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\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Every unitary matrix has an orthonormal
      basis of eigenvectors. That is, for $A \in
      \SU(2)$, there exists $P \in \U(2)$ such
      that $P^{-1} AP \in T$. Then if $Q =
      \frac{1}{\sqrt{\det P}} P \in \SU(2)$,
      $Q^{-1} AQ = P^{-1} AP \in T$, i.e.
      $[A]_{\SU(2)} \cap T \neq \emptyset$.
    \item If $A=  \pm I$ the claim is clear.
      Otherwise
      \begin{align*}
        [A]_{\SU(2)}
        &= [t]_{\SU(2)}
        &&\text{for some $t \in T$} \\
        &= \{gtg^{-1} \st g \in \SU(2)\}
      \end{align*}
      All elements of $gtg^{-1}$ have the same
      eigenvalues as $t$. So if $t' = gtg^{-1} \in
      T$ then $t' \in n\{t^{\pm 1}\}$, i.e.
      $[A]_{\SU(2)} \cap T \subseteq \{t^{\pm
      1}\}$. But if
      \[ s =
      \begin{pmatrix}
        0 & 1 \\
        -1 & 0
      \end{pmatrix}
      \in \SU(2) \]
      then $sts^{-1} = t^{-1}$.
    \item $[A]_{\SU(2)} \mapsto \half \Trace A$ is
      well-defined and injective since conjugate
      matrices have the same trace and if $\half
      \Trace A = \half \Trace B$ for $A, B \in
      \SU(2)$, since $\det A = \det B = 1$, then
      $A$ and $B$ have the same characteristic
      polynomial and hence the same eigenvalues,
      so by (ii) they are conjugate. Moreover
      \[ \half \Trace
      \begin{pmatrix}
        e^{-i\theta} & 0 \\
        0 & e^{-i\theta}
      \end{pmatrix} = \cos \theta \]
      so the image of our map is $[-1, 1)$.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{corollary*}
  A (continuous) class function $f : \SU(2) \to
  \CC$ is determined by its restriction to $T$ and
  $f|_T$ is even, i.e.
  \[ f \left(
  \begin{pmatrix}
    z & 0 \\
    0 & z^{-1}
  \end{pmatrix}
  \right) = f \left(
  \begin{pmatrix}
    z^{-1} & 0 \\
    0 & z
  \end{pmatrix}
  \right) \qquad \forall z \in S^1 .\]
\end{corollary*}
\vspace{-1em}

We'll write
\[ f(z) = f \left(
\begin{pmatrix}
  z & 0 \\
  0 & z^{-1}
\end{pmatrix}
\right) \]
for $z \in S^1$ i.e. identify $T$ and $S^1$.

\begin{notation*}
  \glssymboldefn{ev}{ev}{ev}
  We'll write
  \begin{align*}
    \ZZ[z, z^{-1})^{\text{ev}}
    &= \{f \in \ZZ[z, z^{-1}) \st f(z) =
    f(z^{-1})\} \\
    &= \left\{ \sum a_n z^n \st a_n \in \ZZ, a_n =
    a_{-n} \forall n \in \ZZ\right\}
  \end{align*}
\end{notation*}

\begin{lemma*}
  If $\chichar$ is a \gls{char_rep} of a
  \gls{top_rep} of $\SU(2)$ then $\chichar|_T \in
  \suev\ZZ[z, z^{-1})$.
\end{lemma*}
\vspace{-1em}

It follows that
$\charring(\SU(2)) \lesssim \suev\ZZ[z, z^{-1})$
and we're going to see that it is an equality.

\begin{proof}
  If $V$ is a \gls{top_rep} of $\SU(2)$ and
  $\chichar$ its character, then
  \[ \chichar|_T = \chichar_{\Res_T^{\SU(2)} V} \]
  since every \gls{char_rep} of $T \cong S^1$ lies
  in $\ZZ[z, z^{-1})$ and $\chichar|_T$ is even,
  we're done.
\end{proof}

