Characters of unipotent radicals of standard parabolic subgroups with 3 parts

Abstract

This paper gives explicit constructions of all irreducible representations of unipotent radicals $N_{n_1,n_2,n_3}({\mathbb {F}}_q)$ of the standard parabolic subgroups $P_{n_1,n_2,n_3}({\mathbb {F}}_q)$ of ${\mathrm {GL}}_n({\mathbb {F}}_q),$ corresponding to the ordered partition $(n_1,\ n_2,\ n_3)$ of n. The construction gives a bijection between coadjoint orbits of $N_{n_1,n_2,n_3}({\mathbb {F}}_q)$ and irreducible representations inducing from degree 1 characters in the sense of Boyarchenko’s construction. The result shows that the number of irreducible characters of $N_{n_1,n_2,n_3}({\mathbb {F}}_q)$ with a fixed degree is a polynomial in $q-1$ with nonnegative integer coefficients and verifies analogue conjectures of Higman, Lehrer, and Isaacs.

1 Introduction

As a convention, ${\mathbb {F}}_q$ denotes a finite field of order q. Let ${\mathrm {U}}_n({\mathbb {F}}_q)$ be the unitriangular group consisting of all upper triangular $n\times n$ matrices with ones along the diagonal. Let $N_{n_1,n_2,n_3}({\mathbb {F}}_q)$ denote the unipotent radicals $N_{n_1,n_2,n_3}({\mathbb {F}}_q)$ of the standard parabolic subgroups $P_{n_1,n_2,n_3}({\mathbb {F}}_q)$ of ${\mathrm {GL}}_n({\mathbb {F}}_q),$ corresponding to the ordered partition $(n_1,\ n_2,\ n_3)$ of n.

This paper provides an explicit bijection between the coadjoint orbits of $N_{n_1,n_2,n_3}({\mathbb {F}}_q)$ and its irreducible representations in terms of Boyarchenko’s construction.

The motivation is the extension of conjectures by Higman–Lehrer–Isaacs, originally proposed for the number of irreducible characters of ${\mathrm {U}}_n({\mathbb {F}}_q).$

1.1 Higman, Lehrer, and Isaacs conjectures

Given an algebraic group G, we call $G({\mathbb {F}}_q)$ a q -power degree group if every irreducible representation of $G({\mathbb {F}}_q)$ is of degree a power of q. When $G({\mathbb {F}}_q)$ is a q-power degree group, denote by $N_{G({\mathbb {F}}_q), \ e}$ the number of irreducible representations of $G({\mathbb {F}}_q)$ with degree $q^e.$

In 1960, Higman [10] predicted that the number of conjugacy classes of ${\mathrm {U}}_n({\mathbb {F}}_q)$ is given by a polynomial in q with integer coefficients.

Later, Lehrer, and Isaacs further strengthened Higman conjecture to the followings:

Conjecture 1.1

(Lehrer conjecture, [15]) $N_{{\mathrm {U}}_n({\mathbb {F}}_q), \ e}$ is a polynomial in q with integer coefficients, for all $e\in {\mathbb {Z}}$.

Conjecture 1.2

(Isaacs conjecture, [11]) $N_{{\mathrm {U}}_n({\mathbb {F}}_q), \ e}$ is a polynomial in $q-1$ with nonnegative integer coefficients, for all $e\in {\mathbb {Z}}$.

Definition 1.3

Let G be an algebraic group defined over ${\mathbb {F}}_q$. Then, G is called of Isaacs class if $G({\mathbb {F}}_q)$ is a q-power degree group and $N_{G({\mathbb {F}}_q), \ e}$ is a polynomial in $q-1$ with nonnegative integer coefficients, for all $e\in {\mathbb {Z}}$.

