Download pdf - COHOMOLOGY AND REPRESENTATION THEORY Introduction · COHOMOLOGY AND REPRESENTATION THEORY JON F. CARLSON 0. Introduction In these lectures we shall consider a few aspects of the interaction

COHOMOLOGY AND REPRESENTATION THEORY

JON F. CARLSON∗

0. Introduction

In these lectures we shall consider a few aspects of the interaction between groupcohomology and group representation theory. That interaction has grown tremen-dously in the last thirty years to the point that homological methods are now stan-dard in modular representation theory. The subject is much too large to give acomplete picture in the space of a half semester of lectures. Consequently, we willconcentrate on the methods and results required for one application: the classifica-tion of endotrivial modules. The classification is a statement about modules overgroup algebras and makes almost no mention of homological algebra or cohomology.Yet its proof relies in fundamental ways on the therory of support varieties, on thecomputations of the cohomology rings of extraspecial groups and on several itemsfrom group cohomology.

An outline of the course is the following. In general we assume a basic knowledgeof homological algebra and group representations. The first two or three sections willcover foundational material and be treated mostly as review. In the later sections weencounter some theorems whose proofs, because of time constraints, will be omitedor only sketched.

(1) Modular representations of p-groups.(2) Group cohomology.(3) Support varieties.(4) The cohomology ring of a dihedral group(5) Elementary abelian subgroups in cohomology and representations(6) Quillen’s Dimension Theorem(7) Properties of support varieties(8) The rank of the group of endotrivial modules

Throughout these notes, the symbol k denotes a field of prime characteristic p. Ingeneral, we assume that k is algebraically closed, though for many of the theorems,this restriction is not necessary. All modules are left unital modules unless statedotherwise. The tensor product ⊗ means ⊗k. The k-dual of a kG module or k-vectorspace M is denoted M ∗. All modules will be assumed to be finitely generated.Recall that modules over a finite dimensional algebra satisfy the Krull-SchmidtTheorem. That is, every (finitely generated) module can be written uniquely (up toisomorphism and order of the factors) as a direct sum of indecomposable modules.

Date: April 29, 2005.This research was supported in part by a grant from NSF.

1

2 JON F. CARLSON

For most of the basic material, no references are given. The results can be foundin one or all of the basic text books on the subject [2, 10, 15].

1. Modules over p-groups

In this section we explore the group algebras of p-groups and their representations.All of the material in this section is standard and can be found in almost any textthat deals with modular representation theory. Some of the results of the sectionhold for all finite groups and not just p-groups. The reader who is unfamiliar withsome of the material in this section is encouraged to work through the exercises insome detail.

Throughout this section assume that G is a finite group. We specialize to p-groupslater in the section. First we need some basics on group algebras.

Hopf algebras, tensor products and duals. We have a Hopf algebra structurekG −→ kG × kG given on basis elements by g −→ (g, g). This means that if Mand N are kG-modules, then so is M ⊗ N with the action of g ∈ G defined byg(m× n) = gm⊗ gn for m ∈M and n ∈ N . Likewise we make Homk(M,N) into akG-module by letting (gf)(m) = gf(g−1m) for all f ∈ Homk(M,N) and m ∈M .

Exercise 1.1. Prove that Homk(M,N) ∼= M∗ ⊗N by the map which sends λ ⊗ nto f where f(m) = λ(m)n for all λ ∈ M ∗, n ∈ N and m ∈ M . Show also that theisomorphism is natural in both variables.

Exercise 1.2. Suppose that G = 〈x, y〉 is an elementary abelian group of order 4,and k has characteristic 2. Let M = Mα be the kG-module of dimension 2 for whichthe actions of x and y are given by the matrices

x →

(

1 01 1

)

, y →

(

1 0α 1

)

,

for some element α ∈ k. Find a decomposition of M ⊗ M into a direct sum ofindecomposable modules. Do the same for Mα ⊗Mβ where α and β are differentelement of k.

