\classheader{2012-10-24}


Just as a reminder, the midterm is today, 4:30 to 6:15, in Eckhart 312.

\begin{theorem}[Plancherel, general case]
For $\phi,\psi\in\C\{G\}$,
\[\sum_{g\in G}\phi(g)\overline{\psi(g)}=\sum_{\rho\in\widehat{G}} \frac{\dim(\rho)}{\# G}\tr(\rho(\phi)\rho(\psi)^*).\]
\end{theorem}
\begin{proof}
Let $f=\phi\ast\overline{\psi^\tau}$. Apply the orthogonality relations to $f$:
\[(\phi\ast\overline{\psi^\tau})(e)=\sum_{\rho\in\widehat{G}}\frac{\dim(\rho)}{\#
	G}\tr(\rho(\phi\ast\overline{\psi^\tau})).\]
%(we proved this last time?) 
%Now look at the left side of the above orthogonality relation, and note that it equals the left side of the theorem statement:
%jw 
%\[(\text{LHS of orthogonality})\quad \left.\sum_{y\in G}\phi(xy^{-1})\overline{\psi^\tau}(y)\right|_{x=e} = \sum_{y\in G}\phi(y^{-1})\overline{\psi(y^{-1})}\quad (\text{LHS of theorem}).\]
%Now let's prove that the right sides are equal. The right side of the orthogonality relation is
%\[\sum_{\rho\in\widehat{G}}\frac{\dim(\rho)}{\# G}\tr(\rho(\phi)\rho(\psi)^*).\]
%Consider $\Psi$, defined by \[\Psi=\bigg(\bigoplus_{\rho\in\widehat{G}}\rho\bigg):\C G\to\bigoplus_{\rho\in\widehat{G}}\End_\C(L_\rho)\]
%mapping $f$ to $\operatornamewithlimits{\oplus}\limits_{\rho\in\widehat{G}}\rho(f)$.
By definition of convolution, the LHS of the above orthogonality relation is equal to 
\[\sum_{y\in G}\phi(ey^{-1})\overline{\psi^\tau}(y) = \sum_{y\in G}\phi(y^{-1})\overline{\psi(y^{-1})}\quad (\text{LHS of theorem}).\]
We know that $\rho(\phi * \overline{\psi^\tau}) = \rho(\phi) \rho(\overline{\psi^\tau})$, and by
the last lemma from Lecture 11, $\rho(\overline{\psi^\tau}) = \rho(\psi)^*$. Hence the RHS of
orthogonality coincides with the RHS of the theorem, and we are done.
\end{proof}

%jw
Recall from Wedderburn theory that the map 
\[ \Psi : \C{G} \to \operatornamewithlimits\bigoplus\limits_{\rho \in \widehat G}
\End_\C(L_\rho) : f \mapsto \operatornamewithlimits{\oplus}\limits_{\rho\in
	\widehat{G}}\rho(f) \] is an isomorphism. The inversion theorem gives a formula for the inverse:

\begin{theorem}[Inversion]
For
$a=\operatornamewithlimits{\oplus}\limits_{\rho\in\widehat{G}}a_\rho\in
\operatornamewithlimits\bigoplus\limits_{\rho\in\widehat{G}}\End_\C(L_\rho)$, 
the inverse map $\Psi^{-1}$ is given by
\[\Psi^{-1}(a)(g)=\sum_{\rho\in\widehat{G}}\frac{\dim(\rho)}{\# G}\tr(a_\rho\cdot\rho(g)^{-1}).\]
\end{theorem}
\begin{proof}
WLOG we can assume 
$a=\Psi(f)=\operatornamewithlimits{\oplus}\limits_{\rho\in \widehat{G}}\rho(f)$ for some $f\in\C\{G\}$. 
We need to check\vspace{-0.1in}
\[\sum_{\rho\in\widehat{G}}\frac{\dim(\rho)}{\# G}\tr(\rho(f)\rho(g)^{-1})\stackrel{?}{=}f(g).\]
But the left side is just equal to
\[\sum_{\rho\in\widehat{G}}\frac{\dim(\rho)}{\# G}\tr(\rho(f)\cdot\rho(\mathbf{1}_g)^*)\stackrel{\text{ Plancherel }}{=}\sum_{x\in G}f(x)\overline{\mathbf{1}_g(x)}=f(g).\qedhere\]
\end{proof}
Recall that the character of a representation $\rho$ is defined to be $\chi_\rho(f)=\tr(\rho(g))$.

