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\begin{center}
\textbf{Schur functions and alternating sums}

\textit{Marc A. A. van Leeuwen}

version of 10 November 2006

\textbf{Errata and comments by Darij Grinberg}

\bigskip
\end{center}

The following text is an annotation to Marc A. A. van Leeuwen's paper
\textquotedblleft Schur functions and alternating sums\textquotedblright\ in
its version of 10 November 2006.

This annotation contains corrections of mistakes (or what I believe to be
mistakes) and additional comments (in particular, elaborations of some
arguments that I found insufficiently detailed in the paper). The latter are
printed in \textcolor{blue}{blue}.

Different comments are separated by horizontal lines, like this:

\rule{17cm}{0.3mm}

\textbf{Page 1, Abstract:} \textquotedblleft Our our goal\textquotedblright%
\ should be \textquotedblleft our goal\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 1, \S 0:} Lemma 2.2 actually has appeared in the literature
(although in a far less readable form than in your paper). It appears (in an
equivalent form) in \cite[second equation on p. 29]{DeLeTh}, where it is
attributed to Muir (the reference unfortunately being the whole book
\textquotedblleft A treatise on the theory of determinants\textquotedblright).

\rule{17cm}{0.3mm}

\textbf{Page 1, \S 0:} \textquotedblleft an equally symmetrical doubly
alternating expressions\textquotedblright\ should be either \textquotedblleft
an equally symmetrical doubly alternating expression\textquotedblright\ or
\textquotedblleft equally symmetrical doubly alternating
expressions\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 4, \S 1.2:} \textquotedblleft to transpose semistandard
tableau\textquotedblright\ should be \textquotedblleft to transpose
semistandard tableaux\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 5, Definition 1.2.3:} Replace \textquotedblleft%
$\operatorname*{Tabl}\left(  \lambda/\nu\right)  $\textquotedblright\ by
\textquotedblleft$\operatorname*{Tabl}\left(  \lambda/\mu\right)
$\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 5, \S 1.3:} I would replace \textquotedblleft and for which $\deg
f_{n}$ is bounded\textquotedblright\ by \textquotedblleft and for which the
sequence $\left(  \deg f_{n}\right)  _{n\in\mathbf{N}}$ is
bounded\textquotedblright, to make the wording unambiguous.

\rule{17cm}{0.3mm}

\textbf{Page 6, \S 1.4:} Replace \textquotedblleft be the
sign\textquotedblright\ by \textquotedblleft by the sign\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 6, \S 1.4:} It would be useful to define the notations
$\mathbf{S}_{n}$ and $\varepsilon\left(  \sigma\right)  $ here (both of which
are not really standard):

\begin{itemize}
\item Given $n\in\mathbf{N}$, we let $\mathbf{S}_{n}$ denote the group of all
permutations of the set $\left[  n\right]  =\left\{  0,1,\ldots,n-1\right\}  $.

\item Given $n\in\mathbf{N}$ and $\sigma\in\mathbf{S}_{n}$, we let
$\varepsilon\left(  \sigma\right)  $ denote the sign of $\sigma$.
\end{itemize}

\rule{17cm}{0.3mm}

\textbf{Page 6, \S 1.4:} In the formula \textquotedblleft$a_{\alpha}\left[
X_{\left[  n\right]  }\right]  =\sum_{\sigma\in\mathbf{S}_{n}}\varepsilon
\left(  \sigma\right)  X^{\sigma\cdot\alpha}$\textquotedblright, you are using
the notation \textquotedblleft$\sigma\cdot\alpha$\textquotedblright\ for
$\sigma\left(  \alpha\right)  $. This notation should be defined.

\rule{17cm}{0.3mm}

\textbf{Page 6, \S 1.4:} In (13), replace \textquotedblleft$X_{j}^{\alpha_{j}%
}$\textquotedblright\ by \textquotedblleft$X_{i}^{\alpha_{j}}$%
\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 7:} In the formula for $s_{\left(  3,1\right)  }\left[
X_{\left[  3\right]  }\right]  $, replace \textquotedblleft$\left(
X_{0}-X_{1}\right)  \left(  X_{0}-X_{1}\right)  \left(  X_{1}-X_{2}\right)
$\textquotedblright\ by \textquotedblleft$\left(  X_{0}-X_{1}\right)  \left(
X_{0}-X_{2}\right)  \left(  X_{1}-X_{2}\right)  $\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 7, proof of Proposition 1.4.1:} The sentence \textquotedblleft We
may assume that $\alpha\in\mathbf{N}^{n+1}$ holds, since otherwise both
$s_{\alpha}\left[  X_{\left[  n+1\right]  }\right]  $ and $s_{\alpha}\left[
X_{\left[  n\right]  }\right]  $ are zero by definition.\textquotedblright%
\ should be moved to the very beginning of this proof, because already in the
very first sentence this assumption is being used.

\rule{17cm}{0.3mm}

\begin{details}
\textbf{Page 8, \S 1.4:} The claim that \textquotedblleft The set $\left\{
s_{\lambda}\mid\lambda\in\mathcal{P}\right\}  $ forms a $\mathbf{Z}$-basis of
$\Lambda$\textquotedblright\ is not completely obvious (in my opinion) and
could use a proof. Here is the simplest proof that I am aware of:

\textit{Proof of the fact that the set }$\left\{  s_{\lambda}\mid\lambda
\in\mathcal{P}\right\}  $\textit{ forms a }$\mathbf{Z}$\textit{-basis of
}$\Lambda$\textit{:}

\begin{itemize}
\item For every $n\in\mathbf{N}$, the family $\left(  s_{\lambda}\left[
X_{\left[  n\right]  }\right]  \right)  _{\lambda\in\mathcal{P};\ \lambda
_{n}=0}$ is a $\mathbf{Z}$-basis of $\Lambda_{\left[  n\right]  }$. In
particular, this family is linearly independent. Hence, the family $\left(
s_{\lambda}\right)  _{\lambda\in\mathcal{P}}$ must be linearly independent
(because if this family would satisfy a nontrivial linear dependency relation,
then, for some sufficiently high $n\in\mathbf{N}$, we could apply the
projection $\Lambda\rightarrow\Lambda_{\left[  n\right]  }$ to this relation,
and obtain a nontrivial linear dependency relation for the family $\left(
s_{\lambda}\left[  X_{\left[  n\right]  }\right]  \right)  _{\lambda
\in\mathcal{P};\ \lambda_{n}=0}$; but this would contradict the fact that the
latter family is linearly independent).

\item Now, let us show that the family $\left(  s_{\lambda}\right)
_{\lambda\in\mathcal{P}}$ spans $\Lambda$. Indeed, here are two ways to prove this:

\textit{First proof of the fact that the family }$\left(  s_{\lambda}\right)
_{\lambda\in\mathcal{P}}$\textit{ spans }$\Lambda$\textit{:} We shall use the
notation $\varepsilon\left(  \alpha,\lambda\right)  $ defined in (19).
However, we cannot yet use the equality (19) as its definition, because we
have not yet proven that the set $\left\{  s_{\lambda}\mid\lambda
\in\mathcal{P}\right\}  $ forms a $\mathbf{Z}$-basis of $\Lambda$ (and thus we
have not yet proven that the scalar product $\left\langle \cdot,\cdot
\right\rangle $ is well-defined). We shall instead use a different (but
equivalent) definition of this notation: If $\alpha\in\mathcal{C}$ and
$\lambda\in\mathcal{P}$, then $\varepsilon\left(  \alpha,\lambda\right)  $
shall be defined as follows:

\begin{itemize}
\item If the sequence $\alpha\left[  {}\right]  $ (defined as in \S 1.5) has
two equal entries, then we set $\varepsilon\left(  \alpha,\lambda\right)  =0$.

