On certain degenerate Whittaker Models for cuspidal representations of ${\mathrm{GL}_{k \cdot n}(\mathbb {F}_q)}$

Abstract

Let $\pi$ be an irreducible cuspidal representation of $\mathrm {GL}_{kn}(\mathbb {F}_q)$. Assume that $\pi = \pi _{\theta }$, corresponds to a regular character $\theta$ of $\mathbb {F}_{q^{kn}}^{*}$. We consider the twisted Jacquet module of $\pi$ with respect to a non-degenerate character of the unipotent radical corresponding to the partition $(n,n,\ldots ,n)$ of kn. We show that, as a $\mathrm {GL}_{n}(\mathbb {F}_q)$-representation, this Jacquet module is isomorphic to $\pi _{\theta \upharpoonright _{\mathbb {F}_n^*}} \otimes \mathrm {St}^{\otimes (k-1)}$, where $\mathrm {St}$ is the Steinberg representation of $\mathrm {GL}_{n}(\mathbb {F}_q)$. This generalizes a theorem of D. Prasad, who considered the case $k=2$. We prove and rely heavily on a formidable identity involving q-hypergeometric series and linear algebra.

1 Introduction

Let $\mathbb {F}:=\mathbb {F}_q$ be the finite field of size q. We fix a nontrivial character $\psi _0$ of $\mathbb {F}$. Denote by $\mathbb {F}_m:=\mathbb {F}_{q^m}$ the unique degree m field extension of $\mathbb {F}$. For a positive integer r, we denote the diagonal subgroup of $\left( \mathrm {GL}_{\ell }(\mathbb {F})\right) ^r$ by

$$\begin{aligned} \Delta ^r\left( \mathrm {GL}_{\ell }(\mathbb {F})\right) :=\left\{ \left. \left( g,\ldots ,g\right) \in \left( \mathrm {GL}_{\ell }(\mathbb {F})\right) ^r\ \right| g\in \mathrm {GL}_{\ell }(\mathbb {F}) \right\} . \end{aligned}$$

For a partition $\rho =\left( k_1,k_2,\ldots ,k_s\right)$ of $\ell$, denote by $P_{\rho }$ the corresponding standard parabolic subgroup of $\mathrm {GL}_{\ell }(\mathbb {F})$. Let $M_{\rho }$ and $N_{\rho }$ be the corresponding standard Levi subgroup and unipotent radical.

Fix $k \ge 1$. Let $\rho =(n,n,\ldots ,n)$ be the partition of kn consisting of k parts of size n. In this paper we denote $G:=\mathrm {GL}_{kn}(\mathbb {F})$, $P:=P_{\rho }$, $M:=M_{\rho }$ and $N:=N_{\rho }$. We have the Levi decomposition $P=M\ltimes N$. We write $U\in N$ in the form

$$\begin{aligned} \quad U= \begin{pmatrix} I_{n} &{}\quad X_{1,1} &{}\quad X_{1,2} &{}\quad \cdots &{}X_{1,k-2}&{}\quad X_{1,k-1} \\ 0 &{}\quad I_{n} &{}\quad X_{2,2} &{}\quad \cdots &{}X_{2,k-2}&{}\quad X_{2,k-1} \\ 0&{} 0 &{} I_{n} &{}\quad \cdots &{}\quad X_{3,k-2}&{}\quad X_{3,k-1}\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}&{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad 0&{}\quad \cdots &{}\quad I_n &{}\quad X_{k-1,k-1}\\ 0 &{}\quad 0 &{}\quad 0&{}\quad \cdots &{}\quad 0 &{}\quad I_n \end{pmatrix}, \end{aligned}$$

(1.1)

where the matrices $X_{i,j}$ ($1\le i\le j \le k-1$) are elements of $M_n(\mathbb {F})$.

Definition 1.1

A character $\psi :N\rightarrow \mathbb {C}^*$ is said to be non-degenerate if it is of the form

$$\begin{aligned} \psi \left( U\right) :=\psi _{0}\left( \mathrm {tr}\left( \sum _{i=1}^{k-1} A_i X_{i,i}\right) \right) =\prod _{i=1}^{k-1}\psi _{0}\left( \mathrm {tr}\left( A_i X_{i,i}\right) \right) , \end{aligned}$$

where the matrices $A_i$ are invertible.

Let $\psi :N\rightarrow \mathbb {C}^{*}$ be a non-degenerate character. Let $\pi$ be an irreducible representation of G, acting on a space $V_\pi$. We denote by $V_{\pi _{k,N,\psi }}$ the largest subspace of $V_\pi$, on which N operates through $\psi$, i.e.

$$\begin{aligned} {V_{\pi _{k,N,\psi }}}=\left\{ v\in V_{\pi }\ \left| \ \pi (U)v=\psi (U)v,\ \forall U\in N\right. \right\} . \end{aligned}$$

This is the $\left( N,\psi \right)$-isotypic subspace of $V_{\pi }$ and it is the image of the canonical projection of $V_\pi$ on $V_{\pi _{k,N,\psi }}$ given by

$$\begin{aligned} P_{k,N,\psi }\left( v\right) =\frac{1}{\left| N\right| }\sum _{U\in N}\overline{\psi }\left( U\right) \pi (U)v. \end{aligned}$$

(1.2)

Since M normalizes N, it acts on the characters of N as follows. If $m \in M$, then for all $U \in N$

$$\begin{aligned} (m\cdot \psi )(U)=\psi \left( m^{-1}Um\right) . \end{aligned}$$

We have, for $m\in M$,

$$\begin{aligned} \pi (m)V_{\pi _{k,N,\psi }}=V_{\pi _{k,N,m\cdot \psi }}. \end{aligned}$$

Let us compute the stabilizer of $\psi$ in M. If

$$\begin{aligned} m=\begin{pmatrix} B_{1}&{}\quad 0&{}\quad \cdots &{}\quad 0\\ 0&{}\quad B_{2}&{}\quad \cdots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad &{}\quad \vdots \\ 0&{}\quad 0&{}\quad \cdots &{}\quad B_{k} \end{pmatrix}, \end{aligned}$$

where $B_{i}\in \mathrm {GL}_{n}(\mathbb {F})$ for all $1 \le i \le k$, then

$$\begin{aligned} (m\cdot \psi )(U)=\psi _0 \left( \text {tr}\left( \sum _{i=1}^{k-1}A_i B_i^{-1} X_{i,i}B_{i+1}\right) \right) . \end{aligned}$$

Thus, $m\cdot \psi =\psi$ if and only if $B_i=B_{i+1}$ for all $1 \le i \le k-1$. In other words,

$$\begin{aligned} \text {stab}_M\psi = \Delta ^k\left( \mathrm {GL}_{n}(\mathbb {F})\right) \cong \mathrm {GL}_{n}(\mathbb {F}). \end{aligned}$$

Therefore, $V_{\pi _{k,N,\psi }}$ is a $\mathrm {GL}_{n}(\mathbb {F})$-module. We denote by $\pi _{k,N,\psi }$ the resulting representation of $\mathrm {GL}_{n}(\mathbb {F})$ on $V_{\pi _{k,N,\psi }}$. It is easy to see that by conjugation with an element in the standard Levi subgroup, we may simply take all the $A_i$ to be the identity matrix. The corresponding twisted Jacquet modules are isomorphic. In the rest of the paper we assume $A_i=I_n$ and fix

$$\begin{aligned} \psi \left( U\right) :=\psi _{0}\left( \mathrm {tr}\left( \sum _{i=1}^{k-1} X_{i,i}\right) \right) . \end{aligned}$$

The goal of this paper is to calculate the character of $\pi _{k,N,\psi }$, and to describe it in terms of more familiar representations, for an irreducible, cuspidal representation $\pi =\pi _\theta$ of $\mathrm {GL}_{kn}(\mathbb {F})$, associated to a regular character $\theta$ of $\mathbb {F}_{kn}^*$. The paper generalizes Prasad’s result for the case $k=2$ stated below.

Theorem

[11, Thm. 1] Let $\pi$ be an irreducible cuspidal representation of $\mathrm {GL}_{2n}(\mathbb {F})$ obtained from a character $\theta$ of $\mathbb {F}^*_{2n}$. Then

$$\begin{aligned} \pi _{2,N,\psi } \cong \mathrm {Ind}_{\mathbb {F}^*_n}^{\mathrm {GL}_{n}(\mathbb {F})}\theta \upharpoonright _{\mathbb {F}_n^*}. \end{aligned}$$

(1.3)

Prasad proved this theorem by an explicit calculation of the characters of $\pi _{2,N,\psi }$ and of the induced representation $\mathrm {Ind}_{\mathbb {F}^*_n}^{\mathrm {GL}_{n}(\mathbb {F})}\theta \upharpoonright _{\mathbb {F}_n^*}$. At any element of $\mathrm {GL}_{n}(\mathbb {F})$ the characters are the same. Therefore, the two representations are equivalent.

The methods used in this paper are generalizations of the methods used by the second author in his thesis [7] for the case $k=3$. From the character calculation, done in Theorem 3 below, we are able to describe in Theorem 4$\pi _{k,N,\psi }$ in terms of the representations $\mathrm {Ind}_{\mathbb {F}^*_{\ell }} ^{\mathrm {GL}_{n}(\mathbb {F})} \theta \upharpoonright _{\mathbb {F}^*_{\ell }}$, where $\ell \mid n$. This reduces immediately to Prasad’s result when $k=2$. Furthermore, we give a compact description of $\pi _{k,N,\psi }$ in terms of the Steinberg representation in the following theorem.

Theorem 1

Let $k\ge 1$. Let $\pi _\theta$ be an irreducible cuspidal representation of $\mathrm {GL}_{kn}(\mathbb {F})$ obtained from a character $\theta$ of $\mathbb {F}^*_{kn}$. Then

$$\begin{aligned} \pi _{{k},N,\psi }\cong \pi _{\theta \upharpoonright _{\mathbb {F}^*_n}}\otimes \mathrm {St}^{\otimes (k-1)}, \end{aligned}$$

where $\pi _{\theta \upharpoonright _{\mathbb {F}^*_n}}$ is the irreducible cuspidal representation of $\mathrm {GL}_{n}(\mathbb {F})$ obtained from $\theta \upharpoonright _{\mathbb {F}^*_n}$, and $\mathrm {St}^{\otimes (k-1)}$ is the $(k-1)$-fold tensor product of the Steinberg representation of $\mathrm {GL}_{n}(\mathbb {F})$ with itself.

Note that for $n=1$, Theorem 1 gives $\pi _{k,N ,\psi } \cong \theta \upharpoonright _{\mathbb {F}^*}$, which also follows from Gel’fand–Graev [4] in case of $\mathrm {GL}_{k}(\mathbb {F})$ (cf. [12, Ch. 8.1]).

We are currently investigating an analogous construction for a non-Archimedean local field.

1.1 Structure of the paper

In Sect. 2 we set the background material from several topics that are needed in the paper: linear algebra, representation theory, q-hypergeometric identities and arithmetic identities.

In Sect. 3 we calculate the dimension of $\pi _{k,N,\psi }$. Green’s formula allows us to express the dimension as rather complicated sum. We use q-hypergeometric identities and linear algebra to show that this sum admits the following compact form.

Theorem 2

Let $k\ge 2$. We have

$$\begin{aligned} \mathrm {dim}\left( \pi _{k,N,\psi }\right) =q^{(k-2) \frac{n(n-1)}{2}}\frac{\left| \mathrm {GL}_{n}(\mathbb {F})\right| }{q^n-1}. \end{aligned}$$

In Sect. 4 we compute the character of $\pi _{k,N,\psi }$, denoted by $\Theta _{k,N ,\psi }$. Apart from the tools used in Theorem 2 this requires understanding of some conjugacy classes of $\mathrm {GL}_{n}(\mathbb {F})$. When $d \mid m$, we have an embedding $\mathbb {F}_{d}^{*} \hookrightarrow \mathrm {GL}_{m}(\mathbb {F})$ (see Sect. 2.1). The elements in $\mathrm {GL}_{m}(\mathbb {F})$ conjugate to an element in the image of this embedding are said to come from $\mathbb {F}_{d}$.

Theorem 3

Let $k\ge 2$. Let $g = s \cdot u$ be the Jordan decomposition of an element g in $\mathrm {GL}_{n}(\mathbb {F})$, where s and u are the semisimple part and unipotent part, respectively.

• (I) If s does not come from $\mathbb {F}_n$, then

  $$\begin{aligned} \Theta _{k,N ,\psi }(g) =0. \end{aligned}$$

• (II) If the $u\ne I_n$, then

  $$\begin{aligned} \Theta _{k,N ,\psi }(g) =0. \end{aligned}$$

• (III) Assume that $u=I_n$ and s comes from $\mathbb {F}_d\subseteq \mathbb {F}_n$ and $d\mid n$ is minimal. Let $\lambda$ be an eigenvalue of s which generates $\mathbb {F}_d$ over $\mathbb {F}$. Then,

  $$\begin{aligned} \Theta _{k,N ,\psi }(s) = (-1)^{k(n-{d^\prime })} q^{(k-2)\frac{n({d^\prime }-1)}{2}} \cdot \left[ \sum \limits _{i=0}^{d-1} \theta (\lambda ^{q^i})\right] \cdot \frac{\left| \mathrm {GL}_{{d^\prime }}(\mathbb {F}_{d})\right| }{q^n-1} , \end{aligned}$$

  where $d^\prime =n/d$.

In Sect. 5 we obtain from Theorem 3 and Lemma 2.10 an isomorphism of representation relating between $\pi _{k,N,\psi }$ and $\mathrm {Ind}_{\mathbb {F}^*_{\ell }} ^{\mathrm {GL}_{n}(\mathbb {F})} \theta \upharpoonright _{\mathbb {F}^*_{\ell }}$ for all $\ell \mid n$. We write a|b|c for a|b and b|c. For any $\ell$ dividing n and any $k\ge 2$, let

$$\begin{aligned} a_{k;n,\ell }(q)=\frac{q^\ell -1}{q^n-1} \sum _{m: \, \ell \mid m \mid n} \mu \left( \frac{m}{\ell }\right) (-1)^{k(n- \frac{n}{m})} q^{(k-2)\frac{n}{2} (\frac{n}{m}-1)}, \end{aligned}$$

(1.4)

where $\mu$ is the Möbius function.

Theorem 4

Let $k \ge 2$.

