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\ihead{Math 701, Fall 2026, Lecture 9, version \today}
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\begin{document}
\section*{Math 701 Fall 2026, Lecture 9: The Pieri rules}

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

(GPT-6 was used to create examples and to polish and detail some of the
arguments below. \textbf{The current version contains unverified GPT output.})
\medskip

\setcounter{section}{1}\setcounter{subsection}{7}\setcounter{subsubsection}{0}\setcounter{theo}{11}

\begin{proof}
[End of the proof of Lemma~\ref{lem.schur.stembridge}.]As we recall,
$\mathcal{X}$ denotes the set of all SSYTs $T\in\operatorname*{SSYT}\left(
\lambda/\mu,N\right)  $ that are not $\nu$-Yamanouchi. We have defined a map%
\[
f:\mathcal{X}\rightarrow\mathcal{X}%
\]
as follows: Let $T\in\mathcal{X}$. We let $j$ be the largest violator of $T$
(that is, the largest $j\geq1$ such that $\nu+\operatorname*{cont}\left(
\operatorname{col}_{\geq j}T\right)  $ is \textbf{not} an $N$-partition).
Define the $N$-tuple
\[
b:=\nu+\operatorname*{cont}\left(  \operatorname{col}_{\geq j}T\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{not an }N\text{-partition}\right)  .
\]
We let $k$ be the smallest misstep of $T$ (that is, the smallest $k\in\left[
N-1\right]  $ such that $b_{k}<b_{k+1}$). Now we define the tableau $f\left(
T\right)  $ by keeping columns $j,j+1,j+2,\ldots$ of $T$ unchanged while
applying the Bender--Knuth involution $\beta_{k}$ to the remaining columns
$1,2,\ldots,j-1$ (that is, replacing $\operatorname{col}_{<j}T$ by $\beta
_{k}\left(  \operatorname{col}_{<j}T\right)  $). We recall the five claims
that stood between us and the proof of Lemma \ref{lem.schur.stembridge}:

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

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

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

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

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

Claims 1 and 2 were already proved in Lecture 8. Now we prove the rest:

\begin{proof}
[Proof of Claim 3.]Fix $T\in\mathcal{X}$. We did not change anything in
columns $j,j+1,j+2,\ldots$ of $T$. So $j$ is still a violator of $f\left(
T\right)  $, while none of $j+1,j+2,j+3,\ldots$ is. So $j$ is still the
largest violator of $f\left(  T\right)  $. Furthermore, $k$ is still the
smallest misstep of $f\left(  T\right)  $, because the tuple $\nu
+\operatorname*{cont}\left(  \operatorname{col}_{\geq j}\left(  f\left(
T\right)  \right)  \right)  $ is precisely the tuple $\nu+\operatorname*{cont}%
\left(  \operatorname{col}_{\geq j}T\right)  $ (again since we did not change
anything in columns $j,j+1,j+2,\ldots$ of $T$). Hence, when we apply $f$ again
to $f\left(  T\right)  $, the course is set precisely in the same way as it
was when we applied $f$ to $T$: we again leave columns $j,j+1,j+2,\ldots$
unchanged and apply $\beta_{k}$ to the first $j-1$ columns. Thus, in total,
the first $j-1$ columns of $T$ end up undergoing $\beta_{k}$ twice, which
brings them back to their original state (since $\beta_{k}$ is an involution);
meanwhile, all the remaining columns of $T$ never change at all. Altogether,
we thus conclude that $f\left(  f\left(  T\right)  \right)  =T$. This proves
Claim 3.
\end{proof}

It remains to prove Claims 4 and 5. In these proofs, we will use the following
notation: For any $N$-tuple $\alpha$ and any $i\in\left[  N\right]  $, we let
$\alpha_{i}$ denote the $i$-th entry of $\alpha$. If $\alpha\in\mathbb{N}^{N}$
is an $N$-tuple and $\sigma\in S_{N}$ is a permutation, then $\alpha
\circ\sigma$ denotes the $N$-tuple $\left(  \alpha_{\sigma\left(  1\right)
},\alpha_{\sigma\left(  2\right)  },\ldots,\alpha_{\sigma\left(  N\right)
}\right)  \in\mathbb{N}^{N}$. This gives a right action of the group $S_{N}$
on $\mathbb{N}^{N}$; it satisfies
\[
\left(  \alpha+\beta\right)  \circ\sigma=\alpha\circ\sigma+\beta\circ
\sigma\ \ \ \ \ \ \ \ \ \ \text{for all }\alpha,\beta\in\mathbb{N}^{N}\text{
and }\sigma\in S_{N}.
\]
We shall use this equality without further mention. Note also that if
$\alpha\in\mathbb{N}^{N}$ and $k\in\left[  N-1\right]  $, then the $N$-tuple
$\alpha\circ s_{k}$ is obtained from $\alpha$ by swapping the $k$-th and
$\left(  k+1\right)  $-st entries. In particular, if $\alpha_{k}=\alpha_{k+1}%
$, then $\alpha\circ s_{k}=\alpha$.

\begin{proof}
[Proof of Claim 4.]We must prove that $a_{\nu+\operatorname*{cont}T+\rho
}+a_{\nu+\operatorname*{cont}\left(  f\left(  T\right)  \right)  +\rho}=0$.

