Simple biset functors and double Burnside ring

Serge Bouc, Radu Stancu, and Jacques Thévenaz

Abstract : Let G be a finite group and let k be a field. Our purpose is to investigate the simple modules for the double Burnside ring kB(G, G). It turns out that they are evaluations at G of simple biset functors. For a fixed finite group H, we introduce a suitable bilinear form on kB(G, H) and we prove that the quotient of kB(−, H) by the radical of the bilinear form is a semi-simple functor. This allows for a description of the evaluation of simple functors, hence of simple modules for the double Burnside ring.

AMS Subject Classification : 19A22, 20C20. Key words : biset, Burnside ring, simple module.

1. Introduction

Let G be a finite group. The double Burnside ring B(G, G) of all (G, G)-bisets plays a crucial role in many recent developments of representation theory and homotopy theory. It appears in the theory of biset functors [Bo3], which was successfully used for the solution of important problems in representation theory (e.g. the classification of endo-permutation modules for a p-group [Bo2]). In homotopy theory, a subring of B(G, G) appears in the theory of p-completed classifying spaces [MP, BF], fusion systems and p-local groups [BLO], idempotents associated to fusion systems [Ra, RS].

One of the main issues concerning the ring structure is to understand the simple B(G, G)-modules, since they appear in the semi-simple quotient of B(G, G) by its Jacobson radical. For this, it suffices to work over a field k, and we consider the finite-dimensional k-algebra kB(G, G) = k ⊗_{ℤ} B(G, G). Very little is known in general about the ring structure of kB(G, G). It is known that kB(G, G) is semi-simple only for cyclic groups in suitable characteristic, e.g. in characteristic zero (see Section 6.1 in [Bo3]). For a p-group P and for the subring of kB(P, P) spanned by bisets which are free on one side, Benson and Feshbach [BF] have a description of all simple modules, while Henn and Priddy give a sufficient condition for this subring to be a local ring (and they prove that this condition occurs quite frequently). More recently, for any finite group G, Boltje and Danz [BD] define a ghost ring for the subring of kB(G, G) spanned by bisets which are free on one side (respectively on both sides) and this sheds some new light about the ring structure, especially over a field k of characteristic zero.

The main purpose of this paper is to study the full double Burnside ring kB(G, G) and analyze the simple kB(G, G)-modules, using their connection with simple biset functors. This connection is quite deep since any simple kB(G, G)-module determines uniquely a simple biset functor, and conversely any (k-linear) simple biset functor has an evaluation at G which is a simple kB(G, G)-module (provided it is non-zero). The subring of kB(G, G) generated by the bisets which are free on both sides is also considered briefly.

We work mainly over an algebraically closed field k and consider the finitedimensional k-vector space kB(G, H) = k⊗_{ℤ} B(G, H), where G and H are finite groups and B(G, H) is the Grothendieck group of (G, H)-bisets. Let

k\overline{B}(G, H) = kB(G, H)/kI(G, H) ,

where kI(G, H) is the k-subspace generated by all bisets which factor through a proper subquotient of H. We define a canonical bilinear form on k\overline{B}(G, H) and pass to the quotient by the radical R(G, H) of this form. Allowing G to vary, we obtain a biset functor k\overline{B}(−, H) and a subfunctor R(−, H). We prove that k\overline{B}(−, H)/R(−, H) is semi-simple, more precisely the largest semi-simple quotient of the biset functor k\overline{B}(−, H).

Evaluation at G gives rise to a semi-simple kB(G, G)-module

k\overline{B}(G, H)/R(G, H) ,

with an explicit decomposition into simple summands. This provides the main tool for analyzing simple kB(G, G)-modules, or equivalently, evaluation of simple biset functors. In particular, we obtain a formula for the dimension of the evaluation S_{H,V} (G) of a simple biset functor S_{H,V}, generalizing the formula obtained in [Bo1] for the case where V the trivial module.

After an introductory Section 2, we review in Section 3 the connections between simple modules for kB(G, G) and simple functors. In Section 4, we define the standard quotient k\overline{B}(G, H) of kB(G, H) and prove some of its properties. We prove in Section 5 that any simple kB(G, G)-module has a minimal group attached to it. The main construction of two possible bilinear forms on k\overline{B}(G, H) is performed in Section 6. One form is attached to a given simple kOut(H)-module V, the other corresponds to the largest semi-simple quotient kOut(H)/J(kOut(H)). We describe the quotient k\overline{B}(−, H)/R(−, H), where R(G, H) is the (right) kernel of the bilinear form. By using the bilinear form corresponding to a given simple kOut(H)-module V, we prove in Section 7 a formula for the dimension of the evaluation at G of the simple functor S_{H,V}. In Section 8, by using the other bilinear form, we prove that k\overline{B}(−, H)/R(−, H) is the largest semi-simple quotient of the biset functor k\overline{B}(−, H). The same result may not hold after evaluation at some finite group G and this question is studied in Section 9, where some sufficient conditions are given. In Section 10, we analyze the case when the group H is the trivial group, where more information can be obtained. In particular, a rather large subspace of the Jacobson radical J(k\overline{B}(G, G)) is described. In Section 11, we replace the group H by a fixed subquotient P/Q of the group G and describe some natural ideals of kB(G, G) corresponding to this subquotient. The case of the subring of kB(G, G) generated by the bisets which are free on both sides is treated in Section 12. Finally, several examples are presented in Section 13.

2. Biset functors

We first review some know facts about biset functors and refer to [Bo1] and [Bo3] for more details. Given two finite groups G and H, the Burnside group B(G, H) is the Grothendieck group of the category of finite (G, H)-bisets. Since we are interested in simple modules, it is no loss to work over a field k and we define kB(G, H) = k ⊗_{ℤ} B(G, H). In particular, kB(G, G) is a finite dimensional kalgebra, called the double Burnside ring of G. We do not need to consider the usual double Burnside ring B(G, G) (defined over ℤ) and we work only over the field k. In Section 12, we will consider another version of the double Burnside ring, namely the subring obtained by requiring that the bisets are free on both sides, but for the rest of this paper, we always use the full ring kB(G, G).

A section of a finite group G is a pair (S, T) of subgroups of G such that T is a normal subgroup of S. In that case, the group S/T is called a subquotient of G. We write H ⊑ G when the group H is isomorphic to a subquotient of G and we write H ⊏ G if H ⊑ G and H ≅ G̸ (hence |H| < |G|). We also write N_{G}(S, T) for the normalizer of the section, that is, the set of all g ∈ G such that gSg^{−1} = S and gT g^{−1} = T.

If (S, T) is a section of G, then there are elementary bisets Res_{S}^{G}, Def_{S/T}^{S}, Ind_{S}^{G}, Inf_{S/T}^{S}, and their composites Defres_{S/T}^{G} and Indinf_{S/T}^{G} (see Section 2.3 in [Bo3]). Also any group isomorphism σ : A → B defines a (B, A)-biset Iso_{σ}. Recall the following basic result (see Lemma 3 in [Bo1] or Lemma 2.3.26 in [Bo3]).

2.1. Lemma. Let X and Y be finite groups.

1. Any transitive (X, Y)-biset has the form Indinf_{J/K}^{X} Iso_{σ} Defres_{S/T}^{Y}, where (J, K) is a section of X, (S, T) is a section of Y, and σ : S/T → J/K is a group isomorphism.

2. Let E be the set of triples ((J, K), σ, (S, T)) where (J, K) is a section of X, (S, T) is a section of Y, and σ : S/T → J/K is a group isomorphism. The group X × Y acts by conjugation on E. Then the set of all elements Indinf_{J/K}^{X} Iso_{σ} Defres_{S/T}^{Y} is a k-basis of kB(X, Y), where the triple ((J, K), σ, (S, T)) runs over representatives of (X × Y)-orbits in E.

The biset category kC is the k-linear category whose objects are finite groups, with morphisms Hom_{kC}(H, G) = kB(G, H) (note that a (G, H)-biset is a morphism from H to G). The composition of morphisms is the k-linear extension of the usual products of bisets U ×_{H} V. Recall that, if U is a (G, H)-biset and V is an (H, L)-biset, then U ×_{H} V is a (G, L)-biset in the obvious way.

A biset functor is a k-linear functor from kC to the category k−Mod of kvector spaces and we let F be the category of all such biset functors (an abelian category). We often use a dot for the action of bisets on evaluation of functors, that is, α · x ∈ F (G) whenever F ∈ F, x ∈ F (H), and α ∈ kB(G, H). A subquotient of a functor is a quotient of a subfunctor. Moreover, a sequence of functors

0 ⟶ F_{1} ⟶ F_{2} ⟶ F_{3} ⟶ 0

is exact if and only if, for every finite group G, the evaluation sequence

0 ⟶ F_{1}(G) ⟶ F_{2}(G) ⟶ F_{3}(G) ⟶ 0

is exact.

A biset functor is called simple if it is non-zero and has no proper non-zero subfunctor. The evaluation at a finite group G of a simple functor (and also of a representable functor) is always a finite-dimensional k-vector space, so we shall in fact only deal with functors having this additional property.

For any fixed finite group G, consider the representable functor kB(−, G) (which is a projective functor). Its evaluation at a group X has a natural structure of (kB(X, X), kB(G, G))-bimodule. For any kB(G, G)-module W, we define, following [Bo1], the functor

L_{G,W} = kB(−, G) ⊗_{kB(G,G)} W ,

which satisfies the following adjunction property (see Section 2 in [Bo1]).

2.2. Lemma. Let G be a finite group. The functor

kB(G, G)−Mod ⟶ F , W ↦ L_{G,W}

is left adjoint of the evaluation functor

F ⟶ kB(G, G)−Mod , F ↦ F (G) .

Our next result is a slight extension of the first lemma of [Bo1].

2.3. Lemma. Let G be a finite group, let W be a kB(G, G)-module, and let

J_{G,W} (X) = { ∑_{i} ϕ_{i} ⊗ w_{i} ∈ L_{G,W} (X) | ∀ψ ∈ kB(G, X), ∑_{i}(ψϕ_{i}) · w_{i} = 0} .

1. J_{G,W} is the unique subfunctor of L_{G,W} which is maximal with respect to the condition that it vanishes at G.

2. If W is a simple module, then J_{G,W} is the unique maximal subfunctor of L_{G,W} and L_{G,W} /J_{G,W} is a simple functor.

Proof : Note first that the condition ∑_{i}(ψϕ_{i})·w_{i} = 0 in W is equivalent to the condition that ∑_{i} ϕ_{i} ⊗w_{i} lies in the kernel of L_{G,W} (ψ) : L_{G,W} (X) → L_{G,W} (G), in view of the isomorphism L_{G,W} (G) ≅ W induced by ϕ ⊗ w ↦ ϕw. Therefore

J_{G,W} (X) = ⋂_{ψ∈kB(G,X)} Ker(L_{G,W} (ψ) : L_{G,W} (X) → L_{G,W} (G)) .

It is then easy to check that J_{G,W} is a subfunctor of L_{G,W}. It is also clear that it vanishes at G since J_{G,W} (G) lies in the kernel of id. Finally, if F is a subfunctor of L_{G,W} vanishing at G, then F (X) must lie in the kernel of all maps L_{G,W} (ψ) for ψ ∈ kB(G, X), so that F ⊆ J_{G,W}.

The case where W is simple is explicit in [Bo1] and we just recall the main line of the argument. Any subfunctor F of L_{G,W} either vanishes at G, hence F ⊆ J_{G,W}, or satisfies F (G) = W, hence F = L_{G,W}. Therefore J_{G,W} is the unique maximal subfunctor of L_{G,W} and L_{G,W} /J_{G,W} is a simple functor.

We define S_{G,W} = L_{G,W} /J_{G,W} and we emphasize that this is a simple functor provided W is a simple kB(G, G)-module. This provides the first link between simple kB(G, G)-modules and simple functors. Our next section shows that the connection is much stronger.

3. Simple modules and simple functors

We prove in this section that the link between simple kB(G, G)-modules and simple biset functors is deep. Note that the results of this section work in the more general context presented in Section 3.2 of [Bo3], in particular for inflation functors (corresponding to bisets which are free on one side), for global Mackey functors (corresponding to bisets which are free on both sides), and also for functors defined only on a given class of finite groups. For simplicity, we consider all groups and all bisets. An overview of the case of global Mackey functors, using bisets which are free on both sides, is given in Section 12.

Our first proposition is a special case of the results of Section 4.2 in [Bo3]. We repeat the arguments for convenience.

3.1. Proposition.

1. If G is a finite group and W is a simple kB(G, G)-module, then W is the evaluation at G of a simple biset functor, namely S_{G,W}. Moreover S_{G,W} is the unique simple functor, up to isomorphism, such that its evaluation at G is isomorphic to W as a kB(G, G)-module.

2. If S is a simple biset functor and G is a finite group, then either S(G) = 0 or S(G) is a simple kB(G, G)-module. In the latter case, S ≅ S_{G,W}, where W = S(G).

