\sigma\left( k\right) $, so that $i\neq\sigma\left( k\right) $. Now, by the definition of $c_{i}$, we have% \begin{align*} c_{i} & =\left\{ \begin{array} [c]{l}% v_{j_{i}},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\sigma\left( k\right) ;\\ a\rightharpoonup v_{j_{\sigma\left( k\right) }},\ \ \ \ \ \ \ \ \ \ \text{if }i=\sigma\left( k\right) \end{array} \right. =v_{j_{i}}\ \ \ \ \ \ \ \ \ \ \left( \text{since }i\neq\sigma\left( k\right) \right) \\ & =v_{m-i}\ \ \ \ \ \ \ \ \ \ \left( \text{since }j_{i}=m-i\right) . \end{align*} \par Now, forget that we fixed $i$. We thus have shown that $c_{i}=v_{m-i}$ for every $i\in\mathbb{N}$ such that $i\geq\max\left\{ K,\sigma\left( k\right) +1\right\} $. Hence, $c_{i}=v_{m-i}$ for sufficiently large $i$, qed.) \par We have $c_{i}=v_{m-i}$ for sufficiently large $i$. Hence, the infinite wedge product $c_{0}\wedge c_{1}\wedge c_{2}\wedge...$ is well-defined. Since $\left( c_{0},c_{1},c_{2},...\right) =\left( v_{j_{0}},v_{j_{1}% },...,v_{j_{\sigma\left( k\right) -1}},a\rightharpoonup v_{j_{\sigma\left( k\right) }},v_{j_{\sigma\left( k\right) +1}},v_{j_{\sigma\left( k\right) +2}},...\right) $, we have% \[ c_{0}\wedge c_{1}\wedge c_{2}\wedge...=v_{j_{0}}\wedge v_{j_{1}}% \wedge...\wedge v_{j_{\sigma\left( k\right) -1}}\wedge\left( a\rightharpoonup v_{j_{\sigma\left( k\right) }}\right) \wedge v_{j_{\sigma\left( k\right) +1}}\wedge v_{j_{\sigma\left( k\right) +2}% }\wedge.... \] \par But according to Proposition \ref{prop.semiinfwedge.welldef} \textbf{(f)} (applied to $\left( c_{0},c_{1},c_{2},...\right) $ instead of $\left( b_{0},b_{1},b_{2},...\right) $), the infinite wedge product $c_{\sigma\left( 0\right) }\wedge c_{\sigma\left( 1\right) }\wedge c_{\sigma\left( 2\right) }\wedge...$ is well-defined and satisfies \[ c_{\sigma\left( 0\right) }\wedge c_{\sigma\left( 1\right) }\wedge c_{\sigma\left( 2\right) }\wedge...=\left( -1\right) ^{\sigma}\cdot c_{0}\wedge c_{1}\wedge c_{2}\wedge.... \] \par On the other hand, define a sequence $\left( d_{0},d_{1},d_{2},...\right) $ of elements of $V$ by% \begin{equation} \left( d_{i}=\left\{ \begin{array} [c]{l}% v_{j_{\sigma\left( i\right) }},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq k;\\ a\rightharpoonup v_{j_{\sigma\left( k\right) }},\ \ \ \ \ \ \ \ \ \ \text{if }i=k \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}\right) . \label{pf.glinf.glinfact.welldef.ass1.pf.5.pf.d}% \end{equation} Then, $\left( d_{0},d_{1},d_{2},...\right) =\left( v_{j_{\sigma\left( 0\right) }},v_{j_{\sigma\left( 1\right) }},...,v_{j_{\sigma\left( k-1\right) }},a\rightharpoonup v_{j_{\sigma\left( k\right) }}% ,v_{j_{\sigma\left( k+1\right) }},v_{j_{\sigma\left( k+2\right) }% },...\right) $. \par But every $i\in\mathbb{N}$ satisfies% \begin{align*} c_{\sigma\left( i\right) } & =\left\{ \begin{array} [c]{l}% v_{j_{\sigma\left( i\right) }},\ \ \ \ \ \ \ \ \ \ \text{if }\sigma\left( i\right) \neq\sigma\left( k\right) ;\\ a\rightharpoonup v_{j_{\sigma\left( k\right) }},\ \ \ \ \ \ \ \ \ \ \text{if }\sigma\left( i\right) =\sigma\left( k\right) \end{array} \right. \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }% c_{\sigma\left( i\right) }\right) \\ & =\left\{ \begin{array} [c]{l}% v_{j_{\sigma\left( i\right) }},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq k;\\ a\rightharpoonup v_{j_{\sigma\left( k\right) }},\ \ \ \ \ \ \ \ \ \ \text{if }i=k \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\sigma\left( i\right) \neq\sigma\left( k\right) \text{ is equivalent to }i\neq k\text{ (since }\sigma\text{ is bijective), and since}\\ \sigma\left( i\right) =\sigma\left( k\right) \text{ is equivalent to }i=k\text{ (since }\sigma\text{ is bijective)}% \end{array} \right) \\ & =d_{i}\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.glinf.glinfact.welldef.ass1.pf.5.pf.d})}\right) . \end{align*} Thus, \begin{align*} \left( c_{\sigma\left( 0\right) },c_{\sigma\left( 1\right) }% ,c_{\sigma\left( 2\right) },...\right) & =\left( d_{0},d_{1}% ,d_{2},...\right) \\ & =\left( v_{j_{\sigma\left( 0\right) }},v_{j_{\sigma\left( 1\right) }% },...,v_{j_{\sigma\left( k-1\right) }},a\rightharpoonup v_{j_{\sigma\left( k\right) }},v_{j_{\sigma\left( k+1\right) }},v_{j_{\sigma\left( k+2\right) }},...\right) , \end{align*} so that% \[ c_{\sigma\left( 0\right) }\wedge c_{\sigma\left( 1\right) }\wedge c_{\sigma\left( 2\right) }\wedge...=v_{j_{\sigma\left( 0\right) }}\wedge v_{j_{\sigma\left( 1\right) }}\wedge...\wedge v_{j_{\sigma\left( k-1\right) }}\wedge\left( a\rightharpoonup v_{j_{\sigma\left( k\right) }% }\right) \wedge v_{j_{\sigma\left( k+1\right) }}\wedge v_{j_{\sigma\left( k+2\right) }}\wedge.... \] Hence,% \begin{align*} & v_{j_{\sigma\left( 0\right) }}\wedge v_{j_{\sigma\left( 1\right) }% }\wedge...\wedge v_{j_{\sigma\left( k-1\right) }}\wedge\left( a\rightharpoonup v_{j_{\sigma\left( k\right) }}\right) \wedge v_{j_{\sigma\left( k+1\right) }}\wedge v_{j_{\sigma\left( k+2\right) }% }\wedge...\\ & =c_{\sigma\left( 0\right) }\wedge c_{\sigma\left( 1\right) }\wedge c_{\sigma\left( 2\right) }\wedge...=\left( -1\right) ^{\sigma}% \cdot\underbrace{c_{0}\wedge c_{1}\wedge c_{2}\wedge...}_{=v_{j_{0}}\wedge v_{j_{1}}\wedge...\wedge v_{j_{\sigma\left( k\right) -1}}\wedge\left( a\rightharpoonup v_{j_{\sigma\left( k\right) }}\right) \wedge v_{j_{\sigma\left( k\right) +1}}\wedge v_{j_{\sigma\left( k\right) +2}% }\wedge...}\\ & =\left( -1\right) ^{\sigma}\cdot v_{j_{0}}\wedge v_{j_{1}}\wedge...\wedge v_{j_{\sigma\left( k\right) -1}}\wedge\left( a\rightharpoonup v_{j_{\sigma\left( k\right) }}\right) \wedge v_{j_{\sigma\left( k\right) +1}}\wedge v_{j_{\sigma\left( k\right) +2}}\wedge.... \end{align*} This proves (\ref{pf.glinf.glinfact.welldef.ass1.pf.5}).} Hence, (\ref{pf.glinf.glinfact.welldef.ass1.pf.4}) becomes% \begin{align*} & \left( -1\right) ^{\sigma}\cdot F_{a}\left( v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge...\right) \\ & =\sum\limits_{k\geq0}\underbrace{v_{j_{\sigma\left( 0\right) }}\wedge v_{j_{\sigma\left( 1\right) }}\wedge...\wedge v_{j_{\sigma\left( k-1\right) }}\wedge\left( a\rightharpoonup v_{j_{\sigma\left( k\right) }% }\right) \wedge v_{j_{\sigma\left( k+1\right) }}\wedge v_{j_{\sigma\left( k+2\right) }}\wedge...}_{\substack{=\left( -1\right) ^{\sigma}\cdot v_{j_{0}}\wedge v_{j_{1}}\wedge...\wedge v_{j_{\sigma\left( k\right) -1}% }\wedge\left( a\rightharpoonup v_{j_{\sigma\left( k\right) }}\right) \wedge v_{j_{\sigma\left( k\right) +1}}\wedge v_{j_{\sigma\left( k\right) +2}}\wedge...\\\text{(by (\ref{pf.glinf.glinfact.welldef.ass1.pf.5}))}}}\\ & =\sum\limits_{k\geq0}\left( -1\right) ^{\sigma}\cdot v_{j_{0}}\wedge v_{j_{1}}\wedge...\wedge v_{j_{\sigma\left( k\right) -1}}\wedge\left( a\rightharpoonup v_{j_{\sigma\left( k\right) }}\right) \wedge v_{j_{\sigma\left( k\right) +1}}\wedge v_{j_{\sigma\left( k\right) +2}% }\wedge...\\ & =\left( -1\right) ^{\sigma}\cdot\sum\limits_{k\geq0}v_{j_{0}}\wedge v_{j_{1}}\wedge...\wedge v_{j_{\sigma\left( k\right) -1}}\wedge\left( a\rightharpoonup v_{j_{\sigma\left( k\right) }}\right) \wedge v_{j_{\sigma\left( k\right) +1}}\wedge v_{j_{\sigma\left( k\right) +2}% }\wedge.... \end{align*} Dividing this equality by $\left( -1\right) ^{\sigma}$, we obtain% \begin{align*} & F_{a}\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\right) \\ & =\sum\limits_{k\geq0}v_{j_{0}}\wedge v_{j_{1}}\wedge...\wedge v_{j_{\sigma\left( k\right) -1}}\wedge\left( a\rightharpoonup v_{j_{\sigma\left( k\right) }}\right) \wedge v_{j_{\sigma\left( k\right) +1}}\wedge v_{j_{\sigma\left( k\right) +2}}\wedge...\\ & =\sum\limits_{k\geq0}v_{j_{0}}\wedge v_{j_{1}}\wedge...\wedge v_{j_{k-1}% }\wedge\left( a\rightharpoonup v_{j_{k}}\right) \wedge v_{j_{k+1}}\wedge v_{j_{k+2}}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }k\text{ for }% \sigma\left( k\right) \text{ in the sum (since }\sigma\text{ is bijective)}\right) . \end{align*} Thus, (\ref{pf.glinf.glinfact.welldef.ass1}) is proven in Case 2. \end{verlong} \begin{vershort} We have now proven (\ref{pf.glinf.glinfact.welldef.ass1}) in each of the two Cases 1 and 2, hence in all situations. In other words, Assertion \ref{prop.glinf.glinfact.welldef}.1 is proven. \end{vershort} \begin{verlong} Hence, (\ref{pf.glinf.glinfact.welldef.ass1}) is proven in each of the two Cases 1 and 2. Since these two cases cover all possibilities, this shows that (\ref{pf.glinf.glinfact.welldef.ass1}) always holds. This completes the proof of (\ref{pf.glinf.glinfact.welldef.ass1}). In other words, Assertion \ref{prop.glinf.glinfact.welldef}.1 is proven. \end{verlong} Our next goal is the following assertion: \begin{quote} \textit{Assertion \ref{prop.glinf.glinfact.welldef}.2:} Let $a\in \mathfrak{gl}_{\infty}$. Let $b_{0},b_{1},b_{2},...$ be vectors in $V$ which satisfy% \[ b_{i}=v_{m-i}\ \ \ \ \ \ \ \ \ \ \text{for all sufficiently large }i. \] Then,% \[ F_{a}\left( b_{0}\wedge b_{1}\wedge b_{2}\wedge...\right) =\sum \limits_{k\geq0}b_{0}\wedge b_{1}\wedge...\wedge b_{k-1}\wedge\left( a\rightharpoonup b_{k}\right) \wedge b_{k+1}\wedge b_{k+2}\wedge.... \] \end{quote} \begin{vershort} \textit{Proof of Assertion \ref{prop.glinf.glinfact.welldef}.2 (sketched):} We have $b_{i}=v_{m-i}$ for all sufficiently large $i$. In other words, there exists a $K\in\mathbb{N}$ such that every $i\geq K$ satisfies $b_{i}=v_{m-i}$. Fix such a $K$. We have to prove the equality% \begin{equation} F_{a}\left( b_{0}\wedge b_{1}\wedge b_{2}\wedge...\right) =\sum \limits_{k\geq0}b_{0}\wedge b_{1}\wedge...\wedge b_{k-1}\wedge\left( a\rightharpoonup b_{k}\right) \wedge b_{k+1}\wedge b_{k+2}\wedge.... \label{pf.glinf.glinfact.welldef.ass2.pfshort}% \end{equation} This equality is clearly linear in \textbf{each} of the variables $b_{0}$, $b_{1}$, $...$, $b_{K-1}$ (and also in each of the variables $b_{K}$, $b_{K+1}$, $b_{K+2}$, $...$, but we don't care about them). Hence, in proving it, we can WLOG assume that each of the vectors $b_{0}$, $b_{1}$, $...$, $b_{K-1}$ belongs to the basis $\left( v_{j}\right) _{j\in\mathbb{Z}}$ of $V$.\ \ \ \ \footnote{Note that this assumption is allowed because $b_{0}$, $b_{1}$, $...$, $b_{K-1}$ are \textbf{finitely many} vectors. In contrast, if we wanted to WLOG assume that each of the (infinitely many) vectors $b_{0}$, $b_{1}$, $b_{2}$, $...$ belongs to the basis $\left( v_{j}\right) _{j\in\mathbb{Z}}$ of $V$, then we would have to need more justification for such an assumption.} Assume this. Of course, the remaining vectors $b_{K}$, $b_{K+1}$, $b_{K+2}$, $...$ also belong to the basis $\left( v_{j}\right) _{j\in\mathbb{Z}}$ of $V$ (because every $i\geq K$ satisfies $b_{i}=v_{m-i}$). Hence, all the vectors $b_{0}$, $b_{1}$, $b_{2}$, $...$ belong to the basis $\left( v_{j}\right) _{j\in\mathbb{Z}}$ of $V$. Hence, there exists a sequence $\left( j_{0},j_{1},j_{2},...\right) \in\mathbb{Z}^{\mathbb{N}}$ such that every $i\in\mathbb{N}$ satisfies $b_{i}=v_{j_{i}}$. Therefore, the equality that we need to prove, (\ref{pf.glinf.glinfact.welldef.ass2.pfshort}% ), will immediately follow from Assertion \ref{prop.glinf.glinfact.welldef}.1 once we can show that $\left( j_{0},j_{1},j_{2},...\right) $ is a straying $m$-degression. But the latter is obvious (since every $i\geq K$ satisfies $v_{j_{i}}=b_{i}=v_{m-i}$ and thus $j_{i}=m-i$, so that $j_{i}+i=m$). Hence, (\ref{pf.glinf.glinfact.welldef.ass2.pfshort}) is proven. That is, Assertion \ref{prop.glinf.glinfact.welldef}.2 is proven. \end{vershort} \begin{verlong} Rather than prove this directly, we will show the following assertion first: \textit{Assertion \ref{prop.glinf.glinfact.welldef}.3:} Let $K$ be a nonnegative integer. Let $a\in\mathfrak{gl}_{\infty}$. Let $b_{0},b_{1}% ,b_{2},...$ be vectors in $V$ which satisfy% \begin{equation} b_{i}=v_{m-i}\ \ \ \ \ \ \ \ \ \ \text{for all sufficiently large }i. \label{pf.glinf.glinfact.welldef.ass3.stand}% \end{equation} Assume also that% \begin{equation} b_{i}\in\left\{ v_{z}\ \mid\ z\in\mathbb{Z}\right\} \ \ \ \ \ \ \ \ \ \ \text{for all integers }i\geq K. \label{pf.glinf.glinfact.welldef.ass3.hook}% \end{equation} Then,% \begin{equation} F_{a}\left( b_{0}\wedge b_{1}\wedge b_{2}\wedge...\right) =\sum \limits_{k\geq0}b_{0}\wedge b_{1}\wedge...\wedge b_{k-1}\wedge\left( a\rightharpoonup b_{k}\right) \wedge b_{k+1}\wedge b_{k+2}\wedge.... \label{pf.glinf.glinfact.welldef.ass3.state}% \end{equation} Notice that Assertion \ref{prop.glinf.glinfact.welldef}.3 differs from Assertion \ref{prop.glinf.glinfact.welldef}.2 in the presence of an additional condition (namely, (\ref{pf.glinf.glinfact.welldef.ass3.hook})). This condition will turn out to be harmless\footnote{In fact, it is easy to see that for every fixed sequence $\left( b_{0},b_{1},b_{2},...\right) $ of vectors in $V$ which satisfy $\left( b_{i}=v_{m-i}\text{ for all sufficiently large }i\right) $, there exists a $K\in\mathbb{N}$ for which this condition (\ref{pf.glinf.glinfact.welldef.ass3.hook}) is satisfied. We will explain this in more detail later.}, but it allows us to induct over the variable $K$. \textit{Proof of Assertion \ref{prop.glinf.glinfact.welldef}.3:} We will prove Assertion \ref{prop.glinf.glinfact.welldef}.3 by induction over $K$: \textit{Induction base:} To complete the induction base, we need to show that Assertion \ref{prop.glinf.glinfact.welldef}.3 holds for $K=0$. So, assume that $K=0$. Let $a\in\mathfrak{gl}_{\infty}$. Let $b_{0},b_{1},b_{2},...$ be vectors in $V$ which satisfy (\ref{pf.glinf.glinfact.welldef.ass3.stand}) and (\ref{pf.glinf.glinfact.welldef.ass3.hook}). We need to prove that (\ref{pf.glinf.glinfact.welldef.ass3.state}) holds. We know that (\ref{pf.glinf.glinfact.welldef.ass3.stand}) holds. In other words, there exists an $N\in\mathbb{N}$ such that every $i\geq N$ satisfies $b_{i}=v_{m-i}$. Fix such an $N$. Notice that% \begin{equation} b_{i}\in\left\{ v_{z}\ \mid\ z\in\mathbb{Z}\right\} \ \ \ \ \ \ \ \ \ \ \text{for all }i\in\mathbb{N}. \label{pf.glinf.glinfact.welldef.ass3.pf.indbase}% \end{equation} \footnote{\textit{Proof of (\ref{pf.glinf.glinfact.welldef.ass3.pf.indbase}):} Let $i\in\mathbb{N}$. Then, $i\geq0=K$. Thus (according to (\ref{pf.glinf.glinfact.welldef.ass3.hook})), we have $b_{i}\in\left\{ v_{z}\ \mid\ z\in\mathbb{Z}\right\} $. This proves (\ref{pf.glinf.glinfact.welldef.ass3.pf.indbase}).} For every $i\in\mathbb{N}$, let us define an integer $j_{i}$ as follows: If $i\geq N$, set $j_{i}=m-i$. Otherwise, set $j_{i}$ to be an integer $z\in\mathbb{Z}$ such that $b_{i}=v_{z}$ (such an integer $z$ exists, because (\ref{pf.glinf.glinfact.welldef.ass3.pf.indbase}) shows that $b_{i}\in\left\{ v_{z}\ \mid\ z\in\mathbb{Z}\right\} $). Thus, we have defined an integer $j_{i}$ for every $i\in\mathbb{N}$. It is now rather clear that% \begin{equation} b_{i}=v_{j_{i}}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}. \label{pf.glinf.glinfact.welldef.ass3.pf.indbase.2}% \end{equation} \footnote{\textit{Proof of (\ref{pf.glinf.glinfact.welldef.ass3.pf.indbase.2}% ):} Let $i\in\mathbb{N}$. We need to then show that $b_{i}=v_{j_{i}}$. We distinguish between two cases: \par \textit{Case 1:} We have $i\geq N$. \par \textit{Case 2:} We don't have $i\geq N$. \par Let us first consider Case 1. In this case, $i\geq N$. Hence, $j_{i}=m-i$ (by the definition of $j_{i}$). Thus, $m-i=j_{i}$, so that $v_{m-i}=v_{j_{i}}$. But on the other hand, $b_{i}=v_{m-i}$ (since we know that every $i\geq N$ satisfies $b_{i}=v_{m-i}$). Thus, $b_{i}=v_{m-i}=v_{j_{i}}$. This proves $b_{i}=v_{j_{i}}$ in Case 1. \par Now let us consider Case 2. In this case, we don't have $i\geq N$. Hence, $j_{i}$ is an integer $z\in\mathbb{Z}$ such that $b_{i}=v_{z}$ (by the definition of $j_{i}$). Hence, $b_{i}=v_{j_{i}}$. Thus, $b_{i}=v_{j_{i}}$ is proven in Case 2. \par We have now proven $b_{i}=v_{j_{i}}$ in each of the Cases 1 and 2. Hence, $b_{i}=v_{j_{i}}$ always holds (since Cases 1 and 2 cover all possibilities). Thus, (\ref{pf.glinf.glinfact.welldef.ass3.pf.indbase.2}) is proven.} Also,% \begin{equation} \text{every sufficiently high }k\in\mathbb{N}\text{ satisfies }j_{k}+k=m. \label{pf.glinf.glinfact.welldef.ass3.pf.indbase.3}% \end{equation} \footnote{\textit{Proof of (\ref{pf.glinf.glinfact.welldef.ass3.pf.indbase.3}% ):} For every $k\in\mathbb{N}$ such that $k\geq N$, we have $j_{k}=m-k$ (by the definition of $j_{k}$). In other words, for every $k\in\mathbb{N}$ such that $k\geq N$, we have $j_{k}+k=m$. Thus, for every sufficiently high $k\in\mathbb{N}$, we have $j_{k}+k=m$. This proves (\ref{pf.glinf.glinfact.welldef.ass3.pf.indbase.3}).} Hence, $\left( j_{0},j_{1},j_{2},...\right) $ is a straying $m$-degression. Thus, Assertion \ref{prop.glinf.glinfact.welldef}.1 yields% \[ F_{a}\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\right) =\sum\limits_{k\geq0}v_{j_{0}}\wedge v_{j_{1}}\wedge...\wedge v_{j_{k-1}% }\wedge\left( a\rightharpoonup v_{j_{k}}\right) \wedge v_{j_{k+1}}\wedge v_{j_{k+2}}\wedge.... \] But due to (\ref{pf.glinf.glinfact.welldef.ass3.pf.indbase.2}), we have $\left( b_{0},b_{1},b_{2},...\right) =\left( v_{j_{0}},v_{j_{1}},v_{j_{2}% },...\right) $. Therefore, $b_{0}\wedge b_{1}\wedge b_{2}\wedge...=v_{j_{0}% }\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...$. Hence,% \begin{align*} F_{a}\left( b_{0}\wedge b_{1}\wedge b_{2}\wedge...\right) & =F_{a}\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\right) \\ & =\sum\limits_{k\geq0}\underbrace{v_{j_{0}}\wedge v_{j_{1}}\wedge...\wedge v_{j_{k-1}}\wedge\left( a\rightharpoonup v_{j_{k}}\right) \wedge v_{j_{k+1}% }\wedge v_{j_{k+2}}\wedge...}_{\substack{=b_{0}\wedge b_{1}\wedge...\wedge b_{k-1}\wedge\left( a\rightharpoonup b_{k}\right) \wedge b_{k+1}\wedge b_{k+2}\wedge...\\\text{(since }\left( v_{j_{0}},v_{j_{1}},v_{j_{2}% },...\right) =\left( b_{0},b_{1},b_{2},...\right) \text{)}}}\\ & =\sum\limits_{k\geq0}b_{0}\wedge b_{1}\wedge...\wedge b_{k-1}\wedge\left( a\rightharpoonup b_{k}\right) \wedge b_{k+1}\wedge b_{k+2}\wedge.... \end{align*} In other words, (\ref{pf.glinf.glinfact.welldef.ass3.state}) holds. We have thus shown that Assertion \ref{prop.glinf.glinfact.welldef}.3 holds for $K=0$. This completes the induction base. \textit{Induction step:} Let $\kappa\in\mathbb{N}$. Assume that Assertion \ref{prop.glinf.glinfact.welldef}.3 holds for $K=\kappa$. We now need to prove that Assertion \ref{prop.glinf.glinfact.welldef}.3 holds for $K=\kappa+1$. Let $a\in\mathfrak{gl}_{\infty}$. Let $b_{0},b_{1},b_{2},...$ be vectors in $V$ which satisfy (\ref{pf.glinf.glinfact.welldef.ass3.stand}), and satisfy (\ref{pf.glinf.glinfact.welldef.ass3.hook}) for $K=\kappa+1$. We need to prove that (\ref{pf.glinf.glinfact.welldef.ass3.state}) holds. We know that the vectors $b_{0},b_{1},b_{2},...$ satisfy (\ref{pf.glinf.glinfact.welldef.ass3.hook}) for $K=\kappa+1$. In other words,% \begin{equation} b_{i}\in\left\{ v_{z}\ \mid\ z\in\mathbb{Z}\right\} \ \ \ \ \ \ \ \ \ \ \text{for all integers }i\geq\kappa+1. \label{pf.glinf.glinfact.welldef.ass3.indstep.hook}% \end{equation} We have $b_{\kappa}\in V$, while $\left( v_{j}\right) _{j\in\mathbb{Z}}$ is a basis of the vector space $V$. Thus, we can write the vector $b_{\kappa}$ in the form% \[ b_{\kappa}=\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}v_{\ell}% \] for some family $\left( \beta_{\ell}\right) _{\ell\in\mathbb{Z}}% \in\mathbb{C}^{\mathbb{Z}}$ such that all but finitely many $\ell\in \mathbb{Z}$ satisfy $\beta_{\ell}=0$. Consider this family $\left( \beta_{\ell}\right) _{\ell\in\mathbb{Z}}$. Since all but finitely many $\ell\in\mathbb{Z}$ satisfy $\beta_{\ell}=0$, the sum $\sum\limits_{\ell \in\mathbb{Z}}\beta_{\ell}v_{\ell}$ has only finitely many nonzero addends, and thus can be treated as a finite sum in regard to algebraic transformations. In particular, the following computation is allowed:% \begin{align} & b_{0}\wedge b_{1}\wedge b_{2}\wedge...\nonumber\\ & =b_{0}\wedge b_{1}\wedge...\wedge b_{\kappa-1}\wedge b_{\kappa}\wedge b_{\kappa+1}\wedge b_{\kappa+2}\wedge...\nonumber\\ & =b_{0}\wedge b_{1}\wedge...\wedge b_{\kappa-1}\wedge\left( \sum \limits_{\ell\in\mathbb{Z}}\beta_{\ell}v_{\ell}\right) \wedge b_{\kappa +1}\wedge b_{\kappa+2}\wedge...\ \ \ \ \ \ \ \ \ \ \left( \text{since }b_{\kappa}=\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}v_{\ell}\right) \nonumber\\ & =\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}b_{0}\wedge b_{1}% \wedge...\wedge b_{\kappa-1}\wedge v_{\ell}\wedge b_{\kappa+1}\wedge b_{\kappa+2}\wedge...\label{pf.glinf.glinfact.welldef.ass3.indstep.1}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{due to the multilinearity of the infinite wedge product}\right) .\nonumber \end{align} Now, for every $\ell\in\mathbb{Z}$, let us define a sequence $\left( b_{\ell,0},b_{\ell,1},b_{\ell,2},...\right) $ of vectors in $V$ by% \[ \left( b_{\ell,i}=\left\{ \begin{array} [c]{c}% b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}\right) . \] Then, for every $\ell\in\mathbb{Z}$, we have $\left( b_{\ell,0},b_{\ell ,1},b_{\ell,2},...\right) =\left( b_{0},b_{1},...,b_{\kappa-1},v_{\ell },b_{\kappa+1},b_{\kappa+2},...\right) $. Hence, for every $\ell\in \mathbb{Z}$, we have% \[ b_{\ell,0}\wedge b_{\ell,1}\wedge b_{\ell,2}\wedge...=b_{0}\wedge b_{1}% \wedge...\wedge b_{\kappa-1}\wedge v_{\ell}\wedge b_{\kappa+1}\wedge b_{\kappa+2}\wedge.... \] Thus, (\ref{pf.glinf.glinfact.welldef.ass3.indstep.1}) becomes% \begin{align*} & b_{0}\wedge b_{1}\wedge b_{2}\wedge...\\ & =\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}\underbrace{b_{0}\wedge b_{1}\wedge...\wedge b_{\kappa-1}\wedge v_{\ell}\wedge b_{\kappa+1}\wedge b_{\kappa+2}\wedge...}_{=b_{\ell,0}\wedge b_{\ell,1}\wedge b_{\ell,2}% \wedge...}\\ & =\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}b_{\ell,0}\wedge b_{\ell ,1}\wedge b_{\ell,2}\wedge.... \end{align*} Applying the map $F_{a}$ to this equality, we obtain% \begin{align} & F_{a}\left( b_{0}\wedge b_{1}\wedge b_{2}\wedge...\right) \nonumber\\ & =F_{a}\left( \sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}b_{\ell,0}\wedge b_{\ell,1}\wedge b_{\ell,2}\wedge...\right) =\sum\limits_{\ell\in\mathbb{Z}% }\beta_{\ell}F_{a}\left( b_{\ell,0}\wedge b_{\ell,1}\wedge b_{\ell,2}% \wedge...\right) \label{pf.glinf.glinfact.welldef.ass3.indstep.2}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since the map }F_{a}\text{ is linear, and since the sum }\sum \limits_{\ell\in\mathbb{Z}}\beta_{\ell}b_{\ell,0}\wedge b_{\ell,1}\wedge b_{\ell,2}\wedge...\\ \text{has only finitely many nonzero addends}\\ \text{(because all but finitely many }\ell\in\mathbb{Z}\text{ satisfy }% \beta_{\ell}=0\text{)}% \end{array} \right) .\nonumber \end{align} Now, fix an $\ell\in\mathbb{N}$. The sequence $\left( b_{\ell,0},b_{\ell ,1},b_{\ell,2},...\right) \in V^{\mathbb{N}}$ satisfies% \begin{equation} b_{\ell,i}=v_{m-i}\ \ \ \ \ \ \ \ \ \ \text{for all sufficiently large }i. \label{pf.glinf.glinfact.welldef.ass3.indstep.standk+1}% \end{equation} \footnote{\textit{Proof of (\ref{pf.glinf.glinfact.welldef.ass3.indstep.standk+1}):} We know that the vectors $b_{0},b_{1},b_{2},...$ satisfy (\ref{pf.glinf.glinfact.welldef.ass3.stand}). In other words, there exists an $I\in\mathbb{N}$ such that every integer $i>I$ satisfies $b_{i}=v_{m-i}$. Consider such an $I$. \par Let $i\in\mathbb{N}$ be such that $i>\max\left\{ \kappa,I\right\} $. Then, $i>\max\left\{ \kappa,I\right\} \geq I$, so that $b_{i}=v_{m-i}$ (since we know that every integer $i>I$ satisfies $b_{i}=v_{m-i}$). Also, $i>\max \left\{ \kappa,I\right\} \geq\kappa$, and thus $i\neq\kappa$. By the definition of $b_{\ell,i}$, we have% \begin{align*} b_{\ell,i} & =\left\{ \begin{array} [c]{c}% b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. =b_{i}\ \ \ \ \ \ \ \ \ \ \left( \text{since }i\neq\kappa\right) \\ & =v_{m-i}. \end{align*} \par Now, forget that we fixed $i$. We thus have shown that every $i\in\mathbb{N}$ such that $i>\max\left\{ \kappa,I\right\} $ satisfies $b_{\ell,i}=v_{m-i}$. Thus, every sufficiently large $i$ satisfies $b_{\ell,i}=v_{m-i}$. This proves (\ref{pf.glinf.glinfact.welldef.ass3.indstep.standk+1}).} Moreover, the sequence $\left( b_{\ell,0},b_{\ell,1},b_{\ell,2},...\right) \in V^{\mathbb{N}}$ satisfies \begin{equation} b_{\ell,i}\in\left\{ v_{z}\ \mid\ z\in\mathbb{Z}\right\} \ \ \ \ \ \ \ \ \ \ \text{for all integers }i\geq\kappa. \label{pf.glinf.glinfact.welldef.ass3.indstep.hookk+1}% \end{equation} \footnote{\textit{Proof of (\ref{pf.glinf.glinfact.welldef.ass3.indstep.hookk+1}):} Let $i$ be an integer such that $i\geq\kappa$. \par If $i=\kappa$, then% \begin{align*} b_{\ell,i} & =\left\{ \begin{array} [c]{c}% b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }b_{\ell ,i}\right) \\ & =v_{\ell}\ \ \ \ \ \ \ \ \ \ \left( \text{since }i=\kappa\right) \\ & \in\left\{ v_{z}\ \mid\ z\in\mathbb{Z}\right\} \ \ \ \ \ \ \ \ \ \ \left( \text{since }\ell\in\mathbb{Z}\right) . \end{align*} Hence, if $i=\kappa$, then (\ref{pf.glinf.glinfact.welldef.ass3.indstep.hookk+1}) is true. Thus, for the rest of the proof of (\ref{pf.glinf.glinfact.welldef.ass3.indstep.hookk+1}), we can WLOG assume that $i\neq\kappa$. Assume this. \par Since $i\geq\kappa$ and $i\neq\kappa$, we have $i>\kappa$. Since $i$ and $\kappa$ are integers, this shows that $i\geq\kappa+1$. By the definition of $b_{\ell,i}$, we have% \begin{align*} b_{\ell,i} & =\left\{ \begin{array} [c]{c}% b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. =b_{i}\ \ \ \ \ \ \ \ \ \ \left( \text{since }i\neq\kappa\right) \\ & \in\left\{ v_{z}\ \mid\ z\in\mathbb{Z}\right\} \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.glinf.glinfact.welldef.ass3.indstep.hook}), since }i\geq \kappa+1\right) . \end{align*} This proves (\ref{pf.glinf.glinfact.welldef.ass3.indstep.hookk+1}).} But we have assumed (as the induction hypothesis) that Assertion \ref{prop.glinf.glinfact.welldef}.3 holds for $K=\kappa$. Hence, we can apply Assertion \ref{prop.glinf.glinfact.welldef}.3 to $\left( b_{\ell,0}% ,b_{\ell,1},b_{\ell,2},...\right) $ and $\kappa$ instead of $\left( b_{0},b_{1},b_{2},...\right) $ and $K$ (because we know that the sequence $\left( b_{\ell,0},b_{\ell,1},b_{\ell,2},...\right) $ satisfies (\ref{pf.glinf.glinfact.welldef.ass3.indstep.standk+1}) and (\ref{pf.glinf.glinfact.welldef.ass3.indstep.hookk+1})). As a result, we conclude that% \begin{align} & F_{a}\left( b_{\ell,0}\wedge b_{\ell,1}\wedge b_{\ell,2}\wedge...\right) \nonumber\\ & =\sum\limits_{k\geq0}b_{\ell,0}\wedge b_{\ell,1}\wedge...\wedge b_{\ell,k-1}\wedge\left( a\rightharpoonup b_{\ell,k}\right) \wedge b_{\ell,k+1}\wedge b_{\ell,k+2}\wedge.... \label{pf.glinf.glinfact.welldef.ass3.indstep.7}% \end{align} Now, forget that we have fixed $\ell$. We have thus proven (\ref{pf.glinf.glinfact.welldef.ass3.indstep.7}) for every $\ell\in\mathbb{N}% $. Now, (\ref{pf.glinf.glinfact.welldef.ass3.indstep.2}) becomes% \begin{align} & F_{a}\left( b_{0}\wedge b_{1}\wedge b_{2}\wedge...\right) \nonumber\\ & =\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}\underbrace{F_{a}\left( b_{\ell,0}\wedge b_{\ell,1}\wedge b_{\ell,2}\wedge...\right) }% _{\substack{=\sum\limits_{k\geq0}b_{\ell,0}\wedge b_{\ell,1}\wedge...\wedge b_{\ell,k-1}\wedge\left( a\rightharpoonup b_{\ell,k}\right) \wedge b_{\ell,k+1}\wedge b_{\ell,k+2}\wedge...\\\text{(by (\ref{pf.glinf.glinfact.welldef.ass3.indstep.7}))}}}\nonumber\\ & =\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}\sum\limits_{k\geq0}b_{\ell ,0}\wedge b_{\ell,1}\wedge...\wedge b_{\ell,k-1}\wedge\left( a\rightharpoonup b_{\ell,k}\right) \wedge b_{\ell,k+1}\wedge b_{\ell,k+2}\wedge...\nonumber\\ & =\sum\limits_{k\geq0}\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}b_{\ell ,0}\wedge b_{\ell,1}\wedge...\wedge b_{\ell,k-1}\wedge\left( a\rightharpoonup b_{\ell,k}\right) \wedge b_{\ell,k+1}\wedge b_{\ell,k+2}\wedge.... \label{pf.glinf.glinfact.welldef.ass3.indstep.8}% \end{align} It remains to prove that the right hand side of (\ref{pf.glinf.glinfact.welldef.ass3.indstep.7}) equals the right hand side of (\ref{pf.glinf.glinfact.welldef.ass3.state}). So our next goal is to show that every $k\in\mathbb{N}$ satisfies% \begin{align} & \sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}b_{\ell,0}\wedge b_{\ell ,1}\wedge...\wedge b_{\ell,k-1}\wedge\left( a\rightharpoonup b_{\ell ,k}\right) \wedge b_{\ell,k+1}\wedge b_{\ell,k+2}\wedge...\nonumber\\ & =b_{0}\wedge b_{1}\wedge...\wedge b_{k-1}\wedge\left( a\rightharpoonup b_{k}\right) \wedge b_{k+1}\wedge b_{k+2}\wedge.... \label{pf.glinf.glinfact.welldef.ass3.indstep.13}% \end{align} \textit{Proof of (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13}):} Fix $k\in\mathbb{N}$. We need to prove that (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13}) holds. Let us define a sequence $\left( \widetilde{b}_{0},\widetilde{b}% _{1},\widetilde{b}_{2},...\right) $ of vectors in $V$ by% \[ \left( \widetilde{b}_{i}=\left\{ \begin{array} [c]{c}% b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq k;\\ a\rightharpoonup b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i=k \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}\right) . \] Then, $\left( \widetilde{b}_{0},\widetilde{b}_{1},\widetilde{b}% _{2},...\right) =\left( b_{0},b_{1},...,b_{k-1},a\rightharpoonup b_{k},b_{k+1},b_{k+2},...\right) $. For every $\ell\in\mathbb{Z}$, let us define a sequence $\left( \widetilde{b}_{\ell,0},\widetilde{b}_{\ell,1},\widetilde{b}_{\ell ,2},...\right) $ of vectors in $V$ by% \[ \left( \widetilde{b}_{\ell,i}=\left\{ \begin{array} [c]{c}% b_{\ell,i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq k;\\ a\rightharpoonup b_{\ell,i},\ \ \ \ \ \ \ \ \ \ \text{if }i=k \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}\right) . \] Then, for every $\ell\in\mathbb{Z}$, we have $\left( \widetilde{b}_{\ell ,0},\widetilde{b}_{\ell,1},\widetilde{b}_{\ell,2},...\right) =\left( b_{\ell,0},b_{\ell,1},...,b_{\ell,k-1},a\rightharpoonup b_{\ell,k}% ,b_{\ell,k+1},b_{\ell,k+2},...\right) $. Hence, for every $\ell\in\mathbb{Z}% $, we have% \begin{align} & \widetilde{b}_{\ell,0}\wedge\widetilde{b}_{\ell,1}\wedge\widetilde{b}% _{\ell,2}\wedge...\nonumber\\ & =b_{\ell,0}\wedge b_{\ell,1}\wedge...\wedge b_{\ell,k-1}\wedge\left( a\rightharpoonup b_{\ell,k}\right) \wedge b_{\ell,k+1}\wedge b_{\ell ,k+2}\wedge.... \label{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.0}% \end{align} Furthermore, for every $\ell\in\mathbb{Z}$, define an element $w_{\ell}$ of $V$ by $w_{\ell}=\left\{ \begin{array} [c]{c}% v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa\neq k;\\ a\rightharpoonup v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa=k \end{array} \right. $. Since the sum $\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}v_{\ell }$ has only finitely many nonzero addends, we have% \[ \sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}\left( a\rightharpoonup v_{\ell }\right) =a\rightharpoonup\underbrace{\left( \sum\limits_{\ell\in\mathbb{Z}% }\beta_{\ell}v_{\ell}\right) }_{=b_{\kappa}}=a\rightharpoonup b_{\kappa}. \] Now, the sum $\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}w_{\ell}$ has only finitely many nonzero addends (since all but finitely many $\ell\in\mathbb{Z}$ satisfy $\beta_{\ell}=0$), and equals% \begin{align} \sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}w_{\ell} & =\sum\limits_{\ell \in\mathbb{Z}}\beta_{\ell}\left\{ \begin{array} [c]{c}% v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa\neq k;\\ a\rightharpoonup v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa=k \end{array} \right. \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }w_{\ell}=\left\{ \begin{array} [c]{c}% v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa\neq k;\\ a\rightharpoonup v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa=k \end{array} \right. \text{ for every }\ell\in\mathbb{Z}\right) \nonumber\\ & =\left\{ \begin{array} [c]{c}% \sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}v_{\ell}% ,\ \ \ \ \ \ \ \ \ \ \text{if }\kappa\neq k;\\ \sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}\left( a\rightharpoonup v_{\ell }\right) ,\ \ \ \ \ \ \ \ \ \ \text{if }\kappa=k \end{array} \right. =\left\{ \begin{array} [c]{c}% b_{\kappa},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa\neq k;\\ a\rightharpoonup b_{\kappa},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa=k \end{array} \right. \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}v_{\ell}=b_{\kappa }\text{ if }\kappa\neq k\text{,}\\ \text{and since }\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}\left( a\rightharpoonup v_{\ell}\right) =a\rightharpoonup b_{\kappa}\text{ if }\kappa=k \end{array} \right) \nonumber\\ & =\widetilde{b}_{\kappa}\ \ \ \ \ \ \ \ \ \ \left( \text{since the definition of }\widetilde{b}_{\kappa}\text{ yields }\widetilde{b}_{\kappa }=\left\{ \begin{array} [c]{c}% b_{\kappa},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa\neq k;\\ a\rightharpoonup b_{\kappa},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa=k \end{array} \right. \right) . \label{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.1}% \end{align} Now, fix $\ell\in\mathbb{Z}$. It is easy to see that \begin{equation} \widetilde{b}_{\ell,i}=\left\{ \begin{array} [c]{c}% \widetilde{b}_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ w_{\ell}\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{N}. \label{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.2}% \end{equation} \footnote{\textit{Proof of (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.2}):} Let $i\in\mathbb{N}$. We distinguish between two cases: \par \textit{Case 1:} We have $i\neq k$. \par \textit{Case 2:} We have $i=k$. \par Let us consider Case 1 first. In this case, $i\neq k$. By the definition of $\widetilde{b}_{\ell,i}$, we have% \begin{align} \widetilde{b}_{\ell,i} & =\left\{ \begin{array} [c]{c}% b_{\ell,i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq k;\\ a\rightharpoonup b_{\ell,i},\ \ \ \ \ \ \ \ \ \ \text{if }i=k \end{array} \right. =b_{\ell,i}\ \ \ \ \ \ \ \ \ \ \left( \text{since }i\neq k\right) \nonumber\\ & =\left\{ \begin{array} [c]{c}% b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }b_{\ell ,i}\right) . \label{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.2.pf.1}% \end{align} But the definition of $\widetilde{b}_{i}$ yields $\widetilde{b}_{i}=\left\{ \begin{array} [c]{c}% b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq k;\\ a\rightharpoonup b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i=k \end{array} \right. =b_{i}$ (since $i\neq k$). Thus, $b_{i}=\widetilde{b}_{i}$. Also, if $i=\kappa$, then $\kappa\neq k$ (since $i\neq k$). Hence, if $i=\kappa$, then% \begin{align*} w_{\ell} & =\left\{ \begin{array} [c]{c}% v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa\neq k;\\ a\rightharpoonup v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa=k \end{array} \right. \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }w_{\ell }\right) \\ & =v_{\ell}\ \ \ \ \ \ \ \ \ \ \left( \text{since }\kappa\neq k\right) . \end{align*} Hence, if $i=\kappa$, then $v_{\ell}=w_{\ell}$. Now, (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.2.pf.1}) becomes% \begin{align*} \widetilde{b}_{\ell,i} & =\left\{ \begin{array} [c]{c}% b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. =\left\{ \begin{array} [c]{c}% \widetilde{b}_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ w_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{because we have }b_{i}=\widetilde{b}_{i}\text{ if }i\neq\kappa\text{ (this was proven above),}\\ \text{and because we have }v_{\ell}=w_{\ell}\text{ if }i=\kappa\text{ (this was proven above)}% \end{array} \right) . \end{align*} Thus, (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.2}) is proven in Case 1. \par Let us now consider Case 2. In this case, $i=k$. By the definition $\widetilde{b}_{\ell,i}$, we have% \begin{align} \widetilde{b}_{\ell,i} & =\left\{ \begin{array} [c]{c}% b_{\ell,i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq k;\\ a\rightharpoonup b_{\ell,i},\ \ \ \ \ \ \ \ \ \ \text{if }i=k \end{array} \right. =a\rightharpoonup b_{\ell,i}\ \ \ \ \ \ \ \ \ \ \left( \text{since }i=k\right) \nonumber\\ & =a\rightharpoonup\left\{ \begin{array} [c]{c}% b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }b_{\ell,i}=\left\{ \begin{array} [c]{c}% b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. \text{ (by the definition of }b_{\ell,i}\text{)}\right) \nonumber\\ & =\left\{ \begin{array} [c]{c}% a\rightharpoonup b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ a\rightharpoonup v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. . \label{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.2.pf.2}% \end{align} But the definition of $\widetilde{b}_{i}$ yields $\widetilde{b}_{i}=\left\{ \begin{array} [c]{c}% b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq k;\\ a\rightharpoonup b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i=k \end{array} \right. =a\rightharpoonup b_{i}$ (since $i=k$). Thus, $a\rightharpoonup b_{i}=\widetilde{b}_{i}$. Also, if $i=\kappa$, then $\kappa=k$ (since $i=k$). Hence, if $i=\kappa$, then% \begin{align*} w_{\ell} & =\left\{ \begin{array} [c]{c}% v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa\neq k;\\ a\rightharpoonup v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\kappa=k \end{array} \right. \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }w_{\ell }\right) \\ & =a\rightharpoonup v_{\ell}\ \ \ \ \ \ \ \ \ \ \left( \text{since }% \kappa=k\right) . \end{align*} Hence, if $i=\kappa$, then $a\rightharpoonup v_{\ell}=w_{\ell}$. Now, (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.2.pf.2}) becomes% \begin{align*} \widetilde{b}_{\ell,i} & =\left\{ \begin{array} [c]{c}% a\rightharpoonup b_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ a\rightharpoonup v_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. =\left\{ \begin{array} [c]{c}% \widetilde{b}_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq\kappa;\\ w_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }i=\kappa \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{because we have }a\rightharpoonup b_{i}=\widetilde{b}_{i}\text{ if }i\neq\kappa\text{ (this was proven above),}\\ \text{and because we have }a\rightharpoonup v_{\ell}=w_{\ell}\text{ if }i=\kappa\text{ (this was proven above)}% \end{array} \right) . \end{align*} Thus, (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.2}) is proven in Case 2. \par We have thus proven (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.2}) in each of the Cases 1 and 2. Since these two Cases cover all possibilities, this yields that (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.2}) always holds, qed.} In other words, $\left( \widetilde{b}_{\ell,0},\widetilde{b}% _{\ell,1},\widetilde{b}_{\ell,2},...\right) =\left( \widetilde{b}% _{0},\widetilde{b}_{1},...,\widetilde{b}_{\kappa-1},w_{\ell},\widetilde{b}% _{\kappa+1},\widetilde{b}_{\kappa+2},...\right) $. Thus,% \[ \widetilde{b}_{\ell,0}\wedge\widetilde{b}_{\ell,1}\wedge\widetilde{b}_{\ell ,2}\wedge...=\widetilde{b}_{0}\wedge\widetilde{b}_{1}\wedge...\wedge \widetilde{b}_{\kappa-1}\wedge w_{\ell}\wedge\widetilde{b}_{\kappa+1}% \wedge\widetilde{b}_{\kappa+2}\wedge.... \] Compared with (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.0}), this yields% \begin{align} & b_{\ell,0}\wedge b_{\ell,1}\wedge...\wedge b_{\ell,k-1}\wedge\left( a\rightharpoonup b_{\ell,k}\right) \wedge b_{\ell,k+1}\wedge b_{\ell ,k+2}\wedge...\nonumber\\ & =\widetilde{b}_{0}\wedge\widetilde{b}_{1}\wedge...\wedge\widetilde{b}% _{\kappa-1}\wedge w_{\ell}\wedge\widetilde{b}_{\kappa+1}\wedge\widetilde{b}% _{\kappa+2}\wedge.... \label{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.4}% \end{align} Now, forget that we fixed $\ell$. We thus have proven that (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.4}) holds for every $\ell\in\mathbb{Z}$. Now,% \begin{align*} & \sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}\underbrace{b_{\ell,0}\wedge b_{\ell,1}\wedge...\wedge b_{\ell,k-1}\wedge\left( a\rightharpoonup b_{\ell,k}\right) \wedge b_{\ell,k+1}\wedge b_{\ell,k+2}\wedge...}% _{\substack{=\widetilde{b}_{0}\wedge\widetilde{b}_{1}\wedge...\wedge \widetilde{b}_{\kappa-1}\wedge w_{\ell}\wedge\widetilde{b}_{\kappa+1}% \wedge\widetilde{b}_{\kappa+2}\wedge...\\\text{(by (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.4}))}}}\\ & =\sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}\widetilde{b}_{0}% \wedge\widetilde{b}_{1}\wedge...\wedge\widetilde{b}_{\kappa-1}\wedge w_{\ell }\wedge\widetilde{b}_{\kappa+1}\wedge\widetilde{b}_{\kappa+2}\wedge...\\ & =\widetilde{b}_{0}\wedge\widetilde{b}_{1}\wedge...\wedge\widetilde{b}% _{\kappa-1}\wedge\underbrace{\left( \sum\limits_{\ell\in\mathbb{Z}}% \beta_{\ell}w_{\ell}\right) }_{\substack{=\widetilde{b}_{\kappa}\\\text{(by (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13.pf.1}))}}}\wedge \widetilde{b}_{\kappa+1}\wedge\widetilde{b}_{\kappa+2}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since the infinite wedge product is multilinear, and since the sum}\\ \sum\limits_{\ell\in\mathbb{Z}}\beta_{\ell}w_{\ell}\text{ has only finitely many nonzero addends}% \end{array} \right) \\ & =\widetilde{b}_{0}\wedge\widetilde{b}_{1}\wedge...\wedge\widetilde{b}% _{\kappa-1}\wedge\widetilde{b}_{\kappa}\wedge\widetilde{b}_{\kappa+1}% \wedge\widetilde{b}_{\kappa+2}\wedge...\\ & =\widetilde{b}_{0}\wedge\widetilde{b}_{1}\wedge\widetilde{b}_{2}\wedge...\\ & =b_{0}\wedge b_{1}\wedge...\wedge b_{k-1}\wedge\left( a\rightharpoonup b_{k}\right) \wedge b_{k+1}\wedge b_{k+2}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( \widetilde{b}% _{0},\widetilde{b}_{1},\widetilde{b}_{2},...\right) =\left( b_{0}% ,b_{1},...,b_{k-1},a\rightharpoonup b_{k},b_{k+1},b_{k+2},...\right) \right) . \end{align*} This proves (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13}). Now, (\ref{pf.glinf.glinfact.welldef.ass3.indstep.8}) becomes% \begin{align*} & F_{a}\left( b_{0}\wedge b_{1}\wedge b_{2}\wedge...\right) \\ & =\sum\limits_{k\geq0}\underbrace{\sum\limits_{\ell\in\mathbb{Z}}\beta _{\ell}b_{\ell,0}\wedge b_{\ell,1}\wedge...\wedge b_{\ell,k-1}\wedge\left( a\rightharpoonup b_{\ell,k}\right) \wedge b_{\ell,k+1}\wedge b_{\ell ,k+2}\wedge...}_{\substack{=b_{0}\wedge b_{1}\wedge...\wedge b_{k-1}% \wedge\left( a\rightharpoonup b_{k}\right) \wedge b_{k+1}\wedge b_{k+2}\wedge...\\\text{(by (\ref{pf.glinf.glinfact.welldef.ass3.indstep.13}% ))}}}\\ & =\sum\limits_{k\geq0}b_{0}\wedge b_{1}\wedge...\wedge b_{k-1}\wedge\left( a\rightharpoonup b_{k}\right) \wedge b_{k+1}\wedge b_{k+2}\wedge.... \end{align*} In other words, (\ref{pf.glinf.glinfact.welldef.ass3.state}) holds. Now, forget that we fixed $a$ and $b_{0},b_{1},b_{2},...$. We thus have shown the following result: \[ \left( \begin{array} [c]{c}% \text{If }a\text{ is an element of }\mathfrak{gl}_{\infty}\text{, and }% b_{0},b_{1},b_{2},...\text{ are vectors in }V\text{ which satisfy (\ref{pf.glinf.glinfact.welldef.ass3.stand}),}\\ \text{and satisfy (\ref{pf.glinf.glinfact.welldef.ass3.hook}) for }% K=\kappa+1\text{, then (\ref{pf.glinf.glinfact.welldef.ass3.state}) holds}% \end{array} \right) . \] In other words, we have shown that Assertion \ref{prop.glinf.glinfact.welldef}% .3 holds for $K=\kappa+1$. This completes the inductive proof of Assertion \ref{prop.glinf.glinfact.welldef}.3. \textit{Proof of Assertion \ref{prop.glinf.glinfact.welldef}.2:} By the assumptions, we know that $b_{i}=v_{m-i}$ for all sufficiently large $i$. In other words, there exists a $K\in\mathbb{N}$ such that every $i\geq K$ satisfies $b_{i}=v_{m-i}$. Fix such a $K$. Then, every integer $i\geq K$ satisfies $b_{i}=v_{m-i}\in\left\{ v_{z}\ \mid\ z\in\mathbb{Z}\right\} $ (since $m-i\in\mathbb{Z}$). In other words, (\ref{pf.glinf.glinfact.welldef.ass3.hook}) is satisfied. Thus, we can apply Assertion \ref{prop.glinf.glinfact.welldef}.3, and conclude that \[ F_{a}\left( b_{0}\wedge b_{1}\wedge b_{2}\wedge...\right) =\sum \limits_{k\geq0}b_{0}\wedge b_{1}\wedge...\wedge b_{k-1}\wedge\left( a\rightharpoonup b_{k}\right) \wedge b_{k+1}\wedge b_{k+2}\wedge.... \] This proves Assertion \ref{prop.glinf.glinfact.welldef}.2. \end{verlong} Next, here's something obvious that we are going to use a few times in the proof: \begin{quote} \textit{Assertion \ref{prop.glinf.glinfact.welldef}.4:} Let $f$ and $g$ be two endomorphisms of the $\mathbb{C}$-vector space $\wedge^{\dfrac{\infty}{2},m}% V$. If every $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $ satisfies $f\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =g\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) $, then $f=g$. \end{quote} \begin{vershort} This follows from the fact that $\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) _{\left( i_{0},i_{1},i_{2},...\right) \text{ is an }m\text{-degression}}$ is a basis of the $\mathbb{C}$-vector space $\wedge^{\dfrac{\infty}{2},m}V$. \end{vershort} \begin{verlong} \textit{Proof of Assertion \ref{prop.glinf.glinfact.welldef}.4:} Assume that every $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $ satisfies $f\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =g\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) $. In other words, the two maps $f$ and $g$ are equal to each other on every element of the family $\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) _{\left( i_{0},i_{1},i_{2},...\right) \text{ is an }m\text{-degression}}$. Let us recall the following fact from linear algebra: If $\mathfrak{A}$ and $\mathfrak{B}$ are two $\mathbb{C}$-vector spaces, and $S$ is a basis of $\mathfrak{A}$, and $\mathfrak{f}:\mathfrak{A}\rightarrow\mathfrak{B}$ and $\mathfrak{g}:\mathfrak{A}\rightarrow\mathfrak{B}$ are two $\mathbb{C}$-linear maps such that $f$ and $g$ are equal to each other on every element of $S$, then $\mathfrak{f}=\mathfrak{g}$. Applying this fact to $\mathfrak{A}% =\wedge^{\dfrac{\infty}{2},m}V$, $\mathfrak{B}=\wedge^{\dfrac{\infty}{2},m}V$, $S=\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) _{\left( i_{0},i_{1},i_{2},...\right) \text{ is an }m\text{-degression}}$, $\mathfrak{f}=f$ and $\mathfrak{g}=g$, we conclude that $f=g$. This proves Assertion \ref{prop.glinf.glinfact.welldef}.4. \end{verlong} Next, we notice the following easy fact: \begin{quote} \textit{Assertion \ref{prop.glinf.glinfact.welldef}.5:} Let $a\in \mathfrak{gl}_{\infty}$ and $b\in\mathfrak{gl}_{\infty}$. Let $\lambda \in\mathbb{C}$ and $\mu\in\mathbb{C}$. Then, $\lambda F_{a}+\mu F_{b}% =F_{\lambda a+\mu b}$ in the Lie algebra $\mathfrak{gl}\left( \wedge ^{\dfrac{\infty}{2},m}V\right) $. \end{quote} \begin{vershort} This follows very quickly from the linearity of the definition of $F_{a}$ with respect to $a$ (the details are left to the reader). \end{vershort} \begin{verlong} \textit{Proof of Assertion \ref{prop.glinf.glinfact.welldef}.5:} Both maps $F_{a}$ and $F_{b}$ are $\mathbb{C}$-linear (by their definitions). Hence, the map $\lambda F_{a}+\mu F_{b}$ is $\mathbb{C}$-linear. On the other hand, the map $F_{\lambda a+\mu b}$ is $\mathbb{C}$-linear (by its definition). Now, every $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $ satisfies% \begin{align} & \left( \lambda F_{a}+\mu F_{b}\right) \left( v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...\right) \nonumber\\ & =\lambda\underbrace{F_{a}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}% }\wedge...\right) }_{\substack{=\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}% }\wedge...\wedge v_{i_{k-1}}\wedge\left( a\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\\text{(by the definition of }F_{a}\text{)}}}+\mu\underbrace{F_{b}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) }_{\substack{=\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}% }\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\\text{(by the definition of }F_{b}\text{)}}}\nonumber\\ & =\lambda\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( a\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ +\mu\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge... \label{pf.glinf.glinfact.welldef.ass4.1}% \end{align} and% \begin{align} & F_{\lambda a+\mu b}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right) \nonumber\\ & =\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge\left( \underbrace{\left( \lambda a+\mu b\right) \rightharpoonup v_{i_{k}}}_{=\lambda a\rightharpoonup v_{i_{k}}+\mu b\rightharpoonup v_{i_{k}% }}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }F_{\lambda a+\mu b}\right) \nonumber\\ & =\sum\limits_{k\geq0}\underbrace{v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( \lambda a\rightharpoonup v_{i_{k}}+\mu b\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}% \wedge...}_{\substack{=\lambda v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( a\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...+\mu v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\\\text{(by Proposition \ref{prop.semiinfwedge.welldef} \textbf{(c)})}}}\nonumber\\ & =\sum\limits_{k\geq0}\left( \lambda v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge\left( a\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\right. \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left. +\mu v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}% }\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\right) \nonumber\\ & =\lambda\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( a\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ +\mu\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge.... \label{pf.glinf.glinfact.welldef.ass4.2}% \end{align} Hence, every $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $ satisfies% \begin{align*} & \left( \lambda F_{a}+\mu F_{b}\right) \left( v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...\right) \\ & =\lambda\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( a\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ +\mu\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.glinf.glinfact.welldef.ass4.1})}\right) \\ & =F_{\lambda a+\mu b}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.glinf.glinfact.welldef.ass4.2})}\right) . \end{align*} Hence, Assertion \ref{prop.glinf.glinfact.welldef}.4 (applied to $f=\lambda F_{a}+\mu F_{b}$ and $g=F_{\lambda a+\mu b}$) yields that $\lambda F_{a}+\mu F_{b}=F_{\lambda a+\mu b}$. This proves Assertion \ref{prop.glinf.glinfact.welldef}.5. \end{verlong} Here is something rather simple: \begin{quote} \textit{Assertion \ref{prop.glinf.glinfact.welldef}.6:} Let $i\in\mathbb{Z}$ and $j\in\mathbb{Z}$. Let $m\in\mathbb{Z}$. Let $\left( i_{0},i_{1}% ,i_{2},...\right) $ be a straying $m$-degression. \textbf{(a)} For every $\ell\in\mathbb{N}$, the sequence $\left( i_{0}% ,i_{1},...,i_{\ell-1},i,i_{\ell+1},i_{\ell+2},...\right) $ is a straying $m$-degression. \textbf{(b)} If $j\notin\left\{ i_{0},i_{1},i_{2},...\right\} $, then $F_{E_{i,j}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =0$. \textbf{(c)} If there exists a \textbf{unique} $\ell\in\mathbb{N}$ such that $j=i_{\ell}$, then we have% \[ F_{E_{i,j}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{L-1}}\wedge v_{i}\wedge v_{i_{L+1}}\wedge v_{i_{L+2}}\wedge..., \] where $L$ is the unique $\ell\in\mathbb{N}$ such that $j=i_{\ell}$. \end{quote} The proof of Assertion \ref{prop.glinf.glinfact.welldef}.6 is as straightforward as one would expect: it is a matter of substituting $a=E_{i,j}$ and $b_{k}=v_{i_{k}}$ into Assertion \ref{prop.glinf.glinfact.welldef}.2 and taking care of the few addends which are not $0$. \begin{verlong} \textit{Proof of Assertion \ref{prop.glinf.glinfact.welldef}.6:} By the definition of $E_{i,j}$, we have% \begin{equation} E_{i,j}v_{u}=\delta_{j,u}v_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% u\in\mathbb{Z}. \label{pf.glinf.glinfact.welldef.ass6.pf.Eij}% \end{equation} We have% \begin{equation} v_{i_{i}}=v_{m-i}\ \ \ \ \ \ \ \ \ \ \text{for all sufficiently large }i. \label{pf.glinf.glinfact.welldef.ass6.pf.1}% \end{equation} \footnote{\textit{Proof of (\ref{pf.glinf.glinfact.welldef.ass6.pf.1}):} We know that $\left( i_{0},i_{1},i_{2},...\right) $ is a straying $m$-degression. By the definition of a straying $m$-degression, this rewrites as follows: Every sufficiently high $k\in\mathbb{N}$ satisfies $i_{k}+k=m$. In other words, every sufficiently high $k\in\mathbb{N}$ satisfies $i_{k}=m-k$. Hence, every sufficiently high $k\in\mathbb{N}$ satisfies $v_{i_{k}}=v_{m-k}$. Renaming the variable $k$ as $i$ in this result, we conclude: Every sufficiently high $i\in\mathbb{N}$ satisfies $v_{i_{i}}=v_{m-i}$. This proves (\ref{pf.glinf.glinfact.welldef.ass6.pf.1}).}. Hence, Assertion \ref{prop.glinf.glinfact.welldef}.2 (applied to $a=E_{i,j}$ and $b_{k}% =v_{i_{k}}$) yields% \begin{align} & F_{E_{i,j}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge ...\right) \nonumber\\ & =\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge\underbrace{\left( E_{i,j}\rightharpoonup v_{i_{k}}\right) }_{\substack{=E_{i,j}v_{i_{k}}=\delta_{j,i_{k}}v_{i}\\\text{(by (\ref{pf.glinf.glinfact.welldef.ass6.pf.Eij}), applied to }u=i_{k}\text{)}% }}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\nonumber\\ & =\sum\limits_{k\geq0}\underbrace{v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( \delta_{j,i_{k}}v_{i}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...}_{\substack{=\delta_{j,i_{k}}\cdot v_{i_{0}% }\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i}\wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\\\text{(since the infinite wedge product is multilinear)}}}\nonumber\\ & =\sum\limits_{k\geq0}\delta_{j,i_{k}}\cdot v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge v_{i}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}% }\wedge.... \label{pf.glinf.glinfact.welldef.ass6.pf.2}% \end{align} \textbf{(a)} Let $\ell\in\mathbb{N}$. We know that $\left( i_{0},i_{1}% ,i_{2},...\right) $ is a straying $m$-degression. By the definition of a straying $m$-degression, this rewrites as follows: Every sufficiently high $k\in\mathbb{N}$ satisfies $i_{k}+k=m$. In other words, there exists a $K\in\mathbb{N}$ such that every integer $k\geq K$ satisfies $i_{k}+k=m$. Consider this $k$. Define a sequence $\left( i_{0}^{\prime},i_{1}^{\prime},i_{2}^{\prime },...\right) $ of integers by% \[ \left( i_{k}^{\prime}=\left\{ \begin{array} [c]{c}% i_{k},\ \ \ \ \ \ \ \ \ \ \text{if }k\neq\ell;\\ i,\ \ \ \ \ \ \ \ \ \ \text{if }k=\ell \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N}\right) . \] Then, $\left( i_{0}^{\prime},i_{1}^{\prime},i_{2}^{\prime},...\right) =\left( i_{0},i_{1},...,i_{\ell-1},i,i_{\ell+1},i_{\ell+2},...\right) $. Now, let $k$ be any integer such that $k\geq\max\left\{ \ell+1,K\right\} $. Then, $k\geq\max\left\{ \ell+1,K\right\} \geq K$, so that $i_{k}+k=m$. And since $k\geq\max\left\{ \ell+1,K\right\} \geq\ell+1$, we have $k\in \mathbb{N}$ (since $\ell\in\mathbb{N}$). Moreover, since $k\geq\ell+1>\ell$, we have $k\neq\ell$. Now, by the definition of $i_{k}^{\prime}$, we have% \[ i_{k}^{\prime}=\left\{ \begin{array} [c]{c}% i_{k},\ \ \ \ \ \ \ \ \ \ \text{if }k\neq\ell;\\ i,\ \ \ \ \ \ \ \ \ \ \text{if }k=\ell \end{array} \right. =i_{k}\ \ \ \ \ \ \ \ \ \ \left( \text{since }k\neq\ell\right) , \] so that $i_{k}^{\prime}+k=i_{k}+k=m$. Now, forget that we fixed $k$. We thus have proven that every integer $k$ satisfying $k\geq\max\left\{ \ell+1,K\right\} $ satisfies $i_{k}^{\prime}+k=m$. In other words, the sequence $\left( i_{0}^{\prime},i_{1}^{\prime},i_{2}^{\prime},...\right) $ is a straying $m$-degression. Since $\left( i_{0}^{\prime},i_{1}^{\prime },i_{2}^{\prime},...\right) =\left( i_{0},i_{1},...,i_{\ell-1},i,i_{\ell +1},i_{\ell+2},...\right) $, this rewrites as follows: The sequence $\left( i_{0},i_{1},...,i_{\ell-1},i,i_{\ell+1},i_{\ell+2},...\right) $ is a straying $m$-degression. This proves Assertion \ref{prop.glinf.glinfact.welldef}.6 \textbf{(a)}. \textbf{(b)} Assume that $j\notin\left\{ i_{0},i_{1},i_{2},...\right\} $. Then, \begin{equation} \text{every }k\in\mathbb{N}\text{ satisfies }\delta_{j,i_{k}}=0 \label{pf.glinf.glinfact.welldef.ass6.pf.3}% \end{equation} \footnote{\textit{Proof of (\ref{pf.glinf.glinfact.welldef.ass6.pf.3}):} Let $k\in\mathbb{N}$. Then, $j\neq i_{k}$ (because otherwise, we would have $j=i_{k}\in\left\{ i_{0},i_{1},i_{2},...\right\} $, which contradicts $j\notin\left\{ i_{0},i_{1},i_{2},...\right\} $). Thus, $\delta_{j,i_{k}}% =0$. This proves (\ref{pf.glinf.glinfact.welldef.ass6.pf.3}).}. Now, (\ref{pf.glinf.glinfact.welldef.ass6.pf.2}) becomes% \begin{align*} & F_{E_{i,j}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge ...\right) \\ & =\sum\limits_{k\geq0}\underbrace{\delta_{j,i_{k}}}_{\substack{=0\\\text{(by (\ref{pf.glinf.glinfact.welldef.ass6.pf.3}))}}}\cdot v_{i_{0}}\wedge v_{i_{1}% }\wedge...\wedge v_{i_{k-1}}\wedge v_{i}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}% }\wedge...\\ & =\sum\limits_{k\geq0}0\cdot v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...=0. \end{align*} This proves Assertion \ref{prop.glinf.glinfact.welldef}.6 \textbf{(b)}. \textbf{(c)} Assume that there exists a \textbf{unique} $\ell\in\mathbb{N}$ such that $j=i_{\ell}$. Denote this $\ell$ by $L$. Recall that $L$ is an $\ell\in\mathbb{N}$ such that $j=i_{\ell}$. Hence, $j=i_{L}$. Recall that $L$ is a \textbf{unique} $\ell\in\mathbb{N}$ such that $j=i_{\ell }$. From the uniqueness in this statement, we conclude that there exists no $k\in\mathbb{N}$ satisfying $j=i_{k}$ and $k\neq L$. In other words, no $k\in\mathbb{N}$ satisfying $k\neq L$ can satisfy $j=i_{k}$. In other words, every $k\in\mathbb{N}$ satisfying $k\neq L$ satisfies $j\neq i_{k}$. Hence, \begin{equation} \text{every }k\in\mathbb{N}\text{ satisfying }k\neq L\text{ satisfies }% \delta_{j,i_{k}}=0. \label{pf.glinf.glinfact.welldef.ass6.pf.4}% \end{equation} Now, (\ref{pf.glinf.glinfact.welldef.ass6.pf.2}) becomes% \begin{align*} & F_{E_{i,j}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge ...\right) \\ & =\sum\limits_{k\geq0}\delta_{j,i_{k}}\cdot v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge v_{i}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}% }\wedge...\\ & =\underbrace{\delta_{j,i_{L}}}_{\substack{=1\\\text{(since }j=i_{L}% \text{)}}}\cdot v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{L-1}}\wedge v_{i}\wedge v_{i_{L+1}}\wedge v_{i_{L+2}}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{k\geq0;\\k\neq L}% }\underbrace{\delta_{j,i_{k}}}_{\substack{=0\\\text{(by (\ref{pf.glinf.glinfact.welldef.ass6.pf.4}))}}}\cdot v_{i_{0}}\wedge v_{i_{1}% }\wedge...\wedge v_{i_{k-1}}\wedge v_{i}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}% }\wedge...\\ & =\underbrace{1\cdot v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{L-1}% }\wedge v_{i}\wedge v_{i_{L+1}}\wedge v_{i_{L+2}}\wedge...}_{=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{L-1}}\wedge v_{i}\wedge v_{i_{L+1}}\wedge v_{i_{L+2}}\wedge...}\\ & \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\substack{k\geq0;\\k\neq L}}0\cdot v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...}_{=0}\\ & =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{L-1}}\wedge v_{i}\wedge v_{i_{L+1}}\wedge v_{i_{L+2}}\wedge.... \end{align*} This proves Assertion \ref{prop.glinf.glinfact.welldef}.6 \textbf{(c)}. \end{verlong} Now here is something less obvious: \begin{quote} \textit{Assertion \ref{prop.glinf.glinfact.welldef}.7:} Every $a\in \mathfrak{gl}_{\infty}$ and $b\in\mathfrak{gl}_{\infty}$ satisfy $\left[ F_{a},F_{b}\right] =F_{\left[ a,b\right] }$ in the Lie algebra $\mathfrak{gl}\left( \wedge^{\dfrac{\infty}{2},m}V\right) $. \end{quote} There are two possible approaches to proving Assertion \ref{prop.glinf.glinfact.welldef}.7. \begin{landscape} \textit{First proof of Assertion \ref{prop.glinf.glinfact.welldef}.7 (sketched):} In order to prove Assertion \ref{prop.glinf.glinfact.welldef}.7, it is enough to show that% \begin{equation} \left[ F_{a},F_{b}\right] \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}% }\wedge...\right) =F_{\left[ a,b\right] }\left( v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...\right) \label{pf.glinf.glinfact.welldef.ass7.pf.short.2.goal}% \end{equation} for every $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $. (Indeed, once this is done, $\left[ F_{a},F_{b}\right] =F_{\left[ a,b\right] }$ will follow from Assertion \ref{prop.glinf.glinfact.welldef}.4.) So let $\left( i_{0},i_{1},i_{2},...\right) $ be any $m$-degression. Then,% \begin{align*} & F_{a}\left( F_{b}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right) \right) \\ & =F_{a}\left( \sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }F_{b}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right)\text{ is defined as } \sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge... \right) \\ & =\sum\limits_{k\geq0}F_{a}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\right) \\ & =\sum\limits_{q\geq0}\underbrace{F_{a}\left( v_{i_{0}}\wedge v_{i_{1}% }\wedge...\wedge v_{i_{q-1}}\wedge\left( b\rightharpoonup v_{i_{q}}\right) \wedge v_{i_{q+1}}\wedge v_{i_{q+2}}\wedge...\right) }_{\substack{=\sum \limits_{k\geq0}\left\{ \begin{array} [c]{l}% v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}% \wedge...\wedge v_{i_{q-1}}\wedge\left( a\rightharpoonup v_{i_{q}}\right) \wedge v_{i_{q+1}}\wedge v_{i_{q+2}},\ \ \ \ \ \ \ \ \ \ \text{if }kq \end{array} \right. \\\text{(by an application of Assertion \ref{prop.glinf.glinfact.welldef}.2)}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we renamed the summation index }k\text{ as }q\right) \\ & =\sum\limits_{q\geq0}\sum\limits_{k\geq0}\left\{ \begin{array} [c]{l}% v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}% \wedge...\wedge v_{i_{q-1}}\wedge\left( a\rightharpoonup v_{i_{q}}\right) \wedge v_{i_{q+1}}\wedge v_{i_{q+2}}\wedge...,\ \ \ \ \ \ \ \ \ \ \text{if }kq \end{array} \right. \end{align*}% \begin{align} & =\sum\limits_{q\geq0}\sum\limits_{\substack{k\geq0;\\kq}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{q-1}}\wedge\left( a\rightharpoonup v_{i_{q}}\right) \wedge v_{i_{q+1}}\wedge v_{i_{q+2}}% \wedge...\wedge v_{i_{k-1}}\wedge\left( b\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\nonumber\\ & =\sum\limits_{q\geq0}\sum\limits_{\substack{p\geq0;\\pp}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{p-1}}\wedge\left( a\rightharpoonup v_{i_{p}}\right) \wedge v_{i_{p+1}}\wedge v_{i_{p+2}}% \wedge...\wedge v_{i_{q-1}}\wedge\left( b\rightharpoonup v_{i_{q}}\right) \wedge v_{i_{q+1}}\wedge v_{i_{q+2}}\wedge ...\label{pf.glinf.glinfact.welldef.ass7.pf.short.1}% \end{align} \footnote{In the last step of this computation, we did the following substitutions: \par -- We renamed the index $k$ as $p$ in the second sum. \par -- We renamed the index $q$ as $k$ in the third sum. \par -- We switched the meanings of the indices $p$ and $q$ in the fourth and fifth sums.}. Similarly,% \begin{align} & F_{b}\left( F_{a}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right) \right) \nonumber\\ & =\sum\limits_{q\geq0}\sum\limits_{\substack{p\geq0;\\pp}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{p-1}}\wedge\left( b\rightharpoonup v_{i_{p}}\right) \wedge v_{i_{p+1}}\wedge v_{i_{p+2}}% \wedge...\wedge v_{i_{q-1}}\wedge\left( a\rightharpoonup v_{i_{q}}\right) \wedge v_{i_{q+1}}\wedge v_{i_{q+2}}\wedge ....\label{pf.glinf.glinfact.welldef.ass7.pf.short.2}% \end{align} \end{landscape} Now, let us subtract (\ref{pf.glinf.glinfact.welldef.ass7.pf.short.2}) from (\ref{pf.glinf.glinfact.welldef.ass7.pf.short.1}). I am claiming that the first term on the right hand side of (\ref{pf.glinf.glinfact.welldef.ass7.pf.short.2}) cancels against the third term on the right hand side of (\ref{pf.glinf.glinfact.welldef.ass7.pf.short.1}). Indeed, in order to see this, one needs to check that one can interchange the order of summation in the sum% \[ \sum\limits_{q\geq0}\sum\limits_{\substack{p\geq0;\\pp}}$. This is easy to see (indeed, one must show that \newline$v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{p-1}}\wedge\left( b\rightharpoonup v_{i_{p}}\right) \wedge v_{i_{p+1}}\wedge v_{i_{p+2}}\wedge...\wedge v_{i_{q-1}}\wedge\left( a\rightharpoonup v_{i_{q}}\right) \wedge v_{i_{q+1}}\wedge v_{i_{q+2}}% \wedge...=0$ for all but finitely many \textbf{pairs} $\left( i,j\right) \in\mathbb{N}^{2}$), but not trivial a priori\footnote{Here is a cautionary tale on why one cannot always interchange summation in infinite sums. Define a family $\left( \alpha_{p,q}\right) _{\left( p,q\right) \in\mathbb{N}^{2}}$ of integers by $\alpha_{p,q}=\left\{ \begin{array} [c]{c}% 1,\text{ if }p=q;\\ -1,\text{ if }p=q+1 \end{array} \right. $. Then, every $q\in\mathbb{N}$ satisfies $\sum\limits_{p\geq0}% \alpha_{p,q}=0$. Hence, $\sum\limits_{q\geq0}\sum\limits_{p\geq0}\alpha _{p,q}=0$. On the other hand, every $p\in\mathbb{N}$ satisfies $\sum \limits_{q\geq0}\alpha_{p,q}=\delta_{p,0}$. Hence, $\sum\limits_{p\geq0}% \sum\limits_{q\geq0}\alpha_{p,q}=1\neq0=\sum\limits_{q\geq0}\sum \limits_{p\geq0}\alpha_{p,q}$. So the two summation signs in this situation cannot be interchanged, even though all sums (both inner and outer) converge in the discrete topology. Generally, for a family $\left( \lambda _{p,q}\right) _{\left( p,q\right) \in\mathbb{N}^{2}}$ of elements of an additive group, we are guaranteed to have $\sum\limits_{p\geq0}\sum \limits_{q\geq0}\lambda_{p,q}=\sum\limits_{q\geq0}\sum\limits_{p\geq0}% \lambda_{p,q}$ if the \textbf{double sum }$\sum\limits_{\left( p,q\right) \in\mathbb{N}^{2}}\lambda_{p,q}$ still converges in the discrete topology (this is analogous to Fubini's theorem). But the double sum $\sum \limits_{\left( p,q\right) \in\mathbb{N}^{2}}\alpha_{p,q}$ does not converge in the discrete topology, so $\sum\limits_{p\geq0}\sum\limits_{q\geq0}% \alpha_{p,q}\neq\sum\limits_{q\geq0}\sum\limits_{p\geq0}\alpha_{p,q}$ should not come as a surprise.}. So we know that the first term on the right hand side of (\ref{pf.glinf.glinfact.welldef.ass7.pf.short.2}) cancels against the third term on the right hand side of (\ref{pf.glinf.glinfact.welldef.ass7.pf.short.1}). Similarly, the third term on the right hand side of (\ref{pf.glinf.glinfact.welldef.ass7.pf.short.2}) cancels against the first term on the right hand side of (\ref{pf.glinf.glinfact.welldef.ass7.pf.short.1}). Thus, when we subtract (\ref{pf.glinf.glinfact.welldef.ass7.pf.short.2}) from (\ref{pf.glinf.glinfact.welldef.ass7.pf.short.1}), on the right hand side only the second terms of both equations remain, and we obtain% \begin{align*} & F_{a}\left( F_{b}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right) \right) -F_{b}\left( F_{a}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \right) \\ & =\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge\left( \left( ab\right) \rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ -\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge\left( \left( ba\right) \rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & =\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge\left( \left( ab-ba\right) \rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the multilinearity of the infinite wedge product}\right) \\ & =F_{ab-ba}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge ...\right) =F_{\left[ a,b\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) . \end{align*} This proves (\ref{pf.glinf.glinfact.welldef.ass7.pf.short.2.goal}), and thus Assertion \ref{prop.glinf.glinfact.welldef}.7. Filling the details of this proof is left to the reader. \begin{vershort} \textit{Second proof of Assertion \ref{prop.glinf.glinfact.welldef}.7 (sketched):} Due to Assertion \ref{prop.glinf.glinfact.welldef}.5, the value of $F_{c}$ for $c\in\mathfrak{gl}_{\infty}$ depends $\mathbb{C}$-linearly on $c$. But we must prove the equality $\left[ F_{a},F_{b}\right] =F_{\left[ a,b\right] }$ for all $a\in\mathfrak{gl}_{\infty}$ and $b\in\mathfrak{gl}% _{\infty}$. This equality is $\mathbb{C}$-linear in $a$ and $b$ (since the value of $F_{c}$ for $c\in\mathfrak{gl}_{\infty}$ depends $\mathbb{C}% $-linearly on $c$), so it is enough to show it only when $a$ and $b$ belong to the basis $\left( E_{i,j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}$ of $\mathfrak{gl}_{\infty}$. But in this case, one can check this equality by verifying that every $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $ satisfies% \[ \left[ F_{a},F_{b}\right] \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}% }\wedge...\right) =F_{\left[ a,b\right] }\left( v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...\right) . \] This can be done (using Assertion \ref{prop.glinf.glinfact.welldef}.6) by a straightforward distinction of cases (the cases depend on whether some indices belong to $\left\{ i_{0},i_{1},i_{2},...\right\} $ or not, and whether some indices are equal or not). The reader should not have much of a trouble supplying these arguments, but they are as unenlightening as one would expect. There \textbf{is} a somewhat better way to do this verification (better in the sense that less cases have to be considered) by means of exploiting some symmetry; this relies on checking the following assertion: \end{vershort} \begin{verlong} We will soon show a second proof of Assertion \ref{prop.glinf.glinfact.welldef}.7 in more detail. This proof relies substantially on the following assertion: \end{verlong} \begin{quote} \textit{Assertion \ref{prop.glinf.glinfact.welldef}.8:} Let $r$, $s$, $u$ and $v$ be integers. Let $m\in\mathbb{Z}$. Let $\left( i_{0},i_{1},i_{2}% ,...\right) $ be an $m$-degression. Let $I$ denote the set $\left\{ i_{0},i_{1},i_{2},...\right\} $. \textbf{(a)} If $v\notin I$, then \[ \left( F_{E_{r,s}}F_{E_{u,v}}-\delta_{s,u}F_{E_{r,v}}\right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =0. \] \textbf{(b)} If $s=v$, then \[ \left( F_{E_{r,s}}F_{E_{u,v}}-\delta_{s,u}F_{E_{r,v}}\right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =0. \] \textbf{(c)} Assume that $s\neq v$. Let $\mathbf{w}:\mathbb{Z}\rightarrow \mathbb{Z}$ be the function defined by \[ \left( \mathbf{w}\left( k\right) =\left\{ \begin{array} [c]{c}% r,\ \ \ \ \ \ \ \ \ \ \text{if }k=s;\\ u,\ \ \ \ \ \ \ \ \ \ \text{if }k=v;\\ k,\ \ \ \ \ \ \ \ \ \ \text{otherwise}% \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for all }k\in\mathbb{Z}\right) . \] \footnote{Here, the term $\left\{ \begin{array} [c]{c}% r,\ \ \ \ \ \ \ \ \ \ \text{if }k=s;\\ u,\ \ \ \ \ \ \ \ \ \ \text{if }k=v;\\ k,\ \ \ \ \ \ \ \ \ \ \text{otherwise}% \end{array} \right. $ makes sense, since $s\neq v$.} Then, $\left( \mathbf{w}\left( i_{0}\right) ,\mathbf{w}\left( i_{1}\right) ,\mathbf{w}\left( i_{2}\right) ,...\right) $ is a straying $m$-degression, and satisfies \begin{align*} & \left( F_{E_{r,s}}F_{E_{u,v}}-\delta_{s,u}F_{E_{r,v}}\right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\ & =\left[ s\in I\right] \cdot\left[ v\in I\right] \cdot v_{\mathbf{w}% \left( i_{0}\right) }\wedge v_{\mathbf{w}\left( i_{1}\right) }\wedge v_{\mathbf{w}\left( i_{2}\right) }\wedge.... \end{align*} Here, whenever $\mathcal{A}$ is an assertion, we denote by $\left[ \mathcal{A}\right] $ the integer $\left\{ \begin{array} [c]{l}% 1,\text{ if }\mathcal{A}\text{ is true;}\\ 0,\text{ if }\mathcal{A}\text{ is wrong}% \end{array} \right. $. \end{quote} \begin{vershort} The proof of this assertion, as well as the derivation of Assertion \ref{prop.glinf.glinfact.welldef}.7 from it (Assertion \ref{prop.glinf.glinfact.welldef}.8 must be applied twice), is left to the reader. \end{vershort} \begin{verlong} \textit{Proof of Assertion \ref{prop.glinf.glinfact.welldef}.8:} We know that $\left( i_{0},i_{1},i_{2},...\right) $ is an $m$-degression, hence a strictly decreasing straying $m$-degression (since the $m$-degressions are exactly the strictly decreasing straying $m$-degressions). Since $\left( i_{0},i_{1},i_{2},...\right) $ is strictly decreasing, the numbers $i_{0}$, $i_{1}$, $i_{2}$, $...$ are pairwise distinct. Hence,% \begin{equation} \left\{ i_{0},i_{1},...,i_{w-1},i_{w+1},i_{w+2},...\right\} =I\setminus \left\{ i_{w}\right\} \ \ \ \ \ \ \ \ \ \ \text{for every }w\in\mathbb{N}. \label{pf.glinf.glinfact.welldef.8.pf.allbutone}% \end{equation} \footnote{\textit{Proof of (\ref{pf.glinf.glinfact.welldef.8.pf.allbutone}):} Let $w\in\mathbb{N}$. We have $i_{w}\neq i_{p}$ for every $p\in\mathbb{N}$ satisfying $w\neq p$ (since the numbers $i_{0}$, $i_{1}$, $i_{2}$, $...$ are pairwise distinct). Thus,% \begin{align*} i_{w} & \notin\left\{ i_{p}\ \mid\ \underbrace{p\in\mathbb{N};\ w\neq p}_{\substack{\text{this is equivalent to}\\p\in\left\{ q\in\mathbb{N}% \ \mid\ w\neq q\right\} }}\right\} =\left\{ i_{p}\ \mid\ p\in \underbrace{\left\{ q\in\mathbb{N}\ \mid\ w\neq q\right\} }_{=\left\{ 0,1,...,w-1,w+1,w+2,...\right\} }\right\} \\ & =\left\{ i_{p}\ \mid\ p\in\left\{ 0,1,...,w-1,w+1,w+2,...\right\} \right\} =\left\{ i_{0},i_{1},...,i_{w-1},i_{w+1},i_{w+2},...\right\} . \end{align*} Combined with $\left\{ i_{0},i_{1},...,i_{w-1},i_{w+1},i_{w+2},...\right\} \subseteq\left\{ i_{0},i_{1},i_{2},...\right\} $, this yields that \[ \left\{ i_{0},i_{1},...,i_{w-1},i_{w+1},i_{w+2},...\right\} \subseteq \left\{ i_{0},i_{1},i_{2},...\right\} \setminus\left\{ i_{w}\right\} . \] Combined with% \begin{align*} & \underbrace{\left\{ i_{0},i_{1},i_{2},...\right\} }_{\substack{=\left\{ i_{0},i_{1},...,i_{w-1},i_{w},i_{w+1},i_{w+2},...\right\} \\=\left\{ i_{0},i_{1},...,i_{w-1},i_{w+1},i_{w+2},...\right\} \cup\left\{ i_{w}\right\} }}\setminus\left\{ i_{w}\right\} \\ & =\left( \left\{ i_{0},i_{1},...,i_{w-1},i_{w+1},i_{w+2},...\right\} \cup\left\{ i_{w}\right\} \right) \setminus\left\{ i_{w}\right\} \\ & =\underbrace{\left( \left\{ i_{0},i_{1},...,i_{w-1},i_{w+1}% ,i_{w+2},...\right\} \setminus\left\{ i_{w}\right\} \right) }% _{\subseteq\left\{ i_{0},i_{1},...,i_{w-1},i_{w+1},i_{w+2},...\right\} }% \cup\underbrace{\left( \left\{ i_{w}\right\} \setminus\left\{ i_{w}\right\} \right) }_{=\varnothing}\\ & \subseteq\left\{ i_{0},i_{1},...,i_{w-1},i_{w+1},i_{w+2},...\right\} \cup\varnothing=\left\{ i_{0},i_{1},...,i_{w-1},i_{w+1},i_{w+2},...\right\} , \end{align*} this yields $\left\{ i_{0},i_{1},...,i_{w-1},i_{w+1},i_{w+2},...\right\} =\left\{ i_{0},i_{1},i_{2},...\right\} \setminus\left\{ i_{w}\right\} $. Since $\left\{ i_{0},i_{1},i_{2},...\right\} =I$, this rewrites as $\left\{ i_{0},i_{1},...,i_{w-1},i_{w+1},i_{w+2},...\right\} =I\setminus\left\{ i_{w}\right\} $. Thus, (\ref{pf.glinf.glinfact.welldef.8.pf.allbutone}) is proven.} \textbf{(a)} Assume that $v\notin I$. Thus, $v\notin I=\left\{ i_{0}% ,i_{1},i_{2},...\right\} $. Thus, $F_{E_{u,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =0$ (by Assertion \ref{prop.glinf.glinfact.welldef}.6 \textbf{(b)}, applied to $i=u$ and $j=v$) and $F_{E_{r,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right) =0$ (by Assertion \ref{prop.glinf.glinfact.welldef}.6 \textbf{(b)}, applied to $i=r$ and $j=v$). Hence,% \begin{align*} & \left( F_{E_{r,s}}F_{E_{u,v}}-\delta_{s,u}F_{E_{r,v}}\right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\ & =F_{E_{r,s}}\left( \underbrace{F_{E_{u,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) }_{=0}\right) -\delta _{s,u}\underbrace{F_{E_{r,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}% }\wedge...\right) }_{=0}\\ & =\underbrace{F_{E_{r,s}}\left( 0\right) }_{\substack{=0\\\text{(since }F_{E_{r,s}}\text{ is linear)}}}-\underbrace{\delta_{s,u}0}_{=0}=0-0=0. \end{align*} This proves Assertion \ref{prop.glinf.glinfact.welldef}.8 \textbf{(a)}. \textbf{(b)} Assume that $s=v$. If $v\notin I$, then Assertion \ref{prop.glinf.glinfact.welldef}.8 \textbf{(b)} is obviously true\footnote{\textit{Proof.} Assume that $v\notin I$. Then, due to Assertion \ref{prop.glinf.glinfact.welldef}.8 \textbf{(a)}, we have $\left( F_{E_{r,s}% }F_{E_{u,v}}-\delta_{s,u}F_{E_{r,v}}\right) \left( v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...\right) =0$. In other words, Assertion \ref{prop.glinf.glinfact.welldef}.8 \textbf{(b)} is true, qed.}. Hence, for the rest of the proof of Assertion \ref{prop.glinf.glinfact.welldef}.8 \textbf{(b)}, we can WLOG assume that we don't have $v\notin I$. Assume this. So we have $v\in I$ (since we don't have $v\notin I$). Thus, there exists a \textbf{unique} $\ell\in\mathbb{N}$ such that $v=i_{\ell}$% .\ \ \ \ \footnote{\textit{Proof:} Since $v\in I=\left\{ i_{0},i_{1}% ,i_{2},...\right\} $, there exists at least one $\ell\in\mathbb{N}$ such that $v=i_{\ell}$. \par Now, let $\ell_{1}$ and $\ell_{2}$ be two elements $\ell$ of $\mathbb{N}$ such that $v=i_{\ell}$. Then, $v=i_{\ell_{1}}$ (since $\ell_{1}$ is an element $\ell$ of $\mathbb{N}$ such that $v=i_{\ell}$) and $v=i_{\ell_{2}}$ (similarly). Hence, $i_{\ell_{1}}=v=i_{\ell_{2}}$. If we had $\ell_{1}\neq \ell_{2}$, then we would have $i_{\ell_{1}}\neq i_{\ell_{2}}$ (since the numbers $i_{0}$, $i_{1}$, $i_{2}$, $...$ are pairwise distinct), which would contradict $i_{\ell_{1}}=i_{\ell_{2}}$. Thus, we cannot have $\ell_{1}\neq \ell_{2}$. Hence, $\ell_{1}=\ell_{2}$. \par Now, forget that we fixed $\ell_{1}$ and $\ell_{2}$. We thus have proven that if $\ell_{1}$ and $\ell_{2}$ be two elements $\ell$ of $\mathbb{N}$ such that $v=i_{\ell}$, then $\ell_{1}=\ell_{2}$. In other words, there exists at most one $\ell\in\mathbb{N}$ such that $v=i_{\ell}$. Hence, there exists a \textbf{unique} $\ell\in\mathbb{N}$ such that $v=i_{\ell}$ (since we also know that there exists at least one $\ell\in\mathbb{N}$ such that $v=i_{\ell}$), qed.} Denote this $\ell$ by $L$. Then, $v=i_{L}$ (since $L$ is an $\ell \in\mathbb{N}$ such that $v=i_{\ell}$). Now, define a sequence $\left( i_{0}^{\prime},i_{1}^{\prime},i_{2}^{\prime},...\right) $ by% \[ \left( i_{k}^{\prime}=\left\{ \begin{array} [c]{c}% i_{k},\ \ \ \ \ \ \ \ \ \ \text{if }k\neq L;\\ u,\ \ \ \ \ \ \ \ \ \ \text{if }k=L \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N}\right) . \] Then, $\left( i_{0}^{\prime},i_{1}^{\prime},i_{2}^{\prime},...\right) =\left( i_{0},i_{1},...,i_{L-1},u,i_{L+1},i_{L+2},...\right) $. Thus, \begin{align} \left\{ i_{0}^{\prime},i_{1}^{\prime},i_{2}^{\prime},...\right\} & =\left\{ i_{0},i_{1},...,i_{L-1},u,i_{L+1},i_{L+2},...\right\} \nonumber\\ & =\left\{ u\right\} \cup\underbrace{\left\{ i_{0},i_{1},...,i_{L-1}% ,i_{L+1},i_{L+2},...\right\} }_{\substack{=I\setminus\left\{ i_{L}\right\} \\\text{(by (\ref{pf.glinf.glinfact.welldef.8.pf.allbutone}), applied to }w=L\text{)}}}\nonumber\\ & =\left\{ u\right\} \cup\left( I\setminus\left\{ \underbrace{i_{L}}% _{=v}\right\} \right) =\left\{ u\right\} \cup\left( I\setminus\left\{ v\right\} \right) . \label{pf.glinf.glinfact.welldef.8.pf.b.1}% \end{align} By the definition of $i_{L}^{\prime}$, we have $i_{L}^{\prime}=\left\{ \begin{array} [c]{c}% i_{L},\ \ \ \ \ \ \ \ \ \ \text{if }L\neq L;\\ u,\ \ \ \ \ \ \ \ \ \ \text{if }L=L \end{array} \right. =u$ (since $L=L$). On the other hand, $L$ is the \textbf{unique} $\ell\in\mathbb{N}$ such that $v=i_{\ell}$. Hence, Assertion \ref{prop.glinf.glinfact.welldef}.6 \textbf{(c)} (applied to $u$ and $v$ instead of $i$ and $j$) yields \begin{align} F_{E_{u,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) & =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{L-1}}\wedge v_{u}\wedge v_{i_{L+1}}\wedge v_{i_{L+2}}\wedge...\nonumber\\ & =v_{i_{0}^{\prime}}\wedge v_{i_{1}^{\prime}}\wedge v_{i_{2}^{\prime}}% \wedge...\label{pf.glinf.glinfact.welldef.8.pf.b.2}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{0},i_{1},...,i_{L-1}% ,u,i_{L+1},i_{L+2},...\right) =\left( i_{0}^{\prime},i_{1}^{\prime}% ,i_{2}^{\prime},...\right) \right) .\nonumber \end{align} Moreover, $\left( i_{0},i_{1},...,i_{L-1},u,i_{L+1},i_{L+2},...\right) $ is a straying $m$-degression (by Assertion \ref{prop.glinf.glinfact.welldef}.6 \textbf{(a)}, applied to $u$, $u$ and $L$ instead of $i$, $j$ and $\ell$). In other words, $\left( i_{0}^{\prime},i_{1}^{\prime},i_{2}^{\prime},...\right) $ is a straying $m$-degression (since $\left( i_{0}^{\prime},i_{1}^{\prime },i_{2}^{\prime},...\right) =\left( i_{0},i_{1},...,i_{L-1},u,i_{L+1}% ,i_{L+2},...\right) $). We now distinguish between two cases: \textit{Case 1:} We have $s=u$. \textit{Case 2:} We don't have $s=u$. Let us first consider Case 1. In this case, $s=u$. Hence, $s=u=i_{L}^{\prime}% $. Thus, \begin{equation} \text{there exists at least one }\ell\in\mathbb{N}\text{ satisfying }% s=i_{\ell}^{\prime} \label{pf.glinf.glinfact.welldef.8.pf.b.case1.1}% \end{equation} (namely, $\ell=L$). But there exists at most one $\ell\in\mathbb{N}$ satisfying $s=i_{\ell}^{\prime}$\ \ \ \ \footnote{\textit{Proof.} Let $\ell \in\mathbb{N}$ satisfy $s=i_{\ell}^{\prime}$. We will prove that $\ell=L$. \par Assume (for the sake of contradiction) that $\ell\neq L$. Then, by the definition of $i_{\ell}^{\prime}$, we have $i_{\ell}^{\prime}=\left\{ \begin{array} [c]{c}% i_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }\ell\neq L;\\ u,\ \ \ \ \ \ \ \ \ \ \text{if }\ell=L \end{array} \right. =i_{\ell}$ (since $\ell\neq L$). But since $\ell\in\mathbb{N}$ and $\ell\neq L$, we have $\ell\in\mathbb{N}\setminus\left\{ L\right\} =\left\{ 0,1,...,L-1,L+1,L+2,...\right\} $, so that $i_{L}\in\left\{ i_{0}% ,i_{1},...,i_{L-1},i_{L+1},i_{L+2},...\right\} =I\setminus\left\{ i_{L}\right\} $ (by (\ref{pf.glinf.glinfact.welldef.8.pf.allbutone}), applied to $w=L$). Thus, $s=i_{\ell}^{\prime}=i_{\ell}\in I\setminus\left\{ i_{L}\right\} =I\setminus\left\{ v\right\} $ (since $i_{L}=v$). Consequently, $s\neq v$. This contradicts $s=v$. This contradiction shows that our assumption (that $\ell\neq L$) was wrong. Hence, $\ell=L$. \par Now forget that we fixed $\ell$. We thus have shown that every $\ell \in\mathbb{N}$ satisfying $s=i_{\ell}^{\prime}$ must satisfy $\ell=L$. In other words, every $\ell\in\mathbb{N}$ satisfying $s=i_{\ell}^{\prime}$ must equal $L$. Hence, there exists at most one $\ell\in\mathbb{N}$ satisfying $s=i_{\ell}^{\prime}$, qed.}. Combined with (\ref{pf.glinf.glinfact.welldef.8.pf.b.case1.1}), this yields that there exists a \textbf{unique }$\ell\in\mathbb{N}$ such that $s=i_{\ell}^{\prime}$. This $\ell$ is $L$ (because $s=i_{L}^{\prime}$). Therefore, we can apply Assertion \ref{prop.glinf.glinfact.welldef}.6 \textbf{(c)} to $i_{k}^{\prime}% $, $r$ and $s$ instead of $i_{k}$, $i$ and $j$. As a result, we obtain% \[ F_{E_{r,s}}\left( v_{i_{0}^{\prime}}\wedge v_{i_{1}^{\prime}}\wedge v_{i_{2}^{\prime}}\wedge...\right) =v_{i_{0}^{\prime}}\wedge v_{i_{1}% ^{\prime}}\wedge...\wedge v_{i_{L-1}^{\prime}}\wedge v_{r}\wedge v_{i_{L+1}^{\prime}}\wedge v_{i_{L+2}^{\prime}}\wedge.... \] Thus,% \begin{align} & v_{i_{0}^{\prime}}\wedge v_{i_{1}^{\prime}}\wedge...\wedge v_{i_{L-1}% ^{\prime}}\wedge v_{r}\wedge v_{i_{L+1}^{\prime}}\wedge v_{i_{L+2}^{\prime}% }\wedge...\nonumber\\ & =F_{E_{r,s}}\left( \underbrace{v_{i_{0}^{\prime}}\wedge v_{i_{1}^{\prime}% }\wedge v_{i_{2}^{\prime}}\wedge...}_{\substack{=F_{E_{u,v}}\left( v_{i_{0}% }\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\\text{(by (\ref{pf.glinf.glinfact.welldef.8.pf.b.2}))}}}\right) \nonumber\\ & =F_{E_{r,s}}\left( F_{E_{u,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \right) . \label{pf.glinf.glinfact.welldef.8.pf.b.case1.3}% \end{align} On the other hand, define a sequence $\left( i_{0}^{\prime\prime}% ,i_{1}^{\prime\prime},i_{2}^{\prime\prime},...\right) $ by% \[ \left( i_{k}^{\prime\prime}=\left\{ \begin{array} [c]{c}% i_{k},\ \ \ \ \ \ \ \ \ \ \text{if }k\neq L;\\ r,\ \ \ \ \ \ \ \ \ \ \text{if }k=L \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N}\right) . \] Then, $\left( i_{0}^{\prime\prime},i_{1}^{\prime\prime},i_{2}^{\prime\prime },...\right) =\left( i_{0},i_{1},...,i_{L-1},r,i_{L+1},i_{L+2},...\right) $. By the definition of $i_{L}^{\prime\prime}$, we have $i_{L}^{\prime\prime }=\left\{ \begin{array} [c]{c}% i_{L},\ \ \ \ \ \ \ \ \ \ \text{if }L\neq L;\\ r,\ \ \ \ \ \ \ \ \ \ \text{if }L=L \end{array} \right. =r$ (since $L=L$). But we have% \begin{equation} i_{k}^{\prime\prime}=\left\{ \begin{array} [c]{c}% i_{k}^{\prime},\ \ \ \ \ \ \ \ \ \ \text{if }k\neq L;\\ r,\ \ \ \ \ \ \ \ \ \ \text{if }k=L \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N} \label{pf.glinf.glinfact.welldef.8.pf.b.case1.5}% \end{equation} \footnote{\textit{Proof of (\ref{pf.glinf.glinfact.welldef.8.pf.b.case1.5}):} Let $k\in\mathbb{N}$. \par Notice that $\left\{ \begin{array} [c]{c}% i_{L}^{\prime},\ \ \ \ \ \ \ \ \ \ \text{if }L\neq L;\\ r,\ \ \ \ \ \ \ \ \ \ \text{if }L=L \end{array} \right. =r$ (since $L=L$). Thus, $i_{L}^{\prime\prime}=r=\left\{ \begin{array} [c]{c}% i_{L}^{\prime},\ \ \ \ \ \ \ \ \ \ \text{if }L\neq L;\\ r,\ \ \ \ \ \ \ \ \ \ \text{if }L=L \end{array} \right. $. In other words, (\ref{pf.glinf.glinfact.welldef.8.pf.b.case1.5}) holds in the case when $k=L$. Hence, for the rest of the proof of (\ref{pf.glinf.glinfact.welldef.8.pf.b.case1.5}), we can WLOG assume that $k=L$ does not hold. Let us assume this. \par We have assumed that $k=L$ does not hold. In other words, $k\neq L$. By the definition of $i_{k}^{\prime\prime}$, we have $i_{k}^{\prime\prime}=\left\{ \begin{array} [c]{c}% i_{k},\ \ \ \ \ \ \ \ \ \ \text{if }k\neq L;\\ r,\ \ \ \ \ \ \ \ \ \ \text{if }k=L \end{array} \right. =i_{k}$ (since $k\neq L$). But% \begin{align*} \left\{ \begin{array} [c]{c}% i_{k}^{\prime},\ \ \ \ \ \ \ \ \ \ \text{if }k\neq L;\\ r,\ \ \ \ \ \ \ \ \ \ \text{if }k=L \end{array} \right. & =i_{k}^{\prime}\ \ \ \ \ \ \ \ \ \ \left( \text{since }k\neq L\right) \\ & =\left\{ \begin{array} [c]{c}% i_{k},\ \ \ \ \ \ \ \ \ \ \text{if }k\neq L;\\ u,\ \ \ \ \ \ \ \ \ \ \text{if }k=L \end{array} \right. \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }i_{k}% ^{\prime}\right) \\ & =i_{k}\ \ \ \ \ \ \ \ \ \ \left( \text{since }k\neq L\right) . \end{align*} Compared with $i_{k}^{\prime\prime}=i_{k}$, this yields $i_{k}^{\prime\prime }=\left\{ \begin{array} [c]{c}% i_{k}^{\prime},\ \ \ \ \ \ \ \ \ \ \text{if }k\neq L;\\ r,\ \ \ \ \ \ \ \ \ \ \text{if }k=L \end{array} \right. $. This proves (\ref{pf.glinf.glinfact.welldef.8.pf.b.case1.5}).}. Hence, $\left( i_{0}^{\prime\prime},i_{1}^{\prime\prime},i_{2}^{\prime\prime },...\right) =\left( i_{0}^{\prime},i_{1}^{\prime},...,i_{L-1}^{\prime },r,i_{L+1}^{\prime},i_{L+2}^{\prime},...\right) $. Recall that $L$ is the \textbf{unique} $\ell\in\mathbb{N}$ such that $v=i_{\ell}$. Hence, Assertion \ref{prop.glinf.glinfact.welldef}.6 \textbf{(c)} (applied to $r$ and $v$ instead of $i$ and $j$) yields \begin{align} F_{E_{r,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) & =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{L-1}}\wedge v_{r}\wedge v_{i_{L+1}}\wedge v_{i_{L+2}}\wedge...\nonumber\\ & =v_{i_{0}^{\prime\prime}}\wedge v_{i_{1}^{\prime\prime}}\wedge v_{i_{2}^{\prime\prime}}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{0},i_{1},...,i_{L-1}% ,r,i_{L+1},i_{L+2},...\right) =\left( i_{0}^{\prime\prime},i_{1}% ^{\prime\prime},i_{2}^{\prime\prime},...\right) \right) \nonumber\\ & =v_{i_{0}^{\prime}}\wedge v_{i_{1}^{\prime}}\wedge...\wedge v_{i_{L-1}% ^{\prime}}\wedge v_{r}\wedge v_{i_{L+1}^{\prime}}\wedge v_{i_{L+2}^{\prime}% }\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{0}^{\prime\prime}% ,i_{1}^{\prime\prime},i_{2}^{\prime\prime},...\right) =\left( i_{0}^{\prime },i_{1}^{\prime},...,i_{L-1}^{\prime},r,i_{L+1}^{\prime},i_{L+2}^{\prime },...\right) \right) \nonumber\\ & =F_{E_{r,s}}\left( F_{E_{u,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.glinf.glinfact.welldef.8.pf.b.case1.3})}\right) . \label{pf.glinf.glinfact.welldef.8.pf.b.case1.9}% \end{align} Now,% \begin{align*} & \left( F_{E_{r,s}}F_{E_{u,v}}-\delta_{s,u}F_{E_{r,v}}\right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\ & =F_{E_{r,s}}\left( F_{E_{u,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \right) -\underbrace{\delta_{s,u}}% _{\substack{=1\\\text{(since }s=u\text{)}}}\underbrace{F_{E_{r,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) }% _{\substack{=F_{E_{r,s}}\left( F_{E_{u,v}}\left( v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...\right) \right) \\\text{(by (\ref{pf.glinf.glinfact.welldef.8.pf.b.case1.9}))}}}\\ & =F_{E_{r,s}}\left( F_{E_{u,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \right) -F_{E_{r,s}}\left( F_{E_{u,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \right) \\ & =0. \end{align*} Thus, Assertion \ref{prop.glinf.glinfact.welldef}.8 \textbf{(b)} is proven in Case 1. Let us now consider Case 2. In this case, we don't have $s=u$. Thus, $s\neq u$. Hence, $s\notin\left\{ u\right\} $. Combined with $s=v\notin I\setminus\left\{ v\right\} $, this yields $s\notin\left\{ u\right\} \cup\left( I\setminus\left\{ v\right\} \right) =\left\{ i_{0}^{\prime },i_{1}^{\prime},i_{2}^{\prime},...\right\} $ (by (\ref{pf.glinf.glinfact.welldef.8.pf.b.1})). Hence, Assertion \ref{prop.glinf.glinfact.welldef}.6 \textbf{(b)} (applied to $i_{k}^{\prime}$, $r$ and $s$ instead of $i_{k}$, $i$ and $j$) yields \begin{equation} F_{E_{r,s}}\left( v_{i_{0}^{\prime}}\wedge v_{i_{1}^{\prime}}\wedge v_{i_{2}^{\prime}}\wedge...\right) =0. \label{pf.glinf.glinfact.welldef.8.pf.b.case2.1}% \end{equation} Thus,% \begin{align*} & \left( F_{E_{r,s}}F_{E_{u,v}}-\delta_{s,u}F_{E_{r,v}}\right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\ & =F_{E_{r,s}}\left( \underbrace{F_{E_{u,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) }_{\substack{=v_{i_{0}^{\prime}% }\wedge v_{i_{1}^{\prime}}\wedge v_{i_{2}^{\prime}}\wedge...\\\text{(by (\ref{pf.glinf.glinfact.welldef.8.pf.b.case2.1}))}}}\right) -\underbrace{\delta_{s,u}}_{\substack{=0\\\text{(since }s\neq u\text{)}% }}F_{E_{r,v}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\ & =\underbrace{F_{E_{r,s}}\left( v_{i_{0}^{\prime}}\wedge v_{i_{1}^{\prime}% }\wedge v_{i_{2}^{\prime}}\wedge...\right) }_{\substack{=0\\\text{(by (\ref{pf.glinf.glinfact.welldef.8.pf.b.case2.1}))}}}-\underbrace{0F_{E_{r,v}% }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) }% _{=0}=0-0=0. \end{align*} Thus, Assertion \ref{prop.glinf.glinfact.welldef}.8 \textbf{(b)} is proven in Case 2. Hence, Assertion \ref{prop.glinf.glinfact.welldef}.8 \textbf{(b)} is proven in each of the Cases 1 and 2. Thus, Assertion \ref{prop.glinf.glinfact.welldef}.8 \textbf{(b)} is proven in all cases (since Cases 1 and 2 cover all possible situations). \textbf{(c)} It is easy to see that $\left( \mathbf{w}\left( i_{0}\right) ,\mathbf{w}\left( i_{1}\right) ,\mathbf{w}\left( i_{2}\right) ,...\right) $ is a straying $m$-degression\footnote{\textit{Proof.} Recall that $\left( i_{0},i_{1},i_{2},...\right) $ is a straying $m$-degression. Due to the definition of a straying $m$-degression, this means that every sufficiently high $k\in\mathbb{N}$ satisfies $i_{k}+k=m$. Thus, we know that every sufficiently high $k\in\mathbb{N}$ satisfies $i_{k}+k=m$. In other words, there exists a $K\in\mathbb{N}$ such that every nonnegative integer $k\geq K$ satisfies $i_{k}+k=m$. Consider this $k$. \par Now, let $k$ be a nonnegative integer such that $k\geq\max\left\{ K,m-s+1,m-v+1\right\} $. Then, $k\geq\max\left\{ K,m-s+1,m-v+1\right\} \geq K$. Hence, $i_{k}+k=m$, so that $i_{k}=m-k$. Moreover, $k\geq\max\left\{ K,m-s+1,m-v+1\right\} \geq m-s+1>m-s$, so that $m-km-v$, so that $m-k0}\overline{\mathfrak{a}_{\infty}^{i}}$ is the space of all strictly upper-triangular matrices in $\overline {\mathfrak{a}_{\infty}}$. \end{definition} Note that this was completely analogous to Definition \ref{def.glinf.grade}. \subsection{\texorpdfstring{$\mathfrak{a}_{\infty}$}{a-infinity} and its action on \texorpdfstring{$\wedge^{\dfrac{\infty}{2},m}V$}{the semi-infinite wedge space}} \begin{definition} \label{def.glinf.rho}Let $m\in\mathbb{Z}$. Let $\rho:\mathfrak{gl}_{\infty }\rightarrow\operatorname*{End}\left( \wedge^{\dfrac{\infty}{2},m}V\right) $ be the representation of $\mathfrak{gl}_{\infty}$ on $\wedge^{\dfrac{\infty }{2},m}V$ defined in Definition \ref{def.glinf.semiinfwedge}. \end{definition} The following question poses itself naturally now: Can we extend this representation $\rho$ to a representation of $\overline{\mathfrak{a}_{\infty}% }$ in a reasonable way? This question depends on what we mean by \textquotedblleft reasonable\textquotedblright. One way to concretize this is by noticing that $\overline{\mathfrak{a}_{\infty}}=\bigoplus\limits_{i\in\mathbb{Z}}% \overline{\mathfrak{a}_{\infty}^{i}}$, where $\overline{\mathfrak{a}_{\infty }^{i}}$ is the space of all matrices with nonzero entries only on the $i$-th diagonal. For each $i\in\mathbb{Z}$, the vector space $\overline {\mathfrak{a}_{\infty}^{i}}$ can be given the product topology (i. e., the topology in which a net $\left( s_{z}\right) _{z\in Z}$ of matrices converges to a matrix $s$ if and only if for any $\left( m,n\right) \in\mathbb{Z}^{2}$ satisfying $n-m=i$, the net of the $\left( m,n\right) $-th entries of the matrices $s_{z}$ converge to the $\left( m,n\right) $-th entry of $s$ in the discrete topology). Then, $\mathfrak{gl}_{\infty}^{i}$ in dense in $\overline{\mathfrak{a}_{\infty}^{i}}$ for every $i\in\mathbb{Z}$. We can also make $\wedge^{\dfrac{\infty}{2},m}V$ into a topological space by using the discrete topology. Our question can now be stated as follows: Can we extend $\rho$ by continuity to a representation of $\overline{\mathfrak{a}% _{\infty}}$ (where \textquotedblleft continuous\textquotedblright\ means \textquotedblleft continuous on each $\overline{\mathfrak{a}_{\infty}^{i}}% $\textquotedblright, since we have not defined a topology on the whole space $\overline{\mathfrak{a}_{\infty}}$) ? Answer: Almost, but not precisely. We cannot make $\overline{\mathfrak{a}% _{\infty}}$ act on $\wedge^{\dfrac{\infty}{2},m}V$ in such a way that its action extends $\rho$ continuously, but we can make a central extension of $\overline{\mathfrak{a}_{\infty}}$ act on $\wedge^{\dfrac{\infty}{2},m}V$ in a way that only slightly differs from $\rho$. Let us first see what goes wrong if we try to find an extension of $\rho$ to $\overline{\mathfrak{a}_{\infty}}$ by continuity: For $i\neq0$, a typical element $X\in\overline{\mathfrak{a}_{\infty}^{i}}$ is of the form $X=\sum\limits_{j\in\mathbb{Z}}z_{j}E_{j,j+i}$ with $z_{j}% \in\mathbb{C}$. Now we can define $\rho\left( X\right) v=\sum\limits_{j\in \mathbb{Z}}z_{j}\rho\left( E_{j,j+i}\right) v$ for every $v\in\wedge ^{\dfrac{\infty}{2},m}V$; this sum has only finitely many nonzero addends\footnote{\textit{Proof.} We must prove that, for every $v\in \wedge^{\dfrac{\infty}{2},m}V$, the sum $\sum\limits_{j\in\mathbb{Z}}z_{j}% \rho\left( E_{j,j+i}\right) v$ has only finitely many nonzero addends. It is clearly enough to prove this in the case when $v$ is an elementary semiinfinite wedge. So let us WLOG assume that $v$ is an elementary semiinfinite wedge. In other words, WLOG assume that $v=v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...$ for some $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $. Consider this $m$-degression. By the definition of an $m$-degression, every sufficiently high $k\in\mathbb{N}$ satisfies $i_{k}+k=m$. In other words, there exists a $K\in\mathbb{N}$ such that every integer $k\geq K$ satisfies $i_{k}+k=m$. Consider this $K$. Then, every integer $j\leq i_{K}$ appears in the $m$-degression $\left( i_{0}% ,i_{1},i_{2},...\right) $. \par Now, we have the following two observations: \par \begin{itemize} \item Every integer $j>i_{0}-i$ satisfies $\rho\left( E_{j,j+i}\right) v=0$ (because for every integer $j>i_{0}-i$, we have $j+i>i_{0}$, so that the integer $j+i$ does not appear in the $m$-degression $\left( i_{0},i_{1}% ,i_{2},...\right) $). \par \item Every integer $j\leq i_{K}$ satisfies $\rho\left( E_{j,j+i}\right) v=0$ (because every integer $j\leq i_{K}$ appears in the $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $, and because $i\neq0$). \end{itemize} \par Combining these two observations, we conclude that every sufficiently large integer $j$ satisfies $\rho\left( E_{j,j+i}\right) v=0$ and that every sufficiently small integer $j$ satisfies $\rho\left( E_{j,j+i}\right) v=0$. Hence, only finitely many integers $j$ satisfy $\rho\left( E_{j,j+i}\right) v\neq0$. Thus, the sum $\sum\limits_{j\in\mathbb{Z}}z_{j}\rho\left( E_{j,j+i}\right) v$ has only finitely many nonzero addends, qed.} and thus makes sense. But when $i=0$, we run into a problem with this approach: $\rho\left( \sum\limits_{j\in\mathbb{Z}}z_{j}E_{j,j}\right) v=\sum\limits_{j\in \mathbb{Z}}z_{j}\rho\left( E_{j,j}\right) v$ is an infinite sum which may very well have infinitely many nonzero addends, and thus makes no sense. To fix this problem, we define a map $\widehat{\rho}$ which will be a ``small'' modification of $\rho$: \begin{definition} \label{def.glinf.rhohat.abar}Define a linear map $\widehat{\rho}% :\overline{\mathfrak{a}_{\infty}}\rightarrow\operatorname*{End}\left( \wedge^{\dfrac{\infty}{2},m}V\right) $ by% \begin{align} \widehat{\rho}\left( \left( a_{i,j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}\right) & =\sum\limits_{\left( i,j\right) \in \mathbb{Z}^{2}}a_{i,j}\left\{ \begin{array} [c]{c}% \rho\left( E_{i,j}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0;\\ \rho\left( E_{i,j}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0 \end{array} \right. \label{def.glinf.rhohat.generalcase}\\ & \ \ \ \ \ \ \ \ \ \ \left. \text{for every }\left( a_{i,j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}\in\overline{\mathfrak{a}_{\infty}% }\right. \nonumber \end{align} (where $1$ means the endomorphism $\operatorname*{id}$ of $\wedge ^{\dfrac{\infty}{2},m}V$). Here, the infinite sum $\sum\limits_{\left( i,j\right) \in\mathbb{Z}^{2}}a_{i,j}\left\{ \begin{array} [c]{c}% \rho\left( E_{i,j}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0;\\ \rho\left( E_{i,j}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0 \end{array} \right. $ is well-defined as an endomorphism of $\wedge^{\dfrac{\infty}{2}% ,m}V$, because for every $v\in\wedge^{\dfrac{\infty}{2},m}V$, the sum $\sum\limits_{\left( i,j\right) \in\mathbb{Z}^{2}}a_{i,j}\left\{ \begin{array} [c]{c}% \rho\left( E_{i,j}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0;\\ \rho\left( E_{i,j}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0 \end{array} \right. v$ has only finitely many nonzero addends (as Proposition \ref{prop.glinf.rhohat.welldef} shows). \end{definition} The map $\widehat{\rho}$ just defined does not extend the map $\rho$, but is the unique continuous (in the sense explained above) extension of the map $\widehat{\rho}\mid_{\mathfrak{gl}_{\infty}}$ to $\overline{\mathfrak{a}% _{\infty}}$ as a linear map. The map $\widehat{\rho}\mid_{\mathfrak{gl}% _{\infty}}$ is, in a certain sense, a ``very close approximation to $\rho$'', as can be seen from the following remark: \begin{remark} \label{rmk.glinf.rhohat.abar}From Definition \ref{def.glinf.rhohat.abar}, it follows that% \begin{equation} \widehat{\rho}\left( E_{i,j}\right) =\left\{ \begin{array} [c]{c}% \rho\left( E_{i,j}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0;\\ \rho\left( E_{i,j}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0 \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }\left( i,j\right) \in\mathbb{Z}^{2}. \label{def.glinf.rhohat}% \end{equation} \end{remark} We are not done yet: This map $\widehat{\rho}$ is not a representation of $\overline{\mathfrak{a}_{\infty}}$. We will circumvent this by defining a central extension $\mathfrak{a}_{\infty}$ of $\overline{\mathfrak{a}_{\infty}% }$ for which the map $\widehat{\rho}$ (once suitably extended) will be a representation. But first, let us show a lemma that we owe for the definition of $\widehat{\rho}$: \begin{proposition} \label{prop.glinf.rhohat.welldef}Let $\left( a_{i,j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}\in\overline{\mathfrak{a}_{\infty}}$ and $v\in\wedge^{\dfrac{\infty}{2},m}V$. Then, the sum% \[ \sum\limits_{\left( i,j\right) \in\mathbb{Z}^{2}}a_{i,j}\left\{ \begin{array} [c]{c}% \rho\left( E_{i,j}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0;\\ \rho\left( E_{i,j}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0 \end{array} \right. v \] has only finitely many nonzero addends. \end{proposition} \textit{Proof of Proposition \ref{prop.glinf.rhohat.welldef}.} We know that $v$ is an element of $\wedge^{\dfrac{\infty}{2},m}V$. Hence, $v$ is a $\mathbb{C}$-linear combination of elements of the form $v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...$ with $\left( i_{0},i_{1},i_{2}% ,...\right) $ being an $m$-degression (since $\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) _{\left( i_{0},i_{1}% ,i_{2},...\right) \text{ is an }m\text{-degression}}$ is a basis of $\wedge^{\dfrac{\infty}{2},m}V$). Hence, we can WLOG assume that $v$ is an element of the form $v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...$ with $\left( i_{0},i_{1},i_{2},...\right) $ being an $m$-degression (because the claim of Proposition \ref{prop.glinf.rhohat.welldef} is clearly linear in $v$). Assume this. Then, $v=v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...$ for some $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $. Consider this $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $. By the definition of an $m$-degression, every sufficiently high $k\in\mathbb{N}$ satisfies $i_{k}+k=m$. In other words, there exists a $K\in\mathbb{N}$ such that every integer $k\geq K$ satisfies $i_{k}+k=m$. Consider this $K$. Then, every integer which is less or equal to $i_{K}$ appears in the $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $. For every $\left( i,j\right) \in\mathbb{Z}^{2}$, let $r_{i,j}$ be the map $\left\{ \begin{array} [c]{c}% \rho\left( E_{i,j}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0;\\ \rho\left( E_{i,j}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0 \end{array} \right. \in\operatorname*{End}\left( \wedge^{\dfrac{\infty}{2},m}V\right) $. Then, the sum% \[ \sum\limits_{\left( i,j\right) \in\mathbb{Z}^{2}}a_{i,j}\left\{ \begin{array} [c]{c}% \rho\left( E_{i,j}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0;\\ \rho\left( E_{i,j}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0 \end{array} \right. v \] clearly rewrites as $\sum\limits_{\left( i,j\right) \in\mathbb{Z}^{2}% }a_{i,j}r_{i,j}v$. Hence, in order to prove Proposition \ref{prop.glinf.rhohat.welldef}, we only need to prove that the sum $\sum\limits_{\left( i,j\right) \in\mathbb{Z}^{2}}a_{i,j}r_{i,j}v$ has only finitely many nonzero addends. Since $\left( a_{i,j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}% \in\overline{\mathfrak{a}_{\infty}}$, only finitely many diagonals of the matrix $\left( a_{i,j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}$ are nonzero. In other words, there exists an $M\in\mathbb{N}$ such that% \begin{equation} \left( \text{the }m\text{-th diagonal of the matrix }\left( a_{i,j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}\text{ is zero for every }% m\in\mathbb{Z}\text{ such that }\left\vert m\right\vert \geq M\right) . \label{pf.glinf.rhohat.welldef.M}% \end{equation} Consider this $M$. Now, we have the following three observations: \begin{itemize} \item Every $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $j>\max\left\{ i_{0},0\right\} $ satisfies $r_{i,j}v=0\ \ \ \ $\footnote{\textit{Proof.} Let $\left( i,j\right) \in\mathbb{Z}^{2}$ be such that $j>\max\left\{ i_{0},0\right\} $. Then, $j>i_{0}$ and $j>0$. \par Since $j>i_{0}$, the integer $j$ does not appear in the $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $. Hence, $\rho\left( E_{i,j}\right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =0$. Since $v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...=v$, this rewrites as $\rho\left( E_{i,j}\right) v=0$. \par Since $j>0$, we cannot have $i=j$ and $i\leq0$. Now, $r_{i,j}=\left\{ \begin{array} [c]{c}% \rho\left( E_{i,j}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0;\\ \rho\left( E_{i,j}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0 \end{array} \right. =\rho\left( E_{i,j}\right) $ (since we cannot have $i=j$ and $i\leq0$), so that $r_{i,j}v=\rho\left( E_{i,j}\right) v=0$, qed.} and thus $a_{i,j}\underbrace{r_{i,j}v}_{=0}=0$. \item Every $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $i\leq \min\left\{ i_{K},0\right\} $ satisfies $r_{i,j}v=0$% \ \ \ \ \footnote{\textit{Proof.} Let $\left( i,j\right) \in\mathbb{Z}^{2}$ be such that $i\leq\min\left\{ i_{K},0\right\} $. Then, $i\leq i_{K}$ and $i\leq0$. \par Since $i\leq i_{K}$, the integer $i$ appears in the $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $ (because every integer which is less or equal to $i_{K}$ appears in the $m$-degression $\left( i_{0},i_{1},i_{2}% ,...\right) $). We now must be in one of the following two cases: \par \textit{Case 1:} We have $i\neq j$. \par \textit{Case 2:} We have $i=j$. \par Let us first consider Case 1. In this case, $i\neq j$. Thus, $\rho\left( E_{i,j}\right) v=0$ (because the integer $i$ appears in the $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $, so that after applying $\rho\left( E_{i,j}\right) $ to $v=v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...$, we obtain a wedge in which $v_{i}$ appears twice). On the other hand, $i\neq j$, so that we cannot have $i=j$ and $i\leq0$. Now, $r_{i,j}=\left\{ \begin{array} [c]{c}% \rho\left( E_{i,j}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0;\\ \rho\left( E_{i,j}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0 \end{array} \right. =\rho\left( E_{i,j}\right) $ (since we cannot have $i=j$ and $i\leq0$), and thus $r_{i,j}v=\rho\left( E_{i,j}\right) v=0$. \par Now, let us consider Case 2. In this case, $i=j$. Thus, $r_{i,j}=\left\{ \begin{array} [c]{c}% \rho\left( E_{i,j}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0;\\ \rho\left( E_{i,j}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0 \end{array} \right. =\rho\left( E_{i,j}\right) -1$ (since $i=j$ and $i\leq0$). Since $E_{i,j}=E_{i,i}$ (because $j=i$), this rewrites as $r_{i,j}=\rho\left( E_{i,i}\right) -1$. On the other hand, the integer $i$ appears in the $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $, so that $\rho\left( E_{i,i}\right) v=v$. Hence, from $r_{i,j}=\rho\left( E_{i,i}\right) -1$, we get $r_{i,j}v=\left( \rho\left( E_{i,i}\right) -1\right) v=\underbrace{\rho\left( E_{i,i}\right) v}_{=v}-v=v-v=0$. \par Thus, in each of the cases 1 and 2, we have proven that $r_{i,j}v=0$. Hence, $r_{i,j}v=0$ always holds, qed.} and thus $a_{i,j}\underbrace{r_{i,j}v}% _{=0}=0$. \item Every $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $\left\vert i-j\right\vert \geq M$ satisfies $a_{i,j}=0$\ \ \ \ \footnote{\textit{Proof.} Let $\left( u,v\right) \in\mathbb{Z}^{2}$ be such that $\left\vert u-v\right\vert \geq M$. Then, since $\left\vert v-u\right\vert =\left\vert u-v\right\vert \geq M$, the $\left( v-u\right) $-th diagonal of the matrix $\left( a_{i,j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}$ is zero (by (\ref{pf.glinf.rhohat.welldef.M}), applied to $m=v-u$), and thus $a_{u,v}=0$ (since $a_{u,v}$ is an entry on the $\left( v-u\right) $-th diagonal of the matrix $\left( a_{i,j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}$). We thus have shown that every $\left( u,v\right) \in\mathbb{Z}^{2}$ such that $\left\vert u-v\right\vert \geq M$ satisfies $a_{u,v}=0$. Renaming $\left( u,v\right) $ as $\left( i,j\right) $ in this fact, we obtain: Every $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $\left\vert i-j\right\vert \geq M$ satisfies $a_{i,j}=0$, qed.} and thus $\underbrace{a_{i,j}}% _{=0}r_{i,j}v=0$. \end{itemize} Now, for any $\alpha\in\mathbb{Z}$ and $\beta\in\mathbb{Z}$, let $\left[ \alpha,\beta\right] _{\mathbb{Z}}$ denote the set $\left\{ x\in \mathbb{Z}\ \mid\ \alpha\leq x\leq\beta\right\} $ (this set is finite). It is easy to see that% \begin{equation} \left( \begin{array} [c]{c}% \text{every }\left( i,j\right) \in\mathbb{Z}^{2}\text{ such that }% a_{i,j}r_{i,j}v\neq0\text{ satisfies}\\ \left( i,j\right) \in\left[ \min\left\{ i_{K},0\right\} +1,\max\left\{ i_{0},0\right\} +M-1\right] _{\mathbb{Z}}\times\left[ \min\left\{ i_{K},0\right\} -M+2,\max\left\{ i_{0},0\right\} \right] _{\mathbb{Z}}% \end{array} \right) \label{pf.gli.rhohat.welldef.bnd}% \end{equation} \footnote{\textit{Proof of (\ref{pf.gli.rhohat.welldef.bnd}):} Let $\left( i,j\right) \in\mathbb{Z}^{2}$ be such that $a_{i,j}r_{i,j}v\neq0$. Then, we cannot have $j>\max\left\{ i_{0},0\right\} $ (since every $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $j>\max\left\{ i_{0},0\right\} $ satisfies $a_{i,j}r_{i,j}v=0$, whereas we have $a_{i,j}r_{i,j}v\neq0$). In other words, $j\leq\max\left\{ i_{0},0\right\} $. Also, we cannot have $i\leq\min\left\{ i_{K},0\right\} $ (since every $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $i\leq\min\left\{ i_{K},0\right\} $ satisfies $a_{i,j}r_{i,j}v=0$, whereas we have $a_{i,j}r_{i,j}v\neq0$). Thus, we have $i>\min\left\{ i_{K},0\right\} $, so that $i\geq\min\left\{ i_{K}% ,0\right\} +1$ (since $i$ and $\min\left\{ i_{K},0\right\} $ are integers). Finally, we cannot have $\left\vert i-j\right\vert \geq M$ (since every $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $\left\vert i-j\right\vert \geq M$ satisfies $a_{i,j}r_{i,j}v=0$, whereas we have $a_{i,j}r_{i,j}v\neq 0$). Thus, we have $\left\vert i-j\right\vert 0$. - The upper blocks contain the $i$-th rows for all $i\leq0$; the lower blocks contain the $i$-th rows for all $i>0$. Then, $\alpha\left( A,B\right) =\operatorname*{Tr}\left( -B_{12}% A_{21}+A_{12}B_{21}\right) $. (This trace makes sense because the matrices $A_{12}$, $B_{21}$, $A_{21}$, $B_{12}$ have only finitely many nonzero entries.) \end{theorem} \begin{corollary} \label{cor.japan}The bilinear map $\alpha:\overline{\mathfrak{a}_{\infty}% }\times\overline{\mathfrak{a}_{\infty}}\rightarrow\mathbb{C}$ defined in Theorem \ref{thm.japan} is a $2$-cocycle on $\overline{\mathfrak{a}_{\infty}}$. We define $\mathfrak{a}_{\infty}$ as the $1$-dimensional central extension $\widehat{\overline{\mathfrak{a}_{\infty}}}_{\alpha}$ of $\overline {\mathfrak{a}_{\infty}}$ by $\mathbb{C}$ using this cocycle $\alpha$ (see Definition \ref{def.centex} for what this means). \end{corollary} \begin{definition} \label{def.japan}The $2$-cocycle $\alpha:\overline{\mathfrak{a}_{\infty}% }\times\overline{\mathfrak{a}_{\infty}}\rightarrow\mathbb{C}$ introduced in Corollary \ref{cor.japan} is called the \textit{Japanese cocycle}. \end{definition} The proofs of Theorem \ref{thm.japan} and Corollary \ref{cor.japan} are a homework problem. A few remarks on the Japanese cocycle are in order. It can be explicitly computed by the formula% \begin{align*} & \alpha\left( \left( a_{i,j}\right) _{\left( i,j\right) \in \mathbb{Z}^{2}},\left( b_{i,j}\right) _{\left( i,j\right) \in \mathbb{Z}^{2}}\right) \\ & =-\sum_{\substack{i\leq0;\\j>0}}b_{i,j}a_{j,i}+\sum_{\substack{i\leq 0;\\j>0}}a_{i,j}b_{j,i}=-\sum_{\substack{i>0;\\j\leq0}}a_{i,j}b_{j,i}% +\sum_{\substack{i\leq0;\\j>0}}a_{i,j}b_{j,i}\\ & =\sum_{\left( i,j\right) \in\mathbb{Z}^{2}}a_{i,j}b_{j,i}\left( \left[ j>0\right] -\left[ i>0\right] \right) \ \ \ \ \ \ \ \ \ \ \text{for every }\left( a_{i,j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}},\left( b_{i,j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}\in\overline {\mathfrak{a}_{\infty}}% \end{align*} where we are using the \textit{Iverson bracket notation}\footnote{This is the notation $\left[ \mathcal{S}\right] $ for the truth value of any logical statement $\mathcal{S}$ (that is, $\left[ \mathcal{S}\right] $ denotes the integer $% \begin{cases} 1, & \text{if }\mathcal{S}\text{ is true;}\\ 0, & \text{if }\mathcal{S}\text{ is false}% \end{cases} $).}. The cocycle $\alpha$\ owes its name \textquotedblleft Japanese cocycle\textquotedblright\ to the fact that it (first?) appeared in the work of the Tokyo mathematical physicists Date, Jimbo, Kashiwara and Miwa\footnote{More precisely, it is the skew-symmetric bilinear form $c$ in the following paper: \par \begin{itemize} \item Etsuro Date, Michio Jimbo, Masaki Kashiwara, Tetuji Miwa, \textit{Transformation Groups for Soliton Equations -- Euclidean Lie Algebras and Reduction of the KP Hierarchy}, Publ. RIMS, Kyoto Univ. 18 (1982), pp. 1077--1110. \end{itemize} \par In this paper, the Lie algebras that we are denoting by $\overline {\mathfrak{a}_{\infty}}$ and $\mathfrak{a}_{\infty}$ are called $\mathfrak{pgl}\left( \infty\right) $ and $\mathfrak{gl}\left( \infty\right) $, respectively.}. We are going to prove soon (Proposition \ref{prop.japan.nontr} and Corollary \ref{cor.japan.triv}) that $\alpha$ is a nontrivial $2$-cocycle, but its restriction to $\mathfrak{gl}_{\infty}$ is trivial. This is a strange situation (given that $\mathfrak{gl}_{\infty}$ is a dense Lie subalgebra of $\overline{\mathfrak{a}_{\infty}}$ with respect to a reasonably defined topology), but we will later see the reason for this behavior. \begin{theorem} Let us extend the linear map $\widehat{\rho}:\overline{\mathfrak{a}_{\infty}% }\rightarrow\operatorname*{End}\left( \wedge^{\dfrac{\infty}{2},m}V\right) $ (introduced in Definition \ref{def.glinf.rhohat.abar}) to a linear map $\widehat{\rho}:\mathfrak{a}_{\infty}\rightarrow\operatorname*{End}\left( \wedge^{\dfrac{\infty}{2},m}V\right) $ by setting $\widehat{\rho}\left( K\right) =\operatorname*{id}$. (This makes sense since $\mathfrak{a}_{\infty }=\overline{\mathfrak{a}_{\infty}}\oplus\mathbb{C}K$ as vector spaces.) Then, this map $\widehat{\rho}:\mathfrak{a}_{\infty}\rightarrow\operatorname*{End}% \left( \wedge^{\dfrac{\infty}{2},m}V\right) $ is a representation of $\mathfrak{a}_{\infty}$. Thus, $\wedge^{\dfrac{\infty}{2},m}V$ becomes an $\mathfrak{a}_{\infty}$-module. \end{theorem} \begin{definition} \label{def.ainf.grade2}Since $\mathfrak{a}_{\infty}=\overline{\mathfrak{a}% _{\infty}}\oplus\mathbb{C}K$ as vector space, we can define a grading on $\mathfrak{a}_{\infty}$ as the direct sum of the grading on $\overline {\mathfrak{a}_{\infty}}$ (which was defined in Definition \ref{def.ainf.grade}% ) and the trivial grading on $\mathbb{C}K$ (that is the grading which puts $K$ in degree $0$). This is easily seen to make $\mathfrak{a}_{\infty}$ a $\mathbb{Z}$-graded Lie algebra. We will consider $\mathfrak{a}_{\infty}$ to be $\mathbb{Z}$-graded in this way. \end{definition} \begin{proposition} Let $m\in\mathbb{Z}$. With the grading defined in Definition \ref{def.ainf.grade2}, the $\mathfrak{a}_{\infty}$-module $\wedge ^{\dfrac{\infty}{2},m}V$ is graded. \end{proposition} \begin{corollary} \label{cor.japan.triv}The restriction of $\alpha$ to $\mathfrak{gl}_{\infty }\times\mathfrak{gl}_{\infty}$ is a $2$-coboundary. \end{corollary} \textit{Proof of Corollary \ref{cor.japan.triv}.} Let $J$ be the block matrix $\left( \begin{array} [c]{cc}% 0 & 0\\ 0 & -I_{\infty}% \end{array} \right) \in\overline{\mathfrak{a}_{\infty}}$, where the blocks are separated in the same way as in Theorem \ref{thm.japan}. Define a linear map $f:\mathfrak{gl}_{\infty}\rightarrow\mathbb{C}$ by \[ \left( f\left( A\right) =\operatorname*{Tr}\left( JA\right) \ \ \ \ \ \ \ \ \ \ \text{for any }A\in\mathfrak{gl}_{\infty}\right) \] \footnote{Note that $\operatorname*{Tr}\left( JA\right) $ is well-defined for every $A\in\mathfrak{gl}_{\infty}$, since Remark \ref{rmk.ainf.mult} \textbf{(b)} (applied to $J$ and $A$ instead of $A$ and $B$) yields that $JA\in\mathfrak{gl}_{\infty}$.}. Then, any $A\in\mathfrak{gl}_{\infty}$ and $B\in\mathfrak{gl}_{\infty}$ satisfy $\alpha\left( A,B\right) =f\left( \left[ A,B\right] \right) $. This is because (for any $A\in\mathfrak{gl}% _{\infty}$ and $B\in\mathfrak{gl}_{\infty}$) we can write the matrix $\left[ A,B\right] $ in the form $\left[ A,B\right] =\left( \begin{array} [c]{cc}% \ast & \ast\\ \ast & \left[ A_{22},B_{22}\right] +A_{21}B_{12}-B_{21}A_{12}% \end{array} \right) $ (where asterisks mean blocks which we don't care about), so that $J\left[ A,B\right] =\left( \begin{array} [c]{cc}% 0 & 0\\ \ast & -\left( \left[ A_{22},B_{22}\right] +A_{21}B_{12}-B_{21}% A_{12}\right) \end{array} \right) $ and thus \begin{align*} & \operatorname*{Tr}\left( J\left[ A,B\right] \right) \\ & =-\operatorname*{Tr}\left( \left[ A_{22},B_{22}\right] +A_{21}% B_{12}-B_{21}A_{12}\right) =-\underbrace{\operatorname*{Tr}\left[ A_{22},B_{22}\right] }_{=0}-\underbrace{\operatorname*{Tr}\left( A_{21}B_{12}\right) }_{=\operatorname*{Tr}\left( B_{12}A_{21}\right) }+\underbrace{\operatorname*{Tr}\left( B_{21}A_{12}\right) }% _{=\operatorname*{Tr}\left( A_{12}B_{21}\right) }\\ & =-\operatorname*{Tr}\left( B_{12}A_{21}\right) +\operatorname*{Tr}\left( A_{12}B_{21}\right) =\operatorname*{Tr}\left( -B_{12}A_{21}+A_{12}% B_{21}\right) =\alpha\left( A,B\right) . \end{align*} The proof of Corollary \ref{cor.japan.triv} is thus finished. But note that this proof does not extend to $\overline{\mathfrak{a}_{\infty}}% $, because $f$ does not continuously extend to $\overline{\mathfrak{a}% _{\infty}}$ (for any reasonable notion of continuity). \begin{proposition} \label{prop.japan.nontr}The $2$-cocycle $\alpha$ itself is not a $2$-coboundary. \end{proposition} \textit{Proof of Proposition \ref{prop.japan.nontr}.} Let $T$ be the shift operator defined above. The span $\left\langle T^{j}\ \mid\ j\in \mathbb{Z}\right\rangle $ is an abelian Lie subalgebra of $\overline {\mathfrak{a}_{\infty}}$ (isomorphic to the abelian Lie algebra $\mathbb{C}% \left[ t,t^{-1}\right] $, and to the quotient $\overline{\mathcal{A}}$ of the Heisenberg algebra $\mathcal{A}$ by its central subalgebra $\left\langle K\right\rangle $). Any $2$-coboundary must become zero when restricted onto an abelian Lie subalgebra. But the $2$-cocycle $\alpha$, restricted onto the span $\left\langle T^{j}\ \mid\ j\in\mathbb{Z}\right\rangle $, does not become $0$, since% \[ \alpha\left( T^{i},T^{j}\right) =\left\{ \begin{array} [c]{c}% 0,\ \ \ \ \ \ \ \ \ \ \text{if }i\neq-j;\\ i,\ \ \ \ \ \ \ \ \ \ \text{if }i=-j \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\mathbb{Z}. \] Proposition \ref{prop.japan.nontr} is thus proven. In this proof, we have constructed an embedding $\overline{\mathcal{A}% }\rightarrow\overline{\mathfrak{a}_{\infty}}$ which sends $\overline{a_{j}}$ to $T^{j}$ for every $j\in\mathbb{Z}$. This embedding is crucial to what we are going to do, so let us give it a formal definition: \begin{definition} \label{def.ainf.A}The map% \[ \overline{\mathcal{A}}\rightarrow\overline{\mathfrak{a}_{\infty}% },\ \ \ \ \ \ \ \ \ \ a_{j}\mapsto T^{j}% \] (where $\overline{\mathcal{A}}$ is the quotient of the Heisenberg algebra $\mathcal{A}$ by its central subalgebra $\left\langle K\right\rangle $) is an embedding of Lie algebras. We will regard this embedding as an inclusion, and thus we will regard $\overline{\mathcal{A}}$ as a Lie subalgebra of $\overline{\mathfrak{a}_{\infty}}$. This embedding is easily seen to give rise to an embedding $\mathcal{A}% \rightarrow\mathfrak{a}_{\infty}$ of Lie algebras which sends $K$ to $K$ and sends $a_{j}$ to $T^{j}$ for every $j\in\mathbb{Z}$. This embedding will also be regarded as an inclusion, so that $\mathcal{A}$ will be considered as a Lie subalgebra of $\mathfrak{a}_{\infty}$. \end{definition} It is now easy to see: \begin{proposition} Extend our map $\widehat{\rho}:\overline{\mathfrak{a}_{\infty}}\rightarrow \operatorname*{End}\left( \wedge^{\dfrac{\infty}{2},m}V\right) $ to a map $\mathfrak{a}_{\infty}\rightarrow\operatorname*{End}\left( \wedge ^{\dfrac{\infty}{2},m}V\right) $, also denoted by $\widehat{\rho}$, by setting $\widehat{\rho}\left( K\right) =\operatorname*{id}$. Then, this map $\widehat{\rho}:\mathfrak{a}_{\infty}\rightarrow\operatorname*{End}\left( \wedge^{\dfrac{\infty}{2},m}V\right) $ is a Lie algebra homomorphism, i. e., it makes $\wedge^{\dfrac{\infty}{2},m}V$ into an $\mathfrak{a}_{\infty}% $-module. The element $K$ of $\mathfrak{a}_{\infty}$ acts as $\operatorname*{id}$ on this module. By means of the embedding $\mathcal{A}\rightarrow\mathfrak{a}_{\infty}$, this $\mathfrak{a}_{\infty}$-module gives rise to an $\mathcal{A}$-module $\wedge^{\dfrac{\infty}{2},m}V$, on which $K$ acts as $\operatorname*{id}$. \end{proposition} In Proposition \ref{prop.Lomegam}, we identified $\wedge^{\dfrac{\infty}{2}% ,m}V$ as an irreducible highest-weight $\mathfrak{gl}_{\infty}$-module; similarly, we can identify it as an irreducible highest-weight $\mathfrak{a}% _{\infty}$-module: \begin{proposition} \label{prop.Lomegam.a}Let $m\in\mathbb{Z}$. Let $\overline{\omega}_{m}$ be the $\mathbb{C}$-linear map $\mathfrak{a}_{\infty}\left[ 0\right] \rightarrow \mathbb{C}$ which sends every infinite diagonal matrix $\operatorname*{diag}% \left( ...,d_{-2},d_{-1},d_{0},d_{1},d_{2},...\right) \in\overline {\mathfrak{a}_{\infty}}$ to $\left\{ \begin{array} [c]{c}% \sum\limits_{j=1}^{m}d_{j},\ \ \ \ \ \ \ \ \ \ \text{if }m\geq0;\\ -\sum\limits_{j=m+1}^{0}d_{j},\ \ \ \ \ \ \ \ \ \ \text{if }m<0 \end{array} \right. $, and sends $K$ to $1$. Then, the graded $\mathfrak{a}_{\infty}% $-module $\wedge^{\dfrac{\infty}{2},m}V$ is the irreducible highest-weight representation $L_{\overline{\omega}_{m}}$ of $\mathfrak{a}_{\infty}$ with highest weight $L_{\overline{\omega}_{m}}$. Moreover, $L_{\overline{\omega }_{m}}$ is unitary. \end{proposition} \begin{remark} Note the analogy between the weight $\overline{\omega}_{m}$ in Proposition \ref{prop.Lomegam.a} and the weight $\omega_{m}$ in Proposition \ref{prop.Lomegam}: The weight $\omega_{m}$ in Proposition \ref{prop.Lomegam} sends every diagonal matrix $\operatorname*{diag}\left( ...,d_{-2}% ,d_{-1},d_{0},d_{1},d_{2},...\right) \in\mathfrak{gl}_{\infty}$ to $\sum\limits_{j=-\infty}^{m}d_{j}$. Note that this sum $\sum\limits_{j=-\infty }^{m}d_{j}$ is well-defined (because for a diagonal matrix $\operatorname*{diag}\left( ...,d_{-2},d_{-1},d_{0},d_{1},d_{2},...\right) $ to lie in $\mathfrak{gl}_{\infty}$, it has to satisfy $d_{j}=0$ for all but finitely many $j\in\mathbb{Z}$). \end{remark} In analogy to Corollary \ref{cor.lomegam.unit}, we can also show: \begin{corollary} \label{cor.lomegam.unit.a}For every finite sum $\sum\limits_{i\in\mathbb{Z}% }k_{i}\overline{\omega}_{i}$ with $k_{i}\in\mathbb{N}$, the representation $L_{\sum\limits_{i\in\mathbb{Z}}k_{i}\overline{\omega}_{i}}$ of $\mathfrak{a}% _{\infty}$ is unitary. \end{corollary} \subsection{Virasoro actions on \texorpdfstring{$\wedge^{\dfrac{\infty}{2},m}V$} {the semi-infinite wedge space}} We can also embed the Virasoro algebra $\operatorname*{Vir}$ into $\mathfrak{a}_{\infty}$, and not just in one way, but in infinitely many ways depending on two parameters: \begin{proposition} Let $\alpha\in\mathbb{C}$ and $\beta\in\mathbb{C}$. Let the $\operatorname*{Vir}$-module $V_{\alpha,\beta}$ be defined as in Proposition \ref{prop.Vab.1}. For every $k\in\mathbb{Z}$, let $v_{k}=t^{-k+\alpha}\left( dt\right) ^{\beta}\in V_{\alpha,\beta}$. Here, for any $\ell\in\mathbb{Z}$, the term $t^{\ell+\alpha}\left( dt\right) ^{\beta}$ denotes $t^{\ell}t^{\alpha }\left( dt\right) ^{\beta}$. According to Proposition \ref{prop.Vab.1} \textbf{(b)}, every $m\in\mathbb{Z}$ satisfies% \[ L_{m}v_{k}=\left( k-\alpha-\beta\left( m+1\right) \right) v_{k-m}% \ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{Z}. \] Thus, if we write $L_{m}$ as a matrix with respect to the basis $\left( v_{k}\right) _{k\in\mathbb{Z}}$ of $V_{\alpha,\beta}$, then this matrix lies in $\overline{\mathfrak{a}_{\infty}}$ (in fact, its only nonzero diagonal is the $m$-th one). This defines an injective map $\overline{\varphi_{\alpha,\beta}}% :W\rightarrow\overline{\mathfrak{a}_{\infty}}$, which sends every $L_{m}\in W$ to the matrix representing the action of $L_{m}$ on $V_{\alpha,\beta}$. This map $\overline{\varphi_{\alpha,\beta}}$ is a Lie algebra homomorphism (since the $\operatorname*{Vir}$-module $V_{\alpha,\beta}$ has central charge $0$, i. e., is an $W$-module). Hence, this map $\overline{\varphi_{\alpha,\beta}}$ lifts to an injective map $\widehat{W}\rightarrow\mathfrak{a}_{\infty}$, where $\widehat{W}$ is defined as follows: Let $\widetilde{\alpha}:\overline {\mathfrak{a}_{\infty}}\times\overline{\mathfrak{a}_{\infty}}\rightarrow \mathbb{C}$ be the Japanese cocycle (this cocycle has been called $\alpha$ in Definition \ref{def.japan}, but here we use the letter $\alpha$ for something different), and let $\widetilde{\alpha}^{\prime}:W\times W\rightarrow \mathbb{C}$ be the restriction of this Japanese cocycle $\widetilde{\alpha }:\overline{\mathfrak{a}_{\infty}}\times\overline{\mathfrak{a}_{\infty}% }\rightarrow\mathbb{C}$ to $W\times W$ via the map $\overline{\varphi _{\alpha,\beta}}\times\overline{\varphi_{\alpha,\beta}}:W\times W\rightarrow \overline{\mathfrak{a}_{\infty}}\times\overline{\mathfrak{a}_{\infty}}$. Then, $\widehat{W}$ denotes the central extension of $W$ defined by the $2$-cocycle $\widetilde{\alpha}^{\prime}$. But let us now compute $\widetilde{\alpha}^{\prime}$ and $\widehat{W}$. In fact, from a straightforward calculation (Homework Set 4 exercise 3) it follows that% \[ \widetilde{\alpha}^{\prime}\left( L_{m},L_{n}\right) =\delta_{n,-m}\left( \dfrac{n^{3}-n}{12}c_{\beta}+2nh_{\alpha,\beta}\right) \ \ \ \ \ \ \ \ \ \ \text{for all }n,m\in\mathbb{Z}, \] where \[ c_{\beta}=-12\beta^{2}+12\beta-2\ \ \ \ \ \ \ \ \ \ \text{and}% \ \ \ \ \ \ \ \ \ \ h_{\alpha,\beta}=\dfrac{1}{2}\alpha\left( \alpha +2\beta-1\right) . \] Thus, the $2$-cocycle $\widetilde{\alpha}^{\prime}$ differs from the $2$-cocycle $\omega$ (defined in Theorem \ref{thm.H^2(W)}) merely by a multiplicative factor ($\dfrac{c_{\beta}}{2}$) and a $2$-coboundary (which sends every $\left( L_{m},L_{n}\right) $ to $\delta_{n,-m}\cdot 2nh_{\alpha,\beta}$). Thus, the central extension $\widehat{W}$ of $W$ defined by the $2$-cocycle $\widetilde{\alpha}^{\prime}$ is isomorphic (as a Lie algebra) to the central extension of $W$ defined by the $2$-cocycle $\omega$, that is, to the Virasoro algebra $\operatorname*{Vir}$. This turns the Lie algebra homomorphism $\widehat{W}\rightarrow\mathfrak{a}_{\infty}$ into a homomorphism $\operatorname*{Vir}\rightarrow\mathfrak{a}_{\infty}$. Let us describe this homomorphism explicitly: Let $\widehat{L_{0}}$ be the element $\overline{\varphi_{\alpha,\beta}}\left( L_{0}\right) +h_{\alpha,\beta}K\in\mathfrak{a}_{\infty}$. Then, the linear map% \begin{align*} \operatorname*{Vir} & \rightarrow\mathfrak{a}_{\infty},\\ L_{n} & \mapsto\overline{\varphi_{\alpha,\beta}}\left( L_{n}\right) \ \ \ \ \ \ \ \ \ \ \text{for }n\neq0,\\ L_{0} & \mapsto\widehat{L_{0}},\\ C & \mapsto c_{\beta}K \end{align*} is a Lie algebra homomorphism. Denote this map by $\varphi_{\alpha,\beta}$. By means of this homomorphism, we can restrict the $\mathfrak{a}_{\infty}$-module $\wedge^{\dfrac{\infty}{2},m}V$ to a $\operatorname*{Vir}$-module. Denote this $\operatorname*{Vir}$-module by $\wedge^{\dfrac{\infty}{2},m}V_{\alpha,\beta}% $. Note that $\wedge^{\dfrac{\infty}{2},m}V_{\alpha,\beta}$ is a Virasoro module with central charge $c=c_{\beta}$. This $\wedge^{\dfrac{\infty}{2}% ,m}V_{\alpha,\beta}$ is called the \textit{module of semiinfinite forms}. The vector $\psi_{m}=v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...$ (defined in Definition \ref{def.psim}) has highest degree (namely, $0$). We have $L_{i}\psi_{m}=0$ for $i>0$, and we have $L_{0}\psi_{m}=\dfrac{1}% {2}\left( \alpha-m\right) \left( \alpha+2\beta-1-m\right) \psi_{m}$. (Proof: Homework exercise.) \end{proposition} \begin{corollary} Let $\alpha,\beta\in\mathbb{C}$. We have a homomorphism% \begin{align*} M_{\lambda} & \rightarrow\wedge^{\dfrac{\infty}{2},m}V_{\alpha,\beta},\\ v_{\lambda} & \mapsto\psi_{m}% \end{align*} of Virasoro modules, where% \[ \lambda=\left( \dfrac{1}{2}\left( \alpha-m\right) \left( \alpha +2\beta-1-m\right) ,-12\beta^{2}+12\beta-2\right) . \] \end{corollary} We will see that this is an isomorphism for generic $\lambda$. For concrete $\lambda$ it is not always one, and can have a rather complicated kernel. \subsection{The dimensions of the homogeneous components of \texorpdfstring{$\wedge ^{\dfrac{\infty}{2},m}V$}{the semi-infinite wedge space}} Fix $m\in\mathbb{Z}$. We already know from Definition \ref{def.glinf.wedge.grading} that $\wedge^{\dfrac{\infty}{2},m}V$ is a graded $\mathbb{C}$-vector space. More concretely,% \[ \wedge^{\dfrac{\infty}{2},m}V=\bigoplus\limits_{d\geq0}\left( \wedge ^{\dfrac{\infty}{2},m}V\right) \left[ -d\right] , \] where every $d\geq0$ satisfies% \[ \left( \wedge^{\dfrac{\infty}{2},m}V\right) \left[ -d\right] =\left\langle v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\ \mid\ \sum\limits_{k\geq 0}\left( i_{k}+k-m\right) =d\right\rangle . \] We also know that the $m$-degressions are in a 1-to-1 correspondence with the partitions. This correspondence maps any $m$-degression $\left( i_{0}% ,i_{1},i_{2},...\right) $ to the partition $\left( i_{k}+k-m\right) _{k\geq0}$; this is a partition of the integer $\sum\limits_{k\geq0}\left( i_{k}+k-m\right) $. As a consequence, for every integer $d\geq0$, the $m$-degressions $\left( i_{0},i_{1},i_{2},...\right) $ satisfying $\sum\limits_{k\geq0}\left( i_{k}+k-m\right) =d$ are in 1-to-1 correspondence with the partitions of $d$. Hence, for every integer $d\geq0$, the number of all $m$-degressions $\left( i_{0},i_{1},i_{2},...\right) $ satisfying $\sum\limits_{k\geq0}\left( i_{k}+k-m\right) =d$ equals the number of the partitions of $d$. Thus, for every integer $d\geq0$, we have% \begin{align*} & \dim\left( \left( \wedge^{\dfrac{\infty}{2},m}V\right) \left[ -d\right] \right) \\ & =\left( \text{the number of }m\text{-degressions }\left( i_{0}% ,i_{1},i_{2},...\right) \text{ satisfying}\sum\limits_{k\geq0}\left( i_{k}+k-m\right) =d\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge ...\right) _{\left( i_{0},i_{1},i_{2},...\right) \text{ is an }m\text{-degression satisfying }\sum\limits_{k\geq0}\left( i_{k}+k-m\right) =d}\\ \text{ is a basis of }\left( \wedge^{\dfrac{\infty}{2},m}V\right) \left[ -d\right] \end{array} \right) \\ & =\left( \text{the number of partitions of }d\right) =p\left( d\right) , \end{align*} where $p$ is the partition function. Hence: \begin{proposition} \label{prop.wedge.genfun}Let $m\in\mathbb{Z}$. Every integer $d\geq0$ satisfies $\dim\left( \left( \wedge^{\dfrac{\infty}{2},m}V\right) \left[ -d\right] \right) =p\left( d\right) $, where $p$ is the partition function. As a consequence, in the ring of formal power series $\mathbb{C}% \left[ \left[ q\right] \right] $, we have% \[ \sum\limits_{d\geq0}\dim\left( \left( \wedge^{\dfrac{\infty}{2},m}V\right) \left[ -d\right] \right) q^{d}=\sum\limits_{d\geq0}p\left( d\right) q^{d}=\dfrac{1}{\left( 1-q\right) \left( 1-q^{2}\right) \left( 1-q^{3}\right) \cdots}. \] \end{proposition} \subsection{The Boson-Fermion correspondence} \begin{proposition} \label{prop.wedge.fock}Let $m\in\mathbb{Z}$. Recall the vector $\psi_{m}$ defined in Definition \ref{def.psim}. \textbf{(a)} As an $\mathcal{A}$-module, $\wedge^{\dfrac{\infty}{2},m}V$ is isomorphic to the Fock module $F_{m}$. More precisely, there exists a graded $\mathcal{A}$-module isomorphism $\widetilde{\sigma}_{m}:F_{m}\rightarrow \wedge^{\dfrac{\infty}{2},m}V$ of $\mathcal{A}$-modules such that $\widetilde{\sigma}_{m}\left( 1\right) =\psi_{m}$. \textbf{(b)} As an $\mathcal{A}$-module, $\wedge^{\dfrac{\infty}{2},m}V$ is isomorphic to the Fock module $\widetilde{F}_{m}$. More precisely, there exists a graded $\mathcal{A}$-module isomorphism $\sigma_{m}:\widetilde{F}% _{m}\rightarrow\wedge^{\dfrac{\infty}{2},m}V$ of $\mathcal{A}$-modules such that $\sigma_{m}\left( 1\right) =\psi_{m}$. \end{proposition} \textit{Proof of Proposition \ref{prop.wedge.fock}.} \textbf{(a)} Let us first notice that in the ring $\mathbb{C}\left[ \left[ q\right] \right] $, we have \begin{align*} \sum\limits_{d\geq0}\dim\left( \left( \wedge^{\dfrac{\infty}{2},m}V\right) \left[ -d\right] \right) q^{d} & =\dfrac{1}{\left( 1-q\right) \left( 1-q^{2}\right) \left( 1-q^{3}\right) \cdots}\ \ \ \ \ \ \ \ \ \ \left( \text{by Proposition \ref{prop.wedge.genfun}}\right) \\ & =\sum\limits_{n\geq0}\dim\left( \underbrace{F}_{\substack{=F_{m}% \\\text{(as vector spaces)}}}\left[ -n\right] \right) q^{n}% \ \ \ \ \ \ \ \ \ \ \left( \text{by Definition \ref{def.fock.grad}}\right) \\ & =\sum\limits_{n\geq0}\dim\left( F_{m}\left[ -n\right] \right) q^{n}=\sum\limits_{d\geq0}\dim\left( F_{m}\left[ -d\right] \right) q^{d}. \end{align*} By comparing coefficients, this yields that every integer $d\geq0$ satisfies \begin{equation} \dim\left( \left( \wedge^{\dfrac{\infty}{2},m}V\right) \left[ -d\right] \right) =\dim\left( F_{m}\left[ -d\right] \right) . \label{pf.wedge.fock.dim}% \end{equation} We have $a_{i}\psi_{m}=0$ for all $i>0$ (by degree considerations), and we also have $K\psi_{m}=\psi_{m}$. Besides, it is easy to see that $a_{0}\psi _{m}=m\psi_{m}$\ \ \ \ \footnote{\textit{Proof.} The embedding $\mathcal{A}% \rightarrow\mathfrak{a}_{\infty}$ sends $a_{0}$ to $T^{0}=\mathbf{1}$, where $\mathbf{1}$ denotes the identity matrix in $\mathfrak{a}_{\infty}$. Thus, $a_{0}\psi_{m}=\mathbf{1}\psi_{m}$. (Note that $\mathbf{1}\psi_{m}$ needs not equal $\psi_{m}$ in general, since the action of $\mathfrak{a}_{\infty}$ on $\wedge^{\dfrac{\infty}{2},m}V$ is not an associative algebra action, but just a Lie algebra action.) Recall that $\wedge^{\dfrac{\infty}{2},m}V$ became an $\mathfrak{a}_{\infty}$-module via the map $\widehat{\rho}$, so that $U\psi_{m}=\widehat{\rho}\left( U\right) \psi_{m}$ for every $U\in \mathfrak{a}_{\infty}$. Now,% \begin{align*} a_{0}\psi_{m} & =\mathbf{1}\psi_{m}=\sum\limits_{i\in\mathbb{Z}}E_{i,i}% \psi_{m}\ \ \ \ \ \ \ \ \ \ \left( \text{since }\mathbf{1}=\sum \limits_{i\in\mathbb{Z}}E_{i,i}\right) \\ & =\sum\limits_{i\in\mathbb{Z}}\underbrace{\widehat{\rho}\left( E_{i,i}\right) }_{\substack{=\left\{ \begin{array} [c]{c}% \rho\left( E_{i,i}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=i\text{ and }i\leq0;\\ \rho\left( E_{i,i}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=i\text{ and }i\leq0 \end{array} \right. \\\text{(by the definition of }\widehat{\rho}\text{)}}% }\underbrace{\psi_{m}}_{=v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }U\psi_{m}=\widehat{\rho}\left( U\right) \psi_{m}\text{ for every }U\in\mathfrak{a}_{\infty}\right) \\ & =\sum\limits_{i\in\mathbb{Z}}\left\{ \begin{array} [c]{c}% \rho\left( E_{i,i}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=i\text{ and }i\leq0;\\ \rho\left( E_{i,i}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=i\text{ and }i\leq0 \end{array} \right. \cdot v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...\\ & =\sum\limits_{\substack{i\in\mathbb{Z};\\i>0}}\underbrace{\left\{ \begin{array} [c]{c}% \rho\left( E_{i,i}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=i\text{ and }i\leq0;\\ \rho\left( E_{i,i}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=i\text{ and }i\leq0 \end{array} \right. }_{=\rho\left( E_{i,i}\right) }\cdot v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{i\in\mathbb{Z};\\i\leq 0}}\underbrace{\left\{ \begin{array} [c]{c}% \rho\left( E_{i,i}\right) ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=i\text{ and }i\leq0;\\ \rho\left( E_{i,i}\right) -1,\ \ \ \ \ \ \ \ \ \ \text{if }i=i\text{ and }i\leq0 \end{array} \right. }_{=\rho\left( E_{i,i}\right) -1}\cdot v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...\\ & =\sum\limits_{\substack{i\in\mathbb{Z};\\i>0}}\rho\left( E_{i,i}\right) \cdot v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...+\sum\limits_{\substack{i\in \mathbb{Z};\\i\leq0}}\left( \rho\left( E_{i,i}\right) -1\right) \cdot v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge.... \end{align*} \par Now, we distinguish between two cases: \par \textit{Case 1:} We have $m\geq0$. \par \textit{Case 2:} We have $m<0$. \par In Case 1, we have% \begin{align*} a_{0}\psi_{m} & =\sum\limits_{\substack{i\in\mathbb{Z};\\i>0}}\rho\left( E_{i,i}\right) \cdot v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...+\sum \limits_{\substack{i\in\mathbb{Z};\\i\leq0}}\left( \rho\left( E_{i,i}% \right) -1\right) \cdot v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...\\ & =\sum\limits_{\substack{i\in\mathbb{Z};\\i>0;\ i>m}}\underbrace{\rho\left( E_{i,i}\right) \cdot v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...}% _{\substack{=0\\\text{(since }i\text{ does not appear in the }% m\text{-degression }\left( m,m-1,m-2,...\right) \text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{i\in\mathbb{Z};\\i>0;\ i\leq m}}\underbrace{\rho\left( E_{i,i}\right) \cdot v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...}_{\substack{=v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge ...\\\text{(since }i\text{ appears in the }m\text{-degression }\left( m,m-1,m-2,...\right) \text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{i\in\mathbb{Z};\\i\leq 0}}\underbrace{\left( \rho\left( E_{i,i}\right) -1\right) \cdot v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...}_{\substack{=0\\\text{(since }i\text{ appears in the }m\text{-degression }\left( m,m-1,m-2,...\right) \\\text{and thus we have }\rho\left( E_{i,i}\right) \cdot v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...=v_{m}\wedge v_{m-1}\wedge v_{m-2}% \wedge...\text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since we are in Case 1, so that }% m\geq0\right) \\ & =\underbrace{\sum\limits_{\substack{i\in\mathbb{Z};\\i>0;\ i>m}}0}% _{=0}+\sum\limits_{\substack{i\in\mathbb{Z};\\i>0;\ i\leq m}}\underbrace{v_{m}% \wedge v_{m-1}\wedge v_{m-2}\wedge...}_{=\psi_{m}}+\underbrace{\sum \limits_{\substack{i\in\mathbb{Z};\\i\leq0}}0}_{=0}=\sum \limits_{\substack{i\in\mathbb{Z};\\i>0;\ i\leq m}}\psi_{m}=m\psi_{m}. \end{align*} Hence, $a_{0}\psi_{m}=m\psi_{m}$ is proven in Case 1. In Case 2, the proof of $a_{0}\psi_{m}=m\psi_{m}$ is similar (but instead of splitting the $\sum\limits_{\substack{i\in\mathbb{Z};\\i>0}}$ sum into a $\sum \limits_{\substack{i\in\mathbb{Z};\\i>0;\ i>m}}$ and a $\sum \limits_{\substack{i\in\mathbb{Z};\\i>0;\ i\leq m}}$ sum, we must now split the $\sum\limits_{\substack{i\in\mathbb{Z};\\i\leq0}}$ sum into a $\sum\limits_{\substack{i\in\mathbb{Z};\\i\leq0;\ i>m}}$ and a $\sum \limits_{\substack{i\in\mathbb{Z};\\i\leq0;\ i\leq m}}$ sum). Thus, $a_{0}% \psi_{m}=m\psi_{m}$ holds in both cases 1 and 2. In other words, the proof of $a_{0}\psi_{m}=m\psi_{m}$ is complete.}. Hence, Lemma \ref{lem.V=F.A.gr} (applied to $m$ and $\wedge^{\dfrac{\infty}% {2},m}V$ instead of $\mu$ and $V$) yields that there exists a $\mathbb{Z}% $-graded homomorphism $\widetilde{\sigma}_{m}:F_{m}\rightarrow\wedge ^{\dfrac{\infty}{2},m}V$ of $\mathcal{A}$-modules such that $\widetilde{\sigma }_{m}\left( 1\right) =\psi_{m}$. (An alternative way to prove the existence of this $\widetilde{\sigma}_{m}$ would be to apply Lemma \ref{lem.singvec}, making use of the fact (Proposition \ref{prop.fockverma.A}) that $F_{m}$ is a Verma module for $\mathcal{A}$.) This $\widetilde{\sigma}_{m}$ is injective (since $F_{m}$ is irreducible) and $\mathbb{Z}$-graded. Hence, for every integer $d\geq0$, it induces a homomorphism from $F_{m}\left[ -d\right] $ to $\left( \wedge^{\dfrac {\infty}{2},m}V\right) \left[ -d\right] $. This induced homomorphism must be injective (since $\widetilde{\sigma}_{m}$ was injective), and thus is an isomorphism (since the vector spaces $F_{m}\left[ -d\right] $ and $\left( \wedge^{\dfrac{\infty}{2},m}V\right) \left[ -d\right] $ have the same dimension (by (\ref{pf.wedge.fock.dim})) and are both finite-dimensional). Since this holds for every integer $d\geq0$, this yields that $\widetilde{\sigma}_{m}$ itself must be an isomorphism. This proves Proposition \ref{prop.wedge.fock} \textbf{(a)}. Proposition \ref{prop.wedge.fock} \textbf{(b)} follows from Proposition \ref{prop.wedge.fock} \textbf{(a)} due to Proposition \ref{prop.resc} \textbf{(b)}. Note that Proposition \ref{prop.wedge.fock} is surprising: It gives an isomorphism between a space of polynomials (the Fock space $F_{m}$, also called a \textit{bosonic space}) and a space of wedge products (the space $\wedge^{\dfrac{\infty}{2},m}V$, also called a \textit{fermionic space}); isomorphisms like this are unheard of in finite-dimensional contexts. \begin{definition} We write $\mathcal{B}^{\left( m\right) }$ for the $\mathcal{A}$-module $\widetilde{F}_{m}$. We write $\mathcal{B}$ for the $\mathcal{A}$-module $\bigoplus\limits_{m}\mathcal{B}^{\left( m\right) }=\bigoplus\limits_{m}% \widetilde{F}_{m}$. We write $\mathcal{F}^{\left( m\right) }$ for the $\mathcal{A}$-module $\wedge^{\dfrac{\infty}{2},m}V$. We write $\mathcal{F}$ for the $\mathcal{A}$-module $\bigoplus\limits_{m}\mathcal{F}^{\left( m\right) }$. The isomorphism $\sigma_{m}$ (constructed in Proposition \ref{prop.wedge.fock} \textbf{(b)}) is thus an isomorphism $\mathcal{B}^{\left( m\right) }\rightarrow\mathcal{F}^{\left( m\right) }$. We write $\sigma$ for the $\mathcal{A}$-module isomorphism $\bigoplus\limits_{m}\sigma_{m}% :\mathcal{B}\rightarrow\mathcal{F}$. This $\sigma$ is called the \textit{Boson-Fermion Correspondence}. \end{definition} Note that we can do the same for the Virasoro algebra: If $M_{\lambda}$ is irreducible, then the homomorphism $M_{\lambda}\rightarrow\wedge ^{\dfrac{\infty}{2},m}V_{\alpha,\beta}$ is an isomorphism. And we know that $\operatorname*{Vir}$ is nondegenerate, so $M_{\lambda}$ is irreducible for Weil-generic $\lambda$. \begin{corollary} For generic $\alpha$ and $\beta$, the $\operatorname*{Vir}$-module $\wedge^{\dfrac{\infty}{2},m}V_{\alpha,\beta}$ is irreducible. \end{corollary} But now, back to the Boson-Fermion Correspondence: Both $\mathcal{B}$ and $\mathcal{F}$ are $\mathcal{A}$-modules, and Proposition \ref{prop.wedge.fock} \textbf{(b)} showed us that they are isomorphic as such through the isomorphism $\sigma:\mathcal{B}\rightarrow \mathcal{F}$. However, $\mathcal{F}$ is also an $\mathfrak{a}_{\infty}% $-module, whereas $\mathcal{B}$ is not. But of course, with the isomorphism $\sigma$ being given, we can transfer the $\mathfrak{a}_{\infty}$-module structure from $\mathcal{F}$ to $\mathcal{B}$. The same can be done with the $\mathfrak{gl}_{\infty}$-module structure. Let us explicitly define these: \begin{definition} \textbf{(a)} We make $\mathcal{B}$ into an $\mathfrak{a}_{\infty}$-module by transferring the $\mathfrak{a}_{\infty}$-module structure on $\mathcal{F}$ (given by the map $\widehat{\rho}:\mathfrak{a}_{\infty}\rightarrow \operatorname*{End}\mathcal{F}$) to $\mathcal{B}$ via the isomorphism $\sigma:\mathcal{B}\rightarrow\mathcal{F}$. Note that the $\mathcal{A}$-module $\mathcal{B}$ is a restriction of the $\mathfrak{a}_{\infty}$-module $\mathcal{B}$ (since the $\mathcal{A}$-module $\mathcal{F}$ is the restriction of the $\mathfrak{a}_{\infty}$-module $\mathcal{F}$). We denote the $\mathfrak{a}_{\infty}$-module structure on $\mathcal{B}$ by $\widehat{\rho }:\mathfrak{a}_{\infty}\rightarrow\operatorname*{End}\mathcal{B}$. \textbf{(b)} We make $\mathcal{B}$ into a $\mathfrak{gl}_{\infty}$-module by transferring the $\mathfrak{gl}_{\infty}$-module structure on $\mathcal{F}$ (given by the map $\rho:\mathfrak{gl}_{\infty}\rightarrow\operatorname*{End}% \mathcal{F}$) to $\mathcal{B}$ via the isomorphism $\sigma:\mathcal{B}% \rightarrow\mathcal{F}$. We denote the $\mathfrak{gl}_{\infty}$-module structure on $\mathcal{B}$ by $\rho:\mathfrak{gl}_{\infty}\rightarrow \operatorname*{End}\mathcal{B}$. \end{definition} How do we describe these module structures on $\mathcal{B}$ explicitly (i. e., in formulas?) This question is answered using the so-called \textit{vertex operator construction}. But first, some easier things: \begin{definition} \label{def.createdestroy}Let $m\in\mathbb{Z}$. Let $i\in\mathbb{Z}$. \textbf{(a)} We define the so-called $i$\textit{-th wedging operator} $\widehat{v_{i}}:\mathcal{F}^{\left( m\right) }\rightarrow\mathcal{F}% ^{\left( m+1\right) }$ by% \[ \widehat{v_{i}}\cdot\psi=v_{i}\wedge\psi\ \ \ \ \ \ \ \ \ \ \text{for all }\psi\in\mathcal{F}^{\left( m\right) }. \] Here, $v_{i}\wedge\psi$ is formally defined as follows: Write $\psi$ as a $\mathbb{C}$-linear combination of (well-defined) semiinfinite wedge products $b_{0}\wedge b_{1}\wedge b_{2}\wedge...$ (for instance, elementary semiinfinite wedges); then, $v_{i}\wedge\psi$ is obtained by replacing each such product $b_{0}\wedge b_{1}\wedge b_{2}\wedge...$ by $v_{i}\wedge b_{0}\wedge b_{1}\wedge b_{2}\wedge...$. \textbf{(b)} We define the so-called $i$\textit{-th contraction operator} $\overset{\vee}{v_{i}}:\mathcal{F}^{\left( m\right) }\rightarrow \mathcal{F}^{\left( m-1\right) }$ as follows: For every $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $, we let $\overset{\vee}{v_{i}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right) $ be% \[ \left\{ \begin{array} [c]{l}% 0,\ \ \ \ \ \ \ \ \ \ \text{if }i\notin\left\{ i_{0},i_{1},i_{2},...\right\} ;\\ \left( -1\right) ^{j}v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\wedge v_{i_{j-1}}\wedge v_{i_{j+1}}\wedge v_{i_{j+2}}\wedge ...,\ \ \ \ \ \ \ \ \ \ \text{if }i\in\left\{ i_{0},i_{1},i_{2},...\right\} \end{array} \right. , \] where, in the case $i\in\left\{ i_{0},i_{1},i_{2},...\right\} $, we denote by $j$ the integer $k$ satisfying $i_{k}=i$. Thus, the map $\overset{\vee }{v_{i}}$ is defined on all elementary semiinfinite wedges; we extend this to a map $\mathcal{F}^{\left( m\right) }\rightarrow\mathcal{F}^{\left( m-1\right) }$ by linearity. \end{definition} Note that the somewhat unwieldy definition of $\overset{\vee}{v_{i}}$ can be slightly improved: While it only gave a formula for $m$-degressions, it is easy to see that the same formula holds for straying $m$-degressions: \begin{proposition} Let $m\in\mathbb{Z}$ and $i\in\mathbb{Z}$. Let $\left( i_{0},i_{1}% ,i_{2},...\right) $ be a straying $m$-degression which has no two equal elements. Then,% \begin{align*} & \overset{\vee}{v_{i}}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}% }\wedge...\right) \\ & =\left\{ \begin{array} [c]{l}% 0,\ \ \ \ \ \ \ \ \ \ \text{if }i\notin\left\{ i_{0},i_{1},i_{2},...\right\} ;\\ \left( -1\right) ^{j}v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\wedge v_{i_{j-1}}\wedge v_{i_{j+1}}\wedge v_{i_{j+2}}\wedge ...,\ \ \ \ \ \ \ \ \ \ \text{if }i\in\left\{ i_{0},i_{1},i_{2},...\right\} \end{array} \right. , \end{align*} where, in the case $i\in\left\{ i_{0},i_{1},i_{2},...\right\} $, we denote by $j$ the integer $k$ satisfying $i_{k}=i$. \end{proposition} These operators satisfy the relations% \begin{align*} \widehat{v_{i}}\widehat{v_{j}}+\widehat{v_{j}}\widehat{v_{i}} & =0,\ \ \ \ \ \ \ \ \ \ \overset{\vee}{v_{i}}\overset{\vee}{v_{j}% }+\overset{\vee}{v_{j}}\overset{\vee}{v_{i}}=0,\\ \overset{\vee}{v_{i}}\widehat{v_{j}}+\widehat{v_{j}}\overset{\vee}{v_{i}} & =\delta_{i,j}% \end{align*} for all $i\in\mathbb{Z}$ and $j\in\mathbb{Z}$. \begin{definition} For every $i\in\mathbb{Z}$, define $\xi_{i}=\widehat{v_{i}}$ and $\xi _{i}^{\ast}=\overset{\vee}{v_{i}}$. \end{definition} Then, all $i\in\mathbb{Z}$ and $j\in\mathbb{Z}$ satisfy $\rho\left( E_{i,j}\right) =\xi_{i}\xi_{j}^{\ast}$ and \[ \widehat{\rho}\left( E_{i,j}\right) =\left\{ \begin{array} [c]{c}% \xi_{i}\xi_{j}^{\ast}-1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0,\\ \xi_{i}\xi_{j}^{\ast},\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0 \end{array} \right. . \] The $\xi_{i}$ and $\xi_{i}^{\ast}$ are called \textit{fermionic operators}. So what are the $\xi_{i}$ in terms of $a_{j}$ ? \subsection{The vertex operator construction} We identify the space $\mathbb{C}\left[ z,z^{-1},x_{1},x_{2},...\right] =\bigoplus\limits_{m}z^{m}\mathbb{C}\left[ x_{1},x_{2},...\right] $ with $\mathcal{B}=\bigoplus\limits_{m}\mathcal{B}^{\left( m\right) }$ by means of identifying $z^{m}\mathbb{C}\left[ x_{1},x_{2},...\right] $ with $\mathcal{B}^{\left( m\right) }$ for every $m\in\mathbb{Z}$ (the identification being made through the map% \begin{align*} \mathcal{B}^{\left( m\right) } & \rightarrow z^{m}\mathbb{C}\left[ x_{1},x_{2},...\right] ,\\ p & \mapsto z^{m}\cdot p \end{align*} ). Note also that $z$ (that is, multiplication by $z$) is an isomorphism of $\mathcal{A}_{0}$-modules, but not of $\mathcal{A}$-modules. The Boson-Fermion correspondence goes like this:% \[ \mathcal{F}=\bigoplus\limits_{m}\mathcal{F}^{\left( m\right) }% \overset{\sigma=\bigoplus\limits_{m}\sigma_{m}}{\leftarrow}\mathcal{B}% =\bigoplus\limits_{m}\mathcal{B}^{\left( m\right) }. \] On $\mathcal{F}$ there are operators $\widehat{v_{i}}=\xi_{i}$, $\overset{\vee }{v_{i}}=\xi_{i}^{\ast}$, $\rho\left( E_{i,j}\right) =\xi_{i}\xi_{j}^{\ast}% $, \newline$\widehat{\rho}\left( E_{i,j}\right) =\left\{ \begin{array} [c]{c}% \xi_{i}\xi_{j}^{\ast}-1,\ \ \ \ \ \ \ \ \ \ \text{if }i=j\text{ and }i\leq0,\\ \xi_{i}\xi_{j}^{\ast},\ \ \ \ \ \ \ \ \ \ \text{unless }i=j\text{ and }i\leq0 \end{array} \right. $. By conjugating with the Boson-Fermion correspondence $\sigma$, these operators give rise to operators on $\mathcal{B}$. How do the latter operators look like? \begin{definition} \label{def.euler.XGamma}Introduce the quantum fields% \begin{align*} X\left( u\right) & =\sum\limits_{n\in\mathbb{Z}}\xi_{n}u^{n}\in\left( \operatorname*{End}\mathcal{F}\right) \left[ \left[ u,u^{-1}\right] \right] ,\\ X^{\ast}\left( u\right) & =\sum\limits_{n\in\mathbb{Z}}\xi_{n}^{\ast }u^{-n}\in\left( \operatorname*{End}\mathcal{F}\right) \left[ \left[ u,u^{-1}\right] \right] ,\\ \Gamma\left( u\right) & =\sigma^{-1}\circ X\left( u\right) \circ \sigma\in\left( \operatorname*{End}\mathcal{B}\right) \left[ \left[ u,u^{-1}\right] \right] ,\\ \Gamma^{\ast}\left( u\right) & =\sigma^{-1}\circ X^{\ast}\left( u\right) \circ\sigma\in\left( \operatorname*{End}\mathcal{B}\right) \left[ \left[ u,u^{-1}\right] \right] . \end{align*} Note that $\sigma^{-1}\circ X\left( u\right) \circ\sigma$ is to be read as ``conjugate every term of the power series $X\left( u\right) $ by $\sigma $''; in other words, $\sigma^{-1}\circ X\left( u\right) \circ\sigma$ means $\sum\limits_{n\in\mathbb{Z}}\left( \sigma^{-1}\circ\xi_{n}\circ \sigma\right) u^{n}$. \end{definition} Recall that $\xi_{n}=\widehat{v_{n}}$ sends $\mathcal{F}^{\left( m\right) }$ to $\mathcal{F}^{\left( m+1\right) }$ for any $m\in\mathbb{Z}$ and $n\in\mathbb{Z}$. Thus, every term of the power series $X\left( u\right) =\sum\limits_{n\in\mathbb{Z}}\xi_{n}u^{n}$ sends $\mathcal{F}^{\left( m\right) }$ to $\mathcal{F}^{\left( m+1\right) }$ for any $m\in\mathbb{Z}$. Abusing notation, we will abbreviate this fact by saying that $X\left( u\right) :\mathcal{F}^{\left( m\right) }\rightarrow\mathcal{F}^{\left( m+1\right) }$ for any $m\in\mathbb{Z}$. Similarly, $X^{\ast}\left( u\right) :\mathcal{F}^{\left( m\right) }\rightarrow\mathcal{F}^{\left( m-1\right) }$ for any $m\in\mathbb{Z}$ (since $\xi_{n}^{\ast}=\overset{\vee}{v_{n}}$ sends $\mathcal{F}^{\left( m\right) }$ to $\mathcal{F}^{\left( m-1\right) }$ for any $m\in\mathbb{Z}$ and $n\in\mathbb{Z}$). As a consequence, $\Gamma\left( u\right) :\mathcal{B}^{\left( m\right) }\rightarrow \mathcal{B}^{\left( m+1\right) }$ and $\Gamma^{\ast}\left( u\right) :\mathcal{B}^{\left( m\right) }\rightarrow\mathcal{B}^{\left( m-1\right) }$ for any $m\in\mathbb{Z}$. Now, here is how we can describe $\Gamma\left( u\right) $ and $\Gamma^{\ast }\left( u\right) $ (and therefore the operators $\sigma^{-1}\circ\xi _{n}\circ\sigma$ and $\sigma^{-1}\circ\xi_{n}^{\ast}\circ\sigma$) in terms of $\mathcal{B}$: \begin{theorem} \label{thm.euler}Let $m\in\mathbb{Z}$. On $\mathcal{B}^{\left( m\right) }$, we have% \begin{align*} \Gamma\left( u\right) & =u^{m+1}z\exp\left( \sum\limits_{j>0}% \dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac {a_{j}}{j}u^{-j}\right) ;\\ \Gamma^{\ast}\left( u\right) & =u^{-m}z^{-1}\exp\left( -\sum \limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( \sum \limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) . \end{align*} Here, $\exp A$ means $1+A+\dfrac{A^{2}}{2!}+\dfrac{A^{3}}{3!}+...$ for any $A$ for which this series makes any sense. \end{theorem} Let us explain what we mean by the products $\exp\left( \sum\limits_{j>0}% \dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac {a_{j}}{j}u^{-j}\right) $ and $\exp\left( -\sum\limits_{j>0}\dfrac{a_{-j}% }{j}u^{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{a_{j}}{j}% u^{-j}\right) $ in Theorem \ref{thm.euler}. Why do these products (which are products of exponentials of infinite sums) make any sense? This is easily answered: \begin{itemize} \item For any $v\in\mathcal{B}^{\left( m\right) }$, the term $\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \left( v\right) $ is well-defined and is valued in $\mathcal{B}^{\left( m\right) }\left[ u^{-1}\right] $. (In fact, if we blindly expand% \begin{align*} \exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) & =\sum\limits_{\ell=0}^{\infty}\dfrac{1}{\ell!}\left( -\sum\limits_{j>0}% \dfrac{a_{j}}{j}u^{-j}\right) ^{\ell}\\ & =\sum\limits_{\ell=0}^{\infty}\dfrac{1}{\ell!}\left( -1\right) ^{\ell }\sum\limits_{j_{1},j_{2},...,j_{\ell}\text{ positive integers}}% \dfrac{a_{j_{1}}a_{j_{2}}...a_{j_{\ell}}}{j_{1}j_{2}...j_{\ell}}u^{-\left( j_{1}+j_{2}+...+j_{\ell}\right) }, \end{align*} and apply every term of the resulting power series to $v$, then (for fixed $v$) only finitely many of these terms yield a nonzero result, since $v$ is a polynomial and thus has finite degree, whereas each $a_{j}$ lowers degree by $j$.) \item For any $v\in\mathcal{B}^{\left( m\right) }$, the term $\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) v$ is well-defined and is valued in $\mathcal{B}^{\left( m\right) }\left( \left( u\right) \right) $. (In fact, we have just shown that $\exp\left( -\sum\limits_{j>0}% \dfrac{a_{j}}{j}u^{-j}\right) \left( v\right) \in\mathcal{B}^{\left( m\right) }\left[ u^{-1}\right] $; therefore, applying $\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \in\left( \operatorname*{End}% \left( \mathcal{B}^{\left( m\right) }\right) \right) \left[ \left[ u\right] \right] $ to this gives a well-defined power series in $\mathcal{B}^{\left( m\right) }\left( \left( u\right) \right) $ (because if $\mathfrak{A}$ is an algebra and $\mathfrak{M}$ is an $\mathfrak{A}% $-module, then the application of a power series in $\mathfrak{A}\left[ \left[ u\right] \right] $ to an element of $\mathfrak{M}\left[ u^{-1}\right] $ gives a well-defined element of $\mathfrak{M}\left( \left( u\right) \right) $).) \item For any $v\in\mathcal{B}^{\left( m\right) }$, the term $\exp\left( -\sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) v$ is well-defined and is valued in $\mathcal{B}^{\left( m\right) }\left( \left( u\right) \right) $. (This is proven similarly.) \end{itemize} Thus, the formulas of Theorem \ref{thm.euler} make sense. \begin{remark} Here is some of physicists' intuition for the right hand sides of the equations in Theorem \ref{thm.euler}. [Note: I (=Darij) don't fully understand it, so don't expect me to explain it well.] Consider the quantum field $a\left( u\right) =\sum\limits_{j\in\mathbb{Z}% }a_{j}u^{-j-1}\in U\left( \mathcal{A}\right) \left[ \left[ u,u^{-1}% \right] \right] $ defined in Section \ref{subsect.quantumfields}. Let us work on an informal level, and pretend that integration of series in $U\left( \mathcal{A}\right) \left[ \left[ u,u^{-1}\right] \right] $ is well-defined and behaves similar to that of functions on $\mathbb{R}$. Then, $\int a\left( u\right) du=-\sum\limits_{j\neq0}\dfrac{a_{j}}{j}u^{-j}% +a_{0}\log u$. Exponentiating this \textbf{``in the normal ordering''} (this means we expand the series $\exp\left( -\sum\limits_{j\neq0}\dfrac{a_{j}}% {j}u^{-j}+a_{0}\log u\right) $ and replace all products by their normal ordered versions, i. e., shovel all $a_{m}$ with $m<0$ to the left and all $a_{m}$ with $m>0$ to the right), we obtain% \begin{align*} & \left. :\exp\left( \int a\left( u\right) du\right) :\right. =\left. :\exp\left( -\sum\limits_{j\neq0}\dfrac{a_{j}}{j}u^{-j}+a_{0}\log u\right) :\right. \\ & =\exp\left( \underbrace{-\sum\limits_{j<0}\dfrac{a_{j}}{j}u^{-j}}% _{=\sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}}\right) \cdot\exp\left( a_{0}\log u\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \\ & =\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot \exp\left( a_{0}\log u\right) \cdot\exp\left( -\sum\limits_{j>0}% \dfrac{a_{j}}{j}u^{-j}\right) . \end{align*} But for every $m\in\mathbb{Z}$, we have% \begin{align*} & \Gamma\left( u\right) \\ & =u^{m+1}z\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by Theorem \ref{thm.euler}}\right) \\ & =uz\cdot\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot\underbrace{u^{m}}_{\substack{=\exp\left( m\log u\right) =\exp\left( a_{0}\log u\right) \\\text{(since }a_{0}\text{ acts by }m\text{ on }\mathcal{B}^{\left( m\right) }\text{,}\\\text{and thus }\exp\left( a_{0}\log u\right) =\exp\left( m\log u\right) \text{ on }\mathcal{B}% ^{\left( m\right) }\text{)}}}\cdot\exp\left( -\sum\limits_{j>0}\dfrac {a_{j}}{j}u^{-j}\right) \\ & =uz\cdot\underbrace{\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}% u^{j}\right) \cdot\exp\left( a_{0}\log u\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) }_{=\left. :\exp\left( \int a\left( u\right) du\right) :\right. }\\ & =uz\cdot\left. :\exp\left( \int a\left( u\right) du\right) :\right. . \end{align*} Since the right hand side of this equality does not depend on $m$, we thus have $\Gamma\left( u\right) =uz\left. :\exp\left( \int a\left( u\right) du\right) :\right. $. Hence, we have rewritten half of the statement of Theorem \ref{thm.euler} as the identity $\Gamma\left( u\right) =uz\left. :\exp\left( \int a\left( u\right) du\right) :\right. $ (which holds on all of $\mathcal{B}$). Similarly, the other half of Theorem \ref{thm.euler} rewrites as the identity $\Gamma^{\ast}\left( u\right) =z^{-1}\left. :\exp\left( -\int a\left( u\right) du\right) :\right. $. This is reminiscent of Euler's formula $y=c\exp\left( \int a\left( u\right) du\right) $ for the solution $y$ of the differential equation $y^{\prime}=ay$. \end{remark} Before we can show Theorem \ref{thm.euler}, we state a lemma about the action of $\mathcal{A}$ on $\mathcal{B}$: \begin{lemma} \label{lem.euler.aGamma}For every $j\in\mathbb{Z}$, we have $\left[ a_{j},\Gamma\left( u\right) \right] =u^{j}\Gamma\left( u\right) $ and $\left[ a_{j},\Gamma^{\ast}\left( u\right) \right] =-u^{j}\Gamma^{\ast }\left( u\right) $. \end{lemma} \textit{Proof of Lemma \ref{lem.euler.aGamma}.