Resolutions of Mackey functors^{1}

S. Bouc

Abstract. I will build some standard resolutions for Mackey functors which are projective relative to p-subgroups. Those resolutions are closely related to the poset of p-subgroups. They lead to generalizations of known results on cohomology. They give a way to compute the Cartan matrix for Mackey functors, in terms of p-permutation modules, and to precise the structure of projective Mackey functors. They also provide results on complexes of projective Mackey functors and complexes of p-permutation modules.

1. Introduction

I will build some standard resolutions for Mackey functors which are projective relative to p-subgroups. Those resolutions are closely related to the poset of psubgroups. They lead to generalizations of known results on cohomology. They give a way to compute the Cartan matrix for Mackey functors, in terms of ppermutation modules, and to precise the structure of projective Mackey functors. They also provide results on complexes of projective Mackey functors and complexes of p-permutation modules.

2. Notation

2.1. The Mackey algebra. Most of the results I will need can be found in J.Thévenaz and P.Webb’s paper ([6]) on the structure of Mackey functors : among other possible definitions, a Mackey functor is a module over the Mackey algebra. A possible definition of this algebra is the following :

Definition 2.1. Let G be a finite group, and R be a commutative ring. The Mackey algebra μ_{R}(G) is the algebra generated by the elements t_{H}^{K}, r_{H}^{K}, and c_{g,H}, where H and K are subgroups of G such that H ⊆ K, and g is an element of G, subject to the relations :

t_{K}^{L}t_{H}^{K} = t_{H}^{L} for all H ⊆ K ⊆ L
r_{H}^{K}r_{K}^{L} = r_{H}^{L} for all H ⊆ K ⊆ L
c_{g,^{h}K}c_{h,K} = c_{gh,K} for all g, h, K
t_{H}^{H} = r_{H}^{H} = c_{h,H} for all h, H such that h ∈ H

1991 Mathematics Subject Classification. Primary 19A22 ; Secundary 20J05.

1This is a rough translation of my paper Résolutions de foncteurs de Mackey (in french)

AMS PSPM 63 (1998) 31-83. A few comments have been added in footnotes (date : 2008-11-24)

c_{g,K}t_{H}^{K} = t_{^{g}H}^{^{g}K}c_{g,H} for all g, h, H, K
c_{g,K}r_{K}^{H} = r_{^{g}K}^{^{g}H}c_{g,H} for all g, h, H, K

∑_{H} t_{H}^{H} = ∑_{H} r_{H}^{H} = 1

r_{K}^{H}t_{L}^{H} = ∑_{x∈K\H/L} t_{K∩^{x}L}^{K}r_{K∩^{x}L}^{^{x}L}c_{x,L} for all K ⊆ H ⊇ L (Mackey axiom)

any other product of r_{H}^{K}, t_{H}^{K} and c_{g,H} being equal to zero.

Setting c_{x} = ∑_{H} c_{x,H} for x ∈ G, it follows that c_{x,H} = c_{x}t_{H}^{H}, and that the map x ↦ c_{x} endows μ_{R}(G) with a structure of interior G-algebra (in Puig’s sense [3]).

Proposition (3.4) of [6] proves that the c_{x,1}, hence also the c_{x}, are linearly independent, which allows for an identification of x and c_{x}. With this notation, Proposition (3.2) of [6] proves that μ_{R}(G) admits a basis over R, consisting of the elements t_{^{x}L}^{H}xr_{L}^{K}, where H and K are subgroups of G, where x ∈ K\G/H, and L is a subgroup of H ∩ K^{x} up to conjugation by H ∩ K^{x}.

If H is a subgroup of G, the defining relations of μ_{R}(G) allow to define a morphism from the Burnside ring b_{R}(H) of H with coefficients in R, to the algebra μ_{R}(G), sending the element H/K to the element t_{K}^{H}r_{K}^{H}. I will denote by X ↦ [X] this morphism.

On the other hand, there is a natural morphism z from the Burnside ring of G to the center of the Mackey algebra ([6], Proposition (9.2)), defined by

z(X) = ∑_{K⊆G}[Res_{K}^{G}X]

which allows the use of the idempotents of the Burnside ring to split the Mackey algebra into smaller pieces.

In the remainder of this paper, I will only consider the “characteristic p” case : I will assume that any prime number different from p is invertible in the ring R. Under these conditions (cf [6] Section 9-10), the algebra μ_{R}(G) is Morita-equivalent to a direct product of algebras indexed by the p-perfect subgroups of G (i.e. the subgroups with no non trivial quotient p-group).

More precisely, the algebra μ_{R}(G) is Morita-equivalent to the direct product over the p-perfect subgroups H of G (up to G-conjugation) of the algebras I will denote by μ_{R}^{1}(N_{G}(H)/H) (the category of μ_{R}^{1}(G)-modules is denoted by Mack_{R}(G, 1) in [6]) : the algebra μ_{R}^{1}(G) is the piece of μ_{R}(G) corresponding to the idempotent f_{1}^{G} of the Burnside ring of G. Denoting by \underline{s}_{p}(G) the set of p-subgroups of G, the idempotent f_{1}^{G} of the Burnside ring b_{Q}(G) with rational coefficients is defined par

f_{1}^{G} = ∑_{P ∈\underline{s}_{p}(G)/G} e_{P}^{G}

with

e_{P}^{G} = \frac{1}{|N_{G}(P)|} ∑_{Q⊆P} χ̃]Q, P [|Q|G/Q

where χ̃]Q, P [ is the reduced Euler-Poincaré characteristic of the set of p-subgroups strictly containing Q and strictly contained in P. An easy computation shows then that

f_{1}^{G} = − ∑_{P ∈\underline{s}_{p}(G)/G} \frac{χ̃(s_{p}(N_{G}(P)/P))}{|N_{G}(P)/P |}G/P

with s_{p}(G) = \underline{s}_{p}(G) − \{1\}.

This expression shows that f_{1}^{G} is p-integral, hence f_{1}^{G} in b_{R}(G) (it is indeed well known that χ̃(s_{p}(G)) is divisible by the p-part of the order of G, since e.g. it is the degree of a projective character of G (cf. [4])).

With this notation, the algebra μ_{R}^{1}(G) identifies to z(f_{1}^{G})μ_{R}(G). Another way of seeing this algebra up to Morita equivalence is given by the following lemma :

Lemma 2.2. Let e = ∑_{P ∈\underline{s}_{p}(G)} t_{P}^{P}. Then eμ_{R}(G)e is a subalgebra of μ_{R}^{1}(G), and the inclusion of eμ_{R}(G)e in μ_{R}^{1}(G) is a Morita equivalence.

First I have to check that eμ_{R}(G)e is a subalgebra of μ_{R}^{1}(G) : it suffices to check that z(f_{1}^{G})e = e, which will follow from the equality z(f_{1}^{G})t_{P}^{P} = t_{P}^{P} for all p-subgroup P of G. But it follows from Section 9 of [6] that

z(f_{1}^{G})t_{P}^{P} = [Res_{P}^{G}(f_{1}^{G})]t_{P}^{P} = [Res_{P}^{G}(f_{1}^{G})P/P] = [Res_{P}^{G}(f_{1}^{G})]

and the result will follow, if I know that Res_{P}^{G}f_{1}^{G} = f_{1}^{P}, since the expression of f_{1}^{P} shows that f_{1}^{P} = P/P if P is a p-group. This in turn follows from the

Lemma 2.3. Let H be a subgroup of G. Then Res_{H}^{G}f_{1}^{G} = f_{1}^{H}

The computation can be done inside b_{Q}(G) : the idempotents e_{L}^{G} are such that for all X of b_{Q}(G)

X = ∑_{{L∈G \atop Lmod.G}} |X^{L}|e_{L}^{G}

Then

Res_{H}^{G}e_{P}^{G} = ∑_{{L⊆H \atop Lmod.H}} Res_{H}^{G}e_{P}^{G}.e_{L}^{H} = ∑_{{L⊆H \atop Lmod.H}} |(Res_{H}^{G}e_{P}^{G})^{L}|e_{L}^{H}

and since |(Res_{H}^{G}e_{P}^{G})^{L}| is non zero only if L is a conjugate of P in G, in which case it is equal to 1, I see that Res_{H}^{G}e_{P}^{G} is the sum of the e_{L}^{H}, where L runs through the conjugates of P contained in H, up to H-conjugation. Then Res_{H}^{G}f_{1}^{G} is the sum over the p-subgroups L of H, modulo H, of the e_{L}^{H}. Lemma 2.3, and the first assertion of Lemma 2.2. The second one follows then from

Lemma 2.4. ^{2} Let A be a ring (with identity element), and e be an idempotent of A. The following assertions are equivalent :

1) The inclusion of eAe in A is a Morita equivalence.

2) The two sided ideal of A generated by e is equal to A (i.e. AeA = A).

The inclusion of eAe in A defines a functor of restriction r from the category Amod of left A-modules to eAe-mod, by r(M) = eM. This functor has a left adjoint i, defined by i(M) = A ⊗_{eAe} M. Then ri(M) = eA ⊗_{eAe} M = eAe ⊗_{eAe} M = M and ir(M) identifies with AeM. Hence if 1) holds, then in particular AeA = A, hence 2) holds. Conversely, if 2) holds, then ir(M) = AeM = AeAM = AM = M, hence 1) holds.

To complete the proof of Lemma 2.2, it remains to check that the identity element of μ_{R}^{1}(G), i.e. z(f_{1}^{G}), lies in the two sided ideal of μ_{R}^{1}(G) generated by e. It suffices to show that t_{P}^{H}r_{P}^{H} lies in this ideal, for any subgroup H of G and any p-subgroup P of H. But this is clear, since t_{P}^{H}r_{P}^{H} = t_{P}^{H}er_{P}^{H}. Lemma 2.2 follows.

2This lemma is true, but the proof given here is not quite complete

Remark 2.5. The eμ_{R}(G)e-modules are actually very natural objects : they are exactly the Mackey functors “defined only over p-subgroups ”. Whereas the μ_{R}^{1}(G)-modules are the Mackey functors which are projective relative to p-subgroups ([6] Theorem 9.7). The algebra μ_{R}^{1}(G) identifies with the subalgebra of μ_{R}(G) generated by the x ∈ G, and the r_{Q}^{H} and t_{Q}^{H}, where Q is a p-subgroup of G and H an arbitrary subgroup of G.

On the other hand, Lemma 2.2 has the following consequence :

Corollary 2.6. Let S be a Sylow p-subgroup of G, and M_{∗} be a complex of Mackey functors in Mack_{R}(G, 1). Then M_{∗} is acyclic (resp. split acyclic) if and only if its restriction to S is.

Indeed, a Morita equivalence maps an exact sequence (resp. a split exact sequence) to an exact sequence (resp. to a split exact sequence).

2.2. Another algebra. I will use also here another algebra, that I denote by rμ_{R}(G), defined similarly by considering only p-subgroups of G, and “forgetting” the generators t_{H}^{K} and the Mackey axiom.

More precisely, let A be a free R-module on the set of triples (x, Q, P), where x is an element of G, and P, Q are p-subgroups of G such that Q ⊆ P. I turn A into an algebra by defining the multiplication of the basis elements by

(x, Q, P)(y, R, S) = δ_{P^{y},R}(xy, Q^{y}, S)

It is easy to see that A is an associative algebra. Moreover, let

e = ∑_{P ∈\underline{s}_{p}(G)}(1, P, P)

then

e.(y, Q, R) = ∑_{P ∈\underline{s}_{p}(G)} δ_{P^{y},Q}(y, P^{y}, R) = (y, Q, R)

and

(y, Q, R).e = ∑_{P ∈\underline{s}_{p}(G)} δ_{R,P} (y, Q, P) = (y, Q, R)

hence e is an identity element of A.

Let then I the R-submodule of A generated by the elements of the form u_{x,z,Q,P} = (xz, Q^{z}, P) − (x, Q, P), where z ∈ P. Then I is a two sided ideal of A. Indeed

(y, R, S)u_{x,z,Q,P} = δ_{S^{xz},Q^{z}}(yxz, R^{xz}, P) − δ_{S^{x},Q}(yx, R^{x}, P)
= δ_{S^{x},Q}((yxz, R^{xz}, P) − (yx, R^{x}, P)) = δ_{S^{x},Q}u_{yx,z,R^{x},P}

Similarly

u_{x,z,Q,P} (y, R, S) = δ_{P^{y},R}(xzy, Q^{zy}, S) − δ_{P^{y},R}(xy, Q^{y}, S)
= δ_{P^{y},R}u_{xy,z^{y},Q^{y},S}

which makes sense, since if P^{y} = R, then z^{y} ∈ R ⊆ S.

I will denote by rμ_{R}(G) the quotient algebra A/I. It is easy to find a basis of rμ_{R}(G) :

Lemma 2.7. The elements (x, Q, P), for Q ⊆ P ∈ \underline{s}_{p}(G), and x ∈ G/P, form a basis of rμ_{R}(G).

Lemma 2.7 is just a reformulation of the following :

Lemma 2.8. Let A be an R-algebra, admitting a basis B over R. Let E be an equivalence relation over B such that the elements x − y, with xEy, generate a two sided ideal I of A. Then A/I admits a basis over R, consisting of the images in A/I of the elements of B/E.

Indeed, it is clear that the images of B/E generate A/I. since all the elements of the equivalence class of x ∈ B are congruent to x modulo I. And if ∑_{x∈B/E} r_{x}x ∈ I, with r_{x} ∈ R, then there exists scalars s_{x,y} ∈ R such that

∑_{x∈B/E} r_{x}x = ∑_{xEy} s_{x,y}(x − y)

Then for all x

r_{x} = ∑_{xEy} s_{x,y} − ∑_{yEx} s_{y,x}

Summing this relation for x in the equivalence class of z, I can suppose that r_{z} alone is non zero, hence

r_{z} = ∑_{xEz}(∑_{xEy} s_{x,y} − ∑_{yEx} s_{y,x}) = ∑_{x,yEz} s_{x,y} − ∑_{x,yEz} s_{x,y} = 0

since the relation E is symmetric and transitive. The above lemmas follow.

I can then define an application f from rμ_{R}(G) to eμ_{R}^{1}(G)e by sending (x, Q, P) to xr_{Q}^{P}. It is easy to check that the map f is a morphism of algebras (with identity elements).

Lemma 2.9. The morphism f is injective.

(In the sequel, I will identify rμ_{R}(G) with its image in μ_{R}(G).) To show this, I must check that the elements xr_{Q}^{P}, for x ∈ G/P and Q ⊆ P, are linearly independent in μ_{R}(G) : but if

∑_{x∈G/P,Q,P} r_{x,Q,P} xr_{Q}^{P} = 0 in μ_{R}(G)

then multiplying this relation on the left by t_{R}^{R} and on the right by r_{P}^{P}, for given subgroups R and P, it follows that

∑_{{x∈G/P,Q \atop ^{x}Q=R}} r_{x,Q,P} t_{R}^{R}xr_{Q}^{P} = 0

i.e.

∑_{{x∈G/P \atop R^{x}⊆P}} r_{x,R^{x},P} t_{R}^{R}xr_{R^{x}}^{P} = 0

Then using the above basis of μ_{R}(G), it follows that, for any double coset Rx_{0}P and any subgroup Q_{0} of R^{x_{0}} ∩ P

∑_{{x∈Rx_{0}P/P \atop R^{x}=Q_{0}mod.R^{x_{0}} ∩P}} r_{x,R^{x},P} = 0

i.e.

∑_{{y∈R/(R∩^{x_{0}} P) \atop R^{yx_{0}} =Q_{0}mod.R^{x_{0}} ∩P}} r_{yx_{0},R^{yx_{0}} ,P} = 0

or equivalently

∑_{{y∈R/(R∩^{x_{0}} P) \atop R^{x_{0}} =Q_{0}mod.R^{x_{0}} ∩P}} r_{yx_{0},R^{x_{0}} ,P} = 0

In particular taking Q_{0} = R^{x_{0}} ∩ P, I conclude that r_{x_{0},R^{x_{0}} ∩P,P} = 0, and as this relation is true for all x_{0}, R, P, the above linear independence claim holds, and the injectivity of f follows.

It is also natural to consider here the algebra tμ_{R}(G) defined similarly to rμ_{R}(G) by “forgetting” the generators r_{H}^{K}. This algebra identifies with the subalgebra of μ_{R}(G) generated by the elements t_{Q}^{P}x. It also identifies to the opposite algebra of rμ_{R}(G) : the map sending t_{Q}^{P} to r_{Q}^{P} and c_{x,Q} to c_{x^{−1},Q} defines indeed an anti-isomorphism from tμ_{R}(G) to rμ_{R}(G). The results I will prove here for rμ_{R}(G) will have a “dual” version for tμ_{R}(G), and the previous argument will allow for a single proof for both.

The algebra homomorphism from rμ_{R}(G) to μ_{R}^{1}(G) yields a forgetful functor from the category of Mackey functors which are projective relative to p-subgroups to the category of rμ_{R}(G)-modules. I will denote this functor by R : if M is a Mackey functor, then R(M) is the functor M, for which I only consider evaluations at p-subgroups, and restrictions and conjugation by elements of G.

The functor R admits a left adjoint : let N be a rμ_{R}(G)-module. I set N(P) = r_{P}^{P} N. If H is a subgroup of G, then H acts on ⊕_{P ∈\underline{s}_{p}(H)}N(P), and I set

I(N)(H) = H_{0}(H, ⊕_{P ∈\underline{s}_{p}(H)}N(P))

that I will also denote by (⊕_{P ∈\underline{s}_{p}(H)}N(P))_{H}.

If P is a p-subgroup of H, and v is an element of ⊕_{P ∈\underline{s}_{p}(H)}N(P), I denote by π_{H}(v) its image in I(N)(H).

If H ⊆ K are subgroups of G, and if v ∈ ⊕_{P ∈\underline{s}_{p}(H)}N(P), I set

t_{H}^{K}(π_{H}(v)) = π_{K}(v)

which makes sense since H ⊆ K.

Similarly, if Q is a p-subgroup of K, and v ∈ N(Q), I set

r_{H}^{K}(π_{K}(v)) = ∑_{x∈H\K/Q} π_{H}(r_{H∩^{x}Q}^{^{x}Q}xv)

It is easy to see that the map r_{H}^{K} is well defined.

Finally in the same situation, if x ∈ G, I set

c_{x,K}(π_{K}(v)) = π_{^{x}K}(xv)

With this notation :

Lemma 2.10. The correspondence sending N to I(N) defines a functor from the category of rμ_{R}(G)-modules in Mack_{R}(G, 1), and this functor is left adjoint to the functor R.

I must check that I(N) is a Mackey functor, which is moreover in Mack_{R}(G, 1), and then check the adjunction property. For the first property, the only non obvious points are the transitivity of restrictions and the Mackey axiom. So let H ⊆ K ⊆ L be subgroups of G, and x = π_{L}(n) ∈ I(N)(L), with P ∈ \underline{s}_{p}(L) and n ∈ N(P).

Then

r_{K}^{L}(x) = ∑_{l∈K\L/P} π_{K}(r_{K∩^{l}P}^{^{l}P} ln)

Thus

r_{H}^{K}r_{K}^{L}(x) = ∑_{k∈H\K/K∩^{l}P} ∑_{l∈K\L/P} π_{H}(r_{H∩^{k}(K∩^{l}P)}^{^{k}(K∩^{l}P)}kr_{K∩^{l}P}^{^{l}P} ln)

= ∑_{k∈H\K/K∩^{l}P} ∑_{l∈K\L/P} π_{H}(r_{H∩^{kl}P}^{^{kl}P} kln)

But the map (k, l) ↦ kl is a bijection from the set of pairs (k, l) such that l ∈ K\L/P and k ∈ H\K/K ∩^{l}P to the set H\L/P. Thus indeed r_{H}^{K}r_{K}^{L}(x) = r_{H}^{L}(x), and restrictions are transitive.

For the Mackey axiom, let H ⊆ K ⊇ L be subgroups of G, and x = π_{L}(n) ∈ I(N)(L), with P ∈ \underline{s}_{p}(L). Then t_{L}^{K}(x) = π_{K}(n) hence

r_{H}^{K}t_{L}^{K}(x) = ∑_{k∈H\K/P} π_{H}(r_{H∩^{k}P}^{^{k}P} kn)

On the other hand

∑_{y∈H\K/L} t_{H∩^{y}L}^{H}r_{H∩^{y}L}^{^{y}L}yx = ∑_{y∈H\K/L} t_{H∩^{y}L}^{H}r_{H∩^{y}L}^{^{y}L}π_{^{y}L}(yn)

= ∑_{y∈H\K/L} t_{H∩^{y}L}^{H} ∑_{z∈H∩^{y}L\^{y}L/^{y}P} π_{H∩^{y}L}(r_{H∩^{y}L∩^{zy}P}^{^{zy}P} zyn)

= ∑_{y∈H\K/L} ∑_{z∈H∩^{y}L\^{y}L/^{y}P} π_{H}(r_{H∩^{zy}P}^{^{zy}P} zyn)

Replacing z by z^{y} in this sum, I get

r_{H}^{K}t_{L}^{K}(x) = ∑_{y∈H\K/L} ∑_{z∈H^{y}∩L\L/P} π_{H}(r_{H∩^{yz}P}^{^{yz}P} yzn)

and as above, the map (y, z) ↦ yz induces a bijection from the set of pairs (y, z) such that y ∈ H\K/L and z ∈ H^{y} ∩ L\L/P to the set H\K/P. Then indeed

r_{H}^{K}t_{L}^{K}(x) = ∑_{y∈H\K/L} t_{H∩^{y}L}^{H}r_{H∩^{y}L}^{^{y}L}yx

and the Mackey axiom holds. Thus I(N) is indeed a Mackey functor.

To check that it is in Mack_{R}(G, 1), I must check that z(f_{1}^{G}) acts trivially on it. So let H be a subgroup of G, and x = π_{H}(n) ∈ I(H)(N), with n ∈ N(P). As x = t_{P}^{H}(π_{P} (n)), I have z(f_{1}^{G})x = [f_{1}^{H}]t_{P}^{H}(π_{P} (n)). Then if I know that [f_{1}^{H}]t_{P}^{H} = t_{P}^{H}[Res_{P}^{H}f_{1}^{H}], the proof will be complete, since Res_{P}^{H}f_{1}^{H} = f_{1}^{P} = P/P. But this follows from

Lemma 2.11. Let K ⊆ H be subgroups of G, and X ∈ b(H). Then

[X]t_{K}^{H} = t_{K}^{H}[Res_{K}^{H}X] in μ_{R}(G)

Indeed if X = H/L, then [X] = t_{L}^{H}r_{L}^{H} and

[X]t_{K}^{H} = t_{L}^{H}r_{L}^{H}t_{K}^{H} = ∑_{x∈L\H/K} t_{L∩^{x}K}^{H}r_{L∩^{x}K}^{^{x}K}x = ∑_{x∈L\H/K} xt_{L^{x}∩K}^{H}r_{L^{x}∩K}^{K}

= ∑_{x∈L\H/K} t_{L^{x}∩K}^{H}r_{L^{x}∩K}^{K}

since x ∈ H, which can be written as

[X]t_{K}^{H} = t_{K}^{H} ∑_{x∈L\H/K} t_{L^{x}∩K}^{K}r_{L^{x}∩K}^{K} = t_{K}^{H}[Res_{K}^{H}H/L]

proving Lemma2.11.

