\classheader{2012-10-17}

Let $V$ be a vector space over $k$.

\begin{definition}
A representation of a group $G$ is a group homomorphism $\rho:G\to \GL(V)$. This is equivalent to a linear $G$-action on $V$.
\end{definition}

Observe that this is also equivalent to specifying a $kG$-module structure on $V$. If $\rho$ is a group representation of $G$, then we declare that $a=\sum_{g\in G}c_gg\in kG$ will act on $V$ via
\[\rho(a)=\sum_{g\in G}c_g\rho(g):V\to V\,\]
or in other words, for all $v\in V$,
\[\rho(a)v=\sum_{g\in G}c_g\rho(g)v.\]
(Note that sometimes we'll just write $av$ instead of $\rho(a)v$). Using this view, we have the notions of a subrepresentation, quotient representation, and direct sum of representations. An ``irrep'' is a simple $kG$-module, and an ``intertwiner'' is a morphism of $kG$-modules.

\begin{theorem}[Schur Lemma for Representations]
Given an algebraically closed field $k$, let $V_1,V_2$ be finite-dimensional irreps. Let $f:V_1\to V_2$ be an intertwiner. If $V_1\not\cong V_2$, then $f=0$, and if $V_1\cong V_2$, then $f=\lambda\cdot\id_V$ for some $\lambda\in k$.
\end{theorem}
\begin{proof}
Apply the Schur lemma for algebras to $kG$.
\end{proof}

From now on let $k=\C$. Let $V$ be a vector space over $\C$ with a positive definite hermitian inner product $(\cdot,\cdot)$. 
\begin{definition}
A unitary representation of $G$ in $V$ is a homomorphism $G\to \U(V)\subset\GL(V)$, or in other words, a linear $G$-action on $V$ by isometries.
\end{definition}
\begin{lemma}
Any unitary representation is completely reducible.
\end{lemma}
\begin{proof}
Let $W\subset V$ be a subrepresentation. We have that $V=W\oplus W^\perp$. We need to show that $W^\perp$ is $G$-stable.

Let $x\in W^\perp$. We need to check that $(\rho(g)x,W)=0$. Note a very important fact:
\begin{center}
\fbox{\;$\rho$ is unitary $\iff$ $\rho(g)^*=\rho(g)^{-1}=\rho(g^{-1})$\;}
\end{center}
Thus
\[(\rho(g)x,W)=(x,\rho(g)^*W)=(x,\rho(g^{-1})W\subset(x,W)=0,\]
and thus $W^\perp$ is a subrepresentation of $V$.
\end{proof}

Now let's consider representations of topological groups. Let $G$ be a topological group, and let $V$ be a finite dimensional vector space over $\C$. Then $\GL(V)\subset\End_{\C}(V)\cong \C^{\dim(V)^2}$, and this inclusion is in fact an open embedding.
\begin{definition}
A representation of $G$ in $V$ is a continuous representation of $G$.
\end{definition}
More concretely, a representation of $G$ in $\C^n$ is a homomorphism $\rho:G\to\GL_n(\C)$, with $g\mapsto (\rho_{ij}(g))$, and this is continuous if and only if for each $1\leq i,j\leq n$, $g\mapsto\rho_{ij}(g)$ is a continuous function from $G$ to $\C$.

As we discussed last time, the natural candidate for the group algebra over a topological group is the algebra $C_c(G)$ with convolution as the product. Fix a left-invariant integral $\int$ on $G$. Then any representation $\rho:G\to\GL(V)$ gives $V$ a $(C_c(G),\ast)$-module structure: for $f\in C_c(G)$, define
\[\rho(f)=\int f(g)\rho(g)\in\End_\C(V).\]

(Squirrels try to get into the classroom. Ginzburg relates that he once found a squirrel in his office that he gave nuts to.)

