Some groups have simpler representation theories than others. For example, the representation theory of $S_n$ is simpler and more beautiful than the representation theory of $A_n$ – the characters of $S_n$ all have integer values, and are easily parametrized, while we saw that $A_4$ and $A_5$ have characters that do not extend to $S_4$ and $S_5$, and have more complicated descriptions using roots of unity (cube roots for $A_4$ and fifth roots for $A_5$). It is also true that the conjugacy classes of $S_n$ are easier to describe than the conjugacy classes of $A_n$. The best way to understand the conjugacy classes or the irreducible representations of $A_n$ is probably to first understand the conjugacy classes or representations of $S_n$, then understand how some conjugacy classes or representations split into two when restricted to the smaller group.

Similarly, the representation theory of $\operatorname{GL} (n, \mathbb{F}_q)$ is much simpler and more beautiful than the representation theory of $\operatorname{SL} (n, \mathbb{F}_q)$. This theory was completed in 1956 by J. A. Green in an important paper. We will use Mackey theory to construct one of the four families of irreducible representations when $n = 2$. Everything we do generalized beautifully to general $n$ in Green's theory.

The group $\operatorname{GL} (2, \mathbb{F}_q)$ has irreducible representations of degrees $1, q, q - 1$ and $q + 1$. There are $q - 1$ linear characters, which are the characters $\chi (\det (g))$ where $\chi$ is any linear character of $\mathbb{F}_q^{\times}$; this obviously is a character since $\det : \operatorname{GL} (2, \mathbb{F}_q) \longrightarrow \mathbb{F}_q^{\times}$ is a homomorphism.

There are also $q - 1$ irreducible characters of degree $q$. One irreducible
character, called the *Steinberg character* can be obtained by taking
the permutation character on $\mathbb{P}^1 (\mathbb{F}_q)$ and
subtracting $1$. You already know this character, but we mention that the
Steinberg character can be generalized to $\operatorname{GL} (n, \mathbb{F}_q)$. It
is an irreducible character of degree $q^{\frac{1}{2} n (n - 1)}$ with
remarkable properties. The remaining irreducible characters can be obtained by
multiplying by the $q - 1$ linear characters.

The two families of degrees $q - 1$ and $q + 1$ are "twins.'' They are
parametrized by characters of two important abelian subgroups called
*tori*.

The first torus $T_{\operatorname{split}}$ is the diagonal subgroup of order $(q -
1)^2$ consisting of all matrices of the form
\[ T_{\operatorname{split}} = \left\{ \left(\begin{array}{cc}
t_1 & \\
& t_2
\end{array}\right) {\large{|}} t_1, t_2 \in \mathbb{F}_q^{\times} \right\}
\]
The characters of $T_{\operatorname{split}}$ are of the form $t =
\left(\begin{array}{cc}
t_1 & \\
& t_2
\end{array}\right) \longmapsto \psi (t) = \psi_1 (t_1) \psi_2 (t_2)$, and each
such character parametrizes a representation that we will call $\pi (\psi_1,
\psi_2)$. In order to construct it, we first extend the charater to a larger
subgroup
\[ B = \left\{ \left(\begin{array}{cc}
t_1 & x\\
0 & t_2
\end{array}\right) {\large{|}} t_1, t_2 \in \mathbb{F}_q^{\times}, x \in
\mathbb{F}_q \right\} \]
called the *Borel subgroup*, by the same formula:
\[ \psi \left(\begin{array}{cc}
t_1 & x\\
0 & t_2
\end{array}\right) = \psi_1 (t_1) \psi_2 (t_2), \]
and we define $\pi (\psi_1, \psi_2)$ to be the representation of $\operatorname{GL}
(2, \mathbb{F}_q)$ induced from this. We will deduce its properties from
Mackey's theorem as a typical application.

**Exercise 4.4.1:***
Prove that the degree of $\pi (\psi_1, \psi_2)$ is $q + 1$.
*

The first step is to compute the double cosets $B \backslash G / B$ where $G = \operatorname{GL} (2, \mathbb{F}_q) .$

**Lemma 4.4.1:***
( Bruhat decomposition.) There are two double cosets in $B \backslash G
/ B$. They are
\[ \left\{ \left(\begin{array}{cc}
a & b\\
c & d
\end{array}\right) {\large{|}} c = 0 \right\} = B \left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right) B \]
and
\[ \left\{ \left(\begin{array}{cc}
a & b\\
c & d
\end{array}\right) {\large{|}} c \neq 0 \right\} = B
\left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right) B. \]
*

**Proof. **(Click to Expand/Collapse)

It is obvious that
\[ \left\{ \left(\begin{array}{cc}
a & b\\
c & d
\end{array}\right) {\large{|}} c = 0 \right\} = B = B
\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right) B. \]
For the other double coset, if $c \neq 0$ we can write
\[ \left(\begin{array}{cc}
a & b\\
c & d
\end{array}\right) = \left(\begin{array}{cc}
1 & a / c\\
0 & 1
\end{array}\right) \left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right) \left(\begin{array}{cc}
c & d\\
& - \frac{c^{- 1}}{a d - b c}
\end{array}\right) . \]
This identity shows that these entries form a single double coset $B
\left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right) B$.