• (I) If k is even or n is odd, we have

  $$\begin{aligned} \begin{array}{ll} \pi _{k,N,\psi } \cong&\bigoplus _{\ell \mid n} a_{k;n,\ell }(q) \cdot \mathrm {Ind}_{\mathbb {F}^*_{\ell }} ^{\mathrm {GL}_{n}(\mathbb {F})} \theta \upharpoonright _{\mathbb {F}^*_{\ell }}. \end{array} \end{aligned}$$

  (1.5)

• (II) If k is odd and n is even, we have

  $$\begin{aligned} \bigg (\pi _{k,N,\psi }\oplus \bigoplus _{\ell : \, \ell \mid n, 2 \not \mid \frac{n}{\ell }}(-a_{k;n,\ell }(q))\cdot \mathrm {Ind}_{\mathbb {F}^*_{\ell }} ^{\mathrm {GL}_{n}(\mathbb {F})} \theta \upharpoonright _{\mathbb {F}^*_{\ell }}\bigg ) \cong \bigoplus _{\ell : \, \ell \mid n, 2 \mid \frac{n}{\ell }} a_{k;n,\ell }(q) \cdot \mathrm {Ind}_{\mathbb {F}^*_{\ell }} ^{\mathrm {GL}_{n}(\mathbb {F})} \theta \upharpoonright _{\mathbb {F}^*_{\ell }}. \end{aligned}$$

  (1.6)

We note that the coefficients in Theorem 4 are non-negative integers. Indeed, when $k=2$, it is easily shown (see Lemma 2.10) that $a_{2;n,\ell }(q) = \delta _{\ell ,n}$, which gives (1.3). If $k>2$ we show in Lemma 2.10 that $a_{k;n,\ell }(q)$ is a positive integer, except when k is odd, n is even and $2 \not \mid \frac{n}{\ell }$, in which case $-a_{k;n,\ell }(q)$ is a positive integer.

In Sect. 6 we deduce Theorem 1 from Theorem 3.

2 Preliminaries

2.1 Cuspidal representations

We review the irreducible cuspidal representations of $\mathrm {GL}_{m}(\mathbb {F})$ as in Gel’fand [3, Sect. 6] (originally in Green [5]). Irreducible cuspidal representations of $\mathrm {GL}_{m}(\mathbb {F})$, from which all the other irreducible representations of $\mathrm {GL}_{m}(\mathbb {F})$ are obtained via the process of parabolic induction, are associated to regular characters of $\mathbb {F}_{m}^*$. A multiplicative character $\theta$ of $\mathbb {F}_{m}^*$ is called regular if, under the action of the Galois group of $\mathbb {F}_{m}$ over $\mathbb {F}$, the orbit of $\theta$ consists of m distinct characters of $\mathbb {F}_{m}^*$.

We denote the irreducible cuspidal representation of $\mathrm {GL}_{m}(\mathbb {F})$ associated to a regular character $\theta$ of $\mathbb {F}_{m}^*$ by $\pi _\theta$ and the character of the representation $\pi _\theta$ by $\Theta _\theta$.

Given $a\in \mathbb {F}_{m}$, consider the map $m_a:\mathbb {F}_{m}\rightarrow \mathbb {F}_{m}$, defined by $m_a(x)=ax$. The map $a\mapsto m_a$ is an injective homomorphism of algebras $\mathbb {F}_{m} \hookrightarrow \mathrm {End}_{\mathbb {F}}(\mathbb {F}_{m})$. This way, every element of $\mathbb {F}_{m}^*$ gives rise to a well-defined conjugacy class in $\mathrm {GL}_{m}(\mathbb {F})$. The elements in the conjugacy classes in $\mathrm {GL}_{m}(\mathbb {F})$, which are so obtained from elements of $\mathbb {F}_{m}^*$, are said to come from $\mathbb {F}_{m}^*$.

We summarize the information about the character $\Theta _\theta$ in the following theorem. We refer to the paper [11, Thm. 2] for the statement of this theorem in this explicit form, which is originally due to Green [5, Thm. 14] (cf. [3, 14]).

Theorem 2.1

(Green [5]) Let $\Theta _\theta$ be the character of a cuspidal representation $\pi _\theta$ of $\mathrm {GL}_{m}(\mathbb {F})$ associated to a regular character $\theta$ of $\mathbb {F}_{m}^*$. Let $g = s \cdot u$ be the Jordan decomposition of an element g in $\mathrm {GL}_{m}(\mathbb {F})$ (s is a semisimple element, u is unipotent and s, u commute). If $\Theta _\theta (g) \not = 0$, then the semisimple element s must come from $\mathbb {F}_{m}^*$. Suppose that s comes from $\mathbb {F}_{m}^*$. Let $\lambda$ be an eigenvalue of s in $\mathbb {F}_{m}^*$, and let $t=\mathrm {dim}_{\mathbb {F}_{m}}\ker (g-\lambda I)$. Then

$$\begin{aligned} \Theta _\theta (s\cdot u) = (-1)^{m-1}\left[ \sum \limits _{\alpha =0}^{d-1}\theta (\lambda ^{q^\alpha })\right] (1-q^d)(1-({q^d})^2)\cdots (1-({q^d})^{t-1}) \end{aligned}$$

(2.1)

where $q^d$ is the cardinality of the field generated by $\lambda$ over $\mathbb {F}$, and the summation is over the various distinct Galois conjugates of $\lambda$.

Corollary 2.2

The value $\Theta _\theta (g)$ is determined by the eigenvalue of g and the number of Jordan blocks of g, which, in turn, is determined by $\mathrm {dim}_{\mathbb {F}_{m}}\ker (g-\lambda I)$.

2.2 Characters induced from subfields

The following lemma summarizes the information about the character of $\mathrm {Ind}_{\mathbb {F}_\ell ^*} ^{\mathrm {GL}_{n}(\mathbb {F})} (\theta \upharpoonright _{\mathbb {F}_\ell ^*})$, where $\ell \mid n$ and $\theta$ is a character of $\mathbb {F}_n^*$.

Lemma 2.3

[7, Lem. 2.4] Let $\theta$ be a character of $\mathbb {F}^*_n$. Suppose that $s\in \mathrm {GL}_{n}(\mathbb {F})$ comes from $\mathbb {F}_d\subseteq \mathbb {F}_\ell$ ($d\mid \ell$ is minimal). Let $\lambda$ be an eigenvalue of s in $\mathbb {F}_{d}^*$. Then, the character $\Theta _{\mathrm {Ind}_\ell }$ of $\mathrm {Ind}_{\mathbb {F}_\ell ^*} ^{\mathrm {GL}_{n}(\mathbb {F})} (\theta \upharpoonright _{\mathbb {F}_\ell ^*})$ at s is given by

$$\begin{aligned} \Theta _{\mathrm {Ind}_\ell }(s)= & {} \frac{1}{q^\ell -1}\sum _{\begin{array}{c} g\in \mathrm {GL}_{n}(\mathbb {F})\\ g^{-1}sg\in \mathbb {F}^*_\ell \end{array}}\theta (g^{-1}sg) \end{aligned}$$

(2.2)

$$\begin{aligned}= & {} \frac{\left| \mathrm {GL}_{d^\prime }(\mathbb {F}_d)\right| }{q^\ell -1}\left[ \sum \limits _{i=0}^{d-1}\theta (\lambda ^{q^i})\right] , \end{aligned}$$

(2.3)

where $d^\prime = n/d$, and the last sum is over the various distinct Galois conjugates of $\lambda$. The value of the character $\Theta _{\mathrm {Ind}_\ell }$ at an element of $\mathrm {GL}_{n}(\mathbb {F})$ which does not come from $\mathbb {F}_\ell$ is zero.

Remark 2.4

Recall that in (2.2) $\mathbb {F}_\ell ^*$ is considered a subgroup of $\mathrm {GL}_{n}(\mathbb {F})$ by the injective map $a\mapsto [m_a]$, where $[m_a]$ is the representing matrix of $m_a$ with respect to a fixed basis of $\mathbb {F}_{n}$ over $\mathbb {F}$. Note that the choice of basis for $[m_a]$ does not affect the values of $\Theta _{\mathrm {Ind}_\ell }$.

2.3 On some conjugacy classes of $\mathrm {GL}_{n}(\mathbb {F})$

2.3.1 Analogue of Jordan form

Let $g\in \mathrm {GL}_{n}(\mathbb {F})$ and $g=s\cdot u$ be its Jordan decomposition. Assume that s comes from $\mathbb {F}_d\subseteq \mathbb {F}_n$ ($d\mid n$ is minimal). Let $\lambda \in \mathbb {F}_d^*$ be an eigenvalue of s, which generates the field $\mathbb {F}_d$ over $\mathbb {F}$. Denote by f the characteristic polynomial of $\lambda$ (of degree d), and by $L_f\in \mathrm {GL}_{d}(\mathbb {F})$ the companion matrix of f. For $\ell \ge 1$ we denote

$$\begin{aligned} L_{f,\ell }=\begin{pmatrix} L_f&{}\quad I_d&{}\quad &{}\\ {} &{}\quad L_f&{}\quad &{}\\ {} &{}\quad &{}\quad \ddots &{}\quad I_d\\ {} &{}\quad &{}\quad &{}\quad L_f \end{pmatrix}\in \mathrm {GL}_{\ell \cdot d}(\mathbb {F}). \end{aligned}$$

This is an analogue of a Jordan block. As in [3, 5], there exists $\rho =\left( \ell _1,\ldots ,\ell _r\right)$, a partition of $\frac{n}{d}$, $\ell _1\ge \ell _2\ge \cdots \ge \ell _r$, such that g is conjugate to

$$\begin{aligned} L_{\rho }(f):=\begin{pmatrix} L_{f,\ell _1}&{}\quad &{}\quad &{}\\ {} &{}\quad L_{f,\ell _2}&{}\quad &{}\\ {} &{}\quad &{}\quad \ddots &{}\\ {} &{}\quad &{}\quad &{}\quad L_{f,\ell _r} \end{pmatrix}, \end{aligned}$$

i.e. there exists $R\in \mathrm {GL}_{n}(\mathbb {F})$ such that

$$\begin{aligned} R^{-1}gR=L_\rho (f). \end{aligned}$$

(2.4)

Notice that in case $u=I_n$ (g is semisimple), we have $\rho =(1^{n/d})$ and there exists $R\in \mathrm {GL}_{n}(\mathbb {F})$ such that $R^{-1}gR$ is a block diagonal matrix with $d^\prime =n/d$ times $L_f$ on the diagonal. Otherwise, $\ell _1>1$ and, in particular, there exists $R\in \mathrm {GL}_{n}(\mathbb {F})$ such that the upper $2d\times 2d$ left corner of $R^{-1}gR$ is

$$\begin{aligned} \begin{pmatrix} L_f&{}I_d\\ &{}L_f \end{pmatrix}. \end{aligned}$$

Now, s (and so g) has d different eigenvalues obtained by applying the Frobenius automorphism $\sigma$, which generates the Galois group $\mathrm {Gal}(\mathbb {F}_d/\mathbb {F})$, namely

$$\begin{aligned} \left\{ \lambda , \sigma (\lambda ), \ldots , \sigma ^{d-1} (\lambda ) \right\} =\left\{ \lambda , \lambda ^q, \ldots , \lambda ^{q^{d-1}} \right\} , \end{aligned}$$

all of multiplicity ${d^\prime }=n/d$ in the characteristic polynomial of s. Let $0\ne v_0\in \mathbb {F}_d^d$ satisfy $L_f\cdot v_0=\lambda v_0$. So $L_f\cdot \sigma ^{i}(v_0)=\lambda ^{q^i}\sigma ^{i}(v_0)$, for $0\le i\le d-1$. Hence, $B=\left\{ v_0,\sigma (v_0),\ldots ,\sigma ^{d-1}(v_0)\right\} \subseteq \mathbb {F}_d^d$ is linearly independent over $\mathbb {F}_d$, since its elements are eigenvectors of $L_f$ for different eigenvalues. Let $T\in \mathrm {GL}_{d}(\mathbb {F}_d)$ be the diagonalizing matrix of $L_f$ obtained by B, i.e.

$$\begin{aligned} T^{-1}L_fT=D, \end{aligned}$$

(2.5)

where

$$\begin{aligned} D:=\mathrm {diag}\left( \lambda ,\ldots ,\lambda ^{q^{d-1}}\right) . \end{aligned}$$

Denote by $\Delta ^{d^\prime }\left( T\right)$ the block diagonal matrix with $d^\prime$ times T on the diagonal. Explicitly, the columns of $\Delta ^{d^\prime }\left( T\right)$ are the vectors of the basis

$$\begin{aligned} C=\left\{ v_0(i,j)\right\} _{0\le i\le d-1}^{0\le j\le d^\prime -1}, \end{aligned}$$

(2.6)

whose $(j\cdot d+i)$-th vector is given by

$$\begin{aligned} v_0(i,j)=\begin{pmatrix} \underline{0}_{j\cdot d}\\ \sigma ^{i}(v_0)\\ \underline{0}_{n-(j+1)\cdot d} \end{pmatrix}\in \mathbb {F}_d^n, \end{aligned}$$

where ${0\le i\le d-1}$ and ${0\le j\le d^\prime -1}$. Thus, in case $u=I_n$

$$\begin{aligned} \Delta ^{d^\prime }\left( T^{-1}\right) R^{-1}gR\Delta ^{d^\prime }\left( T\right) =\begin{pmatrix} D&{}\quad &{} \\ &{}\quad \ddots &{} \\ &{}\quad &{}D \end{pmatrix}. \end{aligned}$$

Otherwise

$$\begin{aligned} \Delta ^{d^\prime }\left( T^{-1}\right) R^{-1}gR\Delta ^{d^\prime }\left( T\right) =\begin{pmatrix} D&{}\quad I_d&{}\quad &{}\quad &{} \\ &{}\quad D&{}\quad &{}\quad &{} \\ &{}\quad &{}\quad D&{}\quad *&{}\\ &{}\quad &{}\quad &{}\quad \ddots &{}\quad *\\ &{}\quad &{}\quad &{}\quad &{}\quad D \end{pmatrix}, \end{aligned}$$

where $*$ means either $I_d$ or $0_d$ above the diagonal. We denote

$$\begin{aligned} g_{\rho }:=g_{\rho ,R}=\Delta ^{d^\prime }\left( T^{-1}\right) R^{-1}gR\Delta ^{d^\prime }\left( T\right) . \end{aligned}$$

(2.7)

The matrix $g_{\rho }$ is sometimes referred to as an analogue of the Jordan form of g [3, Sect. 0].

