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Irreducible tensor products of representations of finite quasi-simple

groups of Lie type

Kay Magaard and Pham Huu Tiep^{∗}

Abstract. Let G be a finite quasi-simple group of Lie type defined over a field of characteristic p. We determine whether G can possess irreducible Brauer characters α, β (of degree > 1) in characteristic other than p such that α ⊗ β is irreducible.

1. Introduction

Let G be a finite group, r a fixed prime (or r = 0). Consider the following statement

P(r) : α⊗β is not irreducible for any nonlinear Brauer characters α, β ∈ IBr_{r}(G)

(by IBr_{r}(G) with r = 0 we mean Irr(G)).

One motivation for studying the property P(r) comes from the desire to classify the overgroups of the irreducible subgroups of the finite classical groups G(q). According to Aschbacher’s Theorem [A], if M is a maximal subgroup of G(q), then either M belongs to one of 8 collections C_{i}, 1 ≤ i ≤ 8, of “natural” subgroups of G(q), or M ∈ S, a collection of quasi-simple groups which act irreducibly on the natural module V of G(q). For instance, C_{4} consists of stabilizers of tensor decompositions of V. Conversely, if M ∈ ∪_{i=1}^{8}C_{i}, then the maximality of M is determined in [KL]. Now suppose M ∈ S. Could it happen that M < N < G for some N ∈ C_{4}, in other words, could the (irreducible) representation of M on V be tensor-decomposable?

This problem of tensor-decomposable irreducible modules has been investigated by Isaacs [Is] for solvable groups G. In particular, he conjectures that if the solvable group G has two faithful complex characters α and β such that αβ is irreducible, then G is abelian. He also proved this conjecture in the case where every minimal normal subgroup of G is cyclic.

∗Part of this work was done 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 and support.

The authors are grateful to M. Geck, P. Sin and A. Turull for helpful discussion. It is the first author’s pleasure to thank the SFB 343 at the Universität Bielefeld for its hospitality and support.

The goal of this paper is to find all finite quasi-simple groups with property P(r). It has been shown by Bessenrodt and Kleshchev [BK1] that the symmetric group 𝕊_{n} (n ≥ 5) satisfies P(0). They also classified all complex irreducible characters α, β of degree > 1 of the alternating group 𝔸_{n} (n ≥ 5) such that α ⊗ β is irreducible: it happens if and only if n = a^{2} for some a ≥ 3, one of the characters α, β is the nontrivial irreducible component of the natural permutation character, and the other one corresponds to the partition (a^{a}). More recently, Bessenrodt and Kleshchev have shown in [BK2] that 𝕊_{n} satisfies P(r) for any odd prime r, and that 𝕊_{n} satisfies P(2) if n is odd. The cases of complex characters of 2𝕊_{n} and 2𝔸_{n} are being studied. We note that there are examples of irreducible tensor products of complex spin representations of 𝔸_{9}, 𝔸_{10}. Also, some examples of irreducible tensor products of representations of 𝕊_{n} in characteristic 2 have been found by Gow and Kleshchev [GK].

We investigated the 26 sporadic finite simple groups (and their covers) using the character tables that we found in GAP. We found that M_{11}, J_{1} and 2HS satisfy P(r) for all primes r, and that HN, Ly, He, F i_{23} and J_{4} satisfy P(0) whereas all other sporadic groups do not even satisfy P(0). The list of tensor-decomposable irreducible representations that we found can be reproduced in GAP with minimal effort and is hence not included in this paper. Instead, we list a couple of examples of irreducible tensor products Φ ⊗ Ψ for G = BM, 2 · BM:

\begin{matrix}Φ(1) = 4371, & Ψ(1) = 53936390144, \\ Φ(1) = 96256, & Ψ(1) = 90807234375, \end{matrix}

and for G = M :

\begin{matrix}Φ(1) = 196883, & Ψ(1) = 8980616927734375, \\ Φ(1) = 19360062527, & Ψ(1) = 8980616927734375, \\ Φ(1) = 21296876, & Ψ(1) = 3503434660075044981. \end{matrix}

We focus our attention on the finite quasi-simple groups G of Lie type defined over a field 𝔽_{q} of characteristic p, q = p^{f}, and p ≠ r.

Observe that G cannot satisfy P(p). Indeed, let us view G as the fixed point subgroup G^{F}, where G is a simple algebraic group defined over \overline{𝔽}_{p} and F is the corresponding Frobenius endomorphism. Then the Steinberg tensor product theorem asserts that any irreducible \overline{𝔽}_{p}G-module can be written as the tensor product V_{1} ⊗ V_{2}^{(p)} ⊗ … ⊗ V_{f}^{(p^{f−1})} for some restricted irreducible \overline{𝔽}_{p}G-modules V_{1}, …, V_{f}; and conversely, any such tensor product is irreducible. This remarkable theorem gives us many examples of modules whose tensor product is irreducible. Even more, in combination with a result of Seitz, this theorem allows us to describe all pairs (U, V) of \overline{𝔽}_{p}G-modules such that U ⊗ V is irreducible. For, let

V = V_{1} ⊗ V_{2}^{(p)} ⊗ … ⊗ V_{f}^{(p^{f−1})}, U = U_{1} ⊗ U_{2}^{(p)} ⊗ … ⊗ U_{f}^{(p^{f−1})},

where each U_{i}, V_{i} is restricted and irreducible. Then

U ⊗ V ≃ (U_{1} ⊗ V_{1}) ⊗ (U_{2} ⊗ V_{2})^{(p)} ⊗ … ⊗ (U_{f} ⊗ V_{f})^{(p^{f−1})}.

Suppose U ⊗ V is irreducible. Then for each i, the G-module W_{i} := U_{i} ⊗ V_{i} is irreducible. In particular, the G-module W_{i} is irreducible. Now [S1, (1.6)] states that W_{i} is restricted, and it also gives us the list of all pairs (U_{i}, V_{i}) of restricted modules such that the G-module W_{i} is irreducible. Conversely, suppose that we are given irreducible restricted modules U_{i}, V_{i}, 1 ≤ i ≤ f, such that each W_{i} is irreducible. Since W_{i} is restricted in this case, the G-module W_{i} is irreducible. Hence the G-module U ⊗V is irreducible by the Steinberg tensor product theorem.

The main result of the paper is the following

Theorem 1.1. Let G be a finite quasi-simple group of Lie type defined over a field 𝔽_{q}. Then either G has property P(r) for any r |̸q, or one of the following holds.

(i) q ≤ 3.

(ii) G = Sp_{2n}(q), where q is odd.

(iii) G = Sp_{2n}(q), 2|q.

(iv) G/Z(G) = L_{3}(4).

(v) G = F_{4}(q), 2|q.

(vi) G = ^{2}F_{4}(q), 2|q.

Throughout the paper, r |̸q means that either r = 0, or r does not divide q. We prove more detailed results in the body of the paper. In particular we give examples where property P(r) does not hold. For example, it is shown in Proposition 5.5 that when G is Sp_{2n}(3) or Sp_{2n}(5), then P(0) does not hold. Using GAP one sees that 4L_{3}(4), Sp_{6}(2), Sp_{8}(2), and Sp_{10}(2) do not have property P(0). Thus the cases (i) – (iv) in Theorem 1.1 actually occur. On the other hand, we show that when q is odd and ≥ 7, then Sp_{2n}(q) has property P(0). Moreover we conjecture that Sp_{2n}(q) has property P(r) when q is odd and ≥ 7. We also conjecture, but are not able to show, that none of the cases (v) and (vi) can happen. In particular, the cases (v) and (vi) cannot happen if r = 0, see Proposition 8.4. Furthermore we conjecture that Sp_{2n}(q), with 2|q and q ≠ 2 does have property P(r).

For reader’s convenience, we formulate the main result in the complex case.

Theorem 1.2. Let G be a finite quasi-simple group of Lie type defined over a field 𝔽_{q}. Then either G has property P(0), or one of the following holds.

(i) q ≤ 3.

(ii) G = Sp_{2n}(5).

(iii) G = Sp_{2n}(q), 2|q.

(iv) G/Z(G) = L_{3}(4).

The dual problem of determining whether the product of two conjugacy classes of a finite group could form a single conjugacy class has been investigated by Arad [Ar]. Also, property P(0) is closely related to factorizations of quasi-primitive characters intensively investigated by Ferguson and Turull [FT].

2. Preliminaries

We begin with the following trivial statements.

Lemma 2.1. Let G = H ∗C be a central product of a subgroup H and an abelian subgroup C. Then any irreducible 𝔽H-module V extends to G.

Proof. Let W be a simple submodule of V^{G}. Then W = A ⊗ B, where A is an irreducible H-module and B an irreducible C-module. Since C is abelian, dim(B) = 1. Next, 0 ≠ Hom(W, V^{G}) ≃ Hom(W |_{H}, V), hence W |_{H} = V. ⊓ ⊔

Lemma 2.2. Let V be a G-module. Let H ⊲G and let U be a simple submodule, resp. a simple quotient, of V |_{H}. Suppose U extends to G and G/H is abelian. Then the G-module V has a submodule, resp. a quotient, Ũ, which is an extension of U.