} Let us prove the first formula. Let $j\in\mathbb{Z}$. On the fermionic space $\mathcal{F}$, the element $a_{j}\in\mathcal{A}$ acts as% \begin{align*} \widehat{\rho}\left( T^{j}\right) & =\sum\limits_{i}\widehat{\rho}\left( E_{i,i+j}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }T^{j}% =\sum\limits_{i\in\mathbb{Z}}E_{i,i+j}\right) \\ & =\sum\limits_{i}\left\{ \begin{array} [c]{c}% \xi_{i}\xi_{i+j}^{\ast}-1,\ \ \ \ \ \ \ \ \ \ \text{if }i=i+j\text{ and }% i\leq0,\\ \xi_{i}\xi_{i+j}^{\ast},\ \ \ \ \ \ \ \ \ \ \text{unless }i=i+j\text{ and }i\leq0 \end{array} \right. \end{align*} (since $\widehat{\rho}\left( E_{i,i+j}\right) =\left\{ \begin{array} [c]{c}% \xi_{i}\xi_{i+j}^{\ast}-1,\ \ \ \ \ \ \ \ \ \ \text{if }i=i+j\text{ and }% i\leq0,\\ \xi_{i}\xi_{i+j}^{\ast},\ \ \ \ \ \ \ \ \ \ \text{unless }i=i+j\text{ and }i\leq0 \end{array} \right. $ for every $i\in\mathbb{Z}$). Hence, on $\mathcal{F}$, we have% \begin{align*} \left[ a_{j},X\left( u\right) \right] & =\left[ \sum\limits_{i}\left\{ % \begin{array} [c]{c}% \xi_{i}\xi_{i+j}^{\ast}-1,\ \ \ \ \ \ \ \ \ \ \text{if }i=i+j\text{ and }% i\leq0,\\ \xi_{i}\xi_{i+j}^{\ast},\ \ \ \ \ \ \ \ \ \ \text{unless }i=i+j\text{ and }i\leq0 \end{array} \right. ,X\left( u\right) \right] \\ & =\sum\limits_{i}\left\{ \begin{array} [c]{c}% \left[ \xi_{i}\xi_{i+j}^{\ast}-1,X\left( u\right) \right] ,\ \ \ \ \ \ \ \ \ \ \text{if }i=i+j\text{ and }i\leq0,\\ \left[ \xi_{i}\xi_{i+j}^{\ast},X\left( u\right) \right] ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=i+j\text{ and }i\leq0 \end{array} \right. \\ & =\sum\limits_{i}\left\{ \begin{array} [c]{c}% \left[ \xi_{i}\xi_{i+j}^{\ast},X\left( u\right) \right] ,\ \ \ \ \ \ \ \ \ \ \text{if }i=i+j\text{ and }i\leq0,\\ \left[ \xi_{i}\xi_{i+j}^{\ast},X\left( u\right) \right] ,\ \ \ \ \ \ \ \ \ \ \text{unless }i=i+j\text{ and }i\leq0 \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left[ \xi_{i}\xi_{i+j}^{\ast }-1,X\left( u\right) \right] =\left[ \xi_{i}\xi_{i+j}^{\ast},X\left( u\right) \right] \right) \\ & =\sum\limits_{i}\left[ \xi_{i}\xi_{i+j}^{\ast},X\left( u\right) \right] =\sum\limits_{i}\left[ \xi_{i}\xi_{i+j}^{\ast},\sum\limits_{m}\xi_{m}% u^{m}\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since }X\left( u\right) =\sum\limits_{m}\xi_{m}u^{m}\right) \\ & =\sum\limits_{i}\sum\limits_{m}\underbrace{\left[ \xi_{i}\xi_{i+j}^{\ast },\xi_{m}\right] }_{\substack{=\delta_{m,i+j}\xi_{i}\\\text{(this is easy to check)}}}u^{m}=\sum\limits_{i}\sum\limits_{m}\delta_{m,i+j}\xi_{i}u^{m}\\ & =\sum\limits_{m}\xi_{m-j}u^{m}=u^{j}\underbrace{\sum\limits_{m}\xi _{m-j}u^{m-j}}_{=X\left( u\right) }=u^{j}X\left( u\right) . \end{align*} Conjugating this equation by $\sigma$, we obtain $\left[ a_{j},\Gamma\left( u\right) \right] =u^{j}\Gamma\left( u\right) $. Similarly, we can prove $\left[ a_{j},\Gamma^{\ast}\left( u\right) \right] =-u^{j}\Gamma^{\ast }\left( u\right) $. Lemma \ref{lem.euler.aGamma} is proven. \textit{Proof of Theorem \ref{thm.euler}.} Define an element $\Gamma _{+}\left( u\right) $ of the $\mathbb{C}$-algebra $\left( \operatorname*{End}\mathcal{B}\right) \left[ \left[ u^{-1}\right] \right] $ by $\Gamma_{+}\left( u\right) =\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}% }{j}u^{-j}\right) $. Then,% \begin{align} \left[ a_{i},\Gamma_{+}\left( u\right) \right] & =0\ \ \ \ \ \ \ \ \ \ \text{if }i\geq0;\label{pf.euler.1}\\ \left[ a_{i},\Gamma_{+}\left( u\right) \right] & =u^{i}\Gamma_{+}\left( u\right) \ \ \ \ \ \ \ \ \ \ \text{if }i<0. \label{pf.euler.2}% \end{align} In fact, (\ref{pf.euler.1}) is trivial (because when $i\geq0$, the element $a_{i}$ commutes with $a_{j}$ for every $j>0$, and thus also commutes with $\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) $). To prove (\ref{pf.euler.2}), it is enough to show that $\left[ a_{i},\exp\left( -\dfrac{a_{-i}}{-i}u^{i}\right) \right] =u^{i}\exp\left( -\dfrac{a_{-i}% }{-i}u^{i}\right) $ (since we can write $\Gamma_{+}\left( u\right) $ in the form \[ \Gamma_{+}\left( u\right) =\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}% {j}u^{-j}\right) =\prod\limits_{j>0}\exp\left( -\dfrac{a_{j}}{j}% u^{-j}\right) , \] and it is clear that $a_{i}$ commutes with all terms $-\dfrac{a_{j}}{j}u^{-j}$ for $j\neq-i$). But this is easily checked using the fact that $\left[ a_{i},a_{-i}\right] =i$ and Lemma \ref{lem.powerseries1} (applied to $K=\mathbb{Q}$, $R=\left( \operatorname*{End}\mathcal{B}\right) \left[ \left[ u^{-1}\right] \right] $, $\alpha=a_{i}$, $\beta=a_{-i}$ and $P=\exp\left( -\dfrac{X}{-i}u^{i}\right) $). This completes the proof of (\ref{pf.euler.2}). Since $\Gamma_{+}\left( u\right) $ is an invertible power series in $\left( \operatorname*{End}\mathcal{B}\right) \left[ \left[ u^{-1}\right] \right] $ (because the constant term of $\Gamma_{+}\left( u\right) $ is $1$), it makes sense to speak of the power series $\Gamma_{+}\left( u\right) ^{-1}% \in\left( \operatorname*{End}\mathcal{B}\right) \left[ \left[ u^{-1}\right] \right] $. From (\ref{pf.euler.1}) and (\ref{pf.euler.2}), we can derive the formulas% \begin{align} \left[ a_{i},\Gamma_{+}\left( u\right) ^{-1}\right] & =0\ \ \ \ \ \ \ \ \ \ \text{if }i\geq0;\label{pf.euler.1inv}\\ \left[ a_{i},\Gamma_{+}\left( u\right) ^{-1}\right] & =-u^{i}\Gamma _{+}\left( u\right) ^{-1}\ \ \ \ \ \ \ \ \ \ \text{if }i<0 \label{pf.euler.2inv}% \end{align} (using the standard fact that $\left[ \alpha,\beta^{-1}\right] =-\beta ^{-1}\left[ \alpha,\beta\right] \beta^{-1}$ for any two elements $\alpha$ and $\beta$ of a ring such that $\beta$ is invertible). Now define a map $\Delta\left( u\right) :\mathcal{B}^{\left( m\right) }\rightarrow\mathcal{B}^{\left( m\right) }\left( \left( u\right) \right) $ by $\Delta\left( u\right) =\Gamma\left( u\right) \Gamma_{+}\left( u\right) ^{-1}z^{-1}$. Let us check why this definition makes sense: \begin{itemize} \item For any $v\in\mathcal{B}^{\left( m\right) }$, we have $z^{-1}% v\in\mathcal{B}^{\left( m-1\right) }$, and the term $\Gamma_{+}\left( u\right) ^{-1}z^{-1}v$ is well-defined and is valued in $\mathcal{B}^{\left( m-1\right) }\left[ u^{-1}\right] $.\ \ \ \ \footnote{\textit{Proof.} Recall that $\mathcal{A}$ is a $\mathbb{Z}$-graded Lie algebra, and that $\mathcal{B}$ is a $\mathbb{Z}$-graded $\mathcal{A}$-module concentrated in nonpositive degrees. Let us (for this single proof!) change the $\mathbb{Z}% $-gradings on both $\mathcal{A}$ and $\mathcal{B}$ to their inverses (i. e., switch $\mathcal{A}\left[ N\right] $ with $\mathcal{A}\left[ -N\right] $ for every $N\in\mathbb{Z}$, and switch $\mathcal{B}\left[ N\right] $ with $\mathcal{B}\left[ -N\right] $ for every $N\in\mathbb{Z}$); then, $\mathcal{A}$ remains still a $\mathbb{Z}$-graded Lie algebra, but $\mathcal{B}$ is now a $\mathbb{Z}$-graded $\mathcal{A}$-module concentrated in nonnegative degrees. Moreover, $\mathcal{B}$ is actually a $\mathbb{Z}% $-graded $\operatorname*{End}\nolimits_{\operatorname{hg}}\mathcal{B}$-module concentrated in nonnegative degrees. \par The power series $\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\in\left( \operatorname*{End}\nolimits_{\operatorname{hg}}\mathcal{B}\right) \left[ \left[ u^{-1}\right] \right] $ is now equigraded (since our modified grading on $\mathcal{A}$ has the property that $\deg\left( a_{j}\right) =-j$), so that the power series $\exp\left( \sum\limits_{j>0}\dfrac{a_{j}}% {j}u^{-j}\right) \in\left( \operatorname*{End}\nolimits_{\operatorname{hg}% }\mathcal{B}\right) \left[ \left[ u^{-1}\right] \right] $ is equigraded as well (because a consequence of Proposition \ref{prop.equigraded.basics} \textbf{(b)} is that whenever the exponential of an equigraded power series is well-defined, this exponential is also equigraded). Since% \[ \Gamma_{+}\left( u\right) ^{-1}=\left( \exp\left( -\sum\limits_{j>0}% \dfrac{a_{j}}{j}u^{-j}\right) \right) ^{-1}=\exp\left( \sum\limits_{j>0}% \dfrac{a_{j}}{j}u^{-j}\right) \] (since Corollary \ref{cor.exp(-w)} (applied to $R=\left( \operatorname*{End}% \mathcal{B}\right) \left[ \left[ u^{-1}\right] \right] $, $I=\left( \text{the ideal of }R\text{ consisting of all power series with constant term }1\right) $, and $\gamma=-\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}$) yields $\left( \exp\left( \sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \right) \cdot\left( \exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \right) =1$), this rewrites as follows: The power series $\Gamma_{+}\left( u\right) ^{-1}\in\left( \operatorname*{End}\nolimits_{\operatorname{hg}% }\mathcal{B}\right) \left[ \left[ u^{-1}\right] \right] $ is equigraded. \par Therefore, Proposition \ref{prop.equigraded.fx} \textbf{(c)} (applied to $\operatorname*{End}\nolimits_{\operatorname{hg}}\mathcal{B}$, $\mathcal{B}$, $\Gamma_{+}\left( u\right) ^{-1}$ and $z^{-1}v$ instead of $A$, $M$, $f$ and $x$) yields that $\Gamma_{+}\left( u\right) ^{-1}z^{-1}v$ is a well-defined element of $\mathcal{B}^{\left( m-1\right) }\left[ u^{-1}\right] $, qed.} \item For any $v\in\mathcal{B}^{\left( m\right) }$, the term $\Gamma\left( u\right) \Gamma_{+}\left( u\right) ^{-1}z^{-1}$ is well-defined and is valued in $\mathcal{B}^{\left( m\right) }\left( \left( u\right) \right) $.\ \ \ \ \footnote{\textit{Proof.} We have just shown that $\Gamma_{+}\left( u\right) ^{-1}z^{-1}v\in\mathcal{B}^{\left( m-1\right) }\left[ u^{-1}\right] $. Thus, $\Gamma_{+}\left( u\right) ^{-1}z^{-1}% v\in\mathcal{B}^{\left( m-1\right) }\left[ u^{-1}\right] \subseteq \mathcal{B}\left[ u^{-1}\right] \subseteq\mathcal{B}\left[ u,u^{-1}\right] $. \par Recall that $\mathcal{A}$ is a $\mathbb{Z}$-graded Lie algebra, and that $\mathcal{B}$ and $\mathcal{F}$ are $\mathbb{Z}$-graded $\mathcal{A}$-modules concentrated in nonpositive degrees. Let us (for this single proof!) change the $\mathbb{Z}$-gradings on all of $\mathcal{A}$, $\mathcal{B}$ and $\mathcal{F}$ to their inverses (i. e., switch $\mathcal{A}\left[ N\right] $ with $\mathcal{A}\left[ -N\right] $ for every $N\in\mathbb{Z}$, and switch $\mathcal{B}\left[ N\right] $ with $\mathcal{B}\left[ -N\right] $ for every $N\in\mathbb{Z}$, and switch $\mathcal{F}\left[ N\right] $ with $\mathcal{F}\left[ -N\right] $ for every $N\in\mathbb{Z}$); then, $\mathcal{A}$ remains still a $\mathbb{Z}$-graded Lie algebra, but $\mathcal{B}$ and $\mathcal{F}$ now are $\mathbb{Z}$-graded $\mathcal{A}% $-modules concentrated in nonnegative degrees. Moreover, $\mathcal{B}$ is actually a $\mathbb{Z}$-graded $\operatorname*{End}% \nolimits_{\operatorname{hg}}\mathcal{B}$-module concentrated in nonnegative degrees, and $\mathcal{F}$ is a $\mathbb{Z}$-graded $\operatorname*{End}% \nolimits_{\operatorname{hg}}\mathcal{F}$-module concentrated in nonnegative degrees. \par It is easy to see (from the definition of $X\left( u\right) $) that $X\left( u\right) \in\left( \operatorname*{End}\nolimits_{\operatorname{hg}% }\mathcal{F}\right) \left[ \left[ u,u^{-1}\right] \right] $ is equigraded. As a consequence, $\Gamma\left( u\right) \in\left( \operatorname*{End}\nolimits_{\operatorname{hg}}\mathcal{B}\right) \left[ \left[ u,u^{-1}\right] \right] $ is equigraded (since $\Gamma\left( u\right) =\sigma^{-1}\circ X\left( u\right) \circ\sigma$). Therefore, Proposition \ref{prop.equigraded.fx} \textbf{(b)} (applied to $\operatorname*{End}\nolimits_{\operatorname{hg}}\mathcal{B}$, $\mathcal{B}$, $\Gamma\left( u\right) $ and $\Gamma_{+}\left( u\right) ^{-1}z^{-1}v$ instead of $A$, $M$, $f$ and $x$) yields that $\Gamma\left( u\right) \Gamma_{+}\left( u\right) ^{-1}z^{-1}$ is a well-defined element of $\mathcal{B}\left( \left( u\right) \right) $. This element actually lies in $\mathcal{B}^{\left( m\right) }\left( \left( u\right) \right) $ (since $\Gamma\left( u\right) :\mathcal{B}^{\left( m-1\right) }% \rightarrow\mathcal{B}^{\left( m\right) }$), qed.} \end{itemize} Since $\left[ a_{0},z\right] =z$ and $\left[ a_{i},z\right] =0$ for all $i\neq0$, we have% \begin{equation} \left[ a_{i},\Delta\left( u\right) \right] =\left\{ \begin{array} [c]{c}% 0,\ \ \ \ \ \ \ \ \ \ \text{if }i\leq0;\\ u^{i}\Delta\left( u\right) ,\ \ \ \ \ \ \ \ \ \ \text{if }i>0 \end{array} \right. \label{pf.euler.Del}% \end{equation} (due to (\ref{pf.euler.1inv}), (\ref{pf.euler.2inv}) and Lemma \ref{lem.euler.aGamma}). In particular, $\left[ a_{i},\Delta\left( u\right) \right] =0$ if $i\leq0$. Thus, $\Delta\left( u\right) $ is a homomorphism of $\mathcal{A}_{-}$-modules, where $\mathcal{A}_{-}$ is the Lie subalgebra $\left\langle a_{-1},a_{-2},a_{-3},...\right\rangle $ of $\mathcal{A}$. (Of course, this formulation means that every term of the formal power series $\Delta\left( u\right) $ is a homomorphism of $\mathcal{A}_{-}$-modules.) Consider now the element $z^{m}$ of $z^{m}\mathbb{C}\left[ x_{1}% ,x_{2},...\right] =\mathcal{B}^{\left( m\right) }=\widetilde{F}_{m}$. Also, consider the element $\psi_{m}=v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...$ of $\wedge^{\dfrac{\infty}{2},m}V=\mathcal{F}^{\left( m\right) }$ defined in Definition \ref{def.psim}. By the definition of $\sigma_{m}$, we have $\sigma_{m}\left( z^{m}\right) =\psi_{m}$. (In fact, $z^{m}$ is what was denoted by $1$ in Proposition \ref{prop.wedge.fock}.) From Lemma \ref{lem.F.P1=P}, it is clear that the Fock module $F$ is generated by $1$ as an $\mathcal{A}_{-}$-module (since $\mathcal{A}_{-}=\left\langle a_{-1},a_{-2},a_{-3},...\right\rangle $). Since there exists an $\mathcal{A}% _{-}$-module isomorphism $F\rightarrow\widetilde{F}$ which sends $1$ to $1$ (in fact, the map $\operatorname*{resc}$ of Proposition \ref{prop.resc} is such an isomorphism), this yields that $\widetilde{F}$ is generated by $1$ as an $\mathcal{A}_{-}$-module. Since there exists an $\mathcal{A}_{-}$-module isomorphism $\widetilde{F}\rightarrow\widetilde{F}_{m}$ which sends $1$ to $z^{m}$ (in fact, multiplication by $z^{m}$ is such an isomorphism), this yields that $\widetilde{F}_{m}$ is generated by $z^{m}$ as an $\mathcal{A}% _{-}$-module. Consequently, the $m$-th term of the power series $\Delta\left( u\right) $ is completely determined by $\left( \Delta\left( u\right) \right) \left( z^{m}\right) $ (because we know that $\Delta\left( u\right) $ is a homomorphism of $\mathcal{A}_{-}$-modules). So let us compute $\left( \Delta\left( u\right) \right) \left( z^{m}\right) $. Since $\Delta\left( u\right) :\mathcal{B}^{\left( m\right) }\rightarrow \mathcal{B}^{\left( m\right) }\left( \left( u\right) \right) $, we know that $\left( \Delta\left( u\right) \right) \left( z^{m}\right) $ is an element of $\underbrace{\mathcal{B}^{\left( m\right) }}_{=z^{m}% \widetilde{F}}\left( \left( u\right) \right) =z^{m}\widetilde{F}\left( \left( u\right) \right) $. In other words, $\left( \Delta\left( u\right) \right) \left( z^{m}\right) $ is $z^{m}$ times a Laurent series in $u$ whose coefficients are polynomials in $x_{1},x_{2},x_{3},...$. Denote this Laurent series by $Q$. Thus, $\left( \Delta\left( u\right) \right) \left( z^{m}\right) =z^{m}Q$. For every $i>0$, we have \[ a_{i}\Delta\left( u\right) =\Delta\left( u\right) a_{i}% +\underbrace{\left[ a_{i},\Delta\left( u\right) \right] }% _{\substack{=u^{i}\Delta\left( u\right) \\\text{(by (\ref{pf.euler.Del}))}% }}=\Delta\left( u\right) a_{i}+u^{i}\Delta\left( u\right) , \] so that% \begin{align*} \left( a_{i}\Delta\left( u\right) \right) \left( z^{m}\right) & =\left( \Delta\left( u\right) a_{i}+u^{i}\Delta\left( u\right) \right) \left( z^{m}\right) =\Delta\left( u\right) \underbrace{a_{i}z^{m}% }_{\substack{=0\\\text{(since }a_{i}=\dfrac{\partial}{\partial x_{i}}\text{)}% }}+u^{i}\underbrace{\left( \Delta\left( u\right) \right) \left( z^{m}\right) }_{=z^{m}Q}\\ & =u^{i}z^{m}Q=z^{m}u^{i}Q. \end{align*} Since $\left( a_{i}\Delta\left( u\right) \right) \left( z^{m}\right) =a_{i}\underbrace{\left( \left( \Delta\left( u\right) \right) \left( z^{m}\right) \right) }_{=z^{m}Q}=z^{m}\underbrace{a_{i}}_{=\dfrac{\partial }{\partial x_{i}}}Q=z^{m}\dfrac{\partial Q}{\partial x_{i}}$, this rewrites as $z^{m}\dfrac{\partial Q}{\partial x_{i}}=z^{m}u^{i}Q$. Hence, for every $i>0$, we have $\dfrac{\partial Q}{\partial x_{i}}=u^{i}Q$. Thus, we can write the formal Laurent series $Q$ in the form $Q=f\left( u\right) \exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) $ for some Laurent series $f\left( u\right) \in\mathbb{C}\left( \left( u\right) \right) \ \ \ \ $% .\footnote{This follows from Proposition \ref{prop.euler.recognizing-exp}, applied to $R=\mathbb{C}\left[ u\right] $, $U=\mathbb{C}\left( \left( u\right) \right) $, $\left( \alpha_{1},\alpha_{2},\alpha_{3},...\right) =\left( u^{1},u^{2},u^{3},...\right) $ and $P=Q$.} Thus, \begin{align*} & \left( \Delta\left( u\right) \right) \left( z^{m}\right) \\ & =z^{m}Q=z^{m}f\left( u\right) \exp\left( \sum\limits_{j>0}x_{j}% u^{j}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }Q=f\left( u\right) \exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) \right) \\ & =f\left( u\right) \exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}% u^{j}\right) \left( z^{m}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since each }\dfrac{a_{-j}}{j}\text{ acts as multiplication by }x_{j}\text{ on }\widetilde{F}\right) . \end{align*} In other words, the two maps $\Delta\left( u\right) $ and $f\left( u\right) \exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) $ are equal on $z^{m}$. Since each of these two maps is an $\mathcal{A}_{-}$-module homomorphism\footnote{In fact, we know that $\Delta\left( u\right) $ is an $\mathcal{A}_{-}$-module homomorphism, and it is clear that $f\left( u\right) \exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) $ is an $\mathcal{A}_{-}$-module homomorphism because $\mathcal{A}_{-}$ is an abelian Lie algebra.}, this yields that these two maps must be identical (because $\widetilde{F}_{m}$ is generated by $z^{m}$ as an $\mathcal{A}_{-}$-module). In other words, $\Delta\left( u\right) =f\left( u\right) \exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) $. Since $\Delta\left( u\right) =\Gamma\left( u\right) \Gamma_{+}\left( u\right) ^{-1}z^{-1}$, this becomes $\Gamma\left( u\right) \Gamma_{+}\left( u\right) ^{-1}% z^{-1}=f\left( u\right) \exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}% u^{j}\right) $, so that% \begin{align} \Gamma\left( u\right) & =f\left( u\right) \exp\left( \sum \limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot z\cdot\Gamma_{+}\left( u\right) =f\left( u\right) \exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}% {j}u^{j}\right) \cdot z\cdot\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}% {j}u^{-j}\right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\Gamma_{+}\left( u\right) =\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \right) \nonumber\\ & =f\left( u\right) z\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}% u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}% u^{-j}\right) \label{pf.euler.Gamma-through-f}% \end{align} on $\mathcal{B}^{\left( m\right) }$. It remains to show that $f\left( u\right) =u^{m+1}$. In order to do this, we recall that \begin{align*} \left( \Gamma\left( u\right) \right) \left( z^{m}\right) & =f\left( u\right) z\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot\underbrace{\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \left( z^{m}\right) }_{\substack{=z^{m}\\\text{(because }a_{j}\left( z^{m}\right) =0\text{ for every }j>0\text{)}}}\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.euler.Gamma-through-f})}\right) \\ & =f\left( u\right) z\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}% u^{j}\right) \left( z^{m}\right) =f\left( u\right) z\exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) z^{m}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since each }\dfrac{a_{-j}}{j}\text{ acts as multiplication by }x_{j}\text{ on }\widetilde{F}\right) \\ & =f\left( u\right) \exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) z^{m+1}. \end{align*} On the other hand, back on the fermionic side, for the vector $\psi_{m}% =v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...$, we have% \begin{align*} \left( X\left( u\right) \right) \psi_{m} & =\sum\limits_{n\in\mathbb{Z}% }\widehat{v_{n}}\left( \psi_{m}\right) u^{n}\ \ \ \ \ \ \ \ \ \ \left( \text{since }X\left( u\right) =\sum\limits_{n\in\mathbb{Z}}\underbrace{\xi _{n}}_{=\widehat{v_{n}}}u^{n}=\sum\limits_{n\in\mathbb{Z}}\widehat{v_{n}}% u^{n}\right) \\ & =\sum\limits_{\substack{n\in\mathbb{Z};\\n\leq m}% }\underbrace{\widehat{v_{n}}\left( \psi_{m}\right) }% _{\substack{=0\\\text{(since }n\leq m\text{, so that }v_{n}\\\text{appears in }v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...=\psi_{m}\text{)}}}u^{n}% +\sum\limits_{\substack{n\in\mathbb{Z};\\n\geq m+1}}\widehat{v_{n}}\left( \psi_{m}\right) u^{n}=\sum\limits_{\substack{n\in\mathbb{Z};\\n\geq m+1}}\widehat{v_{n}}\left( \psi_{m}\right) u^{n}. \end{align*} Thus, $\sigma^{-1}\left( \left( X\left( u\right) \right) \psi_{m}\right) =\sigma^{-1}\left( \sum\limits_{\substack{n\in\mathbb{Z};\\n\geq m+1}}\widehat{v_{n}}\left( \psi_{m}\right) u^{n}\right) $. Compared with% \begin{align*} \sigma^{-1}\left( \left( X\left( u\right) \right) \underbrace{\psi_{m}% }_{=\sigma\left( z^{m}\right) }\right) & =\sigma^{-1}\left( \left( X\left( u\right) \right) \left( \sigma\left( z^{m}\right) \right) \right) =\underbrace{\left( \sigma^{-1}\circ X\left( u\right) \circ \sigma\right) }_{=\Gamma\left( u\right) }\left( z^{m}\right) =\left( \Gamma\left( u\right) \right) \left( z^{m}\right) \\ & =f\left( u\right) \exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) z^{m+1}, \end{align*} this yields $\sigma^{-1}\left( \sum\limits_{\substack{n\in\mathbb{Z};\\n\geq m+1}}\widehat{v_{n}}\left( \psi_{m}\right) u^{n}\right) =f\left( u\right) \exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) z^{m+1}$, so that% \begin{equation} \sigma\left( f\left( u\right) \exp\left( \sum\limits_{j>0}x_{j}% u^{j}\right) z^{m+1}\right) =\sum\limits_{\substack{n\in\mathbb{Z};\\n\geq m+1}}\widehat{v_{n}}\left( \psi_{m}\right) u^{n}. \label{pf.euler.compare}% \end{equation} We want to find $f\left( u\right) $ by comparing the sides of this equation. In order to do this, we recall that each space $\mathcal{B}^{\left( i\right) }$ is graded; hence, $\mathcal{B}$ (being the direct sum of the $\mathcal{B}% ^{\left( i\right) }$) is also graded (by taking the direct sum of all the gradings). Also, each space $\mathcal{F}^{\left( i\right) }$ is graded; hence, $\mathcal{F}$ (being the direct sum of the $\mathcal{F}^{\left( i\right) }$) is also graded (by taking the direct sum of all the gradings). Since each $\sigma_{m}$ is a graded map, the direct sum $\sigma=\bigoplus \limits_{m\in\mathbb{Z}}\sigma_{m}$ is also graded. Therefore,% \begin{align} & \sigma\left( 0\text{-th homogeneous component of }f\left( u\right) \exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) z^{m+1}\right) \nonumber\\ & =\left( 0\text{-th homogeneous component of }\sigma\left( f\left( u\right) \exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) z^{m+1}\right) \right) \nonumber\\ & =\left( 0\text{-th homogeneous component of }\sum\limits_{\substack{n\in \mathbb{Z};\\n\geq m+1}}\widehat{v_{n}}\left( \psi_{m}\right) u^{n}\right) \label{pf.euler.compare2}% \end{align} (by (\ref{pf.euler.compare})). Now, for every $n\in\mathbb{Z}$ satisfying $n\geq m+1$, the element $\widehat{v_{n}}\left( \psi_{m}\right) $ equals $v_{n}\wedge v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...$, and thus has degree $-\left( n-m-1\right) $. Hence, for every nonpositive $i\in\mathbb{Z}$, the $i$-th homogeneous component of the sum $\sum\limits_{\substack{n\in \mathbb{Z};\\n\geq m+1}}\widehat{v_{n}}\left( \psi_{m}\right) u^{n}% \in\mathcal{F}$ is $\widehat{v_{m+1-i}}\left( \psi_{m}\right) u^{m+1-i}$. In particular, the $0$-th homogeneous component of $\sum\limits_{\substack{n\in \mathbb{Z};\\n\geq m+1}}\widehat{v_{n}}\left( \psi_{m}\right) u^{n}$ is $\widehat{v_{m+1}}\left( \psi_{m}\right) u^{m+1}=\psi_{m+1}u^{m+1}$ (since $\widehat{v_{m+1}}\left( \psi_{m}\right) =v_{m+1}\wedge v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...=\psi_{m+1}$). Therefore, (\ref{pf.euler.compare2}) becomes% \begin{equation} \sigma\left( 0\text{-th homogeneous component of }f\left( u\right) \exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) z^{m+1}\right) =\psi _{m+1}u^{m+1}. \label{pf.euler.compare3}% \end{equation} On the other hand, the $0$-th homogeneous component of the element $f\left( u\right) \exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) z^{m+1}% \in\mathcal{B}$ is clearly $f\left( u\right) z^{m+1}$ (because $\exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) =1+\left( \text{terms involving at least one }x_{j}\right) $, and every $x_{j}$ lowers the degree). Thus, (\ref{pf.euler.compare3}) becomes $\sigma\left( f\left( u\right) z^{m+1}\right) =\psi_{m+1}u^{m+1}$. Since $\sigma\left( f\left( u\right) z^{m+1}\right) =f\left( u\right) \underbrace{\sigma\left( z^{m+1}\right) }_{=\psi_{m+1}}=f\left( u\right) \psi_{m+1}$, this rewrites as $f\left( u\right) \psi_{m+1}=\psi_{m+1}u^{m+1}$, so that $f\left( u\right) =u^{m+1}% $. Hence, (\ref{pf.euler.Gamma-through-f}) becomes \begin{align*} \Gamma\left( u\right) & =\underbrace{f\left( u\right) }_{=u^{m+1}}% z\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \\ & =u^{m+1}z\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \ \ \ \ \ \ \ \ \ \ \text{on }\mathcal{B}^{\left( m\right) }. \end{align*} This proves one of the equalities of Theorem \ref{thm.euler}. The other is proven similarly. Theorem \ref{thm.euler} is proven. \begin{corollary} \label{cor.euler}Let $m\in\mathbb{Z}$. On $\mathcal{B}^{\left( m\right) }$, we have% \[ \rho\left( \sum\limits_{\left( i,j\right) \in\mathbb{Z}^{2}}u^{i}% v^{-j}E_{i,j}\right) =\sum\limits_{\left( i,j\right) \in\mathbb{Z}^{2}% }u^{i}v^{-j}\xi_{i}\xi_{j}^{\ast}=X\left( u\right) X^{\ast}\left( v\right) , \] thus% \begin{align*} & \sigma^{-1}\circ\rho\left( \sum\limits_{\left( i,j\right) \in \mathbb{Z}^{2}}u^{i}v^{-j}E_{i,j}\right) \circ\sigma\\ & =\sigma^{-1}\circ X\left( u\right) X^{\ast}\left( v\right) \circ \sigma=\Gamma\left( u\right) \Gamma^{\ast}\left( v\right) \\ & =\dfrac{1}{1-\dfrac{v}{u}}\cdot\left( \dfrac{u}{v}\right) ^{m}\exp\left( \sum\limits_{j>0}\dfrac{u^{j}-v^{j}}{j}a_{-j}\right) \exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}-v^{-j}}{j}a_{j}\right) \end{align*} as linear maps from $\mathcal{B}^{\left( m\right) }$ to $\mathcal{B}% ^{\left( m\right) }\left( \left( u,v\right) \right) $. \end{corollary} \begin{remark} It must be pointed out that the term% \[ \dfrac{1}{1-\dfrac{v}{u}}\cdot\left( \dfrac{u}{v}\right) ^{m}\exp\left( \sum\limits_{j>0}\dfrac{u^{j}-v^{j}}{j}a_{-j}\right) \exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}-v^{-j}}{j}a_{j}\right) \] only makes sense as a map from $\mathcal{B}^{\left( m\right) }$ to $\mathcal{B}^{\left( m\right) }\left( \left( u,v\right) \right) $, but not (for example) as a map from $\mathcal{B}^{\left( m\right) }$ to $\mathcal{B}^{\left( m\right) }\left[ \left[ u,u^{-1},v,v^{-1}\right] \right] $ or as an element of $\left( \operatorname*{End}\left( \mathcal{B}^{\left( m\right) }\right) \right) \left[ \left[ u,u^{-1},v,v^{-1}\right] \right] $. Indeed, $1-\dfrac{v}{u}$ is a zero-divisor in $\mathbb{C}\left[ \left[ u,u^{-1},v,v^{-1}\right] \right] $ (since $\left( 1-\dfrac{v}{u}\right) \sum\limits_{k\in\mathbb{Z}}\left( \dfrac{v}{u}\right) ^{k}=0$), so it does not make sense, for example, to multiply a generic element of $\mathcal{B}^{\left( m\right) }\left[ \left[ u,u^{-1},v,v^{-1}\right] \right] $ by $\dfrac{1}{1-\dfrac{v}{u}}$. An element of $\mathcal{B}^{\left( m\right) }\left( \left( u,v\right) \right) $ needs not always be a multiple of $1-\dfrac{v}{u}$, but at least when it is, the quotient is unique. \end{remark} The importance of Corollary \ref{cor.euler} lies in the fact that it gives an easy way to compute the $\rho$-action of $\mathfrak{gl}_{\infty}$ on $\mathcal{B}^{\left( m\right) }$: In fact, for any $p\in\mathbb{Z}$ and $q\in\mathbb{Z}$, the coefficient of $\sigma^{-1}\circ\rho\left( \sum\limits_{i,j}u^{i}v^{-j}E_{i,j}\right) \circ\sigma\in\left( \operatorname*{End}\left( \mathcal{B}^{\left( m\right) }\right) \right) \left[ \left[ u,u^{-1},v,v^{-1}\right] \right] $ before $u^{p}v^{-q}$ is $\sigma^{-1}\circ\rho\left( E_{p,q}\right) \circ\sigma$, and this is exactly the action of $E_{p,q}$ on $\mathcal{B}^{\left( m\right) }$ obtained by transferring the action $\rho$ of $\mathfrak{gl}_{\infty}$ on $\mathcal{F}% ^{\left( m\right) }$ to $\mathcal{B}^{\left( m\right) }$. \textit{Proof of Corollary \ref{cor.euler}.} First of all, we clearly have% \begin{align*} \rho\left( \sum\limits_{\left( i,j\right) \in\mathbb{Z}^{2}}u^{i}% v^{-j}E_{i,j}\right) & =\sum\limits_{\left( i,j\right) \in\mathbb{Z}^{2}% }u^{i}v^{-j}\underbrace{\rho\left( E_{i,j}\right) }_{=\xi_{i}\xi_{j}^{\ast}% }=\sum\limits_{\left( i,j\right) \in\mathbb{Z}^{2}}u^{i}v^{-j}\xi_{i}\xi _{j}^{\ast}\\ & =\underbrace{\left( \sum\limits_{i\in\mathbb{Z}}\xi_{i}u^{i}\right) }_{=\sum\limits_{n\in\mathbb{Z}}\xi_{n}u^{n}=X\left( u\right) }% \underbrace{\sum\limits_{j\in\mathbb{Z}}\xi_{j}^{\ast}v^{-j}}_{=\sum \limits_{n\in\mathbb{Z}}\xi_{n}^{\ast}v^{-n}=X^{\ast}\left( v\right) }=X\left( u\right) X^{\ast}\left( v\right) , \end{align*} so that% \begin{align*} & \sigma^{-1}\circ\rho\left( \sum\limits_{\left( i,j\right) \in \mathbb{Z}^{2}}u^{i}v^{-j}E_{i,j}\right) \circ\sigma\\ & =\sigma^{-1}\circ X\left( u\right) X^{\ast}\left( v\right) \circ \sigma=\Gamma\left( u\right) \Gamma^{\ast}\left( v\right) . \end{align*} It thus only remains to prove that% \[ \Gamma\left( u\right) \Gamma^{\ast}\left( v\right) =\dfrac{1}{1-\dfrac {v}{u}}\cdot\left( \dfrac{u}{v}\right) ^{m}\exp\left( \sum\limits_{j>0}% \dfrac{u^{j}-v^{j}}{j}a_{-j}\right) \exp\left( -\sum\limits_{j>0}% \dfrac{u^{-j}-v^{-j}}{j}a_{j}\right) . \] By Theorem \ref{thm.euler} (applied to $m-1$ instead of $m$), we have% \[ \Gamma\left( u\right) =u^{m}z\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}% {j}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}% u^{-j}\right) \ \ \ \ \ \ \ \ \ \ \text{on }\mathcal{B}^{\left( m-1\right) }. \] By Theorem \ref{thm.euler}, we have% \[ \Gamma^{\ast}\left( v\right) =v^{-m}z^{-1}\exp\left( -\sum\limits_{j>0}% \dfrac{a_{-j}}{j}v^{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{a_{j}% }{j}v^{-j}\right) \ \ \ \ \ \ \ \ \ \ \text{on }\mathcal{B}^{\left( m\right) }. \] Multiplying these two equalities, we obtain% \begin{align*} \Gamma\left( u\right) \Gamma^{\ast}\left( v\right) & =u^{m}v^{-m}% \cdot\exp\left( \sum\limits_{j>0}\dfrac{u^{j}}{j}a_{-j}\right) \exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}}{j}a_{j}\right) \\ & \ \ \ \ \ \ \ \ \ \ \cdot\exp\left( -\sum\limits_{j>0}\dfrac{v^{j}}% {j}a_{-j}\right) \exp\left( \sum\limits_{j>0}\dfrac{v^{-j}}{j}a_{j}\right) \ \ \ \ \ \ \ \ \ \ \text{on }\mathcal{B}^{\left( m\right) }% \end{align*} (since multiplication by $z$ commutes with each of $\exp\left( \sum \limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) $ and $\exp\left( -\sum \limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) $). We wish to ``switch'' the second and the third exponential on the right hand side of this equation (although they don't commute). To do so, we notice that each of $-\sum \limits_{j>0}\dfrac{u^{-j}}{j}a_{j}$ and $-\sum\limits_{j>0}\dfrac{v^{j}}% {j}a_{-j}$ lies in the ring $\left( \operatorname*{End}\left( \mathcal{B}% ^{\left( m\right) }\right) \right) \left[ \left[ u^{-1},v\right] \right] $\ \ \ \ \footnote{This is the ring of formal power series in the indeterminates $u^{-1}$ and $v$ over the ring $\operatorname*{End}\left( \mathcal{B}^{\left( m\right) }\right) $. Note that $\operatorname*{End}% \left( \mathcal{B}^{\left( m\right) }\right) $ is non-commutative, but the ring of formal power series is still defined in the same way as over commutative rings. The indeterminates $u^{-1}$ and $v$ themselves commute with each other and with each element of $\operatorname*{End}\left( \mathcal{B}% ^{\left( m\right) }\right) $.}. Let $I$ be the ideal of the ring $\left( \operatorname*{End}\left( \mathcal{B}^{\left( m\right) }\right) \right) \left[ \left[ u^{-1},v\right] \right] $ consisting of all power series with constant term $0$. This ring $\left( \operatorname*{End}\left( \mathcal{B}^{\left( m\right) }\right) \right) \left[ \left[ u^{-1},v\right] \right] $ is a $\mathbb{Q}$-algebra and is complete and Hausdorff with respect to the $I$-adic topology. Let $\alpha=-\sum \limits_{j>0}\dfrac{u^{-j}}{j}a_{j}$ and $\beta=-\sum\limits_{j>0}\dfrac {v^{j}}{j}a_{-j}$. Clearly, both $\alpha$ and $\beta$ lie in $I$. Also,% \begin{align*} \left[ \alpha,\beta\right] & =\left[ -\sum\limits_{j>0}\dfrac{u^{-j}}% {j}a_{j},-\sum\limits_{j>0}\dfrac{v^{j}}{j}a_{-j}\right] =\sum\limits_{j>0}% \sum\limits_{k>0}\dfrac{u^{-j}v^{k}}{jk}\underbrace{\left[ a_{j}% ,a_{-k}\right] }_{=\delta_{j,k}j}\\ & =\sum\limits_{j>0}\sum\limits_{k>0}\dfrac{u^{-j}v^{k}}{jk}\delta _{j,k}j=\sum\limits_{j>0}\dfrac{u^{-j}v^{j}}{jj}j=\sum\limits_{j>0}\dfrac {1}{j}\left( \dfrac{v}{u}\right) ^{j}=-\log\left( 1-\dfrac{v}{u}\right) \end{align*} is a power series with coefficients in $\mathbb{Q}$, and thus lies in the center of $\left( \operatorname*{End}\left( \mathcal{B}^{\left( m\right) }\right) \right) \left[ \left[ u^{-1},v\right] \right] $, and hence commutes with each of $\alpha$ and $\beta$. Thus, we can apply Lemma \ref{lem.powerseries3} to $K=\mathbb{Q}$ and $R=\left( \operatorname*{End}% \left( \mathcal{B}^{\left( m\right) }\right) \right) \left[ \left[ u^{-1},v\right] \right] $, and obtain $\left( \exp\alpha\right) \cdot\left( \exp\beta\right) =\left( \exp\beta\right) \cdot\left( \exp\alpha\right) \cdot\left( \exp\left[ \alpha,\beta\right] \right) $. Hence,% \begin{align*} & \Gamma\left( u\right) \Gamma^{\ast}\left( v\right) \\ & =u^{m}v^{-m}\cdot\exp\left( \sum\limits_{j>0}\dfrac{u^{j}}{j}% a_{-j}\right) \exp\underbrace{\left( -\sum\limits_{j>0}\dfrac{u^{-j}}% {j}a_{j}\right) }_{=\alpha}\\ & \ \ \ \ \ \ \ \ \ \ \cdot\exp\underbrace{\left( -\sum\limits_{j>0}% \dfrac{v^{j}}{j}a_{-j}\right) }_{=\beta}\exp\left( \sum\limits_{j>0}% \dfrac{v^{-j}}{j}a_{j}\right) \\ & =u^{m}v^{-m}\cdot\exp\left( \sum\limits_{j>0}\dfrac{u^{j}}{j}% a_{-j}\right) \cdot\underbrace{\left( \exp\alpha\right) \cdot\left( \exp\beta\right) }_{=\left( \exp\beta\right) \cdot\left( \exp \alpha\right) \cdot\left( \exp\left[ \alpha,\beta\right] \right) }% \cdot\exp\left( \sum\limits_{j>0}\dfrac{v^{-j}}{j}a_{j}\right) \\ & =\underbrace{u^{m}v^{-m}}_{=\left( \dfrac{u}{v}\right) ^{m}}\cdot \exp\left( \sum\limits_{j>0}\dfrac{u^{j}}{j}a_{-j}\right) \cdot \exp\underbrace{\beta}_{=-\sum\limits_{j>0}\dfrac{v^{j}}{j}a_{-j}}\\ & \ \ \ \ \ \ \ \ \ \ \cdot\exp\underbrace{\alpha}_{=-\sum\limits_{j>0}% \dfrac{u^{-j}}{j}a_{j}}\cdot\underbrace{\exp\left[ \alpha,\beta\right] }_{\substack{=\dfrac{1}{1-\dfrac{v}{u}}\\\text{(since }\left[ \alpha ,\beta\right] =-\log\left( 1-\dfrac{v}{u}\right) \text{)}}}\cdot\exp\left( \sum\limits_{j>0}\dfrac{v^{-j}}{j}a_{j}\right) \\ & =\left( \dfrac{u}{v}\right) ^{m}\cdot\exp\left( \sum\limits_{j>0}% \dfrac{u^{j}}{j}a_{-j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac {v^{j}}{j}a_{-j}\right) \\ & \ \ \ \ \ \ \ \ \ \ \cdot\exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}}% {j}a_{j}\right) \cdot\dfrac{1}{1-\dfrac{v}{u}}\cdot\exp\left( \sum \limits_{j>0}\dfrac{v^{-j}}{j}a_{j}\right) \end{align*}% \begin{align*} & =\dfrac{1}{1-\dfrac{v}{u}}\cdot\left( \dfrac{u}{v}\right) ^{m}% \cdot\underbrace{\exp\left( \sum\limits_{j>0}\dfrac{u^{j}}{j}a_{-j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{v^{j}}{j}a_{-j}\right) }_{\substack{=\exp\left( \sum\limits_{j>0}\dfrac{u^{j}-v^{j}}{j}% a_{-j}\right) \\\text{(by Theorem \ref{thm.exp(u+v)}, applied to }R=\operatorname*{End}\left( \mathcal{B}^{\left( m\right) }\right) \left[ \left[ u,v\right] \right] \text{,}\\I=\left( \text{the ideal of }R\text{ consisting of all power series with constant term }0\right) \text{,}% \\\alpha=\sum\limits_{j>0}\dfrac{u^{j}}{j}a_{-j}\text{ and }\beta =-\sum\limits_{j>0}\dfrac{v^{j}}{j}a_{-j}\text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ \cdot\underbrace{\exp\left( -\sum\limits_{j>0}% \dfrac{u^{-j}}{j}a_{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac {v^{-j}}{j}a_{j}\right) }_{\substack{=\exp\left( -\sum\limits_{j>0}% \dfrac{u^{-j}-v^{-j}}{j}a_{j}\right) \\\text{(by Theorem \ref{thm.exp(u+v)}, applied to }R=\operatorname*{End}\left( \mathcal{B}^{\left( m\right) }\right) \left[ \left[ u^{-1},v^{-1}\right] \right] \text{,}\\I=\left( \text{the ideal of }R\text{ consisting of all power series with constant term }0\right) \text{,}\\\alpha=-\sum\limits_{j>0}\dfrac{u^{-j}}{j}a_{j}\text{ and }\beta=\sum\limits_{j>0}\dfrac{v^{-j}}{j}a_{j}\text{)}}}\\ & =\dfrac{1}{1-\dfrac{v}{u}}\cdot\left( \dfrac{u}{v}\right) ^{m}\exp\left( \sum\limits_{j>0}\dfrac{u^{j}-v^{j}}{j}a_{-j}\right) \exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}-v^{-j}}{j}a_{j}\right) . \end{align*} This proves Corollary \ref{cor.euler}. \subsection{Expliciting \texorpdfstring{$\sigma^{-1}$}{the inverse of the Boson-Fermion correspondence} using Schur polynomials} Next we are going to give an explicit (in as far as one can do) formula for $\sigma^{-1}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) $ for an elementary semiinfinite wedge $v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...$. Before we do so, we need to introduce the notion of \textit{Schur polynomials}. We first define \textit{elementary Schur polynomials}: \subsubsection{Schur polynomials} \begin{Convention} \label{conv.schur.x}In the following, we let $x$ denote the countable family of indeterminates $\left( x_{1},x_{2},x_{3},...\right) $. Thus, for any polynomial $P$ in countably many indeterminates, we write $P\left( x\right) $ for $P\left( x_{1},x_{2},x_{3},...\right) $. \end{Convention} \begin{definition} \label{def.schur.Sk}For every $k\in\mathbb{N}$, let $S_{k}\in\mathbb{Q}\left[ x_{1},x_{2},x_{3},...\right] $ be the coefficient of the power series $\exp\left( \sum\limits_{i\geq1}x_{i}z^{i}\right) \in\mathbb{Q}\left[ x_{1},x_{2},x_{3},...\right] \left[ \left[ z\right] \right] $ before $z^{k}$. Then, obviously,% \begin{equation} \sum\limits_{k\geq0}S_{k}\left( x\right) z^{k}=\exp\left( \sum \limits_{i\geq1}x_{i}z^{i}\right) . \label{def.schur.sk.genfun}% \end{equation} \end{definition} For example, $S_{0}\left( x\right) =1$, $S_{1}\left( x\right) =x_{1}$, $S_{2}\left( x\right) =\dfrac{x_{1}^{2}}{2}+x_{2}$, $S_{3}\left( x\right) =\dfrac{x_{1}^{3}}{6}+x_{1}x_{2}+x_{3}$. Note that the polynomials $S_{k}$ that we just defined are \textbf{not} symmetric polynomials. Instead, they ``represent'' the complete symmetric functions in terms of the $\dfrac{p_{i}}{i}$ (where $p_{i}$ are the power sums). Here is what exactly we mean by this: \begin{definition} \label{def.schur.y}Let $N\in\mathbb{N}$, and let $y$ denote a family of $N$ indeterminates $\left( y_{1},y_{2},...,y_{N}\right) $. Thus, for any polynomial $P$ in $N$ indeterminates, we write $P\left( y\right) $ for $P\left( y_{1},y_{2},...,y_{N}\right) $. \end{definition} \begin{definition} \label{def.schur.hk}For every $k\in\mathbb{N}$, define the $k$\textit{-th complete symmetric function} $h_{k}$ in the variables $y_{1},y_{2},...,y_{N}$ by $h_{k}\left( y_{1},y_{2},...,y_{N}\right) =\sum\limits_{\substack{p_{1}% ,p_{2},...,p_{N}\in\mathbb{N};\\p_{1}+p_{2}+...+p_{N}=k}}y_{1}^{p_{1}}% y_{2}^{p_{2}}...y_{N}^{p_{N}}$. \end{definition} \begin{proposition} \label{prop.schur.hk}In the ring $\mathbb{Q}\left[ y_{1},y_{2},...,y_{N}% \right] \left[ \left[ z\right] \right] $, we have% \[ \sum\limits_{k\geq0}z^{k}h_{k}\left( y\right) =\prod\limits_{j=1}^{N}% \dfrac{1}{1-zy_{j}}. \] \end{proposition} \textit{Proof of Proposition \ref{prop.schur.hk}.} For every $j\in\left\{ 1,2,...,N\right\} $, the sum formula for the geometric series yields $\dfrac{1}{1-zy_{j}}=\sum\limits_{p\in\mathbb{N}}\left( zy_{j}\right) ^{p}=\sum\limits_{p\in\mathbb{N}}y_{j}^{p}z^{p}$. Hence,% \begin{align*} \prod\limits_{j=1}^{N}\dfrac{1}{1-zy_{j}} & =\prod\limits_{j=1}^{N}\left( \sum\limits_{p\in\mathbb{N}}y_{j}^{p}z^{p}\right) =\sum\limits_{p_{1}% ,p_{2},...,p_{N}\in\mathbb{N}}\underbrace{\left( y_{1}^{p_{1}}z^{p_{1}% }\right) \left( y_{2}^{p_{2}}z^{p_{2}}\right) ...\left( y_{N}^{p_{N}% }z^{p_{N}}\right) }_{=y_{1}^{p_{1}}y_{2}^{p_{2}}...y_{N}^{p_{N}}% z^{p_{1}+p_{2}+...+p_{N}}}\\ & =\sum\limits_{p_{1},p_{2},...,p_{N}\in\mathbb{N}}y_{1}^{p_{1}}y_{2}^{p_{2}% }...y_{N}^{p_{N}}z^{p_{1}+p_{2}+...+p_{N}}=\sum\limits_{k\geq0}% \underbrace{\sum\limits_{\substack{p_{1},p_{2},...,p_{N}\in\mathbb{N}% ;\\p_{1}+p_{2}+...+p_{N}=k}}y_{1}^{p_{1}}y_{2}^{p_{2}}...y_{N}^{p_{N}}% }_{=h_{k}\left( y_{1},y_{2},...,y_{N}\right) =h_{k}\left( y\right) }% z^{k}\\ & =\sum\limits_{k\geq0}h_{k}\left( y\right) z^{k}=\sum\limits_{k\geq0}% z^{k}h_{k}\left( y\right) . \end{align*} This proves Proposition \ref{prop.schur.hk}. \begin{definition} Let $N\in\mathbb{N}$. We define a map $\operatorname*{PSE}\nolimits_{N}% :\mathbb{C}\left[ x_{1},x_{2},x_{3},...\right] \rightarrow\mathbb{C}\left[ y_{1},y_{2},...,y_{N}\right] $ as follows: For every polynomial $P\in\left[ x_{1},x_{2},x_{3},...\right] $, let $\operatorname*{PSE}\nolimits_{N}\left( P\right) $ be the result of substituting $x_{j}=\dfrac{y_{1}^{j}+y_{2}% ^{j}+...+y_{N}^{j}}{j}$ for all positive integers $j$ into the polynomial $P$. Clearly, this map $\operatorname*{PSE}\nolimits_{N}$ is a $\mathbb{C}$-algebra homomorphism. \end{definition} (The notation $\operatorname*{PSE}\nolimits_{N}$ is mine and has been chosen as an abbreviation for ``Power Sum Evaluation in $N$ variables''.) \begin{proposition} \label{prop.schur.h_k.as.schur}For every $N\in\mathbb{N}$, we have $h_{k}\left( y\right) =\operatorname*{PSE}\nolimits_{N}\left( S_{k}\left( x\right) \right) $ for each $k\in\mathbb{N}$. \end{proposition} \textit{Proof of Proposition \ref{prop.schur.h_k.as.schur}.} Fix $N\in\mathbb{N}$. We know that $\sum\limits_{k\geq0}S_{k}\left( x\right) z^{k}=\exp\left( \sum\limits_{i\geq1}x_{i}z^{i}\right) $. Since $\operatorname*{PSE}\nolimits_{N}$ is a $\mathbb{C}$-algebra homomorphism, this yields% \begin{align*} \sum\limits_{k\geq0}\operatorname*{PSE}\nolimits_{N}\left( S_{k}\left( x\right) \right) z^{k} & =\exp\left( \sum\limits_{i\geq1}% \operatorname*{PSE}\nolimits_{N}\left( x_{i}\right) z^{i}\right) =\exp\left( \sum\limits_{i\geq1}\sum\limits_{j=1}^{N}\dfrac{y_{j}^{i}}% {i}z^{i}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\operatorname*{PSE}\nolimits_{N}% \left( x_{i}\right) =\dfrac{y_{1}^{i}+y_{2}^{i}+...+y_{N}^{i}}{i}% =\sum\limits_{j=1}^{N}\dfrac{y_{j}^{i}}{i}\right) \\ & =\exp\left( \sum\limits_{j=1}^{N}\sum\limits_{i\geq1}\dfrac{y_{j}^{i}}% {i}z^{i}\right) =\prod\limits_{j=1}^{N}\exp\left( \sum\limits_{i\geq1}% \dfrac{y_{j}^{i}}{i}z^{i}\right) \\ & =\prod\limits_{j=1}^{N}\exp\underbrace{\left( \sum\limits_{i\geq1}% \dfrac{y_{j}^{i}z^{i}}{i}\right) }_{=-\log\left( 1-y_{j}z\right) }% =\prod\limits_{j=1}^{N}\underbrace{\exp\left( -\log\left( 1-y_{j}z\right) \right) }_{=\dfrac{1}{1-y_{j}z}=\dfrac{1}{1-zy_{j}}}\\ & =\prod\limits_{j=1}^{N}\dfrac{1}{1-zy_{j}}=\sum\limits_{k\geq0}z^{k}% h_{k}\left( y\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by Proposition \ref{prop.schur.hk}}\right) . \end{align*} By comparing coefficients in this equality, we conclude that $\operatorname*{PSE}\nolimits_{N}\left( S_{k}\left( x\right) \right) =h_{k}\left( y\right) $ for each $k\in\mathbb{N}$. Proposition \ref{prop.schur.h_k.as.schur} is proven. \begin{definition} \label{def.schur.Slambda}Let $\lambda=\left( \lambda_{1},\lambda _{2},...,\lambda_{m}\right) $ be a partition, so that $\lambda_{1}\geq \lambda_{2}\geq...\geq\lambda_{m}\geq0$ are integers. We define $S_{\lambda}\left( x\right) \in\mathbb{Q}\left[ x_{1},x_{2}% ,x_{3},...\right] $ to be the polynomial% \begin{align*} & \det\left( \begin{array} [c]{ccccc}% S_{\lambda_{1}}\left( x\right) & S_{\lambda_{1}+1}\left( x\right) & S_{\lambda_{1}+2}\left( x\right) & ... & S_{\lambda_{1}+m-1}\left( x\right) \\ S_{\lambda_{2}-1}\left( x\right) & S_{\lambda_{2}}\left( x\right) & S_{\lambda_{2}+1}\left( x\right) & ... & S_{\lambda_{2}+m-2}\left( x\right) \\ S_{\lambda_{3}-2}\left( x\right) & S_{\lambda_{3}-1}\left( x\right) & S_{\lambda_{3}}\left( x\right) & ... & S_{\lambda_{3}+m-3}\left( x\right) \\ ... & ... & ... & ... & ...\\ S_{\lambda_{m}-m+1}\left( x\right) & S_{\lambda_{m}-m+2}\left( x\right) & S_{\lambda_{m}-m+3}\left( x\right) & ... & S_{\lambda_{m}}\left( x\right) \end{array} \right) \\ & =\det\left( \left( S_{\lambda_{i}+j-i}\left( x\right) \right) _{1\leq i\leq m,\ 1\leq j\leq m}\right) , \end{align*} where $S_{j}$ denotes $0$ if $j<0$. (Note that this does not depend on trailing zeroes in the partition; in other words, $S_{\left( \lambda _{1},\lambda_{2},...,\lambda_{m}\right) }\left( x\right) =S_{\left( \lambda_{1},\lambda_{2},...,\lambda_{m},0,0,...,0\right) }\left( x\right) $ for any number of zeroes. This is because any nonnegative integers $m$ and $\ell$, any $m\times m$-matrix $A$, any $m\times\ell$-matrix $B$ and any upper unitriangular $\ell\times\ell$-matrix $C$ satisfy $\det\left( \begin{array} [c]{cc}% A & B\\ 0 & C \end{array} \right) =\det A$.) We refer to $S_{\lambda}\left( x\right) $ as the \textit{bosonic Schur polynomial corresponding to the partition }$\lambda$. \end{definition} To a reader acquainted with the Schur polynomials of combinatorics (and representation theory of symmetric groups), this definition may look familiar, but it should be reminded that our polynomial $S_{\lambda}\left( x\right) $ is \textbf{not a symmetric function per se} (this is why we call it ``bosonic Schur polynomial'' and not just simply ``Schur polynomial''); instead, it can be made into a symmetric function -- and this will, indeed, be the $\lambda $-Schur polynomial known from combinatorics -- by substituting for each $x_{j}$ the term $\dfrac{\left( j\text{-th power sum symmetric function}% \right) }{j}$. We will prove this in Proposition \ref{prop.schur.Schur=schur} (albeit only for finitely many variables). Let us first formulate one of the many definitions of Schur polynomials from combinatorics: \begin{definition} \label{def.schur.schurpoly}Let $\lambda=\left( \lambda_{1},\lambda _{2},...,\lambda_{m}\right) $ be a partition, so that $\lambda_{1}\geq \lambda_{2}\geq...\geq\lambda_{m}\geq0$ are integers. We define $\lambda _{\ell}$ to mean $0$ for all integers $\ell>m$; thus, we obtain a nonincreasing sequence $\left( \lambda_{1},\lambda_{2},\lambda_{3}% ,...\right) $ of nonnegative integers. Let $N\in\mathbb{N}$. The so-called $\lambda$\textit{-Schur module} $V_{\lambda}$ \textit{over }$\operatorname*{GL}\left( N\right) $ is defined to be the $\operatorname*{GL}\left( N\right) $-module $\operatorname*{Hom}% \nolimits_{S_{n}}\left( S^{\lambda},\left( \mathbb{C}^{N}\right) ^{\otimes n}\right) $, where $n$ denotes the number $\lambda_{1}+\lambda_{2}% +...+\lambda_{m}$ and $S^{\lambda}$ denotes the Specht module over the symmetric group $S_{n}$ corresponding to the partition $\lambda$. (The $\operatorname*{GL}\left( N\right) $-module structure on $\operatorname*{Hom}\nolimits_{S_{n}}\left( S^{\lambda},\left( \mathbb{C}^{N}\right) ^{\otimes n}\right) $ is obtained from the $\operatorname*{GL}\left( N\right) $-module structure on $\mathbb{C}^{N}$.) This $\lambda$-Schur module $V_{\lambda}$ is not only a $\operatorname*{GL}% \left( N\right) $-module, but also a $\mathfrak{gl}\left( N\right) $-module. If $\lambda_{N+1}=0$, then $V_{\lambda}$ is irreducible both as a representation of $\operatorname*{GL}\left( N\right) $ and as a representation of $\mathfrak{gl}\left( N\right) $. If $\lambda_{N+1}\neq0$, then $V_{\lambda}=0$. It is known that there exists a unique polynomial $\chi_{\lambda}\in \mathbb{Q}\left[ y_{1},y_{2},...,y_{N}\right] $ (depending both on $\lambda$ and on $N$) such that every diagonal matrix $A=\operatorname*{diag}\left( a_{1},a_{2},...,a_{N}\right) \in\operatorname*{GL}\left( N\right) $ satisfies $\chi_{\lambda}\left( a_{1},a_{2},...,a_{N}\right) =\left( \operatorname*{Tr}\mid_{V_{\lambda}}\right) \left( A\right) $ (where $\left( \operatorname*{Tr}\mid_{V_{\lambda}}\right) \left( A\right) $ means the trace of the action of $A\in\operatorname*{GL}\left( N\right) $ on $V_{\lambda}$ by means of the $\operatorname*{GL}\left( N\right) $-module structure on $V_{\lambda}$). In the language of representation theory, $\chi_{\lambda}$ is thus the character of the $\operatorname*{GL}\left( N\right) $-module $V_{\lambda}$. This polynomial $\chi_{\lambda}$ is called the $\lambda$\textit{-th Schur polynomial in }$N$ \textit{variables}. \end{definition} Now, the relation between the $S_{\lambda}$ and the Schur polynomials looks like this: \begin{proposition} \label{prop.schur.Schur=schur}Let $\lambda=\left( \lambda_{1},\lambda _{2},...,\lambda_{m}\right) $ be a partition. Then, $\chi_{\lambda}\left( y_{1},y_{2},...,y_{N}\right) =\operatorname*{PSE}\nolimits_{N}\left( S_{\lambda}\left( x\right) \right) $. \end{proposition} This generalizes Proposition \ref{prop.schur.h_k.as.schur} (in fact, set $\lambda=\left( k\right) $ and notice that $V_{\lambda}=S^{k}\mathbb{C}^{N}$). \textit{Proof of Proposition \ref{prop.schur.Schur=schur}.} Define $h_{k}$ to mean $0$ for every $k<0$. Proposition \ref{prop.schur.h_k.as.schur} yields $h_{k}\left( y\right) =\operatorname*{PSE}\nolimits_{N}\left( S_{k}\left( x\right) \right) $ for each $k\in\mathbb{N}$. Since $h_{k}\left( y\right) =\operatorname*{PSE}% \nolimits_{N}\left( S_{k}\left( x\right) \right) $ also holds for every negative integer $k$ (since every negative integer $k$ satisfies $h_{k}=0$ and $S_{k}=0$), we thus conclude that% \begin{equation} h_{k}\left( y\right) =\operatorname*{PSE}\nolimits_{N}\left( S_{k}\left( x\right) \right) \ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{Z}. \label{pf.schur.Schur=schur.1}% \end{equation} We know that $\chi_{\lambda}$ is the $\lambda$-th Schur polynomial in $N$ variables. By the first Giambelli formula, this yields that% \begin{align*} & \chi_{\lambda}\left( y_{1},y_{2},...,y_{N}\right) \\ & =\det\underbrace{\left( \begin{array} [c]{ccccc}% h_{\lambda_{1}}\left( y\right) & h_{\lambda_{1}+1}\left( y\right) & h_{\lambda_{1}+2}\left( y\right) & ... & h_{\lambda_{1}+m-1}\left( y\right) \\ h_{\lambda_{2}-1}\left( y\right) & h_{\lambda_{2}}\left( y\right) & h_{\lambda_{2}+1}\left( y\right) & ... & h_{\lambda_{2}+m-2}\left( y\right) \\ h_{\lambda_{3}-2}\left( y\right) & h_{\lambda_{3}-1}\left( y\right) & h_{\lambda_{3}}\left( y\right) & ... & h_{\lambda_{3}+m-3}\left( y\right) \\ ... & ... & ... & ... & ...\\ h_{\lambda_{m}-m+1}\left( y\right) & h_{\lambda_{m}-m+2}\left( y\right) & h_{\lambda_{m}-m+3}\left( y\right) & ... & h_{\lambda_{m}}\left( y\right) \end{array} \right) }_{=\left( h_{\lambda_{i}+j-i}\left( y\right) \right) _{1\leq i\leq m,\ 1\leq j\leq m}}\\ & =\det\left( \left( h_{\lambda_{i}+j-i}\left( y\right) \right) _{1\leq i\leq m,\ 1\leq j\leq m}\right) =\det\left( \left( \operatorname*{PSE}% \nolimits_{N}\left( S_{\lambda_{i}+j-i}\left( x\right) \right) \right) _{1\leq i\leq m,\ 1\leq j\leq m}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.schur.Schur=schur.1})}\right) \\ & =\operatorname*{PSE}\nolimits_{N}\underbrace{\left( \det\left( \left( S_{\lambda_{i}+j-i}\left( x\right) \right) _{1\leq i\leq m,\ 1\leq j\leq m}\right) \right) }_{=S_{\lambda}\left( x\right) }\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\operatorname*{PSE}\nolimits_{N}\text{ is a }\mathbb{C}% \text{-algebra homomorphism, whereas }\det\text{ is a polynomial}\\ \text{(and any }\mathbb{C}\text{-algebra homomorphism commutes with any polynomial)}% \end{array} \right) \\ & =\operatorname*{PSE}\nolimits_{N}\left( S_{\lambda}\left( x\right) \right) . \end{align*} Proposition \ref{prop.schur.Schur=schur} is proven. \subsubsection{The statement of the fact} \begin{theorem} \label{thm.schur}Whenever $\left( i_{0},i_{1},i_{2},...\right) $ is a $0$-degression (see Definition \ref{def.glinf.m-deg} for what this means), we have $\sigma^{-1}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right) =S_{\lambda}\left( x\right) $ where $\lambda=\left( i_{0}+0,i_{1}+1,i_{2}+2,...\right) $. (Note that this $\lambda$ is indeed a partition since $\left( i_{0},i_{1},i_{2},...\right) $ is a $0$-degression.) \end{theorem} We are going to give two proofs of this theorem. The first proof will be covered in Section \ref{subsect.schur.pf1}, whereas the second proof will encompass Section \ref{subsect.schur.pf2}. \subsection{\label{subsect.schur.pf1}Expliciting \texorpdfstring{$\sigma^{-1}$}{the inverse of the Boson-Fermion correspondence} using Schur polynomials: first proof} \subsubsection{\label{subsubsect.powersums}The power sums are algebraically independent} Our first proof of Theorem \ref{thm.schur} will require some lemmata from algebraic combinatorics. First of all: \begin{lemma} \label{lem.schur.algind}Let $N\in\mathbb{N}$. For every positive integer $j$, let $p_{j}$ denote the polynomial $y_{1}^{j}+y_{2}^{j}+...+y_{N}^{j}% \in\mathbb{C}\left[ y_{1},y_{2},...,y_{N}\right] $. Then, the polynomials $p_{1}$, $p_{2}$, $...$, $p_{N}$ are algebraically independent. \end{lemma} In order to prove this fact, we need the following known facts (which we won't prove): \begin{lemma} \label{lem.schur.elsym}Let $N\in\mathbb{N}$. For every $j\in\mathbb{N}$, let $e_{j}$ denote the $j$-th elementary symmetric polynomial $\sum\limits_{1\leq i_{1} \left( \beta_{1},\beta_{2},...,\beta_{N}\right) $ (since $\left( \alpha_{1},\alpha_{2},...,\alpha_{N}\right) \neq\left( \beta_{1},\beta _{2},...,\beta_{N}\right) $). Thus, $\left( \alpha_{1},\alpha_{2}% ,...,\alpha_{N}\right) \in\mathbb{N}^{N}$ is higher than $\left( \beta _{1},\beta_{2},...,\beta_{N}\right) $. Hence, $\lambda_{\alpha_{1},\alpha _{2},...,\alpha_{N}}=0$ (by (\ref{pf.schur.algind.4})), so that $\lambda _{\alpha_{1},\alpha_{2},...,\alpha_{N}}\widetilde{P}_{1}^{\alpha_{1}% }\widetilde{P}_{2}^{\alpha_{2}}...\widetilde{P}_{N}^{\alpha_{N}}=0$ is clearly a $\mathbb{C}$-linear combination of monomials smaller than $T_{1}^{\alpha _{1}}T_{2}^{\alpha_{2}}...T_{N}^{\alpha_{N}}$. Thus, (\ref{pf.schur.algind.5}) holds in Case 1. \par Now, let us consider Case 2. In this case, $\left( \alpha_{1},\alpha _{2},...,\alpha_{N}\right) <\left( \beta_{1},\beta_{2},...,\beta_{N}\right) $, so that $T_{1}^{\alpha_{1}}T_{2}^{\alpha_{2}}...T_{N}^{\alpha_{N}}% 0$ for every $j\in\left\{ 1,2,...,N\right\} $. Let $m\in\mathbb{N}$. Then,% \begin{align*} & \sum\limits_{k=1}^{N}\det\left( \left( y_{i}^{i_{j-1}+\delta_{j,k}% m+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) \\ & =\left( y_{1}^{m}+y_{2}^{m}+...+y_{N}^{m}\right) \det\left( \left( y_{i}^{i_{j-1}+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) . \end{align*} \end{corollary} \textit{Proof of Corollary \ref{cor.schur.det}.} Applying Proposition \ref{prop.schur.det} to $R=\mathbb{C}\left[ y_{1},y_{2},...,y_{N}\right] $, $\left( a_{i,j}\right) _{1\leq i\leq N,\ 1\leq j\leq N}=\left( y_{i}^{i_{j-1}+N}\right) _{1\leq i\leq N,\ 1\leq j\leq N}$ and $b_{i}% =y_{i}^{m}$, we obtain% \begin{align*} & \sum\limits_{k=1}^{N}\det\left( \left( y_{i}^{i_{j-1}+N-1}\left( y_{i}^{m}\right) ^{\delta_{j,k}}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) \\ & =\left( y_{1}^{m}+y_{2}^{m}+...+y_{N}^{m}\right) \det\left( \left( y_{i}^{i_{j-1}+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) . \end{align*} Since any $i\in\left\{ 1,2,...,N\right\} $, $j\in\left\{ 1,2,...,N\right\} $ and $k\in\left\{ 1,2,...,N\right\} $ satisfy $y_{i}^{i_{j-1}+N-1}\left( y_{i}^{m}\right) ^{\delta_{j,k}}=y_{i}^{i_{j-1}+N+\delta_{j,k}m}% =y_{i}^{i_{j-1}+\delta_{j,k}m+N-1}$, this rewrites as% \begin{align*} & \sum\limits_{k=1}^{N}\det\left( \left( y_{i}^{i_{j-1}+\delta_{j,k}% m+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) \\ & =\left( y_{1}^{m}+y_{2}^{m}+...+y_{N}^{m}\right) \det\left( \left( y_{i}^{i_{j-1}+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) . \end{align*} Corollary \ref{cor.schur.det} is proven. Now, to the main proof. \textit{Proof of Theorem \ref{thm.schur}.} Define a $\mathbb{C}$-linear map $\tau:\mathcal{F}^{\left( 0\right) }\rightarrow\mathbb{C}\left[ x_{1}% ,x_{2},x_{3},...\right] $ by% \[ \tau\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =S_{\left( i_{0}+0,i_{1}+1,i_{2}+2,...\right) }\left( x\right) \ \ \ \ \ \ \ \ \ \ \text{for every }0\text{-degression }\left( i_{0}% ,i_{1},i_{2},...\right) . \] (This definition makes sense, because we know that $\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) _{\left( i_{0},i_{1}% ,i_{2},...\right) \text{ is a }0\text{-degression}}$ is a basis of $\wedge^{\dfrac{\infty}{2},0}V=\mathcal{F}^{\left( 0\right) }$.) Our aim is to prove that $\tau=\sigma^{-1}$. \textit{1st step:} First of all, the definition of $\tau$ (applied to the $0$-degression $\left( 0,-1,-2,...\right) $) yields% \[ \tau\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) =S_{\left( 0+0,-1+1,-2+2,...\right) }\left( x\right) =S_{\left( 0,0,0,...\right) }\left( x\right) =1. \] \textit{2nd step:} If $N\in\mathbb{N}$, and $\left( i_{0},i_{1}% ,i_{2},...\right) $ is a straying $0$-degression, then we say that $\left( i_{0},i_{1},i_{2},...\right) $ is $N$\textit{-finished} if the following two conditions (\ref{pf.schur.step2.fin1}) and (\ref{pf.schur.step2.fin2}) hold:% \begin{align} & \left( \text{every integer }k\geq N\text{ satisfies }i_{k}+k=0\right) ;\label{pf.schur.step2.fin1}\\ & \left( \text{each of the integers }i_{0}\text{, }i_{1}\text{, }...\text{, }i_{N-1}\text{ is }>-N\right) . \label{pf.schur.step2.fin2}% \end{align} Now, we claim the following: For any $N\in\mathbb{N}$, and any $N$-finished straying $0$-degression $\left( i_{0},i_{1},i_{2},...\right) $, we have% \begin{equation} \operatorname*{PSE}\nolimits_{N}\left( \tau\left( v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...\right) \right) =\dfrac{\det\left( \left( y_{i}^{i_{j-1}+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) }% {\det\left( \left( y_{i}^{j-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) }. \label{pf.schur.step2}% \end{equation} \textit{Proof of (\ref{pf.schur.step2}):} Let $N\in\mathbb{N}$, and let $\left( i_{0},i_{1},i_{2},...\right) $ be an $N$-finished straying $0$-degression. Since $\left( i_{0},i_{1},i_{2},...\right) $ is $N$-finished, we conclude (by the definition of ``$N$-finished'') that it satisfies the conditions (\ref{pf.schur.step2.fin1}) and (\ref{pf.schur.step2.fin2}). If some two of the integers $i_{0}$, $i_{1}$, $...$, $i_{N-1}$ are equal, then (\ref{pf.schur.step2}) is true.\footnote{\textit{Proof.} Assume that some two of the integers $i_{0}$, $i_{1}$, $...$, $i_{N-1}$ are equal. Then, some two elements of the sequence $\left( i_{0},i_{1},i_{2},...\right) $ are equal, so that $v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...=0$ (by the definition of $v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...$) and thus $\operatorname*{PSE}\nolimits_{N}\left( \tau\left( v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...\right) \right) =\operatorname*{PSE}% \nolimits_{N}\left( 0\right) =0$. Thus, the left hand side of (\ref{pf.schur.step2}) is $0$. On the other hand, the matrix $\left( y_{i}^{i_{j-1}+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}$ has two equal columns (since two of the integers $i_{0}$, $i_{1}$, $...$, $i_{N-1}$ are equal) and thus its determinant vanishes, i. e., we have $\det\left( \left( y_{i}^{i_{j-1}+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) =0$, so that the right hand side of (\ref{pf.schur.step2}) is $0$. \par Thus, both the left hand side and the right hand side of (\ref{pf.schur.step2}% ) are $0$. Hence, (\ref{pf.schur.step2}) is true, qed.} Hence, for the rest of this proof, we assume that no two of the integers $i_{0}$, $i_{1}$, $...$, $i_{N-1}$ are equal. Then, there exists a permutation $\phi$ of the set $\left\{ 0,1,...,N-1\right\} $ such that $i_{\phi^{-1}\left( 0\right) }>i_{\phi^{-1}\left( 1\right) }>...>i_{\phi^{-1}\left( N-1\right) }$. Consider this $\phi$. It is easy to see that $i_{\phi^{-1}\left( 0\right) }>i_{\phi^{-1}\left( 1\right) }>...>i_{\phi^{-1}\left( N-1\right) }>-N$% .\ \ \ \ \footnote{\textit{Proof.} Every $j\in\left\{ 0,1,...,N-1\right\} $ satisfies $\phi^{-1}\left( j\right) \in\phi^{-1}\left( \left\{ 0,1,...,N-1\right\} \right) =\left\{ 0,1,...,N-1\right\} $. Hence, for every $j\in\left\{ 0,1,...,N-1\right\} $, the integer $i_{\phi^{-1}\left( j\right) }$ is one of the integers $i_{0}$, $i_{1}$, $...$, $i_{N-1}$, and therefore $>-N$ (due to (\ref{pf.schur.step2.fin2})). That is, $i_{\phi ^{-1}\left( j\right) }>-N$ for every $j\in\left\{ 0,1,...,N-1\right\} $. Combining this with $i_{\phi^{-1}\left( 0\right) }>i_{\phi^{-1}\left( 1\right) }>...>i_{\phi^{-1}\left( N-1\right) }$, we get $i_{\phi ^{-1}\left( 0\right) }>i_{\phi^{-1}\left( 1\right) }>...>i_{\phi ^{-1}\left( N-1\right) }>-N$.} Let $\pi$ be the finitary permutation of $\mathbb{N}$ which sends every $k\in\mathbb{N}$ to \newline$\left\{ \begin{array} [c]{l}% \phi\left( k\right) ,\ \ \ \ \ \ \ \ \ \ \text{if }k\in\left\{ 0,1,...,N-1\right\} ;\\ k,\ \ \ \ \ \ \ \ \ \ \text{if }k\notin\left\{ 0,1,...,N-1\right\} \end{array} \right. $. Then, $\left( -1\right) ^{\pi}=\left( -1\right) ^{\phi}$; moreover, every $k\in\mathbb{N}$ satisfies% \begin{equation} \pi^{-1}\left( k\right) =\left\{ \begin{array} [c]{l}% \phi^{-1}\left( k\right) ,\ \ \ \ \ \ \ \ \ \ \text{if }k\in\left\{ 0,1,...,N-1\right\} ;\\ k,\ \ \ \ \ \ \ \ \ \ \text{if }k\notin\left\{ 0,1,...,N-1\right\} \end{array} \right. . \label{pf.schur.step2.pf.2}% \end{equation} In particular, every integer $k\geq N$ satisfies $\pi^{-1}\left( k\right) =k$. From (\ref{pf.schur.step2.pf.2}), it is clear that% \begin{equation} \text{every }k\in\left\{ 0,1,...,N-1\right\} \text{ satisfies }\pi ^{-1}\left( k\right) =\phi^{-1}\left( k\right) . \label{pf.schur.step2.pf.4}% \end{equation} Hence, $i_{\pi^{-1}\left( 0\right) }>i_{\pi^{-1}\left( 1\right) }>...>i_{\pi^{-1}\left( N-1\right) }>-N$ (since $i_{\phi^{-1}\left( 0\right) }>i_{\phi^{-1}\left( 1\right) }>...>i_{\phi^{-1}\left( N-1\right) }>-N$). Now, every integer $k\geq N$ satisfies $\pi^{-1}\left( k\right) =k$, thus $i_{\pi^{-1}\left( k\right) }=i_{k}=-k$ (since (\ref{pf.schur.step2.fin1}) yields $i_{k}+k=0$). Hence, $-N=i_{\pi^{-1}\left( N\right) }>i_{\pi ^{-1}\left( N+1\right) }>i_{\pi^{-1}\left( N+2\right) }>...$ (because $-N=-N>-\left( N+1\right) >-\left( N+2\right) >...$). Combined with $i_{\pi^{-1}\left( 0\right) }>i_{\pi^{-1}\left( 1\right) }>...>i_{\pi ^{-1}\left( N-1\right) }>-N$, this becomes% \[ i_{\pi^{-1}\left( 0\right) }>i_{\pi^{-1}\left( 1\right) }>...>i_{\pi ^{-1}\left( N-1\right) }>-N=i_{\pi^{-1}\left( N\right) }>i_{\pi ^{-1}\left( N+1\right) }>i_{\pi^{-1}\left( N+2\right) }>.... \] Thus,% \[ i_{\pi^{-1}\left( 0\right) }>i_{\pi^{-1}\left( 1\right) }>...>i_{\pi ^{-1}\left( N-1\right) }>i_{\pi^{-1}\left( N\right) }>i_{\pi^{-1}\left( N+1\right) }>i_{\pi^{-1}\left( N+2\right) }>.... \] In other words, the sequence $\left( i_{\pi^{-1}\left( 0\right) }% ,i_{\pi^{-1}\left( 1\right) },i_{\pi^{-1}\left( 2\right) },...\right) $ is strictly decreasing. Since every sufficiently high $k\in\mathbb{N}$ satisfies $i_{\pi^{-1}\left( k\right) }+k=0$ (in fact, every $k\geq N$ satisfies $i_{\pi^{-1}\left( k\right) }=-k$ and thus $i_{\pi^{-1}\left( k\right) }+k=0$), this sequence $\left( i_{\pi^{-1}\left( 0\right) },i_{\pi^{-1}\left( 1\right) },i_{\pi^{-1}\left( 2\right) },...\right) $ must thus be a $0$-degression. Hence, by the definition of $\tau$, we have% \[ \tau\left( v_{i_{\pi^{-1}\left( 0\right) }}\wedge v_{i_{\pi^{-1}\left( 1\right) }}\wedge v_{i_{\pi^{-1}\left( 2\right) }}\wedge...\right) =S_{\left( i_{\pi^{-1}\left( 0\right) }+0,i_{\pi^{-1}\left( 1\right) }+1,i_{\pi^{-1}\left( 2\right) }+2,...\right) }\left( x\right) . \] Since $\pi$ is a finitary permutation of $\mathbb{N}$ such that $\left( i_{\pi^{-1}\left( 0\right) },i_{\pi^{-1}\left( 1\right) },i_{\pi ^{-1}\left( 2\right) },...\right) $ is a $0$-degression, it is clear that $\pi$ is the straightening permutation of $\left( i_{0},i_{1},i_{2}% ,...\right) $. Thus, by the definition of $v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...$, we have% \begin{align*} v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge... & =\underbrace{\left( -1\right) ^{\pi}}_{=\left( -1\right) ^{\phi}}v_{i_{\pi^{-1}\left( 0\right) }}\wedge v_{i_{\pi^{-1}\left( 1\right) }}\wedge v_{i_{\pi ^{-1}\left( 2\right) }}\wedge...\\ & =\left( -1\right) ^{\phi}v_{i_{\pi^{-1}\left( 0\right) }}\wedge v_{i_{\pi^{-1}\left( 1\right) }}\wedge v_{i_{\pi^{-1}\left( 2\right) }% }\wedge..., \end{align*} so that% \begin{align} & \operatorname*{PSE}\nolimits_{N}\left( \tau\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \right) \nonumber\\ & =\operatorname*{PSE}\nolimits_{N}\left( \tau\left( \left( -1\right) ^{\phi}v_{i_{\pi^{-1}\left( 0\right) }}\wedge v_{i_{\pi^{-1}\left( 1\right) }}\wedge v_{i_{\pi^{-1}\left( 2\right) }}\wedge...\right) \right) \nonumber\\ & =\left( -1\right) ^{\phi}\operatorname*{PSE}\nolimits_{N}% \underbrace{\left( \tau\left( v_{i_{\pi^{-1}\left( 0\right) }}\wedge v_{i_{\pi^{-1}\left( 1\right) }}\wedge v_{i_{\pi^{-1}\left( 2\right) }% }\wedge...\right) \right) }_{\substack{=S_{\left( i_{\pi^{-1}\left( 0\right) }+0,i_{\pi^{-1}\left( 1\right) }+1,i_{\pi^{-1}\left( 2\right) }+2,...\right) }\left( x\right) \\\text{(by the definition of }\tau\text{, since }\left( i_{\pi^{-1}\left( 0\right) },i_{\pi^{-1}\left( 1\right) },i_{\pi^{-1}\left( 2\right) },...\right) \text{ is a }0\text{-degression)}% }}\nonumber\\ & =\left( -1\right) ^{\phi}\operatorname*{PSE}\nolimits_{N}\left( S_{\left( i_{\pi^{-1}\left( 0\right) }+0,i_{\pi^{-1}\left( 1\right) }+1,i_{\pi^{-1}\left( 2\right) }+2,...\right) }\left( x\right) \right) . \label{pf.schur.step2.pf.4a}% \end{align} Let $\mu$ be the partition $\left( i_{\pi^{-1}\left( 0\right) }% +0,i_{\pi^{-1}\left( 1\right) }+1,i_{\pi^{-1}\left( 2\right) }+2,...\right) $. For every positive integer $\alpha$, let $\mu_{\alpha}$ denote the $\alpha$-th part of the partition $\mu$, so that $\mu=\left( \mu_{1},\mu_{2},\mu_{3},...\right) $. Then, every $j\in\left\{ 1,2,...,N\right\} $ satisfies% \begin{align*} \mu_{j} & =i_{\pi^{-1}\left( j-1\right) }+\left( j-1\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\mu\right) \\ & =i_{\phi^{-1}\left( j-1\right) }+\left( j-1\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since (\ref{pf.schur.step2.pf.4}) (applied to }k=j-1\text{) yields }\pi^{-1}\left( j-1\right) =\phi^{-1}\left( j-1\right) \right) , \end{align*} so that $\mu_{j}+N-j=i_{\phi^{-1}\left( j-1\right) }+\left( j-1\right) +N-j=i_{\phi^{-1}\left( j-1\right) }+N-1$. Hence,% \begin{equation} \det\left( \left( y_{i}^{\mu_{j}+N-j}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) =\det\left( \left( y_{i}^{i_{\phi^{-1}\left( j-1\right) }% +N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) . \label{pf.schur.step2.pf.5}% \end{equation} But the matrix $\left( y_{i}^{i_{\phi^{-1}\left( j-1\right) }+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}$ is obtained from the matrix $\left( y_{i}^{i_{j-1}+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}$ by permuting the columns using the permutation $\phi$. Hence,% \[ \det\left( \left( y_{i}^{i_{\phi^{-1}\left( j-1\right) }+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) =\left( -1\right) ^{\phi}% \det\left( \left( y_{i}^{i_{j-1}+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) \] (since permuting the columns of a matrix changes the determinant by the sign of the permutation). Combining this with (\ref{pf.schur.step2.pf.5}), we obtain% \begin{equation} \det\left( \left( y_{i}^{\mu_{j}+N-j}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) =\left( -1\right) ^{\phi}\det\left( \left( y_{i}^{i_{j-1}% +N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) . \label{pf.schur.step2.pf.7}% \end{equation} Also, by the definition of $\mu$, we have $\mu_{N+1}=i_{\pi^{-1}\left( N\right) }+N=0$ (because $-N=i_{\pi^{-1}\left( N\right) }$), and thus we can apply Theorem \ref{thm.schur.altern} to $\mu$ instead of $\lambda$. This results in \begin{align} \operatorname*{PSE}\nolimits_{N}\left( S_{\mu}\left( x\right) \right) & =\dfrac{\det\left( \left( y_{i}^{\mu_{j}+N-j}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) }{\det\left( \left( y_{i}^{j-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) }\nonumber\\ & =\dfrac{\left( -1\right) ^{\phi}\det\left( \left( y_{i}^{i_{j-1}% +N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) }{\det\left( \left( y_{i}^{j-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) } \label{pf.schur.step2.pf.8}% \end{align} (by (\ref{pf.schur.step2.pf.7})). But (\ref{pf.schur.step2.pf.4a}) becomes% \begin{align*} & \operatorname*{PSE}\nolimits_{N}\left( \tau\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \right) \\ & =\left( -1\right) ^{\phi}\operatorname*{PSE}\nolimits_{N}\left( S_{\left( i_{\pi^{-1}\left( 0\right) }+0,i_{\pi^{-1}\left( 1\right) }+1,i_{\pi^{-1}\left( 2\right) }+2,...\right) }\left( x\right) \right) \\ & =\left( -1\right) ^{\phi}\operatorname*{PSE}\nolimits_{N}\left( S_{\mu }\left( x\right) \right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{\pi^{-1}\left( 0\right) }+0,i_{\pi^{-1}\left( 1\right) }+1,i_{\pi ^{-1}\left( 2\right) }+2,...\right) =\mu\right) \\ & =\left( -1\right) ^{\phi}\dfrac{\left( -1\right) ^{\phi}\det\left( \left( y_{i}^{i_{j-1}+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) }{\det\left( \left( y_{i}^{j-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) }\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.schur.step2.pf.8}% )}\right) \\ & =\dfrac{\det\left( \left( y_{i}^{i_{j-1}+N-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) }{\det\left( \left( y_{i}^{j-1}\right) _{1\leq i\leq N,\ 1\leq j\leq N}\right) }. \end{align*} This proves (\ref{pf.schur.step2}). The proof of the 2nd step is thus complete. \textit{3rd step:} Consider the action of the Heisenberg algebra $\mathcal{A}$ on $\widetilde{F}=\mathcal{B}^{\left( 0\right) }$ and $\wedge^{\dfrac {\infty}{2},0}V=\mathcal{F}^{\left( 0\right) }$. We will now prove that the map $\tau:\wedge^{\dfrac{\infty}{2},0}V\rightarrow\widetilde{F}$ satisfies% \begin{equation} \tau\circ a_{-m}=a_{-m}\circ\tau\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }m. \label{pf.schur.step3}% \end{equation} \textit{Proof of (\ref{pf.schur.step3}):} Let $m$ be a positive integer. Let $\left( i_{0},i_{1},i_{2},...\right) $ be a $0$-degression. By the definition of a $0$-degression, $\left( i_{0},i_{1},i_{2},...\right) $ is a strictly decreasing sequence of integers such that every sufficiently high $k\in\mathbb{N}$ satisfies $i_{k}+k=0$. In other words, there exists an $\ell\in\mathbb{N}$ such that every integer $k\geq\ell$ satisfies $i_{k}+k=0$. Consider this $\ell$. Let $N$ be any integer satisfying $N\geq\ell+m$. Then, it is easy to see that, for every integer $k\geq N$, we have $i_{k}+m=i_{k-m}$. By the definition of the $\mathcal{A}$-module structure on $\wedge ^{\dfrac{\infty}{2},0}V$, the action of $a_{-m}$ on $\wedge^{\dfrac{\infty}% {2},0}V$ is $\widehat{\rho}\left( T^{-m}\right) $, where $T$ is the shift operator. Thus,% \begin{equation} a_{-m}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =\left( \widehat{\rho}\left( T^{-m}\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) . \label{pf.schur.step3.2}% \end{equation} Since $m\neq0$, the matrix $T^{-m}$ has the property that, for every integer $i$, the $\left( i,i\right) $-th entry of $T^{-m}$ is $0$. Hence, Proposition \ref{prop.glinf.ainfact} (applied to $0$, $T^{-m}$ and $v_{i_{k}}$ instead of $m$, $a$ and $b_{k}$) yields% \begin{align*} & \left( \widehat{\rho}\left( T^{-m}\right) \right) \left( v_{i_{0}% }\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\ & =\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge\underbrace{\left( T^{-m}\rightharpoonup v_{i_{k}}\right) }_{=v_{i_{k}+m}}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & =\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge v_{i_{k}+m}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & =\underbrace{\sum\limits_{\substack{k\geq0;\\k 0}y_{s}\dfrac{\partial}{\partial z_{s}}\right) P\left( z\right) =P\left( y+z\right) . \] Here, $y+z$ means the componentwise sum of the sequences $y$ and $z$ (so that $y+z=\left( y_{1}+z_{1},y_{2}+z_{2},y_{3}+z_{3},...\right) $). \end{lemma} Lemma \ref{lem.hirota.newton} is actually a multivariate generalization of the famous Taylor formula% \[ \exp\left( \alpha\dfrac{\partial}{\partial\xi}\right) P\left( \xi\right) =P\left( \alpha+\xi\right) \] which holds for any polynomial $P\in K\left[ \xi\right] $ and any $\alpha\in K$. \textit{Proof of Lemma \ref{lem.hirota.newton}.} Let $A$ be the map% \[ \exp\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}\right) :K\left[ z_{1},z_{2},z_{3},...\right] \rightarrow K\left[ z_{1},z_{2}% ,z_{3},...\right] \] (this is easily seen to be well-defined). Let $B$ be the map% \[ K\left[ z_{1},z_{2},z_{3},...\right] \rightarrow K\left[ z_{1},z_{2}% ,z_{3},...\right] ,\ \ \ \ \ \ \ \ \ \ P\mapsto P\left( y+z\right) . \] We have $A=\exp\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}% }\right) $, so that $A$ is the exponential of a derivation (since $\sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}$ is a derivation). Thus, $A$ is a $K$-algebra homomorphism (since there is a known fact that the exponential of a derivation is a $K$-algebra homomorphism). Combined with the fact that $B$ is a $K$-algebra homomorphism (in fact, $B$ is an evaluation homomorphism), this yields that both $A$ and $B$ are $K$-algebra homomorphisms. Now, let $k$ be a positive integer. We will prove that $Az_{k}=Bz_{k}$. We have \[ \left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}\right) z_{k}=\sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}z_{k}% =y_{k}\underbrace{\dfrac{\partial}{\partial z_{k}}z_{k}}_{=1}+\sum \limits_{\substack{s>0;\\s\neq k}}y_{s}\underbrace{\dfrac{\partial}{\partial z_{s}}z_{k}}_{\substack{=0\\\text{(since }s\neq k\text{)}}}=y_{k}% +\underbrace{\sum\limits_{\substack{s>0;\\s\neq k}}y_{s}0}_{=0}=y_{k}, \] so that% \[ \left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}\right) ^{2}z_{k}=\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}% }\right) \underbrace{\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}\right) z_{k}}_{=y_{k}}=\left( \sum\limits_{s>0}y_{s}\dfrac{\partial }{\partial z_{s}}\right) y_{k}=\sum\limits_{s>0}y_{s}\underbrace{\dfrac {\partial}{\partial z_{s}}y_{k}}_{=0}=0. \] As a consequence, \begin{equation} \text{every integer }i\geq2\text{ satisfies }\left( \sum\limits_{s>0}% y_{s}\dfrac{\partial}{\partial z_{s}}\right) ^{i}z_{k}=0. \label{pf.hirota.newton.1}% \end{equation} Now, since $A=\exp\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}\right) =\sum\limits_{i\in\mathbb{N}}\dfrac{1}{i!}\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}\right) ^{i}$, we have% \begin{align*} Az_{k} & =\sum\limits_{i\in\mathbb{N}}\dfrac{1}{i!}\left( \sum \limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}\right) ^{i}z_{k}\\ & =\underbrace{\dfrac{1}{0!}}_{=1}\underbrace{\left( \sum\limits_{s>0}% y_{s}\dfrac{\partial}{\partial z_{s}}\right) ^{0}}_{=\operatorname*{id}}% z_{k}+\underbrace{\dfrac{1}{1!}}_{=1}\underbrace{\left( \sum\limits_{s>0}% y_{s}\dfrac{\partial}{\partial z_{s}}\right) ^{1}}_{=\sum\limits_{s>0}% y_{s}\dfrac{\partial}{\partial z_{s}}}z_{k}+\sum\limits_{i\geq2}\dfrac{1}% {i!}\underbrace{\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}% }\right) ^{i}z_{k}}_{\substack{=0\\\text{(by (\ref{pf.hirota.newton.1}))}}}\\ & =\underbrace{\operatorname*{id}z_{k}}_{=z_{k}}+\underbrace{\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}\right) z_{k}}_{=y_{k}% }+\underbrace{\sum\limits_{i\geq2}\dfrac{1}{i!}0}_{=0}=z_{k}+y_{k}=y_{k}% +z_{k}. \end{align*} Compared to \begin{align*} Bz_{k} & =z_{k}\left( y+z\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }B\right) \\ & =y_{k}+z_{k}, \end{align*} this yields $Az_{k}=Bz_{k}$. Now, forget that we fixed $k$. We thus have shown that $Az_{k}=Bz_{k}$ for every positive integer $k$. In other words, the maps $A$ and $B$ coincide on the set $\left\{ z_{1},z_{2},z_{3},...\right\} $. Since the set $\left\{ z_{1},z_{2},z_{3},...\right\} $ generates $K\left[ z_{1},z_{2}% ,z_{3},...\right] $ as a $K$-algebra, this yields that the maps $A$ and $B$ coincide on a generating set of the $K$-algebra $K\left[ z_{1},z_{2}% ,z_{3},...\right] $. Since $A$ and $B$ are $K$-algebra homomorphisms, this yields that $A=B$ (because if two $K$-algebra homomorphisms coincide on a $K$-algebra generating set of their domain, then they must be equal). Hence, every $P\in K\left[ z_{1},z_{2},z_{3},...\right] $ satisfies% \[ \underbrace{\exp\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}% }\right) }_{=A=B}\underbrace{P\left( z\right) }_{=P}=BP=P\left( y+z\right) \] (by the definition of $B$). This proves Lemma \ref{lem.hirota.newton}. \subsubsection{\label{subsubsect.GLinfty} \texorpdfstring{$\operatorname*{GL}\left( \infty\right) $}{GL-infinity} and \texorpdfstring{$\operatorname*{M}\left( \infty\right) $}{M-infinity}} We now introduce the groups $\operatorname*{GL}\left( \infty\right) $ and $\operatorname*{M}\left( \infty\right) $ and their actions on $\wedge ^{\dfrac{\infty}{2},m}V$. On the one hand, this will prepare us to the second proof of Theorem \ref{thm.schur}; on the other hand, these group actions are of autonomous interest, and we will meet them again in Subsection \ref{subsubsect.infgrass}. \begin{definition} \label{def.Minf}We let $\operatorname*{M}\left( \infty\right) $ denote the set $\operatorname*{id}+\mathfrak{gl}_{\infty}$. In other words, we let $\operatorname*{M}\left( \infty\right) $ denote the set of all infinite matrices (infinite in both directions) which are equal to the infinite identity matrix $\operatorname*{id}$ in all but finitely many entries. \end{definition} Clearly, $\operatorname*{M}\left( \infty\right) \subseteq\overline {\mathfrak{a}_{\infty}}$ as sets. We notice that: \begin{proposition} \label{prop.Minf.monoid}\textbf{(a)} For every $A\in\operatorname*{M}\left( \infty\right) $ and $B\in\operatorname*{M}\left( \infty\right) $, the matrix $AB$ is well-defined and lies in $\operatorname*{M}\left( \infty\right) $. \textbf{(b)} We have $\operatorname*{id}\in\operatorname*{M}\left( \infty\right) $ (where $\operatorname*{id}$ denotes the infinite identity matrix). \textbf{(c)} The set $\operatorname*{M}\left( \infty\right) $ becomes a monoid under multiplication of matrices. \textbf{(d)} If a matrix $A\in\operatorname*{M}\left( \infty\right) $ is invertible, then its inverse also lies in $\operatorname*{M}\left( \infty\right) $. \textbf{(e)} Denote by $\operatorname*{GL}\left( \infty\right) $ the subset $\left\{ A\in\operatorname*{M}\left( \infty\right) \ \mid\ A\text{ is invertible}\right\} $ of $\operatorname*{M}\left( \infty\right) $. Then, $\operatorname*{GL}\left( \infty\right) $ becomes a group under multiplication of matrices. \end{proposition} \begin{remark} \label{rmk.GLinf}In Proposition \ref{prop.Minf.monoid}, a matrix $A\in\operatorname*{M}\left( \infty\right) $ is said to be \textit{invertible} if there exists an infinite matrix $B$ (with rows and columns indexed by integers) satisfying $AB=BA=\operatorname*{id}$. The matrix $B$ is then called the \textit{inverse} of $A$. Note that we don't a-priori require that $B$ lie in $\operatorname*{M}\left( \infty\right) $, or any other ``finiteness conditions'' for $B$; Proposition \ref{prop.Minf.monoid} \textbf{(d)} shows that these conditions are automatically satisfied. \end{remark} \begin{definition} \label{def.GLinf}Let $\operatorname*{GL}\left( \infty\right) $ denote the group $\operatorname*{GL}\left( \infty\right) $ defined in Proposition \ref{prop.Minf.monoid} \textbf{(e)}. \end{definition} \textit{Proof of Proposition \ref{prop.Minf.monoid}.} \textbf{(a)} Let $A\in\operatorname*{M}\left( \infty\right) $ and $B\in\operatorname*{M}% \left( \infty\right) $. Since $A\in\operatorname*{M}\left( \infty\right) =\operatorname*{id}+\mathfrak{gl}_{\infty}$, there exists an $a\in \mathfrak{gl}_{\infty}$ such that $A=\operatorname*{id}+a$. Consider this $a$. Since $B\in\operatorname*{M}\left( \infty\right) =\operatorname*{id}% +\mathfrak{gl}_{\infty}$, there exists a $b\in\mathfrak{gl}_{\infty}$ such that $B=\operatorname*{id}+b$. Consider this $b$. Since $A=\operatorname*{id}+a$ and $B=\operatorname*{id}+b$, we have $AB=\left( \operatorname*{id}+a\right) \left( \operatorname*{id}+b\right) =\operatorname*{id}+a+b+ab$, which is clearly well-defined (because $a\in\mathfrak{gl}_{\infty}$ and $b\in\mathfrak{gl}_{\infty}$ lead to $ab$ being well-defined) and lies in $\operatorname*{M}\left( \infty\right) $ (since $\underbrace{a}_{\in\mathfrak{gl}_{\infty}}+\underbrace{b}% _{\in\mathfrak{gl}_{\infty}}+\underbrace{ab}_{\substack{\in\mathfrak{gl}% _{\infty}\\\text{(since }a\in\mathfrak{gl}_{\infty}\text{ and }b\in \mathfrak{gl}_{\infty}\text{)}}}\in\mathfrak{gl}_{\infty}+\mathfrak{gl}% _{\infty}+\mathfrak{gl}_{\infty}\subseteq\mathfrak{gl}_{\infty}$ and thus $\operatorname*{id}+a+b+ab\in\operatorname*{id}+\mathfrak{gl}_{\infty }=\operatorname*{M}\left( \infty\right) $). This proves Proposition \ref{prop.Minf.monoid} \textbf{(a)}. \textbf{(b)} Trivial. \textbf{(c)} Follows from \textbf{(a)} and \textbf{(b)}. \textbf{(d)} Let $A\in\operatorname*{M}\left( \infty\right) $ be invertible. Since $A\in\operatorname*{M}\left( \infty\right) =\operatorname*{id}% +\mathfrak{gl}_{\infty}$, there exists an $a\in\mathfrak{gl}_{\infty}$ such that $A=\operatorname*{id}+a$. Consider this $a$. Since $A$ is invertible, there exists an infinite matrix $B$ (with rows and columns indexed by integers) satisfying $AB=BA=\operatorname*{id}$ (according to how we defined ``invertible'' in Remark \ref{rmk.GLinf}). Consider this $B$. This $B$ is the inverse of $A$. Let $b=B-\operatorname*{id}$. Then, $B=\operatorname*{id}+b$. Since $A=\operatorname*{id}+a$ and $B=\operatorname*{id}+b$, we have $AB=\left( \operatorname*{id}+a\right) \left( \operatorname*{id}+b\right) =\operatorname*{id}+a+b+ab$, which is clearly well-defined (because $a\in\mathfrak{gl}_{\infty}$ leads to $ab$ being well-defined). Since $\operatorname*{id}=AB=\operatorname*{id}+ab+a+b$, we have $0=ab+a+b$. Let us introduce two notations that we will use during this proof: \begin{itemize} \item For any infinite matrix $M$ and any pair $\left( i,j\right) $ of integers, let us denote by $M_{i,j}$ the $\left( i,j\right) $-th entry of the matrix $M$. (In particular, for any pair $\left( i,j\right) $ of integers, we denote by $a_{i,j}$ the $\left( i,j\right) $-th entry of the matrix $a$ (not of the matrix $A$ !), and we denote by $b_{i,j}$ the $\left( i,j\right) $-th entry of the matrix $b$ (not of the matrix $B$ !).) \item For any assertion $\mathcal{A}$, let $\left[ \mathcal{A}\right] $ denote the integer $\left\{ \begin{array} [c]{l}% 1,\text{ if }\mathcal{A}\text{ is true;}\\ 0,\text{ if }\mathcal{A}\text{ is wrong}% \end{array} \right. $. \end{itemize} Since $a\in\mathfrak{gl}_{\infty}$, only finitely many entries of the matrix $a$ are nonzero. In particular, this yields that only finitely many columns of the matrix $a$ are nonzero. Hence, there exists a nonnegative integer $N$ such that \begin{equation} \left( \text{for every integer }j\text{ with }\left\vert j\right\vert >N\text{, the }j\text{-th column of }a\text{ is zero}\right) . \label{pf.Minf.monoid.1}% \end{equation} Consider this $N$. Clearly,% \begin{equation} \left( \text{for every }\left( i,j\right) \in\mathbb{Z}^{2}\text{ such that }\left\vert j\right\vert >N\text{, we have }a_{i,j}=0\right) \label{pf.Minf.monoid.1a}% \end{equation} (because for every $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $\left\vert j\right\vert >N$, the $j$-th column of $a$ is zero (by (\ref{pf.Minf.monoid.1})), so that every entry on the $j$-th column of $a$ is zero, so that $a_{i,j}$ is zero (because the element $a_{i,j}$ is the $\left( i,j\right) $-th entry of $a$, hence an entry on the $j$-th column of $a$)). Recall that only finitely many entries of the matrix $a$ are nonzero. In particular, this yields that only finitely many rows of the matrix $a$ are nonzero. Hence, there exists a nonnegative integer $M$ such that \begin{equation} \left( \text{for every integer }i\text{ with }\left\vert i\right\vert >M\text{, the }i\text{-th row of }a\text{ is zero}\right) . \label{pf.Minf.monoid.2}% \end{equation} Consider this $M$. Clearly,% \begin{equation} \left( \text{for every }\left( i,j\right) \in\mathbb{Z}^{2}\text{ such that }\left\vert i\right\vert >M\text{, we have }a_{i,j}=0\right) \label{pf.Minf.monoid.2a}% \end{equation} (because for every $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $\left\vert i\right\vert >M$, the $i$-th row of $a$ is zero (by (\ref{pf.Minf.monoid.2})), so that every entry on the $i$-th row of $a$ is zero, so that $a_{i,j}$ is zero (because the element $a_{i,j}$ is the $\left( i,j\right) $-th entry of $a$, hence an entry on the $i$-th row of $a$)). Let $P=\max\left\{ M,N\right\} $. Clearly, $P\geq M$ and $P\geq N$. It is now easy to see that \begin{equation} \text{any }\left( i,j\right) \in\mathbb{Z}^{2}\text{ satisfies }% a_{i,j}=\left[ \left\vert i\right\vert \leq P\right] \cdot a_{i,j}. \label{pf.Minf.monoid.a}% \end{equation} \footnote{\textit{Proof of (\ref{pf.Minf.monoid.a}):} Let $\left( i,j\right) \in\mathbb{Z}^{2}$. Then, we must be in one of the following three cases: \par \textit{Case 1:} We don't have $\left\vert i\right\vert \leq P$. \par \textit{Case 2:} We have $\left\vert i\right\vert \leq P$. \par Let us consider Case 1 first. In this case, we don't have $\left\vert i\right\vert \leq P$. Thus, $\left[ \left\vert i\right\vert \leq P\right] =0$ and $\left\vert i\right\vert >P$. From $\left\vert i\right\vert >P\geq M$, we conclude that $a_{i,j}=0$ (by (\ref{pf.Minf.monoid.2a})). Compared with $\underbrace{\left[ \left\vert i\right\vert \leq P\right] }_{=0}\cdot a_{i,j}=0$, this yields $a_{i,j}=\left[ \left\vert i\right\vert \leq P\right] \cdot a_{i,j}$. Hence, (\ref{pf.Minf.monoid.a}) is proven in Case 1. \par Finally, let us consider Case 2. In this case, we have $\left\vert i\right\vert \leq P$. Hence, $\left[ \left\vert i\right\vert \leq P\right] =1$. Thus, $\underbrace{\left[ \left\vert i\right\vert \leq P\right] }% _{=1}\cdot a_{i,j}=a_{i,j}$. Hence, (\ref{pf.Minf.monoid.a}) is proven in Case 2. \par Altogether, we have thus proven (\ref{pf.Minf.monoid.a}) in each of the two cases 1 and 2. Since these two cases cover all possibilities, this shows that (\ref{pf.Minf.monoid.a}) always holds. Thus, (\ref{pf.Minf.monoid.a}) is proven.} Similarly,% \begin{equation} \text{any }\left( i,j\right) \in\mathbb{Z}^{2}\text{ satisfies }% a_{i,j}=\left[ \left\vert j\right\vert \leq P\right] \cdot a_{i,j}. \label{pf.Minf.monoid.a'}% \end{equation} Let $b^{\prime}$ be the infinite matrix (with rows and columns indexed by integers) defined by% \begin{equation} \left( b_{i,j}^{\prime}=\left[ \left\vert i\right\vert \leq P\right] \cdot\left[ \left\vert j\right\vert \leq P\right] \cdot b_{i,j}% \ \ \ \ \ \ \ \ \ \ \text{for all }\left( i,j\right) \in\mathbb{Z}% ^{2}\right) . \label{pf.Minf.monoid.bprime}% \end{equation} It is clear that only finitely many entries of $b^{\prime}$ are nonzero\footnote{\textit{Proof.} Let $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $b_{i,j}^{\prime}\neq0$. Then, $\left\vert i\right\vert \leq P$ (because otherwise, we would have $\left[ \left\vert i\right\vert \leq P\right] =0$, so that $b_{i,j}^{\prime}=\underbrace{\left[ \left\vert i\right\vert \leq P\right] }_{=0}\cdot\left[ \left\vert j\right\vert \leq P\right] \cdot b_{i,j}=0$, contradicting to $b_{i,j}^{\prime}\neq0$), so that $i\in\left\{ -P,-P+1,...,P\right\} $, and similarly $j\in\left\{ -P,-P+1,...,P\right\} $. Hence, $\left( i,j\right) \in\left\{ -P,-P+1,...,P\right\} ^{2}$ (since $i\in\left\{ -P,-P+1,...,P\right\} $ and $j\in\left\{ -P,-P+1,...,P\right\} $). \par Now forget that we fixed $\left( i,j\right) $. We thus have showed that every $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $b_{i,j}^{\prime}% \neq0$ satisfies $\left( i,j\right) \in\left\{ -P,-P+1,...,P\right\} ^{2}% $. Since there are only finitely many $\left( i,j\right) \in\left\{ -P,-P+1,...,P\right\} ^{2}$, this yields that there are only finitely many $\left( i,j\right) \in\mathbb{Z}^{2}$ such that $b_{i,j}^{\prime}\neq0$. In other words, there are only finitely many $\left( i,j\right) \in \mathbb{Z}^{2}$ such that the $\left( i,j\right) $-th entry of $b^{\prime}$ is nonzero. In other words, only finitely many entries of $b^{\prime}$ are nonzero, qed.}. In other words, $b^{\prime}\in\mathfrak{gl}_{\infty}$, so that $\operatorname*{id}+b^{\prime}\in\operatorname*{id}+\mathfrak{gl}_{\infty }=\operatorname*{M}\left( \infty\right) $. We will now prove that $A\left( \operatorname*{id}+b^{\prime}\right) =\operatorname*{id}$. For every $\left( i,j\right) \in\mathbb{Z}^{2}$, we have% \begin{align*} & \left( ab^{\prime}+a+b^{\prime}\right) _{i,j}\\ & =\underbrace{\left( ab^{\prime}\right) _{i,j}}_{\substack{=\sum \limits_{k\in\mathbb{Z}}a_{i,k}b_{k,j}^{\prime}\\\text{(by the definition of the}\\\text{product of two matrices)}}}+a_{i,j}+\underbrace{b_{i,j}^{\prime}% }_{\substack{=\left[ \left\vert i\right\vert \leq P\right] \cdot\left[ \left\vert j\right\vert \leq P\right] \cdot b_{i,j}\\\text{(by (\ref{pf.Minf.monoid.bprime}))}}}\\ & =\sum\limits_{k\in\mathbb{Z}}a_{i,k}\underbrace{b_{k,j}^{\prime}% }_{\substack{=\left[ \left\vert k\right\vert \leq P\right] \cdot\left[ \left\vert j\right\vert \leq P\right] \cdot b_{k,j}\\\text{(by (\ref{pf.Minf.monoid.bprime}), applied to}\\k\text{ instead of }i\text{)}% }}+a_{i,j}+\left[ \left\vert i\right\vert \leq P\right] \cdot\left[ \left\vert j\right\vert \leq P\right] \cdot b_{i,j}\\ & =\sum\limits_{k\in\mathbb{Z}}\underbrace{a_{i,k}\left[ \left\vert k\right\vert \leq P\right] }_{\substack{=\left[ \left\vert k\right\vert \leq P\right] \cdot a_{i,k}=a_{i,k}\\\text{(since (\ref{pf.Minf.monoid.a'}) (applied to}\\k\text{ instead of }j\text{) yields }a_{i,k}=\left[ \left\vert k\right\vert \leq P\right] \cdot a_{i,k}\text{)}}}\cdot\left[ \left\vert j\right\vert \leq P\right] \cdot b_{k,j}+\underbrace{a_{i,j}}% _{\substack{=\left[ \left\vert i\right\vert \leq P\right] \cdot a_{i,j}\\\text{(by (\ref{pf.Minf.monoid.a}))}}}+\left[ \left\vert i\right\vert \leq P\right] \cdot\left[ \left\vert j\right\vert \leq P\right] \cdot b_{i,j}\\ & =\sum\limits_{k\in\mathbb{Z}}\underbrace{a_{i,k}}_{\substack{=\left[ \left\vert i\right\vert \leq P\right] \cdot a_{i,k}\\\text{(by (\ref{pf.Minf.monoid.a}), applied to}\\k\text{ instead of }j\text{)}}% }\cdot\left[ \left\vert j\right\vert \leq P\right] \cdot b_{k,j}+\left[ \left\vert i\right\vert \leq P\right] \cdot\underbrace{a_{i,j}}% _{\substack{=\left[ \left\vert j\right\vert \leq P\right] \cdot a_{i,j}\\\text{(by (\ref{pf.Minf.monoid.a'}))}}}+\left[ \left\vert i\right\vert \leq P\right] \cdot\left[ \left\vert j\right\vert \leq P\right] \cdot b_{i,j}\\ & =\sum\limits_{k\in\mathbb{Z}}\left[ \left\vert i\right\vert \leq P\right] \cdot a_{i,k}\cdot\left[ \left\vert j\right\vert \leq P\right] \cdot b_{k,j}+\left[ \left\vert i\right\vert \leq P\right] \cdot\left[ \left\vert j\right\vert \leq P\right] \cdot a_{i,j}+\left[ \left\vert i\right\vert \leq P\right] \cdot\left[ \left\vert j\right\vert \leq P\right] \cdot b_{i,j}\\ & =\left[ \left\vert i\right\vert \leq P\right] \cdot\left[ \left\vert j\right\vert \leq P\right] \cdot\left( \sum\limits_{k\in\mathbb{Z}}% a_{i,k}b_{k,j}+a_{i,j}+b_{i,j}\right) =\left[ \left\vert i\right\vert \leq P\right] \cdot\left[ \left\vert j\right\vert \leq P\right] \cdot \underbrace{\left( \left( ab\right) _{i,j}+a_{i,j}+b_{i,j}\right) }_{\substack{=\left( ab+a+b\right) _{i,j}=0\\\text{(since }ab+a+b=0\text{)}% }}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\left( ab\right) _{i,j}=\sum\limits_{k\in\mathbb{Z}}% a_{i,k}b_{k,j}\text{ (by the definition of the product of two matrices),}\\ \text{so that }\sum\limits_{k\in\mathbb{Z}}a_{i,k}b_{k,j}=\left( ab\right) _{i,j}% \end{array} \right) \\ & =0. \end{align*} Thus, $ab^{\prime}+a+b^{\prime}=0$. Since $A=\operatorname*{id}+a$, we have $A\left( \operatorname*{id}+b^{\prime}\right) =\left( \operatorname*{id}% +a\right) \left( \operatorname*{id}+b^{\prime}\right) =\operatorname*{id}% +\underbrace{ab^{\prime}+a+b^{\prime}}_{=0}=\operatorname*{id}$. We thus have shown that $A\left( \operatorname*{id}+b^{\prime}\right) =\operatorname*{id}$. Now, it is easy to see that the products $B\left( A\left( \operatorname*{id}% +b^{\prime}\right) \right) $ and $\left( BA\right) \left( \operatorname*{id}+b^{\prime}\right) $ are well-defined and satisfy associativity, i. e., we have $B\left( A\left( \operatorname*{id}+b^{\prime }\right) \right) =\left( BA\right) \left( \operatorname*{id}+b^{\prime }\right) $. Now,% \[ B=B\cdot\underbrace{\operatorname*{id}}_{=A\left( \operatorname*{id}% +b^{\prime}\right) }=B\left( A\left( \operatorname*{id}+b^{\prime}\right) \right) =\underbrace{\left( BA\right) }_{=\operatorname*{id}}\left( \operatorname*{id}+b^{\prime}\right) =\operatorname*{id}+b^{\prime}% \in\operatorname*{M}\left( \infty\right) . \] Since $B$ is the inverse of $A$, this yields that the inverse of $A$ lies in $\operatorname*{M}\left( \infty\right) $. This proves Proposition \ref{prop.Minf.monoid} \textbf{(d)}. \textbf{(e)} Follows from \textbf{(c)} and \textbf{(d)}. The proof of Proposition \ref{prop.Minf.monoid} is complete. We now construct a group action of $\operatorname*{GL}\left( \infty\right) $ on $\mathcal{F}^{\left( m\right) }$ that is related to the Lie algebra action $\rho$ of $\mathfrak{gl}_{\infty}$ on $\mathcal{F}^{\left( m\right) }$ in the same way as the action of a Lie group on a representation is usually related to its ``derivative'' action of the corresponding Lie algebra: \begin{definition} \label{def.GLinf.act}Let $m\in\mathbb{Z}$. We define an action $\varrho :\operatorname*{M}\left( \infty\right) \rightarrow\operatorname*{End}\left( \mathcal{F}^{\left( m\right) }\right) $ of the monoid $\operatorname*{M}% \left( \infty\right) $ on the vector space $\mathcal{F}^{\left( m\right) }=\wedge^{\dfrac{\infty}{2},m}V$ as follows: For every $A\in\operatorname*{M}% \left( \infty\right) $ and every $m$-degression $\left( i_{0},i_{1}% ,i_{2},...\right) $, we set% \[ \left( \varrho\left( A\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...\right) =Av_{i_{0}}\wedge Av_{i_{1}}\wedge Av_{i_{2}}\wedge.... \] (This is then extended to the whole $\mathcal{F}^{\left( m\right) }$ by linearity.) It is very easy to see that this is well-defined (because $Av_{k}=v_{k}$ for all sufficiently small $k$) and indeed gives a monoid action. The restriction $\varrho\mid_{\operatorname*{GL}\left( \infty\right) }:\operatorname*{GL}\left( \infty\right) \rightarrow\operatorname*{End}% \left( \mathcal{F}^{\left( m\right) }\right) $ to $\operatorname*{GL}% \left( \infty\right) $ is thus a group action of $\operatorname*{GL}\left( \infty\right) $ on $\mathcal{F}^{\left( m\right) }$. Since we have defined an action of $\operatorname*{M}\left( \infty\right) $ on $\mathcal{F}^{\left( m\right) }$ for every $m\in\mathbb{Z}$, we thus obtain an action of $\operatorname*{M}\left( \infty\right) $ on $\mathcal{F}=\bigoplus\limits_{m\in\mathbb{Z}}\mathcal{F}^{\left( m\right) }$ (namely, the direct sum of the previous actions). This latter action will also be denoted by $\varrho$. \end{definition} Note that the letter $\varrho$ is a capital rho, as opposed to $\rho$ which is the lowercase rho. When $A$ is a matrix in $\operatorname*{M}\left( \infty\right) $, the endomorphism $\varrho\left( A\right) $ of $\mathcal{F}^{\left( m\right) }$ can be seen as an infinite analogue of the endomorphisms $\wedge^{\ell}A$ of $\wedge^{\ell}V$ defined for all $\ell\in\mathbb{N}$. We are next going to give an explicit formula for the action of $\varrho \left( A\right) $ on $\mathcal{F}^{\left( m\right) }$ in terms of (infinite) minors of $A$. The formula will be an infinite analogue of the following well-known formula: \begin{proposition} \label{prop.GLinf.det.fin}Let $P$ be a finite-dimensional $\mathbb{C}$-vector space with basis $\left( e_{1},e_{2},...,e_{n}\right) $, and let $Q$ be a finite-dimensional $\mathbb{C}$-vector space with basis $\left( f_{1}% ,f_{2},...,f_{m}\right) $. Let $\ell\in\mathbb{N}$. Let $f:P\rightarrow Q$ be a linear map, and let $A$ be the $m\times n$-matrix which represents this map $f$ with respect to the bases $\left( e_{1}% ,e_{2},...,e_{n}\right) $ and $\left( f_{1},f_{2},...,f_{m}\right) $ of $P$ and $Q$. Let $i_{1}$, $i_{2}$, $...$, $i_{\ell}$ be integers such that $1\leq i_{1} N$ satisfies $\pi_{m-i}\left( b_{i}\right) =v_{m-i}$. (Such an $N$ exists, since we know that $\pi_{m-i}\left( b_{i}\right) =v_{m-i}$ for all sufficiently large $i$.) Then, we define $b_{0}\wedge b_{1}\wedge b_{2}% \wedge...$ to be the element \[ \pi_{m-N}\left( b_{0}\right) \wedge\pi_{m-N}\left( b_{1}\right) \wedge...\wedge\pi_{m-N}\left( b_{N}\right) \wedge v_{m-N-1}\wedge v_{m-N-2}\wedge v_{m-N-3}\wedge...\in\wedge^{\dfrac{\infty}{2},m}V. \] This element does not depend on the choice of $N$ (according to Proposition \ref{prop.uinf.Vhatwedge.welldef} below). Hence, $b_{0}\wedge b_{1}\wedge b_{2}\wedge...$ is well-defined. \end{definition} The next few propositions state some properties of wedge products of elements of $\widehat{V}$ similar to some properties of wedge products of elements of $V$ stated above. We will not prove them; neither of them is actually difficult to verify. \begin{proposition} \label{prop.uinf.Vhatwedge.welldef}Let $m\in\mathbb{Z}$. Let $b_{0}% ,b_{1},b_{2},...$ be vectors in $\widehat{V}$ which satisfy% \[ \pi_{m-i}\left( b_{i}\right) =v_{m-i}\ \ \ \ \ \ \ \ \ \ \text{for all sufficiently large }i. \] If we pick some $N\in\mathbb{N}$ such that every $i>N$ satisfies $\pi _{m-i}\left( b_{i}\right) =v_{m-i}$, then the element% \[ \pi_{m-N}\left( b_{0}\right) \wedge\pi_{m-N}\left( b_{1}\right) \wedge...\wedge\pi_{m-N}\left( b_{N}\right) \wedge v_{m-N-1}\wedge v_{m-N-2}\wedge v_{m-N-3}\wedge...\in\wedge^{\dfrac{\infty}{2},m}V \] does not depend on the choice of $N$. \end{proposition} \begin{proposition} The wedge product defined in Definition \ref{def.uinf.Vhatwedge} is antisymmetric and multilinear (in the appropriate sense). \end{proposition} \begin{definition} \label{def.uinf.semiinfwedge}Let $m\in\mathbb{Z}$. Define an action of the Lie algebra $\mathfrak{u}_{\infty}$ on the vector space $\wedge^{\dfrac{\infty}% {2},m}V$ by the equation% \[ a\rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right) =\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge\left( a\rightharpoonup v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge... \] for all $a\in\mathfrak{u}_{\infty}$ and all elementary semiinfinite wedges $v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...$ (and by linear extension). \end{definition} \begin{proposition} Let $m\in\mathbb{Z}$. Then, Definition \ref{def.uinf.semiinfwedge} really defines a representation of the Lie algebra $\mathfrak{u}_{\infty}$ on the vector space $\wedge^{\dfrac{\infty}{2},m}V$. \end{proposition} \begin{proposition} Let $m\in\mathbb{Z}$. Let $b_{0},b_{1},b_{2},...$ be vectors in $\widehat{V}$ which satisfy% \[ \pi_{m-i}\left( b_{i}\right) =v_{m-i}\ \ \ \ \ \ \ \ \ \ \text{for all sufficiently large }i. \] Let $a\in\mathfrak{u}_{\infty}$. Then,% \[ a\rightharpoonup\left( b_{0}\wedge b_{1}\wedge b_{2}\wedge...\right) =\sum\limits_{k\geq0}b_{0}\wedge b_{1}\wedge...\wedge b_{k-1}\wedge\left( a\rightharpoonup b_{k}\right) \wedge b_{k+1}\wedge b_{k+2}\wedge.... \] \end{proposition} \begin{definition} \label{def.Uinf.rho}Let $m\in\mathbb{Z}$. Let $\rho:\mathfrak{u}_{\infty }\rightarrow\operatorname*{End}\left( \wedge^{\dfrac{\infty}{2},m}V\right) $ be the representation of $\mathfrak{u}_{\infty}$ on $\wedge^{\dfrac{\infty}% {2},m}V$ defined in Definition \ref{def.uinf.semiinfwedge}. (We denote this representation by the same letter $\rho$ as the representation $\mathfrak{gl}% _{\infty}\rightarrow\operatorname*{End}\left( \wedge^{\dfrac{\infty}{2}% ,m}V\right) $ from Definition \ref{def.glinf.rho}. This is intentional and unproblematic, because both of these representations have the same restriction onto $\mathfrak{u}_{\infty}\cap\mathfrak{gl}_{\infty}$.) \end{definition} \begin{remark} \label{rmk.Uinf.rhorhohat}Let $m\in\mathbb{Z}$. Let $a\in\mathfrak{u}_{\infty }\cap\overline{\mathfrak{a}_{\infty}}$. Then, $\rho\left( a\right) =\widehat{\rho}\left( a\right) $ (where $\rho\left( a\right) $ is defined according to Definition \ref{def.Uinf.rho}, and $\widehat{\rho}\left( a\right) $ is defined according to Definition \ref{def.glinf.rhohat.abar}). \end{remark} \begin{definition} \label{def.Uinf}We let $\operatorname*{U}\left( \infty\right) $ denote the set $\operatorname*{id}+\mathfrak{u}_{\infty}$. In other words, $\operatorname*{U}\left( \infty\right) $ is the set of all upper-triangular infinite matrices (with rows and columns indexed by integers) whose all diagonal entries are $=1$. This set $\operatorname*{U}\left( \infty\right) $ is easily seen to be a group (with respect to matrix multiplication). Inverses in this group can be computed by means of the formula $\left( I_{\infty }+a\right) ^{-1}=\sum\limits_{k=0}^{\infty}a^{k}$ for all $a\in \mathfrak{u}_{\infty}$\ \ \ \ \footnotemark \end{definition} \footnotetext{Here, we are using the fact that, for every $a\in\mathfrak{u}% _{\infty}$, the sum $\sum\limits_{k=0}^{\infty}a^{k}$ converges entrywise (i. e., for every $\left( i,j\right) \in\mathbb{Z}^{2}$, the sum $\sum \limits_{k=0}^{\infty}\left( \text{the }\left( i,j\right) \text{-th entry of }a^{k}\right) $ converges in the discrete topology). Here is why this holds: \par Since $a\in\mathfrak{u}_{\infty}$, we know that the $\left( i,j\right) $-th entry of $a$ is $0$ for all $\left( i,j\right) \in\mathbb{Z}^{2}$ satisfying $i>j-1$. From this, it is easy to conclude (by induction over $k$) that for every $k\in\mathbb{N}$, the $\left( i,j\right) $-th entry of $a^{k}$ is $0$ for all $\left( i,j\right) \in\mathbb{Z}^{2}$ satisfying $i>j-k$. Hence, for every $\left( i,j\right) \in\mathbb{Z}^{2}$, the $\left( i,j\right) $-th entry of $a^{k}$ is $0$ for all nonnegative integers $k$ satisfying $k>j-i$. As a consequence, for every $\left( i,j\right) \in\mathbb{Z}^{2}$, all but finitely many addends of the sum% \[ \sum\limits_{k=0}^{\infty}\left( \text{the }\left( i,j\right) \text{-th entry of }a^{k}\right) \] are $0$. In other words, for every $\left( i,j\right) \in\mathbb{Z}^{2}$, the sum $\sum\limits_{k=0}^{\infty}\left( \text{the }\left( i,j\right) \text{-th entry of }a^{k}\right) $ converges in the discrete topology. Hence, the sum $\sum\limits_{k=0}^{\infty}a^{k}$ converges entrywise, qed.} \begin{definition} \label{def.Uinf.act}Let $m\in\mathbb{Z}$. We define an action $\varrho :\operatorname*{U}\left( \infty\right) \rightarrow\operatorname*{End}\left( \mathcal{F}^{\left( m\right) }\right) $ of the group $\operatorname*{U}% \left( \infty\right) $ on the vector space $\mathcal{F}^{\left( m\right) }=\wedge^{\dfrac{\infty}{2},m}V$ as follows: For every $A\in\operatorname*{U}% \left( \infty\right) $ and every $m$-degression $\left( i_{0},i_{1}% ,i_{2},...\right) $, we set% \[ \left( \varrho\left( A\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...\right) =Av_{i_{0}}\wedge Av_{i_{1}}\wedge Av_{i_{2}}\wedge.... \] (This is then extended to the whole $\mathcal{F}^{\left( m\right) }$ by linearity.) It is very easy to see that this is well-defined (because $\pi_{v_{k}}\left( Av_{k}\right) =v_{k}$ for all sufficiently small $k$) and indeed gives a group action. (We denote this action by the same letter $\varrho$ as the action $\operatorname*{M}\left( \infty\right) \rightarrow\operatorname*{End}\left( \mathcal{F}^{\left( m\right) }\right) $ from Definition \ref{def.GLinf.act}. This is intentional and unproblematic, because both of these actions have the same restriction onto $\operatorname*{U}\left( \infty\right) \cap\operatorname*{M}\left( \infty\right) $.) \end{definition} In analogy to Remark \ref{rmk.GLinf.det}, we have: \begin{remark} \label{rmk.Uinf.det}Let $\left( i_{0},i_{1},i_{2},...\right) $ be an $m$-degression. Let $A\in\operatorname*{U}\left( \infty\right) $. For any $m$-degression $\left( j_{0},j_{1},j_{2},...\right) $, let $A_{j_{0}% ,j_{1},j_{2},...}^{i_{0},i_{1},i_{2},...}$ denote the $\mathbb{N}% \times\mathbb{N}$-matrix defined by% \[ \left( \left( \text{the }\left( u,v\right) \text{-th entry of }% A_{j_{0},j_{1},j_{2},...}^{i_{0},i_{1},i_{2},...}\right) =\left( \text{the }\left( j_{u},i_{v}\right) \text{-th entry of }A\right) \ \ \ \ \ \ \ \ \ \ \text{for every }\left( u,v\right) \in\mathbb{N}% ^{2}\right) . \] (In other words, let $A_{j_{0},j_{1},j_{2},...}^{i_{0},i_{1},i_{2},...}$ denote the matrix which is obtained from $A$ by removing all columns except for the $i_{0}$-th, the $i_{1}$-th, the $i_{2}$-th, etc. ones and removing all rows except for the $j_{0}$-th, the $j_{1}$-th, the $j_{2}$-th, etc. ones, and then inverting the order of the rows, and inverting the order of the columns.) Then, for any $m$-degression $\left( j_{0},j_{1},j_{2},...\right) $, the matrix $\left( A_{j_{0},j_{1},j_{2},...}^{i_{0},i_{1},i_{2},...}\right) ^{T}$ is upper almost-unitriangular, and thus the determinant $\det\left( \left( A_{j_{0},j_{1},j_{2},...}^{i_{0},i_{1},i_{2},...}\right) ^{T}\right) $ makes sense (according to Definition \ref{def.infdet} \textbf{(e)}). We have% \[ \left( \varrho\left( A\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...\right) =\sum\limits_{\left( j_{0},j_{1}% ,j_{2},...\right) \text{ is an }m\text{-degression}}\det\left( \left( A_{j_{0},j_{1},j_{2},...}^{i_{0},i_{1},i_{2},...}\right) ^{T}\right) v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge.... \] \end{remark} The analogy between Remark \ref{rmk.GLinf.det} and Remark \ref{rmk.Uinf.det} is somewhat marred by the fact that the transposed matrix $\left( A_{j_{0},j_{1},j_{2},...}^{i_{0},i_{1},i_{2},...}\right) ^{T}$ is used in Remark \ref{rmk.Uinf.det} instead of the matrix $A_{j_{0},j_{1},j_{2}% ,...}^{i_{0},i_{1},i_{2},...}$. This is merely a technical difference, and if we would have defined the determinant of a \textbf{lower} almost-unitriangular matrix, we could have avoided using the transpose in Remark \ref{rmk.Uinf.det}. \begin{remark} There is a way to ``merge'' $\operatorname*{GL}\left( \infty\right) $ and $\operatorname*{U}\left( \infty\right) $ into a bigger group of infinite matrices. Indeed, let $\operatorname*{M}\nolimits^{\operatorname*{U}}\left( \infty\right) $ the set of all matrices $A\in\operatorname*{U}\left( \infty\right) $ such that all but finitely many among the $\left( i,j\right) \in\mathbb{Z}^{2}$ satisfying $i\geq j$ satisfy $\left( \text{the }\left( i,j\right) \text{-th entry of }A\right) =\delta_{i,j}$. (Note that this condition does not restrict the $\left( i,j\right) $-th entry of $A$ for any $\left( i,j\right) \in\mathbb{Z}^{2}$ satisfying $i >}[d]_{\pi} %& P^{\otimes\ell} \ar@{->>}[d]^{\pi} \\ %\wedge^{\ell} P \ar[r]^{\wedge^{\ell} \left(\exp a\right)} & \wedge^{\ell} P %}}}% %BeginExpansion \xymatrixcolsep{4pc} \xymatrix{ P^{\otimes\ell} \ar[r]^{\left(\exp a\right)^{\otimes\ell}} \ar@{->>}[d]_{\pi} & P^{\otimes\ell} \ar@{->>}[d]^{\pi} \\ \wedge^{\ell} P \ar[r]^{\wedge^{\ell} \left(\exp a\right)} & \wedge^{\ell} P }% %EndExpansion \] commutes. In other words, $\pi\circ\left( \exp a\right) ^{\otimes\ell }=\left( \wedge^{\ell}\left( \exp a\right) \right) \circ\pi$. Thus, \[ \left( \wedge^{\ell}\left( \exp a\right) \right) \circ\pi=\pi\circ\left( \exp a\right) ^{\otimes\ell}=\left( \exp\left( \rho_{P,\ell}\left( a\right) \right) \right) \circ\pi. \] Since the morphism $\pi$ is right-cancellable (since it is surjective), this yields $\wedge^{\ell}\left( \exp a\right) =\exp\left( \rho_{P,\ell}\left( a\right) \right) $. This proves Theorem \ref{thm.finitary.rhoRho}. \end{verlong} \begin{vershort} \textit{Second proof of Theorem \ref{thm.finitary.rhoRho} (sketched).} Since $a$ is nilpotent, it is known that the exponential $\exp a$ is a well-defined unipotent element of $\operatorname*{GL}\left( P\right) $. But for every $\ell\in\mathbb{N}$, the $\ell$-th exterior power of any unipotent element of $\operatorname*{GL}\left( P\right) $ is a unipotent element of $\operatorname*{GL}\left( \wedge^{\ell}P\right) $. Since $\exp a$ is a unipotent element of $\operatorname*{GL}\left( P\right) $, this yields that $\wedge^{\ell}\left( \exp a\right) $ is a unipotent element of $\operatorname*{GL}\left( \wedge^{\ell}P\right) $ for every $\ell \in\mathbb{N}$. Hence, the logarithm $\log\left( \wedge^{\ell}\left( \exp a\right) \right) $ is well-defined for every $\ell\in\mathbb{N}$. \end{vershort} \begin{verlong} \textit{Second proof of Theorem \ref{thm.finitary.rhoRho}.} Since $a$ is nilpotent, it is known that the exponential $\exp a$ is a well-defined unipotent element of $\operatorname*{GL}\left( P\right) $. But for every $\ell\in\mathbb{N}$, the $\ell$-th exterior power of any unipotent element of $\operatorname*{GL}\left( P\right) $ is a unipotent element of $\operatorname*{GL}\left( \wedge^{\ell}P\right) $% \ \ \ \ \footnote{\textit{Proof.} Let $\ell\in\mathbb{N}$. Let $\alpha$ be a unipotent element of $\operatorname*{GL}\left( P\right) $. We must prove that $\wedge^{\ell}\alpha$ is a unipotent element of $\operatorname*{GL}% \left( \wedge^{\ell}P\right) $. \par Since $\alpha$ is unipotent, $\alpha-1\in\operatorname*{End}P$ is nilpotent. That is, there exists an $n\in\mathbb{N}$ such that $\left( \alpha-1\right) ^{n}=0$. Consider this $n$. \par We will denote the identity map $\operatorname*{id}:P\rightarrow P$ by $1$ (since it is the unity of the algebra $\operatorname*{End}P$). \par For every $i\in\left\{ 1,2,...,\ell\right\} $, let $\alpha_{i}$ denote the endomorphism $\underbrace{\alpha\otimes\alpha\otimes...\otimes\alpha }_{i-1\text{ times }\alpha}\otimes\left( \alpha-1\right) \otimes \underbrace{1\otimes1\otimes...\otimes1}_{\ell-i\text{ times }1}$ of $P^{\otimes\ell}$. Then, the endomorphisms $\alpha_{1}$, $\alpha_{2}$, $...$, $\alpha_{n}$ commute with each other (because $\alpha$, $\alpha-1$ and $1$ commute with each other). But for every $i\in\left\{ 1,2,...,\ell\right\} $, we have% \[ \alpha_{i}=\underbrace{\alpha\otimes\alpha\otimes...\otimes\alpha}_{i-1\text{ times }\alpha}\otimes\left( \alpha-1\right) \otimes\underbrace{1\otimes 1\otimes...\otimes1}_{\ell-i\text{ times }1}, \] so that% \begin{align*} \alpha_{i}^{n} & =\left( \underbrace{\alpha\otimes\alpha\otimes ...\otimes\alpha}_{i-1\text{ times }\alpha}\otimes\left( \alpha-1\right) \otimes\underbrace{1\otimes1\otimes...\otimes1}_{\ell-i\text{ times }1 }\right) ^{n}\\ & =\underbrace{\alpha^{n}\otimes\alpha^{n}\otimes...\otimes\alpha^{n}% }_{i-1\text{ times }\alpha^{n}}\otimes\underbrace{\left( \alpha-1\right) ^{n}}_{=0}\otimes\underbrace{1^{n}\otimes1^{n}\otimes...\otimes1^{n}}% _{\ell-i\text{ times }1^{n}}=0, \end{align*} so that $\alpha_{i}$ is nilpotent. \par Meanwhile, it is known that if finitely many nilpotent elements of an algebra commute with each other, then the sum of these elements must also be nilpotent. Applying this result to the nilpotent elements $\alpha_{1}$, $\alpha_{2}$, $...$, $\alpha_{\ell}$ of the algebra $\operatorname*{End}V$ (these elements commute with each other, as we know), we conclude that the sum $\sum\limits_{i=1}^{\ell}\alpha_{i}$ is nilpotent. But every $i\in\left\{ 1,2,...,\ell\right\} $ satisfies% \begin{align*} \alpha_{i} & =\underbrace{\alpha\otimes\alpha\otimes...\otimes\alpha }_{i-1\text{ times }\alpha}\otimes\left( \alpha-1\right) \otimes \underbrace{1\otimes1\otimes...\otimes1}_{\ell-i\text{ times }1}\\ & =\underbrace{\underbrace{\alpha\otimes\alpha\otimes...\otimes\alpha }_{i-1\text{ times }\alpha}\otimes\alpha}_{=\underbrace{\alpha\otimes \alpha\otimes...\otimes\alpha}_{i\text{ times }\alpha}}\otimes \underbrace{1\otimes1\otimes...\otimes1}_{\ell-i\text{ times }1 }% -\underbrace{\alpha\otimes\alpha\otimes...\otimes\alpha}_{i-1\text{ times }\alpha}\otimes\underbrace{1\otimes\underbrace{1\otimes1\otimes...\otimes 1}_{\ell-i\text{ times }1}}_{=\underbrace{1\otimes1\otimes...\otimes1}% _{\ell-i+1\text{ times }1}=\underbrace{1\otimes1\otimes...\otimes1}% _{\ell-\left( i-1\right) \text{ times }1}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the multilinearity of the tensor product}\right) \\ & =\underbrace{\alpha\otimes\alpha\otimes...\otimes\alpha}_{i\text{ times }\alpha}\otimes\underbrace{1\otimes1\otimes...\otimes1}_{\ell-i\text{ times }1}-\underbrace{\alpha\otimes\alpha\otimes...\otimes\alpha}_{i-1\text{ times }\alpha}\otimes\underbrace{1\otimes1\otimes...\otimes1}_{\ell-\left( i-1\right) \text{ times }1}. \end{align*} Hence,% \begin{align*} \sum\limits_{i=1}^{\ell}\alpha_{i} & =\sum\limits_{i=1}^{\ell}\left( \underbrace{\alpha\otimes\alpha\otimes...\otimes\alpha}_{i\text{ times }% \alpha}\otimes\underbrace{1\otimes1\otimes...\otimes1}_{\ell-i\text{ times }% 1}-\underbrace{\alpha\otimes\alpha\otimes...\otimes\alpha}_{i-1\text{ times }\alpha}\otimes\underbrace{1\otimes1\otimes...\otimes1}_{\ell-\left( i-1\right) \text{ times }1}\right) \\ & =\underbrace{\underbrace{\alpha\otimes\alpha\otimes...\otimes\alpha}% _{\ell\text{ times }\alpha}}_{=\alpha^{\otimes\ell}}\otimes \underbrace{\underbrace{1\otimes1\otimes...\otimes1}_{\ell-\ell\text{ times }1}}_{=\left( \text{empty tensor product}\right) }% -\underbrace{\underbrace{\alpha\otimes\alpha\otimes...\otimes\alpha}_{0\text{ times }\alpha}}_{=\left( \text{empty tensor product}\right) }\otimes \underbrace{\underbrace{1\otimes1\otimes...\otimes1}_{\ell-0\text{ times }1}% }_{=1^{\otimes\left( \ell-0\right) }=1^{\otimes\ell}=\operatorname*{id}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the telescope principle}\right) \\ & =\alpha^{\otimes\ell}-\operatorname*{id}. \end{align*} Since we know that $\sum\limits_{i=1}^{\ell}\alpha_{i}$ is nilpotent, this yields that $\alpha^{\otimes\ell}-\operatorname*{id}$ is nilpotent. In other words, $\alpha^{\otimes\ell}$ is unipotent. \par But let $\pi$ be the canonical projection $P^{\otimes\ell}\rightarrow \wedge^{\ell}P$. Then, we have a commutative diagram% \[% %TCIMACRO{\TeXButton{tensor vs wedge functoriality}{\xymatrixcolsep{4pc} %\xymatrix{ %P^{\otimes\ell} \ar[r]^{\alpha^{\otimes\ell}} \ar@{->>}[d]_{\pi} %& P^{\otimes\ell} \ar@{->>}[d]^{\pi} \\ %\wedge^{\ell} P \ar[r]^{\wedge^{\ell} \alpha} & \wedge^{\ell} P %}}}% %BeginExpansion \xymatrixcolsep{4pc} \xymatrix{ P^{\otimes\ell} \ar[r]^{\alpha^{\otimes\ell}} \ar@{->>}[d]_{\pi} & P^{\otimes\ell} \ar@{->>}[d]^{\pi} \\ \wedge^{\ell} P \ar[r]^{\wedge^{\ell} \alpha} & \wedge^{\ell} P }% %EndExpansion \] (since the canonical projection $P^{\otimes\ell}\rightarrow\wedge^{\ell}P$ is functorial). Thus, $\left( \wedge^{\ell}\alpha\right) \circ\pi=\pi \circ\alpha^{\otimes\ell}$. Hence, \[ \left( \wedge^{\ell}\alpha-\operatorname*{id}\right) \circ\pi =\underbrace{\left( \wedge^{\ell}\alpha\right) \circ\pi}_{=\pi\circ \alpha^{\otimes\ell}} - \underbrace{\operatorname*{id}\circ\pi}_{=\pi=\pi \circ\operatorname*{id}} = \pi\circ\alpha^{\otimes\ell} - \pi\circ \operatorname*{id} = \pi\circ\left( \alpha^{\otimes\ell}-\operatorname*{id}% \right) \] (since composition of linear maps is bilinear). \par Now, for every $m\in\mathbb{N}$, we have \begin{equation} \left( \wedge^{\ell}\alpha-\operatorname*{id}\right) ^{m}\circ\pi=\pi \circ\left( \alpha^{\otimes\ell}-\operatorname*{id}\right) ^{m}. \label{pf.finitary.rhoRho.m}% \end{equation} (\textit{Proof of (\ref{pf.finitary.rhoRho.m}):} We will prove (\ref{pf.finitary.rhoRho.m}) by induction over $m$: \par \textit{Induction base:} Comparing $\underbrace{\left( \wedge^{\ell}% \alpha-\operatorname*{id}\right) ^{0}}_{=\operatorname*{id}}\circ\pi=\pi$ and $\pi\circ\underbrace{\left( \alpha^{\otimes\ell}-\operatorname*{id}\right) ^{0}}_{=\operatorname*{id}}=\pi$, we obtain $\left( \wedge^{\ell}% \alpha-\operatorname*{id}\right) ^{0}\circ\pi=\pi\circ\left( \alpha ^{\otimes\ell}-\operatorname*{id}\right) ^{0}$. Thus, (\ref{pf.finitary.rhoRho.m}) holds for $m=0$. This completes the induction base. \par \textit{Induction step:} Let $\mu\in\mathbb{N}$. Assume that (\ref{pf.finitary.rhoRho.m}) holds for $m=\mu$. We must prove that (\ref{pf.finitary.rhoRho.m}) holds for $m=\mu+1$ as well. \par Since (\ref{pf.finitary.rhoRho.m}) holds for $m=\mu$, we have $\left( \wedge^{\ell}\alpha-\operatorname*{id}\right) ^{\mu}\circ\pi=\pi\circ\left( \alpha^{\otimes\ell}-\operatorname*{id}\right) ^{\mu}$. Now,% \begin{align*} \underbrace{\left( \wedge^{\ell}\alpha-\operatorname*{id}\right) ^{\mu+1}% }_{=\left( \wedge^{\ell}\alpha-\operatorname*{id}\right) ^{\mu}\circ\left( \wedge^{\ell}\alpha-\operatorname*{id}\right) }\circ\pi & =\left( \wedge^{\ell}\alpha-\operatorname*{id}\right) ^{\mu}\circ\underbrace{\left( \wedge^{\ell}\alpha-\operatorname*{id}\right) \circ\pi}_{=\pi\circ\left( \alpha^{\otimes\ell}-\operatorname*{id}\right) }=\underbrace{\left( \wedge^{\ell}\alpha-\operatorname*{id}\right) ^{\mu}\circ\pi}_{=\pi \circ\left( \alpha^{\otimes\ell}-\operatorname*{id}\right) ^{\mu}}% \circ\left( \alpha^{\otimes\ell}-\operatorname*{id}\right) \\ & =\pi\circ\underbrace{\left( \alpha^{\otimes\ell}-\operatorname*{id}% \right) ^{\mu}\circ\left( \alpha^{\otimes\ell}-\operatorname*{id}\right) }_{=\left( \alpha^{\otimes\ell}-\operatorname*{id}\right) ^{\mu+1}}=\pi \circ\left( \alpha^{\otimes\ell}-\operatorname*{id}\right) ^{\mu+1}. \end{align*} Thus, (\ref{pf.finitary.rhoRho.m}) holds for $m=\mu+1$. This completes the induction step. The induction proof of (\ref{pf.finitary.rhoRho.m}) is thus complete.) \par But since $\alpha^{\otimes\ell}-\operatorname*{id}$ is nilpotent, there exists some $m\in\mathbb{N}$ such that $\left( \alpha^{\otimes\ell}% -\operatorname*{id}\right) ^{m}=0$. Consider this $m$. By (\ref{pf.finitary.rhoRho.m}), we have $\left( \wedge^{\ell}\alpha -\operatorname*{id}\right) ^{m}\circ\pi=\pi\circ\underbrace{\left( \alpha^{\otimes\ell}-\operatorname*{id}\right) ^{m}}_{=0}=\pi\circ0=0$. Since $\pi$ is right-cancellable (because $\pi$ is a projection and thus surjective), this yields $\left( \wedge^{\ell}\alpha-\operatorname*{id}% \right) ^{m}=0$. Hence, $\wedge^{\ell}\alpha-\operatorname*{id}$ is nilpotent, so that $\wedge^{\ell}\alpha$ is unipotent. \par We have thus shown that $\wedge^{\ell}\alpha$ is a unipotent element of $\operatorname*{GL}\left( \wedge^{\ell}P\right) $ whenever $\alpha$ is a unipotent element of $\operatorname*{GL}\left( P\right) $. In other words, the $\ell$-th exterior power of any unipotent element of $\operatorname*{GL}% \left( P\right) $ is a unipotent element of $\operatorname*{GL}\left( \wedge^{\ell}P\right) $, qed.}. Since $\exp a$ is a unipotent element of $\operatorname*{GL}\left( P\right) $, this yields that $\wedge^{\ell}\left( \exp a\right) $ is a unipotent element of $\operatorname*{GL}\left( \wedge^{\ell}P\right) $ for every $\ell\in\mathbb{N}$. Hence, the logarithm $\log\left( \wedge^{\ell}\left( \exp a\right) \right) $ is well-defined for every $\ell\in\mathbb{N}$. \end{verlong} On the other hand, consider the map $\wedge\left( \exp a\right) :\wedge P\rightarrow\wedge P$. This map is an algebra homomorphism (because generally, if $Q$ and $R$ are two vector spaces, and $f:Q\rightarrow R$ is a linear map, then $\wedge f:\wedge Q\rightarrow\wedge R$ is an algebra homomorphism) and identical with the direct sum $\bigoplus\limits_{\ell\in\mathbb{N}}% \wedge^{\ell}\left( \exp a\right) :\bigoplus\limits_{\ell\in\mathbb{N}% }\wedge^{\ell}P\rightarrow\bigoplus\limits_{\ell\in\mathbb{N}}\wedge^{\ell}P$ of the linear maps $\wedge^{\ell}\left( \exp a\right) :\wedge^{\ell }P\rightarrow\wedge^{\ell}P$. Since $\wedge\left( \exp a\right) =\bigoplus\limits_{\ell\in\mathbb{N}% }\wedge^{\ell}\left( \exp a\right) $, we have $\log\left( \wedge\left( \exp a\right) \right) =\log\left( \bigoplus\limits_{\ell\in\mathbb{N}% }\wedge^{\ell}\left( \exp a\right) \right) =\bigoplus\limits_{\ell \in\mathbb{N}}\log\left( \wedge^{\ell}\left( \exp a\right) \right) $ (because logarithms on direct sums are componentwise).\footnote{Note that the map $\wedge\left( \exp a\right) $ needs not be unipotent, but the logarithm $\log\left( \wedge\left( \exp a\right) \right) $ nevertheless makes sense because the map $\wedge\left( \exp a\right) $ is a direct sum of unipotent maps (and thus is locally unipotent).} As a consequence, every $\ell \in\mathbb{N}$ and every $p_{1},p_{2},...,p_{\ell}\in P$ satisfy $p_{1}\wedge p_{2}\wedge...\wedge p_{\ell}\in\wedge^{\ell}P$ and thus $\left( \log\left( \wedge\left( \exp a\right) \right) \right) \left( p_{1}\wedge p_{2}% \wedge...\wedge p_{\ell}\right) =\left( \log\left( \wedge^{\ell}\left( \exp a\right) \right) \right) \left( p_{1}\wedge p_{2}\wedge...\wedge p_{\ell}\right) $. But it is well-known that if $A$ is an algebra and $f:A\rightarrow A$ is an algebra endomorphism such that $\log f$ is well-defined, then $\log f:A\rightarrow A$ is a derivation. Applied to $A=\wedge P$ and $f=\wedge \left( \exp a\right) $, this yields that $\log\left( \wedge\left( \exp a\right) \right) :\wedge P\rightarrow\wedge P$ is a derivation. But every $p\in P$ satisfies \begin{equation} \left( \log\left( \wedge\left( \exp a\right) \right) \right) \left( p\right) =a\rightharpoonup p, \label{pf.finitary.rhoRho.deg1}% \end{equation} where $p$ is viewed as an element of $\wedge^{1}P\subseteq\wedge P$.\ \ \ \ \footnote{\textit{Proof of (\ref{pf.finitary.rhoRho.deg1}):} Let $p\in P$. Since $\log\left( \wedge\left( \exp a\right) \right) =\bigoplus\limits_{\ell\in\mathbb{N}}\log\left( \wedge^{\ell}\left( \exp a\right) \right) $ and $p\in P=\wedge^{1}P$, we have \[ \left( \log\left( \wedge\left( \exp a\right) \right) \right) \left( p\right) =\left( \log\underbrace{\left( \wedge^{1}\left( \exp a\right) \right) }_{=\exp a}\right) \left( p\right) =\underbrace{\left( \log\left( \exp a\right) \right) }_{=a}\left( p\right) =ap=a\rightharpoonup p. \] This proves (\ref{pf.finitary.rhoRho.deg1}).} Now recall the Leibniz identity for derivations. In its general form, it says that if $A$ is an algebra, $M$ is an $A$-bimodule, and $d:A\rightarrow M$ is a derivation, then every $\ell\in\mathbb{N}$ and every $p_{1},p_{2},...,p_{\ell }\in A$ satisfy \[ d\left( p_{1}p_{2}...p_{\ell}\right) =\sum\limits_{k=1}^{\ell}p_{1}% p_{2}...p_{k-1}d\left( p_{k}\right) p_{k+1}p_{k+2}...p_{\ell}. \] Applying this to $A=\wedge P$, $M=\wedge P$ and $d=\log\left( \wedge\left( \exp a\right) \right) $, we conclude that every $\ell\in\mathbb{N}$ and every $p_{1},p_{2},...,p_{\ell}\in\wedge P$ satisfy% \[ \left( \log\left( \wedge\left( \exp a\right) \right) \right) \left( p_{1}p_{2}...p_{\ell}\right) =\sum\limits_{k=1}^{\ell}p_{1}p_{2}% ...p_{k-1}\left( \log\left( \wedge\left( \exp a\right) \right) \right) \left( p_{k}\right) p_{k+1}p_{k+2}...p_{\ell}% \] (since $\log\left( \wedge\left( \exp a\right) \right) :\wedge P\rightarrow\wedge P$ is a derivation). Thus, every $\ell\in\mathbb{N}$ and every $p_{1},p_{2},...,p_{\ell}\in P$ satisfy \begin{align} \left( \log\left( \wedge\left( \exp a\right) \right) \right) \left( p_{1}p_{2}...p_{\ell}\right) & =\sum\limits_{k=1}^{\ell}p_{1}% p_{2}...p_{k-1}\underbrace{\left( \log\left( \wedge\left( \exp a\right) \right) \right) \left( p_{k}\right) }_{\substack{=a\rightharpoonup p_{k}\\\text{(by (\ref{pf.finitary.rhoRho.deg1}), applied to }p=p_{k}\text{)}% }}p_{k+1}p_{k+2}...p_{\ell}\nonumber\\ & =\sum\limits_{k=1}^{\ell}\underbrace{p_{1}p_{2}...p_{k-1}\left( a\rightharpoonup p_{k}\right) p_{k+1}p_{k+2}...p_{\ell}}_{\substack{=p_{1}% \wedge p_{2}\wedge...\wedge p_{k-1}\wedge\left( a\rightharpoonup p_{k}\right) \wedge p_{k+1}\wedge p_{k+2}\wedge...\wedge p_{\ell }\\\text{(since the multiplication in }\wedge P\text{ is given by the wedge product)}}}\nonumber\\ & =\sum\limits_{k=1}^{\ell}p_{1}\wedge p_{2}\wedge...\wedge p_{k-1}% \wedge\left( a\rightharpoonup p_{k}\right) \wedge p_{k+1}\wedge p_{k+2}\wedge...\wedge p_{\ell}\nonumber\\ & =\left( \rho_{P,\ell}\left( a\right) \right) \left( p_{1}\wedge p_{2}\wedge...\wedge p_{\ell}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{def.finitary.rho})}\right) . \label{pf.finitary.rhoRho.calc}% \end{align} On the other hand, every $\ell\in\mathbb{N}$ and every $p_{1},p_{2}% ,...,p_{\ell}\in P$ satisfy \begin{align*} & \left( \log\left( \wedge\left( \exp a\right) \right) \right) \underbrace{\left( p_{1}p_{2}...p_{\ell}\right) }_{\substack{=p_{1}\wedge p_{2}\wedge...\wedge p_{\ell}\\\text{(since the multiplication in }\wedge P\text{ is given by the wedge product)}}}\\ & =\left( \log\left( \wedge\left( \exp a\right) \right) \right) \left( p_{1}\wedge p_{2}\wedge...\wedge p_{\ell}\right) =\left( \log\left( \wedge^{\ell}\left( \exp a\right) \right) \right) \left( p_{1}\wedge p_{2}\wedge...\wedge p_{\ell}\right) . \end{align*} Compared with (\ref{pf.finitary.rhoRho.calc}), this yields% \[ \left( \rho_{P,\ell}\left( a\right) \right) \left( p_{1}\wedge p_{2}\wedge...\wedge p_{\ell}\right) =\left( \log\left( \wedge^{\ell }\left( \exp a\right) \right) \right) \left( p_{1}\wedge p_{2}% \wedge...\wedge p_{\ell}\right) \] for every $\ell\in\mathbb{N}$ and every $p_{1},p_{2},...,p_{\ell}\in P$. Now fix $\ell\in\mathbb{N}$. We know that \[ \left( \rho_{P,\ell}\left( a\right) \right) \left( p_{1}\wedge p_{2}\wedge...\wedge p_{\ell}\right) =\left( \log\left( \wedge^{\ell }\left( \exp a\right) \right) \right) \left( p_{1}\wedge p_{2}% \wedge...\wedge p_{\ell}\right) \] for every $p_{1},p_{2},...,p_{\ell}\in P$. Since the vector space $\wedge^{\ell}P$ is spanned by elements of the form $p_{1}\wedge p_{2}% \wedge...\wedge p_{\ell}$ with $p_{1},p_{2},...,p_{\ell}\in P$, this yields that the two linear maps $\rho_{P,\ell}\left( a\right) $ and $\log\left( \wedge^{\ell}\left( \exp a\right) \right) $ are equal to each other on a spanning set of the vector space $\wedge^{\ell}P$. Therefore, these two maps must be identical (because if two linear maps are equal to each other on a spanning set of their domain, then they must always be identical). In other words, $\rho_{P,\ell}\left( a\right) =\log\left( \wedge^{\ell}\left( \exp a\right) \right) $. Exponentiating this equality, we obtain $\exp\left( \rho_{P,\ell}\left( a\right) \right) =\wedge^{\ell}\left( \exp a\right) $. This proves Theorem \ref{thm.finitary.rhoRho}. \subsubsection{\label{subsubsect.schur2.reduct}Reduction to fermions} We are now going to reduce Theorem \ref{thm.schur} to a ``purely fermionic'' statement -- a statement (Theorem \ref{thm.schur.fermi}) not involving the bosonic space $\mathcal{B}$ or the Boson-Fermion correspondence $\sigma$ in any way. We will later (Subsection \ref{subsubsect.skewschur}) generalize this statement, and yet later prove the generalization. First, a definition: \begin{definition} Let $\mathbf{R}$ (not to be confused with the field $\mathbb{R}$) be a commutative $\mathbb{Q}$-algebra. We denote by $\mathcal{A}_{\mathbf{R}}$ the Heisenberg algebra defined over the ground ring $\mathbf{R}$ in lieu of $\mathbb{C}$. We denote by $\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }$ the $\mathcal{A}_{\mathbf{R}}$-module $\mathcal{B}^{\left( 0\right) }$ defined over the ground ring $\mathbf{R}$ in lieu of $\mathbb{C}$. We denote by $\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }$ the $\mathcal{A}_{\mathbf{R}}% $-module $\mathcal{F}^{\left( 0\right) }$ defined over the ground ring $\mathbf{R}$ in lieu of $\mathbb{C}$. We denote by $\sigma_{\mathbf{R}}$ the map $\sigma$ defined over the ground ring $\mathbf{R}$ in lieu of $\mathbb{C}% $. (This $\sigma_{\mathbf{R}}$ is thus a graded $\mathcal{A}_{\mathbf{R}}% $-module homomorphism $\mathcal{B}_{\mathbf{R}}\rightarrow\mathcal{F}% _{\mathbf{R}}$, where $\mathcal{B}_{\mathbf{R}}$ and $\mathcal{F}_{\mathbf{R}% }$ are the $\mathcal{A}_{\mathbf{R}}$-modules $\mathcal{B}$ and $\mathcal{F}$ defined over the ground ring $\mathbf{R}$ in lieu of $\mathbb{C}$.) \end{definition} Next, some preparations: \begin{proposition} \label{prop.schur.fermi.welldef}Let $\mathbf{R}$ be a commutative $\mathbb{Q}% $-algebra. Let $y_{1},y_{2},y_{3},...$ be some elements of $\mathbf{R}$. \textbf{(a)} Let $M$ be a $\mathbb{Z}$-graded $\mathcal{A}_{\mathbf{R}}% $-module concentrated in nonpositive degrees (i. e., satisfying $M\left[ n\right] =0$ for all positive integers $n$). The map $\exp\left( y_{1}% a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) :M\rightarrow M$ is well-defined, in the following sense: For every $m\in M$, expanding the expression $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) m$ yields an infinite sum with only finitely many nonzero addends. \textbf{(b)} Let $M$ and $N$ be two $\mathbb{Z}$-graded $\mathcal{A}% _{\mathbf{R}}$-modules concentrated in nonpositive degrees. Let $\eta :M\rightarrow N$ be an $\mathcal{A}_{\mathbf{R}}$-module homomorphism. Then,% \[ \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ\eta=\eta\circ\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) \right) \] as maps from $M$ to $N$. \textbf{(c)} Consider the $\mathbb{Z}$-graded $\mathcal{A}_{\mathbf{R}}% $-module $\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }$. This $\mathbb{Z}% $-graded $\mathcal{A}_{\mathbf{R}}$-module $\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }$ is concentrated in nonpositive degrees. Hence, by Theorem \ref{thm.schur.fermi}, the map $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) :\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }% \rightarrow\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }$ is well-defined. Thus, $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) $ is well-defined for every $0$-degression $\left( i_{0},i_{1},i_{2},...\right) $. \end{proposition} \textit{Proof of Proposition \ref{prop.schur.fermi.welldef}.} \textbf{(a)} Let $m\in M$. We will prove that expanding the expression $\exp\left( y_{1}% a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) m$ yields an infinite sum with only finitely many nonzero terms. Since $M$ is $\mathbb{Z}$-graded, we can write $m$ in the form $m=\sum \limits_{n\in\mathbb{Z}}m_{n}$ for a family $\left( m_{n}\right) _{n\in\mathbb{Z}}$ of elements of $M$ which satisfy $\left( m_{n}\in M\left[ n\right] \text{ for every }n\in\mathbb{Z}\right) $ and $\left( m_{n}=0\text{ for all but finitely many }n\in\mathbb{Z}\right) $. Consider this family $\left( m_{n}\right) _{n\in\mathbb{Z}}$. We know that $m_{n}=0$ for all but finitely many $n\in\mathbb{Z}$. In other words, there exists a finite subset $I$ of $\mathbb{Z}$ such that every $n\in\mathbb{Z}\setminus I$ satisfies $m_{n}=0$. Consider this $I$. Let $s$ be an integer which is smaller than every element of $I$. (Such an $s$ exists since $I$ is finite.) Then,% \begin{equation} fm=0\ \ \ \ \ \ \ \ \ \ \text{for every integer }q\geq-s\text{ and every }f\in U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) \left[ q\right] \label{pf.schur.fermi.welldef.fm=0}% \end{equation} (where $U_{\mathbf{R}}$ means ``enveloping algebra over the ground ring $\mathbf{R}$ '').\ \ \ \ \footnote{\textit{Proof of (\ref{pf.schur.fermi.welldef.fm=0}):} Let $q\geq-s$ be an integer, and let $f\in U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) \left[ q\right] $. Since $s$ is smaller than every element of $I$, we have $s -n$ for every $n\in I$, so that $q+n>0$ for every $n\in I$ and thus $M\left[ q+n\right] =0$ for every $n\in I$ (since $M$ is concentrated in nonpositive degrees). \par Notice that $M$ is a graded $\mathcal{A}_{\mathbf{R}}$-module, thus a graded $U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) $-module. But% \[ m=\sum\limits_{n\in\mathbb{Z}}m_{n}=\sum\limits_{n\in I}m_{n}+\sum \limits_{n\in\mathbb{Z}\setminus I}\underbrace{m_{n}}% _{\substack{=0\\\text{(since }n\in\mathbb{Z}\setminus I\text{)}}% }=\sum\limits_{n\in I}m_{n}+\underbrace{\sum\limits_{n\in\mathbb{Z}\setminus I}0}_{=0}=\sum\limits_{n\in I}m_{n}, \] so that% \[ fm=f\sum\limits_{n\in I}m_{n}=\sum\limits_{n\in I}\underbrace{fm_{n}% }_{\substack{\in M\left[ q+n\right] \\\text{(since }f\in U_{\mathbf{R}% }\left( \mathcal{A}_{\mathbf{R}}\right) \left[ q\right] \text{ and }% m_{n}\in M\left[ n\right] \text{,}\\\text{and since }M\text{ is a graded }U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) \text{-module)}}% }\in\sum\limits_{n\in I}\underbrace{M\left[ q+n\right] }% _{\substack{=0\\\text{(since }n\in I\text{)}}}=\sum\limits_{n\in I}0=0, \] so that $fm=0$, qed.} Expanding the expression $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) m$, we obtain% \begin{align*} & \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) m\\ & =\sum\limits_{i=0}^{\infty}\dfrac{1}{i!}\left( \underbrace{y_{1}% a_{1}+y_{2}a_{2}+y_{3}a_{3}+...}_{=\sum\limits_{j\in\left\{ 1,2,3,...\right\} }y_{j}a_{j}}\right) ^{i}m=\sum\limits_{i=0}^{\infty}% \dfrac{1}{i!}\underbrace{\left( \sum\limits_{j\in\left\{ 1,2,3,...\right\} }y_{j}a_{j}\right) ^{i}}_{=\sum\limits_{\left( j_{1},j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}}y_{j_{1}}y_{j_{2}}...y_{j_{i}}a_{j_{1}% }a_{j_{2}}...a_{j_{i}}}m\\ & =\sum\limits_{i=0}^{\infty}\dfrac{1}{i!}\sum\limits_{\left( j_{1}% ,j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}}y_{j_{1}}% y_{j_{2}}...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}m\\ & =\sum\limits_{\substack{i\in\mathbb{N};\\\left( j_{1},j_{2},...,j_{i}% \right) \in\left\{ 1,2,3,...\right\} ^{i}}}\dfrac{1}{i!}y_{j_{1}}y_{j_{2}% }...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}m. \end{align*} But this infinite sum has only finitely many nonzero addends\footnote{\textit{Proof.} Let $i\in\mathbb{N}$ and $\left( j_{1}% ,j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}$ be such that $\dfrac{1}{i!}y_{j_{1}}y_{j_{2}}...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}% }m\neq0$. Since $a_{j_{k}}\in U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}% }\right) \left[ j_{k}\right] $ for every $k\in\left\{ 1,2,...,i\right\} $, we have \begin{align*} a_{j_{1}}a_{j_{2}}...a_{j_{i}} & \in\left( U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) \left[ j_{1}\right] \right) \left( U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) \left[ j_{2}\right] \right) ...\left( U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) \left[ j_{i}\right] \right) \\ & \subseteq U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) \left[ j_{1}+j_{2}+...+j_{i}\right] , \end{align*} so that% \[ \dfrac{1}{i!}y_{j_{1}}y_{j_{2}}...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}% \in\dfrac{1}{i!}y_{j_{1}}y_{j_{2}}...y_{j_{i}}U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) \left[ j_{1}+j_{2}+...+j_{i}\right] \subseteq U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) \left[ j_{1}+j_{2}+...+j_{i}\right] . \] Hence, if $j_{1}+j_{2}+...+j_{i}\geq-s$, then $\dfrac{1}{i!}y_{j_{1}}y_{j_{2}% }...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}m=0$ (by (\ref{pf.schur.fermi.welldef.fm=0}), applied to $f=\dfrac{1}{i!}y_{j_{1}% }y_{j_{2}}...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}$ and $q=j_{1}% +j_{2}+...+j_{i}$), contradicting $\dfrac{1}{i!}y_{j_{1}}y_{j_{2}}...y_{j_{i}% }a_{j_{1}}a_{j_{2}}...a_{j_{i}}m\neq0$. As a consequence, we cannot have $j_{1}+j_{2}+...+j_{i}\geq-s$. We must thus have $j_{1}+j_{2}+...+j_{i}<-s$. \par Now forget that we fixed $i$ and $\left( j_{1},j_{2},...,j_{i}\right) $. We thus have shown that every $i\in\mathbb{N}$ and $\left( j_{1},j_{2}% ,...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}$ such that $\dfrac {1}{i!}y_{j_{1}}y_{j_{2}}...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}m\neq0$ must satisfy $j_{1}+j_{2}+...+j_{i}<-s$. Since there are only finitely many pairs $\left( i,\left( j_{1},j_{2},...,j_{i}\right) \right) $ of $i\in\mathbb{N}$ and $\left( j_{1},j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}$ satisfying $j_{1}+j_{2}+...+j_{i}<-s$, this yields that there are only finitely many pairs $\left( i,\left( j_{1}% ,j_{2},...,j_{i}\right) \right) $ of $i\in\mathbb{N}$ and $\left( j_{1},j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}$ satisfying $\dfrac{1}{i!}y_{j_{1}}y_{j_{2}}...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}% }m\neq0$. In other words, the infinite sum $\sum\limits_{\substack{i\in \mathbb{N};\\\left( j_{1},j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}}}\dfrac{1}{i!}y_{j_{1}}y_{j_{2}}...y_{j_{i}}a_{j_{1}% }a_{j_{2}}...a_{j_{i}}m$ has only finitely many nonzero addends, qed.}. Thus, we have shown that for every $m\in M$, expanding the expression $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) m$ yields an infinite sum with only finitely many nonzero addends. This proves Proposition \ref{prop.schur.fermi.welldef} \textbf{(a)}. \begin{verlong} \textbf{(b)} Just as in the proof of Proposition \ref{prop.schur.fermi.welldef} \textbf{(a)} above, we can show that% \begin{equation} \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) m=\sum \limits_{\substack{i\in\mathbb{N};\\\left( j_{1},j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}}}\dfrac{1}{i!}y_{j_{1}}y_{j_{2}% }...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}m\ \ \ \ \ \ \ \ \ \ \text{for every }m\in M, \label{pf.schur.fermi.welldef.b.1}% \end{equation} with the infinite sum $\sum\limits_{\substack{i\in\mathbb{N};\\\left( j_{1},j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}}}\dfrac {1}{i!}y_{j_{1}}y_{j_{2}}...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}m$ having only finitely many nonzero addends (for every fixed $m$). Similarly,% \begin{equation} \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) n=\sum \limits_{\substack{i\in\mathbb{N};\\\left( j_{1},j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}}}\dfrac{1}{i!}y_{j_{1}}y_{j_{2}% }...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}n\ \ \ \ \ \ \ \ \ \ \text{for every }n\in N. \label{pf.schur.fermi.welldef.b.2}% \end{equation} Now, let $m\in M$. Since $\eta$ is an $\mathcal{A}_{\mathbf{R}}$-module homomorphism, $\eta$ must also be an $U_{\mathbf{R}}\left( \mathcal{A}% _{\mathbf{R}}\right) $-module homomorphism (since every $\mathcal{A}% _{\mathbf{R}}$-module homomorphism is an $U_{\mathbf{R}}\left( \mathcal{A}% _{\mathbf{R}}\right) $-module homomorphism). Thus, \begin{equation} \eta\left( gm\right) =g\cdot\eta\left( m\right) \ \ \ \ \ \ \ \ \ \ \text{for every }g\in U_{\mathbf{R}}\left( \mathcal{A}% _{\mathbf{R}}\right) . \label{pf.schur.fermi.welldef.b.3}% \end{equation} Since $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) $ is not (in general) an element of $U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) $, we cannot directly apply this to $g=\exp\left( y_{1}a_{1}+y_{2}a_{2}% +y_{3}a_{3}+...\right) $. However, we have% \begin{align*} & \left( \eta\circ\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) \right) \right) \left( m\right) \\ & =\eta\left( \underbrace{\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) m}_{=\sum\limits_{\substack{i\in\mathbb{N};\\\left( j_{1},j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}}}\dfrac {1}{i!}y_{j_{1}}y_{j_{2}}...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}m}\right) \\ & =\eta\left( \sum\limits_{\substack{i\in\mathbb{N};\\\left( j_{1}% ,j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}}}\dfrac{1}% {i!}y_{j_{1}}y_{j_{2}}...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}m\right) \\ & =\sum\limits_{\substack{i\in\mathbb{N};\\\left( j_{1},j_{2},...,j_{i}% \right) \in\left\{ 1,2,3,...\right\} ^{i}}}\dfrac{1}{i!}y_{j_{1}}y_{j_{2}% }...y_{j_{i}}\underbrace{\eta\left( a_{j_{1}}a_{j_{2}}...a_{j_{i}}m\right) }_{\substack{=a_{j_{1}}a_{j_{2}}...a_{j_{i}}\eta\left( m\right) \\\text{(by (\ref{pf.schur.fermi.welldef.b.3}), applied to }g=a_{j_{1}}a_{j_{2}% }...a_{j_{i}}\text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\eta\text{ is }\mathbf{R}\text{-linear, while the infinite sum }\sum\limits_{\substack{i\in\mathbb{N};\\\left( j_{1},j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}}}\dfrac{1}{i!}y_{j_{1}}y_{j_{2}% }...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}m\\ \text{has only finitely many nonzero addends}% \end{array} \right) \\ & =\sum\limits_{\substack{i\in\mathbb{N};\\\left( j_{1},j_{2},...,j_{i}% \right) \in\left\{ 1,2,3,...\right\} ^{i}}}\dfrac{1}{i!}y_{j_{1}}y_{j_{2}% }...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}\eta\left( m\right) \\ & =\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \left( \eta\left( m\right) \right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since (\ref{pf.schur.fermi.welldef.b.2}) (applied to }n=\eta\left( m\right) \text{) yields}\\ \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \left( \eta\left( m\right) \right) =\sum\limits_{\substack{i\in\mathbb{N};\\\left( j_{1},j_{2},...,j_{i}\right) \in\left\{ 1,2,3,...\right\} ^{i}}}\dfrac {1}{i!}y_{j_{1}}y_{j_{2}}...y_{j_{i}}a_{j_{1}}a_{j_{2}}...a_{j_{i}}\eta\left( m\right) \end{array} \right) \\ & =\left( \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ\eta\right) \left( m\right) . \end{align*} Now forget that we fixed $m$. We thus have proven that every $m\in M$ satisfies% \[ \left( \eta\circ\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) \right) \right) \left( m\right) =\left( \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ \eta\right) \left( m\right) . \] In other words, $\eta\circ\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) \right) =\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}% +y_{3}a_{3}+...\right) \right) \circ\eta$. This proves Proposition \ref{prop.schur.fermi.welldef} \textbf{(b)}. \end{verlong} \begin{vershort} \textbf{(b)} In order to prove Proposition \ref{prop.schur.fermi.welldef} \textbf{(b)}, we must clearly show that% \begin{equation} \eta\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) m\right) =\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\eta\left( m\right) \label{pf.schur.fermi.welldef.b.short.1}% \end{equation} for every $m\in M$. Fix $m\in M$. Since $\eta$ is an $\mathcal{A}_{\mathbf{R}}$-module homomorphism, $\eta$ must also be an $U_{\mathbf{R}}\left( \mathcal{A}% _{\mathbf{R}}\right) $-module homomorphism (since every $\mathcal{A}% _{\mathbf{R}}$-module homomorphism is an $U_{\mathbf{R}}\left( \mathcal{A}% _{\mathbf{R}}\right) $-module homomorphism). Thus, \[ \eta\left( gm\right) =g\cdot\eta\left( m\right) \ \ \ \ \ \ \ \ \ \ \text{for every }g\in U_{\mathbf{R}}\left( \mathcal{A}% _{\mathbf{R}}\right) . \] If $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) $ was an element of $U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) $, then we could apply this to $g=\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) $ and conclude (\ref{pf.schur.fermi.welldef.b.short.1}) immediately. Unfortunately, $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) $ is not an element of $U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) $, but this problem is easy to amend: By Proposition \ref{prop.schur.fermi.welldef} \textbf{(a)}, we can find a finite partial sum $g$ of the expanded power series $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) $ satisfying% \begin{align*} \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) m & =gm\ \ \ \ \ \ \ \ \ \ \text{and}\\ \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\eta\left( m\right) & =g\cdot\eta\left( m\right) . \end{align*} Consider such a $g$. Since $g$ is only a finite partial sum, we have $g\in U_{\mathbf{R}}\left( \mathcal{A}_{\mathbf{R}}\right) $, and thus $\eta\left( gm\right) =g\cdot\eta\left( m\right) $. Hence,% \begin{align*} \eta\left( \underbrace{\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}% +...\right) m}_{=gm}\right) & =\eta\left( gm\right) =g\cdot\eta\left( m\right) \\ & =\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\eta\left( m\right) , \end{align*} so that (\ref{pf.schur.fermi.welldef.b.short.1}) is proven. Thus, Proposition \ref{prop.schur.fermi.welldef} \textbf{(b)} is proven. \end{vershort} \textbf{(c)} This is obvious. Let us make a remark which we will only use in the ``finitary'' version of our proof of Theorem \ref{thm.schur.fermi}. First, a definition: \begin{definition} For every commutative ring $\mathbf{R}$, let $\mathcal{A}_{+\mathbf{R}}$ be the Lie algebra $\mathcal{A}_{+}$ defined for the ground ring $\mathbf{R}$ instead of $\mathbb{C}$. \end{definition} Now, it is easy to see that Proposition \ref{prop.schur.fermi.welldef} holds with $\mathcal{A}_{\mathbf{R}}$ replaced by $\mathcal{A}_{+\mathbf{R}}$. We will only use the analogues of parts \textbf{(a)} and \textbf{(b)}: \begin{proposition} \label{prop.schur.fermi.welldef.A+}Let $\mathbf{R}$ be a commutative $\mathbb{Q}$-algebra. Let $y_{1},y_{2},y_{3},...$ be some elements of $\mathbf{R}$. \textbf{(a)} Let $M$ be a $\mathbb{Z}$-graded $\mathcal{A}_{+\mathbf{R}}% $-module concentrated in nonpositive degrees (i. e., satisfying $M\left[ n\right] =0$ for all positive integers $n$). The map $\exp\left( y_{1}% a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) :M\rightarrow M$ is well-defined, in the following sense: For every $m\in M$, expanding the expression $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) m$ yields an infinite sum with only finitely many nonzero addends. \textbf{(b)} Let $M$ and $N$ be two $\mathbb{Z}$-graded $\mathcal{A}% _{+\mathbf{R}}$-modules concentrated in nonpositive degrees. Let $\eta:M\rightarrow N$ be an $\mathcal{A}_{+\mathbf{R}}$-module homomorphism. Then,% \[ \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ\eta=\eta\circ\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) \right) \] as maps from $M$ to $N$. \end{proposition} \textit{Proof of Proposition \ref{prop.schur.fermi.welldef.A+}.} In order to obtain proofs of Proposition \ref{prop.schur.fermi.welldef.A+}, it is enough to simply replace $\mathcal{A}_{\mathbf{R}}$ by $\mathcal{A}_{+\mathbf{R}}$ throughout the proof of parts \textbf{(a)} and \textbf{(b)} of Proposition \ref{prop.schur.fermi.welldef}. Now, let us state the ``fermionic'' version of Theorem \ref{thm.schur}: \begin{theorem} \label{thm.schur.fermi}Let $\mathbf{R}$ be a commutative $\mathbb{Q}$-algebra. Let $y_{1},y_{2},y_{3},...$ be some elements of $\mathbf{R}$. Denote by $y$ the family $\left( y_{1},y_{2},y_{3},...\right) $. Let $\left( i_{0}% ,i_{1},i_{2},...\right) $ be a $0$-degression. The $\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) $-coordinate of $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) $ (this is a well-defined element of $\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }$ due to Proposition \ref{prop.schur.fermi.welldef} \textbf{(c)}) with respect to the basis\footnotemark\ $\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}% \wedge...\right) _{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression}}$ of $\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }$ equals $S_{\left( i_{k}+k\right) _{k\geq0}}\left( y\right) $. (Here, we are using the fact that $\left( i_{k}+k\right) _{k\geq0}$ is a partition for every $0$-degression $\left( i_{0},i_{1},i_{2},...\right) $. This follows from Proposition \ref{prop.glinf.wedge.grading}, applied to $m=0$.) \end{theorem} \footnotetext{Here, ``basis'' means ``$\mathbf{R}$-module basis'', not ``$\mathbb{C}$-vector space basis''.}Let us see how this yields Theorem \ref{thm.schur}: \textit{Proof of Theorem \ref{thm.schur} using Theorem \ref{thm.schur.fermi}.} Fix a $0$-degression $\left( i_{0},i_{1},i_{2},...\right) $; then, $i_{0}>i_{1}>i_{2}>...$ and $v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\in\mathcal{F}^{\left( 0\right) }$. Let $\lambda$ be the partition $\left( i_{0}+0,i_{1}+1,i_{2}+2,...\right) $. Denote the element $\sigma^{-1}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \in\mathcal{B}^{\left( 0\right) }$ by $P\left( x\right) $. We need to show that $P\left( x\right) =S_{\lambda}\left( x\right) $. From now on, we let $y$ denote another countable family of indeterminates $\left( y_{1},y_{2},y_{3},...\right) $ (rather than a finite family like the $\left( y_{1},y_{2},...,y_{N}\right) $ of Definition \ref{def.schur.y}). Thus, whenever $Q$ is a polynomial in countably many indeterminates, $Q\left( y\right) $ will mean $Q\left( y_{1},y_{2},y_{3},...\right) $. Let $\mathbf{R}$ be the polynomial ring $\mathbb{C}\left[ y_{1},y_{2}% ,y_{3},...\right] $. Then, $y$ is a family of elements of $\mathbf{R}$. By the definition of $\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }$, we have $\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }=\mathbf{R}\left[ x_{1}% ,x_{2},x_{3},...\right] $ as a vector space, so that $\mathcal{B}% _{\mathbf{R}}^{\left( 0\right) }=\left( \mathbb{C}\left[ y_{1},y_{2}% ,y_{3},...\right] \right) \left[ x_{1},x_{2},x_{3},...\right] $ as a vector space. Let us denote by $1\in\mathcal{B}^{\left( 0\right) }$ the unity of the algebra $\mathbb{C}\left[ x_{1},x_{2},x_{3},...\right] $. Clearly, $\mathcal{B}^{\left( 0\right) }\subseteq\mathcal{B}_{\mathbf{R}% }^{\left( 0\right) }$, and thus $1\in\mathcal{B}^{\left( 0\right) }\subseteq\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }$. We still let $x$ denote the whole collection of variables $\left( x_{1}% ,x_{2},x_{3},...\right) $. Also, let $x+y$ denote the family $\left( x_{1}+y_{1},x_{2}+y_{2},x_{3}+y_{3},...\right) $ of elements of $\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }$. Recall the $\mathbb{C}$-bilinear form $\left( \cdot,\cdot\right) :F\times F\rightarrow\mathbb{C}$ defined in Proposition \ref{prop.A.contravariantform}. Since $F=\widetilde{F}=\mathcal{B}^{\left( 0\right) }$ (as vector spaces), this form $\left( \cdot,\cdot\right) $ is a $\mathbb{C}$-bilinear form $\mathcal{B}^{\left( 0\right) }\times\mathcal{B}^{\left( 0\right) }\rightarrow\mathbb{C}$. Since the definition of the form did not depend of the ground ring, we can analogously define an $\mathbf{R}$-bilinear form $\left( \cdot,\cdot\right) :\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }\times\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }\rightarrow\mathbf{R}$. The restriction of this latter $\mathbf{R}$-bilinear form $\left( \cdot ,\cdot\right) :\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }\times \mathcal{B}_{\mathbf{R}}^{\left( 0\right) }\rightarrow\mathbf{R}$ to $\mathcal{B}^{\left( 0\right) }\times\mathcal{B}^{\left( 0\right) }$ is clearly the former $\mathbb{C}$-bilinear form $\left( \cdot,\cdot\right) :\mathcal{B}^{\left( 0\right) }\times\mathcal{B}^{\left( 0\right) }\rightarrow\mathbb{C}$; therefore we will use the same notation for these two forms. In the following, elements of $\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }=\mathbf{R}\left[ x_{1},x_{2},x_{3},...\right] $ will be considered as polynomials in the variables $x_{1},x_{2},x_{3},...$ over the ring $\mathbf{R}$, and not as polynomials in the variables $x_{1},x_{2}% ,x_{3},...,y_{1},y_{2},y_{3},...$ over the field $\mathbb{C}$. Hence, for an $R\in\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }$, the notation $R\left( 0,0,0,...\right) $ will mean the result of substituting $0$ for the variables $x_{1},x_{2},x_{3},...$ in $R$ (but the variables $y_{1},y_{2},y_{3},...$ will stay unchanged!). We will abbreviate $R\left( 0,0,0,...\right) $ by $R\left( 0\right) $. Every polynomial $R\in\mathcal{B}^{\left( 0\right) }$ satisfies:% \begin{equation} R\left( 0\right) =\left( \begin{array} [c]{c}% \text{the }\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) \text{-coordinate of }\sigma\left( R\right) \\ \text{with respect to the basis }\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\right) _{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression}}\text{ of }\mathcal{F}^{\left( 0\right) }% \end{array} \right) \label{pf.schur.1}% \end{equation} \footnote{\textit{Proof of (\ref{pf.schur.1}).} Let $R\in\mathcal{B}^{\left( 0\right) }$. Thus, $R\in\mathcal{B}^{\left( 0\right) }=\widetilde{F}$. \par Let $p_{0,\mathcal{B}}$ be the canonical projection of the graded space $\mathcal{B}^{\left( 0\right) }$ onto its $0$-th homogeneous component $\mathcal{B}^{\left( 0\right) }\left[ 0\right] =\mathbb{C}\cdot1$, and let $p_{0,\mathcal{F}}$ be the canonical projection of the graded space $\mathcal{F}^{\left( 0\right) }$ onto its $0$-th homogeneous component $\mathcal{F}^{\left( 0\right) }\left[ 0\right] =\mathbb{C}\psi_{0}$. Since $\sigma_{0}:\mathcal{B}^{\left( 0\right) }\rightarrow\mathcal{F}^{\left( 0\right) }$ is a graded homomorphism, $\sigma_{0}$ commutes with the projections on the $0$-th graded components; in other words, $\sigma_{0}\circ p_{0,\mathcal{B}}=p_{0,\mathcal{F}}\circ\sigma_{0}$. Now, we know that $p_{0,\mathcal{B}}\left( R\right) =R\left( 0\right) \cdot1$ (since $\mathcal{B}=\widetilde{F}=\mathbb{C}\left[ x_{1},x_{2},x_{3},...\right] $), and thus $\left( \sigma_{0}\circ p_{0,\mathcal{B}}\right) \left( R\right) =\sigma_{0}\left( \underbrace{p_{0,\mathcal{B}}\left( R\right) }_{=R\left( 1\right) \cdot1}\right) =\sigma_{0}\left( R\left( 0\right) \cdot1\right) =R\left( 0\right) \cdot\underbrace{\sigma_{0}\left( 1\right) }_{=\psi_{0}% }=R\left( 0\right) \psi_{0}$. \par On the other hand, let $\kappa$ denote the $\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) $-coordinate of $\sigma\left( R\right) $ with respect to the basis $\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}% \wedge...\right) _{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression}}$ of $\mathcal{F}^{\left( 0\right) }$. Then, the projection of $\sigma\left( R\right) $ onto the $0$-th graded component $\mathcal{F}^{\left( 0\right) }\left[ 0\right] $ of $\mathcal{F}^{\left( 0\right) }$ is $\kappa\cdot v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...$ (because the basis $\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}% \wedge...\right) _{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression}}$ of $\mathcal{F}^{\left( 0\right) }$ is a graded basis, and the $0$-th graded component $\mathcal{F}^{\left( 0\right) }\left[ 0\right] $ of $\mathcal{F}^{\left( 0\right) }$ is spanned by $\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) $). In other words, $p_{0,\mathcal{F}}\left( \sigma\left( R\right) \right) =\kappa \cdot\underbrace{v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...}_{=\psi_{0}}% =\kappa\psi_{0}$. Hence,% \[ R\left( 0\right) \psi_{0}=\underbrace{\left( \sigma_{0}\circ p_{0,\mathcal{B}}\right) }_{=p_{0,\mathcal{F}}\circ\sigma_{0}}\left( R\right) =\left( p_{0,\mathcal{F}}\circ\sigma_{0}\right) \left( R\right) =p_{0,\mathcal{F}}\left( \sigma\left( R\right) \right) =\kappa\psi_{0}. \] Thus, $\left( R\left( 0\right) -\kappa\right) \psi_{0}% =\underbrace{R\left( 0\right) \psi_{0}}_{=\kappa\psi_{0}}-\kappa\psi _{0}=\kappa\psi_{0}-\kappa\psi_{0}=0$. \par But $\psi_{0}$ is an element of a basis of $\mathcal{F}^{\left( 0\right) }$ (namely, of the basis $\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}% \wedge...\right) _{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression}}$). Thus, every scalar $\mu\in\mathbb{C}$ satisfying $\mu\psi_{0}=0$ must satisfy $\mu=0$. Applying this to $\mu=R\left( 0\right) -\kappa$, we obtain $R\left( 0\right) -\kappa=0$ (since $\left( R\left( 0\right) -\kappa\right) \psi_{0}=0$). Thus,% \[ R\left( 0\right) =\kappa=\left( \begin{array} [c]{c}% \text{the }\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) \text{-coordinate of }\sigma\left( R\right) \\ \text{with respect to the basis }\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\right) _{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression}}\text{ of }\mathcal{F}^{\left( 0\right) }% \end{array} \right) . \] This proves (\ref{pf.schur.1}).}. Since the proof of (\ref{pf.schur.1}) clearly does not depend on the ground ring, an analogous result holds over the ring $\mathbf{R}$: Every polynomial $R\in\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }$ satisfies% \begin{equation} R\left( 0\right) =\left( \begin{array} [c]{c}% \text{the }\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) \text{-coordinate of }\sigma_{\mathbf{R}}\left( R\right) \\ \text{with respect to the basis }\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\right) _{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression}}\text{ of }\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }% \end{array} \right) \label{pf.schur.1R}% \end{equation} \footnote{Of course, ``basis'' means ``$\mathbf{R}$-module basis'' and no longer ``$\mathbb{C}$-vector space basis'' in this statement.}. On the other hand, for every polynomial $R\in\mathcal{B}^{\left( 0\right) }% $, we can view $R=R\left( x\right) $ as an element of $\mathcal{B}% _{\mathbf{R}}^{\left( 0\right) }$ (since $\mathcal{B}^{\left( 0\right) }\subseteq\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }$), and this way we obtain% \begin{align} & \left( 1,\exp\left( \underbrace{y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}% +...}_{=\sum\limits_{s>0}y_{s}a_{s}}\right) R\left( x\right) \right) =\left( 1,\exp\left( \sum\limits_{s>0}y_{s}a_{s}\right) R\left( x\right) \right) \nonumber\\ & =\left( 1,\exp\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial x_{s}}\right) R\left( x\right) \right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }a_{s}\text{ acts as }\dfrac{\partial}{\partial x_{s}}\text{ on }\mathcal{B}^{\left( 0\right) }\text{ for every }s\geq1\right) \nonumber\\ & =\left( 1,R\left( x+y\right) \right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\exp\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial x_{s}}\right) R\left( x\right) =R\left( x+y\right) \\ \text{by Lemma \ref{lem.hirota.newton} (applied to }R\text{, }\left( x_{1},x_{2},x_{3},...\right) \text{ and }\mathbf{R}\\ \text{ instead of }P\text{, }\left( z_{1},z_{2},z_{3},...\right) \text{ and }K\text{)}% \end{array} \right) \nonumber\\ & =\left( R\left( x+y\right) \right) \left( 0\right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{because the analogue of Proposition \ref{prop.A.contravariantform} \textbf{(b)} for}\\ \text{the ground ring }\mathbf{R}\text{ yields }\left( 1,Q\right) =Q\left( 0\right) \text{ for every }Q\in\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }% \end{array} \right) \nonumber\\ & =R\left( y\right) \label{pf.schur.usingfermi.1}% \end{align} in $\mathbf{R}$. Recall that the map $\sigma_{\mathbf{R}}$ is defined analogously to $\sigma$ but for the ground ring $\mathbf{R}$ instead of $\mathbb{C}$. Thus, $\sigma_{\mathbf{R}}\left( Q\right) =\sigma\left( Q\right) $ for every $Q\in\mathcal{B}^{\left( 0\right) }$. Applied to $Q=P\left( x\right) $, this yields $\sigma_{\mathbf{R}}\left( P\left( x\right) \right) =\sigma\left( P\left( x\right) \right) =v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...$ (since $P\left( x\right) =\sigma^{-1}\left( v_{i_{0}% }\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) $). On the other hand, since $\sigma_{\mathbf{R}}:\mathcal{B}_{\mathbf{R}% }^{\left( 0\right) }\rightarrow\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }$ is an $\mathcal{A}_{\mathbf{R}}$-module homomorphism, and since $\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }$ and $\mathcal{F}_{\mathbf{R}% }^{\left( 0\right) }$ are two $\mathcal{A}_{\mathbf{R}}$-modules concentrated in nonpositive degrees, we can apply Proposition \ref{prop.schur.fermi.welldef} \textbf{(b)} to $\sigma_{\mathbf{R}}$, $\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }$ and $\mathcal{F}_{\mathbf{R}% }^{\left( 0\right) }$ instead of $\eta$, $M$ and $N$. As a result, we obtain% \[ \sigma_{\mathbf{R}}\circ\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) \right) =\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}% +y_{3}a_{3}+...\right) \right) \circ\sigma_{\mathbf{R}}% \] as maps from $\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }$ to $\mathcal{F}% _{\mathbf{R}}^{\left( 0\right) }$. This easily yields% \begin{align} \sigma_{\mathbf{R}}\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) P\left( x\right) \right) & =\underbrace{\left( \sigma_{\mathbf{R}}\circ\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) \right) \right) }_{=\left( \exp\left( y_{1}a_{1}% +y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ\sigma_{\mathbf{R}}}\left( P\left( x\right) \right) \nonumber\\ & =\left( \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ\sigma_{\mathbf{R}}\right) \left( P\left( x\right) \right) \nonumber\\ & =\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot \underbrace{\sigma_{\mathbf{R}}\left( P\left( x\right) \right) }_{=v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...