IRREDUCIBILITY OF ALTERNATING

AND SYMMETRIC SQUARES

Kay Magaard and Gunter Malle

Wayne State University and Universität Heidelberg

We investigate the question when the alternating or symmetric square of an absolutely irreducible projective representation of a non-abelian simple group G is again irreducible. The knowledge of such representations is of importance in the description of the maximal subgroups of simple classical groups of Lie type. We obtain complete results for G an alternating group and for G a projective special linear group when the given representation is in non-defining characteristic. For the proof we exhibit a linear composition factor in the socle of the restriction to a large subgroup of the alternating or symmetric square of a given projective representation V. Assuming irreducibility this shows that the dimension of V has to be very small. A good knowledge of projective representations of small dimension allows to rule out these cases as well.

1. Introduction

Let R = R(ℓ^{f}) be a finite classical group of Lie type. Let G < R be a quasi-simple subgroup acting absolutely irreducibly on the natural module of R, not of Lie type in characteristic ℓ. We study those cases where G has the same number of composition factors on the adjoint module for R as R itself. These embeddings are of importance in the determination of maximal subgroups of the finite classical groups of Lie type.

Let V be the natural module for a classical group R. We will write Λ̃^{2}(V), Σ̃^{2}(V) respectively Ã(V) for the largest irreducible R-sub-quotient of Λ^{2}(V), Sym^{2}(V), V ⊗ V^{∗}. In Table 1.1 we recall the dimension of X̃(V) for certain choices (R, X) with X ∈ {Λ, Σ^{2}, A} (see [3], resp. [1, (8.9), (9.6) and (11.6)]).

In this paper we study quasi-simple subgroups G of classical groups R which act irreducibly on V as well as on X̃(V) with X as in Table 1.1. We say that V is of plustype (for G) if it carries a G-invariant quadratic form, and that V is of minus-type if it carries a bilinear alternating form but no quadratic form. Thus if V is of plus-type for G, we have to consider both Λ̃^{2}(V) and Σ̃^{2}(V), if V is of minus-type, we consider Σ̃^{2}(V), and Λ̃^{2}(V) if ℓ = 2, and if V carries no G-invariant non-degenerate form, we have to consider Ã(V). Previous results in this direction were obtained by the first

Typeset by AMS-TEX

This paper was written while the authors were participating in the special semester on representations of algebraic groups and related finite groups at the Isaac Newton Institute (Cambridge). It is a pleasure to thank the institute for its hospitality.

author for special linear groups [8], and by the second author for arbitrary quasisimple groups in the case ℓ = 0 and X = A [9]. We hope to pursue this investigation for other types of quasi-simple groups G in the future.

1.1. Choices for R and X.

R  X̃ dim(X̃(V))  condition
SL_{m}  Ã m^{2} − 1  ℓ |̸m
Ã m^{2} − 2  ℓ|m
Sp_{m} Σ̃^{2} \frac{1}{2}m(m + 1) ℓ odd
Σ̃^{2} \frac{1}{2}m(m − 1) − 1  ℓ = 2, m ≡ 2 (mod 4)
Σ̃^{2} \frac{1}{2}m(m − 1) − 2  ℓ = 2, m ≡ 0 (mod 4)
SO_{m} Λ̃^{2} \frac{1}{2}m(m − 1) ℓ odd
Λ̃^{2} \frac{1}{2}m(m − 1) − 1  ℓ = 2, m ≡ 2 (mod 4)
Λ̃^{2} \frac{1}{2}m(m − 1) − 2  ℓ = 2, m ≡ 0 (mod 4)
Σ̃^{2} \frac{1}{2}m(m + 1) − 1  ℓ |̸m
Σ̃^{2} \frac{1}{2}m(m + 1) − 2  ℓ|m

2. The alternating groups

For a field F and a partition λ ⊢ n of n let S^{λ} denote the Specht module for the symmetric group 𝔖_{n} over F. It is irreducible if the characteristic of F is prime to n!. We first collect some results on the constituents of alternating and symmetric squares of small Specht modules.

