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\ihead{Math 701, Fall 2026, Lecture 8, version \today}
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\begin{document}
\section*{Math 701 Fall 2026, Lecture 8: Proof of the Littlewood--Richardson
rule}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/26fs}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/26fs/}}}

(GPT-6 was used to create examples and to polish and detail some of the
arguments below.) \medskip

\setcounter{section}{1}\setcounter{subsection}{7}\setcounter{subsubsection}{0}\setcounter{theo}{7}

In order to prove the Littlewood--Richardson rule, we will first show
\textbf{Stembridge's lemma}:

\begin{lemma}
[Stembridge's lemma]\label{lem.schur.stembridge}Let $\lambda/\mu$ be a skew
partition and let $\nu$ be an $N$-partition. Then,%
\[
a_{\nu+\rho}\cdot s_{\lambda/\mu}=\sum_{\substack{T\in\operatorname*{SSYT}%
\left(  \lambda/\mu,N\right)  ;\\T\text{ is }\nu\text{-Yamanouchi}}%
}a_{\nu+\operatorname*{cont}T+\rho}.
\]

\end{lemma}

\begin{sloppypar}
This lemma, due to Stembridge \cite{Stembr02}, will easily yield both the
Littlewood--Richardson rule (Theorem \ref{thm.schur.generalized-lr} in Lecture
7) and the alternant formula for Schur polynomials (Theorem
\ref{thm.schur.bialternant} \textbf{(b)} in Lecture 5). Let me explain how.
The proof of Lemma \ref{lem.schur.stembridge} will be given afterwards.
\end{sloppypar}

First, we recover the alternant formula:

\begin{proof}
[Proof of Theorem \ref{thm.schur.bialternant} \textbf{(b)} using Lemma
\ref{lem.schur.stembridge}.]Let $\lambda$ be a partition of length $\leq N$.
We identify $\lambda$ with the $N$-partition $\left(  \lambda_{1},\lambda
_{2},\ldots,\lambda_{N}\right)  $ formed by its first $N$ entries (since all
its other entries are $0$). Set $\mu=\varnothing$ and $\nu=\varnothing
=(0,0,\ldots,0)$ in Lemma~\ref{lem.schur.stembridge}. We obtain%
\begin{equation}
a_{\rho}\cdot s_{\lambda}=\sum_{\substack{T\in\operatorname*{SSYT}\left(
\lambda,N\right)  ;\\T\text{ is }\varnothing\text{-Yamanouchi}}%
}a_{\operatorname*{cont}T+\rho} \label{pf.thm.schur.bialternant.b.2}%
\end{equation}
(since $s_{\lambda/\varnothing}=s_{\lambda}$ and $\operatorname*{SSYT}\left(
\lambda/\varnothing,N\right)  =\operatorname*{SSYT}\left(  \lambda,N\right)
$). But what are the $\varnothing$-Yamanouchi tableaux $T\in
\operatorname*{SSYT}\left(  \lambda,N\right)  $ ?

Let $T_{0}$ be the tableau of shape $\lambda$ whose entries $T_{0}\left(
i,j\right)  $ are simply given by%
\begin{equation}
T_{0}\left(  i,j\right)  =i\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
i,j\right)  \in Y\left(  \lambda\right)
\label{pf.thm.schur.bialternant.b.T0ij=i}%
\end{equation}
(that is, all entries of the $i$-th row of $T_{0}$ equal $i$ for each $i$).
For example, if $\lambda=\left(  5,3,2\right)  $, then%
\[
T_{0}=\ytableaushort{11111,222,33}\ \ .
\]
It is clear that the tableau $T_{0}$ is semistandard, and that its entries
belong to $\left[  N\right]  $ (since $\ell\left(  \lambda\right)  \leq N$);
hence, $T_{0}\in\operatorname*{SSYT}\left(  \lambda,N\right)  $. We call this
tableau $T_{0}$ the \textbf{minimalistic} tableau of shape $\lambda$, because
its entry in each cell is the smallest that it can be (indeed, it is easy to
see that any SSYT $T$ of shape $\lambda$ must satisfy%
\begin{equation}
T\left(  i,j\right)  \geq i\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
i,j\right)  \in Y\left(  \lambda\right)
\label{pf.thm.schur.bialternant.b.Tijgeqi}%
\end{equation}
\footnote{\textit{Proof.} Let $T$ be an SSYT of shape $\lambda$. Let $\left(
i,j\right)  \in Y\left(  \lambda\right)  $. Since $Y\left(  \lambda\right)  $
is a straight Young diagram, all the $i$ cells $\left(  1,j\right)  ,\ \left(
2,j\right)  ,\ \ldots,\ \left(  i,j\right)  $ must belong to $Y\left(
\lambda\right)  $ as well. The entries in these cells are positive integers
and satisfy $T\left(  1,j\right)  <T\left(  2,j\right)  <\cdots<T\left(
i,j\right)  $ (since the entries of $T$ strictly increase down each column).
Thus, the entry $T\left(  i,j\right)  $ must have $i-1$ distinct smaller
entries above it in its same column; consequently, it must itself be $\geq i$.
That is, $T\left(  i,j\right)  \geq i$, qed.}, and of course the tableau
$T_{0}$ achieves equality in this inequality). Others call it the
\textbf{highest-weight tableau}, since its content is lexicographically larger
than the content of any other SSYT of shape $\lambda$ with entries in $[N]$
(we will prove this later on).

