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\ihead{Math 701, Fall 2026, Lecture 7, version \today}
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\begin{document}
\section*{Math 701 Fall 2026, Lecture 7: Skew Schur polynomials continued}

\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 some of the examples below and to rewrite Remark
\ref{rmk.schur.yamanouchi-votes}.) \medskip

\setcounter{section}{1}\setcounter{subsection}{6}\setcounter{subsubsection}{0}\setcounter{theo}{16}

\subsubsection{The Bender--Knuth involutions (continued)}

We continue where we left off in Lecture 6:

\begin{proof}
[Finishing the proof of Theorem \ref{thm.schur.skew-symmetry}.]We continue
with the notation from Lecture 6. In particular, fix $k\in\left[  N-1\right]
$. We now define a map%
\[
\beta_{k}:\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  \rightarrow
\operatorname*{SSYT}\left(  \lambda/\mu,N\right)
\]
as follows: Given $T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  $, we
perform the following procedure:

\begin{itemize}
\item For each row of $T$, if there are $a$ free $k$'s and $b$ free $\left(
k+1\right)  $'s in this row, then we replace them by $b$ many $k$'s followed
by $a$ many $\left(  k+1\right)  $'s, from left to right in the same cells.
All other entries remain unchanged.
\end{itemize}

We define $\beta_{k}\left(  T\right)  $ to be the result of this procedure. We
still need to prove the following:

\begin{statement}
\textit{Observation 2:} This map $\beta_{k}$ is well-defined, i.e., the
tableau $\beta_{k}\left(  T\right)  $ belongs to $\operatorname*{SSYT}\left(
\lambda/\mu,N\right)  $.
\end{statement}

Before we prove this, let us give an example of $\beta_{k}$.

\begin{example}
\label{exa.schur.bk-action}Let $k=2$ and $N=4$. We use the tableau $T$ from
Example \ref{exa.schur.bk-matched-free} in Lecture 6, of shape $\left(
9,8,6\right)  /\left(  3,1\right)  $. The map $\beta_{2}$ acts as follows:
\begingroup
\definecolor{bkmatched}{gray}{0.85} \ytableausetup{boxsize=1.15em}
\[
T=\begin{ytableau}
\none & \none & \none & 1 & 1 & 1 & *(bkmatched) \mathbf{2} & \mathbf{3} & 4\\
\none & 1 & *(bkmatched) \mathbf{2} & \mathbf{2} & \mathbf{2} & \mathbf{3} & *(bkmatched) \mathbf{3} & 4\\
1 & \mathbf{2} & *(bkmatched) \mathbf{3} & 4 & 4 & 4
\end{ytableau}
\quad\longmapsto\quad\beta_{2}(T)=\begin{ytableau}
\none & \none & \none & 1 & 1 & 1 & *(bkmatched) \mathbf{2} & \mathbf{2} & 4\\
\none & 1 & *(bkmatched) \mathbf{2} & \mathbf{2} & \mathbf{3} & \mathbf{3} & *(bkmatched) \mathbf{3} & 4\\
1 & \mathbf{3} & *(bkmatched) \mathbf{3} & 4 & 4 & 4
\end{ytableau}\ .
\]
\endgroup
As before, the $2$'s and $3$'s are boldfaced, and the matched entries are
shaded. The shaded entries stay unchanged. The free blocks change as follows:
\[%
\begin{array}
[c]{c@{\qquad}c@{\qquad}c}%
\text{row }1 & \text{row }2 & \text{row }3\\[4pt]%
(3)\longmapsto(2) & (2,2,3)\longmapsto(2,3,3) & (2)\longmapsto(3)
\end{array}
\]
All entries other than $2$ and $3$ stay unchanged as well. Thus, the five
$2$'s and four $3$'s in $T$ become four $2$'s and five $3$'s in $\beta_{2}%
(T)$. Notice that in the second row, we replace $223$ by $233$, not by $332$:
the new free block must still be weakly increasing.
\end{example}

\begin{proof}
[Proof of Observation 2.]The entries of $\beta_{k}\left(  T\right)  $ are
clearly in $\left[  N\right]  $, since $k,k+1\in\left[  N\right]  $. It
remains to show that the tableau $\beta_{k}\left(  T\right)  $ is semistandard.

