A Selberg Trace Formula for GL₃(${\mathbb{F}}_{\mathit{p}}$)∖GL₃(${\mathbb{F}}_{\mathit{q}}$)/K

Abstract

In this paper, we prove a discrete analog of the Selberg Trace Formula for the group ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right).$ By considering a cubic extension of the finite field ${\mathbb{F}}_q$, we define an analog of the upper half-space and an action of ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ on it. To compute the orbital sums, we explicitly identify the double coset spaces and fundamental domains in our upper half space. To understand the spectral side of the trace formula, we decompose the induced representation $\rho ={\mathrm{Ind}}_{\Gamma }^{G}1$ for $G={\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ and $\Gamma ={\mathrm{GL}}_3\left({\mathbb{F}}_p\right).$

1. Introduction

The Selberg trace formula [1] is one of the most celebrated mathematical results of the past century. In its original form, it relates the spectrum of the Laplacian on a hyperbolic surface to lengths of closed geodesics on the surface. It can be viewed as a non-Abelian generalization of the well-known Poisson summation formula in Fourier analysis, and has found numerous applications in many fields, for example, in number theory and the theory of automorphic forms (see Sarnak [2]).

The Selberg trace formula deals with surfaces which can be represented as double cosets of the form $\Gamma \setminus {SL}_2\left(\mathbb{R}\right)/{SO}_2\left(\mathbb{R}\right)$, where $\Gamma$ is a discrete subgroup of ${SL}_2\left(\mathbb{R}\right)$. It is a natural question to work out the trace formula for higher-dimensional matrix groups. To quote Dennis Hejhal: “Of course, we won’t really understand the trace formula until it is written down for ${SL}_4\left(\mathbb{Z}\right)$”. While Dorothy Wallace [3] has worked out the explicit details of the trace formula for ${SL}_3\left(\mathbb{Z}\right)\setminus {SL}_3\left(\mathbb{R}\right)/{SO}_3\left(\mathbb{R}\right)$, the four-dimensional case remains mysterious. A further generalization is the so-called Arthur–Selberg trace formula [4], which plays an important role in the Langlands program. In this regard, it should be mentioned that an extension of the Selberg trace formulae to higher-dimensional symmetric spaces $\Gamma \setminus G/K$ over real or complex numbers requires handling quite complicated analytic issues that do not appear in the original two-dimensional case. It is for this reason that study of Selberg trace formulae for higher-dimensional special linear groups over finite fields is a necessary step towards understanding a general trace formula. In this paper, we take a first step in this direction by proving a Selberg trace formula for a three-dimensional Poincaré upper half-plane over finite fields. At the same time, our results show that extending this formula to higher-dimensional upper half-planes for more general matrix groups G would involve complicated calculations.

Selberg-type trace formulae have been derived for discrete spaces such as graphs as well. By realizing a k-regular graph as a quotient of the infinite k-regular tree, Ahumada [5] found such a formula for k-regular graphs. Audrey Terras [6] introduced the concept of finite upper half-planes as a finite analogue of the Poincaré upper half-plane model in which the complex number field is replaced by a quadratic extension of a finite field ${\mathbb{F}}_q$, where $q=p^r$ is a prime power. She constructed a family of Ramanujan graphs [7] using these upper half-planes and developed a representation-theoretic trace formula on ${\mathrm{GL}}_2\left({\mathbb{F}}_p\right)\setminus {\mathrm{GL}}_2\left({\mathbb{F}}_q\right)/K$, where K is the finite analogue of ${SO}_2\left(\mathbb{R}\right)$.

The aim of this paper is to present an analogous trace formula for ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$. Our main result in this paper is as follows. Consider the pre-trace formula

$$\sum _{\pi \in \widehat{G}}m(\pi ,\rho )Tr\left(\widehat{f}\left(\pi \right)\right)=\sum _{\left\{\gamma \right\}\in C_{\Gamma }}\frac{|G_{\gamma }|}{|{\Gamma }_{\gamma }|}{I}_G(f,\gamma ),$$

(1)

where G is a finite group, $\Gamma$ is a subgroup of G, $\rho ={Ind}_{\Gamma }^{G}1$ is the induced representation of the trivial representation of $\Gamma$, and $I_G(f,\gamma )$ is the orbital sum of f at $\gamma$. These terms are defined in Section 3, and a proof of the pre-trace formula is recalled there as well. Note that (1) can be viewed as a representation-theoretic analogue of Selberg’s trace formula [1,6]; the left hand side can be thought of as the spectral side, summing over the irreducible representations of G and their multiplicities in $\rho$, while the right hand side can be interpreted as the geometric side, summing over conjugacy classes in G. In this paper, to obtain a trace formula for the finite upper half-space we take $G={\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ and $\Gamma ={\mathrm{GL}}_3\left({\mathbb{F}}_p\right)$; also, $f:G\to \mathbb{C}$ is chosen to be a $K-$bi-invariant function (K is the stabilizer of $p_0$ as above), i.e., $\forall k,h\in K$ and $\forall x\in G$ we have $f\left(kxh\right)=f\left(x\right)$. One can also think of f as a function on $G/K\simeq {\mathbb{H}}_q$ which is invariant under left-multiplication by elements in K.

The rest of this paper is organized as follows. In Section 2, we define an analogue, denoted ${\mathbb{H}}_q,$ of the upper half-plane for the group ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ by considering a cubic extension of the finite field ${\mathbb{F}}_q$. We define an action of ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ on ${\mathbb{H}}_q$, which sets the stage for our discrete trace formula. In Section 3, we recall a standard pre-trace formula for finite groups. In Section 4, Section 5 and Section 6, we compute each term on the right-hand side of the pre-trace formula (5) (the so-called geometric side) using the explicit description for conjugacy classes of ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ provided in [8]. The conjugacy classes are separated into central, hyperbolic, parabolic, and elliptic terms. Each of these types of terms is explicitly identified. It would be helpful to have a more conceptual understanding of these computations. In Section 7, we provide some simple examples of the application of our formula in which the left-hand side of the pre-trace formula is known a priori. In Section 8 and Section 9, we calculate the left-hand side of the pre-trace formula by computing the character of the induced representation $\rho ={Ind}_{\Gamma }^{G}1$, then using this to compute its decomposition into irreducible representations in the case that $2,3\nmid n$.

2. A Finite Upper Half-Space

To generalize the results of Terras to the ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ case, we must first discuss the generalization of the finite upper half-space. Let p be a prime number and $q=p^n$ for a positive integer n. The definition of the upper half-space ${\mathbb{H}}_q$ comes from considering the cubic extension of ${\mathbb{F}}_q$. The construction was first provided by Martinez in [9]. Here, we describe it explicitly for our special case of ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$.

Lemma 1. 

${\mathbb{F}}_q$ has a cubic nonresidue if and only if $q\equiv 1\left(\mathrm{mod}3\right)$.

Proof. 

Let $q=p^n$. Consider the cube map $x\mapsto x^3:{\mathbb{F}}_q^{\times }\to {\mathbb{F}}_q^{\times }$. This map has as a kernel the subgroup ${\mu }_3$ consisting of the cube roots of unity in ${\mathbb{F}}_q^{\times }$. Let $\alpha$ be a generator of ${\mathbb{F}}_q^{\times }$. Then, if $q\equiv 1\left(\mathrm{mod}3\right)$, we have ${\mu }_3=\{1,{\alpha }^{(q-1)/3},{\alpha }^{2(q-1)/3}\}$, in which case there are $(q-1)/3$ cubic residues. Otherwise, ${\mu }_3=\left\{1\right\}$, meaning that each element of ${\mathbb{F}}_q^{\times }$ is a cubic residue. □

From now on, we assume that $q\equiv 1\left(\mathrm{mod}3\right)$. Take a cubic nonresidue $\delta \in {\mathbb{F}}_q$ and consider the cubic extension ${\mathbb{F}}_q\left(\sqrt[3]{\delta }\right)\simeq {\mathbb{F}}_{q^3}$. A basis for ${\mathbb{F}}_q\left(\sqrt[3]{\delta }\right)/{\mathbb{F}}_q$ is $\{1,{\delta }^{1/3},{\delta }^{2/3}\}$. For an element $\alpha$, we write $({\alpha }_1,{\alpha }_2,{\alpha }_3)$ to denote its components, i.e.,

$$\alpha ={\alpha }_1+{\alpha }_2{\delta }^{1/3}+{\alpha }_3{\delta }^{2/3}.$$

We define

$${\mathbb{H}}_q=\left\{(\alpha ,\beta )\in {\mathbb{F}}_q\left(\sqrt[3]{\delta }\right)\times {\mathbb{F}}_q\left(\sqrt[3]{\delta }\right):\left|\begin{array}{cc}{\alpha }_2& {\alpha }_3\\ {\beta }_2& {\beta }_3\end{array}\right|\ne 0\right\}.$$

Lemma 2. 

The following map defines an action of ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ on ${\mathbb{H}}_q$:

$$\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ r& s& t\end{array}\right](\alpha ,\beta )=\left(\frac{a\alpha +b\beta +c}{r\alpha +s\beta +t},\frac{d\alpha +e\beta +f}{r\alpha +s\beta +t}\right).$$

Proof. 

The compatibility of the action with multiplication in ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ follows from the fact that ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ has an action on the projective space ${\mathbb{P}}^2\left({\mathbb{F}}_{q^3}\right)$. Thus, we need to check that the action is closed in ${\mathbb{H}}_q$. This is a routine verification, and can be found in Appendix A. □

Our distinguished point will be $p_0=({\delta }^{2/3},{\delta }^{1/3})\in {\mathbb{H}}_q$. The action of ${Aff}_3\left({\mathbb{F}}_q\right)\le {\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ is transitive on ${\mathbb{H}}_q$, where

$${Aff}_3\left({\mathbb{F}}_q\right):=\left\{\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ 0& 0& 1\end{array}\right]\right\}.$$

With this action, we calculate that

$$K:=stab\left(p_0\right)=\left\{\left[\begin{array}{ccc}a& c\delta & b\delta \\ b& a& c\delta \\ c& b& a\end{array}\right]\right\}.$$

Indeed, we find that

$$K\stackrel{\sim }{\to }{\mathbb{F}}_q{\left({\delta }^{1/3}\right)}^{\times }:\left[\begin{array}{ccc}a& c\delta & b\delta \\ b& a& c\delta \\ c& b& a\end{array}\right]\mapsto a+b{\delta }^{1/3}+c{\delta }^{2/3}.$$

In addition, we have the homogeneous space ${\mathbb{H}}_q\simeq {\mathrm{GL}}_3\left({\mathbb{F}}_q\right)/K$. As a quick check, we can note that we have $\left|{\mathrm{GL}}_3\right({\mathbb{F}}_q\left)\right|\hspace{0.17em}=(q^3-1)(q^3-q)(q^3-q^2)$ and $\left|K\right|=q^3-1$ (from the canonical isomorphism); thus, we should have $|{\mathbb{H}}_q|$ equal to

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

Indeed, from the definition, we have

$$\left|{\mathbb{H}}_q\right|\hspace{0.17em}=\hspace{0.17em}\left|{\mathrm{GL}}_2\left({\mathbb{F}}_q\right)\right|·q^2=(q^2-1)(q^2-q)q^2=(q^2-1)(q-1)q^3.$$

Remark 1. 