\item If the sequence $\alpha\left[  {}\right]  $ has no two equal entries,
then there exists a unique permutation $\sigma\in\mathbf{S}_{\infty}$ such
that $\sigma\left(  \alpha\left[  {}\right]  \right)  $ is strictly
decreasing. We consider this $\sigma$, and we set $\varepsilon\left(
\alpha,\lambda\right)  =%
\begin{cases}
\varepsilon\left(  \sigma\right)  , & \text{if }\sigma\left(  \alpha\left[
{}\right]  \right)  =\lambda\left[  {}\right]  ;\\
0, & \text{if }\sigma\left(  \alpha\left[  {}\right]  \right)  \neq
\lambda\left[  {}\right]
\end{cases}
$.
\end{itemize}

It is rather clear that, with this definition of $\varepsilon\left(
\alpha,\lambda\right)  $, the equality (20) holds for every $\alpha
\in\mathcal{C}$. Proposition 2.1 and Lemma 2.2 hold. (Their proofs did not
make use of the fact that the set $\left\{  s_{\lambda}\mid\lambda
\in\mathcal{P}\right\}  $ forms a $\mathbf{Z}$-basis of $\Lambda$; they only
used the notation $\varepsilon\left(  \alpha,\lambda\right)  $.)

Let $f\in\Lambda$. Then, we can write $f\left[  X_{\mathbf{N}}\right]  $ in
the form $f\left[  X_{\mathbf{N}}\right]  =\sum_{\alpha\in\mathcal{C}%
}c_{\alpha}X^{\alpha}$ for some family $\left(  c_{\alpha}\right)  _{\alpha
\in\mathcal{C}}\in\mathbf{Z}^{\mathcal{C}}$ of integers. Consider this family
$\left(  c_{\alpha}\right)  _{\alpha\in\mathcal{C}}$. Applying Lemma 2.2 to
$\beta=\left(  0\right)  $, we obtain $fs_{\left(  0\right)  }=\sum_{\alpha
\in\mathcal{C}}c_{\alpha}s_{\alpha+\left(  0\right)  }$ (and the sum on the
right hand side of this equality is effectively finite -- i.e., all but
finitely many of its addends are zero). Compared with $f\underbrace{s_{\left(
0\right)  }}_{=1}=f$, this yields%
\[
f=\sum_{\alpha\in\mathcal{C}}c_{\alpha}\underbrace{s_{\alpha+\left(  0\right)
}}_{=s_{\alpha}}=\sum_{\alpha\in\mathcal{C}}c_{\alpha}\underbrace{s_{\alpha}%
}_{=\sum_{\lambda\in\mathcal{P}}\varepsilon\left(  \alpha,\lambda\right)
s_{\lambda}}=\sum_{\alpha\in\mathcal{C}}c_{\alpha}\left(  \sum_{\lambda
\in\mathcal{P}}\varepsilon\left(  \alpha,\lambda\right)  s_{\lambda}\right)
=\sum_{\alpha\in\mathcal{C}}\sum_{\lambda\in\mathcal{P}}c_{\alpha}%
\varepsilon\left(  \alpha,\lambda\right)  s_{\lambda}.
\]
Thus, $f$ belongs to the $\mathbf{Z}$-linear span of the family $\left(
s_{\lambda}\right)  _{\lambda\in\mathcal{P}}$.

Now, let us forget that we fixed $f$. We thus have shown that every
$f\in\Lambda$ belongs to the $\mathbf{Z}$-linear span of the family $\left(
s_{\lambda}\right)  _{\lambda\in\mathcal{P}}$. In other words, the family
$\left(  s_{\lambda}\right)  _{\lambda\in\mathcal{P}}$ spans $\Lambda$.

\textit{Second proof of the fact that the family }$\left(  s_{\lambda}\right)
_{\lambda\in\mathcal{P}}$\textit{ spans }$\Lambda$\textit{:} For every
$m\in\mathbf{N}$ and every graded ring $A$, we let $A_{\left(  m\right)  }$
denote the $m$-th graded component of the ring $A$. Then, if $n\in\mathbf{N}$
and $m\in\mathbf{N}$ are arbitrary, then%
\begin{equation}
\text{the family }\left(  s_{\lambda}\left[  X_{\left[  n\right]  }\right]
\right)  _{\lambda\in\mathcal{P};\ \lambda_{n}=0;\ \left\vert \lambda
\right\vert =m}\text{ is a }\mathbf{Z}\text{-basis of }\left(  \Lambda
_{\left[  n\right]  }\right)  _{m} \label{p8.10}%
\end{equation}
\footnote{\textit{Proof of (\ref{p8.10}):} Let $n\in\mathbf{N}$ and
$m\in\mathbf{N}$. We know that the family $\left(  s_{\lambda}\left[
X_{\left[  n\right]  }\right]  \right)  _{\lambda\in\mathcal{P};\ \lambda
_{n}=0}$ is a $\mathbf{Z}$-basis of $\Lambda_{\left[  n\right]  }$. Therefore,
for reasons of gradedness, we see that the family $\left(  s_{\lambda}\left[
X_{\left[  n\right]  }\right]  \right)  _{\lambda\in\mathcal{P};\ \lambda
_{n}=0;\ \left\vert \lambda\right\vert =m}$ is a $\mathbf{Z}$-basis of
$\left(  \Lambda_{\left[  n\right]  }\right)  _{m}$. Qed.}.