It remains to check the adjunction property : so let α be a morphism from I(N) to a Mackey functor M. Such a morphism is characterized by specifying, for all subgroup H of G, of a morphism α_{H} of I(N)(H) in M(H), such that

α_{H}t_{K}^{H} = t_{K}^{H}α_{K} , α_{K}r_{K}^{H} = r_{K}^{H}α_{H} , c_{x,H}α_{H} = α_{^{x}H}c_{x,H}

for all K ⊆ H and all x ∈ G.

Then if P is a p-subgroup of G, I define a map β_{P} from N(P) to M(P) by β_{P} = α_{P} π_{P} . If Q is a subgroup of P, and if n ∈ N(P), then

r_{Q}^{P}(π_{P} (n)) = ∑_{x∈Q\P/P} π_{Q}(r_{Q∩^{x}P}^{^{x}P} xn) = π_{Q}(r_{Q}^{P}(n))

hence

r_{Q}^{P}β_{P} (n) = r_{Q}^{P}α_{P} π_{P} (n) = α_{Q}r_{Q}^{P}π_{P} (n) = α_{Q}π_{Q}r_{Q}^{P}(n) = β_{Q}r_{Q}^{P}(n)

Similarly

c_{x,P} β_{P} (n) = c_{x,P} α_{P} π_{P} (n) = α_{^{x}P} c_{x,P} π_{P} (n) = α_{^{x}P} π_{^{x}P} c_{x,P} (n) = β_{^{x}P} c_{x,P} (n)

and β defines a morphism from N to R(M).

Conversely, if β is such a morphism, then for any p-subgroup P of G, I have a morphism β_{P} of N(P) in M(P), such that

r_{Q}^{P}β_{P} = β_{Q}r_{Q}^{P} c_{x,P} β_{P} = β_{^{x}P} c_{x,P}

Then if H is a subgroup of G, I define a map α_{H} from I(N)(H) to M(H) by α_{H}(π_{H}(n)) = t_{P}^{H}β_{P} (n), if P is a p-subgroup of H, and if n ∈ N(P). If K is a subgroup of H, if Q is a subgroup of P, and if n ∈ N(Q), then

α_{H}t_{K}^{H}(π_{K}(n)) = α_{H}(π_{H}(n)) = t_{Q}^{H}β_{Q}(n)

whereas

t_{K}^{H}α_{K}(π_{K}(n)) = t_{K}^{H}t_{Q}^{K}β_{Q}(n)

which proves that α_{H}t_{K}^{H} = t_{K}^{H}α_{K}.

Similarly, if n ∈ N(P), then

α_{K}r_{K}^{H}(π_{H}(n)) = α_{K}( ∑_{x∈K\H/P} π_{K}(r_{K∩^{x}P}^{^{x}P} xn)) = ∑_{x∈K\H/P} t_{K∩^{x}P}^{K} β_{K∩^{x}P} r_{K∩^{x}P}^{^{x}P} xn

= ∑_{x∈K\H/P} t_{K∩^{x}P}^{K} r_{K∩^{x}P}^{^{x}P} β_{^{x}P} xn = ∑_{x∈K\H/P} t_{K∩^{x}P}^{K} r_{K∩^{x}P}^{^{x}P} xβ_{P} n

= r_{K}^{H}t_{P}^{H}β_{P} (n) = r_{K}^{H}α_{H}(π_{H}(n))

which proves that α_{K}r_{K}^{H} = r_{K}^{H}α_{H}.

Finally

c_{x,H}α_{H}π_{H}(n) = c_{x,H}t_{P}^{H}β_{P} (n) = t_{^{x}P}^{^{x}H}c_{x,P} β_{P} (n) = t_{^{x}P}^{^{x}H}β_{^{x}P} c_{x,P} (n) = α_{^{x}H}c_{x,H}(π_{H}(n))

which proves that c_{x,H}α_{H} = α_{^{x}H}c_{x,H}, hence that α defines a morphism from I(N) to M. These correspondences between Hom_{μ_{R}(G)}(I(N), M) and Hom_{rμ_{R}(G)}(N, R(M)) are clearly mutual inverse bijections, and this proves the adjunction property, completing the proof of Lemma 2.10.

3. Examples of rμ_{R}(G)-modules

Let I the R-submodule of rμ_{R}(G) generated by the xr_{Q}^{P}, where x ∈ G, and Q is a proper subgroup of P. Then :

Lemma 3.1. The submodule I is a nilpotent two sided ideal of rμ_{R}(G), and the quotient rμ_{R}(G)/I is isomorphic to the direct product over the p-subgroups P of G up to G-conjugation, of the algebras Ind_{N_{G}(P)/P}^{G} RN_{G}(P)/P.

(about the notion of induced algebra, see [3])

Indeed, if Q is a proper subgroup of P, and if A ⊆ B are p-subgroups of G, and if c ∈ G, then cr_{A}^{B}xr_{Q}^{P} is equal to zero if B^{x} ≠ Q, and to cxr_{A^{x}}^{P} otherwise, and then A^{x} is a proper subgroup of P. Similarly, the product xr_{Q}^{P}cr_{A}^{B} is equal to zero if P^{c} ≠ A, and to xcr_{Q^{c}}^{B} otherwise, and then Q^{c} is a proper subgroup of B.

It is clear moreover that I is nilpotent : indeed, if x_{1}r_{Q_{1}}^{P_{1}}x_{2}r_{Q_{2}}^{P_{2}} … x_{n}r_{Q_{n}}^{P_{n}} ≠ 0, then P_{1}^{x_{2}} = Q_{2}, P_{2}^{x_{3}} = Q_{3}, etc, and the sequence P_{1}^{x_{2}...x_{n}}, P_{2}^{x_{3}...x_{n}}, …, P_{n} is strictly decreasing. Its length is hence bounded by the p-valuation of the order of G.

Then if P is a p-subgroup of G, let e_{P} = ∑_{Q=_{G}P} r_{Q}^{Q}. The elements e_{P} are idempotents of rμ_{R}(G). Moreover e_{P} and e_{R} are orthogonal if P and R are not conjugate in G. Finally ∑_{P mod.G} e_{P} = ∑_{P} r_{P}^{P} is the identity element of rμ_{R}(G).

On the other hand, if the product r_{P}^{P} cr_{A}^{B}r_{P}^{P} is non zero, then P^{c} = A ⊆ B = P, hence the algebra r_{P}^{P} rμ_{R}(G)r_{P}^{P} has a basis consisting of the elements xr_{P}^{P}, for x ∈ N_{G}(P)/P (which form a subset of the basis of rμ_{R}(G) of Lemma 2.7). Moreover, the map x ∈ N_{G}(P)/P ↦ xr_{P}^{P} defines a morphism of algebras α of RN_{G}(P)/P in r_{P}^{P} rμ_{R}(G)r_{P}^{P}, which is hence an isomorphism. It is then clear that the map from Ind_{N_{G}(P)/P}^{G} RN_{G}(P)/P to e_{P} rμ_{R}(G)e_{P}, sending g ⊗ u ⊗ h to gα(u)h is a isomorphism, and this completes the proof of Lemma 3.1.

But if H is a subgroup of G, and A is an interior H-algebra, the inclusion of A in Ind_{H}^{G}A is a Morita equivalence : this follows e.g. from Lemma 2.4. This remark, together with Lemma 3.1, yields a classification of the simple rμ_{R}(G)-modules.

Indeed, let P be a p-subgroup of G and V be an RN_{G}(P)/P -module. I define an rμ_{R}(G)-module N_{P,V} by N_{P,V} = Ind_{N_{G}(P)/P}^{G} V. The element cr_{A}^{B} acts by 0 on N_{P,V} if A is a proper subgroup of B, or if B is not conjugate to P. And if A = B = P^{x}, then cr_{A}^{B}.(y ⊗ v) = 0 if xy ∉ N_{G}(P), and cr_{A}^{B}.(y ⊗ v) = cx^{−1} ⊗ xyv otherwise (note that this element does not depend on the choice of the element x such that B = P^{x}).

This defines a genuine rμ_{R}(G)-module : I must essentially check that Q acts trivially on N_{P,V} (Q) and that

cr_{P^{x}}^{P^{x}}(dr_{P^{y}}^{P^{y}}.(z ⊗ v)) = (cr_{P^{x}}^{P^{x}}dr_{P^{y}}^{P^{y}}).(z ⊗ v)

The left hand side is equal to 0 if yz ∉ N_{G}(P), and to cr_{P^{x}}^{P^{x}}.(dy^{−1} ⊗ yzv) otherwise. This is equal to zero if xdy^{−1} ∉ N_{G}(P), and to cx^{−1} ⊗ xdzv = cx^{−1} ⊗ xdy^{−1}yzv = cdy^{−1} ⊗ yzv otherwise. But the right hand side is equal to zero if P^{xd} ≠ P^{y}, i.e. if xdy^{−1} ∉ N_{G}(P), and to cdr_{P^{y}}^{P^{y}}.(z ⊗ v) otherwise. This is equal to zero if yz ∉ N_{G}(P), and to cdy^{−1} ⊗ yzv otherwise. Hence the two sides are equal.

The definition of N_{P,V} implies that N_{P,V} (Q) = r_{Q}^{Q}N_{P,V} is equal to zero if Q is not conjugate to P in G. And if Q = P^{x}, then N_{P,V} (Q) identifies with x^{−1} ⊗ V. Then if q ∈ Q, and if x^{−1} ⊗ v ∈ N_{P,V} (Q), I have q.(x^{−1} ⊗ v) = qx^{−1} ⊗ v = x^{−1}.^{x}q ⊗ v = x^{−1} ⊗^{x} qv = x^{−1} ⊗ v since ^{x}Q = P and since P acts trivially on V. Thus N_{P,V} is a genuine rμ_{R}(G)-module.

The previous remarks, together with Lemma 3.1, yield the following

Proposition 3.2. The modules N_{P,V}, when P runs through a set of representatives of conjugacy classes of p-subgroups of G, and V through a set of representatives of isomorphism classes of simple RN_{G}(P)/P -modules, form a full set of representatives of isomorphism classes of simple rμ_{R}(G)-modules.

In order to describe the indecomposable projective rμ_{R}(G)-modules, I need a notation : if A and B are subgroups of G, I denote by T_{G}(A, B) the set of x ∈ G such that A^{x} ⊆ B. Let then P be a p-subgroup of G, and V be a RN_{G}(P)/P -module. If Q is a p-subgroup of G, I define an R-submodule of Ind_{N_{G}(P)/P}^{G} V, denoted by L_{P,V} (Q), by

L_{P,V} (Q) = ⊕_{x∈T_{G}(Q,P)/N_{G}(P)}x ⊗ V

and I set L_{P,V} = ⊕_{Q}L_{P,V} (Q).

If Q ⊇ R, then the inclusion of T_{G}(Q, P)/N_{G}(P) in T_{G}(R, P)/N_{G}(P) induces a morphism r_{R}^{Q} from L_{P,V} (Q) to L_{P,V} (R). And if x ∈ G, the natural bijection yN_{G}(P) ↦ xyN_{G}(P) from T_{G}(Q, P)/N_{G}(P) to T_{G}(^{x}Q, P)/N_{G}(P) induces a morphism, that I still denote by x, from L_{P,V} (Q) to L_{P,V} (^{x}Q). Then

Proposition 3.3. These definitions turn L_{P,V} into an rμ_{R}(G)-module, and the functor from RN_{G}(P)/P -Mod to rμ_{R}(G)-Mod sending V to L_{P,V} is left adjoint to the restriction functor induced by the inclusion u ↦ ur_{P}^{P} from RN_{G}(P)/P to rμ_{R}(G).

The action of cr_{A}^{B} on the element x ⊗ v of L_{P,V} (Q) is defined by
cr_{A}^{B}.x ⊗ v = δ_{B,Q}cx ⊗ v

where, when B = Q, the element cx⊗v is seen in L_{P,V} (^{c}A). In particular, if q ∈ Q, then q.(x ⊗ v) = qx ⊗ v = xq^{x} ⊗ v = x ⊗ q^{x}v = x ⊗ v, since Q^{x} ⊆ P. Thus Q acts trivially on L_{P,V} (Q).

Similarly,

dr_{E}^{F}.(cr_{A}^{B}.x ⊗ v) = δ_{F,^{c}A}δ_{B,Q}dcx ⊗ v

with dcx ⊗ v ∈ L_{P,V} (^{d}E) if B = Q and F = ^{c}A.

Since dr_{E}^{F}.cr_{A}^{B} = δ_{F^{c},A}dcr_{E^{c}}^{B}, I have

(dr_{E}^{F}.cr_{A}^{B}).x ⊗ v = δ_{F^{c},A}δ_{B}^{Q}dcx ⊗ v

with dcx ⊗ v ∈ L_{P,V} (^{dc}(E^{c})) if F^{c} = A and B = Q. Thus L_{P,V} is indeed an rμ_{R}(G)-module.

It is clear that the construction sending V to L_{P,V} is functorial in V. If M is an rμ_{R}(G)-module, a morphism α from L_{P,V} to M is defined by specifying, for any p-subgroup Q of G, a morphism α_{Q} from L_{P,V} (Q) to M(Q), such that

α_{R}r_{R}^{Q} = r_{R}^{Q}α_{Q} xα_{Q} = α_{^{x}Q}x

Since L_{P,V} (P) identifies with V as an RN_{G}(P)/P -module, this yields a morphism β = α_{P} from V to M(P). And since the element x ⊗ v of L_{P,V} (Q) is equal to r_{Q}^{^{x}P} (x ⊗ v), I have

α_{Q}(x ⊗ v) = r_{Q}^{^{x}P} α_{^{x}P} (x ⊗ v) = r_{Q}^{^{x}P} xα_{P} (1 ⊗ v) = xr_{Q^{x}}^{P}β(v)

and β determines entirely α.

Conversely, if β is a morphism from V to M(P), if Q is a p-subgroup of G, and if x ⊗ v ∈ L_{P,V} (Q), I set α_{Q}(x ⊗ v) = xr_{Q^{x}}^{P}β(v), which makes sense since Q^{x} ⊆ P.

Then r_{R}^{Q}α_{Q}(x ⊗ v) = r_{R}^{Q}xr_{Q^{x}}^{P}β(v) = r_{R^{x}}^{P}β(v) = α_{R}r_{R}^{Q}(x ⊗ v), hence α_{R}r_{R}^{Q} = r_{R}^{Q}α_{Q}.

On the other hand, yα_{Q}(x ⊗ v) = yxr_{Q^{x}}^{P}β(v), whereas α_{^{y}Q}(y.(x ⊗ v)) = α_{^{y}Q}(yx ⊗ v) = yxr_{(^{y}Q)^{yx}}^{P}β(v), which proves that α is a morphism of rμ_{R}(G)- modules.

Finally it is clear that these correspondences between Hom_{rμ_{R}(G)}(L_{P,V} , M) and Hom_{RN_{G}(P)/P} (V, M(P)) are mutual inverse bijections. The proposition follows.

Corollary 3.4. If E_{V} is a projective cover of the simple RN_{G}(P)/P -module V, then L_{P,E_{V}} is a projective cover of the simple rμ_{R}(G)-module N_{P,V}.

This follows indeed from the isomorphisms

Hom_{rμ_{R}(G)}(L_{P,E_{V}} , M) ≅ Hom_{RN_{G}(P)/P} (E_{V} , M(P))

and N_{P,V} (P) ≅ V : if f is an essential morphism from E_{V} to V, then f defines a morphism L(f) of L_{P,E_{V}} in N_{P,V}, which is also essential : if L_{1} is a submodule of L_{P,E_{V}} which maps onto N_{P,V} by L(f), then L_{1}(P) maps onto V by f, hence L_{1}(P) = E_{V}. And since L_{1}(Q) is the sum for x ∈ T_{G}(Q, P) of xr_{Q^{x}}^{P}(L_{1}(P)), it follows that L_{1}(Q) = L_{P,V} (Q) for all Q, hence that L_{1} = L_{P,V}.

Remark 3.5. The module L_{P,E_{V}} (1) identifies with X = Ind_{N_{G}(P)/P}^{G} E_{V} as an RG-module. In the case where R is a field of characteristic p, it is easy to check that the module L_{P,E_{V}} (Q) identifies with the module

X[Q] = X^{Q}/( ∑_{R⊂Q} T r_{R}^{Q}(X^{R}))

The projective indecomposable rμ_{R}(G)-modules are then in one to one correspondence with p-permutation RG-modules of the form Ind_{N_{G}(P)/P}^{G} E_{V}, where P is a p-subgroup of G and E_{V} is an indecomposable projective RN_{G}(P)/P -module.

The restriction functor induced by the inclusion u ↦ ur_{P}^{P} from RN_{G}(P)/P to rμ_{R}(G) also admits a right adjoint : if V is an RN_{G}(P)/P module, and Q a p-subgroup of G, then Q acts on the right on N_{G}(P)\T_{G}(P, Q), hence on the left on ⊕_{x∈N_{G}(P)\T_{G}(P,Q)}x^{−1} ⊗ V, and I set

L_{P,V}^{o} (Q) = (⊕_{x∈N_{G}(P)\T_{G}(P,Q)}x^{−1} ⊗ V)^{Q}

Then if Q ⊆ R, the inclusion of N_{G}(P)\T_{G}(P, Q) in N_{G}(P)\T_{G}(P, R) allows to define a morphism r_{Q}^{R} from L_{P,V}^{o} (R) to L_{P,V}^{o} (Q), by r_{Q}^{R}(⊕_{x}x^{−1}⊗v_{x}) = ⊕_{P^{x}⊆Q}x^{−1}⊗v_{x}.

Similarly, the natural bijection y ↦ yx^{−1} from N_{G}(P)\T_{G}(P, Q) to N_{G}(P)\T_{G}(P,^{x} Q) allows to define a morphism denoted by x from L_{P,V}^{o} (Q) to L_{P,V}^{o} (^{x}Q), by setting x. ⊕_{y} (y^{−1} ⊗ v_{y}) = ⊕_{y}(yx^{−1})^{−1} ⊗ v_{y} = ⊕_{y}xy^{−1} ⊗ v_{y}.

It is then clear that Q acts trivially on L_{P,V}^{o} (Q). Moreover

dr_{E}^{F}(cr_{A}^{B}.(⊕_{x}x^{−1} ⊗ v_{x})) = δ_{B,Q}dr_{E}^{F}(⊕_{P^{x}⊆A}cx^{−1} ⊗ v_{x})
= δ_{B,Q}δ_{F,^{c}A} ⊕_{P^{x}⊆A,P^{xc^{−1}} ⊆E} dcx^{−1} ⊗ v_{x}

But if F = ^{c}A, then P^{xc^{−1}} ⊆ E implies that P^{x} ⊆ E^{c} ⊆ F^{c} = A, hence

dr_{E}^{F}(cr_{A}^{B}.(⊕_{x}x^{−1} ⊗ v_{x})) = δ_{B,Q}δ_{F,^{c}A} ⊕_{P^{xc^{−1}} ⊆E} dcx^{−1} ⊗ v_{x}

On the other hand, since dr_{E}^{F}cr_{A}^{B} = δ_{F^{c},A}dcr_{E^{c}}^{B}, I also have

(dr_{E}^{F}cr_{A}^{B}) ⊕_{x} x^{−1} ⊗ v_{x} = δ_{B,Q}δ_{F^{c},A} ⊕_{P^{x}⊆E^{c}} dcx^{−1} ⊗ v_{x}

and L_{P,V}^{o} = ⊕_{Q}L_{P,V}^{o} (Q) is indeed an rμ_{R}(G)-module.

This construction is clearly functorial in V. And if α is a morphism from an rμ_{R}(G)-module M to L_{P,V}^{o}, then as L_{P,V}^{o} (P) = 1⊗V, this yields a morphism β from M(P) to V, such that α_{P} = 1 ⊗ β. If Q is a p-subgroup of G, if m ∈ M(Q), and if α_{Q}(m) = ⊕_{x∈N_{G}(P)\T_{G}(P,Q)}x^{−1} ⊗ v_{x}, let x ∈ N_{G}(P)\T_{G}(P, Q). Since r_{P^{x}}^{Q}α_{Q}(m) = x^{−1}⊗v_{x}, I must hence have x^{−1}⊗v_{x} = r_{P^{x}}^{Q}α_{Q}(m) = α_{P^{x}}r_{P^{x}}^{Q}(m) = x^{−1}α_{P} xr_{P^{x}}^{Q}(m), hence v_{x} = β(xr_{P^{x}}^{Q}(m)), and β determine entirely α.

Conversely, if β is given, I define α_{Q} from M(Q) to L_{P,V}^{o} (Q) by

α_{Q}(m) = ⊕_{x∈N_{G}(P)\T_{G}(P,Q)}x^{−1} ⊗ β(xr_{P^{x}}^{Q}(m))

It is easy to see that α is a morphism from M to L_{P,V}^{o}. These two constructions are clearly inverse to each other. Thus

Proposition 3.6. The functor which maps V to L_{P,V}^{o} is right adjoint to the restriction functor from rμ_{R}(G)-Mod to RN_{G}(P)/P -Mod.

This allows of course for a description of injective rμ_{R}(G)-modules.

4. Resolutions

4.1. Projectivity relative to a functor. ^{3} The following definition generalizes the notion of relative projectivity, when R is a functor of restriction to a subgroup.

Definition 4.1. Let C and D be categories, and R a functor from C to D. I will say that an object M of C is projective relative to R (or R-projective) if for any diagram

\begin{aligned}& & M \\ & & ↓ β \\ & α \\ X & → & Y \end{aligned}

such that the morphism R(α) is a split epimorphism, there exists a morphism ϕ from M to X such that αϕ = β.

3The content of this section is very close to the notion of comonad or cotriple in category theory. See e.g. C.A. Weibel. An introduction to homological algebra, Cambridge studies in advanced mathematics 38, Chapter 8.6

The following lemma is a straightforward consequence of the definitions :

Lemma 4.2. a) If M is projective relative to R, and if f : M → N is a split epimorphism, then N is projective relative to R.

b) If M_{1} and M_{2} are projective relative to R, and if N is a coproduct of M_{1} and M_{2}, then N is projective relative to R

The following lemma is a formalization of known results in the classical cases of relative projectivity, when R has a left adjoint :

Lemma 4.3. Let C and D be categories, and R be a functor of C in D, admitting a left adjoint I. Let M be an object of C. The following conditions are equivalent :

(1) The object M is projective relative to R.

(2) The counit morphism IR(M) → M is a split epimorphism.

(3) There exists an object Y of D and a split epimorphism I(Y) → M.

If X is an object of C, if Y is an object of D, and α (resp. β) is a morphism from Y to R(X) (resp. a morphism from I(Y) to X), I denote by α^{∗} (resp. β_{∗}) the morphism from I(Y) to X (resp. from Y to R(X)) associated by adjunction. Then (α^{∗})_{∗} = α and (β_{∗})^{∗} = β.

