\documentclass[numbers=enddot,12pt,final,onecolumn,notitlepage]{scrartcl}%
\usepackage[headsepline,footsepline,manualmark]{scrlayer-scrpage}
\usepackage[all,cmtip]{xy}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{framed}
\usepackage{comment}
\usepackage{color}
\usepackage[breaklinks=True]{hyperref}
\usepackage[sc]{mathpazo}
\usepackage[T1]{fontenc}
\usepackage{needspace}
\usepackage{tabls}
\usepackage{ytableau}
\usepackage{xr}
\usepackage{tikz}
\usepackage{pgfplots}
\usepackage[type={CC}, modifier={zero}, version={1.0},]{doclicense}
%TCIDATA{OutputFilter=latex2.dll}
%TCIDATA{Version=5.50.0.2960}
%TCIDATA{LastRevised=Monday, October 05, 2026 14:57:36}
%TCIDATA{SuppressPackageManagement}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{Language=American English}
%BeginMSIPreambleData
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
%EndMSIPreambleData
\externaldocument{lec01-701}
\externaldocument{lec02-701}
\externaldocument{lec03-701}
\externaldocument{lec04-701}
\externaldocument{lec05-701}
\pgfplotsset{compat=1.13}
\theoremstyle{definition}
\newtheorem{theo}{Theorem}[subsection]
\newtheorem{exer}{Exercise}[subsubsection]
\newenvironment{theorem}[1][]
{\begin{theo}[#1]\begin{leftbar}}
{\end{leftbar}\end{theo}}
\newtheorem{lem}[theo]{Lemma}
\newenvironment{lemma}[1][]
{\begin{lem}[#1]\begin{leftbar}}
{\end{leftbar}\end{lem}}
\newtheorem{prop}[theo]{Proposition}
\newenvironment{proposition}[1][]
{\begin{prop}[#1]\begin{leftbar}}
{\end{leftbar}\end{prop}}
\newtheorem{defi}[theo]{Definition}
\newenvironment{definition}[1][]
{\begin{defi}[#1]\begin{leftbar}}
{\end{leftbar}\end{defi}}
\newtheorem{remk}[theo]{Remark}
\newenvironment{remark}[1][]
{\begin{remk}[#1]\begin{leftbar}}
{\end{leftbar}\end{remk}}
\newtheorem{coro}[theo]{Corollary}
\newenvironment{corollary}[1][]
{\begin{coro}[#1]\begin{leftbar}}
{\end{leftbar}\end{coro}}
\newtheorem{conv}[theo]{Convention}
\newenvironment{convention}[1][]
{\begin{conv}[#1]\begin{leftbar}}
{\end{leftbar}\end{conv}}
\newtheorem{quest}[theo]{Question}
\newenvironment{question}[1][]
{\begin{quest}[#1]\begin{leftbar}}
{\end{leftbar}\end{quest}}
\newtheorem{warn}[theo]{Warning}
\newenvironment{warning}[1][]
{\begin{warn}[#1]\begin{leftbar}}
{\end{leftbar}\end{warn}}
\newtheorem{conj}[theo]{Conjecture}
\newenvironment{conjecture}[1][]
{\begin{conj}[#1]\begin{leftbar}}
{\end{leftbar}\end{conj}}
\newtheorem{exam}[theo]{Example}
\newenvironment{example}[1][]
{\begin{exam}[#1]\begin{leftbar}}
{\end{leftbar}\end{exam}}
\newtheorem{exmp}[exer]{Exercise}
\newenvironment{exercise}[1][]
{\begin{exmp}[#1]\begin{leftbar}}
{\end{leftbar}\end{exmp}}
\newenvironment{statement}{\begin{quote}}{\end{quote}}
\newenvironment{fineprint}{\begin{small}}{\end{small}}
\iffalse
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\newenvironment{convention}[1][Convention]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\newenvironment{question}[1][Question]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\newenvironment{sloppypar}[1][sloppypar]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\fi
\let\sumnonlimits\sum
\let\prodnonlimits\prod
\let\cupnonlimits\bigcup
\let\capnonlimits\bigcap
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
\renewcommand{\bigcup}{\cupnonlimits\limits}
\renewcommand{\bigcap}{\capnonlimits\limits}
\usetikzlibrary{arrows,arrows.meta,decorations.markings}
\setlength\tablinesep{3pt}
\setlength\arraylinesep{3pt}
\setlength\extrarulesep{3pt}
\setlength\textheight{22.5cm}
\setlength\textwidth{14.8cm}
\newenvironment{verlong}{}{}
\newenvironment{vershort}{}{}
\newenvironment{noncompile}{}{}
\excludecomment{verlong}
\includecomment{vershort}
\excludecomment{noncompile}
\newcommand{\defn}[1]{{\color{darkred}\emph{#1}}}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\KK}{\mathbb{K}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\tup}[1]{\left( #1 \right)}
\newcommand{\ive}[1]{\left[ #1 \right]}
\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
\newcommand{\mono}{\hookrightarrow}
\newcommand{\epi}{\twoheadrightarrow}
\newcommand{\iso}{\overset{\cong}{\to}}
\newcommand{\arinj}{\ar@{_{(}->}}
\newcommand{\arinjrev}{\ar@{^{(}->}}
\newcommand{\arsurj}{\ar@{->>}}
\newcommand{\arelem}{\ar@{|->}}
\newcommand{\arback}{\ar@{<-}}
\newcommand{\Ker}{\operatorname{Ker}}
\newcommand{\Coker}{\operatorname{Coker}}
\newcommand{\incpdftexpic}[1]{{\def\svgwidth{\columnwidth} \input{#1} }}
\definecolor{dbluecolor}{rgb}{0.01,0.02,0.7}
\definecolor{dgreencolor}{rgb}{0.2,0.4,0.0}
\definecolor{darkred}{rgb}{0.7,0,0}
\newtheoremstyle{plainsl}
{8pt plus 2pt minus 4pt}
{8pt plus 2pt minus 4pt}
{\slshape}
{0pt}
{\bfseries}
{.}
{5pt plus 1pt minus 1pt}
{}
\theoremstyle{plainsl}
\ihead{Math 701, Fall 2026, Lecture 6, version \today}
\ohead{page \thepage}
\cfoot{}
\begin{document}
\section*{Math 701 Fall 2026, Lecture 6: Skew Schur polynomials}