\begin{itemize}
\item Why are its rows weakly increasing? Consider a row of $T$. In $\beta
_{k}\left(  T\right)  $, some of its free entries may have changed from $k$ to
$k+1$ or vice versa; no other entries changed. But Observation 1 from Lecture
6 shows that the matched $k$'s are to the left of the free $k$'s and the
matched $\left(  k+1\right)  $'s are to the right of the free $\left(
k+1\right)  $'s; thus, all the free entries form a single contiguous block in
our row. By definition, this block stays weakly increasing in $\beta
_{k}\left(  T\right)  $. All the entries in this block are either $k$'s or
$\left(  k+1\right)  $'s in both $T$ and $\beta_{k}\left(  T\right)  $; all
the entries to the left of this block are $\leq k$; all the entries to the
right of this block are $\geq k+1$. Thus, the row is still weakly increasing
in $\beta_{k}\left(  T\right)  $.

\item Why are its columns strictly increasing? Because we only changed free
$k$'s and free $\left(  k+1\right)  $'s, and there are no other $k$'s or
$\left(  k+1\right)  $'s in their columns to mess up the inequalities. (For
instance, changing a free $k$ to a $k+1$ still keeps it smaller than the
neighboring entry to its south (if there is one), since that entry cannot be a
$k+1$ because the $k$ was free.)
\end{itemize}

\noindent So $\beta_{k}\left(  T\right)  $ is semistandard, and Observation 2
is proved.
\end{proof}

\begin{statement}
\textit{Observation 3:} This map $\beta_{k}$ is an involution, i.e., we have
$\beta_{k}\circ\beta_{k}=\operatorname*{id}$.
\end{statement}

\begin{proof}
[Proof of Observation 3.]We must show that
\[
\beta_{k}\left(  \beta_{k}\left(  T\right)  \right)  =T\qquad\text{for each
}T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  .
\]
{}

Fix $T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  $. To compute
$\beta_{k}\left(  T\right)  $ from $T$, we identify the free and matched $k$'s
and $\left(  k+1\right)  $'s, and flip the numbers of free $k$'s and free
$\left(  k+1\right)  $'s in each row. Note that the matched entries of $T$ are
unchanged in $\beta_{k}\left(  T\right)  $ and thus, in particular, remain
matched in $\beta_{k}\left(  T\right)  $ (since their \textquotedblleft
partners\textquotedblright\ also don't change). On the other hand, the free
entries of $T$ are replaced by new entries, which are again free (since all
the other entries in their respective columns are $<k$ or $>k+1$, and thus
don't change). Hence, if a row of $T$ had $a$ free $k$'s and $b$ free $\left(
k+1\right)  $'s, then the same row of $\beta_{k}\left(  T\right)  $ has $b$
free $k$'s and $a$ free $\left(  k+1\right)  $'s.

Now, when we compute $\beta_{k}\left(  \beta_{k}\left(  T\right)  \right)  $
from $\beta_{k}\left(  T\right)  $ (by the same procedure that gave us
$\beta_{k}\left(  T\right)  $ from $T$), we flip the numbers of free $k$'s and
free $\left(  k+1\right)  $'s in each row again. Hence, if a row of $T$ had
$a$ free $k$'s and $b$ free $\left(  k+1\right)  $'s, then the same row of
$\beta_{k}\left(  T\right)  $ has $b$ free $k$'s and $a$ free $\left(
k+1\right)  $'s (as we saw above), and therefore the same row of $\beta
_{k}\left(  \beta_{k}\left(  T\right)  \right)  $ will have $a$ free $k$'s and
$b$ free $\left(  k+1\right)  $'s again, which is the same distribution that
$T$ used to have. Since the other entries in that row don't change at all,
this shows that the whole row is the same in $\beta_{k}\left(  \beta
_{k}\left(  T\right)  \right)  $ as it was in $T$. We have proved this for
every row; thus, $\beta_{k}\left(  \beta_{k}\left(  T\right)  \right)  =T$.
\end{proof}