Since the isomorphism from Hom_{C}(I(−), −) to Hom_{D}(−, R(−)) is an isomorphism of bifunctors, if f ∈ Hom_{D}(Y_{1}, Y_{2}), and if α ∈ Hom_{D}(Y_{2}, R(X)), I have (αf)^{∗} = α^{∗}I(f). Similarly, if g ∈ Hom_{C}(X_{1}, X_{2}), and if β ∈ Hom_{C}(I(Y), X_{1}), then (gβ)_{∗} = R(g)β_{∗}.

Then if 1) holds, the diagram

M
↓ Id_{M}
(Id_{R(M)})^{∗}
IR(M)  →  M

shows that 2) holds, if I know that the morphism u = R((Id_{R(M)})^{∗}) is a split epimorphism. But if I set v = (Id_{IR(M)})_{∗}, I have

uv = ((Id_{R(M)})^{∗}Id_{R(M)})_{∗} = ((Id_{R(M)})^{∗})_{∗} = Id_{R(M)}

It is clear that 2) implies 3). Conversely, if u : I(Y) → M is a split epimorphism, then there exists v : M → I(Y) such that uv = Id_{M}. On the other hand, the morphism I(u_{∗}) is a morphism from I(Y) to IR(M) and

(Id_{R(M)})^{∗}I(u_{∗}) = (Id_{R(M)}u_{∗})^{∗} = u

hence

(Id_{R(M)})^{∗}I(u_{∗})v = uv = Id_{M}

which prove 2). Finally if 2) holds, let τ : M → IR(M) be such that (Id_{R(M)})^{∗}τ = Id_{M}. If I have a diagram

\begin{aligned}& & M \\ & & ↓ β \\ & α \\ X & → & Y \end{aligned}

such that R(α) is a split epimorphism, then there exists a morphism

γ : R(Y) → R(X)

such that R(α)γ = Id_{R(Y)}. Setting then

ϕ = (Id_{R(X)})^{∗}I(γ)IR(β)τ

it follows that

αϕ = α(Id_{R(X)})^{∗}I(γR(β))τ
= (R(α)Id_{R(X)})^{∗}I(γR(β))τ
= (R(α)γR(β))^{∗}τ = (R(β))^{∗}τ
= (R(β)Id_{R(M)})^{∗}τ = β(Id_{R(M)})^{∗}τ = β

which proves 1), and the equivalence of the three conditions.

Lemma 4.4. Let C and D be categories, and R be a functor from C to D, admitting a left adjoint I. Then R is faithful if and only if for any object M of C, the counit morphism from IR(M) to M is an epimorphism.

Indeed, the morphism from IR(M) to M is an epimorphism if and only if for any object X, the morphism

Hom(M, X) → Hom(IR(M), X)

is injective, i.e. by adjunction if and only if the morphism

Hom(M, X) → Hom(R(M), R(X))

is injective.

Remark 4.5. It follows that if R is faithful, then any projective object is projective relative to R.

I suppose now that C and D are abelian categories, that R is faithful and admits a left adjoint I. I denote by K(M) the kernel of the epimorphism from IR(M) to M.

Then K(M) is the quotient of IRK(M) by K(K(M)) = K^{2}(M), which itself is the quotient of IRK^{2}(M) by K(K^{2}(M)) = K^{3}(M), and I can build that way a resolution of M by objects of C of the form I(L), which are projective relative to R. Actually :

Lemma 4.6. Let C and D be abelian categories, and R be a faithful functor from C to D admitting a left adjoint I. Then for any object M of C, there exists a resolution

… → L_{i} → L_{i−1} → … → L_{0} → M → 0

where the L_{i}’s are projective relative to R, such that the complex

… → R(L_{i}) → R(L_{i−1}) → … → R(L_{0}) → R(M) → 0

is exact and split, and such a resolution of M is unique to homotopy.

The existence of such a resolution follows from the previous argument, and from the fact that the morphism RIR(M) → R(M) is a split epimorphism (see the proof of Lemma 4.3).

A standard homological argument shows that if M and N are objects of C, if X_{∗} is a resolution of M by R-projective objects, and Y_{∗} a resolution of N such that the complex R(Y_{∗}) → R(N) is split, any homomorphism of M in N can be lifted to an homomorphism from X_{∗} to Y_{∗}, and that such a lift is unique up to homotopy. The uniqueness of the resolution follows.

Remark 4.7. If M is projective, a resolution of M with the properties of Lemma 4.6 is split, because M is projective relative to R.

4.2. Resolutions of Mackey functors. The functors R and I defined above between the categories rμ_{R}(G)-Mod and Mack_{R}(G, 1) fulfill the assumptions of Lemma 4.6 (the functor R is faithful because it is the composition of a forgetful functor and an equivalence of categories).

Thus, any Mackey functor in Mack_{R}(G, 1) admits a resolution by R-projective functors, whose image by R is split. This means that this resolution can be split by homomorphisms which commute with conjugations by G, and with restrictions (but dot not commute, in general, with transfers). Moreover such a resolution is unique up to homotopy. The additional fact here is that this resolution can be chosen to be finite.

Indeed, let M a Mackey functor. Then K(M)(H) is the image in the quotient (⊕_{P ∈\underline{s}_{p}(H)}M(P))_{H} of the set of sequences n_{P} such that ∑_{P} t_{P}^{H}n_{P} = 0. In particular, if H is a p-subgroup Q of G, this condition is equivalent to n_{Q} = − ∑_{P ⊂Q} t_{P}^{Q}n_{P}. It follows that RK(M)(Q) identifies with

R(M)_{1}(Q) = (⊕_{P ⊂Q}M(P))_{Q}

More generally, let M a rμ_{R}(G)-module : I will set M_{0} = M, and if i is a positive integer, I set

M_{i}(Q) = (⊕_{P_{0}⊂P_{1}...⊂P_{i−1}⊂Q}M(P_{0}))_{Q}

If n ∈ M(P_{0}), I denote by n_{P_{0},P_{1},...,P_{i−1},Q} the corresponding element of (⊕_{P_{0}⊂P_{1}...⊂P_{i−1}⊂Q}M(P_{0})), and π_{Q}(n_{P_{0},P_{1},...,P_{i−1},Q}) its image in M_{i}(Q). If x ∈ G, I set

xπ_{Q}(n_{P_{0},...,P_{i−1},Q}) = π_{^{x}Q}((xn)_{^{x}P_{0},...,^{x}P_{i−1},^{x}Q})

and if S is a subgroup of Q, I set

r_{S}^{Q}π_{Q}(n_{P_{0},...,P_{i−1},Q}) = ∑_{{x∈S\Q/P_{0} \atop S∩^{x}P_{0}≠S∩^{x}P_{1}...≠S∩^{x}P_{i−1}≠S}} π_{S}((xr_{S^{x}∩P_{0}}^{P_{0}}n)_{S∩^{x}P_{0},...,S∩^{x}P_{i−1},S})

Lemma 4.8. The previous definitions turn M_{i} into an rμ_{R}(G)-module.

The only non-trivial point is the transitivity of restrictions. So let T ⊆ R ⊆ Q be p-subgroups of G, and s = (P_{0}, …, P_{i−1}, Q) be an increasing sequence of subgroups of Q. I will say that s is a proper sequence if s is strictly increasing. If A is a subgroup of G, I will denote by A ∩ s the sequence (A ∩ P_{0}, …, A ∩ P_{i−1}, A ∩ Q).

Then if n ∈ M(P_{0}), I have

r_{S}^{Q}(π_{Q}(n_{s})) = ∑_{{x∈S\Q/P_{0} \atop S∩^{x}s proper}} π_{S}((xr_{S^{x}∩P_{0}}^{P_{0}}n)_{S∩^{x}s})

Thus

r_{T}^{S}r_{S}^{Q}(π_{Q}(n_{s})) = ∑_{{x∈S\Q/P_{0} \atop S∩^{x}s proper}} ∑_{{y∈R\S/S∩^{x}P_{0} \atop R∩^{y}(S∩^{x}s) proper}} π_{R}((yr_{R^{y}∩S∩^{x}P_{0}}^{S∩^{x}P_{0}}xr_{S^{x}∩P_{0}}^{P_{0}}n)_{R∩^{y}(S∩^{x}s)})

i.e.

r_{T}^{S}r_{S}^{Q}(π_{Q}(n_{s})) = ∑_{{x∈S\Q/P_{0} \atop S∩^{x}s proper}} ∑_{{y∈R\S/S∩^{x}P_{0} \atop R∩^{yx}s proper}} π_{R}((yxr_{R^{yx}∩P_{0}}^{P_{0}}n)_{R∩^{yx}s})

Then the element z = yx runs through a set of representatives of doubles cosets R\Q/P_{0} such that R ∩^{z}s is proper. Thus

r_{T}^{S}r_{S}^{Q}(π_{Q}(n_{s})) = ∑_{{z∈R\Q/P_{0} \atop R∩^{z}s proper}} π_{R}((zr_{R^{z}∩P_{0}}^{P_{0}}n)_{R∩^{z}s}) = r_{T}^{Q}(π_{Q}(n_{s}))

which proves the lemma.

The previous construction associates to any rμ_{R}(G)-module M some rμ_{R}(G)- modules M_{i}, and moreover M_{i} = 0 if i is greater than (or equal to) the p-valuation of the order of G. If M is a Mackey functor in Mack_{R}(G, 1), and i a non negative integer, I will denote by δ_{i}(M) the functor I(R(M)_{i}), and I will build a resolution of M from these functors. The functor δ_{i}(M) admits the following simple description :

Lemma 4.9. Let M in Mack_{R}(G, 1), and H a subgroup of G. Then

δ_{i}(M)(H) = (⊕_{P_{0}⊂...⊂P_{i}∈\underline{s}_{p}(H)}M(P_{0}))_{H}

If s = (P_{0}, …, P_{i}) is a sequence of p-subgroups of H, and if n ∈ M(P_{0}), I denote by n_{s} the corresponding element of (⊕_{P_{0}⊂...⊂P_{i}∈\underline{s}_{p}(H)}M(P_{0})) if the sequence s is strictly increasing, and I set n_{s} = 0 otherwise. I denote by π_{H}(n_{s}) the image of n_{s} in (⊕_{P_{0}⊂...⊂P_{i}∈\underline{s}_{p}(H)}M(P))_{H}. I can then define a linear map f from δ_{i}(M)(H) to (⊕_{P_{0}⊂...⊂P_{i}∈\underline{s}_{p}(H)}M(P))_{H} by

f(π_{H}π_{Q}(n_{s})) = π_{H}(n_{s∪Q})

which makes sense since if h ∈ H and q ∈ Q, then

f(π_{H}(hπ_{Q}(q.n_{s}))) = f(π_{H}π_{^{h}Q}((hqn)_{^{hq}s})) = π_{H}((hqn)_{^{hq}s∪^{h}Q}))
= π_{H}(h.(qn)_{^{q}s∪Q}) = π_{H}(q.n_{s∪Q}) = π_{H}(n_{s∪Q})

Conversely, I define a linear map g from (⊕_{P_{0}⊂...⊂P_{i}∈\underline{s}_{p}(H)}M(P))_{H} to δ_{i}(M)(H) by

g(π_{H}(n_{s})) = π_{H}π_{sups}(n_{s−sups})

which makes sense because if h ∈ H, then

g(π_{H}(h.n_{s})) = g(π_{H}((hn)_{^{h}s})) = π_{H}π_{^{h}sups}((hn)_{^{h}s−^{h}sups}))
= π_{H}(h.π_{sups}(n_{s−sups})) = π_{H}π_{sups}(n_{s−sups})

It is then clear that f and g are mutual inverse isomorphisms, which proves the lemma.

If K is a subgroup of H, if s = (P_{0}, …, P_{i}) is a sequence of p-subgroups of K, and if n ∈ M(P_{0}), it follows from the definitions that

t_{K}^{H}(π_{K}(n_{s})) = π_{H}(n_{s})

Similarly, if s = (P_{0}, …, P_{i}) is a sequence of p-subgroups of H, and if n ∈ M(P_{0}), then

r_{K}^{H}(π_{H}(n_{s})) = ∑_{{x∈K\H/P_{0} \atop K∩^{x}s proper}} π_{K}((xr_{K^{x}∩P_{0}}^{P_{0}}n)_{K∩^{x}s})

Hence I can identify δ_{0}(M) with IR(M). I will denote by d_{0} the canonical morphism from δ_{0}(M) to M. If i > 0, I define a linear map d_{i} from δ_{i}(M) to δ_{i−1}(M) by

d_{i,H}(π_{H}(n_{s})) = π_{H}((t_{P_{0}}^{P_{1}}n)_{s_{0}} + ∑_{j=1}^{i}(−1)^{j}n_{s_{j}})

where s_{j} denotes the sequence s − {P_{j}}. Then

Theorem 4.10. Let M in Mack_{R}(G, 1). The sequence

0 → … →^{d_{i+1}} δ_{i}(M) →^{d_{i}} … →^{d_{1}} δ_{0}(M) →^{d_{0}} M → 0

is an exact complex of Mackey functors in Mack_{R}(G, 1), and the complex

0 → … →^{R(d_{i+1})} R(δ_{i}(M)) →^{R(d_{i})} … →^{R(d_{1})} R(δ_{0}(M)) →^{R(d_{0})} R(M) → 0

is exact and split.

First I must that the maps d_{i} are morphisms of Mackey functors. This is clear for d_{0}. If i > 0, if H ⊆ K are subgroups of G, if s = (P_{0}, …, P_{i}) is an increasing sequence of p-subgroups of H, and if n ∈ M(P_{0}), then

d_{i,K}(t_{H}^{K}π_{H}(n_{s})) = d_{i,K}(π_{K}(n_{s})) = π_{K}((t_{P_{0}}^{P_{1}}n)_{s_{0}} + ∑_{j=1}^{i}(−1)^{j}n_{s_{j}})

whereas

t_{H}^{K}d_{i,H}(π_{H}(N_{s})) = t_{H}^{K}(π_{H}((t_{P_{0}}^{P_{1}}n)_{s_{0}} + ∑_{j=1}^{i}(−1)^{j}n_{s_{j}}) = π_{K}((t_{P_{0}}^{P_{1}}n)_{s_{0}} + ∑_{j=1}^{i}(−1)^{j}n_{s_{j}})

and the maps d_{i} commute with traces.

Similarly, if s = (P_{0}, …, P_{i}) is an increasing sequence of p-subgroups of K, and if n ∈ M(P_{0}), then

d_{i,H}(r_{H}^{K}π_{K}(n_{s})) = d_{i,H}( ∑_{x∈H\K/P_{0}} π_{H}((xr_{H^{x}∩P_{0}}^{P_{0}}n)_{H∩^{x}s}))

= ∑_{x∈H\K/P_{0}} π_{H}((t_{H∩^{x}P_{0}}^{H∩^{x}P_{1}}xr_{H^{x}∩P_{0}}^{P_{0}}n)_{(H∩^{x}s)_{0}} + ∑_{j=1}^{i}(−1)^{j}(xr_{H^{x}∩P_{0}}^{P_{0}})_{(H∩^{x}s)_{j}}) (1)

On the other hand

r_{H}^{K}d_{i,K}(π_{K}(n_{s})) = r_{H}^{K}π_{K}((t_{P_{0}}^{P_{1}}n)_{s_{0}} + ∑_{j=1}^{i}(−1)^{j}n_{s_{j}})

But if j > 0, then the smallest element of s_{j} is P_{0}, and

r_{H}^{K}(π_{K}(n_{s_{j}})) = ∑_{x∈H\K/P_{0}} π_{H}((xr_{H^{x}∩P_{0}}^{P_{0}}n)_{H∩^{x}s_{j}})

And these terms are equal to the corresponding terms of Equality (1), since for any m of M(P_{0}), I have m_{(H∩^{x}s)_{j}} = m_{H∩^{x}s_{j}}. Similarly

r_{H}^{K}π_{K}((t_{P_{0}}^{P_{1}}n)_{s_{0}}) = ∑_{y∈H\K/P_{1}} π_{H}(yr_{H^{y}∩P_{1}}^{P_{1}}(t_{P_{0}}^{P_{1}}n)_{H∩^{y}s_{0}})

= π_{H} ∑_{y∈H\K/P_{1}}(y ∑_{z∈H^{y}∩P_{1}\P_{1}/P_{0}} t_{H^{y}∩^{z}P_{0}}^{H^{y}∩P_{1}}r_{H^{y}∩^{z}P_{0}}^{^{z}P_{0}}zn)_{H∩^{y}s_{0}})

= ∑_{y∈H\K/P_{1}} ∑_{z∈H^{y}∩P_{1}\P_{1}/P_{0}} π_{H}(t_{H∩^{yz}P_{0}}^{H∩^{y}P_{1}}r_{H∩^{yz}P_{0}}^{^{yz}P_{0}}yzn)_{H∩^{y}s_{0}})

and as z ∈ P_{1} = inf s_{0}, I have ^{y}P_{1} = ^{yz}P_{1} and H ∩^{y}s_{0} = H ∩^{yz}P_{0}, and I can sum over x = yz, which runs through the set H\K/P_{0}. It follows that

r_{H}^{K}π_{K}((t_{P_{0}}^{P_{1}}n)_{s_{0}}) = ∑_{x∈H\K/P_{0}} π_{H}(t_{H∩^{x}P_{0}}^{H∩^{x}P_{1}}r_{H∩^{x}P_{0}}^{^{x}P_{0}}xn)_{H∩^{x}s_{0}})

= ∑_{x∈H\K/P_{0}} π_{H}(t_{H∩^{x}P_{0}}^{H∩^{x}P_{1}}xr_{H^{x}∩P_{0}}^{P_{0}}n)_{H∩^{x}s_{0}})

which is equal to the corresponding term of Equality (1), since (H ∩^{x}s)_{0} = H ∩^{x}s_{0}, which proves that the maps d_{i} commute with restrictions.

Finally it is clear that the d_{i}’s commute with conjugations by elements of G, hence they are indeed morphisms of Mackey functors.

A standard computation shows that d_{i}d_{i+1} = 0 for all i, and the sequence in the statement of Theorem 4.10 is indeed a complex of Mackey functors (in Mack_{R}(G, 1) by definition of functors δ_{i}(M)).

To check that this complex is exact, it suffices to check that its image by R is acyclic, since the functor R is a forgetful functor. I will show that this complex is split and acyclic, as claimed in Theorem 4.10.

Let Q be a p-subgroup of G. If s = (P_{0}, …, P_{i}) is a sequence of subgroups of Q, and if n ∈ M(P_{0}), I set α_{i,Q}(π_{Q}(n_{s})) = π_{Q}(n_{s∪Q}). Similarly, if n ∈ M(Q), I set α_{−1,Q}(n) = π_{Q}(n_{{Q}}). This yields for any i a map from δ_{i}(M)(Q) to δ_{i+1}(M)(Q). This map is actually a morphism from R(δ_{i}(M)) to R(δ_{i+1}(M)) : indeed, if S is a subgroup of Q, then

α_{i,S}r_{S}^{Q}π_{Q}(n_{s}) = α_{i} ∑_{x∈S\Q/P_{0}} π_{S}((xr_{S^{x}∩P_{0}}^{P_{0}}n)_{S∩^{x}s}) = ∑_{x∈S\Q/P_{0}} π_{S}((xr_{S^{x}∩P_{0}}^{P_{0}}n)_{(S∩^{x}s)∪S})

But

r_{S}^{Q}α_{i,Q}π_{Q}(n_{s}) = r_{S}^{Q}π_{Q}(n_{s∪Q}) = ∑_{{x∈S\Q/P_{0} \atop S∩^{x}(s∪Q) proper}} π_{S}((xr_{S^{x}∩P_{0}}^{P_{0}}n)_{S∩^{x}(s∪Q)})

and since for any m of M(S), I have m_{(S∩^{x}s)∪S} = m_{S∩^{x}(s∪Q)} if S ∩ ^{x}(s ∪ Q) is proper, and m_{(S∩^{x}s)∪S} = 0 otherwise, the maps α_{i} commute with restrictions. It is clear on the other hand that they commute with conjugations, hence that this are morphisms from R(δ_{i}(M)) to R(δ_{i+1}(M)).

Finally

d_{i+1,Q}α_{i,Q}π_{Q}(n_{s}) = d_{i+1,Q}π_{Q}(n_{s∪Q}) = π_{Q}((t_{P_{0}}^{P_{1}}n)_{(s∪Q)_{0}} + ∑_{j=1}^{i+1}(−1)^{j}n_{(s∪Q)_{j}})

On the other hand

α_{i−1,Q}d_{i,Q}π_{Q}(n_{s}) = α_{i−1,Q}π_{Q}((t_{P_{0}}^{P_{1}}n)_{s_{0}} + ∑_{j=1}^{i}(−1)^{j}n_{s_{j}})

= π_{Q}((t_{P_{0}}^{P_{1}}n)_{s_{0}∪Q} + ∑_{j=1}^{i}(−1)^{j}n_{s_{j}∪Q})

By difference, it follows that

d_{i+1,Q}α_{i,Q}π_{Q}(n_{s}) − α_{i−1,Q}d_{i,Q}π_{Q}(n_{s}) = (−1)^{i+1}π_{Q}(n_{(s∪Q)_{i+1}}) = (−1)^{i+1}π_{Q}(n_{s})

hence d_{i+1}α_{i} −α_{i−1}d_{i} = (−1)^{i+1}Id, and a suitable change of sign on the α_{i}’s proves the last assertion of the theorem.

4.3. Dual results. I have showed above that the algebra tμ_{R}(G) identifies with the opposite algebra of rμ_{R}(G). The previous results have a translation in terms of this algebra, obtained by replacing restrictions by traces, traces by restrictions, elements of G by their inverses, coinvariants by invariants, and reversing the arrows.

I will denote by T the restriction functor associated to the inclusion from tμ_{R}(G) into μ_{R}(G). The functor T admits a right adjoint, that I denote by J, defined par

J (N)(H) = (⊕_{P ∈\underline{s}_{p}(H)}N(P))^{H}

when N is a tμ_{R}(G)-module and H a subgroup of G.

If K is a subgroup of H, and if ⊕_{P} n_{P} ∈ J (N)(H), then r_{K}^{H}(⊕_{P} n_{P}) = ⊕_{Q}m_{Q}, with m_{Q} = n_{Q} if Q is a subgroup of K, and m_{Q} = 0 otherwise.