\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.) \medskip

\setcounter{section}{1}\setcounter{subsection}{5}

\subsection{Skew Schur polynomials}

\subsubsection{Skew partitions, tableaux and Schur functions}

We recall that the length of a partition $\lambda$ (that is, the \# of its
nonzero entries\footnote{The symbol \textquotedblleft\#\textquotedblright%
\ means \textquotedblleft number\textquotedblright.}) is denoted by
$\ell\left(  \lambda\right)  $.

\begin{definition}
\label{def.schur.partition-containment}Let $\lambda$ and $\mu$ be two partitions.

We say that $\mu$ is \textbf{contained} in $\lambda$ if $Y\left(  \mu\right)
\subseteq Y\left(  \lambda\right)  $. We write \textquotedblleft$\mu
\subseteq\lambda$\textquotedblright\ for this statement. Equivalently,
$\mu\subseteq\lambda$ holds if and only if%
\[
\mu_{i}\leq\lambda_{i}\ \ \ \ \ \ \ \ \ \ \text{for each }i\geq1
\]
(where $\mu_{i}$ means the $i$-th entry of $\mu$, and similarly $\lambda_{i}$
is understood). Here we are again identifying partitions with infinite
sequences (by padding them with infinitely many zeroes), so we have
$\lambda_{i}=0$ for all $i>\ell\left(  \lambda\right)  $.
\end{definition}

\begin{example}
\label{exa.schur.partition-containment}Let $\lambda=\left(  4,3,1\right)  $,
$\mu=\left(  2,1\right)  $ and $\nu=\left(  3,2,2\right)  $. Their Young
diagrams are
\[%
\begin{array}
[c]{cc@{\qquad}cc@{\qquad}c}%
Y\left(  \lambda\right)  &  & Y\left(  \mu\right)  &  & Y\left(  \nu\right)
\\[4pt]%
\ydiagram{4,3,1} &  & \ydiagram{2,1} &  & \ydiagram{3,2,2}
\end{array}
\]
We have $\mu\subseteq\lambda$, since $2\leq4$ and $1\leq3$ and $0\leq1$ (and
$0\leq0$ for all the remaining entries). Likewise, we have $\mu\subseteq\nu$.
However, we don't have $\nu\subseteq\lambda$, since $\nu_{3}=2>1=\lambda_{3}$.
Thus, the cell $\left(  3,2\right)  $ belongs to $Y\left(  \nu\right)  $ but
not to $Y\left(  \lambda\right)  $. Notice that $\left\vert \nu\right\vert
=7<8=\left\vert \lambda\right\vert $; having fewer boxes does not guarantee containment.
\end{example}

\begin{definition}
\label{def.schur.skew-diagram}Let $\lambda$ and $\mu$ be two partitions such
that $\mu\subseteq\lambda$. Then, the pair $\left(  \lambda,\mu\right)  $ is
called a \textbf{skew partition}, and is denoted by $\lambda/\mu$.

