Let $G={\rm GL}(n,\mathbb{F}_q)$ and let $N$ be the nilpotent radical of the standard Borel subgroup, i.e., $N$ is the subgroup of $G$ consisting of upper triangular matrices with 1’s on the diagonal. Given a non-trivial character $\psi: \mathbb{F}_q\rightarrow \mathbb{C}^\times$, we define a character $\psi_N$ of $N$ by

$\psi_N( \begin{pmatrix} 1 & x_{12} & x_{13} & \cdots & x_{1n}\\ & 1 & x_{23} & \cdots & x_{2n}\\ & & 1 & \cdots & \vdots \\ & & & \ddots & \vdots\\ & & & & 1\\ \end{pmatrix}) = \psi(x_{12}+x_{23}+\cdots+x_{n-1, n}).$

This defines a one-dimensional representation of $N$. One can show that the induced representation $\mathcal{G}={\rm Ind}_N^G(\psi_N)$ is multiplicity-free, i.e., when we decompose $\mathcal{G}={\rm Ind}_N^G(\psi_N)$ into a sum of irreducible constituents,

$\mathcal{G}={\rm Ind}_N^G(\psi_N)=\oplus a_i\pi_i,$

we have $a_1=1$ for all $i$. Each constituent $\pi_i$ is called a generic irreducible representation. There are many reasons for calling them generic. One is that no matter which non-trivial character $\psi$ we choose, we obtain the same induced representation up to equivalence. One can see this by looking at the induced character values on each conjugacy class. These values will be polynomials in $q$ and, once simplified, do not depend on the character $\psi$. Moreover, this Gelfand-Graev representation $\mathcal{G}$ contains most irreducible representations of the group.

To see this, we can compare the number of irreducible constituents of $\mathcal{G}$ with the number of irreducible representations of $G$. By a standard result in the representation theory of finite groups, the number of irreducible representations of $G$ is equal to its number of conjugacy classes.

J. A. Green wrote a paper about the irreducible characters of ${\rm GL}(n,\mathbb{F}_q)$ in which he gives a generating function for the number $c(n,q)$ of classes of ${\rm GL}(n,\mathbb{F}_q)$:

Let $p_n$ be the number of partitions of n. We define the partition function

$p(x)=\sum_{n=0}^\infty p_n x^n = 1/(1-x)(1-x^2)\cdots.$

The generating function for the number $c(n,q)$ of classes of ${\rm GL}(n,\mathbb{F}_q)$ is

$\sum_{n=0}^\infty c(n,q)x^n = \prod_{d=1}^\infty p(x^d)^{w(d,q)}$

where

$w(d,q)=\frac{1}{d}\sum_{k|d}\mu(k)q^{d/k}$

if the number of irreducible polynomials $f(t)$ of degree $d$ over $\mathbb{F}_q$.

Comparing $c(n,q)$ with the character inner product $(\mathcal{G},\mathcal{G})$, one can see that most irreducible representations of $G$ are generic.

General symplectic groups

We have a similar story for $G={\rm GSp}(2n,\mathbb{F}_q)$. Again, let $N$ be nilpotent radical of the standard Borel subgroup, i.e., $N$ is the subgroup of $G$ consisting of upper triangular matrices with 1’s on the diagonal. Given a non-trivial character $\psi: \mathbb{F}_q\rightarrow \mathbb{C}^\times$, we define a character $\psi_N$ of $N$ by

$\psi_N( \begin{pmatrix} 1 & x_{12} & x_{13} & \cdots & x_{1n}\\ & 1 & x_{23} & \cdots & x_{2n}\\ & & 1 & \cdots & \vdots \\ & & & \ddots & \vdots\\ & & & & 1\\ \end{pmatrix}) = \psi(x_{12}+x_{23}+\cdots+x_{n-1, n})$

and induce to $G$. The induced representation is multiplicity-free and contains most of the irreducible representations of $G$. The case $n=2$ was made explicit in my thesis.