We shall show that the $N$-tuple $\nu+\operatorname*{cont}\left(  f\left(
T\right)  \right)  +\rho$ is obtained from $\nu+\operatorname*{cont}T+\rho$ by
swapping the $k$-th and $\left(  k+1\right)  $-st entries. Thus, we shall show
that%
\begin{equation}
\nu+\operatorname*{cont}\left(  f\left(  T\right)  \right)  +\rho=\left(
\nu+\operatorname*{cont}T+\rho\right)  \circ s_{k}.
\label{eq.pf.lem.schur.stembridge.shifted-swap}%
\end{equation}


\begin{sloppypar}
By Lemma \ref{lem.schur.alternant-swap} \textbf{(b)}, this will entail
$a_{\nu+\operatorname*{cont}\left(  f\left(  T\right)  \right)  +\rho}%
=-a_{\nu+\operatorname*{cont}T+\rho}$, thus $a_{\nu+\operatorname*{cont}%
T+\rho}+a_{\nu+\operatorname*{cont}\left(  f\left(  T\right)  \right)  +\rho
}=0$, which will prove Claim 4.
\end{sloppypar}

By the definition of $f\left(  T\right)  $, we have
\[
\operatorname{col}_{\geq j}\left(  f\left(  T\right)  \right)
=\operatorname{col}_{\geq j}T
\]
(since only the first $j-1$ columns change from $T$ to $f\left(  T\right)  $).
Moreover, the first $j-1$ columns of $f\left(  T\right)  $ are obtained from
the corresponding columns of $T$ by the Bender--Knuth involution $\beta_{k}$;
thus, $\operatorname{col}_{<j}\left(  f\left(  T\right)  \right)  =\beta
_{k}\left(  \operatorname{col}_{<j}T\right)  $, so that
\[
\operatorname*{cont}\left(  \operatorname{col}_{<j}\left(  f\left(  T\right)
\right)  \right)  =\operatorname*{cont}\left(  \beta_{k}\left(
\operatorname{col}_{<j}T\right)  \right)  =\operatorname*{cont}\left(
\operatorname{col}_{<j}T\right)  \circ s_{k}%
\]
(since the Bender--Knuth involution $\beta_{k}$ swaps the \#s of $k$'s and
$\left(  k+1\right)  $'s in a tableau, while leaving all other \#s unchanged
(by Observation 4 in the proof of Theorem \ref{thm.schur.skew-symmetry});
thus, it acts on the content by swapping the $k$-th and $\left(  k+1\right)
$-st entries). Finally,%
\begin{equation}
\operatorname*{cont}T=\operatorname*{cont}\left(  \operatorname{col}_{\geq
j}T\right)  +\operatorname*{cont}\left(  \operatorname{col}_{<j}T\right)
\label{eq.pf.lem.schur.stembridge.cont+cont}%
\end{equation}
(since each entry of $T$ belongs either to $\operatorname{col}_{\geq j}T$ or
to $\operatorname{col}_{<j}T$).

Now, recall that $b=\nu+\operatorname*{cont}\left(  \operatorname{col}_{\geq
j}T\right)  $. Define also $a:=\operatorname*{cont}\left(  \operatorname{col}%
_{<j}T\right)  $. So%
\[
\nu+\underbrace{\operatorname*{cont}T}_{=\operatorname*{cont}\left(
\operatorname{col}_{\geq j}T\right)  +\operatorname*{cont}\left(
\operatorname{col}_{<j}T\right)  }=\underbrace{\nu+\operatorname*{cont}\left(
\operatorname{col}_{\geq j}T\right)  }_{=b}+\underbrace{\operatorname*{cont}%
\left(  \operatorname{col}_{<j}T\right)  }_{=a}=b+a.
\]
Meanwhile, (\ref{eq.pf.lem.schur.stembridge.cont+cont}) (applied to $f\left(
T\right)  $ instead of $T$) yields%
\begin{align*}
\operatorname*{cont}\left(  f\left(  T\right)  \right)   &
=\operatorname*{cont}\left(  \underbrace{\operatorname{col}_{\geq j}\left(
f\left(  T\right)  \right)  }_{=\operatorname{col}_{\geq j}T}\right)
+\underbrace{\operatorname*{cont}\left(  \operatorname{col}_{<j}\left(
f\left(  T\right)  \right)  \right)  }_{=\operatorname*{cont}\left(
\operatorname{col}_{<j}T\right)  \circ s_{k}}\\
&  =\operatorname*{cont}\left(  \operatorname{col}_{\geq j}T\right)
+\underbrace{\operatorname*{cont}\left(  \operatorname{col}_{<j}T\right)
}_{=a}\circ\,s_{k}=\operatorname*{cont}\left(  \operatorname{col}_{\geq
j}T\right)  +a\circ s_{k}.
\end{align*}
Adding $\nu$ to both sides, we thus obtain%
\begin{equation}
\nu+\operatorname*{cont}\left(  f\left(  T\right)  \right)  =\underbrace{\nu
+\operatorname*{cont}\left(  \operatorname{col}_{\geq j}T\right)  }%
_{=b}+\,a\circ s_{k}=b+a\circ s_{k}%
.\ \ \ \ \ \ \ \ \ \ \label{eq.pf.lem.schur.stembridge.nuft}%
\end{equation}


But $\rho=\left(  N-1,N-2,\ldots,1,0\right)  $ yields $\rho_{k+1}=\rho_{k}-1$.
Recall also that $b_{k+1}=b_{k}+1$, as we saw previously in the construction
of $f\left(  T\right)  $. Thus,%
\[
\left(  b+\rho\right)  _{k+1}=\underbrace{b_{k+1}}_{=b_{k}+1}+\underbrace{\rho
_{k+1}}_{=\rho_{k}-1}=\left(  b_{k}+1\right)  +\left(  \rho_{k}-1\right)
=b_{k}+\rho_{k}=\left(  b+\rho\right)  _{k}%
\]
and therefore
\[
\left(  b+\rho\right)  \circ s_{k}=b+\rho
\]
(since an $N$-tuple $\alpha\in\mathbb{N}^{N}$ that satisfies $\alpha
_{k}=\alpha_{k+1}$ always satisfies $\alpha\circ s_{k}=\alpha$).