Symmetric and self-injective algebras. Let σ : kG −→ k by σ(∑

ag · g) = a1.That is, σ applied to an element of kG returns the coefficient on the identity elementof G. Define a nondegenerate symmetric bilinear form ( , ) : kG× kG −→ k by therule (α, β) = σ(α · β). It can be seen that the form is G-invariant in the sense that(αg, β) = (α, gβ) for all g ∈ G, α, β ∈ kG. The form proves that kG is a symmetricalgebra. That is, there is an isomorphism φ : kG ∼= kG∗ given by φ(α) = (α, ). Aconsequence of this is the following.

Theorem 1.3. The group algebra kG is a self-injective algebra. That is, everyfinitely generated projective module is injective, and conversely, every finitely gen-erated injective module is projective.

Exercise 1.4. Prove the theorem. Show first that finitely generated free modulesare injective using the duality.

COHOMOLOGY AND REPRESENTATION THEORY 3

Module categories. We let mod(kG) denote the category of finitely generatedkG-modules. Let stmod(kG) denote the stable category of kG-modules moduloprojectives. The objects in stmod(kG) are the same as those in mod(kG), but themorphisms from modules M to N are given by

HomkG(M,N) = HomkG(M,N)/PHomkG(M,N)

where PHomkG(M,N) is the set of all homomorphisms from M to N that factorthrough a projective module.

Induction and Frobenius reciprocity. Suppose that H is a subgroup of G. IfM is a kG-module we let MH denote the restriction of M to a kH-module. If someemphasis is required we use the symbol M↓H to denote the restriction. If N is akH-module, then the induced module N ↑G = kG ⊗kH N is a kG-module with theaction of G on the left. Both restriction and induction are functors on the modulecategories. There are two results relating induction and restriction that are veryuseful to us. The first is known as Frobenius Reciprocity.

Theorem 1.5. Let M be a kG-module and N a kH-module. Then

M ⊗N↑G ∼= (MH ⊗N)↑G

The isomorphism is given by the map m⊗ (g⊗n) → g⊗ (g−1m⊗n) for all g ∈ G,m ∈M and n ∈ N . In the other direction, the map sends g⊗(m⊗n) = gm⊗(g⊗n).

The other result is known as the Mackey formula.

Theorem 1.6. Suppose that M is a finitely generated kH-mdoule for H a subgroupof G. Let K be another subgroup of G. Then

(M↑G)↓K∼=

∑

k×H

((x⊗M)K∩xHx−1)↑K =∑

KxH

(x⊗Mx−1Kx∩H)↑G

where the sum is indexed by the double KxH-cosets in G.

Now notice that if H = 1, the identity subgroup and kH is the trivial kH-

module, then k↑GH∼= kG as left kG-modules. A consequence of this and Frobenius

Reciprocity is the following.

Exercise 1.7. Suppose that P is a projective kG-module and that M is any kG-module. Prove that M ⊗ P is projective.

Degree shifting. The notation and ideas of this section are vital for the rest ofthe course. First we recall Schanuel’s Lemma.

Proposition 1.8. Let R be a ring and let M be an R-module. Suppose that P1 andP2 are projective modules and that θ1 : P1 −→ M , θ2 : P2 −→ M are surjectivehomomorphisms. Let Ki be the kernel of θi for i = 1, 2. Then K1 ⊕ P2

∼= K2 ⊕ P1.

If M is a finitely generated kG-module, then there exists a finitely generatedprojective cover θ : P −→ M . That is, P is a projective module of least dimensionsuch that there is a surjective homomophism (theta) onto M . We denote the kernel

4 JON F. CARLSON

of θ by Ω(M). Notice that Ω(M) has no projective submodule, because if Q werea projective submodule of Ω(M), then Q would be a projective and also injectivesubmodule of P . Hence Q would be a direct summand of P , thus contradicting theminimality of P . Moreover, Ω(M) is uniquely defined in the sense that if γ : Q −→M is any surjective homomorphism with Q projective, then by Schanuel’s Lemmathe kernel of γ is Ω(M) ⊕ (proj), where by ⊕ (proj) we mean the direct sum withsome projective module.