The above conjectures are still open. In [22], Pak and Soffer verified Higman’s conjecture for $n\le 16$ and suggested that it probably fails for $n\ge 59$. Similar conjectures are generalized to other finite groups with analogous structures, for example: finite pattern groups, unipotent radical of symplectic or orthogonal groups, etc. Related problems are considered, wild and various tools from different aspects, including geometry, combinatorics, algebra, supercharacter theory, have been developed. For instance, refer to: [2, 3, 5,6,7,8,9, 11, 12, 14, 16,17,18,19,20,21,22,23,24,25], etc.¹

1.2 Pattern groups

Let ${\mathrm {Mat}}_{n\times m}$ denote the set of $n\times m$ matrices and ${\mathrm {Mat}}_n$ the set of $n\times n$ matrices. For a finite set X, denote by $\#X$ the cardinality of X.

Denote by $e_{i,j}$ the matrix unit in ${\mathrm {Mat}}_n$ whose entries are 0 except the (i, j)-entry, in which case is 1. Then $e_{i,j}e_{r,s}=\delta _{jr}e_{i,s}$ where $\delta _{jr}$ is the Kronecker delta.

Let ${\mathfrak {U}}_n$ be the Lie algebra $Lie({\mathrm {U}}_n )$ of ${\mathrm {U}}_n$ and ${\mathfrak {U}}_n^t$ be its dual algebra. We identify ${\mathfrak {U}}_n$ (resp. ${\mathfrak {U}}_n^t$) with the set of upper (resp. lower) triangular nilpotent matrices.

Next, let us recall pattern groups and their associated pattern algebras. Refer to [12] for more details on the finite field case and also [3] for their generalization to p-adic case. Denote by $\Delta _n=\{(i,j)\mid 1\le i< j\le n\}.$ A subset D of $\Delta _n$ is called closed, if D contains (i, k) whenever it contains both (i, j) and (j, k). Then for closed $D\subset \Delta _n,$

$$\begin{aligned} A_D:=\{(n_{i,j})\in {\mathfrak {U}}_n\mid n_{i,j}=0\text { if } (i,j)\notin D\} \end{aligned}$$

is a subalgebra of ${\mathfrak {U}}_n$. Define $G_D:=1+A_D$, which is a subgroup of ${\mathrm {U}}_n$. The algebra $A_D$ (resp. the group $G_D$) is called the pattern algebra (resp. pattern group) corresponding to D.

For a matrix $C=(c_{i,j})\in {\mathfrak {U}}_n$, denote by

$$\begin{aligned} J(C):=\{(i, j)\in \Delta _n \mid c_{i,j}\ne 0\}.\end{aligned}$$

(1.1)

For a subset $X\subset {\mathfrak {U}}_n$, denote by

$$\begin{aligned} J(X):=\bigcup _{C\in X}\{J(C) \}. \end{aligned}$$

(1.2)

Let ${Ind }^{G}$ denote the induction functor. Let F be a self-dual field, that is, a finite field, or ${\mathbb {R}}$, or ${\mathbb {C}}$, or a finite extension of ${\mathbb {Q}}_p$, or a field ${\mathbb {F}}_q((t))$ of formal Laurent series in one variable over a finite field. Let A be a finite-dimensional associative nilpotent algebra over F. Then A inherits a natural topology from F, and $1+A$ becomes a locally compact (Hausdorff) and second countable topological group. Moreover, it is unimodular. If B is an F-subalgebra of A, then $1+B$ can be viewed as a closed subgroup of $1+A.$

In this situation, Boyarchenko confirmed a conjecture proposed by Gutkin in 1973 with the following theorem.

Theorem 1.4

[3, Theorem 1.3] Let $\pi$ be a irreducible unitary representation of $1+A$. Then there exist an F-subalgebra $B\subset A$ and a unitary character $\alpha : 1 + B \mapsto {\mathbb {C}}$ such that $\pi \cong u-{Ind }_{1+B}^{1+A}\alpha ,$ where $u-{Ind }$ denotes the operation of unitary induction.