Observe that the above construction is a natural generalization of the finite upper half-plane defined by Terras [6] (here, δ is a nonsquare in the finite field ${\mathbb{F}}_q$):

$$H_q=\left\{x+y\sqrt{\delta }:x,y\in {\mathbb{F}}_q,y\ne 0\right\}.$$

The condition $y\ne 0$ is equivalent to 1 and $x+y\sqrt{\delta }$ being linearly independent elements of ${\mathbb{F}}_q\left(\sqrt{\delta }\right)$ over ${\mathbb{F}}_q$.

3. Trace Formula

3.1. The Pre-Trace Formula

Our trace formula relies on the interpretation of a known result, namely, the pre-trace formula. We now derive the pre-trace formula and contextualize it in our setting of the group $G={\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$. The idea behind the pre-trace formula is to compute the trace of a specific operator in two different bases. We may construct the relevant information for this operator as follows:

Suppose that G is a finite group and $\Gamma \le G$ is a subgroup. Let $f:G\to \mathbb{C}$ be a complex-valued function on G. For a representation $\kappa :G\to \mathrm{GL}\left(V\right)$ of G (where V is a finite-dimensional complex vector space endowed with a linear G-action), define the Fourier transform of f at $\kappa$ by

$$\widehat{f}\left(\kappa \right)=\sum _{g\in G}f\left(g\right)\kappa \left(g^{-1}\right).$$

It can be observed that the Fourier transform $\widehat{f}\left(\kappa \right)$ is $EndV$-valued. The pre-trace formula is now obtained by computing the trace of $\widehat{f}\left(\rho \right)$ in two different bases with $\rho ={Ind}_{\Gamma }^{G}1$. First, we can decompose $\rho$ as a direct sum of irreducible representations of G to obtain

$$Tr\left(\widehat{f}\left(\rho \right)\right)=\sum _{\pi \in \widehat{G}}m(\pi ,\rho )Tr\left(\widehat{f}\left(\pi \right)\right).$$

(2)

Here, $\widehat{G}$ is the set of irreducible representations of G and $m(\pi ,\rho )$ is the multiplicity of the irreducible representation $\pi \in \widehat{G}$ in $\rho$. Let

$$V=\{f:G\to \mathbb{C}|f\left(\gamma g\right)=f\left(g\right)\forall \gamma \in \Gamma ,g\in G\}=L^2(\Gamma \setminus G).$$

As G acts linearly on V via right-multiplication, the corresponding representation is precisely $\rho ={Ind}_{\Gamma }^{G}1$. More concretely, for $\varphi \in V$, $g,x\in G$, we have

$$\left[\rho \left(g\right)\varphi \right]\left(x\right)=\varphi \left(xg\right).$$

Thus,

$$\left[\widehat{f}\left(\rho \right)\varphi \right]\left(x\right)=\sum _{y\in G}f\left(y\right)\varphi \left(x{y}^{-1}\right)=\sum _{u\in G}f\left(u^{-1}x\right)\varphi \left(u\right).$$

Further, as $\varphi$ is $\Gamma$-invariant,

$$\left[\widehat{f}\left(\rho \right)\varphi \right]\left(x\right)=\sum _{y\in \Gamma \setminus G}\sum _{\gamma \in \Gamma }f\left(y^{-1}\gamma x\right)\varphi \left(y\right).$$

Now, we are ready to compute the trace of $\widehat{f}\left(\rho \right)$ in another way. Consider the indicator basis ${\left\{{\delta }_y\right\}}_{y\in \Gamma \setminus G}$ of V (for $y\in \Gamma \setminus G$, ${\delta }_y$ is the function on $\Gamma \setminus G$ which outputs 1 if the input is y and 0 otherwise). With respect to this basis, $\widehat{f}\left(\rho \right)$ is represented by a matrix where the entry corresponding to $x,y\in \Gamma \setminus G$ is ${\sum }_{\gamma \in \Gamma }f\left(y^{-1}\gamma x\right)$. Therefore,

$$Tr\left[\widehat{f}\left(\rho \right)\right]=\sum _{x\in \Gamma \setminus G}\sum _{\gamma \in \Gamma }f\left(x^{-1}\gamma x\right).$$

(3)

We can rewrite the right-hand side as a sum over conjugacy classes in G. Let:

$${\Gamma }_{\gamma }=\left\{x\in \Gamma |x^{-1}\gamma x=\gamma \right\}=\mathrm{centralizer}\mathrm{of}\gamma \mathrm{in}\Gamma$$

$$G_{\gamma }=\left\{x\in G|x^{-1}\gamma x=\gamma \right\}=\mathrm{centralizer}\mathrm{of}\gamma \mathrm{in}G$$

$$\left\{\gamma \right\}=\left\{x^{-1}\gamma x|x\in G\right\}=\mathrm{conjugacy}\mathrm{class}\mathrm{of}\gamma \mathrm{in}G$$

$$C_{\Gamma }=\mathrm{set}\mathrm{of}\mathrm{conjugacy}\mathrm{classes}\mathrm{in}\Gamma .$$

Note that there is a one-to-one correspondence between the right cosets ${\Gamma }_{\gamma }\setminus \Gamma$ and elements of $\left\{\gamma \right\}$ provided by ${\Gamma }_{\gamma }x\mapsto x^{-1}\gamma x$. Because conjugacy classes partition $\Gamma$, we can rewrite the inner sum as

$$\sum _{x\in \Gamma \setminus G}\sum _{\gamma \in \Gamma }f\left(x^{-1}\gamma x\right)=\sum _{x\in \Gamma \setminus G}\sum _{\left\{\gamma \right\}\in C_{\Gamma }}\sum _{u\in {\Gamma }_{\gamma }\setminus \Gamma }f\left(x^{-1}{u}^{-1}\gamma ux\right).$$

Setting $y=ux$, we can then rewrite this as a sum over right cosets $y\in {\Gamma }_{\gamma }\setminus G$:

$$\sum _{x\in \Gamma \setminus G}\sum _{\gamma \in \Gamma }f\left(x^{-1}\gamma x\right)=\sum _{y\in {\Gamma }_{\gamma }\setminus G}\sum _{\left\{\gamma \right\}\in C_{\Gamma }}f\left(y^{-1}\gamma y\right).$$

Observing that ${\Gamma }_{\gamma }\le G_{\gamma }$, we can write $y=ts$ for $t\in {\Gamma }_{\gamma }\setminus G_{\gamma }$ and $s\in G_{\gamma }\setminus G$:

$$\sum _{x\in \Gamma \setminus G}\sum _{\gamma \in \Gamma }f\left(x^{-1}\gamma x\right)=\sum _{t\in {\Gamma }_{\gamma }\setminus G_{\gamma }}\sum _{s\in G_{\gamma }\setminus G}\sum _{\left\{\gamma \right\}\in C_{\Gamma }}f\left(s^{-1}{t}^{-1}\gamma ts\right)=\sum _{\left\{\gamma \right\}\in C_{\Gamma }}\frac{|G_{\gamma }|}{|{\Gamma }_{\gamma }|}\sum _{s\in G_{\gamma }\setminus G}f\left(s^{-1}\gamma s\right).$$

(4)

We define the orbital sum $I_G(f,\gamma )$ of f at $\gamma$ to be

$$I_G(f,\gamma ):=\sum _{s\in G_{\gamma }\setminus G}f\left(s^{-1}\gamma s\right).$$

The pre-trace formula now follows from combining (2)–(4):

$$\sum _{\pi \in \widehat{G}}m(\pi ,\rho )Tr\left(\widehat{f}\left(\pi \right)\right)=\sum _{\left\{\gamma \right\}\in C_{\Gamma }}\frac{|G_{\gamma }|}{|{\Gamma }_{\gamma }|}{I}_G(f,\gamma )$$

(5)

where $\rho ={Ind}_{\Gamma }^{G}1$ and $I_G(f,\gamma )$ is the orbital sum of f at $\gamma$ as defined above. Note that (5) can be viewed as a representation-theoretic analogue of Selberg’s trace formula [1,6]; the left-hand side can be thought of as the spectral side, summing over the irreducible representations of G and their multiplicities in $\rho$, while the right-hand side can be interpreted as the geometric side, summing over conjugacy classes in G. For further references, see [6] (Ch. 22).

To obtain the trace formula for the finite upper half-space, we take $G={\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ and $\Gamma ={\mathrm{GL}}_3\left({\mathbb{F}}_p\right)$. In addition, $f:G\to \mathbb{C}$ is chosen to be a $K-$bi-invariant function (K is the stabilizer of $p_0$, as above), i.e., $\forall k,h\in K$ and $\forall x\in G$ we have $f\left(kxh\right)=f\left(x\right)$. One can also think of f as a function on $G/K\simeq {\mathbb{H}}_q$ which is invariant under left-multiplication by elements in K.

Remark 2. 

The left-hand side of (5) can be expressed solely in terms of the characters ${\chi }_{\pi }$ of the group G:

$$\begin{array}{cc}\hfill Tr\left[\widehat{f}\left(\pi \right)\right]& =Tr\left[\sum _{g\in G}f\left(g\right)\pi \left(g^{-1}\right)\right]=\sum _{g\in G}f\left(g\right)Tr\left[\pi \left(g^{-1}\right)\right]\hfill \\ & =\sum _{g\in G}f\left(g\right){\chi }_{\pi }\left(g^{-1}\right)=\sum _{g\in G}f\left(g\right)\overline{{\chi }_{\pi }\left(g\right)}=⟨f,{\chi }_{\pi }⟩.\hfill \end{array}$$

Here, $⟨x,y⟩$ denotes the usual Hermitian inner product on the complex vector space $L^2\left(G\right)$:

$$⟨x,y⟩=\sum _{g\in G}x\left(g\right)\overline{y\left(g\right)}$$

.

In Section 4, Section 5 and Section 6, we compute each term in the sum on the right-hand side of the pre-trace formula (5) using the explicit description of conjugacy classes of ${\mathrm{GL}}_3\left({\mathbb{F}}_p\right)$ provided in [8]. The conjugacy classes are separated into the cases of central, hyperbolic, parabolic, and elliptic terms. It is helpful to introduce one more tool that will help us simplify these computations.