For every $n\in\mathbf{N}$, let $\pi_{n}$ be the projection $\Lambda_{\left[
n+1\right]  }\rightarrow\Lambda_{\left[  n\right]  }$ defined by the
substitution $X_{n}:=0$. Then, $\pi_{n}\left(  f\right)  =f\left[
X_{n}:=0\right]  $ for every $n\in\mathbf{N}$ and $f\in\Lambda_{\left[
n+1\right]  }$. Hence, every $n\in\mathbf{N}$ and $\lambda\in\mathcal{P}$
satisfy%
\begin{equation}
\pi_{n}\left(  s_{\lambda}\left[  X_{\left[  n+1\right]  }\right]  \right)
=s_{\lambda}\left[  X_{\left[  n+1\right]  }\right]  \left[  X_{n}:=0\right]
=s_{\lambda}\left[  X_{\left[  n\right]  }\right]  \label{p8.11}%
\end{equation}
(by Proposition 1.4.1, applied to $\alpha=\lambda$). For every $n\in
\mathbf{N}$ and $m\in\mathbf{N}$ satisfying $n\geq m$, it is easy to see that%
\begin{equation}
\text{the map }\pi_{n}\mid_{\left(  \Lambda_{\left[  n+1\right]  }\right)
_{m}}:\left(  \Lambda_{\left[  n+1\right]  }\right)  _{m}\rightarrow
\Lambda_{\left[  n\right]  }\text{ is injective} \label{p8.12}%
\end{equation}
\footnote{\textit{Proof of (\ref{p8.12}):} Let $n\in\mathbf{N}$ and
$m\in\mathbf{N}$ be such that $n\geq m$. The family $\left(  s_{\lambda
}\left[  X_{\left[  n+1\right]  }\right]  \right)  _{\lambda\in\mathcal{P}%
;\ \lambda_{n+1}=0;\ \left\vert \lambda\right\vert =m}$ is a $\mathbf{Z}%
$-basis of $\left(  \Lambda_{\left[  n+1\right]  }\right)  _{m}$ (because of
(\ref{p8.10}), applied to $n+1$ instead of $n$).
\par
Recall that $n\geq m$. Thus, it is easy to see that any $\lambda\in
\mathcal{P}$ satisfying $\left\vert \lambda\right\vert =m$ must also satisfy
$\lambda_{n}=0$. Hence, the family $\left(  s_{\lambda}\left[  X_{\left[
n\right]  }\right]  \right)  _{\lambda\in\mathcal{P};\ \lambda_{n+1}%
=0;\ \left\vert \lambda\right\vert =m}$ is a subfamily of the $\mathbf{Z}%
$-basis $\left(  s_{\lambda}\left[  X_{\left[  n\right]  }\right]  \right)
_{\lambda\in\mathcal{P};\ \lambda_{n}=0}$ of $\Lambda_{\left[  n\right]  }$.
Therefore, this family $\left(  s_{\lambda}\left[  X_{\left[  n\right]
}\right]  \right)  _{\lambda\in\mathcal{P};\ \lambda_{n+1}=0;\ \left\vert
\lambda\right\vert =m}$ is linearly independent.
\par
Now, for every $\lambda\in\mathcal{P}$, we have $\left(  \pi_{n}\mid_{\left(
\Lambda_{\left[  n+1\right]  }\right)  _{m}}\right)  \left(  s_{\lambda
}\left[  X_{\left[  n+1\right]  }\right]  \right)  =\pi_{n}\left(  s_{\lambda
}\left[  X_{\left[  n+1\right]  }\right]  \right)  =s_{\lambda}\left[
X_{\left[  n\right]  }\right]  $ (by (\ref{p8.11})). Therefore, the
$\mathbf{Z}$-linear map $\pi_{n}\mid_{\left(  \Lambda_{\left[  n+1\right]
}\right)  _{m}}$ sends the $\mathbf{Z}$-basis $\left(  s_{\lambda}\left[
X_{\left[  n+1\right]  }\right]  \right)  _{\lambda\in\mathcal{P}%
;\ \lambda_{n+1}=0;\ \left\vert \lambda\right\vert =m}$ of $\left(
\Lambda_{\left[  n+1\right]  }\right)  _{m}$ to the family $\left(
s_{\lambda}\left[  X_{\left[  n\right]  }\right]  \right)  _{\lambda
\in\mathcal{P};\ \lambda_{n+1}=0;\ \left\vert \lambda\right\vert =m}$. Since
the latter family is linearly independent, this shows that the map $\pi
_{n}\mid_{\left(  \Lambda_{\left[  n+1\right]  }\right)  _{m}}$ is injective
(since it sends a $\mathbf{Z}$-basis of its domain to a linearly independent
family). This proves (\ref{p8.12}).}.

Now, let $f\in\Lambda$ be homogeneous. Set $m=\deg f$. We know that the family
$\left(  s_{\lambda}\left[  X_{\left[  m\right]  }\right]  \right)
_{\lambda\in\mathcal{P};\ \lambda_{m}=0;\ \left\vert \lambda\right\vert =m}$
is a $\mathbf{Z}$-basis of $\left(  \Lambda_{\left[  m\right]  }\right)  _{m}$
(by (\ref{p8.10}), applied to $n=m$). Thus, there exists a family $\left(
w_{\lambda}\right)  _{\lambda\in\mathcal{P};\ \lambda_{m}=0}$ of integers such
that%
\begin{equation}
f\left[  X_{\left[  m\right]  }\right]  =\sum_{\lambda\in\mathcal{P}%
;\ \lambda_{m}=0}w_{\lambda}s_{\lambda}\left[  X_{\left[  m\right]  }\right]
\label{p8.13}%
\end{equation}
(since $f\left[  X_{\left[  m\right]  }\right]  \in\left(  \Lambda_{\left[
m\right]  }\right)  _{m}$). Consider this family. We claim that%
\begin{equation}
f\left[  X_{\left[  n\right]  }\right]  =\sum_{\lambda\in\mathcal{P}%
;\ \lambda_{m}=0}w_{\lambda}s_{\lambda}\left[  X_{\left[  n\right]  }\right]
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbf{N}. \label{p8.14}%
\end{equation}
Indeed, we can prove (\ref{p8.14}) as follows: For $n<m$, the equality
(\ref{p8.14}) follows by applying the canonical projection $\Lambda_{\left[
m\right]  }\rightarrow\Lambda_{\left[  n\right]  }$ (defined by the
substitutions $X_{i}:=0$ for all $i\in\left\{  n,n+1,\ldots,m-1\right\}  $) to
(\ref{p8.13}). For $n\geq m$, we can prove (\ref{p8.14}) by induction over
$n$: The induction base ($n=m$) follows from (\ref{p8.13}). For the induction
step, we fix an $n\in\mathbf{N}$ satisfying $n\geq m$, and we assume (as the
induction hypothesis) that%
\begin{equation}
f\left[  X_{\left[  n\right]  }\right]  =\sum_{\lambda\in\mathcal{P}%
;\ \lambda_{m}=0}w_{\lambda}s_{\lambda}\left[  X_{\left[  n\right]  }\right]
; \label{p8.15}%
\end{equation}
our goal is then to prove that%
\begin{equation}
f\left[  X_{\left[  n+1\right]  }\right]  =\sum_{\lambda\in\mathcal{P}%
;\ \lambda_{m}=0}w_{\lambda}s_{\lambda}\left[  X_{\left[  n+1\right]
}\right]  . \label{p8.16}%
\end{equation}
But both $f\left[  X_{\left[  n+1\right]  }\right]  $ and $\sum_{\lambda
\in\mathcal{P};\ \lambda_{m}=0}w_{\lambda}s_{\lambda}\left[  X_{\left[
n+1\right]  }\right]  $ are elements of $\left(  \Lambda_{\left[  n+1\right]
}\right)  _{m}$. We have%
\begin{align*}
\pi_{n}\left(  f\left[  X_{\left[  n+1\right]  }\right]  \right)   &
=f\left[  X_{\left[  n\right]  }\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\pi_{n}\right) \\
&  =\sum_{\lambda\in\mathcal{P};\ \lambda_{m}=0}w_{\lambda}%
\underbrace{s_{\lambda}\left[  X_{\left[  n\right]  }\right]  }%
_{\substack{=\pi_{n}\left(  s_{\lambda}\left[  X_{\left[  n+1\right]
}\right]  \right)  \\\text{(by (\ref{p8.11}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{p8.15})}\right) \\
&  =\sum_{\lambda\in\mathcal{P};\ \lambda_{m}=0}w_{\lambda}\pi_{n}\left(
s_{\lambda}\left[  X_{\left[  n+1\right]  }\right]  \right)  =\pi_{n}\left(
\sum_{\lambda\in\mathcal{P};\ \lambda_{m}=0}w_{\lambda}s_{\lambda}\left[
X_{\left[  n+1\right]  }\right]  \right)  .
\end{align*}
Since the map $\pi_{n}\mid_{\left(  \Lambda_{\left[  n+1\right]  }\right)
_{m}}$ is injective (by (\ref{p8.12})), this shows that $f\left[  X_{\left[
n+1\right]  }\right]  =\sum_{\lambda\in\mathcal{P};\ \lambda_{m}=0}w_{\lambda
}s_{\lambda}\left[  X_{\left[  n+1\right]  }\right]  $ (because both $f\left[
X_{\left[  n+1\right]  }\right]  $ and $\sum_{\lambda\in\mathcal{P}%
;\ \lambda_{m}=0}w_{\lambda}s_{\lambda}\left[  X_{\left[  n+1\right]
}\right]  $ live in $\left(  \Lambda_{\left[  n+1\right]  }\right)  _{m}$). In
other words, (\ref{p8.16}) holds. This completes the induction step, and thus
(\ref{p8.14}) is proven. Now, from (\ref{p8.14}), it follows that
$f=\sum_{\lambda\in\mathcal{P};\ \lambda_{m}=0}w_{\lambda}s_{\lambda}$.
Therefore, $f$ belongs to the $\mathbf{Z}$-linear span of the family $\left(
s_{\lambda}\right)  _{\lambda\in\mathcal{P}}$.

