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\subsection{The character table of $B$}

As a warm-up we compute the \gls{char_table} of
\[ B = \left\{
\begin{pmatrix}
  a & b \\
  c & d
\end{pmatrix}
~\bigg|~ a, d \in \FF_q^\times, b \in \FF_q
\right\} \le G ,\]
a group of order $(q - 1)^2 q$.

Recall
\[ N = \left\{
\begin{pmatrix}
  1 & b \\
  0 & b
\end{pmatrix}
~\bigg|~ b \in \FF_q
\right\} \normalsub B \]
and
\begin{align*}
  B / N
  &\simeq T \simeq \FF_q^\times \FF_q^\times \\
  \begin{pmatrix}
    a & b \\
    0 & d
  \end{pmatrix}
  N
  &\to
  \begin{pmatrix}
    a & 0 \\
    0 & d
  \end{pmatrix}
\end{align*}
The conjugacy classes in $B$ are
\begin{center}
  \begin{tabular}{c|c|c|c}
    Rep & $C_B$ & size of class & \#classes \\
    \hline
    $\begin{pmatrix} \lambda & 0 \\ 0 & \lambda
    \end{pmatrix}$ & $B$ & $1$ & $q - 1$ \\
    $\begin{pmatrix} \lambda & 1 \\ 0 & \lambda
    \end{pmatrix}$ & $ZN$ & $q - 1$ & $q - 1$ \\
    $\begin{pmatrix} \lambda & 0 \\ 0 & \mu
    \end{pmatrix}$ & $T$ & $q$ & $(q - 1)(q - 2)$
  \end{tabular}
\end{center}
Moreover if $\Theta_q : \{\text{\glspl{rep}
$\FF_q^\times \to \CC^\times$}\}$ then $\Theta_q$
is a cyclic group of order $q - 1$ under pointwise
operations since $\FF_q^\times \simeq C_{q - 1}$
and for each pair $\theta, \phi \in \Theta_q$ we
can define a \repdim{1} \gls{rep} of $B$
(factoring through $B / N$) given by
\[ \chichar_{\theta, \phi} \left(
\begin{pmatrix}
  a & b \\
  0 & d
\end{pmatrix}
\right) = \theta(a) \phi(d) \]
giving $(q - 1)^2$ \repdim{1} \glspl{rep}. We
will build the remaining \gls{irred}
\glspl{char_rep} of $B$ by \gls{induction} from $ZN$.
\begin{align*}
  ZN
  &\simeq \FF_q^\times \times (\FF_q, +) \\
  \begin{pmatrix}
    a & b \\
    0 & a
  \end{pmatrix}
  &\mapsto (a, a^{-1} b)
\end{align*}
so given a \repdim{1} \gls{rep} $\gamma : (\FF_q,
+) \to \CC^\times$ and $\theta \in \Theta_q$, we
can define a \repdim{1} \gls{rep} of $ZN$
\begin{align*}
  \rho_{\theta, \gamma} : ZN
  &\to \CC^\times \\
  \begin{pmatrix}
    a & b \\
    0 & a
  \end{pmatrix}
  &\mapsto \theta(a) \gamma(a^{-1} b)
\end{align*}
Now $ZN \normalsub B$, so by
\nameref{mack_irred_crit},
\[ \text{$\Ind_{ZN}^B \rho_{\theta, \gamma}$ is
\gls{irred}} \iff \leftgrep{}^g \rho_{\theta,
\gamma} \neq \rho_{\theta, \gamma} \]
for all $gZN \in B / ZN \setminus ZN / ZN$. Since
\[ \left\{ t_\lambda =
\begin{pmatrix}
  1 & 0 \\
  0 & \lambda
\end{pmatrix}
~\bigg|~ \lambda \in \FF_q^\times
