Low-dimensional Representations of Special Linear Groups in Cross Characteristics

Robert M. Guralnick^{†}

Department of Mathematics University of Southern California Los Angeles, CA 90089-1113

Pham Huu Tiep^{∗}

Department of Mathematics University of Florida Gainesville, FL 32611-8105

(Proc. London Math. Soc. 78 (1999), 116 − 138)

Abstract The low-dimensional projective irreducible representations in cross characteristics of the projective special linear group L_{n}(q) are investigated. If n ≥ 3 and (n, q) ≠ (3, 2), (3, 4), (4, 2), (4, 3), all such representations of the first degree (which is (q^{n} − q)/(q − 1) − κ_{n} with κ_{n} = 0 or 1) and the second degree (which is (q^{n} − 1)/(q − 1)) come from Weil representations. We show that the gap between the second and the third degree is roughly q^{2n−4}.

Key words: Special linear group, low-dimensional representation in cross characteristic, Weil representation

1991 Mathematics Subject Classification: 20C20, 20C33

1 Introduction

Let G(q) be a finite simple group of Lie type over a field of order q, where q is a prime power, and let ℓ(G(q)) (resp. d_{ℂ}(G(q))) denote the smallest integer t > 1 such that G(q) has a projective irreducible representation of degree t over a field of cross characteristic (that is, of characteristic not dividing q), (resp. over the field ℂ of complex numbers). The main result of [20], [25] is a table of lower bounds for ℓ(G(q)). This result turns out to be extremely useful in numerous applications. For reader’s convenience, we reproduce the bounds for L_{n}(q) in Table I.

†E-mail: guralnic@mtha.usc.edu

Partially supported by the NSF.

∗E-mail: tiep@math.ufl.edu

Table I. The Landázuri-Seitz-Zalesskii lower bounds for ℓ(L_{n}(q))

G(q)  Bound  Exceptions
L_{2}(q) (q − 1)/ gcd(2, q − 1) \{ \begin{matrix}ℓ(L_{2}(4)) = 2, \\ ℓ(L_{2}(9)) = 3. \end{matrix}

\begin{matrix}L_{n}(q), \\ n ≥ 3 \end{matrix} (q^{n} − 1)/(q − 1) − n \{ \begin{matrix}ℓ(L_{3}(2)) = 2, \\ ℓ(L_{3}(4)) = 4, \\ ℓ(L_{4}(2)) = 7, \\ ℓ(L_{4}(3)) = 26. \end{matrix}

For the case of complex representations, one can say more about lowdimensional representations, using the Deligne-Lusztig theory. In particular, the low-dimensional projective complex characters of the finite classical groups are determined in [27]; the corresponding representations are investigated by the second author (work in progress). Yet these results have many applications, see for instance [21], [22], [23], [26], [28]. For reader’s convenience we reproduce the relevant result [27] for L_{n}(q). Given a finite group X, d_{j,ℂ}(X) denotes the j−th degree of nontrivial projective irreducible complex representations of X, and N_{j} denotes the number of (equivalence classes of) the corresponding representations.

Table II. Complex characters of the first three degrees of L_{n}(q)

L  (d_{1,ℂ}, N_{1})  (d_{2,ℂ}, N_{2})  (d_{3,ℂ}, N_{3})  Conditions
L_{2}(q) (\frac{q−1}{2}, 2)  (\frac{q+1}{2}, 2)  (q − 1, \frac{q−1}{2})  2 |̸q, q ≠ 9
L_{2}(q) (q − 1, \frac{q}{2})  (q, 1)  (q + 1, \frac{q−2}{2})  2|q, q ≠ 4
L_{2}(4)  (2, 2)  (3, 2)  (4, 2)
L_{2}(9)  (3, 4)  (4, 2)  (5, 2)
L_{3}(q) (q(q + 1), 1) (q^{2} + q + 1, q − 2) ((q − 1)(q^{2} − 1), \frac{q(q+1)}{3}) \{ \begin{matrix}3 |̸(q − 1), \\ q ≠ 2 \end{matrix}
L_{3}(q) (q(q + 1), 1) (q^{2} + q + 1, q − 2)  (\frac{(q−1)(q^{2}−1)}{3}, 6)  \{ \begin{matrix}3|(q − 1), \\ q ≠ 4 \end{matrix}
L_{3}(2)  (3, 2)  (4, 2)  (6, 3)
L_{3}(4)  (6, 2)  (8, 12)  (10, 6)
L_{4}(q) (\frac{q^{4}−q}{q−1} , 1)  (\frac{q^{4}−1}{q−1}, q − 2)  (\frac{(q^{3}−1)(q−1)}{2}, 2)  2 |̸q, q ≠ 3
L_{4}(q) (\frac{q^{4}−q}{q−1} , 1)  (\frac{q^{4}−1}{q−1}, q − 2)  ((q^{3} − 1)(q − 1), \frac{q}{2})  2|q, q ≠ 2
L_{4}(2)  (7, 1)  (8, 1)  (14, 1)
L_{4}(3)  (26, 2)  (39, 1)  (40, 1)
L_{n}(q) (\frac{q^{n}−q}{q−1} , 1)  (\frac{q^{n}−1}{q−1}, q − 2)  (\frac{(q^{n}−1)(q^{n−1}−q^{2})}{(q−1)(q^{2}−1)}, 1)  \{ \begin{matrix}n ≥ 5, q ≥ 3, \\ (n, q) ≠ (6, 3) \end{matrix}
L_{n}(2) (\frac{q^{n}−q}{q−1} , 1) (\frac{(q^{n}−1)(q^{n−1}−q^{2})}{(q−1)(q^{2}−1)}, 1)  (\frac{(q^{n}−1)(q^{n−1}−1)}{(q−1)(q^{2}−1)}, 1)  \{ \begin{matrix}n ≥ 5, n ≠ 6, \\ q = 2 \end{matrix}
L_{6}(2)  (62, 1)  (217, 1)  (588, 1)
L_{6}(3)  (363, 1)  (364, 1)  (6292, 2)

In this paper we concentrate on the projective representations of the projective special linear group L = L_{n}(q). In the “generic” case one has ℓ(L) ≥ (q^{n} − 1)/(q − 1) − n. As mentioned in [25], the first author of the present paper was able to improve this bound to the best possible bound (q^{n}− q)/(q − 1) − κ_{n}. Here and hereafter, if p is a fixed prime integer not dividing q, then κ_{n} is defined to be 1 if p divides (q^{n} − 1)/(q − 1), and 0 otherwise. Furthermore, ζ is defined to be 1 if p divides [\frac{n−1}{2}], and 0 otherwise. For a finite group X, let d_{j,p}(X) denote the j−th degree of nontrivial projective irreducible representations of X over a field of characteristic p. Thus L has d_{1,p} = (q^{n}−q)/(q−1)−κ_{n}, d_{2,p} = (q^{n}−1)/(q−1) in the “generic” case. Up to the very recent time it was an open question if all representations of degree d_{1,p} and d_{2,p} were (irreducible) constituents of the p-modular reduction of the Weil representations (cf. [8], [11], [24], [29], [30]) of SL_{n}(q). It has recently been shown in [9] that the answer is yes for the representations of degree (q^{n} − q)/(q − 1) and (q^{n} − 1)/(q − 1). Furthermore, in various applications, see for instance [9], one needs to show that there is a relatively big gap between d_{2,p} and d_{3,p}, say, d_{3,p} ≥ 2d_{1,p}.

In the 80’s R. Dipper and G. James [3], [4], [13], [14], [15] developed a powerful theory of irreducible representations of GL_{n}(q) in cross characteristics, which can be viewed as a modular analogue of J. A. Green’s theory of irreducible complex representations of GL_{n}(q) [10]. (The blocks of GL_{n}(q) in cross characteristics are described in [6].) Using these results of Dipper and James, we are able to prove the following result. Given an integer m, m_{p′} denotes the p′-part of m.

Theorem 1.1 Let n ≥ 3, q a prime power, p a prime not dividing q and (n, q) ≠ (3, 2), (3, 4), (4, 2), (4, 3), (6, 2), (6, 3). Suppose L = L_{n}(q) has a projective absolutely irreducible representation Φ of degree d over a field of characteristic p, where

1 < d < 𝔡(L) = \{ \begin{matrix}(q − 1)(q^{2} − 1)/(3, q − 1), & if n = 3 \\ (q − 1)(q^{3} − 1)/(2, q − 1), & if n = 4 \\ (q^{n−1} − 1) (\frac{q^{n−2}−q}{q−1} − κ_{n−2}) , & if n ≥ 5 \end{matrix} .

Then Φ is an irreducible constituent of the p-modular reduction of a complex Weil representation of SL_{n}(q). More precisely, either Φ is equivalent to a unique representation of degree (q^{n} − q)/(q − 1) − κ_{n}, or it is equivalent to one of (q − 1)_{p′} − 1 (inequivalent) representations of degree (q^{n} − 1)/(q − 1).

Observe that if n = 3, 4, then SL_{n}(q) has an irreducible representation in any cross characteristic of degree 𝔡(L), which is equal to d_{3,ℂ}, cf. Table II. Thus the bound 𝔡(L) given in Theorem 1.1 is best possible when n = 3, 4. Suppose n ≥ 5. Then 𝔡(L) is slightly less than d_{3,ℂ} (see Table II), but it is a polynomial of the same degree 2n − 4 and with the same leading term q^{2n−4} as of d_{3,ℂ}. Let 𝔟(L) denote the greatest integer t with property: any (projective) absolutely irreducible representation of L = L_{n}(q) of degree d, 1 < d < t, is one of the (q − 1)_{p′} representations listed in Theorem 1.1. (Thus 𝔟(L) = d_{3,p}(L) if (q − 1)_{p′} > 1, and 𝔟(L) = d_{2,p}(L) if (q − 1)_{p′} = 1.) According to Corollary 3.2 and Proposition 5.4, SL_{n}(q) has an irreducible representation of degree

ℓ := \{ \begin{matrix}(\frac{q^{n}−1}{q−1}) (\frac{q^{n−1}−q^{2}}{q^{2}−1} − κ_{n−2}) + κ_{n−2}(κ_{n} + 1) − κ_{n−1}, & if p |̸(q + 1) \\ (\frac{q^{n}−1}{q−1}) (\frac{q^{n−1}−q^{2}}{q^{2}−1} − κ_{n}) + κ_{n}(κ_{n} + 1) − ζ, & if p|(q + 1) \end{matrix} .

Clearly, 𝔡(L) ≤ 𝔟(L) ≤ ℓ. Conjecturally ℓ = 𝔟(L) in the “generic” case. Moreover, ℓ−𝔡(L) is a polynomial of q with leading term q^{2n−6}. The following result improves Theorem 1.1 by giving a lower bound for 𝔟(L), which differs from ℓ by a polynomial of q with leading term q^{2n−8} (when n ≥ 7).

Theorem 1.2 Let n ≥ 5, q a prime power, p a prime not dividing q, (n, q) ≠ (5, 2), (5, 4), (6, 2), (6, 3), (8, 2), (8, 3), and let L = L_{n}(q). Then 𝔟(L) ≥ (q^{n−1} − 1)(\frac{q^{n−2}−1}{q−1} − κ_{n−2}) + 𝔟(L_{n−1}(q)).