Observation 3 shows that $\beta_{k}$ is a bijection (since any involution is a bijection).

\begin{statement}
\textit{Observation 4:} Let $T\in\operatorname*{SSYT}\left(  \lambda
/\mu,N\right)  $. Then, we have the equalities%
\begin{align*}
\left(  \text{\# of }k\text{'s in }\beta_{k}\left(  T\right)  \right)   &
=\left(  \text{\# of }\left(  k+1\right)  \text{'s in }T\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left(  \text{\# of }\left(  k+1\right)  \text{'s in }\beta_{k}\left(
T\right)  \right)   &  =\left(  \text{\# of }k\text{'s in }T\right)
\end{align*}
and furthermore%
\[
\left(  \text{\# of }i\text{'s in }\beta_{k}\left(  T\right)  \right)
=\left(  \text{\# of }i\text{'s in }T\right)  \ \ \ \ \ \ \ \ \ \ \text{for
each }i\notin\left\{  k,k+1\right\}  .
\]

\end{statement}

\begin{proof}
[Proof of Observation 4.]In the proof of Observation 3, we observed that, as
we transform $T$ into $\beta_{k}\left(  T\right)  $, all matched entries stay
matched and remain unchanged, while all free entries remain free (though they
can change their values). Thus, the matched $k$'s in $\beta_{k}\left(
T\right)  $ are precisely the matched $k$'s in $T$. Hence,%
\begin{align*}
\left(  \text{\# of matched }k\text{'s in }\beta_{k}\left(  T\right)  \right)
&  =\left(  \text{\# of matched }k\text{'s in }T\right) \\
&  =\left(  \text{\# of matched }\left(  k+1\right)  \text{'s in }T\right)
\end{align*}
(as we saw above, back in Lecture 6).

Now let $i\geq1$. Recall again from our proof of Observation 3 that when we
transform $T$ into $\beta_{k}\left(  T\right)  $, all free entries remain free
(though they can change their values), while all matched entries remain
matched. Hence, if the $i$-th row of $T$ has $a$ free $k$'s and $b$ free
$\left(  k+1\right)  $'s, then, in $\beta_{k}\left(  T\right)  $, these
entries become $b$ free $k$'s and $a$ free $\left(  k+1\right)  $'s, and there
are no further free $k$'s in the $i$-th row of $\beta_{k}\left(  T\right)  $
(since matched entries of $T$ remain matched); therefore the $i$-th row of
$\beta_{k}\left(  T\right)  $ has $b$ free $k$'s. That is,%
\begin{align*}
&  \left(  \text{\# of free }k\text{'s in the }i\text{-th row of }\beta
_{k}\left(  T\right)  \right) \\
&  =\left(  \text{\# of free }\left(  k+1\right)  \text{'s in the }i\text{-th
row of }T\right)  .
\end{align*}
Summing this equality over all $i\geq1$, we obtain%
\[
\left(  \text{\# of free }k\text{'s in }\beta_{k}\left(  T\right)  \right)
=\left(  \text{\# of free }\left(  k+1\right)  \text{'s in }T\right)  .
\]


