\classheader{2012-11-16}

Let $V$ be a finite-dimensional vector space over $\C$, and consider the tensor power $V^{\otimes n}$. We have two natural actions, $S_n\rightcirclearrow V^{\otimes n}\leftcirclearrow \GL(V)$, where $S_n$ permutes tensorands, and $g\in\GL(V)$ acts by
\[g(v_1\otimes \cdots \otimes v_n)=gv_1\otimes\cdots\otimes gv_n.\]
This gives us maps
\begin{center}
\begin{tikzcd}
S_n \ar{r}{j_1} & \End_\C(V^{\otimes n}) & \GL(V)\ar{l}[swap]{j_2}
\end{tikzcd}
\end{center}
Let $A(S_n)$ be the $\C$-linear span of $\im(j_1)$, and similarly let $A(\GL)$ be the $\C$-linear span of $\im(j_2)$.

\begin{theorem}[Schur-Weyl duality]
$\text{}$

\begin{enumerate}
\item Inside $\End_\C(V^{\otimes n})$, we have $A(S_n)^!=A(\GL)$ and $A(\GL)^!=A(S_n)$.
\item If $\dim(V)\geq n$, then each irrep of $S_n$ occurs in $V^{\otimes n}$, and in particular,
\[V^{\otimes n}=\bigoplus_{\lambda\in\eucal{P}_n}V_\lambda\otimes L_\lambda\]
where the $V_\lambda$ are the Specht modules and the $L_\lambda$ are mutually non-isomorphic $\GL(V)$-irreps.
\end{enumerate}
\end{theorem}

\begin{proof}[Proof of 1]
Clearly, $A(\GL)\subseteq A(S_n)^!$; the non-trivial part is the other inclusion. The argument we will use is an illustration of Lie theory.

Consider $a\in\End_\C(V)$. Then $e^{ta}\in\GL(V)$ for any $t\in\C$, so that $j_2(e^{ta})\in A(\GL)$ for all $t\in\C$. Applying it to a simple tensor,
\begin{align*}
j_2(e^{ta})(v_1\otimes\cdots\otimes v_n)&=e^{ta}(v_1)\otimes\cdots\otimes e^{ta}(v_n)\\
&=(v_1+ta(v_1)+o(t))\otimes\cdots\otimes(v_n+ta(v_n)+o(t))\\
&= (v_1\otimes\cdots\otimes v_n)+t\left(\textstyle\sum\limits_{i=1}^n(v_1\otimes\cdots \otimes av_i\otimes\cdots\otimes v_n)\right)+o(t)
\end{align*}
Taking the derivative with respect to $t$ and evaluating at $0$, 
we see that %jw
\[(a\otimes 1\otimes\cdots\otimes 1)+(1\otimes a\otimes \cdots\otimes 1)+\cdots\in A(\GL)\]
for all $a\in\End(V)$. (Since $A(\GL)$ is a finite dimensional vector space, it is topologically
closed.) By the lemma from last time, the elements of the above form generate the algebra 
$\End_{S_n}(V^{\otimes n})=A(S_n)^!$. 



%This 




Since $\C S_n$ is a semisimple algebra, we have that $A(S_n)$ is a semisimple algebra, and you will show on the homework that this implies $A(S_n)^!$ is semisimple, hence $A(\GL)$ is semisimple, hence $A(S_n)=A(S_n)^{!!}=A(\GL)^!$.
\end{proof}
\begin{proof}[Proof of 2]
We want to show that there is a copy of the regular representation of $S_n$ on $V^{\otimes n}$ if $\dim(V)\geq n$. Let $\{v_1,\ldots,v_d\mid d\geq n\}$ be a $\C$-basis of $V$, and let $s\in S_n$. Then the elements
\[v_{s(1)}\otimes\cdots\otimes v_{s(n)}\in V^{\otimes n}\]
span a copy of $\C S_n$.
\end{proof}