The corresponding result for the remaining groups is tabulated as follows:

Table III. Modular characters of the first three degrees of small groups

L  (d_{1,p}, N_{1})  (d_{2,p}, N_{2})  (d_{3,p}, N_{3})  Conditions
(\frac{q−1}{2}, 2)  (q − 1, \frac{q−1}{4}) (q + 1, \frac{r−1}{2})  2 = p|\frac{q−1}{2}
L_{2}(q),  (\frac{q−1}{2}, 2)  (q − 1, \frac{r−1}{2}) (q + 1, \frac{q−3}{4})  2 = p|\frac{q+1}{2}
2 |̸q,  (\frac{q−1}{2}, 2)  (\frac{q+1}{2}, 2)  (q − 1, \frac{q−1}{2}) 2 ≠ p|(q − 1)
q ≠ 3, 5, 9  (\frac{q−1}{2}, 2)  (\frac{q+1}{2}, 2)  (q − 1, \frac{r}{2})  2 ≠ p|(q + 1)
(\frac{q−1}{2}, 2)  (\frac{q+1}{2}, 2)  (q − 1, \frac{q−1}{2})  p = 0
L_{2}(q),  (q − 1, \frac{q}{2})  (q, 1)  (q + 1, \frac{r−1}{2})  p|(q − 1)
2|q,  (q − 1, \frac{r+1}{2}) (q + 1, \frac{q}{2} − 1)  −  p|(q + 1)
q ≥ 8  (q − 1, \frac{q}{2})  (q, 1)  (q + 1, \frac{q−2}{2})  p = 0
L_{2}(5)  (2, 2)  (4, 1)  −  p = 2
L_{2}(4)  \begin{matrix}(2, 2) \\ (2, 1) \\ (2, 2) \end{matrix}  \begin{matrix}(3, 2) \\ (3, 1) \\ (3, 2) \end{matrix}  \begin{matrix}(4, 1) \\ (4, 1) \\ (4, 2) \end{matrix}  \begin{matrix}p = 3 \\ p = 5 \\ p = 0 \end{matrix}

L_{2}(9)  \begin{matrix}(3, 4) \\ (3, 2) \\ (3, 4) \end{matrix}  \begin{matrix}(4, 2) \\ (4, 2) \\ (4, 2) \end{matrix}  \begin{matrix}(8, 2) \\ (5, 2) \\ (5, 2) \end{matrix}  \begin{matrix}p = 2 \\ p = 5 \\ p = 0 \end{matrix}
L_{3}(2)  \begin{matrix}(3, 2) \\ (2, 1) \end{matrix}  \begin{matrix}(4, 2) \\ (3, 1) \end{matrix}  \begin{matrix}(6, 3) \\ (4, 1) \end{matrix}  \begin{matrix}p = 0, 3 \\ p = 7 \end{matrix}

L_{3}(4)  \begin{matrix}(4, 6) \\ (6, 2) \\ (6, 2) \\ (6, 2) \end{matrix}  \begin{matrix}(6, 3) \\ (8, 6) \\ (8, 12) \\ (8, 12) \end{matrix}  \begin{matrix}(8, 12) \\ (10, 6) \\ (10, 3) \\ (10, 6) \end{matrix}  \begin{matrix}p = 3 \\ p = 5 \\ p = 7 \\ p = 0 \end{matrix}

L_{4}(2)  \begin{matrix}(7, 1) \\ (7, 1) \end{matrix}  \begin{matrix}(8, 1) \\ (8, 1) \end{matrix}  \begin{matrix}(13, 1) \\ (14, 1) \end{matrix}  \begin{matrix}p = 3, 5 \\ p = 0, 7 \end{matrix}
L_{4}(3)  \begin{matrix}(26, 2) \\ (26, 2) \\ (26, 2) \end{matrix}  \begin{matrix}(38, 1) \\ (38, 1) \\ (39, 1) \end{matrix}  \begin{matrix}(208, 2) \\ (40, 1) \\ (40, 1) \end{matrix}  \begin{matrix}p = 2 \\ p = 5 \\ p = 0, 13 \end{matrix}
L_{6}(2)  \begin{matrix}(62 − κ_{6}, 1) \\ (62 − κ_{6}, 1) \end{matrix}  \begin{matrix}(217, 1) \\ (217, 1) \end{matrix}  \begin{matrix}(526 + κ_{6}, 1) \\ (588 − κ_{5}, 1) \end{matrix}  \begin{matrix}p = 3, 5 \\ p = 0, 7, 31 \end{matrix}
L_{6}(3)  (363 − κ_{6}, 1)  (364, 1)  (6292, 2)

In this Table, r is defined to be (q −1)_{2′} if 2 = p|\frac{q−1}{2}, (q +1)_{2′} if 2 = p|\frac{q+1}{2}, (q − 1)_{p′} if 2 ≠ p|(q − 1), and (q + 1)_{p′} if 2 ≠ p|(q + 1).

In what follows, G = GL_{n}(q), p a prime not dividing q, and let (𝕂, R, 𝔽) be a p-modular splitting system for G. Fix a generator ϵ of the multiplicative group 𝔽_{q}^{•} of the field 𝔽_{q} of q elements, and fix an injection ϵ ↦ ϵ̄ from 𝔽_{q}^{•} to ℂ.

2 Modular Representations of GL_{n}(q)

In this section we summarize some results of Dipper and James that we need in the sequel.

If s ∈ \overline{𝔽_{q}}, the degree of s is the degree of the minimal polynomial of s over 𝔽_{q}. Suppose that s is an element of degree d over 𝔽_{q} and let λ be a partition of k (write λ ⊢ k) with dk = n. Let λ^{∗} denote the partition dual to λ. Then there exists a 𝕂G-module S_{𝕂}(s, λ) and an 𝔽G-module S_{𝔽}(s, λ) such that the following hold.

Proposition 2.1 [14] (i) S_{𝕂}(s, λ) is an absolutely irreducible 𝕂G-module, and S_{𝔽}(s, λ) is a p-modular reduction of S_{𝕂}(s, λ).

(ii) S_{𝔽}(s, λ) has a unique maximal submodule S_{𝔽}(s, λ)^{max}.

Let D_{𝔽}(s, λ) = S_{𝔽}(s, λ)/S_{𝔽}(s, λ)^{max}.

Proposition 2.2 [4] (i) S_{𝕂}(s, λ) is isomorphic to S_{𝕂}(s, μ) if and only if λ = μ. D_{𝔽}(s, λ) is isomorphic to D_{𝔽}(s, μ) if and only if λ = μ.

(ii) If s and t are nonzero roots of the same irreducible polynomial over 𝔽_{q}, then S_{𝕂}(s, λ) ≃ S_{𝕂}(t, λ).

(iii) Suppose s and t are nonzero roots of irreducible polynomials over 𝔽_{q} of the same degree, and the p′-parts of s and t are roots of the same polynomial over 𝔽_{q}. Then S_{𝔽}(s, λ) ≃ S_{𝔽}(t, λ) and D_{𝔽}(s, λ) ≃ D_{𝔽}(t, λ).

Proposition 2.3 [3] Suppose that λ ⊢ k and s is a nonzero element of degree d over 𝔽_{q}. Let t be the p′-part of s and suppose that t has degree a over 𝔽_{q}. Then a divides d. Let μ be the partition of dk/a given by μ_{i}^{∗} = (d/a)λ_{i}^{∗} for all i, where λ_{i}^{∗} is the i-th part of the partition λ^{∗} which is dual to λ, and μ_{i}^{∗} is defined similarly. Then D_{𝔽}(s, λ) ≃ D_{𝔽}(t, μ).

In view of these results, we choose one root of each monic irreducible polynomial over 𝔽_{q} (except the polynomial X) and call this set of roots C. Order C suitably. Then the following describes a complete set of inequivalent irreducible F G-modules (F = 𝕂 or 𝔽):

Proposition 2.4 [3] Suppose that s_{1} < s_{2} < … < s_{N} are elements of C. Let d_{i} denote the degree of s_{i} over 𝔽_{q} and let λ^{(i)} ⊢ k_{i} > 0 for 1 ≤ i ≤ N. Suppose that n = d_{1}k_{1} + d_{2}k_{2} + … + d_{N}k_{N}. Then

(i) (S_{𝕂}(s_{1}, λ^{(1)}) ∘ S_{𝕂}(s_{2}, λ^{(2)}) ∘ … ∘ S_{𝕂}(s_{N}, λ^{(N)})) ↑ G is an irreducible 𝕂G-module.

(ii) (D_{𝔽}(s_{1}, λ^{(1)}) ∘ D_{𝔽}(s_{2}, λ^{(2)}) ∘ … ∘ D_{𝔽}(s_{N}, λ^{(N)})) ↑ G is an irreducible 𝔽G-module, provided that s_{1}, …, s_{N} are all p-regular.

In this proposition, the module

(S_{𝕂}(s_{1}, λ^{(1)}) ∘ S_{𝕂}(s_{2}, λ^{(2)}) ∘ … ∘ S_{𝕂}(s_{N}, λ^{(N)})) ↑ G

is constructed as follows. By Proposition 2.1, S_{𝕂}(s_{i}, λ^{(i)}) is an irreducible 𝕂G_{i}-module, where G_{i} = GL_{d_{i}k_{i}}(q). Hence

V := S_{𝕂}(s_{1}, λ^{(1)}) ⊗ … ⊗ S_{𝕂}(s_{N}, λ^{(N)})

is an irreducible 𝕂L-module, where L = G_{1} × … × G_{N}. Next we can view L as a Levi subgroup of a parabolic subgroup P (which is the stabilizer of a flag of type (d_{1}k_{1}, d_{1}k_{1} + d_{2}k_{2}, …, d_{1}k_{1} + d_{2}k_{2} + … d_{N}k_{N})), and V as an irreducible 𝕂P -module, which is trivially acted on by the radical of P and which is denoted by S_{𝕂}(s_{1}, λ^{(1)}) ∘ S_{𝕂}(s_{2}, λ^{(2)}) ∘ … ∘ S_{𝕂}(s_{N}, λ^{(N)}). Finally, we can induce this module from P to G and obtain

(S_{𝕂}(s_{1}, λ^{(1)}) ∘ S_{𝕂}(s_{2}, λ^{(2)}) ∘ … ∘ S_{𝕂}(s_{N}, λ^{(N)})) ↑ G.

The same applies to (D_{𝔽}(s_{1}, λ^{(1)}) ∘ D_{𝔽}(s_{2}, λ^{(2)}) ∘ … ∘ D_{𝔽}(s_{N}, λ^{(N)})) ↑ G. Clearly, this construction has the following properties:

(a) (S_{𝕂}(s_{a_{1}}, λ^{(a_{1})}) ∘ S_{𝕂}(s_{a_{2}}, λ^{(a_{2})}) ∘ … ∘ S_{𝕂}(s_{a_{N}} , λ^{(a_{N})})) ↑ G is isomorphic to (S_{𝕂}(s_{1}, λ^{(1)}) ∘ S_{𝕂}(s_{2}, λ^{(2)}) ∘ … ∘ S_{𝕂}(s_{N}, λ^{(N)})) ↑ G, if {a_{1}, a_{2}, …, a_{N}} = {1, 2, …, N} (cf. [13], Theorem 16.1);

(b) The modules

(S_{𝕂}(s_{1}, λ^{(1)}) ∘ S_{𝕂}(s_{2}, λ^{(2)}) ∘ S_{𝕂}(s_{3}, λ^{(3)})) ↑ G,

(((S_{𝕂}(s_{1}, λ^{(1)}) ∘ S_{𝕂}(s_{2}, λ^{(2)})) ↑ H) ∘ S_{𝕂}(s_{3}, λ^{(3)})) ↑ G,

(S_{𝕂}(s_{1}, λ^{(1)}) ∘ (S_{𝕂}(s_{2}, λ^{(2)}) ∘ S_{𝕂}(s_{3}, λ^{(3)})) ↑ H′) ↑ G,

where G = GL_{d_{1}k_{1}+d_{2}k_{2}+d_{3}k_{3}}(q), H = GL_{d_{1}k_{1}+d_{2}k_{2}}(q), H′ = GL_{d_{2}k_{2}+d_{3}k_{3}}(q), are isomorphic.