1.3 Coadjoint orbits

Given a pattern group $1+A$, we may identify its dual space of the Lie algebra A with $A^t$ via the trace map. The coadjoint action of $g\in 1+A$ on $\alpha \in A^t$ in [13] is given by

$$\begin{aligned} (g\cdot \alpha )(a)={\mathrm {tr}}(\alpha gag^{-1}), \text { for } a\in A. \end{aligned}$$

(1.3)

It can be identified with $\alpha \mapsto [g^{-1}\alpha g]_A$, where $[\cdot ]_A$ is the projection from ${\mathrm {Mat}}_n$ to $A^t.$ More precisely, for $m=(m_{i,j})\in {\mathrm {Mat}}_n$,

$$\begin{aligned} ([m]_A)_{i,j}=m_{i,j} \text { if }(j,i)\in J(A), \text { and } ([m]_A)_{i,j}=0 \text { if }(j,i)\notin J(A) \text { or } i\le j. \end{aligned}$$

Define the action of $1+A$ on $A^t$ by

$$\begin{aligned} g\circ _A \alpha :=[g^{-1}\alpha g]_A\text { for }g\in 1+A\text { and }\alpha \in A^t. \end{aligned}$$

Since $g\cdot \alpha (a)={\mathrm {tr}}([g\alpha g^{-1}]_Aa)$ for all $a\in A$, the action $\circ _A$ of $1+A$ on $A^t$ coincides with the coadjoint action, whose orbits are called coadjoint orbits. The following lemma relates the number of coadjoint orbits to the number of conjugacy classes.

Lemma 1.5

[5, Lemma 4.1] Let A be a finite-dimensional associative nilpotent algebra over ${\mathbb {F}}_q$. Then, the number of coadjoint orbits is the same as the number of conjugacy classes, which is equal to the number of isomorphism classes of irreducible representations of $1+A$.

Next, we recall Mackey’s theorem for finite groups, which provides a nice way to compute the dimension of intertwining operator between two induced representations.

Theorem 1.6

(Mackey’s theorem; [4, Proposition 4.1.2]) Let G be a finite group, $H_i$ its subgroups and $\pi _i$ representations of $H_i,i=1,2$. Denote by

$$\begin{aligned} \mathfrak S=\{ \triangle : G\mapsto {\mathrm {Hom}}_{\mathbb {C}}(\pi _1 , \pi _2 ) |\ \triangle (h_2gh_1)=\pi _2(h_2)\circ \triangle (g)\circ \pi _1(h_1), h_i\in H_i\}.\end{aligned}$$

(1.4)

As a vector space, ${\mathrm {Hom}}_G({Ind }_{H_1}^G\pi _1 , {Ind }_{H_2}^G\pi _2 )$ is isomorphic to $\mathfrak {S}.$

We end this introduction by summarizing the main result, Theorem 2.3 as follows.

For $G=N_{n_1,n_2,n_3}=1+A,$ there is a natural bijection between irreducible characters of $G({\mathbb {F}}_q)$ and its coadjoint orbits in the following sense:

• (1) Each irreducible character of $G({\mathbb {F}}_q)$ is given by ${Ind }_{I_T}\psi _T,$ for some $I_T$ pattern subgroup of $G({\mathbb {F}}_q)$ attached to $T\in {\mathfrak {n}}^t_{n_1,n_2, n_3},$ and the character $\psi _T:I_T\mapsto {\mathbb {C}}^*$ on $I_T$ is given by

  $$\begin{aligned} \psi _T(W):= \psi _q({\mathrm {tr}}[TW]),\text { for }W\in G({\mathbb {F}}_q). \end{aligned}$$

• (2) ${Ind }_{I_T}\psi _T \cong {Ind }_{I_{T'}}\psi _{T'}$ if and only if T and $T'$ belong to the same coadjoint orbits.

The above correspondence provides an efficient way to verify that $N_{n_1,n_2,n_3}({\mathbb {F}}_q)$ is of Isaacs class. It would be interesting to find criterions on finite pattern group $G({\mathbb {F}}_q)$ so that such explicit correspondence holds for $G({\mathbb {F}}_q)$, and to find conceptual ways for computing the number of irreducible characters with a fixed degree. We also wonder if an analogous correspondence could be extended to $N_{n_1,n_2,n_3}(F)$ or other G(F),  where G is a pattern group and F is a self-dual field.

