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\ihead{Errata to ``... the boson-fermion correspondence''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{A combinatorial generalization of the boson-fermion correspondence}

\textit{Thomas Lam}

\texttt{lam - boson-fermion - 0507341v1.pdf}

arXiv preprint
\href{http://arxiv.org/abs/math/0507341v1}{arXiv:math/0507341v1}, version of
17 July 2005

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

I will refer to the results appearing in the article \textquotedblleft A
combinatorial generalization of the boson-fermion
correspondence\textquotedblright\ by the numbers under which they appear in
this article (specifically, in its version of 17 July 2005, posted on arXiv
under the identifier arXiv:math/0507341v1).

\setcounter{section}{10}

\section{Errata}

\begin{itemize}
\item \textbf{Page 1:} Replace \textquotedblleft composition\textquotedblright%
\ by \textquotedblleft weak composition\textquotedblright. (A \textit{weak
composition} is an infinite sequence $\left(  r_{1},r_{2},r_{3},\ldots\right)
$ of nonnegative integers such that only finitely many $i\in\left\{
1,2,3,\ldots\right\}  $ satisfy $r_{i}\neq0$. In contrast, a composition is a
finite sequence of positive integers. The weight of a semistandard tableau $T$
is a weak composition; it is not necessarily a composition, because it can
have a $0$ followed by a positive integer.)

\item \textbf{Page 2:} In \textquotedblleft the definition of a
tableaux\textquotedblright, replace \textquotedblleft
tableaux\textquotedblright\ by \textquotedblleft tableau\textquotedblright.

\item \textbf{Page 2:} In \textquotedblleft a semi-standard Young
tableaux\textquotedblright, replace \textquotedblleft
tableaux\textquotedblright\ by \textquotedblleft tableau\textquotedblright.

\item \textbf{Page 3:} After \textquotedblleft The ring $\Lambda_{K}$ should
be thought of as the ring of formal power series in countably many variables
$x_{1},x_{2},\ldots$, of bounded degree\textquotedblright, add
\textquotedblleft, which are invariant under permutations of $x_{1}%
,x_{2},\ldots$\textquotedblright. (Or else, you should replace
\textquotedblleft thought of as the ring\textquotedblright\ by
\textquotedblleft thought of as a subring\textquotedblright.)

\item \textbf{Page 4, \S 2:} Replace \textquotedblleft a Young tableaux
$T$\textquotedblright\ by \textquotedblleft a Young tableau $T$%
\textquotedblright.

\item \textbf{Page 4, \S 2:} Replace \textquotedblleft a Young tableaux of
skew shape\textquotedblright\ by \textquotedblleft a Young tableau of skew
shape\textquotedblright.

\item \textbf{Page 4, \S 3:} Replace \textquotedblleft$\left[  B_{k}%
,B_{l}\right]  =l\cdot a_{l}\cdot\delta_{k,-l}$\textquotedblright\ by
\textquotedblleft$\left[  B_{k},B_{l}\right]  =k\cdot a_{k}\cdot\delta_{k,-l}%
$\textquotedblright.

\item \textbf{Page 4, \S 3:} Replace \textquotedblleft$a_{l}=-a_{-l}%
$\textquotedblright\ (or else replace \textquotedblleft$\left[  B_{k}%
,B_{l}\right]  =k\cdot a_{k}\cdot\delta_{k,-l}$\textquotedblright\ by
\textquotedblleft$\left[  B_{k},B_{l}\right]  =a_{k}\cdot\delta_{k,-l}%
$\textquotedblright; but this option would require further changes in various
other places).

\item \textbf{Page 5, \S 3:} I think it would be useful to state the fact that
the $H$-module $\Lambda_{K}$ is faithful. This shows that any identity in $H$
can be proven by verifying the corresponding identity for operators on
$\Lambda_{K}$ (which, to a combinatorialist, is far more familiar terrain).

