\classheader{2012-11-14}

Last time we constructed a ring homomorphism from the Grothendieck group $K_\Z=\bigoplus_{n}K_\Z(S_n)$ to the ring $R$ by sending $V^\lambda$ to $s_\lambda$. We showed that for all $\lambda\in\eucal{P}_n$, either $\pm V^\lambda\in\widehat{S_n}$.

It remains to show that this map is an isomorphism. We see that $\ch$ is surjective, since the Schur functions form a $\Z$-basis of $R$, and $s_\lambda\in\im(\ch)$. The map $\ch$ is injective, as it is an isometry.
\begin{corollary}
$\chi_{V^\lambda}(c_\mu)=\langle s_\lambda,p_\mu\rangle_{\text{\emph{Hall}}}$
\end{corollary}
\begin{proof}
We have
\[s_\lambda=\ch(V^\lambda)=\sum_{\mu\in\eucal{P}_n}z_\mu^{-1}\chi_{V^\lambda}(c_\mu)p_\mu,\]
so
\[\langle s_\lambda,p_\nu\rangle=\sum_\mu z_\mu^{-1}\chi_{V^\lambda}(c_\mu)\langle p_\mu,p_\nu\rangle=\chi_{V^\lambda}(c_\mu).\qedhere\]
\end{proof}

There are three main ways partitions come up in what we've been doing.
\begin{center}
\begin{tikzpicture}[scale=1.1]
\node[rounded rectangle,draw] (a) at (0,0) {partitions};
\node[rounded rectangle,draw] (b) at (170:3) {reps of $S_n$};
\node[rounded rectangle,draw] (c) at (270:1.4) {\parbox{1.3in}{\centering nilpotent conjugacy \\classes in $\M_n(\C)$}};
\node[rounded rectangle,draw] (d) at (10:3) {reps of $\GL_n(\C)$};
\draw[thick] (a) to (b);
\draw[thick] (a) to (c);
\draw[thick] (a) to (d);
\draw[thick,<->,>=latex] (b.north) to [out=20,in=160] node[midway,label=above:Schur-Weyl duality] {} (d.north);
\end{tikzpicture}
\end{center}
We define a partial order on partitions. Given $\lambda=(\lambda_1,\ldots),\mu=(\mu_1,\ldots)\in\Z^n$, we say $\lambda\leq \mu$ if $\lambda_1\leq \mu_1$, and $\lambda_1+\lambda_2\leq\mu_1+\mu_2$, $\ldots$, and $\lambda_1+\cdots+\lambda_n\leq\mu_1+\cdots+\mu_n$.

Define $O_\lambda$ to be the nilpotent conjugacy class with Jordan blocks $\lambda_1,\ldots,\lambda_n$.
\begin{proposition}
$\text{}$

\begin{enumerate}
\item $O_\lambda\subseteq \overline{O_\mu}$ $\iff$ $\lambda\leq\mu$.
\item $\lambda\leq\mu$ $\iff$ $\mu^t\leq\lambda^t$.
\item $s(\rho)<\rho$ for all non-identity $s\in S_n$.
\end{enumerate}
\end{proposition}


\begin{proposition}
$\text{}$

\begin{enumerate}
\item $[\Ind_{S_\mu}^{S_n}\triv:V^\lambda]=\begin{cases}
0 & \text{ unless }\mu\leq\lambda,\\
1 & \text{ if }\mu=\lambda.
\end{cases}$
\item $[\Ind_{S_\mu}^{S_n}\sign:V^\lambda]=\begin{cases}
0 & \text{ unless }\mu\leq \lambda^t,\\
1 & \text{ if }\mu=\lambda^t.
\end{cases}$
\item $V^\lambda=V_{\lambda^t}$
\end{enumerate}
\end{proposition}
\begin{proof}
\begingroup
\addtolength{\jot}{1em}
\begin{align*}
V^\lambda & = \sum_{s\in S_n}\text{sign}(s)\cdot\left[\Ind_{S_{\lambda+\rho-s(p)}}^{S_n}(\text{triv})\right]\\
&=\Ind_{S_{\lambda}}^{S_n}(\text{triv}) +\sum_{\mu>\lambda}a_{\lambda\mu}\left[\Ind_{S_\mu}^{S_n}(\text{triv})\right]
\end{align*}
\endgroup
The transition matrix from $\Ind_{S_\mu}^{S_n}$ to $V^\lambda$ is of the form
\[\begin{pmatrix}
1 & * & \cdots &*\\
& 1 & \ddots & \vdots\\
& & \ddots & *\\
& & & 1
\end{pmatrix}\]
with respect to the partial ordering. The inverse of a strict upper triangular matrix is also
	strictly upper triangular, so %jw
\[\Ind_{S_\lambda}^{S_n}(\text{triv})=V^\lambda+\sum_{\mu>\lambda}b_{\lambda\mu}\cdot V^\mu\]
which proves part 1. Part 2 follows from part 1 by applying $\ch$ and $\tau$. %jw

Thus $V^\lambda$ is the unique irrep of $S_n$ which occurs both in $\Ind_{S_\lambda}^{S_n}(\text{triv})$ and 
$\Ind_{S_{\lambda^t}}^{S_n}(\text{sign})$.