Now, let us forget that we fixed $f$. We thus have shown that every
homogeneous $f\in\Lambda$ belongs to the $\mathbf{Z}$-linear span of the
family $\left(  s_{\lambda}\right)  _{\lambda\in\mathcal{P}}$. Therefore,
every $f\in\Lambda$ belongs to the $\mathbf{Z}$-linear span of the family
$\left(  s_{\lambda}\right)  _{\lambda\in\mathcal{P}}$ (since every
$f\in\Lambda$ is a sum of homogeneous elements of $\Lambda$). In other words,
the family $\left(  s_{\lambda}\right)  _{\lambda\in\mathcal{P}}$ spans
$\Lambda$.

\item The family $\left(  s_{\lambda}\right)  _{\lambda\in\mathcal{P}}$ spans
$\Lambda$ and is linearly independent. Hence, the family $\left(  s_{\lambda
}\right)  _{\lambda\in\mathcal{P}}$ is a $\mathbf{Z}$-basis of $\Lambda$, qed.
\end{itemize}
\end{details}

\rule{17cm}{0.3mm}

\textbf{Page 8, \S 1.5:} Remove the comma before \textquotedblleft that
transforms $\alpha\left[  {}\right]  $ into $\lambda\left[  {}\right]
$\textquotedblright\ on the last line of page 8.

\rule{17cm}{0.3mm}

\textbf{Page 10, Proposition 1.5.1:} Replace \textquotedblleft$\mathbf{Z}%
$\textquotedblright\ by \textquotedblleft$\mathbf{N}$\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 12:} Replace \textquotedblleft save intervals\textquotedblright%
\ by \textquotedblleft safe intervals\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 13:} On the second line of page 13, replace \textquotedblleft%
$K_{\lambda/\lambda,\alpha}^{\prime}=K_{\lambda/\lambda,\alpha}^{\prime
}=\left[  \alpha=\left(  0\right)  \right]  $\textquotedblright\ by
\textquotedblleft$K_{\lambda/\lambda,\alpha}^{\prime}=K_{\lambda
/\lambda,\alpha}=\left[  \alpha=\left(  0\right)  \right]  $\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 13, Corollary 2.6:} Replace \textquotedblleft
transposed\textquotedblright\ by \textquotedblleft transpose\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 13, \S 2:} Two lines above (25), replace \textquotedblleft
transposed\textquotedblright\ by \textquotedblleft transpose\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 14, proof of Proposition 3.2:} After \textquotedblleft of equal
shape\textquotedblright, add \textquotedblleft(or, for (30), transposed
shape)\textquotedblright.

\rule{17cm}{0.3mm}

\begin{details}
\textbf{Page 14:} You prove Proposition 3.2 using a reference to Knuth's paper
[Knu]. Let me give an alternative proof of Proposition 3.2, which relies only
on results that you prove in your paper.

\textit{Second proof of Proposition 3.2:} The following proof of Proposition
3.2 uses only results you prove in your paper. Thus, it would make the paper self-contained.

We first notice some basic properties of minimal symmetric functions (whose
proofs are very easy):

\begin{itemize}
\item Every $\nu\in\mathcal{C}$ satisfies%
\begin{equation}
m_{\nu}\left[  X_{\mathbf{N}}\right]  =\sum_{\beta\in\mathcal{C}}\left[
\nu^{+}=\beta^{+}\right]  X^{\beta}. \label{p14.mnu.1}%
\end{equation}


\item Every $\nu\in\mathcal{P}$ satisfies%
\begin{equation}
m_{\nu}\left[  X_{\mathbf{N}}\right]  =\sum_{\alpha\in\mathcal{C}}\left[
\nu=\alpha^{+}\right]  X^{\alpha}. \label{p14.mnu.2}%
\end{equation}


\item For every $\nu\in\mathcal{P}$ and $\alpha\in\mathcal{C}$, we have%
\begin{equation}
\left(  \text{the }X^{\alpha}\text{-coefficient of }m_{\nu}\left[
X_{\mathbf{N}}\right]  \right)  =\left[  \nu=\alpha^{+}\right]  .
\label{p14.mnu.4}%
\end{equation}

