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\begin{document}
\section*{Math 701 Fall 2026, Lecture 5: 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}\setcounter{subsubsection}{1}\setcounter{theo}{4}

We will soon see more examples. First, two notations:

\begin{convention}
The length of a partition $\lambda$ is denoted by $\ell\left(  \lambda\right)
$. As we recall, this is the number of nonzero entries of $\lambda$.
\end{convention}

For instance, $\ell\left(  \left(  2,2,1\right)  \right)  =3$.

\begin{convention}
\label{conv.schur.tuple-arithmetic} We add and subtract $N$-tuples of integers
entrywise. That is, if $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{N}\right)  \in\mathbb{Z}^{N}$ and $\beta=\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{N}\right)  \in\mathbb{Z}^{N}$, then we set%
\begin{align*}
\alpha+\beta &  :=\left(  \alpha_{1}+\beta_{1},\alpha_{2}+\beta_{2}%
,\ldots,\alpha_{N}+\beta_{N}\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\alpha-\beta &  :=\left(  \alpha_{1}-\beta_{1},\alpha_{2}-\beta_{2}%
,\ldots,\alpha_{N}-\beta_{N}\right)  .
\end{align*}
Of course, if $\alpha\in\mathbb{N}^{N}$ and $\beta\in\mathbb{N}^{N}$, then
$\alpha+\beta$ belongs to $\mathbb{N}^{N}$ as well (but $\alpha-\beta$ does
not always).
\end{convention}

Recall that an $N$-partition is a weakly decreasing $N$-tuple of nonnegative
integers. Recall furthermore that $\rho$ is the specific $N$-partition
$\left(  N-1,N-2,\ldots,1,0\right)  $.

\begin{lemma}
\label{lem.schur.rho-shift} If $\lambda\in\mathbb{N}^{N}$ is an $N$-partition,
then $\lambda+\rho$ is a \textbf{strictly} decreasing $N$-tuple of nonnegative integers.

Conversely, if $\alpha\in\mathbb{N}^{N}$ is a strictly decreasing $N$-tuple of
nonnegative integers, then $\alpha-\rho$ is an $N$-partition.

Thus, we obtain a bijection%
\begin{align*}
\left\{  N\text{-partitions}\right\}   &  \rightarrow\left\{  \text{strictly
decreasing }N\text{-tuples }\alpha\in\mathbb{N}^{N}\right\}  ,\\
\lambda &  \mapsto\lambda+\rho
\end{align*}
(with inverse map given by $\alpha\mapsto\alpha-\rho$).
\end{lemma}

\begin{proof}
We shall only prove the \textquotedblleft Conversely\textquotedblright\ claim,
since the rest is (even more) trivial.

So let $\alpha\in\mathbb{N}^{N}$ be a strictly decreasing $N$-tuple of
nonnegative integers. Write $\alpha$ as $\alpha=\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{N}\right)  $; thus, $\alpha_{1}>\alpha_{2}>\cdots
>\alpha_{N}\geq0$. Likewise, write $\rho$ as $\rho=\left(  \rho_{1},\rho
_{2},\ldots,\rho_{N}\right)  $, so that $\rho_{i}=N-i$ for each $i\in\left[
N\right]  $.

We WLOG assume that $N>0$ (else, the claim is trivial).

Now, let $i\in\left[  N-1\right]  $. Then, $\left(  \alpha-\rho\right)
_{i}=\alpha_{i}-\underbrace{\rho_{i}}_{=N-i}=\alpha_{i}-\left(  N-i\right)
=\alpha_{i}-N+i$ and likewise $\left(  \alpha-\rho\right)  _{i+1}=\alpha
_{i+1}-N+\left(  i+1\right)  $. But $\alpha_{1}>\alpha_{2}>\cdots>\alpha_{N}$,
so that $\alpha_{i}>\alpha_{i+1}$, and therefore $\alpha_{i}\geq\alpha
_{i+1}+1$ (since $\alpha_{i}$ and $\alpha_{i+1}$ are integers). Hence,%
\begin{align*}
\left(  \alpha-\rho\right)  _{i}  &  =\underbrace{\alpha_{i}}_{\geq
\alpha_{i+1}+1}-\,N+i\\
&  \geq\alpha_{i+1}+1-N+i=\alpha_{i+1}-N+\left(  i+1\right)  =\left(
\alpha-\rho\right)  _{i+1}.
\end{align*}


