}.)
\end{lemma}
\textit{Proof of Proposition \ref{prop.pbw.U(g)h}}. Let $\left( e_{i}\right)
_{i\in P}$ be a basis of the $k$-module $\mathfrak{h}$, and let $\left(
e_{i}\right) _{i\in Q}$ be a basis of the $k$-module $N$. Assume WLOG that
the sets $P$ and $Q$ are disjoint. Let $I=P\cup Q$. Choose a well-ordering on
$I$ such that every element of $Q$ is smaller than any element of $P$. Then,
$\left( e_{i}\right) _{i\in I}$ is a basis of the $k$-module $\mathfrak{h}%
\oplus N=\mathfrak{g}$. Proposition \ref{thm.pbw.basis} thus yields that
$\left( \overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes e_{i_{n}}%
}\right) _{\substack{n\in\mathbb{N};\ \left( i_{1},i_{2},...,i_{n}\right)
\in I^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n}}}$ is a basis of the $k$-module
$U\left( \mathfrak{g}\right) $.
Let $I^{\ast}$ be the disjoint union of the sets $I^{n}$ for all
$n\in\mathbb{N}$. In other words, let $I^{\ast}$ be the set of all finite
sequences of elements of $I$. In particular, the empty sequence (i. e., the
only element of $I^{0}$) is an element of $I^{\ast}$.
For every two elements $\kappa\in I^{\ast}$ and $\kappa^{\prime}\in I^{\ast}$,
we define an element $\kappa\cdot\kappa^{\prime}\in I^{\ast}$ as follows:
Write $\kappa$ in the form $\kappa=\left( i_{1},i_{2},...,i_{n}\right) $ and
write $\kappa^{\prime}$ in the form $\kappa^{\prime}=\left( j_{1}%
,j_{2},...,j_{m}\right) $; then set $\kappa\cdot\kappa^{\prime}=\left(
i_{1},i_{2},...,i_{n},j_{1},j_{2},....,j_{m}\right) $.
Let $K$ be the set%
\begin{align*}
& \left\{ \left( i_{1},i_{2},...,i_{n}\right) \ \mid\ \left( i_{1}%
,i_{2},...,i_{n}\right) \in I^{\ast};\ i_{1}\leq i_{2}\leq...\leq
i_{n}\right\} \\
& =\bigcup_{n\in\mathbb{N}}\left\{ \left( i_{1},i_{2},...,i_{n}\right)
\ \mid\ \left( i_{1},i_{2},...,i_{n}\right) \in I^{n};\ i_{1}\leq i_{2}%
\leq...\leq i_{n}\right\} .
\end{align*}
In other words, $K$ is the set of all increasing finite sequences of elements
of $I$.
Let us note that every $q\in K\cap Q^{\ast}$ and $p\in K\cap P^{\ast}$ satisfy
$q\cdot p\in K$. (This is because we have chosen a well-ordering on $I$ such
that every element of $Q$ is smaller than any element of $P$.)
For every $\kappa\in I^{\ast}$, define an element $\mathbf{e}_{\kappa}$ of
$U\left( \mathfrak{g}\right) $ by%
\[
\mathbf{e}_{\kappa}=\overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes
e_{i_{n}}},\ \ \ \ \ \ \ \ \ \ \text{where }\left( i_{1},i_{2},...,i_{n}%
\right) \text{ is such that }\kappa=\left( i_{1},i_{2},...,i_{n}\right) .
\]
Then, $\left( \mathbf{e}_{\kappa}\right) _{\kappa\in K}$ is a basis of the
$k$-module $U\left( \mathfrak{g}\right) $ (since this is just another way to
state our knowledge that $\left( \overline{e_{i_{1}}\otimes e_{i_{2}}%
\otimes...\otimes e_{i_{n}}}\right) _{\substack{n\in\mathbb{N};\ \left(
i_{1},i_{2},...,i_{n}\right) \in I^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n}}}$
is a basis of the $k$-module $U\left( \mathfrak{g}\right) $). Thus,
$U\left( \mathfrak{g}\right) =\left\langle \mathbf{e}_{\kappa}\ \mid
\ \kappa\in K\right\rangle $.
Let $X$ be the subset $\left\{ \left( i_{1},i_{2},...,i_{n}\right) \in
K\ \mid\ i_{n}\in P\right\} $ of $K$.
Let $Y$ be the subset $\left\{ \left( i_{1},i_{2},...,i_{n}\right) \in
K\ \mid\ n\leq m\right\} $ of $K$.
Clearly, $X\cap Y$ is the subset $\left\{ \left( i_{1},i_{2},...,i_{n}%
\right) \in K\ \mid\ i_{n}\in P;\ n\leq m\right\} $ of $K$.
\begin{noncompile}
Let $Z$ be the subset $\left\{ \left( i_{1},i_{2},...,i_{n}\right) \in
K\ \mid\ n\leq m-1\right\} $ of $K$.
\end{noncompile}
We notice that $\left( \mathbf{e}_{\kappa}\right) _{\kappa\in Y}$ is a basis
of the $k$-module $U_{\leq m}\left( \mathfrak{g}\right) $ (since this is
just another way to state our knowledge that the family $\left(
\overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes e_{i_{n}}}\right)
_{\substack{n\in\mathbb{N};\ \left( i_{1},i_{2},...,i_{n}\right) \in
I^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n};\ n\leq m}}$ is a basis of the
$k$-module $U_{\leq m}\left( \mathfrak{g}\right) $).
\begin{noncompile}
Applying this very fact to $m-1$ instead of $m$, we see that $\left(
\mathbf{e}_{\kappa}\right) _{\kappa\in Z}$ is a basis of the $k$-module
$U_{\leq\left( m-1\right) }\left( \mathfrak{g}\right) $.
\end{noncompile}
The definition of $\mathbf{e}_{\kappa}$ readily yields%
\begin{equation}
\mathbf{e}_{\kappa}\cdot\mathbf{e}_{\kappa^{\prime}}=\mathbf{e}_{\kappa
\cdot\kappa^{\prime}}\ \ \ \ \ \ \ \ \ \ \text{for every }\kappa\in I^{\ast
}\text{ and }\kappa^{\prime}\in I^{\ast}. \label{pf.pbw.U(g)h.0}%
\end{equation}
Also note that%
\begin{equation}
\mathbf{e}_{\left( i\right) }=e_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in
I, \label{pf.pbw.U(g)h.-1}%
\end{equation}
where $e_{i}$ means $\overline{e_{i}}=\psi\left( e_{i}\right) \in U\left(
\mathfrak{g}\right) $ on the right hand side (this is a slight abuse of
notation, but legitimate in view of the Poincar\'{e}-Birkhoff-Witt theorem).
Now let us show that $U\left( \mathfrak{g}\right) \cdot\mathfrak{h}%
=\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle $. In fact,
\begin{align}
\underbrace{U\left( \mathfrak{g}\right) }_{=\left\langle \mathbf{e}_{\kappa
}\ \mid\ \kappa\in K\right\rangle }\cdot\underbrace{\mathfrak{h}%
}_{\substack{=\left\langle e_{i}\ \mid\ i\in P\right\rangle \\\text{(since
}\left( e_{i}\right) _{i\in P}\text{ is a basis of }\mathfrak{h}\text{)}}}
& =\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in K\right\rangle
\cdot\left\langle e_{i}\ \mid\ i\in P\right\rangle \nonumber\\
& =\left\langle \mathbf{e}_{\kappa}e_{i}\ \mid\ \left( \kappa,i\right) \in
K\times P\right\rangle . \label{pf.pbw.U(g)h.3a}%
\end{align}
Hence, in order to prove that $U\left( \mathfrak{g}\right) \cdot
\mathfrak{h}=\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle
$, it is enough to show that $\left\langle \mathbf{e}_{\kappa}e_{i}%
\ \mid\ \left( \kappa,i\right) \in K\times P\right\rangle =\left\langle
\mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle $. In order to do so, we
must prove that%
\begin{equation}
\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle
\subseteq\left\langle \mathbf{e}_{\kappa}e_{i}\ \mid\ \left( \kappa,i\right)
\in K\times P\right\rangle \label{pf.pbw.U(g)h.1}%
\end{equation}
and%
\begin{equation}
\left\langle \mathbf{e}_{\kappa}e_{i}\ \mid\ \left( \kappa,i\right) \in
K\times P\right\rangle \subseteq\left\langle \mathbf{e}_{\kappa}\ \mid
\ \kappa\in X\right\rangle . \label{pf.pbw.U(g)h.2}%
\end{equation}
\textit{Proof of (\ref{pf.pbw.U(g)h.1}).} Every $\kappa\in X$ can be written
in the form $\left( i_{1},i_{2},...,i_{n}\right) $ for some $n\in\mathbb{N}$
and $\left( i_{1},i_{2},...,i_{n}\right) \in K$ satisfying $i_{n}\in P$.
Thus,%
\begin{align*}
\mathbf{e}_{\kappa} & =\mathbf{e}_{\left( i_{1},i_{2},...,i_{n}\right)
}=\mathbf{e}_{\left( i_{1},i_{2},...,i_{n-1}\right) \cdot\left(
i_{n}\right) }=\underbrace{\mathbf{e}_{\left( i_{1},i_{2},...,i_{n-1}%
\right) }}_{\in U\left( \mathfrak{g}\right) }\underbrace{\mathbf{e}%
_{\left( i_{n}\right) }}_{=e_{i_{n}}\text{ (by (\ref{pf.pbw.U(g)h.-1}))}%
}\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.pbw.U(g)h.0})}\right) \\
& \in U\left( \mathfrak{g}\right) \underbrace{e_{i_{n}}}_{\in
\mathfrak{h}\text{ (since }i_{n}\in P\text{)}}\subseteq U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}.
\end{align*}
We have thus shown that every $\kappa\in X$ satisfies $\mathbf{e}_{\kappa}\in
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}$. Therefore,%
\begin{equation}
\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle \subseteq
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}. \label{pf.pbw.U(g)h.1a}%
\end{equation}
Combined with (\ref{pf.pbw.U(g)h.3a}), this proves (\ref{pf.pbw.U(g)h.1}).
\textit{Proof of (\ref{pf.pbw.U(g)h.2}).} Let $\left( \kappa,i\right) \in
K\times P$ be arbitrary. We only have to prove that $\mathbf{e}_{\kappa}%
e_{i}\in\left\langle \mathbf{e}_{\kappa^{\prime}}\ \mid\ \kappa^{\prime}\in
X\right\rangle $ (because once this is shown for every $\left( \kappa
,i\right) \in K\times P$, it will become clear that $\left\langle
\mathbf{e}_{\kappa}e_{i}\ \mid\ \left( \kappa,i\right) \in K\times
P\right\rangle \subseteq\left\langle \mathbf{e}_{\kappa^{\prime}}%
\ \mid\ \kappa^{\prime}\in X\right\rangle =\left\langle \mathbf{e}_{\kappa
}\ \mid\ \kappa\in X\right\rangle $, and this will prove (\ref{pf.pbw.U(g)h.2})).
Since $\kappa\in K$, we can write $\kappa$ in the form $\left( i_{1}%
,i_{2},...,i_{n}\right) $ for some $n\in\mathbb{N}$ and $\left( i_{1}%
,i_{2},...,i_{n}\right) \in I^{n}$ satisfying $i_{1}\leq i_{2}\leq...\leq
i_{n}$. Let $i_{n+1}=i$. Let $\nu$ be the smallest integer in $\left\{
1,2,...,n+1\right\} $ such that $i_{\nu}\in P$ (such a $\nu$ exists since
$i_{n+1}=i\in P$). Then, $i_{1}$, $i_{2}$, $...$, $i_{\nu-1}$ lie in $Q$
whereas $i_{\nu}$, $i_{\nu+1}$, $...$, $i_{n}$ lie in $P$ (this is because
$i_{1}\leq i_{2}\leq...\leq i_{n}$ and because every element of $Q$ is smaller
than any element of $P$). Since $i_{n+1}=i$ also lies in $P$, we conclude that
all of the elements $i_{\nu}$, $i_{\nu+1}$, $...$, $i_{n+1}$ lie in $P$. This
yields $\left( i_{\nu},i_{\nu+1},...,i_{n+1}\right) \in P^{\ast}$.
Since $i_{1}$, $i_{2}$, $...$, $i_{\nu-1}$ lie in $Q$, we have $\left(
i_{1},i_{2},...,i_{\nu-1}\right) \in Q^{\ast}$. Combined with $\left(
i_{1},i_{2},...,i_{\nu-1}\right) \in K$ (since $i_{1}\leq i_{2}\leq...\leq
i_{\nu-1}$), this yields $\left( i_{1},i_{2},...,i_{\nu-1}\right) \in K\cap
Q^{\ast}$.
Now we have%
\begin{align*}
\mathbf{e}_{\kappa}e_{i} & =\mathbf{e}_{\left( i_{1},i_{2},...,i_{n}%
\right) }e_{i}\ \ \ \ \ \ \ \ \ \ \left( \text{since }\kappa=\left(
i_{1},i_{2},...,i_{n}\right) \right) \\
& =\mathbf{e}_{\left( i_{1},i_{2},...,i_{n}\right) }\underbrace{e_{i_{n+1}%
}}_{=\mathbf{e}_{\left( i_{n+1}\right) }\text{ (by (\ref{pf.pbw.U(g)h.-1}%
))}}\ \ \ \ \ \ \ \ \ \ \left( \text{since }i=i_{n+1}\right) \\
& =\mathbf{e}_{\left( i_{1},i_{2},...,i_{n}\right) }\mathbf{e}_{\left(
i_{n+1}\right) }=\mathbf{e}_{\left( i_{1},i_{2},...,i_{n}\right)
\cdot\left( i_{n+1}\right) }\ \ \ \ \ \ \ \ \ \ \left( \text{by
(\ref{pf.pbw.U(g)h.0})}\right) \\
& =\mathbf{e}_{\left( i_{1},i_{2},...,i_{n+1}\right) }=\mathbf{e}_{\left(
i_{1},i_{2},...,i_{\nu-1}\right) \cdot\left( i_{\nu},i_{\nu+1}%
,...,i_{n+1}\right) }=\mathbf{e}_{\left( i_{1},i_{2},...,i_{\nu-1}\right)
}\mathbf{e}_{\left( i_{\nu},i_{\nu+1},...,i_{n+1}\right) }%
\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.pbw.U(g)h.0})}\right) .
\end{align*}
But we can identify the universal enveloping algebra $U\left( \mathfrak{h}%
\right) $ with a $k$-submodule of $U\left( \mathfrak{g}\right) $ - namely,
with the $k$-submodule $\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in
P^{\ast}\right\rangle $ of $U\left( \mathfrak{g}\right) $. The element
$\mathbf{e}_{\left( i_{\nu},i_{\nu+1},...,i_{n+1}\right) }$ of $U\left(
\mathfrak{g}\right) $ lies in this submodule $U\left( \mathfrak{h}\right) $
(because $\left( i_{\nu},i_{\nu+1},...,i_{n+1}\right) \in P^{\ast}$).
Applying Proposition \ref{prop.pbw.gen} to $\mathfrak{h}$ and $\left(
e_{i}\right) _{i\in P}$ instead of $\mathfrak{g}$ and $\left( e_{i}\right)
_{i\in I}$, we see that $\left( \overline{e_{i_{1}}\otimes e_{i_{2}}%
\otimes...\otimes e_{i_{n}}}\right) _{\substack{n\in\mathbb{N};\ \left(
i_{1},i_{2},...,i_{n}\right) \in P^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n}}}$
is a generating set of the $k$-module $U\left( \mathfrak{h}\right) $. In
other words,%
\begin{align*}
U\left( \mathfrak{h}\right) & =\left\langle \underbrace{\overline
{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes e_{i_{n}}}}_{=\mathbf{e}_{\left(
i_{1},i_{2},...,i_{n}\right) }}\ \mid\ n\in\mathbb{N};\ \left( i_{1}%
,i_{2},...,i_{n}\right) \in P^{n};\ \underbrace{i_{1}\leq i_{2}\leq...\leq
i_{n}}_{\substack{\text{this is equivalent to}\\\left( i_{1},i_{2}%
,...,i_{n}\right) \in K}}\right\rangle \\
& =\left\langle \mathbf{e}_{\left( i_{1},i_{2},...,i_{n}\right) }%
\ \mid\ n\in\mathbb{N};\ \left( i_{1},i_{2},...,i_{n}\right) \in
P^{n};\ \left( i_{1},i_{2},...,i_{n}\right) \in K\right\rangle \\
& =\left\langle \mathbf{e}_{\left( i_{1},i_{2},...,i_{n}\right) }%
\ \mid\ n\in\mathbb{N};\ \left( i_{1},i_{2},...,i_{n}\right) \in K\cap
P^{n}\right\rangle =\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in K\cap
P^{\ast}\right\rangle .
