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\section{Topological Groups}

In this chapter, $k = \CC$. This is important,
because we will be using topological properties of
$\CC$ (contrary to previously, where we normally
are just using the fact that it is algebraically
closed).

\subsection{Definitions and Examples}
\label{sub8_1}

\begin{flashcard}[top-group]
\begin{definition*}[Topological group]
  \glsnoundefn{top_group}{topological
  group}{topological groups}
  \cloze{A \emph{topological group} $G$ is a group $G$
  which also has the structure of a topological
  space such that the multiplication map $G \times
  G \to G$, $(x, y) \mapsto xy$ and the inversion
  map $G \to G$, $x \mapsto x^{-1}$ are both
  continuous.}
\end{definition*}
\end{flashcard}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
  \item $\GL_n(\CC)$ with the subspace topology
    from $\Mat_n(\CC) \simeq \CC^{n^2}$, since
    \[ (AB)_{ij} = \sum_{k = 1}^n A_{ik} B_{kj}
    \qquad \text{and} \qquad A^{-1} =
    \frac{1}{\det A} \adj A \]
    are both continuous. More generally, if $V$ is
    any $\CC$-vector space we can give $\GL(V)$
    the topology that makes the isomorphism
    $\GL(V) \to \GL_n(\CC)$ (given by choosing a
    basis) a homeomorphism. Since conjugation on
    $\GL_n(\CC)$, $X \mapsto P^{-1} XP$ is
    continuous for all $P \in \GL_n(\CC)$, this
    does not depend on the choice of basis.
  \item $G$ finite with discrete topology since
    all amps $G \times G \to G$ and $G \to G$ are
    continuous.
  \item $\On(n) = \{A \to \GL_n(\RR) \st A^\top A =
    I\}$, $\SO(n) = \{A \in \On(n) \st \det A =
    1\}$.
  \item $\U(n) = \{A \in \GL_n(\CC) \st
    \ol{A}^\top A = I\}$, $\SU(n) = \{A \in \U(n)
    \st \det(A) = 1\}$.
    In particular, $\U(1) = S^1 = (\{x \in
    \CC^\times \st |z| = 1\}, \cdot)$.
  \item $*$ (non-examinable) $G$ a profinite group
    such as $\ZZ_p$ the completion of $\ZZ$ with
    respect to $p$-adic metric.
\end{enumerate}

\begin{flashcard}[rep-of-top-group]
\begin{definition*}[Representation of a topological
  group]
  \glsnoundefn{top_rep}{representation}{representations}
  \cloze{A \emph{representation of a
  \gls{top_group}}
  $G$ is a continuous homomorphism
  \[ \rho : G \to \GL(V) \]
  ($V$ a vector space over $\CC$).}
\end{definition*}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item If $G$ is a (finite) group with the
      discrete topology then every function $G \to
      \GL(V)$ is continuous and we recover the old
      definition.
    \item If $X$ is any topological space then
      $\alpha : X \to \GL_n(\CC)$ is continuous
      if and only if $\alpha_{ij} : X \to \CC$,
      $\alpha_{ij}(x) \defeq \alpha(x)_{ij}$ is
      continuous for all $i, j$.
  \end{enumerate}
\end{remark*}

\subsection{Compact groups}

Our most powerful when studying finite groups was
the operator $\frac{1}{|G|} \sum_{g \in G}$. We
want to replace $\sum$ by $\int$.

\begin{flashcard}[hair-integral-defn]
\begin{definition*}[Haar integral]
  \glssymboldefn{Haar_int}{$\int_G$}{$\int_G$}
  \glsnoundefn{Haar_int}{Haar integral}{Haar
  integrals}
  \glssymboldefn{cts_func}{$C(X, Y)$}{$C(X, Y)$}
  \cloze{For $G$ a \gls{top_group} and $C(G, \RR) = \{f
  : G \to \RR \st f \text{ continuous}\}$, a linear
  map $\int_G : \cts(G, \RR) \to \RR$ is called a
  \emph{Haar integral} if
  \begin{enumerate}[(i)]
    \item $\int_G \indicator{G} = 1$ (So $\int_G$
      is normalised so that total volume is $1$).
      \glsadjdefn{ti}{translation
      invariant}{\gls{Haar_int}}
    \item $\int_G f(xg) \dd g = \int_G f(g) \dd g
      = \int_G f(gx) \dd g$ for all $x \in G$ (so
      $\int_G$ is translation invariant).
      (we write $\int_G f(g) \dd g = \int_G f$ and
      $\int_G f(xg) \dd g$ means apply $\int_G$ to $g
      \mapsto f(xg) \in \cts(G, \RR)$).
    \item $\int_G f \ge 0$ if $f(g) \ge 0$ for
      all $g \in G$ (positivity).
  \end{enumerate}
  }
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item If $G$ is finite then $\Gint_G f =
      \frac{1}{|G|} \sum_{g \in G} f(g)$ is a
      \gls{Haar_int}.
    \item If $G = S^1$, $\Gint_G f = \frac{1}{2\pi}
      \int_0^{2\pi} f(e^{i\theta}) \dd \theta$ is
      a \gls{Haar_int}.
  \end{enumerate}
\end{example*}
\vspace{-1em}