Similarly, if ⊕_{Q}m_{Q} ∈ J (N)(K), then t_{K}^{H}(⊕_{Q}m_{Q}) = ⊕_{P} n_{P}, with

n_{P} = ∑_{x∈P \H/K} t_{P ∩^{x}K}^{P}xn_{P^{x}∩K}

If M is a Mackey functor in Mack_{R}(G, 1), I can define similarly the functors ∂^{i}(M) by the formulas

∂^{i}(M)(H) = (⊕_{P_{0}⊂...⊂P_{i}∈\underline{s}_{p}(H)}M(P_{0}))^{H}

With this notation r_{K}^{H}(⊕_{s}n_{s}) = ⊕_{t}m_{t}, with m_{t} = n_{t} if sup t ⊆ K, and m_{t} = 0 otherwise. Similarly, t_{K}^{H}(⊕_{t}m_{t}) = ⊕_{s}n_{s}, with

n_{s} = ∑_{x∈P_{0}\H/K} t_{P_{0}∩^{x}K}^{P_{0}}xm_{s^{x}∩K} where P_{0} = inf s

The differential d̂^{i} from ∂^{i}(M) to ∂^{i+1}(M) is given by d̂^{i}(⊕_{s}n_{s}) = ⊕_{t}m_{t} where

m_{t} = r_{P_{0}}^{P_{1}}n_{s_{0}} + ∑_{j=1}^{i}(−1)^{j}n_{s_{j}}

The dual result to Theorem 4.10 is then the following :

Theorem 4.11. Let M in Mack_{R}(G, 1). The sequence

0 → M →^{d̂^{−1}} ∂^{0}(M) →^{d̂^{0}} … →^{d̂^{i−1}} ∂^{i}(M) →^{d̂^{i}} … → 0

is an exact complex of Mackey functors in Mack_{R}(G, 1), and the complex

0 → T (M) →^{T (d̂^{−1})} T (∂^{0}(M)) →^{T (d̂^{0})} … →^{T (d̂^{i−1})} T (∂^{i}(M)) →^{T (d̂^{i})} … → 0

is exact and split.

5. Applications

5.1. Steinberg Modules. Theorem 4.11 allows to extend to any Mackey functor in Mack_{R}(G, 1) a result of P.Webb expressing homology in terms of Steinberg modules (cf [7]) :

Proposition 5.1. Let M be a Mackey functor in Mack_{R}(G, 1), and H be a subgroup of G. Then

M(H) = − ∑_{P ∈\underline{s}_{p}(H)/N_{G}(H)} Ind_{N_{G}(H,P)/N_{H}(P)}^{N_{G}(H)/H}Hom_{RN_{H}(P)/P} (St_{p}(N_{H}(P)/P), M(P))

in the Green ring of RN_{G}(H)/H-modules.

Remark 5.2. Webb’s Theorem is the case where H = G and M(K) = Ĥ^{n}(K, V)_{p}, where V denotes a Z_{p}G-module and n ∈ Z : in this case, the term corresponding to P = 1 in the above sum is equal to zero. On the other hand, the fact that Ĥ^{n}(−, V)_{p} lies in Mack_{R}(G, 1) is proved in [6] (section [16]).

Recall that the Steinberg module over R of the group G at the prime number p is defined by

St_{p}(G) = −R − ∑_{s}(−1)^{|s|}Ind_{N_{G}(s)}^{G}R

where the summation runs over a set of representatives of G-conjugacy classes of strictly increasing sequences s of non trivial p-subgroups of G, and |s| denotes the cardinality of s. This sum is an element of the Green ring of finitely generated RGmodules. On the other hand, if A, B and C are RG-modules, then Hom_{RG}(A − B, C) is equal by definition to Hom_{RG}(A, C) − Hom_{RG}(B, C).

Proposition 5.1 follows from the fact that, by Theorem 4.11, if H is a subgroup of G, the sequence

0 → M(H) → ∂^{0}(M)(H) → … → ∂^{i}(M)(H) → … → 0

is a split exact sequence of RN_{G}(H)/H-modules. Denoting by Δ_{i}(H) the set of sequences P_{0} ⊂ … ⊂ P_{i} of p-subgroups of H, I have

M(H) = ∑_{i≥0}(−1)^{i}∂^{i}(M)(H) = ∑_{i≥0}(−1)^{i}(⊕_{s∈Δ_{i}(H)}M(inf s))^{H}

But ⊕_{s∈Δ_{i}(H)}M(inf s) is an RN_{G}(H)-module, isomorphic to

⊕_{s∈Δ_{i}(H)/N_{G}(H)}Ind_{N_{G}(H,s)}^{N_{G}(H)}M(inf s)

Moreover, if H is a normal subgroup of the group G, if K is a subgroup of G, and L is a RK-module, the module (Ind_{K}^{G}L)^{H} identifies to to Ind_{HK/H}^{G/H}L^{H∩K} as an RG/H-module. This remark shows that

∂^{i}(M)(H) = ⊕_{s∈Δ_{i}(H)/N_{G}(H)}Ind_{HN_{G}(H,s)/H}^{N_{G}(H)/H}M(inf s)^{N_{H}(s)}

On the other hand, the module St_{p}(N_{H}(P)/P) is a RN_{G}(H, P)-module, isomorphic to

− ∑_{i}(−1)^{i} ∑_{{s∈Δ_{i}(H)/N_{G}(H,P) \atop inf s=P}} Ind_{N_{G}(H,s)}^{N_{G}(H,P)}R

hence the module Hom_{R}(St_{p}(N_{H}(P)/P), M(P)) is isomorphic to

− ∑_{i}(−1)^{i} ∑_{{s∈Δ_{i}(H)/N_{G}(H,P) \atop inf s=P}} Ind_{N_{G}(H,s)}^{N_{G}(H,P)}Hom_{R}(R, M(P))

Another application of the previous remark, for the group N_{G}(H, P) and its normal subgroup N_{H}(P), gives then Hom_{RN_{H}(P)}(St_{p}(N_{H}(P)/P), M(P)) =

− ∑_{i}(−1)^{i} ∑_{{s∈Δ_{i}(H)/N_{G}(H,P) \atop inf s=P}} Ind_{N_{G}(H,s)/N_{H}(s)}^{N_{G}(H,P)/N_{H}(P)}M(P)^{N_{H}(s)}

and the right hand side of the equality of the proposition becomes

∑_{P ∈\underline{s}_{p}(H)/N_{G}(H)} ∑_{i}(−1)^{i} ∑_{{s∈Δ_{i}(H)/N_{G}(H,P) \atop inf s=P}} Ind_{N_{G}(H,s)/N_{H}(s)}^{N_{G}(H)/H}M(P)^{N_{H}(s)}

i.e.

∑_{i}(−1)^{i} ∑_{s∈Δ_{i}(H)/N_{G}(H)} Ind_{N_{G}(H,s)/N_{H}(s)}^{N_{G}(H)/H}M(inf s)^{N_{H}(s)}

and this is equal to ∑_{i}(−1)^{i}∂^{i}(M)(H), which proves the proposition.

5.2. T -injective functors and R-projective functors.

5.2.1. Residues. The definitions and results of Section 3 about the projectivity relative to R can be dualized by reversing the arrows and replacing “left” by “right”, and “projective” by “injective” (in other words, I will say that an object M is injective relative to the functor T if it is projective relative to the dual functor of T for the opposite categories). For the next statement, I will need the following notation : Let M a Mackey functor for the group G, and K be a subgroup of G. I will denote by \overline{M}(K) the quotient

\overline{M}(K) = M(K)/ ∑_{L⊂K} t_{L}^{K}M(L)

and \underline{M}(K) the intersection

\underline{M}(K) = ⋂_{L⊂K} Ker r_{L}^{K}

These definitions of “residues” of M at K are dual to each other : if N is a Mackey functor for the group N_{G}(K)/K, I define (with Thévenaz and Webb cf.[5]) a functor from the category of R(N_{G}(K)/K)-Mackey functors to the category of μ_{R}(G)-modules mapping N to the functor

F (N) = Ind_{N_{G}(K)}^{G}Inf_{N_{G}(K)/K}^{N_{G}(K)}N

Then the functor F a a left adjoint, defined by

M^{K}(L/K) = M(L)/ ∑_{K⊄H⊂L} t_{H}^{L}M(H)

and a right adjoint, defined by

M_{K}(L/K) = ⋂_{K⊄H⊂L} Ker r_{H}^{L}

Then \overline{M}(K) is the value at K/K of M^{K}, and \underline{M}(K) is the value at K/K of M_{K}.

Let then X be an rμ_{R}(G)-module, and P be a p-subgroup of G. It is easy to see that \overline{I(X)}(P) identifies with X(P). If F P_{V} denotes the N_{G}(P)/P -Mackey functor of fixed points on V, defined by F P_{V} (H) = V^{H}, r_{K}^{H}v = v, and t_{K}^{H}(w) = T r_{K}^{H}(v), this implies that

Hom(I(X), Ind_{N_{G}(P)}^{G}Inf_{N_{G}(P)/P}^{N_{G}(P)} F P_{V}) = Hom(I(X)^{P} , F P_{V}) =
Hom_{RN_{G}(P)/P} (X(P), V)

the second equality following from the fact that that the functor sending V to F P_{V} is right adjoint to the evaluation functor at P/P (from the category of RN_{G}(P)/P - Mackey functors to the category of RN_{G}(P)/P -modules)(cf.[5] Proposition 6.1).

Since moreover

Hom(I(X), Ind_{N_{G}(P)}^{G}Inf_{N_{G}(P)/P}^{N_{G}(P)} F P_{V}) = Hom(X, R(Ind_{N_{G}(P)}^{G}Inf_{N_{G}(P)/P}^{N_{G}(P)} F P_{V}))

it follows that the functor mapping V to R(Ind_{N_{G}(P)}^{G}Inf_{N_{G}(P)/P}^{N_{G}(P)} F P_{V}) is right adjoint to the functor sending X to X(P) (from the category of rμ_{R}(RN_{G}(P)/P)- modules to the category of RN_{G}(P)/P -modules), hence that it identifies with the functor sending V to L_{P,V}^{o} (cf. Proposition 3.6). The following proposition shows how to compute the values of an R-projective or a T -injective Mackey functor from its residues :

Proposition 5.3. Let M be a Mackey functor in Mack_{R}(G, 1).

(1) If M is R-projective (e.g. if M is projective), then

M(H) = (⊕_{P ∈\underline{s}_{p}(H)}\overline{M}(P))_{H}

as an RN_{G}(H)/H-module.

(2) If M is T -injective (e.g. if M is injective), then

M(H) = (⊕_{P ∈\underline{s}_{p}(H)}\underline{M}(P))^{H}

as an RN_{G}(H)/H-module.

These two results are dual to each other, hence it suffices to prover the second one, which will be a consequence of the following stronger (but more obscure) result :

Proposition 5.4. Let M be a Mackey functor in Mack_{R}(G, 1). If M is T - injective, then

R(M) = ⨁_{P ∈\underline{s}_{p}(G)/G} L_{P,\underline{M}(P)}^{o} = R( ⨁_{P ∈\underline{s}_{p}(G)/G} Ind_{N_{G}(P)}^{G}Inf_{N_{G}(P)/P}^{N_{G}(P)} F P_{\underline{M}(P)})

Proposition 5.3 follows, since

( ⨁_{P ∈\underline{s}_{p}(G)/G} Ind_{N_{G}(P)}^{G}Inf_{N_{G}(P)/P}^{N_{G}(P)} F P_{\underline{M}(P)})(H) =

⨁_{P ∈\underline{s}_{p}(G)/G} ⨁_{x∈N_{G}(P)\G/H}(Inf_{N_{G}(P)/P}^{N_{G}(P)} F P_{\underline{M}(P)})(N_{G}(P) ∩^{x}H)

= ⨁_{P ∈\underline{s}_{p}(G)/G} ⨁_{x∈N_{G}(P)\T_{G}(P,H)/H} F P_{\underline{M}(P)}(N_{^{x}H}(P)/P)

and using rational coefficients for this computation, this sum is equal to

⨁_{P ∈\underline{s}_{p}(G)} \frac{|N_{G}(P)|}{|G|} ⨁_{x∈T_{G}(P,H)} \frac{|N_{^{x}H}(P)|}{|N_{G}(P)||H|}\underline{M}(P)^{N_{^{x}H}(P)/P} =

⨁_{P ∈\underline{s}_{p}(G),x∈G,P^{x}⊆H} \frac{|N_{H}(P^{x})|}{|G||H|}\underline{M}(P^{x})^{N_{H}(P^{x})/P^{x}}

= ⨁_{P ∈\underline{s}_{p}(H)} \frac{|N_{H}(P)|}{|H|}\underline{M}(P)^{N_{H}(P)} = (⊕_{P ∈\underline{s}_{p}(H)}\underline{M}(P))^{H}

Proposition 5.4 shows that the identification of part 2 of Proposition 5.3 commutes with restrictions.

In order to prove Proposition 5.4, I will first characterize those rμ_{R}(G)-modules which are isomorphic to direct sums of modules of the form L_{P,V_{P}}^{o}. First I notice that I can define \underline{M}(P) when M is an rμ_{R}(G)-module and P is a p-subgroup of G, since this definition uses only restriction maps. Then :

Lemma 5.5. Let M be an rμ_{R}(G)-module. The following conditions are equivalent :

(1) There exists N_{G}(P)/P -modules V_{P} such that M is isomorphic to

⨁_{P ∈\underline{s}_{p}(G)/G} L_{P,V_{P}}^{o}

(2) The module M is isomorphic to

⨁_{P ∈\underline{s}_{p}(G)/G} L_{P,\underline{M}(P)}^{o}

(3) For any p-subgroup P of G, the inclusion of \underline{M}(P) in M(P) is a split injection of RN_{G}(P)/P -modules, and the quotient M(P)/\underline{M}(P) is isomorphic to (\underleftarrow{lim}_{Q⊂P} M(Q))^{P}.

I recall that L_{P,V}^{o} is defined by

L_{P,V}^{o} (Q) = (⊕_{x∈N_{G}(P)\T_{G}(P,Q)}x^{−1} ⊗ V)^{Q}

and

r_{Q}^{R}(⊕_{x}x^{−1} ⊗ v_{x}) = ⊕_{P^{x}⊆Q}x^{−1} ⊗ v_{x}

In particular L_{P,V}^{o} (R) is zero if P is not contained in R up to conjugation, hence \underline{L}_{P,V}^{o} (R) = 0 in this case.

Otherwise, if v = ⊕_{x∈N_{G}(P)\T_{G}(P,R)}x^{−1}⊗v ∈ \underline{L}_{P,V} (R), and if R is not conjugate to P, let y ∈ T_{G}(P, R). Since r_{P^{y}}^{R}(v) = y^{−1} ⊗ v_{y} = 0, it follows that v = 0. This proves that \underline{L}_{P,V}^{o} (R) is zero if R is not conjugate to P, and equal to V otherwise, hence Assertion 1) implies Assertion 2).

It is clear conversely that Assertion 2) implies Assertion 1). The previous argument also shows that Assertion 1) implies the first part of Assertion 3), because if M = L_{P,V}^{o}, then the injection of \underline{M}(R) in M(R) is zero if R is not conjugate to P, and the identity map otherwise.

To prove the second part, I notice that the kernel of the natural morphism from M(R) to (\underleftarrow{lim}_{Q⊂R}M(Q))^{R} is equal to \underline{M}(R). Hence it suffices to show that this

morphism is surjective.

So let M = L_{P,V}^{o}. There is nothing to prove if P is not contained in R up to conjugation. Similarly, if R = P, then (\underleftarrow{lim}_{Q⊂R}M(Q))^{R} = 0, and \underline{M}(R) = M(R)

in this case. Hence I can suppose that R contains P strictly up to conjugation. Then if v ∈ (\underleftarrow{lim}_{Q⊂R}M(Q))^{R}, for any proper subgroup Q of R, I have an element

v_{Q} of M(Q), such that if S ⊆ Q, then r_{S}^{Q}(v_{Q}) = v_{S}. In particular, if x ∈ T_{G}(P, R), then the element v_{P^{x}} can be written x^{−1} ⊗ v_{x}. Since v is fixed by R, the element w = ⊕_{x∈N_{G}(P)\T_{G}(P,R)}x^{−1} ⊗ v_{x} is in M(R), and it is clear that r_{Q}^{R}(w) = v_{Q} for any proper subgroup Q of R, which proves the second part of Assertion 3).

It remains to show that Assertion 3) implies Assertion 2). Let X be the module ⊕_{P ∈\underline{s}_{p}(G)/G}L_{P,\underline{M}(P)}^{o}. I notice first that

Hom(M, X) = ⊕_{P} Hom_{RN_{G}(P)/P} (M(P), \underline{M}(P))

from the adjunction property of the functor V ↦ L_{P,V}^{o}. Then by the first part of Assertion 3), I can choose, for any P modulo G, a section of the inclusion of \underline{M}(P) in M(P), and this choice determines a morphism ϕ from M to X.

This morphism is such that \underline{ϕ}(P) is the identity map for any P. Let then K be the kernel of ϕ, and Q be a minimal p-subgroup of G such that K(Q) ≠ 0. Then by definition of Q,

K(Q) = \underline{K}(Q) = Ker \underline{ϕ}(Q) = 0

and this contradiction proves that ϕ is injective. Let Y the cokernel of ϕ. I postpone the proof of the following lemma :

Lemma 5.6. The functor sending M to \underline{M}(P) is left exact, and its first derived functor D_{P} is given by

D_{P} (M) = (\underleftarrow{lim}_{Q⊂P} M(Q))^{P} /(Image of M(P))

Then the exact sequence 0 → M →^{ϕ} X → Y → 0 yields for any P the exact

sequence 0 → \underline{M}(P) →^{\underline{ϕ}(P)} \underline{X}(P) → \underline{Y} (P) → 0, by the second part of Assertion 3). Since \underline{ϕ}(P) is the identity map, it follows that \underline{Y} (P) = 0 for all P.

Then if Q is minimal such that Y (Q) ≠ 0, I have \underline{Y} (Q) = Y (Q) = 0, and this contradiction proves that Y is zero, hence that ϕ is surjective, and Assertion 2) of Lemma 5.5 follows.

It remains to prove Lemma 5.6. If 0 → X → Y → Z → 0 is an exact sequence of rμ_{R}(G)-modules, it is clear that \underline{X}(P) is a submodule of \underline{Y} (P). And if w is in the kernel of the map \underline{Y} (P) → \underline{Z}(P), then w is an element of Y (P) which maps to 0 in Z(P). Thus w is in X(P) ∩ \underline{Y} (P) = \underline{X}(P), which proves that the functor M ↦ \underline{M}(P) is left exact.

To complete the proof of Lemma 5.6, it remains to see that its first derived functor D_{P} is as claimed. I will denote by d_{P} the functor which maps M to the quotient of (\underleftarrow{lim}_{Q⊂P} M(Q))^{P} by the image of M(P).

I have already observed (in the proof of the implication 1)⇒ 3) of Lemma 5.5), that d_{P} (M) = 0 if M is of the form L_{P,V}^{o}. The adjunction property of the functor V ↦ L_{P,V}^{o} shows that L_{P,V}^{o} is injective if V is, and I have indeed D_{P} (L_{P,V}^{o}) = d_{P} (L_{P,V}^{o}) = 0 in this case.

Let then M be an arbitrary rμ_{R}(G)-module. I choose, for any P modulo G, an injective RN_{G}(P)/P -module I_{P} containing \underline{M}(P). The inclusion map from \underline{M}(P) into I_{P} extends to a morphism from M(P) to I_{P}, and I get this way a morphism from M to X = ⊕_{P} L_{P,I_{P}}^{o}, and X is an injective module. It is easy to see that this morphism is moreover injective.

Let Y the cokernel of this morphism. If I show that any exact sequence 0 → M → X → Y → 0 yields an exact sequence

0 → \underline{M}(P) → \underline{X}(P) → \underline{Y} (P) → d_{P} (M) → d_{P} (X)

I get then an exact sequence

0 → \underline{M}(P) → \underline{X}(P) → \underline{Y} (P) → d_{P} (M) → 0

Since moreover, by definition of D_{P}, I have the exact sequence

0 → \underline{M}(P) → \underline{X}(P) → \underline{Y} (P) → D_{P} (M) → 0

Lemma 5.6 will be proved.

So let 0 → M → X →^{b} Y → 0 be an exact sequence, in which I consider M as a submodule of X, and y be an element of \underline{Y} (P). Then there exists x in X(P) such that y = b(x). Moreover, if Q is a proper subgroup of P, then r_{Q}^{P}x belongs to M(Q).

The sequence (r_{Q}^{P}x)_{Q⊂P} defines an element of (\underleftarrow{lim}_{Q⊂P} M(Q))^{P}, well defined

up to an element of the image of M(P), i.e. an element of d_{P} (M). This element is zero if and only if there exists an element m of M(P) such that r_{Q}^{P}x = r_{Q}^{P}m for all subgroup proper Q of P, i.e. if y is in the image of \underline{X}(P) (because then y = b(x − m), and x − m ∈ \underline{X}(P)). Thus I have built an exact sequence

0 → \underline{M}(P) → \underline{X}(P) → \underline{Y} (P) →^{c} d_{P} (M)

The inclusion a from M to X yields a morphism d_{P} (a) from d_{P} (M) to d_{P} (X). Let m be an element of d_{P} (M). Then m is represented by a sequence (m_{Q})_{Q⊂P} of elements of M(Q) such that r_{S}^{Q}(m_{Q}) = m_{S} if S ⊆ Q ⊂ P, and pm_{Q} = m_{^{p}Q} if p ∈ P. The element m is in the kernel of d_{P} (a) if and only if there exists an element x of X(P) such that m_{Q} = r_{Q}^{P}x for all Q ⊂ P, hence if and only if m is in the image of \underline{Y} (P) (the element m is then the image by c of b(x)).

Hence I have proved that the sequence

0 → \underline{M}(P) → \underline{X}(P) → \underline{Y} (P) →^{c} d_{P} (M) →^{d_{P} (a)} d_{P} (X)

is exact, which proves Lemma 5.6, and completes the proof of Lemma 5.5.

It remains to see how Lemma 5.5 implies Proposition 5.4. So let M be a T - injective Mackey functor in Mack_{R}(G, 1). I will show that R(M) fulfills Condition 3 of Lemma 5.5.

This condition being inherited by direct summands, I can suppose that M is of the form J (X). Let then P be a p-subgroup of G. By definition

J (X)(P) = ( ⨁_{Q⊆P} X(Q))^{P}

If R is a subgroup of P, and if v = (v_{Q}) ∈ J (X)(P), then the Q-component of r_{R}^{P}(v) is equal to v_{Q} (for Q ⊆ R). Hence if v is in \underline{J (X)}(P), then v_{Q} is zero for any subgroup of a proper subgroup of P, i.e. for any proper subgroup of P. Hence the only non zero component of v is its P -component. But it is clear that J (X)(P) splits as

J (X)(P) = X(P) ⨁( ⨁_{Q⊂P} X(Q))^{P}

as an N_{G}(P)/P -module, which proves that the injection of \underline{J (X)}(P) in J (X)(P) is a split.

Similarly, if w ∈ ( lim ; ⟵_{Q⊂P} J (X)(Q))^{P}, then w is a sequence of elements w_{Q} of

J (X)(Q) such that r_{R}^{Q}(w_{Q}) = w_{R} if Q ⊂ P, and x.w_{Q} = w_{^{x}Q} if x ∈ P. Each element w_{Q} is itself defined by a sequence w_{R,Q} of elements of X(R), for R ⊆ Q. Then if R ⊆ S ⊆ Q ⊂ P, I must have r_{S}^{Q}(w_{Q}) = w_{S}, hence w_{R,S} = w_{R,Q}. Setting u_{R} = w_{R,R} for R ⊂ P, and u_{P} = 0, I define indeed a element u = (u_{R}) of J (X)(P) (because w is invariant by P) such that r_{Q}^{P}(u) = w_{Q} for Q ⊂ P. The natural morphism from J (X)(P) to ( lim ; ⟵_{Q⊂P} J (X)(Q))^{P} is hence indeed surjective, which

proves Condition 3 of Lemma 5.5, and Proposition 5.4.