Altogether,%
\begin{align*}
\left(  \underbrace{\nu+\operatorname*{cont}T}_{=b+a}+\,\rho\right)  \circ
s_{k}  &  =\underbrace{\left(  b+a+\rho\right)  }_{=\left(  b+\rho\right)
+a}\circ\,s_{k}=\left(  \left(  b+\rho\right)  +a\right)  \circ s_{k}\\
&  =\underbrace{\left(  b+\rho\right)  \circ s_{k}}_{=b+\rho}+\,a\circ
s_{k}=b+\rho+a\circ s_{k}\\
&  =\underbrace{b+a\circ s_{k}}_{\substack{=\nu+\operatorname*{cont}\left(
f\left(  T\right)  \right)  \\\text{(by (\ref{eq.pf.lem.schur.stembridge.nuft}%
))}}}+\,\rho=\nu+\operatorname*{cont}\left(  f\left(  T\right)  \right)
+\rho,
\end{align*}
which proves (\ref{eq.pf.lem.schur.stembridge.shifted-swap}). As we explained
above, this completes the proof of Claim 4.
\end{proof}

\begin{proof}
[Proof of Claim 5.]Assume that $f\left(  T\right)  =T$. But Claim 4 tells us
that
\[
a_{\nu+\operatorname*{cont}T+\rho}+a_{\nu+\operatorname*{cont}\left(  f\left(
T\right)  \right)  +\rho}=0.
\]
In view of $f\left(  T\right)  =T$, this simplifies to $a_{\nu
+\operatorname*{cont}T+\rho}+a_{\nu+\operatorname*{cont}T+\rho}=0$. That is,
$2a_{\nu+\operatorname*{cont}T+\rho}=0$. If $K=\mathbb{Z}$, then this
immediately implies that $a_{\nu+\operatorname*{cont}T+\rho}=0$, since we can
cancel the factor $2$ in a polynomial ring over $\mathbb{Z}$. So at least we
have proved the equality $a_{\nu+\operatorname*{cont}T+\rho}=0$ for
$K=\mathbb{Z}$. Now, applying the coefficientwise ring morphism $\mathcal{P}%
_{\mathbb{Z}}\rightarrow\mathcal{P}_{K}$ from Proposition
\ref{prop.symp.functoriality} \textbf{(b)} proves the same equality over any
$K$ (since this morphism sends the $\alpha$-alternant $a_{\alpha}$ defined
over $\mathbb{Z}$ to the analogous $\alpha$-alternant $a_{\alpha}$ defined
over $K$ for any $\alpha\in\mathbb{N}^{N}$); thus, Claim 5 is proved. Note
that this argument uses functoriality (in the form of the morphism
$\mathcal{P}_{\mathbb{Z}}\rightarrow\mathcal{P}_{K}$) to avoid the need for
cancelling $2$ in $\mathcal{P}_{K}$; roughly speaking, it transfers this
cancellation to the polynomial ring $\mathcal{P}_{\mathbb{Z}}$, where it can
be easily justified.

Alternatively, we can argue as follows: In the proof of Claim 4, we proved the
equality (\ref{eq.pf.lem.schur.stembridge.shifted-swap}). Since $f\left(
T\right)  =T$, this equality rewrites as%
\[
\nu+\operatorname*{cont}T+\rho=\left(  \nu+\operatorname*{cont}T+\rho\right)
\circ s_{k}.
\]
This means that the tuple $\nu+\operatorname*{cont}T+\rho$ has equal entries
in positions $k$ and $k+1$. Therefore, Lemma \ref{lem.schur.alternant-swap}
\textbf{(a)} gives $a_{\nu+\operatorname*{cont}T+\rho}=0$. So Claim 5 is
proved again. This argument works directly over any $K$.
\end{proof}

\Needspace{18\baselineskip}

\begin{example}
\label{exa.schur.stembridge-fixed-point}Fixed points of $f$ really can occur.
Let $N=2$, $\nu=(0,0)$, $\lambda=(2,1)$ and $\mu=\varnothing$, and consider
the tableau
\[
T=\begin{ytableau}
1 & 2\\
2
\end{ytableau}\ .
\]
Its largest violator is $j=2$, since the last column has content $(0,1)$,
whereas every column restriction starting at $3$ or later is empty. Thus,
$b=(0,1)$ and $k=1$. In the first column, the $1$ and $2$ are matched, so
$\beta_{1}$ leaves that column unchanged. Consequently, $f(T)=T$. Just as
Claim 5 predicts, the corresponding summand $a_{\nu+\operatorname*{cont}%
T+\rho}$ is zero: since $\operatorname*{cont}T=(1,2)$ and $\rho=(1,0)$, we
have
\[
a_{\nu+\operatorname*{cont}T+\rho}=a_{(2,2)}=\det%
\begin{pmatrix}
x_{1}^{2} & x_{1}^{2}\\
x_{2}^{2} & x_{2}^{2}%
\end{pmatrix}
=0.
\]

\end{example}

We can now complete the cancellation. The set $\mathcal{X}$ is finite, since
its tableaux have a fixed finite shape and entries in $[N]$. By Claim 3, it
splits into one-element and two-element orbits of $f$. A one-element orbit
contributes $0$ to the sum $\sum_{T\in\mathcal{X}}a_{\nu+\operatorname*{cont}%
T+\rho}$ by Claim 5, and a two-element orbit contributes $0$ by Claim 4.
Hence,
\[
\sum_{T\in\mathcal{X}}a_{\nu+\operatorname*{cont}T+\rho}=0.
\]
But we proved in Lecture 8 that%
\[
a_{\nu+\rho}\cdot s_{\lambda/\mu}=\sum_{T\in\operatorname*{SSYT}\left(
\lambda/\mu,N\right)  }a_{\nu+\operatorname*{cont}T+\rho}.
\]
Splitting this sum into two parts, according to whether $T$ is $\nu
$-Yamanouchi or not, we find%
\[
a_{\nu+\rho}\cdot s_{\lambda/\mu}=\sum_{\substack{T\in\operatorname*{SSYT}%
\left(  \lambda/\mu,N\right)  ;\\T\text{ is }\nu\text{-Yamanouchi}}%
}a_{\nu+\operatorname*{cont}T+\rho}+\sum_{\substack{T\in\operatorname*{SSYT}%
\left(  \lambda/\mu,N\right)  ;\\T\text{ is not }\nu\text{-Yamanouchi}}%
}a_{\nu+\operatorname*{cont}T+\rho}.
\]
Since%
\begin{align*}
\sum_{\substack{T\in\operatorname*{SSYT}\left(  \lambda/\mu,N\right)
;\\T\text{ is not }\nu\text{-Yamanouchi}}}a_{\nu+\operatorname*{cont}T+\rho}
&  =\sum_{T\in\mathcal{X}}a_{\nu+\operatorname*{cont}T+\rho}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathcal{X}\right) \\
&  =\sum_{\substack{T\in\mathcal{X};\\f\left(  T\right)  =T}%
}\underbrace{a_{\nu+\operatorname*{cont}T+\rho}}_{\substack{=0\\\text{(by
Claim 5)}}}+\sum_{\substack{T\in\mathcal{X};\\f\left(  T\right)  \neq
T}}a_{\nu+\operatorname*{cont}T+\rho}\\
&  =\sum_{\substack{T\in\mathcal{X};\\f\left(  T\right)  \neq T}%
}a_{\nu+\operatorname*{cont}T+\rho}=0
\end{align*}