\end{itemize}

Now, every $\lambda\in\mathcal{P}$ satisfies%
\begin{equation}
s_{\lambda}\left[  X_{\mathbf{N}}\right]  =\sum_{T\in\operatorname*{SST}%
\left(  \lambda\right)  }X^{\operatorname*{wt}\left(  T\right)  }
\label{p14.1}%
\end{equation}
\ \ \ \ \footnote{\textit{Proof of (\ref{p14.1}):} Let $\lambda\in\mathcal{P}%
$. Define a power series $f\left[  X_{\mathbf{N}}\right]  \in\mathbf{Z}\left[
\left[  X_{\mathbf{N}}\right]  \right]  $ by $f\left[  X_{\mathbf{N}}\right]
=\sum_{T\in\operatorname*{SST}\left(  \lambda\right)  }X^{\operatorname*{wt}%
\left(  T\right)  }$. In \S 5, you show that the power series $f\left[
X_{\mathbf{N}}\right]  $ is symmetric. Since this power series is also
homogeneous, this shows that $f\left[  X_{\mathbf{N}}\right]  \in\Lambda$.
Now,%
\[
f\left[  X_{\mathbf{N}}\right]  =\sum_{T\in\operatorname*{SST}\left(
\lambda\right)  }X^{\operatorname*{wt}\left(  T\right)  }=\sum_{\alpha
\in\mathcal{C}}\left\vert \left\{  T\in\operatorname*{SST}\left(
\lambda\right)  \ \mid\ \operatorname*{wt}\left(  T\right)  =\alpha\right\}
\right\vert X^{\alpha}.
\]
Thus, Lemma 2.2 (applied to $\beta=\left(  0\right)  $ and $c_{\alpha
}=\left\vert \left\{  T\in\operatorname*{SST}\left(  \lambda\right)
\ \mid\ \operatorname*{wt}\left(  T\right)  =\alpha\right\}  \right\vert $)
yields%
\begin{align*}
fs_{\left(  0\right)  }  &  =\sum_{\alpha\in\mathcal{C}}\left\vert \left\{
T\in\operatorname*{SST}\left(  \lambda\right)  \ \mid\ \operatorname*{wt}%
\left(  T\right)  =\alpha\right\}  \right\vert \underbrace{s_{\alpha+\left(
0\right)  }}_{=s_{\alpha}}\\
&  =\sum_{\alpha\in\mathcal{C}}\left\vert \left\{  T\in\operatorname*{SST}%
\left(  \lambda\right)  \ \mid\ \operatorname*{wt}\left(  T\right)
=\alpha\right\}  \right\vert s_{\alpha}.
\end{align*}
Compared with $f\underbrace{s_{\left(  0\right)  }}_{=1}=f$, this yields%
\begin{equation}
f=\sum_{\alpha\in\mathcal{C}}\left\vert \left\{  T\in\operatorname*{SST}%
\left(  \lambda\right)  \ \mid\ \operatorname*{wt}\left(  T\right)
=\alpha\right\}  \right\vert s_{\alpha}. \label{p14.2}%
\end{equation}
\par
But in (48), you have proven that $\sum_{T\in\operatorname*{SST}\left(
\lambda\right)  }s_{\operatorname*{wt}\left(  T\right)  }=s_{\lambda}$, so
that%
\[
s_{\lambda}=\sum_{T\in\operatorname*{SST}\left(  \lambda\right)
}s_{\operatorname*{wt}\left(  T\right)  }=\sum_{\alpha\in\mathcal{C}%
}\left\vert \left\{  T\in\operatorname*{SST}\left(  \lambda\right)
\ \mid\ \operatorname*{wt}\left(  T\right)  =\alpha\right\}  \right\vert
s_{\alpha}=f\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{p14.2})}\right)  .
\]
Hence, $s_{\lambda}\left[  X_{\mathbf{N}}\right]  =f\left[  X_{\mathbf{N}%
}\right]  =\sum_{T\in\operatorname*{SST}\left(  \lambda\right)  }%
X^{\operatorname*{wt}\left(  T\right)  }$, qed.} and therefore%
\begin{equation}
\left(  \left(  \text{the }X^{\alpha}\text{-coefficient of }s_{\lambda}\left[
X_{\mathbf{N}}\right]  \right)  =K_{\lambda,\alpha}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\alpha\in\mathcal{C}\right)
\label{p14.5r}%
\end{equation}
\footnote{\textit{Proof of (\ref{p14.5r}):} Let $\lambda\in\mathcal{P}$. Let
$\alpha\in\mathcal{C}$. We have%
\begin{align*}
&  \left(  \text{the }X^{\alpha}\text{-coefficient of }\underbrace{s_{\lambda
}\left[  X_{\mathbf{N}}\right]  }_{\substack{=\sum_{T\in\operatorname*{SST}%
\left(  \lambda\right)  }X^{\operatorname*{wt}\left(  T\right)  }\\\text{(by
(\ref{p14.1}))}}}\right) \\
&  =\left(  \text{the }X^{\alpha}\text{-coefficient of }\sum_{T\in
\operatorname*{SST}\left(  \lambda\right)  }X^{\operatorname*{wt}\left(
T\right)  }\right)  =\sum_{T\in\operatorname*{SST}\left(  \lambda\right)
}\underbrace{\left(  \text{the }X^{\alpha}\text{-coefficient of }%
X^{\operatorname*{wt}\left(  T\right)  }\right)  }_{=\left[  \alpha
=\operatorname*{wt}\left(  T\right)  \right]  }\\
&  =\sum_{T\in\operatorname*{SST}\left(  \lambda\right)  }\left[
\alpha=\operatorname*{wt}\left(  T\right)  \right]  =\sum_{\substack{T\in
\operatorname*{SST}\left(  \lambda\right)  ;\\\alpha=\operatorname*{wt}\left(
T\right)  }}\underbrace{\left[  \alpha=\operatorname*{wt}\left(  T\right)
\right]  }_{\substack{=1\\\text{(since }\alpha=\operatorname*{wt}\left(
T\right)  \text{)}}}+\sum_{\substack{T\in\operatorname*{SST}\left(
\lambda\right)  ;\\\alpha\neq\operatorname*{wt}\left(  T\right)
}}\underbrace{\left[  \alpha=\operatorname*{wt}\left(  T\right)  \right]
}_{\substack{=0\\\text{(since }\alpha\neq\operatorname*{wt}\left(  T\right)
\text{)}}}\\
&  =\sum_{\substack{T\in\operatorname*{SST}\left(  \lambda\right)
;\\\alpha=\operatorname*{wt}\left(  T\right)  }}1+\underbrace{\sum
_{\substack{T\in\operatorname*{SST}\left(  \lambda\right)  ;\\\alpha
\neq\operatorname*{wt}\left(  T\right)  }}0}_{=0}=\sum_{\substack{T\in
\operatorname*{SST}\left(  \lambda\right)  ;\\\alpha=\operatorname*{wt}\left(
T\right)  }}1=\left\vert \left\{  T\in\operatorname*{SST}\left(
\lambda\right)  \ \mid\ \alpha=\operatorname*{wt}\left(  T\right)  \right\}
\right\vert \cdot1\\
&  =\left\vert \underbrace{\left\{  T\in\operatorname*{SST}\left(
\lambda\right)  \ \mid\ \alpha=\operatorname*{wt}\left(  T\right)  \right\}
}_{\substack{=\left\{  T\in\operatorname*{SST}\left(  \lambda\right)
\ \mid\ \operatorname*{wt}\left(  T\right)  =\alpha\right\}
\\=\operatorname*{SST}\left(  \lambda,\alpha\right)  }}\right\vert =\left\vert
\operatorname*{SST}\left(  \lambda,\alpha\right)  \right\vert
=\#\operatorname*{SST}\left(  \lambda,\alpha\right)  =K_{\lambda,\alpha},
\end{align*}
and thus (\ref{p14.5r}) is proven.}. Moreoer, every $\lambda\in\mathcal{P}$
satisfies%
\begin{equation}
s_{\lambda}=\sum_{\nu\in\mathcal{P}}K_{\lambda,\nu}m_{\nu} \label{p14.5}%
\end{equation}
(by (\ref{p14.5r}), since $s_{\lambda}$ is symmetric). Finally, every
$\lambda\in\mathcal{P}$ and $\alpha\in\mathcal{C}$ satisfy%
\begin{equation}
K_{\lambda,\alpha^{+}}=K_{\lambda,\alpha} \label{p14.Kla}%
\end{equation}
(by (\ref{p14.5r}), since $s_{\lambda}$ is symmetric).