\glssymboldefn{vHaar_int}{$\int_G$}{$\int_G$}
Note that for any $\RR$-vector space $V$,
$\Gint_G$ induces a linear map (also called
$\Gint_G$)
\[ \int_G : \cts(G, V) \to V \]
Under the identification $V \simeq V^{**}$ for
$\theta \in V^*$, $f \in \cts(G, V)$,
\[ \theta \left( \vGint_G f \right) = \Gint_G
\theta(f(g)) \dd g \]

More concretely, if $v_1, \ldots, v_n$ is a basis
for $V$ and $f \in \cts(G, V)$ then
\[ f = \sum_{i = 1}^n f_i v_i \]
with $f_i \in \cts(G, \RR)$ and
\[ \vGint_G f = \sum_{i = 1}^n \left( \Gint_G f_i
\right) v_i .\]
This map is also \gls{ti} and sends a
constant function to its only value. Moreover if
$\alpha : V \to W$ is linear and $f \in \cts(G,
V)$, then
\[ \alpha \left( \vGint_G f \right) = \vGint_G
\alpha(f) .\]
In particular if $V$ is a $\CC$-vector space $v
\mapsto iv$ is $\RR$-linear so $\vGint_G : \cts(G,
V) \to V$ is $\CC$-linear.

\begin{theorem*}
  If $G$ is a compact Hausdorff group then there
  is a unique \gls{Haar_int} on $G$.
\end{theorem*}

\begin{proof}
  Omitted.
\end{proof}

\glsadjdefn{cpt}{compact}{\gls{top_group}}
All the examples in \cref{sub8_1} are compact
Hausdorff except $\GL_n(\CC)$ which is not
compact. We'll follow standard practice in this
field and write ``compact'' to mean ``compact and
Hausdorff''.

\begin{flashcard}[weyls-unitary-trick]
\begin{corollary*}[Weyl's unitary trick]
  If $G$ is a \gls{cpt} \gls{top_group} then every
  \gls{top_rep} $(\rho, V)$ of $G$ is
  \gls{uni_rep}.
\end{corollary*}

\begin{proof}
  \cloze{
  As for finite groups, let $\langle \bullet,
  \bullet \rangle$ be an inner product on $V$.
  Then
  \[ (v, w) \defeq \Gint_G \langle \rho(g) v,
  \rho(g) w \rangle \dd g \]
  is the required \Ginv{} inner product. Since,
  for $x \in G$ and $v, w \in V$,
  \begin{align*}
    (\rho(x) v, \rho(x) w)
    &= \Gint_G \langle \rho(gx) v, \rho(gx) w
    \rangle \dd g \\
    &= \Gint_G \langle \rho(g) v, \rho(g) w
    \rangle \dd g
    &&\text{(by $G$-invariance of $\Gint_G$)} \\
    &= (v, w)
  \end{align*}
  Clearly $(\bullet, \bullet)$ is an inner product
  by using $\CC$-linearity of $\Gint_G$ and
  positivity of $\Gint_G$.}
\end{proof}
\end{flashcard}

\begin{remark*}
  It follows that every \gls{cpt} subgroup of
  $\GL_n(\CC)$ is conjugate to a subgroup of
  $\U(n)$.
\end{remark*}

\begin{corollary*}
  All \glspl{top_rep} of a \gls{cpt} group are
  \gls{comp_red}.
\end{corollary*}
\vspace{-1em}

If $G \to \GL(V)$ is a \gls{top_rep} then
$\chichar_\rho \defeq \Trace \rho$ is a continuous
\gls{cls_fn} on $G$.

\begin{flashcard}[dim-Ug-formula]
\begin{lemma*}
  If $U$ is a \gls{top_rep} of $G$ \gls{cpt} then
  \[ \dim U^G = \Gint_G \chichar .\]
\end{lemma*}

\begin{proof}
  \cloze{
  Let $\pi \in \Hom_k(U, U)$ be defined by $\pi =
  \vGint_G \rho \in \Hom_k(U, U)$. If $x \in G$
  then
  \[ \rho(x) \cdot \pi = \rho(x) \vGint_G
  \rho(g) \dd g = \vGint_G \rho(xg) \dd g = \pi \]
  since $\vGint_G$ is \gls{ti}. So $\Im \pi \le
  U^G$. If $u \in U^G$ then
  \[ \pi(u) = \left( \vGint_G \rho(g) \dd g
  \right)(u) = \vGint_G \rho(g) u \dd g = \vGint_G
  u = u \]
  Thus $\pi$ is a projection onto $U^G$. So
  \[ \dim U^G = \Trace \pi = \Trace \left(
  \vGint_G \rho \right) = \Gint_G \Trace \rho =
  \Gint_G \chichar .\]
  }
\end{proof}
\end{flashcard}

\begin{corollary*}[Orthogonality of characters]
  If $G$ is a \gls{cpt} group and $V$, $W$ are
  \gls{irred} \glspl{rep} of $G$ then
  \[ \langle \chichar_V, \chichar_W \rangle_G =
  \begin{cases}
    1 & \text{if $V \simeq W$} \\
    0 & \text{if $V \not\simeq W$}
  \end{cases} \]
  where
  \glssymboldefn{Gip_int}{$\langle f, g
  \rangle_G$}{$\langle f, g \rangle_G$}
  \[ \langle f_1, f_2 \rangle_G = \Gint_G
  \ol{f_1}(g) f_2(g) \dd g \]
\end{corollary*}
\vspace{-1em}

To prove this as in the finite case we use
$\chichar_v(g^{-1}) = \ol{\chichar_V}(g)$. This
holds because $V$ is unitary.