The proof of the faithfulness of the $H$-module $\Lambda_{K}$ is not
difficult: One can first prove that the family
\[
\left(  \overrightarrow{\prod\limits_{i\in\mathbb{Z}\setminus\left\{
0\right\}  }}\left(  B_{i}\right)  ^{a_{i}}\right)  _{\substack{\left(
a_{i}\right)  _{i\in\mathbb{Z}\setminus\left\{  0\right\}  }\text{ is a
family}\\\text{of nonnegative integers}\\\text{such that only finitely
many}\\i\in\mathbb{Z}\setminus\left\{  0\right\}  \text{ satisfy }a_{i}\neq
0}}
\]
(where the $\overrightarrow{\prod\limits_{i\in\mathbb{Z}\setminus\left\{
0\right\}  }}$ symbol signifies a product taken in increasing order, i.e., we
have $\overrightarrow{\prod\limits_{i\in\mathbb{Z}\setminus\left\{  0\right\}
}}\left(  B_{i}\right)  ^{a_{i}}=\cdots\left(  B_{-2}\right)  ^{a_{-2}}\left(
B_{-1}\right)  ^{a_{-1}}\left(  B_{1}\right)  ^{a_{1}}\left(  B_{2}\right)
^{a_{2}}\cdots$) generates the $K$-module $H$ (mainly because the relation
$\left[  B_{k},B_{l}\right]  =k\cdot a_{k}\cdot\delta_{k,-l}$ allows us to
rearrange the terms in any monomial of the form $B_{p_{1}}B_{p_{2}}\cdots
B_{p_{s}}$ into weakly increasing order, at the cost of creating smaller
monomials). Moreover, the actions of the elements of this family on
$\Lambda_{K}$ are linearly independent (as one can easily see as well). Thus,
the $H$-module $\Lambda_{K}$ is faithful.

\item \textbf{Page 5, Lemma 1:} Here the notation $B_{\lambda}$ is being used;
this notation is not defined until later (in \S 4).

\item \textbf{Page 5, Lemma 1:} Replace \textquotedblleft$B_{-k}B_{\lambda
}={k}a_{k}m_{k}(\lambda)B_{\mu}+B_{\lambda}B_{-k}$\textquotedblright\ by
\textquotedblleft$B_{-k}B_{\lambda}=-{k}a_{k}m_{k}(\lambda)B_{\mu}+B_{\lambda
}B_{-k}$\textquotedblright.

\item \textbf{Page 5:} Replace \textquotedblleft the parameters $a_{l}=1$ for
$l\geq1$ and $a_{l}=-1$ for $l\leq-1$\textquotedblright\ by \textquotedblleft
the parameters $a_{l}=1$\textquotedblright.

\item \textbf{Page 5:} Replace \textquotedblleft$\{v_{j}:j\in Z\}$%
\textquotedblright\ by \textquotedblleft$\{v_{j}:j\in\mathbb{Z}\}$%
\textquotedblright.

\item \textbf{Page 5, (5):} Replace \textquotedblleft$v_{i_{0}}\wedge
v_{i_{-1}}\wedge\cdots\wedge v_{i_{j-1}}\wedge v_{i_{j}-k}\wedge v_{i_{j+1}%
}\wedge\cdots$\textquotedblright\ by \textquotedblleft$v_{i_{0}}\wedge
v_{i_{-1}}\wedge\cdots\wedge v_{i_{j+1}}\wedge v_{i_{j}-k}\wedge v_{i_{j-1}%
}\wedge\cdots$\textquotedblright. (The two subscripts \textquotedblleft%
$i_{j-1}$\textquotedblright\ and \textquotedblleft$i_{j+1}$\textquotedblright%
\ have been switched.)

\item \textbf{Page 5, Theorem 3:} After \textquotedblleft$\sigma(v_{i_{0}%
}\wedge v_{i_{-1}}\wedge\cdots)=s_{\lambda}$\textquotedblright, add
\textquotedblleft, where $\lambda=\left(  \lambda_{0},\lambda_{1},\lambda
_{2},\ldots\right)  $\textquotedblright.

\item \textbf{Page 6, \S 3:} Replace \textquotedblleft have been studied
previously\textquotedblright\ by \textquotedblleft has been studied
previously\textquotedblright.