We have proved this for each $i\in\left[  N-1\right]  $. In other words,%
\[
\left(  \alpha-\rho\right)  _{1}\geq\left(  \alpha-\rho\right)  _{2}\geq
\cdots\geq\left(  \alpha-\rho\right)  _{N}.
\]
Since we furthermore have $\left(  \alpha-\rho\right)  _{N}=\alpha
_{N}-\underbrace{\rho_{N}}_{=0}=\alpha_{N}\geq0$, we thus obtain%
\[
\left(  \alpha-\rho\right)  _{1}\geq\left(  \alpha-\rho\right)  _{2}\geq
\cdots\geq\left(  \alpha-\rho\right)  _{N}\geq0.
\]
Hence, $\alpha-\rho$ is a weakly decreasing $N$-tuple of nonnegative integers,
i.e., an $N$-partition. This proves the \textquotedblleft
Conversely\textquotedblright\ claim.
\end{proof}

\begin{example}
\label{exa.schur.alternant-quotients} Assume that $N>0$. Example
\ref{exa.schur.exa2} from Lecture 4 showed that%
\[
a_{\left(  1\right)  +\rho}=a_{\rho}\cdot\left(  x_{1}+x_{2}+\cdots
+x_{N}\right)  .
\]
(Keep in mind that we identify partitions of length $\leq N$ with
$N$-partitions, so that $\left(  1\right)  =\left(  1,0,0,\ldots,0\right)  $
with $N-1$ many zeroes. The $N$-tuple $\alpha$ from Example
\ref{exa.schur.exa2} is what we now call $\left(  1\right)  +\rho$.) Thus,%
\[
a_{\left(  1\right)  +\rho}/a_{\rho}=x_{1}+x_{2}+\cdots+x_{N}.
\]
And of course,%
\[
a_{\left(  {}\right)  +\rho}/a_{\rho}=1.
\]
More generally, it can be shown that%
\[
a_{\left(  n\right)  +\rho}/a_{\rho}=h_{n}\ \ \ \ \ \ \ \ \ \ \text{for each
}n\in\mathbb{N}%
\]
(this can be derived, e.g., by Laplace expansion from \cite[Theorem 1]%
{Nica22}). Furthermore, using the shorthand $\left(  1^{n}\right)  $ for the
$n$-tuple $\left(  1,1,\ldots,1\right)  $, we have%
\[
a_{\left(  1^{n}\right)  +\rho}/a_{\rho}=e_{n}\ \ \ \ \ \ \ \ \ \ \text{for
each }n\in\left\{  0,1,\ldots,N\right\}
\]
(see \cite[1.15]{Prasol94} for a particularly nice proof).

For another example, assume that $N\geq3$. Then, it can be shown that%
\begin{align*}
a_{\left(  2,1\right)  +\rho}/a_{\rho}  &  =\underbrace{\sum_{i<j}\left(
x_{i}^{2}x_{j}+x_{i}x_{j}^{2}\right)  }_{=m_{\left(  2,1\right)  }%
}+2\underbrace{\sum_{i<j<k}x_{i}x_{j}x_{k}}_{=m_{\left(  1,1,1\right)  }}\\
&  =m_{\left(  2,1\right)  }+2m_{\left(  1,1,1\right)  }.
\end{align*}
Do we see any patterns here? How can we generalize this?
\end{example}

Today we shall give a general answer to \textquotedblleft what is
$a_{\lambda+\rho}/a_{\rho}$, combinatorially?\textquotedblright. In the next
few lectures, we will prove this and more.

\subsubsection{Young diagrams, Young tableaux and Schur polynomials}

We first define some combinatorial notions. In \cite[Chapter 7]{21s}, they are
defined for $N$-partitions; here we define them for partitions (of arbitrary
length) instead, thus avoiding an unnecessary dependence on $N$. Recall that
$\left[  k\right]  $ denotes the set $\left\{  1,2,\ldots,k\right\}  $
whenever $k$ is an integer.

\begin{definition}
\label{def.schur.young-diagram} Let $\lambda=\left(  \lambda_{1},\lambda
_{2},\lambda_{3},\ldots\right)  $ be a partition (written as an infinite
sequence by inserting infinitely many zeroes at its end).

The \textbf{Young diagram} of $\lambda$ is defined to be the finite set%
\[
\left\{  \left(  i,j\right)  \ \mid\ i\in\left\{  1,2,3,\ldots\right\}  \text{
and }j\in\left[  \lambda_{i}\right]  \right\}  \subseteq\left\{
1,2,3,\ldots\right\}  ^{2},
\]
that is, the set of all pairs $\left(  i,j\right)  $ of positive integers $i$
and $j$ satisfying $j\leq\lambda_{i}$.