\begin{sloppypar}
\noindent(because the tableaux indexing this sum can be split into unordered
pairs $\left\{  T,f\left(  T\right)  \right\}  $, and each pair contributes
two addends $a_{\nu+\operatorname*{cont}T+\rho}$ and $a_{\nu
+\operatorname*{cont}\left(  f\left(  T\right)  \right)  +\rho}$ that sum to
$0$ by Claim 4), this rewrites as%
\[
a_{\nu+\rho}\cdot s_{\lambda/\mu}=\sum_{\substack{T\in\operatorname*{SSYT}%
\left(  \lambda/\mu,N\right)  ;\\T\text{ is }\nu\text{-Yamanouchi}}%
}a_{\nu+\operatorname*{cont}T+\rho}+0=\sum_{\substack{T\in\operatorname*{SSYT}%
\left(  \lambda/\mu,N\right)  ;\\T\text{ is }\nu\text{-Yamanouchi}}%
}a_{\nu+\operatorname*{cont}T+\rho}.
\]

\end{sloppypar}

This proves Lemma \ref{lem.schur.stembridge} (and thus completes the proofs of
Theorem \ref{thm.schur.generalized-lr} and Theorem \ref{thm.schur.bialternant}
\textbf{(b)} that relied on it).
\end{proof}

\subsection{Intermezzo: Antisymmetric polynomials}

Antisymmetric polynomials are an analogue of symmetric polynomials, much as
antisymmetric matrices are analogous to symmetric matrices. Let us define them:

\begin{definition}
\label{def.symp.antisymmetric}A polynomial $f\in\mathcal{P}$ is said to be
\textbf{antisymmetric} if it satisfies%
\[
\sigma\cdot f=\left(  -1\right)  ^{\sigma}\cdot f\ \ \ \ \ \ \ \ \ \ \text{for
all }\sigma\in S_{N}.
\]
Here, $\left(  -1\right)  ^{\sigma}$ means the sign of the permutation
$\sigma$.
\end{definition}

Again, this condition is equivalent to requiring that $s_{k}\cdot f=-f$ for
all $k\in\left[  N-1\right]  $ (for a similar reason as Lemma
\ref{lem.symp.simple-transpositions} in Lecture 3).

\begin{noncompile}
Indeed, the forward implication is immediate. Conversely, write any $\sigma\in
S_{N}$ as a product of $r$ simple transpositions. Applying them one at a time
gives $\sigma\cdot f=(-1)^{r}f=(-1)^{\sigma}f$. This also covers $N=0$ and
$N=1$, when $S_{N}$ contains only the identity.
\end{noncompile}

A standard example of an antisymmetric polynomial is the Vandermonde
determinant $a_{\rho}=\prod_{i<j}\left(  x_{i}-x_{j}\right)  $. But beware:
Over a field of characteristic $2$, antisymmetric polynomials are the same as
symmetric ones, since $-1=1$. Our definition of antisymmetric polynomials is
not well-adapted to \textquotedblleft characteristic-$2$\textquotedblright%
\ behavior. This can be fixed by switching to a more careful definition (see
Remark \ref{rmk.symp.antisymmetric-arbitrary-ring} below), but we shall
instead fix it by stating our main result (Theorem
\ref{thm.symp.antisymmetric-bases}) under the condition that we are not in
characteristic $2$ (or, more precisely, that $2$ is a regular element of our
base ring $K$). We will mostly need the case $K=\mathbb{Z}$ anyway.

\begin{definition}
\label{def.symp.antisymmetric-module}Let $\mathcal{A}$ be the $K$-module of
all antisymmetric polynomials $f\in\mathcal{P}$. (In general, $\mathcal{A}$ is
not a subring of $\mathcal{P}$: the product of two antisymmetric polynomials
is symmetric, since the two signs cancel. But it is obviously a $K$-submodule.)
\end{definition}

Every alternant $a_{\alpha}$ belongs to $\mathcal{A}$: Indeed, permuting the
variables $x_{1},x_{2},\ldots,x_{N}$ permutes the rows of the determinant
$\det\left(  \left(  x_{i}^{\alpha_{j}}\right)  _{i,j\in\left[  N\right]
}\right)  $ defining $a_{\alpha}$, and hence multiplies it by the sign of the permutation.

\Needspace{13\baselineskip}

\begin{example}
\label{exa.symp.antisymmetric-cautions}Let $N=2$, and write $x,y$ for
$x_{1},x_{2}$.

\begin{enumerate}
\item[\textbf{(a)}] Over $\mathbb{Z}$, the polynomial $x-y$ is antisymmetric,
but $(x-y)^{2}$ is not (swapping $x$ and $y$ leaves this nonzero square
unchanged, rather than multiplying it by $-1$).