2.3.2 Conjugating an arbitrary matrix

We use the notation of Sect. 2.3.1. In particular, we have a fixed $g \in \mathrm {GL}_{n}(\mathbb {F})$ and corresponding R and T as defined in (2.4) and (2.5). Let $A\in M_n(\mathbb {F})$. We study the following conjugation

$$\begin{aligned} A_{\rho }:=A_{\rho ,R}=\Delta ^{d^\prime }\left( T^{-1}\right) R^{-1}AR\Delta ^{d^\prime }\left( T\right) \in M_n(\mathbb {F}_d). \end{aligned}$$

Define $A_R$ by $A_R=R^{-1}AR$, and so $A_\rho =\Delta ^{d^\prime }\left( T^{-1}\right) A_R\Delta ^{d^\prime }\left( T\right)$.

Let $B \in M_n(\mathbb {F}_d)$. Let us represent the vectors $B\cdot v_0(0,m)$, for any $0\le m \le {d^\prime }-1$, as a linear combination of the basis C given in (2.6):

$$\begin{aligned} B\cdot v_0(0,m) = \sum \limits _{\begin{array}{c} 0\le i \le d-1\\ 0\le j \le d^\prime -1 \end{array}} a_{m,i;j} \cdot v_0(i,j),\qquad a_{m,i;j}\in \mathbb {F}_d. \end{aligned}$$

A necessary and sufficient condition for $B \in M_n(\mathbb {F})$ is that for all $0\le m \le {d^\prime }-1,\ 0\le r \le d-1$,

$$\begin{aligned} B\cdot {v_0\left( r,m\right) } = \sum \limits _{\begin{array}{c} 0\le i \le d-1\\ 0\le j \le d^\prime -1 \end{array}} \sigma ^r( a_{m,i;j})\cdot v_0\left( i+r\pmod d,j\right) . \end{aligned}$$

(2.8)

By taking $B = A_R \in M_n(\mathbb {F})$, we get that (2.8) holds for $A_R$. Therefore, $[A_R]_C = A_{\rho }$ is a $d^\prime \times d^\prime$ matrix with entries from $M_d\left( \mathbb {F}_d\right)$. For $0\le m,j\le d^\prime -1$, the m-th row and j-th column of $A_{\rho }$, denoted by $A_{m,j}$, is given by

$$\begin{aligned} A_{m,j}=\left( \sigma ^r\left( a_{m,i-r\pmod d;j}\right) \right) _{0\le i,r \le d-1} , \end{aligned}$$

(2.9)

i.e. $A_{m,j}\in M_d\left( \mathbb {F}_d\right)$ and for ${0\le i,r \le d-1}$, the i-th row and r-th column of $A_{m,j}$ is $\sigma ^r\left( a_{m,i-r\pmod d;j}\right)$. The above discussion can be summarized in the following lemma.

Lemma 2.5

In the above notations, the map $A\mapsto A_\rho$ induces an $\mathbb {F}$-linear isomorphism $M_n(\mathbb {F})\rightarrow M_{n\times d^\prime }(\mathbb {F}_d) \cong \left[ M_{d\times d^\prime }(\mathbb {F}_d)\right] ^{d^\prime }$. It is given by

$$\begin{aligned} A\mapsto \begin{pmatrix} \left( a_{0,i;j}\right) _{\begin{array}{c} 0\le i \le d-1\\ 0\le j \le d^\prime -1 \end{array}}\\ \vdots \\ \left( a_{d^\prime -1,i;j}\right) _{\begin{array}{c} 0\le i \le d-1\\ 0\le j \le d^\prime -1 \end{array}} \end{pmatrix}, \end{aligned}$$

where the $(m\cdot d+i)$-th row and j-th column of the image of A is $a_{m,i;j}\in \mathbb {F}_d$, for $0\le m,j\le d^\prime -1$ and ${0\le i \le d-1}$.

2.3.3 Trace under conjugation

For $g\in \mathrm {GL}_{n}(\mathbb {F})$ and $A\in M_n(\mathbb {F})$ we shall be interested in $\mathrm {tr}\left( g^{-1}A\right)$. We use the notation of Sects. 2.3.1 and 2.3.2. By (2.7), we have

$$\begin{aligned} \mathrm {tr}\left( g^{-1}A\right) =\mathrm {tr}\left( g_{\rho }^{-1}A_\rho \right) . \end{aligned}$$

The inverse of an analogue of a Jordan block of order $d\cdot \ell$ is given by

$$\begin{aligned} \begin{array}{lll} \left( \begin{pmatrix} D&{}I_d&{}\\ {} &{}\ddots &{}I_d\\ &{}&{}D \end{pmatrix}^{-1}\right) _{i,j}={\left\{ \begin{array}{ll} (-1)^{j-i}D^{-j+i-1},&{}i\le j\\ 0,&{}i>j, \end{array}\right. } \end{array} \end{aligned}$$

(2.10)

for $0\le i,j\le \ell$, where the LHS of (2.10) denotes the block matrix in the i-th row and j-th column. We have

$$\begin{aligned} \begin{array}{lll} \mathrm {tr}\left( g_{\rho }^{-1}A_{\rho }\right) &{}=&{} \sum _{m=0}^{d^\prime -1}\mathrm {tr}\left( D^{-1}A_{m,m}+D^{-2}\alpha _m\left( g,D^{-1},A_\rho \right) \right) \\ &{}=&{}\mathrm {tr}\left( \sum _{m=0}^{d^\prime -1} D^{-1}A_{m,m} \right) + \sum _{m=0}^{d^\prime -1} \mathrm {tr}\left( D^{-2}\alpha _m\left( g,D^{-1},A_\rho \right) \right) , \end{array} \end{aligned}$$

(2.11)

where $\alpha _m\left( g,D^{-1},A_\rho \right)$, for $0\le m \le d^\prime -1$ are determined by the analogous Jordan form of g. Notice, that in case g is semisimple, then $\alpha _m\left( g,D^{-1},A_\rho \right) =0$ for all $0\le m \le d^\prime -1$. Otherwise, for $0\le m \le d^\prime -1$, $D^{-2}\alpha _m\left( g,D^{-1},A_\rho \right)$ equals to a sum of terms of the form $(-1)^{\ell }D^{-\ell -1}A_{\ell ,m}$, where $m<\ell \le d^\prime -1$.

By (2.9) we have

$$\begin{aligned} D^{-1}A_{m,m}=\left( \left( \lambda ^{-1}\right) ^{q^r}\sigma ^r\left( a_{m,{i-r\pmod d};m}\right) \right) _{1\le i,r \le d-1}. \end{aligned}$$

So the first sum in the RHS of (2.11) becomes

$$\begin{aligned} \sum _{m=0}^{d^\prime -1}\sum _{r=0}^{d-1}\left( \lambda ^{-1}\right) ^{q^r}\sigma ^r\left( a_{m,0;m}\right) =\sum _{r=0}^{d-1}\sigma ^r\left( \lambda ^{-1}\sum _{m=0}^{d^\prime -1}a_{m,0;m}\right) = \mathrm {Tr}_{\mathbb {F}_{d}/\mathbb {F}}\left( \lambda ^{-1} \cdot \sum \limits _{m=0} ^{d^\prime -1} a_{m,0;m}\right) . \end{aligned}$$

On the other hand, for each $0\le m\le d^\prime -1$, the term $\mathrm {tr}\left( D^{-2}\alpha _m\left( g,D^{-1},A_\rho \right) \right)$ in (2.11) does not depend on the elements $a_{\ell ,0;m}$, where $\ell =m$. Each such term depends only on $\lambda$ and on $a_{\ell ,i,m}$ where $\ell >m$. We summarize the above results in the following lemma.

Lemma 2.6

In the above notations,

$$\begin{aligned} \mathrm {tr}\left( g^{-1}A\right) = \mathrm {Tr}_{\mathbb {F}_{d}/\mathbb {F}}\left( \lambda ^{-1} \cdot \sum \limits _{m=0} ^{d^\prime -1} a_{m,0;m}\right) +\sum _{m=0}^{d^\prime -1}\mathrm {tr}\left( D^{-2}\alpha _m\left( g,D^{-1},A_\rho \right) \right) , \end{aligned}$$

and each summand $\mathrm {tr}\left( D^{-2}\alpha _m\left( g,D^{-1},A_\rho \right) \right)$ is independent of $a_{m,0;m}$ appearing in the first summand, for all $0\le m \le d^\prime -1$.

In case $g=s$ is semisimple we have

$$\begin{aligned} \mathrm {tr}\left( g^{-1}A\right) = \mathrm {Tr}_{\mathbb {F}_{d}/\mathbb {F}}\left( \lambda ^{-1} \cdot \sum \limits _{m=0} ^{d^\prime -1} a_{m,0;m}\right) . \end{aligned}$$

2.4 q-Hypergeometric identity

In order to calculate the dimension of $\pi _{k,N,\psi }$, we need a combinatorial identity related to ranks of triangular block matrices. We first prove a lemma that is a special case of a q-analogue of the Chu–Vandermonde identity, phrased in a manner that we use in the proof of the combinatorial identity. We recall the definition of the q-Pochhammer symbol:

$$\begin{aligned} (a;q)_n = \prod _{i=0}^{n-1}(1-aq^i). \end{aligned}$$

Lemma 2.7

Let $R_q(n,m,r)$ be the number of $n \times m$ matrices of rank r over the finite field of size q (n, m may be 0, with the convention that the empty matrix has rank 0). Let a be an integer greater or equal to $n+m$. Then

$$\begin{aligned} \sum _{r \ge 0} R_q(n,m,r) (q;q)_{a-r} = q^{nm}\frac{(q;q)_{a-n}(q;q)_{a-m}}{(q;q)_{a-n-m}}. \end{aligned}$$

Proof

We start by stating a q-analogue of the Chu–Vandermonde identity [2, Eq. (1.5.2)]:

$$\begin{aligned} \sum _{r =0}^{i} \frac{ (q^{-i};q)_r (b;q)_r }{(c;q)_r (q;q)_r } \left( \frac{cq^i}{b} \right) ^r= \frac{(c/b;q)_i}{(c;q)_i}, \end{aligned}$$

where i is a non-negative integer, and b, c are complex numbers that satisfy $b \ne 0$ and $c \notin \{ q^{-1},\ldots ,q^{-(i-1)} \}$. Choosing $i=n$, $b=q^{-m}$, $c=q^{-a}$, we obtain

$$\begin{aligned} \sum _{r=0}^{n} \frac{ (q^{-n};q)_r (q^{-m};q)_r }{(q^{-a};q)_r (q;q)_r } q^{(n+m-a)r}= \frac{(q^{m-a};q)_n}{(q^{-a};q)_n}. \end{aligned}$$

(2.12)

We have the following formula for $R_q(n,m,r)$ by Landsberg [9]:

$$\begin{aligned} R_q(n,m,r)=\frac{(-1)^r (q^{-n};q)_r (q^{-m};q)_r q^{(n+m)r-\left( {\begin{array}{c}r\\ 2\end{array}}\right) }}{(q;q)_r}. \end{aligned}$$

By expressing the r-th summand of (2.12) as

$$\begin{aligned} \begin{array}{ll} &{}\frac{(-1)^r (q^{-n};q)_r (q^{-m};q)_r q^{(n+m)r-\left( {\begin{array}{c}r\\ 2\end{array}}\right) }}{(q;q)_r} \cdot \frac{q^{-ar+\left( {\begin{array}{c}r\\ 2\end{array}}\right) }}{(-1)^r (q^{-a};q)_r} \\ &{}\quad = R_q(n,m,r) \cdot \frac{q^{-ar+\left( {\begin{array}{c}r\\ 2\end{array}}\right) }}{(-1)^r (q^{-a};q)_r}, \end{array} \end{aligned}$$

we obtain that

$$\begin{aligned} \sum _{r =0}^{n} R_q(n,m,r) \frac{q^{-ar+\left( {\begin{array}{c}r\\ 2\end{array}}\right) } (-1)^r}{(q^{-a};q)_r} =\frac{(q^{m-a};q)_n}{(q^{-a};q)_n}. \end{aligned}$$

(2.13)

The proof is concluded by applying to (2.13) the simple identity

$$\begin{aligned} (q^{-x};q)_y = (-1)^y q^{\left( {\begin{array}{c}y\\ 2\end{array}}\right) -xy} \frac{(q;q)_x}{(q;q)_{x-y}} \end{aligned}$$

with $(x,y) \in \{ (a,n), (a-m,n), (a,r)\}$. $\square$

We now state our main combinatorial identity needed for computing the dimension. Let k be a positive integer. We define the following family of functions.

$$\begin{aligned} f_{k,q}\Big (a;\begin{array}{c} n_1,\ldots , n_k\\ m_1,\ldots ,m_k \end{array}\Big )=\sum _A \left( q;q\right) _{a-\mathrm {rk}A}, \end{aligned}$$

(2.14)

where $\{ n_i\}_{i=1}^{k}, \{ m_j \}_{j=1}^{k}$ are sequences of non-negative integers, a is an integer such that

$$\begin{aligned} a \ge \max \left\{ \sum _{j=1}^{i} n_j + \sum _{j=i}^{k} m_j \mid 1 \le i \le k \right\} \end{aligned}$$

(2.15)

and the sum is over all matrices $A\in M_{(\sum \nolimits _{i=1}^k n_i) \times (\sum \nolimits _{j=1}^k m_j)}(\mathbb {F})$ of the form

$$\begin{aligned} A=\begin{pmatrix} Y_{1,1} &{}\quad Y_{1,2}&{}\quad \cdots &{}Y_{1,k} \\ 0&{}\quad Y_{2,2}&{}\quad \cdots &{}\quad Y_{2,k}\\ \vdots &{}\quad \vdots &{}&{}\quad \vdots \\ 0 &{}\quad 0&{}\quad \cdots &{}\quad Y_{k,k} \end{pmatrix}, \end{aligned}$$

(2.16)

where $Y_{i,j}\in M_{n_i\times m_j}(\mathbb {F})$ for all $1\le i\le j\le k$.