Inductively, we define, Ωn(M) = Ω(Ωn−1(M)) for all natural numbers n > 1. Forthe reasons given Ωn(M) is well defined up to isomorphism. The module M has aninjective hull given by θ : M−→Q where Q a smallest injective (projective) moduleinto which M injects. Then the kernel of θ is denoted Ω−1(M) and has no injective(hence projective) submodules. Iterating, we define Ω−n(M) = Ω−1(Ω−n+1(M)) forn > 0. We let Ω0(M) be the nonprojective part of M , the direct sum of all of thenonprojective indecomposable summands of M .

With the above definitions and some facts that we know about projective modules,we can prove the following very useful result.

Exercise 1.9. Suppose thatM andN are kG-modules andm and n are any integers.Then

(1) Ωm(M) ⊗ Ωn(N) ∼= Ωm+n(M ⊗N) ⊕ (proj), and(2) (Ωn(M))∗ ∼= Ω−n(M∗).

Definition 1.10. A kG-module is an endotrivial module provided its k-endomorphismring is the direct sum of a trivial module and a projective module. That is, M isendotrivial if and only if

Homk(M,M) ∼= M∗ ⊗M ∼= k ⊕ (proj)

.

The previous exercise shows that for any integer n, Ωn(k) is an endotrivial module.

Group algebras of p-groups. Suppose now that G is a p-group. Note that ifx ∈ G then (x − 1)p

n

= xpn

− 1 = 0, if pn is the order of x. Consequently, theaugmentation ideal I(kG) of kG, the ideal generated by all x − 1 for x in G, isgenerated by nilpotent elements. Slightly harder to prove is the following.

Exercise 1.11. LetG be a p-group. Then the augmentation ideal I(G) is a nilpotentideal of codimension 1 in kG. In particular, I(G) is the radical of kG. It containsevery proper ideal of kG, and kG is a local ring.

The exercise implies the following result. The fact that projective modules overa local ring are free modules is well known.

Corollary 1.12. If G is a p-group, then kG is a local ring, and projective kG-modules are free.

Another corollary of the exercise is that kG has only one irreducible module,namely, the trivial module k. Another consequence is that the projectivity of amodule can be established simply by considering the action of a single element. Let


NG =∑

g∈G g be the sum of all of the elements of G. Note that if G = 〈g〉 is a

cyclic group of order pn then NG = (x− 1)pn−1. As usual, |G| denotes the order of

G.

Lemma 1.13. Let M be a kG-module, then

Dim NG ·M ≤1

|G|DimM.

Moreover, we have equality if and only if M is a projective module.

Proof. (Sketch) Note first that kNG ⊆ kG is the unique minimal ideal in kG. Inparticular it is contained in every other nonzero ideal. Also, the ideal of kG generatedby NG has dimension one and is the k-basis for the submodule NG ·M . Let a1, . . . , atbe a basis for NGM . For each i, let bi ∈ M be an element such that NGbi = ai.Then define ψ : kGt −→M by the formula

ψ(α1, . . . , αt) −→

t∑

i=1

αibi.

Now we see that ψ is injective because its kernel is zero. Hence, we have an exactsequence

0 // kGtψ

// M

Because kGt is injective the sequence splits. Therefore, M ∼= kGt⊕M ′ where M ′ isthe cokernel of ψ. The Lemma follows from the fact that NG ·M ′ = 0.

The following is very useful in the later development.

Corollary 1.14. If G = 〈x〉 is a cyclic group of order p, then a kG-module Mis projective if and only the rank of the matrix of the action of x on M is ((p −1)/p) DimM .

