Non-additive exact functors and tensor induction for Mackey functors

Serge Bouc

Abstract. First I will introduce a generalization of the notion of (right)-exact functor between abelian categories to the case of non-additive functors. The main result of this section is an extension theorem: any functor defined on a suitable subcategory can be extended uniquely to a right exact functor defined on the whole category.

Next I use those results to define various functors of generalized tensor induction, associated to finite bisets, between categories attached to finite groups. This includes a definition of tensor induction for Mackey functors, for cohomological Mackey functors, for p-permutation modules and algebras. This also gives a single formalism of bisets for restriction, inflation, and ordinary tensor induction for modules.

Contents

1. Introduction 1 2. Non additive exact functors 3 3. Permutation Mackey functors 16 4. Tensor induction for Mackey functors 22 5. Relations with the functors L_{U} 30 6. Direct product of Mackey functors 35 7. Tensor induction for Green functors 44 8. Cohomological tensor induction 51 9. Tensor induction for p-permutation modules 65 10. Tensor induction for modules 70 References 75

1. Introduction

1 The theory of Mackey functors for a finite group G over a ring R provides a single framework for the various representation theories of G and its subgroups. So it looks like an extension of the notion of RG-module. The usual notions of

1991 Mathematics Subject Classification. Primary 18A22,19A22; Secundary 20J05. Key words and phrases. additive, exact, tensor induction, Burnside, Mackey functor, Green

functor.

1Received by the editor November 20, 1997, and in revised form November 16, 1998.

induction, restriction, inflation, ... for modules, have their analogues for Mackey functors. I will describe here a missing item in that list: tensor induction.

The first part of this paper is actually more general, and not specific to Mackey functors. It introduces an extension of the notion of right exact functor between abelian categories to non-additive functors: the usual definition of right exactness actually implies additivity, so it has to be modified in order to be extended. The main theorem of this general setting concerns the extension of a (non necessarily additive) functor from a suitable sub-category P of an abelian category A to an abelian category B, to a right exact functor from A to B.

Next I apply those results to various constructions of tensor induction. In all those cases, I will consider two finite groups G and H, a finite set U with a H × G^{op}-action (or H-set-G, or biset), and I will define a (generalized) tensor induction associated to U, which will be a functor between categories C(G) and C(H) naturally attached to G and H.

In the first case, the category C(G) is the category of Mackey functors for G. I will apply the extension theorem to the subcategory of “permutation functors”, and this leads to a generalized tensor induction functor T_{U} from Mackey functors for G to Mackey functors for H, associated to a finite biset U. This tensor induction behaves well with respect to composition of functors, tensor product of Mackey functors, and disjoint unions of bisets ^{2}. There is also a kind of binomial formula for the tensor induction of a direct sum.

Next I consider the relations between tensor induction and other functors between categories of Mackey functors, such as induction, restriction, inflation, ... I also define a reasonable notion of direct product of Mackey functors, and study its relations with tensor induction. Finally, I extend those notions to the case of Green functors.

The second case deals with the category C(G) of cohomological Mackey functors for G over a commutative ring R, and uses the subcategory of “permutation cohomological Mackey functors”. There is a generalized tensor induction functor associated to finite biset U, whenever U is “free enough” with respect to R. This cohomological tensor induction is closely related to the tensor induction for Mackey functors.

It leads to the definition of a generalized tensor induction for p-permutation modules and p-permutation algebras: this was the very starting point of that work, in a conversation with Jacques Thévenaz, who asked me about the possibility of such a generalized construction, giving a suitable functorial structure for the Dade group. In our joint recent preprint [BT98], we give an independent exposition of this generalized tensor induction for permutations algebras for p-groups, and use it to solve some open questions about the Dade group.

The third case is the case of the category C(G) of RG modules, using the subcategory of free RG-modules. This leads to a generalized tensor induction associated to a finite right-free biset U. The case U = H, when G is a subgroup of H, is the usual tensor induction from RG-modules to RH-modules. There is no essentially new construction here, since the other cases correspond to restriction and inflation of modules.

2The construction of a tensor induction for Mackey functors with those properties was a question of T. Yoshida at the Seattle AMS conference (Problem 37 in [ACPW98])

It should be noted however that even in those well known cases, the formalism of bisets gives a single natural framework involving restriction, inflation and tensor induction. The classical properties of those construction, such as Mackey formula, transitivity of tensor induction, or composition of inflation and tensor induction, are various aspects of a single simple composition formula. Similarly, the formula giving the tensor induction of a direct sum, which is generally incomplete (and evaluated only “up to terms induced from proper subgroups”), is nothing but a generalization of the binomial formula, and can be written explicitly.

2. Non additive exact functors

2.1. Notations. If M, N and P are objects of an abelian category A, and f : M ⊕ N → P is a morphism, I will denote f by (f ∘ i_{M}, f ∘ i_{N}), where i_{M} and i_{N} are the canonical injections from M and N to M ⊕ N. Similarly, if g : P → M ⊕ N is a morphism, I will denote it by \binom{s_{M} ∘g}{s_{N} ∘g}, where s_{M} and s_{N} are the canonical surjections from M ⊕ N to M and N. With those notations, the usual rules of matrix multiplication apply.

The identity morphism of M will generally be denoted by 1, and by Id_{M} if some precision is needed. The zero morphisms will be denoted by 0.

2.2. Definition. First I observe that the classical definition of an exact functor actually implies additivity:

Lemma 2.1. Let A and B be abelian categories, and F be a functor from A to B, which is not supposed to be additive. Suppose that for any exact sequence in A

M →^{φ} N →^{ψ} L → 0

the associated sequence

F (M) →^{F (φ)} F (N) →^{F (ψ)} F (L) → 0

is exact in B. Then F is additive.

Proof. Note that I don’t suppose that the second exact sequence is the image by F of the first one. In other words, I don’t suppose that F (0) = 0. But it is a consequence of the exactness of the second sequence: indeed, as F (ψ) ∘ F (ψ) = F (ψ ∘ ϕ) = F (0) has to be zero, and as the identity of the zero object factors through any (zero) morphism, the identity of F (0) has to be zero. Hence F maps the zero object to the zero object.

Let M and N be objects of A. Applying the hypothesis to the sequence

M \underrightarrow{\binom{1}{0}} M ⊕ N \underrightarrow{(0, 1)} N → 0

shows that the sequence

F (M) \underrightarrow{F \binom{1}{0}} F (M ⊕ N) \underrightarrow{F (0, 1)} F (N) → 0

is exact. But the morphism F \binom{1}{0} is a split monomorphism, and F (0, 1) is a split epimorphism. This proves that there are inverse isomorphisms

i_{M,N} = \binom{F (1, 0)}{F (0, 1)} : F (M ⊕ N) → F (M) ⊕ F (N)

j_{M,N} = (F \binom{1}{0}, F \binom{0}{1}) : F (M) ⊕ F (N) → F (M ⊕ N)

Now if f, g : M → N are morphisms in A, their sum f +g is obtained by composition

M \underrightarrow{\binom{1}{1}} M ⊕ M \underrightarrow{( \begin{matrix}f & 0 \\ 0 & g \end{matrix} )} N ⊕ N \underrightarrow{(1, 1)} N

Taking images by F gives the commutative diagram

F (M) \underrightarrow{F \binom{1}{1}} F (M ⊕ M) \underrightarrow{F ( \begin{matrix}f & 0 \\ 0 & g \end{matrix} )} F (N ⊕ N) \underrightarrow{F (1, 1)} F (N)
1 ↓ ↑ 1  i_{M,M} ↓ ↑ j_{M,M}  i_{N,N} ↓ ↑ j_{N,N}  1 ↓ ↑ 1
F (M) \underrightarrow{Δ} F (M) ⊕ F (M) \underrightarrow{φ} F (N) ⊕ F (N) \underrightarrow{Σ} F (N)

where the bottom row is obtained through the previous isomorphisms. Thus

Δ = i_{M,M} ∘ F \binom{1}{1} = \binom{F (1, 0)}{F (0, 1)}F \binom{1}{1} = \binom{F (1)}{F (1)} = \binom{1}{1}

and similarly Σ = (1, 1). Moreover

φ = \binom{F (1, 0)}{F (0, 1)}F ( \begin{matrix}f & 0 \\ 0 & g \end{matrix} ) (F \binom{1}{0}, F \binom{0}{1})

It follows that

φ = \binom{F (1, 0)}{F (0, 1)}(F \binom{f}{0}, F \binom{0}{g}) = ( \begin{matrix}F (f) & F (0) \\ F (0) & F (g) \end{matrix} ) = ( \begin{matrix}F (f) & 0 \\ 0 & F (g) \end{matrix} )

Now the composition Σ ∘ φ ∘ Δ is equal to F (f) + F (g). It is also equal to F (f + g), so F is additive.

I will modify the definition of right exactness to extend it to non-additive functors. First I need the following notation:

Notation 2.2. If φ : M → N is a morphism in A, I can build the morphisms (φ, 1) and (0, 1) from M ⊕ N to N. So I have morphisms F (φ, 1) and F (0, 1) in B from F (M ⊕ N) to F (N). I denote by ΔF (φ) their difference

ΔF (φ) = F (φ, 1) − F (0, 1) : F (M ⊕ N) → F (N)

If ψ : N → L is a morphism in A such that ψ ∘ φ = 0, then of course

F (ψ) ∘ ΔF (φ) = F (ψ ∘ (φ, 1)) − F (ψ ∘ (0, 1)) = 0

since ψ ∘ (φ, 1) = (ψ ∘ φ, ψ) = (0, ψ) = ψ ∘ (0, 1). This leads to the following definition:

Definition 2.3. Let F : A → B be a (non-necessarily additive) functor between abelian categories. I will say that F is right exact, if for any exact sequence

M →^{φ} N →^{ψ} L → 0 (2.1)

the associated sequence

F (M ⊕ N) \underrightarrow{ΔF (φ)} F (N) \underrightarrow{F (ψ)} F (L) → 0 (2.2)

is exact.

In particular, a right exact functor maps epimorphims to epimorphisms.

Remark 2.4. If F is additive, then ΔF (φ) = F (φ, 0), so the previous sequence factors as

F (M ⊕ N) \underrightarrow{F (1, 0)} F (M) \underrightarrow{F (φ)} F (N) \underrightarrow{F (ψ)} F (L) → 0

The left morphism is a split epimorphism. So F is right exact for the modified definition if and only if the sequence

F (M) \underrightarrow{F (φ)} F (N) \underrightarrow{F (ψ)} F (L) → 0

is exact, that is if and only if F is right exact in the usual sense. So the new definition is equivalent to the usual one for additive functors.

Remark 2.5. Let P be any object of B. Define F (M) = P for any object M of A, and F (φ) = Id_{P} for any map φ in A. Then F is a (trivial) example of a right exact functor, which is not additive if P is non-zero.

Remark 2.6. A functor F is exact if and only if the sequence (2.2) is exact for any short exact sequence

0 → M →^{φ} N →^{ψ} L → 0 (2.3)

Indeed this is obviously a necessary condition. Conversely, if the sequence (2.2) is exact for any short exact sequence (2.3), then in particular, the functor F maps epimorphisms to epimorphisms. Now if

M′ →^{φ′} N →^{ψ} L → 0

is an exact sequence, denoting by M the cokernel of φ′, then φ′ factors through M as φ′ = ι ∘ σ, where ι is a monomorphism and σ is an epimorphism. Now the

map ( \begin{matrix}σ & 0 \\ 0 & 1 \end{matrix} ) is an epimorphism from M′ ⊕ N to M ⊕ N, and so is F ( \begin{matrix}σ & 0 \\ 0 & 1 \end{matrix} ).

Moreover

ΔF (ι)∘F ( \begin{matrix}σ & 0 \\ 0 & 1 \end{matrix} ) = F ((ι, 1) ( \begin{matrix}σ & 0 \\ 0 & 1 \end{matrix} ) )−F ((0, 1) ( \begin{matrix}σ & 0 \\ 0 & 1 \end{matrix} ) ) = …

… = F (φ′, 1) − F (0, 1) = ΔF (φ′)

So if θ : F (N) → P is any map in B, then θ ∘ ΔF (ι) is zero if and only if θ ∘ ΔF (φ′)

is, since F ( \begin{matrix}σ & 0 \\ 0 & 1 \end{matrix} ) is an epimorphism. Thus ΔF (ι) and ΔF (φ′) have the same

image, and the sequence

F (M′ ⊕ N) \underrightarrow{ΔF (φ′)} F (N) \underrightarrow{F (ψ)} F (L) → 0

is also exact.

Remark 2.7. Suppose that the sequence

M →^{φ} N →^{ψ} L → 0

is exact and split, in the following sense: there exist a morphism α : N → M and a morphism β : L → N such that

ψ ∘ β = 1 φ ∘ α + β ∘ ψ = 1

(note that this will be the case in particular if it is exact, and if M, N and L are projective in A). Then the sequence

F (M ⊕ N) \underrightarrow{ΔF (φ)} F (N) \underrightarrow{F (ψ)} F (L) → 0

is also exact and split: indeed, set

A = −F \binom{−α}{1} : F (N) → F (M ⊕ N)

B = F (β) : F (L) → F (N)

Then it is clear that F (ψ) ∘ B = F (ψ ∘ β) = F (1) = 1, and that

ΔF (φ)∘A+B ∘F (ψ) = −(F (φ, 1)−F (0, 1))∘F \binom{−α}{1}+F (β ∘ψ) = …

… = −F (−φ ∘ α + 1) + F (1) + F (β ∘ ψ) = F (1) = 1

So the condition of the definition of a right exact functor is void on the split exact sequences. In particular, if every exact sequence in A is split, then every functor from A to an abelian category is right exact.

Remark 2.8. Let p = F (0, 1) : F (M ⊕ N) → F (N), and i = F \binom{0}{1} : F (N) →

F (M ⊕N). Then i∘p = F ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ) is an idempotent endomorphism of F (M ⊕N).

Moreover

F (φ, 1)(1 − F ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ) ) = F (φ, 1) − F (0, 1) = ΔF (φ)

So the functor F is right exact if and only if for any short exact sequence (2.1), the sequence

(1 − F ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ))F (M ⊕ N) \underrightarrow{F (φ, 1)} F (N) \underrightarrow{F (ψ)} F (L) → 0

is exact.

2.3. Basic properties.

2.3.1. Composition. The class of right exact functors is closed by composition:

Proposition 2.9. Let F : A → B and G : B → C be right exact functors between abelian categories. Then G ∘ F is right exact.

Proof. Let M →^{φ} N →^{ψ} L → 0 be an exact sequence in A. Since F is right exact, the sequence

F (M ⊕ N) \underrightarrow{ΔF (φ)} F (N) \underrightarrow{F (ψ)} F (L) → 0

is exact. And since G is right exact, the sequence

G(F (M ⊕ N) ⊕ F (N)) \underrightarrow{ΔG(ΔF (φ))} G ∘ F (N) \underrightarrow{G ∘ F (ψ)} G ∘ F (L) → 0

is exact. Moreover

ΔG(ΔF (φ)) = G(F (φ, 1) − F (0, 1), 1) − G(0, 1)

On the other hand, the functor G ∘ F is right exact if and only if the sequence

G ∘ F (M ⊕ N) \underrightarrow{Δ(G ∘ F)(φ)} G ∘ F (N) \underrightarrow{G ∘ F (ψ)} G ∘ F (L) → 0 (2.4)

is exact. Let

D = ΔG(ΔF (φ)) = G(F (φ, 1) − F (0, 1), 1) − G(0, 1)
D′ = Δ(G ∘ F)(φ) = G ∘ F (φ, 1) − G ∘ F (0, 1)

I will show that Im D = Im D′.

Let α be the morphism from F (M ⊕ N) to F (M ⊕ N) ⊕ F (N) defined by

α = \binom{1 − F ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} )}{F (0, 1)}

and let A = G(α). Let β be the morphism from F (M ⊕ N) ⊕ F (N) to F (M ⊕ N) defined by

β = (1 − F ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ) , F \binom{0}{1})

and let B = G(β). Then D′ ∘ B = G(v) − G(v′), where

v = F (φ, 1) ∘ β v′ = F (0, 1) ∘ β

Note that v′ is obtained from v by replacing φ by 0. But

v = (F (φ, 1)(1 − F ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} )), F (φ, 1)F \binom{0}{1}) = (F (φ, 1) − F (0, 1), 1)

So v′ = (0, 1), and

D′ ∘ B = G(F (φ, 1) − F (0, 1), 1) − G(0, 1) = D

Moreover B ∘ A = G(s), with

s = (1 − F ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ) , F \binom{0}{1})\binom{1 − F ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} )}{F (0, 1)} = …

… = 1 − F ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ) + F ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ) = 1

So B ∘ A = 1, and in particular B is an epimorphism. So if θ : G ∘ F (N) → P is any map in B, then

θ ∘ D = 0 ⇔ θ ∘ D′ ∘ B = 0 ⇔ θ ∘ D′ = 0

So D and D′ have the same image, and the sequence (2.4) is exact. This completes the proof of the proposition.

2.3.2. Products and sums. If A and A′ are abelian categories, then their product A × A′ is also abelian. If f : M → N and f′ : M′ → N′ are maps in A and A′, I will denote by [M, M′] and [N, N′] the associated couples in A × A′, and [f, f′] : [M, M′] → [N, N′] the associated morphism. The image of [M, M′] under a functor F will be denoted by F [M, M′] instead of F ([M, M′]).

Proposition 2.10. Let F : A → B and F′ : A′ → B′ be right exact functors between abelian categories. Then

F × F′ : A × A′ → B × B′

is right exact.

Proof. This is obvious, since a product of exact sequences is exact.

Corollary 2.11. Let F, F′ : A → B be right exact functors between abelian categories. Then F ⊕ F′ is right exact.

Proof. The functor F ⊕ F′ factors as

A →^{Δ} A × A \underrightarrow{F × F′} B × B →^{Σ} B

where Δ is the diagonal functor, mapping the object M to [M, M] and the morphism f to [f, f], and Σ is the direct sum functor mapping the object [P, Q] to P ⊕ Q, and the morphism [f, g] to f ⊕ g. Those two functors are obviously additive and exact, so the corollary follows from proposition 2.9.

2.3.3. Pairings. If A, A′, and B are abelian categories, a pairing F : A×A′ → B is just a biadditive functor: for any object M of A, the functor F [M, −] : A′ → B is additive, and for any object M′ of A′, the functor F [−, M′] : A → B is additive. Note that F itself is not additive in general.

Proposition 2.12. Let A, A′, and B be abelian categories, and F : A × A′ → B be a pairing. The following are equivalent:

1. The functor F is right exact.

2. For any objects M of A and M′ of A′, the (additive) functors F [M, −] and F [−, M′] are right exact.

3. For any exact sequences

M →^{φ} N →^{ψ} L → 0 M′ →^{φ′} N′ →^{ψ′} L′ → 0

the sequence

F [M, N′] ⊕ F [N, M′] \underrightarrow{(F [φ, 1], F [1, φ′])} F [N, N′] \underrightarrow{F [ψ, ψ′]} F [L, L′] → 0

is exact.

Proof. Suppose that F is right exact. Fix an object M′ of A′. Now the functor F [−, M′] : M ↦ F [M, M′] factors as

F [−, M′] = F ∘ (Id_{A} × c_{M′}) ∘ Δ

where Δ is the diagonal functor A → A × A as above, and c_{M′} is the constant functor, equal to M′ everywhere. So F [−, M′] is composed of three right exact functors, hence it is right exact by proposition 2.9. A similar argument shows that F [M, −] is right exact for any object M of A, so 1) implies 2) (note that this does not depend on the fact that F is a pairing).

Now if 2) holds, and if

M →^{φ} N →^{ψ} L → 0 M′ →^{φ′} N′ →^{ψ′} L′ → 0

are exact sequences, I have the following commutative diagram

F [M, M′] → F [M, N′] → F [M, L′] → 0
↓ ↓h ↓
F [N, M′] →^{g} F [N, N′] →^{b} F [N, L′] → 0
↓f ↓a ↓d
F [L, M′] →^{e} F [L, N′] →^{c} F [L, L′] → 0
↓  ↓  ↓
0  0  0

where a = F [ψ, 1], … h = F [φ, 1]. The rows and columns of this diagram are exact. To prove that 3) holds, I must show that the sequence

F [M, N′] ⊕ F [N, M′] \underrightarrow{(h, g)} F [N, N′] \underrightarrow{c ∘ a} F [L, L′] → 0

is exact. But c ∘ a is the product of two epimorphisms, so it is an epimorphism. And if θ : F [N, N′] → P is any map in B such that θ ∘ (h, g) = 0, then θ ∘ h = 0 and θ ∘ g = 0. As the middle column is exact, the map θ factors as θ = θ′ ∘ a. Now

0 = θ ∘ g = θ′ ∘ a ∘ g = θ′ ∘ e ∘ f

As f is an epimorphism, this gives θ′ ∘ e = 0, and as the bottom row is exact, the map θ′ factors as θ′ = θ” ∘ c. So θ = θ” ∘ c ∘ a, and the image of (h, g) is the kernel of c ∘ a. So 2) implies 3).

Now suppose 3) holds. Let

M →^{φ} N →^{ψ} L → 0 M′ →^{φ′} N′ →^{ψ′} L′ → 0

be exact sequences in A and A′. The sequence

U = F [M, N′] ⊕ F [N, M′] \underrightarrow{(F [φ, 1], F [1, φ′])} F [N, N′] \underrightarrow{F [ψ, ψ′]} F [L, L′] → 0

is exact, and to prove 1), I must show that the sequence

V = F [M ⊕ N, M′ ⊕ N′] \underrightarrow{ΔF [φ, φ′]} F [N, N′] \underrightarrow{F [ψ, ψ′]} F [L, L′] → 0

is exact. I will set D = (F [φ, 1], F [1, φ′]).

Let i : [M, N′] → [M ⊕ N, M′ ⊕ N′] be the map in A × A′ defined by i = [\binom{1}{0}, \binom{0}{1}]. Similarly, let j : [N, M′] → [M ⊕ N, M′ ⊕ N′] be the map defined by

j = [\binom{0}{1}, \binom{1}{0}]. Let A : U → V defined by A = (F (i), F (j)).

Define moreover k : [M ⊕ N, M′ ⊕ N′] → [M, N′] by k = [(1, 0), (φ′, 1)] and

l : [M ⊕ N, M′ ⊕ N′] → [N, M′] by l = [(0, 1), (1, 0)]. Let B : V → U defined by

B = \binom{F (k)}{F (l)}. Now

D ∘ B = (F [φ, 1], F [1, φ′]) ∘ \binom{F (k)}{F (l)} = …

… = F ([φ, 1][(1, 0), (φ′, 1)])+F ([1, φ′][(0, 1), (1, 0)]) = …
… = F ([(φ, 0), (φ′, 1)]) + F ([(0, 1), (φ′, 0)])

Since F is biadditive, this is also equal to

F ([(φ, 0), (φ′, 0)]) + F ([(φ, 0), (0, 1)]) + F ([(0, 1), (φ′, 0)])

On the other hand, the map D′ = ΔF [φ, φ′] is equal to

ΔF [φ, φ′] = F [(φ, 1), (φ′, 1)] − F [(0, 1), (0, 1)] = …

… = F ([(φ, 0), (φ′, 0)]) + F ([(φ, 0), (0, 1)]) + F ([(0, 1), (φ′, 0)])

Thus D ∘ B = D′. Moreover

B ∘ A = \binom{F (k)}{F (l)}(F (i), F (j)) = ( \begin{matrix}F (k ∘ i) & F (k ∘ j) \\ F (l ∘ i) & F (l ∘ j) \end{matrix} )

But

k ∘ i = [(1, 0), (φ′, 1)][\binom{1}{0}, \binom{0}{1}] = [1, 1]

So F (k ∘ i) = 1. Similarly

k ∘ j = [(1, 0), (φ′, 1)][\binom{0}{1}, \binom{1}{0}] = [0, φ′]

So F (k ∘ j) = 0, since F is biadditive. Moreover

l ∘ i = [(0, 1), (1, 0)][\binom{1}{0}, \binom{0}{1}] = [0, 0]

l ∘ j = [(0, 1), (1, 0)][\binom{0}{1}, \binom{1}{0}] = [1, 1]

Finally, I have

B ∘ A = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) = 1

In particular B is an epimorphism. As D ∘ B = D′, the images of D and D′ are the same. So 3) implies 1), and this completes the proof.

Corollary 2.13. Let [M, N] ↦ M ⊗ N be a (biadditive) tensor product from B ×B to an abelian category C, which is right exact with respect to M and N. Then

1. If F and F′ are right exact functors from A to B, so is F ⊗ F′ : A → C.

2. In particular if A = B = C, then for any positive integer n, define inductively the functor M ↦ M^{⊗n} by

M^{⊗1} = M M^{⊗n} = M^{⊗(n−1)} ⊗ M if n > 1

Then the functor M ↦ M^{⊗n} is right exact.

Proof. The functor F ⊗ F′ is the functor from A to C defined by the composition

A →^{Δ} A × A \underrightarrow{F × F′} B × B →^{⊗} B

So it is exact by proposition 2.9.

If moreover A = B = C, the functor M ↦ M^{⊗n} is right exact by an easy induction argument.

2.4. Extension of functors. The main result concerning right exact functors is the following:

Theorem 2.14. Let P be a full subcategory of an abelian category A with the following properties:

1. The objects of P are projective in A.

2. Any object of A is a quotient of an object of P.

3. If P and Q are objects of P, then so is P ⊕ Q. Then any functor F from P to an abelian category B can be uniquely extended (up to isomorphism of functors) to a right exact functor from A to B.

Proof. First uniqueness is almost obvious: let F_{1} and F_{2} be right exact functors from A to B, and θ be an isomorphism from the restriction of F_{1} to P to the restriction of P. In particular, for any object P of P, there is an isomorphism θ_{P} from F_{1}(P) to F_{2}(P).

Now uniqueness will follow from the following

Proposition 2.15. Let A be an abelian category, and P be a full subcategory of A satisfying the hypothesis of theorem 2.14. Let F_{1} and F_{2} be (non-necessarily additive) right exact functors from A to an abelian category B. Then if θ is a natural transformation from the restriction of F_{1} to P to the restriction of F_{2} to P, there exists a unique natural transformation θ̃ from F_{1} to F_{2} which coincides with θ on P.

Proof. For any object M of A, choose a short exact sequence

Q →^{φ} P →^{ψ} M → 0 (2.5)

with P and Q in P. Such a sequence exists by condition 2). Since F_{1} and F_{2} are right exact, the rows of the following commutative diagram are exact

F_{1}(Q ⊕ P) \underrightarrow{ΔF_{1}(φ)} F_{1}(P) \underrightarrow{F_{1}(ψ)} F_{1}(M) → 0
↓ θ_{Q⊕P}  ↓ θ_{P}
F_{2}(Q ⊕ P) \underrightarrow{ΔF_{2}(φ)} F_{2}(P) \underrightarrow{F_{2}(ψ)} F_{2}(M) → 0

Then there is a unique morphism θ̃_{M} : F_{1}(M) → F_{2}(M) completing this diagram into a commutative one. This morphism does not depend on the choice of the resolution 2.5: indeed, if

Q′ →^{φ′} P′ →^{ψ′} M → 0

is another resolution of M by objects of P, then as ψ′ is an epimorphism and as P is projective, there is a morphism a : P → P′ such that ψ′ ∘ a = ψ. Thus if θ̃′_{M} : F_{1}(M) → F_{2}(M) is built using the second resolution, I have

θ̃′_{M}F_{1}(ψ) = θ̃′_{M}F_{1}(ψ′)F_{1}(a) = F_{2}(ψ′)θ_{P′}F_{1}(a) = …

… = F_{2}(ψ′)F_{2}(a)θ_{P} = F_{2}(ψ)θ_{P} = θ̃_{M}F_{1}(ψ)

So θ̃_{M} = θ̃′_{M} since F_{1}(ψ) is an epimorphism. Moreover if M = P is already in P, I can choose the resolution P →^{0} P →^{1} M → 0 of M, and then

θ̃_{M} = θ̃_{M}F_{1}(1) = F_{2}(1)θ_{P} = θ_{P}

Now it is clear that the maps θ̃_{M} give a well defined natural transformation extending θ, and that this extension is unique.