Proposition 2.5 [4] Suppose s is an element of degree d over 𝔽_{q} and λ ⊢ k with n = kd. The 𝔽G-module S_{𝔽}(s, λ) has the same composition factors as ⊕_{μ⊢k}d_{λμ}D_{𝔽}(s, μ), where Δ(s, k) = (d_{λμ}) is a lower unitriangular matrix of non-negative integers (if the partitions of k are ordered lexicographically), and d_{λμ} ≠ 0 only if μ ⊵ λ. The matrix Δ(s, k) does not depend on the choice of s, and it is equal to the matrix Δ_{k} := Δ(1, k) of decomposition numbers for the unipotent representations of GL_{k}(q^{d}).

Corollary 2.6 Suppose s is an element of degree d over 𝔽_{q} and n = kd. Then the GL_{n}(q)-module S_{𝕂}(s, (k)) is irreducible in any cross characteristic.

Proof. Observe that if μ ⊢ k and μ ⊵ (k), then μ = (k). Now one applies Propositions 2.1 and 2.5. □

A necessary and sufficient condition for an ordinary irreducible representation of GL_{n}(q) to remain irreducible modulo p is given in [16]. Corollary 2.6 shows that GL_{n}(q) has a complex irreducible module which is (absolutely) irreducible in any cross characteristic. A more general example of such a module is given in the next statement.

Corollary 2.7 Let p_{1} < … < p_{N} be pairwise coprime integers and suppose s_{i} is an element of degree p_{i} over 𝔽_{q}, 1 ≤ i ≤ N. Suppose G = GL_{n}(q) with n = ∑_{i=1}^{N} k_{i}p_{i}. Then the G-module

V := (S_{𝕂}(s_{1}, (k_{1})) ∘ S_{𝕂}(s_{2}, (k_{2})) ∘ … ∘ S_{𝕂}(s_{N}, (k_{N}))) ↑ G

is irreducible in any cross characteristic.

Proof. Let p be any prime not dividing q and let t_{i} denote the p′-part of s_{i}. First suppose that t_{i} has degree p_{i} over 𝔽_{q} for all i. By Proposition 2.2 and Corollary 2.6, V mod p is isomorphic to

(D_{𝔽}(t_{1}, (k_{1})) ∘ D_{𝔽}(t_{2}, (k_{2})) ∘ … ∘ D_{𝔽}(t_{N}, (k_{N}))) ↑ G,

and the latter is irreducible by Proposition 2.4. Now suppose that some t_{i} has degree a less than p_{i}, say for i = 1. In this case p divides (q^{p_{1}} −1)/(q−1). (For, a|p_{1} by Proposition 2.3. Let r denote the p-part of s_{1}. Since t_{1} ∈ 𝔽_{q^{a}}^{•} and s_{1} ∈ 𝔽_{q^{p_{1}}}^{•}, one has 𝔽_{q^{a}}^{•} ∋ (s_{1}/t_{1})^{(q^{p_{1}}−1)/(q^{a}−1)} = r^{(q^{p_{1}}−1)/(q^{a}−1)}. But r is a pelement. Hence, if p |̸(q^{p_{1}} −1)/(q −1), then r ∈ 𝔽_{q^{a}}^{•}, yielding s_{1} ∈ 𝔽_{q^{a}}^{•}, i.e. s_{1} has degree at most a, a contradiction.) Assume t_{j} has degree less than p_{j} for some j ≥ 2. Then p divides (q^{p_{j}}−1)/(q−1), and so p divides q^{(p_{1},p_{j})}−1 = q−1, which implies that p divides both p_{1} and p_{j}, a contradiction. Thus t_{j} has degree p_{j} for all j ≥ 2. By Corollary 2.6, S_{𝔽}(s_{1}, (k_{1})) is irreducible, and it is isomorphic to D_{𝔽}(t_{1}, (k_{1}^{p_{1}/a})) by Proposition 2.3. Therefore,

V mod p ≃ (D_{𝔽}(t_{1}, (k_{1}^{p_{1}/a})) ∘ D_{𝔽}(s_{2}, (k_{2})) ∘ … ∘ D_{𝔽}(s_{N}, (k_{N}))) ↑ G,

and so it is irreducible. □

The matrices Δ_{n} with n ≤ 10 are calculated in [15]. The calculation is particularly based on the following two rules (Rules 5.4 and 5.8 of [15]), which will be used later in this paper. Given a prime p not dividing q, we define e to be the least positive integer r such that p divides (q^{r} −1)/(q −1). Observe that 2 ≤ e ≤ p, and e = p if and only if p|(q − 1). Denote Δ_{n} = (d_{λμ}), Δ_{n}^{−1} = (c_{λμ}). For any positive integer m, let 𝕊_{m} be the symmetric group of degree m. If α ⊢ k, β ⊢ l with k + l = n, then let [α] (resp. [β]) denote the corresponding ordinary character of 𝕊_{k} (resp. of 𝕊_{l}). The partition of n whose ith part is α_{i} + β_{i} is denoted by α + β. We can embed 𝕊_{k} × 𝕊_{l} in a natural way in 𝕊_{n}, then induce [α] ⊗ [β] to a character of 𝕊_{n} denoted by [α] · [β]. If

[α] · [β] = ∑_{γ⊢n} a_{αβγ}[γ],

then the non-negative integers a_{αβγ} are called Littlewood-Richardson coefficients.

Proposition 2.8 [15] Let λ, μ ⊢ n.

(i) Let σ ⊢ k, τ ⊢ n − k be the partitions which are determined by the properties that λ^{∗} = σ^{∗}+τ^{∗}, σ is e-regular, and all the parts of τ^{∗} are divisible by e. Then

c_{λμ} = ∑_{α⊢k} ∑_{β⊢n−k} c_{σα}c_{τβ}a_{αβμ} .

(ii) Suppose that λ_{1} = μ_{1} = k. Let λ^{(1)} = (λ_{2}, λ_{3}, …) ⊢ n − k and μ^{(1)} = (μ_{2}, μ_{3}, …) ⊢ n − k. Then d_{λμ} = d_{λ^{(1)}μ^{(1)}}.

In what follows, we denote the character afforded by the module S_{𝕂}(s, λ) by the symbol σ(s, λ), and the Brauer character afforded by the module D_{𝔽}(s, λ) by the symbol δ(s, λ). We need to work with the following partitions of n: α = (n), β = (n − 1, 1), γ = (n − 2, 2) (if n ≥ 4), and δ = (n − 2, 1, 1). We will need the following simple statement:

Lemma 2.9 Suppose t, s_{1}, …, s_{N} ∈ 𝔽_{q}^{•}, λ^{(i)} ⊢ k_{i}, 1 ≤ i ≤ N, and n = ∑_{i=1}^{N} k_{i}. Then

(σ(ts_{1}, λ^{(1)}) ∘ σ(ts_{2}, λ^{(2)}) ∘ … ∘ σ(ts_{N}, λ^{(N)})) ↑ G =
= σ(t, α) · {(σ(s_{1}, λ^{(1)}) ∘ σ(s_{2}, λ^{(2)}) ∘ … ∘ σ(s_{N}, λ^{(N)})) ↑ G} .

Proof. By reciprocity, it suffices to show that σ(t, (k)) · σ(s, λ) = σ(ts, λ), if s, t ∈ 𝔽_{q}^{•} and λ ⊢ k. Observe that σ(t, (k)) is a linear character of GL_{k}(q) afforded by the module M_{𝕂}(t, (k)) (cf. [14]). According to the construction of S_{𝕂}(s, λ) as a submodule of M_{𝕂}(s, (1^{k})) given in [14], one sees that it suffices to prove the desired relation for (the character afforded by) the module M_{𝕂}(s, (1^{k})). Next,

M_{𝕂}(s, (1^{k})) = (M_{𝕂}(s, (1)) ∘ … ∘ M_{𝕂}(s, (1))) ↑ GL_{k}(q).

Finally, we observe that M_{𝕂}(st, (1)) ≃ M_{𝕂}(s, (1)) ⊗ M_{𝕂}(t, (1)) as GL_{1}(q)- modules. (For, we can identify GL_{1}(q) with 𝔽_{q}^{•} = ⟨ϵ⟩. If s = ϵ^{i} and g = ϵ^{j} ∈ 𝔽_{q}^{•}, then g acts on M_{𝕂}(s, (1)) as the scalar ϵ̄^{ij}.) □

3 A Fragment of the Matrix Δ_{n}

Recall that the matrix Δ_{n} consists of decomposition numbers for the unipotent representations of GL_{n}(q), cf. Proposition 2.5. Observe that if λ ⊢ n and λ⊵δ, then λ ∈ {α, β, γ, δ}. Also, α▷β▷γ▷δ.

Proposition 3.1 Suppose n ≥ 4. The 4 × 4 fragment D of the matrix Δ_{n} corresponding to α, β, γ, and to δ is equal to

[ \begin{matrix}1 & 0 & 0 & 0 \\ κ_{n} & 1 & 0 & 0 \\ κ_{n−1} & κ_{n−2} & 1 & 0 \\ 0 & κ_{n} & 0 & 1 \end{matrix} ] ,

if p |̸(q + 1), and

[ \begin{matrix}1 & 0 & 0 & 0 \\ κ_{n} & 1 & 0 & 0 \\ ζ & κ_{n} & 1 & 0 \\ ζ + 1 & κ_{n} & 1 & 1 \end{matrix} ] ,

if p|(q + 1).

Proof. 1) Let d = ord_{p}(q). Observe that either d = e, or p|(q − 1), d = 1, e = p. For any integer b, let b^{∗} = [\frac{b}{d}]. Given two non-negative integers a, b, we say that a contains b to base p (cf. [13]) if and only if

(a) a > 0, a = a_{0} + a_{1}p + … + a_{k}p^{k} with 0 ≤ a_{i} < p, a_{k} ≠ 0, and

(b) b = b_{0} + b_{1}p + … with 0 ≤ b_{i} < p, b_{k} = b_{k+1} = … = 0, and for each i, either b_{i} = 0 or b_{i} = a_{i}. Given two non-negative integers n, m, let f_{p,d}(n, m) = 1 if m = 0, f_{p,d}(n, m) = 1 if (n + 1)^{∗} contains m^{∗} to base p, and either d divides m or d divides (n + 1 − m). Otherwise let f_{p,d}(n, m) = 0.

