\classheader{2012-11-07}

Let $\lambda\in\eucal{P}_d$. Recall the various classes of symmetric polynomials we defined last time:
\begin{center}
\begin{tabular}{c|c}
$e_\lambda$ & elementary\\\hline
$m_\lambda$ & monomial\\\hline
$h_\lambda$ & complete\\\hline
$p_\lambda$ & power sum\\\hline
$s_\lambda$ & Schur
\end{tabular}
\end{center}
Last time, we defined the quantity
\[XY=\prod_{i,j\geq 1}\frac{1}{1-x_iy_j}.\]
\begin{proposition}[2nd Cauchy identity]
\[XY=\sum_\lambda h_\lambda(x)m_\lambda(y)=\sum_\lambda h_\lambda(y)m_\lambda(x).\]
\end{proposition}
\begin{proof}
Recall that the generating function for the complete symmetric polynomials was
\[H(t)=\prod_{i\geq 1}\frac{1}{1-tx_i}.\]
Thus,
\[XY=\prod_jH(y_j)=\prod_j\sum_{r\geq 0}h_r(x)y_j^r=\sum_{\alpha\in\Z_{\geq 0}^n} h_\alpha(x)y^\alpha=\sum_{\lambda}h_{\lambda}(x)m_{\lambda}(y),\]
where in the last equality we grouped the $\alpha$'s according to which $\lambda$ they are a permutation of.
\end{proof}

\begin{proposition}[3rd Cauchy identity]
\[XY=\sum_\lambda s_\lambda(x)s_\lambda(y).\]
\end{proposition}
\begin{proof}
The determinantal formula says that for any $\alpha\in\Z_{\geq 0}^n$,
\[a(x^\alpha)=a(x^\rho)\det(h_{\alpha_i-n+j})_{ij},\]
where $h$'s with negative indices are declared to be 0, and $\rho$ is as we defined last time,
\[\rho=(n-1,n-2,\ldots,1,0).\]
Thus,
\[a(x^\alpha)=a(x^\rho)\sum_{\substack{s\in S_n\\ \beta\in\Z_{\geq 0}^n}}\sign(s)h_{\beta-s(\rho)}(x).\]
Now we use the 2nd identity to see that 
\[a(x^\rho)a(y^\rho)\cdot XY=a(x^\rho)\sum_{\substack{s\in S_n\\ \lambda}} h_\lambda(x)\sign(s)y^{s(p)}m_\lambda(y),\]
where we expanded $a(y^\rho)$ as an alternation explicitly. Then
\[a(x^\rho)a(y^\rho)\cdot XY=a(x^\rho)\sum_{\substack{\alpha\in\Z_{\geq 0}^n\\s\in S_n}} h_\alpha(x)\sign(s)y^{\alpha+s(\rho)}.\]
Letting $\beta=\alpha+s(\rho)$,
\[a(x^\rho)a(y^\rho)\cdot XY=a(x^\rho)\sum_{\beta,s}\sign(s)h_{\beta-s(\rho)}(x)y^\beta,\]
which is precisely the expression in the determinental formula. Thus,
\[a(x^\rho)a(y^\rho)\cdot XY=\sum_{\beta\in\Z_{\geq 0}^n}a(x^\beta)y^\beta=\sum_{\substack{\mu\\ s\in S_n}}a(x^{s(\mu)})y^{s(\mu)}=\sum_{\mu,s}a(x^\mu)\sign(s)y^{s(\mu)},\]
where in the last step we used that $a(x^{s(\mu)})=\sign(s)\cdot a(x^\mu)$. Finally, letting $\mu=\lambda+\rho$, this is equal to
\[\sum_{\mu}a(x^\mu)a(y^\mu)=\sum_{\lambda}a(x^{\lambda+\rho})a(y^{\lambda+\rho}).\]
Dividing both sides by $a(x^\rho)a(y^\rho)$ and applying the definition of the Schur polynomials, we are done.
\end{proof}

The Hall inner product $\langle\,\cdot\,,\,\cdot\,\rangle:R\times R\to\Q$ is defined by $\langle h_\mu,m_\lambda\rangle=\delta_{\lambda\mu}$.

