\classheader{2012-11-12}

Using the notation from last time, $K_Z(S_d)$ is the Grothendieck group of finite-dimensional $S_d$-representations over a commutative ring $Z$.

We can make a commutative ring structure on
\[K_Z=\bigoplus_{d\geq 0}K_Z(S_d)\]
by defining the circle product $\circ:K_Z(S_m)\times K_Z(S_n)\to K_Z(S_{m+n})$ as follows: we consider the natural inclusion $j:S_m\times S_n\to S_{m+n}$, and for any $M$ and $N$ (representations of $S_m$ and $S_n$, respectively), we define
\[M\circ N=\Ind_{S_m\times S_n}^{S_{m+n}}M\boxtimes N.\]
More generally, given $\lambda=(\lambda_1\geq\lambda_2\geq\cdots)\in \eucal{P}_n$, we define
\[S_\lambda=S_{\lambda_1}\times S_{\lambda_2}\times\cdots,\]
and given representations $M_i$ over each $S_{\lambda_i}$, we can form
\[M_1\circ M_2\circ\cdots\circ M_k=\Ind_{S_\lambda}^{S_n}(M_1\boxtimes M_2\boxtimes\cdots\boxtimes M_k).\]
\begin{lemma}
The circle product is associative and commutative, and thus $K_Z$ is a commutative associative ring.
\end{lemma}
\begin{proof}
Let $M$, $N$, and $L$ be representations of $S_m$, $S_n$, and $S_\ell$ respectively. Then directly from the definition,
\[(M\circ N)\circ L=\Ind_{S_m\times S_n\times S_\ell}^{S_{m+n+\ell}}(M\boxtimes N\boxtimes L)=M\circ (N\circ L).\]
Commutativity follows from the fact that inducing from two subgroups which are conjugate to one another gives the same result, so
\[\Ind_{S_m\times S_n}^{S_{m+n}}(M\boxtimes N)=\Ind_{S_n\times S_m}^{S_{m+n}}(N\boxtimes M).\qedhere\]
\end{proof}

\subsection*{Frobenius characteristic}

We define the Frobenius characteristic map
$\ch_n:K_\C(S_n)\xrightarrow{\;\;\chi\;\;}\C\{S_n\}^{S_n}\xrightarrow{\;\;\Psi\;\;}\C\otimes_\Z R$
(where $R$ is the ring of symmetric functions\footnote{Reader believes that everything works out if
 we take $R$ to be $R_\infty$ as defined in Lecture 16}, complexified by tensoring with $\C$) by sending $[M]$ to
\[\frac{1}{n!}\sum_{s\in S_n}\chi_M(s)\cdot p_{\lambda(s)}.\]
Then we let $\ch$ be the sum of all these maps,
\[\ch=\bigg(\bigoplus_{n\geq 0}\ch_n\bigg):K_\C\to \C\otimes_\Z R.\]
\newpage

\begin{theorem}\ 
\begin{enumerate}
\item The map $\ch$ restricts to a ring isomorphism $(K_\Z,\circ)\to R$.
\item The map $\ch$ is an isometry.
\item The map $\ch$ takes $\sign\otimes(-)$ to $\tau$, i.e. $\ch(\sign\otimes M)=\tau(\ch(M))$.
\item The map $\ch$ acts as follows, for all $\lambda\in \eucal{P}_n$:
\begin{itemize}
\item $V_\lambda\longmapsto s_{\lambda^t}$
\item $\Ind_{S_\lambda}^{S_n}(\text{\emph{triv}})\longmapsto h_\lambda$
\item $\Ind_{S_\lambda}^{S_n}(\text{\emph{sign}})\longmapsto e_\lambda$
\end{itemize}
\end{enumerate}
\end{theorem}