Now we claim that $V_\lambda$ occurs in $\Ind_{S_{\lambda^t}}^{S_n}(\text{triv})$ and $\Ind_{S_\lambda}^{S_n}(\text{sign})$. Recall that a $\lambda\in\eucal{P}_n$ corresponds to
\[D_\lambda=D_{\lambda_1}\cdot D_{\lambda_2}\cdots\]
and $s(D_\lambda)=\sign(s)\cdot D_\lambda$ for all $s\in S_\lambda$, so there exists a surjective map $\Ind_{S_\lambda}^{S_n}(\text{sign})\to V_\lambda$, sending 1 to $D_\lambda$. Because $V_\lambda$ is simple, this implies that $V_\lambda$ occurs in $\Ind_{S_\lambda}^{S_n}(\text{sign})$.

In our notation, $D_n=a(x^\rho)$, so that
\[D_\lambda=a(x^{\rho_{\lambda_1}})a(x^{\rho_{\lambda_2}})\cdots\]
\begin{center}
\begin{tikzpicture}[scale=0.6]
\draw[thick] (0,0) rectangle (1,7);
\draw[thick] (0,0) rectangle (1,6);
\draw[thick] (0,0) rectangle (1,5);
\draw[thick] (0,0) rectangle (1,4);
\draw[thick] (0,0) rectangle (1,3);
\draw[thick] (0,0) rectangle (1,2);
\draw[thick] (0,0) rectangle (1,1);
\draw[thick] (0,0) rectangle (2,5);
\draw[thick] (0,0) rectangle (2,4);
\draw[thick] (0,0) rectangle (2,3);
\draw[thick] (0,0) rectangle (2,2);
\draw[thick] (0,0) rectangle (2,1);
\draw[thick] (0,0) rectangle (3,5);
\draw[thick] (0,0) rectangle (3,4);
\draw[thick] (0,0) rectangle (3,3);
\draw[thick] (0,0) rectangle (3,2);
\draw[thick] (0,0) rectangle (3,1);
\draw[thick] (0,0) rectangle (8,2);
\draw[thick] (0,0) rectangle (7,2);
\draw[thick] (0,0) rectangle (6,2);
\draw[thick] (0,0) rectangle (5,2);
\draw[thick] (0,0) rectangle (4,2);
\draw[thick] (0,0) rectangle (10,1);
\node at (-0.7,0.5) {$\lambda_1$};
\node at (-0.7,1.5) {$\lambda_2$};
\node at (0.5,-0.7) {$x_1$};
\node at (1.5,-0.7) {$x_2$};
\node at (-0.7,3.2) {$\vdots$};
\node at (0.5,0.5) {0};
\node at (0.5,1.5) {0};
\node at (0.5,2.5) {0};
\node at (1.5,0.5) {1};
\node at (1.5,1.5) {1};
\node at (1.5,2.5) {1};
\node at (2.5,0.5) {2};
\node at (2.5,1.5) {2};
\node at (2.5,2.5) {2};
\node at (9,0.5) {\small $\lambda_1-1$};
\end{tikzpicture} %\vspace{-0.1in}
\end{center}
Expanding the product and counting along columsn instead of rows (see diagram), we have 
\[D_\lambda=(x_1x_{\lambda_1+1}x_{\lambda_1+\lambda_2+1}\cdots)^0(x_2x_{\lambda_1+2}x_{\lambda_1+\lambda_2+2}\cdots)^1\cdots\]
Therefore $S_{\lambda^t}$ acts trivially on $D_\lambda$ with coordinates transposed, i.e., we get a
nonzero map $j:Ind_{S_{\lambda^t}}^{S_n}(\text{triv})\to V_\lambda$. This shows Part 3.
\end{proof}

\subsection*{Representation theory of $S_n$ and $\GL_m$}

Let $V$ be a finite-dimensional $\C$-vector space. For any $n\geq 1$, we have a natural action
$S_n\lcirclearrowright V^{\otimes n}$. We let $j:\End_\C(V) \to \End_\C(V^{\otimes n})$ be the
Lie algebra action defined by
\[a(v_1\otimes v_2\otimes\cdots\otimes v_n)=(av_1)\otimes v_2\otimes\cdots\otimes v_n+v_1\otimes (av_2)\otimes\cdots\otimes v_n+\cdots\]
\begin{lemma}
$\End_{S_n}(V^{\otimes n})$ is the algebra generated by $\im(j)$.
\end{lemma}
\begin{proof}
The inclusion $\supseteq$ is clear.

Now we will show the other inclusion. First, we claim
\[\End(V^{\otimes n})\cong \End(V)^{\otimes n}\]
\[\supseteq\]
\begin{align*}
\End(V^{\otimes n})&=V^{\otimes n}\otimes (V^{\otimes n})^*\\
&= V^{\otimes n}\otimes (V^*)^{\otimes n}\\
&=(V\otimes V^*)^{\otimes n}\\
&=(\End(V))^{\otimes n}
\end{align*}

Second, we see that
\[\End_{S_n}(V^{\otimes n})=(\End(V^{\otimes n}))^{S_n}=((\End(V))^{\otimes n})^{S_n}.\]

Third, we know from the homework that $(A^{\otimes n})^{S_n}$ is generated by elements of the form
\[(a\otimes 1\otimes\cdots\otimes 1)+(1\otimes a\otimes \cdots\otimes 1)+\cdots\]
for any algebra $A$, and thus this is true in particular for $A=\End(V)$.
\end{proof}