\right\} \]
is a coset representative for $B / ZN$ and
\begin{align*}
  \leftgrep{}^{t_\lambda} \rho_{\theta, \gamma}
  \left(
  \begin{pmatrix}
    a & b \\
    0 & a
  \end{pmatrix}
  \right)
  &= \rho_{\theta, \gamma} \left(
  \begin{pmatrix}
    1 & 0 \\
    0 & \lambda^{-1}
  \end{pmatrix}
  \begin{pmatrix}
    a & b \\
    0 & a
  \end{pmatrix}
  \begin{pmatrix}
    1 & 0 \\
    0 & \lambda
  \end{pmatrix}
  \right) \\
  &= \rho_{\theta, \gamma} \left(
  \begin{pmatrix}
    a & \lambda b \\
    0 & a
  \end{pmatrix}
  \right) \\
  &= \theta(a) \gamma(a^{-1} \lambda b)
\end{align*}
We see that
\begin{align*}
  \leftgrep{}^{t_\lambda} \rho_{\theta, \gamma} =
  \rho_{\theta, \gamma}
  &\iff \gamma(\lambda b) = \gamma(b)
  &&\forall b \in (\FF_q, +) \\
  &\iff \gamma((\lambda - 1) b) = 1
  &&\forall b \in \FF_q \\
  &\iff \gamma = \indicator{\FF_q} \text{ or }
  \lambda = 1
\end{align*}
So $\Ind_{ZN}^B \rho_{\theta, \gamma}$ is
\gls{irred} if and only if $\gamma \neq
\indicator{\FF_q}$.

Now since
\[ (\Ind_{ZN}^B) = \sum_{[g]_{ZN} \subseteq [b]_B}
\frac{|C_B(b)|}{|C_{ZN}(g)|} \chichar(g) .\]
We can compute
\begin{align*}
  (\Ind_{ZN}^B \rho_{\theta, \gamma}) \left(
  \begin{pmatrix}
    \lambda & 0 \\
    0 & \lambda
  \end{pmatrix}
  \right)
  &= \frac{|B|}{|ZN|} \rho_{\theta, \gamma} \left(
  \begin{pmatrix}
    \lambda & 0 \\
    0 & \lambda
  \end{pmatrix}
  \right) = (q - 1)\theta(\lambda) \\
  (\Ind_{ZN}^B \rho_{\theta, \gamma}) \left(
  \begin{pmatrix}
    \lambda & 1 \\
    0 & \lambda
  \end{pmatrix}
  \right)
  &= \sum_{b \in \FF_q^\times} \frac{|BN|}{|BN|}
  \rho_{\theta, \gamma} \left(
  \begin{pmatrix}
    \lambda & b \\
    0 & \lambda
  \end{pmatrix}
  \right) \\
  &= \theta(\lambda) \left( \left( \sum_{b \in
  \FF_q} \gamma(\lambda^{-1} b) \right) - 1
  \right) \\
  &= \theta(\lambda) (q \langle \gamma,
  \indicator{\FF_q} \rangle_{(\FF_q, +)} - 1) \\
  &= \begin{cases}
    -\theta(\lambda) & \gamma \neq \FF_q \\
    (q - 1)\theta(\lambda) & \gamma =
    \indicator{\FF_q}
  \end{cases} \\
  \Ind_{ZN}^B
  &= \left(
  \begin{pmatrix}
    \lambda & 0 \\
    0 & \mu
  \end{pmatrix}
  \right)
  &= 0 \qquad (\lambda \neq \mu)
\end{align*}
Let $\mu_\theta \defeq \Ind_{ZN}^B \rho_{\theta,
\gamma}$ for $\gamma \neq \indicator{\FF_q}$,
noting this does not depend on the choice of
$\gamma$. Then each $\mu_\theta$ is \gls{irred} by
earlier calculation and we have $q - 1$
\gls{irred} \glspl{rep} of $B$ all of degree $q -
1$.