We visualize each pair $\left(  i,j\right)  \in\mathbb{Z}^{2}$ as a square of
size $1$ in the plane, centered at the point with matrix coordinates $\left(
i,j\right)  $. \textbf{Matrix coordinates} are Cartesian coordinates where the
x-axis runs top-to-bottom and the y-axis runs left-to-right. (This is the
coordinate system used to index the entries of a matrix: the usual way to
write down a matrix puts its $\left(  i,j\right)  $-th entry at matrix
coordinates $\left(  i,j\right)  $.)

So a Young diagram is a set of squares, which are called its \textbf{cells} or
\textbf{boxes}. It looks like a table of left-aligned rows, with the $i$-th
row (counted from the top) having $\lambda_{i}$ boxes. (By definition, it has
infinitely many rows, but only finitely many of them are nonempty.) Formally
speaking, a \textbf{cell} (aka \textbf{box}) is just a pair $\left(
i,j\right)  $ of integers.

We denote the Young diagram of $\lambda$ by $Y\left(  \lambda\right)  $. It
contains $\left\vert \lambda\right\vert $ boxes.
\end{definition}

\begin{example}
Here are some partitions $\lambda$ and (underneath them) their Young diagrams
$Y\left(  \lambda\right)  $ (visualized):%
\[%
\begin{tabular}
[c]{|c||c|c|c|c|c|}\hline
$\lambda$ & $\left(  3\right)  $ & $\left(  1,1,1\right)  $ & $\left(
2,1\right)  $ & $\left(  4,2,2,1\right)  $ & $\left(  3,2,1\right)  $\\\hline
$Y\left(  \lambda\right)  $ & $\ydiagram{3}$ & $\ydiagram{1,1,1}$ &
$\ydiagram{2,1}$ & $\ydiagram{4,2,2,1}$ & $\ydiagram{3,2,1}$\\\hline
\end{tabular}
\ \
\]
Formally speaking,%
\begin{align*}
Y\left(  \left(  3\right)  \right)   &  =\left\{  \left(  1,1\right)
,\ \left(  1,2\right)  ,\ \left(  1,3\right)  \right\}  ,\\
Y\left(  \left(  1,1,1\right)  \right)   &  =\left\{  \left(  1,1\right)
,\ \left(  2,1\right)  ,\ \left(  3,1\right)  \right\}  ,\\
Y\left(  \left(  2,1\right)  \right)   &  =\left\{  \left(  1,1\right)
,\ \left(  1,2\right)  ,\ \left(  2,1\right)  \right\}  ,
\end{align*}
and so on. Note that $Y\left(  \left(  {}\right)  \right)  $ is the empty set
$\varnothing$.
\end{example}

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

A \textbf{Young tableau} of \textbf{shape} $Y\left(  \lambda\right)  $ means a
map $T:Y\left(  \lambda\right)  \rightarrow\left\{  1,2,3,\ldots\right\}  $.
We visualize it as a way of filling the boxes of the Young diagram $Y\left(
\lambda\right)  $ with positive integers (one integer per box), simply by
writing into each box $c\in Y\left(  \lambda\right)  $ the corresponding value
$T\left(  c\right)  $.

We often omit the word \textquotedblleft Young\textquotedblright\ in
\textquotedblleft Young tableau\textquotedblright\ (so we just speak of
\textquotedblleft a tableau\textquotedblright). We furthermore write
\textquotedblleft of shape $\lambda$\textquotedblright\ instead of
\textquotedblleft of shape $Y\left(  \lambda\right)  $\textquotedblright.

The plural of the word \textquotedblleft tableau\textquotedblright\ is
\textquotedblleft tableaux\textquotedblright.
\end{definition}

\begin{warning}
\label{warn.schur.young-no-N}In \cite[Chapter 7]{21s}, I define Young tableaux
only for $N$-partitions $\lambda$ (not for all partitions $\lambda$), and I
require their entries to belong to $\left[  N\right]  $ (see \cite[Definition
7.3.5]{21s}). Here I do not impose this requirement, since it would limit us
too much.
\end{warning}

\begin{example}
\label{exa.schur.youngtab-433}Here is a Young tableau of shape $\left(
4,3,3\right)  $:%
\[
\begin{ytableau}
5 & 11 & 8 & 2\\
3 & 1 & 3\\
4 & 5 & 5
\end{ytableau}\ \ .
\]
Formally speaking, this is a map $Y\left(  \left(  4,3,3\right)  \right)
\rightarrow\left\{  1,2,3,\ldots\right\}  $ sending $\left(  1,1\right)  $ to
$5$, sending $\left(  1,2\right)  $ to $11$ and so on.
\end{example}

The visual interpretation of a tableau as a Young diagram filled with numbers
suggests some rather natural concepts to define:

\begin{itemize}
\item The \textbf{entry} of a tableau $T$ in a cell $\left(  i,j\right)  $
means the value $T\left(  i,j\right)  $.

