# Introduction

The group $GL_2(\mathbb{F}_q)$ consists of invertible $2 \times 2$ matrices with entries in the finite field $\mathbb{F}_q$, where $q=p^n$ for some prime $p$ (throughout we exclude the case when $p=2$ because it can be quite difficult). As a $\mathbb{F}_p$-vector space, $\mathbb{F}_q$ has dimension $n$. The Galois group $G(\mathbb{F}_q/\mathbb{F}_p)$ is cyclic and is generated by the Frobenius map.

The field $\mathbb{F}_q$ itself is already pretty complicated, let alone a matrix group over it. In this post we try to follow Fulton-Harris' idea on Representation Theory: A First Course to classify all irreducible representations of $G=GL_2(\mathbb{F}_q)$. To be specific, we are talking about group homomorphisms $\rho:G \to GL(V)$ where $V$ is a $\mathbb{C}$-vector space.

First of all we determine the cardinality of $G=GL_2(\mathbb{F}_q)$. Along the way, we will introduce some important subgroups.

$G \supset B = \left\{\begin{pmatrix}a & b \\ 0 & d\end{pmatrix}:a,d \ne 0\right\} \supset N= \left\{\begin{pmatrix}1 & b \\ 0 & 1\end{pmatrix}\right\}.$

The cardinality of $G$ is determined by the class formula, consider the canonical action on $\mathbb{P}^1(\mathbb{F}_q)$.

First of all, notice that $|\mathbb{P}^1(\mathbb{F}_q)|=q+1$. There are $q^2$ elements in $\mathbb{F}_q \times \mathbb{F}_q$, excluding the zero, we have $q^2-1$ remaining. Since $(r:s)=a(r:s)$ for all $a \in \mathbb{F}_q^\ast$, we divide $q^2-1$ by $|\mathbb{F}_q^\ast|=q-1$ to obtain the cardinality of the projective space.

The action of $G$ on $\mathbb{P}^1(\mathbb{F}_q)$ is defined canonically as follows:

$\begin{pmatrix} a & b \\ c & d \end{pmatrix}(r:s)=(ar+bs:cr+ds).$

In particular, $B$ is the isotropy group of the set $\{(1:0)\}$, because in this case, $(ar+bs:cr+ds)=(a:0)=(1:0)$. There are $(q-1)(q-1)q$ elements of $B$.

Since $G$ clearly acts on $\mathbb{P}^1(\mathbb{F}_q)$ transitively, by the class equation, we have

$|G|=|B||\mathbb{P}^1(\mathbb{F}_q)|=(q-1)^2q(q+1).$

In general, the cardinality of $GL_n(\mathbb{F}_q)$ is $\prod_{k=0}^{n-1}(q^n-q^k)$. One can check this document.

We next consider the diagonal subgroup

$D=\left\{\begin{pmatrix}a & 0 \\ 0 & d\end{pmatrix}:a,d \ne 0 \right\} = \mathbb{F}_q^\ast \times \mathbb{F}_q^\ast.$

Let $\mathbb{F}'=\mathbb{F}_{q^2}$ be the extension of $\mathbb{F}_q$ of degree $2$. We can certainly identify $GL_2(\mathbb{F}_q)$ as the group of $\mathbb{F}_q$-linear invertible automorphisms of $\mathbb{F}'$. Each $h \in (\mathbb{F}')^\ast$ induces a $\mathbb{F}_q$-linear automorphism by multiplication, hence $h$ can be embedded into $GL_2(\mathbb{F}_q)$. The question is, how. Let $K = (\mathbb{F}')^\ast$ be this subgroup. We write down the matrix representation explicitly.

Let  be a generator of the cyclic group $\mathbb{F}_q^\ast$, then $X^2-\varepsilon$ is irreducible in $\mathbb{F}_q[X]$. We therefore have $\mathbb{F}_{q^2} \cong \mathbb{F}_q[X]/(X^2-\varepsilon)$. We see $\{1,X\}$, or more precisely, $\{1,\sqrt\varepsilon\}$, is a basis of $\mathbb{F}_{q^2}$ as a vector space over $\mathbb{F}_q$. We can then identify $(\mathbb{F}')^\ast$ as a subgroup $K$ of $G$ where

$K=\left\{\begin{pmatrix}x & \varepsilon{y} \\ y & x\end{pmatrix}:\text{x and y not identically zero}\right\} \cong (\mathbb{F}')^\ast.$

The isomorphism is given by

$\begin{pmatrix}x & \varepsilon{y} \\ y & x\end{pmatrix} \leftrightarrow \zeta = x + y\sqrt\varepsilon.$

To make $K$ a subgroup of $G$, each entry must be in $\mathbb{F}_q$. That's why we write $\varepsilon$ instead of $\sqrt\varepsilon$ in the definition of $K$.