3.2. Double Cosets and Fundamental Domains

To calculate the orbital sums, it is convenient to identify $G_{\gamma }\setminus G$ with $G_{\gamma }\setminus G/K\times K/Z$ (when possible), where Z is the center of G (the subgroup of diagonal matrices $\{aI:a\in {\mathbb{F}}_q^{\times }\}$). Let ${\left\{t_i\right\}}_{i=1}^l$ be representatives of double cosets in $G_{\gamma }\setminus G/K$, and consider the following map:

$$\begin{array}{cc}\hfill G_{\gamma }\setminus G/K\times K/Z& \to G_{\gamma }\setminus G\hfill \\ \hfill (G_{\gamma }{t}_{i}K,kZ)& \mapsto G_{\gamma }{t}_{i}k.\hfill \end{array}$$

We first show that this map is well-defined and onto. If $z\in Z$, then $G_{\gamma }{t}_{i}kz=G_{\gamma }z{t}_{i}k=G_{\gamma }{t}_{i}k$, because $z\in G_{\gamma }$ (as z commutes with every element of G). Now, let $G_{\gamma }x\in G\setminus G_{\gamma }$; then, because double cosets $G_{\gamma }{t}_{i}K$ partition G and are unions of right cosets $G_{\gamma }y$, we can find a representative $t_i\in G$ and $k\in K$ such that $G_{\gamma }x=G_{\gamma }{t}_{i}k$.

For injectivity, let $t_i,t_j$ be representatives of two double cosets in $G_{\gamma }\setminus G/K$ and let $k,k^{\prime }$ be representatives of two left cosets in $K/Z$. Suppose that $G_{\gamma }{t}_{i}k=G_{\gamma }{t}_j{k}^{\prime }.$ Because double cosets are either disjoint or identical, this is only possible if $t_i=t_j$ while letting $t=t_i$. Thus, we have

$$G_{\gamma }=G_{\gamma }t{k}^{\prime }{k}^{-1}{t}^{-1},$$

(6)

which happens if and only if $t{k}^{\prime }{k}^{-1}{t}^{-1}\in G_{\gamma }$, or equivalently, $k^{\prime }{k}^{-1}\in t{G}_{\gamma }{t}^{-1}\cap K$. This can also be stated as $kH=k^{\prime }H$, where $H=t{G}_{\gamma }{t}^{-1}\cap K$. Note that $Z\le K$ and $Z\le t{G}_{\gamma }{t}^{-1}=G_{t\gamma t^{-1}}$, as Z is the intersection of all centralizers in G; hence, we always have $Z\le K\cap t{G}_{\gamma }{t}^{-1}=H$. If $Z=H$, then (6) is equivalent to $kZ=k^{\prime }Z$, which implies that the above map is injective. In our specific example, the centralizers $G_{\gamma }$ contain no non-central elements conjugate to some matrix in K unless $\gamma$ itself is conjugate to a member of K (this can be seen by, e.g., taking simple representatives $\gamma$ of conjugacy classes in G and looking at the characteristic polynomials and eigenvalues of elements in $G_{\gamma }$, which are preserved by conjugation).

Thus, we have two cases:

Case 1: $\gamma$ is not conjugate to an element of K. Then, we have

$$I_G(f,\gamma )=\sum _{s\in G_{\gamma }\setminus G}f\left(s^{-1}\gamma s\right)=\sum _{t\in G_{\gamma }\setminus G/K}\sum _{k\in K/Z}f\left({\left(tk\right)}^{-1}\gamma tk\right)=|K/Z|\sum _{t\in G_{\gamma }\setminus G/K}f\left(t^{-1}\gamma t\right).$$

The last equality follows from the fact that f is $K-$bi-invariant. We can identify $G_{\gamma }\setminus G/K$ with $G_{\gamma }\setminus {\mathbb{H}}_q$. In order to calculate orbital sums, we find fundamental domains for the action of $G_{\gamma }$ on ${\mathbb{H}}_q$, i.e., subsets of ${\mathbb{H}}_q$ which contain exactly one element from each $G_{\gamma }$-orbit.

Case 2: $\gamma$ is conjugate to an element of K. In this case, the above map is not necessarily injective; however, $G_{\gamma }=G\mathrm{or}K$, depending on whether $\gamma$ is conjugate to a central element of K or not. Then, $G_{\gamma }\setminus G$ is either trivial or can be identified with ${\mathbb{H}}_q$.

4. Central and Hyperbolic Terms

4.1. Central Terms

Let $\gamma =\left[\begin{array}{ccc}a& 0& 0\\ 0& a& 0\\ 0& 0& a\end{array}\right]$ for $a\in {\mathbb{F}}_p^{\times }$. In this case, $G_{\gamma }=G$ and ${\Gamma }_{\gamma }=\Gamma$. Thus, $I_G(f,\gamma )=f\left(p_0\right)$, and the total contribution from such $\gamma$ is

$$\frac{\left|G\right|}{|\Gamma |}f\left(p_0\right)·(p-1)=\frac{(q^3-1)(q^2-1)(q-1)q^3}{(p^3-1)(p^2-1)p^3}f\left(p_0\right).$$

4.2. Hyperbolic Terms of the First Kind

Let $\gamma =\left[\begin{array}{ccc}a& 0& 0\\ 0& a& 0\\ 0& 0& b\end{array}\right],a\ne b\in {\mathbb{F}}_p^{\times }.$ Here,

$$G_{\gamma }=\left\{\left[\begin{array}{ccc}x& y& 0\\ w& z& 0\\ 0& 0& t\end{array}\right]\right\}\simeq {\mathrm{GL}}_2\left({\mathbb{F}}_q\right)\times {\mathbb{F}}_q^{\times }$$

and similarly for ${\Gamma }_{\gamma }$.

A fundamental domain for the action of $G_{\gamma }$ is provided by

$$\begin{array}{c}\hfill G_{\gamma }\setminus {\mathbb{H}}_q=\{(u+{\delta }^{1/3},v+{\delta }^{2/3}):u,v\in {\mathbb{F}}_q\}.\end{array}$$

(7)

We verify this in the next proposition for the sake of illustration. In the next sections, the verification of the fundamental domains is relegated to Appendix A.

Proposition 1. 

A fundamental domain for $G_{\gamma }$ is provided by (7).

Proof. 

First, to show the uniqueness of each representative, suppose that there exists $m\in G_{\gamma }$ such that

$$m·(u_1+{\delta }^{1/3},v_1+{\delta }^{2/3})=(u_2+{\delta }^{1/3},v_2+{\delta }^{2/3}).$$

Then, we deduce that $y=0=w$ and $x=t=z$; thus, $u_1=u_2$ and $v_1=v_2$.

Next, to show the completeness of this fundamental domain, we can take an arbitrary element $(\alpha ,\beta )\in {\mathbb{H}}_q$. We wish to find $m\in G_{\gamma },u,v\in {\mathbb{F}}_q$ such that $m·(u+{\delta }^{1/3},v+{\delta }^{2/3})=(\alpha ,\beta )$. Setting $x={\alpha }_2,y={\alpha }_3,w={\beta }_2,z={\beta }_3$, and $t=1$, it can be seen that $u,v$ is a solution to

$$\begin{array}{c}\hfill \left[\begin{array}{cc}{\alpha }_2& {\alpha }_3\\ {\beta }_2& {\beta }_3\end{array}\right]\left[\begin{array}{c}u\\ v\end{array}\right]=\left[\begin{array}{c}{\alpha }_1\\ {\beta }_1\end{array}\right],\end{array}$$

which has a solution because the left matrix is invertible. □

Thus,

$$\begin{array}{cc}\hfill I_G(f,\gamma )& =\sum _{t\in G_{\gamma }\setminus G/K}\sum _{u\in K/\left\{aI\right\}}f\left({\left(tu\right)}^{-1}\gamma \left(tu\right)\right)\hfill \\ \hfill & =\sum _{t\in G_{\gamma }\setminus {\mathbb{H}}_q}\frac{\left|K\right|}{\left|\right\{aI\left\}\right|}f\left(t^{-1}\gamma t\right)\hfill \\ \hfill & =\frac{q^3-1}{q-1}\sum _{x,y\in {\mathbb{F}}_q}f\left(\left[\begin{array}{ccc}0& 1& -y\\ 1& 0& -x\\ 0& 0& 1\end{array}\right]\gamma \left[\begin{array}{ccc}0& 1& x\\ 1& 0& y\\ 0& 0& 1\end{array}\right]p_0\right)\hfill \\ \hfill & =(q^2+q+1)\sum _{x,y\in {\mathbb{F}}_q}f\left(\left[\begin{array}{ccc}a& 0& (a-b)y\\ 0& a& (a-b)x\\ 0& 0& b\end{array}\right]p_0\right)\hfill \\ \hfill & =(q^2+q+1)\sum _{x,y\in {\mathbb{F}}_q}f\left((x+\frac{a}{b}{\delta }^{2/3},y+\frac{a}{b}{\delta }^{1/3})\right).\hfill \end{array}$$

Let us define the horocycle transform $Hf:{\mathrm{GL}}_2\left({\mathbb{F}}_q\right)\to \mathbb{C}$ of $f\in L^2(K\setminus G/K)$ by

$$Hf\left(\kappa \right):=\sum _{x,y\in {\mathbb{F}}_q}\sum _{\xi \in \left\{\kappa \right\}}f\left(\left[\begin{array}{ccc}{\xi }_{11}& {\xi }_{12}& x\\ {\xi }_{21}& {\xi }_{22}& y\\ 0& 0& 1\end{array}\right]p_0\right),$$

where the inner sum runs over $\xi \in {\mathrm{GL}}_2\left({\mathbb{F}}_q\right)$ in the conjugacy class of $\kappa$. Thus, this hyperbolic term is equal to

$$\begin{array}{cc}\hfill & (q^2+q+1)\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}\frac{(q^2-1)(q^2-q)(q-1)}{(p^2-1)(p^2-p)(p-1)}·Hf\left(\frac{a}{b}I\right)\hfill \\ \hfill & =\frac{(q^3-1)(q^2-1)(q^2-q)}{(p^2-1)(p^2-p)(p-1)}\sum _{\begin{array}{c}a\in {\mathbb{F}}_p^{\times }\\ a\ne 1\end{array}}(p-1)·Hf\left(aI\right)\hfill \\ \hfill & =\frac{(q^3-1)(q^2-1)(q^2-q)}{(p^2-1)(p^2-p)}\sum _{\begin{array}{c}a\in {\mathbb{F}}_p^{\times }\\ a\ne 1\end{array}}Hf\left(aI\right).\hfill \end{array}$$

We have shown the full computation for the orbital sums for this case both to demonstrate the computational technique and to motivate and introduce the horocycle transform. For the next cases, we omit detailed computations for orbital sums; however, some are included in Appendix A for the curious reader.

4.3. Hyperbolic Terms of the Second Kind

Let $\gamma =\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]$, with $a,b,c\in {\mathbb{F}}_p^{\times }$ being distinct. Here, $G_{\gamma }$ is the set of diagonal matrices and

$$\begin{array}{cc}\hfill G_{\gamma }\setminus {\mathbb{H}}_q& =\{(x+{\delta }^{1/3}+y{\delta }^{2/3},r+s{\delta }^{1/3}+{\delta }^{2/3}):ys\ne 1\}\bigsqcup \{(x+{\delta }^{2/3},r+{\delta }^{1/3}+s{\delta }^{2/3}):s\ne 0\}\hfill \\ \hfill & \bigsqcup \{(x+y{\delta }^{1/3}+{\delta }^{2/3},r+{\delta }^{1/3}):y\ne 0\}\bigsqcup \left\{(x+{\delta }^{2/3},y+{\delta }^{1/3})\right\}.\hfill \end{array}$$

We have

$$I_G(f,\gamma )=(q^2+q+1)·Hf\left(\left[\begin{array}{cc}a/c& 0\\ 0& b/c\end{array}\right]\right),$$

where $a,b,c\in {\mathbb{F}}_p^{\times }$ comes from a change of variables outlined in Appendix A.