\begin{corollary}
The element $e_\rho=\frac{\dim(\rho)}{\# G}\sum\limits_{g\in G}\overline{\chi_\rho(g)}\cdot g$ is a central idempotent in $\C G$. Moreover,\vspace{-0.1in}
\[\rho'(e_\rho)=\begin{cases}
\id_{L_\rho}&\text{ if }\rho'\cong \rho,\\
0&\text{ if }\rho'\not\cong\rho.
\end{cases}\]
\end{corollary}
\begin{proof}
Take $a=\oplus_{\rho'}a_{\rho'}$, where $a_{\rho'}=\begin{cases}
\id_{L_\rho}&\text{ if }\rho'\cong \rho,\\
0&\text{ if }\rho'\not\cong\rho.
\end{cases}$. Then
\[\Psi^{-1}(a)(g)\stackrel{\text{ inversion }}{=}\frac{\dim(\rho)}{\# G}\tr(\rho(g^{-1}))\]
which implies that $\Psi^{-1}(a)=\frac{\dim(\rho)}{\# G}\sum\limits_{g\in G}\tr(\rho(g)^*)$. But \[\tr(\rho(g)^*)=\tr(\overline{\rho(g)})=\overline{\tr(\rho(g))}=\overline{\chi_\rho(g)}\]
so we are done.
\end{proof}

Let $\C\{G\}^G$ be the collection of class functions on $G$. 
\begin{theorem}[Classical orthogonality]
The set $\{\chi_\rho\mid\rho\in\widehat{G}\}$ is an orthonormal basis of $\C\{G\}^G$, i.e.
\[(\chi_\rho,\chi_{\rho'})=\begin{cases}\# G & \text{ if }\rho\cong\rho',\\
0 & \text{ if }\rho\not\cong\rho'.\end{cases}\]
\end{theorem}
\begin{proof}
\[(\overline{\chi_\rho},\overline{\chi_\rho})=\sum_{g\in G}\overline{\chi_\rho(g)}\chi_{\rho'}(g)=\left(\frac{\# G}{\dim(\rho)}e_\rho,\frac{\# G}{\dim(\rho')}e_{\rho'}\right)\]
\[\stackrel{\text{ Plancherel }}{=} \sum_{\sigma\in\widehat{G}}\frac{\dim(\sigma)}{\# G}\tr\left(\sigma\left(\frac{\# G}{\dim(\rho)}e_\rho\right)\cdot\sigma\left(\frac{\# G}{\dim(\rho')}e_{\rho'}\right)^*\right)\]
\[=\sum_{\sigma\in\widehat{G}}\frac{\dim(\sigma)}{\# G}\frac{(\# G)^2}{\dim(\rho)\dim(\rho')}\tr\left(\delta_{\sigma\rho}\id_{L_\rho}\cdot\;\delta_{\sigma\rho'}\id_{L_{\rho'}}\right)\]
\[=\begin{cases}
0 & \text{ if }\rho\not\cong\rho',\\
\frac{\dim(\rho)}{\# G}\cdot\frac{(\# G)^2}{\dim(\rho)^2}\dim(\rho)=\#G & \text{ if }\rho\cong\rho'.
\end{cases}\qedhere\]
\end{proof}

\section*{Tensor products}

Let $B$ be a ring. We can consider right $B$-modules, which are equivalent to left $B^{\text{op}}$-modules. If $B$ is commutative, then left is the same as right.
For example, we can think of the collection of column vectors
\[\begin{pmatrix}
B\\ B\\ \vdots\\  B
\end{pmatrix}\]
as a left $\M_n(B)$-module, and the collection of row vectors 
\[\begin{pmatrix}
B & B & \cdots & B
\end{pmatrix}\]
as a right $\M_n(B)$-module.

Let $A$ and $B$ be rings. An $(A,B)$-bimodule is an abelian group $M$ which is a left $A$-module and a right $B$-module such that the $A$-action and $B$-action commute.

\begin{examples}
$\text{}$

\begin{itemize}
\item $A$ is an $(A,A)$-bimodule in the obvious way.
\item The space of $m\times n$ matrices over $A$ is an $(\M_m(A),\M_n(A))$-bimodule.
\end{itemize}
\end{examples}

\begin{definition}
Let $M$ be a right $A$-module, and $N$ be a left $A$-module. Then we define
\[M\otimes_A N=\frac{\{\text{free abelian group on symbols }m\otimes n\}}{\left\langle\begin{array}{c}
(m_1+m_2)\otimes n-m_1\otimes n-m_2\otimes n\\
m\otimes (n_1+n_2) - m\otimes n_1-m\otimes n_2\\
ma\otimes n-m\otimes an
\end{array}\right\rangle}\]
\end{definition}

\begin{comments}
$\text{}$

\begin{itemize}
\item If $A$ is a $k$-algebra, then $M\otimes_A N$ is a quotient of $M\otimes_kN$.
\item $A\otimes_AN=N$ and $M\otimes_AA=M$
\item If we have rings $A$, $B$, and if $M$ is an $(A,B)$-bimodule, then for any left $B$-module $N$, there is still a left $A$-action on $M\otimes_BN$, making it into a left $A$-module.
\item The tensor product $\otimes_A$ has a universal property,
\[\xymatrix@!C=0.9in{ M\times N \ar[dr]_f \ar[r]^{\text{ canonical }}& M\otimes_A N\ar@{-->}[d]^{\exists ! \widetilde{f}}\\ 
& M}\]
where $\widetilde{f}$ is a map of abelian groups and $f$ is a ``middle $A$-linear'' map.
\end{itemize}
\end{comments}