# Conjugacy Classes

At the end of this section one can see a table of the result.

To matrices, conjugacy gives rise to eigenvalues and Jordan canonical form. So we immediately come up with the three following forms:

$a_x = \begin{pmatrix}x & 0 \\ 0 & x \end{pmatrix}, \quad b_x = \begin{pmatrix}x & 1 \\ 0 & x \end{pmatrix}, \quad c_{x,y} = \begin{pmatrix}x & 0 \\ 0 & y \end{pmatrix} (x \ne y).$

For each $x$ and $y$, $a_x$, $b_x$ and $c_{x,y}$ represents three different conjugacy classes respectively, and they do not intersect. We will study these three families of conjugacy classes and see how far we can go. Spoiler: we will miss $\frac{q(q-1)}{2}$ conjugacy classes, which will be found in the subgroup $K$.

Conjugacy classes represented by $a_x$ is the easiest one. Since scalar matrices commutes with any matrix, for any invertible matrix $A$, we have

$A^{-1}\begin{pmatrix}x & 0 \\ 0 & x \end{pmatrix}A=\begin{pmatrix}x & 0 \\ 0 & x \end{pmatrix}A^{-1}A=\begin{pmatrix}x & 0 \\ 0 & x \end{pmatrix}.$

Therefore there is only one element in the conjugacy class represented by $a_x$. Ranging through all $x \ne 0$, we obtain $q-1$ such classes.

For Jordan canonical form like $b_x$, the story is different. Ranging through all $x \ne 0$, we again obtain $q-1$ such classes. Nevertheless, to determine the cardinality of each class, it is unrealistic to work only in the scope of matrices.

Let $\mathcal{C}=(b_x)$ be a conjugacy class. Let $G$ act on $\mathcal{C}$ by conjugation. The action is transitive: for $A,B \in \mathcal{C}$, there are invertible matrices $U$ and $V$ such that $U^{-1}AU=b_x=V^{-1}BV$, and therefore $A=(VU^{-1})^{-1}B(VU^{-1})$.

To determine the cardinality of $\mathcal{C}$, we use the class formula again. Suppose $A=(a_{ij})$ fixes $b_x$, i.e. $A^{-1}b_xA=b_x$, or $b_xA=Ab_x$, then

$Ab_x=\begin{pmatrix} a_{11}x & a_{11}+a_{12}x \\ a_{21}x & a_{21}+a_{22}x \end{pmatrix} = \begin{pmatrix} a_{21}+a_{11}x & a_{22}+a_{12}x \\ a_{21}x & a_{22}x \end{pmatrix} = b_x A.$

The equation above implies that $c=0$ and $a=d$. Therefore the isotropy group of $\{b_x\}$ is

$J=\left\{\begin{pmatrix}a & b \\ 0 & a \end{pmatrix}:a \ne 0\right\}.$

It follows that $|\mathcal{C}|=|G|/|J|=(q^2-q)(q^2-1)/(q^2-q)=q^2-1$.

Let $\mathcal{D}=(c_{x,y})$ be a conjugacy class. Ranging through all $x,y \ne 0$ with $y \ne x$, then divide it by $2$, we obtain $\frac{(q-1)(q-2)}{2}$ conjugacy classes in the same form of $\mathcal{D}$. We divide it by $2$ because $c_{x,y}$ is conjugate to $c_{y,x}$ as they share the same eigenvalues.

We determine the cardinality of $\mathcal{D}$ in the same way as $\mathcal{C}$. The isotropy group of $\{c_{x,y}\}$ is $D$ in the introduction. Therefore $|\mathcal{D}|=|G|/|D|=q^2+q$.

Now let's count how many conjugacy classes we have obtained:

$1(q-1)+(q^2-1)(q-1)+(q^2+q)\frac{(q-1)(q-2)}{2}=\frac{1}{2}(q^4-3q^2+2q).$

We still need to find $\frac{q^2(q-1)^2}{2}$ elements. Look at subgroups we derived in the introduction. Subgroups like $B$, $N$ and $D$ all go down into Jordan canonical form immediately, but $K$ is not the case. Consider

$d_{x,y}=\begin{pmatrix} x & \varepsilon{y} \\ y & x \end{pmatrix}, \quad y \ne 0.$

Then the eigenvalues of $d_{x,y}$ are $x\pm \sqrt\varepsilon y$, none of which lies in $\mathbb{F}_q$. Therefore it has nothing to do with Jordan canonical form. We will explore the remainder of conjugacy classes of $G$ here.