The tableau $T_{0}$ has content $\operatorname*{cont}T_{0}=\lambda$ (since the
number of $i$'s in $T_{0}$ is the number of cells in the $i$-th row of
$Y\left(  \lambda\right)  $, but the latter number is $\lambda_{i}$). It is
furthermore easy to see that this tableau $T_{0}$ is $\varnothing$-Yamanouchi
(indeed, for each $j\geq1$, the tableau $\operatorname{col}_{\geq j}\left(
T_{0}\right)  $ has content $\operatorname*{cont}\left(  \operatorname{col}%
_{\geq j}\left(  T_{0}\right)  \right)  =\left(  \lambda_{1,j},\lambda
_{2,j},\ldots,\lambda_{N,j}\right)  $, where $\lambda_{i,j}:=\max\left\{
\lambda_{i}-j+1,\ 0\right\}  $; and this $N$-tuple is easily seen to be weakly
decreasing because $\lambda$ is). Now I claim:

\begin{statement}
\textit{Claim 1:} The \textbf{only} $\varnothing$-Yamanouchi tableau
$T\in\operatorname*{SSYT}\left(  \lambda,N\right)  $ is $T_{0}$.
\end{statement}

\begin{proof}
[Proof of Claim 1.]Let $T$ be any $\varnothing$-Yamanouchi tableau in
$\operatorname*{SSYT}\left(  \lambda,N\right)  $. We must show that $T=T_{0}$.

We can argue column-by-column (by induction, starting with the rightmost
column) that all entries in each column must be $1,2,\ldots,h$, where $h$ is
the length of the column. Here are the details. Suppose that all columns
strictly to the right of a given column $j$ have already been shown to have
this form. We must show that column $j$ has this form as well.\footnote{Here
is an example of our situation (column $j$ is highlit in gray):%
\[
\begin{ytableau}
? & ? & *(lightgray)? & 1 & 1 & 1 \\
? & ? & *(lightgray)? & 2 & 2 \\
? & ? & *(lightgray)? \\
? & ?
\end{ytableau}\ \ .
\]
}

Let $r$ be the length of column $j+1$ (or $0$ if column $j$ is the rightmost
column), and let $h$ be the length of column $j$. Since the shape of $T$ is
straight, its columns get shorter as we move east; thus, $r\leq h$. For each
$i\leq r$, we have $T(i,j)\geq i$ by (\ref{pf.thm.schur.bialternant.b.Tijgeqi}%
), while weak increase along row $i$ gives $T(i,j)\leq T(i,j+1)=i$ (by
induction). Thus, the first $r$ entries in column $j$ are $1,2,\ldots,r$.

It remains to prove\footnote{Here is an example of our situation (column $j$
is highlit in gray):%
\[
\begin{ytableau}
? & ? & *(lightgray)1 & 1 & 1 & 1 \\
? & ? & *(lightgray)2 & 2 & 2 \\
? & ? & *(lightgray)? \\
? & ?
\end{ytableau}\ \ .
\]
} that the remaining entries in column $j$ must be $r+1,r+2,\ldots,h$. Assume
the contrary. Since these remaining entries of column $j$ are all $\geq r+1$
(because of (\ref{pf.thm.schur.bialternant.b.Tijgeqi})) and are strictly
increasing down this column (since $T$ is semistandard), we conclude that they
must skip one of the integers $r+1,r+2,\ldots,h$; that is, there exists some
$k>r$ such that $k$ is not an entry of column $j$ but $k+1$ is. Now, this $k$
cannot appear in any of the columns $j+1,j+2,j+3,\ldots$ either (since all
entries in these columns are $\leq r$ by induction); hence, $k$ does not
appear in any of the columns $j,j+1,j+2,\ldots$. In other words, the $k$-th
entry of $\operatorname*{cont}\left(  \operatorname{col}_{\geq j}T\right)  $
is $0$. But the $\left(  k+1\right)  $-st entry of $\operatorname*{cont}%
\left(  \operatorname{col}_{\geq j}T\right)  $ is positive (since $k+1$
appears at least in column $j$). Therefore, the $N$-tuple
$\operatorname*{cont}\left(  \operatorname{col}_{\geq j}T\right)  $ cannot be
weakly decreasing (since its $k$-th entry is $0$, but its $\left(  k+1\right)
$-st entry is positive). This contradicts the fact that $\operatorname*{cont}%
\left(  \operatorname{col}_{\geq j}T\right)  $ is weakly decreasing (since $T$
is $\varnothing$-Yamanouchi).