Lemma 2.1. For all n ≥ 9 we have

\begin{aligned}Λ^{2}(S^{n−1,1}) & = S^{n−2,1^{2}}, Σ^{2}(S^{n−1,1}) = S^{n} + S^{n−1,1} + S^{n−2,2}, \\ Λ^{2}(S^{n−2,2}) & = S^{n−2,1^{2}} + S^{n−3,2,1} + S^{n−3,1^{3}} + S^{n−4,3,1}, \\ Σ^{2}(S^{n−2,2}) & = S^{n} + S^{n−1,1} + 2S^{n−2,2} + S^{n−3,3} + S^{n−3,2,1} + S^{n−4,4} + S^{n−4,2^{2}} \\ Λ^{2}(S^{n−2,1^{2}}) & = S^{n−2,1^{2}} + S^{n−3,2,1} + S^{n−3,1^{3}} + S^{n−4,2,1^{2}} \\ Σ^{2}(S^{n−2,1^{2}}) & = S^{n} + S^{n−1,1} + 2S^{n−2,2} + S^{n−3,3} + S^{n−3,2,1} + S^{n−4,2^{2}} + S^{n−4,1^{4}} . \end{aligned}

Proof. It is well-known that the character χ_{λ} of any Specht module S^{λ} can be written as a linear combination of permutation characters 1_{𝔖_{μ}}^{𝔖_{n}} on Young subgroups 𝔖_{μ} for partitions μ ⊵ λ. The decomposition of 1_{𝔖_{μ}}^{𝔖_{n}} ⊗ 1_{𝔖_{μ}}^{𝔖_{n}} where the largest part μ_{1} of μ is at least n − 2 can easily be done using the Mackey formula for tensor products. The formulas for the decomposition of alternating and symmetric squares are then proved by induction. We give the details for the simple case of S^{n−1,1}. By the Mackey formula we have

1_{𝔖_{n−1}}^{𝔖_{n}} ⊗ 1_{𝔖_{n−1}}^{𝔖_{n}} = 1_{𝔖_{n−1}}^{𝔖_{n}} + 1_{𝔖_{n−2}}^{𝔖_{n}} ,

hence since 1_{𝔖_{n−1}}^{𝔖_{n}} = S^{n} + S^{n−1} we obtain

(2.2)  S^{n−1,1} ⊗ S^{n−1,1} = S^{n} + S^{n−1,1} + S^{n−2,2} + S^{n−2,1^{2}} .

The branching rule yields

Λ^{2}(S^{n−1,1})|_{𝔖_{n−1}} = Λ^{2}(S^{n−2,1} + S^{n−1}) = Λ^{2}(S^{n−2,1}) + S^{n−2,1} = S^{n−3,1^{2}} + S^{n−2,1} ,

the last equality by the induction hypothesis, starting from n = 4 where the assertion is easy to check. Now conversely the branching rule shows that Λ^{2}(S^{n−1,1}) can only have been S^{n−2,1^{2}}. Equation (2.2) then also determines the decomposition of the symmetric square. The other cases can be handled similarly. □

For any field F of characteristic ℓ and any ℓ-regular partition λ ⊢ n we denote by D^{λ} the ℓ-modular irreducible F 𝔖_{n}-module indexed by λ. The decomposition of the Specht modules S^{λ} into irreducible constituents D^{μ} is known for small partitions. We will need the following assertions:

\begin{aligned}S^{n−1,1} & = D^{n−1,1} + δ_{1}D^{n}, \\ S^{n−2,2} & = D^{n−2,2} + δ_{2}D^{n−1,1} + δ_{3}D^{n}, \\ S^{n−2,1^{2}} & = D^{n−2,1^{2}} + δ_{1}D^{n−1,1} if ℓ is odd, \end{aligned}

where

δ_{1} := \{ \begin{matrix}1 & if ℓ|n, \\ 0 & else, \end{matrix} δ_{2} := \{ \begin{matrix}1 & if ℓ|n − 2, \\ 0 & else, \end{matrix}

δ_{3} := \{ \begin{matrix}1 & if 2 ≠ ℓ|n − 1, or ℓ = 2, n ≡ 1, 2 (mod 4), \\ 0 & else \end{matrix}

(see [5, Th. 24.1 and 24.15]). Finally, we need some information on irreducible modules of small dimension. For k ≥ 1 we define the set R_{n}(k) to consist of those irreducible 𝔖_{n}-modules D for which there exists an ℓ-regular partition λ ⊢ n with λ_{1} ≥ n − k such that either D ≅ D^{λ} or D ≅ S^{1^{n}} ⊗ D^{λ}. The following is a slight strengthening of a result of James [6, Th. 7] (where the bound (n − 1)(n − 2)/2 was given):

Proposition 2.3. Let F be a field of characteristic ℓ ≥ 0. Let n ≥ 12 if ℓ ≠ 2, respectively n ≥ 17 if ℓ = 2. Then any irreducible F 𝔖_{n}-module D satisfies one of

D ∈ R_{n}(2)  or  dim(D) > (n − 2)(n − 3) .