However, each $k$ in $\beta_{k}\left(  T\right)  $ is either free or matched.
Thus,%
\begin{align*}
&  \left(  \text{\# of }k\text{'s in }\beta_{k}\left(  T\right)  \right) \\
&  =\underbrace{\left(  \text{\# of matched }k\text{'s in }\beta_{k}\left(
T\right)  \right)  }_{=\left(  \text{\# of matched }\left(  k+1\right)
\text{'s in }T\right)  }+\underbrace{\left(  \text{\# of free }k\text{'s in
}\beta_{k}\left(  T\right)  \right)  }_{\substack{=\left(  \text{\# of free
}\left(  k+1\right)  \text{'s in }T\right)  }}\\
&  =\left(  \text{\# of matched }\left(  k+1\right)  \text{'s in }T\right)
+\left(  \text{\# of free }\left(  k+1\right)  \text{'s in }T\right) \\
&  =\left(  \text{\# of }\left(  k+1\right)  \text{'s in }T\right)  .
\end{align*}
This proves the first equality of Observation 4. The second equality is
similar; the third is trivial (since no entries other than $k$ and $k+1$ change).
\end{proof}

\begin{statement}
\textit{Observation 5:} For each $T\in\operatorname*{SSYT}\left(  \lambda
/\mu,N\right)  $, we have%
\[
x_{\beta_{k}\left(  T\right)  }=s_{k}\cdot x_{T}.
\]

\end{statement}

\begin{proof}
[Proof of Observation 5.]We have%
\[
x_{T}=\prod_{i=1}^{N}x_{i}^{\left(  \text{\# of }i\text{'s in }T\right)
}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ x_{\beta_{k}\left(
T\right)  }=\prod_{i=1}^{N}x_{i}^{\left(  \text{\# of }i\text{'s in }\beta
_{k}\left(  T\right)  \right)  }.
\]
These two monomials have largely the same exponents, by Observation 4; only
the exponents on $x_{k}$ and on $x_{k+1}$ get swapped. This is precisely
saying that one monomial turns into the other when we switch the variables
$x_{k}$ and $x_{k+1}$. In other words, the action of $s_{k}$ takes one
monomial into the other. This proves Observation 5.
\end{proof}

Now, by the definition of $s_{\lambda/\mu}$, we have%
\begin{align*}
s_{\lambda/\mu}  &  =\sum_{T\in\operatorname*{SSYT}\left(  \lambda
/\mu,N\right)  }x_{T}=\sum_{T\in\operatorname*{SSYT}\left(  \lambda
/\mu,N\right)  }x_{\beta_{k}\left(  T\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\beta_{k}\left(  T\right)  \text{ for }T\text{,
since }\beta_{k}\text{ is a bijection}\\
\text{(by Observation 3)}%
\end{array}
\right) \\
&  =\sum_{T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  }s_{k}\cdot
x_{T}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Observation 5}\right) \\
&  =s_{k}\cdot\underbrace{\sum_{T\in\operatorname*{SSYT}\left(  \lambda
/\mu,N\right)  }x_{T}}_{=s_{\lambda/\mu}}=s_{k}\cdot s_{\lambda/\mu}.
\end{align*}
That is, $s_{k}\cdot s_{\lambda/\mu}=s_{\lambda/\mu}$. Since we have proved
this for each $k\in\left[  N-1\right]  $, we conclude (by Lemma
\ref{lem.symp.simple-transpositions} from Lecture 3) that $s_{\lambda/\mu}$ is
symmetric. So Theorem \ref{thm.schur.skew-symmetry} is proved.
\end{proof}

\bigskip

As we mentioned before, if $\lambda$ is an $N$-partition and $\mu
=\varnothing=\left(  {}\right)  $, then $s_{\lambda/\mu}=s_{\lambda}$, so we
conclude from Theorem \ref{thm.schur.skew-symmetry} that $s_{\lambda}$ is
symmetric. Thus, Theorem \ref{thm.schur.bialternant} \textbf{(a)} is proved.