The \textbf{skew Young diagram} $Y\left(  \lambda/\mu\right)  $ of
$\lambda/\mu$ is defined to be the set%
\begin{align*}
Y\left(  \lambda\right)  \setminus Y\left(  \mu\right)   &  =\left\{  \left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\in\left[
\lambda_{i}\right]  \setminus\left[  \mu_{i}\right]  \right\} \\
&  =\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}%
\ \mid\ \mu_{i}<j\leq\lambda_{i}\right\}  .
\end{align*}


The original Young diagrams $Y\left(  \lambda\right)  $ will occasionally be
called \textbf{straight Young diagrams} in order to distinguish them from skew ones.

As usual, we will often omit the word \textquotedblleft
Young\textquotedblright.
\end{definition}

\begin{example}
\label{exa.schur.skew-diagram}For $\lambda=\left(  4,3,1\right)  $ and
$\mu=\left(  2,1\right)  $ as above, we have
\[
Y\left(  \lambda/\mu\right)  =\ydiagram{2+2,1+2,1}\ .
\]
The missing boxes at the start of the first two rows are the boxes of
$Y\left(  \mu\right)  $; the remaining boxes of $Y\left(  \lambda\right)  $
stay in their original positions. More explicitly,
\[
Y\left(  \lambda/\mu\right)  =\left\{  (1,3),(1,4),(2,2),(2,3),(3,1)\right\}
.
\]
This diagram has $5=\left\vert \lambda\right\vert -\left\vert \mu\right\vert $ boxes.
\end{example}

Of course, each straight\ Young diagram $Y\left(  \lambda\right)  $ is a skew
Young diagram, namely $Y\left(  \lambda/\left(  {}\right)  \right)  $. We
often denote the empty partition $\left(  {}\right)  $ by $\varnothing$, so we
have $Y\left(  \lambda\right)  =Y\left(  \lambda/\varnothing\right)  $.

Skew Young diagrams (unlike straight ones, i.e., ones of the form $Y\left(
\lambda\right)  $) need not be left-aligned. Nevertheless, their rows and
their columns still don't have \textquotedblleft gaps\textquotedblright%
\ (i.e., any space between two cells in the same row or column is also
occupied by cells). This is a necessary, but not a sufficient condition for a
set of cells to be a skew Young diagram; for example, the set%
\[
\ydiagram{2+1,1+3,2+1}
\]
has this property as well but is not a skew Young diagram. However, we can say
more about skew Young diagrams. The most comfortable way to do so is by
introducing a partial order on $\mathbb{Z}^{2}$:

\begin{definition}
\label{def.schur.nw-order}We define a partial order $\leq$ on the set of all
cells $\left(  i,j\right)  \in\mathbb{Z}^{2}$ by setting%
\[
\left(  a,b\right)  \leq\left(  c,d\right)  \ \ \ \ \ \ \ \ \ \ \text{if and
only if}\ \ \ \ \ \ \ \ \ \ \left(  a\leq c\text{ and }b\leq d\right)  .
\]
This is the \textbf{entrywise partial order} on $\mathbb{Z}^{2}$, and is also
called the \textbf{northwest order} (since cells become smaller as you move northwest).
\end{definition}

For example, $\left(  1,1\right)  \leq\left(  2,4\right)  \leq\left(
3,4\right)  \leq\left(  3,5\right)  $, but the two cells $\left(  1,2\right)
$ and $\left(  2,1\right)  $ are incomparable.

We can now characterize straight and skew Young diagrams in terms of this order:

\begin{lemma}
[northwest-closure of straight Young diagrams]%
\label{lem.schur.northwest-closure}Let $\lambda$ be a partition. Let $c$ and
$d$ be two cells in $\left\{  1,2,3,\ldots\right\}  ^{2}$ such that $c\leq d$
(in the northwest order) and $d\in Y\left(  \lambda\right)  $. Then, $c\in
Y\left(  \lambda\right)  $.
\end{lemma}

\begin{proof}
Write the cells $c$ and $d$ as $c=\left(  i^{\prime},j^{\prime}\right)  $ and
$d=\left(  i,j\right)  $. Then, $\left(  i,j\right)  =d\in Y\left(
\lambda\right)  $, so that $j\in\left[  \lambda_{i}\right]  $. That is,
$j\leq\lambda_{i}$. But $c\leq d$ means that $i^{\prime}\leq i$ and
$j^{\prime}\leq j$ (since $c=\left(  i^{\prime},j^{\prime}\right)  $ and
$d=\left(  i,j\right)  $). Since $\lambda$ is a partition, we have
$\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\cdots$ and thus $\lambda
_{i^{\prime}}\geq\lambda_{i}$ (because $i^{\prime}\leq i$). Now, $j^{\prime
}\leq j\leq\lambda_{i}\leq\lambda_{i^{\prime}}$ (since $\lambda_{i^{\prime}%
}\geq\lambda_{i}$). In other words, $j^{\prime}\in\left[  \lambda_{i^{\prime}%
}\right]  $, so that $\left(  i^{\prime},j^{\prime}\right)  \in Y\left(
\lambda\right)  $. That is, $c\in Y\left(  \lambda\right)  $, and Lemma
\ref{lem.schur.northwest-closure} is proved.
\end{proof}

