%! TEX root = RT.tex
% vim: tw=50
% 25/10/2023 11AM

\begin{corollary*}
  The number of \gls{irred} $\CC$-\glspl{rep} of
  $G$ is the number of conjugacy classes in $G$.
\end{corollary*}

\begin{notation*}
  For $g \in G$ we'll write $[g]_G = \{xgx^{-1}
  \st x \in G\}$ for the conjugacy class
  containing $g$.
\end{notation*}

\begin{corollary*}
  For $g \in G$, $\chichar(g) \in \RR$ for every
  \gls{irred} \gls{char_rep} $\chichar$ if and
  only if $[g]_G = [g^{-1}]_G$.
\end{corollary*}

\begin{proof}
  Since $\chichar(g^{-1}) = \ol{\chichar(g)}$,
  $\chichar(g) \in \RR \iff \chichar(g) =
  \chichar(g^{-1})$. So we can rephrase the
  statement as
  \[ \text{$\chichar(g) = \chichar(g^{-1})$ for
  every \gls{char_rep} $\chichar$} \iff [g]_G =
  [g^{-1}]_G \]
  Since the (\gls{irred}) \gls{char_rep} span
  $\mathcal{C}_G$,
  \[ \text{$\chichar(g) = \chichar(g^{-1})$ for
  every (\gls{irred}) \gls{char_rep}} \iff
  \text{$f(g) = f(g^{-1})$ for every $f \in
  \mathcal{C}_G$} \]
  Since $\indicator{[g]_G}$ is a \gls{cls_fn}
  $f(g) = f(g^{-1})$ for every $f \in
  \mathcal{C}_G$ $\iff$ $[g]_G = [g^{-1}]_G$.
\end{proof}

\subsection{Character Tables}

\begin{flashcard}[char-table-defn]
\begin{definition*}[Character table]
  \glsnoundefn{char_table}{character
  table}{character tables}
  \cloze{
  The \emph{character table} of a finite group is
  defined as follows. We list the conjugacy
  classes of $G$, $[g_1]_G, \ldots, [g_r]_G$ (by
  convention $g_1 = e$ always). We list the
  \gls{irred} \gls{char_rep} of $G$ (over $\CC$)
  $\chichar_1, \ldots, \chichar_r$ (by convention
  $\chichar_1 = \indicator{G}$ the \gls{char_rep}
  of \gls{triv_rep}). Then we write the matrix
  \begin{center}
    \begin{tabular}{c|c|c|c|c|c|c|c}
       & $e$ & $g_2$ & $g_3$ & $\cdots$ & $g_j$ &
      $\cdots$ & $g_r$ \\
      \hline
      $\chichar_1$ & $1$ & $1$ & $1$ & $\cdots$ &
      $1$ & $\cdots$ & $1$ \\
      $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ &
      $\vdots$ & $\vdots$ & $\vdots$ \\
      $\chichar_i$ & $\cdots$ & $\cdots$ &
      $\cdots$ & $\cdots$ & $\chichar_i(g_j)$ & $\cdots$ &
      $\cdots$ \\
      $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ &
      $\vdots$ & $\vdots$ & $\vdots$ \\
      $\chichar_i$ & $\cdots$ & $\cdots$ &
      $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ &
      $\cdots$ \\
    \end{tabular}
  \end{center}
  We sometimes write $|[g_i]_G|$ above $g_i$ and
  sometimes $|\mathcal{C}_G(g_i)|$ (recall
  $|[g_i]_G| |\mathcal{C}_G(g_i)| = |G|$ by
  Orbit-Stabiliser Theorem).}
\end{definition*}
\end{flashcard}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
  \item $C_3 = \langle x \rangle$. Let $\omega =
    e^{2\pi i/3}$. So $\omega^2 = \omega^{-1} =
    \ol{\omega}$.
    \begin{center}
      \begin{tabular}{c|c|c|c}
         & $e$ & $x$ & $x^2$ \\
        \hline
        $\chichar_1$ & $1$ & $1$ & $1$ \\
        $\chichar_2$ & $1$ & $\omega$ &
        $\ol{\omega}$ \\
        $\chichar_3$ & $1$ & $\ol{\omega}$ &
        $\omega$
      \end{tabular}
    \end{center}
    Note the rows are indeed orthogonal with
    respect to $\Gip \langle \bullet, \bullet
    \rangle_G$ and the columns are orthogonal with
    respect to standard Hermitian inner product.
  \item $S_3$. Conjugacy classes are $\{e\}$,
    $\{(12), (13), (23)\}$, $\{(123), (132)\}$. So
    there must be $3$ \gls{irred} \glspl{rep} /
    \gls{char_rep}. $\chichar_1 = \indicator{G}$ is
    the \gls{char_rep} of the \gls{triv_rep}.
    $\chichar_2 = \eps : S_3 \to \{\pm 1\} \subset
    \CC^\times$ (where $\eps$ is the sign of a
    permutation) is a homomorphism and so a
    \repdim{1} \gls{rep} and so a \gls{char_rep}.
    To compute $\chichar_3$ we can use
    orthogonality of
    \glsref[char_rep]{characters}.
    Let $\chichar_3(e) = a$, $\chichar_3((12)) =
    b$, $\chichar_3((123)) = c$. Since every $g$
    in $S_3$ is conjugate to $g^{-1}$ in $S_3$,
    $a, b, c \in \RR$. Then
    \[ 0 = \Gip\langle \indicator{}, \chichar_3
    \rangle_G = \frac{1}{6} (a + 3b + 2c) \]
    \[ 0 = \Gip\langle \eps, \indicator{} \rangle_G
    = \frac{1}{6} (a - 3b + 2c) \]
    which can be solved to give $b = 0$, $a =
    -2c$. Then
    \[ 1 = \Gip\langle \chichar_3, \chichar_3
    \rangle_G = \frac{1}{6}(a^2 + 3b^2 + 2c^2) \]
    So $c^2 = 1$. But $a$ is the dimension of the
    \gls{rep} with \gls{char_rep} $\chichar_3$ so
    $a \ge 1$. So $a = 2$, $c = -1$.
    \begin{center}
      \begin{tabular}{c|ccc}
        $|C_{S_3}(g_i)|$ & $6$ & $2$ & $3$ \\
        $g_i$ & $e$ & $(12)$ & $(123)$ \\
        \hline
        $\indicator{}$ & $1$ & $1$ & $1$ \\
        $\eps$ & $1$ & $-1$ & $1$ \\
        $\chichar_3$ & $2$ & $0$ & $-1$
      \end{tabular}
    \end{center}
    In fact we already know about $\chichar_3$ as
    the \gls{char_rep} of the \gls{rep} of $S_3$
    ($\cong D_6$) on $\RR^2$ induced from the
    symmetries of a triangle.
    Once again the columns are orthogonal and
    their lengths are $1^2 + 1^2 + 2^2 = 6 =
    |C_{S_3}(e)|$, $1^2
    + (-1)^2 + 0^2 = 2 = |C_{S_3}((12))|$, $1^2 + 1^2
    + (-1)^2 = 3 = |C_{S_3}((123))|$.
\end{enumerate}