This contradiction shows that our assumption was wrong. So the entries in
column $j$ must be $1,2,\ldots,h$. This proves the claim by induction.

So the entries of $T$ are given by $T\left(  i,j\right)  =i$. In other words,
$T=T_{0}$. This proves Claim 1.
\end{proof}

Claim 1 yields
\[
\sum_{\substack{T\in\operatorname*{SSYT}\left(  \lambda,N\right)  ;\\T\text{
is }\varnothing\text{-Yamanouchi}}}a_{\operatorname*{cont}T+\rho}%
=a_{\lambda+\rho},
\]
and thus our above equality (\ref{pf.thm.schur.bialternant.b.2}) rewrites as
$a_{\rho}\cdot s_{\lambda}=a_{\lambda+\rho}$. This proves Theorem
\ref{thm.schur.bialternant} \textbf{(b)}, assuming that Lemma
\ref{lem.schur.stembridge} holds.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.schur.generalized-lr} using Lemma
\ref{lem.schur.stembridge}.]\setlength{\emergencystretch}{2em}We have assumed
that Lemma \ref{lem.schur.stembridge} holds, i.e., we have%
\begin{equation}
a_{\nu+\rho}\cdot s_{\lambda/\mu}=\sum_{\substack{T\in\operatorname*{SSYT}%
\left(  \lambda/\mu,N\right)  ;\\T\text{ is }\nu\text{-Yamanouchi}}%
}a_{\nu+\operatorname*{cont}T+\rho}. \label{pf.thm.schur.generalized-lr.1}%
\end{equation}
By Theorem \ref{thm.schur.bialternant} \textbf{(b)}, we have $a_{\nu+\rho
}=a_{\rho}\cdot s_{\nu}$ and $a_{\nu+\operatorname*{cont}T+\rho}=a_{\rho}\cdot
s_{\nu+\operatorname*{cont}T}$ for any $\nu$-Yamanouchi tableau $T\in
\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  $ (the latter because
$\nu+\operatorname*{cont}T$ is an $N$-partition whenever $T$ is $\nu
$-Yamanouchi). Thus, the equality (\ref{pf.thm.schur.generalized-lr.1})
rewrites as%
\begin{equation}
a_{\rho}\cdot s_{\nu}\cdot s_{\lambda/\mu}=\sum_{\substack{T\in
\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  ;\\T\text{ is }%
\nu\text{-Yamanouchi}}}a_{\rho}\cdot s_{\nu+\operatorname*{cont}T}.
\label{pf.thm.schur.generalized-lr.2}%
\end{equation}
Cancelling $a_{\rho}$ would give us precisely the claim of Theorem
\ref{thm.schur.generalized-lr}.

Why can we cancel $a_{\rho}$ ? Well, if $K$ is an integral domain, then the
polynomial ring $\mathcal{P}=K\left[  x_{1},x_{2},\ldots,x_{N}\right]  $ is an
integral domain as well, and the polynomial $a_{\rho}=\prod_{1\leq i<j\leq
N}\left(  x_{i}-x_{j}\right)  $ (the Vandermonde determinant) is nonzero, so
it can be cancelled. In the general case, we can argue as
follows:\footnote{See \cite[Lemma 7.3.38]{21s} for details.}

\begin{itemize}
\item Either argue that Theorem \ref{thm.schur.generalized-lr} is true for
$K=\mathbb{Z}$ (by what we just said, since $\mathbb{Z}$ is an integral
domain) and therefore for arbitrary $K$ (since we know from Proposition
\ref{prop.symp.functoriality} \textbf{(b)} that there is a functorial ring
morphism $\mathcal{P}_{\mathbb{Z}}\rightarrow\mathcal{P}_{K}$ that sends skew
Schur polynomials in $\mathcal{P}_{\mathbb{Z}}$ to skew Schur polynomials in
$\mathcal{P}_{K}$; applying this morphism to the claim of Theorem
\ref{thm.schur.generalized-lr} for $K=\mathbb{Z}$ yields the general case of
Theorem \ref{thm.schur.generalized-lr}).

\item Or argue that the element $a_{\rho}=\prod_{1\leq i<j\leq N}(x_{i}%
-x_{j})\in\mathcal{P}$ is regular (i.e., multiplication by this element is
injective as a map from $\mathcal{P}$ to $\mathcal{P}$) for any $K$. Indeed,
each factor $x_{i}-x_{j}$ is monic as a polynomial in $x_{i}$ over the ring of
polynomials in the other variables. Multiplication by a monic polynomial is
injective over any commutative ring (the leading nonzero coefficient of a
nonzero polynomial stays nonzero in the product). Multiplication by the
product of these factors $x_{i}-x_{j}$ is therefore injective as well. That
is, multiplication by $a_{\rho}$ is injective. Hence, we can cancel $a_{\rho}$
from equalities such as (\ref{pf.thm.schur.generalized-lr.2}).
\end{itemize}

\noindent Either way, Theorem \ref{thm.schur.generalized-lr} is proved
provided that we can prove Lemma \ref{lem.schur.stembridge}.
\end{proof}

So it remains to prove Lemma \ref{lem.schur.stembridge}. We prepare with a
simple observation about alternants:

\begin{lemma}
[Alternants are alternating in $\alpha$]\label{lem.schur.alternant-swap}Let
$\alpha\in\mathbb{N}^{N}$ be an $N$-tuple.

\begin{enumerate}
\item[\textbf{(a)}] If $\alpha$ has two equal entries, then $a_{\alpha}=0$.