2 Irreducible representations of $N_{n_1,n_2, n_3}({\mathbb {F}}_q)$

Let $\mathfrak {n}_ {n_1, n_2, n_3}$ be the pattern algebra corresponding to the closed set

$$\begin{aligned} \{(j_1,n_1+i_1),\ (j_1,n_1+n_2+i_2),\ (n_1+i_1,n_1+n_2+i_2)\ | \ 1\le j_1\le n_1,\ 1\le i_1\le n_2, \end{aligned}$$

$\ 1\le i_2\le n_3\}.$ Then $N_{n_1, n_2, n_3}=1+\mathfrak {n}_ {n_1, n_2, n_3}$ is a pattern group.

Let $n=n_1+n_2+n_3,$ and

$$\begin{aligned} n(X,Y,Z):=\begin{pmatrix}{\mathrm {I}}_{n_1}&{}X&{} Y\\ 0&{}{\mathrm {I}}_{n_2}&{}Z\\ 0&{}0&{}{\mathrm {I}}_{n_3}\end{pmatrix}\in N_{n_1,n_2, n_3}({\mathbb {F}}_q), \end{aligned}$$

where $X\in {\mathrm {Mat}}_{n_1\times n_2}, \ Y\in {\mathrm {Mat}}_{n_1\times n_3},\ Z\in {\mathrm {Mat}}_{n_2\times n_3} .$

For $B\in {\mathrm {Mat}}_{n_2\times n_1}, \ A\in {\mathrm {Mat}}_{n_3\times n_1},\ C\in {\mathrm {Mat}}_{n_3\times n_2} ,$ denote by

$$\begin{aligned} s(A,B, C):=\begin{pmatrix}0&{}0&{}0\\ B&{}0&{}0\\ A&{}C&{}0\end{pmatrix}\in {\mathfrak {n}}^t_{n_1,n_2, n_3}, \end{aligned}$$

where ${\mathfrak {n}}^t_{n_1,n_2, n_3}$ denotes the dual algebra of $N_{n_1,n_2, n_3}.$ Write

$$\begin{aligned} A=(A_1,\ldots ,A_{n_2})=\begin{pmatrix}A^1\\ \vdots \\ A^{n_3}\end{pmatrix}, \end{aligned}$$

where $A_i$(resp. $B_i$ and $C_i$) is the ith column and $A^i$ ( resp. $B^i$ and $C^i$) is the ith row of A (resp. B and C).

Assume that the rank of A is $k\ge 0.$ Let $i_1$(resp. $j_1$) be the smallest index so that $A_{i_1}\ne 0$ (resp. $A^{j_1}\ne 0$). When $k=0$, this is void. Assume that $A_{i_t}$ and $A^{j_t}$ have been defined for $1\le t\le k-1$. Define inductively $i_{t+1}$ (resp. $j_{t+1}$) to be the smallest index so that $A_{i_1},\ldots , A_{i_t}, A_{i_{t+1}}$ (resp. $A^{j_1},\ldots , A^{j_t}, A^{j_{t+1}}$) are linearly independent.

Then

$$\begin{aligned}_{\mathfrak {n}_ {n_1, n_2, n_3}} =[s(A,B+TA,C-AR)]_{\mathfrak {n}_ {n_1, n_2, n_3}}. \end{aligned}$$

Let $M=TA.$ Then

$$\begin{aligned} M^i=\sum _{r=1}^{n_3}t_{i,r}A^r,\ 1\le i\le n_2. \end{aligned}$$

Since $\{A_{i_1},\ldots , A_{i_k}\}$ and $\{A^{j_1},\ldots , A^{j_k}\}$ are linearly independent, by choosing suitable $t_{i,r}$ we can make the $i_1,\ i_2,\ldots , \ i_k$th columns of $B+TA$ are zeros. Symmetrically, we can find R so that the $j_1,\ j_2,\ \ldots , \ j_k$th rows of $C-AR$ are zeros. We claim that s(A, B, C) is in the same coadjoint orbit with a unique matrix of the form s(A, E, F),  where

$$\begin{aligned} E_{i_1}=\cdots = E_{i_k}=0,\text { and }F^{j_1}=\cdots =F^{j_k}=0.\end{aligned}$$

(2.1)