5.2.2. Homomorphisms. Let X be an rμ_{R}(G)-module, and Y be a tμ_{R}(G)- module. I define a tμ_{R}(G)-module h(X, Y) by

h(X, Y)(P) = Hom_{R}(X(P), Y (P))

If P is a subgroup of Q, and if ϕ ∈ h(X, Y)(P), I set

t_{P}^{Q}(ϕ) = t_{P}^{Q}.ϕ.r_{P}^{Q}

If x is a element of G, I define the conjugate of ϕ by x by

x(ϕ) = x.ϕ.x^{−1}

A similar (though more complicated) construction, exists for Mackey functors : if X and Y are G-Mackey functors, I define the Mackey functor H(X, Y) by

H(X, Y)(K) = Hom(Res_{K}^{G}X, Res_{K}^{G}Y)

Hence an element of H(X, Y)(K) is a sequence of homomorphisms ϕ_{L} of X(L) in Y (L), for all subgroups L of K such that

ϕ_{L}.t_{M}^{L} = t_{M}^{L}.ϕ_{M}  ϕ_{M}.r_{M}^{L} = r_{M}^{L}.ϕ_{L}  si M ⊆ L ⊆ K
x.ϕ_{L} = ϕ_{^{x}L}.x if x ∈ K, L ⊆ K

The restriction of such an element ϕ = (ϕ_{L})_{L⊆K} to a subgroup M of K is defined by r_{M}^{K}(ϕ)_{L} = ϕ_{L} for L ⊆ M. The trace of ϕ from K to a subgroup N of G containing K is defined by

t_{K}^{N}(ϕ)_{L} = ∑_{x∈K\N/L} t_{K^{x}∩L}^{L}.x^{−1}.ϕ_{K∩^{x}L}.x.r_{K^{x}∩L}^{L}

Finally, if x ∈ G, then the conjugate of ϕ by x is defined par

(xϕ)_{L} = x.ϕ_{L^{x}}.x^{−1}

Its a rather tedious calculation to check that H(X, Y) is indeed a Mackey functor. This construction plays the same role for Mackey functors as the functor Hom(−, −) plays for G-modules : for example, a Mackey functor X is projective relative to a subgroup K of G if and only if H(X, X)(G) = t_{K}^{G}H(X, X)(K) (Higman’s criterion).

On the other hand, it is easy to see that if X and Y are in Mack_{R}(G, 1), then so does H(X, Y) (this follows from the fact that Res_{K}^{G}f_{1}^{G} = f_{1}^{K}).

The previous two constructions are related by the following lemma :

Lemma 5.7. Let X be an rμ_{R}(G)-module, and Y be a tμ_{R}(G)-module. Then

H(I(X), J (Y)) ≃ J (h(X, Y))

Corollary 5.8. Let X and Y be Mackey functors in Mack_{R}(G, 1). If X is R-projective and if Y is T -injective, then H(X, Y) is T -injective, and in particular

Hom(X, Y) = ⊕_{P ∈\underline{s}_{p}(G)/G}Hom_{RN_{G}(P)/P} (\overline{X}(P), \underline{Y} (P))

Let K be a subgroup of G. There is an obvious restriction functor from the category of rμ_{R}(G)-modules to the category of rμ_{R}(K)-modules : if P is a psubgroup of K, and X a rμ_{R}(G)-module, then Res_{K}^{G}(X)(P) = X(P). It is clear moreover that this functor commutes with the functors I and R : more precisely, if I_{G} (resp. I_{K}) denotes the functor I for the group G (resp. for the group K), then Res_{K}^{G}I_{G} = I_{K}Res_{K}^{G} (here the first Res denotes restriction for Mackey functors, the second one restriction for rμ_{R}(G)-modules).

Similarly, there is a restriction functor, that I still denote by Res_{K}^{G}, from the category of tμ_{R}(G)-modules to the category of tμ_{R}(K)-modules, which commutes with functors J and T .

With this notation

Hom(Res_{K}^{G}I_{G}(X), Res_{K}^{G}J (Y)) = Hom(I_{K}(Res_{K}^{G}X), Res_{K}^{G}J (Y)) =
Hom(Res_{K}^{G}X, R_{K}Res_{K}^{G}J_{G}(Y)) = Hom(Res_{K}^{G}X, R_{K}J_{K}(Res_{K}^{G}Y))

But J_{K}(Res_{K}^{G}Y) is T_{K}-injective, and \underline{J_{K}(Res_{K}^{G})(Y)}(P) = Y (P). Proposition 5.4 shows then that

R_{K}J_{K}(Res_{K}^{G}Y) = ⨁_{P ∈\underline{s}_{p}(K)/K} L_{P,Y (P)}^{o}

and then

Hom(Res_{K}^{G}I_{G}(X), Res_{K}^{G}J (Y)) = ⨁_{P ∈\underline{s}_{p}(K)/K} Hom_{N_{G}(P)/P} (X(P), Y (P))

hence

Hom(Res_{K}^{G}I_{G}(X), Res_{K}^{G}J (Y)) = ( ⨁_{P ∈\underline{s}_{p}(K)} h(X, Y)(P))^{K}

which is also J (h(X, Y))(K). It is easy to describe this isomorphism precisely : if for all p-subgroup P of K, I have a morphism ϕ_{P} from X(P) to Y (P) such that xϕ_{P} x^{−1} = ϕ_{^{x}P} if x ∈ K, and if L is a subgroup of K, I define a morphism ϕ_{L} from (⊕_{P ⊆L}X(P))_{L} to (⊕_{Q⊆L}Y (Q))^{L} by

ϕ_{L}(π_{L}(u))_{Q} = ∑_{x∈P \L/Q} t_{Q∩^{x}P}^{Q} .x.r_{Q^{x}∩P}^{P} .ϕ_{P} (u)

This identification allows to check that the resulting isomorphism is indeed an isomorphism of Mackey functors, which proves Lemma 5.7.

Then if X is R-projective and if Y is T -injective, the functor H(X, Y) is a direct summand of a functor of the form H(I(A), J (B)), isomorphic to J (h(A, B)), hence it is T -injective. To prove the corollary, it remains then to apply Proposition 5.4 and the following lemma :

Lemma 5.9. Let X and Y be G-Mackey functors, and K be a subgroup of G. Then H(X, Y)_{P} identifies with H(X^{P} , Y_{P})

This lemma shows indeed that

\underline{H(X, Y)}(P) = H(X^{P} , Y_{P})(P/P) = Hom_{R}(\overline{X}(P), \underline{Y} (P))

hence that (⊕_{P ∈\underline{s}_{p}(G)}\underline{H(X, Y)}(P))^{G} = ⊕_{P ∈\underline{s}_{p}(G)/G}Hom_{N_{G}(P)/P} (\overline{X}(P), \underline{Y} (P)).

So it remains to prove Lemma 5.9. Let K be a subgroup of G. Then by definition, H(X, Y)(K) is the group of homomorphisms (as K-Mackey functors) from Res_{K}^{G}X to Res_{K}^{G}Y. An element of H(X, Y)(K) is hence a sequence (ϕ_{L}) of homomorphisms from X(L) to Y (L), for L ⊆ K, which commute with restrictions, traces, and elements of K.

If P is a normal subgroup of K, such an element is in H(X, Y)_{P} (K/P) if its restriction to any subgroup M of K not containing P is zero, i.e. if ϕ_{M} = 0 if P ⊈ M.

If L/P is a subgroup of K/P, then r_{M}^{L}ϕ_{L} = ϕ_{M}r_{M}^{L} = 0 if P ⊈ M, and the image of ϕ_{L} is contained in Y_{P} (L/P). Similarly, since ϕ_{L}t_{M}^{L} = t_{L}^{M}ϕ_{M} = 0, the map ϕ_{L} factors through X^{P} (L/P). I define that way for all subgroup L/P of K/P a morphism from X^{P} (L/P) to Y_{P} (L/P). It is clear that these morphisms commute with restrictions, traces and elements of K, hence I have defined a morphism of K/P -Mackey functors from X^{P} to Y_{P}.

This construction can be obviously reversed, and this proves Lemma 5.9.

5.3. Projective functors and Cartan matrix.

5.3.1. Notation and recall. I will suppose in this section that R is a field k, of characteristic p > 0.

Under these conditions, the simple Mackey functors in Mack_{k}(G, 1) are indexed by the G-conjugacy classes of pairs (Q, V), where Q is a p-subgroup of G and V a simple kN_{G}(Q)/Q-module (cf.[5] Theorem 8.3 and [6] Theorem 9.7). The simple functor S_{Q,V}^{G} is defined for Q ≠ 1 by

S_{Q,V}^{G} = Ind_{N_{G}(Q)}^{G}Inf_{N_{G}(Q)/Q}^{N_{G}(Q)}S_{1,V}^{N_{G}(Q)/Q}

and the functor S_{1,V}^{G} is the unique minimal subfunctor of the fixed points functor F P_{V}. The value at a subgroup H is given by S_{1,V}^{G} (H) = T r_{1}^{H}(V).

Denoting by F Q_{V} the functor of coinvariants on V, whose value at H is F Q_{V} (H) = V_{H}, the functor S_{1,V}^{G} is also the quotient of F Q_{V} by its unique maximal subfunctor.

I will denote by P_{Q,V}^{G} the projective cover of the functor S_{Q,V}^{G}. Then P_{Q,V}^{G} is in Mack_{k}(G, 1). Denoting by b_{Q} the Burnside functor for the group Q, the functor P_{Q,V}^{G} is a direct summand of the functor Ind_{Q}^{G}b_{Q}. The module P_{Q,V} (1) is an indecomposable p-permutation kG-module (cf.[6] Theorem 12.7). Conversely, if W is a p-permutation kG-module, there exists a unique projective Mackey functor L in Mack_{k}(G, 1) such that L(1) = W.

Thus the projective Mackey functors in Mack_{k}(G, 1) are characterized by their value at the trivial subgroup of G. The main result of this section will be a formula to compute the Cartan matrix of Mack(G, 1) from the p-permutation modules.

If M is a Mackey functor, I will denote by M^{∗} its dual, defined by M^{∗}(H) = M(H)^{∗}, by t_{H}^{K}(ϕ) = ϕ ∘ r_{H}^{K}, by r_{H}^{K}(ϕ) = ϕ ∘ r_{H}^{K} and x.ϕ = ϕ ∘ x^{−1}. If M and N are Mackey functors, then Hom(M, N) identifies with Hom(N^{∗}, M^{∗}), hence the dual of a projective functor is an injective functor, and the dual of an injective functor is a projective functor.

The dual of the functor S_{Q,V}^{G} being the functor S_{Q,V^{∗}}^{G}, the dual of the functor P_{Q,V}^{G} is the injective hull I_{Q,V^{∗}}^{G} of the functor S_{Q,V^{∗}}^{G}.

If L is a projective Mackey functor, then the natural morphisms from F Q_{L(1)} to L and from L to F P_{L(1)} are respectively injective and surjective (cf.[6] Lemma 12.4). The same is true by duality if L is an injective functor, since the dual of the functor F P_{V} is the functor F Q_{V^{∗}}.

Finally if K is a subgroup of G, the dual of the functor M^{K} is the functor (M^{∗})_{K}.

5.3.2. Residues. The Mackey functors in Mack_{R}(G, 1) which are moreover projective (resp. injective) are R-projective (resp. T -injective). I can apply Corollary 5.8 to these functors (and this argument also holds when R is an arbitrary ring). When R is a field k, the residues projective or injective Mackey functors can be easily computed :

Lemma 5.10. Let M be a projective (resp. injective) functor in Mack_{k}(G, 1), and P a p-subgroup of G. Then \overline{M}(P) (resp. \underline{M}(P)) is isomorphic to M(1)[P].

(for the notation V [P], see Remark 3.5)

It suffices by duality of prove this lemma when M is projective : indeed, if M is injective, then M^{∗} is projective and

\underline{M}(P) = M_{P} (P/P) = ((M^{∗})^{∗})_{P} (P/P) = ((M^{∗})^{P})^{∗}(P/P) = ((M^{∗})^{P} (P/P))^{∗} =

(M^{∗}(1)[P])^{∗} = (M(1)[P]^{∗})^{∗} = M(1)[P]

since if V is a p-permutation module, then V^{∗}[P] = (V [P])^{∗}.

So let M be a projective Mackey functor in Mack_{k}(G, 1), and Q be a p-subgroup of G. The natural morphism α from M(Q) to F P_{M(1)}(Q) = M(1)^{Q} is the map sending v to r_{1}^{Q}v. This morphism is surjective. On the other hand, if v is a relative trace t_{S}^{Q}w from a proper subgroup S of Q, then

r_{1}^{Q}v = ∑_{x∈Q/S} t_{1}^{1}xr_{1}^{S}w = T r_{S}^{Q}(w)

and the morphism α passes down to quotients, giving a surjective morphism \overline{α} from \overline{M}(Q) to M(1)[Q]. I must now show that this morphism is injective. This property being inherited by direct sums and direct summands, I can suppose that M is of the form Ind_{S}^{G}b_{S}, where S is a p-subgroup of G.

In this case, the module M(1) is isomorphic to Ind_{S}^{G}k, and denoting by \overline{T}_{G}(Q, S) the quotient N_{G}(Q)\G/S, I have (cf.[2])

M(1)[Q] = (Ind_{S}^{G}k)[Q] = ∑_{x∈\overline{T}_{G}(Q,S)} Ind_{N_{^{x}S}(Q)/Q}^{N_{G}(Q)/Q}k

On the other hand, if L is a Mackey functor for the group N_{G}(Q)/Q, then

Hom((Ind_{S}^{G}b_{S})^{Q}, L) = Hom(Ind_{S}^{G}b_{S}, Ind_{N_{G}(Q)}^{G}Inf_{N_{G}(Q)/Q}^{N_{G}(Q)}L)

or

Hom((Ind_{S}^{G}b_{S})^{Q}, L) = Hom(Res_{N_{G}(Q)}^{G}Ind_{S}^{G}b_{S}, Inf_{N_{G}(Q)/Q}^{N_{G}(Q)}L)

Moreover

Res_{N_{G}(Q)}^{G}Ind_{S}^{G}b_{S} = ∑_{x∈N_{G}(Q)\G/S} Ind_{N_{G}(Q)∩^{x}S}^{N_{G}(Q)}b_{N_{G}(Q)∩^{x}S}

hence

Hom((Ind_{S}^{G}b_{S})^{Q}, L) = ∑_{x∈N_{G}(Q)\G/S} Hom(b_{N_{G}(Q)∩^{x}S}, Res_{N_{G}(Q)∩^{x}S}^{N_{G}(Q)}Inf_{N_{G}(Q)/Q}^{N_{G}(Q)}L)

Since for any group S and any S-Mackey functor X, the group Hom(b_{S}, X) is isomorphic to X(S), and since (Inf_{N_{G}(Q)/Q}^{N_{G}(Q)}L)(N_{^{x}S}(Q)) is zero if Q is not contained in ^{x}S, i.e. if x ∉ \overline{T}_{G}(Q, S), and equal to L(N_{^{x}S}(Q)/Q) otherwise, I have

Hom((Ind_{S}^{G}b_{S})^{Q}, L) = ∑_{x∈\overline{T}_{G}(Q,S)} L(N_{^{x}S}(Q)/Q)

Finally, for any L, I have

Hom((Ind_{S}^{G}b_{S})^{Q}, L) = ∑_{x∈\overline{T}_{G}(Q,S)} Hom(Ind_{N_{^{x}S}(Q)/Q}^{N_{G}(Q)/Q}b_{N_{^{x}S}(Q)/Q}, L)

But the functor sending M to M^{Q} being left adjoint of an exact functor, it follows that M^{Q} is projective if M is. Then the functors X = (Ind_{S}^{G}b_{S})^{Q} and Y = ∑_{x∈\overline{T}_{G}(Q,S)} Ind_{N_{^{x}S}(Q)/Q}^{N_{G}(Q)/Q}b_{N_{^{x}S}(Q)/Q} are both projective, and such that Hom(X, L) = Hom(Y, L) for any L. Hence they are isomorphic.

So they have the same value at Q/Q, which gives

\overline{M}(Q) = ∑_{x∈\overline{T}_{G}(Q,S)} Ind_{N_{^{x}S}(Q)/Q}^{N_{G}(Q)/Q}k = M(1)[Q]

and completes the proof of Lemma 5.10.

5.3.3. Cartan matrix. I suppose moreover here that the field k is big enough (i.e. that it is a splitting field for all the groups N_{G}(Q)/Q, for Q ∈ \underline{s}_{p}(G)).

Let Q (resp. L) be a p-subgroup of G, and V (resp. W) be a simple kN_{G}(Q)/Qmodule (resp. a simple kN_{G}(L)/L-module). Lemma 5.10 and Corollary 5.8 show that

Hom(P_{Q,V}^{G} , (P_{L,W}^{G})^{∗}) = ⊕_{S∈\underline{s}_{p}(G)/G}Hom_{kN_{G}(S)/S}(P_{Q,V}^{G} (1)[S], (P_{L,W}^{G})^{∗}(1)[S])

On the other hand, in the Grothendieck group of Mackey functors

P_{L,W}^{G} = ∑_{H,M} C_{(L,W),(H,M)}S_{H,M}^{G}

where the sum runs over the indexing pairs of simple Mackey functors in Mack_{k}(G, 1), denoting by C_{(L,W),(H,M)} the coefficient of the Cartan matrix corresponding to the simple functors S_{L,W}^{G} and S_{H,M}^{G}. The dual of S_{H,M}^{G} being S_{H,M^{∗}}^{G}, it follows that

(P_{L,W}^{G})^{∗} = ∑_{H,M} C_{(L,W),(H,M)}S_{H,M^{∗}}^{G}

and since Hom(P_{Q,V}^{G} , S_{H,M^{∗}}^{G}) is zero if the pair (H, M^{∗}) is not conjugate to the pair (Q, V), and one dimensional k (since k is big enough), it follows that

C_{(L,W),(Q,V^{∗})} = ∑_{S∈\underline{s}_{p}(G)/G} dim_{k}Hom_{kN_{G}(S)/S}(P_{Q,V}^{G} (1)[S], (P_{L,W}^{G})^{∗}(1)[S])

But (P_{L,W}^{G})^{∗}(1) = (P_{L,W}^{G} (1))^{∗}, and (P_{L,W}^{G} (1))^{∗} = P_{L,W^{∗}}^{G}(1) (because P_{L,W}^{G} (1) is the Green correspondent of the projective cover of W by Theorem 12.7 of [6], and since the Green correspondence commutes with the duality). Changing V to V^{∗} in the above equality gives

C_{(L,W),(Q,V)} = ∑_{S∈\underline{s}_{p}(G)/G} dim_{k}Hom_{kN_{G}(S)/S}((P_{Q,V}^{G} (1))^{∗}[S], (P_{L,W}^{G} (1))^{∗}[S])

i.e.

C_{(L,W),(Q,V)} = ∑_{S∈\underline{s}_{p}(G)/G} dim_{k}Hom_{kN_{G}(S)/S}(P_{L,W}^{G} (1)[S], P_{Q,V}^{G} (1)[S])

Finally, by linearity, this gives :

Proposition 5.11. Let k be a “big enough” field. If M and N are projective Mackey functors in Mack_{k}(G, 1), then

dim_{k}Hom(M, N) = ∑_{S∈\underline{s}_{p}(G)/G} dim_{k}Hom_{kN_{G}(S)/S}(M(1)[S], N(1)[S])

Corollary 5.12. Let pg(G) the Green ring of p-permutation kG-modules. The bilinear form on pg(G) defined by

< X, Y >_{G}= ∑_{S∈\underline{s}_{p}(G)/G} dim_{k}Hom_{kN_{G}(S)/S}(X[S], Y [S])

is symmetric, positive, and definite. Moreover, if H is a subgroup of G, then

< Ind_{H}^{G}X, Y >_{G}=< X, Res_{H}^{G}Y >_{H}

This corollary follows from the fact that the Cartan matrix of Mack_{k}(G, 1) is symmetric, positive, and definite, and that the corresponding bilinear form satisfies Frobenius reciprocity (moreover, if M is a kH-Mackey functor, then (Ind_{H}^{G}M)(1) = Ind_{H}^{G}M(1)).

5.3.4. Another formula. I suppose here that R is a field k of characteristic p. The notation about the generalized Steinberg modules was introduced in [1].

Proposition 5.13. Let L be a projective Mackey functor in Mack_{k}(G, 1), and X be any Mackey functor for G over k. Then

dim_{k} Hom(L, X) = ∑_{P ∈\underline{s}_{p}(G)/G} dim_{k} Hom_{kN_{G}(P)/P} (St(N_{G}(P)/P, L(1)[P]), X(P))

To prove this proposition, I use the following ingredients :

1) If N is a normal subgroup of G, there exists a natural algebra homomorphism from μ_{k}(G/N) to μ_{k}(G), mapping the element t_{\overline{x}L/N}^{K/N} \overline{x}r_{L/N}^{H/N} to t_{^{x}L}^{K}xr_{L}^{H}. If N is a normal p-subgroup of G, then this morphism maps the algebra μ_{k}^{1}(G/N) in the algebra μ_{k}^{1}(G) : indeed, if P/N is a p-subgroup of G/N, then the image of t_{P/N}^{H/N}

(resp. r_{P/N}^{H/N}) is t_{P}^{H} (resp. r_{P}^{H}), and P is a p-subgroup of G.

2) This morphism induces an exact functor ρ_{G/N}^{G} from the category of G-Mackey functors to the category of G/N-Mackey functors : if M is a G-Mackey functor, then the functor ρ_{G/N}^{G}(M) is such that

ρ_{G/N}^{G}(M)(H/N) = M(H)

Moreover, if N is a normal p-subgroup of G, the functor ρ_{G/N}^{G} maps Mack_{k}(G, 1) in Mack_{k}(G/N, 1).

The functor ρ_{G/N}^{G} admits a left adjoint, denoted by ι_{G/N}^{G}. Since the functor ρ_{G/N}^{G} is exact, the functor ι_{G/N}^{G}(M) is projective if M is. If N is a normal psubgroup of G, and if M is in Mack_{k}(G/N, 1), then ι_{G/N}^{G}(M) is in Mack_{k}(G, 1).

3) If M is a G/N-Mackey functor, then for any kG-module V

Hom(ι_{G/N}^{G}(M), F P_{V}) = Hom(M, ρ_{G/N}^{G}(F P_{V}))

and it is easy to see that ρ_{G/N}^{G}(F P_{V}) identifies with F P_{V^{N}}. Then

Hom(ι_{G/N}^{G}(M), F P_{V}) = Hom_{kG/N}(M(N/N), V^{N}) = Hom_{kG}(M(N/N), V)

On the other hand, since

Hom(ι_{G/N}^{G}(M), F P_{V}) = Hom_{G}(ι_{G/N}^{G}(M)(1), V)

it follows that ι_{G/N}^{G}(M)(1) is isomorphic as a kG-module to the module M(N/N).