To complete the proof of the theorem, I have to prove the existence of an extension F′ of F. For any object of A, I choose an exact sequence (2.5). Since F′ must be right exact, and coincide with F on P, the sequence

F (Q ⊕ P) \underrightarrow{ΔF (φ)} F (P) \underrightarrow{F′(ψ)} F′(M) → 0

must be exact. So I can define F′(M) as the cokernel of ΔF (φ). Of course, I must make this definition functorial with respect to M.

I will show that if f : M → M′ is a morphism in A, then there is a well defined morphism F′_{f} : F′(M) → F′(M′), which is moreover functorial with respect to f: this follows from standard arguments on projective resolutions.

If Q′ →^{φ′} P′ →^{ψ′} M′ → 0 is any exact sequence with P′ and Q′ in P, then there is a map a : P → P′ such that ψ′ ∘ a = f ∘ ψ, because P is projective and ψ′ is an epimorphism. Now ψ′ ∘ a ∘ φ = 0, so as Kerψ′ = Imφ′, and as Q is projective, there is a map b : Q → Q′ such that φ′ ∘ b = a ∘ φ.

Now I have the following diagram with exact rows

F (Q ⊕ P) \underrightarrow{ΔF (φ)} F (P) \underrightarrow{F′(ψ)} F′(M) → 0
F ( \begin{matrix}b & 0 \\ 0 & a \end{matrix} ) ↓  F (a) ↓
F (Q′ ⊕ P′) \underrightarrow{ΔF (φ′)} F (P′) \underrightarrow{F′(ψ′)} F′(M′) → 0

This diagram is commutative since

ΔF (φ′)F ( \begin{matrix}b & 0 \\ 0 & a \end{matrix} ) = (F (φ′, 1)−F (0, 1))F ( \begin{matrix}b & 0 \\ 0 & a \end{matrix} ) = F (φ′∘b, a)−F (0, a) = …

… = F (a∘φ, a)−F (0, a) = F (a)(F (φ, 1)−F (0, 1)) = F (a)∘ΔF (φ)

It follows that there is a morphism F′_{f} : F′(M) → F′(M′) such that F′_{f} ∘ F′(ψ) = F′(ψ′) ∘ F (a).

This morphism does not depend on the choice of a and b, since if a′ : P → P′ is another map such that ψ′ ∘ a′ = f ∘ ψ, then ψ′ ∘ (a − a′) = 0, so there is a map c : P → Q′ such that a − a′ = φ′ ∘ c. Now F \binom{c}{a′} is a map from F (P) to F (Q′ ⊕ P′), and

ΔF (φ′)F \binom{c}{a′} = (F (φ′, 1)−F (0, 1))F \binom{c}{a′} = F (φ′∘c+a′)−F (a′) = F (a)−F (a′)

This proves first that F′_{f} is well defined. So if F = 1, as I can take a = 1 and b = 1, I have F′_{1} = 1. A similar argument shows that F′_{f∘g} = F′_{f} ∘ F′_{g}, so the correspondence M ↦ F′(M) and f ↦ F′_{f} is a functor from A to B.

Moreover if P is in P, then I can choose the following exact sequence for P

P →^{0} P →^{1} P → 0

The associated sequence is

F (P ⊕ P) \underrightarrow{ΔF (0)} F (P) \underrightarrow{F′(1)} F′(P) → 0

As ΔF (0) = 0, I have F′(P) ≃ F (P). Moreover, if f : P → P′ is a morphism in P, then as the diagram

P →^{0} P →^{1} P → 0
f ↓ f ↓ f ↓
P′ →^{0} P′ →^{1} P′ → 0

is commutative, the following commutative diagram

F (P ⊕ P) \underrightarrow{ΔF (0)} F (P) \underrightarrow{1} F (P) → 0
F ( \begin{matrix}f & 0 \\ 0 & f \end{matrix} ) ↓  F (f) ↓  F′(f) ↓
F (P′ ⊕ P′) \underrightarrow{ΔF (0)} F (P′) \underrightarrow{1} F (P′) → 0

shows that F′(f) = F (f), and this induces an isomorphism between F and the restriction of F′ to P. Finally, the diagram

F (P)  \underrightarrow{0} F (P) \underrightarrow{1} F (P) → 0
F \binom{0}{1} ↓  1 ↓  F′_{ψ} ↓
F (Q ⊕ P) \underrightarrow{ΔF (φ)} F (P) \underrightarrow{F′(ψ)} F′(M) → 0

shows that F′_{ψ} = F′(ψ).

It remains to check that the functor F′ is right exact: denote by

Q_{M} →^{φ_{M}} P_{M} →^{ψ_{M}} M → 0

the chosen resolution by objects of P for the object M of A. Suppose moreover as before that if M = P is in P, then this sequence is

P →^{0} P →^{1} P → 0

so that F′(P) can be identified with F (P).

Now let M be any object of A, and consider first an exact sequence in A

Q →^{φ} P →^{ψ} M → 0

with P and Q in P. By the above arguments there are maps a : P → P_{M}, b : Q → Q_{M}, a′ : P_{M} → P and b′ : Q_{M} → Q and a commutative diagram

Q \underrightarrow{φ} P \underrightarrow{ψ} M → 0
b ↓ ↑ b′  a ↓ ↑ a′  1 ↓ ↑ 1
Q_{M}  \underrightarrow{φ_{M}} P_{M}  \underrightarrow{ψ_{M}} M → 0

As ψ ∘ a′ ∘ a = ψ_{M} ∘ a = ψ, there is a map c : P_{M} → Q such that 1 − a′ ∘ a = φ ∘ c. Similarly, there is a map c′ : P → Q_{M} such that 1 − a ∘ a′ = φ_{M} ∘ c′. Now there is a commutative diagram

\begin{aligned}F (Q ⊕ P) & \underrightarrow{ΔF (φ)} & F (P) & \underrightarrow{F′(ψ)} & F′(M) & → 0 \\ F ( \begin{matrix}b & 0 \\ 0 & a \end{matrix} ) ↓ & F (a) ↓ & & 1 ↓ \\ F (Q_{M} ⊕ P_{M}) & \underrightarrow{ΔF (φ_{M})} & F (P_{M}) & \underrightarrow{F′(ψ_{M})} & F′(M) & → 0 \end{aligned} (2.6)

in which the bottom line is exact by construction of F′(M). Now F \binom{c}{a′∘a} is a map from F (P_{M}) to F (Q ⊕ P), and

ΔF (φ) ∘ F \binom{c}{a′ ∘ a} = (F (φ, 1) − F (0, 1)) ∘ F \binom{c}{a′ ∘ a} = …

… = F (φ∘c+a′ ∘a)−F (a′ ∘a) = F (1)−F (a′)∘F (a) = 1−F (a′)∘F (a)

Similarly, the map F \binom{c′}{a∘a′} : F (P) → F (Q_{M} ⊕ P_{M}) is such that

ΔF (φ_{M}) ∘ F \binom{c′}{a ∘ a′} = 1 − F (a) ∘ F (a′)

This shows that F (a) and F (a′) induce mutual inverse isomorphisms between the cokernel of ΔF (φ) and the cokernel of ΔF (φ_{M}), equal to F′(M) by definition. In other words, the top line in 2.6 is exact.

In particular, if ψ : P → M is an epimorphism from an object of P to M, then F′(ψ) is also an epimorphism.

Now let

0 → M →^{φ} N →^{ψ} L → 0

be an arbitrary short exact sequence in A. It is well-known (see [Wei94] Horseshoe lemma 2.2.8) that it is possible to find a resolution

Q′_{N} →^{φ′_{N}} P′_{N} →^{ψ′_{N}} N → 0

by objects of P, such that there is a commutative diagram

0 → Q_{M} → Q′_{N} →^{γ} Q_{L} → 0
↓  φ′_{N} ↓  ↓
0 → P_{M} →^{α} P′_{N} →^{β} P_{L} → 0
↓ ψ′_{N} ↓ ↓
0 → M → N → L → 0

with exact rows and columns (note that this resolution needs not be equal to the prescribed resolution of N).

Then it is easy to check that the following diagram is commutative

F (Q′_{N} ⊕ P′_{N}) \underrightarrow{F ( \begin{matrix}γ & 0 \\ 0 & β \end{matrix} )} F (Q_{L} ⊕ P_{L}) → 0
ΔF (φ′_{N}) ↓  ΔF (φ_{L}) ↓
F (P_{M} ⊕ P′_{N}) \underrightarrow{ΔF (α)}  F (P′_{N})  \underrightarrow{F (β)}  F (P_{L}) → 0
F′ ( \begin{matrix}ψ_{M} & 0 \\ 0 & ψ′_{N} \end{matrix} ) ↓  F′(ψ′_{N}) ↓  F′(ψ_{L}) ↓
F′(M ⊕ N) \underrightarrow{ΔF′(φ)}  F′(N)  \underrightarrow{F′(ψ)}  F′(L)  → 0
↓  ↓  ↓
0  0  0

Moreover, its columns are exact by the above remarks, as well as its two top lines. Now the bottom line is also exact: the map F′(ψ) ∘ F′(ψ′_{N}) = F (ψ′_{L}) ∘ F (β) is an epimorphism, because F (ψ′_{L}) and F (β) are. Hence F′(ψ) is also an epimorphism. And if θ : F′(N) → B is any map in B such that θ ∘ ΔF′(φ) = 0, then

θ ∘ ΔF′(φ) ∘ F′ ( \begin{matrix}ψ_{M} & 0 \\ 0 & ψ′_{N} \end{matrix} ) = θ ∘ F′(ψ′_{N}) ∘ ΔF (α) = 0

As the middle row is exact, there is a map μ : F (P_{L}) → B such that

θ ∘ F′(ψ′_{N}) = μ ∘ F (β)

Now

0 = θ ∘ F′(ψ′_{N}) ∘ ΔF (φ′_{N}) = μ ∘ F (β) ∘ ΔF (φ′_{N}) = μ ∘ ΔF (φ_{L}) ∘ F ( \begin{matrix}γ & 0 \\ 0 & β \end{matrix} )

As the top line is exact, I have

μ ∘ ΔF (φ_{L}) = 0

and as the right column is exact, there is a map λ : F′(L) → B such that μ = λ ∘ F′(ψ_{L}). Thus

θ ∘ F′(ψ′_{N}) = λ ∘ F′(ψ_{L}) ∘ F (β) = λ ∘ F′(ψ) ∘ F′(ψ′_{N})

As F′(ψ′_{N}) is an epimorphism, I have θ = λ∘F′(ψ), so the bottom line of the above diagram is exact, and the functor F′ is right exact. This completes the proof of the theorem.

3. Permutation Mackey functors

3.1. Mackey functors. Let R be a commutative ring, and G be a finite group. I will use Dress definition of Mackey functors for G over R (see [Dre73]):

Definition 3.1. A Mackey functor M for G over R is a bivariant functor from the category G-set of finite G-sets to the category R-Mod of (left) R-modules, satisfying the following conditions:

• If X and Y are finite G-sets, and i_{X} and i_{Y} are the respective inclusions of X and Y into their disjoint union X ⊔ Y, then the following morphisms

M(X) ⊕ M(Y) \underrightarrow{(M_{∗}(i_{X}), M_{∗}(i_{Y}))} M(X ⊔ Y) \underrightarrow{\binom{M^{∗}(i_{X})}{M^{∗}(i_{Y})}} M(X) ⊕ M(Y)

are mutual inverse isomorphisms.

• If

X ⟶^{a} Y
b↓  ↓^{c}
Z ⟶_{d} T

is a cartesian square of finite G-sets, then M_{∗}(b)M^{∗}(a) = M^{∗}(d)M_{∗}(c)

I will mostly consider the case R = ℤ of Mackey functors over the integers, called simply Mackey functors. The Mackey functors for G, and natural transformation of bivariant functors, form an abelian category, denoted by Mack(G).

If M is a Mackey functor for G, and H is a subgroup of G, then M(G/H) is also denoted M(H). If K ⊆ H are subgroups of G, then p_{K}^{H} is the projection G/K → G/H defined by p_{K}^{H}(xK) = xH. The transfer t_{K}^{H} is the map M_{∗}(p_{K}^{H}), and the restriction r_{K}^{H} is the map M^{∗}(p_{K}^{H}).

3.2. Burnside functors. I will denote by b_{G} or b the Burnside Mackey functor for the group G. Its value b(H) for a subgroup H of G is just the Grothendieck group of finite H-sets, for relations given by decomposition in disjoint union.

More generally, its value b(X) on a finite G-set X is the Grothendieck group of the category G-set↓_{X} of finite sets over X (see [Bou97]). If f : X → Y is a morphism in G-set, then b_{∗}(f) : b(X) → b(Y) is defined by composition, and b^{∗}(f) : b(Y) → b(X) by pull-back.

If M is a Mackey functor for the group G, and X is a finite G-set, the Mackey functor M_{X} is defined (see [Web91], [TW95], [Bou97]) on a finite G-set Y by M_{X}(Y) = M(Y × X), and for a morphism f : Y → Z in G-set by M_{X,∗}(f) = M_{∗}(f × Id_{X}) and M_{X}^{∗}(f) = M^{∗}(f × Id_{X}). This construction is functorial with respect to X: if f : X → X′ is a morphism in G-set, then there are obvious morphisms M_{f} : M_{X} → M_{X′} and M^{f} : M_{X′} → M_{X}.

More generally, if X is any G-set (finite or not), and if X = ⊔_{ω∈G\X}ω is its decomposition in (finite) G-orbits, I will set

M_{X} = ⊕_{ω∈G\X} M_{ω}

For a finite G-set X, these two definitions of M_{X} coincide, in that there is a canonical isomorphism between them: if i_{ω} is the inclusion of ω into X, then the sequence (M_{i_{ω}})_{ω∈G\X} is an isomorphism from ⊕_{ω∈G\X} M_{ω} to M_{X}.

Note that with this definition, for any G-set X and any finite G-set Y, there is a natural isomorphism

⊕_{(y,x)∈[G\(Y ×X)]} M(G_{y,x}) → M_{X}(Y) = ⊕_{ω∈G\X} M(Y × ω) (3.7)

mapping the element v ∈ M(G_{y,x}) to the element M_{∗}(m_{y,x})(v) ∈ M(Y × Gx), where m_{y,x} is the map from G/G_{y,x} → Y × Gx given by

m_{y,x}(gG_{y,x}) = (gy, gx)

To avoid the choice of a system of representatives [G\(Y × X)], one can also view the left hand side module of (3.7) as

( ⊕_{(y,x)∈Y ×X} M(G_{y,x}))^{G}

Using this isomorphism, if ψ : Y → Y′ is a morphism of finite G-sets, then the map M_{X,∗}(ψ) becomes the map

( ⊕_{(y,x)∈Y ×X} M(G_{y,x}))^{G} → ( ⊕_{(y′,x)∈Y′×X} M(G_{y′,x}))^{G}

sending (v_{y,x})_{(y,x)∈Y ×X} to (v′_{y′,x})_{(y′,x)∈Y′×X} defined by

v′_{y′,x} = ∑_{y∈G_{y′,x}\ψ^{−1}(y′)} t_{G_{y,x}}^{G_{y′,x}}m_{y,x}

Similarly, the map M_{X}^{∗}(ψ) gives the map

( ⊕_{(y′,x)∈Y′×X} M(G_{y′,x}))^{G} → ( ⊕_{(y,x)∈Y ×X} M(G_{y,x}))^{G}

sending (v′_{y′,x})_{(y′,x)∈Y′×X} to (v_{y,x})_{(y,x)∈Y ×X} defined by

v_{y,x} = r_{G_{y,x}}^{G_{ψ(y),x}}v′_{ψ(y),x}

3.3. Permutation functors.

Definition 3.2. A permutation Mackey functor is a Mackey functor isomorphic to b_{X}, for some G-set X. I denote by P Mack(G) the full subcategory of Mack(G) formed by permutation Mackey functors.

Note that if X is any G-set, and Y is a finite G-set, then b_{X}(Y) is the Grothendieck group of the category of finite G-sets over Y × X.

For a finite G-set X, and any Mackey functor M, it is easy to check (see [Bou97]) that

Hom_{Mack(G)}(b_{X}, M) ≃ M(X) (3.8)

This isomorphism is as follows: if m ∈ M(X), then for any finite G-set Y, the map associated to m from b_{X}(Y) = b(Y × X) → M(Y) is defined by

[figure: Y; X; Z; a; b; ↦ M_{∗}(a)M^{∗}(b)(m) ∈ M(Y)]

In particular, the functor b_{X} is projective. Hence for any G-set X (finite or not), the functor b_{X} is projective, as a direct sum of projectives. If M is any Mackey functor, then

Hom_{Mack(G)}(b_{X}, M) ≃ ∏_{ω∈G\X} M(ω)

3.4. Lefschetz invariants. Let X be a G-set (finite or not). There is another possible interpretation of b_{X}, using Lefschetz invariants. If Y is a finite G-poset, recall (see [Thé87]) that the Lefschetz invariant of Y is the element of b(G) defined by

Λ_{Y}^{G} = − ∑_{s∈G\Sd(Y)}(−1)^{|s|}G/G_{s}

where Sd(Y) is the set of totally ordered non-empty subsets of Y, and |s| is the cardinality of s and G_{s} its stabilizer in G. If H is a subgroup of G, then the (algebraic) number of fixed points of H on Λ_{Y}^{G} is the Euler-Poincaré characteristic of the set of fixed points of H on Y

|(Λ_{Y}^{G})^{H}| = χ(Y^{H})

So two G-posets Y and Z have the same Lefschetz invariant if and only if for any subgroup H of G, the Euler-Poincaré characteristic χ(Y^{H}) and χ(Z^{H}) are equal.

Any G-set X can be viewed as a G-poset with the discrete ordering (x ≤ y ⇔ x = y): when no other poset structure is given, the discrete one will be understood. Similarly, all maps between G-posets will be G-equivariant maps, compatible with the (given or understood) poset structures.

Definition 3.3. If X and Y are G-sets, and (Δ, f) is a (finite or not) G-poset over Y × X, I denote by f_{Y} (resp. f_{X}) the composition of f with the projection onto Y (resp. onto X).

I denote by G-poset↓_{Y,X} the category of G-posets (Z, f) over Y × X such that for any y ∈ Y, the fibre f_{Y}^{−1}(y) is finite. Such a poset is said to have finite fibres over Y. Note that this implies in particular that f^{−1}(y, x) is finite for any (y, x) ∈ Y × X, and that Z is finite if Y is.

I say that two objects (Δ, f) and (Δ′, f′) of G-poset↓_{Y,X} are equivalent (notation (Δ, f) ∼ (Δ′, f′)), if

∀(y, x) ∈ Y × X, Λ_{f^{−1}(y,x)}^{G_{y,x}} = Λ_{f′^{−1}(y,x)}^{G_{y,x}}

I denote by h_{G}(Y, X) the set of equivalence classes of objects of G-poset↓_{Y,X}, modulo this equivalence relation.

Lemma 3.4. Let X be a G-set

1. Let Y be a finite G-set. Then the correspondence which maps the finite G-poset (Z, f) over Y × X to

(Λ_{f^{−1}(y,x)}^{G_{y,x}})_{(y,x)∈Y ×X} ∈ ( ⊕_{(y,x)∈Y ×X} b(G_{y,x}))^{G} ≃ b_{X}(Y)

induces a one to one correspondence θ_{Y,X} between h_{G}(Y, X) and b_{X}(Y).

2. If (Z, f) and (Z′, f′) are finite G-posets over Y × X, then

θ_{Y,X}(Z ⊔ Z′, f ⊔ f′) = θ_{Y,X}(Z, f) + θ_{Y,X}(Z′, f′)

3. If ψ : Y → Y′ is a morphism of finite G-sets, and (Z, f) is a finite G-poset over Y × X, then

b_{X,∗}(ψ)θ_{Y,X}(Z, f) = θ_{Y′,X}(Z, (ψ × Id_{X}) ∘ f)

If (Z′, f′) is a finite G-poset over Y′ × X, let (Z, f_{Y}) be the G-set defined by the pull-back diagram

Z ⟶^{a} Z′
f_{Y} ↓  ↓^{f′_{Y ′}}
Y ⟶_{ψ} Y′

viewed as a sub-G-poset of Y ×Z′. Then setting f = (f_{Y} , f′_{X} ∘a) : Z → Y ×X turns (Z, f) into a finite G-poset over Y × X, and

b_{X}^{∗}θ_{Y′,X}(Z′, f′) = θ_{Y,X}(Z, f)

Proof. The correspondence θ is clearly well-defined and injective, by the very definition of equivalence of posets over Y ×X, since it is clear that for (y, x) ∈ Y ×X and g ∈ G, I have

Λ_{f^{−1}(gy,gx)}^{G_{gy,gx}} = ^{g}(Λ_{f^{−1}(y,x)}^{G_{y,x}})

Conversely, if (β_{y,x}) ∈ ( ⊕_{(y,x)∈(Y ×X)} b(G_{y,x}))^{G}, then for each (y, x) ∈ [G\(Y ×X)], it is possible to find a finite G_{y,x}-poset Z_{y,x} with Lefschetz invariant equal to β_{y,x} (see Lemme 2 of [Bou92]). Of course, if β_{y,x} = 0, I take Z_{y,x} = ∅. Now define

Z = ⊔_{(y,x)∈[G\(Y ×X)]}Ind_{G_{y,x}}^{G}Z_{y,x}

This is a finite G-poset. Let f : Z → Y × X mapping the element (g, z) of Ind_{G_{y,x}}^{G}Z_{y,x} to (gy, gx). Then (Z, f) is a finite G-poset over Y × X, and moreover for (y, x) ∈ [G\(Y × X)], it is clear that

f^{−1}(y, x) = {1} × Z_{y,x} ⊆ Ind_{G_{y,x}}^{G}Z_{y,x}

It is isomorphic to Z_{y,x}, so its Lefschetz invariant is β_{y,x}, and the first assertion follows.

The second assertion is clear, since the Lefschetz invariant of a disjoint union E ⊔ E′ is the sum of the Lefschetz invariant of E and E′.

The third assertion follows from the fact that

(ψ ∘ f_{Y} , f_{X})^{−1}(y′, x) = ⊔_{y∈ψ^{−1}(y′)}(f_{Y} , f_{X})^{−1}(y, x)

so that

Λ_{f′^{−1}(y′,x)}^{G_{y′,x}} = ∑_{y∈G_{y′,x}\ψ^{−1}(y′)} Ind_{G_{y,x}}^{G_{y′,x}}Λ_{f^{−1}(y,x)}^{G_{y,x}}

Moreover the transfer for the Burnside functor is just induction.

Similarly, if (Z′, f′) is a finite G-poset over Y′, let Z be the pull-back of Y and Z′ over Y′:

[figure: Y; Y′; X; Z′; Z; f′_{Y′}; f′_{X}; ψ]

Let f be the map from Z to Y × X defined by this diagram. Now if (y, x) ∈ Y × X, then

f^{−1}(y, x) = f′^{−1}(ψ(y), x)

so

Λ_{f^{−1}(y,x)}^{G_{y,x}} = Res_{G_{y,x}}^{G_{ψ(y),x}}Λ_{f′^{−1}(ψ(y),x)}^{G_{ψ(y),x}}

This completes the proof of the lemma.

Now I have a nice interpretation of b_{X}, and even a little more: observe that in the case of two finite G-sets X and Y, isomorphism (3.8) gives

Hom_{Mack(G)}(b_{Y} , b_{X}) ≃ b(Y × X)

(see also [Bou97]). When Y is infinite, this is no longer true. The correct formulation is the following:

Proposition 3.5. Let G be a finite group, and X, Y be any G-sets. For any object (Δ, f) of G-poset↓_{Y,X} and any finite G-set Z, let b_{Δ,f,Z} be the map sending the finite G-poset (T, g) over Z × Y to the G-poset (U, h) over Z × X defined by the pullback-diagram

[figure: Z; Y; X; T; Δ; U; g_{Z}; f_{Y}; g_{Y}; f_{X}]

Then the map b_{Δ,f,Z} passes down to equivalence classes of finite G-posets over Z × Y, and this defines a morphism of Mackey functors b_{Δ,f} from b_{Y} to b_{X}. Moreover this induces a one to one correspondence

θ_{Y,X} : h_{G}(Y, X) ≃ Hom_{Mack(G)}(b_{Y} , b_{X})

such that θ_{Y,X}(Δ ⊔ Δ′, f ⊔ f′) = θ_{Y,X}(Δ, f) + θ_{Y,X}(Δ′, f′).

Proof. First it is clear that the pull-back U is finite, since g_{Y} (T) is, and f has finite fibres over Y. It is also clear that b_{Δ,f,Z} passes down to equivalence classes of finite G-posets over Z × Y: with the notation of the proposition, if (z, x) ∈ Z × X, then

h^{−1}(z, x) = ⊔_{y∈Y} g^{−1}(z, y) × f^{−1}(y, x)

So

Λ_{h^{−1}(z,x)}^{G_{z,x}} = ∑_{y∈G_{z,x}\Y} Ind_{G_{z,y,x}}^{G_{z,x}}(Res_{G_{z,y,x}}^{G_{z,y}}Λ_{g^{−1}(z,y)}^{G_{z,y}})(Res_{G_{z,y,x}}^{G_{y,x}}Λ_{f^{−1}(y,x)}^{G_{y,x}})

So this only depends on the equivalence class of (T, g) and (Δ, f).

Moreover, the maps b_{Δ,f,Z} define a morphism of Mackey functors: if ψ : Z → Z′ is a morphism of finite G-sets, an element of b_{Y} (Z) can be represented by the equivalence class of a finite G-poset (T, g) over Z × Y. Then b_{Y,∗}(ψ)(T, g) is the class of (T, g′), where g′ = (ψ × Id_{Y}) ∘ g. But b_{Δ,f,Z}(T, g) = (U, h), and

b_{X,∗}(ψ)(U, h) = (U, (ψ × Id_{X}) ∘ h)

On the other hand

b_{Δ,f,Z′}(T, (ψ × Id_{Y}) ∘ g) = (U, h′)

where h′ = (ψ × Id_{X}) ∘ h. This proves that

b_{X,∗} ∘ b_{Δ,f,Z} = b_{Δ,f,Z′} ∘ b_{Y,∗}

Similarly, if now ψ : Z′ → Z is a morphism of finite G-sets, then

b_{Y}^{∗} ∘ b_{Δ,f,Z′} = b_{Δ,f,Z} ∘ b_{X}^{∗}

This is because in the following diagram

[figure: Z′; T_{1}; T_{2}; Z; Y; X; T; Δ; U; g_{Z}; f_{Y}; g_{Y}; f_{X}]

if (T, g) is a finite G-poset over Z × X, if T_{1} is the pull-back of Z′ and T over Z, and T_{2} the pull-back of T_{1} and U over T, then T_{2} is also the pull-back of Z′ and U over Z (this last pull-back involves the composition of two cartesian squares, which is cartesian).