\end{align*}
Therefore, $\mathbf{e}_{\left( i_{\nu},i_{\nu+1},...,i_{n+1}\right) }%
=\sum\limits_{\lambda\in K\cap P^{\ast}}\rho_{\lambda}\mathbf{e}_{\lambda}$
for some scalars $\rho_{\lambda}\in k$ (because $\mathbf{e}_{\left( i_{\nu
},i_{\nu+1},...,i_{n+1}\right) }\in U\left( \mathfrak{h}\right) $).
Consider these scalars $\rho_{\lambda}$. It is easily seen that $\rho_{\left(
{}\right) }=0$, where $\left( {}\right) $ denotes the empty sequence (in
fact, if the map $\varepsilon_{U\left( \mathfrak{g}\right) }:U\left(
\mathfrak{g}\right) \rightarrow k$ is defined as in Proposition
\ref{prop.Uhopf.easy}, then Proposition \ref{prop.Uhopf.easy} \textbf{(b)}
yields that $\varepsilon_{U\left( \mathfrak{g}\right) }\left(
\mathbf{e}_{\left( i_{\nu},i_{\nu+1},...,i_{n+1}\right) }\right) =0$
(because $\left( i_{\nu},i_{\nu+1},...,i_{n+1}\right) $ is not the empty
sequence) on one hand but $\varepsilon_{U\left( \mathfrak{g}\right) }\left(
\sum\limits_{\lambda\in K\cap P^{\ast}}\rho_{\lambda}\mathbf{e}_{\lambda
}\right) =\rho_{\left( {}\right) }$ on the other, so that we obtain
$0=\rho_{\left( {}\right) }$). Therefore, $\mathbf{e}_{\left( i_{\nu
},i_{\nu+1},...,i_{n+1}\right) }=\sum\limits_{\lambda\in K\cap P^{\ast}}%
\rho_{\lambda}\mathbf{e}_{\lambda}$ can be rewritten as%
\[
\mathbf{e}_{\left( i_{\nu},i_{\nu+1},...,i_{n+1}\right) }=\sum
\limits_{\substack{\lambda\in K\cap P^{\ast};\\\lambda\neq\left( {}\right)
}}\rho_{\lambda}\mathbf{e}_{\lambda}%
\]
(here we removed the $\lambda=\left( {}\right) $ addend, since
$\rho_{\left( {}\right) }=0$).
Now,%
\begin{align*}
\mathbf{e}_{\kappa}e_{i} & =\mathbf{e}_{\left( i_{1},i_{2},...,i_{\nu
-1}\right) }\underbrace{\mathbf{e}_{\left( i_{\nu},i_{\nu+1},...,i_{n+1}%
\right) }}_{=\sum\limits_{\substack{\lambda\in K\cap P^{\ast};\\\lambda
\neq\left( {}\right) }}\rho_{\lambda}\mathbf{e}_{\lambda}}=\sum
\limits_{\substack{\lambda\in K\cap P^{\ast};\\\lambda\neq\left( {}\right)
}}\rho_{\lambda}\underbrace{\mathbf{e}_{\left( i_{1},i_{2},...,i_{\nu
-1}\right) }\mathbf{e}_{\lambda}}_{\substack{=\mathbf{e}_{\left( i_{1}%
,i_{2},...,i_{\nu-1}\right) \cdot\lambda}\\\text{(by (\ref{pf.pbw.U(g)h.0}%
))}}}\\
& =\sum\limits_{\substack{\lambda\in K\cap P^{\ast};\\\lambda\neq\left(
{}\right) }}\rho_{\lambda}\mathbf{e}_{\left( i_{1},i_{2},...,i_{\nu
-1}\right) \cdot\lambda}.
\end{align*}
Now, every $\lambda\in K\cap P^{\ast}$ satisfies $\left( i_{1},i_{2}%
,...,i_{\nu-1}\right) \cdot\lambda\in K$ (because $\left( i_{1}%
,i_{2},...,i_{\nu-1}\right) \in K\cap Q^{\ast}$ and $\lambda\in K\cap
P^{\ast}$, and because every $q\in K\cap Q^{\ast}$ and $p\in K\cap P^{\ast}$
satisfy $q\cdot p\in K$). Therefore, every $\lambda\in K\cap P^{\ast}$ such
that $\lambda\neq\left( {}\right) $ satisfies $\left( i_{1},i_{2}%
,...,i_{\nu-1}\right) \cdot\lambda\in X$ (because $\left( i_{1}%
,i_{2},...,i_{\nu-1}\right) \cdot\lambda\in K$ on the one hand, but on the
other hand the last element of the sequence $\left( i_{1},i_{2},...,i_{\nu
-1}\right) \cdot\lambda$ lies in $P$\ \ \ \ {\footnotemark}\footnotetext{This
is because $\lambda\neq\left( {}\right) $ and $\lambda\in P^{\ast}$.}).
Thus,%
\[
\mathbf{e}_{\kappa}e_{i}=\sum\limits_{\substack{\lambda\in K\cap P^{\ast
};\\\lambda\neq\left( {}\right) }}\rho_{\lambda}\mathbf{e}_{\left(
i_{1},i_{2},...,i_{\nu-1}\right) \cdot\lambda}\in\left\langle \mathbf{e}%
_{\kappa^{\prime}}\ \mid\ \kappa^{\prime}\in X\right\rangle
\]
(since $\left( i_{1},i_{2},...,i_{\nu-1}\right) \cdot\lambda\in X$ for every
$\lambda\in K\cap P^{\ast}$ such that $\lambda\neq\left( {}\right) $). This
proves (\ref{pf.pbw.U(g)h.2}).
We could also prove $U_{\leq\left( m-1\right) }\left( \mathfrak{g}\right)
\cdot\mathfrak{h}=\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\cap
Y\right\rangle $ by a similar argument, but this would be a slight overkill.
Instead we only need the weaker result that $\left\langle \mathbf{e}_{\kappa
}\ \mid\ \kappa\in X\cap Y\right\rangle \subseteq U_{\leq\left( m-1\right)
}\left( \mathfrak{g}\right) \cdot\mathfrak{h}$, which can be seen exactly
the same way as we have shown (\ref{pf.pbw.U(g)h.1a}).
Now, combining $U_{\leq\left( m-1\right) }\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\subseteq\left( U\left( \mathfrak{g}\right) \cdot
\mathfrak{h}\right) \cap U_{\leq m}\left( \mathfrak{g}\right) $ (this is
because $\underbrace{U_{\leq\left( m-1\right) }\left( \mathfrak{g}\right)
}_{\subseteq U\left( \mathfrak{g}\right) }\cdot\mathfrak{h}\subseteq
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}$ and $U_{\leq\left(
m-1\right) }\left( \mathfrak{g}\right) \cdot\underbrace{\mathfrak{h}%
}_{\subseteq\mathfrak{g}}\subseteq U_{\leq\left( m-1\right) }\left(
\mathfrak{g}\right) \cdot\mathfrak{g}\subseteq U_{\leq m}\left(
\mathfrak{g}\right) $) with
\begin{align*}
\underbrace{\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right)
}_{=\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle }%
\cap\underbrace{U_{\leq m}\left( \mathfrak{g}\right) }_{=\left\langle
\mathbf{e}_{\kappa}\ \mid\ \kappa\in Y\right\rangle } & =\left\langle
\mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle \cap\left\langle
\mathbf{e}_{\kappa}\ \mid\ \kappa\in Y\right\rangle \\
& =\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\cap Y\right\rangle
\ \ \ \ \ \ \ \ \ \ \left( \text{by Lemma \ref{lem.linalg.int}}\right) \\
& \subseteq U_{\leq\left( m-1\right) }\left( \mathfrak{g}\right)
\cdot\mathfrak{h},
\end{align*}
we obtain $\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right)
\cap U_{\leq m}\left( \mathfrak{g}\right) =U_{\leq\left( m-1\right)
}\left( \mathfrak{g}\right) \cdot\mathfrak{h}$. In other words, Proposition
\ref{prop.pbw.U(g)h} is proven.
\subsection{The kernel of $\operatorname*{gr}\nolimits_{n}\tau
:\operatorname*{gr}\nolimits_{n}\left( \left( \otimes\mathfrak{g}\right)
\diagup\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
\right) \rightarrow\operatorname*{gr}\nolimits_{n}\left( U\left(
\mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) \right) $}
Next we prove a result that generalizes Lemma 4.3 of \cite{calaque1}:
\begin{theorem}
\label{thm.l42}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$n\in\mathbb{N}$. Assume that the inclusion $\mathfrak{h}\hookrightarrow
\mathfrak{g}$ splits \textit{as a }$k$\textit{-module inclusion} (but not
necessarily as an $\mathfrak{h}$-module inclusion). This means that there
exists a $k$-submodule $N$ of $\mathfrak{g}$ such that $\mathfrak{g}%
=\mathfrak{h}\oplus N$. \newline Let us work with the notations introduced in
Theorem \ref{thm.l25} and in Definition \ref{defs.U}. Let also $\psi$ denote
the canonical projection from the $k$-algebra $\otimes\mathfrak{g}$ to the
factor algebra $\left( \otimes\mathfrak{g}\right) \diagup I_{\mathfrak{g}%
}=U\left( \mathfrak{g}\right) $. Clearly, $\psi$ is a $k$-algebra
homomorphism.\newline Let us abbreviate the $k$-submodule $U\left(
\mathfrak{g}\right) \cdot\psi\left( \mathfrak{h}\right) $ of $U\left(
\mathfrak{g}\right) $ by $U\left( \mathfrak{g}\right) \cdot\mathfrak{h}%
$.\newline Let $\rho$ be the canonical $k$-module projection $U\left(
\mathfrak{g}\right) \rightarrow U\left( \mathfrak{g}\right) \diagup\left(
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) $.\newline\textbf{(a)}
There exists one and only one map $\theta:\left( \otimes\mathfrak{g}\right)
\diagup\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
\rightarrow U\left( \mathfrak{g}\right) \diagup\left( U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}\right) $ for which the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%\otimes\mathfrak{g} \ar@{->>}[r]^{\psi} \ar@{->>}[d]_{\zeta} & U\left
%(\mathfrak{g}\right) \ar@{->>}[d]^{\rho} \\
%\left(\otimes\mathfrak{g}\right) \diagup\left(J + \left(\otimes\mathfrak
%{g}\right)\cdot\mathfrak{h}\right) \ar[r]_{\theta} & U\left(\mathfrak{g}%
%\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak{h}\right)
%}} }%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
\otimes\mathfrak{g} \ar@{->>}[r]^{\psi} \ar@{->>}[d]_{\zeta} & U\left
(\mathfrak{g}\right) \ar@{->>}[d]^{\rho} \\
\left(\otimes\mathfrak{g}\right) \diagup\left(J + \left(\otimes\mathfrak
{g}\right)\cdot\mathfrak{h}\right) \ar[r]_{\theta} & U\left(\mathfrak{g}%
\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak{h}\right)
}
%EndExpansion
\label{thm.l42.diag}%
\end{equation}
commutes. Denote this map $\theta$ by $\tau$.\newline\textbf{(b)} This $\tau$
is a surjective $\mathfrak{h}$-module homomorphism. Also, it satisfies
$\tau\circ\zeta=\rho\circ\psi$. In other words, the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%\otimes\mathfrak{g} \ar@{->>}[r]^{\psi} \ar@{->>}[d]_{\zeta} & U\left
%(\mathfrak{g}\right) \ar@{->>}[d]^{\rho} \\
%\left(\otimes\mathfrak{g}\right) \diagup\left(J + \left(\otimes\mathfrak
%{g}\right)\cdot\mathfrak{h}\right) \ar[r]_{\tau} & U\left(\mathfrak{g}%
%\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak{h}\right)
%}} }%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
\otimes\mathfrak{g} \ar@{->>}[r]^{\psi} \ar@{->>}[d]_{\zeta} & U\left
(\mathfrak{g}\right) \ar@{->>}[d]^{\rho} \\
\left(\otimes\mathfrak{g}\right) \diagup\left(J + \left(\otimes\mathfrak
{g}\right)\cdot\mathfrak{h}\right) \ar[r]_{\tau} & U\left(\mathfrak{g}%
\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak{h}\right)
}
%EndExpansion
\label{thm.l42.diag.real}%
\end{equation}
commutes.\newline\textbf{(c)} Every of the four corners of the commutative
square (\ref{thm.l42.diag.real}) is endowed with a filtration - namely as
follows:\newline- The filtration on $\otimes\mathfrak{g}$ is the degree
filtration $\left( \mathfrak{g}^{\otimes\leq n}\right) _{n\geq0}$.\newline-
The filtration on $\left( \otimes\mathfrak{g}\right) \diagup\left(
J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right) $ is the
filtration $\left( F_{n}\right) _{n\geq0}$ defined in Theorem \ref{thm.l25}
\textbf{(b)} by $F_{n}=\zeta\left( \mathfrak{g}^{\otimes\leq n}\right)
$.\newline- The filtration on $U\left( \mathfrak{g}\right) $ is the
filtration $\left( U_{\leq n}\left( \mathfrak{g}\right) \right) _{n\geq0}$
defined in Convention \ref{conv.Un} by $U_{\leq n}\left( \mathfrak{g}\right)
=\psi\left( \mathfrak{g}^{\otimes\leq n}\right) $.\newline- The filtration
on $U\left( \mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) $ is the filtration $\left( W_{n}\right) _{n\geq
0}$ defined by $W_{n}=\left( \tau\circ\zeta\right) \left( \mathfrak{g}%
^{\otimes\leq n}\right) $.\newline Then, $W_{n}=\left( \rho\circ\psi\right)
\left( \mathfrak{g}^{\otimes\leq n}\right) =\rho\left( U_{\leq n}\left(
\mathfrak{g}\right) \right) =\tau\left( F_{n}\right) $. Also, the maps
$\psi$, $\rho$, $\zeta$ and $\tau$ all respect the filtration. Therefore, for
every $n\in\mathbb{N}$, we can apply the functor $\operatorname*{gr}%
\nolimits_{n}$ to the diagram (\ref{thm.l42.diag.real}), and obtain the
commutative diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar@{->>}%
%[r]^{\operatorname*{gr}_n\psi} \ar@{->>}[d]_{\operatorname*{gr}_n\zeta}
%& \operatorname*{gr}_n\left(U\left(\mathfrak{g}\right)\right) \ar@
%{->>}[d]^{\operatorname*{gr}_n\rho} \\
%\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right) \diagup
%\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right)\right
%) \ar[r]_{\operatorname*{gr}_n\tau} & \operatorname*{gr}_n\left(U\left
%(\mathfrak{g}\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak
%{h}\right)\right)
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar@{->>}%
[r]^{\operatorname*{gr}_n\psi} \ar@{->>}[d]_{\operatorname*{gr}_n\zeta}
& \operatorname*{gr}_n\left(U\left(\mathfrak{g}\right)\right) \ar@
{->>}[d]^{\operatorname*{gr}_n\rho} \\
\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right) \diagup
\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right)\right
) \ar[r]_{\operatorname*{gr}_n\tau} & \operatorname*{gr}_n\left(U\left
(\mathfrak{g}\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak
{h}\right)\right)
}%
%EndExpansion
. \label{thm.l42.diag.gr}%
\end{equation}
\newline\textbf{(d)} Assume that both $\mathfrak{h}$ and $N$ are free
$k$-modules. Let $n\in\mathbb{N}$. Then, $\left( \operatorname*{grad}%
\nolimits_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right) \left(
\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\tau\right)
\right) =K_{n}\left( \mathfrak{n}\right) $. (For the definition of
$K_{n}\left( \mathfrak{n}\right) $, see Definition \ref{defs.symm-alg},
applied to $V=\mathfrak{n}$. For the definition of $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}$ and $\omega_{n}$, see Theorem \ref{thm.l25}
\textbf{(c)}.)