\subsection*{Jacobson density}

Let $X$ and $Y$ be sets, and consider $\Map(X,Y)$. We equip this with the pointwise topology, which is defined as follows. For $f\in \Map(X,Y)$, a basis of open neighborhoods of $f$ is given by, for all $n\geq 1$ and $x_1,\ldots,x_n\in X$,
\[\cal{U}_{x_1,\ldots,x_n}=\{f':X\to Y\mid f'(x_i)=f(x_i)\text{ for all }i=1,\ldots,n\}.\]
Thus, $f_i\to f$ if and only if, for any $x\in X$, we have $f_i(x)=f(x)$ for all $i\gg 0$.

Let $A$ be a ring, and let $M$ be an $A$-module; equivalently, we have an action map $A\to\End_{\Z}(M)$. We have
\[A\to A_M:=\im(\text{action})\hookrightarrow A_M^{!!}.\]
\begin{theorem}[Jacobson density]
If $M$ is a semisimple $A$-module, then $A_M$ is pointwise dense in $A_M^{!!}$; in other words, for any $A_M^!$-linear map $f:M\to M$ and any collection $m_1,\ldots,m_n\in M$, there is an $a\in A$ such that $f(m_1)=am_1$ for all $i=1,\ldots,n$.
\end{theorem}
\begin{example}
Let $A=k$, so that $M$ is a vector space over $k$, say with $\dim(M)=n$. Then $k^!=A_M^!=\M_n(k)$, and so $k^{!!}=A_M^{!!}=\M_n(k)^!=k$.
\end{example}
\begin{lemma}
Let $A$ be a ring, and let $L$ be an $A$-module. Let $M=L^n$. Then $A_{L^n}^{!!}\cong A_L^{!!}$.
\end{lemma}

\begin{example}
Let $L=A$. Then $A_L^!=A^\op$, so that $A_L^{!!}=(A^\op)^\op=A$. Thus, the lemma implies that $(A_{A^n})^{!!}=A$, so we see that the fact that $k$ was a field in our earlier example was unnecessary.
\end{example}
\begin{proof}[Proof of Lemma]
First, recall that we proved a long time ago,
\[A_{L^n}^!=\End_A(L^n)=\M_n(\End_A(L))=\M_n(A_L^!).\]
Second, note that, if $L$ is a $B$-module, then $L^n$ has a $\M_n(B)$-module structure, and
\[\End_{\M_n(B)}(L^n)=\{\diag(a,\ldots,a)\mid a\in\underbrace{\End_B(L)}_{:=B_L^!}\}.\]
Third, we therefore have that $(A_{L^n})^{!!}=(\M_n(A_L^!))^!$. We have $B=A_L^!$, and by our second observation,
\[(A_{L^n})^{!!}=(\M_n(A_L^!))^!=\{\diag(a,\ldots,a)\mid a\in(A_L^!)^!=A_L^{!!}\}.\qedhere\]
\end{proof}
\begin{example}
Let $A=\M_d(k)$, and let $L=k^d$. Let $n\geq 1$, so that $L^n=(k^d)^n$ can be identified with the $d\times n$ matrices over $k$. Then $A=\M_d(k)$ acts on the left on $L^n$, and $A^!=\M_n(k)$ acts on the right, and
\[(k^d)^n=(k^n)^d\implies A^{!!}=\M_d(k).\]
\end{example}

Let $L$ be a finite-dimensional vector space over $k$. Let $A=\End_k(L)$, and let $M$ be a finitely generated $A$-module (we know $M\cong L^n$ for some $n$, but we ignore this for now). We have an evaluation map
\[\ev:L\otimes\underbrace{\Hom_A(L,M)}_{L^\lozenge}\xrightarrow{\;\sim\;}M,\]
and
\[\End_k(L\otimes L^\lozenge)=\End(L)\otimes\End(L^\lozenge).\]
Then $A=\End(L)\otimes 1$ and $A^!=1\otimes \End(L^\lozenge)$.