The last assertion of the proposition follows from the fact that if Y is a finite G-set, then by (3.8)

Hom_{Mack(G)}(b_{Y} , b_{X}) = ⊕_{ω∈G\X} Hom_{Mack(G)}(b_{Y} , b_{ω}) ≃ ⊕_{ω∈G\X} b(Y × ω) ≃ b_{X}(Y)

Thus if Y is any G-set

Hom_{Mack(G)}(b_{Y} , b_{X}) = ∏_{ω′∈G\Y} b_{X}(ω′) ≃ ∏_{ω′∈G\Y} h_{G}(ω′, X)

Now a sequence (T_{ω′}) ∈ ∏_{ω′∈G\Y} h_{G}(ω′, X) defines a G-poset T = ⊔_{ω′∈Y \Y} T_{ω′}, with finite fibres over Y. The equivalence class of T depends only on the sequence (T_{ω′}). Conversely, the equivalence class of the G-poset (T, f) over Y × X, with finite fibres over Y, defines the sequence T_{ω′} = f^{−1}(ω′ × X). This completes the proof of the proposition.

Corollary 3.6. Let X, Y and Z be G-sets. If f : b_{Y} → b_{X} is the morphism of Mackey functors defined by the class of the object (Δ, f) of G-poset↓_{Y,X}, and g : b_{Z} → b_{Y} is the morphism defined by the class of the object (Δ′, f′) of G-poset↓_{Z,Y}, then the morphism f ∘ g : b_{Z} → b_{X} is defined by the class of the object Δ” of G-poset↓_{Z,X} defined by the pull-back diagram

[figure: Z; Y; X; Δ′; Δ; Δ”; f′_{Z}; f_{Y}; f′_{Y}; f_{X}]

Proof. This is clear from the proposition. Note that Δ” has finite fibres over Z, if Δ has finite fibres over Y and Δ′ has finite fibres over Z.

3.5. Resolutions. The subcategory of permutation Mackey functors is “big enough”:

Proposition 3.7. For any Mackey functor M for the group G, there exists a G-set X and an epimorphism φ : b_{X} → M.

Proof. Let E be a set of subgroups of G, and for each H ∈ E, let S_{H} be a non-empty subset of M(H). Let

X = ⊔_{H∈E}(G/H) × S_{H}

where G acts trivially on S_{H}. There is a natural morphism θ from b_{X} to M: indeed the orbits of G on X are the sets (G/H) × {s}, for H ∈ E and s ∈ S_{H}. This gives

Hom_{Mack(G)}(b_{X}, M) ≃ ∏_{{H∈E \atop s∈S_{H}}} M(G/H)

The right hand side contains the element (s)_{H∈E,s∈S_{H}}, which corresponds to a morphism θ from b_{X} to M. The element s ∈ M(G/H) defines the morphism from b_{G/H} to M mapping the (discrete) set (T, f) over Z × (G/H) to the element M_{∗}(f_{Z})M^{∗}(f_{G/H})(s). In particular, if Z = T = G/H, and f is the diagonal inclusion, then this element is just s itself. In other words, if I = Im θ, then I(H) contains S_{H}, so it contains the sub-Mackey functor J of M generated by the union of the S_{H}, for H ∈ E (this is the intersection of all the subfunctors N of M such that N(H) ⊇ S_{H} for all H ∈ E (see [TW90] Proposition 2.1)).

The image I of θ is actually equal to J: indeed, for any subgroup K of G, the value of J at K is

J(K) =< t_{L}^{K}(^{x}r_{K^{x}}^{H}s) | H ∈ E, s ∈ S_{H}, x ∈ G, L ⊆ K ∩^{x}H >

But the element t_{L}^{K}(^{x}r_{K^{x}}^{H}s) is the image under θ_{K} of the element (G/L, f) of b_{G/H}(K), where f(gL) = (gK, gxH).

This proves in particular that the morphism θ is an epimorphism if and only if the set ⊔_{H∈E}S_{H} generates M as a Mackey functor. The proposition follows, taking for E the set of all subgroups of G, and S_{H} = M(H) for all H.

4. Tensor induction for Mackey functors

4.1. Bisets. Let G and H be finite groups. A set U is called a biset (more precisely an H-set-G) if the group H × G^{op} acts on U. Equivalently, the group H acts on the left on U, and the group G acts on the right, and those two actions commute.

If U is an H-set-G, I denote by U^{op} the G-set-H which is equal to U as a set, and double action is given for g ∈ G, h ∈ H, and u ∈ U by

g.u.h (in U^{op}) = h^{−1}ug^{−1} (in U)

So let U be a finite H-set-G, and Z be a G-set. Let Hom_{G}(U^{op}, Z) be the set of morphism of G-sets from U^{op} to Z. This is an H-set: if h ∈ H and φ ∈ Hom_{G}(U^{op}, Z), then hφ is the morphism of G-sets from U^{op} to Z defined by

(hφ)(u) = φ(h^{−1}u) ∀u ∈ U, h ∈ H

If now Z is a G-poset, then Hom_{G}(U^{op}, Z) is an H-poset:

φ ≤ ψ in Hom_{G}(U^{op}, Z) ⇔ ∀u ∈ U, φ(u) ≤ ψ(u) in Z

Proposition 4.1. The correspondence X ↦ Hom_{G}(U^{op}, X) induces a functor T_{U} from P Mack(G) to P Mack(H), mapping the object b_{X} to b_{Hom_{G}(U^{op},X)}, and the morphism φ : b_{Y} → b_{X} defined by the class of the object (Δ, f) of G-poset↓_{Y,X} to the morphism defined by the class of the H-poset

T_{U}(Δ, f) = (Hom_{G}(U^{op}, Δ), Hom_{G}(U^{op}, f))

over

Hom_{G}(U^{op}, Y × X) ≃ Hom_{G}(U^{op}, Y) × Hom_{G}(U^{op}, X)

Proof. First I must check that if (Δ, f) is a G-poset over Y × X, with finite fibres over Y, then

(D, F) = T_{U}(Δ, f) = (Hom_{G}(U^{op}, Δ), Hom_{G}(U^{op}, f))

is an H-set over Hom_{G}(U^{op}, Y × X), identified with

Hom_{G}(U^{op}, Y) × Hom_{G}(U^{op}, X)

which has finite fibres over Hom_{G}(U^{op}, Y). Fix φ ∈ Hom_{G}(U^{op}, Y), and look for α ∈ Hom_{G}(U^{op}, Δ) such that f_{Y} ∘ α = φ: then for all u ∈ U, the element α(u) has to be in f_{Y}^{−1}φ(u), which is a finite set by hypothesis. As U is finite, there is only a finite number of possibles choices, hence a finite number of possibles α’s.

Next I must show that the class of T_{U}(Δ, f) depends only on the class of (Δ, f). Fix φ ∈ Hom_{G}(U^{op}, Y) and ψ ∈ Hom_{G}(U^{op}, X), and a subgroup K of H_{φ,ψ}. Then an element α of Hom_{G}(U^{op}, Δ)^{K} such that f ∘α = (φ, ψ) is defined by the following conditions

∀u ∈ U, α(u) ∈ f^{−1}(φ(u), ψ(u)) ∀u ∈ U, ∀g ∈ G, ∀k ∈ K, α(kug^{−1}) = gα(u)

So α defines a sequence of elements α(u) in f^{−1}(φ(u), ψ(u)), for u in a system [K\U/G] of representatives of orbits of K × G^{op} on U. The element α(u) must be invariant by the subgroup of G defined by

G_{K,u} = {g ∈ G | ∃k ∈ K, kug^{−1} = u}

Conversely, if I choose elements a_{u} ∈ (f^{−1}(φ(u), ψ(u)))^{G_{K,u}}, for u ∈ [K\U/G],

then I can define α(v) for any v ∈ U by setting α(v) = ga_{u}, if v = kug^{−1}, with u ∈ [K\U/G], g ∈ G, and k ∈ K.

In other words, there is a bijection

(F^{−1}(φ, ψ))^{K} ≃ ∏_{u∈[K\U/G]} (f^{−1}(φ(u), ψ(u)))^{G_{K,u}}

This is clearly an isomorphism of posets, so

χ ((F^{−1}(φ, ψ))^{K}) = ∏_{u∈[K\U/G]} χ ((f^{−1}(φ(u), ψ(u)))^{G_{K,u}})

In particular, this depends only on the Lefschetz invariants Λ_{f^{−1}(φ(u),ψ(u))}^{G_{φ(u),ψ(u)}} (note that G_{K,u} is a subgroup of G_{φ(u),ψ(u)} if K is a subgroup of H_{φ,ψ}). Since these Euler-characteristics define the Lefschetz invariant Λ_{F^{−1}(φ,ψ)}^{H_{φ,ψ}}, the equivalence class of (D, F) depends only on the class of (Δ, f).

Finally I have to check functoriality: first the identity morphism of b_{X} is associated to the class (X, d_{X}) of the (discrete) set X over X × X, where d_{X}(x) = (x, x). Clearly

T_{U}(X, d_{X}) = (Hom_{G}(U^{op}, X), d_{Hom_{G}(U^{op},X)})

so T_{U} maps the identity to the identity. Moreover, if

[figure: Z; Y; X; Δ′; Δ; Δ”; f′_{Z}; f_{Y}; f′_{Y}; f_{X}]

is a composition of morphisms b_{Z} → b_{Y} → b_{X}, then taking images by the functor Hom_{G}(U^{op}, −) gives the diagram

[figure: Hom_{G}(U^{op}, Z); Hom_{G}(U^{op}, Y); Hom_{G}(U^{op}, X); Hom_{G}(U^{op}, Δ′); Hom_{G}(U^{op}, Δ); Hom_{G}(U^{op}, Δ”)]

and the middle square in this diagram is cartesian, by definition. This proves that T_{U} commutes with composition of morphisms, and this completes the proof of the proposition.

4.2. Tensor induction. Now I can give the definition of tensor induction for Mackey functors:

Definition 4.2. Let G and H be finite groups, and U be a finite H-set-G. I call tensor induction associated to U the right exact functor from Mack(G) to Mack(H) extending the functor T_{U} : P Mack(G) → Mack(H). This extension is again denoted by T_{U}.

Recall that T_{U} is constructed as follows: for a Mackey functor M for G, choose an exact sequence

b_{Y} \underrightarrow{φ} b_{X} \underrightarrow{ψ} M → 0

Then T_{U}(M) is defined as the cokernel of the map ΔT_{U}(φ), so the following sequence is exact

T_{U}(b_{Y} ⊕ b_{X}) \underrightarrow{ΔT_{U}(φ)} T_{U}(b_{X}) \underrightarrow{T_{U}(ψ)} T_{U}(M) → 0

Moreover

T_{U}(b_{Y} ⊕ b_{X}) = T_{U}(b_{Y ⊔X}) = b_{Hom_{G}(U^{op},Y ⊔X)}

Suppose that φ is defined by the G-poset (D, f) over Y × X. Then ΔT_{U}(D, f) is the morphism from b_{Hom_{G}(U^{op},Y ⊔X)} to b_{Hom_{G}(U^{op},X)} defined by the difference

[figure: Hom_{G}(U^{op}, Y ⊔ X); Hom_{G}(U^{op}, X); Hom_{G}(U^{op}, D ⊔ X); −; Hom_{G}(U^{op}, Y ⊔ X); Hom_{G}(U^{op}, X); Hom_{G}(U^{op}, ∅ ⊔ X)]

The left hand term is the image of the poset

[figure: Y ⊔ X; X; D ⊔ X; a; b]

where a is f_{Y} ⊔Id_{X}, and b is f_{X} on D and identity on X. The other one is obtained from it by replacing D by ∅.

Define Hom_{G}(U^{op}, D; X) as the set of G-morphisms α from U to D ⊔ X such that Im α ⊈ X. Then T_{U}(M) is also the cokernel of the map

b_{Hom_{G}(U,Y;X)} → b_{Hom_{G}(U,X)}

defined by the following poset

[figure: Hom_{G}(U^{op}, Y; X); Hom_{G}(U^{op}, X); Hom_{G}(U^{op}, D; X)]

4.2.1. Examples. The case U = ∅. Let U = ∅. Then for any G-set X, the H-set Hom_{G}(U, X) is a one element set. It follows that the functor T_{∅} is the constant functor, equal to b. The case H = G = U. Suppose H = G, and that U is the set G, with double action given by left and right multiplication. Then for any G-set X

Hom_{G}(G, X) ≃ X

It follows that the functor T_{G} is the identity functor. Projectivity. As the functor T_{U} maps permutation functors to permutation functors, it follows that it maps projectives to projectives: if P is any projective Mackey functor for G, then P is a direct summand of some permutation functor b_{X}. Then T_{U}(P) is a direct summand of T_{U}(b_{X}) = b_{Hom_{G}(U^{op},X)}, hence it is projective.

4.3. Composition. The tensor induction T_{U} is functorial with respect to U:

Proposition 4.3. Let G, H and K be finite groups. If U is a finite H-set-G and V is a finite K-set-H, then there is an isomorphism of functors

T_{V} ∘ T_{U} ≃ T_{V ×_{H}U}

Proof. Recall that V ×_{H} U is the quotient V × U, viewed as a K-set-G, by the action of H given by (v, u)h = (vh, h^{−1}u).

Now observe that T_{V ×_{H}U} is right exact by definition, and that T_{V} ∘ T_{U} is right exact by proposition 2.9. Moreover, for any G-set X

(4.9) (T_{V} ∘ T_{U})(b_{X}) = T_{V} (b_{Hom_{G}(U^{op},X)}) = b_{Hom_{H}(V^{op},Hom_{G}(U^{op},X))}

But there is an adjunction

Hom_{H}(V^{op}, Hom_{G}(U^{op}, X)) ≃ Hom_{G}(U^{op} ×_{H} V^{op}, X)

and moreover U^{op}×_{H} V^{op} ≃ (V ×_{H} U)^{op}. Equation 4.9 is easily seen to be functorial with respect to X: in other words, the restrictions of T_{V ×_{H}U} and T_{V} ∘ T_{U} to P Mack(G) are isomorphic. As they are both right exact, proposition follows from theorem 2.14.

4.4. Tensor product. Recall from [Bou97] chapter 1 that if M and N are Mackey functors, then their tensor product is the Mackey functor M ˆ⊗ N defined on the finite G-set X by

(M ˆ⊗ N)(X) = ( ⊕_{Y →^{φ}X} M(Y) ⊗ N(Y))/J

where (Y, φ) is a (finite) G-set over X, and J is the submodule generated by the elements

M_{∗}(f)(m) ⊗ n′ − m ⊗ N^{∗}(f)(n′) ∀m ∈ M(Y), n′ ∈ N(Y′)

and the elements

M^{∗}(f)(m′) ⊗ n − m ⊗ N_{∗}(f)(n) ∀m′ ∈ M(Y′), n ∈ N(Y)

whenever f : (Y, φ) → (Y′, φ′) is a morphism of G-sets over X, which means that φ′ ∘ f = φ. If m ∈ M(Y) and n ∈ N(Y) for a G-set (Y, φ) over X, I denote by [m ⊗ n]_{(Y,φ)} the image of m ⊗ n in (M ˆ⊗ N)(X).

If θ : X → X′ is a morphism of G-sets, then the image of [m ⊗ n]_{(Y,φ)} by (M ˆ⊗ N)_{∗}(θ) is [m ⊗ n]_{(Y,θφ)}. If (Y′, φ′) is a G-set over X′, if m′ ∈ M(Y′) and n′ ∈ N(Y′), then the image of [m′ ⊗ n′]_{(Y′,φ′)} by (M ˆ⊗ N)^{∗}(f) is equal to [M^{∗}(a)(m) ⊗ N^{∗}(a)(n)]_{(Y,φ)}, where (Y, φ) is the pull-back of (Y′, φ′) along θ

Y ⟶^{a} Y′
φ↓  ↓^{φ′}
X ⟶_{θ} X′

Also recall from [Bou97] Lemma 7.2.3 that if X and Y are finite G-sets, there are natural isomorphisms

(M ˆ⊗ N)_{X×Y} ≃ M_{X} ˆ⊗ N_{Y}

Now this isomorphism clearly holds also for infinite sets X and Y, because

M_{X} ˆ⊗ N_{Y} ≃ ( ⊕_{ω∈G\X} M_{ω}) ˆ⊗( ⊕_{ω′∈G\Y} N_{ω′}) ≃ …

… ≃ ⊕_{{ω∈G\X \atop ω′∈G\Y}} M_{ω} ˆ⊗ N_{ω′} ≃ ⊕_{{ω∈G\X \atop ω′∈G\Y}} (M ˆ⊗ N)_{ω×ω′} ≃ (M ˆ⊗ N)_{⊔ω×ω′} ≃ …

… ≃ (M ˆ⊗ N)_{X×Y}

The following proposition states this isomorphism precisely, when M and N are both equal to the Burnside functor (note that b ˆ⊗ b ≃ b by Proposition 2.4.5 of [Bou97]):

Proposition 4.4. Let X and Y be G-sets, and Z be a finite G-set. If (T, φ) is a finite G-set over Z, if (U, f) is a finite G-poset over T × X, and (V, g) is a finite G-poset over T × Y, denote by U._{T} V the fibre product of U and V over T, viewed as a sub-G-poset of U × V. Then U._{T} V is a finite G-poset over Z × X × Y, and the correspondence

(U, f) ⊗ (V, g) ∈ b_{X}(T) ⊗ b_{Y} (T) ↦ U._{T} V ∈ b_{X×Y} (Z)

induces an isomorphism (b_{X} ˆ⊗ b_{Y})(Z) → b_{X×Y} (Z), which is part of an isomorphism of Mackey functors

b_{X} ˆ⊗ b_{Y} → b_{X×Y}

Proof. First I have to specify the morphism h : U._{T} V → Z × X × Y: if (u, v) ∈ U._{T} V, that is (u, v) ∈ U × V and f_{T} (u) = g_{T} (v), then

h(u, v) = (φ ∘ f_{T} (u), f_{X}(u), g_{Y} (v))

Now if (z, x, y) ∈ Z × X × Y, then

h^{−1}(z, x, y) = ⊔_{t∈φ^{−1}(z)}f^{−1}(t, x) × g^{−1}(t, y)

which gives of course

Λ_{h^{−1}(z,x,y)}^{G_{z,x,y}} = ∑_{t∈G_{z,x,y}\φ^{−1}(z)} Ind_{G_{t,x,y}}^{G_{z,x,y}}(Res_{G_{t,x,y}}^{G_{t,x}}Λ_{f^{−1}(t,x)}^{G_{t,x}})(Res_{G_{t,x,y}}^{G_{t,y}}Λ_{g^{−1}(t,y)}^{G_{t,y}})

So the equivalence class of (U._{T} V, h) depends only on the classes of (U, f) and (V, g).

Next I have to check that the submodule of relations J is mapped to zero in b_{X×Y} (Z): so let θ : (T, φ) → (T′, φ′) be a morphism of G-sets over Z. Let (U, f) be a finite G-poset over T × X, which gives a G-poset (U′, f′) over T′ by composition with θ × Id_{X} (note that U = U′). Let (V′, g′) be a finite G-poset over T′, which gives a G-set (V, g) over T by pull-back along θ. I must check that the posets U._{T} V and U′._{T′}V′ over Z × X × Y are equivalent.

They are in fact isomorphic: an element of U._{T} V is an element in U × V, or a triple (u, t, v′) in U × T × V′, such that

θ(t) = g′_{T′}(v′) f_{T} (u) = t

This element is mapped to (φf_{T} (u), f_{X}(u), g′_{Y} (v′)) ∈ Z × X × Y.

On the other hand, an element of U′._{T′}V′ is an element (u, v′) ∈ U × V′, such that θf_{T} (u) = g′_{T′}(v′). It is mapped to (φ′θf_{T} (u), f_{X}(u), g′_{Y} (v′)). Now the map

(u, v′) ∈ U′._{T′}V′ ↦ (u, f_{T} (u), v′) ∈ U._{T} V

is clearly an isomorphism of posets.

So I have a well-defined map from Θ_{Z} : (b_{X} ˆ⊗ b_{Y})(Z) → b_{X×Y} (Z). This is an isomorphism: indeed, let Φ_{Z} be the map from b_{X×Y} (Z) to (b_{X} ˆ⊗ b_{Y})(Z) which sends the finite G-poset (W, f) over Z × X × Y to

[(W, Id_{W} × f_{X}) ⊗ (W, Id_{W} × f_{Y})]_{(W,f_{Z})}

where Id_{W} × f_{X} : W → W × X is defined by (Id_{W} × f_{X})(w) = (w, f_{X}(w)), and the maps f_{Z}, f_{X} and f_{Y} are obtained from f by composition with the projections on Z, X and Y.

It is clear that Θ_{Z} ∘ Φ_{Z} is the identity, since the fibre product W._{W} W is equal to W. Conversely, if (T, φ) is a finite G-set over Z, if (U, f) is a finite G-poset over T × X, and (V, g) is a finite G-poset over T × Y, then

Φ_{Z}∘Θ_{Z}([(U, f)⊗(V, g)]_{(T,φ)})=[(U._{T} V , Id_{U._{T} V} ×h_{X})⊗(U._{T} V , Id_{U._{T} V} ×h_{Y})]_{(U._{T} V ,h_{Z})}

But as (U, f) = b_{X,∗}(f_{T})(U, Id_{U} × f_{X}), up to an element of J, I have

[(U, f) ⊗ (V, g)]_{(T,φ)} = [(U, Id_{U} × f_{X}) ⊗ b_{Y}^{∗} (f_{T})(V, g)]_{(U,φf_{T})}

But b_{Y}^{∗} (f_{T})(V, g) = (U._{T} V, k), where k : U._{T} V → U × Y is the map defined by k(u, v) = (u, g_{Y} (v)). Now again (U._{T} V, k) = b_{Y,∗}(k_{U})(U._{T} V, Id_{U._{T} V} × k_{Y}), so up to J, I have

[(U, f)⊗(V, g)]_{(T,φ)} = [b_{X}^{∗}(k_{U})(U, Id_{U} ×f_{X})⊗(U._{T} V, Id_{U._{T} V} ×k_{Y})]_{(U._{T} V,h_{Z})} = …

… = [(U._{T} V , Id_{U._{T} V} × h_{X}) ⊗ (U._{T} V , Id_{U._{T} V} × h_{Y})]_{(U._{T} V ,h_{Z})}

So Φ_{Z} ∘ Θ_{Z} is also the identity. To complete the proof of the proposition, it suffices to check that the maps Φ_{Z} define a morphism of Mackey functors, which is clear. So the maps Θ_{Z} define also a morphism of Mackey functors, and Θ and Φ are inverse isomorphisms.

Corollary 4.5. Let G and H be finite groups, and U be a finite H-set-G. If M and M′ are Mackey functors for G then

T_{U}(M ˆ⊗ M′) ≃ T_{U}(M) ˆ⊗ T_{U}(M′)

as Mackey functors for H, and this isomorphism is functorial in M and M′.

Proof. I will prove that there is an isomorphism of functors from Mack(G) × Mack(G) to Mack(H)

([M, M′] ↦ T_{U}(M ˆ⊗ M′))  ≃  ([M, M′] ↦ T_{U}(M) ˆ⊗ T_{U}(M′))

Indeed, the previous proposition states an isomorphism of functors between the restriction of those two functors to the subcategory P = P Mack(G) × P Mack(G). But the objects of P are projective in C = Mack(G) × Mack(G), and any object of C is the quotient of some object of P. Moreover P is closed under direct sums.

The left hand side functor above is right exact, because it is composed of T_{U}, which is right exact, and of the tensor product, which is right exact (as any tensor product over a ring, which it actually is. See [Bou97] Chapter 1). The right hand side is composed of T_{U} × T_{U}, which is right exact, and of the tensor product. So both sides are right exact, and isomorphic when restricted to P. Hence they are isomorphic, by theorem 2.14.

4.5. Disjoint unions. The tensor induction T_{U} also behaves well with respect to disjoints unions of bisets:

Proposition 4.6. Let G and H be finite groups. If U and U′ are finite Hsets-G, then for any Mackey functor M for G

T_{U⊔U′}(M) ≃ T_{U}(M) ˆ⊗ T_{U′}(M)

and this isomorphism is functorial in M.

Proof. Here again, both sides are right exact functors in M. So it suffices to check that their restrictions to P Mack(G) are isomorphic.

But for any G-set X

Hom_{G}((U ⊔U′)^{op}, X) = Hom_{G}(U^{op} ⊔U′^{op}, X) ≃ Hom_{G}(U^{op}, X)×Hom_{G}(U′^{op}, X)

This gives an isomorphism

T_{U⊔U′}(b_{X}) ≃ b_{Hom_{G}(U^{op},X)×Hom_{G}(U′^{op},X)} ≃ b_{Hom_{G}(U^{op},X)} ˆ⊗ b_{Hom_{G}(U′^{op},X)} = …

… = T_{U}(b_{X}) ˆ⊗ T_{U′}(b_{X})

which is easily seen to be functorial in X. So the proposition follows from theorem 2.14.

4.6. Direct sums. One can think of T_{U}(M) as a sort of U-th power of M. Here is the corresponding “binomial identity”:

Proposition 4.7. Let G and H be finite groups, and U be a finite H-set-G. For any Mackey functors M and M′ for G

T_{U}(M ⊕ M′) ≃ ⊕_{V ⊆U V G−invariant V mod.H} Ind_{H_{V}}^{H} T_{V} (M) ˆ⊗ T_{U−V} (M′)

(where H_{V} = {h ∈ H | hV = V }), and this isomorphism is functorial in M and M′.

Proof. Both sides define functors from Mack(G) × Mack(G) to Mack(H). The left hand side is clearly right exact, since it is composed of T_{U} and of the direct sum, which is (additive!) exact. The right hand side is a direct sum of induced of tensor products of tensor inductions. As induction is an exact functor, the right hand side is also right exact. So again, it suffices to check that both sides are isomorphic when restricted to P Mack(G) × P Mack(G).

But if X and X′ are G-sets, then there is a bijection

Hom_{G}(U^{op}, X ⊔ X′) ≃ ⨆_{V ⊆U V G−invariant} Hom_{G}(V^{op}, X) × Hom_{G}((U − V)^{op}, X′)

Indeed any φ : U → X ⊔ X′ is determined by V = φ^{−1}(X), which is a sub-G-set of U, and its restrictions to V and U − V, which are G-morphisms from V to X and U − V to X′ respectively.