Proof. 1) We claim that, if W is an arbitrary irreducible constituent of the Gmodule U^{G}, then W |_{H} ≃ U. Indeed, since U is G-invariant, U^{G}|_{H} is the sum of |G/H| copies of U. Hence, W |_{H} is a direct sum of copies of U. By [Hup, (V.17.5)], W = W_{1} ⊗ W_{2}, where W_{1} is a projective representation of G and W_{2} a projective representation of G/H. We can choose W_{1} to be an extension Û of U to G. Since the factor sets in W_{1} and W_{2} are inverse sets, W_{2} is also a linear representation of G/H. But G/H is abelian, hence dim W_{2} = 1 and W is an extension of U.

2) Consider the case where U is a submodule of V |_{H}. Then 0 ≠ Hom(U, V |_{H}) ≃ Hom(U^{G}, V). Thus V has a submodule V_{1} which is a nonzero quotient of U^{G}. By 1), V_{1} has a submodule which is an extension of U.

3) Consider the case where U is a quotient of V |_{H}. Then 0 ≠ Hom(V |_{H}, U) ≃ Hom(V, U^{G}). Thus V has a quotient V_{1} which is a nonzero submodule of U^{G}. By 1), V_{1} has a quotient which is an extension of U. ⊓ ⊔

Observe that one cannot drop the assumption G/H is abelian in Lemma 2.2. For example, one can have examples where V is the tensor product of an extension of U and an irreducible G/H-module of dimension > 1, and yet V is irreducible.

We are going to consider representations of a finite group G over an algebraically closed field 𝔽 of characteristic r. Usually, G contains a p-subgroup of special type for a prime p ≠ r, that is, [Q, Q] = Z(Q) = Φ(Q) has exponent p and [x, Q] = Z(W) for all x ∈ Q \ Z(Q). More concretely, if p is odd then Q is the set

{[X, Y, z]|X, Y ∈ 𝔽_{q}^{m}, z ∈ 𝔽_{q}}

(for a natural integer m and a power q = p^{f}) endowed with the following group operation

[X, Y, z] ∘ [X′, Y′, z′] = [X + X′, Y + Y′, z + z′ + ∑_{i=1}^{m}(x_{i}y′_{i} − x′_{i}y_{i})],

if

\begin{matrix}X = (x_{1}, …, x_{m}), Y = (y_{1}, …, y_{m}), \\ X′ = (x′_{1}, …, x′_{m}), Y′ = (y′_{1}, …, y′_{m}). \end{matrix}

Clearly, [Q, Q] = Z(Q) = Φ(Q) coincides with the subgroup {t_{c} := [0, 0, c] | c ∈ 𝔽_{q}}. Moreover, [x, Q] = Z(Q) for any x ∈ Q \ Z(Q), hence Q is of special type. If G is a finite group of Lie type defined over a field of characteristic p, then we can select Q such that P := N_{G}(Q) is a “big” parabolic subgroup of G.

Let Q be a p-group of special type. For any 𝔽Q-module V, let

Ω(V) = {λ ∈ Irr(Z(Q)) \ {1} | λ occurs in V |_{Z(Q)}}.

For any λ ∈ Ω(V), let V_{λ} be the λ-eigenspace for Z(Q) on V, and let

d_{λ}(V) = dim(V_{λ})/\sqrt{|Q/Z(Q)|}.

Lemma 2.3. Let G be a finite group with a p-subgroup Q of special type and char(𝔽) ≠ p. Let C ≤ N_{G}(Q) ∩ C_{G}(Z(Q)). Suppose that V and W are 𝔽Gmodules such that there is a λ ∈ IBr_{r}(Z(Q))\{1} with λ ∈ Ω(V) and λ^{−1} ∈ Ω(W). Then V ⊗ W has a C-submodule F of dimension d_{λ}(V)d_{λ^{−1}}(W), on which Q acts trivially.

Proof. Since char(𝔽) ≠ p, we can write

V = ( ⨁_{μ∈Ω(V)} V_{μ}) ⊕ V_{1}, W = ( ⨁_{μ∈Ω(W)} W_{μ}) ⊕ W_{1}.

But C centralizes Z(Q), hence C preserves each of the summands in these decompositions. In particular, U := V_{λ} ⊗ W_{λ^{−1}} is a direct summand of the C-module V ⊗ W. Let α be the unique irreducible character (over 𝔽) of Q whose restriction to Z(Q) contains λ. Then the character of Q on V_{λ}, respectively on W_{λ^{−1}}, equals d_{λ}(V)α, respectively d_{λ^{−1}}(W)ᾱ. It is easy to see that αᾱ is the regular character of Q/Z(Q). Hence Q acts on U with character d_{λ}(V)d_{λ^{−1}}(W) ∑_{β∈IBr_{r}(Q/Z(Q))} β. Let F be the fixed point subspace of Q on U. Then clearly dim(F) = d_{λ}(V)d_{λ^{−1}}(W), F is C-invariant, and Q acts trivially on F. ⊓ ⊔

A key ingredient of our proof is the following

Proposition 2.4. Let G be a finite group with a p-subgroup Q of special type and char(𝔽) ≠ p. Let Q ≤ C = N_{G}(Q) ∩ C_{G}(Z(Q)). Suppose that V and W are 𝔽G-modules such that there is a λ ∈ IBr_{r}(Z(Q)) \ {1} with λ ∈ Ω(V) and λ^{−1} ∈ Ω(W). Suppose in addition that G has a subgroup N ≤ N_{G}(Q) such that N/Q = C/Q ∗ B for some abelian subgroup B. Then V ⊗ W has an N-submodule T of dimension at most d_{λ}(V)d_{λ^{−1}}(W). In particular, if the G-module V ⊗ W is irreducible, then

dim(V) · dim(W) ≤ (G : N)d_{λ}(V)d_{λ^{−1}}(W).

Proof. By Lemma 2.3, the C-module V ⊗ W has a submodule F of dimension d_{λ}(V)d_{λ^{−1}}(W), on which Q acts trivially. Let F′ be a simple C-submodule of F.

By Lemma 2.1, F′ extends to N. By Lemma 2.2, the N-module V ⊗ W has a submodule T, which is an extension of F′. If the G-module V ⊗ W is irreducible, then 0 ≠ Hom_{N}(T, (V ⊗W)|_{N}) ≃ Hom_{G}(T^{G}, V ⊗W), whence V ⊗W is a quotient of T^{G}, and so

dim(V) · dim(W) ≤ dim(T^{G}) ≤ (G : N)d_{λ}(V)d_{λ^{−1}}(W).
⊓ ⊔

Extending the proof of [S2, (1.4)] and [Hup, (V.17.5)], one can prove the following supplement to Lemma 2.3, which we will use as an induction hypothesis in what follows.

Lemma 2.5. Keep the notation and assumption of Lemma 2.3. In addition, suppose that

(i) C is a semi-direct product of Q by A × B;

(ii) A acts faithfully on Q and A is perfect;

(iii) B has trivial Schur multiplier;

(iv) (|Z(Q)|, |Q/Z(Q)|) ≠ (3, 9), (9, 81). Then the C-module T mentioned in Lemma 2.3 is actually the tensor product of two C/Q-modules, of dimension d_{λ}(V) and d_{λ^{−1}}(W), respectively. □

Lemma 2.6. Let G be a finite group with property P(r). Let H be a normal subgroup of G of prime index t, and let γ, δ ∈ IBr_{r}(H) be characters of degree > 1 such that

(i) γ extends to G;

(ii) For a fixed h ∈ G \ H, if δ^{h} ≠ δ, then γ ⊗ δ^{h} ≠ γ ⊗ δ. Then the H-character γ ⊗ δ is not irreducible.

Proof. Consider the inertia group I of δ. Then H ≤ I ≤ G. If I = G, then δ extends to G by [F, 3.2.14], and so γ ⊗ δ is reducible by the condition P(r) for G. Suppose I < G. Then I = H, and δ has exactly t distinct conjugates δ_{i}, i = 1, …, t, and δ^{G}|_{H} = ∑_{i=1}^{t} δ_{i}. Let γ′ denote an extension of γ to G. Suppose γ ⊗δ is irreducible. Then γ ⊗δ_{i}, i = 1, …, t, are all irreducible and distinct, hence the G-character γ′ ⊗ δ^{G} is irreducible, again contradicting the condition P(r) for G. ⊓ ⊔

3. Linear Groups

Let G = GL_{n}(q), S = SL_{n}(q) with q = p^{f}. It is easy to check that all covers of L_{2}(q) with q > 3 and q ≠ 9 have property P(r) for all r. On the other hand, the universal cover of L_{2}(9) ≃ Sp_{4}(2)′ does not have property P(r) for any r. In what follows we assume that n ≥ 3, (n, q) ≠ (3, 2), (3, 4), (4, 2), (4, 3). Let 𝔽_{q}^{n} be the natural module for G, with a basis (e_{1}, …, e_{n}). We consider the parabolic subgroup P = St_{S}(⟨e_{2}⟩, ⟨e_{2}, …, e_{n}⟩_{𝔽_{q}}) of S. Then Q = O_{p}(P) is a group of order q^{2n−3} of special type. The center Z(Q) has order q and contains q−1 transvections

t_{c}, where c ∈ 𝔽_{q}^{•} and t_{c} has matrix diag (( \begin{matrix}1 & 0 \\ c & 1 \end{matrix} ) , I_{n−2}) in the given basis of

V. It is easy to see that Q has a complement

{diag(a, b, X) | a, b ∈ 𝔽_{q}^{•}, X ∈ GL_{n−2}(q), ab · det(X) = 1} ,

and all the transvections t_{c} are conjugate in P. Our assumption on (n, q) guarantees that S is the universal cover of P SL_{n}(q).

