The unipotent modules of ${{\mathrm {GL}}}_{n}({{\mathbb {F}}}_{q})$ via tableaux

Abstract

We construct the irreducible unipotent modules of the finite general linear groups from actions on tableaux. Our approach is analogous to that of James (Bull Lond Math Soc 8:229–232, 1976) for the symmetric groups, answering an open question as to whether such a construction exists. We show that our modules are isomorphic to those previously constructed by James (Representations of general linear groups, London Mathematical Society Lecture Note Series, vol. 94. Cambridge University Press, Cambridge, 1984. doi:10.1017/CBO9780511661921) , although the two presentations are quite different. Key to our construction are the generalized Gelfand–Graev representations of Kawanaka (Generalized Gel’fand-Graev representations and Ennola duality. In: Algebraic groups and related topics (Kyoto/Nagoya, 1983), advanced studies in pure math., vol. 6, pp. 175–206. North-Holland, Amsterdam 1985).

1 Introduction

A standard construction of the irreducible representations of the symmetric group $S_{n}$, initially due to James [4], uses the action of $S_{n}$ on Young tableaux to define a module for each integer partition of n. In characteristic 0, these modules are (up to isomorphism) all of the irreducible $S_{n}$-modules; in other characteristics, the irreducible modules appear as quotients of these modules.

There is a principle in representation theory that information about the finite general linear group ${{\mathrm {GL}}}_{n}({{\mathbb {F}}}_{q})$ can be related to that of $S_{n}$ by “setting $q=1$.” In particular, there is a collection of irreducible representations of ${{\mathrm {GL}}}_{n}({{\mathbb {F}}}_{q})$, known as “unipotent representations,” that one would expect to behave like the irreducible representations of $S_{n}$. James [6], constructs the unipotent modules of ${{\mathrm {GL}}}_{n}({{\mathbb {F}}}_{q})$ over any field containing a nontrivial $p\text {th}$ root of unity (where p is the characteristic of ${{\mathbb {F}}}_{q}$). The construction is quite different from the tableaux approach for the symmetric group, and in particular the proofs do not translate to proofs for the symmetric group. James [6] asks in the introduction to whether an alternative construction exists that is more along the lines of that of the symmetric group.

Our approach is to label the boxes of Young diagrams with elements of ${{\mathbb {F}}}_{q}^{n}$ rather than by integers. The group ${{\mathrm {GL}}}_{n}({{\mathbb {F}}}_{q})$ acts on these objects; we use this action to define the irreducible unipotent modules. Our construction is a natural analogue to that of the symmetric group, providing a positive answer to James’ question. A key tool in our approach is the use of the “generalized Gelfand–Graev representations,” initially constructed by Kawanaka [8] and recently studied by Thiem and the author [1].

James shows in [5] that the irreducible module of $S_{n}$ corresponding to the partition $\lambda$ has a basis indexed by the standard Young tableaux of shape $\lambda$. It is still an open problem to determine bases for the unipotent irreducible modules of ${{\mathrm {GL}}}_{n}({{\mathbb {F}}}_{q})$ (see, for instance, [2]). In providing a new construction of the irreducible unipotent modules, we hope to shed some light on this question.

In Sect. 2, we cover necessary background material on partitions and the finite general linear groups. We provide motivation for our construction in Sect. 3, in particular explaining why the generalized Gelfand–Graev representations are the right tool for constructing unipotent modules. Our construction is in Sect. 4, and in Sect. 5 we look at some further directions for this line of research.

2 Preliminaries

2.1 Partitions and tableaux

Let n be a positive integer; a partition of n is a sequence $\lambda = (\lambda _{1},\lambda _{2},\ldots ,\lambda _{k})$ of positive integers with $\lambda _{1} \ge \lambda _{2} \ge \cdots \ge \lambda _{k}$ and $\lambda _{1}+\lambda _{2}+\cdots +\lambda _{k} = n$. We write $\lambda \vdash n$ to indicate that $\lambda$ is a partition of n.

There is a partial order on the set of partitions of n with $\lambda \succeq \mu$ if and only if

$$\begin{aligned} \sum _{i=1}^k\lambda _i \ge \sum _{i=1}^k\mu _i \end{aligned}$$

for all k (setting $\lambda _{i}=0$ if $\lambda$ has fewer than i parts). This order is called the dominance order on partitions.

To each partition $\lambda$, we associate a Young diagram, which is a left-justified array of blocks such that the number of blocks in the $i\hbox {th}$ row is $\lambda _{i}$.

Example 1

Let $\lambda = (4,3,1,1)$; then the Young diagram of shape $\lambda$ is

The conjugate of a partition $\lambda$, denoted $\lambda '$, is the partition defined by $\lambda _{i}' = |\{j \mid \lambda _{j} \ge i\}|$. Note that the Young diagram of $\lambda '$ is obtained from that of $\lambda$ by reflection about the diagonal.

If $\lambda$ is a partition of n, a tableau of shape $\lambda$ is a filling of the Young diagram of shape $\lambda$ by the integers from 1 through n, each appearing exactly once. We say that a tableau is standard if the entries increase along rows and columns.