Ranging through all $x$ and $y \ne 0$, then divide it by $2$, we obtain $\frac{q(q-1)}{2}$ conjugacy classes. We divide it by $2$ because $d_{x,y}$ and $d_{x,-y}$ are conjugate by any

$\begin{pmatrix} a & -\varepsilon{c} \\ c & -a \end{pmatrix}.$

Now let $\mathcal{E}=(d_{x,y})$ be a conjugacy class. Notice the isotropy group of $d_{x,y}$ is $K$, so we can obtain the cardinality $|\mathcal{E}|=|G|/|K|=(q-1)^2q(q+1)/(q^2-1)=q^2-q$. Now our search is complete.

Representative Number of Elements in Class Number of Classes
$a_x=\begin{pmatrix}x & 0 \\ 0 & x\end{pmatrix}$ $1$ $q-1$
$b_x = \begin{pmatrix}x & 1 \\ 0 & x \end{pmatrix}$ $q^2-1$ $q-1$
$c_{x,y} = \begin{pmatrix}x & 0 \\ 0 & y \end{pmatrix} (x \ne y)$ $q^2+q$ $\frac{(q-1)(q-2)}{2}$
$d_{x,y}=\begin{pmatrix} x & \varepsilon{y} \\ y & x \end{pmatrix}, y \ne 0$ $q^2-q$ $\frac{q(q-1)}{2}$

These matrices would not frequently appear in the remainder of the post because it will mess up the format.

# Irreducible Representations

There are $q-1+q-1+\frac{(q-1)(q-2)+q(q-1)}{2}=q^2-1$ conjugacy classes, so we need to find $q^2-1$ irreducible representations. Of course we cannot list down all of them. We will instead classify them with certain reasonings. A character table can be found in the next section.

Some computations are omitted because if not, this section would be unreadable. However, the author of this post has checked most of them on paper. The reader should find it easy to compute by themselves. For completed computation, one refers to this note. Note however the classification is a little bit different from here. Reading this first may help you to get the hang of it.

## Attempt 1 - Subrepresentations of the Permutation

Recall how we find irreducible representations of $\mathfrak{S}_3$: consider permutation of a basis in a vector space of dimension $3$. We do a similar thing here. Let $G$ acts on $\mathbb{P}^1(\mathbb{F}_q)$ by permutation. This induces a $q+1$ dimensional representation $W$ because $\mathbb{P}^1(\mathbb{F}_q)$ has $q+1$ elements. It contains the trivial representation $U$. Let $V$ be the complement of $U$, i.e. $W=U \oplus V$, then $V$ has dimension $q$. Now we determine the character of $V$. Since $\chi_V=\chi_W-\chi_U=\chi_W-1$, we only need to calculate $\chi_W$, i.e. to see fixed points of the permutation on each conjugacy class.

1. $\chi_W(a_x)=q+1$. It fixes every point.

2. $\chi_W(b_x)=1$. It only fixes one point: $(1:0)$.

3. $\chi_W(c_{x,y})=2$. It fixes two points: $(1:0)$ and $(0:1)$.

4. $\chi_W(d_{x,y})=0$. If $d_{x,y}$ fixes $(a:b)$, then $a^2=\varepsilon b^2$, and this cannot happen.

Therefore we have

$a_x$ $b_x$ $c_{x,y}$ $d_{x,y}$
$\chi_V$ $q$ $0$ $1$ $-1$

We see, $(\chi_V,\chi_V)=1$. Therefore $V$ is irreducible and we cannot decompose $W$ further. We have to find different approaches.

## Attempt 2 - Pontryagin Dual

The Pontryagin dual of a group $H$ is defined to be

$\hat{H}=\operatorname{Hom}(H,S^1).$

If $H$ admits a topology, we may want to eliminate non-continuous homomorphisms but it's not our concern here because we only care about finite groups now, which admits discrete topology. Notice that if $H$ is finite and cyclic, then $\hat{H} \cong H$. We will use this fact right now.