\begin{remark}
Let $j:H\hookrightarrow G$ be a group embedding. Let $E$ and $F$ be representations of $H$ and $G$, respectively. Then there is a canonical isomorphism
\[\Hom_G(\Ind_H^GE,F)\cong \Hom_H(E,j^*F).\]
Thus, for any class function $f\in \C\{G\}^G$,
\[\langle\chi_{\Ind_H^G E},f\rangle_G=\langle \chi_E,f|_H\rangle_H.\]
This is the key ingredient in the proof of the first statement; the rest is just the definitions.
\end{remark}

\begin{proof}[Proof of 1 (ring homomorphism)]
Let $M$ and $N$ be representations of $S_m$ and $S_n$, respectively. Then
\begingroup
\addtolength{\jot}{1ex}
\begin{align*}
\ch(M\circ N)&=\Psi(\chi(\Ind_{S_m\times S_n}^{S_{m+n}}(M\boxtimes N)))\\
&=\langle \chi_{\Ind_{S_m\times S_n}^{S_{m+n}}(M\boxtimes N)},\psi\rangle_{S_{m+n}}\\
\text{(by remark) }&=\langle \chi_{M\boxtimes N},\psi|_{S_m\times S_n}\rangle_{S_m\times S_n}\\
&=\frac{1}{m!}\frac{1}{n!}\sum_{\substack{s\in S_m\\ s'\in S_n}}\chi_{M\boxtimes N}(s\times s')\cdot\psi(s\times s')\\
&=\frac{1}{m!n!}\sum_{s,s'}\chi_M(s)\chi_N(s')p_{\lambda(s\times s')}.\\
\intertext{Note that, for any $\mu\in\eucal{P}_m$ and $\nu\in\eucal{P}_n$, we have $p_{\mu\sqcup \nu}=p_\mu\cdot p_\nu$. The cycle type $\lambda(s\times s')$ is just $\lambda(s)\sqcup\lambda(s')$, and therefore $p_{\lambda(s\times s')}=p_{\lambda(s)}\cdot p_{\lambda(s')}$.}
&=\frac{1}{m!n!}\sum_{s,s'}\chi_M(s)\chi_N(s')p_{\lambda(s)}p_{\lambda(s')}\\
&=\left[\frac{1}{m!}\sum_{s\in S_m}\chi_M(s)p_{\lambda(s)}\right]\left[\frac{1}{n!}\sum_{s'\in S_n}\chi_N(s')p_{\lambda(s')}\right]
\end{align*}
\endgroup
so we have shown $\ch$ is a ring homomorphism. We will show it is an isomorphism next lecture.
\end{proof}
\begin{proof}[Proof of 2]
We showed that $\chi$ was an isometry last time, so it will suffice to show that $\Psi$ is an isometry because $\ch=\chi\circ \Psi$.