The total contribution from the second hyperbolic term is

$$\frac{1}{6}(q^2+q+1)\frac{{(q-1)}^3}{{(p-1)}^3}\sum _{\begin{array}{c}a,b,c\in {\mathbb{F}}_p^{\times }\\ a\ne b\ne c\end{array}}Hf\left(\left[\begin{array}{cc}a/c& 0\\ 0& b/c\end{array}\right]\right)=\frac{1}{6}(q^2+q+1)\frac{{(q-1)}^3}{{(p-1)}^2}\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\setminus \left\{1\right\}\\ a\ne b\end{array}}Hf\left(\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]\right)$$

$$=\frac{1}{3}(q^3-1)\frac{{(q-1)}^2}{{(p-1)}^2}\sum _{\begin{array}{c}\{a,b\}\subseteq {\mathbb{F}}_p^{\times }\setminus \left\{1\right\}\end{array}}Hf\left(\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]\right).$$

5. Parabolic Terms

5.1. Parabolic Terms of the First Kind

Let $\gamma =\left[\begin{array}{ccc}a& 0& a\\ 0& a& 0\\ 0& 0& a\end{array}\right],a\in {\mathbb{F}}_p^{\times }$. In this case,

$$G_{\gamma }=\left\{\left[\begin{array}{ccc}d& y& x\\ 0& c& b\\ 0& 0& d\end{array}\right]:c,d\in {\mathbb{F}}_q^{\times },b,x,y\in {\mathbb{F}}_q\right\}.$$

Proposition 2. 

A fundamental domain for $G_{\gamma }$ is

$$G_{\gamma }\setminus {\mathbb{H}}_q=\{(u{\delta }^{1/3},v{\delta }^{1/3}+{\delta }^{2/3}):u,v\in {\mathbb{F}}_q,u\ne 0\}\bigsqcup \{({\delta }^{1/3}+u{\delta }^{2/3},{\delta }^{1/3}):u\in {\mathbb{F}}_q^{\times }\}.$$

Proof. 

See Appendix A. □

Following the computations outlined in Appendix A, the orbital sum in this case is

$$I_G(f,\gamma )=(q^2+q+1)\left(Hf\left(I\right)-f\left(p_0\right)\right).$$

In total, the contribution from the parabolic terms of the first kind is

$$\frac{q^3{(q-1)}^2}{p^3{(p-1)}^2}(q^2+q+1)\left(Hf\left(I\right)-f\left(p_0\right)\right)·(p-1)=\frac{q^3(q^3-1)(q-1)}{p^3(p-1)}\left(Hf\left(I\right)-f\left(p_0\right)\right).$$

5.2. Parabolic Terms of the Second Kind

Let $\gamma =\left[\begin{array}{ccc}a& a& 0\\ 0& a& a\\ 0& 0& a\end{array}\right],a\in {\mathbb{F}}_p^{\times }$. We find that

$$G_{\gamma }=B=\left\{\left[\begin{array}{ccc}z& y& x\\ 0& c& b\\ 0& 0& d\end{array}\right]:d,c,z\in {\mathbb{F}}_q^{\times },b,x,y\in {\mathbb{F}}_q\right\}.$$

Here, B is the Borel subgroup of upper-triangular matrices.

Proposition 3. 

A fundamental domain for B is

$$B\setminus {\mathbb{H}}_q=\{({\delta }^{1/3},v{\delta }^{1/3}+{\delta }^{2/3}):v\in {\mathbb{F}}_q\}\bigsqcup \left\{({\delta }^{2/3},{\delta }^{1/3})\right\}.$$

Proof. 

See Appendix A. □

We find that the orbital sum (see Appendix A) is

$$I_G(f,\gamma )=(q^2+q+1)\left(f\left({\delta }^{2/3}+{\delta }^{1/3},{\delta }^{1/3}+1\right)+\sum _{v\in {\mathbb{F}}_q}f\left((1-v){\delta }^{2/3}-v^2{\delta }^{1/3}+1,{\delta }^{2/3}+(v+1){\delta }^{1/3}\right)\right).$$

Note that the orbital sum is independent of a; therefore, we denote each $I_G(f,\gamma )$ by ${\tilde{I}}_G\left(f\right)$. Thus, in total, the contribution of these parabolic terms is

$$\frac{q^3{(q-1)}^2}{p^3{(p-1)}^2}{\tilde{I}}_G\left(f\right)·(p-1)=\frac{q^3{(q-1)}^2}{p^3(p-1)}{\tilde{I}}_G\left(f\right).$$

5.3. Parabolic Terms of the Third Kind

Let $\gamma =\left[\begin{array}{ccc}a& 0& 0\\ a& a& 0\\ 0& 0& b\end{array}\right]$ with $a\ne b\in {\mathbb{F}}_p^{\times }$. Here,

$$G_{\gamma }=\left\{\left[\begin{array}{ccc}x& 0& 0\\ y& x& 0\\ 0& 0& z\end{array}\right]:x,z\in {\mathbb{F}}_q^{\times },y\in {\mathbb{F}}_q\right\}.$$

Again, by a change of variables, we find the following orbital sum (see Appendix A):

$$I_G(f,\gamma )=(q^2+q+1)·Hf\left(\left[\begin{array}{cc}a/b& 1\\ 0& a/b\end{array}\right]\right).$$

The overall contribution of these terms is

$$\begin{array}{cc}\hfill \sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}\frac{q{(q-1)}^2}{p{(p-1)}^2}(q^2+q+1)·Hf\left(\left[\begin{array}{cc}a/b& 1\\ 0& a/b\end{array}\right]\right)& =\frac{q{(q-1)}^2}{p{(p-1)}^2}(q^2+q+1)\sum _{\begin{array}{c}c\in {\mathbb{F}}_p^{\times }\\ c\ne 1\end{array}}(p-1)·Hf\left(\left[\begin{array}{cc}c& 1\\ 0& c\end{array}\right]\right)\hfill \\ \hfill & =\frac{q(q-1)(q^3-1)}{p(p-1)}\sum _{\begin{array}{c}c\in {\mathbb{F}}_p^{\times }\\ c\ne 1\end{array}}Hf\left(\left[\begin{array}{cc}c& 1\\ 0& c\end{array}\right]\right).\hfill \end{array}$$

6. Elliptic Terms

6.1. Elliptic Terms of the First Kind

Irreducible $\gamma$. The characteristic polynomial of such a $\gamma$ is irreducible in ${\mathbb{F}}_p\left[t\right]$. Recall that $q=p^n$. Depending on whether $n\equiv 0\left(\mathrm{mod}3\right)$, there are two cases to consider.

• $n\equiv 0\left(\mathrm{mod}3\right)$. In this case, we have the tower of field extensions ${\mathbb{F}}_q/{\mathbb{F}}_{p^3}/{\mathbb{F}}_p$. Thus, $\gamma$ is similar to a diagonal matrix in ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$. However, the same is not the case in ${\mathrm{GL}}_3\left({\mathbb{F}}_p\right)$, which is why this case is different from the second hyperbolic term. Therefore, $G_{\gamma }$ is the subgroup of diagonal matrices over ${\mathbb{F}}_q$, while ${\Gamma }_{\gamma }$ is K with entries in ${\mathbb{F}}_p$, i.e., $\left|{\Gamma }_{\gamma }\right|=p^3-1$. Because $G_{\gamma }$ is the same, we can reuse the fundamental domain for $G_{\gamma }$ from the earlier second hyperbolic case.
  Suppose that $\gamma$ is similar to $\left(\begin{array}{ccc}{\xi }_1& 0& 0\\ 0& {\xi }_2& 0\\ 0& 0& {\xi }_3\end{array}\right)$ in ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$, where ${\xi }_1,{\xi }_2,{\xi }_3\in {\mathbb{F}}_{p^3}\setminus {\mathbb{F}}_p$. Similar to the second hyperbolic term, the orbital sum will be

  $$I_G(f,\gamma )=|K/Z|·Hf\left(\left[\begin{array}{cc}{\xi }_1/{\xi }_3& 0\\ 0& {\xi }_2/{\xi }_3\end{array}\right]\right).$$
  Because ${\xi }_1,{\xi }_2,{\xi }_3$ are Galois conjugates, without loss of generality, ${\xi }_1={\xi }_3^p$ and ${\xi }_2={\xi }_3^{p^2}$. Therefore, the total contribution of the first elliptic terms is

  $$\frac{1}{3}·\frac{{(q-1)}^3}{p^3-1}·\frac{q^3-1}{q-1}\sum _{\xi \in {\mathbb{F}}_{p^3}^{\times }\setminus {\mathbb{F}}_p^{\times }}Hf\left(\left[\begin{array}{cc}{\xi }^{p-1}& 0\\ 0& {\xi }^{p^2-1}\end{array}\right]\right).$$
  Note that the factor of $\frac{1}{3}$ here comes from the fact that each $\gamma$ is counted three times in the sum.
• $n\not\equiv 0\left(\mathrm{mod}3\right)$. In this case, there is no cubic extension intermediate to ${\mathbb{F}}_q/{\mathbb{F}}_p$; thus, ${\mathbb{F}}_q\left({\delta }^{1/3}\right)$ is the minimal field (in the sense of containment, of course, not size) over which the characteristic polynomial of $\gamma$ has a root. Suppose that an eigenvalue of $\gamma$ is ${\alpha }_1+{\alpha }_2{\delta }^{1/3}+{\alpha }_3{\delta }^{2/3}$ with eigenvector $x_1+x_2{\delta }^{1/3}+x_3{\delta }^{2/3}$, where ${\alpha }_i\in {\mathbb{F}}_q$ and $x_i\in {\mathbb{F}}_q^3$. Then,

  $$\begin{array}{cc}\hfill \gamma x_1& ={\alpha }_1{x}_1+{\alpha }_3\delta x_2+{\alpha }_2\delta x_3,\hfill \\ \hfill \gamma x_2& ={\alpha }_2{x}_1+{\alpha }_1{x}_2+{\alpha }_3\delta x_3,\hfill \\ \hfill \gamma x_3& ={\alpha }_3{x}_1+{\alpha }_2{x}_2+{\alpha }_1{x}_3.\hfill \end{array}$$
  Thus, ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ when viewed in $\gamma$ is similar to a matrix in K. Conversely, non-diagonal matrices in K provide irreducible elements in ${\mathrm{GL}}_3\left({\mathbb{F}}_p\right)$. Further, in this case $G_{\gamma }$ is precisely K; thus, $G_{\gamma }\setminus G={\mathbb{H}}_q$. After identifying elements $y\in {\mathbb{H}}_q$ with elements $y=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ 0& 0& 1\end{array}\right]\in {Aff}_3\left({\mathbb{F}}_q\right)$ and writing $\gamma =\left[\begin{array}{ccc}\alpha & \epsilon \delta & \beta \delta \\ \beta & \alpha & \epsilon \delta \\ \epsilon & \beta & \alpha \end{array}\right]$, we can compute

  $$\begin{array}{cc}\hfill I_G(f,\gamma )& =\sum _{y\in {\mathbb{H}}_q}f\left(y^{-1}\gamma y\right)\hfill \\ \hfill & =\sum _{a,b,c,d,e,f\in {\mathbb{F}}_q}f\left(\frac{1}{bd-ae}\left[\begin{array}{ccc}-e& b& ce-bf\\ d& -a& af-cd\\ 0& 0& bd-ae\end{array}\right]\left[\begin{array}{ccc}\alpha & \epsilon \delta & \beta \delta \\ \beta & \alpha & \epsilon \delta \\ \epsilon & \beta & \alpha \end{array}\right]\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ 0& 0& 1\end{array}\right]p_0\right)\hfill \\ \hfill & =\sum _{a,b,c,d,e,f\in {\mathbb{F}}_q}f\left(\frac{1}{bd-ae}\left[\begin{array}{ccc}{x}_{11}& x_{12}& x_{13}\\ x_{21}& x_{22}& x_{23}\\ x_{31}& x_{32}& x_{33}\end{array}\right]p_0\right),\hfill \end{array}$$

  where the entries of the matrix, by column, are as follows:

  ◇

  $x_{11}=\alpha (bd-ae)+\beta (ab+cde-bdf)+\epsilon (ace-abf-de\delta )$

  ◇

  $x_{21}=\beta (adf-a^2-c{d}^2)+\epsilon (a^{2}f-acd+d^2\delta )$

  ◇

  $x_{31}=\beta d+\epsilon a$

  ◇

  $x_{12}=\beta (c{e}^2-bef+b^2)+\epsilon (bce-b^{2}f-e^2\delta )$

  ◇

  $x_{22}=\alpha (bd-ae)+\beta (aef-cde-ab)+\epsilon (abf-bcd+de\delta )$

  ◇

  $x_{32}=\beta e+\epsilon b$

  ◇

  $x_{13}=\beta (cef+bc-b{f}^2-e\delta )$

  ◇

  $x_{23}=\beta (a{f}^2-cdf-ac+d\delta )+\epsilon (acf-a{c}^2+df\delta -a\delta )$

  ◇

  $x_{33}=\alpha +\beta f+\epsilon c$.

  Unfortunately, there does not seem to be an obvious way to simplify this expression further. When computing this case for a specific field ${\mathbb{F}}_q$, it seems that applying this formula may not be more helpful than computing directly.

6.2. Elliptic Terms of the Second Kind

$\gamma =\left[\begin{array}{ccc}a& b& 0\\ c& d& 0\\ 0& 0& e\end{array}\right]$, where $g=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is irreducible and $e\in {\mathbb{F}}_p^{\times }$. In addition,

$${\Gamma }_{\gamma }=\left\{\left[\begin{array}{ccc}a& b\xi & 0\\ b& a& 0\\ 0& 0& c\end{array}\right]:c(a^2-b^2\xi )\ne 0\right\},$$

where $\xi$ is a nonsquare in ${\mathbb{F}}_p$.

Again, depending on whether $n\equiv 0\left(\mathrm{mod}2\right)$, i.e., whether n is even or odd, there are two cases to consider.

• n is even. In this case, ${\mathbb{F}}_p\subset {\mathbb{F}}_{p^2}\subseteq {\mathbb{F}}_q$ and $\gamma$ is diagonalizable in ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$. $G_{\gamma }$ is a subgroup of diagonal matrices over ${\mathbb{F}}_q$, meaning that the fundamental domain $G_{\gamma }\setminus {\mathbb{H}}_q$ is the same as for the second hyperbolic case.
  Suppose that $\gamma$ is similar to $\left(\begin{array}{ccc}{\eta }_1& 0& 0\\ 0& {\eta }_2& 0\\ 0& 0& e\end{array}\right)$ in G, where ${\eta }_1,{\eta }_2\in {\mathbb{F}}_{p^2}\setminus {\mathbb{F}}_p$ and $e\in {\mathbb{F}}_p^{\times }$. Then, (analogously to the second hyperbolic case) the orbital sum will be

  $$I_G(f,\gamma )=|K/Z|·Hf\left(\left[\begin{array}{cc}{\eta }_1/e& 0\\ 0& {\eta }_2/e\end{array}\right]\right).$$
  As ${\eta }_1,{\eta }_2$ are Galois conjugates, without loss of generality we may assume that ${\eta }_1={\eta }_2^p$. Therefore, the total contribution of the second elliptic terms will be

  $$\frac{1}{2}·\frac{{(q-1)}^3}{(p-1)(p^2-1)}·\frac{q^3-1}{q-1}\sum _{e\in {\mathbb{F}}_p^{\times }}\sum _{\eta \in {\mathbb{F}}_{p^2}^{\times }\setminus {\mathbb{F}}_p^{\times }}Hf\left(\left[\begin{array}{cc}\eta /e& 0\\ 0& {\eta }^p/e\end{array}\right]\right)$$

  $$=\frac{1}{2}·\frac{{(q-1)}^3}{(p^2-1)}·\frac{q^3-1}{q-1}\sum _{\eta \in {\mathbb{F}}_{p^2}^{\times }\setminus {\mathbb{F}}_p^{\times }}Hf\left(\left[\begin{array}{cc}\eta & 0\\ 0& {\eta }^p\end{array}\right]\right).$$
• n is odd. In this case, $\gamma$ is not diagonalizable in ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$. In general, it will be conjugate to an element of the following form:

  $$\left[\begin{array}{ccc}k& l\xi & 0\\ l& k& 0\\ 0& 0& m\end{array}\right]$$

  where $k\in {\mathbb{F}}_p$, $m,l,\xi \in {\mathbb{F}}_p^{\times }$ such that $\xi$ is a nonsquare in ${\mathbb{F}}_p$ (this ensures that the determinant $m(k^2-l^2\xi )$ is nonzero). In this case, we have

  $$G_{\gamma }=\left\{\left[\begin{array}{ccc}a& b\xi & 0\\ b& a& 0\\ 0& 0& c\end{array}\right]:a,b,c\in {\mathbb{F}}_q,c(a^2-b^2\xi )\ne 0\right\}.$$

  We seek a fundamental domain for this specific $G_{\gamma }$.

  Proposition 4. 

  A fundamental domain for $G_{\gamma }$ is provided by

  $$G_{\gamma }\setminus {\mathbb{H}}_q=\left\{\left(x+u{\delta }^{1/3}+v{\delta }^{2/3},y+{\delta }^{1/3}\right):v\in {\mathbb{F}}_q^{\times },x,y,u\in {\mathbb{F}}_q\right\}.$$

  The corresponding orbital sum is

  $$I_G(f,\gamma )=\frac{(q^3-1)(q^2-1)}{(p-1)(p^2-1)}Hf\left(\left[\begin{array}{cc}k/m& l\xi /m\\ l/m& k/m\end{array}\right]\right).$$

  The total contribution from the second elliptic terms in this case is

  $$\frac{(q^3-1)(q^2-1)}{(p^2-1)}\sum _{\begin{array}{c}s\in {\mathbb{F}}_p\\ t\in {\mathbb{F}}_p^{\times }\end{array}}Hf\left(\left[\begin{array}{cc}s& t\xi \\ t& s\end{array}\right]\right).$$

  A detailed proof of both the proposition and the orbital sum calculation can be found in Appendix A.

7. Examples

Throughout this section, subscripts under multiplicities indicate which group is the domain of the corresponding representations. We set $\rho ={Ind}_{\Gamma }^{G}1$ and $\kappa ={Ind}_K^{G}1$.

Example 1. 

Let $f:G\to \mathbb{C}$ be the constant one function. The right-hand side of the trace formula is $\left|G\right|$, while the left-hand side (by orthogonality of characters) becomes

$$\sum _{\pi \in \widehat{G}}m{(\pi ,\rho )}_G⟨f,{\chi }_{\pi }⟩=\sum _{\pi \in \widehat{G}}m{(\pi ,\rho )}_G⟨{\chi }_1,{\chi }_{\pi }⟩=\left|G\right|·m{(1,\rho )}_G.$$

Thus, the trace formula says $\left|G\right|·m(1,\rho )=\left|G\right|$, which is a manifestation of Frobenius reciprocity:

$$m{(1,\rho )}_G=\frac{1}{\left|G\right|}{⟨1,{Ind}_{\Gamma }^{G}1⟩}_G=\frac{1}{|\Gamma |}{⟨{Res}_{\Gamma }^{G}1,1⟩}_{\Gamma }=m{(1,1)}_{\Gamma }=1.$$

Example 2. 

Let $f:{\mathbb{H}}_q\to \mathbb{C}$ be the indicator function ${\delta }_{p_0}$, which is 1 on $p_0$ and zero otherwise. It is $K-$bi-invariant because $stab\left(p_0\right)=K$. On the left-hand side of the trace formula, we have

$$\begin{array}{cc}\hfill \sum _{\pi \in \widehat{G}}m{(\pi ,\rho )}_G⟨f,{\chi }_{\pi }⟩& =\sum _{\pi \in \widehat{G}}m{(\pi ,\rho )}_G\sum _{g\in G}f(g·p_0){\chi }_{\pi }\left(g^{-1}\right)=\sum _{\pi \in \widehat{G}}m{(\pi ,\rho )}_G\sum _{k\in K}{\chi }_{\pi }\left(k^{-1}\right)\hfill \\ \hfill & =\sum _{\pi \in \widehat{G}}m{(\pi ,\rho )}_G{⟨1,{\chi }_{\pi }⟩}_K=\left|K\right|·\sum _{\pi \in \widehat{G}}m{(\pi ,\rho )}_G·m{(1,{Res}_K^G\pi )}_K\hfill \\ \hfill & =\left|K\right|·\sum _{\pi \in \widehat{G}}m{(\pi ,\rho )}_G·m{(\pi ,\kappa )}_G.\hfill \end{array}$$

On the right-hand side, the only nonzero terms are the central terms and the first elliptic terms when n is not divisible by 3. If that is the case, we claim that for a non-central γ in K we have $I_G(f,\gamma )=3.$

Suppose that γ corresponds to multiplication by $\xi \in {\mathbb{F}}_q{\left({\delta }^{1/3}\right)}^{\times }$ under the isomorphism $K\simeq {\mathbb{F}}_q{\left({\delta }^{1/3}\right)}^{\times }$. When viewed as an element of ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$, the eigenvalues of γ are the Galois conjugates of ξ, which are ξ, ${\xi }^q$, and ${\xi }^{q^2}$ in our setting. This means that the conjugates of γ lying in K are precisely ${\gamma }^q$ and ${\gamma }^{q^2}$; hence, there will be exactly three non-zero terms in each orbital sum $I_G(f,\gamma )$.

