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\subsection{Worked example: $S^1$}

\textbf{Goal:} Understand the \glspl{top_rep} of
$S^1$. Since
\[ f \mapsto \frac{1}{2\pi} \int_0^{2\pi}
f(e^{i\theta}) \dd \theta \]
is a \gls{Haar_int}, these \glspl{top_rep} are all
unitary and hence \gls{comp_red}. So it is enough
to understand the \gls{irred} (unitary)
\glspl{top_rep} of $S^1$.

By \nameref{schurs_lemma} all such have degree
$1$, i.e. we have a correspondence
\[ \{\text{\gls{irred} \glspl{top_rep} of $S^1$}\}
\simbijto \{\text{continuous group homomorphisms
$S^1 \to S^1$}\} .\]
Since $\RR \to S^1$, $x \mapsto e^{2 \pi ix}$
induces an isomorphism of \glspl{top_group}
\begin{align*}
  \RR / \ZZ
  &\simeqto S_1 \\
  \left\{\substack{\text{continuous group}\\
  \text{homomorphisms $S^1 \to
  S^1$}}\right\}
  &\simbijto
  \left\{\substack{\text{continuous group}\\
  \text{homomorphisms $\theta :
  R \to S^1$}} ~\Big|~ \ker\theta \ge \ZZ\right\}
\end{align*}

\textbf{Fact:} If $f : \RR \to S^1$ is a
continuous function with $f(0) = 1$ there is a
unique continuous function $\alpha : \RR \to \RR$
such that $\alpha(0) = 0$ and $f(x) = e^{2\pi i
\alpha(x)}$ for all $x \in \RR$.
\begin{center}
\begin{tikzcd}
   & \RR \ar[d, "e^{2\pi i x}"] \\
  \RR \ar[ur, "\alpha", dashed] \ar[r, "f"]
   & S^1
\end{tikzcd}
\end{center}
\begin{proof}[Sketch proof]
  On small intervals we can define $\alpha(x) =
  \frac{1}{2\pi i \log x}$ and we choose the
  branch of $\log$ so that $\alpha(0) = 0$ and
  $\alpha$ is continuous.
\end{proof}

\begin{lemma*}
  If $\theta : \RR \to S^1$ is a continuous group
  homomorphism there is $\psi : \RR \to \RR$ a
  continuous group homomorphism such that
  $\theta(x) = e^{2\pi i \psi(x)}$ for all $x \in
  \RR$.
\end{lemma*}

\begin{proof}
  Our fact uniquely determines $\psi : \RR \to
  \RR$ continuous function such that $\psi(0) = 0$
  and $\theta(x) = e^{2\pi i \psi(x)}$. We must
  show $\psi$ is a group homomorphism. To this end
  we consider
  \begin{align*}
    \Delta : \RR^2
    &\to \RR \\
    \Delta(a, b)
    &= \psi(a + b) - \psi(a) - \psi(b)
  \end{align*}
  We must show $\Delta \equiv 0$. It is easy to
  see that $\Delta$ is continuous. Also,
  \[ e^{2\pi i \Delta(a, b)} = \theta(a + b)
  \theta(a)^{-1} \theta(b)^{-1} = 1 \]
  so $\Delta$ takes values in $\ZZ$. So as $\RR^2$
  is connected, $\Delta$ is constant ($\ZZ$ is
  discrete). But $\Delta(0, 0) = 0$, so $\Delta
  \equiv 0$ as required.
\end{proof}

\begin{lemma*}
  If $\psi : (\RR, +) \to (\RR, +)$ is a
  continuous group homomorphism then $\exists
  \lambda \in \RR$ such that $\psi(x) = \lambda x$
  for all $x \in \RR$.
\end{lemma*}

\begin{proof}
  Let $\lambda = \psi(1)$. Then $\psi(n) = \lambda
  n$ for all $n \in \ZZ$ ($\psi$ is a
  homomorphism). So
  \[ m \psi \left( \frac{n}{m} \right) = \psi(n) =
  \lambda n \]
  for all $\frac{n}{m} \in \QQ$ ($\psi$ is a
  homomorphism), i.e. $\psi(x) = \lambda x$ for
  all $x \in \QQ$. But $\QQ$ is dense in $\RR$, so
  $\psi(x) = \lambda x$ for all $x \in \RR$.
\end{proof}