Proof. Let first ℓ ≠ 2. Define f(n) := (n−2)(n−3) for n ≥ 13 and f(12) := 88. Then 2f(n) > f(n + 2) for all n ≥ 12. Also, from the known decomposition matrices [5] it follows that for n ∈ {12, 13} and F of characteristic different from 2 the irreducible F 𝔖_{n}-modules D either lie in R_{n}(2) or satisfy dim(D) > f(n). (𝔖_{12} has an 89dimensional 5-modular irreducible module not in R_{n}(2).) Moreover, by [6, Table 1] we have dim(D) > f(n) for all D ∈ R_{n}(4) \ R_{n}(2) and n ≥ 12. But then by Lemma 4 in [6] the first part follows.

If F has characteristic 2 we define f(15) = f(16) = 126, f(n) = (n − 2)(n − 3) for n ≥ 17. Then the same argument can be used since for n ∈ {15, 16} the irreducible F 𝔖_{n}-modules D either lie in R_{n}(2) or satisfy dim(D) > 126. □

A corresponding result for the projective representations of the alternating group was proved by Wagner [11]:

Proposition 2.4 (Wagner). Let F be a field of characteristic ℓ ≠ 2, n ≥ 8, and V an absolutely irreducible faithful F [2.𝔄_{n}]-module. Then dim(V) is divisible by 2^{⌊ \frac{n−s−1}{2} ⌋} where s is the number of 1’s in the 2-adic expansion of n.

Lemma 2.5. Let G be a covering group of the alternating group 𝔄_{n}, n ≥ 8, V a faithful F G-module with F of characteristic ℓ ≥ 0. Then X̃(V), for X̃ ∈ {Λ̃^{2}, Σ̃^{2}, Ã}, contains one of the constituents of the permutation character 1_{𝔄_{n−4}}^{𝔄_{n}}.

Proof. Let first V be an absolutely irreducible faithful 𝔄_{n}-module and ℓ ≠ 2. We consider the restriction of V to a natural subgroup H = 𝔄_{4} and write V_{1}, V_{2}, V_{3} for the non-trivial isotypic components of V under the elementary abelian Sylow 2subgroup of H. Since ℓ ≠ 2 these are direct summands of V which are cyclically permuted by the elements of order 3 in H. Thus V_{1}, V_{2}, V_{3} are equivalent modules for the centralizer 𝔄_{n−4} of H. Hence X(V) contains at least three trivial 𝔄_{n−4}- composition factors in the socle. By our above considerations this implies that X̃(V) contains a constituent of the permutation module 1_{𝔄_{n−4}}^{𝔄_{n}} as claimed.

In the case ℓ = 2 we let σ_{1}, σ_{2} ∈ G = 𝔄_{n} be 3-cycles which generate an 𝔄_{4}. Let V_{ω}, V_{ω^{2}} denote the non-trivial eigenspaces of σ_{1} on V. These are modules for the centralizer 𝔄_{n−3} of σ_{1}. Let W_{1}, …, W_{k} denote the composition factors in the socle of the restriction of V_{ω} to the centralizer 𝔄_{n−4} of H := ⟨σ_{1}, σ_{2}⟩. Assume first that V is not self-dual, so X = A. If k ≥ 2, or if k = 1 and W_{1} is equivalent to the socle of V_{ω^{2}}, X(V) contains at least four trivial 𝔄_{n−4}-composition factors in the socle. If the socle of V_{ω} ⊕ V_{ω^{2}} consists of two non-equivalent 𝔄_{n−4}-modules then W_{1} has to be σ_{2}-invariant, so V_{ω} is G-invariant, which is a contradiction. So now assume that V is self-dual. If one of the W_{i} is self-dual, or if two of them are equivalent, or if k ≥ 3 then X(V) has sufficiently many trivial 𝔄_{n−4}-composition factors in the socle and X̃(V) contains a constituent of 1_{𝔄_{n−4}}^{𝔄_{n}}. If k = 1 and W_{1} is not self-dual then V_{ω} is ⟨σ_{1}, σ_{2}, 𝔄_{n−3}⟩ = G-invariant, so V = V_{ω}, which is a contradiction. Thus finally assume that k = 2 and W_{1}, W_{2} are non-equivalent and not self-dual. Then the conjugate of W_{1} by σ_{2} is another composition factor of the socle of V as 𝔄_{n−4}- module. Since V_{ω} cannot be G-invariant, W_{2} is dual to W_{1} so that again X(V) has three trivial 𝔄_{n−4}-composition factors in the socle.