Now Theorem 20.6 of [13] says that for 0 ≤ j ≤ m ≤ n/2, the composition multiplicity of D_{𝔽}(1, (n − j, j)) in S_{𝔽}(1, (n − m, m)) is f_{p,d}(n − 2j, m − j). We will apply this Theorem for m = 1, 2.

2) By Proposition 2.5, d_{λλ} = 1 and d_{λμ} = 0 unless μ ⊵ λ. In other words, D is lower unitriangular. Due to 1), d_{βα} = f_{p,d}(n, 1) = 0 or 1. If d = 1, i.e. p|(q − 1), then f_{p,d}(n, 1) = 1 if and only if n + 1 contains 1 to base p, i.e. p|n. The latter is equivalent to the condition that p divides (q^{n} − 1)/(q − 1). If d > 1, then f_{p,d}(n, 1) = 1 if and only if d divides n, i.e. p divides (q^{n} − 1)/(q − 1). We conclude that d_{βα} = κ_{n}.

Next we have d_{γα} = f_{p,d}(n, 2) = 0 or 1. Suppose d = 1. Then f_{p,d}(n, 2) = 1 if and only if n + 1 contains 2 to base p. If p ≥ 3, this is equivalent to the condition that p|(n − 1), i.e. p divides (q^{n−1} − 1)/(q − 1). If p = 2, then n + 1 contains 2 to base 2 precisely when p divides [\frac{n−1}{2}]. Next suppose d = 2, i.e. p divides q +1 but not q −1. Then f_{p,d}(n, 2) = 1 if and only if [\frac{n+1}{2}] contains 1 to base p, i.e. p divides [\frac{n−1}{2}]. Finally, suppose d ≥ 3. Then f_{p,d}(n, 2) = 1 if and only if d divides n − 1, i.e. p divides (q^{n−1} − 1)/(q − 1). We conclude that d_{γα} = κ_{n−1} if p |̸(q + 1), and d_{γα} = ζ if p|(q + 1).

Again due to 1), d_{γβ} = f_{p,d}(n−2, 1) = 0 or 1. If d = 1, then f_{p,d}(n−2, 1) = 1 if and only if n − 1 contains 1 to base p, i.e. p|(n − 2). The latter is equivalent to the condition that p divides (q^{n−2} − 1)/(q − 1). If d > 1, then f_{p,d}(n−2, 1) = 1 if and only if d divides n−2, i.e. p divides (q^{n−2}−1)/(q−1). We conclude that d_{γβ} = κ_{n−2}. Observe that κ_{n−2} = κ_{n} when p|(q + 1).

3) Set γ^{(1)} = (2), δ^{(1)} = (1, 1). By Proposition 2.8 (ii), d_{δγ} is equal to d_{δ}_{(1)}_{γ}_{(1)}, hence it is 1 if p|(q + 1) and 0 otherwise.

It remains to determine d_{δα} and d_{δβ}. First we suppose that e ≠ 2, i.e. p |̸(q + 1), and n ≥ 4. (The case n = 3 can be read off from [15]). In

this case, α, β, and δ are e-regular. By Theorem 6.22 of [15], d_{δα} = 0, and d_{δβ} = κ_{n}.

4) Finally, we compute d_{δα} and d_{δβ} for the case e = 2, i.e. p|(q + 1), and n ≥ 4. In this case, δ is e-singular. Write δ^{∗} = (3, 1^{n−3}) = (1^{n−2}) + (2) = σ^{∗} + τ^{∗} with σ = (n − 2), τ = (1^{2}). Denote ν = (2). Observe that

Δ_{2} = [ \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} ], hence Δ_{2}^{−1} = [ \begin{matrix}1 & 0 \\ −1 & 1 \end{matrix} ]. Applying Proposition 2.8 (i), we

obtain

c_{δμ} = ∑_{α⊢n−2} ∑_{β⊢2} c_{σα}c_{τβ}a_{αβμ} = ∑_{α⊢n−2} (c_{σα}a_{ατμ} − c_{σα}a_{ανμ})

for any μ ⊢ n. But Δ_{n−2}^{−1} is lower unitriangular, hence

c_{δμ} = a_{στμ} − a_{σνμ}.

By Example 14.4 of [12],

[σ] · [ν] = [n − 2] · [2] = [n] + [n − 1, 1] + [n − 2, 2] = [α] + [β] + [γ].

Next, [1] · [1] = [2] + [1^{2}]. Writing down all semistandard tableaux of type (n − 2, 1^{2}), we obtain by Young’s rule that

[n − 2] · [1] · [1] = [n − 2, 1^{2}] + [n − 2, 2] + 2[n − 1, 1] + [n],

hence

[σ] · [τ] = [n − 2] · [1] · [1] − [n − 2] · [2] = [n − 2, 1^{2}] + [n − 1, 1] = [β] + [δ].

Consequently, (c_{δα}, c_{δβ}, c_{δγ}, c_{δδ}) = (−1, 0, −1, 1). This is the fourth row of D^{−1}. On the other hand, we already know the first three rows of D. Hence the fourth row of D is (ζ + 1, κ_{n}, 1, 1). □

Corollary 3.2 (i) dim D_{𝔽}(1, α) = 1.

(ii) dim D_{𝔽}(1, β) = (q^{n} − q)/(q − 1) − κ_{n}.

(iii) dim D_{𝔽}(1, γ) is equal to

(\frac{q^{n} − 1}{q − 1}) (\frac{q^{n−1} − q^{2}}{q^{2} − 1} − κ_{n−2}) + κ_{n−2}(κ_{n} + 1) − κ_{n−1}

if p |̸(q + 1), and

(\frac{q^{n} − 1}{q − 1}) (\frac{q^{n−1} − q^{2}}{q^{2} − 1} − κ_{n}) + κ_{n}(κ_{n} + 1) − ζ

if p|(q + 1).

(iv) dim D_{𝔽}(1, δ) is equal to

(\frac{q^{n} − q}{q − 1} ) (\frac{q^{n} − q^{2}}{q^{2} − 1} − κ_{n}) + κ_{n}^{2}

if p |̸(q + 1), and

\frac{(q^{n} − 1)(q^{n−1} − 1)}{q^{2} − 1}

if p|(q + 1). □

Let n ≥ 3. The irreducible Weil modules of G = GL_{n}(q) are exactly the modules S_{𝕂}(s, β) with s ∈ 𝔽_{q}^{•} (of degree (q^{n} − q)/(q − 1)), and (S_{𝕂}(s, (1)) ∘ S_{𝕂}(t, (n − 1))) ↑ G with s, t ∈ 𝔽_{q}^{•}, s ≠ t (of degree (q^{n} − 1)/(q − 1)). The S_{𝕂}(s, β)’s restricted to S = SL_{n}(q) give us just one irreducible module, whose character is denoted by τ_{n,q}^{0}, and the modules (S_{𝕂}(s, (1)) ∘ S_{𝕂}(t, (n − 1))) ↑ G restricted to S give us q −2 irreducible modules, whose characters are denoted by τ_{n,q}^{i}, where s/t = ϵ^{i}, 1 ≤ i ≤ q − 2. As an easy application of the Dipper-James theory, we have the following statement (see also [9]).

Proposition 3.3 Let n ≥ 3 and p a prime not dividing q. All characters are considered as SL_{n}(q)-characters.

(i) The irreducible Brauer character δ(1, β) has degree (q^{n}−q)/(q−1)−κ_{n}. Furthermore, τ_{n,q}^{0} mod p = δ(1, β) + κ_{n} · 1_{S}.

(ii) Suppose 1 ≤ j ≤ q − 2. The reduction modulo p of any τ_{n,q}^{j} is equal either to some τ_{n,q}^{i} mod p with 1 ≠ ϵ^{i} being p-regular, or to 1_{S} + τ_{n,q}^{0} mod p. The ((q − 1)_{p′} − 1) characters τ_{n,q}^{i} mod p with 1 ≠ ϵ^{i} being p-regular are distinct irreducible Brauer characters.

Proof. (i) The decomposition numbers for the G-module S_{𝔽}(1, β) are shown in the second row of Δ_{n}, hence S_{𝔽}(1, β) has at most two composition factors, the trivial one with multiplicity κ_{n} and D_{𝔽}(1, β) with multiplicity 1. In particular, dim D_{𝔽}(1, β) = (q^{n} −q)/(q −1)−κ_{n}. It is easy (using for example ℓ(L_{n}(q))) to show that the S-module D_{𝔽}(1, β) is irreducible.

(ii) Let s, t ∈ 𝔽_{q}^{•}, s/t = ϵ^{i}. Let W = ⟨e_{1}, …, e_{n}⟩_{𝔽_{q}} be the natural 𝔽_{q}S-module. Let P = Stab_{S}(⟨e_{1}⟩_{𝔽_{q}}), and let U(ϵ^{l}) = ⟨v⟩_{F} denote the 1dimensional P -module with the following P -action: if g ∈ P and g(e_{1}) = ϵ^{j}e_{1}, then g(v) = ϵ̄^{lj}, and F = 𝕂 or 𝔽. Clearly, the restriction to S of (S_{𝕂}(s, (1)) ∘ S_{𝕂}(t, (n − 1))) ↑ G is V (s, t) := U(s/t) ↑ S. In particular, this restriction depends only on s/t.

Let s′ and t′ denote the p′-parts of s and t. By Proposition 2.2, the restriction to S of (S_{𝔽}(s, (1)) ∘ S_{𝔽}(t, (n − 1))) ↑ G is equal to the restriction to S of (S_{𝔽}(s′, (1)) ∘ S_{𝔽}(t′, (n − 1))) ↑ G and hence equal to V (s′/t′). If s′ = t′, then V (s′/t′) affords the character 1_{S} + τ_{n,q}^{0} mod p. If s′ ≠ t′, then (S_{𝔽}(s′, (1)) ∘ S_{𝔽}(t′, (n − 1))) ↑ G is irreducible by Proposition 2.4, hence V (s′/t′) is completely reducible. But due to Mackey’s formula (cf. [5]), dim Hom_{𝔽S}(V (s′/t′), V (s′/t′)) = 1, therefore V (s′/t′) is irreducible. Mackey’s formula also shows that dim Hom_{𝔽S}(V (r), V (r′)) = 0, whenever r, r′ ∈ 𝔽_{q}^{•} are p-regular and different. □

4 Small Groups

In this section we treat the case of small groups L = L_{n}(q), that is, the ones listed in Table III or n = 3.

The cases L = L_{2}(5), L_{2}(9), L_{3}(2), L_{3}(4), L_{4}(2) and L_{4}(3) can be read off from [17]. The cases L_{6}(2) and L_{6}(3) will be handled later, cf. Proposition 6.3. The desired information about L_{2}(q) with q ≥ 11 can be obtained from [2] or in a similar fashion.

Now we can suppose that L = L_{3}(q) with q ≠ 2, 4. Then (see e.g. Theorem 5.1.4 of [18]) S = SL_{3}(q) is the universal cover of L. Thus it suffices to deal with representations of S. But first we need to find the degrees of all irreducible modular representations of G = GL_{3}(q). According to Proposition 2.4, we have to deal with the following cases.

1) V = D_{𝔽}(s, λ) with λ ⊢ 3 and s ∈ 𝔽_{q}^{•} is p-regular.