Since $G$ can be pretty big, it is not realistic to study all eigenvalues of representations. Instead, we consider the Pontryagin dual of $\mathbb{F}_q^\ast$, which is again a finite cyclic group. For each of the $q-1$ elements in $\hat{H}$, $\alpha:\mathbb{F}_q^\ast \to S^1$, we have a one-dimensional representation $U_\alpha$ of $G$ defined by

$\chi_{U_\alpha}(g)=\alpha(\det(g)), \quad g \in G.$

Note the trivial representation is one of the $U_\alpha$, once one realises that $\alpha$ defined by $\alpha(x)=1$ for all $x \in \mathbb{F}_q^\ast$ is also a homomorphism into $S^1$.

Tensoring $U_\alpha$ with $V$, we obtain another family of irreducible representations $\{V_\alpha = V \otimes U_\alpha\}$. Note $V$ is one of the $V_\alpha$. The character table of them are easily computed.

$a_x$ $b_x$ $c_{x,y}$ $d_{x,y}$
$U_\alpha$ $\alpha(x)^2$ $\alpha(x)^2$ $\alpha(x)\alpha(y)$ $\alpha(x^2-\varepsilon y^2)$
$V_\alpha$ $q\alpha(x)^2$ $0$ $\alpha(x)\alpha(y)$ $-\alpha(x^2-\varepsilon y^2)$

## Attempt 3 - Pontryagin Dual x Pontryagin Dual

We have successfully determined $2(q-1)$ irreducible representations, still $(q-1)^2$ of them to be found. We now make use of those subgroups we have determined. For each $\alpha,\beta \in \widehat{\mathbb{F}_q^\ast}$, we have a new character of a representation:

\begin{aligned} \gamma_{\alpha,\beta}:B \to B/N \cong D \cong \mathbb{F}^\ast \times \mathbb{F}^\ast &\xrightarrow{(\alpha,\beta)}\mathbb{C}^\ast \times \mathbb{C}^\ast \xrightarrow{\times}\mathbb{C}^\ast \\ \begin{pmatrix} a & b \\ 0 & d \end{pmatrix} &\mapsto \alpha(a)\beta(b) \end{aligned}

Let $W'_{\alpha,\beta}$ be the representation of $B$ with character $\gamma_{\alpha,\beta}$, and let $W_{\alpha,\beta}=\operatorname{Ind}_B^G W_{\alpha,\beta}'$. We can quite easily (no, with a lot of dirty computation) write down the character table of $W_{\alpha,\beta}$.

$a_x$ $b_x$ $c_x$ $d_x$
$W_{\alpha,\beta}$ $(a+1)\alpha(x)\beta(x)$ $\alpha(x)\beta(x)$ $\alpha(x)\beta(y)+\alpha(y)\beta(x)$ $0$

If $\alpha=\beta$, then $W_{\alpha,\beta}=U_\alpha \oplus V_\beta$ so not irreducible. However, if $\alpha \ne \beta$, then $W_{\alpha,\beta} \cong W_{\beta,\alpha}$ is irreducible, if one computes $(\chi,\chi)$. The dimension of $W_{\alpha,\beta}$ is $[G:B]=q+1$, and there are $\frac{1}{2}(q-1)(q-2)$ of them. Still there are $\frac{1}{2}q(q-1)$ irreducible representations to be found.

## Attempt 4 - Pontryagin Dual of K

We haven't used this subgroup yet so we first explore it in the same vein as attempt 3. We consider the the dual of $\mathbb{F}' \cong K$. Each $\varphi:(\mathbb{F}')^\ast \to \mathbb{C}^\ast$ also determines a representation on $K$ with character $\varphi$. One immediately think about $\operatorname{Ind}_K^G(\varphi)$, for which we simply write $\operatorname{Ind}(\varphi)$. It is easy to compute the character table so far.

$a_x$ $b_x$ $c_{x,y}$ $d_{x,y}$
$\operatorname{Ind}(\varphi)$ $q(q-1)\varphi(x)$ $0$ $0$ $\varphi(\zeta)+\varphi(\zeta)^q$

We put $\zeta=x+y\sqrt\varepsilon \in K = (\mathbb{F}')^\ast$. Since $\operatorname{Ind}(\varphi) \cong \operatorname{Ind}(\varphi^q)$, we obtain $\frac{1}{2}q(q-1)$ representations out of this, with the restriction that $\varphi^q \ne \varphi$. Nevertheless, we have $(\chi,\chi)=q$ if $\varphi^q=\varphi$, $(\chi,\chi)=q-1$ if $\varphi^q \ne \varphi$. We still need to work on it from different direction.