Let V be any irreducible 𝔽S-module of dimension > 1. Then the subgroup {diag(a^{−1}, 1, a, I_{n−3}) | a ∈ 𝔽_{q}^{•}} acts transitively on the set IBr_{r}(Z(Q))\{1}, hence Ω(V) = IBr_{r}(Z(Q))\{1}, and all d_{λ}(V) are the same. We begin with the following estimate

Lemma 3.1. In the above notation,

d_{λ}(V) ≤ \frac{dim(V)}{q^{n−1} − 1}.

If d_{λ}(V) = 1 and n ≥ 5, then V is one of Weil modules.

Proof. Let R = St_{S}(e_{2}) and let A = O_{p}(R). Then A is an elementary abelian group of order q^{n−1}. Let V (μ) be the homogeneous component of V |_{A} corresponding to μ ∈ IBr_{r}(A). Fix a nontrivial μ ∈ IBr_{r}(A). Then dim(V) = dim(V (1))+(q^{n−1} −1) dim(V (μ)). In particular, dim(V (μ)) ≤ dim(V)/(q^{n−1} −1). Let χ be the Brauer character of V. Then χ(t_{1}) = dim(V (1)) − dim(V (μ)), and so dim(V (μ)) = (dim(V) − χ(t_{1})/q^{n−1}. In the meantime,

dim(V) = dim(V_{1}) + q^{n−2}(q − 1)d_{λ}(V), χ(t_{1}) = dim(V_{0}) − q^{n−2}d_{λ}(V),

whence

d_{λ}(V) = \frac{dim(V) − χ(t_{1})}{q^{n−1}} = dim(V (μ)) ≤ \frac{dim(V)}{q^{n−1} − 1}.

Next suppose that d_{λ}(V) = 1. Then dim V (μ) = 1 and dim(V) = dim(V (1)) + (q^{n−1}−1). Observe that V (1) = C_{V} (A). Now we can look at the opposite parabolic subgroup R′ and its radical A′ of order q^{n−1}. Since S = ⟨A, A′⟩ (cf. [GM]) and V is irreducible and non-linear, C_{V} (A) ∩ C_{V} (A′) = 0. The above computation shows that dim(C_{V} (A)) = (dim(V)+(q^{n−1} −1)χ(t_{1}))/q^{n−1}. Similarly, if t′ ∈ A′ is a nonidentity element, then dim(C_{V} (A′)) = (dim(V) + (q^{n−1} − 1)χ(t′))/q^{n−1}. Observe that t_{1} and t′ are S-conjugate, as n > 2. Hence, dim(C_{V} (A)) = dim(C_{V} (A′)) and so dim(V) ≥ 2 dim(C_{V} (A)). We conclude that dim(V) ≤ 2(q^{n−1} − 1). It remains to apply the main result of [GT]. ⊓ ⊔

Observe that Lemma 3.1 remains true for an 𝔽G-module V.

Proposition 3.2. Suppose n ≥ 2. Then the groups SL_{n}(q) with q ≥ 4, and GL_{n}(q) with q ≥ 3, have property P(r) for any r |̸q.

Proof. The cases (3, 4) and (4, 3) can be checked using GAP. Thus we may assume that n ≥ 3, and (n, q) ≠ (3, 4), (4, 3). First we assume that G = GL_{n}(q), q ≥ 3, has irreducible 𝔽-representations V, W of dimension > 1 such that V ⊗ W is irreducible. Pick any λ ∈ Ω(V). Consider the parabolic subgroup P̂ = St_{G}(⟨e_{2}⟩, ⟨e_{2}, …, e_{n}⟩_{𝔽_{p}}) of G. Then Q = O_{p}(P̂), and Q has a complement

D := {diag(a, b, X) | a, b ∈ 𝔽_{q}^{•}, X ∈ GL_{n−2}(q)}

in P̂. Consider the following subgroups

A := {diag(a, a, X) | a ∈ 𝔽_{q}^{•}, X ∈ GL_{n−2}(q)} , B := {diag(1, a, I_{n−2}) | a ∈ 𝔽_{q}^{•}}

of D, and let C = Q : A. Clearly, C normalizes Q and centralizes Z(Q). Moreover, P̂ = CB, and B centralizes C/Q. Applying Proposition 2.4, we get dim(V) · dim(W) ≤ (G : P̂) · d_{λ}(V) · d_{λ^{−1}}(W). Combined with Lemma 3.1, this inequality yields

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

a contradiction since q ≥ 3.

Next we assume that S = SL_{n}(q), q ≥ 4, has irreducible 𝔽-representations V, W of dimension > 1 such that V ⊗ W is irreducible. Consider the complement

D := {diag(a, b, X) | a, b ∈ 𝔽_{q}^{•}, X ∈ GL_{n−2}(q), ab · det(X) = 1}

of Q in P. Consider the following subgroups

A := {diag(a, a, X) | a ∈ 𝔽_{q}^{•}, det(X) = a^{−2}} , B := {diag(a, a^{−1}, I_{n−2}) | a ∈ 𝔽_{q}^{•}}

of D, and let C = Q : A. Clearly, C normalizes Q and centralizes Z(Q), and B centralizes C/Q. By Proposition 2.4, dim(V) · dim(W) ≤ (S : BC) · d_{λ}(V) · d_{λ^{−1}}(W). Combined with Lemma 3.1, this inequality yields

1 ≤ \frac{(q^{n} − 1)(q^{n−1} − 1) · gcd(2, q − 1)}{(q − 1)^{2} · (q^{n−1} − 1)^{2}},

a contradiction since q ≥ 4. ⊓ ⊔

Proposition 3.3. The groups SL_{n}(2), n ≥ 5, has property P(r) for any r ≠ 2.

Proof. We proceed by induction on n ≥ 5. The cases n = 5, 6 can be checked directly using GAP.

Consider the case n ≥ 7 and assume the contrary: S = SL_{n}(2) has irreducible 𝔽-representations V, W of dimension > 1 such that V ⊗ W is irreducible. We pick any λ ∈ Ω(V) and distinguish two subcases.

1) d_{λ}(V), d_{λ^{−1}}(W) > 1.

By Lemma 2.5, P has a submodule T inside V ⊗ W, where T is acted on trivially by Q and it is the tensor product of two P/Q-modules of dimension d_{λ}(V) and d_{λ^{−1}}(W). Since P/Q ≃ SL_{n−2}(2), by induction hypothesis T is not irreducible. Hence we can choose a composition factor U of the P -module T such that dim(U) ≤ dim(T)/2. Moreover, we can choose U as a submodule or a quotient of T. In the former case,

0 ≠ Hom(U, (V ⊗ W)|_{P}) ≃ Hom(U^{S}, V ⊗ W),

hence V ⊗ W is a quotient of U^{S}. In the latter case,

0 ≠ Hom((V ⊗ W)|_{P} , U) ≃ Hom(V ⊗ W, U^{S}),

hence V ⊗ W is a submodule of U^{S}.

Let L be a Levi subgroup of P. Then L ≃ SL_{n−2}(2), and U is actually an L-module. Observe that U^{S} is the Harish-Chandra induction R_{L}^{S}(U) (where we use the parabolic subgroup P containing L). We introduce two more parabolic subgroups. The first one, P_{1} = St_{S}(⟨e_{1}⟩, ⟨e_{1}, e_{2}, …, e_{n−1}⟩) is obviously conjugate to P. We can view U as a module U_{1} for a Levi subgroup L_{1} of P_{1}, and then U^{S} = R_{L}^{S}(U) ≃ R_{L_{1}}^{S}(U_{1}). The second one is P_{2} = St_{S}(⟨e_{1}⟩, ⟨e_{1}, e_{2}⟩) having L_{2} := L as a Levi subgroup. Let S = {(1, 2), (2, 3), …, (n−1, n)} be the standard set of generators for the Weyl group of S = SL_{n}(2). Since

P_{1} = \{[ \begin{matrix}1 & ∗ & ∗ \\ 0 & X & ∗ \\ 0 & 0 & 1 \end{matrix} ] | X ∈ SL_{n−2}(2)\} , P_{2} = \{[ \begin{matrix}1 & ∗ & ∗ \\ 0 & 1 & ∗ \\ 0 & 0 & X \end{matrix} ] | X ∈ SL_{n−2}(2)\} ,

P_{1} is the standard parabolic subgroup corresponding to the subset

I_{1} = {(2, 3), (3, 4), …, (n − 2, n − 1)},

and P_{2} is the standard parabolic subgroup corresponding to the subset

I_{2} = {(3, 4), (4, 5), …, (n − 1, n)}

of S. Clearly, I_{1} and I_{2} are conjugate via the cycle w = (2, 3, 4, …, n − 1, n) of the Weyl group of S. Now the independence property of the Harish-Chandra induction (cf. [HL], [DD]) implies that R_{L_{1}}^{S}(U_{1}) ≃ R_{L_{2}}^{S}(U_{2}), where U_{2} = ^{w}U_{1} (the w-conjugate of U_{1}). Thus U^{S} is isomorphic to the Harish-Chandra induction (1 ∘ 1 ∘ U_{2}) ↑ S from P_{2} to S of the P_{2}/O_{p}(P_{2})-module U_{2} in the notation of [GT].