Assume that $C-AR=F,$ and $C-AR'=F'$ such that $F^{j_1}=\cdots =F^{j_k}={F'}^{j_1}=\cdots ={F'}^{j_k}=0.$ Let $Q=R'-R.$ Then $AQ=F-F',$ whose $j_v$th row are zeros, $\text { for all } 1\le v\le k.$ Let

$$\begin{aligned}A=\begin{pmatrix}A^1\\ \vdots \\ A^{n_3}\end{pmatrix}, \text { and }Q=(Q_1,\ldots , Q_{n_2}). \end{aligned}$$

Since the $j_v$th row of AQ is $(A^{j_v}Q_1, \ A^{j_v}Q_2,\ \ldots , A^{j_v}Q_{n_2}),$

$$\begin{aligned} A^{j_v}Q_i=0,\text { for all } 1\le v\le k,\ 1\le i \le n_2. \end{aligned}$$

Then

$$\begin{aligned} A^jQ_i=0,\text { for all } 1\le j\le n_3,\ 1\le i \le n_2, \end{aligned}$$

since the row space of A is generated by $\{A^{j_1}\ldots , A^{j_k}\}.$ Therefore, $AQ=0$ and $F=F'$. By symmetry, there is also a unique E satisfying Eq. (2.1).

Let

$$\begin{aligned} S:=\{s(A,E,F)\ \ \ A,\ E,\ F \text { satisfying Eq.} 2.1 \}. \end{aligned}$$

Then, S is a complete set of coadjoint orbits with no repeating elements corresponding to same orbits.

Let

$$\begin{aligned} N_A:=\{n(X,Y,Z) | \ AX=0 \}. \end{aligned}$$

Fix a nontrivial additive character $\psi _q$ of ${\mathbb {F}}_q$ and define

$$\begin{aligned} \chi _{s(A,E, F)}:N_{n_1,n_2,n_3}({\mathbb {F}}_q)\mapsto {\mathbb {C}}^* \end{aligned}$$

by $\chi _{s(A,E, F)} (n(X,Y,Z))=\psi _q({\mathrm {tr}}[EX]+{\mathrm {tr}}[AY]+{\mathrm {tr}}[FZ]) .$ Then $\chi _{s(A,E, F)}$ is a character of $N_A({\mathbb {F}}_q)$, since

$$\begin{aligned}&\chi _{s(A,E, F)} (n(X,Y,Z)n(X',Y',Z'))\\&\quad =\chi _{s(A,E, F)} (n(X+X',Y+Y'+XZ',Z+Z'))\\&\quad =\psi _q({\mathrm {tr}}[E(X+X')]+{\mathrm {tr}}[A(Y+Y')]+{\mathrm {tr}}[F(Z+Z')])\\&\quad =\chi _{s(A,E, F)}(n(X,Y,Z))\chi _{s(A,E, F)} (n(X',Y',Z')). \end{aligned}$$

Lemma 2.1

For $s(A,E,F)\in S,$ ${Ind }_{N_A({\mathbb {F}}_q)}^{N_{n_1,n_2,n_3}({\mathbb {F}}_q)}\chi _{s(A,E, F)}$ is irreducible of degree $q^{kn_2}.$ Moreover, given $s(A,E,F),\ s(A',E',F')\in S,$

$$\begin{aligned}&{Ind }_{N_A({\mathbb {F}}_q)}^{N_{n_1,n_2,n_3}({\mathbb {F}}_q)}\chi _{s(A,E, F)}\cong {Ind }_{N_{A'}({\mathbb {F}}_q)}^{N_{n_1,n_2,n_3}({\mathbb {F}}_q)}\chi _{s(A',E', F')} \\&\quad \text { if and only if } (A,E,F)=(A',E', F'). \end{aligned}$$

Proof

First, the double coset

$$\begin{aligned} N_A({\mathbb {F}}_q)n(R,V,W) N_A({\mathbb {F}}_q)=N_A({\mathbb {F}}_q)n(R,0,0) N_A({\mathbb {F}}_q), \end{aligned}$$

and

$$\begin{aligned}&n(R,0,0)n(X,Y,Z)n(R,0,0)^{-1}\\ \nonumber&\quad =n(X,Y+RZ,Z)\in n(R,0,0)N_A({\mathbb {F}}_q)n(R,0,0)^{-1}\cap N_A({\mathbb {F}}_q), \end{aligned}$$