By [15], the matrix Δ_{3} is [ \begin{matrix}1 \\ κ_{3} & 1 \\ κ_{2} & κ_{3} & 1 \end{matrix} ]. Hence, by Proposition 2.5,

δ(s, β) = σ(s, β) − κ_{3}, δ(s, γ) = σ(s, γ) − κ_{3} · σ(s, β) + κ_{3}^{2} − κ_{2}.

Thus, dim D_{𝔽}(s, λ) equals 1 if λ = α = (3), and q^{2} + q − κ_{3} if λ = β = (2, 1). Furthermore, dim D_{𝔽}(s, γ) is equal to q^{3} − 1 if p|(q + 1), (q − 1)(q^{2} − 1) if p|(q^{2} + q + 1), and q^{3} otherwise.

2) V = (D_{𝔽}(s, λ) ∘ D_{𝔽}(t, (1))) ↑ G with λ ⊢ 2, and s, t ∈ 𝔽_{q}^{•} are p-regular. Interchanging λ and (1) if necessary, we may assume that s < t (relative to the fixed order on C). If λ = (2), then dim V = q^{2} + q + 1. If λ = (1^{2}), then dim V = q(q^{2} + q + 1).

3) V = (D_{𝔽}(r, (1)) ∘ D_{𝔽}(s, (1)) ∘ D_{𝔽}(t, (1))) ↑ G with r, s, t ∈ 𝔽_{q}^{•}, r < s < t, and r, s, t are p-regular. Then dim V = (q + 1)(q^{2} + q + 1).

4) V = D_{𝔽}(s, (1)) with s a p-regular element of degree 3 over 𝔽_{q}. Then dim V = (q − 1)(q^{2} − 1).

5) V = (D_{𝔽}(s, (1)) ∘ D_{𝔽}(t, (1))) ↑ G with p-regular elements s, t, where t ∈ 𝔽_{q}^{•} and s has degree 2 over 𝔽_{q}. Then dim V = q^{3} − 1.

Thus (d_{1,p}, d_{2,p}, d_{3,p}) = (q^{2} + q − κ_{3}, q^{2} + q + 1, (q − 1)(q^{2} − 1)) for G = GL_{3}(q).

Proof of Theorem 1.1 for n = 3. First suppose that (3, q − 1) = 1. Then G = S × Z(G) with Z(G) ≃ ℤ_{q−1}. Hence S has the same degrees of irreducible Brauer characters as G does, and we are done. Next suppose 3|(q − 1) (and q ≥ 7), and ψ ∈ IBr_{p}(G). If 1 < ψ(1) ≤ q^{2} + q + 1, then due to the description above, ψ is the p-modular reduction of an irreducible Weil character of G, hence, according to Proposition 3.3, ψ_{S} is irreducible, and it is one of (q − 1)_{p′} Brauer characters of degree q^{2} + q − κ_{3} or q^{2} + q + 1. If ψ(1) > q^{2} + q + 1, then ψ ≥ (q − 1)(q^{2} − 1). But G = (S ∗ Z(G)) · ℤ_{3}, hence every irreducible constituent of ψ_{S} has degree ≥ (q − 1)(q^{2} − 1)/3.

Finally, we show that S has an irreducible Brauer character of degree (q−1)(q^{2}−1)/3. For, S has a χ ∈ Irr(S) with χ(1) = (q−1)(q^{2}−1)/3 (see e.g. [27]). Extend χ to an irreducible character χ̃ of S∗Z(G). Clearly, φ := χ̃ ↑ G has degree (q − 1)(q^{2} − 1), and it is irreducible, hence φ = S_{𝕂}(s, (1)) for an element s of degree 3 over 𝔽_{q}. By Corollary 2.6, φ mod p is irreducible, hence χ mod p ∈ IBr_{p}(S).

5 Some Induced Modules

From now on we suppose that n ≥ 4. Let G = GL_{n}(q). We begin with the following simple statement

Lemma 5.1 (i) (S_{𝕂}(1, (1)) ∘ S_{𝕂}(1, (n − 1))) ↑ G ≃ S_{𝕂}(1, α) ⊕ S_{𝕂}(1, β). (ii) (S_{𝕂}(1, (2)) ∘ S_{𝕂}(1, (n − 2))) ↑ G ≃ S_{𝕂}(1, α) ⊕ S_{𝕂}(1, β) ⊕ S_{𝕂}(1, γ). (iii) (S_{𝕂}(1, (1)) ∘ S_{𝕂}(1, (1)) ∘ S_{𝕂}(1, (n − 2))) ↑ G ≃ S_{𝕂}(1, α) ⊕ S_{𝕂}(1, β) ⊕

S_{𝕂}(1, β) ⊕ S_{𝕂}(1, γ) ⊕ S_{𝕂}(1, δ).

Proof. Let W = ⟨e_{1}, e_{2}, …, e_{n}⟩_{𝔽_{q}} denote the natural 𝔽_{q}G-module. Let

P_{1} = Stab_{G} (⟨e_{1}⟩_{𝔽_{q}}) , P_{2} = Stab_{G} (⟨e_{1}, e_{2}⟩_{𝔽_{q}}) , R = Stab_{G} (⟨e_{1}⟩, ⟨e_{1}, e_{2}⟩/⟨e_{1}⟩) .

Then the characters afforded by the modules mentioned in the Lemma are α := 1_{P_{1}} ↑ G, β := 1_{P_{2}} ↑ G, and γ := 1_{R} ↑ G. By Mackey’s formula, (α, α)_{G} = |P_{1}\G/P_{1}| = 2. On the other hand, obviously α has constituents σ(1, α) and σ(1, β), hence α = σ(1, α) + σ(1, β). Next,

(β, β)_{G} = |P_{2}\G/P_{2}| = 3, (α, β)_{G} = |P_{1}\G/P_{2}| = 2.

Observe that β has a constituent equal to σ(1, γ) (cf. §§10, 11 of [13]). Hence β = σ(1, α) + σ(1, β) + σ(1, γ). Finally,

(γ, γ)_{G} = |R\G/R| = 7, (α, γ)_{G} = |P_{1}\G/R| = 3, (β, γ)_{G} = |P_{2}\G/R| = 4,

and γ has a constituent equal to σ(1, δ) (cf. §§10, 11 of [13]). Consequently, γ = σ(1, α) + 2σ(1, β) + σ(1, γ) + σ(1, δ). (Actually, we could omit this calculation by referring to the Littlewood-Richardson rule, since the arising Littlewood-Richardson coefficients have been computed in p. 4) of the proof of Proposition 3.1.) □

In this section we find the p-modular reduction of the modules of type

V = (U ∘ S_{𝕂}(t, μ)) ↑ G,

where μ = (n − 2), t ∈ 𝔽_{q}^{•}, U is an irreducible 𝕂GL_{2}(q)-module.

Proposition 5.2 Keep the above assumptions. Suppose the p-modular reduction of V has a composition factor W such that dim W < (q^{n} − 1)(q^{n−1} − 1)/(q^{2} − 1). Then W is isomorphic to one of the following modules:

(i) D_{𝔽}(a, α), D_{𝔽}(a, β), and D_{𝔽}(a, γ), where a ∈ 𝔽_{q}^{•} is p-regular; (ii) (D_{𝔽}(a, (1)) ∘ D_{𝔽}(b, (n − 1))) ↑ G, where a, b ∈ 𝔽_{q}^{•} are p-regular and

a ≠ b;

(iii) (D_{𝔽}(a, (2)) ∘ D_{𝔽}(b, (n − 2))) ↑ G, where a, b ∈ 𝔽_{q}^{•} are p-regular and a ≠ b.

Proof. By Proposition 2.4, U is either S_{𝕂}(s, λ) with an element s of degree d over 𝔽_{q} and λ ⊢ 2/d, or (S_{𝕂}(r, (1)) ∘ S_{𝕂}(s, (1))) ↑ GL_{2}(q) with r, s ∈ 𝔽_{q}^{•}. Let V′ denote the p-modular reduction of V.

1) First we consider the former case: U ≃ S_{𝕂}(s, λ). Let s′ and t′ denote the p′-parts of s and t. By Propositions 2.1 and 2.2, V′ is isomorphic to (S_{𝔽}(s′, λ) ∘ S_{𝔽}(t′, μ)) ↑ G, if s′ and s have the same degree over 𝔽_{q}. Clearly, S_{𝔽}(t′, μ) = D_{𝔽}(t′, μ). We distinguish the following subcases.

1a) λ = (2).

Clearly, S_{𝔽}(s′, λ) = D_{𝔽}(s′, λ). If s′ ≠ t′, then V′ is irreducible by Proposition 2.4, whence we arrive at (iii). Suppose s′ = t′. Using Lemma 2.9, without loss of generality we may assume that s′ = t′ = 1. By Lemma 5.1, (S_{𝕂}(1, (2)) ∘ S_{𝕂}(1, (n − 2))) ↑ G affords the character σ(1, α)+σ(1, β)+ σ(1, γ). By Proposition 2.5, any composition factor of V′ is isomorphic to one of the modules listed in (i).

1b) λ = (1^{2}).

By Proposition 3.1, S_{𝔽}(s′, λ) has only D_{𝔽}(s′, λ) (of dimension q or q − 1) and possibly D_{𝔽}(s′, (2)) as composition factors. Because of 1) one can discard D_{𝔽}(s′, (2)). If s′ ≠ t′, then (D_{𝔽}(s′, λ) ∘ D_{𝔽}(t′, μ)) ↑ G is an irreducible module of dimension at least

(q − 1) · \frac{(q^{n} − 1)(q^{n−1} − 1)}{(q − 1)(q^{2} − 1)} = \frac{(q^{n} − 1)(q^{n−1} − 1)}{q^{2} − 1}.

Hence s′ = t′. Using Lemma 2.9, we may assume that s′ = t′ = 1. Since

S_{𝕂}(1, (2)) ⊕ S_{𝕂}(1, (1^{2})) ≃ (S_{𝕂}(1, (1)) ∘ S_{𝕂}(1, (1))) ↑ GL_{2}(q),

the module (S_{𝕂}(1, λ) ∘ S_{𝕂}(1, μ)) ↑ G is a composition factor of

(S_{𝕂}(1, (1)) ∘ S_{𝕂}(1, (1)) ∘ S_{𝕂}(1, (n − 2))) ↑ G.

Therefore, by Lemma 5.1 and Proposition 2.5, any composition factor of V′ is isomorphic either to one of the modules listed in (i), or to D_{𝔽}(1, δ). By Corollary 3.2, W cannot be D_{𝔽}(1, δ).

1c) λ = (1).

In this case, dim S_{𝕂}(s, λ) = q − 1. If s′ has degree 2 over 𝔽_{q}, then S_{𝔽}(s, λ) ≃ S_{𝔽}(s′, λ) ≃ D_{𝔽}(s′, λ) is irreducible. Moreover, by Proposition 2.4 V′ is an irreducible module of dimension (q^{n} −1)(q^{n−1} −1)/(q^{2} −1). Hence s′ has degree 1 over 𝔽_{q}. In this situation p|(q + 1), and D_{𝔽}(s, λ) ≃ D_{𝔽}(s′, (1^{2})) is an irreducible module of dimension q − 1. Thus S_{𝔽}(s, λ) ≃ D_{𝔽}(s′, (1^{2})), whence we arrive at the case 1b).