Finally, assume that G = 𝔄̃_{n} = 2.𝔄_{n} and ℓ ≠ 2. Here we restrict to a cyclic subgroup generated by an element σ of order 4 in a natural subgroup H = 𝔄̃_{4} and denote by V_{i}, V_{−i} the eigenspaces of σ for the primitive fourth roots of unity. These are direct summands of V which are interchanged by elements in the Sylow 2-subgroup of H, hence they are equivalent modules for the centralizer 𝔄̃_{n−4} of H. Thus X(V) contains at least one trivial composition factor in the socle. If V is of plus-type, then so are the V_{i} as 𝔄̃_{n−4}-modules, and Σ^{2}(V) contains at least three trivial composition factors in the socle. Thus we obtain that X̃(V) contains a constituent of 1𝔄̃_{n} ; 𝔄̃_{n−4} as claimed. □

Proposition 2.6. Let G be a covering group of the alternating group 𝔄_{n}, n ≥ 7, V an absolutely irreducible faithful F G-module with F of characteristic ℓ ≥ 0. Then X̃(V) is reducible for X̃ ∈ {Λ̃^{2}, Σ̃^{2}, Ã} unless (G, ℓ, dim(V), X̃) are as in Table 2.7.

2.7. Examples in alternating groups 𝔄_{n}, n ≥ 7.

G  ℓ dim(V)  Λ̃^{2} Σ̃^{2} Ã
𝔄_{n}  ℓ |̸n n − 1  ×
𝔄_{n}  ℓ|n n − 2  × ×
3.𝔄_{7}  5  3  ×
𝔄_{7}  2  4  ×
2.𝔄_{7}  7  4  ×
2.𝔄_{7} ≠ 2, 7  4  ×
6.𝔄_{7} ≠ 2, 3  6  ×
𝔄_{8}  2  4  ×
2.𝔄_{8}  ≠ 2  8  ×
𝔄_{9}  2  8  ×
2.𝔄_{9}  3  8  ×
2.𝔄_{9} ≠ 2, 3  8  × ×
2.𝔄_{10}  5  8  × ×

Proof. The Brauer character tables for 2.𝔄_{n}, n ≤ 13, are known (see [5]) and the assertion can be checked, so we may assume n ≥ 14.

We first consider irreducible representations V of 𝔄_{n}. Then X̃(V) is a constituent of W := 1_{𝔄_{n−4}}^{𝔄_{n}} by Lemma 2.5. The ordinary constituents of W are Specht modules S^{λ} for partitions λ ⊢ n with λ_{1} ≥ n − 4. The largest degree of such a constituent is dim(S^{n−4,2,1^{2}}) = n(n − 2)(n − 3)(n − 5)/8. On the other hand by Prop. 2.3 either dim(V) > (n − 2)(n − 3) or V ∈ R_{n}(2), or n ≤ 16 and ℓ = 2. Since

\frac{1}{8}n(n − 2)(n − 3)(n − 5) < \frac{1}{8}(n − 2)(n − 3)((n − 2)(n − 3) − 2) − 2

it follows that either V is a constituent of the restriction to 𝔄_{n} of some D^{λ} with λ_{1} ≥ n − 2, or n ≤ 16, ℓ = 2. For these modules the assertion easily follows from Lemma 2.1 and the decompositions into irreducibles of S^{λ} collected above. If ℓ = 2, n ∈ {14, 15, 16}, the degrees of the constituents of 1_{𝔄_{n−4}}^{𝔄_{n}} can be computed explicitly using [4]; they are collected in Table 2.8. The degrees in line 3 correspond to the examples in Table 2.7. The only other entries of the form m^{2} − ϵ, m(m − 1)/2 − ϵ for integral m and ϵ ∈ {1, 2} are

208 = 21 · 20/2 − 2, 560 = 34 · 33/2 − 1, 624 = 25^{2} − 1.