\item \textbf{Page 6, \S 4:} Remove the \textquotedblleft and $U_{k}%
:=\sum_{\lambda\vdash k}z_{\lambda}^{-1}B_{-\lambda}$\textquotedblright\ part
from \textquotedblleft Let $D_{k}:=\sum_{\lambda\vdash k}z_{\lambda}%
^{-1}B_{\lambda}$ and $U_{k}:=\sum_{\lambda\vdash k}z_{\lambda}^{-1}%
B_{-\lambda}$\textquotedblright. (It is entirely unnecessary, since the same
definition of $U_{k}$ appears again in the next paragraph.)

\item \textbf{Page 6, \S 4:} In \textquotedblleft$\langle v_{s},v_{s}^{\prime
}\rangle=\delta_{ss^{\prime}}$\textquotedblright, replace \textquotedblleft%
$v_{s}^{\prime}$\textquotedblright\ by \textquotedblleft$v_{s^{\prime}}%
$\textquotedblright.

\item \textbf{Page 6, \S 4:} Replace \textquotedblleft Define the generating
functions
\[
F_{s/t}^{V}(x_{1},x_{2},\ldots)=F_{s/t}(x_{1},x_{2},\ldots):=\sum_{\alpha
}x^{\alpha}\langle U_{\alpha_{l}}U_{\alpha_{l-1}}\cdots U_{\alpha_{1}}\cdot
t,s\rangle,
\]
where the sum is over all compositions $\alpha=(\alpha_{1},\alpha_{2}%
,\ldots,\alpha_{l})$. Similarly define
\[
G_{s/t}^{V}(x_{1},x_{2},\ldots)=G_{s/t}(x_{1},x_{2},\ldots)=\sum_{\alpha
}x^{\alpha}\langle D_{\alpha_{l}}D_{\alpha_{l-1}}\cdots D_{\alpha_{1}}\cdot
s,t\rangle.
\]
\textquotedblright\ by \textquotedblleft Define the generating functions
\[
F_{s/t}^{V}(x_{1},x_{2},\ldots)=F_{s/t}(x_{1},x_{2},\ldots):=\sum_{\alpha
}x^{\alpha}\langle\cdots U_{\alpha_{3}}U_{\alpha_{1}}U_{\alpha_{1}}\cdot
v_{t},v_{s}\rangle,
\]
where the sum is over all weak compositions $\alpha=(\alpha_{1},\alpha
_{2},\alpha_{3},\ldots)$. Similarly define
\[
G_{s/t}^{V}(x_{1},x_{2},\ldots)=G_{s/t}(x_{1},x_{2},\ldots)=\sum_{\alpha
}x^{\alpha}\langle\cdots D_{\alpha_{3}}D_{\alpha_{2}}D_{\alpha_{1}}\cdot
v_{s},v_{t}\rangle.
\]
\textquotedblright

I have made two changes here: First, the \textquotedblleft$t$%
\textquotedblright\ and \textquotedblleft$s$\textquotedblright\ on the right
hand sides have been replaced by \textquotedblleft$v_{t}$\textquotedblright%
\ and \textquotedblleft$v_{s}$\textquotedblright, respectively (since $t$ and
$s$ themselves are not vectors, but just elements of the indexing set $S$).
Second, instead of summing over all compositions $\alpha=(\alpha_{1}%
,\alpha_{2},\ldots,\alpha_{l})$, I am summing over all weak compositions,
since we need to allow $0$'s followed by positive integers here, and also
since the \textquotedblleft$l$\textquotedblright\ is confusing (if you sum
over finite sequences, do you count $\left(  2,0,3\right)  $ and $\left(
2,0,3,0\right)  $ as two distinct finite sequences?).

