\classheader{2012-11-09}

Recall some notation from last time: we let $\{1,\ldots,n\}=I_1\sqcup \cdots\sqcup I_k$ be a partition, which collectively we will call $I$. This naturally corresponds to $\lambda$, a partition in the sense we've been using, by letting $\lambda_j=\# I_j$. We also defined a more general Vandemonde polynomial
\[D_I=D_{I_1}\cdots D_{I_k},\]
with
\[d_\lambda:=\deg(D_I)=\sum\frac{\lambda_i(\lambda_i-1)}{2}.\]
We then defined
\[V_\lambda=kS_n\cdot D_I\subset k^{d_\lambda}[x_1,\ldots,x_n],\]
a representation which is called the Specht $S_n$-module ({\small Reader comment: this is actually what most references call the Specht module of $\lambda^t$. The module $V^\lambda = V_{\lambda^t}$ we later construct is the Specht module for $\lambda$}). The subgroup
\[S_I=S_{I_1}\times\cdots\times S_{I_k}\subset S_n\]
is called a Young subgroup.
\begin{theorem}
$\text{}$

\begin{enumerate}
\item $V_\lambda$ is irreducible.
\item $V_\lambda\not\cong V_\mu$ for $\lambda\neq \mu$.
\item Any irrep of $S_n$ is isomorphic to some $V_\lambda$.
\end{enumerate}
\end{theorem}
\begin{proof}[Proof]
We'll assume that $k\subseteq\C$, though it holds in more generality. It will suffice to show that $V_\lambda$ is simple for $k=\C$, since extending the field can only cause it to split more.

Suppose $V_\lambda=V\oplus V'$ is a non-trivial $S_n$-stable decomposition. Then $\pr_V\in \End_\C(V_\lambda)$. To reach a contradiction, we will show that every intertwiner $f:V_\lambda\to V_\lambda$ is a scalar operator. Let $f$ be an intertwiner; then
\[f(D_I)=D'\in k^{d_\lambda}[x_1,\ldots,x_n]\]
for some $D'$. We have $s(D_I)=\sign(s)\cdot D_I$ for all $s\in S_I$, so $s(D')=\sign(s)\cdot D'$ for all $s\in S_I$ because $f$ is an intertwiner. Thus, for all $i,j\in I_r$ for any $r$, we have that $s_{ij}(D')=-D'$, so $D'$ vanishes on the set
\[\{(a_1,\ldots,a_n)\in\C^n\mid a_i=a_j\},\]
hence $(x_i-x_j)\mid D'$, and because we are working in a UFD,
\[\bigg(\prod_{p=1}^k\prod_{i,j\in I_p}(x_i-x_j)\bigg)\mid D',\]
so that $D_I\mid D'$. But $D_I$ and $D'$ are homogeneous polynomials of degree $d_\lambda$, so $D'=c\cdot D_I$ for some $c\in\C$. Thus $f(D_I)=c\cdot D_I$, and so for any $a\in \C S_n$, we have 
\[f(aD_I)=a\cdot f(D_I)=c\cdot a\cdot D_I,\]
and hence $f=c\cdot\id$.

%jw
The same argument shows that there are no interwiners between $V_\lambda$ and $V_\mu$ if $\lambda
\ne \mu$. 

The conjugacy classes of $S_n$ are indexed by $\eucal{P}_n$. Thus, the number of conjugacy classes is 
$\#\eucal{P}_n = \#\widehat{S_n}$. This proves (3).
\end{proof}

\begin{corollary}\leavevmode\vspace{-0.1in}
\begin{enumerate}
\item $[k^d[x_1,\ldots,x_n]:V_\lambda]=\begin{cases}
1 & \text{ if }d=d_\lambda,\\
0 & \text{ if }d<d_\lambda.
\end{cases}$
\item The representation $V_\lambda$ is defined over $\Q$.
\item $V_\lambda\cong V_\lambda^*$.
\end{enumerate}
\end{corollary}
\begin{proof}
There exists an $S_n$-invariant, positive definite $\Q$-bilinear form $\beta$ on $V_\lambda$ (considered as a vector space over $\Q$), and the map from $V\to V^*$ sending $v$ to $\beta(-,v)$ is a $\Q$-linear function (we can't do this over $\C$ because it'd be skew-linear).
\end{proof}