2. Group cohomology

We assume here that the reader is familiar with basic homological constructionssuch as chain complexes and chain maps, homology, homotopy of maps, the Kunnethformula for tensor product of chain complexes and the functors Ext and Tor. Inparticular, we often make use of the fact that Ext is the derived functor of Hom aswell as being the set of equivalence classes of extensions. Our main concern here is toestablish some facts about group cohomology and the relations with representationtheory. Most of this relationship is expressed in the notions of support varieties.The notation is as before. In particular, assume that G is a finite group and k is afield of prime characteristic p.

Group cohomology. Suppose that M and N are kG-modules and that

(P∗, ε) : . . . // P2// P1

// P0ε

// M // 0

6 JON F. CARLSON

is a projective resolution of M . Recall that for n > 0, ExtnkG(M,N) is defined to bethe (co)homology of the cochain complex HomkG(P∗, N). That is,

ExtnkG(M,N) = Hn(HomkG(P∗, N)).

We also have that

ExtnkG(M,N) = Hn(HomkG(M,Q∗)).

Where Q∗ is an injective resolution of N . Moreover, the Ext functor can be becomputed in the stable category as

ExtnkG(M,N) ∼= HomkG(Ωn(M), N) ∼= HomkG(M,Ω−n(N)).

The group cohomology of G with coefficients in a module M is defined as

Hn(G,M) ∼= ExtnkG(k,M).

With the canonical isomorphism HomkG(M,N) ∼= HomkG(k,M∗ ⊗N) we can provethat

ExtnkG(M,N) ∼= Hn(G,M∗ ⊗N).

Minimal Resolutions. Let M be a finitely generated kG-module. A projectiveresolution

(P∗, ε) : . . . // P2// P1

// P0ε

// M // 0

of M is a minimal projective resolution if ∂(Pn) ⊆ Rad(Pn−1) for every n > 0.We construct the minimal projective resolution of M by taking projective covers.That is, let ε : P0 → M be a projective cover of M . Then the kernel of ε is Ω(M)which has no projective submodules. Now take a projective cover P1 of Ω(M) andcontinue. Some useful properties of minimal resolutions are itemized in the followingproposition.

Proposition 2.1. Let (P∗, ε) be a projective resolution of a finitely generated kG-module M . The following are equivalent statements.

(1) (P∗, ε) is a minimal projective resolution of M .(2) If S is a simple kG-module, then for all n > 0

HomkG(Pn, S) = ExtnkG(M,S).

(3) If S is a simple kG-module, then for every n ≥ 0 the cohomology map

∂∗ : Hom(Pn, S) −→ Hom(Pn+1, S)

is the zero map.(4) Let (Q∗, ε

′) be any projective resolution of M . Then the chain map µ :(Q∗, ε

′) −→ (P∗, ε) that lifts the identity map on M is surjective.(5) Let (Q∗, ε

′) be any projective resolution of M . Then any chain map ν∗ :(P∗, ε) −→ (Q∗, ε

′) that lifts the identity on M is injective.

Exercise 2.2. Prove the proposition.

The usual example of a minimal resolution is for a cyclic group.


Example 2.3. Suppose that G = 〈x|xpn

= 1〉 is a cyclic p-group, written multi-plicatively. Let NG =

∑

g∈G g ∈ kG be the sum of the elements in G. Notice that

NG = (x− 1)pn−1. Then we have a periodic projective resolution (X∗, ε) of the form

. . . NG// X3

x−1// X2

NG// X1

x−1// X0

ε// k // 0

where for every i, Xi∼= kG. That is, the boundary map on Xi for i odd is

multiplication by x − 1, while for i even it is multiplication by NG. The ex-actness of the resolution can be checked from the observation that the elements1, x− 1, (x− 1)2, . . . , (x− 1)p

n−2,NG form a k-basis for the free k-module Xi forevery i. This resolution may be constructed for any commutative ring of coefficientsk.

Exercise 2.4. Let G denote the quaternion group of order 8, given by

G = 〈x, y|x2 = y2 = (xy)2, x4 = 1〉.

Note that yxy−1 = x−1 and that x2 is the unique element of order 2 in G. Showthat the trivial kG-module k has a minimal projective resolution of the form

. . . // KG2 // kG2 // kG // kG // kG2 // kG2 // kG // k // 0

which repeats after every four steps.

