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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 polish and detail some of the arguments below.
\textbf{Unchecked GPT output included currently; to be fixed.}) \medskip

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

Recall:

\begin{definition}
\label{def.schur.yamanouchi-recall}Let $\lambda/\mu$ be a skew partition, and
let $\nu$ be an $N$-partition. A tableau $T\in\operatorname*{SSYT}\left(
\lambda/\mu,N\right)  $ is said to be $\nu$\textbf{-Yamanouchi} if it has the
property that for each positive integer $j$, the $N$-tuple $\nu
+\operatorname*{cont}\left(  \operatorname{col}_{\geq j}T\right)  $ is an
$N$-partition (i.e., weakly decreasing).
\end{definition}

We want to prove:

\begin{theorem}
[Zelevinsky's generalized Littlewood--Richardson rule, in Yamanouchi form for
finite $N$]\label{thm.schur.generalized-lr-recall}\ \ 

Let $\lambda/\mu$ be a skew partition and let $\nu$ be an $N$-partition. Then,%
\[
s_{\nu}\cdot s_{\lambda/\mu}=\sum_{\substack{T\in\operatorname*{SSYT}\left(
\lambda/\mu,N\right)  ;\\T\text{ is }\nu\text{-Yamanouchi}}}s_{\nu
+\operatorname*{cont}T}.
\]

\end{theorem}

Instead of proving this theorem, we will first show 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}

But before we show this, let me explain why this lemma implies both the LR
(Littlewood--Richardson) rule and the $s_{\lambda}\cdot a_{\rho}%
=a_{\lambda+\rho}$ theorem (for $\ell(\lambda)\leq N$) that we have long been
trying to prove.

To get the latter, we just specialize Stembridge's lemma:

\begin{proof}
[Proof of the bialternant identity using Stembridge's lemma.]Let $\lambda$ be
a partition of length $\leq N$, identified with an $N$-partition by padding it
with zeroes. Set $\mu=\varnothing$ and $\nu=\varnothing=(0,0,\ldots,0)$ in
Lemma~\ref{lem.schur.stembridge}. We obtain%
\[
a_{\rho}\cdot s_{\lambda}=\sum_{\substack{T\in\operatorname*{SSYT}\left(
\lambda,N\right)  ;\\T\text{ is }\varnothing\text{-Yamanouchi}}%
}a_{\operatorname*{cont}T+\rho}.
\]
But what must a straight tableau $T\in\operatorname*{SSYT}\left(
\lambda,N\right)  $ satisfy in order to be $\varnothing$-Yamanouchi? We can
argue column-by-column (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 column
$j$ have already been shown to have this form. 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 is straight, $r\leq h$. For each $i\leq
r$, strict increase down column $j$ gives $T(i,j)\geq i$, while weak increase
along row $i$ gives $T(i,j)\leq T(i,j+1)=i$. Thus, the first $r$ entries in
column $j$ are $1,2,\ldots,r$.

All entries in the columns to the right are $\leq r$. Therefore, the remaining
entries in column $j$ must be $r+1,r+2,\ldots,h$: otherwise the content of
$\operatorname{col}_{\geq j}T$ would have a zero coordinate with index $>r$
followed by a nonzero coordinate. This would contradict its being weakly
decreasing. This argument also covers the rightmost column, where $r=0$, and
proves the claim by induction. If $\lambda$ is empty, there are no columns to check.

So $T$ must be the tableau of shape $\lambda$ whose entries $T\left(
i,j\right)  $ are simply given by $T\left(  i,j\right)  =i$. This is the
tableau I call the \textbf{minimalistic tableau} of shape $\lambda$; 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). Conversely, this
minimalistic tableau really is $\varnothing$-Yamanouchi: the $i$-th entry of
$\operatorname*{cont}(\operatorname{col}_{\geq j}T)$ is $\max\{\lambda
_{i}-j+1,0\}$, and these entries are weakly decreasing in $i$. Its entries
belong to $[N]$ since $\ell(\lambda)\leq N$, and its content is
$\operatorname*{cont}T=\lambda$. Hence,%
\[
\sum_{\substack{T\in\operatorname*{SSYT}\left(  \lambda,N\right)  ;\\T\text{
is }\varnothing\text{-Yamanouchi}}}a_{\operatorname*{cont}T+\rho}%
=a_{\lambda+\rho},
\]
and thus our above equality shows that $a_{\rho}\cdot s_{\lambda}%
=a_{\lambda+\rho}$, provided that Stembridge's lemma holds.
\end{proof}