For any $\lambda,\mu\in\eucal{P}_n$, we take the Hall inner product
\[\langle \Psi(\mathbf{1}_{C_\lambda}),\Psi(\mathbf{1}_{C_\mu})\rangle_{\text{Hall}}.\]
We know that $\Psi(\mathbf{1}_{C_\lambda})=z_\lambda^{-1}\cdot p_\lambda$, so we compute
\begin{align*}
\langle \Psi(\mathbf{1}_{C_\lambda}),\Psi(\mathbf{1}_{C_\mu})\rangle_{\text{Hall}} & = \langle z_\lambda^{-1}\cdot p_\lambda,z_\mu^{-1}\cdot p_\mu\rangle 
=z_\lambda^{-1}z_\mu^{-1} \langle p_\lambda, p_\mu \rangle \\ 
&=\begin{cases}
z_\lambda^{-1} & \text{ if }\lambda=\mu,\\
0 & \text{ if }\lambda\neq \mu 
\end{cases} = \langle \mathbf{1}_{C_\lambda}, \mathbf{1}_{C_\mu} \rangle 
\qedhere
\end{align*}
\end{proof}
%jw
\begin{proof}[Proof of 3]
Write the character $\chi_M=\sum_{\lambda\in\eucal{P}_n}\chi(C_\lambda)\cdot\mathbf{1}_{C_\lambda}$
so that 
\[\ch(M)=\sum_{\lambda\in\eucal{P}_n}\chi(C_\lambda)\cdot z_{\lambda}^{-1}\cdot p_\lambda.\]
We have
$\tau(p_d)=(-1)^{d-1}\cdot p_d$ and taking product, $\tau(p_\lambda)=(-1)^{\sum(\lambda_i-1)}\cdot
p_\lambda$.
Thus,
\[\tau(\ch(M))=\sum_{\lambda\in\eucal{P}_n}\chi(C_{\lambda})\cdot z_{\lambda}^{-1}(-1)^{\sum \lambda_i-1}\cdot p_\lambda\]
Note that
\[\sign(\text{cycle of length }d)=(-1)^{d-1},\]
so
\[\sign|_{C_\lambda}=(-1)^{\sum (\lambda_i -1)}.\]
Thus
\[\chi_{\sign\otimes M}=\sum_{\lambda\in\eucal{P}_n}\chi_{\sign\otimes M}(C_\lambda)\mathbf{1}_{C_\lambda}=\sum_{\lambda}\chi_M(C_\lambda)\cdot(-1)^{\sum(\lambda_i-1)}\mathbf{1}_{C_\lambda},\]
and therefore
\[\ch(\sign\otimes M)=\sum_{\lambda}\chi(C_\lambda)(-1)^{\sum
	\lambda_i-1}\Psi(\mathbf{1}_{C_\lambda})=\sum_{\lambda}\chi(C_\lambda)(-1)^{\sum
		(\lambda_i-1)}z_\lambda^{-1} p_\lambda\]
which is exactly the expression we wanted.
\end{proof}
\begin{proof}[Proof of 4]
Let's compute the Frobenius character of the trivial representation.
\[\ch(\text{triv}_n)=\sum_{\lambda\in\eucal{P}_n}z_{\lambda}^{-1}p_\lambda=h_n\in R.\]
For the sign representation, we just get
\[\ch(\sign)=\tau(h_n)=e_n.\]
Using the fact that $\ch$ is a homomorphism (which we just proved),
\begin{align*}
\ch(\Ind_{S_\lambda}^{S_n}(\text{triv}))&=\ch(\text{triv}_{\lambda_1}\circ\text{triv}_{\lambda_2}\circ\cdots)\\
&=\ch(\text{triv}_{\lambda_1})\ch(\text{triv}_{\lambda_2})\cdots\\
&=h_{\lambda_1}h_{\lambda_2}\cdots\\
&=h_\lambda.\qedhere
\end{align*}

For any $\lambda\in\eucal{P}_n$, we define an alternating sum in the Grothendieck group,
\begin{align*}
V^\lambda&=\sum_{s\in S_n}\sign(s)\left[\Ind_{S_{\lambda+\rho-s(\rho)}}^{S_n}(\text{triv})\right]
\end{align*}
so that
\begin{align*}
\ch(V^\lambda)&=\sum_{s\in S_n}\sign(s)\ch(\Ind\cdots)\\
&=\sum_{s\in S_n}\sign(s)\cdot h_{\lambda+\rho-s(\rho)}\overset{\substack{\text{determinantal}\\ \text{identity}}}{=}s_\lambda.
\end{align*}
Therefore
\[\langle V^\lambda,V^\mu\rangle=\langle \ch(V^\lambda),\ch(V^{\mu})\rangle_{\text{Hall}}=\langle s_\lambda,s_\mu\rangle_{\text{Hall}}=\delta_{\lambda\mu},\]
and in particular,
\[\langle V^\lambda,V^\lambda\rangle=1\]
which implies that either $V^{\lambda}$ or $-V^{\lambda}$ (this sign is in the Grothendieck group) is an actual 
irrep. %You'll show that the correct sign is $+$ on the homework.
Next time we will see that $V_\lambda = V^{\lambda^t}$, which finishes the proof.
\end{proof}