Thus the \gls{char_table} of $B$ is
\begin{center}
  \begin{tabular}{c|c|c|c|c}
     & $\begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}$
    & $\begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}$
    & $\begin{pmatrix} \lambda & 0 \\ 0 & \mu
    \end{pmatrix}$ \\
    \hline
    $\chichar_{\theta, \phi}$ & $\theta(\lambda)
    \phi(\lambda)$ & $\theta(\lambda)
    \phi(\lambda)$ & $\theta(\lambda)\phi(\mu)$ &
    $\theta, \phi \in \Theta_q$ \\
    \hline
    $\mu_\theta$ & $(q - 1)\theta(\lambda)$ &
    $-\theta(\lambda)$ & $0$ & $\theta \in
    \Theta_q$
  \end{tabular}
\end{center}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item The $0$ in the bottom right corner
      appears in $(q - 1)$ rows and $(q - 1)(q -
      2)$ columns. They are all forced to be $0$
      by a Lemma from \cref{sub7_4}, since order
      of the conjugacy class is $q$ and the
      dimension of an \gls{irred} \gls{rep} are
      coprime and these elements don't act by
      scalars since the \glspl{rep} are
      \gls{ff_rep} and the elements are not in the
      centre.
    \item
      \[ B = Z \times \left\{
      \begin{pmatrix}
        a & b \\
        0 & 1
      \end{pmatrix}
      ~\bigg|~ a \in \FF_q^\times, b \in \FF_q
      \right\} \]
      and the second factor is a \gls{frob_gp}
      group, so \es[10]{3} tells us that
      \glspl{char_rep} of the second factor arise
      essentially as we have constructed them.
  \end{enumerate}
\end{remark*}

\subsection{The character table of $G$}

As $\det : G \to \FF_q^\times$ is a surjective
group homomorphism, for each $\theta \in
\Theta_q$, $\theta \circ \det : G \to \CC^\times$
is a distinct \repdim{1} \gls{rep} of $G$. We get
$q - 1$ in all. Next we'll do induction from $B$.
Define
\[ s =
\begin{pmatrix}
  0 & 1 \\
  1 & 0
\end{pmatrix}
\in G ,\]
and note that
\[
\begin{pmatrix}
  a & b \\
  0 & d
\end{pmatrix}
\begin{pmatrix}
  0 & 1 \\
  1 & 0
\end{pmatrix}
\begin{pmatrix}
  1 & \beta \\
  0 & 1
\end{pmatrix}
=
\begin{pmatrix}
  b & a + b\beta \\
  d & \beta d
\end{pmatrix}
\]
These elements are all disinct. Hence
\[ |BsN| = q|B| = |G \setminus B| \]
($G / B$ has order $q + 1$). Thus $BsN = BsB$ and
$G = B \ci BsB$ (Bruhat decomposition) and $B
\leftquot G / B = \{B, BsB\}$. By the proof of
\nameref{mack_irred_crit},
\[ \langle \Ind_B^G \chichar, \Ind_B^G \chichar
\rangle_G = \langle \chichar, \chichar \rangle_B +
\langle \Res_{B \cap \leftgrep{}^s B}^B \chichar,
\Res_{B \cap \leftgrep{}^s B}^{\leftgrep{}^s B}
\leftgrep{}^s \chichar \rangle_{B \cap \leftgrep{}^s B}
\]
Now
\[ s
\begin{pmatrix}
  a & b \\
  0 & d
\end{pmatrix}
s^{-1} =
\begin{pmatrix}
  d & 0 \\
  b & a
\end{pmatrix} \]
So $B \cap \leftgrep^{s} B = T$ and
\[ \Gip\langle \Ind_B^G \chichar, \Ind_B^G \chichar