Example 2

Let $\lambda = (4,3,1,1)$, and let

then T and $T'$ are both tableaux of shape $\lambda$, but $T'$ is not standard as the pairs (7, 8) and (6, 9) violate the row-increasing and column-increasing conditions.

2.2 The finite general linear groups

Let q be a power of a prime, and let ${{\mathbb {F}}}_q$ be the finite field with q elements. We are interested in $G = {{\mathrm {GL}}}_n({{\mathbb {F}}}_q)$, the group of invertible $n \times n$ matrices with entries in ${{\mathbb {F}}}_q$.

Let $\lambda$ be a partition of n, and let T be the row-reading tableau of shape $\lambda$. We define

$$\begin{aligned} P_\lambda&= \{g \in G \mid g_{ij} = 0 \text { if } i \text { is strictly below } j \text { in } T\} \text { and} \\ U_\lambda&= \{g \in G \mid g_{ii} = 1 \text { and } g_{ij} = 0 \text { unless } i = j \text { or }i \text { is strictly above } j \text { in } T\}. \end{aligned}$$

Example 3

Let $\lambda = (4,3,1,1)$; then

and we have

$$\begin{aligned} P_\lambda = \left\{ \left( \begin{array}{ccccccccc} * &{} * &{} * &{} * &{} * &{} * &{} * &{} * &{} * \\ * &{} * &{} * &{} * &{} * &{} * &{} * &{} * &{} * \\ * &{} * &{} * &{} * &{} * &{} * &{} * &{} * &{} * \\ * &{} * &{} * &{} * &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} * \end{array} \right) \in {{\mathrm {GL}}}_n({{\mathbb {F}}}_q)\right\} \end{aligned}$$

and

$$\begin{aligned} U_\lambda = \left\{ \left( \begin{array}{ccccccccc} 1 &{} 0 &{} 0 &{} 0 &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 1 &{} 0 &{} 0 &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 0 &{} 1 &{} 0 &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 1 &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 \end{array}\right) \in {{\mathrm {GL}}}_n({{\mathbb {F}}}_q)\right\} . \end{aligned}$$

We also define

$$\begin{aligned} P_\lambda ^- = (P_\lambda )^\textit{t} \quad \text {and} \quad U_\lambda ^- = (U_\lambda )^\textit{t} \end{aligned}$$

to be the transposes of $P_\lambda$ and $U_\lambda$.

Remark 1

The groups $P_\lambda$ and $P_\lambda ^-$ are parabolic subgroups of G, with unipotent radicals $U_\lambda$ and $U_\lambda ^-$. In particular, the group of upper triangular matrices in G given by

$$\begin{aligned} B_n({{\mathbb {F}}}_q) = P_{(1^n)} \end{aligned}$$

is a Borel subgroup of G with unipotent radical

$$\begin{aligned} {\mathrm {UT}}_n({{\mathbb {F}}}_q) = U_{(1^n)}. \end{aligned}$$

We say that an irreducible module of G (over some field) is unipotent if it is a composition factor of ${\mathrm {Ind}}_{B_n({{\mathbb {F}}}_q)}^G({\mathbbm {1}})$.

3 Motivation behind the construction

In this section, we motivate our construction in the specific case where we are taking our representations over the field of complex numbers. None of the results of this section are necessary for the construction; however they provide insight as to why our approach makes sense. First, we recall some facts about the character theory of $S_n$ and ${{\mathrm {GL}}}_n({{\mathbb {F}}}_q)$ over ${\mathbb {C}}$ and the connection to symmetric functions.

The irreducible complex characters of $S_n$ are indexed by the partitions of n; let $\psi ^\lambda$ denote the irreducible character of $S_n$ corresponding to $\lambda$, as in [9]. Note that $\psi ^{(n)}$ is the trivial character and $\psi ^{(1^n)}$ is the sign character (denoted $\epsilon$). The following proposition follows from the Pieri rules (see [9, I.5]).

Proposition 1

Let $\lambda$ be a partition of n, and let $W_\lambda$ be a Young subgroup of $S_n$ of shape $\lambda$. Then

$$\begin{aligned} {\mathrm {Ind}}_{W_\lambda }^{S_n}({\mathbbm {1}}) = \sum _{\mu \succeq \lambda } K_{\mu \lambda }\psi ^\mu \quad \text {and}\quad {\mathrm {Ind}}_{W_\lambda }^{S_n}(\epsilon ) = \sum _{\mu ' \succeq \lambda } K_{\mu '\lambda }\psi ^\mu , \end{aligned}$$

where the $K_{\mu \lambda }$ are the Kostka numbers (see [9, I.6.4]).