\begin{lemma}
[convexity of skew Young diagrams]\label{lem.schur.skew-convexity}Let
$\lambda/\mu$ be a skew partition. Let $c,d,e$ be three cells in
$\mathbb{Z}^{2}$ such that $c\leq d\leq e$ (in the northwest order) and $c\in
Y\left(  \lambda/\mu\right)  $ and $e\in Y\left(  \lambda/\mu\right)  $. Then,
$d\in Y\left(  \lambda/\mu\right)  $.
\end{lemma}

\begin{proof}
First of all, $c\in Y\left(  \lambda/\mu\right)  \subseteq\left\{
1,2,3,\ldots\right\}  ^{2}$. That is, both coordinates of the cell $c$ are
$\geq1$ (where the \textquotedblleft coordinates\textquotedblright\ of a cell
$\left(  i,j\right)  $ are $i$ and $j$, as you would expect from the picture).
But $c\leq d$ shows that the coordinates of the cell $c$ are $\leq$ to the
respective coordinates of the cell $d$. Since the former coordinates are
$\geq1$, this entails that the latter are $\geq1$ as well. In other words,
$d\in\left\{  1,2,3,\ldots\right\}  ^{2}$.

From $d\in\left\{  1,2,3,\ldots\right\}  ^{2}$ and $d\leq e$ and $e\in
Y\left(  \lambda/\mu\right)  =Y\left(  \lambda\right)  \setminus Y\left(
\mu\right)  \subseteq Y\left(  \lambda\right)  $, we obtain $d\in Y\left(
\lambda\right)  $ (by Lemma \ref{lem.schur.northwest-closure}, applied to $d$
and $e$ instead of $c$ and $d$).

Now, we shall prove that $d\notin Y\left(  \mu\right)  $. Indeed, assume the
contrary. Then, $d\in Y\left(  \mu\right)  $. From $c\in\left\{
1,2,3,\ldots\right\}  ^{2}$ and $c\leq d$ and $d\in Y\left(  \mu\right)  $, we
then obtain $c\in Y\left(  \mu\right)  $ (by Lemma
\ref{lem.schur.northwest-closure}, applied to $\mu$ instead of $\lambda$). But
this contradicts $c\in Y\left(  \lambda/\mu\right)  =Y\left(  \lambda\right)
\setminus Y\left(  \mu\right)  $. This contradiction shows that our assumption
was false. Hence, $d\notin Y\left(  \mu\right)  $ is proved.

Combining $d\in Y\left(  \lambda\right)  $ with $d\notin Y\left(  \mu\right)
$, we obtain $d\in Y\left(  \lambda\right)  \setminus Y\left(  \mu\right)
=Y\left(  \lambda/\mu\right)  $, qed.
\end{proof}

It turns out that, among finite subsets of $\left\{  1,2,3,\ldots\right\}
^{2}$, Lemma \ref{lem.schur.northwest-closure} and Lemma
\ref{lem.schur.skew-convexity} actually characterize straight Young diagrams
and skew Young diagrams, respectively. This is easy to prove for straight
Young diagrams (see \cite[Proposition 5.1.7 \textbf{(a)}]{24s} for a proof);
the version for skew ones will not be needed.

Note that a Young diagram $Y\left(  \lambda\right)  $ uniquely determines the
partition $\lambda$ (see \cite[Proposition 5.1.7 \textbf{(c)}]{24s} for a
proof), but a skew Young diagram $Y\left(  \lambda/\mu\right)  $ does not
uniquely determine $\lambda$ and $\mu$. For example, $\left(  3,2\right)
/\left(  3\right)  \neq\left(  2,2\right)  /\left(  2\right)  $, but
\[
Y\left(  \left(  3,2\right)  /\left(  3\right)  \right)  =Y\left(  \left(
2,2\right)  /\left(  2\right)  \right)  =\left\{  (2,1),(2,2)\right\}  .
\]


Let us now define Young tableaux of skew shapes. This is done in the exact
same way as for straight shapes:

\begin{definition}
\label{def.schur.skew-tableau}Let $\lambda/\mu$ be a skew partition.

A \textbf{Young tableau} of shape $Y\left(  \lambda/\mu\right)  $ (or just of
shape $\lambda/\mu$) means a map $T:Y\left(  \lambda/\mu\right)
\rightarrow\left\{  1,2,3,\ldots\right\}  $, that is, a way to fill the boxes
of $Y\left(  \lambda/\mu\right)  $ with positive integers (one per box). Such
Young tableaux are often called \textbf{skew Young tableaux}. Again, we omit
\textquotedblleft Young\textquotedblright.