\item[\textbf{(b)}] If $\beta$ is an $N$-tuple obtained from $\alpha$ by
swapping two entries, then $a_{\beta}=-a_{\alpha}$.

\item[\textbf{(c)}] Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{N}\right)  \in\mathbb{N}^{N}$ be any $N$-tuple. Let $\sigma\in S_{N}$
be a permutation. As before, let $\alpha\circ\sigma$ denote the $N$-tuple
$\left(  \alpha_{\sigma\left(  1\right)  },\alpha_{\sigma\left(  2\right)
},\ldots,\alpha_{\sigma\left(  N\right)  }\right)  $. Then, $a_{\alpha
\circ\sigma}=\left(  -1\right)  ^{\sigma}a_{\alpha}$, where $\left(
-1\right)  ^{\sigma}$ denotes the sign of a permutation $\sigma$.
\end{enumerate}
\end{lemma}

\begin{proof}
Recall that $a_{\alpha}=\det\left(  \left(  x_{i}^{\alpha_{j}}\right)
_{i,j\in\left[  N\right]  }\right)  $, and use the fact that the determinant
of a matrix is alternating in its columns. (See \cite[Lemma 7.3.39]{21s} for
detailed proofs of parts \textbf{(a)} and \textbf{(b)}. Part \textbf{(c)}
follows from \cite[second equality in Corollary 6.4.15]{21s}.)
\end{proof}

\begin{proof}
[Beginning of the proof of Lemma~\ref{lem.schur.stembridge}.]\leavevmode


\nobreak\noindent\renewcommand{\qedsymbol}{} This proof is due to Stembridge
\cite{Stembr02}, although Gasharov gave a less elementary proof in 1998 based
on some of the same ideas \cite{Gashar98}.

We will use the notation $\beta_{i}$ for the $i$-th entry of any tuple $\beta$.

Recall that any $\sigma\in S_{N}$ and any $f,g\in\mathcal{P}$ satisfy%
\begin{equation}
\sigma\cdot\left(  fg\right)  =\left(  \sigma\cdot f\right)  \cdot\left(
\sigma\cdot g\right)  . \label{pf.lem.schur.stembridge.sifg}%
\end{equation}


