%! TEX root = RT.tex
% vim: tw=50
% 24/11/2023 11AM

\begin{proposition*}[Cletsch-Gordon rule]
  For $n, m \in \NN$,
  \[ V_n \tprod V_m \simeq V_{n + m} \oplus V_{n +
  m - 2} \oplus \cdots \oplus V_{|n - m|} \]
\end{proposition*}

\begin{proof}
  Without loss of generality $n \ge m$. Then
  \begin{align*}
    (\chichar_n \chichar_m)(z)
    &= \left( \frac{z^{n + 1} - z^{-(n + 1)}}{z -
    z^{-1}} \right) (z^m + z^{m - 1} + \cdots +
    z^{-m}) \\
    &= \sum_{j = 0}^m \left( \frac{z^{n + m + 1 -
    2j} - z^{-n + m + 1 - 2j}}{z - z^{-1}}
    \right) \\
    &= \sum_{j = 0}^m \chichar_{n + m - 2j}(z)
  \end{align*}
\end{proof}

\subsection{Representations of $\SO(3)$}

\begin{proposition*}
  The action of $\SU(2)$ on the 3D $\RR$-normed
  vector space
  \[ \left\{
  \begin{pmatrix}
    a & b \\
    -\ol{b} & \ol{a}
  \end{pmatrix}
  ~\bigg|~ a + \ol{a} = 0
  \right\} \subseteq M_2(\CC) \]
  with norm $\|A\|^2 = \det A$ by conjugation
  induces an isomorphism of \glspl{top_group}
  \[ \frac{\SU(2)}{\{\pm I\}} \to \SO(3) .\]
\end{proposition*}

\begin{proof}
  See \es[4]{4} (for more hints see lecturers
  notes from 2012).
\end{proof}

\begin{corollary*}
  Every \gls{irred} \gls{top_rep} of $\SO(3)$ is
  of the form $V_{2n}$ for some $n \ge 0$.
\end{corollary*}

\begin{proof}
  It follows from the previous proposition that
  the \gls{irred} \gls{rep} of $\SO(3)$ correspond
  to \gls{irred} \glspl{rep} of $\SU(2)$ whose
  kernel contains $\pm I$. But it is easy to see
  that $-I$ acts on $V_n$ by $(-1)^n$.
\end{proof}

\newpage
\section{Character table of $\GL_2(\FF_q)$}

\subsection{$\FF_q$}

Let $p > 2$ be a prime, $q = p^a$ a power of $p$
for some $a > 0$ and let $\FF_q$ be the field with
$q$ elements. We know that $\FF_q^\times \simeq
C_{q - 1}$ and $\FF_q^\times \to \FF_q^\times$, $x
\mapsto x^2$ is a homomorphism with kernel $\{\pm
1\}$. Thus half the elements are squares and half
are not. Moreover, $x \mapsto x^{\frac{q - 1}{2}}$
sends squares to $1$ and non-squares to $-1$. Let
$\eps \in \FF_q^\times$ be a fixed non-square. So
$\eps^{\frac{q - 1}{2}} = -1$ and let
\[ \FF_{q^2} = \{a + b\sqrt{\eps} \mid a, b \in
\FF_q\} \]
the field extension of $\FF_q$ (with respect to
the obvious operations) of order $q^2$.

Every element of $\FF_q$ has a square root in
$\FF_q^2$, since if $\lambda \in \FF_q$ is a
non-square then $\frac{\lambda}{\eps}$ is a
square, $\mu^2$ say, and $(\sqrt{\eps} \mu)^2 =
\eps \mu^2 = \lambda$. Thus every quadratic
polynomial with coefficients in $\FF_q$ factorises
over $\FF_q^2$. Notice $(a + b\sqrt{\eps})^q = a^q
+ b^q \eps^{\frac{q - 1}{2}} = a - b\sqrt{\eps}$
since $p \mid {q \choose i}$ for $0 < i < q$.