\item \textbf{Page 6, \S 4:} For a combinatorialist reader, it would be
helpful to explain how the various elements of $H$ you have defined act on
$\Lambda_{K}$. Namely, the map $H\rightarrow\operatorname*{End}\nolimits_{K}%
\left(  \Lambda_{K}\right)  $ given by the action of $H$ on $\Lambda_{K}$
sends%
\begin{align*}
B_{-k}  &  \mapsto a_{k}p_{k}\ \ \ \ \ \ \ \ \ \ \text{for every }k\geq1;\\
B_{k}  &  \mapsto k\dfrac{\partial}{\partial p_{k}}=p_{k}^{\perp
}\ \ \ \ \ \ \ \ \ \ \text{for every }k\geq1;\\
D_{k}  &  \mapsto h_{k}^{\perp}\ \ \ \ \ \ \ \ \ \ \text{for every }k\geq0;\\
D_{\lambda}  &  \mapsto h_{\lambda}^{\perp}\ \ \ \ \ \ \ \ \ \ \text{for every
partition }\lambda.
\end{align*}
It also sends%
\begin{align*}
U_{k}  &  \mapsto h_{k}\ \ \ \ \ \ \ \ \ \ \text{for every }k\geq0;\\
U_{\lambda}  &  \mapsto h_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every
partition }\lambda;\\
S_{\lambda}  &  \mapsto s_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every
partition }\lambda
\end{align*}
when all $a_{l}$ are $1$.

\item \textbf{Page 7, \S 4:} I do not understand what you mean by
\textquotedblleft The element $\Omega(H_{-},X)\cdot v_{b}\in V\hat{\otimes
}\Lambda_{K}(X)$ depends only on the choice of $v_{b}$.\textquotedblright.

\item \textbf{Page 7, \S 5:} In \textquotedblleft By Proposition~2, there is a
canonical map of $H$-modules $\phi:H\cdot b\rightarrow\Lambda_{K}$ sending
$v_{b}\mapsto1$.\textquotedblright, replace \textquotedblleft$H\cdot
b$\textquotedblright\ by \textquotedblleft$H\cdot v_{b}$\textquotedblright.
More importantly, I do not understand how this follows from Proposition 2. How
do we know that $H\cdot v_{b}$ is irreducible?

Here is the simplest proof I can find for the existence of a canonical
$H$-module homomorphism $\phi:H\cdot v_{b}\rightarrow\Lambda_{K}$:

We use the following fact:

\textbf{Proposition 2a.} Let $V$ be a representation of $H$. Let $v\in V$ be a
highest weight vector. Then, there exists a unique $H$-module homomorphism
$\phi:K\left[  B_{-1},B_{-2},\ldots\right]  \rightarrow V$ such that
$\phi\left(  1\right)  =v$.

Proposition 2a is relatively well-known (it appeared in Pavel Etingof's class
on infinite-dimensional Lie algebras), and I am sure that your proof of
Proposition 2 uses Proposition 2a as an intermediate step.

Now, applying Proposition 2a to $H\cdot v_{b}$ and $v_{b}$ instead of $V$ and
$v$, we conclude that there exists a unique $H$-module homomorphism
$\phi:K\left[  B_{-1},B_{-2},\ldots\right]  \rightarrow H\cdot v_{b}$ such
that $\phi\left(  1\right)  =v_{b}$. This homomorphism $\phi$ is injective
(because it is nonzero, and because $K\left[  B_{-1},B_{-2},\ldots\right]  $
is an irreducible $H$-module) and surjective (because it satisfies
$H\cdot\underbrace{v_{b}}_{=\phi\left(  1\right)  }=H\cdot\phi\left(
1\right)  =\phi\left(  H\cdot1\right)  \subseteq\phi\left(  K\left[
B_{-1},B_{-2},\ldots\right]  \right)  $), and thus an isomorphism. Hence, the
inverse $\Phi$ of $\phi$ is an $H$-module isomorphism $\phi:H\cdot
v_{b}\rightarrow\Lambda_{K}$.