But 𝔄_{14} does not have 2-modular irreducible representations of degree 21 or 34, nor does 𝔄_{15} have such of degree 25, as restriction to 𝔄_{13} shows.

2.8. Dimensions of some 𝔖_{n}-modules D^{λ}, ℓ = 2.

\begin{matrix}λ & n = 14 & n = 15 & n = 16 \\ n & 1 & 1 & 1 \\ n − 1, 1 & 12 & 14 & 14 \\ n − 2, 2 & 64 & 90 & 90 \\ n − 3, 3 & 208 & 336 & 336 \\ n − 3, 2, 1 & 560 & 624 & 896 \\ n − 4, 4 & 364 & 910 & 910 \\ n − 4, 3, 1 & 1300 & 1300 & 2548 \end{matrix}

In the case of faithful 2.𝔄_{n}-modules Lemma 2.5 and the result of Wagner (Prop. 2.4) give a contradiction unless n ∈ {14, 15}, dim(V) ∈ {32, 64}. We first consider n = 15. If ℓ ≠ 13 the smallest dimension of a faithful 2.𝔖_{13}-module is 64, while for ℓ = 13 the smallest degree of a faithful 2.𝔄_{12}-module is 32. Thus restriction to these subgroups shows that in fact dim(V) ≥ 64 if n = 15. Now from [5, Th. 24.1 and 24.15] and Table 1 in [6] it follows that no irreducible constituent of W has the same degree as X̃(V). So finally assume that n = 14. In any characteristic, any faithful irreducible 2.𝔄_{11}-module of dimension less than 100 has dimension 16, and there are at most two inequivalent ones. Thus if dim(V) = 64 we obtain enough trivial 2.𝔄_{11}-composition factors in the socle of X(V) to conclude that X̃(V) is a constituent of 1_{𝔄_{11}}^{𝔄_{14}}. This contradicts dim(V) = 64. Thus dim(V) = 32. Again from [6, Table 1] and [5, Th. 24.1 and 24.15] we deduce ℓ = 3. From the known tables it can be checked that the smallest faithful 2.𝔄_{13}-modules (of dimension 32) are of minus-type, while the 32-dimensional faithful 2.𝔖_{12}-module is of plus-type. So V does not carry a nondegenerate form. The tensor product of a 32-dimensional faithful 2.𝔄_{13}-module with itself contains a constituent of degree 66, so dim(Ã(V)) ≤ 924, which gives the final contradiction. □

3. The linear groups

Partial results in the case of linear groups had already been obtained in [8]. It turns out that the methods used there extend to our more general situation.

Proposition 3.1. Let G be a covering group of L_{2}(q) or L_{3}(q), ℓ a prime with gcd(q, ℓ) = 1 or ℓ = 0 and V an absolutely irreducible faithful F G-module. Then X(V) is reducible for X ∈ {Λ̃^{2}, Σ̃^{2}, Ã} unless (G, ℓ, dim(V), X) are as in Table 3.2.

Proof. Let first S = L_{2}(q). The cases q ≤ 13 can be treated using the tables in the modular Atlas [7]. Hence assume that q ≥ 16. Then, by [10] the minimal degree of a faithful irreducible projective representation of L_{2}(q) in characteristic ℓ |̸q is (q − 1)/ gcd(2, q − 1). On the other hand, the largest degree of a complex projective irreducible character for S is q+1. For the case of the alternating square, irreducibility of X(V) gives the inequality \frac{1}{8}(q − 1)(q − 3) − 2 ≤ q + 1 which is not satisfied for q ≥ 16. For Σ^{2}(V) and A(V) the bounds restrict q even more.

The groups L_{3}(q) for q ≤ 4 can be handled using [7]. So assume now that S = L_{3}(q), q ≥ 5. Then the minimal degree of a faithful irreducible projective representation of S is at least q^{2} − 1 [10], while the largest degree of a complex projective irreducible character for S is (q^{2} + q + 1)(q + 1). The irreducibility of the alternating square thus forces the inequality

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

which does not hold for q ≥ 5. □

3.2. Examples in linear groups.

G  ℓ dim(V)  Λ̃^{2} Σ̃^{2} Ã
2.L_{2}(4) ≠ 2, 5  2  ×
L_{2}(4)  ≠ 2, 5  3  × ×
L_{2}(7)  ≠ 2, 7  3  ×
3.L_{2}(9)  ≠ 3  3  ×
L_{2}(9)  2  4  ×
2.L_{2}(9) ≠ 2, 3  4  ×
L_{2}(9)  ≠ 2, 3  5  ×
L_{2}(13)  2  6  ×
4_{2}.L_{3}(4)  3  4  ×
2.L_{3}(4)  3  6  × ×
6.L_{3}(4) ≠ 2, 3  6  ×
4_{1}.L_{3}(4)  ≠ 2  8  ×
2.L_{3}(4)  7  10  ×
L_{4}(2)  ≠ 2  7  ×