\end{theorem}
The condition that both $\mathfrak{h}$ and $N$ are free in \textbf{(d)} is
somewhat restrictive - we will partly lift it in Subsection
\ref{subsect.goodwillie}.
\textit{Proof of Theorem \ref{thm.l42}.} \textbf{(a)} Let $J_{0}$ denote the
$k$-submodule
\[
\left\langle v\otimes w-w\otimes v-\left[ v,w\right] \ \mid\ \left(
v,w\right) \in\mathfrak{g}\times\mathfrak{h}\right\rangle
\]
of $\otimes\mathfrak{g}$. Then, $J=\left( \otimes\mathfrak{g}\right) \cdot
J_{0}\cdot\left( \otimes\mathfrak{g}\right) $ (by Proposition
\ref{prop.l25a} \textbf{(b)}). On the other hand,%
\begin{align*}
J_{0} & =\left\langle v\otimes w-w\otimes v-\left[ v,w\right]
\ \mid\ \left( v,w\right) \in\mathfrak{g}\times\mathfrak{h}\right\rangle \\
& \subseteq\left\langle v\otimes w-w\otimes v-\left[ v,w\right]
\ \mid\ \left( v,w\right) \in\mathfrak{g}\times\mathfrak{g}\right\rangle
\end{align*}
(since $\mathfrak{g}\times\mathfrak{h}\subseteq\mathfrak{g}\times\mathfrak{g}%
$). Now
\begin{align*}
J & =\left( \otimes\mathfrak{g}\right) \cdot\underbrace{J_{0}}%
_{\subseteq\left\langle v\otimes w-w\otimes v-\left[ v,w\right]
\ \mid\ \left( v,w\right) \in\mathfrak{g}\times\mathfrak{g}\right\rangle
}\cdot\left( \otimes\mathfrak{g}\right) \\
& \subseteq\left( \otimes\mathfrak{g}\right) \cdot\left\langle v\otimes
w-w\otimes v-\left[ v,w\right] \ \mid\ \left( v,w\right) \in
\mathfrak{g}\times\mathfrak{g}\right\rangle \cdot\left( \otimes
\mathfrak{g}\right) =I_{\mathfrak{g}}.
\end{align*}
With the help of%
\begin{align*}
\psi\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
& =\psi\left( \underbrace{J}_{\subseteq I_{\mathfrak{g}}}\right)
+\underbrace{\psi\left( \left( \otimes\mathfrak{g}\right) \cdot
\mathfrak{h}\right) }_{\substack{=\psi\left( \otimes\mathfrak{g}\right)
\cdot\psi\left( \mathfrak{h}\right) \\\text{(since }\psi\text{ is a
}k\text{-algebra}\\\text{homomorphism)}}}\subseteq\underbrace{\psi\left(
I_{\mathfrak{g}}\right) }_{\substack{=0\text{ (since }\psi\text{ is
the}\\\text{projection on }\left( \otimes\mathfrak{g}\right) \diagup
I_{\mathfrak{g}}\text{)}}}+\underbrace{\psi\left( \otimes\mathfrak{g}\right)
}_{\subseteq U\left( \mathfrak{g}\right) }\cdot\psi\left( \mathfrak{h}%
\right) \\
& \subseteq0+U\left( \mathfrak{g}\right) \cdot\psi\left( \mathfrak{h}%
\right) =U\left( \mathfrak{g}\right) \cdot\psi\left( \mathfrak{h}\right)
=U\left( \mathfrak{g}\right) \cdot\mathfrak{h},
\end{align*}
we obtain%
\[
\left( \rho\circ\psi\right) \left( J+\left( \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right) =\rho\left( \underbrace{\psi\left( J+\left(
\otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right) }_{\subseteq U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}}\right) \subseteq\rho\left( U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}\right) =0
\]
(since $\rho$ is the projection on $U\left( \mathfrak{g}\right)
\diagup\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) $). In
other words, $\left( \rho\circ\psi\right) \left( J+\left( \otimes
\mathfrak{g}\right) \cdot\mathfrak{h}\right) =0$.
Thus, by the homomorphism theorem, the map $\rho\circ\psi:\otimes
\mathfrak{g}\rightarrow U\left( \mathfrak{g}\right) \diagup\left( U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}\right) $ factors through the factor
map $\zeta:\otimes\mathfrak{g}\rightarrow\left( \otimes\mathfrak{g}\right)
\diagup\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
$. In other words, there exists one and only one map $\theta:\left(
\otimes\mathfrak{g}\right) \diagup\left( J+\left( \otimes\mathfrak{g}%
\right) \cdot\mathfrak{h}\right) \rightarrow U\left( \mathfrak{g}\right)
\diagup\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) $
satisfying $\theta\circ\zeta=\rho\circ\psi$, and this map $\theta$ is a
$k$-module homomorphism. Since $\theta\circ\zeta=\rho\circ\psi$ is equivalent
to the commutativity of the diagram (\ref{thm.l42.diag}), this result rewrites
as follows: There exists one and only one map $\theta:\left( \otimes
\mathfrak{g}\right) \diagup\left( J+\left( \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right) \rightarrow U\left( \mathfrak{g}\right)
\diagup\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) $ for
which the diagram (\ref{thm.l42.diag}) commutes, and this map $\theta$ is a
$k$-module homomorphism. This proves Theorem \ref{thm.l42} \textbf{(a)}, and
even a part of Theorem \ref{thm.l42} \textbf{(b) }(namely, the part saying
that $\tau$ is a $k$-module homomorphism).
\textbf{(b)} We have already proven that $\tau$ is a $k$-module homomorphism.
Now, we need only show that $\tau$ is surjective and an $\mathfrak{h}$-module homomorphism.
By the definition of $\tau$, the diagram (\ref{thm.l42.diag}) commutes for
$\theta=\tau$. That is, $\tau\circ\zeta=\rho\circ\psi$. Since the maps $\rho$
and $\psi$ are surjective (because $\rho$ and $\psi$ are projections), their
composition $\rho\circ\psi$ is surjective. Thus, $\tau\circ\zeta=\rho\circ
\psi$ is surjective, so that $\tau$ must too be surjective.
We know that $\zeta$ is surjective, and that $\tau\circ\zeta$ is an
$\mathfrak{h}$-module homomorphism (since $\tau\circ\zeta=\rho\circ\psi$, and
since both $\rho$ and $\psi$ are $\mathfrak{h}$-module homomorphisms).
Applying Lemma \ref{lem.surj} to $\otimes\mathfrak{g}$, $\left(
\otimes\mathfrak{g}\right) \diagup\left( J+\left( \otimes\mathfrak{g}%
\right) \cdot\mathfrak{h}\right) $, $U\left( \mathfrak{g}\right)
\diagup\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) $,
$\zeta$ and $\tau$ instead of $A$, $B$, $C$, $f$ and $g$, we thus obtain that
$\tau$ is an $\mathfrak{h}$-module homomorphism.
This completes the proof of Theorem \ref{thm.l42} \textbf{(b)}.
\textbf{(c)} It is pretty much trivial that $\left( W_{n}\right) _{n\geq0}$
is indeed a filtration of $U\left( \mathfrak{g}\right) \diagup\left(
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) $ (all we use here is
that $\tau\circ\zeta$ is surjective). Now, $W_{n}=\left( \tau\circ
\zeta\right) \left( \mathfrak{g}^{\otimes\leq n}\right) $ yields
$W_{n}=\left( \rho\circ\psi\right) \left( \mathfrak{g}^{\otimes\leq
n}\right) $ because of $\tau\circ\zeta=\rho\circ\psi$. Also, $W_{n}=\left(
\tau\circ\zeta\right) \left( \mathfrak{g}^{\otimes\leq n}\right)
=\tau\left( \underbrace{\zeta\left( \mathfrak{g}^{\otimes\leq n}\right)
}_{=F_{n}}\right) =\tau\left( F_{n}\right) $ and $W_{n}=\left( \rho
\circ\psi\right) \left( \mathfrak{g}^{\otimes\leq n}\right) =\rho\left(
\underbrace{\psi\left( \mathfrak{g}^{\otimes\leq n}\right) }_{=U_{\leq
n}\left( \mathfrak{g}\right) }\right) =\rho\left( U_{\leq n}\left(
\mathfrak{g}\right) \right) $. It is now absolutely obvious that the maps
$\psi$, $\rho$, $\zeta$ and $\tau$ all respect the filtration. Theorem
\ref{thm.l42} \textbf{(c)} is now proven.
\textbf{(d)} Assume that both $\mathfrak{h}$ and $N$ are free $k$-modules.
This, of course, yields that $\mathfrak{g}=\mathfrak{h}\oplus N$ must also be
free, and thus $\mathfrak{g}$ satisfies the $n$-PBW condition (by Theorem
\ref{thm.pbw} \textbf{(a)}). Thus, Proposition \ref{prop.pbw} \textbf{(b)}
yields $\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\psi\right)
=\operatorname*{grad}\nolimits_{\mathfrak{g},n}\left( K_{n}\left(
\mathfrak{g}\right) \right) $. In other words, $\operatorname*{grad}%
\nolimits_{\mathfrak{g},n}^{-1}\left( \operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\psi\right) \right) =K_{n}\left(
\mathfrak{g}\right) $ (since $\operatorname*{grad}\nolimits_{\mathfrak{g},n}$
is an isomorphism).
On the other hand, the construction of $K_{n}\left( V\right) $ for a
$k$-module $V$ yields that whenever $f:A\rightarrow B$ is a surjective
$k$-linear map between two $k$-modules $A$ and $B$, then $f^{\otimes n}\left(
K_{n}\left( A\right) \right) =K_{n}\left( B\right) $. Applying this to
$A=\mathfrak{g}$, $B=\mathfrak{n}$ and $f=\pi$, we obtain $\pi^{\otimes
n}\left( K_{n}\left( \mathfrak{g}\right) \right) =K_{n}\left(
\mathfrak{n}\right) $.
Our goal is to show that $\left( \operatorname*{grad}\nolimits_{\mathfrak{n}%
,n}^{-1}\circ\omega_{n}\right) \left( \operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\tau\right) \right) =K_{n}\left(
\mathfrak{n}\right) $. This will be done once we have proven that%
\begin{equation}
\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\tau\right)
=\left( \operatorname*{gr}\nolimits_{n}\zeta\right) \left(
\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\psi\right)
\right) . \label{pf.l42.1}%
\end{equation}
In fact, once (\ref{pf.l42.1}) is shown, we have%
\begin{align*}
& \left( \operatorname*{grad}\nolimits_{\mathfrak{n},n}^{-1}\circ\omega
_{n}\right) \left( \underbrace{\operatorname*{Ker}\left( \operatorname*{gr}%
\nolimits_{n}\tau\right) }_{=\left( \operatorname*{gr}\nolimits_{n}%
\zeta\right) \left( \operatorname*{Ker}\left( \operatorname*{gr}%
\nolimits_{n}\psi\right) \right) }\right) \\
& =\left( \operatorname*{grad}\nolimits_{\mathfrak{n},n}^{-1}\circ\omega
_{n}\right) \left( \left( \operatorname*{gr}\nolimits_{n}\zeta\right)
\left( \operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\psi\right)
\right) \right) =\left( \operatorname*{grad}\nolimits_{\mathfrak{n},n}%
^{-1}\circ\underbrace{\omega_{n}\circ\operatorname*{gr}\nolimits_{n}\zeta
}_{\substack{=\operatorname*{gr}\nolimits_{n}\left( \otimes\pi\right)
\\\text{(since the diagram}\\\text{(\ref{thm.l25.c.diag'}) commutes)}%
}}\right) \left( \operatorname*{Ker}\left( \operatorname*{gr}%
\nolimits_{n}\psi\right) \right) \\
& =\underbrace{\left( \operatorname*{grad}\nolimits_{\mathfrak{n},n}%
^{-1}\circ\operatorname*{gr}\nolimits_{n}\left( \otimes\pi\right) \right)
}_{\substack{=\pi^{\otimes n}\circ\operatorname*{grad}\nolimits_{\mathfrak{g}%
,n}^{-1}\\\text{(this follows easily from the}\\\text{definitions of
}\operatorname*{grad}\nolimits_{\mathfrak{g},n}\text{ and }%
\operatorname*{grad}\nolimits_{\mathfrak{n},n}\text{)}}}\left(
\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\psi\right)
\right) =\left( \pi^{\otimes n}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{g},n}^{-1}\right) \left( \operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\psi\right) \right) \\
& =\pi^{\otimes n}\left( \underbrace{\operatorname*{grad}%
\nolimits_{\mathfrak{g},n}^{-1}\left( \operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\psi\right) \right) }_{=K_{n}\left(
\mathfrak{g}\right) }\right) =\pi^{\otimes n}\left( K_{n}\left(
\mathfrak{g}\right) \right) =K_{n}\left( \mathfrak{n}\right) ,
\end{align*}
which proves Theorem \ref{thm.l42} \textbf{(d)}. So, in order to complete the
proof of Theorem \ref{thm.l42} \textbf{(d)}, the only thing we need to do is
verify (\ref{pf.l42.1}).
According to Proposition \ref{prop.9l} (applied to the diagram
(\ref{thm.l42.diag.gr}) instead of the diagram (\ref{prop.9l.diag})), the
equality (\ref{pf.l42.1}) will follow once we can show that
$\operatorname*{gr}\nolimits_{n}\zeta$ is surjective and that%
\begin{equation}
\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\rho\right)
\subseteq\left( \operatorname*{gr}\nolimits_{n}\psi\right) \left(
\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\zeta\right)
\right) . \label{pf.l42.2}%
\end{equation}
But it is easy to see that $\operatorname*{gr}\nolimits_{n}\zeta$ is
surjective (and it was proven in detail during the proof of Proposition
\ref{prop.l25.Ker-alpha} \textbf{(e)}). Thus, we only have to prove
(\ref{pf.l42.2}) now.
Due to the commutative diagram (\ref{thm.l25.c.diag'}), and due to the fact
that $\omega_{n}$ is an isomorphism, we have $\operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\zeta\right) =\operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\left( \otimes\pi\right) \right) $.
Let $s\in\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}%
\rho\right) $ be arbitrary. Then, $s\in\operatorname*{gr}\nolimits_{n}\left(
U\left( \mathfrak{g}\right) \right) =\left( U_{\leq n}\left(
\mathfrak{g}\right) \right) \diagup\left( U_{\leq\left( n-1\right)
}\left( \mathfrak{g}\right) \right) $, so that there exists some $S\in
U_{\leq n}\left( \mathfrak{g}\right) $ such that $s=\overline{S}%
$.\ \ \ \ {\footnotemark}\footnotetext{Henceforth until the end of this proof
of Theorem \ref{thm.l42}, the term $\overline{S}$ always denotes the residue
class of $S$ modulo $U_{\leq\left( n-1\right) }\left( \mathfrak{g}\right)
$. Similarly, the terms $\overline{\rho\left( S\right) }$ and $\overline
{S-S^{\prime}}$ will have to be interpreted.} Consider this $S$. Since
$s\in\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\rho\right) $,
we have $\left( \operatorname*{gr}\nolimits_{n}\rho\right) \left( s\right)
=0$. But $s=\overline{S}$ shows that $\left( \operatorname*{gr}%
\nolimits_{n}\rho\right) \left( s\right) =\left( \operatorname*{gr}%
\nolimits_{n}\rho\right) \left( \overline{S}\right) =\overline{\rho\left(
S\right) }$, so that $\left( \operatorname*{gr}\nolimits_{n}\rho\right)
\left( s\right) =0$ becomes $\overline{\rho\left( S\right) }=0$. In other
words, $\rho\left( S\right) \in W_{n-1}=\rho\left( U_{\leq\left(
n-1\right) }\left( \mathfrak{g}\right) \right) $. This means that there
exists some $S^{\prime}\in U_{\leq\left( n-1\right) }\left( \mathfrak{g}%
\right) $ such that $\rho\left( S\right) =\rho\left( S^{\prime}\right) $.
Consider this $S^{\prime}$. Then, $\rho\left( S\right) =\rho\left(
S^{\prime}\right) $ yields $0=\rho\left( S\right) -\rho\left( S^{\prime
}\right) =\rho\left( S-S^{\prime}\right) $, so that $S-S^{\prime}%
\in\operatorname*{Ker}\rho=U\left( \mathfrak{g}\right) \cdot\mathfrak{h}$.