Next we can write

(1 ∘ 1 ∘ U_{2}) ↑ S ≃ (((1 ∘ 1) ↑ SL_{2}(2)) ∘ U_{2}) ↑ S.

But (1 ∘ 1) ↑ SL_{2}(2) is a reducible 3-dimensional module. Therefore, any composition factor of U^{S} has dimension at most (2/3) dim(U^{S}).

The above arguments show that dim(V ⊗W) ≤ (1/3)(S : P) dim(T). Combined with Lemma 3.1, this yields

1 ≤ \frac{1}{3} · \frac{(2^{n} − 1)(2^{n−1} − 1)}{(2^{n−1} − 1)^{2}},

a contradiction.

2) Either d_{λ}(V) or d_{λ^{−1}}(W) = 1.

For definiteness let us suppose d_{λ}(V) = 1. The above argument with Harish-Chandra induction shows that dim(V ⊗W) ≤ (2/3)·(S : P)d_{λ}(V)d_{λ^{−1}}(W). Recall that d_{λ}(V) = 1, and d_{λ^{−1}}(W) ≤ dim(W)/(2^{n−1} − 1) by Lemma 3.1. By [GT], dim(V) ≥ 2^{n} − 3. We come to the inequality

2^{n} − 3 ≤ \frac{2}{3} · \frac{(2^{n} − 1)(2^{n−1} − 1)}{2^{n−1} − 1},

again a contradiction. ⊓ ⊔

Proposition 3.4. Let n ≥ 3 and let V, W be two nontrivial irreducible representations of SL_{n}(3) in characteristic r ≠ 3 such that V ⊗W is absolutely irreducible. Then r > 0, n is even, and at least one of the modules V, W is a Weil module.

Proof. We proceed by induction on n ≥ 3. The cases n = 3, 4 can be checked directly using GAP. So we assume that n ≥ 5. Pick any λ ∈ Ω(V) and consider two cases.

1) d_{λ}(V), d_{λ^{−1}}(W) > 1.

Let P′ be the (unique) normal subgroup of type Q : SL_{n−2}(3) in P. By Lemma 2.5, P′ has a submodule T inside V ⊗W, where T is acted on trivially by Q and it is the tensor product of two P′/Q-modules of dimension d_{λ}(V) and d_{λ^{−1}}(W). Since P′/Q ≃ SL_{n−2}(3), by induction hypothesis T is not irreducible. Hence we can choose a composition factor L of the P′-module T such that dim(L) ≤ dim(T)/2. Moreover, we can choose L as a submodule or a quotient of T. Arguing as in the proof of Proposition 3.3, we see that V ⊗ W is either a quotient or a submodule of L^{S}.

Consider the subgroup D := {diag(a, a, I_{n−2}) | a ∈ 𝔽_{q}^{•}} in P. Since D centralizes P′/Q, L extends to a P′D-module of the same dimension, whence L^{P′D} consists of two constituents of the same dimension. Hence we arrive at the bound dim(V ⊗ W) ≤ (1/2)(S : P′D) dim(T). Combined with Lemma 3.1, this yields

1 ≤ \frac{1}{2} · \frac{(3^{n} − 1)(3^{n−1} − 1)}{2(3^{n−1} − 1)^{2}},

a contradiction.

2) Either d_{λ}(V) or d_{λ^{−1}}(W) = 1. Clearly, the case n is odd follows from Proposition 3.2, since GL_{n}(3) = SL_{n}(3) × ℤ_{2} when n is odd.

Now we assume that n is even. Let α, resp. β, be the character afforded by V, resp. W. For definiteness we assume that d_{λ}(V) = 1, whence α is one of the two Weil characters of S by Lemma 3.1. By the virtue of Proposition 3.2, we may also assume that β is not extendible to G. In the rest of the proof we assume that r = 0.

First we consider the case α(1) = (3^{n} − 1)/3. For any g ∈ S let d(g) denote the dimension of the fixed point subspace of g on the natural module of S. Assume g is unipotent. Then d(g) ≥ 1. Following the proof of [TZ, Lemma 4.1], one can show that α(g) = (3^{d(g)} − 1)/2 ≥ 1. Also, it is easy to show that α extends to G. By our assumption, β^{G} is an irreducible complex character of G, and β^{G}|_{S} = β + β′ for some G-conjugate β′ of β. Let T be a Sylow 3-subgroup of S. By a result of Zelevinsky [Z], there is a character λ of T such that the multiplicity of λ in β^{G}|_{T} is one. From this it follows that λ enters exactly one of the two characters β|_{T} and β′|_{T}, whence β|_{T} ≠ β′|_{T}. Thus there is a unipotent element g ∈ S such that β(g) ≠ β′(g). Since α(g) ≥ 1, α ⊗ β ≠ α ⊗ β′. We have seen that the pair (α, β) satisfies the asumption of Lemma 2.6 (with H = S). By Lemma 2.6 and Proposition 3.2, α ⊗ β cannot be irreducible, a contradiction.

Next we consider the case α(1) = (3^{n} − 3)/2. Let R = A : L denote the first maximal parabolic subgroup of S with A := ℤ_{3}^{n−1} and L := GL_{n−1}(3). Then α = (1_{R})^{S} − 1_{S}, whence (α|_{R}, 1_{R})_{R} = 1. We claim that β|_{R} is irreducible. (For, suppose that β|_{R} is reducible: β|_{R} = γ + γ′ for some R-characters γ and γ′ with γ(1) ≤ β(1)/2. Then (α ⊗ β, γ^{S})_{S} = (α|_{R} ⊗ β|_{R}, γ)_{R} ≥ 1, but α ⊗ β is irreducible by assumption, hence

\frac{3^{n} − 3}{2}β(1) = α(1)β(1) ≤ γ^{S}(1) = (S : R)γ(1) ≤ (S : R)β(1)/2 = \frac{3^{n} − 1}{4}β(1),

a contradiction.) Next, the transitivity of L on Irr(A)\{1_{A}} and the irreducibility of β|_{R} imply that (3^{n−1} − 1)|β(1). Let ℓ be a primitive prime divisor (cf. [Zs]) of 3^{n−1} − 1, and let ℓ^{a} be the largest power of ℓ dividing 3^{n−1} − 1. Then ℓ^{2a} divides (α ⊗ β)(1) and does not divide |S|. Therefore α ⊗ β cannot be irreducible, again a contradiction. ⊓ ⊔

Thus we obtain a complete result in the complex case:

Theorem 3.5. Suppose n ≥ 2. Then the groups SL_{n}(q) and GL_{n}(q) have property P(0). □

4. Unitary Groups

Let G = U_{n}(q), S = SU_{n}(q) with q = p^{f}, n ≥ 3, (n, q) ≠ (3, 2), (4, 2), (4, 3), (6, 2). Let 𝔽_{q^{2}}^{n} be the natural module for G, endowed with a Hermitian form with Gram

matrix diag (( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) , I_{n−2}) in a basis (e_{1}, …, e_{n}). We consider the parabolic

subgroup P = St_{S}(⟨e_{1}⟩_{𝔽_{q^{2}}} ) of S. Then Q = O_{p}(P) is a group of order q^{2n−3} of special type. The centre Z(Q) has order q and contains q − 1 transvections t_{c}, where c ∈ 𝔽_{q^{2}}^{•}, c + c^{q} = 0. It is easy to see that Q has a complement

{diag(a, a^{−q}, X) | a ∈ 𝔽_{q^{2}}^{•}, X ∈ U_{n−2}(q), det(X) = a^{q−1}} ,

and all the transvections t_{c} are conjugate in P. Our assumption on (n, q) guarantees that S is the universal cover of P SU_{n}(q).

Let V be any irreducible 𝔽S-module of dimension > 1. Since the subgroup {diag(a, a^{−q}, a^{q−1}, I_{n−3}) | a ∈ 𝔽_{q^{2}}^{•}} acts transitively on IBr_{r}(Z(Q)) \ {1}, Ω(V) =

IBr_{r}(Z(Q))\{1} and all d_{λ}(V) are the same. In particular, d_{λ}(V) ≤ dim(V)/q^{n−2}(q− Proposition 4.1. Suppose n ≥ 3. Then the groups SU_{n}(q) with q ≥ 4, and U_{n}(q) with q ≥ 3, have property P(r) for any r |̸q.