The total contribution of the first elliptic terms is

$$\frac{1}{3}·\frac{q^3-1}{p^3-1}\sum _{\begin{array}{c}\gamma \in K\\ \gamma \notin Z\end{array}}{I}_G(f,\gamma )=\frac{(q^3-1)(q^3-q)}{p^3-1}.$$

Thus, we obtain the following cases:

Case 1: n is divisible by 3; then, we have

$$\sum _{\pi \in \widehat{G}}m{(\pi ,\rho )}_G·m{(\pi ,\kappa )}_G=\frac{q^3(q^2-1)(q-1)}{p^3(p^2-1)(p^3-1)}.$$

Case 2: n is not divisible by 3; then, we have

$$\sum _{\pi \in \widehat{G}}m{(\pi ,\rho )}_G·m{(\pi ,\kappa )}_G=\frac{q^3(q^2-1)(q-1)}{p^3(p^2-1)(p^3-1)}+\frac{q^3-q}{p^3-1}.$$

Remark 3. 

Note that the sum on the left-hand side can also be written as

$$\sum _{\pi \in \widehat{G}}m{(\pi ,\rho )}_G·m{(\pi ,\kappa )}_G=\frac{1}{\left|G\right|}⟨{\chi }_{\rho },{\chi }_{\kappa }⟩.$$

8. Character of the Representation $\mathit{\rho }={\mathbf{Ind}}_{\mathbf{\Gamma }}^{\mathit{G}}\mathbf{1}$

Having effectively computed the right-hand side of the pre-trace formula (5), we now turn to the left-hand side. For this, we require the character ${\chi }_{\rho }$ of $\rho$, which we now calculate. We use the usual formula for the character of an induced representation (see [6] (Ch. 16)):

$${\chi }_{\rho }\left(\gamma \right)=\frac{1}{|\Gamma |}\sum _{x\in G}{\mathbb{1}}_{\Gamma }\left(x\gamma x^{-1}\right).$$

We treat each conjugacy class. By $H_{1,q},H_{2,q},P_{1,q},∆,E_{2,q}$, we denote the size of various types of conjugacy classes in ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$. Given a set S, we use ${\mathbb{1}}_S$ to denote the indicator function for S.

• Central class: $\gamma =\left[\begin{array}{ccc}a& 0& 0\\ 0& a& 0\\ 0& 0& a\end{array}\right],a\in {\mathbb{F}}_q^{\times }.$ Because $\gamma \in Z\left(G\right)$, we have $x\gamma x^{-1}=\gamma$; thus,

  $${\chi }_{\rho }\left(\gamma \right)=\frac{\left|G\right|}{|\Gamma |}{\mathbb{1}}_{{\mathbb{F}}_p}\left(a\right).$$
• First hyperbolic class: $\gamma =\left[\begin{array}{ccc}a& 0& 0\\ 0& a& 0\\ 0& 0& b\end{array}\right]$ with $a\ne b\in {\mathbb{F}}_q^{\times }.$ Here, $\gamma$ is similar to an element of $\Gamma$ only when $a,b$ are roots of a polynomial over ${\mathbb{F}}_p$. This occurs only when $a,b\in {\mathbb{F}}_p^{\times }$; for instance, this can be seen by considering the determinant $Det\left(\gamma \right)=a^{2}b$. As this is the constant coefficient of ${\mathrm{char}}_{\gamma }\in {\mathbb{F}}_p\left[x\right]$, we have $Det\left(\gamma \right)\in {\mathbb{F}}_p$, which implies that $Det\left(\gamma \right)$ is invariant under the Frobenius automorphism (as $Gal({\mathbb{F}}_q/{\mathbb{F}}_p)$ is generated by the Frobenius). Because ${\mathbb{F}}_q/{\mathbb{F}}_p$ is Galois, if $a,b\notin {\mathbb{F}}_p$, then they must be Galois conjugates, i.e., $a^p=b$ and $b^p=a$. However, this forces $a^{2}b={\left(a^{2}b\right)}^p=b^{2}a,$ i.e., $a=b$, which is a contradiction. Thus, by a simple application of the orbit-stabilizer theorem, it follows that

  $${\chi }_{\rho }\left(\gamma \right)=\frac{\left|G\right|}{|\Gamma |}\frac{H_{1,p}}{H_{1,q}}{\mathbb{1}}_{\Gamma }\left(\gamma \right).$$
• Second hyperbolic class: $\gamma =\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]$ with $a,b,c\in {\mathbb{F}}_q^{\times }$ distinct. There are three cases which allow $\gamma$ to be similar to an element of $\Gamma$: either all three are Galois conjugates in ${\mathbb{F}}_{p^3}/{\mathbb{F}}_p$ (only possible if $3\mid n$), or $a,b$ are Galois conjugates in ${\mathbb{F}}_{p^2}/{\mathbb{F}}_p$ and $c\in {\mathbb{F}}_p$ (only possible if $2\mid n$), or all three belong to ${\mathbb{F}}_p$. These cases give rise to the three terms in the sum:

  $${\chi }_{\rho }\left(\gamma \right)=\frac{\left|G\right|}{|\Gamma |·H_{2,q}}\left(E_{2,p}·{\mathbb{1}}_{{\mathbb{F}}_{p^3}\setminus {\mathbb{F}}_p}\left(a\right){\delta }_b\left(a^p\right){\delta }_c\left(a^{p^2}\right)+E_{1,p}·{\mathbb{1}}_{{\mathbb{F}}_{p^2}\setminus {\mathbb{F}}_p}\left(a\right){\delta }_b\left(a^p\right){\mathbb{1}}_{{\mathbb{F}}_p}\left(c\right)+H_{2,p}·{\mathbb{1}}_{\Gamma }\left(\gamma \right)\right).$$
• First parabolic class: $\gamma =\left[\begin{array}{ccc}a& 1& 0\\ 0& a& 0\\ 0& 0& a\end{array}\right],a\in {\mathbb{F}}_q^{\times }$. If $a\notin {\mathbb{F}}_p$ were the root of a polynomial over ${\mathbb{F}}_p$, its Galois conjugate $a^p\ne a$ would also be a root, which is clearly not the case. Thus,

  $${\chi }_{\rho }\left(\gamma \right)=\frac{\left|G\right|}{|\Gamma |}\frac{P_{1,p}}{P_{1,q}}{\mathbb{1}}_{{\mathbb{F}}_p}\left(a\right).$$
• Second parabolic class: $\gamma =\left[\begin{array}{ccc}a& 1& 0\\ 0& a& 1\\ 0& 0& a\end{array}\right],a\in {\mathbb{F}}_q^{\times }.$ Similar to the previous class,

  $${\chi }_{\rho }\left(\gamma \right)=\frac{\left|G\right|}{|\Gamma |}\frac{P_{2,p}}{P_{2,q}}{\mathbb{1}}_{{\mathbb{F}}_p}\left(a\right).$$
• Third parabolic class: $\gamma =\left[\begin{array}{ccc}a& 1& 0\\ 0& a& 0\\ 0& 0& b\end{array}\right]$ with $a\ne b\in {\mathbb{F}}_q^{\times }$. Following a similar argument as for the first hyperbolic class, we have

  $${\chi }_{\rho }\left(g\right)=\frac{\left|G\right|}{|\Gamma |}\frac{P_{3,p}}{P_{3,q}}{\mathbb{1}}_{\Gamma }\left(g\right).$$
• First elliptic class: $\gamma =\left[\begin{array}{ccc}w& 0& 0\\ 0& w^q& 0\\ 0& 0& r\end{array}\right]$ with $w\in {\mathbb{F}}_{q^2}\setminus {\mathbb{F}}_q$ and $r\in {\mathbb{F}}_q^{\times }$. Here, $\gamma$ is similar to an element of $\Gamma$ if and only if $w,w^q$ are roots of a degree-2 irreducible polynomial over ${\mathbb{F}}_p$. However, if $2\mid n$, then ${\mathbb{F}}_{p^2}$ is an intermediate extension in ${\mathbb{F}}_q/{\mathbb{F}}_p$, meaning that the roots of any degree-2 polynomial over ${\mathbb{F}}_p$ are included in ${\mathbb{F}}_q$. Thus, ${\chi }_{\rho }\left(\gamma \right)=0$ if n is divisible by 2; otherwise,

  $${\chi }_{\rho }\left(\gamma \right)=\frac{\left|G\right|}{|\Gamma |}\frac{E_{1,p}}{E_{1,q}}{\mathbb{1}}_{{\mathbb{F}}_{p^2}\setminus {\mathbb{F}}_p}\left(w\right){\mathbb{1}}_{{\mathbb{F}}_p}\left(r\right).$$
• Second elliptic class: $\gamma =\left[\begin{array}{ccc}w& 0& 0\\ 0& w^q& 0\\ 0& 0& w^{q^2}\end{array}\right],w\in {\mathbb{F}}_q\left({\delta }^{1/3}\right)\setminus {\mathbb{F}}_q$. As for the previous class, ${\chi }_{\rho }\left(\gamma \right)=0$ if n is divisible by 3; otherwise,

  $${\chi }_{\rho }\left(\gamma \right)=\frac{\left|G\right|}{|\Gamma |}\frac{E_{2,p}}{E_{2,q}}{\mathbb{1}}_{{\mathbb{F}}_{p^3}\setminus {\mathbb{F}}_p}\left(w\right).$$

9. Decomposition of $\mathit{\rho }$ When $\mathbf{2},\mathbf{3}\nmid \mathit{N}$

In this section, we compute the decomposition of $\rho$ when neither 2 nor 3 divide n; thus, elliptic elements in ${\mathrm{GL}}_3\left({\mathbb{F}}_p\right)$ remain elliptic in ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$.

We use the character table for ${\mathrm{GL}}_3\left({\mathbb{F}}_q\right)$ in Terras [6] (pp. 383–384) and Steinberg [10]. As usual, to calculate the multiplicity $m(\pi ,\rho )$ of an irreducible representation $\pi$ present in $\rho$, we compute the inner product of their respective characters. By $H_1,H_2,P_1\cdots ,E_2$ we denote the size of different types of conjugacy classes in ${\mathrm{GL}}_3\left({\mathbb{F}}_p\right)$, while by ${Nm}_{E/F}$ we denote the norm of the field extension $E/F$.

1.