Fix a group $G$ and a commutative ring $Z$ (which in most cases will in fact be $\Z$).
\begin{definition}
The Grothendieck group $K_Z(G)$ is defined as follows. Take $M$, the free $Z$-module with basis consisting of isomorphism classes $[V]$ of finite-dimensional, completely reducible $G$-representations $V$, and quotient $M$ by the relation $[V\oplus V']-[V]-[V']$.
\end{definition}
Note that, for example, if $V=L_1^{m_1}\oplus\cdots\oplus L_k^{m_k}$ where $L_i\in\widehat{G}$, we have
\[[V]=m_1[L_1]+\cdots+m_k[L_k].\]
Thus, we could also think of $K_Z(G)$ as being the free $Z$-module with basis $[L_i]$, for each $L_i\in\widehat{G}$. Note that any element of $K_Z(G)$ can be put in the form $[M]-[N]$ by looking at the terms with positive or negative coefficients and grouping them.

For an example, note that if $G=\{e\}$, then $K_Z(\{e\})=\Z$, where $[V]\leftrightarrow\dim(V)$.

Let $G$ be a finite group. We define an inner product $K_Z(G)\times K_Z(G)\to Z$ by 
\[\langle [L],[L']\rangle=\begin{cases}
1 & \text{ if }L\cong L',\\
0 & \text{ if }L\not\cong L'.
\end{cases}\]
Alternatively, we can define
\[\langle [M],[N]\rangle=\dim(\Hom_G(M,N)).\]
Moreover, $K_Z(G)$ has a commutative ring structure, by $[M][N]=[M\otimes N]$. This is valid because $M\otimes (N_1\oplus N_2)\cong (M\otimes N_1)\oplus(M\otimes N_2)$. The unit for the multiplication is the trivial representation.

Define $\chi:K_\C(G)\to \C\{G\}^G$ by sending $[M]$ to $\chi_M$. We know that $\chi_{M\otimes N}=\chi_M\cdot\chi_N$, so that $\chi$ is a ring homomorphism.
\begin{proposition}
$\chi$ is in fact an isometric ring isomorphism.
\end{proposition}
\begin{proof}
This follows directly from the orthogonality relations for irreducible characters, and the fact that the irreducible characters are a basis for the space of class functions.
\end{proof}

Now let $G=S_d$, and let $R^d$ be the homogeneous symmetric functions of degree $d$ (in a large, unspecified number of variables). We define $\psi:S_d\to R^d$ by $\psi(s)=p_{\lambda(s)}$, where $\lambda(s)$ is the cycle type of $s$. Now we define a map on the space of class functions, $\Psi:\C\{S_d\}^{S_d}\to R^d$, by
\[\Psi(f)=\frac{1}{d!}\sum_{s\in S_d}f(s)\psi(s)\]
We often refer to this as $\langle f,\psi\rangle_{S_d}$, even though $\psi$ is not the same kind of object as $f$.

\begin{lemma}
If $s\in C_\lambda$ (i.e., $s$ has cycle type $\lambda$) then
\[\#(\text{\emph{centralizer of }}s)=z_\lambda.\]
\end{lemma}

This then implies that $\# C_\lambda=\frac{\#S_d}{\#(\text{centralizer})}=\frac{d!}{z_\lambda}$.

\begin{corollary}
$\Psi(\mathbf{1}_{C_\lambda})=z_\lambda^{-1}p_\lambda$
\end{corollary}

A key map we will talk about next class is the Frobenius characteristic map
\[\mathrm{ch}:K_\C(S_d)\xrightarrow{\;\;\chi\;\;}\C\{S_d\}^{S_d}\xrightarrow{\;\;\Psi\;\;}R^d\]
defined by sending $[M]$ to $\langle \chi_M,\psi\rangle_{S_d}$.