For any $\beta\in\mathbb{N}^{N}$, we have%
\begin{align}
a_{\beta} &  =\det\left(  \left(  x_{i}^{\beta_{j}}\right)  _{i,j\in\left[
N\right]  }\right)  =\det\left(  \left(  x_{j}^{\beta_{i}}\right)
_{i,j\in\left[  N\right]  }\right)  \qquad\left(
\begin{array}
[c]{c}%
\text{since }\det\left(  A^{T}\right)  =\det A\\
\text{for any matrix }A
\end{array}
\right)  \nonumber\\
&  =\sum_{\sigma\in S_{N}}\underbrace{\left(  -1\right)  ^{\sigma}}_{\text{the
sign of }\sigma}\underbrace{x_{\sigma\left(  1\right)  }^{\beta_{1}}%
x_{\sigma\left(  2\right)  }^{\beta_{2}}\cdots x_{\sigma\left(  N\right)
}^{\beta_{N}}}_{\substack{=\sigma\cdot\left(  x_{1}^{\beta_{1}}x_{2}%
^{\beta_{2}}\cdots x_{N}^{\beta_{N}}\right)  \\=\sigma\cdot x^{\beta}%
}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the definition of}\\
\text{a determinant}%
\end{array}
\right)  \nonumber\\
&  =\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\sigma\cdot x^{\beta
}.\label{pf.lem.schur.stembridge.ab=}%
\end{align}
Applying this to $\beta=\nu+\rho$, we find%
\[
a_{\nu+\rho}=\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\sigma\cdot
x^{\nu+\rho}.
\]
Multiplying this by $s_{\lambda/\mu}$, we find%
\begin{align*}
a_{\nu+\rho}\cdot s_{\lambda/\mu} &  =\sum_{\sigma\in S_{N}}\left(  -1\right)
^{\sigma}\left(  \sigma\cdot x^{\nu+\rho}\right)  \cdot\underbrace{s_{\lambda
/\mu}}_{\substack{=\sigma\cdot s_{\lambda/\mu}\\\text{(since }s_{\lambda/\mu
}\text{ is symmetric)}}}\\
&  =\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\underbrace{\left(
\sigma\cdot x^{\nu+\rho}\right)  \cdot\left(  \sigma\cdot s_{\lambda/\mu
}\right)  }_{\substack{=\sigma\cdot\left(  x^{\nu+\rho}s_{\lambda/\mu}\right)
\\\text{(by (\ref{pf.lem.schur.stembridge.sifg}))}}}\\
&  =\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\sigma\cdot\left(
x^{\nu+\rho}s_{\lambda/\mu}\right)  .
\end{align*}
Since%
\begin{align*}
x^{\nu+\rho}s_{\lambda/\mu} &  =x^{\nu+\rho}\sum_{T\in\operatorname*{SSYT}%
\left(  \lambda/\mu,N\right)  }x^{\operatorname*{cont}T}=\sum_{T\in
\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  }\underbrace{x^{\nu+\rho
}x^{\operatorname*{cont}T}}_{=x^{\nu+\operatorname*{cont}T+\rho}}\\
&  =\sum_{T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  }%
x^{\nu+\operatorname*{cont}T+\rho},
\end{align*}
we can rewrite this as%
\begin{align}
a_{\nu+\rho}\cdot s_{\lambda/\mu} &  =\sum_{\sigma\in S_{N}}\left(  -1\right)
^{\sigma}\sigma\cdot\sum_{T\in\operatorname*{SSYT}\left(  \lambda
/\mu,N\right)  }x^{\nu+\operatorname*{cont}T+\rho}\nonumber\\
&  =\sum_{T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  }%
\ \ \underbrace{\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\sigma\cdot
x^{\nu+\operatorname*{cont}T+\rho}}_{\substack{=a_{\nu+\operatorname*{cont}%
T+\rho}\\\text{(by (\ref{pf.lem.schur.stembridge.ab=}), applied}%
\\\text{backwards to }\beta=\nu+\operatorname*{cont}T+\rho\text{)}%
}}\nonumber\\
&  =\sum_{T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  }%
a_{\nu+\operatorname*{cont}T+\rho}%
.\label{pf.lem.schur.stembridge.before-cancel}%
\end{align}
This is \textbf{almost} the sum we want; we just want the right-hand side to
be%
\begin{equation}
\sum_{\substack{T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)
;\\T\text{ is }\nu\text{-Yamanouchi}}}a_{\nu+\operatorname*{cont}T+\rho
}\label{pf.lem.schur.stembridge.goal}%
\end{equation}
instead. All that remains is to prove that these two sums are equal, i.e.,
that the non-$\nu$-Yamanouchi tableaux in the sum in
(\ref{pf.lem.schur.stembridge.before-cancel}) cancel out. To be precise, we
must show that%
\begin{equation}
\sum_{\substack{T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)
;\\T\text{ is not }\nu\text{-Yamanouchi}}}a_{\nu+\operatorname*{cont}T+\rho
}=0.\label{pf.lem.schur.stembridge.bad-sum}%
\end{equation}
But we will achieve this by cancelling addends that are equal-up-to-sign and
throwing out addends that are zero.

The cancellation, as so often, is produced by a sign-reversing involution. Let
$\mathcal{X}$ be the set of all $T\in\operatorname*{SSYT}\left(  \lambda
/\mu,N\right)  $ that are not $\nu$-Yamanouchi. We shall find an involution
\[
f:\mathcal{X}\rightarrow\mathcal{X}%
\]
with the properties that

\begin{enumerate}
\item if a tableau $T\in\mathcal{X}$ satisfies $f\left(  T\right)  =T$, then
$a_{\nu+\operatorname*{cont}T+\rho}=0$;

\item for all $T\in\mathcal{X}$, we have $a_{\nu+\operatorname*{cont}T+\rho
}+a_{\nu+\operatorname*{cont}\left(  f\left(  T\right)  \right)  +\rho}=0$.
\end{enumerate}

If such an involution $f$ is found, then the sum
(\ref{pf.lem.schur.stembridge.bad-sum}) will break up into addends that are
simply $0$ (the ones corresponding to $T$ that satisfy $f\left(  T\right)
=T$) and pairs of addends that cancel one another (the ones corresponding to
all the other $T$'s and their respective $f\left(  T\right)  $'s). As a
consequence, the whole sum (\ref{pf.lem.schur.stembridge.bad-sum}) will
therefore be $0$. (This argument is spelt out in more detail in \cite[end of
proof of Lemma 7.3.34]{21s}.)

To define our involution $f$, let us analyze the non-$\nu$-Yamanouchi tableaux
$T\in\mathcal{X}$ more closely. Let $T\in\mathcal{X}$ be such a tableau. Then,
there exists at least one $j\geq1$ such that $\nu+\operatorname*{cont}\left(
\operatorname{col}_{\geq j}T\right)  $ is \textbf{not} an $N$-partition (since
$T$ is not $\nu$-Yamanouchi). Any such $j$ will be called a \textbf{violator}
of $T$. Note that $T$ has only finitely many violators (since
$\operatorname{col}_{\geq j}T$ is empty for all sufficiently large $j$, so its
content is the zero $N$-tuple and therefore $\nu+\operatorname*{cont}%
(\operatorname{col}_{\geq j}T)=\nu$, which is an $N$-partition).