If we don't have $\mu\subseteq\lambda$, then, by convention, there are no
Young tableaux of shape $\lambda/\mu$.

The notions of \textquotedblleft\textbf{semistandard}\textquotedblright\ and
\textquotedblleft\textbf{standard}\textquotedblright\ are defined for skew
Young tableaux in the same way as for straight ones: \textquotedblleft
semistandard\textquotedblright\ means \textquotedblleft weakly increasing
along rows, strictly increasing down columns\textquotedblright, whereas
\textquotedblleft standard\textquotedblright\ additionally requires that the
entries are $1,2,\ldots,\left\vert Y\left(  \lambda/\mu\right)  \right\vert $
with no repetition. Again, we abbreviate \textquotedblleft semistandard Young
tableau\textquotedblright\ as \textquotedblleft SSYT\textquotedblright.

We again use the notation $\operatorname*{SSYT}\left(  \lambda/\mu\right)  $
for the set of all SSYTs of shape $Y\left(  \lambda/\mu\right)  $, and the
notation $\operatorname*{SSYT}\left(  \lambda/\mu,k\right)  $ for the set of
those SSYTs of shape $Y\left(  \lambda/\mu\right)  $ whose entries belong to
$\left[  k\right]  $.
\end{definition}

\begin{lemma}
\label{lem.schur.skew-tableau-monotonicity}Let $\lambda$ and $\mu$ be two
partitions. Let $T$ be an SSYT of shape $\lambda/\mu$. Then:

\begin{enumerate}
\item[\textbf{(a)}] If two cells $c,d\in Y\left(  \lambda/\mu\right)  $
satisfy $c\leq d$ (in the northwest order), then $T\left(  c\right)  \leq
T\left(  d\right)  $.

\item[\textbf{(b)}] If two cells $c,d\in Y\left(  \lambda/\mu\right)  $
satisfy $c\leq d$ but do not lie in the same row, then $T\left(  c\right)
<T\left(  d\right)  $.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof sketch.] We outline the proofs; details can be found in \cite[\S B.10,
detailed proof of Lemma 7.3.17]{21s} (for $N$-partitions instead of
partitions, but this does not affect the proof). \medskip

\textbf{(a)} Let $c,d\in Y\left(  \lambda/\mu\right)  $ be two cells that
satisfy $c\leq d$. Thus, we can walk from $c$ to $d$ by a sequence of steps,
each of which is either an east-step (going from a cell $\left(  i,j\right)  $
to the cell $\left(  i,j+1\right)  $) or a south-step (going from a cell
$\left(  i,j\right)  $ to the cell $\left(  i+1,j\right)  $). Each of the
intermediate cells $p$ we traverse during this walk satisfies $c\leq p\leq d$
and thus belongs to $Y\left(  \lambda/\mu\right)  $ (by Lemma
\ref{lem.schur.skew-convexity}, applied to $p$ and $d$ instead of $d$ and
$e$); thus, it has an entry in $T$. The definition of an SSYT ensures that the
entries in these intermediate cells on our path increase weakly as we traverse
the path (since the entries of SSYT weakly increase along rows and strictly
increase down columns); thus, in particular, the entry $T\left(  c\right)  $
(in our starting cell) must be $\leq$ to the entry $T\left(  d\right)  $ (in
our ending cell). This proves part \textbf{(a)}. \medskip

\textbf{(b)} Proceed as in the proof of part \textbf{(a)} above, but now
additionally observe that our walk must contain at least one south-step (since
$c$ and $d$ do not lie in the same row), and the entries of $T$ increase
strictly as we make a south-step (since they strictly increase down columns).
Thus, the final inequality $T\left(  c\right)  \leq T\left(  d\right)  $ must
be strict, and this proves part \textbf{(b)}.
\end{proof}

The following two definitions are direct calques of Definition
\ref{def.schur.tableau-monomial} and Definition \ref{def.schur.polynomial}:

\begin{definition}
\label{def.schur.skew-tableau-monomial}Let $\lambda/\mu$ be a skew partition.
Let $T$ be any skew tableau of shape $\lambda/\mu$ whose entries belong to
$\left[  N\right]  $. Then, we define the corresponding monomial%
\[
x_{T}:=\prod_{c\text{ is a cell of }Y\left(  \lambda/\mu\right)  }x_{T\left(
c\right)  }=\prod_{\left(  i,j\right)  \in Y\left(  \lambda/\mu\right)
}x_{T\left(  i,j\right)  }=\prod_{k=1}^{N}x_{k}^{\left(  \text{\# of
}k\text{'s in }T\right)  }%
\]
in the indeterminates $x_{1},x_{2},\ldots,x_{N}$. (Of course,
\textquotedblleft\# of $k$'s in $T$\textquotedblright\ means the number of
boxes of $T$ with a $k$ in them.)
\end{definition}

\begin{definition}
\label{def.schur.skew-polynomial}Let $\lambda/\mu$ be a skew partition. Then,
the \textbf{(skew) Schur polynomial} $s_{\lambda/\mu}\in\mathcal{P}$ is
defined by%
\[
s_{\lambda/\mu}:=\sum_{T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)
}x_{T}.
\]