Proposition 2.8

Let $k \ge 1$. For any sequences of non-negative integers, $\{ n_i\}_{i=1}^{k}$ and $\{ m_j \}_{j=1}^{k}$, and for any integer a satisfying (2.15), we have

$$\begin{aligned} f_{k,q}\Big (a;\begin{array}{c} n_1,\ldots , n_k\\ m_1,\ldots ,m_k \end{array}\Big )=q^{\sum \limits _{1\le i\le j\le k}n_i m_j}\cdot \frac{\prod \nolimits _{i=0}^{k}\left( q;q\right) _{a-\sum \nolimits _{j=1}^{k-i}n_j-\sum \nolimits _{j=k-i+1}^{k}m_j}}{\prod \nolimits _{i=1}^{k}\left( q;q\right) _{a-\sum \nolimits _{j=1}^{k-i+1}n_j-\sum \nolimits _{j=k-i+1}^{k}m_j}}. \end{aligned}$$

(2.17)

Proof

We use the following notation:

$$\begin{aligned} I_{r,n,m} = \begin{pmatrix} I_r&{}0_{m-r}\\ 0_{n-r}&{}0 \end{pmatrix}, \quad (r\le \min \{n,m\}). \end{aligned}$$

(2.18)

We prove the proposition by induction on k. Let $k=1$. Then

$$\begin{aligned} f_{1,q}\Big (a;\begin{array}{c} n\\ m \end{array}\Big )=\sum _{A\in M_{n\times m}(\mathbb {F})} \left( q;q\right) _{a-\mathrm {rk}A}=\sum _{r\ge 0} R_q(n,m,r) \left( q;q\right) _{a-r}. \end{aligned}$$

By Lemma 2.7 we find that

$$\begin{aligned} f_1\Big (a;\begin{array}{c} n\\ m \end{array}\Big )=q^{nm}\frac{(q;q)_{a-n}(q;q)_{a-m}}{(q;q)_{a-n-m}}, \end{aligned}$$

as needed. We now perform the induction step, i.e. assume that (2.17) holds for $k-1$ in place of k, and prove it for k. We split the sum defining $f_{k,q}\Big (a;\begin{array}{c} n_1,\ldots , n_k\\ m_1,\ldots ,m_k \end{array}\Big )$ as follows:

$$\begin{aligned} f_{k,q}\Big (a;\begin{array}{c} n_1,\ldots , n_k\\ m_1,\ldots ,m_k \end{array}\Big ) = \sum _{\begin{array}{c} Y_{i,i}\in M_{n_i\times m_i}(\mathbb {F})\\ 1\le i \le k \end{array}} \sum _{\begin{array}{c} Y_{i,j}\in M_{n_i\times m_j}(\mathbb {F})\\ 1\le i<j\le k \end{array}} \left( q;q\right) _{a-\mathrm {rk}A}. \end{aligned}$$

(2.19)

In the inner sum of (2.19) the ranks of $Y_{i,i}$ are fixed for all $1\le i \le k$, so we set $r_i=\mathrm {rk}(Y_{i,i})$. There exist invertible matrices $E_i,C_{i}$ such that $Y_{i,i}=E_i I_{r_i,n_i,m_i}C_{i}$, for all $1\le i \le k$. So, one can write A in the inner sum of (2.19) as $\mathrm {diag}\left( E_1,\ldots ,E_{k}\right) \cdot \widetilde{A}\cdot \mathrm {diag}\left( C_1,\ldots ,C_{k}\right) ,$ where

$$\begin{aligned} \widetilde{A}= \begin{pmatrix} I_{r_1,n_1,m_1} &{} \widetilde{Y}_{1,2}&{}\quad \cdots &{}\quad \widetilde{Y}_{1,k} \\ 0 &{} I_{r_2,n_2,m_2}&{}\quad \cdots &{}\quad \widetilde{Y}_{2,k} \\ \vdots &{}\vdots &{}\quad \ddots &{}\quad \vdots \\ 0 &{}0&{}\quad \cdots &{}\quad I_{r_{k},n_k,m_k}\\ \end{pmatrix} \end{aligned}$$

(2.20)

and $\widetilde{Y}_{i,j}=E_{i}^{-1}Y_{i,j}C_{j}^{-1}$ for all $1\le i<j\le k$. Together with the fact that rank is invariant under elementary operations, (2.19) becomes

$$\begin{aligned} f_{k,q}\Big (a;\begin{array}{c} n_1,\ldots , n_k\\ m_1,\ldots ,m_k \end{array}\Big ) = \sum _{\begin{array}{c} \forall 1\le i \le k: \\ r_i \ge 0 \end{array}} \prod _{i=1}^{k} R_q(n_i,m_i,r_i) \sum _{\widetilde{A}} \left( q;q\right) _{a-\mathrm {rk}\widetilde{A}}, \end{aligned}$$

(2.21)

where the inner sum is over matrices $\widetilde{A}$ of the form (2.20). We can use Gaussian elimination operations on $\widetilde{Y}_{i,j}$ for all $1\le i< j\le k$ (which do not affect the rank of $\widetilde{A}$) as follows: the first $r_i$ rows of each $\widetilde{Y}_{i,j}$ are being canceled by the pivot elements in $I_{r_{i},n}$ (using elementary row operations) and the first $r_j$ columns of each $\widetilde{Y}_{i,j}$ are being canceled by the pivot elements in $I_{r_j,n}$ (using elementary column operations). Formally, the composition of these elementary operations maps the sequence of matrices $\{ \widetilde{Y}_{i,j} \}_{1\le i<j\le k}$$\mathbb {F}$-linearly to a sequence of matrices

$$\begin{aligned} \Big \{ \widehat{\widetilde{Y}}_{i,j} = \begin{pmatrix} 0&{}0\\ 0&{}Z_{i,j} \end{pmatrix} \Big \}_{1 \le i <j \le k}, \end{aligned}$$

(2.22)

where $Z_{i,j}\in M_{(n_i-r_i)\times (m_j-r_{j})}(\mathbb {F})$. This linear map is a projection by construction. Its kernel is of size $q^{\sum _{t=1}^{k-1}r_t \sum _{\ell =t+1}^{k} m_{\ell }+\sum _{t=2}^{k}r_{t} \sum _{\ell =1}^{t-1}(n_{\ell }-r_{\ell })}$. The dimension of the kernel corresponds to the number of elements which we canceled. Equation (2.21) becomes

$$\begin{aligned} \begin{array}{ll} f_{k,q}\Big (a;\begin{array}{c} n_1,\ldots , n_k\\ m_1,\ldots ,m_k \end{array}\Big ) &{}= \sum \limits _{\begin{array}{c} \forall 1 \le i \le k: \\ r_i\ge 0 \end{array}} \prod \limits _{i=1}^{k} R_q(n_i,m_i,r_i) q^{\sum \nolimits _{t=1}^{k-1}r_t\sum \nolimits _{\ell =t+1}^k m_\ell +\sum \nolimits _{t=2}^k r_t\sum \nolimits _{\ell =1}^{t-1}\left( n_\ell -r_\ell \right) }\\ &{}\quad \cdot \sum \limits _{\widehat{\widetilde{A}}} \left( q;q\right) _{a-\mathrm {rk}\widehat{\widetilde{A}}}, \end{array} \end{aligned}$$

(2.23)

where the inner sum is over matrices of the form

$$\begin{aligned} \widehat{\widetilde{A}}= \begin{pmatrix} I_{r_1,n_1,m_1} &{} \widehat{\widetilde{Y}}_{1,2}&{} \cdots &{} \widehat{\widetilde{Y}}_{1,k} \\ 0 &{} I_{r_2,n_2,m_2}&{} \cdots &{}\widehat{\widetilde{Y}}_{2,k} \\ \vdots &{}\vdots &{}\ddots &{}\vdots \\ 0 &{}0&{} \cdots &{}I_{r_{k},n_k,m_k}\\ \end{pmatrix}, \end{aligned}$$

and $\widehat{\widetilde{Y}}_{i,j}$ are as defined in (2.22). Note that $\mathrm {rk}\widehat{\widetilde{A}}=\sum _{j=1}^k r_j+\mathrm {rk}Z,$ where

$$\begin{aligned} Z= \begin{pmatrix} Z_{1,2} &{} \cdots &{} Z_{1,k} \\ \vdots &{}\ddots &{}\vdots \\ 0 &{}\cdots &{}Z_{k-1,k}\\ \end{pmatrix}. \end{aligned}$$

Hence, from (2.23) we obtain the following recursive relation:

$$\begin{aligned} \begin{array}{lll} f_{k,q}\Big (a;\begin{array}{c} n_1,\ldots , n_k\\ m_1,\ldots ,m_k \end{array}\Big ) &{}=&{} \sum \limits _{\begin{array}{c} \forall 1 \le i \le k: \\ r_i\ge 0 \end{array}}\prod \limits _{i=1}^{k} R_q(n_i,m_i,r_i) q^{\sum \nolimits _{t=1}^{k-1}r_t\sum \nolimits _{\ell =t+1}^k m_\ell +\sum \nolimits _{t=2}^k r_t\sum \nolimits _{\ell =1}^{t-1}\left( n_\ell -r_\ell \right) } \\ &{}&{} \cdot f_{k-1,q}\Big (a-\sum \limits _{j=1}^k r_j;\begin{array}{c} n_1-r_1,\ldots , n_{k-1}-r_{k-1}\\ m_2-r_2,\ldots ,m_k-r_k \end{array}\Big ). \end{array} \end{aligned}$$

(2.24)

Plugging the induction assumption in (2.24) we get that $f_{k,q}\Big (a;\begin{array}{c} n_1,\ldots , n_k\\ m_1,\ldots ,m_k \end{array}\Big )$ equals

$$\begin{aligned} \begin{array}{ll} &{} \sum \limits _{\begin{array}{c} \forall 1 \le i \le k: \\ r_i\ge 0 \end{array}} \prod \limits _{i=1}^{k}R_q(n_i,m_i,r_i) q^{\sum \nolimits _{t=1}^{k-1}r_t\sum \nolimits _{\ell =t+1}^k m_\ell +\sum \nolimits _{t=2}^k r_t\sum \nolimits _{\ell =1}^{t-1}\left( n_\ell -r_\ell \right) } \\ &{} \quad \cdot q^{\sum \nolimits _{1\le i\le j\le k-1}\left( n_i-r_i\right) \cdot \left( m_{j+1}-r_{j+1}\right) }\cdot \frac{\prod \nolimits _{i=0}^{k-1}\left( q;q\right) _{a-\sum \nolimits _{j=1}^{k}r_j-\sum \nolimits _{j=1}^{k-1-i}\left( n_j-r_j\right) -\sum \nolimits _{j=k-i}^{k-1}\left( m_{j+1}-r_{j+1}\right) }}{\prod \nolimits _{i=1}^{k-1}\left( q;q\right) _{a-\sum \nolimits _{j=1}^{k}r_j-\sum \nolimits _{j=1}^{k-i}\left( n_j-r_j\right) -\sum \nolimits _{j=k-i}^{k-1}\left( m_{j+1}-r_{j+1}\right) }}. \end{array} \end{aligned}$$

(2.25)

Rearranging (2.25), we see that the sum over $r_1,\ldots ,r_k$ may be written as a product over k sums, where the i-th sum is over $r_i$:

$$\begin{aligned} \begin{array}{lll} f_{k,q}\Big (a;\begin{array}{c} n_1,\ldots , n_k\\ m_1,\ldots ,m_k \end{array}\Big ) &{}=&{} \frac{q^{\sum \nolimits _{1\le i\le j\le k-1}n_i m_{j+1}}}{\prod \nolimits _{i=1}^{k-1}\left( q;q\right) _{a-\sum \nolimits _{j=1}^{k-i}n_j-\sum \nolimits _{j=k-i}^{k-1}m_{j+1}}} \\ &{}&{}\cdot {} \prod \nolimits _{i=1}^{k}\Big (\sum _{r_i \ge 0} R_q(n_{i},m_{i},r_{i})\left( q;q\right) _{a-r_{i}-\sum \nolimits _{j=1}^{i-1}n_j-\sum \nolimits _{j=i}^{k-1}m_{j+1}}\Big ). \end{array} \end{aligned}$$

(2.26)

Using Lemma 2.7 we substitute each inner sum of (2.26) with

$$\begin{aligned} q^{n_{i}\cdot m_{i}}\frac{(q;q)_{a-\sum \nolimits _{j=1}^{i}n_j-\sum \nolimits _{j=i}^{k-1}m_{j+1}}(q;q)_{s-\sum \nolimits _{j=1}^{i-1}n_j-\sum \nolimits _{j=i-1}^{k-1}m_{j+1}}}{(q;q)_{a-\sum \nolimits _{j=1}^{i}n_j-\sum \nolimits _{j=i-1}^{k-1}m_{j+1}}}, \end{aligned}$$

and by simplifying we complete the induction step and obtain the desired identity. $\square$

Remark 2.9

Solomon [13] proved a relation between the following two quantities: the number of placements of k non-attacking rooks on a $n \times n$ chessboard, counted with certain weights depending on q, and the number of matrices in $M_{n \times n}(\mathbb {F})$ of rank k. Haglund generalized Solomon’s result to any “Ferrers board” [6, Thm. 1], which means that the number of matrices of the form (2.16) over $\mathbb {F}$ of rank k is related to the q-rook polynomial $R_k(B,q)$, where B is a certain Ferrers board associated with (2.16). For the definition of a Ferrers board and $R_k(B,q)$, see the introduction to the paper by Garsia and Remmel [1]. In particular, Proposition 2.8 may be deduced from a result of Garcia and Remmel on q-rook polynomials, see [6, Cor. 2]. Our proof of Proposition 2.8 is direct and so we believe it is more accessible. More importantly, the ideas used in the proof reappear in the proofs of Theorems 2 and 3.

2.5 Arithmetic properties of certain polynomials

For any d dividing n and any $k \ge 2$, let

$$\begin{aligned} a_{k;n,d}(x)=\frac{x^{d}-1}{x^{n}-1} \sum _{m: \, d \mid m \mid n} \mu \left( \frac{m}{d} \right) (-1)^{k(n- \frac{n}{m})} x^{(k-2)\frac{n}{2} (\frac{n}{m}-1)} \in \mathbb {Q}(x), \end{aligned}$$

(2.27)

where $\mu : \mathbb {N} \rightarrow \mathbb {C}$ is the Möbius function, defined by $\mu (1)=1$ and

$$\begin{aligned} \mu (n) = {\left\{ \begin{array}{ll} 0 &{} \text { if } \,p^2 \mid n \text { for some prime }p, \\ (-1)^m &{} \text { if }\,n=p_1 p_2 \ldots p_m, \text { where } p_i \text { are distinct primes}. \end{array}\right. } \end{aligned}$$

We recall the following properties of $\mu$ [8, Ch. 2].