Now keeping track of the action of H gives the isomorphism of H-sets

Hom_{G}(U^{op}, X ⊔ X′) ≃ ⨆_{V ⊆U V G−invariant V mod.H} Ind_{H_{V}}^{H} Hom_{G}(V^{op}, X) × Hom_{G}((U − V)^{op}, X′) (4.10)

But if K is a subgroup of G, and Z is a K-set, it is easy to see that

b_{Ind_{K}^{G}Z} ≃ Ind_{K}^{G}b_{Z}

This follows by taking direct sums from the case of a finite set Z. In that case for any G-set Y

b_{Ind_{K}^{G}Z}(Y) = b(Y × Ind_{K}^{G}Z) = b(Ind_{K}^{G}(Res_{K}^{G}Y.Z)) = …

… = (Res_{K}^{G}b)(Res_{K}^{G}Y.Z) = b_{Z}(Res_{K}^{G}Y) = (Ind_{K}^{G})b_{Z}(Y)

Now equation 4.10 gives the following isomorphism

b_{Ind_{G}(U^{op},X⊔X′)} = ⊕_{V ⊆U V G−invariant V mod.H} Ind_{H_{V}}^{H} b_{Hom_{G}(V^{op},X)×Hom_{G}((U−V)^{op},X′)} ≃ …

… ≃ ⊕_{V ⊆U V G−invariant V mod.H} Ind_{H_{V}}^{H} (b_{Hom_{G}(V^{op},X)} ˆ⊗ b_{Hom_{G}((U−V)^{op},X′)})

This isomorphism is functorial in X and X′, and this completes the proof of the proposition.

5. Relations with the functors L_{U}

5.1. Construction of L_{U}. If G, H, and K are groups, if U is an H-set-G and V is a K-set-H, then in [Bou96a], I defined the product V ∘_{H} U by

V ∘_{H} U = {(v, u) ∈ V × U | ∀h ∈ H, vh = v ⇒ ∃g ∈ G, hu = ug}/H

This is a (generally strict) sub-K-set-G of V ×_{H} U.

In particular, if U is finite, the product X ↦ U ∘_{G} X is a functor from G-set to H-set, and those functors are exactly those preserving disjoint unions and cartesian squares ([Bou96a] Théorème 1). By composition, they give functors M ↦ M ∘ U from Mack(H) to Mack(G).

Those functors between Mackey categories have left and right adjoints, described in [Bou97] Chapter 9. I denote by L_{U} the left adjoint of the functor M ↦ M ∘ U. So L_{U} is a functor from Mack(G) to Mack(H). Let me recall some notations and definitions:

Definition 5.1. If X is a finite H-set, let D_{U}(X) be the following category:

• The objects of D_{U}(X) are the finite H-sets over X × (U/G). If (Y, f) is such an object, I denote by f_{X} the X-component of f, and by f_{U} its U/Gcomponent. I denote by U._{f}Y or U.Y the fibre product of U and Y over U/G. It is an H-set-G, with action given by h(u, y)g = (hug, hy). To simplify the notations, I view H\U.Y as a left G-set (it should really be denoted (H\U.Y)^{op}).

• A morphism α : (Y, f) → (Y′, f′) in D_{U}(X) is a morphism of H-sets over X × (U/G), such that moreover the morphism U.α : U.Y → U.Y′ is injective on each H-orbit, that is

∀(u, y) ∈ U.Y, ∀h ∈ H, hu = u, α(hy) = α(y) ⇒ hy = y

If (Y, f) is an object of D_{U}(X), then a G-set (Z, a) over H\U.Y is said to be ν-disjoint if the following condition holds

∀(z, u, y) ∈ Z × U × Y, a(z) = H(u, y) ⇒ (u, z) ∉ U ∘_{G} Z

(this last condition means that there exists an g ∈ G with ug = u but gz ≠ z). If moreover M is a Mackey functor for G, I set

Q_{U}(M)(Y, f) = M(H\U.Y)/ ∑_{(Z,a)} M_{∗}(a)M(Z)

where the sum runs through the G-sets (Z, a) over H\U.Y which are ν-disjoint. Then Q_{U}(M) is a functor on D_{U}(X), with values in abelian groups. I set

L_{U}(M)(X) = lim ; ⟶_{(Y,f)∈D_{U} (X)} Q_{U}(M)(Y, f)

If (Y, f) is an object of D_{U}(X) and if m ∈ M(H\U.Y), I denote by m_{(Y,f)} the image of m in L_{U}(M)(X).

The correspondence X ↦ L_{U}(M)(X) can be turned into a Mackey functor for H, denoted by L_{U}(M).

Morever, if θ : M → M′ is a morphism of Mackey functors for G, then set

L_{U}(θ)(m_{(Y,f)}) = θ_{H\U.Y} (m)_{(Y,f)}

This gives a well defined morphism L_{U}(θ) of Mackey functors for H from L_{U}(M) to L_{U}(M′). More precisely:

Theorem 5.2. ([Bou97] Theorem 9.5.2) Let G and H be finite groups, and U be a finite G-set-H. The correspondence

M ↦ L_{U}(M)
θ ∈ Hom_{Mack(G)}(M, M′) ↦ L_{U}(θ) ∈ Hom_{Mack(H)}(L_{U}(M), L_{U}(M′))

is a functor from Mack(G) to Mack(H), which is left adjoint to the functor N ↦ N ∘ U.

Corollary 5.3. The functor L_{U} is right exact.

Proof. This is because it is additive and has a right adjoint: see for example [Wei94] Theorem 2.6.1.

5.2. Examples.

5.2.1. Induction and restriction. Let H be a subgroup of G. If U = G, viewed as an H-set-G, then the functor N ↦ N ∘ U is the induction functor Ind_{H}^{G}, and the functor L_{U} is the restriction functor Res_{H}^{G}. If U = G is viewed as a G-set-H, then the functor N ↦ N ∘ U is the restriction functor Res_{H}^{G}, and the functor L_{U} is the induction functor Ind_{H}^{G}.

5.2.2. Inflation. Let N be a normal subgroup of G, and H = G/N. If U = H is viewed as an H-set-G, then the functor N ↦ N ∘U is the inflation functor Inf_{H}^{G} (see [TW90],[TW95]). The functor L_{U} will be denoted by M ↦ M^{N} (it is denoted M^{+} in [TW95]).

5.2.3. Coinflation. Let N be a normal subgroup of G, and H = G/N. If U = H is viewed as a G-set-H, then the functor N ∘ U will be denoted by ρ_{H}^{G} (it is the functor β^{!} of [TW95] section 5). The functor L_{U} will be denoted by ι_{H}^{G} (its existence is mentioned in [TW95] and it is called β_{!}).

The functor ι_{H}^{G} can be described as follows (see [Bou97] 9.9.3): if X is a G-set, let D_{H}^{G}(X) be the following category:

• The objects of D_{H}^{G}(X) are the G-sets over X.

• A morphism α : (Y, f) → (Y′, f′) in D_{H}^{G}(X) is a morphism of G-sets over X, which is moreover injective when restricted to every orbit of N on Y.

Then if M is a Mackey functor for H, the correspondence (Y, f) ↦ M(N\Y) is a functor on D_{H}^{G}(X), and

ι_{H}^{G}(M)(X) = lim ; ⟶_{(Y,f)∈D_{H}^{G}(X)} M(N\Y)

5.3. The case U/G = •.

Lemma 5.4. Let G and H be finite groups, and U be a finite H-set-G. Then for any G-set X

L_{U}(b_{X}) = b_{U∘_{G}X}

Proof. This follows from the case of a finite G-set X, which is a remark at the end of Chapter 9 of [Bou97]: indeed for any Mackey functor N for H

Hom_{Mack(H)}(L_{U}(b_{X}), N) = Hom_{Mack(G)}(b_{X}, N ∘U) = (N ∘U)(X) = …

… = N(U ∘_{G} X) = Hom_{Mack(H)}(b_{U∘_{G}X}, N)

The lemma follows, because all those isomorphisms are natural in N.

Proposition 5.5. Let G and H be finite groups, and U be a finite H-set-G. The following conditions are equivalent:

1. The functor T_{U} is additive.

2. The functors T_{U} and L_{U} are isomorphic.

3. The group G is transitive on U.

Proof. First if T_{U} is additive, then for any G-sets X and Y

T_{U}(b_{X⊔Y}) ≃ b_{Hom_{G}(U^{op},X⊔Y)} ≃ T_{U}(b_{X} ⊕b_{Y}) ≃ T_{U}(b_{X})⊕T_{U}(b_{Y}) ≃ …

… ≃ b_{Hom_{G}(U^{op},X)} ⊕ b_{Hom_{G}(U^{op},Y)} ≃ b_{Hom_{G}(U^{op},X)⊔Hom_{G}(U^{op},Y)}

As the rank of the evaluation at {1} of b_{X} is the cardinality of X, this implies in particular, if X and Y are finite trivial G-sets, that

(|X| + |Y |)^{|U/G|} = |X|^{|U/G|} + |Y |^{|U/G|}

which forces |U/G| = 1. So 3) holds.

Obviously 2) implies 1), since the functors L_{U} are always additive.

Now if 3) holds, I claim that Hom_{G}(U^{op}, X) ≃ U ∘_{G} X for any G-set X: choose u in U, take φ ∈ Hom_{G}(U^{op}, X), and consider the couple

θ(φ) = (u, φ(u)) ∈ U ×_{G} X

This does not depend on the choice of u, since by 3) any other choice is in the G-orbit of u, say equal to ug, for g ∈ G, and since

(ug, φ(ug)) = (ug, g^{−1}φ(u)) = (u, φ(u))

Moreover if g ∈ G fixes u on the right, then

gφ(u) = φ(ug^{−1}) = φ(u)

This proves that θ is a map from Hom_{G}(U^{op}, X) to U ∘_{G} X. It is a morphism of H-sets, since for h ∈ H

θ(hφ) = (u, (hφ)(u)) = (u, φ(h^{−1}u))

Choose g ∈ G such that h^{−1}u = ug^{−1} (or ug = hu), which is possible by 3). Then

(u, φ(h^{−1}u)) = (u, φ(ug^{−1})) = (u, gφ(u)) = (ug, φ(u)) = (hu, φ(u)) = hθ(φ)

Conversely, if (u, x) ∈ U ∘_{G} X, define a map θ′(u, x) : U → X by

θ′(u, x)(u′) = g^{−1}x if g ∈ G, u′ = ug

This map is well-defined by 3), and because (u, x) is in U ∘_{G} X. It is a map of H-sets: if h ∈ H, then choose g_{0} ∈ G with hu = ug_{0}. Then

(hθ′(u, x))(u′) = θ′(u, x)(h^{−1}u′) = θ′(u, x)(h^{−1}ug) = θ′(u, x)(ug_{0}^{−1}g) = g^{−1}g_{0}x

whereas θ′(hu, x)(u′) is given by

θ′(hu, x)(u′) = θ′(ug_{0}, x)(u′) = g^{−1}g_{0}x

since u′ = (ug_{0})g_{0}^{−1}g.

Now θ′θ(φ) = θ′(u, φ(u)). But for u′ ∈ U

θ′(u, φ(u))(u′) = g^{−1}φ(u) if g ∈ G, u′ = ug

So θ′(u, φ(u))(u′) = φ(ug) = φ(u′), and the map θ′ ∘ θ is the identity.

Conversely

θθ′(u, x) = (u, θ′(u, x)(u)) = (u, x)

So θ and θ′ are mutually inverse isomorphisms.

Now for any G-set

T_{U}(b_{X}) = b_{Hom_{G}(U^{op},X)} ≃ b_{U∘_{G}X} ≃ L_{U}(b_{X})

This is natural in X. As T_{U} and L_{U} are right exact, they are isomorphic by theorem 2.14. So 2) holds, and this completes the proof of the proposition.

Corollary 5.6. Let G and H be finite groups, and U be a finite H-set-G. Then for any Mackey functor M for G

1. If K is a subgroup of H

Res_{K}^{H}T_{U}(M) ≃ T_{Res_{K×G^{op}}^{H×G^{op}} U}(M)

2. If N is a normal subgroup of H, then

(T_{U}(M))^{N} ≃ T_{N\U}(M)

3. If N is a normal subgroup of a group K, and H = K/N, then

ι_{H}^{K}T_{U}(M) ≃ T_{Inf_{H×G^{op}}^{K×G^{op}} U}(M)

Moreover, these isomorphisms are natural in M.

Proof. This is straightforward, using proposition 4.3, with V = H, viewed as a K-set-H in case 1), and V = H viewed as a K-set-H in case 2), and V = H viewed as an H-set-K in case 3).

5.4. The products T_{V} ∘ L_{U}. The previous corollary gives examples of composition L_{V} ∘ T_{U}. The next proposition computes the composition T_{V} ∘ L_{U}:

Proposition 5.7. Let G, H and K be finite groups. Let U be a finite H-set-G, and V be a finite K-set-H. If f : V^{op} → U/G is a morphism of H-sets, let V._{f,H}U denote the quotient set of the fibre product of V and U over U/G, by the right action of H given by (v, u)h = (vh, h^{−1}u). If K_{f} is the stabilizer of f in K, then V._{f,H}U is a finite K_{f}-set-G by k.(v, u).g = (kv, ug). Then for any Mackey functor M for G

T_{V} ∘ L_{U}(M) ≃ ⊕_{f∈K\Hom_{H}(V^{op},U/G)} Ind_{K_{f}}^{K} T_{V._{f,H}U}(M)

and this isomorphism is natural in M.

Proof. Both sides are right exact functors. So it suffices to check that their restrictions to P Mack(G) coincide.

So let M = b_{X}, for some G-set X. The left hand side is

T_{V} ∘ L_{U}(b_{X}) ≃ T_{V} (b_{U∘_{G}X}) = b_{Hom_{H}(V^{op},U∘_{G}X)}

Now as there is always a (unique) map p_{X} from X to the trivial G-set •, there is a natural map υ_{X} = U ∘_{G}p_{X} from U ∘_{G}X to U ∘_{G}• ≃ U/G, given by (u, x) ↦ uG. So the functor X ↦ (U ∘_{G} X, υ_{X}) is a functor from the category G-Set of (arbitrary) G-sets, to the category H-Set↓_{U/G} of (arbitrary) H-sets over U/G. Conversely, if (Y, f) is an H-set over U/G, then the set H\U._{f}Y is a G-set, and moreover:

Proposition 5.8. Let G and H be finite groups, and U be a finite H-set-G. The functor (Y, f) ↦ H\U._{f}Y from H-Set↓_{U/G} to G-Set is left adjoint to the functor X ↦ (U ∘_{G} X, υ_{X}).

Proof. The corresponding statement for finite sets is proposition 9.1.1 of [Bou97]. The proof extends verbatim to arbitrary sets: the unit of this adjunction is denoted by ν (it it the reason for the word ν-disjoint in definition 5.1), and defined for an H-set (Y, f) over U/G by the morphism

ν_{(Y,f)} : Y → U ∘_{G} (H\U._{f}Y)

given by ν_{(Y,f)}(y) = (u, H(u, y)) if f(y) = uG.

The counit of this adjunction is denoted by η. It is defined for a G-set Z by the map

η_{Z} : H\U._{υ_{Z}} (U ∘_{G} Z) → Z

defined by η_{Z}(H(u′, (u, z)) = h^{−1}z if h ∈ H is such that u′ = uh.

This result means that there is a one to one correspondence between the set of morphisms of H-sets φ from V^{op} to U ∘_{G} X, such that υ ∘ φ is a given morphism f from V^{op} to U/G, and the set of morphisms of G-sets from H\U._{f}V to X. In other words

Hom_{H}(V^{op}, U ∘_{G} X) = ⨆_{f∈Hom_{H}(V^{op},U/G)} Hom_{G}(H\U._{f}V, X)

Keeping track of the action of K gives

Hom_{H}(V^{op}, U ∘_{G} X) = ⨆_{f∈K\Hom_{H}(V^{op},U/G)} Ind_{K_{f}}^{K} Hom_{G}(H\U._{f}V, X)

Thus

b_{Hom_{H}(V^{op},U∘_{G}X)} ≃ ⊕_{f∈K\Hom_{H}(V^{op},U/G)} Ind_{K_{f}}^{K} b_{Hom_{G}(H\U._{f} V,X)}

This isomorphism is natural in X. This completes the proof, since with the notations of the proposition 5.7, the map (u, v) ↦ (v, u) induces an isomorphism

H\U._{f}V ≃ V._{f,H}U

of K_{f}-sets-G.

Proposition 5.9. Let G and H be finite groups, and U be a finite H-set-G. Let X be a G-set and M be a Mackey functor for G. Then there is an isomorphism

T_{U}(M_{X}) ≃ T_{U}(M)_{Hom_{G}(U^{op},X)}

natural in M.

Proof. Here again its enough to check the isomorphism for permutation functors. But if Y is a G-set, it is clear that (b_{Y})_{X} is isomorphic to b_{Y ×X}. As

Hom_{G}(U^{op}, Y × X) ≃ Hom_{G}(U^{op}, Y) × Hom_{G}(U^{op}, X)

it follows that

T_{U}((b_{Y})_{X}) ≃ (b_{Hom_{G}(U^{op},Y)})_{Hom_{G}(U^{op},X)}

so the result holds for permutation functors, and the proposition follows.i

Remark 5.10. This proposition is also a consequence of proposition 5.7 in the case of a finite G-set X: let X̃ = G × X, viewed as a G-set-G by g.(g′, x).g′′ = (gg′g′′, g.x). Then one can prove that the functor M ↦ M_{X} is isomorphic to the functor LX̃: this is because it is self adjoint, and moreover for any G-set Y, one has X̃ ∘_{G} Y ≃ X × Y.

6. Direct product of Mackey functors

6.1. Definition. There is a reasonable definition of the direct product of Mackey functors:

Definition 6.1. Let G and H be finite groups. If M is a Mackey functor for G, and N a Mackey functor for H, I denote by M ⊠ N, and I call direct product of M and N, the Mackey functor for G × H defined by

M ⊠ N = ι_{G}^{G×H}(M) ˆ⊗ ι_{H}^{G×H}(N)

6.2. Examples.

6.2.1. Direct product of permutation functors. The direct product of two permutation Mackey functors is a permutation Mackey functor:

Lemma 6.2. Let G and H be finite groups. If X is a G-set and Y is an H-set, then

b_{X} ⊠ b_{Y} ≃ b_{Inf_{G}^{G×H}X×Inf_{H}^{G×H}Y}

Proof. This follows from the fact that

ι_{G}^{G×H}(b_{X}) = b_{Inf_{H}^{G×H}X}

by lemma 5.4, and from the isomorphism

b_{Inf_{H}^{G×H}X} ˆ⊗ b_{Inf_{H}^{G×H}Y} ≃ b_{Inf_{H}^{G×H}X×Inf_{H}^{G×H}Y}

6.2.2. Extension of coefficients. As a special case of direct product of Mackey functors, I will look at the case G = {1}: the category Mack({1}) is equivalent to the category of abelian groups, because a Mackey functor for the trivial group is completely determined by its value on the trivial set, which is just an abelian group. Conversely, any abelian group A defines a Mackey functor for the trivial group, still denoted by A, which value A(X) on the finite set X is equal to the group A^{(X)} of functions from X to A. If f : X → Y is a map of finite sets, then A^{∗}(f) : A(Y) → A(X) is the composition with f, and A_{∗}(f) : A(X) → A(Y) is defined by

A_{∗}(f)(α)(y) = ∑_{x∈f^{−1}(y)} α(x)

Proposition 6.3. Let G be a finite group. If M is a Mackey functor for G, and A is an abelian group, then after identification of {1}×G with G, for any finite G-set X

(A ⊠ M)(X) = A⊗_{ℤ}M(X)

and for any map of finite G-sets f : X → Y

(A ⊠ M)_{∗}(f) = A⊗_{ℤ}M_{∗}(f) (A ⊠ M)^{∗}(f) = A⊗_{ℤ}M^{∗}(f)

In other words, the functor A ⊠ M is just the functor M, “with coefficients in A”.

Proof. First consider the case A = ℤ. This is just the Burnside functor for the trivial group. But ι_{{1}}^{G}(b) = b by lemma 5.4, and I have

ℤ ⊠ M = b ˆ⊗ ι_{G}^{G}(M) = b ˆ⊗ M ≃ M

More generally, if S is any set, then the free abelian group ℤ^{(S)} is the functor b_{S} for the trivial group. So

ι_{{1}}^{G}(ℤ^{(S)}) = b_{Inf_{{1}}^{G}S} ˆ⊗ M ≃ ⊕_{s∈S} M

So its value on any G-set X is ℤ^{(S)} ⊗ M(X). Now for a fixed Mackey functor M, the functor A ↦ A ⊠ M is clearly additive and right exact. Choosing a resolution of A by free ℤ-modules

ℤ^{(T)} → ℤ^{(S)} → A → 0

leads for any G-set X to the exact sequence

(ℤ^{(T)} ⊠ M)(X) → (ℤ^{(S)} ⊠ M)(X) → (A ⊠ M)(X) → 0

This sequence is isomorphic to

ℤ^{(T)} ⊗ M(X) → ℤ^{(S)} ⊗ M(X) → (A ⊠ M)(X) → 0

This proves that (A ⊠ M)(X) ≃ A ⊗ M(X), and this isomorphism is natural in M and A. The proposition follows.

6.3. Associativity. The direct product of Mackey functors is associative:

Proposition 6.4. Let G, H, and K be finite groups. If M is a Mackey functor for G, if N is a Mackey functor for H and P is a Mackey functor for P, then after identification of (G × H) × K with G × (H × K), there is an isomorphism

(M ⊠ N) ⊠ P ≃ M ⊠ (N ⊠ P)

which is natural in M, N and P.

Proof. Indeed, by definition

(M ⊠ N) ⊠ P = ι_{G×H}^{G×H×K}(ι_{G}^{G×H}(M) ˆ⊗ ι_{H}^{G×H}(N)) ˆ⊗ ι_{K}^{G×H×K}(P)

Let U = G × H, viewed as a (G × H × K)-set-(G × H). Then

ι_{G×H}^{G×H×K}(ι_{G}^{G×H}(M) ˆ⊗ ι_{H}^{G×H}(N)) = T_{U}(ι_{G}^{G×H}(M) ˆ⊗ ι_{H}^{G×H}(N))

By proposition 4.4, this is also

T_{U}(ι_{G}^{G×H}(M)) ˆ⊗ T_{U}(ι_{H}^{G×H}(N))

Let V = G, wiewed as a (G × H)-set-G. Then

T_{U}(ι_{G}^{G×H}(M)) = T_{U} ∘ T_{V} (M) = T_{U∘_{G×H}V} (M)

But U ∘_{G×H} V is the set G, viewed as a (G × H × K)-set-G. Thus

T_{U∘_{G×H}V} (M) = ι_{G}^{G×H×K}(M)

A similar argument shows that

T_{U}(ι_{H}^{G×H}(N)) = ι_{H}^{G×H×K}(N)

Finally

(M ⊠ N) ⊠ P ≃ (ι_{G}^{G×H×K}(M) ˆ⊗ ι_{H}^{G×H×K}(N)) ˆ⊗ ι_{K}^{G×H×K}(N)

On the other hand

M ⊠ (N ⊠ P) ≃ ι_{G}^{G×H×K}(M) ˆ⊗ (ι_{H}^{G×H×K}(N) ˆ⊗ ι_{K}^{G×H×K}(N))

Now proposition follows from the associativity of tensor product (see [Bou97] Proposition 1.9.1), and from the naturality of the above isomorphisms.

6.4. Tensor induction of direct products.

Proposition 6.5. Let G, H, and K be finite groups, and U be a finite K-set- (G × H). Then U/H is a finite K-set-G and U/G is a finite K-set-H. If M is a Mackey functor for G and N is a Mackey functor for H, then

T_{U}(M ⊠ N) ≃ T_{U/H}(M) ˆ⊗ T_{U/G}(N)

and this isomorphism is natural in M and N.

Proof. Let V = G, viewed as a (G × H)-set-G, and W = H, viewed as a (G × H)-set-H. Then by proposition 4.4

T_{U}(M ⊠ N) = T_{U}(T_{V} (M) ˆ⊗ T_{W} (N)) = (T_{U} ∘ T_{V} (M)) ˆ⊗ (T_{U} ∘ T_{W} (N))

But moreover

T_{U} ∘ T_{V} (M) ≃ T_{U×_{G×H}V} (M)

and as U ×_{G×H} V ≃ U/H as K-set-H, this gives

T_{U} ∘ T_{V} (M) ≃ T_{U/H}(M)

A similar argument shows that

T_{U} ∘ T_{W} (N) ≃ T_{U/G}(N)

and the proposition follows.

6.5. Tensor product from direct product. The tensor product of Mackey functors can be recovered from the direct product, by diagonal restriction:

Proposition 6.6. Let G be a finite group, identified with the diagonal subgroup Δ(G) of G × G. Then if M and N are Mackey functors for G

Res_{Δ(G)}^{G×G}(M ⊠ N) ≃ M ˆ⊗ N

and this isomorphism is natural in M and N.

Proof. Indeed the restriction to Δ(G) can be viewed as the tensor induction T_{U} for the set U = G × G, viewed as a Δ(G)-set-(G × G). Now by proposition 6.5

T_{U}(M ⊠ N) = T_{U/G}(M) ˆ⊗ T_{U/G}(N)

Here the first U/G is relative to the second factor of G × G, and the second is relative to the first factor. Anyway, those two G-sets-G are isomorphic to G, with its left and right action by multiplication. So T_{U/G} is the identity functor, and the proposition follows.

6.6. Identification of the direct product. The direct product of Mackey functors can be computed as follows:

Proposition 6.7. Let G and H be finite groups, and X be a finite (G×H)-set. If M is a Mackey functor for G, and N is a Mackey functor for H, then

(M ⊠ N)(X) ≃ ( ⊕_{Y →^{φ}X} M(H\Y) ⊗ N(G\Y))/J (6.11)

where (Y, φ) is a finite (G × H)-set over X, and J is the submodule generated by

M_{∗}(H\f)(m) ⊗ n′ − m ⊗ N^{∗}(G\f)(n′) ∀m ∈ M(H\Y), n′ ∈ N(G\Y′) whenever f : (Y, φ) → (Y′, φ′) is a morphism of (G × H)-sets over X, which is injective when restricted to every H-orbit, and by the elements

M^{∗}(H\f)(m′) ⊗ n − m′ ⊗ N_{∗}(G\f)(n) ∀m′ ∈ M(H\Y′), n ∈ N(G\Y)

whenever f : (Y, φ) → (Y′, φ′) is a morphism of (G × H)-sets over X, which is injective when restricted to every G-orbit

Proof. Let K = G × H. Recall that for a K-set Y, the value of ι_{G}^{K}(M)(Y) is

ι_{G}^{K}(M)(Y) = lim ; ⟶_{(Z,f)∈D_{G}^{K}(Y)} M(H\Z)

where D_{G}^{K}(Y) is the category with K-sets over Y as objects, the morphisms being the morphisms of K-sets over Y which are injective when restricted to each orbit of H. If m ∈ M(H\Z), denote by m_{(Z,f)} its image in ι_{G}^{K}(Y).

Let

(M ⊠′ N)(X) = ( ⊕_{Y →^{φ}X} M(H\Y) ⊗ N(G\Y))/J

If (Y, φ) is a finite K-set over X, if m ∈ M(H\Y), and n ∈ N(G\Y), I denote by [m ⊠ n]_{(Y,φ)} the image of m ⊗ n in (M ⊠′ N)(X).