Proof : The first claim of (1) is an immediate consequence of Lemma 2.3 above, because S_{G,W} (G) = L_{G,W} (G) ≅ W. The second claim of (1) is a consequence of (2).

For the proof of (2), suppose that S(G) ≠ 0. Let M be a non-zero kB(G, G)- submodule of S(G) and let i : M → S(G) be the inclusion. By the adjunction of Lemma 2.2, this corresponds to a non-zero morphism θ : L_{G,M} → S. Since S is simple, θ must be surjective, that is, surjective on every evaluation. Thus θ(G) : L_{G,M}(G) → S(G) is surjective, but this is just θ(G) = i : M → S(G). This proves that M = S(G). Consequently S(G) is simple. Letting now W = S(G), we obtain a surjective morphism θ : L_{G,W} → S. But S_{G,W} is the unique simple quotient of L_{G,W}, by Lemma 2.3 above, so S ≅ S_{G,W}.

The fact that a simple kB(G, G)-module determines uniquely a simple biset functor shows that the study of simple kB(G, G)-modules is, in some sense, equivalent to the study of simple biset functors. So we have, in some sense, enriched the structure of objects we are working with. On the other hand, the evaluations of simple functors are unfortunately not easy to determine. One of our main results in Section 7 will actually give some answer to this question.

A simple functor S has many realizations S ≅ S_{G,W}, one for each G such that S(G) ≠ 0. But recall that, in order to obtain a parametrization, one can do better, as follows (see Section 4 in [Bo1] or Section 4.3 in [Bo3]).

3.2. Proposition. Let S be a simple biset functor, let H be a group of minimal order such that S(H) ≠ 0, and let V = S(H).

1. H is unique up to isomorphism and S ≅ S_{H,V}.

2. Let kI(H, H) be the ideal of kB(H, H) generated by all bisets which factor through a proper subquotient of H, so that kB(H, H)/kI(H, H) ≅ kOut(H) (where Out(H) denotes the group of outer automorphisms of H). Then kI(H, H) acts by zero on V and V is a kOut(H)-module.

3. If S(G) ≠ 0 for some finite group G, then H ⊑ G.

This provides a parametrization of simple functors by pairs (H, V) where H is a finite group and V is a simple kOut(H)-module. In the next sections of this paper, we shall often fix a finite group H and a simple kOut(H)-module V. Then we shall work with the simple functor S_{H,V} and consider finite groups G such that W = S_{H,V} (G) ≠ 0 (so that in fact S_{H,V} ≅ S_{G,W}).

3.3. Corollary. If G is a finite group, the number of isomorphism classes of simple kB(G, G)-modules is equal to the number of isomorphism classes of pairs (H, V), where H is finite group such that H ⊑ G and V is a simple kOut(H)-module, subject to the condition that S_{H,V} (G) ≠ 0.

Proof : This follows immediately from Propositions 3.1 and 3.2.

It follows from this proposition that the question of the vanishing of evaluation of simple functors is a crucial issue for the description of simple kB(G, G)- modules. Whenever S_{H,V} (G) = 0, the pair (H, V) must not be counted. This question of the vanishing of evaluation of simple functors will be considered in another paper [BST].

Another useful fact is the following :

3.4. Proposition. Let S be a simple biset functor and let G be a group such that S(G) ≠ 0. Then S is generated by S(G), that is, S(X) = kB(X, G)S(G) for all finite groups X. More precisely, if 0 ≠ u ∈ S(G), then S(X) = kB(X, G) · u.

Proof : Given 0 ≠ u ∈ S(G), let S′(X) = kB(X, G) · u for all finite groups X. This clearly defines a non-zero subfunctor S′ of S, so S′ = S by simplicity of S.

The connection between simple functors and simple evaluations extends further, as follows.

3.5. Proposition. Let S be a simple biset functor and let G be a finite group such that S(G) ≠ 0. Let F be any biset functor. Then the following are equivalent:

1. S is isomorphic to a subquotient of F.

2. The simple kB(G, G)-module S(G) is isomorphic to a subquotient of the kB(G, G)-module F (G).

Proof : It is clear that (1) implies (2). Suppose that (2) holds and let W_{1}, W_{2} be submodules of F (G) such that W_{2} ⊂ W_{1} and W_{1}/W_{2} ≅ S(G). For i ∈ {1, 2}, let F_{i} be the subfunctor of F generated by W_{i}. Explicitly, for any finite group X, F_{i}(X) = kB(X, G) · W_{i} ⊆ F (X). Then F_{i}(G) = W_{i} and (F_{1}/F_{2})(G) = W_{1}/W_{2} ≅ S(G). The isomorphism S(G) → (F_{1}/F_{2})(G) induces, by the adjunction of Lemma 2.2, a non-zero morphism θ : L_{G,S(G)} → F_{1}/F_{2}.

Since S(G) is simple, L_{G,S(G)} has a unique maximal subfunctor J_{G,S(G)}, by Lemma 2.3, and L_{G,S(G)}/J_{G,S(G)} ≅ S_{G,S(G)} = S, by Proposition 3.1. Let F′_{1} = θ(L_{G,S(G)}) and F′_{2} = θ(J_{G,S(G)}). Since θ ≠ 0, we obtain

F′_{1}/F′_{2} ≅ L_{G,S(G)}/J_{G,S(G)} ≅ S_{G,S(G)} = S ,

showing that S is isomorphic to a subquotient of F.

Note that, in the proof above, we have F_{2} ⊆ F′_{2} ⊆ F′_{1} ⊆ F_{1} ⊆ F. Moreover (F′_{1}/F′_{2})(G) ≅ S(G) ≅ (F_{1}/F_{2})(G), so that we have (F_{1}/F′_{1})(G) = 0 and (F′_{2}/F_{2})(G) = 0. But in general we don’t have F′_{i} = F_{i}.

4. The standard quotient of a representable functor

In this section, we construct the biset functor k\overline{B}(−, H), which plays a central role in the rest of the paper. Let us fix a finite group H and consider the representable functor kB(−, H). For every finite group X, define

I(X, H) := ∑_{K⊏H} B(X, K)B(K, H) .

This is a subgroup of the abelian group B(X, H). Extending scalars to k, we obtain the k-subspace

kI(X, H) := ∑_{K⊏H} kB(X, K)B(K, H) .

4.1. Lemma. kI(−, H) is a subfunctor of kB(−, H).

Proof : We have

B(Y, X)I(X, H) = B(Y, X) ∑_{K⊏H} B(X, K)B(K, H)
= ∑_{K⊏H} B(Y, X)B(X, K)B(K, H)
⊆ ∑_{K⊏H} B(Y, K)B(K, H) = I(Y, H) .

The idea of passing to the quotient by all morphisms factorizing below H has been widely used (e.g. in Section 4 of [We2]). We define the standard quotient of kB(−, H) to be

k\overline{B}(−, H) = kB(−, H)/kI(−, H) .

This terminology will be motivated below. In particular kI(H, H) is an ideal of the double Burnside ring kB(H, H) and k\overline{B}(H, H) ≅ kOut(H), as already noticed in Proposition 3.2.

We first note the following elementary result.

4.2. Lemma. Let H and X be finite groups. Then \overline{B}(X, H) ≠ 0 if and only if H ⊑ X.

Proof : If \overline{B}(X, H) ≠ 0, there exists a transitive (X, H)-biset U such that its class in \overline{B}(X, H) is non-zero. But by Lemma 2.1, we can write

U = Indinf_{J/K}^{X}Iso_{σ}Defres_{S/T}^{H} ,

where (S, T) is a section of H, (J, K) is a section of X, and σ : S/T → J/K is an isomorphism. If S/T was a proper subquotient of H, then we would have Defres_{S/T}^{H} ∈ I(S/T, H), hence U ∈ I(X, H), and the class of U would be zero in \overline{B}(X, H). Therefore S = H and T = 1, so that U = Indinf_{J/K}^{X}Iso_{σ} where σ : H → J/K is an isomorphism. This proves that H ⊑ X.

Conversely, if H ⊑ X, then there is an isomorphism σ : H → J/K where (J, K) is a section of X, and U = Indinf_{J/K}^{X}Iso_{σ} defines a non-zero element in \overline{B}(X, H), by Lemma 2.1.

For any finite group X, the evaluation k\overline{B}(X, H) has a natural structure of right kOut(H)-module, because the right action of kI(H, H) is zero. For any left kOut(H)-module V, we define the functor

\overline{L}_{H,V} = k\overline{B}(−, H) ⊗_{kOut(H)} V ,

or also \overline{L}_{H,V} ≅ k\overline{B}(−, H) ⊗_{kB(H,H)} V if V is viewed as a kB(H, H)-module. Applying the functor − ⊗_{kB(H,H)} V to the exact sequence

0 → kI(X, H) → kB(X, H) → k\overline{B}(X, H) → 0 ,

we see that \overline{L}_{H,V} (X) is a quotient of L_{H,V} (X) (quotient by the image of kI(X, H) ⊗_{kB(H,H)} V). We call \overline{L}_{H,V} the standard quotient of L_{H,V}. Note that if V = kOut(H), then \overline{L}_{H,kOut(H)} = k\overline{B}(−, H).

Following Section 4 of [We2], we define, for any biset functor F and any finite group H, the restriction kernel

\underline{F} (H) = ⋂_{K⊏H, ϕ∈B(K,H)} Ker(F (ϕ)) .

Clearly kI(H, H) acts by zero on \underline{F} (H), so that \underline{F} (H) is a kOut(H)-module. We also refer to [Ya] for a use of restriction kernels. Similarly to the adjunction of Lemma 2.2, we have the following analogous property for the standard quotient.

4.3. Lemma. Let H be a finite group. The functor

kOut(H)−Mod ⟶ F , V ↦ \overline{L}_{H,V}

is left adjoint of the functor

F ⟶ kOut(H)−Mod , F ↦ \underline{F} (H) .

Proof : Let g : V → \underline{F} (H) be a kOut(H)-linear map. Composing with the inclusion i : \underline{F} (H) → F (H) and applying the adjunction of Lemma 2.2, we obtain a morphism θ : L_{H,V} → F such that θ(H) = i ∘ g. Now fix K ⊏ H and let α ∈ B(X, K) and β ∈ B(K, H), so that αβ is among the generators of kI(X, H). Then we obtain

θ(X)L_{H,V} (αβ) = F (αβ)θ(H) = F (α)F (β)ig = 0 ,

because F (β)i = 0 by definition of \underline{F} (H). It follows that θ(X) is zero on the image of kI(X, H) ⊗_{kB(H,H)} V in L_{H,V} (X) = kB(X, H) ⊗_{kB(H,H)} V. Therefore θ(X) induces a morphism \overline{θ}(X) : \overline{L}_{H,V} (X) → F (X). Then \overline{θ} : \overline{L}_{H,V} → F is the morphism corresponding to g under the required adjunction.

In Section 5 of [We2], Webb defines a biset functor Δ_{H,V} and proves that V ↦ Δ_{H,V} is left adjoint of the functor F ↦ \underline{F} (H). Therefore we have in fact

Δ_{H,V} ≅ \overline{L}_{H,V} .

One of the main results of Webb asserts that the functors Δ_{H,V}, where V is simple, are the standard objects in a highest weight category structure on F (over a field of characteristic zero). Thus we see that the functors \overline{L}_{H,V}, hence also the functors k\overline{B}(−, H), deserve to be called standard. This explains our terminology of standard quotient.

Webb also shows that, if V is a simple kOut(H)-module, then the simple functor S_{H,V} is a quotient of Δ_{H,V}. We now show that this works also for any kOut(H)-module V, using our definition of the corresponding functor S_{H,V}.

4.4. Proposition. Let H be a finite group and V a kOut(H)-module.

1. The image of kI(X, H) ⊗_{kB(H,H)} V in kB(X, H) ⊗_{kB(H,H)} V = L_{H,V} (X) is contained in J_{H,V} (X).

2. The quotient morphism ε : kB(−, H) → k\overline{B}(−, H) induces a surjective morphism ε : L_{H,V} → \overline{L}_{H,V} and an isomorphism

\overline{ε} : S_{H,V} = L_{H,V} /J_{H,V} ⟶^{∼} \overline{L}_{H,V} /\overline{J}_{H,V} ,

where \overline{J}_{H,V} = ε(J_{H,V}).

3. If V is a simple module, then \overline{J}_{H,V} is the unique maximal subfunctor of \overline{L}_{H,V}.

Proof : Let α ∈ kI(X, H). For any ψ ∈ kB(H, X), we have ψα ∈ kI(H, H), hence ψα · v = 0 for any v ∈ V. This shows that α ⊗ v ∈ J_{H,V} (X), proving the first part. The other parts follow immediately.

4.5. Remark. By copying the proof of Lemma 2.3, one easily obtains that

\overline{J}_{H,V} (X) = \{ ∑_{i} \overline{ϕ}_{i} ⊗ v_{i} ∈ \overline{L}_{G,W} (X) | ∀ψ ∈ kB(H, X), ∑_{i}(\overline{ψϕ}_{i})v_{i} = 0\} ,

and that \overline{J}_{H,V} is the unique subfunctor of \overline{L}_{H,V} which is maximal with respect to the condition that it vanishes at H.