• The divisor sum $\sum _{d \mid n} \mu (d)$ is given by

  $$\begin{aligned} \sum _{d \mid n} \mu (d) = \delta _{1,n}. \end{aligned}$$

  (2.28)

• The Möbius function is multiplicative.

Lemma 2.10

Let $k \ge 2$. The following hold.

• (I) For any $d\mid n$, $a_{k;n,d}(x)$ is a polynomial in $\mathbb {Z}[x]$. Furthermore, in case $d \notin \{ n, \frac{n}{2} \}$, $a_{k;n,d}(x)$ is divisible by $x^d-1$. In the remaining cases we have

  $$\begin{aligned} a_{k;n,d}(x) = {\left\{ \begin{array}{ll} (-1)^{k(n-1)} &{} \text {if }\,d=n,\\ \frac{x^{\frac{(k-2)n}{2}}+(-1)^{k+1}}{x^{\frac{n}{2}}+1} &{} \text {if }\,d= \frac{n}{2}. \end{array}\right. } \end{aligned}$$

  (2.29)

• (II) If $k > 2$ we have $\deg \left( a_{k;n,d}\right) = \frac{(n(k-2)-2d)(n-d)}{2d}$, and $a_{k;n,d}$ has leading coefficient $(-1)^{k(n-\frac{n}{d})}$. If $k=2$, we have $a_{k;n,d} = \delta _{n,d}$.

• (III) Assume $k>2$. For any prime power q, $a_{k;n,d}(q)$ is a non-zero integer. Its sign equals the sign of $(-1)^{k(n-\frac{n}{d})}$, i.e. it is a positive integer unless k is odd, n is even and $2 \not \mid \frac{n}{d}$.

Proof

We begin by proving the first part of the lemma. If $d \in \{n, \frac{n}{2}\}$, a short calculation reveals that (2.29) holds. From now on we assume that $d \notin \{n, \frac{n}{2} \}$. We shall show that

$$\begin{aligned} x^n-1 \mid \sum _{m: \, d \mid m \mid n} \mu \left( \frac{m}{d}\right) (-1)^{k(n- \frac{n}{m})} x^{(k-2)\frac{n}{2} (\frac{n}{m}-1)} \end{aligned}$$

(2.30)

in $\mathbb {Q}[x]$, which implies that $a_{k;n,d}(x)$ is a polynomial divisible by $x^d-1$. Gauss’s lemma, applied to (2.30), implies that $a_{k;n,d}(x)\in \mathbb {Z}[x]$. We now prove (2.30).

Let z be a root of unity of order dividing n. Assume first that n is odd or that k is even. Then for all $m\mid n$ we have

$$\begin{aligned} z^{(k-2)\frac{n}{2}(\frac{n}{m}-1)} = (z^n)^{(k-2)\frac{\frac{n}{m}-1}{2}}= 1. \end{aligned}$$

Hence, using (2.28),

$$\begin{aligned} \sum _{m: \, d \mid m \mid n} \mu \left( \frac{m}{d} \right) (-1)^{k(n- \frac{n}{m})} z^{(k-2)\frac{n}{2} (\frac{n}{m}-1)} = \sum _{m: \, d \mid m \mid n} \mu \left( \frac{m}{d} \right) =\sum _{a: \, a \mid \frac{n}{d}} \mu (a) = \delta _{d,n} = 0. \end{aligned}$$

(2.31)

Now we assume instead that n is even and k is odd. We are led to consider two cases.

• If $z^{\frac{n}{2}} = -1$ then for all $m\mid n$ we have,

  $$\begin{aligned} z^{(k-2)\frac{n}{2}(\frac{n}{m}-1)} = (-1)^{\frac{n}{m}-1}. \end{aligned}$$

  Hence, using (2.28),

  $$\begin{aligned}&\sum _{m: \, d \mid m \mid n} \mu \left( \frac{m}{d}\right) (-1)^{k(n- \frac{n}{m})} z^{(k-2)\frac{n}{2} (\frac{n}{m}-1)} = -\sum _{m: \, d \mid m \mid n} \mu \left( \frac{m}{d}\right) \nonumber \\&\quad = -\sum _{a \mid \frac{n}{d}} \mu (a) = -\delta _{d,n} = 0. \end{aligned}$$

  (2.32)

• If $z^{\frac{n}{2}}=1$ then for all $m\mid n$ we have,

  $$\begin{aligned} z^{(k-2)\frac{n}{2}(\frac{n}{m}-1)} = 1. \end{aligned}$$

  Hence,

  $$\begin{aligned}&\sum _{m: \, d \mid m \mid n} \mu \left( \frac{m}{d}\right) (-1)^{k(n- \frac{n}{m})} z^{(k-2)\frac{n}{2} (\frac{n}{m}-1)}\nonumber \\&\quad = \sum _{m: \, d \mid m \mid n} \mu \left( \frac{m}{d}\right) (-1)^{\frac{n}{m}} = \sum _{a \mid \frac{n}{d}} \mu (a) (-1)^{\frac{n}{ad}} \nonumber \\&\quad = \sum _{\begin{array}{c} a \mid \frac{n}{d} \\ 2 \mid \frac{n}{ad} \end{array}} \mu (a) - \sum _{\begin{array}{c} a \mid \frac{n}{d} \\ 2 \not \mid \frac{n}{ad} \end{array}} \mu (a) \nonumber \\&\quad = {\left\{ \begin{array}{ll} 0-\sum _{a \mid \frac{n}{d}}\mu (a) &{}\text { if }\, 2 \not \mid \frac{n}{d} \\ \sum _{a \mid \frac{n}{2d}} \mu (a) - \sum _{\begin{array}{c} a \mid \frac{n}{d}\\ 2 \mid a \end{array}} \mu (2 \cdot \frac{a}{2}) &{} \text { if }\, 2 \mid \frac{n}{d}, 4 \not \mid \frac{n}{d} \\ \sum _{a \mid \frac{n}{2d}} \mu (a) - \sum _{\begin{array}{c} a \mid \frac{n}{d} \\ 2 \not \mid \frac{n}{ad} \end{array}} \mu (4\cdot \frac{a}{4})&\text { if }\, 4 \mid \frac{n}{d}\end{array}\right. } \nonumber \\&\quad = {\left\{ \begin{array}{ll} -\delta _{d,n}&{} \text { if }\,2 \not \mid \frac{n}{d} \\ \delta _{2d,n} - \mu (2) \delta _{2d,n} &{} \text { if }\,2 \mid \frac{n}{d}, 4 \not \mid \frac{n}{d} \\ \delta _{2d,n} &{} \text { if }\,4 \mid \frac{n}{d}\end{array}\right. } \nonumber \\&\quad = 0. \end{aligned}$$

  (2.33)

Equations (2.31), (2.32) and (2.33) show that the RHS of (2.30) vanishes on each root of the separable polynomial $x^n-1$, which establishes (2.30). This concludes the proof of the first part of the lemma.

The second part of the lemma for $k>2$ follows from the observation that the numerator of $a_{k;n,d}(x)$ has degree $d + (k-2)\frac{n}{2}(\frac{n}{d}-1)$ (arising from the term corresponding to $m=d$) and leading coefficient equal to $(-1)^{k(n-\frac{n}{d})}$, while the denominator of $a_{k;n,d}(x)$ has degree n and leading coefficient equal to 1.

When $k=2$, all terms in the sum in (2.27) are constants, and we have

$$\begin{aligned} a_{2;n,d}(x)=\frac{x^{d}-1}{x^{n}-1} \sum _{m: \, d \mid m \mid n} \mu \left( \frac{m}{d} \right) = \frac{x^d-1}{x^n-1}\delta _{n,d} = \delta _{n,d}. \end{aligned}$$

We now turn to the third part of the lemma. Since $a_{k;n,d}(x)$ has integer coefficients, $a_{k;n,d}(q)$ is an integer. We now determine its sign when $k>2$, and in particular show that it is non-zero.

Since $q^d-1$, $q^n-1$, $q^{\frac{n}{2}}$ are positive, we deal with the expression

$$\begin{aligned} \begin{array}{ll} \widetilde{a}_{k;n,d}(q) :&{}= \frac{q^n-1}{q^d-1}q^{(k-2)\frac{n}{2}}\cdot a_{k;n,d}(q) \\ {} &{}= \sum _{m: \, d\mid m \mid n} \mu \left( \frac{m}{d}\right) (-1)^{k(n-\frac{n}{m})} (q^{(k-2)\frac{n}{2}})^{\frac{n}{m}}\\ {} &{}= \sum _{a \mid \frac{n}{d}} \mu (a) (-1)^{k(n-\frac{n}{ad})} (q^{(k-2)\frac{n}{2}})^{\frac{n}{ad}}, \end{array} \end{aligned}$$

whose sign is the same as the sign of $a_{k;n,d}(q)$. If $d=n$ then

$$\begin{aligned} (-1)^{k(n-\frac{n}{d})} \widetilde{a}_{k;n,d}(q) = q^{(k-2)\frac{n}{2}} >0. \end{aligned}$$

If $d=\frac{n}{2}$ then

$$\begin{aligned} (-1)^{k(n-\frac{n}{d})} \widetilde{a}_{k;n,d}(q) = (q^{(k-2)\frac{n}{2}})^{2} +(-1)^{k+1} q^{(k-2)\frac{n}{2}} > 0. \end{aligned}$$

If $\frac{n}{d} \ge 3$, we set $t = q^{(k-2)\frac{n}{2}}$. Then, $t \ge 2^{\frac{3}{2}} >2$ and

$$\begin{aligned} \begin{array}{lll} (-1)^{k(n-\frac{n}{d})} \widetilde{a}_{k;n,d}(q) &{}\ge &{} (q^{(k-2)\frac{n}{2}})^{\frac{n}{d}} - \sum _{1 \le i \le \frac{n}{2d}} (q^{(k-2)\frac{n}{2}})^i \ge (q^{(k-2)\frac{n}{2}})^{\frac{n}{d}} - \frac{(q^{(k-2)\frac{n}{2}})^{\frac{n}{2d}}}{1-q^{-(k-2)\frac{n}{2}}} \\ &{}=&{} (q^{(k-2)\frac{n}{2}})^{\frac{n}{2d}} \left( (q^{(k-2)\frac{n}{2}})^{\frac{n}{2d}} - \frac{1}{1-q^{-(k-2)\frac{n}{2}}}\right) \\ &{}\ge &{} (q^{(k-2)\frac{n}{2}})^{\frac{n}{2d}} \left( (q^{(k-2)\frac{n}{2}})^{\frac{3}{2}} - \frac{1}{1-q^{-(k-2)\frac{n}{2}}}\right) \\ &{}=&{} \frac{(q^{(k-2)\frac{n}{2}})^{\frac{n}{2d}}}{1-q^{-(k-2)\frac{n}{2}}} \left( t^{\frac{1}{2}}(t-1) - 1\right) > 0. \end{array} \end{aligned}$$

$\square$

Remark 2.11

The polynomials $a_{k;n,d}(x)$ may be expressed using the necklace polynomials (see Moreau [10]), defined by

$$\begin{aligned} M_n(x) = \frac{1}{n} \sum _{d \mid n} \mu (d) x^\frac{n}{d}. \end{aligned}$$

Indeed,

$$\begin{aligned} a_{k;n,d}(x) = \frac{x^d-1}{x^n-1} \cdot \left( \frac{(-1)^n}{x^\frac{n}{2}} \right) ^{k-2} \cdot M_{\frac{n}{d}}\left( \left( -x^\frac{n}{2} \right) ^{k-2}\right) . \end{aligned}$$

3 Calculation of the dimension of $\pi _{k,N,\psi }$

Here we prove Theorem 2. Recall that $\Theta _\theta$ is the character of the irreducible cuspidal representation $\pi _\theta$ associated to a regular character $\theta$ of $\mathbb {F}_{n}^*$. Given $U\in N$, we write it in the notation of (1.1). From (1.2),

$$\begin{aligned} \begin{array}{lll} \mathrm {dim}\left( \pi _{k,N,\psi }\right) &{}=&{}\frac{1}{|N|}\sum \limits _{U\in N}\Theta _\theta (U)\overline{\psi }(U)= \frac{1}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}}\sum \limits _{U\in N}\Theta _\theta \left( U\right) \overline{\psi }\left( U\right) \\ &{}=&{}\frac{1}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}} \sum \limits _{\begin{array}{c} X_{i,i}\in M_n(\mathbb {F})\\ 1\le i \le k-1 \end{array}} \sum \limits _{\begin{array}{c} X_{i,j}\in M_n\left( \mathbb {F}\right) \\ 1\le i< j \le k-1 \end{array}}\Theta _\theta \left( U\right) \overline{\psi }\left( U\right) . \end{array} \end{aligned}$$

The character $\psi \left( U\right) =\psi \left( X_{1,1},\ldots ,X_{k-1,k-1}\right)$ is determined by the traces of $X_{i,i}$, $1\le i\le k-1$. Hence,

$$\begin{aligned} \begin{array}{l} \mathrm {dim}\left( \pi _{k,N,\psi }\right) =\frac{1}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}} \sum \limits _{\begin{array}{c} X_{i,i}\in M_n(\mathbb {F})\\ 1\le i \le k-1 \end{array}} \overline{\psi }\left( U\right) \sum \limits _{\begin{array}{c} X_{i,j}\in M_n\left( \mathbb {F}\right) \\ 1 \le i < j \le k-1 \end{array}}\Theta _\theta \left( U\right) . \end{array} \end{aligned}$$

(3.1)