\item \textbf{Page 7, Theorem 6:} According to what you wrote in the mail, the
assumptions of Theorem 6 need to be changed. Namely, instead of assuming that
$v_{b}\in V$ is a highest weight vector of $H$, you want to assume that
\begin{equation}
\left\langle B_{l}x,v_{b}\right\rangle =0\ \ \ \ \ \ \ \ \ \ \text{for every
}x\in V\text{ and }l<0. \label{p7.miss-ass}%
\end{equation}
This assumption is used in deriving $\langle B_{\lambda}B_{l}\cdot v_{s}%
,v_{b}\rangle=ka_{k}m_{k}(\lambda)\langle B_{\mu}v_{s},v_{b}\rangle$ from
Lemma 1 in the proof of Theorem 6. (Namely, we have
\begin{align*}
&  \left\langle B_{\lambda}\underbrace{B_{l}}_{\substack{=B_{-k}\\\text{(since
}l=-k\text{)}}}\cdot v_{s},v_{b}\right\rangle \\
&  =\left\langle \underbrace{B_{\lambda}B_{-k}}_{=B_{-k}B_{\lambda}%
+ka_{k}m_{k}\left(  \lambda\right)  B_{\mu}}\cdot v_{s},v_{b}\right\rangle
=\left\langle \left(  B_{-k}B_{\lambda}+ka_{k}m_{k}\left(  \lambda\right)
B_{\mu}\right)  \cdot v_{s},v_{b}\right\rangle \\
&  =\underbrace{\left\langle B_{-k}B_{\lambda}\cdot v_{s},v_{b}\right\rangle
}_{\substack{=0\\\text{(by (\ref{p7.miss-ass}), applied to }-k\text{ and
}B_{\lambda}\cdot v_{s}\\\text{instead of }l\text{ and }x\text{)}}%
}+ka_{k}m_{k}\left(  \lambda\right)  \left\langle B_{\mu}\cdot v_{s}%
,v_{b}\right\rangle =ka_{k}m_{k}\left(  \lambda\right)  \left\langle B_{\mu
}\cdot v_{s},v_{b}\right\rangle .
\end{align*}
)

\item \textbf{Page 8, proof of Theorem 6:} In \textquotedblleft$\sum
_{c}\langle B_{l}\cdot v_{s},v_{c}\rangle\left(  \sum_{\lambda}z_{\lambda
}^{-1}\langle B_{\lambda}\cdot v_{c},v_{b}\rangle\right)  $\textquotedblright,
add a \textquotedblleft$p_{\lambda}$\textquotedblright\ before the
\textquotedblleft$\langle B_{\lambda}\cdot v_{c},v_{b}\rangle$%
\textquotedblright.

\item \textbf{Page 8, \S 5:} I find it worthwhile to mention that the map
$\Phi:V\rightarrow\Lambda_{K}$ constructed in Theorem 6 is a $K\left[
B_{1},B_{2},\ldots\right]  $-modules even if we don't require
(\ref{p7.miss-ass}) to hold. (Indeed, this follows from the second part of the
proof of Theorem 6, that begins with \textquotedblleft Now suppose
$k>0$\textquotedblright; this part does not use (\ref{p7.miss-ass}).)

\item \textbf{Page 8, \S 5:} Shouldn't \textquotedblleft a different action of
$H$ on $\Lambda_{K}$\textquotedblright\ be \textquotedblleft a different
action of $H$ on $V$\textquotedblright\ ? Anyway, I fear I don't fully
understand this paragraph, and it appears to me that it might use some
additional assumptions.

\item \textbf{Page 8, \S 6:} In \textquotedblleft under the map $\kappa
:\Lambda_{K}\rightarrow K$ given by $\kappa(p_{k})=a_{k}$\textquotedblright,
replace \textquotedblleft map\textquotedblright\ by \textquotedblleft%
$K$-algebra homomorphism\textquotedblright.

\item \textbf{Page 9, Theorem 7:} In \textquotedblleft$\langle U_{k}\cdot
s,t\rangle$\textquotedblright, replace \textquotedblleft$s$\textquotedblright%
\ and \textquotedblleft$t$\textquotedblright\ by \textquotedblleft$v_{s}%
$\textquotedblright\ and \textquotedblleft$v_{t}$\textquotedblright,
respectively. Do the analogous replacements in the other three equalities.

