\classheader{2012-10-15}

\subsection*{Integration on topological groups}


Let $X$ be a locally compact topological space, i.e. any point has a compact neighborhood.

Let $C(X)$ be the space of continuous functions $X\to\C$. Let $C_c(X)$ be the subspace of $C(X)$ consisting of functions with compact support.

An integral on $X$ is a linear functional $\int:C_c(X)\to\C$ satisfying
\begin{itemize}
\item If $f(x)\geq 0$ for all $x$, then $\int f\geq 0$, and $\int f =0\iff f=0$.
\item Continuity: for every compact $K \subseteq X$, there exists a constant $C_K \geq 0$ such that for all $f \in C_c(X)$ with $\supp(f) \subseteq K$, $|\int_K f| \leq C_K \cdot\max\limits_{x \in K} |f(x)|$.
\end{itemize}
If $X$ is actually compact, then clearly $C_c(X) = C(X)$ and $\vol(X) = \int 1$. Moreover, Fubini's theorem guarantees that
\[ \int_Y\left(\int_X f(x,y)\,dx\right)dy=\int_X\left(\int_Y f(x,y)\,dy\right)dx.\]


A topological group $G$ is a group that is a topological space such that multiplication $m: G \times G \to G$ and inversion $i: G \to G$ are continuous. 

\begin{examples}
$(\R,+)$, $(\C^\times,\cdot)$, $(\S^1,\cdot)$, $(\GL_n(\R),\cdot)$, $(\U(\R^n),\cdot)$
\end{examples}

When we say that a topological group $G$ acts on a topological space $X$, we require that the action map is continuous.

Given a function $f:X\to\C$ and a $g\in G$, define $g^*f(x) = f(g^{-1} x)$. We say that an integral on $X$ is $G$-invariant if $\int_X g^*(f) = \int_X f$ for all $f \in C_c(X)$ and for all $g \in G$. Alternatively, when thinking about measures, we say that a measure $\mu$ is $G$-invariant if for all $S\subseteq X$ and $g\in G$, we have $\vol(gS)=\vol(S)$.

\begin{theorem}[Haar]
Any locally compact topological group $G$ has a left-invariant integral which is unique up to a constant factor.
\end{theorem}

\begin{examples}
Some examples of integrals which can be obtained this way:
\begin{enumerate}
\item $(\R, +)$ with $dx$,
\item $(\R^+, \cdot)$ with $d\mu = \frac{dx}{x}$,
\item $(S^1, \theta)$ with $d\theta$.
\end{enumerate}
\end{examples}

Using differential forms (on Lie groups say), this theorem is obvious.

Note that a left-invariant integral is not necessarily right-invariant.

\begin{proposition}
If $G$ is a compact group, then $G$ is unimodular (i.e. a left-invariant integral on $G$ is automatically right-invariant).
\end{proposition}
\begin{proof}
Let $\int_L$ be a left-invariant integral on a compact group $G$. Let $f \in C(G)$. Then define $\phi: G \to \C$ by $\phi(g) =\int_L f(xg) dx$. Notice that $\phi(g)$ is also a left-invariant integral. Therefore, there is a constant $c(g) \in \C$ such that 
\[ \int_L f(xg) dx  = c(g) \int_L f(x).\]
Then $c(g)$ has the following properties,
\begin{enumerate}
\item $g \to c(g)$ is continuous on $G$ (to see this, just plug in $f=1$).
\item $c(g) > 0$ for all $g$.
\item $g \to c(g)$ is a group homomorphism into the multiplicative group of $\R$.
\end{enumerate}
Thus, $c(g) = 1$ since the image of $c: G \to \R^+$ is a compact subgroup of $\R^+$, hence $1$.
\end{proof}


\subsection*{The group algebra}
If $G$ is a finite group and $k$ is a field, then $kG$  is a $k$-vector space with basis $g \in G$ and with obvious multiplication. Alternatively, the group algebra $k\{ G \}$ is the set of functions $G \to k$ with convolution and addition, i.e.
\[ (\phi * \psi )(x) = \sum_{g \in G} \phi(xg^{-1})\psi(g) \]
\begin{proposition}
For a finite group $G$, $k\{G\} = kG$.
\end{proposition}
\begin{proof}
The elements $1_g$ form a basis of $k\{G\}$, and $1_g * 1_h = 1_{gh}$.
\end{proof}

Now let $G$ be a locally compact topological group with left-invariant integral $\int$, and let $k$ be a topological field. For $\phi, \psi \in C_c(G)$, define\footnote{A reader suggests that if $G$ is not unimodular, convolution should be defined using $\phi(y)
\psi(y^{-1} x)$, not the other way around, making the action on a
representation a left action.}
\[ (\phi * \psi)(x) = \int_G \phi(xy^{-1})\psi(y) dy. \]

\begin{comments}
$\text{}$

\begin{enumerate}
\item Any discrete group is locally compact (discrete topology), but then $C_c(G)$ only includes functions with finite support.
\item If $G$ is not discrete, then $1_g$ is not a continuous function.
\item If $G$ is discrete (e.g. if $G$ is finite), the unit of the group algebra is the function $1_e$. If $G$ is not discrete, then in fact there is no unit!
\end{enumerate}
\end{comments}