\begin{lemma}
Let $\{u_\lambda\}$ and $\{v_\lambda\}$ be a pair of bases of $R$. Then the following are equivalent:
\begin{enumerate}
\item $\langle u_\lambda,v_\mu\rangle=\delta_{\lambda\mu}$
\item $\sum_{\lambda} u_\lambda(x)v_\lambda(y)=XY$.
\end{enumerate}
\end{lemma}
\begin{proof}
Expanding in the basis,
\[u_\lambda=\sum_{\nu}a_{\lambda\nu}h_\nu,\qquad v_\mu=\sum_{\sigma} b_{\mu\sigma}m_\sigma.\]
Then 1 is equivalent to
\[\sum_\nu a_{\lambda\nu}b_{\mu\nu}=\delta_{\lambda\mu}\]
and 2 is equivalent to
\[\sum_{\lambda}u_\lambda(x)v_\lambda(y)=XY\overset{\text{Cauchy 2}}{=}\sum_\nu h_\nu(x)m_\nu(y).\]
But both of these are equivalent to the claim that the matrix of $a$'s is inverse to the matrix of
$b$'s, i.e.,
\[\sum_\lambda a_{\lambda\nu}b_{\lambda\sigma}=\delta_{\nu\sigma}.\qedhere\]
\end{proof}
Applying the 1st Cauchy identity, 
\[XY=\sum_{\lambda}\frac{1}{z_\lambda}p_\lambda(x)p_\lambda(y)\]
which we proved last time, we get as a corollary of this lemma that
\[\langle p_\lambda,p_\mu\rangle=z_\lambda\cdot\delta_{\lambda\mu}.\]
Another corollary is that $\langle s_\lambda,s_\mu\rangle=\delta_{\lambda\mu}$ (by the 3rd Cauchy
		identity, and the lemma). Lastly, we get as a corollary that
$\langle\,\cdot\,,\,\cdot\,\rangle$ is a symmetric positive definite bilinear form 
(this is important because it is defined in an \textit{a priori} non-symmetric way).

Now we define an involution $\tau:R\to R$.
\begin{definition}
Define $\tau:R\to R$ by $\tau(e_r)=h_r$. Because these generate $R$ as an algebra, this then implies that $\tau(e_\lambda)=h_\lambda$ for all $\lambda\in\eucal{P}_n$.
\end{definition}
\begin{proposition}
The map $\tau$ is an involution, and $\tau$ respects $\langle\,\cdot\,,\,\cdot\,\rangle$.
\end{proposition}
\begin{proof}
The action of $\tau$ is as a matrix $T$, and Newton's identities tell us that
\[\sum_{r=0}^n(-1)^re_rh_{n-r}=0.\]
Because $T^2(e_r)=T(h_r)$ and because Newton's identities are symmetric under switching $e$'s and $h$'s, we must have that $T^2(e_r)=e_r$, and hence $T^2=\id$.

You have equations expressing $p_r$ in terms of $e_\lambda$'s and $h_\lambda$'s from last time.
%, and you can check that they are of the form
%\[p_r=\langle e_\lambda,\,\cdot\,\rangle,\qquad p_r=\langle h_\lambda,\,\cdot\,\rangle.\]
%jw
These equations and induction imply that $\tau(p_r)=(-1)^{r-1}p_r$. Thus,
\[\langle \tau(p_\lambda),\tau(p_\mu)\rangle=\begin{cases}
\pm \langle p_\lambda,p_\mu\rangle = 0 & \text{ if }\lambda\neq \mu,\\
\langle p_\lambda,p_\lambda\rangle & \text{ if }\lambda=\mu.
\end{cases}\qedhere\]
\end{proof}

\subsection*{Young diagrams}

The Young diagram of a partition $\lambda=(\lambda_1,\lambda_2,\ldots)$ with $|\lambda|=n$ is
\begin{center}
\begin{tikzpicture}[scale=0.4]
\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 (9,1);
\node at (-0.7,0.5) {$\lambda_1$};
\node at (-0.7,1.5) {$\lambda_2$};
\node at (-0.7,3.2) {$\vdots$};
\end{tikzpicture} %\vspace{-0.1in}
\end{center}
Thus, the total number of blocks in the diagram is $n$.

There is a natural involution on Young diagrams, $\lambda\mapsto \lambda^t$, defined by interchanging rows and columns.

There is another determinantal formula, which is symmetric with respect to this involution:
\[s_\lambda=\det(e_{\lambda_i^t-i+j}).\]
As a corollary, we get that $\tau(s_\lambda)=s_{\lambda^t}$ (apply involution to first determinantal formula; $e$'s and $h$'s are interchanged, and we flip $\lambda$).

Now we'll start relating this to the representation theory of $S_n$.

Fix a field $k$, and consider the polynomial ring $k[x_1,\ldots,x_n]$.

Let $\{1,\ldots,n\}=I_1\sqcup \cdots\sqcup I_k$ be a partition with $\# I_j=\lambda_j$, which collectively we will call $I$. This naturally corresponds to $\lambda$, a partition in the sense we've been using. Then we define
\[S_I=S_{I_1}\times\cdots\times S_{I_k}\subset S_n.\]
We define $D_I$ to be a product of certain Vandermonde determinants of various sizes, \[D_I=D_{I_1}\times\cdots\times D_{I_k},\] where
\[D_{I_m}:=\prod_{i,j\in I_m}(x_i-x_j).\]

\begin{theorem}
Let $\operatorname{char}(k)=0$, and let $I$ be a partition as above. Then $V_\lambda:=kS_n\cdot D_I$ is an irrep of $S_n$, and in fact $\{V_\lambda\mid \lambda\in \eucal{P}_n\}\cong\widehat{S_n}$.
\end{theorem}