4) If M = A − B is a virtual projective kG-module, I denote by F Q_{M} the Mackey virtual projective functor F Q_{A} − F Q_{B}, in the Green ring of projective Mackey functors in Mack_{k}(G, 1). If X and Y are virtual projective Mackey functors in Mack_{k}(G, 1) such that X(1) = Y (1) in the ring of Green of kG-modules, then X = Y. The previous considerations prove the following lemma :

Lemma 5.14. Let L be a projective Mackey functor in Mack_{k}(G, 1). Then

L = ∑_{P ∈\underline{s}_{p}(G)/G} Ind_{N_{G}(P)}^{G}ι_{N_{G}(P)/P}^{N_{G}(P)} (F Q_{St(N_{G}(P)/P,L(1)[P])})

in the Green ring of projective functors in Mack_{k}(G, 1).

Indeed, the module L(1) can be written (cf.[1])

L(1) = ∑_{P ∈\underline{s}_{p}(G)/G} Ind_{N_{G}(P)}^{G}St(N_{G}(P)/P, L(1)[P])

Then both sides of the equality of the lemma are virtual projective Mackey functors in Mack_{k}(G, 1), which have the same value at the trivial subgroup. Hence they are equal.

The proposition follows, applying the functor Hom(−, X) to both sides.

Remark 5.15. It is possible to compute effectively the functor ι_{G/N}^{G} : let M be a kG/N-Mackey functor, and K be a subgroup of G. Let ω_{N}(K) the set of subgroups of K ordered by the relation

L ⪯ L′ ⇔ \{ \begin{matrix}L ⊆ L′ \\ L ∩ N = L′ ∩ N \end{matrix}

I denote by

lim ; ⟶_{L∈ω_{N} (K)} M(LN/N)

the quotient of ⊕_{L⊆K}M(LN/N) by the submodule generated by elements of the form t_{LN/N}^{L′N/N}m − m, for L ⪯ L′ and m ∈ M(LN/N).

The group K acts on lim ; ⟶_{L∈ω_{N} (K)} M(LN/N), and then

ι_{H}^{G}(M)(K) = ( lim ; ⟶_{L∈ω_{N} (K)} M(LN/N))_{K}

is the largest quotient on which K acts trivially. If L is a subgroup of K, and m an element of M(LN/N), I denote by m_{L}^{K} the image of m in ι_{H}^{G}(M)(K).

If K ⊆ K′, then t_{K}^{K′} (resp. r_{K}^{K′}) is the map from ι_{H}^{G}(M)(K) to ι_{H}^{G}(M)(K′) (resp. from ι_{H}^{G}(M)(K′) to ι_{H}^{G}(M)(K)) defined by

t_{K}^{K′}(m_{L}^{K}) = m_{L}^{K′} r_{K}^{K′}(m′_{L′}^{K′}) = ∑_{x∈K\K′/L′} (^{x}r_{(K^{x}∩T he)N/N}^{L′N/N}m′)_{K∩^{x}L′}^{K}

Finally if x ∈ G, then ^{x}(m_{L}^{K}) = (^{x}m)_{^{x}L}^{^{x}K}.

6. Projective functors and image of I

I will try to see in this section under which conditions a projective Mackey functor in Mack(G, 1) lies in the image of I. I will suppose that R is a complete local ring, with residue field of characteristic p.

6.1. Finite projective resolutions. I will state the following equivalence :

Theorem 6.1. Let X be an rμ_{R}(G)-module. The following conditions are equivalent :

(1) The functor I(X) is projective.

(2) The module X has a finite projective resolution.

The proof of this theorem requires a series of preliminary results.

Lemma 6.2. Let X and Y be projective functors in Mack_{R}(G, 1). The morphism ϕ ↦ ϕ(1) of evaluation at 1 from Hom(X, Y) to Hom_{RG}(X(1), Y (1)) is surjective. Moreover ϕ is a split injective (resp. a split surjective) if and only if ϕ(1) is a split injective (resp. a split surjective).

I have already recalled that the natural morphism from Y to F P_{Y (1)} is surjective if Y is projective. Let then f in Hom_{RG}(X(1), Y (1)). It yields by adjunction a morphism F of X in F P_{Y (1)}. Then the diagram

Y
↓
X →^{F} F P_{Y (1)}

can be completed to a commutative diagram by a morphism ϕ from X to Y, because the vertical arrow is surjective, and the functor X is projective. It is clear that ϕ(1) is equal to f, which proves the first assertion.

It is clear moreover that if ϕ is a split injection, then ϕ(1) is also split injective. Conversely, if ϕ(1) is a split injection, then there exists a morphism f from Y (1) to X(1) such that fϕ(1) = Id_{X(1)}. The morphism f can be lifted to a morphism f̃ from Y to X, and then the morphism ψ = f̃ϕ is an endomorphism of X such that ψ(1) = Id. Then some power of ψ is a Fitting element^{4}. Its image I is then a direct summand of X, hence I is a projective Mackey functor in Mack(G, 1) such that I(1) = X(1), which proves that I is isomorphic to X, hence that ψ = f̃ϕ is invertible, and ϕ is a split injection. The case of a split surjection is similar.

4This probably requires X to be finitely generated

Lemma 6.3. Let X and Y be projective functors in Mack_{R}(G, 1), and ϕ be a morphism from X to Y. The following conditions are equivalent :

(1) The morphism ϕ is a split injection (resp. a split surjection).

(2) For any p subgroup P of G, the morphism \overline{ϕ}(P) of \overline{X}(P) in \overline{Y} (P) is injective (resp. surjective).

Since one can lift projective objects and morphisms between them from k to R, I can assume that R is a field k of characteristic p. In this case, the module \overline{X}(P) identifies with X(1)[P], and the morphism \overline{ϕ}(P) to ϕ(1)[P]. I will prove simultaneously Lemma 6.3 and the following proposition :

Proposition 6.4. Let M and N be p-permutation kG-modules, and f be a morphism from M to N. The following conditions are equivalent :

(1) The morphism f is a split injection (resp. a split surjection).

(2) For any p subgroup P of G, the morphism f[P] of M[P] in N[P] is injective (resp. surjective).

Indeed let X, Y and ϕ be as in Lemma 6.3. It is clear that if ϕ is a split surjection, then for any P, the morphism \overline{ϕ}(P) is (split) surjective. Conversely, if \overline{ϕ}(P) is surjective for all P, then ϕ(1) = \overline{ϕ}(1) is surjective. Let Q be a p-subgroup such that ϕ(R) is surjective for all proper subgroup R of Q. Let moreover v ∈ Y (P).

Since \overline{ϕ}(P) is surjective, there exists elements v_{R} ∈ Y (R), for R ⊂ Q, and an element w ∈ X(Q) such that

v = ϕ(Q)(w) + ∑_{R⊂Q} t_{R}^{Q}v_{R}

Then each v_{R} can be written v_{R} = ϕ(R)(w_{R}) since ϕ(R) is surjective, and then

v = ϕ(Q)(W) + ∑_{R⊂Q} t_{R}^{Q}ϕ(R)(w_{R}) = ϕ(Q)(w) + ∑_{R⊂Q} ϕ(Q)(t_{R}^{Q}w_{R})

which proves that ϕ(Q) is surjective for all Q. Then ϕ is surjective, hence ϕ is a split surjection since Y is projective. This shows the equivalence of conditions of Lemma 6.3 in the surjective case.

Then let M, N and f be as in Proposition 6.4. If f is a split surjection, it is clear that f[P] is (split) surjective for all P. Conversely, if f[P] is surjective for all P, let L_{M} (resp. L_{N}) the projective Mackey functor in Mack_{k}(G, 1) such that L_{M}(1) = M (resp. L_{N}(1) = N). The morphism f can be extended to a morphism ϕ of L_{M} in L_{N}, such that ϕ(1) = f. The hypothesis implies that \overline{ϕ}(P), which identifies with f[P], is surjective for all P. Then ϕ is a split surjection, and f = ϕ(1) is also split surjective. Which proves the equivalence of conditions of Proposition 6.4 in the surjective case.

But by duality, this proves the equivalence of the conditions of Proposition 6.4 in the injective case^{5}.

5This proof is not correct for infinitely generated modules. This was fixed in Le complexe de chaînes d’un G-complexe simplicial acyclique J. Alg. 220 (1999) 415-436

Let then again X, Y, and ϕ be as in Lemma 6.3. It is clear that if ϕ is a split injection, then \overline{ϕ}(P) is (split) injective for all P. Conversely, if \overline{ϕ}(P) is injective for all P, let f = ϕ(1). Then f is a morphism from X(1) to Y (1) such that f[P] is injective for all P, hence f is a split injection, and Lemma 6.2 implies that ϕ is a split injection, which completes the proof of the Lemma 6.3 and Proposition 6.4.

By the same argument as above about lifting of projective modules, it suffices to prove Theorem 6.1 in the case where R is a filed k of characteristic p.

If M is a kG-module, and P a p-subgroup of G, I will denote by Br_{P} the projection morphism from M^{P} to M[P]. If X is an rμ_{R}(G)-module, as the image of r_{1}^{P} is contained in X(1)^{P}, I can abuse notation, and denote by Br_{P} r_{1}^{P} the composite map from X(P) to X(1)[P]. With this notation :

Proposition 6.5. Let X be an rμ_{k}(G)-module. The following conditions are equivalent :

(1) The module X has a finite projective resolution.

(2) The module X(1) is a p-permutation kG-module, and for any p-subgroup P of G, the map Br_{P} r_{1}^{P} is a isomorphism from X(P) to X(1)[P].

I will show that Assertion 1 of the proposition implies Assertion 2 by induction on the length of a finite projective resolution of X.

I already observed (in Section 2) that Assertion 2 is true if X is projective : indeed, if P is a p-subgroup of G, if E is a projective kN_{G}(P)/P -module projective, and if X = L_{P,E}, then

X(Q) = ⊕_{x∈T_{G}(Q,P)/N_{G}(P)}x ⊗ E

On the other hand, the module X(1) identifies with Ind_{N_{G}(P)}^{G}E. Hence an element v of X(1)^{Q} can be written ∑_{x∈G/P} x⊗v_{x}, the sequence v_{x} being such that v_{σ_{q}(x)} = h_{q,x}v_{x}, denoting by x ↦ σ_{q}(x) the permutation of G/P induced by q, and h_{q,x} the element of N_{G}(P) defined by qx = σ_{q}(x)h_{q,x}. In particular, the element v_{x} is invariant by Q ∩^{x}P, and v can also be written

v = ∑_{x∈Q\G/N_{G}(P)} T r_{Q∩^{x}P}^{Q} (x ⊗ v_{x})

It is then clear that r_{1}^{Q}X(Q) is a supplement in X(1)^{Q} of the kernel of Br_{Q}.

So let X be an rμ_{k}(G)-module having a finite projective resolution. There exists a projective rμ_{k}(G)-module L and an rμ_{k}(G)-module Y having a strictly shorter finite projective resolution than the one of X, and an exact sequence

0 → Y →^{i} L → X → 0

The induction hypothesis implies that Y (1) is a p-permutation module, and that Br_{P} r_{1}^{P} is an isomorphism from Y (P) to Y (1)[P]. Then the commutative diagram

0 →  Y (P)  →^{i(P)} L(P)
↓ Br_{P} r_{1}^{P}  ↓ Br_{P} r_{1}^{P}
Y (1)[P] →^{i(1)[P]} L(1)[P]

where the vertical arrows are isomorphisms, shows that the map i(1) is an injection from Y (1) in L(1) such that i(1)[P] is injective for all P. Thus i(1) is a split injection by Proposition 6.4, and the sequence

0 → Y (1) →^{i(1)} L(1) → X(1) → 0

is split exact. In particular, the module X(1) is a direct summand of L(1). Hence it is a p-permutation module. Then the completed diagram

0 →  Y (P)  →^{i(P)}  L(P)  →  X(P)  → 0
↓ Br_{P} r_{1}^{P}  ↓ Br_{P} r_{1}^{P}  ↓ Br_{P} r_{1}^{P}
0 → Y (1)[P] →^{i(1)[P]} L(1)[P] → X(1)[P] → 0

shows that the vertical arrow on the right is an isomorphism, which proves Assertion 2 of the proposition.

Now I will show that Assertion 2 of the proposition implies Assertion 1. First a notation : if X is an rμ_{k}(G)-module, I denote by Supp(X) the set of subgroups P such that X(P) ≠ 0, and \overline{Supp}(X) its “downwards closure” for inclusion, i.e. the set of subgroups which are contained in a element of Supp(X).

Let X be an rμ_{k}(G)-module with the properties of Assertion 2 of the proposition. I will proceed by induction on the cardinality of \overline{Supp}(X).

There is nothing to prove if this cardinality is zero, since then X is zero, hence projective. I postpone the proof of the following lemma :

Lemma 6.6. Let X be an rμ_{k}(G)-module, and for any p-subgroup P of G, let E_{P} be a projective cover of the kN_{G}(P)/P -module

X(P)/( ∑_{Q⊃P} r_{P}^{Q}X(Q))

Then ⊕_{P ∈\underline{s}_{p}(G)/G}L_{P,E_{P}} is a projective cover of X.

The properties of Assertion 2 show indeed that if P is a maximal element of Supp(X) (or equivalently of \overline{Supp}(X)), then the module X(P) is a p-permutation kN_{G}(P)/P -module. Moreover, if P ⊂ Q ⊆ N_{G}(P), then X(P)[Q/P], isomorphic to X(1)[P][Q/P], hence to X(1)[Q], hence to X(Q), is zero. Then X(P) is a projective kN_{G}(P)/P -module. Since moreover r_{P}^{Q}X(Q) is zero if Q strictly contains P, I see that X(P) is equal to E_{P} for any maximal element P of Supp(X).

Let then L = ⊕_{P} L_{P,E_{P}} be a projective cover of X, and Y the kernel of a surjection s from L to X. Since L_{P,E_{P}} (P) = E_{P} = X(P), I see that Y (P) is zero if P is maximal in Supp(X). Moreover L(Q) is zero if Q is not in \overline{Supp(X)} : indeed, the support of L_{P,E_{P}} is the set of subgroups of G which are contained in P up to conjugation, if E_{P}, hence X(P) are non zero.

Hence the cardinality of \overline{Supp(Y)} is strictly smaller than the cardinality of \overline{Supp(X)}. On the other hand, the commutative diagram

L(P) →^{s(P)}  X(P)  → 0
↓ Br_{P} r_{1}^{P}  ↓ Br_{P} r_{1}^{P}
L(1)[P] →^{s(1)[P]}  X(1)[P]

shows as above that s is a split surjection. Similarly, the completed diagram

0 →  Y (P)  →  L(P)  →^{s(P)}  X(P)  → 0
↓ Br_{P} r_{1}^{P}  ↓ Br_{P} r_{1}^{P}  ↓ Br_{P} r_{1}^{P}
0 → Y (1)[P] → L(1)[P] →^{s(1)[P]} X(1)[P] → 0

shows then that Y has the properties of Assertion 2. The induction hypothesis implies that Y has a finite projective resolution. It follows that X has a finite projective resolution, and this completes the proof of Proposition 6.5.

Remark 6.7. The previous proof also shows that if p^{n} is the p-part of the order of G, then any rμ_{k}(G)-module having a finite projective resolution has a resolution of length smaller than n. The length of minimal finite projective resolutions is hence bounded^{6}.

I must now prove to Lemma 6.6. Recall that the simple rμ_{k}(G)-modules, that I have denoted by N_{P,V}, are indexed by pairs {P, V }, where P is a p-subgroup of G and V a simple kN_{G}(P)/P -module. They are defined by N_{P,V} (Q) = 0 if Q is not conjugate to P, and N_{P,V} (P) = V. It is then easy to see that for all rμ_{k}(G)-module X

Hom(X, N_{P,V}) = Hom_{kN_{G}(P)/P} (X(P)/ ∑_{Q⊃P} r_{P}^{Q}X(Q), V)

which shows that, denoting by J(X) the radical of X,

J(X)(P) = J(X(P)) + ∑_{Q⊃P} r_{P}^{Q}X(Q)

Lemma 6.6 follows easily.

I can now prove Theorem 6.1 : let X be an rμ_{k}(G)-module such that I(X) is projective. Then, since X(1) = I(X)(1), the module X(1) is a p-permutation module. On the other hand, the module \overline{I(X)}(P) identifies with X(1)[P] if I(X) is projective, by Lemma 5.10. But for any X, it also identifies with X(P). Moreover the diagram

X(P) → I(X)(P)
↓ r_{1}^{P}  ↓ r_{1}^{P}
X(1) → I(X)(1)

is commutative, hence X fulfills the conditions of Assertion 2 of Proposition 6.5. Thus X has a finite projective resolution, which proves Assertion 2 of the theorem.

To show that Assertion 2 of the theorem implies Assertion 1, I proceed by induction on the length of a finite projective resolution of X. If X is projective, then I(X) is projective, because the functor I is left adjoint to an exact functor.

If X has a finite projective resolution, then there exists a projective rμ_{k}(G)- module L, an rμ_{k}(G)-module Y having a finite projective resolution, strictly shorter than the one of X, and an exact sequence

0 → Y →^{i} L → X → 0

Then I(Y) and I(L) are projective Mackey functors. Moreover, since \overline{I(i)}(P) = i(P) is injective for all P, the morphism I(i) is a split injection by Lemma 6.3. Since the functor I is right exact, it follows that the sequence

0 → I(Y) →^{I(i)} I(L) → I(X) → 0

is split exact, hence the functor I(X) is projective, which completes the proof of Theorem 6.1.

6In other words, the finitistic dimension of rμ_{k}(G) is at most equal to n

6.2. Examples. Let M be a p-permutation module. There exists a unique projective Mackey functor L_{M} in Mack_{k}(G, 1), such that L_{M}(1) = M. Proposition 6.5 and Lemma 5.10 give a necessary and sufficient condition for the existence of an rμ_{k}(G)-module X such that L_{M} is isomorphic to I(X) : indeed in this case, the module X(P) must be isomorphic to \overline{L_{M}}(P), hence to M[P]. Hence there must exist maps r_{Q}^{P}, defined for Q ⊆ P, from M[P] to M[Q], such that

• r_{S}^{Q}r_{Q}^{P} = r_{S}^{P} if S ⊆ Q ⊆ P.
• r_{P}^{P} = Id for all P ∈ \underline{s}_{p}(G).
• xr_{Q}^{P}x^{−1} = r_{^{x}Q}^{^{x}P} for any x ∈ G and any Q ⊆ P.

• r_{1}^{P} is injective and its image is a supplement of KerBr_{P} in M^{P} = M[1]^{P} Conversely, if such maps exist, then they define an rμ_{k}(G)-module X such that I(X) is projective, and moreover I(X)(1) = X(1) = M, hence I(X) is isomorphic to L_{M}.

6.2.1. Permutation modules. A simple example of this situation is the case where M is a permutation module : indeed, if B is a G-stable basis of M, then the inclusion of B^{P} in B^{Q} yields the required map r_{Q}^{P}. For example, if M = k, the functor L_{M} is the functor b_{p}(G) of Mack_{k}(G, 1) associated Burnside to the Burnside functor (i.e. the subfunctor of the functor of Burnside such that b_{p}(H) is generated by the elements H/P, where P is a p-subgroup of H). The associated rμ_{k}(G)-module X is such that X(P) = k for all P, the maps r_{Q}^{P} being identity maps, as well as the conjugations by the elements of G. In other words, the module X is isomorphic to R(F P_{k}). Thus

Proposition 6.8. The module R(F P_{k}) has a finite projective resolution.

I will now give other examples of this situation.

6.2.2. Some indecomposable p-permutation modules. Recall (cf.[2]) that the indecomposable p-permutation modules can be indexed by the pairs (P, E), where P is a p-subgroup of G and E is an indecomposable projective kN_{G}(P)/P -module : the module M(P, E) corresponding to the pair (P, E) is the unique indecomposable p-permutation module with vertex P such that M(P, E)[P] = E. The multiplicity of M(P, E) as a direct summand of a p-permutation module N is equal to the multiplicity of E as a direct summand of N[P]. This multiplicity is given by the following :

Lemma 6.9. Let N be a kG-module, and E be a projective kG-module. The multiplicity of E as a direct summand of N is equal to

dim_{k}T r_{1}^{G}Hom_{k}(N, E/J(E))/dim_{k}End_{kG}(E/J(E))

Indeed dim_{k} T r_{1}^{G}Hom_{k}(N, E/J(E)) is the dimension of the space of kG-homo-morphisms f from N to E/J(E) which factor through a projective module, i.e. which can be written f = πg, where π is the projection from E to E/J(E) (since E is a projective cover of E/J(E)). Since π is essential, the morphism g is then surjective if f is non zero, hence surjective. Then E is a direct summand of N. Conversely, if N can be written N = E^{n} ⊕ M, where the module M has no direct summand isomorphic to E, then dim_{k}T r_{1}^{G}Hom_{k}(M, E/J(E)) is equal to zero, and dim_{k}T r_{1}^{G}Hom_{k}(N, E/J(E)) is equal to ndim_{k}T r_{1}^{G}Hom_{k}(E, E/J(E)), hence to ndim_{k}End_{kG}(E/J(E)).

Proposition 6.10. Let P (resp. Q) be a p-subgroup of G, and E (resp. F) be an indecomposable projective kN_{G}(P)/P -module (resp. kN_{G}(Q)/Q)-module) If M(P, E) is direct summand of Ind_{N_{G}(Q)}^{G}F, then there exists an element x ∈ G such that Q^{x} ∩ O_{p}(N_{G}(P)) = P, and in particular

Q ∩ O_{p}(G) ⊆ ^{x}P ⊆ Q

Corollary 6.11. If Q ⊆ O_{p}(G), then the module Ind_{N_{G}(Q)}^{G}F is indecomposable (equal to M(Q, F)), and the functor L_{M(Q,F)} is isomorphic to I(L_{Q,F}).

Indeed (cf. [1] Lemme 3), if N = Ind_{N_{G}(Q)}^{G}F, then

N[P] = ∑_{x∈N_{G}(P)\T_{G}(P,Q)/N_{G}(Q)} Ind_{N_{G}(P,^{x}Q)/P}^{N_{G}(P)/P} ^{x}F

moreover assuming for simplicity x = 1, hence P ⊂ Q

dim_{k} T r_{1}^{N_{G}(P)/P} Hom_{k}(Ind_{N_{G}(P,Q)/P}^{N_{G}(P)/P} F, E/J(E)) = …

… = dim_{k} T r_{1}^{N_{G}(P,Q)/P} Hom_{k}(F, E/J(E))

But N_{Q}(P)/P acts trivially on F, hence

dim_{k} T r_{1}^{N_{G}(P,Q)/P} Hom_{k}(F, E/J(E)) = …

… = dim_{k} T r_{N_{Q}(P)/P}^{N_{G}(P,Q)/P} Hom_{N_{Q}(P)/P} (F, T r_{1}^{N_{Q}(P)/P} (E/J(E))

But O_{p}(N_{G}(P)/P) acts trivially on the simple kN_{G}(P)/P module E/J(E). This expression is hence equal to zero if N_{Q}(P) ∩ O_{p}(N_{G}(P)) is different from P, hence if Q ∩ O_{p}(N_{G}(P)) ≠ P.