$\alpha$ is a character of ${\mathbb{F}}_q^{\times }$. Then,

$$\begin{array}{cc}\hfill m(\alpha ,\rho )& =\frac{1}{|\Gamma |}(\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a^3\right)+H_1\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}\alpha \left(a^{2}b\right)+\frac{H_2}{6}\sum _{\begin{array}{c}a,b,c\in {\mathbb{F}}_p^{\times }\\ a,b,c\mathrm{distinct}\end{array}}\alpha \left(abc\right)+P_1\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a^3\right)+P_2\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a^3\right)\hfill \\ \hfill & +P_3\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}\alpha \left(a^{2}b\right)+\frac{E_1}{2}\sum _{\begin{array}{c}w\in {\mathbb{F}}_{p^2}\setminus {\mathbb{F}}_p\\ r\in {\mathbb{F}}_p^{\times }\end{array}}\alpha \left(r{Nm}_{{\mathbb{F}}_{q^2}/{\mathbb{F}}_q}\left(w\right)\right)+\frac{E_2}{3}\sum _{\begin{array}{c}w\in {\mathbb{F}}_{p^3}\setminus {\mathbb{F}}_p\end{array}}\alpha \left({Nm}_{{\mathbb{F}}_{q^3}/{\mathbb{F}}_q}\left(w\right)\right)).\hfill \end{array}$$

We can simplify these sums on the basis of three cases:

• ${\alpha }^3{\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$. Each of the sums is zero; thus, $m(\alpha ,\rho )=0$.
• ${\alpha }^3{\mid }_{{\mathbb{F}}_p^{\times }}=1$ but $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$. A short calculation yields

  $$m(\alpha ,\rho )=\frac{1}{|\Gamma |}\left(1-H_1+\frac{H_2}{3}+P_1+P_2-P_3-\frac{E_2}{3}\right)(p-1)=0.$$
• $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}=1$. Simply,

  $$m(\alpha ,\rho )=\frac{1}{|\Gamma |}\sum _{\left\{g\right\}\in C_{\Gamma }}\left|\left\{g\right\}\right|=1.$$

2.

${\pi }_{\alpha }$, where $\alpha$ is a character of ${\mathbb{F}}_q^{\times }$.

$$\begin{array}{cc}\hfill m({\pi }_{\alpha },\rho )& =\frac{1}{|\Gamma |}((q^2+q)\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a^3\right)+(q+1)H_1\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}\alpha \left(a^{2}b\right)+\frac{H_2}{3}\sum _{\begin{array}{c}a,b,c\in {\mathbb{F}}_p^{\times }\\ a,b,c\mathrm{distinct}\end{array}}\alpha \left(abc\right)+q{P}_1\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a^3\right)\hfill \\ \hfill & +0+P_3\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}\alpha \left(a^{2}b\right)+0-\frac{E_2}{3}\sum _{\begin{array}{c}w\in {\mathbb{F}}_{p^3}\setminus {\mathbb{F}}_p\end{array}}\alpha \left({Nm}_{{\mathbb{F}}_{q^3}/{\mathbb{F}}_q}\left(w\right)\right)).\hfill \end{array}$$

• ${\alpha }^3{\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$. Here, all the sums are again zero, leading to $m({\pi }_{\alpha },\rho )=0$.
• ${\alpha }^3{\mid }_{{\mathbb{F}}_p^{\times }}=1$, but $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$.

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha },\rho )& =\frac{1}{|\Gamma |}\left((q^2+q)-(q+1)H_1+\frac{2}{3}{H}_2+q{P}_1-P_3+\frac{E_2}{3}\right)(p-1)\hfill \\ \hfill & =\frac{(q-p)(q-p^2)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$
• $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}=1$.

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha },\rho )& =\frac{1}{|\Gamma |}((q^2+q)(p-1)+(q+1)H_1(p-1)(p-2)+(p-1)(p-2)(p-3)\frac{H_2}{3}\hfill \\ \hfill & +q{P}_1(p-1)+(p-1)(p-2)P_3-\frac{E_2}{3}(p^3-p))\hfill \\ \hfill & =\frac{(q-p)(q+p^5-2p^2)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$
  As a sanity check, showing that this is actually an integer is not very hard.

3.

${\pi }_{\alpha }^{\prime },$ where $\alpha$ is a character of ${\mathbb{F}}_q^{\times }$.

$$\begin{array}{cc}\hfill m({\pi }_{\alpha }^{\prime },\rho )& =\frac{1}{|\Gamma |}(q^3\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a^3\right)+q{H}_1\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}\alpha \left(a^{2}b\right)+\frac{H_2}{6}\sum _{\begin{array}{c}a,b,c\in {\mathbb{F}}_p^{\times }\\ a,b,c\mathrm{distinct}\end{array}}\alpha \left(abc\right)+0+0+0\hfill \\ \hfill & -\frac{E_1}{2}\sum _{\begin{array}{c}w\in {\mathbb{F}}_{p^2}\setminus {\mathbb{F}}_p\\ r\in {\mathbb{F}}_p^{\times }\end{array}}\alpha \left(r{Nm}_{{\mathbb{F}}_{q^2}/{\mathbb{F}}_q}\left(w\right)\right)+\frac{E_2}{3}\sum _{\begin{array}{c}w\in {\mathbb{F}}_{p^3}\setminus {\mathbb{F}}_p\end{array}}\alpha \left({Nm}_{{\mathbb{F}}_{q^3}/{\mathbb{F}}_q}\left(w\right)\right)).\hfill \end{array}$$

• ${\alpha }^3{\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$. $m({\pi }_{\alpha },\rho )=0$.
• ${\alpha }^3{\mid }_{{\mathbb{F}}_p^{\times }}=1$, but $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$.

  $$\begin{array}{c}\hfill m({\pi }_{\alpha }^{\prime },\rho )=\frac{1}{|\Gamma |}\left(q^3-q{H}_1+\frac{1}{3}{H}_2-\frac{E_2}{3}\right)(p-1)=\frac{(q-p)(q-p^2)(q+p^2+p)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\end{array}$$
• $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}=1$.

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha }^{\prime },\rho )& =\frac{1}{|\Gamma |}(q^3(p-1)+q(p-1)(p-2)H_1+(p-1)(p-2)(p-3)\frac{H_2}{6}\hfill \\ \hfill & -(p^2-p)(p-1)\frac{E_1}{2}+(p^3-p)\frac{E_2}{3})\hfill \\ \hfill & =\frac{(q-p)(q^2+pq+p^5-p^4-p^3-p^2)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$

4.

${\pi }_{\alpha ,\beta }$ for distinct characters $\alpha$ and $\beta$ of ${\mathbb{F}}_q^{\times }$.

$$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta },\rho )& =\frac{1}{|\Gamma |}((q^2+q+1)\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a^2\right)\beta \left(a\right)+H_1\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}((q+1)\alpha \left(ab\right)\beta \left(a\right)+\alpha \left(a^2\right)\beta \left(b\right))\hfill \\ \hfill & +\frac{H_2}{6}\sum _{\begin{array}{c}a,b,c\in {\mathbb{F}}_p^{\times }\\ a,b,c\mathrm{distinct}\end{array}}(\beta \left(a\right)\alpha \left(bc\right)+\beta \left(b\right)\alpha \left(ac\right)+\beta \left(c\right)\alpha \left(ab\right))+(q+1)P_1\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a^2\right)\beta \left(a\right)\hfill \\ \hfill & +P_2\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a^2\right)\beta \left(a\right)+P_3\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}(\alpha \left(ab\right)\beta \left(a\right)+\alpha \left(a^2\right)\beta \left(b\right))\hfill \\ \hfill & +\frac{E_1}{2}\sum _{\begin{array}{c}w\in {\mathbb{F}}_{p^2}\setminus {\mathbb{F}}_p\\ r\in {\mathbb{F}}_p^{\times }\end{array}}\alpha \left({Nm}_{{\mathbb{F}}_{q^2}/{\mathbb{F}}_q}\left(w\right)\right)\beta \left(r\right)+0).\hfill \end{array}$$

• ${\alpha }^2\beta {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$. $m({\pi }_{\alpha ,\beta },\rho )=0$.
• ${\alpha }^2\beta {\mid }_{{\mathbb{F}}_p^{\times }}=1$, but ${\alpha }^2{\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$ and $\beta {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$.

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta },\rho )& =\frac{1}{|\Gamma |}((q^2+q+1)(p-1)-(q+2)(p-1)H_1+(p-1)H_2\hfill \\ \hfill & +(q+1)(p-1)P_1+(p-1)P_2-2(p-1)P_3)\hfill \\ \hfill & =\frac{(q-p)(q-p^2)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$
• ${\alpha }^2{\mid }_{{\mathbb{F}}_p^{\times }}=1$ and $\beta {\mid }_{{\mathbb{F}}_p^{\times }}=1$, but $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$.

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta },\rho )& =\frac{1}{|\Gamma |}((q^2+q+1)(p-1)-(p-1)(q-p+3)H_1-(p-1)(p-3)\frac{H_2}{2}\hfill \\ \hfill & +(q+1)(p-1)P_1+(p-1)P_2+(p-1)(p-3)P_3-{(p-1)}^2\frac{E_1}{2})\hfill \\ \hfill & =\frac{(q-p)(q-p^2)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$
• $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}=1$ and $\beta {\mid }_{{\mathbb{F}}_p^{\times }}=1$.

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta },\rho )& =\frac{1}{|\Gamma |}((q^2+q+1)(p-1)+(q+2)(p-1)(p-2)H_1+(p-1)(p-2)(p-3)\frac{H_2}{2}\hfill \\ \hfill & +(q+1)(p-1)P_1+(p-1)P_2+2(p-1)(p-2)P_3+(p-1)(p^2-p)\frac{E_1}{2})\hfill \\ \hfill & =\frac{q(q+p^5-2p^2-p)+p^3(p^5-2p^3-p^2+3)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$

5.

${\pi }_{\alpha ,\beta }^{\prime }$ for distinct characters $\alpha$ and $\beta$ of ${\mathbb{F}}_q^{\times }$.

$$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta }^{\prime },\rho )& =\frac{1}{|\Gamma |}(q(q^2+q+1)\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a^2\right)\beta \left(a\right)+H_1\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}((q+1)\alpha \left(ab\right)\beta \left(a\right)+q\alpha \left(a^2\right)\beta \left(b\right))\hfill \\ \hfill & +\frac{H_2}{6}\sum _{\begin{array}{c}a,b,c\in {\mathbb{F}}_p^{\times }\\ a,b,c\mathrm{distinct}\end{array}}(\beta \left(a\right)\alpha \left(bc\right)+\beta \left(b\right)\alpha \left(ac\right)+\beta \left(c\right)\alpha \left(ab\right))+q{P}_1\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a^2\right)\beta \left(a\right)\hfill \\ \hfill & +0+P_3\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}\alpha \left(ab\right)\beta \left(a\right)-\frac{E_1}{2}\sum _{\begin{array}{c}w\in {\mathbb{F}}_{p^2}\setminus {\mathbb{F}}_p\\ r\in {\mathbb{F}}_p^{\times }\end{array}}\alpha \left({Nm}_{{\mathbb{F}}_{q^2}/{\mathbb{F}}_q}\left(w\right)\right)\beta \left(r\right)+0).\hfill \end{array}$$

• ${\alpha }^2\beta {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$. $m({\pi }_{\alpha ,\beta }^{\prime },\rho )=0$.
• ${\alpha }^2\beta {\mid }_{{\mathbb{F}}_p^{\times }}=1$, but ${\alpha }^2{\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$ and $\beta {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$.

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta }^{\prime },\rho )& =\frac{1}{|\Gamma |}(q(q^2+q+1)(p-1)-(2q+1)(p-1)H_1+(p-1)H_2\hfill \\ \hfill & +q(p-1)P_1-(p-1)P_3)\hfill \\ \hfill & =\frac{(q-p)(q-p^2)(q+p^2+p+1)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$
• ${\alpha }^2{\mid }_{{\mathbb{F}}_p^{\times }}=1$ and $\beta {\mid }_{{\mathbb{F}}_p^{\times }}=1$, but $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$.