Proof. The case (4, 3) can be checked directly using GAP, hence we may assume that (n, q) ≠ (4, 3). First we assume that G = U_{n}(q), q ≥ 3, has irreducible 𝔽-representations V, W of dimension > 1 such that V ⊗ W is irreducible. Pick any λ ∈ Ω(V) and consider the parabolic subgroup P̂ = St_{G}(⟨e_{1}⟩_{𝔽_{q^{2}}} ) of G. Then Q = O_{p}(P̂), and Q has a complement

D := {diag(a, a^{−q}, X) | a ∈ 𝔽_{q^{2}}^{•}, X ∈ U_{n−2}(q)}

in P̂. Consider the following subgroups

A := {diag(1, 1, X) | X ∈ U_{n−2}(q)} , B := {diag(a, a^{−q}, I_{n−2}) | a ∈ 𝔽_{q^{2}}^{•}}

of D, and let C = Q : A. Clearly, C normalizes Q and centralizes Z(Q). Moreover, P̂ = CB, and B centralizes C/Q. Hence dim(V) · dim(W) ≤ (G : P̂) · d_{λ}(V) · d_{λ^{−1}}(W) by Proposition 2.4. Thus

q^{2n−4}(q − 1)^{2} ≤ \frac{dim(V) · dim(W)}{d_{λ}(V) · d_{λ^{−1}}(W)} ≤ \frac{(q^{n} − (−1)^{n})(q^{n−1} − (−1)^{n−1})}{q^{2} − 1},

a contradiction since q ≥ 3 and n ≥ 3.

Next we assume that S = SU_{n}(q), q ≥ 4, has irreducible 𝔽-representations V, W of dimension > 1 such that V ⊗ W is irreducible. Consider the complement

D := {diag(a, a^{−q}, X) | a ∈ 𝔽_{q^{2}}^{•}, X ∈ U_{n−2}(q), det(X) = a^{q−1}}

of Q in P. Consider the following subgroups

A := {diag(a, a, X) | a^{q+1} = 1, det(X) = a^{−2}} , B := {diag(a, a^{−1}, I_{n−2}) | a^{q−1} = 1}

of D, and let C = Q : A. Again, C normalizes Q and centralizes Z(Q), and B centralizes C/Q. Hence dim(V) · dim(W) ≤ (S : BC) · d_{λ}(V) · d_{λ^{−1}}(W) by Proposition 2.4. Thus

q^{2n−4}(q −1)^{2} ≤ \frac{dim(V) · dim(W)}{d_{λ}(V) · d_{λ^{−1}}(W)} ≤ \frac{(q^{n} − (−1)^{n})(q^{n−1} − (−1)^{n−1})}{q^{2} − 1} ·gcd(2, q −1),

a contradiction since q ≥ 4 and n ≥ 3. ⊓ ⊔

5. Symplectic Groups in Odd Characteristic

Let G = CSp_{2n}(q), S = Sp_{2n}(q) with q = p^{f}, n ≥ 2, p > 2. Let 𝔽_{q}^{2n} be the natural module for G, endowed with a symplectic form with Gram matrix

diag (( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) , ( \begin{matrix}0 & I_{n−2} \\ −I_{n−2} & 0 \end{matrix} )) in a basis (e_{1}, …, e_{2n}). We consider the

parabolic subgroup P = St_{S}(⟨e_{1}⟩_{𝔽_{q}}) of S. Then Q = O_{p}(P) is a group of order q^{2n−1} of special type. The centre Z(Q) has order q and contains q−1 transvections

t_{c}, where c ∈ 𝔽_{q}^{•} and t_{c} has matrix (( \begin{matrix}1 & c \\ 0 & 1 \end{matrix} ) , I_{2n−2}) in the given basis. It is

easy to see that Q has a complement

{diag(a, a^{−1}, X) | a ∈ 𝔽_{q}^{•}, X ∈ Sp_{2n−2}(q)}

in P. Our assumption on (n, q) guarantees that S is the universal cover of P Sp_{2n}(q).

The transvections t_{c} are conjugate in G, but they lie in two conjugacy classes in S. We can identify IBr_{r}(Z(Q)) with 𝔽_{q} by assigning to each a ∈ 𝔽_{q} the character

λ_{a} : t_{c} ↦ ϵ^{Tr_{𝔽q /𝔽p} (ac)},

where ϵ = exp(2πi/p). Let Ω_{+}, resp. Ω_{−}, be the set of characters corresponding to the squares, resp. non-squares, in 𝔽_{q}^{•}. Let V be any irreducible 𝔽S-module of dimension > 1. Then Ω(V) is either Ω_{+}, Ω_{−}, or Ω := IBr_{r}(Z(Q))\{1} = Ω_{+} ∪Ω_{−}. In any case, d_{λ}(V) ≤ 2 dim(V)/q^{n−1}(q − 1) if λ ∈ Ω(V).

If V is extendible to G, then clearly Ω(V) = Ω and all d_{λ}(V) are the same. Actually we can say more about the case where Ω(V) ≠ Ω.

Lemma 5.1. Let V be an 𝔽S-module. Let S_{n−1} ≃ Sp_{2n−2}(q) be the pointwise stabilizer of the 2-subspace ⟨e_{2n−1}, e_{2n}⟩_{𝔽_{q}}. Suppose that Ω(V) ≠ Ω. Then Ω(U) ≠ Ω for any irreducible constituent U of V |_{S_{n−1}}.

Proof. Clearly Z(Q) < S_{n−1}. The assumption on Ω(V) means that the set of nontrivial characters of Z(Q) occuring in V |_{Z(Q)} is either ∅, Ω_{+}, or Ω_{−}. The same is true for U|_{Z(Q)}, where U is any irreducible constituent of V |_{S_{1}}. This forces Ω(U) to be different from Ω. ⊓ ⊔

Corollary 5.2. Let V be an irreducible 𝔽S-module of characteristic r = 0 such that Ω(V) ≠ Ω. Then V is either trivial, or one of Weil modules of S.

Proof. We proceed by induction on n ≥ 1. If n = 1 then our statement can be checked directly. Suppose n > 1. By Lemma 5.1 and by induction hypothesis, any irreducible constituent U of V |_{S_{n−1}} is either trivial, or one of Weil modules of S_{n−1}. This implies that V itself is either trivial or one of Weil modules of S_{n}, according to the main result of [TZ]. ⊓ ⊔

In view of Corollary 5.2 we call a nontrivial 𝔽S-module such that Ω(V) ≠ Ω a module of Weil-type. We remark that it seems very like that a module of Weil-type is in fact a Weil module.

Proposition 5.3. Suppose n ≥ 2, p > 2, p ≠ char(𝔽). Let V, W be irreducible 𝔽S-modules of dimension > 1 such that V ⊗ W is irreducible.

(i) If q ≥ 7, then both of V, W are Weil-type modules.

(ii) If q = 5, then at least one of V, W is a Weil-type module, and both of them are not invariant under CSp_{2n}(q).

Proof. (i) Assume the contrary, say W is not a Weil-type module. Then Ω(W) = Ω and so we can choose a λ ∈ Ω(V) such that λ^{−1} ∈ Ω(W). Consider the complement

D := {diag(a, a^{−1}, X) | a ∈ 𝔽_{q}^{•}, X ∈ Sp_{2n−2}(q)}

of Q in P. Consider the following subgroups

A := {diag(1, 1, X) | X ∈ Sp_{2n−2}(q)} , B := {diag(a, a^{−1}, I_{n−2}) | a ∈ 𝔽_{q}^{•}}

of D. Clearly, C = Q : A normalizes Q and centralizes Z(Q). Also, P = BC and B centralizes C/Q. Hence, dim(V) · dim(W) ≤ (S : P) · d_{λ}(V) · d_{λ^{−1}}(W) by Proposition 2.4. Thus

q^{2n−2}(q − 1)^{2}/4 ≤ \frac{dim(V) · dim(W)}{d_{λ}(V) · d_{λ^{−1}}(W)} ≤ \frac{q^{2n} − 1}{q − 1},

a contradiction since q ≥ 7.

(ii) First we consider the case where Ω(V) = Ω(W) = Ω. Then we can choose a λ ∈ Ω(V) such that dim(W_{λ^{−1}}) is smallest among all dim(W_{μ}), μ ∈ Ω(W). Clearly, dim(V)/d_{λ}(V) ≥ q^{n−1}(q − 1)/2 and dim(W)/d_{λ^{−1}}(W) ≥ q^{n−1}(q − 1). Using the same arguments as in (i) we come to the inequality

q^{2n−2}(q − 1)^{2}/2 ≤ \frac{dim(V) · dim(W)}{d_{λ}(V) · d_{λ^{−1}}(W)} ≤ \frac{q^{2n} − 1}{q − 1},

a contradiction since q = 5.

Next we consider the case where W is invariant under CSp_{2n}(q). Then Ω(W) = Ω and dim(W_{μ}) is the same for all μ ∈ Ω(W), hence dim(W)/d_{λ^{−1}}(W) ≥ q^{n−1}(q− 1). Now we can use the same arguments as in (i) and again come to the contradiction that

q^{2n−2}(q − 1)^{2}/2 ≤ \frac{dim(V) · dim(W)}{d_{λ}(V) · d_{λ^{−1}}(W)} ≤ \frac{q^{2n} − 1}{q − 1}.