\item[\textbf{(b)}] Over $K=\mathbb{Z}/4\mathbb{Z}$, the nonzero constant
polynomial $2$ is antisymmetric, because $2=-2$ in $K$. Yet it is not a
$K$-linear combination of alternants $a_{\alpha}$; nor is it divisible by
$a_{\rho}=x-y$. This explains why we require the regularity of $2$ in Theorem
\ref{thm.symp.antisymmetric-bases} below (at least in parts \textbf{(a)},
\textbf{(b)} and \textbf{(c)}; part \textbf{(d)} does not actually need this requirement).
\end{enumerate}
\end{example}

We can now state the antisymmetric analogue of Theorem \ref{thm.symp.m-basis}:

\begin{theorem}
\label{thm.symp.antisymmetric-bases}Assume that $2$ is regular in $K$ (that
is, any $u\in K$ satisfying $2u=0$ must itself be $0$). Then:

\begin{enumerate}
\item[\textbf{(a)}] The family $\left(  a_{\alpha}\right)  _{\alpha
\in\mathbb{N}^{N}\text{ is strictly decreasing}}$ is a basis of the $K$-module
$\mathcal{A}$.

\item[\textbf{(b)}] Each antisymmetric polynomial $f\in\mathcal{A}$ satisfies%
\[
f=\sum_{\substack{\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{N}\right)  \in\mathbb{N}^{N}\\\text{is strictly decreasing}}%
}\ \ \underbrace{\left(  \left[  x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\cdots
x_{N}^{\alpha_{N}}\right]  f\right)  }_{\text{the }x_{1}^{\alpha_{1}}%
x_{2}^{\alpha_{2}}\cdots x_{N}^{\alpha_{N}}\text{-coefficient of }f}a_{\alpha
}.
\]
Or, to put this more compactly:%
\begin{equation}
f=\sum_{\substack{\alpha\in\mathbb{N}^{N}\\\text{is strictly decreasing}%
}}\left(  \left[  x^{\alpha}\right]  f\right)  a_{\alpha}.
\label{eq.thm.symp.antisymmetric-bases.b.claim}%
\end{equation}


\item[\textbf{(c)}] The map
\begin{align*}
\mathcal{S}  &  \rightarrow\mathcal{A},\\
f  &  \mapsto a_{\rho}f
\end{align*}
is a $K$-module isomorphism. (In particular, $\mathcal{A}=a_{\rho}\mathcal{S}$.)

\item[\textbf{(d)}] The family $\left(  s_{\lambda}\right)  _{\lambda\text{ is
an }N\text{-partition}}$ is a basis of the $K$-module $\mathcal{S}$.
\end{enumerate}
\end{theorem}

\begin{remark}
\label{rmk.symp.antisymmetric-arbitrary-ring}Part \textbf{(d)} of Theorem
\ref{thm.symp.antisymmetric-bases} holds even if we don't assume anything
about $K$ (and we will prove this later). For parts \textbf{(a)}, \textbf{(b)}
and \textbf{(c)}, if we want $K$ to be arbitrary, we need to adapt the
definition of \textquotedblleft antisymmetric\textquotedblright\ to
additionally require that $f$ becomes $0$ whenever two of the $N$ variables
are set equal. Antisymmetric polynomials $f$ that satisfy this additional
requirement are called \emph{alternating}; if we replace $\mathcal{A}$ by the
$K$-module of alternating polynomials, then parts \textbf{(a)}, \textbf{(b)}
and \textbf{(c)} of Theorem \ref{thm.symp.antisymmetric-bases} still hold (see
\cite[(SYM.6.1), (SYM.6.6) and (SYM.6.7)]{LLPT}). We will not need these
subtleties, since the case most important to us is the case $K=\mathbb{Z}$;
almost all combinatorially defined symmetric polynomials are defined over
$\mathbb{Z}$ and can be mapped into $\mathcal{P}_{K}$ for any $K$ by functoriality.
\end{remark}

Parts \textbf{(a)} and \textbf{(b)} of Theorem
\ref{thm.symp.antisymmetric-bases} are analogues of respective parts of
Theorem \ref{thm.symp.m-basis} in Lecture 4; their proofs are mostly similar,
but somewhat complicated by signs and by monomials with equal exponents. Due
to these complications, we shall give all the details. First, recall
Definition \ref{def.symp.monomial-sort} in Lecture 4. Note that if
$a\in\mathbb{N}^{N}$ is an $N$-tuple whose entries are distinct, then
$\operatorname*{sort}a$ is a \textbf{strictly} decreasing $N$-tuple. The
following is an analogue of Lemma \ref{lem.symp.sort-a} \textbf{(b)} in
Lecture 4:

\begin{lemma}
\label{lem.asymp.sort-a}Assume that $2$ is regular in $K$ (that is, any $u\in
K$ satisfying $2u=0$ must itself be $0$). Let $f\in\mathcal{P}$ be a
polynomial. Then:

\begin{enumerate}
\item[\textbf{(a)}] The polynomial $f$ is antisymmetric (i.e., belongs to
$\mathcal{A}$) if and only if all $a\in\mathbb{N}^{N}$ satisfy%
\[
\left[  x^{a}\right]  f=\left(  -1\right)  ^{\sigma}\cdot\left[
x^{\operatorname*{sort}a}\right]  f,
\]
where $\sigma\in S_{N}$ is such that $\operatorname*{sort}a=a\circ\sigma$.
(Here,  $a\circ\sigma$ denotes the $N$-tuple $\left(  a_{\sigma\left(
1\right)  },a_{\sigma\left(  2\right)  },\ldots,a_{\sigma\left(  N\right)
}\right)  $, where $a=\left(  a_{1},a_{2},\ldots,a_{N}\right)  $.)