\bigskip

\subsection{The Littlewood--Richardson rule}

What happens when we multiply two Schur polynomials? Here are two examples:%
\begin{align*}
s_{\left(  2,1\right)  }s_{\left(  1,1\right)  }  &  =s_{\left(
2,1,1,1\right)  }+s_{\left(  2,2,1\right)  }+s_{\left(  3,1,1\right)
}+s_{\left(  3,2\right)  };\\
s_{\left(  2,1\right)  }s_{\left(  2,1\right)  }  &  =s_{\left(  2,2,2\right)
}+2s_{\left(  3,2,1\right)  }+s_{\left(  3,3\right)  }+s_{\left(
4,1,1\right)  }\\
&  \qquad+s_{\left(  4,2\right)  }+s_{\left(  2,2,1,1\right)  }+s_{\left(
3,1,1,1\right)  }.
\end{align*}
In both cases, we have expanded the product as a $\mathbb{Z}$-linear
combination of Schur polynomials. The fact that such an expansion always
exists is not too surprising; indeed, the family $\left(  s_{\lambda}\right)
_{\lambda\text{ is an }N\text{-partition}}$ is a basis of $\mathcal{S}$, as we
will later show. That the same expansion works for every $N$ is also
unsurprising once we get to the $N\rightarrow\infty$ case (in the next
chapter). But the fact that there are no negative coefficients in this
expansion requires explanation! Historically, this expansion appeared in
Schur's and Weyl's work on representations of $\operatorname*{GL}\left(
N\right)  $, which explained the nonnegativity of its coefficients in terms of
representation theory. We shall instead explain the nonnegativity of the
coefficients via an entirely combinatorial formula known as the
\textbf{Littlewood--Richardson rule}. This rule was the holy grail of
algebraic combinatorics for decades in the mid-20th century, but since
Stembridge's proof in 2002 \cite{Stembr02}, it has become so accessible that
we can prove it as one of our first properties of Schur polynomials.

To state it, we will need a bit more notation.

\begin{definition}
\label{def.schur.content}Let $\lambda$ and $\mu$ be two partitions. Let
$T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  $. We define the
\textbf{content} of $T$ to be the $N$-tuple $\operatorname*{cont}T=\left(
a_{1},a_{2},\ldots,a_{N}\right)  $, where%
\[
a_{i}=\left(  \text{\# of }i\text{'s in }T\right)  =\left\vert \left\{  c\in
Y\left(  \lambda/\mu\right)  \ \mid\ T\left(  c\right)  =i\right\}
\right\vert .
\]

\end{definition}

For example, if $N=5$, then%
\[
\operatorname*{cont}\ \begin{ytableau}
1 & 1 & 2 & 4\\
3 & 3
\end{ytableau}=\left(  2,1,2,1,0\right)  .
\]


Note that
\[
x_{T}=x^{\operatorname*{cont}T}\ \ \ \ \ \ \ \ \ \ \text{for each }%
T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  ,
\]
where, for any $N$-tuple $\left(  a_{1},a_{2},\ldots,a_{N}\right)
\in\mathbb{N}^{N}$, we set $x^{\left(  a_{1},a_{2},\ldots,a_{N}\right)
}=\prod_{i=1}^{N}x_{i}^{a_{i}}$.

\begin{definition}
\label{def.schur.column-restriction}Let $\lambda/\mu$ be a skew partition. Let
$T$ be a tableau of shape $\lambda/\mu$. Let $j$ be a positive integer. Then,
$\operatorname{col}_{\geq j}T$ means the restriction of $T$ to columns
$j,j+1,j+2,\ldots$ (that is, the result of removing the first $j-1$ columns
from $T$). The remaining cells retain their original coordinates. Thus,%
\[
\operatorname{col}_{\geq j}T=T\mid_{\left\{  \left(  u,v\right)  \in Y\left(
\lambda/\mu\right)  \ \mid\ v\geq j\right\}  }.
\]
This is again a skew tableau (of shape $\widetilde{\lambda}/\widetilde{\mu}$,
where $\widetilde{\lambda}=\left(  \widetilde{\lambda}_{1},\widetilde{\lambda
}_{2},\ldots,\widetilde{\lambda}_{M}\right)  $ and $\widetilde{\mu}=\left(
\widetilde{\mu}_{1},\widetilde{\mu}_{2},\ldots,\widetilde{\mu}_{M}\right)  $
are two partitions given by $M=\ell\left(  \lambda\right)  $ and
$\widetilde{\lambda}_{i}=\max\left\{  j-1,\lambda_{i}\right\}  $ and
$\widetilde{\mu}_{i}=\max\left\{  j-1,\mu_{i}\right\}  $ for all $1\leq
i\leq\ell\left(  \lambda\right)  $).
\end{definition}

For example, $\operatorname{col}_{\geq1}T=T$, whereas $\operatorname{col}%
_{\geq j}T$ is empty for any sufficiently large $j$.