⊓ ⊔

The same proof as above shows

Proposition 5.4. Suppose n ≥ 1, p > 2, and q = p^{f} ≥ 5. Then CSp_{2n}(q) has property P(r) for any r. □

In view of Proposition 5.3, one has to study the tensor product of Weil modules of S = Sp_{2n}(q). Recall that S has four complex irreducible Weil modules, whose characters are denoted by ξ, ξ^{∗} (of degree (q^{n} + 1)/2), and η, η^{∗} (of degree (q^{n} − 1)/2). If r ≠ 2, p, then the reduction modulo r of these four characters are absolutely irreducible, and distinct (look at the character values at transvections). If r = 2, then ξ(mod 2) = 1 + η(mod 2), ξ^{∗}(mod 2) = 1 + η^{∗}(mod 2), and η(mod 2) and η^{∗}(mod 2) are absolutely irreducible and distinct. If V is any irreducible module, then let S^{2}(V), resp. ∧^{2}(V), be the quotient (of codimension 0 or 1) of Sym^{2}(V), resp. of V ∧ V, by the fixed point subspace. If V is not self-dual, then let A(V) be the natural submodule of codimension 1 in V ⊗ V^{∗}. Let

M = {V ⊗ W, S^{2}(V), ∧^{2}(V), A^{2}(V) | V, W Weil modules of Sp_{2n}(q)} .

Proposition 5.5. In the above notation, assume n ≥ 2.

(i) Suppose q ≥ 11 and r ≠ p. Then no M ∈ M can be irreducible.

(ii) Suppose q = 7, 9. Then the only modules M ∈ M which can be irreducible are S^{2}(η), S^{2}(η^{∗}), ∧^{2}(ξ), and ∧^{2}(ξ^{∗}). Moreover, these modules are irreducible if r = 0.

(iii) Suppose q = 5. Then the only modules M ∈ M which can be irreducible are S^{2}(ξ) = S^{2}(ξ^{∗}), ∧^{2}(ξ), ∧^{2}(ξ^{∗}), S^{2}(η), S^{2}(η^{∗}), ∧^{2}(η) = ∧^{2}(η^{∗}), ξη = ξ^{∗}η^{∗}. Moreover, these modules are irreducible if r = 0.

(iii) Suppose q = 3. Then the only modules M ∈ M which can be irreducible are A(ξ), A(η), ξη^{∗}, ξ^{∗}η, S^{2}(ξ) = S^{2}(ξ^{∗}), ∧^{2}(ξ), ∧^{2}(ξ^{∗}), S^{2}(η), S^{2}(η^{∗}), ∧^{2}(η) = ∧^{2}(η^{∗}), ξη = ξ^{∗}η^{∗}. Moreover, these modules are irreducible if r = 0.

Proof. The discussion before this proposition shows that it is enough to prove this proposition for complex Weil modules.

We can embed S naturally in A := GL_{2n}(q). The group A has a so-called (reducible) rational Weil character τ of degree q^{2n}, and τ|_{S} is just the permutation character of S acting on the natural module 𝔽_{q}^{2n}, in particular, (τ|_{S}, τ|_{S})_{S} = 2q+2. Furthermore, τ_{S} is the sum of twice the trivial character, an irreducible character of degree (q^{n} − 1)(q^{n} + q)/2(q − 1) (of multiplicity 1), an irreducible character of degree (q^{n} + 1)(q^{n} − q)/2(q − 1) (of multiplicity 1), two irreducible characters of degree (q^{2n} − 1)/2(q − 1) (each of multiplicity 1), and (q − 3)/2 irreducible characters of degree (q^{2n} − 1)/(q − 1) (each of multiplicity 2) (cf. [T]).

We can also embed S naturally in B := U_{2n}(q). The group B has a so-called (reducible) ℚ-valued Weil character σ of degree q^{2n}, and (σ|_{S}, σ|_{S})_{S} = 2q + 2. Furthermore, σ_{S} is the sum of an irreducible character of degree (q^{n} + 1)(q^{n} + q)/2(q + 1) (of multiplicity 1), an irreducible character of degree (q^{n} − 1)(q^{n} − q)/2(q+1) (of multiplicity 1), two irreducible characters of degree (q^{2n}−1)/2(q+1) (each of multiplicity 1), and (q−1)/2 irreducible characters of degree (q^{2n}−1)/(q+ 1) (each of multiplicity 2) (cf. [T]).

We distinguish two cases.

1) q ≡ 1(mod4). In this case the characters ξ, ξ^{∗} are real, and the characters η, η^{∗} are real-valued but have Schur index 2 over ℝ.

One can show that τ|_{S} = (ξ + η)^{2} = (ξ^{∗} + η^{∗})^{2} and σ|_{S} = (ξ + η)(ξ^{∗} + η^{∗}) (cf. [T]). The biggest dimension of irreducible constituents of σ_{S} is (q^{2n} − 1)/(q + 1), which is less than (q^{n} − 1)^{2}/4, hence all the characters ξξ^{∗}, ηη^{∗}, ξη^{∗}, and ξ^{∗}η are not irreducible.

Next we suppose q ≥ 9. Then the biggest dimension of irreducible constituents of τ_{S} is (q^{2n} − 1)/(q − 1), which is less than the dimension of S^{2}(ξ) and ξη. Also, the dimension of ∧^{2}(η) cannot be equal to the dimension of any irreducible constituents of τ|_{S}. Finally, the only case when the dimension of S^{2}(η) and ∧^{2}(ξ) is equal to the dimension of an irreducible constituent of τ|_{S} is q = 9. In fact, if q = 9, then one can work out the following decomposition of the characters in M in terms of the irreducible constituents of τ|_{S}:

ξη = \overline{\frac{q^{2n} − 1}{q − 1}} + \overline{\frac{q^{2n} − 1}{q − 1}}, S^{2}(ξ) = \overline{\frac{q^{2n} − 1}{2(q − 1)}} + \overline{\frac{(q^{n} − 1)(q^{n} + q)}{2(q − 1)}},

∧^{2}(ξ), S^{2}(η) = \overline{\frac{q^{2n} − 1}{q − 1}}, ∧^{2} (η) = \overline{\frac{q^{2n} − 1}{2(q − 1)}} + \overline{\frac{(q^{n} + 1)(q^{n} − q)}{2(q − 1)}},

where \overline{d} denotes an irreducible character of degree d.

Finally, suppose q = 5. Then we have the obvious decomposition

τ_{S} = (ξ + η)^{2} = S^{2}(ξ) + ∧^{2}(ξ) + S^{2}(η) + ∧^{2}(η) + 2ξη + 2 · 1_{S}.

Since (τ_{S}, τ_{S})_{S} = 12, each summand in this decomposition is irreducible. Here are the degrees of these summands:

(ξη)(1) = \frac{q^{2n} − 1}{q − 1}, S^{2}(ξ)(1) = \frac{(q^{n} − 1)(q^{n} + q)}{2(q − 1)},

∧^{2}(ξ)(1) = S^{2}(η)(1) = \frac{q^{2n} − 1}{2(q − 1)}, ∧^{2} (η)(1) = \frac{(q^{n} + 1)(q^{n} − q)}{2(q − 1)}.

2) q ≡ 3(mod4). Then ξ^{∗} = \overline{ξ}, η^{∗} = \overline{η}, and both ξ, η are non-real.

One can show that τ|_{S} = (ξ+η)(ξ^{∗}+η^{∗}) and σ|_{S} = (ξ+η)^{2} = (ξ^{∗}+η^{∗})^{2} (cf. [T]). First we look at τ|_{S}. Suppose q ≥ 7. Then the biggest dimension of irreducible constituents of τ_{S} is (q^{2n}−1)/(q−1), which is less than (q^{n}−1)(q^{n}−3)/4, hence all the characters A(ξ), A(η), ξη^{∗}, and ξ^{∗}η are not irreducible. Next suppose q = 3. Then we have the obvious decomposition

τ_{S} = A(ξ) + A(η) + ξη^{∗} + ξ^{∗}η + 2 · 1_{S}.

Since (τ_{S}, τ_{S})_{S} = 8, each summand in this decomposition is irreducible. Here are the degrees of these summands:

A(ξ)(1) = \frac{(q^{n} − 1)(q^{n} + q)}{2(q − 1)}, A(η)(1) = \frac{(q^{n} + 1)(q^{n} − q)}{2(q − 1)},

(ξη^{∗})(1) = (ξ^{∗}η)(1) = \frac{q^{2n} − 1}{2(q − 1)}.

Next we consider σ|_{S} and suppose q ≥ 7. Then the biggest dimension of irreducible constituents of σ_{S} is (q^{2n} − 1)/(q + 1), which is less than the dimension of S^{2}(ξ) and ξη. Also, the dimension of ∧^{2}(η) cannot be equal to the dimension of any irreducible constituents of σ|_{S}. Finally, the only case when the dimension of S^{2}(η) and ∧^{2}(ξ) is equal to the dimension of an irreducible constituent of σ|_{S} is q = 7. In fact, if q = 7, then one can work out the following decomposition of the characters in M in terms of the irreducible constituents of σ|_{S}:

ξη = \overline{\frac{q^{2n} − 1}{q + 1}} + \overline{\frac{q^{2n} − 1}{q + 1}}, S^{2}(ξ) = \overline{\frac{q^{2n} − 1}{2(q + 1)}} + \overline{\frac{(q^{n} + 1)(q^{n} + q)}{2(q + 1)}},

∧^{2}(ξ), S^{2}(η) = \overline{\frac{q^{2n} − 1}{q + 1}}, ∧^{2} (η) = \overline{\frac{q^{2n} − 1}{2(q + 1)}} + \overline{\frac{(q^{n} − 1)(q^{n} − q)}{2(q + 1)}}.