5. The minimal group of a simple kB(G, G)-module

In this section, we fix a finite group G and we let W be a simple kB(G, G)- module. A minimal group for W is a group P of minimal order subject to the condition that B(G, P)B(P, G)W ≠ 0. Note that composition of bisets defines a homomorphism B(G, P)B(P, G) → B(G, G), so that B(G, P)B(P, G) acts on W. Note also that the image of this homomorphism is a two-sided ideal of B(G, G), generated by all the (G, G)-bisets which factor through P, so that B(G, P)B(P, G)W is a kB(G, G)-submodule of W (hence either 0 or W by simplicity of W). This notion of minimality only involves the double Burnside ring, but it has a very useful interpretation in terms of biset functors, as follows.

5.1. Proposition. Let G be a finite group, let W be a simple kB(G, G)- module, and let P be a minimal group for W.

1. P is unique up to isomorphism and P ⊑ G.

2. Let (H, V) be a pair which parametrizes the simple functor S_{G,W}, that is, S_{G,W} = S_{H,V}, where H is a group of minimal order such that S_{G,W} (H) ≠ 0 and V is a simple kOut(H)-module (see Proposition 3.2). Then H is a minimal group for W. Moreover V = S_{G,W} (H) and W = S_{G,W} (G).

Proof : We first give a proof using the simple functor S = S_{G,W} = S_{H,V}, where (H, V) is as in statement (2), so in particular H ⊑ G. We have W = S(G) and B(P, G)W = B(P, G)S(G) makes sense because bisets can be applied to this evaluation. If B(G, P)B(P, G)W ≠ 0, then B(P, G)S(G) ≠ 0, hence S(P) ≠ 0. This implies that H is isomorphic to a subquotient of P, by Proposition 3.2. On the other hand, by Proposition 3.4, we have

B(G, H)B(H, G)W = B(G, H)B(H, G)S(G) = B(G, H)S(H) = S(G) ≠ 0 .

By minimality of P, this implies that the subquotient H of P must be isomorphic to P. This proves both statements.

We now sketch a second proof of the first statement, using only bisets. Let α ∈ B(G, P) and β ∈ B(P, G) be transitive bisets such that αβW ≠ 0. By minimality of P, αβ cannot factor through a proper subquotient of P. Therefore, by Lemma 2.1, α = Indinf_{J/K}^{G}Iso_{σ} for some section (J, K) of G and β = Iso_{ρ}Defres_{S/T}^{G} for some section (S, T) of G. If now P′ is another minimal group for W, then

B(G, P′)B(P′, G)B(G, P)B(P, G)W = W

and we get transitive bisets α′ ∈ B(G, P′) and β′ ∈ B(P′, G) such that

α′β′αβW ≠ 0 .

Then we decompose the (P′, P)-biset β′α as a sum of transitive bisets and we obtain at least one transitive summand γ ∈ B(P′, P) such that α′γβW ≠ 0. But γ factorizes through subquotients of P and P′ (by Lemma 2.1), so by minimality of P and P′, the only possibility is that γ is an isomorphism. Therefore P ≅ P′, as required.

We now establish the link bewteen minimal groups and standard quotients.

5.2. Proposition. Let G be a finite group, let W be a simple kB(G, G)- module, and let P be a minimal group for W.

1. W is isomorphic to a quotient of the kB(G, G)-module k\overline{B}(G, P).

2. If W is isomorphic to a quotient of the kB(G, G)-module k\overline{B}(G, H) for some finite group H, then H is isomorphic to a subquotient of P.

Proof : By Proposition 5.1, P is a minimal group for S_{G,W} and we have S_{G,W} = S_{P,V}, where V is a simple kOut(P)-module. Moreover, there are surjective morphisms

k\overline{B}(−, P) ⟶ \overline{L}_{P,V} ⟶ S_{P,V} ,

where the first morphism maps ϕ ∈ k\overline{B}(X, P) to ϕ⊗v ∈ \overline{L}_{P,V} (X) for some fixed v ∈ V, and the second comes from Proposition 4.4. By evaluating at G, we see that W = S_{G,W} (G) = S_{P,V} (G) is isomorphic to a quotient of the kB(G, G)- module k\overline{B}(G, P), proving (1).

Assume now that W = S_{P,V} (G) is isomorphic to a quotient of the kB(G, G)- module k\overline{B}(G, H) for some H. By Proposition 3.5, S_{P,V} is isomorphic to a subquotient of k\overline{B}(−, H). By evaluating at P, we see that V = S_{P,V} (P) is isomorphic to a subquotient of the kB(G, G)-module k\overline{B}(P, H). Therefore k\overline{B}(P, H) ≠ 0, hence H ⊑ P by Proposition 4.2, proving (2).

The above results show that it is worth considering the simple quotients of k\overline{B}(G, H) in order to find the simple kB(G, G)-modules. Every such simple module must appear, possibly for several finite groups H. Our next task is to study further semi-simple quotients of k\overline{B}(G, H).

6. Bilinear forms on standard quotients

In this section, we introduce one of the main constructions of this paper. We fix a finite group H and consider a quotient algebra E of kOut(H) with corresponding k-algebra map \overline{π} : kOut(H) → E, which we compose with the quotient map kB(H, H) → kOut(H) to get

π : kB(H, H) ⟶ E .

We assume that E is a symmetric algebra with symmetrizing form τ : E → k. This means that τ is k-linear and induces a symmetric non-degenerate bilinear form on E given by

(a, b) ↦ τ(ab) .

We have in mind two cases which we shall consider later :

1. k is algebraically closed (or at least large enough for the group Out(H)), E = End_{k}(V) where V is a simple kOut(H)-module, \overline{π} : kOut(H) → End_{k}(V) is the natural surjection, and τ = τ_{V} is the trace map on End_{k}(V). Note that if V is self-dual, then τ_{V} satisfies the additional condition τ_{V} (\overline{π}(s^{−1})) = τ_{V} (\overline{π}(s)) for all s ∈ Out(H).

2. k is algebraically closed (or at least large enough for the group Out(H)), E = kOut(H)/J(kOut(H)) ≅ ∏_{i=1}^{r} End_{k}(V_{i}), where V_{1}, …, V_{r} are the simple kOut(H)-modules, \overline{π} : kOut(H) → kOut(H)/J(kOut(H)) is the quotient map, and τ = ∑_{i=1}^{r} τ_{V_{i}}, the sum of the trace maps on each End_{k}(V_{i}). Note that τ satisfies the additional condition τ(\overline{π}(s^{−1})) = τ(\overline{π}(s)) for all s ∈ Out(H). If the characteristic of k does not divide |Out(H)|, then E = kOut(H), \overline{π} = id, and we could also choose for τ the ordinary map for group algebras (coefficient of 1), because any symmetrizing form would do.

With such data, we can construct a bilinear form on k\overline{B}(X, H), for any finite group X. Recall that any (X, H)-biset U has an opposite U^{op}, which is an (H, X)-biset (see Section 2.3 in [Bo3]), and this extends to a k-linear map kB(X, H) → kB(H, X), α ↦ α^{op}, such that (αβ)^{op} = β^{op}α^{op}. Then we define

<−, −>_{X} : k\overline{B}(X, H) × k\overline{B}(X, H) ⟶ k , <\overline{ϕ}, \overline{ψ}>_{X} = τπ(ϕ^{op}ψ) ,

where ϕ, ψ ∈ kB(X, H) are representatives of \overline{ϕ} and \overline{ψ} respectively.

6.1. Lemma. Let H and X be finite groups and let π : kB(H, H) → E and τ : E → k be as above. Let R(X, H) be the right kernel of the form <−, −>_{X}, that is, the set of all elements \overline{ψ} ∈ k\overline{B}(X, H) such that <\overline{ϕ}, \overline{ψ}>_{X} = 0 for all \overline{ϕ} ∈ k\overline{B}(X, H),

1. The map <−, −>_{X} is well-defined and k-bilinear.

2. If τ satisfies the condition τ(\overline{π}(s^{−1})) = τ(\overline{π}(s)) for all s ∈ Out(H) (e.g. in case (2) above), then the bilinear form <−, −>_{X} is symmetric.

3. We have <\overline{α^{op}} \overline{β}, \overline{ψ}>_{X} = <\overline{β}, \overline{α} \overline{ψ}>_{Y} for all \overline{α} ∈ k\overline{B}(Y, X), \overline{β} ∈ k\overline{B}(Y, H), \overline{ψ} ∈ k\overline{B}(X, H).

4. R(−, H) is a subfunctor of k\overline{B}(−, H).

5. We have <\overline{ϕ}, \overline{ψ} \overline{γ}>_{X} = <\overline{ϕ} \overline{γ}^{op}, \overline{ψ}>_{X} for all \overline{ϕ}, \overline{ψ} ∈ k\overline{B}(X, H) and \overline{γ} ∈ k\overline{B}(H, H) = kOut(H).

6. R(X, H) is a right kOut(H)-submodule of k\overline{B}(X, H).

7. R(H, H) = Ker(\overline{π}).

Proof : (1) Replace ψ ∈ kB(X, H) by ψ′ = ψ + χ, where χ ∈ kI(X, H). Then, for all ϕ ∈ kB(X, H),

π(ϕ^{op}ψ′) = π(ϕ^{op}ψ) + π(ϕ^{op}χ) = π(ϕ^{op}ψ) ,

because ϕ^{op}χ ∈ kI(H, H), hence π(ϕ^{op}χ) = 0. A similar argument applies if we modify ϕ by an element of kI(X, H) and this shows that the form is well-defined. It is obvious that it is bilinear.

(2) Let δ = ϕ^{op}ψ ∈ kB(H, H), so that ψ^{op}ϕ = δ^{op}. The map π factors through kOut(H) and the passage to opposite bisets induces on kOut(H) the map s ↦ s^{−1} for each s ∈ Out(H). Since we have τ(\overline{π}(s^{−1})) = τ(\overline{π}(s)) by our assumption, we obtain τπ(δ) = τπ(δ^{op}), and the symmetry follows.

(3) This follows immediately from the equality (α^{op}β)^{op}ψ = β^{op}(αψ).

(4) Let \overline{ψ} ∈ R(X, H), so that <\overline{ϕ}, \overline{ψ}>_{X} = 0 for all \overline{ϕ} ∈ k\overline{B}(X, H). Apply the morphism α ∈ kB(Y, X). Then, for all \overline{β} ∈ k\overline{B}(Y, H), we have <\overline{β}, \overline{α} \overline{ψ}>_{Y} = <\overline{α^{op}} \overline{β}, \overline{ψ}>_{X} = 0. Therefore α · \overline{ψ} ∈ R(Y, H). This shows that R(−, H) is a subfunctor of k\overline{B}(−, H).

(5) Using the symmetry of the form τ, we obtain

<\overline{ϕ}, \overline{ψ} \overline{γ}>_{X} = τπ(ϕ^{op}ψγ) = τπ(γϕ^{op}ψ) = τπ((ϕγ^{op})^{op}ψ) = <\overline{ϕ} \overline{γ}^{op}, \overline{ψ}>_{X} .

(6) Let \overline{ψ} ∈ R(X, H), and let \overline{γ} ∈ k\overline{B}(H, H) = kOut(H). Then, for all \overline{ϕ} ∈ k\overline{B}(X, H),

<\overline{ϕ}, \overline{ψ} \overline{γ}>_{X} = <\overline{ϕ} \overline{γ}^{op}, \overline{ψ}>_{X} = 0

This proves that \overline{ψ} \overline{γ} ∈ R(X, H).

(7) Let \overline{γ} ∈ k\overline{B}(H, H) = kOut(H). Then \overline{γ} ∈ R(H, H) if and only if τπ(ϕ^{op}γ) = 0 for all ϕ ∈ kB(H, H). But this gives τ(π(ϕ^{op})π(γ)) = 0 for all ϕ, hence π(γ) = 0 by non-degeneracy of the bilinear form on E induced by τ. Therefore \overline{γ} ∈ R(H, H) if and only if \overline{γ} ∈ Ker(\overline{π}).

If necessary, it could be possible to define the bilinear form directly on kB(X, H) instead of its quotient k\overline{B}(X, H). Then kI(X, H) would be in the kernel of the form and the form would induce the one defined above.

We are interested in the quotient k\overline{B}(−, H)/R(−, H) and we first determine its structure.

6.2. Theorem. Let H be a finite group, let π : kB(H, H) → E and τ : E → k be as above. Let R(−, H) be the subfunctor of k\overline{B}(−, H) defined in Lemma 6.1. Then the quotient functor k\overline{B}(−, H)/R(−, H) is isomorphic to S_{H,E}.