**Lemma 4.4.2:***
Let $\chi (\psi_1, \psi_2)$ and $\chi (\phi_1, \phi_2)$ be the
characters of $\pi (\psi_1, \psi_2)$ and $\pi (\phi_1, \phi_2)$,
respectively, where $\psi_i$ and $\phi_i$ are characters of
$\mathbb{F}_q^{\times}$. Then
\[ \left\langle \chi (\psi_1, \psi_2), \chi (\phi_1, \phi_2) \right\rangle =
\left\{ \begin{array}{cc}
1 & \text{if $\phi_1 = \psi_1$ and $\phi_2 = \psi_2,$}\\
0 & \text{otherwise;}
\end{array} \right\} + \left\{ \begin{array}{cc}
1 & \text{if $\phi_1 = \psi_2$ and $\phi_2 = \psi_1,$}\\
0 & \text{otherwise;}
\end{array} \right\} . \]
*

**Proof. **(Click to Expand/Collapse)

By the Corollary to the Geometric Form of Mackey's Theorem, the inner
product is equal to the dimension of the vector space of function $\Delta$
on $G$ such that

It is obvious from the previous Lemma that such a function is determined by
the values

since any $g \in G$ is of the form
\[ \left(\begin{array}{cc}
t_1 & x\\
0 & t_2
\end{array}\right) w \left(\begin{array}{cc}
u_1 & y\\
0 & u_2
\end{array}\right), \]
where $w = I$ or $\left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right)$. Thus the dimension of the space of such $\Delta$ is at
most two. But it could be less, since there is a consistency relation that
needs to be satisfied in order for $\Delta (w)$ to be nonzero for $w = I$ or
$\left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right)$.

Indeed, we have
\[ \left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right) = \left(\begin{array}{cc}
t_1 & \\
& t_2
\end{array}\right) \left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right) \left(\begin{array}{cc}
t_1^{- 1} & \\
& t^{- 1}_2
\end{array}\right), \]
so (4.4.1) implies that for all $t_1$ and $t_2$ in
$\mathbb{F}_q^{\times}$ we have
\[ \Delta \left( \left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right) \right) = \phi_1 (t_1) \phi_2 (t_2) \psi_1 (t_1)^{- 1}
\psi_2 (t_2)^{- 1} \Delta \left( \left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right) \right) . \]
If $\phi_1 \neq \psi_1$, we can choose $t_1$ so that $\phi_1 (t_1) \psi_1
(t_1)^{- 1} \neq 1$, and then choose $t_2 = 1$; we see that $\Delta (I)$
must equal zero. Similarly if $\phi_2 \neq \psi_2$ we prove that $\Delta (I)
= 0$. This proves (4.4.3). One must also check that if $\phi_1 =
\psi_1$ and $\phi_2 = \psi_2$ then (4.4.1) with $g = I$
gives a consistent $\Delta$ supported in this double coset, but we leave
this to the reader.

Indeed, we have
\[ \left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right) = \left(\begin{array}{cc}
t_1 & \\
& t_2
\end{array}\right) \left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right) \left(\begin{array}{cc}
t_2^{- 1} & \\
& t^{- 1}_1
\end{array}\right), \]
whence
\[ \Delta \left( \left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right) \right) = \phi_1 (t_1) \phi_2 (t_2) \psi_1 (t_2)^{- 1}
\psi_2 (t_1)^{- 1} \Delta \left( \left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right) \right), \]
from which one deduces (4.4.4), analogously to the proof of
(4.4.3). One must also check that if $\phi_1 = \psi_2$ and
$\phi_2 = \psi_1$ then (4.4.1) with $g =
\left(\begin{array}{cc}
& 1\\
1 &
\end{array}\right)$ gives a consistent $\Delta$ supported in this double
coset, and again we leave this to the reader.

\[ \label{deltabconsistent} \Delta \left( \left(\begin{array}{cc} t_1 & x\\ 0 & t_2 \end{array}\right) g \left(\begin{array}{cc} u_1 & y\\ 0 & u_2 \end{array}\right) \right) = \phi_1 (t_1) \phi_2 (t_2) \Delta (g) \psi_1 (u_1) \psi_2 (u_2) . \] | (4.4.1) |

\[ \label{twodeltav} \Delta \left( \left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right) \right), \hspace{2em} \Delta \left( \left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right) \right) \] | (4.4.2) |

First, we show that

\[ \label{deltaicond} \Delta \left( \left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right) \right) = 0\qquad\text{unless}\qquad \phi_1 = \psi_1\quad\text{and}\quad \phi_2 = \psi_2 . \] | (4.4.3) |

Next we prove that

\[ \label{deltawcond}\Delta \left( \left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right) \right) = 0\qquad\text{unless}\qquad\phi_1 = \psi_2\quad\text{and}\quad \phi_2 = \psi_1. \] | (4.4.4) |

To summarize, $\Delta$ is determined by its values (4.4.2). There are however necessary and sufficient conditions as to whether these values are nonzero. Taking these into account one obtains the Lemma.