\begin{proposition*}[Column orthogonality]
  If $G$ is a finite group and $\chichar_1$,
  \ldots, $\chichar_r$ is a complete list of its
  \gls{irred} \glsref[char_rep]{characters} then
  for $g, h \in G$,
  \[ \sum_{i = 1}^r
  \ol{\chichar_i(g)}\chichar_i(h) = \begin{cases}
    0 & \text{if $[g]_G \neq [h]_G$} \\
    |C_G(g)| & \text{if $[g]_G = [h]_G$}
  \end{cases} \]
  In particular
  \[ \sum_{i = 1}^ (\dim V_i)^2 = \sum_{i = 1}^r
  \chichar_i(e)^2 = |G| \]
  where $V_i$ is a \gls{rep} such that
  $\chichar_{V_i} = \chichar_i$.
\end{proposition*}

\begin{proof}
  Let $X$ be the \gls{char_table} viewed as a
  matrix $X_{ij} = \chichar_i(g_j)$ and $D$ be the
  diagonal (real) matrix with $D_{ii} =
  |C_G(g_i)|$. Orthogonality of characters gives
  \begin{align*}
    \delta_{ij}
    &= \langle \chichar_i, \chichar_j
    \rangle_G \\
    &= \sum_k \frac{1}{|C_G(g_k)|}
    \ol{\chichar_i(g_k)} \chichar_j(g_k) \\
    &= \sum_k \frac{1}{D_{kk}} \ol{\chichar_{ik}}
    \chichar_{jk} \\
    &= (\ol{X} D^{-1} X^\top)_{ij}
  \end{align*}
  So $\ol{X} D^{-1} X^\top I$ since $X$ is square,
  $X$ is invertible and $\ol{D^{-1} X^\top} =
  X^{-1}$ so $D = \ol{X^\top} X$ since $D$ is
  real, i.e. $\sum_k \ol{\chichar_k(g_i)}
  \chichar_k(g_j) = D_{ij} = \delta_{ij}
  |C_G(g_i)|$.
\end{proof}

\subsection{Permutation Representations}

If recall that if a group $G$ acts on a finite set
$X$, $\CC X = \{f : X \to \CC\}$ is a \gls{rep}
via
\[ (g \cdot f)(x) = f(g^{-1} x) \]
or equivalently $g \cdot \delta_x = \delta_g x
~\forall g \in G, x \in X$.

\begin{lemma*}
  If $\chichar$ is the \gls{char_rep} of $\CC X$
  then $\chichar(g) = |\{x \in X \st gx = x\}|$.
\end{lemma*}

\begin{proof}
  If $X = \{x_1, \ldots, x_d\}$ then with respect
  to the basis $\delta_{x_1}, \ldots,
  \delta_{x_d}$ the matrix of $g$ has $i$-th
  column with a $1$ in entry $j$ and $0$
  elsewhere if $g \cdot x_i = x_j$. So $i$-th
  column contributes $1$ to the trace if $i = j$
  and $0$ otherwise.
\end{proof}