Let's write
\[ O_x = \left\{A \in \SU(2) \st \half \Trace A = x\right\} \]
for $x \in [-1, 1)$. These are the conjugacy
classes of $\SU(2)$, $O_1 = \{I\}$, $O_{-1} =
\{-I\}$ and for $-1 < x < 1$. There is some
(unique) $\theta \in (0, \pi)$ such that
$\cos\theta = x$ and
\[ O_x = \left\{
\begin{pmatrix}
  a & b \\
  -\ol{b} & \ol{a}
\end{pmatrix}
~\bigg|~ (\Im a)^2 + |b|^2 = 1 - x^2 = \sin^2
\theta \right\} \]
(since $\Re a = x$). That is $O_x$ is a $2$-sphere
of radius $|\sin \theta|$.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/asdfasdfasdf.png}
\end{center}
Thus if $f$ is a class function on $\SU(2)$, since
$f$ is constant on $O_{\cos\theta}$,
\begin{align*}
  \Gint_{\SU(2)} f(g) \dd g
  &= \frac{1}{2\pi^2} \int_0^\pi \left[
  \Gint_{O_{\cos\theta}} f(e^{i\theta}) \right]
  \dd \theta \\
  &= \frac{1}{2\pi^2} \int_0^\pi 4\pi \sin^2\theta
  f(e^{i\theta}) \dd \theta \\
  &= \frac{1}{\pi} \int_0^{2\pi} f(e^{i\theta})
  \sin^2 \theta \dd \theta
\end{align*}
since $f$ is even.
Note this is normalised correctly since
\[ \frac{1}{\pi} \int_0^{2\pi} \sin^2\theta \dd
\theta = \frac{\pi}{\pi} = 1 .\]
So we can compute $\langle f, g\rangle_{\SU(2)}$
on class functiosn (and so characters) as
\[ \langle f, g\rangle_{\SU(2)} = \frac{1}{\pi}
\int_0^{2\pi} \ol{f}(e^{i\theta}) g(e^{i\theta})
\sin^2\theta \dd \theta \]

\subsection{Representations of SU$(2)$}

Let $V_n$ be the $\CC$-vector space of homogeneous
polynomials in $x$ and $y$ of degree $n$. So
\[ V_n = \bigoplus_{i = 1}^n \CC x^i y^{n - i} \]
has dimension $n + 1$. $\GL_2(\CC)$ acts on $V_n$
via
\begin{align*}
  \rho_n : \GL_2(\CC)
  &\to \GL(V_n) \\
  \rho_n \left(
  \begin{pmatrix}
    a & b \\
    c & d
  \end{pmatrix}
  \right) (f(x, y))
  &= f(ax + cy, bx + dy)
\end{align*}
i.e.
\[ \rho_n \left(
\begin{pmatrix}
  a & b \\
  c & d
\end{pmatrix}
\right) x^i y^i = (ax + cy)^i (bx + dy)^j .\]

\begin{example*}
  $V_0 = \CC$ is the \gls{triv_rep}.

  $V_1 = \CC x \oplus \CC y$ is the natural
  \gls{rep} of $\GL_2(\CC)$ on $\CC^2$ with
  respect to basis $x, y$.

  $V_2 = \CC x^2 \oplus \CC xy \oplus \CC y^2$ and
  with respect to this basis,
  \[ \rho_2 \left(
  \begin{pmatrix}
    a & b \\
    c & d
  \end{pmatrix}
  \right) = \left(
  \begin{pmatrix}
    a^2 & ab & b^2 \\
    2ac & ad + bc & 2bd \\
    c^2 & cd & d^2
  \end{pmatrix} \right) \]

  In general $V_n \simeq S^n V_1$ as
  \glspl{top_rep} of $\GL_2(\CC)$.
\end{example*}
\vspace{-1em}

Since $\SU(2)$ is a subgroup of $\GL_2(\CC)$ we
can view these $V_n$ as \glspl{rep} of $\SU(2)$ by
restriction. In fact we'll see the $V_n$ are
precisely the \gls{irred} \glspl{top_rep} of
$\SU(2)$.