On the other hand, $S\in U_{\leq n}\left( \mathfrak{g}\right) $ and
$S^{\prime}\in U_{\leq\left( n-1\right) }\left( \mathfrak{g}\right)
\subseteq U_{\leq n}\left( \mathfrak{g}\right) $ lead to $S-S^{\prime}\in
U_{\leq n}\left( \mathfrak{g}\right) $ (since $U_{\leq n}\left(
\mathfrak{g}\right) $ is a $k$-module). Combining this with $S-S^{\prime}\in
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}$, we obtain%
\begin{align*}
S-S^{\prime} & \in U_{\leq n}\left( \mathfrak{g}\right) \cap\left(
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) =U_{\leq\left(
n-1\right) }\left( \mathfrak{g}\right) \cdot\mathfrak{h}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{by Proposition \ref{prop.pbw.U(g)h},
applied to }n\text{ instead of }m\right) \\
& =\psi\left( \mathfrak{g}^{\otimes\leq\left( n-1\right) }\cdot
\mathfrak{h}\right) .
\end{align*}
Thus,
\begin{align*}
\overline{S-S^{\prime}} & \in\overline{\psi\left( \mathfrak{g}^{\otimes
\leq\left( n-1\right) }\cdot\mathfrak{h}\right) }\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, }\overline{\psi\left( \mathfrak{g}^{\otimes\leq\left(
n-1\right) }\cdot\mathfrak{h}\right) }\text{ means the image of }\psi\left(
\mathfrak{g}^{\otimes\leq\left( n-1\right) }\cdot\mathfrak{h}\right) \\
\text{under the canonical projection }U_{\leq n}\left( \mathfrak{g}\right)
\rightarrow\operatorname{gr}_{n}\left( U\left( \mathfrak{g}\right) \right)
\end{array}
\right) \\
& =\left( \operatorname*{gr}\nolimits_{n}\psi\right) \left(
\underbrace{\overline{\mathfrak{g}^{\otimes\leq\left( n-1\right) }%
\cdot\mathfrak{h}}}_{\subseteq\operatorname*{Ker}\left( \operatorname*{gr}%
\nolimits_{n}\left( \otimes\pi\right) \right) =\operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\zeta\right) }\right) \\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, }\overline{\mathfrak{g}^{\otimes\leq\left( n-1\right) }%
\cdot\mathfrak{h}}\text{ means the image of }\mathfrak{g}^{\otimes\leq\left(
n-1\right) }\cdot\mathfrak{h}\\
\text{under the canonical projection }\mathfrak{g}^{\otimes\leq n}%
\rightarrow\operatorname{gr}_{n}\left( \otimes\mathfrak{g}\right)
\end{array}
\right) \\
& \subseteq\left( \operatorname*{gr}\nolimits_{n}\psi\right) \left(
\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\zeta\right)
\right) .
\end{align*}
Now,%
\begin{align*}
s & =\overline{S}=\overline{S-S^{\prime}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }S^{\prime}\in U_{\leq\left( n-1\right) }\left( \mathfrak{g}%
\right) \right) \\
& \in\left( \operatorname*{gr}\nolimits_{n}\psi\right) \left(
\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\zeta\right)
\right) .
\end{align*}
Since this is shown for every $s\in\operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\rho\right) $, we thus conclude that
$\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\rho\right)
\subseteq\left( \operatorname*{gr}\nolimits_{n}\psi\right) \left(
\operatorname*{Ker}\left( \operatorname*{gr}\nolimits_{n}\zeta\right)
\right) $. Thus, (\ref{pf.l42.2}) is shown. This completes the proof of
Theorem \ref{thm.l42} \textbf{(d)}.
\subsection{The associated graded object of $U\left( \mathfrak{g}\right)
\diagup\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) $}
Here an important consequence of Theorem \ref{thm.l42}:
\begin{corollary}
\label{cor.l42}Consider the situation of Theorem \ref{thm.l42} \textbf{(d)}.
Let $n\in\mathbb{N}$.\newline Consider the canonical projection
$\operatorname{sym}_{\mathfrak{n},n}:\mathfrak{n}^{\otimes n}\rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$.\newline\textbf{(a)} The map
$\operatorname{gr}_{n}\tau:\operatorname{gr}_{n}\left( \left( \otimes
\mathfrak{g}\right) \diagup\left( J+\left( \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right) \right) \rightarrow\operatorname{gr}_{n}\left(
U\left( \mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) \right) $ is surjective.\newline\textbf{(b)} There
exists one and only one $k$-module homomorphism $\Xi:\operatorname{gr}%
_{n}\left( U\left( \mathfrak{g}\right) \diagup\left( U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}\right) \right) \rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$ for which the diagram%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ \operatorname{gr}_n\left(\left(\otimes \mathfrak g\right)\diagup \left(J+\left(\otimes \mathfrak g\right)\cdot \mathfrak h\right)\right) \ar[r]^{\operatorname*{gr}_n\tau} \ar[d]_{\omega_n}^{\cong} & \operatorname*{gr}_n\left( U\left(\mathfrak g\right) \diagup \left(U\left(\mathfrak g\right)\cdot\mathfrak h\right) \right) \ar[dd]^{\Xi} \\ \operatorname{gr}_n\left(\otimes\mathfrak n\right) \ar[d]_{\operatorname*{grad}_{\mathfrak n,n}^{-1}}^{\cong} & \\ \mathfrak n^{\otimes n} \ar[r]_{\operatorname*{sym}_{\mathfrak n,n}} & \operatorname*{Sym}^n \mathfrak n }
\label{cor.l42.diag}%
\end{equation}
commutes.\newline\textbf{(c)} Let us denote this map $\Xi$ by $\Theta_{n}$.
Then, $\Theta_{n}$ is an $\mathfrak{h}$-module isomorphism $\operatorname{gr}%
_{n}\left( U\left( \mathfrak{g}\right) \diagup\left( U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}\right) \right) \rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$ which satisfies $\operatorname{sym}%
_{\mathfrak{n},n}\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega
_{n}=\Theta_{n}\circ\operatorname{gr}_{n}\tau$.\newline\textbf{(d)} Let $\pi$
be the canonical projection $\mathfrak{g}\rightarrow\mathfrak{g}%
\diagup\mathfrak{h}=\mathfrak{n}$. The diagram%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ \operatorname{gr}_n\left(\otimes \mathfrak g\right) \ar[r]^-{\operatorname*{gr}_n\left(\rho\circ\psi\right)} \ar[d]_{\operatorname*{gr}_n\left(\otimes \pi\right)} & \operatorname*{gr}_n\left( U\left(\mathfrak g\right) \diagup \left(U\left(\mathfrak g\right)\cdot\mathfrak h\right) \right) \ar[dd]^{\Theta_n} \\ \operatorname{gr}_n\left(\otimes\mathfrak n\right) \ar[d]_{\operatorname*{grad}_{\mathfrak n,n}^{-1}}^{\cong} & \\ \mathfrak n^{\otimes n} \ar[r]_{\operatorname*{sym}_{\mathfrak n,n}} & \operatorname*{Sym}^n \mathfrak n }
\label{cor.l42.diag2}%
\end{equation}
commutes.
\end{corollary}
\textit{Proof of Corollary \ref{cor.l42}.} \textbf{(a)} This follows from
Proposition \ref{prop.gr.surj}, because $\tau\left( F_{n}\right) =W_{n}$.
\textbf{(b)} We have $\operatorname{Ker}\operatorname{sym}_{\mathfrak{n}%
,n}=K_{n}\left( \mathfrak{n}\right) $ (since $\operatorname{sym}%
_{\mathfrak{n},n}$ is the projection from $\mathfrak{n}^{\otimes n}$ on
$\mathfrak{n}^{\otimes n}\diagup K_{n}\left( \mathfrak{n}\right) $).
Since $\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}$ is a
$k$-module isomorphism (which is because $\operatorname{grad}_{\mathfrak{n}%
,n}^{-1}$ and $\omega_{n}$ are $k$-module isomorphisms), we have
\begin{align*}
\operatorname{Ker}\left( \operatorname{sym}_{\mathfrak{n},n}\circ
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right) & =\left(
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right) ^{-1}\left(
\underbrace{\operatorname{Ker}\operatorname{sym}_{\mathfrak{n},n}}%
_{=K_{n}\left( \mathfrak{n}\right) }\right) =\left( \operatorname{grad}%
_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right) ^{-1}\left( K_{n}\left(
\mathfrak{n}\right) \right) \\
& =\operatorname{Ker}\left( \operatorname{gr}_{n}\tau\right)
\end{align*}
(because Theorem \ref{thm.l42} \textbf{(d)} says that $\left(
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right) \left(
\operatorname{Ker}\left( \operatorname{gr}_{n}\tau\right) \right)
=K_{n}\left( \mathfrak{n}\right) $, and because $\operatorname{grad}%
_{\mathfrak{n},n}^{-1}\circ\omega_{n}$ is an isomorphism). In particular, this
yields that $\operatorname{Ker}\left( \operatorname{gr}_{n}\tau\right)
\subseteq\operatorname{Ker}\left( \operatorname{sym}_{\mathfrak{n},n}%
\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right) $. Since
$\operatorname{gr}_{n}\tau$ is surjective, the homomorphism theorem thus
yields that there exists one and only one $k$-module homomorphism
$\Xi:\operatorname{gr}_{n}\left( U\left( \mathfrak{g}\right) \diagup\left(
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) \right)
\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ which satisfies $\Xi
\circ\operatorname{gr}_{n}\tau=\operatorname{sym}_{\mathfrak{n},n}%
\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}$. In other
words, there exists one and only one $k$-module homomorphism $\Xi
:\operatorname{gr}_{n}\left( U\left( \mathfrak{g}\right) \diagup\left(
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) \right)
\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ for which the diagram
(\ref{cor.l42.diag}) commutes. This proves Corollary \ref{cor.l42}
\textbf{(b)}.
\textbf{(c)} The map $\operatorname{sym}_{\mathfrak{n},n}\circ
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}:\operatorname{gr}%
_{n}\left( \left( \otimes\mathfrak{g}\right) \diagup\left( J+\left(
\otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right) \right) \rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$ is surjective (since $\operatorname{sym}%
_{\mathfrak{n},n}$ is surjective while $\operatorname{grad}_{\mathfrak{n}%
,n}^{-1}$ and $\omega_{n}$ are isomorphisms). Since $\operatorname{sym}%
_{\mathfrak{n},n}\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega
_{n}=\Theta_{n}\circ\operatorname{gr}_{n}\tau$ (because the map $\Theta_{n}$
is defined as the map $\Xi$ for which the diagram (\ref{cor.l42.diag})
commutes), this yields that the map $\Theta_{n}\circ\operatorname{gr}_{n}\tau$
is surjective. Hence, the map $\Theta_{n}$ is surjective.
On the other hand, let $i$ be an arbitrary element of $\operatorname{Ker}%
\Theta_{n}$. Then, we can write $i$ in the form $i=\left( \operatorname{gr}%
_{n}\tau\right) \left( i^{\prime}\right) $ for some $i^{\prime}%
\in\operatorname{gr}_{n}\left( \left( \otimes\mathfrak{g}\right)
\diagup\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
\right) $ (since $\operatorname{gr}_{n}\tau$ is surjective). Now,
$i\in\operatorname{Ker}\Theta_{n}$ yields $\Theta_{n}\left( i\right) =0$, so
that%
\[
\left( \underbrace{\operatorname{sym}_{\mathfrak{n},n}\circ
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}}_{=\Theta_{n}%
\circ\operatorname{gr}_{n}\tau}\right) \left( i^{\prime}\right) =\left(
\Theta_{n}\circ\operatorname{gr}_{n}\tau\right) \left( i^{\prime}\right)
=\Theta_{n}\left( \underbrace{\left( \operatorname{gr}_{n}\tau\right)
\left( i^{\prime}\right) }_{=i}\right) =\Theta_{n}\left( i\right) =0,
\]
thus $i^{\prime}\in\operatorname{Ker}\left( \operatorname{sym}_{\mathfrak{n}%
,n}\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right)
=\operatorname{Ker}\left( \operatorname{gr}_{n}\tau\right) $, so that
$0=\left( \operatorname{gr}_{n}\tau\right) \left( i^{\prime}\right) =i$.
We have thus shown that every $i\in\operatorname{Ker}\Theta_{n}$ satisfies
$i=0$. Thus, $\operatorname{Ker}\Theta_{n}=0$, so that $\Theta_{n}$ is injective.
Since $\Theta_{n}$ is surjective and injective and a $k$-module homomorphism,
we conclude that $\Theta_{n}$ is a $k$-module isomorphism.
We know that $\operatorname{gr}_{n}\tau$ is a surjective $\mathfrak{h}$-module
homomorphism (since $\operatorname{gr}_{n}\tau$ is surjective according to
part \textbf{(a)}, and is an $\mathfrak{h}$-module homomorphism since $\tau$
is an $\mathfrak{h}$-module homomorphism (by Theorem \ref{thm.l42}
\textbf{(b)})), and we know that $\Theta_{n}\circ\operatorname{gr}_{n}\tau$ is
an $\mathfrak{h}$-module homomorphism (since $\Theta_{n}\circ\operatorname{gr}%
_{n}\tau=\operatorname{sym}_{\mathfrak{n},n}\circ\operatorname{grad}%
_{\mathfrak{n},n}^{-1}\circ\omega_{n}$, and since all of the maps
$\operatorname{sym}_{\mathfrak{n},n}$, $\operatorname{grad}_{\mathfrak{n}%
,n}^{-1}$ and $\omega_{n}$ are $\mathfrak{h}$-module homomorphisms). Applying
Lemma \ref{lem.surj} to $\operatorname{gr}_{n}\left( \otimes\mathfrak{g}%
\diagup\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
\right) $, $\operatorname{gr}_{n}\left( U\left( \mathfrak{g}\right)
\diagup\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right)
\right) $, $\operatorname{Sym}^{n}\mathfrak{n}$, $\operatorname{gr}_{n}\tau$
and $\Theta_{n}$ instead of $A$, $B$, $C$, $f$ and $g$, we thus conclude that
$\Theta_{n}$ is an $\mathfrak{h}$-module homomorphism. Combining this with the
fact that $\Theta_{n}$ is a $k$-module isomorphism, we conclude that
$\Theta_{n}$ is an $\mathfrak{h}$-module isomorphism. This proves Corollary
\ref{cor.l42} \textbf{(c)}.
\textbf{(d)} By the functoriality of $\operatorname{gr}_{n}$, we have
$\operatorname{gr}_{n}\left( \tau\circ\zeta\right) =\operatorname{gr}%
_{n}\tau\circ\operatorname{gr}_{n}\zeta$. On the other hand, the commutative
diagram (\ref{thm.l25.c.diag'}) yields $\omega_{n}\circ\operatorname{gr}%
_{n}\zeta=\operatorname{gr}_{n}\left( \otimes\pi\right) $. Now,%
\begin{align*}
\Theta_{n}\circ\operatorname{gr}_{n}\left( \underbrace{\rho\circ\psi
}_{\substack{=\tau\circ\zeta\\\text{(by Theorem \ref{thm.l42} \textbf{(b)})}%
}}\right) & =\Theta_{n}\circ\underbrace{\operatorname{gr}_{n}\left(
\tau\circ\zeta\right) }_{=\operatorname{gr}_{n}\tau\circ\operatorname{gr}%
_{n}\zeta}=\underbrace{\Theta_{n}\circ\operatorname{gr}_{n}\tau}%
_{\substack{=\operatorname{sym}_{\mathfrak{n},n}\circ\operatorname{grad}%
_{\mathfrak{n},n}^{-1}\circ\omega_{n}\\\text{(by Corollary \ref{cor.l42}
\textbf{(c)})}}}\circ\operatorname{gr}_{n}\zeta\\
& =\operatorname{sym}_{\mathfrak{n},n}\circ\operatorname{grad}_{\mathfrak{n}%
,n}^{-1}\circ\underbrace{\omega_{n}\circ\operatorname{gr}_{n}\zeta
}_{=\operatorname{gr}_{n}\left( \otimes\pi\right) }=\operatorname{sym}%
_{\mathfrak{n},n}\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ
\operatorname{gr}_{n}\left( \otimes\pi\right) .