\item The $u$\textbf{-th row} of a tableau $T$ means the sequence of all
entries of $T$ in the cells $\left(  i,j\right)  $ with $i=u$, read from left
to right.

\item The $v$\textbf{-th column} of a tableau $T$ means the sequence of all
entries of $T$ in the cells $\left(  i,j\right)  $ with $j=v$, read from top
to bottom.

\item The \textbf{boxes} (or \textbf{cells}) of a tableau $T$ are just the
boxes of its shape $Y\left(  \lambda\right)  $.

\item Two boxes are said to be \textbf{adjacent} if they have an edge in
common in the picture, i.e., if one of them has the form $\left(  i,j\right)
$ and the other has the form $\left(  i+1,j\right)  $ or $\left(
i,j+1\right)  $.

\item The words \textquotedblleft north\textquotedblright, \textquotedblleft
west\textquotedblright, etc. mean what they would mean in the picture: e.g.,
the box $\left(  2,4\right)  $ lies one step north and three steps west of
$\left(  3,7\right)  $.
\end{itemize}

Some Young tableaux are nicer than others:

\begin{definition}
\label{def.schur.semistandard} Let $\lambda$ be a partition.

A Young tableau $T$ of shape $\lambda$ is said to be \textbf{semistandard} if
its entries

\begin{itemize}
\item increase weakly along each row from left to right (i.e., we have
$T\left(  i,j\right)  \leq T\left(  i,j+1\right)  $ whenever $\left(
i,j\right)  $ and $\left(  i,j+1\right)  $ belong to $Y\left(  \lambda\right)
$), and

\item increase strictly down each column from top to bottom (i.e., we have
$T\left(  i,j\right)  <T\left(  i+1,j\right)  $ whenever $\left(  i,j\right)
$ and $\left(  i+1,j\right)  $ belong to $Y\left(  \lambda\right)  $).
\end{itemize}

We let $\operatorname*{SSYT}\left(  \lambda\right)  $ denote the set of all
semistandard Young tableaux of shape $\lambda$. Furthermore, for any
$k\in\mathbb{N}$, we let $\operatorname*{SSYT}\left(  \lambda,k\right)  $
denote the set of all semistandard Young tableaux of shape $\lambda$ whose
entries belong to $\left[  k\right]  $. We also abbreviate the word
\textquotedblleft semistandard Young tableau\textquotedblright\ as
\textbf{\textquotedblleft SSYT\textquotedblright.}
\end{definition}

\begin{definition}
\label{def.schur.standard} Furthermore, a Young tableau $T$ of shape $\lambda$
is said to be \textbf{standard} if it is semistandard and its entries are
$1,2,\ldots,\left\vert Y\left(  \lambda\right)  \right\vert $ with no
repetition. (Keep in mind: $\left\vert Y\left(  \lambda\right)  \right\vert $
is the number of boxes in $Y\left(  \lambda\right)  $.)
\end{definition}

\begin{example}
\label{exa.schur.tableau-types}Here are five tableaux of shape $\left(
3,2\right)  $:
\[%
\begin{array}
[c]{c@{\qquad}c@{\qquad}c@{\qquad}c@{\qquad}c}%
\begin{ytableau}1 & 3 & 2\\2 & 4\end{ytableau} &
\begin{ytableau}1 & 2 & 3\\2 & 2\end{ytableau} &
\begin{ytableau}1 & 1 & 2\\2 & 3\end{ytableau} &
\begin{ytableau}1 & 2 & 4\\3 & 6\end{ytableau} &
\begin{ytableau}1 & 2 & 4\\3 & 5\end{ytableau}\\[4pt]%
T_{1} & T_{2} & T_{3} & T_{4} & T_{5}%
\end{array}
\]


\begin{enumerate}
\item[\textbf{(a)}] The tableaux $T_{1}$ and $T_{2}$ are \textbf{not
semistandard}. In $T_{1}$, the first row decreases from $3$ to $2$ (although
the columns increase strictly). In $T_{2}$, the rows increase weakly, but the
second column has equal entries.