As

(M ⊠ N)(X) = (ι_{G}^{K}(M) ˆ⊗ ι_{H}^{K}(N))(X) = ( ⊕_{Y →^{φ}X} ι_{G}^{K}(M)(H\Y) ⊗ ι_{H}^{K}(N)(G\Y))/K

for a suitable submodule K, I can try to define a morphism Φ from (M ⊠′ N)(X) to (M ⊠ N)(X) by

Φ([m ⊠ n]_{(Y,φ)}) = [m_{(Y,Id_{Y} )} ⊗ n_{(Y,Id_{Y} )}]_{(Y,φ)}

This is possible if J is mapped to zero. So let f : (Y, φ) ↦ (Y′, φ′) be a morphism of K-sets over X, which is injective on every H-orbit. Let m ∈ M(H\Y) and n′ ∈ N(G\Y′). Now

Φ([m ⊠ N^{∗}(G\f)(n′)]_{(Y,φ)}) = [m_{(Y,Id_{Y} )} ⊗ N^{∗}(G\f)(n′)_{(Y,Id_{Y} )}]_{(Y,φ)}

But N^{∗}(G\f)(n′)_{(Y,Id_{Y} )} = ι_{H}^{K}(N)^{∗}(f)(n′_{(Y′,Id_{Y ′})}), so in (M ⊠ N)(X), I have

Φ([m ⊠ N^{∗}(G\f)(n′)]_{(Y,φ)}) = [ι_{G}^{K}(M)_{∗}(f)(m_{(Y,Id_{Y} )}) ⊠ n′_{(Y′,Id′_{Y} )}]_{(Y′,φ′)}

As ι_{G}^{K}(M)_{∗}(f)(m_{(Y,Id_{Y} )}) = m_{(Y,f)} and as f is supposed to be injective on every H-orbit, it is a morphism in D_{G}^{K}(X) from (Y, f) to (Y′, Id_{Y′}). Thus in ι_{G}^{K}(Y′), I have

m_{(Y,f)} = M_{∗}(H\f)(m)_{(Y′,Id_{Y ′})}

This gives

Φ([m⊠N^{∗}(G\f)(n′)]_{(Y,φ)}) = [M_{∗}(H\f)(m)_{(Y′,Id_{Y ′})}⊗n′_{(Y′,Id_{Y ′})}]_{(Y′,φ′)} = …

… = Φ([M_{∗}(H\f)(m) ⊠ n′]_{(Y′,φ′)})

A similar argument obtained by reversing the roles of M and N shows that Φ is well-defined.

I will now define a map Θ from (M ⊠ N)(X) to (M ⊠′ N)(X): let (Y, φ) be a K-set over X. If (Z, f) is an object of D_{G}^{K}(Y), if m ∈ M(H\Z), if (T, g) is an object of D_{H}^{K}(Y) and N ∈ N(G\T), I build the fibre product Z.T

Z.T ⟶^{π_{Z}} Z
π_{T} ↓  ↓^{f}
T ⟶_{g} Y

and then I set

Θ([m_{(Z,f)} ⊗ n_{(T,g)}]_{(Y,φ)}) = [M^{∗}(H\π_{Z})(m) ⊠ N^{∗}(G\π_{T})(n)]_{(Z.T,φ∘f∘π_{Z})}

This map is also well-defined: I must check that Θ is compatible with the inductive limits in the definition of ι, and that K is mapped to zero.

Suppose first that I replace m_{(Z,f)} by M_{∗}(H\α)(m)_{(Z′,f′)}, for some morphism α : (Z, f) → (Z′, f′) in D_{G}^{K}(Y). Then

Θ([M_{∗}(H\α)(m)_{(Z′,f′)} ⊗ n_{(T,g)}]_{(Y,φ)}) = …

… = [M^{∗}(H\π_{Z′})M_{∗}(H\α)(m) ⊠ N^{∗}(G\π_{T})(n)]_{(Z′.T,φ∘f′∘π_{Z′})}

There is a commutative diagram

[figure: T; Y; Z′.T; Z′; Z.T; Z; π_{Z}; π_{Z′}; g; α; α.T; π_{T}; π_{T}; f; f′]

Now it is clear that the square

Z.T ⟶^{π_{Z}} Z
α.T ↓  ↓^{α}
Z′.T ⟶_{π_{Z′}} Z′

is cartesian, so the square

H\Z.T ⟶^{H\π_{Z}} H\Z
H\α.T ↓  ↓^{H\α}
H\Z′.T ⟶_{H\π_{H\Z′}} H\Z′

is also cartesian by Lemma 9.3.3 of [Bou97], because α is injective on the H-orbits. Now it follows that

M^{∗}(H\π_{Z′})M_{∗}(H\α) = M_{∗}(H\α.T)M^{∗}(H\π_{Z})

Moreover, the morphism α.T : Z.T → Z′.T is injective on every H-orbit, because α is. Thus in (M ⊠′ N)(X), I have

[M_{∗}(H\α.T)M^{∗}(H\π_{Z}) ⊠ N^{∗}(G\π_{T})(n)]_{(Z′.T,φ∘f′∘π_{Z′})} = …

… = [M^{∗}(H\π_{Z}) ⊠ N^{∗}(G\α.T)N^{∗}(G\π_{T})(n)]_{(Z.T,φ∘f∘π_{Z})}

But

N^{∗}(G\α.T)N^{∗}(G\π_{T})(n) = N^{∗}(G\π_{T})(n)

This proves that

Θ([M_{∗}(H\α)(m)_{(Z′,f′)} ⊗ n_{(T,g)}]_{(Y,φ)}) = Θ([m_{(Z,f)} ⊗ n_{(T,g)}]_{(Y,φ)})

and a similar argument shows that Θ is compatible with the relations coming from the inductive limits.

Now suppose that α : (Y, φ) → (Y′, φ′) is a morphism of K-sets over X. Let (Z, f) be an object of D_{G}^{K}(Y) and m ∈ M(H\Z). Let (T′, g′) be an object in D_{H}^{K}(Y′), and n′ ∈ N(G\T). In (M ⊠ N)(X), I have

[m_{(Z,f)} ⊗ ι_{H}^{K}(N)^{∗}(α)(n′_{(T′,g′)})]_{(Y,φ)} = [ι_{G}^{K}(M)_{∗}(α)(m_{(Z,f)}) ⊗ n′_{(T′,g′)}]_{(Y′,φ′)} (6.12)

I have to check that both sides have the same image in (M ⊠′ N)(X). But

ι_{G}^{K}(M)_{∗}(α)(m_{(Z,f)}) = m_{(Z,α∘f)}

Moreover, if the square

T ⟶^{a} T′
g↓  ↓^{g′}
Y ⟶_{α} Y′

is cartesian, then

ι_{H}^{K}(N)^{∗}(α)(n′_{(T′,g′)}) = N^{∗}(G\a)(n′)_{(T,g)}

The image by Θ of the right hand side of (6.12) is

[M^{∗}(H\π_{Z})(m) ⊠ N^{∗}(G\π_{T})N^{∗}(G\a)(n′)]_{(Z.T,φ∘g∘π_{T})} (6.13)

Now the square

Z.T ⟶^{a∘π_{T}} T′
π_{Z} ↓  ↓^{g′}
Z ⟶_{α∘f} Y′

is cartesian, because it is composed of two cartesian squares. So the image by Θ of the left hand side of (6.12) is

[M^{∗}(H\π_{Z})(m) ⊠ N^{∗}(G\(a ∘ π_{T}))(n′)]_{(Z.T,φ′∘g′∘a∘π_{T})}

This is equal to (6.13). So Θ is well-defined.

It is clear that Θ ∘ Φ is the identity: indeed, if (Y, φ) is a K-set over X, if m ∈ M(H\Y) and n ∈ N(G\Y), then

Θ ∘ Φ([m ⊠ n]_{(Y,φ)}) = Θ([m_{(Y,Id)} ⊗ n_{(Y,Id)}]_{(Y,φ)}) = [m ⊠ n]_{(Y,φ)}

since Y.Y = Y. Moreover, if (Z, f) and (T, g) are any K-sets over Y, if m ∈ M(H\Z) and n ∈ N(H\T), then in (M ⊠ N)(X)

[m_{(Z,f)} ⊗ n_{(T,g)}]_{(Y,φ)} = [ι_{G}^{K}(M)_{∗}(f)(m_{(Z,Id)}) ⊗ n_{(T,g)}]_{(Y,φ)} = …

… = [m_{(Z,Id)} ⊗ ι_{H}^{K}(N)^{∗}(f)(n_{(T,g)})]_{(Z,φ∘f)} = …
… = [m_{(Z,Id)} ⊗ N^{∗}(G\π_{T})(n)_{(Z.T,π_{Z})}]_{(Z,φ∘f)} = …
… = [m_{(Z,Id)}⊗ι_{H}^{K}(N)_{∗}(π_{Z})(N^{∗}(G\π_{T})(n)_{(Z.T,Id_{Z.T})})]_{(Z,φ∘f)} = …

… = [M^{∗}(H\π_{Z})(m)_{(Z.T,Id_{Z.T})} ⊗ N^{∗}(G\π_{T})(n)_{(Z.T,Id_{Z.T})}]_{(Y,φ)}

Thus Φ ∘ Θ is also the identity, and this proves the proposition.

From now on, I will identify (M ⊠ N)(X) with the right hand side of (6.11).

Corollary 6.8. If θ : X → X′ is a morphism of (G × H)-sets, then

(M ⊠ N)_{∗}(θ)([m ⊠ n]_{(Y,φ)}) = [m ⊠ n]_{(Y,θ∘φ)}

If (Y′, φ′) is a (G × H)-set over X′, if m′ ∈ M(H\Y′) and n′ ∈ N(G\Y′), then

(M ⊠ N)^{∗}([m′ ⊠ n′]_{(Y′,φ′)}) = [M^{∗}(H\a)(m′) ⊠ N^{∗}(G\a)(n′)]_{(Y,φ)}

where Y, φ, and a are defined by the cartesian square

Y ⟶^{a} Y′
φ↓  ↓^{φ′}
X ⟶_{θ} X′

Proof. This follows from straightforward translations, using the isomorphisms Φ and Θ of the proposition.

Remark 6.9. Those identifications give an explicit form of the isomorphisms of proposition 6.3

(A ⊠ M)(X) ≃ A ⊗ M(X)

Just map a ⊗ m ∈ A ⊗ M(X) to [A^{∗}(p_{X})(a) ⊠ m]_{(X,Id_{X})} ∈ (A ⊠ M)(X), where p_{X} : X → • is the only possible map. Conversely, map [α ⊗ m]_{(Y,φ)} to

∑_{y∈G\Y} α(y) ⊗ M_{∗}(φ)M_{∗}(i_{y})M^{∗}(i_{y})(m)

where i_{y} : G/G_{y} → Y is defined by i_{y}(gG_{y}) = gy.

Remark 6.10. Recall from [Bou96b] Section 3 the following notations: if L is a subgroup of G × H, denote by G′ = p_{1}(L) (resp. H′ = p_{2}(L)) the projection of L on G (resp. on H). Let

k_{1}(L) = {g ∈ G | (g, 1) ∈ L} k_{2}(L) = {h ∈ H | (1, h) ∈ L}

Then k_{j}(L) \underline{◁} p_{j}(L) for j = 1, 2, and the quotients K =p_{1}(L)/k_{1}(L) and p_{2}(L)/k_{2}(L) are isomorphic. Let s : G′ → K be the canonical surjection. Then there exists a surjection t : H′ → K such that

L = {(g, h) ∈ G′ × H′ | s(g) = t(h)}

The previous identification of M ⊠N gives the following evaluation at the subgroup L of G × H:

(M ⊠ N)(L) = ( ⊕_{K⊆L} M(p_{1}(K)) ⊗ N(p_{2}(K))) /J

where J is the submodule generated by

t_{p_{1}(K′)}^{p_{1}(K)}m ⊗ n − m ⊗ r_{p_{2}(K′)}^{p_{2}(K)}n for K′ ⊆ K, k_{2}(K′) = k_{2}(K)

r_{p_{1}(K′)}^{p_{1}(K)}m ⊗ n − m ⊗ t_{p_{2}(K′)}^{p_{2}(K)}n for K′ ⊆ K, k_{1}(K′) = k_{1}(K)

xm ⊗ yn − m ⊗ n for (x, y) ∈ L

6.7. Wreath products. Let G be a group, and n be a positive integer. I denote by G ≀ S_{n} the wreath product of G with the symmetric group S_{n}: it is the semi-direct product of G^{n} and S_{n}, for the action of S_{n} by permutations of the factors of G^{n}. There is a more intrinsic presentation of those wreath products:

Proposition 6.11. Let G be a group, and E be a free right G-set. If n = |E/G| is finite, then Aut_{G}(E) ≃ G ≀ S_{n}.

Proof. Choose a system of representatives Ω of E/G. Then as a right G-set

E ≃ Ω × G

where G acts trivially on Ω. Now if θ ∈ Aut_{G}(Ω × G)

∀ω ∈ Ω, ∀g ∈ G, θ(ω, g) = θ(ω, 1)g

so there exists a map γ : G → Ω and a permutation σ of Ω, such that

θ(ω, g) = (σ(ω), γ(ω)g)

Conversely, if γ ∈ G^{Ω} and σ is a bijection of Ω, this formula defines a G-auto-morphism of Ω × G. If |Ω| = n, this states a bijection between Aut_{G}(Ω × G) and G ≀ S_{n}, which is easily seen to be a group isomorphism.

Definition 6.12. If G is a finite group, and n is a non-negative integer, define M^{⊠n} as the Mackey functor for G^{n} equal to the Burnside functor for the trivial group if n = 0, and to M^{n−1} ⊠ M if n ≥ 1.

Note in particular that M^{⊠1} ≃ M.

Proposition 6.13. Let G be a finite group, and E be a finite free right G-set. Then if |E/G| = n, and if H = Aut_{G}(E), the group H is isomorphic to G ≀ S_{n}, and the set E is a finite H-set-G. Then for any Mackey functor M for G

Res_{G^{n}}^{H}T_{E}(M) ≃ M^{⊠n}

Proof. It follows from corollary 5.6 that

Res_{G^{n}}T_{E}(M) = T_{Res_{G^{n}×G^{op}} E}(M)

Now the restriction of E to G^{n} × G^{op} is the disjoint union of n-copies of G: more precisely, if G(i) is the i^{th}-factor of G^{n}

Res_{G^{n}×G^{op}}E = ⨆_{i=1}^{n} Inf_{G(i)×G^{op}}^{G^{n}×G^{op}}G

It follows that

Res_{G^{n}}T_{E}(M) ≃ ˆ⊗_{i=1}^{n} T_{Inf_{G(i)×G^{op}}^{G^{n}×G^{op}} G}(M)

But moreover

T_{Inf_{G(i)×G^{op}}^{G^{n}×G^{op}} G}(M) = ι_{G(i)}^{G^{n}}(M)

An easy induction argument shows that

M^{⊠n} = ι_{G(1)}^{G^{n}}(M) ˆ⊗ … ˆ⊗ ι_{G(n)}^{G^{n}}(M)

and this completes the proof of the proposition.

Remark 6.14. Suppose that H is a subgroup of G, of index n. Let E = G viewed as a free right H-set. Then |E/H| = n. Moreover, the group G acts on the left on E by multiplication, and this action commutes with the right action of H. This gives a morphism G → Aut_{H}(E) ≃ H ≀ S_{n}: this is the Frobenius morphism.

In that case, the functor T_{E} certainly deserves the name of tensor induction, and could be denoted by Ten_{H}^{G}.

7. Tensor induction for Green functors

7.1. Green functors. Recall the following definition (see [Bou97] Chapter 2)

Definition 7.1. Let R be a commutative ring. A Green functor A (over R) for the group G is a Mackey functor (over R) endowed for any G-sets X and Y of bilinear maps

A(X) × A(Y) → A(X × Y)

denoted by (a, b) ↦ a × b which are bifunctorial, associative, and unitary, in the following sense:

• (Bifunctoriality) If f : X → X′ and g : Y → Y′ are morphisms of G-sets, then the squares

A(X) × A(Y) ⟶^{×} A(X × Y)
A_{∗}(f)×A_{∗}(g)↓  ↓^{A_{∗}(f×g)}
A(X′) × A(Y′) ⟶_{×} A(X′ × Y′)
A(X) × A(Y) ⟶^{×} A(X × Y)
A^{∗}(f)×A^{∗}(g)↑  ↑^{A^{∗}(f×g)}
A(X′) × A(Y′) ⟶_{×} A(X′ × Y′)

are commutative.

• (Associativity) If X, Y and Z are G-sets, then the square

A(X) × A(Y) × A(Z) ⟶^{Id_{A(X)}×(×)} A(X) × A(Y × Z)
(×)×Id_{A(Z)}↓  ↓^{×}
A(X × Y) × A(Z)  ⟶_{×}  A(X × Y × Z)

is commutative, up to identifications (X×Y)×Z ≃ X×Y ×Z ≃ X×(Y ×Z).

• (Unitarity) If • denotes the G-set with one element, there exists an element ε ∈ A(•) such that for any G-set X and for any a ∈ A(X)

A_{∗}(p_{X})(a × ε) = a = A_{∗}(q_{X})(ε × a)

denoting by p_{X} (resp. q_{X}) the (bijective) projection from X × • (resp. from • × X) to X.

If A and B are Green functors for the group G, a morphism f (of Green functors) from A to B is a morphism of Mackey functors such that for any G-sets X and Y, the square

A(X) × A(Y) ⟶^{×} A(X × Y)
f_{X}×f_{Y} ↓  ↓^{f_{X×Y}}
B(X) × B(Y) ⟶_{×} B(X × Y)

is commutative. If moreover f_{•} : A(•) → B(•) maps the unit of A to the unit of B, then I will say that f is unitary.

Equivalently, a Green functor A (over ℤ) can be defined as a Mackey functor equipped with morphisms

μ : A ˆ⊗ A → A ε : b → A

satisfying the usual conditions of associativity and unitarity: the following diagrams are commutative

\begin{matrix}A ˆ⊗ A ˆ⊗ A & ⟶^{1 ˆ⊗ μ} & A ˆ⊗ A \\ ^{μ ˆ⊗ 1}↓ & & ↓^{μ} \\ A ˆ⊗ A & ⟶_{μ} & A \end{matrix} [figure: b ˆ⊗ A; A; A ˆ⊗ A; A ˆ⊗ b; \underrightarrow{ε ˆ⊗ 1}; \underleftarrow{1 ˆ⊗ ε}; η; μ; η] (7.14)

Here η is any of the canonical isomorphisms b ˆ⊗ A ≃ A ≃ A ˆ⊗ b.

Similarly, if A is a Green functor for G, an A-module M is a Mackey functor equipped with a morphism μ : A ˆ⊗ M → M, such that the following diagrams are commutative

\begin{matrix}A ˆ⊗ A ˆ⊗ M & ⟶^{1 ˆ⊗ μ} & A ˆ⊗ M \\ ^{μ ˆ⊗ 1}↓ & & ↓^{μ} \\ A ˆ⊗ M & ⟶_{μ} & M \end{matrix} [figure: b ˆ⊗ M; M; A ˆ⊗ M; \underrightarrow{ε ˆ⊗ 1}; η; μ] (7.15)

I will denote by A-Mod the category of A-modules.

7.2. Tensor induction. Tensor induction maps Green functors to Green functors:

Proposition 7.2. Let G and H be finite groups, and U be a finite H-set-G.

1. If A is a Green functor for G, then T_{U}(A) is a Green functor for H.

2. If M is an A-module, then T_{U}(M) is a T_{U}(A)-module.

3. The correspondence M ↦ T_{U}(M) is a (generally non-additive) right exact functor from A-Mod to T_{U}(A)-Mod.

Proof. This is clear, since T_{U} commute with tensor products, and maps the Burnside functor to the Burnside functor. So the image by T_{U} of the diagrams (7.14) give commutative diagrams

T_{U}(A) ˆ⊗ T_{U}(A) ˆ⊗ T_{U}(A) ⟶^{1 ˆ⊗ T_{U} (μ)} T_{U}(A) ˆ⊗ T_{U}(A)
T_{U} (μ) ˆ⊗ 1↓  ↓^{T_{U} (μ)}
T_{U}(A) ˆ⊗ T_{U}(A)  ⟶_{T_{U} (μ)}  T_{U}(A)

[figure: b ˆ⊗ T_{U}(A); T_{U}(A); T_{U}(A) ˆ⊗ T_{U}(A); T_{U}(A) ˆ⊗ b; \underrightarrow{T_{U}(ε) ˆ⊗ 1}; \underleftarrow{1 ˆ⊗ T_{U}(ε)}; η; T_{U}(μ); η]

So T_{U}(A) is a Green functor.

Similarly, if M is an A-module, then the images of the diagrams (7.15) give a structure of T_{U}(A) module on T_{U}(M). By definition, the correspondence M ↦ T_{U}(M) is a right exact functor.

Remark 7.3. There is a similar result for the functors L_{U} (see [Bou97] Proposition 10.3.2). In the case U/G = •, I have seen that T_{U} = L_{U}. One can check that the two structures of Green functor coincide in this case.

Remark 7.4. If G and H are finite groups, if A is a Green functor for G and B is a Green functor for H, then A ⊠ B is a Green functor for G × H: this follows from the fact that ι_{G}^{G×H}(A) and ι_{H}^{G×H}(B) are, and that the tensor product of Green functors is a Green functor (see [Bou97] Proposition 6.3.1).

7.3. Examples.

7.3.1. Extension of coefficients. Let G and H be finite groups. If R is a commutative ring, then a Mackey functor over R for G is just a module over the Green functor

R ⊠ b ≃ ι_{{1}}^{G}(R) ˆ⊗ b ≃ ι_{{1}}^{G}(R)

If U is a finite G-set-H, then T_{U}(M) is a module over the Green functor

T_{U}(R ⊠ b) ≃ T_{U/G}(R) ˆ⊗ T_{U/{1}}(b) ≃ T_{U/G}(R)

This functor T_{U/G}(R) depends only on R and U/G, which is just a finite H-set. More generally, if A is an abelian group, and V is an H-set, the functor T_{V} (A) is a Mackey functor for H, that may be quite complicated, even if A is cyclic:

Proposition 7.5. Let n be an positive integer, and Ω be a set of cardinality n. If H is a finite group, and V is a finite H-set, then T_{V} (Z/nZ) is the quotient of the Burnside functor b for H by the subfunctor generated by the elements Ω^{W} = Hom_{{1}}(W, Ω) ∈ b(H_{W}), where W is a non-empty subset of V.

Proof. Let 0 → Z →^{n} Z → Z/nZ → 0 be the standard resolution. As Mackey functors over the trivial group, it can be viewed as

0 → b_{•} →^{n} b_{•} → Z/nZ → 0

The {1}-poset over • × • defining the morphism n : b_{•} → b_{•} is just the discrete set Ω with n elements. Now T_{V} (Z/nZ) is defined by the following exact sequence

b_{Hom_{{1}}(V,•;•)} → b_{Hom_{{1}}(V,•)} → T_{V} (Z/nZ) → 0

This is also

b_{Hom_{{1}}(V,•;•)} → b → T_{V} (Z/nZ) → 0

Evaluation at the trivial H-set gives

b(Hom_{{1}}(V, •; •)) → b(•) → T_{V} (Z/nZ)(•) → 0

Now an element φ of Hom_{{1}}(V, •; •) is completely determined by the preimage W of the first point, which must be non-empty. If K is a subgroup of H, then φ in invariant by K if and only if W is. So the H-sets H/K → Hom_{{1}}(V, •; •) are in one to one correspondence with the K-invariant non-empty subsets of V, and they generate b(Hom_{{1}}(V, •; •)) as K runs through the subgroups of H. To find the image of these elements in b(•), I have to build pull-backs

T ⟶ Hom_{{1}}(V, Ω; •)
↓  ↓^{a}
H/K ⟶_{φ} Hom_{{1}}(V, •; •)

Now if φ is associated to the subset W of V, the pull-back T is isomorphic to

T ≃ Ind_{K}^{H}a^{−1}(φ(K))

Moreover a^{−1}(φ(K)) is the set of maps β : V → Ω ⊔ •, such that β(W) ⊆ Ω, and β(V − W) = •. So it is isomorphic to Hom_{{1}}(W, Ω). Thus

Ind_{K}^{H}a^{−1}(φ(K)) = Ind_{K}^{H}Res_{K}^{H_{W}} Ω^{W}

Now the proposition follows, since for any subgroup L of H, the module

T_{V} (L) = (Res_{L}^{H}T_{V})(•) = T_{Res_{L}^{H}V} (•)

can be computed by the previous procedure.

Example 7.6. Suppose H = ℤ/pℤ, for a prime p, and V = H, acted on by multiplication. Then if ∅ ≠ W ⊆ V, either H_{W} is trivial, or H_{W} = H and W = V. This proves that T_{V} (ℤ/nℤ)(H) is the quotient of b(H), generated by H/H and H/{1}, by the submodule generated by the elements

nH/H + \frac{n^{p} − n}{p}H/1 and nH/1

whereas T_{V} (ℤ/nℤ)(1) is the quotient of b(1) ≃ ℤ by n{1}/{1}. So the evaluation of T_{V} (ℤ/nℤ) at the trivial subgroup is always ℤ/nℤ. Its evaluation at H is the quotient of ℤ^{2} by the submodule generated by (n, d) and (0, n), where d is the g.c.d of n and \frac{n^{p}−n}{p}, equal to n if n is prime to p, and to n/p if p | n. So if n is prime to p, the functor T_{V} is just ℤ/nℤ ⊠ b. But if p | n, then

T_{V} (ℤ/nℤ)(H) ≃ ℤ/(n/p)ℤ × ℤ/(np)ℤ

This example shows that T_{U}(R) need not be a Mackey functor over R, even if R = ℤ/pℤ for a prime p. However, there is a simple case:

Corollary 7.7. Let n be a positive integer, and H be a finite group. If V is a (non-empty) finite H-set, then

1. If n is prime to the order of H, then

T_{V} (ℤ/nℤ) ≃ ℤ/nℤ ⊠ b

2. Let h be the least integer such that if H_{1} ⊂ H_{2} … ⊂ H_{h} = H is a strictly increasing sequence of subgroups of G, then H_{1} is of order prime to n. Then T_{V} (ℤ/nℤ) is a Mackey functor over ℤ/n^{h}ℤ.

Proof. The first assertion is proved by induction on the order of H. If H is trivial, this is true, because T_{U}(ℤ/nℤ)(1) is a tensor product of copies of ℤ/nℤ, thus isomorphic to ℤ/nℤ. This is because the functor T_{•} is the identity functor for the trivial group. Note next that Ω^{W_{1}⊔W_{2}} = Ω^{W_{1}} × Ω^{W_{2}} in b(H). Now if W is any orbit of H on V, then Ω^{W} can be written as

Ω^{W} = nH/H + ∑_{f} H/H_{f} (7.16)

where the H_{f} are proper subgroups of H, equal to stabilizers of non-constant functions f from W to Ω. Moreover, the cyclic group C = ℤ/nℤ acts on Ω^{W}, and this action commutes with the action of H. The stabilizer of f ∈ Ω^{W} in C × H is the direct product D × K of its projections on C and H, because C and H have relatively prime orders. Now D stabilizes f, so D = 1.