Proposition 3.3. Let G be a covering group of L_{n}(q), n ≥ 4, q ≠ 2, ℓ a prime with gcd(q, ℓ) = 1 or ℓ = 0 and V an absolutely irreducible faithful F G-module. Then X̃(V) is reducible for X̃ ∈ {Λ̃^{2}, Σ̃^{2}, Ã} unless (G, ℓ, dim(V), X̃) are as in Table 3.2.

Proof. Using the modular Atlas to treat L_{4}(3) we may assume that V is a module for G = SL_{n}(q), and q ≥ 4 for n = 4. We now essentially follow the argument in [8]. Let P_{1} denote the stabilizer in G of a 1-dimensional subspace of the natural module. Thus P_{1} is a maximal parabolic subgroup of G, with Levi decomposition P_{1} = U_{1}: G_{1} where the Levi factor G_{1} = GL_{n−1}(q) acts on the elementary abelian group U_{1} of order q^{n−1} as it does on its natural module. Upon restriction to U_{1} the module V splits as

V |_{U_{1}} = ⨁_{λ∈Hom(U_{1},F^{×})} V_{λ} .

Moreover G_{1} permutes transitively the spaces V_{λ} for λ ≠ 1.

We fix 1 ≠ λ ∈ Hom(U_{1}, F^{×}) and denote by P_{2} its stabilizer in P_{1}. Thus P_{2} = U_{1}: U_{2}: G_{2} where U_{2}: G_{2} is the stabilizer of λ in G_{1} = GL_{n−1}(q), |U_{2}| = q^{n−2} and G_{2} ≅ GL_{n−2}(q). Let ⟨λ⟩ denote the set of q − 1 non-trivial elements of Hom(U_{1}, F^{×}) stabilized by P_{2}. We consider two cases.

Case 1: First assume that U_{2} acts nontrivially on V_{λ}. We write

V_{λ}|_{U_{2}} = ⨁_{μ∈Hom(U_{2},F^{×})} V_{λ,μ} .

By our assumption V_{λ,μ} ≠ 0 for some (and hence all) μ ≠ 1. The stabilizer P_{3} in P_{2} of such a V_{λ,μ} is of the shape U_{1}: U_{2}: U_{3}: GL_{n−3}(q), with U_{3} elementary abelian of order q^{n−3}. Let ⟨μ⟩ denote the set of non-trivial elements in the centralizer in Hom(U_{2}, F^{×}) of GL_{n−3}(q), of order q − 1, and

V_{λ,⟨μ⟩} := ⨁_{μ′∈⟨μ⟩} V_{λ,μ′} .

Thus the stabilizer H_{3} of V_{λ,⟨μ⟩} is of the form U_{1}: U_{2}: U_{3}: (GL_{n−3}(q) × (q − 1)). Now W := X(V_{λ,⟨μ⟩}) is a non-trivial H_{3}-invariant subspace of X(V). Assuming that X̃(V) is irreducible this gives the upper bound

dim(X̃(V)) ≤ [G : H_{3}] dim(W) ≤ (q^{n} − 1)(q^{n−1} − 1)(q^{n−2} − 1) dim(X̃(V_{λ,μ})) .

Since |Hom(U_{i}, F^{×})| = |U_{i}| = q^{n−i} we have the lower bound dim(V) ≥ (q^{n−1} − 1) × (q^{n−2} − 1)d_{λ,μ} where d_{λ,μ} := dim(V_{λ,μ}), so

(3.4) dim(X̃(V)) ≥ \frac{1}{2}(q^{n−1} − 1)(q^{n−2} − 1)d_{λ,μ}((q^{n−1} − 1)(q^{n−2} − 1)d_{λ,μ} − 1) − 2 .

These two inequalities yield a contradiction for all (n, q).