Cohomology products. When we think of the cohomology as Ext*kG in terms of

extensions, then it is convenient to think of the products as compositions of sequences(splices). However, there are several other methods of defining the products, someof which are peculiar to group cohomology. Happily, the various definitions areequivalent and each gives us some insight into the multiplicative structure.

With the product structure, Ext*kG(k, k) = H∗(G, k) is a graded (associative)

ring. We are able to show that the multiplication is graded commutative, in thatζγ = (−1)deg(ζ) deg(γ)γζ. If M is a kG-module, then Ext*

kG(M,M) is also a ring andassociative. As long as M is finitely generated as a module, its cohomology ring isfinitely generated as an algebra. However, it is not always true that Ext*

kG(M,M)is commutative or graded commutative.

Yoneda splices and compositions of chain maps. Suppose that L,M and Nare kG-modules and that ζ ∈ ExtnkG(M,L), γ ∈ ExtmkG(N,M) for n > 0, m > 0.Then there are exact sequences

E : 0 // L // Bn−1// · · · // B0

ν// M // 0

and

E ′ : 0 // Mµ

// Cm−1// · · · // C0

// N // 0

which represent ζ and γ respectively. The Yoneda splice or Yoneda composite of thetwo sequences is a sequence E E ′ of length n +m given as

0 // L // Bn−1// · · · // B0

µν// Cm−1

// · · · // C0// N // 0

8 JON F. CARLSON

Then the product of ζγ is defined to be the class of the extension E E ′ inExtmkG(N,L).

Exercise 2.5. Prove that the above product is well defined.

Now suppose that (P∗, ε) and (Q∗, ε′) are projective resolutions of the kG-modules

M and L, respectively. Let Un(M,L) denote the homotopy classes of chain maps ofdegree −n from P∗ to Q∗, n > 0.

Proposition 2.6. Un(M,L) ∼= ExtnkG(M,L).

Let f : Pn −→ L be a cocycle representing a cohomology element γ ∈ ExtnkG(M,L).Then let the image of γ be the chain map µ of the following commutative diagram:

· · · // Pn+1∂n+1

//

µ1

Pn∂n

//

µ0

f

""DD

D

D

D

D

D

D

D

Pn−1// · · ·

· · · // Q1∂1

// Q0ε′

// L // 0.

So in addition to the Yoneda splice of sequences, we also have a well defined producton cohomology

ExtmkG(M,L) ⊗ ExtnkG(N,M) → Extm+nkG (N,L)

which is given by composing the chain maps.

Proposition 2.7. The two products defined on cohomology by Yoneda splice and bythe composition of chain maps coincide.

We should note that the same result could have been completed with injectiveresolutions. That is, the cohomology product defined by the Yoneda splice operationon exact sequences coincides with the operation which we could define by takingcompositions of chain maps on injective resolutions of the modules. Another viewof cohomology products is the following.

Products in the stable category. Suppose we look at the cohomology as theHom functor in the stable category. Recall that, for any m and any kG-modules Mand N , we have ExtmkG(N,M) ∼= HomkG(Ωm(N),M) ∼= HomkG(Ωm+n(N),Ωn(M)).We can define a product

ExtnkG(M,L)⊗ExtmkG(N,M) ∼=

HomkG(Ωn(M), L) ⊗ HomkG(Ωn+m(N),Ωn(M))

−→ HomkG(Ωn+m(N), L) ∼= Extn+mkG (N,L),

where the middle map is composition of homomorphisms in the stable category. Wecan prove the following.

Proposition 2.8. The product on cohomology given by the composition of maps inthe stable category coincides with the product defined by Yoneda splice of sequences.