For example,
\[
\operatorname{col}_{\geq3}\ \begin{ytableau}
\none & 1 & 1 & 2\\
1 & 2 & 3\\
3 & 4
\end{ytableau}=\begin{ytableau}
\none & \none & 1 & 2\\
\none & \none & 3
\end{ytableau}\ .
\]


The final ingredient for the Littlewood--Richardson rule is a property of some
semistandard tableaux known as $\nu$\textbf{-Yamanouchiness}:

\begin{definition}
\label{def.schur.yamanouchi}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 $j$,%
\[
\text{the }N\text{-tuple }\nu+\operatorname*{cont}\left(  \operatorname{col}%
_{\geq j}T\right)  \text{ is an }N\text{-partition (i.e., weakly decreasing).}%
\]

\end{definition}

\begin{remark}
\label{rmk.schur.yamanouchi-votes}Here is a memorable metaphor for $\nu$-Yamanouchiness:

Imagine an election with $N$ candidates labelled $1,2,\ldots,N$. The entries
of $T$ are votes waiting to be counted, with each entry $k$ representing one
vote for candidate $k$. Some votes have already been counted: candidate $i$
starts with $\nu_{i}$ votes.

We count the votes in $T$ one column at a time, starting with the rightmost
column and working our way to the left. For example, a column with entries $2$
and $4$ contributes one vote to candidate $2$ and one to candidate $4$. Each
column contributes at most one vote to any given candidate, since its entries
are strictly increasing. After counting all votes from any single column, we
take a break before we move on to the next column.

The tableau $T$ is $\nu$-Yamanouchi exactly when, during each break, candidate
$1$ has at least as many votes as candidate $2$, candidate $2$ has at least as
many as candidate $3$, and so on. This ordering holds at the start because
$\nu$ is an $N$-partition. (Ties are allowed.)
\end{remark}

\Needspace{12\baselineskip}

\begin{example}
\label{exa.schur.yamanouchi}\ \ 

\begin{enumerate}
\item[\textbf{(a)}] Let $N=3$ and $\nu=\varnothing=\left(  0,0,0\right)  $.
Which of the following seven tableaux are $\nu$-Yamanouchi?
\[
\renewcommand{\arraystretch}{1.1}%
\begin{array}
[c]{@{}ccc@{\qquad}ccc@{\qquad}c}%
T_{1}=\begin{ytableau} \none & 1 & 1\\ 2 & 2 \end{ytableau} &  &  &
T_{2}=\begin{ytableau} \none & 1 & 1\\ 2 & 3 \end{ytableau} &  &  &
T_{3}=\begin{ytableau} \none & 1 & 1\\ 1 & 2 \end{ytableau}\\[12pt]%
T_{4}=\begin{ytableau} \none & 1\\ 1\\ 2 \end{ytableau} &  &  & T_{5}%
=\begin{ytableau} \none & 1\\ 1\\ 3 \end{ytableau} &  &  & T_{6}%
=\begin{ytableau} \none & \none & 1 & 1\\ \none & 1 & 2 & 2\\ 2 & 3 \end{ytableau}\\[12pt]%
T_{7}=\begin{ytableau} \none & 1 & 2\\ 2 & 2 \end{ytableau} &  &  &  &  &  &
\end{array}
\]