}\nonumber\\ & =\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) . \label{pf.schur.usingfermi.4}% \end{align} But (\ref{pf.schur.usingfermi.1}) (applied to $R=P$) yields% \[ \left( 1,\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) P\left( x\right) \right) =P\left( y\right) , \] so that% \begin{align*} & P\left( y\right) \\ & =\left( 1,\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) P\left( x\right) \right) \\ & =\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) P\left( x\right) \right) \left( 0\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{because the analogue of Proposition \ref{prop.A.contravariantform} \textbf{(b)} for}\\ \text{the ground ring }\mathbf{R}\text{ yields }\left( 1,Q\right) =Q\left( 0\right) \text{ for every }Q\in\mathcal{B}_{\mathbf{R}}^{\left( 0\right) }% \end{array} \right) \\ & =\left( \begin{array} [c]{c}% \text{the }\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) \text{-coordinate of }\sigma_{\mathbf{R}}\left( \exp\left( y_{1}a_{1}% +y_{2}a_{2}+y_{3}a_{3}+...\right) P\left( x\right) \right) \\ \text{with respect to the basis }\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\right) _{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression}}\text{ of }\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }% \end{array} \right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.schur.1R}), applied to }R=\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) P\left( x\right) \right) \\ & =\left( \begin{array} [c]{c}% \text{the }\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) \text{-coordinate of}\\ \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\ \text{with respect to the basis }\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\right) _{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression}}\text{ of }\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }% \end{array} \right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.schur.usingfermi.4})}\right) \\ & =S_{\left( i_{k}+k\right) _{k\geq0}}\left( y\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by Theorem \ref{thm.schur.fermi}}\right) \\ & =S_{\lambda}\left( y\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{k}+k\right) _{k\geq0}=\left( i_{0}+0,i_{1}+1,i_{2}+2,...\right) =\lambda\right) . \end{align*} Substituting $x_{i}$ for $y_{i}$ in this equation, we obtain $P\left( x\right) =S_{\lambda}\left( x\right) $ (since both $P$ and $S_{\lambda}$ are polynomials in $\mathbb{C}\left[ x_{1},x_{2},x_{3},...\right] $). Thus,% \[ S_{\lambda}\left( x\right) =P\left( x\right) =\sigma^{-1}\left( v_{i_{0}% }\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) . \] This proves Theorem \ref{thm.schur}. \subsubsection{\label{subsubsect.skewschur}Skew Schur polynomials} Rather than prove Theorem \ref{thm.schur.fermi} directly, let us formulate and verify a stronger statement which will be in no way harder to prove. First, we need a definition: \begin{definition} \label{def.skewschur}Let $\lambda$ and $\mu$ be two partitions. \textbf{(a)} We write $\mu\subseteq\lambda$ if every $i\in\left\{ 1,2,3,...\right\} $ satisfies $\lambda_{i}\geq\mu_{i}$, where the partitions $\lambda$ and $\mu$ have been written in the forms $\lambda=\left( \lambda_{1},\lambda_{2},\lambda_{3},...\right) $ and $\mu=\left( \mu_{1}% ,\mu_{2},\mu_{3},...\right) $. \textbf{(b)} We define a polynomial $S_{\lambda\diagup\mu}\left( x\right) \in\mathbb{Q}\left[ x_{1},x_{2},x_{3},...\right] $ as follows: Write $\lambda$ and $\mu$ in the forms $\lambda=\left( \lambda_{1},\lambda _{2},...,\lambda_{m}\right) $ and $\mu=\left( \mu_{1},\mu_{2},...,\mu _{m}\right) $ for some $m\in\mathbb{N}$. Then, let $S_{\lambda\diagup\mu }\left( x\right) $ be the polynomial% \begin{align*} & \det\left( \begin{array} [c]{ccccc}% S_{\lambda_{1}-\mu_{1}}\left( x\right) & S_{\lambda_{1}-\mu_{2}+1}\left( x\right) & S_{\lambda_{1}-\mu_{3}+2}\left( x\right) & ... & S_{\lambda _{1}-\mu_{m}+m-1}\left( x\right) \\ S_{\lambda_{2}-\mu_{1}-1}\left( x\right) & S_{\lambda_{2}-\mu_{2}}\left( x\right) & S_{\lambda_{2}-\mu_{3}+1}\left( x\right) & ... & S_{\lambda _{2}-\mu_{m}+m-2}\left( x\right) \\ S_{\lambda_{3}-\mu_{1}-2}\left( x\right) & S_{\lambda_{3}-\mu_{2}-1}\left( x\right) & S_{\lambda_{3}-\mu_{3}}\left( x\right) & ... & S_{\lambda _{3}-\mu_{m}+m-3}\left( x\right) \\ ... & ... & ... & ... & ...\\ S_{\lambda_{m}-\mu_{1}-m+1}\left( x\right) & S_{\lambda_{m}-\mu_{2}% -m+2}\left( x\right) & S_{\lambda_{m}-\mu_{3}-m+3}\left( x\right) & ... & S_{\lambda_{m}-\mu_{m}}\left( x\right) \end{array} \right) \\ & =\det\left( \left( S_{\lambda_{i}-\mu_{j}+j-i}\left( x\right) \right) _{1\leq i\leq m,\ 1\leq j\leq m}\right) , \end{align*} where $S_{j}$ denotes $0$ if $j<0$. (Note that this does not depend on the choice of $m$ (that is, increasing $m$ at the cost of padding the partitions $\lambda$ and $\mu$ with trailing zeroes does not change the value of $\det\left( \left( S_{\lambda_{i}-\mu_{j}+j-i}\left( x\right) \right) _{1\leq i\leq m,\ 1\leq j\leq m}\right) $). This is because any nonnegative integers $m$ and $\ell$, any $m\times m$-matrix $A$, any $m\times\ell$-matrix $B$ and any upper unitriangular $\ell\times\ell$-matrix $C$ satisfy $\det\left( \begin{array} [c]{cc}% A & B\\ 0 & C \end{array} \right) =\det A$.) We refer to $S_{\lambda\diagup\mu}\left( x\right) $ as the \textit{bosonic Schur polynomial corresponding to the skew partition }$\lambda\diagup\mu$. \end{definition} Before we formulate the strengthening of Theorem \ref{thm.schur.fermi}, three remarks: \begin{remark} \label{rmk.skewschur.empty}Let $\varnothing$ denote the partition $\left( 0,0,0,...\right) $. For every partition $\lambda$, we have $\varnothing \subseteq\lambda$ and $S_{\lambda\diagup\varnothing}\left( x\right) =S_{\lambda}\left( x\right) $. \end{remark} \begin{remark} \label{rmk.skewschur.0}Let $\lambda$ and $\mu$ be two partitions. Then, $S_{\lambda\diagup\mu}\left( x\right) =0$ unless $\mu\subseteq\lambda$. \end{remark} \begin{remark} \label{rmk.skewschur.infdet}Recall that in Definition \ref{def.infdet} \textbf{(c)}, we defined the notion of an ``upper almost-unitriangular'' $\mathbb{N}\times\mathbb{N}$-matrix. In the same way, we can define the notion of an ``upper almost-unitriangular'' $\left\{ 1,2,3,...\right\} \times\left\{ 1,2,3,...\right\} $-matrix. In Definition \ref{def.infdet} \textbf{(e)}, we defined the determinant of an upper almost-unitriangular $\mathbb{N}\times\mathbb{N}$-matrix. Analogously, we can define the determinant of an upper almost-unitriangular $\left\{ 1,2,3,...\right\} \times\left\{ 1,2,3,...\right\} $-matrix. Let $\lambda=\left( \lambda_{1},\lambda_{2},\lambda_{3},...\right) $ and $\mu=\left( \mu_{1},\mu_{2},\mu_{3},...\right) $ be two partitions. Then, the $\left\{ 1,2,3,...\right\} \times\left\{ 1,2,3,...\right\} $-matrix $\left( S_{\lambda_{i}-\mu_{j}+j-i}\left( x\right) \right) _{\left( i,j\right) \in\left\{ 1,2,3,...\right\} ^{2}}$ is upper almost-unitriangular, and we have% \begin{align} S_{\lambda\diagup\mu}\left( x\right) & =\det\left( \left( S_{\lambda _{i}-\mu_{j}+j-i}\left( x\right) \right) _{\left( i,j\right) \in\left\{ 1,2,3,...\right\} ^{2}}\right) \label{rmk.skewschur.infdet.1}\\ & =\det\left( \begin{array} [c]{cccc}% S_{\lambda_{1}-\mu_{1}}\left( x\right) & S_{\lambda_{1}-\mu_{2}+1}\left( x\right) & S_{\lambda_{1}-\mu_{3}+2}\left( x\right) & ...\\ S_{\lambda_{2}-\mu_{1}-1}\left( x\right) & S_{\lambda_{2}-\mu_{2}}\left( x\right) & S_{\lambda_{2}-\mu_{3}+1}\left( x\right) & ...\\ S_{\lambda_{3}-\mu_{1}-2}\left( x\right) & S_{\lambda_{3}-\mu_{2}-1}\left( x\right) & S_{\lambda_{3}-\mu_{3}}\left( x\right) & ...\\ ... & ... & ... & ... \end{array} \right) .\nonumber \end{align} \end{remark} All of the above three remarks follow easily from Definition \ref{def.skewschur}. Now, let us finally give the promised strengthening of Theorem \ref{thm.schur.fermi}: \begin{theorem} \label{thm.schur.fermi.skew}Let $\mathbf{R}$ be a commutative $\mathbb{Q}% $-algebra. Let $y_{1},y_{2},y_{3},...$ be some elements of $\mathbf{R}$. Denote by $y$ the family $\left( y_{1},y_{2},y_{3},...\right) $. Let $\left( i_{0},i_{1},i_{2},...\right) $ be a $0$-degression. Recall that $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) $ is a well-defined element of $\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }$ due to Proposition \ref{prop.schur.fermi.welldef} \textbf{(c)}. Recall also that $\left( j_{k}+k\right) _{k\geq0}$ is a partition for every $0$-degression $\left( j_{0},j_{1},j_{2},...\right) $ (this follows from Proposition \ref{prop.glinf.wedge.grading}, applied to $0$ and $\left( j_{0},j_{1}% ,j_{2},...\right) $ instead of $m$ and $\left( i_{0},i_{1},i_{2},...\right) $). In particular, $\left( i_{k}+k\right) _{k\geq0}$ is a partition. We have% \begin{align} & \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \nonumber\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge.... \label{thm.schur.fermi.skew.eq}% \end{align} (Note that the sum on the right hand side of (\ref{thm.schur.fermi.skew.eq}) is a finite sum, since only finitely many $0$-degressions $\left( j_{0}% ,j_{1},j_{2},...\right) $ satisfy $\left( j_{k}+k\right) _{k\geq0}% \subseteq\left( i_{k}+k\right) _{k\geq0}$.) \end{theorem} Before we prove this, let us see how this yields Theorem \ref{thm.schur.fermi}: \textit{Proof of Theorem \ref{thm.schur.fermi} using Theorem \ref{thm.schur.fermi.skew}.} Remark \ref{rmk.skewschur.empty} (applied to $\lambda=\left( i_{k}+k\right) _{k\geq0}$) yields $\varnothing \subseteq\left( i_{k}+k\right) _{k\geq0}$ and $S_{\left( i_{k}+k\right) _{k\geq0}\diagup\varnothing}\left( x\right) =S_{\left( i_{k}+k\right) _{k\geq0}}\left( x\right) $. By substituting $y$ for $x$ in the equality $S_{\left( i_{k}+k\right) _{k\geq0}\diagup\varnothing}\left( x\right) =S_{\left( i_{k}+k\right) _{k\geq0}}\left( x\right) $, we conclude $S_{\left( i_{k}+k\right) _{k\geq0}\diagup\varnothing}\left( y\right) =S_{\left( i_{k}+k\right) _{k\geq0}}\left( y\right) $. Theorem \ref{thm.schur.fermi.skew} yields that (\ref{thm.schur.fermi.skew.eq}) holds. On the other hand, every $0$-degression $\left( j_{0},j_{1},j_{2},...\right) $ satisfying $\left( j_{k}+k\right) _{k\geq0}\not \subseteq \left( i_{k}+k\right) _{k\geq0}$ must satisfy% \begin{equation} S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}% }\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...=0 \label{pf.schur.fermi.notcontained}% \end{equation} \ \ \ \ \footnote{\textit{Proof.} Let $\left( j_{0},j_{1},j_{2},...\right) $ be a $0$-degression satisfying $\left( j_{k}+k\right) _{k\geq0}% \not \subseteq \left( i_{k}+k\right) _{k\geq0}$. We know that $\left( i_{k}+k\right) _{k\geq0}$ and $\left( j_{k}+k\right) _{k\geq0}$ are partitions. Thus, Remark \ref{rmk.skewschur.0} (applied to $\lambda=\left( i_{k}+k\right) _{k\geq0}$ and $\mu=\left( j_{k}+k\right) _{k\geq0}$) yields that $S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( x\right) =0$ unless $\left( j_{k}+k\right) _{k\geq 0}\subseteq\left( i_{k}+k\right) _{k\geq0}$. Since we don't have $\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}$ (because by assumption, we have $\left( j_{k}+k\right) _{k\geq0}\not \subseteq \left( i_{k}+k\right) _{k\geq0}$), we thus know that $S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( x\right) =0$. Substituting $y$ for $x$ in this equation, we obtain $S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}% }\left( y\right) =0$, so that $S_{\left( i_{k}+k\right) _{k\geq0}% \diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}% }\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...=0$, qed.}. Hence, each of the addends of the infinite sum $\sum\limits_{\substack{\left( j_{0},j_{1}% ,j_{2},...\right) \text{ a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\not \subseteq \left( i_{k}+k\right) _{k\geq0}}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...$ equals $0$. Thus, the infinite sum $\sum\limits_{\substack{\left( j_{0},j_{1}% ,j_{2},...\right) \text{ a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\not \subseteq \left( i_{k}+k\right) _{k\geq0}}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...$ is well-defined and equals $0$. We thus have% \begin{equation} 0=\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\not \subseteq \left( i_{k}+k\right) _{k\geq0}}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge.... \label{pf.schur.fermi.onlysubdiagrams}% \end{equation} Adding this equality to (\ref{thm.schur.fermi.skew.eq}), we obtain% \begin{align*} & \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left( j_{0},j_{1}% ,j_{2},...\right) \text{ a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\not \subseteq \left( i_{k}+k\right) _{k\geq0}}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\\ & =\sum\limits_{\left( j_{0},j_{1},j_{2},...\right) \text{ a }% 0\text{-degression}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge.... \end{align*} Hence, the $\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) $-coordinate of $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) $ with respect to the basis $\left( v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}% \wedge...\right) _{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression}}$ of $\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }$ equals \begin{align*} S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( -k+k\right) _{k\geq0}% }\left( y\right) & =S_{\left( i_{k}+k\right) _{k\geq0}\diagup \varnothing}\left( y\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( -k+k\right) _{k\geq0}=\left( 0\right) _{k\geq0}=\left( 0,0,0,...\right) =\varnothing\right) \\ & =S_{\left( i_{k}+k\right) _{k\geq0}}\left( y\right) . \end{align*} This proves Theorem \ref{thm.schur.fermi} using Theorem \ref{thm.schur.fermi.skew}. \subsubsection{\label{subsubsect.schur2}Proof of Theorem \ref{thm.schur.fermi.skew} using \texorpdfstring{$\operatorname*{U}\left( \infty\right) $}{U-infinity}} One final easy lemma: \begin{lemma} \label{lem.schur.fermi.skew.toeplitzmatrix}For every $n\in\mathbb{Z}$, let $c_{n}$ be an element of $\mathbb{C}$. Assume that $c_{n}=0$ for every negative $n\in\mathbb{Z}$. Consider the shift operator $T:V\rightarrow V$ of Definition \ref{def.shiftoperator}. Then, $\sum\limits_{k\geq0}c_{k}% T^{k}=\left( c_{j-i}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}$. \end{lemma} \begin{vershort} The proof of this lemma is immediate from the definition of $T$. \end{vershort} \begin{verlong} \textit{Proof of Lemma \ref{lem.schur.fermi.skew.toeplitzmatrix}.} By Definition \ref{def.shiftoperator}, we have $Tv_{i+1}=v_{i}$ for every $i\in\mathbb{Z}$. Substituting $i-1$ for $i$ in this equality, we obtain: $Tv_{i}=v_{i-1}$ for every $i\in\mathbb{Z}$. Using this, we can readily find that% \begin{equation} T^{k}v_{i}=v_{i-k}\ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N}\text{ and }i\in\mathbb{Z}. \label{pf.schur.fermi.skew.toeplitzmatrix.1}% \end{equation} \footnote{\textit{Proof of (\ref{pf.schur.fermi.skew.toeplitzmatrix.1}):} We will prove (\ref{pf.schur.fermi.skew.toeplitzmatrix.1}) by induction over $k$. \par \textit{Induction base:} For $k=0$, we have $\underbrace{T^{k}}_{=T^{0}% =\operatorname*{id}}v_{i}=\operatorname*{id}v_{i}=v_{i}=v_{i-0}=v_{i-k}$ (since $0=k$) for every $i\in\mathbb{Z}$. In other words, (\ref{pf.schur.fermi.skew.toeplitzmatrix.1}) is true for $k=0$. This completes the induction base. \par \textit{Induction step:} Let $\ell\in\mathbb{N}$. Assume that (\ref{pf.schur.fermi.skew.toeplitzmatrix.1}) holds for $k=\ell$. We must prove that (\ref{pf.schur.fermi.skew.toeplitzmatrix.1}) also holds for $k=\ell+1$. \par Let $i\in\mathbb{Z}$. Since (\ref{pf.schur.fermi.skew.toeplitzmatrix.1}) holds for $k=\ell$, we can apply (\ref{pf.schur.fermi.skew.toeplitzmatrix.1}) to $\ell$ and $i-1$ instead of $k$ and $i$, and obtain $T^{\ell}v_{i-1}% =v_{i-1-\ell}$. Now, $T^{\ell+1}=T^{\ell}T$, so that $T^{\ell+1}v_{i}=T^{\ell }\underbrace{Tv_{i}}_{=v_{i-1}}=T^{\ell}v_{i-1}=v_{i-1-\ell}=v_{i-\left( \ell+1\right) }$. Now, forget that we fixed $i$. We thus have proven that $T^{\ell+1}v_{i}=v_{i-\left( \ell+1\right) }$ for every $i\in\mathbb{Z}$. Thus, (\ref{pf.schur.fermi.skew.toeplitzmatrix.1}) holds for $k=\ell+1$. This completes the induction step. The induction proof of (\ref{pf.schur.fermi.skew.toeplitzmatrix.1}) is thus finished.} Hence, every $k\in\mathbb{N}$ and $i\in\mathbb{Z}$ satisfy% \begin{align*} & \left( \text{the }i\text{-th column of the matrix }\left( \delta _{v-u,k}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}\right) \\ & =\sum\limits_{u\in\mathbb{Z}}\delta_{i-u,k}v_{u}=\sum \limits_{\substack{u\in\mathbb{Z};\\u\neq i-k}}\underbrace{\delta_{i-u,k}% }_{\substack{=0\\\text{(since }i-u\neq k\\\text{(since }u\neq i-k\text{))}% }}v_{u}+\underbrace{\delta_{i-\left( i-k\right) ,k}}% _{\substack{=1\\\text{(since }i-\left( i-k\right) =k\text{)}}}v_{i-k}\\ & =\underbrace{\sum\limits_{\substack{u\in\mathbb{Z};\\u\neq i-k}}0v_{u}% }_{=0}+v_{i-k}=v_{i-k}=T^{k}v_{i}\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.schur.fermi.skew.toeplitzmatrix.1})}\right) \\ & =\left( \text{the }i\text{-th column of the matrix }T^{k}\right) . \end{align*} Thus, every $k\in\mathbb{N}$ satisfies $\left( \delta_{v-u,k}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}=T^{k}$. Hence, $\sum\limits_{k\geq 0}c_{k}\underbrace{\left( \delta_{v-u,k}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}}_{=T^{k}}=\sum\limits_{k\geq0}c_{k}T^{k}$, so that% \[ \sum\limits_{k\geq0}c_{k}T^{k}=\sum\limits_{k\geq0}c_{k}\left( \delta _{v-u,k}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}=\left( \sum\limits_{k\geq0}c_{k}\delta_{v-u,k}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}. \] But recall that $c_{n}=0$ for every negative $n\in\mathbb{Z}$. In other words, $c_{k}=0$ for every negative $k\in\mathbb{Z}$. Hence, $c_{k}\delta_{v-u,k}=0$ for every negative $k\in\mathbb{Z}$ and every $\left( u,v\right) \in\mathbb{Z}^{2}$. Thus, the sum $\sum\limits_{k<0}c_{k}\delta_{v-u,k}$ is well-defined and equals $0$ for every $\left( u,v\right) \in\mathbb{Z}^{2}$. Hence, in $\overline{\mathfrak{a}_{\infty}}$, we have $\left( \sum \limits_{k<0}c_{k}\delta_{v-u,k}\right) _{\left( u,v\right) \in \mathbb{Z}^{2}}=\left( 0\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}=0$. Now,% \begin{align*} \sum\limits_{k\geq0}c_{k}T^{k} & =\underbrace{\sum\limits_{k\geq0}c_{k}% T^{k}}_{=\left( \sum\limits_{k\geq0}c_{k}\delta_{v-u,k}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}}+\underbrace{0}_{=\left( \sum\limits_{k<0}% c_{k}\delta_{v-u,k}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}}\\ & =\left( \sum\limits_{k\geq0}c_{k}\delta_{v-u,k}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}+\left( \sum\limits_{k<0}c_{k}\delta _{v-u,k}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}=\left( \sum\limits_{k\geq0}c_{k}\delta_{v-u,k}+\sum\limits_{k<0}c_{k}\delta _{v-u,k}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}. \end{align*} But since every $\left( u,v\right) \in\mathbb{Z}^{2}$ satisfies% \begin{align*} \sum\limits_{k\geq0}c_{k}\delta_{v-u,k}+\sum\limits_{k<0}c_{k}\delta_{v-u,k} & =\sum\limits_{k\in\mathbb{Z}}c_{k}\delta_{v-u,k}=\sum \limits_{\substack{k\in\mathbb{Z};\\k\neq v-u}}c_{k}\underbrace{\delta _{v-u,k}}_{\substack{=0\\\text{(since }v-u\neq k\text{)}}}+c_{v-u}% \underbrace{\delta_{v-u,v-u}}_{=1}\\ & =\underbrace{\sum\limits_{\substack{k\in\mathbb{Z};\\k\neq v-u}}c_{k}% 0}_{=0}+c_{v-u}=c_{v-u}, \end{align*} this rewrites as $\sum\limits_{k\geq0}c_{k}T^{k}=\left( \underbrace{\sum \limits_{k\geq0}c_{k}\delta_{v-u,k}+\sum\limits_{k<0}c_{k}\delta_{v-u,k}% }_{=c_{v-u}}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}=\left( c_{v-u}\right) _{\left( u,v\right) \in\mathbb{Z}^{2}}=\left( c_{j-i}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}$ (here, we renamed $\left( u,v\right) $ as $\left( i,j\right) $). This proves Lemma \ref{lem.schur.fermi.skew.toeplitzmatrix}. \end{verlong} We now give a proof of Theorem \ref{thm.schur.fermi.skew} using the actions $\rho:\mathfrak{u}_{\infty}\rightarrow\operatorname*{End}\left( \wedge^{\dfrac{\infty}{2},m}V\right) $ and $\varrho:\operatorname*{U}\left( \infty\right) \rightarrow\operatorname*{GL}\left( \wedge^{\dfrac{\infty}% {2},m}V\right) $ introduced in Subsection \ref{subsubsect.Uinf} and their properties. \textit{First proof of Theorem \ref{thm.schur.fermi.skew}.} In order to simplify notation, we assume that $\mathbf{R}=\mathbb{C}$. (All the arguments that we will make in the following are independent of the ground ring, as long as the ground ring is a commutative $\mathbb{Q}$-algebra. Therefore, we are actually allowed to assume that $\mathbf{R}=\mathbb{C}$.) Since we assumed that $\mathbf{R}=\mathbb{C}$, we have $\mathcal{A}_{\mathbf{R}}=\mathcal{A}$ and $\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }=\mathcal{F}^{\left( 0\right) }$. Now consider the shift operator $T:V\rightarrow V$ of Definition \ref{def.shiftoperator}. As a matrix in $\overline{\mathfrak{a}_{\infty}}$, this $T$ is the matrix which has $1$'s on the diagonal right above the main one, and $0$'s everywhere else. The embedding $\mathcal{A}\rightarrow \mathfrak{a}_{\infty}$ that we are using to define the action of $\mathcal{A}$ on $\mathcal{F}^{\left( 0\right) }$ sends $a_{j}$ to $T^{j}$ for every $j\in\mathbb{Z}$. Thus, every positive integer $j$ satisfies% \begin{align*} a_{j}\mid_{\mathcal{F}^{\left( 0\right) }} & =T^{j}\mid_{\mathcal{F}% ^{\left( 0\right) }}=\widehat{\rho}\left( T^{j}\right) =\rho\left( T^{j}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by Remark \ref{rmk.Uinf.rhorhohat}, applied to }m=0\text{ and }a=T^{j}\text{ (since }T^{j}\in\mathfrak{u}_{\infty }\cap\overline{\mathfrak{a}_{\infty}}\text{)}\right) . \end{align*} Since $y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...=\sum\limits_{j\geq1}y_{j}a_{j}$, we have% \begin{align*} \left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \mid_{\mathcal{F}% ^{\left( 0\right) }} & =\left( \sum\limits_{j\geq1}y_{j}a_{j}\right) \mid_{\mathcal{F}^{\left( 0\right) }}=\sum\limits_{j\geq1}y_{j}% \underbrace{\left( a_{j}\mid_{\mathcal{F}^{\left( 0\right) }}\right) }_{=\rho\left( T^{j}\right) }=\sum\limits_{j\geq1}y_{j}\rho\left( T^{j}\right) \\ & =\rho\left( \sum\limits_{j\geq1}y_{j}T^{j}\right) . \end{align*} Here, we have used the fact that $\sum\limits_{j\geq1}y_{j}T^{j}% \in\mathfrak{u}_{\infty}$ (this ensures that $\rho\left( \sum\limits_{j\geq 1}y_{j}T^{j}\right) $ is well-defined). On the other hand, substituting $y$ for $x$ in (\ref{def.schur.sk.genfun}), we obtain% \[ \sum\limits_{k\geq0}S_{k}\left( y\right) z^{k}=\exp\left( \sum \limits_{i\geq1}y_{i}z^{i}\right) \ \ \ \ \ \ \ \ \ \ \text{in }% \mathbb{C}\left[ \left[ z\right] \right] . \] Substituting $T$ for $z$ in this equality, we obtain $\sum\limits_{k\geq 0}S_{k}\left( y\right) T^{k}=\exp\left( \sum\limits_{i\geq1}y_{i}% T^{i}\right) $. Thus,% \begin{equation} \exp\left( \sum\limits_{j\geq1}y_{j}T^{j}\right) =\exp\left( \sum \limits_{i\geq1}y_{i}T^{i}\right) =\sum\limits_{k\geq0}S_{k}\left( y\right) T^{k}=\left( S_{j-i}\left( y\right) \right) _{\left( i,j\right) \in\mathbb{Z}^{2}} \label{pf.schur.fermi.skew.1}% \end{equation} (by Lemma \ref{lem.schur.fermi.skew.toeplitzmatrix}, applied to $c_{n}% =S_{n}\left( y\right) $ (since $S_{n}\left( y\right) =0$ for every negative $n\in\mathbb{Z}$)). Now,% \begin{align*} & \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \mid_{\mathcal{F}^{\left( 0\right) }}\\ & =\exp\underbrace{\left( \left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}% +...\right) \mid_{\mathcal{F}^{\left( 0\right) }}\right) }_{=\rho\left( \sum\limits_{j\geq1}y_{j}T^{j}\right) }\\ & =\exp\left( \rho\left( \sum\limits_{j\geq1}y_{j}T^{j}\right) \right) =\varrho\left( \exp\left( \sum\limits_{j\geq1}y_{j}T^{j}\right) \right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since Theorem \ref{thm.Uinf.rhoRho} (applied to }a=\sum\limits_{j\geq 1}y_{j}T^{j}\text{) yields}\\ \varrho\left( \exp\left( \sum\limits_{j\geq1}y_{j}T^{j}\right) \right) =\exp\left( \rho\left( \sum\limits_{j\geq1}y_{j}T^{j}\right) \right) \end{array} \right) \\ & =\varrho\left( \left( S_{j-i}\left( y\right) \right) _{\left( i,j\right) \in\mathbb{Z}^{2}}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.schur.fermi.skew.1})}\right) . \end{align*} Denote the matrix $\left( S_{j-i}\left( y\right) \right) _{\left( i,j\right) \in\mathbb{Z}^{2}}\in\operatorname*{U}\left( \infty\right) $ by $A$. Thus, we have% \[ \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \mid_{\mathcal{F}^{\left( 0\right) }}=\varrho\left( \underbrace{\left( S_{j-i}\left( y\right) \right) _{\left( i,j\right) \in\mathbb{Z}^{2}}% }_{=A}\right) =\varrho\left( A\right) . \] Hence,% \begin{align} & \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \nonumber\\ & =\left( \varrho\left( A\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =\sum\limits_{\left( j_{0}% ,j_{1},j_{2},...\right) \text{ is a }0\text{-degression}}\det\left( \left( A_{j_{0},j_{1},j_{2},...}^{i_{0},i_{1},i_{2},...}\right) ^{T}\right) v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge ...\label{pf.schur.fermi.skew.3}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by Remark \ref{rmk.Uinf.det}, applied to }m=0\right) .\nonumber \end{align} But a close look at the matrix $\left( A_{j_{0},j_{1},j_{2},...}^{i_{0}% ,i_{1},i_{2},...}\right) ^{T}$ proves that% \begin{equation} \det\left( \left( A_{j_{0},j_{1},j_{2},...}^{i_{0},i_{1},i_{2},...}\right) ^{T}\right) =S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \ \ \ \ \ \ \ \ \ \ \text{for every }0\text{-degression }\left( j_{0},j_{1},j_{2},...\right) \label{pf.schur.fermi.skew.2}% \end{equation} \footnote{\textit{Proof of (\ref{pf.schur.fermi.skew.2}):} Let $\left( j_{0},j_{1},j_{2},...\right) $ be a $0$-degression. Since $A=\left( S_{j-i}\left( y\right) \right) _{\left( i,j\right) \in\mathbb{Z}^{2}}$, we have $A_{j_{0},j_{1},j_{2},...}^{i_{0},i_{1},i_{2},...}=\left( S_{i_{v}-j_{u}}\left( y\right) \right) _{\left( u,v\right) \in \mathbb{N}^{2}}$, so that $\left( A_{j_{0},j_{1},j_{2},...}^{i_{0}% ,i_{1},i_{2},...}\right) ^{T}=\left( S_{i_{v}-j_{u}}\left( y\right) \right) _{\left( v,u\right) \in\mathbb{N}^{2}}$. But define two partitions $\lambda$ and $\mu$ by $\lambda=\left( i_{k}+k\right) _{k\geq0}$ and $\mu=\left( j_{k}+k\right) _{k\geq0}$. Write the partitions $\lambda$ and $\mu$ in the forms $\lambda=\left( \lambda_{1},\lambda_{2},\lambda _{3},...\right) $ and $\mu=\left( \mu_{1},\mu_{2},\mu_{3},...\right) $. Then, $\lambda_{v}=i_{v-1}+\left( v-1\right) $ for every $v\in\left\{ 1,2,3,...\right\} $, and $\mu_{u}=j_{u-1}+\left( u-1\right) $ for every $u\in\left\{ 1,2,3,...\right\} $. Thus, for every $\left( u,v\right) \in\left\{ 1,2,3,...\right\} ^{2}$, we have% \begin{equation} \underbrace{\lambda_{v}}_{=i_{v-1}+\left( v-1\right) }-\underbrace{\mu_{u}% }_{=j_{u-1}+\left( u-1\right) }+u-v=\left( i_{v-1}+\left( v-1\right) \right) -\left( j_{u-1}+\left( u-1\right) \right) +u-v=i_{v-1}-j_{u-1}. \label{pf.schur.fermi.skew.2.pf.1}% \end{equation} But (\ref{rmk.skewschur.infdet.1}) yields $S_{\lambda\diagup\mu}\left( x\right) =\det\left( \left( S_{\lambda_{i}-\mu_{j}+j-i}\left( x\right) \right) _{\left( i,j\right) \in\left\{ 1,2,3,...\right\} ^{2}}\right) $. Substituting $y$ for $x$ in this equality, we obtain% \begin{align*} S_{\lambda\diagup\mu}\left( y\right) & =\det\left( \left( S_{\lambda _{i}-\mu_{j}+j-i}\left( y\right) \right) _{\left( i,j\right) \in\left\{ 1,2,3,...\right\} ^{2}}\right) =\det\left( \left( S_{\lambda_{v}-\mu _{u}+u-v}\left( y\right) \right) _{\left( v,u\right) \in\left\{ 1,2,3,...\right\} ^{2}}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }\left( v,u\right) \text{ for }\left( i,j\right) \right) \\ & =\det\left( \left( S_{i_{v-1}-j_{u-1}}\left( y\right) \right) _{\left( v,u\right) \in\left\{ 1,2,3,...\right\} ^{2}}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.schur.fermi.skew.2.pf.1}% )}\right) \\ & =\det\left( \underbrace{\left( S_{i_{v}-j_{u}}\left( y\right) \right) _{\left( v,u\right) \in\mathbb{N}^{2}}}_{=\left( A_{j_{0},j_{1},j_{2}% ,...}^{i_{0},i_{1},i_{2},...}\right) ^{T}}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }\left( v,u\right) \text{ for }\left( v-1,u-1\right) \right) \\ & =\det\left( \left( A_{j_{0},j_{1},j_{2},...}^{i_{0},i_{1},i_{2}% ,...}\right) ^{T}\right) . \end{align*} Since $\lambda=\left( i_{k}+k\right) _{k\geq0}$ and $\mu=\left( j_{k}+k\right) _{k\geq0}$, this rewrites as $S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) =\det\left( \left( A_{j_{0},j_{1},j_{2},...}^{i_{0},i_{1},i_{2},...}\right) ^{T}\right) $. This proves (\ref{pf.schur.fermi.skew.2}).}. Now, (\ref{pf.schur.fermi.skew.3}) becomes% \begin{align*} & \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \cdot\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\ & =\sum\limits_{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression}}\underbrace{\det\left( \left( A_{j_{0},j_{1},j_{2}% ,...}^{i_{0},i_{1},i_{2},...}\right) ^{T}\right) }_{\substack{=S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \\\text{(by (\ref{pf.schur.fermi.skew.2}))}}}v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\\ & =\sum\limits_{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge...\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\substack{\left( j_{0}% ,j_{1},j_{2},...\right) \text{ a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\not \subseteq \left( i_{k}+k\right) _{k\geq0}% }}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq 0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge ...}_{\substack{=0\\\text{(by (\ref{pf.schur.fermi.onlysubdiagrams}))}}}\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge.... \end{align*} This proves Theorem \ref{thm.schur.fermi.skew}. We can now combine the above to obtain a proof of Theorem \ref{thm.schur}: \textit{Second proof of Theorem \ref{thm.schur}.} We have proven Theorem \ref{thm.schur.fermi} using Theorem \ref{thm.schur.fermi.skew}. Since we know that Theorem \ref{thm.schur.fermi.skew} holds, this yields that Theorem \ref{thm.schur.fermi} holds. This, in turn, entails that Theorem \ref{thm.schur} holds (since we have proven Theorem \ref{thm.schur} using Theorem \ref{thm.schur.fermi}). \subsubsection{\label{subsubsect.schur2.finitary}``Finitary'' proof of Theorem \ref{thm.schur.fermi.skew}} The above second proof of Theorem \ref{thm.schur} had the drawback of requiring a slew of new notions (those of $\mathfrak{u}_{\infty}$, of $\operatorname*{U}\left( \infty\right) $, of the determinant of an almost upper-triangular matrix etc.) and of their properties (Proposition \ref{prop.uinf.Vhatwedge.welldef}, Remark \ref{rmk.Uinf.det}, Theorem \ref{thm.Uinf.rhoRho} and others). We will now give a proof of Theorem \ref{thm.schur} which is more or less equivalent to the second proof of Theorem \ref{thm.schur} shown above, but avoiding these new notions. It will eschew using infinite matrices other than those in $\overline{\mathfrak{a}% _{\infty}}$, and instead work with finite objects most of the time. Since we already know how to derive Theorem \ref{thm.schur} from Theorem \ref{thm.schur.fermi.skew}, we only need to verify Theorem \ref{thm.schur.fermi.skew}. Let us first introduce some finite-dimensional subspaces of the vector space $V$: \begin{definition} \label{def.finitary.Valphabeta}Let $\alpha$ and $\beta$ be integers such that $\alpha-1\leq\beta$. \textbf{(a)} Then, $V_{\left] \alpha,\beta\right] }$ will denote the vector subspace of $V$ spanned by the vectors $v_{\alpha+1}$, $v_{\alpha+2}$, $...$, $v_{\beta}$. It is clear that $\left( v_{\alpha+1},v_{\alpha+2},...,v_{\beta }\right) $ is a basis of this vector space $V_{\left] \alpha,\beta\right] }$, so that $\dim\left( V_{\left] \alpha,\beta\right] }\right) =\beta-\alpha$. \textbf{(b)} Let $T_{\left] \alpha,\beta\right] }$ be the endomorphism of the vector space $V_{\left] \alpha,\beta\right] }$ defined by% \[ \left( T_{\left] \alpha,\beta\right] }\left( v_{i}\right) =\left\{ \begin{array} [c]{l}% v_{i-1},\ \ \ \ \ \ \ \ \ \ \text{if }i>\alpha+1;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }i=\alpha+1 \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{ \alpha+1,\alpha +2,...,\beta\right\} \right) . \] \textbf{(c)} We let $\mathcal{A}_{+}$ be the Lie subalgebra $\left\langle a_{1},a_{2},a_{3},...\right\rangle $ of $\mathcal{A}$. This Lie subalgebra $\mathcal{A}_{+}$ is abelian. We define an $\mathcal{A}_{+}$-module structure on the vector space $V_{\left] \alpha,\beta\right] }$ by letting $a_{i}$ act as $T_{\left] \alpha,\beta\right] }^{i}$ for every positive integer $i$. (This is well-defined, since the powers of $T_{\left] \alpha,\beta\right] }$ commute, just as the elements of $\mathcal{A}_{+}$.) Thus, for every $\ell \in\mathbb{N}$, the $\ell$-th exterior power $\wedge^{\ell}\left( V_{\left] \alpha,\beta\right] }\right) $ is canonically equipped with an $\mathcal{A}_{+}$-module structure. \textbf{(d)} For every $\ell\in\mathbb{N}$, let $R_{\ell,\left] \alpha ,\beta\right] }:\wedge^{\ell}\left( V_{\left] \alpha,\beta\right] }\right) \rightarrow\wedge^{\dfrac{\infty}{2},\alpha+\ell}V$ be the linear map defined by% \[ \left( \begin{array} [c]{r}% R_{\ell,\left] \alpha,\beta\right] }\left( b_{1}\wedge b_{2}\wedge...\wedge b_{\ell}\right) =b_{1}\wedge b_{2}\wedge...\wedge b_{\ell}\wedge v_{\alpha }\wedge v_{\alpha-1}\wedge v_{\alpha-2}\wedge...\\ \ \ \ \ \ \ \ \ \ \ \text{for any }b_{1},b_{2},...,b_{\ell}\in V_{\left] \alpha,\beta\right] }% \end{array} \right) . \] \end{definition} \begin{remark} \label{rmk.finitary.Valphabeta}Let $\alpha$ and $\beta$ be integers such that $\alpha-1\leq\beta$. \textbf{(a)} The $\left( \beta-\alpha\right) $-tuple $\left( v_{\beta },v_{\beta-1},...,v_{\alpha+1}\right) $ is a basis of this vector space $V_{\left] \alpha,\beta\right] }$. With respect to this basis, the endomorphism $T_{\left] \alpha,\beta\right] }$ of $V_{\left] \alpha ,\beta\right] }$ is represented by the $\left( \beta-\alpha\right) \times\left( \beta-\alpha\right) $ matrix $\left( \begin{array} [c]{cccccc}% 0 & 0 & 0 & ... & 0 & 0\\ 1 & 0 & 0 & ... & 0 & 0\\ 0 & 1 & 0 & ... & 0 & 0\\ 0 & 0 & 1 & ... & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 & 0 \end{array} \right) $. \textbf{(b)} We have $T_{\left] \alpha,\beta\right] }^{\beta-\alpha}=0$. \textbf{(c)} For every sequence $\left( y_{1},y_{2},y_{3},...\right) $ of elements of $\mathbb{C}$, the endomorphism $\sum\limits_{i=1}^{\infty}% y_{i}T_{\left] \alpha,\beta\right] }^{i}$ of $V_{\left] \alpha ,\beta\right] }$ is well-defined and nilpotent. \textbf{(d)} For every sequence $\left( y_{1},y_{2},y_{3},...\right) $ of elements of $\mathbb{C}$, the endomorphism $\exp\left( \sum\limits_{i=1}% ^{\infty}y_{i}T_{\left] \alpha,\beta\right] }^{i}\right) $ of $V_{\left] \alpha,\beta\right] }$ is well-defined. \textbf{(e)} For every sequence $\left( y_{1},y_{2},y_{3},...\right) $ of elements of $\mathbb{C}$, the endomorphism $\exp\left( y_{1}a_{1}+y_{2}% a_{2}+y_{3}a_{3}+...\right) $ of $V_{\left] \alpha,\beta\right] }$ is well-defined. \textbf{(f)} Every $j\in\mathbb{N}$ satisfies% \begin{equation} T_{\left] \alpha,\beta\right] }^{j}v_{u}=\left\{ \begin{array} [c]{l}% v_{u-j},\ \ \ \ \ \ \ \ \ \ \text{if }u-j>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u-j\leq\alpha \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }u\in\left\{ \alpha +1,\alpha+2,...,\beta\right\} . \label{pf.finitary.Valphabeta.R.Tabjvu}% \end{equation} \textbf{(g)} For every $n\in\mathbb{Z}$, let $c_{n}$ be an element of $\mathbb{C}$. Assume that $c_{n}=0$ for every negative $n\in\mathbb{Z}$. Then, the sum $\sum\limits_{k\geq0}c_{k}T_{\left] \alpha,\beta\right] }^{k}$ is a well-defined endomorphism of $V_{\left] \alpha,\beta\right] }$, and the matrix representing this endomorphism with respect to the basis $\left( v_{\beta},v_{\beta-1},...,v_{\alpha+1}\right) $ of $V_{\left] \alpha ,\beta\right] }$ is $\left( c_{i-j}\right) _{\left( i,j\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}$. \end{remark} \begin{vershort} \textit{Proof of Remark \ref{rmk.finitary.Valphabeta}.} Parts \textbf{(a)} through \textbf{(f)} of Remark \ref{rmk.finitary.Valphabeta} are trivial, and part \textbf{(g)} is just the finitary analogue of Lemma \ref{lem.schur.fermi.skew.toeplitzmatrix} and proven in the same way. \end{vershort} \begin{verlong} \textit{Proof of Remark \ref{rmk.finitary.Valphabeta}.} \textbf{(a)} We know that $\left( v_{\alpha+1},v_{\alpha+2},...,v_{\beta}\right) $ is a basis of this vector space $V_{\left] \alpha,\beta\right] }$. Thus, $\left( v_{\beta},v_{\beta-1},...,v_{\alpha+1}\right) $ also is a basis of this vector space $V_{\left] \alpha,\beta\right] }$. With respect to this basis, the endomorphism $T_{\left] \alpha,\beta\right] }$ of $V_{\left] \alpha,\beta\right] }$ is represented by the $\left( \beta-\alpha\right) \times\left( \beta-\alpha\right) $ matrix $\left( \begin{array} [c]{cccccc}% 0 & 0 & 0 & ... & 0 & 0\\ 1 & 0 & 0 & ... & 0 & 0\\ 0 & 1 & 0 & ... & 0 & 0\\ 0 & 0 & 1 & ... & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 & 0 \end{array} \right) $ (this follows readily from the definition of $T_{\left] \alpha,\beta\right] }$). This proves Remark \ref{rmk.finitary.Valphabeta} \textbf{(a)}. \textbf{(b)} We know (from Remark \ref{rmk.finitary.Valphabeta} \textbf{(a)}) that the endomorphism $T_{\left] \alpha,\beta\right] }$ of $V_{\left] \alpha,\beta\right] }$ is represented by the $\left( \beta-\alpha\right) \times\left( \beta-\alpha\right) $ matrix $\left( \begin{array} [c]{cccccc}% 0 & 0 & 0 & ... & 0 & 0\\ 1 & 0 & 0 & ... & 0 & 0\\ 0 & 1 & 0 & ... & 0 & 0\\ 0 & 0 & 1 & ... & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 & 0 \end{array} \right) $. This matrix is a strictly lower-triangular $\left( \beta -\alpha\right) \times\left( \beta-\alpha\right) $ matrix, and therefore its $\left( \beta-\alpha\right) $-th power is $0$\ \ \ \ \footnote{Here, we are using the following simple fact from linear algebra: If $n$ is a nonnegative integer, and $M$ is a strictly lower-triangular $n\times n$-matrix, then $M^{n}=0$.}. That is, $T_{\left] \alpha,\beta\right] }^{\beta-\alpha}=0$. This proves Remark \ref{rmk.finitary.Valphabeta} \textbf{(b)}. \textbf{(c)} Let $\left( y_{1},y_{2},y_{3},...\right) $ be a sequence of elements of $\mathbb{C}$. We have $T_{\left] \alpha,\beta\right] }% ^{\beta-\alpha}=0$. Thus, $T_{\left] \alpha,\beta\right] }$ is nilpotent, so that the endomorphism $\sum\limits_{i=1}^{\infty}y_{i}T_{\left] \alpha ,\beta\right] }^{i}$ is well-defined. We have $\sum\limits_{i=1}^{\infty}y_{i}\underbrace{T_{\left] \alpha ,\beta\right] }^{i}}_{=T_{\left] \alpha,\beta\right] }^{i-1}\circ T_{\left] \alpha,\beta\right] }}=\left( \sum\limits_{i=1}^{\infty}% y_{i}T_{\left] \alpha,\beta\right] }^{i-1}\right) \circ T_{\left] \alpha,\beta\right] }$ (here, the endomorphism $\sum\limits_{i=1}^{\infty }y_{i}T_{\left] \alpha,\beta\right] }^{i-1}$ is well-defined, since $T_{\left] \alpha,\beta\right] }$ is nilpotent), so that% \begin{align*} \left( \sum\limits_{i=1}^{\infty}y_{i}T_{\left] \alpha,\beta\right] }% ^{i}\right) ^{\beta-\alpha} & =\left( \left( \sum\limits_{i=1}^{\infty }y_{i}T_{\left] \alpha,\beta\right] }^{i-1}\right) \circ T_{\left] \alpha,\beta\right] }\right) ^{\beta-\alpha}=\left( \sum\limits_{i=1}% ^{\infty}y_{i}T_{\left] \alpha,\beta\right] }^{i-1}\right) ^{\beta-\alpha }\circ\underbrace{T_{\left] \alpha,\beta\right] }^{\beta-\alpha}}_{=0}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\sum\limits_{i=1}^{\infty}% y_{i}T_{\left] \alpha,\beta\right] }^{i-1}\text{ and }T_{\left] \alpha,\beta\right] }\text{ commute}\right) \\ & =0, \end{align*} so that the endomorphism $\sum\limits_{i=1}^{\infty}y_{i}T_{\left] \alpha,\beta\right] }^{i}$ is nilpotent. This proves Remark \ref{rmk.finitary.Valphabeta} \textbf{(c)}. \textbf{(d)} Let $\left( y_{1},y_{2},y_{3},...\right) $ be a sequence of elements of $\mathbb{C}$. By Remark \ref{rmk.finitary.Valphabeta} \textbf{(c)}, the endomorphism $\sum\limits_{i=1}^{\infty}y_{i}T_{\left] \alpha,\beta\right] }^{i}$ is nilpotent. Thus, the endomorphism $\exp\left( \sum\limits_{i=1}^{\infty}y_{i}T_{\left] \alpha,\beta\right] }^{i}\right) $ of $V_{\left] \alpha,\beta\right] }$ is well-defined. This proves Remark \ref{rmk.finitary.Valphabeta} \textbf{(d)}. \textbf{(e)} Let $\left( y_{1},y_{2},y_{3},...\right) $ be a sequence of elements of $\mathbb{C}$. We know that the endomorphism $\exp\left( \sum\limits_{i=1}^{\infty}y_{i}T_{\left] \alpha,\beta\right] }^{i}\right) $ is well-defined. Since $T_{\left] \alpha,\beta\right] }^{i}$ is the action of $a_{i}$ on $V_{\left] \alpha,\beta\right] }$ for every positive integer $i$, this endomorphism rewrites as \[ \exp\left( \sum\limits_{i=1}^{\infty}y_{i}\underbrace{T_{\left] \alpha ,\beta\right] }^{i}}_{=a_{i}}\right) =\exp\left( \underbrace{\sum \limits_{i=1}^{\infty}y_{i}a_{i}}_{=y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}% +...}\right) =\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) . \] Hence, the endomorphism $\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) $ of $V_{\left] \alpha,\beta\right] }$ is well-defined. This proves Remark \ref{rmk.finitary.Valphabeta} \textbf{(e)}. \textbf{(f)} We will prove (\ref{pf.finitary.Valphabeta.R.Tabjvu}) by induction over $j$: \textit{Induction base:} For every $u\in\left\{ \alpha+1,\alpha +2,...,\beta\right\} $, we have $\underbrace{T_{\left] \alpha,\beta\right] }^{0}}_{=\operatorname*{id}}v_{u}=\operatorname*{id}\left( v_{u}\right) =v_{u}$ and $\left\{ \begin{array} [c]{l}% v_{u-0},\ \ \ \ \ \ \ \ \ \ \text{if }u-0>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u-0\leq\alpha \end{array} \right. =\left\{ \begin{array} [c]{l}% v_{u},\ \ \ \ \ \ \ \ \ \ \text{if }u>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u\leq\alpha \end{array} \right. =v_{u}$ (since $u>\alpha$). Thus, for every $u\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} $, we have $T_{\left] \alpha ,\beta\right] }^{0}v_{u}=v_{u}=\left\{ \begin{array} [c]{l}% v_{u-0},\ \ \ \ \ \ \ \ \ \ \text{if }u-0>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u-0\leq\alpha \end{array} \right. $. This proves (\ref{pf.finitary.Valphabeta.R.Tabjvu}) for $j=0$. This completes the induction base. \textit{Induction step:} Let $J\in\mathbb{N}$. Assume that (\ref{pf.finitary.Valphabeta.R.Tabjvu}) holds for $j=J$. We now must prove that (\ref{pf.finitary.Valphabeta.R.Tabjvu}) also holds for $j=J+1$. Since (\ref{pf.finitary.Valphabeta.R.Tabjvu}) holds for $j=J$, we have% \begin{equation} T_{\left] \alpha,\beta\right] }^{J}v_{u}=\left\{ \begin{array} [c]{l}% v_{u-J},\ \ \ \ \ \ \ \ \ \ \text{if }u-J>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u-J\leq\alpha \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for every }u\in\left\{ \alpha +1,\alpha+2,...,\beta\right\} . \label{pf.finitary.Valphabeta.g.1}% \end{equation} Now let $u\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} $ be arbitrary. We will prove that% \begin{equation} T_{\left] \alpha,\beta\right] }^{J+1}v_{u}=\left\{ \begin{array} [c]{l}% v_{u-\left( J+1\right) },\ \ \ \ \ \ \ \ \ \ \text{if }u-\left( J+1\right) >\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u-\left( J+1\right) \leq\alpha \end{array} \right. . \label{pf.finitary.Valphabeta.g.2}% \end{equation} In order to do so, we distinguish between two cases: \textit{Case 1:} We have $u>\alpha+1$. \textit{Case 2:} We have $u\leq\alpha+1$. First, consider Case 1. In this case, $u>\alpha+1$. But the definition of $T_{\left] \alpha,\beta\right] }$ yields $T_{\left] \alpha,\beta\right] }\left( v_{u}\right) =\left\{ \begin{array} [c]{l}% v_{u-1},\ \ \ \ \ \ \ \ \ \ \text{if }u>\alpha+1;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u=\alpha+1 \end{array} \right. =v_{u-1}$ (since $u>\alpha+1$). Since $T_{\left] \alpha ,\beta\right] }^{J+1}=T_{\left] \alpha,\beta\right] }^{J}\circ T_{\left] \alpha,\beta\right] }$, we have% \begin{align*} T_{\left] \alpha,\beta\right] }^{J+1}v_{u} & =\left( T_{\left] \alpha,\beta\right] }^{J}\circ T_{\left] \alpha,\beta\right] }\right) \left( v_{u}\right) =T_{\left] \alpha,\beta\right] }^{J}\left( \underbrace{T_{\left] \alpha,\beta\right] }\left( v_{u}\right) }% _{=v_{u-1}}\right) =T_{\left] \alpha,\beta\right] }^{J}v_{u-1}\\ & =\left\{ \begin{array} [c]{l}% v_{u-1-J},\ \ \ \ \ \ \ \ \ \ \text{if }u-1-J>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u-1-J\leq\alpha \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{according to (\ref{pf.finitary.Valphabeta.g.1}), applied to }u-1\text{ instead of }u\right) \\ & =\left\{ \begin{array} [c]{l}% v_{u-\left( J+1\right) },\ \ \ \ \ \ \ \ \ \ \text{if }u-\left( J+1\right) >\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u-\left( J+1\right) \leq\alpha \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }u-1-J=u-\left( J+1\right) \right) . \end{align*} This proves (\ref{pf.finitary.Valphabeta.g.2}) in the Case 1. Now, let us consider Case 2. In this case, $u\leq\alpha+1$. Thus, $u=\alpha+1$ (since $u\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} $). Hence, $u-\left( J+1\right) =\left( \alpha+1\right) -\left( J+1\right) =\alpha-\underbrace{J}_{\geq0}\leq\alpha$, so that $\left\{ \begin{array} [c]{l}% v_{u-\left( J+1\right) },\ \ \ \ \ \ \ \ \ \ \text{if }u-\left( J+1\right) >\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u-\left( J+1\right) \leq\alpha \end{array} \right. =0$. But the definition of $T_{\left] \alpha,\beta\right] }$ yields $T_{\left] \alpha,\beta\right] }\left( v_{u}\right) =\left\{ \begin{array} [c]{l}% v_{u-1},\ \ \ \ \ \ \ \ \ \ \text{if }u>\alpha+1;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u=\alpha+1 \end{array} \right. =0$ (since $u=\alpha+1$). Since $T_{\left] \alpha,\beta\right] }^{J+1}=T_{\left] \alpha,\beta\right] }^{J}\circ T_{\left] \alpha ,\beta\right] }$, we have% \begin{align*} T_{\left] \alpha,\beta\right] }^{J+1}v_{u} & =\left( T_{\left] \alpha,\beta\right] }^{J}\circ T_{\left] \alpha,\beta\right] }\right) \left( v_{u}\right) =T_{\left] \alpha,\beta\right] }^{J}\left( \underbrace{T_{\left] \alpha,\beta\right] }\left( v_{u}\right) }% _{=0}\right) =T_{\left] \alpha,\beta\right] }^{J}\left( 0\right) \\ & =0=\left\{ \begin{array} [c]{l}% v_{u-\left( J+1\right) },\ \ \ \ \ \ \ \ \ \ \text{if }u-\left( J+1\right) >\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }u-\left( J+1\right) \leq\alpha \end{array} \right. . \end{align*} This proves (\ref{pf.finitary.Valphabeta.g.2}) in the Case 2. We have thus proven (\ref{pf.finitary.Valphabeta.g.2}) in each of the Cases 1 and 2. Thus, (\ref{pf.finitary.Valphabeta.g.2}) always holds. We have thus proven (\ref{pf.finitary.Valphabeta.g.2}) for every $u\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} $. In other words, (\ref{pf.finitary.Valphabeta.R.Tabjvu}) holds for $j=J+1$. This completes the induction step. Thus, (\ref{pf.finitary.Valphabeta.R.Tabjvu}) is proven by induction over $j$. In other words, Remark \ref{rmk.finitary.Valphabeta} \textbf{(f)} is proven. \textbf{(g)} Since $T_{\left] \alpha,\beta\right] }^{\beta-\alpha}=0$, the endomorphism $T_{\left] \alpha,\beta\right] }$ is nilpotent. Thus, the infinite sum $\sum\limits_{k\geq0}c_{k}T_{\left] \alpha,\beta\right] }^{k}$ converges (with respect to the discrete topology), and hence is a well-defined endomorphism of $V_{\left] \alpha,\beta\right] }$. Let us identify every endomorphism of the vector space $V_{\left] \alpha,\beta\right] }$ with the matrix representing this endomorphism with respect to the basis $\left( v_{\beta},v_{\beta-1},...,v_{\alpha+1}\right) $ of $V_{\left] \alpha,\beta\right] }$. Then, every $k\in\mathbb{N}$ and $i\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} $ satisfy% \begin{align*} & \left( \delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}\cdot v_{i}\\ & =\left( \text{the }\left( \beta+1-i\right) \text{-th column of the matrix }\left( \delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }v_{i}\text{ is the }\left( \beta+1-i\right) \text{-th element of the basis }\left( v_{\beta}% ,v_{\beta-1},...,v_{\alpha+1}\right) \right) \\ & =\sum\limits_{u\in\left\{ 1,2,...,\beta-\alpha\right\} }% \underbrace{\delta_{u-\left( \beta+1-i\right) ,k}}_{\substack{=\delta _{i-\left( \beta+1-u\right) ,k}\\\text{(since }u-\left( \beta+1-i\right) =i-\left( \beta+1-u\right) \text{)}}}v_{\beta+1-u}=\sum\limits_{u\in\left\{ 1,2,...,\beta-\alpha\right\} }\delta_{i-\left( \beta+1-u\right) ,k}% v_{\beta+1-u}\\ & =\sum\limits_{u\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} }% \delta_{i-u,k}v_{u}\ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }u\text{ for }\beta+1-u\right) \\ & =\sum\limits_{\substack{u\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} ;\\u\neq i-k}}\underbrace{\delta_{i-u,k}}_{\substack{=0\\\text{(since }i-u\neq k\\\text{(since }u\neq i-k\text{))}}}v_{u}+\sum\limits_{\substack{u\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} ;\\u=i-k}}\underbrace{\delta_{i-u,k}% }_{\substack{=1\\\text{(since }i-u=k\\\text{(since }u=i-k\text{))}}}v_{u}\\ & =\underbrace{\sum\limits_{\substack{u\in\left\{ \alpha+1,\alpha +2,...,\beta\right\} ;\\u\neq i-k}}0v_{u}}_{=0}+\sum\limits_{\substack{u\in \left\{ \alpha+1,\alpha+2,...,\beta\right\} ;\\u=i-k}}v_{u}=\sum \limits_{\substack{u\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} ;\\u=i-k}}v_{u}\\ & =\left\{ \begin{array} [c]{l}% v_{i-k},\ \ \ \ \ \ \ \ \ \ \text{if }i-k\in\left\{ \alpha+1,\alpha +2,...,\beta\right\} ;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{otherwise}% \end{array} \right. \\ & =\left\{ \begin{array} [c]{l}% v_{i-k},\ \ \ \ \ \ \ \ \ \ \text{if }\left( i-k>\alpha\text{ and }% i-k\leq\beta\right) ;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{otherwise}% \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }i-k\in\left\{ \alpha +1,\alpha+2,...,\beta\right\} \text{ is equivalent to }\left( i-k>\alpha \text{ and }i-k\leq\beta\right) \right) \\ & =\left\{ \begin{array} [c]{l}% v_{i-k},\ \ \ \ \ \ \ \ \ \ \text{if }i-k>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{otherwise}% \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\left( i-k>\alpha\text{ and }i-k\leq\beta\right) \text{ is equivalent to }i-k>\alpha\\ \text{(because }i-k\leq\beta\text{ is automatically satisfied (since }% i\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} \\ \text{yields }i\leq\beta\text{, while }k\in\mathbb{N}\text{ yields }% k\geq0\text{, so that }\underbrace{i}_{\leq\beta}-\underbrace{k}_{\geq0}% \leq\beta-0=\beta\text{))}% \end{array} \right) \\ & =\left\{ \begin{array} [c]{l}% v_{i-k},\ \ \ \ \ \ \ \ \ \ \text{if }i-k>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }i-k\leq\alpha \end{array} \right. . \end{align*} and% \[ T_{\left] \alpha,\beta\right] }^{k}v_{i}=\left\{ \begin{array} [c]{l}% v_{i-k},\ \ \ \ \ \ \ \ \ \ \text{if }i-k>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }i-k\leq\alpha \end{array} \right. . \] (by (\ref{pf.finitary.Valphabeta.R.Tabjvu}), applied to $k$ and $i$ instead of $j$ and $u$). Hence, every $k\in\mathbb{N}$ and $i\in\left\{ \alpha +1,\alpha+2,...,\beta\right\} $ satisfy% \[ \left( \delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}\cdot v_{i}=\left\{ \begin{array} [c]{l}% v_{i-k},\ \ \ \ \ \ \ \ \ \ \text{if }i-k>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }i-k\leq\alpha \end{array} \right. =T_{\left] \alpha,\beta\right] }^{k}v_{i}. \] Thus, $\left( \delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}=T_{\left] \alpha,\beta\right] }^{k}$ for every $k\in\mathbb{N}$. Hence,% \begin{equation} \sum\limits_{k\geq0}c_{k}\underbrace{T_{\left] \alpha,\beta\right] }^{k}% }_{=\left( \delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}}=\sum\limits_{k\geq0}c_{k}\left( \delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta -\alpha\right\} ^{2}}=\left( \sum\limits_{k\geq0}c_{k}\delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}. \label{pf.finitary.Valphabeta.g.7}% \end{equation} On the other hand, recall that $c_{n}=0$ for every negative $n\in\mathbb{Z}$. In other words, $c_{k}=0$ for every negative $k\in\mathbb{Z}$. Hence, $c_{k}\delta_{u-v,k}=0$ for every negative $k\in\mathbb{Z}$ and every $\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}$. Thus, the sum $\sum\limits_{k<0}c_{k}\delta_{u-v,k}$ is well-defined and equals $0$ for every $\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}$. Hence, in $\left( \sum\limits_{k<0}c_{k}\delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}=\left( 0\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}=0$. In other words, $0=\left( \sum\limits_{k<0}c_{k}\delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}$. Adding this equality to (\ref{pf.finitary.Valphabeta.g.7}), we obtain% \begin{align*} \sum\limits_{k\geq0}c_{k}T_{\left] \alpha,\beta\right] }^{k} & =\left( \sum\limits_{k\geq0}c_{k}\delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}+\left( \sum\limits_{k<0}% c_{k}\delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta -\alpha\right\} ^{2}}\\ & =\left( \sum\limits_{k<0}c_{k}\delta_{u-v,k}+\sum\limits_{k<0}c_{k}% \delta_{u-v,k}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta -\alpha\right\} ^{2}}. \end{align*} But since every $\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}$ satisfies% \begin{align*} \sum\limits_{k\geq0}c_{k}\delta_{u-v,k}+\sum\limits_{k<0}c_{k}\delta_{u-v,k} & =\sum\limits_{k\in\mathbb{Z}}c_{k}\delta_{u-v,k}=\sum \limits_{\substack{k\in\mathbb{Z};\\k\neq u-v}}c_{k}\underbrace{\delta _{u-v,k}}_{\substack{=0\\\text{(since }u-v\neq k\text{)}}}+c_{u-v}% \underbrace{\delta_{u-v,u-v}}_{=1}\\ & =\underbrace{\sum\limits_{\substack{k\in\mathbb{Z};\\k\neq u-v}}c_{k}% 0}_{=0}+c_{u-v}=c_{u-v}, \end{align*} this rewrites as $\sum\limits_{k\geq0}c_{k}T_{\left] \alpha,\beta\right] }^{k}=\left( \underbrace{\sum\limits_{k<0}c_{k}\delta_{u-v,k}+\sum \limits_{k<0}c_{k}\delta_{u-v,k}}_{=c_{u-v}}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}=\left( c_{u-v}\right) _{\left( u,v\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}=\left( c_{i-j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}$ (here, we renamed $\left( u,v\right) $ as $\left( i,j\right) $). In other words, the matrix representing the endomorphism $\sum\limits_{k\geq0}c_{k}T_{\left] \alpha,\beta\right] }^{k}$ of $V_{\left] \alpha,\beta\right] }$ with respect to the basis $\left( v_{\beta},v_{\beta-1},...,v_{\alpha+1}\right) $ of $V_{\left] \alpha,\beta\right] }$ is $\left( c_{i-j}\right) _{\left( i,j\right) \in\mathbb{Z}^{2}}$. This proves Remark \ref{rmk.finitary.Valphabeta} \textbf{(g)}. \end{verlong} This completes the proof of Remark \ref{rmk.finitary.Valphabeta}. A less trivial observation is the following: \begin{proposition} \label{prop.finitary.Valphabeta.R}Let $\alpha$ and $\beta$ be integers such that $\alpha-1\leq\beta$. Let $\ell\in\mathbb{N}$. Then, $R_{\ell,\left] \alpha,\beta\right] }:\wedge^{\ell}\left( V_{\left] \alpha,\beta\right] }\right) \rightarrow\mathcal{F}^{\left( \alpha+\ell\right) }$ is an $\mathcal{A}_{+}$-module homomorphism (where the $\mathcal{A}_{+}$-module structure on $\mathcal{F}^{\left( \alpha+\ell\right) }$ is obtained by restricting the $\mathcal{A}$-module structure on $\mathcal{F}^{\left( \alpha+\ell\right) }$). \end{proposition} \textit{Proof of Proposition \ref{prop.finitary.Valphabeta.R}.} Let $T$ be the shift operator defined in Definition \ref{def.shiftoperator}. Let $j$ be a positive integer. Then, for every integer $i\leq0$, the $\left( i,i\right) $-th entry of the matrix $T^{j}$ is $0$ (in fact, the matrix $T^{j}$ has only zeroes on its main diagonal). Moreover, from the definition of $T$, it follows quickly that% \begin{equation} T^{j}v_{u}=v_{u-j}\ \ \ \ \ \ \ \ \ \ \text{for every }u\in\mathbb{Z}. \label{pf.finitary.Valphabeta.R.Tjvu}% \end{equation} \begin{verlong} (Actually, (\ref{pf.finitary.Valphabeta.R.Tjvu}) follows from (\ref{pf.schur.fermi.skew.toeplitzmatrix.1}), applied to $j$ and $u$ instead of $k$ and $i$.) \end{verlong} Let $\left( i_{0},i_{1},...,i_{\ell-1}\right) $ be an $\ell$-tuple of elements of $\left\{ \alpha+1,\alpha+2,...,\beta\right\} $ such that $i_{0}>i_{1}>...>i_{\ell-1}$. We will prove that% \begin{equation} a_{j}\rightharpoonup\left( R_{\ell,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) =R_{\ell,\left] \alpha,\beta\right] }\left( a_{j}\rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) . \label{pf.finitary.Valphabeta.R.1}% \end{equation} Indeed, let us extend the $\ell$-tuple $\left( i_{0},i_{1},...,i_{\ell -1}\right) $ to a sequence $\left( i_{0},i_{1},i_{2},...\right) $ of integers by setting $\left( i_{k}=\ell+\alpha-k\text{ for every }k\in\left\{ \ell,\ell+1,\ell+2,...\right\} \right) $. Then, $\left( i_{0},i_{1}% ,i_{2},...\right) =\left( i_{0},i_{1},...,i_{\ell-1},\alpha,\alpha -1,\alpha-2,...\right) $. As a consequence, the sequence $\left( i_{0}% ,i_{1},i_{2},...\right) $ is strictly decreasing (since $i_{0}>i_{1}% >...>i_{\ell-1}>\alpha>\alpha-1>\alpha-2>...$) and hence an $\left( \alpha+\ell\right) $-degression. Note that \begin{equation} v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...=0\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N}\text{ satisfying }k\geq\ell\label{pf.finitary.Valphabeta.R.2}% \end{equation} \footnote{\textit{Proof of (\ref{pf.finitary.Valphabeta.R.2}):} Let $k\in\mathbb{N}$ satisfy $k\geq\ell$. Then, $k+j\geq\ell$ as well (since $j$ is positive), so that $i_{k+j}=\ell+\alpha-\left( k+j\right) =\underbrace{\left( \ell+\alpha-k\right) }_{\substack{=i_{k}\\\text{(since }k\geq\ell\text{)}}}-j=i_{k}-j$. Thus, the sequence $\left( i_{0}% ,i_{1},...,i_{k-1},i_{k}-j,i_{k+1},i_{k+2},...\right) $ has two equal terms (since $k+j\neq k$ (due to $j$ being positive)). Thus, $v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...=0$. This proves (\ref{pf.finitary.Valphabeta.R.2}).}. Also,% \begin{equation} v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...=0\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N}\text{ satisfying }k<\ell\text{ and }i_{k}-j\leq\alpha. \label{pf.finitary.Valphabeta.R.3}% \end{equation} \footnote{\textit{Proof of (\ref{pf.finitary.Valphabeta.R.3}):} Let $k\in\mathbb{N}$ satisfy $k<\ell$ and $i_{k}-j\leq\alpha$. Every integer $\leq\alpha$ is contained in the sequence $\left( i_{0},i_{1},i_{2}% ,...\right) $ (since $\left( i_{0},i_{1},i_{2},...\right) =\left( i_{0},i_{1},...,i_{\ell-1},\alpha,\alpha-1,\alpha-2,...\right) $). Since $i_{k}-j\leq\alpha$, this yields that the integer $i_{k}-j$ is contained in the sequence $\left( i_{0},i_{1},i_{2},...\right) $. Hence, there exists a $p\in\mathbb{N}$ such that $i_{p}=i_{k}-j$. Consider this $p$. \par Since $k<\ell$, we have $i_{k}\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} $, so that $i_{k}>\alpha\geq i_{k}-j=i_{p}$, and hence $i_{k}\neq i_{p}$. Thus, $k\neq p$. Since $i_{k}-j=i_{p}$ and $k\neq p$, the sequence $\left( i_{0},i_{1},...,i_{k-1},i_{k}-j,i_{k+1},i_{k+2},...\right) $ has two equal terms. Thus, $v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...=0$, and this proves (\ref{pf.finitary.Valphabeta.R.3}).} \begin{vershort} The definition of $R_{\ell,\left] \alpha,\beta\right] }$ yields% \begin{align*} & R_{\ell,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}% }\wedge...\wedge v_{i_{\ell-1}}\right) \\ & =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\wedge v_{\alpha }\wedge v_{\alpha-1}\wedge v_{\alpha-2}\wedge...\\ & =v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge ...\ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{0},i_{1},...,i_{\ell -1},\alpha,\alpha-1,\alpha-2,...\right) =\left( i_{0},i_{1},i_{2}% ,...\right) \right) , \end{align*} so that% \begin{align} & a_{j}\rightharpoonup\left( R_{\ell,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) \nonumber\\ & =a_{j}\rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}% }\wedge...\right) =\left( \widehat{\rho}\left( T^{j}\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }a_{j}\mid_{\mathcal{F}^{\left( \alpha+\ell\right) }}=T^{j}\mid_{\mathcal{F}^{\left( \alpha+\ell\right) }% }=\widehat{\rho}\left( T^{j}\right) \right) \nonumber\\ & =\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge\underbrace{\left( T^{j}\rightharpoonup v_{i_{k}}\right) }_{\substack{=T^{j}v_{i_{k}}=v_{i_{k}-j}\\\text{(by (\ref{pf.finitary.Valphabeta.R.Tjvu}), applied to }u=i_{k}\text{)}}}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{by Proposition \ref{prop.glinf.ainfact}, applied to }\left( b_{0}% ,b_{1},b_{2},...\right) =\left( v_{i_{0}},v_{i_{1}},v_{i_{2}},...\right) \text{ and }a=T^{j}\\ \text{(since for every integer }i\leq0\text{, the }\left( i,i\right) \text{-th entry of }T^{j}\text{ is }0\text{).}% \end{array} \right) \nonumber\\ & =\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge....\nonumber \end{align} The sum on the right hand side of this equation is infinite, but lots of its terms vanish: Namely, all its terms with $k\geq\ell$ vanish (because of (\ref{pf.finitary.Valphabeta.R.2})), and all its terms with $k<\ell$ and $i_{k}-j\leq\alpha$ vanish (because of (\ref{pf.finitary.Valphabeta.R.3})). We can thus replace the $\sum\limits_{k\geq0}$ sign by a $\sum \limits_{\substack{k\geq0;\\k<\ell;\\i_{k}-j>\alpha}}$ sign, and obtain% \begin{align} & a_{j}\rightharpoonup\left( R_{\ell,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) \nonumber\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell;\\i_{k}-j>\alpha}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge.... \label{pf.finitary.Valphabeta.R.oneside.short}% \end{align} On the other hand, by the definition of the $\mathcal{A}_{+}$-module $\wedge^{\ell}\left( V_{\left] \alpha,\beta\right] }\right) $, we have% \begin{align*} & a_{j}\rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \\ & =\sum\limits_{k=0}^{\ell-1}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\underbrace{\left( a_{j}\rightharpoonup v_{i_{k}}\right) }_{\substack{=T_{\left] \alpha,\beta\right] }^{j}v_{i_{k}}\\\text{(since }a_{j}\text{ acts as }T_{\left] \alpha,\beta\right] }^{j}\\\text{on }V_{\left] \alpha,\beta\right] }\text{)}}}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\wedge v_{i_{\ell-1}}\\ & =\sum\limits_{k=0}^{\ell-1}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( T_{\left] \alpha,\beta\right] }^{j}v_{i_{k}% }\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\wedge v_{i_{\ell-1}}. \end{align*} Applying the linear map $R_{\ell,\left] \alpha,\beta\right] }$ to this equation, we obtain \begin{align*} & R_{\ell,\left] \alpha,\beta\right] }\left( a_{j}\rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) \\ & =\sum\limits_{k=0}^{\ell-1}\underbrace{R_{\ell,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( T_{\left] \alpha,\beta\right] }^{j}v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\wedge v_{i_{\ell-1}}\right) }% _{\substack{=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( T_{\left] \alpha,\beta\right] }^{j}v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\wedge v_{i_{\ell-1}}\wedge v_{\alpha}\wedge v_{\alpha-1}\wedge v_{\alpha-2}\wedge...\\\text{(by the definition of }% R_{\ell,\left] \alpha,\beta\right] }\text{)}}}\\ & =\underbrace{\sum\limits_{k=0}^{\ell-1}}_{=\sum\limits_{\substack{k\geq 0;\\k<\ell}}}\underbrace{v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge\left( T_{\left] \alpha,\beta\right] }^{j}v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\wedge v_{i_{\ell-1}}\wedge v_{\alpha }\wedge v_{\alpha-1}\wedge v_{\alpha-2}\wedge...}_{\substack{=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( T_{\left] \alpha ,\beta\right] }^{j}v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}% }\wedge...\\\text{(since }\left( i_{0},i_{1},...,i_{\ell-1},\alpha ,\alpha-1,\alpha-2,...\right) =\left( i_{0},i_{1},i_{2},...\right) \text{)}}}\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell}}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge\underbrace{\left( T_{\left] \alpha ,\beta\right] }^{j}v_{i_{k}}\right) }_{\substack{=\left\{ \begin{array} [c]{l}% v_{i_{k}-j},\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j\leq\alpha \end{array} \right. \\\text{(by (\ref{pf.finitary.Valphabeta.R.Tabjvu}), applied to }u=i_{k}\text{)}}}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell}}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge\left\{ \begin{array} [c]{l}% v_{i_{k}-j},\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j\leq\alpha \end{array} \right. \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell;\\i_{k}-j>\alpha}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge.... \end{align*} Compared with (\ref{pf.finitary.Valphabeta.R.oneside.short}), this yields% \[ a_{j}\rightharpoonup\left( R_{\ell,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) =R_{\ell,\left] \alpha,\beta\right] }\left( a_{j}\rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) . \] We have thus proven (\ref{pf.finitary.Valphabeta.R.1}). \end{vershort} \begin{verlong} The definition of $R_{\ell,\left] \alpha,\beta\right] }$ yields% \begin{align*} & R_{\ell,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}% }\wedge...\wedge v_{i_{\ell-1}}\right) \\ & =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\wedge v_{\alpha }\wedge v_{\alpha-1}\wedge v_{\alpha-2}\wedge...\\ & =v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge ...\ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{0},i_{1},...,i_{\ell -1},\alpha,\alpha-1,\alpha-2,...\right) =\left( i_{0},i_{1},i_{2}% ,...\right) \right) , \end{align*} so that% \begin{align} & a_{j}\rightharpoonup\left( R_{\ell,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) \nonumber\\ & =a_{j}\rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}% }\wedge...\right) =\left( \widehat{\rho}\left( T^{j}\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }a_{j}\mid_{\mathcal{F}^{\left( \alpha+\ell\right) }}=T^{j}\mid_{\mathcal{F}^{\left( \alpha+\ell\right) }% }=\widehat{\rho}\left( T^{j}\right) \right) \nonumber\\ & =\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge\underbrace{\left( T^{j}\rightharpoonup v_{i_{k}}\right) }_{\substack{=T^{j}v_{i_{k}}=v_{i_{k}-j}\\\text{(by (\ref{pf.finitary.Valphabeta.R.Tjvu}), applied to }u=i_{k}\text{)}}}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{by Proposition \ref{prop.glinf.ainfact}, applied to }\left( b_{0}% ,b_{1},b_{2},...\right) =\left( v_{i_{0}},v_{i_{1}},v_{i_{2}},...\right) \text{ and }a=T^{j}\\ \text{(since for every integer }i\leq0\text{, the }\left( i,i\right) \text{-th entry of }T^{j}\text{ is }0\text{).}% \end{array} \right) \nonumber\\ & =\sum\limits_{k\geq0}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\nonumber\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell}}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{k\geq0;\\k\geq\ell }}\underbrace{v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...}% _{\substack{=0\\\text{(due to (\ref{pf.finitary.Valphabeta.R.2}))}% }}\nonumber\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell}}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\nonumber\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell;\\i_{k}-j>\alpha}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{k\geq0;\\k<\ell;\\i_{k}% -j\leq\alpha}}\underbrace{v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...}% _{\substack{=0\\\text{(due to (\ref{pf.finitary.Valphabeta.R.3}))}% }}\nonumber\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell;\\i_{k}-j>\alpha}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge.... \label{pf.finitary.Valphabeta.R.oneside}% \end{align} On the other hand, by the definition of the $\mathcal{A}_{+}$-module $\wedge^{\ell}\left( V_{\left] \alpha,\beta\right] }\right) $, we have% \begin{align*} & a_{j}\rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \\ & =\sum\limits_{k=0}^{\ell-1}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\underbrace{\left( a_{j}\rightharpoonup v_{i_{k}}\right) }_{\substack{=T_{\left] \alpha,\beta\right] }^{j}v_{i_{k}}\\\text{(since }a_{j}\text{ acts as }T_{\left] \alpha,\beta\right] }^{j}\\\text{on }V_{\left] \alpha,\beta\right] }\text{)}}}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\wedge v_{i_{\ell-1}}\\ & =\sum\limits_{k=0}^{\ell-1}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( T_{\left] \alpha,\beta\right] }^{j}v_{i_{k}% }\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\wedge v_{i_{\ell-1}}, \end{align*} so that% \begin{align*} & R_{\ell,\left] \alpha,\beta\right] }\left( a_{j}\rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) \\ & =R_{\ell,\left] \alpha,\beta\right] }\left( \sum\limits_{k=0}^{\ell -1}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( T_{\left] \alpha,\beta\right] }^{j}v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\wedge v_{i_{\ell-1}}\right) \\ & =\sum\limits_{k=0}^{\ell-1}\underbrace{R_{\ell,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( T_{\left] \alpha,\beta\right] }^{j}v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\wedge v_{i_{\ell-1}}\right) }% _{\substack{=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( T_{\left] \alpha,\beta\right] }^{j}v_{i_{k}}\right) \wedge v_{i_{k+1}% }\wedge v_{i_{k+2}}\wedge...\wedge v_{i_{\ell-1}}\wedge v_{\alpha}\wedge v_{\alpha-1}\wedge v_{\alpha-2}\wedge...\\\text{(by the definition of }% R_{\ell,\left] \alpha,\beta\right] }\text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }R_{\ell,\left] \alpha ,\beta\right] }\text{ is linear}\right) \\ & =\underbrace{\sum\limits_{k=0}^{\ell-1}}_{=\sum\limits_{\substack{k\geq 0;\\k<\ell}}}\underbrace{v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}% }\wedge\left( T_{\left] \alpha,\beta\right] }^{j}v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\wedge v_{i_{\ell-1}}\wedge v_{\alpha }\wedge v_{\alpha-1}\wedge v_{\alpha-2}\wedge...}_{\substack{=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\left( T_{\left] \alpha ,\beta\right] }^{j}v_{i_{k}}\right) \wedge v_{i_{k+1}}\wedge v_{i_{k+2}% }\wedge...\\\text{(since }\left( i_{0},i_{1},...,i_{\ell-1},\alpha ,\alpha-1,\alpha-2,...\right) =\left( i_{0},i_{1},i_{2},...\right) \text{)}}}\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell}}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge\underbrace{\left( T_{\left] \alpha ,\beta\right] }^{j}v_{i_{k}}\right) }_{\substack{=\left\{ \begin{array} [c]{l}% v_{i_{k}-j},\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j\leq\alpha \end{array} \right. \\\text{(by (\ref{pf.finitary.Valphabeta.R.Tabjvu}), applied to }u=i_{k}\text{)}}}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell}}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{k-1}}\wedge\left\{ \begin{array} [c]{l}% v_{i_{k}-j},\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j\leq\alpha \end{array} \right. \wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell;\\i_{k}-j>\alpha}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge\underbrace{\left\{ \begin{array} [c]{l}% v_{i_{k}-j},\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j\leq\alpha \end{array} \right. }_{\substack{=v_{i_{k}-j}\\\text{(since }i_{k}-j>\alpha\text{)}% }}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{k\geq0;\\k<\ell;\\i_{k}% -j\leq\alpha}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}% \wedge\underbrace{\left\{ \begin{array} [c]{l}% v_{i_{k}-j},\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j>\alpha;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{if }i_{k}-j\leq\alpha \end{array} \right. }_{\substack{=0\\\text{(since }i_{k}-j\leq\alpha\text{)}}}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell;\\i_{k}-j>\alpha}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\substack{k\geq 0;\\k<\ell;\\i_{k}-j\leq\alpha}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge0\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge...}_{=0}\\ & =\sum\limits_{\substack{k\geq0;\\k<\ell;\\i_{k}-j>\alpha}}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{k-1}}\wedge v_{i_{k}-j}\wedge v_{i_{k+1}}\wedge v_{i_{k+2}}\wedge.... \end{align*} Compared with (\ref{pf.finitary.Valphabeta.R.oneside}), this yields% \[ a_{j}\rightharpoonup\left( R_{\ell,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) =R_{\ell,\left] \alpha,\beta\right] }\left( a_{j}\rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) . \] We have thus proven (\ref{pf.finitary.Valphabeta.R.1}). \end{verlong} Now, forget that we fixed $j$ and $\left( i_{0},i_{1},...,i_{\ell-1}\right) $. We have thus proven the equality (\ref{pf.finitary.Valphabeta.R.1}) for every positive integer $j$ and every $\ell$-tuple $\left( i_{0}% ,i_{1},...,i_{\ell-1}\right) $ of elements of $\left\{ \alpha+1,\alpha +2,...,\beta\right\} $ such that $i_{0}>i_{1}>...>i_{\ell-1}$. \begin{vershort} Since $\mathcal{A}_{+}=\left\langle a_{1},a_{2},a_{3},...\right\rangle $ and since $\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) _{\beta\geq i_{0}>i_{1}>...>i_{\ell-1}\geq\alpha+1}$ is a basis of the vector space $\wedge^{\ell}\left( V_{\left] \alpha,\beta\right] }\right) $, this yields (by linearity) that $x\rightharpoonup\left( R_{\ell,\left] \alpha,\beta\right] }\left( w\right) \right) =R_{\ell,\left] \alpha ,\beta\right] }\left( x\rightharpoonup w\right) $ holds for every $x\in\mathcal{A}_{+}$ and $w\in\wedge^{\ell}\left( V_{\left] \alpha ,\beta\right] }\right) $. Thus, $R_{\ell,\left] \alpha,\beta\right] }$ is an $\mathcal{A}_{+}$-module homomorphism. This proves Proposition \ref{prop.finitary.Valphabeta.R}. \end{vershort} \begin{verlong} From the above, we can easily obtain that% \begin{equation} x\rightharpoonup\left( R_{\ell,\left] \alpha,\beta\right] }\left( w\right) \right) =R_{\ell,\left] \alpha,\beta\right] }\left( x\rightharpoonup w\right) \ \ \ \ \ \ \ \ \ \ \text{for every }% x\in\mathcal{A}_{+}\text{ and }w\in\wedge^{\ell}\left( V_{\left] \alpha,\beta\right] }\right) . \label{pf.finitary.Valphabeta.R.7}% \end{equation} \footnote{\textit{Proof of (\ref{pf.finitary.Valphabeta.R.7}):} Let $x\in\mathcal{A}_{+}$ and $w\in\wedge^{\ell}\left( V_{\left] \alpha ,\beta\right] }\right) $. Since $x\in\mathcal{A}_{+}=\left\langle a_{1},a_{2},a_{3},...\right\rangle $, there exists a sequence $\left( \lambda_{1},\lambda_{2},\lambda_{3},...\right) $ of elements of $\mathbb{C}$ such that $x=\sum\limits_{j=1}^{\infty}\lambda_{j}a_{j}$ and such that all but finitely many positive integers $j$ satisfy $\lambda_{j}=0$. Consider this sequence $\left( \lambda_{1},\lambda_{2},\lambda_{3},...\right) $. \par We know that $\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}% }\right) _{\beta\geq i_{0}>i_{1}>...>i_{\ell-1}\geq\alpha+1}$ is a basis of the vector space $\wedge^{\ell}\left( V_{\left] \alpha,\beta\right] }\right) $ (because $\left( v_{\beta},v_{\beta-1},...,v_{\alpha+1}\right) $ is a basis of the vector space $V_{\left] \alpha,\beta\right] }$). Since $w\in\wedge^{\ell}\left( V_{\left] \alpha,\beta\right] }\right) $, there must thus exist a family $\left( \alpha_{\left( i_{0},i_{1},...,i_{\ell -1}\right) }\right) _{\beta\geq i_{0}>i_{1}>...>i_{\ell-1}\geq\alpha+1}$ of elements of $\mathbb{C}$ such that $w=\sum\limits_{\beta\geq i_{0}% >i_{1}>...>i_{\ell-1}\geq\alpha+1}\alpha_{\left( i_{0},i_{1},...,i_{\ell -1}\right) }v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}$. Consider this family $\left( \alpha_{\left( i_{0},i_{1},...,i_{\ell -1}\right) }\right) _{\beta\geq i_{0}>i_{1}>...>i_{\ell-1}\geq\alpha+1}$. \par Since $x=\sum\limits_{j=1}^{\infty}\lambda_{j}a_{j}$ and $w=\sum \limits_{\beta\geq i_{0}>i_{1}>...>i_{\ell-1}\geq\alpha+1}\alpha_{\left( i_{0},i_{1},...,i_{\ell-1}\right) }v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}$, we have% \begin{align*} x\rightharpoonup\left( R_{\ell,\left] \alpha,\beta\right] }\left( w\right) \right) & =\left( \sum\limits_{j=1}^{\infty}\lambda_{j}% a_{j}\right) \rightharpoonup\left( R_{\ell,\left] \alpha,\beta\right] }\left( \sum\limits_{\beta\geq i_{0}>i_{1}>...>i_{\ell-1}\geq\alpha+1}% \alpha_{\left( i_{0},i_{1},...,i_{\ell-1}\right) }v_{i_{0}}\wedge v_{i_{1}% }\wedge...\wedge v_{i_{\ell-1}}\right) \right) \\ & =\sum\limits_{j=1}^{\infty}\lambda_{j}\sum\limits_{\beta\geq i_{0}% >i_{1}>...>i_{\ell-1}\geq\alpha+1}\alpha_{\left( i_{0},i_{1},...,i_{\ell -1}\right) }\underbrace{a_{j}\rightharpoonup\left( R_{\ell,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) \right) }_{\substack{=R_{\ell,\left] \alpha ,\beta\right] }\left( a_{j}\rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}% }\wedge...\wedge v_{i_{\ell-1}}\right) \right) \\\text{(by (\ref{pf.finitary.Valphabeta.R.1}))}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since the action of }\mathcal{A}% _{+}\text{ is bilinear, and }R_{\ell,\left] \alpha,\beta\right] }\text{ is linear}\right) \\ & =\sum\limits_{j=1}^{\infty}\lambda_{j}\sum\limits_{\beta\geq i_{0}% >i_{1}>...>i_{\ell-1}\geq\alpha+1}\alpha_{\left( i_{0},i_{1},...,i_{\ell -1}\right) }R_{\ell,\left] \alpha,\beta\right] }\left( a_{j}% \rightharpoonup\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}% }\right) \right) \\ & =R_{\ell,\left] \alpha,\beta\right] }\left( \underbrace{\left( \sum\limits_{j=1}^{\infty}\lambda_{j}a_{j}\right) }_{=x}\rightharpoonup \underbrace{\left( \sum\limits_{\beta\geq i_{0}>i_{1}>...>i_{\ell-1}% \geq\alpha+1}\alpha_{\left( i_{0},i_{1},...,i_{\ell-1}\right) }v_{i_{0}% }\wedge v_{i_{1}}\wedge...\wedge v_{i_{\ell-1}}\right) }_{=w}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since the action of }\mathcal{A}% _{+}\text{ is bilinear, and }R_{\ell,\left] \alpha,\beta\right] }\text{ is linear}\right) \\ & =R_{\ell,\left] \alpha,\beta\right] }\left( x\rightharpoonup w\right) . \end{align*} This proves (\ref{pf.finitary.Valphabeta.R.7}).} As a consequence, $R_{\ell,\left] \alpha,\beta\right] }:\wedge^{\ell}\left( V_{\left] \alpha,\beta\right] }\right) \rightarrow\mathcal{F}^{\left( \alpha +\ell\right) }$ is an $\mathcal{A}_{+}$-module homomorphism. This proves Proposition \ref{prop.finitary.Valphabeta.R}. \end{verlong} Now, we can turn to the promised proof: \textit{Second proof of Theorem \ref{thm.schur.fermi.skew}.} In order to simplify notation, we assume that $\mathbf{R}=\mathbb{C}$. (All the arguments that we will make in the following are independent of the ground ring, as long as the ground ring is a commutative $\mathbb{Q}$-algebra. Therefore, we are actually allowed to assume that $\mathbf{R}=\mathbb{C}$.) Since we assumed that $\mathbf{R}=\mathbb{C}$, we have $\mathcal{A}_{\mathbf{R}}=\mathcal{A}$ and $\mathcal{F}_{\mathbf{R}}^{\left( 0\right) }=\mathcal{F}^{\left( 0\right) }$. Since $\left( i_{0},i_{1},i_{2},...\right) $ is a $0$-degression, every sufficiently high $k\in\mathbb{N}$ satisfies $i_{k}+k=0$. In other words, there exists some $K\in\mathbb{N}$ such that every $k\in\mathbb{N}$ satisfying $k\geq K$ satisfies $i_{k}+k=0$. Consider this $K$. WLOG assume that $K>0$ (else, replace $K$ by $K+1$). Since every $k\in\mathbb{N}$ satisfying $k\geq K$ satisfies $i_{k}+k=0$ and thus $i_{k}=-k$, we have $\left( i_{0}% ,i_{1},i_{2},...\right) =\left( i_{0},i_{1},i_{2},...,i_{K-1},-K,-\left( K+1\right) ,-\left( K+2\right) ,...\right) =\left( i_{0},i_{1}% ,i_{2},...,i_{K-1},-K,-K-1,-K-2,...\right) $. In particular, $i_{K}=-K$. Let $\alpha=i_{K}$ and $\beta=i_{0}$. Since $\left( i_{0},i_{1}% ,i_{2},...\right) $ is a $0$-degression, we have $i_{0}>i_{1}>i_{2}>...$. Thus, $i_{0}>i_{1}>i_{2}>...>i_{K-1}>i_{K}$. In other words, $i_{0}\geq i_{0}>i_{1}>i_{2}>...>i_{K-1}>i_{K}$. Since $i_{0}=\beta$ and $i_{K}=\alpha$, this rewrites as $\beta\geq i_{0}>i_{1}>i_{2}>...>i_{K-1}>\alpha$. Thus, the integers $i_{0}$, $i_{1}$, $i_{2}$, $...$, $i_{K-1}$ lie in the set $\left\{ \alpha+1,\alpha+2,...,\beta\right\} $. Hence, the vectors $v_{i_{0}}$, $v_{i_{1}}$, $...$, $v_{i_{K-1}}$ lie in the vector space $V_{\left] \alpha,\beta\right] }$. Thus, the definition of the map $R_{K,\left] \alpha,\beta\right] }$ (defined according to Definition \ref{def.finitary.Valphabeta} \textbf{(d)}) yields% \begin{align} R_{K,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{K-1}}\right) & =v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{K-1}}\wedge v_{\alpha}\wedge v_{\alpha-1}\wedge v_{\alpha-2}\wedge...\nonumber\\ & =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\wedge v_{-K}\wedge v_{-K-1}\wedge v_{-K-2}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\alpha=i_{K}=-K\right) \nonumber\\ & =v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge... \label{pf.finitary.Valphabeta.R.8}% \end{align} (since $\left( i_{0},i_{1},i_{2},...,i_{K-1},-K,-K-1,-K-2,...\right) =\left( i_{0},i_{1},i_{2},...\right) $). For every $p\in\left\{ 1,2,...,K\right\} $, define an integer $\widetilde{i}% _{p}$ by $\widetilde{i}_{p}=\beta+1-i_{p-1}$. Subtracting the chain of inequalities $\beta\geq i_{0}>i_{1}>i_{2}% >...>i_{K-1}>\alpha$ from $\beta+1$, we obtain $\beta+1-\beta\leq\beta +1-i_{0}<\beta+1-i_{1}<\beta+1-i_{2}<...<\beta+1-i_{K-1}<\beta+1-\alpha$. Since $\beta+1-i_{p-1}=\widetilde{i}_{p}$ for every $p\in\left\{ 1,2,...,K\right\} $, this rewrites as $\beta+1-\beta\leq\widetilde{i}% _{1}<\widetilde{i}_{2}<\widetilde{i}_{3}<...<\widetilde{i}_{K}<\beta+1-\alpha$. This simplifies to $1\leq\widetilde{i}_{1}<\widetilde{i}_{2}<\widetilde{i}% _{3}<...<\widetilde{i}_{K}<\beta+1-\alpha$. Since $\widetilde{i}_{K}$ and $\beta+1-\alpha$ are integers, we obtain $\widetilde{i}_{K}\leq\beta-\alpha$ from $\widetilde{i}_{K}<\beta+1-\alpha$. Thus, $1\leq\widetilde{i}% _{1}<\widetilde{i}_{2}<\widetilde{i}_{3}<...<\widetilde{i}_{K}\leq\beta -\alpha$. On the other hand, substituting $y$ for $x$ in (\ref{def.schur.sk.genfun}), we obtain% \[ \sum\limits_{k\geq0}S_{k}\left( y\right) z^{k}=\exp\left( \sum \limits_{i\geq1}y_{i}z^{i}\right) \ \ \ \ \ \ \ \ \ \ \text{in }% \mathbb{C}\left[ \left[ z\right] \right] . \] Substituting $T_{\left] \alpha,\beta\right] }$ for $z$ in this equality, we obtain \begin{equation} \sum\limits_{k\geq0}S_{k}\left( y\right) T_{\left] \alpha,\beta\right] }^{k}=\exp\left( \sum\limits_{i\geq1}y_{i}T_{\left] \alpha,\beta\right] }^{i}\right) . \label{pf.schur.fermi.skew.finitary.1}% \end{equation} From Remark \ref{rmk.finitary.Valphabeta}, we know that the endomorphisms $\exp\left( \sum\limits_{i=1}^{\infty}y_{i}T_{\left] \alpha,\beta\right] }^{i}\right) $ and \newline$\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) $ of $V_{\left] \alpha,\beta\right] }$ are well-defined. Denote the endomorphism \newline$\exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}% a_{3}+...\right) $ of $V_{\left] \alpha,\beta\right] }$ by $f$. Then, \begin{align*} f & =\exp\left( \underbrace{y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...}% _{=\sum\limits_{i=1}^{\infty}y_{i}a_{i}}\right) =\exp\left( \sum \limits_{i=1}^{\infty}y_{i}\underbrace{a_{i}}_{\substack{=T_{\left] \alpha,\beta\right] }^{i}\\\text{(since the action of }a_{i}\text{ on }V_{\left] \alpha,\beta\right] }\\\text{was defined to be }T_{\left] \alpha,\beta\right] }^{i}\text{)}}}\right) =\exp\left( \sum\limits_{i=1}% ^{\infty}y_{i}T_{\left] \alpha,\beta\right] }^{i}\right) \\ & =\exp\left( \sum\limits_{i\geq1}y_{i}T_{\left] \alpha,\beta\right] }% ^{i}\right) =\sum\limits_{k\geq0}S_{k}\left( y\right) T_{\left] \alpha,\beta\right] }^{k}\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.schur.fermi.skew.finitary.1})}\right) . \end{align*} Note that $S_{n}\left( y\right) \in\mathbb{C}$ for every $n\in\mathbb{Z}$ (since we assumed that $\mathbf{R}=\mathbb{C}$). Also note that $S_{n}\left( y\right) =0$ for every negative $n\in\mathbb{Z}$ (since $S_{n}=0$ for every negative $n$). Hence, according to Remark \ref{rmk.finitary.Valphabeta} \textbf{(g)} (applied to $c_{n}=S_{n}\left( y\right) $), the sum $\sum\limits_{k\geq0}S_{k}\left( y\right) T_{\left] \alpha,\beta\right] }^{k}$ is a well-defined endomorphism of $V_{\left] \alpha,\beta\right] }$, and the matrix representing this endomorphism with respect to the basis $\left( v_{\beta},v_{\beta-1},...,v_{\alpha+1}\right) $ of $V_{\left] \alpha,\beta\right] }$ is $\left( S_{i-j}\left( y\right) \right) _{\left( i,j\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}$. Denote this matrix $\left( S_{i-j}\left( y\right) \right) _{\left( i,j\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}$ by $A$. Let $n=\beta-\alpha$, and denote the basis $\left( v_{\beta},v_{\beta-1},...,v_{\alpha+1}\right) $ of $V_{\left] \alpha,\beta\right] }$ by $\left( e_{1},e_{2},...,e_{n}% \right) $. Then,% \begin{equation} e_{k}=v_{\beta+1-k}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{ 1,2,...,n\right\} . \label{pf.schur.fermi.skew.finitary.2}% \end{equation} As a consequence,% \begin{equation} e_{\beta+1-k}=v_{k}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} \label{pf.schur.fermi.skew.finitary.2a}% \end{equation} (because for every $k\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} $, we can apply (\ref{pf.schur.fermi.skew.finitary.2}) to $\beta+1-k$ instead of $k$, and thus obtain $e_{\beta+1-k}=v_{\beta+1-\left( \beta+1-k\right) }=v_{k}$). We have shown that the matrix representing the endomorphism $\sum \limits_{k\geq0}S_{k}\left( y\right) T_{\left] \alpha,\beta\right] }^{k}$ of $V_{\left] \alpha,\beta\right] }$ with respect to the basis $\left( v_{\beta},v_{\beta-1},...,v_{\alpha+1}\right) $ of $V_{\left] \alpha ,\beta\right] }$ is $\left( S_{i-j}\left( y\right) \right) _{\left( i,j\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}}$. Since $\sum\limits_{k\geq0}S_{k}\left( y\right) T_{\left] \alpha,\beta\right] }^{k}=f$, $\left( v_{\beta},v_{\beta-1},...,v_{\alpha+1}\right) =\left( e_{1},e_{2},...,e_{n}\right) $, and $\left( S_{i-j}\left( y\right) \right) _{\left( i,j\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}% }=A$, this rewrites as follows: The matrix representing the endomorphism $f$ of $V_{\left] \alpha,\beta\right] }$ with respect to the basis $\left( e_{1},e_{2},...,e_{n}\right) $ of $V_{\left] \alpha,\beta\right] }$ is $A$. In other words, $A$ is the $n\times n$-matrix which represents the map $f$ with respect to the bases $\left( e_{1},e_{2},...,e_{n}\right) $ and $\left( e_{1},e_{2},...,e_{n}\right) $ of $V_{\left] \alpha,\beta\right] }$ and $V_{\left] \alpha,\beta\right] }$. Therefore, we can apply Proposition \ref{prop.GLinf.det.fin} to $n$, $V_{\left] \alpha,\beta\right] }$, $V_{\left] \alpha,\beta\right] }$, $\left( e_{1},e_{2},...,e_{n}% \right) $, $\left( e_{1},e_{2},...,e_{n}\right) $, $K$, $f$, $A$ and $\left( \widetilde{i}_{1},\widetilde{i}_{2},\widetilde{i}_{3}% ,...,\widetilde{i}_{K}\right) $ instead of $m$, $P$, $Q$, $\left( e_{1},e_{2},...,e_{n}\right) $, $\left( f_{1},f_{2},...,f_{m}\right) $, $\ell$, $f$, $A$ and $\left( i_{1},i_{2},...,i_{\ell}\right) $. As a result, we obtain% \begin{equation} \left( \wedge^{K}\left( f\right) \right) \left( e_{\widetilde{i}_{1}% }\wedge e_{\widetilde{i}_{2}}\wedge...\wedge e_{\widetilde{i}_{K}}\right) =\sum\limits_{\substack{j_{1}\text{, }j_{2}\text{, }...\text{, }j_{K}\text{ are }K\text{ integers;}\\1\leq j_{1} \beta+1-\left( \beta+1-j_{1}\right) >...>\beta+1-\left( \beta+1-j_{K-1}\right) \geq \beta+1-\left( \beta+\alpha\right) . \] This simplifies to $\beta\geq j_{0}>j_{1}>...>j_{K-1}\geq\alpha+1$ (since $\beta+1-1=\beta$, since $\beta+1-\left( \beta+1-j_{p}\right) =j_{p}$ for every $p\in\left\{ 0,1,...,K-1\right\} $, and since $\beta+1-\left( \beta+\alpha\right) =\alpha+1$). Thus, $\beta\geq j_{0}$, so that $j_{0}% \leq\beta$. Also $-K>-\left( K+1\right) >-\left( K+2\right) >...$. Since $j_{k}=-k$ for every $k\geq K$, this rewrites as $j_{K}>j_{K+1}>j_{K+2}>...$. Also, since $j_{k}=-k$ for every $k\geq K$, we have $j_{K}=-K$. Compared with $i_{K}=-K$, this yields $j_{K}=i_{K}=\alpha$. Thus, $j_{K-1}\geq \alpha+1>\alpha=j_{K}$. Combined with $j_{0}>j_{1}>...>j_{K-1}$, this yields $j_{0}>j_{1}>...>j_{K}$. Combined with $j_{K}>j_{K+1}>j_{K+2}>...$, this yields $j_{0}>j_{1}>j_{2}>...$. Thus, $\left( j_{0},j_{1},j_{2},...\right) $ is a strictly decreasing sequence of integers. Since we know that $\left( j_{k}=-k\text{ for every }k\geq K\right) $, this yields that $\left( j_{0},j_{1},j_{2},...\right) $ is a $0$-degression. Recall also that $j_{0}\leq\beta$. This proves Assertion 1. \par \textit{Proof of Assertion 2:} Assume that $\left( j_{0},j_{1},j_{2}% ,...\right) $ is a $0$-degression satisfying $j_{0}\leq\beta$. Since $\left( j_{0},j_{1},j_{2},...\right) $ is a $0$-degression, we have $j_{0}% >j_{1}>...>j_{K-1}>j_{K}$. \par Recall that $\left( j_{k}=-k\text{ for every }k\geq K\right) $. Applied to $k=K$, this yields $j_{K}=-K$. Compared with $i_{K}=-K$, this yields $j_{K}=i_{K}=\alpha$. Thus, $j_{K-1}>j_{K}=\alpha$. Since $j_{K-1}$ and $\alpha$ are integers, this yields $j_{K-1}\geq\alpha+1$. Combined with $j_{0}>j_{1}>...>j_{K-1}$, this becomes $j_{0}>j_{1}>...>j_{K-1}\geq\alpha+1$. Combined with $\beta\geq j_{0}$ (since $j_{0}\leq\beta$), this becomes $\beta\geq j_{0}>j_{1}>...>j_{K-1}\geq\alpha+1$. Subtracting this chain of inequalities from $\beta+1$, we obtain $\beta+1-1\leq\beta+1-j_{0}% <\beta+1-j_{1}<...<\beta+1-j_{K-1}\leq\beta+1-\left( \alpha+1\right) $. Since $\beta+1-1=\beta$ and $\beta+1-\left( \alpha+1\right) =\beta-\alpha$, this simplifies to $1\leq\beta+1-j_{0}<\beta+1-j_{1}<...<\beta+1-j_{K-1}% \leq\beta-\alpha$. This proves Assertion 2. \par Now, both Assertions 1 and 2 are proven. Combining these two assertions, we conclude that $1\leq\beta+1-j_{0}<\beta+1-j_{1}<...<\beta+1-j_{K-1}\leq \beta-\alpha$ holds if and only if $\left( j_{0},j_{1},j_{2},...\right) $ is a $0$-degression satisfying $j_{0}\leq\beta$, qed.}. Hence, we can replace the sum sign $\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \in\mathbb{Z}^{\mathbb{N}}\text{;}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\1\leq\beta+1-j_{0}<\beta+1-j_{1}<...<\beta+1-j_{K-1}\leq \beta-\alpha}}$ by $\sum\limits_{\substack{\left( j_{0},j_{1},j_{2}% ,...\right) \in\mathbb{Z}^{\mathbb{N}}\text{;}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression such that }j_{0}\leq\beta}}$ in (\ref{pf.schur.fermi.skew.finitary.32}). Hence, (\ref{pf.schur.fermi.skew.finitary.32}) becomes% \begin{align} & R_{K,\left] \alpha,\beta\right] }\left( \left( \exp\left( y_{1}% a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \right) \nonumber\\ & =\underbrace{\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \in\mathbb{Z}^{\mathbb{N}}\text{;}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\1\leq\beta+1-j_{0}<\beta+1-j_{1}<...<\beta+1-j_{K-1}\leq \beta-\alpha}}}_{=\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \in\mathbb{Z}^{\mathbb{N}}\text{;}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\\left( j_{0},j_{1},j_{2},...\right) \text{ is a }% 0\text{-degression such that }j_{0}\leq\beta}}=\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\j_{k}% =-k\text{ for every }k\geq K\text{;}\\j_{0}\leq\beta}}}\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \det\left( A_{\beta+1-j_{0}% ,\beta+1-j_{1},...,\beta+1-j_{K-1}}^{\widetilde{i}_{1},\widetilde{i}% _{2},...,\widetilde{i}_{K}}\right) v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}% }\wedge...\nonumber\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\j_{0}% \leq\beta}}\det\left( A_{\beta+1-j_{0},\beta+1-j_{1},...,\beta+1-j_{K-1}% }^{\widetilde{i}_{1},\widetilde{i}_{2},...,\widetilde{i}_{K}}\right) v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge.... \label{pf.schur.fermi.skew.finitary.35}% \end{align} But it is easily revealed that% \begin{equation} \det\left( A_{\beta+1-j_{0},\beta+1-j_{1},...,\beta+1-j_{K-1}}^{\widetilde{i}% _{1},\widetilde{i}_{2},...,\widetilde{i}_{K}}\right) =S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \label{pf.schur.fermi.skew.finitary.36}% \end{equation} for any $0$-degression $\left( j_{0},j_{1},j_{2},...\right) $ satisfying $\left( j_{k}=-k\text{ for every }k\geq K\right) $ and $j_{0}\leq\beta $\ \ \ \ \footnote{\textit{Proof of (\ref{pf.schur.fermi.skew.finitary.36}):} Let $\left( j_{0},j_{1},j_{2},...\right) $ be a $0$-degression satisfying $\left( j_{k}=-k\text{ for every }k\geq K\right) $ and $j_{0}\leq\beta$. By the definition of the matrix $A$, we have $A=\left( S_{i-j}\left( y\right) \right) _{\left( i,j\right) \in\left\{ 1,2,...,\beta-\alpha\right\} ^{2}% }$. Hence,% \begin{align*} & A_{\beta+1-j_{0},\beta+1-j_{1},...,\beta+1-j_{K-1}}^{\widetilde{i}% _{1},\widetilde{i}_{2},...,\widetilde{i}_{K}}\\ & =\left( S_{\left( \beta+1-j_{u-1}\right) -\widetilde{i}_{v}}\left( y\right) \right) _{\left( u,v\right) \in\left\{ 1,2,...,K\right\} ^{2}% }=\left( S_{\left( \beta+1-j_{u-1}\right) -\left( \beta+1-i_{v-1}\right) }\left( y\right) \right) _{\left( u,v\right) \in\left\{ 1,2,...,K\right\} ^{2}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\widetilde{i}_{v}=\beta +1-i_{v-1}\text{ (by the definition of }\widetilde{i}_{v}\text{) for every }v\in\left\{ 1,2,...,K\right\} \right) \\ & =\left( S_{i_{v-1}-j_{u-1}}\left( y\right) \right) _{\left( u,v\right) \in\left\{ 1,2,...,K\right\} ^{2}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( \beta+1-j_{u-1}\right) -\left( \beta+1-i_{v-1}\right) =i_{v-1}-j_{u-1}\text{ for every }\left( u,v\right) \in\left\{ 1,2,...,K\right\} ^{2}\right) . \end{align*} \par Now, define two partitions $\lambda$ and $\mu$ by $\lambda=\left( i_{k}+k\right) _{k\geq0}$ and $\mu=\left( j_{k}+k\right) _{k\geq0}$. Write the partitions $\lambda$ and $\mu$ in the forms $\lambda=\left( \lambda _{1},\lambda_{2},\lambda_{3},...\right) $ and $\mu=\left( \mu_{1},\mu _{2},\mu_{3},...\right) $. Then, $\lambda_{v}=i_{v-1}+\left( v-1\right) $ for every $v\in\left\{ 1,2,3,...\right\} $, and $\mu_{u}=j_{u-1}+\left( u-1\right) $ for every $u\in\left\{ 1,2,3,...\right\} $. Thus, for every $\left( u,v\right) \in\left\{ 1,2,...,K\right\} ^{2}$, we have% \begin{equation} \underbrace{\lambda_{v}}_{=i_{v-1}+\left( v-1\right) }-\underbrace{\mu_{u}% }_{=j_{u-1}+\left( u-1\right) }+u-v=\left( i_{v-1}+\left( v-1\right) \right) -\left( j_{u-1}+\left( u-1\right) \right) +u-v=i_{v-1}-j_{u-1}. \label{pf.schur.fermi.skew.finitary.37}% \end{equation} \par But every integer $v\geq K+1$ satisfies $\lambda_{v}=i_{v-1}+\left( v-1\right) =0$ (because every integer $v\geq K+1$ satisfies $v-1\geq K$, and therefore $i_{v-1}+\left( v-1\right) =0$ (due to the fact that $\left( i_{k}+k=0\text{ for every }k\geq K\right) $, applied to $k=v-1$). Hence, the partition $\left( \lambda_{1},\lambda_{2},\lambda_{3},...\right) $ can be written in the form $\left( \lambda_{1},\lambda_{2},...,\lambda_{K}\right) $. \par Also, every integer $u\geq K+1$ satisfies $\mu_{u}=j_{u-1}+\left( u-1\right) =0$ (because every integer $u\geq K+1$ satisfies $u-1\geq K$, and therefore $j_{u-1}+\left( u-1\right) =0$ (due to the fact that $\left( j_{k}% +k=0\text{ for every }k\geq K\right) $, applied to $k=u-1$). Hence, the partition $\left( \mu_{1},\mu_{2},\mu_{3},...\right) $ can be written in the form $\left( \mu_{1},\mu_{2},...,\mu_{K}\right) $. \par Since $\lambda=\left( \lambda_{1},\lambda_{2},\lambda_{3},...\right) =\left( \lambda_{1},\lambda_{2},...,\lambda_{K}\right) $ and $\mu=\left( \mu_{1},\mu_{2},\mu_{3},...\right) =\left( \mu_{1},\mu_{2},...,\mu _{K}\right) $, the definition of $S_{\lambda\diagup\mu}\left( x\right) $ yields $S_{\lambda\diagup\mu}\left( x\right) =\det\left( \underbrace{\left( S_{\lambda_{i}-\mu_{j}+j-i}\left( x\right) \right) _{1\leq i\leq K,\ 1\leq j\leq K}}_{=\left( S_{\lambda_{i}-\mu_{j}+j-i}\left( x\right) \right) _{\left( i,j\right) \in\left\{ 1,2,...,K\right\} ^{2}}% }\right) =\det\left( \left( S_{\lambda_{i}-\mu_{j}+j-i}\left( x\right) \right) _{\left( i,j\right) \in\left\{ 1,2,...,K\right\} ^{2}}\right) =\left( \left( S_{\lambda_{v}-\mu_{u}+u-v}\left( y\right) \right) _{\left( v,u\right) \in\left\{ 1,2,...,K\right\} ^{2}}\right) $ (here, we substituted $\left( v,u\right) $ for $\left( i,j\right) $). Substituting $y$ for $x$ in this equality, we obtain% \begin{align*} S_{\lambda\diagup\mu}\left( y\right) & =\det\left( \left( S_{\lambda _{v}-\mu_{u}+u-v}\left( y\right) \right) _{\left( v,u\right) \in\left\{ 1,2,...,K\right\} ^{2}}\right) \\ & =\det\left( \underbrace{\left( S_{i_{v-1}-j_{u-1}}\left( y\right) \right) _{\left( v,u\right) \in\left\{ 1,2,...,K\right\} ^{2}}% }_{=A_{\beta+1-j_{0},\beta+1-j_{1},...,\beta+1-j_{K-1}}^{\widetilde{i}% _{1},\widetilde{i}_{2},...,\widetilde{i}_{K}}}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.schur.fermi.skew.finitary.37}% )}\right) \\ & =\det\left( A_{\beta+1-j_{0},\beta+1-j_{1},...,\beta+1-j_{K-1}% }^{\widetilde{i}_{1},\widetilde{i}_{2},...,\widetilde{i}_{K}}\right) . \end{align*} Thus, $\det\left( A_{\beta+1-j_{0},\beta+1-j_{1},...,\beta+1-j_{K-1}% }^{\widetilde{i}_{1},\widetilde{i}_{2},...,\widetilde{i}_{K}}\right) =S_{\lambda\diagup\mu}\left( y\right) =S_{\left( i_{k}+k\right) _{k\geq 0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) $ (since $\lambda=\left( i_{k}+k\right) _{k\geq0}$ and $\mu=\left( j_{k}+k\right) _{k\geq0}$). This proves (\ref{pf.schur.fermi.skew.finitary.36}).}. Therefore, (\ref{pf.schur.fermi.skew.finitary.35}) becomes% \begin{align} & R_{K,\left] \alpha,\beta\right] }\left( \left( \exp\left( y_{1}% a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \right) \nonumber\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\j_{0}% \leq\beta}}\underbrace{\det\left( A_{\beta+1-j_{0},\beta+1-j_{1}% ,...,\beta+1-j_{K-1}}^{\widetilde{i}_{1},\widetilde{i}_{2},...,\widetilde{i}% _{K}}\right) }_{\substack{=S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \\\text{(by (\ref{pf.schur.fermi.skew.finitary.36}))}}}v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\nonumber\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\j_{0}% \leq\beta}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}% }\wedge.... \label{pf.schur.fermi.skew.finitary.42}% \end{align} \end{verlong} But Proposition \ref{prop.finitary.Valphabeta.R} (applied to $K$ instead of $\ell$) yields that $R_{K,\left] \alpha,\beta\right] }:\wedge^{K}\left( V_{\left] \alpha,\beta\right] }\right) \rightarrow\mathcal{F}^{\left( \alpha+K\right) }$ is an $\mathcal{A}_{+}$-module homomorphism. Since $\underbrace{\alpha}_{=i_{K}=-K}+K=-K+K=0$, this rewrites as follows: $R_{K,\left] \alpha,\beta\right] }:\wedge^{K}\left( V_{\left] \alpha ,\beta\right] }\right) \rightarrow\mathcal{F}^{\left( 0\right) }$ is an $\mathcal{A}_{+}$-module homomorphism. \begin{vershort} Now, $\mathcal{A}_{+}$ is a graded Lie subalgebra of $\mathcal{A}$, and it is easy to define a grading on the $\mathcal{A}_{+}$-module $\wedge^{K}\left( V_{\left] \alpha,\beta\right] }\right) $ such that $\wedge^{K}\left( V_{\left] \alpha,\beta\right] }\right) $ is concentrated in nonpositive degrees.\footnote{Indeed, let us define a grading on the vector space $V_{\left] \alpha,\beta\right] }$ by setting the degree of $v_{i}$ to be $\alpha+1-i$ for every $i\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} $. Then, the vector space $V_{\left] \alpha,\beta\right] }$ is concentrated in nonpositive degrees, so that its $K$-th exterior power $\wedge^{K}\left( V_{\left] \alpha,\beta\right] }\right) $ is also concentrated in nonpositive degrees. On the other hand, $V_{\left] \alpha,\beta\right] }$ is a graded $\mathcal{A}_{+}$-module (this is very easy to check), so that its $K$-th exterior power $\wedge^{K}\left( V_{\left] \alpha,\beta\right] }\right) $ is also a graded $\mathcal{A}_{+}$-module.} Thus, applying Proposition \ref{prop.schur.fermi.welldef.A+} \textbf{(b)} to $\mathbb{C}$, $\wedge^{K}\left( V_{\left] \alpha,\beta\right] }\right) $, $\mathcal{F}% ^{\left( 0\right) }$ and $R_{K,\left] \alpha,\beta\right] }$ instead of $\mathbf{R}$, $M$, $N$ and $\eta$, we obtain% \[ \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ R_{K,\left] \alpha,\beta\right] }=R_{K,\left] \alpha,\beta\right] }\circ\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \] as maps from $\wedge^{K}\left( V_{\left] \alpha,\beta\right] }\right) $ to $\mathcal{F}^{\left( 0\right) }$. Hence,% \begin{align*} & \left( \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ R_{K,\left] \alpha,\beta\right] }\right) \left( v_{i_{0}% }\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \\ & =\left( R_{K,\left] \alpha,\beta\right] }\circ\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \\ & =R_{K,\left] \alpha,\beta\right] }\left( \left( \exp\left( y_{1}% a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \right) \\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\j_{0}% \leq\beta}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}% }\wedge...\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.schur.fermi.skew.finitary.42.short})}\right) . \end{align*} Compared with% \begin{align*} & \left( \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ R_{K,\left] \alpha,\beta\right] }\right) \left( v_{i_{0}% }\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \\ & =\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \underbrace{\left( R_{K,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \right) }_{\substack{=v_{i_{0}% }\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\\\text{(by (\ref{pf.finitary.Valphabeta.R.8}))}}}\\ & =\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) , \end{align*} this becomes% \begin{align} & \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \nonumber\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\j_{0}% \leq\beta}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}% }\wedge...\nonumber\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}\\j_{k}=-k\text{ for every }k\geq K\text{;}% \\j_{0}\leq\beta}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge... \label{pf.schur.fermi.skew.finitary.48.short}% \end{align} (here, we deprived the sum of all addends for which $\left( j_{k}+k\right) _{k\geq0}\not \subseteq \left( i_{k}+k\right) _{k\geq0}$, because (\ref{pf.schur.fermi.notcontained}) shows that all such addends are $0$). \end{vershort} \begin{verlong} Let us define a grading on the vector space $V_{\left] \alpha,\beta\right] }$ by setting the degree of $v_{i}$ to be $\alpha+1-i$ for every $i\in\left\{ \alpha+1,\alpha+2,...,\beta\right\} $. Then, the vector space $V_{\left] \alpha,\beta\right] }$ is concentrated in nonpositive degrees, so that its $K$-th exterior power $\wedge^{K}\left( V_{\left] \alpha,\beta\right] }\right) $ is also concentrated in nonpositive degrees. On the other hand, $\mathcal{A}_{+}$ is a graded Lie subalgebra of $\mathcal{A}$, and $V_{\left] \alpha,\beta\right] }$ is a graded $\mathcal{A}_{+}$-module (this is very easy to check), so that its $K$-th exterior power $\wedge^{K}\left( V_{\left] \alpha,\beta\right] }\right) $ is also a graded $\mathcal{A}_{+}$-module. Now, applying Proposition \ref{prop.schur.fermi.welldef.A+} \textbf{(b)} to $\mathbb{C}$, $\wedge^{K}\left( V_{\left] \alpha,\beta\right] }\right) $, $\mathcal{F}^{\left( 0\right) }$ and $R_{K,\left] \alpha,\beta\right] }$ instead of $\mathbf{R}$, $M$, $N$ and $\eta$, we obtain% \[ \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ R_{K,\left] \alpha,\beta\right] }=R_{K,\left] \alpha,\beta\right] }\circ\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \] as maps from $\wedge^{K}\left( V_{\left] \alpha,\beta\right] }\right) $ to $\mathcal{F}^{\left( 0\right) }$. Hence,% \begin{align*} & \left( \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ R_{K,\left] \alpha,\beta\right] }\right) \left( v_{i_{0}% }\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \\ & =\left( R_{K,\left] \alpha,\beta\right] }\circ\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \\ & =R_{K,\left] \alpha,\beta\right] }\left( \left( \exp\left( y_{1}% a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \right) \\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\j_{0}% \leq\beta}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}% }\wedge...\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.schur.fermi.skew.finitary.42})}\right) . \end{align*} Compared with% \begin{align*} & \left( \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \circ R_{K,\left] \alpha,\beta\right] }\right) \left( v_{i_{0}% }\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \\ & =\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \underbrace{\left( R_{K,\left] \alpha,\beta\right] }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{K-1}}\right) \right) }_{\substack{=v_{i_{0}% }\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\\\text{(by (\ref{pf.finitary.Valphabeta.R.8}))}}}\\ & =\left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) , \end{align*} this becomes% \begin{align} & \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \nonumber\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\j_{0}% \leq\beta}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}% }\wedge...\nonumber\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}\\j_{k}=-k\text{ for every }k\geq K\text{;}% \\j_{0}\leq\beta}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left( j_{0},j_{1}% ,j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\not \subseteq \left( i_{k}+k\right) _{k\geq0}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\j_{0}\leq\beta}}\underbrace{S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...}_{\substack{=0\\\text{(by (\ref{pf.schur.fermi.notcontained}), since }\left( j_{0},j_{1},j_{2}% ,...\right) \text{ is a }0\text{-degression}\\\text{satisfying }\left( j_{k}+k\right) _{k\geq0}\not \subseteq \left( i_{k}+k\right) _{k\geq 0}\text{)}}}\nonumber\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}\\j_{k}=-k\text{ for every }k\geq K\text{;}% \\j_{0}\leq\beta}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge...+\underbrace{\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\not \subseteq \left( i_{k}+k\right) _{k\geq 0}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\j_{0}\leq\beta}}0}% _{=0}\nonumber\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}\\j_{k}=-k\text{ for every }k\geq K\text{;}% \\j_{0}\leq\beta}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge.... \label{pf.schur.fermi.skew.finitary.48}% \end{align} \end{verlong} \begin{vershort} But for any $0$-degression $\left( j_{0},j_{1},j_{2},...\right) $ satisfying $\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}$, we automatically have $\left( j_{k}=-k\text{ for every }k\geq K\right) $ and $j_{0}\leq\beta$ (this is very easy to see). Hence, we can replace the summation sign $\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}% \subseteq\left( i_{k}+k\right) _{k\geq0}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\j_{0}\leq\beta}}$ on the right hand side of (\ref{pf.schur.fermi.skew.finitary.48.short}) by a $\sum \limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}}}$ sign. Thus, (\ref{pf.schur.fermi.skew.finitary.48.short}) simplifies to% \begin{align*} & \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge.... \end{align*} This proves Theorem \ref{thm.schur.fermi.skew}. \end{vershort} \begin{verlong} But for any $0$-degression $\left( j_{0},j_{1},j_{2},...\right) $ satisfying $\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}$, we automatically have $\left( j_{k}=-k\text{ for every }k\geq K\right) $ and $j_{0}\leq\beta$\ \ \ \ \footnote{\textit{Proof.} Let $\left( j_{0}% ,j_{1},j_{2},...\right) $ be a $0$-degression satisfying $\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}$. Since $\left( j_{k}+k\right) _{k\geq0}$ is a partition, every $k\geq0$ satisfies $j_{k}+k\geq0$. \par Now recall that $\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}$. In other words, $j_{k}+k\leq i_{k}+k$ for every $k\geq0$. In other words, $j_{k}\leq i_{k}$ for every $k\geq0$. Applied to $k=0$, this yields $j_{0}\leq i_{0}=\beta$. \par Now, let $k\in\mathbb{N}$ satisfy $k\geq K$. Then, $j_{k}\leq i_{k}$ (since $k\geq0$) and $i_{k}=-k$ (since $k\geq K$). Hence, $j_{k}\leq i_{k}=-k$. Combined with $j_{k}\geq-k$ (since $j_{k}+k=0$), this yields $j_{k}=-k$. Now forget that we fixed $k$. We thus have proven that $\left( j_{k}=-k\text{ for every }k\geq K\right) $. \par Altogether, we now know that $\left( j_{k}=-k\text{ for every }k\geq K\right) $ and $j_{0}\leq\beta$, qed.}. Hence, we can replace the summation sign $\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}\\j_{k}=-k\text{ for every }k\geq K\text{;}% \\j_{0}\leq\beta}}$ on the right hand side of (\ref{pf.schur.fermi.skew.finitary.48}) by a $\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}}}$ sign. Thus, (\ref{pf.schur.fermi.skew.finitary.48}) becomes% \begin{align*} & \left( \exp\left( y_{1}a_{1}+y_{2}a_{2}+y_{3}a_{3}+...\right) \right) \left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) \\ & =\underbrace{\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}% \subseteq\left( i_{k}+k\right) _{k\geq0}\\j_{k}=-k\text{ for every }k\geq K\text{;}\\j_{0}\leq\beta}}}_{=\sum\limits_{\substack{\left( j_{0}% ,j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}}% }}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq 0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}% \wedge...\\ & =\sum\limits_{\substack{\left( j_{0},j_{1},j_{2},...\right) \text{ is a }0\text{-degression;}\\\left( j_{k}+k\right) _{k\geq0}\subseteq\left( i_{k}+k\right) _{k\geq0}}}S_{\left( i_{k}+k\right) _{k\geq0}\diagup\left( j_{k}+k\right) _{k\geq0}}\left( y\right) \cdot v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge.... \end{align*} This proves Theorem \ref{thm.schur.fermi.skew}. \end{verlong} \subsection{Applications to integrable systems} Let us show how these things can be applied to partial differential equations. \begin{Convention} If $v$ is a function in several variables $x_{1}$, $x_{2}$, $...$, $x_{k}$, then, for every $i\in\left\{ 1,2,...,k\right\} $, the derivative of $v$ by the variable $x_{i}$ will be denoted by $\partial_{x_{i}}v$ and by $v_{x_{i}}% $. In other words, $\partial_{x_{i}}v=v_{x_{i}}=\dfrac{\partial}{\partial x_{i}}v$. (For example, if $v$ is a function in two variables $x$ and $t$, then $v_{t}$ will mean the derivative of $v$ by $t$.) \end{Convention} The PDE (partial differential equation) we will be concerned with is the \textbf{Korteweg-de Vries equation} (abbreviated as \textbf{KdV equation}% )\textbf{:} This is the equation $u_{t}=\dfrac{3}{2}uu_{x}+\dfrac{1}{4}% u_{xxx}$ for a function $u\left( t,x\right) $.\ \ \ \ \footnote{There seems to be no consistent definition of the KdV equation across literature. We defined the KdV equation as $u_{t}=\dfrac{3}{2}uu_{x}+\dfrac{1}{4}u_{xxx}$ because this is the form most suited to our approach. Some other authors, instead, define the KdV equation as $v_{t}=v_{xxx}+6vv_{x}$ for a function $v\left( t,x\right) $. Others define it as $w_{t}+ww_{x}+w_{xxx}=0$ for a function $w\left( t,x\right) $. Yet others define it as $q_{t}% +q_{xxx}+6qq_{x}=0$ for a function $q\left( t,x\right) $. These equations are not literally equivalent, but can be transformed into each other by very simple substitutions. In fact, for a function $u\left( t,x\right) $, we have the following equivalence of assertions:% \begin{align*} & \ \left( \text{the function }u\left( t,x\right) \text{ satisfies the equation }u_{t}=\dfrac{3}{2}uu_{x}+\dfrac{1}{4}u_{xxx}\right) \\ & \Longleftrightarrow\ \left( \text{the function }v\left( t,x\right) :=u\left( 4t,x\right) \text{ satisfies the equation }v_{t}=v_{xxx}% +6vv_{x}\right) \\ & \Longleftrightarrow\ \left( \text{the function }w\left( t,x\right) :=6u\left( -4t,x\right) \text{ satisfies the equation }w_{t}+ww_{x}% +w_{xxx}=0\right) \\ & \Longleftrightarrow\ \left( \text{the function }q\left( t,x\right) :=u\left( -4t,x\right) \text{ satisfies the equation }q_{t}+q_{xxx}% +6qq_{x}=0\right) . \end{align*} } We will discuss several interesting solutions of this equation. Here is the most basic family of solutions:% \[ u\left( t\right) =\dfrac{2a^{2}}{\cosh^{2}\left( a\left( x+a^{2}t\right) \right) }\ \ \ \ \ \ \ \ \ \ \left( \text{for }a\text{ being arbitrary but fixed}\right) . \] These are so-called ``traveling wave solutions''. It is a peculiar kind of wave: it has only one bump; it is therefore called a \textit{soliton} (or \textit{solitary wave}). Such waves never occur in linear systems. Note that when we speak of ``wave'', we are imagining a time-dependent 2-dimensional graph with the x-axis showing $t$, the y-axis showing $u\left( t\right) $, and the time parameter being $x$. So when we speak of ``traveling wave'', we mean that it is a wave for any fixed time $x$ and ``travels'' when $x$ moves. The first to study this kind of waves was J. S. Russell in 1834, describing the motion of water in a shallow canal (tsunami waves are similar). The first models for these waves were found by Korteweg-de Vries in 1895. The term $\dfrac{1}{4}u_{xxx}$ in the Korteweg-de Vries equation $u_{t}% =\dfrac{3}{2}uu_{x}+\dfrac{1}{4}u_{xxx}$ is called the \textit{dispersion term}. \textbf{Exercise:} Solve the equation $u_{t}=\dfrac{3}{2}uu_{x}$. (Note that the waves solving this equation develop shocks, in contrast to those solving the Korteweg-de Vries equation.) The Korteweg-de Vries equation is famous for having lots of explicit solutions (unexpectedly for a nonlinear partial differential equation). We will construct some of them using infinite-dimensional Lie algebras. (There are many other ways to construct solutions. In some sense, every field of mathematics is related to some of its solutions.) We will also study the \textbf{Kadomtsev-Petviashvili equation} (abbreviated as \textbf{KP equation}) \[ u_{yy}=\left( u_{t}-\dfrac{3}{2}uu_{x}-\dfrac{1}{4}u_{xxx}\right) _{x}% \] (or, after some rescaling, $\dfrac{3}{4}\partial_{y}^{2}u=\partial_{x}\left( \partial_{t}u-\dfrac{3}{2}u\partial_{x}u-\dfrac{1}{4}\partial_{x}^{3}u\right) $) on a function $u\left( t,x,y\right) $. We will obtain functions which solve this equation (among others). We are going to use the \textit{infinite Grassmannian} for this. First, recall what the \textit{finite Grassmannian} is: \subsubsection{The finite Grassmannian} \begin{definition} Let $k$ and $n$ be integers satisfying $0\leq k\leq n$. Let $V$ be the $\mathbb{C}$-vector space $\mathbb{C}^{n}$. Let $\left( v_{1},v_{2}% ,...,v_{n}\right) $ be the standard basis of $\mathbb{C}^{n}$. Recall that $\wedge^{k}V$ is a representation of $\operatorname*{GL}\left( V\right) $ with a highest-weight vector $v_{1}\wedge v_{2}\wedge...\wedge v_{k}$. Denote by $\Omega$ the orbit of $v_{1}\wedge v_{2}\wedge...\wedge v_{k}$ under $\operatorname*{GL}\left( V\right) $. \end{definition} \begin{proposition} Let $k$ and $n$ be integers satisfying $0\leq k\leq n$. We have $\Omega =\left\{ x\in\wedge^{k}V\text{ nonzero}\ \mid\ x=x_{1}\wedge x_{2}% \wedge...\wedge x_{k}\text{ for some }x_{i}\in V\right\} $. Also, $x_{1}\wedge x_{2}\wedge...\wedge x_{k}\neq0$ if and only if $x_{1}$, $x_{2}$, $...$, $x_{k}$ are linearly independent. \end{proposition} \textit{Proof.} Very easy. \begin{definition} Let $V$ be a $\mathbb{C}$-vector space. Let $k$ be a nonnegative integer. The $k$\textit{-Grassmannian of }$V$ is defined to be the set of all $k$-dimensional vector subspaces of $V$. This set is denoted by $\operatorname*{Gr}\left( k,V\right) $. \end{definition} When $V$ is a finite-dimensional $\mathbb{C}$-vector space, there is a way to define the structure of a projective variety on the Grassmannian $\operatorname*{Gr}\left( k,V\right) $. While we won't ever need the existence of this structure, we will need the so-called Pl\"{u}cker embedding which is the main ingredient in defining this structure:\footnote{In the following definition (and further below), we use the notation $\mathbb{P}% \left( W\right) $ for the projective space of a $\mathbb{C}$-vector space $W$. This projective space is defined to be the quotient set $\left( W \setminus\left\{ 0\right\} \right) / \sim$, where $\sim$ is the proportionality relation (i.e., two vectors $w_{1}$ and $w_{2}$ in $W \setminus\left\{ 0\right\} $ satisfy $w_{1} \sim w_{2}$ if and only if they are linearly dependent).} \begin{definition} Let $k$ and $n$ be integers satisfying $0\leq k\leq n$. Let $V$ be the $\mathbb{C}$-vector space $\mathbb{C}^{n}$. The \textit{Pl\"{u}cker embedding} (corresponding to $n$ and $k$) is defined as the map% \begin{align*} \operatorname*{Pl}:\operatorname*{Gr}\left( k,V\right) & \rightarrow \mathbb{P}\left( \wedge^{k}V\right) ,\\ \left( \begin{array} [c]{c}% k\text{-dimensional subspace of }V\\ \text{with basis }x_{1},x_{2},...,x_{k}% \end{array} \right) & \mapsto\left( \begin{array} [c]{c}% \text{projection of}\\ x_{1}\wedge x_{2}\wedge...\wedge x_{k}\in\wedge^{k}V\diagdown\left\{ 0\right\} \\ \text{on }\mathbb{P}\left( \wedge^{k}V\right) \end{array} \right) . \end{align*} It is easy to see that this is well-defined (i. e., that the projection of $x_{1}\wedge x_{2}\wedge...\wedge x_{k}\in\wedge^{k}V\diagdown\left\{ 0\right\} $ on $\mathbb{P}\left( \wedge^{k}V\right) $ does not depend on the choice of basis $x_{1},x_{2},...,x_{k}$). The image of this map is $\operatorname*{Im}\operatorname*{Pl}=\Omega\diagup\left( \text{scalars}% \right) $. \end{definition} \begin{proposition} \label{prop.plu.injective}This map $\operatorname*{Pl}$ is injective. \end{proposition} \textit{Proof of Proposition \ref{prop.plu.injective}.} Proving Proposition \ref{prop.plu.injective} boils down to showing that if $\lambda$ is a complex number and $v_{1}$, $v_{2}$, $...$, $v_{k}$, $w_{1}$, $w_{2}$, $...$, $w_{k}$ are any vectors in a vector space $U$ satisfying $v_{1}\wedge v_{2}% \wedge...\wedge v_{k}=\lambda\cdot w_{1}\wedge w_{2}\wedge...\wedge w_{k}% \neq0$, then the vector subspace of $U$ spanned by the vectors $v_{1}$, $v_{2}$, $...$, $v_{k}$ is identical with the vector subspace of $U$ spanned by the vectors $w_{1}$, $w_{2}$, $...$, $w_{k}$. This is a well-known fact. The details are left to the reader. Thus, $\operatorname*{Gr}\left( k,V\right) \cong\Omega\diagup\left( \text{scalars}\right) $. (For algebraic geometers: $\Omega$ is the total space of the determinant bundle on $\operatorname*{Gr}\left( k,V\right) $ (but only the nonzero elements).) We are now going to describe the image $\operatorname*{Im}\operatorname*{Pl}$ by algebraic equations. These equations go under the name \textit{Pl\"{u}cker relations}. First, we define (in analogy to Definition \ref{def.createdestroy}) ``wedging'' and ``contraction'' operators on the exterior algebra of $V$: \begin{definition} \label{def.createdestroy.fin}Let $n\in\mathbb{N}$. Let $k\in\mathbb{Z}$. Let $V$ be the vector space $\mathbb{C}^{n}$. Let $\left( v_{1},v_{2}% ,...,v_{n}\right) $ be the standard basis of $V$. Let $i\in\left\{ 1,2,...,n\right\} $. \textbf{(a)} We define the so-called $i$\textit{-th wedging operator} $\widehat{v_{i}}:\wedge^{k}V\rightarrow\wedge^{k+1}V$ by% \[ \widehat{v_{i}}\cdot\psi=v_{i}\wedge\psi\ \ \ \ \ \ \ \ \ \ \text{for all }\psi\in\wedge^{k}V. \] \textbf{(b)} We define the so-called $i$\textit{-th contraction operator} $\overset{\vee}{v_{i}}:\wedge^{k}V\rightarrow\wedge^{k-1}V$ as follows: For every $k$-tuple $\left( i_{1},i_{2},...,i_{k}\right) $ of integers satisfying $1\leq i_{1} r$, so that $e_{i}^{\ast}\left( e_{j}\right) =0$ for all $j\in\left\{ 1,2,...,r\right\} $, so that $e_{i}^{\ast}\left( E^{\prime}\right) =0$ (since $\left( e_{1}% ,e_{2},...,e_{r}\right) $ is a basis of $E^{\prime\perp}$), so that $e_{i}^{\ast}\in\left( E^{\prime\perp}\right) ^{\perp}=E^{\prime}$). The vectors $\widehat{e_{i}}\tau$ for $i\in\left\{ m+1,m+2,...,n\right\} $ are linearly independent (because if some linear combination of them was zero, then some linear combination of the $e_{i}$ with $i\in\left\{ m+1,m+2,...,n\right\} $ would lie in $\left\{ v\in V\ \mid\ \widehat{v}% \tau=0\right\} =E$, but this contradicts the fact that $\left( e_{1}% ,e_{2},...,e_{m}\right) $ is a basis of $E$). Hence, the vectors $\widehat{e_{i}}\tau$ for $i\in\left\{ m+1,m+2,...,r\right\} $ are linearly independent. We defined $S$ using the basis $\left( v_{1},v_{2},...,v_{n}\right) $ of $V$ by the formula $S=\sum\limits_{i=1}^{n}\widehat{v_{i}}\otimes\overset{\vee }{v_{i}}$. Since $S$ did not depend on the basis, we get the same $S$ if we define it using the basis $\left( e_{1},e_{2},...,e_{n}\right) $. Thus, we have $S=\sum\limits_{i=1}^{n}\widehat{e_{i}}\otimes\overset{\vee}{e_{i}}$. Hence,% \begin{align*} S\left( \tau\otimes\tau\right) & =\sum\limits_{i=1}^{m}% \underbrace{\widehat{e_{i}}\tau}_{\substack{=0\\\text{(since }i\in\left\{ 1,2,...,m\right\} \text{)}}}\otimes\overset{\vee}{e_{i}^{\ast}}\tau +\sum\limits_{i=m+1}^{r}\widehat{e_{i}}\tau\otimes\overset{\vee}{e_{i}^{\ast}% }\tau+\sum\limits_{i=r+1}^{n}\widehat{e_{i}}\tau\otimes \underbrace{\overset{\vee}{e_{i}^{\ast}}\tau}_{\substack{=0\\\text{(since }i\in\left\{ r+1,r+2,...,n\right\} \text{)}}}\\ & =\sum\limits_{i=m+1}^{r}\widehat{e_{i}}\tau\otimes\overset{\vee }{e_{i}^{\ast}}\tau. \end{align*} Thus, $S\left( \tau\otimes\tau\right) =0$ rewrites as $\sum\limits_{i=m+1}% ^{r}\widehat{e_{i}}\tau\otimes\overset{\vee}{e_{i}^{\ast}}\tau=0$. But since the vectors $\widehat{e_{i}}\tau$ for $i\in\left\{ m+1,m+2,...,r\right\} $ are linearly independent, this yields that $\overset{\vee}{e_{i}^{\ast}}% \tau=0$ for any $i\in\left\{ m+1,m+2,...,r\right\} $. Thus, for every $i\in\left\{ m+1,m+2,...,r\right\} $, we have $e_{i}^{\ast}\in\left\{ f\in V^{\ast}\ \mid\ \overset{\vee}{f}\tau=0\right\} =E^{\prime}$, so that $e_{i}^{\ast}\left( E^{\prime\perp}\right) =0$. But on the other hand, for every $i\in\left\{ m+1,m+2,...,r\right\} $, we have $e_{i}\in E^{\prime \perp}$ (since $\left( e_{1},e_{2},...,e_{r}\right) $ is a basis of $E^{\prime\perp}$, and since $i\leq r$). Thus, for every $i\in\left\{ m+1,m+2,...,r\right\} $, we have $1=e_{i}^{\ast}\left( \underbrace{e_{i}% }_{\in E^{\prime\perp}}\right) \in e_{i}^{\ast}\left( E^{\prime\perp }\right) =0$. This is a contradiction unless there are no $i\in\left\{ m+1,m+2,...,r\right\} $ at all. So we conclude that there are no $i\in\left\{ m+1,m+2,...,r\right\} $ at all. In other words, $m=r$. Thus, $\dim E=m=r=\dim\left( E^{\prime\perp }\right) $. Combined with $E\subseteq E^{\prime\perp}$, this yields $E=E^{\prime\perp}$. Now, recall that $\left( e_{i_{1}}\wedge e_{i_{2}}\wedge...\wedge e_{i_{k}% }\right) _{1\leq i_{1} m=r$, so that $i\in\left\{ r+1,r+2,...,n\right\} $. As we saw above, this yields $\overset{\vee}{e_{i}^{\ast}}\tau=0$. Thus, \begin{align*} 0 & =\underbrace{\overset{\vee}{e_{i}^{\ast}}}_{=\overset{\vee}{e_{i}}}% \tau=\overset{\vee}{e_{i}}\tau=\sum\limits_{1\leq i_{1} i_{1}% >...>i_{N+m}\geq-N}$ of $\wedge^{N+m+1}\left( V_{N}\right) $ are sent by $j_{N}^{\left( m\right) }$ to pairwise distinct elements of the basis $\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) _{\left( i_{0},i_{1},i_{2},...\right) \text{ is an }m\text{-degression}}$ of $\mathcal{F}^{\left( m\right) }$). \end{definition} In the terminology of Definition \ref{def.finitary.Valphabeta}, the map $j_{N}^{\left( m\right) }$ that we have just defined is the map $R_{N+m+1,\left] -N-1,N\right] }$. Our definitions of $j_{N}^{\left( m\right) }$ and of $i_{N}$ satisfy reasonable compatibilities: \begin{proposition} \label{prop.plu.inf.iNjmN}Let $N\in\mathbb{N}$ and $m\in\mathbb{Z}$. For any $u\in\wedge^{N+m+1}\left( V_{N}\right) $ and $A\in\operatorname*{M}\left( V_{N}\right) $, we have% \[ i_{N}\left( A\right) \cdot j_{N}^{\left( m\right) }\left( u\right) =j_{N}^{\left( m\right) }\left( Au\right) . \] (Here, of course, $i_{N}\left( A\right) \cdot j_{N}^{\left( m\right) }\left( u\right) $ stands for $\left( \varrho\left( i_{N}\left( A\right) \right) \right) \left( j_{N}^{\left( m\right) }\left( u\right) \right) $.) \end{proposition} \textit{Proof of Proposition \ref{prop.plu.inf.iNjmN}.} Let $A\in \operatorname*{M}\left( V_{N}\right) $ and $u\in\wedge^{N+m+1}\left( V_{N}\right) $. We must prove the equality $i_{N}\left( A\right) \cdot j_{N}^{\left( m\right) }\left( u\right) =j_{N}^{\left( m\right) }\left( Au\right) $. Since this equality is linear in $u$, we can WLOG assume that $u$ is an element of the basis $\left( v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{N+m}}\right) _{N\geq i_{0}>i_{1}>...>i_{N+m}\geq-N}$ of $\wedge^{N+m+1}\left( V_{N}\right) $. Assume this. Then, there exists an $N+m+1$-tuple $\left( i_{0},i_{1},...,i_{N+m}\right) $ of integers such that $N\geq i_{0}>i_{1}>...>i_{N+m}\geq-N$ and $u=v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{N+m}}$. Consider this $N+m+1$-tuple. By the definition of $i_{N}\left( A\right) $, we have \begin{equation} \left( i_{N}\left( A\right) \cdot v_{k}=Av_{k}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{ -N,-N+1,...,N\right\} \right) \label{pf.plu.inf.iNjmN.in}% \end{equation} and \begin{equation} \left( i_{N}\left( A\right) \cdot v_{k}=v_{k}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{Z}\diagdown\left\{ -N,-N+1,...,N\right\} \right) . \label{pf.plu.inf.iNjmN.out}% \end{equation} Note that every $\ell\in\left\{ 0,1,...,N+m\right\} $ satisfies $i_{\ell}% \in\left\{ -N,-N+1,...,N\right\} $ (since $N\geq i_{0}>i_{1}>...>i_{N+m}% \geq-N$ and thus $N\geq i_{\ell}\geq-N$) and thus \begin{equation} i_{N}\left( A\right) \cdot v_{i_{\ell}}=Av_{i_{\ell}} \label{pf.plu.inf.iNjmN.in2}% \end{equation} (by (\ref{pf.plu.inf.iNjmN.in}), applied to $k=i_{\ell}$). Also, every positive integer $r$ satisfies $-N-r\in\mathbb{Z}\diagdown\left\{ -N,-N+1,...,N\right\} $ and thus% \begin{equation} i_{N}\left( A\right) \cdot v_{-N-r}=v_{-N-r} \label{pf.plu.inf.iNjmN.out2}% \end{equation} (by (\ref{pf.plu.inf.iNjmN.out}), applied to $k=-N-r$). Now, since $u=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}$, we have% \begin{align*} j_{N}^{\left( m\right) }\left( u\right) & =j_{N}^{\left( m\right) }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\right) \\ & =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge... \end{align*} (by the definition of $j_{N}^{\left( m\right) }$), so that% \begin{align} & i_{N}\left( A\right) \cdot j_{N}^{\left( m\right) }\left( u\right) \nonumber\\ & =i_{N}\left( A\right) \cdot\left( v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}% \wedge...\right) \nonumber\\ & =\underbrace{i_{N}\left( A\right) \cdot v_{i_{0}}\wedge i_{N}\left( A\right) \cdot v_{i_{1}}\wedge...\wedge i_{N}\left( A\right) \cdot v_{i_{N+m}}}_{\substack{=Av_{i_{0}}\wedge Av_{i_{1}}\wedge...\wedge Av_{i_{N+m}}\\\text{(because every }\ell\in\left\{ 0,1,...,N+m\right\} \text{ satisfies }i_{N}\left( A\right) \cdot v_{i_{\ell}}=Av_{i_{\ell}% }\text{ (by (\ref{pf.plu.inf.iNjmN.in2})))}}}\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \wedge\underbrace{i_{N}\left( A\right) \cdot v_{-N-1}\wedge i_{N}\left( A\right) \cdot v_{-N-2}\wedge i_{N}\left( A\right) \cdot v_{-N-3}\wedge...}_{\substack{=v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...\\\text{(because every positive integer }r\text{ satisfies }i_{N}\left( A\right) \cdot v_{-N-r}=v_{-N-r}\text{ (by (\ref{pf.plu.inf.iNjmN.out2})))}}}\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of the action }% \varrho:\operatorname*{M}\left( \infty\right) \rightarrow\operatorname*{End}% \left( \mathcal{F}^{\left( m\right) }\right) \right) \nonumber\\ & =Av_{i_{0}}\wedge Av_{i_{1}}\wedge...\wedge Av_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge.... \label{pf.plu.inf.iNjmN.left}% \end{align} On the other hand, since $u=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}$, we have $Au=Av_{i_{0}}\wedge Av_{i_{1}}\wedge...\wedge Av_{i_{N+m}}$, so that% \begin{align*} j_{N}^{\left( m\right) }\left( Au\right) & =j_{N}^{\left( m\right) }\left( Av_{i_{0}}\wedge Av_{i_{1}}\wedge...\wedge Av_{i_{N+m}}\right) \\ & =Av_{i_{0}}\wedge Av_{i_{1}}\wedge...\wedge Av_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge... \end{align*} (by the definition of $j_{N}^{\left( m\right) }$). Compared with (\ref{pf.plu.inf.iNjmN.left}), this yields $i_{N}\left( A\right) \cdot j_{N}^{\left( m\right) }\left( u\right) =j_{N}^{\left( m\right) }\left( Au\right) $. This proves Proposition \ref{prop.plu.inf.iNjmN}. An important property of the maps $j_{N}^{\left( m\right) }$ is that their images (for fixed $m$ and varying $N$) cover (not just span, but actually cover) all of $\mathcal{F}^{\left( m\right) }$: \begin{proposition} \label{prop.plu.inf.cover}Let $m\in\mathbb{Z}$. \textbf{(a)} We have% \[ j_{0}^{\left( m\right) }\left( \wedge^{0+m+1}\left( V_{0}\right) \right) \subseteq j_{1}^{\left( m\right) }\left( \wedge^{1+m+1}\left( V_{1}\right) \right) \subseteq j_{2}^{\left( m\right) }\left( \wedge^{2+m+1}\left( V_{2}\right) \right) \subseteq.... \] \textbf{(b)} For every $Q\in\mathbb{N}$, we have $\mathcal{F}^{\left( m\right) }=\bigcup\limits_{\substack{N\in\mathbb{N};\\N\geq Q}}j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) $. \end{proposition} Actually, the ``$N\geq Q$'' in Proposition \ref{prop.plu.inf.cover} \textbf{(b)} doesn't have much effect since Proposition \ref{prop.plu.inf.cover} \textbf{(a)} yields $\bigcup\limits_{\substack{N\in \mathbb{N};\\N\geq Q}}j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) =\bigcup\limits_{N\in\mathbb{N}}j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) $; but we prefer to put it in because it is needed in our application. \textit{Proof of Proposition \ref{prop.plu.inf.cover}.} \textbf{(a)} Let $N\in\mathbb{N}$. From the definitions of $j_{N}$ and $j_{N+1}$, it is easy to see that% \[ j_{N}^{\left( m\right) }\left( b_{0}\wedge b_{1}\wedge...\wedge b_{N+m}\right) =j_{N+1}^{\left( m\right) }\left( b_{0}\wedge b_{1}% \wedge...\wedge b_{N+m}\wedge v_{-N-1}\right) \] for any $b_{0},b_{1},...,b_{N+m}\in V_{N}$. Due to linearity, this yields that $j_{N}^{\left( m\right) }\left( a\right) =j_{N+1}^{\left( m\right) }\left( a\wedge v_{-N-1}\right) $ for any $a\in\wedge^{N+m+1}\left( V_{N}\right) $. Hence, $j_{N}^{\left( m\right) }\left( a\right) =j_{N+1}^{\left( m\right) }\left( a\wedge v_{-N-1}\right) \in j_{N+1}^{\left( m\right) }\left( \wedge^{\left( N+1\right) +m+1}\left( V_{N+1}\right) \right) $ for any $a\in\wedge^{N+m+1}\left( V_{N}\right) $. In other words, $j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) \subseteq j_{N+1}^{\left( m\right) }\left( \wedge^{\left( N+1\right) +m+1}\left( V_{N+1}\right) \right) $. We thus have proven that every $N\in\mathbb{N}$ satisfies% \[ j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) \subseteq j_{N+1}^{\left( m\right) }\left( \wedge^{\left( N+1\right) +m+1}\left( V_{N+1}\right) \right) . \] In other words,% \[ j_{0}^{\left( m\right) }\left( \wedge^{0+m+1}\left( V_{0}\right) \right) \subseteq j_{1}^{\left( m\right) }\left( \wedge^{1+m+1}\left( V_{1}\right) \right) \subseteq j_{2}^{\left( m\right) }\left( \wedge^{2+m+1}\left( V_{2}\right) \right) \subseteq.... \] Proposition \ref{prop.plu.inf.cover} \textbf{(a)} is proven. \textbf{(b)} We need three notations: \begin{itemize} \item For any $m$-degression $\mathbf{i}$, define a nonnegative integer $\operatorname*{exting}\left( \mathbf{i}\right) $ as the largest $k\in\mathbb{N}$ satisfying $i_{k}+k\neq m$\ \ \ \ \footnote{If no such $k$ exists, then we set $\operatorname*{exting}\left( \mathbf{i}\right) $ to be $0$.}, where $\mathbf{i}$ is written in the form $\left( i_{0},i_{1}% ,i_{2},...\right) $. (Such a largest $k$ indeed exists, because (by the definition of an $m$-degression) every sufficiently high $k\in\mathbb{N}$ satisfies $i_{k}+k=m$.) \item For any $m$-degression $\mathbf{i}$, define an integer $\operatorname*{head}\left( \mathbf{i}\right) $ by $\operatorname*{head}% \left( \mathbf{i}\right) =i_{0}$, where $\mathbf{i}$ is written in the form $\left( i_{0},i_{1},i_{2},...\right) $. \item For any $m$-degression $\mathbf{i}$, define an element $v_{\mathbf{i}}$ of $\mathcal{F}^{\left( m\right) }$ by $v_{\mathbf{i}}=v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...$, where $\mathbf{i}$ is written in the form $\left( i_{0},i_{1},i_{2},...\right) $. \end{itemize} Thus, $\left( v_{\mathbf{i}}\right) _{\mathbf{i}\text{ is an }% m\text{-degression}}=\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right) _{\left( i_{0},i_{1},i_{2},...\right) \text{ is an }m\text{-degression}}$. Since \newline$\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) _{\left( i_{0},i_{1},i_{2},...\right) \text{ is an }m\text{-degression}}$ is a basis of the vector space $\mathcal{F}^{\left( m\right) }$, we thus conclude that $\left( v_{\mathbf{i}}\right) _{\mathbf{i}\text{ is an }m\text{-degression}}$ is a basis of the vector space $\mathcal{F}^{\left( m\right) }$. Now we prove a simple fact:% \begin{equation} \left( \begin{array} [c]{c}% \text{If }\mathbf{i}\text{ is an }m\text{-degression, and }P\text{ is an integer such that }\\ P\geq\max\left\{ 0,\operatorname*{exting}\left( \mathbf{i}\right) -m,\operatorname*{head}\left( \mathbf{i}\right) \right\} \text{, then }v_{\mathbf{i}}\in j_{P}^{\left( m\right) }\left( \wedge^{P+m+1}\left( V_{P}\right) \right) \end{array} \right) . \label{pf.plu.inf.cover.1}% \end{equation} \textit{Proof of (\ref{pf.plu.inf.cover.1}):} Let $\mathbf{i}$ be an $m$-degression, and $P$ be an integer such that $P\geq\max\left\{ 0,\operatorname*{exting}\left( \mathbf{i}\right) -m,\operatorname*{head}% \left( \mathbf{i}\right) \right\} $. Write $\mathbf{i}$ in the form $\left( i_{0},i_{1},i_{2},...\right) $. Then, $\operatorname*{exting}\left( \mathbf{i}\right) $ is the largest $k\in\mathbb{N}$ satisfying $i_{k}+k\neq m$ (by the definition of $\operatorname*{exting}\left( \mathbf{i}\right) $). Hence,% \begin{equation} \text{every }k\in\mathbb{N}\text{ such that }k>\operatorname*{exting}\left( \mathbf{i}\right) \text{ satisfies }i_{k}+k=m. \label{pf.plu.inf.cover.2}% \end{equation} Since $P\geq\max\left\{ 0,\operatorname*{exting}\left( \mathbf{i}\right) -m,\operatorname*{head}\left( \mathbf{i}\right) \right\} \geq0$, the map $j_{P}^{\left( m\right) }$ and the space $V_{P}$ are well-defined. Since $P\geq\max\left\{ 0,\operatorname*{exting}\left( \mathbf{i}\right) -m,\operatorname*{head}\left( \mathbf{i}\right) \right\} \geq \operatorname*{exting}\left( \mathbf{i}\right) -m$, we have $P+m\geq \operatorname*{exting}\left( \mathbf{i}\right) \geq0$. Now,% \begin{equation} \text{every positive integer }\ell\text{ satisfies }i_{P+m+\ell}% =-P-\ell\label{pf.plu.inf.cover.3}% \end{equation} \footnote{\textit{Proof of (\ref{pf.plu.inf.cover.3}):} Let $\ell\in \mathbb{N}$ be a positive integer. Then, $P+m+\underbrace{\ell}_{>0}% >P+m\geq\operatorname*{exting}\left( \mathbf{i}\right) $. Hence, (\ref{pf.plu.inf.cover.2}) (applied to $k=P+m+\ell$) yields $i_{P+m+\ell }+P+m+\ell=m$. In other words, $i_{P+m+\ell}=-P-\ell$. This proves (\ref{pf.plu.inf.cover.3}).}. Applied to $\ell=1$, this yields $i_{P+m+1}% =-P-1$. Notice also that $P\geq\max\left\{ 0,\operatorname*{exting}\left( \mathbf{i}\right) -m,\operatorname*{head}\left( \mathbf{i}\right) \right\} \geq\operatorname*{head}\left( \mathbf{i}\right) =i_{0}$ (by the definition of $\operatorname*{head}\left( \mathbf{i}\right) $). Now it is easy to see that% \begin{equation} \text{every }k\in\mathbb{N}\text{ such that }k\leq P+m\text{ satisfies }v_{i_{k}}\in V_{P}. \label{pf.plu.inf.cover.4}% \end{equation} \footnote{\textit{Proof of (\ref{pf.plu.inf.cover.4}):} Let $k\in\mathbb{N}$ be such that $k\leq P+m$. Thus, $k i_{1}>i_{2}>...$. As a consequence, $i_{0}\geq i_{k}$ (since $0\leq k$) and $i_{k}>i_{P+m+1}$ (since $k
i_{P+m+1}=-P-1$, we have $i_{k}\geq-P$ (since both $i_{k}$ and $-P$ are integers). Combining $P\geq i_{0}\geq i_{k}$ with $i_{k}\geq-P$, we obtain $P\geq i_{k}\geq-P$. Hence, $v_{i_{k}}\in\left\langle v_{-P}% ,v_{-P+1},...,v_{P}\right\rangle =V_{P}$ (because $V_{P}$ is defined as $\left\langle v_{-P},v_{-P+1},...,v_{P}\right\rangle $). This proves (\ref{pf.plu.inf.cover.4}).} Hence, $v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{P+m}}\in\wedge^{P+m+1}\left( V_{P}\right) $. Now, by the definition of $j_{P}^{\left( m\right) }$, we have% \begin{align*} & j_{P}^{\left( m\right) }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{P+m}}\right) \\ & =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{P+m}}\wedge \underbrace{v_{-P-1}\wedge v_{-P-2}\wedge v_{-P-3}\wedge...}% _{\substack{=v_{i_{P+m+1}}\wedge v_{i_{P+m+2}}\wedge v_{i_{P+m+3}}% \wedge...\\\text{(because every positive integer }\ell\\\text{satisfies }-P-\ell=i_{P+m+\ell}\text{ (by (\ref{pf.plu.inf.cover.3})))}}}\\ & =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{P+m}}\wedge v_{i_{P+m+1}% }\wedge v_{i_{P+m+2}}\wedge v_{i_{P+m+3}}\wedge...=v_{i_{0}}\wedge v_{i_{1}% }\wedge v_{i_{2}}\wedge...=v_{\mathbf{i}}% \end{align*} (since $v_{\mathbf{i}}$ was defined as $v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...$). Thus, $v_{\mathbf{i}}=j_{P}^{\left( m\right) }\left( \underbrace{v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{P+m}}}_{\in \wedge^{P+m+1}\left( V_{P}\right) }\right) \in j_{P}^{\left( m\right) }\left( \wedge^{P+m+1}\left( V_{P}\right) \right) $. This proves (\ref{pf.plu.inf.cover.1}). Now, fix an arbitrary $Q\in\mathbb{N}$. Let $w$ be any element of $\mathcal{F}^{\left( m\right) }$. Since $\left( v_{\mathbf{i}}\right) _{\mathbf{i}\text{ is an }m\text{-degression}}$ is a basis of $\mathcal{F}^{\left( m\right) }$, we can write $w$ as a linear combination of elements of the family $\left( v_{\mathbf{i}}\right) _{\mathbf{i}\text{ is an }m\text{-degression}}$. Since every linear combination contains only finitely many vectors, this yields that we can write $w$ as a linear combination of \textbf{finitely many} elements of the family $\left( v_{\mathbf{i}}\right) _{\mathbf{i}\text{ is an }m\text{-degression}% }$. In other words, there exists a finite set $S$ of $m$-degressions such that $w$ is a linear combination of the family $\left( v_{\mathbf{i}}\right) _{\mathbf{i}\in S}$. Consider this $S$. Since $w$ is a linear combination of the family $\left( v_{\mathbf{i}}\right) _{\mathbf{i}\in S}$, we can find a scalar $\lambda_{\mathbf{i}}\in\mathbb{C}$ for every $\mathbf{i}\in S$ such that $w=\sum\limits_{\mathbf{i}\in S}\lambda_{\mathbf{i}}v_{\mathbf{i}}$. Consider these scalars $\lambda_{\mathbf{i}}$. Let \[ P=\max\left\{ Q,\max\left\{ \max\left\{ 0,\operatorname*{exting}\left( \mathbf{j}\right) -m,\operatorname*{head}\left( \mathbf{j}\right) \right\} \ \mid\ \mathbf{j}\in S\right\} \right\} \] (where the maximum of the empty set is to be understood as $0$). Then, first of all, $P\geq Q$. Second, every $\mathbf{i}\in S$ satisfies% \begin{align*} P & =\max\left\{ Q,\max\left\{ \max\left\{ 0,\operatorname*{exting}% \left( \mathbf{j}\right) -m,\operatorname*{head}\left( \mathbf{j}\right) \right\} \ \mid\ \mathbf{j}\in S\right\} \right\} \\ & \geq\max\left\{ \max\left\{ 0,\operatorname*{exting}\left( \mathbf{j}\right) -m,\operatorname*{head}\left( \mathbf{j}\right) \right\} \ \mid\ \mathbf{j}\in S\right\} \\ & \geq\max\left\{ 0,\operatorname*{exting}\left( \mathbf{i}\right) -m,\operatorname*{head}\left( \mathbf{i}\right) \right\} \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\max\left\{ 0,\operatorname*{exting}\left( \mathbf{i}\right) -m,\operatorname*{head}\left( \mathbf{i}\right) \right\} \text{ is an element of the set}\\ \left\{ \max\left\{ 0,\operatorname*{exting}\left( \mathbf{j}\right) -m,\operatorname*{head}\left( \mathbf{j}\right) \right\} \ \mid \ \mathbf{j}\in S\right\} \text{ (because }\mathbf{i}\in S\text{),}\\ \text{and the maximum of a set is }\geq\text{ to any element of this set}% \end{array} \right) \end{align*} and thus $v_{\mathbf{i}}\in j_{P}^{\left( m\right) }\left( \wedge ^{P+m+1}\left( V_{P}\right) \right) $ (by (\ref{pf.plu.inf.cover.1})). Hence,% \begin{align*} w & =\sum\limits_{\mathbf{i}\in S}\lambda_{\mathbf{i}}% \underbrace{v_{\mathbf{i}}}_{\in j_{P}^{\left( m\right) }\left( \wedge^{P+m+1}\left( V_{P}\right) \right) }\in\sum\limits_{\mathbf{i}\in S}\lambda_{\mathbf{i}}j_{P}^{\left( m\right) }\left( \wedge^{P+m+1}\left( V_{P}\right) \right) \subseteq j_{P}^{\left( m\right) }\left( \wedge^{P+m+1}\left( V_{P}\right) \right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }j_{P}^{\left( m\right) }\left( \wedge^{P+m+1}\left( V_{P}\right) \right) \text{ is a vector space}\right) \\ & \subseteq\bigcup\limits_{\substack{N\in\mathbb{N};\\N\geq Q}}j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }P\geq Q\right) . \end{align*} Now, forget that we fixed $w$. We thus have proven that every $w\in \mathcal{F}^{\left( m\right) }$ satisfies $w\in\bigcup \limits_{\substack{N\in\mathbb{N};\\N\geq Q}}j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) $. Thus, $\mathcal{F}^{\left( m\right) }\subseteq\bigcup\limits_{\substack{N\in\mathbb{N};\\N\geq Q}% }j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) $. Combined with the obvious inclusion $\bigcup \limits_{\substack{N\in\mathbb{N};\\N\geq Q}}j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) \subseteq\mathcal{F}^{\left( m\right) }$, this yields $\mathcal{F}^{\left( m\right) }=\bigcup \limits_{\substack{N\in\mathbb{N};\\N\geq Q}}j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) $. Proposition \ref{prop.plu.inf.cover} \textbf{(b)} is thus proven. \begin{verlong} And a corollary of Proposition \ref{prop.plu.inf.cover} (that we won't need): \begin{corollary} \label{cor.plu.inf.cover.tensor}Let $m\in\mathbb{Z}$. We have $\mathcal{F}% ^{\left( m\right) }\otimes\mathcal{F}^{\left( m\right) }=\bigcup \limits_{N\in\mathbb{N}}\left( j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) \otimes j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) \right) $. \end{corollary} To prove this, we need the following lemma: \begin{lemma} \label{lem.plu.inf.cover.tensor}Let $W$ be a vector space, and $\left( W_{n}\right) _{n\in\mathbb{N}}$ a family of vector subspaces of $W$ such that $W_{0}\subseteq W_{1}\subseteq W_{2}\subseteq...$ and $W=\bigcup \limits_{n\in\mathbb{N}}W_{n}$. Let $U$ be a vector space, and $\left( U_{n}\right) _{n\in\mathbb{N}}$ a family of vector subspaces of $U$ such that $U_{0}\subseteq U_{1}\subseteq U_{2}\subseteq...$ and $U=\bigcup \limits_{n\in\mathbb{N}}U_{n}$. Then, $U\otimes W=\bigcup\limits_{n\in \mathbb{N}}\left( U_{n}\otimes W_{n}\right) $. \end{lemma} \textit{Proof of Lemma \ref{lem.plu.inf.cover.tensor}.} Let $t\in U\otimes W$ be arbitrary. Since $t$ is a tensor, we can write $t$ in the form $t=\sum\limits_{i=1}^{m}u_{i}\otimes w_{i}$ for some $m\in\mathbb{N}$, some elements $u_{1}$, $u_{2}$, $...$, $u_{m}$ of $U$, and some elements $w_{1}$, $w_{2}$, $...$, $w_{m}$ of $W$. Consider this $u$, these $u_{1}$, $u_{2}$, $...$, $u_{m}$ and these $w_{1}$, $w_{2}$, $...$, $w_{m}$. For every $i\in\left\{ 1,2,...,m\right\} $, there exists some $\alpha_{i}% \in\mathbb{N}$ such that $u_{i}\in U_{\alpha_{i}}$ (since $u_{i}\in U=\bigcup\limits_{n\in\mathbb{N}}U_{n}$). Consider this $\alpha_{i}$. For every $i\in\left\{ 1,2,...,m\right\} $, there exists some $\beta_{i}% \in\mathbb{N}$ such that $w_{i}\in W_{\beta_{i}}$ (since $w_{i}\in W=\bigcup\limits_{n\in\mathbb{N}}W_{n}$). Consider this $\beta_{i}$. Let $N=\max\left( \left\{ \alpha_{1},\alpha_{2},...,\alpha_{m}\right\} \cup\left\{ \beta_{1},\beta_{2},...,\beta_{m}\right\} \right) $. Then, every $i\in\left\{ 1,2,...,m\right\} $ satisfies $\alpha_{i}\in\left\{ \alpha_{1},\alpha_{2},...,\alpha_{m}\right\} \subseteq\left\{ \alpha _{1},\alpha_{2},...,\alpha_{m}\right\} \cup\left\{ \beta_{1},\beta _{2},...,\beta_{m}\right\} $, so that% \begin{align*} \alpha_{i} & \leq\max\left( \left\{ \alpha_{1},\alpha_{2},...,\alpha _{m}\right\} \cup\left\{ \beta_{1},\beta_{2},...,\beta_{m}\right\} \right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since any element of a finite set is }\leq\text{ to the maximum of this set}\right) \\ & =N \end{align*} and thus $U_{\alpha_{i}}\subseteq U_{N}$ (since $U_{0}\subseteq U_{1}\subseteq U_{2}\subseteq...$), so that $u_{i}\in U_{\alpha_{i}}\subseteq U_{N}$. Similarly, every $i\in\left\{ 1,2,...,m\right\} $ satisfies $w_{i}\in W_{N}% $. Thus,% \begin{align*} t & =\sum\limits_{i=1}^{m}\underbrace{u_{i}}_{\in U_{N}}\otimes \underbrace{w_{i}}_{\in W_{N}}\in\sum\limits_{i=1}^{m}U_{N}\otimes W_{N}\subseteq U_{N}\otimes W_{N}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }U_{N}\otimes W_{N}\text{ is a }k\text{-vector space}\right) \\ & \subseteq\bigcup\limits_{n\in\mathbb{N}}\left( U_{n}\otimes W_{n}\right) . \end{align*} Now, forget that we fixed $t$. We thus have proven that every $t\in U\otimes W$ satisfies $t\in\bigcup\limits_{n\in\mathbb{N}}\left( U_{n}\otimes W_{n}\right) $. In other words, $U\otimes W\subseteq\bigcup\limits_{n\in \mathbb{N}}\left( U_{n}\otimes W_{n}\right) $. Combined with the obvious inclusion $\bigcup\limits_{n\in\mathbb{N}}\left( U_{n}\otimes W_{n}\right) \subseteq U\otimes W$, this yields $U\otimes W=\bigcup\limits_{n\in\mathbb{N}% }\left( U_{n}\otimes W_{n}\right) $, so that Lemma \ref{lem.plu.inf.cover.tensor} is proven. \textit{Proof of Corollary \ref{cor.plu.inf.cover.tensor}.} Proposition \ref{prop.plu.inf.cover} \textbf{(b)} (applied to $Q=0$) yields \[ \mathcal{F}^{\left( m\right) }=\bigcup\limits_{\substack{N\in\mathbb{N}% ;\\N\geq0}}j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) =\bigcup\limits_{N\in\mathbb{N}}j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) . \] Proposition \ref{prop.plu.inf.cover} \textbf{(a)} yields $j_{0}^{\left( m\right) }\left( \wedge^{0+m+1}\left( V_{0}\right) \right) \subseteq j_{1}^{\left( m\right) }\left( \wedge^{1+m+1}\left( V_{1}\right) \right) \subseteq j_{2}^{\left( m\right) }\left( \wedge^{2+m+1}\left( V_{2}\right) \right) \subseteq...$. Thus, Lemma \ref{lem.plu.inf.cover.tensor} (applied to $W=\mathcal{F}^{\left( m\right) }$, $W_{i}=j_{i}^{\left( m\right) }\left( \wedge^{i+m+1}\left( V_{1}\right) \right) $, $U=\mathcal{F}^{\left( m\right) }$ and $U_{i}=j_{i}^{\left( m\right) }\left( \wedge^{i+m+1}\left( V_{1}\right) \right) $) yields $\mathcal{F}^{\left( m\right) }\otimes\mathcal{F}% ^{\left( m\right) }=\bigcup\limits_{N\in\mathbb{N}}\left( j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) \otimes j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) \right) $. This proves Corollary \ref{cor.plu.inf.cover.tensor}. \end{verlong} What comes next is almost a carbon copy of Definition \ref{def.createdestroy.fin}: \begin{definition} \label{def.plu.inf.createdestroy}Let $N\in\mathbb{N}$. Let $k\in\mathbb{Z}$. Let $i\in\left\{ -N,-N+1,...,N\right\} $. \textbf{(a)} We define the so-called $i$\textit{-th wedging operator} $\widehat{v_{i}^{\left( N\right) }}:\wedge^{k}\left( V_{N}\right) \rightarrow\wedge^{k+1}\left( V_{N}\right) $ by% \[ \widehat{v_{i}^{\left( N\right) }}\cdot\psi=v_{i}\wedge\psi \ \ \ \ \ \ \ \ \ \ \text{for all }\psi\in\wedge^{k}\left( V_{N}\right) . \] \textbf{(b)} We define the so-called $i$\textit{-th contraction operator} $\overset{\vee}{v_{i}^{\left( N\right) }}:\wedge^{k}\left( V_{N}\right) \rightarrow\wedge^{k-1}\left( V_{N}\right) $ as follows: For every $k$-tuple $\left( i_{1},i_{2},...,i_{k}\right) $ of integers satisfying $N\geq i_{1}>i_{2}>...>i_{k}\geq-N$, we let $\overset{\vee }{v_{i}^{\left( N\right) }}\left( v_{i_{1}}\wedge v_{i_{2}}\wedge...\wedge v_{i_{k}}\right) $ be% \[ \left\{ \begin{array} [c]{l}% 0,\ \ \ \ \ \ \ \ \ \ \text{if }i\notin\left\{ i_{1},i_{2},...,i_{k}\right\} ;\\ \left( -1\right) ^{j-1}v_{i_{1}}\wedge v_{i_{2}}\wedge...\wedge v_{i_{j-1}% }\wedge v_{i_{j+1}}\wedge v_{i_{j+2}}\wedge...\wedge v_{i_{k}}% ,\ \ \ \ \ \ \ \ \ \ \text{if }i\in\left\{ i_{1},i_{2},...,i_{k}\right\} \end{array} \right. , \] where, in the case $i\in\left\{ i_{1},i_{2},...,i_{k}\right\} $, we denote by $j$ the integer $\ell$ satisfying $i_{\ell}=i$. Thus, the map $\overset{\vee}{v_{i}^{\left( N\right) }}$ is defined on a basis of the vector space $\wedge^{k}\left( V_{N}\right) $; we extend this to a map $\wedge^{k}\left( V_{N}\right) \rightarrow\wedge^{k-1}\left( V_{N}\right) $ by linearity. Note that, for every negative $\ell\in\mathbb{Z}$, we understand $\wedge ^{\ell}\left( V_{N}\right) $ to mean the zero space. \end{definition} Also: \begin{definition} For every $N\in\mathbb{N}$ and $k\in\left\{ 1,2,...,2N+1\right\} $, let $\Omega_{N}^{\left( k\right) }$ denote the orbit of $v_{N}\wedge v_{N-1}\wedge...\wedge v_{N-k+1}$ under the action of $\operatorname*{GL}% \left( V_{N}\right) $. \end{definition} The following lemma, then, is an easy corollary of Theorem \ref{thm.plu}: \begin{lemma} \label{lem.plu.inf.plu}Let $N\in\mathbb{N}$ and $k\in\mathbb{Z}$. Let $S_{N}^{\left( k\right) }=\sum\limits_{i=-N}^{N}\widehat{v_{i}^{\left( N\right) }}\otimes\overset{\vee}{v_{i}^{\left( N\right) }}:\wedge ^{k}\left( V_{N}\right) \otimes\wedge^{k}\left( V_{N}\right) \rightarrow\wedge^{k+1}\left( V_{N}\right) \otimes\wedge^{k-1}\left( V_{N}\right) $. \textbf{(a)} This map $S_{N}^{\left( k\right) }$ does not depend on the choice of the basis of $V_{N}$, and is $\operatorname*{GL}\left( V_{N}\right) $-invariant. In other words, for \textbf{any} basis $\left( w_{N},w_{N-1},...,w_{-N}\right) $ of $V_{N}$, we have $S_{N}^{\left( k\right) }=\sum\limits_{i=-N}^{N}\widehat{w_{i}^{\left( N\right) }}% \otimes\overset{\vee}{w_{i}^{\left( N\right) }}$ (where the maps $\widehat{w_{i}^{\left( N\right) }}$ and $\overset{\vee}{w_{i}^{\left( N\right) }}$ are defined just as $\widehat{v_{i}^{\left( N\right) }}$ and $\overset{\vee}{v_{i}^{\left( N\right) }}$, but with respect to the basis $\left( w_{N},w_{N-1},...,w_{-N}\right) $). \textbf{(b)} Let $k\in\left\{ 1,2,...,2N+1\right\} $. A nonzero element $\tau\in\wedge^{k}\left( V_{N}\right) $ belongs to $\Omega_{N}^{\left( k\right) }$ if and only if $S_{N}^{\left( k\right) }\left( \tau\otimes \tau\right) =0$. \textbf{(c)} The map $S_{N}^{\left( k\right) }$ is $\operatorname*{M}\left( V_{N}\right) $-invariant. \end{lemma} \textit{Proof of Lemma \ref{lem.plu.inf.plu}.} If we set $n=2N+1$ in Theorem \ref{thm.plu}, and do the following renaming operations: \begin{itemize} \item rename the standard basis $\left( v_{1},v_{2},...,v_{n}\right) $ as $\left( v_{N},v_{N-1},...,v_{-N}\right) $; \item rename the vector space $V$ as $V_{N}$; \item rename the map $S$ as $S_{N}^{\left( k\right) }$; \item rename the basis $\left( w_{1},w_{2},...,w_{n}\right) $ as $\left( w_{N},w_{N-1},...,w_{-N}\right) $; \item rename the maps $\widehat{v_{i}}$ as $\widehat{v_{i}^{\left( N\right) }}$; \item rename the maps $\overset{\vee}{v_{i}}$ as $\overset{\vee}{v_{i}% ^{\left( N\right) }}$; \item rename the maps $\widehat{w_{i}}$ as $\widehat{w_{i}^{\left( N\right) }}$; \item rename the maps $\overset{\vee}{w_{i}}$ as $\overset{\vee}{w_{i}% ^{\left( N\right) }}$; \item rename the set $\Omega$ as $\Omega_{N}^{\left( k\right) }$; \end{itemize} then what we obtain is exactly the statement of Lemma \ref{lem.plu.inf.plu}. Thus, Lemma \ref{lem.plu.inf.plu} is proven. The maps $S_{N}^{\left( k\right) }$ have their own compatibility relation with the $j_{N}^{\left( m\right) }$: \begin{lemma} \label{lem.plu.inf.S.comp}Let $N\in\mathbb{N}$ and $m\in\mathbb{Z}$. Define the notation $S_{N}^{\left( N+m+1\right) }$ as in Lemma \ref{lem.plu.inf.plu}. Then,% \[ \left( j_{N}^{\left( m+1\right) }\otimes j_{N}^{\left( m-1\right) }\right) \circ S_{N}^{\left( N+m+1\right) }=S\circ\left( j_{N}^{\left( m\right) }\otimes j_{N}^{\left( m\right) }\right) . \] \end{lemma} \textit{Proof of Lemma \ref{lem.plu.inf.S.comp}.} Define the maps $\widehat{v_{i}^{\left( N\right) }}$ and $\overset{\vee}{v_{i}^{\left( N\right) }}$ (for all $i\in\left\{ -N,-N+1,...,N\right\} $) as in Definition \ref{def.plu.inf.createdestroy}. Define the maps $\widehat{v_{i}}$ and $\overset{\vee}{v_{i}}$ (for all $i\in\mathbb{Z}$) as in Definition \ref{def.createdestroy}. \textbf{a)} Let us first show that% \begin{equation} j_{N}^{\left( m+1\right) }\circ\widehat{v_{i}^{\left( N\right) }% }=\widehat{v_{i}}\circ j_{N}^{\left( m\right) }\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{ N,N-1,...,-N\right\} . \label{pf.plu.inf.S.comp.a}% \end{equation} \textit{Proof of (\ref{pf.plu.inf.S.comp.a}):} Let $i\in\left\{ N,N-1,...,-N\right\} $. In order to prove (\ref{pf.plu.inf.S.comp.a}), it is clearly enough to show that $\left( j_{N}^{\left( m+1\right) }% \circ\widehat{v_{i}^{\left( N\right) }}\right) \left( u\right) =\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) $ for every $u\in\wedge^{N+m+1}\left( V_{N}\right) $. So let $u$ be any element of $\wedge^{N+m+1}\left( V_{N}\right) $. We must prove the equality $\left( j_{N}^{\left( m+1\right) }\circ\widehat{v_{i}% ^{\left( N\right) }}\right) \left( u\right) =\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) $. Since this equality is linear in $u$, we can WLOG assume that $u$ is an element of the basis $\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\right) _{N\geq i_{0}>i_{1}>...>i_{N+m}\geq-N}$ of $\wedge^{N+m+1}\left( V_{N}\right) $. Assume this. Then, there exists an $N+m+1$-tuple $\left( i_{0},i_{1}% ,...,i_{N+m}\right) $ of integers such that $N\geq i_{0}>i_{1}>...>i_{N+m}% \geq-N$ and $u=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}$. Consider this $N+m+1$-tuple. Comparing% \begin{align*} \left( j_{N}^{\left( m+1\right) }\circ\widehat{v_{i}^{\left( N\right) }% }\right) \left( u\right) & =j_{N}^{\left( m+1\right) }% \underbrace{\left( \widehat{v_{i}^{\left( N\right) }}\left( u\right) \right) }_{\substack{=v_{i}\wedge u\\\text{(by the definition of }\widehat{v_{i}^{\left( N\right) }}\text{)}}}=j_{N}^{\left( m+1\right) }\left( v_{i}\wedge\underbrace{u}_{=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}}\right) \\ & =j_{N}^{\left( m+1\right) }\left( v_{i}\wedge v_{i_{0}}\wedge v_{i_{1}% }\wedge...\wedge v_{i_{N+m}}\right) \\ & =v_{i}\wedge v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }j_{N}^{\left( m+1\right) }\right) \end{align*} with% \begin{align*} \left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) & =\widehat{v_{i}}\left( j_{N}^{\left( m\right) }\left( u\right) \right) =v_{i}\wedge j_{N}^{\left( m\right) }\left( \underbrace{u}_{=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\widehat{v_{i}}\right) \\ & =v_{i}\wedge\underbrace{j_{N}^{\left( m\right) }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\right) }_{\substack{=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...\\\text{(by the definition of }j_{N}^{\left( m\right) }\text{)}}}\\ & =v_{i}\wedge v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge..., \end{align*} we obtain $\left( j_{N}^{\left( m+1\right) }\circ\widehat{v_{i}^{\left( N\right) }}\right) \left( u\right) =\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) $. This is exactly what we needed to prove in order to complete the proof of (\ref{pf.plu.inf.S.comp.a}). The proof of (\ref{pf.plu.inf.S.comp.a}) is thus finished. \textbf{b)} Let us next show that% \begin{equation} j_{N}^{\left( m+1\right) }\circ\overset{\vee}{v_{i}^{\left( N\right) }% }=\overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }% \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{ N,N-1,...,-N\right\} . \label{pf.plu.inf.S.comp.b}% \end{equation} \textit{Proof of (\ref{pf.plu.inf.S.comp.b}):} Let $i\in\left\{ N,N-1,...,-N\right\} $. In order to prove (\ref{pf.plu.inf.S.comp.b}), it is clearly enough to show that $\left( j_{N}^{\left( m+1\right) }% \circ\overset{\vee}{v_{i}^{\left( N\right) }}\right) \left( u\right) =\left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) $ for every $u\in\wedge^{N+m+1}\left( V_{N}\right) $. So let $u$ be any element of $\wedge^{N+m+1}\left( V_{N}\right) $. We must prove the equality $\left( j_{N}^{\left( m+1\right) }\circ\overset{\vee }{v_{i}^{\left( N\right) }}\right) \left( u\right) =\left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) $. Since this equality is linear in $u$, we can WLOG assume that $u$ is an element of the basis $\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\right) _{N\geq i_{0}>i_{1}>...>i_{N+m}\geq-N}$ of $\wedge ^{N+m+1}\left( V_{N}\right) $. Assume this. Then, there exists an $N+m+1$-tuple $\left( i_{0},i_{1},...,i_{N+m}\right) $ of integers such that $N\geq i_{0}>i_{1}>...>i_{N+m}\geq-N$ and $u=v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{N+m}}$. Consider this $N+m+1$-tuple. Let $\left( j_{0},j_{1},j_{2},...\right) $ be the sequence $\left( i_{0},i_{1},...,i_{N+m},-N-1,-N-2,-N-3,...\right) $. From $u=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}$, we obtain% \begin{align} j_{N}^{\left( m\right) }\left( u\right) & =j_{N}^{\left( m\right) }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\right) =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }j_{N}^{\left( m\right) }\right) \nonumber\\ & =v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{0},i_{1},...,i_{N+m}% ,-N-1,-N-2,-N-3,...\right) =\left( j_{0},j_{1},j_{2},...\right) \right) . \label{pf.plu.inf.S.comp.b.0}% \end{align} We distinguish between two cases: \textit{Case 1:} We have $i\notin\left\{ i_{0},i_{1},...,i_{N+m}\right\} $. \textit{Case 2:} We have $i\in\left\{ i_{0},i_{1},...,i_{N+m}\right\} $. Let us first consider Case 1. In this case, from $u=v_{i_{0}}\wedge v_{i_{1}% }\wedge...\wedge v_{i_{N+m}}$, we obtain% \begin{align*} & \overset{\vee}{v_{i}^{\left( N\right) }}\left( u\right) \\ & =\overset{\vee}{v_{i}^{\left( N\right) }}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\right) \\ & =\left\{ \begin{array} [c]{l}% 0,\ \ \ \ \ \ \ \ \ \ \text{if }i\notin\left\{ i_{0},i_{1},...,i_{N+m}% \right\} ;\\ \left( -1\right) ^{j-1}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\left( j-1\right) -1}}\wedge v_{i_{\left( j-1\right) +1}}\wedge v_{i_{\left( j-1\right) +2}}\wedge...\wedge v_{i_{N+m}}% ,\ \ \ \ \ \ \ \ \ \ \text{if }i\in\left\{ i_{0},i_{1},...,i_{N+m}\right\} \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\overset{\vee }{v_{i}^{\left( N\right) }}\right) , \end{align*} where, in the case $i\in\left\{ i_{0},i_{1},...,i_{N+m}\right\} $, we denote by $j$ the integer $\ell$ satisfying $i_{\ell-1}=i$.\ \ \ \ \footnote{If you are wondering where the $-1$ (for example, in $i_{\ell-1}$ and in $i_{\left( j-1\right) -1}$) comes from: It comes from the fact that the indexing of our $N+m+1$-tuple $\left( v_{i_{0}},v_{i_{1}},...,v_{i_{N+m}}\right) $ begins with $0$, and not with $1$ as in Definition \ref{def.createdestroy.fin}.} Since $i\notin\left\{ i_{0},i_{1},...,i_{N+m}\right\} $ (because we are in Case 1), this simplifies to% \[ \overset{\vee}{v_{i}^{\left( N\right) }}\left( u\right) =0. \] On the other hand, combining $i\notin\left\{ -N-1,-N-2,-N-3,...\right\} $ (which is because $i\in\left\{ N,N-1,...,-N\right\} $) with $i\notin\left\{ i_{0},i_{1},...,i_{N+m}\right\} $ (which is because we are in Case 1), we obtain% \begin{align*} i & \notin\left\{ i_{0},i_{1},...,i_{N+m}\right\} \cup\left\{ -N-1,-N-2,-N-3,...\right\} \\ & =\left\{ i_{0},i_{1},...,i_{N+m},-N-1,-N-2,-N-3,...\right\} =\left\{ j_{0},j_{1},j_{2},...\right\} \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{0},i_{1},...,i_{N+m}% ,-N-1,-N-2,-N-3,...\right) =\left( j_{0},j_{1},j_{2},...\right) \right) . \end{align*} Now,% \begin{align*} \left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) & =\overset{\vee}{v_{i}}\left( j_{N}^{\left( m\right) }\left( u\right) \right) =\overset{\vee}{v_{i}}\left( v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge...\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }j_{N}^{\left( m\right) }\left( u\right) =v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\text{ by (\ref{pf.plu.inf.S.comp.b.0})}\right) \\ & =\left\{ \begin{array} [c]{l}% 0,\ \ \ \ \ \ \ \ \ \ \text{if }i\notin\left\{ j_{0},j_{1},j_{2},...\right\} ;\\ \left( -1\right) ^{j}v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}% \wedge...\wedge v_{j_{j-1}}\wedge v_{j_{j+1}}\wedge v_{j_{j+2}}\wedge ...,\ \ \ \ \ \ \ \ \ \ \text{if }i\in\left\{ j_{0},j_{1},j_{2},...\right\} \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\overset{\vee }{v_{i}}\right) , \end{align*} where, in the case $i\in\left\{ j_{0},j_{1},j_{2},...