(2.2)

$\text { for all }n(X,Y,Z)\in N_A({\mathbb {F}}_q).$ If $N_A({\mathbb {F}}_q)n(R,0,0) N_A({\mathbb {F}}_q)$ is an admissible coset for $(\chi _{s(A,E, F)}, \chi _{s(A,E, F)})$ in the sense of Theorem 1.6, then by Eq. (2.2)

$$\begin{aligned}&\chi _{s(A,E, F)}(n(X,Y+RZ,Z))\\&\quad =\chi _{s(A,E, F)}(n(X,Y ,Z)),\text { for all }n(X,Y,Z)\in N_A({\mathbb {F}}_q). \end{aligned}$$

That is, $\psi _q({\mathrm {tr}}[ARZ])=1$ for all $Z\in {\mathrm {Mat}}_{n_2,n_3},$ which implies that $AR=0$ and $N_A({\mathbb {F}}_q)n(R,0,0) N_A({\mathbb {F}}_q)=N_A({\mathbb {F}}_q).$ By Theorem 1.6,

$$\begin{aligned} \dim {\mathrm {Hom}}_{ N_{n_1,n_2,n_3}}\left( {Ind }_{N_A({\mathbb {F}}_q)}^{N_{n_1,n_2,n_3}({\mathbb {F}}_q)}\chi _{s(A,E, F)},\ {Ind }_{N_A({\mathbb {F}}_q)}^{N_{n_1,n_2,n_3}({\mathbb {F}}_q)}\chi _{s(A,E, F)}\right) =1, \end{aligned}$$

and ${Ind }_{N_A({\mathbb {F}}_q)}^{N_{n_1,n_2,n_3}({\mathbb {F}}_q)}\chi _{s(A,E, F)}$ is irreducible with dimension

$$\begin{aligned} \dim {Ind }_{N_A({\mathbb {F}}_q)}^{N_{n_1,n_2,n_3}({\mathbb {F}}_q)}\chi _{s(A,E, F)}= \frac{\#N_{n_1,n_2,n_3}({\mathbb {F}}_q)}{\#N_A({\mathbb {F}}_q)}= q^{kn_2} .\end{aligned}$$

Given $s(A,E,F),\ s(A',E',F')\in S,$ if $N_{A'}({\mathbb {F}}_q)n(R,0,0) N_A({\mathbb {F}}_q)$ is an admissible coset for $(\chi _{s(A,E, F)}, \chi _{s(A',E', F')})$ in the sense of Theorem 1.6, then

$$\begin{aligned}&\chi _{s(A',E', F')}(n(R,0,0)n(X,Y,Z)n(R,0,0)^{-1})\\&\quad =\chi _{s(A',E', F')}(n(X,Y+RZ,Z))\\&\quad =\psi _q({\mathrm {tr}}[E'X]+{\mathrm {tr}}[A'(Y+RZ)]+{\mathrm {tr}}[F'Z])\\&\quad =\chi _{s(A,E, F)}(n(X,Y ,Z))\\&\quad =\psi _q({\mathrm {tr}}[EX]+{\mathrm {tr}}[AY]+{\mathrm {tr}}[FZ]), \end{aligned}$$

$\text { for all }n(X,Y,Z)\in N_A({\mathbb {F}}_q)\cap N_{A'}({\mathbb {F}}_q).$ That is,

$$\begin{aligned} A=A',\ \psi _q({\mathrm {tr}}[(E-E')X])=1,\text { for all }AX=0 \text { and }A'R+F'=F. \end{aligned}$$