2) Now we suppose that U ≃ (S_{𝕂}(r, (1)) ∘ S_{𝕂}(s, (1))) ↑ GL_{2}(q). Let r′, s′ denote the p′-parts of r and s. Then the p-modular reduction of V is isomorphic to

V′ = (D_{𝔽}(r′, (1)) ∘ D_{𝔽}(s′, (1)) ∘ D_{𝔽}(t′, μ)) ↑ G.

If r′, s′, t′ are pairwise different, then V′ is irreducible by Proposition 2.4, and dim V′ = (q^{n} −1)(q^{n−1} −1)/(q −1)^{2}. We are left with the following subcases.

2a) r′ ≠ s′ = t′.

By Lemma 2.9 we may assume that s′ = t′ = 1. Observe that

S_{𝕂}(1, (n − 1)) ⊕ S_{𝕂}(1, (n − 2, 1)) ≃ (S_{𝕂}(1, (1)) ∘ S_{𝕂}(1, (n − 2))) ↑ GL_{n−1}(q)

by Lemma 5.1. Hence W is a composition factor either of

(S_{𝔽}(r′, (1)) ∘ S_{𝔽}(1, (n − 1))) ↑ G,

or

(S_{𝔽}(r′, (1)) ∘ S_{𝔽}(1, (n − 2, 1))) ↑ G.

In the former case we arrive at one of the modules listed in (ii). In the latter case, W is isomorphic either to (D_{𝔽}(r′, (1)) ∘ D_{𝔽}(1, (n − 1))) ↑ G, or (D_{𝔽}(r′, (1)) ∘ D_{𝔽}(1, (n − 2, 1))) ↑ G. The first module belongs to list (ii). The second module has dimension at least

(\frac{q^{n} − 1}{q − 1}) · (\frac{q^{n−1} − q}{q − 1} − 1) > \frac{(q^{n} − 1)(q^{n−1} − 1)}{q^{2} − 1}

by Corollary 3.2.

2b) s′ ≠ r′ = t′.

Interchanging r′ and s′, we come back to 2a).

2c) r′ = s′ = t′.

By Lemma 2.9, we may assume that r′ = 1. Thus we come back to the module (D_{𝔽}(1, (1)) ∘ D_{𝔽}(1, (1)) ∘ D_{𝔽}(1, (n − 2))) ↑ G already handled in 1b).

2d) r′ = s′ ≠ t′.

By Lemma 2.9, we may assume that r′ = 1. Since

(S_{𝕂}(1, (1)) ∘ S_{𝕂}(1, (1))) ↑ GL_{2}(q) ≃ S_{𝕂}(1, (2)) ⊕ S_{𝕂}(1, (1^{2})),

we can apply the results of 1a) and 1b). □

Next we determine the restriction of the modules listed in Proposition 5.2 to S = SL_{n}(q). By Proposition 3.3, the modules D_{𝔽}(a, α), D_{𝔽}(a, β), where a ∈ 𝔽_{q}^{•} is p-regular, and (D_{𝔽}(a, (1)) ∘ D_{𝔽}(b, (n − 1))) ↑ G, where a, b ∈ 𝔽_{q}^{•} are p-regular and a ≠ b, are irreducible S-modules.

Proposition 5.3 Let n ≥ 4. Let M denote the restriction of the module (D_{𝔽}(a, (2)) ∘ D_{𝔽}(b, (n − 2))) ↑ G to S = SL_{n}(q), where a, b ∈ 𝔽_{q}^{•} are p-regular and a ≠ b. Then one of the following holds.

(i) M is irreducible.

(ii) n = 4, q is odd, p is odd, b = −a, and M is a direct sum of two G-conjugate irreducible modules.

Proof. Keep the notation of the proof of Lemma 5.1, and set P = P_{2} ∩ S. Suppose c = a/b = ϵ^{j}. Define ĉ: 𝔽_{q}^{•} → 𝔽^{•} by ĉ(ϵ^{i}) = ϵ̄^{ij}. Let N = ⟨v⟩_{𝔽} denote the 1-dimensional 𝔽P -module with the following action by any element g =

[ \begin{matrix}X & ∗ \\ 0 & Y \end{matrix} ] in P (X ∈ GL_{2}(q) and Y ∈ GL_{n−2}(q)): g(v) = ĉ(det X)v. Then it

is clear that M = N ↑ S. By Clifford’s Theorem M is completely reducible, and by Mackey’s formula,

dim Hom_{𝔽S}(M, M) = ∑_{z∈T} dim Hom_{𝔽Q^{z}}(N^{z} ↓ Q^{z}, N ↓ Q^{z}),

where T is a complete set of representatives of double cosets in P \S/P and Q^{z} = P^{z} ∩ P. We can take T = {1, x, y} with

x : e_{1} ↦ e_{1}, e_{2} ↦ e_{3} ↦ −e_{2}, e_{i} ↦ e_{i} for i ≥ 4,
y : e_{1} ↔ e_{3}, e_{2} ↔ e_{4}, e_{i} ↦ e_{i} for i ≥ 5.

A typical element g ∈ Q^{x} is of the form [ \begin{matrix}\begin{matrix}a_{1} & b_{1} & c_{1} \\ 0 & b_{2} & 0 \\ 0 & 0 & c_{3} \end{matrix} & ∗ \\ 0 & X \end{matrix} ] with a_{1}, b_{1},

c_{1}, b_{2} and c_{3} in 𝔽_{q}, X ∈ GL_{n−3}(q) and a_{1} · b_{2} · c_{3} · det X = 1. Then g acts on N and N^{x} as scalars ĉ(a_{1}b_{2}) and ĉ(a_{1}c_{3}), respectively. Since n ≥ 3, we can choose g ∈ Q^{x} such that a_{1} is p-regular, ĉ(a_{1}) ≠ 1, b_{2} = 1 and c_{3} = a_{1}^{−1}, and so ĉ(a_{1}b_{2}) ≠ ĉ(a_{1}c_{3}). This shows that Hom_{𝔽Q^{x}}(N^{x} ↓ Q^{x}, N ↓ Q^{x}) = 0.

Clearly, Hom_{𝔽Q^{y}}(N^{y} ↓ Q^{y}, N ↓ Q^{y}) = 0 or 𝔽. Suppose that Hom_{𝔽Q^{y}}(N^{y} ↓

Q^{y}, N ↓ Q^{y}) = 𝔽. A typical element g ∈ Q^{y} is of the form [ \begin{matrix}\begin{matrix}A & 0 \\ 0 & B \end{matrix} & ∗ \\ 0 & C \end{matrix} ]

with A, B ∈ GL_{2}(q), C ∈ GL_{n−4}(q) and det A · det B · det C = 1. Then g acts on N and N^{y} as scalars ĉ(det A) and ĉ(det B), respectively. If n ≥ 5, we can choose g ∈ Q^{y} such that det A is p-regular, ĉ(det A) ≠ 1, det B = 1, and so ĉ(det A) ≠ ĉ(det B). This shows that Hom_{𝔽Q^{y}}(N^{y} ↓ Q^{y}, N ↓ Q^{y}) = 0. Hence n = 4. Now we can choose g with det A being any p-regular element in 𝔽_{q}^{•} and det B = det A^{−1}. This shows that c = −1, and so p and q are odd.

We have shown that: either dim Hom_{𝔽S}(M, M) = 1, or c = −1 and dim Hom_{𝔽S}(M, M) = 2. In the former case M is irreducible. In the latter case M is a direct sum of two G-conjugate irreducible submodules. □

Now we want to consider the S-restriction of D_{𝔽}(s, γ), s ∈ 𝔽_{q}^{•}. Lemma 2.9 shows that the S-modules D_{𝔽}(s, γ) and D_{𝔽}(1, γ) are isomorphic. Hence if suffices to study the S-module D_{𝔽}(1, γ).

Proposition 5.4 Suppose n ≥ 5. Then the S-module D_{𝔽}(1, γ) is irreducible.

Proof. Observe that the statement is trivial in the case (n, q) = (5, 2), (5, 3), where G = S × Z(G). Henceforth we may assume that (n, q) ≠ (5, 2), (5, 3).

1) Keep the notation of the proof of Lemma 5.1, and let

W′ = ⟨e_{2}, e_{3}, …, e_{n}⟩_{𝔽_{q}}, T = Stab_{S}(W′), Q = O_{q}(T).

Then Q is an elementary abelian group of order q^{n−1} and T/Q ≃ GL_{n−1}(q) acts transitively on Q \ {1} and on Irr(Q)^{#} := Irr(Q) \ {1_{Q}}. By Lemma 5.1, the module S_{𝕂}(1, γ) affords the S-character σ(1, γ) = β − α, in the notation of the proof of the Lemma. Take any non-identity element g ∈ Q. Without

loss of generality we may assume that g = [ \begin{matrix}1 \\ 1 & 1 \\ & & . & . & . \\ & & & & & 1 \end{matrix} ]. Hence

α(g) = \frac{q^{n−1} − 1}{q − 1}, β(g) = \frac{(q^{n−1} − 1)(q^{n−2} − 1)}{(q − 1)(q^{2} − 1)} + q^{n−2}.

This implies that

α ↓ Q = (1 + \frac{q^{n−1} − 1}{q − 1}) 1_{Q} + ∑_{ρ∈Irr(Q)^{#}} ρ,

β ↓ Q = (\frac{q^{n−1} − 1}{q − 1} + \frac{(q^{n−1} − 1)(q^{n−2} − 1)}{(q − 1)(q^{2} − 1)}) 1_{Q} + \frac{q^{n−2} − 1}{q − 1} ∑_{ρ∈Irr(Q)^{#}} ρ.

Recall that α is afforded by the permutation module of S on 1-spaces in W, and β is afforded by the permutation module of S on 2-spaces in W. Observe that Q acts trivially and T acts transitively on the set of 1-spaces inside W′. This implies by Lemma 5.1 applied to T/Q that the 1_{Q}-homogeneous part of α ↓ T affords the T/Q-character 2 + σ(1, (n − 2, 1)). Similarly, Q acts trivially and T acts transitively on the set of 2-spaces inside W′. By Lemma 5.1, this action affords the T/Q-character 1+σ(1, (n−2, 1))+σ(1, (n−3, 2)). Consequently, the 1_{Q}-homogeneous part of σ(1, γ) ↓ T contains the T/Qcharacter σ(1, (n − 3, 2)), of degree d := (q^{n−1} − 1)(q^{n−2} − q^{2})/(q − 1)(q^{2} − 1).

2) Consider the “lowest” constituent δ := δ(1, (n − 3, 2)) of σ(1, (n − 3, 2)) mod p. By Corollary 3.2, δ(1) ≥ (q^{n−1}−1)(q^{n−2}−2q^{2}+1)/(q−1)(q^{2}−1). The discussion in 1) ensures that δ is a constituent of the 1_{Q}-homogeneous part of σ(1, γ) ↓ T. Also, since n ≥ 5 and (n, q) ≠ (5, 2), (5, 3), we have that

(δ(1, γ) ↓ Q, 1_{Q})_{Q} ≤ (σ(1, γ) ↓ Q, 1_{Q})_{Q} = ((β − α) ↓ Q, 1_{Q})_{Q} =

= \frac{(q^{n−1} − 1)(q^{n−2} − 1)}{(q − 1)(q^{2} − 1)} − 1 < 2δ(1),

and

((σ(1, γ) − δ(1, γ)) ↓ Q, 1_{Q})_{Q} ≤ \frac{q^{n−1} − 1}{q − 1} + 1 < δ(1).