\item[\textbf{(b)}] If $f$ is antisymmetric, and if $a\in\mathbb{N}^{N}$ is an
$N$-tuple with two equal entries, then%
\[
\left[  x^{a}\right]  f=0.
\]

\end{enumerate}
\end{lemma}

\begin{proof}
TODO.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.symp.antisymmetric-bases}.]\textbf{(a)},
\textbf{(b)} This is similar to proving that $\left(  m_{\lambda}\right)
_{\lambda\text{ is an }N\text{-partition}}$ is a basis of $\mathcal{S}$ (see
Theorem \ref{thm.symp.m-basis} \textbf{(a)} in Lecture 4). The only
complication now is that signs appear, and we must rule out monomials with
equal exponents.

Linear independence of $\left(  a_{\alpha}\right)  _{\alpha\in\mathbb{N}%
^{N}\text{ is strictly decreasing}}$ is easy: Each
\[
a_{\alpha}=\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\sigma\cdot
x^{\alpha}%
\]
contains the monomial $x^{\alpha}=x_{1}^{\alpha_{1}}\cdots x_{N}^{\alpha_{N}}$
with coefficient $1$ (nothing can cancel it because $\alpha$ is strictly
decreasing, so every nonidentity permutation of its entries changes the
tuple), and contains none of the $x^{\beta}$ for other strictly decreasing
$N$-tuples $\beta$. So any linear combination $\sum c_{\alpha}a_{\alpha}$ will
\textquotedblleft remember\textquotedblright\ the coefficient $c_{\alpha}$ as
its $x^{\alpha}$-coefficient; in particular, if it is $0$, then all
$c_{\alpha}$'s must be $0$.

It remains to show that this family $\left(  a_{\alpha}\right)  _{\alpha
\in\mathbb{N}^{N}\text{ is strictly decreasing}}$ spans $\mathcal{A}$. So we
fix an antisymmetric polynomial $f\in\mathcal{A}$, and we try to decompose it
as a linear combination of $\left(  a_{\alpha}\right)  _{\alpha\in
\mathbb{N}^{N}\text{ is strictly decreasing}}$. Namely, we shall show that%
\[
f=\sum_{\alpha\in\mathbb{N}^{N}\text{ is strictly decreasing}}\left(  \left[
x^{\alpha}\right]  f\right)  a_{\alpha}.
\]
Only finitely many coefficients in this sum are nonzero, since $f$ is a
polynomial. Both sides are antisymmetric, because every $a_{\alpha}$ is
antisymmetric. They agree in all coefficients of the strictly decreasing
monomials (that is, monomials whose exponent tuples are strictly decreasing),
by the coefficient calculation just used for linear independence.

Here is why this also gives equality of the coefficients of all monomials with
distinct exponents. Each such monomial is $m=\sigma\cdot x^{\alpha}$ for some
strictly decreasing $\alpha$ and some $\sigma\in S_{N}$. For any antisymmetric
polynomial $g$, we have
\[
[x^{\alpha}]g=[m](\sigma\cdot g)=(-1)^{\sigma}[m]g, \qquad\text{so}%
\qquad[m]g=(-1)^{\sigma}[x^{\alpha}]g.
\]
Applying this to both sides proves that their $m$-coefficients agree.

It remains to handle those monomials whose exponent tuples have repeated
entries, such as $x^{2}yz^{2}$. The RHS of the above equality does not contain
any such monomials, since none of these $a_{\alpha}$'s contains them. We must
show that the LHS does not contain them either. In other words, we must show
that if $\alpha\in\mathbb{N}^{N}$ has two equal entries -- say, $\alpha
_{i}=\alpha_{j}$ for $i<j$ -- then $\left[  x^{\alpha}\right]  f=0$.

But $f$ is antisymmetric, so the transposition $t_{i,j}\in S_{N}$ that swaps
$i$ with $j$ satisfies $t_{i,j}\cdot f=\left(  -1\right)  ^{t_{i,j}}\cdot
f=-f$. But this transposition $t_{i,j}$ does not change the tuple $\alpha$
(since $\alpha_{i}=\alpha_{j}$), and thus does not change the monomial
$x^{\alpha}$. So
\[
\left[  x^{\alpha}\right]  f=\left[  x^{\alpha}\right]  \underbrace{\left(
t_{i,j}\cdot f\right)  }_{=-f}=\left[  x^{\alpha}\right]  \left(  -f\right)
=-\left[  x^{\alpha}\right]  f.
\]
Therefore, $2\left[  x^{\alpha}\right]  f=0$. Since $2$ is regular in $K$,
this entails $\left[  x^{\alpha}\right]  f=0$. Thus all coefficients agree,
which proves the desired expansion and completes part \textbf{(a)} and
\textbf{(b)}. \medskip

\textbf{(c)} Well-definedness: If $f\in\mathcal{S}$, then $a_{\rho}%
f\in\mathcal{A}$, since
\[
\sigma\cdot\left(  a_{\rho}f\right)  =\left(  \sigma\cdot a_{\rho}\right)
\left(  \sigma\cdot f\right)  =(-1)^{\sigma}a_{\rho}f
\]
for every $\sigma\in S_{N}$.

$K$-linearity is even more obvious.

\begin{sloppypar}
Injectivity follows from the regularity of $a_{\rho}$, proved in Lecture 8
while deducing Theorem \ref{thm.schur.generalized-lr} from Stembridge's lemma.
Recall the argument: each factor $x_{i}-x_{j}$ is monic in $x_{i}$, so
multiplication by it is injective; hence multiplication by their product
$a_{\rho}$ is injective as well.
\end{sloppypar}

Surjectivity: Every $f\in\mathcal{A}$ is a linear combination of $\left(
a_{\alpha}\right)  _{\alpha\in\mathbb{N}^{N}\text{ is strictly decreasing}}$
by part \textbf{(a)}. But each strictly decreasing $N$-tuple $\alpha$ can be
written uniquely as $\lambda+\rho$ for an $N$-partition $\lambda$ (by Lemma
\ref{lem.schur.rho-shift}), and furthermore we then have $a_{\alpha
}=a_{\lambda+\rho}=a_{\rho}s_{\lambda}$ (by Theorem
\ref{thm.schur.bialternant} \textbf{(b)}). So each $a_{\alpha}$ lies in the
image of the map; therefore the map is surjective. \medskip