\right\} $, we denote by $j$ the integer $k$ satisfying $j_{k}=i$. Since $i\notin\left\{ j_{0},j_{1},j_{2},...\right\} $, this simplifies to% \[ \left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) =0. \] Compared with% \[ \left( j_{N}^{\left( m+1\right) }\circ\overset{\vee}{v_{i}^{\left( N\right) }}\right) \left( u\right) =j_{N}^{\left( m+1\right) }\underbrace{\left( \overset{\vee}{v_{i}^{\left( N\right) }}\left( u\right) \right) }_{=0}=0, \] this yields $\left( j_{N}^{\left( m+1\right) }\circ\overset{\vee }{v_{i}^{\left( N\right) }}\right) \left( u\right) =\left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) $. We have thus proven $\left( j_{N}^{\left( m+1\right) }% \circ\overset{\vee}{v_{i}^{\left( N\right) }}\right) \left( u\right) =\left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) $ in Case 1. Next, let us consider Case 2. In this case, $i\in\left\{ i_{0},i_{1}% ,...,i_{N+m}\right\} $, so there exists an $\ell\in\left\{ 0,1,...,N+m\right\} $ such that $i_{\ell}=i$. Denote this $\ell$ by $\kappa$. Then, $i_{\kappa}=i$. Clearly, \begin{align} & \left( i_{0},i_{1},...,i_{\kappa-1},i_{\kappa+1},i_{\kappa+2}% ,...,i_{N+m},-N-1,-N-2,-N-3,...\right) \nonumber\\ & =\left( \text{result of removing the }\kappa+1\text{-th term from the sequence} \phantom{\dfrac{\dfrac{I}{I}}{I}} \right. \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left. \underbrace{\left( i_{0},i_{1},...,i_{N+m},-N-1,-N-2,-N-3,...\right) }_{=\left( j_{0}% ,j_{1},j_{2},...\right) }\right) \nonumber\\ & =\left( \text{result of removing the }\kappa+1\text{-th term from the sequence }\left( j_{0},j_{1},j_{2},...\right) \right) \nonumber\\ & =\left( j_{0},j_{1},...,j_{\kappa-1},j_{\kappa+1},j_{\kappa+2},...\right) . \label{pf.plu.inf.S.comp.b.2.triv}% \end{align} From $u=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}$, we obtain% \begin{align*} & \overset{\vee}{v_{i}^{\left( N\right) }}\left( u\right) \\ & =\overset{\vee}{v_{i}^{\left( N\right) }}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\right) \\ & =\left\{ \begin{array} [c]{l}% 0,\ \ \ \ \ \ \ \ \ \ \text{if }i\notin\left\{ i_{0},i_{1},...,i_{N+m}% \right\} ;\\ \left( -1\right) ^{j-1}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\left( j-1\right) -1}}\wedge v_{i_{\left( j-1\right) +1}}\wedge v_{i_{\left( j-1\right) +2}}\wedge...\wedge v_{i_{N+m}}% ,\ \ \ \ \ \ \ \ \ \ \text{if }i\in\left\{ i_{0},i_{1},...,i_{N+m}\right\} \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\overset{\vee }{v_{i}^{\left( N\right) }}\right) , \end{align*} where, in the case $i\in\left\{ i_{0},i_{1},...,i_{N+m}\right\} $, we denote by $j$ the integer $\ell$ satisfying $i_{\ell-1}=i$.\ \ \ \ \footnote{If you are wondering where the $-1$ (for example, in $i_{\ell-1}$ and in $i_{\left( j-1\right) -1}$) comes from: It comes from the fact that the indexing of our $N+m+1$-tuple $\left( v_{i_{0}},v_{i_{1}},...,v_{i_{N+m}}\right) $ begins with $0$, and not with $1$ as in Definition \ref{def.createdestroy.fin}.} Since $i\in\left\{ i_{0},i_{1},...,i_{N+m}\right\} $, this simplifies to% \[ \overset{\vee}{v_{i}^{\left( N\right) }}\left( u\right) =\left( -1\right) ^{j-1}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\left( j-1\right) -1}}\wedge v_{i_{\left( j-1\right) +1}}\wedge v_{i_{\left( j-1\right) +2}}\wedge...\wedge v_{i_{N+m}}, \] where we denote by $j$ the integer $\ell$ satisfying $i_{\ell-1}=i$. Since the integer $\ell$ satisfying $i_{\ell-1}=i$ is $\kappa+1$ (because $i_{\left( \kappa+1\right) -1}=i_{\kappa}=i$), this rewrites as% \begin{align*} \overset{\vee}{v_{i}^{\left( N\right) }}\left( u\right) & =\left( -1\right) ^{\left( \kappa+1\right) -1}v_{i_{0}}\wedge v_{i_{1}}% \wedge...\wedge v_{i_{\left( \left( \kappa+1\right) -1\right) -1}}\wedge v_{i_{\left( \left( \kappa+1\right) -1\right) +1}}\wedge v_{i_{\left( \left( \kappa+1\right) -1\right) +2}}\wedge...\wedge v_{i_{N+m}}\\ & =\left( -1\right) ^{\kappa}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\kappa-1}}\wedge v_{i_{\kappa+1}}\wedge v_{i_{\kappa+2}}\wedge...\wedge v_{i_{N+m}}% \end{align*} (since $\left( \kappa+1\right) -1=\kappa$). Thus,% \begin{align} & \left( j_{N}^{\left( m+1\right) }\circ\overset{\vee}{v_{i}^{\left( N\right) }}\right) \left( u\right) \nonumber\\ & =j_{N}^{\left( m+1\right) }\left( \underbrace{\overset{\vee }{v_{i}^{\left( N\right) }}\left( u\right) }_{=\left( -1\right) ^{\kappa}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\kappa-1}}\wedge v_{i_{\kappa+1}}\wedge v_{i_{\kappa+2}}\wedge...\wedge v_{i_{N+m}}}\right) \nonumber\\ & =j_{N}^{\left( m+1\right) }\left( \left( -1\right) ^{\kappa}v_{i_{0}% }\wedge v_{i_{1}}\wedge...\wedge v_{i_{\kappa-1}}\wedge v_{i_{\kappa+1}}\wedge v_{i_{\kappa+2}}\wedge...\wedge v_{i_{N+m}}\right) \nonumber\\ & =\left( -1\right) ^{\kappa}v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{\kappa-1}}\wedge v_{i_{\kappa+1}}\wedge v_{i_{\kappa+2}}\wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }j_{N}^{\left( m+1\right) }\right) \nonumber\\ & =\left( -1\right) ^{\kappa}v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}% }\wedge...\wedge v_{j_{\kappa-1}}\wedge v_{j_{\kappa+1}}\wedge v_{j_{\kappa +2}}\wedge...\label{pf.plu.inf.S.comp.b.2}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since (\ref{pf.plu.inf.S.comp.b.2.triv}) yields}\\ \left( i_{0},i_{1},...,i_{\kappa-1},i_{\kappa+1},i_{\kappa+2},...,i_{N+m}% ,-N-1,-N-2,-N-3,...\right) \\ =\left( j_{0},j_{1},...,j_{\kappa-1},j_{\kappa+1},j_{\kappa+2},...\right) \end{array} \right) .\nonumber \end{align} On the other hand, \begin{align*} i & \in\left\{ i_{0},i_{1},...,i_{N+m}\right\} \subseteq\left\{ i_{0},i_{1},...,i_{N+m},-N-1,-N-2,-N-3,...\right\} =\left\{ j_{0}% ,j_{1},j_{2},...\right\} \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{0},i_{1},...,i_{N+m}% ,-N-1,-N-2,-N-3,...\right) =\left( j_{0},j_{1},j_{2},...\right) \right) . \end{align*} Moreover, the integer $k$ satisfying $j_{k}=i$ is $\kappa$% \ \ \ \ \footnote{because% \begin{align*} j_{\kappa} & =i_{\kappa}\ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{0},i_{1},...,i_{N+m},-N-1,-N-2,-N-3,...\right) =\left( j_{0},j_{1}% ,j_{2},...\right) \text{ and }\kappa\in\left\{ 0,1,...,N+m\right\} \right) \\ & =i \end{align*} }. Now, \begin{align*} \left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) & =\overset{\vee}{v_{i}}\left( j_{N}^{\left( m\right) }\left( u\right) \right) =\overset{\vee}{v_{i}}\left( v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge...\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }j_{N}^{\left( m\right) }\left( u\right) =v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\text{ by (\ref{pf.plu.inf.S.comp.b.0})}\right) \\ & =\left\{ \begin{array} [c]{l}% 0,\ \ \ \ \ \ \ \ \ \ \text{if }i\notin\left\{ j_{0},j_{1},j_{2},...\right\} ;\\ \left( -1\right) ^{j}v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}% \wedge...\wedge v_{j_{j-1}}\wedge v_{j_{j+1}}\wedge v_{j_{j+2}}\wedge ...,\ \ \ \ \ \ \ \ \ \ \text{if }i\in\left\{ j_{0},j_{1},j_{2},...\right\} \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\overset{\vee }{v_{i}}\right) , \end{align*} where, in the case $i\in\left\{ j_{0},j_{1},j_{2},...\right\} $, we denote by $j$ the integer $k$ satisfying $j_{k}=i$. Since $i\in\left\{ j_{0}% ,j_{1},j_{2},...\right\} $, this simplifies to% \[ \left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) =\left( -1\right) ^{j}v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}% }\wedge...\wedge v_{j_{j-1}}\wedge v_{j_{j+1}}\wedge v_{j_{j+2}}\wedge..., \] where we denote by $j$ the integer $k$ satisfying $j_{k}=i$. Since the integer $k$ satisfying $j_{k}=i$ is $\kappa$, this rewrites as% \[ \left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) =\left( -1\right) ^{\kappa}v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\wedge v_{j_{\kappa-1}}\wedge v_{j_{\kappa+1}}\wedge v_{j_{\kappa+2}}\wedge.... \] Compared with (\ref{pf.plu.inf.S.comp.b.2}), this yields $\left( j_{N}^{\left( m+1\right) }\circ\widehat{v_{i}^{\left( N\right) }}\right) \left( u\right) =\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) $. This is exactly what we needed to prove in order to complete the proof of (\ref{pf.plu.inf.S.comp.b}). The proof of (\ref{pf.plu.inf.S.comp.b}) is thus finished. \textbf{c)} Let us next show that% \begin{equation} \widehat{v_{i}}\circ j_{N}^{\left( m\right) }=0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{ -N-1,-N-2,-N-3,...\right\} . \label{pf.plu.inf.S.comp.c}% \end{equation} \textit{Proof of (\ref{pf.plu.inf.S.comp.c}):} Let $i\in\left\{ -N-1,-N-2,-N-3,...\right\} $. In order to prove (\ref{pf.plu.inf.S.comp.c}), it is clearly enough to show that $\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) =0$ for every $u\in\wedge^{N+m+1}\left( V_{N}\right) $. So let $u$ be any element of $\wedge^{N+m+1}\left( V_{N}\right) $. We must prove the equality $\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) =0$. Since this equality is linear in $u$, we can WLOG assume that $u$ is an element of the basis $\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\right) _{N\geq i_{0}>i_{1}% >...>i_{N+m}\geq-N}$ of $\wedge^{N+m+1}\left( V_{N}\right) $. Assume this. Then, there exists an $N+m+1$-tuple $\left( i_{0},i_{1},...,i_{N+m}\right) $ of integers such that $N\geq i_{0}>i_{1}>...>i_{N+m}\geq-N$ and $u=v_{i_{0}% }\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}$. Consider this $N+m+1$-tuple. The vector $v_{i}$ occurs twice in the semiinfinite wedge $v_{i}\wedge v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...$ (namely, it occurs once in the very beginning of this wedge, and then it occurs again in the $v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...$ part (because $i\in\left\{ -N-1,-N-2,-N-3,...\right\} $)). Hence, the semiinfinite wedge $v_{i}\wedge v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...$ equals $0$ (since a semiinfinite wedge in which a vector occurs more than once must always be equal to $0$). Now,% \begin{align*} \left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) & =\widehat{v_{i}}\left( j_{N}^{\left( m\right) }\left( u\right) \right) =v_{i}\wedge j_{N}^{\left( m\right) }\left( \underbrace{u}_{=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\widehat{v_{i}}\right) \\ & =v_{i}\wedge\underbrace{j_{N}^{\left( m\right) }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\right) }_{\substack{=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...\\\text{(by the definition of }j_{N}^{\left( m\right) }\text{)}}}\\ & =v_{i}\wedge v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...\\ & =0\ \ \ \ \ \ \ \ \ \ \left( \text{as we proved above}\right) . \end{align*} This is exactly what we needed to prove in order to complete the proof of (\ref{pf.plu.inf.S.comp.c}). The proof of (\ref{pf.plu.inf.S.comp.c}) is thus finished. \textbf{d)} Let us now show that \begin{equation} \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }% =0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{ N+1,N+2,N+3,...\right\} . \label{pf.plu.inf.S.comp.d}% \end{equation} \textit{Proof of (\ref{pf.plu.inf.S.comp.d}):} Let $i\in\left\{ N+1,N+2,N+3,...\right\} $. In order to prove (\ref{pf.plu.inf.S.comp.b}), it is clearly enough to show that $\left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) =0$ for every $u\in \wedge^{N+m+1}\left( V_{N}\right) $. So let $u$ be any element of $\wedge^{N+m+1}\left( V_{N}\right) $. We must prove the equality $\left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) =0$. Since this equality is linear in $u$, we can WLOG assume that $u$ is an element of the basis $\left( v_{i_{0}% }\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\right) _{N\geq i_{0}% >i_{1}>...>i_{N+m}\geq-N}$ of $\wedge^{N+m+1}\left( V_{N}\right) $. Assume this. Then, there exists an $N+m+1$-tuple $\left( i_{0},i_{1},...,i_{N+m}% \right) $ of integers such that $N\geq i_{0}>i_{1}>...>i_{N+m}\geq-N$ and $u=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}$. Consider this $N+m+1$-tuple. Notice that $i\in\left\{ N+1,N+2,N+3,...\right\} $, so that $i\notin\left\{ N,N-1,...,-N\right\} $ and $i\notin\left\{ N,N-1,N-2,...\right\} $. Since $N\geq i_{0}>i_{1}>...>i_{N+m}\geq-N$, we have $\left\{ i_{0}% ,i_{1},...,i_{N+m}\right\} \subseteq\left\{ N,N-1,...,-N\right\} $ and thus $i\notin\left\{ i_{0},i_{1},...,i_{N+m}\right\} $ (because $i\notin\left\{ N,N-1,...,-N\right\} $). Let $\left( j_{0},j_{1},j_{2},...\right) $ be the sequence $\left( i_{0},i_{1},...,i_{N+m},-N-1,-N-2,-N-3,...\right) $. Then,% \begin{align*} \left\{ j_{0},j_{1},j_{2},...\right\} & =\left\{ i_{0},i_{1}% ,...,i_{N+m},-N-1,-N-2,-N-3,...\right\} \\ & =\underbrace{\left\{ i_{0},i_{1},...,i_{N+m}\right\} }_{\subseteq\left\{ N,N-1,...,-N\right\} }\cup\left\{ -N-1,-N-2,-N-3,...\right\} \\ & \subseteq\left\{ N,N-1,...,-N\right\} \cup\left\{ -N-1,-N-2,-N-3,...\right\} =\left\{ N,N-1,N-2,...\right\} . \end{align*} Thus, $i\notin\left\{ j_{0},j_{1},j_{2},...\right\} $ (since $i\notin% \left\{ N,N-1,N-2,...\right\} $). From $u=v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}$, we obtain% \begin{align} j_{N}^{\left( m\right) }\left( u\right) & =j_{N}^{\left( m\right) }\left( v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\right) =v_{i_{0}}\wedge v_{i_{1}}\wedge...\wedge v_{i_{N+m}}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }j_{N}^{\left( m\right) }\right) \nonumber\\ & =v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( i_{0},i_{1},...,i_{N+m}% ,-N-1,-N-2,-N-3,...\right) =\left( j_{0},j_{1},j_{2},...\right) \right) , \label{pf.plu.inf.S.comp.d.0}% \end{align} so that% \begin{align*} \left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) & =\overset{\vee}{v_{i}}\left( j_{N}^{\left( m\right) }\left( u\right) \right) =\overset{\vee}{v_{i}}\left( v_{j_{0}}\wedge v_{j_{1}% }\wedge v_{j_{2}}\wedge...\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }j_{N}^{\left( m\right) }\left( u\right) =v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}\wedge...\text{ by (\ref{pf.plu.inf.S.comp.d.0})}\right) \\ & =\left\{ \begin{array} [c]{l}% 0,\ \ \ \ \ \ \ \ \ \ \text{if }i\notin\left\{ j_{0},j_{1},j_{2},...\right\} ;\\ \left( -1\right) ^{j}v_{j_{0}}\wedge v_{j_{1}}\wedge v_{j_{2}}% \wedge...\wedge v_{j_{j-1}}\wedge v_{j_{j+1}}\wedge v_{j_{j+2}}\wedge ...,\ \ \ \ \ \ \ \ \ \ \text{if }i\in\left\{ j_{0},j_{1},j_{2},...\right\} \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\overset{\vee }{v_{i}}\right) , \end{align*} where, in the case $i\in\left\{ j_{0},j_{1},j_{2},...\right\} $, we denote by $j$ the integer $k$ satisfying $j_{k}=i$. Since $i\notin\left\{ j_{0},j_{1},j_{2},...\right\} $, this simplifies to $\left( \overset{\vee }{v_{i}}\circ j_{N}^{\left( m\right) }\right) \left( u\right) =0$. This is exactly what we needed to prove in order to complete the proof of (\ref{pf.plu.inf.S.comp.d}). The proof of (\ref{pf.plu.inf.S.comp.d}) is thus finished. \textbf{e)} Now it is the time to draw conclusions. We have $S=\sum\limits_{i\in\mathbb{Z}}\widehat{v_{i}}\otimes\overset{\vee }{v_{i}}$ (by the definition of $S$). Thus,% \begin{align*} & S\circ\left( j_{N}^{\left( m\right) }\otimes j_{N}^{\left( m\right) }\right) =\left( \sum\limits_{i\in\mathbb{Z}}\widehat{v_{i}}\otimes \overset{\vee}{v_{i}}\right) \circ\left( j_{N}^{\left( m\right) }\otimes j_{N}^{\left( m\right) }\right) =\sum\limits_{i\in\mathbb{Z}}% \underbrace{\left( \widehat{v_{i}}\otimes\overset{\vee}{v_{i}}\right) \circ\left( j_{N}^{\left( m\right) }\otimes j_{N}^{\left( m\right) }\right) }_{=\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \otimes\left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) }\\ & =\sum\limits_{i\in\mathbb{Z}}\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \otimes\left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) \\ & =\sum\limits_{i=-\infty}^{-N-1}\underbrace{\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) }_{\substack{=0\\\text{(by (\ref{pf.plu.inf.S.comp.c}))}}}\otimes\left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) +\sum\limits_{i=-N}^{N}\underbrace{\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) }_{\substack{=j_{N}% ^{\left( m+1\right) }\circ\widehat{v_{i}^{\left( N\right) }}\\\text{(by (\ref{pf.plu.inf.S.comp.a}))}}}\otimes\underbrace{\left( \overset{\vee }{v_{i}}\circ j_{N}^{\left( m\right) }\right) }_{\substack{=j_{N}^{\left( m+1\right) }\circ\overset{\vee}{v_{i}^{\left( N\right) }}\\\text{(by (\ref{pf.plu.inf.S.comp.b}))}}}+\sum\limits_{i=N+1}^{\infty}\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \otimes \underbrace{\left( \overset{\vee}{v_{i}}\circ j_{N}^{\left( m\right) }\right) }_{\substack{=0\\\text{(by (\ref{pf.plu.inf.S.comp.d}))}}}\\ & =\underbrace{\sum\limits_{i=-\infty}^{-N-1}0\otimes\left( \overset{\vee }{v_{i}}\circ j_{N}^{\left( m\right) }\right) }_{=0}+\sum\limits_{i=-N}% ^{N}\underbrace{\left( j_{N}^{\left( m+1\right) }\circ\widehat{v_{i}% ^{\left( N\right) }}\right) \otimes\left( j_{N}^{\left( m+1\right) }\circ\overset{\vee}{v_{i}^{\left( N\right) }}\right) }_{=\left( j_{N}^{\left( m+1\right) }\otimes j_{N}^{\left( m-1\right) }\right) \circ\left( \widehat{v_{i}^{\left( N\right) }}\otimes\overset{\vee }{v_{i}^{\left( N\right) }}\right) }+\underbrace{\sum\limits_{i=N+1}% ^{\infty}\left( \widehat{v_{i}}\circ j_{N}^{\left( m\right) }\right) \otimes0}_{=0}\\ & =\sum\limits_{i=-N}^{N}\left( j_{N}^{\left( m+1\right) }\otimes j_{N}^{\left( m-1\right) }\right) \circ\left( \widehat{v_{i}^{\left( N\right) }}\otimes\overset{\vee}{v_{i}^{\left( N\right) }}\right) =\left( j_{N}^{\left( m+1\right) }\otimes j_{N}^{\left( m-1\right) }\right) \circ\left( \sum\limits_{i=-N}^{N}\widehat{v_{i}^{\left( N\right) }}% \otimes\overset{\vee}{v_{i}^{\left( N\right) }}\right) . \end{align*} But since $S_{N}^{\left( N+m+1\right) }=\sum\limits_{i=-N}^{N}% \widehat{v_{i}^{\left( N\right) }}\otimes\overset{\vee}{v_{i}^{\left( N\right) }}$ (by the definition of $S_{N}^{\left( N+m+1\right) }$), this rewrites as% \[ S\circ\left( j_{N}^{\left( m\right) }\otimes j_{N}^{\left( m\right) }\right) =\left( j_{N}^{\left( m+1\right) }\otimes j_{N}^{\left( m-1\right) }\right) \circ\underbrace{\left( \sum\limits_{i=-N}% ^{N}\widehat{v_{i}^{\left( N\right) }}\otimes\overset{\vee}{v_{i}^{\left( N\right) }}\right) }_{=S_{N}^{\left( N+m+1\right) }}=\left( j_{N}^{\left( m+1\right) }\otimes j_{N}^{\left( m-1\right) }\right) \circ S_{N}^{\left( N+m+1\right) }. \] This proves Lemma \ref{lem.plu.inf.S.comp}. Now we can finally come to proving Theorem \ref{thm.plu.inf}: \textit{Proof of Theorem \ref{thm.plu.inf}.} Let $\varrho^{\prime }:\operatorname*{M}\left( \infty\right) \rightarrow\operatorname*{End}% \left( \mathcal{F}\otimes\mathcal{F}\right) $ be the action of the monoid $\operatorname*{M}\left( \infty\right) $ on the tensor product of the $\operatorname*{M}\left( \infty\right) $-module $\mathcal{F}$ with itself. Clearly,% \[ \varrho^{\prime}\left( M\right) =\varrho\left( M\right) \otimes \varrho\left( M\right) \ \ \ \ \ \ \ \ \ \ \text{for every }M\in \operatorname*{M}\left( \infty\right) \] (because this is how one defines the tensor product of two modules over a monoid). \textbf{(c)} Let $m\in\mathbb{Z}$. Let $M\in\operatorname*{M}\left( \infty\right) $. Let $v\in\mathcal{F}^{\left( m\right) }$ and $w\in\mathcal{F}^{\left( m\right) }$. We are going to prove that $\left( S\circ\varrho^{\prime}\left( M\right) \right) \left( v\otimes w\right) =\left( \varrho^{\prime}\left( M\right) \circ S\right) \left( v\otimes w\right) $. Since $M\in\operatorname*{M}\left( \infty\right) =\bigcup\limits_{N\in \mathbb{N}}i_{N}\left( \operatorname*{M}\left( V_{N}\right) \right) $ (by Remark \ref{rmk.plu.inf.iN} \textbf{(c)}), there exists an $R\in\mathbb{N}$ such that $M\in i_{R}\left( \operatorname*{M}\left( V_{R}\right) \right) $. Consider this $R$. Since $v\in\mathcal{F}^{\left( m\right) }=\bigcup\limits_{\substack{N\in \mathbb{N};\\N\geq R}}j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) $ (by Proposition \ref{prop.plu.inf.cover} \textbf{(b)}, applied to $Q=R$), there exists some $T\in\mathbb{N}$ such that $T\geq R$ and $v\in j_{T}^{\left( m\right) }\left( \wedge^{T+m+1}\left( V_{T}\right) \right) $. Consider this $T$. Since $w\in\mathcal{F}^{\left( m\right) }=\bigcup\limits_{\substack{N\in \mathbb{N};\\N\geq T}}j_{N}^{\left( m\right) }\left( \wedge^{N+m+1}\left( V_{N}\right) \right) $ (by Proposition \ref{prop.plu.inf.cover} \textbf{(b)}, applied to $Q=T$), there exists some $P\in\mathbb{N}$ such that $P\geq T$ and $w\in j_{P}^{\left( m\right) }\left( \wedge^{P+m+1}\left( V_{P}\right) \right) $. Consider this $P$. There exists a $w^{\prime}% \in\wedge^{P+m+1}\left( V_{P}\right) $ such that $w=j_{P}^{\left( m\right) }\left( w^{\prime}\right) $ (because $w\in j_{P}^{\left( m\right) }\left( \wedge^{P+m+1}\left( V_{P}\right) \right) $). Consider this $w^{\prime}$. Applying Proposition \ref{prop.plu.inf.cover} \textbf{(a)}, we get $j_{0}^{\left( m\right) }\left( \wedge^{0+m+1}\left( V_{0}\right) \right) \subseteq j_{1}^{\left( m\right) }\left( \wedge^{1+m+1}\left( V_{1}\right) \right) \subseteq j_{2}^{\left( m\right) }\left( \wedge^{2+m+1}\left( V_{2}\right) \right) \subseteq...$. Thus, $j_{T}^{\left( m\right) }\left( \wedge^{T+m+1}\left( V_{T}\right) \right) \subseteq j_{P}^{\left( m\right) }\left( \wedge^{P+m+1}\left( V_{P}\right) \right) $ (since $T\leq P$), so that $v\in j_{T}^{\left( m\right) }\left( \wedge^{T+m+1}\left( V_{T}\right) \right) \subseteq j_{P}^{\left( m\right) }\left( \wedge^{P+m+1}\left( V_{P}\right) \right) $. Hence, there exists a $v^{\prime}\in\wedge^{P+m+1}\left( V_{P}\right) $ such that $v=j_{P}^{\left( m\right) }\left( v^{\prime}\right) $. Consider this $v^{\prime}$. Since $v=j_{P}^{\left( m\right) }\left( v^{\prime }\right) $ and $w=j_{P}^{\left( m\right) }\left( w^{\prime}\right) $, we have% \begin{equation} v\otimes w=j_{P}^{\left( m\right) }\left( v^{\prime}\right) \otimes j_{P}^{\left( m\right) }\left( w^{\prime}\right) =\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) \left( v^{\prime }\otimes w^{\prime}\right) . \label{pf.plu.inf.-20}% \end{equation} Since $R\leq T\leq P$, we have $i_{R}\left( \operatorname*{M}\left( V_{R}\right) \right) \subseteq i_{P}\left( \operatorname*{M}\left( V_{P}\right) \right) $ (since Remark \ref{rmk.plu.inf.iN} \textbf{(b)} yields $i_{0}\left( \operatorname*{M}\left( V_{0}\right) \right) \subseteq i_{1}\left( \operatorname*{M}\left( V_{1}\right) \right) \subseteq i_{2}\left( \operatorname*{M}\left( V_{2}\right) \right) \subseteq...$). Thus, $M\in i_{R}\left( \operatorname*{M}\left( V_{R}\right) \right) \subseteq i_{P}\left( \operatorname*{M}\left( V_{P}\right) \right) $. In other words, there exists an $A\in\operatorname*{M}\left( V_{P}\right) $ such that $M=i_{P}\left( A\right) $. Consider this $A$. In the following, we will write the action of $\operatorname*{M}\left( \infty\right) $ on $\mathcal{F}$ as a left action. In other words, we will abbreviate $\left( \varrho\left( N\right) \right) u$ by $Nu$, wherever $N\in\operatorname*{M}\left( \infty\right) $ and $u\in\mathcal{F}$. Similarly, we will write the action of $\operatorname*{M}\left( \infty\right) $ on $\mathcal{F}\otimes\mathcal{F}$ (this action is obtained by tensoring the $\operatorname*{M}\left( \infty\right) $-module $\mathcal{F}$ with itself); this action satisfies $\varrho^{\prime}\left( A\right) =\varrho\left( A\right) \otimes\varrho\left( A\right) $. Let us also denote by $\varrho$ the action of the monoid $\operatorname*{M}% \left( V_{N}\right) $ on $\wedge\left( V_{N}\right) $. Moreover, let us denote by $\varrho^{\prime}$ the action of the monoid $\operatorname*{M}% \left( V_{N}\right) $ on $\wedge\left( V_{N}\right) \otimes\wedge\left( V_{N}\right) $ (this action is obtained by tensoring the $\operatorname*{M}% \left( V_{N}\right) $-module $\wedge\left( V_{N}\right) $ with itself). We notice that every $\ell\in\mathbb{Z}$ satisfies% \begin{equation} \left( \varrho\left( M\right) \right) \circ j_{P}^{\left( \ell\right) }=j_{P}^{\left( \ell\right) }\circ\left( \varrho\left( A\right) \right) . \label{pf.plu.inf.-10}% \end{equation} \footnote{\textit{Proof of (\ref{pf.plu.inf.-10}):} Let $\ell\in\mathbb{Z}$. Every $u\in\mathcal{F}^{\left( \ell\right) }$ satisfies% \begin{align*} \left( \left( \varrho\left( M\right) \right) \circ j_{P}^{\left( \ell\right) }\right) \left( u\right) & =\left( \varrho\left( M\right) \right) \left( j_{P}^{\left( \ell\right) }\left( u\right) \right) =\underbrace{M}_{=i_{P}\left( A\right) }\cdot j_{P}^{\left( \ell\right) }\left( u\right) =i_{P}\left( A\right) \cdot j_{P}^{\left( \ell\right) }u=j_{P}^{\left( \ell\right) }\underbrace{\left( Au\right) }_{=\left( \varrho\left( A\right) \right) u}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by Proposition \ref{prop.plu.inf.iNjmN}, applied to }P\text{ and }\ell\text{ instead of }N\text{ and }m\right) \\ & =j_{P}^{\left( \ell\right) }\left( \left( \varrho\left( A\right) \right) u\right) =\left( j_{P}^{\left( \ell\right) }\circ\left( \varrho\left( A\right) \right) \right) \left( u\right) . \end{align*} Thus, $\left( \varrho\left( M\right) \right) \circ j_{P}^{\left( \ell\right) }=j_{P}^{\left( \ell\right) }\circ\left( \varrho\left( A\right) \right) $, so that (\ref{pf.plu.inf.-10}) is proven.} Applying Lemma \ref{lem.plu.inf.S.comp} to $N=P$, we obtain \begin{equation} \left( j_{P}^{\left( m+1\right) }\otimes j_{P}^{\left( m-1\right) }\right) \circ S_{P}^{\left( P+m+1\right) }=S\circ\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) . \label{pf.plu.inf.1}% \end{equation} On the other hand, the map $S_{P}^{\left( P+m+1\right) }$ is $\operatorname*{M}\left( \infty\right) $-invariant (by Lemma \ref{lem.plu.inf.plu} \textbf{(c)}, applied to $N=P$ and $k=P+m+1$), so that% \[ S_{P}^{\left( P+m+1\right) }\circ\left( \varrho^{\prime}\left( A\right) \right) =\left( \varrho^{\prime}\left( A\right) \right) \circ S_{P}^{\left( P+m+1\right) }. \] Since $\varrho^{\prime}\left( A\right) =\varrho\left( A\right) \otimes\varrho\left( A\right) $, this rewrites as \begin{equation} S_{P}^{\left( P+m+1\right) }\circ\left( \varrho\left( A\right) \otimes\varrho\left( A\right) \right) =\left( \varrho\left( A\right) \otimes\varrho\left( A\right) \right) \circ S_{P}^{\left( P+m+1\right) }. \label{pf.plu.inf.2}% \end{equation} Comparing% \begin{align*} & S\circ\underbrace{\left( \varrho^{\prime}\left( M\right) \right) }_{=\varrho\left( M\right) \otimes\varrho\left( M\right) }\circ\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) \\ & =S\circ\underbrace{\left( \varrho\left( M\right) \otimes\varrho\left( M\right) \right) \circ\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) }_{=\left( \left( \varrho\left( M\right) \right) \circ j_{P}^{\left( m\right) }\right) \otimes\left( \left( \varrho\left( M\right) \right) \circ j_{P}^{\left( m\right) }\right) }\\ & =S\circ\left( \underbrace{\left( \left( \varrho\left( M\right) \right) \circ j_{P}^{\left( m\right) }\right) }_{\substack{=j_{P}^{\left( m\right) }\circ\left( \varrho\left( A\right) \right) \\\text{(by (\ref{pf.plu.inf.-10}), applied to }\ell=m\text{)}}}\otimes\underbrace{\left( \left( \varrho\left( M\right) \right) \circ j_{P}^{\left( m\right) }\right) }_{\substack{=j_{P}^{\left( m\right) }\circ\left( \varrho\left( A\right) \right) \\\text{(by (\ref{pf.plu.inf.-10}), applied to }% \ell=m\text{)}}}\right) \\ & =S\circ\underbrace{\left( \left( j_{P}^{\left( m\right) }\circ\left( \varrho\left( A\right) \right) \right) \otimes\left( j_{P}^{\left( m\right) }\circ\left( \varrho\left( A\right) \right) \right) \right) }_{=\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) \circ\left( \varrho\left( A\right) \otimes\varrho\left( A\right) \right) }=\underbrace{S\circ\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) }_{\substack{=\left( j_{P}^{\left( m+1\right) }\otimes j_{P}^{\left( m-1\right) }\right) \circ S_{P}^{\left( P+m+1\right) }\\\text{(by (\ref{pf.plu.inf.1}))}}}\circ\left( \varrho\left( A\right) \otimes\varrho\left( A\right) \right) \\ & =\left( j_{P}^{\left( m+1\right) }\otimes j_{P}^{\left( m-1\right) }\right) \circ\underbrace{S_{P}^{\left( P+m+1\right) }\circ\left( \varrho\left( A\right) \otimes\varrho\left( A\right) \right) }_{\substack{=\left( \varrho\left( A\right) \otimes\varrho\left( A\right) \right) \circ S_{P}^{\left( P+m+1\right) }\\\text{(by (\ref{pf.plu.inf.2}% ))}}}=\left( j_{P}^{\left( m+1\right) }\otimes j_{P}^{\left( m-1\right) }\right) \circ\left( \varrho\left( A\right) \otimes\varrho\left( A\right) \right) \circ S_{P}^{\left( P+m+1\right) }% \end{align*} with% \begin{align*} & \underbrace{\left( \varrho^{\prime}\left( M\right) \right) }% _{=\varrho\left( M\right) \otimes\varrho\left( M\right) }\circ \underbrace{S\circ\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) }_{\substack{=\left( j_{P}^{\left( m+1\right) }\otimes j_{P}^{\left( m-1\right) }\right) \circ S_{P}^{\left( P+m+1\right) }\\\text{(by (\ref{pf.plu.inf.1}))}}}\\ & =\underbrace{\left( \varrho\left( M\right) \otimes\varrho\left( M\right) \right) \circ\left( j_{P}^{\left( m+1\right) }\otimes j_{P}^{\left( m-1\right) }\right) }_{=\left( \left( \varrho\left( M\right) \right) \circ j_{P}^{\left( m+1\right) }\right) \otimes\left( \left( \varrho\left( M\right) \right) \circ j_{P}^{\left( m-1\right) }\right) }\circ S_{P}^{\left( P+m+1\right) }\\ & =\left( \underbrace{\left( \left( \varrho\left( M\right) \right) \circ j_{P}^{\left( m+1\right) }\right) }_{\substack{=j_{P}^{\left( m+1\right) }\circ\left( \varrho\left( A\right) \right) \\\text{(by (\ref{pf.plu.inf.-10}), applied to }\ell=m+1\text{)}}}\otimes \underbrace{\left( \left( \varrho\left( M\right) \right) \circ j_{P}^{\left( m-1\right) }\right) }_{\substack{=j_{P}^{\left( m-1\right) }\circ\left( \varrho\left( A\right) \right) \\\text{(by (\ref{pf.plu.inf.-10}), applied to }\ell=m-1\text{)}}}\right) \circ S_{P}^{\left( P+m+1\right) }\\ & =\underbrace{\left( \left( j_{P}^{\left( m+1\right) }\circ\left( \varrho\left( A\right) \right) \right) \otimes\left( j_{P}^{\left( m-1\right) }\circ\left( \varrho\left( A\right) \right) \right) \right) }_{=\left( j_{P}^{\left( m+1\right) }\otimes j_{P}^{\left( m-1\right) }\right) \circ\left( \varrho\left( A\right) \otimes\varrho\left( A\right) \right) }\circ S_{P}^{\left( P+m+1\right) }\\ & =\left( j_{P}^{\left( m+1\right) }\otimes j_{P}^{\left( m-1\right) }\right) \circ\left( \varrho\left( A\right) \otimes\varrho\left( A\right) \right) \circ S_{P}^{\left( P+m+1\right) }, \end{align*} we obtain% \begin{equation} S\circ\left( \varrho^{\prime}\left( M\right) \right) \circ\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) =\left( \varrho^{\prime}\left( M\right) \right) \circ S\circ\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) . \label{pf.plu.inf.14}% \end{equation} Now, \begin{align*} & \left( S\circ\left( \varrho^{\prime}\left( M\right) \right) \right) \underbrace{\left( v\otimes w\right) }_{\substack{=\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) \left( v^{\prime }\otimes w^{\prime}\right) \\\text{(by (\ref{pf.plu.inf.-20}))}}}\\ & =\left( S\circ\left( \varrho^{\prime}\left( M\right) \right) \right) \left( \left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) \left( v^{\prime}\otimes w^{\prime}\right) \right) =\underbrace{\left( S\circ\left( \varrho^{\prime}\left( M\right) \right) \circ\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) \right) }_{\substack{=\left( \varrho^{\prime}\left( M\right) \right) \circ S\circ\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) \\\text{(by (\ref{pf.plu.inf.14}))}}}\left( v^{\prime }\otimes w^{\prime}\right) \\ & =\left( \left( \varrho^{\prime}\left( M\right) \right) \circ S\circ\left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) \right) \left( v^{\prime}\otimes w^{\prime}\right) =\left( \left( \varrho^{\prime}\left( M\right) \right) \circ S\right) \underbrace{\left( \left( j_{P}^{\left( m\right) }\otimes j_{P}^{\left( m\right) }\right) \left( v^{\prime}\otimes w^{\prime}\right) \right) }_{\substack{=v\otimes w\\\text{(by (\ref{pf.plu.inf.-20}))}}}\\ & =\left( \left( \varrho^{\prime}\left( M\right) \right) \circ S\right) \left( v\otimes w\right) . \end{align*} Now forget that we fixed $v$ and $w$. We thus have proven that $\left( S\circ\varrho^{\prime}\left( M\right) \right) \left( v\otimes w\right) =\left( \varrho^{\prime}\left( M\right) \circ S\right) \left( v\otimes w\right) $ for every $v\in\mathcal{F}^{\left( m\right) }$ and $w\in\mathcal{F}^{\left( m\right) }$. In other words, the two maps $S\circ\varrho^{\prime}\left( M\right) $ and $\varrho^{\prime}\left( M\right) \circ S$ are equal to each other on every pure tensor in $\mathcal{F}^{\left( m\right) }\otimes\mathcal{F}^{\left( m\right) }$. Thus, these two maps must be identical (on $\mathcal{F}^{\left( m\right) }\otimes\mathcal{F}^{\left( m\right) }$). In other words, $S\circ \varrho^{\prime}\left( M\right) =\varrho^{\prime}\left( M\right) \circ S$. Now forget that we fixed $M$. We have proven that $S\circ\varrho^{\prime }\left( M\right) =\varrho^{\prime}\left( M\right) \circ S$ for every $M\in\operatorname*{M}\left( \infty\right) $. In other words, $S$ is $\operatorname*{M}\left( \infty\right) $-invariant. This proves Theorem \ref{thm.plu.inf} \textbf{(c)}. \textbf{(a)} Theorem \ref{thm.plu.inf} \textbf{(a)} follows from Theorem \ref{thm.plu.inf} \textbf{(c)} since $\operatorname*{GL}\left( \infty\right) \subseteq\operatorname*{M}\left( \infty\right) $. \textbf{(b)} $\Longrightarrow:$ Assume that $\tau\in\Omega$. We want to prove that $S\left( \tau\otimes\tau\right) =0$. Since $\Omega=\operatorname*{GL}\left( \infty\right) \cdot\psi_{0}$, we have $\tau\in\Omega=\operatorname*{GL}\left( \infty\right) \cdot\psi_{0}$. In other words, there exists $A\in\operatorname*{GL}\left( \infty\right) $ such that $\tau=A\psi_{0}$. Consider this $A$. It is easy to see that% \begin{equation} \overset{\vee}{v_{i}}\left( \psi_{0}\right) =0\ \ \ \ \ \ \ \ \ \ \text{for every integer }i>0. \label{pf.plu.inf.b1}% \end{equation} \footnote{\textit{Proof of (\ref{pf.plu.inf.b1}):} Let $i>0$ be an integer. Then,% \begin{align*} \overset{\vee}{v_{i}}\left( \psi_{0}\right) & =\overset{\vee}{v_{i}% }\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }\psi_{0}=v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) \\ & =\left\{ \begin{array} [c]{l}% 0,\ \ \ \ \ \ \ \ \ \ \text{if }i\notin\left\{ 0,-1,-2,...\right\} ;\\ \left( -1\right) ^{j}v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\wedge v_{-\left( j-1\right) }\wedge v_{-\left( j+1\right) }\wedge v_{-\left( j+2\right) }\wedge...,\ \ \ \ \ \ \ \ \ \ \text{if }i\in\left\{ 0,-1,-2,...\right\} \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\overset{\vee }{v_{i}}\right) , \end{align*} where, in the case $i\in\left\{ 0,-1,-2,...\right\} $, we denote by $j$ the integer $k$ satisfying $-k=i$. Since $i\notin\left\{ 0,-1,-2,...\right\} $ (because $i>0$), this simplifies to $\overset{\vee}{v_{i}}\left( \psi _{0}\right) =0$. This proves (\ref{pf.plu.inf.b1}).} Also,% \begin{equation} \widehat{v_{i}}\left( \psi_{0}\right) =0\ \ \ \ \ \ \ \ \ \ \text{for every integer }i\leq0. \label{pf.plu.inf.b2}% \end{equation} \footnote{\textit{Proof of (\ref{pf.plu.inf.b2}):} Let $i\leq0$ be an integer. Since $\psi_{0}=v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...$, we have% \[ \widehat{v_{i}}\left( \psi_{0}\right) =\widehat{v_{i}}\left( v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...\right) =v_{i}\wedge v_{0}\wedge v_{-1}\wedge v_{-2}\wedge... \] (by the definition of $\widehat{v_{i}}$). But the semiinfinite wedge $v_{i}\wedge v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...$ contains the vector $v_{i}$ twice (in fact, it contains the vector $v_{i}$ once in its very beginning, and once again in its $v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...$ part (since $i\leq0$)), and thus must equal $0$ (since any semiinfinite wedge which contains a vector more than once must equal $0$). We thus have \[ \widehat{v_{i}}\left( \psi_{0}\right) =v_{i}\wedge v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...=0. \] This proves (\ref{pf.plu.inf.b2}).} Since $S=\sum\limits_{i\in\mathbb{Z}}\widehat{v_{i}}\otimes\overset{\vee }{v_{i}}$, we have% \begin{align*} S\left( \psi_{0}\otimes\psi_{0}\right) & =\sum\limits_{i\in\mathbb{Z}% }\underbrace{\left( \widehat{v_{i}}\otimes\overset{\vee}{v_{i}}\right) \left( \psi_{0}\otimes\psi_{0}\right) }_{=\widehat{v_{i}}\left( \psi _{0}\right) \otimes\overset{\vee}{v_{i}}\left( \psi_{0}\right) }% =\sum\limits_{i\in\mathbb{Z}}\widehat{v_{i}}\left( \psi_{0}\right) \otimes\overset{\vee}{v_{i}}\left( \psi_{0}\right) \\ & =\sum\limits_{\substack{i\in\mathbb{Z};\\i\leq0}}\underbrace{\widehat{v_{i}% }\left( \psi_{0}\right) }_{\substack{=0\\\text{(by (\ref{pf.plu.inf.b2}))}% }}\otimes\overset{\vee}{v_{i}}\left( \psi_{0}\right) +\sum \limits_{\substack{i\in\mathbb{Z};\\i>0}}\widehat{v_{i}}\left( \psi _{0}\right) \otimes\underbrace{\overset{\vee}{v_{i}}\left( \psi_{0}\right) }_{\substack{=0\\\text{(by (\ref{pf.plu.inf.b1}))}}}\\ & =\underbrace{\sum\limits_{\substack{i\in\mathbb{Z};\\i\leq0}}0\otimes \overset{\vee}{v_{i}}\left( \psi_{0}\right) }_{=0}+\underbrace{\sum \limits_{\substack{i\in\mathbb{Z};\\i>0}}\widehat{v_{i}}\left( \psi _{0}\right) \otimes0}_{=0}=0. \end{align*} Now, since $\tau=A\psi_{0}$, we have $\tau\otimes\tau=A\psi_{0}\otimes A\psi_{0}=A\left( \psi_{0}\otimes\psi_{0}\right) $, so that% \begin{align*} S\left( \tau\otimes\tau\right) & =S\left( A\left( \psi_{0}\otimes \psi_{0}\right) \right) \\ & =A\cdot\underbrace{S\left( \psi_{0}\otimes\psi_{0}\right) }% _{=0}\ \ \ \ \ \ \ \ \ \ \left( \text{since }S\text{ is }\operatorname*{M}% \left( \infty\right) \text{-linear (by Theorem \ref{thm.plu.inf} \textbf{(c)})}\right) \\ & =A\cdot0=0. \end{align*} This proves the $\Longrightarrow$ direction of Theorem \ref{thm.plu.inf} \textbf{(b)}. $\Longleftarrow:$ Let $\tau\in\mathcal{F}^{\left( 0\right) }$ be such that $S\left( \tau\otimes\tau\right) =0$. We want to prove that $\tau\in\Omega$. Since $\tau\in\mathcal{F}^{\left( 0\right) }=\bigcup\limits_{\substack{N\in \mathbb{N};\\N\geq0}}j_{N}^{\left( 0\right) }\left( \wedge^{N+0+1}\left( V_{N}\right) \right) $ (by Proposition \ref{prop.plu.inf.cover} \textbf{(b)}, applied to $m=0$ and $Q=0$), there exists some $N\in\mathbb{N}$ such that $N\geq0$ and $\tau\in j_{N}^{\left( 0\right) }\left( \wedge^{N+0+1}\left( V_{N}\right) \right) $. Consider this $N$. Lemma \ref{lem.plu.inf.S.comp} (applied to $m=0$) yields \begin{equation} \left( j_{N}^{\left( 1\right) }\otimes j_{N}^{\left( -1\right) }\right) \circ S_{N}^{\left( N+1\right) }=S\circ\left( j_{N}^{\left( 0\right) }\otimes j_{N}^{\left( 0\right) }\right) . \label{pf.plu.inf.b5}% \end{equation} Recall that the map $j_{N}^{\left( m\right) }$ is injective for every $m\in\mathbb{Z}$. In particular, the maps $j_{N}^{\left( 1\right) }$ and $j_{N}^{\left( -1\right) }$ are injective, so that the map $j_{N}^{\left( 1\right) }\otimes j_{N}^{\left( -1\right) }$ is also injective. But $\tau\in j_{N}^{\left( 0\right) }\left( \wedge^{N+0+1}\left( V_{N}\right) \right) =j_{N}^{\left( 0\right) }\left( \wedge^{N+1}\left( V_{N}\right) \right) $. In other words, there exists some $\tau^{\prime}% \in\wedge^{N+1}\left( V_{N}\right) $ such that $\tau=j_{N}^{\left( 0\right) }\left( \tau^{\prime}\right) $. Consider this $\tau^{\prime}$. Since $\tau=j_{N}^{\left( 0\right) }\left( \tau^{\prime}\right) $, we have $\tau\otimes\tau=j_{N}^{\left( 0\right) }\left( \tau^{\prime}\right) \otimes j_{N}^{\left( 0\right) }\left( \tau^{\prime}\right) =\left( j_{N}^{\left( 0\right) }\otimes j_{N}^{\left( 0\right) }\right) \left( \tau^{\prime}\otimes\tau^{\prime}\right) $, so that% \begin{align*} S\left( \tau\otimes\tau\right) & =S\left( \left( j_{N}^{\left( 0\right) }\otimes j_{N}^{\left( 0\right) }\right) \left( \tau^{\prime }\otimes\tau^{\prime}\right) \right) =\underbrace{\left( S\circ\left( j_{N}^{\left( 0\right) }\otimes j_{N}^{\left( 0\right) }\right) \right) }_{\substack{=\left( j_{N}^{\left( 1\right) }\otimes j_{N}^{\left( -1\right) }\right) \circ S_{N}^{\left( N+1\right) }\\\text{(by (\ref{pf.plu.inf.b5}))}}}\left( \tau^{\prime}\otimes\tau^{\prime}\right) \\ & =\left( \left( j_{N}^{\left( 1\right) }\otimes j_{N}^{\left( -1\right) }\right) \circ S_{N}^{\left( N+1\right) }\right) \left( \tau^{\prime}\otimes\tau^{\prime}\right) =\left( j_{N}^{\left( 1\right) }\otimes j_{N}^{\left( -1\right) }\right) \left( S_{N}^{\left( N+1\right) }\left( \tau^{\prime}\otimes\tau^{\prime}\right) \right) . \end{align*} Compared with $S\left( \tau\otimes\tau\right) =0$, this yields $\left( j_{N}^{\left( 1\right) }\otimes j_{N}^{\left( -1\right) }\right) \left( S_{N}^{\left( N+1\right) }\left( \tau^{\prime}\otimes\tau^{\prime}\right) \right) =0$. Since $j_{N}^{\left( 1\right) }\otimes j_{N}^{\left( -1\right) }$ is injective, this yields $S_{N}^{\left( N+1\right) }\left( \tau^{\prime}\otimes\tau^{\prime}\right) =0$. But Lemma \ref{lem.plu.inf.plu} \textbf{(b)} (applied to $N+1$ and $\tau^{\prime}$ instead of $k$ and $\tau$) yields that $\tau^{\prime}$ belongs to $\Omega_{N}^{\left( N+1\right) }$ if and only if $S_{N}^{\left( N+1\right) }\left( \tau^{\prime}\otimes \tau^{\prime}\right) =0$. Since we know that $S_{N}^{\left( N+1\right) }\left( \tau^{\prime}\otimes\tau^{\prime}\right) =0$, we can thus conclude that $\tau^{\prime}$ belongs to $\Omega_{N}^{\left( N+1\right) }$. Since $\Omega_{N}^{\left( N+1\right) }$ is the orbit of $v_{N}\wedge v_{N-1}% \wedge...\wedge v_{N-\left( N+1\right) +1}$ under the action of $\operatorname*{GL}\left( V_{N}\right) $ (this is how $\Omega_{N}^{\left( N+1\right) }$ was defined), this yields that $\tau^{\prime}$ belongs to the orbit of $v_{N}\wedge v_{N-1}\wedge...\wedge v_{N-\left( N+1\right) +1}$ under the action of $\operatorname*{GL}\left( V_{N}\right) $. In other words, there exists some $A\in\operatorname*{GL}\left( V_{N}\right) $ such that $\tau^{\prime}=A\cdot\left( v_{N}\wedge v_{N-1}\wedge...\wedge v_{N-\left( N+1\right) +1}\right) $. Consider this $A$. We have $\tau^{\prime}=A\cdot\left( v_{N}\wedge v_{N-1}\wedge...\wedge \underbrace{v_{N-\left( N+1\right) +1}}_{=v_{0}}\right) =A\cdot\left( v_{N}\wedge v_{N-1}\wedge...\wedge v_{0}\right) $. There clearly exists an invertible linear map $B\in\operatorname*{GL}\left( V_{N}\right) $ which sends $v_{N}$, $v_{N-1}$, $...$, $v_{0}$ to $v_{0}$, $v_{-1}$, $...$, $v_{-N}$, respectively\footnote{\textit{Proof.} Since $\left( v_{N},v_{N-1},...,v_{-N}\right) $ is a basis of $V_{N}$, there exists a linear map $B\in\operatorname*{End}\left( V_{N}\right) $ which sends $v_{i}$ to $\left\{ \begin{array} [c]{c}% v_{i-N},\ \ \ \ \ \ \ \ \ \ \text{if }i\geq0;\\ v_{-i},\ \ \ \ \ \ \ \ \ \ \text{if }i<0 \end{array} \right. $ for every $i\in\left\{ N,N-1,...,-N\right\} $. This linear map $B$ is invertible (since it permutes the elements of the basis $\left( v_{N},v_{N-1},...,v_{-N}\right) $ of $V_{N}$), and thus lies in $\operatorname*{GL}\left( V_{N}\right) $, and it clearly sends $v_{N}$, $v_{N-1}$, $...$, $v_{0}$ to $v_{0}$, $v_{-1}$, $...$, $v_{-N}$, respectively. Qed.}. Pick such a $B$. Then, $B\cdot\left( v_{N}\wedge v_{N-1}% \wedge...\wedge v_{0}\right) =v_{0}\wedge v_{-1}\wedge...\wedge v_{-N}$ (since $B$ sends $v_{N}$, $v_{N-1}$, $...$, $v_{0}$ to $v_{0}$, $v_{-1}$, $...$, $v_{-N}$, respectively), so that $B^{-1}\cdot\left( v_{0}\wedge v_{-1}\wedge...\wedge v_{-N}\right) =v_{N}\wedge v_{N-1}\wedge...\wedge v_{0}$ and thus% \[ A\underbrace{B^{-1}\cdot\left( v_{0}\wedge v_{-1}\wedge...\wedge v_{-N}\right) }_{=v_{N}\wedge v_{N-1}\wedge...\wedge v_{0}}=A\cdot\left( v_{N}\wedge v_{N-1}\wedge...\wedge v_{0}\right) =\tau^{\prime}. \] Let $M=i_{N}\left( AB^{-1}\right) $. Then, $M=i_{N}\underbrace{\left( AB^{-1}\right) }_{\in\operatorname*{GL}\left( V_{N}\right) }\in i_{N}\left( \operatorname*{GL}\left( V_{N}\right) \right) \subseteq \operatorname*{GL}\left( \infty\right) $. Also,% \begin{align*} j_{N}^{\left( 0\right) }\left( v_{0}\wedge v_{-1}\wedge...\wedge v_{-N}\right) & =v_{0}\wedge v_{-1}\wedge...\wedge v_{-N}\wedge v_{-N-1}\wedge v_{-N-2}\wedge v_{-N-3}\wedge...\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }j_{N}^{\left( 0\right) }\right) \\ & =v_{0}\wedge v_{-1}\wedge v_{-2}\wedge...=\psi_{0}. \end{align*} Now,% \begin{align*} & \underbrace{M}_{=i_{N}\left( AB^{-1}\right) }\cdot\underbrace{\psi_{0}% }_{=j_{N}^{\left( 0\right) }\left( v_{0}\wedge v_{-1}\wedge...\wedge v_{-N}\right) }\\ & =i_{N}\left( AB^{-1}\right) \cdot j_{N}^{\left( 0\right) }\left( v_{0}\wedge v_{-1}\wedge...\wedge v_{-N}\right) =j_{N}^{\left( 0\right) }\left( \underbrace{AB^{-1}\cdot\left( v_{0}\wedge v_{-1}\wedge...\wedge v_{-N}\right) }_{=\tau^{\prime}}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by Proposition \ref{prop.plu.inf.iNjmN}, applied to }0\text{, }AB^{-1}\text{ and }v_{0}\wedge v_{-1}\wedge...\wedge v_{-N}\text{ instead of }m\text{, }A\text{ and }u\right) \\ & =j_{N}^{\left( 0\right) }\left( \tau^{\prime}\right) =\tau. \end{align*} Thus, $\tau=\underbrace{M}_{\in\operatorname*{GL}\left( \infty\right) }% \cdot\psi_{0}\in\operatorname*{GL}\left( \infty\right) \cdot\psi_{0}=\Omega $. This proves the $\Longleftarrow$ direction of Theorem \ref{thm.plu.inf} \textbf{(b)}. \subsubsection{The semiinfinite Grassmannian} Denote $\Omega\diagup\mathbb{C}^{\times}$ by $\operatorname*{Gr}$; this is called the \textit{semiinfinite Grassmannian}. Think of the space $V$ as $\mathbb{C}\left[ t,t^{-1}\right] $ (by identifying $v_{i}$ with $t^{-i}$). Then, $\left\langle v_{0},v_{-1}% ,v_{-2},...\right\rangle =\mathbb{C}\left[ t\right] $. \textbf{Exercise:} Then, $\operatorname*{Gr}$ is the set% \[ \left\{ E\subseteq V\ \text{subspace\ }\mid\ \left( \begin{array} [c]{c}% E\supseteq t^{N}\mathbb{C}\left[ t\right] \text{ for sufficiently large }N\text{, and}\\ \dim\left( E\diagup t^{N}\mathbb{C}\left[ t\right] \right) =N\text{ for sufficiently large }N \end{array} \right) \right\} . \] \footnote{Here, ``subspace'' means ``$\mathbb{C}$-vector subspace''.} (Note that when the relations $E\supseteq t^{N}\mathbb{C}\left[ t\right] $ and $\dim\left( E\diagup t^{N}\mathbb{C}\left[ t\right] \right) =N$ hold for \textit{some} $N$, it is easy to see that they also hold for all greater $N$.) We can also replace $\mathbb{C}\left[ t,t^{-1}\right] $ with $\mathbb{C}% \left( \left( t\right) \right) $ (the formal Laurent series), and then \[ \operatorname*{Gr}=\left\{ E\subseteq V\text{ subspace}\ \mid\ \left( \begin{array} [c]{c}% E\supseteq t^{N}\mathbb{C}\left[ \left[ t\right] \right] \text{ for sufficiently large }N\text{, and}\\ \dim\left( E\diagup t^{N}\mathbb{C}\left[ \left[ t\right] \right] \right) =N\text{ for sufficiently large }N \end{array} \right) \right\} . \] For any $E\in\operatorname*{Gr}$, there exists some $N\in\mathbb{N}$ such that $t^{N}\mathbb{C}\left[ t\right] \subseteq E\subseteq t^{-N}\mathbb{C}\left[ t\right] $, so that the quotient $E\diagup t^{N}\mathbb{C}\left[ t\right] \subseteq t^{-N}\mathbb{C}\left[ t\right] \diagup t^{N}\mathbb{C}\left[ t\right] \cong\mathbb{C}^{2N}$. Thus, $\operatorname*{Gr}=\bigcup\limits_{N\geq1}\operatorname*{Gr}\left( N,2N\right) $ (a nested union). (By a variation of this construction, $\operatorname*{Gr}=\bigcup\limits_{N\geq1}\bigcup\limits_{M\geq 1}\operatorname*{Gr}\left( N,N+M\right) $.) \subsubsection{The preimage of the Grassmannian under the Boson-Fermion correspondence: the Hirota bilinear relations} Now, how do we actually use these things to find solutions to the Kadomtsev-Petviashvili equations and other integrable systems? By Theorem \ref{thm.plu.inf} \textbf{(b)}, the elements of $\Omega$ are exactly the nonzero elements $\tau$ of $\mathcal{F}^{\left( 0\right) }$ satisfying $S\left( \tau\otimes\tau\right) =0$. We might wonder what happens to these elements under the Boson-Fermion correspondence $\sigma$: how can their preimages under $\sigma$ be described? In other words, can we find a necessary and sufficient condition for a polynomial $\tau\in\mathcal{B}% ^{\left( 0\right) }$ to satisfy $\sigma\left( \tau\right) \in\Omega$ (without using $\sigma$ in this very condition)? Recall the power series $X\left( u\right) =\sum\limits_{i\in\mathbb{Z}}% \xi_{i}u^{i}$ and $X^{\ast}\left( u\right) =\sum\limits_{i\in\mathbb{Z}}% \xi_{i}^{\ast}u^{-i}$ defined in Definition \ref{def.euler.XGamma}. These power series ``act'' on the fermionic space $\mathcal{F}$. The word ``act'' has been put in inverted commas here because it is not the power series but their coefficients which really act on $\mathcal{F}$, whereas the power series themselves only map elements of $\mathcal{F}$ to elements of $\mathcal{F}% \left( \left( u\right) \right) $. This, actually, is an important observation:% \begin{equation} \text{every }\omega\in\mathcal{F}\text{ satisfies }X\left( u\right) \omega\in\mathcal{F}\left( \left( u\right) \right) \text{ and }X^{\ast }\left( u\right) \omega\in\mathcal{F}\left( \left( u\right) \right) . \label{KdV.F((u))}% \end{equation} \footnote{\textit{Proof of (\ref{KdV.F((u))}):} Let $\omega\in\mathcal{F}$. Since $X\left( u\right) =\sum\limits_{i\in\mathbb{Z}}\xi_{i}u^{i}$, we have $X\left( u\right) \omega=\sum\limits_{i\in\mathbb{Z}}\xi_{i}\left( \omega\right) u^{i}\in\mathcal{F}\left( \left( u\right) \right) $, because every sufficiently small $i\in\mathbb{Z}$ satisfies $\xi_{i}\left( \omega\right) =0$ (this is easy to see). On the other hand, since $X^{\ast }\left( u\right) =\sum\limits_{i\in\mathbb{Z}}\xi_{i}^{\ast}u^{-i}$, we have $X^{\ast}\left( u\right) =\sum\limits_{i\in\mathbb{Z}}\xi_{i}^{\ast}\left( \omega\right) u^{-i}\in\mathcal{F}\left( \left( u\right) \right) $, since every sufficiently high $i\in\mathbb{Z}$ satisfies $\xi_{i}^{\ast}\left( \omega\right) =0$ (this, again, is easy to see). This proves (\ref{KdV.F((u))}).} Let $\tau\in\mathcal{B}^{\left( 0\right) }$ be arbitrary. We want to find an equivalent form for the equation $S\left( \sigma\left( \tau\right) \otimes\sigma\left( \tau\right) \right) =0$ which does not refer to $\sigma$. Let us give two definitions first: \begin{definition} \label{def.OMEGA}Let $A$ and $B$ be two $\mathbb{C}$-vector spaces, and let $u$ be a symbol. Then, the map% \begin{align*} A\left( \left( u\right) \right) \times B\left( \left( u\right) \right) & \rightarrow\left( A\otimes B\right) \left( \left( u\right) \right) ,\\ \left( \sum\limits_{i\in\mathbb{Z}}a_{i}u^{i},\sum\limits_{i\in\mathbb{Z}% }b_{i}u^{i}\right) & \mapsto\sum\limits_{i\in\mathbb{Z}}\left( \sum\limits_{j\in\mathbb{Z}}a_{j}\otimes b_{i-j}\right) u^{i}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{where all }a_{i}\text{ lie in }A\text{ and all }b_{i}\text{ lie in }B\right) \end{align*} is well-defined (in fact, it is easy to see that for any Laurent series $\sum\limits_{i\in\mathbb{Z}}a_{i}u^{i}\in A\left( \left( u\right) \right) $ with all $a_{i}$ lying in $A$, any Laurent series $\sum\limits_{i\in \mathbb{Z}}b_{i}u^{i}\in B\left( \left( u\right) \right) $ with all $b_{i}$ lying in $B$, and any integer $i\in\mathbb{Z}$, the sum $\sum \limits_{j\in\mathbb{Z}}a_{j}\otimes b_{i-j}$ has only finitely many addends and vanishes if $i$ is small enough) and $\mathbb{C}$-bilinear. Hence, it induces a $\mathbb{C}$-linear map% \begin{align*} A\left( \left( u\right) \right) \otimes B\left( \left( u\right) \right) & \rightarrow\left( A\otimes B\right) \left( \left( u\right) \right) ,\\ \left( \sum\limits_{i\in\mathbb{Z}}a_{i}u^{i}\right) \otimes\left( \sum\limits_{i\in\mathbb{Z}}b_{i}u^{i}\right) & \mapsto\sum\limits_{i\in \mathbb{Z}}\left( \sum\limits_{j\in\mathbb{Z}}a_{j}\otimes b_{i-j}\right) u^{i}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{where all }a_{i}\text{ lie in }A\text{ and all }b_{i}\text{ lie in }B\right) . \end{align*} This map will be denoted by $\Omega_{A,B,u}$. \end{definition} More can be said about the map $\Omega_{A,B,u}$: It factors as a composition of the canonical projection $A\left( \left( u\right) \right) \otimes B\left( \left( u\right) \right) \rightarrow A\left( \left( u\right) \right) \otimes_{\mathbb{C}\left( \left( u\right) \right) }B\left( \left( u\right) \right) $ with a $\mathbb{C}\left( \left( u\right) \right) $-linear map $A\left( \left( u\right) \right) \otimes _{\mathbb{C}\left( \left( u\right) \right) }B\left( \left( u\right) \right) \rightarrow\left( A\otimes B\right) \left( \left( u\right) \right) $. We won't need this in the following. What we will need is the following observation: \begin{remark} \label{rmk.OMEGA.linear}Let $A$ and $B$ be two $\mathbb{C}$-algebras, and let $u$ be a symbol. Then, the map $\Omega_{A,B,u}$ is $A\otimes B$-linear. \end{remark} \begin{definition} Let $A$ be a $\mathbb{C}$-vector space, and let $u$ be a symbol. Then, $\operatorname*{CT}\nolimits_{u}:A\left( \left( u\right) \right) \rightarrow A$ will denote the map which sends every Laurent series $\sum\limits_{i\in\mathbb{Z}}a_{i}u^{i}\in A\left( \left( u\right) \right) $ (where all $a_{i}$ lie in $A$) to $a_{0}\in A$. The image of a Laurent series $\alpha$ under $\operatorname*{CT}\nolimits_{u}$ will be called the \textbf{constant term} of $\alpha$. The map $\operatorname*{CT}\nolimits_{u}$ is clearly $A$-linear. \end{definition} This notion of ``constant term'' we have thus defined for Laurent series is, of course, completely analogous to the one used for polynomials and formal power series. The label $\operatorname*{CT}\nolimits_{u}$ is an abbreviation for ``constant term with respect to the variable $u$''. Now, for every $\omega\in\mathcal{F}^{\left( 0\right) }$ and $\rho \in\mathcal{F}^{\left( 0\right) }$, we have% \begin{equation} S\left( \omega\otimes\rho\right) =\operatorname*{CT}\nolimits_{u}\left( \Omega_{\mathcal{F},\mathcal{F},u}\left( X\left( u\right) \omega\otimes X^{\ast}\left( u\right) \rho\right) \right) . \label{KdV.S=CT}% \end{equation} \footnote{\textit{Proof of (\ref{KdV.S=CT}):} Let $\omega\in\mathcal{F}% ^{\left( 0\right) }$ and $\rho\in\mathcal{F}^{\left( 0\right) }$. Since $X\left( u\right) =\sum\limits_{i\in\mathbb{Z}}\xi_{i}u^{i}$ and $X^{\ast }\left( u\right) =\sum\limits_{i\in\mathbb{Z}}\xi_{i}^{\ast}u^{-i}% =\sum\limits_{i\in\mathbb{Z}}\xi_{-i}^{\ast}u^{i}$ (here, we substituted $-i$ for $i$ in the sum), we have% \[ X\left( u\right) \omega\otimes X^{\ast}\left( u\right) \rho=\left( \sum\limits_{i\in\mathbb{Z}}\xi_{i}u^{i}\right) \omega\otimes\left( \sum\limits_{i\in\mathbb{Z}}\xi_{-i}^{\ast}u^{i}\right) \rho=\left( \sum\limits_{i\in\mathbb{Z}}\xi_{i}\left( \omega\right) u^{i}\right) \otimes\left( \sum\limits_{i\in\mathbb{Z}}\xi_{-i}^{\ast}\left( \rho\right) u^{i}\right) , \] so that% \begin{align*} & \Omega_{\mathcal{F},\mathcal{F},u}\left( X\left( u\right) \omega\otimes X^{\ast}\left( u\right) \rho\right) \\ & =\Omega_{\mathcal{F},\mathcal{F},u}\left( \left( \sum\limits_{i\in \mathbb{Z}}\xi_{i}\left( \omega\right) u^{i}\right) \otimes\left( \sum\limits_{i\in\mathbb{Z}}\xi_{-i}^{\ast}\left( \rho\right) u^{i}\right) \right) =\sum\limits_{i\in\mathbb{Z}}\left( \sum\limits_{j\in\mathbb{Z}}% \xi_{j}\left( \omega\right) \otimes\xi_{-\left( i-j\right) }^{\ast}\left( \rho\right) \right) u^{i}% \end{align*} (by the definition of $\Omega_{\mathcal{F},\mathcal{F},u}$). Thus (by the definition of $\operatorname*{CT}\nolimits_{u}$) we have% \begin{align*} & \operatorname*{CT}\nolimits_{u}\left( \Omega_{\mathcal{F},\mathcal{F}% ,u}\left( X\left( u\right) \omega\otimes X^{\ast}\left( u\right) \rho\right) \right) \\ & =\sum\limits_{j\in\mathbb{Z}}\xi_{j}\left( \omega\right) \otimes \xi_{-\left( 0-j\right) }^{\ast}\left( \rho\right) =\sum\limits_{j\in \mathbb{Z}}\xi_{j}\left( \omega\right) \otimes\xi_{j}^{\ast}\left( \rho\right) =\sum\limits_{i\in\mathbb{Z}}\underbrace{\xi_{i}}% _{=\widehat{v_{i}}}\left( \omega\right) \otimes\underbrace{\xi_{i}^{\ast}% }_{=\overset{\vee}{v_{i}}}\left( \rho\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }i\text{ for }j\text{ in the sum}\right) \\ & =\sum\limits_{i\in\mathbb{Z}}\widehat{v_{i}}\left( \omega\right) \otimes\overset{\vee}{v_{i}}\left( \rho\right) =\underbrace{\left( \sum\limits_{i\in\mathbb{Z}}\widehat{v_{i}}\otimes\overset{\vee}{v_{i}% }\right) }_{\substack{=S\\\text{(because this is how }S\text{ was defined)}% }}\left( \omega\otimes\rho\right) =S\left( \omega\otimes\rho\right) , \end{align*} so that (\ref{KdV.S=CT}) is proven.} Now, let $\tau\in\mathcal{B}^{\left( 0\right) }$. Due to (\ref{KdV.F((u))}) (applied to $\omega=\sigma\left( \tau\right) $), we have $X\left( u\right) \sigma\left( \tau\right) \in\mathcal{F}\left( \left( u\right) \right) $ and $X^{\ast}\left( u\right) \sigma\left( \tau\right) \in\mathcal{F}% \left( \left( u\right) \right) $. Now, let us abuse notation and denote by $\sigma$ the map from $\mathcal{B}% \left( \left( u\right) \right) $ to $\mathcal{F}\left( \left( u\right) \right) $ which is canonically induced by the Boson-Fermion correspondence $\sigma:\mathcal{B}\rightarrow\mathcal{F}$. Then, of course, this new map $\sigma:\mathcal{B}\left( \left( u\right) \right) \rightarrow \mathcal{F}\left( \left( u\right) \right) $ is also an isomorphism. Then, the equalities $\Gamma\left( u\right) =\sigma^{-1}\circ X\left( u\right) \circ\sigma$ and $\Gamma^{\ast}\left( u\right) =\sigma^{-1}\circ X^{\ast }\left( u\right) \circ\sigma$ (from Definition \ref{def.euler.XGamma}) are not just abbreviations for termwise equalities (as we explained them back in Definition \ref{def.euler.XGamma}), but also hold literally (if we interpret $\sigma$ to mean our isomorphism $\sigma:\mathcal{B}\left( \left( u\right) \right) \rightarrow\mathcal{F}\left( \left( u\right) \right) $ rather than the original Boson-Fermion correspondence $\sigma:\mathcal{B}% \rightarrow\mathcal{F}$). As a consequence, $\sigma\circ\Gamma\left( u\right) =X\left( u\right) \circ\sigma$ and $\sigma\circ\Gamma^{\ast }\left( u\right) =X^{\ast}\left( u\right) \circ\sigma$. Thus,% \[ \sigma\left( \Gamma\left( u\right) \tau\right) =\underbrace{\left( \sigma\circ\Gamma\left( u\right) \right) }_{=X\left( u\right) \circ \sigma}\tau=\left( X\left( u\right) \circ\sigma\right) \tau=X\left( u\right) \sigma\left( \tau\right) \] and% \[ \sigma\left( \Gamma^{\ast}\left( u\right) \tau\right) =\underbrace{\left( \sigma\circ\Gamma^{\ast}\left( u\right) \right) }_{=X^{\ast}\left( u\right) \circ\sigma}\tau=\left( X^{\ast}\left( u\right) \circ \sigma\right) \tau=X^{\ast}\left( u\right) \sigma\left( \tau\right) , \] so that% \[ \underbrace{X\left( u\right) \sigma\left( \tau\right) }_{=\sigma\left( \Gamma\left( u\right) \tau\right) }\otimes\underbrace{X^{\ast}\left( u\right) \sigma\left( \tau\right) }_{=\sigma\left( \Gamma^{\ast}\left( u\right) \tau\right) }=\sigma\left( \Gamma\left( u\right) \tau\right) \otimes\sigma\left( \Gamma^{\ast}\left( u\right) \tau\right) =\left( \sigma\otimes\sigma\right) \left( \Gamma\left( u\right) \tau\otimes \Gamma^{\ast}\left( u\right) \tau\right) . \] Now, \begin{align*} S\left( \sigma\left( \tau\right) \otimes\sigma\left( \tau\right) \right) & =\operatorname*{CT}\nolimits_{u}\left( \Omega_{\mathcal{F},\mathcal{F}% ,u}\underbrace{\left( X\left( u\right) \sigma\left( \tau\right) \otimes X^{\ast}\left( u\right) \sigma\left( \tau\right) \right) }_{=\left( \sigma\otimes\sigma\right) \left( \Gamma\left( u\right) \tau\otimes \Gamma^{\ast}\left( u\right) \tau\right) }\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{KdV.S=CT}), applied to }% \omega=\sigma\left( \tau\right) \text{ and }\rho=\sigma\left( \tau\right) \right) \\ & =\operatorname*{CT}\nolimits_{u}\left( \Omega_{\mathcal{F},\mathcal{F}% ,u}\left( \left( \sigma\otimes\sigma\right) \left( \Gamma\left( u\right) \tau\otimes\Gamma^{\ast}\left( u\right) \tau\right) \right) \right) \\ & =\underbrace{\left( \operatorname*{CT}\nolimits_{u}\circ\Omega _{\mathcal{F},\mathcal{F},u}\circ\left( \sigma\otimes\sigma\right) \right) }_{\substack{=\left( \sigma\otimes\sigma\right) \circ\operatorname*{CT}% \nolimits_{u}\circ\Omega_{\mathcal{B},\mathcal{B},u}\\\text{(since }\operatorname*{CT}\nolimits_{u}\text{ and }\Omega_{A,B,u}\text{ are functorial)}}}\left( \Gamma\left( u\right) \tau\otimes\Gamma^{\ast}\left( u\right) \tau\right) \\ & =\left( \left( \sigma\otimes\sigma\right) \circ\operatorname*{CT}% \nolimits_{u}\circ\Omega_{\mathcal{B},\mathcal{B},u}\right) \left( \Gamma\left( u\right) \tau\otimes\Gamma^{\ast}\left( u\right) \tau\right) \\ & =\left( \sigma\otimes\sigma\right) \left( \operatorname*{CT}% \nolimits_{u}\left( \Omega_{\mathcal{B},\mathcal{B},u}\left( \Gamma\left( u\right) \tau\otimes\Gamma^{\ast}\left( u\right) \tau\right) \right) \right) . \end{align*} Therefore, the equation $S\left( \sigma\left( \tau\right) \otimes \sigma\left( \tau\right) \right) =0$ is equivalent to \newline$\left( \sigma\otimes\sigma\right) \left( \operatorname*{CT}\nolimits_{u}\left( \Omega_{\mathcal{B},\mathcal{B},u}\left( \Gamma\left( u\right) \tau \otimes\Gamma^{\ast}\left( u\right) \tau\right) \right) \right) =0$. This latter equation, in turn, is equivalent to $\operatorname*{CT}\nolimits_{u}% \left( \Omega_{\mathcal{B},\mathcal{B},u}\left( \Gamma\left( u\right) \tau\otimes\Gamma^{\ast}\left( u\right) \tau\right) \right) =0$ (since $\sigma\otimes\sigma$ is an isomorphism\footnote{because $\sigma$ is an isomorphism}). This, in turn, is equivalent to $\left( z^{-1}\otimes z\right) \cdot\operatorname*{CT}\nolimits_{u}\left( \Omega_{\mathcal{B}% ,\mathcal{B},u}\left( \Gamma\left( u\right) \tau\otimes\Gamma^{\ast}\left( u\right) \tau\right) \right) =0$ (because $z^{-1}\otimes z$ is an invertible element of $\mathcal{B}\otimes\mathcal{B}$). Since% \begin{align*} & \left( z^{-1}\otimes z\right) \cdot\operatorname*{CT}\nolimits_{u}\left( \Omega_{\mathcal{B},\mathcal{B},u}\left( \Gamma\left( u\right) \tau \otimes\Gamma^{\ast}\left( u\right) \tau\right) \right) \\ & =\operatorname*{CT}\nolimits_{u}\left( \underbrace{\left( z^{-1}\otimes z\right) \cdot\Omega_{\mathcal{B},\mathcal{B},u}\left( \Gamma\left( u\right) \tau\otimes\Gamma^{\ast}\left( u\right) \tau\right) }_{\substack{=\Omega_{\mathcal{B},\mathcal{B},u}\left( \left( z^{-1}\otimes z\right) \left( \Gamma\left( u\right) \tau\otimes\Gamma^{\ast}\left( u\right) \tau\right) \right) \\\text{(since }\Omega_{\mathcal{B}% ,\mathcal{B},u}\text{ is }\mathcal{B}\otimes\mathcal{B}\text{-linear)}% }}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\operatorname*{CT}\nolimits_{u}% \text{ is }\mathcal{B}\otimes\mathcal{B}\text{-linear (by Remark \ref{rmk.OMEGA.linear})}\right) \\ & =\operatorname*{CT}\nolimits_{u}\left( \Omega_{\mathcal{B},\mathcal{B}% ,u}\underbrace{\left( \left( z^{-1}\otimes z\right) \left( \Gamma\left( u\right) \tau\otimes\Gamma^{\ast}\left( u\right) \tau\right) \right) }_{=z^{-1}\Gamma\left( u\right) \tau\otimes z\Gamma^{\ast}\left( u\right) \tau}\right) \\ & =\operatorname*{CT}\nolimits_{u}\left( \Omega_{\mathcal{B},\mathcal{B}% ,u}\left( z^{-1}\Gamma\left( u\right) \tau\otimes z\Gamma^{\ast}\left( u\right) \tau\right) \right) , \end{align*} this is equivalent to $\operatorname*{CT}\nolimits_{u}\left( \Omega _{\mathcal{B},\mathcal{B},u}\left( z^{-1}\Gamma\left( u\right) \tau\otimes z\Gamma^{\ast}\left( u\right) \tau\right) \right) =0$. Let us combine what we have proven: We have proven the equivalence of assertions \begin{equation} \left( S\left( \sigma\left( \tau\right) \otimes\sigma\left( \tau\right) \right) =0\right) \ \Longleftrightarrow\ \left( \operatorname*{CT}% \nolimits_{u}\left( \Omega_{\mathcal{B},\mathcal{B},u}\left( z^{-1}% \Gamma\left( u\right) \tau\otimes z\Gamma^{\ast}\left( u\right) \tau\right) \right) =0\right) . \label{pf.hirota.firstrewriting}% \end{equation} Now, let us simplify $\operatorname*{CT}\nolimits_{u}\left( \Omega _{\mathcal{B},\mathcal{B},u}\left( z^{-1}\Gamma\left( u\right) \tau\otimes z\Gamma^{\ast}\left( u\right) \tau\right) \right) $. For this, we recall that $\mathcal{B}^{\left( 0\right) }=\widetilde{F}% =\mathbb{C}\left[ x_{1},x_{2},x_{3},...\right] $. Thus, the elements of $\mathcal{B}^{\left( 0\right) }$ are polynomials in the countably many indeterminates $x_{1}$, $x_{2}$, $x_{3}$, $...$. We are going to interpret the elements of $\mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }$ as polynomials in ``twice as many'' indeterminates; by this we mean the following: \begin{Convention} \label{conv.hirota.x'x''}Let $\left( x_{1}^{\prime},x_{2}^{\prime}% ,x_{3}^{\prime},...\right) $ and $\left( x_{1}^{\prime\prime},x_{2}% ^{\prime\prime},x_{3}^{\prime\prime},...\right) $ be two countable families of new symbols. We denote the family $\left( x_{1}^{\prime},x_{2}^{\prime },x_{3}^{\prime},...\right) $ by $x^{\prime}$, and we denote the family $\left( x_{1}^{\prime\prime},x_{2}^{\prime\prime},x_{3}^{\prime\prime },...\right) $ by $x^{\prime\prime}$. Thus, if $P\in\mathbb{C}\left[ x_{1},x_{2},x_{3},...\right] $, we will denote by $P\left( x^{\prime }\right) $ the polynomial $P\left( x_{1}^{\prime},x_{2}^{\prime}% ,x_{3}^{\prime},...\right) $, and we will denote by $P\left( x^{\prime \prime}\right) $ the polynomial $P\left( x_{1}^{\prime\prime},x_{2}% ^{\prime\prime},x_{3}^{\prime\prime},...\right) $. The $\mathbb{C}$-linear map% \begin{align*} \mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) } & \rightarrow\mathbb{C}\left[ x_{1}^{\prime},x_{1}^{\prime\prime},x_{2}% ^{\prime},x_{2}^{\prime\prime},x_{3}^{\prime},x_{3}^{\prime\prime},...\right] ,\\ P\otimes Q & \mapsto P\left( x^{\prime}\right) Q\left( x^{\prime\prime }\right) \end{align*} is a $\mathbb{C}$-algebra isomorphism. By means of this isomorphism, we are going to identify $\mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }$ with $\mathbb{C}\left[ x_{1}^{\prime},x_{1}^{\prime\prime}% ,x_{2}^{\prime},x_{2}^{\prime\prime},x_{3}^{\prime},x_{3}^{\prime\prime },...\right] $. \end{Convention} Another convention: \begin{Convention} \label{conv.hirota.P(y)}For any $P\in\mathcal{B}^{\left( 0\right) }\left( \left( u\right) \right) $ and any family $\left( y_{1},y_{2}% ,y_{3},...\right) $ of pairwise commuting elements of a $\mathbb{C}$-algebra $A$, we define an element $P\left( y_{1},y_{2},y_{3},...\right) $ of $A\left( \left( u\right) \right) $ as follows: Write $P$ in the form $P=\sum\limits_{i\in\mathbb{Z}}P_{i}\cdot u^{i}$ for some $P_{i}\in \mathcal{B}^{\left( 0\right) }$, and set $P\left( y_{1},y_{2}% ,y_{3},...\right) =\sum\limits_{i\in\mathbb{Z}}P_{i}\left( y_{1},y_{2}% ,y_{3},...\right) \cdot u^{i}$. (In words, $P\left( y_{1},y_{2}% ,y_{3},...\right) $ is defined by substituting $y_{1},y_{2},y_{3},...$ for the variables $x_{1},x_{2},x_{3},...$ in $P$ while keeping the variable $u$ unchanged). \end{Convention} Now, let us notice that: \begin{lemma} \label{lem.hirota.PQ}For any $P\in\mathcal{B}^{\left( 0\right) }\left( \left( u\right) \right) $ and $Q\in\mathcal{B}^{\left( 0\right) }\left( \left( u\right) \right) $, we have% \[ \Omega_{\mathcal{B},\mathcal{B},u}\left( P\otimes Q\right) =P\left( x^{\prime}\right) \cdot Q\left( x^{\prime\prime}\right) \] (where $P\left( x^{\prime}\right) $ and $Q\left( x^{\prime\prime}\right) $ are to be understood according to Convention \ref{conv.hirota.P(y)} and Convention \ref{conv.hirota.x'x''}, and where $\mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }$ is identified with $\mathbb{C}% \left[ x_{1}^{\prime},x_{1}^{\prime\prime},x_{2}^{\prime},x_{2}^{\prime \prime},x_{3}^{\prime},x_{3}^{\prime\prime},...\right] $ according to Convention \ref{conv.hirota.x'x''}). \end{lemma} \textit{Proof of Lemma \ref{lem.hirota.PQ}.} Let $P\in\mathcal{B}^{\left( 0\right) }\left( \left( u\right) \right) $ and $Q\in\mathcal{B}^{\left( 0\right) }\left( \left( u\right) \right) $. Write $P$ in the form $P=\sum\limits_{i\in\mathbb{Z}}P_{i}\cdot u^{i}$ for some $P_{i}\in \mathcal{B}^{\left( 0\right) }$. Write $Q$ in the form $Q=\sum \limits_{i\in\mathbb{Z}}Q_{i}\cdot u^{i}$ for some $Q_{i}\in\mathcal{B}% ^{\left( 0\right) }$. Since $P=\sum\limits_{i\in\mathbb{Z}}P_{i}\cdot u^{i}$ and $Q=\sum\limits_{i\in\mathbb{Z}}Q_{i}\cdot u^{i}$, we have \begin{align*} \Omega_{\mathcal{B},\mathcal{B},u}\left( P\otimes Q\right) & =\Omega_{\mathcal{B},\mathcal{B},u}\left( \left( \sum\limits_{i\in \mathbb{Z}}P_{i}\cdot u^{i}\right) \otimes\left( \sum\limits_{i\in \mathbb{Z}}Q_{i}\cdot u^{i}\right) \right) \\ & =\sum\limits_{i\in\mathbb{Z}}\left( \sum\limits_{j\in\mathbb{Z}% }\underbrace{P_{j}\otimes Q_{i-j}}_{\substack{=P_{j}\left( x^{\prime}\right) \cdot Q_{i-j}\left( x^{\prime\prime}\right) \\\text{(due to our identification of}\\\mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}% ^{\left( 0\right) }\text{ with}\\\mathbb{C}\left[ x_{1}^{\prime}% ,x_{1}^{\prime\prime},x_{2}^{\prime},x_{2}^{\prime\prime},x_{3}^{\prime}% ,x_{3}^{\prime\prime},...\right] \text{)}}}\right) u^{i}% \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\Omega_{\mathcal{B}% ,\mathcal{B},u}\right) \\ & =\sum\limits_{i\in\mathbb{Z}}\left( \sum\limits_{j\in\mathbb{Z}}% P_{j}\left( x^{\prime}\right) \cdot Q_{i-j}\left( x^{\prime\prime}\right) \right) u^{i}% \end{align*} and% \begin{align*} P\left( x^{\prime}\right) \cdot Q\left( x^{\prime\prime}\right) & =\underbrace{\left( \sum\limits_{i\in\mathbb{Z}}P_{i}\cdot u^{i}\right) \left( x^{\prime}\right) }_{\substack{=\sum\limits_{i\in\mathbb{Z}}% P_{i}\left( x^{\prime}\right) \cdot u^{i}=\sum\limits_{j\in\mathbb{Z}}% P_{j}\left( x^{\prime}\right) \cdot u^{j}\\\text{(here, we renamed }i\text{ as }j\text{)}}}\cdot\underbrace{\left( \sum\limits_{i\in\mathbb{Z}}Q_{i}\cdot u^{i}\right) \left( x^{\prime\prime}\right) }_{=\sum\limits_{i\in \mathbb{Z}}Q_{i}\left( x^{\prime\prime}\right) \cdot u^{i}}\\ & =\left( \sum\limits_{j\in\mathbb{Z}}P_{j}\left( x^{\prime}\right) \cdot u^{j}\right) \cdot\left( \sum\limits_{i\in\mathbb{Z}}Q_{i}\left( x^{\prime\prime}\right) \cdot u^{i}\right) =\sum\limits_{j\in\mathbb{Z}}% \sum\limits_{i\in\mathbb{Z}}P_{j}\left( x^{\prime}\right) \cdot u^{j}\cdot Q_{i}\left( x^{\prime\prime}\right) \cdot u^{i}\\ & =\sum\limits_{j\in\mathbb{Z}}\sum\limits_{i\in\mathbb{Z}}P_{j}\left( x^{\prime}\right) \cdot\underbrace{u^{j}\cdot Q_{i-j}\left( x^{\prime\prime }\right) \cdot u^{i-j}}_{=Q_{i-j}\left( x^{\prime\prime}\right) \cdot u^{i}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }i-j\text{ for }i\text{ in the second sum}\right) \\ & =\sum\limits_{j\in\mathbb{Z}}\sum\limits_{i\in\mathbb{Z}}P_{j}\left( x^{\prime}\right) \cdot Q_{i-j}\left( x^{\prime\prime}\right) \cdot u^{i}=\sum\limits_{i\in\mathbb{Z}}\left( \sum\limits_{j\in\mathbb{Z}}% P_{j}\left( x^{\prime}\right) \cdot Q_{i-j}\left( x^{\prime\prime}\right) \right) u^{i}=\Omega_{\mathcal{B},\mathcal{B},u}\left( P\otimes Q\right) . \end{align*} This proves Lemma \ref{lem.hirota.PQ}. Now, Theorem \ref{thm.euler} (applied to $m=0$) yields% \begin{align} \Gamma\left( u\right) & =uz\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}% {j}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}% u^{-j}\right) \ \ \ \ \ \ \ \ \ \ \text{and}\label{pf.hirota.2}\\ \Gamma^{\ast}\left( u\right) & =z^{-1}\exp\left( -\sum\limits_{j>0}% \dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{a_{j}% }{j}u^{-j}\right) \label{pf.hirota.3}% \end{align} on $\mathcal{B}^{\left( 0\right) }$. Thus,% \begin{align*} z^{-1}\Gamma\left( u\right) \tau & =z^{-1}uz\exp\left( \sum\limits_{j>0}% \dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac {a_{j}}{j}u^{-j}\right) \tau\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.hirota.2})}\right) \\ & =u\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot \exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \tau\\ & =u\exp\left( \sum\limits_{j>0}\dfrac{jx_{j}}{j}u^{j}\right) \cdot \exp\left( -\sum\limits_{j>0}\dfrac{\left( \dfrac{\partial}{\partial x_{j}% }\right) }{j}u^{-j}\right) \tau\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }a_{j}\text{ acts as }\dfrac{\partial}{\partial x_{j}}\text{ on }\widetilde{F}\text{ for every }j>0\text{,}\\ \text{ and since }a_{-j}\text{ acts as }jx_{j}\text{ on }\widetilde{F}\text{ for every }j>0 \end{array} \right) \\ & =u\exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}}u^{-j}\right) \tau, \end{align*} so that% \begin{align} \left( z^{-1}\Gamma\left( u\right) \tau\right) \left( x^{\prime}\right) & =\left( u\exp\left( \sum\limits_{j>0}x_{j}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}}u^{-j}\right) \tau\right) \left( x^{\prime}\right) \nonumber\\ & =u\exp\left( \sum\limits_{j>0}x_{j}^{\prime}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}% u^{-j}\right) \left( \tau\left( x^{\prime}\right) \right) . \label{pf.hirota.5}% \end{align} Also,% \begin{align*} z\Gamma^{\ast}\left( u\right) \tau & =zz^{-1}\exp\left( -\sum \limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( \sum \limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \tau\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.hirota.3})}\right) \\ & =\exp\left( -\sum\limits_{j>0}\dfrac{a_{-j}}{j}u^{j}\right) \cdot \exp\left( \sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \tau\\ & =\exp\left( -\sum\limits_{j>0}\dfrac{jx_{j}}{j}u^{j}\right) \cdot \exp\left( \sum\limits_{j>0}\dfrac{\left( \dfrac{\partial}{\partial x_{j}% }\right) }{j}u^{-j}\right) \tau\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }a_{j}\text{ acts as }\dfrac{\partial}{\partial x_{j}}\text{ on }\widetilde{F}\text{ for every }j>0\text{, and }\\ \text{since }a_{-j}\text{ acts as }jx_{j}\text{ on }\widetilde{F}\text{ for every }j>0 \end{array} \right) \\ & =\exp\left( -\sum\limits_{j>0}x_{j}u^{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}}u^{-j}\right) \tau, \end{align*} so that% \begin{align} \left( z\Gamma^{\ast}\left( u\right) \tau\right) \left( x^{\prime\prime }\right) & =\left( \exp\left( -\sum\limits_{j>0}x_{j}u^{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}% }u^{-j}\right) \tau\right) \left( x^{\prime\prime}\right) \nonumber\\ & =\exp\left( -\sum\limits_{j>0}x_{j}^{\prime\prime}u^{j}\right) \cdot \exp\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}}u^{-j}\right) \left( \tau\left( x^{\prime\prime }\right) \right) . \label{pf.hirota.6}% \end{align} Now,% \begin{align} & \Omega_{\mathcal{B},\mathcal{B},u}\left( z^{-1}\Gamma\left( u\right) \tau\otimes z\Gamma^{\ast}\left( u\right) \tau\right) \nonumber\\ & =\left( z^{-1}\Gamma\left( u\right) \tau\right) \left( x^{\prime }\right) \cdot\left( z\Gamma^{\ast}\left( u\right) \tau\right) \left( x^{\prime\prime}\right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by Lemma \ref{lem.hirota.PQ}, applied to }P=z^{-1}\Gamma\left( u\right) \tau\text{ and }Q=z\Gamma^{\ast}\left( u\right) \tau\right) \nonumber\\ & =u\exp\left( \sum\limits_{j>0}x_{j}^{\prime}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}% u^{-j}\right) \left( \tau\left( x^{\prime}\right) \right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \cdot\exp\left( -\sum\limits_{j>0}x_{j}^{\prime\prime }u^{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial }{\partial x_{j}^{\prime\prime}}u^{-j}\right) \left( \tau\left( x^{\prime\prime}\right) \right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.hirota.5}) and (\ref{pf.hirota.6})}\right) \nonumber\\ & =u\exp\left( \sum\limits_{j>0}x_{j}^{\prime}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}x_{j}^{\prime\prime}u^{j}\right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \cdot\exp\left( -\sum\limits_{j>0}\dfrac{1}{j}% \dfrac{\partial}{\partial x_{j}^{\prime}}u^{-j}\right) \left( \tau\left( x^{\prime}\right) \right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{1}% {j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}}u^{-j}\right) \left( \tau\left( x^{\prime\prime}\right) \right) . \label{pf.hirota.7}% \end{align} We are going to rewrite the right hand side of this equality. First of all, notice that Theorem \ref{thm.exp(u+v)} (applied to $R=\left( \mathcal{B}% ^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }\right) \left( \left( u\right) \right) $, \newline$I=\left( \text{closure of the ideal of }R\text{ generated by }x_{j}^{\prime}\text{ and }x_{j}^{\prime\prime}\text{ with }j\text{ ranging over all positive integers}\right) $, $\alpha =\sum\limits_{j>0}x_{j}^{\prime}u^{j}$ and $\beta=-\sum\limits_{j>0}% x_{j}^{\prime\prime}u^{j}$) yields% \[ \exp\left( \sum\limits_{j>0}x_{j}^{\prime}u^{j}+\left( -\sum\limits_{j>0}% x_{j}^{\prime\prime}u^{j}\right) \right) =\exp\left( \sum\limits_{j>0}% x_{j}^{\prime}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}x_{j}% ^{\prime\prime}u^{j}\right) . \] Thus,% \begin{align} \exp\left( \sum\limits_{j>0}x_{j}^{\prime}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}x_{j}^{\prime\prime}u^{j}\right) & =\exp \underbrace{\left( \sum\limits_{j>0}x_{j}^{\prime}u^{j}+\left( -\sum\limits_{j>0}x_{j}^{\prime\prime}u^{j}\right) \right) }_{=\sum \limits_{j>0}u^{j}\left( x_{j}^{\prime}-x_{j}^{\prime\prime}\right) }\nonumber\\ & =\exp\left( \sum\limits_{j>0}u^{j}\left( x_{j}^{\prime}-x_{j}% ^{\prime\prime}\right) \right) . \label{pf.hirota.8}% \end{align} Now, let us recall a very easy fact: If $\phi$ is an endomorphism of a vector space $V$, and $v$ is a vector in $V$ such that $\phi v=0$, then $\left( \exp\phi\right) v$ is well-defined (in the sense that the power series $\sum\limits_{n\geq0}\dfrac{1}{n!}\phi^{n}v$ converges) and satisfies $\left( \exp\phi\right) v=v$. Applying this fact to $V=\left( \mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }\right) \left[ u,u^{-1}\right] $, $\phi=\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial }{\partial x_{j}^{\prime\prime}}u^{-j}$ and $v=\tau\left( x^{\prime}\right) $, we see that $\exp\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial }{\partial x_{j}^{\prime\prime}}u^{-j}\right) \left( \tau\left( x^{\prime }\right) \right) $ is well-defined and satisfies \begin{equation} \exp\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}}u^{-j}\right) \left( \tau\left( x^{\prime}\right) \right) =\tau\left( x^{\prime}\right) \label{pf.hirota.9}% \end{equation} (since $\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}}u^{-j}\right) \left( \tau\left( x^{\prime}\right) \right) =\sum\limits_{j>0}\dfrac{1}{j}\underbrace{\dfrac{\partial}{\partial x_{j}^{\prime\prime}}\left( \tau\left( x^{\prime}\right) \right) }% _{=0}u^{-j}=0$). The same argument (with $x_{j}^{\prime}$ and $x_{j}% ^{\prime\prime}$ switching places) shows that $\exp\left( \sum\limits_{j>0}% \dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}u^{-j}\right) \left( \tau\left( x^{\prime\prime}\right) \right) $ is well-defined and satisfies% \begin{equation} \exp\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}u^{-j}\right) \left( \tau\left( x^{\prime\prime}\right) \right) =\tau\left( x^{\prime\prime}\right) . \label{pf.hirota.10}% \end{equation} Now,% \begin{align} \exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}}{j}\left( \dfrac{\partial }{\partial x_{j}^{\prime}}-\dfrac{\partial}{\partial x_{j}^{\prime\prime}% }\right) \right) & =\exp\left( \left( -\sum\limits_{j>0}\dfrac{1}% {j}\dfrac{\partial}{\partial x_{j}^{\prime}}u^{-j}\right) +\sum \limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}}% u^{-j}\right) \nonumber\\ & =\exp\left( -\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}u^{-j}\right) \circ\exp\left( \sum\limits_{j>0}\dfrac{1}% {j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}}u^{-j}\right) \label{pf.hirota.11}% \end{align} \footnote{Here, the last equality sign follows from Theorem \ref{thm.exp(u+v)}% , applied to \begin{align*} R & =\left( \begin{array} [c]{c}% \text{closure of the }\mathbb{C}\left[ u,u^{-1}\right] \text{-subalgebra of }\operatorname*{End}\nolimits_{\mathbb{C}\left[ u,u^{-1}\right] }\left( \left( \mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }\right) \left[ u,u^{-1}\right] \right) \\ \text{generated by }\dfrac{\partial}{\partial x_{j}^{\prime}}\text{ and }\dfrac{\partial}{\partial x_{j}^{\prime\prime}}\text{ with }j\text{ ranging over all positive integers}% \end{array} \right) ,\\ I & =\left( \begin{array} [c]{c}% \text{closure of the ideal of }R\text{ generated by }\dfrac{\partial}{\partial x_{j}^{\prime}}\text{ and }\dfrac{\partial}{\partial x_{j}^{\prime\prime}% }\text{ with}\\ j\text{ ranging over all positive integers}% \end{array} \right) ,\\ \alpha & =-\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}u^{-j},\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \beta =\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}% }u^{-j}. \end{align*} } and similarly% \begin{equation} \exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}}{j}\left( \dfrac{\partial }{\partial x_{j}^{\prime}}-\dfrac{\partial}{\partial x_{j}^{\prime\prime}% }\right) \right) =\exp\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial }{\partial x_{j}^{\prime\prime}}u^{-j}\right) \circ\exp\left( -\sum \limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}% u^{-j}\right) . \label{pf.hirota.12}% \end{equation} But since $-\sum\limits_{j>0}\dfrac{u^{-j}}{j}\left( \dfrac{\partial }{\partial x_{j}^{\prime}}-\dfrac{\partial}{\partial x_{j}^{\prime\prime}% }\right) $ is a derivation (from $\left( \mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }\right) \left[ u,u^{-1}\right] $ to \newline$\left( \mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }\right) \left[ u,u^{-1}\right] $), its exponential $\exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}}{j}\left( \dfrac{\partial}{\partial x_{j}^{\prime}}-\dfrac{\partial}{\partial x_{j}^{\prime\prime}}\right) \right) $ is a $\mathbb{C}$-algebra homomorphism (since exponentials of derivations are $\mathbb{C}$-algebra homomorphisms), so that% \begin{align*} & \exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}}{j}\left( \dfrac{\partial }{\partial x_{j}^{\prime}}-\dfrac{\partial}{\partial x_{j}^{\prime\prime}% }\right) \right) \left( \tau\left( x^{\prime}\right) \tau\left( x^{\prime\prime}\right) \right) \\ & =\underbrace{\exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}}{j}\left( \dfrac{\partial}{\partial x_{j}^{\prime}}-\dfrac{\partial}{\partial x_{j}^{\prime\prime}}\right) \right) }_{\substack{=\exp\left( -\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}% u^{-j}\right) \circ\exp\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial }{\partial x_{j}^{\prime\prime}}u^{-j}\right) \\\text{(by (\ref{pf.hirota.11}% ))}}}\left( \tau\left( x^{\prime}\right) \right) \\ & \ \ \ \ \ \ \ \ \ \ \cdot\underbrace{\exp\left( -\sum\limits_{j>0}% \dfrac{u^{-j}}{j}\left( \dfrac{\partial}{\partial x_{j}^{\prime}}% -\dfrac{\partial}{\partial x_{j}^{\prime\prime}}\right) \right) }_{\substack{=\exp\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial }{\partial x_{j}^{\prime\prime}}u^{-j}\right) \circ\exp\left( -\sum \limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}% u^{-j}\right) \\\text{(by (\ref{pf.hirota.12}))}}}\left( \tau\left( x^{\prime\prime}\right) \right) \\ & =\left( \exp\left( -\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial }{\partial x_{j}^{\prime}}u^{-j}\right) \circ\exp\left( \sum\limits_{j>0}% \dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}}u^{-j}\right) \right) \left( \tau\left( x^{\prime}\right) \right) \\ & \ \ \ \ \ \ \ \ \ \ \cdot\left( \exp\left( \sum\limits_{j>0}\dfrac{1}% {j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}}u^{-j}\right) \circ \exp\left( -\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}u^{-j}\right) \right) \left( \tau\left( x^{\prime\prime }\right) \right) \\ & =\exp\left( -\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}u^{-j}\right) \underbrace{\left( \exp\left( \sum \limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}}% u^{-j}\right) \left( \tau\left( x^{\prime}\right) \right) \right) }_{\substack{=\tau\left( x^{\prime}\right) \\\text{(by (\ref{pf.hirota.9}% ))}}}\\ & \ \ \ \ \ \ \ \ \ \ \cdot\exp\left( \sum\limits_{j>0}\dfrac{1}{j}% \dfrac{\partial}{\partial x_{j}^{\prime\prime}}u^{-j}\right) \underbrace{\left( \exp\left( -\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial }{\partial x_{j}^{\prime}}u^{-j}\right) \left( \tau\left( x^{\prime\prime }\right) \right) \right) }_{\substack{=\tau\left( x^{\prime\prime}\right) \\\text{(by (\ref{pf.hirota.10}))}}}\\ & =\exp\left( -\sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}u^{-j}\right) \left( \tau\left( x^{\prime}\right) \right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}}u^{-j}\right) \left( \tau\left( x^{\prime\prime }\right) \right) . \end{align*} Hence, (\ref{pf.hirota.7}) becomes% \begin{align} & \Omega_{\mathcal{B},\mathcal{B},u}\left( z^{-1}\Gamma\left( u\right) \tau\otimes z\Gamma^{\ast}\left( u\right) \tau\right) \nonumber\\ & =u\underbrace{\exp\left( \sum\limits_{j>0}x_{j}^{\prime}u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}x_{j}^{\prime\prime}u^{j}\right) }_{\substack{=\exp\left( \sum\limits_{j>0}u^{j}\left( x_{j}^{\prime}% -x_{j}^{\prime\prime}\right) \right) \\\text{(by (\ref{pf.hirota.8}))}% }}\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \cdot\underbrace{\exp\left( -\sum\limits_{j>0}% \dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime}}u^{-j}\right) \left( \tau\left( x^{\prime}\right) \right) \cdot\exp\left( \sum\limits_{j>0}% \dfrac{1}{j}\dfrac{\partial}{\partial x_{j}^{\prime\prime}}u^{-j}\right) \left( \tau\left( x^{\prime\prime}\right) \right) }_{=\exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}}{j}\left( \dfrac{\partial}{\partial x_{j}^{\prime}}-\dfrac{\partial}{\partial x_{j}^{\prime\prime}}\right) \right) \left( \tau\left( x^{\prime}\right) \tau\left( x^{\prime\prime }\right) \right) }\nonumber\\ & =u\exp\left( \sum\limits_{j>0}u^{j}\left( x_{j}^{\prime}-x_{j}% ^{\prime\prime}\right) \right) \cdot\exp\left( -\sum\limits_{j>0}% \dfrac{u^{-j}}{j}\left( \dfrac{\partial}{\partial x_{j}^{\prime}}% -\dfrac{\partial}{\partial x_{j}^{\prime\prime}}\right) \right) \left( \tau\left( x^{\prime}\right) \tau\left( x^{\prime\prime}\right) \right) . \label{pf.hirota.15}% \end{align} Thus, (\ref{pf.hirota.firstrewriting}) rewrites as% \begin{align} & \left( S\left( \sigma\left( \tau\right) \otimes\sigma\left( \tau\right) \right) =0\right) \nonumber\\ & \Longleftrightarrow\ \left( \operatorname*{CT}\nolimits_{u}\left( u\exp\left( \sum\limits_{j>0}u^{j}\left( x_{j}^{\prime}-x_{j}^{\prime\prime }\right) \right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}}{j}\left( \dfrac{\partial}{\partial x_{j}^{\prime}}-\dfrac{\partial}{\partial x_{j}^{\prime\prime}}\right) \right) \left( \tau\left( x^{\prime}\right) \tau\left( x^{\prime\prime}\right) \right) \right) =0\right) . \label{pf.hirota.secondrewriting}% \end{align} This already gives a criterion for a $\tau\in\mathcal{B}^{\left( 0\right) }$ to satisfy $\sigma\left( \tau\right) \in\Omega$, but it is yet a rather messy one. We are going to simplify it in the following. First, we do a substitution of variables: \begin{Convention} Let $\left( y_{1},y_{2},y_{3},...\right) $ be a sequence of new symbols. We identify the $\mathbb{C}$-algebra $\mathbb{C}\left[ x_{1},y_{1},x_{2}% ,y_{2},x_{3},y_{3},...\right] $ with the $\mathbb{C}$-algebra $\mathbb{C}% \left[ x_{1}^{\prime},x_{1}^{\prime\prime},x_{2}^{\prime},x_{2}^{\prime \prime},x_{3}^{\prime},x_{3}^{\prime\prime},...\right] =\mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }$ by the following substitution:% \begin{align*} x_{j}^{\prime} & =x_{j}-y_{j}\ \ \ \ \ \ \ \ \ \ \text{for every }j>0;\\ x_{j}^{\prime\prime} & =x_{j}+y_{j}\ \ \ \ \ \ \ \ \ \ \text{for every }j>0. \end{align*} If we define the sum and the difference of two sequences by componentwise addition resp. subtraction, then this rewrites as follows:% \begin{align*} x^{\prime} & =x-y;\\ x^{\prime\prime} & =x+y. \end{align*} \end{Convention} It is now easy to see that% \[ x_{j}^{\prime}-x_{j}^{\prime\prime}=-2y_{j}\ \ \ \ \ \ \ \ \ \ \text{for every }j>0, \] and% \[ \dfrac{\partial}{\partial x_{j}^{\prime}}-\dfrac{\partial}{\partial x_{j}^{\prime\prime}}=-\dfrac{\partial}{\partial y_{j}}% \ \ \ \ \ \ \ \ \ \ \text{for every }j>0 \] (where $\dfrac{\partial}{\partial x_{j}^{\prime}}$ and $\dfrac{\partial }{\partial x_{j}^{\prime\prime}}$ mean differentiation over the variables $x_{j}^{\prime}$ and $x_{j}^{\prime\prime}$ in the polynomial ring $\mathbb{C}\left[ x_{1}^{\prime},x_{1}^{\prime\prime},x_{2}^{\prime}% ,x_{2}^{\prime\prime},x_{3}^{\prime},x_{3}^{\prime\prime},...\right] $, whereas $\dfrac{\partial}{\partial y_{j}}$ means differentiation over the variable $y_{j}$ in the polynomial ring $\mathbb{C}\left[ x_{1},y_{1}% ,x_{2},y_{2},x_{3},y_{3},...\right] $). As a consequence,% \begin{align*} & u\exp\left( \sum\limits_{j>0}u^{j}\underbrace{\left( x_{j}^{\prime}% -x_{j}^{\prime\prime}\right) }_{=-2y_{j}}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{u^{-j}}{j}\underbrace{\left( \dfrac{\partial }{\partial x_{j}^{\prime}}-\dfrac{\partial}{\partial x_{j}^{\prime\prime}% }\right) }_{=-\dfrac{\partial}{\partial y_{j}}}\right) \left( \tau\left( \underbrace{x^{\prime}}_{=x-y}\right) \tau\left( \underbrace{x^{\prime \prime}}_{=x+y}\right) \right) \\ & =u\exp\left( -2\sum\limits_{j>0}u^{j}y_{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{u^{-j}}{j}\dfrac{\partial}{\partial y_{j}}\right) \left( \tau\left( x-y\right) \tau\left( x+y\right) \right) . \end{align*} Hence, (\ref{pf.hirota.secondrewriting}) rewrites as% \begin{align} & \left( S\left( \sigma\left( \tau\right) \otimes\sigma\left( \tau\right) \right) =0\right) \nonumber\\ & \Longleftrightarrow\ \left( \operatorname*{CT}\nolimits_{u}\left( u\exp\left( -2\sum\limits_{j>0}u^{j}y_{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{u^{-j}}{j}\dfrac{\partial}{\partial y_{j}}\right) \left( \tau\left( x-y\right) \tau\left( x+y\right) \right) \right) =0\right) . \label{pf.hirota.thirdrewriting}% \end{align} To simplify this even further, a new notation is needed: \begin{definition} \label{def.hirota.A(P,f,g)}Let $K$ be a commutative ring. Let $\left( x_{1},x_{2},x_{3},...\right) $, $\left( z_{1},z_{2},z_{3},...\right) $, and $\left( w_{1},w_{2},w_{3},...\right) $ be three disjoint families of indeterminates. Denote by $x$ the family $\left( x_{1},x_{2},x_{3}% ,...\right) $, and denote by $z$ the family $\left( z_{1},z_{2}% ,z_{3},...\right) $. \textbf{(a)} For any polynomial $r\in K\left[ x_{1},x_{2},x_{3}% ,...,z_{1},z_{2},z_{3},...\right] $, let $r\mid_{z=0}$ denote the polynomial in $K\left[ x_{1},x_{2},x_{3},...\right] $ obtained by substituting $\left( 0,0,0,...\right) $ for $\left( z_{1},z_{2},z_{3},...\right) $ in $P$. \textbf{(b)} Consider the differential operators $\dfrac{\partial}{\partial z_{1}},\dfrac{\partial}{\partial z_{2}},\dfrac{\partial}{\partial z_{3}},...$ on $K\left[ x_{1},x_{2},x_{3},...,z_{1},z_{2},z_{3},...\right] $. For any power series $P\in K\left[ \left[ w_{1},w_{2},w_{3},...\right] \right] $, let $P\left( \partial_{z}\right) $ mean the value of $P$ when applied to the family $\left( \dfrac{\partial}{\partial z_{1}},\dfrac{\partial}{\partial z_{2}},\dfrac{\partial}{\partial z_{3}},...\right) $ (that is, the result of substituting $\dfrac{\partial}{\partial z_{j}}$ for each $w_{j}$ in $P$). This value is a well-defined differential operator on $K\left[ x_{1},x_{2}% ,x_{3},...,z_{1},z_{2},z_{3},...\right] $ (due to Remark \ref{rmk.hirota.welldef} below). \textbf{(c)} For any power series $P\in K\left[ \left[ w_{1},w_{2}% ,w_{3},...\right] \right] $ and any two polynomials $f\in K\left[ x_{1},x_{2},x_{3},...\right] $ and $g\in K\left[ x_{1},x_{2},x_{3}% ,...\right] $, define a polynomial $A\left( P,f,g\right) \in K\left[ x_{1},x_{2},x_{3},...\right] $ by \[ A\left( P,f,g\right) =\left( P\left( \partial_{z}\right) \left( f\left( x-z\right) g\left( x+z\right) \right) \right) \mid_{z=0}. \] \end{definition} \begin{remark} \label{rmk.hirota.welldef}Let $K$ be a commutative ring. Let $\left( x_{1},x_{2},x_{3},...\right) $, $\left( z_{1},z_{2},z_{3},...\right) $, and $\left( w_{1},w_{2},w_{3},...\right) $ be three disjoint families of indeterminates. Let $P\in K\left[ \left[ w_{1},w_{2},w_{3},...\right] \right] $ be a power series. Then, if we apply the power series $P$ to the family $\left( \dfrac{\partial}{\partial z_{1}},\dfrac{\partial}{\partial z_{2}},\dfrac{\partial}{\partial z_{3}},...\right) $, we obtain a well-defined endomorphism of $K\left[ x_{1},x_{2},x_{3},...,z_{1},z_{2}% ,z_{3},...\right] $. \end{remark} \textit{Proof of Remark \ref{rmk.hirota.welldef}.} Let $\mathbb{N}% _{\operatorname*{fin}}^{\left\{ 1,2,3,...\right\} }$ be defined as in Convention \ref{conv.fin}. Write the power series $P$ in the form \[ P=\sum\limits_{\left( i_{1},i_{2},i_{3},...\right) \in\mathbb{N}% _{\operatorname*{fin}}^{\left\{ 1,2,3,...\right\} }}\lambda_{\left( i_{1},i_{2},i_{3},...\right) }w_{1}^{i_{1}}w_{2}^{i_{2}}w_{3}^{i_{3}}... \] for $\lambda_{\left( i_{1},i_{2},i_{3},...\right) }\in K$. Then, if we apply the power series $P$ to the family $\left( \dfrac{\partial}{\partial z_{1}% },\dfrac{\partial}{\partial z_{2}},\dfrac{\partial}{\partial z_{3}% },...\right) $, we obtain% \[ \sum\limits_{\left( i_{1},i_{2},i_{3},...\right) \in\mathbb{N}% _{\operatorname*{fin}}^{\left\{ 1,2,3,...\right\} }}\lambda_{\left( i_{1},i_{2},i_{3},...\right) }\left( \dfrac{\partial}{\partial z_{1}% }\right) ^{i_{1}}\left( \dfrac{\partial}{\partial z_{2}}\right) ^{i_{2}% }\left( \dfrac{\partial}{\partial z_{3}}\right) ^{i_{3}}.... \] In order to prove that this is a well-defined endomorphism of $K\left[ x_{1},x_{2},x_{3},...,z_{1},z_{2},z_{3},...\right] $, we must prove that for every $r\in K\left[ x_{1},x_{2},x_{3},...,z_{1},z_{2},z_{3},...\right] $, the sum% \[ \sum\limits_{\left( i_{1},i_{2},i_{3},...\right) \in\mathbb{N}% _{\operatorname*{fin}}^{\left\{ 1,2,3,...\right\} }}\lambda_{\left( i_{1},i_{2},i_{3},...\right) }\left( \left( \dfrac{\partial}{\partial z_{1}}\right) ^{i_{1}}\left( \dfrac{\partial}{\partial z_{2}}\right) ^{i_{2}}\left( \dfrac{\partial}{\partial z_{3}}\right) ^{i_{3}}...\right) r \] is well-defined, i. e., has only finitely many nonzero addends. But this is clear, because only finitely many $\left( i_{1},i_{2},i_{3},...\right) \in\mathbb{N}_{\operatorname*{fin}}^{\left\{ 1,2,3,...\right\} }$ satisfy $\left( \left( \dfrac{\partial}{\partial z_{1}}\right) ^{i_{1}}\left( \dfrac{\partial}{\partial z_{2}}\right) ^{i_{2}}\left( \dfrac{\partial }{\partial z_{3}}\right) ^{i_{3}}...\right) r\neq0$ \ \ \ \ \footnote{This is because $r$ is a polynomial, so that only finitely many variables occur in $r$, and the degrees of the monomials of $r$ are bounded from above.}. Hence, we have proven that the sum $\sum\limits_{\left( i_{1},i_{2},i_{3}% ,...\right) \in\mathbb{N}_{\operatorname*{fin}}^{\left\{ 1,2,3,...\right\} }}\lambda_{\left( i_{1},i_{2},i_{3},...\right) }\left( \dfrac{\partial }{\partial z_{1}}\right) ^{i_{1}}\left( \dfrac{\partial}{\partial z_{2}% }\right) ^{i_{2}}\left( \dfrac{\partial}{\partial z_{3}}\right) ^{i_{3}% }...$ is a well-defined endomorphism of $K\left[ x_{1},x_{2},x_{3}% ,...,z_{1},z_{2},z_{3},...\right] $. Since this sum is the result of applying the power series $P$ to the family $\left( \dfrac{\partial}{\partial z_{1}% },\dfrac{\partial}{\partial z_{2}},\dfrac{\partial}{\partial z_{3}% },...\right) $, we thus conclude that applying the power series $P$ to the family $\left( \dfrac{\partial}{\partial z_{1}},\dfrac{\partial}{\partial z_{2}},\dfrac{\partial}{\partial z_{3}},...\right) $ yields a well-defined endomorphism of $K\left[ x_{1},x_{2},x_{3},...,z_{1},z_{2},z_{3},...\right] $. Remark \ref{rmk.hirota.welldef} is proven. \textbf{Example:} If $P\left( w\right) =w_{1}$ (the first variable), then% \[ A\left( P,f,g\right) =\left( \dfrac{\partial}{\partial z_{1}}\left( f\left( x-z\right) g\left( x+z\right) \right) \right) \mid_{z=0}% =-\dfrac{\partial f}{\partial x_{1}}g+\dfrac{\partial g}{\partial x_{1}}f. \] \begin{lemma} For any three polynomials $P,f,g$, we have $A\left( P,f,g\right) =A\left( P_{-},g,f\right) $, where $P_{-}\left( w\right) =P\left( -w\right) $. \end{lemma} \begin{corollary} \label{cor.hirota.odd}For any two polynomials $P$ and $f$, we have $A\left( P,f,f\right) =0$ if $P$ is odd. \end{corollary} This is clear from the definition. We now state the so-called \textit{Hirota bilinear relations}, which are a simplified version of (\ref{pf.hirota.thirdrewriting}): \begin{theorem} [Hirota bilinear relations]\label{thm.hirota}Let $\tau\in\mathcal{B}^{\left( 0\right) }$ be a nonzero vector. Let $\left( y_{1},y_{2},y_{3},...\right) $ and $\left( w_{1},w_{2},w_{3},...\right) $ be two families of new symbols. Let $\widetilde{w}$ denote the sequence $\left( \dfrac{w_{1}}{1},\dfrac {w_{2}}{2},\dfrac{w_{3}}{3},...\right) $. Define the elementary Schur polynomials $S_{k}$ as in Definition \ref{def.schur.Sk}. Then, $\sigma\left( \tau\right) \in\Omega$ if and only if% \begin{equation} A\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}w_{s}\right) ,\tau,\tau\right) =0, \label{thm.hirota.eqn}% \end{equation} where the term $A\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}% w_{s}\right) ,\tau,\tau\right) $ is to be interpreted by applying Definition \ref{def.hirota.A(P,f,g)} \textbf{(c)} to $K=\mathbb{C}\left[ \left[ y_{1},y_{2},y_{3},...\right] \right] $ (since $\sum\limits_{j=0}^{\infty }S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}w_{s}\right) \in\left( \mathbb{C}\left[ \left[ y_{1},y_{2},y_{3},...\right] \right] \right) \left[ \left[ w_{1}% ,w_{2},w_{3},...\right] \right] $ and $\tau\in\mathcal{B}^{\left( 0\right) }=\mathbb{C}\left[ x_{1},x_{2},x_{3},...\right] \subseteq\left( \mathbb{C}\left[ \left[ y_{1},y_{2},y_{3},...\right] \right] \right) \left[ x_{1},x_{2},x_{3},...\right] $). \end{theorem} Before we prove this, we need a simple lemma about polynomials: \begin{lemma} \label{lem.hirota.y+z}Let $K$ be a commutative $\mathbb{Q}$-algebra. Let $\left( y_{1},y_{2},y_{3},...\right) $ and $\left( z_{1},z_{2}% ,z_{3},...\right) $ be two sequences of new symbols. Denote the sequence $\left( y_{1},y_{2},y_{3},...\right) $ by $y$. Denote the sequence $\left( z_{1},z_{2},z_{3},...\right) $ by $z$. Denote by $\widetilde{\partial_{y}}$ the sequence $\left( \dfrac{1}{1}\dfrac{\partial}{\partial y_{1}},\dfrac {1}{2}\dfrac{\partial}{\partial y_{2}},\dfrac{1}{3}\dfrac{\partial}{\partial y_{3}},...\right) $ of endomorphisms of $\left( K\left[ \left[ y_{1}% ,y_{2},y_{3},...\right] \right] \right) \left[ z_{1},z_{2},z_{3}% ,...\right] $. Denote by $\widetilde{\partial_{z}}$ the sequence $\left( \dfrac{1}{1}\dfrac{\partial}{\partial z_{1}},\dfrac{1}{2}\dfrac{\partial }{\partial z_{2}},\dfrac{1}{3}\dfrac{\partial}{\partial z_{3}},...\right) $ of endomorphisms of $\left( K\left[ \left[ y_{1},y_{2},y_{3},...\right] \right] \right) \left[ z_{1},z_{2},z_{3},...\right] $. Let $P$ and $Q$ be two elements of $K\left[ w_{1},w_{2},w_{3},...\right] $ (where $\left( w_{1},w_{2},w_{3},...\right) $ is a further sequence of new symbols). Then,% \[ Q\left( \widetilde{\partial_{y}}\right) \left( P\left( y+z\right) \right) =Q\left( \widetilde{\partial_{z}}\right) \left( P\left( y+z\right) \right) . \] \end{lemma} \textit{Proof of Lemma \ref{lem.hirota.y+z}.} Let $D$ be the $K$-subalgebra of $\operatorname*{End}\left( \left( K\left[ \left[ y_{1},y_{2}% ,y_{3},...\right] \right] \right) \left[ z_{1},z_{2},z_{3},...\right] \right) $ generated by $\dfrac{\partial}{\partial y_{1}},\dfrac{\partial }{\partial y_{2}},\dfrac{\partial}{\partial y_{3}},...,\dfrac{\partial }{\partial z_{1}},\dfrac{\partial}{\partial z_{2}},\dfrac{\partial}{\partial z_{3}},...$. Then, clearly, $D$ is a commutative $K$-algebra (since its generators commute), and all elements of the sequences $\widetilde{\partial _{y}}$ and $\widetilde{\partial_{z}}$ lie in $D$ (since $\widetilde{\partial _{y}}=\left( \dfrac{1}{1}\dfrac{\partial}{\partial y_{1}},\dfrac{1}{2}% \dfrac{\partial}{\partial y_{2}},\dfrac{1}{3}\dfrac{\partial}{\partial y_{3}% },...\right) $ and $\widetilde{\partial_{z}}=\left( \dfrac{1}{1}% \dfrac{\partial}{\partial z_{1}},\dfrac{1}{2}\dfrac{\partial}{\partial z_{2}% },\dfrac{1}{3}\dfrac{\partial}{\partial z_{3}},...\right) $). Let $I$ be the ideal of $D$ generated by $\dfrac{\partial}{\partial y_{i}% }-\dfrac{\partial}{\partial z_{i}}$ with $i$ ranging over the positive integers. Then, $\dfrac{\partial}{\partial y_{i}}-\dfrac{\partial}{\partial z_{i}}\in I$ for every positive integer $i$. Hence, every positive integer $i$ satisfies $\dfrac{1}{i}\dfrac{\partial}{\partial y_{i}}\equiv\dfrac{1}% {i}\dfrac{\partial}{\partial z_{i}}\operatorname{mod}I$ (since $\dfrac{1}% {i}\dfrac{\partial}{\partial y_{i}}-\dfrac{1}{i}\dfrac{\partial}{\partial z_{i}}=\dfrac{1}{i}\underbrace{\left( \dfrac{\partial}{\partial y_{i}}% -\dfrac{\partial}{\partial z_{i}}\right) }_{\in I}\in I$). In other words, for every positive integer $i$, the $i$-th element of the sequence $\widetilde{\partial_{y}}$ is congruent to the $i$-th element of the sequence $\widetilde{\partial_{z}}$ modulo $I$ (since the $i$-th element of the sequence $\widetilde{\partial_{y}}$ is $\dfrac{1}{i}\dfrac{\partial}{\partial y_{i}}$, while the $i$-th element of the sequence $\widetilde{\partial_{z}}$ is $\dfrac{1}{i}\dfrac{\partial}{\partial z_{i}}$). Thus, each element of the sequence $\widetilde{\partial_{y}}$ is congruent to the corresponding element of the sequence $\widetilde{\partial_{z}}$ modulo $I$. Hence, $Q\left( \widetilde{\partial_{y}}\right) \equiv Q\left( \widetilde{\partial_{z}% }\right) \operatorname{mod}I$ (since $Q$ is a polynomial, and $I$ is an ideal). Hence, \begin{align*} & Q\left( \widetilde{\partial_{y}}\right) -Q\left( \widetilde{\partial _{z}}\right) \in I\\ & =\left( \text{ideal of }D\text{ generated by }\dfrac{\partial}{\partial y_{i}}-\dfrac{\partial}{\partial z_{i}}\text{ with }i\text{ ranging over the positive integers}\right) . \end{align*} In other words, $Q\left( \widetilde{\partial_{y}}\right) -Q\left( \widetilde{\partial_{z}}\right) $ is a $D$-linear combinations of terms of the form $\dfrac{\partial}{\partial y_{i}}-\dfrac{\partial}{\partial z_{i}}$ with $i$ ranging over the positive integers. Thus, we can write $Q\left( \widetilde{\partial_{y}}\right) -Q\left( \widetilde{\partial_{z}}\right) $ in the form $Q\left( \widetilde{\partial_{y}}\right) -Q\left( \widetilde{\partial_{z}}\right) =\sum\limits_{i>0}d_{i}\circ\left( \dfrac{\partial}{\partial y_{i}}-\dfrac{\partial}{\partial z_{i}}\right) $, where each $d_{i}$ is an element of $D$, and all but finitely many $i>0$ satisfy $d_{i}=0$. Consider these $d_{i}$. But it is easy to see that% \begin{equation} \text{every positive integer }i\text{ satisfies }\left( \dfrac{\partial }{\partial y_{i}}-\dfrac{\partial}{\partial z_{i}}\right) \left( P\left( y+z\right) \right) =0. \label{pf.hirota.y+z.1}% \end{equation} \footnote{\textit{Proof of (\ref{pf.hirota.y+z.1}):} Let $i$ be a positive integer. Let us identify $\mathbb{C}\left[ w_{1},w_{2},w_{3},...\right] $ with $\left( \mathbb{C}\left[ w_{1},w_{2},...,w_{i-1},w_{i+1},w_{i+2}% ,...\right] \right) \left[ w_{i}\right] $. Then, $P\in\mathbb{C}\left[ w_{1},w_{2},w_{3},...\right] =\left( \mathbb{C}\left[ w_{1},w_{2}% ,...,w_{i-1},w_{i+1},w_{i+2},...\right] \right) \left[ w_{i}\right] $, so that we can write $P$ as a polynomial in the variable $w_{i}$ over the ring $\mathbb{C}\left[ w_{1},w_{2},...,w_{i-1},w_{i+1},w_{i+2},...\right] $. In other words, we can write $P$ in the form $P=\sum\limits_{n\in\mathbb{N}}% p_{n}w_{i}^{n}$, where every $n\in\mathbb{N}$ satisfies $p_{n}\in \mathbb{C}\left[ w_{1},w_{2},...,w_{i-1},w_{i+1},w_{i+2},...\right] $ and all but finitely many $n\in\mathbb{N}$ satisfy $p_{n}=0$. Consider these $p_{n}$. \par Let $n\in\mathbb{N}$ be arbitrary. Consider $p_{n}\in\mathbb{C}\left[ w_{1},w_{2},...,w_{i-1},w_{i+1},w_{i+2},...\right] $ as an element of $\mathbb{C}\left[ w_{1},w_{2},w_{3},...\right] $ (by means of the canonical embedding $\mathbb{C}\left[ w_{1},w_{2},...,w_{i-1},w_{i+1},w_{i+2}% ,...\right] \subseteq\mathbb{C}\left[ w_{1},w_{2},w_{3},...\right] $). Then, $p_{n}$ is a polynomial in which the variable $w_{i}$ does not occur. Hence, $p_{n}\left( y+z\right) $ is a polynomial in which neither of the variables $y_{i}$ and $z_{i}$ occur. Thus, $\dfrac{\partial}{\partial y_{i}% }\left( p_{n}\left( y+z\right) \right) =0$ and $\dfrac{\partial}{\partial z_{i}}\left( p_{n}\left( y+z\right) \right) =0$. \par On the other hand, it is very easy to check that $\dfrac{\partial}{\partial y_{i}}\left( y_{i}+z_{i}\right) ^{n}=\dfrac{\partial}{\partial z_{i}}\left( y_{i}+z_{i}\right) ^{n}$ (in fact, this is obvious in the case when $n=0$, and in every other case follows from $\dfrac{\partial}{\partial y_{i}}\left( y_{i}+z_{i}\right) ^{n}=n\left( y_{i}+z_{i}\right) ^{n-1}$ and $\dfrac{\partial}{\partial z_{i}}\left( y_{i}+z_{i}\right) ^{n}=n\left( y_{i}+z_{i}\right) ^{n-1}$). Now, by the Leibniz rule,% \begin{align*} \dfrac{\partial}{\partial y_{i}}\left( p_{n}\left( y+z\right) \cdot\left( y_{i}+z_{i}\right) ^{n}\right) & =\underbrace{\left( \dfrac{\partial }{\partial y_{i}}\left( p_{n}\left( y+z\right) \right) \right) }_{=0=\dfrac{\partial}{\partial z_{i}}\left( p_{n}\left( y+z\right) \right) }\cdot\left( y_{i}+z_{i}\right) ^{n}+p_{n}\left( y+z\right) \cdot\underbrace{\dfrac{\partial}{\partial y_{i}}\left( y_{i}+z_{i}\right) ^{n}}_{=\dfrac{\partial}{\partial z_{i}}\left( y_{i}+z_{i}\right) ^{n}}\\ & =\left( \dfrac{\partial}{\partial z_{i}}\left( p_{n}\left( y+z\right) \right) \right) \cdot\left( y_{i}+z_{i}\right) ^{n}+p_{n}\left( y+z\right) \cdot\dfrac{\partial}{\partial z_{i}}\left( y_{i}+z_{i}\right) ^{n}. \end{align*} Compared with% \[ \dfrac{\partial}{\partial z_{i}}\left( p_{n}\left( y+z\right) \cdot\left( y_{i}+z_{i}\right) ^{n}\right) =\left( \dfrac{\partial}{\partial z_{i}% }\left( p_{n}\left( y+z\right) \right) \right) \cdot\left( y_{i}% +z_{i}\right) ^{n}+p_{n}\left( y+z\right) \cdot\dfrac{\partial}{\partial z_{i}}\left( y_{i}+z_{i}\right) ^{n}% \] (this follows from the Leibniz rule), this yields% \begin{equation} \dfrac{\partial}{\partial y_{i}}\left( p_{n}\left( y+z\right) \cdot\left( y_{i}+z_{i}\right) ^{n}\right) =\dfrac{\partial}{\partial z_{i}}\left( p_{n}\left( y+z\right) \cdot\left( y_{i}+z_{i}\right) ^{n}\right) . \label{pf.hirota.y+z.2}% \end{equation} \par Now, forget that we fixed $n\in\mathbb{N}$. We have shown that every $n\in\mathbb{N}$ satisfies (\ref{pf.hirota.y+z.2}). Now, since $P=\sum \limits_{n\in\mathbb{N}}p_{n}w_{i}^{n}$, we have $P\left( y+z\right) =\sum\limits_{n\in\mathbb{N}}p_{n}\left( y+z\right) \cdot\left( y_{i}% +z_{i}\right) ^{n}$, so that% \begin{align*} & \left( \dfrac{\partial}{\partial y_{i}}-\dfrac{\partial}{\partial z_{i}% }\right) \left( P\left( y+z\right) \right) \\ & =\left( \dfrac{\partial}{\partial y_{i}}-\dfrac{\partial}{\partial z_{i}% }\right) \left( \sum\limits_{n\in\mathbb{N}}p_{n}\left( y+z\right) \cdot\left( y_{i}+z_{i}\right) ^{n}\right) \\ & =\sum\limits_{n\in\mathbb{N}}\underbrace{\dfrac{\partial}{\partial y_{i}% }\left( p_{n}\left( y+z\right) \cdot\left( y_{i}+z_{i}\right) ^{n}\right) }_{\substack{=\dfrac{\partial}{\partial z_{i}}\left( p_{n}\left( y+z\right) \cdot\left( y_{i}+z_{i}\right) ^{n}\right) \\\text{(by (\ref{pf.hirota.y+z.2}))}}}-\sum\limits_{n\in\mathbb{N}}% \dfrac{\partial}{\partial z_{i}}\left( p_{n}\left( y+z\right) \cdot\left( y_{i}+z_{i}\right) ^{n}\right) \\ & =\sum\limits_{n\in\mathbb{N}}\dfrac{\partial}{\partial z_{i}}\left( p_{n}\left( y+z\right) \cdot\left( y_{i}+z_{i}\right) ^{n}\right) -\sum\limits_{n\in\mathbb{N}}\dfrac{\partial}{\partial z_{i}}\left( p_{n}\left( y+z\right) \cdot\left( y_{i}+z_{i}\right) ^{n}\right) =0. \end{align*} This proves (\ref{pf.hirota.y+z.1}).} Thus,% \begin{align*} & Q\left( \widetilde{\partial_{y}}\right) \left( P\left( y+z\right) \right) -Q\left( \widetilde{\partial_{z}}\right) \left( P\left( y+z\right) \right) \\ & =\underbrace{\left( Q\left( \widetilde{\partial_{y}}\right) -Q\left( \widetilde{\partial_{z}}\right) \right) }_{=\sum\limits_{i>0}d_{i}% \circ\left( \dfrac{\partial}{\partial y_{i}}-\dfrac{\partial}{\partial z_{i}% }\right) }\left( P\left( y+z\right) \right) =\sum\limits_{i>0}\left( d_{i}\circ\left( \dfrac{\partial}{\partial y_{i}}-\dfrac{\partial}{\partial z_{i}}\right) \right) \left( P\left( y+z\right) \right) \\ & =\sum\limits_{i>0}d_{i}\underbrace{\left( \left( \dfrac{\partial }{\partial y_{i}}-\dfrac{\partial}{\partial z_{i}}\right) \left( P\left( y+z\right) \right) \right) }_{\substack{=0\\\text{(by (\ref{pf.hirota.y+z.1}))}}}=\sum\limits_{i>0}\underbrace{d_{i}\left( 0\right) }_{=0}=0. \end{align*} In other words, $Q\left( \widetilde{\partial_{y}}\right) \left( P\left( y+z\right) \right) =Q\left( \widetilde{\partial_{z}}\right) \left( P\left( y+z\right) \right) $. This proves Lemma \ref{lem.hirota.y+z}. \textit{Proof of Theorem \ref{thm.hirota}.} We introduce a new family of indeterminates $\left( z_{1},z_{2},z_{3},...\right) $. Denote this family by $z$. (This $z$ has nothing to do with the element $z$ of $\mathcal{B}$. It is best to forget about $\mathcal{B}$ here, and only think about $\mathcal{B}% ^{\left( 0\right) }=\mathbb{C}\left[ x_{1},x_{2},x_{3},...\right] $.) Denote by $\widetilde{\partial_{z}}$ the sequence $\left( \dfrac{1}{1}% \dfrac{\partial}{\partial z_{1}},\dfrac{1}{2}\dfrac{\partial}{\partial z_{2}% },\dfrac{1}{3}\dfrac{\partial}{\partial z_{3}},...\right) $. Denote by $\widetilde{\partial_{y}}$ the sequence $\left( \dfrac{1}{1}% \dfrac{\partial}{\partial y_{1}},\dfrac{1}{2}\dfrac{\partial}{\partial y_{2}% },\dfrac{1}{3}\dfrac{\partial}{\partial y_{3}},...\right) $. Also, let $-2y$ be the sequence $\left( -2y_{1},-2y_{2},-2y_{3},...\right) $. Then,% \begin{align} \sum\limits_{k=0}^{\infty}S_{k}\left( -2y\right) u^{k} & =\sum \limits_{k\geq0}S_{k}\left( -2y\right) u^{k}=\exp\left( \sum\limits_{i\geq 1}-2y_{i}u^{i}\right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{def.schur.sk.genfun}), with }-2y\text{ substituted for }x\text{ and }u\text{ substituted for }z\right) \nonumber\\ & =\exp\left( \sum\limits_{j\geq1}-2y_{j}u^{j}\right) =\exp\left( -2\sum\limits_{j>0}u^{j}y_{j}\right) \label{pf.hirota.29}% \end{align} and% \begin{align} \sum\limits_{k=0}^{\infty}S_{k}\left( \widetilde{\partial_{y}}\right) u^{-k} & =\sum\limits_{k\geq0}S_{k}\left( \widetilde{\partial_{y}}\right) u^{-k}=\exp\left( \sum\limits_{i\geq1}\dfrac{1}{i}\dfrac{\partial}{\partial y_{i}}u^{-i}\right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{def.schur.sk.genfun}), with }\widetilde{\partial_{y}}\text{ substituted for }x\text{ and }u^{-1}\text{ substituted for }z\right) \nonumber\\ & =\exp\left( \sum\limits_{j\geq1}\dfrac{1}{j}\dfrac{\partial}{\partial y_{j}}u^{-j}\right) =\exp\left( \sum\limits_{j>0}\dfrac{u^{-j}}{j}% \dfrac{\partial}{\partial y_{j}}\right) . \label{pf.hirota.30}% \end{align} Applying Lemma \ref{lem.hirota.newton} to $K=\left( \mathbb{C}\left[ \left[ y_{1},y_{2},y_{3},...\right] \right] \right) \left[ x_{1},x_{2}% ,x_{3},...\right] $ and $P=\tau\left( x+z\right) \tau\left( x-z\right) $, we obtain% \begin{equation} \exp\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}\right) \left( \tau\left( x+z\right) \tau\left( x-z\right) \right) =\tau\left( x+y+z\right) \tau\left( x-y-z\right) . \label{pf.hirota.31}% \end{equation} Now,% \begin{align*} & \operatorname*{CT}\nolimits_{u}\left( u\exp\left( -2\sum\limits_{j>0}% u^{j}y_{j}\right) \exp\left( \sum\limits_{j>0}\dfrac{u^{-j}}{j}% \dfrac{\partial}{\partial y_{j}}\right) \left( \tau\left( x-y\right) \tau\left( x+y\right) \right) \right) \\ & =\operatorname*{CT}\nolimits_{u}\left( u\underbrace{\exp\left( -2\sum\limits_{j>0}u^{j}y_{j}\right) }_{\substack{=\sum\limits_{k=0}^{\infty }S_{k}\left( -2y\right) u^{k}\\\text{(by (\ref{pf.hirota.29}))}% }}\underbrace{\exp\left( \sum\limits_{j>0}\dfrac{u^{-j}}{j}\dfrac{\partial }{\partial y_{j}}\right) }_{\substack{=\sum\limits_{k=0}^{\infty}S_{k}\left( \widetilde{\partial_{y}}\right) u^{-k}\\\text{(by (\ref{pf.hirota.30}))}% }}\left( \tau\left( x+y+z\right) \tau\left( x-y-z\right) \right) \right) \mid_{z=0}\\ & =\operatorname*{CT}\nolimits_{u}\left( u\left( \sum\limits_{k=0}^{\infty }S_{k}\left( -2y\right) u^{k}\right) \left( \sum\limits_{k=0}^{\infty }S_{k}\left( \widetilde{\partial_{y}}\right) u^{-k}\right) \left( \tau\left( x+y+z\right) \tau\left( x-y-z\right) \right) \right) \mid_{z=0}\\ & =\sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) \underbrace{S_{j+1}% \left( \widetilde{\partial_{y}}\right) \left( \tau\left( x+y+z\right) \tau\left( x-y-z\right) \right) }_{\substack{=S_{j+1}\left( \widetilde{\partial_{z}}\right) \left( \tau\left( x+y+z\right) \tau\left( x-y-z\right) \right) \\\text{(by Lemma \ref{lem.hirota.y+z}, applied to}\\K=\mathbb{C}\left[ x_{1},x_{2},x_{3},...\right] \text{, }P=\tau\left( x+w\right) \tau\left( x-w\right) \text{ and }Q=S_{j+1}\left( w\right) \text{)}}}\mid_{z=0}\\ & =\sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{\partial_{z}}\right) \underbrace{\left( \tau\left( x+y+z\right) \tau\left( x-y-z\right) \right) }_{\substack{=\exp\left( \sum \limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}}\right) \left( \tau\left( x+z\right) \tau\left( x-z\right) \right) \\\text{(by (\ref{pf.hirota.31}% ))}}}\mid_{z=0}\\ & =\sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{\partial_{z}}\right) \exp\left( \sum\limits_{s>0}y_{s}% \dfrac{\partial}{\partial z_{s}}\right) \left( \tau\left( x+z\right) \tau\left( x-z\right) \right) \mid_{z=0}. \end{align*} Compared with the fact that (by the definition of $A\left( \sum \limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}% \right) \exp\left( \sum\limits_{s>0}y_{s}w_{s}\right) ,\tau,\tau\right) $) we have% \begin{align*} & A\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}w_{s}\right) ,\tau,\tau\right) \\ & =\underbrace{\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}% w_{s}\right) \right) \left( \partial_{z}\right) }_{=\sum\limits_{j=0}% ^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{\partial_{z}% }\right) \exp\left( \sum\limits_{s>0}y_{s}\dfrac{\partial}{\partial z_{s}% }\right) }\left( \tau\left( x+z\right) \tau\left( x-z\right) \right) \mid_{z=0}\\ & =\sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{\partial_{z}}\right) \exp\left( \sum\limits_{s>0}y_{s}% \dfrac{\partial}{\partial z_{s}}\right) \left( \tau\left( x+z\right) \tau\left( x-z\right) \right) \mid_{z=0}, \end{align*} this yields% \begin{align*} & \operatorname*{CT}\nolimits_{u}\left( u\exp\left( -2\sum\limits_{j>0}% u^{j}y_{j}\right) \exp\left( \sum\limits_{j>0}\dfrac{u^{-j}}{j}% \dfrac{\partial}{\partial y_{j}}\right) \left( \tau\left( x-y\right) \tau\left( x+y\right) \right) \right) \\ & =A\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}w_{s}\right) ,\tau,\tau\right) . \end{align*} Hence, (\ref{pf.hirota.thirdrewriting}) rewrites as follows:% \[ \left( S\left( \sigma\left( \tau\right) \otimes\sigma\left( \tau\right) \right) =0\right) \ \Longleftrightarrow\ \left( A\left( \sum \limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}% \right) \exp\left( \sum\limits_{s>0}y_{s}w_{s}\right) ,\tau,\tau\right) =0\right) . \] Since $S\left( \sigma\left( \tau\right) \otimes\sigma\left( \tau\right) \right) =0$ is equivalent to $\sigma\left( \tau\right) \in\Omega$ (by Theorem \ref{thm.plu.inf} \textbf{(b)}, applied to $\sigma\left( \tau\right) $ instead of $\tau$), this rewrites as follows:% \[ \left( \sigma\left( \tau\right) \in\Omega\right) \ \Longleftrightarrow \ \left( A\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}% w_{s}\right) ,\tau,\tau\right) =0\right) . \] This proves Theorem \ref{thm.hirota}. Theorem \ref{thm.hirota} tells us that a nonzero $\tau\in\mathcal{B}^{\left( 0\right) }$ satisfies $\sigma\left( \tau\right) \in\Omega$ if and only if it satisfies the equation (\ref{thm.hirota.eqn}). The left hand side of this equation is a power series with respect to the variables $y_{1},y_{2}% ,y_{3},...$. A power series is $0$ if and only if each of its coefficients is $0$. Hence, the equation (\ref{thm.hirota.eqn}) holds if and only if for each monomial in $y_{1},y_{2},y_{3},...$, the coefficient of the left hand side of (\ref{thm.hirota.eqn}) in front of this monomial is $0$. Thus, the equation (\ref{thm.hirota.eqn}) is equivalent to \textbf{a system of infinitely many equations}, one for each monomial in $y_{1},y_{2},y_{3},...$. We don't know of a good way to describe these equations (without using the variables $y_{1},y_{2},y_{3},...$), but we can describe the equations corresponding to the simplest among our monomials: the monomials of degree $0$ and those of degree $1$. In the following, we consider $\left( \mathbb{C}\left[ \left[ y_{1}% ,y_{2},y_{3},...\right] \right] \right) \left[ x_{1},x_{2},x_{3}% ,...\right] $ as a subring of \newline$\left( \mathbb{C}\left[ x_{1}% ,x_{2},x_{3},...\right] \right) \left[ \left[ y_{1},y_{2},y_{3}% ,...\right] \right] $. For every commutative ring $K$, every element $T$ of $K\left[ \left[ y_{1},y_{2},y_{3},...\right] \right] $ and any monomial\footnote{When we say ``monomial'', we mean a monomial without coefficient.} $\mathfrak{m}$ in the variables $y_{1},y_{2},y_{3},...$, we denote by $T\left[ \mathfrak{m}\right] $ the coefficient of the monomial $\mathfrak{m}$ in the power series $T$. (For example, $\left( \exp\left( x_{2}y_{2}\right) \right) \left[ y_{2}^{3}\right] =\dfrac{x_{2}^{3}}{6}$; note that $K=\mathbb{C}\left[ x_{1},x_{2},x_{3},...\right] $ in this example, so that $x_{2}$ counts as a constant!) For every $P\in\left( \mathbb{C}\left[ \left[ y_{1},y_{2},y_{3},...\right] \right] \right) \left[ \left[ w_{1},w_{2},w_{3},...\right] \right] $ and every monomial $\mathfrak{m}$ in the variables $y_{1},y_{2},y_{3},...$, we have \begin{equation} \left( A\left( P,\tau,\tau\right) \right) \left[ \mathfrak{m}\right] =A\left( P\left[ \mathfrak{m}\right] ,\tau,\tau\right) . \label{pf.hirota.66}% \end{equation} \footnote{\textit{Proof.} We have $P=\sum\limits_{\substack{\mathfrak{n}\text{ is a monomial}\\\text{in }y_{1},y_{2},y_{3},...}}P\left[ \mathfrak{n}\right] \cdot\mathfrak{n}$. Since the map% \begin{align*} \left( \mathbb{C}\left[ \left[ y_{1},y_{2},y_{3},...\right] \right] \right) \left[ \left[ w_{1},w_{2},w_{3},...\right] \right] & \rightarrow\left( \mathbb{C}\left[ \left[ y_{1},y_{2},y_{3},...\right] \right] \right) \left[ x_{1},x_{2},x_{3},...\right] ,\\ Q & \mapsto A\left( Q,\tau,\tau\right) \end{align*} is $\mathbb{C}\left[ \left[ y_{1},y_{2},y_{3},...\right] \right] $-linear, we have \[ A\left( \sum\limits_{\substack{\mathfrak{n}\text{ is a monomial}\\\text{in }y_{1},y_{2},y_{3},...}}P\left[ \mathfrak{n}\right] \cdot\mathfrak{n}% ,\tau,\tau\right) =\sum\limits_{\substack{\mathfrak{n}\text{ is a monomial}\\\text{in }y_{1},y_{2},y_{3},...}}A\left( P\left[ \mathfrak{n}% \right] ,\tau,\tau\right) \cdot\mathfrak{n}. \] But $P=\sum\limits_{\substack{\mathfrak{n}\text{ is a monomial}\\\text{in }y_{1},y_{2},y_{3},...}}P\left[ \mathfrak{n}\right] \cdot\mathfrak{n}$ shows that \[ A\left( P,\tau,\tau\right) =A\left( \sum\limits_{\substack{\mathfrak{n}% \text{ is a monomial}\\\text{in }y_{1},y_{2},y_{3},...}}P\left[ \mathfrak{n}\right] \cdot\mathfrak{n},\tau,\tau\right) =\sum \limits_{\substack{\mathfrak{n}\text{ is a monomial}\\\text{in }y_{1}% ,y_{2},y_{3},...}}A\left( P\left[ \mathfrak{n}\right] ,\tau,\tau\right) \cdot\mathfrak{n}, \] so that the coefficient of $A\left( P,\tau,\tau\right) $ before $\mathfrak{m}$ equals $A\left( P\left[ \mathfrak{m}\right] ,\tau ,\tau\right) $. Since we denoted the coefficient of $A\left( P,\tau ,\tau\right) $ before $\mathfrak{m}$ by $\left( A\left( P,\tau,\tau\right) \right) \left[ \mathfrak{m}\right] $, this rewrites as $\left( A\left( P,\tau,\tau\right) \right) \left[ \mathfrak{m}\right] =A\left( P\left[ \mathfrak{m}\right] ,\tau,\tau\right) $, qed.} Now, let us describe the equations that are obtained from (\ref{thm.hirota.eqn}) by taking coefficients before monomials of degree $0$ and $1$: \textbf{Monomials of degree }$0$\textbf{:} The only monomial of degree $0$ in $y_{1},y_{2},y_{3},...$ is $1$. We have% \begin{align*} & \left( A\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}% w_{s}\right) ,\tau,\tau\right) \right) \left[ 1\right] \\ & =A\left( \underbrace{\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum \limits_{s>0}y_{s}w_{s}\right) \right) \left[ 1\right] }_{=S_{1}\left( \widetilde{w}\right) =w_{1}},\tau,\tau\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.hirota.66}), applied to }P=\sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}w_{s}\right) \text{ and }\mathfrak{m}=1\right) \\ & =A\left( w_{1},\tau,\tau\right) =0\ \ \ \ \ \ \ \ \ \ \left( \text{by Corollary \ref{cor.hirota.odd}, since }w_{1}\text{ is odd}\right) . \end{align*} Therefore, if we take coefficients with respect to the monomial $1$ in the equation (\ref{pf.hirota.66}), we obtain a tautology. \textbf{Monomials of degree }$1$\textbf{:} This will be more interesting. The monomials of degree $1$ in $y_{1},y_{2},y_{3},...$ are $y_{1},y_{2},y_{3}% ,...$. Let $r$ be a positive integer. We have% \begin{align} & \left( A\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}% w_{s}\right) ,\tau,\tau\right) \right) \left[ y_{r}\right] \nonumber\\ & =A\left( \underbrace{\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum \limits_{s>0}y_{s}w_{s}\right) \right) \left[ y_{r}\right] }% _{\substack{=-2S_{r+1}\left( \widetilde{w}\right) +w_{1}w_{r}\\\text{(by easy computations)}}},\tau,\tau\right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.hirota.66}), applied to }P=\sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}w_{s}\right) \text{ and }\mathfrak{m}=y_{r}\right) \nonumber\\ & =A\left( -2S_{r+1}\left( \widetilde{w}\right) +w_{1}w_{r},\tau ,\tau\right) . \label{pf.hirota.69}% \end{align} Denote the polynomial $-2S_{r+1}\left( \widetilde{w}\right) +w_{1}w_{r}$ by $T_{r}\left( w\right) $. Then, (\ref{pf.hirota.69}) rewrites as% \begin{equation} \left( A\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}% w_{s}\right) ,\tau,\tau\right) \right) \left[ y_{r}\right] =A\left( T_{r}\left( w\right) ,\tau,\tau\right) . \label{pf.hirota.70}% \end{equation} We have $T_{1}\left( w\right) =w_{2}$, $T_{2}\left( w\right) =-\dfrac{w_{1}^{3}}{3}-\dfrac{2w_{3}}{3}$ and $T_{3}\left( w\right) =\dfrac{w_{1}w_{3}}{3}-\dfrac{w_{4}}{2}-\dfrac{w_{2}^{2}}{4}-\dfrac{w_{1}^{4}% }{12}-\dfrac{w_{1}^{2}w_{2}}{2}$. Since $T_{1}\left( w\right) $ and $T_{2}\left( w\right) $ are odd, we have $A\left( T_{1}\left( w\right) ,\tau,\tau\right) =0$ and $A\left( T_{2}\left( w\right) ,\tau,\tau\right) =0$ (by Corollary \ref{cor.hirota.odd}). Therefore, taking coefficients with respect to the monomials $y_{1}$ and $y_{2}$ in the equation (\ref{pf.hirota.66}) yields tautologies. However, $T_{3}\left( w\right) $ is \textbf{not odd}. Applying (\ref{pf.hirota.70}) to $r=3$, we obtain% \begin{align*} & \left( A\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}% w_{s}\right) ,\tau,\tau\right) \right) \left[ y_{3}\right] \\ & =A\left( T_{3}\left( w\right) ,\tau,\tau\right) =A\left( \dfrac {w_{1}w_{3}}{3}-\dfrac{w_{4}}{2}-\dfrac{w_{2}^{2}}{4}-\dfrac{w_{1}^{4}}% {12}-\dfrac{w_{1}^{2}w_{2}}{2},\tau,\tau\right) \\ & =A\left( \dfrac{w_{1}w_{3}}{3}-\dfrac{w_{2}^{2}}{4}-\dfrac{w_{1}^{4}}% {12},\tau,\tau\right) +\underbrace{A\left( -\dfrac{w_{4}}{2}-\dfrac {w_{1}^{2}w_{2}}{2},\tau,\tau\right) }_{\substack{=0\\\text{(by Corollary \ref{cor.hirota.odd}, since}\\-\dfrac{w_{4}}{2}-\dfrac{w_{1}^{2}w_{2}}% {2}\text{ is odd)}}}\\ & =A\left( \dfrac{w_{1}w_{3}}{3}-\dfrac{w_{2}^{2}}{4}-\dfrac{w_{1}^{4}}% {12},\tau,\tau\right) =\left( \left( \dfrac{\dfrac{\partial}{\partial z_{1}}\dfrac{\partial}{\partial z_{3}}}{3}-\dfrac{\left( \dfrac{\partial }{\partial z_{2}}\right) ^{2}}{4}-\dfrac{\left( \dfrac{\partial}{\partial z_{1}}\right) ^{4}}{12}\right) \left( \tau\left( x-z\right) \tau\left( x+z\right) \right) \right) \mid_{z=0}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }A\left( \dfrac{w_{1}w_{3}}{3}-\dfrac{w_{2}^{2}}{4}-\dfrac{w_{1}^{4}}{12},\tau ,\tau\right) \right) \\ & =\dfrac{1}{12}\left( \left( 4\dfrac{\partial}{\partial z_{1}}% \dfrac{\partial}{\partial z_{3}}-3\left( \dfrac{\partial}{\partial z_{2}% }\right) ^{2}-\left( \dfrac{\partial}{\partial z_{1}}\right) ^{4}\right) \left( \tau\left( x-z\right) \tau\left( x+z\right) \right) \right) \mid_{z=0}\\ & =\dfrac{1}{12}\left( \left( 4\dfrac{\partial}{\partial w_{1}}% \dfrac{\partial}{\partial w_{3}}-3\left( \dfrac{\partial}{\partial w_{2}% }\right) ^{2}-\left( \dfrac{\partial}{\partial w_{1}}\right) ^{4}\right) \left( \tau\left( x-w\right) \tau\left( x+w\right) \right) \right) \mid_{w=0}. \end{align*} Since $\dfrac{\partial}{\partial w_{j}}=\partial_{w_{j}}$ for every $j$, we rewrite this as% \begin{align*} & \left( A\left( \sum\limits_{j=0}^{\infty}S_{j}\left( -2y\right) S_{j+1}\left( \widetilde{w}\right) \exp\left( \sum\limits_{s>0}y_{s}% w_{s}\right) ,\tau,\tau\right) \right) \left[ y_{3}\right] \\ & =\dfrac{1}{12}\left( \left( 4\partial_{w_{1}}\partial_{w_{3}}% -3\partial_{w_{2}}^{2}-\partial_{w_{1}}^{4}\right) \left( \tau\left( x-w\right) \tau\left( x+w\right) \right) \right) \mid_{w=0}. \end{align*} Hence, taking coefficients with respect to the monomial $y_{3}$ in the equation (\ref{thm.hirota.eqn}), we obtain% \[ \dfrac{1}{12}\left( \left( 4\partial_{w_{1}}\partial_{w_{3}}-3\partial _{w_{2}}^{2}-\partial_{w_{1}}^{4}\right) \left( \tau\left( x-w\right) \tau\left( x+w\right) \right) \right) \mid_{w=0}=0. \] In other words, \begin{equation} \left( \partial_{w_{1}}^{4}+3\partial_{w_{2}}^{2}-4\partial_{w_{1}}% \partial_{w_{3}}\right) \left( \tau\left( x-w\right) \tau\left( x+w\right) \right) \mid_{w=0}=0. \label{KdV.star}% \end{equation} This does not yet look like a PDE in any usual form. We will now transform it into one. We make the substitution $x_{1}=x$, $x_{2}=y$, $x_{3}=t$, $x_{m}=c_{m}$ for $m\geq4$. Here, $x$, $y$, $t$ and $c_{m}$ (for $m\geq4$) are new symbols (in particularly, $x$ and $y$ no longer denote the sequences $\left( x_{1}% ,x_{2},x_{3},...\right) $ and $\left( y_{1},y_{2},y_{3},...\right) $). Let $u=2\partial_{x}^{2}\log\tau$. \begin{proposition} \label{prop.KdV.computation}The polynomial $\tau\left( x,y,t,c_{4}% ,c_{5},...\right) $ satisfies (\ref{KdV.star}) if and only if the function $u$ satisfies the KP equation \[ \dfrac{3}{4}\partial_{y}^{2}u=\partial_{x}\left( \partial_{t}u-\dfrac{3}% {2}u\partial_{x}u-\dfrac{1}{4}\partial_{x}^{3}u\right) \] (where $c_{4}$, $c_{5}$, $c_{6}$, $...$ are considered as constants). \end{proposition} \textit{Proof of Proposition \ref{prop.KdV.computation}.} Optional homework exercise. Thus, we know that any element $\tau$ of $\Omega$ gives rise to a solution of the KP equation (namely, the solution is $2\partial_{x}^{2}\log\left( \sigma^{-1}\left( \tau\right) \right) $). Two elements of $\Omega$ differing from each other by a scalar factor yield one and the same solution of the KP equation. Hence, any element of $\operatorname*{Gr}$ gives rise to a solution of the KP equation. Since we know how to produce elements of $\operatorname*{Gr}$, we thus know how to produce solutions of the KP equation! This does not give \textbf{all} solutions, and in fact we cannot even hope to find all solutions explicitly (since they depend on boundary conditions, and these can be arbitrarily nonexplicit), but we will use this to find a dense subset of them (in an appropriate sense). The KP equation is not the KdV (Korteweg-de Vries) equation; but if we have a solution of the KP equation which does not depend on $y$, then this solution satisfies $\partial_{t}u-\dfrac{3}{2}u\partial_{x}u-\dfrac{1}{4}\partial _{x}^{3}u=\operatorname*{const}$, and with some work it gives rise to a solution of the KdV equation (under appropriate decay-at-infinity conditions). The equations corresponding to the coefficients of the monomials $y_{4}$, $y_{5}$, $...$ in (\ref{pf.hirota.66}) correspond to the \textit{KP hierarchy} of higher-order PDEs. There is no point in writing them up explicitly; they become more and more complicated. \begin{corollary} \label{cor.KdV.schursols}Let $\lambda$ be a partition. Then, $2\partial _{x}^{2}\log\left( S_{\lambda}\left( x,y,t,c_{4},c_{5},...\right) \right) $ is a solution of the KP equation (and of the whole KP hierarchy), where $c_{4}$, $c_{5}$, $c_{6}$, $...$ are considered as constants. \end{corollary} \textit{Proof of Corollary \ref{cor.KdV.schursols}.} Write $\lambda$ in the form $\lambda=\left( \lambda_{0},\lambda_{1},\lambda_{2},...\right) $. Let $\left( i_{0},i_{1},i_{2},...\right) $ be the sequence defined by $i_{k}=\lambda_{k}-k$ for every $k\in\mathbb{N}$. Then, $\left( i_{0}% ,i_{1},i_{2},...\right) $ is a $0$-degression, and we know that the elementary semiinfinite wedge $v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...$ is in $\Omega$. But Theorem \ref{thm.schur} yields $\sigma ^{-1}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\right) =S_{\lambda}\left( x\right) $ (since $\lambda=\left( i_{0}+0,i_{1}% +1,i_{2}+2,...\right) $), so that $\sigma\left( S_{\lambda}\left( x\right) \right) =v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}\wedge...\in\Omega$. Hence, the function $2\partial_{x}^{2}\log\left( S_{\lambda}\left( x,y,t,c_{4}% ,c_{5},...\right) \right) $ satisfies the KP equation (and the whole KP hierarchy). This proves Corollary \ref{cor.KdV.schursols}. \subsubsection{\textbf{[unfinished]} \texorpdfstring{$n$}{n}-soliton solutions of KdV} Now we will construct other solutions of the KdV equations (which are called multisoliton solutions). We will identify the $\mathcal{A}$-modules $\mathcal{B}^{\left( 0\right) }$ and $\mathcal{F}^{\left( 0\right) }$ along the Boson-Fermion correspondence $\sigma$. \begin{definition} Define a quantum field $\Gamma\left( u,v\right) \in\left( \operatorname*{End}\left( \mathcal{B}^{\left( 0\right) }\right) \right) \left[ \left[ u,u^{-1},v,v^{-1}\right] \right] $ by% \begin{equation} \Gamma\left( u,v\right) =\exp\left( \sum\limits_{j\geq1}\dfrac{u^{j}-v^{j}% }{j}a_{-j}\right) \exp\left( -\sum\limits_{j\geq1}\dfrac{u^{-j}-v^{-j}}% {j}a_{j}\right) . \label{n-soliton.Gamma(u,v).def1}% \end{equation} \end{definition} It is possible to rewrite the equality (\ref{n-soliton.Gamma(u,v).def1}) in the following form:% \begin{equation} \Gamma\left( u,v\right) =u\left. :\Gamma\left( u\right) \Gamma^{\ast }\left( v\right) :\right. . \label{n-soliton.Gamma(u,v).def2}% \end{equation} However, before we can make sense of this equality (\ref{n-soliton.Gamma(u,v).def2}), we need to explain what we mean by $\left. :\Gamma\left( u\right) \Gamma^{\ast}\left( v\right) :\right. $. Theorem \ref{thm.euler} (applied to $m=-1$ and to $m=0$) yields that% \begin{equation} \Gamma\left( u\right) =z\exp\left( \sum\limits_{j>0}\dfrac{a_{-j}}{j}% u^{j}\right) \cdot\exp\left( -\sum\limits_{j>0}\dfrac{a_{j}}{j}% u^{-j}\right) \ \ \ \ \ \ \ \ \ \ \text{on }\mathcal{B}^{\left( -1\right) } \label{pf.n-soliton.0}% \end{equation} and% \begin{equation} \Gamma^{\ast}\left( u\right) =z^{-1}\exp\left( -\sum\limits_{j>0}% \dfrac{a_{-j}}{j}u^{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{a_{j}% }{j}u^{-j}\right) \ \ \ \ \ \ \ \ \ \ \text{on }\mathcal{B}^{\left( 0\right) }. \label{pf.n-soliton.1}% \end{equation} Renaming $u$ as $v$ in (\ref{pf.n-soliton.1}), we obtain% \begin{equation} \Gamma^{\ast}\left( v\right) =z^{-1}\exp\left( -\sum\limits_{j>0}% \dfrac{a_{-j}}{j}v^{j}\right) \cdot\exp\left( \sum\limits_{j>0}\dfrac{a_{j}% }{j}v^{-j}\right) \ \ \ \ \ \ \ \ \ \ \text{on }\mathcal{B}^{\left( 0\right) }. \label{pf.n-soliton.2}% \end{equation} If we now extend the ``normal ordered product'' which we have defined on $U\left( \mathcal{A}\right) $ to a ``normal ordered multiplication map'' $U\left( \mathcal{A}\right) \left[ z\right] \left[ \left[ u,u^{-1}% \right] \right] \times U\left( \mathcal{A}\right) \left[ z\right] \left[ \left[ v,v^{-1}\right] \right] \rightarrow U\left( \mathcal{A}% \right) \left[ z\right] \left[ \left[ u,u^{-1},v,v^{-1}\right] \right] $ [...] [This isn't really that easy to formalize, and this formalization is wrong.] [According to Etingof, one can put these power series on a firm footing by defining a series $\gamma\in\left( \operatorname*{Hom}\left( A,B\right) \right) \left[ \left[ u,u^{-1}\right] \right] $ (where $A$ and $B$ are two \textbf{graded} vector spaces) to be ``sampled-rational'' if every homogeneous $w\in A$ and every homogeneous $f\in B^{\ast}$ satisfy $\left\langle f,\gamma w\right\rangle \in\mathbb{C}\left( u\right) $. Sampled-rational power series form a torsion-free $\mathbb{C}\left( u\right) $-module\footnote{But I don't think the composition of any two sampled-rational power series is sampled-rational. Ideas?}. And limits are defined sample-wise (see below). But it probably needs some explanations how $\mathbb{C}\left( u\right) $ is embedded in $\mathbb{C}\left[ \left[ u,u^{-1}\right] \right] $ (or what it means for an element of $\mathbb{C}% \left[ \left[ u,u^{-1}\right] \right] $ to be a rational function).] We will use the following notation, generalizing Definition \ref{def.OMEGA}: \begin{definition} Let $A$ and $B$ be two $\mathbb{C}$-vector spaces, and let $\left( u_{1},u_{2},...,u_{\ell}\right) $ be a sequence of distinct symbols. For every $\ell$-tuple $\mathbf{i}\in\mathbb{Z}^{\ell}$, define a monomial $\mathbf{u}^{\mathbf{i}}\in\mathbb{C}\left( \left( u_{1},u_{2},...,u_{\ell }\right) \right) $ by $\mathbf{u}^{\mathbf{i}}=u_{1}^{i_{1}}u_{2}^{i_{2}% }...u_{\ell}^{i_{\ell}}$, where the $\ell$-tuple $\mathbf{i}$ is written in the form $\mathbf{i}=\left( i_{1},i_{2},...,i_{\ell}\right) $. Then, the map% \begin{align*} A\left( \left( u_{1},u_{2},...,u_{\ell}\right) \right) \times B\left( \left( u_{1},u_{2},...,u_{\ell}\right) \right) & \rightarrow\left( A\otimes B\right) \left( \left( u_{1},u_{2},...,u_{\ell}\right) \right) ,\\ \left( \sum\limits_{\mathbf{i}\in\mathbb{Z}^{\ell}}a_{\mathbf{i}}% \mathbf{u}^{\mathbf{i}},\sum\limits_{\mathbf{i}\in\mathbb{Z}^{\ell}% }b_{\mathbf{i}}\mathbf{u}^{\mathbf{i}}\right) & \mapsto\sum \limits_{\mathbf{i}\in\mathbb{Z}^{\ell}}\left( \sum\limits_{\mathbf{j}% \in\mathbb{Z}^{\ell}}a_{\mathbf{j}}\otimes b_{\mathbf{i}-\mathbf{j}}\right) \mathbf{u}^{\mathbf{i}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{where all }a_{\mathbf{i}}\text{ lie in }A\text{ and all }b_{\mathbf{i}}\text{ lie in }B\right) \end{align*} is well-defined (in fact, it is easy to see that for any Laurent series $\sum\limits_{\mathbf{i}\in\mathbb{Z}^{\ell}}a_{\mathbf{i}}\mathbf{u}% ^{\mathbf{i}}\in A\left( \left( u_{1},u_{2},...,u_{\ell}\right) \right) $ with all $a_{\mathbf{i}}$ lying in $A$, any Laurent series $\sum \limits_{\mathbf{i}\in\mathbb{Z}^{\ell}}b_{\mathbf{i}}\mathbf{u}^{\mathbf{i}% }\in B\left( \left( u_{1},u_{2},...,u_{\ell}\right) \right) $ with all $b_{\mathbf{i}}$ lying in $B$, and any $\ell$-tuple $\mathbf{i}\in \mathbb{Z}^{\ell}$, the sum $\sum\limits_{\mathbf{j}\in\mathbb{Z}^{\ell}% }a_{\mathbf{j}}\otimes b_{\mathbf{i}-\mathbf{j}}$ has only finitely many addends and vanishes if any coordinate of $\mathbf{i}$ is small enough) and $\mathbb{C}$-bilinear. Hence, it induces a $\mathbb{C}$-linear map% \begin{align*} A\left( \left( u_{1},u_{2},...,u_{\ell}\right) \right) \otimes B\left( \left( u_{1},u_{2},...,u_{\ell}\right) \right) & \rightarrow\left( A\otimes B\right) \left( \left( u_{1},u_{2},...,u_{\ell}\right) \right) ,\\ \left( \sum\limits_{\mathbf{i}\in\mathbb{Z}^{\ell}}a_{\mathbf{i}}% \mathbf{u}^{\mathbf{i}}\right) \otimes\left( \sum\limits_{\mathbf{i}% \in\mathbb{Z}^{\ell}}b_{\mathbf{i}}\mathbf{u}^{\mathbf{i}}\right) & \mapsto\sum\limits_{\mathbf{i}\in\mathbb{Z}^{\ell}}\left( \sum \limits_{\mathbf{j}\in\mathbb{Z}^{\ell}}a_{\mathbf{j}}\otimes b_{\mathbf{i}% -\mathbf{j}}\right) \mathbf{u}^{\mathbf{i}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{where all }a_{\mathbf{i}}\text{ lie in }A\text{ and all }b_{\mathbf{i}}\text{ lie in }B\right) . \end{align*} This map will be denoted by $\Omega_{A,B,\left( u_{1},u_{2},...,u_{\ell }\right) }$. Clearly, when $\ell=1$, this map $\Omega_{A,B,\left( u_{1}\right) }$ is identical with the map $\Omega_{A,B,u_{1}}$ defined in Definition \ref{def.OMEGA}. \end{definition} \begin{proposition} \label{prop.KdV.grassm}If $\tau\in\Omega$ and $a\in\mathbb{C}$, then% \[ \left( 1+a\Gamma\left( u,v\right) \right) \tau\in\Omega_{u,v}, \] where% \[ \Omega_{u,v}=\left\{ \tau\in\mathcal{B}^{\left( 0\right) }\left( \left( u,v\right) \right) \ \mid\ S\left( \tau\otimes\tau\right) =0\right\} . \] (Here, the $S$ really means not the map $S:\mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }\rightarrow\mathcal{B}^{\left( 1\right) }\otimes\mathcal{B}^{\left( -1\right) }$ itself, but rather the map $\left( \mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }\right) \left( \left( u,v\right) \right) \rightarrow\left( \mathcal{B}^{\left( 1\right) }\otimes\mathcal{B}^{\left( -1\right) }\right) \left( \left( u,v\right) \right) $ it induces. And $\tau \otimes\tau$ means not $\tau\otimes\tau\in\mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }$ but rather $\Omega_{\mathcal{B}% ^{\left( 0\right) },\mathcal{B}^{\left( 0\right) },\left( u,v\right) }\left( \tau\otimes\tau\right) \in\left( \mathcal{B}^{\left( 0\right) }\otimes\mathcal{B}^{\left( 0\right) }\right) \left( \left( u,v\right) \right) $.) \end{proposition} \begin{corollary} For any $a^{(1)},a^{(2)},...,a^{(n)}\in\mathbb{C}$, we have \begin{align*} & \left( 1+a^{(1)}\Gamma\left( u_{1},v_{1}\right) \right) \left( 1+a^{(2)}\Gamma\left( u_{2},v_{2}\right) \right) ...\left( 1+a^{(n)}% \Gamma\left( u_{n},v_{n}\right) \right) \mathbf{1}\\ & \in\Omega \end{align*} (in fact, in an appropriate $\Omega_{u_{1},v_{1},u_{2},v_{2},...}$ rather than in $\Omega$ itself). \end{corollary} \textit{Idea of proof of Proposition.} We will prove $\Gamma\left( u,v\right) ^{2}=0$, but we will have to make sense of a term like $\Gamma\left( u,v\right) ^{2}$ in order to define this. Thus, $1+a\Gamma \left( u,v\right) $ will become $\exp\left( a\Gamma\left( u,v\right) \right) $. We will formalize this proof later. But first, here is the punchline of this: \begin{proposition} Let $a^{(1)},a^{(2)},...,a^{(n)}\in\mathbb{C}$. If $\tau=\left( 1+a^{(1)}\Gamma\left( u_{1},v_{1}\right) \right) \left( 1+a^{(2)}% \Gamma\left( u_{2},v_{2}\right) \right) ...\left( 1+a^{(n)}\Gamma\left( u_{n},v_{n}\right) \right) \mathbf{1}$, then $2\partial_{x}^{2}\log\tau$ is given by a convergent series and defines a solution of KP depending on the parameters $a^{(i)}$, $u_{i}$ and $v_{i}$. \end{proposition} This solution is called an $n$\textit{-soliton solution}. For $n=1$, we have% \[ \tau=\left( 1+a\Gamma\left( u,v\right) \right) \mathbf{1}=1+a\exp\left( \left( u-v\right) x+\left( u^{2}-v^{2}\right) y+\left( u^{3}% -v^{3}\right) t+\left( u^{4}-v^{4}\right) c_{4}+...\right) . \] Absorb the $c_{i}$ parameters into a single constant $c$, which can be absorbed into $a$. So we get% \[ \tau=1+a\exp\left( \left( u-v\right) x+\left( u^{2}-v^{2}\right) y+\left( u^{3}-v^{3}\right) t\right) . \] This $\tau$ satisfies \[ 2\partial_{x}^{2}\log\tau=\dfrac{\left( u-v\right) ^{2}}{2}\dfrac{1}% {\cosh^{2}\left( \dfrac{1}{2}\left( \left( u-v\right) x+\left( u^{2}-v^{2}\right) y+\left( u^{3}-v^{3}\right) \tau\right) \right) }. \] Call this function $U$. To make it independent of $y$ (so we get a solution of KdV equation), we set $v=-u$, and this becomes% \[ U=\dfrac{2u^{2}}{\cosh^{2}\left( ux+u^{3}t\right) }. \] This is exactly the soliton solution of KdV. But let us now give the promised proof of Proposition \ref{prop.KdV.grassm}. \textit{Proof of Proposition \ref{prop.KdV.grassm}.} Recall that $\Gamma\left( u,v\right) =u\left. :\Gamma\left( u\right) \Gamma^{\ast }\left( v\right) :\right. $. We can show: \begin{lemma} \label{lem.KdV.GG}We have% \[ \Gamma\left( u\right) \Gamma\left( v\right) =\left( u-v\right) \cdot\left. :\Gamma\left( u\right) \Gamma\left( v\right) :\right. \] and% \[ \Gamma\left( u\right) \Gamma^{\ast}\left( v\right) =\dfrac{1}{u-v}\left. :\Gamma\left( u\right) \Gamma^{\ast}\left( v\right) :\right. \] and% \[ \Gamma^{\ast}\left( u\right) \Gamma\left( v\right) =\dfrac{1}{u-v}\left. :\Gamma^{\ast}\left( u\right) \Gamma\left( v\right) :\right. \] and% \[ \Gamma^{\ast}\left( u\right) \Gamma^{\ast}\left( v\right) =\left( u-v\right) \cdot\left. :\Gamma^{\ast}\left( u\right) \Gamma^{\ast}\left( v\right) :\right. . \] \end{lemma} \textit{Proof of Lemma \ref{lem.KdV.GG}.} We have% \[ ...\exp\left( \sum\limits_{j>0}\dfrac{a_{j}}{j}u^{-j}\right) \exp\left( \sum\limits_{k>0}\dfrac{a_{-k}}{k}v^{k}\right) ... \] and we have to switch these two terms. We get something like% \[ \exp\left( -\log\left( 1-\dfrac{v}{u}\right) \right) =\dfrac{1}% {1-\dfrac{v}{u}}=\dfrac{u}{u-v}. \] Etc. We can generalize this: If $\varepsilon=1$ or $\varepsilon=-1$, we can define $\Gamma_{\varepsilon}$ by $\Gamma_{+1}=\Gamma$ and $\Gamma_{-1}=\Gamma^{\ast}% $. Then, \begin{proposition} We have% \[ \Gamma_{\varepsilon_{1}}\left( u_{1}\right) \Gamma_{\varepsilon_{2}}\left( u_{2}\right) ...\Gamma_{\varepsilon_{n}}\left( u_{n}\right) =\prod \limits_{i