By Corollary 2.2, the value $\Theta _\theta (U)$ is determined by $\mathrm {dim}_{\mathbb {F}_{kn}}\ker (U-I)$ which is in turn determined by $\mathrm {rank}_{\mathbb {F}_{kn}}(U-I)$. In the inner sum of (3.1) set $r_i=\mathrm {rk}\left( X_{i,i}\right)$ for $1\le i \le k-1$. We write $I_{r,n}:=I_{r,n,n}$ as defined in (2.18). There exist invertible matrices $E_i,C_{i+1}$ such that $X_{i,i}=E_i I_{r_i,n}C_{i+1}$. So, one can write $U$ in the inner sum of (3.1) as $I_{kn}$ plus

$$\begin{aligned} \mathrm {diag}\left( E_1,\ldots ,E_{k-1},I_n\right) \begin{pmatrix} 0 &{}\quad I_{r_1,n} &{}\quad \cdots &{}\quad \widetilde{X}_{1,k-2}&{}\quad \widetilde{X}_{1,k-1} \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad \widetilde{X}_{2,k-2}&{}\quad \widetilde{X}_{2,k-1} \\ \vdots &{}\quad \vdots &{}&{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad I_{r_{k-1},n}\\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 \end{pmatrix} \mathrm {diag}\left( I_n,C_2,\ldots ,C_{k}\right) , \end{aligned}$$

where $\widetilde{X}_{i,j}=E_{i}^{-1}X_{i,j}C_{j+1}^{-1}$ for all $1 \le i< j \le k-1$.Together with the fact that rank is invariant under elementary operations, we now have

$$\begin{aligned} \begin{array}{ll} \mathrm {dim}\left( \pi _{k,N,\psi }\right) = &{} \frac{1}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}} \sum \limits _{\begin{array}{c} X_{i,i}\in M_n(\mathbb {F})\\ 1\le i \le k-1 \end{array}} \overline{\psi }\left( U\right) \\ &{}\cdot \sum \limits _{\begin{array}{c} \widetilde{X}_{i,j}\in M_n\left( \mathbb {F}\right) \\ 1 \le i < j \le k-1 \end{array}}\Theta _\theta \left( I_{kn}+ \begin{pmatrix} 0 &{}\quad I_{r_1,n} &{}\quad \cdots &{}\quad \widetilde{X}_{1,k-2}&{}\quad \widetilde{X}_{1,k-1} \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad \widetilde{X}_{2,k-2}&{}\quad \widetilde{X}_{2,k-1} \\ \vdots &{}\quad \vdots &{}&{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad I_{r_{k-1},n}\\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 \end{pmatrix}\right) . \end{array} \end{aligned}$$

(3.2)

As in the proof of Proposition 2.8, we can use Gaussian elimination operations on $\widetilde{X}_{i,j}$ for all $1 \le i <j \le k-1$ (which do not affect the rank nor dimension of the kernel of the matrix minus $I_{kn}$, and the number of Jordan blocks is not affected as well) in such a way that the sequence of matrices $\{ \widetilde{X}_{i,j} \}_{1\le i<j\le k-1}$ is mapped $\mathbb {F}$-linearly to a sequence of matrices

$$\begin{aligned} \Big \{ \widehat{\widetilde{X}}_{i,j} = \begin{pmatrix} 0&{}0\\ 0&{}Y_{i,j} \end{pmatrix} \Big \}_{1 \le i <j \le k-1}, \end{aligned}$$

where $Y_{i,j}\in M_{(n-r_i)\times (n-r_{j})}(\mathbb {F})$. The kernel of this mapping is of size $q^{\sum _{i=1}^{k-2}r_i(k-i-1)n+\sum _{i=2}^{k-1}r_i\sum _{j=1}^{i-1}(n-r_j)}$. The dimension of the kernel corresponds to the number of elements which we cancel. Equation (3.2) becomes

$$\begin{aligned} \begin{array}{lll} \mathrm {dim}\left( \pi _{k,N,\psi }\right) &{}=&{}\frac{1}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}} \sum \limits _{\begin{array}{c} X_{i,i}\in M_n(\mathbb {F})\\ 1\le i \le k-1 \end{array}} \overline{\psi }\left( U\right) q^{\sum \nolimits _{i=1}^{k-2}r_i(k-i-1)n+\sum \nolimits _{i=2}^{k-1}r_i\sum \nolimits _{j=1}^{i-1}(n-r_j)}\\ &{}&{} \cdot \sum \limits _{\begin{array}{c} Y_{i,j}\in M_{(n-r_i)\times (n-r_{j})}(\mathbb {F})\\ 1 \le i <j \le k-1 \end{array}}\Theta _\theta \left( g\right) , \end{array} \end{aligned}$$

(3.3)

where

$$\begin{aligned} g= I_{kn}+\begin{pmatrix} 0 &{}\quad I_{r_1,n} &{}\quad \cdots &{}\quad \widehat{\widetilde{X}}_{1,k-2}&{}\quad \widehat{\widetilde{X}}_{1,k-1} \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad \widehat{\widetilde{X}}_{2,k-2}&{}\quad \widehat{\widetilde{X}}_{2,k-1} \\ \vdots &{}\quad \vdots &{}&{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad I_{r_{k-1},n}\\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 \end{pmatrix} \end{aligned}$$

Using the character formula (2.1), we can calculate $\Theta _\theta (g)$. In this case $m=kn$, $g=s \cdot u$ where $s=I_{kn}$, so $\lambda =1$ and

$$\begin{aligned} t=\mathrm {dim}\ \mathrm {ker}(g-I)=kn-\mathrm {rk}(g-I)=kn-\sum _{i=1}^{k-1}r_i -\mathrm {rk}A, \end{aligned}$$

where

$$\begin{aligned} A=\begin{pmatrix} Y_{1,2} &{} \cdots &{}Y_{1,k-1} \\ \vdots &{}&{}\vdots \\ 0 &{} \cdots &{}Y_{k-2,k-1} \end{pmatrix},\quad 1 \le i < j \le k-1. \end{aligned}$$

(3.4)

So,

$$\begin{aligned} \Theta _\theta \left( g\right) ={}&(-1)^{kn-1}(1-q)(1-q^2)\cdots (1-q^{kn-\sum \nolimits _{i=1}^{k-1}r_i -\mathrm {rk}A-1})\nonumber \\ ={}&(-1)^{kn-1}(q;q)_{kn-\sum \nolimits _{i=1}^{k-1}r_i -\mathrm {rk}A-1}. \end{aligned}$$

Equation (3.3) can now be written as

$$\begin{aligned} \begin{array}{ll} \mathrm {dim}\left( \pi _{k,N,\psi }\right) =\frac{1}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}} &{}\sum \limits _{\begin{array}{c} X_{i,i}\in M_n(\mathbb {F})\\ 1\le i \le k-1 \end{array}} \overline{\psi }\left( U\right) q^{\sum \nolimits _{i=1}^{k-2}r_i(k-i-1)n+\sum \nolimits _{i=2}^{k-1}r_i\sum \nolimits _{j=1}^{i-1}(n-r_j)}\\ &{}\quad \cdot (-1)^{kn-1}\sum \limits _{A}(q;q)_{kn-\sum \nolimits _{i=1}^{k-1}r_i -\mathrm {rk}A-1}, \end{array} \end{aligned}$$

(3.5)

where the inner sum is over all matrices of the form (3.4) and by the definition (2.14) it is equal to

$$\begin{aligned} f_{k-2,q}\Big (kn-\sum _{i=1}^{k-1}r_i-1;\begin{array}{c} n-r_1,\ldots , n-r_{k-2}\\ n-r_2,\ldots ,n-r_{k-1} \end{array}\Big ). \end{aligned}$$

By applying Proposition 2.8 we replace the inner sum in (3.5) by

$$\begin{aligned} q^{\sum \limits _{1\le i\le j\le k-2}(n-r_i)\cdot (n-r_{j+1})}\cdot \frac{\prod \nolimits _{i=0}^{k-2}\left( q;q\right) _{kn-\sum \nolimits _{j=1}^{k-1}r_j-1-\sum \nolimits _{j=1}^{k-2-i}(n-r_j)-\sum \nolimits _{j=k-i-1}^{k-2}(n-r_{j+1})}}{\prod \nolimits _{i=1}^{k-2}\left( q;q\right) _{kn-\sum \nolimits _{j=1}^{k-1}r_j-1-\sum \nolimits _{j=1}^{k-i-1}(n-r_j)-\sum \nolimits _{j=k-i-1}^{k-2}(n-r_{j+1})}}, \end{aligned}$$

which equals

$$\begin{aligned} q^{\sum \limits _{1\le i\le j\le k-2}(n-r_i)\cdot (n-r_{j+1})}\cdot \frac{\prod \nolimits _{i=1}^{k-1}\left( q;q\right) _{2n-1-r_i}}{\left( (q;q)_{n-1} \right) ^{(k-2)}}. \end{aligned}$$

Now (3.5) becomes

$$\begin{aligned} \mathrm {dim}\left( \pi _{k,N,\psi }\right) =\frac{(-1)^{kn-1}}{\left( (q;q)_{n-1}\right) ^{\left( k-2\right) }q^{(k-1)n^{2}}} \sum _{\begin{array}{c} X_{i,i}\in M_n(\mathbb {F})\\ 1\le i \le k-1 \end{array}}\prod _{i=1}^{k-1} \overline{\psi _{0}}\left( \mathrm {tr}\left( X_{i,i}\right) \right) \left( q;q\right) _{2n-1-r_i}. \end{aligned}$$

(3.6)

Changing the order of sum and product in (3.6) we get that

$$\begin{aligned} \mathrm {dim}\left( \pi _{k,N,\psi }\right) =\frac{(-1)^{kn-1}}{\left( (q;q)_{n-1}\right) ^{\left( k-2\right) }q^{(k-1)n^{2}}} \prod _{i=1}^{k-1} \sum _{X_{i,i}\in M_n(\mathbb {F})} \overline{\psi _{0}}\left( \mathrm {tr}\left( X_{i,i}\right) \right) \left( q;q\right) _{2n-1-r_i}. \end{aligned}$$

(3.7)

From Sect. 5 of [11], each inner sum in (3.7) is equal to

$$\begin{aligned} \sum _{X_{i,i}\in M_n(\mathbb {F})} \overline{\psi _{0}}\left( \mathrm {tr}\left( X_{i,i}\right) \right) \left( q;q\right) _{2n-1-r_i} = (-1)^{n} \cdot q^{n^2} \cdot q^{\left( {\begin{array}{c}n\\ 2\end{array}}\right) } (q;q)_{n-1}. \end{aligned}$$

(3.8)

Plugging (3.8) in (3.7), we obtain

$$\begin{aligned} \mathrm {dim}\left( \pi _{k,N,\psi }\right) = q^{(k-1)\left( {\begin{array}{c}n\\ 2\end{array}}\right) } (-1)^{n-1}(q;q)_{n-1} =q^{(k-2)\left( {\begin{array}{c}n\\ 2\end{array}}\right) } \frac{|\mathrm {GL}_{n}(\mathbb {F})|}{q^n-1}, \end{aligned}$$

as needed. $\square$

4 Calculation of the character $\Theta _{k,N,\psi }$

In this section we prove Theorem 3. Namely, we calculate $\Theta _{k,N,\psi }$. From now on we use the following notations:

$$\begin{aligned} \begin{array}{llllll} h_{g;U} = \begin{pmatrix} g &{}\quad X_{1,1} &{}\quad X_{1,2} &{}\quad \cdots &{}\quad X_{1,k-2}&{}\quad X_{1,k-1} \\ 0 &{}\quad g &{}\quad X_{2,2} &{}\quad \cdots &{}\quad X_{2,k-2}&{}\quad X_{2,k-1} \\ 0&{}\quad 0 &{}\quad g &{}\quad \cdots &{}\quad X_{3,k-2},&{}\quad X_{3,k-1}\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}&{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad 0&{}\quad \cdots &{}\quad g &{}\quad X_{k-1,k-1}\\ 0 &{}\quad 0 &{}\quad 0&{}\quad \cdots &{}\quad 0 &{}\quad g \end{pmatrix}, \end{array} \end{aligned}$$

where $U$ (and so $X_{i,j}$) is as in (1.1). Note that $h_{I_n,U}=U$. We also define

$$\begin{aligned} \Delta ^r\left( g\right) = \mathrm {diag}\left( g,\ldots ,g\right) \in \Delta ^r\left( \mathrm {GL}_{n}(\mathbb {F})\right) , \qquad g\in \mathrm {GL}_{n}(\mathbb {F}). \end{aligned}$$

By definition,

$$\begin{aligned} \Theta _{k,N,\psi }\left( g\right)&=\mathrm{{tr}}\left( \pi _{k,N,\psi }\left( g\right) \right) = \mathrm{{tr}}\left( \pi (\Delta ^k(g)){\upharpoonright _{V_{k,N,\psi }}}\right) \nonumber \\&= \mathrm{{tr}}\left( \pi (\Delta ^k(g))\circ P_{k,N,\psi }\right) . \end{aligned}$$

(4.1)

Substituting (1.2) into (4.1) we have

$$\begin{aligned} \begin{array}{lll} \Theta _{k,N,\psi }\left( g\right) &{}=&{} \mathrm {tr}\left( \frac{1}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}}\sum \limits _{U\in N}\pi \left[ \Delta ^k(g)\ \cdot U\right] \overline{\psi }\left( U\right) \right) \\ &{}=&{} \frac{1}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}}\sum \limits _{U\in N}\mathrm {tr}\left( \pi \left[ \Delta ^k(g)\cdot U\right] \right) \overline{\psi }\left( U\right) . \end{array} \end{aligned}$$

(4.2)

Now we perform the change of variables

$$\begin{aligned} X_{i,j} \mapsto g^{-1}X_{i,j}, \qquad 1 \le i \le j \le k-1 \end{aligned}$$

in (4.2) and obtain

$$\begin{aligned} \Theta _{k,N,\psi }\left( g\right) = \frac{1}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}}\sum _{U\in N}\Theta _{\theta }\left( h_{g;U}\right) \overline{\psi }\left( g^{-1}X_{1,1},\ldots ,g^{-1}X_{k-1,k-1}\right) . \end{aligned}$$

(4.3)

In parts Sects. 4.1, 4.2 and 4.3 we prove parts (I), (II) and (III) of Theorem 3, respectively.