Moreover, if c ∈ C and f ∈ Ω^{W}, then the stabilizer of c.f in H is equal to the stabilizer of f, and c.f cannot be in the orbit of f under H if c ≠ 1. This proves that the coefficient of H/H_{f} in the sum (7.16) is a multiple of n, so T_{V} (ℤ/nℤ)(H) is the quotient of b(H) by a submodule I contained in nb(H). Moreover, by induction hypothesis, the element nH_{f}/H_{f} is zero in T_{V} (ℤ/nℤ)(H_{f}). As

nH/H_{f} = Ind_{H_{f}}^{H} (nH_{f}/H_{f})

this element is zero in T_{V} (ℤ/nℤ)(H). So nH/H = 0 in T_{V} (ℤ/nℤ)(H), and nb(H) ⊆ I. So

T_{V} (ℤ/nℤ)(H) = b(H)/nb(H) ≃ (ℤ/nℤ ⊠ b)(H)

This proves the first assertion.

The second assertion follows by induction on h: the case h = 1 is the first assertion. All I have to show then is that n^{h}T_{V} (ℤ/nℤ)(H) = 0. But this is clear, since multiplying equation (7.16) by n^{h−1} gives

n^{h}H/H + ∑_{f} n^{h−1}H/H_{f} = 0 in T_{V} (H)

By induction hypothesis, for each f in this sum, I have n^{h−1}H_{f}/H_{f} = 0 in T_{V} (H_{f}), so n^{h}H/H = 0, as required.

7.3.2. Direct product of Green functors. If A and B are Green functors for the group G, then A ˆ⊗ B is also a Green functor for G (see [Bou97] 6.3). The product on A ˆ⊗ B follows from associativity and commutativity of tensor product of Mackey functors, which give the following morphisms

(A ˆ⊗ B) ˆ⊗(A ˆ⊗ B) →^{≃} A ˆ⊗ B ˆ⊗ A ˆ⊗ B →^{≃} A ˆ⊗ A ˆ⊗ B ˆ⊗ B → A ˆ⊗ B

Similarly, the unit of A ˆ⊗ B follows from the isomorphism b ˆ⊗ b ≃ b

b ≃ b ˆ⊗ b →^{ε_{A} ˆ⊗ ε_{B}} A ˆ⊗ B

The following lemma deals with similar morphisms between directs products:

Lemma 7.8. Let G and H be finite groups.

1. If A and B are Mackey functors for G, if C and D are Mackey functors for H, then there are isomorphisms

(A ˆ⊗ B) ⊠ (C ˆ⊗ D) ≃ (A ⊠ C) ˆ⊗(B ⊠ D)

of Mackey functors for G × H, which are natural in A, B, C, and D.

2. There is an isomorphism

b ⊠ b ≃ b

of Mackey functors for G × H.

Proof. The first assertion follows from associativity and commutativity of tensor product, and of corollary 4.5, since

(A ˆ⊗ B) ⊠ (C ˆ⊗ D) = ι_{G}^{G×H}(A ˆ⊗ B) ˆ⊗ ι_{H}^{G×H}(C ˆ⊗ D) ≃ …

… ≃ ι_{G}^{G×H}(A) ˆ⊗ ι_{G}^{G×H}(B) ˆ⊗ ι_{H}^{G×H}(C) ˆ⊗ ι_{H}^{G×H}(D) ≃ …

… ≃ ι_{G}^{G×H}(A) ˆ⊗ ι_{H}^{G×H}(C) ˆ⊗ ι_{G}^{G×H}(B) ˆ⊗ ι_{H}^{G×H}(D) ≃ (A⊠C) ˆ⊗(B ⊠D)

The second assertion is also clear, since

b ⊠ b = ι_{G}^{G×H}(b) ˆ⊗ ι_{H}^{G×H}(b) ≃ b ˆ⊗ b ≃ b

Note that in b ⊠ b ≃ b, the first b is the Burnside functor for G, the second is the Burnside functor for H, and the third is the Burnside functor for G × H.

Now if A is a Green functor for G, and H is a Green functor for H, then the morphisms

(A ⊠ B) ˆ⊗(A ⊠ B) →^{≃} (A ˆ⊗ A) ⊠ (B ˆ⊗ B) → A ⊠ B

b ≃ b ⊠ b \underrightarrow{ε_{A} ⊠ ε_{B}} A ⊠ B

turn A ⊠ B into a Green functor for G × H. Similarly, if M is an A-module and N is a B-module, then M ⊠ N has a natural structure of A ⊠ B-module. Moreover, the isomorphisms of corollary 4.5 and proposition 6.5 are isomorphisms of Green functors:

Proposition 7.9. Let G, H, and K be finite groups.

1. If A is a Green functor for G, if B is a Green functor for H, and U is a finite K-set-(G × H), then there is an isomorphism of Green functors

T_{U}(A ⊠ B) ≃ T_{U/H}(A) ˆ⊗ T_{U/G}(B)

2. If A and B are Green functors for G, and if U is a finite H-set-G, then there there is an isomorphism of Green functors

T_{U}(A ˆ⊗ B) ≃ T_{U}(A) ˆ⊗ T_{U}(B)

3. If A is a Green functor for G, and U and U′ are finite H-sets-G, then there there is an isomorphism of Green functors

T_{U⊔U′}(A) ≃ T_{U}(A) ˆ⊗ T_{U′}(A)

4. If A is a Green functor for G, and if U is a finite H-set-G, then there is an isomorphism of Green functors

T_{U}(A^{op}) ≃ T_{U}(A)^{op}

Proof. The two first assertions follows from the definition of the Green functor structure on T_{U}(A⊠B) and T_{U}(A ˆ⊗ B). The third one follows from the the following diagram

T_{U⊔U′}(A) ˆ⊗ T_{U⊔U′}(A)  \underrightarrow{≃} T_{U⊔U′}(A ˆ⊗ A)

\begin{matrix}↓ \\ T_{U}(A) ˆ⊗ T_{U′}(A) ˆ⊗ T_{U}(A) ˆ⊗ T_{U′}(A) \\ ↓ \\ T_{U}(A) ˆ⊗ T_{U}(A) ˆ⊗ T_{U′}(A) ˆ⊗ T_{U′}(A) \\ ↓ \\ T_{U}(A ˆ⊗ A) ˆ⊗ T_{U′}(A ˆ⊗ A) \\ ↓ \end{matrix}  ↓ T_{U⊔U′}(p)

T_{U}(A) ⊔ T_{U′}(A)  \overrightarrow{≃} T_{U⊔U′}(A)

where p : A ˆ⊗ A → A is the product of A. This diagram is commutative, because for any G-sets X and Y, the following diagram is commutative

b_{Hom_{(}U⊔U′,X)} ˆ⊗ b_{Hom_{G}(U⊔U′,X)}  \underrightarrow{≃}b_{Hom_{G}(U⊔U′,X×Y)}
↓

b_{Hom_{G}(U,X)} ˆ⊗ b_{Hom_{G}(U′,X)} ˆ⊗ b_{Hom_{G}(U,Y)} ˆ⊗ b_{Hom_{G}(U′,Y)}

↓

b_{Hom_{G}(U,X)} ˆ⊗ b_{Hom_{G}(U,Y)} ˆ⊗ b_{Hom_{G}(U′,X)} ˆ⊗ b_{Hom_{G}(U′,Y)} ↓

↓ Id

b_{Hom_{G}(U,X×Y)} ˆ⊗ b_{Hom_{G}(U′,X×Y)}  \overrightarrow{≃}b_{Hom_{G}(U⊔U′,X×Y)}

For last assertion, recall that the product on A^{op} can be defined by

A ˆ⊗ A →^{σ} A ˆ⊗ A → A

where σ denotes the natural isomorphism expressing the commutativity of the tensor product. Now the assertion follows from the following commutative diagram, where M and N are Mackey functors for G

T_{U}(M ˆ⊗ N) ⟶^{≃} T_{U}(M) ˆ⊗ T_{U}(N)
T_{U} (σ)↓  ↓^{σ}
T_{U}(N ˆ⊗ M) ⟶_{≃} T_{U}(N) ˆ⊗ T_{U}(M)

To check the commutativity of this diagram, it suffices to suppose M = b_{X} and N = b_{Y}, for some G-sets X and Y. In that case, it follows from the commutativity of the following square

Hom_{G}(U^{op}, X × Y) ⟶^{≃} Hom_{G}(U^{op}, X) × Hom_{G}(U^{op}, Y)
Hom_{G}(U^{op},s)↓  ↓^{s}
Hom_{G}(U^{op}, Y × X ⟶_{≃} Hom_{G}(U^{op}, Y) × Hom_{G}(U^{op}, Y)

where s is the bijection exchanging the factors of a product of sets.

8. Cohomological tensor induction

8.1. Cohomological Mackey functors. Let G be a finite group, and R be a commutative ring. A Mackey functor M for G over R is called cohomological if for any subgroups K ⊆ H of G, the composition t_{K}^{H}r_{K}^{H} is multiplication by the index [H : K]. This is equivalent to say that M is a module over the Green functor F P_{R} (see [TW95] Proposition 16.3). The functor F P_{V} is defined more generally for an RG-module V and a finite G-set X by

F P_{V} (X) = Hom_{RG}(RX, V)

where RX is the permutation RG-module associated to X (see [Bou97] 4.5.2). Clearly F P_{R} ≃ R × F P_{ℤ}, so a Mackey functor over R is cohomological if and only if it is cohomological as a Mackey functor (over ℤ).

Lemma 8.1. Let G be a finite group, and M be a Mackey functor for G. Then M admits a biggest cohomological quotient M^{coh}, given by

M^{coh} = F P_{ℤ} ˆ⊗ M

Proof. This follows from the alternative description of the tensor product of Mackey functors (see [Bou97] Proposition 1.5.1): if M and N are Mackey functors for G, if H is a subgroup of G, then

(N ˆ⊗ M)(H) ≃ ( ⊕_{K⊆H} N(K) ⊗ M(K))/J

where J is the submodule generated by the elements

t_{L}^{K}n ⊗ m − n ⊗ r_{L}^{K}m for L ⊆ K ⊆ H, n ∈ N(L), m ∈ M(K)
r_{L}^{K}n ⊗ m − n ⊗ t_{L}^{K}m for L ⊆ K ⊆ H, n ∈ N(K), m ∈ M(L)
hn ⊗ hm − n ⊗ m for K ⊆ H, n ∈ N(K), m ∈ M(K), h ∈ H

Moreover, if K is a subgroup of G, then F P_{ℤ}(K) ≃ ℤ. Restriction maps are identity, and transfers are multiplication by the index. Thus

(F P_{ℤ} ˆ⊗ M)(H) ≃ ( ⊕_{K⊆H} ℤ ⊗ M(K))/J

where J is generated by

1 ⊗ m − 1 ⊗ t_{L}^{K}m for L ⊆ K ⊆ H, m ∈ M(L)
1 ⊗ r_{L}^{K}m − [K : L]1 ⊗ m for L ⊆ K ⊆ H, m ∈ M(K)
1 ⊗ m − 1 ⊗ hm for K ⊆ H, m ∈ M(K), h ∈ H

Now the element 1 ⊗ m, for K ⊆ H, and m ∈ M(K), is equal to 1 ⊗ t_{K}^{H}(m) in (F P_{ℤ} ⊗ M)(H). This proves that (F P_{ℤ} ⊗ M)(H) is the quotient of M(H) by the submodule generated by the elements t_{K}^{H}r_{K}^{H}m − [H : K]m, for K ⊆ H and m ∈ M(K). Thus F P_{ℤ} ˆ⊗ M is a cohomological quotient of M. Moreover, it is clear that any morphism from M to a cohomological functor must factor through F P_{ℤ} ˆ⊗ M, and the lemma follows.

Lemma 8.2. Let G be a finite group, and X be a G-set. Then

(b_{X})^{coh} ≃ F P_{ℤX}

Proof. This follows from the previous lemma, and from the isomorphisms

(b_{X})^{coh} = F P_{ℤ} ˆ⊗ b_{X} ≃ (F P_{ℤ} ˆ⊗ b)_{X} ≃ (F P_{ℤ})_{X}

Moreover the isomorphism (F P_{ℤ})_{X} ≃ F P_{ℤ} follows from the case of a finite G-set X: in that case indeed, for a finite G-set Y

(F P_{ℤ})_{X}(Y) = F P_{ℤ}(Y ×X) = Hom_{ℤG}(ℤ(Y ×X), ℤ) ≃ Hom_{ℤG}(ℤY ⊗ℤX, ℤ) ≃ …

… ≃ Hom_{ℤG}(ℤY, Hom_{ℤ}(ℤX, ℤ)) ≃ Hom_{ℤG}(ℤY, ℤX) = F P_{ℤX}(Y)

since ℤX is self-dual. Those isomorphisms are moreover functorial in Y.

8.2. Tensor induction of cohomological Mackey functors. The image of cohomological Mackey functors by tensor induction is often zero:

Proposition 8.3. Let G and H be finite groups, and U be a finite H-set-G. The following conditions are equivalent:

1. There exists a cohomological Mackey functor M such that T_{U}(M) is nonzero.

2. The functor T_{U}(F P_{ℤ}) is non-zero.

3. There exists a prime number p such that for any u ∈ U, the stabilizer G_{u} of u in G is a p-group.

Proof. If M is cohomological, then M is a module over F P_{ℤ}. Thus T_{U}(M) is a module over T_{U}(F P_{ℤ}). So if T_{U}(F P_{ℤ}) = 0, then T_{U}(M) is a module over the zero Green functor, so it is zero. Thus 1) implies 2).

Now suppose 2) holds. Decompose U as a disjoint union of transitive bisets U_{i}, for i ∈ I. Then

T_{U}(F P_{ℤ}) ≃ ⊗_{i∈I} T_{U_{i}}(F P_{ℤ})

Each U_{i} is a transitive biset, isomorphic to (H × G)/L_{i}, for some subgroup L_{i} of H × G. Choose an index i, and denote by H′ = p_{1}(L_{i}) (resp. G′ = p_{2}(L_{i})) the projection of L_{i} on H (resp. on G). Let

k_{1}(L_{i}) = {h ∈ H | (h, 1) ∈ L_{i}} N = k_{2}(L_{i}) = {g ∈ G | (1, g) ∈ L_{i}}

Let K = p_{2}(L_{i})/k_{2}(L_{i}), and s : G′ → K be the canonical surjection. Then there exists a surjection t : H′ → K such that

L_{i} = {(h, g) ∈ H′ × G′ | t(h) = s(g)}

Moreover, if

K_{1} = {(h, t(h) | h ∈ H′)} ⊆ H × K K_{2} = {(s(g), g) | g ∈ G′} ⊆ K × G

then (see [Bou96b] Lemme 3)

U_{i} = (G × H)/L_{i} ≃ (H × K)/K_{1} ×_{K} (K × G)/K_{2}

Thus the functor T_{U_{i}} factors as

T_{U_{i}} = T_{(H×K)/K_{1}} ∘ T_{(K×G)/K_{2}}

Moreover G is transitive on (K × G)/K_{2} (since the projection of K_{2} on K is K itself), so

T_{(K×G)/K_{2}} = L_{(K×G)/K_{2}}

Thus for any Mackey functor M for G

T_{(K×G)/K_{2}}(M) = (Res_{G′}^{G}M)^{N}

Note that the stabilizer in G of the element u = (h, g)L_{i} of U_{i} is equal to ^{g}N. Now if M = F P_{ℤ}, then Res_{G′}^{G}M = F P_{ℤ}. Moreover, for any subgroup K′ = S/N of K = G′/N

(F P_{ℤ})^{N}(K′) = F P_{ℤ}(S)/J

where J is generated by the submodules t_{T}^{S} F P_{ℤ}(T), for subgroups T of S not containing N. In other words

(Res_{G′}^{G}F P_{ℤ})^{N}(S/N) = ℤ/gcd{[S : T] | N ⊈ T ⊂ S}ℤ

Now if N is not a p-group for some p, then the Sylow subgroups T of S (for various primes), do not contain N, and have relatively prime index in S. So if N is not a p-group for some prime p, then

(Res_{G′}^{G}F P_{ℤ})^{N} = 0

and it follows that T_{U_{i}}(F P_{ℤ}) = 0 in that case. Moreover, if N is a non-trivial p-group, then the q-Sylow subgroups of S for q ≠ p do not contain N, and the greatest common divisor of their indexes in S is a power of p (dividing the order

of S). In particular, the functor (Res_{G′}^{G}F P_{ℤ})^{N} is a Mackey functor over ℤ/p^{a}ℤ, where p^{a} is the p-part of the order of G. Thus if N is a non-trivial p-group, corollary 7.7 shows that T_{U_{i}}(F P_{ℤ}) is a Mackey functor over ℤ/p^{m}ℤ, for some power p^{m} of p.

But the tensor product of a Mackey functor over ℤ/p^{m}ℤ and a Mackey functor over ℤ/q^{l}ℤ is zero if p and q are relatively prime (since the tensor product of abelian groups of relatively prime orders is zero). So if 2) holds, then there exists a prime p such that for any u ∈ U, the stabilizer of u in G is either trivial, or a non-trivial p-group. So 3) holds.

I will admit for a while that 3) implies 1): this will be a consequence of proposition 8.11.

8.3. Cohomological tensor induction. The fixed point functor F P_{R} plays for cohomological Mackey functors over R the role of the Burnside functor for Mackey functors. One can try to translate in the category Comack_{R}(G) of cohomological Mackey functors over R the definitions of tensor induction for Mackey functors, in order to define a “cohomological tensor induction”.

Recall the following theorem:

Theorem 8.4. (Thévenaz-Webb [TW95] Theorem 16.5). Let G be a finite group and R be a commutative ring.

1. A Mackey functor M for G over R is cohomological if and only if it is isomorphic to a quotient of a fixed point functor F P_{V}, where V is an RGmodule, which can be chosen to be a permutation module.

2. A cohomological Mackey functor is projective in Comack_{R}(G) if and only if it is isomorphic to a fixed point functor F P_{V}, where V is a direct summand of a permutation module for G.

I will also use the following:

Lemma 8.5. Let G be a finite group. If X and Y are G-sets, then

Hom_{Comack_{R}(G)}(F P_{RX}, F P_{RY}) ≃ Hom_{RG}(RX, RY)

Proof. This is a direct consequence of the adjunction property of the functors F P_{V} (see [TW90] 6.1): the functor V ↦ F P_{V} is right adjoint to the functor of evaluation at {1}. Moreover F P_{RX}(1) ≃ RX.

Definition 8.6. A permutation cohomological Mackey functor for G over R is a Mackey functor isomorphic to F P_{RX}, for some G-set X. The full subcategory of Comack_{R}(G) consisting of permutation cohomological Mackey functors will be denoted by P Comack_{R}(G).

Thus the category P Comack_{R}(G) can be described as follows: the objects of P Comack_{R}(G) are the G-sets. A morphism in P Comack_{R}(G) from Y to X is a morphism of RG-modules from RY to RX, or equivalently a matrix m(x, y) of coefficients in R, indexed by X ×Y, which is G-invariant (that is m(gx, gy) = m(x, y) for all g ∈ G and (x, y) ∈ X × Y), such that for all y ∈ Y, the coefficient m(x, y) is zero except for a finite number of x ∈ X. The composition of morphisms is given by matrix multiplication.

The previous theorem shows that the subcategory P = P Comack_{R}(G) of C = Comack_{R}(G) satisfies the hypothesis of theorem 2.14: it is formed of projective objects, any object of C is a quotient of some object of P, and P is closed by direct sums. So any functor from P to an abelian category can be uniquely extended to a right exact functor defined on C.

Definition 8.7. (see [Bou96a] 3.1) Let R be a commutative ring, and G and H be finite groups. If U is an H-set-G, I will say that U is free-R if for any u ∈ U, the prime factors of the order of the stabilizer G_{u} of u in G are equal to zero in R.

Since two distinct primes cannot vanish in R if R is non-zero, there are only two cases left if U is free-R:

• Either R has prime characteristic p > 0, and all the groups G_{u} are p-groups (the set U will be called free-p in that case).

• Or R has characteristic zero or non-prime, and all the groups G_{u} are trivial (so the set U is right free).

Proposition 8.8. Let R be a commutative ring. Let G and H be finite groups, and U be a finite free-R H-set-G. If m(x, y) is a matrix defining a morphism in Comack_{R}(G) from Y to X, if φ : U^{op} → X and ψ : U^{op} → Y are morphisms of G-sets, define

m̃(φ, ψ) = ∏_{u∈U/G} m(φ(u), ψ(u))

(this does not depend of the choice of representatives of U/G). Then the correspondence

X ↦ Hom_{G}(U^{op}, X) m ↦ m̃

is a functor T_{U}^{R coh} from P Comack_{R}(G) to P Comack_{R}(H).

Proof. First I must check that m̃ defines a morphism in Comack_{R}(H) from Hom_{G}(U^{op}, Y) to Hom_{G}(U^{op}, X): for a given ψ, the coefficient

m̃(φ, ψ) = ∏_{u∈U/G} m(φ(u), ψ(u))

is non-zero if and only if for any u ∈ U, the coefficient m(φ(u), ψ(u)) is non-zero. So for each u, there is only a finite number of possible values for φ(u). As U is finite, there is a finite number of φ such that m̃(φ, ψ) ≠ 0.

Next I have to check that m̃ is H-invariant. But this is clear, since for h ∈ H

m̃(hφ, hψ) = ∏_{u∈U/G} m(φ(h^{−1}u), ψ(h^{−1}u)) = m̃(φ, ψ)

since the image by h^{−1} of a system of representatives of U/G is another system of representatives.

Now of course, if m represents the identity morphism, then m̃(φ, ψ) is non-zero if and only if φ(u) = ψ(u) for all u ∈ U, or equivalently if φ = ψ. So m̃ is the identity morphism.

Finally, let Z be another G-set, and p be a matrix representing a morphism in Comack_{R}(G) from Z to Y. The product matrix m.p is defined by

(m.p)(x, z) = ∑_{y∈Y} m(x, y)p(y, z)

Let θ ∈ Hom_{G}(U^{op}, Z), and φ ∈ Hom_{G}(U^{op}, X). Then

m̃.p(φ, θ) = ∏_{u∈U/G} (∑_{y∈Y} m(φ(u), y)p(y, θ(u))) (8.17)

Now for a given u ∈ U/G

∑_{y∈Y} m(φ(u), y)p(y, θ(u)) = ∑_{{y∈G_{u}\Y \atop g∈G_{u}/G_{u,y}}} m(φ(u), gy)p(gy, θ(u)) = …

… = ∑_{{y∈G_{u}\Y \atop g∈G_{u}/G_{u,y}}} m(φ(ug), y)p(y, θ(ug)) = ∑_{y∈G_{u}\Y} [G_{u} : G_{u,y}]m(φ(u), y)p(y, θ(u))

As U is free-R, the coefficient [G_{u} : G_{u,y}] is zero unless G_{u,y} = G_{u}, or equivalently y ∈ Y^{G_{u}}.

Now expanding the product in equation (8.17) is equivalent to choosing for each u ∈ U/G an element y_{u} ∈ Y^{G_{u}}. This in turn is equivalent to defining a G-morphism ψ from U^{op} to Y (by ψ(u′) = gy_{u} if u ∈ U/G, g ∈ G, and u′ = ug^{−1}). This gives finally

m̃.p(φ, θ) = ∑_{ψ∈Hom_{G}(U^{op},Y)} ∏_{u∈U/G} m(θ(u), ψ(u))p(ψ(u), φ(u)) = …

… = ∑_{ψ∈Hom_{G}(U^{op},Y)} m̃(θ, ψ)p̃(ψ, φ)

This proves that m ↦ m̃ is multiplicative, and the proposition follows.

Definition 8.9. Let R be a commutative ring, and G and H be finite groups. If U is a finite free-R H-set-G, I will call cohomological tensor induction the unique right exact functor extending the functor T_{U}^{R coh} from P Comack_{R}(G) to Comack_{R}(H). This extension will still be denoted by T_{U}^{R coh}.

Recall that if M is a cohomological Mackey functor for G over R, then T_{U}^{R coh}(M) is obtained by choosing a resolution of M

F P_{RY} →^{φ} F P_{RX} →^{ψ} M → 0

by permutation cohomological Mackey functors. Then T_{U}^{R coh}(M) is defined by the exact sequence

T_{U}^{R coh}(F P_{RY} ⊕ F P_{RX}) \underrightarrow{ΔT_{U}^{R coh}(φ)} T_{U}^{R coh}(F P_{RX}) ⟶ T_{U}^{R coh}(M) → 0

8.4. Extension of coefficients. A cohomological Mackey functor M for G over R is a module over F P_{R}. If f : R → R′ is a morphism of commutative rings, then the induced morphism F P_{R} → F P_{R′} turns F P_{R′} into a right F P_{R}-module. I can then consider the functor R′ ⊠_{R} M defined by

R′ ⊠_{R} M = F P_{R′} ˆ⊗_{F P_{R}} M

(see [Bou97] 6.6 for the definition of this tensor product over F P_{R}). It is a module over F P_{R′}, or equivalently a cohomological Mackey functor over R′. Its evaluation on a G-set X is simply given by tensoring with R′

(R′ ⊠_{R} M)(X) = R′⊗_{R}M(X)

If M admits a resolution

F P_{RY} →^{φ} F P_{RX} →^{ψ} M → 0

then the sequence

R′ ⊠_{R} F P_{RY} \underrightarrow{R′ ⊠_{R} φ} R′ ⊠_{R} F P_{RX} \underrightarrow{R′ ⊠_{R} ψ} R′ ⊠_{R} M → 0

is still exact, because the functor R′⊗_{R}− is right exact. But clearly, for any G-set X

R′ ⊠_{R} F P_{RX} = F P_{R′X}

So I have a resolution of R′M by permutation cohomological Mackey functors over R′.

Now if H is another finite group, and U is a finite free-R H-set-G, then U is also free-R′, since the morphism f is unitary. By definition T_{U}^{R′ coh}(R′ ⊠_{R} M) is the cokernel of

T_{U}^{R′ coh}(F P_{R′Y} ⊕ F P_{R′X}) \underrightarrow{ΔT_{U}^{R′ coh}(φ)} T_{U}^{R′ coh}(F P_{R′X}) (8.18)

But moreover

T_{U}^{R′ coh}(F P_{R′X}) = F P_{R′Hom_{G}(U^{op},X)} = …
… = R′ ⊠_{R} F P_{RHom_{G}(U^{op},X)} = R′ ⊠_{R} T_{U}^{R coh}(F P_{RX})

So T_{U}^{R′ coh}(R′ ⊠_{R} M) is also the cokernel of

R′ ⊠_{R} T_{U}^{R coh}(F P_{RY} ⊕ F P_{RX}) \underrightarrow{R′ ⊠_{R} ΔT_{U}^{R coh}(φ)} R′ ⊠_{R} T_{U}^{R coh}(F P_{RX})

And this proves the following

Lemma 8.10. Let f : R → R′ be a unitary morphism of commutative rings. If G and H are finite groups, if U is a finite free-R H-set-G, then U is also free-R′, and for any cohomological Mackey functor M for G over R, there is an isomorphism

T_{U}^{R′ coh}(R′ ⊠_{R} M) ≃ R′ ⊠_{R} T_{U}^{R coh}(M)

which is natural in M.