\begin{flashcard}[S1-reps-thm]
\begin{theorem*}[Representations of $S^1$]
  \cloze{Every \gls{irred} \gls{top_rep} of $S^1$ is
  \repdim{1} and is of the form $z \mapsto z^n$
  for some $n \in \ZZ$.}
\end{theorem*}  

\begin{proof}
  \cloze{
  We've already seen that if $\rho : S^1 \to
  \GL_d(\CC)$ is an \gls{irred} \gls{top_rep} then $d
  = 1$ and $\rho(S^1) \le S^1$. Moreover $\rho$
  induces a continuous homomorphism $\theta : \RR
  \to S^1$ given by $\theta(x) = \rho(e^{2\pi i
  x})$. By the last two lemmas, there exists
  $\lambda \in \RR$ such that $\theta(x) = e^{2\pi
  i \lambda x}$ for all $x \in \RR$. Since
  $\theta(1) = \rho(e^{2\pi i}) = \rho(1) = 1$, we
  deduce $e^{2\pi i \lambda} = 1$, i.e. $\lambda
  \in \ZZ$. So $\rho(e^{2\pi i x}) = (e^{2\pi i
  x})^\lambda$ for $\lambda \in \ZZ$.
  }
\end{proof}
\end{flashcard}

The theorem says that the ``\gls{char_table}'' of
$S^1$ has rows given by $\chichar_n$ for $n \in \ZZ$,
$\chichar_n(z) = z^n$.

(unitary \repdim{1} \glsref[char_rep]{characters}
of $\ZZ$ are all of the form $n \mapsto
e^{in\theta}$ for some $e^{i\theta} \in S^1$).

\begin{notation*}
  $\ZZ[z, z^{-1}) = \left\{ \sum_{n \in \ZZ} a_n
  z^n ~\bigg|~ a_n \in \ZZ, \sum_{n \in \ZZ} |a_n|
  < \infty\right\}j$ a ring under natual
  operations.
\end{notation*}
\vspace{-1em}

If $V$ is any \gls{top_rep} of $S^1$ then it
decomposes as a direct sum of \repdim{1}
\glspl{subrep} and its character $\chichar_V =
\sum_{n \in \ZZ} a_n z^n$ with all $a_n \ge 0$ and
$\sum a_n = \dim V$, where as usual $a_n$ is the
number of copies of $(z \mapsto z^n)$ in $V$.

So
\[ \charring(S^1) = \{\chichar - \chichar' \st
\text{$\chichar$, $\chichar'$ are
\glsref[char_rep]{characters} of $S^1$}\} = \ZZ[z,
z^{-1}) \]

By orthogonality of characters,
\begin{align*}
  \langle \chichar_n, \chichar_m \rangle_{S^1}
  &= \frac{1}{2\pi} \int_0^{2\pi} e^{2\pi i (m -
  n)\theta} \dd \theta = d_{m, n} \\
  a_n
  &= \langle \chichar_n, \chichar_V \rangle_{S^1}
  = \frac{1}{2\pi} \int_0^{2\pi}
  \chichar_V(e^{i\phi}) e^{-in\phi} \dd \phi
\end{align*}
and so
\[ \chichar_V(e^{i\theta}) = \sum_{n \in \ZZ}
\left( \frac{1}{2\pi} \int_0^{2\pi}
\chichar_V(e^{i\phi}) e^{-in\phi} \dd \phi \right)
e^{in\phi} \]
So Fourier decomposition of $\chichar_V$
decomposes $\chichar_V$ into \gls{irred}
\glsref[char_rep]{characters} and the FOurier mode
is the multiplicity.