4.1 Character at $g = s\cdot u$ such that the semisimple part s does not come from $\mathbb {F}_n$

Let $g = s\cdot u$. Assume that the semisimple part s does not come from $\mathbb {F}_n$. The semisimple part of $h_{g;U}$ is $\Delta ^k(s)$, which also does not come from $\mathbb {F}_n$. By Theorem 2.1, we have $\Theta _\theta \left( h_{g;U}\right) =0$. Hence, by (4.3) $\Theta _{k,N,\psi }\left( g\right) =0$. $\square$

4.2 Character calculation at a non-semisimple element

Assume that s comes from $\mathbb {F}_d\subseteq \mathbb {F}_n$ and $d\mid n$ is minimal. In addition, $d<n$ since g is not semisimple. Let $\lambda \in \mathbb {F}_d^*$ be an eigenvalue of s which generates the field $\mathbb {F}_d$ over $\mathbb {F}$. We use the notation of Sect. 2.3. Thus, there exist $R\in \mathrm {GL}_{n}(\mathbb {F})$ and $\rho$ a partition of $d^\prime =n/d$ such that $R^{-1}gR=L_\rho (f)$. There exists $\Delta ^{d^\prime }\left( T\right) \in \mathrm {GL}_{n}(\mathbb {F}_d)$ such that

$$\begin{aligned} g_{\rho }=\Delta ^{d^\prime }\left( T^{-1}\right) R^{-1}gR\Delta ^{d^\prime }\left( T\right) , \end{aligned}$$

the analogue of the Jordan form of g. Recall that by Lemma 2.5, the map

$$\begin{aligned} A\mapsto A_\rho :=A_{\rho ,R}=\Delta ^{d^\prime }\left( T^{-1}\right) R^{-1}AR\Delta ^{d^\prime }\left( T\right) \end{aligned}$$

induces an isomorphism. By the notation of Sect. 2.3.2 we have for each

$$\begin{aligned} X_{a,b}, \qquad \forall 1 \le a \le b \le k-1, \end{aligned}$$

the corresponding isomorphism of Lemma 2.5

$$\begin{aligned} X_{a,b}\mapsto (X_{a,b})_{\rho }= \begin{pmatrix} \left( x^{(a,b)}_{0,i;j}\right) _{\begin{array}{c} 0\le i \le d-1\\ 0\le j \le d^\prime -1 \end{array}}\\ \vdots \\ \left( x^{(a,b)}_{d^\prime -1,i;j}\right) _{\begin{array}{c} 0\le i \le d-1\\ 0\le j \le d^\prime -1 \end{array}} \end{pmatrix}. \end{aligned}$$

Note that

$$\begin{aligned} \Delta ^k\left( \Delta ^{d^\prime }\left( T^{-1}\right) \right) \Delta ^k\left( R^{-1}\right) h_{g;U} \Delta ^k\left( R\right) \Delta ^k\left( \Delta ^{d^\prime }\left( T\right) \right) =h_{g_{\rho };U_{\rho }}, \end{aligned}$$

(4.4)

where $U_{\rho }$ is the element of N with $(X_{a,b})_{\rho }$ instead of $X_{a,b}$. From (4.4) we obtain

$$\begin{aligned} \mathrm {rk}\left( h_{g-\lambda I_n;U} \right) =\mathrm {rk}\left( h_{g_{\rho }-\lambda I_n;U_\rho } \right) . \end{aligned}$$

We prove that $\mathrm {rk}\left( h_{g-\lambda I_n;U} \right)$ (which by Corollary 2.2 determines the value of $\Theta _\theta \left( h_{g;U} \right)$) is independent of $x^{(k-1,k-1)}_{1,0;1}\in \mathbb {F}_d$. The matrix $h_{g_\rho -\lambda I_n;U_\rho }$ is of the form

$$\begin{aligned} h_{g_{\rho }-\lambda I_n;U_\rho }= \begin{pmatrix} g_\rho -\lambda I_n &{}\quad (X_{1,1})_\rho &{}\quad \cdots &{}\quad (X_{1,k-2})_{\rho }&{}\quad (X_{1,k-1})_{\rho } \\ 0 &{}\quad g_\rho -\lambda I_n &{}\quad \cdots &{}\quad (X_{2,k-2})_{\rho }&{}\quad (X_{2,k-1})_{\rho } \\ \vdots &{}\quad \vdots &{}&{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad g_\rho -\lambda I_n &{}\quad (X_{k-1,k-1})_{\rho }\\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad \boxed {g_\rho -\lambda I_n} \end{pmatrix}. \end{aligned}$$

(4.5)

Consider the boxed block in (4.5). The $2d\times 2d$ upper left block of the boxed matrix $g_\rho -\lambda I_n$ is of the form

$$\begin{aligned} \begin{pmatrix} 0&{}&{}&{}\quad \boxed {1}&{}&{}&{}\\ {} &{}\quad \lambda ^q-\lambda &{}&{}&{}1&{}&{}\\ &{}&{}\quad \ddots &{}&{}&{}1&{}\\ &{}&{}&{}\quad \lambda ^{q^{d-1}}-\lambda &{}&{}&{}1\\ &{}&{}&{}\quad 0&{}&{}&{}\\ &{}&{}&{}&{}\quad \lambda ^q-\lambda &{}&{}\\ &{}&{}&{}&{}&{}\quad \ddots &{}\\ &{}&{}&{}&{}&{}&{}\quad \lambda ^{q^{d-1}}-\lambda \end{pmatrix} \end{aligned}$$

(4.6)

Let $Z:=X_{k-1,k-1}$, $Z_{\rho } := (X_{k-1,k-1})_{\rho }$ and $z_{m,i;j}:=x^{(k-1,k-1)}_{m,i;j}$. One can eliminate the $(d+1)$-th column in $Z_\rho$ by the boxed 1 from (4.6), i.e. all the elements $\left\{ z_{m,i;1}\right\} _{\begin{array}{c} 0\le i\le d-1 \\ 0\le m\le d^\prime -1 \end{array}}$. In particular, $z_{1,0;1} = x^{(k-1,k-1)}_{1,0;1}$ is eliminated. Now, by Lemma 2.6, (4.3) can be written as

$$\begin{aligned} \begin{array}{lll} \Theta _{k,N,\psi }(g) &{}=&{}\frac{1}{q^ { \left( {\begin{array}{c}k\\ 2\end{array}}\right) n^2 }} \sum \limits _{U\in N} \Theta _\theta \left( h_{g;U} \right) \cdot \prod _{i=1}^{k-2}\overline{\psi _0}\left( g^{-1}X_{i,i}\right) \\ &{}\quad \cdot &{} \overline{\psi }_0 \left( \mathrm {Tr}_{\mathbb {F}_{d}/\mathbb {F}}\left( \lambda ^{-1} \cdot \sum \limits _{m=0} ^{d^\prime -1} z_{m,0;m}\right) +\mathrm {tr}\left( D^{-2}\alpha \left( g,D^{-1},Z_\rho \right) \right) \right) . \end{array} \end{aligned}$$

(4.7)

By Lemma 2.5, going over $Z\in M_n(\mathbb {F})$ is equivalent to going over $\left( z_{m,i;j}\right) _{\begin{array}{c} 0\le i \le d-1\\ 0\le j,m \le d^\prime -1 \end{array}}$, $z_{m,i;j}\in \mathbb {F}_d$. We have just shown that $\Theta _\theta \left( h_{g;U} \right)$ is independent of $z_{1,0;1}$, and by Lemma 2.6$\mathrm {tr}\left( D^{-2}\alpha \left( g,D^{-1},Z_\rho \right) \right)$ in (4.7) is also independent of $z_{1,0;1}$. Thus, we may write (4.7) as the following double sum, where the inner sum is over $z_{1,0;1}$ and the outer sum is over the rest of the coordinates of U:

$$\begin{aligned} \begin{array}{lll} \Theta _{k,N,\psi }(g) &{}=&{}\frac{1}{q^ { \left( {\begin{array}{c}k\\ 2\end{array}}\right) n^2 }} \sum \limits _{ \begin{array}{c} X_{i,j}\in N, (i,j) \ne (k-1,k-1) \\ z_{m,i;j} \in \mathbb {F}_d, (m,i,j) \ne (1,0,1) \end{array}} \Theta _\theta \left( h_{g;U} \right) \cdot \prod \limits _{i=1}^{k-2}\overline{\psi _0}\left( g^{-1}X_{i,i}\right) \\ &{}\quad \cdot &{} \overline{\psi }_0 \left( \mathrm {tr}\left( D^{-2}\alpha \left( g,D^{-1},Z_\rho \right) \right) \right) \cdot \overline{\psi }_0 \left( \mathrm {Tr}_{\mathbb {F}_{d}/\mathbb {F}}\left( \lambda ^{-1} \cdot \sum \limits _{\begin{array}{c} 0 \le m \le d^\prime -1 \\ m \ne 1 \end{array}} z_{m,0;m}\right) \right) \\ &{}\quad \cdot &{} \sum \limits _{z_{1,0;1} \in \mathbb {F}_d} \overline{\psi _0} \left( \mathrm {Tr}_{\mathbb {F}_{d}/\mathbb {F}}\left( \lambda ^{-1} \cdot z_{1,0;1}\right) \right) . \end{array} \end{aligned}$$

Since $\overline{\psi }_0\circ \mathrm {Tr}_{\mathbb {F}_{d}/\mathbb {F}}$ is a nontrivial character, we have

$$\begin{aligned} \sum \limits _{z_{1,0;1}\in \mathbb {F}_d}\overline{\psi }_0 \left( \mathrm {Tr}_{\mathbb {F}_{d}/\mathbb {F}}\left( \lambda ^{-1} \cdot z_{1,0;1}\right) \right) =0. \end{aligned}$$

Thus, $\Theta _{k,N ,\psi }(g) =0$. $\square$

4.3 Character calculation at a semisimple element

Here we use (4.3) to calculate the value of $\Theta _{k,N,\psi }(g)$ for $g=s$ where s is semisimple element which comes from a subfield of $\mathbb {F}_{n}$ ($u=I_n$). Again, we use the notation of Sect. 2.3. Thus, there exist $R\in \mathrm {GL}_{n}(\mathbb {F})$, $\rho$ a partition of n / d and $\Delta ^{d^\prime }\left( T\right) \in \mathrm {GL}_{n}(\mathbb {F}_d)$ such that

$$\begin{aligned} s_{\rho }=\Delta ^{d^\prime }\left( T^{-1}\right) R^{-1}sR\Delta ^{d^\prime }\left( T\right) , \end{aligned}$$

(4.8)

the analogue of the Jordan form of s. We also use the notation of Sect. 2.3.2, and in particular define $(X_{a,b})_{\rho }$ as in Sect. 4.2.

Let $\lambda \in \mathbb {F}_n^*$ be an eigenvalue of s. If $\lambda \in \mathbb {F}^*$ then $s=\lambda I$, and we have by (4.3)

$$\begin{aligned} \Theta _{k,N,\psi }(\lambda I) ={}\frac{1}{q^ { \left( {\begin{array}{c}k\\ 2\end{array}}\right) n^2 }} \sum \limits _{U\in N} \Theta _\theta \left( h_{\lambda I;U} \right) \overline{\psi }\left( \lambda ^{-1}X_{1,1}, \ldots ,\lambda ^{-1}X_{k-1,k-1}\right) . \end{aligned}$$

By the change of variables

$$\begin{aligned} X_{i,j}\mapsto \lambda X_{i,j}, \end{aligned}$$

we get

$$\begin{aligned} \Theta _{k,N,\psi }\left( \lambda I\right) = \frac{1}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}}\sum \limits _{U\in N}\Theta _{\theta }\left( \lambda h_{I;U}\right) \overline{\psi }\left( X_{1,1},\ldots ,X_{k-1,k-1}\right) . \end{aligned}$$

By Theorem 2.1, we have $\Theta _{\theta }\left( \lambda \cdot h_{I;U}\right) =\theta (\lambda )\Theta _{\theta }\left( h_{I;U}\right)$, and so

$$\begin{aligned} \Theta _{k,N,\psi }\left( \lambda I\right) =\theta (\lambda )\Theta _{k,N,\psi }\left( I\right) =\theta (\lambda )\mathrm {dim}\left( \pi _{k,N,\psi }\right) . \end{aligned}$$

By Theorem 2, this proves the case $\lambda \in \mathbb {F}^*$.

If $\lambda \in \mathbb {F}^*_{d} \subseteq \mathbb {F}^*_{n}$ is an eigenvalue of s and $1<d\mid n$ is such that $\mathbb {F}_d$ is generated by $\lambda$ over $\mathbb {F}$, we have by (4.3)

$$\begin{aligned} \Theta _{k,N,\psi }\left( s\right) = \frac{1}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}}\sum _{U\in N}\Theta _{\theta } \left( h_{s;U}\right) \overline{\psi }\left( s^{-1}X_{1,1},\ldots ,s^{-1}X_{k-1,k-1}\right) . \end{aligned}$$

(4.9)

In order to compute $\Theta _{\theta }(h_{s;U})$, we need to find conditions for $X_{i,j}$, such that $h_{s;U}$ will have a fixed number of Jordan blocks. This is equivalent to saying that $h_{s;U}-\lambda I_{kn}$ will have a given kernel dimension, or a given rank. Rank and trace are invariant under conjugation, so let us denote by $h_{s_{\rho },U_{\rho }}$, the matrix $h_{s;U}$ conjugated by $\Delta ^k(R)\Delta ^k\left( \Delta ^{d^\prime }\left( T\right) \right)$, where R and T are defined by s in (4.8):

$$\begin{aligned} h_{s_\rho ;U_\rho } := \Delta ^k\left( \Delta ^{d^\prime }\left( T^{-1}\right) \right) \Delta ^k\left( R^{-1}\right) h_{s;U} \Delta ^k(R)\Delta ^k\left( \Delta ^{d^\prime }\left( T\right) \right) . \end{aligned}$$

We have a matrix in $\mathrm {GL}_{kn}(\mathbb {F}_d)$ and our goal is to find out how many matrices of the form

$$\begin{aligned} h_{s_\rho ;U_\rho } -\lambda I_{kn}= h_{s_{\rho }-\lambda I_n;U_\rho }, \end{aligned}$$

where U varies, have a given rank $\ell$.