Products by Hopf algebra structure. Let (P∗, ε) and (P ′∗, ε

′) be projectiveresolutions of M and M ′ respectively. Then Hn(P∗) = 0 unless n = 0 in whichcase H0(P∗) = M , and similarly for P ′

∗. So we have that Hn(P∗ ⊗ P ′∗) = 0 unless

n = 0, and H0((P∗ ⊗ P ′∗)) = M ⊗M ′ by the Kunneth Tensor Formula. Suppose

the cohomology classes ζ ∈ ExtnkG(M,N) and γ ∈ ExtmkG(M ′, N ′) are represented bycocyles f : Pn → N and f ′ : P ′

m → N ′. Then we have a cocycle f ⊗ f ′ : Pn ⊗ P ′m →

N⊗N ′. Note that Pn⊗P′m is a direct summand of (P⊗P ′)n+m =

∑n+mi=1 Pi⊗P

′n+m−i.

Hence we can consider f ⊗ f ′ : (P ⊗ P ′)n+m → N ⊗N ′ as the cocycle with supportPn ⊗ P ′

m as above. Then the class of f ⊗ f ′ is the outer product of ζ and γ.In the situation that M ∼= M ′ ∼= k, then we have a cohomology product

H∗(G,N) ⊗ H∗(G,N ′) // H∗(G,N ⊗N ′).

Suppose that ζ ∈ Hm(G,N) and ζ ′ ∈ Hn(G,N ′) are represented by cocycles f :Pm −→ N and f ′ : Pn −→ N ′, where (P∗, ε) is a projective resolution of k. Thenthe product ζ⊗ ζ ′ is represented by the cocycle µ (f ⊗f ′) where µ : P∗ −→ P∗⊗P∗

is a diagonal approximation, a chain map that lifts the identity on k. In topology,µ is often called the Alexander-Whitney map. If (P∗, ε) is the bar resolution, thenµ can be given a very explicit form.

Another method for defining the outer product is via the tensor product of com-plexes. That is, let P∗ be a projective resolution of M and ζ ∈ ExtnkG(M,N) berepresented by an exact sequence

E : 0 // N // Bn−1// · · · // B0

η′// M // 0.

Then we have a chain map µ as in the diagram:

· · · // Pn+1//

µn+1

Pn //

µn

Pn−1//

µn−1

// P0//

µ0

M // 0

E : 0 // N // Bn−1// · · · // B0

// M // 0

That is, if we let B be the complex

B : 0 // N // Bn−1// · · · // B0

// 0,

then we have a chain map µ : P∗ −→ B whose induced map on homology is theidentity (on M).

Similarly, if P ′∗ is a resolution of M ′, γ ∈ ExtmkG(M ′, N ′) is represented by a

sequence

E ′ : 0 // N ′ // Cm−1// · · · // C0

η′// M ′ // 0

and we let C be the complex

C : 0 // N ′ // Cm−1// · · · // C0

// 0,

then H0(C) = M ′ and there is a chain map ν : P ′∗ → C that induces the identity

on homology. Then (B ⊗ C)∗ is a complex with the property that Hn(B ⊗ C) = 0 if

10 JON F. CARLSON

n 6= 0 and H0(B ⊗ C) = M ⊗M ′. Therefore we have an exact sequence

U : 0 // N ⊗N ′ // (Bn−1 ⊗N ′) ⊕ (N ⊗ Cm−1) // · · ·

// B0 ⊗ C0

η⊗η′// M ⊗M ′ // 0.

That is, the sequence is the augmented complex of B ⊗ C.

Proposition 2.9. The sequence U represents the cohomology class ζ ⊗ γ in thecohomology group Extm+n

kG (M ⊗M ′, N ⊗N ′).

Exercise 2.10. Verify that all of the above products are equivalent. In other words,fill in the details in the above discussion.

Commutativity of products. Suppose that (P∗, ε) is a projective resolution ofthe trivial module k.

Proposition 2.11. The map µ : (P ⊗ P )∗ −→ (P ⊗ P )∗ given by µ(x ⊗ y) =(−1)deg(x) deg(y)y ⊗ x is a chain map that lifts the identity on k.