Proof : Consider the exact sequence 0 → Ker(\overline{π}) → kOut(H) → E → 0. Tensor this sequence with k\overline{B}(X, H), where X is a finite group. Since tensoring is right exact and since k\overline{B}(X, H) ⊗_{kOut(H)} kOut(H) ≅ k\overline{B}(X, H), we obtain

0 ⟶ k\overline{B}(X, H) · Ker(\overline{π}) ⟶ k\overline{B}(X, H) ⟶ k\overline{B}(X, H) ⊗_{kOut(H)} E ⟶ 0 ,

where the first map is just the inclusion map. Note that k\overline{B}(X, H)⊗_{kOut(H)}E = \overline{L}_{H,E}(X). Now it is obvious that k\overline{B}(X, H) · Ker(\overline{π}) is contained in the kernel R(X, H) of the bilinear form. Therefore we obtain the exact sequence

0 ⟶ k\overline{B}(X, H) · Ker(\overline{π}) ⟶ R(X, H) ⟶ R(X, H) ⊗_{kOut(H)} E ⟶ 0 .

We claim that R(X, H) ⊗_{kOut(H)} E is equal to \overline{J}_{H,E}(X), as defined in Proposition 4.4. It then follows that

k\overline{B}(X, H)/R(X, H) ≅ (k\overline{B}(X, H) ⊗_{kOut(H)} E)/(R(X, H) ⊗_{kOut(H)} E)
= \overline{L}_{H,E}(X)/\overline{J}_{H,E}(X)
≅ S_{H,E}(X) ,

by Proposition 4.4, proving the first statement.

Now we prove the claim. Any element of k\overline{B}(X, H)⊗_{kOut(H)}E can be written \overline{ϕ} ⊗ 1_{E} with ϕ ∈ kB(X, H), because \overline{π} : kOut(H) → E is surjective. Then we have

\overline{ϕ} ∈ R(X, H) ⇔ <\overline{α}, \overline{ϕ}>_{X} = 0 , ∀α ∈ kB(X, H)
⇔ τπ(βϕ) = 0 , ∀β ∈ kB(H, X)
⇔ τ(π(γ)π(βϕ)) = 0 , ∀β ∈ kB(H, X) , ∀γ ∈ kB(H, H)
⇔ π(βϕ) = 0 , ∀β ∈ kB(H, X)
⇔ \overline{βϕ} · 1_{E} = 0 , ∀β ∈ kB(H, X)
⇔ \overline{ϕ} ⊗ 1_{E} ∈ \overline{J}_{H,E}(X) ,

where we used the non-degeneracy of the bilinear form on E induced by τ. This completes the proof.

6.3. Corollary. With the notation of Theorem 6.2, if E is semi-simple as a kOut(H)-module, then the biset functor k\overline{B}(−, H)/R(−, H) is semi-simple.

Proof : This follows from the observation that S_{H,V ⊕W} ≅ S_{H,V} ⊕ S_{H,W} and that S_{H,V} is a simple functor if V is a simple kOut(H)-module.

7. Evaluation of simple functors

In this section, we use the results of Section 6 to determine the dimension of the evaluation of a simple functor S_{H,V}. In general, the explicit computation of an evaluation S_{H,V} (G) is not easy. There are general procedures for the determination of such evaluations (see Theorem 4.3.20 in [Bo3] and also Section 11.2 in [We2]) and there are some special cases where the evaluation is known in detail, for instance when k has characteristic zero, H is a b-group, and V = k is the trivial module (see Section 7.2.4 in [Bo1] and Theorem 5.5.4 in [Bo3]).

Here we prove that the dimension of an evaluation S_{H,V} (G) can be obtained as the rank of a bilinear form, a result which had been known for a long time in the case where V = k, the trivial kOut(H)-module (see Section 8.2 in [Bo1]), but remained open in the general case. In Section 11.2 of [We2], Webb mentions a bilinear form for computing dim(S_{H,V} (G)) (although this is made explicit only if dim(V) = 1), but the form is defined on another vector space. However, the result sketched by Webb is quite similar to our theorem below.

The construction of the simple functor S_{H,V} uses a quotient of L_{H,V} or \overline{L}_{H,V}, hence one of the tensor products

L_{H,V} = kB(−, H) ⊗_{kB(H,H)} V or \overline{L}_{H,V} = k\overline{B}(−, H) ⊗_{kOut(H)} V .

The advantage of Theorem 6.2 is that we obtain S_{H,E} directly as a quotient of k\overline{B}(−, H), without needing to tensor with E. This implies that the dimension of S_{H,E}(G) can be computed by a direct use of the bilinear form and its kernel. This allows for a determination of the dimension of the evaluation of simple functors.

As before, we fix a finite group H and we assume that k is algebraically closed (or at least large enough for the group Out(H)). We assume that V is a simple kOut(H)-module, \overline{π} : kOut(H) → E is the natural surjection onto E = End_{k}(V), and τ = τ_{V} is the trace map on End_{k}(V). In this situation, we have the following.

7.1. Theorem. Let V be a simple kOut(H)-module, let \overline{π} : kOut(H) → E be the natural surjection onto E = End_{k}(V), and let τ = τ_{V} be the trace map on End_{k}(V). Let <−, −>_{G} be the corresponding bilinear form on k\overline{B}(G, H), as defined in Section 6. If G is a finite group such that H ⊑ G, then

dim(S_{H,V} (G)) = \frac{rank <−, −>_{G}}{dim(V)} .

Proof : As a kOut(H)-module, we have End_{k}(V) ≅ m · V, where m = dim(V) and m · V denotes the direct sum of m copies of V. Moreover, the functor W ↦ S_{H,W} is additive in W. Therefore, by Theorem 6.2, we obtain

k\overline{B}(G, H)/R(G, H) ≅ S_{H,E}(G) ≅ m · S_{H,V} (G)

where R(G, H) is the right kernel of the bilinear form. Now the dimension of the left hand side is the rank of the bilinear form <−, −>_{G} on k\overline{B}(G, H). The result follows.

We now describe a procedure for computing the dimension of S_{H,V} (G), for any finite group G. First we need a basis of k\overline{B}(G, H). We consider all sections (S, T) of G such that S/T ≅ H, up to G-conjugation. Then Indinf_{S/T}^{G} Iso_{σ} is a basis element of k\overline{B}(G, H), where σ : H → S/T is an isomorphism. But we have Indinf_{S/T}^{G} Iso_{σ} = Indinf_{S/T}^{G} Iso_{ρ} whenever ρ = Conj_{g} σ, where Conj_{g} denotes conjugation by some element g ∈ N_{G}(S, T).

7.2. Lemma. Let H and G be finite groups. The set of elements of the form Indinf_{S/T}^{G} Iso_{σ}, where (S, T) runs over all sections of G such that S/T ≅ H, up to G-conjugation, and where σ : H → S/T runs over all isomorphisms up to left composition by N_{G}(S, T), is a k-basis of k\overline{B}(G, H).

Proof : This easily follows from Lemma 2.1 and the analysis above.

Now we consider the form <−, −>_{G} on k\overline{B}(G, H). Let α = Indinf_{S/T}^{G} Iso_{σ} and β = Indinf_{J/K}^{G} Iso_{ρ}, where σ : H → S/T and ρ : H → J/K are isomorphisms. Then

<α, β>_{G} = τ\overline{π}(α^{op}β)

and we have α^{op} = Iso_{σ^{−1}}Defres_{S/T}^{G}. The generalized Mackey formula (see Proposition A1 in [BT1] and Lemma 2.5 in [BT2]) tells us how to decompose the biset Defres_{S/T}^{G} Indinf_{J/K}^{G} as a sum of transitive bisets (using butterflies, as defined in [BT2]). Many terms factor through a smaller subquotient and are therefore zero in k\overline{B}(G, H). We are left with terms involving conjugates of (J, K) which are linked to (S, T) and each of them produces an element of Out(H) (see [BT2] for details about linked sections). Computing τ\overline{π} on such a term x is just computing the character value τ_{V} (x). All this is, at least in principle, easy to compute, using some standard computer software of group theory. Note that there is an effective method for computing the evaluation S_{H,V} (G) of a simple functor (see Theorem 4.3.20 in [Bo3]), but this requires the knowledge of the kOut(H)-module V rather than merely its character.

8. Semi-simple quotients

We continue with the notation of Section 6 and consider another special case. We fix a finite group H and we suppose that k is algebraically closed (or at least large enough for the group Out(H)). We assume now that the k-algebra E is the largest semi-simple quotient of kOut(H), that is,

E = kOut(H)/J(kOut(H)) ≅ ∏_{i=1}^{r} End_{k}(V_{i}) ,

where V_{1}, …, V_{r} are the simple kOut(H)-modules. The symmetrizing form τ : E → k is the sum of all trace maps τ = ∑_{i=1}^{r} τ_{V_{i}}. If k has characteristic 0 (or a prime not dividing |Out(H)|), then E = kOut(H) and we could also take the ordinary map τ for group algebras (coefficient of 1). Then, for each finite group X, we have a symmetric bilinear form <−, −>_{X} on k\overline{B}(X, H) with kernel R(X, H), and this defines a subfunctor R(−, H) of k\overline{B}(−, H).

8.1. Theorem. With the notation above, the following holds.

1. The semi-simple functor k\overline{B}(−, H)/R(−, H) is isomorphic to

k\overline{B}(−, H)/R(−, H) ≅ ⨁_{i=1}^{r} S_{H,End_{k}(V_{i})} ≅ ⨁_{i=1}^{r} m_{i} · S_{H,V_{i}} ,

where m_{i} = dim(V_{i}) and m_{i} · S_{H,V_{i}} denotes the direct sum of m_{i} copies of S_{H,V_{i}}.

2. k\overline{B}(−, H)/R(−, H) is the largest semi-simple quotient of the biset functor k\overline{B}(−, H). In other words, R(−, H) is the Jacobson radical of k\overline{B}(−, H).

It should be noted that, if each evaluation of a biset functor F is finitedimensional, the functor F itself may not have finite length. In particular, there are examples of biset functors which have no maximal subfunctors at all. Here, the statement of the theorem shows that k\overline{B}(−, H) has finitely many maximal subfunctors and that it has a Jacobson radical R(−, H) such that k\overline{B}(−, H)/R(−, H) has finite length.

Proof : The first part follows immediately from Theorem 6.2, together with the following isomorphisms of kOut(H)-modules :

E = kOut(H)/J(kOut(H)) ≅ ⨁_{i=1}^{r} End_{k}(V_{i}) ≅ ⨁_{i=1}^{r} m_{i} · V_{i} .

To prove part (2), we assume that a simple functor S is isomorphic to a quotient of k\overline{B}(−, H), and we write S = S_{J,W} where J is a group of minimal order such that S(J) ≠ 0 and W is a simple kOut(J)-module. Thus we obtain a non-zero surjective morphism

kB(−, H) ⟶ \overline{k}B(−, H) ⟶ S_{J,W} ,

which must correspond to an element of S_{J,W} (H) by Yoneda’s lemma. Therefore S_{J,W} (H) ≠ 0, hence J ⊑ H by Proposition 3.2. On the other hand, we must have \overline{k}B(J, H) ≠ 0, because of the surjection

\overline{k}B(J, H) → S_{J,W} (J) = W ≠ 0 .

It follows that H ⊑ J by Lemma 4.2. Therefore we obtain J ≅ H.

Now S = S_{H,W} is a quotient of k\overline{B}(−, H). If it was not a quotient of k\overline{B}(−, H)/R(−, H), we would obtain a semi-simple quotient of k\overline{B}(−, H) isomorphic to

S ⊕ k\overline{B}(−, H)/R(−, H) .

Then, on evaluation at H, we would obtain a semi-simple quotient of kOut(H) isomorphic to

S_{H,W} (H) ⊕ kOut(H)/R(H, H) = W ⊕ kOut(H)/J(kOut(H)) ,

because R(H, H) = Ker(\overline{π}) = J(kOut(H)) by Lemma 6.1 and our choice of E. But kOut(H) cannot have such a semi-simple quotient, by definition of the Jacobson radical. So this is impossible and it follows that k\overline{B}(−, H)/R(−, H) is the largest semi-simple quotient of k\overline{B}(−, H).

Since the evaluation of a simple functor is a simple module, we obtain the following corollary.

8.2. Corollary. With the same notation as above, let G be a finite group. The kB(G, G)-module k\overline{B}(G, H)/R(G, H) is semi-simple, isomorphic to

k\overline{B}(G, H)/R(G, H) ≅ ⨁_{i=1}^{r} m_{i} · S_{H,V_{i}}(G) ,

where m_{i} = dim(V_{i}) and m_{i} · S_{H,V_{i}}(G) denotes the direct sum of m_{i} copies of S_{H,V_{i}}(G).

Proof : This follows immediately from Theorem 8.1 by taking evaluation at G.

We warn the reader that some of the evaluations S_{H,V_{i}}(G) might be zero. This is an important issue for understanding simple kB(G, G)-modules and counting them (see [BST]).