First, notice that by the invariance of rank under elementary row and column operations on $h_{s_{\rho }-\lambda I_n;U_\rho }$, we can use the nonzero elements on the diagonal of $s_\rho -\lambda I_n$ to cancel the corresponding elements of $(X_{a,b})_\rho$. These elementary operations map the sequence of matrices $\{(X_{a,b})_\rho \}_{1 \le a \le b \le k-1}$$\mathbb {F}_d$-linearly to the sequence

$$\begin{aligned} \left\{ (\widehat{X}_{a,b})_\rho = \begin{pmatrix} x^{(a,b)}_{0,0;0} &{}\quad \cdots &{}\quad x^{(a,b)}_{{d^\prime -1},0;0} \\ \vdots &{}\quad \ddots &{}\quad \vdots \\ x^{(a,b)}_{0,0;d^\prime -1} &{}\quad \cdots &{}\quad x^{(a,b)}_{{d^\prime -1},0;d^\prime -1} \end{pmatrix} \in M_{d^\prime }\left( \mathbb {F}_d\right) \right\} _{1 \le a \le b \le k-1}. \end{aligned}$$

The dimension of the kernel of this map is $\left( {\begin{array}{c}k\\ 2\end{array}}\right) (n-d^\prime )d^\prime$, corresponding to the number of elements we canceled. Hence, the number of matrices $h_{s_{\rho }-\lambda I_n;U_\rho }$ of rank $\ell$ is $(q^{d})^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) (n-{d^\prime }){d^\prime }}$ times the number of matrices of the form

$$\begin{aligned} A:=\begin{pmatrix} (\widehat{X}_{1,1})_\rho &{} \cdots &{}(\widehat{X}_{1,k-2})_{\rho }&{}(\widehat{X}_{1,k-1})_{\rho } \\ 0 &{} \cdots &{}(\widehat{X}_{2,k-2})_{\rho }&{}(\widehat{X}_{2,k-1})_{\rho } \\ \vdots &{}&{}\vdots &{}\vdots \\ 0 &{} \cdots &{} 0 &{}(\widehat{X}_{k-1,k-1})_{\rho } \end{pmatrix}\in M_{(k-1)d^\prime }(\mathbb {F}_d) \end{aligned}$$

(4.10)

of rank $\ell -k(n-d^\prime )$. Using the character formula (2.1), we can calculate $\Theta _\theta (h_{s;U})$. In this case $m=kn$, $g=h_{s;U}$ and

$$\begin{aligned} t=\mathrm {dim}\ \mathrm {ker}(h_{s;U}-I)=kn-\mathrm {rk}(h_{s;U}-I)=kn-k(n-d^\prime ) -\mathrm {rk}A=kd^\prime -\mathrm {rk}A. \end{aligned}$$

Thus

$$\begin{aligned} \Theta _\theta \left( h_{s;U}\right)= & {} (-1)^{kn-1}\left[ \sum \limits _{i=0}^{d-1} \theta (\lambda ^{q^i})\right] (1-q^d)(1-(q^d)^2)\cdots (1-(q^d)^{kd^\prime -\mathrm {rk}A-1})\nonumber \\= & {} (-1)^{kn-1}\left[ \sum \limits _{i=0}^{d-1} \theta (\lambda ^{q^i})\right] (q^d;q^d)_{kd^\prime -\mathrm {rk}A-1}. \end{aligned}$$

(4.11)

Now, by (4.11) and Lemma 2.6, (4.9) can be written as

$$\begin{aligned} \begin{array}{ll} \Theta _{k,N,\psi }\left( s\right) = \frac{(-1)^{kn-1}(q^{d})^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) (n-{d^\prime }){d^\prime }}}{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) n^{2}}} &{}\left[ \sum \limits _{i=0}^{d-1} \theta (\lambda ^{q^i})\right] \sum _{A}(q^d;q^d)_{kd^\prime -\mathrm {rk}A-1}\\ {} &{}\cdot \prod \limits _{i=1}^{k-1} \overline{\psi }_0 \left( \mathrm {Tr}_{\mathbb {F}_{d}/\mathbb {F}}\left( \lambda ^{-1} \cdot \sum \limits _{m=0} ^{d^\prime -1} x^{(i,i)}_{m,0;m}\right) \right) , \end{array} \end{aligned}$$

(4.12)

where the sum is over matrices A as in (4.10). By the character formula (2.1), the RHS of (4.12) is $(-1)^{k(n-d^\prime )}\left[ \sum \nolimits _{i=0}^{d-1} \theta (\lambda ^{q^i})\right]$ times the RHS of (3.1), when one replaces n with ${d^\prime }$, q with $q^d$ and $\psi _0$ with

$$\begin{aligned} \psi ^{\prime }_0:\mathbb {F}_d \rightarrow \mathbb {C}^{*}, \quad \psi ^{\prime }_0(x)=\psi _0 \Big ( \mathrm {Tr}_{\mathbb {F}_{d}/\mathbb {F}}(\lambda ^{-1}x) \Big ). \end{aligned}$$

Thus, the RHS of (4.12) is equal to $\mathrm {dim}\left( \pi _{k,N,\psi }\right)$ (which is calculated in Theorem 2) after the substitution of $n,q,\psi _0$ with the relevant values. Hence,

$$\begin{aligned} \begin{array}{ll} \Theta _{k,N,\psi }\left( s\right) =(-1)^{k(n-d^\prime )}\left[ \sum \limits _{i=0}^{d-1} \theta (\lambda ^{q^i})\right] (q^d)^{(k-2) \frac{d^\prime (d^\prime -1)}{2}}\frac{\left| \mathrm {GL}_{d^\prime }(\mathbb {F}_d)\right| }{q^n-1}, \end{array} \end{aligned}$$

as desired. $\square$

5 Proof of Theorem 4

Notice first that by part (III) of Lemma 2.10, the coefficients in both (1.5) and (1.6) are positive integers, unless $k=2$ in which case they may also be zero.

Representations of a finite group are equivalent if the corresponding characters coincide. Hence, both parts of the theorem are equivalent to

$$\begin{aligned} \forall g \in \mathrm {GL}_{n}(\mathbb {F}): \, \Theta _{k;N,\psi }(g) = \sum _{\ell \mid n} a_{k;n,\ell }(q) \cdot \Theta _{\mathrm {Ind}_\ell }(g), \end{aligned}$$

(5.1)

where $\Theta _{\mathrm {Ind}_\ell }$ is the character of $\mathrm {Ind}_{\mathbb {F}_\ell ^*} ^{\mathrm {GL}_{n}(\mathbb {F})} (\theta \upharpoonright _{\mathbb {F}_\ell ^*})$. We prove now (5.1) for any $g\in \mathrm {GL}_{n}(\mathbb {F})$. If g is not semisimple or does not come from $\mathbb {F}_n$ then the LHS of (5.1) is zero by parts (I) and (II) of Theorem 3. The RHS of (5.1) is also zero on such elements by Lemma 2.3.

Let g be a semisimple element, which comes from $\mathbb {F}_d\subseteq \mathbb {F}_n$ and $d\mid n$ is minimal. Let $\lambda$ be an eigenvalue of s, which generates $\mathbb {F}_d$ over $\mathbb {F}$. For such g, part (III) of Theorem 3 and Lemma 2.3 imply that (5.1) is equivalent to

$$\begin{aligned}&(-1)^{k(n-{d^\prime })} \left[ \sum \limits _{i=0}^{d-1} \theta (\lambda ^{q^i})\right] q^{(k-2)\frac{n({d^\prime }-1)}{2}}\cdot \frac{\left| \mathrm {GL}_{{d^\prime }}(\mathbb {F}_{d})\right| }{q^n-1}\nonumber \\&\quad = \sum _{\ell :\ d\mid \ell \mid n} a_{k;n,\ell }(q) \frac{\left| \mathrm {GL}_{d^\prime }(\mathbb {F}_d)\right| }{q^\ell -1}\left[ \sum \limits _{i=0}^{d-1}\theta (\lambda ^{q^i})\right] , \end{aligned}$$

(5.2)

where $d^\prime =n/d$. The following identity, which we now prove, establishes (5.2):

$$\begin{aligned} \frac{(-1)^{k(n-{d^\prime })} q^{(k-2)\frac{n({d^\prime }-1)}{2}}}{q^n-1}= \sum _{\ell :\ d\mid \ell \mid n} \frac{a_{k;n,\ell }(q)}{q^\ell -1}. \end{aligned}$$

(5.3)

Using (1.4), the RHS of (5.3) is

$$\begin{aligned} \sum _{\ell :\ d\mid \ell \mid n} \sum _{m: \, \ell \mid m \mid n} \frac{\mu \left( \frac{m}{\ell }\right) (-1)^{k(n- \frac{n}{m})} q^{(k-2)\frac{n}{2} (\frac{n}{m}-1)}}{q^n-1}. \end{aligned}$$

(5.4)

We simplify (5.4) using (2.28) as follows:

$$\begin{aligned} \begin{array}{lll} \sum \limits _{\ell :\ d\mid \ell \mid n} \sum \limits _{m: \, \ell \mid m \mid n} &{}&{}\frac{\mu \left( \frac{m}{\ell }\right) (-1)^{k(n- \frac{n}{m})} q^{(k-2)\frac{n}{2} (\frac{n}{m}-1)}}{q^n-1}\\ &{}=&{} \sum \limits _{m: \, d \mid m \mid n} \frac{(-1)^{k(n-\frac{n}{m})} q^{(k-2)\frac{n}{2} (\frac{n}{m}-1)}}{q^n-1} \sum \limits _{\ell :\ d\mid \ell \mid m} \mu \left( \frac{m}{\ell }\right) \\ &{}=&{}\sum \limits _{m: \, d \mid m \mid n} (-1)^{k(n-\frac{n}{m})} \frac{ q^{(k-2)\frac{n}{2} (\frac{n}{m}-1)}}{q^n-1}\delta _{d,m}\\ &{}=&{} (-1)^{k(n-\frac{n}{d})} \frac{ q^{(k-2)\frac{n}{2} (\frac{n}{d}-1)}}{q^n-1}, \end{array} \end{aligned}$$

which is the LHS of (5.3). Hence the proof is complete. $\square$

6 Proof of Theorem 1

Representations of a finite group are equivalent if the corresponding characters coincide. Hence, the theorem is equivalent to

$$\begin{aligned} \forall g \in \mathrm {GL}_{n}(\mathbb {F}): \,\quad \Theta _{k,N,\psi }(g) = \Theta _{\theta \upharpoonright _{\mathbb {F}^*_n}}(g) \cdot \left( \mathrm {St}(g)\right) ^{k-1}, \end{aligned}$$

(6.1)

where we use the notation $\mathrm {St}$ also for the character of the Steinberg representation. We prove now (6.1) for any $g\in \mathrm {GL}_{n}(\mathbb {F})$.

We first prove (6.1) for $k=1$. Note that $N=\left\{ I_n\right\}$ and so

$$\begin{aligned} V_{\pi _{1,N,\psi }}=\left\{ v\in V_{\pi _\theta }\ \left| \ \pi (I_n)v=v\right. \right\} =V_{\pi _\theta } . \end{aligned}$$

Hence $\pi _{1,N,\psi }(g) = \pi _{\theta }(g)$ as needed.

Now assume $k \ge 2$. If the semisimple part s of g does not come from $\mathbb {F}_n$, or g is not semisimple, then $\Theta _{k,N,\psi }(g)=0$ by Theorem 3. From Theorem 2.1, we have $\Theta _{\theta \upharpoonright _{\mathbb {F}^*_n}}(g) = 0$. Hence, (6.1) is proved in that case.

Otherwise, $g=s$ is a semisimple element which comes from $\mathbb {F}_d \subseteq \mathbb {F}_n$ and $d \mid n$ is minimal. We begin by calculating the character value $\mathrm {St}(g)$. For any prime p, let $m_p$ be the p-part of m. By [12], Thm. 6.5.9],

$$\begin{aligned} \mathrm {St}(g)=\varepsilon _{\mathrm {GL}_{n}} \varepsilon _{C(g)^{\circ }} \left| C(g)^{\mathbb {F}} \right| _{\mathrm {char}(\mathbb {F})}, \end{aligned}$$

where $\varepsilon _G$ is $(-1)$ to the power of the $\mathbb {F}$-rank of G, C(g) is the centralizer of g in $\mathrm {GL}_{n}(\overline{\mathbb {F}})$, $C(g)^{\circ }$ is its identity component and $C(g)^{\mathbb {F}}$ is the subgroup of $\mathbb {F}$-rational points in C(g). The $\mathbb {F}$-rank of $\mathrm {GL}_{n}$ is n. Let $\rho = \left( 1,1,\ldots , 1\right)$, a partition of $d^\prime =\frac{n}{d}$ and let f be the characteristic polynomial of s. By Sect. 2.3.1, the centralizer $C(g)^{\mathbb {F}}$ is isomorphic to $C(L_{f,\rho })^{\mathbb {F}}$, which in turn is isomorphic to $\mathrm {GL}_{d^\prime }(\mathbb {F}_{d})$ (cf. [5], Lem. 2.4] and the discussion preceding it). Thus, $\varepsilon _{C(g)^{\circ }} =\varepsilon _{\mathrm {GL}_{d^\prime }}= (-1)^{d^\prime }$ and

$$\begin{aligned} \left| C(g)^{\mathbb {F}} \right| =q^{\sum _{i=1}^{d^\prime } d(d^\prime -i)} \prod _{k=1}^{d^\prime }\left( q^{d k}-1\right) , \quad \left| C(g)^{\mathbb {F}} \right| _{\mathrm {char}(\mathbb {F})} = q^{\frac{n(d^{\prime }-1)}{2}}. \end{aligned}$$

The discussion shows that

$$\begin{aligned} \mathrm {St}(g)=(-1)^{n-d^\prime }q^{ \frac{n(d^\prime -1)}{2}}. \end{aligned}$$

(6.2)

By Theorem 2.1,

$$\begin{aligned} \Theta _{\theta \upharpoonright _{\mathbb {F}^*_n}}(g)= & {} (-1)^{n-1}\left[ \sum \limits _{\alpha =0}^{d-1}\theta (\lambda ^{q^\alpha })\right] (1-q^d)(1-({q^d})^2)\cdots (1-({q^d})^{d^\prime -1})\nonumber \\= & {} (-1)^{n-d^\prime } \left[ \sum \limits _{\alpha =0}^{d-1}\theta (\lambda ^{q^\alpha })\right] (q^d-1)(q^{2d}-1)\cdots (q^{n-d} -1) \frac{q^n-1}{q^n-1}\nonumber \\= & {} (-1)^{n-d^\prime } \left[ \sum \limits _{\alpha =0}^{d-1}\theta (\lambda ^{q^\alpha }) \right] \frac{\left| \mathrm {GL}_{{d^\prime }}(\mathbb {F}_{d}) \right| }{(q^n-1) q^{\frac{n(d^\prime -1)}{2}}}, \end{aligned}$$

(6.3)

where $\lambda$ is an eigenvalue of g. By Theorem 3

$$\begin{aligned} \Theta _{k,N,\psi }(g) = (-1)^{k(n-{d^\prime })} q^{(k-2)\frac{n({d^\prime }-1)}{2}} \cdot \left[ \sum \limits _{i=0}^{d-1} \theta (\lambda ^{q^i})\right] \cdot \frac{\left| \mathrm {GL}_{{d^\prime }}(\mathbb {F}_{d})\right| }{q^n-1}. \end{aligned}$$

(6.4)

Multiplying (6.3) by (6.2) raised to the $(k-1)$-th power, we get (6.4) as needed. $\square$