\rangle_G = \langle \chichar, \chichar \rangle_B +
\langle \chichar|_T, \leftgrep{}^s \chichar|_T
\rangle_T \]
where
\[ (\leftgrep{}^s \chichar) \left(
\begin{pmatrix}
  a & 0 \\
  0 & d
\end{pmatrix}
\right) = \chichar \left(
\begin{pmatrix}
  d & 0 \\
  0 & a
\end{pmatrix}
\right) \]
Thus $W_{\theta, \gamma} \defeq \Ind_B^G
\chichar_{\theta, \phi}$ is \gls{irred} if $\theta
\neq \phi \in \Theta_q$. These are called
\emph{principal series representations}. We can
also compute $W_{\theta, \theta}$ has $2$ distinct
\gls{irred} summands:
\[ \langle \Ind_B^G \mu_\theta, \Ind_B^G
\mu_\theta \rangle = 1 + \frac{1}{|T|}
\sum_{\lambda \in \FF_q^\times} |(q - 1)
\theta(\lambda)|^2 = 1 + \frac{(q - 1)^3}{(q -
1)^2} = q .\]
For any \gls{char_rep} $\chichar$ of $B$,
\begin{align*}
  (\Ind_B^G \chichar)(g)
  &= \sum_{[b]_B \subseteq [g]_G}
  \frac{|C_G(g)|}{|C_B(b)|} \chichar(b) \\
  (\Ind_B^G \chichar) \left(
  \begin{pmatrix}
    \lambda & 0 \\
    0 & \lambda
  \end{pmatrix}
  \right)
  &= (q + 1) \chichar \left(
  \begin{pmatrix}
    \lambda & 0 \\
    0 & \lambda
  \end{pmatrix}
  \right) \\
  (\Ind_B^G \chichar) \left(
  \begin{pmatrix}
    \lambda & 1 \\
    0 & \lambda
  \end{pmatrix}
  \right)
  &= \chichar \left(
  \begin{pmatrix}
    \lambda & 1 \\
    0 & \lambda
  \end{pmatrix}
  \right) \\
  (\Ind_B^G \chichar) \left(
  \begin{pmatrix}
    \lambda & 0 \\
    0 & \mu
  \end{pmatrix}
  \right)
  &= \chichar \left(
  \begin{pmatrix}
    \lambda & 0 \\
    0 & \mu
  \end{pmatrix}
  \right) + \chichar \left(
  \begin{pmatrix}
    \mu & 0 \\
    0 & \lambda
  \end{pmatrix}
  \right) \\
  (\Ind_B^G \chichar) \left(
  \begin{pmatrix}
    \lambda & \eps \mu \\
    \mu & \lambda
  \end{pmatrix}
  \right)
  &= 0
\end{align*}
Notice $W_{\theta, \phi} \simeq W_{\phi, \theta}$
so so get ${q - 1 \choose 2}$ principal series
representations. Also, $W_{\theta, \theta} =
\chichar_\theta \tensrepprod W_{\indicator{},
\indicator{}}$ and
\[ W_{\indicator{}, \indicator{}} = \Ind_B^G
\indicator{} = \CC (G / B) \]
is a permutation \gls{rep}. Thus $W_{\indicator{},
\indicator{}} = \CC \oplus V_1$ with
$V_{\indicator{}}$ an (explicit) \gls{irred}
\gls{rep} of degree $q$ (the \emph{Steinberg
representation}) and $W_{\theta, \theta} =
\chichar_\theta \oplus V_\theta$ where $V_\theta =
\theta \tensrepprod V_{\indicator{}}$ (a twisted
Steinbery representation). We've explicitly
constructed $(q - 1) + {q - 1 \choose 2} + (q -
1)$ \gls{irred} \glspl{rep}. We have ${q \choose
2}$ \gls{irred} \glspl{rep} to go. It will turn
out that they are indexed by \gls{irred}
\glspl{rep} of $K$ such that $\varphi \neq
\varphi^q$ up to $\varphi \leftrightarrow
\varphi^q$. We won't explicitly construct these
\glspl{rep}, just their
\glsref[char_rep]{characters}.