Thus cohomological tensor induction commutes with extension of coefficients. In particular, if p is a prime, if U is free-p, then for any cohomological Mackey functor M over 𝔽_{p}

T_{U}^{R coh}(R ⊠_{𝔽_{p}} M) ≃ R ⊠_{𝔽_{p}} T_{U}^{𝔽_{p} coh}(M)

8.5. Comparison. If M is a cohomological Mackey functor for G over a commutative ring R, and if U is finite free-R H-set-G, I can build T_{U}^{R coh}(M). But as M is also a Mackey functor, I can build T_{U}(M). Those two constructions are different, but closely related:

Proposition 8.11. Let G and H be finite groups, and R be a commutative ring. If U is a finite free-R H-set-G, then for any Mackey functor M for G, the functor R ⊠ M^{coh} is a cohomological Mackey functor over R, and there are isomorphisms

T_{U}^{R coh}(R ⊠ M^{coh}) ≃ R ⊠ T_{U}(M)^{coh} ≃ F P_{R} ˆ⊗ T_{U}(M) (8.19)

which are moreover natural in M.

In particular, if M is a cohomological Mackey functor, and if U is right free, then

T_{U}^{ℤ coh}(M) = T_{U}(M)^{coh} (8.20)

Similarly, if p is a prime, if U is free-p, then for any cohomological Mackey functor M over 𝔽_{p}

T_{U}^{𝔽_{p} coh}(M) ≃ 𝔽_{p} ⊠ T_{U}(M)^{coh} ≃ F P_{𝔽_{p}} ˆ⊗ T_{U}(M) (8.21)

Proof. The right hand side isomorphism in (8.19) is clear, since

R ⊠ T_{U}(M)^{coh} ≃ R ⊠ (F P_{ℤ} ˆ⊗ T_{U}(M)) ≃ F P_{R} ˆ⊗ T_{U}(M)

Now both sides of the left hand side isomorphism in (8.19) are functors from Mack(G) to Comack_{R}(H), which are clearly right exact, since they are composed of right exact functors. So it suffices to check that their restrictions to P Mack(G) are isomorphic. But if M = b_{X}, for some G-set X, then M^{coh} = F P_{ℤX}, and

R ⊠ M^{coh} = F P_{RX}

It follows that

T_{U}^{coh}(R ⊠ M^{coh}) ≃ F P_{RHom_{G}(U^{op},X)}

On the other hand, the functor T_{U}(M) is b_{Hom_{G}(U^{op},X)}, thus

R ⊠ T_{U}(M)^{coh} ≃ R ⊠ F P_{ℤHom_{G}(U^{op},X)} ≃ F P_{RHom_{G}(U^{op},X)}

So the functors M ↦ T_{U}^{coh}(R ⊠ M^{coh}) and M ↦ R ⊠ T_{U}(M)^{coh} coincide on objects of P Mack(G). To see that the restrictions of those functors to P Mack(G) are isomorphic, I have to look at their actions on morphisms.

Let me first give this action for the functor Q : M ↦ M^{coh} from Mack(G) to Comack_{ℤ}(G). Lemma 8.2 shows that it maps P mack(G) to P Comack_{ℤ}(G). Now if X and Y are G-sets, and if a is a morphism in P Mack(G) from Y to X, represented by a poset (Δ, f) over Y × X with finite fibres over Y, I want to describe the morphism Q(a). This is a G invariant matrix indexed by X × Y, with coefficients in ℤ, which is obtained by evaluation at the trivial subgroup of the morphism F P_{ℤY} → F P_{ℤX} deduced from a. Since for any Mackey functor M, the evaluations of M and M^{coh} at {1} coincide, the morphism Q(a) is also the evaluation of a at {1}, and its expression will follow from the explicit description of the isomorphism

b_{X}(1) ≃ ℤX

But an element of b_{X}(1) ≃ h_{G}(G/1, X) is an equivalence class of finite G-posets over (G/1) × X. If (Z, f) is such a poset, then as the stabilizer G_{g,x} of any element (g, x) of G × X is trivial, I have b(G_{g,x}) ≃ ℤ, and the Lefchetz invariant Λ_{f^{−1}(g,x)}^{G_{g,x}}

corresponds to χ(f^{−1}(g, x)) under this isomorphism. Now the isomorphism of lemma 3.4 shows that the map

q(Z, f) = ∑_{x∈X} χ(f^{−1}(1, x))x

induces the required isomorphism h_{G}(G/1, X) ≃ b_{X}(1) ≃ ℤX. The inverse isomorphism maps x_{0} ∈ X to the poset (Z, f) defined by

Z = G/1(= G) ∀g ∈ G, f(g) = (g, gx_{0}) ∈ (G/1) × X

were G is given the discrete ordering (clearly f^{−1}(1, x) is empty if x ≠ x_{0}, and f^{−1}(1, x_{0}) = {1} is a singleton).

Now if y ∈ Y is a basis element of ℤY, it can be viewed as the element

[figure: G/1; Y; G/1; Id; m_{y}]

of b_{Y} (G/1), where m_{y}(g) = gy for g ∈ G. Its image by a in b_{X}(1) is given by pull-back

[figure: G/1; Y; X; G/1; Δ; U; Id; f_{Y}; m_{y}; f_{X}]

Denote by φ the map U → (G/1) × X in this diagram. I can write

Q(a)(y) = ∑_{x∈X} χ(φ^{−1}(1, x))x

But moreover

φ^{−1}(1, x) = {(g, δ) ∈ G × Δ | g = 1, gy = f_{Y} (δ), f_{X}(δ) = x} ≃ f^{−1}(y, x)

In other words

Q(a)(y) = ∑_{x∈X} χ(f^{−1}(y, x))x

and the (x, y) entry in the matrix Q(a) is χ(f^{−1}(y, x)).

It follows from the definition of the functor T_{U}^{R coh} that the morphism

T_{U}^{R coh}(R ⊠ Q(a)) : T_{U}^{R coh}(R ⊠ (b_{Y})^{coh}) → T_{U}^{R coh}(R ⊠ (b_{X})^{coh})

is defined by the following matrix of coefficients in R

α(φ, ψ) = ∏_{u∈U/G} (χ(f^{−1}(ψ(u), φ(u))))

for φ ∈ Hom_{G}(U^{op}, X) and ψ ∈ Hom_{G}(U^{op}, Y).

On the other hand, let F denote the morphism

Hom_{G}(U^{op}, f) : Hom_{G}(U^{op}, Δ) → Hom_{G}(U^{op}, Y) × Hom_{G}(U^{op}, X)

It represents a morphism from T_{U}(b_{Y}) to T_{U}(b_{X}). The associated morphism from R ⊠ T_{U}(b_{Y})^{coh} to R ⊠ T_{U}(b_{X})^{coh} is given by the matrix

β(φ, ψ) = χ(F^{−1}(ψ, φ))

But I have shown in the proof of proposition 4.1 that for any subgroup K of H_{ψ,φ}

χ(F^{−1}(ψ, φ)^{K}) = ∏_{u∈[K\U/G]} χ ((f^{−1}(ψ(u), φ(u)))^{G_{K,u}})

where

G_{K,u} = {g ∈ G | ∃k ∈ K, ku = ug}

In the case K = {1}, I have G_{K,u} = G_{u}, which is a p-group if R is of prime characteristic p, and trivial otherwise. In both cases, I have

χ ((f^{−1}(ψ(u), φ(u)))^{G_{u}}) = χ(f^{−1}(ψ(u), φ(u))) (in R)

Hence the matrices α and β are equal. This shows that the restrictions of the functors M ↦ T_{U}^{coh}(R ⊠ M^{coh}) and M ↦ R ⊠ T_{U}(M)^{coh} to P Mack(G) are isomorphic. Since they are both right exact, those two functors are isomorphic, and this completes the proof of proposition 8.19.

Thus if M is cohomological and U is right free, then M^{coh} = M and (8.20) follows from (8.19) for R = ℤ.

Now if M is a cohomological Mackey functor over 𝔽_{p}, then M^{coh} = M, because M is cohomological, and 𝔽_{p}⊠M = M since M is a functor over 𝔽_{p}. So (8.21) follows from (8.19) in the case R = 𝔽_{p}. This completes the proof of proposition 8.11.

As a consequence, I can now prove the missing implication 3) ⇒ 1) in proposition 8.3

Corollary 8.12. Let G and H be finite groups. If p is a prime, and U is a finite free-p H-set-G, then F P_{𝔽_{p}} is a cohomological Mackey functor, and T_{U}(F P_{𝔽_{p}}) ≠ 0.

Proof. Indeed

F P_{𝔽_{p}} ˆ⊗ T_{U}(F P_{𝔽_{p}}) ≃ T_{U}^{𝔽_{p} coh}(F P_{𝔽_{p}}) = F P_{𝔽_{p}} ≠ 0

so in particular T_{U}(F P_{𝔽_{p}}) ≠ 0.

8.6. Composition and graded bisets. In this section, I will explain how to compose those functors T_{U}^{R coh}. Since cohomological tensor induction seems quite compatible with tensor induction of Mackey functors, one could expect a formula similar to proposition 4.3. However, one must be very careful about the meaning of such a formula.

The context is the following: there are three finite groups G, H, and K, a finite biset U, wich is an H-set-G, and a finite biset V, which is a K-set-H. The problem is to compare the functors T_{V}^{R coh} ∘ T_{U}^{R coh} and T_{V ×_{H}U}^{R coh}. Let X be a G-set. Then

T_{V}^{R coh} ∘ T_{U}^{R coh}(F P_{RX}) = T_{V}^{R coh}(F P_{RHom_{G}(U^{op},X)}) = F P_{RHom_{H}(V^{op},Hom_{G}(U^{op},X))}

As usual, the canonical isomorphism

(8.22) Hom_{H}(V^{op}, Hom_{G}(U^{op}, X)) ≃ Hom_{G}((V × U)^{op}, X)

shows that T_{V}^{R coh} ∘ T_{U}^{R coh}(F P_{RX}) and T_{V ×_{H}U}^{R coh}(F P_{RX}) are isomorphic. So the restrictions of T_{V}^{R coh} ∘ T_{U}^{R coh} and T_{V ×_{H}U}^{R coh} to P Comack_{R}(G) are isomorphic on objects. However, they are not isomorphic as functors, which means that the previous isomorphism is badly behaved with respect to morphisms.

To see this, let m : Y → X be a morphism in P Comack_{R}(G), i.e. a G-invariant matrix indexed by X × Y, such that for any y ∈ Y, there is only a finite number of x ∈ X for which m(x, y) ≠ 0. Using isomorphism 8.22, it follows from the definitions that for φ ∈ Hom_{G}((V ×_{H} U)^{op}, X) and ψ ∈ Hom_{G}((V ×_{H} U)^{op}, Y ), I have

T_{V}^{R coh} ∘ T_{U}^{R coh}(m)(φ, ψ) = ∏_{v∈V/H} ∏_{u∈U/G} m(φ(v, u), ψ(v, u)) (8.23)

On the other hand

T_{V ×_{H}U}^{R coh}(m)(φ, ψ) = ∏_{(v,u)∈(V ×_{H}U)/G} m(φ(v, u), ψ(v, u)) (8.24)

As the sets V/H × U/G and (V ×_{H} U)/G are not isomorphic in general, it follows that the expressions in 8.23 and 8.24 are not equal. To see how much they differ, fix sets [U/G] and [V/H] of orbits representatives, and consider the map

θ : [V/H] × [U/G] → (V ×_{H} U)/G , θ(v_{0}, u_{0}) = (v_{0}, u_{0})G .

This map is surjective, since if (v, u) ∈ V ×_{H} U, there exist v_{0} ∈ [V/H] (unique) and h ∈ H such that v = v_{0}h. Now in V ×_{H} U, I have (v, u) = (v_{0}, hu), and there exist u_{0} ∈ [U/G] (unique) and g ∈ G such that hu = u_{0}g. Then clearly

θ(v_{0}, u_{0}) = (v_{0}, u_{0})G = (v_{0}, u_{0}g)G = (v_{0}, hu)G = (v_{0}h, u)G = (v, u)G .

Now two pairs (v_{0}, u_{0}) and (v_{1}, u_{1}) have the same image under θ if and only if there exist h ∈ H and g ∈ G such that

v_{1} = v_{0}h  and  hu_{1} = u_{0}g .

The first equality gives v_{1} = v_{0} = v_{0}h, since v_{1} and v_{0} are in the set of representatives [V/H]. The second one gives u_{1} = h^{−1}u_{0}g. In other words u_{1}G is in the left orbit of u_{0}G under the right stabilizer H_{v_{0}} of v_{0}. So the inverse image θ^{−1}((v, u)G) has cardinality [H_{v} : H_{v,uG}], where

H_{v} = {h ∈ H | vh = v} H_{v,uG} = {h ∈ H | vh = v, ∃g ∈ G, hu = ug}

and I can rewrite equation 8.23 as

(T_{V}^{R coh} ∘ T_{U}^{R coh})(m)(φψ) = ∏_{(v,u)∈[(V ×_{H}U)/G]} (m(φ(v, u), ψ(v, u)))^{[H_{v}:H_{v,uG}]} (8.25)

In particular, this shows that the functors T_{V}^{R coh}∘T_{U}^{R coh} and T_{V ×_{H}U}^{R coh} are isomorphic if and only if elevation at the power [H_{v} : H_{v,uG}] is the identity endomorphism of R, for any (v, u) ∈ V ×_{H} U. This will be the case in any of the following situations:

• The group H acts freely on V, i.e. H_{v} = {1} for all v ∈ V. This will be true by hypothesis if R has characteristic 0 or non-prime.

• The group G acts transitively on U, since in that case, the set U/G is a single point.

• The ground ring is R = 𝔽_{p}.

In general, one can pass from 8.24 to 8.25 by elevating each term m(φ(u), ψ(u)) to the power [H_{v} : H_{v,uG}]. This is always equal to 1 if R has characteristic 0 or non-prime, and a power of p if R has prime characteristic p. In any case, elevation at the power [H_{v} : H_{v,uG}] is an endomorphism of the ring R.

So this suggests to use such ring endomorphisms to define functors between categories of cohomological Mackey functors, and leads to the following definition:

Definition 8.13. An End(R)-graded H-set-G, or graded biset for short, is a couple (U, a) where U is a finite free-R H-set-G, and a is a function from H\U/G to End(R).

If (U′, a′), then the disjoint union of (U, a) and (U′, a′) is the graded biset defined by

(U, a) ⊔ (U′, a′) = (U ⊔ U′, a ⊔ a′)

where a ⊔ a′ is the obvious function H\(U ⊔ U′)/G → End(R).

If K is another group, and (V, b) is an End(R)-graded K-set-H, then the product (V, b) ×_{H} (U, a) is the End(R)-graded K-set-G defined by

(V, b) ×_{H} (U, a) = (V ×_{H} U, b ×_{H} a)

where b ×_{H} a is the function from K\(V ×_{H} U)/G to End(R) defined by

(b ×_{H} a)(v, u) = b(v) ∘ a(u) ∘ π_{v,u}

where π_{v,u} is elevation at the power [H_{v} : H_{v,uG}].

Definition 8.14. If (U, a) is an End(R)-graded H-set-G, and if X is a G-set, let

T_{(U,a)}(X) = Hom_{G}(U^{op}, X)

If m is a morphism in P Comack_{R}(G) from Y to X, let

T_{(U,a)}(m)(φ, ψ) = ∏_{u∈U/G} a(u) [m(φ(u), ψ(u))]

for φ ∈ Hom_{G}(U^{op}, X) and ψ ∈ Hom_{G}(U^{op}, Y).

Clearly, if I take for a the constant function equal to the identity endomorphism of R, then the functor T_{(U,a)} is just the previous functor T_{U}^{R coh}.

Lemma 8.15. The correspondence T_{(U,a)} is a functor from P Comack_{R}(G) to P Comack_{R}(H).

Proof. Let X, Y and Z be G-sets, and let m : Y → X and n : Z → Y be morphisms in P Comack_{R}(G). Then by definition, for φ ∈ Hom_{G}(U^{op}, X) and θ ∈ Hom_{G}(U^{op}, Z)

T_{(U,a)}(m ∘ n)(φ, θ) = ∏_{u∈U/G} a(u) [(m ∘ n)(φ(u), θ(u))] (8.26)

Moreover, for u ∈ U

(m ∘ n)(φ(u), θ(u)) = ∑_{y∈Y} m(φ(u), y)n(y, θ(u))

This can also be written as

(m ∘ n)(φ(u), θ(u)) = ∑_{{y∈G_{u}\Y \atop g∈G_{u}/G_{u,y}}} m(φ(u), gy)n(gy, θ(u)) = …

… = ∑_{y∈G_{u}\Y} [G_{u} : G_{u,y}]m(φ(u), y)n(y, θ(u)) = ∑_{y∈Y^{Gu}} m(φ(u), y)n(y, θ(u))

Now expanding the product in 8.26 is equivalent to choosing an element y ∈ Y^{G_{u}} for all u ∈ U/G. This in turn is equivalent to choosing a G-morphism from U^{op} to Y. Finally, this gives

T_{(U,a)}(m∘n)(φ, θ) = ∑_{ψ∈Hom_{G}(U^{op},Y)} ∏_{u∈U/G} a(u) [m(φ(u), ψ(u))n(ψ(u), θ(u))] = …

… = ∑_{ψ∈Hom_{G}(U^{op},Y)} ∏_{u∈U/G} a(u) [m(φ(u), ψ(u))] ∏_{u∈U/G} a(u) [n(φ(u), ψ(u))] = …

… = ∑_{ψ∈Hom_{G}(U^{op},Y)} T_{(U,a)}(m)(φ, ψ)T_{(U,a)}(n)(ψ, θ)

It proves that T_{(U,a)} is multiplicative on morphisms. Moreover, it is clear that if m is the identity morphism, then

T_{(U,a)}(φ, ψ) = ∏_{u∈U/G} a(u)(δ_{φ(u),ψ(u)})

where δ is a Kronecker symbol. This is non-zero if and only if φ = ψ, and in this case it is equal to 1.

It follows that T_{(U,a)} maps the identity morphism of X to the identity morphism of T_{(U,a)}(X). This completes the proof of the lemma.

Notation 8.16. I will denote by T_{(U,a)}^{R coh} the unique extension of the functor T_{(U,a)} to a right exact functor from Comack_{R}(G) to Comack_{R}(H). This will be called the cohomological tensor induction associated to the graded biset (U, a).

8.7. Properties of cohomological tensor induction.

8.7.1. Tensor product over R. Before stating properties of cohomological tensor induction associated to a graded biset, I have to extend the notion of tensor product to cohomological Mackey functors over R: first note that cohomological Mackey functors over R are also Mackey functors over R, or (R ⊠ b)-modules. Since R ⊠ b is a commutative Green functor, any left (R⊠b)-module is also a right (R⊠b)-module. Thus if M and N are Mackey functors over R, I can define

M ˆ⊗_{R}N = M ˆ⊗_{R⊠b} N

(see [Bou97] 6.6 for the definition of the tensor product in the right hand side). With this definition, it is easy to check that for any subgroup H of G

(M ˆ⊗_{R} N)(H) ≃ ( ⊕_{K⊆H} M(K) ˆ⊗_{R} N(K))/J

where J is the R-submodule generated by

t_{L}^{K}n ⊗ m − n ⊗ r_{L}^{K}m for L ⊆ K ⊆ H, n ∈ N(L), m ∈ M(K)
r_{L}^{K}n ⊗ m − n ⊗ t_{L}^{K}m for L ⊆ K ⊆ H, n ∈ N(K), m ∈ M(L)
hn ⊗ hm − n ⊗ m for K ⊆ H, n ∈ N(K), m ∈ M(K), h ∈ H

In other words, to compute M ˆ⊗_{R} N, replace ⊗ by ⊗_{R}, and “submodule” by Rsubmodule in the formulae for M ˆ⊗ N.

Lemma 8.17. Let G be a finite group, and R be a commutative ring. If M and N are cohomological Mackey functors for G over R, then so is M ˆ⊗_{R} N.

Proof. Indeed M ˆ⊗_{R} N is a Mackey functor over F P_{R} ˆ⊗_{R} F P_{R}. Now the lemma follows from the following isomorphism of Green functors

F P_{R} ˆ⊗_{R} F P_{R} ≃ F P_{R}

which in turn follows from the previous identification of M ˆ⊗_{R} N(H): indeed, if H is a subgroup of G, then

(F P_{R} ˆ⊗_{R} F P_{R})(H) = ( ⊕_{K⊆H} R ˆ⊗_{R} R)/J

where J is the R-submodule generated by

[K : L][1 ⊗ 1]_{L} − [1 ⊗ 1]_{K} for L ⊆ K ⊆ H
[1 ⊗ 1]_{^{h}K} − [1 ⊗ 1]_{K} for K ⊆ H, h ∈ H

So for any L ⊆ H, I have [1 ⊗ 1]_{L} = [H : L][1 ⊗ 1]_{H}, and this proves that (F P_{R} ˆ⊗_{R} F P_{R})(H) ≃ R.

Remark 8.18. The corresponding statement for direct product is false: if G and H are finite groups, if M is a cohomological Mackey functor for G over R and N is a cohomological Mackey functor for H over R, then in general M ⊠ N is not a cohomological Mackey functor: this is because the functor ι_{G}^{G×H} does not preserve cohomological Mackey functors.

8.7.2. Properties.

Theorem 8.19. Let R be a commutative ring.

1. If G, H and K are finite groups, if (U, a) is an End(R)-graded H-set-G and (V, b) is an End(R)-graded K-set-H, then there is an isomorphism of functors

T_{(V,b)}^{R coh} ∘ T_{(U,a)}^{R coh} ≃ T_{(V,b)×_{H}(U,a)}^{R coh}

2. If G and H are finite groups, if M and N are cohomological Mackey functors for G over R, and if (U, a) is an End(R)-graded H-set-G, then

T_{(U,a)}^{R coh}(M ˆ⊗_{R} N) ≃ T_{(U,a)}^{R coh}(M) ˆ⊗_{R} T_{(U,a)}^{R coh}(N)

3. If G and H are finite groups, if (U, a) and (U′, a′) are End(R)-graded Hsets-G, and M is a cohomological Mackey functor for G over R, then

T_{(U,a)⊔(U′,a′)}^{R coh}(M) ≃ T_{(U,a)}^{R coh}(M) ˆ⊗_{R} T_{(U′,a′)}^{R coh}(M)

4. If G and H are finite groups, if (U, a) is an End(R)-graded H-set-G, and if M and M′ are cohomological Mackey functors for G, then

T_{(U,a)}^{R coh}(M ⊕ M′) = ⊕_{V ⊆U V invariant by G V mod. H} Ind_{H_{V}}^{H} T_{(V,a_{|V} )}^{R coh}(M) ˆ⊗_{R} T_{(U−V,a_{|(U−V )})}^{R coh}(M′)

5. If G and H are finite groups, if (U, a) is an End(R)-graded H-set-G, if X is a G-set and M a cohomological Mackey functor for G, then M_{X} is a cohomological Mackey functor for G, and

T_{(U,a)}^{R coh}(M_{X}) ≃ T_{(U,a)}^{R coh}(M)_{Hom_{G}(U^{op},X)}

Moreover, all these isomorphisms are natural.

Proof. Assertion 1) should be clear from the discussion of composition of functors T_{U}^{R coh}. The other assertions state isomorphisms between right exact functors, so it is enough to check these isomorphisms on the restrictions to the corresponding subcategories of permutation cohomological Mackey functors, and this is easy: for instance, assertion 2) is a consequence of the natural isomorphism

F P_{RX} ˆ⊗ F P_{RY} ≃ F P_{R(X×Y)}

and of the bijection Hom_{G}(U^{op}, X × Y) ≃ Hom_{G}(U^{op}, X) × Hom_{G}(U^{op}, Y).

8.7.3. Cohomological Green functors. A cohomogical Green functor A for G over R is a Green functor over R, which is cohomological as a Mackey functor. Equivalently, it is a Green functor over R, such that the unit morphism ε : b → A factors through b → F P_{ℤ}. This is also equivalent to say that A is a Green functor, with a (unitary) morphism of Green functors F P_{R} → A. In particular, any Amodule is a cohomological Mackey functor for G over R.

Lemma 8.20. Let R be a commutative ring. Let G and H be finite groups, and (U, a) be an End(R)-graded H-set-G. If A is a cohomological Green functor for G over R, then T_{(U,a)}^{R coh}(A) is a cohomological Green functor for H over R. If M is an A-module, then T_{(U,a)}^{R coh}(M) is a T_{(U,a)}^{R coh}(A)-module, and the correspondence M ↦ T_{(U,a)}^{R coh}(M) is a functor from A-Mod to T_{(U,a)}^{R coh}(A)-Mod.

Proof. The images by T_{(U,a)}^{R coh} of the commutative diagrams

\begin{matrix}A ˆ⊗_{R} A ˆ⊗_{R} A & ⟶^{1 ˆ⊗_{R} μ} & A ˆ⊗_{R} A \\ ^{μ ˆ⊗_{R} 1}↓ & & ↓^{μ} \\ A ˆ⊗_{R} A & ⟶_{μ} & A \end{matrix} [figure: F P_{R} ˆ⊗_{R} A; A; A ˆ⊗_{R} A; A ˆ⊗_{R} F P_{R}; \underrightarrow{ε ˆ⊗_{R} 1}; \underleftarrow{1 ˆ⊗_{R} ε}; η; μ; η]

give a Green functor structure on T_{(U,a)}^{R coh}(A), and a unitary morphism of Green functors T_{(U,a)}^{R coh}(F P_{R}) = F P_{R} → T_{(U,a)}^{R coh}(A), so T_{(U,a)}^{R coh}(A) is a cohomological Green functor. The other assertions of the lemma are trivial.

Proposition 8.21. Let R be a commutative ring, and G and H be finite groups.

1. If A and B are cohomological Green functor for G over R, and if (U, a) is an End(R)-graded H-set-G, then there is an isomorphism of cohomological Green functors

T_{(U,a)}^{R coh}(A ˆ⊗_{R} B) ≃ T_{(U,a)}^{R coh}(A) ˆ⊗_{R} T_{(U,a)}^{R coh}(B)

2. If A is a cohomological Green functor for G over R, and (U, a) and (U′, a′) are End(R)-graded H-sets-G, then there is an isomorphism of Green functors

T_{(U,a)⊔(U′,a′)}^{R coh}(A) ≃ T_{(U,a)}^{R coh}(A) ˆ⊗_{R} T_{(U′,a′)}^{R coh}(A)

3. If A is a cohomological Green functor for G over R, and (U, a) is an End(R)- graded H-set-G then there is an isomorphism of Green functors

T_{(U,a)}^{R coh}(A^{op}) ≃ T_{(U,a)}^{R coh}(A)^{op}

Proof. The proof of these assertions is similar to the proof of the corresponding assertions of proposition 7.9.

9. Tensor induction for p-permutation modules

9.1. Definition. Let k be a field of characteristic p. If G and H are finite groups, and if (U, a) is an End(k)-graded H-set-G, I have built the functor T_{(U,a)}^{k coh} from Comack_{k}(G) to Comack_{k}(H). This functor maps permutation cohomogical Mackey functors for G over k to permutation cohomological Mackey functors for H. Hence it maps projective cohomological Mackey functors for G to projective cohomological Mackey functors for H.