\end{definition}

\begin{example}
\label{exa.schur.skew-22-1}Assume that $N\geq3$. The skew diagram $Y\left(
\left(  2,2\right)  /\left(  1\right)  \right)  $ and a tableau of this shape
$T$ look as follows:%
\[
Y\left(  \left(  2,2\right)  /\left(  1\right)  \right)
=\ydiagram{1+1,2}\ \ ,\qquad T=\begin{ytableau}
\none & i\\
j & k
\end{ytableau}\ .
\]
Such a tableau $T$ belongs to $\operatorname{SSYT}\left(  \left(  2,2\right)
/\left(  1\right)  ,N\right)  $ if and only if $i,j,k\in\left[  N\right]  $
and $i<k$ and $j\leq k$. Its monomial is $x_{T}=x_{i}x_{j}x_{k}$. Thus,
\begin{align*}
s_{\left(  2,2\right)  /\left(  1\right)  } &  =\sum_{i<k;\ j\leq k}x_{i}%
x_{j}x_{k}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{where all summation indices are}\\
\text{understood to belong to }\left[  N\right]
\end{array}
\right)  \\
&  =2\sum_{u<v<w}x_{u}x_{v}x_{w}+\sum_{u<v}x_{u}^{2}x_{v}+\sum_{u<v}x_{u}%
x_{v}^{2}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by a computation similar to that done in}\\
\text{Example \ref{exa.schur.rows-columns-21} \textbf{(c)} (exercise!)}%
\end{array}
\right)  \\
&  =2m_{\left(  1,1,1\right)  }+m_{\left(  2,1\right)  }.
\end{align*}


This is the same as $s_{\left(  2,1\right)  }$ from Example
\ref{exa.schur.rows-columns-21} \textbf{(c)}. We will later see why this is
not a coincidence.
\end{example}

\begin{example}
\label{exa.schur.skew-three-independent}The skew diagram $Y\left(  \left(
3,2,1\right)  /\left(  2,1\right)  \right)  $ has three boxes, lying in three
different rows and three different columns:
\[
Y\left(  \left(  3,2,1\right)  /\left(  2,1\right)  \right)
=\ydiagram{2+1,1+1,1}\ \ ,\qquad T=\begin{ytableau}
\none & \none & i\\
\none & j\\
k
\end{ytableau}\ .
\]
Thus, every tableau of shape $Y\left(  \left(  3,2,1\right)  /\left(
2,1\right)  \right)  $ is automatically semistandard (there are no row or
column inequalities to check); each of its entries $i,j,k$ can be chosen
independently from $\left[  N\right]  $. We obtain
\[
s_{\left(  3,2,1\right)  /\left(  2,1\right)  }=\sum_{i,j,k\in\lbrack N]}%
x_{i}x_{j}x_{k}=\left(  x_{1}+x_{2}+\cdots+x_{N}\right)  ^{3}=e_{1}%
^{3}=s_{\left(  1\right)  }^{3}.
\]

\end{example}

Now we generalize the symmetry of $s_{\lambda}$ (Theorem
\ref{thm.schur.bialternant} \textbf{(a)}) as follows:

\begin{theorem}
\label{thm.schur.skew-symmetry}Let $\lambda/\mu$ be any skew partition. Then,
the polynomial $s_{\lambda/\mu}$ is symmetric.
\end{theorem}

Setting $\mu=\varnothing=\left(  {}\right)  $, this recovers the symmetry of
$s_{\lambda}$.

We will next prove Theorem \ref{thm.schur.skew-symmetry} bijectively, using
the so-called \textbf{Bender--Knuth involutions}.

\begin{remark}
\label{rmk.schur.skew-degree}The polynomial $s_{\lambda/\mu}$ is clearly
homogeneous of degree $\left\vert Y\left(  \lambda/\mu\right)  \right\vert
=\left\vert \lambda\right\vert -\left\vert \mu\right\vert $, since each
tableau $T$ of shape $Y\left(  \lambda/\mu\right)  $ has $\left\vert Y\left(
\lambda/\mu\right)  \right\vert $ entries.
\end{remark}

\subsubsection{The Bender--Knuth involutions}

We begin the proof of Theorem \ref{thm.schur.skew-symmetry} now; we will
conclude it in Lecture 7.

\begin{proof}
[Beginning of the proof of Theorem \ref{thm.schur.skew-symmetry}%
.]\let\savedqedsymbol\qedsymbol
\renewcommand{\qedsymbol}{} For each $k\in\left[  N-1\right]  $, let $s_{k}\in
S_{N}$ be the transposition that swaps $k$ with $k+1$. As we know (by Lemma
\ref{lem.symp.simple-transpositions} from Lecture 3), it suffices to prove
that $s_{k}\cdot s_{\lambda/\mu}=s_{\lambda/\mu}$ for each $k\in\left[
N-1\right]  $.