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta },\rho )& =\frac{1}{|\Gamma |}(q(q^2+q+1)(p-1)+(p-1)(qp-3q-1)H_1-(p-1)(p-3)\frac{H_2}{2}\hfill \\ \hfill & +q(p-1)P_1-(p-1)P_3+{(p-1)}^2\frac{E_1}{2})\hfill \\ \hfill & =\frac{(q-p)(q(q+p+1)+p^2(p^3-p^2-p-2))}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$
• $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}=1$ and $\beta {\mid }_{{\mathbb{F}}_p^{\times }}=1$.

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta },\rho )& =\frac{1}{|\Gamma |}(q(q^2+q+1)(p-1)+(2q+1)(p-1)(p-2)H_1+(p-1)(p-2)(p-3)\frac{H_2}{2}\hfill \\ \hfill & +q(p-1)P_1+(p-1)(p-2)P_3-(p-1)(p^2-p)\frac{E_1}{2})\hfill \\ \hfill & =\frac{(q-p)(q(q+p+1)+p^2(2p-3)(p^2+p+1))}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$

6.

${\pi }_{\alpha ,\beta ,\gamma }$ for distinct characters $\alpha ,\beta ,\gamma$ of ${\mathbb{F}}_q^{\times }$.

$$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta ,\gamma },\rho )& =\frac{1}{|\Gamma |}((q+1)(q^2+q+1)\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a\right)\beta \left(a\right)\gamma \left(a\right)+H_2\sum _{\begin{array}{c}a,b,c\in {\mathbb{F}}_p^{\times }\\ a,b,c\mathrm{distinct}\end{array}}\alpha \left(a\right)\beta \left(b\right)\gamma \left(c\right)\hfill \\ \hfill & +(q+1)H_1\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}(\alpha \left(b\right)\beta \left(a\right)\gamma \left(a\right)+\beta \left(b\right)\gamma \left(a\right)\alpha \left(a\right)+\gamma \left(b\right)\alpha \left(a\right)\beta \left(a\right))\hfill \\ \hfill & +(2q+1)P_1\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a\right)\beta \left(a\right)\gamma \left(a\right)+P_2\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a\right)\beta \left(a\right)\gamma \left(a\right)\hfill \\ \hfill & +P_3\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}(\alpha \left(b\right)\beta \left(a\right)\gamma \left(a\right)+\beta \left(b\right)\gamma \left(a\right)\alpha \left(a\right)+\gamma \left(b\right)\alpha \left(a\right)\beta \left(a\right))+0+0).\hfill \end{array}$$

• $\alpha \beta \gamma {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$. $m({\pi }_{\alpha ,\beta ,\gamma },\rho )=0$.
• $\alpha \beta \gamma {\mid }_{{\mathbb{F}}_p^{\times }}=1$, but none of $\alpha ,\beta ,$ or $\gamma$ are restricted to the trivial character on ${\mathbb{F}}_p^{\times }.$

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta ,\gamma },\rho )& =\frac{1}{|\Gamma |}((q+1)(q^2+q+1)(p-1)+2(p-1)H_2-3(q+1)(p-1)H_1\hfill \\ \hfill & +(2q+1)(p-1)P_1+(p-1)P_2-3(p-1)P_3)\hfill \\ \hfill & =\frac{(q-p)(q-p^2)(q+p^2+p+2)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$
• $\alpha \beta {\mid }_{{\mathbb{F}}_p^{\times }}=1$ and $\gamma {\mid }_{{\mathbb{F}}_p^{\times }}=1$, but $\alpha$ and $\beta$ are nontrivial on ${\mathbb{F}}_p^{\times }$. Note that because $m({\pi }_{\alpha ,\beta ,\gamma },\rho )$ is symmetric in $\alpha ,\beta ,\gamma$, the calculation here also holds for the other two permutations of this case.

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta ,\gamma },\rho )& =\frac{1}{|\Gamma |}((q+1)(q^2+q+1)(p-1)-(p-1)(p-3)H_2+(q+1)(p-1)(p-4)H_1\hfill \\ \hfill & +(2q+1)(p-1)P_1+(p-1)P_2+(p-1)(p-4)P_3)\hfill \\ \hfill & =\frac{(q-p)(q(q+p+2)+p^2(p^3-p^2-p-3))}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$
• All three of $\alpha ,\beta ,\gamma$ are restricted to the trivial character on ${\mathbb{F}}_p^{\times }$.

  $$\begin{array}{cc}\hfill m({\pi }_{\alpha ,\beta ,\gamma },\rho )& =\frac{1}{|\Gamma |}((q+1)(q^2+q+1)(p-1)+(p-1)(p-2)(p-3)H_2+3(q+1)(p-1)(p-2)H_1\hfill \\ \hfill & +(2q+1)(p-1)P_1+(p-1)P_2+3(p-1)(p-2)P_3)\hfill \\ \hfill & =\frac{(q(q^2+2q+3p^5-p^4-p^3-6p^2-2p)+p^3(p^5-4p^3+p+6))}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$

7.

${\rho }_{\alpha ,\nu }$ for $\alpha$ as a character of ${\mathbb{F}}_q^{\times }$ and $\nu$ as a character of ${\mathbb{F}}_{q^2}^{\times }$ such that ${\nu }^q\ne \nu$ (i.e., it is nondecomposable).

$$\begin{array}{cc}\hfill m({\rho }_{\alpha ,\nu },\rho )& =\frac{1}{|\Gamma |}((q-1)(q^2+q+1)\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a\right)\nu \left(a\right)+(q-1)H_1\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}\alpha \left(b\right)\nu \left(a\right)+0\hfill \\ \hfill & -P_1\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a\right)\nu \left(a\right)-P_2\sum _{a\in {\mathbb{F}}_p^{\times }}\alpha \left(a\right)\nu \left(a\right)-P_3\sum _{\begin{array}{c}a,b\in {\mathbb{F}}_p^{\times }\\ a\ne b\end{array}}\alpha \left(b\right)\nu \left(a\right)\hfill \\ \hfill & -\frac{E_1}{2}\sum _{\begin{array}{c}w\in {\mathbb{F}}_{p^2}\setminus {\mathbb{F}}_p\\ r\in {\mathbb{F}}_p^{\times }\end{array}}\alpha \left(r\right)(\nu \left(w\right)+\nu \left(w^p\right))+0).\hfill \end{array}$$

Note that the last nonzero sum can be written as $-E_1{\sum }_{\begin{array}{c}w\in {\mathbb{F}}_{p^2}\setminus {\mathbb{F}}_p\\ r\in {\mathbb{F}}_p^{\times }\end{array}}\alpha \left(r\right)\nu \left(w\right).$

• $\alpha \nu {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$. $m({\rho }_{\alpha ,\nu },\rho )=0$.
• $\alpha \nu {\mid }_{{\mathbb{F}}_p^{\times }}=1$, but $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$ and $\nu {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$.

  $$\begin{array}{cc}\hfill m({\rho }_{\alpha ,\nu },\rho )& =\frac{1}{|\Gamma |}((q-1)(q^2+q+1)(p-1)-(q-1)(p-1)H_1\hfill \\ \hfill & -(p-1)P_1-(p-1)P_2+(p-1)P_3)\hfill \\ \hfill & =\frac{(q-p)(q-p^2)(q+p^2+p)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$
• $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}=1$, $\nu {\mid }_{{\mathbb{F}}_{p^3}^{\times }}\ne 1$, but $\nu {\mid }_{{\mathbb{F}}_p^{\times }}=1$.

  $$\begin{array}{cc}\hfill m({\rho }_{\alpha ,\nu },\rho )& =\frac{1}{|\Gamma |}((q-1)(q^2+q+1)(p-1)+(q-1)(p-1)(p-2)H_1\hfill \\ \hfill & -(p-1)P_1-(p-1)P_2-(p-1)(p-2)P_3+{(p-1)}^2{E}_1)\hfill \\ \hfill & =\frac{(q-p)(q(q+p)+p^2(p^3-p^2-p-1))}{{(p-1)}^2{p}^3(p+1)(p^2+p+1)}.\hfill \end{array}$$
• $\alpha {\mid }_{{\mathbb{F}}_p^{\times }}=1$ and $\nu {\mid }_{{\mathbb{F}}_{p^3}^{\times }}=1$.

  $$\begin{array}{cc}\hfill m({\rho }_{\alpha ,\nu },\rho )& =\frac{1}{|\Gamma |}((q-1)(q^2+q+1)(p-1)+(q-1)(p-1)(p-2)H_1\hfill \\ \hfill & -(p-1)P_1-(p-1)P_2-(p-1)(p-2)P_3-(p-1)(p^2-p)E_1)\hfill \\ \hfill & =\frac{q(q^2+p^5-p^4-p^3-2p^2)+p^4(-p^4+2p+1)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$

8.

${\sigma }_{\mu }$ for $\mu$ as a character of ${\mathbb{F}}_{q^3}^{\times }$ such that ${\mu }^q\ne \mu$.

$$\begin{array}{cc}\hfill m({\sigma }_{\mu },\rho )& =\frac{1}{|\Gamma |}({(q-1)}^2(q+1)\sum _{a\in {\mathbb{F}}_p^{\times }}\mu \left(a\right)+0+0-(q-1)P_1\sum _{a\in {\mathbb{F}}_p^{\times }}\mu \left(a\right)+P_2\sum _{a\in {\mathbb{F}}_p^{\times }}\mu \left(a\right)+0\hfill \\ \hfill & +0+\frac{E_2}{3}\sum _{w\in {\mathbb{F}}_{p^3}\setminus {\mathbb{F}}_p}(\mu \left(w\right)+\mu \left(w^p\right)+\mu \left(w^{p^2}\right))).\hfill \end{array}$$

• $\mu {\mid }_{{\mathbb{F}}_p^{\times }}\ne 1$. $m({\sigma }_{\mu },\rho )=0$.
• $\mu {\mid }_{{\mathbb{F}}_{p^3}^{\times }}\ne 1$, but $\mu {\mid }_{{\mathbb{F}}_p^{\times }}=1$.

  $$\begin{array}{cc}\hfill m({\sigma }_{\mu },\rho )& =\frac{1}{|\Gamma |}\left({(q-1)}^2(q+1)(p-1)-(q-1)(p-1)P_1+(p-1)P_2-(p-1)E_2\right)\hfill \\ \hfill & =\frac{(q-p)(q-p^2)(q+p^2+p-1)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$
• $\mu {\mid }_{{\mathbb{F}}_{p^3}^{\times }}=1$.

  $$\begin{array}{cc}\hfill m({\sigma }_{\mu },\rho )& =\frac{1}{|\Gamma |}\left({(q-1)}^2(q+1)(p-1)-(q-1)(p-1)P_1+(p-1)P_2+(p^3-p)E_2\right)\hfill \\ \hfill & =\frac{q(q^2-q-p^4-p^3+p)+p^4(p^4-p^2+1)}{p^3{(p-1)}^2(p+1)(p^2+p+1)}.\hfill \end{array}$$