If $V=\C^n$, then
\[\rho(f)_{ij}=\int f(g)\rho_{ij}(g),\]
and
\[\rho(f)v=\int f(g)(\rho(g)v).\]
\begin{lemma}
Let $\rho:G\to\GL(V)$ be a continuous representation, and let $W\subset V$ be a $C_c(G)$-stable subspace. Then in fact $W$ is $G$-stable, so it is a subrepresentation.
\end{lemma}
(Note that unless $G$ is discrete, we have $1_g\notin C_c(G)$ for all $g\in G$, so this isn't obvious.)
\begin{proof}
Given $g\in G$, we could recover the action of the group element using a delta function,
\[\int \delta_{g_0}(g)\rho(g)=\rho(g_0).\]
Since we aren't doing functional analysis, we can't really use delta functions, but we will try to approximate them anyway.

Let $U_0$ be a compact neighborhood of $g\in G$. Because $U_0$ is compact, there is some $C$ such that
\[\left|\int_{U_0}f\right|\leq C\cdot\max_{x\in U_0}|f(x)|.\]
For any $\epsilon>0$, find an open neighborhood $U_\epsilon$ of $g$ and a function $\phi_\epsilon$ such that\footnote{Multiple readers have suggested that $G$ needs to be locally compact and Hausdorff in order to prove that such a function exists.}
\begin{enumerate}
\item $|\rho(x)-\rho(g)|\leq \epsilon$ for all $x\in U_\epsilon$
\item $\supp(\phi_\epsilon)=$ a compact subset of $U_\epsilon$
\item $\phi_\epsilon\geq 0$
\item $\int \phi_\epsilon=1$
\end{enumerate}
The standard way of thinking about this is of course

\begin{center}
\begin{tikzpicture}[scale=0.5]
\begin{scope}
\clip (-5,0) rectangle (5,3.5);
\draw [style={thick,line cap=round}] plot [smooth,samples=20] coordinates {(-4,0.0) (-1,0.25) (0,3) (1,0.25) (4,0.00)};
\end{scope}
\draw[thick] (-4,0) -- (4,0);
\node[fill,circle,inner sep=0pt,outer sep=0pt,minimum size=0.15cm] at (0,0) [label={270:$g$}] {};
\node at (5,0) {$G$};
\node at (0,3.5) {$\phi_\epsilon$};
\end{tikzpicture}
\end{center}

We claim that
\[\left|\int \phi_\epsilon(x)\rho(x)-\rho(g)\right|\leq \epsilon C_0.\]
To see this, note that
\[\left|\int \phi_\epsilon(x)\rho(x)-\rho(g)\right|=\left|\int_G\rho_\epsilon(x)\rho(x)-\int_G\phi_\epsilon(x)\rho(g)\right|\leq\int \phi_\epsilon(x)|\rho(x)-\rho(g)|\leq C\epsilon\]
Finally,
\[\lim_{n\to\infty}\int\phi_{1/n}(x)\rho(x)=\rho(g).\]
Therefore $W$ is $\rho(g)$-stable.
\end{proof}

Given a $G$-action on $X$, we let $X^G$ be the set of $G$-fixed points.


\newcommand{\Av}{\operatorname{Av}}
\begin{lemma}[Averaging Lemma]
Let $G$ be a compact group, and let $\rho:G\to\GL(V)$ be a representation. Then the map $V\to V$ defined by
\[v\mapsto \frac{1}{\vol(G)}\int_G\rho(g)v\]
is a projection to $V^G\subseteq V$. In particular, given a continuous $G$-action on $X$ and a function $f\in C(X)$, we can define $\Av(f)$ by
\[x\mapsto \int_Gf(gx)\;dg=\Av(f)(x),\]
which is a $G$-invariant function on $X$.
\end{lemma}
\begin{proof}
We need to check that
\begin{enumerate}
\item If $v\in V^G$, then $\frac{1}{\vol(G)}\int \rho(g)v=v$.
\item For all $v\in V$, we have $\frac{1}{\vol(G)}\int \rho(g)v\in V^G$.
\end{enumerate}
For 1, note that if $\rho(g)v=v$ for all $g$, then
\[\frac{1}{\vol(G)}\int v=\frac{1}{\vol(G)}\cdot\vol(G)v=v.\]
For 2, we have for any $h\in G$ that
\[\rho(h)\int_G\rho(g)v=\int_G\rho(h)\rho(g)v=\int_G\rho(hg)v\]
which, because our measure is left-invariant, is equal to
\[\int \rho(g)v.\qedhere\]
\end{proof}