\begin{proof}
[Proof of the LR rule.]We have assumed that Stembridge's lemma holds, i.e., we
have%
\[
a_{\nu+\rho}\cdot s_{\lambda/\mu}=\sum_{\substack{T\in\operatorname*{SSYT}%
\left(  \lambda/\mu,N\right)  ;\\T\text{ is }\nu\text{-Yamanouchi}}%
}a_{\nu+\operatorname*{cont}T+\rho}.
\]
By what we just proved, $a_{\nu+\rho}=a_{\rho}\cdot s_{\nu}$ and
$a_{\nu+\operatorname*{cont}T+\rho}=a_{\rho}\cdot s_{\nu+\operatorname*{cont}%
T}$ (the latter applies because $\nu+\operatorname*{cont}T$ is an
$N$-partition whenever $T$ is $\nu$-Yamanouchi). Thus, this equality rewrites
as%
\[
a_{\rho}\cdot s_{\nu}\cdot s_{\lambda/\mu}=\sum_{\substack{T\in
\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  ;\\T\text{ is }%
\nu\text{-Yamanouchi}}}a_{\rho}\cdot s_{\nu+\operatorname*{cont}T}.
\]
Cancelling $a_{\rho}$ would give us precisely the LR rule.

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)  $ (Vandermonde determinant) is nonzero, so it
can be cancelled. In the general case, we can argue as follows:

\begin{itemize}
\item Either argue that the LR rule is true for $K=\mathbb{Z}$ (which is an
integral domain) and therefore for arbitrary $K$ (since there is a functorial
morphism $\mathcal{P}_{\mathbb{Z}}\rightarrow\mathcal{P}_{K}$ that sends skew
Schur polynomials in $\mathcal{P}_{\mathbb{Z}}$ to skew Schur polynomials in
$\mathcal{P}_{K}$).

\item Or argue that multiplication by $a_{\rho}=\prod_{1\leq i<j\leq N}%
(x_{i}-x_{j})$ is injective 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 even over a ring
with zero divisors: the leading nonzero coefficient of a nonzero polynomial
stays nonzero in the product. Multiplication by the product of these factors
is therefore injective as well.
\end{itemize}

Either way, the LR rule is proved provided that we can prove Stembridge's lemma.
\end{proof}

So it remains to prove Stembridge's lemma. We prepare with a simple
observation about alternants:

\begin{lemma}
\label{lem.schur.alternant-swap}Let $\alpha\in\mathbb{N}^{N}$.

\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 $a_{\beta}=-a_{\alpha}$.
\end{enumerate}
\end{lemma}

\begin{proof}
Recall that $a_{\alpha}=\det\left(  x_{i}^{\alpha_{j}}\right)  $ and use the
fact that the determinant of a matrix is alternating in its columns.
\end{proof}

\begin{proof}
[Beginning of the proof of Lemma~\ref{lem.schur.stembridge}.]%
\renewcommand{\qedsymbol}{} This proof is due to
\href{https://doi.org/10.37236/1666}{Stembridge (2002)}. Gasharov gave a less
elementary proof in 1998 based on some of the same ideas.

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

Keep in mind that $\sigma\cdot\left(  fg\right)  =\left(  \sigma\cdot
f\right)  \cdot\left(  \sigma\cdot g\right)  $ for any $\sigma\in S_{N}$ and
$f,g\in\mathcal{P}$.

For any $\beta\in\mathbb{N}^{N}$, we have%
\begin{align*}
a_{\beta}  &  =\det\left(  x_{i}^{\beta_{j}}\right)  =\det\left(  x_{j}%
^{\beta_{i}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\det\left(
A^{T}\right)  =\det A\right) \\
&  =\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}}}\\
&  =\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\sigma\cdot x^{\beta}.
\end{align*}
So%
\[
a_{\nu+\rho}=\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\sigma\cdot
x^{\nu+\rho}.
\]
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)  }_{=\sigma\cdot\left(  x^{\nu+\rho}s_{\lambda/\mu}\right)  }\\
&  =\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\sigma\cdot\left(
x^{\nu+\rho}s_{\lambda/\mu}\right)  .
\end{align*}
Since%
\[
x^{\nu+\rho}s_{\lambda/\mu}=x^{\nu+\rho}\sum_{T\in\operatorname*{SSYT}\left(
\lambda/\mu,N\right)  }x^{\operatorname*{cont}T}=\sum_{T\in
\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  }x^{\nu+\operatorname*{cont}%
T+\rho},
\]
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}\\
&  =\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 the above }a_{\beta}=\sum_{\sigma\in S_{N}}\left(
-1\right)  ^{\sigma}\sigma\cdot x^{\beta}\\\text{formula, applied backwards)}%
}}\\
&  =\sum_{T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  }%
a_{\nu+\operatorname*{cont}T+\rho}.
\end{align*}
This is \textbf{almost} the sum we want; we want%
\[
\sum_{\substack{T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)
;\\T\text{ is }\nu\text{-Yamanouchi}}}a_{\nu+\operatorname*{cont}T+\rho}.
\]
All that remains is to prove that these two sums are equal, i.e., that the
non-$\nu$-Yamanouchi tableaux in the above sum cancel out. To be precise, we
must show that%
\[
\sum_{\substack{T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)
;\\T\text{ is not }\nu\text{-Yamanouchi}}}a_{\nu+\operatorname*{cont}T+\rho
}=0.
\]
But we will achieve this by cancellation and throwing out zero addends.