We can use this to prove the graded-commutativity in the ring H∗(G, k).

Theorem 2.12. Suppose that ζ ∈ Hn(G, k) and γ ∈ Hm(G, k). Then ζγ =(−1)mnγζ.

Cyclic group cohomology. Now we consider the special case of a cyclic group.

Proposition 2.13. Let G = 〈x|xpn

= 1〉 be a cyclic group. Then

H∗(G, k) ∼=

k[ζ] if pn = 2 (deg ζ = 1)k[η, ζ]/(η2) if pn > 2 (deg η = 1, deg ζ = 2).

Exercise 2.14. Verify the formula by composing chain maps for the products.

Cohomology rings of elementary abelian groups. Suppose thatG = 〈x1, . . . , xn〉is an elementary abelian group of order pn. Then G ∼= 〈x1〉 × · · · × 〈xn〉 is a direct

product of cyclic groups of order p. Let (X(i)∗ , εi) be a minimal projective k〈xi〉-

resolution of k as before. We can see that (P∗, ε) = (X (1), ε1) ⊗ · · · ⊗ (X (n), εn) isa minimal projective kG-resolution of k. We also have the product formula on thecohomology. Recall the notation Λ(η1, η2, . . . , ηn) for the exterior algebra generatedby η1, . . . , ηn.

Proposition 2.15. Let G = 〈x1, . . . , xn〉 be an elementary abelian group of orderpn and let k be a field of characteristic p. Then

H∗(G, k) ∼=

k[ζ1, . . . , ζn] if p = 2

k[ζ1, . . . , ζn] ⊗ Λ(η1, . . . , ηn) if p > 2.

If p = 2, then each ζi occurs in degree 1. If p > 2, then each ηi is in degree 1 whileeach ζi is in degree 2. We have relations η2

i = 0 and ηiηj = −ηjηi.


Finite generation. The following is a theorem that will be of great use to us. Theproof is far too complicated to be presented in the time alloted.

Theorem 2.16. (Evens, Venkov) The cohomology ring H∗(G, k) is a finitely gener-ated k-algebra. Moreover, if M is a finitely generated kG-module then H∗(G,M) isa finitely generated module over the ring H∗(G, k).

3. Support Varieties

Because of the Evens-Venkov Theorem on finite generation of group cohomologywe can consider the geometry of the cohomology rings and the geometry of coho-mology modules. The theorem tells us that cohomology rings are always quotientsof finitely generated free graded commutative algebras by finitely generated ideals.In the case that p = 2, the graded commutative rings are actually commutative, andthe cohomology ring H∗(G, k) is a quotient of a polynomial ring. When p is odd,H∗(G, k)/Rad(H∗(G, k)) is a polynomial ring because every element in odd degree isin the radical. Note also that Rad(H∗(G, k)) is in every prime or maximal ideal sinceit is a nilpotent ideal. Consequently, the collections of maximal ideals for H∗(G, k)is the same as that for H∗(G, k)/Rad(H∗(G, k)).

Examples of cohomology rings As an example, consider the cohomology ring ofa dihedral 2-group in the case that the coefficient ring k is a field of characteristic2. It has the form.

H∗(G, k) ∼= k[z, y, x]/(zy).

It is the quotient of the polynomial ring k[z.y.x] by the ideal generated by zy, andit looks like the coordinate ring of the union of two planes (the z-x plane and they-x plane) that intersect in a line. The cohomology generators z, y, x are in degrees1, 1, 2.

For the semi-dihedral group of order 16, the cohomology ring has the form

H∗(G, k) = k[z, y, x, w]/(z3, zy, zx, x2 + y2w).