\item[\textbf{(b)}] The tableaux $T_{3}$ and $T_{4}$ are \textbf{semistandard
but not standard}. Their rows increase weakly and their columns increase
strictly. However, $T_{3}$ has repeated entries, while $T_{4}$ has entries
$1,2,3,4,6$ instead of $1,2,3,4,5$. Thus, distinct entries alone do not make a
semistandard tableau standard.

\item[\textbf{(c)}] The tableau $T_{5}$ is \textbf{standard}: it is
semistandard and contains each of $1,2,3,4,5$ exactly once.
\end{enumerate}
\end{example}

\begin{definition}
\label{def.schur.tableau-monomial} Let $T$ be a Young tableau of shape
$\lambda$ whose entries belong to $\left[  N\right]  $. Then, we define the
corresponding monomial%
\[
x_{T}:=\prod_{c\text{ is a cell of }Y\left(  \lambda\right)  }x_{T\left(
c\right)  }=\prod_{\left(  i,j\right)  \in Y\left(  \lambda\right)
}x_{T\left(  i,j\right)  }=\prod_{k=1}^{N}x_{k}^{\left(  \text{\# of
}k\text{'s in }T\right)  }%
\]
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}

%Added worked example: semistandardness and the tableau monomial.
\Needspace{10\baselineskip}

\begin{example}
\label{exa.schur.tableau-monomial} Let $N=3$. Consider the tableau
\[
T=\begin{ytableau}
1 & 1 & 3\\
2 & 3
\end{ytableau},\qquad\text{with corresponding monomial }x_{T}=x_{1}x_{1}%
x_{3}x_{2}x_{3}=x_{1}^{2}x_{2}x_{3}^{2}.
\]
The rows increase weakly and the columns increase strictly, so this tableau of
shape $\left(  3,2\right)  $ is semistandard. It is not standard, since some
entries repeat.
\end{example}

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

\end{definition}

\begin{example}
\label{exa.schur.rows-columns-21} \ 

\begin{enumerate}
\item[\textbf{(a)}] Assume that $N>0$. Let $n\in\mathbb{N}$. Then,%
\[
s_{\left(  n\right)  }=h_{n},
\]
because an SSYT of shape $\left(  n\right)  $ with entries in $\left[
N\right]  $ is just a weakly increasing $n$-tuple of elements of $\left[
N\right]  $ (the entries of its single row).

\item[\textbf{(b)}] Let $n\in\left\{  0,1,\ldots,N\right\}  $. Then,%
\[
s_{\left(  1^{n}\right)  }=e_{n},
\]
because an SSYT of shape $\left(  1^{n}\right)  $ with entries in $\left[
N\right]  $ is just a strictly increasing $n$-tuple of elements of $\left[
N\right]  $ (the entries of its single column).

\item[\textbf{(c)}] Assume that $N\geq3$. An SSYT of shape $\left(
2,1\right)  $ with entries in $\left[  N\right]  $ is a tableau of the form
\[
\begin{ytableau}
i & j\\
k
\end{ytableau}\qquad\text{with }i,j,k\in\left[  N\right]  ,\quad i\leq j\text{
and }i<k.
\]
Thus, we have\allowdisplaybreaks[4]
\begin{align*}
s_{\left(  2,1\right)  }  &  =\sum_{i\leq j;\ i<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) \\
&  =\sum_{\substack{i\leq j;\ i<k;\\j<k}}x_{i}x_{j}x_{k}+\sum_{\substack{i\leq
j;\ i<k;\\j\geq k}}x_{i}x_{j}x_{k}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split the sum}\\
\text{according to whether }j<k\text{ or }j\geq k
\end{array}
\right) \\
&  =\sum_{i\leq j<k}x_{i}x_{j}x_{k}+\sum_{i<k\leq j}x_{i}x_{j}x_{k}\\
&  =\sum_{i<j<k}x_{i}x_{j}x_{k}+\sum_{i=j<k}x_{i}x_{j}x_{k}+\sum_{i<k<j}%
x_{i}x_{j}x_{k}+\sum_{i<k=j}x_{i}x_{j}x_{k}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split the second sum}\\
\text{according to whether }k<j\text{ or }k=j\text{,}\\
\text{and the first sum according to}\\
\text{whether }i<j\text{ or }i=j
\end{array}
\right) \\
&  =\sum_{u<v<w}x_{u}x_{v}x_{w}+\sum_{u<v}x_{u}^{2}x_{v}+\sum_{u<v<w}%
\underbrace{x_{u}x_{w}x_{v}}_{=x_{u}x_{v}x_{w}}+\sum_{u<v}x_{u}x_{v}^{2}\\
&  =2\underbrace{\sum_{u<v<w}x_{u}x_{v}x_{w}}_{=m_{\left(  1,1,1\right)  }%
}+\underbrace{\sum_{u<v}x_{u}^{2}x_{v}+\sum_{u<v}x_{u}x_{v}^{2}}_{=m_{\left(
2,1\right)  }}\\
&  =2m_{\left(  1,1,1\right)  }+m_{\left(  2,1\right)  }.
\end{align*}