Thus the roots of an irreducible quadratic over
$\FF_q$ are of the form $\lambda, \lambda^q$
($\lambda \mapsto \lambda^q$ is like complex
conjugation).

\subsection{$\GL_2(\FF_q)$ and its conjugacy
classes}

We want to compute the \gls{char_table} of the
group
\[ \GL_2(\FF_q) = \left\{
\begin{pmatrix}
  a & b \\
  c & d
\end{pmatrix}
~\bigg|~ a, b, c, d \in \FF_q, ad - bc \neq 0
\right\} .\]
The order of $\GL_2(\FF_q)$ is equal to the number
of bases for $\FF_q^2$ over $\FF_q$. This is $(q^2
- 1)(q^2 - q) = q(q - 1)^2(q + 1)$. First we
compute the conjugacy classes of $\GL_2(\FF_q)
\eqdef G$. We know from linear algebra (rational
canonical form) that for $A \in G$, $[A]_G$ is
determined by $m_A(x)$, the minimal polynomial and
$\deg m_A(x) \le 2$ (Cayley-Hamilton). Moreover
$m_A(0) \neq 0$.

There are $4$ cases:
\begin{enumerate}[\bfseries Case 1:]
  \item $m_A(x) = (x - \lambda)$ for some $\lambda
    \in \FF_q^\times$. Then $A = \lambda I$ so
    $C_G(A) = G$ and $|[A]_G| = |\{\lambda I\}| =
    1$. There are $q - 1$ such classes -- one for
    each $\lambda$.
  \item $m_A(x) = (x - \lambda)^2$ for some
    $\lambda \in \FF_q^\times$. Then
    \[ [A]_G = \left[
    \begin{pmatrix}
      \lambda & 1 \\
      0 & \lambda
    \end{pmatrix}
    \right]_G \]
    Now
    \[ C_G \left(
    \begin{pmatrix}
      \lambda & 1 \\
      0 & \lambda
    \end{pmatrix}
    \right) = \left\{
    \begin{pmatrix}
      a & b \\
      0 & a
    \end{pmatrix}
    ~\bigg|~ a, b \in \FF_q, a \neq 0
    \right\} \]
    (compute!). So
    \[ |[A]_G| = \frac{q(q - 1)^2 (q + 1)}{(q -
    1)q} = (q - 1)(q + 1) .\]
    There are $q - 1$ such classes -- one for each
    $\lambda$.
  \item $m_A(x) = (x - \lambda)(x - \mu)$ for
    $\lambda, \mu \in \FF_q^\times$ distinct. So
    \[ [A]_G = \left[
    \begin{pmatrix}
      \lambda & 0 \\
      0 & \mu
    \end{pmatrix}
    \right] =
    \left[
    \begin{pmatrix}
      \mu & 0 \\
      0 & \lambda
    \end{pmatrix}
    \right]_G \]
    Moreover
    \[ C_G \left(
    \begin{pmatrix}
      \lambda & 0 \\
      0 & \mu
    \end{pmatrix}
    \right)
    = \left\{
    \begin{pmatrix}
      a & 0 \\
      0 & d
    \end{pmatrix}
    ~\bigg|~ a, d \in \FF_q^\times
    \right\} \eqdef T \]
    so
    \[ |[A]_G| = \frac{q(q - 1)^2(q + 1)}{(q -
    1)^2} = q(q + 1) \]
    There are ${q - 1 \choose 2}$ such classes --
    one for each pair $\lambda, \mu$.
  \item $m_A(x)$ is irreducible over $\FF_q$ of
    degree $2$. So for some $\alpha \in \FF_{q^2}
    \setminus \FF_q$, $\alpha = \lambda + \mu
    \sqrt{\eps}$ for some $\lambda, \mu \in
    \FF_q$, $\mu \neq 0$,
    \begin{align*}
      m_A(x)
      &= (x - \alpha)(x - \alpha^q) \\
      &= (x^2 - (\alpha + \alpha^q) x + \alpha
      \alpha^q) \\