\end{align*}
In other words, the diagram (\ref{cor.l42.diag2}) commutes. This proves
Corollary \ref{cor.l42} \textbf{(d)}.
Note that Corollary \ref{cor.l42} can be used to prove Theorem \ref{thm.pbw}
\textbf{(a)}. This is not particularly surprising and not particularly useful,
as we have used Theorem \ref{thm.pbw} \textbf{(a)} in our proof of Corollary
\ref{cor.l42}. But let us give the proof for the sake of completeness:
\textit{Proof of Theorem \ref{thm.pbw} \textbf{(a)} using Corollary
\ref{cor.l42}:} Let $n\in\mathbb{N}$. Let $\mathfrak{g}$ be a $k$-Lie algebra
which is a free $k$-module. Let $\mathfrak{h}=0$ and $N=\mathfrak{g}$. Then,
$\mathfrak{h}$ is a Lie subalgebra of $\mathfrak{g}$ such that $\mathfrak{g}%
=\mathfrak{h}\oplus N$. Moreover, both $\mathfrak{h}$ and $N$ are free
$k$-modules. Moreover, the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$
splits \textit{as a }$k$\textit{-module inclusion}. Thus, we can apply
Corollary \ref{cor.l42} \textbf{(d)} and obtain that the diagram
(\ref{cor.l42.diag2}) commutes (where we are using the notations of Corollary
\ref{cor.l42}, of course). But since we are in a situation where
$\mathfrak{h}=0$ and $\mathfrak{n}=\mathfrak{g}\diagup\mathfrak{h}%
=\mathfrak{g}$, this diagram simplifies to%
\[
\xymatrixcolsep{7pc}\xymatrix{
\operatorname{gr}_n\left(\otimes \mathfrak g\right) \ar[r]^-{\operatorname*{gr}_n\psi} \ar[d]_{\operatorname*{id}} & \operatorname*{gr}_n\left( U\left(\mathfrak g\right) \right) \ar[dd]^{\Theta_n} \\
\operatorname{gr}_n\left(\otimes\mathfrak g\right) \ar[d]_{\operatorname*{grad}_{\mathfrak g,n}^{-1}}^{\cong} &
\\ \mathfrak g^{\otimes n} \ar[r]_{\operatorname*{sym}_{\mathfrak g,n}} & \operatorname*{Sym}^n \mathfrak g }
\]
(since $\pi=\operatorname{id}$, $\otimes\pi=\operatorname{id}$,
$\operatorname{gr}_{n}\left( \otimes\pi\right) =\operatorname{id}$ and
$\rho=\operatorname{id}$). Thus, $\Theta_{n}\circ\operatorname{gr}_{n}%
\psi=\operatorname{sym}_{\mathfrak{g},n}\circ\operatorname{grad}%
_{\mathfrak{g},n}^{-1}$. Thus, $\Theta_{n}\circ\operatorname{gr}_{n}\psi
\circ\operatorname{grad}_{\mathfrak{g},n}=\operatorname{sym}_{\mathfrak{g},n}%
$, so that%
\[
\operatorname{sym}_{\mathfrak{g},n}=\Theta_{n}\circ
\underbrace{\operatorname{gr}_{n}\psi\circ\operatorname{grad}_{\mathfrak{g}%
,n}}_{\substack{=\operatorname{PBW}_{\mathfrak{g},n}\circ\operatorname{sym}%
_{\mathfrak{g},n}\\\text{(since this is how we defined }\operatorname{PBW}%
_{\mathfrak{g},n}\text{)}}}=\Theta_{n}\circ\operatorname{PBW}_{\mathfrak{g}%
,n}\circ\operatorname{sym}_{\mathfrak{g},n}.
\]
Since $\operatorname{sym}_{\mathfrak{g},n}$ is surjective, this yields
$\operatorname{id}=\Theta_{n}\circ\operatorname{PBW}_{\mathfrak{g},n}$. Thus,
$\operatorname{PBW}_{\mathfrak{g},n}$ is the inverse of the $\mathfrak{h}%
$-module isomorphism $\Theta_{n}$ (here, we are using the fact that
$\Theta_{n}$ is an $\mathfrak{h}$-module isomorphism; this follows from
Corollary \ref{cor.l42} \textbf{(c)}). This yields that $\operatorname{PBW}%
_{\mathfrak{g},n}$ itself is an $\mathfrak{h}$-module isomorphism. In other
words, $\mathfrak{g}$ satisfies the $n$-PBW condition, and our proof of
Theorem \ref{thm.pbw} \textbf{(a)} is complete. As already explained, this
proof does not take us far, as the proof of Corollary \ref{cor.l42} given
above made substantial use of Theorem \ref{thm.pbw} \textbf{(a)}; but at least
it shows that Corollary \ref{cor.l42} is indeed a generalization of Theorem
\ref{thm.pbw} \textbf{(a)}.
\subsection{The splitting of the filtration}
The next theorem encompasses a part of \cite[Theorem 4.5]{calaque1}, and
(together with the theorems we have already proven) will (almost) complete the
proof of the main result of \cite{calaque1} ("almost" because the converse
direction will still be missing, but it is rather easy and straightforward).
\begin{theorem}
\label{thm.l45}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Assume that
the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$ splits \textit{as a
}$k$\textit{-module inclusion} (but not necessarily as an $\mathfrak{h}%
$-module inclusion). This means that there exists a $k$-submodule $N$ of
$\mathfrak{g}$ such that $\mathfrak{g}=\mathfrak{h}\oplus N$.\newline Assume
that both $\mathfrak{h}$ and $N$ are free $k$-modules.\newline Let us consider
the $\mathfrak{g}$-module $U\left( \mathfrak{g}\right) \ \ \ \ $%
.{\footnotemark}\newline Let us also use the notations introduced in Theorem
\ref{thm.l25}, in particularly the $\mathfrak{h}$-module $\mathfrak{n}%
$.\newline We also consider the $\mathfrak{h}$-module $\operatorname{Sym}%
^{n}\mathfrak{n}$ (see Definition \ref{defs.symm-alg.g} for its definition)
for every $n\in\mathbb{N}$.\newline\textbf{Assume that this }$\mathfrak{h}%
$\textbf{-module }$\mathfrak{n}$ \textbf{is actually the restriction of some
}$\left( \mathfrak{g},\mathfrak{h}\right) $\textbf{-semimodule to
}$\mathfrak{h}$\textbf{.}\newline Now let us assume the following statement,
which we call the \textbf{symmetric splitting assumption: The canonical
projection }$\operatorname{sym}_{\mathfrak{n},n}:\mathfrak{n}^{\otimes
n}\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$\textbf{ (defined in
Definition \ref{defs.symm-alg}) splits as an }$\mathfrak{h}$\textbf{-module
projection for every }$n\in\mathbb{N}$\textbf{.\newline}(Note that the
symmetric splitting assumption is automatically satisfied in the case when
every positive integer is invertible in the ring $k$, because in this case we
can split the projection $\operatorname{sym}_{\mathfrak{n},n}:\mathfrak{n}%
^{\otimes n}\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ by the map
$\operatorname{Sym}^{n}\mathfrak{n}\rightarrow\mathfrak{n}^{\otimes n}$ which
sends $\overline{v_{1}\otimes v_{2}\otimes...\otimes v_{n}}$ to $\dfrac{1}%
{n!}\sum\limits_{\sigma\in S_{n}}v_{\sigma\left( 1\right) }\otimes
v_{\sigma\left( 2\right) }\otimes...\otimes v_{\sigma\left( n\right) }$
for every $\left( v_{1},v_{2},...,v_{n}\right) \in\mathfrak{n}^{n}$. But
this is not the only case in which the symmetric splitting assumption
holds.)\newline Define a filtration $\left( W_{n}\right) _{n\geq0}$ on
$U\left( \mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) $ as in Theorem \ref{thm.l42} \textbf{(c)}.\newline
Then, the filtration $\left( W_{n}\right) _{n\geq0}$ is an $\mathfrak{h}%
$-split $\mathfrak{h}$-module filtration.\newline More precisely, we can
construct a splitting for the $\mathfrak{h}$-module projection $W_{n}%
\rightarrow W_{n}\diagup W_{n-1}$ for every $n\geq1$ as follows:\newline
Consider the map $\mathfrak{h}$-module isomorphism $\Theta_{n}%
:\operatorname{gr}_{n}\left( U\left( \mathfrak{g}\right) \diagup\left(
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) \right)
\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ from Corollary \ref{cor.l42}
\textbf{(c)}. Since $\operatorname{gr}_{n}\left( U\left( \mathfrak{g}%
\right) \diagup\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}%
\right) \right) =W_{n}\diagup W_{n-1}$, this $\Theta_{n}$ is thus an
$\mathfrak{h}$-module isomorphism $W_{n}\diagup W_{n-1}\rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$.\newline Let $\mathbf{I}_{n}%
:\operatorname{Sym}^{n}\mathfrak{n}\rightarrow\mathfrak{n}^{\otimes n}$ be an
$\mathfrak{h}$-module homomorphism such that $\operatorname{sym}%
_{\mathfrak{n},n}\circ\mathbf{I}_{n}=\operatorname*{id}$. (In other words, let
$\mathbf{I}_{n}:\operatorname{Sym}^{n}\mathfrak{n}\rightarrow\mathfrak{n}%
^{\otimes n}$ be an $\mathfrak{h}$-module homomorphism which splits the
projection $\operatorname{sym}_{\mathfrak{n},n}:\mathfrak{n}^{\otimes
n}\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$. Such an $\mathbf{I}_{n}$
exists due to the symmetric splitting assumption. If every positive integer is
invertible in the ring $k$, then we can even find a \textit{canonical}
$\mathbf{I}_{n}$.)\newline Since the filtration $\left( F_{n}\right)
_{n\geq0}$ is $\mathfrak{h}$-split (by Theorem \ref{thm.l37ba}), the exact
sequence $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%0 \ar[r] & F_{n-1} \ar[r]^-{\text{inclusion}} & F_n \ar[r]^-{\text
%{projection}} & F_n \diagup F_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
0 \ar[r] & F_{n-1} \ar[r]^-{\text{inclusion}} & F_n \ar[r]^-{\text
{projection}} & F_n \diagup F_{n-1} \ar[r] & 0
}%
%EndExpansion
$ is $\mathfrak{h}$-split. So there exists some $\mathfrak{h}$-module
homomorphism $\vartheta_{n}:F_{n}\diagup F_{n-1}\rightarrow F_{n}$ which
splits the projection $F_{n}\rightarrow F_{n}\diagup F_{n-1}$. Consider this
$\vartheta_{n}$.\newline Let $\tau\mid_{F_{n}}^{W_{n}}$ denote the map
$F_{n}\rightarrow W_{n}$ obtained by restricting $\tau$ to $F_{n}$ (since we
know that $\tau\left( F_{n}\right) =W_{n}$).\newline Then, the map $\left(
\tau\mid_{F_{n}}^{W_{n}}\right) \circ\vartheta_{n}\circ\omega_{n}^{-1}%
\circ\operatorname*{grad}\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}%
\circ\Theta_{n}^{-1}:W_{n}\diagup W_{n-1}\rightarrow W_{n}$ is an
$\mathfrak{h}$-module homomorphism which splits the projection $W_{n}%
\rightarrow W_{n}\diagup W_{n-1}$.
\end{theorem}
\begin{vershort}
\footnotetext{Again, let us remind ourselves that "the $\mathfrak{g}$-module
$U\left( \mathfrak{g}\right) $" is to be understood in accordance to Remark
\ref{rmk.g-mod-U(g).short} here.}
\end{vershort}
\begin{verlong}
\footnotetext{Again, let us remind ourselves that "the $\mathfrak{g}$-module
$U\left( \mathfrak{g}\right) $" is to be understood in accordance to Remark
\ref{rmk.g-mod-U(g).long} here.}
\end{verlong}
\textit{Proof of Theorem \ref{thm.l45}.} First, let us make sure that the map
$\left( \tau\mid_{F_{n}}^{W_{n}}\right) \circ\vartheta_{n}\circ\omega
_{n}^{-1}\circ\operatorname*{grad}\nolimits_{\mathfrak{n},n}\circ
\mathbf{I}_{n}\circ\Theta_{n}:W_{n}\diagup W_{n-1}\rightarrow W_{n}$ is
well-defined at all. This is since $\Theta_{n}:W_{n}\diagup W_{n-1}%
\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ (in fact, we know that
$\Theta_{n}:\operatorname{gr}_{n}\left( U\left( \mathfrak{g}\right)
\diagup\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right)
\right) \rightarrow\operatorname{Sym}^{n}\mathfrak{n}$, but by definition of
$\operatorname*{gr}\nolimits_{n}$ we have $\operatorname{gr}_{n}\left(
U\left( \mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) \right) =W_{n}\diagup W_{n-1}$), since
$\mathbf{I}_{n}:\operatorname{Sym}^{n}\mathfrak{n}\rightarrow\mathfrak{n}%
^{\otimes n}$, since $\operatorname*{grad}\nolimits_{\mathfrak{n}%
,n}:\mathfrak{n}^{\otimes n}\rightarrow\operatorname*{gr}\nolimits_{n}\left(
\otimes\mathfrak{n}\right) $, since $\omega_{n}^{-1}:\operatorname*{gr}%
\nolimits_{n}\left( \otimes\mathfrak{n}\right) \rightarrow F_{n}\diagup
F_{n-1}$ (because $\omega_{n}:F_{n}\diagup F_{n-1}\rightarrow
\operatorname*{gr}\nolimits_{n}\left( \otimes\mathfrak{n}\right) $), since
$\vartheta_{n}:F_{n}\diagup F_{n-1}\rightarrow F_{n}$ and since $\tau
\mid_{F_{n}}^{W_{n}}:F_{n}\rightarrow W_{n}$.
Second, this map $\left( \tau\mid_{F_{n}}^{W_{n}}\right) \circ\vartheta
_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}\nolimits_{\mathfrak{n}%
,n}\circ\mathbf{I}_{n}\circ\Theta_{n}$ is indeed an $\mathfrak{h}$-module
homomorphism, since it is a composition of six $\mathfrak{h}$-module homomorphisms.
We must now prove that this map $\left( \tau\mid_{F_{n}}^{W_{n}}\right)
\circ\vartheta_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}$ splits the
projection $W_{n}\rightarrow W_{n}\diagup W_{n-1}$. In fact, let us denote the
projection $W_{n}\rightarrow W_{n}\diagup W_{n-1}$ by $\Gamma_{n}$. Let us
denote the projection $F_{n}\rightarrow F_{n}\diagup F_{n-1}$ by $\Gamma
_{n}^{\prime}$. Then, $\Gamma_{n}\circ\left( \tau\mid_{F_{n}}^{W_{n}}\right)
=\operatorname*{gr}\nolimits_{n}\tau\circ\Gamma_{n}^{\prime}$ follows easily
from the definition of $\operatorname*{gr}\nolimits_{n}$. On the other hand,
$\Gamma_{n}^{\prime}\circ\vartheta_{n}=\operatorname*{id}$ (since
$\vartheta_{n}$ is defined as a splitting of the projection $F_{n}\rightarrow
F_{n}\diagup F_{n-1}$, but that projection $F_{n}\rightarrow F_{n}\diagup
F_{n-1}$ is $\Gamma_{n}^{\prime}$). Thus,
\begin{align*}
& \Gamma_{n}\circ\left( \left( \tau\mid_{F_{n}}^{W_{n}}\right)
\circ\vartheta_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}\right) \\
& =\underbrace{\Gamma_{n}\circ\left( \tau\mid_{F_{n}}^{W_{n}}\right)
}_{=\operatorname*{gr}\nolimits_{n}\tau\circ\Gamma_{n}^{\prime}}\circ
\vartheta_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}\\
& =\operatorname*{gr}\nolimits_{n}\tau\circ\underbrace{\Gamma_{n}^{\prime
}\circ\vartheta_{n}}_{=\operatorname*{id}}\circ\omega_{n}^{-1}\circ
\operatorname*{grad}\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ
\Theta_{n}=\operatorname*{gr}\nolimits_{n}\tau\circ\omega_{n}^{-1}%
\circ\operatorname*{grad}\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}%
\circ\Theta_{n}\\
& =\Theta_{n}^{-1}\circ\operatorname{sym}_{\mathfrak{n},n}\circ
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\underbrace{\omega_{n}%
\circ\omega_{n}^{-1}}_{=\operatorname*{id}}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\operatorname{sym}_{\mathfrak{n},n}\circ\operatorname{grad}%
_{\mathfrak{n},n}^{-1}\circ\omega_{n}=\Theta_{n}\circ\operatorname{gr}_{n}%
\tau\text{ (by Corollary \ref{cor.l42} \textbf{(c)}),}\\
\text{and thus }\operatorname{gr}_{n}\tau=\Theta_{n}^{-1}\circ
\operatorname{sym}_{\mathfrak{n},n}\circ\operatorname{grad}_{\mathfrak{n}%
,n}^{-1}\circ\omega_{n}\text{ (since }\Theta_{n}\text{ is an isomorphism)}%
\end{array}
\right) \\
& =\Theta_{n}^{-1}\circ\operatorname{sym}_{\mathfrak{n},n}\circ
\underbrace{\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}}_{=\operatorname*{id}}\circ\mathbf{I}_{n}%
\circ\Theta_{n}=\Theta_{n}^{-1}\circ\underbrace{\operatorname{sym}%
_{\mathfrak{n},n}\circ\mathbf{I}_{n}}_{=\operatorname*{id}}\circ\Theta
_{n}=\Theta_{n}^{-1}\circ\Theta_{n}=\operatorname*{id}.