Answer: $T_{1}$, $T_{3}$, $T_{4}$ and $T_{6}$ are; $T_{2}$, $T_{5}$ and
$T_{7}$ are not. Here is a column-by-column check. For the sake of brevity, we
set
\[
C_{j}(T):=\nu+\operatorname*{cont}\left(  \operatorname{col}_{\geq j}T\right)
=\operatorname*{cont}\left(  \operatorname{col}_{\geq j}T\right)
\]
(since $\nu=\varnothing$). We can compute these easily by starting on the very
right (none of our tableaux goes beyond column $4$, so we have $C_{j}\left(
T\right)  =\operatorname*{cont}\left(  \operatorname{col}_{\geq j}T\right)
=0$ for all $j\geq5$) and adding one column at a time (since $C_{j}\left(
T\right)  =C_{j+1}\left(  T\right)  +\operatorname*{cont}\left(
\operatorname{col}_{j}T\right)  $, where $\operatorname{col}_{j}T$ is the
$j$-th column of $T$). Thus we obtain the following contents. The boldfaced
tuples are the ones that are not weakly decreasing:
\[
\renewcommand{\arraystretch}{1.15}%
\begin{array}
[c]{c|cccc}
& C_{4}(T) & C_{3}(T) & C_{2}(T) & C_{1}(T)\\\hline
T_{1} & (0,0,0) & (1,0,0) & (2,1,0) & (2,2,0)\\
T_{2} & (0,0,0) & (1,0,0) & \mathbf{(2,0,1)} & (2,1,1)\\
T_{3} & (0,0,0) & (1,0,0) & (2,1,0) & (3,1,0)\\
T_{4} & (0,0,0) & (0,0,0) & (1,0,0) & (2,1,0)\\
T_{5} & (0,0,0) & (0,0,0) & (1,0,0) & \mathbf{(2,0,1)}\\
T_{6} & (1,1,0) & (2,2,0) & (3,2,1) & (3,3,1)\\
T_{7} & (0,0,0) & \mathbf{(0,1,0)} & \mathbf{(1,2,0)} & \mathbf{(1,3,0)}%
\end{array}
\]
For $j\geq5$, all these contents are $(0,0,0)$. Thus, the table checks every
condition in the definition. In particular, $T_{2}$ fails at $j=2$ because
there is a $3$ but no $2$ in its last two columns; $T_{5}$ fails at $j=1$ for
similar reasons. The tableau $T_{7}$ already fails at $j=3$, since its last
column contains a $2$ but no $1$. Notice that the full content of $T_{2}$ is
weakly decreasing; checking just the full content is therefore not enough.

Note that any semistandard tableau $T$ that is $\varnothing$-Yamanouchi will
automatically be $\nu$-Yamanouchi for any $N$-partition $\nu$ (because the
entrywise sum of two $N$-partitions is an $N$-partition).

\item[\textbf{(b)}] Now let us vary $\nu$. What $\nu$ should you take to make
$T_{2}$ and $T_{5}$ become $\nu$-Yamanouchi? Try to find a $\nu$ that is as
small in size as possible.

We can take $\nu=\left(  1,1,0\right)  $. After adding each nonempty column,
from right to left, the tuples $\nu+\operatorname*{cont}\left(
\operatorname{col}_{\geq j}T\right)  $ are
\[%
\begin{array}
[c]{c|ccc}%
T_{2} & (2,1,0) & (3,1,1) & (3,2,1)\\
T_{5} & (2,1,0) & (3,1,1) &
\end{array}
\]
All these tuples are weakly decreasing, as is the initial tuple $\nu$. This
proves that both tableaux are $\nu$-Yamanouchi.

Why is size $2$ minimal? For $T_{2}$ at $j=2$, and for $T_{5}$ at $j=1$, the
condition requires $\nu+(2,0,1)$ to be weakly decreasing. Thus, $\nu_{2}%
\geq\nu_{3}+1\geq1$. Since $\nu$ itself is weakly decreasing, we also have
$\nu_{1}\geq\nu_{2}\geq1$. Therefore, $\left\vert \nu\right\vert \geq2$, and
our choice $\nu=\left(  1,1,0\right)  $ attains this bound for each of the two tableaux.