      &= (x^2 - (\Trace A) x + \det A)
    \end{align*}
    Then
    \[ [A]_G = \left[
    \begin{pmatrix}
      \lambda & \eps \mu \\
      \mu & \lambda
    \end{pmatrix}
    \right]_G = \left[
    \begin{pmatrix}
      \lambda & -\eps\mu \\
      -\mu & \lambda
    \end{pmatrix}
    \right]_G \]
    Since these matrices have trace $2\lambda =
    \alpha + \alpha^q$ and $\det \lambda^2 - \eps
    \mu^2 = \alpha \alpha^q$. Now
    \[ C_g \left(
    \begin{pmatrix}
      \lambda & \eps \mu \\
      \mu & \lambda
    \end{pmatrix}
    \right) = \left\{
    \begin{pmatrix}
      a & \eps b \\
      b & a
    \end{pmatrix}
    ~\bigg|~ a, b \in \FF_q, a^2 - \eps b^2 \neq 0
    \right\} \eqdef K \]
    If $a^2 - \eps^2 b = 0$, then $a^2 = \eps b^2$
    so $\eps$ is a square or $a = b = 0$. So $|K|
    = q^2 - 1$ and
    \[ |[A]_G| = \frac{q(q - 1)^2(q + 1)}{(q -
    1)(q + 1)} = q(q - 1) .\]
    There are $\frac{q(q - 1)}{2} = {q \choose 2}$ such classes --
    one for each pair $\{\alpha, \alpha^q\}
    \subset \FF_{q^2} \setminus \FF_q$.
\end{enumerate}
In summary:
\begin{center}
  \begin{tabular}{c|c|c|c}
    Rep $A$ & $C_G$ & $|[A]_G|$ & \# classes \\
    \hline
    $\lambda I$ & $G$ & $1$ & $q - 1$ \\
    $\begin{pmatrix}
      \lambda & 1 \\ 0 & \lambda
    \end{pmatrix}$
    & $\left\{ \begin{pmatrix}
      a & b \\ 0 & a
    \end{pmatrix}\right\}$
    & $(q - 1)(q + 1)$
    & $q - 1$ \\
    $\begin{pmatrix}
      \lambda & 0 \\ 0 & \mu
    \end{pmatrix}$
    & $T$
    & $q(q + 1)$
    & ${q - 1 \choose 2}$ \\
    $\begin{pmatrix}
      \lambda & \eps\mu \\ \mu & \lambda
    \end{pmatrix}$
    & $K$
    & $q(q - 1)$
    & ${q \choose 2}$
  \end{tabular}
\end{center}
The groups $T$ and $K$ are both called \emph{maximal
tori}, i.e. they are maximal subgroups such that
they are conjugate to a diagonal subgroup in
$\GL_2(\FF)$ for some $\FF / \FF_q$. $T$ is called
\emph{split} and $K$ is called \emph{non-split}.

Some other important subgroups of $G$ are
\begin{itemize}
  \item The subgroup of scalar matrices (the
    centre of $G$):
    \[ Z = \{\lambda I \st \lambda \in
    \FF_q^\times\} .\]
  \item A Sylow-$p$-subgroup of $G$
    \[ N = \left\{
    \begin{pmatrix}
      1 & b \\
      0 & 1
    \end{pmatrix}
    ~\bigg|~ b \in \FF_q
    \right\} \]
    so
    \[ ZN = \left\{
    \begin{pmatrix}
      a & b \\
      0 & a
    \end{pmatrix}
    ~\bigg|~ a, b \in \FF_q, a \neq 0
    \right\} \]
  \item A \emph{Borel subgroup} of $G$
    \[ B = \left\{
    \begin{pmatrix}
      a & b \\
      0 & d
    \end{pmatrix}
    ~\bigg|~ a, d \in \FF_q^\times, b \in \FF_q
    \right\} \]
\end{itemize}
Then $N \normalsub B$ and $B / N \simeq T \simeq
\FF_q^\times \times \FF_q^\times \simeq C_{q - 1}
\times C_{q - 1}$.