Case 2: Thus we may assume that U_{2} acts trivially on V_{λ}. Case 2a: First we treat the subcase that d_{λ} := dim(V_{λ}) ≥ 2. The subspace W := X(V_{λ}) of X(V) is P_{2}-invariant by definition and has trivial U_{2}-action by assumption. Since the kernel Ũ_{1} < U_{1} of λ (of order q^{n−2}) acts trivially on W, this is a P̃_{2} := Ũ_{1}: U_{2}: G_{2}-submodule of X(V). Our assumptions force W ≠ 0, so if X̃(V) is irreducible it must be a constituent of W G ; P̃_{2}. Now P̃_{2} lies in a maximal parabolic

subgroup P of G of the form Ũ_{1}: U_{2}: (SL_{2}(q) ∘ SL_{n−2}(q)) : (q − 1). Since any irreducible complex character of GL_{2}(q) has degree at most q + 1 we see that X̃(V) is the constituent of the induced of a P -module of dimension at most (q + 1) dim(W). This gives the bound

dim(X̃(V)) ≤ \frac{(q^{n} − 1)(q^{n−1} − 1)}{(q − 1)^{2}} dim(X(V_{λ})) .

On the other hand clearly dim(V) ≥ (q^{n−1} − 1)d_{λ}, so for example

(3.5)  dim(Λ̃^{2}(V)) ≥ \frac{1}{2}(q^{n−1} − 1)d_{λ}((q^{n−1} − 1)d_{λ} − 1) − 2

in the case X = Λ^{2}. Comparing these bounds we obtain a contradiction unless q = 3, d_{λ} ≤ 3, X = Σ^{2}. In the latter case we may assume n ≥ 5. Then the derived group of G_{2} has to act trivially on V_{λ}, hence V_{λ} has only linear G_{2}-constituents. Replacing V_{λ} by one of these, the previous argument goes through. So X̃(V) can not have been irreducible. Case 2b: Now assume that d_{λ} = 1. Then the same argument as before can be applied to the submodules W := V_{λ} ⊗ V_{λ′}, where λ′ ∈ Hom(U_{1}, F^{×}) different from λ and from 1 is stabilized by P_{2}, yielding the bound

dim(X̃(V)) ≤ \frac{(q^{n} − 1)(q^{n−1} − 1)}{(q − 1)^{2}} .

By [10] we have dim(V) ≥ (q^{n} − 1)/(q − 1) − n ≥ q^{n−1} + q^{n−2} for (n, q) ≠ {(4, 3)}. The ensuing inequality is not satisfied for n ≥ 4. This completes the proof. □

To treat the linear groups over 𝔽_{2} we introduce some more notation. For a partition λ ⊢ n let χ_{λ} denote the (complex irreducible) unipotent character of SL_{n}(q) indexed by λ. For a Levi subgroup L of G = SL_{n}(q) we write R_{L}^{G} (respectively ^{∗}R_{L}^{G}) for the operation of Harish-Chandra induction (restriction).

Lemma 3.6. Let n ≥ 4. The unipotent character χ_{n−2,1^{2}} of SL_{n}(2) is a constituent of both Λ^{2}(χ_{n−1,1}) and Σ^{2}(χ_{n−1,1}).

Proof. We proceed by induction on n. The assertion is true for SL_{4}(2) ≅ 𝔄_{8}. Let P = U: L be the maximal parabolic subgroup of G = SL_{n}(2) with L = GL_{n−1}(2) and denote by V a ℂG-module affording the character χ_{n−1,1}. Harish-Chandra induction and restriction of unipotent characters χ_{λ} of SL_{n}(q) behave precisely as ordinary induction and restriction of Specht modules S^{λ} for the symmetric group 𝔖_{n}. Thus V |_{P} contains an L-module with character χ_{n−2,1} on the centralizer V_{1} of U. Now X(V)|_{P} = X(V |_{P}) for X ∈ {Λ^{2}, Σ^{2}}, hence by the inductive assumption we have that X(V)|_{P} contains a submodule with character χ_{n−3,1^{2}} on the centralizer X(V)_{1} of U. In other words ⟨^{∗}R_{L}^{G}(X(χ_{n−1,1})), χ_{n−3,1^{2}}⟩ ≠ 0. By the adjointness of Harish-Chandra induction and restriction this implies that X(χ_{n−1,1}) contains a constituent of R_{L}^{G}(χ_{n−3,1^{2}}). By the Littlewood-Richardson rule we have

R_{L}^{G}(χ_{n−3,1^{2}}) = χ_{n−2,1^{2}} + χ_{n−3,2,1} + χ_{n−3,1^{3}}.