So we fix some $k\in\left[  N-1\right]  $. We will construct a bijection%
\[
\beta_{k}:\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  \rightarrow
\operatorname*{SSYT}\left(  \lambda/\mu,N\right)
\]
(called the $k$\textbf{-th Bender--Knuth involution}) that

\begin{itemize}
\item interchanges the \# of $k$'s with the \# of $\left(  k+1\right)  $'s in
a tableau (i.e., if a given tableau $T\in\operatorname*{SSYT}\left(
\lambda/\mu,N\right)  $ has $i$ many $k$'s and $j$ many $\left(  k+1\right)
$'s, then its image $\beta_{k}\left(  T\right)  $ will have $j$ many $k$'s and
$i$ many $\left(  k+1\right)  $'s), and

\item leaves all other entries of the tableau unchanged.
\end{itemize}

Once such a bijection $\beta_{k}$ is constructed, we will easily conclude that%
\[
x_{\beta_{k}\left(  T\right)  }=s_{k}\cdot x_{T}\ \ \ \ \ \ \ \ \ \ \text{for
all }T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  ,
\]
and thus applying $s_{k}$ to $s_{\lambda/\mu}=\sum_{T\in\operatorname*{SSYT}%
\left(  \lambda/\mu,N\right)  }x_{T}$ will merely permute the addends of this
sum, showing that $s_{k}\cdot s_{\lambda/\mu}=s_{\lambda/\mu}$.

In order to construct the map $\beta_{k}$, we analyze the $k$'s and the
$\left(  k+1\right)  $'s in a semistandard tableau more closely. Let
$T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)  $. We focus on the
$k$'s and the $\left(  k+1\right)  $'s in $T$. An entry $k$ in $T$ will be
called \textbf{matched} if there is a $k+1$ in the same column (thus
necessarily in the neighboring cell to its south\footnote{Why? Because
\par
\begin{itemize}
\item if this $k+1$ were in a cell further north than the $k$, then we would
immediately get a contradiction to the \textquotedblleft increase strictly
down each column\textquotedblright\ condition for a semistandard tableau;
\par
\item if this $k+1$ were more than one row further south than the $k$, then
there would be at least one further entry between the $k$ and the $k+1$; then
(again by the \textquotedblleft increase strictly down each
column\textquotedblright\ condition) this entry would have to be
simultaneously larger than $k$ and smaller than $k+1$, which is impossible.
\end{itemize}
\par
So the only place the $k+1$ can be is in the neighboring cell to the south of
the $k$.}). An entry $k+1$ in $T$ will be called \textbf{matched} if there is
a $k$ in the same column (thus necessarily in the neighboring cell to its
north\footnote{for a similar reason as in the previous sentence}). All other
$k$'s and $\left(  k+1\right)  $'s in $T$ will be called \textbf{free}.

\Needspace{15\baselineskip}

\begin{example}
\label{exa.schur.bk-matched-free}Let $k=2$ and $N=4$. Consider the following
SSYT of shape $\left(  9,8,6\right)  /\left(  3,1\right)  $ (we have set all
the $k$'s and the $\left(  k+1\right)  $'s in boldface for convenience):
\[
\begingroup\definecolor{bkmatched}{gray}{0.85}T=\begin{ytableau}
\none & \none & \none & 1 & 1 & 1 & *(bkmatched) \mathbf{2} & \mathbf{3} & 4\\
\none & 1 & *(bkmatched) \mathbf{2} & \mathbf{2} & \mathbf{2} & \mathbf{3} & *(bkmatched) \mathbf{3} & 4\\
1 & \mathbf{2} & *(bkmatched) \mathbf{3} & 4 & 4 & 4
\end{ytableau}\ .\endgroup
\]
The shaded boxes contain the matched entries. There are two matched pairs: one
in column $3$ and one in column $7$. The free $2$'s are in cells $\left(
2,4\right)  $, $\left(  2,5\right)  $ and $\left(  3,2\right)  $; the free
$3$'s are in cells $\left(  1,8\right)  $ and $\left(  2,6\right)  $. In
particular, there are three free $2$'s and two free $3$'s. The entries $1$ and
$4$ are neither matched nor free, since these terms refer only to $k$ and
$k+1$.
\end{example}

Clearly,%
\[
\left(  \text{\# of matched }k\text{'s}\right)  =\left(  \text{\# of matched
}\left(  k+1\right)  \text{'s}\right)  ,
\]
since matched entries come in pairs of adjacent entries (each matched $k$ has
a matched $k+1$ in its neighboring cell to its south, and each matched $k+1$
has a matched $k$ in its neighboring cell to its north; this gives a 1-to-1
correspondence between the matched $k$'s and the matched $\left(  k+1\right)
$'s). Each column of $T$ that contains a matched entry will contain a matched
$k$, a matched $k+1$ and no other $k$'s or $\left(  k+1\right)  $'s (since
entries in a column strictly increase).