The Kostka numbers $K_{\mu \lambda }$ satisfy that $K_{\lambda \lambda }=1$ and $K_{\mu \lambda }=0$ unless $\mu \succeq \lambda$. In particular, if $W_\lambda$ is a Young subgroup of shape $\lambda$ and $W_{\lambda '}$ is a Young subgroup of shape $\lambda '$, we have

$$\begin{aligned} \langle {\mathrm {Ind}}_{W_\lambda }^{S_n}({\mathbbm {1}}), {\mathrm {Ind}}_{W_{\lambda '}}^{S_n}(\epsilon ) \rangle = 1, \end{aligned}$$

and the common irreducible constituent is $\psi ^\lambda$. James [4] constructs an irreducible submodule $V^\lambda$ of ${\mathrm {Ind}}_{W_\lambda }^{S_n}({\mathbbm {1}})$ in such a way that $V^\lambda$ contains a one-dimensional subspace on which $W_{\lambda '}$ acts as $\epsilon$. In other words,

$$\begin{aligned} \langle {\mathrm {Ind}}_{W_\lambda }^{S_n}({\mathbbm {1}}), V^\lambda \rangle \ne 0 \end{aligned}$$

and

$$\begin{aligned} \langle {\mathrm {Ind}}_{W_{\lambda '}}^{S_n}(\epsilon ), V^\lambda \rangle = \langle \epsilon , {\mathrm {Res}}_{W_{\lambda '}}^{S_n}(V^\lambda ) \rangle \ne 0. \end{aligned}$$

This forces $V^\lambda$ to be a module affording the character $\psi ^\lambda$.

In order to apply a similar approach to construct the irreducible unipotent modules of ${{\mathrm {GL}}}_n({{\mathbb {F}}}_q)$, we need to find modules that are analogous to ${\mathrm {Ind}}_{W_\lambda }^{S_n}({\mathbbm {1}})$ and ${\mathrm {Ind}}_{W_{\lambda '}}^{S_n}(\epsilon )$. The irreducible unipotent characters of ${{\mathrm {GL}}}_n({{\mathbb {F}}}_q)$ are indexed by partitions of n; let $\chi ^\lambda$ be the irreducible unipotent character corresponding to $\lambda$, as in [3, 9]. Note that $\chi ^{(1^n)}$ is the trivial character and $\chi ^{(n)}$ is the Steinberg character. As the trivial character is indexed by the transpose of (n) [rather than by (n) as with the symmetric group], we should expect our results to be transposed to some extent. The next proposition follows from the Pieri rules.

Proposition 2

Let $P_\lambda$ be the parabolic subgroup of G of shape $\lambda$; then we have

$$\begin{aligned} {\mathrm {Ind}}_{P_\lambda }^G({\mathbbm {1}}) = \sum _{\mu ' \succeq \lambda } K_{\mu '\lambda }\chi ^\mu . \end{aligned}$$

Both James’ construction and our construction use ${\mathrm {Ind}}_{P_{\lambda '}}^G({\mathbbm {1}})$ as the analogue of ${\mathrm {Ind}}_{W_\lambda }^{S_n}({\mathbbm {1}})$.

Let $\varPsi ^\lambda$ be the degenerate Gelfand–Graev character corresponding to the partition $\lambda$, as in [11, Section 12.1]. The character $\varPsi ^\lambda$ is obtained by inducing a linear character of ${\mathrm {UT}}_n({{\mathbb {F}}}_q)$ that is trivial on certain root subgroups determined by the partition $\lambda$.

Proposition 3

([11, Section 12.1]) If $\lambda$ and $\mu$ are partitions of n, we have that

$$\begin{aligned} \langle \varPsi ^\lambda , \chi ^\mu \rangle = K_{\mu \lambda }. \end{aligned}$$

It follows that $\langle {\mathrm {Ind}}_{P_{\lambda '}}^G({\mathbbm {1}}),\varPsi ^\lambda \rangle = 1$; the construction of James [6] uses $\varPsi ^\lambda$ as the analogue of ${\mathrm {Ind}}_{W_{\lambda '}}^{S_n}(\epsilon )$.

Kawanaka [8], constructs the generalized Gelfand–Graev characters of a reductive group over a finite field, with each character associated with a nilpotent orbit of the corresponding Lie algebra. In the case of ${{\mathrm {GL}}}_n({{\mathbb {F}}}_q)$, the nilpotent orbits are indexed by the partitions of n; let $\varGamma ^\lambda$ be the generalized Gelfand–Graev character indexed by the partition $\lambda$. Thiem and the author [1] show that $\varGamma ^\lambda$ can be obtained by inducing a linear character from $U_{\lambda '}$ and calculate the multiplicities of the unipotent characters in $\varGamma ^\lambda$.

Theorem 1

([1]) Let $\lambda$ and $\mu$ be partitions of n; then

$$\begin{aligned} \langle \varGamma ^\lambda , \chi ^\mu \rangle = K_{\mu \lambda }(q), \end{aligned}$$

where $K_{\mu \lambda }(q)$ is the Kostka polynomial (see [9, III.6]).

It follows that $\langle {\mathrm {Ind}}_{P_{\lambda '}}^G({\mathbbm {1}}),\varGamma ^\lambda \rangle = 1$. Furthermore, note that $K_{\mu \lambda }(1) = K_{\mu \lambda }$; by setting $q=1$, we see that $\varGamma ^\lambda$ is another analogue of ${\mathrm {Ind}}_{W_{\lambda '}}^{S_n}(\epsilon )$. We construct an irreducible submodule $S^\lambda$ of ${\mathrm {Ind}}_{P_{\lambda '}}^G({\mathbbm {1}})$ that contains a one-dimensional submodule on which $U_{\lambda '}$ acts as the appropriate linear character. This forces $S^\lambda$ to be a module affording the irreducible unipotent character $\chi ^\lambda$.