\textbf{(d)} The isomorphism from part \textbf{(c)} sends $s_{\lambda}$ to
$a_{\lambda+\rho}$. By Lemma \ref{lem.schur.rho-shift}, as $\lambda$ runs
through all $N$-partitions, $\lambda+\rho$ runs through all strictly
decreasing $N$-tuples in $\mathbb{N}^{N}$, without repetitions. Thus, the
$s_{\lambda}$ are exactly the preimages of the basis in part \textbf{(a)}, and
hence form a basis of $\mathcal{S}$.
\end{proof}

In particular, any symmetric polynomial defined over $\mathbb{Z}$ (for
example, any product of known symmetric polynomials such as $h_{5}p_{7}%
e_{3}s_{\left(  3,2\right)  /\left(  1\right)  }$) can be expanded in the
Schur basis $\left(  s_{\lambda}\right)  _{\lambda\text{ is an }%
N\text{-partition}}$ -- first over the ring $\mathbb{Z}$, then (by
functoriality) over any $K$. This is called the \textbf{Schur expansion} of
that polynomial. We will now try to find Schur expansions of many common
symmetric polynomials.

\begin{example}
\label{exa.symp.antisymmetric-schur-expansion}Let $N=2$, $K=\mathbb{Z}$, and
write $x,y$ for $x_{1},x_{2}$. The antisymmetric polynomial
\[
F=2x^{3}-x^{2}y+xy^{2}-2y^{3}
\]
has the alternant expansion
\[
F=2a_{(3,0)}-a_{(2,1)},
\]
since $a_{(3,0)}=x^{3}-y^{3}$ and $a_{(2,1)}=x^{2}y-xy^{2}$. Notice that the
coefficients $2$ and $-1$ can be read off as $[x^{3}]F$ and $[x^{2}y]F$. Now
$\rho=(1,0)$, so dividing by $a_{\rho}=x-y$ gives
\[
2x^{2}+xy+2y^{2}=2s_{(2)}-s_{(1,1)}.
\]
Indeed, $s_{(2)}=h_{2}=x^{2}+xy+y^{2}$ and $s_{(1,1)}=e_{2}=xy$. Thus, a
symmetric polynomial can have nonnegative monomial coefficients but a negative
coefficient in its Schur expansion.
\end{example}

The easiest way to do this is by finding rules for multiplying a Schur
polynomial by another polynomial -- mainly by $h_{n}$, $e_{n}$ or $p_{n}$.

\subsection{The Pieri rules}

\begin{definition}
\label{def.schur.strips}Let $\lambda/\mu$ be a skew partition, and let
$n\in\mathbb{N}$.

\begin{enumerate}
\item[\textbf{(a)}] We say that $\lambda/\mu$ is a \textbf{horizontal strip}
if $Y\left(  \lambda/\mu\right)  $ has no two cells in the same column.

\item[\textbf{(b)}] We say that $\lambda/\mu$ is a \textbf{vertical strip} if
$Y\left(  \lambda/\mu\right)  $ has no two cells in the same row.

\item[\textbf{(c)}] We say that $\lambda/\mu$ is a \textbf{horizontal }%
$n$\textbf{-strip} if $\lambda/\mu$ is a horizontal strip of size $n$ (that
is, $\left\vert Y\left(  \lambda/\mu\right)  \right\vert =n$).

\item[\textbf{(d)}] We say that $\lambda/\mu$ is a \textbf{vertical }%
$n$\textbf{-strip} if $\lambda/\mu$ is a vertical strip of size $n$ (that is,
$\left\vert Y\left(  \lambda/\mu\right)  \right\vert =n$).
\end{enumerate}
\end{definition}

\begin{example}
\label{exa.schur.strips}All four skew partitions below have inner partition
$\mu=(2,1)$ and size $3$: \ytableausetup{boxsize=1.2em}
\[%
\begin{array}
[c]{c@{\qquad}c}%
(4,2)/(2,1) & (2,2,1,1)/(2,1)\\[4pt]%
\begin{ytableau}\none&\none&{}&{}\\\none&{}\end{ytableau} &
\begin{ytableau}\none&\none\\\none&{}\\{}\\{}\end{ytableau}\\[8pt]%
\text{horizontal only} & \text{vertical only}\\[12pt]%
(3,2,1)/(2,1) & (3,3)/(2,1)\\[4pt]%
\begin{ytableau}\none&\none&{}\\\none&{}\\{}\end{ytableau} &
\begin{ytableau}\none&\none&{}\\\none&{}&{}\end{ytableau}\\[8pt]%
\text{both horizontal and vertical} & \text{neither}%
\end{array}
\]
\ytableausetup{boxsize=normal} In the first diagram, the occupied columns are
$2,3,4$, but two cells are in row $1$. In the second, the occupied rows are
$2,3,4$, but two cells are in column $1$. In the third, no row or column
contains two cells. In the fourth, row $2$ and column $3$ each contain two cells.

In particular, a horizontal strip need not lie in a single row, and a vertical
strip need not lie in a single column. The empty skew diagram is both a
horizontal $0$-strip and a vertical $0$-strip.
\end{example}

\Needspace{8\baselineskip}

\begin{proposition}
\label{prop.schur.strips-characterization}Let $\lambda$ and $\mu$ be two
$N$-partitions with $\mu\subseteq\lambda$. Then:

\begin{enumerate}
\item[\textbf{(a)}] The skew partition $\lambda/\mu$ is a horizontal strip if
and only if%
\[
\lambda_{1}\geq\mu_{1}\geq\lambda_{2}\geq\mu_{2}\geq\cdots\geq\lambda_{N}%
\geq\mu_{N}.
\]


\item[\textbf{(b)}] The skew partition $\lambda/\mu$ is a vertical strip if
and only if%
\[
\mu_{i}\leq\lambda_{i}\leq\mu_{i}+1\qquad\text{for each }i\in[N].
\]

\end{enumerate}
\end{proposition}

\begin{proof}
\textbf{(a)} Since $\mu\subseteq\lambda$, the displayed inequalities amount to
$\lambda_{i+1}\leq\mu_{i}$ for every $i\in[N-1]$.