Conjugacy classes of ${\rm GSp}(4,\mathbb{F}_q)$

The conjugacy classes can be found by using a paper by Wall to compute the classes of ${\rm SO}(5, \mathbb{F}_q)\cong {\rm PGSp}(4,\mathbb{F}_q):={\rm GSp}(4,\mathbb{F}_q)/Z$, where $Z$ denotes the center of $G$. The list of conjugacy classes of ${\rm PGSp}(4,\mathbb{F}_q)$ is then used to determine the classes of ${\rm GSp}(4,\mathbb{F}_q)$. Let’s investigate how the class representatives of ${\rm PGSp}(4,\mathbb{F}_q)$ lead to representatives for the classes of ${\rm GSp}(4,\mathbb{F}_q)$.

Consider the natural projection map from ${\rm GSp}(4,\mathbb{F}_q)$ to ${\rm PGSp}(4,\mathbb{F}_q)$ given by

${\rm GSp}(4,\mathbb{F}_q) \longrightarrow {\rm PGSp}(4, \mathbb{F}_q), \text{ } g \mapsto \overline{g}.$

Let $g,h\in {\rm GSp}(4, \mathbb{F}_q)$. If $g=xhx^{-1}$, then, by taking multipliers on each side, it is clear that the multiplier of $g$ is equal to the multiplier of $h$. Moreover, under the projection map, we have

$\overline{g}=\overline{xhx^{-1}}=\overline{x}\cdot\overline{h}\cdot \overline{x^{-1}}.$

So if two elements are conjugate in ${\rm GSp}(4,\mathbb{F}_q)$, they must be conjugate in ${\rm PGSp}(4,\mathbb{F}_q)$. The list of class representatives in ${\rm PGSp}(4,\mathbb{F}_q)$, when pulled back to ${\rm GSp}(4,\mathbb{F}_q)$, hit class representatives of all the conjugacy classes of ${\rm GSp}(4,\mathbb{F}_q)$. Suppose now that two elements $g,h\in {\rm GSp}(4, \mathbb{F}_q)$ are conjugate in ${\rm PGSp}(4,\mathbb{F}_q)$, i.e. $\overline{g}=\overline{x}\cdot\overline{h}\cdot\overline{x^{-1}}$, for some $\overline{x}\in {\rm PGSp}(4, \mathbb{F}_q)$. Then, for some $\gamma^i\in \mathbb{F}_q^\times$,

$g = \begin{pmatrix} \gamma^i & & & \\ & \gamma^i & & \\ & & \gamma^i & \\ & & & \gamma^i \\ \end{pmatrix}x h x^{-1}.$

Taking multipliers on both sides of the equation above, we have $\lambda(g)=\gamma^{2i}\lambda(h)$. So if the multiplier of $g$ is a square, then the multiplier of $h$ is a square and if the multiplier of $g$ is a non-square, then the multiplier of $h$ is a non-square. Write $g$ and $h$ in the following way

$g = \begin{pmatrix} 1 & & & \\ & 1 & & \\ & & \gamma^{i_g} & \\ & & & \gamma^{i_g} \\ \end{pmatrix}\cdot\begin{pmatrix} \gamma^{j_g} & & & \\ & \gamma^{j_g} & & \\ & & \gamma^{j_g} & \\ & & & \gamma^{j_g} \\ \end{pmatrix}\cdot g',$
$h = \begin{pmatrix} 1 & & & \\ & 1 & & \\ & & \gamma^{i_h} & \\ & & & \gamma^{i_h} \\ \end{pmatrix}\cdot\begin{pmatrix} \gamma^{j_h} & & & \\ & \gamma^{j_h} & & \\ & & \gamma^{j_h} & \\ & & & \gamma^{j_h} \\ \end{pmatrix}\cdot h',$

with $g',h'\in {\rm Sp}(4, F_q)$, $i_g,i_h\in\{0,1\}$, and $j_g,j_h\in T_3$. So $\lambda(g)=\gamma^{i_g+2j_g}, \lambda(h)=\gamma^{i_h+2j_h}$. If $g$ and $h$ are conjugate, then $i_g=i_h$. Then $\gamma^{2j_h} = \gamma^{2j_h}$, or $({\gamma^{j_g}})^2 = ({\gamma^{j_h}})^2$. So $\lambda(h)=\pm\lambda(g)$, i.e., the multipliers can only differ by a minus sign.