Let's compute $\chichar_{V_n}|_T$ of $(\rho_n,
V_n)$.
\[ \rho_n \left(
\begin{pmatrix}
  z & 0 \\
  0 & z^{-1}
\end{pmatrix}
\right) (z^i y^j) = (zx)^i (z^{-1} y)^j
= z^{i - j} x^i y^j \]
So for each $0 \le j \le n$, $\CC x^j y^{n - j}$ is
a \repdim{1} \gls{top_rep} of $T$ with
\gls{char_rep} $z^{2j - n}$ and
\[ \chichar_{V_n}(z) = z^n + z^{n - 2} + z^{n - 4}
+ \cdots + z^{2 - n} + z^{-n} = \frac{z^{n + 1} -
z^{-(n + 1)}}{z - z^{-1}} \in \suev\ZZ[z, z^{-1}) \]

\begin{theorem*}
  Each $V_n$ is \gls{irred} as a \gls{top_rep} of
  $\SU(2)$.
\end{theorem*}

\begin{proof}
  Let $0 \neq W \le V_n$ be $\SU(2)$-invariant. We
  must show $W = V_n$. $W$ is also $T$-invariant
  as $\Res_T^{\SU(2)} V_n = \bigoplus_{j = 0}^n
  \CC x^j y^{n - j}$ is as direct sum of
  non-isomorphic \repdim{1} \glspl{subrep}. ($*$)
  $W$ has a basis that is a subset of $\{x^j y^{n
  - j} \st 0 \le j \le n\}$ (uniqueness of
  isotypical decomposition). Thus $x^j y^{n - j}
  \in W$ for some $0 \le j \le n$.
  \[ \frac{1}{\sqrt{2}}
  \begin{pmatrix}
    1 & 1 \\
    -1 & 1
  \end{pmatrix}
  x^j y{n - j} = \frac{1}{\sqrt{2}} (x - y)^j (x +
  y)^{n - j} \in W
  \]
  so by ($*$), $x^n \in W$. Repeat the same
  calculation for $j = n$, we get $(x - y)^n \in
  W$. So $x^i y^{n - i} \in W$ for all $0 \le i
  \le n$ and $W = V_n$.
\end{proof}

\textbf{Exercise} (Alternative proof)\textbf{:}
Show
\[ \langle \chichar_{V_n}, \chichar_{V_n}
\rangle_{\SU(2)} = \frac{1}{2\pi} \int_0^{2\pi}
\left( \frac{e^{(n + 1) i\theta} - e^{-(n +
1)i\theta}}{e^{i\theta} - e^{-i\theta}}
\right)^2 \sin^2\theta \dd \theta = 1 \]

\begin{theorem*}
  Every \gls{irred} \gls{top_rep} of $\SU(2)$ is
  isomorphic to $V_n$ for some $n\ge 0$.
\end{theorem*}

\begin{proof}
  Let $V$ be an \gls{irred} \gls{top_rep} of
  $\SU(2)$, $\chichar_V |_T \in \suev\ZZ[z,
  z^{-1})$. Let $\chichar_n = \chichar_{V_n}|_T$
  for $n \ge 0$, so $\chichar_0 = 1$, $\chichar_1
  = z + z^{-1}$, $\chichar_2 = z^2 + 1 + z^{-2}$
  etc. It is easy to see that there exists
  $\lambda_0, \lambda_1, \ldots, \lambda_m \in
  \ZZ$ such that $\chichar_V |_T = \sum_{i = 0}^n
  \lambda_i \chichar_i$. By
  \nameref{orthog_of_chars},
  \[ \lambda_i = \langle \chichar_V,
  \chichar_{V_i} \rangle_{\SU(2)} = \begin{cases}
    1 & \text{if $V \simeq V_i$} \\
    0 & \text{if $V \not\simeq V_i$}
  \end{cases} \]
  Since $\chichar_V \neq 0$ there is some $i$ such
  that $\lambda_i = 1$ and $\chichar_V =
  \chichar_{V_i}$.
\end{proof}

We want to understand $\tensprod$ for
\glspl{top_rep} of $\SU(2)$. We know
\[ \chichar_{V \tensprod W} = \chichar_V
\chichar_W \]
for \glspl{rep} $V$ and $W$ of any group $G$.
Let's compute some examples:
\[ \chichar_{V_1 \tensprod V_2} (z) = (z +
z^{-1})^2 = z^2 + 2 + z^{-2} = \chichar_{V_2} +
\chichar_{V_0} \]
so $\chichar_{V_1} \tensprod \chichar_{V_1} \simeq
V_2 \oplus V_0$.
\[ \chichar_{V_1 \tensprod V_2}(z) = (z +
z^{-1})(z^2 + 1 + z^{-2}) = z^3 + z + z^{-1} + z +
z^{-1} + z^{-3} = \chichar_{V_3}(z) +
\chichar_{V_1}(z) \]
so $V_1 \tensprod V_2 \simeq V_3 \oplus V_1$.