\end{enumerate}
\end{example}

%Added worked example: the eight tableaux contributing to s_(2,1).


\Needspace{14\baselineskip}

\begin{example}
\label{exa.schur.21-three-variables} Let $N=3$, and write $x,y,z$ for
$x_{1},x_{2},x_{3}$. The eight tableaux $T\in\operatorname{SSYT}\left(
\left(  2,1\right)  ,3\right)  $ are shown below, with their respective
monomials $x_{T}$ underneath:
\[%
\begin{array}
[c]{c@{\qquad}c@{\qquad}c@{\qquad}c}%
\begin{ytableau}1 & 1\\2\end{ytableau} &
\begin{ytableau}1 & 1\\3\end{ytableau} &
\begin{ytableau}1 & 2\\2\end{ytableau} &
\begin{ytableau}1 & 3\\3\end{ytableau}\\[4pt]%
x^{2}y & x^{2}z & xy^{2} & xz^{2}\\[12pt]
&  &  & \\
\begin{ytableau}2 & 2\\3\end{ytableau} &
\begin{ytableau}2 & 3\\3\end{ytableau} &
\begin{ytableau}1 & 2\\3\end{ytableau} &
\begin{ytableau}1 & 3\\2\end{ytableau}\\[4pt]%
y^{2}z & yz^{2} & xyz & xyz
\end{array}
\]
Hence,
\[
s_{\left(  2,1\right)  }=x^{2}y+x^{2}z+xy^{2}+xz^{2}+y^{2}z+yz^{2}+2xyz.
\]
In particular, the $xyz$-coefficient $2$ comes from the two different
semistandard tableaux that use $1,2,3$ exactly once. These are precisely the
standard tableaux of shape $\left(  2,1\right)  $.
\end{example}

These examples seem to suggest that $s_{\lambda}$ is always symmetric. Not
only that, but it also resembles the $a_{\lambda+\rho}/a_{\rho}$ that we have
seen above. Both of these are true, and we will prove them next week:

\begin{theorem}
\label{thm.schur.bialternant} Let $\lambda$ be a partition. Then:

\begin{enumerate}
\item[\textbf{(a)}] The polynomial $s_{\lambda}$ is symmetric.

\item[\textbf{(b)}] If $\ell\left(  \lambda\right)  \leq N$ (that is, if
$\lambda$ is an $N$-partition), then
\[
a_{\lambda+\rho}=a_{\rho}\cdot s_{\lambda}%
\]
(where $\lambda+\rho$ is the entrywise sum of the two $N$-partitions
$\lambda,\rho\in\mathbb{N}^{N}\subseteq\mathbb{Z}^{N}$).
\end{enumerate}
\end{theorem}

In particular, Conjecture \ref{conj.schur.pos} from Lecture 4 will follow from
this, since $s_{\lambda}$ obviously has nonnegative coefficients.

We will prove Theorem \ref{thm.schur.bialternant} \textbf{(a)} soon; but first
we will generalize our Schur polynomials $s_{\lambda}$ to the \textbf{skew
Schur polynomials} $s_{\lambda/\mu}$.

\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[Nica22]{Nica22}\href{https://arxiv.org/abs/2212.13624v2}{Bogdan Nica,
\textit{On an identity of Sylvester}, arXiv:2212.13624v2, Expositiones
Mathematicae, 2023.}

\bibitem[Prasol94]{Prasol94}V. Prasolov, \textit{Problems and Theorems in
Linear Algebra}, 1994, translated by Dimitry Leites.\newline\url{https://staff.math.su.se/mleites/books/prasolov-1994-problems.pdf}
\end{thebibliography}


\end{document}