It is possible that an element $g$ of ${\rm GSp}(4,\mathbb{F}_q)$ is conjugate to $-g$. An example is

$\begin{pmatrix} 1 & & & \\ & 1 & & \\ & & -1 & \\ & & & -1 \\ \end{pmatrix}.\begin{pmatrix} & & 1 & \\ & & & 1 \\ 1 & & & \\ & 1 & & \\ \end{pmatrix}.\begin{pmatrix} 1 & & & \\ & 1 & & \\ & & -1 & \\ & & & -1 \\ \end{pmatrix}$
$=\begin{pmatrix} & & -1 & \\ & & & -1 \\ -1 & & & \\ & -1 & & \\ \end{pmatrix}.$

The centralizers are somewhat affected when we pull back our representatives in ${\rm PGSp}(4,\mathbb{F}_q)$ to ${\rm GSp}(4,\mathbb{F}_q)$. There are two types of pullbacks. The first type consists of elements $g$ such that $g\neq x(-g)x^{-1}$ for any $x\in {\rm GSp}(4, \mathbb{F}_q)$. The second type consists of elements $g$ such that $g=x(-g)x^{-1}$ for some $x\in {\rm GSp}(4, \mathbb{F}_q)$. Let ${\rm Cent_{PGSp}}(\overline{g})$ denote the centralizer of $\overline{g}$ in ${\rm PGSp}(4,\mathbb{F}_q)$ and let ${\rm Cent_{GSp}}(g)$ denote the centralizer of $g$ in ${\rm GSp}(4,\mathbb{F}_q)$.

Type 1

Let $g\in {\rm GSp}(4,\mathbb{F}_q)$ be of the first type, i.e., $g$ is not conjugate to $-g$. Let $\overline{h}\in {\rm Cent_{PGSp}}(\overline{g})$. Then $\overline{g} = \overline{h}\cdot \overline{g}\cdot \overline{h^{-1}}$. When pulled back to ${\rm GSp}(4,\mathbb{F}_q)$, $g = z_0hgh^{-1}$, with $z_0=\pm I$. $z_0\neq -I$ since $g$ is not conjugate to $-g$. So $z_0=I$, $g=hgh^{-1}$, and $h\in {\rm Cent_{GSp}}(g)$. We get a short exact sequence

$1 \longrightarrow Z \longrightarrow {\rm Cent_{GSp}}(g) \longrightarrow {\rm Cent_{PGSp}}(\overline{g}) \longrightarrow 1.$

Therefore $\# {\rm Cent_{GSp}}(g) = (q-1)\cdot \# {\rm Cent_{PGSp}}(\overline{g})$.

Type 2

Let $g\in {\rm GSp}(4,\mathbb{F}_q)$ be of the second type, i.e., $g$ is conjugate to $-g$. Define the set $S_g=\{h\in {\rm GSp}(4, F_q) : hgh^{-1}=-g\}$. Fix $s_0\in S_g$. $S_g$ is not a group, but there is a bijection of sets $S_g \longrightarrow {\rm Cent_{GSp}}(g)$, given by the map $h\mapsto s_0h$. Given $\overline{h}\in {\rm Cent_{PGSp}}(\overline{g})$, either $h\in {\rm Cent_{GSp}}(g)$ or $h\in S_g$. $S_g\sqcup {\rm Cent_{GSp}}(g)$ maps onto ${\rm Cent_{PGSp}}(\overline{g})$ via the projection map. Moreover, ${\rm Cent'_{GSp}}(g) := S_g\sqcup {\rm Cent_{GSp}}(g)$ is a group with respect to matrix multiplication and the projection map is a group homomorphism. ${\rm Cent_{GSp}}(g)$ is a subgroup of ${\rm Cent'_{GSp}}(g)$ of index 2. We get a short exact sequence