\left\vert u_{2}\right\vert >...>\left\vert u_{n}\right\vert $. \end{proposition} \begin{corollary} The matrix elements of $\Gamma_{\varepsilon_{1}}\left( u_{1}\right) \Gamma_{\varepsilon_{2}}\left( u_{2}\right) ...\Gamma_{\varepsilon_{n}% }\left( u_{n}\right) $ (this means expressions of the form $\left( w^{\ast },\Gamma_{\varepsilon_{1}}\left( u_{1}\right) \Gamma_{\varepsilon_{2}% }\left( u_{2}\right) ...\Gamma_{\varepsilon_{n}}\left( u_{n}\right) w\right) $ with $w\in\mathcal{B}^{\left( 0\right) }$ and $w^{\ast}% \in\mathcal{B}^{\left( 0\right) \ast}$ (where $^{\ast}$ means restricted dual); a priori, these are only series) are series which converge to rational functions of the form% \[ P\left( u\right) \cdot\prod\limits_{i 0$. \begin{proposition} We have $\det\nolimits_{n}\left( c,h\right) =0$ if and only if there exists a singular vector $w\neq0$ in $M_{c,h}$ of degree $\leq n$ and $>0$. In particular, if $\det\nolimits_{n}\left( c,h\right) =0$, then $\det\nolimits_{n+1}\left( c,h\right) =0$. \end{proposition} In fact, we will see that $\det\nolimits_{n+1}$ is divisible by $\det \nolimits_{n}$. \textit{Proof of Proposition.} Apparently this is supposed to follow from something we did. We recall examples:% \begin{align*} \det\nolimits_{1} & =2h,\\ \det\nolimits_{2} & =2h\left( 16h^{2}+2hc-10h+c\right) . \end{align*} Also recall that $M_{c,h}$ is irreducible if and only if every positive $n$ satisfies $\det\nolimits_{n}\left( c,h\right) \neq0$. \begin{proposition} Let $\left( c,h\right) \in\mathbb{R}^{2}$. If $M_{c,h}$ is unitary, then $\det\nolimits_{n}\left( c,h\right) >0$ for all positive $n$. More generally, if $L_{c,h}\left[ n\right] \cong M_{c,h}\left[ n\right] $ for some positive $n$, and $L_{c,h}$ is unitary, then $\det\nolimits_{n}% \left( c,h\right) >0$. \end{proposition} \textit{Proof of Proposition.} A positive-definite Hermitian matrix has positive determinant. \begin{theorem} \label{thm.kac.leader}Fix $c$. Regard $\det\nolimits_{m}\left( c,h\right) $ as a polynomial in $h$. Then,% \[ \det\nolimits_{m}\left( c,h\right) =K\cdot h^{\sum\limits_{\substack{r,s\geq 1;\\rs\leq m}}p\left( m-rs\right) }+\left( \text{lower terms}\right) \] for some nonzero constant $K$ (which depends on the choice of the basis). \end{theorem} \textit{Proof.} We computed before the leading term of $\det\nolimits_{m}$ for any graded Lie algebra. $\left( L_{-k}^{m_{k}}...L_{-1}^{m_{1}}v_{\lambda},L_{-k}^{n_{k}}% ...L_{-1}^{n_{1}}v_{\lambda}\right) $: the main contribution to the leading term comes from diagonal. What degree in $h$ do we get? If $\mu$ is a partition of $m$, we can write $m=1k_{1}\left( \mu\right) +2k_{2}\left( \mu\right) +...$, where $k_{i}\left( \mu\right) $ is the number of times $i$ occurs in $\mu$. $\left( L_{-\ell}^{k_{\ell}}...L_{-1}^{k_{1}}v,L_{-\ell}^{k_{\ell}}% ...L_{-1}^{k_{1}}v\right) =\left( v,L_{1}^{k_{1}}...L_{\ell}^{k_{\ell}% }L_{-\ell}^{k_{\ell}}...L_{-1}^{k_{1}}v\right) $. So $\mu$ contributes $k_{1}+...+k_{\ell}$ to the exponent of $h$. So we conclude that the total exponent of $h$ is $\sum\limits_{\mu\vdash m}\sum\limits_{i}k_{i}\left( \mu\right) $. The rest is easy combinatorics: Let $m\left( r,s\right) $ denote the number of partitions of $m$ in which $r$ occurs exactly $s$ times. Then, $m\left( r,s\right) =p\left( m-rs\right) -p\left( m-r\left( s+1\right) \right) $. Thus, with $m$ and $r$ fixed,% \begin{align*} \sum\limits_{s}sm\left( r,s\right) & =\sum\limits_{s}s\left( p\left( m-rs\right) -p\left( m-r\left( s+1\right) \right) \right) \\ & =\sum\limits_{s}sp\left( m-rs\right) -\sum\limits_{s}sp\left( m-r\left( s+1\right) \right) \\ & =\sum\limits_{s}sp\left( m-rs\right) -\sum\limits_{s}\left( s-1\right) p\left( m-rs\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }s-1\text{ for }s\text{ in the second sum}\right) \\ & =\sum\limits_{s}\underbrace{\left( s-\left( s-1\right) \right) }% _{=1}p\left( m-rs\right) =\sum\limits_{s}p\left( m-rs\right) . \end{align*} So our job is to show that $\sum\limits_{\mu\vdash m}\sum\limits_{i}% k_{i}\left( \mu\right) =\sum\limits_{\substack{r,s\geq1;\\rs\leq m}}sm\left( r,s\right) $. But $\sum\limits_{\substack{s\geq1;\\s\leq m}}sm\left( r,s\right) $ is the total number of occurrences of $r$ in all partitions of $m$. Summed over $r$, it yields the total number of parts of all partitions of $m$. But this is also $\sum\limits_{\mu\vdash m}\sum \limits_{i}k_{i}\left( \mu\right) $, qed. We now quote a theorem which was proved independently by Kac and Feigin-Fuchs: \begin{theorem} Suppose $rs\leq m$. Then, if% \[ h=h_{r,s}\left( c\right) :=\dfrac{1}{48}\left( \left( 13-c\right) \left( r^{2}+s^{2}\right) +\sqrt{\left( c-1\right) \left( c-25\right) }\left( r^{2}-s^{2}\right) -24rs-2+2c\right) , \] then $\det\nolimits_{m}\left( c,h\right) =0$. (This is true for any of the branches of the square root.) \end{theorem} \begin{theorem} \label{thm.kac.thm1}If $h=h_{r,s}\left( c\right) $, then $M_{c,h}$ has a nonzero singular vector in degree $1\leq d\leq rs$. \end{theorem} \begin{theorem} [Kac, also proved by Feigin-Fuchs]\label{thm.kac.thm2}We have% \[ \det\nolimits_{m}\left( c,h\right) =K_{m}\cdot\prod \limits_{\substack{r,s\geq1;\\rs\leq m}}\left( h-h_{r,s}\left( c\right) \right) ^{p\left( m-rs\right) }, \] where $K_{m}$ is some constant. Note that we should choose the same branch of the square root in $\sqrt{\left( c-1\right) \left( c-25\right) }$ for $h_{r,s}$ and $h_{s,r}$. The square roots ``cancel out'' and give way to a polynomial in $h$ and $c$. \end{theorem} To prove these, we will use the following lemma: \begin{lemma} \label{lem.kac.linalg}Let $A\left( t\right) $ be a polynomial in one variable $t$ with values in $\operatorname*{End}V$, where $V$ is a finite-dimensional vector space. Suppose that $\dim\operatorname*{Ker}\left( A\left( 0\right) \right) \geq n$. Then, $\det\left( A\left( t\right) \right) $ is divisible by $t^{n}$. \end{lemma} \textit{Proof of Lemma \ref{lem.kac.linalg}.} Pick a basis $e_{1}% ,e_{2},...,e_{m}$ of $V$ such that the first $n$ vectors $e_{1},e_{2}% ,...,e_{n}$ are in $\operatorname*{Ker}\left( A\left( 0\right) \right) $. Then, the matrix of $A\left( t\right) $ in this basis has first $n$ columns divisible by $t$, so that its determinant $\det\left( A\left( t\right) \right) $ is divisible by $t^{n}$. \textit{Proof of Theorem \ref{thm.kac.thm2}.} Let $A=A\left( h\right) $ be the matrix of the contravariant form in degree $m$, considered as a polynomial in $h$. If $h=h_{r,s}\left( c\right) $, we have a singular vector $w$ in degree $1\leq d\leq rs$ (by Theorem \ref{thm.kac.thm1}), which generates a Verma submodule $M_{c,h^{\prime}}\subseteq M_{c,h}$ (by Homework Set 3 problem 1) (the $c$ is the same since $c$ is central and thus acts by the same number on all vectors). So $M_{c,h}\left[ m\right] \supseteq M_{c,h^{\prime}}\left[ m-d\right] $. We also have $\dim\left( M_{c,h^{\prime}}\left[ m-d\right] \right) =p\left( m-d\right) \geq p\left( m-rs\right) $ (since $d\leq rs$) and $M_{c,h^{\prime}}\left[ m-d\right] \subseteq\operatorname*{Ker}\left( \cdot,\cdot\right) $ (when $h=h_{r,s}$). Hence, $\dim\left( \operatorname*{Ker}\left( \cdot,\cdot\right) \right) \geq p\left( m-rs\right) $. By Lemma \ref{lem.kac.linalg}, this yields that $\det \nolimits_{m}\left( c,h\right) $ is divisible by $\left( h-h_{r,s}\left( c\right) \right) ^{p\left( m-rs\right) }$. But it is easy to see that for Weil-generic $c$, the $h-h_{r,s}\left( c\right) $ are different, so that $\det\nolimits_{m}\left( c,h\right) $ is divisible by $\prod\limits_{\substack{r,s\geq1;\\rs\leq m}}\left( h-h_{r,s}\left( c\right) \right) ^{p\left( m-rs\right) }$. But by Theorem \ref{thm.kac.leader}, the leading term of $\det\nolimits_{m}\left( c,h\right) $ is $K\cdot h^{\sum\limits_{\substack{r,s\geq1;\\rs\leq m}}p\left( m-rs\right) }$, which has exactly the same degree. So $\det\nolimits_{m}\left( c,h\right) $ is a constant multiple of $\prod\limits_{\substack{r,s\geq1;\\rs\leq m}}\left( h-h_{r,s}\left( c\right) \right) ^{p\left( m-rs\right) }$. Theorem \ref{thm.kac.thm2} is proven. We will not prove Theorem \ref{thm.kac.thm1}, since we do not have the tools for that. \begin{corollary} The module $M_{c,h}$ is irreducible if and only if $\left( c,h\right) $ does not lie on the lines \[ h-h_{r,r}\left( c\right) =0\ \Longleftrightarrow\ h+\left( r^{2}-1\right) \left( c-1\right) /24=0 \] and quadrics (in fact, hyperbolas if we are over $\mathbb{R}$)% \begin{align*} & \ \left( h-h_{r,s}\left( c\right) \right) \left( h-h_{s,r}\left( c\right) \right) =0\\ & \Longleftrightarrow\ \left( h-\dfrac{\left( r-s\right) ^{2}}{4}\right) ^{2}+\dfrac{h}{24}\left( c-1\right) \left( r^{2}+s^{2}-2\right) +\dfrac {1}{576}\left( r^{2}-1\right) \left( s^{2}-1\right) \left( c-1\right) ^{2}\\ & \ \ \ \ \ \ \ \ \ \ +\dfrac{1}{48}\left( c-1\right) \left( r-s\right) ^{2}\left( rs+1\right) =0. \end{align*} \end{corollary} \begin{corollary} \label{cor.kac.irred}\textbf{(1)} Let $h\geq0$ and $c\geq1$. Then, $L_{c,h}$ is unitary. \textbf{(2)} Let $h>0$ and $c>1$. Then, $M_{c,h}\cong L_{c,h}$, so that $M_{c,h}$ is irreducible. \end{corollary} \textit{Proof of Corollary \ref{cor.kac.irred}.} \textbf{(2)} Lines and hyperbolas do not pass through the region. For part \textbf{(1)} we need a lemma: \begin{lemma} Let $\mathfrak{g}$ be a graded Lie algebra (with $\dim\mathfrak{g}_{i}% \neq\infty$) with a real structure $\dag$. Let $U\subseteq\mathfrak{g}% _{0\mathbb{R}}^{\ast}$ be the set of all $\lambda$ such that $L_{\lambda}$ is unitary. Then, $U$ is closed in the usual metric. [\textbf{Note:} This lemma possibly needs additional assumptions, like the assumption that the map $\dag$ reverses the degree (i. e., every $j\in\mathbb{Z}$ satisfies $\dag\left( \mathfrak{g}_{j}\right) \subseteq\mathfrak{g}_{-j}$) and that $\mathfrak{g}_{0}$ is an abelian Lie algebra.] \end{lemma} \textit{Proof of Lemma.} It follows from the fact that if $\left( A_{n}\right) $ is a sequence of positive definite Hermitian matrices, and $\lim\limits_{n\rightarrow\infty}A_{n}=A_{\infty}$, then $A_{\infty}$ is nonnegative definite. Okay, sorry, we are not going to use this lemma; we will derive the special case we need. Now I claim that if $h>0$ and $c>1$, then $L_{c,h}=M_{c,h}$ is unitary. We know this is true for some points of this region (namely, the ones ``above the zigzag line''). Then everything follows from the fact that if $A\left( t\right) $ is a continuous family of nondegenerate Hermitian matrices parametrized by $t\in\left[ 0,1\right] $ such that $A\left( 0\right) >0$, then $A\left( t\right) >0$ for every $t$. (This fact is because the signature of a nondegenerate Hermitian matrix is a continuous map to a discrete set, and thus constant on connected components.) e. g., consider $M_{1,h}$ as a limit of $M_{1+\dfrac{1}{n},h}$ (this is irreducible for large $n$). So the matrix of the form in $M_{1,h}\left[ m\right] $ is a limit of the matrices for $M_{1+\dfrac{1}{n},h}\left[ m\right] $. So the matrix for $M_{1,h}\left[ m\right] $ is $\geq0$. But kernel lies in $J_{1,h}\left[ m\right] $, so the form on $L_{1,h}\left[ m\right] =\left( M_{1,h}\diagup J_{1,h}\right) \left[ m\right] $ is strictly positive. By analyzing the Kac curves, we can show (although \textit{we} will \textit{not} show) that in the region $0\leq c<1$, there are only countably many points where we possibly can have unitarity: $c\left( m\right) =1-\dfrac{6}{\left( m+2\right) \left( m+3\right) };$ $h_{r,s}\left( m\right) =\dfrac{\left( \left( m+3\right) r-\left( m+2\right) s\right) ^{2}-1}{4\left( m+2\right) \left( m+3\right) }$ with $1\leq r\leq s\leq m+1$. for $m\geq0$. In fact we will show that at these points we indeed have unitary representations. \begin{proposition} \textbf{(1)} If $c\geq0$ and $L_{c,h}$ is unitary, then $h=0$. \textbf{(2)} We have $L_{0,h}=M_{0,h}$ if and only if $h\neq\dfrac{m^{2}% -1}{24}$ for all $m\geq0$. \textbf{(3)} We have $L_{1,h}=M_{1,h}$ if and only if $h\neq\dfrac{m^{2}}{24}$ for all $m\geq0$. \end{proposition} \textit{Proof.} \textbf{(2)} and \textbf{(3)} follow immediately from the Kac determinant formula. For \textbf{(1)}, just compute $\det\left( \begin{array} [c]{cc}% \left( L_{-N}^{2}v,L_{-N}^{2}v\right) & \left( L_{-N}^{2}v,L_{-2N}v\right) \\ \left( L_{-2N}v,L_{-N}^{2}v\right) & \left( L_{-2N}v,L_{-2N}v\right) \end{array} \right) =4N^{3}h^{2}\left( 8h-5N\right) $ (this is $<0$ for high enough $N$ as long as $h\neq0$), so that the only possibility for unitarity is $h=0$. \section{Affine Lie algebras} \subsection{Introducing \texorpdfstring{$\protect\widehat{\mathfrak{gl}_{n}}$}{gl-n-hat}} \begin{definition} \label{def.glnhat}Let $V$ denote the vector representation of $\mathfrak{gl}% _{\infty}$ defined in Definition \ref{def.glinf.V}. Let $n$ be a positive integer. Consider $L\mathfrak{gl}_{n}=\mathfrak{gl}% _{n}\left[ t,t^{-1}\right] $; this is the loop algebra of the Lie algebra $\mathfrak{gl}_{n}$. This loop algebra clearly acts on $\mathbb{C}^{n}\left[ t,t^{-1}\right] $ (by $\left( At^{i}\right) \rightharpoonup\left( wt^{j}\right) =Awt^{i+j}$ for all $A\in\mathfrak{gl}_{n}$, $w\in \mathbb{C}^{n}$, $i\in\mathbb{Z}$ and $j\in\mathbb{Z}$). But we can identify the vector space $\mathbb{C}^{n}\left[ t,t^{-1}\right] $ with $V$ as follows: Let $\left( e_{1},e_{2},...,e_{n}\right) $ be the standard basis of $\mathbb{C}^{n}$. Then we identify $e_{i}t^{k}\in\mathbb{C}^{n}\left[ t,t^{-1}\right] $ with $v_{i-kn}\in V$ for every $i\in\left\{ 1,2,...,n\right\} $ and $k\in\mathbb{Z}$. The action of $L\mathfrak{gl}_{n}$ on $\mathbb{C}^{n}\left[ t,t^{-1}\right] $ now becomes an action of $L\mathfrak{gl}_{n}$ on $V$. Hence, $L\mathfrak{gl}_{n}$ maps into $\operatorname*{End}V$. More precisely, $L\mathfrak{gl}_{n}$ maps into $\overline{\mathfrak{a}_{\infty}}\subseteq\operatorname*{End}V$. Here is a direct way to construct this mapping: Let $a\left( t\right) \in L\mathfrak{gl}_{n}$ be a Laurent polynomial with coefficients in $\mathfrak{gl}_{n}$. Write $a\left( t\right) $ in the form $a\left( t\right) =\sum\limits_{k\in\mathbb{Z}}a_{k}t^{k}$ with all $a_{k}$ lying in $\mathfrak{gl}_{n}$. Then, let $\operatorname*{Toep}\nolimits_{n}% \left( a\left( t\right) \right) $ be the matrix% \[ \left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & a_{0} & a_{1} & a_{2} & ...\\ ... & a_{-1} & a_{0} & a_{1} & ...\\ ... & a_{-2} & a_{-1} & a_{0} & ...\\ ... & ... & ... & ... & ... \end{array} \right) \in\overline{\mathfrak{a}_{\infty}}. \] Formally speaking, this matrix is defined as the matrix whose $\left( ni+\alpha,nj+\beta\right) $-th entry equals the $\left( \alpha,\beta\right) $-th entry of the $n\times n$ matrix $a_{j-i}$ for all $i\in\mathbb{Z}$, $j\in\mathbb{Z}$, $\alpha\in\left\{ 1,2,...,n\right\} $ and $\beta \in\left\{ 1,2,...,n\right\} $. In other words, this is the block matrix consisting of infinitely many $n\times n$-blocks such that the ``$i$-th block diagonal'' is filled with $a_{i}$'s for every $i\in\mathbb{Z}$. We thus have defined a map $\operatorname*{Toep}\nolimits_{n}:L\mathfrak{gl}% _{n}\rightarrow\overline{\mathfrak{a}_{\infty}}$. This map $\operatorname*{Toep}\nolimits_{n}$ is injective, and is exactly the map $L\mathfrak{gl}_{n}\rightarrow\overline{\mathfrak{a}_{\infty}}$ we obtain from the above action of $L\mathfrak{gl}_{n}$ on $V$. In particular, this map $\operatorname*{Toep}\nolimits_{n}$ is a Lie algebra homomorphism. In the following, we will often regard the injective map $\operatorname*{Toep}% \nolimits_{n}$ as an inclusion, i. e., we will identify any $a\left( t\right) \in L\mathfrak{gl}_{n}$ with its image $\operatorname*{Toep}% \nolimits_{n}\left( a\left( t\right) \right) \in\overline{\mathfrak{a}% _{\infty}}$. \end{definition} Note that I chose the notation $\operatorname*{Toep}\nolimits_{n}$ because of the notion of Toeplitz matrices. For any $a\left( t\right) \in L\mathfrak{gl}_{n}$, the matrix $\operatorname*{Toep}\nolimits_{n}\left( a\left( t\right) \right) $ can be called an infinite ``block-Toeplitz'' matrix. If $n=1$, then $\operatorname*{Toep}\nolimits_{1}\left( a\left( t\right) \right) $ is an actual infinite Toeplitz matrix. \begin{example} Since $\mathfrak{gl}_{1}$ is a $1$-dimensional abelian Lie algebra, we can identify $L\mathfrak{gl}_{1}$ with the Lie algebra $\overline{\mathcal{A}}$. The image $\operatorname*{Toep}\nolimits_{1}\left( L\mathfrak{gl}_{1}\right) $ is the abelian Lie subalgebra $\left\langle T^{j}\ \mid\ j\in\mathbb{Z}% \right\rangle $ of $\overline{\mathfrak{a}_{\infty}}$ (where $T$ is the shift operator) and is isomorphic to $\overline{\mathcal{A}}$. \end{example} It is easy to see that: \begin{proposition} \label{prop.Toep.alg}Let $n$ be a positive integer. Define an associative algebra structure on $L\mathfrak{gl}_{n}=\mathfrak{gl}_{n}\left[ t,t^{-1}\right] $ by% \[ \left( at^{i}\right) \cdot\left( bt^{j}\right) =abt^{i+j}% \ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathfrak{gl}_{n}\text{, }% b\in\mathfrak{gl}_{n}\text{, }i\in\mathbb{Z}\text{ and }j\in\mathbb{Z}. \] Then, $\operatorname*{Toep}\nolimits_{n}$ is not only a Lie algebra homomorphism, but also a homomorphism of associative algebras. \end{proposition} \textit{Proof of Proposition \ref{prop.Toep.alg}.} Let $a\left( t\right) \in L\mathfrak{gl}_{n}$ and $b\left( t\right) \in L\mathfrak{gl}_{n}$. Write $a\left( t\right) $ in the form $a\left( t\right) =\sum\limits_{k\in \mathbb{Z}}a_{k}t^{k}$ with all $a_{k}$ lying in $\mathfrak{gl}_{n}$. Write $b\left( t\right) $ in the form $b\left( t\right) =\sum\limits_{k\in \mathbb{Z}}b_{k}t^{k}$ with all $b_{k}$ lying in $\mathfrak{gl}_{n}$. By the definition of $\operatorname*{Toep}\nolimits_{n}$, we have% \begin{align*} \operatorname*{Toep}\nolimits_{n}\left( a\left( t\right) \right) & =\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & a_{0} & a_{1} & a_{2} & ...\\ ... & a_{-1} & a_{0} & a_{1} & ...\\ ... & a_{-2} & a_{-1} & a_{0} & ...\\ ... & ... & ... & ... & ... \end{array} \right) \\ \text{and}\ \ \ \ \ \ \ \ \ \ \operatorname*{Toep}\nolimits_{n}\left( b\left( t\right) \right) & =\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & b_{0} & b_{1} & b_{2} & ...\\ ... & b_{-1} & b_{0} & b_{1} & ...\\ ... & b_{-2} & b_{-1} & b_{0} & ...\\ ... & ... & ... & ... & ... \end{array} \right) . \end{align*} Hence,% \begin{align} & \operatorname*{Toep}\nolimits_{n}\left( a\left( t\right) \right) \cdot\operatorname*{Toep}\nolimits_{n}\left( b\left( t\right) \right) \nonumber\\ & =\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & a_{0} & a_{1} & a_{2} & ...\\ ... & a_{-1} & a_{0} & a_{1} & ...\\ ... & a_{-2} & a_{-1} & a_{0} & ...\\ ... & ... & ... & ... & ... \end{array} \right) \cdot\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & b_{0} & b_{1} & b_{2} & ...\\ ... & b_{-1} & b_{0} & b_{1} & ...\\ ... & b_{-2} & b_{-1} & b_{0} & ...\\ ... & ... & ... & ... & ... \end{array} \right) \nonumber\\ & =\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & \sum\limits_{k\in\mathbb{Z}}a_{k-\left( -1\right) }b_{-1-k} & \sum\limits_{k\in\mathbb{Z}}a_{k-\left( -1\right) }b_{0-k} & \sum \limits_{k\in\mathbb{Z}}a_{k-\left( -1\right) }b_{1-k} & ...\\ ... & \sum\limits_{k\in\mathbb{Z}}a_{k-0}b_{-1-k} & \sum\limits_{k\in \mathbb{Z}}a_{k-0}b_{0-k} & \sum\limits_{k\in\mathbb{Z}}a_{k-0}b_{1-k} & ...\\ ... & \sum\limits_{k\in\mathbb{Z}}a_{k-1}b_{-1-k} & \sum\limits_{k\in \mathbb{Z}}a_{k-1}b_{0-k} & \sum\limits_{k\in\mathbb{Z}}a_{k-1}b_{1-k} & ...\\ ... & ... & ... & ... & ... \end{array} \right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the rule for multiplying block matrices}\right) \nonumber\\ & =\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{\left( -1\right) +\left( -1\right) -k} & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{\left( -1\right) +0-k} & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{\left( -1\right) +1-k} & ...\\ ... & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{0+\left( -1\right) -k} & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{0+0-k} & \sum\limits_{k\in\mathbb{Z}}% a_{k}b_{0+1-k} & ...\\ ... & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{1+\left( -1\right) -k} & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{1+0-k} & \sum\limits_{k\in\mathbb{Z}}% a_{k}b_{1+1-k} & ...\\ ... & ... & ... & ... & ... \end{array} \right) \label{pf.Toep.alg.1}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since any }\left( i,j\right) \in\mathbb{Z}^{2}\text{ satisfies }\sum\limits_{k\in\mathbb{Z}}a_{k-i}% b_{j-k}=\sum\limits_{k\in\mathbb{Z}}a_{k}b_{i+j-k}\right) .\nonumber \end{align} On the other hand, multiplying $a\left( t\right) =\sum\limits_{k\in \mathbb{Z}}a_{k}t^{k}$ and $b\left( t\right) =\sum\limits_{k\in\mathbb{Z}% }b_{k}t^{k}$, we obtain% \[ a\left( t\right) \cdot b\left( t\right) =\left( \sum\limits_{k\in \mathbb{Z}}a_{k}t^{k}\right) \cdot\left( \sum\limits_{k\in\mathbb{Z}}% b_{k}t^{k}\right) =\sum\limits_{i\in\mathbb{Z}}\left( \sum\limits_{k\in \mathbb{Z}}a_{k}b_{i-k}\right) t^{i}% \] (by the definition of the product of two Laurent polynomials), so that% \[ \operatorname*{Toep}\nolimits_{n}\left( a\left( t\right) \cdot b\left( t\right) \right) =\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{\left( -1\right) +\left( -1\right) -k} & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{\left( -1\right) +0-k} & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{\left( -1\right) +1-k} & ...\\ ... & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{0+\left( -1\right) -k} & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{0+0-k} & \sum\limits_{k\in\mathbb{Z}}% a_{k}b_{0+1-k} & ...\\ ... & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{1+\left( -1\right) -k} & \sum\limits_{k\in\mathbb{Z}}a_{k}b_{1+0-k} & \sum\limits_{k\in\mathbb{Z}}% a_{k}b_{1+1-k} & ...\\ ... & ... & ... & ... & ... \end{array} \right) \] (by the definition of $\operatorname*{Toep}\nolimits_{n}$). Compared with (\ref{pf.Toep.alg.1}), this yields $\operatorname*{Toep}\nolimits_{n}\left( a\left( t\right) \right) \cdot\operatorname*{Toep}\nolimits_{n}\left( b\left( t\right) \right) =\operatorname*{Toep}\nolimits_{n}\left( a\left( t\right) \cdot b\left( t\right) \right) $. Now forget that we fixed $a\left( t\right) $ and $b\left( t\right) $. We thus have proven that every $a\left( t\right) \in L\mathfrak{gl}_{n}$ and $b\left( t\right) \in L\mathfrak{gl}_{n}$ satisfy $\operatorname*{Toep}% \nolimits_{n}\left( a\left( t\right) \right) \cdot\operatorname*{Toep}% \nolimits_{n}\left( b\left( t\right) \right) =\operatorname*{Toep}% \nolimits_{n}\left( a\left( t\right) \cdot b\left( t\right) \right) $. Combined with the fact that $\operatorname*{Toep}\nolimits_{n}\left( 1\right) =\operatorname*{id}$ (this is very easy to prove), this yields that $\operatorname*{Toep}\nolimits_{n}$ is a homomorphism of associative algebras. Hence, $\operatorname*{Toep}\nolimits_{n}$ is also a homomorphism of Lie algebras. Proposition \ref{prop.Toep.alg} is proven. Recall that the Lie algebra $\overline{\mathfrak{a}_{\infty}}$ has a central extension $\mathfrak{a}_{\infty}$, which equals $\overline{\mathfrak{a}% _{\infty}}\oplus\mathbb{C}K$ as a vector space but has its Lie bracket defined using the cocycle $\alpha$. \begin{proposition} \label{prop.ainf.alphaomega}Let $\alpha:\overline{\mathfrak{a}_{\infty}}% \times\overline{\mathfrak{a}_{\infty}}\rightarrow\mathbb{C}$ be the Japanese cocycle. Let $n\in\mathbb{N}$. Let $\omega:L\mathfrak{gl}_{n}\times L\mathfrak{gl}% _{n}\rightarrow\mathbb{C}$ be the $2$-cocycle on $L\mathfrak{gl}_{n}$ which is defined by% \begin{equation} \omega\left( a\left( t\right) ,b\left( t\right) \right) =\sum \limits_{k\in\mathbb{Z}}k\operatorname*{Tr}\left( a_{k}b_{-k}\right) \ \ \ \ \ \ \ \ \ \ \text{for all }a\left( t\right) \in L\mathfrak{gl}% _{n}\text{ and }b\left( t\right) \in L\mathfrak{gl}_{n} \label{prop.ainf.alphaomega.form}% \end{equation} (where we write $a\left( t\right) $ in the form $a\left( t\right) =\sum\limits_{i\in\mathbb{Z}}a_{i}t^{i}$ with $a_{i}\in\mathfrak{gl}_{n}$, and where we write $b\left( t\right) $ in the form $b\left( t\right) =\sum\limits_{i\in\mathbb{Z}}b_{i}t^{i}$ with $b_{i}\in\mathfrak{gl}_{n}$). Then, the restriction of the Japanese cocycle $\alpha:\overline{\mathfrak{a}% _{\infty}}\times\overline{\mathfrak{a}_{\infty}}\rightarrow\mathbb{C}$ to $L\mathfrak{gl}_{n}\times L\mathfrak{gl}_{n}$ is the $2$-cocycle $\omega$. \end{proposition} \begin{remark} The $2$-cocycle $\omega$ in Proposition \ref{prop.ainf.alphaomega} coincides with the cocycle $\omega$ defined in Definition \ref{def.loop} in the case when $\mathfrak{g}=\mathfrak{gl}_{n}$ and $\left( \cdot,\cdot\right) $ is the form $\mathfrak{gl}_{n}\times\mathfrak{gl}_{n}\rightarrow\mathbb{C}% ,\ \left( a,b\right) \mapsto\operatorname*{Tr}\left( ab\right) $. The $1$-dimensional central extension $\widehat{\mathfrak{gl}_{n}}_{\omega}$ induced by this $2$-cocycle $\omega$ (by the procedure shown in Definition \ref{def.loop}) will be denoted by $\widehat{\mathfrak{gl}_{n}}$ in the following. Note that $\widehat{\mathfrak{gl}_{n}}=L\mathfrak{gl}_{n}% \oplus\mathbb{C}K$ as a vector space. Note that the equality (\ref{prop.ainf.alphaomega.form}) can be rewritten in the suggestive form% \[ \omega\left( a\left( t\right) ,b\left( t\right) \right) =\operatorname*{Res}\nolimits_{t=0}\operatorname*{Tr}\left( da\left( t\right) b\left( t\right) \right) \ \ \ \ \ \ \ \ \ \ \text{for all }a\left( t\right) \in L\mathfrak{gl}_{n}\text{ and }b\left( t\right) \in L\mathfrak{gl}_{n}% \] (as long as the ``matrix-valued differential form'' $da\left( t\right) b\left( t\right) $ is understood correctly). \end{remark} \textit{Proof of Proposition \ref{prop.ainf.alphaomega}.} We need to prove that $\alpha\left( a\left( t\right) ,b\left( t\right) \right) =\omega\left( a\left( t\right) ,b\left( t\right) \right) $ for any $a\left( t\right) \in L\mathfrak{gl}_{n}$ and $b\left( t\right) \in L\mathfrak{gl}_{n}$ (where, of course, we consider $a\left( t\right) $ and $b\left( t\right) $ as elements of $\overline{\mathfrak{a}_{\infty}}$ in the term $\alpha\left( a\left( t\right) ,b\left( t\right) \right) $). Write $a\left( t\right) $ in the form $a\left( t\right) =\sum \limits_{k\in\mathbb{Z}}a_{k}t^{k}$ with all $a_{k}$ lying in $\mathfrak{gl}% _{n}$. Write $b\left( t\right) $ in the form $b\left( t\right) =\sum\limits_{k\in\mathbb{Z}}b_{k}t^{k}$ with all $b_{k}$ lying in $\mathfrak{gl}_{n}$. In the following, for any integers $u$ and $v$, the $\left( u,v\right) $-th \textit{block} of a matrix will mean the submatrix obtained by leaving only the rows numbered $un+1$, $un+2$, $...$, $\left( u+1\right) n$ and the columns numbered $vn+1$, $vn+2$, $...$, $\left( v+1\right) n$. (This, of course, makes sense only when the matrix has such rows and such columns.) By the definition of our embedding $\operatorname*{Toep}\nolimits_{n}\left( a\left( t\right) \right) :L\mathfrak{gl}_{n}\rightarrow\overline {\mathfrak{a}_{\infty}}$, we have% \begin{align*} a\left( t\right) & =\operatorname*{Toep}\nolimits_{n}\left( a\left( t\right) \right) =\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & a_{0} & a_{1} & a_{2} & ...\\ ... & a_{-1} & a_{0} & a_{1} & ...\\ ... & a_{-2} & a_{-1} & a_{0} & ...\\ ... & ... & ... & ... & ... \end{array} \right) \ \ \ \ \ \ \ \ \ \ \text{and}\\ b\left( t\right) & =\operatorname*{Toep}\nolimits_{n}\left( b\left( t\right) \right) =\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & b_{0} & b_{1} & b_{2} & ...\\ ... & b_{-1} & b_{0} & b_{1} & ...\\ ... & b_{-2} & b_{-1} & b_{0} & ...\\ ... & ... & ... & ... & ... \end{array} \right) , \end{align*} where the matrices $\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & a_{0} & a_{1} & a_{2} & ...\\ ... & a_{-1} & a_{0} & a_{1} & ...\\ ... & a_{-2} & a_{-1} & a_{0} & ...\\ ... & ... & ... & ... & ... \end{array} \right) $ and $\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & b_{0} & b_{1} & b_{2} & ...\\ ... & b_{-1} & b_{0} & b_{1} & ...\\ ... & b_{-2} & b_{-1} & b_{0} & ...\\ ... & ... & ... & ... & ... \end{array} \right) $ are understood as block matrices made of $n\times n$ blocks. In order to compute $\alpha\left( a\left( t\right) ,b\left( t\right) \right) $, let us write these two infinite matrices $a\left( t\right) $ and $b\left( t\right) $ as $2\times2$ block matrices \textbf{made of infinite blocks each}, where the blocks are separated as follows: - The left blocks contain the $j$-th columns for all $j\leq0$; the right blocks contain the $j$-th columns for all $j>0$. - The upper blocks contain the $i$-th rows for all $i\leq0$; the lower blocks contain the $i$-th rows for all $i>0$. Written like this, the matrix $a\left( t\right) $ takes the form $\left( \begin{array} [c]{cc}% A_{11} & A_{12}\\ A_{21} & A_{22}% \end{array} \right) $ with% \begin{align*} A_{11} & =\left( \begin{array} [c]{cccc}% ... & ... & ... & ...\\ ... & a_{0} & a_{1} & a_{2}\\ ... & a_{-1} & a_{0} & a_{1}\\ ... & a_{-2} & a_{-1} & a_{0}% \end{array} \right) ,\ \ \ \ \ \ \ \ \ \ A_{12}=\left( \begin{array} [c]{cccc}% ... & ... & ... & ...\\ a_{3} & a_{4} & a_{5} & ...\\ a_{2} & a_{3} & a_{4} & ...\\ a_{1} & a_{2} & a_{3} & ... \end{array} \right) ,\\ A_{21} & =\left( \begin{array} [c]{cccc}% ... & a_{-3} & a_{-2} & a_{-1}\\ ... & a_{-4} & a_{-3} & a_{-2}\\ ... & a_{-5} & a_{-4} & a_{-3}\\ ... & ... & ... & ... \end{array} \right) ,\ \ \ \ \ \ \ \ \ \ A_{22}=\left( \begin{array} [c]{cccc}% a_{0} & a_{1} & a_{2} & ...\\ a_{-1} & a_{0} & a_{1} & ...\\ a_{-2} & a_{-1} & a_{0} & ...\\ ... & ... & ... & ... \end{array} \right) , \end{align*} and the matrix $b\left( t\right) $ takes the form $\left( \begin{array} [c]{cc}% B_{11} & B_{12}\\ B_{21} & B_{22}% \end{array} \right) $ with similarly-defined blocks $B_{11}$, $B_{12}$, $B_{21}$ and $B_{22}$. By the definition of $\alpha$ given in Theorem \ref{thm.japan}, we now have $\alpha\left( a\left( t\right) ,b\left( t\right) \right) =\operatorname*{Tr}\left( -B_{12}A_{21}+A_{12}B_{21}\right) $. We now need to compute $B_{12}A_{21}$ and $A_{12}B_{21}$ in order to simplify this. Now, since $B_{12}=\left( \begin{array} [c]{cccc}% ... & ... & ... & ...\\ b_{3} & b_{4} & b_{5} & ...\\ b_{2} & b_{3} & b_{4} & ...\\ b_{1} & b_{2} & b_{3} & ... \end{array} \right) $ and $A_{21}=\left( \begin{array} [c]{cccc}% ... & a_{-3} & a_{-2} & a_{-1}\\ ... & a_{-4} & a_{-3} & a_{-2}\\ ... & a_{-5} & a_{-4} & a_{-3}\\ ... & ... & ... & ... \end{array} \right) $, the matrix $B_{12}A_{21}$ is a matrix whose rows and columns are indexed by nonpositive integers, and whose $\left( i,j\right) $-th block equals $\sum\limits_{k\in\mathbb{Z};\ k>0}b_{k-\left( i+1\right) }a_{-k+\left( j+1\right) }$ for any pair of negative integers $i$ and $j$. Similarly, the matrix $A_{12}B_{21}$ is a matrix whose rows and columns are indexed by nonpositive integers, and whose $\left( i,j\right) $-th block equals $\sum\limits_{k\in\mathbb{Z};\ k>0}a_{k-\left( i+1\right) }b_{-k+\left( j+1\right) }$ for any pair of negative integers $i$ and $j$. Thus, the matrix $-B_{12}A_{21}+A_{12}B_{21}$ is a matrix whose rows and columns are indexed by nonpositive integers, and whose $\left( i,j\right) $-th block equals $-\sum\limits_{k\in\mathbb{Z};\ k>0}b_{k-\left( i+1\right) }a_{-k+\left( j+1\right) }+\sum\limits_{k\in\mathbb{Z};\ k>0}a_{k-\left( i+1\right) }b_{-k+\left( j+1\right) }$ for any pair of negative integers $i$ and $j$. But since $\operatorname*{Tr}\left( -B_{12}A_{21}+A_{12}% B_{21}\right) $ is clearly the sum of the traces of the $\left( i,i\right) $-th blocks of the matrix $-B_{12}A_{21}+A_{12}B_{21}$ over all negative integers $i$, we thus have% \begin{align*} \operatorname*{Tr}\left( -B_{12}A_{21}+A_{12}B_{21}\right) & =\sum\limits_{i\in\mathbb{Z};\ i<0}\operatorname*{Tr}\left( -\sum \limits_{k\in\mathbb{Z};\ k>0}b_{k-\left( i+1\right) }a_{-k+\left( i+1\right) }+\sum\limits_{k\in\mathbb{Z};\ k>0}a_{k-\left( i+1\right) }b_{-k+\left( i+1\right) }\right) \\ & =\sum\limits_{i\in\mathbb{Z};\ i\leq0}\operatorname*{Tr}\left( -\sum\limits_{k\in\mathbb{Z};\ k>0}b_{k-i}a_{-k+i}+\sum\limits_{k\in \mathbb{Z};\ k>0}a_{k-i}b_{-k+i}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }i\text{ for }i+1\right) \\ & =\sum\limits_{i\in\mathbb{Z};\ i\geq0}\operatorname*{Tr}\left( -\sum\limits_{k\in\mathbb{Z};\ k>0}b_{k+i}a_{-k-i}+\sum\limits_{k\in \mathbb{Z};\ k>0}a_{k+i}b_{-k-i}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }i\text{ for }-i\text{ in the first sum}\right) . \end{align*} We are now going to split the first sum on the right hand side and get the $\operatorname*{Tr}$ out of it. To see that this is allowed, we notice that each of the infinite sums $\sum\limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2};\\i\geq0;\ k>0}}b_{k+i}a_{-k-i}$ and $\sum \limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2};\\i\geq 0;\ k>0}}a_{k+i}b_{-k-i}$ converges with respect to the discrete topology\footnote{\textit{Proof.} Since $\sum\limits_{k\in\mathbb{Z}}% a_{k}t^{k}=a\left( t\right) \in L\mathfrak{gl}_{n}$, only finitely many $k\in\mathbb{Z}$ satisfy $a_{k}\neq0$. Hence, there exists some $N\in \mathbb{Z}$ such that every $\nu\in\mathbb{Z}$ satisfying $\nu -N$ satisfies $-k-i=-\underbrace{\left( k+i\right) }_{>-N} 0$ satisfy $b_{k+i}a_{-k-i}=0$ (because it is clear that all but finitely many pairs $\left( i,k\right) \in\mathbb{Z}^{2}$ such that $i\geq0$ and $k>0$ satisfy $k+i>-N$). In other words, the sum $\sum\limits_{\substack{\left( i,k\right) \in\mathbb{Z}% ^{2};\\i\geq0;\ k>0}}b_{k+i}a_{-k-i}$ converges with respect to the discrete topology. A similar argument shows that the sum $\sum \limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2};\\i\geq 0;\ k>0}}a_{k+i}b_{-k-i}$ converges with respect to the discrete topology.}. Hence, we can transform these sums as we please: For example,% \begin{align} & \sum\limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2};\\i\geq 0;\ k>0}}b_{k+i}a_{-k-i}\nonumber\\ & =\sum\limits_{\substack{\ell\in\mathbb{Z};\\\ell>0}}\sum \limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2};\\i\geq 0;\ k>0;\\k+i=\ell}}\underbrace{b_{k+i}}_{\substack{=b_{\ell}\\\text{(since }k+i=\ell\text{)}}}\underbrace{a_{-k-i}}_{\substack{=a_{-\left( k+i\right) }=a_{-\ell}\\\text{(since }k+i=\ell\text{)}}}\ \ \ \ \ \ \ \ \ \ \left( \text{since }k+i>0\text{ for all }i\geq0\text{ and }k>0\right) \nonumber\\ & =\sum\limits_{\substack{\ell\in\mathbb{Z};\\\ell>0}}\underbrace{\sum \limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2};\\i\geq 0;\ k>0;\\k+i=\ell}}b_{\ell}a_{-\ell}}_{\substack{=\ell b_{\ell}a_{-\ell }\\\text{(since there exist exactly }\ell\text{ pairs }\left( i,k\right) \in\mathbb{Z}^{2}\\\text{satisfying }i\geq0\text{, }k>0\text{ and }% k+i=\ell\text{)}}}=\sum\limits_{\substack{\ell\in\mathbb{Z};\\\ell>0}}\ell b_{\ell}a_{-\ell}=\sum\limits_{\substack{k\in\mathbb{Z};\\k>0}}kb_{k}a_{-k} \label{pf.ainf.alphaomega.sum1}% \end{align} (here, we renamed the summation index $\ell$ as $k$) and similarly \begin{equation} \sum\limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2};\\i\geq 0;\ k>0}}a_{k+i}b_{-k-i}=\sum\limits_{\substack{k\in\mathbb{Z};\\k>0}% }ka_{k}b_{-k}. \label{pf.ainf.alphaomega.sum2}% \end{equation} The equality (\ref{pf.ainf.alphaomega.sum1}) becomes% \begin{align*} \sum\limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2};\\i\geq 0;\ k>0}}b_{k+i}a_{-k-i} & =\sum\limits_{\substack{k\in\mathbb{Z}% ;\\k>0}}kb_{k}a_{-k}=\sum\limits_{\substack{k\in\mathbb{Z};\\k<0}}\left( -k\right) b_{-k}a_{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }k\text{ for }-k\text{ in the first sum}\right) \\ & =-\sum\limits_{\substack{k\in\mathbb{Z};\\k<0}}kb_{-k}a_{k}, \end{align*} so that \begin{equation} \sum\limits_{\substack{k\in\mathbb{Z};\\k<0}}kb_{-k}a_{k}=-\sum \limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2};\\i\geq 0;\ k>0}}b_{k+i}a_{-k-i}. \label{pf.ainf.alphaomega.sum3}% \end{equation} But \begin{align*} & \omega\left( a\left( t\right) ,b\left( t\right) \right) \\ & =\sum\limits_{k\in\mathbb{Z}}k\operatorname*{Tr}\left( a_{k}b_{-k}\right) =\sum\limits_{\substack{k\in\mathbb{Z};\\k<0}}k\underbrace{\operatorname*{Tr}% \left( a_{k}b_{-k}\right) }_{=\operatorname*{Tr}\left( b_{-k}a_{k}\right) }+\underbrace{0\operatorname*{Tr}\left( a_{0}b_{-0}\right) }_{=0}% +\sum\limits_{\substack{k\in\mathbb{Z};\\k>0}}k\operatorname*{Tr}\left( a_{k}b_{-k}\right) \\ & =\underbrace{\sum\limits_{\substack{k\in\mathbb{Z};\\k<0}% }k\operatorname*{Tr}\left( b_{-k}a_{k}\right) }_{=\operatorname*{Tr}\left( \sum\limits_{\substack{k\in\mathbb{Z};\\k<0}}kb_{-k}a_{k}\right) }+\underbrace{\sum\limits_{\substack{k\in\mathbb{Z};\\k>0}}k\operatorname*{Tr}% \left( a_{k}b_{-k}\right) }_{=\operatorname*{Tr}\left( \sum \limits_{\substack{k\in\mathbb{Z};\\k>0}}ka_{k}b_{-k}\right) }\\ & =\operatorname*{Tr}\left( \underbrace{\sum\limits_{\substack{k\in \mathbb{Z};\\k<0}}kb_{-k}a_{k}}_{\substack{=-\sum\limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2};\\i\geq0;\ k>0}}b_{k+i}a_{-k-i}\\\text{(by (\ref{pf.ainf.alphaomega.sum3}))}}}\right) +\operatorname*{Tr}\left( \underbrace{\sum\limits_{\substack{k\in\mathbb{Z};\\k>0}}ka_{k}b_{-k}% }_{\substack{=\sum\limits_{\substack{\left( i,k\right) \in\mathbb{Z}% ^{2};\\i\geq0;\ k>0}}a_{k+i}b_{-k-i}\\\text{(by (\ref{pf.ainf.alphaomega.sum2}% ))}}}\right) \\ & =\operatorname*{Tr}\left( -\sum\limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2};\\i\geq0;\ k>0}}b_{k+i}a_{-k-i}\right) +\operatorname*{Tr}% \left( \sum\limits_{\substack{\left( i,k\right) \in\mathbb{Z}^{2}% ;\\i\geq0;\ k>0}}a_{k+i}b_{-k-i}\right) \\ & =\operatorname*{Tr}\left( -\sum\limits_{i\in\mathbb{Z};\ i\geq0}% \sum\limits_{k\in\mathbb{Z};\ k>0}b_{k+i}a_{-k-i}\right) +\operatorname*{Tr}% \left( \sum\limits_{i\in\mathbb{Z};\ i\geq0}\sum\limits_{k\in\mathbb{Z}% ;\ k>0}a_{k+i}b_{-k-i}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we have unfolded our single sums into double sums}\right) \\ & =\sum\limits_{i\in\mathbb{Z};\ i\geq0}\operatorname*{Tr}\left( -\sum\limits_{k\in\mathbb{Z};\ k>0}b_{k+i}a_{-k-i}\right) +\sum \limits_{i\in\mathbb{Z};\ i\geq0}\operatorname*{Tr}\left( \sum\limits_{k\in \mathbb{Z};\ k>0}a_{k+i}b_{-k-i}\right) \\ & =\sum\limits_{i\in\mathbb{Z};\ i\geq0}\operatorname*{Tr}\left( -\sum\limits_{k\in\mathbb{Z};\ k>0}b_{k+i}a_{-k-i}+\sum\limits_{k\in \mathbb{Z};\ k>0}a_{k+i}b_{-k-i}\right) =\operatorname*{Tr}\left( -B_{12}A_{21}+A_{12}B_{21}\right) \\ & =\alpha\left( a\left( t\right) ,b\left( t\right) \right) . \end{align*} Thus, $\alpha\left( a\left( t\right) ,b\left( t\right) \right) =\omega\left( a\left( t\right) ,b\left( t\right) \right) $ is proven, so we have verified Proposition \ref{prop.ainf.alphaomega}. Note that Proposition \ref{prop.ainf.alphaomega} gives a new proof of Proposition \ref{prop.japan.nontr}. This proof (whose details are left to the reader) uses two easy facts: \begin{itemize} \item If $\sigma:\mathfrak{g}\times\mathfrak{g}\rightarrow\mathbb{C}$ is a $2$-coboundary on a Lie algebra $\mathfrak{g}$, and $\mathfrak{h}$ is a Lie subalgebra of $\mathfrak{g}$, then $\sigma\mid_{\mathfrak{h}\times \mathfrak{h}}$ must be a $2$-coboundary on $\mathfrak{h}$. \item For any positive integer $n$, the $2$-cocycle $\omega$ of Proposition \ref{prop.ainf.alphaomega} is not a $2$-coboundary. \end{itemize} But if we look closely at this argument, we see that it is not a completely new proof; it is a direct generalization of the proof of Proposition \ref{prop.japan.nontr} that we gave above. In fact, in the particular case when $n=1$, our embedding of $L\mathfrak{gl}_{n}$ into $\overline {\mathfrak{a}_{\infty}}$ becomes the canonical injection of the abelian Lie subalgebra $\left\langle T^{j}\ \mid\ j\in\mathbb{Z}\right\rangle $ into $\overline{\mathfrak{a}_{\infty}}$ (where $T$ is as in the proof of Proposition \ref{prop.japan.nontr}), and we see that what we just did was generalizing that abelian Lie subalgebra. \begin{definition} Due to Proposition \ref{prop.ainf.alphaomega}, the restriction of the $2$-cocycle $\alpha$ to $L\mathfrak{gl}_{n}\times L\mathfrak{gl}_{n}$ is the $2$-cocycle $\omega$. Thus, the $1$-dimensional central extension of $L\mathfrak{gl}_{n}$ determined by the $2$-cocycle $\omega$ canonically injects into the $1$-dimensional central extension of $\overline {\mathfrak{a}_{\infty}}$ determined by the $2$-cocycle $\alpha$. If we recall that the $1$-dimensional central extension of $L\mathfrak{gl}_{n}$ by the $2$-cocycle $\omega$ is $\widehat{\mathfrak{gl}_{n}}$ whereas the $1$-dimensional central extension of $\overline{\mathfrak{a}_{\infty}}$ determined by the $2$-cocycle $\alpha$ is $\mathfrak{a}_{\infty}$, we can rewrite this as follows: We have an injection $\widehat{\mathfrak{gl}_{n}% }\rightarrow\mathfrak{a}_{\infty}$ which lifts the inclusion $L\mathfrak{gl}% _{n}\subseteq\overline{\mathfrak{a}_{\infty}}$ and sends $K$ to $K$. We denote this inclusion map $\widehat{\mathfrak{gl}_{n}}\rightarrow\mathfrak{a}% _{\infty}$ by $\widehat{\operatorname*{Toep}\nolimits_{n}}$, but we will often consider it as an inclusion. Similarly, we can get an inclusion $\widehat{\mathfrak{sl}_{n}}\subseteq \mathfrak{a}_{\infty}$ which lifts the inclusion $L\mathfrak{sl}_{n}% \subseteq\overline{\mathfrak{a}_{\infty}}$. So $\mathcal{B}^{\left( m\right) }\cong\mathcal{F}^{\left( m\right) }$ is a module over $\widehat{\mathfrak{gl}_{n}}$ and $\widehat{\mathfrak{sl}_{n}}$ at level $1$ (this means that $K$ acts as $1$). \end{definition} \begin{corollary} \label{cor.glnhat.div}There is a Lie algebra isomorphism $\widehat{\phi }:\mathcal{A}\rightarrow\widehat{\mathfrak{gl}_{1}}$ which sends $K$ to $K$ and sends $a_{m}$ to $T^{m}\in\widehat{\mathfrak{gl}_{1}}$ for every $m\in\mathbb{Z}$. (Here, we are considering the injection $\widehat{\mathfrak{gl}_{1}}\rightarrow\mathfrak{a}_{\infty}$ as an inclusion, so that $\widehat{\mathfrak{gl}_{1}}$ is identified with the image of this inclusion.) \end{corollary} \textit{Proof of Corollary \ref{cor.glnhat.div}.} There is an obvious Lie algebra isomorphism $\phi:\overline{\mathcal{A}}\rightarrow L\mathfrak{gl}% _{1}$ which sends $a_{m}$ to $t^{m}\in L\mathfrak{gl}_{1}$ for every $m\in\mathbb{Z}$. This isomorphism $\phi$ is easily seen to satisfy% \begin{equation} \omega\left( \phi\left( x\right) ,\phi\left( y\right) \right) =\omega^{\prime}\left( x,y\right) \ \ \ \ \ \ \ \ \ \ \text{for all }% x\in\overline{\mathcal{A}}\text{ and }y\in\overline{\mathcal{A}}\text{,} \label{pf.glinhat.div.1}% \end{equation} where $\omega:L\mathfrak{gl}_{1}\times L\mathfrak{gl}_{1}\rightarrow \mathbb{C}$ is the $2$-cocycle on $L\mathfrak{gl}_{1}$ defined in Proposition \ref{prop.ainf.alphaomega}, and $\omega^{\prime}:\overline{\mathcal{A}}% \times\overline{\mathcal{A}}\rightarrow\mathbb{C}$ is the $2$-cocycle on $\overline{\mathcal{A}}$ defined by% \[ \omega^{\prime}\left( a_{k},a_{\ell}\right) =k\delta_{k,-\ell}% \ \ \ \ \ \ \ \ \ \ \text{for all }k\in\mathbb{Z}\text{ and }\ell\in \mathbb{Z}. \] Thus, the Lie algebra isomorphism $\phi:\overline{\mathcal{A}}\rightarrow L\mathfrak{gl}_{1}$ gives rise to an isomorphism $\widehat{\phi}$ from the extension of $\overline{\mathcal{A}}$ defined by the $2$-cocycle $\omega^{\prime}$ to the extension of $L\mathfrak{gl}_{1}$ defined by the $2$-cocycle $\omega$. Since the extension of $\overline{\mathcal{A}}$ defined by the $2$-cocycle $\omega^{\prime}$ is $\mathcal{A}$, while the extension of $L\mathfrak{gl}_{1}$ defined by the $2$-cocycle $\omega$ is $\widehat{\mathfrak{gl}_{1}}$, this rewrites as follows: The Lie algebra isomorphism $\phi:\overline{\mathcal{A}}\rightarrow L\mathfrak{gl}_{1}$ gives rise to an isomorphism $\widehat{\phi}:\mathcal{A}\rightarrow \widehat{\mathfrak{gl}_{1}}$. This isomorphism $\widehat{\phi}$ sends $K$ to $K$, and sends $a_{m}$ to $t^{m}\in\widehat{\mathfrak{gl}_{1}}$ for every $m\in\mathbb{Z}$. Since $t^{m}$ corresponds to $T^{m}$ under our inclusion $\widehat{\mathfrak{gl}_{1}}\rightarrow\mathfrak{a}_{\infty}$ (in fact, $\operatorname*{Toep}\nolimits_{1}\left( t^{m}\right) =T^{m}$), this shows that $\widehat{\phi}$ sends $a_{m}$ to $T^{m}\in\widehat{\mathfrak{gl}_{1}}$ for every $m\in\mathbb{Z}$. Corollary \ref{cor.glnhat.div} is thus proven. \begin{proposition} \label{prop.glnhat.T}Let $n$ be a positive integer. Consider the shift operator $T$. Let us regard the injections $\overline{\mathfrak{a}_{\infty}% }\rightarrow\mathfrak{a}_{\infty}$, $L\mathfrak{gl}_{n}\rightarrow \overline{\mathfrak{a}_{\infty}}$ and $\widehat{\mathfrak{gl}_{n}}% \rightarrow\mathfrak{a}_{\infty}$ as inclusions, so that $L\mathfrak{gl}_{n}$, $\widehat{\mathfrak{gl}_{n}}$ and $\mathfrak{a}_{\infty}$ all become subspaces of $\mathfrak{a}_{\infty}$. \textbf{(a)} For every $m\in\mathbb{Z}$, we have $T^{m}\in L\mathfrak{gl}% _{n}\subseteq\widehat{\mathfrak{gl}_{n}}$. \textbf{(b)} We have $\widehat{\mathfrak{gl}_{1}}\subseteq \widehat{\mathfrak{gl}_{n}}$. Hence, the Lie algebra isomorphism $\widehat{\phi}:\mathcal{A}\rightarrow\widehat{\mathfrak{gl}_{1}}$ constructed in Corollary \ref{cor.glnhat.div} induces a Lie algebra injection $\mathcal{A}\rightarrow\widehat{\mathfrak{gl}_{n}}$ (which sends every $a\in\mathcal{A}$ to $\widehat{\phi}\left( a\right) \in \widehat{\mathfrak{gl}_{n}}$). The restriction of the $\widehat{\mathfrak{gl}% _{n}}$-module $\mathcal{F}^{\left( m\right) }$ by means of this injection is the $\mathcal{A}$-module $\mathcal{F}^{\left( m\right) }$ that we know. \end{proposition} \textit{First proof of Proposition \ref{prop.glnhat.T}.} \textbf{(a)} We recall that $T=\left( \begin{array} [c]{cccccc}% ... & ... & ... & ... & ... & ...\\ ... & 0 & 1 & 0 & 0 & ...\\ ... & 0 & 0 & 1 & 0 & ...\\ ... & 0 & 0 & 0 & 1 & ...\\ ... & 0 & 0 & 0 & 0 & ...\\ ... & ... & ... & ... & ... & ... \end{array} \right) $ (this is the matrix which has $1$'s on the $1$-st diagonal and $0$'s everywhere else). Clearly, $T\in\overline{\mathfrak{a}_{\infty}}$. We want to prove that $T$ lies in $L\mathfrak{gl}_{n}\subseteq\overline {\mathfrak{a}_{\infty}}$. Let $a_{0}=\left( \begin{array} [c]{ccccc}% 0 & 1 & 0 & ... & 0\\ 0 & 0 & 1 & ... & 0\\ 0 & 0 & 0 & ... & 0\\ ... & ... & ... & ... & ...\\ 0 & 0 & 0 & ... & 0 \end{array} \right) $ (this is the $n\times n$ matrix which has $1$'s on the $1$-st diagonal and $0$'s everywhere else). Let $a_{1}=\left( \begin{array} [c]{ccccc}% 0 & 0 & 0 & ... & 0\\ 0 & 0 & 0 & ... & 0\\ 0 & 0 & 0 & ... & 0\\ ... & ... & ... & ... & ...\\ 0 & 0 & 0 & ... & 0\\ 1 & 0 & 0 & ... & 0 \end{array} \right) $ (this is the $n\times n$ matrix which has a $1$ in its lowermost leftmost corner, and $0$'s everywhere else). Then, $T=\operatorname*{Toep}\nolimits_{n}\left( a_{0}+ta_{1}\right) $. Thus, for every $m\in\mathbb{N}$, we have% \begin{align*} T^{m} & =\left( \operatorname*{Toep}\nolimits_{n}\left( a_{0}% +ta_{1}\right) \right) ^{m}=\operatorname*{Toep}\nolimits_{n}\left( \left( a_{0}+ta_{1}\right) ^{m}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{because of Proposition \ref{prop.Toep.alg}}\right) \\ & \in\operatorname*{Toep}\nolimits_{n}\left( L\mathfrak{gl}_{n}\right) =L\mathfrak{gl}_{n}\ \ \ \ \ \ \ \ \ \ \left( \text{since we regard }\operatorname*{Toep}\nolimits_{n}\text{ as an inclusion}\right) . \end{align*} Since it is easy to see that $T^{-1}\in L\mathfrak{gl}_{n}$ as well\footnote{This is analogous to $T\in L\mathfrak{gl}_{n}$ (because $T^{-1}$ is the matrix which has $1$'s on the $\left( -1\right) $-st diagonal and $0$'s everywhere else).}, a similar argument yields that $\left( T^{-1}\right) ^{m}\in L\mathfrak{gl}_{n}$ for all $m\in\mathbb{N}$. In other words, $T^{-m}\in L\mathfrak{gl}_{n}$ for all $m\in\mathbb{N}$. In other words, $T^{m}\in\mathfrak{gl}_{n}$ for all nonpositive integers $m$. Combined with the fact that $T^{m}\in L\mathfrak{gl}_{n}$ for all $m\in\mathbb{N}$, this yields that $T^{m}\in L\mathfrak{gl}_{n}$ for all $m\in\mathbb{Z}$. Since $L\mathfrak{gl}_{n}\subseteq\widehat{\mathfrak{gl}_{n}}$, we thus have $T^{m}\in L\mathfrak{gl}_{n}\subseteq\widehat{\mathfrak{gl}_{n}}$ for all $m\in\mathbb{Z}$. This proves Proposition \ref{prop.glnhat.T} \textbf{(a)}. \textbf{(b)} For every $a\left( t\right) \in L\mathfrak{gl}_{1}$, we have $\operatorname*{Toep}\nolimits_{1}\left( a\left( t\right) \right) \in\left\langle T^{j}\ \mid\ j\in\mathbb{Z}\right\rangle $% \ \ \ \ \footnote{\textit{Proof.} Let $a\left( t\right) \in L\mathfrak{gl}% _{1}$. Write $a\left( t\right) $ in the form $\sum\limits_{i\in\mathbb{Z}% }a_{i}t^{i}$ with $a_{i}\in\mathfrak{gl}_{1}$. Then, of course, the $a_{i}$ are scalars (since $\mathfrak{gl}_{1}=\mathbb{C}$). By the definition of $\operatorname*{Toep}\nolimits_{1}$, we have% \[ \operatorname*{Toep}\nolimits_{1}\left( a\left( t\right) \right) =\left( \begin{array} [c]{ccccc}% ... & ... & ... & ... & ...\\ ... & a_{0} & a_{1} & a_{2} & ...\\ ... & a_{-1} & a_{0} & a_{1} & ...\\ ... & a_{-2} & a_{-1} & a_{0} & ...\\ ... & ... & ... & ... & ... \end{array} \right) =\sum\limits_{i\in\mathbb{Z}}a_{i}T^{i}\in\left\langle T^{j}% \ \mid\ j\in\mathbb{Z}\right\rangle , \] qed.}. Thus, $\operatorname*{Toep}\nolimits_{1}\left( L\mathfrak{gl}% _{1}\right) \subseteq\left\langle T^{j}\ \mid\ j\in\mathbb{Z}\right\rangle $. Since we are considering $\operatorname*{Toep}\nolimits_{1}$ as an inclusion, this becomes $L\mathfrak{gl}_{1}\subseteq\left\langle T^{j}\ \mid \ j\in\mathbb{Z}\right\rangle $. Combined with $\left\langle T^{j}\ \mid \ j\in\mathbb{Z}\right\rangle \subseteq L\mathfrak{gl}_{n}$ (because every $m\in\mathbb{Z}$ satisfies $T^{m}\in L\mathfrak{gl}_{n}$ (according to Proposition \ref{prop.glnhat.T} \textbf{(a)})), this yields $L\mathfrak{gl}% _{1}\subseteq L\mathfrak{gl}_{n}$. Thus, $\widehat{\mathfrak{gl}_{1}}% \subseteq\widehat{\mathfrak{gl}_{n}}$. Hence, the Lie algebra isomorphism $\widehat{\phi}:\mathcal{A}\rightarrow \widehat{\mathfrak{gl}_{1}}$ constructed in Corollary \ref{cor.glnhat.div} induces a Lie algebra injection $\mathcal{A}\rightarrow\widehat{\mathfrak{gl}% _{n}}$ (which sends every $a\in\mathcal{A}$ to $\widehat{\phi}\left( a\right) \in\widehat{\mathfrak{gl}_{n}}$). This injection is exactly the embedding $\mathcal{A}\rightarrow\mathfrak{a}_{\infty}$ constructed in Definition \ref{def.ainf.A} (apart from the fact that its target is $\widehat{\mathfrak{gl}_{n}}$ rather than $\mathfrak{a}_{\infty}$). Hence, the restriction of the $\widehat{\mathfrak{gl}_{n}}$-module $\mathcal{F}^{\left( m\right) }$ by means of this injection is the $\mathcal{A}$-module $\mathcal{F}^{\left( m\right) }$ that we know\footnote{because both the $\widehat{\mathfrak{gl}_{n}}$-module $\mathcal{F}^{\left( m\right) }$ and the $\mathcal{A}$-module $\mathcal{F}^{\left( m\right) }$ were defined as restrictions of the $\mathfrak{a}_{\infty}$-module $\mathcal{F}^{\left( m\right) }$}. This proves Proposition \ref{prop.glnhat.T} \textbf{(b)}. Our inclusions $L\mathfrak{gl}_{n}\subseteq\overline{\mathfrak{a}_{\infty}}$ and $\widehat{\mathfrak{gl}_{n}}\subseteq\mathfrak{a}_{\infty}$ can be somewhat refined: For any positive integers $n$ and $N$ satisfying $n\mid N$, we have $L\mathfrak{gl}_{n}\subseteq L\mathfrak{gl}_{N}$ and $\widehat{\mathfrak{gl}_{n}}\subseteq\widehat{\mathfrak{gl}_{N}}$. Let us formulate this more carefully without abuse of notation: \begin{proposition} \label{prop.glnhat.div}Let $n$ and $N$ be positive integers such that $n\mid N$. Then, the inclusion $\operatorname*{Toep}\nolimits_{n}:L\mathfrak{gl}% _{n}\rightarrow\overline{\mathfrak{a}_{\infty}}$ factors through the inclusion $\operatorname*{Toep}\nolimits_{N}:L\mathfrak{gl}_{N}\rightarrow \overline{\mathfrak{a}_{\infty}}$. More precisely: Let $d=\dfrac{N}{n}$. Let $\operatorname*{Toep}\nolimits_{n,N}:L\mathfrak{gl}% _{n}\rightarrow L\mathfrak{gl}_{N}$ be the map which sends every $a\left( t\right) \in L\mathfrak{gl}_{n}$ to% \[ \sum\limits_{\ell\in\mathbb{Z}}\underbrace{\left( \begin{array} [c]{ccccc}% a_{\left( j-i\right) d} & a_{\left( j-i\right) d+1} & a_{\left( j-i\right) d+2} & ... & a_{\left( j-i\right) d+\left( d-1\right) }\\ a_{\left( j-i\right) d-1} & a_{\left( j-i\right) d} & a_{\left( j-i\right) d+1} & ... & a_{\left( j-i\right) d+\left( d-2\right) }\\ a_{\left( j-i\right) d-2} & a_{\left( j-i\right) d-1} & a_{\left( j-i\right) d} & ... & a_{\left( j-i\right) d+\left( d-3\right) }\\ ... & ... & ... & ... & ...\\ a_{\left( j-i\right) d-\left( d-1\right) } & a_{\left( j-i\right) d-\left( d-2\right) } & a_{\left( j-i\right) d-\left( d-3\right) } & ... & a_{\left( j-i\right) d}% \end{array} \right) }_{\substack{\text{this is an }N\times N\text{-matrix constructed as a }d\times d\text{-block matrix}\\\text{consisting of }n\times n\text{-blocks; one can formally define this matrix}\\\text{as the }N\times N\text{-matrix whose }\left( nI+\alpha,nJ+\beta\right) \text{-th entry equals}\\\text{the }\left( \alpha,\beta\right) \text{-th entry of }a_{\left( j-i\right) d+J-I}\text{ for all }I\in\left\{ 0,1,...,d-1\right\} \text{,}\\J\in\left\{ 0,1,...,d-1\right\} \text{, }\alpha\in\left\{ 1,2,...,n\right\} \text{ and }\beta\in\left\{ 1,2,...,n\right\} }}t^{\ell}\in L\mathfrak{gl}_{N}% \] (where we write $a\left( t\right) $ in the form $a\left( t\right) =\sum\limits_{i\in\mathbb{Z}}a_{i}t^{i}$ with $a_{i}\in\mathfrak{gl}_{n}$). \textbf{(a)} We have $\operatorname*{Toep}\nolimits_{N}\circ \operatorname*{Toep}\nolimits_{n,N}=\operatorname*{Toep}\nolimits_{n}$. In other words, we can regard $\operatorname*{Toep}\nolimits_{n,N}$ as an inclusion map $L\mathfrak{gl}_{n}\rightarrow L\mathfrak{gl}_{N}$ which forms a commutative triangle with the inclusion maps $\operatorname*{Toep}% \nolimits_{n}:L\mathfrak{gl}_{n}\rightarrow\overline{\mathfrak{a}_{\infty}}$ and $\operatorname*{Toep}\nolimits_{N}:L\mathfrak{gl}_{N}\rightarrow \overline{\mathfrak{a}_{\infty}}$. In other words, if we consider $L\mathfrak{gl}_{n}$ and $L\mathfrak{gl}_{N}$ as Lie subalgebras of $\overline{\mathfrak{a}_{\infty}}$ (by means of the injections $\operatorname*{Toep}\nolimits_{n}:L\mathfrak{gl}_{n}\rightarrow \overline{\mathfrak{a}_{\infty}}$ and $\operatorname*{Toep}\nolimits_{N}% :L\mathfrak{gl}_{N}\rightarrow\overline{\mathfrak{a}_{\infty}}$), then $L\mathfrak{gl}_{n}\subseteq L\mathfrak{gl}_{N}$. \textbf{(b)} If we consider $\operatorname*{Toep}\nolimits_{n,N}$ as an inclusion map $L\mathfrak{gl}_{n}\rightarrow L\mathfrak{gl}_{N}$, then the $2$-cocycle $\omega:L\mathfrak{gl}_{n}\times L\mathfrak{gl}_{n}\rightarrow \mathbb{C}$ defined in Proposition \ref{prop.ainf.alphaomega} is the restriction of the similarly-defined $2$-cocycle $\omega:L\mathfrak{gl}% _{N}\times L\mathfrak{gl}_{N}\rightarrow\mathbb{C}$ (we also call it $\omega$ because it is constructed similarly) to $L\mathfrak{gl}_{n}\times L\mathfrak{gl}_{n}$. As a consequence, the inclusion map $\operatorname*{Toep}% \nolimits_{n,N}:L\mathfrak{gl}_{n}\rightarrow L\mathfrak{gl}_{N}$ induces a Lie algebra injection $\widehat{\operatorname*{Toep}\nolimits_{n,N}% }:\widehat{\mathfrak{gl}_{n}}\rightarrow\widehat{\mathfrak{gl}_{N}}$ which satisfies $\widehat{\operatorname*{Toep}\nolimits_{N}}\circ \widehat{\operatorname*{Toep}\nolimits_{n,N}}=\widehat{\operatorname*{Toep}% \nolimits_{n}}$. Thus, this injection $\widehat{\operatorname*{Toep}% \nolimits_{n,N}}$ forms a commutative triangle with the inclusion maps $\widehat{\operatorname*{Toep}\nolimits_{n}}:\widehat{\mathfrak{gl}_{n}% }\rightarrow\mathfrak{a}_{\infty}$ and $\widehat{\operatorname*{Toep}% \nolimits_{N}}:\widehat{\mathfrak{gl}_{N}}\rightarrow\mathfrak{a}_{\infty}$. In other words, if we consider $\widehat{\mathfrak{gl}_{n}}$ and $\widehat{\mathfrak{gl}_{N}}$ as Lie subalgebras of $\mathfrak{a}_{\infty}$ (by means of the injections $\widehat{\operatorname*{Toep}\nolimits_{n}% }:\widehat{\mathfrak{gl}_{n}}\rightarrow\mathfrak{a}_{\infty}$ and $\widehat{\operatorname*{Toep}\nolimits_{N}}:\widehat{\mathfrak{gl}_{N}% }\rightarrow\mathfrak{a}_{\infty}$), then $\widehat{\mathfrak{gl}_{n}% }\subseteq\widehat{\mathfrak{gl}_{N}}$. \end{proposition} \textit{Proof of Proposition \ref{prop.glnhat.div}.} \textbf{(a)} The proof of Proposition \ref{prop.glnhat.div} \textbf{(a)} is completely straightforward. (One has to show that the $\left( Ni+nI+\alpha,Nj+nJ+\beta\right) $-th entry of $\left( \operatorname*{Toep}\nolimits_{N}\circ\operatorname*{Toep}% \nolimits_{n,N}\right) \left( a\left( t\right) \right) $ equals the $\left( Ni+nI+\alpha,Nj+nJ+\beta\right) $-th entry of $\operatorname*{Toep}% \nolimits_{n}\left( a\left( t\right) \right) $ for every $a\left( t\right) \in L\mathfrak{gl}_{n}$, every $i\in\mathbb{Z}$, every $j\in\mathbb{Z}$, every $I\in\left\{ 0,1,...,d-1\right\} $, $J\in\left\{ 0,1,...,d-1\right\} $, $\alpha\in\left\{ 1,2,...,n\right\} $ and $\beta \in\left\{ 1,2,...,n\right\} $.) \textbf{(b)} The $2$-cocycle $\omega:L\mathfrak{gl}_{n}\times L\mathfrak{gl}% _{n}\rightarrow\mathbb{C}$ defined in Proposition \ref{prop.ainf.alphaomega} is the restriction of the similarly-defined $2$-cocycle $\omega:L\mathfrak{gl}% _{N}\times L\mathfrak{gl}_{N}\rightarrow\mathbb{C}$ to $L\mathfrak{gl}% _{n}\times L\mathfrak{gl}_{n}$. (This is because both of these $2$-cocycles are restrictions of the Japanese cocycle $\alpha:\overline{\mathfrak{a}% _{\infty}}\times\overline{\mathfrak{a}_{\infty}}\rightarrow\mathbb{C}$, as shown in Proposition \ref{prop.ainf.alphaomega}.) This proves Proposition \ref{prop.glnhat.div}. Note that Proposition \ref{prop.glnhat.div} can be used to derive Proposition \ref{prop.glnhat.T}: \textit{Second proof of Proposition \ref{prop.glnhat.T}.} \textbf{(a)} For every $m\in\mathbb{Z}$, we have $T^{m}\in\widehat{\mathfrak{gl}_{1}}$ (because the Lie algebra isomorphism $\widehat{\phi}$ constructed in Corollary \ref{cor.glnhat.div} satisfies $\phi\left( a_{m}\right) =T^{m}$, so that $T^{m}\in\phi\left( a_{m}\right) \in\widehat{\mathfrak{gl}_{1}}$). Thus, for every $m\in\mathbb{Z}$, we have $T^{m}\in\widehat{\mathfrak{gl}_{1}}% \cap\overline{\mathfrak{a}_{\infty}}=L\mathfrak{gl}_{1}$. Due to Proposition \ref{prop.glnhat.div} \textbf{(a)}, we have $L\mathfrak{gl}% _{1}\subseteq L\mathfrak{gl}_{n}$ (since $1\mid n$). Thus, for every $m\in\mathbb{Z}$, we have $T^{m}\in L\mathfrak{gl}_{1}\subseteq L\mathfrak{gl}% _{n}\subseteq\widehat{\mathfrak{gl}_{n}}$. This proves Proposition \ref{prop.glnhat.T} \textbf{(a)}. \textbf{(b)} Due to Proposition \ref{prop.glnhat.div} \textbf{(b)}, we have $\widehat{\mathfrak{gl}_{1}}\subseteq\widehat{\mathfrak{gl}_{n}}$ (since $1\mid n$). Hence, the Lie algebra isomorphism $\widehat{\phi}:\mathcal{A}% \rightarrow\widehat{\mathfrak{gl}_{1}}$ constructed in Corollary \ref{cor.glnhat.div} induces a Lie algebra injection $\mathcal{A}% \rightarrow\widehat{\mathfrak{gl}_{n}}$ (which sends every $a\in\mathcal{A}$ to $\widehat{\phi}\left( a\right) \in\widehat{\mathfrak{gl}_{n}}$). Formally speaking, this injection is the map $\widehat{\operatorname*{Toep}% \nolimits_{1,n}}\circ\widehat{\phi}:\mathcal{A}\rightarrow \widehat{\mathfrak{gl}_{n}}$ (because the injection $\widehat{\mathfrak{gl}% _{1}}\rightarrow\widehat{\mathfrak{gl}_{n}}$ is $\widehat{\operatorname*{Toep}% \nolimits_{1,n}}$). Therefore, the restriction of the $\widehat{\mathfrak{gl}% _{n}}$-module $\mathcal{F}^{\left( m\right) }$ by means of this injection is% \begin{align*} & \left( \text{the restriction of the }\widehat{\mathfrak{gl}_{n}% }\text{-module }\mathcal{F}^{\left( m\right) }\text{ by means of the injection }\widehat{\operatorname*{Toep}\nolimits_{1,n}}\circ\widehat{\phi }:\mathcal{A}\rightarrow\widehat{\mathfrak{gl}_{n}}\right) \\ & =\left( \text{the restriction of the }\mathfrak{a}_{\infty}\text{-module }\mathcal{F}^{\left( m\right) }\text{ by means of the injection }\widehat{\operatorname*{Toep}\nolimits_{n}}\circ\widehat{\operatorname*{Toep}% \nolimits_{1,n}}\circ\widehat{\phi}:\mathcal{A}\rightarrow\mathfrak{a}% _{\infty}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{because the }\widehat{\mathfrak{gl}_{n}}\text{-module }\mathcal{F}% ^{\left( m\right) }\text{ itself was the restriction of the }\mathfrak{a}% _{\infty}\text{-module }\mathcal{F}^{\left( m\right) }\\ \text{by means of the injection }\widehat{\operatorname*{Toep}\nolimits_{n}% }:\widehat{\mathfrak{gl}_{n}}\rightarrow\mathfrak{a}_{\infty}% \end{array} \right) \\ & =\left( \text{the restriction of the }\mathfrak{a}_{\infty}\text{-module }\mathcal{F}^{\left( m\right) }\text{ by means of the injection }\widehat{\operatorname*{Toep}\nolimits_{1}}\circ\widehat{\phi}:\mathcal{A}% \rightarrow\mathfrak{a}_{\infty}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\widehat{\operatorname*{Toep}\nolimits_{n}}\circ \widehat{\operatorname*{Toep}\nolimits_{1,n}}=\widehat{\operatorname*{Toep}% \nolimits_{1}}\\ \text{(by Proposition \ref{prop.glnhat.div} \textbf{(b)}, applied to }n\text{ and }1\text{ instead of }N\text{ and }n\text{)}% \end{array} \right) \\ & =\left( \begin{array} [c]{c}% \text{the restriction of the }\mathfrak{a}_{\infty}\text{-module }% \mathcal{F}^{\left( m\right) }\text{ by means of the}\\ \text{embedding }\mathcal{A}\rightarrow\mathfrak{a}_{\infty}\text{ constructed in Definition \ref{def.ainf.A}}% \end{array} \right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{because }\widehat{\operatorname*{Toep}\nolimits_{1}}\circ\widehat{\phi }:\mathcal{A}\rightarrow\mathfrak{a}_{\infty}\text{ is exactly the}\\ \text{embedding }\mathcal{A}\rightarrow\mathfrak{a}_{\infty}\text{ constructed in Definition \ref{def.ainf.A}}% \end{array} \right) \\ & =\left( \text{the }\mathcal{A}\text{-module }\mathcal{F}^{\left( m\right) }\text{ that we know}\right) . \end{align*} This proves Proposition \ref{prop.glnhat.T} \textbf{(b)}. \subsection{The semidirect product \texorpdfstring{$\protect\widetilde{\mathfrak{gl}_{n}}$}{gl-n-tilde} and its representation theory} \subsubsection{Extending affine Lie algebras by derivations} Now we give a definition pertaining to general affine Lie algebras: \begin{definition} \label{def.gwave}If $\widehat{\mathfrak{g}}=L\mathfrak{g}\oplus\mathbb{C}K$ is an affine Lie algebra (the $\oplus$ sign here only means a direct sum of vector spaces, not a direct sum of Lie algebras), then there exists a unique linear map $d:\widehat{\mathfrak{g}}\rightarrow\widehat{\mathfrak{g}}$ such that $d\left( a\left( t\right) \right) =ta^{\prime}\left( t\right) $ for every $a\left( t\right) \in L\mathfrak{g}$ (so that $d\left( at^{\ell }\right) =\ell at^{\ell}$ for every $a\in\mathfrak{g}$ and $\ell\in \mathbb{N}$) and $d\left( K\right) =0$. This linear map $d$ is a derivation (as can be easily checked). Thus, the abelian Lie algebra $\mathbb{C}d$ (a one-dimensional Lie algebra) acts on the Lie algebra $\widehat{\mathfrak{g}}$ by derivations (in the obvious way, with $d$ acting as $d$). Thus, a semidirect product $\mathbb{C}d\ltimes\widehat{\mathfrak{g}}$ is well-defined (according to Definition \ref{def.semidir.lielie}). Set $\widetilde{\mathfrak{g}}=\mathbb{C}d\ltimes\widehat{\mathfrak{g}}$. Clearly, $\widetilde{\mathfrak{g}}=\mathbb{C}d\oplus\widehat{\mathfrak{g}}$ as vector space. The Lie algebra $\widetilde{\mathfrak{g}}$ is graded by taking the grading of $\widehat{\mathfrak{g}}$ and additionally giving $d$ the degree $0$. \end{definition} One can wonder which $\widehat{\mathfrak{g}}$-modules can be extended to $\widetilde{\mathfrak{g}}$-modules. This can't be generally answered, but here is a partial uniqueness result: \begin{lemma} \label{lem.gwave.uniqueder}Let $\mathfrak{g}$ be a Lie algebra, and $d$ be the unique derivation $\widehat{\mathfrak{g}}\rightarrow\widehat{\mathfrak{g}}$ constructed in Definition \ref{def.gwave}. Let $M$ be a $\widehat{\mathfrak{g}% }$-module, and $v$ an element of $M$ such that $M$ is generated by $v$ as a $\widehat{\mathfrak{g}}$-module. Then, there exists \textbf{at most one} extension of the $\widehat{\mathfrak{g}}$-representation on $M$ to $\widetilde{\mathfrak{g}}$ such that $dv=0$. \end{lemma} \textit{Proof of Lemma \ref{lem.gwave.uniqueder}.} Let $\rho_{1}% :\widetilde{\mathfrak{g}}\rightarrow\operatorname*{End}M$ and $\rho _{2}:\widetilde{\mathfrak{g}}\rightarrow\operatorname*{End}M$ be two extensions of the $\widehat{\mathfrak{g}}$-representation on $M$ to $\widetilde{\mathfrak{g}}$ such that $\rho_{1}\left( d\right) v=0$ and $\rho_{2}\left( d\right) v=0$. If we succeed in showing that $\rho_{1}% =\rho_{2}$, then Lemma \ref{lem.gwave.uniqueder} will be proven. Let $U$ be the subset $\left\{ u\in M\ \mid\ \rho_{1}\left( d\right) u=\rho_{2}\left( d\right) u\right\} $ of $M$. Clearly, $U$ is a vector subspace of $M$. Also, $v\in U$ (since $\rho_{1}\left( d\right) v=0=\rho _{2}\left( d\right) v$). We will now show that $U$ is a $\widehat{\mathfrak{g}}$-submodule of $M$. In fact, since $\rho_{1}$ is an action of $\widetilde{\mathfrak{g}}$ on $M$, every $m\in M$ and every $\alpha\in\widehat{\mathfrak{g}}$ satisfy% \[ \left( \rho_{1}\left( d\right) \right) \left( \rho_{1}\left( \alpha\right) m\right) -\left( \rho_{1}\left( \alpha\right) \right) \left( \rho_{1}\left( d\right) m\right) =\rho_{1}\left( \left[ d,\alpha\right] \right) m. \] Since $\rho_{1}\left( \alpha\right) m=\alpha\rightharpoonup m$ (because the action $\rho_{1}$ extends the $\widehat{\mathfrak{g}}$-representation on $M$) and $\left[ d,\alpha\right] =d\left( \alpha\right) $ (by the definition of the Lie bracket on the semidirect product $\widetilde{\mathfrak{g}}% =\mathbb{C}d\ltimes\widehat{\mathfrak{g}}$), this rewrites as follows: Every $m\in M$ and every $\alpha\in\widehat{\mathfrak{g}}$ satisfy% \[ \left( \rho_{1}\left( d\right) \right) \left( \alpha\rightharpoonup m\right) -\left( \rho_{1}\left( \alpha\right) \right) \left( \rho _{1}\left( d\right) m\right) =\rho_{1}\left( d\left( \alpha\right) \right) m. \] Since $\left( \rho_{1}\left( \alpha\right) \right) \left( \rho_{1}\left( d\right) m\right) =\alpha\rightharpoonup\left( \rho_{1}\left( d\right) m\right) $ (again because the action $\rho_{1}$ extends the $\widehat{\mathfrak{g}}$-representation on $M$) and $\rho_{1}\left( d\left( \alpha\right) \right) m=\left( d\left( \alpha\right) \right) \rightharpoonup m$ (for the same reason), this further rewrites as follows: Every $m\in M$ and every $\alpha\in\widehat{\mathfrak{g}}$ satisfy% \begin{equation} \left( \rho_{1}\left( d\right) \right) \left( \alpha\rightharpoonup m\right) -\alpha\rightharpoonup\left( \rho_{1}\left( d\right) m\right) =\left( d\left( \alpha\right) \right) \rightharpoonup m. \label{pf.gwave.uniqueder.1}% \end{equation} Now, let $m\in U$ and $\alpha\in\widehat{\mathfrak{g}}$ be arbitrary. Then, $\rho_{1}\left( d\right) m=\rho_{2}\left( d\right) m$ (by the definition of $U$, since $m\in U$), but we have% \[ \left( \rho_{1}\left( d\right) \right) \left( \alpha\rightharpoonup m\right) =\alpha\rightharpoonup\left( \rho_{1}\left( d\right) m\right) +\left( d\left( \alpha\right) \right) \rightharpoonup m \] (by (\ref{pf.gwave.uniqueder.1})) and% \[ \left( \rho_{2}\left( d\right) \right) \left( \alpha\rightharpoonup m\right) =\alpha\rightharpoonup\left( \rho_{2}\left( d\right) m\right) +\left( d\left( \alpha\right) \right) \rightharpoonup m \] (similarly). Hence,% \begin{align*} \left( \rho_{1}\left( d\right) \right) \left( \alpha\rightharpoonup m\right) & =\alpha\rightharpoonup\underbrace{\left( \rho_{1}\left( d\right) m\right) }_{=\rho_{2}\left( d\right) m}+\left( d\left( \alpha\right) \right) \rightharpoonup m\\ & =\alpha\rightharpoonup\left( \rho_{2}\left( d\right) m\right) +\left( d\left( \alpha\right) \right) \rightharpoonup m=\left( \rho_{2}\left( d\right) \right) \left( \alpha\rightharpoonup m\right) , \end{align*} so that $\alpha\rightharpoonup m\in U$ (by the definition of $U$). Now forget that we fixed $m\in U$ and $\alpha\in\widehat{\mathfrak{g}}$. We thus have showed that $\alpha\rightharpoonup m\in U$ for every $m\in U$ and $\alpha\in\widehat{\mathfrak{g}}$. In other words, $U$ is a $\widehat{\mathfrak{g}}$-submodule of $M$. Since $v\in U$, this yields that $U$ is a $\widehat{\mathfrak{g}}$-submodule of $M$ containing $v$, and thus must be the whole $M$ (since $M$ is generated by $v$ as a $\widehat{\mathfrak{g}}$-module). Thus, $M=U=\left\{ u\in M\ \mid\ \rho _{1}\left( d\right) u=\rho_{2}\left( d\right) u\right\} $. Hence, every $u\in M$ satisfies $\rho_{1}\left( d\right) u=\rho_{2}\left( d\right) u$. Thus, $\rho_{1}\left( d\right) =\rho_{2}\left( d\right) $. Combining $\rho_{1}\mid_{\widehat{\mathfrak{g}}}=\rho_{2}\mid _{\widehat{\mathfrak{g}}}$ (because both $\rho_{1}$ and $\rho_{2}$ are extensions of the $\widehat{\mathfrak{g}}$-representation on $M$, and thus coincide on $\widehat{\mathfrak{g}}$) and $\rho_{1}\mid_{\mathbb{C}d}=\rho _{2}\mid_{\mathbb{C}d}$ (because $\rho_{1}\left( d\right) =\rho_{2}\left( d\right) $), we obtain $\rho_{1}=\rho_{2}$ (because the vector space $\widetilde{\mathfrak{g}}=\mathbb{C}d\ltimes\widehat{\mathfrak{g}}$ is generated by $\mathbb{C}d$ and $\widehat{\mathfrak{g}}$, and thus two linear maps which coincide on $\mathbb{C}d$ and on $\widehat{\mathfrak{g}}$ must be identical). Thus, as we said above, Lemma \ref{lem.gwave.uniqueder} is proven. \subsubsection{\texorpdfstring{$\protect\widetilde{\mathfrak{gl}_{n}}$} {gl-n-tilde}} Applying Definition \ref{def.gwave} to $\mathfrak{g}=\mathfrak{gl}_{n}$, we obtain a Lie algebra $\widetilde{\mathfrak{gl}_{n}}$. We want to study its highest weight theory. \begin{Convention} For the sake of disambiguation, let us, in the following, use $E_{i,j}% ^{\mathfrak{gl}_{n}}$ to denote the elementary matrices of $\mathfrak{gl}_{n}$ (these are defined for $\left( i,j\right) \in\left\{ 1,2,...,n\right\} ^{2}$), and use $E_{i,j}^{\mathfrak{gl}_{\infty}}$ to denote the elementary matrices of $\mathfrak{gl}_{\infty}$ (these are defined for $\left( i,j\right) \in\mathbb{Z}^{2}$). \end{Convention} \begin{definition} We can make $L\mathfrak{gl}_{n}$ into a graded Lie algebra by setting $\deg E_{i,j}^{\mathfrak{gl}_{n}}=j-i$ (this, so far, is the standard grading on $\mathfrak{gl}_{n}$) and $\deg t=n$. Consequently, $\widehat{\mathfrak{gl}% _{n}}=\mathbb{C}K\oplus\mathfrak{gl}_{n}$ (this is just a direct sum of vector spaces) becomes a graded Lie algebra with $\deg K=0$, and $\widetilde{\mathfrak{gl}_{n}}=\mathbb{C}d\oplus\widehat{\mathfrak{gl}_{n}}$ (again, this is only a direct sum of vector spaces) becomes a graded Lie algebra with $\deg d=0$. \end{definition} The triangular decomposition of $\widetilde{\mathfrak{gl}_{n}}$ is $\widetilde{\mathfrak{gl}_{n}}=\widetilde{\mathfrak{n}_{-}}\oplus \widetilde{\mathfrak{h}}\oplus\widetilde{\mathfrak{n}_{+}}$. Here, $\widetilde{\mathfrak{h}}=\mathbb{C}K\oplus\mathbb{C}d\oplus\mathfrak{h}$ where $\mathfrak{h}$ is the Lie algebra of diagonal $n\times n$ matrices (in other words, $\mathfrak{h}=\left\langle E_{1,1}^{\mathfrak{gl}_{n}}% ,E_{2,2}^{\mathfrak{gl}_{n}},...,E_{n,n}^{\mathfrak{gl}_{n}}\right\rangle $). Further, $\widetilde{\mathfrak{n}_{+}}=\mathfrak{n}_{+}\oplus t\mathfrak{gl}% _{n}\left[ t\right] $ (where $\mathfrak{n}_{+}$ is the Lie algebra of strictly upper-triangular matrices) and $\widetilde{\mathfrak{n}_{-}% }=\mathfrak{n}_{-}\oplus t^{-1}\mathfrak{gl}_{n}\left[ t^{-1}\right] $ (where $\mathfrak{n}_{-}$ is the Lie algebra of strictly lower-triangular matrices). \begin{definition} For every $m\in\mathbb{Z}$, define the weight $\widetilde{\omega}_{m}% \in\widetilde{\mathfrak{h}}^{\ast}$ by% \begin{align*} \widetilde{\omega}_{m}\left( E_{i,i}^{\mathfrak{gl}_{n}}\right) & =\left\{ \begin{array} [c]{c}% 1,\text{ if }i\leq\overline{m};\\ 0,\text{ if }i>\overline{m}% \end{array} \right. +\dfrac{m-\overline{m}}{n}\ \ \ \ \ \ \ \ \ \ \text{for all }% i\in\left\{ 1,2,...,n\right\} ;\\ \widetilde{\omega}_{m}\left( K\right) & =1;\\ \widetilde{\omega}_{m}\left( d\right) & =0, \end{align*} where $\overline{m}$ is the remainder of $m$ modulo $n$ (that is, the element of $\left\{ 0,1,...,n-1\right\} $ satisfying $m\equiv\overline {m}\operatorname{mod}n$). \end{definition} Note that we can rewrite the definition of $\widetilde{\omega}_{m}\left( E_{i,i}^{\mathfrak{gl}_{n}}\right) $ as% \begin{align*} & \widetilde{\omega}_{m}\left( E_{i,i}^{\mathfrak{gl}_{n}}\right) \\ & =\left\{ \begin{array} [c]{c}% \left( \text{the number of all }j\in\mathbb{Z}\text{ such that }j\equiv i\operatorname{mod}n\text{ and }1\leq j\leq m\right) ,\ \ \ \ \ \ \ \ \ \ \text{if }m\geq0;\\ -\left( \text{the number of all }j\in\mathbb{Z}\text{ such that }j\equiv i\operatorname{mod}n\text{ and }m \overline{m}% \end{array} \right. +\dfrac{m-\overline{m}}{n}\right) }_{=\widetilde{\omega}_{m}\left( E_{i,i}^{\mathfrak{gl}_{n}}\right) }\psi_{m}=\widetilde{\omega}_{m}\left( E_{i,i}^{\mathfrak{gl}_{n}}\right) \psi_{m}, \] where $\overline{m}$ is the element of $\left\{ 0,1,...,n-1\right\} $ satisfying $m\equiv\overline{m}\operatorname{mod}n$. Thus, we have checked that $\widetilde{\mathfrak{n}_{+}}\psi_{m}=0$ and $x\psi_{m}=\widetilde{\omega}_{m}\left( x\right) \psi_{m}$ for every $x\in\widetilde{\mathfrak{h}}$. Thus, $\psi_{m}$ is a singular vector of weight $\widetilde{\omega}_{m}$. In other words, $\psi_{m}\in \operatorname*{Sing}\nolimits_{\widetilde{\omega}_{m}}\left( \mathcal{F}% ^{\left( m\right) }\right) $. By Lemma \ref{lem.singvec}, we thus have a canonical isomorphism% \begin{align*} \operatorname*{Hom}\nolimits_{\widetilde{\mathfrak{gl}_{n}}}\left( M_{\widetilde{\omega}_{m}}^{+},\mathcal{F}^{\left( m\right) }\right) & \rightarrow\operatorname*{Sing}\nolimits_{\widetilde{\omega}_{m}}\left( \mathcal{F}^{\left( m\right) }\right) ,\\ \phi & \mapsto\phi\left( v_{\widetilde{\omega}_{m}}^{+}\right) . \end{align*} Thus, since $\psi_{m}\in\operatorname*{Sing}\nolimits_{\widetilde{\omega}_{m}% }\left( \mathcal{F}^{\left( m\right) }\right) $, there exists a $\widetilde{\mathfrak{gl}_{n}}$-module homomorphism $\phi:M_{\widetilde{\omega }_{m}}^{+}\rightarrow\mathcal{F}^{\left( m\right) }$ such that $\phi\left( v_{\widetilde{\omega}_{m}}^{+}\right) =\psi_{m}$. Consider this $\phi$. Since $\mathcal{F}^{\left( m\right) }$ is generated by $\psi_{m}$ as a $\widehat{\mathfrak{gl}_{n}}$-module (this was proven in the proof of Proposition \ref{prop.glwave.F}), it is clear that $\mathcal{F}^{\left( m\right) }$ is generated by $\psi_{m}$ as a $\widetilde{\mathfrak{gl}_{n}}% $-module as well. Thus, $\phi$ must be surjective (because $\psi_{m}% =\phi\left( v_{\widetilde{\omega}_{m}}^{+}\right) \in\phi\left( M_{\widetilde{\omega}_{m}}^{+}\right) $). Hence, $\mathcal{F}^{\left( m\right) }$ is (isomorphic to) a quotient of the $\widetilde{\mathfrak{gl}% _{n}}$-module $M_{\widetilde{\omega}_{m}}^{+}$. In other words, $\mathcal{F}% ^{\left( m\right) }$ is a highest-weight module with highest weight $\widetilde{\omega}_{m}$. Combined with the irreducibility of $\mathcal{F}% ^{\left( m\right) }$, this proves Proposition \ref{prop.glwave.F.irr}. \subsubsection{The \texorpdfstring{$\protect\widetilde{\mathfrak{gl}_{n}}$}{gl-n-tilde}-module \texorpdfstring{$\mathcal{B}^{\left( m\right) }$}{structure on the bosonic Fock space}} By applying the Boson-Fermion correspondence $\sigma$ to Proposition \ref{prop.glwave.F}, we obtain: \begin{proposition} \label{prop.glwave.B}Let $m\in\mathbb{Z}$. Let $\psi_{m}^{\prime}$ be the element $\sigma^{-1}\left( v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...\right) \in\mathcal{B}^{\left( m\right) }$ (the highest-weight vector of $\mathcal{B}^{\left( m\right) }$). There exists a unique extension of the $\widehat{\mathfrak{gl}_{n}}% $-representation on $\mathcal{B}^{\left( m\right) }$ to $\widetilde{\mathfrak{gl}_{n}}$ such that $d\psi_{m}^{\prime}=0$. The action of $d$ in this extension is given by% \[ d\left( \sigma^{-1}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}}% \wedge...\right) \right) =\left( \sum\limits_{k\geq0}\left( \left\lceil \dfrac{m-k}{n}\right\rceil -\left\lceil \dfrac{i_{k}}{n}\right\rceil \right) \right) \cdot\sigma^{-1}\left( v_{i_{0}}\wedge v_{i_{1}}\wedge v_{i_{2}% }\wedge...\right) \] for every $m$-degression $\left( i_{0},i_{1},i_{2},...\right) $. \end{proposition} By applying the Boson-Fermion correspondence $\sigma$ to Proposition \ref{prop.glwave.F.irr}, we obtain: \begin{proposition} Let $m\in\mathbb{Z}$. Let $\psi_{m}^{\prime}$ be the element $\sigma ^{-1}\left( v_{m}\wedge v_{m-1}\wedge v_{m-2}\wedge...\right) \in \mathcal{B}^{\left( m\right) }$ (the highest-weight vector of $\mathcal{B}% ^{\left( m\right) }$). \textbf{(a)} The $\widehat{\mathfrak{gl}_{n}}$-module $\mathcal{B}^{\left( m\right) }$ is irreducible. \textbf{(b)} Let $\widehat{\rho}\mid_{\widetilde{\mathfrak{gl}_{n}}% }:\widetilde{\mathfrak{gl}_{n}}\rightarrow\operatorname*{End}\left( \mathcal{B}^{\left( m\right) }\right) $ denote the unique extension of the $\widehat{\mathfrak{gl}_{n}}$-representation on $\mathcal{B}^{\left( m\right) }$ to $\widetilde{\mathfrak{gl}_{n}}$ such that $d\psi_{m}^{\prime }=0$. (This is well-defined due to Proposition \ref{prop.glwave.B}.) The $\widetilde{\mathfrak{gl}_{n}}$-module $\left( \mathcal{B}^{\left( m\right) },\widehat{\rho}\mid_{\widetilde{\mathfrak{gl}_{n}}}\right) $ is irreducible with highest weight $\widetilde{\omega}_{m}$. \end{proposition} \subsubsection{\texorpdfstring{$\protect\widetilde{\mathfrak{sl}_{n}}$} {sl-n-tilde} and its action on \texorpdfstring{$\mathcal{B}^{\left( m\right) }$}{the bosonic Fock space}} We have $\left[ I_{n}t,\widehat{\mathfrak{sl}_{n}}\right] =0$ in the Lie algebra $\widehat{\mathfrak{gl}_{n}}$ (this is because $\left[ I_{n}% t,L\mathfrak{sl}_{n}\right] =0$ in the Lie algebra $L\mathfrak{gl}_{n}$, and because $\omega\left( I_{n}t,L\mathfrak{sl}_{n}\right) =0$ where the $2$-cocycle $\omega$ is the one defined in Proposition \ref{prop.ainf.alphaomega}). Since $I_{n}t\in\widehat{\mathfrak{gl}_{n}}$ acts on $\mathcal{F}$ by the operator $\widehat{\operatorname*{Toep}\nolimits_{n}% }\left( I_{n}t\right) =T^{n}$ (more precisely, by the action of $T^{n}$ on $\mathcal{F}$, but let us abbreviate this by $T^{n}$ here), this yields that the action of $T^{n}$ on $\mathcal{F}$ is an $\widehat{\mathfrak{sl}_{n}}% $-module homomorphism. Thus, the action of $T^{n}$ on $\mathcal{B}$ also is an $\widehat{\mathfrak{sl}_{n}}$-module homomorphism. As a consequence, the restriction to $\widehat{\mathfrak{sl}_{n}}$ of the representation $\mathcal{B}^{\left( m\right) }$ is not irreducible. But $\psi_{m}^{\prime}$ is still a highest-weight vector with highest weight $\widetilde{\omega}_{m}$. Let us look at how this representation $\mathcal{B}^{\left( m\right) }$ decomposes. \begin{definition} Let $h_{i}=E_{i,i}^{\mathfrak{gl}_{n}}-E_{i+1,i+1}^{\mathfrak{gl}_{n}}$ for $i\in\left\{ 1,2,...,n-1\right\} $, and let $h_{0}=K-h_{1}-h_{2}% -...-h_{n-1}$. Then, $\left( h_{0},h_{1},...,h_{n-1},d\right) $ is a basis of $\widetilde{\mathfrak{h}}\cap\widetilde{\mathfrak{sl}_{n}}$ (which is the $0$-th homogeneous component of $\widetilde{\mathfrak{sl}_{n}}$). \end{definition} \begin{definition} For every $m\in\mathbb{Z}$, define the weight $\omega_{m}\in\left( \widetilde{\mathfrak{h}}\cap\widetilde{\mathfrak{sl}_{n}}\right) ^{\ast}$ to be the restriction $\widetilde{\omega}_{m}\mid_{\widetilde{\mathfrak{h}}% \cap\widetilde{\mathfrak{sl}_{n}}}$ of $\widetilde{\omega}_{m}$ to the $0$-th homogeneous component of $\widetilde{\mathfrak{sl}_{n}}$. \end{definition} This weight $\omega_{m}$ does not depend on $m$ but only depends on the residue class of $m$ modulo $n$. In fact, it satisfies% \begin{align*} \omega_{m}\left( h_{i}\right) & =\widetilde{\omega}_{m}\left( h_{i}\right) =\left\{ \begin{array} [c]{c}% 1,\text{ if }i\equiv m\operatorname{mod}n;\\ 0,\text{ if }i\not \equiv m\operatorname{mod}n \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{ 0,1,...,n-1\right\} ;\\ \omega_{m}\left( d\right) & =\widetilde{\omega}_{m}\left( d\right) =0. \end{align*} \begin{definition} Let $\mathcal{A}^{\left( n\right) }$ be the Lie subalgebra $\left\langle K\right\rangle +\left\langle a_{ni}\ \mid\ i\in\mathbb{Z}\right\rangle $ of $\mathcal{A}$. \end{definition} Note that the map% \begin{align*} \mathcal{A} & \rightarrow\mathcal{A}^{\left( n\right) },\\ a_{i} & \mapsto a_{ni}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\mathbb{Z},\\ K & \mapsto nK \end{align*} is a Lie algebra isomorphism. But we still consider $\mathcal{A}^{\left( n\right) }$ as a Lie subalgebra of $\mathcal{A}$, and we won't identify it with $\mathcal{A}$ via this isomorphism. Since $\mathcal{A}^{\left( n\right) }$ is a Lie subalgebra of $\mathcal{A}$, both $\mathcal{A}$-modules $\mathcal{F}$ and $\mathcal{B}$ become $\mathcal{A}^{\left( n\right) }$-modules. Let us consider the direct sum $\widehat{\mathfrak{sl}_{n}}\oplus \mathcal{A}^{\left( n\right) }$ of Lie algebras. Let us denote by $K_{1}$ the element $\left( K,0\right) $ of $\widehat{\mathfrak{sl}_{n}}% \oplus\mathcal{A}^{\left( n\right) }$ (where the $K$ means the element $K$ of $\widehat{\mathfrak{sl}_{n}}$), and let us denote by $K_{2}$ the element $\left( 0,K\right) $ of $\widehat{\mathfrak{sl}_{n}}\oplus\mathcal{A}% ^{\left( n\right) }$ (where the $K$ means the element $K$ of $\mathcal{A}% ^{\left( n\right) }$). Note that both elements $K_{1}=\left( K,0\right) $ and $K_{2}=\left( 0,K\right) $ lie in the center of $\widehat{\mathfrak{sl}% _{n}}\oplus\mathcal{A}^{\left( n\right) }$; hence, so does their difference $K_{1}-K_{2}=\left( K,-K\right) $. Thus, $\left\langle K_{1}-K_{2}% \right\rangle $ (the $\mathbb{C}$-linear span of the set $\left\{ K_{1}% -K_{2}\right\} $) is an ideal of $\widehat{\mathfrak{sl}_{n}}\oplus \mathcal{A}^{\left( n\right) }$. Thus, $\left( \widehat{\mathfrak{sl}_{n}% }\oplus\mathcal{A}^{\left( n\right) }\right) \diagup\left( K_{1}% -K_{2}\right) $ is a Lie algebra. \begin{proposition} The Lie algebras $\widehat{\mathfrak{gl}_{n}}$ and $\left( \widehat{\mathfrak{sl}_{n}}\oplus\mathcal{A}^{\left( n\right) }\right) \diagup\left( K_{1}-K_{2}\right) $ are isomorphic. More precisely, the maps% \begin{align*} \left( \widehat{\mathfrak{sl}_{n}}\oplus\mathcal{A}^{\left( n\right) }\right) \diagup\left( K_{1}-K_{2}\right) & \rightarrow \widehat{\mathfrak{gl}_{n}},\\ \overline{\left( At^{\ell},0\right) } & \mapsto At^{\ell}% \ \ \ \ \ \ \ \ \ \ \text{for every }A\in\mathfrak{sl}_{n}\text{ and }\ell \in\mathbb{Z},\\ \overline{\left( 0,a_{n\ell}\right) } & \mapsto\operatorname*{id}% \nolimits_{n}t^{\ell}\ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in\mathbb{Z},\\ \overline{K_{1}}=\overline{K_{2}} & \mapsto K \end{align*} and% \begin{align*} \widehat{\mathfrak{gl}_{n}} & \rightarrow\left( \widehat{\mathfrak{sl}_{n}% }\oplus\mathcal{A}^{\left( n\right) }\right) \diagup\left( K_{1}% -K_{2}\right) ,\\ At^{\ell} & \mapsto\overline{\left( \left( A-\dfrac{1}{n}\left( \operatorname*{Tr}A\right) \cdot\operatorname*{id}\nolimits_{n}\right) t^{\ell},\left( \dfrac{1}{n}\operatorname*{Tr}A\right) a_{n\ell}\right) }\ \ \ \ \ \ \ \ \ \ \text{for every }A\in\mathfrak{gl}_{n}\text{ and }\ell \in\mathbb{Z},\\ K & \mapsto\overline{K_{1}}=\overline{K_{2}}. \end{align*} are mutually inverse isomorphisms of Lie algebras. \end{proposition} The proof of this proposition is left to the reader (it is completely straightforward). This isomorphism $\widehat{\mathfrak{gl}_{n}}\cong\left( \widehat{\mathfrak{sl}_{n}}\oplus\mathcal{A}^{\left( n\right) }\right) \diagup\left( K_{1}-K_{2}\right) $ allows us to consider any $\widehat{\mathfrak{gl}_{n}}$-module as an $\left( \widehat{\mathfrak{sl}% _{n}}\oplus\mathcal{A}^{\left( n\right) }\right) \diagup\left( K_{1}% -K_{2}\right) $-module, i. e., as an $\widehat{\mathfrak{sl}_{n}}% \oplus\mathcal{A}^{\left( n\right) }$-module on which $K_{1}$ and $K_{2}$ act the same way. In particular, $\mathcal{F}$ and $\mathcal{B}$ become $\widehat{\mathfrak{sl}_{n}}\oplus\mathcal{A}^{\left( n\right) }$-modules. Of course, the actions of the two addends $\widehat{\mathfrak{sl}_{n}}$ and $\mathcal{A}^{\left( n\right) }$ on $\mathcal{F}$ and $\mathcal{B}$ are exactly the actions of $\widehat{\mathfrak{sl}_{n}}$ and $\mathcal{A}^{\left( n\right) }$ on $\mathcal{F}$ and $\mathcal{B}$ that result from the canonical inclusions $\widehat{\mathfrak{sl}_{n}}\subseteq\widehat{\mathfrak{gl}_{n}% }\subseteq\mathfrak{a}_{\infty}$ and $\mathcal{A}^{\left( n\right) }\subseteq\mathcal{A}\cong\widehat{\mathfrak{gl}_{1}}\subseteq\mathfrak{a}% _{\infty}$. (This is clear for the action of $\widehat{\mathfrak{sl}_{n}}$, and is very easy to see for the action of $\mathcal{A}^{\left( n\right) }$.) We checked above that the action of $T^{n}$ on $\mathcal{B}$ is an $\widehat{\mathfrak{sl}_{n}}$-module homomorphism. This easily generalizes: For every integer $i$, the action of $T^{ni}$ on $\mathcal{B}$ is an $\widehat{\mathfrak{sl}_{n}}$-module homomorphism.\footnote{\textit{Proof.} Let $i$ be an integer. We have $\left[ I_{n}t^{i},\widehat{\mathfrak{sl}_{n}% }\right] =0$ in the Lie algebra $\widehat{\mathfrak{gl}_{n}}$ (this is because $\left[ I_{n}t^{i},L\mathfrak{sl}_{n}\right] =0$ in the Lie algebra $L\mathfrak{gl}_{n}$, and because $\omega\left( I_{n}t^{i},L\mathfrak{sl}% _{n}\right) =0$ where the $2$-cocycle $\omega$ is the one defined in Proposition \ref{prop.ainf.alphaomega}). Since $I_{n}t^{i}\in \widehat{\mathfrak{gl}_{n}}$ acts on $\mathcal{F}$ by the operator $\widehat{\operatorname*{Toep}\nolimits_{n}}\left( I_{n}t^{i}\right) =T^{ni}$ (more precisely, by the action of $T^{ni}$ on $\mathcal{F}$, but let us abbreviate this by $T^{ni}$ here), this yields that the action of $T^{ni}$ on $\mathcal{F}$ is an $\widehat{\mathfrak{sl}_{n}}$-module homomorphism. Thus, the action of $T^{ni}$ on $\mathcal{B}$ also is an $\widehat{\mathfrak{sl}_{n}}$-module homomorphism.} Thus, the subspace $\mathcal{B}_{0}^{\left( m\right) }=\left\{ v\in\mathcal{B}^{\left( m\right) }\ \mid\ T^{ni}v=0\text{ for all }i>0\right\} $ of $\mathcal{B}% ^{\left( m\right) }$ is an $\widehat{\mathfrak{sl}_{n}}$-submodule. Recalling that $\mathcal{B}^{\left( m\right) }=\mathbb{C}\left[ x_{1}% ,x_{2},x_{3},...\right] $, with $T^{ni}$ acting as $ni\dfrac{\partial }{\partial x_{ni}}$, we have $\mathcal{B}_{0}^{\left( m\right) }% \cong\mathbb{C}\left[ x_{j}\ \mid\ n\nmid j\right] $. \begin{theorem} \label{thm.B0m}This $\mathcal{B}_{0}^{\left( m\right) }$ is an irreducible $\widehat{\mathfrak{sl}_{n}}$-module (or $\widetilde{\mathfrak{sl}_{n}}% $-module; this doesn't matter) with highest weight $\omega_{\overline{m}}$ (this means that $\mathcal{B}_{0}^{\left( m\right) }\cong L_{\omega _{\overline{m}}}$) and depends only on $\overline{m}$ (the remainder of $m$ modulo $n$) rather than on $m$. Moreover, $\mathcal{B}^{\left( m\right) }\cong\mathcal{B}_{0}^{\left( m\right) }\otimes\widetilde{F}_{m}$, where $\widetilde{F}_{m}$ is the appropriate Fock module over $\mathcal{A}^{\left( n\right) }$. \end{theorem} \textit{Proof of Theorem \ref{thm.B0m}.} We clearly have such a decomposition as vector spaces, $\widetilde{F}_{m}=\mathbb{C}\left[ x_{n},x_{2n}% ,x_{3n},...\right] $. Each of the two Lie algebras acts in its own factor: $\mathcal{A}^{\left( n\right) }$ acts in $\widetilde{F}_{m}$, and $\widehat{\mathfrak{gl}_{n}}$ commutes with $\mathcal{A}^{\left( n\right) }% $. Since the tensor product is irreducible, each factor is irreducible, so that $\mathcal{B}_{0}^{\left( m\right) }$ is irreducible. \subsubsection{\textbf{[unfinished]} Classification of unitary highest-weight \texorpdfstring{$\protect\widehat{\mathfrak{sl}_{n}}$}{sl-n-hat}-modules} We can now classify unitary highest-weight representations of $\widehat{\mathfrak{sl}_{n}}$: \begin{proposition} The highest-weight representation $L_{\omega_{m}}$ is unitary for each $m\in\left\{ 0,1,...,n-1\right\} $. \end{proposition} \textit{Proof.} The contravariant Hermitian form on $L_{\omega_{m}}$ is the restriction of the form on $\mathcal{B}^{\left( m\right) }$. \begin{corollary} If $k_{0},k_{1},...,k_{n-1}$ are nonnegative integers, then $L_{k_{0}% \omega_{0}+k_{1}\omega_{1}+...+k_{n-1}\omega_{n-1}}$ is unitary (of level $k_{0}+k_{1}+...+k_{n-1}$). \end{corollary} \textit{Proof.} The tensor product $L_{\omega_{0}}^{\otimes k_{0}}\otimes L_{\omega_{1}}^{\otimes k_{1}}\otimes...\otimes L_{\omega_{n-1}}^{\otimes k_{n-1}}$ is unitary (being a tensor product of unitary representations), and thus is a direct sum of irreducible representations. Clearly, $L_{k_{0}% \omega_{0}+k_{1}\omega_{1}+...+k_{n-1}\omega_{n-1}}$ is a summand of this module, and thus also unitary, qed. \begin{theorem} \label{thm.sln.unitaries}These $L_{k_{0}\omega_{0}+k_{1}\omega_{1}% +...+k_{n-1}\omega_{n-1}}$ (with $k_{0},k_{1},...,k_{n-1}$ being nonnegative integers) are the only unitary highest-weight representations of $\widehat{\mathfrak{sl}_{n}}$. \end{theorem} To prove this, first a lemma: \begin{lemma} \label{lem.sl2.unitaries}Consider the antilinear $\mathbb{R}$-antiinvolution $\dag:\mathfrak{sl}_{2}\rightarrow\mathfrak{sl}_{2}$ defined by $e^{\dag}=f$, $f^{\dag}=e$ and $h^{\dag}=h$. Let $\lambda\in\mathfrak{h}^{\ast}$. We identify the function $\lambda\in\mathfrak{h}^{\ast}$ with the value $\lambda\left( h\right) \in\mathbb{C}$. Then, $L_{\lambda}$ is a unitary representation of $\mathfrak{sl}_{2}$ if and only if $\lambda\in\mathbb{Z}% _{+}$. \end{lemma} \textit{Proof of Lemma \ref{lem.sl2.unitaries}.} Assume that $L_{\lambda}$ is a unitary representation of $\mathfrak{sl}_{2}$. Let $v_{\lambda}=v_{\lambda }^{+}$. Since $L_{\lambda}$ is unitary, the form $\left( \cdot,\cdot\right) $ is positive definite, so that $\left( v_{\lambda},v_{\lambda}\right) >0$. Every $n\in\mathbb{N}$ satisfies \[ \left( f^{n}v_{\lambda},f^{n}v_{\lambda}\right) =n!\overline{\lambda}\left( \overline{\lambda}-1\right) ...\left( \overline{\lambda}-n+1\right) \left( v_{\lambda},v_{\lambda}\right) \] (the proof of this is analogous to the proof of (\ref{exa.sl2.bilinform}), but uses $e^{\dag}=f$). Since $\left( \cdot,\cdot\right) $ is positive definite, we must have $\left( f^{n}v_{\lambda},f^{n}v_{\lambda}\right) \geq0$ for every $n\in\mathbb{N}$. Thus, every $n\in\mathbb{N}$ satisfies $n!\overline {\lambda}\left( \overline{\lambda}-1\right) ...\left( \overline{\lambda }-n+1\right) \left( v_{\lambda},v_{\lambda}\right) =\left( f^{n}% v_{\lambda},f^{n}v_{\lambda}\right) \geq0$, so that $\overline{\lambda }\left( \overline{\lambda}-1\right) ...\left( \overline{\lambda }-n+1\right) \geq0$ (since $\left( v_{\lambda},v_{\lambda}\right) >0$). Applied to $n=1$, this yields $\overline{\lambda}\geq0$, so that $\overline{\lambda}\in\mathbb{R}$ and thus $\lambda\in\mathbb{R}$. Hence, $\overline{\lambda}\geq0$ becomes $\lambda\geq0$. Every $n\in\mathbb{N}$ satisfies $\lambda\left( \lambda-1\right) ...\left( \lambda-n+1\right) =\overline{\lambda}\left( \overline{\lambda}-1\right) ...\left( \overline{\lambda}-n+1\right) \geq0$. Thus, $\lambda\in \mathbb{Z}_{+}$ (otherwise, $\lambda\left( \lambda-1\right) ...\left( \lambda-n+1\right) $ would alternate in sign for each sufficiently large $n$). This proves one direction of Lemma \ref{lem.sl2.unitaries}. The converse direction is classical and easy. Lemma \ref{lem.sl2.unitaries} is proven. \begin{corollary} Let $\lambda\in\mathbb{C}$. If $\mathfrak{g}$ is a Lie algebra with antilinear $\mathbb{R}$-antiinvolution $\dag$ and $\mathfrak{sl}_{2}$ is a Lie subalgebra of $\mathfrak{g}$, and if $\dag\mid_{\mathfrak{sl}_{2}}$ sends $e,f,h$ to $f,e,h$, and if $V$ is a unitary representation of $\mathfrak{g}$, and if some $v\in V$ satisfies $ev=0$ and $hv=\lambda v$, then $\lambda\in\mathbb{Z}_{+}$. \end{corollary} \textit{Proof of Theorem \ref{thm.sln.unitaries}.} For every $i\in\left\{ 0,1,...,n-1\right\} $, we have an $\mathfrak{sl}_{2}$-subalgebra:% \begin{align*} h_{i} & =\left\{ \begin{array} [c]{c}% E_{i,i}-E_{i+1,i+1},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq0;\\ K+E_{n,n}-E_{1,1},\ \ \ \ \ \ \ \ \ \ \text{if }i=0 \end{array} \right. ,\\ e_{i} & =\left\{ \begin{array} [c]{c}% E_{i,i+1},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq0;\\ E_{n,1}t,\ \ \ \ \ \ \ \ \ \ \text{if }i=0 \end{array} \right. ;\\ f_{i} & =\left\{ \begin{array} [c]{c}% E_{i+1,i},\ \ \ \ \ \ \ \ \ \ \text{if }i\neq0;\\ E_{1,n}t^{-1},\ \ \ \ \ \ \ \ \ \ \text{if }i=0 \end{array} \right. \end{align*} \footnote{Here, $E_{i,j}$ means $E_{i,j}^{\mathfrak{gl}_{n}}$.} (these form an $\mathfrak{sl}_{2}$-triple, as can be easily checked). These satisfy $e_{i}^{\dag}=f_{i}$, $f_{i}^{\dag}=e_{i}$ and $h_{i}^{\dag}=h_{i}$. Thus, if $L_{\lambda}$ is a unitary representation of $\widehat{\mathfrak{sl}_{n}}$, then $\lambda\left( h_{i}\right) \in\mathbb{Z}_{+}$. But $\omega_{i}$ are a basis for the weights, and namely the dual basis to the basis of the $h_{i}$. Thus, $\lambda=\sum\limits_{i=0}^{n-1}\lambda\left( h_{i}\right) \omega_{i}% $. Hence, $\lambda=\sum\limits_{i=0}^{n-1}k_{i}\omega_{i}$ with $k_{i}% \in\mathbb{Z}_{+}$. Qed. \begin{remark} Relation between $\widehat{\mathfrak{sl}_{n}}$-modules and $\mathfrak{sl}_{n}$-modules: Let $L_{\lambda}$ be a unitary $\widehat{\mathfrak{sl}_{n}}$-module, with $\lambda=k_{0}\omega_{0}+k_{1}\omega_{1}+...+k_{n-1}\omega_{n-1}$. Then, $U\left( \mathfrak{sl}_{n}\right) v_{\lambda}=L_{\overline{\lambda}}$ where $\overline{\lambda}=k_{1}\omega_{1}+k_{2}\omega_{2}+...+k_{n-1}% \omega_{n-1}$ is a weight for $\mathfrak{sl}_{n}$. And if the level of $L_{\lambda}$ was $k$, then we must have $k_{1}+k_{2}+...+k_{n-1}\leq k$. \end{remark} \subsection{The Sugawara construction} We will now study the Sugawara construction. It constructs a $\operatorname*{Vir}$ action on a $\widehat{\mathfrak{g}}$-module (under some conditions), and it generalizes the action of $\operatorname*{Vir}$ on the $\mu$-Fock representation $F_{\mu}$ (that was constructed in Proposition \ref{prop.fockvir.answer2}). \begin{definition} \label{def.sugawara}Let $\mathfrak{g}$ be a finite-dimensional $\mathbb{C}% $-Lie algebra equipped with a $\mathfrak{g}$-invariant symmetric bilinear form $\left( \cdot,\cdot\right) $. (This form needs not be nondegenerate; it is even allowed to be $0$.) Consider the $2$-cocycle $\omega:\mathfrak{g}\left[ t,t^{-1}\right] \times\mathfrak{g}\left[ t,t^{-1}\right] \rightarrow\mathbb{C}$ defined by% \[ \omega\left( a,b\right) =\operatorname*{Res}\nolimits_{t=0}\left( a^{\prime},b\right) dt\ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathfrak{g}% \left[ t,t^{-1}\right] \text{ and }b\in\mathfrak{g}\left[ t,t^{-1}\right] . \] (This is the $2$-cocycle $\omega$ in Definition \ref{def.loop}. We just slightly rewrote the definition.) Also consider the affine Lie algebra $\widehat{\mathfrak{g}}=\mathfrak{g}\left[ t,t^{-1}\right] \oplus \mathbb{C}K$ defined through this cocycle $\omega$. Let $\operatorname*{Kil}$ denote the Killing form on $\mathfrak{g}$, defined by \[ \operatorname*{Kil}\left( a,b\right) =\operatorname*{Tr}\left( \operatorname*{ad}\left( a\right) \cdot\operatorname*{ad}\left( b\right) \right) \ \ \ \ \ \ \ \ \ \ \text{for all }a,b\in\mathfrak{g}. \] An element $k\in\mathbb{C}$ is said to be \textit{non-critical for }$\left( \mathfrak{g},\left( \cdot,\cdot\right) \right) $ if and only if the form $k\cdot\left( \cdot,\cdot\right) +\dfrac{1}{2}\operatorname*{Kil}$ is nondegenerate. \end{definition} \begin{definition} \label{def.sugawara.M}Let $M$ be a $\widehat{\mathfrak{g}}$-module. We say that $M$ is \textit{admissible} if for every $v\in M$, there exists some $N\in\mathbb{N}$ such that every integer $n\geq N$ and every $a\in\mathfrak{g}$ satisfy $at^{n}\cdot v=0$. If $k\in\mathbb{C}$, then we say that $M$ is \textit{of level }$k$ if $K\mid_{M}=k\cdot\operatorname*{id}$. \end{definition} \begin{proposition} \label{prop.WtoDerg}Let $\mathfrak{g}$ be a finite-dimensional $\mathbb{C}% $-Lie algebra equipped with a $\mathfrak{g}$-invariant symmetric bilinear form $\left( \cdot,\cdot\right) $. Consider the affine Lie algebra $\widehat{\mathfrak{g}}$ defined as in Definition \ref{def.sugawara}. \textbf{(a)} Then, there is a natural homomorphism $\eta _{\widehat{\mathfrak{g}}}:W\rightarrow\operatorname*{Der}\widehat{\mathfrak{g}% }$ of Lie algebras given by \[ \left( \eta_{\widehat{\mathfrak{g}}}\left( f\partial\right) \right) \left( g,\alpha\right) =\left( fg^{\prime},0\right) \ \ \ \ \ \ \ \ \ \ \text{for all }f\in\mathbb{C}\left[ t,t^{-1}\right] \text{, }g\in\mathfrak{g}\left[ t,t^{-1}\right] \text{ and }\alpha \in\mathbb{C}. \] \textbf{(b)} There also is a natural homomorphism $\widetilde{\eta }_{\widehat{\mathfrak{g}}}:\operatorname*{Vir}\rightarrow\operatorname*{Der}% \widehat{\mathfrak{g}}$ of Lie algebras given by \[ \left( \widetilde{\eta}_{\widehat{\mathfrak{g}}}\left( f\partial+\lambda K\right) \right) \left( g,\alpha\right) =\left( fg^{\prime},0\right) \ \ \ \ \ \ \ \ \ \ \text{for all }f\in\mathbb{C}\left[ t,t^{-1}\right] \text{, }g\in\mathfrak{g}\left[ t,t^{-1}\right] \text{, }\lambda \in\mathbb{C}\text{ and }\alpha\in\mathbb{C}. \] This homomorphism $\widetilde{\eta}_{\widehat{\mathfrak{g}}}$ is simply the extension of the homomorphism $\eta_{\widehat{\mathfrak{g}}}:W\rightarrow \operatorname*{Der}\widehat{\mathfrak{g}}$ to $\operatorname*{Vir}$ by means of requiring that $\widetilde{\eta}_{\widehat{\mathfrak{g}}}\left( K\right) =0$. This homomorphism $\widetilde{\eta}_{\widehat{\mathfrak{g}}}$ makes $\widehat{\mathfrak{g}}$ a $\operatorname*{Vir}$-module on which $\operatorname*{Vir}$ acts by derivations. Therefore, a Lie algebra $\operatorname*{Vir}\ltimes\widehat{\mathfrak{g}}$ is defined (according to Definition \ref{def.semidir.lielie}). \end{proposition} The proof of Proposition \ref{prop.WtoDerg} is left to the reader. (A proof of Proposition \ref{prop.WtoDerg} \textbf{(a)} can be obtained by carefully generalizing the proof of Lemma \ref{lem.WtoDerA}. Actually, Proposition \ref{prop.WtoDerg} \textbf{(a)} generalizes Lemma \ref{lem.WtoDerA}, since (as we will see in Remark \ref{rmk.sugawara.fockvir}) the Lie algebra $\widehat{\mathfrak{g}}$ generalizes $\mathcal{A}$.) The following theorem is one of the most important facts about affine Lie algebras: \begin{theorem} [Sugawara construction]\label{thm.sugawara}Let us work in the situation of Definition \ref{def.sugawara}. Let $k\in\mathbb{C}$ be non-critical for $\left( \mathfrak{g},\left( \cdot,\cdot\right) \right) $. Let $M$ be an admissible $\widehat{\mathfrak{g}}$-module of level $k$. Let $B\subseteq\mathfrak{g}$ be a basis orthonormal with respect to the form $k\left( \cdot,\cdot\right) +\dfrac{1}{2}\operatorname*{Kil}$. For every $x\in\mathfrak{g}$ and $n\in\mathbb{Z}$, let us denote by $x_{n}$ the element $xt^{n}\in\widehat{\mathfrak{g}}$. For every $x\in\mathfrak{g}$, every $m\in\mathbb{Z}$ and $\ell\in\mathbb{Z}$, define the ``normal ordered product'' $\left. :x_{m}x_{\ell}:\right. $ in $U\left( \widehat{\mathfrak{g}}\right) $ by% \[ \left. :x_{m}x_{\ell}:\right. =\left\{ \begin{array} [c]{c}% x_{m}x_{\ell},\ \ \ \ \ \ \ \ \ \ \text{if }m\leq\ell;\\ x_{\ell}x_{m},\ \ \ \ \ \ \ \ \ \ \text{if }m>\ell \end{array} \right. . \] For every $n\in\mathbb{Z}$, define an endomorphism $L_{n}$ of $M$ by% \[ L_{n}=\dfrac{1}{2}\sum\limits_{a\in B}\sum\limits_{m\in\mathbb{Z}}\left. :a_{m}a_{n-m}:\right. . \] \textbf{(a)} This endomorphism $L_{n}$ is indeed well-defined. In other words, for every $n\in\mathbb{Z}$, every $a\in B$ and every $v\in M$, the sum $\sum\limits_{m\in\mathbb{Z}}\left. :a_{m}a_{n-m}:\right. v$ converges in the discrete topology (i. e., has only finitely many nonzero addends). \textbf{(b)} For every $n\in\mathbb{Z}$, the endomorphism $L_{n}$ does not depend on the choice of the orthonormal basis $B$. \textbf{(c)} The endomorphisms $L_{n}$ for $n\in\mathbb{Z}$ give rise to a $\operatorname*{Vir}$-representation on $M$ with central charge% \[ c=k\cdot\sum\limits_{a\in B}\left( a,a\right) . \] \textbf{(d)} These formulas (for $L_{n}$ and $c$) extend the action of $\widehat{\mathfrak{g}}$ on $M$ to an action of $\operatorname*{Vir}% \ltimes\widehat{\mathfrak{g}}$, so they satisfy $\left[ L_{n},a_{m}\right] =-ma_{n+m}$ and $\left[ L_{n},K\right] =0$. \textbf{(e)} We have $\left[ L_{n},a_{m}\right] =-ma_{n+m}$ for any $a\in\mathfrak{g}$ and any integers $n$ and $m$. \end{theorem} \begin{remark} \label{rmk.sugawara.fockvir}We have already encountered an example of this construction: namely, the example where $\mathfrak{g}$ is the trivial Lie algebra $\mathbb{C}$, where $\left( \cdot,\cdot\right) :\mathfrak{g}% \times\mathfrak{g}\rightarrow\mathbb{C}$ is the bilinear form $\left( x,y\right) \mapsto xy$, where $k=1$, and where $M$ is the $\widehat{\mathfrak{g}}$-module $F_{\mu}$. (To make sense of this, notice that when $\mathfrak{g}$ is the trivial Lie algebra $\mathbb{C}$, the affine Lie algebra $\widehat{\mathfrak{g}}$ is canonically isomorphic to the Heisenberg algebra $\mathcal{A}$, through an isomorphism $\widehat{\mathfrak{g}% }\rightarrow\mathcal{A}$ which takes $t^{n}$ to $a_{n}$ and $K$ to $K$.) In this example, the operators $L_{n}$ defined in Theorem \ref{thm.sugawara} are exactly the operators $L_{n}$ defined in Definition \ref{def.fockvir}. \end{remark} Before we prove Theorem \ref{thm.sugawara}, we formulate a number of lemmas. First, an elementary lemma on Killing forms of finite-dimensional Lie algebras: \begin{lemma} \label{lem.sugawara.Kil}Let $\mathfrak{g}$ be a finite-dimensional Lie algebra. Denote by $\operatorname*{Kil}$ the Killing form of $\mathfrak{g}$. Let $n\in\mathbb{N}$ and $p_{1},p_{2},...,p_{n}\in\mathfrak{g}$ and $q_{1},q_{2},...,q_{n}\in\mathfrak{g}$ be such that the tensor $\sum \limits_{i=1}^{n}p_{i}\otimes q_{i}\in\mathfrak{g}\otimes\mathfrak{g}$ is $\mathfrak{g}$-invariant. Then, $\sum\limits_{i=1}^{n}\left[ \left[ b,p_{i}\right] ,q_{i}\right] =\sum\limits_{i=1}^{n}\operatorname*{Kil}% \left( b,p_{i}\right) q_{i}$ for every $b\in\mathfrak{g}$. \end{lemma} Here, we are using the following notation: \begin{remark} Let $\mathfrak{g}$ be a Lie algebra. An element $m$ of a $\mathfrak{g}$-module $M$ is said to be $\mathfrak{g}$\textit{-invariant} if and only if it satisfies $\left( x\rightharpoonup m=0\text{ for every }x\in\mathfrak{g}% \right) $. We regard $\mathfrak{g}$ as a $\mathfrak{g}$-module by means of the adjoint action of $\mathfrak{g}$ (that is, we set $x\rightharpoonup m=\left[ x,m\right] $ for every $x\in\mathfrak{g}$ and $m\in\mathfrak{g}$); thus, $\mathfrak{g}\otimes\mathfrak{g}$ becomes a $\mathfrak{g}$-module as well. Explicitly, the action of $\mathfrak{g}$ on $\mathfrak{g}\otimes \mathfrak{g}$ is given by% \[ x\rightharpoonup\left( \sum\limits_{i=1}^{n}p_{i}\otimes q_{i}\right) =\sum\limits_{i=1}^{n}\left[ x,p_{i}\right] \otimes q_{i}+\sum \limits_{i=1}^{n}p_{i}\otimes\left[ x,q_{i}\right] \] for every tensor $\sum\limits_{i=1}^{n}p_{i}\otimes q_{i}\in\mathfrak{g}% \otimes\mathfrak{g}$. Hence, a tensor $\sum\limits_{i=1}^{n}p_{i}\otimes q_{i}\in\mathfrak{g}\otimes\mathfrak{g}$ is $\mathfrak{g}$-invariant if and only if every $x\in\mathfrak{g}$ satisfies $\sum\limits_{i=1}^{n}\left[ x,p_{i}\right] \otimes q_{i}+\sum\limits_{i=1}^{n}p_{i}\otimes\left[ x,q_{i}\right] =0$. In other words, a tensor $\sum\limits_{i=1}^{n}% p_{i}\otimes q_{i}\in\mathfrak{g}\otimes\mathfrak{g}$ is $\mathfrak{g}% $-invariant if and only if every $x\in\mathfrak{g}$ satisfies $\sum \limits_{i=1}^{n}\left[ p_{i},x\right] \otimes q_{i}=-\sum\limits_{i=1}% ^{n}p_{i}\otimes\left[ q_{i},x\right] $. \end{remark} \textit{Proof of Lemma \ref{lem.sugawara.Kil}.} Let $\left( c_{1}% ,c_{2},...,c_{m}\right) $ be a basis of the vector space $\mathfrak{g}$, and let $\left( c_{1}^{\ast},c_{2}^{\ast},...,c_{m}^{\ast}\right) $ be the dual basis of $\mathfrak{g}^{\ast}$. Then, every $i\in\left\{ 1,2,...,n\right\} $ satisfies \[ \operatorname*{Kil}\left( b,p_{i}\right) =\operatorname*{Tr}\left( \left( \operatorname*{ad}b\right) \circ\left( \operatorname*{ad}p_{i}\right) \right) =\sum\limits_{j=1}^{m}c_{j}^{\ast}\left( \left( \left( \operatorname*{ad}b\right) \circ\left( \operatorname*{ad}p_{i}\right) \right) \left( c_{j}\right) \right) =\sum\limits_{j=1}^{m}c_{j}^{\ast }\left( \left[ b,\left[ p_{i},c_{j}\right] \right] \right) . \] Hence,% \begin{align*} \sum\limits_{i=1}^{n}\operatorname*{Kil}\left( b,p_{i}\right) q_{i} & =\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}c_{j}^{\ast}\left( \left[ b,\left[ p_{i},c_{j}\right] \right] \right) q_{i}=\sum\limits_{j=1}% ^{m}\sum\limits_{i=1}^{n}c_{j}^{\ast}\left( \left[ b,\left[ p_{i}% ,c_{j}\right] \right] \right) q_{i}\\ & =-\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{n}c_{j}^{\ast}\left( \left[ b,p_{i}\right] \right) \left[ q_{i},c_{j}\right] \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\sum\limits_{i=1}^{n}p_{i}\otimes q_{i}\text{ is } \mathfrak{g}% \text{-invariant, so that}\\ \sum\limits_{i=1}^{n}\left[ p_{i},c_{j}\right] \otimes q_{i}=-\sum \limits_{i=1}^{n}p_{i}\otimes\left[ q_{i},c_{j}\right] \text{ for every }j\in\left\{ 1,2,...,m\right\} \text{, and thus}\\ \sum\limits_{i=1}^{n}c_{j}^{\ast}\left( \left[ b,\left[ p_{i},c_{j}\right] \right] \right) q_{i}=-\sum\limits_{i=1}^{n}c_{j}^{\ast}\left( \left[ b,p_{i}\right] \right) \left[ q_{i},c_{j}\right] \text{ for every }% j\in\left\{ 1,2,...,m\right\} \end{array} \right) \\ & =-\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{n}\left[ q_{i},c_{j}^{\ast }\left( \left[ b,p_{i}\right] \right) c_{j}\right] =-\sum\limits_{i=1}% ^{n}\left[ q_{i},\underbrace{\sum\limits_{j=1}^{m}c_{j}^{\ast}\left( \left[ b,p_{i}\right] \right) c_{j}}_{\substack{=\left[ b,p_{i}\right] \\\text{(since }\left( c_{1}^{\ast},c_{2}^{\ast},...,c_{m}^{\ast}\right) \text{ is the dual basis}\\\text{to the basis }\left( c_{1},c_{2}% ,...,c_{m}\right) \text{)}}}\right] \\ & =-\sum\limits_{i=1}^{n}\left[ q_{i},\left[ b,p_{i}\right] \right] =\sum\limits_{i=1}^{n}\left[ \left[ b,p_{i}\right] ,q_{i}\right] , \end{align*} which proves Lemma \ref{lem.sugawara.Kil}. Here comes another lemma on $\mathfrak{g}$-invariant bilinear forms: \begin{lemma} \label{lem.sugawara.Kil2}Let $\mathfrak{g}$ be a finite-dimensional $\mathbb{C}$-Lie algebra equipped with a $\mathfrak{g}$-invariant symmetric bilinear form $\left\langle \cdot,\cdot\right\rangle $. Let $B\subseteq \mathfrak{g}$ be a basis orthonormal with respect to the form $\left\langle \cdot,\cdot\right\rangle $. \textbf{(a)} Then, the tensor $\sum\limits_{a\in B}a\otimes a$ is $\mathfrak{g}$-invariant in $\mathfrak{g}\otimes\mathfrak{g}$. \textbf{(b)} Let $B^{\prime}$ also be a basis of $\mathfrak{g}$ orthonormal with respect to the form $\left\langle \cdot,\cdot\right\rangle $. Then, $\sum\limits_{a\in B}a\otimes a=\sum\limits_{a\in B^{\prime}}a\otimes a$. \end{lemma} \textit{Proof of Lemma \ref{lem.sugawara.Kil2}.} The bilinear form $\left\langle \cdot,\cdot\right\rangle $ is nondegenerate (since it has an orthonormal basis). \textbf{(a)} For every $v\in\mathfrak{g}$, let $v^{\ast}:\mathfrak{g}% \rightarrow\mathbb{C}$ be the $\mathbb{C}$-linear map which sends every $w\in\mathfrak{g}$ to $\left\langle v,w\right\rangle $. Then, $\mathfrak{g}% ^{\ast}=\left\{ v^{\ast}\ \mid\ v\in\mathfrak{g}\right\} $ (since the form $\left\langle \cdot,\cdot\right\rangle $ is nondegenerate). Let $b\in\mathfrak{g}$. We will now prove that $h\left( \sum\limits_{a\in B}\left( \left[ b,a\right] \otimes a+a\otimes\left[ b,a\right] \right) \right) =0$ for every $h\in\left( \mathfrak{g}\otimes\mathfrak{g}\right) ^{\ast}$. In fact, let $h\in\left( \mathfrak{g}\otimes\mathfrak{g}\right) ^{\ast}$. Since $\mathfrak{g}$ is finite-dimensional, we have $\left( \mathfrak{g}% \otimes\mathfrak{g}\right) ^{\ast}=\mathfrak{g}^{\ast}\otimes\mathfrak{g}% ^{\ast}$, so that $h\in\mathfrak{g}^{\ast}\otimes\mathfrak{g}^{\ast}$. We can WLOG assume that $h=f_{1}\otimes f_{2}$ for some $f_{1}\in\mathfrak{g}^{\ast}$ and $f_{2}\in\mathfrak{g}^{\ast}$ (because every tensor in $\mathfrak{g}% ^{\ast}\otimes\mathfrak{g}^{\ast}$ is a $\mathbb{C}$-linear combination of pure tensors, and the assertion which we want to prove (namely, the equality $h\left( \sum\limits_{a\in B}\left( \left[ b,a\right] \otimes a+a\otimes\left[ b,a\right] \right) \right) =0$) is $\mathbb{C}$-linear in $h$). Assume this. Since $f_{1}\in\mathfrak{g}^{\ast}=\left\{ v^{\ast}\ \mid\ v\in \mathfrak{g}\right\} $, there exists some $v_{1}\in\mathfrak{g}$ such that $f_{1}=v_{1}^{\ast}$. Consider this $v_{1}$. Since $f_{2}\in\mathfrak{g}^{\ast}=\left\{ v^{\ast}\ \mid\ v\in \mathfrak{g}\right\} $, there exists some $v_{2}\in\mathfrak{g}$ such that $f_{2}=v_{2}^{\ast}$. Consider this $v_{2}$. Since $B$ is an orthonormal basis with respect to $\left\langle \cdot ,\cdot\right\rangle $, we have $\sum\limits_{a\in B}a\left\langle \left[ b,v_{2}\right] ,a\right\rangle =\left[ b,v_{2}\right] $ and $\sum \limits_{a\in B}\left\langle \left[ b,v_{1}\right] ,a\right\rangle a=\left[ b,v_{1}\right] $. Now, $h=\underbrace{f_{1}}_{=v_{1}^{\ast}}\otimes\underbrace{f_{2}}% _{=v_{2}^{\ast}}=v_{1}^{\ast}\otimes v_{2}^{\ast}$, so that% \begin{align*} & h\left( \sum\limits_{a\in B}\left( \left[ b,a\right] \otimes a+a\otimes\left[ b,a\right] \right) \right) \\ & =\left( v_{1}^{\ast}\otimes v_{2}^{\ast}\right) \left( \sum\limits_{a\in B}\left( \left[ b,a\right] \otimes a+a\otimes\left[ b,a\right] \right) \right) \\ & =\sum\limits_{a\in B}\left( \underbrace{v_{1}^{\ast}\left( \left[ b,a\right] \right) }_{\substack{=\left\langle v_{1},\left[ b,a\right] \right\rangle \\\text{(by the definition of }v_{1}^{\ast}\text{)}}% }\cdot\underbrace{v_{2}^{\ast}\left( a\right) }_{\substack{=\left\langle v_{2},a\right\rangle \\\text{(by the definition of }v_{2}^{\ast}\text{)}% }}+\underbrace{v_{1}^{\ast}\left( a\right) }_{\substack{=\left\langle v_{1},a\right\rangle \\\text{(by the definition of }v_{1}^{\ast}\text{)}% }}\cdot\underbrace{v_{2}^{\ast}\left( \left[ b,a\right] \right) }_{\substack{=\left\langle v_{2},\left[ b,a\right] \right\rangle \\\text{(by the definition of }v_{2}^{\ast}\text{)}}}\right) \\ & =\sum\limits_{a\in B}\left( \underbrace{\left\langle v_{1},\left[ b,a\right] \right\rangle }_{\substack{=-\left\langle \left[ b,v_{1}\right] ,a\right\rangle \\\text{(since }\left\langle \cdot,\cdot\right\rangle \text{ is invariant)}}}\cdot\left\langle v_{2},a\right\rangle +\left\langle v_{1},a\right\rangle \cdot\underbrace{\left\langle v_{2},\left[ b,a\right] \right\rangle }_{\substack{=-\left\langle \left[ b,v_{2}\right] ,a\right\rangle \\\text{(since }\left\langle \cdot,\cdot\right\rangle \text{ is invariant)}}}\right) \\ & =\sum\limits_{a\in B}\left( -\left\langle \left[ b,v_{1}\right] ,a\right\rangle \cdot\left\langle v_{2},a\right\rangle -\left\langle v_{1},a\right\rangle \cdot\left\langle \left[ b,v_{2}\right] ,a\right\rangle \right) \\ & =-\underbrace{\sum\limits_{a\in B}\left\langle \left[ b,v_{1}\right] ,a\right\rangle \cdot\left\langle v_{2},a\right\rangle }_{=\left\langle v_{2},\sum\limits_{a\in B}\left\langle \left[ b,v_{1}\right] ,a\right\rangle a\right\rangle }-\underbrace{\sum\limits_{a\in B}\left\langle v_{1}% ,a\right\rangle \cdot\left\langle \left[ b,v_{2}\right] ,a\right\rangle }_{=\left\langle v_{1},\sum\limits_{a\in B}a\left\langle \left[ b,v_{2}\right] ,a\right\rangle \right\rangle }\\ & =-\left\langle v_{2},\underbrace{\sum\limits_{a\in B}\left\langle \left[ b,v_{1}\right] ,a\right\rangle a}_{=\left[ b,v_{1}\right] }\right\rangle -\left\langle v_{1},\underbrace{\sum\limits_{a\in B}a\left\langle \left[ b,v_{2}\right] ,a\right\rangle }_{=\left[ b,v_{2}\right] }\right\rangle \\ & =-\underbrace{\left\langle v_{2},\left[ b,v_{1}\right] \right\rangle }_{\substack{=\left\langle \left[ b,v_{1}\right] ,v_{2}\right\rangle \\\text{(since }\left\langle \cdot,\cdot\right\rangle \text{ is symmetric)}% }}-\underbrace{\left\langle v_{1},\left[ b,v_{2}\right] \right\rangle }_{\substack{=-\left\langle \left[ b,v_{1}\right] ,v_{2}\right\rangle \\\text{(since }\left\langle \cdot,\cdot\right\rangle \text{ is invariant)}% }}=-\left\langle \left[ b,v_{1}\right] ,v_{2}\right\rangle -\left( -\left\langle \left[ b,v_{1}\right] ,v_{2}\right\rangle \right) =0. \end{align*} We thus have proven that $h\left( \sum\limits_{a\in B}\left( \left[ b,a\right] \otimes a+a\otimes\left[ b,a\right] \right) \right) =0$ for every $h\in\left( \mathfrak{g}\otimes\mathfrak{g}\right) ^{\ast}$. Consequently, $\sum\limits_{a\in B}\left( \left[ b,a\right] \otimes a+a\otimes\left[ b,a\right] \right) =0$. Hence, we have shown that $\sum\limits_{a\in B}\left( \left[ b,a\right] \otimes a+a\otimes\left[ b,a\right] \right) =0$ for every $b\in \mathfrak{g}$. In other words, the tensor $\sum\limits_{a\in B}a\otimes a$ is $\mathfrak{g}$-invariant. Lemma \ref{lem.sugawara.Kil2} \textbf{(a)} is proven. \textbf{(b)} For every $a\in B$ and $b\in B^{\prime}$, let $\xi_{a,b}$ be the $b$-coordinate of $a$ with respect to the basis $B^{\prime}$. Then, every $a\in B$ satisfies $a=\sum\limits_{b\in B^{\prime}}\xi_{a,b}b$. Thus, $\left( \xi_{a,b}\right) _{\left( a,b\right) \in B\times B^{\prime}}$ (this is a matrix whose rows and columns are indexed by elements of $B$ and $B^{\prime}$, respectively) is the matrix which represents the change of bases from $B^{\prime}$ to $B$ (or from $B$ to $B^{\prime}$, depending on how you define the matrix representing a change of basis). Since both $B$ and $B^{\prime}$ are two orthonormal bases with respect to the same bilinear form $\left\langle \cdot,\cdot\right\rangle $, this matrix must thus be orthogonal. Hence, every $b\in B^{\prime}$ and $b^{\prime}\in B^{\prime}$ satisfy $\sum\limits_{a\in B}\xi_{a,b}\xi_{a,b^{\prime}}=\delta_{b,b^{\prime}}$ (where $\delta _{b,b^{\prime}}$ is the Kronecker delta of $b$ and $b^{\prime}$). Now, since every $a\in B$ satisfies $a=\sum\limits_{b\in B^{\prime}}\xi_{a,b}b$ and $a=\sum\limits_{b\in B^{\prime}}\xi_{a,b}b=\sum\limits_{b^{\prime}\in B^{\prime}}\xi_{a,b^{\prime}}b^{\prime}$ (here, we renamed $b$ as $b^{\prime}$ in the sum), we have% \begin{align*} & \sum\limits_{a\in B}\underbrace{a}_{=\sum\limits_{b\in B^{\prime}}\xi _{a,b}b}\otimes\underbrace{a}_{=\sum\limits_{b^{\prime}\in B^{\prime}}% \xi_{a,b^{\prime}}b^{\prime}}\\ & =\sum\limits_{a\in B}\left( \sum\limits_{b\in B^{\prime}}\xi _{a,b}b\right) \otimes\left( \sum\limits_{b^{\prime}\in B^{\prime}}% \xi_{a,b^{\prime}}b^{\prime}\right) =\sum\limits_{a\in B}\sum\limits_{b\in B^{\prime}}\sum\limits_{b^{\prime}\in B^{\prime}}\xi_{a,b}\xi_{a,b^{\prime}% }b\otimes b^{\prime}\\ & =\sum\limits_{b\in B^{\prime}}\sum\limits_{b^{\prime}\in B^{\prime}% }\underbrace{\sum\limits_{a\in B}\xi_{a,b}\xi_{a,b^{\prime}}}_{=\delta _{b,b^{\prime}}}b\otimes b^{\prime}=\sum\limits_{b\in B^{\prime}% }\underbrace{\sum\limits_{b^{\prime}\in B^{\prime}}\delta_{b,b^{\prime}% }b\otimes b^{\prime}}_{=b\otimes b}=\sum\limits_{b\in B^{\prime}}b\otimes b\\ & =\sum\limits_{a\in B^{\prime}}a\otimes a\ \ \ \ \ \ \ \ \ \ \left( \text{here, we renamed }b\text{ as }a\text{ in the sum}\right) . \end{align*} This proves Lemma \ref{lem.sugawara.Kil2} \textbf{(b)}. As a consequence of this lemma, we get: \begin{lemma} \label{lem.sugawara.Kil3}Let $\mathfrak{g}$ be a finite-dimensional $\mathbb{C}$-Lie algebra equipped with a $\mathfrak{g}$-invariant symmetric bilinear form $\left( \cdot,\cdot\right) $. Denote by $\operatorname*{Kil}$ the Killing form of $\mathfrak{g}$. Let $B\subseteq\mathfrak{g}$ be a basis orthonormal with respect to the form $k\left( \cdot,\cdot\right) +\dfrac {1}{2}\operatorname*{Kil}$. Let $b\in\mathfrak{g}$. \textbf{(a)} We have $\sum\limits_{a\in B}\left( \left[ b,a\right] \otimes a+a\otimes\left[ b,a\right] \right) =0$. \textbf{(b)} We have $\dfrac{1}{2}\sum\limits_{a\in B}\left[ \left[ b,a\right] ,a\right] +k\sum\limits_{a\in B}\left( b,a\right) a=b$. \textbf{(c)} We have $\left( \left[ b,a\right] ,a\right) =0$ for every $a\in\mathfrak{g}$. \end{lemma} \textit{Proof of Lemma \ref{lem.sugawara.Kil3}.} The basis $B$ is orthonormal with respect to a symmetric $\mathfrak{g}$-invariant bilinear form (namely, the form $k\left( \cdot,\cdot\right) +\dfrac{1}{2}\operatorname*{Kil}$). As a consequence, the tensor $\sum\limits_{a\in B}a\otimes a$ is $\mathfrak{g}% $-invariant in $\mathfrak{g}\otimes\mathfrak{g}$ (by Lemma \ref{lem.sugawara.Kil2} \textbf{(a)}, applied to $\left\langle \cdot ,\cdot\right\rangle =k\left( \cdot,\cdot\right) +\dfrac{1}{2}% \operatorname*{Kil}$). In other words, $\sum\limits_{a\in B}\left( \left[ b,a\right] \otimes a+a\otimes\left[ b,a\right] \right) =0$. This proves Lemma \ref{lem.sugawara.Kil3} \textbf{(a)}. \textbf{(b)} If $\left\langle \cdot,\cdot\right\rangle $ is any nondegenerate inner product\footnote{By ``inner product'', we mean a symmetric bilinear form.} on a finite-dimensional vector space $V$ and $B$ is an orthonormal basis with respect to that product, then any vector $b\in V$ is equal to $\sum\limits_{a\in B}\left\langle b,a\right\rangle a$. Applying this fact to the inner product $\left\langle \cdot,\cdot\right\rangle =k\left( \cdot ,\cdot\right) +\dfrac{1}{2}\operatorname*{Kil}$ on the vector space $V=\mathfrak{g}$, we conclude that $b=k\sum\limits_{a\in B}\left( b,a\right) a+\dfrac{1}{2}\sum\limits_{a\in B}\operatorname*{Kil}\left( b,a\right) a$. Now, applying Lemma \ref{lem.sugawara.Kil} to the $\mathfrak{g}$-invariant tensor $\sum\limits_{a\in B}a\otimes a$ in lieu of $\sum\limits_{i=1}^{n}% p_{i}\otimes q_{i}$, we see that $\sum\limits_{a\in B}\left[ \left[ b,a\right] ,a\right] =\sum\limits_{a\in B}\operatorname*{Kil}\left( b,a\right) a$. Hence,% \[ b=k\sum\limits_{a\in B}\left( b,a\right) a+\dfrac{1}{2}\underbrace{\sum \limits_{a\in B}\operatorname*{Kil}\left( b,a\right) a}_{=\sum\limits_{a\in B}\left[ \left[ b,a\right] ,a\right] }=\dfrac{1}{2}\sum\limits_{a\in B}\left[ \left[ b,a\right] ,a\right] +k\sum\limits_{a\in B}\left( b,a\right) a. \] This proves Lemma \ref{lem.sugawara.Kil3} \textbf{(b)}. \begin{vershort} \textbf{(c)} Every $c\in\mathfrak{g}$ satisfies $\left( \left[ a,b\right] ,c\right) +\left( b,\left[ a,c\right] \right) =0$ (due to the $\mathfrak{g}$-invariance of $\left( \cdot,\cdot\right) $). Applying this to $c=a$, we obtain $\left( \left[ a,b\right] ,a\right) +\left( b,\left[ a,a\right] \right) =0$. Since $\left[ a,a\right] =0$ and $\left[ a,b\right] =-\left[ b,a\right] $, this rewrites as $\left( -\left[ b,a\right] ,a\right) +\left( b,0\right) =0$. This simplifies to $-\left( \left[ b,a\right] ,a\right) =0$. Thus, $\left( \left[ b,a\right] ,a\right) =0$. This proves Lemma \ref{lem.sugawara.Kil3} \textbf{(c)}. \end{vershort} \begin{verlong} \textbf{(c)} Every $c\in\mathfrak{g}$ satisfies $\left( \left[ a,b\right] ,c\right) +\left( b,\left[ a,c\right] \right) =0$ (due to the $\mathfrak{g}$-invariance of $\left( \cdot,\cdot\right) $). Applying this to $c=a$, we obtain $\left( \left[ a,b\right] ,a\right) +\left( b,\left[ a,a\right] \right) =0$. Thus,% \begin{align*} 0 & =\left( \underbrace{\left[ a,b\right] }_{\substack{=-\left[ b,a\right] \\\text{(since the Lie bracket}\\\text{is antisymmetric)}% }},a\right) +\left( b,\underbrace{\left[ a,a\right] }% _{\substack{=0\\\text{(since the Lie bracket}\\\text{is antisymmetric)}% }}\right) =\underbrace{\left( -\left[ b,a\right] ,a\right) }% _{\substack{=-\left( \left[ b,a\right] ,a\right) \\\text{(since the form }\left( \cdot,\cdot\right) \\\text{is bilinear)}}}+\underbrace{\left( b,0\right) }_{\substack{=0\\\text{(since the form }\left( \cdot ,\cdot\right) \\\text{is bilinear)}}}\\ & =-\left( \left[ b,a\right] ,a\right) +0=-\left( \left[ b,a\right] ,a\right) . \end{align*} Adding $\left( \left[ b,a\right] ,a\right) $ to this equality, we obtain $\left( \left[ b,a\right] ,a\right) =0$. This proves Lemma \ref{lem.sugawara.Kil3} \textbf{(c)}. \end{verlong} \begin{noncompile} Here is a \textit{different proof of Lemma \ref{lem.sugawara.Kil3} \textbf{(c)} (which I had written before I found the trivial proof above):} Every $c\in\mathfrak{g}$ satisfies $\left( \left[ b,a\right] ,c\right) +\left( a,\left[ b,c\right] \right) =0$ (due to the $\mathfrak{g}% $-invariance of $\left( \cdot,\cdot\right) $). Applying this to $c=a$, we obtain $\left( \left[ b,a\right] ,a\right) +\left( a,\left[ b,a\right] \right) =0$. Since the form $\left( \cdot,\cdot\right) $ is symmetric, this rewrites as $2\left( \left[ b,a\right] ,a\right) =0$. Thus, $\left( \left[ b,a\right] ,a\right) =0$. This proves Lemma \ref{lem.sugawara.Kil3} \textbf{(c)}. \end{noncompile} Next, we formulate the analogue of Remark \ref{rmk.fockvir.normal.mn}: \begin{remark} \label{rmk.sugawara.normal.mn}Let $x\in\mathfrak{g}$. If $m$ and $n$ are integers such that $m\neq-n$, then $\left. :x_{m}x_{n}:\right. =x_{m}x_{n}$. (This is because $\left[ x_{m},x_{n}\right] =0$ in $\widehat{\mathfrak{g}}$ when $m\neq-n$.) \end{remark} In analogy to Remark \ref{rmk.fockvir.normal.comm} \textbf{(a)}, we have commutativity of normal ordered products: \begin{remark} \label{rmk.sugawara.normal.comm}Let $x\in\mathfrak{g}$. Any $m\in\mathbb{Z}$ and $n\in\mathbb{Z}$ satisfy $\left. :x_{m}x_{n}:\right. =\left. :x_{n}x_{m}:\right. $. \end{remark} Also, here is a simple way to rewrite the definition of $\left. :x_{m}% x_{n}:\right. $: \begin{remark} \label{rmk.sugawara.normal.max}Let $x\in\mathfrak{g}$. Any $m\in\mathbb{Z}$ and $n\in\mathbb{Z}$ satisfy $\left. :x_{m}x_{n}:\right. =x_{\min\left\{ m,n\right\} }x_{\max\left\{ m,n\right\} }$. \end{remark} Generalizing Remark \ref{rmk.fockvir.normal.K}, we have: \begin{remark} \label{rmk.sugawara.normal.K}Let $x\in\mathfrak{g}$. Let $m$ and $n$ be integers. \textbf{(a)} Then, $\left. :x_{m}x_{n}:\right. =x_{m}x_{n}+n\left[ m>0\right] \delta_{m,-n}\left( x,x\right) K$. Here, when $\mathfrak{A}$ is an assertion, we denote by $\left[ \mathfrak{A}\right] $ the truth value of $\mathfrak{A}$ (that is, the number $\left\{ \begin{array} [c]{c}% 1\text{, if }\mathfrak{A}\text{ is true;}\\ 0\text{, if }\mathfrak{A}\text{ is false }% \end{array} \right. $). \textbf{(b)} For any $y\in U\left( \widehat{\mathfrak{g}}\right) $, we have $\left[ y,\left. :x_{m}x_{n}:\right. \right] =\left[ y,x_{m}x_{n}\right] $ in $U\left( \widehat{\mathfrak{g}}\right) $ (where $\left[ \cdot ,\cdot\right] $ denotes the commutator in $U\left( \widehat{\mathfrak{g}% }\right) $). \end{remark} The proof of this is left to the reader (it follows very quickly from the definitions). Next, here is a completely elementary lemma: \begin{lemma} \label{lem.telescope}Let $G$ be an abelian group (written additively). Whenever $\left( u_{m}\right) _{m\in\mathbb{Z}}\in G^{\mathbb{Z}}$ is a family of elements of $G$, and $\mathcal{A}\left( m\right) $ is an assertion for every $m\in\mathbb{Z}$, let us abbreviate the sum $\sum \limits_{\substack{m\in\mathbb{Z};\\\mathcal{A}\left( m\right) }}u_{m}$ (if this sum is well-defined) by $\sum\limits_{\mathcal{A}\left( m\right) }% u_{m}$. (For instance, we will abbreviate the sum $\sum\limits_{\substack{m\in \mathbb{Z};\\3\leq m\leq7}}u_{m}$ by $\sum\limits_{3\leq m\leq7}u_{m}$.) For any integers $\alpha$ and $\beta$ such that $\alpha\leq\beta$, for any nonnegative integer $N$, and for any family $\left( u_{m}\right) _{m\in\mathbb{Z}}\in G^{\mathbb{Z}}$ of elements of $G$, we have% \[ \sum\limits_{\left\vert m-\beta\right\vert \leq N}u_{m}-\sum \limits_{\left\vert m-\alpha\right\vert \leq N}u_{m}=-\sum\limits_{\alpha -N\leq m<\beta-N}u_{m}+\sum\limits_{\alpha+N n-m$. \par Let us first consider Case 1. In this case, $m\leq n-m$. Hence, every $a\in\mathfrak{g}$ satisfies $\left. :a_{m}a_{n-m}:\right. =a_{m}a_{n-m}$. Thus, \begin{align*} \sum\limits_{a\in B}\left. :a_{m}a_{n-m}:\right. & =\sum\limits_{a\in B}a_{m}a_{n-m}=\sum\limits_{a\in B^{\prime}}\underbrace{a_{m}a_{n-m}% }_{=\left. :a_{m}a_{n-m}:\right. }\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.sugawara.basisind.pf.1}), applied to }u=m\text{ and }v=n-m\right) \\ & =\sum\limits_{a\in B^{\prime}}\left. :a_{m}a_{n-m}:\right. . \end{align*} This proves $\sum\limits_{a\in B}\left. :a_{m}a_{n-m}:\right. =\sum \limits_{a\in B^{\prime}}\left. :a_{m}a_{n-m}:\right. $ in Case 1. \par Let us now consider Case 2. In this case, $m>n-m$. Hence, every $a\in \mathfrak{g}$ satisfies $\left. :a_{m}a_{n-m}:\right. =a_{n-m}a_{m}$. Thus, \begin{align*} \sum\limits_{a\in B}\left. :a_{m}a_{n-m}:\right. & =\sum\limits_{a\in B}a_{n-m}a_{m}=\sum\limits_{a\in B^{\prime}}\underbrace{a_{n-m}a_{m}% }_{=\left. :a_{m}a_{n-m}:\right. }\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.sugawara.basisind.pf.1}), applied to }u=n-m\text{ and }v=m\right) \\ & =\sum\limits_{a\in B^{\prime}}\left. :a_{m}a_{n-m}:\right. . \end{align*} This proves $\sum\limits_{a\in B}\left. :a_{m}a_{n-m}:\right. =\sum \limits_{a\in B^{\prime}}\left. :a_{m}a_{n-m}:\right. $ in Case 2. \par Hence, $\sum\limits_{a\in B}\left. :a_{m}a_{n-m}:\right. =\sum\limits_{a\in B^{\prime}}\left. :a_{m}a_{n-m}:\right. $ is proven in each of the cases 1 and 2. Thus, $\sum\limits_{a\in B}\left. :a_{m}a_{n-m}:\right. =\sum\limits_{a\in B^{\prime}}\left. :a_{m}a_{n-m}:\right. $ always holds (since cases 1 and 2 cover all possibilities), qed.}. Hence,% \begin{align*} L_{n} & =\dfrac{1}{2}\sum\limits_{a\in B}\sum\limits_{m\in\mathbb{Z}}\left. :a_{m}a_{n-m}:\right. =\dfrac{1}{2}\sum\limits_{m\in\mathbb{Z}}% \underbrace{\sum\limits_{a\in B}\left. :a_{m}a_{n-m}:\right. }% _{=\sum\limits_{a\in B^{\prime}}\left. :a_{m}a_{n-m}:\right. }=\dfrac{1}% {2}\sum\limits_{m\in\mathbb{Z}}\sum\limits_{a\in B^{\prime}}\left. :a_{m}a_{n-m}:\right. \\ & =\dfrac{1}{2}\sum\limits_{a\in B^{\prime}}\sum\limits_{m\in\mathbb{Z}% }\left. :a_{m}a_{n-m}:\right. . \end{align*} Thus, (\ref{pf.sugawara.basisind.1}) is proven. As we said, this completes the proof of Theorem \ref{thm.sugawara} \textbf{(b)}. \textbf{(c)} \textit{1st step:} Let us first show that% \begin{equation} \left[ b_{r},L_{n}\right] =rb_{n+r}\ \ \ \ \ \ \ \ \ \ \text{for every }% b\in\mathfrak{g}\text{ and any integers }r\text{ and }n. \label{pf.sugawara.step1}% \end{equation} \textit{Proof of (\ref{pf.sugawara.step1}):} Let $b\in\mathfrak{g}$, $r\in\mathbb{Z}$ and $n\in\mathbb{Z}$. We must be careful here with infinite sums, since not even formal algebra allows us to manipulate infinite sums like $\sum\limits_{m\in\mathbb{Z}% }\left[ b,a\right] _{r+m}a_{n-m}$ (for good reasons: these are divergent in every meaning of this word). While we were working in the Heisenberg algebra $\mathcal{A}$ (which can be written as $\widehat{\mathfrak{g}}$ for $\mathfrak{g}$ being the trivial Lie algebra $\mathbb{C}$), these infinite sums made sense due to all of their addends being $0$ (since $\left[ b,a\right] =0$ for all $a$ and $b$ lying in the trivial Lie algebra $\mathbb{C}$). But this was an exception rather than the rule, and now we need to take care. Let us first assume that $r\geq0$. Since% \begin{align*} L_{n} & =\dfrac{1}{2}\sum\limits_{a\in B}\underbrace{\sum\limits_{m\in \mathbb{Z}}\left. :a_{m}a_{n-m}:\right. }_{=\lim\limits_{N\rightarrow\infty }\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left. :a_{m}a_{n-m}:\right. }=\dfrac{1}{2}\sum\limits_{a\in B}\lim \limits_{N\rightarrow\infty}\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left. :a_{m}a_{n-m}:\right. \\ & =\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum\limits_{a\in B}% \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left. :a_{m}% a_{n-m}:\right. , \end{align*} we have% \begin{align} & \left[ b_{r},L_{n}\right] \nonumber\\ & =\left[ b_{r},\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum \limits_{a\in B}\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left. :a_{m}a_{n-m}:\right. \right] \nonumber\\ & =\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum\limits_{a\in B}% \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\underbrace{\left[ b_{r},\left. :a_{m}a_{n-m}:\right. \right] }% _{\substack{_{\substack{=\left[ b_{r},a_{m}a_{n-m}\right] }}\\\text{(by Remark \ref{rmk.sugawara.normal.K} \textbf{(b)}, applied to}\\a\text{, }% b_{r}\text{ and }n-m\text{ instead of }x\text{, }y\text{ and }n\text{)}% }}\nonumber\\ & =\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum\limits_{a\in B}% \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\underbrace{\left[ b_{r},a_{m}a_{n-m}\right] }_{=\left[ b_{r},a_{m}\right] a_{n-m}% +a_{m}\left[ b_{r},a_{n-m}\right] }\nonumber\\ & =\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum\limits_{a\in B}% \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \underbrace{\left[ b_{r},a_{m}\right] }_{\substack{=\left[ b,a\right] _{r+m}+K\omega\left( b_{r},a_{m}\right) \\\text{(by (\ref{pf.sugawara.lie}% ))}}}a_{n-m}+a_{m}\underbrace{\left[ b_{r},a_{n-m}\right] }% _{\substack{=\left[ b,a\right] _{n+r-m}+K\omega\left( b_{r},a_{n-m}\right) \\\text{(by (\ref{pf.sugawara.lie}))}}}\right) \nonumber\\ & =\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum\limits_{a\in B}% \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+K\omega\left( b_{r},a_{m}\right) a_{n-m}% +a_{m}\left[ b,a\right] _{n+r-m}+a_{m}K\omega\left( b_{r},a_{n-m}\right) \right) \nonumber\\ & =\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum\limits_{a\in B}% \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}+K\omega\left( b_{r},a_{m}\right) a_{n-m}+a_{m}K\omega\left( b_{r},a_{n-m}\right) \right) . \label{pf.sugawara.b.1}% \end{align} Now fix $a\in B$. We now notice that for any $N\in\mathbb{N}$, the sum $\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}K\omega\left( b_{r},a_{m}\right) a_{n-m}$ (in $\operatorname*{End}M$) has at most one nonzero addend (because $\omega\left( b_{r},a_{m}\right) $ can be nonzero for at most one integer $m$ (namely, for $m=-r$)). Hence, this sum $\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}K\omega\left( b_{r},a_{m}\right) a_{n-m}$ converges for any $N\in\mathbb{N}$. For sufficiently high $N$, this sum does have an addend for $m=-r$, and all other addends of this sum are $0$ (since $\omega\left( b_{r},a_{m}\right) =0$ whenever $m\neq-r$), so that the value of this sum is $\underbrace{K}% _{\substack{=k\\\text{(since }K\text{ acts as}\\k\cdot\operatorname*{id}\text{ on }M\text{)}}}\underbrace{\omega\left( b_{r},a_{-r}\right) }% _{\substack{=r\left( b,a\right) \\\text{(by the definition of }% \omega\text{)}}}\underbrace{a_{n-\left( -r\right) }}_{=a_{n+r}}=kr\left( b,a\right) a_{n+r}$. We thus have shown that the sum $\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}K\omega\left( b_{r},a_{m}\right) a_{n-m}$ converges for all $N\in\mathbb{N}$, and satisfies% \begin{equation} \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}K\omega\left( b_{r},a_{m}\right) a_{n-m}=kr\left( b,a\right) a_{n+r}% \ \ \ \ \ \ \ \ \ \ \text{for sufficiently high }N\text{.} \label{pf.sugawara.b.2a}% \end{equation} Similarly, we see that the sum $\sum\limits_{\left\vert m-\dfrac{n}% {2}\right\vert \leq N}a_{m}K\omega\left( b_{r},a_{n-m}\right) $ converges for all $N\in\mathbb{N}$, and satisfies% \begin{equation} \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}a_{m}K\omega\left( b_{r},a_{n-m}\right) =a_{n+r}kr\left( b,a\right) \ \ \ \ \ \ \ \ \ \ \text{for sufficiently high }N\text{.} \label{pf.sugawara.b.2b}% \end{equation} Finally, for all $N\in\mathbb{N}$, the sum $\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}% a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}\right) $ converges\footnote{\textit{Proof.} Let $N\in\mathbb{N}$. The sum $\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}+K\omega\left( b_{r},a_{m}\right) a_{n-m}+a_{m}K\omega\left( b_{r},a_{n-m}\right) \right) $ converges (because it appears on the right hand side of (\ref{pf.sugawara.b.1})), and the sums $\sum\limits_{\left\vert m-\dfrac{n}% {2}\right\vert \leq N}K\omega\left( b_{r},a_{m}\right) a_{n-m}$ and $\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}a_{m}K\omega\left( b_{r},a_{n-m}\right) $ converge (as we have just seen). Hence, the sum% \begin{align*} & \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}% +K\omega\left( b_{r},a_{m}\right) a_{n-m}+a_{m}K\omega\left( b_{r}% ,a_{n-m}\right) \right) \right. \\ & \ \ \ \ \ \ \ \ \ \ \left. -K\omega\left( b_{r},a_{m}\right) a_{n-m}-a_{m}K\omega\left( b_{r},a_{n-m}\right) \right) \end{align*} converges as well (since it is obtained by subtracting the latter two sums from the former sum componentwise). But this sum clearly simplifies to $\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}\right) $. Hence, the sum $\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}\right) $ converges, qed.}. Since the sums $\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}K\omega\left( b_{r},a_{m}\right) a_{n-m}$, $\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}a_{m}K\omega\left( b_{r},a_{n-m}\right) $ and \newline$\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}\right) $ converge for every $N\in\mathbb{N}$, we have% \begin{align*} & \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}+K\omega\left( b_{r},a_{m}\right) a_{n-m}+a_{m}K\omega\left( b_{r},a_{n-m}\right) \right) \\ & =\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}\right) +\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}K\omega\left( b_{r},a_{m}\right) a_{n-m}\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}a_{m}K\omega\left( b_{r},a_{n-m}\right) \end{align*} for every $N\in\mathbb{N}$. Hence, for every sufficiently high $N\in \mathbb{N}$, we have% \begin{align} & \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}+K\omega\left( b_{r},a_{m}\right) a_{n-m}+a_{m}K\omega\left( b_{r},a_{n-m}\right) \right) \nonumber\\ & =\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}\right) +\underbrace{\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}% K\omega\left( b_{r},a_{m}\right) a_{n-m}}_{\substack{=kr\left( b,a\right) a_{n+r}\text{ for sufficiently high }N\\\text{(by (\ref{pf.sugawara.b.2a}))}% }}\nonumber\\ & \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\left\vert m-\dfrac{n}% {2}\right\vert \leq N}a_{m}K\omega\left( b_{r},a_{n-m}\right) }% _{\substack{=a_{n+r}kr\left( b,a\right) \text{ for sufficiently high }N\\\text{(by (\ref{pf.sugawara.b.2a}))}}}\nonumber\\ & =\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}\right) +\underbrace{kr\left( b,a\right) a_{n+r}+a_{n+r}kr\left( b,a\right) }_{=2rk\cdot\left( b,a\right) a_{n+r}}\nonumber\\ & =\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}\right) +2rk\cdot\left( b,a\right) a_{n+r}. \label{pf.sugawara.b.3.sufficiently}% \end{align} Now, forget that we fixed $a$. The equality (\ref{pf.sugawara.b.1}) becomes% \begin{align} & \left[ b_{r},L_{n}\right] \nonumber\\ & =\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum\limits_{a\in B}\underbrace{\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}% +K\omega\left( b_{r},a_{m}\right) a_{n-m}+a_{m}K\omega\left( b_{r}% ,a_{n-m}\right) \right) }_{\substack{=\sum\limits_{\left\vert m-\dfrac{n}% {2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}\right) +2rk\cdot\left( b,a\right) a_{n+r}\\\text{for sufficiently high }N\text{ (by (\ref{pf.sugawara.b.3.sufficiently}))}% }}\nonumber\\ & =\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum\limits_{a\in B}\left( \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \left[ b,a\right] _{r+m}a_{n-m}+a_{m}\left[ b,a\right] _{n+r-m}\right) +2rk\cdot\left( b,a\right) a_{n+r}\right) \nonumber\\ & =\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \sum\limits_{a\in B}\left[ b,a\right] _{r+m}a_{n-m}+\sum\limits_{a\in B}a_{m}\left[ b,a\right] _{n+r-m}\right) +rk\sum\limits_{a\in B}\left( b,a\right) a_{n+r}. \label{pf.sugawara.b.6}% \end{align} But since $\sum\limits_{a\in B}\left( \left[ b,a\right] \otimes a+a\otimes\left[ b,a\right] \right) =0$ (by Lemma \ref{lem.sugawara.Kil3} \textbf{(a)}), we have \newline$\sum\limits_{a\in B}\left( \left[ b,a\right] _{\ell}\otimes a_{s}+a_{\ell}\otimes\left[ b,a\right] _{s}\right) =0$ for any two integers $\ell$ and $s$. In particular, every $m\in\mathbb{Z}$ satisfies $\sum\limits_{a\in B}\left( \left[ b,a\right] _{m}\otimes a_{n+r-m}+a_{m}\otimes\left[ b,a\right] _{n+r-m}\right) =0$. Hence, every $m\in\mathbb{Z}$ satisfies $\sum\limits_{a\in B}\left( \left[ b,a\right] _{m}a_{n+r-m}+a_{m}\left[ b,a\right] _{n+r-m}\right) =0$, so that $\sum\limits_{a\in B}\left[ b,a\right] _{m}a_{n+r-m}+\sum\limits_{a\in B}a_{m}\left[ b,a\right] _{n+r-m}=0$ and thus $\sum\limits_{a\in B}% a_{m}\left[ b,a\right] _{n+r-m}=-\sum\limits_{a\in B}\left[ b,a\right] _{m}a_{n+r-m}$. Hence, (\ref{pf.sugawara.b.6}) becomes% \begin{align} & \left[ b_{r},L_{n}\right] \nonumber\\ & =\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \sum\limits_{a\in B}\left[ b,a\right] _{r+m}a_{n-m}+\underbrace{\sum\limits_{a\in B}a_{m}\left[ b,a\right] _{n+r-m}}_{=-\sum\limits_{a\in B}\left[ b,a\right] _{m}% a_{n+r-m}}\right) +rk\sum\limits_{a\in B}\left( b,a\right) a_{n+r}% \nonumber\\ & =\dfrac{1}{2}\lim\limits_{N\rightarrow\infty}\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \sum\limits_{a\in B}\left[ b,a\right] _{r+m}a_{n-m}-\sum\limits_{a\in B}\left[ b,a\right] _{m}a_{n+r-m}\right) +rk\sum\limits_{a\in B}\left( b,a\right) a_{n+r}. \label{pf.sugawara.b.9}% \end{align} We will now transform the limit in this equation: In fact,% \begin{align*} & \lim\limits_{N\rightarrow\infty}\underbrace{\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left( \sum\limits_{a\in B}\left[ b,a\right] _{r+m}a_{n-m}-\sum\limits_{a\in B}\left[ b,a\right] _{m}a_{n+r-m}\right) }_{=\sum\limits_{a\in B}\left( \sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left[ b,a\right] _{r+m}a_{n-m}% -\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left[ b,a\right] _{m}a_{n+r-m}\right) }\\ & =\lim\limits_{N\rightarrow\infty}\sum\limits_{a\in B}\left( \underbrace{\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left[ b,a\right] _{r+m}a_{n-m}}_{\substack{=\sum\limits_{\left\vert m-r-\dfrac {n}{2}\right\vert \leq N}\left[ b,a\right] _{m}a_{n+r-m}\\\text{(here, we substituted }m-r\text{ for }m\text{ in the sum)}}}-\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left[ b,a\right] _{m}a_{n+r-m}\right) \\ & =\lim\limits_{N\rightarrow\infty}\sum\limits_{a\in B}\underbrace{\left( \sum\limits_{\left\vert m-r-\dfrac{n}{2}\right\vert \leq N}\left[ b,a\right] _{m}a_{n+r-m}-\sum\limits_{\left\vert m-\dfrac{n}{2}\right\vert \leq N}\left[ b,a\right] _{m}a_{n+r-m}\right) }_{\substack{=-\sum\limits_{\dfrac{n}% {2}-N\leq m<\dfrac{n}{2}+r-N}\left[ b,a\right] _{m}a_{n+r-m}+\sum \limits_{\dfrac{n}{2}+N \dfrac{n}{2}+N\geq\mathbf{M}$ and thus $\left[ b,a\right] t^{m}\cdot w=0$ (by (\ref{pf.sugawara.foot.2a}), applied to $m$ and $\left[ b,a\right] $ instead of $\mathbf{m}$ and $a$), thus $a_{n+r-m}\underbrace{\left[ b,a\right] _{m}}_{=\left[ b,a\right] t^{m}}w=a_{n+r-m}\underbrace{\left[ b,a\right] t^{m}\cdot w}_{=0}=0$. Hence, $\sum\limits_{\dfrac{n}{2}+N 0}}\alpha$ be the half-sum of all positive roots. The \textit{dual Coxeter number} $h^{\vee}$ of $\mathfrak{g}$ is defined by $h^{\vee}=1+\left( \theta,\rho\right) $. It is easy to show that $h^{\vee}$ is a positive integer. \end{definition} \begin{definition} \label{def.standform}Let $\mathfrak{g}$ be a simple finite-dimensional Lie algebra. The \textit{standard form} on $\mathfrak{g}$ will mean the scalar multiple of the Killing form under which $\left( \alpha,\alpha\right) $ (under the inverse form on $\mathfrak{g}^{\ast}$) equals $2$ for long roots $\alpha$. (We do not care to define what a long root is, but it is enough to say that the maximal root $\theta$ is a long root, and this is clearly enough to define the standard form.) (The \textit{inverse form} of a nondegenerate bilinear form $\left( \cdot,\cdot\right) $ on $\mathfrak{g}$ means the bilinear form on $\mathfrak{g}^{\ast}=\mathfrak{h}^{\ast}\oplus\mathfrak{n}_{+}^{\ast}% \oplus\mathfrak{n}_{-}^{\ast}$ obtained by dualizing the bilinear form $\left( \cdot,\cdot\right) $ on $\mathfrak{g}=\mathfrak{h}\oplus \mathfrak{n}_{+}\oplus\mathfrak{n}_{-}$ using itself.) We are going to denote the standard form by $\left( \cdot,\cdot\right) $. \end{definition} \begin{lemma} \label{lem.dualcox}Let $B$ be an orthonormal basis of $\mathfrak{g}$ with respect to the standard form. Let $C=\sum\limits_{a\in B}a^{2}\in U\left( \mathfrak{g}\right) $. This element $C$ is known to be central in $U\left( \mathfrak{g}\right) $ (this is easily checked), and is called the \textit{quadratic Casimir}. Then: \textbf{(1)} For every $\lambda\in\mathfrak{h}^{\ast}$, the element $C\in U\left( \mathfrak{g}\right) $ acts on $L_{\lambda}$ by $\left( \lambda,\lambda+2\rho\right) \cdot\operatorname*{id}$. (Here, $L_{\lambda}$ means $L_{\lambda}^{+}$, but actually can be replaced by any highest-weight module with highest weight $\lambda$.) \textbf{(2)} The element $C\in U\left( \mathfrak{g}\right) $ acts on the adjoint representation $\mathfrak{g}$ by $2h^{\vee}\cdot\operatorname*{id}$. \end{lemma} \textit{Proof of Lemma \ref{lem.dualcox}.} If $\left( b_{i}\right) _{i\in I}$ is any basis of $\mathfrak{g}$, and $\left( b_{i}^{\ast}\right) _{i\in I}$ is the dual basis of $\mathfrak{g}$ with respect to the standard form $\left( \cdot,\cdot\right) $, then% \begin{equation} C=\sum\limits_{i\in I}b_{i}b_{i}^{\ast}. \label{pf.dualcox.Csum}% \end{equation} \footnote{This is a well-known property of the quadratic Casimir.} \textbf{(1)} Let $\lambda\in\mathfrak{h}^{\ast}$. Let us refine the triangular decomposition $\mathfrak{g}=\mathfrak{h}% \oplus\mathfrak{n}_{+}\oplus\mathfrak{n}_{-}$ to the weight space decomposition $\mathfrak{g}=\mathfrak{h}\oplus\left( \bigoplus\limits_{\alpha >0}\mathfrak{g}_{\alpha}\right) \oplus\left( \bigoplus\limits_{\alpha <0}\mathfrak{g}_{\alpha}\right) $, where $\mathfrak{g}_{\alpha}% =\mathbb{C}e_{\alpha}$ for roots $\alpha>0$, and $\mathfrak{g}_{-\alpha }=\mathbb{C}f_{\alpha}$ for roots $\alpha>0$. (This is standard theory of simple Lie algebras.) Normalize the $f_{\alpha}$ in such a way that $\left( e_{\alpha},f_{\alpha}\right) =1$. As usual, denote $h_{\alpha}=\left[ e_{\alpha},f_{\alpha}\right] $ for every root $\alpha>0$. Fix an orthonormal basis $\left( x_{i}\right) _{i\in\left\{ 1,2,...,r\right\} }$ of $\mathfrak{h}$. Clearly, $\left( x_{i}\right) _{i\in\left\{ 1,2,...,r\right\} }\cup\left( e_{\alpha}\right) _{\alpha >0}\cup\left( f_{\alpha}\right) _{\alpha>0}$ (where the index $\alpha$ runs over positive roots only) is a basis of $\mathfrak{g}$. Since% \begin{align*} \left( e_{\alpha},x_{i}\right) & =\left( f_{\alpha},x_{i}\right) =0\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{ 1,2,...,r\right\} \text{ and roots }\alpha>0;\\ \left( e_{\alpha},f_{\beta}\right) & =0\ \ \ \ \ \ \ \ \ \ \text{for any two distinct roots }\alpha>0\text{ and }\beta>0;\\ \left( e_{\alpha},e_{\gamma}\right) & =\left( f_{\alpha},f_{\gamma }\right) =0\ \ \ \ \ \ \ \ \ \ \text{for any roots }\alpha>0\text{ and }\gamma>0;\\ \left( x_{i},x_{j}\right) & =\delta_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{ 1,2,...,r\right\} \text{ and }j\in\left\{ 1,2,...,r\right\} ;\\ \left( e_{\alpha},f_{\alpha}\right) & =\left( f_{\alpha},e_{\alpha }\right) =1\ \ \ \ \ \ \ \ \ \ \text{for any root }\alpha>0\text{,}% \end{align*} we see that $\left( x_{i}\right) _{i\in\left\{ 1,2,...,r\right\} }% \cup\left( f_{\alpha}\right) _{\alpha>0}\cup\left( e_{\alpha}\right) _{\alpha>0}$ is the dual basis to this basis $\left( x_{i}\right) _{i\in\left\{ 1,2,...,r\right\} }\cup\left( e_{\alpha}\right) _{\alpha >0}\cup\left( f_{\alpha}\right) _{\alpha>0}$ with respect to the standard form $\left( \cdot,\cdot\right) $. Thus, (\ref{pf.dualcox.Csum}) yields% \[ C=\sum\limits_{i=1}^{r}x_{i}^{2}+\sum\limits_{\alpha>0}\left( f_{\alpha }e_{\alpha}+e_{\alpha}f_{\alpha}\right) , \] so that (denoting $v_{\lambda}^{+}$ by $v_{\lambda}$) we have% \begin{align*} Cv_{\lambda} & =\sum\limits_{i=1}^{r}\underbrace{x_{i}^{2}v_{\lambda}% }_{=\lambda\left( x_{i}\right) ^{2}v_{\lambda}}+\sum\limits_{\alpha >0}\left( f_{\alpha}e_{\alpha}+e_{\alpha}f_{\alpha}\right) v_{\lambda }=\underbrace{\sum\limits_{i=1}^{r}\lambda\left( x_{i}\right) ^{2}% }_{=\left( \lambda,\lambda\right) }v_{\lambda}+\sum\limits_{\alpha>0}\left( f_{\alpha}\underbrace{e_{\alpha}v_{\lambda}}_{=0}+\underbrace{e_{\alpha }f_{\alpha}}_{=f_{\alpha}e_{\alpha}+\left[ e_{\alpha},f_{\alpha}\right] }v_{\lambda}\right) \\ & =\left( \lambda,\lambda\right) v_{\lambda}+\sum\limits_{\alpha>0}\left( f_{\alpha}e_{\alpha}+\left[ e_{\alpha},f_{\alpha}\right] \right) v_{\lambda}\\ & =\left( \lambda,\lambda\right) v_{\lambda}+\sum\limits_{\alpha >0}f_{\alpha}\underbrace{e_{\alpha}v_{\lambda}}_{=0}+\sum\limits_{\alpha >0}\underbrace{\left[ e_{\alpha},f_{\alpha}\right] }_{=h_{\alpha}}% v_{\lambda}\\ & =\left( \lambda,\lambda\right) v_{\lambda}+\sum\limits_{\alpha >0}\underbrace{h_{\alpha}v_{\lambda}}_{=\lambda\left( h_{\alpha}\right) v_{\lambda}}=\left( \lambda,\lambda\right) v_{\lambda}+\sum\limits_{\alpha >0}\underbrace{\lambda\left( h_{\alpha}\right) }_{=\left( \lambda ,\alpha\right) }v_{\lambda}\\ & =\left( \lambda,\lambda\right) v_{\lambda}+\sum\limits_{\alpha>0}\left( \lambda,\alpha\right) v_{\lambda}=\underbrace{\left( \left( \lambda ,\lambda\right) +\sum\limits_{\alpha>0}\left( \lambda,\alpha\right) \right) }_{\substack{=\left( \lambda,\lambda+\sum\limits_{\alpha>0}% \alpha\right) =\left( \lambda,\lambda+2\rho\right) \\\text{(since }% \sum\limits_{\alpha>0}\alpha=2\rho\text{)}}}v_{\lambda}=\left( \lambda ,\lambda+2\rho\right) v_{\lambda}. \end{align*} Thus, every $a\in U\left( \mathfrak{g}\right) $ satisfies \begin{align*} Cav_{\lambda} & =a\underbrace{Cv_{\lambda}}_{=\left( \lambda,\lambda +2\rho\right) v_{\lambda}}\ \ \ \ \ \ \ \ \ \ \left( \text{since }C\text{ is central in }U\left( \mathfrak{g}\right) \right) \\ & =\left( \lambda,\lambda+2\rho\right) av_{\lambda}. \end{align*} Hence, $C$ acts as $\left( \lambda,\lambda+2\rho\right) \cdot \operatorname*{id}$ on $L_{\lambda}$ (because every element of $L_{\lambda}$ has the form $av_{\lambda}$ for some $a\in U\left( \mathfrak{g}\right) $). This proves Lemma \ref{lem.dualcox} \textbf{(1)}. \textbf{(2)} We have $\mathfrak{g}=L_{\theta}$, and thus Lemma \ref{lem.dualcox} \textbf{(1)} yields% \[ C\mid_{L_{\theta}}=\left( \theta,\theta+2\rho\right) =\underbrace{\left( \theta,\theta\right) }_{=2}+2\left( \theta,\rho\right) =2+2\left( \theta,\rho\right) =2h^{\vee}. \] This proves Lemma \ref{lem.dualcox} \textbf{(2)}. Here is a little table of dual Coxeter numbers, depending on the root system type of $\mathfrak{g}$: For $A_{n-1}$, we have $h^{\vee}=n$. For $B_{n}$, we have $h^{\vee}=2n-1$. For $C_{n}$, we have $h^{\vee}=n+1$. For $D_{n}$, we have $h^{\vee}=2n-2$. For $E_{6}$, we have $h^{\vee}=12$. For $E_{7}$, we have $h^{\vee}=18$. For $E_{8}$, we have $h^{\vee}=30$. For $F_{4}$, we have $h^{\vee}=9$. For $G_{2}$, we have $h^{\vee}=4$. Every Lie theorist is supposed to remember these by heart. \begin{lemma} \label{lem.dualcox.kil}Let $\mathfrak{g}$ be a simple finite-dimensional Lie algebra. Then,% \[ \operatorname*{Kil}\left( a,b\right) =2h^{\vee}\cdot\left( a,b\right) \ \ \ \ \ \ \ \ \ \ \text{for any }a,b\in\mathfrak{g}. \] \end{lemma} \textit{Proof of Lemma \ref{lem.dualcox.kil}.} Let $B$ be an orthonormal basis of $\mathfrak{g}$ with respect to the standard form. Define the quadratic Casimir $C=\sum\limits_{a\in B}a^{2}$ as in Lemma \ref{lem.dualcox}. Then,% \[ \operatorname*{Tr}\nolimits_{\mathfrak{g}}\left( C\right) =\sum\limits_{a\in B}\underbrace{\operatorname*{Tr}\nolimits_{\mathfrak{g}}\left( a^{2}\right) }_{=\operatorname*{Tr}\left( \left( \operatorname*{ad}a\right) \circ\left( \operatorname*{ad}a\right) \right) =\operatorname*{Kil}\left( a,a\right) }=\sum\limits_{a\in B}\operatorname*{Kil}\left( a,a\right) . \] Comparing this with% \begin{align*} \operatorname*{Tr}\nolimits_{\mathfrak{g}}\left( C\right) & =2h^{\vee }\underbrace{\operatorname*{Tr}\nolimits_{\mathfrak{g}}\left( \operatorname*{id}\right) }_{=\dim\mathfrak{g}}\ \ \ \ \ \ \ \ \ \ \left( \text{since }C\mid_{\mathfrak{g}}=2h^{\vee}\operatorname*{id}\text{ by Lemma \ref{lem.dualcox} \textbf{(2)}}\right) \\ & =2h^{\vee}\underbrace{\dim\mathfrak{g}}_{\substack{=\left\vert B\right\vert =\sum\limits_{a\in B}1=\sum\limits_{a\in B}\left( a,a\right) \\\text{(since every }a\in B\text{ satisfies }\left( a,a\right) =1\text{)}}}=2h^{\vee}% \sum\limits_{a\in B}\left( a,a\right) , \end{align*} we obtain $\sum\limits_{a\in B}\operatorname*{Kil}\left( a,a\right) =2h^{\vee}\sum\limits_{a\in B}\left( a,a\right) $. Since $\operatorname*{Kil}$ is a scalar multiple of $\left( \cdot,\cdot\right) $ (because there is only one $\mathfrak{g}$-invariant symmetric bilinear form on $\mathfrak{g}$ up to scaling), this yields $\operatorname*{Kil}=2h^{\vee}% \cdot\left( \cdot,\cdot\right) $ (because $\sum\limits_{a\in B}% \underbrace{\left( a,a\right) }_{=1}=\sum\limits_{a\in B}1=\left\vert B\right\vert \neq0$). Lemma \ref{lem.dualcox.kil} is proven. So let us now look at the Sugawara construction when $\mathfrak{g}$ is simple finite-dimensional. First of all, $k$ is non-critical if and only if $k\neq-h^{\vee}$. (The value $k=-h^{\vee}$ is called the \textit{critical level}.) If $B^{\prime}$ is an orthonormal basis under $\left( \cdot,\cdot\right) $ (rather than under $k\left( \cdot,\cdot\right) +\dfrac{1}{2}% \operatorname*{Kil}=\left( k+h^{\vee}\right) \left( \cdot,\cdot\right) $), then we have% \begin{align} L_{n} & =\dfrac{1}{2\left( k+h^{\vee}\right) }\sum\limits_{a\in B^{\prime }}\sum\limits_{m\in\mathbb{Z}}\left. :a_{m}a_{n-m}:\right. \ \ \ \ \ \ \ \ \ \ \text{and}\nonumber\\ c & =\dfrac{k}{k+h^{\vee}}\underbrace{\sum\limits_{a\in B^{\prime}}\left( a,a\right) }_{\substack{=\left\vert B^{\prime}\right\vert \\\text{(since }\left( a,a\right) =1\text{ for every }a\in B^{\prime}\text{)}}}=\dfrac {k}{k+h^{\vee}}\underbrace{\left\vert B^{\prime}\right\vert }_{=\dim \mathfrak{g}}=\dfrac{k\dim\mathfrak{g}}{k+h^{\vee}}. \label{thm.sugawara.simple.c}% \end{align} In particular, this induces an internal grading on any $\widehat{\mathfrak{g}% }$-module which is a quotient of $M_{\lambda}^{+}$ by eigenvalues of $L_{0}$, whenever $\lambda$ is a weight of $\widehat{\mathfrak{g}}$. This is a grading by complex numbers, since eigenvalues of $L_{0}$ are not necessarily integers. (Note that this does not work for general admissible modules in lieu of quotients of $M_{\lambda}^{+}$.) What happens at the critical level $k=-h^{\vee}$ ? The above formulas with $k+h^{\vee}$ in the denominators clearly don't work at this level anymore. We can, however, remove the denominators, i. e., consider the operators% \[ T_{n}=\dfrac{1}{2}\sum\limits_{a\in B^{\prime}}\sum\limits_{m\in\mathbb{Z}% }\left. :a_{m}a_{n-m}:\right. . \] Then, the same calculations as we did in the proof of Theorem \ref{thm.sugawara} tell us that these $T_{n}$ satisfy $\left[ T_{n}% ,a_{m}\right] =0$ and $\left[ T_{n},T_{m}\right] =0$; they are thus central ``elements'' of $U\left( \widehat{\mathfrak{g}}\right) $ (except that they are not actually elements of $U\left( \widehat{\mathfrak{g}}\right) $, but of some completion of $U\left( \widehat{\mathfrak{g}}\right) $ acting on admissible modules). For any complex numbers $\gamma_{1},\gamma_{2},\gamma_{3},...$, we can construct a $\widehat{\mathfrak{g}}$-module $M_{\lambda}\diagup\left( \sum\limits_{m\geq1}\left( \left( T_{m}-\gamma_{m}\right) M_{\lambda }\right) \right) $, which does not have a grading. So, at the critical level, we do not automatically get gradings on quotients of $M_{\lambda}$ anymore. This is one reason why representations at the critical level are considered more difficult than those at non-critical levels. \subsection{The Sugawara construction and unitarity} We now will show that the Sugawara construction preserves unitarity: \begin{proposition} Consider the situation of Theorem \ref{thm.sugawara}. If $M$ is a unitary admissible module for $\widehat{\mathfrak{g}}$, then $M$ is a unitary $\operatorname*{Vir}\ltimes\widehat{\mathfrak{g}}$-module. (We recall that the Virasoro algebra had its unitary structure given by $L_{n}^{\dag}=L_{-n}$.) \end{proposition} But for $M$ to be unitary for $\widehat{\mathfrak{g}}$, we need $k\in \mathbb{Z}_{+}$ (this is easy to prove; we proved it for $\mathfrak{sl}_{n}$, and the general case is similar). Since for $k=0$, there is only the trivial representation, we really must require $k\geq1$ to get something interesting. And since $c=\dfrac{k\dim\mathfrak{g}}{k+h^{\vee}}$, the $c$ is then $\geq1$, since $\dim\mathfrak{g}\geq1+h^{\vee}$. These modules are already known to us to be unitary, so this construction does not help us in constructing new unitary modules. But there is a way to amend this by a variation of the Sugawara construction: the Goddard-Kent-Olive construction. \subsection{The Goddard-Kent-Olive construction (a.k.a. the coset construction)} \begin{definition} \label{def.goddardkentolive}Let $\mathfrak{g}$ and $\mathfrak{p}$ be two finite-dimensional Lie algebras such that $\mathfrak{g}\supseteq\mathfrak{p}$. Let $\left( \cdot,\cdot\right) $ be a $\mathfrak{g}$-invariant form (possibly degenerate) on $\mathfrak{g}$. We can restrict this form to $\mathfrak{p}$, and obtain a $\mathfrak{p}$-invariant form on $\mathfrak{p}$. Construct an affine Lie algebra $\widehat{\mathfrak{g}}$ as in Definition \ref{def.sugawara} using the $\mathfrak{g}$-invariant form $\left( \cdot,\cdot\right) $ on $\mathfrak{g}$, and similarly construct an affine Lie algebra $\widehat{\mathfrak{p}}$ using the restriction of this form to $\mathfrak{p}$. Then, $\widehat{\mathfrak{g}}\supseteq\widehat{\mathfrak{p}}$. Choose a level $k$ which is non-critical for both $\mathfrak{g}$ and $\mathfrak{p}$. Let $M$ be an admissible $\widehat{\mathfrak{g}}$-module at level $k$. Then, $M$ automatically becomes an admissible $\widehat{\mathfrak{p}}$-module at level $k$. Hence, on $M$, we have two Virasoro actions: one which is obtained from the $\widehat{\mathfrak{g}}$-action, and one which is obtained from the $\widehat{\mathfrak{p}}$-action. We will denote these actions by $\left( L_{i}^{\mathfrak{g}}\right) _{i\in\mathbb{Z}}$ and $\left( L_{i}% ^{\mathfrak{p}}\right) _{i\in\mathbb{Z}}$, respectively (that is, for every $i\in\mathbb{Z}$, we denote by $L_{i}^{\mathfrak{g}}$ the action of $L_{i}% \in\operatorname*{Vir}$ obtained from the $\widehat{\mathfrak{g}}$-module structure on $M$, and we denote by $L_{i}^{\mathfrak{p}}$ the action of $L_{i}\in\operatorname*{Vir}$ obtained from the $\widehat{\mathfrak{p}}% $-module structure on $M$), and we will denote their central charges by $c_{\mathfrak{g}}$ and $c_{\mathfrak{p}}$, respectively. \end{definition} \begin{theorem} \label{thm.goddardkentolive}Consider the situation of Definition \ref{def.goddardkentolive}. Let $L_{i}=L_{i}^{\mathfrak{g}}-L_{i}% ^{\mathfrak{p}}$ for all $i\in\mathbb{Z}$. \textbf{(a)} Then, $\left( L_{i}\right) _{i\in\mathbb{Z}}$ is a $\operatorname*{Vir}$-action on $M$ with central charge $c=c_{\mathfrak{g}% }-c_{\mathfrak{p}}$. \textbf{(b)} Also, $\left[ L_{n},\widehat{p}\right] =0$ for all $\widehat{p}\in\widehat{\mathfrak{p}}$ and $n\in\mathbb{Z}$. \textbf{(c)} Moreover, $\left[ L_{n},L_{m}^{\mathfrak{p}}\right] =0$ for all $n\in\mathbb{Z}$ and $m\in\mathbb{Z}$. \end{theorem} \textit{Proof of Theorem \ref{thm.goddardkentolive}.} \textbf{(b)} Let $n\in\mathbb{Z}$. Every $p\in\mathfrak{p}$ and $m\in\mathbb{Z}$ satisfy% \[ \left[ \underbrace{L_{n}}_{=L_{n}^{\mathfrak{g}}-L_{n}^{\mathfrak{p}}}% ,p_{m}\right] =\underbrace{\left[ L_{n}^{\mathfrak{g}},p_{m}\right] }_{\substack{=-mp_{n+m}\\\text{(by Theorem \ref{thm.sugawara} \textbf{(e)}% ,}\\\text{applied to }p\text{ instead of }a\text{)}}}-\underbrace{\left[ L_{n}^{\mathfrak{p}},p_{m}\right] }_{\substack{=-mp_{n+m}\\\text{(by Theorem \ref{thm.sugawara} \textbf{(e)},}\\\text{applied to }p\text{ and }% \mathfrak{p}\text{ instead of }a\text{ and }\mathfrak{g}\text{)}}}=\left( -mp_{m+n}\right) -\left( -mp_{m+n}\right) =0. \] Combined with the fact that $\left[ L_{n},K\right] =0$ (this is trivial, since $K$ acts as $k\cdot\operatorname*{id}$ on $M$), this yields that $\left[ L_{n},\widehat{p}\right] =0$ for all $\widehat{p}\in \widehat{\mathfrak{p}}$ and $n\in\mathbb{Z}$ (because every $\widehat{p}% \in\widehat{\mathfrak{p}}$ is a $\mathbb{C}$-linear combination of terms of the form $p_{m}$ (with $p\in\mathfrak{p}$ and $m\in\mathbb{Z}$) and $K$). Thus, Theorem \ref{thm.goddardkentolive} \textbf{(b)} is proven. \textbf{(c)} Let $n\in\mathbb{Z}$. We recall that $L_{n}^{\mathfrak{p}}$ was defined by $L_{n}^{\mathfrak{p}}=\dfrac{1}{2}\sum\limits_{a\in B}% \sum\limits_{m\in\mathbb{Z}}\left. :a_{m}a_{n-m}:\right. $, where $B$ is an orthonormal basis of $\mathfrak{p}$ with respect to a certain bilinear form on $\mathfrak{p}$. Thus, $L_{n}^{\mathfrak{p}}$ is a sum of products of elements of $\widehat{\mathfrak{p}}$ (or, more precisely, their actions on $M$). Now, let $m\in\mathbb{Z}$. We have just seen that $L_{n}^{\mathfrak{p}}$ is a sum of products of elements of $\widehat{\mathfrak{p}}$ (or, more precisely, their actions on $M$). Similarly, $L_{m}^{\mathfrak{p}}$ is a sum of products of elements of $\widehat{\mathfrak{p}}$ (or, more precisely, their actions on $M$). Since we know that $L_{n}$ commutes with every element of $\widehat{\mathfrak{p}}$ (due to Theorem \ref{thm.goddardkentolive} \textbf{(b)}), this yields that $L_{n}$ commutes with $L_{m}^{\mathfrak{p}}$. In other words, $\left[ L_{n},L_{m}^{\mathfrak{p}}\right] =0$. Theorem \ref{thm.goddardkentolive} \textbf{(c)} is thus established. \textbf{(a)} Any $n\in\mathbb{Z}$ and $m\in\mathbb{Z}$ satisfy% \begin{align*} & \left[ L_{n},\underbrace{L_{m}}_{=L_{m}^{\mathfrak{g}}-L_{m}% ^{\mathfrak{p}}}\right] \\ & =\left[ L_{n},L_{m}^{\mathfrak{g}}-L_{m}^{\mathfrak{p}}\right] =\left[ L_{n},L_{m}^{\mathfrak{g}}\right] -\underbrace{\left[ L_{n},L_{m}% ^{\mathfrak{p}}\right] }_{\substack{=0\\\text{(by Theorem \ref{thm.goddardkentolive} \textbf{(c)})}}}=\left[ \underbrace{L_{n}}% _{=L_{n}^{\mathfrak{g}}-L_{n}^{\mathfrak{p}}},L_{m}^{\mathfrak{g}}\right] \\ & =\left[ L_{n}^{\mathfrak{g}}-L_{n}^{\mathfrak{p}},L_{m}^{\mathfrak{g}% }\right] =\left[ L_{n}^{\mathfrak{g}},L_{m}^{\mathfrak{g}}\right] -\underbrace{\left[ L_{n}^{\mathfrak{p}},L_{m}^{\mathfrak{g}}\right] }_{\substack{=\left[ L_{n}^{\mathfrak{p}},L_{m}^{\mathfrak{g}}-L_{m}% ^{\mathfrak{p}}\right] +\left[ L_{n}^{\mathfrak{p}},L_{m}^{\mathfrak{p}% }\right] \\\text{(since }L_{m}^{\mathfrak{g}}=\left( L_{m}^{\mathfrak{g}% }-L_{m}^{\mathfrak{p}}\right) +L_{m}^{\mathfrak{p}}\text{)}}}\\ & =\left[ L_{n}^{\mathfrak{g}},L_{m}^{\mathfrak{g}}\right] -\left[ L_{n}^{\mathfrak{p}},\underbrace{L_{m}^{\mathfrak{g}}-L_{m}^{\mathfrak{p}}% }_{=L_{m}}\right] -\left[ L_{n}^{\mathfrak{p}},L_{m}^{\mathfrak{p}}\right] \\ & =\underbrace{\left[ L_{n}^{\mathfrak{g}},L_{m}^{\mathfrak{g}}\right] }_{\substack{=\left( n-m\right) L_{n+m}^{\mathfrak{g}}-\dfrac{n^{3}-n}% {12}c_{\mathfrak{g}}\delta_{n,-m}\\\text{(by Theorem \ref{thm.sugawara} \textbf{(c)})}}}-\underbrace{\left[ L_{n}^{\mathfrak{p}},L_{m}\right] }_{=-\left[ L_{m},L_{n}^{\mathfrak{p}}\right] }-\underbrace{\left[ L_{n}^{\mathfrak{p}},L_{m}^{\mathfrak{p}}\right] }_{\substack{=\left( n-m\right) L_{n+m}^{\mathfrak{p}}-\dfrac{n^{3}-n}{12}c_{\mathfrak{p}}% \delta_{n,-m}\\\text{(by Theorem \ref{thm.sugawara} \textbf{(c)}% ,}\\\text{applied to }\mathfrak{p}\text{ instead of }\mathfrak{g}\text{)}}}\\ & =\left( \left( n-m\right) L_{n+m}^{\mathfrak{g}}-\dfrac{n^{3}-n}% {12}c_{\mathfrak{g}}\delta_{n,-m}\right) +\left[ L_{m},L_{n}^{\mathfrak{p}% }\right] -\left( \left( n-m\right) L_{n+m}^{\mathfrak{p}}-\dfrac{n^{3}% -n}{12}c_{\mathfrak{p}}\delta_{n,-m}\right) \\ & =\left( n-m\right) \underbrace{\left( L_{n+m}^{\mathfrak{g}}% -L_{n+m}^{\mathfrak{p}}\right) }_{=L_{n+m}}-\dfrac{n^{3}-n}{12}\left( c_{\mathfrak{g}}-c_{\mathfrak{p}}\right) \delta_{n,-m}+\underbrace{\left[ L_{m},L_{n}^{\mathfrak{p}}\right] }_{\substack{=0\\\text{(by Theorem \ref{thm.goddardkentolive} \textbf{(c)},}\\\text{applied to }m\text{ and }n\text{ instead of }n\text{ and }m\text{)}}}\\ & =\left( n-m\right) L_{n+m}-\dfrac{n^{3}-n}{12}\left( c_{\mathfrak{g}% }-c_{\mathfrak{p}}\right) \delta_{n,-m}. \end{align*} Hence, $\left( L_{i}\right) _{i\in\mathbb{Z}}$ is a $\operatorname*{Vir}% $-action on $M$ with central charge $c=c_{\mathfrak{g}}-c_{\mathfrak{p}}$. Theorem \ref{thm.goddardkentolive} \textbf{(a)} is thus proven. This completes the proof of Theorem \ref{thm.goddardkentolive}. \begin{example} Let $\mathfrak{a}$ be a simple finite-dimensional Lie algebra. Let $\mathfrak{g}=\mathfrak{a}\oplus\mathfrak{a}$, and let $\mathfrak{p}% =\mathfrak{a}_{\operatorname*{diag}}\subseteq\mathfrak{a}\oplus\mathfrak{a}$ (where $\mathfrak{a}_{\operatorname*{diag}}$ denotes the Lie subalgebra $\left\{ \left( x,x\right) \ \mid\ x\in\mathfrak{a}\right\} $ of $\mathfrak{a}\oplus\mathfrak{a}$). Consider the standard form $\left( \cdot,\cdot\right) $ on $\mathfrak{a}$. Define a symmetric bilinear form on $\mathfrak{a}\oplus\mathfrak{a}$ as the direct sum of the standard forms on $\mathfrak{a}$ and $\mathfrak{a}$. Let $V^{\prime}$ and $V^{\prime\prime}$ be admissible $\widehat{\mathfrak{a}}% $-modules at levels $k^{\prime}$ and $k^{\prime\prime}$. Theorem \ref{thm.sugawara} endows these vector spaces $V^{\prime}$ and $V^{\prime \prime}$ with $\operatorname*{Vir}$-module structures. These $\operatorname*{Vir}$-module structures have central charges $c_{\mathfrak{a}% }^{\prime}=\dfrac{k^{\prime}\dim\mathfrak{a}}{k^{\prime}+h^{\vee}}$ and $c_{\mathfrak{a}}^{\prime\prime}=\dfrac{k^{\prime\prime}\dim\mathfrak{a}% }{k^{\prime\prime}+h^{\vee}}$, respectively (by (\ref{thm.sugawara.simple.c}% )). Let $\left( L_{i}^{\prime}\right) _{i\in\mathbb{Z}}$ and $\left( L_{i}^{\prime\prime}\right) _{i\in\mathbb{Z}}$ denote the actions of $\operatorname*{Vir}$ on these modules. Then, $V^{\prime}\otimes V^{\prime\prime}$ is an admissible $\widehat{\mathfrak{g}}$-module at level $k^{\prime}+k^{\prime\prime}$. Thus, by Theorem \ref{thm.sugawara}, this vector space $V^{\prime}\otimes V^{\prime\prime}$ becomes a $\operatorname*{Vir}$-module. The action $\left( L_{i}^{\mathfrak{g}}\right) _{i\in\mathbb{Z}}$ of $\operatorname*{Vir}$ on this $\operatorname*{Vir}$-module $V^{\prime}\otimes V^{\prime\prime}$ is given by $L_{i}^{\mathfrak{g}}=L_{i}^{\prime}+L_{i}^{\prime\prime}$ (or, more precisely, $L_{i}^{\mathfrak{g}}=L_{i}^{\prime}\otimes\operatorname*{id}% +\operatorname*{id}\otimes L_{i}^{\prime\prime}$). The central charge $c_{\mathfrak{g}}$ of this $\operatorname*{Vir}$-module $V^{\prime}\otimes V^{\prime\prime}$ is \[ c_{\mathfrak{g}}=c_{\mathfrak{a}}^{\prime}+c_{\mathfrak{a}}^{\prime\prime }=\dfrac{k^{\prime}\dim\mathfrak{a}}{k^{\prime}+h^{\vee}}+\dfrac {k^{\prime\prime}\dim\mathfrak{a}}{k^{\prime\prime}+h^{\vee}}. \] Since $\widehat{\mathfrak{p}}=\widehat{\mathfrak{a}}$ acts on $V^{\prime }\otimes V^{\prime\prime}$ by diagonal action, we also get a $\operatorname*{Vir}$-module structure $\left( L_{i}^{\mathfrak{p}}\right) _{i\in\mathbb{Z}}$ on $V^{\prime}\otimes V^{\prime\prime}$ by applying Theorem \ref{thm.sugawara} to $\mathfrak{p}$ instead of $\mathfrak{g}$. The central charge of this $\operatorname*{Vir}$-module is \[ c_{\mathfrak{p}}=\dfrac{k^{\prime}+k^{\prime\prime}}{k^{\prime}+k^{\prime \prime}+h^{\vee}}\dim\mathfrak{a}% \] (since the level of the $\widehat{\mathfrak{p}}$-module $V^{\prime}\otimes V^{\prime\prime}$ is $k^{\prime}+k^{\prime\prime}$). Thus, the central charge $c$ of the $\operatorname*{Vir}$-action on $V^{\prime}\otimes V^{\prime\prime}$ given by Theorem \ref{thm.goddardkentolive} is% \begin{align*} c & =c_{\mathfrak{a}}^{\prime}+c_{\mathfrak{a}}^{\prime\prime}% -c_{\mathfrak{p}}=\dfrac{k^{\prime}\dim\mathfrak{a}}{k^{\prime}+h^{\vee}% }+\dfrac{k^{\prime\prime}\dim\mathfrak{a}}{k^{\prime\prime}+h^{\vee}}% -\dfrac{k^{\prime}+k^{\prime\prime}}{k^{\prime}+k^{\prime\prime}+h^{\vee}}% \dim\mathfrak{a}\\ & =\left( \dfrac{k^{\prime}}{k^{\prime}+h^{\vee}}+\dfrac{k^{\prime\prime}% }{k^{\prime\prime}+h^{\vee}}-\dfrac{k^{\prime}+k^{\prime\prime}}{k^{\prime }+k^{\prime\prime}+h^{\vee}}\right) \dim\mathfrak{a}. \end{align*} We can use this construction to obtain, for every positive integer $m$, a unitary representation of $\operatorname*{Vir}$ with central charge $1-\dfrac{6}{\left( m+2\right) \left( m+3\right) }$: In fact, let $\mathfrak{a}=\mathfrak{sl}_{2}$, so that $h^{\vee}=2$, and let $k^{\prime}=1$ and $k^{\prime\prime}=m$. Then,% \[ c=3\left( \dfrac{1}{3}+\dfrac{m}{m+2}-\dfrac{m+1}{m+3}\right) =1-\dfrac {6}{\left( m+2\right) \left( m+3\right) }. \] So we get unitary representations of $\operatorname*{Vir}$ with central charge $c$ for these values of $c$. \end{example} \subsection{\label{subsect.prelims}Preliminaries to simple and Kac-Moody Lie algebras} Our next goal is defining and studying the Kac-Moody Lie algebras. Before we do this, however, we will recollect some properties of simple finite-dimensional Lie algebras (which are, in some sense, the prototypical Kac-Moody Lie algebras); and yet before that, we show some general results from the theory of Lie algebras which will be used in our later proofs. [This whole Section \ref{subsect.prelims} is written by Darij and aims at covering the gap between introductory courses in Lie algebras and Etingof's class. It states some folklore facts about Lie algebras which will be used later.] \subsubsection{A basic property of \texorpdfstring{$\mathfrak{sl}_{2}$}{sl-2}-modules} We begin with a lemma from the representation theory of $\mathfrak{sl}_{2}$: \begin{lemma} \label{lem.serre-gen.sl2}Let $e$, $f$ and $h$ mean the classical basis elements of $\mathfrak{sl}_{2}$. Let $\lambda\in\mathbb{C}$. We consider any $\mathfrak{sl}_{2}$-module as a $U\left( \mathfrak{sl}_{2}\right) $-module. \textbf{(a)} Let $V$ be an $\mathfrak{sl}_{2}$-module. Let $x\in V$ be such that $ex=0$ and $hx=\lambda x$. Then, every $n\in\mathbb{N}$ satisfies $e^{n}f^{n}x=n!\lambda\left( \lambda-1\right) ...\left( \lambda-n+1\right) x$. \textbf{(b)} Let $V$ be an $\mathfrak{sl}_{2}$-module. Let $x\in V$ be such that $fx=0$ and $hx=\lambda x$. Then, every $n\in\mathbb{N}$ satisfies $f^{n}e^{n}x=n!\lambda\left( \lambda+1\right) ...\left( \lambda+n-1\right) x$. \textbf{(c)} Let $V$ be a finite-dimensional $\mathfrak{sl}_{2}$-module. Let $x$ be a nonzero element of $V$ satisfying $ex=0$ and $hx=\lambda x$. Then, $\lambda\in\mathbb{N}$ and $f^{\lambda+1}x=0$. \end{lemma} \textit{Proof of Lemma \ref{lem.serre-gen.sl2}.} \textbf{(a)} \textit{1st step:} We will see that% \begin{equation} hf^{m}x=\left( \lambda-2m\right) f^{m}x\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N}. \label{pf.serre-gen.sl2.1}% \end{equation} \textit{Proof of (\ref{pf.serre-gen.sl2.1}):} We will prove (\ref{pf.serre-gen.sl2.1}) by induction over $m$: \textit{Induction base:} For $m=0$, we have $hf^{m}x=hf^{0}x=hx=\lambda x$ and $\left( \lambda-2m\right) f^{m}x=\left( \lambda-2\cdot0\right) f^{0}x=\lambda x$, so that $hf^{m}x=\left( \lambda-2m\right) f^{m}x$ holds for $m=0$. In other words, (\ref{pf.serre-gen.sl2.1}) holds for $m=0$. This completes the induction base. \textit{Induction step:} Let $M\in\mathbb{N}$. Assume that (\ref{pf.serre-gen.sl2.1}) holds for $m=M$. We must then prove that (\ref{pf.serre-gen.sl2.1}) holds for $m=M+1$ as well. Since (\ref{pf.serre-gen.sl2.1}) holds for $m=M$, we have $hf^{M}x=\left( \lambda-2M\right) f^{M}x$. Now,% \begin{align*} h\underbrace{f^{M+1}}_{=ff^{M}}x & =\underbrace{hf}_{=fh+\left[ h,f\right] }f^{M}x=\left( fh+\left[ h,f\right] \right) f^{M}x=f\underbrace{hf^{M}% x}_{=\left( \lambda-2M\right) f^{M}x}+\underbrace{\left[ h,f\right] }_{=-2f}f^{M}x\\ & =\left( \lambda-2M\right) \underbrace{ff^{M}}_{=f^{M+1}}% x-2\underbrace{ff^{M}}_{=f^{M+1}}x=\left( \lambda-2M\right) f^{M+1}% x-2f^{M+1}x\\ & =\underbrace{\left( \lambda-2M-2\right) }_{=\lambda-2\left( M+1\right) }f^{M+1}x=\left( \lambda-2\left( M+1\right) \right) f^{M+1}x. \end{align*} Thus, (\ref{pf.serre-gen.sl2.1}) holds for $m=M+1$ as well. This completes the induction step. The induction proof of (\ref{pf.serre-gen.sl2.1}) is thus complete. \textit{2nd step:} We will see that% \begin{equation} ef^{m}x=m\left( \lambda-m+1\right) f^{m-1}x\ \ \ \ \ \ \ \ \ \ \text{for every positive }m\in\mathbb{N}. \label{pf.serre-gen.sl2.2}% \end{equation} \textit{Proof of (\ref{pf.serre-gen.sl2.2}):} We will prove (\ref{pf.serre-gen.sl2.2}) by induction over $m$: \textit{Induction base:} For $m=1$, we have% \[ ef^{m}x=\underbrace{ef^{1}}_{=ef=\left[ e,f\right] +fe}x=\left( \left[ e,f\right] +fe\right) x=\underbrace{\left[ e,f\right] }_{=h}% x+f\underbrace{ex}_{=0}=hx+f0=hx=\lambda x \] and $m\left( \lambda-m+1\right) f^{m-1}x=1\underbrace{\left( \lambda -1+1\right) }_{=\lambda}\underbrace{f^{1-1}}_{=1}x=\lambda x$, so that $ef^{m}x=m\left( \lambda-m+1\right) f^{m-1}x$ holds for $m=1$. In other words, (\ref{pf.serre-gen.sl2.2}) holds for $m=1$. This completes the induction base. \textit{Induction step:} Let $M\in\mathbb{N}$ be positive. Assume that (\ref{pf.serre-gen.sl2.2}) holds for $m=M$. We must then prove that (\ref{pf.serre-gen.sl2.2}) holds for $m=M+1$ as well. Since (\ref{pf.serre-gen.sl2.2}) holds for $m=M$, we have $ef^{M}x=M\left( \lambda-M+1\right) f^{M-1}x$. Now,% \begin{align*} e\underbrace{f^{M+1}}_{=ff^{M}}x & =\underbrace{ef}_{=fe+\left[ e,f\right] }f^{M}x=\left( fe+\left[ e,f\right] \right) f^{M}x=f\underbrace{ef^{M}% }_{=M\left( \lambda-M+1\right) f^{M-1}x}x+\underbrace{\left[ e,f\right] }_{=h}f^{M}x\\ & =M\left( \lambda-M+1\right) \underbrace{ff^{M-1}}_{=f^{M}}% x+\underbrace{hf^{M}x}_{\substack{=\left( \lambda-2M\right) f^{M}% x\\\text{(by (\ref{pf.serre-gen.sl2.1}), applied to }m=M\text{)}}}\\ & =M\left( \lambda-M+1\right) f^{M}x+\left( \lambda-2M\right) f^{M}x=\underbrace{\left( M\left( \lambda-M+1\right) +\left( \lambda-2M\right) \right) }_{=\left( M+1\right) \left( \lambda-\left( M+1\right) +1\right) }f^{M}x\\ & =\left( M+1\right) \left( \lambda-\left( M+1\right) +1\right) f^{M}x. \end{align*} Thus, (\ref{pf.serre-gen.sl2.2}) holds for $m=M+1$ as well. This completes the induction step. The induction proof of (\ref{pf.serre-gen.sl2.2}) is thus complete. \textit{3rd step:} We will see that% \begin{equation} e^{n}f^{n}x=n!\lambda\left( \lambda-1\right) ...\left( \lambda-n+1\right) x\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}. \label{pf.serre-gen.sl2.3}% \end{equation} \textit{Proof of (\ref{pf.serre-gen.sl2.3}):} We will prove (\ref{pf.serre-gen.sl2.3}) by induction over $n$: \textit{Induction base:} For $n=0$, we have $e^{n}f^{n}x=e^{0}f^{0}x=x$ and $n!\lambda\left( \lambda-1\right) ...\left( \lambda-n+1\right) x=\underbrace{0!}_{=1}\underbrace{\lambda\left( \lambda-1\right) ...\left( \lambda-0+1\right) }_{=\left( \text{empty product}\right) =1}x=x$, so that $e^{n}f^{n}x=n!\lambda\left( \lambda-1\right) ...\left( \lambda-n+1\right) x$ holds for $n=0$. In other words, (\ref{pf.serre-gen.sl2.3}) holds for $n=0$. This completes the induction base. \textit{Induction step:} Let $N\in\mathbb{N}$. Assume that (\ref{pf.serre-gen.sl2.3}) holds for $n=N$. We must then prove that (\ref{pf.serre-gen.sl2.3}) holds for $n=N+1$ as well. Since (\ref{pf.serre-gen.sl2.3}) holds for $n=N$, we have $e^{N}% f^{N}x=N!\lambda\left( \lambda-1\right) ...\left( \lambda-N+1\right) x$. Now,% \begin{align*} \underbrace{e^{N+1}}_{=e^{N}e}f^{N+1}x & =e^{N}\underbrace{ef^{N+1}% x}_{\substack{=\left( N+1\right) \left( \lambda-\left( N+1\right) +1\right) f^{\left( N+1\right) -1}x\\\text{(by (\ref{pf.serre-gen.sl2.2}), applied to }m=N+1\text{)}}}=\left( N+1\right) \left( \lambda-\left( N+1\right) +1\right) e^{N}\underbrace{f^{\left( N+1\right) -1}}_{=f^{N}% }x\\ & =\left( N+1\right) \left( \lambda-\left( N+1\right) +1\right) \underbrace{e^{N}f^{N}x}_{=N!\lambda\left( \lambda-1\right) ...\left( \lambda-N+1\right) x}\\ & =\left( N+1\right) \left( \lambda-\left( N+1\right) +1\right) \cdot N!\lambda\left( \lambda-1\right) ...\left( \lambda-N+1\right) x\\ & =\underbrace{\left( \left( N+1\right) \cdot N!\right) }_{=\left( N+1\right) !}\cdot\underbrace{\left( \lambda\left( \lambda-1\right) ...\left( \lambda-N+1\right) \right) \cdot\left( \lambda-\left( N+1\right) +1\right) }_{=\lambda\left( \lambda-1\right) ...\left( \lambda-\left( N+1\right) +1\right) }x\\ & =\left( N+1\right) !\lambda\left( \lambda-1\right) ...\left( \lambda-\left( N+1\right) +1\right) x. \end{align*} Thus, (\ref{pf.serre-gen.sl2.3}) holds for $n=N+1$ as well. This completes the induction step. The induction proof of (\ref{pf.serre-gen.sl2.3}) is thus complete. Lemma \ref{lem.serre-gen.sl2} \textbf{(a)} immediately follows from (\ref{pf.serre-gen.sl2.3}). \textbf{(b)} The proof of Lemma \ref{lem.serre-gen.sl2} \textbf{(b)} is analogous to the proof of Lemma \ref{lem.serre-gen.sl2} \textbf{(a)}. \textbf{(c)} By assumption, $\dim V<\infty$. Now, the endomorphism $h\mid_{V}$ of $V$ has at most $\dim V$ distinct eigenvalues (since an endomorphism of any finite-dimensional vector space $W$ has at most $\dim W$ distinct eigenvalues). From this, it is easy to conclude that $f^{\dim V}% x=0$\ \ \ \ \footnote{\textit{Proof.} Assume the opposite. Then, $f^{\dim V}x\neq0$. \par Now, let $m\in\left\{ 0,1,...,\dim V\right\} $ be arbitrary. We will prove that $\lambda-2m$ is an eigenvalue of $h\mid_{V}$. \par In fact, $m\leq\dim V$, so that $f^{\dim V-m}\left( f^{m}x\right) =f^{\dim V-m+m}x=f^{\dim V}x\neq0$ and thus $f^{m}x\neq0$. Since $hf^{m}x=\left( \lambda-2m\right) f^{m}x$ (by (\ref{pf.serre-gen.sl2.1})), this yields that $f^{m}x$ is a nonzero eigenvector of $h\mid_{V}$ with eigenvalue $\lambda-2m$. Thus, $\lambda-2m$ is an eigenvalue of $h\mid_{V}$. \par Now forget that we fixed $m$. Thus, we have proven that $\lambda-2m$ is an eigenvalue of $h\mid_{V}$ for every $m\in\left\{ 0,1,...,\dim V\right\} $. Thus we have found $\dim V+1$ pairwise distinct eigenvalues of $h\mid_{V}$. This contradicts the fact that $h\mid_{V}$ has at most $\dim V$ distinct eigenvalues. This contradiction shows that our assumption was wrong, qed.}. Thus, there exists a smallest $m\in\mathbb{N}$ satisfying $f^{m}x=0$. Denote this $m$ by $u$. Then, $f^{u}x=0$. Since $f^{0}x=x\neq0$, this $u$ is $\neq0$, so that $f^{u-1}x$ is well-defined. Moreover, $f^{u-1}x\neq0$ (since $u$ is the smallest $m\in\mathbb{N}$ satisfying $f^{m}x=0$). Lemma \ref{lem.serre-gen.sl2} \textbf{(a)} (applied to $n=u$) yields $e^{u}f^{u}x=u!\lambda\left( \lambda-1\right) ...\left( \lambda-u+1\right) x$. Since $e^{u}\underbrace{f^{u}x}_{=0}=0$, this rewrites as $u!\lambda \left( \lambda-1\right) ...\left( \lambda-u+1\right) x=0$. Since $\operatorname*{char}\mathbb{C}=0$, we can divide this equation by $u!$, and obtain $\lambda\left( \lambda-1\right) ...\left( \lambda-u+1\right) x=0$. Since $x\neq0$, this yields $\lambda\left( \lambda-1\right) ...\left( \lambda-u+1\right) =0$. Thus, one of the numbers $\lambda$, $\lambda-1$, $...$, $\lambda-u+1$ must be $0$. In other words, $\lambda\in\left\{ 0,1,...,u-1\right\} $. Hence, $\lambda\in\mathbb{N}$ and $\lambda\leq u-1$. Applying (\ref{pf.serre-gen.sl2.1}) to $m=u-1$, we obtain $hf^{u-1}x=\left( \lambda-2\left( u-1\right) \right) f^{u-1}x$. Denote $\lambda-2\left( u-1\right) $ by $\mu$. Then, $hf^{u-1}x=\underbrace{\left( \lambda-2\left( u-1\right) \right) }_{=\mu}f^{u-1}x=\mu f^{u-1}x$. Also, $ff^{u-1}% x=f^{u}x=0$. Thus, we can apply Lemma \ref{lem.serre-gen.sl2} \textbf{(b)} to $\mu$, $f^{u-1}x$ and $u-1$ instead of $\lambda$, $x$ and $n$. Thus, we obtain% \[ f^{u-1}e^{u-1}f^{u-1}x=\left( u-1\right) !\mu\left( \mu+1\right) ...\left( \mu+\left( u-1\right) -1\right) f^{u-1}x. \] But $\mu=\underbrace{\lambda}_{\leq u-1}-2\left( u-1\right) \leq\left( u-1\right) -2\left( u-1\right) =-\left( u-1\right) $, so that each of the integers $\mu$, $\mu+1$, $...$, $\mu+\left( u-1\right) -1$ is nonzero. Thus, their product $\mu\left( \mu+1\right) ...\left( \mu+\left( u-1\right) -1\right) $ also is $\neq0$. Combined with $\left( u-1\right) !\neq0$, this yields $\left( u-1\right) !\mu\left( \mu+1\right) ...\left( \mu+\left( u-1\right) -1\right) \neq0$. Combined with $f^{u-1}x\neq0$, this yields $\left( u-1\right) !\mu\left( \mu+1\right) ...\left( \mu+\left( u-1\right) -1\right) f^{u-1}x\neq0$. Thus,% \[ f^{u-1}e^{u-1}f^{u-1}x=\left( u-1\right) !\mu\left( \mu+1\right) ...\left( \mu+\left( u-1\right) -1\right) f^{u-1}x\neq0, \] so that $e^{u-1}f^{u-1}x\neq0$. But Lemma \ref{lem.serre-gen.sl2} \textbf{(a)} (applied to $n=u-1$) yields $e^{u-1}f^{u-1}x=\left( u-1\right) !\lambda\left( \lambda-1\right) ...\left( \lambda-\left( u-1\right) +1\right) x$. Thus, \[ \left( u-1\right) !\lambda\left( \lambda-1\right) ...\left( \lambda-\left( u-1\right) +1\right) x=e^{u-1}f^{u-1}x\neq0. \] Hence, $\lambda\left( \lambda-1\right) ...\left( \lambda-\left( u-1\right) +1\right) \neq0$. Hence, $\dbinom{\lambda}{u-1}=\dfrac{1}{\left( u-1\right) !}\underbrace{\lambda\left( \lambda-1\right) ...\left( \lambda-\left( u-1\right) +1\right) }_{\neq0}\neq0$, so that $u-1\leq \lambda$ (because otherwise, we would have $\dbinom{\lambda}{u-1}=0$, contradicting $\dbinom{\lambda}{u-1}\neq0$). Combined with $u-1\geq\lambda$, this yields $u-1=\lambda$. Thus, $u=\lambda+1$. Hence, $f^{u}x=0$ rewrites as $f^{\lambda+1}x=0$. This proves Lemma \ref{lem.serre-gen.sl2} \textbf{(c)}. \subsubsection{\texorpdfstring{$Q$}{Q}-graded Lie algebras} The following generalization of the standard definition of a $\mathbb{Z}% $-graded Lie algebra suggests itself: \begin{definition} \label{def.Q-graded.lie}Let $Q$ be an abelian group, written additively. \textbf{(a)} A $Q$\textit{-graded vector space} will mean a vector space $V$ equipped with a family $\left( V\left[ \alpha\right] \right) _{\alpha\in Q}$ of vector subspaces $V\left[ \alpha\right] $ of $V$ (indexed by elements of $Q$) satisfying $V=\bigoplus\limits_{\alpha\in Q}V\left[ \alpha\right] $. For every $\alpha\in Q$, the subspace $V\left[ \alpha\right] $ is called the $\alpha$\textit{-th homogeneous component} of the $Q$-graded vector space $V$. The family $\left( V\left[ \alpha\right] \right) _{\alpha\in Q}$ is called a $Q$\textit{-grading} on the vector space $V$. \textbf{(b)} A $Q$\textit{-graded Lie algebra} will mean a Lie algebra $\mathfrak{g}$ equipped with a family $\left( \mathfrak{g}\left[ \alpha\right] \right) _{\alpha\in Q}$ of vector subspaces $\mathfrak{g}% \left[ \alpha\right] $ of $\mathfrak{g}$ (indexed by elements of $Q$) satisfying $\mathfrak{g}=\bigoplus\limits_{\alpha\in Q}\mathfrak{g}\left[ \alpha\right] $ and satisfying \[ \left[ \mathfrak{g}\left[ \alpha\right] ,\mathfrak{g}\left[ \beta\right] \right] \subseteq\mathfrak{g}\left[ \alpha+\beta\right] \ \ \ \ \ \ \ \ \ \ \text{for all }\alpha,\beta\in Q\text{.