4 Construction of the modules

The construction in this section is motivated by the construction of the irreducible representations of the symmetric group (see [4, 5, 10]). Many of the results are similar to those found in [6], which is not surprising as our constructions produce isomorphic modules.

Let p be the characteristic of ${{\mathbb {F}}}_q$, and let ${\mathbb {K}}$ be a field that contains a nontrivial $p\text {th}$ root of unity (in particular, the characteristic of ${\mathbb {K}}$ cannot be p). For the remained of the paper, fix a nontrivial homomorphism $\theta :{{\mathbb {F}}}_q^+ \rightarrow {\mathbb {K}}^\times$.

Definition 1

Let $\lambda$ be a partition of n, and let T be a filling of the Young diagram of shape $\lambda$ with linearly independent elements of ${\mathbb {F}}_q^n$. We call T an ${\mathbb {F}}_q^n$ -tableau.

Note that G acts on the set of ${\mathbb {F}}_q^n$-tableaux by left multiplication of the entries (considered as column vectors). If T is an ${\mathbb {F}}_q^n$-tableau of shape $\lambda$, we obtain an ordered basis ${\mathcal {B}}(T)$ of ${\mathbb {F}}_q^n$ by numbering the entries of T from top to bottom and then left to right.

Example 4

Let $\lambda = (4,2^2,1)$; then the ordered basis ${\mathcal {B}}(T) = \{v_1,\ldots ,v_9\}$ corresponds to the tableau

To each ${{\mathbb {F}}}_q^n$-tableaux T, we associate two subgroups of G, given by

$$\begin{aligned}&U(T) = \{g \in G \mid g \cdot v_i -v_i \in {\mathbb {F}}_q\text {-span}\{v_j \mid v_j \text { is strictly left of }v_i \text { in }T\}\text { for all }i\} \text { and} \\&P(T) = \{g \in G \mid g \cdot v_i \in {\mathbb {F}}_q\text {-span}\{v_j \mid v_j \text { is nonstrictly right of }v_i \text { in }T\}\text { for all }i\}. \end{aligned}$$

If ${{\mathcal {B}}}(T)$ is the standard ordered basis of ${\mathbb {F}}_q^n$, then $U(T) = U_{\lambda '}$ and $P(T)=P_{\lambda '}^-$.

Example 5

Let $\lambda = (4,2^2,1)$, and let ${\mathcal {B}}(T)$ be the standard ordered basis. Then

and we have

$$\begin{aligned} U(T) = \left\{ \left( \begin{array}{ccccccccc} 1 &{} 0 &{} 0 &{} 0 &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 1 &{} 0 &{} 0 &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 0 &{} 1 &{} 0 &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 1 &{} * &{} * &{} * &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} * &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} * \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 \end{array} \right) \right\} \end{aligned}$$

and

$$\begin{aligned} P(T) = \left\{ \left( \begin{array}{ccccccccc} * &{} * &{} * &{} * &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ * &{} * &{} * &{} * &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ * &{} * &{} * &{} * &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ * &{} * &{} * &{} * &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ * &{} * &{} * &{} * &{} * &{} * &{} * &{} 0 &{} 0 \\ * &{} * &{} * &{} * &{} * &{} * &{} * &{} 0 &{} 0 \\ * &{} * &{} * &{} * &{} * &{} * &{} * &{} 0 &{} 0 \\ * &{} * &{} * &{} * &{} * &{} * &{} * &{} * &{} 0 \\ * &{} * &{} * &{} * &{} * &{} * &{} * &{} * &{} * \end{array} \right) \right\} . \end{aligned}$$

If T is the ${{\mathbb {F}}}_q^n$-tableau corresponding to the ordered basis ${\mathcal {B}}(T)=\{v_1,v_2,\ldots ,v_n\}$, let X(T) be the set of pairs (i, j) such that $v_i$ lies in the box directly to the left of $v_j$ in T. In Example 5,

$$\begin{aligned} X(T) = \{(1,5),(2,6),(3,7),(5,8),(8,9)\}. \end{aligned}$$

Define a linear character $\psi _T$ of U(T) by

$$\begin{aligned} \psi _T(u) = \theta \left( \sum _{(i,j) \in X(T)} \text {the coefficient of } v_i \text { in }uv_j \right) , \end{aligned}$$

where $\theta$ is the fixed nontrivial homomorphism from ${{\mathbb {F}}}_q^+$ to ${\mathbb {K}}^\times$.

Remark 2

The groups P(T) and U(T) are analogous to the row-stabilizer $R_T$ and the column-stabilizer $C_T$ of the symmetric group; the linear character $\psi _T$ is analogous to the sign character (see [4, 5, 10]).

Consider the permutation G-module

$$\begin{aligned} {\mathbb {K}}\text {-span}\{T \mid T \text { is an } {{\mathbb {F}}}_q^n\text {-tableau of shape }\lambda \}; \end{aligned}$$

this module is isomorphic to the left regular module of G. We define

$$\begin{aligned} m_T&= \sum _{p \in P(T)} pT \quad \text {and} \\ e_T&= \sum _{u \in U(T)} \psi _T(u^{-1})m_{uT}. \end{aligned}$$

The following proposition is easy to verify directly.