In view of part (2) of Theorem 8.1, the obvious question is to know whether or not k\overline{B}(G, H)/R(G, H) is the largest semi-simple quotient of k\overline{B}(G, H). In other words, when is the inclusion

J(k\overline{B}(G, H)) ⊆ R(G, H)

an equality J(k\overline{B}(G, H)) = R(G, H) ? We shall see in Section 9 that the answer is negative in general, but positive in some specific cases, for instance when G is abelian.

Corollary 8.2 produces simple modules for the double Burnside ring kB(G, G) by means of an easily computable quotient k\overline{B}(G, H)/R(G, H) of a rather straightforward module k\overline{B}(G, H). This is enough for the determination of the Jacobson radical of kB(G, G), in view of the following result.

8.3. Proposition. Let G be a finite group. The Jacobson radical J(kB(G, G)) of the k-algebra kB(G, G) is equal to the kernel of the action of kB(G, G) on the semi-simple module

⨁_{H⊑G} k\overline{B}(G, H)/R(G, H) .

In other words, every simple kB(G, G)-module appears in one of the modules k\overline{B}(G, H)/R(G, H), for some subquotient H ⊑ G.

Proof : Let α ∈ kB(G, G). Then α ∈ J(kB(G, G)) if and only if α acts by zero on every simple module S_{H,V_{i}}(G), where H ⊑ G and V_{1}, …, V_{r} are the simple kOut(H)-modules. Equivalently, for every subquotient H ⊑ G, α acts by zero on ⨁_{i=1}^{r} m_{i} · S_{H,V_{i}}(G), that is, on k\overline{B}(G, H)/R(G, H) by Corollary 8.2.

8.4. Remark. It is interesting to note that S_{H,V_{i}}(G) (provided it is non-zero) appears at least m_{i} = dim(V_{i}) times as a quotient of k\overline{B}(G, H), hence also as a quotient of

⨁_{H⊑G} k\overline{B}(G, H)/R(G, H) .

On the other hand, if k is a splitting field for kB(G, G), the simple module S_{H,V_{i}}(G) must appear n_{i} times as a quotient of kB(G, G), where n_{i} = dim(S_{H,V_{i}}(G)). The comparison between m_{i} and n_{i} is not straightforward. In case k\overline{B}(G, H) is a generated by a single element (hence a quotient of kB(G, G)), then we must have n_{i} ≥ m_{i}. This happens in suitable examples (see Remark 9.3 and Proposition 9.4). But in general, it seems difficult to have specific information on the dimension of evaluations, apart from the general result of Theorem 7.1.

9. Further results on evaluations

As in the previous section, H denotes a finite group, E = kOut(H)/J(kOut(H)), and R(G, H) is the kernel of the corresponding bilinear form on k\overline{B}(G, H), where G is a finite group (with H ⊑ G). In this section, we examine a few cases where the equality J(k\overline{B}(−, H)) = R(−, H) remains true on evaluation at G. We first give a sufficient condition.

9.1. Proposition. With the notation above, suppose that there exists α ∈ kB(H, G) and β ∈ kB(G, H) such that αβ ≡ id mod kI(H, H). Then

J(k\overline{B}(G, H)) = R(G, H)

and the only simple quotients of k\overline{B}(G, H) (as kB(G, G)-module) are the simple modules S_{H,V} (G) (provided they are non-zero), where V is a simple kOut(H)- module.

Proof : Let W be a simple quotient of k\overline{B}(G, H) as kB(G, G)-module and let P be a minimal group for W. By Proposition 5.1, W = S_{P,V} (G) for some simple kOut(P)-module V. By Proposition 5.2, H ⊑ P. Now by assumption

k\overline{B}(G, H) = k\overline{B}(G, H)\overline{α}\overline{β} = k\overline{B}(G, H)\overline{α}\overline{β}\overline{α}\overline{β} ⊆ kB(G, G)βα · \overline{β} ⊆ k\overline{B}(G, H) , hence kB(G, G)βα · \overline{β} = k\overline{B}(G, H). Let w_{0} be the image of \overline{β} ∈ k\overline{B}(G, H) via the quotient map k\overline{B}(G, H) → W. Then βα·w_{0} ≠ 0, otherwise the whole image of k\overline{B}(G, H) would be zero. But this can be viewed as successive actions of α and β in the functor S_{P,V}, because W = S_{P,V} (G). We deduce that α · w_{0} ≠ 0 in S_{P,V} (H). Therefore S_{P,V} (H) ≠ 0, hence P ⊑ H by Proposition 3.2. It follows that P ≅ H and so W = S_{H,V} (G). Therefore, every simple module appearing as a quotient of k\overline{B}(G, H) must be indexed by H. We are left with the question of the multiplicities.

By Corollary 8.2, we know that k\overline{B}(G, H)/R(G, H) is semi-simple, with S_{H,V} (G) appearing m times, where m = dim(V). We need to prove that the multiplicity of S_{H,V} (G) as a quotient of k\overline{B}(G, H) is exactly m. From this, the equality J(k\overline{B}(G, H)) = R(G, H) will follow. Now the multiplicity \overline{m} of W = S_{H,V} (G) as a quotient of k\overline{B}(G, H) is equal to

\overline{m} = dim Hom_{kB(G,G)}(k\overline{B}(G, H), W)

and there is a k-linear map

f : V ⟶ Hom_{kB(G,G)}(k\overline{B}(G, H), W) , v ↦ f(v) ,

where f(v) maps \overline{γ} ∈ k\overline{B}(G, H) to γ · v. This makes sense in the functor S_{H,V}, because S_{H,V} (H) = V and S_{H,V} (G) = W, and it only depends on \overline{γ}, because S_{H,V} vanishes on proper subquotients of H. The map f is non-zero because there is some γ ∈ kB(G, H) and v ∈ V such that γ · v ≠ 0, by Proposition 3.4. We shall prove that f is an isomorphism, so that m = dim(V) must be equal to \overline{m}, as required.

Now the map f is kOut(H)-linear, with respect to the left action on the target given by

(\overline{ϕ} · q)(\overline{γ}) = q(\overline{γ}\overline{ϕ}) ,

where \overline{ϕ} ∈ kOut(H), q ∈ Hom_{kB(G,G)}(k\overline{B}(G, H), W), and \overline{γ} ∈ k\overline{B}(G, H). By simplicity of V as a kOut(H)-module, the map f is injective. To prove the surjectivity of f, we let q ∈ Hom_{kB(G,G)}(k\overline{B}(G, H), W). By our assumption, we have

q(\overline{γ}) = q(\overline{γ} \overline{α}\overline{β}) = γα q(\overline{β})

by kB(G, G)-linearity of q. Therefore q(\overline{γ}) = γ · v, where v = α q(\overline{β}), and so q = f(v), proving the surjectivity.

9.2. Remark. If k has characteristic zero or prime to |Out(H)|, then we can avoid the argument of the second part of the proof and prove directly that \overline{m} = m, as follows. We already know that k\overline{B}(G, H) has a semi-simple quotient with S_{H,V} (G) appearing m times, where m = dim(V). If the multiplicity of S_{H,V} (G) is \overline{m} ≥ m, then S_{H,V} appears \overline{m} times as a composition factor of k\overline{B}(−, H), by repeated applications of Proposition 3.5. Then, on evaluation at H, the module V = S_{H,V} (H) appears at least \overline{m} times as a composition factor of k\overline{B}(H, H) = kOut(H). But since kOut(H) is semi-simple by Maschke’s theorem, the module V appears exactly m = dim(V) times as a composition factor of kOut(H). Therefore \overline{m} ≤ m.

9.3. Remark. The assumption of Proposition 9.1 implies that k\overline{B}(G, H) is generated by a single element as a kB(G, G)-module, because kB(G, G)βα·\overline{β} = k\overline{B}(G, H), as we have seen in the proof. However, there are many examples where k\overline{B}(G, H) is not a cyclic kB(G, G)-module (e.g. Example 13.5 below). It seems to be an interesting question to understand when this happens. It clearly depends on the various ways H is realized as a subquotient of the group G.

We now show that the situation of Proposition 9.1 is rather common, so that the equality J(k\overline{B}(G, H)) = R(G, H) often occurs.

9.4. Proposition. The assumption of Proposition 9.1 is satisfied in each of the following cases.

1. H is isomorphic to a quotient of G.

2. G is abelian.

3. H is isomorphic to a subgroup Z of G such that N_{G}(Z) = ZC_{G}(Z), and |N_{G}(Z) : Z| is non-zero in k.

4. H is isomorphic to a central subgroup Z of G, and |G : Z| is non-zero in k.

Proof : (1) If H = G/N, then Def_{G/N}^{G} Inf_{G/N}^{G} = id_{G/N}.

(2) Any subgroup of an abelian group G is isomorphic to a quotient of G, by using the isomorphism between G and its dual. It follows that any subquotient of G is isomorphic to a quotient of G. Thus part (1) applies.

(4) is a special case of (3). Note that, in characteristic zero, (2) can also be proved by using (4).

We are left with a proof of (3). Let α = |N_{G}(Z) : Z|^{−1}Res_{Z}^{G} and β = Ind_{Z}^{G}. By the Mackey formula, we have

Res_{Z}^{G} Ind_{Z}^{G} = ∑_{g∈[Z\G/Z]} Ind_{Z∩^{g}Z}^{Z} Res_{Z∩^{g}Z}^{^{g}Z} Conj_{g}
≡ ∑_{g∈[N_{G}(Z)/Z]} Ind_{Z}^{Z} Res_{Z}^{Z} Conj_{g} mod I(Z, Z)
≡ |N_{G}(Z) : Z| · id mod I(Z, Z) ,

because Conj_{g} is the identity for g ∈ ZC_{G}(Z) = N_{G}(Z). It follows that αβ ≡ id mod kI(Z, Z).

9.5. Remark. Statement (3) is in fact a special case of a more general, but more technical, result. Suppose that H is isomorphic to a subquotient S/T of G, where (S, T) is a section of G such that T ≤ Φ(S) and the image of N_{G}(S, T) in Out(S) is trivial. Suppose also that |N_{G}(S, T) : S| is non-zero in k. Then the assumption of Proposition 9.1 is satisfied. The proof uses the idempotents e_{S}^{G} of the ordinary Burnside ring B(G) and their images ẽ_{S}^{G} ∈ B(G, G) defined in Section 2.5 of [Bo3]. We let

α = |N_{G}(S, T) : S|^{−1}Defres_{S/T}^{G} ẽ_{S}^{G} and β = ẽ_{S}^{G} Indinf_{S/T}^{G} .

Then the actual computation is similar in spirit to the one used in the proof above, but more involved. One needs the generalized Mackey formula (Proposition A1 in [BT1] and Lemma 2.5 in [BT2]), which tells us how to decompose the biset Defres_{S/T}^{G} Indinf_{S/T}^{G} as a sum of transitive bisets, using butterflies, as defined in [BT2]. The condition that T is contained in the Frattini subgroup

Φ(S) implies that ẽ_{S}^{G} Indinf_{S/T}^{G} = Indinf_{S/T}^{G} ẽ_{S/T}^{S/T} and this is used to show that many terms in the the sum lie in fact in I(S/T, S/T). The only remaining terms are conjugations by elements g ∈ N_{G}(S, T). The assumption on the action of this group ensures that such a conjugation is the identity and the result follows.

9.6. Remark. Yet another case where we obtain the equality J(k\overline{B}(G, H)) = R(G, H) appears when H = N_{G}(T)/T and T is an expansive subgroup of G, as defined in Section 6.4 of [Bo3]. We consider the elements

α = f_{1}^{N_{G}(T)/T} Defres_{N_{G}(T)/T}^{G} and β = Indinf_{S/T}^{G} f_{1}^{N_{G}(T)/T} ,

where f_{1}^{N_{G}(T)/T} is also defined in Section 6.4 of [Bo3]. Then α and β satisfy the property αβ ≡ id mod kI(H, H) and Proposition 9.1 applies again.

We have seen various cases where J(k\overline{B}(G, H)) = R(G, H), but we shall see in Section 13 several examples where this is not so.

10. The case of the trivial group

We assume now that the finite group H is trivial. Then k\overline{B}(G, 1) = kB(G, 1) = kB(G) for any finite group G, where B(G) denotes the ordinary Burnside ring of G. Viewed as biset functors, we have

k\overline{B}(−, 1) = kB(−, 1) = kB(−) = L_{1,k} = \overline{L}_{1,k} .

Moreover, this has a unique simple quotient S_{1,k} by Lemma 2.3 (or also by Theorem 8.1). For simplicity, we assume that k is a field of characteristic zero. In this special case, our bilinear form on k\overline{B}(G, 1) = kB(G) was already considered in Section 7.2 of [Bo1]. The kernel R(G, 1) of the bilinear form is equal to the kernel of the surjective map

q : kB(G) → kR_{ℚ}(G) ,

where R_{ℚ} is the functor of rational representations. Here q denotes the k-linear extension of the natural homomorphism q : B(G) → R_{ℚ}(G) mapping a G-set X to the permutation ℚG-module q(X) = ℚX. It follows that S_{1,k} ≅ kR_{ℚ}. In particular the dimension of S_{1,k}(G) is the number of conjugacy classes of cyclic subgroups of G.