}% \] In this case, $Q$ is called the \textit{root lattice} of this $Q$-graded Lie algebra $\mathfrak{g}$. (This does not mean that $Q$ actually has to be a lattice of roots of $\mathfrak{g}$, or that $Q$ must be related in any way to the roots of $\mathfrak{g}$.) Clearly, any $Q$-graded Lie algebra is a $Q$-graded vector space. Thus, the notion of the $\alpha$-th homogeneous component of a $Q$-graded Lie algebra makes sense for every $\alpha\in Q$. \end{definition} \begin{Convention} Whenever $Q$ is an abelian group, $\alpha$ is an element of $Q$, and $V$ is a $Q$-graded vector space or a $Q$-graded Lie algebra, we will denote the $\alpha$-th homogeneous component of $V$ by $V\left[ \alpha\right] $. \end{Convention} In the context of a $Q$-graded vector space (or Lie algebra) $V$, one often writes $V_{\alpha}$ instead of $V\left[ \alpha\right] $ for the $\alpha$-th homogeneous component of $V$. This notation, however, can sometimes be misunderstood. When a group homomorphism from $Q$ to $\mathbb{Z}$ is given, a $Q$-graded Lie algebra canonically becomes a $\mathbb{Z}$-graded Lie algebra: \begin{proposition} \label{prop.Q-graded.principal}Let $Q$ be an abelian group. Let $\ell :Q\rightarrow\mathbb{Z}$ be a group homomorphism. Let $\mathfrak{g}$ be a $Q$-graded Lie algebra. \textbf{(a)} For every $m\in\mathbb{Z}$, the internal direct sum $\bigoplus\limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =m}}\mathfrak{g}\left[ \alpha\right] $ is well-defined. \textbf{(b)} Denote this internal direct sum $\bigoplus \limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =m}}\mathfrak{g}% \left[ \alpha\right] $ by $\mathfrak{g}_{\left[ m\right] }$. Then, the Lie algebra $\mathfrak{g}$ equipped with the grading $\left( \mathfrak{g}% _{\left[ m\right] }\right) _{m\in\mathbb{Z}}$ is a $\mathbb{Z}$-graded Lie algebra. (This grading $\left( \mathfrak{g}_{\left[ m\right] }\right) _{m\in\mathbb{Z}}$ is called the \textit{principal grading} on $\mathfrak{g}$ induced by the given $Q$-grading on $\mathfrak{g}$ and the map $\ell$.) \end{proposition} \begin{vershort} The proof of this proposition is straightforward and left to the reader. \end{vershort} \begin{verlong} \textit{Proof of Proposition \ref{prop.Q-graded.principal}.} \textbf{(a)} Since $\mathfrak{g}$ is $Q$-graded, we have \[ \mathfrak{g}=\bigoplus\limits_{\alpha\in Q}\mathfrak{g}\left[ \alpha\right] =\bigoplus\limits_{m\in\mathbb{Z}}\bigoplus\limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =m}}\mathfrak{g}\left[ \alpha\right] . \] Thus, the internal direct sum $\bigoplus\limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =m}}\mathfrak{g}\left[ \alpha\right] $ is defined for every $m\in\mathbb{Z}$. This proves Proposition \ref{prop.Q-graded.principal} \textbf{(a)}. \textbf{(b)} We have \[ \mathfrak{g}=\bigoplus\limits_{m\in\mathbb{Z}}\underbrace{\bigoplus \limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =m}}\mathfrak{g}% \left[ \alpha\right] }_{=\mathfrak{g}_{\left[ m\right] }}=\bigoplus \limits_{m\in\mathbb{Z}}\mathfrak{g}_{\left[ m\right] }. \] Also, every $m_{1}\in\mathbb{Z}$ and every $m_{2}\in\mathbb{Z}$ satisfy% \[ \mathfrak{g}_{\left[ m_{1}\right] }\mathfrak{g}_{\left[ m_{2}\right] }\subseteq\mathfrak{g}_{\left[ m_{1}+m_{2}\right] }% \] \footnote{\textit{Proof.} Let $m_{1}\in\mathbb{Z}$ and $m_{2}\in\mathbb{Z}$. Then, the definition of $\mathfrak{g}_{\left[ m_{1}\right] }$ yields \begin{align*} \mathfrak{g}_{\left[ m_{1}\right] } & =\bigoplus\limits_{\substack{\alpha \in Q;\\\ell\left( \alpha\right) =m_{1}}}\mathfrak{g}\left[ \alpha\right] =\sum\limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =m_{1}% }}\mathfrak{g}\left[ \alpha\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since direct sums are sums}\right) \\ & =\sum\limits_{\substack{\beta\in Q;\\\ell\left( \beta\right) =m_{1}% }}\mathfrak{g}\left[ \beta\right] \ \ \ \ \ \ \ \ \ \ \left( \text{here, we renamed the index }\alpha\text{ as }\beta\right) . \end{align*} Also, the definition of $\mathfrak{g}_{\left[ m_{2}\right] }$ yields% \begin{align*} \mathfrak{g}_{\left[ m_{2}\right] } & =\bigoplus\limits_{\substack{\alpha \in Q;\\\ell\left( \alpha\right) =m_{2}}}\mathfrak{g}\left[ \alpha\right] =\sum\limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =m_{2}% }}\mathfrak{g}\left[ \alpha\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since direct sums are sums}\right) \\ & =\sum\limits_{\substack{\gamma\in Q;\\\ell\left( \gamma\right) =m_{2}% }}\mathfrak{g}\left[ \gamma\right] \ \ \ \ \ \ \ \ \ \ \left( \text{here, we renamed the index }\alpha\text{ as }\gamma\right) . \end{align*} Finally, the definition of $\mathfrak{g}_{\left[ m_{1}+m_{2}\right] }$ yields% \[ \mathfrak{g}_{\left[ m_{1}+m_{2}\right] }=\bigoplus\limits_{\substack{\alpha \in Q;\\\ell\left( \alpha\right) =m_{1}+m_{2}}}\mathfrak{g}\left[ \alpha\right] =\sum\limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =m_{1}+m_{2}}}\mathfrak{g}\left[ \alpha\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since direct sums are sums}\right) . \] \par Every $\beta\in Q$ and $\gamma\in Q$ such that $\ell\left( \beta\right) =m_{1}$ and $\ell\left( \gamma\right) =m_{2}$ satisfy% \begin{align*} \ell\left( \beta+\gamma\right) & =\underbrace{\ell\left( \beta\right) }_{=m_{1}}+\underbrace{\ell\left( \gamma\right) }_{=m_{2}}% \ \ \ \ \ \ \ \ \ \ \left( \text{since }\ell\text{ is a }\mathbb{Z}% \text{-module homomorphism}\right) \\ & =m_{1}+m_{2}. \end{align*} Thus, every $\beta\in Q$ and $\gamma\in Q$ such that $\ell\left( \beta\right) =m_{1}$ and $\ell\left( \gamma\right) =m_{2}$ satisfy% \begin{equation} \mathfrak{g}\left[ \beta+\gamma\right] \subseteq\sum \limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =m_{1}+m_{2}% }}\mathfrak{g}\left[ \alpha\right] =\mathfrak{g}_{\left[ m_{1}% +m_{2}\right] }. \label{pf.Q-graded.principal.1}% \end{equation} \par Since $\mathfrak{g}_{\left[ m_{1}\right] }=\sum\limits_{\substack{\beta\in Q;\\\ell\left( \beta\right) =m_{1}}}\mathfrak{g}\left[ \beta\right] $ and $\mathfrak{g}_{\left[ m_{2}\right] }=\sum\limits_{\substack{\gamma\in Q;\\\ell\left( \gamma\right) =m_{2}}}\mathfrak{g}\left[ \gamma\right] $, we have% \begin{align*} \left[ \mathfrak{g}_{\left[ m_{1}\right] },\mathfrak{g}_{\left[ m_{2}\right] }\right] & =\left[ \sum\limits_{\substack{\beta\in Q;\\\ell\left( \beta\right) =m_{1}}}\mathfrak{g}\left[ \beta\right] ,\sum\limits_{\substack{\gamma\in Q;\\\ell\left( \gamma\right) =m_{2}% }}\mathfrak{g}\left[ \gamma\right] \right] =\sum\limits_{\substack{\beta\in Q;\\\ell\left( \beta\right) =m_{1}}}\sum\limits_{\substack{\gamma\in Q;\\\ell\left( \gamma\right) =m_{2}}}\underbrace{\left[ \mathfrak{g}\left[ \beta\right] ,\mathfrak{g}\left[ \gamma\right] \right] }% _{\substack{\subseteq\mathfrak{g}\left[ \beta+\gamma\right] \\\text{(since }\mathfrak{g}\text{ is a }Q\text{-graded Lie algebra)}}}\\ & \subseteq\sum\limits_{\substack{\beta\in Q;\\\ell\left( \beta\right) =m_{1}}}\sum\limits_{\substack{\gamma\in Q;\\\ell\left( \gamma\right) =m_{2}}}\underbrace{\mathfrak{g}\left[ \beta+\gamma\right] }% _{\substack{\subseteq\mathfrak{g}_{\left[ m_{1}+m_{2}\right] }\\\text{(by (\ref{pf.Q-graded.principal.1}), since}\\\ell\left( \beta\right) =m_{1}\text{ and }\ell\left( \gamma\right) =m_{2}\text{)}}}\subseteq \sum\limits_{\substack{\beta\in Q;\\\ell\left( \beta\right) =m_{1}}% }\sum\limits_{\substack{\gamma\in Q;\\\ell\left( \gamma\right) =m_{2}% }}\mathfrak{g}_{\left[ m_{1}+m_{2}\right] }\\ & \subseteq\mathfrak{g}_{\left[ m_{1}+m_{2}\right] }% \ \ \ \ \ \ \ \ \ \ \left( \text{since }\mathfrak{g}_{\left[ m_{1}% +m_{2}\right] }\text{ is a vector space}\right) , \end{align*} qed.}. Combined with the fact that $\mathfrak{g}=\bigoplus\limits_{m\in \mathbb{Z}}\mathfrak{g}_{\left[ m\right] }$, this yields that the Lie algebra $\mathfrak{g}$ equipped with the family $\left( \mathfrak{g}_{\left[ m\right] }\right) _{m\in\mathbb{Z}}$ is a $\mathbb{Z}$-graded Lie algebra. This proves Proposition \ref{prop.Q-graded.principal}. \end{verlong} \subsubsection{A few lemmas on generating subspaces of Lie algebras} We proceed with some facts about generating sets of Lie algebras (free or not): \begin{lemma} \label{lem.generation.1}Let $\mathfrak{g}$ be a Lie algebra, and let $T$ be a vector subspace of $\mathfrak{g}$. Assume that $\mathfrak{g}$ is generated by $T$ as a Lie algebra. Let $U$ be a vector subspace of $\mathfrak{g}$ such that $T\subseteq U$ and $\left[ T,U\right] \subseteq U$. Then, $U=\mathfrak{g}$. \end{lemma} Notice that Lemma \ref{lem.generation.1} is not peculiar to Lie algebras. A similar result holds (for instance) if ``Lie algebra'' is replaced by ``commutative nonunital algebra'' and ``$\left[ T,U\right] $'' is replaced by ``$TU$''. The following proof is written merely for the sake of completeness; intuitively, Lemma \ref{lem.generation.1} should be obvious from the observation that all iterated Lie brackets of elements of $T$ can be written as linear combinations of Lie brackets of the form $\left[ t_{1},\left[ t_{2},\left[ ...,\left[ t_{k-1},t_{k}\right] \right] \right] \right] $ (with $t_{1},t_{2},...,t_{k}\in T$) by applying the Jacobi identity iteratively. \textit{Proof of Lemma \ref{lem.generation.1}.} Define a sequence $\left( T_{n}\right) _{n\geq1}$ of vector subspaces of $\mathfrak{g}$ recursively as follows: Let $T_{1}=T$, and for every positive integer $n$, set $T_{n+1}% =\left[ T,T_{n}\right] $. We have% \begin{equation} \left[ T_{i},T_{j}\right] \subseteq T_{i+j}\ \ \ \ \ \ \ \ \ \ \text{for any positive integers }i\text{ and }j\text{.} \label{pf.generation.1.additivity}% \end{equation} \footnote{\textit{Proof of (\ref{pf.generation.1.additivity}):} We will prove (\ref{pf.generation.1.additivity}) by induction over $i$. \par \textit{Induction base:} For any positive integer $j$, we have $T_{j+1}% =\left[ T,T_{j}\right] $ (by the definition of $T_{j+1}$) and thus $\left[ \underbrace{T_{1}}_{=T},T_{j}\right] =\left[ T,T_{j}\right] =T_{j+1}% =T_{1+j}$. In other words, (\ref{pf.generation.1.additivity}) holds for $i=1$. This completes the induction base. \par \textit{Induction step:} Let $k$ be a positive integer. Assume that (\ref{pf.generation.1.additivity}) is proven for $i=k$. We now will prove (\ref{pf.generation.1.additivity}) for $i=k+1$. \par Since (\ref{pf.generation.1.additivity}) is proven for $i=k$, we have% \begin{equation} \left[ T_{k},T_{j}\right] \subseteq T_{k+j}\ \ \ \ \ \ \ \ \ \ \text{for any positive integer }j\text{.} \label{pf.generation.1.additivity.2}% \end{equation} \par Now, let $j$ be a positive integer. Then, $T_{k+j+1}=\left[ T,T_{k+j}\right] $ (by the definition of $T_{k+j+1}$) and $T_{j+1}=\left[ T,T_{j}\right] $ (by the definition of $T_{j+1}$). Now, any $x\in T$, $y\in T_{k}$ and $z\in T_{j}$ satisfy% \begin{align*} \left[ \left[ x,y\right] ,z\right] & =-\underbrace{\left[ \left[ y,z\right] ,x\right] }_{=-\left[ x,\left[ y,z\right] \right] }-\left[ \underbrace{\left[ z,x\right] }_{=-\left[ x,z\right] },y\right] \ \ \ \ \ \ \ \ \ \ \left( \text{by the Jacobi identity}\right) \\ & =\underbrace{-\left( -\left[ x,\left[ y,z\right] \right] \right) }_{=\left[ x,\left[ y,z\right] \right] }-\underbrace{\left[ -\left[ x,z\right] ,y\right] }_{=-\left[ \left[ x,z\right] ,y\right] =\left[ y,\left[ x,z\right] \right] }=\left[ \underbrace{x}_{\in T},\left[ \underbrace{y}_{\in T_{k}},\underbrace{z}_{\in T_{j}}\right] \right] -\left[ \underbrace{y}_{\in T_{k}},\left[ \underbrace{x}_{\in T}% ,\underbrace{z}_{\in T_{j}}\right] \right] \\ & \in\left[ T,\underbrace{\left[ T_{k},T_{j}\right] }_{\substack{\subseteq T_{k+j}\\\text{(by (\ref{pf.generation.1.additivity.2}))}}}\right] +\left[ T_{k},\underbrace{\left[ T,T_{j}\right] }_{=T_{j+1}}\right] \subseteq \underbrace{\left[ T,T_{k+j}\right] }_{=T_{k+j+1}}+\underbrace{\left[ T_{k},T_{j+1}\right] }_{\substack{\subseteq T_{k+j+1}\\\text{(by (\ref{pf.generation.1.additivity.2}), applied to}\\j+1\text{ instead of }j\text{)}}}\\ & \subseteq T_{k+j+1}+T_{k+j+1}\subseteq T_{k+j+1}\ \ \ \ \ \ \ \ \ \ \left( \text{since }T_{k+j+1}\text{ is a vector space}\right) \\ & =T_{\left( k+1\right) +j}. \end{align*} Hence, $\left[ \left[ T,T_{k}\right] ,T_{j}\right] \subseteq T_{\left( k+1\right) +j}$ (since $T_{\left( k+1\right) +j}$ is a vector space). Since $\left[ T,T_{k}\right] =T_{k+1}$ (by the definition of $T_{k+1}$), this rewrites as $\left[ T_{k+1},T_{j}\right] \subseteq T_{\left( k+1\right) +j}$. Since we have proven this for every positive integer $j$, we have thus proven (\ref{pf.generation.1.additivity}) for $i=k+1$. The induction step is thus complete. This finishes the proof of (\ref{pf.generation.1.additivity}).} Now, let $S$ be the vector subspace $\sum\limits_{i\geq1}T_{i}$ of $\mathfrak{g}$. Then, every positive integer $k$ satisfies $T_{k}\subseteq S$. In particular, $T_{1}\subseteq S$. Since $S=\sum\limits_{i\geq1}T_{i}$ and $S=\sum\limits_{i\geq1}T_{i}=\sum\limits_{j\geq1}T_{j}$, we have% \[ \left[ S,S\right] =\left[ \sum\limits_{i\geq1}T_{i},\sum\limits_{j\geq 1}T_{j}\right] =\sum\limits_{i\geq1}\sum\limits_{j\geq1}\underbrace{\left[ T_{i},T_{j}\right] }_{\substack{\subseteq T_{i+j}\subseteq S\\\text{(since every positive}\\\text{integer }k\text{ satisfies }T_{k}\subseteq S\text{)}% }}\subseteq\sum\limits_{i\geq1}\sum\limits_{j\geq1}S\subseteq S \] (since $S$ is a vector space). Thus, $S$ is a Lie subalgebra of $\mathfrak{g}% $. Since $T=T_{1}\subseteq S$, this yields that $S$ is a Lie subalgebra of $\mathfrak{g}$ containing $T$ as a subset. Since the smallest Lie subalgebra of $\mathfrak{g}$ containing $T$ as a subset is $\mathfrak{g}$ itself (because $\mathfrak{g}$ is generated by $T$ as a Lie algebra), this yields that $S\supseteq\mathfrak{g}$. In other words, $S=\mathfrak{g}$. Now, it is easy to see that% \begin{equation} T_{i}\subseteq U\text{ for every positive integer }i. \label{pf.generation.2}% \end{equation} \footnote{\textit{Proof of (\ref{pf.generation.2}):} We will prove (\ref{pf.generation.2}) by induction over $i$. \par \textit{Induction base:} We have $T_{1}=T\subseteq U$. Thus, (\ref{pf.generation.2}) holds for $i=1$. This completes the induction base. \par \textit{Induction step:} Let $k$ be a positive integer. Assume that (\ref{pf.generation.2}) holds for $i=k$. We now will prove (\ref{pf.generation.2}) for $i=k+1$. \par Since (\ref{pf.generation.2}) holds for $i=k$, we have $T_{k}\subseteq U$. Since $T_{k+1}=\left[ T,T_{k}\right] $ (by the definition of $T_{k+1}$), we have $T_{k+1}=\left[ T,\underbrace{T_{k}}_{\subseteq U}\right] \subseteq\left[ T,U\right] \subseteq U$. In other words, (\ref{pf.generation.2}) holds for $i=k+1$. This completes the induction step. Thus, (\ref{pf.generation.2}) is proven.} Hence,% \[ \mathfrak{g}=S=\sum\limits_{i\geq1}\underbrace{T_{i}}_{\subseteq U}% \subseteq\sum\limits_{i\geq1}U\subseteq U \] (since $U$ is a vector space). Thus, $U=\mathfrak{g}$, and this proves Lemma \ref{lem.generation.1}. The next result is related: \begin{theorem} \label{thm.FreeLie.grading1}Let $\mathfrak{g}$ be a $\mathbb{Z}$-graded Lie algebra. Let $T$ be a vector subspace of $\mathfrak{g}\left[ 1\right] $ such that $\mathfrak{g}$ is generated by $T$ as a Lie algebra. Then, $T=\mathfrak{g}\left[ 1\right] $. \end{theorem} \begin{vershort} The proof of this theorem proceeds by defining the sequence $\left( T_{n}\right) _{n\geq1}$ as in the proof of Lemma \ref{lem.generation.1}, and showing that $T_{i}\subseteq\mathfrak{g}\left[ i\right] $ for every positive integer $i$. The details are left to the reader. \end{vershort} \begin{verlong} \textit{Proof of Theorem \ref{thm.FreeLie.grading1}.} Define a sequence $\left( T_{n}\right) _{n\geq1}$ of vector subspaces of $\mathfrak{g}$ recursively as follows: Let $T_{1}=T$, and for every positive integer $n$, set $T_{n+1}=\left[ T,T_{n}\right] $. Let $S$ be the vector subspace $\sum\limits_{i\geq1}T_{i}$ of $\mathfrak{g}$. Just as in the proof of Lemma \ref{lem.generation.1}, we can see that $S=\mathfrak{g}$. Let $\pi_{1}$ be the canonical projection from the graded vector space $\mathfrak{g}$ on its $1$-th homogeneous component $\mathfrak{g}\left[ 1\right] $. Then, $\pi_{1}$ sends every homogeneous component of $\mathfrak{g}$ other than $\mathfrak{g}\left[ 1\right] $ to $0$. In other words, $\pi_{1}$ sends $\mathfrak{g}\left[ i\right] $ to $0$ for every integer $i\neq1$. In other words,% \begin{equation} \text{every integer }i\neq1\text{ satisfies }\pi_{1}\left( \mathfrak{g}% \left[ i\right] \right) =0. \label{pf.FreeLie.grading1}% \end{equation} On the other hand, since $\pi_{1}$ is a projection on $\mathfrak{g}\left[ 1\right] $, we have $\mathfrak{g}\left[ 1\right] =\pi_{1}\left( \mathfrak{g}\right) $. Since $\pi_{1}$ is a projection on $\mathfrak{g}\left[ 1\right] $, we have% \begin{equation} \pi_{1}\left( x\right) =x\ \ \ \ \ \ \ \ \ \ \text{for every }% x\in\mathfrak{g}\left[ 1\right] . \label{pf.FreeLie.grading0}% \end{equation} Now, it is easy to see that% \begin{equation} T_{i}\subseteq\mathfrak{g}\left[ i\right] \ \ \ \ \ \ \ \ \ \ \text{for every positive integer }i. \label{pf.FreeLie.grading2}% \end{equation} \footnote{\textit{Proof of (\ref{pf.FreeLie.grading2}):} We will prove (\ref{pf.FreeLie.grading2}) by induction over $i$. \par \textit{Induction base:} We have $T_{1}=T\subseteq\mathfrak{g}\left[ 1\right] $. Thus, (\ref{pf.FreeLie.grading2}) is proven for $i=1$. This completes the induction base. \par \textit{Induction step:} Let $n$ be a positive integer. Assume that (\ref{pf.FreeLie.grading2}) holds for $i=n$. We now must prove that (\ref{pf.FreeLie.grading2}) also holds for $i=n+1$. \par Since (\ref{pf.FreeLie.grading2}) holds for $i=n$, we have $T_{n}% \subseteq\mathfrak{g}\left[ n\right] $. By the definition of $T_{n+1}$, we have \begin{align*} T_{n+1} & =\left[ \underbrace{T}_{\subseteq\mathfrak{g}\left[ 1\right] },\underbrace{T_{n}}_{\subseteq\mathfrak{g}\left[ n\right] }\right] \subseteq\left[ \mathfrak{g}\left[ 1\right] ,\mathfrak{g}\left[ n\right] \right] \subseteq\mathfrak{g}\left[ 1+n\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since }\mathfrak{g}\text{ is a }\mathbb{Z}\text{-graded Lie algebra}\right) \\ & =\mathfrak{g}\left[ n+1\right] . \end{align*} Thus, (\ref{pf.FreeLie.grading2}) also holds for $i=n+1$. This completes the induction step. The induction proof of (\ref{pf.FreeLie.grading2}) is thus complete.} Hence, for every positive integer $i\geq2$, we have% \[ \pi_{1}\left( \underbrace{T_{i}}_{\subseteq\mathfrak{g}\left[ i\right] }\right) \subseteq\pi_{1}\left( \mathfrak{g}\left[ i\right] \right) =0 \] (by (\ref{pf.FreeLie.grading1}), since $i\neq1$). Hence,% \begin{align*} \mathfrak{g}\left[ 1\right] & =\pi_{1}\left( \underbrace{\mathfrak{g}% }_{=S=\sum\limits_{i\geq1}T_{i}}\right) =\pi_{1}\left( \sum\limits_{i\geq 1}T_{i}\right) =\sum\limits_{i\geq1}\pi_{1}\left( T_{i}\right) =\pi _{1}\left( T_{1}\right) +\sum\limits_{i\geq2}\underbrace{\pi_{1}\left( T_{i}\right) }_{\substack{=0\\\text{(since }i\geq2\text{)}}}\\ & =\pi_{1}\left( T_{1}\right) +\underbrace{\sum\limits_{i\geq2}0}_{=0}% =\pi_{1}\left( T_{1}\right) =\left\{ \underbrace{\pi_{1}\left( x\right) }_{\substack{=x\\\text{(by (\ref{pf.FreeLie.grading0}), since}\\x\in T_{1}=T\subseteq\mathfrak{g}\left[ 1\right] \text{)}}}\ \mid\ x\in T_{1}\right\} =\left\{ x\ \mid\ x\in T_{1}\right\} =T_{1}=T. \end{align*} This proves Theorem \ref{thm.FreeLie.grading1}. \end{verlong} Generating subspaces can help in proving that Lie algebra homomorphisms are $Q$-graded: \begin{proposition} \label{prop.generation.Q-gr}Let $\mathfrak{g}$ and $\mathfrak{h}$ be two $Q$-graded Lie algebras. Let $T$ be a $Q$-graded vector subspace of $\mathfrak{g}$. Assume that $\mathfrak{g}$ is generated by $T$ as a Lie algebra. Let $f:\mathfrak{g}\rightarrow\mathfrak{h}$ be a Lie algebra homomorphism. Assume that $f\mid_{T}:T\rightarrow\mathfrak{h}$ is a $Q$-graded map. Then, the map $f$ is $Q$-graded. \end{proposition} \begin{vershort} The proof of this is left to the reader. \end{vershort} \begin{verlong} \textit{Proof of Proposition \ref{prop.generation.Q-gr}.} For every $\alpha\in Q$, let $P_{\alpha}$ be the vector subspace $\left( \mathfrak{g}\left[ \alpha\right] \right) \cap f^{-1}\left( \mathfrak{h}\left[ \alpha\right] \right) $ of $\mathfrak{g}$. Then,% \begin{equation} \left[ P_{\alpha},P_{\beta}\right] \subseteq P_{\alpha+\beta}% \ \ \ \ \ \ \ \ \ \ \text{for any }\alpha\in Q\text{ and }\beta\in Q. \label{pf.generation.Q-gr.PaPb}% \end{equation} \footnote{\textit{Proof of (\ref{pf.generation.Q-gr.PaPb}):} Let $\alpha\in Q$ and $\beta\in Q$. Let $x\in P_{\alpha}$ and $y\in P_{\beta}$. Then, $x\in P_{\alpha}=\left( \mathfrak{g}\left[ \alpha\right] \right) \cap f^{-1}\left( \mathfrak{h}\left[ \alpha\right] \right) $ (by the definition of $P_{\alpha}$), so that $x\in\mathfrak{g}\left[ \alpha\right] $ and $x\in f^{-1}\left( \mathfrak{h}\left[ \alpha\right] \right) $. Also, $y\in P_{\beta}=\left( \mathfrak{g}\left[ \beta\right] \right) \cap f^{-1}\left( \mathfrak{h}\left[ \beta\right] \right) $ (by the definition of $P_{\beta}$), so that $y\in\mathfrak{g}\left[ \beta\right] $ and $y\in f^{-1}\left( \mathfrak{h}\left[ \beta\right] \right) $. From $x\in f^{-1}\left( \mathfrak{h}\left[ \alpha\right] \right) $, we obtain $f\left( x\right) \in\mathfrak{h}\left[ \alpha\right] $. From $y\in f^{-1}\left( \mathfrak{h}\left[ \beta\right] \right) $, we get $f\left( y\right) \in\mathfrak{h}\left[ \beta\right] $. Now,% \begin{align*} f\left( \left[ x,y\right] \right) & =\left[ \underbrace{f\left( x\right) }_{\in\mathfrak{h}\left[ \alpha\right] },\underbrace{f\left( y\right) }_{\in\mathfrak{h}\left[ \beta\right] }\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since }f\text{ is a Lie algebra homomorphism}\right) \\ & \in\left[ \mathfrak{h}\left[ \alpha\right] ,\mathfrak{h}\left[ \beta\right] \right] \subseteq\mathfrak{h}\left[ \alpha+\beta\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since }\mathfrak{h}\text{ is a }Q\text{-graded Lie algebra}\right) , \end{align*} and thus $\left[ x,y\right] \in f^{-1}\left( \mathfrak{h}\left[ \alpha+\beta\right] \right) $. Combined with% \[ \left[ \underbrace{x}_{\in\mathfrak{g}\left[ \alpha\right] },\underbrace{y}% _{\in\mathfrak{g}\left[ \beta\right] }\right] \in\left[ \mathfrak{g}% \left[ \alpha\right] ,\mathfrak{g}\left[ \beta\right] \right] \subseteq\mathfrak{g}\left[ \alpha+\beta\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since }\mathfrak{g}\text{ is a }Q\text{-graded Lie algebra}\right) ; \] this yields $\left[ x,y\right] \in\left( \mathfrak{g}\left[ \alpha +\beta\right] \right) \cap f^{-1}\left( \mathfrak{h}\left[ \alpha +\beta\right] \right) $. But since $P_{\alpha+\beta}=\left( \mathfrak{g}% \left[ \alpha+\beta\right] \right) \cap f^{-1}\left( \mathfrak{h}\left[ \alpha+\beta\right] \right) $ (by the definition of $P_{\alpha+\beta}$), this rewrites as $\left[ x,y\right] \in P_{\alpha+\beta}$. \par Now forget that we fixed $x$ and $y$. We thus have proven that every $x\in P_{\alpha}$ and $y\in P_{\beta}$ satisfy $\left[ x,y\right] \in P_{\alpha+\beta}$. Since $P_{\alpha+\beta}$ is a vector space, this yields $\left[ P_{\alpha},P_{\beta}\right] \subseteq P_{\alpha+\beta}$. This proves (\ref{pf.generation.Q-gr.PaPb}).} Now, let $P$ be the vector subspace $\sum\limits_{\alpha\in Q}P_{\alpha}$ of $\mathfrak{g}$. Then,% \begin{equation} P_{\alpha}\subseteq P\text{ for every }\alpha\in Q\text{.} \label{pf.generation.Q-gr.tauto}% \end{equation} But since $P=\sum\limits_{\alpha\in Q}P_{\alpha}$ and $P=\sum\limits_{\alpha \in Q}P_{\alpha}=\sum\limits_{\beta\in Q}P_{\beta}$ (here, we renamed the summation index $\alpha$ as $\beta$), we have% \begin{align*} \left[ P,P\right] & =\left[ \sum\limits_{\alpha\in Q}P_{\alpha}% ,\sum\limits_{\beta\in Q}P_{\beta}\right] =\sum\limits_{\alpha\in Q}% \sum\limits_{\beta\in Q}\underbrace{\left[ P_{\alpha},P_{\beta}\right] }_{\substack{\subseteq P_{\alpha+\beta}\\\text{(by (\ref{pf.generation.Q-gr.PaPb}))}}}\ \ \ \ \ \ \ \ \ \ \left( \text{since the Lie bracket is bilinear}\right) \\ & \subseteq\sum\limits_{\alpha\in Q}\sum\limits_{\beta\in Q}% \underbrace{P_{\alpha+\beta}}_{\substack{\subseteq P\\\text{(by (\ref{pf.generation.Q-gr.tauto}), applied to}\\\alpha+\beta\text{ instead of }\alpha\text{)}}}\subseteq\sum\limits_{\alpha\in Q}\sum\limits_{\beta\in Q}P\subseteq P\ \ \ \ \ \ \ \ \ \ \left( \text{since }P\text{ is a vector space}\right) . \end{align*} As a consequence, $P$ is a Lie subalgebra of $\mathfrak{g}$. Since $T$ is a $Q$-graded vector subspace, we have $T=\bigoplus\limits_{\alpha \in Q}T\left[ \alpha\right] $, and every $\alpha\in Q$ satisfies $T\left[ \alpha\right] \subseteq\mathfrak{g}\left[ \alpha\right] $. Now, $T\subseteq P$\ \ \ \ \footnote{\textit{Proof.} Let $\alpha\in Q$. Let $x\in T\left[ \alpha\right] $. Then, $\left( f\mid_{T}\right) \left( x\right) \in\mathfrak{h}\left[ \alpha\right] $ (since $f\mid_{T}$ is $Q$-graded). Thus, $f\left( x\right) =\left( f\mid_{T}\right) \left( x\right) \in\mathfrak{h}\left[ \alpha\right] $, so that $x\in f^{-1}\left( \mathfrak{h}\left[ \alpha\right] \right) $. Combined with $x\in T\left[ \alpha\right] \subseteq\mathfrak{g}\left[ \alpha\right] $, this yields $x\in\left( \mathfrak{g}\left[ \alpha\right] \right) \cap f^{-1}\left( \mathfrak{h}\left[ \alpha\right] \right) =P_{\alpha}$. \par Now forget that we fixed $x$. We thus have proven that every $x\in T\left[ \alpha\right] $ satisfies $x\in P_{\alpha}$. In other words, $T\left[ \alpha\right] \subseteq P_{\alpha}$. \par Now forget that we fixed $\alpha$. We thus have proven that $T\left[ \alpha\right] \subseteq P_{\alpha}$ for every $\alpha\in Q$. But \begin{align*} T & =\bigoplus\limits_{\alpha\in Q}T\left[ \alpha\right] =\sum \limits_{\alpha\in Q}\underbrace{T\left[ \alpha\right] }_{\subseteq P_{\alpha}}\ \ \ \ \ \ \ \ \ \ \left( \text{since direct sums are sums}\right) \\ & \subseteq\sum\limits_{\alpha\in Q}P_{\alpha}=P, \end{align*} qed.}. Hence, $P$ is a Lie subalgebra of $\mathfrak{g}$ containing $T$ as a subset. But since every Lie subalgebra of $\mathfrak{g}$ containing $T$ as a subset must be $\mathfrak{g}$ (because $\mathfrak{g}$ is generated by $T$ as a Lie algebra), this yields that $P=\mathfrak{g}$. Now, let $\beta\in Q$ be arbitrary. Let $x\in\mathfrak{g}\left[ \beta\right] $. Then, $x\in\mathfrak{g}\left[ \beta\right] \subseteq\mathfrak{g}% =P=\sum\limits_{\alpha\in Q}P_{\alpha}=P_{\beta}+\sum\limits_{\substack{\alpha \in Q;\\\alpha\neq\beta}}P_{\alpha}$. Hence, there exist some $y\in P_{\beta}$ and some $z\in\sum\limits_{\substack{\alpha\in Q;\\\alpha\neq\beta}}P_{\alpha }$ such that $x=y+z$. Consider these $y$ and $z$. From $x=y+z$, we obtain $x-y=z$. But since $\mathfrak{g}$ is $Q$-graded, we have \[ \mathfrak{g}=\bigoplus\limits_{\alpha\in Q}\mathfrak{g}\left[ \alpha\right] =\left( \mathfrak{g}\left[ \beta\right] \right) \oplus \underbrace{\bigoplus\limits_{\substack{\alpha\in Q;\\\alpha\neq\beta }}\mathfrak{g}\left[ \alpha\right] }_{\substack{=\sum \limits_{\substack{\alpha\in Q;\\\alpha\neq\beta}}\mathfrak{g}\left[ \alpha\right] \\\text{(since direct sums are sums)}}}=\left( \mathfrak{g}% \left[ \beta\right] \right) \oplus\left( \sum\limits_{\substack{\alpha\in Q;\\\alpha\neq\beta}}\mathfrak{g}\left[ \alpha\right] \right) . \] Thus, the internal direct sum $\left( \mathfrak{g}\left[ \beta\right] \right) \oplus\left( \sum\limits_{\substack{\alpha\in Q;\\\alpha\neq\beta }}\mathfrak{g}\left[ \alpha\right] \right) $ is well-defined, so that $\left( \mathfrak{g}\left[ \beta\right] \right) \cap\left( \sum \limits_{\substack{\alpha\in Q;\\\alpha\neq\beta}}\mathfrak{g}\left[ \alpha\right] \right) =0$. By the definition of $P_{\beta}$, we have $P_{\beta}=\left( \mathfrak{g}% \left[ \beta\right] \right) \cap f^{-1}\left( \mathfrak{h}\left[ \beta\right] \right) \subseteq\mathfrak{g}\left[ \beta\right] $. Hence, $y\in P_{\beta}\subseteq\mathfrak{g}\left[ \beta\right] $. Combined with $x\in\mathfrak{g}\left[ \beta\right] $, this yields $x-y\in\mathfrak{g}% \left[ \beta\right] -\mathfrak{g}\left[ \beta\right] \subseteq \mathfrak{g}\left[ \beta\right] $ (since $\mathfrak{g}\left[ \beta\right] $ is a vector space). Since $x-y=z$, this rewrites as $z\in\mathfrak{g}\left[ \beta\right] $. Combined with $z\in\sum\limits_{\substack{\alpha\in Q;\\\alpha\neq\beta}}\underbrace{P_{\alpha}}_{\substack{\subseteq \mathfrak{g}\left[ \alpha\right] \\\text{(this is proven in the same}\\\text{way as we showed }P_{\beta}\subseteq\mathfrak{g}\left[ \beta\right] \text{)}}}\subseteq\sum\limits_{\substack{\alpha\in Q;\\\alpha\neq\beta}}\mathfrak{g}\left[ \alpha\right] $, this yields $z\in\left( \mathfrak{g}\left[ \beta\right] \right) \cap\left( \sum\limits_{\substack{\alpha\in Q;\\\alpha\neq\beta}}\mathfrak{g}\left[ \alpha\right] \right) =0$. Hence, $z=0$. Thus, $x-y=z=0$, so that $x=y\in P_{\beta}=\left( \mathfrak{g}\left[ \beta\right] \right) \cap f^{-1}\left( \mathfrak{h}\left[ \beta\right] \right) \subseteq f^{-1}\left( \mathfrak{h}\left[ \beta\right] \right) $, hence $f\left( x\right) \in\mathfrak{h}\left[ \beta\right] $. Now forget that we fixed $x$. We thus have proven that $f\left( x\right) \in\mathfrak{h}\left[ \beta\right] $ for every $x\in\mathfrak{g}\left[ \beta\right] $. In other words, $f\left( \mathfrak{g}\left[ \beta\right] \right) \subseteq\mathfrak{h}\left[ \beta\right] $. Now forget that we fixed $\beta$. We thus have shown that $f\left( \mathfrak{g}\left[ \beta\right] \right) \subseteq\mathfrak{h}\left[ \beta\right] $ for every $\beta\in Q$.\ In other words, the map $f$ is $Q$-graded. This proves Proposition \ref{prop.generation.Q-gr}. \end{verlong} Next, a result on free Lie algebras: \begin{proposition} \label{prop.Ufree}Let $V$ be a vector space. We let $\operatorname*{FreeLie}V$ denote the free Lie algebra on the vector space $V$ (not on the set $V$), and let $T\left( V\right) $ denote the tensor algebra of $V$. Then, there exists a canonical algebra isomorphism $U\left( \operatorname*{FreeLie}V\right) \rightarrow T\left( V\right) $, which commutes with the canonical injections of $V$ into $U\left( \operatorname*{FreeLie}V\right) $ and into $T\left( V\right) $. \end{proposition} We are going to prove Proposition \ref{prop.Ufree} by combining the universal properties of the universal enveloping algebra, the free Lie algebra, and the tensor algebra. Let us first formulate these properties. First, the universal property of the universal enveloping algebra: \begin{theorem} \label{thm.universal.U}Let $\mathfrak{g}$ be a Lie algebra. We denote by $\iota_{\mathfrak{g}}^{U}:\mathfrak{g}\rightarrow U\left( \mathfrak{g}% \right) $ the canonical map from $\mathfrak{g}$ into $U\left( \mathfrak{g}% \right) $. (This map $\iota_{\mathfrak{g}}^{U}$ is injective by the Poincar\'{e}-Birkhoff-Witt theorem, but this is not relevant to the current theorem.) For any algebra $B$ and any Lie algebra homomorphism $f:\mathfrak{g}% \rightarrow B$ (where the Lie algebra structure on $B$ is defined by the commutator of the multiplication of $B$), there exists a unique algebra homomorphism $F:U\left( \mathfrak{g}\right) \rightarrow B$ satisfying $f=F\circ\iota_{\mathfrak{g}}^{U}$. \end{theorem} Next, the universal property of the free Lie algebra: \begin{theorem} \label{thm.universal.FreeLie}Let $V$ be a vector space. We denote by $\iota_{V}^{\operatorname*{FreeLie}}:V\rightarrow\operatorname*{FreeLie}V$ the canonical map from $V$ into $\operatorname*{FreeLie}V$. (The construction of $\operatorname*{FreeLie}V$ readily shows that this map $\iota_{V}% ^{\operatorname*{FreeLie}}$ is injective.) For any Lie algebra $\mathfrak{h}$ and any linear map $f:V\rightarrow\mathfrak{h}$, there exists a unique Lie algebra homomorphism $F:\operatorname*{FreeLie}V\rightarrow\mathfrak{h}$ satisfying $f=F\circ\iota_{V}^{\operatorname*{FreeLie}}$. \end{theorem} Finally, the universal property of the tensor algebra: \begin{theorem} \label{thm.universal.tensor}Let $V$ be a vector space. We denote by $\iota _{V}^{T}:V\rightarrow T\left( V\right) $ the canonical map from $V$ into $T\left( V\right) $. (This map $\iota_{V}^{T}$ is known to be injective.) For any algebra $B$ and any linear map $f:V\rightarrow B$, there exists a unique algebra homomorphism $F:T\left( V\right) \rightarrow B$ satisfying $f=F\circ\iota_{V}^{T}$. \end{theorem} \textit{Proof of Proposition \ref{prop.Ufree}.} The algebra $T\left( V\right) $ canonically becomes a Lie algebra (by defining the Lie bracket on $T\left( V\right) $ as the commutator of the multiplication). Similarly, the algebra $U\left( \operatorname*{FreeLie}V\right) $ becomes a Lie algebra. Applying Theorem \ref{thm.universal.FreeLie} to $\mathfrak{h}=T\left( V\right) $ and $f=\iota_{V}^{T}$, we obtain that there exists a unique Lie algebra homomorphism $F:\operatorname*{FreeLie}V\rightarrow T\left( V\right) $ satisfying $\iota_{V}^{T}=F\circ\iota_{V}^{\operatorname*{FreeLie}}$. Denote this Lie algebra homomorphism $F$ by $h$. Then, $h:\operatorname*{FreeLie}% V\rightarrow T\left( V\right) $ is a Lie algebra homomorphism satisfying $\iota_{V}^{T}=h\circ\iota_{V}^{\operatorname*{FreeLie}}$. Applying Theorem \ref{thm.universal.U} to $\mathfrak{g}% =\operatorname*{FreeLie}V$, $B=T\left( V\right) $ and $f=h$, we obtain that there exists a unique algebra homomorphism $F:U\left( \operatorname*{FreeLie}% V\right) \rightarrow T\left( V\right) $ satisfying $h=F\circ\iota _{\operatorname*{FreeLie}V}^{U}$. Denote this algebra homomorphism $F$ by $\alpha$. Then, $\alpha:U\left( \operatorname*{FreeLie}V\right) \rightarrow T\left( V\right) $ is an algebra homomorphism satisfying $h=\alpha\circ \iota_{\operatorname*{FreeLie}V}^{U}$. Applying Theorem \ref{thm.universal.tensor} to $B=U\left( \operatorname*{FreeLie}V\right) $ and $f=\iota_{\operatorname*{FreeLie}V}% ^{U}\circ\iota_{V}^{\operatorname*{FreeLie}}$, we obtain that there exists a unique algebra homomorphism $F:T\left( V\right) \rightarrow U\left( \operatorname*{FreeLie}V\right) $ satisfying $\iota_{\operatorname*{FreeLie}% V}^{U}\circ\iota_{V}^{\operatorname*{FreeLie}}=F\circ\iota_{V}^{T}$. Denote this algebra homomorphism $F$ by $\beta$. Then, $\beta:T\left( V\right) \rightarrow U\left( \operatorname*{FreeLie}V\right) $ is an algebra homomorphism satisfying $\iota_{\operatorname*{FreeLie}V}^{U}\circ\iota _{V}^{\operatorname*{FreeLie}}=\beta\circ\iota_{V}^{T}$. Both $\alpha$ and $\beta$ are algebra homomorphisms, and therefore Lie algebra homomorphisms. Also, $\iota_{\operatorname*{FreeLie}V}^{U}$ is a Lie algebra homomorphism. We have% \[ \beta\circ\underbrace{\alpha\circ\iota_{\operatorname*{FreeLie}V}^{U}}% _{=h}\circ\iota_{V}^{\operatorname*{FreeLie}}=\beta\circ\underbrace{h\circ \iota_{V}^{\operatorname*{FreeLie}}}_{=\iota_{V}^{T}}=\beta\circ\iota_{V}% ^{T}=\iota_{\operatorname*{FreeLie}V}^{U}\circ\iota_{V}% ^{\operatorname*{FreeLie}}% \] and% \[ \alpha\circ\underbrace{\beta\circ\iota_{V}^{T}}_{=\iota _{\operatorname*{FreeLie}V}^{U}\circ\iota_{V}^{\operatorname*{FreeLie}}% }=\underbrace{\alpha\circ\iota_{\operatorname*{FreeLie}V}^{U}}_{=h}\circ \iota_{V}^{\operatorname*{FreeLie}}=h\circ\iota_{V}^{\operatorname*{FreeLie}% }=\iota_{V}^{T}. \] Now, applying Theorem \ref{thm.universal.FreeLie} to $\mathfrak{h}=U\left( \operatorname*{FreeLie}V\right) $ and $f=\iota_{\operatorname*{FreeLie}V}% ^{U}\circ\iota_{V}^{\operatorname*{FreeLie}}$, we obtain that there exists a unique Lie algebra homomorphism $F:\operatorname*{FreeLie}V\rightarrow U\left( \operatorname*{FreeLie}V\right) $ satisfying $\iota _{\operatorname*{FreeLie}V}^{U}\circ\iota_{V}^{\operatorname*{FreeLie}}% =F\circ\iota_{V}^{\operatorname*{FreeLie}}$. Thus, any two Lie algebra homomorphisms $F:\operatorname*{FreeLie}V\rightarrow U\left( \operatorname*{FreeLie}V\right) $ satisfying $\iota_{\operatorname*{FreeLie}% V}^{U}\circ\iota_{V}^{\operatorname*{FreeLie}}=F\circ\iota_{V}% ^{\operatorname*{FreeLie}}$ must be equal. Since $\beta\circ\alpha\circ \iota_{\operatorname*{FreeLie}V}^{U}$ and $\iota_{\operatorname*{FreeLie}% V}^{U}$ are two such Lie algebra homomorphisms (because we know that $\beta\circ\alpha\circ\iota_{\operatorname*{FreeLie}V}^{U}\circ\iota _{V}^{\operatorname*{FreeLie}}=\iota_{\operatorname*{FreeLie}V}^{U}\circ \iota_{V}^{\operatorname*{FreeLie}}$ and clearly $\iota _{\operatorname*{FreeLie}V}^{U}\circ\iota_{V}^{\operatorname*{FreeLie}}% =\iota_{\operatorname*{FreeLie}V}^{U}\circ\iota_{V}^{\operatorname*{FreeLie}}% $), this yields that $\beta\circ\alpha\circ\iota_{\operatorname*{FreeLie}% V}^{U}$ and $\iota_{\operatorname*{FreeLie}V}^{U}$ must be equal. In other words,% \[ \beta\circ\alpha\circ\iota_{\operatorname*{FreeLie}V}^{U}=\iota _{\operatorname*{FreeLie}V}^{U}. \] Next, applying Theorem \ref{thm.universal.U} to $\mathfrak{g}% =\operatorname*{FreeLie}V$, $B=U\left( \operatorname*{FreeLie}V\right) $ and $f=\iota_{\operatorname*{FreeLie}V}^{U}$, we obtain that there exists a unique algebra homomorphism $F:U\left( \operatorname*{FreeLie}V\right) \rightarrow U\left( \operatorname*{FreeLie}V\right) $ satisfying $\iota _{\operatorname*{FreeLie}V}^{U}=F\circ\iota_{\operatorname*{FreeLie}V}^{U}$. Thus, any two algebra homomorphisms $F:U\left( \operatorname*{FreeLie}% V\right) \rightarrow U\left( \operatorname*{FreeLie}V\right) $ satisfying $\iota_{\operatorname*{FreeLie}V}^{U}=F\circ\iota_{\operatorname*{FreeLie}% V}^{U}$ must be equal. Since $\beta\circ\alpha$ and $\operatorname*{id}% \nolimits_{U\left( \operatorname*{FreeLie}V\right) }$ are two such algebra homomorphisms (because $\beta\circ\alpha\circ\iota_{\operatorname*{FreeLie}% V}^{U}=\iota_{\operatorname*{FreeLie}V}^{U}$ and $\operatorname*{id}% \nolimits_{U\left( \operatorname*{FreeLie}V\right) }\circ\iota _{\operatorname*{FreeLie}V}^{U}=\iota_{\operatorname*{FreeLie}V}^{U}$), this yields that $\beta\circ\alpha$ and $\operatorname*{id}\nolimits_{U\left( \operatorname*{FreeLie}V\right) }$ must be equal. Thus,% \[ \beta\circ\alpha=\operatorname*{id}\nolimits_{U\left( \operatorname*{FreeLie}% V\right) }. \] On the other hand, applying Theorem \ref{thm.universal.tensor} to $B=T\left( V\right) $ and $f=\iota_{V}^{T}$, we obtain that there exists a unique algebra homomorphism $F:T\left( V\right) \rightarrow T\left( V\right) $ satisfying $\iota_{V}^{T}=F\circ\iota_{V}^{T}$. Therefore, any two algebra homomorphisms $F:T\left( V\right) \rightarrow T\left( V\right) $ satisfying $\iota_{V}^{T}=F\circ\iota_{V}^{T}$ must be equal. Since $\alpha\circ\beta$ and $\operatorname*{id}\nolimits_{T\left( V\right) }$ are two such algebra homomorphisms (because we know that $\alpha\circ\beta \circ\iota_{V}^{T}=\iota_{V}^{T}$ and $\operatorname*{id}\nolimits_{T\left( V\right) }\circ\iota_{V}^{T}=\iota_{V}^{T}$), this yields that $\alpha \circ\beta$ and $\operatorname*{id}\nolimits_{T\left( V\right) }$ must be equal. In other words, $\alpha\circ\beta=\operatorname*{id}\nolimits_{T\left( V\right) }$. Combined with $\beta\circ\alpha=\operatorname*{id}% \nolimits_{U\left( \operatorname*{FreeLie}V\right) }$, this yields that $\alpha$ and $\beta$ are mutually inverse, and thus $\alpha$ and $\beta$ are algebra isomorphisms. Hence, $\alpha:U\left( \operatorname*{FreeLie}V\right) \rightarrow T\left( V\right) $ is a canonical algebra isomorphism. Also, $\alpha$ commutes with the canonical injections of $V$ into $U\left( \operatorname*{FreeLie}V\right) $ and into $T\left( V\right) $, because% \[ \underbrace{\alpha\circ\iota_{\operatorname*{FreeLie}V}^{U}}_{=h}\circ \iota_{V}^{\operatorname*{FreeLie}}=h\circ\iota_{V}^{\operatorname*{FreeLie}% }=\iota_{V}^{T}. \] Hence, there exists a canonical algebra isomorphism $U\left( \operatorname*{FreeLie}V\right) \rightarrow T\left( V\right) $, which commutes with the canonical injections of $V$ into $U\left( \operatorname*{FreeLie}V\right) $ and into $T\left( V\right) $ (namely, $\alpha$). Proposition \ref{prop.Ufree} is proven. \begin{verlong} There is a special version of Proposition \ref{prop.Ufree} available for graded vector spaces: \begin{proposition} \label{prop.Ufree.gr}Let $Q$ be an abelian group. Let $V$ be a $Q$-graded vector space. Let us use the notations of Proposition \ref{prop.Ufree}. Then, the canonical algebra isomorphism $U\left( \operatorname*{FreeLie}V\right) \rightarrow T\left( V\right) $ constructed in Proposition \ref{prop.Ufree} is an isomorphism of $Q$\textbf{-graded} algebras. \end{proposition} One way to prove Proposition \ref{prop.Ufree.gr} is to scatter the word ``graded'' across the proof of Proposition \ref{prop.Ufree}; of course, we would need the graded analogues of Theorems \ref{thm.universal.tensor}, \ref{thm.universal.U} and \ref{thm.universal.FreeLie} for this to work. Here is a slightly different way, which will require only the graded version of Theorem \ref{thm.universal.tensor}: \begin{theorem} \label{thm.universal.tensor.gr}Let $Q$ be an abelian group. Let $V$ be a $Q$-graded vector space. We denote by $\iota_{V}^{T}:V\rightarrow T\left( V\right) $ the canonical map from $V$ into $T\left( V\right) $. (This map $\iota_{V}^{T}$ is known to be injective.) Let $B$ be any $Q$-graded algebra, and $f:V\rightarrow B$ be any $Q$-graded linear map. According to Theorem \ref{thm.universal.tensor}, there exists a unique algebra homomorphism $F:T\left( V\right) \rightarrow B$ satisfying $f=F\circ\iota_{V}^{T}$. This homomorphism $F$ is $Q$-graded. \end{theorem} \textit{Proof of Theorem \ref{thm.universal.tensor.gr}.} Consider the unique algebra homomorphism $F:T\left( V\right) \rightarrow B$ satisfying $f=F\circ\iota_{V}^{T}$. We need to show that this $F$ is $Q$-graded. For every $n\in\mathbb{N}$ and every $\left( q_{1},q_{2},...,q_{n}\right) \in Q^{n}$ and every $w\in\left( V\left[ q_{1}\right] \right) \otimes\left( V\left[ q_{2}\right] \right) \otimes...\otimes\left( V\left[ q_{n}\right] \right) $, we have% \begin{equation} F\left( w\right) \in B\left[ q_{1}+q_{2}+...+q_{n}\right] . \label{pf.universal.tensor.gr.1}% \end{equation} \textit{Proof of (\ref{pf.universal.tensor.gr.1}):} We will treat the map $\iota_{V}^{T}$ as an inclusion map, so that $\iota_{V}^{T}\left( x\right) =x$ for every $x\in V$. Let $n\in\mathbb{N}$ and $\left( q_{1},q_{2},...,q_{n}\right) \in Q^{n}$. We need to prove the relation (\ref{pf.universal.tensor.der.gr.1}) for all $w\in\left( V\left[ q_{1}\right] \right) \otimes\left( V\left[ q_{2}\right] \right) \otimes...\otimes\left( V\left[ q_{n}\right] \right) $. In order to achieve this, it is enough to prove the relation (\ref{pf.universal.tensor.der.gr.1}) for all pure tensors $w\in\left( V\left[ q_{1}\right] \right) \otimes\left( V\left[ q_{2}\right] \right) \otimes...\otimes\left( V\left[ q_{n}\right] \right) $ (because every tensor is a linear combination of pure tensors, but the relation (\ref{pf.universal.tensor.gr.1}) is linear in $w$). Thus, we can assume WLOG that $w$ is a pure tensor. Assume this. Then, there exists a $\left( v_{1},v_{2},...,v_{n}\right) \in\left( V\left[ q_{1}\right] \right) \times\left( V\left[ q_{2}\right] \right) \times...\times\left( V\left[ q_{n}\right] \right) $ such that $w=v_{1}\otimes v_{2}\otimes...\otimes v_{n}$. Consider this $\left( v_{1},v_{2},...,v_{n}\right) $. We have $w=v_{1}\otimes v_{2}\otimes...\otimes v_{n}=v_{1}v_{2}...v_{n}$ (since the multiplication on the algebra $T\left( V\right) $ is given by the tensor product). On the other hand, every $i\in\left\{ 1,2,...,n\right\} $ satisfies $v_{i}\in V\left[ q_{i}\right] \subseteq V$, and thus $\iota _{V}^{T}\left( v_{i}\right) =v_{i}$ (since $\iota_{V}^{T}\left( x\right) =x$ for every $x\in V$), so that $\underbrace{f}_{=F\circ\iota_{V}^{T}}\left( v_{i}\right) =\left( F\circ\iota_{V}^{T}\right) \left( v_{i}\right) =F\left( \underbrace{\iota_{V}^{T}\left( v_{i}\right) }_{=v_{i}}\right) =F\left( v_{i}\right) $. Thus, $\left( f\left( v_{1}\right) ,f\left( v_{2}\right) ,...,f\left( v_{n}\right) \right) =\left( F\left( v_{1}\right) ,F\left( v_{2}\right) ,...,F\left( v_{n}\right) \right) $, so that $f\left( v_{1}\right) f\left( v_{2}\right) ...f\left( v_{n}\right) =F\left( v_{1}\right) F\left( v_{2}\right) ...F\left( v_{n}\right) $. Hence,% \begin{align*} F\left( w\right) & =F\left( v_{1}v_{2}...v_{n}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }w=v_{1}v_{2}...v_{n}\right) \\ & =F\left( v_{1}\right) F\left( v_{2}\right) ...F\left( v_{n}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }F\text{ is an algebra homomorphism}% \right) \\ & =f\left( v_{1}\right) f\left( v_{2}\right) ...f\left( v_{n}\right) . \end{align*} But every $i\in\left\{ 1,2,...,n\right\} $ satisfies $f\left( v_{i}\right) \in B\left[ q_{i}\right] $ (because $v_{i}\in V\left[ q_{i}\right] $ and since $f$ is $Q$-graded). Hence, $\left( f\left( v_{1}\right) ,f\left( v_{2}\right) ,...,f\left( v_{n}\right) \right) \in\left( B\left[ q_{1}\right] \right) \times\left( B\left[ q_{2}\right] \right) \times...\times\left( B\left[ q_{n}\right] \right) $. Thus, \[ f\left( v_{1}\right) f\left( v_{2}\right) ...f\left( v_{n}\right) \in\left( B\left[ q_{1}\right] \right) \left( B\left[ q_{2}\right] \right) ...\left( B\left[ q_{n}\right] \right) \subseteq B\left[ q_{1}+q_{2}+...+q_{n}\right] \] (since $B$ is a graded algebra). Altogether, we now have% \[ F\left( w\right) =f\left( v_{1}\right) f\left( v_{2}\right) ...f\left( v_{n}\right) \in B\left[ q_{1}+q_{2}+...+q_{n}\right] . \] This proves (\ref{pf.universal.tensor.gr.1}). Now, let $q\in Q$. By the definition of the grading on a tensor product, we have% \[ V^{\otimes n}\left[ q\right] =\bigoplus\limits_{\substack{\left( q_{1},q_{2},...,q_{n}\right) \in Q^{n};\\q_{1}+q_{2}+...+q_{n}=q}}\left( V\left[ q_{1}\right] \right) \otimes\left( V\left[ q_{2}\right] \right) \otimes...\otimes\left( V\left[ q_{n}\right] \right) \] for every $n\in\mathbb{N}$. On the other hand, $T\left( V\right) =\bigoplus\limits_{n\in\mathbb{N}}V^{\otimes n}$ (where the direct sum is a direct sum of graded vector spaces), so that% \begin{align*} \left( T\left( V\right) \right) \left[ q\right] & =\left( \bigoplus\limits_{n\in\mathbb{N}}V^{\otimes n}\right) \left[ q\right] =\bigoplus\limits_{n\in\mathbb{N}}\underbrace{V^{\otimes n}\left[ q\right] }_{=\bigoplus\limits_{\substack{\left( q_{1},q_{2},...,q_{n}\right) \in Q^{n};\\q_{1}+q_{2}+...+q_{n}=q}}\left( V\left[ q_{1}\right] \right) \otimes\left( V\left[ q_{2}\right] \right) \otimes...\otimes\left( V\left[ q_{n}\right] \right) }\\ & =\bigoplus\limits_{n\in\mathbb{N}}\bigoplus\limits_{\substack{\left( q_{1},q_{2},...,q_{n}\right) \in Q^{n};\\q_{1}+q_{2}+...+q_{n}=q}}\left( V\left[ q_{1}\right] \right) \otimes\left( V\left[ q_{2}\right] \right) \otimes...\otimes\left( V\left[ q_{n}\right] \right) \\ & =\sum\limits_{n\in\mathbb{N}}\sum\limits_{\substack{\left( q_{1}% ,q_{2},...,q_{n}\right) \in Q^{n};\\q_{1}+q_{2}+...+q_{n}=q}}\left( V\left[ q_{1}\right] \right) \otimes\left( V\left[ q_{2}\right] \right) \otimes...\otimes\left( V\left[ q_{n}\right] \right) \end{align*} (since direct sums are sums). Thus,% \begin{align*} & F\left( \left( T\left( V\right) \right) \left[ q\right] \right) =F\left( \sum\limits_{n\in\mathbb{N}}\sum\limits_{\substack{\left( q_{1},q_{2},...,q_{n}\right) \in Q^{n};\\q_{1}+q_{2}+...+q_{n}=q}}\left( V\left[ q_{1}\right] \right) \otimes\left( V\left[ q_{2}\right] \right) \otimes...\otimes\left( V\left[ q_{n}\right] \right) \right) \\ & =\sum\limits_{n\in\mathbb{N}}\sum\limits_{\substack{\left( q_{1}% ,q_{2},...,q_{n}\right) \in Q^{n};\\q_{1}+q_{2}+...+q_{n}=q}% }\underbrace{F\left( \left( V\left[ q_{1}\right] \right) \otimes\left( V\left[ q_{2}\right] \right) \otimes...\otimes\left( V\left[ q_{n}\right] \right) \right) }_{\substack{\subseteq B\left[ q_{1}% +q_{2}+...+q_{n}\right] \\\text{(since (\ref{pf.universal.tensor.gr.1}) yields that }F\left( w\right) \in B\left[ q_{1}+q_{2}+...+q_{n}\right] \\\text{for every }w\in\left( V\left[ q_{1}\right] \right) \otimes\left( V\left[ q_{2}\right] \right) \otimes...\otimes\left( V\left[ q_{n}\right] \right) \text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }F\text{ is linear}\right) \\ & \subseteq\sum\limits_{n\in\mathbb{N}}\sum\limits_{\substack{\left( q_{1},q_{2},...,q_{n}\right) \in Q^{n};\\q_{1}+q_{2}+...+q_{n}=q}}B\left[ \underbrace{q_{1}+q_{2}+...+q_{n}}_{=q}\right] =\sum\limits_{n\in\mathbb{N}% }\sum\limits_{\substack{\left( q_{1},q_{2},...,q_{n}\right) \in Q^{n};\\q_{1}+q_{2}+...+q_{n}=q}}B\left[ q\right] \subseteq B\left[ q\right] \end{align*} (since $B\left[ q\right] $ is a vector space). Now forget that we fixed $q$. We thus have shown that $F\left( \left( T\left( V\right) \right) \left[ q\right] \right) \subseteq B\left[ q\right] $. The map $F$ therefore is $Q$-graded. Theorem \ref{thm.universal.tensor.gr} is thus proven. \textit{Proof of Proposition \ref{prop.Ufree.gr}.} Define the maps $\alpha$ and $\beta$ as in the proof of Proposition \ref{prop.Ufree}. Then, $\beta$ is the unique algebra homomorphism $F:T\left( V\right) \rightarrow U\left( \operatorname*{FreeLie}V\right) $ satisfying $\iota_{\operatorname*{FreeLie}% V}^{U}\circ\iota_{V}^{\operatorname*{FreeLie}}=F\circ\iota_{V}^{T}$. Hence, Theorem \ref{thm.universal.tensor.gr} (applied to $B=U\left( \operatorname*{FreeLie}V\right) $ and $f=\iota_{\operatorname*{FreeLie}V}% ^{U}\circ\iota_{V}^{\operatorname*{FreeLie}}$) yields that this homomorphism $\beta$ is $Q$-graded. Also, we have seen in the proof of Proposition \ref{prop.Ufree} that $\beta$ is an algebra isomorphism, and that $\alpha$ and $\beta$ are mutually inverse. Since $\alpha$ and $\beta$ are mutually inverse, we have $\alpha=\beta^{-1}$, so that $\alpha$ is the inverse of a $Q$-graded algebra isomorphism (because $\beta$ is a $Q$-graded algebra isomorphism). Thus, $\alpha$ itself is $Q$-graded (because any inverse of a $Q$-graded isomorphism must itself be $Q$-graded). Since $\alpha$ is the canonical algebra isomorphism $U\left( \operatorname*{FreeLie}V\right) \rightarrow T\left( V\right) $ constructed in Proposition \ref{prop.Ufree}, this yields that the canonical algebra isomorphism $U\left( \operatorname*{FreeLie}V\right) \rightarrow T\left( V\right) $ constructed in Proposition \ref{prop.Ufree} is $Q$-graded. Hence, the canonical algebra isomorphism $U\left( \operatorname*{FreeLie}V\right) \rightarrow T\left( V\right) $ constructed in Proposition \ref{prop.Ufree} is an isomorphism of $Q$-graded algebras. Proposition \ref{prop.Ufree.gr} is proven. \end{verlong} \subsubsection{Universality of the tensor algebra with respect to derivations} Next, let us notice that the universal property of the tensor algebra (Theorem \ref{thm.universal.tensor}) has an analogue for derivations in lieu of algebra homomorphisms: \begin{theorem} \label{thm.universal.tensor.der}Let $V$ be a vector space. We denote by $\iota_{V}^{T}:V\rightarrow T\left( V\right) $ the canonical map from $V$ into $T\left( V\right) $. (This map $\iota_{V}^{T}$ is known to be injective.) For any $T\left( V\right) $-bimodule $M$ and any linear map $f:V\rightarrow M$, there exists a unique derivation $F:T\left( V\right) \rightarrow M$ satisfying $f=F\circ\iota_{V}^{T}$. \end{theorem} It should be noticed that ``derivation'' means ``$\mathbb{C}$-linear derivation'' here. Before we prove this theorem, let us extend its uniqueness part a bit: \begin{proposition} \label{prop.derivation.unique}Let $A$ be an algebra. Let $M$ be an $A$-bimodule, and $d:A\rightarrow M$ and $e:A\rightarrow M$ two derivations. Let $S$ be a subset of $A$ which generates $A$ as an algebra. Assume that $d\mid_{S}=e\mid_{S}$. Then, $d=e$. \end{proposition} \textit{Proof of Proposition \ref{prop.derivation.unique}.} Let $U$ be the subset $\operatorname*{Ker}\left( d-e\right) $ of $A$. Clearly, $U$ is a vector space (since $d-e$ is a linear map (since $d$ and $e$ are linear)). It is known that any derivation $f:A\rightarrow M$ satisfies $f\left( 1\right) =0$. Applying this to $f=d$, we get $d\left( 1\right) =0$. Similarly, $e\left( 1\right) =0$. Thus, $\left( d-e\right) \left( 1\right) =\underbrace{d\left( 1\right) }_{=0}-\underbrace{e\left( 1\right) }_{=0}=0$, so that $1\in\operatorname*{Ker}\left( d-e\right) =U$. Now let $b\in U$ and $c\in U$. Since $b\in U=\operatorname*{Ker}\left( d-e\right) $, we have $\left( d-e\right) \left( b\right) =0$. Thus, $d\left( b\right) -e\left( b\right) =\left( d-e\right) \left( b\right) =0$, so that $d\left( b\right) =e\left( b\right) $. Similarly, $d\left( c\right) =e\left( c\right) $. Now, since $d$ is a derivation, the Leibniz formula yields $d\left( bc\right) =d\left( b\right) \cdot c+b\cdot d\left( c\right) $. Similarly, $e\left( bc\right) =e\left( b\right) \cdot c+b\cdot e\left( c\right) $. Hence,% \begin{align*} \left( d-e\right) \left( bc\right) & =\underbrace{d\left( bc\right) }_{=d\left( b\right) \cdot c+b\cdot d\left( c\right) }% -\underbrace{e\left( bc\right) }_{=e\left( b\right) \cdot c+b\cdot e\left( c\right) }=\left( \underbrace{d\left( b\right) }_{=e\left( b\right) }\cdot c+b\cdot\underbrace{d\left( c\right) }_{=e\left( c\right) }\right) -\left( e\left( b\right) \cdot c+b\cdot e\left( c\right) \right) \\ & =\left( e\left( b\right) \cdot c+b\cdot e\left( c\right) \right) -\left( e\left( b\right) \cdot c+b\cdot e\left( c\right) \right) =0. \end{align*} In other words, $bc\in\operatorname*{Ker}\left( d-e\right) =U$. Now forget that we fixed $b$ and $c$. We have thus showed that any $b\in U$ and $c\in U$ satisfy $bc\in U$. Combined with the fact that $U$ is a vector space and that $1\in U$, this yields that $U$ is a subalgebra of $A$. Since $S\subseteq U$ (because every $s\in S$ satisfies% \[ \left( d-e\right) \left( s\right) =\underbrace{d\left( s\right) }_{=\left( d\mid_{S}\right) \left( s\right) }-\underbrace{e\left( s\right) }_{=\left( e\mid_{S}\right) \left( s\right) }% =\underbrace{\left( d\mid_{S}\right) }_{=e\mid_{S}}\left( s\right) -\left( e\mid_{S}\right) \left( s\right) =\left( e\mid_{S}\right) \left( s\right) -\left( e\mid_{S}\right) \left( s\right) =0 \] and thus $s\in\operatorname*{Ker}\left( d-e\right) =U$), this yields that $U$ is a subalgebra of $A$ containing $S$ as a subset. But since the smallest subalgebra of $A$ containing $S$ as a subset is $A$ itself (because $S$ generates $A$ as an algebra), this yields that $U\supseteq A$. Hence, $A\subseteq U=\operatorname*{Ker}\left( d-e\right) $, so that $d-e=0$ and thus $d=e$. Proposition \ref{prop.derivation.unique} is proven. \textit{Proof of Theorem \ref{thm.universal.tensor.der}.} For any $n\in\mathbb{N}$, we can define a linear map $\Phi_{n}:V^{\otimes n}\rightarrow M$ by the equation% \begin{equation} \left( \begin{array} [c]{r}% \Phi_{n}\left( v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right) =\sum\limits_{k=1}^{n}v_{1}\cdot v_{2}\cdot...\cdot v_{k-1}\cdot f\left( v_{k}\right) \cdot v_{k+1}\cdot v_{k+2}\cdot...\cdot v_{n}\\ \text{for all }v_{1},v_{2},...,v_{n}\in V \end{array} \right) \label{pf.universal.tensor.der.Phin}% \end{equation} (by the universal property of the tensor product, since the term $\sum\limits_{k=1}^{n}v_{1}\cdot v_{2}\cdot...\cdot v_{k-1}\cdot f\left( v_{k}\right) \cdot v_{k+1}\cdot v_{k+2}\cdot...\cdot v_{n}$ is clearly multilinear in $v_{1}$, $v_{2}$, $...$, $v_{n}$). Define this map $\Phi_{n}$. Let $\Phi$ be the map $\bigoplus\limits_{n\in\mathbb{N}}\Phi_{n}% :\bigoplus\limits_{n\in\mathbb{N}}V^{\otimes n}\rightarrow M$. Then, every $n\in\mathbb{N}$ and every $v_{1},v_{2},...,v_{n}$ satisfy% \begin{align} \Phi\left( v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right) & =\Phi _{n}\left( v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right) \nonumber\\ & =\sum\limits_{k=1}^{n}v_{1}\cdot v_{2}\cdot...\cdot v_{k-1}\cdot f\left( v_{k}\right) \cdot v_{k+1}\cdot v_{k+2}\cdot...\cdot v_{n}. \label{pf.universal.tensor.der.Phi}% \end{align} Since $\bigoplus\limits_{n\in\mathbb{N}}V^{\otimes n}=T\left( V\right) $, the map $\Phi$ is a map from $T\left( V\right) $ to $M$. We will now prove that $\Phi$ is a derivation. In fact, in order to prove this, we must show that% \begin{equation} \Phi\left( ab\right) =\Phi\left( a\right) \cdot b+a\cdot\Phi\left( b\right) \ \ \ \ \ \ \ \ \ \ \text{for any }a\in T\left( V\right) \text{ and }b\in T\left( V\right) . \label{pf.universal.tensor.der.Phider}% \end{equation} \textit{Proof of (\ref{pf.universal.tensor.der.Phider}):} Every element of $T\left( V\right) $ is a linear combination of elements of $V^{\otimes n}$ for various $n\in\mathbb{N}$ (because $T\left( V\right) =\bigoplus \limits_{n\in\mathbb{N}}V^{\otimes n}$). Meanwhile, every element of $V^{\otimes n}$ for any $n\in\mathbb{N}$ is a linear combination of pure tensors. Combining these two observations, we see that every element of $T\left( V\right) $ is a linear combination of pure tensors. We need to prove the equation (\ref{pf.universal.tensor.der.Phider}) for all $a\in T\left( V\right) $ and $b\in T\left( V\right) $. Since this equation is linear in each of $a$ and $b$, we can WLOG assume that $a$ and $b$ are pure tensors (since every element of $T\left( V\right) $ is a linear combination of pure tensors). Assume this. Then, $a$ is a pure tensor, so that there exists an $n\in\mathbb{N}$ and some $v_{1},v_{2},...,v_{n}\in V$ satisfying $a=v_{1}\otimes v_{2}\otimes...\otimes v_{n}$. Consider this $n$ and these $v_{1},v_{2},...,v_{n}$. Also, $b$ is a pure tensor, so that there exists an $m\in\mathbb{N}$ and some $w_{1},w_{2},...,w_{m}\in V$ satisfying $b=w_{1}\otimes w_{2}\otimes...\otimes w_{m}$. Consider this $m$ and these $w_{1},w_{2},...,w_{m}$. By (\ref{pf.universal.tensor.der.Phi}) (applied to $m$ and $w_{1}% ,w_{2},...,w_{m}$ instead of $n$ and $v_{1},v_{2},...,v_{n}$), we have% \begin{align*} \Phi\left( w_{1}\otimes w_{2}\otimes...\otimes w_{m}\right) & =\sum\limits_{k=1}^{m}w_{1}\cdot w_{2}\cdot...\cdot w_{k-1}\cdot f\left( w_{k}\right) \cdot w_{k+1}\cdot w_{k+2}\cdot...\cdot w_{m}\\ & =\sum\limits_{k=n+1}^{n+m}w_{1}\cdot w_{2}\cdot...\cdot w_{k-n-1}\cdot f\left( w_{k-n}\right) \cdot w_{k-n+1}\cdot w_{k-n+2}\cdot...\cdot w_{m}% \end{align*} (here, we substituted $k-n$ for $k$ in the sum). Let $\left( u_{1},u_{2},...,u_{n+m}\right) $ be the $\left( n+m\right) $-tuple $\left( v_{1},v_{2},...,v_{n},w_{1},w_{2},...,w_{m}\right) $. Then, \[ u_{1}\otimes u_{2}\otimes...\otimes u_{n+m}=\underbrace{v_{1}\otimes v_{2}\otimes...\otimes v_{n}}_{=a}\otimes\underbrace{w_{1}\otimes w_{2}% \otimes...\otimes w_{m}}_{=b}=a\otimes b=ab. \] By (\ref{pf.universal.tensor.der.Phi}) (applied to $n+m$ and $u_{1}% ,u_{2},...,u_{n+m}$ instead of $n$ and $v_{1},v_{2},...,v_{n}$), we have% \begin{align*} & \Phi\left( u_{1}\otimes u_{2}\otimes...\otimes u_{n+m}\right) \\ & =\sum\limits_{k=1}^{n+m}u_{1}\cdot u_{2}\cdot...\cdot u_{k-1}\cdot f\left( u_{k}\right) \cdot u_{k+1}\cdot u_{k+2}\cdot...\cdot u_{n+m}\\ & =\sum\limits_{k=1}^{n}\underbrace{u_{1}\cdot u_{2}\cdot...\cdot u_{k-1}\cdot f\left( u_{k}\right) \cdot u_{k+1}\cdot u_{k+2}\cdot...\cdot u_{n+m}}_{\substack{=v_{1}\cdot v_{2}\cdot...\cdot v_{k-1}\cdot f\left( v_{k}\right) \cdot v_{k+1}\cdot v_{k+2}\cdot...\cdot v_{n}\cdot w_{1}\cdot w_{2}\cdot...\cdot w_{m}\\\text{(since }\left( u_{1},u_{2},...,u_{n+m}% \right) =\left( v_{1},v_{2},...,v_{n},w_{1},w_{2},...,w_{m}\right) \text{ and }k\leq n\text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{k=n+1}^{n+m}\underbrace{u_{1}\cdot u_{2}\cdot...\cdot u_{k-1}\cdot f\left( u_{k}\right) \cdot u_{k+1}\cdot u_{k+2}\cdot...\cdot u_{n+m}}_{\substack{=v_{1}\cdot v_{2}\cdot...\cdot v_{n}\cdot w_{1}\cdot w_{2}\cdot...\cdot w_{k-n-1}\cdot f\left( w_{k-n}\right) \cdot w_{k-n+1}\cdot w_{k-n+2}\cdot...\cdot w_{m}% \\\text{(since }\left( u_{1},u_{2},...,u_{n+m}\right) =\left( v_{1}% ,v_{2},...,v_{n},w_{1},w_{2},...,w_{m}\right) \text{ and }k>n\text{)}}}\\ & =\sum\limits_{k=1}^{n}v_{1}\cdot v_{2}\cdot...\cdot v_{k-1}\cdot f\left( v_{k}\right) \cdot v_{k+1}\cdot v_{k+2}\cdot...\cdot v_{n}\cdot \underbrace{w_{1}\cdot w_{2}\cdot...\cdot w_{m}}_{=w_{1}\otimes w_{2}% \otimes...\otimes w_{m}=b}\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{k=n+1}^{n+m}\underbrace{v_{1}\cdot v_{2}\cdot...\cdot v_{n}}_{=v_{1}\otimes v_{2}\otimes...\otimes v_{n}=a}\cdot w_{1}\cdot w_{2}\cdot...\cdot w_{k-n-1}\cdot f\left( w_{k-n}\right) \cdot w_{k-n+1}\cdot w_{k-n+2}\cdot...\cdot w_{m}\\ & =\sum\limits_{k=1}^{n}v_{1}\cdot v_{2}\cdot...\cdot v_{k-1}\cdot f\left( v_{k}\right) \cdot v_{k+1}\cdot v_{k+2}\cdot...\cdot v_{n}\cdot b\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{k=n+1}^{n+m}a\cdot w_{1}\cdot w_{2}% \cdot...\cdot w_{k-n-1}\cdot f\left( w_{k-n}\right) \cdot w_{k-n+1}\cdot w_{k-n+2}\cdot...\cdot w_{m}\\ & =\underbrace{\left( \sum\limits_{k=1}^{n}v_{1}\cdot v_{2}\cdot...\cdot v_{k-1}\cdot f\left( v_{k}\right) \cdot v_{k+1}\cdot v_{k+2}\cdot...\cdot v_{n}\right) }_{\substack{=\Phi\left( v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right) \\\text{(by (\ref{pf.universal.tensor.der.Phi}))}}}\cdot b\\ & \ \ \ \ \ \ \ \ \ \ +a\cdot\underbrace{\left( \sum\limits_{k=n+1}% ^{n+m}w_{1}\cdot w_{2}\cdot...\cdot w_{k-n-1}\cdot f\left( w_{k-n}\right) \cdot w_{k-n+1}\cdot w_{k-n+2}\cdot...\cdot w_{m}\right) }_{=\Phi\left( w_{1}\otimes w_{2}\otimes...\otimes w_{m}\right) }\\ & =\Phi\left( \underbrace{v_{1}\otimes v_{2}\otimes...\otimes v_{n}}% _{=a}\right) \cdot b+a\cdot\Phi\left( \underbrace{w_{1}\otimes w_{2}% \otimes...\otimes w_{m}}_{=b}\right) =\Phi\left( a\right) \cdot b+a\cdot\Phi\left( b\right) . \end{align*} Thus, (\ref{pf.universal.tensor.der.Phider}) is proven. Now that we know that $\Phi$ satisfies (\ref{pf.universal.tensor.der.Phider}), we conclude that $\Phi$ is a derivation. Next, notice that every $v\in V$ satisfies $\iota_{V}^{T}\left( v\right) =v$ (since $\iota_{V}^{T}$ is just the inclusion map). Hence, every $v\in V$ satisfies% \begin{align*} \left( \Phi\circ\iota_{V}^{T}\right) \left( v\right) & =\Phi\left( \underbrace{\iota_{V}^{T}\left( v\right) }_{\substack{=v}}\right) =\Phi\left( v\right) \\ & =\sum\limits_{k=1}^{1}f\left( v\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.universal.tensor.der.Phi}), applied to }n=1\text{ and }% v_{1}=v\right) \\ & =f\left( v\right) . \end{align*} Thus, $\Phi\circ\iota_{V}^{T}=f$. So we know that $\Phi$ is a derivation satisfying $f=\Phi\circ\iota_{V}^{T}$. Thus, we have shown that there exists a derivation $F:T\left( V\right) \rightarrow M$ satisfying $f=F\circ\iota_{V}^{T}$ (namely, $F=\Phi$). In order to complete the proof of Theorem \ref{thm.universal.tensor.der}, we only need to check that this derivation is unique. In other words, we need to check that whenever a derivation $F:T\left( V\right) \rightarrow M$ satisfies $f=F\circ\iota_{V}^{T}$, we must have $F=\Phi$. Let us prove this now. Let $F:T\left( V\right) \rightarrow M$ be any derivation satisfying $f=F\circ\iota_{V}^{T}$. Then, every $v\in V$ satisfies% \begin{align*} \left( F\mid_{V}\right) \left( v\right) & =F\left( \underbrace{v}% _{=\iota_{V}^{T}\left( v\right) }\right) =F\left( \iota_{V}^{T}\left( v\right) \right) =\underbrace{\left( F\circ\iota_{V}^{T}\right) }% _{=f=\Phi\circ\iota_{V}^{T}}\left( v\right) =\left( \Phi\circ\iota_{V}% ^{T}\right) \left( v\right) \\ & =\Phi\left( \underbrace{\iota_{V}^{T}\left( v\right) }_{=v}\right) =\Phi\left( v\right) =\left( \Phi\mid_{V}\right) \left( v\right) . \end{align*} Thus, $F\mid_{V}=\Phi\mid_{V}$. Proposition \ref{prop.derivation.unique} (applied to $A=T\left( V\right) $, $d=F$, $e=\Phi$ and $S=V$) thus yields $F=\Phi$ (since $V$ generates $T\left( V\right) $ as an algebra). This completes the proof of Theorem \ref{thm.universal.tensor.der} (as we have seen above). \begin{verlong} We record a graded version of Theorem \ref{thm.universal.tensor.der}: \begin{theorem} \label{thm.universal.tensor.der.gr}Let $Q$ be an abelian group. Let $V$ be a $Q$-graded vector space. We denote by $\iota_{V}^{T}:V\rightarrow T\left( V\right) $ the canonical map from $V$ into $T\left( V\right) $. (This map $\iota_{V}^{T}$ is known to be injective and $Q$-graded.) For any $Q$-graded $T\left( V\right) $-bimodule $M$ and any $Q$-graded linear map $f:V\rightarrow M$, there exists a unique $Q$-graded derivation $F:T\left( V\right) \rightarrow M$ satisfying $f=F\circ\iota_{V}^{T}$. \end{theorem} \textit{Proof of Theorem \ref{thm.universal.tensor.der.gr}.} We will treat the map $\iota_{V}^{T}$ as an inclusion map, so that $\iota_{V}^{T}\left( x\right) =x$ for every $x\in V$. Define the map $\Phi$ as in the proof of Theorem \ref{thm.universal.tensor.der}. As we have seen in the proof of Theorem \ref{thm.universal.tensor.der}, this map $\Phi$ is a derivation $T\left( V\right) \rightarrow M$ satisfying $f=\Phi\circ\iota_{V}^{T}$. We will now prove that $\Phi$ is $Q$-graded. For every $Q$-graded vector space $W$ and every $q\in Q$, let $\pi_{q}^{W}$ be the canonical projection from $W$ to the $q$-th homogeneous component $W\left[ q\right] $. Of course, for every $Q$-graded vector space $W$ and every $w\in W$, we have% \begin{equation} w=\sum\limits_{q\in Q}\pi_{q}^{W}\left( w\right) . \label{pf.universal.tensor.der.gr.0}% \end{equation} Let us notice that for every $Q$-graded vector space $W$ and any two distinct elements $p$ and $q$ of $Q$, we have% \begin{equation} \pi_{q}^{W}\left( W\left[ p\right] \right) =0 \label{pf.universal.tensor.der.gr.0a}% \end{equation} (because $\pi_{q}^{W}$ is the projection of the $Q$-graded vector space $W$ onto its $q$-th homogeneous component $W\left[ q\right] $, while $W\left[ p\right] $ is a homogeneous component of $W$ distinct from $W\left[ q\right] $). For every $q\in Q$, let $\Phi_{q}$ be the linear map $\left( T\left( V\right) \right) \left[ q\right] \rightarrow M\left[ q\right] $ defined by% \[ \left( \Phi_{q}\left( x\right) =\pi_{q}^{M}\left( \Phi\left( x\right) \right) \ \ \ \ \ \ \ \ \ \ \text{for every }x\in\left( T\left( V\right) \right) \left[ q\right] \right) . \] Then, the direct sum $\bigoplus\limits_{q\in Q}\Phi_{q}:\bigoplus\limits_{q\in Q}\left( T\left( V\right) \right) \left[ q\right] \rightarrow \bigoplus\limits_{q\in Q}M\left[ q\right] $ is a $Q$-graded linear map from $T\left( V\right) $ to $M$ (since $\bigoplus\limits_{q\in Q}\left( T\left( V\right) \right) \left[ q\right] =T\left( V\right) $ and $\bigoplus \limits_{q\in Q}M\left[ q\right] =M$). Denote this map $\bigoplus \limits_{q\in Q}\Phi_{q}$ by $\Phi^{\prime}$. Every $r\in Q$ and every $x\in\left( T\left( V\right) \right) \left[ r\right] $ satisfy% \begin{align} \Phi^{\prime}\left( x\right) & =\left( \bigoplus\limits_{q\in Q}% \Phi\left[ q\right] \right) \left( x\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }\Phi^{\prime}=\bigoplus\limits_{q\in Q}\Phi\left[ q\right] \right) \nonumber\\ & =\Phi_{r}\left( x\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }% x\in\left( T\left( V\right) \right) \left[ r\right] \right) \nonumber\\ & =\pi_{r}^{M}\left( \Phi\left( x\right) \right) \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\Phi_{r}\right) . \label{pf.universal.tensor.der.gr.1}% \end{align} Now, every $r\in Q$ and every $x\in V\left[ r\right] $ satisfy% \begin{align*} \left( \Phi^{\prime}\circ\iota_{V}^{T}\right) \left( x\right) & =\Phi^{\prime}\left( \underbrace{\iota_{V}^{T}\left( x\right) }% _{=x}\right) =\Phi^{\prime}\left( x\right) \\ & =\pi_{r}^{M}\left( \Phi\left( \underbrace{x}_{=\iota_{V}^{T}\left( x\right) }\right) \right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.universal.tensor.der.gr.1}) (since }x\in V\left[ r\right] \subseteq\left( T\left( V\right) \right) \left[ r\right] \text{)}\right) \\ & =\pi_{r}^{M}\left( \underbrace{\Phi\left( \iota_{V}^{T}\left( x\right) \right) }_{=\left( \Phi\circ\iota_{V}^{T}\right) \left( x\right) }\right) =\pi_{r}^{M}\left( \underbrace{\left( \Phi\circ\iota_{V}% ^{T}\right) }_{=f}\left( x\right) \right) =\pi_{r}^{M}\left( f\left( x\right) \right) =f\left( x\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }f\left( x\right) \in M\left[ r\right] \text{ (because }f\text{ is }Q\text{-graded and }x\in V\left[ r\right] \text{), and thus}\\ \text{the projection }\pi_{r}^{M}\text{ onto }M\left[ r\right] \text{ leaves }f\left( x\right) \text{ invariant}% \end{array} \right) . \end{align*} Thus, $\Phi^{\prime}\circ\iota_{V}^{T}=f$. We will next show that $\Phi^{\prime}$ is a derivation. Indeed, in order to prove this, we must show that% \begin{equation} \Phi^{\prime}\left( ab\right) =\Phi^{\prime}\left( a\right) \cdot b+a\cdot\Phi^{\prime}\left( b\right) \ \ \ \ \ \ \ \ \ \ \text{for any }a\in T\left( V\right) \text{ and }b\in T\left( V\right) . \label{pf.universal.tensor.der.gr.2}% \end{equation} \textit{Proof of (\ref{pf.universal.tensor.der.gr.2}):} We need to prove the equation (\ref{pf.universal.tensor.der.gr.2}) for all $a\in T\left( V\right) $ and $b\in T\left( V\right) $. Since this equation is linear in each of $a$ and $b$, we can WLOG assume that $a$ and $b$ are homogeneous (since every element of $T\left( V\right) $ is a linear combination of homogeneous elements). Assume this. Then, $a$ is homogeneous, so there exists a $p\in Q$ such that $a\in\left( T\left( V\right) \right) \left[ p\right] $. Consider this $p$. Since $b$ is homogeneous, there exists an $r\in Q$ such that $b\in\left( T\left( V\right) \right) \left[ r\right] $. Consider this $r$. Since $a\in\left( T\left( V\right) \right) \left[ p\right] $ and $b\in\left( T\left( V\right) \right) \left[ r\right] $, we have $ab\in\left( T\left( V\right) \right) \left[ p+r\right] $ (since $T\left( V\right) $ is a $Q$-graded algebra), so that% \begin{align*} \Phi^{\prime}\left( ab\right) & =\pi_{p+r}^{M}\left( \underbrace{\Phi \left( ab\right) }_{\substack{=\Phi\left( a\right) \cdot b+a\cdot \Phi\left( b\right) \\\text{(since }\Phi\text{ is a derivation)}}}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.universal.tensor.der.gr.1}), applied to }ab\text{ and }p+r\text{ instead of }x\text{ and }r\right) \\ & =\pi_{p+r}^{M}\left( \Phi\left( a\right) \cdot b+a\cdot\Phi\left( b\right) \right) =\pi_{p+r}^{M}\left( \Phi\left( a\right) \cdot b\right) +\pi_{p+r}^{M}\left( a\cdot\Phi\left( b\right) \right) . \end{align*} Now, it is easy to see that $\pi_{p+r}^{M}\left( \Phi\left( a\right) \cdot b\right) =\pi_{p}^{M}\left( \Phi\left( a\right) \right) \cdot b$\ \ \ \ \footnote{\textit{Proof.} Applying (\ref{pf.universal.tensor.der.gr.0}) to $W=M$ and $w=\Phi\left( a\right) $, we obtain $\Phi\left( a\right) =\sum\limits_{q\in Q}\pi_{q}^{M}\left( \Phi\left( a\right) \right) $, so that% \begin{align*} \pi_{p+r}^{M}\left( \Phi\left( a\right) \cdot b\right) & =\pi_{p+r}% ^{M}\left( \sum\limits_{q\in Q}\pi_{q}^{M}\left( \Phi\left( a\right) \right) \cdot b\right) =\sum\limits_{q\in Q}\pi_{p+r}^{M}\left( \pi_{q}% ^{M}\left( \Phi\left( a\right) \right) \cdot b\right) \\ & =\pi_{p+r}^{M}\left( \pi_{p}^{M}\left( \Phi\left( a\right) \right) \cdot b\right) +\sum\limits_{\substack{q\in Q;\\q\neq p}}\pi_{p+r}^{M}\left( \pi_{q}^{M}\left( \Phi\left( a\right) \right) \cdot b\right) . \end{align*} \par Now, for every $q\in Q$, we have $\underbrace{\pi_{q}^{M}\left( \Phi\left( a\right) \right) }_{\substack{\in M\left[ q\right] \\\text{(since }\pi _{q}^{M}\text{ is a}\\\text{projection on }M\left[ q\right] \text{)}}% }\cdot\underbrace{b}_{\in\left( T\left( V\right) \right) \left[ r\right] }\in\left( M\left[ q\right] \right) \cdot\left( \left( T\left( V\right) \right) \left[ r\right] \right) \subseteq M\left[ q+r\right] $ (since $M$ is a graded $T\left( V\right) $-bimodule). For every $q\in Q$ satisfying $q+r\neq p+r$, we have $\pi_{p+r}^{W}\left( M\left[ q+r\right] \right) =0$ (by (\ref{pf.universal.tensor.der.gr.0a}), applied to $M$, $p+r$ and $q+r$ instead of $W$, $q$ and $p$). Thus,% \[ \sum\limits_{\substack{q\in Q;\\q\neq p}}\pi_{p+r}^{M}\left( \underbrace{\pi _{q}^{M}\left( \Phi\left( a\right) \right) \cdot b}_{\in M\left[ q+r\right] }\right) \in\sum\limits_{\substack{q\in Q;\\q\neq p}% }\underbrace{\pi_{p+r}^{M}\left( M\left[ q+r\right] \right) }% _{\substack{=0\\\text{(since }q+r\neq p+r\\\text{(because }q\neq p\text{))}% }}=\sum\limits_{\substack{q\in Q;\\q\neq p}}0=0, \] so that $\sum\limits_{\substack{q\in Q;\\q\neq p}}\pi_{p+r}^{M}\left( \pi _{q}^{M}\left( \Phi\left( a\right) \right) \cdot b\right) =0$. Hence,% \[ \pi_{p+r}^{M}\left( \Phi\left( a\right) \cdot b\right) =\pi_{p+r}% ^{M}\left( \pi_{p}^{M}\left( \Phi\left( a\right) \right) \cdot b\right) +\underbrace{\sum\limits_{\substack{q\in Q;\\q\neq p}}\pi_{p+r}^{M}\left( \pi_{q}^{M}\left( \Phi\left( a\right) \right) \cdot b\right) }_{=0}% =\pi_{p+r}^{M}\left( \pi_{p}^{M}\left( \Phi\left( a\right) \right) \cdot b\right) . \] On the other hand, $\underbrace{\pi_{p}^{M}\left( \Phi\left( a\right) \right) }_{\substack{\in M\left[ p\right] \\\text{(since }\pi_{p}^{M}\text{ is a}\\\text{projection on }M\left[ p\right] \text{)}}}\cdot\underbrace{b}% _{\in\left( T\left( V\right) \right) \left[ r\right] }\in\left( M\left[ p\right] \right) \cdot\left( \left( T\left( V\right) \right) \left[ r\right] \right) \subseteq M\left[ p+r\right] $ (since $M$ is a graded $T\left( V\right) $-bimodule). Thus, $\pi_{p+r}^{M}\left( \pi _{p}^{M}\left( \Phi\left( a\right) \right) \cdot b\right) =\pi_{p}% ^{M}\left( \Phi\left( a\right) \right) \cdot b$ (because $\pi_{p+r}^{M}$ is a projection on $M\left[ p+r\right] $ and thus leaves every element in $M\left[ p+r\right] $ fixed). Hence,% \[ \pi_{p+r}^{M}\left( \Phi\left( a\right) \cdot b\right) =\pi_{p+r}% ^{M}\left( \pi_{p}^{M}\left( \Phi\left( a\right) \right) \cdot b\right) =\pi_{p}^{M}\left( \Phi\left( a\right) \right) \cdot b, \] qed.}. Since $\pi_{p}^{M}\left( \Phi\left( a\right) \right) =\Phi^{\prime }\left( a\right) $ (because (\ref{pf.universal.tensor.der.gr.1}) (applied to $a$ and $p$ instead of $x$ and $r$) yields $\Phi^{\prime}\left( a\right) =\pi_{p}^{M}\left( \Phi\left( a\right) \right) $), this rewrites as $\pi_{p+r}^{M}\left( \Phi\left( a\right) \cdot b\right) =\Phi^{\prime }\left( a\right) \cdot b$. The same argument (but with the right action of $T\left( V\right) $ on $M$ replaced by left action) shows that $\pi _{p+r}^{M}\left( b\cdot\Phi\left( a\right) \right) =b\cdot\Phi^{\prime }\left( a\right) $. If we apply this equality to $a$, $b$, $p$ and $r$ in lieu of $b$, $a$, $r$ and $p$, we obtain $\pi_{r+p}^{M}\left( a\cdot \Phi\left( b\right) \right) =a\cdot\Phi^{\prime}\left( b\right) $. In other words, $\pi_{p+r}^{M}\left( a\cdot\Phi\left( b\right) \right) =a\cdot\Phi^{\prime}\left( b\right) $. Thus,% \begin{align*} \Phi^{\prime}\left( ab\right) & =\underbrace{\pi_{p+r}^{M}\left( \Phi\left( a\right) \cdot b\right) }_{=\Phi^{\prime}\left( a\right) \cdot b}+\underbrace{\pi_{p+r}^{M}\left( a\cdot\Phi\left( b\right) \right) }_{=a\cdot\Phi^{\prime}\left( b\right) }\\ & =\Phi^{\prime}\left( a\right) \cdot b+a\cdot\Phi^{\prime}\left( b\right) . \end{align*} This proves (\ref{pf.universal.tensor.der.gr.2}). From (\ref{pf.universal.tensor.der.gr.2}), it becomes clear that $\Phi ^{\prime}$ is a derivation. Since $\Phi^{\prime}$ also is $Q$-graded and satisfies $f=\Phi^{\prime}\circ\iota_{V}^{T}$, we thus conclude that there exists a $Q$-graded derivation $F:T\left( V\right) \rightarrow M$ satisfying $f=F\circ\iota_{V}^{T}$ (namely, $F=\Phi$). Combining this with the fact that there exists \textbf{at most one} $Q$-graded derivation $F:T\left( V\right) \rightarrow M$ satisfying $f=F\circ\iota_{V}^{T}$% \ \ \ \ \footnote{\textit{Proof.} Theorem \ref{thm.universal.tensor.der} yields that there exists a unique derivation $F:T\left( V\right) \rightarrow M$ satisfying $f=F\circ\iota_{V}^{T}$. Hence, there exists at most one derivation $F:T\left( V\right) \rightarrow M$ satisfying $f=F\circ\iota _{V}^{T}$. In particular, there exists at most one $Q$-graded derivation $F:T\left( V\right) \rightarrow M$ satisfying $f=F\circ\iota_{V}^{T}$, qed.}, we conclude that there exists \textbf{a unique} $Q$-graded derivation $F:T\left( V\right) \rightarrow M$ satisfying $f=F\circ\iota_{V}^{T}$. Theorem \ref{thm.universal.tensor.der.gr} is proven. \end{verlong} We will later use a corollary of Proposition \ref{prop.derivation.unique}: \begin{corollary} \label{cor.derivation.unique.ihg}Let $A$ be an algebra. Let $B$ be a subalgebra of $A$. Let $C$ be a subalgebra of $B$. Let $d:A\rightarrow A$ be a derivation of the algebra $A$. Let $S$ be a subset of $C$ which generates $C$ as an algebra. Assume that $d\left( S\right) \subseteq B$. Then, $d\left( C\right) \subseteq B$. \end{corollary} \textit{Proof of Corollary \ref{cor.derivation.unique.ihg}.} Since $C\subseteq B\subseteq A$, the vector spaces $A$ and $B$ become $C$-modules. Let $\pi:A\rightarrow A\diagup B$ be the canonical projection. Clearly, $\pi$ is a $C$-module homomorphism, and satisfies $\operatorname*{Ker}\pi=B$. Let $d^{\prime}:C\rightarrow A\diagup B$ be the restriction of the map $\pi\circ d:A\rightarrow A\diagup B$ to $C$. It is easy to see that $d^{\prime }:C\rightarrow A\diagup B$ is a derivation\footnote{\textit{Proof.} Every $x\in C$ and $y\in C$ satisfy% \begin{align*} d^{\prime}\left( xy\right) & =\left( \pi\circ d\right) \left( xy\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }d^{\prime}\text{ is the restriction of }\pi\circ d\text{ to }C\right) \\ & =\pi\left( \underbrace{d\left( xy\right) }_{\substack{=d\left( x\right) \cdot y+x\cdot d\left( y\right) \\\text{(since }d\text{ is a derivation)}}}\right) =\pi\left( d\left( x\right) \cdot y+x\cdot d\left( y\right) \right) \\ & =\underbrace{\pi\left( d\left( x\right) \cdot y\right) }% _{\substack{=\pi\left( d\left( x\right) \right) \cdot y\\\text{(since }% \pi\text{ is a }C\text{-module}\\\text{homomorphism)}}}+\underbrace{\pi\left( x\cdot d\left( y\right) \right) }_{\substack{=x\cdot\pi\left( d\left( y\right) \right) \\\text{(since }\pi\text{ is a }C\text{-module}% \\\text{homomorphism)}}}\\ & =\underbrace{\pi\left( d\left( x\right) \right) }_{\substack{=\left( \pi\circ d\right) \left( x\right) =d^{\prime}\left( x\right) \\\text{(since }d^{\prime}\text{ is the restriction of}\\\pi\circ d\text{ to }C\text{, and since }x\in C\text{)}}}\cdot y+x\cdot\underbrace{\pi\left( d\left( y\right) \right) }_{\substack{=\left( \pi\circ d\right) \left( y\right) =d^{\prime}\left( y\right) \\\text{(since }d^{\prime}\text{ is the restriction of}\\\pi\circ d\text{ to }C\text{, and since }y\in C\text{)}% }}=d^{\prime}\left( x\right) \cdot y+x\cdot d^{\prime}\left( y\right) . \end{align*} Thus, $d^{\prime}$ is a derivation, qed.}. On the other hand, $0:C\rightarrow A\diagup B$ is a derivation as well. Every $s\in S$ satisfies% \begin{align*} \left( d^{\prime}\mid_{S}\right) \left( s\right) & =d^{\prime}\left( s\right) =\left( \pi\circ d\right) \left( s\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }d^{\prime}\text{ is the restriction of }\pi\circ d\text{ to }C\right) \\ & =\pi\left( d\left( s\right) \right) =0\ \ \ \ \ \ \ \ \ \ \left( \text{since }d\left( \underbrace{s}_{\in S}\right) \in d\left( S\right) \subseteq B=\operatorname*{Ker}\pi\right) \\ & =0\left( s\right) =\left( 0\mid_{S}\right) \left( s\right) . \end{align*} Thus, $d^{\prime}\mid_{S}=0\mid_{S}$. Proposition \ref{prop.derivation.unique} (applied to $C$, $A\diagup M$, $d^{\prime}$ and $0$ instead of $A$, $M$, $d$ and $e$) therefore yields that $d^{\prime}=0$ on $C$. But since $d^{\prime}$ is the restriction of $\pi\circ d$ to $C$, we have $d^{\prime}=\left( \pi\circ d\right) \mid_{C}$. Thus, $\left( \pi\circ d\right) \mid _{C}=d^{\prime}=0$, so that $\left( \pi\circ d\right) \left( C\right) =0$. Thus, $\pi\left( d\left( C\right) \right) =\left( \pi\circ d\right) \left( C\right) =0$, so that $d\left( C\right) \subseteq \operatorname*{Ker}\pi=B$. Corollary \ref{cor.derivation.unique.ihg} is therefore proven. \begin{corollary} \label{cor.derivation.Lie.semidir}Let $\mathfrak{g}$ be a Lie algebra. Let $\mathfrak{h}$ be a vector space equipped with both a Lie algebra structure and a $\mathfrak{g}$-module structure. Assume that $\mathfrak{g}$ acts on $\mathfrak{h}$ by derivations. Consider the semidirect product $\mathfrak{g}% \ltimes\mathfrak{h}$ defined as in Definition \ref{def.semidir.lielie} \textbf{(b)}. Consider $\mathfrak{g}$ as a Lie subalgebra of $\mathfrak{g}% \ltimes\mathfrak{h}$. Consider $\mathfrak{g}\ltimes\mathfrak{h}$ as a Lie subalgebra of $U\left( \mathfrak{g}\ltimes\mathfrak{h}\right) $ (where the Lie bracket on $U\left( \mathfrak{g}\ltimes\mathfrak{h}\right) $ is defined as the commutator of the multiplication). Consider $\mathfrak{h}$ as a Lie subalgebra of $\mathfrak{g}\ltimes\mathfrak{h}$, whence $U\left( \mathfrak{h}\right) $ becomes a subalgebra of $U\left( \mathfrak{g}% \ltimes\mathfrak{h}\right) $. Then, $\left[ \mathfrak{g},U\left( \mathfrak{h}\right) \right] \subseteq U\left( \mathfrak{h}\right) $ (as subsets of $U\left( \mathfrak{g}% \ltimes\mathfrak{h}\right) $). \end{corollary} \textit{Proof of Corollary \ref{cor.derivation.Lie.semidir}.} Let $x\in\mathfrak{g}$. Define a map $\xi:U\left( \mathfrak{g}\ltimes \mathfrak{h}\right) \rightarrow U\left( \mathfrak{g}\ltimes\mathfrak{h}% \right) $ by% \[ \left( \xi\left( y\right) =\left[ x,y\right] \ \ \ \ \ \ \ \ \ \ \text{for every }y\in U\left( \mathfrak{g}\ltimes \mathfrak{h}\right) \right) . \] Then, $\xi$ is clearly a derivation of the algebra $U\left( \mathfrak{g}% \ltimes\mathfrak{h}\right) $. We are identifying $\mathfrak{g}$ with a Lie subalgebra of $\mathfrak{g}% \ltimes\mathfrak{h}$. Clearly, $x\in\mathfrak{g}$ corresponds to $\left( x,0\right) \in\mathfrak{g}\ltimes\mathfrak{h}$ under this identification. We are also identifying $\mathfrak{h}$ with a Lie subalgebra of $\mathfrak{g}% \ltimes\mathfrak{h}$. Every $y\in\mathfrak{h}$ corresponds to $\left( 0,y\right) \in\mathfrak{g}\ltimes\mathfrak{h}$ under this identification. Thus, every $y\in\mathfrak{h}$ satisfies% \begin{align*} \left[ \underbrace{x}_{=\left( x,0\right) },\underbrace{y}_{=\left( 0,y\right) }\right] & =\left[ \left( x,0\right) ,\left( 0,y\right) \right] =\left( \underbrace{\left[ x,0\right] }_{=0},\underbrace{\left[ 0,y\right] }_{=0}+x\rightharpoonup y-\underbrace{0\rightharpoonup0}% _{=0}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of the Lie bracket on }\mathfrak{g}\ltimes\mathfrak{h}\right) \\ & =\left( 0,x\rightharpoonup y\right) =x\rightharpoonup y\in\mathfrak{h}. \end{align*} Hence, $\xi\left( y\right) =\left[ x,y\right] \in\mathfrak{h}$ for every $y\in\mathfrak{h}$. Thus, $\xi\left( \mathfrak{h}\right) \subseteq \mathfrak{h}\subseteq U\left( \mathfrak{h}\right) $. Now, we notice that the subset $\mathfrak{h}$ of $U\left( \mathfrak{h}% \right) $ generates $U\left( \mathfrak{h}\right) $ as an algebra. Thus, Corollary \ref{cor.derivation.unique.ihg} (applied to $A=U\left( \mathfrak{g}\ltimes\mathfrak{h}\right) $, $B=U\left( \mathfrak{h}\right) $, $C=U\left( \mathfrak{h}\right) $, $d=\xi$ and $S=\mathfrak{h}$) yields $\xi\left( U\left( \mathfrak{h}\right) \right) \subseteq U\left( \mathfrak{h}\right) $. Hence, every $u\in U\left( \mathfrak{h}\right) $ satisfies $\xi\left( u\right) \in U\left( \mathfrak{h}\right) $. But since $\xi\left( u\right) =\left[ x,u\right] $ (by the definition of $\xi$), this yields that every $u\in U\left( \mathfrak{h}\right) $ satisfies $\left[ x,u\right] \in U\left( \mathfrak{h}\right) $. Now forget that we fixed $x$. We thus have shown that every $x\in\mathfrak{g}$ and every $u\in U\left( \mathfrak{h}\right) $ satisfy $\left[ x,u\right] \in U\left( \mathfrak{h}\right) $. Thus, $\left[ \mathfrak{g},U\left( \mathfrak{h}\right) \right] \subseteq U\left( \mathfrak{h}\right) $ (since $U\left( \mathfrak{h}\right) $ is a vector space). This proves Corollary \ref{cor.derivation.Lie.semidir}. \subsubsection{Universality of the free Lie algebra with respect to derivations} Both Theorem \ref{thm.universal.tensor.der} and Proposition \ref{prop.derivation.unique} have analogues pertaining to Lie algebras in lieu of (associative) algebras.\footnote{Notice that the Lie-algebraic analogue of a derivation from an algebra $A$ into an $A$-bimodule is a $1$-cocycle from a Lie algebra $\mathfrak{g}$ into a $\mathfrak{g}$-module.} We are going to formulate both of these analogues, but we start with that of Proposition \ref{prop.derivation.unique}, since it is the one we will find utile in our study of Kac-Moody Lie algebras: \begin{proposition} \label{prop.derivation.Lie.unique}Let $\mathfrak{g}$ be a Lie algebra. Let $M$ be a $\mathfrak{g}$-module, and $d:\mathfrak{g}\rightarrow M$ and $e:\mathfrak{g}\rightarrow M$ two $1$-cocycles. Let $S$ be a subset of $\mathfrak{g}$ which generates $\mathfrak{g}$ as a Lie algebra. Assume that $d\mid_{S}=e\mid_{S}$. Then, $d=e$. \end{proposition} \begin{vershort} The proof of Proposition \ref{prop.derivation.Lie.unique} is analogous to that of Proposition \ref{prop.derivation.unique}. \end{vershort} \begin{verlong} The proof of Proposition \ref{prop.derivation.Lie.unique} is analogous to that of Proposition \ref{prop.derivation.unique}. Here are its details: \textit{Proof of Proposition \ref{prop.derivation.Lie.unique}.} Let $U$ be the subset $\operatorname*{Ker}\left( d-e\right) $ of $\mathfrak{g}$. Clearly, $U$ is a vector space (since $d-e$ is a linear map (since $d$ and $e$ are linear)). Let $b\in U$ and $c\in U$. Since $b\in U=\operatorname*{Ker}\left( d-e\right) $, we have $\left( d-e\right) \left( b\right) =0$. Thus, $d\left( b\right) -e\left( b\right) =\left( d-e\right) \left( b\right) =0$, so that $d\left( b\right) =e\left( b\right) $. Similarly, $d\left( c\right) =e\left( c\right) $. Now, since $d$ is a $1$-cocycle, we have $d\left( \left[ b,c\right] \right) =\left[ d\left( b\right) ,c\right] +\left[ b,d\left( c\right) \right] $ (by the definition of $1$-cocycles). Similarly, $e\left( \left[ b,c\right] \right) =\left[ e\left( b\right) ,c\right] +\left[ b,e\left( c\right) \right] $. Hence,% \begin{align*} \left( d-e\right) \left( \left[ b,c\right] \right) & =\underbrace{d\left( \left[ b,c\right] \right) }_{=\left[ d\left( b\right) ,c\right] +\left[ b,d\left( c\right) \right] }% -\underbrace{e\left( \left[ b,c\right] \right) }_{=\left[ e\left( b\right) ,c\right] +\left[ b,e\left( c\right) \right] }=\left( \left[ \underbrace{d\left( b\right) }_{=e\left( b\right) },c\right] +\left[ b,\underbrace{d\left( c\right) }_{=e\left( c\right) }\right] \right) -\left( \left[ e\left( b\right) ,c\right] +\left[ b,e\left( c\right) \right] \right) \\ & =\left( \left[ e\left( b\right) ,c\right] +\left[ b,e\left( c\right) \right] \right) -\left( \left[ e\left( b\right) ,c\right] +\left[ b,e\left( c\right) \right] \right) =0. \end{align*} In other words, $\left[ b,c\right] \in\operatorname*{Ker}\left( d-e\right) =U$. Now forget that we fixed $b$ and $c$. We have thus showed that any $b\in U$ and $c\in U$ satisfy $\left[ b,c\right] \in U$. Combined with the fact that $U$ is a vector space, this yields that $U$ is a Lie subalgebra of $\mathfrak{g}$. Since $S\subseteq U$ (because every $s\in S$ satisfies% \[ \left( d-e\right) \left( s\right) =\underbrace{d\left( s\right) }_{=\left( d\mid_{S}\right) \left( s\right) }-\underbrace{e\left( s\right) }_{=\left( e\mid_{S}\right) \left( s\right) }% =\underbrace{\left( d\mid_{S}\right) }_{=e\mid_{S}}\left( s\right) -\left( e\mid_{S}\right) \left( s\right) =\left( e\mid_{S}\right) \left( s\right) -\left( e\mid_{S}\right) \left( s\right) =0 \] and thus $s\in\operatorname*{Ker}\left( d-e\right) =U$), this yields that $U$ is a Lie subalgebra of $\mathfrak{g}$ containing $S$ as a subset. But since the smallest Lie subalgebra of $\mathfrak{g}$ containing $S$ as a subset is $\mathfrak{g}$ itself (because $S$ generates $\mathfrak{g}$ as a Lie algebra), this yields that $U\supseteq\mathfrak{g}$. Hence, $\mathfrak{g}% \subseteq U=\operatorname*{Ker}\left( d-e\right) $, so that $d-e=0$ and thus $d=e$. Proposition \ref{prop.derivation.Lie.unique} is proven. \end{verlong} \begin{verlong} An analogue of Theorem \ref{thm.universal.tensor.der.gr} for Lie algebras can also be given, and is left to the reader. \end{verlong} We record a corollary of Proposition \ref{prop.derivation.Lie.unique}: \begin{corollary} \label{cor.derivation.Lie.unique.ihg}Let $\mathfrak{g}$ be a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $\mathfrak{i}$ be a Lie subalgebra of $\mathfrak{h}$. Let $d:\mathfrak{g}\rightarrow\mathfrak{g}$ be a derivation of the Lie algebra $\mathfrak{g}$. Let $S$ be a subset of $\mathfrak{i}$ which generates $\mathfrak{i}$ as a Lie algebra. Assume that $d\left( S\right) \subseteq\mathfrak{h}$. Then, $d\left( \mathfrak{i}% \right) \subseteq\mathfrak{h}$. \end{corollary} \begin{vershort} This corollary is analogous to Corollary \ref{cor.derivation.unique.ihg}, and proven accordingly. \end{vershort} \begin{verlong} This corollary is analogous to Corollary \ref{cor.derivation.unique.ihg}, and proven accordingly: \textit{Proof of Corollary \ref{cor.derivation.Lie.unique.ihg}.} We regard $\mathfrak{g}$ as a $\mathfrak{g}$-module by means of the adjoint action. Since $\mathfrak{i}\subseteq\mathfrak{h}\subseteq\mathfrak{g}$, the $\mathfrak{g}$-module $\mathfrak{g}$ thus becomes an $\mathfrak{i}$-module. We also regard $\mathfrak{h}$ as an $\mathfrak{h}$-module by means of the adjoint action. Since $\mathfrak{i}\subseteq\mathfrak{h}$, the $\mathfrak{h}% $-module $\mathfrak{h}$ thus becomes an $\mathfrak{i}$-module. Let $\pi:\mathfrak{g}\rightarrow\mathfrak{g}\diagup\mathfrak{h}$ be the canonical projection. Clearly, $\pi$ is an $\mathfrak{i}$-module homomorphism, and satisfies $\operatorname*{Ker}\pi=\mathfrak{h}$. Let $d^{\prime }:\mathfrak{i}\rightarrow\mathfrak{g}\diagup\mathfrak{h}$ be the restriction of the map $\pi\circ d:\mathfrak{g}\rightarrow\mathfrak{g}\diagup\mathfrak{h}$ to $\mathfrak{i}$. It is easy to see that $d^{\prime}:\mathfrak{i}% \rightarrow\mathfrak{g}\diagup\mathfrak{h}$ is a $1$% -cocycle\footnote{\textit{Proof.} Every $x\in\mathfrak{i}$ and $y\in \mathfrak{i}$ satisfy% \begin{align*} d^{\prime}\left( \left[ x,y\right] \right) & =\left( \pi\circ d\right) \left( \left[ x,y\right] \right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }d^{\prime}\text{ is the restriction of }\pi\circ d\text{ to }\mathfrak{i}% \right) \\ & =\pi\left( \underbrace{d\left( \left[ x,y\right] \right) }_{\substack{=\left[ d\left( x\right) ,y\right] +\left[ x,d\left( y\right) \right] \\\text{(since }d\text{ is a derivation)}}}\right) =\pi\left( \underbrace{\left[ d\left( x\right) ,y\right] }_{=-\left[ y,d\left( x\right) \right] }+\left[ x,d\left( y\right) \right] \right) \\ & =\pi\left( -\underbrace{\left[ y,d\left( x\right) \right] }_{\substack{=y\rightharpoonup\left( d\left( x\right) \right) \\\text{(since }\mathfrak{g}\text{ is a }\mathfrak{g}\text{-module}\\\text{by the adjoint action)}}}+\underbrace{\left[ x,d\left( y\right) \right] }_{\substack{=x\rightharpoonup\left( d\left( y\right) \right) \\\text{(since }\mathfrak{g}\text{ is a }\mathfrak{g}\text{-module}\\\text{by the adjoint action)}}}\right) \\ & =\pi\left( -y\rightharpoonup\left( d\left( x\right) \right) +x\rightharpoonup\left( d\left( y\right) \right) \right) \\ & =-\underbrace{\pi\left( y\rightharpoonup\left( d\left( x\right) \right) \right) }_{\substack{=y\rightharpoonup\left( \pi\left( d\left( x\right) \right) \right) \\\text{(since }\pi\text{ is an }\mathfrak{i}% \text{-module homomorphism)}}}+\underbrace{\pi\left( x\rightharpoonup\left( d\left( y\right) \right) \right) }_{\substack{=x\rightharpoonup\left( \pi\left( d\left( y\right) \right) \right) \\\text{(since }\pi\text{ is an }\mathfrak{i}\text{-module homomorphism)}}}\\ & =-y\rightharpoonup\left( \underbrace{\pi\left( d\left( x\right) \right) }_{\substack{=\left( \pi\circ d\right) \left( x\right) =d^{\prime}\left( x\right) \\\text{(since }d^{\prime}\text{ is the restriction of}\\\pi\circ d\text{ to }\mathfrak{i}\text{, and since }% x\in\mathfrak{i}\text{)}}}\right) +x\rightharpoonup\left( \underbrace{\pi \left( d\left( y\right) \right) }_{\substack{=\left( \pi\circ d\right) \left( y\right) =d^{\prime}\left( y\right) \\\text{(since }d^{\prime }\text{ is the restriction of}\\\pi\circ d\text{ to }\mathfrak{i}\text{, and since }y\in\mathfrak{i}\text{)}}}\right) \\ & =-y\rightharpoonup\left( d^{\prime}\left( x\right) \right) +x\rightharpoonup\left( d^{\prime}\left( y\right) \right) =x\rightharpoonup\left( d^{\prime}\left( y\right) \right) -y\rightharpoonup\left( d^{\prime}\left( x\right) \right) . \end{align*} Thus, $d^{\prime}$ is a $1$-cocycle, qed.}. On the other hand, $0:\mathfrak{i}% \rightarrow\mathfrak{g}\diagup\mathfrak{h}$ is a $1$-cocycle as well. Every $s\in S$ satisfies% \begin{align*} \left( d^{\prime}\mid_{S}\right) \left( s\right) & =d^{\prime}\left( s\right) =\left( \pi\circ d\right) \left( s\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }d^{\prime}\text{ is the restriction of }\pi\circ d\text{ to }\mathfrak{i}\right) \\ & =\pi\left( d\left( s\right) \right) =0\ \ \ \ \ \ \ \ \ \ \left( \text{since }d\left( \underbrace{s}_{\in S}\right) \in d\left( S\right) \subseteq\mathfrak{h}=\operatorname*{Ker}\pi\right) \\ & =0\left( s\right) =\left( 0\mid_{S}\right) \left( s\right) . \end{align*} Thus, $d^{\prime}\mid_{S}=0\mid_{S}$. Proposition \ref{prop.derivation.Lie.unique} (applied to $\mathfrak{i}$, $\mathfrak{g}% \diagup\mathfrak{h}$, $d^{\prime}$ and $0$ instead of $\mathfrak{g}$, $M$, $d$ and $e$) therefore yields that $d^{\prime}=0$ on $\mathfrak{i}$. But since $d^{\prime}$ is the restriction of $\pi\circ d$ to $\mathfrak{i}$, we have $d^{\prime}=\left( \pi\circ d\right) \mid_{\mathfrak{i}}$. Thus, $\left( \pi\circ d\right) \mid_{\mathfrak{i}}=d^{\prime}=0$, so that $\left( \pi\circ d\right) \left( \mathfrak{i}\right) =0$. Thus, $\pi\left( d\left( \mathfrak{i}\right) \right) =\left( \pi\circ d\right) \left( \mathfrak{i}\right) =0$, so that $d\left( \mathfrak{i}\right) \subseteq\operatorname*{Ker}\pi=\mathfrak{h}$. Corollary \ref{cor.derivation.Lie.unique.ihg} is therefore proven. \end{verlong} Let us now state the analogue of Proposition \ref{prop.derivation.unique} in the Lie-algebraic setting: \begin{theorem} \label{thm.universal.FreeLie.der}Let $V$ be a vector space. We denote by $\iota_{V}^{\operatorname*{FreeLie}}:V\rightarrow\operatorname*{FreeLie}V$ the canonical map from $V$ into $\operatorname*{FreeLie}V$. (This map $\iota _{V}^{\operatorname*{FreeLie}}$ is easily seen to be injective.) For any $\operatorname*{FreeLie}V$-module $M$ and any linear map $f:V\rightarrow M$, there exists a unique $1$-cocycle $F:\operatorname*{FreeLie}V\rightarrow M$ satisfying $f=F\circ\iota_{V}^{\operatorname*{FreeLie}}$. \end{theorem} \begin{vershort} Although we will not use this theorem anywhere in the following, let us briefly discuss how it is proven. Theorem \ref{thm.universal.FreeLie.der} cannot be proven as directly as we proved Theorem \ref{thm.universal.tensor.der}. Instead, a way to prove Theorem \ref{thm.universal.FreeLie.der} is by using the following lemma: \end{vershort} \begin{verlong} Although we will not use this theorem anywhere in the following, let us discuss how it is proven. Theorem \ref{thm.universal.FreeLie.der} cannot be proven as directly as we proved Theorem \ref{thm.universal.tensor.der}. Instead, a way to prove Theorem \ref{thm.universal.FreeLie.der} is by using the following lemma: \end{verlong} \begin{lemma} \label{lem.semidir.coc-to-deriv}Let $\mathfrak{g}$ be a Lie algebra. Let $M$ be a $\mathfrak{g}$-module. Define the semidirect product $\mathfrak{g}\ltimes M$ as in Definition \ref{def.semidir}. Let $\varphi:\mathfrak{g}\rightarrow M$ be a linear map. Then, $\varphi:\mathfrak{g}\rightarrow M$ is a $1$-cocycle if and only if the map% \[ \mathfrak{g}\rightarrow\mathfrak{g}\ltimes M,\ \ \ \ \ \ \ \ \ \ x\mapsto \left( x,\varphi\left( x\right) \right) \] is a Lie algebra homomorphism. \end{lemma} \begin{vershort} This lemma helps reducing Theorem \ref{thm.universal.FreeLie.der} to Theorem \ref{thm.universal.FreeLie}. We leave the details of this proof (both of the lemma and of Theorem \ref{thm.universal.FreeLie.der}) to the reader. \end{vershort} \begin{verlong} We will use this lemma to reduce Theorem \ref{thm.universal.FreeLie.der} to Theorem \ref{thm.universal.FreeLie}; let us, however, first establish the lemma itself: \textit{Proof of Lemma \ref{lem.semidir.coc-to-deriv}.} Let $\Phi$ denote the map \[ \mathfrak{g}\rightarrow\mathfrak{g}\ltimes M,\ \ \ \ \ \ \ \ \ \ x\mapsto \left( x,\varphi\left( x\right) \right) . \] Clearly, the map $\Phi$ is linear. We will now prove the following two assertions: \textit{Assertion }$\mathfrak{K}_{1}$\textit{:} If $\varphi:\mathfrak{g}% \rightarrow M$ is a $1$-cocycle, then $\Phi$ is a Lie algebra homomorphism. \textit{Assertion }$\mathfrak{K}_{2}$\textit{:} If $\Phi$ is a Lie algebra homomorphism, then $\varphi:\mathfrak{g}\rightarrow M$ is a $1$-cocycle. \textit{Proof of Assertion }$\mathfrak{K}_{1}$\textit{:} Assume that $\varphi:\mathfrak{g}\rightarrow M$ is a $1$-cocycle. By the definition of ``$1$-cocycle'', this means that \[ \varphi\left( \left[ a,b\right] \right) =a\rightharpoonup\varphi\left( b\right) -b\rightharpoonup\varphi\left( a\right) \ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathfrak{g}\text{ and }b\in \mathfrak{g}. \] Let $a\in\mathfrak{g}$ and $b\in\mathfrak{g}$. By the definition of $\Phi$, we have $\Phi\left( a\right) =\left( a,\varphi\left( a\right) \right) $ and $\Phi\left( b\right) =\left( b,\varphi\left( b\right) \right) $. Thus,% \[ \left[ \underbrace{\Phi\left( a\right) }_{=\left( a,\varphi\left( a\right) \right) },\underbrace{\Phi\left( b\right) }_{=\left( b,\varphi\left( b\right) \right) }\right] =\left[ \left( a,\varphi \left( a\right) \right) ,\left( b,\varphi\left( b\right) \right) \right] =\left( \left[ a,b\right] ,a\rightharpoonup\varphi\left( b\right) -b\rightharpoonup\varphi\left( a\right) \right) \] (by the definition of the semidirect product $\mathfrak{g}\ltimes M$). On the other hand, the definition of $\Phi$ yields \[ \Phi\left( \left[ a,b\right] \right) =\left( \left[ a,b\right] ,\underbrace{\varphi\left( \left[ a,b\right] \right) }_{=a\rightharpoonup \varphi\left( b\right) -b\rightharpoonup\varphi\left( a\right) }\right) =\left( \left[ a,b\right] ,a\rightharpoonup\varphi\left( b\right) -b\rightharpoonup\varphi\left( a\right) \right) =\left[ \Phi\left( a\right) ,\Phi\left( b\right) \right] . \] Now forget that we fixed $a$ and $b$. We thus have shown that every $a\in\mathfrak{g}$ and $b\in\mathfrak{g}$ satisfy $\Phi\left( \left[ a,b\right] \right) =\left[ \Phi\left( a\right) ,\Phi\left( b\right) \right] $. Since $\Phi$ is linear, this yields that $\Phi$ is a Lie algebra homomorphism. This proves Assertion $\mathfrak{K}_{1}$. \textit{Proof of Assertion }$\mathfrak{K}_{2}$\textit{:} Assume that $\Phi$ is a Lie algebra homomorphism. Thus,% \[ \Phi\left( \left[ a,b\right] \right) =\left[ \Phi\left( a\right) ,\Phi\left( b\right) \right] \ \ \ \ \ \ \ \ \ \ \text{for all }% a\in\mathfrak{g}\text{ and }b\in\mathfrak{g}. \] Now let $a\in\mathfrak{g}$ and $b\in\mathfrak{g}$. By the definition of $\Phi $, we have the three equalities $\Phi\left( a\right) =\left( a,\varphi \left( a\right) \right) $, $\Phi\left( b\right) =\left( b,\varphi\left( b\right) \right) $ and $\Phi\left( \left[ a,b\right] \right) =\left( \left[ a,b\right] ,\varphi\left( \left[ a,b\right] \right) \right) $. Thus,% \begin{align*} \left( \left[ a,b\right] ,\varphi\left( \left[ a,b\right] \right) \right) & =\Phi\left( \left[ a,b\right] \right) =\left[ \underbrace{\Phi\left( a\right) }_{=\left( a,\varphi\left( a\right) \right) },\underbrace{\Phi\left( b\right) }_{=\left( b,\varphi\left( b\right) \right) }\right] \\ & =\left[ \left( a,\varphi\left( a\right) \right) ,\left( b,\varphi\left( b\right) \right) \right] =\left( \left[ a,b\right] ,a\rightharpoonup\varphi\left( b\right) -b\rightharpoonup\varphi\left( a\right) \right) \end{align*} (by the definition of the semidirect product $\mathfrak{g}\ltimes M$). Hence, $\varphi\left( \left[ a,b\right] \right) =a\rightharpoonup\varphi\left( b\right) -b\rightharpoonup\varphi\left( a\right) $. Now forget that we fixed $a$ and $b$. We thus have shown that% \[ \varphi\left( \left[ a,b\right] \right) =a\rightharpoonup\varphi\left( b\right) -b\rightharpoonup\varphi\left( a\right) \ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathfrak{g}\text{ and }b\in \mathfrak{g}. \] By the definition of ``$1$-cocycle'', this means exactly that $\varphi :\mathfrak{g}\rightarrow M$ is a $1$-cocycle (since we already know that $\varphi$ is linear). We have therefore shown that $\varphi:\mathfrak{g}% \rightarrow M$ is a $1$-cocycle. This proves Assertion $\mathfrak{K}_{2}$. Combining Assertion $\mathfrak{K}_{1}$ with Assertion $\mathfrak{K}_{2}$, we conclude that $\varphi:\mathfrak{g}\rightarrow M$ is a $1$-cocycle if and only if $\Phi$ is a Lie algebra homomorphism. Since $\Phi$ is the map% \[ \mathfrak{g}\rightarrow\mathfrak{g}\ltimes M,\ \ \ \ \ \ \ \ \ \ x\mapsto \left( x,\varphi\left( x\right) \right) , \] this rewrites as follows: $\varphi:\mathfrak{g}\rightarrow M$ is a $1$-cocycle if and only if the map% \[ \mathfrak{g}\rightarrow\mathfrak{g}\ltimes M,\ \ \ \ \ \ \ \ \ \ x\mapsto \left( x,\varphi\left( x\right) \right) \] is a Lie algebra homomorphism. This proves Lemma \ref{lem.semidir.coc-to-deriv}. \textit{Proof of Theorem \ref{thm.universal.FreeLie.der}.} Fix a $\operatorname*{FreeLie}V$-module $M$ and any linear map $f:V\rightarrow M$. Let $\mathfrak{g}=\operatorname*{FreeLie}V$. Let $\pi_{1}:\mathfrak{g}\ltimes M\rightarrow\mathfrak{g}$ be the canonical projection on the first addend. Then, $\pi_{1}$ is known to be a Lie algebra homomorphism. Let $\pi _{2}:\mathfrak{g}\ltimes M\rightarrow M$ be the canonical projection on the second addend. Regard the canonical injection $\iota_{V}^{\operatorname*{FreeLie}% }:V\rightarrow\operatorname*{FreeLie}V=\mathfrak{g}$ as an inclusion. Let $\psi$ be the map% \[ V\rightarrow\mathfrak{g}\ltimes M,\ \ \ \ \ \ \ \ \ \ x\mapsto\left( x,f\left( x\right) \right) . \] This map $\psi$ is clearly linear. Thus, Theorem \ref{thm.universal.FreeLie} (applied to $\mathfrak{g}\ltimes M$ and $\psi$ instead of $\mathfrak{h}$ and $f$) yields that there exists a unique Lie algebra homomorphism $F:\operatorname*{FreeLie}V\rightarrow\mathfrak{g}\ltimes M$ satisfying $\psi=F\circ\iota_{V}^{\operatorname*{FreeLie}}$. Denote this $F$ by $\Psi$. Thus, $\Psi:\operatorname*{FreeLie}V\rightarrow\mathfrak{g}\ltimes M$ is a Lie algebra homomorphism satisfying $\psi=\Psi\circ\iota_{V}% ^{\operatorname*{FreeLie}}$. Since $\pi_{1}$ and $\Psi$ are Lie algebra homomorphisms, their composition $\pi_{1}\circ\Psi:\operatorname*{FreeLie}V\rightarrow\mathfrak{g}$ must also be a Lie algebra homomorphism. Note that $\operatorname*{id}% \nolimits_{\mathfrak{g}}$ also is a Lie algebra homomorphism from $\operatorname*{FreeLie}V$ to $\mathfrak{g}$ (since $\mathfrak{g}% =\operatorname*{FreeLie}V$). Every $x\in V$ satisfies $\iota_{V}^{\operatorname*{FreeLie}}\left( x\right) =x$ (since we regard the map $\iota_{V}^{\operatorname*{FreeLie}}$ as an inclusion). But every $x\in V$ satisfies% \[ \left( \pi_{1}\circ\psi\right) \left( x\right) =\pi_{1}\left( \underbrace{\psi\left( x\right) }_{\substack{=\left( x,f\left( x\right) \right) \\\text{(by the definition of }\psi\text{)}}}\right) =\pi_{1}\left( x,f\left( x\right) \right) =x \] (since $\pi_{1}:\mathfrak{g}\ltimes M\rightarrow\mathfrak{g}$ was defined as the canonical projection on the first addend). Thus, every $x\in V$ satisfies $\left( \pi_{1}\circ\psi\right) \left( x\right) =x=\iota_{V}% ^{\operatorname*{FreeLie}}\left( x\right) $. Hence, $\pi_{1}\circ\psi =\iota_{V}^{\operatorname*{FreeLie}}$. Applying Theorem \ref{thm.universal.FreeLie} to $\mathfrak{g}$ and $\pi _{1}\circ\psi$ instead of $\mathfrak{h}$ and $f$, we conclude that there exists a unique Lie algebra homomorphism $F:\operatorname*{FreeLie}% V\rightarrow\mathfrak{g}$ satisfying $\pi_{1}\circ\psi=F\circ\iota _{V}^{\operatorname*{FreeLie}}$. Hence, if $F_{1}$ and $F_{2}$ are two Lie algebra homomorphisms $\operatorname*{FreeLie}V\rightarrow\mathfrak{g}$ satisfying $\pi_{1}\circ\psi=F_{1}\circ\iota_{V}^{\operatorname*{FreeLie}}$ and $\pi_{1}\circ\psi=F_{2}\circ\iota_{V}^{\operatorname*{FreeLie}}$, then we must have $F_{1}=F_{2}$. Applying this to $F_{1}=\pi_{1}\circ\Psi$ and $F_{2}=\operatorname*{id}\nolimits_{\mathfrak{g}}$ (since $\pi_{1}% \circ\underbrace{\psi}_{=\Psi\circ\iota_{V}^{\operatorname*{FreeLie}}}=\pi _{1}\circ\Psi\circ\iota_{V}^{\operatorname*{FreeLie}}$ and $\pi_{1}\circ \psi=\iota_{V}^{\operatorname*{FreeLie}}=\operatorname*{id}% \nolimits_{\mathfrak{g}}\circ\iota_{V}^{\operatorname*{FreeLie}}$), we obtain $\pi_{1}\circ\Psi=\operatorname*{id}\nolimits_{\mathfrak{g}}$. Now, let $\varphi$ denote the linear map $\pi_{2}\circ\Psi :\operatorname*{FreeLie}V\rightarrow M$. Since $\operatorname*{FreeLie}% V=\mathfrak{g}$, this map $\varphi$ is a linear map $\mathfrak{g}\rightarrow M$. We shall now show that this map $\varphi$ is a $1$-cocycle satisfying $f=\varphi\circ\iota_{V}^{\operatorname*{FreeLie}}$. Indeed, every $x\in\mathfrak{g}$ satisfies% \[ \Psi\left( x\right) =\left( x,\varphi\left( x\right) \right) \] \footnote{\textit{Proof.} Let $x\in\mathfrak{g}$. Then, $x\in\mathfrak{g}% =\operatorname*{FreeLie}V$, so that $\Psi\left( x\right) $ is an element of $\mathfrak{g}\ltimes M$. Hence, we can write $\Psi\left( x\right) $ in the form $\Psi\left( x\right) =\left( y,z\right) $ for some $y\in\mathfrak{g}$ and $z\in M$ (by the definition of $\mathfrak{g}\ltimes M$). Consider these $y$ and $z$. Since $\Psi\left( x\right) =\left( y,z\right) $, we have $\pi_{1}\left( \Psi\left( x\right) \right) =\pi_{1}\left( y,z\right) =y$ (since $\pi_{1}:\mathfrak{g}\ltimes M\rightarrow\mathfrak{g}$ is the canonical projection on the first addend) and $\pi_{2}\left( \Psi\left( x\right) \right) =\pi_{2}\left( y,z\right) =z$ (since $\pi_{2}:\mathfrak{g}\ltimes M\rightarrow M$ is the canonical projection on the second addend). But now, $y=\pi_{1}\left( \Psi\left( x\right) \right) =\underbrace{\left( \pi _{1}\circ\Psi\right) }_{=\operatorname*{id}\nolimits_{\mathfrak{g}}}\left( x\right) =x$ and $z=\pi_{2}\left( \Psi\left( x\right) \right) =\underbrace{\left( \pi_{2}\circ\Psi\right) }_{=\varphi}\left( x\right) $, so we have $\Psi\left( x\right) =\left( \underbrace{y}_{=x},\underbrace{z}% _{=\varphi\left( x\right) }\right) =\left( x,\varphi\left( x\right) \right) $, qed.}. In other words, $\Psi$ is the map \[ \mathfrak{g}\rightarrow\mathfrak{g}\ltimes M,\ \ \ \ \ \ \ \ \ \ x\mapsto \left( x,\varphi\left( x\right) \right) . \] Since we know that $\Psi$ is a Lie algebra homomorphism, we thus conclude that the map \[ \mathfrak{g}\rightarrow\mathfrak{g}\ltimes M,\ \ \ \ \ \ \ \ \ \ x\mapsto \left( x,\varphi\left( x\right) \right) \] is a Lie algebra homomorphism. Hence, $\varphi:\mathfrak{g}\rightarrow M$ is a $1$-cocycle (because Lemma \ref{lem.semidir.coc-to-deriv} yields that $\varphi:\mathfrak{g}\rightarrow M$ is a $1$-cocycle if and only if the map% \[ \mathfrak{g}\rightarrow\mathfrak{g}\ltimes M,\ \ \ \ \ \ \ \ \ \ x\mapsto \left( x,\varphi\left( x\right) \right) \] is a Lie algebra homomorphism). In other words, $\varphi :\operatorname*{FreeLie}V\rightarrow M$ is a $1$-cocycle (since $\mathfrak{g}% =\operatorname*{FreeLie}V$). Every $x\in V$ satisfies% \begin{align*} \left( \underbrace{\varphi}_{=\pi_{2}\circ\Psi}\circ\iota_{V}% ^{\operatorname*{FreeLie}}\right) \left( x\right) & =\left( \pi_{2}% \circ\underbrace{\Psi\circ\iota_{V}^{\operatorname*{FreeLie}}}_{=\psi}\right) \left( x\right) =\left( \pi_{2}\circ\psi\right) \left( x\right) =\pi _{2}\left( \underbrace{\psi\left( x\right) }_{\substack{=\left( x,f\left( x\right) \right) \\\text{(by the definition of }\psi\text{)}}}\right) \\ & =\pi_{2}\left( x,f\left( x\right) \right) =f\left( x\right) \end{align*} (since $\pi_{2}:\mathfrak{g}\ltimes M\rightarrow M$ was defined as the canonical projection on the second addend). Thus, $\varphi\circ\iota _{V}^{\operatorname*{FreeLie}}=f$. Altogether, we know that $\varphi:\operatorname*{FreeLie}V\rightarrow M$ is a $1$-cocycle satisfying $f=\varphi\circ\iota_{V}^{\operatorname*{FreeLie}}$. We thus conclude that there exists a $1$-cocycle $F:\operatorname*{FreeLie}% V\rightarrow M$ satisfying $f=F\circ\iota_{V}^{\operatorname*{FreeLie}}$ (namely, $F=\varphi$). Now, let us prove the uniqueness of such an $F$. Let $F$ be a $1$-cocycle $\operatorname*{FreeLie}V\rightarrow M$ satisfying $f=F\circ\iota_{V}^{\operatorname*{FreeLie}}$. We will prove that $F=\varphi$. Recall that the free Lie algebra $\operatorname*{FreeLie}V$ is generated by its subset $V$ as a Lie algebra. Every $x\in V$ satisfies% \[ \underbrace{f}_{=F\circ\iota_{V}^{\operatorname*{FreeLie}}}\left( x\right) =\left( F\circ\iota_{V}^{\operatorname*{FreeLie}}\right) \left( x\right) =F\left( \underbrace{\iota_{V}^{\operatorname*{FreeLie}}\left( x\right) }_{=x}\right) =F\left( x\right) =\left( F\mid_{V}\right) \left( x\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }x\in V\right) \] and% \[ \underbrace{f}_{=\varphi\circ\iota_{V}^{\operatorname*{FreeLie}}}\left( x\right) =\left( \varphi\circ\iota_{V}^{\operatorname*{FreeLie}}\right) \left( x\right) =\varphi\left( \underbrace{\iota_{V}% ^{\operatorname*{FreeLie}}\left( x\right) }_{=x}\right) =\varphi\left( x\right) =\left( \varphi\mid_{V}\right) \left( x\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }x\in V\right) . \] Thus, every $x\in V$ satisfies $\left( F\mid_{V}\right) \left( x\right) =f\left( x\right) =\left( \varphi\mid_{V}\right) \left( x\right) $. Hence, $F\mid_{V}=\varphi\mid_{V}$. Thus, Proposition \ref{prop.derivation.Lie.unique} (applied to $\operatorname*{FreeLie}V$, $F$, $\varphi$ and $V$ instead of $\mathfrak{g}$, $d$, $e$ and $S$) yields that $F=\varphi$. Now forget that we fixed $F$. We thus have proven that every $1$-cocycle $F:\operatorname*{FreeLie}V\rightarrow M$ satisfying $f=F\circ\iota _{V}^{\operatorname*{FreeLie}}$ must satisfy $F=\varphi$. Hence, there exists at most one $1$-cocycle $F:\operatorname*{FreeLie}V\rightarrow M$ satisfying $f=F\circ\iota_{V}^{\operatorname*{FreeLie}}$. Combined with the fact that there exists a $1$-cocycle $F:\operatorname*{FreeLie}V\rightarrow M$ satisfying $f=F\circ\iota_{V}^{\operatorname*{FreeLie}}$ (this fact was proven above), this yields that there exists a unique $1$-cocycle $F:\operatorname*{FreeLie}V\rightarrow M$ satisfying $f=F\circ\iota _{V}^{\operatorname*{FreeLie}}$. This proves Theorem \ref{thm.universal.FreeLie.der}. \end{verlong} An alternative way to prove Theorem \ref{thm.universal.FreeLie.der} is the following: Apply Theorem \ref{thm.universal.tensor.der} to construct a derivation $F:T\left( V\right) \rightarrow M$ (of algebras) satisfying $f=F\circ\iota_{V}^{T}$, and then identify $\operatorname*{FreeLie}V$ with a Lie subalgebra of $T\left( V\right) $ (because Proposition \ref{prop.Ufree} $U\left( \operatorname*{FreeLie}V\right) \cong T\left( V\right) $, and because the Poincar\'{e}-Birkhoff-Witt theorem entails an injection $\operatorname*{FreeLie}V\rightarrow U\left( \operatorname*{FreeLie}V\right) $). Then, restricting the derivation $F:T\left( V\right) \rightarrow M$ to $\operatorname*{FreeLie}V$, we obtain a $1$-cocycle $\operatorname*{FreeLie}% V\rightarrow M$ with the required properties. The uniqueness part of Theorem \ref{thm.universal.FreeLie.der} is easy (and follows from Proposition \ref{prop.derivation.Lie.unique} below). This proof of Theorem \ref{thm.universal.FreeLie.der} has the disadvantage that it makes use of the Poincar\'{e}-Birkhoff-Witt theorem, which does not generalize to the case of Lie algebras over rings (whereas Theorem \ref{thm.universal.FreeLie.der} does generalize to this case). \subsubsection{Derivations from grading} The following simple lemma will help us defining derivations on Lie algebras: \begin{lemma} \label{lem.deriv.grading}Let $Q$ be an abelian group. Let $s\in \operatorname*{Hom}\left( Q,\mathbb{C}\right) $ be a group homomorphism. Let $\mathfrak{n}$ be a $Q$-graded Lie algebra. Let $\eta:\mathfrak{n}% \rightarrow\mathfrak{n}$ be a linear map satisfying% \begin{equation} \eta\left( x\right) =s\left( w\right) \cdot x\ \ \ \ \ \ \ \ \ \ \text{for every }w\in Q\text{ and every }x\in\mathfrak{n}\left[ w\right] . \label{lem.deriv.grading.1}% \end{equation} Then, $\eta$ is a derivation (of Lie algebras). \end{lemma} \textit{Proof of Lemma \ref{lem.deriv.grading}.} In order to prove that $\eta$ is a derivation, we need to check that \begin{equation} \eta\left( \left[ a,b\right] \right) =\left[ \eta\left( a\right) ,b\right] +\left[ a,\eta\left( b\right) \right] \ \ \ \ \ \ \ \ \ \ \text{for any }a\in\mathfrak{n}\text{ and }b\in \mathfrak{n}. \label{pf.deriv.grading.1}% \end{equation} Let us prove the equation (\ref{pf.deriv.grading.1}). Since this equation is linear in each of $a$ and $b$, we can WLOG assume that $a$ and $b$ are homogeneous (because any element of $\mathfrak{n}$ is a sum of homogeneous elements). So, assume this. We will write the binary operation of the group $Q$ as addition. Since $a$ is homogeneous, we have $a\in\mathfrak{n}\left[ u\right] $ for some $u\in Q$. Consider this $u$. Since $b$ is homogeneous, we have $b\in\mathfrak{n}\left[ v\right] $ for some $v\in Q$. Fix this $v$. Thus, $\left[ a,b\right] \in\mathfrak{n}\left[ u+v\right] $ (since $a\in\mathfrak{n}\left[ u\right] $ and $b\in\mathfrak{n}\left[ v\right] $ and since $\mathfrak{n}$ is $Q$-graded). Thus, (\ref{lem.deriv.grading.1}) (applied to $x=a+b$ and $w=u+v$) yields $\eta\left( \left[ a,b\right] \right) =\underbrace{s\left( u+v\right) }_{\substack{=s\left( u\right) +s\left( v\right) \\\text{(since }s\text{ is a group}\\\text{homomorphism)}% }}\cdot\left[ a,b\right] =\left( s\left( u\right) +s\left( v\right) \right) \cdot\left[ a,b\right] $. On the other hand, (\ref{lem.deriv.grading.1}) (applied to $x=a$ and $w=u$) yields $\eta\left( a\right) =s\left( u\right) \cdot a$. Also, (\ref{lem.deriv.grading.1}) (applied to $x=b$ and $y=v$) yields $\eta\left( b\right) =s\left( v\right) \cdot b$. Now,% \[ \left[ \underbrace{\eta\left( a\right) }_{=s\left( u\right) \cdot a},b\right] +\left[ a,\underbrace{\eta\left( b\right) }_{=s\left( v\right) \cdot b}\right] =s\left( u\right) \cdot\left[ a,b\right] +s\left( v\right) \cdot\left[ a,b\right] =\left( s\left( u\right) +s\left( v\right) \right) \cdot\left[ a,b\right] =\eta\left( \left[ a,b\right] \right) . \] This proves (\ref{pf.deriv.grading.1}). Now that (\ref{pf.deriv.grading.1}) is proven, we conclude that $\eta$ is a derivation. Lemma \ref{lem.deriv.grading} is proven. \subsubsection{The commutator of derivations} The following proposition is the classical analogue of Proposition \ref{prop.commutator.derivs} for algebras in lieu of Lie algebras: \begin{proposition} \label{prop.commutator.derivs.alg}Let $A$ be an algebra. Let $f:A\rightarrow A$ and $g:A\rightarrow A$ be two derivations of $A$. Then, $\left[ f,g\right] $ is a derivation of $A$. (Here, the Lie bracket is to be understood as the Lie bracket on $\operatorname*{End}A$, so that we have $\left[ f,g\right] =f\circ g-g\circ f$.) \end{proposition} The proof of this is completely analogous to that of Proposition \ref{prop.commutator.derivs}. Moreover, by the same argument, the following slight generalization of Proposition \ref{prop.commutator.derivs.alg} can be shown: \begin{proposition} \label{prop.commutator.derivs.alg.2}Let $A$ be a subalgebra of an algebra $B$. Let $f:A\rightarrow B$ and $g:B\rightarrow B$ be two derivations such that $g\left( A\right) \subseteq A$. Then, $f\circ\left( g\mid_{A}\right) -g\circ f:A\rightarrow B$ is a derivation. \end{proposition} \begin{verlong} \textit{Proof of Proposition \ref{prop.commutator.derivs.alg.2}.} Let $a\in A$ and $b\in A$. Since $f$ is a derivation, we have $f\left( ab\right) =f\left( a\right) \cdot b+a\cdot f\left( b\right) $. Thus,% \begin{align*} \left( g\circ f\right) \left( ab\right) & =g\left( \underbrace{f\left( ab\right) }_{=f\left( a\right) \cdot b+a\cdot f\left( b\right) }\right) =g\left( f\left( a\right) \cdot b+a\cdot f\left( b\right) \right) \\ & =\underbrace{g\left( f\left( a\right) \cdot b\right) }% _{\substack{=g\left( f\left( a\right) \right) \cdot b+f\left( a\right) \cdot g\left( b\right) \\\text{(since }g\text{ is a derivation)}% }}+\underbrace{g\left( a\cdot f\left( b\right) \right) }% _{\substack{=g\left( a\right) \cdot f\left( b\right) +a\cdot g\left( f\left( b\right) \right) \\\text{(since }g\text{ is a derivation)}}}\\ & =\underbrace{g\left( f\left( a\right) \right) }_{=\left( g\circ f\right) \left( a\right) }\cdot b+f\left( a\right) \cdot g\left( b\right) +g\left( a\right) \cdot f\left( b\right) +a\cdot \underbrace{g\left( f\left( b\right) \right) }_{=\left( g\circ f\right) \left( b\right) }\\ & =\left( g\circ f\right) \left( a\right) \cdot b+f\left( a\right) \cdot g\left( b\right) +g\left( a\right) \cdot f\left( b\right) +a\cdot\left( g\circ f\right) \left( b\right) . \end{align*} Let us notice that $f\left( g\left( a\right) \right) $ and $f\left( g\left( b\right) \right) $ are well-defined (since $g\left( \underbrace{a}% _{\in A}\right) \in g\left( A\right) \subseteq A$ and $g\left( \underbrace{b}_{\in A}\right) \in g\left( A\right) \subseteq A$). Since $g$ is a derivation, we have $g\left( ab\right) =g\left( a\right) \cdot b+a\cdot g\left( b\right) $. Thus,% \begin{align*} \left( f\circ\left( g\mid_{A}\right) \right) \left( ab\right) & =f\left( \underbrace{\left( g\mid_{A}\right) \left( ab\right) }_{=g\left( ab\right) =g\left( a\right) \cdot b+a\cdot g\left( b\right) }\right) =f\left( g\left( a\right) \cdot b+a\cdot g\left( b\right) \right) \\ & =\underbrace{f\left( g\left( a\right) \cdot b\right) }% _{\substack{=f\left( g\left( a\right) \right) \cdot b+g\left( a\right) \cdot f\left( b\right) \\\text{(since }g\text{ is a derivation)}% }}+\underbrace{f\left( a\cdot g\left( b\right) \right) }% _{\substack{=f\left( a\right) \cdot g\left( b\right) +a\cdot f\left( g\left( b\right) \right) \\\text{(since }g\text{ is a derivation)}}}\\ & =f\left( \underbrace{g\left( a\right) }_{=\left( g\mid_{A}\right) \left( a\right) }\right) \cdot b+g\left( a\right) \cdot f\left( b\right) +f\left( a\right) \cdot g\left( b\right) +a\cdot f\left( \underbrace{g\left( b\right) }_{=\left( g\mid_{A}\right) \left( b\right) }\right) \\ & =\underbrace{f\left( \left( g\mid_{A}\right) \left( a\right) \right) }_{=\left( f\circ\left( g\mid_{A}\right) \right) \left( a\right) }\cdot b+g\left( a\right) \cdot f\left( b\right) +f\left( a\right) \cdot g\left( b\right) +a\cdot\underbrace{f\left( \left( g\mid_{A}\right) \left( b\right) \right) }_{=\left( f\circ\left( g\mid_{A}\right) \right) \left( b\right) }\\ & =\left( f\circ\left( g\mid_{A}\right) \right) \left( a\right) \cdot b+g\left( a\right) \cdot f\left( b\right) +f\left( a\right) \cdot g\left( b\right) +a\cdot\left( f\circ\left( g\mid_{A}\right) \right) \left( b\right) . \end{align*} Thus,% \begin{align*} & \left( f\circ\left( g\mid_{A}\right) -g\circ f\right) \left( ab\right) \\ & =\underbrace{\left( f\circ\left( g\mid_{A}\right) \right) \left( ab\right) }_{=\left( f\circ\left( g\mid_{A}\right) \right) \left( a\right) \cdot b+g\left( a\right) \cdot f\left( b\right) +f\left( a\right) \cdot g\left( b\right) +a\cdot\left( f\circ\left( g\mid _{A}\right) \right) \left( b\right) }-\underbrace{\left( g\circ f\right) \left( ab\right) }_{=\left( g\circ f\right) \left( a\right) \cdot b+f\left( a\right) \cdot g\left( b\right) +g\left( a\right) \cdot f\left( b\right) +a\cdot\left( g\circ f\right) \left( b\right) }\\ & =\left( \left( f\circ\left( g\mid_{A}\right) \right) \left( a\right) \cdot b+g\left( a\right) \cdot f\left( b\right) +f\left( a\right) \cdot g\left( b\right) +a\cdot\left( f\circ\left( g\mid_{A}\right) \right) \left( b\right) \right) \\ & \ \ \ \ \ \ \ \ \ \ -\left( \left( g\circ f\right) \left( a\right) \cdot b+f\left( a\right) \cdot g\left( b\right) +g\left( a\right) \cdot f\left( b\right) +a\cdot\left( g\circ f\right) \left( b\right) \right) \\ & =\underbrace{\left( f\circ\left( g\mid_{A}\right) \right) \left( a\right) \cdot b-\left( g\circ f\right) \left( a\right) \cdot b}_{=\left( \left( f\circ\left( g\mid_{A}\right) \right) \left( a\right) -\left( g\circ f\right) \left( a\right) \right) \cdot b}+\underbrace{a\cdot\left( f\circ\left( g\mid_{A}\right) \right) \left( b\right) -a\cdot\left( g\circ f\right) \left( b\right) }_{=a\cdot\left( \left( f\circ\left( g\mid_{A}\right) \right) \left( b\right) -\left( g\circ f\right) \left( b\right) \right) }\\ & =\underbrace{\left( \left( f\circ\left( g\mid_{A}\right) \right) \left( a\right) -\left( g\circ f\right) \left( a\right) \right) }_{=\left( f\circ\left( g\mid_{A}\right) -g\circ f\right) \left( a\right) }\cdot b+a\cdot\underbrace{\left( \left( f\circ\left( g\mid _{A}\right) \right) \left( b\right) -\left( g\circ f\right) \left( b\right) \right) }_{=\left( f\circ\left( g\mid_{A}\right) -g\circ f\right) \left( b\right) }\\ & =\left( f\circ\left( g\mid_{A}\right) -g\circ f\right) \left( a\right) \cdot b+a\cdot\left( f\circ\left( g\mid_{A}\right) -g\circ f\right) \left( b\right) . \end{align*} We have thus proven that any $a\in A$ and $b\in A$ satisfy $\left( f\circ\left( g\mid_{A}\right) -g\circ f\right) \left( ab\right) =\left( f\circ\left( g\mid_{A}\right) -g\circ f\right) \left( a\right) \cdot b+a\cdot\left( f\circ\left( g\mid_{A}\right) -g\circ f\right) \left( b\right) $. In other words, $f\circ\left( g\mid_{A}\right) -g\circ f$ is a derivation. This proves Proposition \ref{prop.commutator.derivs.alg.2}. \textit{Proof of Proposition \ref{prop.commutator.derivs.alg}.} Applying Proposition \ref{prop.commutator.derivs.alg.2} to $B=A$, we obtain that $f\circ\left( g\mid_{A}\right) -g\circ f:A\rightarrow A$ is a derivation (since $g\left( A\right) \subseteq A$). In other words, $\left[ f,g\right] $ is a derivation (since $f\circ\underbrace{\left( g\mid_{A}\right) }% _{=g}-g\circ f=f\circ g-g\circ f=\left[ f,g\right] $). Hence, Proposition \ref{prop.commutator.derivs.alg} is proven. \end{verlong} \protect\begin{noncompile} \subsubsection{Extending degree-zero invariant forms on graded Lie algebras} In this subsection we will collect some results on how to construct a degree-$0$ invariant bilinear form on a graded Lie algebra if we are given its values on the first few homogeneous components and we know that these homogeneous components generate the whole Lie algebra. Our main results will be the following two theorems: \begin{theorem} [...][ext] \end{theorem} \begin{theorem} [...][symmetry gives symmetry] \end{theorem} These two theorems will be later applied to Kac-Moody algebras, on which it will allow the construction of an invariant form. Nevertheless I (Darij) am going to prove them in full generality (at least, the greatest generality known to me) and full detail, seeing that most references known to me offer neither. However, I don't know of any application of these theorems beyond the case of Kac-Moody algebras. We start with the easiest part: the uniqueness of the extended form. We state it in probably the most general form: \begin{proposition} \label{prop.bilext.uni1}Let $\mathfrak{g}$ be a $\mathbb{Z}$-graded Lie algebra over a field $k$. Let $M$ and $N$ be two $\mathbb{Z}$-graded $\mathfrak{g}$-modules. Let $K\in\mathbb{N}$ be a nonnegative integer. Assume that every integer $n>K$ satisfies% \begin{equation} N\left[ -n\right] =\sum\limits_{i=1}^{n-1}\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -n+i\right] . \label{prop.bilext.uni1.gen1}% \end{equation} \footnotemark\ For every $n\in\left\{ 1,2,...,K\right\} $, let $\beta_{n}$ be a $k$-bilinear form $M\left[ n\right] \times N\left[ -n\right] \rightarrow k$. Then, there exists \textbf{at most} one sequence $\left( \gamma_{1},\gamma_{2},\gamma_{3},...\right) $ of maps satisfying the following three properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}.3: \textit{Property \ref{prop.bilext.uni1}.1:} For every positive integer $n$, the map $\gamma_{n}$ is a $k$-bilinear form $M\left[ n\right] \times N\left[ -n\right] \rightarrow k$. \textit{Property \ref{prop.bilext.uni1}.2:} We have $\gamma_{n}=\beta_{n}$ for every $n\in\left\{ 1,2,...,K\right\} $. \textit{Property \ref{prop.bilext.uni1}.3:} Every positive integer $n$, every $i\in\left\{ 1,2,...,n-1\right\} $, every $x\in\mathfrak{g}\left[ -i\right] $, every $a\in M\left[ n\right] $ and every $b\in N\left[ -n+i\right] $ satisfy% \begin{equation} \gamma_{n-i}\left( x\rightharpoonup a,b\right) +\gamma_{n}\left( a,x\rightharpoonup b\right) =0. \label{prop.bilext.uni1.main}% \end{equation} \end{proposition} \footnotetext{Here and in the following, we are using the notation introduced in Definition \ref{def.liesubspace}.}\ This proposition is rather straightforward to prove by induction: \begin{vershort} \textit{Proof of Proposition \ref{prop.bilext.uni1}.} Let $\left( \delta _{1},\delta_{2},\delta_{3},...\right) $ and $\left( \varepsilon _{1},\varepsilon_{2},\varepsilon_{3},...\right) $ be two sequences $\left( \gamma_{1},\gamma_{2},\gamma_{3},...\right) $ of maps satisfying the properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}.3. All we need to prove is that $\left( \delta _{1},\delta_{2},\delta_{3},...\right) =\left( \varepsilon_{1},\varepsilon _{2},\varepsilon_{3},...\right) $. That is, all we need to prove is that $\delta_{n}=\varepsilon_{n}$ for every positive integer $n$. We are going to prove this by strong induction over $n$. This means that we fix a positive integer $n$ and are going to show that $\delta_{n}% =\varepsilon_{n}$ assuming that $\delta_{n^{\prime}}=\varepsilon_{n^{\prime}}$ holds for every positive integer $n^{\prime} K$. This entails that (\ref{prop.bilext.uni1.gen1}) holds. In order to prove that $\delta_{n}=\varepsilon_{n}$, we have to show that $\delta_{n}\left( a,b\right) =\varepsilon_{n}\left( a,b\right) $ for every $a\in M\left[ n\right] $ and $b\in N\left[ -n\right] $. So let $a\in M\left[ n\right] $ and $b\in N\left[ -n\right] $ be arbitrary. Then,% \[ b\in N\left[ -n\right] =\sum\limits_{i=1}^{n-1}\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -n+i\right] \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{prop.bilext.uni1.gen1})}\right) \] In other words, $b$ is a sum of elements of $\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -n+i\right] $ for varying $i\in\left\{ 1,2,...,n-1\right\} $. Since the identity that we want to prove (that is, $\delta_{n}\left( a,b\right) =\varepsilon_{n}\left( a,b\right) $) is $k$-linear in $b$, we can therefore WLOG assume that $b\in\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -n+i\right] $. Assume this. Then, $b$ is a $k$-linear combination of elements of the form $z\rightharpoonup u$ with $z\in\mathfrak{g}\left[ -i\right] $ and $u\in N\left[ -n+i\right] $. Again, by linearity, we can WLOG assume that $b$ is such an element (rather than only a $k$-linear combination). Assume this. So we have $b=z\rightharpoonup u$ for some $z\in\mathfrak{g}\left[ -i\right] $ and $u\in N\left[ -n+i\right] $. Considering this $z$ and this $u$, we have $\delta_{n}\left( a,b\right) =\delta_{n}\left( a,z\rightharpoonup u\right) $. But we know that $\left( \delta_{1},\delta_{2},\delta_{3},...\right) $ is a sequence $\left( \gamma_{1},\gamma_{2},\gamma_{3},...\right) $ of maps satisfying Property \ref{prop.bilext.uni1}.3. Hence, we can apply Property \ref{prop.bilext.uni1}.3 to $\delta_{j}$, $z$ and $u$ instead of $\gamma_{j}$, $x$ and $b$, and this yields $\delta_{n-i}\left( z\rightharpoonup a,u\right) +\delta_{n}\left( a,z\rightharpoonup u\right) =0$. Thus, $\delta_{n}\left( a,z\rightharpoonup u\right) =-\delta_{n-i}\left( z\rightharpoonup a,u\right) $. Altogether, $\delta_{n}\left( a,b\right) =\delta_{n}\left( a,z\rightharpoonup u\right) =-\delta_{n-i}\left( z\rightharpoonup a,u\right) $. The same argument, applied to $\varepsilon_{j}$ instead of $\delta_{j}$, shows that $\varepsilon_{n}\left( a,b\right) =-\varepsilon _{n-i}\left( z\rightharpoonup a,u\right) $. But recall the induction hypothesis saying that $\delta_{n^{\prime}}=\varepsilon_{n^{\prime}}$ holds for every positive integer $n^{\prime} K$. Let us consider Case 1 first. In this case, $\mathbf{n}\leq K$. Thus, $\mathbf{n}\in\left\{ 1,2,...,K\right\} $. Hence, we can apply Property \ref{prop.bilext.uni1}.2 to $\left( \gamma_{1},\gamma_{2},\gamma _{3},...\right) =\left( \delta_{1},\delta_{2},\delta_{3},...\right) $ and $n=\mathbf{n}$ (because we know that $\left( \delta_{1},\delta_{2},\delta _{3},...\right) $ is a sequence $\left( \gamma_{1},\gamma_{2},\gamma _{3},...\right) $ of maps satisfying Property \ref{prop.bilext.uni1}.2). As a consequence, we obtain $\delta_{\mathbf{n}}=\beta_{\mathbf{n}}$. The same argument (but with $\delta_{i}$ replaced by $\varepsilon_{i}$) yields that $\varepsilon_{\mathbf{n}}=\beta_{\mathbf{n}}$. Hence, $\delta_{\mathbf{n}% }=\beta_{\mathbf{n}}=\varepsilon_{\mathbf{n}}$. We have thus proven $\delta_{\mathbf{n}}=\varepsilon_{\mathbf{n}}$ in Case 1. Now, let us consider Case 2. In this case, $\mathbf{n}>K$. Let $a\in M\left[ \mathbf{n}\right] $. We are first going to show that \begin{equation} \delta_{\mathbf{n}}\left( a,b\right) =\varepsilon_{\mathbf{n}}\left( a,b\right) \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{ 1,2,...,\mathbf{n}-1\right\} \text{ and }b\in\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -\mathbf{n}+i\right] \text{.} \label{pf.bilext.uni1.1}% \end{equation} \textit{Proof of (\ref{pf.bilext.uni1.1}):} Let $i\in\left\{ 1,2,...,\mathbf{n}-1\right\} $ and $b\in\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -\mathbf{n}+i\right] $. By the definition of $\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -\mathbf{n}+i\right] $, we know that $\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -\mathbf{n}+i\right] $ is the $k$-linear span of all elements of the form $y\rightharpoonup u$ with $y\in\mathfrak{g}\left[ -i\right] $ and $u\in N\left[ -\mathbf{n}+i\right] $. Hence, the elements of $\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -\mathbf{n}+i\right] $ are $k$-linear combinations of elements of the form $z\rightharpoonup u$ with $z\in \mathfrak{g}\left[ -i\right] $ and $u\in N\left[ -\mathbf{n}+i\right] $. Since $b$ is an element of $\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -\mathbf{n}+i\right] $, this yields that $b$ is a $k$-linear combination of elements of the form $z\rightharpoonup u$ with $z\in \mathfrak{g}\left[ -i\right] $ and $u\in M\left[ -\mathbf{n}+i\right] $. In other words, there exist an $L\in\mathbb{N}$, some elements $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{L}$ of $k$, some elements $z_{1}$, $z_{2}$, $...$, $z_{L}$ of $\mathfrak{g}\left[ -i\right] $ and some elements $u_{1}$, $u_{2}$, $...$, $u_{L}$ of $M\left[ -\mathbf{n}+i\right] $ such that $b=\sum\limits_{\ell=1}^{L}\lambda_{\ell}z_{\ell}\rightharpoonup u_{\ell}$. Consider this $L$, these $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{L}$, these $z_{1}$, $z_{2}$, $...$, $z_{L}$, and these $u_{1}$, $u_{2}$, $...$, $u_{L}$. Since $b=\sum\limits_{\ell=1}^{L}\lambda_{\ell}z_{\ell}\rightharpoonup u_{\ell}$, we have \begin{align} \delta_{\mathbf{n}}\left( a,b\right) & =\delta_{\mathbf{n}}\left( a,\sum\limits_{\ell=1}^{L}\lambda_{\ell}z_{\ell}\rightharpoonup u_{\ell }\right) =\sum\limits_{\ell=1}^{L}\lambda_{\ell}\underbrace{\delta _{\mathbf{n}}\left( a,z_{\ell}\rightharpoonup u_{\ell}\right) }% _{\substack{=-\delta_{\mathbf{n}-i}\left( z_{\ell}\rightharpoonup a,u_{\ell }\right) \\\text{(by (\ref{pf.bilext.uni1.prop1delta}), applied to }\mathbf{n}\text{, }i\text{, }z_{\ell}\text{, }a\text{ and }u_{\ell }\\\text{instead of }n\text{, }i\text{, }x\text{, }a\text{ and }b\text{)}% }}\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\delta_{\mathbf{n}}\text{ is }k\text{-bilinear}\right) .\nonumber\\ & =\sum\limits_{\ell=1}^{L}\left( -\delta_{\mathbf{n}-i}\left( z_{\ell }\rightharpoonup a,u_{\ell}\right) \right) . \label{pf.bilext.uni1.3}% \end{align} The same argument (but with $\delta_{j}$ replaced by $\varepsilon_{j}$) yields% \begin{equation} \varepsilon_{\mathbf{n}}\left( a,b\right) =\sum\limits_{\ell=1}^{L}\left( -\varepsilon_{\mathbf{n}-i}\left( z_{\ell}\rightharpoonup a,u_{\ell}\right) \right) . \label{pf.bilext.uni1.4}% \end{equation} But since $i\in\left\{ 1,2,...,\mathbf{n}-1\right\} $, we have $i>0$, so that $\mathbf{n}-i<\mathbf{n}$. Hence, (\ref{pf.bilext.uni1.claim}) holds for $n=\mathbf{n}-i$ (because we have assumed that (\ref{pf.bilext.uni1.claim}) holds whenever $n<\mathbf{n}$). In other words, $\delta_{\mathbf{n}% -i}=\varepsilon_{\mathbf{n}-i}$. Now, (\ref{pf.bilext.uni1.3}) becomes% \[ \delta_{\mathbf{n}}\left( a,b\right) =\sum\limits_{\ell=1}^{L}\left( -\underbrace{\delta_{\mathbf{n}-i}}_{=\varepsilon_{\mathbf{n}-i}}\left( z_{\ell}\rightharpoonup a,u_{\ell}\right) \right) =\sum\limits_{\ell=1}% ^{L}\left( -\varepsilon_{\mathbf{n}-i}\left( z_{\ell}\rightharpoonup a,u_{\ell}\right) \right) =\varepsilon_{\mathbf{n}}\left( a,b\right) \] (by (\ref{pf.bilext.uni1.4})). This proves (\ref{pf.bilext.uni1.1}). Now, let $b$ be any element of $N\left[ -\mathbf{n}\right] $. Since $\mathbf{n}>K$, we can apply (\ref{prop.bilext.uni1.gen1}) to $n=\mathbf{n}$. As a result, we obtain% \[ N\left[ -\mathbf{n}\right] =\sum\limits_{i=1}^{\mathbf{n}-1}\mathfrak{g}% \left[ -i\right] \rightharpoonup N\left[ -\mathbf{n}+i\right] . \] Hence,% \[ b\in N\left[ -\mathbf{n}\right] =\sum\limits_{i=1}^{\mathbf{n}% -1}\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -\mathbf{n}% +i\right] . \] Thus, there exists an $\left( \mathbf{n}-1\right) $-tuple $\left( b_{1},b_{2},...,b_{\mathbf{n}-1}\right) $ of elements of $N$ such that $\left( \text{every }i\in\left\{ 1,2,...,\mathbf{n}-1\right\} \text{ satisfies }b_{i}\in\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -\mathbf{n}+i\right] \right) $ and $b=\sum\limits_{i=1}^{\mathbf{n}-1}b_{i}% $. Consider this $\left( \mathbf{n}-1\right) $-tuple $\left( b_{1}% ,b_{2},...,b_{\mathbf{n}-1}\right) $. Since $b=\sum\limits_{i=1}% ^{\mathbf{n}-1}b_{i}$, we have% \begin{align*} \delta_{\mathbf{n}}\left( a,b\right) & =\delta_{\mathbf{n}}\left( a,\sum\limits_{i=1}^{\mathbf{n}-1}b_{i}\right) =\sum\limits_{i=1}% ^{\mathbf{n}-1}\underbrace{\delta_{\mathbf{n}}\left( a,b_{i}\right) }_{\substack{=\varepsilon_{\mathbf{n}}\left( a,b_{i}\right) \\\text{(by (\ref{pf.bilext.uni1.1}) (applied to }b_{i}\text{ instead of }b\text{)}% \\\text{(since }b_{i}\in\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -\mathbf{n}+i\right] \text{))}}}\ \ \ \ \ \ \ \ \ \ \left( \text{since }\delta_{\mathbf{n}}\text{ is }k\text{-linear}\right) \\ & =\sum\limits_{i=1}^{\mathbf{n}-1}\varepsilon_{\mathbf{n}}\left( a,b_{i}\right) =\varepsilon_{\mathbf{n}}\left( a,\underbrace{\sum \limits_{i=1}^{\mathbf{n}-1}b_{i}}_{=b}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }\varepsilon_{\mathbf{n}}\text{ is }k\text{-linear}\right) \\ & =\varepsilon_{\mathbf{n}}\left( a,b\right) . \end{align*} Now, forget that we fixed $a$ and $b$. We thus have shown that $\delta _{\mathbf{n}}\left( a,b\right) =\varepsilon_{\mathbf{n}}\left( a,b\right) $ for every $a\in M\left[ \mathbf{n}\right] $ and $b\in N\left[ -\mathbf{n}\right] $. In other words, $\delta_{\mathbf{n}}=\varepsilon _{\mathbf{n}}$. We have thus proven $\delta_{\mathbf{n}}=\varepsilon _{\mathbf{n}}$ in Case 2. Hence, $\delta_{\mathbf{n}}=\varepsilon_{\mathbf{n}}$ is proven in each of the Cases 1 and 2. Since the Cases 1 and 2 cover all possibilities, this yields that $\delta_{\mathbf{n}}=\varepsilon_{\mathbf{n}}$ always holds. In other words, (\ref{pf.bilext.uni1.claim}) holds when $n=\mathbf{n}$. This completes the induction step. The induction proof of (\ref{pf.bilext.uni1.claim}) is thus complete. From (\ref{pf.bilext.uni1.claim}), we immediately obtain $\left( \delta _{1},\delta_{2},\delta_{3},...\right) =\left( \varepsilon_{1},\varepsilon _{2},\varepsilon_{3},...\right) $. Now, forget that we fixed $\left( \delta_{1},\delta_{2},\delta_{3}% ,...\right) $ and $\left( \varepsilon_{1},\varepsilon_{2},\varepsilon _{3},...\right) $. We thus have shown that if $\left( \delta_{1},\delta _{2},\delta_{3},...\right) $ and $\left( \varepsilon_{1},\varepsilon _{2},\varepsilon_{3},...\right) $ are two sequences $\left( \gamma _{1},\gamma_{2},\gamma_{3},...\right) $ of maps satisfying the properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}% .3, then $\left( \delta_{1},\delta_{2},\delta_{3},...\right) =\left( \varepsilon_{1},\varepsilon_{2},\varepsilon_{3},...\right) $. In other words, any two sequences $\left( \gamma_{1},\gamma_{2},\gamma_{3},...\right) $ of maps satisfying the properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}.3 must be equal. In other words, there exists at most one sequence $\left( \gamma_{1},\gamma_{2}% ,\gamma_{3},...\right) $ of maps satisfying the properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}% .3. This proves Proposition \ref{prop.bilext.uni1}. \end{verlong} We will now supplement this proposition with a corresponding existence statement, albeit one that requires more assumptions: \begin{proposition} \label{prop.bilext.1}Let $\mathfrak{g}$ be a $\mathbb{Z}$-graded Lie algebra over a field $k$. Let $M$ and $N$ be two $\mathbb{Z}$-graded $\mathfrak{g}% $-modules. Let $K\in\mathbb{N}$ be a nonnegative integer. Assume that every integer $n>K$ satisfies% \begin{equation} N\left[ -n\right] =\sum\limits_{i=1}^{n-1}\mathfrak{g}\left[ -i\right] \rightharpoonup N\left[ -n+i\right] . \label{prop.bilext.1.gen1}% \end{equation} Assume further that every integer $n>K$ satisfies% \begin{equation} M\left[ n\right] =\sum\limits_{i=1}^{n-1}\mathfrak{g}\left[ i\right] \rightharpoonup M\left[ n-i\right] . \label{prop.bilext.1.gen2}% \end{equation} For every $i\in\left\{ 1,2,...,K\right\} $, let $\beta_{i}$ be a $k$-bilinear form $M\left[ i\right] \times N\left[ -i\right] \rightarrow k$. Assume that every $n\in\left\{ 1,2,...,K\right\} $, every $i\in\left\{ 1,2,...,n-1\right\} $, every $x\in\mathfrak{g}\left[ -i\right] $, every $a\in M\left[ n\right] $ and every $b\in N\left[ -n+i\right] $ satisfy% \begin{equation} \beta_{n-i}\left( x\rightharpoonup a,b\right) +\beta_{n}\left( a,x\rightharpoonup b\right) =0. \label{prop.bilext.1.main}% \end{equation} Then, there exists \textbf{one and only one} sequence $\left( \gamma _{1},\gamma_{2},\gamma_{3},...\right) $ of maps satisfying the three properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}.3. (See Proposition \ref{prop.bilext.uni1} for the statements of these three properties.) \end{proposition} Before we prove this, we will need two lemmas: \begin{lemma} \label{lem.bilext.1.1}Let $p\in\mathbb{N}$. Let $k$ be a field. Let $W$ be a $k$-vector space. Let $V_{1}$, $V_{2}$, $...$, $V_{p}$ be $k$-vector spaces. For every $i\in\left\{ 1,2,...,p\right\} $, let $f_{i}:V_{i}\rightarrow W$ be a $k$-linear map. We denote by $\sum\limits_{i=1}^{p}f_{i}$ the map $\bigoplus\limits_{i=1}^{p}V_{i}\rightarrow W$ obtained from the maps $f_{i}:V_{i}\rightarrow W$ through the universal property of the direct sum $\bigoplus\limits_{i=1}^{p}V_{i}$. (This is the map which sends every $\left( v_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }\in\bigoplus\limits_{i=1}% ^{p}V_{i}$ to $\sum\limits_{i=1}^{p}f_{i}\left( v_{i}\right) $.) Then,% \[ \left( \sum\limits_{i=1}^{p}f_{i}\right) \left( \bigoplus\limits_{i=1}% ^{p}V_{i}\right) =\sum\limits_{i=1}^{p}f_{i}\left( V_{i}\right) . \] \end{lemma} Lemma \ref{lem.bilext.1.1} is just a basic fact from linear algebra. \begin{verlong} \textit{Proof of Lemma \ref{lem.bilext.1.1}.} Let $w\in\left( \sum \limits_{i=1}^{p}f_{i}\right) \left( \bigoplus\limits_{i=1}^{p}V_{i}\right) $. Then, there exists a $v\in\bigoplus\limits_{i=1}^{p}V_{i}$ such that $w=\left( \sum\limits_{i=1}^{p}f_{i}\right) \left( v\right) $. Consider this $v$. Since $v\in\bigoplus\limits_{i=1}^{p}V_{i}$, we know that $v$ can be written in the form $v=\left( v_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }$, where $v_{1}$, $v_{2}$, $...$, $v_{p}$ are vectors in $V_{1}$, $V_{2}$, $...$, $V_{p}$, respectively. Consider these $v_{1}$, $v_{2}$, $...$, $v_{p}$. Now,% \begin{align*} w & =\left( \sum\limits_{i=1}^{p}f_{i}\right) \left( v\right) =\left( \sum\limits_{i=1}^{p}f_{i}\right) \left( \left( v_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }v=\left( v_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }\right) \\ & =\sum\limits_{i=1}^{p}f_{i}\left( \underbrace{v_{i}}_{\in V_{i}}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\sum\limits_{i=1}% ^{p}f_{i}\right) \\ & \in\sum\limits_{i=1}^{p}f_{i}\left( V_{i}\right) . \end{align*} Now, forget that we fixed $w$. We thus have shown that $\left( \sum \limits_{i=1}^{p}f_{i}\right) \left( \bigoplus\limits_{i=1}^{p}V_{i}\right) \subseteq\sum\limits_{i=1}^{p}f_{i}\left( V_{i}\right) $. On the other hand, let $u\in\sum\limits_{i=1}^{p}f_{i}\left( V_{i}\right) $. Then, there exists a family $\left( u_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }$ of elements of $W$ which satisfies $u=\sum \limits_{i=1}^{p}u_{i}$, $\left( u_{i}\in f_{i}\left( V_{i}\right) \text{ for every }i\in\left\{ 1,2,...,p\right\} \right) $ and $\left( \text{all but finitely many }i\in\left\{ 1,2,...,p\right\} \text{ satisfy }% u_{i}=0\right) $\ \ \ \ \footnote{The third of these conditions is a tautology, of course.}. Consider such a family $\left( u_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }$. We know that for every $i\in\left\{ 1,2,...,p\right\} $, there exists an $x\in V_{i}$ such that $u_{i}% =f_{i}\left( x\right) $ (because $u_{i}\in f_{i}\left( V_{i}\right) $). Denote this $x$ by $x_{i}$. Then, for every $i\in\left\{ 1,2,...,p\right\} $, we have $x_{i}\in V_{i}$ and $u_{i}=f_{i}\left( x_{i}\right) $. But $\bigoplus\limits_{i=1}^{p}V_{i}=\prod\limits_{i=1}^{p}V_{i}$ (because a direct sum of finitely many $k$-vector spaces is the same as their direct product) and $\left( x_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }% \in\prod\limits_{i=1}^{p}V_{i}$ (since $x_{i}\in V_{i}$ for every $i\in\left\{ 1,2,...,p\right\} $). Hence, \[ \left( x_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }\in\prod \limits_{i=1}^{p}V_{i}=\bigoplus\limits_{i=1}^{p}V_{i}, \] so that $\left( \sum\limits_{i=1}^{p}f_{i}\right) \left( \left( x_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }\right) $ is well-defined. By the definition of $\sum\limits_{i=1}^{p}f_{i}$, we have% \[ \left( \sum\limits_{i=1}^{p}f_{i}\right) \left( \left( x_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }\right) =\sum\limits_{i=1}^{p}f_{i}\left( x_{i}\right) , \] so that% \[ u=\sum\limits_{i=1}^{p}\underbrace{u_{i}}_{=f_{i}\left( x_{i}\right) }% =\sum\limits_{i=1}^{p}f_{i}\left( x_{i}\right) =\left( \sum\limits_{i=1}% ^{p}f_{i}\right) \left( \left( x_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }\right) \in\left( \sum\limits_{i=1}^{p}f_{i}\right) \left( \bigoplus\limits_{i=1}^{p}V_{i}\right) . \] Now, forget that we fixed $u$. We thus have shown that every $u\in \sum\limits_{i=1}^{p}f_{i}\left( V_{i}\right) $ satisfies $u\in\left( \sum\limits_{i=1}^{p}f_{i}\right) \left( \bigoplus\limits_{i=1}^{p}% V_{i}\right) $. In other words, $\sum\limits_{i=1}^{p}f_{i}\left( V_{i}\right) \subseteq\left( \sum\limits_{i=1}^{p}f_{i}\right) \left( \bigoplus\limits_{i=1}^{p}V_{i}\right) $. Combining this with $\left( \sum\limits_{i=1}^{p}f_{i}\right) \left( \bigoplus\limits_{i=1}^{p}% V_{i}\right) \subseteq\sum\limits_{i=1}^{p}f_{i}\left( V_{i}\right) $, we obtain $\left( \sum\limits_{i=1}^{p}f_{i}\right) \left( \bigoplus \limits_{i=1}^{p}V_{i}\right) =\sum\limits_{i=1}^{p}f_{i}\left( V_{i}\right) $. This proves Lemma \ref{lem.bilext.1.1}. \end{verlong} \begin{lemma} \label{lem.bilext.1.2}Let $\mathfrak{g}$ be a $\mathbb{Z}$-graded Lie algebra over a field $k$. Let $M$ and $N$ be two $\mathbb{Z}$-graded $\mathfrak{g}% $-modules. Let $\mathbf{n}$ be a nonnegative integer. Let $\left( \gamma _{1},\gamma_{2},...,\gamma_{\mathbf{n}-1}\right) $ be a sequence of maps such that the two properties \ref{prop.bilext.uni1}.1 and \ref{prop.bilext.uni1}.3 are satisfied for all $n\leq\mathbf{n}-1$. [...] \end{lemma} \textit{Proof of Lemma \ref{lem.bilext.1.2}.} [...] \textit{Proof of Proposition \ref{prop.bilext.1}.} From Proposition \ref{prop.bilext.uni1}, we know that there exists \textbf{at most} one sequence $\left( \gamma_{1},\gamma_{2},\gamma_{3},...\right) $ of maps satisfying the three properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}.3. It thus remains to show that there exists \textbf{at least} one such sequence. So let us construct such a sequence. We will construct a sequence $\left( \gamma_{1},\gamma_{2},\gamma _{3},...\right) $ of maps satisfying the three properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}% .3 recursively, more precisely by strong induction. This means that we will assume that $\mathbf{n}$ is a positive integer and we are given a sequence $\left( \gamma_{1},\gamma_{2},...,\gamma_{\mathbf{n}-1}\right) $ of maps satisfying the three properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}.3 for all $n\leq \mathbf{n}-1$\ \ \ \ \footnote{When I say ``for all $n\leq\mathbf{n}-1$'', I don't mean to remove the conditions imposed on $n$ in the properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}% .3, but I just mean to add the extra condition that $n\leq\mathbf{n}-1$ to the existing conditions. So, for example, saying that Property \ref{prop.bilext.uni1}.2 holds for all $n\leq\mathbf{n}-1$ is equivalent to saying that $\gamma_{n}=\beta_{n}$ for every $n\in\left\{ 1,2,...,K\right\} $ satisfying $n\leq\mathbf{n}-1$, but not to saying that $\gamma_{n}=\beta _{n}$ for all positive integers $n$ whatsoever satisfying $n\leq\mathbf{n}% -1$.}, and we will show how to construct a new map $\gamma_{\mathbf{n}}$ which extends this sequence to a sequence $\left( \gamma_{1},\gamma_{2}% ,...,\gamma_{\mathbf{n}}\right) $ of maps satisfying the three properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}% .3 for all $n\leq\mathbf{n}$. This will allow recursively constructing maps $\gamma_{1}$, $\gamma_{2}$, $\gamma_{3}$, $...$ in such a way that the resulting sequence $\left( \gamma_{1},\gamma_{2},\gamma_{3},...\right) $ satisfies the three properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}.3 for all $n$. This will show that there exists at least one such sequence. So, let us perform the induction step in the recursive construction of the sequence $\left( \gamma_{1},\gamma_{2},\gamma_{3},...\right) $. Let $\mathbf{n}$ be a positive integer. Assume that we are given a sequence $\left( \gamma_{1},\gamma_{2},...,\gamma_{\mathbf{n}-1}\right) $ of maps satisfying the three properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}.3 for all $n\leq \mathbf{n}-1$. We need to construct a new map $\gamma_{\mathbf{n}}$ which extends this sequence to a sequence $\left( \gamma_{1},\gamma_{2}% ,...,\gamma_{\mathbf{n}}\right) $ of maps satisfying the three properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}% .3 for all $n\leq\mathbf{n}$. Let us define the map $\gamma_{\mathbf{n}}:M\left[ \mathbf{n}\right] \times N\left[ -\mathbf{n}\right] \rightarrow k$ as follows: Let us distinguish between two cases: \textit{Case 1:} We have $\mathbf{n}\in\left\{ 1,2,...,K\right\} $. \textit{Case 2:} We have $\mathbf{n}\notin\left\{ 1,2,...,K\right\} $. Let us consider Case 1 first. In this case, $\mathbf{n}\in\left\{ 1,2,...,K\right\} $. Thus, $\beta_{\mathbf{n}}$ is a well-defined map $M\left[ \mathbf{n}\right] \times N\left[ -\mathbf{n}\right] \rightarrow k$. Hence, we can just define $\gamma_{\mathbf{n}}$ to be $\beta_{\mathbf{n}}% $. Define $\gamma_{\mathbf{n}}$ this way. Thus, $\gamma_{\mathbf{n}}$ is defined in Case 1. (We will later prove that the sequence $\left( \gamma _{1},\gamma_{2},...,\gamma_{\mathbf{n}}\right) $ satisfies the three properties \ref{prop.bilext.uni1}.1, \ref{prop.bilext.uni1}.2 and \ref{prop.bilext.uni1}.3 for all $n\leq\mathbf{n}$.) Let us now consider Case 2. In this case, $\mathbf{n}\notin\left\{ 1,2,...,K\right\} $, so that $\mathbf{n}>K$ (since $\mathbf{n}$ is a positive integer). Let $i\in\left\{ 1,2,...,\mathbf{n}-1\right\} $. Then, $\mathbf{n}% -i\in\left\{ 1,2,...,\mathbf{n}-1\right\} $. In other words, $\mathbf{n}-i$ is a positive integer satisfying $\mathbf{n}-i\leq\mathbf{n}-1$. Hence, Property \ref{prop.bilext.uni1}.1 is satisfied for $n=\mathbf{n}-i$ (because we have assumed that Property \ref{prop.bilext.uni1}.1 is satisfied for all $n\leq\mathbf{n}-1$). Hence, we can apply Property \ref{prop.bilext.uni1}.1 to $n=\mathbf{n}-i$. We thus conclude that the map $\gamma_{\mathbf{n}-i}$ is a $k$-bilinear form $M\left[ \mathbf{n}-i\right] \times N\left[ -\left( \mathbf{n}-i\right) \right] \rightarrow k$. In other words, the map $\gamma_{\mathbf{n}-i}$ is a $k$-bilinear form $M\left[ \mathbf{n}-i\right] \times N\left[ -\mathbf{n}+i\right] \rightarrow k$. Define a map $\xi_{i}:\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] \rightarrow\operatorname*{Hom}\left( M\left[ \mathbf{n}\right] ,k\right) $ as follows: For every $\left( x,b\right) \in\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] $, define $\xi_{i}\left( x,b\right) $ to be the map% \begin{align*} M\left[ \mathbf{n}\right] & \rightarrow k,\\ a & \mapsto\gamma_{\mathbf{n}-i}\left( x\rightharpoonup a,b\right) . \end{align*} This is well-defined\footnote{\textit{Proof.} For every $\left( x,b\right) \in\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] $, the map% \begin{align*} M\left[ \mathbf{n}\right] & \rightarrow k,\\ a & \mapsto\gamma_{\mathbf{n}-i}\left( x\rightharpoonup a,b\right) \end{align*} is $k$-linear (since $\gamma_{\mathbf{n}-i}$ and the action of $\mathfrak{g}$ on $M$ are both $k$-bilinear), and hence is an element of $\operatorname*{Hom}% \left( M\left[ \mathbf{n}\right] ,k\right) $. Hence, $\xi_{i}$ is well-defined.}. Thus, we have defined a map $\xi_{i}$. For every $\left( x,b\right) \in\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] $, we know that $\xi_{i}\left( x,b\right) $ is the map% \begin{align*} M\left[ \mathbf{n}\right] & \rightarrow k,\\ a & \mapsto\gamma_{\mathbf{n}-i}\left( x\rightharpoonup a,b\right) \end{align*} (because this is how we have defined $\xi_{i}\left( x,b\right) $). In other words, for every $\left( x,b\right) \in\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] $, we have% \begin{equation} \left( \xi_{i}\left( x,b\right) \right) \left( a\right) =\gamma _{\mathbf{n}-i}\left( x\rightharpoonup a,b\right) \ \ \ \ \ \ \ \ \ \ \text{for every }a\in M\left[ \mathbf{n}\right] . \label{pf.bilext.1.xi_i}% \end{equation} The map $\xi_{i}$ is $k$-bilinear\footnote{\textit{Proof.} \textbf{a)} Let $x_{1}$ and $x_{2}$ be elements of $\mathfrak{g}\left[ -i\right] $. Let $\lambda_{1}$ and $\lambda_{2}$ be elements of $k$. Let $b$ be an element of $N\left[ -\mathbf{n}+i\right] $. Since $x_{1}\in\mathfrak{g}\left[ -i\right] $ and $x_{2}\in\mathfrak{g}\left[ -i\right] $, we have $\lambda_{1}x_{1}+\lambda_{2}x_{2}\in\mathfrak{g}\left[ -i\right] $, so that $\left( \lambda_{1}x_{1}+\lambda_{2}x_{2},b\right) \in\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] $. Thus, every $a\in M\left[ \mathbf{n}\right] $ satisfies% \begin{align} & \left( \xi_{i}\left( \lambda_{1}x_{1}+\lambda_{2}x_{2},b\right) \right) \left( a\right) \nonumber\\ & =\gamma_{\mathbf{n}-i}\left( \underbrace{\left( \lambda_{1}x_{1}% +\lambda_{2}x_{2}\right) \rightharpoonup a}_{=\lambda_{1}x_{1}\rightharpoonup a+\lambda_{2}x_{2}\rightharpoonup a},b\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.bilext.1.xi_i}), applied to }x=\lambda_{1}x_{1}+\lambda _{2}x_{2}\right) \nonumber\\ & =\gamma_{\mathbf{n}-i}\left( \lambda_{1}x_{1}\rightharpoonup a+\lambda _{2}x_{2}\rightharpoonup a,b\right) =\lambda_{1}\gamma_{\mathbf{n}-i}\left( x_{1}\rightharpoonup a,b\right) +\lambda_{2}\gamma_{\mathbf{n}-i}\left( x_{2}\rightharpoonup a,b\right) \label{pf.bilext.1.1}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\gamma_{\mathbf{n}-i}\text{ is a }k\text{-bilinear map}\right) .\nonumber \end{align} On the other hand, every $a\in M\left[ \mathbf{n}\right] $ satisfies \begin{equation} \left( \xi_{i}\left( x_{1},b\right) \right) \left( a\right) =\gamma_{\mathbf{n}-i}\left( x_{1}\rightharpoonup a,b\right) \label{pf.bilext.1.1b}% \end{equation} (by (\ref{pf.bilext.1.xi_i}), applied to $x=x_{1}$) and \begin{equation} \left( \xi_{i}\left( x_{2},b\right) \right) \left( a\right) =\gamma_{\mathbf{n}-i}\left( x_{2}\rightharpoonup a,b\right) \label{pf.bilext.1.1c}% \end{equation} (by (\ref{pf.bilext.1.xi_i}), applied to $x=x_{2}$). Hence, every $a\in M\left[ \mathbf{n}\right] $ satisfies \begin{align*} & \left( \xi_{i}\left( \lambda_{1}x_{1}+\lambda_{2}x_{2},b\right) \right) \left( a\right) \\ & =\lambda_{1}\underbrace{\gamma_{\mathbf{n}-i}\left( x_{1}\rightharpoonup a,b\right) }_{\substack{=\left( \xi_{i}\left( x_{1},b\right) \right) \left( a\right) \\\text{(by (\ref{pf.bilext.1.1b}))}}}+\lambda _{2}\underbrace{\gamma_{\mathbf{n}-i}\left( x_{2}\rightharpoonup a,b\right) }_{\substack{=\left( \xi_{i}\left( x_{2},b\right) \right) \left( a\right) \\\text{(by (\ref{pf.bilext.1.1c}))}}}\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.bilext.1.1})}\right) \\ & =\lambda_{1}\left( \xi_{i}\left( x_{1},b\right) \right) \left( a\right) +\lambda_{2}\left( \xi_{i}\left( x_{2},b\right) \right) \left( a\right) =\left( \lambda_{1}\xi_{i}\left( x_{1},b\right) +\lambda_{2}% \xi_{i}\left( x_{2},b\right) \right) \left( a\right) . \end{align*} In other words, $\xi_{i}\left( \lambda_{1}x_{1}+\lambda_{2}x_{2},b\right) =\lambda_{1}\xi_{i}\left( x_{1},b\right) +\lambda_{2}\xi_{i}\left( x_{2},b\right) $. \par Now, forget that we fixed $x_{1}$, $x_{2}$, $\lambda_{1}$, $\lambda_{2}$ and $b$. We thus have proven that $\xi_{i}\left( \lambda_{1}x_{1}+\lambda _{2}x_{2},b\right) =\lambda_{1}\xi_{i}\left( x_{1},b\right) +\lambda_{2}% \xi_{i}\left( x_{2},b\right) $ for all $x_{1}\in\mathfrak{g}\left[ -i\right] $, $x_{2}\in\mathfrak{g}\left[ -i\right] $, $\lambda_{1}\in k$, $\lambda_{2}\in k$ and $b\in N\left[ -\mathbf{n}+i\right] $. In other words, the map $\xi_{i}$ is $k$-linear in its first argument. \par \textbf{b)} Now, let $b_{1}$ and $b_{2}$ be elements of $N\left[ -\mathbf{n}+i\right] $. Let $\mu_{1}$ and $\mu_{2}$ be elements of $k$. Let $x$ be an element of $\mathfrak{g}\left[ -i\right] $. Since $b_{1}\in N\left[ -\mathbf{n}+i\right] $ and $b_{2}\in N\left[ -\mathbf{n}+i\right] $, we have $\mu_{1}b_{1}+\mu_{2}b_{2}\in N\left[ -\mathbf{n}+i\right] $, so that $\left( x,\mu_{1}b_{1}+\mu_{2}b_{2}\right) \in\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] $. Thus, every $a\in M\left[ \mathbf{n}\right] $ satisfies% \begin{align} & \left( \xi_{i}\left( x,\mu_{1}b_{1}+\mu_{2}b_{2}\right) \right) \left( a\right) \nonumber\\ & =\gamma_{\mathbf{n}-i}\left( x\rightharpoonup a,\mu_{1}b_{1}+\mu_{2}% b_{2}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.bilext.1.xi_i}), applied to }b=\mu_{1}b_{1}+\mu_{2}b_{2}\right) \nonumber\\ & =\mu_{1}\gamma_{\mathbf{n}-i}\left( x\rightharpoonup a,b_{1}\right) +\mu_{2}\gamma_{\mathbf{n}-i}\left( x\rightharpoonup a,b_{2}\right) \label{pf.bilext.1.2}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\gamma_{\mathbf{n}-i}\text{ is a }k\text{-bilinear map}\right) .\nonumber \end{align} On the other hand, every $a\in M\left[ \mathbf{n}\right] $ satisfies \begin{equation} \left( \xi_{i}\left( x,b_{1}\right) \right) \left( a\right) =\gamma_{\mathbf{n}-i}\left( x\rightharpoonup a,b_{1}\right) \label{pf.bilext.1.2b}% \end{equation} (by (\ref{pf.bilext.1.xi_i}), applied to $b=b_{1}$) and \begin{equation} \left( \xi_{i}\left( x,b_{2}\right) \right) \left( a\right) =\gamma_{\mathbf{n}-i}\left( x\rightharpoonup a,b_{2}\right) \label{pf.bilext.1.2c}% \end{equation} (by (\ref{pf.bilext.1.xi_i}), applied to $b=b_{2}$). Hence, every $a\in M\left[ \mathbf{n}\right] $ satisfies \begin{align*} & \left( \xi_{i}\left( x,\mu_{1}b_{1}+\mu_{2}b_{2}\right) \right) \left( a\right) \\ & =\mu_{1}\underbrace{\gamma_{\mathbf{n}-i}\left( x\rightharpoonup a,b_{1}\right) }_{\substack{=\left( \xi_{i}\left( x,b_{1}\right) \right) \left( a\right) \\\text{(by (\ref{pf.bilext.1.2b}))}}}+\mu_{2}% \underbrace{\gamma_{\mathbf{n}-i}\left( x\rightharpoonup a,b_{2}\right) }_{\substack{=\left( \xi_{i}\left( x,b_{2}\right) \right) \left( a\right) \\\text{(by (\ref{pf.bilext.1.2c}))}}}\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.bilext.1.2})}\right) \\ & =\mu_{1}\left( \xi_{i}\left( x,b_{1}\right) \right) \left( a\right) +\mu_{2}\left( \xi_{i}\left( x,b_{2}\right) \right) \left( a\right) =\left( \mu_{1}\xi_{i}\left( x,b_{1}\right) +\mu_{2}\xi_{i}\left( x,b_{2}\right) \right) \left( a\right) . \end{align*} In other words, $\xi_{i}\left( x,\mu_{1}b_{1}+\mu_{2}b_{2}\right) =\mu _{1}\xi_{i}\left( x,b_{1}\right) +\mu_{2}\xi_{i}\left( x,b_{2}\right) $. \par Now, forget that we fixed $b_{1}$, $b_{2}$, $\mu_{1}$, $\mu_{2}$ and $x$. We thus have proven that $\xi_{i}\left( x,\mu_{1}b_{1}+\mu_{2}b_{2}\right) =\mu_{1}\xi_{i}\left( x,b_{1}\right) +\mu_{2}\xi_{i}\left( x,b_{2}\right) $ for all $b_{1}\in N\left[ -\mathbf{n}+i\right] $, $b_{2}\in N\left[ -\mathbf{n}+i\right] $, $\mu_{1}\in k$, $\mu_{2}\in k$ and $x\in \mathfrak{g}\left[ -i\right] $. In other words, the map $\xi_{i}$ is $k$-linear in its second argument. \par \textbf{c)} Now, we know that the map $\xi_{i}$ is $k$-linear in its first argument and $k$-linear in its second argument. Hence, the map $\xi_{i}$ is $k$-bilinear, qed.}. Hence, by the universal property of the tensor product, this map $\xi_{i}:\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] \rightarrow\operatorname*{Hom}\left( M\left[ \mathbf{n}\right] ,k\right) $ gives rise to a $k$-linear map $\Xi _{i}:\mathfrak{g}\left[ -i\right] \otimes N\left[ -\mathbf{n}+i\right] \rightarrow\operatorname*{Hom}\left( M\left[ \mathbf{n}\right] ,k\right) $ satisfying% \[ \left( \Xi_{i}\left( x\otimes b\right) =\xi_{i}\left( x,b\right) \ \ \ \ \ \ \ \ \ \ \text{for every }\left( x,b\right) \in\mathfrak{g}% \left[ -i\right] \times N\left[ -\mathbf{n}+i\right] \right) . \] Consider this map $\Xi_{i}$. Then, every $x\in\mathfrak{g}\left[ -i\right] $ and $b\in N\left[ -\mathbf{n}+i\right] $ satisfy% \[ \Xi_{i}\left( x\otimes b\right) =\xi_{i}\left( x,b\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( x,b\right) \in \mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] \text{ (since }x\in\mathfrak{g}\left[ -i\right] \text{ and }b\in N\left[ -\mathbf{n}+i\right] \text{)}\right) . \] Thus, every $x\in\mathfrak{g}\left[ -i\right] $, $b\in N\left[ -\mathbf{n}+i\right] $ and $a\in M\left[ \mathbf{n}\right] $ satisfy% \begin{equation} \left( \underbrace{\Xi_{i}\left( x\otimes b\right) }_{=\xi_{i}\left( x,b\right) }\right) \left( a\right) =\left( \xi_{i}\left( x,b\right) \right) \left( a\right) =\gamma_{\mathbf{n}-i}\left( x\rightharpoonup a,b\right) \label{pf.bilext.1.Xi_i}% \end{equation} (by (\ref{pf.bilext.1.xi_i})). Now, forget that we fixed $i$. We thus have defined a $k$-linear map $\Xi _{i}:\mathfrak{g}\left[ -i\right] \otimes N\left[ -\mathbf{n}+i\right] \rightarrow\operatorname*{Hom}\left( M\left[ \mathbf{n}\right] ,k\right) $ for every $i\in\left\{ 1,2,...,\mathbf{n}-1\right\} $. Consider the map $\sum\limits_{i=1}^{\mathbf{n}-1}\Xi_{i}$ (defined in the same way as the map $\sum\limits_{i=1}^{p}f_{i}$ in Lemma \ref{lem.bilext.1.1}). Denote this map by $\Xi$. Thus, $\Xi=\sum\limits_{i=1}^{\mathbf{n}-1}\Xi_{i}$. On the other hand, let $i\in\left\{ 1,2,...,\mathbf{n}-1\right\} $ again. Define a map $\omega_{i}:\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] \rightarrow N\left[ -\mathbf{n}\right] $ by% \begin{equation} \left( \omega_{i}\left( x,b\right) =x\rightharpoonup b\ \ \ \ \ \ \ \ \ \ \text{for all }\left( x,b\right) \in\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] \right) . \label{pf.bilext.1.omega_i}% \end{equation} This is well-defined\footnote{\textit{Proof.} For every $\left( x,b\right) \in\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] $, the element $x\rightharpoonup b$ lies in $N\left[ -\mathbf{n}\right] $ (because we have $\left( x,b\right) \in\mathfrak{g}\left[ -i\right] \times N\left[ -\mathbf{n}+i\right] $, so that $x\in\mathfrak{g}\left[ -i\right] $ and $b\in N\left[ -\mathbf{n}+i\right] $, so that \begin{align*} \underbrace{x}_{\in\mathfrak{g}\left[ -i\right] }\rightharpoonup \underbrace{b}_{\in N\left[ -\mathbf{n}+i\right] } & \in\mathfrak{g}% \left[ -i\right] \rightharpoonup N\left[ -\mathbf{n}+i\right] \subseteq N\left[ \underbrace{\left( -i\right) +\left( -\mathbf{n}+i\right) }_{=-\mathbf{n}}\right] \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }N\text{ is a }\mathbb{Z}% \text{-graded }\mathfrak{g}\text{-module}\right) \\ & =N\left[ -\mathbf{n}\right] \end{align*} ). Hence, $\omega_{i}$ is well-defined, qed.