\end{align*}
In other words, the map $\left( \tau\mid_{F_{n}}^{W_{n}}\right)
\circ\vartheta_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}$ splits the
projection $W_{n}\rightarrow W_{n}\diagup W_{n-1}$ (because $\Gamma_{n}$ is
the projection $W_{n}\rightarrow W_{n}\diagup W_{n-1}$). Therefore, the exact
sequence $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%0 \ar[r] & W_{n-1} \ar[r]^-{\text{inclusion}} & W_n \ar[r]^-{\text
%{projection}} & W_n \diagup W_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
0 \ar[r] & W_{n-1} \ar[r]^-{\text{inclusion}} & W_n \ar[r]^-{\text
{projection}} & W_n \diagup W_{n-1} \ar[r] & 0
}%
%EndExpansion
$ is $\mathfrak{h}$-split (since the map $\left( \tau\mid_{F_{n}}^{W_{n}%
}\right) \circ\vartheta_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}$ is an
$\mathfrak{h}$-module homomorphism). Since we have proven this for each
$n\in\mathbb{N}$, we thus conclude that the filtration $\left( W_{n}\right)
_{n\geq0}$ is $\mathfrak{h}$-split.
This proves Theorem \ref{thm.l45}.
\subsection{Non-canonical isomorphisms}
Corollary \ref{cor.l42} \textbf{(c)} gave us a canonical $\mathfrak{h}$-module
isomorphism $\operatorname{gr}_{n}\left( U\left( \mathfrak{g}\right)
\diagup\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right)
\right) \rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ for every
$n\in\mathbb{N}$. Therefore,
\[
\bigoplus\limits_{n\in\mathbb{N}}\operatorname{gr}_{n}\left( U\left(
\mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) \right) \cong\bigoplus\limits_{n\in\mathbb{N}%
}\operatorname{Sym}^{n}\mathfrak{n}=\operatorname*{Sym}\mathfrak{n}%
\]
by a canonical isomorphism. Now, the standard intuition for the direct sum
$\bigoplus\limits_{n\in\mathbb{N}}\operatorname{gr}_{n}V$ (where $V$ is some
filtered $k$-module) is that this sum is a kind of "approximation" for $V$,
which is usually simpler than $V$ itself (for example, $U\left(
\mathfrak{g}\right) $ is generally a non-commutative algebra, while
$\bigoplus\limits_{n\in\mathbb{N}}\operatorname{gr}_{n}\left( U\left(
\mathfrak{g}\right) \right) $ is a commutative one). So now, knowing that
the "approximation" $\bigoplus\limits_{n\in\mathbb{N}}\operatorname{gr}%
_{n}\left( U\left( \mathfrak{g}\right) \diagup\left( U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}\right) \right) $ of $U\left(
\mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) $ is isomorphic to $\operatorname*{Sym}\mathfrak{n}%
$, we can ask ourselves what $U\left( \mathfrak{g}\right) \diagup\left(
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) $ itself is isomorphic
to. It turns out that in the situation of Theorem \ref{thm.l45}, we have the
same answer, but the isomorphism is not canonical anymore:
\begin{proposition}
\label{prop.l49}Consider the situation of Theorem \ref{thm.l45}.\newline We
have $U\left( \mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}%
\right) \cdot\mathfrak{h}\right) \cong\operatorname*{Sym}\mathfrak{n}$ as
$\mathfrak{h}$-modules.\newline More precisely, there is an $\mathfrak{h}%
$-module isomorphism $U\left( \mathfrak{g}\right) \diagup\left( U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}\right) \rightarrow\operatorname*{Sym}%
\mathfrak{n}$ which respects the filtration such that the inverse of this
isomorphism also respects the filtration.
\end{proposition}
\textit{Proof of Proposition \ref{prop.l49}.} We know from Theorem
\ref{thm.l45} that the filtration $\left( W_{n}\right) _{n\geq0}$ is
$\mathfrak{h}$-split. Thus, Proposition \ref{prop.split.filtr-g-2} (applied to
$U\left( \mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) $, $\left( W_{n}\right) _{n\geq0}$ and
$\mathfrak{h}$ instead of $V$, $\left( V_{n}\right) _{n\geq0}$ and
$\mathfrak{g}$) yields that there exists a bifiltered $\mathfrak{h}$-module
isomorphism $U\left( \mathfrak{g}\right) \diagup\left( U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}\right) \rightarrow\bigoplus
\limits_{p\in\mathbb{N}}\operatorname*{gr}\nolimits_{p}\left( U\left(
\mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) \right) $.
On the other hand, every $p\in\mathbb{N}$ satisfies $\operatorname*{gr}%
\nolimits_{p}\left( U\left( \mathfrak{g}\right) \diagup\left( U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}\right) \right) \cong%
\operatorname*{Sym}\nolimits^{p}\mathfrak{n}$ as $\mathfrak{h}$-modules
(because Corollary \ref{cor.l42} \textbf{(c)} (applied to $p$ instead of $n$)
shows that $\Theta_{p}$ is an $\mathfrak{h}$-module isomorphism
$\operatorname{gr}_{p}\left( U\left( \mathfrak{g}\right) \diagup\left(
U\left( \mathfrak{g}\right) \cdot\mathfrak{h}\right) \right)
\rightarrow\operatorname{Sym}^{p}\mathfrak{n}$). Therefore, there exists a
bifiltered $\mathfrak{h}$-module isomorphism $\bigoplus\limits_{p\in
\mathbb{N}}\operatorname*{gr}\nolimits_{p}\left( U\left( \mathfrak{g}%
\right) \diagup\left( U\left( \mathfrak{g}\right) \cdot\mathfrak{h}%
\right) \right) \rightarrow\bigoplus\limits_{p\in\mathbb{N}}%
\operatorname*{Sym}\nolimits^{p}\mathfrak{n}=\operatorname*{Sym}\mathfrak{n}$.
Composing this isomorphism with our bifiltered $\mathfrak{h}$-module
isomorphism $U\left( \mathfrak{g}\right) \diagup\left( U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}\right) \rightarrow\bigoplus
\limits_{p\in\mathbb{N}}\operatorname*{gr}\nolimits_{p}\left( U\left(
\mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) \right) $, we obtain a bifiltered $\mathfrak{h}%
$-module isomorphism $U\left( \mathfrak{g}\right) \diagup\left( U\left(
\mathfrak{g}\right) \cdot\mathfrak{h}\right) \rightarrow\operatorname*{Sym}%
\mathfrak{n}$. By the definition of "bifiltered", this is an isomorphism which
respects the filtration such that the inverse of this isomorphism also
respects the filtration. This proves Proposition \ref{prop.l49}.
Even if the (somewhat restrictive) conditions of Theorem \ref{thm.l45} are not
satisfied, we can still obtain a $k$\textit{-module} isomorphy $U\left(
\mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) \cong\operatorname*{Sym}\mathfrak{n}$ under somewhat
weaker conditions:
\begin{proposition}
\label{prop.l50}Consider the situation of Theorem \ref{thm.l42}. Assume that
$\mathfrak{h}$ and $N$ are free $k$-modules.\newline We have $U\left(
\mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) \cong\operatorname*{Sym}\mathfrak{n}$ as
$k$-modules.\newline More precisely, there is a $k$-module isomorphism
$U\left( \mathfrak{g}\right) \diagup\left( U\left( \mathfrak{g}\right)
\cdot\mathfrak{h}\right) \rightarrow\operatorname*{Sym}\mathfrak{n}$ which
respects the filtration such that the inverse of this isomorphism also
respects the filtration.
\end{proposition}
\textit{Proof of Proposition \ref{prop.l50}.} The idea of the proof of
Proposition \ref{prop.l50} is to proceed as in the proof of Proposition
\ref{prop.l49}, but to read "$k$-split", "$k$-module" and "$k$-module
homomorphism" instead of each "$\mathfrak{h}$-split", "$\mathfrak{h}$-module"
and "$\mathfrak{h}$-module homomorphism", respectively. Do the same
modifications in the proof of Theorem \ref{thm.l45}.
Of course, it is not really that easy, because Theorem \ref{thm.l45} has some
more conditions than we have assumed in Proposition \ref{prop.l50}. Here is
how to deal with them:
\begin{itemize}
\item Since we read "$k$-module" instead of "$\mathfrak{h}$-module", the
symmetric splitting assumption (which is needed for Theorem \ref{thm.l45} to
hold) now takes the following form: "The canonical projection
$\operatorname{sym}_{\mathfrak{n},n}:\mathfrak{n}^{\otimes n}\rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$ (defined in Definition \ref{defs.symm-alg}%
) splits as a $k$-module projection for every $n\in\mathbb{N}$." But this is
obviously satisfied, because $\mathfrak{n}\cong N$ (as $k$-modules) is a free
$k$-module.
\item The assumption that the $\mathfrak{h}$-module $\mathfrak{n}$ be the
restriction of some $\left( \mathfrak{g},\mathfrak{h}\right) $-semimodule to
$\mathfrak{h}$ is not granted anymore. Fortunately, we only use it to make
sure that the filtration $\left( F_{n}\right) _{n\geq0}$ is $\mathfrak{h}%
$-split. Since we read "$k$-split" instead of "$\mathfrak{h}$-split", we
therefore just need a new argument for why the filtration $\left(
F_{n}\right) _{n\geq0}$ is $k$-split. We have more or less done this in
Section \ref{sect.L25} already: From Proposition \ref{prop.l25.Ker-alpha}
\textbf{(b)} and \textbf{(d)}, the map $\overline{\varphi}$ (constructed in
Proposition \ref{prop.l25.Ker-alpha} \textbf{(a)}) is a bifiltered isomorphism
$\left( \otimes\mathfrak{g}\right) \diagup\left( J+\left( \otimes
\mathfrak{g}\right) \cdot\mathfrak{h}\right) \rightarrow\otimes N$. Since
the filtration $\left( N^{\otimes\leq n}\right) _{n\geq0}$ of $\otimes N$ is
$k$-split, we can now easily conclude that so is the filtration $\left(
F_{n}\right) _{n\geq0}$ of $\left( \otimes\mathfrak{g}\right)
\diagup\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
$.
\end{itemize}
This proves Proposition \ref{prop.l50}.
\section{\label{sect.gen}Generalizations, improvements and analogues}
While the results we gave above were already somewhat more general than those
of \cite{calaque1}, some of them can be extended even further, and/or shown to
have analogues. These extensions and analogues have never been studied in
detail, and neither am I going to do so in the present paper, but I will
discuss them in brief in this Section \ref{sect.gen}.
\subsection{\label{subsect.general}When $\mathfrak{g}$ is not a Lie algebra}
The results of Sections \ref{sect.L25} and \ref{sect.split} can be
substantially generalized once we notice the following: All results of
Sections \ref{sect.L25} and \ref{sect.split} were formulated in the following
setup:%
\begin{align*}
& \mathfrak{g}\text{ is a }k\text{-Lie algebra, and }\mathfrak{h}\text{ is a
Lie subalgebra of }\mathfrak{g}\text{.}\\
& \text{We consider }\mathfrak{g}\text{ as an }\mathfrak{h}\text{-module (by
restricting the }\mathfrak{g}\text{-module }\mathfrak{g}\text{).}%
\end{align*}
However, we have never used this setup in its full glory in Sections
\ref{sect.L25} and \ref{sect.split}, and everything done in these Sections can
be extended to the case when this setup is replaced by the following one:%
\begin{align*}
& \mathfrak{h}\text{ is a }k\text{-Lie algebra, and }\mathfrak{g}\text{ is an
}\mathfrak{h}\text{-module which}\\
& \text{happens to contain the }\mathfrak{h}\text{-module }\mathfrak{h}\text{
as an }\mathfrak{h}\text{-submodule.}%
\end{align*}
For example, Theorem \ref{thm.l25} takes the following form in this case:
\begin{theorem}
\label{thm.l25.gen}Let $k$ be a commutative ring. Let $\mathfrak{h}$ be a
$k$-Lie algebra, and let $\mathfrak{g}$ be an $\mathfrak{h}$-module (not
necessarily a Lie algebra itself!). Assume that the $\mathfrak{h}$-module
$\mathfrak{h}$ itself (this $\mathfrak{h}$-module $\mathfrak{h}$ is defined
according to Definition \ref{defs.g-mod-g}, applied to $\mathfrak{h}$ instead
of $\mathfrak{g}$) is an $\mathfrak{h}$-submodule of $\mathfrak{g}$.\newline
Assume that the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$ splits
\textit{as a }$k$\textit{-module inclusion} (but not necessarily as an
$\mathfrak{h}$-module inclusion).\newline Let $J$ be the two-sided ideal%
\[
\left( \otimes\mathfrak{g}\right) \cdot\left\langle v\otimes w-w\otimes
v+w\rightharpoonup v\ \mid\ \left( v,w\right) \in\mathfrak{g}\times
\mathfrak{h}\right\rangle \cdot\left( \otimes\mathfrak{g}\right)
\]
of the $k$-algebra $\otimes\mathfrak{g}$.\newline Let $\mathfrak{n}%
=\mathfrak{g}\diagup\mathfrak{h}$. This $\mathfrak{n}$ is an $\mathfrak{h}%
$-module (because both $\mathfrak{g}$ and $\mathfrak{h}$ are $\mathfrak{h}%
$-modules).\newline Let $\pi:\mathfrak{g}\rightarrow\mathfrak{n}$ be the
canonical projection with kernel $\mathfrak{h}$. Obviously, $\pi$ is an
$\mathfrak{h}$-module homomorphism. Thus, $\otimes\pi:\otimes\mathfrak{g}%
\rightarrow\otimes\mathfrak{n}$ is also an $\mathfrak{h}$-module homomorphism
(according to Proposition \ref{prop.(X)-hom-is-hom}).\newline We consider
$\mathfrak{h}$ as an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$ by
means of the embedding $\mathfrak{h}\hookrightarrow\mathfrak{g}\hookrightarrow
\otimes\mathfrak{g}$.\newline\textbf{(a)} Both $J$ and $\left( \otimes
\mathfrak{g}\right) \cdot\mathfrak{h}$ are $\mathfrak{h}$-submodules of
$\otimes\mathfrak{g}$. Thus, $\left( \otimes\mathfrak{g}\right)
\diagup\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
$ is an $\mathfrak{h}$-module. Let $\zeta:\otimes\mathfrak{g}\rightarrow
\left( \otimes\mathfrak{g}\right) \diagup\left( J+\left( \otimes
\mathfrak{g}\right) \cdot\mathfrak{h}\right) $ be the canonical projection.