Now, we have%
\begin{equation}
f=\sum_{\lambda\in\mathcal{P}}\left\langle h_{\lambda}\mid f\right\rangle
m_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every }f\in\Lambda\label{p14.21}%
\end{equation}
\footnote{\textit{Proof of (\ref{p14.21}):} Let $f\in\Lambda$. We need to
prove the equality (\ref{p14.21}). This equality is $\mathbf{Z}$-linear in
$f$. Hence, we can WLOG assume that $f$ belongs to the basis $\left(
s_{\lambda}\right)  _{\lambda\in\mathcal{P}}$ of the $\mathbf{Z}$-module
$\Lambda$. Assume this. Thus, $f=s_{\mu}$ for some $\mu\in\mathcal{P}$.
Consider this $\mu$.
\par
For every $\nu\in\mathcal{P}$, we have $K_{\mu/\left(  0\right)  ,\nu
}=\left\langle h_{\nu}\mid s_{\mu/\left(  0\right)  }\right\rangle $ (by the
definition of $K_{\mu/\left(  0\right)  ,\nu}$). Since $\mu/\left(  0\right)
=\mu$, this equality rewrites as%
\begin{equation}
K_{\mu,\nu}=\left\langle h_{\nu}\mid s_{\mu}\right\rangle . \label{p14.22}%
\end{equation}
Now,
\begin{align}
\sum_{\lambda\in\mathcal{P}}\left\langle h_{\lambda}\mid f\right\rangle
m_{\lambda}  &  =\sum_{\nu\in\mathcal{P}}\left\langle h_{\nu}\mid
\underbrace{f}_{=s_{\mu}}\right\rangle m_{\nu}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we renamed the summation index }\lambda\text{ as }\nu\right)
\nonumber\\
&  =\sum_{\nu\in\mathcal{P}}\underbrace{\left\langle h_{\nu}\mid s_{\mu
}\right\rangle }_{\substack{=K_{\mu,\nu}\\\text{(by (\ref{p14.22}))}}}m_{\nu
}=\sum_{\nu\in\mathcal{P}}K_{\mu,\nu}m_{\nu}. \label{p14.23}%
\end{align}
On the other hand, $f=s_{\mu}=\sum_{\nu\in\mathcal{P}}K_{\mu,\nu}m_{\nu}$ (by
(\ref{p14.5}), applied to $\lambda=\mu$). Comparing this with (\ref{p14.23}),
we obtain $f=\sum_{\lambda\in\mathcal{P}}\left\langle h_{\lambda}\mid
f\right\rangle m_{\lambda}$. This proves (\ref{p14.21}).}. Hence, the two
bases $\left(  h_{\lambda}\right)  _{\lambda\in\mathcal{P}}$ and $\left(
m_{\lambda}\right)  _{\lambda\in\mathcal{P}}$ of $\Lambda$ are dual to each
other with respect to the bilinear form $\left\langle \cdot\mid\cdot
\right\rangle $.

Now, as you show right before Proposition 3.2, every $\alpha\in\mathcal{C}$
and $\beta\in\mathcal{C}$ satisfy%
\begin{equation}
\left\langle h_{\alpha}\mid e_{\beta}\right\rangle =\sum_{\lambda
\in\mathcal{P}}K_{\lambda,\alpha}K_{\lambda,\beta}^{\prime} \label{p14.41}%
\end{equation}
and%
\begin{equation}
\left\langle h_{\alpha}\mid h_{\beta}\right\rangle =\sum_{\lambda
\in\mathcal{P}}K_{\lambda,\alpha}K_{\lambda,\beta}. \label{p14.42}%
\end{equation}


Now, fix $\beta\in\mathcal{C}$. Then, (\ref{p14.21}) (applied to $f=h_{\beta}%
$) yields%

\begin{align*}
h_{\beta}  &  =\sum_{\lambda\in\mathcal{P}}\left\langle h_{\lambda}\mid
h_{\beta}\right\rangle m_{\lambda}=\sum_{\nu\in\mathcal{P}}%
\underbrace{\left\langle h_{\nu}\mid h_{\beta}\right\rangle }_{\substack{=\sum
_{\lambda\in\mathcal{P}}K_{\lambda,\nu}K_{\lambda,\beta}\\\text{(by
(\ref{p14.42}), applied to }\alpha=\nu\text{)}}}m_{\nu}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\lambda\text{ as }\nu\right) \\
&  =\sum_{\nu\in\mathcal{P}}\left(  \sum_{\lambda\in\mathcal{P}}K_{\lambda
,\nu}K_{\lambda,\beta}\right)  m_{\nu},
\end{align*}
so that%
\begin{align*}
h_{\beta}\left[  X_{\mathbf{N}}\right]   &  =\sum_{\nu\in\mathcal{P}}\left(
\sum_{\lambda\in\mathcal{P}}K_{\lambda,\nu}K_{\lambda,\beta}\right)
\underbrace{m_{\nu}\left[  X_{\mathbf{N}}\right]  }_{\substack{=\sum
_{\alpha\in\mathcal{C}}\left[  \nu=\alpha^{+}\right]  X^{\alpha}\\\text{(by
(\ref{p14.mnu.2}))}}}=\sum_{\nu\in\mathcal{P}}\left(  \sum_{\lambda
\in\mathcal{P}}K_{\lambda,\nu}K_{\lambda,\beta}\right)  \left(  \sum
_{\alpha\in\mathcal{C}}\left[  \nu=\alpha^{+}\right]  X^{\alpha}\right) \\
&  =\sum_{\alpha\in\mathcal{C}}\underbrace{\sum_{\nu\in\mathcal{P}}\left(
\sum_{\lambda\in\mathcal{P}}K_{\lambda,\nu}K_{\lambda,\beta}\right)  \left[
\nu=\alpha^{+}\right]  }_{=\sum_{\lambda\in\mathcal{P}}K_{\lambda,\alpha^{+}%
}K_{\lambda,\beta}}X^{\alpha}=\sum_{\alpha\in\mathcal{C}}\left(  \sum
_{\lambda\in\mathcal{P}}\underbrace{K_{\lambda,\alpha^{+}}}%
_{\substack{=K_{\lambda,\alpha}\\\text{(by (\ref{p14.Kla}))}}}K_{\lambda
,\beta}\right)  X^{\alpha}\\
&  =\sum_{\alpha\in\mathcal{C}}\left(  \sum_{\lambda\in\mathcal{P}}%
K_{\lambda,\alpha}K_{\lambda,\beta}\right)  X^{\alpha}.
\end{align*}
Hence,%
\[
\sum_{\alpha\in\mathcal{C}}\left(  \sum_{\lambda\in\mathcal{P}}K_{\lambda
,\alpha}K_{\lambda,\beta}\right)  X^{\alpha}=h_{\beta}\left[  X_{\mathbf{N}%
}\right]  =\sum_{\alpha\in\mathcal{C}}\sharp\mathcal{M}_{\alpha,\beta
}X^{\alpha}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (10)}\right)  .
\]
Comparing coefficients in this equality, we conclude that%
\[
\sum_{\lambda\in\mathcal{P}}K_{\lambda,\alpha}K_{\lambda,\beta}=\sharp
\mathcal{M}_{\alpha,\beta}.
\]
This proves (29). Similarly, we can prove (30) (by applying (\ref{p14.21}) to
$f=e_{\beta}$ instead of $f=h_{\beta}$, and by using (\ref{p14.41}) instead of
(\ref{p14.42})). The proof of Proposition 3.2 is complete.

(Alternatively, it is also possible to prove Proposition 3.2 by deriving it
from the identities (33) and (34), which are more widespread than Proposition
3.2 and which can be proven in various ways, not only using the RSK algorithm.)
\end{details}

\rule{17cm}{0.3mm}

\textbf{Page 15:} \textquotedblleft the the image\textquotedblright\ should be
\textquotedblleft the image\textquotedblright.