}. We have thus defined a map $\omega_{i}$. [...] According to Lemma \ref{lem.bilext.1.1} (applied to $p=\mathbf{n}-1$, $W=\operatorname*{Hom}\left( M\left[ \mathbf{n}\right] ,k\right) $, $V_{i}=\mathfrak{g}\left[ -i\right] \otimes N\left[ -\mathbf{n}+i\right] $ and $f_{i}=\Xi_{i}$), we have% \[ \left( \sum\limits_{i=1}^{\mathbf{n}-1}\Xi_{i}\right) \left( \bigoplus \limits_{i=1}^{\mathbf{n}-1}V_{i}\right) =\sum\limits_{i=1}^{\mathbf{n}-1}% \Xi_{i}\left( V_{i}\right) \] [Quote]Let $p\in\mathbb{N}$. Let $k$ be a field. Let $W$ be a $k$-vector space. Let $V_{1}$, $V_{2}$, $...$, $V_{p}$ be $k$-vector spaces. For every $i\in\left\{ 1,2,...,p\right\} $, let $f_{i}:V_{i}\rightarrow W$ be a $k$-linear map. We denote by $\sum\limits_{i=1}^{p}f_{i}$ the map $\bigoplus\limits_{i=1}^{p}V_{i}\rightarrow W$ obtained from the maps $f_{i}:V_{i}\rightarrow W$ through the universal property of the direct sum $\bigoplus\limits_{i=1}^{p}V_{i}$. (This is the map which sends every $\left( v_{i}\right) _{i\in\left\{ 1,2,...,p\right\} }\in\bigoplus\limits_{i=1}% ^{p}V_{i}$ to $\sum\limits_{i=1}^{p}f_{i}\left( v_{i}\right) $.) Then,% \[ \left( \sum\limits_{i=1}^{p}f_{i}\right) \left( \bigoplus\limits_{i=1}% ^{p}V_{i}\right) =\sum\limits_{i=1}^{p}f_{i}\left( V_{i}\right) . \] [...] Let $a\in M\left[ \mathbf{n}\right] $ and $b\in N\left[ -\mathbf{n}\right] $. [...] [...] [...] \end{noncompile} \subsection{Simple Lie algebras: a recollection} The Kac-Moody Lie algebras form a class of Lie algebras which contains all simple finite-dimensional and all affine Lie algebras, but also many more. Before we start studying them, let us recall some facts about simple Lie algebras: Let $\mathfrak{g}$ be a finite-dimensional simple Lie algebra over $\mathbb{C}$. A \textit{Cartan subalgebra} of $\mathfrak{g}$ means a maximal commutative Lie subalgebra which consists of semisimple\footnote{An element of a Lie algebra is said to be \textit{semisimple} if and only if its action on the adjoint representation is a semisimple operator.} elements. There are usually many Cartan subalgebras of $\mathfrak{g}$, but they are all conjugate under the action of the corresponding Lie group $G$ (which satisfies $\mathfrak{g}=\operatorname*{Lie}G$, and can be defined as the connected component of the identity in the group $\operatorname*{Aut}\mathfrak{g}$). Thus, there is no loss of generality in picking one such subalgebra. So pick a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$. We denote the dimension $\dim\mathfrak{h}$ by $n$ and also by $\operatorname*{rank}\mathfrak{g}$. This dimension $\dim\mathfrak{h}=\operatorname*{rank}\mathfrak{g}$ is called the \textit{rank} of $\mathfrak{g}$. The restriction of the Killing form on $\mathfrak{g}$ to $\mathfrak{h}\times\mathfrak{h}$ is a nondegenerate symmetric bilinear form on $\mathfrak{h}$. For every $\alpha\in\mathfrak{h}^{\ast}$, we can define a vector subspace $\mathfrak{g}_{\alpha}$ of $\mathfrak{g}$ by \[ \mathfrak{g}_{\alpha}=\left\{ a\in\mathfrak{g}\ \mid\ \left[ h,a\right] =\alpha\left( h\right) a\text{ for all }h\in\mathfrak{h}\right\} . \] It can be shown that $\mathfrak{g}_{0}=\mathfrak{h}$. Now we let $\Delta$ be the finite subset $\left\{ \alpha\in\mathfrak{h}^{\ast}\diagdown\left\{ 0\right\} \ \mid\ \mathfrak{g}_{\alpha}\neq0\right\} $ of $\mathfrak{h}% ^{\ast}\diagdown\left\{ 0\right\} $. Then, $\mathfrak{g}=\mathfrak{h}% \oplus\bigoplus\limits_{\alpha\in\Delta}\mathfrak{g}_{\alpha}$ (as a direct sum of vector spaces). The subset $\Delta$ is called the \textit{root system} of $\mathfrak{g}$. The elements of $\Delta$ are called the \textit{roots} of $\mathfrak{g}$. It is known that for each $\alpha\in\Delta$, the vector space $\mathfrak{g}_{\alpha}$ is one-dimensional and can be written as $\mathfrak{g}_{\alpha}=\mathbb{C}e_{\alpha}$ for some particular $e_{\alpha }\in\mathfrak{g}_{\alpha}$. We want to use the decomposition $\mathfrak{g}=\mathfrak{h}\oplus \bigoplus\limits_{\alpha\in\Delta}\mathfrak{g}_{\alpha}$ in order to construct a triangular decomposition of $\mathfrak{g}$. This can be done with the grading which we constructed in Proposition \ref{prop.grad.g}, but let us do it again now, with more elementary means: Fix an $\overline{h}\in\mathfrak{h}$ such that every $\alpha\in\Delta$ satisfies $\alpha\left( \overline {h}\right) \in\mathbb{R}\diagdown\left\{ 0\right\} $ (it can be seen that such $\overline{h}$ exists). Define $\Delta_{+}=\left\{ \alpha\in\Delta \ \mid\ \alpha\left( \overline{h}\right) >0\right\} $ and $\Delta _{-}=\left\{ \alpha\in\Delta\ \mid\ \alpha\left( \overline{h}\right) <0\right\} $. Then, $\Delta$ is the union of two disjoint subsets $\Delta _{+}$ and $\Delta_{-}$, and we have $\Delta_{+}=-\Delta_{-}$. The triangular decomposition of $\mathfrak{g}$ is now defined as $\mathfrak{g}=\mathfrak{n}% _{-}\oplus\mathfrak{h}\oplus\mathfrak{n}_{+}$, where $\mathfrak{n}% _{-}=\bigoplus\limits_{\alpha\in\Delta_{-}}\mathfrak{g}_{\alpha}$ and $\mathfrak{n}_{+}=\bigoplus\limits_{\alpha\in\Delta_{+}}\mathfrak{g}_{\alpha}% $. This decomposition depends on the choice of $\overline{h}$ (and $\mathfrak{h}$, of course). The elements of $\Delta_{+}$ are called \textit{positive roots} of $\mathfrak{g}$, and the elements of $\Delta_{-}$ are called \textit{negative roots} of $\mathfrak{g}$. If $\alpha$ is a root of $\mathfrak{g}$, then we write $\alpha>0$ if $\alpha$ is a positive root, and we write $\alpha<0$ if $\alpha$ is a negative root. Let us now construct the grading on $\mathfrak{g}$ which yields this triangular decomposition $\mathfrak{g}=\mathfrak{n}_{-}\oplus\mathfrak{h}% \oplus\mathfrak{n}_{+}$. This grading was already constructed in Proposition \ref{prop.grad.g}, but now we are going to do this in detail: We define the \textit{simple roots} of $\mathfrak{g}$ as the elements of $\Delta_{+}$ which cannot be written as sums of more than one element of $\Delta_{+}$. It can be shown that there are exactly $n$ of these simple roots, and they form a basis of $\mathfrak{h}^{\ast}$. Denote these simple roots as $\alpha_{1}$, $\alpha_{2}$, $...$, $\alpha_{n}$. Every root $\alpha\in\Delta_{+}$ can now be written in the form $\alpha=\sum \limits_{i=1}^{n}k_{i}\left( \alpha\right) \alpha_{i}$ for a unique $n$-tuple $\left( k_{1}\left( \alpha\right) ,k_{2}\left( \alpha\right) ,...,k_{n}\left( \alpha\right) \right) $ of nonnegative integers. For all $\alpha,\beta\in\Delta$ with $\alpha+\beta\notin\Delta\cup\left\{ 0\right\} $, we have $\left[ \mathfrak{g}_{\alpha},\mathfrak{g}_{\beta }\right] =0$. For all $\alpha,\beta\in\mathfrak{h}^{\ast}$, we have $\left[ \mathfrak{g}_{\alpha},\mathfrak{g}_{\beta}\right] \subseteq\mathfrak{g}% _{\alpha+\beta}$. In particular, for every $\alpha\in\mathfrak{h}^{\ast}$, we have $\left[ \mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}\right] \subseteq\mathfrak{h}$. Better yet, we can show that for every $\alpha \in\Delta$, there exists some nonzero $h_{\alpha}\in\mathfrak{h}$ such that $\left[ \mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}\right] =\mathbb{C}% h_{\alpha}$. For every $i\in\left\{ 1,2,...,n\right\} $, pick a generator $e_{i}$ of the vector space $\mathfrak{g}_{\alpha_{i}}$ and a generator $f_{i}$ of the vector space $\mathfrak{g}_{-\alpha_{i}}$. It is possible to normalize $e_{i}$ and $f_{i}$ in such a way that $\left[ h_{i},e_{i}\right] =2e_{i}$ and $\left[ h_{i},f_{i}\right] =-2f_{i}$, where $h_{i}=\left[ e_{i},f_{i}\right] $. This $h_{i}$ will, of course, lie in $\mathfrak{h}$ and be a scalar multiple of $h_{\alpha_{i}}$. We can normalize $h_{\alpha_{i}}$ in such a way that $h_{i}=h_{\alpha_{i}}$. We suppose that all these normalizations are done. Then: \begin{proposition} \label{prop.serre-gen.1}With the notations introduced above, we have: \textbf{(a)} The family $\left( h_{1},h_{2},...,h_{n}\right) $ is a basis of $\mathfrak{h}$. \textbf{(b)} For any $i$ and $j$ in $\left\{ 1,2,...,n\right\} $, denote $\alpha_{j}\left( h_{i}\right) $ by $a_{i,j}$. The Lie algebra $\mathfrak{g}$ is generated (as a Lie algebra) by the elements $e_{i}$, $f_{i}$ and $h_{i}$ with $i\in\left\{ 1,2,...,n\right\} $ (a total of $3n$ elements), and the following relations hold:% \begin{align*} \left[ h_{i},h_{j}\right] & =0\ \ \ \ \ \ \ \ \ \ \text{for all }% i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ h_{i},e_{j}\right] & =\alpha_{j}\left( h_{i}\right) e_{j}% =a_{i,j}e_{j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ h_{i},f_{j}\right] & =-\alpha_{j}\left( h_{i}\right) f_{j}=a_{i,j}f_{j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ e_{i},f_{j}\right] & =\delta_{i,j}h_{i}% \ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} . \end{align*} (This does not mean that no more relations hold. In fact, additional relations, the so-called Serre relations, do hold in $\mathfrak{g}$; we will see these relations later, in Theorem \ref{thm.serre-gen.2}.) The $n\times n$ matrix $A=\left( a_{i,j}\right) _{1\leq i,j\leq n}$ is called the \textit{Cartan matrix} of $\mathfrak{g}$. Let $\left( \cdot,\cdot\right) $ denote the standard form on $\mathfrak{g}$ (defined in Definition \ref{def.standform}). Then, $\left( \cdot ,\cdot\right) $ is a nonzero scalar multiple of the Killing form on $\mathfrak{g}$ (since any two nonzero invariant symmetric bilinear forms on $\mathfrak{g}$ are scalar multiples of each other). Hence, the restriction of $\left( \cdot,\cdot\right) $ to $\mathfrak{h}\times\mathfrak{h}$ is nondegenerate (since the restriction of the Killing form to $\mathfrak{h}% \times\mathfrak{h}$ is nondegenerate). Thus, this restriction gives rise to a vector space isomorphism $\mathfrak{h}\rightarrow\mathfrak{h}^{\ast}$. This isomorphism sends $h_{i}$ to $\alpha_{i}^{\vee}=\dfrac{2\alpha_{i}}{\left( \alpha_{i},\alpha_{i}\right) }$ for every $i$ (where we denote by $\left( \cdot,\cdot\right) $ not only the standard form, but also the inverse form of its restriction to $\mathfrak{h}$). Thus, $a_{i,j}=\alpha_{j}\left( h_{i}\right) =\dfrac{2\left( \alpha_{j},\alpha_{i}\right) }{\left( \alpha_{i},\alpha_{i}\right) }$ for all $i$ and $j$. (Note that the latter equality would still hold if $\left( \cdot,\cdot\right) $ would mean the Killing form rather than the standard form.) The elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ are called \textit{Chevalley generators} of $\mathfrak{g}$. \textbf{Properties of the matrix }$A$\textbf{:} \textbf{1)} We have $a_{i,i}=2$ for all $i\in\left\{ 1,2,...,n\right\} $. \textbf{2)} Any two distinct $i\in\left\{ 1,2,...,n\right\} $ and $j\in\left\{ 1,2,...,n\right\} $ satisfy $a_{i,j}\leq0$ and $a_{i,j}% \in\mathbb{Z}$. Also, $a_{i,j}=0$ if and only if $a_{j,i}=0$. \textbf{3)} The matrix $A$ is indecomposable (i. e., if conjugation of $A$ by a permutation matrix brings $A$ into a block-diagonal form $\left( \begin{array} [c]{cc}% A_{1} & 0\\ 0 & A_{2}% \end{array} \right) $, then either $A_{1}$ or $A_{2}$ is a $0\times0$ matrix). \textbf{4)} The matrix $A$ is positive. Here is what we mean by this: There exists a diagonal $n\times n$ matrix $D$ with positive diagonal entries such that $DA$ is a symmetric and positive definite matrix. \end{proposition} \begin{theorem} An $n\times n$ matrix $A=\left( a_{i,j}\right) _{1\leq i,j\leq n}$ satisfies the four properties \textbf{1)}, \textbf{2)}, \textbf{3)} and \textbf{4)} of Proposition \ref{prop.serre-gen.1} if and only if it is a Cartan matrix of a simple Lie algebra. \end{theorem} Such matrices (and thus, simple finite-dimensional Lie algebras) can be encoded by so-called \textit{Dynkin diagrams}. The \textit{Dynkin diagram} of a simple Lie algebra $\mathfrak{g}$ is defined as the graph with vertex set $\left\{ 1,2,...,n\right\} $, and the following rules for drawing edges\footnote{The notion of a graph we are using here is slightly different from the familiar notions of a graph in graph theory, since this graph can have both directed and undirected edges.}: \begin{itemize} \item If $a_{i,j}=0$, then the vertices $i$ and $j$ are not connected by any edge (directed or undirected). \item If $a_{i,j}=a_{j,i}=-1$, then the vertices $i$ and $j$ are connected by exactly one edge, and this edge is undirected. \item If $a_{i,j}=-2$ and $a_{j,i}=-1$, then the vertices $i$ and $j$ are connected by two directed edges from $j$ to $i$ (and no other edges). \item If $a_{i,j}=-3$ and $a_{j,i}=-1$, then the vertices $i$ and $j$ are connected by three directed edges from $j$ to $i$ (and no other edges). \end{itemize} Here is a classification of simple finite-dimensional Lie algebras by their Dynkin diagrams: $A_{n}=\mathfrak{sl}\left( n+1\right) $ for $n\geq1$; the Dynkin diagram is $% %TCIMACRO{\TeXButton{x}{\xymatrix{ %\circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{}[r]|-{...} & \circ\ar@ %{-}[r] & \circ\ar@{-}[r] & \circ}}}% %BeginExpansion \xymatrix{ \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{}[r]|-{...} & \circ\ar@ {-}[r] & \circ\ar@{-}[r] & \circ}% %EndExpansion $ (with $n$ nodes). $B_{n}=\mathfrak{so}\left( 2n+1\right) $ for $n\geq2$; the Dynkin diagram is $% %TCIMACRO{\TeXButton{x}{\xymatrix{ %\circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{}[r]|-{...} & \circ\ar@ %{-}[r] & \circ\ar@{=>}[r] & \circ}}}% %BeginExpansion \xymatrix{ \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{}[r]|-{...} & \circ\ar@ {-}[r] & \circ\ar@{=>}[r] & \circ}% %EndExpansion $ (with $n$ nodes, only the last edge being directed and double). (Note that $\mathfrak{so}\left( 3\right) \cong\mathfrak{sl}\left( 2\right) $.) $C_{n}=\mathfrak{sp}\left( 2n\right) $ for $n\geq2$; the Dynkin diagram is $% %TCIMACRO{\TeXButton{x}{\xymatrix{ %\circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{}[r]|-{...} & \circ\ar@ %{-}[r] & \circ\ar@{<=}[r] & \circ}}}% %BeginExpansion \xymatrix{ \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{}[r]|-{...} & \circ\ar@ {-}[r] & \circ\ar@{<=}[r] & \circ}% %EndExpansion $ (with $n$ nodes, only the last edge being directed and double). (Note that $\mathfrak{sp}\left( 2\right) \cong\mathfrak{sl}\left( 2\right) $ and $\mathfrak{sp}\left( 4\right) \cong\mathfrak{so}\left( 5\right) $.) $D_{n}=\mathfrak{so}\left( 2n\right) $ for $n\geq4$; the Dynkin diagram is $% %TCIMACRO{\TeXButton{x}{\xymatrix{ %& & & & & \circ\\ %\circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{}[r]|-{...} & \circ\ar@ %{-}[r] & \circ\ar@{-}[ru] \ar@{-}[rd] \\ %& & & & & \circ}}}% %BeginExpansion \xymatrix{ & & & & & \circ\\ \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{}[r]|-{...} & \circ\ar@ {-}[r] & \circ\ar@{-}[ru] \ar@{-}[rd] \\ & & & & & \circ}% %EndExpansion $ (with $n$ nodes). (Note that $\mathfrak{so}\left( 4\right) \cong% \mathfrak{sl}\left( 2\right) \oplus\mathfrak{sl}\left( 2\right) $ and $\mathfrak{so}\left( 6\right) \cong\mathfrak{sl}\left( 4\right) $.) Exceptional Lie algebras: $E_{6}$; the Dynkin diagram is $% %TCIMACRO{\TeXButton{x}{\xymatrix{ %& & \circ\ar@{-}[d] & & \\ %\circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ %}}}% %BeginExpansion \xymatrix{ & & \circ\ar@{-}[d] & & \\ \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ}% %EndExpansion $. $E_{7}$; the Dynkin diagram is $% %TCIMACRO{\TeXButton{x}{\xymatrix{ %& & \circ\ar@{-}[d] & & & \\ %\circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}% %[r] & \circ\ar@{-}[r] & \circ}}}% %BeginExpansion \xymatrix{ & & \circ\ar@{-}[d] & & & \\ \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}% [r] & \circ\ar@{-}[r] & \circ}% %EndExpansion $. $E_{8}$; the Dynkin diagram is $% %TCIMACRO{\TeXButton{x}{\xymatrix{ %& & \circ\ar@{-}[d] & & & & \\ %\circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}% %[r] & \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ}}}% %BeginExpansion \xymatrix{ & & \circ\ar@{-}[d] & & & & \\ \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{-}% [r] & \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ}% %EndExpansion $. $F_{4}$; the Dynkin diagram is $% %TCIMACRO{\TeXButton{x}{\xymatrix{ %\circ\ar@{-}[r] & \circ\ar@{=>}[r] & \circ\ar@{-}[r] & \circ}}}% %BeginExpansion \xymatrix{ \circ\ar@{-}[r] & \circ\ar@{=>}[r] & \circ\ar@{-}[r] & \circ}% %EndExpansion $. $G_{2}$; the Dynkin diagram is $% %TCIMACRO{\TeXButton{x}{\xymatrix{ %\circ& \circ\ar@3{->}[l] %}}}% %BeginExpansion \xymatrix{ \circ& \circ\ar@3{->}[l] }% %EndExpansion $. Now to the Serre relations, which we have not yet written down: \begin{theorem} \label{thm.serre-gen.2}Let $\mathfrak{g}$ be a simple finite-dimensional Lie algebra. Use the notations introduced in Proposition \ref{prop.serre-gen.1}. \textbf{(a)} Let $i$ and $j$ be two distinct elements of $\left\{ 1,2,...,n\right\} $. Then, in $\mathfrak{g}$, we have $\left( \operatorname*{ad}\left( e_{i}\right) \right) ^{1-a_{i,j}}e_{j}=0$ and $\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}% =0$. These relations (totalling up to $2n\left( n-1\right) $ relations, because there are $n\left( n-1\right) $ pairs $\left( i,j\right) $ of distinct elements of $\left\{ 1,2,...,n\right\} $) are called the \textit{Serre relations} for $\mathfrak{g}$. \textbf{(b)} Combined with the relations% \begin{equation} \left\{ \begin{array} [c]{l}% \left[ h_{i},h_{j}\right] =0\ \ \ \ \ \ \ \ \ \ \text{for all }% i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ h_{i},e_{j}\right] =a_{i,j}e_{j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ h_{i},f_{j}\right] =-a_{i,j}f_{j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ e_{i},f_{j}\right] =\delta_{i,j}h_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} \end{array} \right. \label{nonserre-relations}% \end{equation} of Proposition \ref{prop.serre-gen.1}, the Serre relations form a set of defining relations for $\mathfrak{g}$. This means that, if $\widetilde{\mathfrak{g}}$ denotes the quotient Lie algebra \[ \operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) , \] then $\widetilde{\mathfrak{g}}\diagup\left( \text{Serre relations}\right) \cong\mathfrak{g}$. (Here, $\operatorname*{FreeLie}\left( h_{i},f_{i}% ,e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) $ denotes the free Lie algebra with $3n$ generators $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$.) \end{theorem} \begin{remark} If $\mathfrak{g}\cong\mathfrak{sl}_{2}$, then $\mathfrak{g}$ has no Serre relations (because $n=1$), and thus the claim of Theorem \ref{thm.serre-gen.2} \textbf{(b)} rewrites as $\widetilde{\mathfrak{g}}\cong\mathfrak{g}$ (where $\widetilde{\mathfrak{g}}$ is defined as in Theorem \ref{thm.serre-gen.2}). But in all other cases, the Lie algebra $\widetilde{\mathfrak{g}}$ is infinite-dimensional, and while it clearly projects onto $\mathfrak{g}$, it is much bigger than $\mathfrak{g}$. \end{remark} We will give a partial proof of Theorem \ref{thm.serre-gen.2}: We will only prove part \textbf{(a)}. \textit{Proof of Theorem \ref{thm.serre-gen.2} \textbf{(a)}.} Define a $\mathbb{C}$-linear map% \begin{align*} \Phi_{i}:\mathfrak{sl}_{2} & \rightarrow\mathfrak{g},\\ e & \mapsto e_{i},\\ f & \mapsto f_{i},\\ h & \mapsto h_{i}. \end{align*} Since $\left[ e_{i},f_{i}\right] =h_{i}$, $\left[ h_{i},e_{i}\right] =2e_{i}$ and $\left[ h_{i},f_{i}\right] =-2f_{i}$, this map $\Phi_{i}$ is a Lie algebra homomorphism. But $\mathfrak{g}$ is a $\mathfrak{g}$-module (by the adjoint representation of $\mathfrak{g}$), and thus becomes an $\mathfrak{sl}_{2}$-module by means of $\Phi_{i}:\mathfrak{sl}_{2}\rightarrow\mathfrak{g}$. This $\mathfrak{sl}_{2}% $-module satisfies% \[ ef_{j}=\underbrace{\left( \Phi_{i}\left( e\right) \right) }_{=e_{i}}% f_{j}=\left( \operatorname*{ad}\left( e_{i}\right) \right) f_{j}=\left[ e_{i},f_{j}\right] =0\ \ \ \ \ \ \ \ \ \ \left( \text{since }i\neq j\right) \] and% \[ hf_{j}=\underbrace{\left( \Phi_{i}\left( h\right) \right) }_{=h_{i}}% f_{j}=\left( \operatorname*{ad}\left( h_{i}\right) \right) f_{j}=\left[ h_{i},f_{j}\right] =-a_{i,j}f_{j}. \] Hence, Lemma \ref{lem.serre-gen.sl2} \textbf{(c)} (applied to $V=\mathfrak{g}% $, $\lambda=-a_{i,j}$ and $x=f_{j}$) yields that $-a_{i,j}\in\mathbb{N}$ and $f^{-a_{i,j}+1}f_{j}=0$. Since% \[ f^{-a_{i,j}+1}f_{j}=f^{1-a_{i,j}}f_{j}=\left( \underbrace{\Phi_{i}\left( f\right) }_{=f_{i}}\right) ^{1-a_{i,j}}f_{j}=\left( \operatorname*{ad}% \left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}, \] this rewrites as $\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}=0$. Similarly, $\left( \operatorname*{ad}\left( e_{i}\right) \right) ^{1-a_{i,j}}e_{j}=0$. Theorem \ref{thm.serre-gen.2} \textbf{(a)} is thus proven. As we said, we are not going to prove Theorem \ref{thm.serre-gen.2} \textbf{(b)} here. \subsection{\textbf{[unfinished]} Kac-Moody Lie algebras: definition and construction} Now forget about our simple Lie algebra $\mathfrak{g}$. Let us first define the notion of contragredient Lie algebras by axioms; we will construct these algebras later. \begin{definition} \label{def.contragredient}Suppose that $A=\left( a_{i,j}\right) _{1\leq i,j\leq n}$ is any $n\times n$ matrix of complex numbers. Let $Q$ be the free abelian group generated by $n$ symbols $\alpha_{1}$, $\alpha_{2}$, $...$, $\alpha_{n}$ (that is, $Q=\mathbb{Z}\alpha_{1}% \oplus\mathbb{Z}\alpha_{2}\oplus...\oplus\mathbb{Z}\alpha_{n}$). These symbols are just symbols, not weights of any Lie algebra (at the moment). We write the group $Q$ additively. A \textit{contragredient Lie algebra} corresponding to $A$ is a $Q$-graded $\mathbb{C}$-Lie algebra $\mathfrak{g}$ which is (as a Lie algebra) generated by some elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ which satisfy the following three conditions: \textbf{(1)} These elements satisfy the relations (\ref{nonserre-relations}). \textbf{(2)} The vector space $\mathfrak{g}\left[ 0\right] $ has $\left( h_{1},h_{2},...,h_{n}\right) $ as a $\mathbb{C}$-vector space basis, and we have $\mathfrak{g}\left[ \alpha_{i}\right] =\mathbb{C}e_{i}$ and $\mathfrak{g}\left[ -\alpha_{i}\right] =\mathbb{C}f_{i}$ for all $i\in\left\{ 1,2,...,n\right\} $. \textbf{(3)} Every nonzero $Q$-graded ideal in $\mathfrak{g}$ has a nonzero intersection with $\mathfrak{g}\left[ 0\right] $. (Here, we are using the notation $\mathfrak{g}\left[ \alpha\right] $ for the $\alpha$-th homogeneous component of the $Q$-graded Lie algebra $\mathfrak{g}% $, just as in Definition \ref{def.Q-graded.lie}.) Just as in the case of $\mathbb{Z}$-graded Lie algebras, we will denote $\mathfrak{g}\left[ 0\right] $ by $\mathfrak{h}$. \end{definition} Note that the condition \textbf{(3)} is satisfied for simple finite-dimensional Lie algebras $\mathfrak{g}$ (graded by their weight spaces, where $Q$ is the root lattice\footnote{in the meaning which this word has in the theory of simple Lie algebras} of $\mathfrak{g}$, and $A$ is the Cartan matrix); hence, simple finite-dimensional Lie algebras (graded by their weight spaces) are contragredient. \begin{theorem} \label{thm.g(A).exuni}Let $A=\left( a_{i,j}\right) _{1\leq i,j\leq n}$ be a (fixed) $n\times n$ matrix of complex numbers. \textbf{(a)} Then, there exists a unique (up to $Q$-graded isomorphism respecting the generators $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$) contragredient Lie algebra $\mathfrak{g}$ corresponding to $A$. \textbf{(b)} If $A$ is a Cartan matrix, then the contragredient Lie algebra $\mathfrak{g}$ corresponding to $A$ is finite-dimensional and simple. \end{theorem} \begin{definition} Let $A$ be an $n\times n$ matrix of complex numbers. Then, the unique (up to isomorphism) contragredient Lie algebra $\mathfrak{g}$ corresponding to $A$ is denoted by $\mathfrak{g}\left( A\right) $. \end{definition} The proof of Theorem \ref{thm.g(A).exuni} rests upon the following fact: \begin{theorem} \label{thm.gtilde}Let $A=\left( a_{i,j}\right) _{1\leq i,j\leq n}$ be an $n\times n$ matrix of complex numbers. Let $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ be $3n$ distinct symbols (which are, a priori, new and unrelated to the vectors $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ in Definition \ref{def.contragredient}). Let $\widetilde{\mathfrak{g}}$ be the quotient Lie algebra \[ \operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) . \] (Here, $\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) $ denotes the free Lie algebra with $3n$ generators $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$.) By abuse of notation, we will denote the projections of the elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ onto the quotient Lie algebra $\widetilde{\mathfrak{g}}$ by the same letters $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$. Let $Q$ be the free abelian group generated by $n$ symbols $\alpha_{1}$, $\alpha_{2}$, $...$, $\alpha_{n}$ (that is, $Q=\mathbb{Z}\alpha_{1}% \oplus\mathbb{Z}\alpha_{2}\oplus...\oplus\mathbb{Z}\alpha_{n}$). These symbols are just symbols, not weights of any Lie algebra (at the moment). \textbf{(a)} We can make $\widetilde{\mathfrak{g}}$ uniquely into a $Q$-graded Lie algebra by setting% \[ \deg\left( e_{i}\right) =\alpha_{i},\ \ \ \ \ \ \ \ \ \ \deg\left( f_{i}\right) =-\alpha_{i}\ \ \ \ \ \ \ \ \ \ \text{and }\deg\left( h_{i}\right) =0\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{ 1,2,...,n\right\} . \] \textbf{(b)} Let $\widetilde{\mathfrak{n}}_{+}=\operatorname*{FreeLie}\left( e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) $ (this means the free Lie algebra with $n$ generators $e_{1}$, $e_{2}$, $...$, $e_{n}$). Let $\widetilde{\mathfrak{n}}_{-}=\operatorname*{FreeLie}\left( f_{i}% \ \mid\ i\in\left\{ 1,2,...,n\right\} \right) $ (this means the free Lie algebra with $n$ generators $f_{1}$, $f_{2}$, $...$, $f_{n}$). Let $\widetilde{\mathfrak{h}}$ be the free vector space with basis $h_{1},h_{2},...,h_{n}$. Consider $\widetilde{\mathfrak{h}}$ as an abelian Lie algebra. Then, we have well-defined canonical Lie algebra homomorphisms $\iota _{+}:\widetilde{\mathfrak{n}}_{+}\rightarrow\widetilde{\mathfrak{g}}$ and $\iota_{-}:\widetilde{\mathfrak{n}}_{-}\rightarrow\widetilde{\mathfrak{g}}$ given by sending the generators $e_{1}$, $e_{2}$, $...$, $e_{n}$ (in the case of $\iota_{+}$), respectively, $f_{1}$, $f_{2}$, $...$, $f_{n}$ (in the case of $\iota_{-}$) to the corresponding generators $e_{1}$, $e_{2}$, $...$, $e_{n}$ (in the case of $\iota_{+}$), respectively, $f_{1}$, $f_{2}$, $...$, $f_{n}$ (in the case of $\iota_{-}$). Moreover, we have a well-defined linear map $\iota_{0}:\widetilde{\mathfrak{h}}\rightarrow\widetilde{\mathfrak{g}}$ given by sending the generators $h_{1}$, $h_{2}$, $...$, $h_{n}$ to $h_{1}$, $h_{2}$, $...$, $h_{n}$, respectively. These maps $\iota_{+}$, $\iota_{-}$ and $\iota_{0}$ are injective Lie algebra homomorphisms. \textbf{(c)} We have $\widetilde{\mathfrak{g}}=\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \oplus\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \oplus\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. \textbf{(d)} Both $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \oplus\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ and $\iota _{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \oplus\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ are Lie subalgebras of $\widetilde{\mathfrak{g}}$. \textbf{(e)} The $0$-th homogeneous component of $\widetilde{\mathfrak{g}}$ (in the $Q$-grading) is $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. That is, $\widetilde{\mathfrak{g}}\left[ 0\right] =\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. Moreover,% \[ \bigoplus_{\substack{\alpha\text{ is a }\mathbb{Z}\text{-linear combination}% \\\text{of }\alpha_{1}\text{, }\alpha_{2}\text{, }...\text{, }\alpha_{n}\text{ with nonnegative}\\\text{coefficients; }\alpha\neq0}}\widetilde{\mathfrak{g}% }\left[ \alpha\right] =\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \] and% \[ \bigoplus_{\substack{\alpha\text{ is a }\mathbb{Z}\text{-linear combination}% \\\text{of }\alpha_{1}\text{, }\alpha_{2}\text{, }...\text{, }\alpha_{n}\text{ with nonpositive}\\\text{coefficients; }\alpha\neq0}}\widetilde{\mathfrak{g}% }\left[ \alpha\right] =\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) . \] \textbf{(f)} There exists an involutive Lie algebra automorphism of $\widetilde{\mathfrak{g}}$ which sends $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ to $f_{1}$, $f_{2}$, $...$, $f_{n}$, $e_{1}$, $e_{2}$, $...$, $e_{n}$, $-h_{1}$, $-h_{2}$, $...$, $-h_{n}$, respectively. \textbf{(g)} Every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\widetilde{\mathfrak{g}}\left[ \alpha_{i}\right] =\mathbb{C}e_{i}$ and $\widetilde{\mathfrak{g}}\left[ -\alpha_{i}\right] =\mathbb{C}f_{i}$. \textbf{(h)} Let $I$ be the sum of all $Q$-graded ideals in $\widetilde{\mathfrak{g}}$ which have zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. Then, $I$ itself is a $Q$-graded ideal in $\widetilde{\mathfrak{g}}$ which has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. \textbf{(i)} Let $\mathfrak{g}=\widetilde{\mathfrak{g}}\diagup I$. Clearly, $\mathfrak{g}$ is a $Q$-graded Lie algebra. The projections of the elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\widetilde{\mathfrak{g}}$ on the quotient Lie algebra $\widetilde{\mathfrak{g}}\diagup I=\mathfrak{g}$ will still be denoted by $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}% $, $h_{2}$, $...$, $h_{n}$. Then, $\mathfrak{g}$ is a contragredient Lie algebra corresponding to $A$. \end{theorem} \begin{definition} Let $A$ be an $n\times n$ matrix of complex numbers. Then, the Lie algebra $\widetilde{\mathfrak{g}}$ defined in Theorem \ref{thm.gtilde} is denoted by $\widetilde{\mathfrak{g}}\left( A\right) $. \end{definition} \textit{Proof of Theorem \ref{thm.gtilde}.} First of all, for the sake of clarity, let us make a convention: In the following proof, the word ``Lie derivation'' will always mean ``derivation of Lie algebras'', whereas the word ``derivation'' without the word ``Lie'' directly in front of it will always mean ``derivation of algebras''. The only exception to this will be the formulation ``$\mathfrak{a}$ acts on $\mathfrak{b}$ by derivations'' where $\mathfrak{a}$ and $\mathfrak{b}$ are two Lie algebras; this formulation has been defined in Definition \ref{def.semidir.lielie} \textbf{(a)}. \bigskip \begin{vershort} \textbf{(f)} The relations% \[ \left\{ \begin{array} [c]{l}% \left[ -h_{i},-h_{j}\right] =0\ \ \ \ \ \ \ \ \ \ \text{for all }% i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ -h_{i},f_{j}\right] =a_{i,j}f_{j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ -h_{i},e_{j}\right] =-a_{i,j}e_{j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ f_{i},e_{j}\right] =\delta_{i,j}\left( -h_{i}\right) \ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} \end{array} \right. \] are satisfied in $\widetilde{\mathfrak{g}}$ (since they are easily seen to be equivalent to the relations (\ref{nonserre-relations}), and the relations (\ref{nonserre-relations}) are satisfied in $\widetilde{\mathfrak{g}}$ by the definition of $\widetilde{\mathfrak{g}}$). Hence, we can define a Lie algebra homomorphism \[ \omega:\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) \rightarrow\widetilde{\mathfrak{g}}% \] by requiring% \[ \left\{ \begin{array} [c]{c}% \omega\left( e_{i}\right) =f_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} ;\\ \omega\left( f_{i}\right) =e_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} ;\\ \omega\left( h_{i}\right) =-h_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} \end{array} \right. . \] Consider this $\omega$. Since $\operatorname*{FreeLie}\left( h_{i}% ,f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) =\widetilde{\mathfrak{g}}$, this homomorphism $\omega$ is a Lie algebra endomorphism of $\widetilde{\mathfrak{g}}$. It is easy to see that the Lie algebra homomorphisms $\omega^{2}$ and $\operatorname*{id}$ are equal on the generators $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\widetilde{\mathfrak{g}}$. Hence, these must be identical, i. e., we have $\omega^{2}=\operatorname*{id}$. Thus, $\omega$ is an involutive Lie algebra automorphism of $\widetilde{\mathfrak{g}% }$, and as we know from its definition, it sends $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ to $f_{1}$, $f_{2}$, $...$, $f_{n}$, $e_{1}$, $e_{2}$, $...$, $e_{n}$, $-h_{1}$, $-h_{2}$, $...$, $-h_{n}$, respectively. This proves Theorem \ref{thm.gtilde} \textbf{(f)}. \end{vershort} \begin{verlong} \textbf{(f)} Let us notice that the relations (\ref{nonserre-relations}) are equivalent to the relations% \begin{equation} \left\{ \begin{array} [c]{l}% \left[ -h_{i},-h_{j}\right] =0\ \ \ \ \ \ \ \ \ \ \text{for all }% i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ -h_{i},f_{j}\right] =a_{i,j}f_{j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ -h_{i},e_{j}\right] =-a_{i,j}e_{j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ f_{i},e_{j}\right] =\delta_{i,j}\left( -h_{i}\right) \ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} \end{array} \right. \label{nonserre-relations2}% \end{equation} \footnote{\textit{Proof.} We will show that the assertion \begin{equation} \left( \left[ e_{i},f_{j}\right] =\delta_{i,j}h_{i}\text{ for all }% i,j\in\left\{ 1,2,...,n\right\} \right) \label{pf.gtilde.equiv.1}% \end{equation} is equivalent to the assertion \begin{equation} \left( \left[ f_{i},e_{j}\right] =\delta_{i,j}\left( -h_{i}\right) \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} \right) . \label{pf.gtilde.equiv.2}% \end{equation} \par If (\ref{pf.gtilde.equiv.1}) holds, then (\ref{pf.gtilde.equiv.2}) holds as well (because if (\ref{pf.gtilde.equiv.1}) holds, then any $i,j\in\left\{ 1,2,...,n\right\} $ satisfy% \begin{align*} -\left[ f_{i},e_{j}\right] & =\left[ e_{j},f_{i}\right] =\underbrace{\delta_{j,i}}_{=\delta_{i,j}=\left\{ \begin{array} [c]{c}% 1,\text{ if }i=j;\\ 0,\text{ if }i\neq j \end{array} \right. }h_{j}\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.gtilde.equiv.1}), applied to }i\text{ and }j\text{ instead of }j\text{ and }i\right) \\ & =\left\{ \begin{array} [c]{c}% 1,\text{ if }i=j;\\ 0,\text{ if }i\neq j \end{array} \right. h_{j}=\left\{ \begin{array} [c]{c}% h_{j},\text{ if }i=j;\\ 0,\text{ if }i\neq j \end{array} \right. =\left\{ \begin{array} [c]{c}% h_{i},\text{ if }i=j;\\ 0,\text{ if }i\neq j \end{array} \right. \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }j=i\text{ in the case when }i=j\right) \\ & =\underbrace{\left\{ \begin{array} [c]{c}% 1,\text{ if }i=j;\\ 0,\text{ if }i\neq j \end{array} \right. }_{=\delta_{i,j}}h_{i}=\delta_{i,j}h_{i}% \end{align*} and thus $\left[ f_{i},e_{j}\right] =-\delta_{i,j}h_{i}=\delta_{i,j}\left( -h_{i}\right) $). Similarly, if (\ref{pf.gtilde.equiv.2}) holds, then (\ref{pf.gtilde.equiv.1}) holds as well. Thus, the assertion (\ref{pf.gtilde.equiv.1}) is equivalent to the assertion (\ref{pf.gtilde.equiv.2}). In other words, the fourth of the four relations (\ref{nonserre-relations}) is equivalent to the fourth of the four relations (\ref{nonserre-relations2}). But it is easy to see that the second of the four relations (\ref{nonserre-relations}) is equivalent to the third of the four relations (\ref{nonserre-relations2}). Similarly, the third of the four relations (\ref{nonserre-relations}) is equivalent to the second of the four relations (\ref{nonserre-relations2}). Finally, the first of the four relations (\ref{nonserre-relations}) is equivalent to the first of the four relations (\ref{nonserre-relations2}). Altogether, we thus conclude that the relations (\ref{nonserre-relations}) are equivalent to the relations (\ref{nonserre-relations2}), qed.}. Hence, \begin{align*} \widetilde{\mathfrak{g}} & =\operatorname*{FreeLie}\left( h_{i},f_{i}% ,e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) \\ & =\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations2})}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since the relations (\ref{nonserre-relations}) are equivalent to the relations (\ref{nonserre-relations2})}\right) . \end{align*} Hence, the relations (\ref{nonserre-relations2}) are satisfied in $\widetilde{\mathfrak{g}}$. Now, we can define a Lie algebra homomorphism \[ \Omega:\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \rightarrow\widetilde{\mathfrak{g}}% \] by requiring% \begin{equation} \left\{ \begin{array} [c]{c}% \Omega\left( e_{i}\right) =f_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} ;\\ \Omega\left( f_{i}\right) =e_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} ;\\ \Omega\left( h_{i}\right) =-h_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} \end{array} \right. \label{pf.gtilde.OMEGA}% \end{equation} (because we can define a Lie algebra homomorphism from a Lie algebra by arbitrarily choosing its values on the free generators). Define this $\Omega$. Then, $\Omega$ sends the relations (\ref{nonserre-relations}) to the relations (\ref{nonserre-relations2}). Since the relations (\ref{nonserre-relations2}) are satisfied in $\widetilde{\mathfrak{g}}$, this yields that the Lie algebra homomorphism $\Omega$ factors through the factor Lie algebra% \[ \operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) . \] In other words, there exists a unique Lie algebra homomorphism% \[ \omega:\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) \rightarrow\widetilde{\mathfrak{g}}% \] satisfying $\omega\circ\pi=\Omega$, where \begin{align*} \pi: & \operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \\ & \rightarrow\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid \ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) \end{align*} is the canonical projection. Consider this $\omega$. Since \newline% $\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) =\widetilde{\mathfrak{g}}$, this $\omega$ is a Lie algebra homomorphism from $\widetilde{\mathfrak{g}}$ to $\widetilde{\mathfrak{g}}$. Due to $\omega\circ\pi=\Omega$ and because of (\ref{pf.gtilde.OMEGA}), we have% \begin{equation} \left\{ \begin{array} [c]{c}% \omega\left( e_{i}\right) =f_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} ;\\ \omega\left( f_{i}\right) =e_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} ;\\ \omega\left( h_{i}\right) =-h_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} \end{array} \right. . \label{pf.gtilde.omega}% \end{equation} Thus, $\omega$ sends $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ to $f_{1}$, $f_{2}$, $...$, $f_{n}$, $e_{1}$, $e_{2}$, $...$, $e_{n}$, $-h_{1}$, $-h_{2}$, $...$, $-h_{n}% $, respectively. The elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ generate $\widetilde{\mathfrak{g}}$ as a Lie algebra\footnote{This is because $\widetilde{\mathfrak{g}% }=\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) $.}. In other words, the subset $\left\{ e_{1},e_{2},...,e_{n},f_{1},f_{2},...,f_{n},h_{1},h_{2},...,h_{n}\right\} $ generates $\widetilde{\mathfrak{g}}$ as a Lie algebra. The maps $\omega^{2}$ and $\operatorname*{id}$ are equal to each other on the set $\left\{ e_{1},e_{2},...,e_{n},f_{1},f_{2},...,f_{n},h_{1},h_{2}% ,...,h_{n}\right\} $\ \ \ \ \footnote{\textit{Proof.} Let $x\in\left\{ e_{1},e_{2},...,e_{n},f_{1},f_{2},...,f_{n},h_{1},h_{2},...,h_{n}\right\} $. We will prove that $\omega^{2}\left( x\right) =\operatorname*{id}\left( x\right) $. \par Indeed, since $x\in\left\{ e_{1},e_{2},...,e_{n},f_{1},f_{2},...,f_{n}% ,h_{1},h_{2},...,h_{n}\right\} =\left\{ e_{1},e_{2},...,e_{n}\right\} \cup\left\{ f_{1},f_{2},...,f_{n}\right\} \cup\left\{ h_{1},h_{2}% ,...,h_{n}\right\} $, we must be in one of the three following three cases: \par \textit{Case 1:} We have $x\in\left\{ e_{1},e_{2},...,e_{n}\right\} $. \par \textit{Case 2:} We have $x\in\left\{ f_{1},f_{2},...,f_{n}\right\} $. \par \textit{Case 3:} We have $x\in\left\{ h_{1},h_{2},...,h_{n}\right\} $. \par Let us first consider Case 1. In this case, $x\in\left\{ e_{1},e_{2}% ,...,e_{n}\right\} $. Thus, there exists an $i\in\left\{ 1,2,...,n\right\} $ such that $x=e_{i}$. Consider this $i$. From $x=e_{i}$, we obtain $\omega\left( x\right) =\omega\left( e_{i}\right) =f_{i}$ and thus $\omega^{2}\left( x\right) =\omega\left( \underbrace{\omega\left( x\right) }_{=f_{i}}\right) =\omega\left( f_{i}\right) =e_{i}% =x=\operatorname*{id}\left( x\right) $. Thus, $\omega^{2}\left( x\right) =\operatorname*{id}\left( x\right) $ is proven in Case 1. \par Let us next consider Case 2. In this case, $x\in\left\{ f_{1},f_{2}% ,...,f_{n}\right\} $. Thus, there exists an $i\in\left\{ 1,2,...,n\right\} $ such that $x=f_{i}$. Consider this $i$. From $x=f_{i}$, we obtain $\omega\left( x\right) =\omega\left( f_{i}\right) =e_{i}$ and thus $\omega^{2}\left( x\right) =\omega\left( \underbrace{\omega\left( x\right) }_{=e_{i}}\right) =\omega\left( e_{i}\right) =f_{i}% =x=\operatorname*{id}\left( x\right) $. Thus, $\omega^{2}\left( x\right) =\operatorname*{id}\left( x\right) $ is proven in Case 2. \par Let us first consider Case 3. In this case, $x\in\left\{ h_{1},h_{2}% ,...,h_{n}\right\} $. Thus, there exists an $i\in\left\{ 1,2,...,n\right\} $ such that $x=h_{i}$. Consider this $i$. From $x=h_{i}$, we obtain $\omega\left( x\right) =\omega\left( h_{i}\right) =-h_{i}$ and thus $\omega^{2}\left( x\right) =\omega\left( \underbrace{\omega\left( x\right) }_{=-h_{i}}\right) =\omega\left( -h_{i}\right) =-\underbrace{\omega\left( h_{i}\right) }_{=-h_{i}}=-\left( -h_{i}\right) =h_{i}=x=\operatorname*{id}\left( x\right) $. Thus, $\omega^{2}\left( x\right) =\operatorname*{id}\left( x\right) $ is proven in Case 3. \par Hence, the equality $\omega^{2}\left( x\right) =\operatorname*{id}\left( x\right) $ is proven in each of the three Cases 1, 2 and 3. Since these three cases cover all possibilities, this yields that $\omega^{2}\left( x\right) =\operatorname*{id}\left( x\right) $ always holds. \par Now forget that we fixed $x$. Thus, we have shown that $\omega^{2}\left( x\right) =\operatorname*{id}\left( x\right) $ for every $x\in\left\{ e_{1},e_{2},...,e_{n},f_{1},f_{2},...,f_{n},h_{1},h_{2},...,h_{n}\right\} $. In other words, the maps $\omega^{2}$ and $\operatorname*{id}$ are equal to each other on the set $\left\{ e_{1},e_{2},...,e_{n},f_{1},f_{2}% ,...,f_{n},h_{1},h_{2},...,h_{n}\right\} $, qed.}. Since this set $\left\{ e_{1},e_{2},...,e_{n},f_{1},f_{2},...,f_{n},h_{1},h_{2},...,h_{n}\right\} $ generates $\widetilde{\mathfrak{g}}$ as a Lie algebra, this yields that the maps $\omega^{2}$ and $\operatorname*{id}$ are equal to each other on a generating set of the Lie algebra $\widetilde{\mathfrak{g}}$. We also know that $\omega^{2}$ and $\operatorname*{id}$ are Lie algebra homomorphisms (since $\omega$ is a Lie algebra homomorphism). Now, it is well-known that if two Lie algebra homomorphisms from a Lie algebra $\mathfrak{i}$ to another Lie algebra are equal to each other on a generating set of the Lie algebra $\mathfrak{i}$, then these two homomorphisms must be identical. Applied to our two Lie algebra homomorphisms $\omega^{2}$ and $\operatorname*{id}$ (which, as we know, are equal to each other on a generating set of the Lie algebra $\widetilde{\mathfrak{g}}$), we conclude that the two homomorphisms $\omega^{2}$ and $\operatorname*{id}$ must be identical. In other words, $\omega^{2}=\operatorname*{id}$. Hence, $\omega$ is an involutive automorphism of the Lie algebra $\widetilde{\mathfrak{g}}$. Thus, there exists an involutive Lie algebra automorphism of $\widetilde{\mathfrak{g}}$ which sends $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ to $f_{1}$, $f_{2}$, $...$, $f_{n}$, $e_{1}$, $e_{2}$, $...$, $e_{n}$, $-h_{1}$, $-h_{2}$, $...$, $-h_{n}$, respectively (namely, $\omega$). This proves Theorem \ref{thm.gtilde} \textbf{(f)}. \end{verlong} \bigskip \textbf{(a)} In order to define a $Q$-grading on a free Lie algebra, it is enough to choose the degrees of its free generators. Thus, we can define a $Q$-grading on the Lie algebra $\operatorname*{FreeLie}\left( h_{i}% ,f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) $ by setting% \[ \deg\left( e_{i}\right) =\alpha_{i},\ \ \ \ \ \ \ \ \ \ \deg\left( f_{i}\right) =-\alpha_{i}\ \ \ \ \ \ \ \ \ \ \text{and }\deg\left( h_{i}\right) =0\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{ 1,2,...,n\right\} . \] The relations (\ref{nonserre-relations}) are homogeneous with respect to this $Q$-grading; hence, the quotient Lie algebra $\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) $ inherits the $Q$-grading from $\operatorname*{FreeLie}\left( h_{i}% ,f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) $. Since this quotient Lie algebra is $\widetilde{\mathfrak{g}}$, we thus have constructed a $Q$-grading on $\widetilde{\mathfrak{g}}$ which satisfies% \begin{equation} \deg\left( e_{i}\right) =\alpha_{i},\ \ \ \ \ \ \ \ \ \ \deg\left( f_{i}\right) =-\alpha_{i}\ \ \ \ \ \ \ \ \ \ \text{and }\deg\left( h_{i}\right) =0\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{ 1,2,...,n\right\} . \label{pf.gtilde.a.1}% \end{equation} Since this grading is clearly the only one to satisfy (\ref{pf.gtilde.a.1}) (because $\widetilde{\mathfrak{g}}$ is generated as a Lie algebra by $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$), this proves Theorem \ref{thm.gtilde} \textbf{(a)}. \bigskip \textbf{(b)} \underline{\textit{1st step: Definitions and identifications.}} Let $N_{+}$ be the free vector space with basis $e_{1},e_{2},...,e_{n}$. Since $\widetilde{\mathfrak{n}}_{+}=\operatorname*{FreeLie}\left( e_{i}\ \mid \ i\in\left\{ 1,2,...,n\right\} \right) $, we then have a canonical isomorphism $\widetilde{\mathfrak{n}}_{+}\cong\operatorname*{FreeLie}\left( N_{+}\right) $ (where $\operatorname*{FreeLie}\left( N_{+}\right) $ means the free Lie algebra over the vector space (not the set) $N_{+}$). We identify $\widetilde{\mathfrak{n}}_{+}$ with $\operatorname*{FreeLie}\left( N_{+}\right) $ along this isomorphism. Due to the construction of the free Lie algebra, we have a canonical injection $N_{+}\rightarrow \operatorname*{FreeLie}\left( N_{+}\right) =\widetilde{\mathfrak{n}}_{+}$. We will regard this injection as an inclusion (so that $N_{+}\subseteq \widetilde{\mathfrak{n}}_{+}$). By Proposition \ref{prop.Ufree} (applied to $V=N_{+}$), there exists a canonical algebra isomorphism $U\left( \operatorname*{FreeLie}\left( N_{+}\right) \right) \rightarrow T\left( N_{+}\right) $. We identify $U\left( \widetilde{\mathfrak{n}}_{+}\right) =U\left( \operatorname*{FreeLie}\left( N_{+}\right) \right) $ with $T\left( N_{+}\right) $ along this isomorphism. Let $N_{-}$ be the free vector space with basis $f_{1},f_{2},...,f_{n}$. Since $\widetilde{\mathfrak{n}}_{-}=\operatorname*{FreeLie}\left( f_{i}\ \mid \ i\in\left\{ 1,2,...,n\right\} \right) $, we then have a canonical isomorphism $\widetilde{\mathfrak{n}}_{-}\cong\operatorname*{FreeLie}\left( N_{-}\right) $ (where $\operatorname*{FreeLie}\left( N_{-}\right) $ means the free Lie algebra over the vector space (not the set) $N_{-}$). We identify $\widetilde{\mathfrak{n}}_{-}$ with $\operatorname*{FreeLie}\left( N_{-}\right) $ along this isomorphism. Due to the construction of the free Lie algebra, we have a canonical injection $N_{-}\rightarrow \operatorname*{FreeLie}\left( N_{-}\right) =\widetilde{\mathfrak{n}}_{-}$. We will regard this injection as an inclusion (so that $N_{-}\subseteq \widetilde{\mathfrak{n}}_{-}$). By Proposition \ref{prop.Ufree} (applied to $V=N_{-}$), there exists a canonical algebra isomorphism $U\left( \operatorname*{FreeLie}\left( N_{-}\right) \right) \rightarrow T\left( N_{-}\right) $. We identify $U\left( \widetilde{\mathfrak{n}}_{-}\right) =U\left( \operatorname*{FreeLie}\left( N_{-}\right) \right) $ with $T\left( N_{-}\right) $ along this isomorphism. A consequence of the Poincar\'{e}-Birkhoff-Witt theorem says that for any Lie algebra $\mathfrak{i}$, the canonical map $\mathfrak{i}\rightarrow U\left( \mathfrak{i}\right) $ is injective. Thus, the canonical map $\widetilde{\mathfrak{n}}_{+}\rightarrow U\left( \widetilde{\mathfrak{n}}% _{+}\right) $ and the canonical map $\widetilde{\mathfrak{n}}_{-}\rightarrow U\left( \widetilde{\mathfrak{n}}_{-}\right) $ are injective. We will therefore regard these maps as inclusions. Let us identify the group $Q$ with $\mathbb{Z}^{n}$ by means of identifying $\alpha_{i}$ with the column vector $e_{i}=\left( \underbrace{0,0,...,0}% _{i-1\text{ zeroes}},1,\underbrace{0,0,...,0}_{n-i\text{ zeroes}}\right) ^{T}$ for every $i\in\left\{ 1,2,...,n\right\} $. As a consequence, for every $i\in\left\{ 1,2,...,n\right\} $, the row vector $e_{i}^{T}A$ is an element of the group $\operatorname*{Hom}\left( Q,\mathbb{C}\right) $ of group homomorphisms from $Q$ to $\mathbb{C}$. Thus, for every $w\in Q$ and every $i\in\left\{ 1,2,...,n\right\} $, the product $e_{i}^{T}Aw$ is a complex number. \bigskip \underline{\textit{2nd step: Defining a }$Q$\textit{-grading on }% $\widetilde{\mathfrak{n}}_{-}$\textit{.}} Let us define a $Q$-grading on the vector space $N_{-}$ by setting% \[ \deg\left( f_{i}\right) =-\alpha_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }% i\in\left\{ 1,2,...,n\right\} . \] (This is well-defined since $\left( f_{1},f_{2},...,f_{n}\right) $ is a basis of $N_{-}$.) Then, the free Lie algebra $\operatorname*{FreeLie}\left( N_{-}\right) =\widetilde{\mathfrak{n}}_{-}$ canonically becomes a $Q$-graded Lie algebra, and the grading on this Lie algebra also satisfies% \[ \deg\left( f_{i}\right) =-\alpha_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }% i\in\left\{ 1,2,...,n\right\} . \] (This grading clearly makes the map $\iota_{-}$ graded. We will not use this fact, however.) We will later use this grading to define certain Lie derivations $\eta_{1}$, $\eta_{2}$, $...$, $\eta_{n}$ of the Lie algebra $\widetilde{\mathfrak{n}}_{-}$. \bigskip \underline{\textit{3rd step: Defining an }$\widetilde{\mathfrak{h}}% $\textit{-module }$\widetilde{\mathfrak{n}}_{-}$\textit{.}} For every $i\in\left\{ 1,2,...,n\right\} $, let us define a linear map $\eta_{i}:\widetilde{\mathfrak{n}}_{-}\rightarrow\widetilde{\mathfrak{n}}_{-}$ by setting% \begin{equation} \left( \eta_{i}\left( x\right) =\left( e_{i}^{T}Aw\right) \cdot x\ \ \ \ \ \ \ \ \ \ \text{for every }w\in Q\text{ and every }x\in \widetilde{\mathfrak{n}}_{-}\left[ w\right] \right) . \label{pf.gtilde.b.eta.def}% \end{equation} This map $\eta_{i}$ is well-defined (because in order to define a linear map from a $Q$-graded vector space, it is enough to define it linearly on every homogeneous component) and graded (because it multiplies any homogeneous element of $\widetilde{\mathfrak{n}}_{-}$ by a scalar). Actually, $\eta_{i}$ acts as a scalar on each homogeneous component of $\widetilde{\mathfrak{n}% }_{-}$. Moreover, for every $i\in\left\{ 1,2,...,n\right\} $, Lemma \ref{lem.deriv.grading} (applied to $s=e_{i}^{T}A$, $\mathfrak{n}% =\widetilde{\mathfrak{n}}_{-}$ and $\eta=\eta_{i}$) yields that $\eta_{i}$ is a Lie derivation. That is, $\eta_{i}\in\operatorname*{Der}\left( \widetilde{\mathfrak{n}}_{-}\right) $. One can directly see that% \begin{equation} \eta_{i}\left( f_{j}\right) =-a_{i,j}f_{j}\ \ \ \ \ \ \ \ \ \ \text{for any }i\in\left\{ 1,2,...,n\right\} \text{ and }j\in\left\{ 1,2,...,n\right\} \label{pf.gtilde.b.eta.fj}% \end{equation} \footnote{\textit{Proof of (\ref{pf.gtilde.b.eta.fj}):} Let $i\in\left\{ 1,2,...,n\right\} $ and $j\in\left\{ 1,2,...,n\right\} $. By the definition of our grading on $\widetilde{\mathfrak{n}}_{-}$, we have $\deg\left( f_{j}\right) =-\underbrace{\alpha_{j}}_{=e_{j}}=-e_{j}$, so that $f_{j}% \in\widetilde{\mathfrak{n}}_{-}\left[ -e_{j}\right] $. Hence, (\ref{pf.gtilde.b.eta.def}) (applied to $x=f_{j}$ and $w=-\alpha_{j}$) yields $\eta_{i}\left( f_{j}\right) =\left( e_{i}^{T}A\left( -e_{j}\right) \right) \cdot f_{j}=-\underbrace{\left( e_{i}^{T}Ae_{j}\right) }_{=a_{i,j}% }\cdot f_{j}=-a_{i,j}f_{j}$. This proves (\ref{pf.gtilde.b.eta.fj}).}. [Note that, while we defined the $\eta_{i}$ using the grading, there is also an alternative way to define them, by applying Theorem \ref{thm.universal.FreeLie.der}.] It is easy to see that \begin{equation} \left[ \eta_{i},\eta_{j}\right] =0\ \ \ \ \ \ \ \ \ \ \text{for all }% i\in\left\{ 1,2,...,n\right\} \text{ and }j\in\left\{ 1,2,...,n\right\} \label{pf.gtilde.b.eta.commute}% \end{equation} (since each of the maps $\eta_{i}$ and $\eta_{j}$ acts as a scalar on each homogeneous component of $\widetilde{\mathfrak{n}}_{-}$). Define a linear map $\Xi:\widetilde{\mathfrak{h}}\rightarrow \operatorname*{Der}\left( \widetilde{\mathfrak{n}}_{-}\right) $ by% \[ \left( \Xi\left( h_{i}\right) =\eta_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{ 1,2,...,n\right\} \right) \] (this map is well-defined, since $\left( h_{1},h_{2},...,h_{n}\right) $ is a basis of $\widetilde{\mathfrak{h}}$). Then, $\Xi$ is a Lie algebra homomorphism (this follows from (\ref{pf.gtilde.b.eta.commute})), and thus makes $\widetilde{\mathfrak{n}}_{-}$ into an $\widetilde{\mathfrak{h}}$-module on which $\widetilde{\mathfrak{h}}$ acts by derivations. Thus, a Lie algebra $\widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}$ is well-defined (according to Definition \ref{def.semidir.lielie}). Both Lie algebras $\widetilde{\mathfrak{h}}$ and $\widetilde{\mathfrak{n}}_{-}$ canonically inject (by Lie algebra homomorphisms) into this Lie algebra $\widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}$. Therefore, both $\widetilde{\mathfrak{h}}$ and $\widetilde{\mathfrak{n}}_{-}$ will be considered as Lie subalgebras of $\widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}$. In the Lie algebra $\widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}% }_{-}$, every $i\in\left\{ 1,2,...,n\right\} $ and $j\in\left\{ 1,2,...,n\right\} $ satisfy% \begin{align} \left[ h_{i},f_{j}\right] & =h_{i}\rightharpoonup f_{j}% \ \ \ \ \ \ \ \ \ \ \left( \text{where }\rightharpoonup\text{ denotes the action of }\widetilde{\mathfrak{h}}\text{ on }\widetilde{\mathfrak{n}}% _{-}\right) \nonumber\\ & =\underbrace{\left( \Xi\left( h_{i}\right) \right) }_{=\eta_{i}}\left( f_{j}\right) =\eta_{i}\left( f_{j}\right) =-a_{i,j}f_{j}% \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.gtilde.b.eta.fj})}\right) . \label{pf.gtilde.b.semidir.ij}% \end{align} From (\ref{nonserre-relations}), we see that the same relation is satisfied in the Lie algebra $\widetilde{\mathfrak{g}}$. Since $\widetilde{\mathfrak{n}}_{-}$ is a Lie subalgebra of $\widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}$, the universal enveloping algebra $U\left( \widetilde{\mathfrak{n}}_{-}\right) $ is a subalgebra of $U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}% }_{-}\right) $. This makes $U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ into a $U\left( \widetilde{\mathfrak{n}}_{-}\right) $-bimodule. Since $U\left( \widetilde{\mathfrak{n}}_{-}\right) =T\left( N_{-}\right) $, this means that $U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}% _{-}\right) $ is a $T\left( N_{-}\right) $-bimodule. \bigskip \underline{\textit{4th step: Defining an action of }$\widetilde{\mathfrak{g}}$ \textit{on }$U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}% }_{-}\right) $\textit{.}} We are going to construct an action of the Lie algebra $\widetilde{\mathfrak{g}}$ on $U\left( \widetilde{\mathfrak{h}}% \ltimes\widetilde{\mathfrak{n}}_{-}\right) $ (but not by derivations). First, let us define some further maps. Let $\iota_{N_{-}}^{T}:N_{-}\rightarrow T\left( N_{-}\right) $ be the canonical inclusion map. Notice that we are regarding $\iota_{N_{-}}^{T}$ as an inclusion. \begin{vershort} For every $i\in\left\{ 1,2,...,n\right\} $, let us define a derivation\footnote{Here, by ``derivation'', we mean a derivation of algebras, not of Lie algebras.} $\varepsilon_{i}:U\left( \widetilde{\mathfrak{n}}% _{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ by requiring that% \[ \left( \varepsilon_{i}\left( f_{j}\right) =\delta_{i,j}h_{i}% \ \ \ \ \ \ \ \ \ \ \text{for every }j\in\left\{ 1,2,...,n\right\} \right) . \] \footnote{Why is this well-defined? We know that $U\left( \widetilde{\mathfrak{n}}_{-}\right) =T\left( N_{-}\right) $. Hence, (by Theorem \ref{thm.universal.tensor.der}, applied to $V=N_{-}$ and $M=U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $) we can lift any linear map $f:N_{-}\rightarrow U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ to a derivation $U\left( \widetilde{\mathfrak{n}}_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $. Taking $f$ equal to the linear map $N_{-}\rightarrow U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ which sends every $f_{j}$ to $\delta_{i,j}h_{i}$, we obtain a derivation $U\left( \widetilde{\mathfrak{n}% }_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ which sends every $f_{j}$ to $\delta_{i,j}h_{i}$. This is why $\varepsilon_{i}$ is well-defined.} \end{vershort} \begin{verlong} For every $i\in\left\{ 1,2,...,n\right\} $, let $\varepsilon_{i}^{\prime }:N_{-}\rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ be the linear map defined by% \[ \left( \varepsilon_{i}^{\prime}\left( f_{j}\right) =\delta_{i,j}% h_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }j\in\left\{ 1,2,...,n\right\} \right) \] (this map is well-defined, since $\left( f_{1},f_{2},...,f_{n}\right) $ is a basis of $N_{-}$). For every $i\in\left\{ 1,2,...,n\right\} $, there exists a unique derivation\footnote{Here, by ``derivation'', we mean a derivation of algebras, not of Lie algebras.} $F:T\left( N_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ satisfying $\varepsilon_{i}^{\prime}=F\circ\iota_{N_{-}}^{T}$ (according to Theorem \ref{thm.universal.tensor.der}, applied to $V=N_{-}$, $M=U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ and $f=\varepsilon_{i}^{\prime}$). Denote this derivation by $\varepsilon_{i}$. Thus, for every $i\in\left\{ 1,2,...,n\right\} $, the map $\varepsilon _{i}:T\left( N_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ is a derivation satisfying $\varepsilon_{i}^{\prime}=\varepsilon_{i}\circ\iota_{N_{-}}^{T}$. Since $T\left( N_{-}\right) =U\left( \widetilde{\mathfrak{n}}_{-}\right) $, this map $\varepsilon_{i}$ is thus a derivation $U\left( \widetilde{\mathfrak{n}% }_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $. Clearly, every $i\in\left\{ 1,2,...,n\right\} $ and $j\in\left\{ 1,2,...,n\right\} $ satisfy% \begin{equation} \varepsilon_{i}\left( f_{j}\right) =\delta_{i,j}h_{i} \label{pf.gtilde.b.epsilon.fj}% \end{equation} \footnote{\textit{Proof of (\ref{pf.gtilde.b.epsilon.fj}):} Let $i\in\left\{ 1,2,...,n\right\} $ and $j\in\left\{ 1,2,...,n\right\} $. Then, $\iota_{N_{-}}^{T}\left( f_{j}\right) =f_{j}$ (since we regard $\iota _{N_{-}}^{T}$ as an inclusion). But since $\varepsilon_{i}^{\prime }=\varepsilon_{i}\circ\iota_{N_{-}}^{T}$, we have $\varepsilon_{i}^{\prime }\left( f_{j}\right) =\left( \varepsilon_{i}\circ\iota_{N_{-}}^{T}\right) \left( f_{j}\right) =\varepsilon_{i}\left( \underbrace{\iota_{N_{-}}% ^{T}\left( f_{j}\right) }_{=f_{j}}\right) =\varepsilon_{i}\left( f_{j}\right) $. Thus, $\varepsilon_{i}\left( f_{j}\right) =\varepsilon _{i}^{\prime}\left( f_{j}\right) =\delta_{i,j}h_{i}$. This proves (\ref{pf.gtilde.b.epsilon.fj}).}. \end{verlong} Let $\rho:U\left( \widetilde{\mathfrak{n}}_{-}\right) \otimes U\left( \widetilde{\mathfrak{h}}\right) \rightarrow U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ be the vector space homomorphism defined by% \[ \rho\left( \alpha\otimes\beta\right) =\alpha\beta \ \ \ \ \ \ \ \ \ \ \text{for all }\alpha\in U\left( \widetilde{\mathfrak{n}% }_{-}\right) \text{ and }\beta\in U\left( \widetilde{\mathfrak{h}}\right) \] (this is clearly well-defined). Since $\widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}=\widetilde{\mathfrak{n}}_{-}\oplus \widetilde{\mathfrak{h}}$ as vector spaces, Corollary \ref{cor.U(X)U} (applied to $k=\mathbb{C}$, $\mathfrak{c}=\widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}$, $\mathfrak{a}=\widetilde{\mathfrak{n}}_{-}$ and $\mathfrak{b}=\widetilde{\mathfrak{h}}$) yields that $\rho$ is an isomorphism of filtered vector spaces, of left $U\left( \widetilde{\mathfrak{n}}% _{-}\right) $-modules and of right $U\left( \widetilde{\mathfrak{h}}\right) $-modules. Thus, $\rho^{-1}$ also is an isomorphism of filtered vector spaces, of left $U\left( \widetilde{\mathfrak{n}}_{-}\right) $-modules and of right $U\left( \widetilde{\mathfrak{h}}\right) $-modules. \begin{vershort} For every $i\in\left\{ 1,2,...,n\right\} $, define a linear map $E_{i}:U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}% _{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ by% \begin{equation} \left( E_{i}\left( u_{-}u_{0}\right) =\varepsilon_{i}\left( u_{-}\right) u_{0}\ \ \ \ \ \ \ \ \ \ \text{for every }u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) \text{ and }u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) \right) . \label{pf.gtilde.b.Ei.short}% \end{equation} \footnote{Why is this well-defined? Since $\rho$ is an isomorphism, we have $U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) \cong U\left( \widetilde{\mathfrak{n}}_{-}\right) \otimes U\left( \widetilde{\mathfrak{h}}\right) $. In order to define a linear map $U\left( \widetilde{\mathfrak{n}}_{-}\right) \otimes U\left( \widetilde{\mathfrak{h}% }\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $, we just need to take a bilinear map $U\left( \widetilde{\mathfrak{n}}_{-}\right) \times U\left( \widetilde{\mathfrak{h}}\right) \rightarrow U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ and apply the universal property of the tensor product. Taking% \[ U\left( \widetilde{\mathfrak{n}}_{-}\right) \times U\left( \widetilde{\mathfrak{h}}\right) \rightarrow U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) ,\ \ \ \ \ \ \ \ \ \ \left( u_{-},u_{0}\right) \mapsto\varepsilon_{i}\left( u_{-}\right) u_{0}% \] as this bilinear map, we obtain (by the universal property) a linear map $U\left( \widetilde{\mathfrak{n}}_{-}\right) \otimes U\left( \widetilde{\mathfrak{h}}\right) \rightarrow U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ which sends $u_{-}\otimes u_{0}$ to $\varepsilon_{i}\left( u_{-}\right) u_{0}$ for every $u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) $ and $u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) $. Composing this map with the isomorphism $\rho^{-1}:U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}% }_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{n}}_{-}\right) \otimes U\left( \widetilde{\mathfrak{h}}\right) $, we obtain a linear map $U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}% }_{-}\right) $ which sends $u_{-}u_{0}$ to $\varepsilon_{i}\left( u_{-}\right) u_{0}$ for every $u_{-}\in U\left( \widetilde{\mathfrak{n}}% _{-}\right) $ and $u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) $. Therefore, $E_{i}$ is well-defined.} \end{vershort} \begin{verlong} Let $\mu:U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}% _{-}\right) \otimes U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ be the multiplication map of the algebra $U\left( \widetilde{\mathfrak{h}}% \ltimes\widetilde{\mathfrak{n}}_{-}\right) $. We consider $U\left( \widetilde{\mathfrak{h}}\right) $ as a subalgebra of $U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ (since $\widetilde{\mathfrak{h}}$ is a Lie subalgebra of $\widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}$). For every $i\in\left\{ 1,2,...,n\right\} $, define a linear map $E_{i}:U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}% _{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ by $E_{i}=\mu\circ\left( \varepsilon_{i}\otimes\operatorname*{id}\right) \circ\rho^{-1}$. Then, $E_{i}$ is a right $U\left( \widetilde{\mathfrak{h}}\right) $-module homomorphism (because all of $\mu$, $\varepsilon_{i}\otimes\operatorname*{id}$ and $\rho^{-1}$ are right $U\left( \widetilde{\mathfrak{h}}\right) $-module homomorphisms). Also, every $u_{-}\in U\left( \widetilde{\mathfrak{n}}% _{-}\right) $ and $u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) $ satisfy% \begin{equation} E_{i}\left( u_{-}u_{0}\right) =\varepsilon_{i}\left( u_{-}\right) u_{0} \label{pf.gtilde.b.Ei}% \end{equation} \footnote{\textit{Proof.} Let $u_{-}\in U\left( \widetilde{\mathfrak{n}}% _{-}\right) $ and $u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) $. Since $E_{i}=\mu\circ\left( \varepsilon_{i}\otimes\operatorname*{id}\right) \circ\rho^{-1}$, we have \begin{align*} E_{i}\left( u_{-}u_{0}\right) & =\left( \mu\circ\left( \varepsilon _{i}\otimes\operatorname*{id}\right) \circ\rho^{-1}\right) \left( u_{-}u_{0}\right) =\left( \mu\circ\left( \varepsilon_{i}\otimes \operatorname*{id}\right) \right) \left( \underbrace{\rho^{-1}\left( u_{-}u_{0}\right) }_{\substack{=u_{-}\otimes u_{0}\\\text{(since the definition of }\rho\text{ yields}\\\rho\left( u_{-}\otimes u_{0}\right) =u_{-}u_{0}\text{)}}}\right) \\ & =\left( \mu\circ\left( \varepsilon_{i}\otimes\operatorname*{id}\right) \right) \left( u_{-}\otimes u_{0}\right) =\mu\left( \underbrace{\left( \varepsilon_{i}\otimes\operatorname*{id}\right) \left( u_{-}\otimes u_{0}\right) }_{=\varepsilon_{i}\left( u_{-}\right) \otimes u_{0}}\right) =\mu\left( \varepsilon_{i}\left( u_{-}\right) \otimes u_{0}\right) =\varepsilon_{i}\left( u_{-}\right) u_{0}% \end{align*} (since $\mu$ is the multiplication map), qed.}. \end{verlong} For every $i\in\left\{ 1,2,...,n\right\} $, define a linear map $F_{i}:U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}% _{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ by% \[ \left( F_{i}\left( u\right) =f_{i}u\ \ \ \ \ \ \ \ \ \ \text{for every }u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}% _{-}\right) \right) . \] Clearly, $F_{i}$ is a right $U\left( \widetilde{\mathfrak{h}}\right) $-module homomorphism. For every $i\in\left\{ 1,2,...,n\right\} $, define a linear map $H_{i}:U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}% _{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ by% \[ \left( H_{i}\left( u\right) =h_{i}u\ \ \ \ \ \ \ \ \ \ \text{for every }u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}% _{-}\right) \right) . \] Clearly, $H_{i}$ is a right $U\left( \widetilde{\mathfrak{h}}\right) $-module homomorphism. Our next goal is to prove the relations% \begin{equation} \left\{ \begin{array} [c]{l}% \left[ H_{i},H_{j}\right] =0\ \ \ \ \ \ \ \ \ \ \text{for all }% i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ H_{i},E_{j}\right] =a_{i,j}E_{j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ H_{i},F_{j}\right] =-a_{i,j}F_{j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} ;\\ \left[ E_{i},F_{j}\right] =\delta_{i,j}H_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,n\right\} \end{array} \right. \label{pf.gtilde.NONSERRE}% \end{equation} in $\operatorname*{End}\left( U\left( \widetilde{\mathfrak{h}}% \ltimes\widetilde{\mathfrak{n}}_{-}\right) \right) $. Once these relations are proven, it will follow that a Lie algebra homomorphism $\widetilde{\mathfrak{g}}\rightarrow\operatorname*{End}\left( U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right) $ mapping $h_{i}$, $e_{i}$, $f_{i}$ to $H_{i}$, $E_{i}$, $F_{i}$ for all $i$ exists (and is unique), and this map will make $U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ into a $\widetilde{\mathfrak{g}}$-module. This $\widetilde{\mathfrak{g}}$-module structure will then yield Theorem \ref{thm.gtilde} \textbf{(b)} by a rather simple argument. But we must first verify (\ref{pf.gtilde.NONSERRE}). \bigskip \underline{\textit{5th step: Verifying the relations (\ref{pf.gtilde.NONSERRE}% ).}} We will verify the four relations (\ref{pf.gtilde.NONSERRE}) one after the other: \bigskip \textit{Proof of the relation }$\left[ H_{i},H_{j}\right] =0$ \textit{for all } $i,j\in\left\{ 1,2,...,n\right\} $\textit{:} Let $i$ and $j$ be two elements of $\left\{ 1,2,...,n\right\} $. Every $u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ satisfies% \begin{align*} \underbrace{\left[ H_{i},H_{j}\right] }_{=H_{i}\circ H_{j}-H_{j}\circ H_{i}% }u & =\left( H_{i}\circ H_{j}-H_{j}\circ H_{i}\right) \left( u\right) =H_{i}\underbrace{\left( H_{j}u\right) }_{\substack{=h_{j}u\\\text{(by the definition}\\\text{of }H_{j}\text{)}}}-H_{j}\underbrace{\left( H_{i}u\right) }_{\substack{=h_{i}u\\\text{(by the definition}\\\text{of }H_{i}\text{)}}}\\ & =\underbrace{H_{i}\left( h_{j}u\right) }_{\substack{=h_{i}\left( h_{j}u\right) \\\text{(by the definition}\\\text{of }H_{i}\text{)}% }}-\underbrace{H_{j}\left( h_{i}u\right) }_{\substack{=h_{j}\left( h_{i}u\right) \\\text{(by the definition}\\\text{of }H_{j}\text{)}}}\\ & =h_{i}\left( h_{j}u\right) -h_{j}\left( h_{i}u\right) =\underbrace{\left( h_{i}h_{j}-h_{j}h_{i}\right) }_{\substack{=\left[ h_{i},h_{j}\right] =0\\\text{in }U\left( \widetilde{\mathfrak{h}}% \ltimes\widetilde{\mathfrak{n}}_{-}\right) \\\text{(since }\left[ h_{i},h_{j}\right] =0\text{ in }\widetilde{\mathfrak{h}}\text{)}}}u=0. \end{align*} Thus, $\left[ H_{i},H_{j}\right] =0$. Now forget that we fixed $i$ and $j$. We have thus proven the relation $\left[ H_{i},H_{j}\right] =0$ for all $i,j\in\left\{ 1,2,...,n\right\} $. \bigskip \textit{Proof of the relation }$\left[ H_{i},E_{j}\right] =a_{i,j}E_{j}$ \textit{for all } $i,j\in\left\{ 1,2,...,n\right\} $\textit{:} This will be the most difficult among the four relations that we must prove. Applying Corollary \ref{cor.derivation.Lie.semidir} to $\widetilde{\mathfrak{h}}$ and $\widetilde{\mathfrak{n}}_{-}$ instead of $\mathfrak{g}$ and $\mathfrak{h}$, we obtain $\left[ \widetilde{\mathfrak{h}% },U\left( \widetilde{\mathfrak{n}}_{-}\right) \right] \subseteq U\left( \widetilde{\mathfrak{n}}_{-}\right) $ in $U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) $. Let us consider $U\left( \widetilde{\mathfrak{n}}_{-}\right) $ as $\widetilde{\mathfrak{n}}_{-}$-module via the adjoint action. Then, $\widetilde{\mathfrak{n}}_{-}\subseteq U\left( \widetilde{\mathfrak{n}}% _{-}\right) $ as $\widetilde{\mathfrak{n}}_{-}$-modules. Let $i$ be any element of $\left\{ 1,2,...,n\right\} $. Define a map $\zeta_{i}:U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}% }_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ by% \[ \left( \zeta_{i}\left( u\right) =\left[ h_{i},u\right] \ \ \ \ \ \ \ \ \ \ \text{for every }u\in U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right) . \] Clearly, $\zeta_{i}$ is a derivation of algebras. Now, using the relation $\left[ \widetilde{\mathfrak{h}},U\left( \widetilde{\mathfrak{n}}_{-}\right) \right] \subseteq U\left( \widetilde{\mathfrak{n}}_{-}\right) $, it is easy to check that $\zeta _{i}\left( U\left( \widetilde{\mathfrak{n}}_{-}\right) \right) \subseteq U\left( \widetilde{\mathfrak{n}}_{-}\right) $% \ \ \ \ \footnote{\textit{Proof.} Every $u\in U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ satisfies% \[ \zeta_{i}\left( u\right) =\left[ \underbrace{h_{i}}_{\in \widetilde{\mathfrak{h}}},\underbrace{u}_{\in U\left( \widetilde{\mathfrak{n}% }_{-}\right) }\right] \in\left[ \widetilde{\mathfrak{h}},U\left( \widetilde{\mathfrak{n}}_{-}\right) \right] \subseteq U\left( \widetilde{\mathfrak{n}}_{-}\right) . \] In other words, $\zeta_{i}\left( U\left( \widetilde{\mathfrak{n}}% _{-}\right) \right) \subseteq U\left( \widetilde{\mathfrak{n}}_{-}\right) $, qed.}. Now, let $j\in\left\{ 1,2,...,n\right\} $ be arbitrary. Recall that $\zeta_{i}:U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}% }_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ and $\varepsilon_{j}:U\left( \widetilde{\mathfrak{n}}_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ are derivations satisfying $\zeta_{i}\left( U\left( \widetilde{\mathfrak{n}}% _{-}\right) \right) \subseteq U\left( \widetilde{\mathfrak{n}}_{-}\right) $. Thus, Proposition \ref{prop.commutator.derivs.alg.2} (applied to $U\left( \widetilde{\mathfrak{n}}_{-}\right) $, $U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) $, $\varepsilon_{j}$ and $\zeta_{i}$ instead of $A$, $B$, $f$ and $g$) yields that $\varepsilon _{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon_{j}:U\left( \widetilde{\mathfrak{n}}% _{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ is a derivation (of algebras). On the other hand, $-a_{i,j}\varepsilon_{j}:U\left( \widetilde{\mathfrak{n}% }_{-}\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ is also a derivation (of algebras), since $\varepsilon_{j}$ is a derivation. We will now prove that% \begin{equation} \varepsilon_{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}% }_{-}\right) }\right) -\zeta_{i}\circ\varepsilon_{j}=-a_{i,j}\varepsilon _{j}. \label{pf.gtilde.dercomm1}% \end{equation} \footnote{Note that the term $\varepsilon_{j}\circ\left( \zeta_{i}% \mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) $ in this equality is well-defined because $\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) \left( U\left( \widetilde{\mathfrak{n}}_{-}\right) \right) =\zeta_{i}\left( U\left( \widetilde{\mathfrak{n}}_{-}\right) \right) \subseteq U\left( \widetilde{\mathfrak{n}}_{-}\right) $.} This will bring us very close to the proof of the relation $\left[ H_{i},E_{j}\right] =a_{i,j}E_{j}$. \begin{vershort} \textit{Proof of (\ref{pf.gtilde.dercomm1}):} Every $k\in\left\{ 1,2,...,n\right\} $ satisfies% \begin{align*} & \left( \left( \varepsilon_{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon _{j}\right) \mid_{N_{-}}\right) \left( f_{k}\right) \\ & =\varepsilon_{j}\left( \underbrace{\zeta_{i}\left( f_{k}\right) }_{\substack{=\left[ h_{i},f_{k}\right] \\\text{(by the definition of }% \zeta_{i}\text{)}}}\right) -\underbrace{\zeta_{i}\left( \varepsilon _{j}\left( f_{k}\right) \right) }_{\substack{=\left[ h_{i},\varepsilon _{j}\left( f_{k}\right) \right] \\\text{(by the definition of }\zeta _{i}\text{)}}}\\ & =\varepsilon_{j}\left( \underbrace{\left[ h_{i},f_{k}\right] }_{\substack{=-a_{i,k}f_{k}\\\text{(by (\ref{pf.gtilde.b.semidir.ij}), applied to}\\k\text{ instead of }j\text{)}}}\right) -\left[ h_{i}% ,\underbrace{\varepsilon_{j}\left( f_{k}\right) }_{\substack{=\delta _{j,k}h_{j}\\\text{(by the definition of }\varepsilon_{j}\text{)}}}\right] \\ & =\underbrace{\varepsilon_{j}\left( -a_{i,k}f_{k}\right) }_{=-a_{i,k}% \varepsilon_{j}\left( f_{k}\right) }-\underbrace{\left[ h_{i},\delta _{j,k}h_{j}\right] }_{\substack{=0\\\text{(since }\widetilde{\mathfrak{h}% }\text{ is an abelian Lie algebra)}}}=-a_{i,k}\underbrace{\varepsilon _{j}\left( f_{k}\right) }_{\substack{=\delta_{j,k}h_{j}\\\text{(by the definition of }\varepsilon_{j}\text{)}}}\\ & =-\underbrace{a_{i,k}\delta_{j,k}}_{=a_{i,j}\delta_{j,k}}h_{j}% =-a_{i,j}\underbrace{\delta_{j,k}h_{j}}_{\substack{=\varepsilon_{j}\left( f_{k}\right) \\\text{(by the definition of }\varepsilon_{j}\text{)}% }}=-a_{i,j}\varepsilon_{j}\left( f_{k}\right) =\left( \left( -a_{i,j}\varepsilon_{j}\right) \mid_{N_{-}}\right) \left( f_{k}\right) . \end{align*} In other words, the maps $\left( \varepsilon_{j}\circ\left( \zeta_{i}% \mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}% \circ\varepsilon_{j}\right) \mid_{N_{-}}$ and $\left( -a_{i,j}% \varepsilon_{j}\right) \mid_{N_{-}}$ are equal to each other on each of the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ of $N_{-}$. Since $\left( f_{1},f_{2},...,f_{n}\right) $ is a basis of $N_{-}$, this yields that \[ \left( \varepsilon_{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon _{j}\right) \mid_{N_{-}}=\left( -a_{i,j}\varepsilon_{j}\right) \mid_{N_{-}}% \] (because the maps $\left( \varepsilon_{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon _{j}\right) \mid_{N_{-}}$ and $\left( -a_{i,j}\varepsilon_{j}\right) \mid_{N_{-}}$ are linear). Hence, since the set $N_{-}$ generates $U\left( \widetilde{\mathfrak{n}}_{-}\right) $ as an algebra (because $U\left( \widetilde{\mathfrak{n}}_{-}\right) =T\left( N_{-}\right) $), Proposition \ref{prop.derivation.unique} (applied to $U\left( \widetilde{\mathfrak{n}% }_{-}\right) $, $N_{-}$, $U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $, $\varepsilon_{j}\circ\left( \zeta _{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta _{i}\circ\varepsilon_{j}$ and $-a_{i,j}\varepsilon_{j}$ instead of $A$, $S$, $M$, $d$ and $e$) yields that $\varepsilon_{j}\circ\left( \zeta_{i}% \mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}% \circ\varepsilon_{j}=-a_{i,j}\varepsilon_{j}$ (since $\varepsilon_{j}% \circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon_{j}$ and $-a_{i,j}\varepsilon_{j}$ are derivations). This proves (\ref{pf.gtilde.dercomm1}). \end{vershort} \begin{verlong} \textit{Proof of (\ref{pf.gtilde.dercomm1}):} Every $k\in\left\{ 1,2,...,n\right\} $ satisfies% \begin{align*} & \left( \left( \varepsilon_{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon _{j}\right) \mid_{N_{-}}\right) \left( f_{k}\right) \\ & =\left( \varepsilon_{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon _{j}\right) \left( f_{k}\right) \\ & =\varepsilon_{j}\left( \underbrace{\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) \left( f_{k}\right) }_{\substack{=\zeta_{i}\left( f_{k}\right) =\left[ h_{i},f_{k}\right] \\\text{(by the definition of }\zeta_{i}\text{)}}}\right) -\underbrace{\zeta _{i}\left( \varepsilon_{j}\left( f_{k}\right) \right) }% _{\substack{=\left[ h_{i},\varepsilon_{j}\left( f_{k}\right) \right] \\\text{(by the definition of }\zeta_{i}\text{)}}}\\ & =\varepsilon_{j}\left( \underbrace{\left[ h_{i},f_{k}\right] }_{\substack{=-a_{i,k}f_{k}\\\text{(by (\ref{pf.gtilde.b.semidir.ij}), applied to}\\k\text{ instead of }j\text{)}}}\right) -\left[ h_{i}% ,\underbrace{\varepsilon_{j}\left( f_{k}\right) }_{\substack{=\delta _{j,k}h_{j}\\\text{(by (\ref{pf.gtilde.b.epsilon.fj}), applied to}\\j\text{ and }k\text{ instead of }i\text{ and }j\text{)}}}\right] \\ & =\underbrace{\varepsilon_{j}\left( -a_{i,k}f_{k}\right) }_{=-a_{i,k}% \varepsilon_{j}\left( f_{k}\right) }-\underbrace{\left[ h_{i},\delta _{j,k}h_{j}\right] }_{\substack{=0\\\text{(since }\widetilde{\mathfrak{h}% }\text{ is an abelian Lie algebra)}}}=-a_{i,k}\underbrace{\varepsilon _{j}\left( f_{k}\right) }_{\substack{=\delta_{j,k}h_{j}\\\text{(by (\ref{pf.gtilde.b.epsilon.fj}), applied to}\\j\text{ and }k\text{ instead of }i\text{ and }j\text{)}}}\\ & =-a_{i,k}\underbrace{\delta_{j,k}}_{=\left\{ \begin{array} [c]{c}% 1,\text{ if }j=k\\ 0,\text{ if }j\neq k \end{array} \right. }h_{j}=-\underbrace{a_{i,k}\left\{ \begin{array} [c]{c}% 1,\text{ if }j=k\\ 0,\text{ if }j\neq k \end{array} \right. }_{=\left\{ \begin{array} [c]{c}% a_{i,k}\cdot1,\text{ if }j=k\\ a_{i,k}\cdot0,\text{ if }j\neq k \end{array} \right. }h_{j}\\ & =-\left\{ \begin{array} [c]{c}% a_{i,k}\cdot1,\text{ if }j=k\\ a_{i,k}\cdot0,\text{ if }j\neq k \end{array} \right. h_{j}=-\left\{ \begin{array} [c]{c}% a_{i,j}\cdot1,\text{ if }j=k\\ a_{i,k}\cdot0,\text{ if }j\neq k \end{array} \right. h_{j}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }a_{i,k}=a_{i,j}\text{ if }j=k\text{ (because }k=j\text{ if }j=k\text{)}\right) \\ & =-\underbrace{\left\{ \begin{array} [c]{c}% a_{i,j}\cdot1,\text{ if }j=k\\ a_{i,j}\cdot0,\text{ if }j\neq k \end{array} \right. }_{=a_{i,j}\cdot\left\{ \begin{array} [c]{c}% 1,\text{ if }j=k\\ 0,\text{ if }j\neq k \end{array} \right. }h_{j}\ \ \ \ \ \ \ \ \ \ \left( \text{since }a_{i,k}\cdot 0=a_{i,j}\cdot0\text{ if }j\neq k\right) \\ & =-a_{i,j}\cdot\underbrace{\left\{ \begin{array} [c]{c}% 1,\text{ if }j=k\\ 0,\text{ if }j\neq k \end{array} \right. }_{=\delta_{j,k}}h_{j}=-a_{i,j}\delta_{j,k}h_{j}% \end{align*} and% \[ \left( \left( -a_{i,j}\varepsilon_{j}\right) \mid_{N_{-}}\right) \left( f_{k}\right) =\left( -a_{i,j}\varepsilon_{j}\right) \left( f_{k}\right) =-a_{i,j}\underbrace{\varepsilon_{j}\left( f_{k}\right) }_{\substack{=\delta _{j,k}h_{j}\\\text{(by (\ref{pf.gtilde.b.epsilon.fj}), applied to}\\j\text{ and }k\text{ instead of }i\text{ and }j\text{)}}}=-a_{i,j}\delta_{j,k}h_{j}. \] Hence, every $k\in\left\{ 1,2,...,n\right\} $ satisfies% \[ \left( \left( \varepsilon_{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon _{j}\right) \mid_{N_{-}}\right) \left( f_{k}\right) =-a_{i,j}\delta _{j,k}h_{j}=\left( \left( -a_{i,j}\varepsilon_{j}\right) \mid_{N_{-}% }\right) \left( f_{k}\right) . \] In other words, the maps $\left( \varepsilon_{j}\circ\left( \zeta_{i}% \mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}% \circ\varepsilon_{j}\right) \mid_{N_{-}}$ and $\left( -a_{i,j}% \varepsilon_{j}\right) \mid_{N_{-}}$ are equal to each other on each of the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ of $N_{-}$. Since $\left( f_{1},f_{2},...,f_{n}\right) $ is a basis of $N_{-}$, this yields that \[ \left( \varepsilon_{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon _{j}\right) \mid_{N_{-}}=\left( -a_{i,j}\varepsilon_{j}\right) \mid_{N_{-}}% \] (because the maps $\left( \varepsilon_{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon _{j}\right) \mid_{N_{-}}$ and $\left( -a_{i,j}\varepsilon_{j}\right) \mid_{N_{-}}$ are linear). Hence, since the set $N_{-}$ generates $U\left( \widetilde{\mathfrak{n}}_{-}\right) $ as an algebra (because $U\left( \widetilde{\mathfrak{n}}_{-}\right) =T\left( N_{-}\right) $), Proposition \ref{prop.derivation.unique} (applied to $U\left( \widetilde{\mathfrak{n}% }_{-}\right) $, $N_{-}$, $U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $, $\varepsilon_{j}\circ\left( \zeta _{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta _{i}\circ\varepsilon_{j}$ and $-a_{i,j}\varepsilon_{j}$ instead of $A$, $S$, $M$, $d$ and $e$) yields that $\varepsilon_{j}\circ\left( \zeta_{i}% \mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}% \circ\varepsilon_{j}=-a_{i,j}\varepsilon_{j}$ (since $\varepsilon_{j}% \circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon_{j}$ and $-a_{i,j}\varepsilon_{j}$ are derivations). This proves (\ref{pf.gtilde.dercomm1}). \end{verlong} Now, we will show that \begin{equation} \left[ h_{i},\varepsilon_{j}\left( u_{-}\right) \right] -\varepsilon _{j}\left( \left[ h_{i},u_{-}\right] \right) =a_{i,j}\varepsilon _{j}\left( u_{-}\right) \ \ \ \ \ \ \ \ \ \ \text{for every }u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) . \label{pf.gtilde.dercomm2}% \end{equation} \textit{Proof of (\ref{pf.gtilde.dercomm2}):} Let $u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) $. Then,% \begin{align*} \left( \varepsilon_{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon _{j}\right) \left( u_{-}\right) & =\varepsilon_{j}\left( \underbrace{\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}% _{-}\right) }\right) \left( u_{-}\right) }_{\substack{=\zeta_{i}\left( u_{-}\right) =\left[ h_{i},u_{-}\right] \\\text{(by the definition of }\zeta_{i}\text{)}}}\right) -\underbrace{\zeta_{i}\left( \varepsilon _{j}\left( u_{-}\right) \right) }_{\substack{=\left[ h_{i},\varepsilon _{j}\left( u_{-}\right) \right] \\\text{(by the definition of }\zeta _{i}\text{)}}}\\ & =\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) -\left[ h_{i},\varepsilon_{j}\left( u_{-}\right) \right] . \end{align*} Comparing this with% \[ \underbrace{\left( \varepsilon_{j}\circ\left( \zeta_{i}\mid_{U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right) -\zeta_{i}\circ\varepsilon _{j}\right) }_{\substack{=-a_{i,j}\varepsilon_{j}\\\text{(by (\ref{pf.gtilde.dercomm1}))}}}\left( u_{-}\right) =-a_{i,j}\varepsilon _{j}\left( u_{-}\right) , \] we obtain $-a_{i,j}\varepsilon_{j}\left( u_{-}\right) =\varepsilon _{j}\left( \left[ h_{i},u_{-}\right] \right) -\left[ h_{i},\varepsilon _{j}\left( u_{-}\right) \right] $. In other words, $\left[ h_{i}% ,\varepsilon_{j}\left( u_{-}\right) \right] -\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) =a_{i,j}\varepsilon_{j}\left( u_{-}\right) $. This proves (\ref{pf.gtilde.dercomm2}). Now, let us finally prove that $\left[ H_{i},E_{j}\right] =a_{i,j}E_{j}$. \begin{vershort} Indeed, let $u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) $ and $u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) $. Then, $\left[ \underbrace{h_{i}}_{\in\mathfrak{h}},\underbrace{u_{-}}_{\in U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right] \in\left[ \widetilde{\mathfrak{h}},U\left( \widetilde{\mathfrak{n}}_{-}\right) \right] \subseteq U\left( \widetilde{\mathfrak{n}}_{-}\right) $. Thus, (\ref{pf.gtilde.b.Ei.short}) (applied to $\left[ h_{i},u_{-}\right] $ and $j$ instead of $u_{-}$ and $i$) yields% \[ E_{j}\left( \left[ h_{i},u_{-}\right] u_{0}\right) =\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}. \] On the other hand, $\underbrace{h_{i}}_{\in\widetilde{\mathfrak{h}}% }\underbrace{u_{0}}_{\in U\left( \widetilde{\mathfrak{h}}\right) }% \in\widetilde{\mathfrak{h}}U\left( \widetilde{\mathfrak{h}}\right) \subseteq U\left( \widetilde{\mathfrak{h}}\right) $. Hence, (\ref{pf.gtilde.b.Ei.short}) (applied to $h_{i}u_{0}$ and $j$ instead of $u_{0}$ and $i$) yields% \[ E_{j}\left( u_{-}h_{i}u_{0}\right) =\varepsilon_{j}\left( u_{-}\right) h_{i}u_{0}. \] But $\underbrace{h_{i}u_{-}}_{=\left[ h_{i},u_{-}\right] +u_{-}h_{i}}% u_{0}=\left[ h_{i},u_{-}\right] u_{0}+u_{-}h_{i}u_{0}$, so that \begin{align*} E_{j}\left( h_{i}u_{-}u_{0}\right) & =E_{j}\left( \left[ h_{i}% ,u_{-}\right] u_{0}+u_{-}h_{i}u_{0}\right) =\underbrace{E_{j}\left( \left[ h_{i},u_{-}\right] u_{0}\right) }_{=\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}}+\underbrace{E_{j}\left( u_{-}h_{i}% u_{0}\right) }_{=\varepsilon_{j}\left( u_{-}\right) h_{i}u_{0}}\\ & =\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}% +\varepsilon_{j}\left( u_{-}\right) h_{i}u_{0}. \end{align*} On the other hand,% \begin{align*} & \underbrace{\left[ H_{i},E_{j}\right] }_{=H_{i}\circ E_{j}-E_{j}\circ H_{i}}\left( u_{-}u_{0}\right) =\left( H_{i}\circ E_{j}-E_{j}\circ H_{i}\right) \left( u_{-}u_{0}\right) \\ & =H_{i}\left( \underbrace{E_{j}\left( u_{-}u_{0}\right) }% _{\substack{=\varepsilon_{j}\left( u_{-}\right) u_{0}\\\text{(by (\ref{pf.gtilde.b.Ei.short}), applied}\\\text{to }j\text{ instead of }i\text{)}}}\right) -E_{j}\left( \underbrace{H_{i}\left( u_{-}u_{0}\right) }_{\substack{=h_{i}u_{-}u_{0}\\\text{(by the definition of }H_{i}\text{)}% }}\right) \\ & =\underbrace{H_{i}\left( \varepsilon_{j}\left( u_{-}\right) u_{0}\right) }_{\substack{=h_{i}\varepsilon_{j}\left( u_{-}\right) u_{0}\\\text{(by the definition of }H_{i}\text{)}}}-\underbrace{E_{j}\left( h_{i}u_{-}u_{0}\right) }_{\substack{=\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}+\varepsilon_{j}\left( u_{-}\right) h_{i}u_{0}\\\text{(as we saw above)}}}\\ & =h_{i}\varepsilon_{j}\left( u_{-}\right) u_{0}-\left( \varepsilon _{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}+\varepsilon_{j}\left( u_{-}\right) h_{i}u_{0}\right) =h_{i}\varepsilon_{j}\left( u_{-}\right) u_{0}-\varepsilon_{j}\left( u_{-}\right) h_{i}u_{0}-\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}\\ & =\left( \underbrace{h_{i}\varepsilon_{j}\left( u_{-}\right) -\varepsilon_{j}\left( u_{-}\right) h_{i}}_{=\left[ h_{i},\varepsilon _{j}\left( u_{-}\right) \right] }-\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) \right) u_{0}=\underbrace{\left( \left[ h_{i},\varepsilon_{j}\left( u_{-}\right) \right] -\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) \right) }_{\substack{=a_{i,j}% \varepsilon_{j}\left( u_{-}\right) \\\text{(by (\ref{pf.gtilde.dercomm2}))}% }}u_{0}\\ & =a_{i,j}\underbrace{\varepsilon_{j}\left( u_{-}\right) u_{0}% }_{\substack{=E_{j}\left( u_{-}u_{0}\right) \\\text{(since (\ref{pf.gtilde.b.Ei.short}) (applied to }j\text{ instead of }i\text{)}% \\\text{yields }E_{j}\left( u_{-}u_{0}\right) =\varepsilon_{j}\left( u_{-}\right) u_{0}\text{)}}}=a_{i,j}E_{j}\left( u_{-}u_{0}\right) . \end{align*} Now, forget that we fixed $u_{-}$ and $u_{0}$. We thus have proven that% \[ \left[ H_{i},E_{j}\right] \left( u_{-}u_{0}\right) =\left( a_{i,j}% E_{j}\right) \left( u_{-}u_{0}\right) \ \ \ \ \ \ \ \ \ \ \text{for all }u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) \text{ and }u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) . \] Since the vector space $U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ is generated by products $u_{-}u_{0}$ with $u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) $ and $u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) $ (this is because $\rho:U\left( \widetilde{\mathfrak{n}}_{-}\right) \otimes U\left( \widetilde{\mathfrak{h}% }\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ is an isomorphism), this yields that $\left[ H_{i},E_{j}\right] =a_{i,j}E_{j}$. \end{vershort} \begin{verlong} Indeed, let $u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) $ and $u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) $. Then, $\left[ \underbrace{h_{i}}_{\in\mathfrak{h}},\underbrace{u_{-}}_{\in U\left( \widetilde{\mathfrak{n}}_{-}\right) }\right] \in\left[ \widetilde{\mathfrak{h}},U\left( \widetilde{\mathfrak{n}}_{-}\right) \right] \subseteq U\left( \widetilde{\mathfrak{n}}_{-}\right) $. Thus, (\ref{pf.gtilde.b.Ei}) (applied to $\left[ h_{i},u_{-}\right] $ and $j$ instead of $u_{-}$ and $i$) yields% \begin{equation} E_{j}\left( \left[ h_{i},u_{-}\right] u_{0}\right) =\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}. \label{pf.gtilde.dercomm3}% \end{equation} On the other hand, $\underbrace{h_{i}}_{\in\widetilde{\mathfrak{h}}% }\underbrace{u_{0}}_{\in U\left( \widetilde{\mathfrak{h}}\right) }% \in\widetilde{\mathfrak{h}}U\left( \widetilde{\mathfrak{h}}\right) \subseteq U\left( \widetilde{\mathfrak{h}}\right) $. Hence, (\ref{pf.gtilde.b.Ei}) (applied to $h_{i}u_{0}$ and $j$ instead of $u_{0}$ and $i$) yields% \begin{equation} E_{j}\left( u_{-}h_{i}u_{0}\right) =\varepsilon_{j}\left( u_{-}\right) h_{i}u_{0}. \label{pf.gtilde.dercomm4}% \end{equation} But $\underbrace{h_{i}u_{-}}_{=\left[ h_{i},u_{-}\right] +u_{-}h_{i}}% u_{0}=\left[ h_{i},u_{-}\right] u_{0}+u_{-}h_{i}u_{0}$, so that \begin{align} E_{j}\left( h_{i}u_{-}u_{0}\right) & =E_{j}\left( \left[ h_{i}% ,u_{-}\right] u_{0}+u_{-}h_{i}u_{0}\right) =\underbrace{E_{j}\left( \left[ h_{i},u_{-}\right] u_{0}\right) }_{\substack{=\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}\\\text{(by (\ref{pf.gtilde.dercomm3}))}% }}+\underbrace{E_{j}\left( u_{-}h_{i}u_{0}\right) }_{\substack{=\varepsilon _{j}\left( u_{-}\right) h_{i}u_{0}\\\text{(by (\ref{pf.gtilde.dercomm4}))}% }}\nonumber\\ & =\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}% +\varepsilon_{j}\left( u_{-}\right) h_{i}u_{0}. \label{pf.gtilde.dercomm5}% \end{align} On the other hand, $\rho\left( u_{-}\otimes u_{0}\right) =u_{-}u_{0}$ (by the definition of $\rho$), and% \begin{align*} & \left( \left[ H_{i},E_{j}\right] \circ\rho\right) \left( u_{-}\otimes u_{0}\right) \\ & =\underbrace{\left[ H_{i},E_{j}\right] }_{=H_{i}\circ E_{j}-E_{j}\circ H_{i}}\left( \underbrace{\rho\left( u_{-}\otimes u_{0}\right) }% _{=u_{-}u_{0}}\right) =\left( H_{i}\circ E_{j}-E_{j}\circ H_{i}\right) \left( u_{-}u_{0}\right) \\ & =H_{i}\left( \underbrace{E_{j}\left( u_{-}u_{0}\right) }% _{\substack{=\varepsilon_{j}\left( u_{-}\right) u_{0}\\\text{(by (\ref{pf.gtilde.b.Ei}), applied}\\\text{to }j\text{ instead of }i\text{)}% }}\right) -E_{j}\left( \underbrace{H_{i}\left( u_{-}u_{0}\right) }_{\substack{=h_{i}u_{-}u_{0}\\\text{(by the definition of }H_{i}\text{)}% }}\right) \\ & =\underbrace{H_{i}\left( \varepsilon_{j}\left( u_{-}\right) u_{0}\right) }_{\substack{=h_{i}\varepsilon_{j}\left( u_{-}\right) u_{0}\\\text{(by the definition of }H_{i}\text{)}}}-\underbrace{E_{j}\left( h_{i}u_{-}u_{0}\right) }_{\substack{=\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}+\varepsilon_{j}\left( u_{-}\right) h_{i}u_{0}\\\text{(by (\ref{pf.gtilde.dercomm5}))}}}\\ & =h_{i}\varepsilon_{j}\left( u_{-}\right) u_{0}-\left( \varepsilon _{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}+\varepsilon_{j}\left( u_{-}\right) h_{i}u_{0}\right) =h_{i}\varepsilon_{j}\left( u_{-}\right) u_{0}-\varepsilon_{j}\left( u_{-}\right) h_{i}u_{0}-\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) u_{0}\\ & =\left( \underbrace{h_{i}\varepsilon_{j}\left( u_{-}\right) -\varepsilon_{j}\left( u_{-}\right) h_{i}}_{=\left[ h_{i},\varepsilon _{j}\left( u_{-}\right) \right] }-\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) \right) u_{0}=\underbrace{\left( \left[ h_{i},\varepsilon_{j}\left( u_{-}\right) \right] -\varepsilon_{j}\left( \left[ h_{i},u_{-}\right] \right) \right) }_{\substack{=a_{i,j}% \varepsilon_{j}\left( u_{-}\right) \\\text{(by (\ref{pf.gtilde.dercomm2}))}% }}u_{0}\\ & =a_{i,j}\underbrace{\varepsilon_{j}\left( u_{-}\right) u_{0}% }_{\substack{=E_{j}\left( u_{-}u_{0}\right) \\\text{(since (\ref{pf.gtilde.b.Ei}) (applied to }j\text{ instead of }i\text{)}% \\\text{yields }E_{j}\left( u_{-}u_{0}\right) =\varepsilon_{j}\left( u_{-}\right) u_{0}\text{)}}}=a_{i,j}E_{j}\left( \underbrace{u_{-}u_{0}% }_{=\rho\left( u_{-}\otimes u_{0}\right) }\right) =a_{i,j}E_{j}\left( \rho\left( u_{-}\otimes u_{0}\right) \right) \\ & =\left( a_{i,j}E_{j}\circ\rho\right) \left( u_{-}\otimes u_{0}\right) . \end{align*} Now, forget that we fixed $u_{-}$ and $u_{0}$. We thus have proven that% \[ \left( \left[ H_{i},E_{j}\right] \circ\rho\right) \left( u_{-}\otimes u_{0}\right) =\left( a_{i,j}E_{j}\circ\rho\right) \left( u_{-}\otimes u_{0}\right) \ \ \ \ \ \ \ \ \ \ \text{for all }u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) \text{ and }u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) . \] In other words, the two maps $\left[ H_{i},E_{j}\right] \circ\rho$ and $a_{i,j}E_{j}\circ\rho$ are equal to each other on each pure tensor in the tensor product $U\left( \widetilde{\mathfrak{n}}_{-}\right) \otimes U\left( \widetilde{\mathfrak{h}}\right) $. Since these two maps are linear, this yields that these two maps must be identical (because whenever two linear maps from a tensor product are equal to each other on each pure tensor, these maps must be identical). In other words, $\left[ H_{i},E_{j}\right] \circ \rho=a_{i,j}E_{j}\circ\rho$. Since we can cancel the $\rho$ from this equation (because $\rho$ is an isomorphism), this yields $\left[ H_{i},E_{j}\right] =a_{i,j}E_{j}$. \end{verlong} Now forget that we fixed $i$ and $j$. We have thus proven the relation $\left[ H_{i},E_{j}\right] =a_{i,j}E_{j}$ for all $i,j\in\left\{ 1,2,...,n\right\} $. \bigskip \textit{Proof of the relation }$\left[ H_{i},F_{j}\right] =-a_{i,j}F_{j}$ \textit{for all }$i,j\in\left\{ 1,2,...,n\right\} $\textit{:} \begin{vershort} The proof of the relation $\left[ H_{i},F_{j}\right] =-a_{i,j}F_{j}$ for all $i,j\in\left\{ 1,2,...,n\right\} $ is analogous to the above-given proof of the relation $\left[ H_{i},H_{j}\right] =0$ for all $i,j\in\left\{ 1,2,...,n\right\} $ (except that this time, instead of using the equality $\left[ h_{i},h_{j}\right] =0$ in $\widetilde{\mathfrak{h}}$, need to apply the equality $\left[ h_{i},f_{j}\right] =-a_{i,j}f_{j}$ in $\widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}$; the latter equality is a consequence of (\ref{pf.gtilde.b.semidir.ij})). \end{vershort} \begin{verlong} Let $i$ and $j$ be two elements of $\left\{ 1,2,...,n\right\} $. Every $u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $ satisfies% \begin{align*} \underbrace{\left[ H_{i},F_{j}\right] }_{=H_{i}\circ F_{j}-F_{j}\circ H_{i}% }u & =\left( H_{i}\circ F_{j}-F_{j}\circ H_{i}\right) \left( u\right) =H_{i}\underbrace{\left( F_{j}u\right) }_{\substack{=f_{j}u\\\text{(by the definition}\\\text{of }F_{j}\text{)}}}-F_{j}\underbrace{\left( H_{i}u\right) }_{\substack{=h_{i}u\\\text{(by the definition}\\\text{of }H_{i}\text{)}}}\\ & =\underbrace{H_{i}\left( f_{j}u\right) }_{\substack{=h_{i}\left( f_{j}u\right) \\\text{(by the definition}\\\text{of }H_{i}\text{)}% }}-\underbrace{F_{j}\left( h_{i}u\right) }_{\substack{=f_{j}\left( h_{i}u\right) \\\text{(by the definition}\\\text{of }F_{j}\text{)}}}\\ & =h_{i}\left( f_{j}u\right) -f_{j}\left( h_{i}u\right) =\underbrace{\left( h_{i}f_{j}-f_{j}h_{i}\right) }_{\substack{=\left[ h_{i},f_{j}\right] =-a_{i,j}f_{j}\\\text{in }U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) \\\text{(since (\ref{pf.gtilde.b.semidir.ij}) yields }\left[ h_{i},f_{j}\right] =-a_{i,j}f_{j}\text{ in }\widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\text{)}}}u\\ & =-a_{i,j}f_{j}u \end{align*} and $-a_{i,j}F_{j}u=-a_{i,j}f_{j}u$ (because the definition of $F_{j}$ yields $F_{j}u=f_{j}u$). Thus, every $u\in U\left( \widetilde{\mathfrak{h}}% \ltimes\widetilde{\mathfrak{n}}_{-}\right) $ satisfies $\left[ H_{i}% ,F_{j}\right] u=-a_{i,j}F_{j}u$. Hence, $\left[ H_{i},F_{j}\right] =-a_{i,j}F_{j}$. Now forget that we fixed $i$ and $j$. We have thus proven the relation $\left[ H_{i},F_{j}\right] =-a_{i,j}F_{j}$ for all $i,j\in\left\{ 1,2,...,n\right\} $. \end{verlong} \bigskip \textit{Proof of the relation }$\left[ E_{i},F_{j}\right] =\delta_{i,j}% H_{i}$ \textit{for all }$i,j\in\left\{ 1,2,...,n\right\} $\textit{:} \begin{vershort} Let $i$ and $j$ be two elements of $\left\{ 1,2,...,n\right\} $. Let $u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) $ and $u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) $. Since $f_{j}\in \widetilde{\mathfrak{n}}_{-}$ and $u_{-}\in U\left( \widetilde{\mathfrak{n}% }_{-}\right) $, we have $f_{j}u_{-}\in\widetilde{\mathfrak{n}}_{-}\cdot U\left( \widetilde{\mathfrak{n}}_{-}\right) \subseteq U\left( \widetilde{\mathfrak{n}}_{-}\right) $. Thus, we can apply (\ref{pf.gtilde.b.Ei.short}) to $f_{j}u_{-}$ instead of $u_{-}$, and obtain% \begin{align*} E_{i}\left( f_{j}u_{-}u_{0}\right) & =\underbrace{\varepsilon_{i}\left( f_{j}u_{-}\right) }_{\substack{=\varepsilon_{i}\left( f_{j}\right) u_{-}+f_{j}\varepsilon_{i}\left( u_{-}\right) \\\text{(since }% \varepsilon_{i}\text{ is a derivation)}}}u_{0}=\left( \varepsilon_{i}\left( f_{j}\right) u_{-}+f_{j}\varepsilon_{i}\left( u_{-}\right) \right) u_{0}\\ & =\underbrace{\varepsilon_{i}\left( f_{j}\right) }_{\substack{=\delta _{i,j}h_{i}\\\text{(by the definition of }\varepsilon_{i}\text{)}}}u_{-}% u_{0}+f_{j}\varepsilon_{i}\left( u_{-}\right) u_{0}=\delta_{i,j}h_{i}% u_{-}u_{0}+f_{j}\varepsilon_{i}\left( u_{-}\right) u_{0}. \end{align*} \end{vershort} \begin{verlong} Let $i$ and $j$ be two elements of $\left\{ 1,2,...,n\right\} $. Let $u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) $ and $u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) $. Since $f_{j}\in \widetilde{\mathfrak{n}}_{-}$ and $u_{-}\in U\left( \widetilde{\mathfrak{n}% }_{-}\right) $, we have $f_{j}u_{-}\in\widetilde{\mathfrak{n}}_{-}\cdot U\left( \widetilde{\mathfrak{n}}_{-}\right) \subseteq U\left( \widetilde{\mathfrak{n}}_{-}\right) $. Thus, we can apply (\ref{pf.gtilde.b.Ei}) to $f_{j}u_{-}$ instead of $u_{-}$, and obtain% \begin{align} E_{i}\left( f_{j}u_{-}u_{0}\right) & =\underbrace{\varepsilon_{i}\left( f_{j}u_{-}\right) }_{\substack{=\varepsilon_{i}\left( f_{j}\right) u_{-}+f_{j}\varepsilon_{i}\left( u_{-}\right) \\\text{(since }% \varepsilon_{i}\text{ is a derivation)}}}u_{0}=\left( \varepsilon_{i}\left( f_{j}\right) u_{-}+f_{j}\varepsilon_{i}\left( u_{-}\right) \right) u_{0}\nonumber\\ & =\underbrace{\varepsilon_{i}\left( f_{j}\right) }_{\substack{=\delta _{i,j}h_{i}\\\text{(by (\ref{pf.gtilde.b.epsilon.fj}))}}}u_{-}u_{0}% +f_{j}\varepsilon_{i}\left( u_{-}\right) u_{0}\nonumber\\ & =\delta_{i,j}h_{i}u_{-}u_{0}+f_{j}\varepsilon_{i}\left( u_{-}\right) u_{0}. \label{pf.gtilde.dercomm7}% \end{align} \end{verlong} \begin{vershort} But% \begin{align*} \underbrace{\left[ E_{i},F_{j}\right] }_{=E_{i}\circ F_{j}-F_{j}\circ E_{i}% }\left( u_{-}u_{0}\right) & =\left( E_{i}\circ F_{j}-F_{j}\circ E_{i}\right) \left( u_{-}u_{0}\right) \\ & =E_{i}\left( \underbrace{F_{j}\left( u_{-}u_{0}\right) }% _{\substack{=f_{j}u_{-}u_{0}\\\text{(by the definition of }F_{j}\text{)}% }}\right) -F_{j}\left( \underbrace{E_{i}\left( u_{-}u_{0}\right) }_{\substack{=\varepsilon_{i}\left( u_{-}\right) u_{0}\\\text{(by (\ref{pf.gtilde.b.Ei.short}))}}}\right) \\ & =\underbrace{E_{i}\left( f_{j}u_{-}u_{0}\right) }_{\substack{=\delta _{i,j}h_{i}u_{-}u_{0}+f_{j}\varepsilon_{i}\left( u_{-}\right) u_{0}% \\\text{(as we have proven above)}}}-\underbrace{F_{j}\left( \varepsilon _{i}\left( u_{-}\right) u_{0}\right) }_{\substack{=f_{j}\varepsilon _{i}\left( u_{-}\right) u_{0}\\\text{(by the definition of }F_{j}\text{)}% }}\\ & =\delta_{i,j}h_{i}u_{-}u_{0}+f_{j}\varepsilon_{i}\left( u_{-}\right) u_{0}-f_{j}\varepsilon_{i}\left( u_{-}\right) u_{0}=\delta_{i,j}h_{i}% u_{-}u_{0}\\ & =\delta_{i,j}\underbrace{h_{i}u_{-}u_{0}}_{\substack{=H_{i}\left( u_{-}u_{0}\right) \\\text{(since the definition of }H_{i}\text{ yields}\\H_{i}\left( u_{-}u_{0}\right) =h_{i}u_{-}u_{0}\text{)}}% }=\delta_{i,j}H_{i}\left( u_{-}u_{0}\right) . \end{align*} Now, forget that we fixed $u_{-}$ and $u_{0}$. We thus have proven that% \[ \left[ E_{i},F_{j}\right] \left( u_{-}u_{0}\right) =\delta_{i,j}% H_{i}\left( u_{-}u_{0}\right) \ \ \ \ \ \ \ \ \ \ \text{for all }u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) \text{ and }u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) \text{.}% \] Since the vector space $U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ is generated by products $u_{-}u_{0}$ with $u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) $ and $u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) $ (this is because $\rho:U\left( \widetilde{\mathfrak{n}}_{-}\right) \otimes U\left( \widetilde{\mathfrak{h}% }\right) \rightarrow U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ is an isomorphism), this yields that $\left[ E_{i},F_{j}\right] =\delta_{i,j}H_{i}$. \end{vershort} \begin{verlong} But $\rho\left( u_{-}\otimes u_{0}\right) =u_{-}u_{0}$ (by the definition of $\rho$), and% \begin{align*} & \left( \left[ E_{i},F_{j}\right] \circ\rho\right) \left( u_{-}\otimes u_{0}\right) \\ & =\underbrace{\left[ E_{i},F_{j}\right] }_{=E_{i}\circ F_{j}-F_{j}\circ E_{i}}\left( \underbrace{\rho\left( u_{-}\otimes u_{0}\right) }% _{=u_{-}u_{0}}\right) =\left( E_{i}\circ F_{j}-F_{j}\circ E_{i}\right) \left( u_{-}u_{0}\right) \\ & =E_{i}\left( \underbrace{F_{j}\left( u_{-}u_{0}\right) }% _{\substack{=f_{j}u_{-}u_{0}\\\text{(by the definition of }F_{j}\text{)}% }}\right) -F_{j}\left( \underbrace{E_{i}\left( u_{-}u_{0}\right) }_{\substack{=\varepsilon_{i}\left( u_{-}\right) u_{0}\\\text{(by (\ref{pf.gtilde.b.Ei}))}}}\right) \\ & =\underbrace{E_{i}\left( f_{j}u_{-}u_{0}\right) }_{\substack{=\delta _{i,j}h_{i}u_{-}u_{0}+f_{j}\varepsilon_{i}\left( u_{-}\right) u_{0}% \\\text{(by (\ref{pf.gtilde.dercomm7}))}}}-\underbrace{F_{j}\left( \varepsilon_{i}\left( u_{-}\right) u_{0}\right) }_{\substack{=f_{j}% \varepsilon_{i}\left( u_{-}\right) u_{0}\\\text{(by the definition of }% F_{j}\text{)}}}\\ & =\delta_{i,j}h_{i}u_{-}u_{0}+f_{j}\varepsilon_{i}\left( u_{-}\right) u_{0}-f_{j}\varepsilon_{i}\left( u_{-}\right) u_{0}=\delta_{i,j}% h_{i}\underbrace{u_{-}u_{0}}_{=\rho\left( u_{-}\otimes u_{0}\right) }\\ & =\delta_{i,j}\underbrace{h_{i}\rho\left( u_{-}\otimes u_{0}\right) }_{\substack{=H_{i}\left( \rho\left( u_{-}\otimes u_{0}\right) \right) \\\text{(since the definition of }H_{i}\text{ yields}\\H_{i}\left( \rho\left( u_{-}\otimes u_{0}\right) \right) =h_{i}\rho\left( u_{-}\otimes u_{0}\right) \text{)}}}=\delta_{i,j}H_{i}\left( \rho\left( u_{-}\otimes u_{0}\right) \right) \\ & =\left( \delta_{i,j}H_{i}\circ\rho\right) \left( u_{-}\otimes u_{0}\right) . \end{align*} Now, forget that we fixed $u_{-}$ and $u_{0}$. We thus have proven that% \[ \left( \left[ E_{i},F_{j}\right] \circ\rho\right) \left( u_{-}\otimes u_{0}\right) =\left( \delta_{i,j}H_{i}\circ\rho\right) \left( u_{-}\otimes u_{0}\right) \ \ \ \ \ \ \ \ \ \ \text{for all }u_{-}\in U\left( \widetilde{\mathfrak{n}}_{-}\right) \text{ and }u_{0}\in U\left( \widetilde{\mathfrak{h}}\right) . \] In other words, the two maps $\left[ E_{i},F_{j}\right] \circ\rho$ and $\delta_{i,j}H_{i}\circ\rho$ are equal to each other on each pure tensor in the tensor product $U\left( \widetilde{\mathfrak{n}}_{-}\right) \otimes U\left( \widetilde{\mathfrak{h}}\right) $. Since these two maps are linear, this yields that these two maps must be identical (because whenever two linear maps from a tensor product are equal to each other on each pure tensor, these maps must be identical). In other words, $\left[ E_{i},F_{j}\right] \circ\rho=\delta_{i,j}H_{i}\circ\rho$. Since we can cancel the $\rho$ from this equation (because $\rho$ is an isomorphism), this yields $\left[ E_{i},F_{j}\right] =\delta_{i,j}H_{i}$. \end{verlong} Now forget that we fixed $i$ and $j$. We have thus proven the relation $\left[ E_{i},F_{j}\right] =\delta_{i,j}H_{i}$ for all $i,j\in\left\{ 1,2,...,n\right\} $. \bigskip Altogether, we have thus verified all four relations (\ref{pf.gtilde.NONSERRE}% ). Now, let us define a Lie algebra homomorphism $\xi^{\prime}% :\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \rightarrow\operatorname*{End}\left( U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right) $ by the relations% \[ \left\{ \begin{array} [c]{c}% \xi^{\prime}\left( e_{i}\right) =E_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{ 1,2,...,n\right\} ;\\ \xi^{\prime}\left( f_{i}\right) =F_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{ 1,2,...,n\right\} ;\\ \xi^{\prime}\left( h_{i}\right) =H_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{ 1,2,...,n\right\} \end{array} \right. . \] This $\xi^{\prime}$ is clearly well-defined (because a Lie algebra homomorphism from a free Lie algebra over a set can be defined by arbitrarily choosing its values at the elements of this set). This homomorphism $\xi^{\prime}$ clearly maps the four relations (\ref{nonserre-relations}) to the four relations (\ref{pf.gtilde.NONSERRE}). Since we know that the four relations (\ref{pf.gtilde.NONSERRE}) are satisfied in $\operatorname*{End}% \left( U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}% _{-}\right) \right) $, we conclude that the homomorphism $\xi^{\prime}$ factors through the Lie algebra $\operatorname*{FreeLie}\left( h_{i}% ,f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) =\widetilde{\mathfrak{g}}$. In other words, there exists a Lie algebra homomorphism $\xi:\widetilde{\mathfrak{g}}\rightarrow\operatorname*{End}% \left( U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}% _{-}\right) \right) $ such that% \[ \left\{ \begin{array} [c]{c}% \xi\left( e_{i}\right) =E_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} ;\\ \xi\left( f_{i}\right) =F_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} ;\\ \xi\left( h_{i}\right) =H_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} \end{array} \right. . \] Consider this $\xi$. Clearly, the Lie algebra homomorphism $\xi :\widetilde{\mathfrak{g}}\rightarrow\operatorname*{End}\left( U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right) $ makes the vector space $U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $ into a $\widetilde{\mathfrak{g}}$-module. \bigskip \underline{\textit{6th step: Proving the injectivity of }$\iota_{-}$% \textit{.}} We are very close to proving Theorem \ref{thm.gtilde} \textbf{(b)} now. Let $\xi_{-}$ be the map $\xi\circ\iota_{-}:\widetilde{\mathfrak{n}}% _{-}\rightarrow\operatorname*{End}\left( U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right) $. Then, $\xi_{-}$ is a Lie algebra homomorphism (since $\xi$ and $\iota_{-}$ are Lie algebra homomorphisms). Every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\underbrace{\xi_{-}}% _{=\xi\circ\iota_{-}}\left( f_{i}\right) =\left( \xi\circ\iota_{-}\right) \left( f_{i}\right) =\xi\left( \underbrace{\iota_{-}\left( f_{i}\right) }_{\substack{=f_{i}\\\text{(by the definition of }\iota_{-}\text{)}}}\right) =\xi\left( f_{i}\right) =F_{i}$ (by the definition of $\xi$). Let $\mathfrak{s}$ be the subset% \[ \left\{ s\in\widetilde{\mathfrak{n}}_{-}\ \mid\ \left( \xi_{-}\left( s\right) \right) \left( u\right) =su\text{ for all }u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right\} \] of $\widetilde{\mathfrak{n}}_{-}$. Every $i\in\left\{ 1,2,...,n\right\} $ satisfies% \[ \underbrace{\left( \xi_{-}\left( f_{i}\right) \right) }_{=F_{i}}\left( u\right) =F_{i}\left( u\right) =f_{i}u\ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }F_{i}\right) \] for all $u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}% }_{-}\right) $, and therefore% \[ f_{i}\in\left\{ s\in\widetilde{\mathfrak{n}}_{-}\ \mid\ \left( \xi _{-}\left( s\right) \right) \left( u\right) =su\text{ for all }u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right\} =\mathfrak{s}. \] In other words, $\mathfrak{s}$ contains the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$. \begin{vershort} On the other hand, it is very easy to see that $\mathfrak{s}$ is a Lie subalgebra of $\widetilde{\mathfrak{n}}_{-}$. (In fact, all that is needed to prove this is knowing that $\xi_{-}$ is a Lie algebra homomorphism. The details are left to the reader.) \end{vershort} \begin{verlong} Now, we will show that $\mathfrak{s}$ is a Lie subalgebra of $\widetilde{\mathfrak{n}}_{-}$. \textit{Proof that $\mathfrak{s}$ is a Lie subalgebra of }% $\widetilde{\mathfrak{n}}_{-}$\textit{:} The reader can easily verify that $\mathfrak{s}$ is a vector subspace of $\widetilde{\mathfrak{n}}_{-}$ (because for any fixed $u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $, the equation $\left( \xi_{-}\left( s\right) \right) \left( u\right) =su$ is linear in $s$). Now, let $s_{1}\in\mathfrak{s}$ and $s_{2}\in\mathfrak{s}$ be arbitrary. Then,% \[ s_{1}\in\mathfrak{s}=\left\{ s\in\widetilde{\mathfrak{n}}_{-}\ \mid\ \left( \xi_{-}\left( s\right) \right) \left( u\right) =su\text{ for all }u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right\} , \] so that% \begin{equation} \left( \xi_{-}\left( s_{1}\right) \right) \left( u\right) =s_{1}% u\ \ \ \ \ \ \ \ \ \ \text{for all }u\in U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) . \label{pf.gtilde.b.step6.1}% \end{equation} Similarly,% \begin{equation} \left( \xi_{-}\left( s_{2}\right) \right) \left( u\right) =s_{2}% u\ \ \ \ \ \ \ \ \ \ \text{for all }u\in U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) . \label{pf.gtilde.b.step6.2}% \end{equation} Now, since $\xi_{-}$ is a Lie algebra homomorphism, we have $\xi_{-}\left( \left[ s_{1},s_{2}\right] \right) =\left[ \xi_{-}\left( s_{1}\right) ,\xi_{-}\left( s_{2}\right) \right] =\xi_{-}\left( s_{1}\right) \circ \xi_{-}\left( s_{2}\right) -\xi_{-}\left( s_{2}\right) \circ\xi_{-}\left( s_{1}\right) $. Thus, every $u\in U\left( \widetilde{\mathfrak{h}}% \ltimes\widetilde{\mathfrak{n}}_{-}\right) $ satisfies% \begin{align*} \left( \xi_{-}\left( \left[ s_{1},s_{2}\right] \right) \right) \left( u\right) & =\left( \xi_{-}\left( s_{1}\right) \circ\xi_{-}\left( s_{2}\right) -\xi_{-}\left( s_{2}\right) \circ\xi_{-}\left( s_{1}\right) \right) \left( u\right) \\ & =\left( \xi_{-}\left( s_{1}\right) \right) \left( \underbrace{\left( \xi_{-}\left( s_{2}\right) \right) \left( u\right) }_{\substack{=s_{2}% u\\\text{(by (\ref{pf.gtilde.b.step6.2}))}}}\right) -\left( \xi_{-}\left( s_{2}\right) \right) \left( \underbrace{\left( \xi_{-}\left( s_{1}\right) \right) \left( u\right) }_{\substack{=s_{1}u\\\text{(by (\ref{pf.gtilde.b.step6.1}))}}}\right) \\ & =\underbrace{\left( \xi_{-}\left( s_{1}\right) \right) \left( s_{2}u\right) }_{\substack{=s_{1}s_{2}u\\\text{(by (\ref{pf.gtilde.b.step6.1}% ), applied to}\\s_{2}u\text{ instead of }u\text{)}}}-\underbrace{\left( \xi_{-}\left( s_{2}\right) \right) \left( s_{1}u\right) }% _{\substack{=s_{2}s_{1}u\\\text{(by (\ref{pf.gtilde.b.step6.2}), applied to}\\s_{1}u\text{ instead of }u\text{)}}}\\ & =s_{1}s_{2}u-s_{2}s_{1}u=\underbrace{\left( s_{1}s_{2}-s_{2}s_{1}\right) }_{=\left[ s_{1},s_{2}\right] }u=\left[ s_{1},s_{2}\right] u. \end{align*} Hence,% \[ \left[ s_{1},s_{2}\right] \in\left\{ s\in\widetilde{\mathfrak{n}}_{-}% \ \mid\ \left( \xi_{-}\left( s\right) \right) \left( u\right) =su\text{ for all }u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}% }_{-}\right) \right\} =\mathfrak{s}. \] Now, forget that we fixed $s_{1}$ and $s_{2}$. We thus have proven that every $s_{1}\in\mathfrak{s}$ and $s_{2}\in\mathfrak{s}$ satisfy $\left[ s_{1}% ,s_{2}\right] \in\mathfrak{s}$. Since $\mathfrak{s}$ is a vector subspace of $\widetilde{\mathfrak{n}}_{-}$, this yields that $\mathfrak{s}$ is a Lie subalgebra of $\widetilde{\mathfrak{n}}_{-}$. \end{verlong} But now recall that $\widetilde{\mathfrak{n}}_{-}=\operatorname*{FreeLie}% \left( f_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) $. Hence, the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ generate $\widetilde{\mathfrak{n}% }_{-}$ as a Lie algebra. Thus, every Lie subalgebra of $\widetilde{\mathfrak{n}}_{-}$ which contains the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ must be $\widetilde{\mathfrak{n}}_{-}$ itself. Since we know that $\mathfrak{s}$ is a Lie subalgebra of $\widetilde{\mathfrak{n}}_{-}$ and contains the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$, this yields that $\mathfrak{s}$ must be $\widetilde{\mathfrak{n}}_{-}$ itself. In other words, $\mathfrak{s}=\widetilde{\mathfrak{n}}_{-}$. Now, let $s^{\prime}\in\widetilde{\mathfrak{n}}_{-}$ be such that $\iota _{-}\left( s^{\prime}\right) =0$. Then, $\underbrace{\xi_{-}}_{=\xi \circ\iota_{-}}\left( s^{\prime}\right) =\left( \xi\circ\iota_{-}\right) \left( s^{\prime}\right) =\xi\left( \underbrace{\iota_{-}\left( s^{\prime }\right) }_{=0}\right) =\xi\left( 0\right) =0$. But since% \[ s^{\prime}\in\widetilde{\mathfrak{n}}_{-}=\mathfrak{s}=\left\{ s\in \widetilde{\mathfrak{n}}_{-}\ \mid\ \left( \xi_{-}\left( s\right) \right) \left( u\right) =su\text{ for all }u\in U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right\} , \] we have $\left( \xi_{-}\left( s^{\prime}\right) \right) \left( u\right) =s^{\prime}u$ for all $u\in U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $. Applied to $u=1$, this yields $\left( \xi_{-}\left( s^{\prime}\right) \right) \left( 1\right) =s^{\prime}% \cdot1=s^{\prime}$. Compared with $\underbrace{\left( \xi_{-}\left( s^{\prime}\right) \right) }_{=0}\left( 1\right) =0$, this yields $s^{\prime}=0$. Now forget that we fixed $s^{\prime}$. We have thus shown that every $s^{\prime}\in\widetilde{\mathfrak{n}}_{-}$ such that $\iota_{-}\left( s^{\prime}\right) =0$ must satisfy $s^{\prime}=0$. In other words, $\iota _{-}$ is injective. \bigskip \underline{\textit{7th step: Proving the injectivity of }$\iota_{0}$% \textit{.}} \begin{vershort} A similar, but even simpler, argument shows that $\iota_{0}$ is injective. Again, the reader can fill in the details. \end{vershort} \begin{verlong} The proof of the injectivity of $\iota_{0}$ is similar but even simpler. Let $\xi_{0}$ be the map $\xi\circ\iota_{0}:\widetilde{\mathfrak{h}% }\rightarrow\operatorname*{End}\left( U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right) $. Every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\underbrace{\xi_{0}}% _{=\xi\circ\iota_{0}}\left( h_{i}\right) =\left( \xi\circ\iota_{0}\right) \left( h_{i}\right) =\xi\left( \underbrace{\iota_{0}\left( h_{i}\right) }_{\substack{=h_{i}\\\text{(by the definition of }\iota_{0}\text{)}}}\right) =\xi\left( h_{i}\right) =H_{i}$ (by the definition of $\xi$). Let $\mathfrak{t}$ be the subset% \[ \left\{ s\in\widetilde{\mathfrak{h}}\ \mid\ \left( \xi_{0}\left( s\right) \right) \left( u\right) =su\text{ for all }u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right\} \] of $\widetilde{\mathfrak{h}}$. Every $i\in\left\{ 1,2,...,n\right\} $ satisfies% \[ \underbrace{\left( \xi_{0}\left( h_{i}\right) \right) }_{=H_{i}}\left( u\right) =H_{i}\left( u\right) =h_{i}u\ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }H_{i}\right) \] for all $u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}% }_{-}\right) $, and therefore% \[ h_{i}\in\left\{ s\in\widetilde{\mathfrak{h}}\ \mid\ \left( \xi_{0}\left( s\right) \right) \left( u\right) =su\text{ for all }u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right\} =\mathfrak{t}. \] In other words, $\mathfrak{t}$ contains the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$. On the other hand, it is easy to see that $\mathfrak{t}$ is a vector subspace of $\widetilde{\mathfrak{h}}$ (because for any fixed $u\in U\left( \widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}}_{-}\right) $, the equation $\left( \xi_{0}\left( s\right) \right) \left( u\right) =su$ is linear in $s$). But now recall that $\widetilde{\mathfrak{h}}$ is the free vector space with basis $h_{1},h_{2},...,h_{n}$. Thus, the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$ span the vector space $\widetilde{\mathfrak{h}}$. Thus, every vector subspace of $\widetilde{\mathfrak{h}}$ which contains the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$ must be $\widetilde{\mathfrak{h}}$ itself. Since we know that $\mathfrak{t}$ is a vector subspace of $\widetilde{\mathfrak{h}}$ and contains the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$, this yields that $\mathfrak{t}$ must be $\widetilde{\mathfrak{h}}$ itself. In other words, $\mathfrak{t}=\widetilde{\mathfrak{h}}$. Now, let $s^{\prime}\in\widetilde{\mathfrak{h}}$ be such that $\iota _{0}\left( s^{\prime}\right) =0$. Then, $\underbrace{\xi_{0}}_{=\xi \circ\iota_{0}}\left( s^{\prime}\right) =\left( \xi\circ\iota_{0}\right) \left( s^{\prime}\right) =\xi\left( \underbrace{\iota_{0}\left( s^{\prime }\right) }_{=0}\right) =\xi\left( 0\right) =0$. But since% \[ s^{\prime}\in\widetilde{\mathfrak{h}}=\mathfrak{t}=\left\{ s\in \widetilde{\mathfrak{h}}\ \mid\ \left( \xi_{0}\left( s\right) \right) \left( u\right) =su\text{ for all }u\in U\left( \widetilde{\mathfrak{h}% }\ltimes\widetilde{\mathfrak{n}}_{-}\right) \right\} , \] we have $\left( \xi_{0}\left( s^{\prime}\right) \right) \left( u\right) =s^{\prime}u$ for all $u\in U\left( \widetilde{\mathfrak{h}}\ltimes \widetilde{\mathfrak{n}}_{-}\right) $. Applied to $u=1$, this yields $\left( \xi_{0}\left( s^{\prime}\right) \right) \left( 1\right) =s^{\prime}% \cdot1=s^{\prime}$. Compared with $\underbrace{\left( \xi_{0}\left( s^{\prime}\right) \right) }_{=0}\left( 1\right) =0$, this yields $s^{\prime}=0$. Now forget that we fixed $s^{\prime}$. We have thus shown that every $s^{\prime}\in\widetilde{\mathfrak{h}}$ such that $\iota_{0}\left( s^{\prime }\right) =0$ must satisfy $s^{\prime}=0$. In other words, $\iota_{0}$ is injective. \end{verlong} \bigskip \underline{\textit{8th step: Proving the injectivity of }$\iota_{+}$% \textit{.}} \begin{vershort} We have proven the injectivity of the maps $\iota_{-}$ and $\iota_{0}$ above. The proof of the injectivity of the map $\iota_{+}$ is analogous to the above proof of the injectivity of the map $\iota_{-}$. (Alternately, the injectivity of $\iota_{+}$ can be obtained from that of $\iota_{-}$ using the involutive Lie algebra automorphism constructed in Theorem \ref{thm.gtilde} \textbf{(f)}.) \end{vershort} \begin{verlong} We have proven the injectivity of the maps $\iota_{-}$ and $\iota_{0}$ above. The proof of the injectivity of the map $\iota_{+}$ is very similar to the above proof of the injectivity of the map $\iota_{-}$. More precisely, in order to obtain a proof of the injectivity of the map $\iota_{+}$, it is enough to apply the following changes to our above proof of the injectivity of the map $\iota_{-}$: \begin{itemize} \item replace every $e_{i}$ by $f_{i}$, and simultaneously replace every $f_{i}$ by $e_{i}$; \item replace every $h_{i}$ by $-h_{i}$ (this is allowed since $\widetilde{\mathfrak{h}}$ is the free vector space with basis $-h_{1}% ,-h_{2},...,-h_{n}$); \item replace $\widetilde{\mathfrak{n}}_{-}$ by $\widetilde{\mathfrak{n}}_{+}$; \item replace every reference to (\ref{nonserre-relations}) by a reference to (\ref{nonserre-relations2}) (this is allowed since we know that the relations (\ref{nonserre-relations2}) are satisfied in $\widetilde{\mathfrak{g}}$). \end{itemize} We have thus proven that the maps $\iota_{+}$, $\iota_{-}$ and $\iota_{0}$ are injective. Moreover, $\iota_{+}$ and $\iota_{-}$ are Lie algebra homomorphisms (by definition), and $\iota_{0}$ is a Lie algebra homomorphism as well (because any $i,j\in\left\{ 1,2,...,n\right\} $ satisfy $\left[ h_{i}% ,h_{j}\right] =0$ in $\widetilde{\mathfrak{h}}$ (since $\widetilde{\mathfrak{h}}$ is an abelian Lie algebra) and $\left[ h_{i}% ,h_{j}\right] =0$ in $\widetilde{\mathfrak{g}}$ (due to the relations (\ref{nonserre-relations}))). This completes the proof of Theorem \ref{thm.gtilde} \textbf{(b)}. \end{verlong} \bigskip \textbf{(c)} \underline{\textit{1st step: The existence of the direct sum in question.}} Define a relation $\leq$ on $Q$ by positing that two $n$-tuples $\left( \lambda_{1},\lambda_{2},...,\lambda_{n}\right) \in\mathbb{Z}^{n}$ and $\left( \mu_{1},\mu_{2},...,\mu_{n}\right) \in\mathbb{Z}^{n}$ satisfy $\lambda_{1}\alpha_{1}+\lambda_{2}\alpha_{2}+...+\lambda_{n}\alpha_{n}\leq \mu_{1}\alpha_{1}+\mu_{2}\alpha_{2}+...+\mu_{n}\alpha_{n}$ if and only if every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\lambda_{i}\leq\mu_{i}$. It is clear that this relation $\leq$ is a non-strict partial order. Define $\geq$ to be the opposite of $\leq$. Define $>$ and $<$ to be the strict versions of the relations $\geq$ and $\leq$, respectively; thus, any $\alpha\in Q$ and $\beta\in Q$ satisfy $\alpha>\beta$ if and only if $\left( \alpha\neq\beta\text{ and }\alpha\geq\beta\right) $. The elements $\alpha$ of $Q$ satisfying $\alpha>0$ are exactly the nonzero sums $\lambda_{1}\alpha_{1}+\lambda_{2}\alpha_{2}+...+\lambda_{n}\alpha_{n}$ with $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{n}$ being nonnegative integers. The elements $\alpha$ of $Q$ satisfying $\alpha<0$ are exactly the nonzero sums $\lambda_{1}\alpha_{1}+\lambda_{2}\alpha_{2}+...+\lambda _{n}\alpha_{n}$ with $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{n}$ being nonpositive integers. Let $\widetilde{\mathfrak{g}}\left[ <0\right] =\bigoplus \limits_{\substack{\alpha\in Q;\\\alpha<0}}\widetilde{\mathfrak{g}}\left[ \alpha\right] $ and $\widetilde{\mathfrak{g}}\left[ >0\right] =\bigoplus\limits_{\substack{\alpha\in Q;\\\alpha>0}}\widetilde{\mathfrak{g}% }\left[ \alpha\right] $. Then, $\widetilde{\mathfrak{g}}\left[ 0\right] $, $\widetilde{\mathfrak{g}}\left[ <0\right] $ and $\widetilde{\mathfrak{g}% }\left[ >0\right] $ are $Q$-graded Lie subalgebras of $\widetilde{\mathfrak{g}}$ (this is easy to see from the fact that $\widetilde{\mathfrak{g}}$ is a $Q$-graded Lie algebra). It is easy to see that the (internal) direct sum $\widetilde{\mathfrak{g}% }\left[ >0\right] \oplus\widetilde{\mathfrak{g}}\left[ <0\right] \oplus\widetilde{\mathfrak{g}}\left[ 0\right] $ is well-defined.\footnote{\textit{Proof.} We have $\widetilde{\mathfrak{g}% }=\bigoplus\limits_{\alpha\in Q}\widetilde{\mathfrak{g}}\left[ \alpha\right] $ (since $\widetilde{\mathfrak{g}}$ is $Q$-graded). But every $\alpha\in Q$ satisfies \textbf{exactly one} of the four assertions $\alpha>0$, $\alpha<0$, $\alpha=0$ and $\left( \text{neither }\alpha<0\text{ nor }\alpha>0\text{ nor }\alpha=0\right) $. Thus,% \begin{align*} \bigoplus\limits_{\alpha\in Q}\widetilde{\mathfrak{g}}\left[ \alpha\right] & =\underbrace{\left( \bigoplus\limits_{\substack{\alpha\in Q;\\\alpha >0}}\widetilde{\mathfrak{g}}\left[ \alpha\right] \right) }% _{=\widetilde{\mathfrak{g}}\left[ >0\right] }\oplus\underbrace{\left( \bigoplus\limits_{\substack{\alpha\in Q;\\\alpha<0}}\widetilde{\mathfrak{g}% }\left[ \alpha\right] \right) }_{=\widetilde{\mathfrak{g}}\left[ <0\right] }\oplus\underbrace{\left( \bigoplus\limits_{\substack{\alpha\in Q;\\\alpha=0}}\widetilde{\mathfrak{g}}\left[ \alpha\right] \right) }_{=\widetilde{\mathfrak{g}}\left[ 0\right] }\oplus\left( \bigoplus \limits_{\substack{\alpha\in Q;\\\text{neither }\alpha<0\\\text{nor }% \alpha>0\text{ nor }\alpha=0}}\widetilde{\mathfrak{g}}\left[ \alpha\right] \right) \\ & =\widetilde{\mathfrak{g}}\left[ >0\right] \oplus\widetilde{\mathfrak{g}% }\left[ <0\right] \oplus\widetilde{\mathfrak{g}}\left[ 0\right] \oplus\left( \bigoplus\limits_{\substack{\alpha\in Q;\\\text{neither }% \alpha<0\\\text{nor }\alpha>0\text{ nor }\alpha=0}}\widetilde{\mathfrak{g}% }\left[ \alpha\right] \right) . \end{align*} Thus, the (internal) direct sum $\widetilde{\mathfrak{g}}\left[ >0\right] \oplus\widetilde{\mathfrak{g}}\left[ <0\right] \oplus\widetilde{\mathfrak{g}% }\left[ 0\right] $ is well-defined (because it is a partial sum of the direct sum $\widetilde{\mathfrak{g}}\left[ >0\right] \oplus \widetilde{\mathfrak{g}}\left[ <0\right] \oplus\widetilde{\mathfrak{g}% }\left[ 0\right] \oplus\left( \bigoplus\limits_{\substack{\alpha\in Q;\\\text{neither }\alpha<0\\\text{nor }\alpha>0\text{ nor }\alpha =0}}\widetilde{\mathfrak{g}}\left[ \alpha\right] \right) $).} Every $i\in\left\{ 1,2,...,n\right\} $ satisfies \begin{align*} f_{i} & \in\widetilde{\mathfrak{g}}\left[ -\alpha_{i}\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since }\deg\left( f_{i}\right) =-\alpha_{i}\right) \\ & \subseteq\bigoplus\limits_{\substack{\alpha\in Q;\\\alpha<0}% }\widetilde{\mathfrak{g}}\left[ \alpha\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since }-\alpha_{i}<0\right) \\ & =\widetilde{\mathfrak{g}}\left[ <0\right] . \end{align*} Hence, the Lie algebra $\widetilde{\mathfrak{g}}\left[ <0\right] $ contains the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$. But now, recall that $\widetilde{\mathfrak{n}}_{-}=\operatorname*{FreeLie}\left( f_{i}\ \mid \ i\in\left\{ 1,2,...,n\right\} \right) $. Hence, the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ of $\widetilde{\mathfrak{n}}_{-}$ generate $\widetilde{\mathfrak{n}}_{-}$ as a Lie algebra. Thus, the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ of $\widetilde{\mathfrak{g}}$ generate $\iota _{-}\left( \widetilde{\mathfrak{n}}_{-}\right) $ as a Lie algebra (because the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ of $\widetilde{\mathfrak{g}}$ are the images of the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ of $\widetilde{\mathfrak{n}}_{-}$ under the map $\iota_{-}$). Thus, every Lie subalgebra of $\widetilde{\mathfrak{g}}$ which contains the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ must contain $\iota_{-}\left( \widetilde{\mathfrak{n}% }_{-}\right) $ as a subset. Since we know that $\widetilde{\mathfrak{g}% }\left[ <0\right] $ is a Lie subalgebra of $\widetilde{\mathfrak{g}}$ and contains the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$, this yields that $\widetilde{\mathfrak{g}}\left[ <0\right] $ must contain $\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) $ as a subset. In other words, $\iota _{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \subseteq \widetilde{\mathfrak{g}}\left[ <0\right] $. Similarly (by considering the elements $e_{1}$, $e_{2}$, $...$, $e_{n}$ instead of $f_{1}$, $f_{2}$, $...$, $f_{n}$), we can show $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \subseteq\widetilde{\mathfrak{g}}\left[ >0\right] $. \begin{vershort} A similar argument proves $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) \subseteq\widetilde{\mathfrak{g}}\left[ 0\right] $. \end{vershort} \begin{verlong} Finally, every $i\in\left\{ 1,2,...,n\right\} $ satisfies $h_{i}% \in\widetilde{\mathfrak{g}}\left[ 0\right] $ (since $\deg\left( h_{i}\right) =0$). In other words, the vector space $\widetilde{\mathfrak{g}% }\left[ 0\right] $ contains the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$. But now, recall that $\widetilde{\mathfrak{h}}$ is the free vector space with basis $h_{1},h_{2},...,h_{n}$. Thus, the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\widetilde{\mathfrak{h}}$ span the vector space $\widetilde{\mathfrak{h}}$. Consequently, the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\widetilde{\mathfrak{g}}$ span the vector space $\iota _{0}\left( \widetilde{\mathfrak{h}}\right) $ (because the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\widetilde{\mathfrak{g}}$ are the images of the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\widetilde{\mathfrak{h}}$ under the map $\iota_{0}$). Hence, every vector subspace of $\widetilde{\mathfrak{g}% }$ which contains the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$ must contain $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ as a subset. Since we know that $\widetilde{\mathfrak{g}}\left[ 0\right] $ is a vector subspace of $\widetilde{\mathfrak{g}}$ and contains the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$, this yields that $\widetilde{\mathfrak{g}}\left[ 0\right] $ must contain $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ as a subset. In other words, $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) \subseteq \widetilde{\mathfrak{g}}\left[ 0\right] $. \end{verlong} Since the internal direct sum $\widetilde{\mathfrak{g}}\left[ <0\right] \oplus\widetilde{\mathfrak{g}}\left[ >0\right] \oplus\widetilde{\mathfrak{g}% }\left[ 0\right] $ is well-defined, the internal direct sum $\iota _{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \oplus\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \oplus\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ must also be well-defined (because $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \subseteq \widetilde{\mathfrak{g}}\left[ >0\right] $, $\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \subseteq\widetilde{\mathfrak{g}}\left[ <0\right] $ and $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) \subseteq\widetilde{\mathfrak{g}}\left[ 0\right] $). We now must prove that this direct sum is $\widetilde{\mathfrak{g}}$. \bigskip \underline{\textit{2nd step: Identifications.}} Since the maps $\iota_{+}$, $\iota_{-}$ and $\iota_{0}$ are injective Lie algebra homomorphisms, and since their images are linearly disjoint (because the direct sum $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \oplus\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \oplus\iota _{0}\left( \widetilde{\mathfrak{h}}\right) $ is well-defined), we can regard these maps $\iota_{+}$, $\iota_{-}$ and $\iota_{0}$ as inclusions of Lie algebras. Let us do this from now on. Thus, $\widetilde{\mathfrak{n}}_{+}$, $\widetilde{\mathfrak{n}}_{-}$ and $\widetilde{\mathfrak{h}}$ are Lie subalgebras of $\widetilde{\mathfrak{g}}$. The identification of $\widetilde{\mathfrak{n}}_{-}$ with the Lie subalgebra $\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) $ of $\widetilde{\mathfrak{g}}$ eliminates the need of distinguishing between the elements $f_{i}$ of $\widetilde{\mathfrak{n}}_{-}$ and the elements $f_{i}$ of $\widetilde{\mathfrak{g}}$ (because for every $i\in\left\{ 1,2,...,n\right\} $, the element $f_{i}$ of $\widetilde{\mathfrak{g}}$ is the image of the element $f_{i}$ of $\widetilde{\mathfrak{n}}_{-}$ under the map $\iota_{-}$, and since we regard this map $\iota_{-}$ as inclusion, these two elements $f_{i}$ are therefore equal). Similarly, we don't have to distinguish between the elements $e_{i}$ of $\widetilde{\mathfrak{n}}_{+}$ and the elements $e_{i}$ of $\widetilde{\mathfrak{g}}$, nor is it necessary to distinguish between the elements $h_{i}$ of $\widetilde{\mathfrak{h}}$ and the elements $h_{i}$ of $\widetilde{\mathfrak{g}}$. Since we regard the maps $\iota_{+}$, $\iota_{-}$ and $\iota_{0}$ as inclusions, we have $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) =\widetilde{\mathfrak{n}}_{+}$, $\iota_{-}\left( \widetilde{\mathfrak{n}}% _{-}\right) =\widetilde{\mathfrak{n}}_{-}$ and $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =\widetilde{\mathfrak{h}}$. Hence, $\iota _{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \oplus\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \oplus\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =\widetilde{\mathfrak{n}}_{+}\oplus \widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}$. This shows that the internal direct sums $\widetilde{\mathfrak{n}}_{-}\oplus \widetilde{\mathfrak{h}}$ and $\widetilde{\mathfrak{n}}_{+}\oplus \widetilde{\mathfrak{h}}$ are well-defined (since they are partial sums of the direct sum $\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{n}}% _{-}\oplus\widetilde{\mathfrak{h}}$). \bigskip \underline{\textit{3rd step: Proving that }$\widetilde{\mathfrak{n}}_{-}% \oplus\widetilde{\mathfrak{h}}$\textit{ is a Lie subalgebra of }% $\widetilde{\mathfrak{g}}$\textit{.}} We now will prove part \textbf{(d)} of Theorem \ref{thm.gtilde} before we come back and finish the proof of part \textbf{(c)}. \begin{vershort} Indeed, let us first prove that $\left[ \widetilde{\mathfrak{h}% },\widetilde{\mathfrak{n}}_{-}\right] \subseteq\widetilde{\mathfrak{n}}_{-}$. In fact, in order to prove this, it is enough to show that $\left[ h_{i},\widetilde{\mathfrak{n}}_{-}\right] \subseteq\widetilde{\mathfrak{n}% }_{-}$ for every $i\in\left\{ 1,2,...,n\right\} $ (since the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\widetilde{\mathfrak{h}}$ span the vector space $\widetilde{\mathfrak{h}}$). So let $i\in\left\{ 1,2,...,n\right\} $. Let $\xi_{i}:\widetilde{\mathfrak{g}}\rightarrow\widetilde{\mathfrak{g}}$ be the map defined by% \[ \left( \xi_{i}\left( x\right) =\left[ h_{i},x\right] \ \ \ \ \ \ \ \ \ \ \text{for any }x\in\widetilde{\mathfrak{g}}\right) . \] Then, $\xi_{i}$ is a Lie derivation of the Lie algebra $\widetilde{\mathfrak{g}}$. On the other hand, the subset $\left\{ f_{1},f_{2},...,f_{n}\right\} $ of $\widetilde{\mathfrak{n}}_{-}$ generates $\widetilde{\mathfrak{n}}_{-}$ as a Lie algebra (since the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ of $\widetilde{\mathfrak{n}}_{-}$ generate $\widetilde{\mathfrak{n}}_{-}$ as a Lie algebra), and we can easily check that $\xi_{i}\left( \left\{ f_{1},f_{2},...,f_{n}\right\} \right) \subseteq\widetilde{\mathfrak{n}}_{-}$\ \ \ \ \footnote{\textit{Proof.} For every $j\in\left\{ 1,2,...,n\right\} $, we have% \begin{align*} \xi_{i}\left( f_{j}\right) & =\left[ h_{i},f_{j}\right] \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\xi_{i}\right) \\ & =-a_{i,j}\underbrace{f_{j}}_{\in\widetilde{\mathfrak{n}}_{-}}% \ \ \ \ \ \ \ \ \ \ \left( \text{by the relations (\ref{nonserre-relations}% )}\right) \\ & \in-a_{i,j}\widetilde{\mathfrak{n}}_{-}\subseteq\widetilde{\mathfrak{n}% }_{-}. \end{align*} Thus, $\xi_{i}\left( \left\{ f_{1},f_{2},...,f_{n}\right\} \right) \subseteq\widetilde{\mathfrak{n}}_{-}$, qed.}. Hence, Corollary \ref{cor.derivation.Lie.unique.ihg} (applied to $\widetilde{\mathfrak{g}}$, $\widetilde{\mathfrak{n}}_{-}$, $\widetilde{\mathfrak{n}}_{-}$, $\xi_{i}$ and $\left\{ f_{1},f_{2},...,f_{n}\right\} $ instead of $\mathfrak{g}$, $\mathfrak{h}$, $\mathfrak{i}$, $d$ and $S$) yields that $\xi_{i}\left( \widetilde{\mathfrak{n}}_{-}\right) \subseteq\widetilde{\mathfrak{n}}_{-}$. But by the definition of $\xi_{i}$, we have $\xi_{i}\left( \widetilde{\mathfrak{n}}_{-}\right) =\left[ h_{i},\widetilde{\mathfrak{n}% }_{-}\right] $. Hence, $\left[ h_{i},\widetilde{\mathfrak{n}}_{-}\right] =\xi_{i}\left( \widetilde{\mathfrak{n}}_{-}\right) \subseteq \widetilde{\mathfrak{n}}_{-}$. Now forget that we fixed $i$. We thus have proven that $\left[ h_{i},\widetilde{\mathfrak{n}}_{-}\right] \subseteq \widetilde{\mathfrak{n}}_{-}$ for every $i\in\left\{ 1,2,...,n\right\} $. As explained above, this yields $\left[ \widetilde{\mathfrak{h}}% ,\widetilde{\mathfrak{n}}_{-}\right] \subseteq\widetilde{\mathfrak{n}}_{-}$. \end{vershort} \begin{verlong} Indeed, we know that both $\widetilde{\mathfrak{n}}_{-}$ and $\widetilde{\mathfrak{h}}$ are Lie subalgebras of $\widetilde{\mathfrak{g}}$. Thus, $\left[ \widetilde{\mathfrak{n}}_{-},\widetilde{\mathfrak{n}}% _{-}\right] \subseteq\widetilde{\mathfrak{n}}_{-}$ and $\left[ \widetilde{\mathfrak{h}},\widetilde{\mathfrak{h}}\right] \subseteq \widetilde{\mathfrak{h}}$. We will now show that $\left[ \widetilde{\mathfrak{h}},\widetilde{\mathfrak{n}}_{-}\right] \subseteq \widetilde{\mathfrak{n}}_{-}$. \textit{Proof of }$\left[ \widetilde{\mathfrak{h}},\widetilde{\mathfrak{n}% }_{-}\right] \subseteq\widetilde{\mathfrak{n}}_{-}$\textit{:} Let $i\in\left\{ 1,2,...,n\right\} $. Let $\xi_{i}:\widetilde{\mathfrak{g}% }\rightarrow\widetilde{\mathfrak{g}}$ be the map defined by% \[ \left( \xi_{i}\left( x\right) =\left[ h_{i},x\right] \ \ \ \ \ \ \ \ \ \ \text{for any }x\in\widetilde{\mathfrak{g}}\right) . \] Then, $\xi_{i}$ is a Lie derivation of the Lie algebra $\widetilde{\mathfrak{g}}$. On the other hand, the subset $\left\{ f_{1},f_{2},...,f_{n}\right\} $ of $\widetilde{\mathfrak{n}}_{-}$ generates $\widetilde{\mathfrak{n}}_{-}$ as a Lie algebra (since the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ of $\widetilde{\mathfrak{n}}_{-}$ generate $\widetilde{\mathfrak{n}}_{-}$ as a Lie algebra), and we can easily check that $\xi_{i}\left( \left\{ f_{1},f_{2},...,f_{n}\right\} \right) \subseteq\widetilde{\mathfrak{n}}_{-}$\ \ \ \ \footnote{\textit{Proof.} For every $j\in\left\{ 1,2,...,n\right\} $, we have% \begin{align*} \xi_{i}\left( f_{j}\right) & =\left[ h_{i},f_{j}\right] \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\xi_{i}\right) \\ & =-a_{i,j}\underbrace{f_{j}}_{\in\widetilde{\mathfrak{n}}_{-}}% \ \ \ \ \ \ \ \ \ \ \left( \text{by the relations (\ref{nonserre-relations}% )}\right) \\ & \in-a_{i,j}\widetilde{\mathfrak{n}}_{-}\subseteq\widetilde{\mathfrak{n}% }_{-}. \end{align*} Thus, $\left\{ \xi_{i}\left( f_{1}\right) ,\xi_{i}\left( f_{2}\right) ,...,\xi_{i}\left( f_{n}\right) \right\} \subseteq\widetilde{\mathfrak{n}% }_{-}$. Since $\left\{ \xi_{i}\left( f_{1}\right) ,\xi_{i}\left( f_{2}\right) ,...,\xi_{i}\left( f_{n}\right) \right\} =\xi_{i}\left( \left\{ f_{1},f_{2},...,f_{n}\right\} \right) $, this rewrites as $\xi _{i}\left( \left\{ f_{1},f_{2},...,f_{n}\right\} \right) \subseteq \widetilde{\mathfrak{n}}_{-}$, qed.}. Hence, Corollary \ref{cor.derivation.Lie.unique.ihg} (applied to $\widetilde{\mathfrak{g}}$, $\widetilde{\mathfrak{n}}_{-}$, $\widetilde{\mathfrak{n}}_{-}$, $\xi_{i}$ and $\left\{ f_{1},f_{2},...,f_{n}\right\} $ instead of $\mathfrak{g}$, $\mathfrak{h}$, $\mathfrak{i}$, $d$ and $S$) yields that $\xi_{i}\left( \widetilde{\mathfrak{n}}_{-}\right) \subseteq\widetilde{\mathfrak{n}}_{-}$. But% \begin{align*} \xi_{i}\left( \widetilde{\mathfrak{n}}_{-}\right) & =\left\{ \underbrace{\xi_{i}\left( x\right) }_{=\left[ h_{i},x\right] }\mid x\in\widetilde{\mathfrak{n}}_{-}\right\} =\left\{ \left[ h_{i},x\right] \mid x\in\widetilde{\mathfrak{n}}_{-}\right\} =\left[ h_{i}% ,\widetilde{\mathfrak{n}}_{-}\right] =\left[ h_{i},\widetilde{\mathfrak{n}% }_{-}\right] \mathbb{C}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left[ h_{i}% ,\widetilde{\mathfrak{n}}_{-}\right] \text{ is a vector space}\right) \\ & =\left[ h_{i}\mathbb{C},\widetilde{\mathfrak{n}}_{-}\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since the Lie bracket is bilinear}\right) . \end{align*} Thus, $\left[ h_{i}\mathbb{C},\widetilde{\mathfrak{n}}_{-}\right] \subseteq\widetilde{\mathfrak{n}}_{-}$. Now, forget that we fixed $i$. We thus have shown that $\left[ h_{i}\mathbb{C},\widetilde{\mathfrak{n}}_{-}\right] \subseteq\widetilde{\mathfrak{n}}_{-}$ for every $i\in\left\{ 1,2,...,n\right\} $. But the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\widetilde{\mathfrak{h}% }$ span the vector space $\widetilde{\mathfrak{h}}$. Thus, $\widetilde{\mathfrak{h}}=\sum\limits_{i=1}^{n}\left( h_{i}\mathbb{C}\right) $, so that% \begin{align*} \left[ \widetilde{\mathfrak{h}},\widetilde{\mathfrak{n}}_{-}\right] & =\left[ \sum\limits_{i=1}^{n}\left( h_{i}\mathbb{C}\right) ,\ \widetilde{\mathfrak{n}}_{-}\right] =\sum\limits_{i=1}^{n}% \underbrace{\left[ h_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] }_{\subseteq\widetilde{\mathfrak{n}}_{-}}\ \ \ \ \ \ \ \ \ \ \left( \text{since the Lie bracket is bilinear}\right) \\ & \subseteq\sum\limits_{i=1}^{n}\widetilde{\mathfrak{n}}_{-}\subseteq \widetilde{\mathfrak{n}}_{-}. \end{align*} This proves $\left[ \widetilde{\mathfrak{h}},\widetilde{\mathfrak{n}}% _{-}\right] \subseteq\widetilde{\mathfrak{n}}_{-}$. \end{verlong} \begin{vershort} Now, $\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}% =\widetilde{\mathfrak{n}}_{-}+\widetilde{\mathfrak{h}}$, so that% \begin{align*} \left[ \widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}% },\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}\right] & =\left[ \widetilde{\mathfrak{n}}_{-}+\widetilde{\mathfrak{h}}% ,\widetilde{\mathfrak{n}}_{-}+\widetilde{\mathfrak{h}}\right] \\ & =\underbrace{\left[ \widetilde{\mathfrak{n}}_{-},\widetilde{\mathfrak{n}% }_{-}\right] }_{\substack{\subseteq\widetilde{\mathfrak{n}}_{-}\\\text{(since }\widetilde{\mathfrak{n}}_{-}\text{ is a Lie algebra)}}}+\underbrace{\left[ \widetilde{\mathfrak{n}}_{-},\widetilde{\mathfrak{h}}\right] }_{=-\left[ \widetilde{\mathfrak{h}},\widetilde{\mathfrak{n}}_{-}\right] \subseteq\left[ \widetilde{\mathfrak{h}},\widetilde{\mathfrak{n}}_{-}\right] \subseteq \widetilde{\mathfrak{n}}_{-}}+\underbrace{\left[ \widetilde{\mathfrak{h}% },\widetilde{\mathfrak{n}}_{-}\right] }_{\subseteq\widetilde{\mathfrak{n}% }_{-}}+\underbrace{\left[ \widetilde{\mathfrak{h}},\widetilde{\mathfrak{h}% }\right] }_{\substack{\subseteq\widetilde{\mathfrak{h}}\\\text{(since }\widetilde{\mathfrak{h}}\text{ is a Lie algebra)}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since the Lie bracket is bilinear}\right) \\ & \subseteq\underbrace{\widetilde{\mathfrak{n}}_{-}+\widetilde{\mathfrak{n}% }_{-}+\widetilde{\mathfrak{n}}_{-}}_{\subseteq\widetilde{\mathfrak{n}}_{-}% }+\widetilde{\mathfrak{h}}\subseteq\widetilde{\mathfrak{n}}_{-}% +\widetilde{\mathfrak{h}}=\widetilde{\mathfrak{n}}_{-}\oplus \widetilde{\mathfrak{h}}. \end{align*} Thus, $\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}$ is a Lie subalgebra of $\widetilde{\mathfrak{g}}$. \end{vershort} \begin{verlong} Now, $\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}% =\widetilde{\mathfrak{n}}_{-}+\widetilde{\mathfrak{h}}$ (since direct sums are sums), so that \begin{align*} \left[ \widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}% },\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}\right] & =\left[ \widetilde{\mathfrak{n}}_{-}+\widetilde{\mathfrak{h}}% ,\widetilde{\mathfrak{n}}_{-}+\widetilde{\mathfrak{h}}\right] =\underbrace{\left[ \widetilde{\mathfrak{n}}_{-},\widetilde{\mathfrak{n}}% _{-}\right] }_{\subseteq\widetilde{\mathfrak{n}}_{-}}+\underbrace{\left[ \widetilde{\mathfrak{n}}_{-},\widetilde{\mathfrak{h}}\right] }_{=-\left[ \widetilde{\mathfrak{h}},\widetilde{\mathfrak{n}}_{-}\right] \subseteq\left[ \widetilde{\mathfrak{h}},\widetilde{\mathfrak{n}}_{-}\right] \subseteq \widetilde{\mathfrak{n}}_{-}}+\underbrace{\left[ \widetilde{\mathfrak{h}% },\widetilde{\mathfrak{n}}_{-}\right] }_{\subseteq\widetilde{\mathfrak{n}% }_{-}}+\underbrace{\left[ \widetilde{\mathfrak{h}},\widetilde{\mathfrak{h}% }\right] }_{\subseteq\widetilde{\mathfrak{h}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since the Lie bracket is bilinear}\right) \\ & \subseteq\underbrace{\widetilde{\mathfrak{n}}_{-}+\widetilde{\mathfrak{n}% }_{-}+\widetilde{\mathfrak{n}}_{-}}_{\subseteq\widetilde{\mathfrak{n}}_{-}% }+\widetilde{\mathfrak{h}}\subseteq\widetilde{\mathfrak{n}}_{-}% +\widetilde{\mathfrak{h}}=\widetilde{\mathfrak{n}}_{-}\oplus \widetilde{\mathfrak{h}}. \end{align*} Thus, $\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}$ is a Lie subalgebra of $\widetilde{\mathfrak{g}}$. \end{verlong} (Note that the map $\left( \iota_{-},\iota_{0}\right) :\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}\rightarrow \widetilde{\mathfrak{g}}$ is actually a Lie algebra isomorphism from the semidirect product $\widetilde{\mathfrak{h}}\ltimes\widetilde{\mathfrak{n}% }_{-}$ (which was constructed during our proof of Theorem \ref{thm.gtilde} \textbf{(b)}) to $\widetilde{\mathfrak{g}}$. But we will not need this fact, so we will not prove it either.) So we have shown that $\widetilde{\mathfrak{n}}_{-}\oplus \widetilde{\mathfrak{h}}$ is a Lie subalgebra of $\widetilde{\mathfrak{g}}$. A similar argument (but with $\widetilde{\mathfrak{n}}_{-}$ replaced by $\widetilde{\mathfrak{n}}_{+}$, and with $f_{j}$ replaced by $e_{j}$, and with $-a_{i,j}$ replaced by $a_{i,j}$) shows that $\widetilde{\mathfrak{n}}% _{+}\oplus\widetilde{\mathfrak{h}}$ is a Lie subalgebra of $\widetilde{\mathfrak{g}}$. We now know that $\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}$ and $\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}}$ are Lie subalgebras of $\widetilde{\mathfrak{g}}$. Since $\widetilde{\mathfrak{n}}% _{-}=\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) $, $\widetilde{\mathfrak{n}}_{+}=\iota_{+}\left( \widetilde{\mathfrak{n}}% _{+}\right) $ and $\widetilde{\mathfrak{h}}=\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $, this rewrites as follows: $\iota _{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \oplus\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ and $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \oplus\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ are Lie subalgebras of $\widetilde{\mathfrak{g}}$. This proves Theorem \ref{thm.gtilde} \textbf{(d)}. \bigskip \underline{\textit{4th step: Finishing the proof of Theorem \ref{thm.gtilde} \textbf{(c)}.}} We know that the internal direct sum $\widetilde{\mathfrak{n}}_{+}% \oplus\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}$ makes sense. Denote this direct sum $\widetilde{\mathfrak{n}}_{+}\oplus \widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}$ as $V$. We know that $V$ is a vector subspace of $\widetilde{\mathfrak{g}}$. We need to prove that $V=\widetilde{\mathfrak{g}}$. \begin{vershort} Let $N$ be the vector subspace of $\widetilde{\mathfrak{g}}$ spanned by the $3n$ elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$. Then, $\widetilde{\mathfrak{g}}$ is generated by $N$ as a Lie algebra (because the elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ generate $\widetilde{\mathfrak{g}}$ as a Lie algebra). \end{vershort} \begin{verlong} Let $N$ be the vector subspace of $\widetilde{\mathfrak{g}}$ spanned by the $3n$ elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$. Then, $\widetilde{\mathfrak{g}}$ is generated by $N$ as a Lie algebra (because the elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ generate $\widetilde{\mathfrak{g}}$ as a Lie algebra\footnote{This is because $\widetilde{\mathfrak{g}}=\operatorname*{FreeLie}\left( h_{i}% ,f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) $.}). \end{verlong} We will now prove that $\left[ N,V\right] \subseteq V$. Indeed, since $N=\sum\limits_{i=1}^{n}\left( e_{i}\mathbb{C}\right) +\sum\limits_{i=1}^{n}\left( f_{i}\mathbb{C}\right) +\sum\limits_{i=1}% ^{n}\left( h_{i}\mathbb{C}\right) $ (because $N$ is the vector subspace of $\widetilde{\mathfrak{g}}$ spanned by the $3n$ elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$) and $V=\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{n}}% _{-}\oplus\widetilde{\mathfrak{h}}=\widetilde{\mathfrak{n}}_{+}% +\widetilde{\mathfrak{n}}_{-}+\widetilde{\mathfrak{h}}$ (since direct sums are sums), we have% \begin{align} \left[ N,V\right] & =\left[ \sum\limits_{i=1}^{n}\left( e_{i}% \mathbb{C}\right) +\sum\limits_{i=1}^{n}\left( f_{i}\mathbb{C}\right) +\sum\limits_{i=1}^{n}\left( h_{i}\mathbb{C}\right) ,\ \widetilde{\mathfrak{n}}_{+}+\widetilde{\mathfrak{n}}_{-}% +\widetilde{\mathfrak{h}}\right] \nonumber\\ & \subseteq\sum\limits_{i=1}^{n}\left[ e_{i}\mathbb{C}% ,\ \widetilde{\mathfrak{n}}_{+}\right] +\sum\limits_{i=1}^{n}\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] +\sum\limits_{i=1}% ^{n}\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{h}}\right] \nonumber\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{i=1}^{n}\left[ f_{i}\mathbb{C}% ,\ \widetilde{\mathfrak{n}}_{+}\right] +\sum\limits_{i=1}^{n}\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] +\sum\limits_{i=1}% ^{n}\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{h}}\right] \nonumber\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{i=1}^{n}\left[ h_{i}\mathbb{C}% ,\ \widetilde{\mathfrak{n}}_{+}\right] +\sum\limits_{i=1}^{n}\left[ h_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] +\sum\limits_{i=1}% ^{n}\left[ h_{i}\mathbb{C},\ \widetilde{\mathfrak{h}}\right] \label{pf.gtilde.c.1}% \end{align} (since the Lie bracket is bilinear). We will now prove that each summand of each of the nine sums on the right hand side of (\ref{pf.gtilde.c.1}) is $\subseteq V$. \textit{Proof that every }$i\in\left\{ 1,2,...,n\right\} $\textit{ satisfies }$\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{+}\right] \subseteq V$\textit{:} For every $i\in\left\{ 1,2,...,n\right\} $, we have $e_{i}\in \widetilde{\mathfrak{n}}_{+}$ and thus $e_{i}\mathbb{C}\subseteq \widetilde{\mathfrak{n}}_{+}$, so that% \begin{align*} \left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{+}\right] & \subseteq\left[ \widetilde{\mathfrak{n}}_{+},\ \widetilde{\mathfrak{n}}% _{+}\right] \subseteq\widetilde{\mathfrak{n}}_{+}\ \ \ \ \ \ \ \ \ \ \left( \text{since }\widetilde{\mathfrak{n}}_{+}\text{ is a Lie algebra}\right) \\ & \subseteq\widetilde{\mathfrak{n}}_{+}+\widetilde{\mathfrak{n}}% _{-}+\widetilde{\mathfrak{h}}=V. \end{align*} We have thus proven that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{+}\right] \subseteq V$. \textit{Proof that every }$i\in\left\{ 1,2,...,n\right\} $\textit{ satisfies }$\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] \subseteq V$\textit{:} \begin{vershort} Let $i\in\left\{ 1,2,...,n\right\} $. Define a map $\psi_{i}% :\widetilde{\mathfrak{g}}\rightarrow\widetilde{\mathfrak{g}}$ by% \[ \left( \psi_{i}\left( x\right) =\left[ e_{i},x\right] \ \ \ \ \ \ \ \ \ \ \text{for every }x\in\widetilde{\mathfrak{g}}\right) . \] Then, $\psi_{i}$ is a Lie derivation of the Lie algebra $\widetilde{\mathfrak{g}}$. On the other hand, the subset $\left\{ f_{1},f_{2},...,f_{n}\right\} $ of $\widetilde{\mathfrak{n}}_{-}$ generates $\widetilde{\mathfrak{n}}_{-}$ as a Lie algebra (since the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ of $\widetilde{\mathfrak{n}}_{-}$ generate $\widetilde{\mathfrak{n}}_{-}$ as a Lie algebra), and we can easily check that $\psi_{i}\left( \left\{ f_{1},f_{2},...,f_{n}\right\} \right) \subseteq\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}% $\ \ \ \ \footnote{\textit{Proof.} For every $j\in\left\{ 1,2,...,n\right\} $, we have% \begin{align*} \psi_{i}\left( f_{j}\right) & =\left[ e_{i},f_{j}\right] \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\psi_{i}\right) \\ & =\delta_{i,j}\underbrace{h_{i}}_{\in\widetilde{\mathfrak{h}}}% \ \ \ \ \ \ \ \ \ \ \left( \text{by the relations (\ref{nonserre-relations}% )}\right) \\ & \in\widetilde{\mathfrak{h}}\subseteq\widetilde{\mathfrak{n}}_{-}% \oplus\widetilde{\mathfrak{h}}. \end{align*} Thus, $\psi_{i}\left( \left\{ f_{1},f_{2},...,f_{n}\right\} \right) \subseteq\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}$, qed.}. Hence, Corollary \ref{cor.derivation.Lie.unique.ihg} (applied to $\widetilde{\mathfrak{g}}$, $\widetilde{\mathfrak{n}}_{-}\oplus \widetilde{\mathfrak{h}}$, $\widetilde{\mathfrak{n}}_{-}$, $\psi_{i}$ and $\left\{ f_{1},f_{2},...,f_{n}\right\} $ instead of $\mathfrak{g}$, $\mathfrak{h}$, $\mathfrak{i}$, $d$ and $S$) yields that $\psi_{i}\left( \widetilde{\mathfrak{n}}_{-}\right) \subseteq\widetilde{\mathfrak{n}}% _{-}\oplus\widetilde{\mathfrak{h}}$ (since $\widetilde{\mathfrak{n}}_{-}% \oplus\widetilde{\mathfrak{h}}$ is a Lie subalgebra of $\widetilde{\mathfrak{g}}$). But by the definition of $\psi_{i}$, we have% \begin{align*} \psi_{i}\left( \widetilde{\mathfrak{n}}_{-}\right) & =\left[ e_{i},\widetilde{\mathfrak{n}}_{-}\right] =\left[ e_{i}% ,\widetilde{\mathfrak{n}}_{-}\right] \mathbb{C}\ \ \ \ \ \ \ \ \ \ \left( \text{since }\left[ e_{i},\widetilde{\mathfrak{n}}_{-}\right] \text{ is a vector space}\right) \\ & =\left[ e_{i}\mathbb{C},\widetilde{\mathfrak{n}}_{-}\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since the Lie bracket is bilinear}\right) . \end{align*} Thus, $\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] \subseteq\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}% \subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{n}}_{-}% \oplus\widetilde{\mathfrak{h}}=V$. Now, forget that we fixed $i$. We thus have shown that $\left[ e_{i}\mathbb{C},\widetilde{\mathfrak{n}}_{-}\right] \subseteq V$ for every $i\in\left\{ 1,2,...,n\right\} $. \end{vershort} \begin{verlong} Let $i\in\left\{ 1,2,...,n\right\} $. Define a map $\psi_{i}% :\widetilde{\mathfrak{g}}\rightarrow\widetilde{\mathfrak{g}}$ by% \[ \left( \psi_{i}\left( x\right) =\left[ e_{i},x\right] \ \ \ \ \ \ \ \ \ \ \text{for every }x\in\widetilde{\mathfrak{g}}\right) . \] Then, $\psi_{i}$ is a Lie derivation of the Lie algebra $\widetilde{\mathfrak{g}}$. On the other hand, the subset $\left\{ f_{1},f_{2},...,f_{n}\right\} $ of $\widetilde{\mathfrak{n}}_{-}$ generates $\widetilde{\mathfrak{n}}_{-}$ as a Lie algebra (since the elements $f_{1}$, $f_{2}$, $...$, $f_{n}$ of $\widetilde{\mathfrak{n}}_{-}$ generate $\widetilde{\mathfrak{n}}_{-}$ as a Lie algebra), and we can easily check that $\psi_{i}\left( \left\{ f_{1},f_{2},...,f_{n}\right\} \right) \subseteq\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}% $\ \ \ \ \footnote{\textit{Proof.} For every $j\in\left\{ 1,2,...,n\right\} $, we have% \begin{align*} \psi_{i}\left( f_{j}\right) & =\left[ e_{i},f_{j}\right] \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\psi_{i}\right) \\ & =\delta_{i,j}\underbrace{h_{i}}_{\in\widetilde{\mathfrak{h}}}% \ \ \ \ \ \ \ \ \ \ \left( \text{by the relations (\ref{nonserre-relations}% )}\right) \\ & \in\delta_{i,j}\widetilde{\mathfrak{h}}\subseteq\widetilde{\mathfrak{h}% }\ \ \ \ \ \ \ \ \ \ \left( \text{since }\widetilde{\mathfrak{h}}\text{ is a vector space}\right) \\ & \subseteq\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}. \end{align*} Thus, $\left\{ \psi_{i}\left( f_{1}\right) ,\psi_{i}\left( f_{2}\right) ,...,\psi_{i}\left( f_{n}\right) \right\} \subseteq\widetilde{\mathfrak{n}% }_{-}\oplus\widetilde{\mathfrak{h}}$. Since $\left\{ \psi_{i}\left( f_{1}\right) ,\psi_{i}\left( f_{2}\right) ,...,\psi_{i}\left( f_{n}\right) \right\} =\psi_{i}\left( \left\{ f_{1},f_{2},...,f_{n}% \right\} \right) $, this rewrites as $\psi_{i}\left( \left\{ f_{1}% ,f_{2},...,f_{n}\right\} \right) \subseteq\widetilde{\mathfrak{n}}_{-}% \oplus\widetilde{\mathfrak{h}}$, qed.