But a projective cohomological Mackey functor M for G over k is isomorphic to F P_{V}, where V is a direct summand of a permutation kG-module, or p-permutation kG-module (see [Bro85]): this is a kG-module admitting an S-invariant k-basis, for some p-Sylow subgroup S of G. Note that the module V is the evaluation of F P_{V} at {1}. So if V is a p-permutation kG-module, then T_{(U,a)}^{k coh}(F P_{V}) is isomorphic to F P_{W}, for some p-permutation kG-module W. This leads to the following:

Definition 9.1. Let kG-p-Mod denote the full subcategory of the category of kG-modules formed by p-permutation kG-modules (note that this is generally not an abelian category).

The composed functor

V ↦ T_{(U,a)}^{k coh}(F P_{V})(1)

from kG-p-Mod to kH-p-Mod will be called tensor induction for p-permutation modules, and denoted by T_{(U,a)}^{k per}.

If a is the constant function equal to the identity endomorphism of k, then T_{(U,a)}^{k per} will simply be denoted T_{U}^{k per}.

9.2. Properties. Evaluation at the trivial subgroup of the isomorphisms of theorem 8.19 gives the following

Proposition 9.2. Let p be a prime number and k be a field of characteristic p.

1. If G, H, and K are finite groups, if (U, a) is an End(k)-graded H-set-G and (V, b) is an End(k)-graded K-set-H, then

T_{(V,b)}^{k per} ∘ T_{(U,a)}^{k per} = T_{(V,b)×_{H}(U,a)}^{k per}

2. If G and H are finite groups, if M and N are p-permutation kG-modules, and if U is an End(k)-graded H-set-G, then

T_{(U,a)}^{k per}(M ⊗_{k} N) ≃ T_{(U,a)}^{k per}(M) ⊗_{k} T_{(U,a)}^{k per}(N)

3. If G and H are finite groups, if (U, a) and (U′, a′) are End(k)-graded Hsets-G, and M is a p-permutation kG-module, then

T_{(U,a)⊔(U′,a′)}^{k per}(M) ≃ T_{(U,a)}^{k per}(M) ⊗_{k} T_{(U′,a′)}^{k per}(M)

4. If G and H are finite groups, if (U, a) is an End(k)-graded H-set-G, and if M and M′ are p-permutation kG-modules, then

T_{(U,a)}^{k per}(M ⊕ M′) = ⊕_{V ⊆U V invariant by G V mod. H} Ind_{H_{V}}^{H} T_{(V,a_{|V} )}^{k per}(M) ⊗_{k} T_{(U−V,a_{|U−V} )}^{k per}(M′)

Moreover, all these isomorphisms are natural.

9.3. p-permutation algebras.

Definition 9.3. A p-permutation kG-algebra is a k-algebra, with an action of G, admitting an S-invariant k-basis, for some p-Sylow subgroup S of G.

Proposition 9.4. Let p be a prime number and k be a ring of characteristic p. Let G and H be finite groups.

1. If (U, a) is an End(k)-graded H-set-G, and A is a p-permutation kG-algebra, then T_{(U,a)}^{k per}(A) is a p-permutation kH-algebra.

2. If (U, a) is an End(k)-graded H-set-G, and A and B are p-permutation kGalgebras, then there is an isomorphism of p-permutation k-algebras

T_{(U,a)}^{k per}(A⊗_{k}B) ≃ T_{(U,a)}^{k per}(A)⊗_{k}T_{(U,a)}^{k per}(B)

3. If (U, a) and (U′, a′) are End(k)-graded H-sets-G, and A is a p-permutation kG-algebra, then there is an isomorphism of p-permutation kH-algebras

T_{(U,a)⊔(U′,a′)}^{k per}(A) ≃ T_{(U,a)}^{k per}(A)⊗_{k}T_{(U′,a′)}^{k per}(A)

4. If (U, a) is an End(k)-graded H-set-G, and A is a p-permutation kG-algebra,

then

T_{(U,a)}^{k per}(A^{op}) ≃ T_{(U,a)}^{k per}(A)^{op}

Proof: Since A is a kG-algebra, then F P_{A} is a cohomological Green functor for G over k, so T_{(U,a)}^{k coh}(F P_{A}) is a cohomological Green functor for H over k. But as A is a p-permutation kG-module, the functor T_{(U,a)}^{k coh}(F P_{A}) is isomorphic to F P_{B},

for B = T_{U}^{k per}(A). Now evaluation at the trivial subgroup of the product for the Green functor F P_{B} gives a p-permutation kH-algebra structure on B. The first assertion follows. The other ones are evaluations at the trivial subgroup of the corresponding assertions of proposition 8.21.

.

9.4. Identification. Let p be a prime number, and k be a field of characteristic p. Let G and H be finite groups, and U be a finite free-p H-set-G (for simplicity, I will not handle the case of a graded biset in this section). It is possible to give an explicit description of the functor T_{U}^{k per}.

Notation 9.5. If M is a kG-module, denote by {M} the underlying G-set. Consider the H-set Hom_{G}(U^{op}, {M}) as a set of functions from U to M. If x ∈ k and f ∈ Hom_{G}(U^{op}, {M}), define xf ∈ Hom_{G}(U^{op}, {M}) by

(xf)(u) = xf(u) ∀u ∈ U

If λ ∈ Hom_{G}(U^{op}, {k}), set

π(λ) = ∏_{u∈U/G} λ(u)

This does not depend on the choice of U/G. Define λ ∗ f ∈ Hom_{G}(U^{op}, {M}) by

(λ ∗ f)(u) = λ(u)f(u) ∀u ∈ U

If f and f′ are elements of Hom_{G}(U^{op}, {M}), let <f + f′> ∈ Hom_{G}(U^{op}, {M}) denote the sum of f and f′, defined by

<f + f′>(u) = f(u) + f′(u) ∀u ∈ U

If V ⊆ U is a G-subset of U, define the element [f, f′]_{V} of Hom_{G}(U^{op}, {M}) by

[f, f′]_{V} (u) = \{ \begin{matrix}f(u) & if u ∈ V \\ f′(u) & if u ∈ U − V \end{matrix}

If P is a p-subgroup of G, define the Brauer quotient M[P] by

M[P] = M^{P} /( ∑_{Q⊂P} T r_{Q}^{P}M^{Q})

Let Br_{P} denote the natural projection M^{P} → M[P]. If f ∈ Hom_{G}(U, {M}), then in particular for any u ∈ U, the element f(u) is in M^{G_{u}}, so I can consider Br_{G_{u}}(f(u)). Recall finally that if X is a G-set, then for any p-group P, the module

(kX)[P] is isomorphic to k(X^{P}).

Definition 9.6. If M is a kG-module, define

t_{U}(M) = kHom_{G}(U^{op}, {M})/J

where J is the subspace generated by the elements

(λ ∗ f) − π(λ)f for λ ∈ Hom_{G}(U^{op}, {k}), f ∈ Hom_{G}(U^{op}, {M})

<f + f′> − ∑_{V ⊆U V G-invariant}[f, f′]_{V} for f, f′ ∈ Hom_{G}(U^{op}, {M})

and by all the elements f ∈ Hom_{G}(U^{op}, {M}) such that there exists u ∈ U with Br_{G_{u}}(f(u)) = 0.

Lemma 9.7. The correspondence M ↦ t_{U}(M) is a functor from kG-Mod to kH-Mod.

Proof. If φ : M → M′ is a morphism of kG-modules, then φ induces a morphism {φ} of the underlying G-sets, hence a morphism

Φ = Hom_{G}(U^{op}, {φ}) : Hom_{G}(U^{op}, {M}) → Hom_{G}(U^{op}, {M′})

Now if λ ∈ Hom_{G}(U^{op}, {k}) and f ∈ Hom_{G}(U^{op}, {M}), then for any u ∈ U

Φ(λ ∗ f)(u) = φ(λ(u)f(u)) = λ(u)φ(f(u))

so Φ(λ ∗ f) = λ ∗ Φ(f). Similarly Φ(xf) = xΦ(f) if x ∈ k. Thus

Φ(λ ∗ f − π(λ)f) = λ ∗ Φ(f) − π(λ)Φ(f)

Similarly

Φ(<f + f′>)(u) = φ(f(u) + f′(u)) = φ(f(u)) + φ(f′(u))

so Φ(<f + f′>) =< Φ(f) + Φ(f′) >. And if V is a G-subset of U, then

Φ([f, f′]_{V})(u) = φ([f, f′](u)) = \{ \begin{matrix}φ(f(u)) & if u ∈ V \\ φ(f′(u)) & if u ∈ U − V \end{matrix}

Thus Φ([f, f′]_{V}) = [Φ(f), Φ(f′)]_{V}.

Finally, if there exists u ∈ U such that Br_{G_{u}}(f(u)) = 0, then as φ commutes with Br_{G_{u}}

Br_{G_{u}}(Φ(f)(u)) = φBr_{G_{u}}(f(u)) = 0

This proves that Φ passes down to a quotient map t_{U}(φ) : t_{U}(M) → t_{U}(M′), and the lemma follows.

Proposition 9.8. Let p be a prime number, and k be a field of characteristic p. Let G and H be finite groups, and U be a finite free-p H-set-G. Then the functor T_{U}^{k per} is isomorphic to the restriction of t_{U} to kG-p-Mod.

Proof. Suppose that M = kX, for some G-set X. Then by definition

T_{U}^{k per}(M) ≃ (F P_{kHom_{G}(U^{op},X)})(1) ≃ kHom_{G}(U^{op}, X)

Now Hom_{G}(U^{op}, X) ≃ ∏_{u∈U/G} X^{G_{u}}, and as kX^{G_{u}} ≃ (kX)[G_{u}], I have

T_{U}^{k per}(M) ≃ ⊗_{u∈U/G}(kX)[G_{u}]

where ⊗ = ⊗_{k}. Now the proposition follows from the following:

Lemma 9.9. Let p be a prime number, and k be a field of characteristic p. Let G and H be finite groups, and U be a finite free-p H-set-G. Then for any kG-module M, there is an isomorphism of vector spaces

t_{U}(M) ≃ ⊗_{u∈U/G} M[G_{u}]

(where ⊗ = ⊗_{k}) which is natural in M.

Proof. There is a natural map

t_{U}(M) = kHom_{G}(U^{op}, {M})/J → ⊗_{u∈U/G} M[G_{u}]

sending f ∈ Hom_{G}(U^{op}, {M}) to ⊗_{u∈U/G} Br_{G_{u}}(f(u)). Conversely, I can map ⊗_{u∈U/G} Br_{G_{u}}(m_{u}) to the image of the element f ∈ Hom_{G}(U^{op}, {M}) defined by f(u′) = gm_{u}, if u′ = ug^{−1} for u ∈ U/G and g ∈ G. Those maps are well-defined, and inverse to each other.

So the restrictions of T_{U}^{k per} and t_{U} to the category of permutation modules are isomorphic.

Now if M is any p-permutation kG-module, there is an exact sequence

F P_{kY} →^{φ} F P_{kX} →^{ψ} F P_{M} → 0 (9.27)

for suitable G-sets X and Y. By definition, the sequence

T_{U}^{k coh}(F P_{k(Y ⊔X)}) \underrightarrow{ΔT_{U}^{k coh}(φ)} T_{U}^{k coh}(F P_{kX}) → T_{U}^{k coh}(F P_{M}) → 0

is exact, and evaluation at the trivial subgroup gives an exact sequence

T_{U}^{k per}(k(Y ⊔ X)) \underrightarrow{ΔT_{U}^{k per}(φ)} T_{U}^{k per}(kX) → T_{U}^{k per}(M) → 0

The left part of this sequence is isomorphic to

t_{U}(k(Y ⊔ X)) \underrightarrow{Δt_{U}(φ)} t_{U}(kX)

So it suffices to prove that the sequence

t_{U}(k(Y ⊔ X)) \underrightarrow{Δt_{U}(φ)} t_{U}(kX) → t_{U}(M) → 0 (9.28)

is exact if M is a p-permutation module. And this sequence is exact if and only if it is an exact sequence of k-vector spaces.

But if M is a p-permutation kG-module, the morphism ψ in the sequence (9.27) is a split epimorphism, since F P_{M} is a projective cohomological Mackey functor. Thus for any u ∈ U, the sequence

(kY)[G_{u}] \underrightarrow{φ[G_{u}]}(kX)[G_{u}] \underrightarrow{ψ[G_{u}]} M[G_{u}] → 0

is exact. Hence the direct product of those sequences for u ∈ S = U/G

((kY)[G_{u}])_{u∈S} \underrightarrow{(φ[G_{u}])_{u∈S}} ((kX)[G_{u}])_{u∈S} \underrightarrow{(ψ[G_{u}])_{u∈S}} (M[G_{u}])_{u∈S} → 0 (9.29)

is exact. Now up to the isomorphism of the lemma, the sequence (9.28) becomes

⊗_{u∈U/G} ((kY)[G_{u}] ⊕ (kX)[G_{u}]) → ⊗_{u∈U/G}(kX)[G_{u}] → ⊗_{u∈U/G} M[G_{u}] → 0

It is exact, since the sequence (9.29) is exact, and the functor

(L_{u})_{u∈U/G} ↦ ⊗_{u∈U/G} L_{u}

is right exact. So the sequence (9.28) is exact, and this completes the proof of the proposition.

Remark 9.10. For u ∈ U, let

H_{(u)} = {h ∈ H | ∃g ∈ G, hug = u}

Then keeping track of the action of H in lemma 9.9 gives the isomorphims of kHmodules

t_{U}(M) ≃ ⊗_{u∈H\U/G} ten_{H_{(u)}}^{H}M[G_{u}]

where ten_{H_{(u)}}^{H} is the ordinary tensor induction for kH-modules (described in the next section). The module M[G_{u}] is a kH_{(u)}-module via the following action

h.m = g.m if h ∈ H, g ∈ G, hug^{−1} = u

10. Tensor induction for modules

10.1. Definition. Let R be a commutative ring. If G is a finite group, let RG-FMod denote the full subcategory of RG-Mod consisting of free RG-modules. Since a free RG-module is always the permutation module associated to a free Gset, this category can also be seen as the category of free permutation modules: its objects are the free G-sets, and a morphism from Y to X is a G-invariant matrix m(x, y), indexed by X × Y, with coefficients in R, and such that for any y ∈ Y, the coefficient m(x, y) is zero except for a finite number of x. Composition of morphisms is given by matrix multiplication.

Let G and H be finite groups, and U be a finite right free H-set-G. Then if X is a free G-set, the H-set Hom_{G}(U^{op}, X) need not be free in general. But I can consider the RH-module

t_{U}(X) = RHom_{G}(U^{op}, X)

If m : Y → X is a morphism in RG-FMod, then for φ ∈ Hom_{G}(U^{op}, X) and ψ ∈ Hom_{G}(U^{op}, Y), define

m̃(φ, ψ) = ∏_{u∈U/G} m(φ(u), ψ(u))

This matrix defines a morphism of RH-modules from t_{U}(Y) to t_{U}(X): for a given ψ, there is only a finite number of φ such that m̃(φ, ψ) is non-zero, and moreover m̃ is H-invariant (see the proof of proposition 8.8). Actually:

Lemma 10.1. Let R be a commutative ring, and G and H be finite groups. Let U be a finite right free H-set-G. Then the correspondence

X ↦ t_{U}(X) = RHom_{G}(U^{op}, X) m ↦ m̃

is a functor t_{U} from RG-FMod to RH-Mod.

Proof. Clearly if m is the identity matrix, so is m̃. I have only to check that the correspondence m ↦ m̃ is multiplicative.

But if Z is another free G-set, and p is a morphism from Z to Y, then the morphism m ∘ p : Z → X is represented by the product matrix m.p defined for x ∈ X and z ∈ Z by

(m.p)(x, z) = ∑_{y∈Y} m(x, y)p(y, z)

Now if θ ∈ Hom_{G}(U^{op}, Z) and φ ∈ Hom_{G}(U^{op}, X)

m̃.p(φ, θ) = ∏_{u∈U/G} (∑_{y∈Y} m(φ(u), y)p(y, θ(u)))

Now expanding this product is equivalent to choosing a sequence (y_{u})_{u∈U/G} of elements of Y. But as U^{op} is a free G-set, this is equivalent to choosing a Gmorphism ψ from U^{op} to Y. Thus

m̃.p(φ, θ) = ∑_{ψ∈Hom_{G}(U^{op},Y)} ∏_{u∈U/G} m(φ(u), ψ(u))p(ψ(u), θ(u)) = …

… = ∑_{ψ∈Hom_{G}(U^{op},Y)} m̃(φ, ψ)m̃(ψ, θ)

So m ↦ m̃ is multiplicative, and the lemma follows.

Note that the subcategory P = RG-FMod of RG-Mod satisfies the hypothesis of theorem 2.14: its objects are projective, it is closed under directs sums, and any RG-module is a quotient of a free RG-module. This leads to the following:

Definition 10.2. Let R be a commutative ring. If G and H are finite groups, and U is a finite right free H-set-G, I call tensor induction associated to U the unique right exact functor from RG-Mod to RH-Mod extending the functor t_{U} from RG-FMod to RH-Mod. This extension is still denoted by t_{U}.

10.2. Identification. The module t_{U}(M) can be computed as follows:

Proposition 10.3. Let R be a commutative ring. Let G and H be finite groups, and U be a finite right free H-set-G. Then for any RG-module M, there is an isomorphism

t_{U}(M) ≃ RHom_{G}(U^{op}, {M})/J

where J is the R-submodule generated by the elements

(λ ∗ f) − π(λ)f for λ ∈ Hom_{G}(U^{op}, {k}), f ∈ Hom_{G}(U^{op}, {M})

< f + f′ > − ∑_{V ⊆U V G-invariant}[f, f′]_{V} for f, f′ ∈ Hom_{G}(U^{op}, {M})

This isomorphism is moreover natural in M.

Remark 10.4. This shows that the notation t_{U} is coherent with the definition 9.6 given for p-permutation kG-modules, when k is a field of characteristic p: indeed, if U is right free, then G_{u} = {1} for all u ∈ U, so all morphisms Br_{G_{u}} are equal to identity, and their kernels are zero. Moreover, if f(u) = 0 for some u, then setting λ(v) = 0 on the orbit of u, and λ(v) = 1 elsewhere, I have f = λ ∗ f and π(λ) = 0, so f is in the submodule J of the proposition.

Proof. It is easy to see that if M is an RG-module, then

RHom_{G}(U^{op}, {M})/J ≃ ⊗_{u∈U/G} M

like in lemma 9.9. This proves that the functor

M ↦ RHom_{G}(U^{op}, {M})/J

is right exact. Now if M is a permutation module RX, the previous isomorphism gives

RHom_{G}(U^{op}, {M})/J ≃ ⊗_{u∈U/G} RX ≃ RHom_{G}(U^{op}, X) = t_{U}(M)

So both functors in the proposition are right exact functors, which are isomorphic when restricted to P = RG-FMod. The proposition follow from theorem 2.14.

Remark 10.5. This proposition proves that if M = RX is any permutation module (not only for a free X), then

t_{U}(M) ≃ RHom_{G}(U^{op}, X)

10.3. Properties.

Theorem 10.6. Let R be a commutative ring.

1. If G, H, and K are finite groups, if U is a finite right free H-set-G and V is a finite right free K-set-H, then V ×_{H} U is a finite right free K-set-G and

t_{V} ∘ t_{U} = t_{V ×_{H}U}

2. If G, H and K are finite groups, if U is a finite right free K-set-(G × H), if M is an RG-module and N is an RH-module, then

t_{U}(M ⊠_{R} N) = t_{U/H}(M) ⊗_{R} t_{U/G}(N)

where M ⊠_{R} N denotes the tensor product M ⊗_{R} N, viewed as an R(G×H)- module.

3. If G and H are finite groups, if M and N are RG-modules, and if U is a finite right free H-set-G, then

t_{U}(M ⊗_{R} N) ≃ t_{U}(M) ⊗_{R} t_{U}(N)

4. If G and H are finite groups, if U and U′ are finite right free H-sets-G, and M is an RG-module, then

t_{U⊔U′}(M) ≃ t_{U}(M) ⊗_{R} t_{U′}(M)

5. If G and H are finite groups, if U is a finite right free H-set-G, and if M and M′ are RG-modules, then

t_{U}(M ⊕ M′) = ⊕_{V ⊆U V invariant by G V mod. H} Ind_{H_{V}}^{H} t_{V} (M) ⊗_{R} t_{U−V} (M′)

6. If G, H, and K are finite groups, if U is a finite right free H-set-G, and V is a finite right free K-set-H, then for any RG-module M

t_{V} (RU ⊗_{RG} M) ≃ ⊕_{f∈K\Hom_{H}(V^{op},U/G)} Ind_{K_{f}}^{K} t_{V._{f,H}U}(M)

Moreover, all these isomorphisms are natural.

Proof. All those assertions state isomorphisms between right exact functors. So it suffices to check that the restrictions to RG-FMod are isomorphic, and this is straightforward: let me just give the details for the last one, where the notations refer to proposition 5.7. If X is a G-set, then

RU ⊗_{RG} RX ≃ R(U ×_{G} X) ≃ R(U ∘_{G} X)

since U is right free. Now I have seen in proposition 5.7 that

Hom_{H}(V^{op}, U ∘_{G} X) = ⨆_{f∈K\Hom_{H}(V^{op},U/G)} Ind_{K_{f}}^{K} Hom_{G}(H\U._{f}V, X)

It follows that

t_{V} (RU ⊗_{RG} M) ≃ ⊕_{f∈K\Hom_{H}(V^{op},U/G)} Ind_{K_{f}}^{K} RHom_{G}(H\U._{f}V, X)

Moreover H\U._{f}V and V._{f,H}U are isomorphic as K_{f}-sets-G. This proves the desired isomorphism for free permutation modules, and the assertion follows from right exactness.

10.4. Comparison. Tensor induction for Mackey functors and for modules are compatible:

Proposition 10.7. Let G and H be finite groups.

1. If U is a finite right free H-set-G, then for any Mackey functor M for G, there is an isomorphim of ℤG-modules

T_{U}(M)(1) ≃ t_{U}(M(1))

where t_{U} denotes tensor induction for ℤG-modules.

2. If R is a commutative ring, and U is a finite right free H-set-G, then for any cohomological Mackey functor M for G over R, there is an isomorphism of RG-modules

T_{U}^{R coh}(M)(1) ≃ t_{U}(M(1))

where t_{U} denotes tensor induction for RG-modules.

Those isomorphisms are moreover natural in M.

Proof. Both assertions state isomorphisms between right exact functors (from from Mack(G) to ℤG-Mod for the first one, and from Comack_{R}(G) to RG-Mod for the second). So it suffices to check the isomorphism of the restrictions to P Mack(G) for the first assertion, and P Comack_{R}(G) for the second.

But if M = b_{X} for some G-set X, then

T_{U}(M)(1) = (b_{Hom_{G}(U^{op},X)})(1) ≃ ℤHom_{G}(U^{op}, X) ≃ t_{U}(ℤX)

Similarly, if M = F P_{RX} for some G-set X, then

T_{U}^{R^{coh}}(M)(1) = (F P_{RHom_{G}(U^{op},X)})(1) = RHom_{G}(U^{op}, X) = t_{U}(X)

Now proposition follows from theorem 2.14.

10.5. Examples.

10.5.1. Restriction. Let H be a subgroup of G, and U = G, viewed as an Hset-G. Then if X is a free G-set, it is clear that Hom_{G}(U^{op}, X) is isomorphic to the restriction of X to H. Since

Res_{H}^{G}RX = R(Res_{H}^{G}X)

the functors Res_{H}^{G} and t_{U} are isomorphic when restricted to RG-FMod. Since they are both right exact, they are isomorphic, and t_{U} is just the restriction functor in that case.

10.5.2. Inflation. Let N be a normal subgroup of G and H = G/N. Let U = H, viewed as a right free G-set-H. If X is an H-set, then Hom_{H}(U^{op}, X) is isomorphic to Inf_{H}^{G}X. Thus

t_{U}(X) = RInf_{H}^{G}X = Inf_{H}^{G}(RX)

Here again, the functors t_{U} and Inf_{H}^{G} are right exact functors, with isomorphic restrictions to RG-FMod. So the functor t_{U} is just the inflation functor in that case.

10.5.3. Ordinary tensor induction. Let H be a subgroup of G, and U = G, viewed as a G-set-H. Then for any H-set X

t_{U}(X) = RHom_{H}(U^{op}, X) ≃ R(X^{[G:H]}) = (RX)^{⊗[G:H]}

So the functor t_{U} is isomorphic to the (ordinary) tensor induction functor ten_{H}^{G} (see [Ben91] 3.15).

In particular, theorem 10.6 5) and 6) gives explicit formulas for ten_{H}^{G}(M ⊕ N), and ten_{H}^{G}(Ind_{K}^{H}M) (compare with [Ben91] Proposition 3.15.2), whereas Mackey formula for tensor induction, transitivity of tensor induction, formulas for composition of tensor induction with restriction and inflation follow from theorem 10.6 1).

References

[ACPW98] A. Adem, J. Carlson, S. Priddy, and P. Webb (eds.), Group representations: Cohomology, Group Actions and Topology, vol. 63, AMS Proceedings of Symposia in pure mathematics, Janvier 1998.

[Ben91] D.J. Benson, Representations and cohomology I, Cambridge studies in advanced mathematics, vol. 30, Cambridge University Press, 1991.

[Bou92] Serge Bouc, Exponentielle et modules de Steinberg, J. of Algebra 150 (1992), no. 1, 118–157.

[Bou96a] Serge Bouc, Construction de foncteurs entre catégories de G-ensembles, J. of Algebra 183 (1996), no. 0239, 737–825.

[Bou96b] Serge Bouc, Foncteurs d’ensembles munis d’une double action, J. of Algebra 183 (1996), no. 0238, 664–736.

[Bou97] Serge Bouc, Green functors and G-sets, Lecture Notes in Mathematics, vol. 1671, Springer, October 1997.

[Bro85] Michel Broué, On Scott modules and p-permutation modules: an approach through the Brauer morphism, Proc.AMS 93 (1985), no. 3, 401–408.

[BT98] Serge Bouc and Jacques Thévenaz, The group of endo-permutation modules, Preprint, June 1998.

[Dre73] A.W.M Dress, Contributions to the theory of induced representations, Lecture Notes in Mathematics, vol. 342, 183–240, Lecture Notes in Mathematics, Springer-Verlag, 1973, pp. 183–240.

[Thé87] Jacques Thévenaz, Permutation representation arising from simplicial complexes,

J.Combin. Theory 46 (1987), 122–155, Ser.A.

[TW90] Jacques Thévenaz and Peter Webb, Simple Mackey functors, Proceedings of the 2nd International group theory conference Bressanone 1989, Rend. Circ. Mat. Palermo, vol. 23, 1990, Serie II, pp. 299–319.

[TW95] Jacques Thévenaz and Peter Webb, The structure of Mackey functors, Trans. Amer. Math. Soc. 347 (1995), no. 6, 1865–1961.

[Web91] Peter Webb, A split exact sequence for Mackey functors, Comment. Math. Helv. 66 (1991), 34–69.

[Wei94] Charles A. Weibel, An introduction to homological algebra, Cambridge studies in advanced mathematics, vol. 38, Cambridge University Press, 1994.

S. Bouc, Equipe des groupes finis, UFR de Mathématiques, Université Paris 7-Denis Diderot, 2 place Jussieu, 75251. Paris Cedex 05, France

E-mail address: bouc@math.jussieu.fr