Then, $\zeta$ is an $\mathfrak{h}$-module homomorphism.\newline\textbf{(b)}
For every $n\in\mathbb{N}$, let $F_{n}$ be the $\mathfrak{h}$-submodule
$\zeta\left( \mathfrak{g}^{\otimes\leq n}\right) $ of $\left(
\otimes\mathfrak{g}\right) \diagup\left( J+\left( \otimes\mathfrak{g}%
\right) \cdot\mathfrak{h}\right) $.\ \ \ \ {\footnotemark} Also define an
$\mathfrak{h}$-submodule $F_{-1}$ of $\left( \otimes\mathfrak{g}\right)
\diagup\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
$ by $F_{-1}=0$. Then, $\left( F_{n}\right) _{n\geq0}$ is an $\mathfrak{h}%
$-module filtration of $\left( \otimes\mathfrak{g}\right) \diagup\left(
J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right) $ and
satisfies $F_{n}\diagup F_{n-1}\cong\mathfrak{n}^{\otimes n}$ as
$\mathfrak{h}$-modules for every $n\in\mathbb{N}$.\newline\textbf{(c)} Let
$n\in\mathbb{N}$. There exists one and only one $k$-module homomorphism
$\Omega_{n}:F_{n}\diagup F_{n-1}\rightarrow\operatorname*{gr}\nolimits_{n}%
\left( \otimes\mathfrak{n}\right) $ for which the diagram%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{9pc}
%\xymatrix{
%\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
%*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
%\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
%\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
%) \right) = F_n \diagup F_{n-1} \ar[r]_-{\Omega_n} & \operatorname*{gr}%
%_n\left(\otimes\mathfrak{n}\right)
%}}}%
%BeginExpansion
\xymatrixcolsep{9pc}
\xymatrix{
\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
) \right) = F_n \diagup F_{n-1} \ar[r]_-{\Omega_n} & \operatorname*{gr}%
_n\left(\otimes\mathfrak{n}\right)
}%
%EndExpansion
\]
commutes. Denote this homomorphism $\Omega_{n}$ by $\omega_{n}$. Then,
$\omega_{n}$ is an $\mathfrak{h}$-module isomorphism, and the diagram%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{9pc}
%\xymatrix{
%\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
%*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
%\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
%\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
%) \right) = F_n \diagup F_{n-1} \ar[r]_-{\omega_n} & \operatorname*{gr}%
%_n\left(\otimes\mathfrak{n}\right)
%}}}%
%BeginExpansion
\xymatrixcolsep{9pc}
\xymatrix{
\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
) \right) = F_n \diagup F_{n-1} \ar[r]_-{\omega_n} & \operatorname*{gr}%
_n\left(\otimes\mathfrak{n}\right)
}%
%EndExpansion
\]
commutes.\newline Applying Definition \ref{defs.filtr.grad} to $\mathfrak{n}$
and $n$ instead of $V$ and $p$, we obtain a map $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}:\mathfrak{n}^{\otimes n}\rightarrow
\operatorname*{gr}\nolimits_{n}\left( \otimes\mathfrak{n}\right) $.
According to Proposition \ref{prop.filtr.grad.g} (applied to $\mathfrak{h}$,
$n$ and $\mathfrak{n}$ instead of $\mathfrak{g}$, $p$ and $V$), this map
$\operatorname*{grad}\nolimits_{\mathfrak{n},n}$ is a canonical $\mathfrak{h}%
$-module isomorphism. Thus, its inverse $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}^{-1}$ is an $\mathfrak{h}$-module isomorphism as
well. The composition $\operatorname*{grad}\nolimits_{\mathfrak{n},n}%
^{-1}\circ\omega_{n}:F_{n}\diagup F_{n-1}\rightarrow\mathfrak{n}^{\otimes n}$
is an $\mathfrak{h}$-module isomorphism (because $\omega_{n}$ and
$\operatorname*{grad}\nolimits_{\mathfrak{n},n}^{-1}$ are $\mathfrak{h}%
$-module isomorphisms).
\end{theorem}
\footnotetext{In fact, $\zeta\left( \mathfrak{g}^{\otimes\leq n}\right) $ is
indeed an $\mathfrak{h}$-submodule (because $\zeta$ is an $\mathfrak{h}%
$-module homomorphism and because $\mathfrak{g}^{\otimes\leq n}$ is an
$\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$).}
This is indeed a generalization of Theorem \ref{thm.l25}, because in the
situation of Theorem \ref{thm.l25}, we have:%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{the ideal }J\text{ defined in Theorem \ref{thm.l25} is identical}\\
\text{with the ideal }J\text{ defined in Theorem \ref{thm.l25.gen}}%
\end{array}
\right) . \label{thm.l25.gen.deriv.1}%
\end{equation}
\begin{vershort}
This is easy to see (and left to the reader).
\end{vershort}
\begin{verlong}
\textit{Proof of (\ref{thm.l25.gen.deriv.1}).} Every $\left( v,w\right)
\in\mathfrak{g}\times\mathfrak{h}$ satisfies $\left[ v,w\right] =-\left[
w,v\right] $ (due to (\ref{[]-3})) and $w\rightharpoonup v=\left[
w,v\right] $ (due to (\ref{prop.g-is-g-module.form}), applied to $w$ and $v$
instead of $v$ and $w$). Thus, every $\left( v,w\right) \in\mathfrak{g}%
\times\mathfrak{h}$ satisfies
\begin{align}
v\otimes w-w\otimes v-\underbrace{\left[ v,w\right] }_{=-\left[ w,v\right]
} & =v\otimes w-w\otimes v+\underbrace{\left[ w,v\right] }%
_{=w\rightharpoonup v}\nonumber\\
& =v\otimes w-w\otimes v+w\rightharpoonup v. \label{pf.l25.deriv.J}%
\end{align}
Now,%
\begin{align*}
& \left( \text{the ideal }J\text{ defined in Theorem \ref{thm.l25.gen}%
}\right) \\
& =\left( \otimes\mathfrak{g}\right) \cdot\left\langle \underbrace{v\otimes
w-w\otimes v+w\rightharpoonup v}_{=v\otimes w-w\otimes v-\left[ v,w\right]
}\ \mid\ \left( v,w\right) \in\mathfrak{g}\times\mathfrak{h}\right\rangle
\cdot\left( \otimes\mathfrak{g}\right) \\
& =\left( \otimes\mathfrak{g}\right) \cdot\left\langle v\otimes w-w\otimes
v-\left[ v,w\right] \ \mid\ \left( v,w\right) \in\mathfrak{g}%
\times\mathfrak{h}\right\rangle \cdot\left( \otimes\mathfrak{g}\right) \\
& =\left( \text{the ideal }J\text{ defined in Theorem \ref{thm.l25}}\right)
.
\end{align*}
This proves (\ref{thm.l25.gen.deriv.1}).
\end{verlong}
To give a proof of Theorem \ref{thm.l25.gen}, we just have to repeat the proof
of Theorem \ref{thm.l25} that we gave in Section \ref{sect.L25} (including all
the auxiliary facts we showed in Section \ref{sect.L25}) up to the following changes:
\begin{itemize}
\item Replace the words "$\mathfrak{g}$-module" by "$\mathfrak{h}$-module".
\item Replace the words "$\mathfrak{g}$-algebra" by "$\mathfrak{h}$-algebra".
\item Replace the words "$\mathfrak{g}$-submodule" by "$\mathfrak{h}$-submodule".
\item Whenever a term of the form $\left[ x,y\right] $ for some
$x\in\mathfrak{g}$ and $y\in\mathfrak{g}$ appears in Section \ref{sect.L25},
proceed by the following rules:\newline- If $x$ is known to lie in
$\mathfrak{h}$, replace this term by $x\rightharpoonup y$.\newline- If $y$ is
known to lie in $\mathfrak{h}$, replace this term by $-y\rightharpoonup
x$.\newline(If both $x$ and $y$ are known to lie in $\mathfrak{h}$, then it
does not matter which of these two rules is being followed, because
$x\rightharpoonup y=\left[ x,y\right] =-\left[ y,x\right]
=-y\rightharpoonup x$ for any $x\in\mathfrak{h}$ and $y\in\mathfrak{h}%
$.)\newline Fortunately, all terms of the form $\left[ x,y\right] $ that
appear in Section \ref{sect.L25} have either $x$ or $y$ lying in
$\mathfrak{h}$, so that after these replacements, no terms of the form
$\left[ x,y\right] $ remain anymore.
\end{itemize}
Most results in Section \ref{sect.(g,h)} can be generalized as soon as we
extend Definition \ref{defs.semimod} (the definition of a $\left(
\mathfrak{g},\mathfrak{h}\right) $-semimodule) as follows:
\begin{definition}
\label{defs.semimod.gen}Let $k$ be a commutative ring. Let $\mathfrak{h}$ be a
$k$-Lie algebra, and let $\mathfrak{g}$ be an $\mathfrak{h}$-module (not
necessarily a Lie algebra itself!). Assume that the $\mathfrak{h}$-module
$\mathfrak{h}$ itself (this $\mathfrak{h}$-module $\mathfrak{h}$ is defined
according to Definition \ref{defs.g-mod-g}, applied to $\mathfrak{h}$ instead
of $\mathfrak{g}$) is an $\mathfrak{h}$-submodule of $\mathfrak{g}$.\newline
Let $V$ be a $k$-module. Let $\mu:\mathfrak{g}\times V\rightarrow V$ be a
$k$-bilinear map. We say that $\left( V,\mu\right) $ is a $\left(
\mathfrak{g},\mathfrak{h}\right) $\textit{-semimodule} if and only if
\begin{equation}
\left( \mu\left( a\rightharpoonup b,v\right) =\mu\left( a,\mu\left(
b,v\right) \right) -\mu\left( b,\mu\left( a,v\right) \right) \text{ for
every }a\in\mathfrak{h}\text{, }b\in\mathfrak{g}\text{ and }v\in V\right) .
\label{defs.semimod.gen.mu}%
\end{equation}
If $\left( V,\mu\right) $ is a $\left( \mathfrak{g},\mathfrak{h}\right)
$-semimodule, then the $k$-bilinear map $\mu:\mathfrak{g}\times V\rightarrow
V$ is called the \textit{Lie action} of the $\left( \mathfrak{g}%
,\mathfrak{h}\right) $-semimodule $V$.\newline Often, when the map $\mu$ is
obvious from the context, we abbreviate the term $\mu\left( a,v\right) $ by
$a\rightharpoonup v$ for any $a\in\mathfrak{g}$ and $v\in V$. Using this
notation, the relation (\ref{defs.semimod.gen.mu}) rewrites as%
\[
\left( \left( a\rightharpoonup b\right) \rightharpoonup v=a\rightharpoonup
\left( b\rightharpoonup v\right) -b\rightharpoonup\left( a\rightharpoonup
v\right) \text{ for every }a\in\mathfrak{h}\text{, }b\in\mathfrak{g}\text{
and }v\in V\right) .
\]
\newline Also, an abuse of notation allows us to write "$V$ is a $\left(
\mathfrak{g},\mathfrak{h}\right) $-semimodule" instead of "$\left(
V,\mu\right) $ is a $\left( \mathfrak{g},\mathfrak{h}\right) $-semimodule"
if the map $\mu$ is clear from the context or has not been introduced
yet.\newline Besides, when $\left( V,\mu\right) $ is a $\left(
\mathfrak{g},\mathfrak{h}\right) $-semimodule, we will say that $\mu$ is a
$\left( \mathfrak{g},\mathfrak{h}\right) $-\textit{semimodule structure on}
$V$. In other words, if $V$ is a $k$-module, then a $\left( \mathfrak{g}%
,\mathfrak{h}\right) $\textit{-semimodule structure on }$V$ means a map
$\mu:\mathfrak{g}\times V\rightarrow V$ such that $\left( V,\mu\right) $ is
a $\left( \mathfrak{g},\mathfrak{h}\right) $-semimodule. (Thus, in order to
make a $k$-module into a $\left( \mathfrak{g},\mathfrak{h}\right)
$-semimodule, we must define a $\left( \mathfrak{g},\mathfrak{h}\right)
$-semimodule structure on it.)
\end{definition}
Theorem \ref{thm.l37ba} generalizes as follows:
\begin{theorem}
\label{thm.l37ba.gen}Let $k$ be a commutative ring. Let $\mathfrak{h}$ be a
$k$-Lie algebra, and let $\mathfrak{g}$ be an $\mathfrak{h}$-module (not
necessarily a Lie algebra itself!). Assume that the $\mathfrak{h}$-module
$\mathfrak{h}$ itself (this $\mathfrak{h}$-module $\mathfrak{h}$ is defined
according to Definition \ref{defs.g-mod-g}, applied to $\mathfrak{h}$ instead
of $\mathfrak{g}$) is an $\mathfrak{h}$-submodule of $\mathfrak{g}$.\newline
Assume that the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$ splits
\textit{as a }$k$\textit{-module inclusion} (but not necessarily as an
$\mathfrak{h}$-module inclusion).\newline Let $J$ be the two-sided ideal%
\[
\left( \otimes\mathfrak{g}\right) \cdot\left\langle v\otimes w-w\otimes
v+w\rightharpoonup v\ \mid\ \left( v,w\right) \in\mathfrak{g}\times
\mathfrak{h}\right\rangle \cdot\left( \otimes\mathfrak{g}\right)
\]
of the $k$-algebra $\otimes\mathfrak{g}$.\newline Let $\mathfrak{n}%
=\mathfrak{g}\diagup\mathfrak{h}$. This $\mathfrak{n}$ is an $\mathfrak{h}%
$-module (because both $\mathfrak{g}$ and $\mathfrak{h}$ are $\mathfrak{h}%
$-modules). \textbf{Assume that this }$\mathfrak{h}$\textbf{-module
}$\mathfrak{n}$ \textbf{is actually the restriction of some }$\left(
\mathfrak{g},\mathfrak{h}\right) $\textbf{-semimodule to }$\mathfrak{h}$
(where "$\left( \mathfrak{g},\mathfrak{h}\right) $-semimodule" is to be
understood according to Definition \ref{defs.semimod.gen}).\newline Let
$\pi:\mathfrak{g}\rightarrow\mathfrak{n}$ be the canonical projection with
kernel $\mathfrak{h}$. Obviously, $\pi$ is an $\mathfrak{h}$-module
homomorphism. Thus, $\otimes\pi:\otimes\mathfrak{g}\rightarrow\otimes
\mathfrak{n}$ is also an $\mathfrak{h}$-module homomorphism (according to
Proposition \ref{prop.(X)-hom-is-hom}).\newline We consider $\mathfrak{h}$ as
an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$ by means of the embedding
$\mathfrak{h}\hookrightarrow\mathfrak{g}\hookrightarrow\otimes\mathfrak{g}%
$.\newline\textbf{(a)} Both $J$ and $\left( \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$ are $\mathfrak{h}$-submodules of $\otimes\mathfrak{g}$.
Thus, $\left( \otimes\mathfrak{g}\right) \diagup\left( J+\left(
\otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right) $ is an $\mathfrak{h}%
$-module. Let $\zeta:\otimes\mathfrak{g}\rightarrow\left( \otimes
\mathfrak{g}\right) \diagup\left( J+\left( \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right) $ be the canonical projection. Then, $\zeta$ is an
$\mathfrak{h}$-module homomorphism.\newline\textbf{(b)} For every
$n\in\mathbb{N}$, let $F_{n}$ be the $\mathfrak{h}$-submodule $\zeta\left(
\mathfrak{g}^{\otimes\leq n}\right) $ of $\left( \otimes\mathfrak{g}\right)
\diagup\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
$. (That $F_{n}$ indeed is an $\mathfrak{h}$-submodule was proven in Theorem
\ref{thm.l25} already.) Then, $\left( F_{n}\right) _{n\geq0}$ is an
$\mathfrak{h}$-module filtration of $\left( \otimes\mathfrak{g}\right)
\diagup\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
$ and satisfies $F_{n}\cong\mathfrak{n}^{\otimes\leq n}$ as $\mathfrak{h}%
$-modules for every $n\in\mathbb{N}$.\newline\textbf{(c)} There exists an
$\mathfrak{h}$-module isomorphism $\left( \otimes\mathfrak{g}\right)
\diagup\left( J+\left( \otimes\mathfrak{g}\right) \cdot\mathfrak{h}\right)
\rightarrow\otimes\mathfrak{n}$ such that for every $n\in\mathbb{N}$, the
image of $F_{n}$ under this isomorphism is $\mathfrak{n}^{\otimes\leq n}%
$.\newline\textbf{(d)} The filtration $\left( F_{n}\right) _{n\geq0}$ of
$\left( \otimes\mathfrak{g}\right) \diagup\left( J+\left( \otimes
\mathfrak{g}\right) \cdot\mathfrak{h}\right) $ is $\mathfrak{h}$-split.