$1 \longrightarrow Z \longrightarrow {\rm Cent'_{\rm GSp}}(g) \longrightarrow {\rm Cent_{\rm PGSp}}(\overline{g}) \longrightarrow 1$

Then $\# {\rm Cent'_{GSp}}(g) = (q-1) \cdot \# {\rm Cent_{\rm PGSp}}(\overline{g})$. Also, $2\cdot \# {\rm Cent_{\rm GSp}}(g) = \# {\rm Cent'_{\rm GSp}}(g)$. So ${\rm Cent_{\rm GSp}}(g) = \dfrac{q-1}{2} \cdot \# {\rm Cent_{\rm PGSp}}(\overline{g})$.

Thus, given a class representative $\overline{g}\in {\rm PGSp}(4, \mathbb{F}_q)$, we pull it back to ${\rm GSp}(4,\mathbb{F}_q)$. Then, we determine if $g$ is of Type 1 or Type 2. If it is of Type 1, then there are $q-1$ conjugacy classes $zg$, for $z\in Z$, each of order $\dfrac{\# {\rm GSp}(4,\mathbb{F}_q)}{(q-1)\cdot\# {\rm Cent}_{\rm PGSp}(g)}$. If the pullback is of Type 2, then there are $\dfrac{q-1}{2}$ conjugacy classes $z_i g$, with

$z_i = \begin{pmatrix} \gamma^i & & & \\ & \gamma^i & & \\ & & \gamma^i & \\ & & & \gamma^i \\ \end{pmatrix},\, i\in T_2$

Each of these classes is of order

$\dfrac{\# {\rm GSp}(4, \mathbb{F}_q)}{\frac{(q-1)}{2}\cdot \# {\rm Cent_{\rm PGSp}}}(g).$

Some of the conjugacy classes of ${\rm GSp}(4,\mathbb{F}_q)$ are given in the following table. Let $T_3=\{1,2,...,q-1\}$.

 Notation Class representative Number of classes Order of centralizer $A_1(k)$, $k\in T_3$ $\begin{pmatrix} \gamma^k & & & \\ & \gamma^k & & \\ & & \gamma^k & \\ & & & \gamma^k \\ \end{pmatrix}$ $q-1$ $\# {\rm GSp}(4, \mathbb{F}_q)$ $A_2(k)$, $k\in T_3$ $\begin{pmatrix} \gamma^k & & & \\ & \gamma^k & \gamma^k & \\ & & \gamma^k & \\ & & & \gamma^k \\ \end{pmatrix}$ $q-1$ $q^4(q^2-1)(q-1)$ $A_{31}(k)$, $k\in T_3$ $\begin{pmatrix} \gamma^k & & & \gamma^k \\ & \gamma^k & -\gamma^k & \\ & & \gamma^k & \\ & & & \gamma^k \\ \end{pmatrix}$ $q-1$ $2q^3(q-1)^2$ $A_{32}(k)$, $k\in T_3$ $\begin{pmatrix} \gamma^k & & & \gamma^{k+1} \\ & \gamma^k & -\gamma^k & \\ & & \gamma^k & \\ & & & \gamma^k \\ \end{pmatrix}$ $q-1$ $2q^3(q^2-1)$ $A_5(k)$, $k\in T_3$ $\begin{pmatrix} \gamma^k & \gamma^k & -\gamma^k & \\ & \gamma^k & -\gamma^k & \\ & & \gamma^k & -\gamma^k \\ & & & \gamma^k \\ \end{pmatrix}$ $q-1$ $q^2(q-1)$

We find that there are $(q^2+2q+4)(q-1)$ conjugacy classes.

The conjugacy classes of $N$

Every element $g\in N_{\rm GSp(4)}$ can be written uniquely in the form

$g = \begin{pmatrix} 1 & & & \\ & 1 & x & \\ & & 1 & \\ & & & 1 \\ \end{pmatrix}. \begin{pmatrix} 1 & \lambda & \mu & \kappa \\ & 1 & & \mu \\ & & 1 & -\lambda \\ & & & 1 \\ \end{pmatrix},$

with $x, \kappa, \lambda, \mu \in \mathbb{F}_q$. The order of $N$ is therefore $q^4$. The multiplier of the matrix $g$ given above is $\lambda(g)=1$. The conjugacy classes of $N = N_{\rm GSp(4)}$ are listed in the following table.