Since N_{G}(P) normalizes P O_{p}(G), I have N_{P O_{p}(G)}(P) ⊆ O_{p}(N_{G}(P)), hence I must have Q ∩ N_{P O_{p}(G)}(P) = P, i.e. Q ∩ P O_{p}(G) = P, or Q ∩ O_{p}(G) ⊆ P, which proves the proposition.

The first part of the corollary follows, since the only indecomposable summand with vertex Q of N is M(Q, F), with multiplicity 1. On the other hand the functor I(L_{Q,F}) is projective, and its value at the trivial subgroup is Ind_{N_{G}(Q)}^{G}F = M(Q, F).

Proposition 6.12. The following conditions are equivalent :

(1) For any p-subgroup of G and any indecomposable projective kN_{G}(P)/P - module, the module Ind_{N_{G}(P)}^{G}E is indecomposable.

(2) For any indecomposable p-permutation module M, there exists a p-subgroup P and an indecomposable projective kN_{G}(P)/P -module E such that M is isomorphic to Ind_{N_{G}(P)}^{G}E.

(3) The group G has a normal Sylow p-subgroup.

(4) Any rμ_{k}(G)-module having a finite projective resolution is projective.

If Assertion 1) holds, as M(P, E) is a direct summand of Ind_{N_{G}(P)}^{G}E, which is indecomposable, these modules are isomorphic, and Assertion 2) holds.

It is clear that Assertion 2 implies Assertion 3 : indeed, if the module k is isomorphic to Ind_{N_{G}(P)}^{G}E, then [G : N_{G}(P)] dim_{k} E = 1 by consideration of dimensions, hence P is a normal subgroup of G. Since p does not divide the dimension of the projective kN_{G}(P)/P -module E, the group N_{G}(P)/P is a p’-group, and P is a Sylow p-subgroup of G.

Similarly, if G has a normal Sylow p-subgroup, equal to O_{p}(G), then all the p-subgroups of G are contained in O_{p}(G), and Assertion 1) holds by the previous proposition.

Assertion 4) implies Assertion 3), since if R(F P_{k}) has a finite projective resolution, it is a projective rμ_{k}(G)-module, indecomposable since R(F P_{k})(1) = k. Hence there exists P and E such that R(F P_{k}) is isomorphic to L_{P,E}. Then k is isomorphic to Ind_{N_{G}(P)}^{G}E, and G has a normal Sylow p-subgroup by the above argument.

To show that Assertion 1) implies Assertion 4), I will use the following lemma :

Lemma 6.13. Let P be a p-subgroup of G and E be a projective kN_{G}(P)/P - module such that the module Ind_{N_{G}(P)}^{G}E is indecomposable. Let X be an rμ_{k}(G)- module having a finite projective resolution. If M(P, E) = Ind_{N_{G}(P)}^{G}E is a direct summand of X(1), then there exists a submodule of X isomorphic to L_{P,E}.

Let indeed α be a split injection from Ind_{N_{G}(P)}^{G}E into X(1) : the map α is determined by a N_{G}(P)-homomorphism from E to X(1), i.e. by a N_{G}(P)/P - homomorphism β from E to X(1)^{P}. Composing this map with the projection Br_{P} onto X(1)[P], and next with the isomorphism σ, inverse of Br_{P} r_{1}^{P}, I get the map ϕ = σBr_{P} β from E to X(P), which gives by adjunction a morphism Φ from L_{P,E} to X. The morphism Φ(1) is defined by

Φ(1)(x ⊗ e) = xr_{1}^{P} σBr_{P} β(e)

whereas α is defined by

α(x ⊗ e) = xβ(e)

Let then γ be a morphism from X(1) to Ind_{N_{G}(P)}^{G}E such that γα = Id. In particular, I have γβ(e) = 1 ⊗ e. The morphism γΦ(1) is an endomorphism of the indecomposable module Ind_{N_{G}(P)}^{G}E. Hence it is hence invertible or nilpotent. But

γΦ(1)(1 ⊗ e) = γr_{1}^{P} σBr_{P} β(e)

and

Br_{P} r_{1}^{P} σBr_{P} β(e) = Br_{P} β(e)

by definition of σ. Thus r_{1}^{P} σBr_{P} β(e) − β(e) ∈ KerBr_{P}, hence

γΦ(1)(1 ⊗ e) − 1 ⊗ e ∈ KerBr_{P}

i.e.

γΦ(1)[P] = Id_{E}

proving that γΦ(1) is not nilpotent, hence that it is invertible. Thus Φ(1) is a split injection, which proves that Φ(1)[Q] is injective for all Q, hence that Φ is injective, which proves the lemma.

Then if the conditions of Assertion 1) of the proposition hold, and if X is a minimal counter example to Assertion 4), let M(P, E) = Ind_{N_{G}(P)}^{G}E be an indecomposable direct summand of X(1), and L be a submodule of X isomorphic to L_{P,E}. The quotient Y of X by L has a finite projective resolution : indeed, a module Y has a finite projective resolution if and only if there exists an integer n such that Ext^{m}(Y, Z) = 0 for all Z and any m ≥ n. Then if

0 → A → B → C → 0

is an exact sequence, the associated long exact sequence of Ext groups shows that if two of the modules A, B, C have a finite projective resolution, then so does the third.

Then the minimality of X implies that Y is projective, hence direct summand of X, which is hence a direct sum of two projective modules, hence projective. This contradicts the hypothesis on X, and completes the proof of the proposition.

6.2.3. Quasi trivial intersection. I will say that a p-subgroup P of G is a quasi trivial intersection subgroup if for any x ∈ G, either P = P^{x}, or P ∩ P^{x} ⊆ O_{p}(G). In the case where O_{p}(G) = (1), a quasi trivial intersection subgroup is a trivial intersection subgroup.

Proposition 6.14. Let P be a quasi trivial intersection p-subgroup of G, and M be an indecomposable p-permutation kG-module with vertex P. Then there exists an rμ_{k}(G)-module X such that I(X) = L_{M}.

Indeed, if M(Q, F) is a direct summand of Ind_{N_{G}(P)}^{G}E, then there exists x ∈ G such that P^{x} ∩ O_{p}(N_{G}(Q)) = Q. Then if Q ⊈ O_{p}(G), the group Q is contained in a unique conjugate P^{x} of P, hence N_{G}(Q) normalizes P^{x}, which proves that N_{P^{x}}(Q) ⊆ O_{p}(N_{G}(Q)). Then it follows that P^{x} = Q. Hence the module Ind_{N_{G}(P)}^{G}E is the direct sum of M(P, E) and indecomposable modules with vertex contained in O_{p}(G).

Let then M be an indecomposable p-permutation kG-module with vertex P. I denote by E the projective kN_{G}(P)/P -module M[P]. If P is contained in O_{p}(G), then Ind_{N_{G}(P)}^{G}E is indecomposable, equal to M(P, E), and the rμ_{k}(G)-module X = L_{P,E} is such that I(X) is projective, and X(1) = Ind_{N_{G}(P)}^{G}E = M(P, E). Thus X is a solution to the question for the module M = M(P, E).

If P is not contained in O_{p}(G), let X be a minimal quotient of L_{P,E} such that X has a finite projective resolution, and that X(P) = E. Such a quotient exists, since L_{P,E}(P) = E.

Since X is a quotient of L_{P,E}, and since X has a finite projective resolution, I know that X(1) is a direct summand of Ind_{N_{G}(P)}^{G}E = L_{P,E}(1).

Then if M(Q, F) is an indecomposable direct summand of X(1), either Q is conjugate to P, or Q is contained in O_{p}(G). In this case, the module Ind_{N_{G}(Q)}^{G}F is indecomposable, equal to M(Q, F), and the split injection from M(Q, F) to X(1) yields an injection from L_{Q,F} into X, hence an exact sequence

0 → L_{Q,F} → X → Y → 0

Then Y is a quotient of L_{P,E}, since X is, and moreover Y (P) is equal to E, since L_{Q,F} (P) = 0. Since Y has a finite projective resolution, this contradicts the hypothesis on X, which proves that Q is conjugate to P, hence that F = E, and that X(1) is indecomposable with vertex P, such that X(1)[P] = E. Hence X is a solution to the question, which proves the proposition.

Proposition 6.15. If the Sylow p-subgroups of G are trivial intersection psubgroups, then :

(1) For any p-subgroup P of G and any projective kN_{G}(P)/P -module E, there exists a projective kN_{G}(P ∩ O_{p}(G))/(P ∩ O_{p}(G))-module F such that

Ind_{N_{G}(P)}^{G}E = M(P, E) ⊕ Ind_{N_{G}(P ∩O_{p}(G))}^{G}F

(2) For any p-permutation kG-module M, there exists an rμ_{k}(G)-module X such that I(X) = L_{M}.

Indeed, if the Sylow p-subgroups of G/O_{p}(G) are trivial intersection p-subgroups, then they are quasi trivial intersection p-subgroups. If P is a p-subgroup of G, not contained in O_{p}(G), if S is a Sylow p-subgroup of G containing P, then S is the only Sylow subgroup of G containing P.

Then if M(Q, F) is an indecomposable direct summand of Ind_{N_{G}(P)}^{G}E, there exists x ∈ G such that P^{x} ∩ O_{p}(N_{G}(Q)) = Q, and P^{x} ∩ O_{p}(G) ⊆ Q. If Q is not contained in O_{p}(G), then S^{x} is the only Sylow p-subgroup of G containing Q. In particular N_{G}(Q) ⊆ N_{G}(S^{x}), and N_{S^{x}}(Q) ⊆ O_{p}(N_{G}(Q)). Thus N_{P^{x}}(Q) ⊆ O_{p}(N_{G}(Q)), which proves that Q = P^{x}. And if Q is contained in O_{p}(G), then Q = P^{x} ∩ O_{p}(G), which proves Assertion 1).

Let then M be a p-permutation indecomposable module with vertex P, and E = M[P]. Let as above X be a minimal quotient of L_{P,E} having a finite projective resolution, and such that X(P) = E.

In particular, the module X(1) is a direct summand of Ind_{N_{G}(P)}^{G}E. Thus if P ⊆ O_{p}(G), then X(1) is equal to Ind_{N_{G}(P)}^{G}E = M(P, E), and X fulfills the conditions of Assertion 2).

If P is not contained in O_{p}(G), let M(Q, F) be an indecomposable direct summand of X(1). Then M(Q, F) is a direct summand of Ind_{N_{G}(P)}^{G}E. Thus if Q is not conjugate to P, then Q ⊆ O_{p}(G), and M(Q, F) = Ind_{N_{G}(Q)}^{G}F is a direct summand of X(1). The module L_{Q,F} is then a submodule of X, and the quotient Y is a quotient of L_{P,E} strictly smaller than X, having a finite projective resolution, and such that Y (P) = X(P) = E, since L_{Q,F} (P) = 0. This contradicts the definition of X, and proves that Q = P, hence that F = E, and that X(1) is indecomposable. Since X(1)[P] = X(P) = E, this prove that X(1) is isomorphic to M, and Assertion 2) follows.

6.2.4. The case of cyclic groups. Let M is a p-permutation indecomposable module with vertex P, then M is a direct summand of Ind_{P}^{G}k. This is equivalent to saying that there exists a P -invariant linear form ϕ on M and a vector f of M^{P}, such that

Id_{M} = T r_{P}^{G}(ϕ ⊗ f)

denoting by ϕ ⊗ f the endomorphism of M defined by

(ϕ ⊗ f)(v) = ϕ(v)f

Let then Q be a p-subgroup of G. I define an endomorphism α_{Q} of M by

α_{Q}(v) = ∑_{x∈T_{G}(Q,P)/P} ϕ(x^{−1}v)xf

so that α_{1} is the identity map. Then :

Lemma 6.16. With this notation :

(1) If Q ∈ \underline{s}_{p}(G) and x ∈ G, then xα_{Q}x^{−1} = α_{^{x}Q}.

(2) The image of α_{Q} is contained in M^{Q}, and its kernel contains [Q, M].

(3) If A ⊇ B ⊆ C, then α_{A}α_{B}α_{C} = α_{A}α_{C}

(4) In particular, for all p-subgroup Q of G, I have

α_{Q}^{3} = α_{Q}^{2}

and α_{Q}^{2} is a projector whose image is isomorphic to M[Q].

Assertion 1) follows from the fact that T_{G}(^{x}Q, P) = T_{G}(Q, P)x. For Assertion 2), I observe that if q ∈ Q, then

qα_{Q}(v) = q ∑_{x∈T_{G}(Q,P)/P} ϕ(x^{−1}v)xf = ∑_{x∈T_{G}(Q,P)/P} ϕ(x^{−1}v)xq^{x}f = α_{Q}(v)

since f is P -invariant. The second part of Assertion 2) follows then from Assertion 1), since α_{Q}(qv) = qα_{Q}(v) = α_{Q}(v). Under the assumptions of Assertion 3), let v ∈ M^{B}. Since α_{1} is the identity map, I can write

v = ∑_{x∈B\G/P} ∑_{y∈B/B∩^{x}P} ϕ(x^{−1}y^{−1}v)yxv

i.e., since v is B-invariant

v = ∑_{x∈B\G/P} ϕ(x^{−1}v)T r_{B∩^{x}P}^{B} (xv)

But α_{A}(T r_{B∩^{x}P}^{B} (xv)) = 0 if B ⊈ ^{x}P, and hence

α_{A}(v) = α_{A}α_{B}(v) si v ∈ M^{B}

and Assertion 3) follows, as the image of α_{C} is contained in M^{C}, hence in M^{B}.

The first part of Assertion 4) follows, taking A = B = C = Q. The second one follows from the fact that for v ∈ M^{Q}, the vector v − α_{Q}(v) is in the kernel of Br_{Q}, which is contained in the kernel of α_{Q}.

In the case where P is cyclic, this gives :

Proposition 6.17. Let P a cyclic p-subgroup of G, and M be an indecomposable p-permutation kG-module with vertex P. Then there exists an rμ_{k}(G)-module X such that I(X) = L_{M}.

If the Sylow p-subgroups of G are cyclic, then for any indecomposable p-permuta-tion module M there exists an rμ_{k}(G)-module X such that I(X) = L_{M}.

With the previous notation, I set, for any p-subgroup Q of G

β_{Q} = α_{Φ^{n}(Q)}α_{Φ^{n−1}(Q)} … α_{Φ(Q)}α_{Q}^{2}

where Φ(Q) is the Frattini subgroup of Q, the integer n being chosen such that Φ^{n+1}(Q) = (1) (the definition of β_{Q} does not depend on the choice of such an integer n, since α_{1} = Id). It is clear that if x ∈ G, then

xβ_{Q}x^{−1} = β_{^{x}Q}

Moreover, if S is a subgroup of Q, there exists k such that S = Φ^{k}(Q), and

α_{S}β_{Q} = α_{Φ^{k}(Q)}α_{Φ^{n}(Q)}α_{Φ^{n−1}(Q)} … α_{Φ(Q)}α_{Q}^{2}

hence

α_{S}β_{Q} = α_{Φ^{k}(Q)}α_{Φ^{k+1}(Q)} … α_{Q}^{2}

which gives

β_{S}β_{Q} = β_{Q}

It follows that β_{Q} is a projector. Moreover, it is clear that β_{Q}α_{Q} = β_{Q}, and that α_{Q}β_{Q} = α_{Q}^{2} (it is the case S = Q, hence k = 0, of the above equality). In particular, the projectors β_{Q} and α_{Q}^{2} have the same kernel, hence images isomorphic to M[Q].

Moreover if S is a subgroup of Q, then β_{S}β_{Q} = β_{Q}, and the image of β_{Q} is contained in the image of β_{S}.

Denoting by r_{S}^{Q} this inclusion, I get the required maps from M[Q] to M[S] I define that way an rμ_{k}(G)-module X such that I(X) is projective and I(X)(1) = M, hence I(X) is isomorphic to L_{M}, which proves the proposition. The corollary follows trivially.

6.2.5. A counter example. Let G be the group symmetric S_{5}, and k be the field with two elements^{7}. I will build a p-permutation kG-module M for which there is no rμ_{k}(G)-module X such that I(X) is isomorphic to L_{M}.

Let M be a 6-dimensional vector space over k. The standard generators of S_{5} act on M by the following correspondence ρ :

ρ((12)) = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ) ρ((23)) = ( \begin{matrix}0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} )

ρ((34)) = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ) ρ((45)) = ( \begin{matrix}1 & 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} )

Then ρ((12)), ρ((23)), and ρ((34)) act on the basis of M via the following permutations

ρ((12)) ↦ (23)(45)  ρ((23)) ↦ (12)(56)  ρ((34)) ↦ (24)(35)

It is then clear that the restriction of ρ to the subgroup S_{4} generated by (12), (23), and (34) is a permutation representation of S_{4}.

An elementary computation shows that ρ((45)) commutes with ρ((12) and ρ((23)), and that the product ρ((34))ρ((45)) has order 3. It follows that ρ is a representation of S_{5}, and that M is a kS_{5} module.

Since the restriction of M to S_{4} is a permutation module, and since S_{4} contains a Sylow 2-subgroup of S_{5}, the module M is a 2-permutation kS_{5}-module.

Let S the Sylow 2-subgroup of S_{5} generated by (12), (34) and (13)(24). Then S stabilizes the canonical basis B = {e_{1}, …, e_{6}} of M. Since

(13)(24) = (23)(34)(12)(23)

the element (13)(24) of S acts on B by the permutation

(12)(56)(24)(35)(23)(45)(12)(56) = (16)(34)

The orbit of S on B are hence {e_{1}, e_{6}} and {e_{2}, e_{3}, e_{4}, e_{5}}. The stabilizer of e_{1} in S is the group P generated by (12) and (34), and the stabilizer of e_{2} in S is the group T generated by (13)(24). It follows that

Res_{S}^{S_{5}}M = Ind_{P}^{S} k ⊕ Ind_{T}^{S} k

7so p = 2 here

The indecomposable direct summands of M are hence of vertex contained in P or in T up to conjugation. The module M has no projective direct summand, since its dimension is smaller than the 2-part of the order of S_{5}. On the other hand, the module M[T] has a basis in one to one correspondence with B^{T} = {e_{2}, e_{5}}. Since N_{S_{5}}(T)/T has order 4, it follows that M[T] has no kN_{S_{5}}(T)/T -projective direct summand. Hence the module M only has indecomposable direct summands with non trivial vertex contained in P.

The module M[P] has a basis in one to one correspondence with B^{P}, consisting of the vectors e_{1} and e_{6}. The normalizer of P in S_{5} is equal to S, and the group S switches e_{1} and e_{6}. It follows that the module M[P] is kN_{S_{5}}(P)/P -indecomposable and projective, hence that M has a direct summand isomorphic to M(P, E_{k}), where E_{k} denotes a projective cover of the trivial kN_{S_{5}}(P)/P -trivial.

Let then Q be the subgroup of P generated by (12). The module M[Q] has a basis in one to one correspondence with B^{Q}, i.e. {e_{1}, e_{6}}. The normalizer of Q, isomorphic to Z/2Z × S_{3}, is generated by Q and the elements (34) and (45), which act trivially on the quotient of M by the subspace generated by {e_{2}, e_{3}, e_{4}, e_{5}}. Thus N_{S_{5}}(Q)/Q acts trivially on M[Q], hence this kN_{S_{5}}(Q)/Q-module has no projective direct summand. It follows that M has no direct summand with vertex Q.

The only proper non trivial subgroup of P which is not conjugate to Q is the group U generated by (12)(34), which acts on B by the permutation (25)(34). Thus M[U] has dimension 2. Since N_{S_{5}}(U)/U has order 4, the module M[U] has no projective direct summand as kN_{S_{5}}(U)/U-module, and the module M has no direct summand with vertex U.

The module M is hence indecomposable of vertex P. Since M[P] is the projective cover of the trivial module, the module M is the Scott module of S_{5} for the subgroup P (cf. [2]).

Then if it is possible of find suitable restriction maps, in particular, the map r_{1}^{P} must be injective, as well as that the map r_{Q}^{P}, since r_{1}^{P} = r_{1}^{Q}r_{Q}^{P}. Since M[P] and M[Q] have the same dimension, the map r_{Q}^{P} must be surjective. Then r_{1}^{P} and r_{1}^{Q} have the same image W, which must be a subspace of M^{P} invariant by the group H generated by N_{S_{5}}(P) and N_{S_{5}}(Q).

The group H is a transitive subgroup of S_{5} which contains a transposition. Thus H = S_{5}. On the other hand, a vector v of M^{P} is of the form

v = ( \begin{matrix}a \\ b \\ b \\ b \\ b \\ c \end{matrix} )

The image of v by ρ((23)) is the vector

ρ((23))(v) = ( \begin{matrix}b \\ a \\ b \\ b \\ c \\ b \end{matrix} )

Thus if v ∈ W, then ρ((23))(v) ∈ W and then a = b = c. Hence the only subspace of M^{P} which is invariant by S_{5} has dimension 1, generated by the vector

v = ( \begin{matrix}1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{matrix} )

It is the module M^{S_{5}}. The space W cannot have dimension 2, hence it is impossible to find an rμ_{k}(G)-module X such that I(X) = L_{M}.

7. Complexes of projective Mackey functors

I suppose here that the ring R is a complete local ring, whose residue field k has characteristic p.

7.1. Split complexes. Let A be a ring and L^{∗} be a complex of A-modules, whose differential d has degree 1. In [7], Webb gives the following definition of a split complex : the complex L^{∗} is split if there exists maps α_{n} from L^{n} to L^{n−1} such that for all n,

d^{n}α_{n+1}d^{n} = d^{n}

Lemma 7.1. The following conditions are equivalent :

(1) The complex L^{∗} is split.

(2) The complex L^{∗} is homotopic to a complex with zero differentials.

Indeed, if M^{∗} is a complex with zero differentials, and if f is a homotopy equivalence from L^{∗} to M^{∗}, with inverse g, then there exists maps α_{n} from L^{n} to L^{n−1} such that

Id − g_{n}f_{n} = d^{n−1}α_{n} + α_{n+1}d^{n}

Then d^{n}α_{n+1}d^{n} = d^{n}−d^{n}g_{n}f_{n}, and d^{n}g_{n} is zero since g is a morphism of complexes and since the differential of M^{∗} is zero.

Conversely, if L^{∗} is split, let M^{n} = H^{n}(L^{∗}) its n-th homology group. I consider M^{∗} as a complex with zero differentials. Let Z^{n} be the kernel of d^{n}, let i_{n} be the injection from Z^{n} to L^{n}, and p_{n} the projection from Z^{n} to M^{n}. The image of Id − α_{n+1}d^{n} is contained in Z^{n} by hypothesis, and I set

f_{n} = p_{n}(Id − α_{n+1}d^{n})

Conversely, the map Id − d^{n−1}α_{n} sends Z^{n} inside itself, and its restriction to Z^{n} factors through M^{n}, i.e.