We conlude that the 1_{Q}-homogeneous part of δ(1, γ) ↓ T contains the T/Qcharacter δ with composition multiplicity 1.

3) Now we can finish the proof of the Proposition. By Clifford’s Theorem, the S-module D_{𝔽}(1, γ) is the direct sum of, say t, irreducible, G-conjugate submodules V_{i}, 1 ≤ i ≤ t. One may assume that the 1_{Q}-homogeneous part of V_{1} ↓ T has a composition factor affording the Brauer character δ. We can choose the set {h_{λ} := diag(λ, 1, …, 1) | λ ∈ 𝔽_{q}^{•}} as a complete set of representatives of cosets of S\G. Clearly, h_{λ} normalizes Q and centralizes T/Q. This implies that for all i, the 1_{Q}-homogeneous part of V_{i} ↓ T has a composition factor affording the Brauer character δ. According to the conclusion of 2), t = 1. In other words, the S-module D_{𝔽}(1, γ) is irreducible. □

Proposition 5.5 Suppose n = 4. Then the S-module D_{𝔽}(1, γ) is the sum of at most (2, q − 1) G-conjugate irreducible submodules.

Proof. The statement is trivial in the case q is even, because then G = S × Z(G). Henceforth we may assume that q is odd. Observe that now G = ((S ∗ Z(G)) · ℤ_{2,(q−1)/2)}) · ℤ_{2}. This implies by Clifford’s Theorem that the S-module D_{𝔽}(1, γ) is the direct sum of, say t, irreducible G-conjugate submodules V_{i}, 1 ≤ i ≤ t, and t|4. If t|2 we are done. Assume that t = 4. If p |̸(q + 1), then dim D_{𝔽}(1, γ) = q^{2}(q^{2} + 1) − κ_{3}, with κ_{3} = 0 or 1. Thus dim D_{𝔽}(1, γ) is not divisible by t = 4, a contradiction. Hence p|(q + 1). In this case dim D_{𝔽}(1, γ) = (q − 1)(q^{3} − 1). Fix a generator ω of 𝔽_{q^{4}}^{•} and let g = diag(ω, ω^{q}, ω^{q^{2}}, ω^{q^{3}}) ∈ G, h = g^{2(q^{2}−1)} ∈ S. Then h is p-regular, and in the notation of the proof of Lemma 5.1 we have α(h) = β(h) = 0. In particular, if δ denotes the Brauer character afforded by D_{𝔽}(1, γ), then δ(h) = 2. On the other hand, observe that G = ∪_{j=0}^{q−2}g^{j}S, implying that the Brauer characters of the V_{i}’s take the same value at h. Therefore δ(h) = 2 is divisible by t = 4, a contradiction. □

Remark 5.6 If G = GL_{4}(3) and p = 2, then the S-module D_{𝔽}(1, γ) (of dimension 52) splits into two irreducible submodules of dimension 26. In general, it is shown in [7] that the decomposition numbers for unipotent characters of GL_{n}(q) and SL_{n}(q) are the same (in particular, the S-module D_{𝔽}(1, γ) is irreducible), provided that p |̸(n, q − 1).

Observe that dim D_{𝔽}(1, γ) < (q^{n} − 1)(q^{n−1} − 1)/(q − 1)(q^{2} − 1), and G/S ≃ ℤ_{q−1}. Hence, if W is an irreducible G-module of dimension at least (q^{n} − 1)(q^{n−1} − 1)/(q^{2} − 1), then the dimension of any composition factor of the S-module W ↓ S is bigger than dim D_{𝔽}(1, γ). Together with Propositions 3.3, 5.2, 5.3, 5.4, and 5.5, this observation leads to the following consequence:

Corollary 5.7 Let V = (U ∘ S_{𝕂}(t, μ)) ↑ G, where μ = (n − 2), t ∈ 𝔽_{q}^{•}, U is an irreducible 𝕂GL_{2}(q)-module. Let Y denote an arbitrary composition factor of the S-restriction of the p-modular reduction of V. Then one of the following holds.

(i) Y is the trivial module.

(ii) dim Y = (q^{n} − q)/(q − 1) − κ_{n} or (q^{n} − 1)/(q − 1), and the Brauer character of Y is one of the characters listed in Proposition 3.3.

(iii) n ≥ 5 and Y ≃ D_{𝔽}(1, γ).

(iv) n ≥ 5 and dim Y > dim D_{𝔽}(1, γ).

(v) n = 4 and dim Y ≥ (q − 1)(q^{3} − 1)/(2, q − 1). □

6 Proof of Theorem 1.1 for the Case n ≥ 4

Suppose L = L_{n}(q) with n ≥ 4, q = r^{f}, r a prime, and (n, q) ≠ (4, 2). Then S = SL_{n}(q) is the universal cover of L (cf. [18]). Hence we can suppose that the representation Φ mentioned in Theorem 1.1 is a linear representation of S.

Let G = GL_{n}(q) as above and denote

H = {diag(a, a, X) | a ∈ 𝔽_{q}^{•}, X ∈ GL_{n−2}(q), det X = a^{−2}}

and let ℓ_{p}(H) denote the smallest integer t > 1 such that H has a linear irreducible representation of degree t over a field of characteristic p. Clearly, (q^{n−2} − 1)/(q − 1) ≥ ℓ_{p}(H) ≥ ℓ(L_{n−2}(q)).

Proposition 6.1 Suppose n ≥ 4 and (n, q) ≠ (4, 2), (4, 3). Let V be an irreducible 𝔽SL_{n}(q)-module such that dim V < (q^{n−1} − 1)ℓ_{p}(H). Then V satisfies one of the conclusions of Corollary 5.7.

Proof. 1) Let W = ⟨e_{1}, e_{2}, …, e_{n}⟩_{𝔽_{q}} denote the natural 𝔽_{q}S-module. Let P = Stab_{S} (⟨e_{1}⟩_{𝔽_{q}}), Q = O_{r}(P). Then P/Q ≃ GL_{n−1}(q) acts transitively on Q \ {1} and on Irr(Q)^{#} := Irr(Q) \ {1_{Q}}. As (p, q) = 1, V decomposes as [V, Q] ⊕ C_{V} (Q), and [V, Q] is a P -submodule which affords Q-character k ∑_{ρ∈Irr(Q)^{#}} ρ for an integer k. By our assumption k ≤ dim V/(q^{n−1} − 1) < ℓ_{p}(H). Observe that the subgroup

R = \{g = [ \begin{matrix}\begin{matrix}a & b \\ 0 & a \end{matrix} & ∗ \\ 0 & X \end{matrix} ] | a ∈ 𝔽_{q}^{•}, b ∈ 𝔽_{q}, X ∈ GL_{n−2}(q), det X = a^{−2}\}

centralizes a character ρ ∈ Irr(Q)^{#}. (For, any element f ∈ Q can be written as

f_{a_{2},... ,a_{n}} : e_{1} ↦ e_{1}, e_{i} ↦ e_{i} + a_{i}e_{1}, for i ≥ 2,

where a_{i} ∈ 𝔽_{q}. We can fix a primitive r-th root ω of unity and choose ρ to be the map

f_{a_{2},... ,a_{n}} ↦ ω^{Tr_{𝔽q/𝔽r}(a_{2})}.

Now the typical element g ∈ R acts on Q as follows: gf_{a_{2},a_{3},... ,a_{n}}g^{−1} = f_{a_{2},a′_{3},... ,a′_{n}}, hence R centralizes the character ρ.) Since (P : R) = q^{n−1} − 1 = |Irr(Q)^{#}|, R = Stab_{P} (ρ). Let A denote the ρ-homogeneous part of [V, Q] ↓ P. Then A is an R-module of dimension k < ℓ_{p}(H), and [V, Q] ↓ P = A ↑ P. Observe that H embeds naturally in R.

2) Consider the normal subgroup

Q_{2} = \{[ \begin{matrix}1 & & ∗ & … & ∗ \\ & 1 & ∗ & … & ∗ \\ 0 & 0 & 1 \\ 0 & 0 & & ... \\ 0 & 0 & & & 1 \end{matrix} ]\}

of R. This is the direct product of two normal elementary abelian subgroups

of order q^{n−2} each. The first one, \{[ \begin{matrix}1 & & ∗ & … & ∗ \\ & 1 & 0 & … & 0 \\ 0 & 0 & 1 \\ 0 & 0 & & ... \\ 0 & 0 & & & 1 \end{matrix} ]\}, is just the sub-

group {f_{0,a_{3},... ,a_{n}} | a_{3}, …, a_{n} ∈ 𝔽_{q}}; it is contained in Ker(ρ). The elements (and the nontrivial linear characters) of the second subgroup,

\{[ \begin{matrix}1 & & 0 & … & 0 \\ & 1 & ∗ & … & ∗ \\ 0 & 0 & 1 \\ 0 & 0 & & ... \\ 0 & 0 & & & 1 \end{matrix} ]\} ,

are transitively permuted by H. Since dim A = k < ℓ_{p}(H) ≤ q^{n−2} − 1, the second subgroup also acts trivially on A. Thus Q_{2} acts trivially on A, and so A is actually acted on by R/Q_{2} ≃ ℤ_{r}^{f} × H.

Let Ψ denote the representation of H on A. Since dim A < ℓ_{p}(H), Ψ(H) is isomorphic to a subgroup of upper triangular matrices, whence it is solvable. Since (n, q) ≠ (4, 2), (4, 3), [H, H] ≃ SL_{n−2}(q) is perfect, therefore [H, H] acts trivially on A.

We also have

dim Hom_{𝔽S}(A ↑ S, V) = dim Hom_{𝔽S}((A ↑ P) ↑ S, V) =

= dim Hom_{𝔽S}(([V, Q] ↓ P) ↑ S, V) = dim Hom_{𝔽P}([V, Q] ↓ P, V ↓ P) ≠ 0.

3) Now we choose an irreducible factor-module V′ of V ↑ G, i.e.

Hom_{𝔽G}(V ↑ G, V′) ≠ 0.

Then Hom_{𝔽S}(V, V′ ↓ S) ≠ 0, i.e. V is a submodule of V′ ↓ S. Since V is irreducible and

Hom_{𝔽S}(A ↑ S, V) ≠ 0, Hom_{𝔽S}(V, V′ ↓ S) ≠ 0,

one has that Hom_{𝔽S}(A ↑ S, V′ ↓ S) ≠ 0, whence Hom_{𝔽G}(A ↑ G, V′) ≠ 0. Denoting P_{2} = Stab_{G} (⟨e_{1}, e_{2}⟩_{𝔽_{q}}) and B = A ↑ P_{2}, we see that V′ is a factormodule of B ↑ G. The module B is acted on trivially by Q_{2} and [H, H], therefore every composition factor of B ↑ G is isomorphic to a module of the form (U ∘ S_{𝔽}(t, μ)) ↑ G, where μ = (n − 2), t ∈ 𝔽_{q}^{•}, U is an irreducible 𝔽GL_{2}(q)-module. Consequently, V satisfies one of the conclusions of Corollary 5.7. □

Corollary 6.2 Let n ≥ 3 and (n, q) ≠ (3, 2), (3, 4), (4, 2), (4, 3). Then d_{1,p}(L_{n}(q)) = (q^{n} − q)/(q − 1) − κ_{n}.