\item \textbf{Page 9, Theorem 7:} I am not sure what assumptions this theorem
needs. Obviously, (\ref{p7.miss-ass}) is required to apply Theorem 6, but we
might need more for \textquotedblleft the comments immediately after
it\textquotedblright.

\item \textbf{Page 9, proof of Lemma 8:} The notation \textquotedblleft%
$\lambda\cup\mu$\textquotedblright\ should be defined. (Here is a simple
definition: If $\alpha$ and $\beta$ are two partitions, then $\alpha\cup\beta$
will denote the partition obtained by sorting the sequence $\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{l\left(  \alpha\right)  },\beta_{1},\beta
_{2},\ldots,\beta_{l\left(  \beta\right)  }\right)  $ in weakly decreasing order.)

\item \textbf{Page 9, proof of Lemma 8:} Replace both appearances of
\textquotedblleft$\theta$\textquotedblright\ by \textquotedblleft$\kappa
$\textquotedblright.

\item \textbf{Page 9, Theorem 9:} In order to obtain the first equality of
Theorem 9 from the second equality, you again use the assumption
(\ref{p7.miss-ass}). (Indeed, this assumption is what guarantees that
$F_{b/s}=\delta_{b,s}$ for all $s\in S$.)

\item \textbf{Page 10, proof of Theorem 9:} The letter \textquotedblleft%
$i$\textquotedblright\ is used in two different meanings here: Once as a bound
variable in \textquotedblleft$h_{k}\left\langle a_{i}\right\rangle
$\textquotedblright\ (shorthand for \textquotedblleft$h_{k}\left\langle
a_{1},a_{2},\ldots\right\rangle $\textquotedblright), and another time as a
bound variable in the products \textquotedblleft$\prod\limits_{i,j\geq
1}^{\infty}$\textquotedblright.

\item \textbf{Page 11:} Replace \textquotedblleft Suppose further that $B_{k}$
and $B_{l}$ commute\textquotedblright\ by \textquotedblleft Suppose further
that $B_{k}^{\prime}$ and $B_{l}^{\prime}$ commute\textquotedblright.

\item \textbf{Page 11:} Add a whitespace in \textquotedblleft Let$D_{k}%
^{\prime}:=\sum_{\lambda\vdash k}z_{\lambda}^{-1}B_{\lambda}^{\prime}%
$\textquotedblright.

\item \textbf{Page 11:} Replace \textquotedblleft$F_{s/t}^{\prime}(x_{1}%
,x_{2},\ldots):=\sum_{\alpha}x^{\alpha}\langle U_{\alpha_{l}}^{\prime
}U_{\alpha_{l-1}}^{\prime}\cdots U_{\alpha_{1}}^{\prime}\cdot t,s\rangle
$\textquotedblright\ by \textquotedblleft$F_{s/t}^{\prime}(x_{1},x_{2}%
,\ldots):=\sum_{\alpha}x^{\alpha}\langle\cdots U_{\alpha_{3}}^{\prime
}U_{\alpha_{2}}^{\prime}U_{\alpha_{1}}^{\prime}\cdot v_{t},v_{s}\rangle
$\textquotedblright.

\item \textbf{Page 12, \S 8.1:} Replace \textquotedblleft$h_{i}\langle
a_{i}\rangle$\textquotedblright\ by \textquotedblleft$h_{k}\langle
a_{i}\rangle$\textquotedblright.

\item \textbf{Page 12, \S 8.2:} Add a semicolon before \textquotedblleft
otherwise\textquotedblright.

\item \textbf{Page 12, \S 8.3:} Replace \textquotedblleft$\tilde{B}_{k}\cdot
v_{1}\otimes v_{2}=(B_{k}\cdot v_{1})\otimes v_{2}+v_{1}\cdot(B_{k}\cdot
v_{2})$\textquotedblright\ by \textquotedblleft$\tilde{B}_{k}\cdot\left(
v_{1}\otimes v_{2}\right)  =(B_{k}\cdot v_{1})\otimes v_{2}+v_{1}\otimes
(B_{k}\cdot v_{2})$\textquotedblright\ (notice that I've made two changes here).
\end{itemize}


\end{document}