First, we claim that $E=E'.$ Let

$$\begin{aligned} W=E-E'=(W_1,\ldots ,W_{n_1}) =\begin{pmatrix}W^1\\ \vdots \\ W^{n_1} \end{pmatrix}, \end{aligned}$$

where $W_i$ is the ith column and $W^i$ the ith row of W. By Eq. (2.1), then $W_{i_1}=\cdots =W_{i_k}= 0^t.$ If $W^r,$ for some r,  is not in the vector space generated by the rows of A, then there exists $X\in {\mathrm {Mat}}_{n_1\times n_2}$ such that $AX=0$ and $\psi _q({\mathrm {tr}}[WX])\ne 1$. Hence, we may assume that all row vectors of W are in the vector space generated by the rows of A. Since $W_{i_1}=\cdots =W_{i_k}= 0^t,$ by the same argument as in showing the uniqueness of representatives $s(A,E,F)\in S,$ we will see that $W=0$ and $E=E'$.

Note that $F=F'$ by the same argument as showing uniqueness of representatives of the form s(A, E, F) with E and F satisfying Eq. (2.1). Then

$$\begin{aligned}&{Ind }_{N_A({\mathbb {F}}_q)}^{N_{n_1,n_2,n_3}({\mathbb {F}}_q)}\chi _{s(A,E, F)}\cong {Ind }_{N_{A'}({\mathbb {F}}_q)}^{N_{n_1,n_2,n_3}({\mathbb {F}}_q)}\chi _{s(A',E', F')} \\&\qquad \text { if and only if } (A,E,F)=(A',E', F'). \end{aligned}$$

$\square$

The following formula is known to experts, but it is not easy to find earlier references. Here we cite a relatively recent paper by Alexander and Fisher in 1966.

Lemma 2.2

[1, Theorem 2] The number of $m\times n$ matrices over ${\mathbb {F}}_q$ with rank k is given by

$$\begin{aligned} c_k:=\left( \prod _{j=1}^k\,\left( q^m-q^{j-1}\right) \right) \,\left( \sum _{\begin{array}{c} {j_1,\ldots ,j_k\in \mathbb {Z}_{\ge 0}}\\ {j_1+j_2+\ldots +j_k\le n-k} \end{array}}\,q^{\sum _{i=1}^k\,i\,j_i}\right) , \end{aligned}$$

Let $\mu =\min \{n_1,n_3\}.$

Theorem 2.3

The set of irreducible representations of $N_n({\mathbb {F}}_q)$ is given by $\{{Ind }_{N_A}^{N_n}\chi _{s(A,B, C)}\ |\ A,\ B,\ C\text { satisfying }Eq. 2.1\}.$ Moreover, the set of degrees of irreducible representations of pattern group $N_n({\mathbb {F}}_q)$ is $\{q^{kn_2}\ | \ 0\le k\le \mu \}$, and there are $c_k$ degree $q^{kn_2}$ irreducible representations. Hence, $N_{n_1,n_2,n_3}({\mathbb {F}}_q)$ is of Isaacs class.²

Proof

Notations follow the above. For any fixed rank k matrix $A\in {\mathrm {Mat}}_{n_3\times n_1},$ there are $q^{(n_1-k)n_2+(n_3-k)n_2}$ coadjoint orbits with representatives of the form s(A, E, F) satisfying Eq. (2.1). Since the degree of ${Ind }_{N_A({\mathbb {F}}_q)}^{N_{n_1,n_2,n_3}({\mathbb {F}}_q)}\chi _{s(A,B, C)}$ is

$$\frac{\#N_{n_1,n_2,n_3}({\mathbb {F}}_q)}{\#N_A({\mathbb {F}}_q)}=q^{kn_2},$$

we have $q^{(n_1-k)n_2+(n_3-k)n_2}c_k$ degree $q^{kn_2}$ irreducible representations. $\square$

Note that

$$\begin{aligned}&\sum _{k=0}^\mu q^{2kn_2}q^{(n_1-k)n_2+(n_3-k)n_2} c_k = q^{n_1n_2+n_2n_3}\sum _{k=0}^\mu c_k\\&\qquad =q^{n_1n_2+n_2n_3+n_3n_1}=\#N_{n_1,n_2,n_3}({\mathbb {F}}_q). \end{aligned}$$

It verifies that the order of a finite group equals the sum of squares of degrees of its irreducible character.