 Notation Class representative Number of classes Order of centralizer $NA_1$ $\begin{pmatrix} 1 & & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \\ \end{pmatrix}$ $1$ $q^4$ $NA_2^1(k)$, $k\in T_3$ $\begin{pmatrix} 1 & & & \\ & 1 & \gamma^k & \\ & & 1 & \\ & & & 1 \\ \end{pmatrix}$ $q-1$ $q^3$ $NA_2^2(k)$, $k\in T_3$ $\begin{pmatrix} 1 & & & \gamma^k \\ & 1 & & \\ & & 1 & \\ & & & 1 \\ \end{pmatrix}$ $q-1$ $q^4$ $NA_{31}^1(i ,j , \kappa)$, $i,j\in T_3$ $\gamma^{2i}-\gamma^j\kappa = \gamma^{2n}$ for some $n\in T_3$ $\begin{pmatrix} 1 & & \gamma^i & \kappa \\ & 1 & \gamma^j & \gamma^i \\ & & 1 & \\ & & & 1 \\ \end{pmatrix}$ $\dfrac{(q-1)^2}{2}$ $q^3$ $NA_{31}^2(k)$, $k\in T_3$ $\begin{pmatrix} 1 & \gamma^k & & \\ & 1 & & \\ & & 1 & -\gamma^k \\ & & & 1 \\ \end{pmatrix}$ $q-1$ $q^2$ $NA_{31}^3(K)$, $k\in T_3$ $\begin{pmatrix} 1 & & \gamma^k & \\ & 1 & & \gamma^k \\ & & 1 & \\ & & & 1 \\ \end{pmatrix}$ $q-1$ $q^3$ $NA_{32}(i ,j , \kappa)$, $i,j\in T_3$ $\gamma^{2i}-\gamma^j\kappa = \gamma^{2n+1}$ for some $n\in T_3$ $\begin{pmatrix} 1 & & \gamma^i & \kappa \\ & 1 & \gamma^j & \gamma^i \\ & & 1 & \\ & & & 1 \\ \end{pmatrix}$ $\dfrac{(q-1)^2}{2}$ $q^3$ $NA_5(i,j)$, $i,j\in T_3$ $\begin{pmatrix} 1 & \gamma^i & & \\ & 1 & \gamma^j & -\gamma^{i+j} \\ & & 1 & -\gamma^i \\ & & & 1 \\ \end{pmatrix}$ $(q-1)^2$ $q^2$

Generic representations of ${\rm GSp}(4,\mathbb{F}_q)$

The character table of $\mathcal{G}$ is

 Conjugacy class $\mathcal{G}$ character value $A_1(q-1)$ $(q^4-1)(q^2-1)(q-1)$ $A_2(q-1)$ $-(q^2-1)(q-1)$ $A_{31}(q-1)$ $-(q^2-1)(q-1)$ $A_{32}(q-1)$ $-(q^2-1)(q-1)$ $A_5(q-1)$ $q-1$

We compute that there are $q^2(q-1)$ irreducible generic representation, which is the majority of the $(q^2+2q+4)(q-1)$ total irreducible representations.

References

J. Breeding II, Irreducible non-cuspidal characters of ${\rm GSp}(4,\mathbb{F}_q)$, Ph.D. thesis, University of Oklahoma, Norman, OK, 2011.

D. Bump, Lie Groups, Springer, 2004.

W. Fulton and J. Harris, Representation Theory, A First Course, Springer, 2004.

J.A. Green, The Characters of the Finite General Linear Groups, Transactions of the American Mathematical Society, 80, 2 (1955), 402–447.

G.E. Wall, On the Conjugacy Classes in the Unitary, Symplectic and Orthogonal Groups, Journal of the Australian Mathematical Society, 3 (1963), 1–62.