\end{theorem}
Again, the proof of this theorem is a repetition of the proof of Theorem
\ref{thm.l37ba} with the same replacements as we had to do to obtain a proof
of Theorem \ref{thm.l25.gen}.
The results of Section \ref{sect.pbw} probably cannot be generalized in a
similar fashion.
A further surprise seems to be that the proofs of Theorems \ref{thm.l25.gen}
and \ref{thm.l37ba.gen} apparently never use the axioms (\ref{[]-1}) and
(\ref{[]-2}) for the $k$-Lie algebra $\mathfrak{h}$, instead only using%
\[
\left[ u,\left[ v,w\right] \right] =\left[ \left[ u,v\right] ,w\right]
+\left[ v,\left[ u,w\right] \right] \ \ \ \ \ \ \ \ \ \ \text{for all
}u\in\mathfrak{h}\text{, }v\in\mathfrak{h}\text{ and }w\in\mathfrak{h}.
\]
"Apparently" because I have not had enough time to check that this indeed is
the case. If it is, this means that Theorems \ref{thm.l25.gen} and
\ref{thm.l37ba.gen} extend to Leibniz algebras in lieu of Lie algebras. I am
not aware of a similar extension of the Poincar\'{e}-Birkhoff-Witt theorem.
\subsection{\label{subsect.super}The case of Lie superalgebras}
The notion of a \textit{Lie superalgebra} (also known under the name
\textit{super Lie algebra} and studied in \cite{scheunert}, \cite{deligne}) is
one of the most well-understood generalizations of that of a Lie algebra.
While classification results for Lie superalgebras are significantly harder
than their non-super counterparts, most "purely algebraic" properties of Lie
algebras tend to have their analogues for Lie superalgebras, which usually are
even proven in more or less the same manner. This has to do with the fact that
Lie superalgebras are just Lie algebras in the category of super-$k$-modules;
however there is also a much more pedestrian approach to proving properties of
Lie superalgebras by re-reading the proofs of the corresponding facts about
Lie algebras and adding signs via the Koszul rule.
Different sources disagree about the correct way to define the notion of a Lie
superalgebra. This might have to do with the fact that the primary interest
lies in Lie superalgebras over a field of characteristic $0$ (rather than an
arbitrary field, let alone a commutative ring), and all the definitions of a
Lie superalgebra are equivalent to each other if we are over a field of
characteristic $0$. As I am interested in the general case, let me give the
following definition of a Lie superalgebra (which is one of the most
restrictive ones, but not as restrictive as \cite[Definition 8.1.1]%
{neisendorfer}):
\begin{definition}
\label{defs.super.liealg}Let $k$ be a commutative ring. A $k$\textit{-Lie
superalgebra} will mean a super-$k$-module $\mathfrak{g}$ (see Definition
\ref{defs.super.mod} below) together with a $k$-bilinear map $\beta
:\mathfrak{g}\times\mathfrak{g}\rightarrow\mathfrak{g}$ satisfying the
conditions%
\begin{align}
& \left( \beta\left( v,v\right) =0\text{ for every }v\in\mathfrak{g}%
_{0}\right) ;\label{defs.super.liealg.1}\\
& \left(
\begin{array}
[c]{c}%
\left( -1\right) ^{i\ell}\beta\left( u,\beta\left( v,w\right) \right)
+\left( -1\right) ^{ji}\beta\left( v,\beta\left( w,u\right) \right)
+\left( -1\right) ^{\ell j}\beta\left( w,\beta\left( u,v\right) \right)
=0\\
\text{ for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{, }j\in\mathbb{Z}%
\diagup2\mathbb{Z}\text{ and }\ell\in\mathbb{Z}\diagup2\mathbb{Z}\\
\text{and every }u\in\mathfrak{g}_{i}\text{, }v\in\mathfrak{g}_{j}\text{ and
}w\in\mathfrak{g}_{\ell}%
\end{array}
\right) ;\label{defs.super.liealg.2}\\
& \left(
\begin{array}
[c]{c}%
\beta\left( v,w\right) =-\left( -1\right) ^{ij}\beta\left( w,v\right) \\
\text{for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{ and }j\in
\mathbb{Z}\diagup2\mathbb{Z}\text{ and every }v\in\mathfrak{g}_{i}\text{ and
}w\in\mathfrak{g}_{j}%
\end{array}
\right) ;\label{defs.super.liealg.3}\\
& \left( \beta\left( v,\beta\left( v,v\right) \right) =0\text{ for every
}v\in\mathfrak{g}_{1}\right) ;\label{defs.super.liealg.4}\\
& \left( \beta\left( \mathfrak{g}_{i}\times\mathfrak{g}_{j}\right)
\subseteq\mathfrak{g}_{i+j}\text{ for every }i\in\mathbb{Z}\diagup
2\mathbb{Z}\text{ and }j\in\mathbb{Z}\diagup2\mathbb{Z}\right) .
\label{defs.super.liealg.5}%
\end{align}
\newline This $k$-bilinear map $\beta:\mathfrak{g}\times\mathfrak{g}%
\rightarrow\mathfrak{g}$ will be called the \textit{Lie bracket} of the
$k$-Lie superalgebra $\mathfrak{g}$. We will often use the square brackets
notation for $\beta$, which means that we are going to abbreviate
$\beta\left( v,w\right) $ by $\left[ v,w\right] $ for any $v\in
\mathfrak{g}$ and $w\in\mathfrak{g}$. Using this notation, the equations
(\ref{defs.super.liealg.1}), (\ref{defs.super.liealg.2}),
(\ref{defs.super.liealg.3}), (\ref{defs.super.liealg.4}) and
(\ref{defs.super.liealg.5}) rewrite as%
\begin{align}
& \left( \left[ v,v\right] =0\text{ for every }v\in\mathfrak{g}%
_{0}\right) ;\label{defs.super.liealg.[1]}\\
& \left(
\begin{array}
[c]{c}%
\left( -1\right) ^{i\ell}\left[ u,\left[ v,w\right] \right] +\left(
-1\right) ^{ji}\left[ v,\left[ w,u\right] \right] +\left( -1\right)
^{\ell j}\left[ w,\left[ u,v\right] \right] =0\\
\text{ for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{, }j\in\mathbb{Z}%
\diagup2\mathbb{Z}\text{ and }\ell\in\mathbb{Z}\diagup2\mathbb{Z}\\
\text{and every }u\in\mathfrak{g}_{i}\text{, }v\in\mathfrak{g}_{j}\text{ and
}w\in\mathfrak{g}_{\ell}%
\end{array}
\right) ;\label{defs.super.liealg.[2]}\\
& \left(
\begin{array}
[c]{c}%
\left[ v,w\right] =-\left( -1\right) ^{ij}\left[ w,v\right] \\
\text{for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{ and }j\in
\mathbb{Z}\diagup2\mathbb{Z}\text{ and every }v\in\mathfrak{g}_{i}\text{ and
}w\in\mathfrak{g}_{j}%
\end{array}
\right) ;\label{defs.super.liealg.[3]}\\
& \left( \left[ v,\left[ v,v\right] \right] =0\text{ for every }%
v\in\mathfrak{g}_{1}\right) ;\label{defs.super.liealg.[4]}\\
& \left( \left[ \mathfrak{g}_{i},\mathfrak{g}_{j}\right] \subseteq
\mathfrak{g}_{i+j}\text{ for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{ and
}j\in\mathbb{Z}\diagup2\mathbb{Z}\right) \label{defs.super.liealg.[5]}%
\end{align}
(where $\left[ \mathfrak{g}_{i},\mathfrak{g}_{j}\right] $ means the
$k$-linear span $\left\langle \left[ v,w\right] \ \mid\ \left( v,w\right)
\in\mathfrak{g}_{i}\times\mathfrak{g}_{j}\right\rangle $).\newline The
equation (\ref{defs.super.liealg.2}) (or its equivalent version
(\ref{defs.super.liealg.[2]})) is called the \textit{super-Jacobi identity}.
\end{definition}
Here we have used the following definition:
\begin{definition}
\label{defs.super.mod}Let $k$ be a commutative ring. A \textit{super-}%
$k$\textit{-module} will mean a $k$-module $V$ together with a pair $\left(
V_{0},V_{1}\right) $ of $k$-submodules of $V$ such that $V=V_{0}\oplus V_{1}%
$.\newline Here, $0$ and $1$ are considered not as integers, but as elements
of $\mathbb{Z}\diagup2\mathbb{Z}$ (so that $1+1=0$). This sounds like a
useless requirement, but it helps us in handling super-$k$-modules
notationally; for example, the equation (\ref{defs.super.liealg.5}) would not
make sense if $0$ and $1$ would be considered as integers (because in the case
$i=1$ and $j=1$, we would have $i+j=2$, but there is no such thing as
$\mathfrak{g}_{2}$ unless $2$ is treated as an element of $\mathbb{Z}%
\diagup2\mathbb{Z}$).\newline The $k$-submodule $V_{0}$ of $V$ is called the
\textit{even part of }$V$. The $k$-submodule $V_{1}$ of $V$ is called the
\textit{odd part of }$V$.\newline An element of $V$ is said to be
\textit{homogeneous} if it lies in $V_{0}$ or in $V_{1}$.
\end{definition}
\begin{condition}
\label{conv.super.conv}We are going to use the notation $V_{0}$ as a universal
notation for the even part of a super-$k$-module $V$. This means that whenever
we have some super-$k$-module $V$ (it needs not be actually called $V$; I only
refer to it by $V$ here in this Convention), the even part of $V$ will be
called $V_{0}$.\newline Similarly, we are going to use the notation $V_{1}$ as
a universal notation for the odd part of a super-$k$-module $V$.
\end{condition}
\begin{remark}
Our Definition \ref{defs.super.liealg} differs from some definitions of a Lie
superalgebra given in literature by having the axioms
(\ref{defs.super.liealg.1}) and (\ref{defs.super.liealg.4}). These axioms are
indeed dispensable when one is only interested in the case of $k$ being a
field of characteristic $0$ (or, more generally, of $k$ being a commutative
ring in which $2$ and $3$ are invertible){\footnotemark}. However, for the
sake of generality, we keep these axioms in.
\end{remark}
\footnotetext{In fact,
\par
\begin{itemize}
\item axiom (\ref{defs.super.liealg.1}) follows from axiom
(\ref{defs.super.liealg.3}) if $2$ is invertible in $k$;
\par
\item axiom (\ref{defs.super.liealg.4}) follows from axiom
(\ref{defs.super.liealg.2}) if $3$ is invertible in $k$.
\end{itemize}
}
Just as the notion of Lie algebras gives birth to that of $\mathfrak{g}%
$-modules, we can define the notion of a $\mathfrak{g}$\textit{-supermodule}
(or just $\mathfrak{g}$\textit{-module}) over a Lie superalgebra
$\mathfrak{g}$:
\begin{definition}
\label{defs.super.module}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a Lie superalgebra. (According to Convention \ref{conv.super.conv}, this
automatically entails that we denote by $\mathfrak{g}_{0}$ the even part of
$\mathfrak{g}$, and denote by $\mathfrak{g}_{1}$ the odd part of
$\mathfrak{g}$.)\newline Let $V$ be a $k$-supermodule. (According to
Convention \ref{conv.super.conv}, this automatically entails that we denote by
$V_{0}$ the even part of $V$, and denote by $V_{1}$ the odd part of
$V$.)\newline Let $\mu:\mathfrak{g}\times V\rightarrow V$ be a $k$-bilinear
map. We say that $\left( V,\mu\right) $ is a $\mathfrak{g}$%
\textit{-supermodule} if and only if
\begin{equation}
\left(
\begin{array}
[c]{c}%
\mu\left( \left[ a,b\right] ,v\right) =\mu\left( a,\mu\left( b,v\right)
\right) -\left( -1\right) ^{ij}\mu\left( b,\mu\left( a,v\right) \right)
\\
\text{for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{, }j\in\mathbb{Z}%
\diagup2\mathbb{Z}\text{ and every }a\in\mathfrak{g}_{i}\text{, }%
b\in\mathfrak{g}_{j}\text{ and }v\in V
\end{array}
\right) \label{defs.super.module.1}%
\end{equation}
and%
\[
\left( \mu\left( \mathfrak{g}_{i}\times V_{j}\right) \subseteq
V_{i+j}\text{ for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{ and }%
j\in\mathbb{Z}\diagup2\mathbb{Z}\right) .
\]
If $\left( V,\mu\right) $ is a $\mathfrak{g}$-supermodule, then the
$k$-bilinear map $\mu:\mathfrak{g}\times V\rightarrow V$ is called the
\textit{Lie action} of the $\mathfrak{g}$-supermodule $V$.
\end{definition}
(This definition seems to be agreed on in most references. I have not seen any
conflicting definitions as in the case of Definition \ref{defs.super.liealg}.)
While I have not checked the details, I am convinced that all results of
Sections \ref{sect.L25}, \ref{sect.(g,h)} and \ref{sect.split} (and Subsection
\ref{subsect.general}) carry over to Lie superalgebras (and Lie supermodules)
as long as $2$ is invertible in the ground ring $k$. Even the invertibility of
$2$ might actually be redundant for most of these results (and it seems that
the reason for its redundancy is the fact that most of the results still hold
for Leibniz algebras - but, as I already said, this is not thoroughly
checked). As for Section \ref{sect.pbw}, trouble might come from the
Poincar\'{e}-Birkhoff-Witt theorem (Theorem \ref{thm.pbw}), whose validity in
the Lie superalgebra case has not been studied to the extent it has been
studied in the original, Lie algebraic case. However, there \textit{are} two
known results:
\begin{theorem}
\label{thm.pbw.super}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie superalgebra. Let $n\in\mathbb{N}$.\newline\textbf{(a)} If
$\mathfrak{g}_{0}$ and $\mathfrak{g}_{1}$ are free $k$-modules, and if $2$ is
invertible in the ring $k$, then $\mathfrak{g}$ satisfies the $n$-PBW
condition.\newline\textbf{(b)} If $k$ is a $\mathbb{Q}$-algebra, then
$\mathfrak{g}$ satisfies the $n$-PBW condition.
\end{theorem}
A proof of Theorem \ref{thm.pbw.super} \textbf{(b)} was given in \cite[Part 1,
Chapter 1, \S 1.3.7]{deligne} and \cite[\S 2.5]{petracci}; a proof of Theorem
\ref{thm.pbw.super} \textbf{(a)} can be found in \cite[\S 2.3, Theorem
1]{scheunert}. Note that whoever claims that $3$ must be invertible in the
ring $k$ in order for Theorem \ref{thm.pbw.super} \textbf{(a)} to hold is
probably using a definition of Lie superalgebra which does not contain the
axiom (\ref{defs.super.liealg.4}). \textit{However,} even having the axiom
(\ref{defs.super.liealg.1}) does not prevent us from having to require the
invertibility of $2$, unless we replace our definition of a Lie superalgebra
by a significantly more complicated one (\cite[Definition 8.1.1]%
{neisendorfer}), in which case we can indeed drop the invertibility of $2$
(\cite[Theorem 8.2.2]{neisendorfer}). Having said this, we are \textit{not}
going to use \cite[Definition 8.1.1]{neisendorfer} as the definition of a Lie
superalgebra in this paper; instead we will keep understanding a Lie
superalgebra according to Definition \ref{defs.super.liealg}. As a
consequence, we will not be able to get rid of the condition that $2$ be
invertible in $k$ in the Poincar\'{e}-Birkhoff-Witt theorem and its consequences.
The correct analogue of Theorem \ref{thm.pbw.basis} now says:
\begin{theorem}
\label{thm.pbw.basis.super}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie superalgebra. Assume that $2$ is invertible in the ring $k$. Also
assume that the $k$-module $\mathfrak{g}$ has a basis $\left( e_{i}\right)
_{i\in I}$, where $I$ is a totally ordered set, and where $e_{i}$ is
homogeneous for every $i\in I$. Then,
\[
\left( \overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes e_{i_{n}}%
}\right) _{\substack{n\in\mathbb{N};\ \left( i_{1},i_{2},...,i_{n}\right)
\in I^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n};\\\text{every }p\text{ which
satisfies }\left( e_{i_{p}}\in\mathfrak{g}_{1}\text{ and }e_{i_{p+1}}%
\in\mathfrak{g}_{1}\right) \text{ satisfies }i_{p}