But the degrees

\begin{aligned}χ_{n−3,2,1}(1) & = (2^{n} − 1)(2^{n} − 16)(2^{n−2} − 1)/7, \\ χ_{n−3,1^{3}}(1) & = (2^{n} − 2)(2^{n} − 4)(2^{n} − 8)/21, \end{aligned}

of the second and third possible constituent of X(V) are larger than dim(X(V)). This shows the assertion. □

Proposition 3.7. Let G be a covering group of L_{n}(2), n ≥ 5, ℓ ≠ 2 and V an absolutely irreducible faithful F G-module. Then X̃(V) is reducible for X̃ ∈ {Λ̃^{2}, Σ̃^{2}, Ã}.

Proof. As above we may assume that V is a module for G = SL_{n}(q). We follow the proof of the preceding proposition. The argument in Case 1 goes through unless d_{λ,μ} := dim(V_{λ,μ}) = 1 and X = Λ^{2}, when W = 0. In this case let 0 ≠ μ′ ∈ Hom(U_{2}, F^{×}) be different from μ. Then W := V_{λ,μ} ⊗ V_{λ,μ′} is a 1-dimensional P_{4} := U_{1}: U_{2}: U_{3}: U_{4}: GL_{n−4}(2)-module. The maximal dimension of an irreducible GL_{4}(2) ≅ 𝔄_{8}-module is 70, so all constituents of the induction of W from P_{4} to a maximal parabolic subgroup of type [2^{4n−16}]: (GL_{4}(2) ∘ GL_{n−4}(2)) have dimension at most 70. This forces

dim(Λ^{2}(V)) ≤ 70\frac{(2^{n} − 1)(2^{n−1} − 1)(2^{n−2} − 1)(2^{n−3} − 1)}{(2^{2} − 1)(2^{3} − 1)(2^{4} − 1)}

which is a contradiction to the lower bound (3.4).

In Case 2, when U_{2} acts trivially on V_{λ}, let P_{3} = U_{1}: U_{2}: U_{3}: GL_{n−3}(2) denote the stabilizer of a pair λ, λ′ of non-trivial elements of Hom(U_{1}, F^{×}). We write V_{λ} = ⊕_{μ}V_{λ,μ} for the decomposition of V_{λ} into isotypic components for U_{3}. Now the submodule ⊕_{μ}(V_{λ,μ} ⊗ V_{λ′,μ}) of X(V) has trivial action by the unipotent radical of P_{3} while the Levi factor acts by permuting the factors. Thus X̃(V) is a constituent of the permutation module for this subgroup. Since the largest dimension of an irreducible GL_{3}(2)-character is 8, we obtain the estimate

dim(X̃(V)) ≤ \frac{(2^{n} − 1)(2^{n−1} − 1)(2^{n−2} − 1)}{21}.

On the other hand, if GL_{n−2}(2) acts non-trivially on V_{λ}, by [10] we get the lower bound

dim(V) ≥ (2^{n−1} − 1)(2^{n−2} − n + 1)

if n ≥ 7, dim(V) ≥ 217 respectively if n = 6, dim(V) ≥ 30 if n = 5. This yields a contradiction. Hence V_{λ} cannot be a faithful GL_{n−2}(2)-module, and replacing V_{λ} by a 1-dimensional submodule in its socle yields a contradiction to the upper bound (3.5) unless d_{λ} = 1 and X = Λ. Since G is generated by two conjugates of U_{1} we must have dim(V_{1}) ≤ dim(V)/2, so dim(V) ≤ 2(2^{n−1} − 1) = 2^{n} − 2.

By [2, Th. 1.1] this implies that V is contained in the ℓ-modular reduction of a characteristic 0 module with character χ_{n−1,1} and χ_{n−1,1}(1) − 1 ≤ dim(V) ≤ χ_{n−1,1}(1). But now the dimension of the submodule of X(χ_{n−1,1}) exhibited in Lemma 3.6 is too small for X̃(V) to be irreducible. This final contradiction completes the proof. □

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K. M. : Department of Mathematics, Wayne State University, Detroit, Michigan 48202, U.S.A.

   E-mail address: kaym@math.wayne.edu

G. M. : IWR, Im Neuenheimer Feld 368, Universität Heidelberg, D–69120 Heidelberg, Germany

   E-mail address: malle@urania.iwr.uni-heidelberg.de