Lemma 1

Let T be any ${{\mathbb {F}}}_q^n$-tableau.

• 1. For all $g \in G$, we have $U(gT) = gU(T)g^{-1}$ and $P(gT) = gP(T)g^{-1}$.

• 2. For all $g \in G$, we have $g \cdot m_T = m_{g \cdot T}$ and $g \cdot e_T = e_{gT}$.

• 3. For all $u \in U(T)$ and $g \in G$, we have $\psi _{gT}(gug^{-1}) = \psi _T(u)$.

• 4. For all $p \in P(T)$, we have $m_{pT} = m_T$.

• 5. For all $u \in U(T)$, we have $e_{uT} = \psi _T(u)e_T$.

Let

$$\begin{aligned} M^\lambda = {\mathbb {K}}\text {-span}\{m_T \mid T \text { is an }{{\mathbb {F}}}_q^n\text {-tableau of shape }\lambda \} \end{aligned}$$

and

$$\begin{aligned} S^\lambda = {\mathbb {K}}\text {-span}\{e_T \mid T \text { is an }{{\mathbb {F}}}_q^n\text {-tableau of shape }\lambda \}; \end{aligned}$$

by part (2) of Lemma 1, $M^\lambda$ and $S^\lambda$ are both G-modules.

Remark 3

The module $M^\lambda$ is isomorphic to the permutation representation of G on the set of $\lambda$-flags. This means that our module $M^\lambda$ is isomorphic to the module $M_{\lambda '}$ of James (as in [6, 10.1]). Note that

$$\begin{aligned} M^\lambda \cong {\mathrm {Ind}}_{P_{\lambda '}^-}^G({\mathbbm {1}}) \end{aligned}$$

and, in the case ${\mathbb {K}}={\mathbb {C}}$, we have that

$$\begin{aligned} {\mathrm {Ind}}_{U(T)}^{{{\mathrm {GL}}}_n({{\mathbb {F}}}_q)}(\psi _T) = \varGamma ^\lambda \end{aligned}$$

(this follows from the elementary construction of the generalized Gelfand–Graev representations in [1]). In particular, when ${\mathbb {K}}={\mathbb {C}}$ the module $S^\lambda$ satisfies the desired properties from Sect. 3.

Let W be the group of permutation matrices of G. The Bruhat decomposition allows us to write each element $g \in G$ in the form $g = uwb$, where $u \in {\mathrm {UT}}_n({{\mathbb {F}}}_q)$, $w \in W$, and $b \in B^-_n({{\mathbb {F}}}_q)$. For a partition $\lambda$, let $W_\lambda = W \cap P_\lambda$ be the Young subgroup of shape $\lambda$.

Lemma 2

Let T and $T'$ be ${{\mathbb {F}}}_q^n$-tableaux of shape $\lambda$. Then $U(T) \cap P(T') = \{1\}$ if and only if there exist $u \in U(T)$ and $p \in P(T')$ with $uT = pT'$.

Proof

By Lemma 1, we only need to consider the case where T corresponds to the standard ordered basis. In this case, $U(T) = U_{\lambda '}$ and $P(T) = P_{\lambda '}^-$. Let $H={\mathrm {UT}}_n({\mathbb {F}}_q)\cap P(T)$; then

$$\begin{aligned} H \cong {\mathrm {UT}}_{\lambda _1'}({\mathbb {F}}_q) \times \cdots \times {\mathrm {UT}}_{\lambda '_k}({\mathbb {F}}_q) \end{aligned}$$

and ${\mathrm {UT}}_n({{\mathbb {F}}}_q)=U(T)\rtimes H$.

Let $T' = gT$, and let $g = vwb$, with $v \in {\mathrm {UT}}_n({{\mathbb {F}}}_q)$, $w \in W$, and $b \in B^-_n({{\mathbb {F}}}_q)$. Then

$$\begin{aligned} U(T) \cap P(T') = U(T) \cap gP(T)g^{-1} =U(T) \cap vwP(T)w^{-1}v^{-1}. \end{aligned}$$

As ${\mathrm {UT}}_n({{\mathbb {F}}}_q)$ normalizes U(T), we have that $U(T) \cap vwP(T)w^{-1}v^{-1} = \{1\}$ if and only if $U(T) \cap w P(T)w^{-1} = \{1\}$; this occurs exactly when $w \in W_{\lambda '}$. We now have

$$\begin{aligned} g \in U(T)vwP(T) = U(T)vP(T) = U(T)P(T), \end{aligned}$$

as $W_{\lambda '}$ and H are both contained in P(T) and $v \in U(T) \rtimes H$. Write $g = u\tilde{p}$; then

$$\begin{aligned} T'&= gT \\ g\tilde{p}^{-1}g^{-1} T'&= g\tilde{p}^{-1}T \\ g\tilde{p}^{-1}g^{-1} T'&= uT; \end{aligned}$$

set $p = g\tilde{p}^{-1}g^{-1}$. Conversely, suppose that $uT = pT'$, with $u \in U(T)$ and $p \in P(T')$; then $u = pg = g\tilde{p}$ for some $\tilde{p} \in P(T)$. It follows that $g \in U(T)P(T)$; hence $U(T) \cap P(T') = \{1\}$.