Now we extend further the analysis by introducing the two-sided ideal I(G) of kB(G, G) generated by all the (G, G)-bisets which factor through the trivial group. The method finds its origin in the proof of Proposition 6.1.5 in [Bo3]. Thus I(G) is generated as a k-vector space by the (G, G)-bisets

(G × G)/(A × B) = Indinf_{A/A}^{G} Iso_{σ} Defres_{B/B}^{G} ,

where A and B are subgroups of G and σ : B/B → A/A is the obvious unique isomorphism (which we shall ignore in the sequel, for simplicity). It is useful to view G/A as a (G, 1)-biset and B\G as a (1, G)-biset, that is,

G/A = Indinf_{A/A}^{G} and B\G = (G/B)^{op} = Defres_{B/B}^{G} .

With this point of view, we obtain

(G × G)/(A × B) = (G/A) · (G/B)^{op} .

More generally, any α ∈ kB(G) can be viewed as an element of kB(G, 1) and then

αβ^{op} ∈ I(G) , for any α, β ∈ kB(G) .

This applies in particular to the primitive idempotents e_{A}^{G} of the Burnside ring kB(G), which form a k-basis of kB(G), and we obtain generators e_{A}^{G} (e_{B}^{G})^{op} of I(G) as a vector space. Here A and B run over all subgroups of G up to conjugation.

The action of I(G) on S_{1,k}(G) = kR_{ℚ}(G) is described in the following result. 10.1. Lemma. Let k be a field of characteristic zero and let I(G) be the two-sided ideal of kB(G, G) defined above.

1. If H is a non-trivial finite group, the action of I(G) on S_{H,V} (G) is zero, for any kOut(H)-module V.

2. Let α, β ∈ kB(G) and let M be a ℚG-module. The action of αβ^{op} on [M] ∈ kR_{ℚ}(G) is given by :

αβ^{op} · [M] = (q(β) | M)_{G} · q(α) ,

where (− | −)_{G} denotes the ordinary scalar product of ℚG-modules.

3. If either A or B is non-cyclic, then the action of e_{A}^{G} (e_{B}^{G})^{op} on kR_{ℚ}(G) is zero.

4. If A and B run over the set of cyclic subgroups up to G-conjugation, then the images of e_{A}^{G} (e_{B}^{G})^{op} in End_{k}(kR_{ℚ}(G)) are k-linearly independent. So are the images of G/A · (G/B)^{op}, where A and B run over the same set.

Proof : (1) The action of an element of I(G) factors through the trivial group. But S_{H,V} (1) = 0 if H ≠ 1 since H has minimal order such that S_{H,V} (H) ≠ 0.

(2) Taking basis elements, we can assume that α = G/A = Indinf_{A/A}^{G} and β = G/B, so β^{op} = Defres_{B/B}^{G}. Then

G/A · (G/B)^{op}(M) = Indinf_{A/A}^{G} Defres_{B/B}^{G}(M) = Indinf_{A/A}^{G}(M^{B})

because the deflation of a ℚB-module is obtained by taking B-fixed points (actually cofixed points, but this is the same in characteristic 0). Then M^{B} is just the direct sum of dim(M^{B}) copies of ℚ (a module for the trivial group), and then this is induced from A to G. So we obtain

G/A · (G/B)^{op}(M) = dim(M^{B}) · ℚ[G/A]
= (ℚ | Res_{B}^{G}(M))_{B} · q(G/A)
= (ℚ[G/B] | M)_{G} · q(G/A)
= (q(G/B) | M)_{G} · q(G/A) ,

as was to be shown.

(3) If A is non-cyclic, then it is well-known that q(e_{A}^{G}) = 0 (because the restriction of e_{A}^{G} to any cyclic subgroup of G is zero, hence the same holds for q(e_{A}^{G})). So the action of e_{A}^{G} (e_{B}^{G})^{op} is zero by (2). The same holds if B is non-cyclic, because q(e_{B}^{G}) = 0 in that case.

(4) Suppose that ∑_{A,B} λ_{A,B} G/A · (G/B)^{op} acts by zero on kR_{ℚ}(G), where A and B run over the set of cyclic subgroups up to G-conjugation and where λ_{A,B} ∈ k. Then by (2), for every x ∈ kR_{ℚ}(G), we have

∑_{A} ∑_{B} λ_{A,B} (ℚ[G/B] | x) · ℚ[G/A] = 0 .

Since the modules ℚ[G/A] form a k-basis of kR_{ℚ}(G) by Artin’s induction theorem, we obtain

( ∑_{B} λ_{A,B} ℚ[G/B] | x) = 0 .

But this holds for all x and the scalar product is non-degenerate. Therefore

∑_{B} λ_{A,B} ℚ[G/B] = 0 .

Again the modules ℚ[G/B] form a basis, so λ_{A,B} = 0. This shows one of the statement. The other statement follows because the idempotents e_{A}^{G} generate the same subspace of the Burnside ring kB(G) as the elements G/A (where A runs over cyclic subgroups up to conjugation).

Now we can prove that a large part of the ideal I(G) lies in the Jacobson radical J(kB(G, G)).

10.2. Theorem. Let k be a field of characteristic zero and let I(G) be the ideal of kB(G, G) defined above. Let I_{c}(G) be the k-subspace of I(G) generated by all elements (G/A) · (G/B)^{op} such that A and B are cyclic subgroups of G.

1. kB(G, G) = I_{c}(G) ⊕ M, where M denotes the maximal two-sided ideal of kB(G, G) which is the kernel of the action of kB(G, G) on S_{1,k}(G) = kR_{ℚ}(G).

2. I(G) = I_{c}(G) ⊕ (I(G) ∩ J(kB(G, G))).

3. dim (I(G) ∩ J(kB(G, G))) = b(G)^{2} − c(G)^{2}, where b(G) is the number of conjugacy classes of subgroups of G and c(G) is the number of conjugacy classes of cyclic subgroups of G.

Proof : (1) Since kR_{ℚ}(G) = S_{1,k}(G) is a simple kB(G, G)-module, the natural k-algebra homomorphism

r : kB(G, G) ⟶ End_{k}(kR_{ℚ}(G))

is surjective by the density theorem. Part (4) of Lemma 10.1 shows that there are c(G)^{2} elements of I_{c}(G) whose images under r are k-linearly independent. But kR_{ℚ}(G) has dimension c(G) (by Artin’s induction theorem), so dim (End_{k}(kR_{ℚ}(G))) = c(G)^{2}. It follows that the restriction of r to I_{c}(G) is an isomorphism of k-vector spaces and (1) follows (because Ker(r) = M).

(2) Let I′(G) be the k-subspace of I(G) generated by all elements e_{A}^{G} (e_{B}^{G})^{op}, where either A or B is non-cyclic. Then I(G) = I_{c}(G) ⊕ I′(G) because the elements e_{A}^{G} (e_{B}^{G})^{op} where A and B are both cyclic generate I_{c}(G). By parts (1) and (3) of Lemma 10.1, I′(G) acts by zero on every simple kB(G, G)-module, hence I′(G) ⊆ J(kB(G, G)). Since I_{c}(G) acts faithfully on one simple module (by part (4) of Lemma 10.1 ), it follows that I′(G) = I(G) ∩ J(kB(G, G)).

(3) It is clear that dim(I(G)) = b(G)^{2} and dim(I_{c}(G)) = c(G)^{2}.

We now recover a result proved in Section 6.1 of [Bo3] (but the result in [Bo3] is more precise).

10.3. Corollary. If k is a field of characteristic zero and if G is a non cyclic group, then kB(G, G) is not semi-simple.

Proof : With the notation of Theorem 10.2, we have b(G) > c(G) because G is not cyclic. Therefore I(G) ∩ J(kB(G, G)) ≠ 0 and kB(G, G) is not semisimple.

11. Left ideals in the double Burnside ring

In order to understand the ring structure of the double Burnside ring kB(G, G), it is useful to have information on some naturally defined left ideals associated to sections of G. Many of the previous results used a finite group H and the space kB(G, H), or its quotient k\overline{B}(G, H). Now we work with an incarnation of H as a subquotient of G by fixing a section (P, Q) of G such that P/Q = H. Then the group N_{G}(P, Q) comes into play, or more precisely the image Γ_{G}(P, Q) of N_{G}(P, Q) in Out(P/Q).

We compose with Defres_{P/Q}^{G} in order to obtain elements of kB(G, G), that is, we define

K_{(P,Q)} = { α · Defres_{P/Q}^{G} | α ∈ kB(G, P/Q) }
K_{<(P,Q)} = { α · Defres_{P/Q}^{G} | α ∈ kI(G, P/Q) }
\overline{K}_{(P,Q)} = K_{(P,Q)}/K_{<(P,Q)} .

Clearly, K_{(P,Q)} and K_{<(P,Q)} are left ideals of kB(G, G) and hence \overline{K}_{(P,Q)} is a kB(G, G)-module. Moreover, K_{(P,Q)} and \overline{K}_{(P,Q)} only depend on the Gconjugacy class of the section (P, Q). Using Lemma 7.2, one can find a k-basis of \overline{K}_{(P,Q)} consisting of elements of the form Indinf_{J/K}^{G} Iso_{σ} Defres_{P/Q}^{G}, where (J, K) runs over sections of G and σ : P/Q → J/K is an isomorphism, up to left composition by N_{G}(J, K) and right composition by N_{G}(P, Q).

It is possible to filter the whole ring kB(G, G) by left ideals in such a way that each successive quotient is isomorphic to \overline{K}_{(P,Q)} for some section (P, Q), as follows. We consider the set X of all conjugacy classes of sections of G, and we let

∅ = X_{0} ⊂ X_{1} ⊂ … ⊂ X_{N−1} ⊂ X_{N} = X

be a filtration of X such that, for each 1 ≤ i ≤ N,

X_{i} = X_{i−1} ∪ {[(P_{i}, Q_{i})]} ,

where [(P_{i}, Q_{i})] is the conjugacy class of a section (P_{i}, Q_{i}) such that |X/Y | ≤ |P_{i}/Q_{i}| for all [(X, Y)] ∈ X_{i−1}. It is clear that such a filtration always exists. Then we define

K_{X_{i}} = ∑_{[(X,Y)]∈X_{i}} K_{(X,Y)} ,

and we obtain a filtration

0 = K_{X_{0}} ⊂ K_{X_{1}} ⊂ … ⊂ K_{X_{N−1}} ⊂ K_{X_{N}} = kB(G, G) .

The following result is easy and its proof is left to the reader. 11.1. Lemma. With the notation above, K_{X_{i}}/K_{X_{i−1}} ≅ \overline{K}_{(P_{i},Q_{i})}.

This result shows that the module structure of \overline{K}_{(P,Q)} is relevant for the understanding of the ring structure of kB(G, G). We shall describe a semi-simple quotient of \overline{K}_{(P,Q)} when k has characteristic zero (but this simplifying assumption could be dropped with a little more work). We first need the following description of \overline{K}_{(P,Q)}. 11.2. Proposition. Let G be a finite group and let (P, Q) be a section of G.

1. The homomorphism kB(G, P/Q) → K_{(P,Q)}, α ↦ α · Defres_{P/Q}^{G}, induces an isomorphism

μ : kB(G, P/Q)_{N_{G}(P,Q)} ⟶^{∼} K_{(P,Q)}

where the left hand side denotes the quotient of kB(G, P/Q) by the right action of N_{G}(P, Q) by conjugation (cofixed points).

2. The isomorphism μ induces an isomorphism

\overline{μ} : k\overline{B}(G, P/Q)_{Γ_{G}(P,Q)} = k\overline{B}(G, P/Q)_{N_{G}(P,Q)} ⟶^{∼} \overline{K}_{(P,Q)}

where Γ_{G}(P, Q) is the image of N_{G}(P, Q) in Out(P/Q).

3. The isomorphism \overline{μ} induces an isomorphism

k\overline{B}(G, P/Q) ⊗_{kOut(P/Q)} k[Out(P/Q)/Γ_{G}(P, Q)] ⟶^{∼} \overline{K}_{(P,Q)} .

Proof : (1) If g ∈ N_{G}(P, Q), then

α Conj_{g} Defres_{P/Q}^{G} = α Defres_{P/Q}^{G} Conj_{g} = α Defres_{P/Q}^{G} ,

so that the surjective homomorphism kB(G, P/Q) → K_{(P,Q)} passes to the quotient kB(G, P/Q)_{N_{G}(P,Q)} of kB(G, P/Q). Then the induced map μ is easily seen to be injective (by Lemma 2.1).