To ensure that $T_{7}$ is $\nu$-Yamanouchi, we can take $\nu=\left(
2,0,0\right)  $. The successive tuples $\nu+\operatorname*{cont}\left(
\operatorname{col}_{\geq j}T\right)  $, again after adding each column from
right to left, are
\[
(2,1,0),\qquad(3,2,0),\qquad(3,3,0),
\]
so this choice works. It also has the smallest possible size: the full content
of $T_{7}$ is $(1,3,0)$, so any suitable $\nu$ must satisfy $\nu_{1}+1\geq
\nu_{2}+3$, or equivalently $\nu_{1}\geq\nu_{2}+2$. Hence $\left\vert
\nu\right\vert \geq2$.
\end{enumerate}
\end{example}

We can finally state the Littlewood--Richardson rule:

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

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}

Some comments are warranted:

\begin{itemize}
\item The term $s_{\nu+\operatorname*{cont}T}$ in the above sum is always
well-defined: Indeed, if $T$ is $\nu$-Yamanouchi, then
\[
\nu+\operatorname*{cont}T=\nu+\operatorname*{cont}\left(  \operatorname{col}%
_{\geq1}T\right)
\]
is an $N$-partition.

\item By picking $\mu=\varnothing$ in Theorem \ref{thm.schur.generalized-lr},
we get a formula for $s_{\nu}\cdot s_{\lambda}$, that is, for any product of
two Schur polynomials as a sum of Schur polynomials.
\end{itemize}

\begin{noncompile}
We will actually prove something slightly different:%
\[
a_{\nu+\rho}\cdot s_{\lambda/\mu}\overset{?}{=}\sum_{\substack{T\in
\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  ;\\T\text{ is }%
\nu\text{-Yamanouchi}}}a_{\nu+\operatorname*{cont}T+\rho}.
\]
Once this formula is proved, we will apply it twice -- once to get $a_{\rho
}s_{\lambda}=a_{\lambda+\rho}$ for $\ell(\lambda)\leq N$, and again to get the
Littlewood--Richardson rule.
\end{noncompile}

Before we get to the proof of Theorem \ref{thm.schur.generalized-lr}, let us
first get a simple fact out of our way:

\begin{proposition}
\label{prop.schur.too-many-rows}If $\lambda$ is a partition of length $>N$,
then $s_{\lambda}=0$.
\end{proposition}

\begin{proof}
Let $\lambda$ be a partition of length $>N$. Note that the length of $\lambda$
is precisely the number of cells in the first column of $Y\left(
\lambda\right)  $. Hence, the first column of $Y\left(  \lambda\right)  $ has
more than $N$ cells (since $\lambda$ has length $>N$). Therefore, any SSYT of
shape $\lambda$ must have more than $N$ distinct entries in its first column
(since the \textquotedblleft strictly increasing down each
column\textquotedblright\ condition in the definition of an SSYT forces all
the entries in a given column to be distinct). But this is impossible if all
entries of the SSYT are to belong to $\left[  N\right]  $ (since we cannot
find more than $N$ distinct numbers in $\left[  N\right]  $). Thus, there
exists no SSYT of shape $\lambda$ whose entries belong to $\left[  N\right]
$. In other words, the set $\operatorname*{SSYT}\left(  \lambda,N\right)  $ is
empty. Thus, $s_{\lambda}$ is an empty sum (according to its definition) and
equals $0$.
\end{proof}

In contrast, a skew Schur polynomial $s_{\lambda/\mu}$ can be nonzero even if
$\lambda$ and $\mu$ are arbitrarily long. For instance, for any positive
integer $k$, setting $\lambda=\left(  k,k-1,\ldots,2,1\right)  $ and
$\mu=\left(  k-1,k-2,\ldots,1\right)  $, we can fill the $k$ cells of
$\lambda/\mu$ with $1$'s, and this always produces a tableau $T\in
\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  $ for any $N\geq1$, so that
$s_{\lambda/\mu}$ contains at least an $x_{1}^{k}$ term.

\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[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}