## Attempt 5 - Making Use of All of Them

Let's try to tensor what we have found. It is easy to see that $V_\alpha \otimes U_\gamma = V_{\alpha\gamma}$ and $W_{\alpha,\beta} \otimes U_{\gamma}=W_{\alpha\gamma,\beta\gamma}$. So we cannot find anything new here. But tensoring $W_{\alpha,\beta}$ and $V_\alpha$ gives us something quite different. We see for $\alpha \ne 1$,

$a_x$ $b_x$ $c_{x,y}$ $d_{x,y}$
$V \otimes W_{\alpha,1}$ $q(q+1)\alpha(x)$ $0$ $\alpha(x)+\alpha(y)$ $0$

Let $\varphi \in \widehat{(\mathbb{F}')^\ast}$ be a homomorphism such that $\varphi|_{\mathbb{F}_{q}^\ast}=\alpha$. Computing inner products gives us

\begin{aligned} (\chi_{V \otimes W_{\alpha,1}},\chi_{W_{\alpha,1}})&=2, \\ (\chi_{V \otimes W_{\alpha,1}},\chi_{V \otimes W_{\alpha,1}})&=q+3, \\ (\chi_{\operatorname{Ind}(\varphi)},\chi_{W_{\alpha,1}}) &= 1, \\ (\chi_{\operatorname{Ind}(\varphi)},\chi_{V \otimes W_{\alpha,1}}) &= q. \end{aligned}

We see $W_{\alpha,1}$ is contained in the representation determined by $\operatorname{Ind}(\varphi)$. We also see $W_{\alpha,1}$ is contained in $V \otimes W_{\alpha,1}$. Besides, $\operatorname{Ind}(\varphi)$ and $V \otimes W_{\alpha,1}$ has a lot of subrepresentations in common. The first guess would be that $\operatorname{Ind}(\varphi)$ is a subrepresentation of $V \otimes W_{\alpha,1}$. So, maybe we can find something we have been missing here. For this reason, consider the virtual character

$\chi_\varphi=\chi_{V \otimes W_{\alpha,1}}-\chi_{W_{\alpha,1}}-\chi_{\operatorname{Ind}(\varphi)}.$

We can compute that $(\chi_\varphi,\chi_\varphi)=1$. To see this is actually a real character, we compute the character table.

$a_x$ $b_x$ $c_{x,y}$ $d_{x,y}$
$\chi_\varphi$ $(q-1)\alpha(x)$ $-\alpha(x)$ $0$ $\varphi(\zeta)+\varphi(\zeta)^q$

It follows that $\chi_\varphi(1)=q-1>0$. Hence $\chi_\varphi$ is irreducible. Since each $\chi_\varphi$ is determined by $\varphi$ with $\varphi^q \ne \varphi$, and there are $\frac{1}{2}q(q-1)$ of such $\varphi$, we have actually determined all of the irreducible characters. They are denoted by $X_\varphi$.

# Character Table

$GL_2(\mathbb{F}_q)$ $a_x$ $b_x$ $c_{x,y}$ $d_{x,y}\leftrightarrow\zeta$ $\dim$
$U_\alpha$ $\alpha(x^2)$ $\alpha(x^2)$ $\alpha(xy)$ $\alpha(\zeta^q)$ $1$
$V_\alpha$ $q\alpha(x^2)$ $0$ $\alpha(xy)$ $-\alpha(\zeta^q)$ $q$
$W_{\alpha,\beta}$ $(q+1)\alpha(x)\beta(x)$ $\alpha(x)\beta(x)$ $\alpha(x)\beta(y)+\alpha(y)\beta(x)$ $0$ $q+1$
$X_\varphi$ $(q-1)\varphi(x)$ $-\varphi(x)$ $0$ $-(\varphi(\zeta)+\varphi(\zeta^q))$ $q-1$

A few remarks are in order. We can call these four classes of irreducible representations in the following way (excerpted from this document):

1. $U_\alpha$: $1$-dimensional representations. There are $q-1$ of them.

2. $V_\alpha=V \otimes U_\alpha$: $q$-dimensional representations. Here $V$ is also called Steinberg representation. There are $q-1$ of them.

3. $W_{\alpha,\beta}$: ($q+1$)-dimensional irreducible principle series. There are $\frac{1}{2}(q-1)(q-2)$ of them. Some authors may also treat $q$-dimensional representations as principle series.

4. $X_\varphi$: irreducible cuspidal representations or complementary series representations. There are $\frac{1}{2}q(q-1)$ of them. A representation is cuspidal if the Jacquet module is trivial.

Irreducible Representations of GL_2(F_q)

https://desvl.xyz/2022/08/12/rep-gl2-fq/

Desvl

2022-08-12

2022-08-12