(2) Taking cofixed points is right exact, so we have a commutative diagram

kB(G, P/Q)  ⟶  k\overline{B}(G, P/Q)  ⟶ 0
↓  ↓
kB(G, P/Q)_{N_{G}(P,Q)} ⟶ k\overline{B}(G, P/Q)_{N_{G}(P,Q)} ⟶ 0

↓ ≅
K_{(P,Q)}  ⟶  \overline{K}_{(P,Q)}  ⟶ 0

The kernel of the map in the first row is I(G, P/Q). The kernel of the map in the second row is the image I′ of I(G, P/Q) in kB(G, P/Q)_{N_{G}(P,Q)}. Now the image of I′ in K_{(P,Q)} is the image of I(G, P/Q) in K_{(P,Q)}, namely K_{<(P,Q)} by definition. Therefore the vertical isomorphism induces an isomorphism

kB(G, P/Q)_{N_{G}(P,Q)}/I′ ⟶^{∼} K_{(P,Q)}/K_{<(P,Q)} ,

that is, an isomorphism

k\overline{B}(G, P/Q)_{N_{G}(P,Q)} ⟶^{∼} \overline{K}_{(P,Q)} ,

as required. The equality k\overline{B}(G, P/Q)_{N_{G}(P,Q)} = k\overline{B}(G, P/Q)_{Γ_{G}(P,Q)} is clear, since N_{G}(P, Q) acts via its image Γ_{G}(P, Q).

(3) For the right kOut(P/Q)-module k\overline{B}(G, P/Q), taking cofixed points under Γ_{G}(P, Q) is the same as tensoring with k[Out(P/Q)/Γ_{G}(P, Q)].

11.3. Remark. Actually, we do not need to extend scalars to k in Proposition 11.2. The result holds for similarly defined ideals in B(G, G).

For simplicity, we assume now that the field k has characteristic zero and is algebraically closed (or at least large enough for the group Out(P/Q)). Then fixed points and cofixed points are isomorphic (by Maschke’s theorem), so we consider the subspace of fixed points k\overline{B}(G, P/Q)^{Γ_{G}(P,Q)}. As in Section 8, we consider the bilinear form <−, −>_{G} on k\overline{B}(G, P/Q) associated with the usual symmetrizing form τ on the group algebra kOut(P/Q). The bilinear form <−, −>_{G} restricts to the subspace k\overline{B}(G, P/Q)^{Γ_{G}(P,Q)}, hence defines a bilinear form, still denoted <−, −>_{G}, on \overline{K}_{(P,Q)} (by part (2) of Proposition 11.2). We let R′ be the kernel of this bilinear form on \overline{K}_{(P,Q)}.

Now we can prove a result analogous to Corollary 8.2, where we replace k\overline{B}(G, H) by its ‘incarnation’ \overline{K}_{(P,Q)} inside the double Burnside ring. It turns out that the semi-simple module kOut(H) has to be replaced by the module W = k[Out(P/Q)/Γ_{G}(P, Q)].

11.4. Theorem. Assume k is algebraically closed of characteristic zero. Let G be a finite group, let (P, Q) be a section of G, let K_{(P,Q)} and K_{<(P,Q)} be the ideals of kB(G, G) defined above, and let \overline{K}_{(P,Q)} = K_{(P,Q)}/K_{<(P,Q)}. Let also R′ be the kernel of the bilinear form <−, −>_{G} on \overline{K}_{(P,Q)}. Then \overline{K}_{(P,Q)}/R′ is a semi-simple kB(G, G)-module, isomorphic to

\overline{K}_{(P,Q)}/R′ ≅ S_{P/Q,W} (G) ≅ ⨁_{j=1}^{q} m_{j} · S_{P/Q,W_{j}}(G) ,

where W = k[Out(P/Q)/Γ_{G}(P, Q)] and W = ⨁_{j=1}^{q} m_{j} · W_{j} is the decomposition of the kOut(P/Q)-module W as a direct sum of simple modules.

Proof : We write H = P/Q. Since k has characteristic zero, J(kOut(H)) = 0. By Theorem 8.1, we have a short exact sequence of biset functors

0 ⟶ R(−, H) ⟶ k\overline{B}(−, H) ⟶ S_{H,kOut(H)} ⟶ 0 ,

with a semi-simple right-hand side

S_{H,kOut(H)} ≅ ⨁_{i=1}^{r} n_{i} · S_{H,V_{i}} ,

where kOut(H) = ⨁_{i=1}^{r} n_{i} · V_{i} is the decomposition of kOut(H) as a direct sum of simple modules. Since k has characteristic zero, tensoring with W is exact and we obtain a short exact sequence of biset functors

0 ⟶ R(−, H) ⊗_{kOut(H)} W  ⟶  k\overline{B}(−, H) ⊗_{kOut(H)} W
⟶  S_{H,kOut(H)} ⊗_{kOut(H)} W ⟶ 0 .

Evaluating at G, the middle term is isomorphic to \overline{K}_{(P,Q)}, by Proposition 11.2. We claim that the kernel on the left hand side (evaluated at G) is isomorphic to the kernel R′ of the bilinear form. Then it follows that the right hand side (evaluated at G) is

S_{H,kOut(H)}(G) ⊗_{kOut(H)} W ≅ \overline{K}_{(P,Q)}/R′ .

We shall return to this at the end of the proof, but we first prove the claim. We note that tensoring with W is the same as taking (co)fixed points under Γ, where Γ = Γ_{G}(P, Q). Moreover

k\overline{B}(G, H) = k\overline{B}(G, H)^{Γ} ⊕ U ,

where U is the kernel of (right) multiplication by e_{Γ} = \frac{1}{|Γ|} ∑_{γ∈Γ} γ. This direct sum is orthogonal with respect to the bilinear form, because, if \overline{ϕ} ∈ k\overline{B}(G, H)^{Γ} and \overline{α} ∈ U, we have

<\overline{ϕ}, \overline{α}>_{G} = <\overline{ϕ} e_{Γ}, \overline{α}>_{G} = <\overline{ϕ}, \overline{α} e_{Γ}>_{G} = 0 ,

using part (5) of Lemma 6.1 and the fact that e_{Γ}^{op} = e_{Γ}. Since the direct sum is orthogonal, the kernel of the bilinear form decomposes as

R(G, H) = (R(G, H)∩k\overline{B}(G, H)^{Γ})⊕(R(G, H)∩U) = R(G, H)^{Γ}⊕(R(G, H)∩U) .

Thus the kernel of the bilinear form restricted to k\overline{B}(G, H)^{Γ} is R(G, H)^{Γ}. In other words, in terms of tensor products, the kernel of the bilinear form restricted to k\overline{B}(G, H) ⊗_{kOut(H)} W is R(G, H) ⊗_{kOut(H)} W. This completes the proof of the claim.

We now return to the isomorphism

S_{H,kOut(H)}(G) ⊗_{kOut(H)} W ≅ \overline{K}_{(P,Q)}/R′

and we analyze the left-hand side. We do this by using again the biset functor S_{H,kOut(H)} ⊗_{kOut(H)} W. Since tensoring with W is taking cofixed points, hence a quotient, S_{H,kOut(H)} ⊗_{kOut(H)} W is a quotient of S_{H,kOut(H)} and is therefore semi-simple again. So we only need to know the simple functors which occur and their multiplicity. This can be achieved by evaluating at H because all the simple functors which appear are indexed by H. So we evaluate at H and we obtain

S_{H,kOut(H)}(H) ⊗_{kOut(H)} W = kOut(H) ⊗_{kOut(H)} W ≅ W = S_{H,W} (H) .

This forces the isomorphism

S_{H,kOut(H)} ⊗_{kOut(H)} W ≅ S_{H,W} = ⨁_{j=1}^{q} m_{j} · S_{H,W_{j}}

because we know that the left-hand side must be a direct sum of simple functors indexed by H. Evaluating now at G, we obtain the desired isomorphism

\overline{K}_{(P,Q)}/R′ ≅ S_{H,kOut(H)}(G) ⊗_{kOut(H)} W ≅ S_{H,W} (G) ,

and the proof is complete.

11.5. Corollary. With the same assumptions, suppose moreover that G is abelian. Then R′(G, P/Q) = J(\overline{K}_{(P,Q)}) and the corresponding semi-simple quotient is

\overline{K}_{(P,Q)}/R′(G, P/Q) ≅ k\overline{B}(G, P/Q)/R(G, P/Q) ≅ ⨁_{i=1}^{r} m_{i} · S_{P/Q,V_{i}}(G) ,

where kOut(P/Q) = ⨁_{i=1}^{r} m_{i} · V_{i} is the decomposition of kOut(P/Q) as a direct sum of simple modules (so m_{i} = dim(V_{i})).

Proof : Since G is abelian, the action of N_{G}(P, Q) on P/Q is trivial, that is, Γ_{G}(P, Q) = 1. Therefore

\overline{K}_{(P,Q)} ≅ k\overline{B}(G, P/Q) and \overline{K}_{(P,Q)}/R′(G, P/Q) ≅ k\overline{B}(G, P/Q)/R(G, P/Q) .

Now by Propositions 9.1 and 9.4 (using again the fact that G is abelian), R(G, P/Q) is the Jacobson radical of k\overline{B}(G, P/Q), and hence R′(G, P/Q) is the Jacobson radical of \overline{K}_{(P,Q)}. The explicit description of the semi-simple quotient follows now from Theorem 11.4.

12. The bifree double Burnside ring

In this section, we explain briefly that the results of this paper can be adapted to the case of the bifree double Burnside ring kA(G, G). We then recover some special cases of results of Webb [We2]. Let kA(G, H) be the subspace of kB(G, H) generated by all the (G, H)-bisets which are free on the left and on the right. In other words, we do not allow for inflation and deflation maps, so that kA(G, H) is generated by all bisets of the form Ind_{A}^{G} Iso_{σ} Res_{S}^{H}, where A is a subgroup of G, S is a subgroup of H, and σ : S → A is a group isomorphism. In particular kA(G, G) is a subring of kB(G, G).

The version used in homotopy theory is the subring of kB(G, G) generated by all the (G, G)-bisets which are free on one side, so that deflation is not allowed but inflation is still defined. Unfortunately, our methods do not seem to apply to this subring, because we use symmetry, in particular in the definition of the bilinear forms where opposite bisets are used (and the opposite of inflation is deflation). Therefore, we only consider here the bifree version kA(G, G). We could work with a more general version of the double Burnside ring by using, as in [We2], a family of groups X (respectively Y) allowed as normal subgroups involved in deflation (respectively inflation), and our methods would work in case X = Y. But for simplicity, we only consider the case X = Y = {1}, corresponding to bifree bisets.

By restricting to bifree bisets only, the category of biset functors becomes the category of global Mackey functors. All the methods of this paper can be easily adapted and therefore our results hold in this different context. Throughout our arguments, it suffices to replace the sections of a group G by the subgroups of G, so that the relation H ⊑ G now means that H is isomorphic to a subgroup of G. The definition of the standard quotient k\overline{A}(X, H) is similar to that of k\overline{B}(X, H) and we have k\overline{A}(X, H) ≠ 0 if and only if H is isomorphic to a subgroup of X. Then the construction of the two kinds of bilinear forms on k\overline{A}(X, H) follow in the same manner and all the results of Sections 6 – 8 hold. Note that the procedure for computing the dimension of S_{H,V} (G) described at the end of Section 7 becomes easier, for it only involves the Mackey formula instead of its generalized version.

In this context of bifree bisets, the following new feature appears in characteristic zero.

12.1. Proposition. Assume k is a field of characteristic zero. Let X and H be finite groups, and consider the symmetric bilinear form on k\overline{A}(X, H) defined by

<\overline{α}, \overline{β}>_{X} = τπ(α^{op}β) ,

where α, β ∈ kA(X, H) are representatives of \overline{α} and \overline{β} respectively, π denotes the quotient map π : kA(H, H) → kOut(H), and τ is the usual symmetrizing form on kOut(H) (coefficient of 1). Then this bilinear form is non-degenerate.

Proof : The standard quotient k\overline{A}(X, H) has a basis consisting of the elements Ind_{A}^{X} Iso_{σ}, where A is a subgroup of X and σ : H → A is a group isomorphism. Moreover A must be considered up to X-conjugation and Iso_{σ} up to right composition with inner automorphisms of H and left composition with conjugations by elements of N_{X}(A). We compute the matrix of the bilinear form with respect to this basis and we show that it is a diagonal matrix.

Let A and B be subgroups of X and let σ : H → A and ρ : H → B be group isomorphisms, considered up to conjugation as above. By the Mackey formula, we have

<Ind_{A}^{X} Iso_{σ} , Ind_{B}^{X} Iso_{ρ}>_{X} = τπ(Iso_{σ}^{−1} Res_{A}^{X} Ind_{B}^{X} Iso_{ρ})

= ∑_{g∈[A\X/B]} τπ(Iso_{σ}^{−1} Ind_{A∩^{g}B}^{A} Res_{A∩^{g}B}^{^{g}B} Conj_{g} Iso_{ρ}) .