**Theorem 4.4.1:***
The representation $\pi (\psi_1, \psi_2)$ is irreducible unless $\psi_1 =
\psi_2$. The representations $\pi (\psi_1, \psi_2)$ and $\pi (\phi_1,
\phi_2)$, with $\psi_1 \neq \psi_2$ and $\phi_1 \neq \phi_2$ are isomorphic
if and only if either $\psi_1 = \phi_1$ and $\psi_2 = \phi_2$ or $\psi_1 =
\phi_2$ and $\psi_2 = \phi_1$.
*

**Proof. **(Click to Expand/Collapse)

If $\psi_1 \neq \psi_2$ then by Lemma 4.4.2 we have
\[ \left\langle \chi (\psi_1, \psi_2), \chi (\psi_1, \psi_2) \right\rangle =
1 \]
and so this character is irreducible. The Lemma also makes plain the
criterion for $\pi (\psi_1, \psi_2)$ and $\pi (\phi_1, \phi_2)$, when
irreducible, to be isomorphic.

This gives us $\frac{1}{2} (q - 1) (q - 2)$ irreducible representations of degree $q + 1$, parametrized by characters of $T_{\operatorname{split}}$, which is a group with $(q - 1)^2$ elements. The torus $T_{\operatorname{split}}$ has a "twin'' that we will call $T_{\operatorname{nonsplit}}$, which is an abelian group with $q^2 - 1$ elements. Of its $q^2 - 1$ linear characters, $q^2 - q$ characters correspond to irreduccible representations, which have degree $q - 1$. However, as with the $q + 1$ dimensional representations, two characters correspond to the same irreducible representation, so there are only $\frac{1}{2} (q^2 - q)$ irreducible of degree $q - 1$.

The torus $T_{\operatorname{nonsplit}}$ can be constructed as follows. There is a field $\mathbb{F}_{q^2}$ with $q^2$ elements, and it is a vector space of dimension 2 over $\mathbb{F}_q$. The group $\mathbb{F}_{q^2}^{\times}$ acts on $\mathbb{F}_{q^2}$ by multiplication, and choosing a basis $x_1, x_2$ of $\mathbb{F}_{q^2}$ as a vector space over $\mathbb{F}_q$, the elements of $\mathbb{F}_{q^2}^{\times}$ can be realized as matrices. Thus if $\alpha \in \mathbb{F}_{q^2}^{\times}$ we can write \begin{eqnarray*} \alpha x_1 = a x_1 + b x_2, & & \\ \alpha x_2 = c x_1 + d x_2, & & \end{eqnarray*} and we can take $T_{\operatorname{nonsplit}}$ to be the set of all such matrices $\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)$. Of course, it depends on the choice of basis but only up to conjugacy.

The nature of the parametrization of irreducible representations of $\operatorname{GL}
(2, \mathbb{F}_q)$ by linear characters of $T$, where $T$ is either
$T_{\operatorname{split}}$ or $T_{\operatorname{nonsplit}}$ can be described as follows. If
$g \in G = \operatorname{GL} (2, \mathbb{F}_q)$ has *distinct eigenvalues*,
and if $\psi$ is a linear character of either $T_{\operatorname{split}}$ or
$T_{\operatorname{nonsplit}}$ parametrizing the irreducible representation of $G$
with character $\chi$, then $\chi (g) = 0$ unless $g$ has a conjugate in $T$.
If it does, it has two conjugates $\gamma_1$ and $\gamma_2$ in $T$. For
example, if $T = T_{\operatorname{split}}$ and one conjugate is
\[ \gamma_1 = \left(\begin{array}{cc}
t_1 & \\
& t_2
\end{array}\right), \]
the other is
\[ \gamma_2 = \left(\begin{array}{cc}
t_2 & \\
& t_1
\end{array}\right) . \]
Again, if $T = T_{\operatorname{nonsplit}}$ and one conjugate is $\gamma_1$, the
other is $\gamma_2 = \gamma_1^q$. Then
\[ \chi (g) = \left\{ \begin{array}{ll}
\psi (\gamma_1) + \psi (\gamma_2) & \text{if $T = T_{\operatorname{split}}$},\\
- \psi (\gamma_1) - \psi (\gamma_2) & \text{if $T =
T_{\operatorname{nonsplit}}$} .
\end{array} \right. \]
This recipe only tells us the character values for most elements of $G$.

Although the analogy and relationship between the two classes of representations of $G$ is very deep, there is an important difference. The $q-1$-dimensional representations cannot directly be constructed by induction. One approach to constructing them is to take one of the generalized characters $\chi$ constructed by Green's Theorem 4.1.2 and write it as $\chi_1 - \chi_2$ where $\chi_1$ is one of the irreducibles of degree $q + 1$, and $\chi_2$ is one of the ones of degree $q - 1$. If $\chi_1$ is already constructed, this yields $\chi_2$. Another approach, due to Deligne and Lusztig, constructs the representations as acting on certain cohomology groups.