Lemma 3

Suppose that T and $T'$ are ${{\mathbb {F}}}_q^n$-tableaux of shape $\lambda$ that satisfy $U(T) \cap P(T') \ne \{1\}$; then there exists $g \in U(T) \cap P(T')$ with $\psi _T(g) \ne 1$.

Proof

Once again, it suffices to consider the case where T is the tableau corresponding to the standard ordered basis. As $U(T)\cap P(T') \ne \{1\}$, we have

$$\begin{aligned} U(T)\cap P(T') = U(T)\cap vwP(T)w^{-1}v^{-1} = v(U(T)\cap wP(T)w^{-1})v^{-1} \end{aligned}$$

for some $w \in W- W_{\lambda '}$ and $v \in {\mathrm {UT}}_n({{\mathbb {F}}}_q)$. As $w \notin W_{\lambda '}$, there must be at least one pair $(i,j) \in X(T)$ (that is, with $v_{i}$ directly left of $v_{j}$ in T), but with $v_{i}$ appearing nonstrictly right of $v_{j}$ in wT. If $e_{ij}$ denotes the matrix with a 1 in the ith row and jth column and 0’s elsewhere, then $1+\alpha e_{ij} \in U(T)\cap wP(T)w^{-1}$ for all $\alpha \in {\mathbb {F}}_q$; as $\theta$ is nontrivial, for some $\alpha \in {\mathbb {F}}_q$, we have $\psi _T(1+\alpha e_{ij}) \ne 1$.

By the construction of $\psi _T$, we have that $\psi _T(1+\alpha e_{ij}) = \psi _T(v(1+\alpha e_{ij})v^{-1})$ for all $v \in {\mathrm {UT}}_n({{\mathbb {F}}}_q)$. Let $g = v(1+\alpha e_{ij})v^{-1}$.

For an ${{\mathbb {F}}}_q^n$-tableau T, define an element $k_T \in {\mathbb {K}}G$ by

$$\begin{aligned} k_T = \sum _{u \in U(T)} \psi _T(u^{-1})u. \end{aligned}$$

The following two lemmas describe how $k_T$ acts on $m_{T'}$ for certain ${{\mathbb {F}}}_q^n$-tableaux $T'$.

Lemma 4

Let T and $T'$ be ${{\mathbb {F}}}_q^n$-tableaux of the same shape; then $k_Tm_{T'} \in {\mathbb {K}}e_T$.

Proof

First suppose that $U(T) \cap P(T') = \{1\}$; by Lemma 2, there exists $u~\in ~U(T)$ and $p \in P(T')$ with $uT = pT'$. By Lemma 1, we have that

$$\begin{aligned} k_Tm_{T'} = k_T m_{pT'} = k_T m_{uT} = \psi _T(u)k_Tm_T = \psi _T(u)e_T. \end{aligned}$$

Suppose $U(T) \cap P(T') \ne \{1\}$; by Lemma 3, there exists $g \in U(T) \cap P(T')$ with $\psi _T(g) \ne 1$. We have that

$$\begin{aligned} k_{T}m_{T'} = k_T m_{gT'} = \psi _T(g)k_Tm_{T'}, \end{aligned}$$

hence $k_Tm_{T'} = 0$.

Lemma 5

Let $\lambda$ and $\mu$ be partitions of n. If T is an ${{\mathbb {F}}}_q^n$-tableaux of shape $\lambda$ and $T'$ is an ${{\mathbb {F}}}_q^n$-tableaux of shape $\mu$, then $k_Tm_{T'}=0$ unless $\mu \succeq \lambda$.

Proof

Suppose that $\mu \not \succeq \lambda$. It suffice to show that this is true when T is the ${{\mathbb {F}}}_q^n$-tableaux of shape $\lambda$ corresponding to the standard ordered basis of ${{\mathbb {F}}}_q^n$. Let $g \in G$ be such that $g^{-1}T'$ is the ${{\mathbb {F}}}_q^n$-tableaux of shape $\mu$ corresponding to the standard ordered basis of ${{\mathbb {F}}}_q^n$. We have that

$$\begin{aligned} U(T) \cap P(T') = U(T) \cap g P(g^{-1}T')g^{-1} = v(U(T) \cap w P(g^{-1}T')w^{-1})v^{-1}, \end{aligned}$$

where $v \in {\mathrm {UT}}_n({{\mathbb {F}}}_q)$, $w \in W$, and $vwP(g^{-1}T') = gP(g^{-1}T')$.

As $\mu \not \succeq \lambda$, there must be at least one pair $(i,j) \in X(T)$ (that is, with $v_{i}$ directly left of $v_{j}$ in T), but with $v_i$ nonstrictly right of $v_j$ in $wg^{-1}T'$. Then $1+\alpha e_{ij} \in U(T)\cap wP(g^{-1}T')w^{-1}$ for all $\alpha \in {\mathbb {F}}_q$; as $\theta$ is nontrivial, for some $\alpha \in {\mathbb {F}}_q$, we have $\psi _T(1+\alpha e_{ij}) \ne 1$.