Proof. By the results of §4, it suffices to consider the case n ≥ 4. Let Φ be an irreducible projective representation of L = L_{n}(q) over 𝔽 of degree d, 1 < d < 2(q^{n−1} − 1). Then Φ corresponds to an 𝔽SL_{n}(q)-module V. Clearly, ℓ_{p}(H) ≥ 2. Hence V satisfies the assumption, and so the conclusion of Proposition 6.1. Applying Corollary 5.7, one has that dim V ≥ (q^{n} − q)/(q − 1) − κ_{n}, and equality can actually be attained. □

Now we are in a position to complete the proof of Theorem 1.1. The representation Φ mentioned in the Theorem corresponds to an 𝔽SL_{n}(q)- module V. If n = 4 and q ≥ 4, then ℓ_{p}(H) ≥ (q−1)/(2, q−1), whence 𝔡(L) ≤ (q^{n−1}−1)ℓ_{p}(H). If n ≥ 5 and (n, q) ≠ (5, 2), (5, 4), (6, 2), (6, 3), then ℓ_{p}(H) ≥ d_{1,p}(L_{n−2}(q)) = (q^{n−2} − q)/(q − 1) − κ_{n−2} by Corollary 6.2, hence 𝔡(L) ≤ (q^{n−1} − 1)ℓ_{p}(H). The case (n, q) = (5, 2) can be done using [17]. Finally, if (n, q) = (5, 4), then ℓ_{p}(H) = ℓ_{p}(GL_{3}(4)) = (q^{n−2} − q)/(q − 1) − κ_{n−2} ≥ 19 (though ℓ(L_{3}(4)) = 4), and so we again have 𝔡(L) ≤ (q^{n−1} − 1)ℓ_{p}(H). Thus V satisfies the assumption and hence the conclusion of Proposition 6.1. By Corollary 5.7, dim V = (q^{n} − q)/(q − 1) − κ_{n} or (q^{n} − 1)/(q − 1), and V comes from the p-modular reduction of irreducible Weil characters of SL_{n}(q), as stated.

Theorem 1.1 has been proved. □

Proposition 6.3 (i) SL_{4}(q) has an irreducible complex representation of degree (q − 1)(q^{3} − 1)/(2, q − 1), whose reduction is irreducible in any cross characteristic.

(ii) Suppose q = 2, 3. Then Theorem 1.1 is true for SL_{6}(q), with 𝔡(L) replaced by (q^{3} − 1)(q^{5} − 1). SL_{6}(q) has an irreducible complex representation of degree (q^{3} − 1)(q^{5} − 1), whose reduction is irreducible in any cross characteristic.

Proof. (i) The cases q = 2, 3 can be read off from [17]. Suppose q > 3. Then G has a complex irreducible module V of dimension (q −1)(q^{3} −1), whose Srestriction is the sum of (2, q−1) irreducible submodules V_{i}, 1 ≤ i ≤ (2, q−1), of dimension (q − 1)(q^{3} − 1)/(2, q − 1) each (cf. Chap. 13 of [19]). It is easy to show that V ≃ S_{𝕂}(s, (2)) for some element s of degree 2 over 𝔽_{q}. Hence V is irreducible in any cross characteristic, according to Corollary 2.6. Finally, we show that the S-modules V_{i} are irreducible in any cross characteristic. Assume the contrary. Then q must be odd, and some composition factor of V_{i} mod p has dimension equal to (q − 1)(q^{3} − 1)/4, contrary to Theorem 1.1.

(ii) Now suppose (n, q) = (6, 2) or (6, 3). Then ℓ_{p}(H) = q^{3} − 1. The same arguments as above show that Theorem 1.1 is true for S, with 𝔡(L) replaced by (q^{3} − 1)(q^{5} − 1).

One can show that G has a complex irreducible module V of dimension (q − 1)(q^{3} − 1)(q^{5} − 1), whose S-restriction is the sum of q − 1 irreducible submodules V_{i}, 1 ≤ i ≤ q − 1, of dimension (q^{3} − 1)(q^{5} − 1) each (cf. [27]). It is easy to show that V ≃ S_{𝕂}(s, (3)) for some element s of degree 2 over 𝔽_{q}. By Corollary 2.6, V is irreducible in any cross characteristic. Now Clifford’s Theorem ensures that the S-modules V_{i} mod p are irreducible. □

7 Concluding Remarks

As mentioned in §1, the bound 𝔡(L) is best possible when n = 3, 4; namely, 𝔡(L) = 𝔟(L). In this case Theorem 1.1 gives the precise values of d_{1,p}, d_{2,p}, and d_{3,p}, except for (q, p) with (q − 1)_{p′} = 1 (which happens exactly when either p = 2, and q = 9 or q a Fermat prime, or q = 2, or p a Mersenne prime and q = p + 1).

If n ≥ 5, then Theorem 1.1 gives a good bound for 𝔟(L). The techniques of the paper could be exploited further to give the precise values of 𝔟(L), d_{3,p} and possibly d_{4,p} as well. Presumably 𝔟(L) = d_{3,p} = dim D_{𝔽}(1, γ) in the “generic” case. The outline of the arguments would be as follows. Set

𝔞(L) = \{ \begin{matrix}(q − 1)(q^{2} − 1)(q^{4} − 1)/(3, q − 1), & if n = 5 \\ (q − 1)(q^{3} − 1)(q^{5} − 1)/(2, q − 1), & if n = 6 \\ (q^{n−1} − 1)(q^{n−2} − 1), & if n ≥ 7 \end{matrix} ,

and we want to find all (projective) irreducible representations Φ of L_{n}(q) over 𝔽 such that 1 < d := deg Φ < 𝔞(L). (We certainly need to exclude a few small groups, as we did in Theorem 1.1. For safety, let us suppose that (n, q) ≠ (5, 2), (5, 4), (6, 2), (6, 3), (8, 2), (8, 3).) Keep the notation of the proof of Proposition 6.1. Then the same arguments as used in that proof show that V, the SL_{n}(q)-module corresponding to Φ, is isomorphic to a constituent of A ↑ S, where A is an 𝔽R-module of dimension less than q^{n−2} − 1. The last inequality guarantees that A is acted on trivially by Q_{2}. At this place, two ways of improvement are possible.

The first way is as follows. Applying Theorem 1.1 to SL_{n−2}(q), we see that (a submodule of) A is (a factor-module of) an induced 1-dimensional module from the first parabolic subgroup of R/Q_{2}. Now one would need to determine the 7×7 fragment of the matrix Δ_{n} corresponding to all partitions λ ⊢ n with λ ⊵ (n − 3, 1, 1, 1), as we did in §3. Then one would expand §5 to cover all modules of the form

(U ∘ S_{𝕂}(s, (1)) ∘ S_{𝕂}(t, (n − 3))) ↑ G,

where s, t ∈ 𝔽_{q}^{•} and U an irreducible GL_{2}(q)-module, and determine the Srestriction of all irreducible 𝔽G-modules arising in this way. This would give us the exact value of 𝔟(L).

The second way is to observe that, since dim A < 𝔞(L)/(q^{n−1} − 1) ≤ 𝔡(L_{n−2}(q)), all composition factors of the T -module A are either trivial, or constituents of the p-modular reduction of complex Weil representations of T := [H, H] ≃ SL_{n−2}(q). It is so because of Theorem 1.1 applied to T. If τ denotes the Brauer character of any of these composition factors and t is any transvection in T, then (τ(1), τ(t)) is either (1, 1), or (\frac{q^{n−2}−1}{q−1}, \frac{q^{n−3}−1}{q−1}),

or (\frac{q^{n−2}−q}{q−1} − κ_{n−2}, \frac{q^{n−3}−q}{q−1} − κ_{n−2}). Since n ≥ 5, τ(t) > 0. Let φ denote the Brauer character afforded by V, M ≃ GL_{n−1}(q) denote a Levi subgroup of P and s a transvection in M. If W′ denotes the natural 𝔽_{q}M-module, then s acts on any s-invariant hyperplane of W′ (respectively, on any factor-space of W′ by an s-invariant 1-space) as the identity element or a transvection. Hence the [V, Q]-part of φ(s) is positive. On the other hand, if s′ is a transvection in Q, then the [V, Q]-part of φ(s′) is −(dim[V, Q])/(q^{n−1} − 1) < 0. This argument shows that C := C_{V} (Q) ≠ 0. Using this observation, we show that

𝔟(L) ≥ (q^{n−1} − 1)(\frac{q^{n−2} − 1}{q − 1} − κ_{n−2}) + 𝔟(L_{n−1}(q)),

proving Theorem 1.2. It suffices to show: if V does not satisfy any of the conclusions of Corollary 5.7, then dim V ≥ (q^{n−1}−1)(\frac{q^{n−2}−1}{q−1} −κ_{n−2})+𝔟(L_{n−1}(q)).

For, suppose all composition factors of the T -module A are trivial. Then arguing as in the proof of Proposition 6.1 we see that T acts trivially on A. Let V′ denote an irreducible factor-module of V ↑ G and let B = A ↑ P_{2} as in the notation of the proof of Proposition 6.1. Then V′ is a factor-module of B ↑ G. Moreover, every composition factor of B is of the form U ∘ S_{𝔽}(t, μ), where μ = (n − 2), t ∈ 𝔽_{q}^{•}, U is an irreducible 𝔽GL_{2}(q)-module. Hence V satisfies one of the conclusions of Corollary 5.7. Thus we may assume that not all composition factors of the T -module A are trivial. By Corollary 6.2, dim[V, Q] ≥ (q^{n−1} − 1)(\frac{q^{n−2}−1}{q−1} − κ_{n−2}).

Similarly, suppose that dim C < 𝔟(L_{n−1}(q)). Then all composition factors of the M′-module C are either of dimension 1 or constituents of the Weil modules, where M′ = [M, M] ≃ SL_{n−1}(q). The same is true for the P_{1}/Qmodule D, where D = C ↑ P_{1}, and P_{1} = Stab_{G}(⟨e_{1}⟩_{𝔽_{q}}) (it is clear that D is trivially acted on by Q). Choosing V′ as above, we see that V′ is a factor-module of D ↑ G. Arguing as in the proof of Proposition 5.2, one again sees that V satisfies one of the conclusions of Corollary 5.7. Hence dim C ≥ 𝔟(L_{n−1}(q)). Consequently,

dim V = dim[V, Q] + dim C_{V} (Q) ≥ (q^{n−1} − 1)(\frac{q^{n−2} − 1}{q − 1} − κ_{n−2}) + 𝔟(L_{n−1}(q)).

Theorem 1.2 has been proved. □

Remark 7.1 Keep the above notation. One might ask which conditions on a given GL_{n}(q)-module V would guarantee that C_{V} (Q) ≠ 0 (in this case one could use the induced characters from P to G to decsribe all such V). It is certainly the case if (q^{n−1} − 1) does not divide dim V. However, if (q^{n−1} − 1)| dim V then it may happen that C_{V} (Q) = 0. For example, suppose an element s has degree n over 𝔽_{q}. Then the G-module M_{𝔽}(s, (1)) = S_{𝔽}(s, (1)) (of dimension (q − 1)(q^{2} − 1) … (q^{n−1} − 1)) is irreducible in any cross characteristic by Corollary 2.6, and yet the Q-character afforded by this module does not contain the principal character 1_{Q}.

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