}. Hence, Corollary \ref{cor.derivation.Lie.unique.ihg} (applied to $\widetilde{\mathfrak{g}}$, $\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}$, $\widetilde{\mathfrak{n}}_{-}$, $\psi_{i}$ and $\left\{ f_{1},f_{2}% ,...,f_{n}\right\} $ instead of $\mathfrak{g}$, $\mathfrak{h}$, $\mathfrak{i}$, $d$ and $S$) yields that $\psi_{i}\left( \widetilde{\mathfrak{n}}_{-}\right) \subseteq\widetilde{\mathfrak{n}}% _{-}\oplus\widetilde{\mathfrak{h}}$ (since $\widetilde{\mathfrak{n}}_{-}% \oplus\widetilde{\mathfrak{h}}$ is a Lie subalgebra of $\widetilde{\mathfrak{g}}$). But% \begin{align*} \psi_{i}\left( \widetilde{\mathfrak{n}}_{-}\right) & =\left\{ \underbrace{\psi_{i}\left( x\right) }_{=\left[ e_{i},x\right] }\mid x\in\widetilde{\mathfrak{n}}_{-}\right\} =\left\{ \left[ e_{i},x\right] \mid x\in\widetilde{\mathfrak{n}}_{-}\right\} =\left[ e_{i}% ,\widetilde{\mathfrak{n}}_{-}\right] =\left[ e_{i},\widetilde{\mathfrak{n}% }_{-}\right] \mathbb{C}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left[ e_{i}% ,\widetilde{\mathfrak{n}}_{-}\right] \text{ is a vector space}\right) \\ & =\left[ e_{i}\mathbb{C},\widetilde{\mathfrak{n}}_{-}\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since the Lie bracket is bilinear}\right) . \end{align*} Thus, $\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] \subseteq\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}% \subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{n}}_{-}% \oplus\widetilde{\mathfrak{h}}=V$. Now, forget that we fixed $i$. We thus have shown that $\left[ e_{i}\mathbb{C},\widetilde{\mathfrak{n}}_{-}\right] \subseteq V$ for every $i\in\left\{ 1,2,...,n\right\} $. \end{verlong} \textit{Proof that every }$i\in\left\{ 1,2,...,n\right\} $\textit{ satisfies }$\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{h}}\right] \subseteq V$\textit{:} Every $i\in\left\{ 1,2,...,n\right\} $ satisfies $e_{i}\mathbb{C}% \subseteq\widetilde{\mathfrak{n}}_{+}$ (since $e_{i}\in\widetilde{\mathfrak{n}% }_{+}$). Thus, every $i\in\left\{ 1,2,...,n\right\} $ satisfies% \begin{align*} \left[ \underbrace{e_{i}\mathbb{C}}_{\subseteq\widetilde{\mathfrak{n}}% _{+}\subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}}% },\ \underbrace{\widetilde{\mathfrak{h}}}_{\subseteq\widetilde{\mathfrak{n}% }_{+}\oplus\widetilde{\mathfrak{h}}}\right] & \subseteq\left[ \widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}}% ,\ \widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}}\right] \\ & \subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}% }\ \ \ \ \ \ \ \ \ \ \left( \text{since }\widetilde{\mathfrak{n}}_{+}% \oplus\widetilde{\mathfrak{h}}\text{ is a Lie subalgebra of }% \widetilde{\mathfrak{g}}\right) \\ & \subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{n}}% _{-}\oplus\widetilde{\mathfrak{h}}=V. \end{align*} \textit{Proof that every }$i\in\left\{ 1,2,...,n\right\} $\textit{ satisfies }$\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{+}\right] \subseteq V$\textit{:} \begin{vershort} We have proven above that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] \subseteq V$. An analogous argument (or an invocation of the automorphism guaranteed by Theorem \ref{thm.gtilde} \textbf{(f)}) shows that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\left[ f_{i}\mathbb{C}% ,\ \widetilde{\mathfrak{n}}_{+}\right] \subseteq V$. \end{vershort} \begin{verlong} We have proven above that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] \subseteq V$. A similar argument (but with $\widetilde{\mathfrak{n}}_{-}$ replaced by $\widetilde{\mathfrak{n}}_{+}$, and with $e_{i}$ replaced by $f_{i}$, and with $f_{j}$ replaced by $e_{j}$, and with $h_{i}$ replaced by $-h_{i}$, and with (\ref{nonserre-relations}) replaced by (\ref{nonserre-relations2})) shows that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\left[ f_{i}% \mathbb{C},\ \widetilde{\mathfrak{n}}_{+}\right] \subseteq V$. \end{verlong} \textit{Proof that every }$i\in\left\{ 1,2,...,n\right\} $\textit{ satisfies }$\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] \subseteq V$\textit{:} We have proven above that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{+}\right] \subseteq V$. A similar argument (but with $\widetilde{\mathfrak{n}}_{+}$ replaced by $\widetilde{\mathfrak{n}}_{-}$, and with $e_{i}$ replaced by $f_{i}$) shows that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] \subseteq V$. \textit{Proof that every }$i\in\left\{ 1,2,...,n\right\} $\textit{ satisfies }$\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{h}}\right] \subseteq V$\textit{:} We have proven above that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{h}}\right] \subseteq V$. A similar argument (but with $\widetilde{\mathfrak{n}}_{+}$ replaced by $\widetilde{\mathfrak{n}}_{-}$, and with $e_{i}$ replaced by $f_{i}$) shows that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{h}}\right] \subseteq V$. \textit{Proof that every }$i\in\left\{ 1,2,...,n\right\} $\textit{ satisfies }$\left[ h_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{+}\right] \subseteq V$\textit{:} Every $i\in\left\{ 1,2,...,n\right\} $ satisfies $h_{i}\mathbb{C}% \subseteq\widetilde{\mathfrak{h}}$ (since $h_{i}\in\widetilde{\mathfrak{h}}$). Thus, every $i\in\left\{ 1,2,...,n\right\} $ satisfies% \begin{align*} \left[ \underbrace{h_{i}\mathbb{C}}_{\subseteq\widetilde{\mathfrak{n}}% _{+}\subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}}% },\ \underbrace{\widetilde{\mathfrak{n}}_{+}}_{\subseteq \widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}}}\right] & \subseteq\left[ \widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}% },\ \widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}}\right] \\ & \subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}% }\ \ \ \ \ \ \ \ \ \ \left( \text{since }\widetilde{\mathfrak{n}}_{+}% \oplus\widetilde{\mathfrak{h}}\text{ is a Lie subalgebra of }% \widetilde{\mathfrak{g}}\right) \\ & \subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{n}}% _{-}\oplus\widetilde{\mathfrak{h}}=V. \end{align*} \textit{Proof that every }$i\in\left\{ 1,2,...,n\right\} $\textit{ satisfies }$\left[ h_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{-}\right] \subseteq V$\textit{:} We have proven above that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\left[ h_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{+}\right] \subseteq V$. A similar argument (but with $\widetilde{\mathfrak{n}}_{+}$ replaced by $\widetilde{\mathfrak{n}}_{-}$) shows that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $\left[ h_{i}\mathbb{C}% ,\ \widetilde{\mathfrak{n}}_{-}\right] \subseteq V$. \textit{Proof that every }$i\in\left\{ 1,2,...,n\right\} $\textit{ satisfies }$\left[ h_{i}\mathbb{C},\ \widetilde{\mathfrak{h}}\right] \subseteq V$\textit{:} Every $i\in\left\{ 1,2,...,n\right\} $ satisfies $h_{i}\mathbb{C}% \subseteq\widetilde{\mathfrak{h}}$ (since $h_{i}\in\widetilde{\mathfrak{h}}$). Thus, every $i\in\left\{ 1,2,...,n\right\} $ satisfies% \begin{align*} \left[ \underbrace{h_{i}\mathbb{C}}_{\subseteq\widetilde{\mathfrak{n}}% _{+}\subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}}% },\ \underbrace{\widetilde{\mathfrak{h}}}_{\subseteq\widetilde{\mathfrak{n}% }_{+}\oplus\widetilde{\mathfrak{h}}}\right] & \subseteq\left[ \widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}}% ,\ \widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}}\right] \\ & \subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{h}% }\ \ \ \ \ \ \ \ \ \ \left( \text{since }\widetilde{\mathfrak{n}}_{+}% \oplus\widetilde{\mathfrak{h}}\text{ is a Lie subalgebra of }% \widetilde{\mathfrak{g}}\right) \\ & \subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{n}}% _{-}\oplus\widetilde{\mathfrak{h}}=V. \end{align*} Thus, we have proven that every $i\in\left\{ 1,2,...,n\right\} $ satisfies the nine relations $\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}% _{+}\right] \subseteq V$, $\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}% }_{-}\right] \subseteq V$, $\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{h}% }\right] \subseteq V$, $\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{n}% }_{+}\right] \subseteq V$, $\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{n}% }_{-}\right] \subseteq V$, $\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{h}% }\right] \subseteq V$, $\left[ h_{i}\mathbb{C},\ \widetilde{\mathfrak{n}% }_{+}\right] \subseteq V$, $\left[ h_{i}\mathbb{C},\ \widetilde{\mathfrak{n}% }_{-}\right] \subseteq V$, and $\left[ h_{i}\mathbb{C}% ,\ \widetilde{\mathfrak{h}}\right] \subseteq V$. Thus, (\ref{pf.gtilde.c.1}) becomes% \begin{align*} \left[ N,V\right] & \subseteq\sum\limits_{i=1}^{n}\underbrace{\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{+}\right] }_{\subseteq V}% +\sum\limits_{i=1}^{n}\underbrace{\left[ e_{i}\mathbb{C}% ,\ \widetilde{\mathfrak{n}}_{-}\right] }_{\subseteq V}+\sum\limits_{i=1}% ^{n}\underbrace{\left[ e_{i}\mathbb{C},\ \widetilde{\mathfrak{h}}\right] }_{\subseteq V}\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{i=1}^{n}\underbrace{\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{+}\right] }_{\subseteq V}% +\sum\limits_{i=1}^{n}\underbrace{\left[ f_{i}\mathbb{C}% ,\ \widetilde{\mathfrak{n}}_{-}\right] }_{\subseteq V}+\sum\limits_{i=1}% ^{n}\underbrace{\left[ f_{i}\mathbb{C},\ \widetilde{\mathfrak{h}}\right] }_{\subseteq V}\\ & \ \ \ \ \ \ \ \ \ \ +\sum\limits_{i=1}^{n}\underbrace{\left[ h_{i}\mathbb{C},\ \widetilde{\mathfrak{n}}_{+}\right] }_{\subseteq V}% +\sum\limits_{i=1}^{n}\underbrace{\left[ h_{i}\mathbb{C}% ,\ \widetilde{\mathfrak{n}}_{-}\right] }_{\subseteq V}+\sum\limits_{i=1}% ^{n}\underbrace{\left[ h_{i}\mathbb{C},\ \widetilde{\mathfrak{h}}\right] }_{\subseteq V}\\ & \subseteq\sum\limits_{i=1}^{n}V+\sum\limits_{i=1}^{n}V+\sum\limits_{i=1}% ^{n}V+\sum\limits_{i=1}^{n}V+\sum\limits_{i=1}^{n}V+\sum\limits_{i=1}% ^{n}V+\sum\limits_{i=1}^{n}V+\sum\limits_{i=1}^{n}V+\sum\limits_{i=1}^{n}V\\ & \subseteq V \end{align*} (since $V$ is a vector space). This proves $\left[ N,V\right] \subseteq V$. Moreover, \begin{align*} N & =\sum\limits_{i=1}^{n}\underbrace{\left( e_{i}\mathbb{C}\right) }_{\substack{\subseteq V\\\text{(since }e_{i}\in\widetilde{\mathfrak{n}}% _{+}\subseteq\widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{n}}% _{-}\oplus\widetilde{\mathfrak{h}}=V\text{)}}}+\sum\limits_{i=1}% ^{n}\underbrace{\left( f_{i}\mathbb{C}\right) }_{\substack{\subseteq V\\\text{(since }f_{i}\in\widetilde{\mathfrak{n}}_{-}\subseteq \widetilde{\mathfrak{n}}_{+}\oplus\widetilde{\mathfrak{n}}_{-}\oplus \widetilde{\mathfrak{h}}=V\text{)}}}+\sum\limits_{i=1}^{n}\underbrace{\left( h_{i}\mathbb{C}\right) }_{\substack{\subseteq V\\\text{(since }h_{i}% \in\widetilde{\mathfrak{h}}\subseteq\widetilde{\mathfrak{n}}_{+}% \oplus\widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}=V\text{)}}}\\ & \subseteq\sum\limits_{i=1}^{n}V+\sum\limits_{i=1}^{n}V+\sum\limits_{i=1}% ^{n}V\subseteq V \end{align*} (since $V$ is a vector space). So we know that $N$ and $V$ are vector subspaces of $\widetilde{\mathfrak{g}}$ such that $\widetilde{\mathfrak{g}}$ is generated by $N$ as a Lie algebra and such that $N\subseteq V$ and $\left[ N,V\right] \subseteq V$. Hence, Lemma \ref{lem.generation.1} (applied to $\widetilde{\mathfrak{g}}$, $N$ and $V$ instead of $\mathfrak{g}$, $T$ and $U$) yields $V=\widetilde{\mathfrak{g}}$. Thus, $\widetilde{\mathfrak{g}}=V=\widetilde{\mathfrak{n}}_{+}\oplus \widetilde{\mathfrak{n}}_{-}\oplus\widetilde{\mathfrak{h}}=\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \oplus\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \oplus\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ (since $\widetilde{\mathfrak{n}}_{-}% =\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) $, $\widetilde{\mathfrak{n}}_{+}=\iota_{+}\left( \widetilde{\mathfrak{n}}% _{+}\right) $ and $\widetilde{\mathfrak{h}}=\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $). This proves Theorem \ref{thm.gtilde} \textbf{(c)}. \bigskip \textbf{(d)} During the proof of Theorem \ref{thm.gtilde} \textbf{(c)}, we have already proven Theorem \ref{thm.gtilde} \textbf{(d)}. \bigskip \textbf{(e)} We will use the notations we introduced in our proof of Theorem \ref{thm.gtilde} \textbf{(d)}. During this proof, we have shown that $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \subseteq \widetilde{\mathfrak{g}}\left[ >0\right] $, $\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \subseteq\widetilde{\mathfrak{g}}\left[ <0\right] $ and $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) \subseteq\widetilde{\mathfrak{g}}\left[ 0\right] $. Also, we know that $\widetilde{\mathfrak{g}}=\iota_{+}\left( \widetilde{\mathfrak{n}}% _{+}\right) \oplus\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \oplus\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. Finally, we know that the internal direct sum $\widetilde{\mathfrak{g}}\left[ >0\right] \oplus\widetilde{\mathfrak{g}}\left[ <0\right] \oplus\widetilde{\mathfrak{g}% }\left[ 0\right] $ is well-defined. \begin{vershort} Now, a simple fact from linear algebra says the following: If $U_{1}$, $U_{2}% $, $U_{3}$, $V_{1}$, $V_{2}$, $V_{3}$ are six vector subspaces of a vector space $V$ satisfying the four relations $U_{1}\subseteq V_{1}$, $U_{2}% \subseteq V_{2}$, $U_{3}\subseteq V_{3}$ and $V=U_{1}\oplus U_{2}\oplus U_{3}% $, and if the internal direct sum $V_{1}\oplus V_{2}\oplus V_{3}$ is well-defined, then we must have $U_{1}=V_{1}$, $U_{2}=V_{2}$ and $U_{3}=V_{3}$. \end{vershort} \begin{verlong} Now, a simple fact from linear algebra says the following: If $U_{1}$, $U_{2}% $, $U_{3}$, $V_{1}$, $V_{2}$, $V_{3}$ are six vector subspaces of a vector space $V$ satisfying the four relations $U_{1}\subseteq V_{1}$, $U_{2}% \subseteq V_{2}$, $U_{3}\subseteq V_{3}$ and $V=U_{1}\oplus U_{2}\oplus U_{3}% $, and if the internal direct sum $V_{1}\oplus V_{2}\oplus V_{3}$ is well-defined, then we must have $U_{1}=V_{1}$, $U_{2}=V_{2}$ and $U_{3}=V_{3}% $\ \ \ \ \footnote{\textit{Proof.} Let $U_{1}$, $U_{2}$, $U_{3}$, $V_{1}$, $V_{2}$, $V_{3}$ be six vector subspaces of a vector space $V$ satisfying the four relations $U_{1}\subseteq V_{1}$, $U_{2}\subseteq V_{2}$, $U_{3}\subseteq V_{3}$ and $V=V_{1}\oplus V_{2}\oplus V_{3}$. Assume that the internal direct sum $V_{1}\oplus V_{2}\oplus V_{3}$ is well-defined. \par Let $x\in V_{1}$. Then, $x\in V_{1}\subseteq V=U_{1}\oplus U_{2}\oplus U_{3}$. Hence, there exist $x_{1}\in U_{1}$, $x_{2}\in U_{2}$ and $x_{3}\in U_{3}$ such that $x=x_{1}+x_{2}+x_{3}$. Consider these $x_{1}$, $x_{2}$ and $x_{3}$. Since $x=x_{1}+x_{2}+x_{3}$, we have $x-x_{1}=\underbrace{x_{2}}_{\in U_{2}\subseteq V_{2}}+\underbrace{x_{3}}_{\in U_{3}\subseteq V_{3}}\subseteq V_{2}+V_{3}$. Combined with $\underbrace{x}_{\in V_{1}}-\underbrace{x_{1}% }_{\in U_{1}\subseteq V_{1}}\in V_{1}-V_{1}\subseteq V_{1}$, this yields \[ x-x_{1}\in\left( V_{2}+V_{3}\right) \cap V_{1}. \] \par On the other hand, the internal direct sum $V_{1}\oplus V_{2}\oplus V_{3}$ is well-defined. This direct sum rewrites as $V_{1}\oplus V_{2}\oplus V_{3}=\underbrace{V_{2}\oplus V_{3}}_{\substack{=V_{2}+V_{3}\\\text{(since direct sums are sums)}}}\oplus V_{1}=\left( V_{2}+V_{3}\right) \oplus V_{1}% $. Hence, the direct sum $\left( V_{2}+V_{3}\right) \oplus V_{1}$ is well-defined. Thus, $\left( V_{2}+V_{3}\right) \cap V_{1}=0$. \par But we have $x-x_{1}\in\left( V_{2}+V_{3}\right) \cap V_{1}$. In view of $\left( V_{2}+V_{3}\right) \cap V_{1}=0$, this rewrites as $x-x_{1}\in0$. Hence, $x-x_{1}=0$, so that $x=x_{1}\in U_{1}$. \par Now forget that we fixed $x$. We have thus proven that $x\in U_{1}$ for every $x\in V_{1}$. In other words, $V_{1}\subseteq U_{1}$. Combined with $U_{1}\subseteq V_{1}$, this yields $U_{1}=V_{1}$. Similarly, $U_{2}=V_{2}$ and $U_{3}=V_{3}$, qed.}. \end{verlong} If we apply this fact to $\widetilde{\mathfrak{g}}$, $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) $, $\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) $, $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $, $\widetilde{\mathfrak{g}}\left[ >0\right] $, $\widetilde{\mathfrak{g}}\left[ <0\right] $, $\widetilde{\mathfrak{g}}\left[ 0\right] $ instead of $V$, $U_{1}$, $U_{2}$, $U_{3}$, $V_{1}$, $V_{2}$, $V_{3}$, then we obtain that $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) =\widetilde{\mathfrak{g}}\left[ >0\right] $, $\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) =\widetilde{\mathfrak{g}}\left[ <0\right] $ and $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =\widetilde{\mathfrak{g}}\left[ 0\right] $ (because we know that $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) $, $\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) $, $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $, $\widetilde{\mathfrak{g}}\left[ >0\right] $, $\widetilde{\mathfrak{g}}\left[ <0\right] $, $\widetilde{\mathfrak{g}}\left[ 0\right] $ are six vector subspaces of $\widetilde{\mathfrak{g}}$ satisfying the four relations $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \subseteq\widetilde{\mathfrak{g}}\left[ >0\right] $, $\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \subseteq\widetilde{\mathfrak{g}}\left[ <0\right] $, $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) \subseteq\widetilde{\mathfrak{g}}\left[ 0\right] $ and $\widetilde{\mathfrak{g}}=\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) \oplus\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) \oplus\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $, and we know that the internal direct sum $\widetilde{\mathfrak{g}}\left[ >0\right] \oplus\widetilde{\mathfrak{g}% }\left[ <0\right] \oplus\widetilde{\mathfrak{g}}\left[ 0\right] $ is well-defined). So we have proven that $\widetilde{\mathfrak{g}}\left[ 0\right] =\iota _{0}\left( \widetilde{\mathfrak{h}}\right) $. In other words, the $0$-th homogeneous component of $\widetilde{\mathfrak{g}}$ (in the $Q$-grading) is $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. On the other hand, we have proven that $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) =\widetilde{\mathfrak{g}}\left[ >0\right] $. Thus,% \[ \iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) =\widetilde{\mathfrak{g}% }\left[ >0\right] =\bigoplus\limits_{\substack{\alpha\in Q;\\\alpha >0}}\widetilde{\mathfrak{g}}\left[ \alpha\right] =\bigoplus \limits_{\substack{\alpha\text{ is a }\mathbb{Z}\text{-linear combination}% \\\text{of }\alpha_{1}\text{, }\alpha_{2}\text{, }...\text{, }\alpha_{n}\text{ with nonnegative}\\\text{coefficients; }\alpha\neq0}}\widetilde{\mathfrak{g}% }\left[ \alpha\right] \] (since an element $\alpha\in Q$ satisfies $\alpha>0$ if and only if $\alpha$ is a $\mathbb{Z}$-linear combination of $\alpha_{1}$, $\alpha_{2}$, $...$, $\alpha_{n}$ with nonnegative coefficients such that $\alpha\neq0$). \begin{vershort} Similarly, $\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) =\bigoplus\limits_{\substack{\alpha\text{ is a }\mathbb{Z}\text{-linear combination}\\\text{of }\alpha_{1}\text{, }\alpha_{2}\text{, }...\text{, }\alpha_{n}\text{ with nonpositive}\\\text{coefficients; }\alpha\neq 0}}\widetilde{\mathfrak{g}}\left[ \alpha\right] $. \end{vershort} \begin{verlong} Also, we have proven that $\iota_{-}\left( \widetilde{\mathfrak{n}}% _{-}\right) =\widetilde{\mathfrak{g}}\left[ <0\right] $. Hence,% \[ \iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) =\widetilde{\mathfrak{g}% }\left[ <0\right] =\bigoplus\limits_{\substack{\alpha\in Q;\\\alpha <0}}\widetilde{\mathfrak{g}}\left[ \alpha\right] =\bigoplus \limits_{\substack{\alpha\text{ is a }\mathbb{Z}\text{-linear combination}% \\\text{of }\alpha_{1}\text{, }\alpha_{2}\text{, }...\text{, }\alpha_{n}\text{ with nonpositive}\\\text{coefficients; }\alpha\neq0}}\widetilde{\mathfrak{g}% }\left[ \alpha\right] \] (since an element $\alpha\in Q$ satisfies $\alpha<0$ if and only if $\alpha$ is a $\mathbb{Z}$-linear combination of $\alpha_{1}$, $\alpha_{2}$, $...$, $\alpha_{n}$ with nonpositive coefficients such that $\alpha\neq0$). \end{verlong} This completes the proof of Theorem \ref{thm.gtilde} \textbf{(e)}. \bigskip \textbf{(g)} Define a $\mathbb{Z}$-linear map $\ell:Q\rightarrow\mathbb{Z}$ by% \[ \left( \ell\left( \alpha_{i}\right) =1\text{ for every }i\in\left\{ 1,2,...,n\right\} \right) . \] (This is well-defined since $Q$ is a free abelian group with generators $\alpha_{1}$, $\alpha_{2}$, $...$, $\alpha_{n}$.) Then, $\ell$ is a group homomorphism. We will use the notations we introduced in our proof of Theorem \ref{thm.gtilde} \textbf{(c)}. As shown in the proof of Theorem \ref{thm.gtilde} \textbf{(e)}, we have $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) =\widetilde{\mathfrak{g}}\left[ >0\right] $, $\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) =\widetilde{\mathfrak{g}}\left[ <0\right] $ and $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =\widetilde{\mathfrak{g}}\left[ 0\right] $. Just as in the proof of Theorem \ref{thm.gtilde} \textbf{(c)}, we will regard the maps $\iota_{+}$, $\iota_{-}$ and $\iota_{0}$ as inclusions. Thus, $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) =\widetilde{\mathfrak{n}}_{+}$, $\iota_{-}\left( \widetilde{\mathfrak{n}}% _{-}\right) =\widetilde{\mathfrak{n}}_{-}$ and $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =\widetilde{\mathfrak{h}}$. From the proof of Theorem \ref{thm.gtilde} \textbf{(c)}, we know that $\widetilde{\mathfrak{g}}\left[ 0\right] $, $\widetilde{\mathfrak{g}}\left[ <0\right] $ and $\widetilde{\mathfrak{g}}\left[ >0\right] $ are $Q$-graded Lie subalgebras of $\widetilde{\mathfrak{g}}$. Since $\widetilde{\mathfrak{g}% }\left[ 0\right] =\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =\widetilde{\mathfrak{h}}$, $\widetilde{\mathfrak{g}}\left[ <0\right] =\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) =\widetilde{\mathfrak{n}}_{-}$ and $\widetilde{\mathfrak{g}}\left[ >0\right] =\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) =\widetilde{\mathfrak{n}}_{+}$, this rewrites as follows: $\widetilde{\mathfrak{h}}$, $\widetilde{\mathfrak{n}}_{-}$ and $\widetilde{\mathfrak{n}}_{+}$ are $Q$-graded Lie subalgebras of $\widetilde{\mathfrak{g}}$. Fix $i\in\left\{ 1,2,...,n\right\} $. Since $\alpha_{i}>0$, the space $\widetilde{\mathfrak{g}}\left[ \alpha_{i}\right] $ is an addend in the direct sum $\bigoplus\limits_{\substack{\alpha\in Q;\\\alpha>0}% }\widetilde{\mathfrak{g}}\left[ \alpha\right] $ (namely, the addend for $\alpha=\alpha_{i}$). Hence, $\widetilde{\mathfrak{g}}\left[ \alpha _{i}\right] \subseteq\bigoplus\limits_{\substack{\alpha\in Q;\\\alpha >0}}\widetilde{\mathfrak{g}}\left[ \alpha\right] =\widetilde{\mathfrak{g}% }\left[ >0\right] =\widetilde{\mathfrak{n}}_{+}$. But since $\widetilde{\mathfrak{n}}_{+}$ is a $Q$-graded vector subspace of $\widetilde{\mathfrak{g}}$, we have $\widetilde{\mathfrak{n}}_{+}\left[ \alpha_{i}\right] =\left( \widetilde{\mathfrak{g}}\left[ \alpha_{i}\right] \right) \cap\widetilde{\mathfrak{n}}_{+}=\widetilde{\mathfrak{g}}\left[ \alpha_{i}\right] $ (since $\widetilde{\mathfrak{g}}\left[ \alpha _{i}\right] \subseteq\widetilde{\mathfrak{n}}_{+}$). \begin{vershort} Now, $\widetilde{\mathfrak{n}}_{+}$ is a $Q$-graded Lie algebra, and $\ell$ is a group homomorphism. Hence, we can apply Proposition \ref{prop.Q-graded.principal} to $\widetilde{\mathfrak{n}}_{+}$ instead of $\widetilde{\mathfrak{g}}$. Applying Proposition \ref{prop.Q-graded.principal} \textbf{(a)} to $\widetilde{\mathfrak{n}}_{+}$ instead of $\mathfrak{g}$, we see that for every $m\in\mathbb{Z}$, the internal direct sum $\bigoplus \limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =m}% }\widetilde{\mathfrak{n}}_{+}\left[ \alpha\right] $ is well-defined. Denote this internal direct sum $\bigoplus\limits_{\substack{\alpha\in Q;\\\ell \left( \alpha\right) =m}}\widetilde{\mathfrak{n}}_{+}\left[ \alpha\right] $ by $\widetilde{\mathfrak{n}}_{+\left[ m\right] }$. Applying Proposition \ref{prop.Q-graded.principal} \textbf{(b)} to $\widetilde{\mathfrak{n}}_{+}$ instead of $\mathfrak{g}$, we see that the Lie algebra $\widetilde{\mathfrak{n}}_{+}$ equipped with the grading $\left( \widetilde{\mathfrak{n}}_{+\left[ m\right] }\right) _{m\in\mathbb{Z}}$ is a $\mathbb{Z}$-graded Lie algebra. Let $N_{+}$ be the free vector space with basis $e_{1},e_{2},...,e_{n}$. Since $\widetilde{\mathfrak{n}}_{+}=\operatorname*{FreeLie}\left( e_{i}\ \mid \ i\in\left\{ 1,2,...,n\right\} \right) $, we then have a canonical isomorphism $\widetilde{\mathfrak{n}}_{+}\cong\operatorname*{FreeLie}\left( N_{+}\right) $ (where $\operatorname*{FreeLie}\left( N_{+}\right) $ means the free Lie algebra over the vector space (not the set) $N_{+}$). We identify $\widetilde{\mathfrak{n}}_{+}$ with $\operatorname*{FreeLie}\left( N_{+}\right) $ along this isomorphism. Due to the construction of the free Lie algebra, we have a canonical injection $N_{+}\rightarrow \operatorname*{FreeLie}\left( N_{+}\right) =\widetilde{\mathfrak{n}}_{+}$. We will regard this injection as an inclusion (so that $N_{+}\subseteq \widetilde{\mathfrak{n}}_{+}$). Since $\widetilde{\mathfrak{n}}_{+}=\operatorname*{FreeLie}\left( N_{+}\right) $, it is clear that $\widetilde{\mathfrak{n}}_{+}$ is generated by $N_{+}$ as a Lie algebra. Clearly, $e_{j}\in\widetilde{\mathfrak{n}}_{+}\left[ \alpha_{j}\right] \subseteq\widetilde{\mathfrak{n}}_{+\left[ 1\right] }$ for every $j\in\left\{ 1,2,...,n\right\} $. Thus, $N_{+}\subseteq \widetilde{\mathfrak{n}}_{+\left[ 1\right] }$. Combining this with the fact that $\widetilde{\mathfrak{n}}_{+}$ is generated by $N_{+}$ as a Lie algebra, we see that we can apply Theorem \ref{thm.FreeLie.grading1} to the Lie algebra $\widetilde{\mathfrak{n}}_{+}$ (with the $\mathbb{Z}$-grading $\left( \widetilde{\mathfrak{n}}_{+\left[ m\right] }\right) _{m\in\mathbb{Z}}$, not with the original $Q$-grading) and $N_{+}$ instead of the Lie algebra $\mathfrak{g}$ and $T$. As a result, we obtain $N_{+}=\widetilde{\mathfrak{n}% }_{+\left[ 1\right] }$. Since $\widetilde{\mathfrak{g}}\left[ \alpha _{i}\right] =\widetilde{\mathfrak{n}}_{+}\left[ \alpha_{i}\right] \subseteq\widetilde{\mathfrak{n}}_{+\left[ 1\right] }=N_{+}$, we have $\widetilde{\mathfrak{g}}\left[ \alpha_{i}\right] =N_{+}\left[ \alpha _{i}\right] $ (since $N_{+}$ is a $Q$-graded subspace of $\widetilde{\mathfrak{g}}$). But $N_{+}\left[ \alpha_{i}\right] =\mathbb{C}e_{i}$ (this is clear from the fact that $N_{+}$ has basis $e_{1},e_{2},...,e_{n}$, and each of the vectors in this basis has a different degree in the $Q$-grading). Hence, $\widetilde{\mathfrak{g}}\left[ \alpha _{i}\right] =N_{+}\left[ \alpha_{i}\right] =\mathbb{C}e_{i}$. A similar argument (with $-\ell$ taking the role of $\ell$) shows that $\widetilde{\mathfrak{g}}\left[ -\alpha_{i}\right] =\mathbb{C}f_{i}$. This proves Theorem \ref{thm.gtilde} \textbf{(g)}. \end{vershort} \begin{verlong} Now, $\widetilde{\mathfrak{n}}_{+}$ is a $Q$-graded Lie algebra, and $\ell$ is a group homomorphism. Hence, we can apply Proposition \ref{prop.Q-graded.principal} to $\widetilde{\mathfrak{n}}_{+}$ instead of $\widetilde{\mathfrak{g}}$. Applying Proposition \ref{prop.Q-graded.principal} \textbf{(a)} to $\widetilde{\mathfrak{n}}_{+}$ instead of $\mathfrak{g}$, we see that for every $m\in\mathbb{Z}$, the internal direct sum $\bigoplus \limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =m}% }\widetilde{\mathfrak{n}}_{+}\left[ \alpha\right] $ is well-defined. Denote this internal direct sum $\bigoplus\limits_{\substack{\alpha\in Q;\\\ell \left( \alpha\right) =m}}\widetilde{\mathfrak{n}}_{+}\left[ \alpha\right] $ by $\widetilde{\mathfrak{n}}_{+\left[ m\right] }$. Applying Proposition \ref{prop.Q-graded.principal} \textbf{(b)} to $\widetilde{\mathfrak{n}}_{+}$ instead of $\mathfrak{g}$, we see that the Lie algebra $\widetilde{\mathfrak{n}}_{+}$ equipped with the grading $\left( \widetilde{\mathfrak{n}}_{+\left[ m\right] }\right) _{m\in\mathbb{Z}}$ is a $\mathbb{Z}$-graded Lie algebra. Denote this $\mathbb{Z}$-graded Lie algebra by $\widetilde{\mathfrak{n}}_{+}^{\operatorname*{principal}}$. Then, $\widetilde{\mathfrak{n}}_{+}^{\operatorname*{principal}}\left[ m\right] =\widetilde{\mathfrak{n}}_{+\left[ m\right] }$ for every $m\in\mathbb{Z}$. Applied to $m=1$, this yields $\widetilde{\mathfrak{n}}_{+}% ^{\operatorname*{principal}}\left[ 1\right] =\widetilde{\mathfrak{n}% }_{+\left[ 1\right] }$. We have $\widetilde{\mathfrak{n}}_{+\left[ 1\right] }=\bigoplus \limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =1}% }\widetilde{\mathfrak{n}}_{+}\left[ \alpha\right] $ (by the definition of $\widetilde{\mathfrak{n}}_{+\left[ 1\right] }$). For every $j\in\left\{ 1,2,...,n\right\} $, the vector space $\widetilde{\mathfrak{n}}_{+}\left[ \alpha_{j}\right] $ is an addend in the direct sum $\bigoplus\limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =1}}\widetilde{\mathfrak{n}}_{+}\left[ \alpha\right] $ (namely, the addend for $\alpha=\alpha_{j}$), because $\ell\left( \alpha _{j}\right) =1$. Thus, for every $j\in\left\{ 1,2,...,n\right\} $, we have $\widetilde{\mathfrak{n}}_{+}\left[ \alpha_{j}\right] \subseteq \bigoplus\limits_{\substack{\alpha\in Q;\\\ell\left( \alpha\right) =1}}\widetilde{\mathfrak{n}}_{+}\left[ \alpha\right] =\widetilde{\mathfrak{n}}_{+\left[ 1\right] }$. Applied to $j=i$, this yields $\widetilde{\mathfrak{n}}_{+}\left[ \alpha_{i}\right] \subseteq \widetilde{\mathfrak{n}}_{+\left[ 1\right] }$. Let $N_{+}$ be the free vector space with basis $e_{1},e_{2},...,e_{n}$. Since $\widetilde{\mathfrak{n}}_{+}=\operatorname*{FreeLie}\left( e_{i}\ \mid \ i\in\left\{ 1,2,...,n\right\} \right) $, we then have a canonical isomorphism $\widetilde{\mathfrak{n}}_{+}\cong\operatorname*{FreeLie}\left( N_{+}\right) $ (where $\operatorname*{FreeLie}\left( N_{+}\right) $ means the free Lie algebra over the vector space (not the set) $N_{+}$). We identify $\widetilde{\mathfrak{n}}_{+}$ with $\operatorname*{FreeLie}\left( N_{+}\right) $ along this isomorphism. Due to the construction of the free Lie algebra, we have a canonical injection $N_{+}\rightarrow \operatorname*{FreeLie}\left( N_{+}\right) =\widetilde{\mathfrak{n}}_{+}$. We will regard this injection as an inclusion (so that $N_{+}\subseteq \widetilde{\mathfrak{n}}_{+}$). Since $\widetilde{\mathfrak{n}}_{+}=\operatorname*{FreeLie}\left( N_{+}\right) $, it is clear that $\widetilde{\mathfrak{n}}_{+}$ is generated by $N_{+}$ as a Lie algebra. In other words, $\widetilde{\mathfrak{n}}% _{+}^{\operatorname*{principal}}$ is generated by $N_{+}$ as a Lie algebra (since $\widetilde{\mathfrak{n}}_{+}^{\operatorname*{principal}}% =\widetilde{\mathfrak{n}}_{+}$ as Lie algebra). Since $N_{+}$ is the free vector space with basis $e_{1},e_{2},...,e_{n}$, we have $N_{+}=\bigoplus\limits_{j=1}^{n}e_{j}\mathbb{C}$. For every $j\in\left\{ 1,2,...,n\right\} $, we have $\deg\left( e_{j}\right) =\alpha_{j}$ (by the definition of the $Q$-grading on $\widetilde{\mathfrak{g}}$) and thus $e_{j}\in\widetilde{\mathfrak{n}}% _{+}\left[ \alpha_{j}\right] $ (since $e_{j}\in\widetilde{\mathfrak{n}}_{+}% $). Hence, for every $j\in\left\{ 1,2,...,n\right\} $, we have $e_{j}\mathbb{C}\subseteq\widetilde{\mathfrak{n}}_{+}\left[ \alpha _{j}\right] $. Thus, for every $j\in\left\{ 1,2,...,n\right\} $, the vector subspace $e_{j}\mathbb{C}$ of $\widetilde{\mathfrak{n}}_{+}$ lies entirely within one homogeneous component of $\widetilde{\mathfrak{n}}_{+}$. Thus, for every $j\in\left\{ 1,2,...,n\right\} $, the vector subspace $e_{j}% \mathbb{C}$ is a $Q$-graded vector subspace of $\widetilde{\mathfrak{n}}_{+}$. Therefore, the internal direct sum $\bigoplus\limits_{j=1}^{n}e_{j}\mathbb{C}$ also is a $Q$-graded vector subspace of $\widetilde{\mathfrak{n}}_{+}$ (because any internal direct sum of $Q$-graded vector subspaces of a $Q$-graded vector space must itself be a $Q$-graded vector subspace). Since $\bigoplus\limits_{j=1}^{n}e_{j}\mathbb{C}=N_{+}$, this means that $N_{+}$ is a $Q$-graded vector subspace of $\widetilde{\mathfrak{n}}_{+}$. For every $j\in\left\{ 1,2,...,n\right\} $ satisfying $j\neq i$, we have $\left( e_{j}\mathbb{C}\right) \left[ \alpha_{i}\right] =0$ (because \begin{align*} \left( e_{j}\mathbb{C}\right) \left[ \alpha_{i}\right] & =\underbrace{\left( e_{j}\mathbb{C}\right) }_{\subseteq \widetilde{\mathfrak{n}}_{+}\left[ \alpha_{j}\right] }\cap\left( \widetilde{\mathfrak{n}}_{+}\left[ \alpha_{i}\right] \right) \subseteq \left( \widetilde{\mathfrak{n}}_{+}\left[ \alpha_{j}\right] \right) \cap\left( \widetilde{\mathfrak{n}}_{+}\left[ \alpha_{i}\right] \right) =0\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }j\neq i\text{, so that }\alpha_{j}\neq\alpha_{i}\text{, and thus the }\alpha_{j}\text{-th and the}\\ \alpha_{i}\text{-th homogeneous components of }\widetilde{\mathfrak{n}}% _{+}\text{ are linearly}\\ \text{disjoint, i. e., they satisfy }\left( \widetilde{\mathfrak{n}}% _{+}\left[ \alpha_{j}\right] \right) \cap\left( \widetilde{\mathfrak{n}% }_{+}\left[ \alpha_{i}\right] \right) =0 \end{array} \right) \end{align*} and thus $\left( e_{j}\mathbb{C}\right) \left[ \alpha_{i}\right] =0$). Moreover, we know that for every $j\in\left\{ 1,2,...,n\right\} $, we have $e_{j}\mathbb{C}\subseteq\widetilde{\mathfrak{n}}_{+}\left[ \alpha _{j}\right] $. Applied to $j=i$, this yields $e_{i}\mathbb{C}\subseteq \widetilde{\mathfrak{n}}_{+}\left[ \alpha_{i}\right] $. Now, $\left( e_{i}\mathbb{C}\right) \left[ \alpha_{i}\right] =\left( e_{i}% \mathbb{C}\right) \cap\left( \widetilde{\mathfrak{n}}_{+}\left[ \alpha _{i}\right] \right) =e_{i}\mathbb{C}$ (since $e_{i}\mathbb{C}\subseteq \widetilde{\mathfrak{n}}_{+}\left[ \alpha_{i}\right] $). Now, \begin{align*} N_{+}\left[ \alpha_{i}\right] & =\left( \bigoplus\limits_{j=1}^{n}% e_{j}\mathbb{C}\right) \left[ \alpha_{i}\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since }N_{+}=\bigoplus\limits_{j=1}^{n}e_{j}\mathbb{C}\right) \\ & =\bigoplus\limits_{j=1}^{n}\left( \left( e_{j}\mathbb{C}\right) \left[ \alpha_{i}\right] \right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }e_{j}\mathbb{C}\text{ is a }Q\text{-graded vector subspace of }\widetilde{\mathfrak{n}}_{+}\text{ for every }j\in\left\{ 1,2,...,n\right\} \right) \\ & =\bigoplus\limits_{j\in\left\{ 1,2,...,n\right\} }\left( \left( e_{j}\mathbb{C}\right) \left[ \alpha_{i}\right] \right) =\left( \underbrace{\left( e_{i}\mathbb{C}\right) \left[ \alpha_{i}\right] }_{=e_{i}\mathbb{C}}\right) \oplus\bigoplus\limits_{\substack{j\in\left\{ 1,2,...,n\right\} ;\\j\neq i}}\left( \underbrace{\left( e_{j}% \mathbb{C}\right) \left[ \alpha_{i}\right] }_{\substack{=0\\\text{(since }j\neq i\text{)}}}\right) \\ & =\left( e_{i}\mathbb{C}\right) \oplus\underbrace{\bigoplus \limits_{\substack{j\in\left\{ 1,2,...,n\right\} ;\\j\neq i}}0}_{=0}% =e_{i}\mathbb{C}. \end{align*} On the other hand,% \begin{align*} N_{+} & =\bigoplus\limits_{j=1}^{n}e_{j}\mathbb{C}=\sum\limits_{j=1}% ^{n}\underbrace{e_{j}\mathbb{C}}_{\subseteq\widetilde{\mathfrak{n}}_{+}\left[ \alpha_{j}\right] \subseteq\widetilde{\mathfrak{n}}_{+\left[ 1\right] }% }\ \ \ \ \ \ \ \ \ \ \left( \text{since direct sums are sums}\right) \\ & \subseteq\sum\limits_{j=1}^{n}\widetilde{\mathfrak{n}}_{+\left[ 1\right] }\subseteq\widetilde{\mathfrak{n}}_{+\left[ 1\right] }% =\widetilde{\mathfrak{n}}_{+}^{\operatorname*{principal}}\left[ 1\right] . \end{align*} Since $\widetilde{\mathfrak{n}}_{+}^{\operatorname*{principal}}$ is generated by $N_{+}$ as a Lie algebra, this yields that we can apply Theorem \ref{thm.FreeLie.grading1} to $\widetilde{\mathfrak{n}}_{+}% ^{\operatorname*{principal}}$ and $N_{+}$ instead of $\mathfrak{g}$ and $T$. As a result, we obtain that $N_{+}=\widetilde{\mathfrak{n}}_{+}% ^{\operatorname*{principal}}\left[ 1\right] =\widetilde{\mathfrak{n}% }_{+\left[ 1\right] }$. Now,% \[ \widetilde{\mathfrak{g}}\left[ \alpha_{i}\right] =\widetilde{\mathfrak{n}% }_{+}\left[ \alpha_{i}\right] \subseteq\widetilde{\mathfrak{n}}_{+\left[ 1\right] }=N_{+}, \] so that% \begin{align*} \widetilde{\mathfrak{g}}\left[ \alpha_{i}\right] & =\left( \widetilde{\mathfrak{g}}\left[ \alpha_{i}\right] \right) \cap N_{+}% =N_{+}\left[ \alpha_{i}\right] \ \ \ \ \ \ \ \ \ \ \left( \text{since }N_{+}\text{ is a }Q\text{-graded vector subspace of }\widetilde{\mathfrak{g}% }\right) \\ & =e_{i}\mathbb{C}. \end{align*} Now, notice that $Q$ is a free abelian group with generators $-\alpha_{1}$, $-\alpha_{2}$, $...$, $-\alpha_{n}$. Hence, we can prove $\widetilde{\mathfrak{g}}\left[ -\alpha_{i}\right] =f_{i}\mathbb{C}$ by means of making the following modifications to the above proof of $\widetilde{\mathfrak{g}}\left[ \alpha_{i}\right] =e_{i}\mathbb{C}$: \begin{itemize} \item Replace every $>$ sign by a $<$ sign, and vice versa. \item Replace every $\alpha_{j}$ by $-\alpha_{j}$ (this includes replacing every $\alpha_{i}$ by $-\alpha_{i}$). \item Replace every $\widetilde{\mathfrak{n}}_{+}$ by $\widetilde{\mathfrak{n}% }_{-}$ and vice versa. \item Replace every $\iota_{+}$ by $\iota_{-}$ and vice versa. \item Replace every $e_{j}$ by $f_{j}$ (this includes replacing every $e_{i}$ by $f_{i}$). \item Replace every $N_{+}$ by $N_{-}$. \item Replace $\widetilde{\mathfrak{n}}_{+}^{\operatorname*{principal}}$ by $\widetilde{\mathfrak{n}}_{-}^{\operatorname*{principal}}$. \end{itemize} Thus we have proven both $\widetilde{\mathfrak{g}}\left[ \alpha_{i}\right] =e_{i}\mathbb{C}$ and $\widetilde{\mathfrak{g}}\left[ -\alpha_{i}\right] =f_{i}\mathbb{C}$. This proves Theorem \ref{thm.gtilde} \textbf{(g)}. \end{verlong} \bigskip \textbf{(h)} It is clear that $I$ (being a sum of $Q$-graded ideals) is a $Q$-graded ideal. We only need to prove that $I$ has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. Let $\pi_{0}:\widetilde{\mathfrak{g}}\rightarrow\widetilde{\mathfrak{g}% }\left[ 0\right] $ be the canonical projection from the $Q$-graded vector space $\widetilde{\mathfrak{g}}$ on its $0$-th homogeneous component $\widetilde{\mathfrak{g}}\left[ 0\right] $. \begin{vershort} For every $Q$-graded vector subspace $M$ of $\widetilde{\mathfrak{g}}$, we have $\pi_{0}\left( M\right) =M\cap\left( \widetilde{\mathfrak{g}}\left[ 0\right] \right) $ (this is just an elementary property of $Q$-graded vector spaces). Since $\widetilde{\mathfrak{g}}\left[ 0\right] =\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ (by Theorem \ref{thm.gtilde} \textbf{(e)}), this rewrites as follows: For every $Q$-graded vector subspace $M$ of $\widetilde{\mathfrak{g}}$, we have $\pi_{0}\left( M\right) =M\cap\iota _{0}\left( \widetilde{\mathfrak{h}}\right) $. Thus, every $Q$-graded ideal $\mathfrak{i}$ of $\widetilde{\mathfrak{g}}$ which has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ satisfies $\pi_{0}\left( \mathfrak{i}\right) =\mathfrak{i}\cap\iota_{0}\left( \widetilde{\mathfrak{h}% }\right) =0$. Therefore, the sum $I$ of all such ideals also satisfies $\pi_{0}\left( I\right) =0$ (since $\pi_{0}$ is linear). But since $\pi _{0}\left( I\right) =I\cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ (because for every $Q$-graded vector subspace $M$ of $\widetilde{\mathfrak{g}}$, we have $\pi_{0}\left( M\right) =M\cap\iota _{0}\left( \widetilde{\mathfrak{h}}\right) $), this rewrites as $I\cap \iota_{0}\left( \widetilde{\mathfrak{h}}\right) =0$. In other words, $I$ has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. Theorem \ref{thm.gtilde} \textbf{(h)} is proven. \end{vershort} \begin{verlong} For every $Q$-graded vector subspace $M$ of $\widetilde{\mathfrak{g}}$, we have% \begin{equation} M\cap\left( \widetilde{\mathfrak{g}}\left[ 0\right] \right) =\pi _{0}\left( M\right) \label{pf.gtilde.g.1}% \end{equation} \footnote{\textit{Proof of (\ref{pf.gtilde.g.1}):} Let $M$ be a $Q$-graded vector subspace of $\widetilde{\mathfrak{g}}$. Then, we have $M=\bigoplus \limits_{\alpha\in Q}M\left[ \alpha\right] $, and every $\alpha\in Q$ satisfies $M\left[ \alpha\right] =M\cap\left( \widetilde{\mathfrak{g}% }\left[ \alpha\right] \right) $. \par Now, $M=\bigoplus\limits_{\alpha\in Q}M\left[ \alpha\right] =\sum \limits_{\alpha\in Q}M\left[ \alpha\right] $ (since direct sums are sums), so that% \[ \pi_{0}\left( M\right) =\pi_{0}\left( \sum\limits_{\alpha\in Q}M\left[ \alpha\right] \right) =\sum\limits_{\alpha\in Q}\pi_{0}\left( M\left[ \alpha\right] \right) =\pi_{0}\left( M\left[ 0\right] \right) +\sum\limits_{\substack{\alpha\in Q;\\\alpha\neq0}}\pi_{0}\left( M\left[ \alpha\right] \right) . \] \par But now, let $\alpha\in Q$ be such that $\alpha\neq0$. Then, $0$ and $\alpha$ are two distinct elements of $Q$. Whenever $\beta$ and $\gamma$ are two distinct elements of $Q$, the projection from the $Q$-graded vector space $\widetilde{\mathfrak{g}}$ on its $\beta$-th homogeneous component $\widetilde{\mathfrak{g}}\left[ \beta\right] $ sends $\widetilde{\mathfrak{g}}\left[ \gamma\right] $ to zero. Applied to $\beta=0$ and $\gamma=\alpha$, this yields that the projection from the $Q$-graded vector space $\widetilde{\mathfrak{g}}$ on its $0$-th homogeneous component $\widetilde{\mathfrak{g}}\left[ 0\right] $ sends $\widetilde{\mathfrak{g}}\left[ \alpha\right] $ to $0$ (since $0$ and $\alpha$ are two distinct elements of $Q$). Since the projection from the $Q$-graded vector space $\widetilde{\mathfrak{g}}$ on its $0$-th homogeneous component $\widetilde{\mathfrak{g}}\left[ 0\right] $ is the map $\pi_{0}$, this rewrites as follows: The map $\pi_{0}$ sends $\widetilde{\mathfrak{g}% }\left[ \alpha\right] $ to $0$. Thus, $\pi_{0}\left( \widetilde{\mathfrak{g}}\left[ \alpha\right] \right) =0$. Since $M\left[ \alpha\right] \subseteq\widetilde{\mathfrak{g}}\left[ \alpha\right] $, we have $\pi_{0}\left( M\left[ \alpha\right] \right) \subseteq\pi_{0}\left( \widetilde{\mathfrak{g}}\left[ \alpha\right] \right) =0$, so that $\pi _{0}\left( M\left[ \alpha\right] \right) =0$. \par Now forget that we fixed $\alpha$. We thus have shown that every $\alpha\in Q$ such that $\alpha\neq0$ satisfies $\pi_{0}\left( M\left[ \alpha\right] \right) =0$. Hence, $\sum\limits_{\substack{\alpha\in Q;\\\alpha\neq 0}}\underbrace{\pi_{0}\left( M\left[ \alpha\right] \right) }_{=0}% =\sum\limits_{\substack{\alpha\in Q;\\\alpha\neq0}}0=0$. \par On the other hand, $\pi_{0}$ is a projection on $\widetilde{\mathfrak{g}% }\left[ 0\right] $, and therefore leaves every subset of $\widetilde{\mathfrak{g}}\left[ 0\right] $ invariant. Since $M\left[ 0\right] $ is a subset of $\widetilde{\mathfrak{g}}\left[ 0\right] $, this yields that $\pi_{0}$ leaves $M\left[ 0\right] $ invariant. Hence, $\pi _{0}\left( M\left[ 0\right] \right) =M\left[ 0\right] $. \par Now,% \[ \pi_{0}\left( M\right) =\underbrace{\pi_{0}\left( M\left[ 0\right] \right) }_{=M\left[ 0\right] }+\underbrace{\sum\limits_{\substack{\alpha\in Q;\\\alpha\neq0}}\pi_{0}\left( M\left[ \alpha\right] \right) }% _{=0}=M\left[ 0\right] =M\cap\left( \widetilde{\mathfrak{g}}\left[ 0\right] \right) \] (since every $\alpha\in Q$ satisfies $M\left[ \alpha\right] =M\cap\left( \widetilde{\mathfrak{g}}\left[ \alpha\right] \right) $). This proves (\ref{pf.gtilde.g.1}).}. As a consequence, every $Q$-graded vector subspace $M$ of $\widetilde{\mathfrak{g}}$ satisfies% \begin{equation} M\cap\underbrace{\iota_{0}\left( \widetilde{\mathfrak{h}}\right) }_{\substack{=\widetilde{\mathfrak{g}}\left[ 0\right] \\\text{(by Theorem \ref{thm.gtilde} \textbf{(e)})}}}=M\cap\left( \widetilde{\mathfrak{g}}\left[ 0\right] \right) =\pi_{0}\left( M\right) \label{pf.gtilde.g.2}% \end{equation} (by (\ref{pf.gtilde.g.1})). Now, $I$ is the sum of all $Q$-graded ideals in $\widetilde{\mathfrak{g}}$ which have zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}% }\right) $. In other words,% \begin{align*} I & =\sum\limits_{\substack{\mathfrak{i}\text{ is a }Q\text{-graded ideal in }\widetilde{\mathfrak{g}}\\\text{such that }\mathfrak{i}\cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =0}}\mathfrak{i}=\sum \limits_{\substack{\mathfrak{i}\text{ is a }Q\text{-graded ideal in }\widetilde{\mathfrak{g}}\\\text{such that }\pi_{0}\left( \mathfrak{i}% \right) =0}}\mathfrak{i}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{because every }Q\text{-graded ideal }\mathfrak{i}\text{ in }\widetilde{\mathfrak{g}}\text{ satisfies }\mathfrak{i}\cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =\pi_{0}\left( \mathfrak{i}\right) \\ \text{(by (\ref{pf.gtilde.g.2}), applied to }M=\mathfrak{i}\text{)}% \end{array} \right) . \end{align*} Thus,% \[ \pi_{0}\left( I\right) =\pi_{0}\left( \sum\limits_{\substack{\mathfrak{i}% \text{ is a }Q\text{-graded ideal in }\widetilde{\mathfrak{g}}\\\text{such that }\pi_{0}\left( \mathfrak{i}\right) =0}}\mathfrak{i}\right) =\sum\limits_{\substack{\mathfrak{i}\text{ is a }Q\text{-graded ideal in }\widetilde{\mathfrak{g}}\\\text{such that }\pi_{0}\left( \mathfrak{i}% \right) =0}}\underbrace{\pi_{0}\left( \mathfrak{i}\right) }_{=0}% =\sum\limits_{\substack{\mathfrak{i}\text{ is a }Q\text{-graded ideal in }\widetilde{\mathfrak{g}}\\\text{such that }\pi_{0}\left( \mathfrak{i}% \right) =0}}0=0. \] But $I$ is $Q$-graded. Thus, (\ref{pf.gtilde.g.2}) (applied to $M=I$) yields $I\cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =\pi_{0}\left( I\right) =0$. In other words, $I$ has zero intersection with $\iota _{0}\left( \widetilde{\mathfrak{h}}\right) $. This proves Theorem \ref{thm.gtilde} \textbf{(h)}. \end{verlong} \bigskip \textbf{(i)} First, we notice that the Lie algebra $\widetilde{\mathfrak{g}}$ is generated by its elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ (since \[ \widetilde{\mathfrak{g}}=\operatorname*{FreeLie}\left( h_{i},f_{i}% ,e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) \] ). Hence, the Lie algebra $\mathfrak{g}$ is generated by its elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ as well (since $\mathfrak{g}=\widetilde{\mathfrak{g}}\diagup I$). In order to prove that $\mathfrak{g}$ is a contragredient Lie algebra corresponding to $A$, we must prove that it satisfies the conditions \textbf{(1)}, \textbf{(2)} and \textbf{(3)} of Definition \ref{def.contragredient}. \textit{Proof of condition \textbf{(1)}:} The relations (\ref{nonserre-relations}) are satisfied in $\widetilde{\mathfrak{g}}$ (by the definition of $\widetilde{\mathfrak{g}}$ as the quotient Lie algebra $\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) $) and thus also in $\mathfrak{g}$ (since $\mathfrak{g}$ is a quotient Lie algebra of $\widetilde{\mathfrak{g}}$). This proves condition \textbf{(1)} for our $Q$-graded Lie algebra $\mathfrak{g}$. \textit{Proof of condition \textbf{(2)}:} By Theorem \ref{thm.gtilde} \textbf{(e)}, we have $\widetilde{\mathfrak{g}}\left[ 0\right] =\iota _{0}\left( \widetilde{\mathfrak{h}}\right) $. We know that $h_{1}% ,h_{2},...,h_{n}$ is a basis of the vector space $\widetilde{\mathfrak{h}}$ (since $\widetilde{\mathfrak{h}}$ was defined as the free vector space with basis $h_{1},h_{2},...,h_{n}$). Since $\iota_{0}$ is injective, this yields that $h_{1},h_{2},...,h_{n}$ is a basis of $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ (because we identify the images of the vectors $h_{1}$, $h_{2}$, $...$, $h_{n}$ under $\iota_{0}$ with $h_{1}$, $h_{2}$, $...$, $h_{n}$). Thus, in particular, the vectors $h_{1}% ,h_{2},...,h_{n}$ in $\widetilde{\mathfrak{g}}$ span the vector space $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =\widetilde{\mathfrak{g}% }\left[ 0\right] $. As a consequence, the vectors $h_{1}$, $h_{2}$, $...$, $h_{n}$ in $\mathfrak{g}$ span the vector space $\mathfrak{g}\left[ 0\right] $ (because $\mathfrak{g}=\widetilde{\mathfrak{g}}\diagup I$). The vectors $h_{1}$, $h_{2}$, $...$, $h_{n}$ in $\mathfrak{g}$ are linearly independent\footnote{\textit{Proof.} Let $\left( \lambda_{1},\lambda _{2},...,\lambda_{n}\right) \in\mathbb{C}^{n}$ be such that $\lambda_{1}% h_{1}+\lambda_{2}h_{2}+...+\lambda_{n}h_{n}=0$ in $\mathfrak{g}$. Then, $\lambda_{1}h_{1}+\lambda_{2}h_{2}+...+\lambda_{n}h_{n}\in I$ in $\widetilde{\mathfrak{g}}$ (since $\mathfrak{g}=\widetilde{\mathfrak{g}% }\diagup I$). Combined with $\lambda_{1}h_{1}+\lambda_{2}h_{2}+...+\lambda _{n}h_{n}\in\widetilde{\mathfrak{g}}\left[ 0\right] =\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $, this yields $\lambda_{1}h_{1}+\lambda _{2}h_{2}+...+\lambda_{n}h_{n}\in I\cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =0$ (since Theorem \ref{thm.gtilde} \textbf{(h)} yields that $I$ has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $). Thus, $\lambda_{1}h_{1}+\lambda_{2}% h_{2}+...+\lambda_{n}h_{n}=0$ in $\iota_{0}\left( \widetilde{\mathfrak{h}% }\right) $. Since $h_{1},h_{2},...,h_{n}$ is a basis of $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $, this yields $\lambda_{1}=\lambda _{2}=...=\lambda_{n}=0$. \par Now forget that we fixed $\left( \lambda_{1},\lambda_{2},...,\lambda _{n}\right) $. We have thus shown that every $\left( \lambda_{1},\lambda _{2},...,\lambda_{n}\right) \in\mathbb{C}^{n}$ such that $\lambda_{1}% h_{1}+\lambda_{2}h_{2}+...+\lambda_{n}h_{n}=0$ in $\mathfrak{g}$ satisfies $\lambda_{1}=\lambda_{2}=...=\lambda_{n}=0$. In other words, the vectors $h_{1}$, $h_{2}$, $...$, $h_{n}$ in $\mathfrak{g}$ are linearly independent, qed.}. Hence, $h_{1},h_{2},...,h_{n}$ is a basis of the vector space $\mathfrak{g}\left[ 0\right] $ (since the vectors $h_{1}$, $h_{2}$, $...$, $h_{n}$ in $\mathfrak{g}$ span the vector space $\mathfrak{g}\left[ 0\right] $ and are linearly independent). In other words, the vector space $\mathfrak{g}\left[ 0\right] $ has $\left( h_{1},h_{2},...,h_{n}\right) $ as a $\mathbb{C}$-vector space basis. Let $i\in\left\{ 1,2,...,n\right\} $. Theorem \ref{thm.gtilde} \textbf{(g)} yields $\widetilde{\mathfrak{g}}\left[ \alpha_{i}\right] =\mathbb{C}e_{i}$. Projecting this onto $\widetilde{\mathfrak{g}}\diagup I=\mathfrak{g}$, we obtain $\mathfrak{g}\left[ \alpha_{i}\right] =\mathbb{C}e_{i}$ (since the projection of $e_{i}$ onto $\mathfrak{g}$ is also called $e_{i}$). Similarly, $\mathfrak{g}\left[ -\alpha_{i}\right] =\mathbb{C}f_{i}$. Condition \textbf{(2)} is thus verified for our $Q$-graded Lie algebra $\mathfrak{g}$. \textit{Proof of condition \textbf{(3)}:} Let $J$ be a nonzero $Q$-graded ideal in $\mathfrak{g}$. Assume that $J\cap\left( \mathfrak{g}\left[ 0\right] \right) =0$. Recall that $I$ has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. That is, $I\cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =0$. Let $\operatorname*{proj}:\widetilde{\mathfrak{g}}\rightarrow \widetilde{\mathfrak{g}}\diagup I=\mathfrak{g}$ be the canonical projection. Then, $\operatorname*{proj}$ is a $Q$-graded Lie algebra homomorphism, so that $\operatorname*{proj}\nolimits^{-1}\left( J\right) $ is a $Q$-graded ideal of $\widetilde{\mathfrak{g}}$ (since $J$ is a $Q$-graded ideal of $\mathfrak{g}$). Also, $\operatorname*{Ker}\operatorname*{proj}=I$ (since $\operatorname*{proj}$ is the canonical projection $\widetilde{\mathfrak{g}% }\rightarrow\widetilde{\mathfrak{g}}\diagup I$). Let $x\in\operatorname*{proj}\nolimits^{-1}\left( J\right) \cap\iota _{0}\left( \widetilde{\mathfrak{h}}\right) $. Then, $x\in \operatorname*{proj}\nolimits^{-1}\left( J\right) $ and $x\in\iota _{0}\left( \widetilde{\mathfrak{h}}\right) $. Since $x\in \operatorname*{proj}\nolimits^{-1}\left( J\right) $, we have $\operatorname*{proj}\left( x\right) \in J$. Since $x\in\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =\widetilde{\mathfrak{g}}\left[ 0\right] $ (by Theorem \ref{thm.gtilde} \textbf{(e)}), we have $\operatorname*{proj}% \left( x\right) \in\mathfrak{g}\left[ 0\right] $ (since $\operatorname*{proj}$ is $Q$-graded). Combined with $\operatorname*{proj}% \left( x\right) \in J$, this yields $\operatorname*{proj}\left( x\right) \in J\cap\left( \mathfrak{g}\left[ 0\right] \right) =0$, so that $\operatorname*{proj}\left( x\right) =0$, thus $x\in\operatorname*{Ker}% \operatorname*{proj}=I$. Combined with $x\in\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $, this yields $x\in I\cap\left( \iota _{0}\left( \widetilde{\mathfrak{h}}\right) \right) =0$, so that $x=0$. Forget that we fixed $x$. We thus have proven that every $x\in \operatorname*{proj}\nolimits^{-1}\left( J\right) \cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ satisfies $x=0$. Hence, $\operatorname*{proj}\nolimits^{-1}\left( J\right) \cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =0$. Thus, $\operatorname*{proj}% \nolimits^{-1}\left( J\right) $ is a $Q$-graded ideal in $\widetilde{\mathfrak{g}}$ which has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. Hence,% \[ \operatorname*{proj}\nolimits^{-1}\left( J\right) \subseteq\left( \text{sum of all }Q\text{-graded ideals in }\widetilde{\mathfrak{g}}\text{ which have zero intersection with }\iota_{0}\left( \widetilde{\mathfrak{h}}\right) \right) =I. \] Now let $y\in J$ be arbitrary. Since $y\in J\subseteq\mathfrak{g}% =\widetilde{\mathfrak{g}}\diagup I$, there exists a $y^{\prime}\in \widetilde{\mathfrak{g}}$ such that $y=\operatorname*{proj}\left( y^{\prime }\right) $. Consider this $y$. Since $\operatorname*{proj}\left( y^{\prime }\right) =y\in J$, we have $y^{\prime}\in\operatorname*{proj}\nolimits^{-1}% \left( J\right) \subseteq I=\operatorname*{Ker}\operatorname*{proj}$, so that $\operatorname*{proj}\left( y^{\prime}\right) =0$. Thus, $y=\operatorname*{proj}\left( y^{\prime}\right) =0$. Now, forget that we fixed $y$. We thus have proven that every $y\in J$ satisfies $y=0$. Thus, $J=0$, contradicting to the fact that $J$ is nonzero. This contradiction shows that our assumption (that $J\cap\left( \mathfrak{g}\left[ 0\right] \right) =0$) was wrong. In other words, $J\cap\left( \mathfrak{g}\left[ 0\right] \right) \neq0$. Now forget that we fixed $J$. We thus have proven that every nonzero $Q$-graded ideal $J$ in $\mathfrak{g}$ satisfies $J\cap\left( \mathfrak{g}% \left[ 0\right] \right) \neq0$. In other words, every nonzero $Q$-graded ideal in $\mathfrak{g}$ has a nonzero intersection with $\mathfrak{g}\left[ 0\right] $. This proves that Condition \textbf{(3)} holds for our $Q$-graded Lie algebra $\mathfrak{g}$. Now that we have checked all three conditions \textbf{(1)}, \textbf{(2)} and \textbf{(3)} for our $Q$-graded Lie algebra $\mathfrak{g}$, we conclude that $\mathfrak{g}$ indeed is a contragredient Lie algebra corresponding to $A$. Theorem \ref{thm.gtilde} \textbf{(i)} is proven. \bigskip \textit{Proof of Theorem \ref{thm.g(A).exuni}.} \textbf{(a)} Let the $Q$-graded Lie algebra $\mathfrak{g}$ be defined as in Theorem \ref{thm.gtilde}. According to Theorem \ref{thm.gtilde} \textbf{(i)}, this $\mathfrak{g}$ is a contragredient Lie algebra corresponding to $A$. Thus, there exists at least one contragredient Lie algebra corresponding to $A$, namely this $\mathfrak{g}$. Now, it only remains to prove that it is the only such Lie algebra (up to isomorphism). In other words, it remains to prove that whenever $\mathfrak{g}^{\prime}$ is a contragredient Lie algebra corresponding to $A$, then there exists a $Q$-graded Lie algebra isomorphism $\mathfrak{g}% \rightarrow\mathfrak{g}^{\prime}$ which sends the generators $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\mathfrak{g}$ to the respective generators $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\mathfrak{g}^{\prime}$. So let $\mathfrak{g}^{\prime}$ be a contragredient Lie algebra. Then, condition \textbf{(1)} of Definition \ref{def.contragredient} is satisfied for $\mathfrak{g}^{\prime}$. Thus, the relations (\ref{nonserre-relations}) are satisfied in $\mathfrak{g}^{\prime}$. Define a Lie algebra homomorphism $\psi:\widetilde{\mathfrak{g}}% \rightarrow\mathfrak{g}^{\prime}$ by% \[ \left\{ \begin{array} [c]{c}% \psi\left( e_{i}\right) =e_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} ;\\ \psi\left( f_{i}\right) =f_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} ;\\ \psi\left( h_{i}\right) =h_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }% i\in\left\{ 1,2,...,n\right\} \end{array} \right. . \] This $\psi$ is well-defined because the relations (\ref{nonserre-relations}) are satisfied in $\mathfrak{g}^{\prime}$ (and because $\widetilde{\mathfrak{g}% }=\operatorname*{FreeLie}\left( h_{i},f_{i},e_{i}\ \mid\ i\in\left\{ 1,2,...,n\right\} \right) \diagup\left( \text{the relations (\ref{nonserre-relations})}\right) $). Since the Lie algebra $\mathfrak{g}^{\prime}$ is generated by its elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ (by the definition of a contragredient Lie algebra), the homomorphism $\psi$ is surjective (since all of the elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ clearly lie in the image of $\psi$). \begin{vershort} Since $\mathfrak{g}^{\prime}$ is a contragredient Lie algebra, the condition \textbf{(2)} of Definition \ref{def.contragredient} is satisfied for $\mathfrak{g}^{\prime}$. In other words, the vector space $\mathfrak{g}% ^{\prime}\left[ 0\right] $ has $\left( h_{1},h_{2},...,h_{n}\right) $ as a $\mathbb{C}$-vector space basis, and we have $\mathfrak{g}^{\prime}\left[ \alpha_{i}\right] =\mathbb{C}e_{i}$ and $\mathfrak{g}^{\prime}\left[ -\alpha_{i}\right] =\mathbb{C}f_{i}$ for all $i\in\left\{ 1,2,...,n\right\} $. This yields that the elements $e_{i}$, $f_{i}$ and $h_{i}$ of $\mathfrak{g}^{\prime}$ satisfy% \[ \deg\left( e_{i}\right) =\alpha_{i},\ \ \ \ \ \ \ \ \ \ \deg\left( f_{i}\right) =-\alpha_{i}\ \ \ \ \ \ \ \ \ \ \text{and }\deg\left( h_{i}\right) =0\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{ 1,2,...,n\right\} . \] Of course, the elements $e_{i}$, $f_{i}$ and $h_{i}$ of $\widetilde{\mathfrak{g}}$ satisfy the same relations (because of the definition of the $Q$-grading on $\widetilde{\mathfrak{g}}$). As a consequence, it is easy to see that Lie algebra homomorphism $\psi$ is $Q$-graded\footnote{\textit{Proof.} Let $T$ be the vector subspace of $\widetilde{\mathfrak{g}}$ spanned by the elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$. Then, $\widetilde{\mathfrak{g}}$ is generated by $T$ as a Lie algebra (because $\widetilde{\mathfrak{g}}$ is generated by the elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ as a Lie algebra). Due to the relations% \[ \deg\left( e_{i}\right) =\alpha_{i},\ \ \ \ \ \ \ \ \ \ \deg\left( f_{i}\right) =-\alpha_{i}\ \ \ \ \ \ \ \ \ \ \text{and }\deg\left( h_{i}\right) =0\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{ 1,2,...,n\right\} \] holding both in $\widetilde{\mathfrak{g}}$ and in $\mathfrak{g}^{\prime}$, it is clear that the map $\psi\mid_{T}$ is $Q$-graded. Proposition \ref{prop.generation.Q-gr} (applied to $\widetilde{\mathfrak{g}}$, $\mathfrak{g}^{\prime}$ and $\psi$ instead of $\mathfrak{g}$, $\mathfrak{h}$ and $f$) now yields that $\psi$ is $Q$-graded, qed.}. As a consequence, $\operatorname*{Ker}\psi$ is a $Q$-graded Lie ideal of $\widetilde{\mathfrak{g}}$. \end{vershort} \begin{verlong} Since $\mathfrak{g}^{\prime}$ is a contragredient Lie algebra, the condition \textbf{(2)} of Definition \ref{def.contragredient} is satisfied for $\mathfrak{g}^{\prime}$. In other words, the vector space $\mathfrak{g}% ^{\prime}\left[ 0\right] $ has $\left( h_{1},h_{2},...,h_{n}\right) $ as a $\mathbb{C}$-vector space basis, and we have $\mathfrak{g}^{\prime}\left[ \alpha_{i}\right] =\mathbb{C}e_{i}$ and $\mathfrak{g}^{\prime}\left[ -\alpha_{i}\right] =\mathbb{C}f_{i}$ for all $i\in\left\{ 1,2,...,n\right\} $. This yields that the elements $e_{i}$, $f_{i}$ and $h_{i}$ of $\mathfrak{g}^{\prime}$ satisfy% \[ \deg\left( e_{i}\right) =\alpha_{i},\ \ \ \ \ \ \ \ \ \ \deg\left( f_{i}\right) =-\alpha_{i}\ \ \ \ \ \ \ \ \ \ \text{and }\deg\left( h_{i}\right) =0\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{ 1,2,...,n\right\} . \] Of course, the elements $e_{i}$, $f_{i}$ and $h_{i}$ of $\widetilde{\mathfrak{g}}$ satisfy the same relations (because of the definition of the $Q$-grading on $\widetilde{\mathfrak{g}}$). As a consequence, it is easy to see that Lie algebra homomorphism $\psi$ is $Q$-graded\footnote{\textit{Proof.} Let $T$ be the vector subspace of $\widetilde{\mathfrak{g}}$ spanned by the elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$. Then, $\widetilde{\mathfrak{g}}$ is generated by $T$ as a Lie algebra (because $\widetilde{\mathfrak{g}}$ is generated by the elements $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ as a Lie algebra). Due to the relations% \[ \deg\left( e_{i}\right) =\alpha_{i},\ \ \ \ \ \ \ \ \ \ \deg\left( f_{i}\right) =-\alpha_{i}\ \ \ \ \ \ \ \ \ \ \text{and }\deg\left( h_{i}\right) =0\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{ 1,2,...,n\right\} \] holding in $\widetilde{\mathfrak{g}}$, the subspace $T$ is of $\widetilde{\mathfrak{g}}$ is $Q$-graded, and its homogeneous components are% \[ T\left[ \alpha\right] =\left\{ \begin{array} [c]{l}% \mathbb{C}h_{1}+\mathbb{C}h_{2}+...+\mathbb{C}h_{n}% ,\ \ \ \ \ \ \ \ \ \ \text{if }\alpha=0;\\ \mathbb{C}e_{i},\ \ \ \ \ \ \ \ \ \ \text{if }\alpha=\alpha_{i}\text{ for some }i\in\left\{ 1,2,...,n\right\} ;\\ \mathbb{C}f_{i},\ \ \ \ \ \ \ \ \ \ \text{if }\alpha=-\alpha_{i}\text{ for some }i\in\left\{ 1,2,...,n\right\} ;\\ 0,\ \ \ \ \ \ \ \ \ \ \text{otherwise}% \end{array} \right. \ \ \ \ \ \ \ \ \ \ \text{for all }\alpha\in Q. \] Using this fact, it is straightforward to check that $\psi\left( T\left[ \alpha\right] \right) \subseteq\mathfrak{g}^{\prime}\left[ \alpha\right] $ for every $\alpha\in Q$. In other words, $\left( \psi\mid_{T}\right) \left( T\left[ \alpha\right] \right) \subseteq\mathfrak{g}^{\prime}\left[ \alpha\right] $ for every $\alpha\in Q$. Thus, the map $\psi\mid_{T}$ is $Q$-graded. Proposition \ref{prop.generation.Q-gr} (applied to $\widetilde{\mathfrak{g}}$, $\mathfrak{g}^{\prime}$ and $\psi$ instead of $\mathfrak{g}$, $\mathfrak{h}$ and $f$) now yields that $\psi$ is $Q$-graded, qed.}. As a consequence, $\operatorname*{Ker}\psi$ is a $Q$-graded Lie ideal of $\widetilde{\mathfrak{g}}$. \end{verlong} Define $\widetilde{\mathfrak{h}}$, $I$ and $\iota_{0}$ as in Theorem \ref{thm.gtilde}. Then, $\widetilde{\mathfrak{h}}$ is the free vector space with basis $h_{1},h_{2},...,h_{n}$. Thus, the vector space $\widetilde{\mathfrak{h}}$ is spanned by $h_{1},h_{2},...,h_{n}$. As a consequence, the vector space $\iota_{0}\left( \widetilde{\mathfrak{h}% }\right) $ is spanned by $h_{1},h_{2},...,h_{n}$ (since $\iota_{0}$ maps the elements $h_{1},h_{2},...,h_{n}$ of $\widetilde{\mathfrak{h}}$ to the elements $h_{1},h_{2},...,h_{n}$ of $\widetilde{\mathfrak{g}}$). Now, it is easy to see that $\left( \operatorname*{Ker}\psi\right) \cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =0$\ \ \ \ \footnote{\textit{Proof.} Let $x\in\left( \operatorname*{Ker}\psi\right) \cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. Then, $x\in\operatorname*{Ker}\psi$ and $x\in\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. Since $x\in\iota _{0}\left( \widetilde{\mathfrak{h}}\right) $, there exist some elements $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{n}$ of $\mathbb{C}$ such that $x=\lambda_{1}h_{1}+\lambda_{2}h_{2}+...+\lambda_{n}h_{n}$ (since the vector space $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ is spanned by $h_{1},h_{2},...,h_{n}$). Consider these $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{n}$. Since $x\in\operatorname*{Ker}\psi$, we have $\psi\left( x\right) =0$, so that% \begin{align*} 0 & =\psi\left( x\right) =\psi\left( \lambda_{1}h_{1}+\lambda_{2}% h_{2}+...+\lambda_{n}h_{n}\right) =\lambda_{1}\psi\left( h_{1}\right) +\lambda_{2}\psi\left( h_{2}\right) +...+\lambda_{n}\psi\left( h_{n}\right) \\ & =\lambda_{1}h_{1}+\lambda_{2}h_{2}+...+\lambda_{n}h_{n}% \ \ \ \ \ \ \ \ \ \ \left( \text{since }\psi\left( h_{i}\right) =h_{i}\text{ for every }i\in\left\{ 1,2,...,n\right\} \right) \end{align*} in $\mathfrak{g}^{\prime}$. But since the elements $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\mathfrak{g}^{\prime}$ are linearly independent (because the vector space $\mathfrak{g}^{\prime}\left[ 0\right] $ has $\left( h_{1},h_{2},...,h_{n}\right) $ as a $\mathbb{C}$-vector space basis), this yields that $\lambda_{1}=\lambda_{2}=...=\lambda_{n}=0$. Thus, $x=\lambda _{1}h_{1}+\lambda_{2}h_{2}+...+\lambda_{n}h_{n}$ becomes $x=0h_{1}% +0h_{2}+...+0h_{n}=0$. \par Now forget that we fixed $x$. We thus have seen that every $x\in\left( \operatorname*{Ker}\psi\right) \cap\iota_{0}\left( \widetilde{\mathfrak{h}% }\right) $ satisfies $x=0$. In other words, $\left( \operatorname*{Ker}% \psi\right) \cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =0$, qed.}. Hence, $\operatorname*{Ker}\psi$ is a $Q$-graded Lie ideal of $\widetilde{\mathfrak{g}}$ which has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $. But $I$ is the sum of all $Q$-graded ideals in $\widetilde{\mathfrak{g}}$ which have zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}% }\right) $. Thus, every $Q$-graded ideal of $\widetilde{\mathfrak{g}}$ which has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ must be a subset of $I$. Since $\operatorname*{Ker}\psi$ is a $Q$-graded Lie ideal of $\widetilde{\mathfrak{g}}$ which has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $, this yields that $\operatorname*{Ker}\psi\subseteq I$. We will now prove the reverse inclusion, i. e., we will show that $I\subseteq\operatorname*{Ker}\psi$. We know that $I$ is $Q$-graded (by Theorem \ref{thm.gtilde} \textbf{(h)}). Since $\psi$ is $Q$-graded, this yields that $\psi\left( I\right) $ is a $Q$-graded vector subspace of $\mathfrak{g}^{\prime}$. On the other hand, since $I$ is $Q$-graded, we have $I\left[ 0\right] =I\cap\underbrace{\left( \widetilde{\mathfrak{g}}\left[ 0\right] \right) }_{\substack{=\iota _{0}\left( \widetilde{\mathfrak{h}}\right) \\\text{(by Theorem \ref{thm.gtilde} \textbf{(e)})}}}=I\cap\iota_{0}\left( \widetilde{\mathfrak{h}}\right) =0$ (since Theorem \ref{thm.gtilde} \textbf{(h)} yields that $I$ has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $). Since $\mathfrak{g}^{\prime}$ is a contragredient Lie algebra, the condition \textbf{(3)} of Definition \ref{def.contragredient} is satisfied for $\mathfrak{g}^{\prime}$. In other words, every nonzero $Q$-graded ideal in $\mathfrak{g}^{\prime}$ has a nonzero intersection with $\mathfrak{g}^{\prime }\left[ 0\right] $. Since $I$ is an ideal of $\widetilde{\mathfrak{g}}$, the image $\psi\left( I\right) $ is an ideal of $\mathfrak{g}^{\prime}$ (because $\psi$ is a surjective homomorphism of Lie algebras, and because the image of an ideal under a \textbf{surjective} homomorphism of Lie algebras must always be an ideal of the target Lie algebra). Assume that $\psi\left( I\right) \neq0$. Clearly, $\psi\left( I\right) $ is $Q$-graded (since $I$ is $Q$-graded (by Theorem \ref{thm.gtilde} \textbf{(h)}) and since $\psi$ is $Q$-graded). Thus, $\psi\left( I\right) $ is a nonzero $Q$-graded ideal in $\mathfrak{g}^{\prime}$. Thus, $\psi\left( I\right) $ has a nonzero intersection with $\mathfrak{g}^{\prime}\left[ 0\right] $ (because every nonzero $Q$-graded ideal in $\mathfrak{g}^{\prime}$ has a nonzero intersection with $\mathfrak{g}^{\prime}\left[ 0\right] $). In other words, $\psi\left( I\right) \cap\left( \mathfrak{g}^{\prime}\left[ 0\right] \right) \neq0$. The following is a known and easy fact from linear algebra: If $A$ and $B$ are two $Q$-graded vector spaces, and $\Phi:A\rightarrow B$ is a $Q$-graded linear map, then $\Phi\left( A\left[ \beta\right] \right) =\left( \Phi\left( A\right) \right) \left[ \beta\right] $ for every $\beta\in Q$. Applying this fact to $A=I$, $B=\mathfrak{g}^{\prime}$, $\Phi=\psi$ and $\beta=0$, we obtain $\psi\left( I\left[ 0\right] \right) =\left( \psi\left( I\right) \right) \left[ 0\right] $. But since $I\left[ 0\right] =0$, this rewrites as $\psi\left( 0\right) =\left( \psi\left( I\right) \right) \left[ 0\right] $. Hence, $\left( \psi\left( I\right) \right) \left[ 0\right] =\psi\left( 0\right) =0$. But since $\psi\left( I\right) $ is a $Q$-graded vector subspace of $\mathfrak{g}^{\prime}$, we have $\psi\left( I\right) \cap\left( \mathfrak{g}^{\prime}\left[ 0\right] \right) =\left( \psi\left( I\right) \right) \left[ 0\right] =0$. This contradicts the fact that $\psi\left( I\right) \cap\left( \mathfrak{g}^{\prime}\left[ 0\right] \right) \neq0$. Hence, our assumption (that $\psi\left( I\right) \neq0$) must have been wrong. In other words, $\psi\left( I\right) =0$, so that $I\subseteq \operatorname*{Ker}\psi$. Combined with $\operatorname*{Ker}\psi\subseteq I$, this yields $I=\operatorname*{Ker}\psi$. Since the $Q$-graded Lie algebra homomorphism $\psi:\widetilde{\mathfrak{g}% }\rightarrow\mathfrak{g}^{\prime}$ is surjective, it factors (according to the homomorphism theorem) through a $Q$-graded Lie algebra isomorphism $\widetilde{\mathfrak{g}}\diagup\left( \operatorname*{Ker}\psi\right) \rightarrow\mathfrak{g}^{\prime}$. Since $\widetilde{\mathfrak{g}}% \diagup\underbrace{\left( \operatorname*{Ker}\psi\right) }_{=I}% =\widetilde{\mathfrak{g}}\diagup I=\mathfrak{g}$, this means that $\psi$ factors through a $Q$-graded Lie algebra isomorphism $\mathfrak{g}% \rightarrow\mathfrak{g}^{\prime}$. This $Q$-graded Lie algebra isomorphism $\mathfrak{g}\rightarrow\mathfrak{g}^{\prime}$ clearly sends the generators $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\mathfrak{g}$ to the respective generators $e_{1}% $, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}% $, $...$, $h_{n}$ of $\mathfrak{g}^{\prime}$. We have thus proven that there exists a $Q$-graded Lie algebra isomorphism $\mathfrak{g}\rightarrow\mathfrak{g}^{\prime}$ which sends the generators $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ of $\mathfrak{g}$ to the respective generators $e_{1}% $, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}% $, $...$, $h_{n}$ of $\mathfrak{g}^{\prime}$. This completes the proof of Theorem \ref{thm.g(A).exuni} \textbf{(a)}. \textbf{(b)} Let $A$ be the Cartan matrix of a simple finite-dimensional Lie algebra. Clearly it is enough to prove that this Lie algebra is a contragredient Lie algebra corresponding to $A$, that is, is generated by $e_{1}$, $e_{2}$, $...$, $e_{n}$, $f_{1}$, $f_{2}$, $...$, $f_{n}$, $h_{1}$, $h_{2}$, $...$, $h_{n}$ as a Lie algebra and satisfies the conditions \textbf{(1)}, \textbf{(2)} and \textbf{(3)} of Definition \ref{def.contragredient}. But this follows from the standard theory of roots of simple finite-dimensional Lie algebras\footnote{For instance, condition \textbf{(3)} follows from the fact that the Lie algebra in question is simple and thus contains no ideals other than $0$ and itself.}. Theorem \ref{thm.g(A).exuni} \textbf{(b)} is thus proven. \begin{remark} Let $A=\left( a_{i,j}\right) _{1\leq i,j\leq n}$ be a complex $n\times n$ matrix such that every $i\in\left\{ 1,2,...,n\right\} $ satisfies $a_{i,i}=2$. One can show that the Lie algebra $\mathfrak{g}\left( A\right) $ is finite-dimensional if and only if $A$ is the Cartan matrix of a semisimple finite-dimensional Lie algebra. (In this case, $\mathfrak{g}\left( A\right) $ is exactly this semisimple Lie algebra, and the ideal $I$ of Theorem \ref{thm.gtilde} is generated by the left hand sides $\left( \operatorname*{ad}\left( e_{i}\right) \right) ^{1-a_{i,j}}e_{j}$ and $\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}$ of the Serre relations.) \end{remark} [...] [Add something about the total degree on $\widetilde{\mathfrak{g}}$, since this will later be used for the bilinear form. $\widetilde{\mathfrak{g}% }\left[ \operatorname*{tot}0\right] =\widetilde{\mathfrak{g}}\left[ 0\right] =\widetilde{\mathfrak{h}}$, $\widetilde{\mathfrak{g}}\left[ \operatorname*{tot}<0\right] ...$, $\widetilde{\mathfrak{g}}\left[ 1\right] =...$] \begin{remark} \label{rmk.g(A+A)}Let $A_{1}$ and $A_{2}$ be two square complex matrices. As usual, we denote by $A_{1}\oplus A_{2}$ the block-diagonal matrix $\left( \begin{array} [c]{cc}% A_{1} & 0\\ 0 & A_{2}% \end{array} \right) $. Then, $\mathfrak{g}\left( A_{1}\oplus A_{2}\right) \cong\mathfrak{g}\left( A_{1}\right) \oplus\mathfrak{g}\left( A_{2}\right) $ as Lie algebras naturally. \end{remark} \textit{Proof of Remark \ref{rmk.g(A+A)} (sketched).} Say $A_{1}$ is an $\ell\times\ell$ matrix, and $A_{2}$ is an $m\times m$ matrix. Let $n=\ell+m$ and $A=A_{1}\oplus A_{2}$. Introduce the notations $\widetilde{\mathfrak{g}}$, $\widetilde{\mathfrak{h}}$, $\widetilde{\mathfrak{n}}_{+}$, $\widetilde{\mathfrak{n}}_{-}$, $\iota_{0}$, $\iota_{+}$, $\iota_{-}$ and $I$ as in Theorem \ref{thm.gtilde}. Let $\mathfrak{j}_{+}$ be the ideal of the Lie algebra $\widetilde{\mathfrak{n}}_{+}$ generated by all elements of the form $\left[ e_{i},e_{j}\right] $ with $i\in\left\{ 1,2,...,\ell\right\} $ and $j\in\left\{ \ell+1,\ell+2,...,n\right\} $. Let $\mathfrak{j}_{-}$ be the ideal of the Lie algebra $\widetilde{\mathfrak{n}}_{-}$ generated by all elements of the form $\left[ f_{i},f_{j}\right] $ with $i\in\left\{ 1,2,...,\ell\right\} $ and $j\in\left\{ \ell+1,\ell+2,...,n\right\} $. Prove that $\iota_{+}\left( \mathfrak{j}_{+}\right) $ and $\iota_{-}\left( \mathfrak{j}_{-}\right) $ are actually $Q$-graded ideals of $\widetilde{\mathfrak{g}}$ (and not only of $\iota_{+}\left( \widetilde{\mathfrak{n}}_{+}\right) $ and $\iota_{-}\left( \widetilde{\mathfrak{n}}_{-}\right) $), so that both $\iota_{+}\left( \mathfrak{j}_{+}\right) $ and $\iota_{-}\left( \mathfrak{j}_{-}\right) $ are subsets of $I$. For every $i\in\left\{ 1,2\right\} $, let $\widetilde{\mathfrak{g}}_{i}$ be the Lie algebra constructed analogously to $\widetilde{\mathfrak{g}}$ but for the matrix $A_{i}$ instead of $A$. Notice that $\widetilde{\mathfrak{g}}\diagup\left( \iota_{+}\left( \mathfrak{j}% _{+}\right) +\iota_{-}\left( \mathfrak{j}_{-}\right) \right) \cong\widetilde{\mathfrak{g}}_{1}\oplus\widetilde{\mathfrak{g}}_{2}$. Conclude the proof by noticing that if $J$ is a $Q$-graded ideal in $\widetilde{\mathfrak{g}}$ which has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $, and $K$ is the sum of all $Q$-graded ideals in $\widetilde{\mathfrak{g}}\diagup J$ which have zero intersection with the projection of $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ on $\widetilde{\mathfrak{g}}\diagup J$, then $\left( \widetilde{\mathfrak{g}% }\diagup J\right) \diagup K\cong\widetilde{\mathfrak{g}}\diagup I=\mathfrak{g}$. The details are left to the reader. \subsection{\textbf{[unfinished]} Kac-Moody algebras for generalized Cartan matrices} For general $A$, we do not know much about $\mathfrak{g}\left( A\right) $; its definition was not even constructive (find that $I$ !). It is not known in general how to obtain generators for $I$. But for some particular cases -- not only Cartan matrices of semisimple Lie algebras --, things behave well. Here is the most important such case: \begin{definition} An $n\times n$ matrix $A=\left( a_{i,j}\right) _{1\leq i,j\leq n}$ of complex numbers is said to be a \textit{generalized Cartan matrix} if it satisfies: \textbf{(1)} We have $a_{i,i}=2$ for all $i\in\left\{ 1,2,...,n\right\} $. \textbf{(2)} For every $i$ and $j$, the number $a_{i,j}$ is a nonpositive integer. Also, $a_{i,j}=0$ if and only if $a_{j,i}=0$. \textbf{(3)} The matrix $A$ is symmetrizable, i. e., there exists a diagonal matrix $D>0$ such that $\left( DA\right) ^{T}=DA$. \end{definition} Note that a Cartan matrix is the same as a generalized Cartan matrix $A$ with $DA>0$. \begin{example} Let $A=\left( \begin{array} [c]{cc}% 2 & -m\\ -1 & 2 \end{array} \right) $ for $m\geq1$. This matrix $A$ is a generalized Cartan matrix, since $\left( \begin{array} [c]{cc}% 1 & 0\\ 0 & m \end{array} \right) \left( \begin{array} [c]{cc}% 2 & -m\\ -1 & 2 \end{array} \right) =\left( \begin{array} [c]{cc}% 2 & -m\\ -m & 2m \end{array} \right) $. Note that $\det\left( \begin{array} [c]{cc}% 2 & -m\\ -1 & 2 \end{array} \right) =4-m$. For $m=1$, we have $\mathfrak{g}\left( A\right) \cong A_{2}=\mathfrak{sl}% _{3}$. For $m=2$, we have $\mathfrak{g}\left( A\right) \cong B_{2}\cong C_{2}% \cong\mathfrak{sp}_{4}\cong\mathfrak{so}_{5}$. For $m=3$, we have $\mathfrak{g}\left( A\right) \cong G_{2}$. For $m\geq4$, the Lie algebra $\mathfrak{g}\left( A\right) $ is infinite-dimensional. For $m=4$, it is a twisted version of $\widehat{\mathfrak{sl}_{2}}$, called $A_{2}^{2}$. For $m\geq5$, the Lie algebra $\mathfrak{g}\left( A\right) $ is big (in the sense of having exponential growth). This strange behaviour is related to the behaviour of the $m$-subspaces problem (finite for $m\leq3$, tame for $m=4$, wild for $m\geq5$). More generally, Kac-Moody algebras are related to representation theory of quivers. \end{example} \begin{definition} A \textit{symmetrizable Kac-Moody algebra} is a Lie algebra of the form $\mathfrak{g}\left( A\right) $ for a generalized Cartan matrix $A$. \end{definition} \begin{theorem} [Gabber-Kac]\label{thm.g(A).gabber-kac}If $A$ is a generalized Cartan matrix, then the ideal $I\subseteq\widetilde{\mathfrak{g}}\left( A\right) $ is generated by the Serre relations (where the notation $I$ comes from Theorem \ref{thm.gtilde}). \end{theorem} \textit{Partial proof of Theorem \ref{thm.g(A).gabber-kac}.} Proving this theorem requires showing two assertions: first, that the Serre relations are contained in $I$; second, that they actually generate $I$. We will only prove the first of these two assertions. Set $I_{+}=I\cap\widetilde{\mathfrak{n}}_{+}$ and $I_{-}=I\cap \widetilde{\mathfrak{n}}_{-}$. Denote $\widetilde{\mathfrak{g}}\left( A\right) $ by $\widetilde{\mathfrak{g}}$ as in Theorem \ref{thm.gtilde}. We know (from Theorem \ref{thm.gtilde} \textbf{(h)}) that $I$ is a $Q$-graded ideal in $\widetilde{\mathfrak{g}}$ which has zero intersection with $\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ (where the notations are those of Theorem \ref{thm.gtilde}). Since $\widetilde{\mathfrak{g}}\left[ 0\right] =\iota_{0}\left( \widetilde{\mathfrak{h}}\right) $ (by Theorem \ref{thm.gtilde} \textbf{(e)}), this rewrites as follows: $I$ is a $Q$-graded ideal in $\widetilde{\mathfrak{g}}$ which has zero intersection with $\widetilde{\mathfrak{g}}\left[ 0\right] $. Thus, $I=I_{+}\oplus I_{-}$. Let us show that $\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}\in I_{-}$. To do that, it is sufficient to show that $\left[ e_{k},\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}\right] =0$ for all $k$. (If we grade $\widetilde{\mathfrak{g}}$ by setting $\deg\left( f_{i}\right) =-1$, $\deg\left( e_{i}\right) =1$ and $\deg\left( h_{i}\right) =0$ (this is called the \textit{principal grading}), then $f_{k}$ can only lower degree, so that the Lie ideal generated by $\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}$ will lie entirely in negative degrees, and thus $\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}$ will lie in $I_{-}$.) \textit{Case 1:} We have $k\neq i,j$. This case is clear since $e_{k}$ commutes with $f_{i}$ and $f_{j}$ (by our relations). \textit{Case 2:} We have $k=j$. In this case, \begin{align*} \left[ e_{k},\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}\right] & =\left[ e_{j},\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}\right] =\left( \operatorname*{ad}% \left( f_{i}\right) \right) ^{1-a_{i,j}}\left( \left[ e_{j},f_{j}\right] \right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since }\operatorname*{ad}\left( f_{i}\right) \text{ and }\operatorname*{ad}\left( e_{j}\right) \text{ commute, due to }i\neq j\right) \\ & =\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}% h_{j}. \end{align*} We now distinguish between two cases according to whether $a_{i,j}$ is $=0$ or $<0$: \textit{Case 2a:} We have $a_{i,j}=0$. Then, $a_{j,i}=0$ by the definition of generalized Cartan matrices. Thus, $\left[ f_{i},h_{j}\right] =-\left[ h_{j},f_{i}\right] =-a_{j,i}f_{i}=0$, and we are done. \textit{Case 2b:} We have $a_{i,j}<0$. Then, $1-a_{i,j}\geq2$. Now, $\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{2}h_{j}=\left( \operatorname*{ad}\left( f_{i}\right) \right) \left( cf_{i}\right) =0$ for some constant $c$. \textit{Case 3:} We have $k=i$. Let $\left( \mathfrak{sl}_{2}\right) _{i}=\left\langle e_{i},f_{i},h_{i}\right\rangle $. Let $M$ be the $\left( \mathfrak{sl}_{2}\right) _{i}$-submodule in $\widetilde{\mathfrak{g}}\left( A\right) $ generated by $f_{j}$. We have $\left[ h_{i},f_{j}\right] =-a_{i,j}f_{j}=mf_{j}$, where $m=-a_{i,j}\geq0$. Together with $\left[ e_{i},f_{j}\right] =0$, this shows that $f_{j}=:v$ is a highest-weight vector of $M$ with weight $m$. Thus, $f_{i}^{m+1}v=\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}$ is a singular vector for $\left( \mathfrak{sl}_{2}\right) _{i}$ (by representation theory of $\mathfrak{sl}_{2}$\ \ \ \ \footnote{What we are using is the following: Consider the module $M_{\lambda}=\mathbb{C}% \left[ f\right] v$ over $\mathfrak{sl}_{2}$. Then, $ef^{n}v=n\left( \lambda-n+1\right) f^{n-1}v$. Thus, when $n=m+1$ and $\lambda=m$, we get $ef^{n}v=0$.}). So much for our part of the proof of Theorem \ref{thm.g(A).gabber-kac}. Of course, simple Lie algebras are Kac-Moody algebras. The next class of Kac-Moody algebras we are interested in is the \textit{affine Lie algebras}: \begin{remark} Let $\sigma\in S_{n}$ be a permutation, and $A$ be an $n\times n$ complex matrix. Then, $\mathfrak{g}\left( A\right) \cong\mathfrak{g}\left( \sigma A\sigma^{-1}\right) $. \end{remark} \begin{definition} A generalized Cartan matrix $A$ is said to be \textit{indecomposable} if it cannot be written in the form $\sigma\left( A_{1}\oplus A_{2}\right) \sigma^{-1}$ for some permutation $\sigma$ and nontrivial square matrices $A_{1}$ and $A_{2}$. Due to the above remark and to Remark \ref{rmk.g(A+A)}, we need to only consider indecomposable generalized Cartan matrices. \end{definition} \begin{definition} A generalized Cartan matrix $A$ is said to be \textit{affine} if $DA\geq0$ but $DA\not > 0$ (thus, $\det\left( DA\right) =0$). \end{definition} \begin{definition} If $A$ is an affine generalized Cartan matrix, then $\mathfrak{g}\left( A\right) $ is called an \textit{affine Kac-Moody algebra}. \end{definition} Now let $A$ be the (usual) Cartan matrix of a simple Lie algebra, and let $\mathfrak{g}=\mathfrak{g}\left( A\right) $ be this simple Lie algebra. Let $L\mathfrak{g}=\mathfrak{g}\left[ t,t^{-1}\right] $, and let $\widehat{\mathfrak{g}}=L\mathfrak{g}\oplus\mathbb{C}K$ as defined long ago. \begin{theorem} This $\widehat{\mathfrak{g}}$ is an affine Kac-Moody algebra with generalized Cartan matrix $\widetilde{A}$ whose $\left( 1,1\right) $-entry is $2$ and whose submatrix obtained by omitting the first row and the first column is $A$. (We do not yet say what the remaining entries are.) \end{theorem} \textit{Proof of Theorem.} Let $\mathfrak{h}$ be the Cartan subalgebra of $\mathfrak{g}$. Let $r=\dim\mathfrak{h}$; thus, $r$ is the rank of $\mathfrak{g}$. Let $\left( h_{1},h_{2},...,h_{r}\right) $ be a corresponding basis of $\mathfrak{h}$, and let $e_{i},f_{i}$ be standard generators for every $i\in\left\{ 1,2,...,r\right\} $. Let $\theta$ be the maximal root. Let us now define elements $e_{0}=f_{\theta}\cdot t$, $f_{0}=e_{\theta}\cdot t^{-1}$ and $h_{0}=\left[ e_{0},f_{0}\right] =-h_{\theta}% +\underbrace{\left( f_{\theta},e_{\theta}\right) }_{=1\text{ (due to our normalization)}}K=K-h_{\theta}$ of $\widehat{\mathfrak{g}}$ (the commutator is computed in $\widehat{\mathfrak{g}}$, not in $L\mathfrak{g}$). Add these elements to our system of generators. Why do we then get a system of generators of $\widehat{\mathfrak{g}}$ ? First, $h_{i}$ for $i\in\left\{ 0,1,...,r\right\} $ are a basis of $\widehat{\mathfrak{h}}=\mathfrak{h}\oplus\mathbb{C}K$. Also, $\mathfrak{g}t^{0}$ is generated by $e_{i},f_{i},h_{i}$ for $i\in\left\{ 1,2,...,r\right\} $. Now, $\mathfrak{g}t^{1}$ is an irreducible $\mathfrak{g}$-module with lowest-weight vector $f_{\theta}\cdot t$. $\Longrightarrow$ $U\left( \mathfrak{g}\right) \cdot f_{\theta }t=\mathfrak{g}t$. Now, $\mathfrak{g}t$ generates $\mathfrak{g}t\mathbb{C}% \left[ t\right] $ (since $\left[ \mathfrak{g},\mathfrak{g}\right] =\mathfrak{g}$). Similarly, $U\left( \mathfrak{g}\right) \cdot e_{\theta }t^{-1}=\mathfrak{g}t^{-1}$, and $\mathfrak{g}t^{-1}$ generates $\mathfrak{g}% t^{-1}\mathbb{C}\left[ t^{-1}\right] $. $\Longrightarrow$ our $e_{i}$, $f_{i}$, $h_{i}$ (including $i=0$) generate all of $\widehat{\mathfrak{g}}$. Now to the relations. $\left[ h_{i},h_{j}\right] =0$ is clear for all $\left( i,j\right) \in\left\{ 0,1,...,r\right\} ^{2}$. We have $\left[ h_{0},e_{0}\right] =\left[ K-h_{\theta},f_{\theta}t\right] =-\left[ h_{\theta},f_{\theta}\right] t=2f_{\theta}t=2e_{0}$. We have $\left[ h_{0},f_{0}\right] =-2f_{0}$ similarly. We have $\left[ e_{0},f_{0}\right] =h_{0}$. We have $\left[ h_{0},e_{i}\right] =\left[ K-h_{\theta},e_{i}\right] =-\alpha_{i}\left( h_{\theta}\right) e_{i}=-\left( \alpha_{i}% ,\theta\right) e_{i}$ $\Longrightarrow$ $a_{0,i}=-\left( \alpha_{i}% ,\theta\right) =\left( \text{some nonpositive integer}\right) $. We have $\left[ h_{0},f_{i}\right] =\left( \alpha_{i},\theta\right) f_{i}% $, same argument. We have $\left[ h_{i},e_{0}\right] =\left[ h_{i},f_{\theta}t\right] =-\theta\left( h_{i}\right) f_{\theta}t=-\theta\left( h_{i}\right) e_{0}=-\left( \alpha_{i}^{\vee},\theta\right) e_{0}$ (where $\alpha _{i}^{\vee}=\dfrac{2\alpha_{i}}{\left( \alpha_{i},\alpha_{i}\right) }$) $\Longrightarrow$ $a_{i,0}=-\left( \alpha_{i}^{\vee},\theta\right) $. We have $\left[ h_{i},f_{0}\right] =\left( \alpha_{i}^{\vee},\theta\right) f_{0}$, same argument. We have $\left[ e_{0},f_{i}\right] =\left[ f_{\theta}t,f_{i}\right] =0$. We have $\left[ e_{i},f_{0}\right] =\left[ e_{i},e_{\theta}t^{-1}\right] =0$. Thus, all basic relations are satisfied. Now let us define a grading: $\widehat{Q}=Q\oplus\mathbb{Z}\delta$, where $Q$ is the root lattice of $\mathfrak{g}$. Define $\alpha_{0}=\delta-\theta$. $\delta\mid_{\widehat{\mathfrak{h}}}=0$. So if we think of $\alpha_{0}$ as an element of $\widehat{\mathfrak{h}}^{\ast}$, then $\alpha_{0},\alpha _{1},...,\alpha_{r}$ is neither linearly independent nor spanning. So the direct sum $Q\oplus\mathbb{Z}\delta$ is an external direct sum, not an internal one!! $\widehat{Q}$-grading: $\deg\left( e_{i}\right) =\alpha_{i}$, $\deg\left( f_{i}\right) =-\alpha_{i}$ and $\deg\left( h_{i}\right) =0$ for $i=0,1,...,r$. Also $\deg\left( at^{k}\right) =\deg a+k\delta$ (so, so to speak, ``$\deg t=\delta$''). So we have $\widehat{\mathfrak{g}}\left[ 0\right] =\widehat{\mathfrak{h}}$ and $\widehat{\mathfrak{g}}\left[ \alpha_{i}\right] =\left\langle e_{i}\right\rangle $ and $\widehat{\mathfrak{g}}\left[ -\alpha_{i}\right] =\left\langle f_{i}\right\rangle $. Note (which we won't use): $\left[ h,a\right] =\alpha\left( h\right) a$, $a\in\widehat{\mathfrak{g}}\left[ \alpha\right] $ ``if you define things this way''. The only thing we now have to do is to show that $I=0$ in $\widehat{\mathfrak{g}}$. Let $\overline{I}$ be the projection of $I$ to $L\mathfrak{g}% =\widehat{\mathfrak{g}}\diagup\left( K\right) $. Clearly, $\overline{I}% \cap\mathfrak{h}=0$. We must prove that $\overline{I}=0$. But there is a \textbf{claim} that any $\widehat{Q}$-graded ideal in $L\mathfrak{g}$ is $0$ or $L\mathfrak{g}$. (\textit{Proof:} If $J$ is a $\widehat{Q}$-graded ideal of $L\mathfrak{g}$ different from $0$, then there exists a nonzero $a\in\mathfrak{g}$ and an $m\in\mathbb{Z}$ such that $at^{m}\in J$. But $at^{m}$ generates $L\mathfrak{g}$ under the action of $L\mathfrak{g}$, since $\left[ bt^{n-m},at^{m}\right] =\left[ b,a\right] t^{n}$ and $\mathfrak{g}=\left[ \mathfrak{g},\mathfrak{g}\right] $.) Proof of Theorem complete. Let us show how Dynkin diagrams look like for these affine Kac-Moody algebras. Consider the case of $A_{n-1}=\mathfrak{sl}_{n}$. Then, $\theta=\left( 1,0,0,...,0,-1\right) $. Also, $\alpha_{1}=\left( 1,-1,0,0,...,0\right) $, $\alpha_{2}=\left( 0,1,-1,0,0,...,0\right) $, $...$, $\alpha_{n-1}=\left( 0,0,...,0,1,-1\right) $. Also, $\alpha=\alpha^{\vee}$ for all simple roots $\alpha$. We thus have $\left( \theta,\alpha_{i}\right) =1$ if $\alpha \in\left\{ 1,n-1\right\} $ and $=0$ otherwise. The Dynkin diagram of $\widehat{A_{n-1}}=A_{n-1}^{1}=\widehat{\mathfrak{sl}_{n}}$ (these are just three notations for one and the same thing) is thus $% %TCIMACRO{\TeXButton{x}{\xymatrix{ %\circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{}[r]|-{...} & \circ\ar@ %{-}[r] & \circ\ar@{-}[r] & \circ}}}% %BeginExpansion \xymatrix{ \circ\ar@{-}[r] & \circ\ar@{-}[r] & \circ\ar@{}[r]|-{...} & \circ\ar@ {-}[r] & \circ\ar@{-}[r] & \circ}% %EndExpansion $ with a cyclically connected dot underneath. The case $n=2$ is special: double link. $\circ=\circ$ double link. Now let us consider other types. Suppose that $\theta$ is a fundamental weight, i. e., satisfies $\left( \theta,\alpha_{i}^{\vee}\right) =1$ for some $i$ and satisfies $\left( \theta,\alpha_{i}^{\vee}\right) =0$ for all other $i$. (This happens for a lot of simple Lie algebras.) To get $\widehat{D_{n}}=\widehat{\mathfrak{so}_{2n}}$, need to attach a new vertex to the second vertex from the left. To get $\widehat{C_{n}}=\widehat{\mathfrak{sp}_{2n}}$, need to attach a new vertex \textbf{doubly-linked} to the first vertex from the left. (The arrow points to the right, i. e., to the $C_{n}$ diagram.) For $\widehat{G_{2}}$, attach a vertex on the left (where the arrow points to the right). For $\widehat{F_{4}}$, attach a vertex on the left (where the arrow points to the right). For $\widehat{E_{6}}$, attach a vertex to the ``bottom'' (the vertex off the line). For $\widehat{E_{7}}$, attach a vertex to the short leg (to make the graph symmetric). For $\widehat{E_{8}}$, attach a vertex to the long leg. These are untwisted affine Lie algebras ($\widehat{\mathfrak{g}}$). There are also twisted ones: $A_{2}^{2}$ with Cartan matrix $\left( \begin{array} [c]{cc}% 2 & -4\\ -1 & 2 \end{array} \right) $ and Dynkin diagram $\circ\left( 4\text{ arrows pointing rightward}\right) \circ$. We will not discuss this kind of Lie algebras here. \subsection{\textbf{[unfinished]} Representation theory of \texorpdfstring{$\mathfrak{g}\left( A\right) $}{g(A)}} We will now work out the representation theory of $\mathfrak{g}\left( A\right) $. Let us start with the case of $\mathfrak{g}\left( A\right) $ being finite-dimensional. In contrast with usual courses on Lie algebras, we will not restrict ourselves to finite-dimensional representations. We define a Category $\mathcal{O}$ which is analogous but (in its details) somewhat different from the one we defined above. In future, we will use only the new definition. \begin{definition} \label{def.O}The objects of \textit{category }$\mathcal{O}$ will be $\mathfrak{g}$-modules $M$ such that: \textbf{1)} The module $M$ is $\mathfrak{h}$-diagonalizable. By this we mean that $M=\bigoplus\limits_{\mu\in\mathfrak{h}^{\ast}}M\left[ \mu\right] $ (where $M\left[ \mu\right] $ means the $\mu$-weight space of $M$), and every $\mu\in\mathfrak{h}^{\ast}$ satisfies $\dim\left( M\left[ \mu\right] \right) <\infty$. \textbf{2)} Let $\operatorname*{Supp}M$ denote the set of all $\mu \in\mathfrak{h}^{\ast}$ such that $M\left[ \mu\right] \neq0$. Then, there exist finitely many $\lambda_{1},\lambda_{2},...,\lambda_{n}\in\mathfrak{h}% ^{\ast}$ such that $\operatorname*{Supp}M\subseteq D\left( \lambda _{1}\right) \cup D\left( \lambda_{2}\right) \cup...\cup D\left( \lambda_{n}\right) $, where for every $\lambda\in\mathfrak{h}^{\ast}$, we denote by $D\left( \lambda\right) $ the subset% \[ \left\{ \lambda-k_{1}\alpha_{1}-k_{2}\alpha_{2}-...-k_{r}\alpha_{r}% \ \mid\ \left( k_{1},k_{2},...,k_{r}\right) \in\mathbb{N}^{r}\right\} \ \ \ \ \ \ \ \ \ \ \text{of }\mathfrak{h}^{\ast}. \] The \textit{morphisms of category }$\mathcal{O}$ will be $\mathfrak{g}$-module homomorphisms. \end{definition} Examples of modules in Category $\mathcal{O}$ are Verma modules $M_{\lambda }=M_{\lambda}^{+}$ and their irreducible quotients $L_{\lambda}$ (and all of their quotients). Category $\mathcal{O}$ is an abelian category (in our case, this simply means it is closed under taking subquotients and direct sums). \begin{definition} Let $M\in\mathcal{O}$ be a $\mathfrak{g}$-module. Then, the \textit{formal character} of $M$ denotes the sum $\operatorname*{ch}M=\sum\limits_{\mu \in\mathfrak{h}^{\ast}}\dim\left( M\left[ \mu\right] \right) e^{\mu}$. Here $\mathbb{C}\left[ \mathfrak{h}^{\ast}\right] $ denotes the group algebra of the additive group $\mathfrak{h}^{\ast}$, where this additive group $\mathfrak{h}^{\ast}$ is written multiplicatively and every $\mu \in\mathfrak{h}^{\ast}$ is renamed as $e^{\mu}$. Where does this sum $\sum\limits_{\mu\in\mathfrak{h}^{\ast}}\dim\left( M\left[ \mu\right] \right) e^{\mu}$ lie? Let $\Gamma$ be a coset of $Q$ (the root lattice) in $\mathfrak{h}^{\ast}$. Then, let $R_{\Gamma}$ denote the space $\lim\limits_{\mu\in\Gamma}e^{\mu }\mathbb{C}\left[ \left[ e^{-\alpha_{1}},e^{-\alpha_{2}},...,e^{-\alpha_{r}% }\right] \right] $ (this is a union, but not a disjoint union, since $R_{\mu}\subseteq R_{\mu+\alpha_{i}}$ for all $i$ and $\mu$). Let $R=\bigoplus\limits_{\Gamma\in\mathfrak{h}^{\ast}\diagup Q}R_{\Gamma}$. This $R$ is a ring. We view $\operatorname*{ch}M$ as an element of $R$. \end{definition} Now, for an example, let us compute the formal character $\operatorname*{ch}% \left( M_{\lambda}\right) $ of the Verma module $M_{\lambda}=U\left( \mathfrak{n}_{-}\right) v_{\lambda}$. Recall that $U\left( \mathfrak{n}_{-}\right) $ has a Poincar\'{e}-Birkhoff-Witt basis consisting of all elements of the form $f_{\alpha^{\left( 1\right) }}^{m_{1}}f_{\alpha^{\left( 2\right) }}% ^{m_{2}}...f_{\alpha^{\left( \ell\right) }}^{m_{\ell}}$ where $\alpha ^{\left( 1\right) },\alpha^{\left( 2\right) },...,\alpha^{\left( \ell\right) }$ are all positive roots of $\mathfrak{g}$, and $\ell =\dim\left( \mathfrak{n}_{-}\right) $. The weight of this element $f_{\alpha^{\left( 1\right) }}^{m_{1}}f_{\alpha^{\left( 2\right) }}% ^{m_{2}}...f_{\alpha^{\left( \ell\right) }}^{m_{\ell}}$ is $-\left( m_{1}\alpha^{\left( 1\right) }+m_{2}\alpha^{\left( 2\right) }+...+m_{\ell }\alpha^{\left( \ell\right) }\right) $. Thus, the weight of $f_{\alpha ^{\left( 1\right) }}^{m_{1}}f_{\alpha^{\left( 2\right) }}^{m_{2}% }...f_{\alpha^{\left( \ell\right) }}^{m_{\ell}}v_{\lambda}$ is $\lambda-\left( m_{1}\alpha^{\left( 1\right) }+m_{2}\alpha^{\left( 2\right) }+...+m_{\ell}\alpha^{\left( \ell\right) }\right) $. Thus, $\dim\left( M_{\lambda}\left[ \lambda-\beta\right] \right) $ is the number of partitions of $\beta$ into positive roots. We denote this by $p\left( \beta\right) $, and call $p$ the \textit{Kostant partition function}. Now, it is very easy (using geometric series) to see that% \[ \sum\limits_{\beta\in Q_{+}}p\left( \beta\right) e^{-\beta}=\prod \limits_{\substack{\alpha\text{ root;}\\a>0}}\dfrac{1}{1-e^{-\alpha}}. \] Thus,% \[ \operatorname*{ch}\left( M_{\lambda}\right) =\sum\limits_{\beta\in Q_{+}% }p\left( \beta\right) e^{\lambda-\beta}=e^{\lambda}\underbrace{\sum \limits_{\beta\in Q_{+}}p\left( \beta\right) e^{-\beta}}_{=\prod \limits_{\substack{\alpha\text{ root;}\\a>0}}\dfrac{1}{1-e^{-\alpha}}% }=e^{\lambda}\prod\limits_{\substack{\alpha\text{ root;}\\a>0}}\dfrac {1}{1-e^{-\alpha}}. \] \textbf{Example:} Let $\mathfrak{g}=\mathfrak{sl}_{2}$. Then,% \[ \operatorname*{ch}\left( M_{\lambda}\right) =\dfrac{e^{\lambda}% }{1-e^{-\alpha}}=e^{\lambda}+e^{\lambda-\alpha}+e^{\lambda-2\alpha}+.... \] Classically, one identifies weights of $\mathfrak{sl}_{2}$ with elements of $\mathbb{C}$ (by $\omega_{1}\mapsto1$ and thus $\alpha\mapsto2$). Write $x$ for $e^{\omega_{1}}$. Then,% \[ \operatorname*{ch}\left( M_{\lambda}\right) =\dfrac{x^{\lambda}}{1-x^{-2}% }=x^{\lambda}+x^{\lambda-2}+x^{\lambda-4}+.... \] The quotient $L_{\lambda}$ has weights $\lambda$, $\lambda-2$, $...$, $-\lambda$ and thus satisfies% \[ \operatorname*{ch}\left( L_{\lambda}\right) =x^{\lambda}+x^{\lambda -2}+...+x^{-\lambda}=\dfrac{x^{\lambda+1}-x^{-\lambda-1}}{x-x^{-1}}. \] Back to the general case of finite-dimensional $\mathfrak{g}\left( A\right) $. First of all, category $\mathcal{O}$ has tensor products, and they make it into a tensor category. \begin{proposition} \textbf{1)} We have $\operatorname*{ch}\left( M_{1}\otimes M_{2}\right) =\operatorname*{ch}\left( M_{1}\right) \cdot\operatorname*{ch}\left( M_{2}\right) $. \textbf{2)} If $N\subseteq M$ are both in $\mathcal{O}$, then $\operatorname*{ch}M=\operatorname*{ch}N+\operatorname*{ch}\left( M\diagup N\right) $. \end{proposition} \textit{Proof of Proposition.} \textbf{1)} \[ \left( M_{1}\otimes M_{2}\right) \left[ \mu\right] =\bigoplus \limits_{\mu_{1}+\mu_{2}=\mu}M_{1}\left[ \mu_{1}\right] \otimes M_{2}\left[ \mu_{2}\right] . \] \textbf{2)} \[ \left( M\diagup N\right) \left[ \mu\right] =M\left[ \mu\right] \diagup N\left[ \mu\right] . \] Now, let us generalize to the case of Kac-Moody Lie algebras (or $\mathfrak{g}\left( A\right) $ for general $A$). Here we run into troubles: For example, for $\widehat{\mathfrak{sl}_{2}}$, we have $M_{\lambda}=U\left( \widetilde{\mathfrak{n}}_{-}\right) v_{\lambda}$, and the vectors $ht^{-1}v_{\lambda},ht^{-2}v_{\lambda},...$ all have weight $\lambda$ with respect to $\widehat{\mathfrak{h}}=\left\langle h_{0},h_{1}\right\rangle $ with $h_{1}=h,$ $h_{0}=K-h$. This yields that weight spaces are infinite-dimensional, and we cannot define characters. Let us work around this by adding derivations. Assume that $A$ is an $r\times r$ complex matrix. Let $\mathfrak{g}% _{\operatorname*{ext}}\left( A\right) =\mathfrak{g}\left( A\right) \oplus\bigoplus\limits_{i=1}^{r}\mathbb{C}D_{i}$ with new relations% \begin{align*} \left[ D_{i},D_{j}\right] & =0\ \ \ \ \ \ \ \ \ \ \text{for all }i,j;\\ \left[ D_{i},e_{j}\right] & =0\ \ \ \ \ \ \ \ \ \ \text{for all }i\neq j;\\ \left[ D_{i},f_{j}\right] & =0\ \ \ \ \ \ \ \ \ \ \text{for all }i\neq j;\\ \left[ D_{i},h_{j}\right] & =0\ \ \ \ \ \ \ \ \ \ \text{for all }i\neq j;\\ \left[ D_{i},e_{i}\right] & =e_{i};\\ \left[ D_{i},f_{i}\right] & =-f_{i};\\ \left[ D_{i},h_{i}\right] & =0. \end{align*} Note that this definition is equivalent to making $\mathfrak{g}% _{\operatorname*{ext}}\left( A\right) $ a semidirect product, so there is no cancellation here. We have $\mathfrak{g}_{\operatorname*{ext}}\left( A\right) =\mathfrak{n}% _{+}\oplus\mathfrak{h}_{\operatorname*{ext}}\oplus\mathfrak{n}_{-}$ where $\mathfrak{h}_{\operatorname*{ext}}=\mathbb{C}^{r}\oplus\mathfrak{h}$ (here the $\mathbb{C}^{r}$ is spanned by the $\mathbb{C}D_{i}$). Consider $\alpha_{i}$ as maps $\mathfrak{h}_{\operatorname*{ext}}% \rightarrow\mathbb{C}$ given by $\alpha_{i}\left( h_{j}\right) =a_{j,i}$ and $\alpha_{i}\left( D_{j}\right) =\delta_{i,j}$. Then, for every $h\in\mathfrak{h}_{\operatorname*{ext}}$, we have $\left[ h,e_{i}\right] =\alpha_{i}\left( h\right) e_{i}$ and $\left[ h,f_{i}\right] =-\alpha_{i}\left( h\right) f_{i}$. Let $F=Q\otimes_{\mathbb{Z}}\mathbb{C}$ and $P=\mathfrak{h}^{\ast}\oplus F$. Let $\varphi:P\rightarrow\mathfrak{h}_{\operatorname*{ext}}^{\ast}$ be given by $\varphi\left( h_{i}^{\ast}\right) \left( D_{j}\right) =0$, $\varphi\left( h_{i}^{\ast}\right) \left( h_{j}\right) =\delta_{i,j}$, $\varphi\left( \alpha_{i}\right) \left( D_{j}\right) =\delta_{i,j}$, $\varphi\left( \alpha_{i}\right) \left( h_{j}\right) =a_{j,i}$. Easy to see $\varphi$ is an iso. Now the trouble disappears. Do the same as for simple Lie algebras. Now weights lie in $\mathfrak{h}_{\operatorname*{ext}}^{\ast}$. Annoying fact: Now, even when $A$ is a Cartan matrix and $\mathfrak{g}$ is simple finite-dimensional, this is not the same as the usual theory [what?]. But it is equivalent. Namely: Suppose $\chi\in\mathfrak{h}% _{\operatorname*{ext}}^{\ast}$. Let $\mathcal{O}_{\chi}$ be the category of modules whose weights lie in $\chi+F$. Therefore, $\mathcal{O}=\bigoplus \limits_{\chi\in\mathfrak{h}^{\ast}}\mathcal{O}_{\chi}$. \begin{proposition} If $\chi_{1}-\chi_{2}\in\operatorname*{Im}\left( F\rightarrow\mathfrak{h}% ^{\ast}\right) $, then $\mathcal{O}_{\chi_{1}}\cong\mathcal{O}_{\chi_{2}}$. \end{proposition} (See Feigin-Zelevinsky paper for proof.) If $A$ is invertible (in particular, for simple $\mathfrak{g}$), all $\mathcal{O}_{\chi}$ are the same and we just have a single category $\mathcal{O}$ (which is the category $\mathcal{O}$ we defined). Affine case: $\operatorname*{Coker}\left( F\rightarrow\mathfrak{h}^{\ast }\right) $ is $1$-dimensional, so $\chi$ has one essential parameter (namely, the image $k$ of $\chi$ in this $\operatorname*{Coker}$). So we get a $1$-parameter category of categories, $\mathcal{O}\left( k\right) $, parametrized by a complex number $k$. In our old approach to $\widehat{\mathfrak{g}}$, this $k$ is the level of representations (i. e., the eigenvalue of the action of $K$). So we did not get anything new, but we have got a uniform way to treat all cases of this kind. \subsection{\textbf{[unfinished]} Invariant bilinear forms} Now let us start developing the theory of invariant bilinear forms on $\mathfrak{g}\left( A\right) $ and $\widetilde{\mathfrak{g}}\left( A\right) $. [We denote $\mathfrak{g}\left[ \alpha\right] $ as $\mathfrak{g}_{\alpha}$.] Let $A$ be an indecomposable complex matrix. We want to see when we can have nontrivial nonzero invariant symmetric bilinear forms on $\widetilde{\mathfrak{g}}\left( A\right) $ and $\mathfrak{g}\left( A\right) $. Let us only care about forms of degree $0$, which means that they send $\mathfrak{g}_{\alpha}\times\mathfrak{g}_{\beta}$ to $0$ unless $\alpha+\beta=0$. It also sounds like a good goal to have the forms nondegenerate, but this cannot always be reached. Let us impose the weaker condition that, if $e_{i}$ and $f_{i}$ denote generators of $\mathfrak{g}% _{\alpha_{i}}$ and $\mathfrak{g}_{-\alpha_{i}}$, respectively, then $\left( e_{i},f_{i}\right) =d_{i}$ for some $d_{i}\neq0$. These conditions already force some properties upon $\mathfrak{g}\left( A\right) $: First, \[ \left( h_{i},h_{j}\right) =\left( h_{i},\left[ e_{j},f_{j}\right] \right) =-\left( \left[ h_{i},f_{j}\right] ,e_{j}\right) =a_{i,j}\left( f_{j},e_{j}\right) =a_{i,j}d_{j}, \] so that the symmetry of our form (and the condition $d_{i}\neq0$) enforces $a_{i,j}d_{j}=a_{j,i}d_{i}$. Thus, if $D$ denotes the matrix $\operatorname*{diag}\left( d_{1},d_{2},...,d_{r}\right) $, then $\left( AD\right) ^{T}=AD$. This means that $A$ is symmetrizable. (Our definition of ``symmetrizable'' spoke of $DA$ instead of $AD$, but this is simply a matter of replacing $D$ by $D^{-1}$.) \begin{lemma} Let $A$ be an indecomposable symmetrizable matrix. Then, there is a unique diagonal matrix $D$ satisfying $\left( AD\right) ^{T}=AD$ up to scaling. \end{lemma} This lemma is purely combinatorial and more or less trivial. \begin{proposition} Let $A$ be an indecomposable symmetrizable matrix. Then, there is at most one invariant symmetric bilinear form of degree $0$ on $\widetilde{\mathfrak{g}% }\left( A\right) $ up to scaling. \end{proposition} Note that the degree in ``degree $0$'' is the degree with respect to $Q$-grading; this is a tuple. \textit{Proof of Proposition.} Let $B$ be such a form. Then, we can view $B$ as a $\mathfrak{g}$-module homomorphism $B^{\vee}:\mathfrak{g}\rightarrow \mathfrak{g}^{\ast}$. If we fix $d_{i}$ (uniquely up to scaling, as we know from Lemma), then we know $B^{\vee}\left( h_{i}\right) $, $B^{\vee}\left( f_{i}\right) $ and $B^{\vee}\left( e_{i}\right) $ (because the form is of degree $0$, and thus the linear maps $B^{\vee}\left( h_{i}\right) $, $B^{\vee}\left( f_{i}\right) $ and $B^{\vee}\left( e_{i}\right) $ are determined by what they do to the corresponding elements of the corresponding degree). But $\mathfrak{g}$ is generated as a $\mathfrak{g}$-module by $e_{i},f_{i},h_{i}$, so $B$ is uniquely determined if it exists. Proposition is proven. \begin{theorem} Let $A$ be a symmetrizable matrix. Then, there is a nonzero invariant bilinear symmetric form of degree $0$ on $\widetilde{\mathfrak{g}}\left( A\right) $. (We know from the previous proposition that this form is unique up to scaling if $A$ is indecomposable.) \end{theorem} \textit{Proof of Theorem (incomplete, as we will skip some steps).} First, fix the $d_{i}$. Then, we can calculate the form by% \begin{align*} & \left( \underbrace{\left[ e_{i_{1}},\left[ e_{i_{2}},...\left[ e_{i_{n-1}},e_{i_{n}}\right] ...\right] \right] }_{\in\mathfrak{g}_{\alpha }},\underbrace{\left[ f_{j_{1}},\left[ f_{j_{2}},...\left[ f_{j_{n-1}% },f_{j_{n}}\right] ...\right] \right] }_{\in\mathfrak{g}_{-\alpha}}\right) \\ & =-\left( \left[ e_{i_{1}},...\right] ,\underbrace{\left[ \left[ e_{i_{2}},\left[ e_{i_{3}},...\left[ e_{i_{n-1}},e_{i_{n}}\right] ...\right] \right] ,\left[ f_{j_{1}},\left[ f_{j_{2}},...\left[ f_{j_{n-1}},f_{j_{n}}\right] ...\right] \right] \right] }_{\in \mathfrak{g}_{-\alpha}}\right) \\ & +... \end{align*} induction on $\alpha$. For details and well-definedness, see page 51 of the Feigin-Zelevinsky paper. Also, $\widetilde{\mathfrak{g}}\left( A\right) $ has such a form by pullback. As usual, denote these forms by $\left( \cdot,\cdot\right) $. \begin{proposition} The kernel $I$ of the canonical projection $\widetilde{\mathfrak{g}}\left( A\right) \rightarrow\mathfrak{g}\left( A\right) $ is a subset of $\operatorname*{Ker}\left( \left( \cdot,\cdot\right) \right) $. \end{proposition} \textit{Proof of Proposition.} We defined the form $\left( \cdot ,\cdot\right) $ on $\widetilde{\mathfrak{g}}\left( A\right) \times \widetilde{\mathfrak{g}}\left( A\right) $ as the pullback of the form $\left( \cdot,\cdot\right) :\mathfrak{g}\left( A\right) \times \mathfrak{g}\left( A\right) \rightarrow\mathbb{C}$ through the canonical projection $\widetilde{\mathfrak{g}}\left( A\right) \times \widetilde{\mathfrak{g}}\left( A\right) \rightarrow\mathfrak{g}\left( A\right) \times\mathfrak{g}\left( A\right) $. Thus, it is clear that the kernel of the former form contains the kernel of the canonical projection $\widetilde{\mathfrak{g}}\left( A\right) \rightarrow\mathfrak{g}\left( A\right) $. Proposition proven. \begin{lemma} \textbf{1)} The center $Z$ of $\mathfrak{g}\left( A\right) $ is contained in $\mathfrak{h}$, and is% \[ Z=\left\{ \sum\limits_{i}\beta_{i}h_{i}\ \mid\ \beta_{i}\in\mathbb{C}\text{ for all }i\text{, and }\sum\limits_{i}\beta_{i}a_{i,j}=0\text{ for all }j\right\} . \] \textbf{2)} If $A$ is an indecomposable symmetrizable matrix, and $A\neq0$, then any graded proper ideal in $\mathfrak{g}\left( A\right) $ is contained in $Z$. \textbf{3)} If $a_{i,i}\neq0$ for all $i$, then $\left[ \mathfrak{g}\left( A\right) ,\mathfrak{g}\left( A\right) \right] =\mathfrak{g}\left( A\right) $. \end{lemma} \textit{Proof of Lemma.} \textbf{1)} Let $z$ be a nonzero central element of $\mathfrak{g}\left( A\right) $. We can WLOG assume that $z$ is homogeneous. Then, $\mathbb{C}z$ is a graded nonzero ideal of $\mathfrak{g}\left( A\right) $, so that $\deg z$ must be $0$, and thus $z\in\mathfrak{h}$. If $z=\sum\limits_{i}\beta_{i}h_{i}$, then every $j$ satisfies $0=\left[ z,e_{j}\right] =\left[ \sum\limits_{i}\beta_{i}h_{i},e_{j}\right] =\left( \sum\limits_{i}\beta_{i}a_{i,j}\right) e_{j}$, so that $\sum\limits_{i}% \beta_{i}a_{i,j}=0$. This proves that $Z\subseteq\left\{ \sum\limits_{i}\beta_{i}h_{i}% \ \mid\ \beta_{i}\in\mathbb{C}\text{ for all }i\text{, and }\sum \limits_{i}\beta_{i}a_{i,j}=0\text{ for all }j\right\} $. The reverse inclusion is easy to see (using $\left[ h_{i},f_{j}\right] =-a_{i,j}f_{j}$). \textbf{2)} Let $I\neq0$ be a graded ideal. Then, $I\cap\mathfrak{h}\neq0$. So $I=I_{+}\oplus I_{0}\oplus I_{-}$ with $I_{0}$ being a nonzero subspace of $\mathfrak{h}$. Assume $I\not \subseteq Z$. Then we claim that $I_{+}\neq0$ or $I_{-}\neq0$. (In fact, otherwise, we would have $I_{+}=0$ and $I_{-}=0$, so that $I\subseteq\mathfrak{h}$, so that there exists some $h\in I\subseteq \mathfrak{h}$ with $h\notin Z$, so that $\left[ h,e_{j}\right] =\lambda e_{j}$ for some $j$ and some $\lambda\neq0$, so that $e_{j}\in I_{+}$, contradicting $I_{+}=0$ and $I_{-}=0$.) Let $\mathfrak{G}$ be the subset $\left\{ e_{1},e_{2},...,e_{n},f_{1}% ,f_{2},...,f_{n},h_{1},h_{2},...,h_{n}\right\} $ of $\mathfrak{g}\left( A\right) $. As we know, this subset $\mathfrak{G}$ generates the Lie algebra $\mathfrak{g}\left( A\right) $. So let us WLOG assume $I_{+}\neq0$. Then there exists a nonzero $a\in I_{+}\left[ \alpha\right] $ for some $\alpha\neq0$. Set $J$ be the ideal generated by $a$. In other words, $J=U\left( \mathfrak{g}\left( A\right) \right) \cdot a$. This $J$ is a graded ideal. Thus, $J\cap\mathfrak{h}\neq0$. Hence, there exists $x\in U\left( \mathfrak{g}\left( A\right) \right) $ such that $x\rightharpoonup a\in\mathfrak{h}$ and $x\rightharpoonup a\neq0$. We can WLOG assume that $x$ has degree $-\alpha$ and is a product of some elements of the set $\mathfrak{G}$ (with repetitions allowed). Of course, this product is nonempty (otherwise, $a$ itself would be in $I_{0}$, not in $I_{+}% $), and hence (by splitting off its first factor) can be written as $\xi \cdot\eta$ with $\xi$ being an element of the set $\mathfrak{G}$ and $\eta$ being a product of elements of $\mathfrak{G}$. Consider these $\xi$ and $\eta $. We assume WLOG that $\eta$ is a product of elements of $\mathfrak{G}$ with a minimum possible number of factors. Then, $\xi\notin\left\{ h_{1}% ,h_{2},...,h_{n}\right\} $ (because otherwise, we could replace $x$ by $\eta $, and would then, by splitting off the first factor, obtain a new $\eta$ with an even smaller number of factors). So we have either $\xi=e_{i}$ for some $i$, or $\xi=f_{i}$ for some $i$. Let us WLOG assume that we are in the first case, i. e., we have $\xi=e_{i}$ for some $i$. Let $y=\eta\rightharpoonup a$. Then, $y\in I$ (since $a\in I$ and since $I$ is an ideal) and \begin{align*} \left[ \xi,y\right] & =\xi\rightharpoonup\underbrace{y}_{=\eta \rightharpoonup a}\ \ \ \ \ \ \ \ \ \ \left( \text{since }\xi\in \mathfrak{G}\subseteq\mathfrak{g}\left( A\right) \right) \\ & =\xi\rightharpoonup\left( \eta\rightharpoonup a\right) =\underbrace{\left( \xi\cdot\eta\right) }_{=x}\rightharpoonup a=x\rightharpoonup a\in\mathfrak{h}% \end{align*} and $\left[ \xi,y\right] =x\rightharpoonup a\neq0$. Since $\xi=e_{i}% \in\mathfrak{g}_{\alpha_{i}}$ and $y$ is homogeneous, this yields that $y\in\mathfrak{g}_{-\alpha_{i}}$. Thus, $y=\chi\cdot f_{i}$ for some $\chi \in\mathbb{C}$. This $\chi$ is nonzero, since $y$ is nonzero (since $\left[ \xi,y\right] \neq0$). Since $y=\chi\cdot f_{i}$, we have $\left[ e_{i},y\right] =\chi \cdot\underbrace{\left[ e_{i},f_{i}\right] }_{=h_{i}}=\chi h_{i}$. Since $\left[ e_{i},y\right] \in I$ (because $I$ is an ideal and $y\in I$), this becomes $\chi h_{i}\in I$, so that $h_{i}\in I$ (since $\chi$ is nonzero). Moreover, since $\chi\cdot f_{i}=y\in I$, we have $f_{i}\in I$ (since $\chi$ is nonzero). Altogether, we now know that $h_{i}\in I$ and $f_{i}\in I$. If $A$ is an $1\times1$ matrix, then $a_{i,i}\neq0$ (since $A\neq0$), so that $e_{i}=\dfrac{\left[ h_{i},e_{i}\right] }{a_{i,i}}\in I$ (because $h_{i}\in I$). Hence, if $A$ is an $1\times1$ matrix, then all of $e_{i}$, $f_{i}$ and $h_{i}$ lie in $I$, so that $I=\mathfrak{g}\left( A\right) $ (because there exists only one $i$). If the size of $A$ is $>1$, there exists some $j\neq i$ such that $a_{i,j}% \neq0$ and $a_{j,i}\neq0$ (since $A$ is indecomposable and symmetrizable), so that $e_{j}=\dfrac{\left[ h_{i},e_{j}\right] }{a_{i,j}}\in I$ (since $h_{i}\in I$), furthermore $f_{j}=-\dfrac{\left[ h_{i},f_{j}\right] }{a_{i,j}}\in I$, therefore $h_{j}=\left[ e_{j},f_{j}\right] \in I$, and finally $e_{i}=\dfrac{\left[ h_{j},e_{i}\right] }{a_{j,i}}\in I$. And for every $k\neq i$ with $a_{i,k}\neq0$ and $a_{k,i}\neq0$, we similarly get $h_{k}$, $f_{k}$, $e_{k}$ $\in I$ etc.. By repeating this argument, we conclude that $e_{\ell},f_{\ell},h_{\ell}\in I$ for all $\ell$ (since $A$ is indecomposable). That is, $\mathfrak{G}\subseteq I$. Since $\mathfrak{G}$ is a generating set of the Lie algebra $\mathfrak{g}\left( A\right) $, this entails $I=\mathfrak{g}\left( A\right) $. \textbf{3)} If $a_{i,i}\neq0$, then the relations (\ref{nonserre-relations}) imply that all generators are in $\left[ \mathfrak{g}\left( A\right) ,\mathfrak{g}\left( A\right) \right] $. Qed. \begin{proposition} Assume that $A$ is symmetrizable. We have $\operatorname*{Ker}\left( \left( \cdot,\cdot\right) \mid_{\mathfrak{g}\left( A\right) }\right) =Z\left( \mathfrak{g}\left( A\right) \right) $. \end{proposition} \textit{Proof of Proposition.} Assume WLOG that $A$ is indecomposable. \textbf{1)} $1\times1$ case, $A=0$ trivial: $\left[ e,f\right] =h$, $\left[ h,e\right] =\left[ h,f\right] =0$, $\left( e,f\right) =1$. Then the kernel of this form is a graded ideal and is not $\mathfrak{g}\left( A\right) $. Hence, it must be contained in $Z$ by the lemma. But $Z\subseteq\operatorname*{Ker}\left( \left( \cdot,\cdot\right) \mid_{\mathfrak{g}\left( A\right) }\right) $ is easy (because $\left( \sum\limits_{i}\beta_{i}h_{i},h_{j}\right) =\sum\limits_{i}\beta_{i}% a_{i,j}d_{j}=0$). Let $F=Q\otimes_{\mathbb{Z}}\mathbb{C}=\bigoplus_{i=1}^{r}\mathbb{C}\alpha _{i}$. Define $\gamma:F\rightarrow\mathfrak{h}$ isomorphism by $\gamma\left( \alpha_{i}\right) =d_{i}^{-1}h_{i}=:h_{\alpha_{i}}$. Extend by linearity: $\gamma\left( \alpha\right) $ will be called $h_{\alpha}$, $\alpha\in F$. \textbf{Claim:} $\left( h_{\alpha},h\right) =\overline{\alpha}\left( h\right) $, where $\overline{\alpha}$ is the image of $\alpha$ in $\mathfrak{h}^{\ast}$. Proof: $\left( h_{\alpha_{i}},h_{j}\right) =d_{i}^{-1}\left( h_{i}% ,h_{j}\right) =d_{i}^{-1}a_{i,j}d_{j}=d_{i}^{-1}a_{j,i}d_{i}=a_{j,i}% =\overline{\alpha_{i}}\left( h_{j}\right) \ \ \ \ $ ($\left[ h_{j}% ,e_{i}\right] =a_{j,i}e_{i}$). \begin{proposition} If $x\in\mathfrak{g}_{\alpha}$ and $y\in\mathfrak{g}_{-\alpha}$, then $\left[ x,y\right] =\left( x,y\right) h_{\alpha}$. \end{proposition} \textit{Proof of Proposition.} By induction over $\left\vert \alpha\right\vert $, where $\left\vert \alpha\right\vert $ means the sum of the coordinates of $\alpha$. \textit{Base:} $\left\vert \alpha\right\vert =1$, $\alpha=\alpha_{i}$. Want to prove $\left[ e_{i},f_{i}\right] =^{?}\left( e_{i},f_{i}\right) h_{\alpha_{i}}$. But $\left[ e_{i},f_{i}\right] =h_{i}$ and $\left( e_{i},f_{i}\right) h_{\alpha_{i}}=d_{i}d_{i}^{-1}h_{i}$, so we are done with the base. \textit{Step:} For $x\in\mathfrak{g}_{\alpha-\alpha_{i}}$ and $y\in \mathfrak{g}_{\alpha-\alpha_{j}}$, we have \begin{align*} & \left[ \left[ e_{i},x\right] ,\left[ f_{j},y\right] \right] \\ & =\left[ \left[ e_{i},\left[ f_{j},y\right] \right] ,x\right] +\left[ e_{i},\left[ x,\left[ f_{j},y\right] \right] \right] \\ & =-\left( \left[ e_{i},\left[ f_{j},y\right] \right] ,x\right) h_{\alpha-\alpha_{i}}+\left( e_{i},\left[ x,\left[ f_{j},y\right] \right] \right) h_{\alpha_{i}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the induction assumption}\right) \\ & =\left( \left[ f_{j},y\right] ,\left[ e_{i},x\right] \right) \left( h_{\alpha-\alpha_{i}}+h_{\alpha_{i}}\right) =\left( \left[ e_{i},x\right] ,\left[ f_{j},y\right] \right) h_{\alpha}. \end{align*} Induction step complete. Proposition proven. \begin{corollary} If we give $\mathfrak{g}\left( A\right) $ the principal $\mathbb{Z}$-grading (so that $\mathfrak{g}\left( A\right) \left[ n\right] =\bigoplus \limits_{\substack{\alpha\in Q;\\\left\vert \alpha\right\vert =n}% }\mathfrak{g}\left( A\right) \left[ \alpha\right] $), then $\mathfrak{g}% \left( A\right) $ is a nondegenerate Lie algebra. \end{corollary} \textit{Proof.} If $\lambda\in\mathfrak{h}^{\ast}$ is such that $\lambda \left( h_{\alpha}\right) \neq0$, then $\lambda\left( \left[ x,y\right] \right) $ is a nondegenerate form $\mathfrak{g}_{\alpha}\times\mathfrak{g}% _{-\alpha}\rightarrow\mathbb{C}$. Qed. Recall $P=\mathfrak{h}^{\ast}\oplus F\cong\mathfrak{h}_{\operatorname*{ext}% }^{\ast}$. $\left( \cdot,\cdot\right) $ on $P$: $\left( \underbrace{\varphi}% _{\in\mathfrak{h}^{\ast}}\oplus\underbrace{\alpha}_{\in F},\underbrace{\psi }_{\in\mathfrak{h}^{\ast}}\oplus\underbrace{\beta}_{\in F}\right) =\psi\left( h_{\alpha}\right) +\varphi\left( h_{\beta}\right) +\left( h_{\alpha},h_{\beta}\right) $ $\left( h_{\alpha_{i}},h_{\alpha_{j}}\right) =d_{i}^{-1}d_{j}^{-1}\left( h_{i},h_{j}\right) =d_{i}^{-1}d_{j}^{-1}a_{i,j}d_{j}=d_{i}^{-1}a_{i,j}$. Basis $h_{\alpha_{i}}^{\ast}\in\mathfrak{h}^{\ast}$, $\alpha_{i}\in F$ $\Longrightarrow$ matrix of the form $\left( \begin{array} [c]{cc}% 0 & 1\\ 1 & D^{-1}A \end{array} \right) $. Inverse form on $\mathfrak{h}_{\operatorname*{ext}}$: dual basis: $h_{\alpha_{i}},D_{i}$. $\left( D_{i},D_{j}\right) =0$, $\left( D_{i},h_{\alpha_{j}}\right) =\delta_{i,j}$, $\left( h_{\alpha_{i}},h_{\alpha_{j}}\right) =d_{i}% ^{-1}a_{i,j}$. \begin{proposition} The form on $\mathfrak{g}_{\operatorname*{ext}}\left( A\right) =\mathfrak{g}\left( A\right) \oplus\mathbb{C}D_{1}\oplus\mathbb{C}% D_{2}\oplus...\oplus\mathbb{C}D_{r}$ defined by this is a nondegenerate symmetric invariant form. \end{proposition} \subsection{\textbf{[unfinished]} Casimir element} We now define the Casimir element. The problem with the classical ``sum of squares of orthonormal basis'' construction which works well in the finite-dimensional case is that now we are infinite-dimensional and such a sum needs to be defined. Note that it will be a generalization of the $L_{0}$ of the Sugawara construction. Define $\rho\in\mathfrak{h}^{\ast}$ by $\rho\left( h_{i}\right) =\dfrac{a_{i,i}}{2}$ (in the Kac-Moody case, this becomes $\rho\left( h_{i}\right) =1$). $\left( \rho,\rho\right) =0$. Case of a finite-dimensional simple Lie algebra: $\Delta=\sum\limits_{a\in B}a^{2}=\sum\limits_{i=1}^{r}x_{i}^{2}+2h_{\rho}+2\sum\limits_{\alpha >0}f_{\alpha}e_{\alpha}$ where $\left( x_{i}\right) _{i=1,...,r}$ is an orthonormal basis of $\mathfrak{h}$. In the infinite-dimensional case, we fix a basis $\left( e_{\alpha}% ^{i}\right) _{i}$ of $\mathfrak{g}_{\alpha}$ for every $\alpha$, and a dual basis $\left( f_{\alpha}^{i}\right) _{i}$ of $\mathfrak{g}_{-\alpha}$ under the inner product. Then define $\Delta_{+}=2\sum\limits_{\alpha>0}% \sum\limits_{i}f_{\alpha}^{i}e_{\alpha}^{i}$ and $\Delta_{0}=\sum \limits_{j}x_{j}^{2}+2h_{\rho}$ (where $\left( x_{j}\right) $ is an orthonormal basis of $\mathfrak{h}_{\operatorname*{ext}}$). We set $\Delta=\Delta_{+}+\Delta_{0}$. Note that $\Delta_{+}$ is an infinite sum and not in $U\left( \mathfrak{g}% \left( A\right) \right) $. But it becomes finite after applying to any vector in a module in category $\mathcal{O}$. \begin{theorem} \textbf{1)} The operator $\Delta$ commutes with $\mathfrak{g}\left( A\right) $. \textbf{2)} We have $\Delta\mid_{M_{\lambda}}=\left( \lambda,\lambda +2\rho\right) \operatorname*{id}$. \end{theorem} \textit{Proof of Theorem.} Let us first prove \textbf{2)} using \textbf{1)}: \textbf{2)} We have $\Delta v_{\lambda}=\Delta_{0}v_{\lambda}=\left( \sum\limits_{j}\lambda\left( x_{j}\right) ^{2}+2\lambda\left( h_{\rho }\right) \right) v_{\lambda}=\left( \left( \lambda,\lambda\right) +2\left( \lambda,\rho\right) \right) v_{\lambda}=\left( \lambda ,\lambda+2\rho\right) v_{\lambda}$. From \textbf{1)}, we see that every $a\in U\left( \mathfrak{g}\left( A\right) \right) $ satisfies $\Delta av_{\lambda}=a\Delta v_{\lambda }=\left( \lambda,\lambda+2\rho\right) av_{\lambda}$. This proves \textbf{2)} since $M_{\lambda}=U\left( \mathfrak{g}\left( A\right) \right) v_{\lambda }$. \textbf{1)} We need to show that $\left[ \Delta,e_{i}\right] =\left[ \Delta,f_{i}\right] =0$. Let us prove $\left[ \Delta,e_{i}\right] =0$ (the proof of $\left[ \Delta,f_{i}\right] =0$ is similar). We have $\left[ \Delta_{0},e_{i}\right] =\left[ \sum x_{j}^{2}+2h\rho ,e_{i}\right] =\sum x_{j}\left[ x_{j},e_{i}\right] +\sum\left[ x_{j}% ,e_{i}\right] x_{j}+2\left( \alpha_{i},\rho\right) e_{i}$ $=\sum x_{j}\underbrace{\alpha_{i}\left( x_{j}\right) }_{=\left( h_{\alpha_{i}},x_{j}\right) }e_{i}+\sum\alpha_{i}\left( x_{j}\right) e_{i}x_{j}+2\left( \alpha_{i},\rho\right) e_{i}$ $=2h_{\alpha_{i}}e_{i}-\sum\underbrace{\alpha_{i}\left( x_{j}\right) }_{=\left( \alpha_{i},\alpha_{i}\right) e_{i}}\alpha_{i}\left( x_{j}\right) e_{i}+2\left( \alpha_{i},\rho\right) e_{i}=2h_{\alpha}e_{i}$ $\Longrightarrow$ Our job is to show $\left[ \Delta_{+},e_{i}\right] =-2h_{\alpha_{i}}e_{i}$. But $\left[ \Delta_{+},e_{i}\right] =2\sum\limits_{\alpha>0}f_{\alpha}% ^{j}\left[ e_{\alpha}^{j},e_{i}\right] +2\underbrace{\sum\limits_{\alpha >0}\left[ f_{\alpha}^{j},e_{i}\right] e_{\alpha}^{j}}_{\substack{\text{for }\alpha=\alpha_{i}\text{ the addend is}\\-2h_{\alpha_{i}}e_{i}\\\text{because }f_{\alpha_{i}}=d_{i}^{-1}f_{i}\text{, }e_{\alpha_{i}}=e_{i}\text{,}\\\left[ d_{i}^{-1}f_{i},e_{i}\right] e_{i}=-d_{i}^{-1}h_{i}e_{i}=-h_{\alpha_{i}}% e_{i}}}$. So we need to show that% \[ \sum\limits_{\alpha>0}f_{\alpha}^{j}\left[ e_{\alpha}^{j},e_{i}\right] +2\sum\limits_{\substack{\alpha>0;\\\alpha\neq\alpha_{i}}}\left[ f_{\alpha }^{j},e_{i}\right] e_{\alpha}^{j}=0. \] For this it is enough to check% \[ \sum\limits_{\alpha>0}f_{\alpha}^{j}\otimes\left[ e_{\alpha}^{j}% ,e_{i}\right] +2\sum\limits_{\substack{\alpha>0;\\\alpha\neq\alpha_{i}% }}\left[ f_{\alpha}^{j},e_{i}\right] \otimes e_{\alpha}^{j}=0. \] For this it is enough to check that $\left[ e_{i},e_{\alpha}^{k}\right] =\sum\left( e_{\beta}^{k},\left[ f_{\alpha}^{j},e_{i}\right] \right) e_{\alpha}^{j}$. This is somehow obvious. Proof complete. \textbf{Exercise:} for $\widehat{\mathfrak{g}}$ (affine), $\Delta=\left( k+h^{\vee}\right) \left( L_{0}-d\right) $ (Sugawara). \subsection{\textbf{[unfinished]} Preparations for the Weyl-Kac character formula} Let $A$ be a symmetrizable generalized Cartan matrix, WLOG indecomposable. We consider the Kac-Moody algebra $\mathfrak{g}=\mathfrak{g}\left( A\right) \subseteq\mathfrak{g}_{\operatorname*{ext}}\left( A\right) $. \begin{proposition} The Serre relations $\left( \operatorname*{ad}\left( e_{i}\right) \right) ^{1-a_{i,j}}e_{j}=\left( \operatorname*{ad}\left( f_{i}\right) \right) ^{1-a_{i,j}}f_{j}=0$ hold in $\mathfrak{g}\left( A\right) $. \end{proposition} This is a part of Theorem \ref{thm.g(A).gabber-kac} (actually, the part that we proved above). \begin{definition} Let $A$ be an associative algebra (with $1$, as always). Let $V$ be an $A$-module. \textbf{(a)} Let $v\in V$. Then, the vector $v$ is said to be \textit{of finite type} if $\dim\left( Av\right) <\infty$. \textbf{(b)} The $A$-module $V$ is said to be \textit{locally finite} if every $v\in V$ is of finite type. \end{definition} It is very easy to check that: \begin{proposition} \label{prop.locfin.equiv}Let $A$ be an associative algebra (with $1$, as always). Let $V$ be an $A$-module. Then, $V$ is locally finite if and only if $V$ is a sum of finite-dimensional $A$-modules. \end{proposition} \textit{Proof of Proposition \ref{prop.locfin.equiv} (sketched).} $\Longrightarrow:$ Assume that $V$ is locally finite. Then, for every $v\in V$, we have $\dim\left( Av\right) <\infty$ (since $v$ is of finite type), so that $Av$ is a finite-dimensional $A$-module. Thus, $V=\sum\limits_{v\in V}Av$ is a sum of finite-dimensional $A$-modules. $\Longleftarrow:$ Assume that $V$ is a sum of finite-dimensional $A$-modules. Then, for every $v\in V$, the vector $v$ belongs to a sum of \textbf{finitely many} finite-dimensional $A$-modules. But such a sum is finite-dimensional as well. As a consequence, for every $v\in V$, the vector $v$ belongs to a finite-dimensional $A$-module, and thus $\dim\left( Av\right) <\infty$, so that $v$ is of finite type. Thus, $V$ is locally finite. Proposition \ref{prop.locfin.equiv} is proven. \begin{Convention} If $\mathfrak{g}$ is a Lie algebra, then ``locally finite'' and ``of finite type'' with respect to $\mathfrak{g}$ mean locally finite resp. of finite type with respect to $U\left( \mathfrak{g}\right) $. \end{Convention} In the following, let $A=U\left( \mathfrak{g}\right) $ for $\mathfrak{g}% =\mathfrak{g}\left( A\right) $. \begin{definition} Let $V$ be a $\mathfrak{g}\left( A\right) $-module. We say that $V$ is \textit{integrable} if $V$ is locally finite under the $\mathfrak{sl}_{2}% $-subalgebra $\left( \mathfrak{sl}_{2}\right) _{i}=\left\langle e_{i}% ,f_{i},h_{i}\right\rangle $ for every $i\in\left\{ 1,2,...,r\right\} $. \end{definition} To motivate the terminology ``integrable'', let us notice: \begin{proposition} If $V$ is a $\mathfrak{sl}_{2}$-module, then $V$ is locally finite if and only if $V$ is isomorphic to a direct sum $\bigoplus\limits_{n=0}^{\infty}% W_{n}\otimes V_{n}$, where $W_{n}$ are vector spaces and $V_{n}$ is the irreducible representation of $\mathfrak{sl}_{2}$ of highest weight $n$ (so that $\dim\left( V_{n}\right) =n+1$) for every $n\in\mathbb{N}$. (In such a direct sum, we have $W_{n}\cong\operatorname*{Hom}\nolimits_{\mathfrak{sl}% _{2}}\left( V_{n},V\right) $.) Locally-finite $\mathfrak{sl}_{2}$-modules can be lifted to modules over the \textbf{algebraic group} $\operatorname*{SL}\nolimits_{2}\left( \mathbb{C}\right) $. \end{proposition} Since lifting is called ``integrating'' (in analogy to geometry, where an action of a Lie group gives rise to an action of the corresponding of the Lie algebra by ``differentiation'', and thus the converse operation, when it makes sense, is called ``integration''), the last sentence of this proposition explains the name ``integrable''. \begin{proposition} \label{prop.weylkac.gint}The $\mathfrak{g}$-module $\mathfrak{g}% =\mathfrak{g}\left( A\right) $ itself is integrable. \end{proposition} The proof of this proposition is based on the following lemma: \begin{lemma} \label{lem.weylkac.fintypfintyp}Let $\mathfrak{a}$ be a Lie algebra, and $\mathfrak{b}$ be another Lie algebra. Assume that we are given a Lie algebra homomorphism $\mathfrak{b}\rightarrow\operatorname*{Der}\mathfrak{a}$; this makes $\mathfrak{a}$ into a $\mathfrak{b}$-module. Then, if $x,y\in \mathfrak{a}$ are of finite type for $\mathfrak{b}$, then so is $\left[ x,y\right] $. \end{lemma} \textit{Proof of Lemma \ref{lem.weylkac.fintypfintyp}.} In $\mathfrak{a}$ (not in $U\left( \mathfrak{a}\right) $), we have% \[ U\left( \mathfrak{b}\right) \cdot\left[ x,y\right] \subseteq\left[ \underbrace{U\left( \mathfrak{b}\right) \cdot x}_{\text{finite dimensional}% },\underbrace{U\left( \mathfrak{b}\right) \cdot y}_{\text{finite dimensional}}\right] . \] Hence, $U\left( \mathfrak{b}\right) \cdot\left[ x,y\right] $ is finite-dimensional. Hence, $\left[ x,y\right] $ is of finite type for $\mathfrak{b}$. Lemma \ref{lem.weylkac.fintypfintyp} is proven. \textit{Proof of Proposition \ref{prop.weylkac.gint}.} We know that $e_{i}$ is of finite type under $\left( \mathfrak{sl}_{2}\right) _{i}$ (in fact, $e_{i}$ generates a $3$-dimensional representation of $\left( \mathfrak{sl}% _{2}\right) _{i}$), and that $e_{j}$ is of finite type under $\left( \mathfrak{sl}_{2}\right) _{i}$ for every $j\neq i$ (in fact, $e_{j}$ generates a representation of dimension $1-a_{i,j}$). The same applies to $f_{j}$, and hence also to $h_{j}$ (by Lemma \ref{lem.weylkac.fintypfintyp}). Hence (again using Lemma \ref{lem.weylkac.fintypfintyp}), the whole $\mathfrak{g}\left( A\right) $ is locally finite under $\left( \mathfrak{sl}_{2}\right) _{i}$. [Fix some stuff here.] Proposition \ref{prop.weylkac.gint} is proven. \begin{proposition} If $V$ is a $\mathfrak{g}\left( A\right) $-module, then $V$ is integrable if and only if there exists a generating family $\left( v_{\alpha}\right) _{\alpha\in\mathfrak{A}}$ of the $\mathfrak{g}\left( A\right) $-module $V$ such that each $v_{\alpha}$ is of finite type under $\left( \mathfrak{sl}% _{2}\right) _{i}$ for each $i$. \end{proposition} Note that this proposition could just as well be formulated for every Lie algebra $\mathfrak{g}$ instead of $\mathfrak{g}\left( A\right) $. \textit{Proof of Proposition.} $\Longleftarrow:$ Let $v\in V$. We need to show that $v$ is of finite type under $\left( \mathfrak{sl}_{2}\right) _{i}$ for all $i$. Pick some $i\in\left\{ 1,2,...,r\right\} $. Let $\mathfrak{g}=\mathfrak{g}% \left( A\right) $. Fix some $i$. Then, there exist $i_{1},i_{2},...,i_{m}\in\mathfrak{A}$ such that $v\in U\left( \mathfrak{g}\right) \cdot v_{i_{1}}+U\left( \mathfrak{g}\right) \cdot v_{i_{2}}+...+U\left( \mathfrak{g}\right) \cdot v_{i_{m}}$. WLOG assume that $i_{1}=1$, $i_{2}=2$, $...$, $i_{m}=m$, and denote the $\mathfrak{g}$-submodule $U\left( \mathfrak{g}\right) \cdot v_{1}+U\left( \mathfrak{g}\right) \cdot v_{2}+...+U\left( \mathfrak{g}% \right) \cdot v_{m}$ of $V$ by $V^{\prime}$. Then, $v\in U\left( \mathfrak{g}\right) \cdot v_{i_{1}}+U\left( \mathfrak{g}\right) \cdot v_{i_{2}}+...+U\left( \mathfrak{g}\right) \cdot v_{i_{m}}=U\left( \mathfrak{g}\right) \cdot v_{1}+U\left( \mathfrak{g}\right) \cdot v_{2}+...+U\left( \mathfrak{g}\right) \cdot v_{m}=V^{\prime}\subseteq V$. Pick a finite-dimensional $\left( \mathfrak{sl}_{2}\right) _{i}% $-subrepresentation $W$ of $V^{\prime}$ such that $v_{1},v_{2},...,v_{m}\in W$. (This is possible because $v_{1},v_{2},...,v_{m}$ are of finite type under $\left( \mathfrak{sl}_{2}\right) _{i}$.) Then we have a surjective homomorphism of $\left( \mathfrak{sl}_{2}\right) _{i}$-modules $U\left( \mathfrak{g}\right) \otimes W\rightarrow V^{\prime}$ (namely, the homomorphism sending $x\otimes w$ to $xw$), where $\mathfrak{g}$ acts on $U\left( \mathfrak{g}\right) $ by adjoint action, and where $\left( \mathfrak{sl}_{2}\right) _{i}$ acts on $U\left( \mathfrak{g}\right) $ by restricting the $\mathfrak{g}$-action on $U\left( \mathfrak{g}\right) $ to $\left( \mathfrak{sl}_{2}\right) _{i}$. So it suffices to show that $U\left( \mathfrak{g}\right) $ is integrable for the adjoint action of $\mathfrak{g}$. But by the symmetrization map (which is an isomorphism by PBW), we have $U\left( \mathfrak{g}\right) \cong S\left( \mathfrak{g}% \right) =\bigoplus\limits_{m\in\mathbb{N}}S^{m}\left( \mathfrak{g}\right) $ (as $\mathfrak{g}$-modules) (this is true for every Lie algebra over a field of characteristic $0$). Since $S^{m}\left( \mathfrak{g}\right) $ injects into $\mathfrak{g}^{\otimes m}$, and since $\mathfrak{g}^{\otimes m}$ is integrable (because $\mathfrak{g}$ is (in fact, it is easy to see that if $X$ and $Y$ are locally finite $\mathfrak{a}$-modules, then so is $X\otimes Y$)), this yields that $U\left( \mathfrak{g}\right) $ is integrable. Hence, $U\left( \mathfrak{g}\right) \otimes W$ is a locally finite $\left( \mathfrak{sl}_{2}\right) _{i}$-module, and thus $V^{\prime}$ (being a quotient module of $U\left( \mathfrak{g}\right) \otimes W$) is a locally finite $\left( \mathfrak{sl}_{2}\right) _{i}$-module also as well. Hence, $v$ (being an element of $V^{\prime}$) is of finite type under $\left( \mathfrak{sl}_{2}\right) _{i}$. $\Longrightarrow:$ Trivial (take all vectors of $V$ as generators). Proposition proven. \begin{corollary} Let $L_{\lambda}$ be the irreducible highest-weight module for $\mathfrak{g}% \left( A\right) $. Then, $L_{\lambda}$ is integrable if and only if for every $i\in\left\{ 1,2,...,r\right\} $, the value $\lambda\left( h_{i}\right) $ is a nonnegative integer. \end{corollary} \textit{Proof of Corollary.} $\Longrightarrow:$ Assume that $L_{\lambda}$ is integrable. Consider the element $v_{\lambda}$ of $L_{\lambda}$. Since $L_{\lambda}$ is integrable, we know that $v_{\lambda}$ is of finite type under $\left( \mathfrak{sl}_{2}\right) _{i}$. In other words, $U\left( \left( \mathfrak{sl}_{2}\right) _{i}\right) v_{\lambda}$ is a finite-dimensional $\left( \mathfrak{sl}_{2}\right) _{i}$-module. Also, we know that $v_{\lambda}\neq0$, $e_{i}v_{\lambda}=0$ and $h_{i}v_{\lambda }=\lambda\left( h_{i}\right) v_{\lambda}$. Hence, Lemma \ref{lem.serre-gen.sl2} \textbf{(c)} (applied to $\left( \mathfrak{sl}% _{2}\right) _{i}$, $e_{i}$, $h_{i}$, $f_{i}$, $U\left( \left( \mathfrak{sl}_{2}\right) _{i}\right) v_{\lambda}$, $v_{\lambda}$ and $\lambda\left( h_{i}\right) $ instead of $\mathfrak{sl}_{2}$, $e$, $h$, $f$, $V$, $x$ and $\lambda$) yields that $\lambda\left( h_{i}\right) \in\mathbb{N}$ and $f_{i}^{\lambda\left( h_{i}\right) +1}v_{\lambda}=0$. In particular, $\lambda\left( h_{i}\right) $ is a nonnegative integer. $\Longleftarrow:$ We have% \begin{align*} e_{i}f_{i}^{\lambda\left( h_{i}\right) +1}v_{\lambda} & =\left( \lambda\left( h_{i}\right) +1\right) \underbrace{\left( \lambda\left( h_{i}\right) -\left( \lambda\left( h_{i}\right) +1\right) +1\right) }_{=0}f_{i}^{\lambda\left( h_{i}\right) }v_{\lambda}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the formula }e_{i}f_{i}^{m}v_{\lambda }=m\left( \lambda\left( h_{i}\right) -m+1\right) f_{i}^{m-1}v_{\lambda }\right) \\ & =0. \end{align*} Hence, $f_{i}^{\lambda\left( h_{i}\right) +1}v_{\lambda}$ must also be zero (since otherwise, this vector would generate a proper graded submodule). This implies that $v_{\lambda}$ generates a finite-dimensional $\left( \mathfrak{sl}_{2}\right) _{i}$-module of dimension $\lambda\left( h_{i}\right) +1$ with basis $\left( v_{\lambda},f_{i}v_{\lambda}% ,...,f_{i}^{\lambda\left( h_{i}\right) }v_{\lambda}\right) $. Hence, $v_{\lambda}$ is of finite type with respect to $\left( \mathfrak{sl}% _{2}\right) _{i}$. By the previous proposition, this yields that $L_{\lambda}$ is integrable. Proof of Corollary complete. \begin{remark} Assume that for every $i\in\left\{ 1,2,...,r\right\} $, the value $\lambda\left( h_{i}\right) $ is a nonnegative integer. Then, the relations $f_{i}^{\lambda\left( h_{i}\right) +1}v_{\lambda}=0$ are defining for $L_{\lambda}$. \end{remark} We will not prove this now, but this will follow from things we do later (from the main theorem for the character formula). \begin{definition} A weight $\lambda$ for which all $\lambda\left( h_{i}\right) $ are nonnegative integers is called \textit{integral} (for $\mathfrak{g}\left( A\right) $ or for $\mathfrak{g}_{\operatorname*{ext}}\left( A\right) $). \end{definition} Now, our next goal is to compute the character of $L_{\lambda}$ for any dominant integral weight $\lambda$. For finite-dimensional simple Lie algebras, these $L_{\lambda}$ are exactly the finite-dimensional irreducible representations, and their characters can be computed by the well-known Weyl character formula. So our goal is to generalize this formula. The Weyl character formula involves a summation over the Weyl group. So, first of all, we need to define a ``Weyl group'' for Kac-Moody Lie algebras. \subsection{\textbf{[unfinished]} Weyl group} \begin{definition} Consider $P=\mathfrak{h}^{\ast}\oplus F$. We know that there is a nondegenerate form $\left( \cdot,\cdot\right) $ on $P$, and we have $\dim P=2r$. Let $i\in\left\{ 1,2,...,r\right\} $. Let $r_{i}:P\rightarrow P$ be the map given by $r_{i}\left( \chi\right) =\chi-\chi\left( h_{i}\right) \alpha_{i}$. \end{definition} Note that $r_{i}$ is an involution, since% \[ r_{i}^{2}\left( \chi\right) =\chi-\chi\left( h_{i}\right) \alpha_{i}% -\chi\left( h_{i}\right) \alpha_{i}+\chi\left( h_{i}\right) \underbrace{\alpha_{i}\left( h_{i}\right) }_{=2}\alpha_{i}=\chi \] for every $\chi\in P$. Since $r_{i}\left( \alpha_{i}\right) =-\alpha_{i}$, this yields $\det\left( r_{i}\right) =-1$. Easy to check that $\left( r_{i}x,r_{i}y\right) =\left( x,y\right) $ for all $x,y\in P$. \begin{proposition} Let $V$ be an integrable $\mathfrak{g}\left( A\right) $-module. Then, for each $i\in\left\{ 1,2,...,r\right\} $ and any $\mu\in P$, we have an isomorphism $V\left[ \mu\right] \rightarrow V\left[ r_{i}\mu\right] $. In particular, $\dim\left( V\left[ \mu\right] \right) =\dim\left( V\left[ r_{i}\mu\right] \right) $. \end{proposition} \textit{Proof of Proposition.} We have $r_{i}\mu=\mu-\mu\left( h_{i}\right) \alpha_{i}$. Since $V$ is integrable for $\left( \mathfrak{sl}_{2}\right) _{i}$, we know that $\mu\left( h_{i}\right) $ is an integer. We have $\left( r_{i}\mu\right) \left( h_{i}\right) =-\mu\left( h_{i}\right) $. Hence, we can assume WLOG that $\mu\left( h_{i}\right) $ is nonnegative (because otherwise, we can switch $\mu$ with $r_{i}\mu$, and it will change sign). Then we have $f_{i}^{\mu\left( h_{i}\right) }:V\left[ \mu\right] \rightarrow V\left[ r_{i}\mu\right] $. I claim that $f_{i}^{\mu\left( h_{i}\right) }$ is an isomorphism. This follows from: \begin{lemma} If $V$ is a locally finite $\mathfrak{sl}_{2}$-module, then $f^{m}:V\left[ m\right] \rightarrow V\left[ -m\right] $ is an isomorphism. \end{lemma} \begin{definition} The \textit{Weyl group of} $\mathfrak{g}\left( A\right) $ is defined as the subgroup of $\operatorname*{GL}\left( P\right) $ generated by the $r_{i}$. This Weyl group is denoted by $W$. The elements $r_{i}$ are called \textit{simple reflections}. \end{definition} We will not prove: \begin{remark} The Weyl group $W$ is finite if and only if $A$ is a Cartan matrix (of a finite-dimensional Lie algebra). \end{remark} \begin{proposition} \label{prop.weylkac.prop0}\textbf{1)} The form $\left( \cdot,\cdot\right) $ on $P$ is $W$-invariant. \textbf{2)} There exists an isomorphism $V\left[ \mu\right] \rightarrow V\left[ w\mu\right] $ for every $\mu\in P$, $w\in W$ and any integrable $V$. \textbf{3)} The set of roots $R$ is $W$-invariant. (We recall that a \textit{root} means a nonzero element $\alpha\in F=Q\otimes_{\mathbb{Z}% }\mathbb{C}$ such that $\mathfrak{g}_{\alpha}\neq0$. We consider $F$ as a subspace of $P$.) \textbf{4)} We have $r_{i}\left( \alpha_{i}\right) =-\alpha_{i}$. Moreover, $r_{i}$ induces a permutation of all positive roots except for $\alpha_{i}$. \end{proposition} \textit{Proof of Proposition.} \textbf{1)} and \textbf{2)} follow easily from the corresponding statement for generators proven above. \textbf{3)} By part \textbf{2)}, the set of weights $P\left( V\right) $ of an integrable $\mathfrak{g}$-module $V$ is $W$-invariant. (Here, ``weight'' means a weight whose weight subspace is nonzero.) Applied to $V=\mathfrak{g}$, this implies \textbf{3)} (since $P\left( \mathfrak{g}\right) =0\cup R$). \textbf{4)} Proving $r_{i}\left( \alpha_{i}\right) =-\alpha_{i}$ is straightforward. Now for the other part: Any positive root can be written as $\alpha=\sum_{i}k_{i}\alpha_{i}$ where all $k_{i}$ are $\geq0$ and $\sum_{i}k_{i}>0$. Thus, for such a root, $r_{i}\left( \alpha\right) =\alpha-\alpha\left( h_{i}\right) \alpha_{i}=\sum_{j\neq i}k_{j}\alpha_{j}+\left( k_{i}% -\alpha\left( h_{i}\right) \right) \alpha_{i}$. If there exists a $j\neq i$ such that $k_{j}>0$, then $r_{i}\left( \alpha\right) $ must be a positive root (since there is no such thing as a partly-negative-partly-positive root). Alternative: $k_{j}=0$ for all $j\neq i$. But then $\alpha=k_{i}\alpha_{i}$, so that $k_{i}=1$ (because a positive multiple of a simple root is not a root, unless we are multiplying with $1$), but this is the case we excluded (``except for $\alpha_{i}$''). Proposition proven. \subsection{\textbf{[unfinished]} The Weyl-Kac character formula} \begin{theorem} [Kac]\label{thm.weylkac.weylkac}Denote by $P_{+}$ the set $\left\{ \chi\in P\ \mid\ \chi\left( h_{i}\right) \in\mathbb{N}\text{ for all }i\in\left\{ 1,2,...,r\right\} \right\} $. Let $\chi$ be a dominant integral weight of $\mathfrak{g}\left( A\right) $. (This means that $\chi\left( h_{i}\right) $ is a nonnegative integer for every $i\in\left\{ 1,2,...,r\right\} $.) Let $V$ be an integrable highest-weight $\mathfrak{g}_{\operatorname*{ext}}\left( A\right) $-module with highest weight $\chi$. Then: \textbf{(1)} The $\mathfrak{g}$-module $V$ is isomorphic to $L_{\chi}$. (In other words, the $\mathfrak{g}$-module $V$ is irreducible.) \textbf{(2)} The character of $V$ is% \[ \operatorname*{ch}\left( V\right) =\dfrac{\sum\limits_{w\in W}\det\left( w\right) \cdot e^{w\left( \chi+\rho\right) -\rho}}{\prod\limits_{\alpha >0}\left( 1-e^{-\alpha}\right) ^{\dim\left( \mathfrak{g}_{\alpha}\right) }}\ \ \ \ \ \ \ \ \ \ \text{in }R. \] Here, we recall that $R$ is the ring $\lim\limits_{\lambda\in P_{+}}% e^{\lambda}\mathbb{C}\left[ \left[ e^{-\alpha_{1}},e^{-\alpha_{2}% },...,e^{-\alpha_{r}}\right] \right] $ (note that this term increases when $\lambda$ is changed to $\lambda+\alpha_{i}$) in which the characters are defined. Here, $\rho$ is the element of $\mathfrak{h}^{\ast}$ satisfying $\rho\left( h_{i}\right) =1$ (as defined above). Since $\mathfrak{h}^{\ast}\subseteq P$, this $\rho$ becomes an element of $P$. Note that $\det\left( w\right) $ is always $1$ or $-1$ (and, in fact, equals $\left( -1\right) ^{k}$, where $w$ is written in the form $w=r_{i_{1}% }r_{i_{2}}...r_{i_{k}}$). \end{theorem} Part \textbf{(2)} of this theorem is called the \textit{Weyl-Kac character formula}. We want to prove this theorem. Since $\chi$ is a dominant integral weight, we have $\chi\in P_{+}$. Some comments on the theorem: First of all, part \textbf{(2)} implies part \textbf{(1)}, since both $V$ and $L_{\chi}$ satisfy the conditions of the Theorem and thus (according to part \textbf{(2)}) share the same character, but we also have a surjective homomorphism $\varphi:V\rightarrow L_{\chi}$, so (because of the characters being the same) it is an isomorphism. Thus, we only need to bother about proving part \textbf{(2)}. Secondly, let us remark that the theorem yields $L_{\lambda}=M_{\lambda }\diagup\left\langle f_{i}^{\lambda\left( h_{i}\right) +1}v_{\lambda}% \ \mid\ i\in\left\{ 1,2,...,r\right\} \right\rangle $ for all dominant integral weights $\lambda$. Indeed, denote $M_{\lambda}\diagup\left\langle f_{i}^{\lambda\left( h_{i}\right) +1}v_{\lambda}\ \mid\ i\in\left\{ 1,2,...,r\right\} \right\rangle $ by $L_{\lambda}^{\prime}$. Then, $L_{\lambda}^{\prime}$ is integrable (as we showed above more or less; more precisely, we showed that $L_{\lambda}$ was integrable, but this proof went exactly through proving that $L_{\lambda}^{\prime}$ is integrable), so that the theorem is still applicable to $L_{\lambda}^{\prime}$ and we obtain $L_{\lambda}^{\prime}\cong L_{\lambda}$. Our third remark: In the case of a simple finite-dimensional Lie algebra $\mathfrak{g}$, we have% \[ \operatorname*{ch}\left( M_{\lambda}\right) =\dfrac{e^{\lambda}}% {\prod\limits_{\alpha>0}\left( 1-e^{-\alpha}\right) }. \] The denominator can be rewritten $\prod\limits_{\alpha>0}\left( 1-e^{-\alpha }\right) ^{\dim\left( \mathfrak{g}_{\alpha}\right) }$, since $\dim\left( \mathfrak{g}_{\alpha}\right) =1$ for all roots $\alpha$. In the case of Kac-Moody Lie algebras $\mathfrak{g}=\mathfrak{g}\left( A\right) $, we can use similar arguments to show that% \[ \operatorname*{ch}\left( M_{\lambda}\right) =\dfrac{e^{\lambda}}% {\prod\limits_{\alpha>0}\left( 1-e^{-\alpha}\right) ^{\dim\left( \mathfrak{g}_{\alpha}\right) }}. \] So the Weyl-Kac character formula can be written as% \[ \operatorname*{ch}\left( V\right) =\sum\limits_{w\in W}\det\left( w\right) \cdot\operatorname*{ch}\left( M_{w\left( \chi+\rho\right) -\rho}\right) . \] This formula can be proven using the BGG\footnote{Bernstein-Gelfand-Gelfand} resolution (in fact, it is obtained as the Euler character of that resolution), but we will take a different route here. Another remark before we prove the formula. The Weyl-Kac character formula has the following corollary: \begin{corollary} [Weyl-Kac denominator formula]We have $\prod\limits_{\alpha>0}\left( 1-e^{-\alpha}\right) ^{\dim\left( \mathfrak{g}_{\alpha}\right) }% =\sum\limits_{w\in W}\det\left( w\right) \cdot e^{w\rho-\rho}$. \end{corollary} \textit{Proof of Corollary (using Weyl-Kac character formula).} Set $\chi=0$. Then $L_{\chi}=\mathbb{C}$, so that $\operatorname*{ch}\left( L_{\chi }\right) =1$ but on the other hand $\operatorname*{ch}\left( L_{\chi }\right) =\dfrac{\sum\limits_{w\in W}\det\left( w\right) \cdot e^{w\rho-\rho}}{\prod\limits_{\alpha>0}\left( 1-e^{-\alpha}\right) ^{\dim\left( \mathfrak{g}_{\alpha}\right) }}$. Thus, $\prod\limits_{\alpha >0}\left( 1-e^{-\alpha}\right) ^{\dim\left( \mathfrak{g}_{\alpha}\right) }=\sum\limits_{w\in W}\det\left( w\right) \cdot e^{w\rho-\rho}$. To prove the Weyl-Kac character formula, we will have to show several lemmas. \begin{lemma} \label{lem.weylkac.1}Let $\chi\in P_{+}$. \textbf{(1)} Then, $W\chi\subseteq D\left( \chi\right) $ (where, as we recall, $D\left( \chi\right) $ denotes the set $\left\{ \chi-\sum_{i}% k_{i}\alpha_{i}\ \mid\ k_{i}\in\mathbb{N}\text{ for all }i\right\} $. \textbf{(2)} If $D\subseteq D\left( \chi\right) $ is a $W$-invariant subset, then $D\cap P_{+}\neq\varnothing$. \end{lemma} \textit{Proof of Lemma \ref{lem.weylkac.1}.} \textbf{(1)} Consider $L_{\chi}$. Since $L_{\chi}$ is integrable, the set $P\left( L_{\chi}\right) $ is $W$-invariant, so that $W\chi\subseteq P\left( L_{\chi}\right) $. But $P\left( L_{\chi}\right) \subseteq D\left( \chi\right) $, since any weight of $L_{\chi}$ is $\chi$ minus a sum of positive roots. Part \textbf{(1)} is proven. \textbf{(2)} Let $\psi\in D$. Pick $w\in W$ such that $x-w\psi=\sum_{i}% k_{i}\alpha_{i}$ with nonnegative integers $k_{i}$ and minimal $\sum_{i}k_{i}% $. We claim that this $w$ satisfies $w\psi\in P_{+}$. This, of course, will prove part \textbf{(2)}. To prove $w\psi\in P_{+}$, assume that $w\psi\notin P_{+}$. Then, there exists an $i$ such that $\left( w\psi,\alpha_{i}\right) =d_{i}^{-1}\left( w\psi\right) \left( h_{i}\right) <0$. (Note that all the $d_{i}$ are $>0$.) Then, $r_{i}w\psi=w\psi-\left( w\psi\right) \left( h_{i}\right) \alpha _{i}$, so that $\chi-r_{i}w\psi=\chi-w\psi+\left( w\psi\right) \left( h_{i}\right) \alpha_{i}=\sum_{j}k_{j}\alpha_{j}+\left( w\psi\right) \left( h_{i}\right) \alpha_{i}=\sum_{j}k_{j}^{\prime}\alpha_{j}$ and $\sum_{j}% k_{j}^{\prime}=\sum_{j}k_{j}+\left( w\psi\right) \left( h_{i}\right) <\sum_{j}k_{j}$. This contradicts the minimality in our choice of $w$. Part \textbf{(2)} is thus proven. \begin{corollary} \label{cor.weylkac.2}Let $w\in W$ satisfy $w\neq1$. Then, there exists $i$ such that $w\alpha_{i}<0$. (By $w\alpha_{i}<0$ we mean that $w\alpha_{i}$ is a negative root.) \end{corollary} \textit{Proof of Corollary \ref{cor.weylkac.2}.} Choose $\chi\in P_{+}$ such that $w\chi\neq\chi$. (Such a $\chi$ always exists, due to the definition of $P_{+}$). Then, $w^{-1}\chi=\chi-\sum k_{i}\alpha_{i}$ for some $k_{i}% \in\mathbb{N}$ (by Lemma \ref{lem.weylkac.1} \textbf{(1)}). Hence,% \[ \chi=ww^{-1}\chi=w\chi-\sum k_{i}w\alpha_{i}=\left( \chi-\sum k_{i}^{\prime }\alpha_{i}\right) -\sum k_{i}w\alpha_{i}. \] Thus, $\sum k_{i}^{\prime}\alpha_{i}+\sum k_{i}w\alpha_{i}=0$. But $\sum k_{i}^{\prime}>0$, so there must exist an $i$ such that $w\alpha_{i}<0$. Corollary \ref{cor.weylkac.2} is proven. \begin{proposition} \label{prop.weylkac.3}Let $\varphi,\psi\in P$ be such that $\varphi\left( h_{i}\right) >0$ and $\psi\left( h_{i}\right) \geq0$ for each $i$. Let $w\in W$. Then, $w\varphi=\psi$ if and only if $\varphi=\psi$ and $w=1$. \end{proposition} \textit{Proof of Proposition \ref{prop.weylkac.3}.} For every $i$, we have $\varphi\left( h_{i}\right) >0$ if and only if $\left( \varphi,\alpha _{i}\right) >0$. Now suppose that there exists a $w\neq1$ such that $w\varphi=\psi$. Then, by Corollary \ref{cor.weylkac.2}, there exists an $i$ such that $w\alpha_{i}<0$. Then, $\left( \varphi,\alpha_{i}\right) >0$ but $\left( \varphi,\alpha_{i}\right) =\left( w^{-1}\psi,\alpha_{i}\right) =\left( \psi,w\alpha_{i}\right) \leq0$. This is a contradiction. Proposition \ref{prop.weylkac.3} is proven. Next, notice that $W$ acts on $R$. \begin{proposition} \label{prop.weylkac.4}Let $K$ denote the Weyl-Kac denominator $\prod \limits_{\alpha>0}\left( 1-e^{-\alpha}\right) ^{\dim\left( \mathfrak{g}% _{\alpha}\right) }$. Then, $w\cdot K=\det\left( w\right) \cdot K$ for every $w\in W$. \end{proposition} \textit{Proof of Proposition \ref{prop.weylkac.4}.} We can WLOG take $w=r_{i}$ (since $\det$ is multiplicative). Then,% \begin{align*} r_{i}K & =e^{r_{i}\rho}\prod\limits_{\alpha>0}\left( 1-e^{-r_{i}\alpha }\right) ^{\dim\left( \mathfrak{g}_{\alpha}\right) }=e^{r_{i}\rho}\left( 1-e^{+\alpha_{i}}\right) ^{\dim\left( \mathfrak{g}_{\alpha_{i}}\right) }\prod\limits_{\substack{\alpha>0;\\\alpha\neq\alpha_{i}}}\left( 1-e^{-\alpha}\right) ^{\dim\left( \mathfrak{g}_{\alpha}\right) }\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by Proposition \ref{prop.weylkac.prop0}% }\right) \\ & =e^{r_{i}\rho}\left( 1-e^{+\alpha_{i}}\right) \prod \limits_{\substack{\alpha>0;\\\alpha\neq\alpha_{i}}}\left( 1-e^{-\alpha }\right) ^{\dim\left( \mathfrak{g}_{\alpha}\right) }% \ \ \ \ \ \ \ \ \ \ \left( \text{since }\dim\left( \mathfrak{g}_{\alpha_{i}% }\right) =1\right) \\ & =\dfrac{e^{r_{i}\rho}\left( 1-e^{+\alpha_{i}}\right) }{e^{\rho}\left( 1-e^{-\alpha_{i}}\right) }\cdot K. \end{align*} Thus, we must only prove that $\dfrac{e^{r_{i}\rho}\left( 1-e^{+\alpha_{i}% }\right) }{e^{\rho}\left( 1-e^{-\alpha_{i}}\right) }=-1$. But this is very easy: We have $r_{i}\rho=\rho-\underbrace{\rho\left( h_{i}\right) }_{=1}\alpha_{i}=\rho-\alpha_{i}$, so that \[ \dfrac{e^{r_{i}\rho}\left( 1-e^{+\alpha_{i}}\right) }{e^{\rho}\left( 1-e^{-\alpha_{i}}\right) }=\dfrac{e^{\rho-\alpha_{i}}\left( 1-e^{+\alpha _{i}}\right) }{e^{\rho}\left( 1-e^{-\alpha_{i}}\right) }=\dfrac {e^{-\alpha_{i}}\left( 1-e^{+\alpha_{i}}\right) }{1-e^{-\alpha_{i}}}% =\dfrac{e^{-\alpha_{i}}-1}{1-e^{-\alpha_{i}}}=-1. \] Proposition \ref{prop.weylkac.4} is proven. \begin{proposition} \label{prop.weylkac.5}Let $\mu,\nu\in P_{+}$ be such that $\mu\in D\left( \nu\right) $ and $\mu\neq\nu$. Then, $\left( \nu+\rho\right) ^{2}-\left( \mu+\rho\right) ^{2}>0$. Here, $\lambda^{2}$ is defined to mean the inner product $\left( \lambda,\lambda\right) $. \end{proposition} \textit{Proof of Proposition \ref{prop.weylkac.5}.} We have $\nu-\mu =\sum\limits_{i}k_{i}\alpha_{i}$ for some $k_{i}\geq0$ (since $\mu\in D\left( \nu\right) $). There exists an $i$ such that $k_{i}>0$ (because $\mu\neq\nu $). Now,% \[ \left( \nu+\rho\right) ^{2}-\left( \mu+\rho\right) ^{2}=\left( \nu -\mu,\mu+\nu+2\rho\right) =\sum\limits_{i}k_{i}\left( \alpha_{i},\mu +\nu+2\rho\right) . \] But now use $\left( \alpha_{i},\mu\right) \geq0$ (since $\mu\in P_{+}$), also $\left( \alpha_{i},\nu\right) \geq0$ (since $\nu\in P_{+}$) and $\left( \alpha_{i},\rho\right) =d_{i}^{-1}>0$ to conclude that this is $>0$ (since there exists an $i$ such that $k_{i}>0$). Proposition \ref{prop.weylkac.5} is proven. \begin{proposition} \label{prop.weylkac.6}Suppose that $V$ is a $\mathfrak{g}_{\operatorname*{ext}% }\left( A\right) $-module from Category $\mathcal{O}$ such that the Casimir $C$ satisfies $\Delta\mid_{V}=\gamma\cdot\operatorname*{id}$. Then, $\operatorname*{ch}\left( V\right) =\sum c_{\lambda}\operatorname*{ch}% \left( M_{\lambda}\right) $, where the sum is over all $\lambda$ satisfying $\left( \lambda,\lambda+2\rho\right) =\gamma$, and $c_{\lambda}\in \mathbb{Z}$ are some integers. \end{proposition} \textit{Proof of Proposition \ref{prop.weylkac.6}.} The expansion is built inductively as follows: Suppose $P\left( V\right) \subseteq D\left( \lambda_{1}\right) \cup D\left( \lambda_{2}\right) \cup...\cup D\left( \lambda_{m}\right) $ for some weights $\lambda_{1},\lambda_{2},...,\lambda_{m}$. Assume that this is a minimal such union. Then, $\lambda_{i}+\alpha_{j}\notin P\left( V\right) $ for any $i,j$. Let $d_{i}=\dim\left( V\left[ \lambda_{i}\right] \right) $. Then, we have a homomorphism $\varphi:\bigoplus_{i}d_{i}M_{\lambda_{i}}\rightarrow V$ which is an isomorphism in weight $\lambda_{i}$. Let $K=\operatorname*{Ker}\varphi$. Let $C=\operatorname*{Coker}\varphi$. Clearly, both $K$ and $C$ lie in Category $\mathcal{O}$. We have an exact sequence $0\rightarrow K\rightarrow \bigoplus_{i}d_{i}M_{\lambda_{i}}\rightarrow V\rightarrow C\rightarrow0$. Since the alternating sum of characters in an exact sequence is $0$, this yields $\operatorname*{ch}V=\sum_{i}d_{i}\operatorname*{ch}\left( M_{\lambda_{i}}\right) -\operatorname*{ch}K+\operatorname*{ch}C$. Now we claim that $\Delta\mid_{M_{\lambda_{i}}}=\left( \lambda_{i}% ,\lambda_{i}+2\rho\right) =\gamma$ if $d_{i}\neq0$. (Otherwise, a homomorphism $\varphi$ could not exist.) Also, $\Delta\mid_{K}=\Delta\mid_{C}=\gamma$. But if $\mu\in P\left( K\right) \cup P\left( C\right) $, then for some $i$, we have $\lambda_{i}-\mu=\sum k_{j}\alpha_{j}$ with $\sum k_{j}\geq1$. Next step: $\sum k_{i}\geq2$. Etc. If we run this procedure indefinitely, eventually every weight in this cone will be exhausted. Then we apply the procedure to $K$ and $C$, and then to their $K$ and $C$ etc.. \textit{Proof of Weyl-Kac character formula.} According to Proposition \ref{prop.weylkac.6}, we have% \[ \operatorname*{ch}\left( V\right) =\sum_{\psi\in D\left( \chi\right) }c_{\psi}\operatorname*{ch}\left( M_{\psi}\right) \ \ \ \ \ \ \ \ \ \ \text{with }c_{\chi}=1. \] We will now need: \begin{corollary} \label{cor.weylkac.7}If $c_{\psi}\neq0$, then $\left( \psi+\rho\right) ^{2}=\left( \chi+\rho\right) ^{2}$. \end{corollary} \textit{Proof of Corollary \ref{cor.weylkac.7}.} This follows from Proposition \ref{prop.weylkac.6}. \begin{lemma} \label{lem.weylkac.8}If $\psi+\rho=w\left( \chi+\rho\right) $, then $c_{\psi}=\det\left( w\right) \cdot c_{\chi}$. \end{lemma} \textit{Proof of Lemma \ref{lem.weylkac.8}.} We have $wK=\left( \det w\right) \cdot K$ and $w\cdot\operatorname*{ch}V=\operatorname*{ch}V$. Hence, $w\left( K\cdot\operatorname*{ch}V\right) =\left( \det w\right) \cdot\left( K\operatorname*{ch}V\right) $. But since $\operatorname*{ch}% \left( M_{\psi}\right) =\dfrac{\sum c_{\psi}e^{\psi+\rho}}{K}$, we have $K\operatorname*{ch}V=\sum\limits_{\psi\in D\left( \chi\right) }c_{\psi }e^{\psi+\rho}=\left( \det w\right) \cdot\sum\limits_{\psi\in D\left( \chi\right) }c_{\psi}e^{\psi+\rho}$. (If $\psi+\rho=w\left( \chi +\rho\right) $.) Thus, $c_{\psi}=\left( \det w\right) \cdot c_{\chi}$. \begin{lemma} \label{lem.weylkac.9}Let $D=\left\{ \psi\ \mid\ c_{\psi-\rho}\neq0\right\} $. Then, $D=W\left( \chi+\rho\right) $. \end{lemma} \textit{Proof of Lemma \ref{lem.weylkac.9}.} We have $W\left( \chi +\rho\right) \subseteq D$ by Lemma \ref{lem.weylkac.8}. Also, $D$ is $W$-invariant since $V$ is integrable. Suppose $D\neq W\left( \chi+\rho\right) $. Then, $\left( D\diagdown W\left( \chi+\rho\right) \right) \cap P_{+}\neq\varnothing$ by Lemma \ref{lem.weylkac.1} \textbf{(2)}. Take some $\beta\in\left( D\diagdown W\left( \chi+\rho\right) \right) \cap P_{+}$. Then, $\beta-\rho\in D\left( \chi\right) $, so that $\left( \chi+\rho,\chi+\rho\right) -\left( \beta,\beta\right) >0$ (by Proposition \ref{prop.weylkac.5}). Thus, $\beta$ cannot occur in the sum (by Corollary \ref{cor.weylkac.7}). Punchline: $\operatorname*{ch}V=\sum_{w\in W}\dfrac{\left( \det w\right) \cdot e^{w\left( \chi+\rho\right) }}{K}$. This is exactly the Weyl-Kac character formula. \subsection{\textbf{[unfinished]} ...} [...] \section{\textbf{[unfinished]} ...} [...] [747l22.pdf] KZ equations, consistent (define a flat connection) $\mathfrak{g}$ simple Lie algebra $V_{1},V_{2},...,V_{N}$ representations of $\mathfrak{g}$ from Category $\mathcal{O}$. $\mathbb{C}_{0}^{N}=\mathbb{C}^{N}\diagdown\left\{ z_{i}=z_{j}\right\} $ $U\subseteq\mathbb{C}_{0}^{N}$ simply connected open set $F\left( z_{1},...,z_{N}\right) \in\left( V_{1}\otimes V_{2}\otimes ...\otimes V_{N}\right) \left[ \nu\right] $ holomorphic function in $z_{1},...,z_{N}$ for a fixed weight $\nu$. $x\in\mathbb{C}$ [or was it $\kappa\in\mathbb{C}$ ?] $\dfrac{\partial F}{\partial z_{i}}-\overline{h}\sum\limits_{i\neq j}% \dfrac{\Omega_{i,j}}{z_{i}-z_{j}}F$ where $\Omega_{i,j}:V_{1}\otimes V_{2}\otimes...\otimes V_{N}\rightarrow V_{1}\otimes V_{2}\otimes...\otimes V_{N}$ $\Omega\in\left( S^{2}\mathfrak{g}\right) ^{\mathfrak{g}}$ Consistent means: setting $\nabla_{i}=\dfrac{\partial}{\partial z_{i}% }-\overline{h}\sum\limits_{i\neq j}\dfrac{\Omega_{i,j}}{z_{i}-z_{j}}$, we have $\left[ \nabla_{i},\nabla_{j}\right] =0$. Consistent systems are known to have locally unique-and-existent solutions. Why is this in our course? The reason is that these equations arise in the representation theory of affine Lie algebras. Interpretation of KZ equations in terms of $\widehat{\mathfrak{g}}$: Consider $L\mathfrak{g}$, $\widehat{\mathfrak{g}}$, $\widetilde{\mathfrak{g}% }=\widehat{\mathfrak{g}}\rtimes\mathbb{C}d$. Define Weyl modules: \begin{definition} Let $\lambda\in P_{+}$ be a dominant integral weight for a simple finite-dimensional Lie algebra $\mathfrak{g}$. Let $L_{\lambda}$ be an irreducible finite-dimensional representation of $\mathfrak{g}$ with highest weight $\lambda$. Let us extend $L_{\lambda}$ to a $\mathfrak{g}\left[ t\right] \oplus\mathbb{C}K$-module by making $t\mathfrak{g}\left[ t\right] $ act by $0$ and $K$ act by some scalar $k$ (that is, $K\mid_{L_{\lambda}% }=k\cdot\operatorname*{id}$ for some $k\in\mathbb{C}$). Denote this $\mathfrak{g}\left[ t\right] \oplus\mathbb{C}K$-module by $L_{\lambda}^{\left( k\right) }$. Then, we define a $\widehat{\mathfrak{g}}% $-module $V_{\lambda,k}=U\left( \widehat{\mathfrak{g}}\right) \otimes _{U\left( \mathfrak{g}\left[ t\right] \oplus\mathbb{C}K\right) }% L_{\lambda}^{\left( k\right) }$. This module is called a \textit{Weyl module} for $\widehat{\mathfrak{g}}$ at level $k$. \end{definition} By the PBW theorem, we immediately see that $U\left( \widehat{\mathfrak{g}% }\right) \cong U\left( t^{-1}\mathfrak{g}\left[ t^{-1}\right] \right) \otimes U\left( \mathfrak{g}\left[ t\right] \oplus\mathbb{C}K\right) $ and thus $V_{\lambda,k}\cong U\left( t^{-1}\mathfrak{g}\left[ t^{-1}\right] \right) \otimes L_{\lambda}$ (canonically, but only as vector spaces). Assuming that $k\neq-h^{\vee}$, we can extend $V_{\lambda,k}$ to $\widetilde{\mathfrak{g}}$ by letting $d$ act as $-L_{0}$ (from Sugawara construction). \begin{definition} If $V$ is a $\mathfrak{g}$-module, then $V\left[ z,z^{-1}\right] $ is an $L\mathfrak{g}$-module, and in fact a $\widehat{\mathfrak{g}}$-module where $K$ acts by $0$. It extends to $\widetilde{\mathfrak{g}}$ by setting $d=z\dfrac{\partial}{\partial z}$. More generally: Can set $d\left( vz^{n}\right) =\left( n-\Delta\right) vz^{n}$ for any fixed $\Delta\in\mathbb{C}$. Call this module $z^{-\Delta}V\left[ z,z^{-1}\right] $. \end{definition} \begin{lemma} If $k\notin\mathbb{Q}$, then $V_{\lambda,k}$ is irreducible. \end{lemma} \textit{Proof of Lemma.} Assume $V_{\lambda,k}$ is reducible. This $V_{\lambda,k}$ is a highest-weight module. So, it must have a singular vector in degree $\ell>0$. Let $C$ be the Casimir for $\widetilde{\mathfrak{g}}$. We know $C=L_{0}-\deg$ (where $\deg$ returns the positive degree). Assume that $w$ (our singular vector) lives in an irr. repr. of $\mathfrak{g}% $. Singular vector means $a\left( m\right) w=0$ for all $m>0$. Here $a\left( m\right) $ means $at^{m}$. $C\mid_{V_{\lambda,k}}=\dfrac{\left( \lambda,\lambda+2\rho\right) }{2\left( k+h^{\vee}\right) }$ $Cw=\left( \dfrac{\left( \mu,\mu+2\rho\right) }{2\left( k+h^{\vee}\right) }-\ell\right) w$ $L_{0}=\dfrac{1}{2\left( k+h^{\vee}\right) }\sum_{i\in\mathbb{Z}}\sum_{a\in B}:a\left( i\right) a\left( -i\right) :\ =\dfrac{1}{2\left( k+h^{\vee }\right) }\left( \sum_{a\in B}a\left( 0\right) ^{2}+2\sum_{a\in B}% \sum_{m\geq1}a\left( -m\right) a\left( m\right) \right) $ where $a\left( m\right) =at^{m}$. $\Longrightarrow$ $\underbrace{\left( \lambda,\lambda+2\rho\right) =\left( \mu,\mu+2\rho\right) }_{\in\mathbb{Z}}-2\ell\left( k+h^{\vee}\right) $ $\Longrightarrow$ $k=-h^{\vee}+\dfrac{\left( \lambda,\lambda+2\rho\right) -\left( \mu,\mu+2\rho\right) }{2\ell}\in\mathbb{Q}$. $\Longrightarrow$ contradiction. \begin{corollary} If $k\notin\mathbb{Q}$, then $V_{\lambda,k}^{\ast}$ (restricted dual) is $U\left( \widehat{\mathfrak{g}}\right) \otimes_{U\left( \mathfrak{g}\left[ t^{-1}\right] \oplus\mathbb{C}K\right) }L_{\lambda}^{\ast\left( -k\right) }$. (Here, $L_{\lambda}^{\ast\left( -k\right) }$ means $L_{\lambda}^{\ast}$ with $K$ acting as $-k$.) \end{corollary} \textit{Proof of Corollary.} From Frobenius reciprocity, we have a homomorphism $\varphi:U\left( \widehat{\mathfrak{g}}\right) \otimes _{U\left( \mathfrak{g}\left[ t^{-1}\right] \oplus\mathbb{C}K\right) }L_{\lambda}^{\ast\left( -k\right) }\rightarrow V_{\lambda,k}^{\ast}$ which is $\operatorname*{id}$ in degree $0$. In fact, Frobenius reciprocity tells us that% \[ \operatorname*{Hom}\nolimits_{\widehat{\mathfrak{g}}}\left( U\left( \widehat{\mathfrak{g}}\right) \otimes_{U\left( \mathfrak{g}\left[ t^{-1}\right] \oplus\mathbb{C}K\right) }L_{\lambda}^{\ast\left( -k\right) },M\right) \cong\operatorname*{Hom}\nolimits_{\mathfrak{g}\left[ t^{-1}\right] \oplus\mathbb{C}K}\left( L_{\lambda}^{\ast\left( -k\right) },M\right) , \] which, in the case $M=V_{\lambda,k}^{\ast}$, becomes [...]. Because $V_{\lambda,k}$ is irreducible (here we are using $k\notin\mathbb{Q}% $), $V_{\lambda,k}^{\ast}$ is irreducible as well, this homomorphism $\varphi$ is surjective. This $\varphi$ also preserves grading, and the characters are equal. $\Longrightarrow$ $\varphi$ is an isomorphism. \begin{corollary} $\operatorname*{Hom}\nolimits_{\widetilde{\mathfrak{g}}}\left( V_{\lambda ,k}\otimes V_{\nu,k}^{\ast},z^{-\Delta}V\left[ z,z^{-1}\right] \right) \cong\operatorname*{Hom}\nolimits_{\mathfrak{g}}\left( L_{\lambda}\otimes L_{\nu}^{\ast},V\right) $ if $\Delta=\Delta\left( \lambda\right) -\Delta\left( \nu\right) $. \end{corollary} \textit{Proof of Corollary.} Frobenius reciprocity as for the previous corollary. (Skip.) [...] We now cite a classical theorem on ODEs. \begin{theorem} Let $N\in\mathbb{N}$. Let $A\left( z\right) =A_{0}+A_{1}z+A_{2}z^{2}+...$ be a holomorphic function on $\left\{ z\in\mathbb{C}\ \mid\ \left\vert z\right\vert <1\right\} $ with values in $\operatorname*{M}\nolimits_{N}% \left( \mathbb{C}\right) $. Assume that for any eigenvalues $\lambda$ and $\mu$ of $A_{0}$ such that $\lambda\neq\mu$, one has $\lambda-\mu \notin\mathbb{Z}$. Then, the ODE $z\dfrac{dF}{dz}=A\left( z\right) F$ (which, of course, is equivalent to $\dfrac{dF}{dz}=\dfrac{A\left( z\right) }{z}F$) has a matrix solution of the form $F\left( z\right) =\left( 1+B_{1}z+B_{2}z^{2}+...\right) z^{A_{0}}$ such that the power series $1+B_{1}z+B_{2}z^{2}+...$ converges for $\left\vert z\right\vert <1$. Here, $z^{A_{0}}$ means $\exp\left( A_{0}\log z\right) $ (on $\mathbb{C}% \diagdown\mathbb{R}_{\leq0}$). \end{theorem} \begin{remark} This is a development of the following basic theorem: If we are given an ODE $\dfrac{dF}{dz}=C\left( z\right) F$ with $C\left( z\right) $ holomorphic, then there exists a holomorphic $F$ satisfying this equation and having the form $F=1+O\left( z\right) $ (the so-called fundamental equation). \end{remark} \textit{Proof of Theorem.} Plug in the solution $F\left( z\right) $ in the above formula:% \[ \left( \sum\limits_{n\geq1}nB_{n}z^{n}\right) z^{A_{0}}+\left( 1+\sum\limits_{n\geq1}B_{n}z^{n}\right) A_{0}z^{A_{0}}=\left( A_{0}% +A_{1}z+A_{2}z^{2}+...\right) \left( 1+B_{1}z+B_{2}z^{2}+...\right) z^{A_{0}}. \] Cancel $z^{A_{0}}$ from this to obtain% \[ \sum\limits_{n\geq1}nB_{n}z^{n}+\left( 1+\sum\limits_{n\geq1}B_{n}% z^{n}\right) A_{0}=\left( A_{0}+A_{1}z+A_{2}z^{2}+...\right) \left( 1+B_{1}z+B_{2}z^{2}+...\right) . \] This is the system of recursive equations% \[ nB_{n}-A_{0}B_{n}+B_{n}A_{0}=A_{1}B_{n-1}+A_{2}B_{n-2}+...+A_{n-1}B_{1}% +A_{n}. \] This rewrites as% \[ \left( n-\operatorname*{ad}A_{0}\right) \left( B_{n}\right) =A_{1}% B_{n-1}+A_{2}B_{n-2}+...+A_{n-1}B_{1}+A_{n}. \] The operator $n-\operatorname*{ad}A_{0}:\operatorname*{M}\nolimits_{N}\left( \mathbb{C}\right) \rightarrow\operatorname*{M}\nolimits_{N}\left( \mathbb{C}\right) $ is invertible (because eigenvalues of this operator are $n-\left( \lambda-\mu\right) $ for $\lambda$ and $\mu$ being eigenvalues of $A_{0}$, and because of the condition that for any eigenvalues $\lambda$ and $\mu$ of $A_{0}$ such that $\lambda\neq\mu$, one has $\lambda-\mu \notin\mathbb{Z}$). Hence, we can use the above equation to recursively compute $B_{n}$ for all $n$. This implies that a solution in the formal sense exists. We also need to estimate radius of convergence. [...] The following generalizes our theorem to several variables: \begin{theorem} Let $m\in\mathbb{N}$ and $N\in\mathbb{N}$. For every $i\in\left\{ 1,2,...,m\right\} $, let $A_{i}\left( \xi_{1},\xi_{2},...,\xi_{m}\right) $ be a holomorphic on $\left\{ \left( \xi_{1},\xi_{2},...,\xi_{m}\right) \ \mid\ \left\vert \xi_{j}\right\vert <1\text{ for all }j\right\} $ with values in $\operatorname*{M}\nolimits_{N}\left( \mathbb{C}\right) $. Consider the system of differential equations $\xi_{i}\dfrac{dF}{d\xi_{i}% }=A_{i}\left( \xi\right) F$ for all $i\in\left\{ 1,2,...,m\right\} $ on a single function $F:\mathbb{C}^{m}\rightarrow\operatorname*{M}\nolimits_{N}% \left( \mathbb{C}\right) $. Assume \[ \left[ \xi_{i}\dfrac{d}{d\xi_{i}}-A_{i},\xi_{j}\dfrac{d}{d\xi_{j}}% -A_{j}\right] =0\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{ 1,2,...,m\right\} \] (this is called a \textit{consistency condition}, aka a zero curvature equation). Then, $\left[ A_{i}\left( 0\right) ,A_{j}\left( 0\right) \right] =0$ for all $i,j\in\left\{ 1,2,...,m\right\} $, and thus the matrices $A_{i}\left( 0\right) $ for all $i$ can be simultaneously trigonalized. Under this trigonalization, let $\lambda_{i,1}$, $\lambda_{i,2}% $, $...$, $\lambda_{i,N}$ be the diagonal entries of $A_{i}\left( 0\right) $. Assume that the condition \[ \left( \lambda_{1,k}-\lambda_{1,\ell},\lambda_{2,k}-\lambda_{2,\ell },...,\lambda_{m,k}-\lambda_{m,\ell}\right) \notin\mathbb{Z}^{m}\diagdown0 \] holds for all $k$ and $\ell$. [...] \end{theorem} \end{document}