Let $j$ be the \textbf{largest} violator of $T$. Define the two $N$-tuples
\begin{align*}
b  &  :=\nu+\operatorname*{cont}\left(  \operatorname{col}_{\geq j}T\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{not an }N\text{-partition}\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\\
c  &  :=\nu+\operatorname*{cont}\left(  \operatorname{col}_{\geq j+1}T\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{an }N\text{-partition, since }j\text{ is
largest}\right)  .
\end{align*}
Since $b$ is not an $N$-partition, there exists a $k\in\left[  N-1\right]  $
such that $b_{k}<b_{k+1}$. Call such a $k$ a \textbf{misstep} of $T$. Let $k$
be the \textbf{smallest} misstep of $T$. Thus,%
\begin{align*}
b_{1}  &  \geq b_{2}\geq\cdots\geq b_{k}<b_{k+1}%
\ \ \ \ \ \ \ \ \ \ \text{whereas}\\
c_{1}  &  \geq c_{2}\geq\cdots\geq c_{k}\geq c_{k+1}\geq\cdots\geq
c_{N}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }c\text{ is an }%
N\text{-partition}\right)  .
\end{align*}
\Needspace{10\baselineskip}

\begin{example}
\label{exa.schur.stembridge-violators}Let $N=4$, $\nu=(1,1,0,0)$,
$\lambda=(4,4)$ and $\mu=\varnothing$. Consider the SSYT
\[
T=\begin{ytableau}
1 & 2 & 2 & 3\\
2 & 3 & 4 & 4
\end{ytableau}\ .
\]
For each positive integer $r$, write
\[
C_{r}:=\nu+\operatorname*{cont}\left(  \operatorname{col}_{\geq r}T\right)  .
\]
Starting with $\nu$ and adding the columns of $T$ from right to left, we get
\[%
\begin{array}
[c]{c|ccccc}%
r & \geq5 & 4 & 3 & 2 & 1\\\hline
C_{r} & (1,1,0,0) & (1,1,1,1) & \mathbf{(1,2,1,2)} & \mathbf{(1,3,2,2)} &
\mathbf{(2,4,2,2)}%
\end{array}
\]
(the boldfaced tuples are not weakly decreasing). Thus, the violators of $T$
are $1,2,3$, and we choose the largest one: $j=3$. This gives
\[
b=C_{3}=(1,2,1,2),\qquad c=C_{4}=(1,1,1,1).
\]
Now $b_{1}<b_{2}$ and $b_{3}<b_{4}$, whereas $b_{2}\geq b_{3}$. Thus, the
missteps are $1$ and $3$, and we choose the smallest one: $k=1$. Notice the
distinction: $j$ is a column number in $T$, whereas $k$ is an index of an
inequality between consecutive entries of $b$.
\end{example}

We want to define $f\left(  T\right)  $ to be again a non-$\nu$-Yamanouchi
SSYT of shape $\lambda/\mu$ with the same largest violator $j$ and the same
smallest misstep $k$, but we want it to satisfy $a_{\nu+\operatorname*{cont}%
T+\rho}+a_{\nu+\operatorname*{cont}\left(  f\left(  T\right)  \right)  +\rho
}=0$. The simplest way to achieve the latter equality is to have
$\nu+\operatorname*{cont}T+\rho$ and $\nu+\operatorname*{cont}\left(  f\left(
T\right)  \right)  +\rho$ differ by a switch of two entries -- let's even make
it two consecutive entries, in order for the respective entries of $\rho$ to
differ only by $1$.

It stands to reason that $f\left(  T\right)  $ should agree with $T$ in all
columns $j,j+1,j+2,\ldots$. Indeed, if this is the case, then it is automatic
that $f\left(  T\right)  $ still has largest violator $j$ and smallest misstep
$k$ (since $\nu+\operatorname*{cont}\left(  \operatorname{col}_{\geq
i}T\right)  $ remains unchanged from $T$ to $f\left(  T\right)  $ for all
$i\geq j$, and therefore $j$ remains the largest violator; moreover, $k$
remains the smallest misstep, since $\nu+\operatorname*{cont}\left(
\operatorname{col}_{\geq j}T\right)  $ remains unchanged). So let us try to
define the tableau $f\left(  T\right)  $ in such a way that it differs from
$T$ only in the first $j-1$ columns.

Let $\operatorname{col}_{<j}T$ be the tableau obtained from $T$ by removing
all the columns $j,j+1,j+2,\ldots$ (and leaving all remaining columns in their
places). This is again an SSYT of skew shape (namely, of shape $\overline
{\lambda}/\overline{\mu}$, where $\overline{\lambda}_{i}=\min\left\{
j-1,\lambda_{i}\right\}  $ and $\overline{\mu}_{i}=\min\left\{  j-1,\mu
_{i}\right\}  $ for each $i$). (If $j=1$, this is the empty tableau.)