If $\lambda_{i+1}>\mu_{i}$ for some $i\in[N-1]$, then both cells $(i,\mu
_{i}+1)$ and $(i+1,\mu_{i}+1)$ belong to $Y(\lambda/\mu)$: we use $\lambda
_{i}\geq\lambda_{i+1}>\mu_{i}$ and $\mu_{i+1}\leq\mu_{i}$. Thus, the diagram
has two cells in the same column, so it is not a horizontal strip.

Conversely, assume that $\lambda_{i+1}\leq\mu_{i}$ for all $i\in[N-1]$. If
$(r,t)$ and $(s,t)$ were two cells of $Y(\lambda/\mu)$ in the same column,
with $r<s$, then
\[
\mu_{r}<t\leq\lambda_{s}\leq\lambda_{r+1}\leq\mu_{r},
\]
a contradiction. Hence the diagram is a horizontal strip. \medskip

\textbf{(b)} Row $i$ of $Y(\lambda/\mu)$ has $\lambda_{i}-\mu_{i}$ cells.
Thus, every row has at most one cell if and only if $0\leq\lambda_{i}-\mu
_{i}\leq1$ for each $i\in[N]$, which is precisely the stated condition.
\end{proof}

\begin{theorem}
[Pieri rules]\label{thm.schur.pieri}Let $n\in\mathbb{N}$. Let $\mu$ be an
$N$-partition. Then:

\begin{enumerate}
\item[\textbf{(a)}] We have
\[
h_{n}s_{\mu}=\sum_{\substack{\lambda\text{ is an }N\text{-partition;}%
\\\lambda/\mu\text{ is a horizontal }n\text{-strip}}}s_{\lambda}.
\]


\item[\textbf{(b)}] We have
\[
e_{n}s_{\mu}=\sum_{\substack{\lambda\text{ is an }N\text{-partition;}%
\\\lambda/\mu\text{ is a vertical }n\text{-strip}}}s_{\lambda}.
\]

\end{enumerate}
\end{theorem}

\Needspace{33\baselineskip}

\begin{example}
\label{exa.schur.pieri-21}Let $N\geq4$ and $\mu=(2,1)$. We illustrate the
Pieri rules for $n=2$. In the diagrams below, the shaded cells belong to
$Y(\mu)$ and the unshaded cells are the two added cells. \begingroup
\definecolor{pieribase}{gray}{0.88}

\begin{enumerate}
\item[\textbf{(a)}] The horizontal $2$-strips give the four outer partitions
\[%
\begin{array}
[c]{c@{\qquad}c@{\qquad}c@{\qquad}c}%
(4,1) & (3,2) & (3,1,1) & (2,2,1)\\[6pt]%
\begin{ytableau}*(pieribase){}&*(pieribase){}&{}&{}\\*(pieribase){}\end{ytableau} &
\begin{ytableau}*(pieribase){}&*(pieribase){}&{}\\*(pieribase){}&{}\end{ytableau} &
\begin{ytableau}*(pieribase){}&*(pieribase){}&{}\\*(pieribase){}\\{}\end{ytableau} &
\begin{ytableau}*(pieribase){}&*(pieribase){}\\*(pieribase){}&{}\\{}\end{ytableau}
\end{array}
\]
Thus, the first Pieri rule gives
\[
h_{2}s_{(2,1)}=s_{(4,1)}+s_{(3,2)}+s_{(3,1,1)}+s_{(2,2,1)}.
\]
For instance, $(4,1)/(2,1)$ is allowed because its two new cells lie in
columns $3$ and $4$, even though they lie in the same row.

\item[\textbf{(b)}] The vertical $2$-strips give
\[%
\begin{array}
[c]{c@{\qquad}c@{\qquad}c@{\qquad}c}%
(3,2) & (3,1,1) & (2,2,1) & (2,1,1,1)\\[6pt]%
\begin{ytableau}*(pieribase){}&*(pieribase){}&{}\\*(pieribase){}&{}\end{ytableau} &
\begin{ytableau}*(pieribase){}&*(pieribase){}&{}\\*(pieribase){}\\{}\end{ytableau} &
\begin{ytableau}*(pieribase){}&*(pieribase){}\\*(pieribase){}&{}\\{}\end{ytableau} &
\begin{ytableau}*(pieribase){}&*(pieribase){}\\*(pieribase){}\\{}\\{}\end{ytableau}
\end{array}
\]
Hence, the second Pieri rule gives
\[
e_{2}s_{(2,1)}=s_{(3,2)}+s_{(3,1,1)}+s_{(2,2,1)}+s_{(2,1,1,1)}.
\]
Here, $(2,1,1,1)/(2,1)$ is allowed because its two new cells lie in distinct
rows, even though both lie in column $1$.
\end{enumerate}

\endgroup


For $N=3$, the horizontal expansion is unchanged, while the term
$s_{(2,1,1,1)}$ disappears from the vertical expansion by Proposition
\ref{prop.schur.too-many-rows}.
\end{example}

\begin{thebibliography}{9999}                                                                                             %


\bibitem[LLPT]{LLPT}Dan Laksov, Alain Lascoux, Piotr Pragacz and Anders
Thorup,
\href{https://alain-lascoux.github.io/files/laksov_lascoux_pragacz_thorup__the_LLPT_notes__version_20180328.pdf}{\textit{The
LLPT Notes}}, edited by A.~Thorup, 1995; version of 28 March 2018, with
corrections and additions suggested by D.~Grinberg.
\end{thebibliography}


\end{document}