\rule{17cm}{0.3mm}

\begin{details}
\textbf{Page 15:} You write: \textquotedblleft From the proof of proposition
2.1 we see that if $\lambda\in\mathbf{N}^{l}$, then one needs to consider only
permutations $\sigma\in\mathbf{S}_{l}$\textquotedblright. By this, you mean
that if $\lambda\in\mathbf{N}^{l}$ and if $\sigma\in\mathbf{S}_{\infty}$ is a
permutation satisfying $\sigma\left(  \lambda\left[  {}\right]  \right)
\geq\mu\left[  {}\right]  $, then
\begin{equation}
\sigma\in\mathbf{S}_{l}. \label{p15.1}%
\end{equation}


I do not see how (\ref{p15.1}) is supposed to follow from the proof of
Proposition 2.1. Here is, instead, my proof of (\ref{p15.1}):

\textit{Proof of (\ref{p15.1}):} Let $\lambda\in\mathbf{N}^{l}$, and let
$\sigma\in\mathbf{S}_{\infty}$ be a permutation satisfying $\sigma\left(
\lambda\left[  {}\right]  \right)  \geq\mu\left[  {}\right]  $. We need to
prove (\ref{p15.1}). Indeed, assume the contrary (for the sake of
contradiction). Thus, $\sigma\notin\mathbf{S}_{l}$. Thus, there exists some
$i\in\left\{  l,l+1,l+2,\ldots\right\}  $ such that $\sigma\left(  i\right)
\neq i$. Let $j$ be the \textbf{largest} such $i$. Thus, $j$ is an element of
$\left\{  l,l+1,l+2,\ldots\right\}  $ and satisfies $\sigma\left(  j\right)
\neq j$. From $j\in\left\{  l,l+1,l+2,\ldots\right\}  $, we obtain $j\geq l$.

Now, we have $\sigma\left(  j\right)  <j$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Thus, $\sigma\left(  j\right)  \geq j$. Combined with
$\sigma\left(  j\right)  \neq j$, this yields $\sigma\left(  j\right)  >j$.
Now, $\sigma$ is injective (since $\sigma\in\mathbf{S}_{\infty}$). Thus, from
$\sigma\left(  j\right)  \neq j$, we obtain $\sigma\left(  \sigma\left(
j\right)  \right)  \neq\sigma\left(  j\right)  $. Also, $\sigma\left(
j\right)  \geq j\geq l$, so that $\sigma\left(  j\right)  \in\left\{
l,l+1,l+2,\ldots\right\}  $.
\par
But recall that $j$ is the \textbf{largest} $i\in\left\{  l,l+1,l+2,\ldots
\right\}  $ such that $\sigma\left(  i\right)  \neq i$. Hence, every
$i\in\left\{  l,l+1,l+2,\ldots\right\}  $ such that $\sigma\left(  i\right)
\neq i$ satisfies $i\leq j$. Applying this to $i=\sigma\left(  j\right)  $, we
obtain $\sigma\left(  j\right)  \leq j$ (since $\sigma\left(  j\right)
\in\left\{  l,l+1,l+2,\ldots\right\}  $ and $\sigma\left(  \sigma\left(
j\right)  \right)  \neq\sigma\left(  j\right)  $). This contradicts
$\sigma\left(  j\right)  >j$. This contradiction proves that our assumption
was wrong, qed.}. Now, set $i=\sigma\left(  j\right)  $. Thus, $\sigma
^{-1}\left(  i\right)  =j$.

We have $\lambda\in\mathbf{N}^{l}$, and thus $\lambda_{w}=0$ for every
$w\in\mathbf{N}$ satisfying $w\geq l$. Applying this to $w=j$, we obtain
$\lambda_{j}=0$. The definition of $\lambda\left[  j\right]  $ now yields
$\lambda\left[  j\right]  =\underbrace{\lambda_{j}}_{=0}-1-j=-1-j$, so that%
\begin{align*}
-1-j  &  =\lambda\left[  \underbrace{j}_{=\sigma^{-1}\left(  i\right)
}\right]  =\lambda\left[  \sigma^{-1}\left(  i\right)  \right]  =\left(
\lambda\left[  {}\right]  \right)  _{\sigma^{-1}\left(  i\right)  }=\left(
\sigma\left(  \lambda\left[  {}\right]  \right)  \right)  _{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\sigma\left(
\lambda\left[  {}\right]  \right)  \right) \\
&  \geq\left(  \mu\left[  {}\right]  \right)  _{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sigma\left(  \lambda\left[  {}\right]  \right)  \geq\mu\left[
{}\right]  \right) \\
&  =\mu\left[  i\right]  =\underbrace{\mu_{i}}_{\geq0}-1-\underbrace{i}%
_{=\sigma\left(  j\right)  <j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\mu\left[  i\right]  \right) \\
&  >-1-j.
\end{align*}
This is absurd. This contradiction shows that our assumption was wrong. Hence,
(\ref{p15.1}) is proven.
\end{details}

\rule{17cm}{0.3mm}

\begin{details}
\textbf{Page 16, \S 3:} You claim that \textquotedblleft we may apply the
automorphism $\omega$ to equation (37)\textquotedblright\ to obtain the
equality (40). This is correct, but in my opinion could use a bit more
justification. You are tacitly using the fact that%
\begin{equation}
\omega\left(  s_{\lambda/\mu}\right)  =s_{\lambda^{\mathbf{t}}/\mu
^{\mathbf{t}}}\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\mathcal{P}\text{
and }\mu\in\mathcal{P}. \label{p18.1}%
\end{equation}


The proof of (\ref{p18.1}) is simple, but not something I would leave to the reader:

\textit{Proof of (\ref{p18.1}):} The basis $\left(  s_{\lambda}\right)
_{\lambda\in\mathcal{P}}$ of the $\mathbf{Z}$-module $\Lambda$ is orthonormal
with respect to the scalar product $\left\langle \cdot\mid\cdot\right\rangle $
(because of the definition of this scalar product). In other words, we have%
\begin{equation}
\left\langle s_{\lambda}\mid s_{\mu}\right\rangle =\left[  \lambda=\mu\right]
\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\mathcal{P}\text{ and }\mu
\in\mathcal{P}. \label{p18.2}%
\end{equation}
Furthermore,%
\begin{equation}
\left\langle \omega\left(  f\right)  \mid\omega\left(  g\right)  \right\rangle
=\left\langle f\mid g\right\rangle \ \ \ \ \ \ \ \ \ \ \text{for any }%
f\in\Lambda\text{ and }g\in\Lambda\label{p18.invariant}%
\end{equation}
\footnote{\textit{Proof of (\ref{p18.invariant}):} The equality $\left\langle
\omega\left(  f\right)  \mid\omega\left(  g\right)  \right\rangle
=\left\langle f\mid g\right\rangle $ is $\mathbf{Z}$-linear in each of $f$ and
$g$. Thus, it suffices to prove it in the case when $f$ and $g$ belong to the
basis $\left(  s_{\lambda}\right)  _{\lambda\in\mathcal{P}}$ of the
$\mathbf{Z}$-module $\Lambda$. But in this case, it boils down to the identity
$\left[  \alpha^{\mathbf{t}}=\beta^{\mathbf{t}}\right]  =\left[  \alpha
=\beta\right]  $ for any $\alpha\in\mathcal{P}$ and $\beta\in\mathcal{P}$
(because of (\ref{p18.2})), which identity is obvious.}.

Recall that $\omega$ is a ring morphism (since $\omega$ coincides with the
ring morphism $\Lambda\rightarrow\Lambda$ that sends $h_{i}\mapsto e_{i}$ for
all $i>0$). Thus,%
\begin{equation}
s_{\mu^{\mathbf{t}}}s_{\nu^{\mathbf{t}}}=\omega\left(  s_{\mu}s_{\nu}\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }\mu\in\mathcal{P}\text{ and }\nu
\in\mathcal{P} \label{p18.rimo-cons}%
\end{equation}
\footnote{\textit{Proof of (\ref{p18.rimo-cons}):} Let $\mu\in\mathcal{P}$ and
$\nu\in\mathcal{P}$. Since $\omega$ is a ring morphism, we have $\omega\left(
s_{\mu}s_{\nu}\right)  =\underbrace{\omega\left(  s_{\mu}\right)
}_{\substack{=s_{\mu^{\mathbf{t}}}\\\text{(by the definition of }%
\omega\text{)}}}\underbrace{\omega\left(  s_{\nu}\right)  }_{\substack{=s_{\nu
^{\mathbf{t}}}\\\text{(by the definition of }\omega\text{)}}}=s_{\mu
^{\mathbf{t}}}s_{\nu^{\mathbf{t}}}$. This proves (\ref{p18.rimo-cons}).}.