Before we define $f\left(  T\right)  $, let us make some observations. The
$N$-tuples $c=\nu+\operatorname*{cont}\left(  \operatorname{col}_{\geq
j+1}T\right)  $ and $b=\nu+\operatorname*{cont}\left(  \operatorname{col}%
_{\geq j}T\right)  $ differ only in the fact that the $j$-th column of $T$ is
counted in $b$ but not in $c$. Hence, for each $i\in\left[  N\right]  $, we
have%
\[
c_{i}\leq b_{i}\leq c_{i}+1
\]
(since the $j$-th column of $T$ contains at most $1$ copy of $i$). So, in
particular,%
\[
c_{k}\leq b_{k}\leq c_{k}+1\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ c_{k+1}\leq b_{k+1}\leq c_{k+1}+1.
\]
But as we have seen above,%
\[
c_{k}\geq c_{k+1}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ b_{k}%
<b_{k+1}.
\]
These inequalities give the chain
\[
c_{k}\leq b_{k}<b_{k+1}\leq c_{k+1}+1\leq c_{k}+1\ \ \ \ \ \ \ \ \ \ \left(
\text{since }c_{k}\geq c_{k+1}\right)  .
\]
Since all entries here are integers, this entails that all $\leq$ signs must
be $=$ signs (because there are no integers lying strictly between $c_{k}$ and
$c_{k}+1$), and therefore%
\[
c_{k}=c_{k+1},\ \ \ \ \ \ \ \ \ \ b_{k}=c_{k},\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ b_{k+1}=c_{k+1}+1.
\]
Hence, the $j$-th column of $T$ has no $k$ (since $b_{k}=c_{k}$) and one $k+1$
(since $b_{k+1}=c_{k+1}+1$). Moreover,%
\[
b_{k+1}=\underbrace{c_{k+1}}_{=c_{k}=b_{k}}+\,1=b_{k}+1.
\]


Now, we define the tableau $f\left(  T\right)  $ by keeping columns
$j,j+1,j+2,\ldots$ of $T$ unchanged while applying the Bender--Knuth
involution $\beta_{k}$ (from the proof of Theorem
\ref{thm.schur.skew-symmetry} in Lecture 7) to the remaining columns
$1,2,\ldots,j-1$ (that is, replacing $\operatorname{col}_{<j}T$ by $\beta
_{k}\left(  \operatorname{col}_{<j}T\right)  $). We must now prove five claims:

\begin{enumerate}
\item The resulting tableau $f\left(  T\right)  $ is still semistandard.

\item This tableau $f\left(  T\right)  $ again belongs to $\mathcal{X}$.

\item The resulting map $f$ is an involution, i.e., we have $f\left(  f\left(
T\right)  \right)  =T$.

\item We have $a_{\nu+\operatorname*{cont}T+\rho}+a_{\nu+\operatorname*{cont}%
\left(  f\left(  T\right)  \right)  +\rho}=0$.

\item If $f\left(  T\right)  =T$, then $a_{\nu+\operatorname*{cont}T+\rho}=0$.
\end{enumerate}

\Needspace{10\baselineskip}

\begin{example}
\label{exa.schur.stembridge-involution}Continue Example
\ref{exa.schur.stembridge-violators}, where $j=3$ and $k=1$. We apply
$\beta_{1}$ only to the first two columns:
\[
\operatorname{col}_{<3}T=\begin{ytableau}
1 & \mathbf{2}\\
2 & 3
\end{ytableau}\quad\longmapsto\quad\beta_{1}\left(  \operatorname{col}%
_{<3}T\right)  =\begin{ytableau}
1 & \mathbf{1}\\
2 & 3
\end{ytableau}\ .
\]
The $1$ and $2$ in the first column are matched and stay unchanged. The only
free $1$ or $2$ in $\operatorname{col}_{<j}T$ is the $2$ in cell $(1,2)$, so
it becomes a $1$. The entry $3$ is unaffected. Restoring the last two columns,
we obtain \begingroup\definecolor{stembridgefixed}{gray}{0.88}
\[
T=\begin{ytableau}
1 & \mathbf{2} & *(stembridgefixed) 2 & *(stembridgefixed) 3\\
2 & 3 & *(stembridgefixed) 4 & *(stembridgefixed) 4
\end{ytableau}\quad\longmapsto\quad f(T)=\begin{ytableau}
1 & \mathbf{1} & *(stembridgefixed) 2 & *(stembridgefixed) 3\\
2 & 3 & *(stembridgefixed) 4 & *(stembridgefixed) 4
\end{ytableau}\ .
\]
\endgroup The shaded columns are left unchanged; the boldfaced entry is the
only entry that changes. Both tableaux are semistandard. Since columns $3$ and
$4$ agree, the largest violator is still $j=3$ and the smallest misstep is
still $k=1$. Applying the procedure again changes the free $1$ back to $2$, so
$f(f(T))=T$ in this example.

The other violators need not stay the same: for $f(T)$, the tallies at $r=2$
and $r=1$ are $(2,2,2,2)$ and $(3,3,2,2)$, respectively. Thus, $f(T)$ has only
one violator, namely $3$.