Now we focus on the rows. We claim the following:

\begin{statement}
\textit{Observation 1:} Each row of $T$ can be subdivided into the following
six blocks (which may be empty):%
\[%
\begin{array}
[c]{@{}cc@{\quad}cc@{\quad}cc@{\quad}cc@{\quad}cc@{\quad}c}%
\text{entries} &  & \text{matched} &  & \text{free} &  & \text{free} &  &
\text{matched} &  & \text{entries}\\
<k &  & k\text{'s} &  & k\text{'s} &  & (k+1)\text{'s} &  & (k+1)\text{'s} &
& >k+1
\end{array}
\ \ ,
\]
appearing in this order from left to right. (That is, the row starts with
entries $<k$, if it has any; then, as we move right, come the matched $k$'s;
then come the free $k$'s; then the free $\left(  k+1\right)  $'s; then the
matched $\left(  k+1\right)  $'s; and then come the entries $>k+1$ at last.)
\end{statement}

\begin{proof}
[Proof of Observation 1.]\let\qedsymbol\savedqedsymbol The entries in a row
weakly increase from left to right; thus, the row has the form%
\[%
\begin{array}
[c]{@{}cc@{\quad}cc@{\quad}cc@{\quad}c}%
\text{entries} &  &  &  &  &  & \text{entries}\\
<k &  & k\text{'s} &  & (k+1)\text{'s} &  & >k+1
\end{array}
\ \ .
\]
It remains to check that all matched $k$'s come before all free $k$'s, and
that all free $\left(  k+1\right)  $'s come before all matched $\left(
k+1\right)  $'s. In other words, we must check that there cannot be free $k$'s
to the left of matched $k$'s (in the same row), nor can there be matched
$\left(  k+1\right)  $'s to the left of free $\left(  k+1\right)  $'s.

We start by proving the former. If there was a free $k$ to the left of a
matched $k$, then these two $k$'s along with their neighboring cells to the
south would form a configuration of the form%
\[%
\begin{tabular}
[c]{|c|c|c|}\cline{1-1}\cline{3-3}%
\multicolumn{1}{|c|}{free $k$} & $\qquad\qquad$ & matched $k$\\\cline{1-1}%
\cline{3-3}%
\multicolumn{1}{|c|}{$?$} &  & $k+1$\\\cline{1-1}\cline{3-3}%
\end{tabular}
\]
(where the two displayed columns need not be adjacent). The \textquotedblleft%
$?$\textquotedblright\ is an actual cell of $Y\left(  \lambda/\mu\right)  $
(by Lemma \ref{lem.schur.skew-convexity}) and thus must be filled with $k+1$
(since whatever entry lies in this cell must be $>k$ but simultaneously $\leq
k+1$), thus rendering the $k$ above it matched; this contradicts the freeness
of the left $k$. So we have shown that there cannot be free $k$'s to the left
of matched $k$'s (in the same row).

For a similar reason, there cannot be matched $\left(  k+1\right)  $'s to the
left of free $\left(  k+1\right)  $'s. Indeed, if there were, then we would
get a configuration
\[%
\begin{tabular}
[c]{|c|c|c|}\cline{1-1}\cline{3-3}%
\multicolumn{1}{|c|}{$k$} & $\qquad\qquad$ & $?$\\\cline{1-1}\cline{3-3}%
\multicolumn{1}{|c|}{matched $k+1$} &  & free $k+1$\\\cline{1-1}\cline{3-3}%
\end{tabular}
\ \ ,
\]
where the \textquotedblleft$?$\textquotedblright\ would again be an actual
cell (by Lemma \ref{lem.schur.skew-convexity}) and would be filled with $k$
(since its entry would be $\geq k$ but $<k+1$), which would render the $k+1$
underneath it matched and cause a contradiction again.

So Observation 1 is proved.
\end{proof}

This completes the first step of the construction. It remains to define
$\beta_{k}$ using the free entries and verify its required properties.
\end{proof}

\textbf{New starting time: 10:58}.

\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[24s]{24s}Darij Grinberg, \textit{An introduction to the symmetric
group algebra [Math 701, Spring 2024 lecture notes]}, 12 June 2026. \newline\url{https://www.cip.ifi.lmu.de/~grinberg/t/24s/sga.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}