Every $f\in\Lambda$ satisfies%
\begin{equation}
f=\sum_{\lambda\in\mathcal{P}}\underbrace{\left\langle f\mid s_{\lambda
}\right\rangle }_{=\left\langle s_{\lambda}\mid f\right\rangle }s_{\lambda
}=\sum_{\lambda\in\mathcal{P}}\left\langle s_{\lambda}\mid f\right\rangle
s_{\lambda}=\sum_{\nu\in\mathcal{P}}\left\langle s_{\nu}\mid f\right\rangle
s_{\nu} \label{p18.3}%
\end{equation}
(here, we have renamed the summation index $\lambda$ as $\nu$).

Let $\lambda\in\mathcal{P}$ and $\mu\in\mathcal{P}$. The definition of
$\omega$ yields $\omega\left(  s_{\lambda}\right)  =s_{\lambda^{\mathbf{t}}}$.
Applying (\ref{p18.3}) to $f=s_{\lambda/\mu}$, we obtain%
\begin{equation}
s_{\lambda/\mu}=\sum_{\nu\in\mathcal{P}}\underbrace{\left\langle s_{\nu}\mid
s_{\lambda/\mu}\right\rangle }_{\substack{=\left\langle s_{\mu}s_{\nu}\mid
s_{\lambda}\right\rangle \\\text{(since }\left\langle s_{\mu}s_{\nu}\mid
s_{\lambda}\right\rangle =\left\langle s_{\nu}\mid s_{\lambda/\mu
}\right\rangle \\\text{(by (17), applied to }f=s_{\nu}\text{))}}}s_{\nu}%
=\sum_{\nu\in\mathcal{P}}\left\langle s_{\mu}s_{\nu}\mid s_{\lambda
}\right\rangle s_{\nu}. \label{p18.onepart}%
\end{equation}
The same argument (applied to $\lambda^{\mathbf{t}}$ and $\mu^{\mathbf{t}}$
instead of $\lambda$ and $\mu$) yields%
\begin{align*}
s_{\lambda^{\mathbf{t}}/\mu^{\mathbf{t}}}  &  =\sum_{\nu\in\mathcal{P}%
}\left\langle s_{\mu^{\mathbf{t}}}s_{\nu}\mid s_{\lambda^{\mathbf{t}}%
}\right\rangle s_{\nu}=\sum_{\nu\in\mathcal{P}}\left\langle \underbrace{s_{\mu
^{\mathbf{t}}}s_{\nu^{\mathbf{t}}}}_{\substack{=\omega\left(  s_{\mu}s_{\nu
}\right)  \\\text{(by (\ref{p18.rimo-cons}))}}}\mid\underbrace{s_{\lambda
^{\mathbf{t}}}}_{=\omega\left(  s_{\lambda}\right)  }\right\rangle
\underbrace{s_{\nu^{\mathbf{t}}}}_{\substack{=\omega\left(  s_{\nu}\right)
\\\text{(since }\omega\left(  s_{\nu}\right)  =s_{\nu^{\mathbf{t}}}\\\text{(by
the definition of }\omega\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\nu^{\mathbf{t}}\text{ for }\nu\text{ in the
sum,}\\
\text{because the map }\mathcal{P}\rightarrow\mathcal{P},\ \nu\mapsto
\nu^{\mathbf{t}}\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum_{\nu\in\mathcal{P}}\underbrace{\left\langle \omega\left(  s_{\mu
}s_{\nu}\right)  \mid\omega\left(  s_{\lambda}\right)  \right\rangle
}_{\substack{=\left\langle s_{\mu}s_{\nu}\mid s_{\lambda}\right\rangle
\\\text{(by (\ref{p18.invariant}))}}}\omega\left(  s_{\nu}\right) \\
&  =\sum_{\nu\in\mathcal{P}}\left\langle s_{\mu}s_{\nu}\mid s_{\lambda
}\right\rangle \omega\left(  s_{\nu}\right)  =\omega\left(  \underbrace{\sum
_{\nu\in\mathcal{P}}\left\langle s_{\mu}s_{\nu}\mid s_{\lambda}\right\rangle
s_{\nu}}_{\substack{=s_{\lambda/\mu}\\\text{(by (\ref{p18.onepart}))}%
}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\omega\text{ is
}\mathbf{Z}\text{-linear}\right) \\
&  =\omega\left(  s_{\lambda/\mu}\right)  .
\end{align*}
This proves (\ref{p18.1}).
\end{details}

\rule{17cm}{0.3mm}

\textbf{Page 18:} Replace \textquotedblleft by a single factor $h_{j}%
$\textquotedblright\ by \textquotedblleft by a single factor $h_{\beta_{j}}%
$\textquotedblright\ (or something like this, but not $h_{j}$, since $j$
already means something different here).

\rule{17cm}{0.3mm}

\textbf{Page 19:} In the paragraph that begins with \textquotedblleft Now
suppose to the contrary\textquotedblright\ and ends with \textquotedblleft of
the sequence $\left(  \mu^{\mathbf{t}}+\operatorname*{col}\left(  M\right)
\right)  \left[  {}\right]  $\textquotedblright, every appearance of
\textquotedblleft$\beta^{\left(  i\right)  }$\textquotedblright\ should be
replaced by \textquotedblleft$\beta^{\left(  i+1\right)  }$\textquotedblright,
and every appearance of \textquotedblleft$\beta^{\left(  i-1\right)  }%
$\textquotedblright\ should be replaced by \textquotedblleft$\beta^{\left(
i\right)  }$\textquotedblright. Also, the \textquotedblleft$\alpha_{i-1}%
$\textquotedblright\ should be replaced by \textquotedblleft$\alpha_{i}%
$\textquotedblright. (The reason for this is that the claim that
\textquotedblleft$\beta^{\left(  i\right)  }$ was obtained by adding the
binary composition $M_{i}$ to the partition $\beta^{\left(  i-1\right)  }%
$\textquotedblright\ is false; the addition of $M_{i}$ rather turns
$\beta^{\left(  i\right)  }$ into $\beta^{\left(  i+1\right)  }$.)

\rule{17cm}{0.3mm}

\textbf{Page 25:} You claim that \textquotedblleft Conversely any entry $m+1$
for which this condition in terms of $\alpha^{\prime}$ holds, and for which
$k$ is minimal, cannot have an entry $m$ in the same column\textquotedblright.
This is true, but I find this rather nontrivial.

\rule{17cm}{0.3mm}

\begin{thebibliography}{999999}                                                                                           %


\bibitem[DeLeTh]{DeLeTh}J. D\'{e}sarm\'{e}nien, B. Leclerc, J.-Y. Thibon,
\textit{Hall-Littlewood functions and Kostka-Foulkes polynomials in
representation theory}, S\'{e}minaire Lotharingien de Combinatoire [electronic
only] (1994), Volume: 32.
\end{thebibliography}


\end{document}