Finally, suppose q = 3. Then we have the obvious decomposition

σ_{S} = (ξ + η)^{2} = S^{2}(ξ) + ∧^{2}(ξ) + S^{2}(η) + ∧^{2}(η) + 2ξη.

Since (σ_{S}, σ_{S})_{S} = 8, each summand in this decomposition is irreducible. Here are the degrees of these summands:

(ξη)(1) = \frac{q^{2n} − 1}{q + 1}, S^{2}(ξ)(1) = \frac{(q^{n} + 1)(q^{n} + q)}{2(q + 1)},

∧^{2}(ξ)(1) = S^{2}(η)(1) = \frac{q^{2n} − 1}{2(q + 1)}, ∧^{2} (η)(1) = \frac{(q^{n} − 1)(q^{n} − q)}{q + 1}.

⊓ ⊔

Propositions 5.3 and 5.5 altogether imply

Corollary 5.6. Suppose n ≥ 2, p > 2, and q ≥ 7. Then the group Sp_{2n}(q) has property P(0). □

We expect that Corollary 5.6 holds for all r since we expect Corollary 5.2 to hold for all r.

6. Orthogonal Groups In Characteristic 2

Let S = Ω_{2n}^{ϵ}(q) with q = 2^{f} > 2, n ≥ 4, ϵ = ±. Let 𝔽_{q}^{2n} be the natural module for S, endowed with a non-degenerate quadratic form, whose bilinear form has

Gram matrix diag (( \begin{matrix}0 & I_{2} \\ I_{2} & 0 \end{matrix} ) , ( \begin{matrix}0 & I_{n−2} \\ I_{n−2} & 0 \end{matrix} )) in a basis (e_{1}, …, e_{2n}). We

may assume that ⟨e_{1}, e_{2}⟩_{𝔽_{q}} is totally singular. Consider the parabolic subgroup P = St_{S}(⟨e_{1}, e_{2}⟩_{𝔽_{q}}) of S. Then Q = O_{p}(P) is a group of order q^{4n−7} of special type. The center Z(Q) has order q and contains q − 1 elements t_{c}, where c ∈ 𝔽_{q}^{•}. It is easy to see that Q has a complement

D := {diag (A, ^{t}A^{−1}, X) | A ∈ GL_{2}(q), X ∈ Ω_{2n−4}^{ϵ}(q)}

in P, and all the elements t_{c} are conjugate in P. Our assumption on (n, q) guarantees that S is the universal cover of Ω_{2n}^{ϵ}(q).

Let V be any irreducible 𝔽S-module of dimension > 1. Since the subgroup {diag (a, 1, a^{−1}, 1, I_{2n−4}) | a ∈ 𝔽_{q}^{•}} acts transitively on IBr_{r}(Z(Q)) \ {1}, Ω(V) = IBr_{r}(Z(Q))\{1}, and all d_{λ}(V) are the same. In particular, d_{λ}(V) ≤ dim(V)/q^{2n−4}(q− 1).

Proposition 6.1. Suppose q = 2^{f} ≥ 4 and n ≥ 4. Then the group Ω_{2n}^{ϵ}(q) has property P(r) for any r ≠ 2.

Proof. Assume that S = Ω_{2n}^{ϵ}(q) has irreducible 𝔽-representations V, W of dimension > 1 such that V ⊗ W is irreducible. Fix a λ ∈ Ω(V) and consider the subgroup

B := {diag (aI_{2}, a^{−1}I_{2}, I_{2n−4}) | a ∈ 𝔽_{q}^{•}}

in P. Also, let C be the (unique) normal subgroup of type Q : (SL_{2}(q)×Ω_{2n−4}^{ϵ}(q)) in P. Then C normalizes Q and centralizes Z(Q). Furthermore, P = BC, and B centralizes C/Q. Hence dim(V) · dim(W) ≤ (S : P) · d_{λ}(V) · d_{λ^{−1}}(W) by Proposition 2.4. Thus

q^{4n−8}(q − 1)^{2} ≤ \frac{dim(V) · dim(W)}{d_{λ}(V) · d_{λ^{−1}}(W)} ≤ \frac{(q^{n} − ϵ)(q^{2n−2} − 1)(q^{n−2} + ϵ)}{(q − 1)(q^{2} − 1)},

a contradiction since q ≥ 4 and n ≥ 4. ⊓ ⊔

7. Orthogonal Groups In Odd Characteristic

Let S = Spin_{n}^{ϵ}(q) with q = p^{f} ≥ 5, p > 2 a prime, n ≥ 7, ϵ = ±. Let 𝔽_{q}^{n} be the natural module for S̄ := Ω_{n}^{ϵ}(q), endowed with a non-degenerate quadratic form, whose

bilinear form has Gram matrix diag (( \begin{matrix}0 & I_{2} \\ I_{2} & 0 \end{matrix} ) , J) in a basis (e_{1}, …, e_{n}) for a

non-degenerate diagonal matrix J. We may assume that ⟨e_{1}, e_{2}⟩_{𝔽_{q}} is totally singular. Consider the parabolic subgroup P̄ = StS̄(⟨e_{1}, e_{2}⟩_{𝔽_{q}}) of S̄. Then Q = O_{p}(P̄) is a group of order q^{2n−7} of special type. Namely, Q is the subgroup

\{[A, B] := ( \begin{matrix}I_{2} & B & −^{t}A · J \\ 0 & I_{2} & 0 \\ 0 & A & I_{n−4} \end{matrix} ) | \begin{matrix}A ∈ M_{n−4,2}(𝔽_{q}), B ∈ M_{2,2}(𝔽_{q}), \\ B + ^{t}B + ^{t}AJA = 0 \end{matrix} \} .

Each matrix [A, B] ∈ Q can be identified with the triple [X, Y, z], where A =

diag(X, −2J^{−1}Y) and B = z ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )− \frac{1}{2} ^{t}AJA, X, Y ∈ 𝔽_{q}^{n−4} and z ∈ 𝔽_{q}. This

identification shows that Q is isomorphic to the group described before Lemma 2.3 with m = n − 4. The centre Z(Q) has order q and contains q − 1 elements t_{c}, where c ∈ 𝔽_{q}^{•}. Let GL_{2}^{•2}(q) be the subgroup of all A ∈ GL_{2}(q) with det(A) being a square. By [KL, 2.7.2], the element diag(A, ^{t}A^{−1}, I_{n−4}), A ∈ GL_{2}(q), belongs to Ω_{n}^{ϵ}(q) if and only if det(A) is a square. Hence Q has a complement D in P̄, which is isomorphic to (GL_{2}^{•2}(q) × Ω_{n−4}^{ϵ}(q)) · 2, and all the elements t_{c} are conjugate in P. Let C be the (unique) normal subgroup of type SL_{2}(q) × Ω_{n−4}^{ϵ}(q) in D and let P̄′ = Q : C.

Let P and P′ be complete inverse images of P̄ and P̄′ in S. Then P is the parabolic subgroup corresponding to the second node of the Dynkin diagram of S. S and P certainly contain a subgroup N ≃ Spin_{n−4}^{ϵ}(q) (which corresponds to the subdiagram containing all but the first and the second nodes). If a root α corresponds to the first node, then S and P also contain the subgroup M (generated by the root subgroups) corresponding to ±α. It is known that M ≃ SL_{2}(q), since the rank of S is bigger than 1. Clearly [M, N] = 1 and M ∩ N = 1 (as S is of simply connected type). Since M and N are perfect, they are contained in P′. Also, we can lift Q to O_{p}(P), which we also denote by Q. Now it is clear that P′ = Q : (M × N). Observe that M × N acts trivially on Z(Q).

Let V be any irreducible 𝔽S-module of dimension > 1. Since the complete inverse image of the subgroup {diag (a, 1, a^{−1}, 1, I_{n−4}) | a ∈ 𝔽_{q}^{•}} in S permutes acts transitively on IBr_{r}(Z(Q)) \ {1}, Ω(V) = IBr_{r}(Z(Q)) \ {1} and all d_{λ}(V) are the same. In particular, d_{λ}(V) ≤ dim(V)/q^{n−4}(q − 1).

Proposition 7.1. Suppose q = p^{f} ≥ 5 and n ≥ 7. Then the group Spin_{n}^{ϵ}(q) has property P(r) for any r ≠ p.