\begin{remark*}
  In fact by the theory of Fourier series any
  continuous function on $S^1$ can be approximated
  uniformly by a finite $\CC$-linear combination
  of $\chichar_n$. Moreover the $\chichar_n$ form
  a complete orthonormal set in the Hilbert space
  \[ \Lp[2](S^1) = \left\{f : S^1 \to \CC ~\bigg|~
  \int_0^{2\pi} |f(e^{i\theta})|^2 \dd \theta <
  \infty\right\} / \sim \]
  of square integrable functions on $S^1$, i.e.
  every function in $\Lp[2](S^1)$ has a unique
  expression as
  \[ f(e^{i\theta}) = \sum_{n \in \ZZ} \left(
  \frac{1}{2\pi} \int_0^{2\pi} f(e^{i\phi})
  e^{-in\phi} \dd \phi \right) e^{in\phi} \]
  converging with respect to the norm $\|f\|^2 =
  \int_0^{2\pi} |f(e^{i\theta})|^2 \dd \theta$. We
  can phrase this as
  \[ \Lp[2](S^1) = \hat{\bigoplus}_{n \in \ZZ} \CC
  \chichar_n \]
  ($\hat{\bigoplus}$ means complete direct sum),
  which is an analogue of
  \[ \CC G = \bigoplus_{V \in \Irr(G)} (\dim V) V \]
  for finite groups (cf
  \href{https://en.wikipedia.org/wiki/Peter\%E2\%80\%93Weyl_theorem}{Peter
  Weyl Theorem}).
\end{remark*}

\subsection{Second worked example $\SU(2)$}

Recall $\SU(2) = \{A \in \GL_2(\CC) \st
\ol{A}^\top A = I, \det A = 1\}$. If
\[ A =
\begin{pmatrix}
  a & b \\
  c & d
\end{pmatrix} \in \SU(2) \]
then as $\det A = 1$,
\[ A^{-1} =
\begin{pmatrix}
  d & -b \\
  -c & a
\end{pmatrix}
=
\begin{pmatrix}
  \ol{a} & \ol{c} \\
  \ol{b} & \ol{d}
\end{pmatrix}
\]
Thus $d = \ol{a}$ and $c = -\ol{b}$. Moreover,
$|a|^2 + |b|^2 = 1$. In this way,
\[ \SU(2) = \left\{
\begin{pmatrix}
  a & b \\
  -\ol{b} & \ol{a}
\end{pmatrix}
~\Bigg|~ a, b \in \CC, |a|^2 + |b|^2 = 1
\right\} \]
which is homeomorphic to $S^3 \subset \RR^4 \simeq
\CC^2$. More precisely if
\[ \HH = \RR \cdot \SU(2) = \left\{
\begin{pmatrix}
  z & w \\
  -\ol{w} & \ol{z}
\end{pmatrix}
\right\} \subset M_2(\CC) \]
Then $\|A\|^2 = \det A$ defines a norm on
$\HH \simeq \RR^4$ and $\SU(2)$ is the unit sphere
in $\HH$ with respect to this norm. If $A \in
\SU(2)$,
\[ \|AX\| = \|X\| = \|XA\| \qquad \forall X \in
\HH \]
(since $\det A = 1$). So $\SU(2)$ acts on $\HH$ on
both left and right by isometries. So after
normalisation, usual integration on $S^3$ defines
a \gls{Haar_int} on $\SU(2)$, i.e.
\[ \Gint_{\SU(2)} f = \frac{1}{2\pi^2} \Gint_{S^3} f \]
Here $\frac{1}{2\pi^2}$ is the volume of $S^3$ in
$\RR^4$ with respect to usual measure. We now try
to understand conjugacy classes in $\SU(2)$. Let
\[ T = \left\{
\begin{pmatrix}
  z & 0 \\
  0 & z^{-1}
\end{pmatrix}
~\Bigg|~ z \in S^1
\right\} \le \SU(2) \]

\begin{proposition*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Every conjugacy clsas in $\SU(2)$
      contains an element of $T$
    \item More precisely, if $O$ is a conjugacy
      class in $\SU(2)$, $O \cap T = \{t,
      t^{-1}\}$ for some $t \in T$. If $t =
      t^{-1}$, $t = \pm I$ and $O = \{t\}$.
    \item There is a continuous bijection
      \begin{align*}
        \{\text{conjugacy classes in $\SU(2)$}\}
        &\to [-1, 1) \\
        A
        &\mapsto \half \Trace A
      \end{align*}
  \end{enumerate}
\end{proposition*}