Let $\mathcal{X}$ be the set of all $T\in\operatorname*{SSYT}\left(
\lambda/\mu,N\right)  $ that are not $\nu$-Yamanouchi. We must find an
involution
\[
f:\mathcal{X}\rightarrow\mathcal{X}%
\]
with the properties that

\begin{enumerate}
\item if $f\left(  T\right)  =T$, 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 above 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 together sum up to $0$ (the
ones corresponding to all the other $T$'s and their respective $f\left(
T\right)  $'s).

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} a partition. 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
$\nu+\operatorname*{cont}(\operatorname{col}_{\geq j}T)=\nu$, which is a partition).

Let $j$ be the \textbf{largest} violator of $T$. Set
\begin{align*}
b  &  :=\nu+\operatorname*{cont}\left(  \operatorname{col}_{\geq j}T\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{not a partition}\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\\
c  &  :=\nu+\operatorname*{cont}\left(  \operatorname{col}_{\geq j+1}T\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{a partition, since }j\text{ is
largest}\right)  .
\end{align*}
Since $b$ is not a 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}.
\end{align*}
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.

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 nothing has changed that would affect these conditions). So let us
try to obtain $f\left(  T\right)  $ from $T$ by only changing the first $j-1$
columns. We consider the tableau $\operatorname{col}_{<j}T$ obtained from $T$
by removing all the columns $j,j+1,j+2,\ldots$, without moving the remaining
cells. This is again an SSYT of skew shape: its outer and inner partitions
have entries $\min(\lambda_{i},j-1)$ and $\min(\mu_{i},j-1)$, respectively. If
$j=1$, this is the empty tableau.

Before we define $f\left(  T\right)  $, let us make some observations. The
$N$-tuples $c$ and $b$ have the property that for each $i\in\left[  N\right]
$, we have%
\[
c_{i}\leq b_{i}\leq c_{i}+1
\]
(since $b$ differs from $c$ only in that it includes the $j$-th column of $T$,
but this column can contain 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%
\[
c_{k}\geq c_{k+1}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ b_{k}%
<b_{k+1}.
\]
These six inequalities (actually four of them suffice) give the chain
\[
c_{k}\leq b_{k}<b_{k+1}\leq c_{k+1}+1\leq c_{k}+1.
\]
Since all entries are integers, this entails that%
\[
c_{k}=c_{k+1},\ \ \ \ \ \ \ \ \ \ b_{k}=c_{k},\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ b_{k+1}=c_{k+1}+1.
\]
The latter two equalities show that the $j$-th column of $T$ has no $k$ and
one $k+1$. In particular, $b_{k+1}=b_{k}+1$.

Now, we define $f\left(  T\right)  $ by keeping columns $j,j+1,j+2,\ldots$ of
$T$ unchanged while replacing $\operatorname{col}_{<j}T$ by $\beta_{k}\left(
\operatorname{col}_{<j}T\right)  $, where $\beta_{k}$ is the Bender--Knuth
involution. We must prove four claims:

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

\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 $a_{\nu+\operatorname*{cont}T+\rho}=0$.

\item This is an involution, i.e., we have $f\left(  f\left(  T\right)
\right)  =T$.
\end{enumerate}

\smallskip\noindent\textit{Proof of Claim 1.} If $j=1$, no entries change, so
the claim is immediate. Thus, assume $j>1$. If $f\left(  T\right)  $ were not
semistandard, then we would 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 the skew diagram
(because all the other inequalities are guaranteed either by the
semistandardness of $T$ or by the semistandardness of $\beta_{k}\left(
\operatorname{col}_{<j}T\right)  $). Since $T\left(  i,j-1\right)  \leq
T\left(  i,j\right)  $ and $\left(  f\left(  T\right)  \right)  \left(
i,j\right)  =T\left(  i,j\right)  $, 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 $k+1$. But the $j$-th column of $T$ has no $k$, so $T\left(  i,j\right)
\neq k$. Therefore, the inequality $T\left(  i,j-1\right)  \leq T\left(
i,j\right)  $ could not have become false as $T\left(  i,j-1\right)  $
increased from $k$ to $k+1$. Thus, $f\left(  T\right)  $ is semistandard.
Since its columns $j,j+1,\ldots$ are unchanged, its largest violator is still
$j$ and its smallest misstep is still $k$. In particular, $f(T)\in\mathcal{X}%
$, so the map $f:\mathcal{X}\to\mathcal{X}$ is well-defined.

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


\end{document}