Proof. Assume that S = Spin_{n}^{ϵ}(q) has irreducible 𝔽-representations V, W of dimension > 1 such that V ⊗ W is irreducible. Fix a λ ∈ Ω(V) and consider the subgroup L of S generated by the root subgroups corresponding to the first and second nodes of the Dynkin diagram of S. Then L ≃ SL_{3}(q), and L contains M. Inside L we can choose a torus B ≃ ℤ_{q−1} which centralizes M and N. By Proposition 2.4, dim(V) · dim(W) ≤ (S : P′B) · d_{λ}(V) · d_{λ^{−1}}(W). Also, P/(P′B) = ℤ_{2}. Consequently,

q^{4k−8}(q − 1)^{2} ≤ \frac{dim(V) · dim(W)}{d_{λ}(V) · d_{λ^{−1}}(W)} ≤ \frac{2(q^{k} − ϵ)(q^{2k−2} − 1)(q^{k−2} + ϵ)}{(q − 1)(q^{2} − 1)}

if n = 2k, and

q^{4k−6}(q − 1)^{2} ≤ \frac{dim(V) · dim(W)}{d_{λ}(V) · d_{λ^{−1}}(W)} ≤ \frac{2(q^{2k} − 1)(q^{2k−2} − 1)}{(q − 1)(q^{2} − 1)}

if n = 2k + 1. Both of these inequalities cannot hold since q ≥ 5 and n ≥ 7. ⊓ ⊔

8. Exceptional Groups of Lie Type

In this section we consider exceptional groups G of Lie type, i.e. of type ^{2}B_{2}, ^{3}D_{4}, G_{2}, ^{2}G_{2}, F_{4}, ^{2}F_{4}, E_{6}, ^{2}E_{6}, E_{7}, or E_{8}, defined over a finite field 𝔽_{q}. Let ℓ(G) denote the smallest integer t > 1 such that G has a projective irreducible representation of degree t over a field of characteristic not dividing q. Lower bounds for ℓ(G) were found by Landázuri and Seitz in [LS] and improved by Seitz and Zalesskii in [SZ]. Recently, the Landázuri-Seitz-Zalesskii bounds have been improved in a number of cases, cf. [GT], [Hof1], [Hof2]. Here we would like to exhibit the exact value of ℓ(G) in the case G = G_{2}(q) and q is divisible by 3.

Proposition 8.1. Suppose that G = G_{2}(q), where 3|q. Then ℓ(G) = q^{2}(q^{2} + 1) if q > 3, and 14 if q = 3.

Proof. The statement is well-known if q = 3, cf. [Atlas] and [JLPW]. Suppose q = 3^{f} > 3 and r is any prime other than 2 and 3. The Brauer trees for cyclic rblocks of G are described in [Sh]. All the non-cyclic r-blocks have maximal defect, and the decompostion numbers for these blocks are computed in [H]. These two results readily imply that φ(1) ≥ q^{4} +q^{2} +1 for all nontrivial φ ∈ IBr_{r}(G). Finally, the 2-modular characters of G are determined in [HS], from which it follows that the smallest degree of a nontrivial 2-modular character is q^{2}(q^{2} + 1). ⊓ ⊔

Observe that ℓ(G_{2}(q)) = q(q^{2} − 1) if 3 |̸q (and q ≠ 2, 4).

Remark 8.2. After the paper had been submitted, we learned from Prof. G. Hiss that Proposition 8.1 was already proved in his Habilitationsschrift. We are thankful to Prof. Hiss for this remark.

Proposition 8.3. Let G be the universal cover of a finite simple exceptional group of Lie type, defined over a field 𝔽_{q}. Then G has property P(r) for any r |̸q, except possibly for the following cases:

(i) q ≤ 3;

(ii) G is of type F_{4} or ^{2}F_{4}, and 2|q.

Proof. 1) First we consider the case where G is of type G_{2} (and 3|q), ^{2}G_{2}, or 2B_{2}. The subcase of ^{2}B_{2}(8) can be checked by hand using [Atlas] and [JLPW]. In all remaining subcases G is simple, and ℓ(G) = q^{2}(q^{2} + 1) (by Proposition 8.1), q(q−1), and (q−1)\sqrt{q/2}, respectively. On the other hand, the largest degree of any irreducible complex character (and therefore of any irreducible Brauer character in cross characteristic) of G is (q + 1)(q^{2} + q + 1)(q^{3} + 1), (q^{2} − 1)(q + \sqrt{3q} + 1), and (q − 1)(q + \sqrt{2q} + 1), respectively. Since this degree is clearly smaller than ℓ(G)^{2}, G has property P(r).

2) Now we suppose that G is the universal cover of an exceptional group of Lie type E_{8}(q), E_{7}(q), E_{6}(q), ^{2}E_{6}(q), F_{4}(q) (and q is odd), ^{3}D_{4}(q), or G_{2}(q) (and 3 |̸q), respectively. Then G has a subgroup Q of special type, of order q^{1+56}, q^{1+32}, q^{1+20}, q^{1+20}, q^{1+14}, q^{1+8}, and q^{1+4}, respectively (cf. for instance Lemma 4.3 of [LS]). In all cases Z(Q) is an elementary abelian group of order q. If P = N_{G}(Q) and P′ = C_{P} (Z(Q)), then P/Q contains an index gcd(2, q − 1) subgroup (P′/Q) ∗ A, where A = ℤ_{q−1}. Correspondingly, P′/Q is the (simply connected) group of type E_{7}(q), Spin_{12}^{+}(q), SL_{6}(q), SU_{6}(q), Sp_{6}(q), SL_{2}(q^{3}), and SL_{2}(q). Also, P acts transitively on Z(Q) \ {1} and on IBr_{r}(Z(Q)) \ {1} (to see this observe that every long root group is contained in an SL_{3}(q)-subgroup). Let V and W be any two 𝔽G-modules of dimension > 1 such that V ⊗ W is irreducible. Then the same arguments as the ones we used for SL_{n}(q) in the proof of Proposition 3.2 show that

dim(V) · dim(W) ≤ (G : P) · d_{λ}(V) · d_{λ^{−1}}(W) · gcd(2, q − 1).

But dim(V)/d_{λ}(V) and dim(W)/d_{λ^{−1}}(W) are at least (q − 1)\sqrt{|Q|/q}. Hence we get the following inequality

(q − 1)^{2} · |Q| ≤ q · (G : P) · gcd(2, q − 1),

which cannot happen in any of the above cases when q ≥ 4. ⊓ ⊔

Proposition 8.4. Let q = 2^{f} > 2 and G = F_{4}(q) or ^{2}F_{4}(q). Then G has property P(0).

Proof. Suppose α, β ∈ Irr(G) are of degree > 1 and such that γ := α ⊗ β is irreducible. Let ϕ_{m} be the mth cyclotomic polynomial in q.

1) Suppose that G is of type F_{4}. By [S2, (2.1)], γ(1) ≤ (q^{12} − 1)(q^{8} − 1)(q^{6} − 1)(q^{2} − 1)/(q − 1)^{4}. On the other hand, α(1) ≥ ℓ(G) ≥ q^{7}(q^{3} − 1)(q − 1)/2. Hence

β(1) ≤ \frac{2(q^{12} − 1)(q^{8} − 1)(q^{6} − 1)(q^{2} − 1)}{q^{7}(q^{3} − 1)(q − 1)^{5}} < ϕ_{1}ϕ_{3}ϕ_{4}ϕ_{6}ϕ_{8}ϕ_{12},

since q ≥ 4. The five smallest nontrivial degrees d_{i}, 1 ≤ i ≤ 5, of complex irreducible characters of G are computed in [Lu], and the fifth degree d_{5} is exactly ϕ_{1}ϕ_{3}ϕ_{4}ϕ_{6}ϕ_{8}ϕ_{12}. Hence β(1) is one of the four smallest degrees. The same thing is true for α(1).

Suppose that α(1) = d_{4} = q^{2}ϕ_{3}^{2}ϕ_{6}^{2}ϕ_{12}. Since |G| is not divisible by ϕ_{12}^{2} and d_{2} is divisible by ϕ_{12}, β(1) is neither d_{2} nor d_{4}. Since |G| is not divisible by ϕ_{6}^{3} and d_{3} is divisible by ϕ_{6}, β(1) cannot be d_{3}. Finally, since |G| is not divisible by ϕ_{3}^{3} and d_{1} is divisible by ϕ_{3}, β(1) cannot be d_{1}, a contradiction.

Thus α(1) is one of the three smallest degrees. In particular, α(1) is divisible by ϕ_{8}. By symmetry, the same thing is true for β(1). Hence γ(1) and |G| are divisible by ϕ_{8}^{2}, again a contradiction.

2) Next suppose that G = ^{2}F_{4}(q). The three smallest nontrivial degrees d_{i}, 1 ≤ i ≤ 3, of complex irreducible characters of G are computed in [Lu], where d_{3} = ϕ_{1}ϕ_{4}^{2}ϕ_{12} if q ≥ 32, and d_{3} = 13778800 if q = 8. By [S2, (2.1)], γ(1) ≤ (q^{6} + 1)(q^{4} − 1)(q^{3} + 1)(q − 1)/(q − \sqrt{2q} + 1)^{2}. On the other hand, α(1) ≥ ℓ(G) ≥ q^{4}(q − 1)\sqrt{q/2}. Hence

β(1) ≤ \frac{(q^{6} + 1)(q^{4} − 1)(q^{3} + 1)(q − 1)}{(q − \sqrt{2q} + 1)^{2}q^{4}(q − 1)\sqrt{q/2}} < d_{3},

since q ≥ 8. This implies that β(1) is d_{1} or d_{2}. The same thing is true for α(1). But d_{1} and d_{2} are divisible by ϕ_{6}, hence γ(1) and |G| are divisible by ϕ_{6}^{2}, a contradiction. ⊓ ⊔

One can show that if G is as in Proposition 8.4 and α, β ∈ IBr_{r}(G) are nontrivial characters such that α ⊗ β is irreducible, then both α and β belong to unipotent blocks.

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