Id − d^{n−1}α_{n} = h_{n}p_{n}

and I denote by g_{n} the composition of h_{n} and i_{n}. It is then clear that if u = p_{n}(v) ∈ M^{n}

f_{n}g_{n}(u) = p_{n}(Id − α_{n+1}d^{n})i_{n}h_{n}p_{n}(v) = p_{n}(Id − α_{n+1}d^{n})(Id − d^{n−1}α_{n})(v)

i.e., since d^{n}(v) = 0

f_{n}g_{n}(u) = p_{n}(v − d^{n−1}α_{n}(v)) = p_{n}(v) = u

Conversely, if w ∈ L^{n}, then

g_{n}f_{n}(w) = i_{n}h_{n}p_{n}(Id − α_{n+1}d^{n})(w) = i_{n}(Id − d^{n−1}α_{n})(Id − α_{n+1}d^{n})(w)

or

g_{n}f_{n} = Id − α_{n+1}d^{n} − d^{n−1}(α_{n} − α_{n}α_{n+1}d^{n})

and setting β_{n} = α_{n} − α_{n}α_{n+1}d^{n}, I have indeed

g_{n}f_{n} = Id − β_{n+1}d^{n} − d^{n−1}β_{n}

which shows that L^{∗} is homotopic to M^{∗}.

It follows from this lemma that L^{∗} is homotopic to the zero complex if and only if L^{∗} is acyclic and split in the sense of the above definition.

Corollary 7.2. Let L^{∗} be a complex of A-modules. If for any integer n, there exists a complex M^{∗} (possibly depending on n) homotopic to L^{∗} and such that M^{n} = 0, then L^{∗} is homotopic to the zero complex.

Indeed in those conditions, the complex L^{∗} is certainly acyclic, since H^{n}(L^{∗}) = H^{n}(M^{∗}) = 0. Moreover, if f is a homotopy equivalence from L^{∗} to M^{∗}, with inverse g, then there exists maps α_{m} from L^{m} to L^{m−1} and β_{m} from M^{m} to M^{m−1} such that for all m

Id − g_{m}f_{m} = α_{m+1}d^{m} + d^{m−1}α_{m}

In particular, I see that d^{n} − d^{n}g_{n}f_{n} = d^{n}α_{n+1}d^{n}, and since M^{n} = 0, I have d^{n}g_{n} = 0. Then the complex L^{∗} is split, hence it is homotopic to the zero complex.

7.2. Split complexes of projective Mackey functors. Lemma 6.3 and Proposition 6.4 have the following generalization :

Proposition 7.3. Let L^{∗} be a complex of projective functors in Mack_{R}(G, 1), such that there exists n with L^{n} = 0. The following conditions :

(1) The complex L^{∗} is acyclic and split.

(2) For any p-subgroup P of G, the complex \overline{L}^{∗}(P) is acyclic.

It is clear that Assertion 1) implies Assertion 2). Conversely, I can e.g. suppose that L^{−1} is zero. Lemma 6.3 shows then that d^{0} is a split injection, which amounts to saying that the complex L^{∗} is homotopic to the complex

… L^{−2} →^{d^{−2}} 0 → 0 → 0 → L^{1}/L^{0} →^{d′^{1}} L^{2} → …

Lemma 6.3 also shows that d^{−3} is a split surjection. By induction, it follows that the complex L^{∗} is homotopic to a complex with an arbitrary number of consecutive zero modules around L_{0}, and the result follows by Corollary 7.2.

Similarly :

Proposition 7.4. Let M^{∗} a complex of p-permutation RG-modules, such that there exists n with M^{n} = 0. The following conditions are equivalent :

(1) The complex M^{∗} is acyclic and split.

(2) For any p-subgroup of G, the complex M^{∗}[P] is acyclic.

If K^{∗} and L^{∗} are complexes of modules over an arbitrary ring, to any morphism f from K^{∗} to L^{∗} is associated a third complex, the cone of f, denoted by C^{∗}(K, L, f) with the following properties :

• The complex C^{∗}(K, L, f) is acyclic if and only if H^{n}(f) is a isomorphism for all n (i.e. if f is a quasi isomorphism)

• The complex C^{∗}(K, L, f) is acyclic and split if and only if f is a homotopy equivalence.

• The module C^{n}(K, L, f) is the direct sum of K^{n} and L^{n−1}.

The third property shows that if K^{∗} and L^{∗} are complexes of Mackey functors in Mack_{R}(G, 1), then so is C^{∗}(K, L, f). Similarly, if K^{∗} and L^{∗} are complexes of ppermutations modules, then so is C^{∗}(K, L, f). The above propositions and remarks yield then the following two propositions :

Proposition 7.5. Let K^{∗} and L^{∗} be complexes of projective Mackey functors in Mack_{R}(G, 1), such that there exists n with K^{n} = L^{n} = 0 and K^{n−1} = L^{n−1} = 0. Let f be a morphism from K^{∗} to L^{∗}. The following conditions are equivalent :

(1) The morphism f is a homotopy equivalence.

(2) For any p-subgroup P of G, the morphism \overline{f}(P) is a quasi isomorphism.

Proposition 7.6. Let M^{∗} and N^{∗} be complexes of p-permutation kG-modules, such that there exists m with M^{n} = N^{n} = 0 and M^{n−1} = N^{n−1} = 0. Let f be a morphism from M^{∗} to N^{∗}. The following conditions are equivalent :

(1) The morphism f is a homotopy equivalence.

(2) For any p-subgroup P of G, the morphism f[P] is a quasi isomorphism.

7.3. Complexes of p-permutations modules. I will try to see how to extend the previous results when the hypothesis “for all P ” is replaced by “for all non trivial P ” : the consequence of this change will be the replacement of zero modules by projective modules, of zero Mackey functors by projective Mackey functors with trivial vertex, of split acyclic complexes by complexes homotopic to complexes of projective modules.

I suppose that R is a field k. I call p-permutations complex a complex of ppermutations kG-modules.

I will try to find here under which conditions such a complex is homotopic to a complex of projective kG-modules. First, I can suppose that G is a p-group

Lemma 7.7. Let L^{∗} be a complex of finitely generated kG-modules, and S be a Sylow p-subgroup of G. The following conditions are equivalent :

(1) The complex L^{∗} is homotopic to a complex of projective kG-modules.

(2) The restriction of L^{∗} to S is homotopic to a complex of projective kSmodules.

It is clear that Assertion 1) implies Assertion 2). Conversely, let E^{∗} be a complex of projective kS-modules, and a be a homotopy equivalence from E^{∗} to Res_{S}^{G}L^{∗}, with inverse b. Then the complex Ind_{S}^{G}E^{∗} is a complex of projective kG-modules, and by adjunction, the morphism a and b yield morphisms A and B between Ind_{S}^{G}E^{∗} and L^{∗}. It is easy to see that the morphism AB is homotopic to [G : S]Id. The lemma follows then from the following lemma :

Lemma 7.8. Let M^{∗} and L^{∗} be complexes of finitely generated kG-modules. Let A be a morphism from M^{∗} to L^{∗} and B be a morphism from L^{∗} to M^{∗} such that AB is homotopic to the identity of L^{∗}. Then L^{∗} is homotopic to a direct summand of M^{∗}.

Indeed, let L_{1}^{∗} = ∩_{n}(AB)^{n}(L^{∗}), and L_{2}^{∗} = ∑_{n} Ker(AB)^{n}. Then L_{1}^{∗} and L_{2}^{∗} are subcomplexes of L^{∗}, invariant by AB, and L^{∗} identifies with the direct sum of the complexes L_{1}^{∗} and L_{2}^{∗}. The restriction of AB to L_{1}^{∗} and L_{2}^{∗} is homotopic to the identity. Since in each degree n, the restriction of AB to L_{2}^{n} is nilpotent, it follows that for any n, there exists a complex K^{∗} homotopic to L_{2}^{∗}, such that K^{n} = 0. Then the complex L_{2}^{∗} is homotopic to the zero complex, and the complex L^{∗} is homotopic to the complex L_{1}^{∗}.

On the other hand, the complex L_{1}^{∗} is isomorphic to the complex B(L_{1}^{∗}), which is homotopic to ∩_{n}(BA)^{n}(M^{∗}), which is direct summand of M^{∗}. This proves the lemma.

The previous lemmas allow to show the

Proposition 7.9. Let L^{∗} be a complex of p-permutation kG-modules, such that there exists an integer n for which L^{n} is projective. The following conditions are equivalent :

(1) The complex L^{∗} is homotopic to a complex of projective kG-modules.

(2) For any non trivial p-subgroup P of G, the complex L^{∗}[P] is acyclic.

It is clear that Assertion 1) implies Assertion 2). To prove the converse, I need a notation :

If X and Y are kG-modules, I will denote by J_{kG}(X, Y) the set of kG-homomor-phisms f from X to Y such that for any kG-homomorphism g from Y to X, the morphism gf is nilpotent.

For example, J_{kG}(X, X) is the Jacobson radical of End_{kG}(X). It is easy to see on the other hand that this definition is additive with respect to X and Y : the canonical isomorphism between Hom_{kG}(X, Y ⊕Z) and Hom_{kG}(X, Y)⊕Hom_{kG}(X, Z) induces indeed an isomorphism between J_{kG}(X, Y ⊕Z) and J_{kG}(X, Y)⊕J_{kG}(X, Z). Similarly, if X^{∗} and Y^{∗} are the respective duals of X and Y, then ϕ ∈ J_{kG}(X, Y) if and only if ϕ^{∗} ∈ J_{kG}(Y^{∗}, X^{∗}) : indeed, the morphism ψ^{∗}ϕ^{∗} is nilpotent if and only if the morphism ϕ^{∗}ψ^{∗} = (ψϕ)^{∗} is.

By the previous lemma, I can suppose that G is a p-group. Let then n be an arbitrary integer, and ϕ be a morphism from L^{n+1} to L^{n}, such that ϕd^{n} is not nilpotent. The diagram

… L^{n−1} →  L^{n}  →^{d^{n}} L^{n+1} → L^{n+2} …
0 ↓  ϕd^{n} ↓  ↙^{ϕ}  ↓ d^{n}ϕ  ↓ 0
… L^{n−1} →  L^{n}  →^{d^{n}} L^{n+1} → L^{n+2} …

defines then a endomorphism γ of L^{∗}, homotopic to 0. I can then replace L^{∗} by its direct summand ∩_{m}(Id + γ)^{m}(L^{∗}), which is homotopic to it. The only modules modified by this operation are L^{n} and L^{n+1}, which are replaced by some direct summands L′^{n} and L′^{n+1}. The differentials d^{n−1}, d^{n} and d^{n+1} become respectively d′^{n−1}, d′^{n} and d′^{n+1}. If d^{n−1} ∈ J_{kG}(L^{n−1}, L^{n}), then d′^{n−1} ∈ J_{kG}(L′^{n−1}, L′^{n}). Similarly, if d^{n+1} ∈ J_{kG}(L^{n+1}, L^{n+2}), then d′^{n+1} ∈ J_{kG}(L′^{n+1}, L′^{n+2}).

Hence I can suppose that d_{n} ∈ J_{kG}(L^{n}, L^{n+1}) for all n.

Let then l be an integer such that L^{l} is a projective kG-module. If one of the modules L^{n}, for n > l, is not projective, let n be the smallest integer greater than l such that L^{n} is not projective.

Let moreover Q be a maximal subgroup of G which is a vortex of some indecomposable direct summand of L^{n}. The only indecomposable p-permutation kG-module with vertex Q is Ind_{Q}^{G}k, because G is a p-group. The module M is then the sum of its part A with vertex Q, isomorphic to a sum of copies of Ind_{Q}^{G}k, and a module X such that X[Q] = 0. On the other hand, the module L^{n+1} is the sum of its part B of vortex Q, of a module Z such that Z[Q] = 0, and of a module Y having only indecomposable summands with vertex strictly bigger than Q.

Then d^{n} can be represented by a matrix

( \begin{matrix}ϕ_{B,A} & ϕ_{B,X} \\ ϕ_{Y,A} & ϕ_{Y,X} \\ ϕ_{Z,A} & ϕ_{Z,X} \end{matrix} )

The morphism d^{n}[Q] can then be represented by the matrix

( \begin{matrix}ϕ_{B,A}[Q] \\ ϕ_{Y,A}[Q] \end{matrix} )

Then if Q ≠ (1), as the complex L^{∗}[Q] is acyclic, and as L^{n−1}[Q] = 0, the map d^{n}[Q] must be injective.

The morphism ϕ_{B,A} is a matrix with coefficients in End_{kG}(Ind_{Q}^{G}k), whose every coefficient is actually in J(End_{kG}(Ind_{Q}^{G}k)), hence maps the socle of Ind_{Q}^{G}k (isomorphic to k) to zero, since ϕ ∈ J_{kG}(M, N). Thus ϕ_{B,A} maps the socle of A to zero.

But this socle maps \underline{onto} the socle of A[Q]. Thus ϕ_{Y,A}[Q] restricted to this socle must be injective, hence ϕ_{Y,A}[Q] must be injective. Then A[Q], which is projective as an N_{G}(Q)/Q module, maps injectively into Y [Q], hence A[Q] is a direct summand of Y [Q]. But since Y only has direct summands with vertex strictly bigger than Q, this is impossible, and this contradiction proves that Q = (1), hence that L^{n} is projective.

Thus L^{n} is projective for all n ≥ l. I can then apply the same argument to the dual complex Hom_{k}(L^{∗}, k) to show that L_{n} is projective for all n ≤ l, hence that L^{∗} is a complex of projective modules. The proposition follows.

It yields the following corollary, which is a slightly more precise version of a precise a result of Webb (cf.[8]) :

Corollary 7.10. Let Δ be a simplicial complex on which acts the group G. The following conditions are equivalent :

(1) The chain complex C(Δ) of Δ over k is homotopic to a complex of projective kG-modules.

(2) For any subgroup P of order p of G, the set Δ^{P} is acyclic modulo p.

Indeed, a classical argument shows that if Assertion 2) holds, then Δ^{P} is acyclic modulo p for all non trivial p-subgroup P of G. On the other hand, the complex C(Δ)[P] is the chain complex of Δ^{P}.

7.4. Complexes of projective Mackey functors. I suppose here again that R is a field k of characteristic p. Propositions 7.4 and 7.6 have the following consequence :

Proposition 7.11. Let X^{∗} be a complex of projective functors in Mack_{k}(G, 1), such that there exists an integer n for which X^{n} = X^{n+1} = 0. The following conditions are equivalent :

(1) There exists a complex L^{∗} of projective kG-modules such that X^{∗} is homotopic to the complex F Q_{L^{∗}}.

(2) For any non trivial p-subgroup P of G, the complex \overline{X}^{∗}(P) is acyclic.

(3) The complex X^{∗}(1) is homotopic to a complex of projective modules.

(4) The complex X^{∗} is homotopic to the complex F Q_{X^{∗}(1)}.

Indeed, if M is a kG-module and Y is a Mackey functor of the form F Q_{M}, it is easy to see that \overline{Y} (P) is zero when P is non trivial, because the traces t_{R}^{P} are surjective for R ≠ P. It is then clear that Assertion 1) implies the Assertion 2). Similarly, Assertion 4) implies Assertion 2).

Similarly, Assertion 2) implies that the complex X^{∗}(1) fulfills the hypothesis of Assertion 2) of Proposition 7.9. Then there exists a complex L^{∗} of projective kG-modules such that X^{∗}(1) is homotopic to L^{∗}. Thus Assertion 2) implies Assertion 3).

It is clear that Assertion 3) implies Assertion 2), since \overline{X}^{∗}(P) identifies with X^{∗}(1)[P].

Let then L^{∗} be a complex of projective modules homotopic to the complex X^{∗}(1). Up to replacing L^{∗} by a direct summand, I can suppose that L^{n} = L^{n+1} = 0. Let then a be a homotopy equivalence from L^{∗} to X^{∗}(1). By adjunction, the morphism a yields a morphism A from the complex F Q_{L^{∗}} to the complex X^{∗} : the morphism A^{n}(H) from (L^{n})_{H} to X^{n}(H) is given by

A^{n}(v) = t_{1}^{H}a^{n}(v)

The complex X^{∗}(1)[P] is homotopic to the complex \overline{X}^{∗}(P), and to the complex L^{∗}[P], which is zero if P ≠ (1). The complex \overline{X}^{∗}(P) is hence acyclic and split in this case. The complex \overline{F Q_{L}^{∗}}(P) is zero if P ≠ (1). Then the morphism \overline{A}(P) is trivially a quasi isomorphism if P ≠ (1). Since the morphism \overline{A}(1) is equal to A(1), hence to a, it is a homotopy equivalence, hence a quasi isomorphism.

Then the complexes F Q_{L^{∗}} and X^{∗} fulfill the hypotheses of Assertion 2 of Proposition 7.5, and A is a homotopy equivalence, which proves that Assertion 3) implies Assertion 1). Hence Assertions 1), 2) and 3) are equivalent.

Then if Assertion 1) is true, i.e. if X^{∗} is homotopic to the complex F Q_{L^{∗}}, the complex X^{∗}(1) is homotopic to the complex F Q_{L^{∗}}(1), i.e. to the complex L^{∗}, hence the complex F Q_{X^{∗}(1)} is homotopic to the complex F Q_{L^{∗}}, hence also to the complex X^{∗}. Thus Assertion 1) implies Assertion 4), and this completes the proof of the proposition.

I will use this result to give a slightly more precise version of a theorem of Webb (cf.[8] Theorem 1) in a particular case. For this, I need a notation : if the group G acts on the set X, then X can be decomposed in a disjoint union of G-orbits, which are G-sets of the form G/H. Then if M is a Mackey functor, Webb denotes by M_{X} the Mackey functor defined by

M_{X ∐ Y} = M_{X} + M_{Y} M_{G/H} = Ind_{H}^{G}Res_{H}^{G}M

On the other hand, if Δ is a simplicial complex, then Δ_{i} denotes the set of simplices of Δ of dimension i (or of cardinal i + 1). Webb’s Theorem is then the following : Theorem (Webb) : Let Δ be a simplicial complex on which acts the group G. Let Y ⊂ X be sets of subgroups of G, closed under inclusion and G-conjugation. Suppose that :

(1) For any simplex s of Δ, the vertices of s are in distinct G-orbits.

(2) The functor M is projective relative to X.

(3) For any H ∈ X − Y, the complex Δ^{H} is contractible.

(4) For any H ∈ Y, the module M(H) is zero. Then there exists a split exact sequence of Mackey functors

0 → M → M_{Δ_{0}} → … → M_{Δ_{i}} → … → 0

I will consider here the case where X is the set of p-subgroups of G, and Y consists only of the trivial subgroup of G. Then a Mackey functor projective relative to X is a functor in Mack_{k}(G, 1). Denoting by C_{i}(Δ) the kG-module with k-basis Δ_{i},

Proposition 7.12. Let M be a Mackey functor in Mack_{k}(G, 1), and Δ be a simplicial complex on which G acts. If :

(1) For any simplex s of Δ, the vertices of s are in distinct G-orbits,

(2) For any p-subgroup P of order p of G, the complex Δ^{P} is acyclic modulo p,

then the complex

0 → M → M_{Δ_{0}} → … → M_{Δ_{i}} → … → 0

is homotopic to the complex

0 → F Q_{M(1)} → F Q_{M(1)⊗C_{0}(Δ)} → … → F Q_{M(1)⊗C_{i}(Δ)} → … → 0

To prove this proposition, I will apply Proposition 7.11 to the complex b^{∗}

0 → (b_{p}) → (b_{p})_{Δ_{0}} → … → (b_{p})_{Δ_{i}} → … → 0

where I denote by b_{p} the functor z(f_{1}^{G})b_{G}. It is a projective functor in Mack_{k}(G, 1), and since the above complex is finite, it vanishes in two consecutive degrees. The value at (1) of this complex is the chain complex of Δ, which is homotopic to a complex of projective modules by Corollary 7.10. Proposition 7.11 now shows that the complex b^{∗} is homotopic to the complex F Q_{b^{∗}(1)} = F Q_{C^{∗}(Δ)}.

But for any Mackey functor M in Mack_{k}(G, 1), the functor H(b_{p}, M) is isomorphic to the functor M. Then the functor H((b_{p})_{Δ_{i}}, M), isomorphic to H(b_{p}, M_{Δ_{i}}), is isomorphic to M_{Δ_{i}}. Then if N denotes the dual functor of M, the complex

0 ← H((b_{p}), N) ← H((b_{p})_{Δ_{0}}, N) ← … ← H((b_{p})_{Δ_{i}}, N) ← … ← 0

is isomorphic to the complex

0 ← N ← N_{Δ_{0}} ← … ← N_{Δ_{i}} ← … ← 0

and this complex is homotopic to the complex H(F Q_{C_{∗}(Δ)}, N). It is clear on the other hand that for any kG-module V, the functor H(F Q_{V} , N) is isomorphic to the functor F P_{Hom_{k}(V,N(1))}. Since the dual of this functor is the functor F Q_{Hom_{k}(N(1),V)}, and since Hom(N(1), C_{i}(Δ)) is isomorphic to C_{i}(Δ) ⊗ M(1), the dual of the above complex, which is the complex

0 → M → M_{Δ_{0}} → … → M_{Δ_{i}} → … → 0

is indeed homotopic to the complex

0 → F Q_{M(1)} → F Q_{M(1)⊗C_{0}(Δ)} → … → F Q_{M(1)⊗C_{i}(Δ)} → … → 0

which proves the proposition.

Remark 7.13. Hypothesis 1) plays no role in the proof, but it is necessary to build the complex of the proposition (cf.[8]).

References

[1] Serge Bouc, Projecteurs dans l’anneau de Burnside, projecteurs dans l’anneau de Green, et modules de Steinberg généralisés, J. of Algebra 139 (1991), no. 2, 395–445.

[2] Michel Broué, On Scott modules and p-permutation modules : an approach through the Brauer morphism, Proc.AMS 93 (1985), no. 3, 401–408.

[3] Lluis Puig, Pointed groups and construction of characters, Math. Z. 176 (1981), 265–292.

[4] Daniel Quillen, Homotopy properties of the poset of non-trivial p-subgroups, Adv. in Maths 28 (1978), no. 2, 101–128.

[5] 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.

[6] , The structure of Mackey functors, Trans. Amer. Math. Soc. 347 (1995), no. 6, 1865– 1961.

[7] Peter Webb, Subgroup complexes, The Arcata conference on representations of finite groups, vol. 47, AMS Proceedings of Symposia in pure mathematics, 1987.

[8] , A split exact sequence for Mackey functors, Comment. Math. Helv. 66 (1991), 34–69.

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@mathp7.jussieu.fr

Current address : S. Bouc, CNRS-LAMFA, 33 rue St Leu, 80039 - Amiens, France

    E-mail address: serge.bouc@u-picardie.fr