By the construction of $\psi _T$, we have that $\psi _T(1+\alpha e_{ij}) = \psi _T(v(1+\alpha e_{ij})v^{-1})$ for all $v \in {\mathrm {UT}}_n({{\mathbb {F}}}_q)$. In other words, there is an element of $x \in U(T) \cap P(T')$ with $\psi _T(x) \ne 1$. Then

$$\begin{aligned} k_{T}m_{T'} = k_T m_{xT'} = \psi _T(x)k_Tm_{T'}, \end{aligned}$$

hence $k_{T}m_{T'} = 0$.

One consequence of Lemma 4 is the following proposition.

Proposition 4

The module $S^\lambda$ is indecomposable.

Proof

Suppose that $S^\lambda = A \oplus B$ is a decomposition of $S^\lambda$ into a direct sum of G-modules. Let T be any ${{\mathbb {F}}}_q^n$-tableaux of shape $\lambda$; by Lemma 4,

$$\begin{aligned} k_TS^\lambda \subseteq {\mathbb {K}}e_T. \end{aligned}$$

At the same time,

$$\begin{aligned} k_{T}e_T&= \sum _{u \in U(T)} \psi _T(u^{-1})ue_T \\&= \sum _{u \in U(T)} \psi _T(u^{-1})\psi _T(u)e_T \\&= |U(T)|e_T. \end{aligned}$$

As the characteristic of ${\mathbb {K}}$ does not divide |U(T)|, we have that in fact

$$\begin{aligned} {\mathbb {K}}e_T = k_TS^\lambda = k_T A \oplus k_T B, \end{aligned}$$

and $e_T \in k_T A$ or $e_T \in k_T B$. We may assume that $e_T \in k_T A$; as A is a ${\mathbb {K}}G$-module, $e_T \in A$. But $e_T$ generates $S^\lambda$ as a ${\mathbb {K}}G$-module; hence $A = S^\lambda$.

There is an immediate corollary of Proposition 4.

Corollary 1

If $char ({\mathbb {K}})$ does not divide |G|, then $S^\lambda$ is irreducible. In particular, if ${\mathbb {K}}$ has characteristic 0, then $S^\lambda$ is irreducible.

Given an ${{\mathbb {F}}}_q^n$-tableau T, let $\overline{T}$ be the ${{\mathbb {F}}}_q^n$-tableau obtained by replacing the entry $v_i$ with $-v_i$ if $v_i$ is in an odd column of T and fixing the entry $v_i$ if $v_i$ is in an even column of T. For example, if

Lemma 6

For all ${{\mathbb {F}}}_q^n$-tableaux T, we have that $U(\overline{T}) = U(T)$, $\overline{T}\in P(T)\cdot T$, and $\psi _{\overline{T}}(u) = \psi _T(u^{-1})$.

Proof

The first two claims are trivial. For the third, recall that

$$\begin{aligned} \psi _{T}(u) = \theta \left( \sum _{(i,j) \in X(T)} \text {the coefficient of } v_i \text { in }uv_j \right) , \end{aligned}$$

where once again X(T) is the set of pairs (i, j) such that $v_i$ lies in the box directly to the left of $v_j$ in T. If $(i,j) \in X(T)$, then exactly one of $v_i$ or $v_j$ has its sign switched in $\overline{T}$. It follows that

$$\begin{aligned} \psi _{\overline{T}}(u) = \theta \left( -\sum _{(i,j) \in X(T)} \text {the coefficient of } v_i \text { in }uv_j \right) = \psi _{T}(u^{-1}). \end{aligned}$$

We define a G-invariant bilinear form on $M^\lambda$ by

$$\begin{aligned}{}[m_T,m_{T'}] = \delta _{m_T,m_{T'}}, \end{aligned}$$

and extending by linearity. We remark that $\delta _{m_T,m_{T'}}$ is not the same as $\delta _{T,T'}$, as $m_T = m_{T'}$ exactly when $T' \in P(T) \cdot T$.

Proposition 5

Let V be a submodule of $M^\lambda$; then either $S^\lambda \subseteq V$ or $V ~\subseteq ~(S^\lambda )^\perp$.

Proof

First suppose that there exists $x \in V$ and an ${{\mathbb {F}}}_q^n$-tableau T such that $k_T x \ne 0$; then by Lemma 4, we have $e_T \in V$; hence $S^\lambda \subseteq V$.

Conversely suppose that for all $x \in V$ and all ${{\mathbb {F}}}_q^n$-tableaux T, we have $k_T x = 0$. Then

$$\begin{aligned}{}[x,e_T]&= \sum _{u \in U(T)}\psi _T(u^{-1})[x,um_T] \\&= \sum _{u \in U(T)}\psi _T(u^{-1})[u^{-1}x,m_T] \\&= [k_{\overline{T}}x, m_T] \\&= [0,m_T] = 0, \end{aligned}$$

as the bilinear form is G-invariant. It follows that $V \subseteq (S^\lambda )^\perp$.

When constructing the irreducible representations of the symmetric groups, it is possible to have $S^\lambda \subseteq (S^\lambda )^\perp$; that will not be the case, however, with the finite general linear groups.