Finally, let us check the cancellation of the corresponding alternants. The
contents are
\[
\operatorname*{cont}T=(1,3,2,2),\qquad\operatorname*{cont}(f(T))=(2,2,2,2).
\]
Since $\rho=(3,2,1,0)$ and $\nu=(1,1,0,0)$, we obtain
\begin{align*}
\nu+\operatorname*{cont}T+\rho &  =(5,6,3,2),\\
\nu+\operatorname*{cont}(f(T))+\rho &  =(6,5,3,2).
\end{align*}
These tuples differ by swapping their first two entries. Hence, Lemma
\ref{lem.schur.alternant-swap} \textbf{(b)} gives $a_{(5,6,3,2)}%
=-a_{(6,5,3,2)}$, and finally we have%
\[
a_{\nu+\operatorname*{cont}T+\rho}+a_{\nu+\operatorname*{cont}(f(T))+\rho
}=a_{(5,6,3,2)}+a_{(6,5,3,2)}=0.
\]

\end{example}

\begin{proof}
[Proof of Claim 1.]Assume the contrary. Thus, $T$ is semistandard but
$f\left(  T\right)  $ is not. If $j=1$, then no entries change from $T$ to
$f\left(  T\right)  $, so this is clearly impossible. Thus, we must have
$j>1$. Since $f\left(  T\right)  $ is not semistandard, we must have $\left(
f\left(  T\right)  \right)  \left(  i,j-1\right)  >\left(  f\left(  T\right)
\right)  \left(  i,j\right)  $ for some $i$ for which both cells $(i,j-1)$ and
$(i,j)$ belong to $Y\left(  \lambda/\mu\right)  $ (because all the other
semistandardness inequalities for $f\left(  T\right)  $ are guaranteed either
by the semistandardness of $T$ or by the semistandardness of $\beta_{k}\left(
\operatorname{col}_{<j}T\right)  $). But $T\left(  i,j-1\right)  \leq T\left(
i,j\right)  $ (since $T$ is semistandard). So this inequality $T\left(
i,j-1\right)  \leq T\left(  i,j\right)  $ must have flipped to $\left(
f\left(  T\right)  \right)  \left(  i,j-1\right)  >\left(  f\left(  T\right)
\right)  \left(  i,j\right)  $.

Since $\left(  f\left(  T\right)  \right)  \left(  i,j\right)  =T\left(
i,j\right)  $ (because $f\left(  T\right)  $ only differs from $T$ in the
first $j-1$ columns), this can only happen if the $\left(  i,j-1\right)  $-th
entry has increased from $T$ to $f\left(  T\right)  $. Since this entry is
affected only by $\beta_{k}$ applied to $\operatorname{col}_{<j}T$, this means
that it would have increased from $k$ to $k+1$. That is, we must have
$T\left(  i,j-1\right)  =k$ and $\left(  f\left(  T\right)  \right)  \left(
i,j-1\right)  =k+1$. Combining the original row inequality with the assumed
failure of the new one, we obtain%
\[
k=T\left(  i,j-1\right)  \leq T\left(  i,j\right)  =\left(  f\left(  T\right)
\right)  \left(  i,j\right)  <\left(  f\left(  T\right)  \right)  \left(
i,j-1\right)  =k+1.
\]
Hence, $T\left(  i,j\right)  =k$, since this entry is an integer.

But the $j$-th column of $T$ has no $k$ (as we saw above), so $T\left(
i,j\right)  \neq k$. This contradiction shows that our assumption was false,
and so Claim 1 is proved.
\end{proof}

\begin{proof}
[Proof of Claim 2.]As we have said above, $f\left(  T\right)  $ agrees with
$T$ in all columns $j,j+1,j+2,\ldots$. Thus, $f\left(  T\right)  $ still has
largest violator $j$, hence in particular has a violator; that is, $f\left(
T\right)  $ is not $\nu$-Yamanouchi. Therefore, $f\left(  T\right)
\in\mathcal{X}$ (since Claim 1 already shows that $f\left(  T\right)  $ is
semistandard). This proves Claim 2.
\end{proof}

We will finish the other three claims next time.
\end{proof}

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[21s]{21s}Darij Grinberg, \textit{An Introduction to Algebraic
Combinatorics (Math 531, Winter 2024 lecture notes)}, 22 July 2026.\newline\url{http://www.cip.ifi.lmu.de/~grinberg/t/21s/lecs.pdf}

\bibitem[Gashar98]{Gashar98}%
\href{https://doi.org/10.1006/eujc.1998.0212}{Vesselin Gasharov, \textit{A
Short Proof of the Littlewood--Richardson Rule}, Europ. J. Combinatorics
(1998) \textbf{19}, pp. 451--453}.

\bibitem[Stembr02]{Stembr02}\href{https://doi.org/10.37236/1666}{John R.
Stembridge, \textit{A Concise Proof of the Littlewood-Richardson Rule},
Electronic Journal of Combinatorics \textbf{9} (2002), \#N5}
\end{thebibliography}


\end{document}