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.

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