This is zero if B is not conjugate to A, because then A∩^{g}B is a proper subgroup of A and π is zero on bisets which factorize through a proper subgroup of H. Thus we assume now that B is conjugate to A, and in fact B = A without loss of generality. Then we obtain

<Ind_{A}^{X} Iso_{σ} , Ind_{A}^{X} Iso_{ρ}>_{X} =

= ∑_{g∈[A\X/A]} τπ(Iso_{σ}^{−1} Ind_{A∩^{g}A}^{A} Res_{A∩^{g}A}^{^{g}A} Conj_{g} Iso_{ρ})

= ∑_{g∈[N_{X}(A)/A]} τπ(Iso_{σ}^{−1} Conj_{g} Iso_{ρ})

= ∑_{g∈[N_{X}(A)/AC_{X}(A)]} |AC_{X}(A) : A| τπ(Iso_{σ}^{−1} Conj_{g} Iso_{ρ}) .

The term indexed by g is zero if π(σ^{−1} Conj_{g} ρ) is not the identity element of Out(H), so at most one term of the sum is non-zero. More precisely, the sum is zero unless there exists g ∈ N_{X}(A) such that π(σ^{−1} Conj_{g} ρ) = 1 in Out(H), that is, σ = Conj_{g} ρ Conj_{h} for some h ∈ H. But this means that we have the same basis element

Ind_{A}^{X} Iso_{σ} = Ind_{A}^{X} Conj_{g} ρ Conj_{h} = Ind_{A}^{X} Iso_{ρ} .

This shows that the matrix is diagonal. Moreover, for every basis element Ind_{A}^{X} Iso_{σ}, the corresponding diagonal entry is |AC_{X}(A) : A|. Since k has characteristic zero, the bilinear form is non-degenerate.

The non-degeneracy of the bilinear form implies the following special case of results of Webb [We2].

12.2. Corollary. Assume k is a field of characteristic zero.

1. For every finite group H, the biset functor k\overline{A}(−, H) is semi-simple.

2. For every finite groups H and G, the kA(G, G)-module k\overline{A}(G, H) is semisimple.

3. The k-algebra kA(G, G) is semi-simple.

Proof : (1) Since the kernel of the bilinear form on k\overline{A}(X, H) is 0 for each X, the functor k\overline{A}(−, H) is semi-simple (Corollary 6.3).

(2) This follows form (1) and evaluation at G.

(3) We only sketch the argument. It is not hard to prove that kA(G, G) acts faithfully on ∏_{H≤G} k\overline{A}(G, H). Since this module is semi-simple by (2), it follows that the radical J(kA(G, G)) must be zero.

This corollary is a special case of more general results of Webb, who proved that the category of all bifree biset functors is semi-simple in characteristic zero (see Theorem 9.5 in [We2]). In particular the algebra kA(G, G) is semi-simple, because it is the endomorphism algebra of the representable functor kA(−, G). So we see that Corollary 12.2 is a special case of Webb’s results.

Also, note that the question of vanishing evaluations, which is not easy in the general case of simple biset functors (see [BST]), has a direct solution for the category of global Mackey functors. Indeed, Webb proved an explicit formula giving the evaluation of simple functors (see Theorem 2.6 in [We1]).

13. Examples

We illustrate the results of this paper by a few examples, which have been worked out either by hand or by computer calculations using [GAP]. We first start with a very small example, where most computations can be made over ℤ.

13.1. Example. Let p be a prime and let G = C_{p} be a cyclic group of order p. Then \overline{B}(G, G) = ℤOut(G) ≅ ℤC_{p−1} and there is a ring homomorphism

ζ : B(G, G) ⟶ ℤOut(G) ≅ ℤC_{p−1} .

On the other hand \overline{B}(G, 1) = B(G, 1) is free abelian on the two elements Inf_{G/G}^{G} and Ind_{1}^{G}. The action of B(G, G) on B(G, 1) is easy to describe and this yields a ring homomorphism

η : B(G, G) ⟶ End_{ℤ}(B(G, 1)) ≅ M_{2}(ℤ)

whose image is generated by (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix}1 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix}1 & p \\ 0 & 0 \end{matrix}), (\begin{matrix}0 & 0 \\ 1 & 1 \end{matrix}), and (\begin{matrix}0 & 0 \\ 1 & p \end{matrix}).

Then we obtain an injective ring homomorphism

ζ × η : B(G, G) ⟶ ℤC_{p−1} × M_{2}(ℤ)

which provides an explicit representation of B(G, G).

Extending scalars to a field k whose characteristic does not divide p − 1, we get an isomorphism

ζ × η : kB(G, G) ⟶^{∼} kC_{p−1} × M_{2}(k)

and we see that kB(G, G) is semi-simple. If k contains (p − 1)-th roots of unity, then kC_{p−1} decomposes further as a product of copies of k, so kB(G, G) has p − 1 simple modules of dimension 1 and one simple module of dimension 2.

In our next two examples, the double Burnside ring is not semi-simple, but we have a clear description of the Jacobson radical.

13.2. Example. Let D_{8} be the dihedral group of order 8 and let k be a field of characteristic different from 2 and 3. By using Theorem 7.1 (or its forerunner, Proposition 4.4.6 in [Bo3], in the case of the trivial module k), the dimension of S_{H,V} (D_{8}) can be easily computed. Writing (H, V) in the first line, we get the following values for dim(S_{H,V} (D_{8})):

\begin{matrix}(1, k) & (C_{2}, k) & (C_{4}, k) & (C_{4}, ε) & (V_{4}, k) & (V_{4}, ε) & (V_{4}, 2) & (D_{8}, k) & (D_{8}, ε) \\ 5 & 11 & 1 & 0 & 3 & 1 & 4 & 1 & 1 \end{matrix}

Here ε denotes the sign representation of Out(H), the group V_{4} denotes the Klein four group, and 2 denotes the two-dimensional representation of Out(V_{4}) (which is the symmetric group of order 6). Summing up the squares of the dimensions, we find dim(kB(D_{8}, D_{8})/J) = 175, where J = J(kB(D_{8}, D_{8})). On the other hand, the basis of Lemma 2.1 yields dim(kB(D_{8}, D_{8})) = 214, hence dim(J) = 214−175 = 39. But we know, by Theorem 10.2, that the ideal I(G)∩J has dimension b(D_{8})^{2} − c(D_{8})^{2} = 8^{2} − 5^{2} = 39. It follows that I(G) ∩ J = J, so the only contribution to the Jacobson radical comes from the trivial group.

13.3. Example. Let Q_{8} be the quaternion group of order 8 and let k be a field of characteristic different from 2 and 3. This example is similar to the previous one and the computations show again that I(G) ∩ J = J, so the only contribution to the Jacobson radical comes from the trivial group.

We now move to a slightly more involved case.

13.4. Example. Let G = A_{4} and H = C_{3} and suppose that k is a field of characteristic different from 2. Then k\overline{B}(A_{4}, C_{3}) has dimension 4, its radical J has dimension 2 and coincides with the kernel R(A_{4}, C_{3}) of the bilinear form (by Proposition 9.4). Moreover

k\overline{B}(A_{4}, C_{3})/J = S_{C_{3},k_{+}}(A_{4})⊕S_{C_{3},k_{−}}(A_{4}) and J = S_{A_{4},k_{+}}(A_{4})⊕S_{A_{4},k_{−}}(A_{4}) ,

where k_{+} denotes the trivial representation of the cyclic group Out(C_{3}) of order 2 and k_{−} denotes the sign representation. All the evaluations above are 1-dimensional. This also shows, not surprisingly, that the Jacobson radical of kB(G, G) is not entirely contained in the ideal I(G) considered in Section 10.

Our next examples are concerned with the fact that J(k\overline{B}(G, H)) is not necessarily equal to R(G, H), though many cases where equality holds have been seen in Section 9.

13.5. Example. Let G = A_{5} be the alternating group on 5 letters, let H = C_{3} be the cyclic group of order 3, and suppose that k is a field of characteristic different from 2. The vector space k\overline{B}(A_{5}, C_{3}) has dimension 3, with basis Ind_{C_{3}}^{A_{5}}, Indinf_{A_{4}/V_{4}}^{A_{5}}, Indinf_{A_{4}/V_{4}}^{A_{5}}Iso_{σ}, where V_{4} is the Klein four-group and σ is the non-trivial group automorphism of the cyclic group of order 3 (see Lemma 7.2 and note that Ind_{C_{3}}^{A_{5}}Iso_{σ} ≅ Ind_{C_{3}}^{A_{5}}.)

We have E = kOut(C_{3}) ≅ k_{+} × k_{−}, so there are two simple functors S_{C_{3},k_{+}} and S_{C_{3},k_{−}} indexed by C_{3}. The direct computation of the bilinear form shows that R(A_{5}, C_{3}) has dimension 2, generated by the differences of basis elements, so k\overline{B}(A_{5}, C_{3})/R(A_{5}, C_{3}) is one-dimensional. We have in fact

k\overline{B}(A_{5}, C_{3})/R(A_{5}, C_{3}) ≅ S_{C_{3},k_{+}}(A_{5}) ⊕ S_{C_{3},k_{−}}(A_{5}) = S_{C_{3},k_{+}}(A_{5}) = k ,

because actually S_{C_{3},k_{+}}(A_{5}) = k and S_{C_{3},k_{−}}(A_{5}) = 0 in this specific example.

But further computations show that J(k\overline{B}(A_{5}, C_{3})) is only one-dimensional and that we have to take into account the simple functors S_{A_{4},V} indexed by the larger group A_{4}. We have to consider the simple kB(A_{5}, A_{5})-modules S_{A_{4},V} (A_{5}), where V = k_{+} or V = k_{−}, the two simple modules for the cyclic group Out(A_{4}) of order 2. It turns out that we get a semi-simple quotient of dimension 2

k\overline{B}(A_{5}, C_{3})/J(k\overline{B}(A_{5}, C_{3})) ≅ S_{C_{3},k_{+}}(A_{5}) ⊕ S_{A_{4},k_{−}}(A_{5}) ,

with an extra factor indexed by A_{4}. Moreover, the Jacobson radical has dimension 1 and satisfies J(k\overline{B}(A_{5}, C_{3})) ≅ S_{A_{4},k_{+}}(A_{5}).

13.6. Example. There are numerous other examples where the radical J = J(k\overline{B}(G, H)) of the module M = k\overline{B}(G, H) is not equal to the kernel R = R(G, H) of the bilinear form. In such cases, there are additional simple quotients, namely the factors of R/J, which are indexed by groups larger than H. We just list a few such examples.

G = GL(3, 2) and H = C_{3}. Then dim(M) = 5, dim(R) = 4, dim(J) = 3. G = SL(2, 7) and H = C_{3}. Then dim(M) = 8, dim(R) = 6, dim(J) = 4.

Here, M/R = S_{C_{3},k}(G), with dimension 2, but S_{C_{3},k_{−}}(G) = 0. There is one additional simple quotient R/J, of dimension 2, indexed by the group C_{7} ⋊ C_{3}. Moreover, J is the direct sum of two 2-dimensional simple modules, one indexed by A_{4}, and the other by C_{7} ⋊ C_{3}.

G = Ã_{5} (perfect group of order 120) and H = C_{3}. Then dim(M) = 4, dim(R) = 2, dim(J) = 1. The additional simple quotient R/J turns out to be indexed by the group A_{4}.

G = Ã_{5} and H = C_{4}. Then dim(M) = 5, dim(R) = 4, dim(J) = 2. Here, M/R = S_{C_{4},k}(G), with dimension 1, but S_{C_{4},k_{−}}(G) = 0. There are 2 additional simple quotients appearing in R/J, one indexed by the group C_{5} ⋊ C_{4}, and the other by the group C_{3} ⋊ C_{4}.

G = P SL(2, 11) and H = C_{3}. Then dim(M) = 4, dim(R) = 2, dim(J) = 1. The additional simple quotient R/J turns out to be indexed by the group A_{4}.

G = P SL(2, 11) and H = C_{5}. Then dim(M) = 6, dim(R) = 4, dim(J) = 2. Here, the evaluations S_{C_{5},V} (G) vanish for two of the kOut(C_{5})-modules V, while the other two appear in the quotient M/R.

G = P SL(2, 8) and H = C_{7}. Then dim(M) = 9, dim(R) = 6, dim(J) = 3. There are 3 additional simple quotients appearing in R/J, indexed by the group (C_{2})^{3} ⋊ C_{7} (the normalizer of a Sylow 2-subgroup of G).

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Serge Bouc, CNRS-LAMFA, Université de Picardie - Jules Verne, 33, rue St Leu, F-80039 Amiens Cedex 1, France.

serge.bouc@u-picardie.fr

Radu Stancu, CNRS-LAMFA, Université de Picardie - Jules Verne, 33, rue St Leu, F-80039 Amiens Cedex 1, France.

radu.stancu@u-picardie.fr

Jacques Thévenaz, Section de mathématiques, EPFL, Station 8, CH-1015 Lausanne, Switzerland.

Jacques.Thevenaz@epfl.ch