Lemma 7

If T is any ${{\mathbb {F}}}_q^n$-tableaux, then $e_T \notin (S^\lambda )^\perp$, hence $S^\lambda \not \subseteq (S^\lambda )^\perp$.

Proof

We have that

$$\begin{aligned}{}[e_T,e_{\overline{T}}]&= \sum _{u_1,u_2 \in U(T)} \psi _T(u_1^{-1})\psi _{\overline{T}}(u_2^{-1})[m_{u_1T},m_{u_2\overline{T}}] \\&= \sum _{u \in U(T)} \psi _T(u^{-1})\psi _{\overline{T}}(u^{-1}) \\&= |U(T)|, \end{aligned}$$

which is not 0 as $\text {char}({\mathbb {K}})$ does not divide q.

The following corollary is an immediate consequence of Proposition 5.

Corollary 2

We have the following.

• 1. $S^\lambda \cap (S^\lambda )^\perp$ is the unique maximal submodule of $S^\lambda$.

• 2. The G-module $S^\lambda /(S^\lambda \cap (S^\lambda )^\perp )$ is irreducible.

Define $D^\lambda = S^\lambda {/}(S^\lambda \cap (S^\lambda )^\perp )$.

Proposition 6

Let $\lambda$ and $\mu$ be partitions of n; then we have the following.

• 1. If $D^\lambda$ is a composition factor of $M^\mu$, then $\lambda \preceq \mu$.

• 2. $D^\lambda$ is a composition factor of $M^\lambda$.

• 3. If $D^\lambda \cong D^\mu$, then $\lambda = \mu$.

Proof

Suppose that $D^\lambda$ is a composition factor of $M^\mu$; then we have a nonzero G-module homomorphism

$$\begin{aligned} \varphi : D^\lambda \rightarrow M^\mu {/}V \end{aligned}$$

for some submodule V of $M^\mu$. As the elements $e_T+S^\lambda \cap (S^\lambda )^\perp$ generate $D^\lambda$, for some ${{\mathbb {F}}}_q^n$-tableau T we have $\varphi (e_T+S^\lambda \cap (S^\lambda )^\perp ) \ne 0$. Note that $k_Te_T$ is a nonzero multiple of $e_T$; hence $k_T\varphi (e_T+S^\lambda \cap (S^\lambda )^\perp ) \ne 0$. By Lemma 5, $\lambda \preceq \mu$.

Claim (2) follows directly from the definition of $D^\lambda$, and (3) is an immediate consequence of (1) and (2).

James [6] constructs a collection of irreducible modules of G, one for each partition of n. James denotes these modules by $D_\lambda$ and shows the following.

• 1. The modules $D_\lambda$ are the unipotent modules of G. In other words, these modules are exactly the composition factors of ${\mathrm {Ind}}_B^G({\mathbbm {1}})$, up to isomorphism.

• 2. If $D_\lambda \cong D_\mu$, then $\lambda = \mu$.

• 3. The module $D_\lambda$ is a composition factor of ${\mathrm {Ind}}_{P_\lambda }^G({\mathbbm {1}})$.

• 4. Every composition factor of ${\mathrm {Ind}}_{P_\mu }^G({\mathbbm {1}})$ is isomorphic to $D_\lambda$ for some $\lambda \succeq \mu$.

Note that the modules $D_\lambda$ are uniquely characterized (up to isomorphism) by properties (2)–(4). As $M^\lambda \cong {\mathrm {Ind}}_{P_{\lambda '}^-}^G({\mathbbm {1}})$ and ${\mathrm {Ind}}_{P_{\lambda '}^-}^G({\mathbbm {1}}) \cong {\mathrm {Ind}}_{P_{\lambda '}}^G({\mathbbm {1}})$ (see [6, 14.7]), by Proposition 6 we have the following.

Corollary 3

We have that $D^\lambda \cong D_{\lambda '}$. In particular, the $D^\lambda$ are the irreducible unipotent modules of G.

Remark 4

The indexing of our modules and those of James differs by the transpose of the partition. Our indexing is chosen to match the convention of Green [3] and MacDonald [9].

5 Further directions

There are a number of questions raised by our construction; a few of particular interest are listed below.

• Even in the case that ${\mathbb {K}}={\mathbb {C}}$, it is still an open problem to find bases for the irreducible unipotent modules. James addresses this question in [6], and a number of papers (for example [2]) contribute partial results. By using a construction that looks more like that of the symmetric group, where bases are known, one might be able to make further progress.

• The finite orthogonal, symplectic, and unitary groups also have unipotent representations; however explicit constructions of the modules are not known. For these groups, there are not enough degenerate Gelfand–Graev representations to distinguish unipotent representations; however by using the generalized Gelfand–Graev representations it might be possible to construct the unipotent modules.

• Most of the irreducible representations of ${{\mathrm {GL}}}_n({{\mathbb {F}}}_q)$ are not unipotent. In [7], James uses the degenerate Gelfand–Graev representations to construct all of the irreducible modules of ${{\mathrm {GL}}}_n({{\mathbb {F}}}_q)$. By modifying our construction, we might be able to obtain more elementary constructions of these modules using the generalized Gelfand–Graev representations.