This time the generators z, y, x, w are in degrees 1, 1, 3, and 4.The cohomology ring of the extraspecial group of order 27 and exponent 3 is

generated by elements z, y, x, w, v, u, t, s, r in degrees 1, 1, 2, 2, 2, 2, 3, 3, 6, subjectto the relations.

yz, xy − uz, wz + uz, wy + vz, vy + uz,

wx+ uv − tz − sy, w2 + uw + sz, vx + uw + sz,

vw + uv − sy, v2 + uw, ux− uv + sy, ty − sz,

uvz − tu+ sv, u2z − tv − sw, tw + tu, sx− sv,

tvz − suz, tuz − st, tuy − suz, svz − st,

tuw + tu2, tu2y − stv, stx− stv

(and some of these may be redundant). Note that the commutativity relations,which are not trivial, are not in the above list.

The fact that the cohomology ring H∗(G, k) is a finitely generated k-algebra meansthat it is noetherian. We let VG(k) denote the maximal ideal spectrum of H∗(G, k).

12 JON F. CARLSON

For G an elementary abelian p-group of p-rank n, VG(k) ∼= kn. That is, everymaximal ideal of H∗(G, k) is the kernel of a homomorphism H∗(G, k) −→ k whichis given by evaluating a polynomial in H∗(G, k)/Rad(H∗(G, k)) at a point α =(α1, . . . , αn) in kn. This, of course, depends on k being algebraically closed. For thedihedral group given above, VG(k) is the union of two planes joined along a line. Inany case, VG(k) is homogeneous affine variety.

Definitions and properties. For kG-modules M and N , a consequence of theEvens-Venkov Theorem is that Ext∗kG(M,N) is a finitely generated module overH∗(G, k) ∼= Ext∗kG(k, k). For any kG-module M , let J(M) be the annihilator inH∗(G, k) of the cohomology ring Ext∗kG(M,M). We can take J(M) to be the anni-hilator in H∗(G, k) of the identity element IdM . Then we let VG(M) = VG(J(M))be the closed subset of VG(k) consisting of all maximal ideals that contain J(M).Some of the properties of support varieties are given in the following.

Theorem 3.1. Let L,M and N be kG-modules.

(1) VG(M) = 0 if and only if M is projective.(2) If 0 → L → M → N → 0 is exact then the variety of any one of L,M or

N is contained in the union of the varieties of the other two. Moreover, ifVG(L) ∩ VG(N) = 0, then the sequence splits.

(3) VG(M ⊗N) = VG(M) ∩ VG(N).(4) VG(Ωn(M)) = VG(M) = VG(M∗) where M∗ = Homk(M, k) is the k-dual

of M .(5) If VG(M) = V1∪V2 where V1 and V2 are non-zero closed subsets of VG(k) and

V1 ∩ V2 = 0, then M ∼= M1 ⊕M2 where VG(M1) = V1 and VG(M2) = V2.(6) A nonprojective module M is periodic (i.e. for some n > 0, Ωn(M) ∼=

Ω0(M)) if and only if its variety VG(M) is a union of lines through theorigin in VG(k).

Proof. Recall that Ext∗kG(M,M) is finitely generated as a module over H∗(G, k). Soif VG(M) = 0, then for n sufficiently large ExtnkG(M,M) = 0. Now H∗(G, k)acts on Ext∗kG(N,M) through its action on Ext∗kG(M,M). Consequently for anymodule N , ExtnkG(N,M) = 0 for n sufficiently large. This implies that M hasfinite projective dimension. But as kG is a self-injective ring, M must be projective.This proves (1).

Part (2) is a consequence of the long exact sequence on cohomology. Statement(3) is rather complicated. We will try to indicate why it is true later. For (4) weneed only notice that Ext∗kG(M,M) ∼= Ext∗kG(Ωn(M),Ωn(M)), not only as ringsbut as modules over H∗(G, k). Moreover, Ext∗kG(M∗,M∗) is anti-isomorphic toExt∗kG(M,M). The proof of Statement (5) is again more complicated than we areprepared to handle. The last statement is a consequence of the following moregeneral development.


Complexity. We define the complexity of a module M to be the least integer s ≥ 0such that

limn→∞

(DimPn)/ns = 0.

It can also be defined as the least s ≥ 0 such that

limn→∞

(DimExt∗kG(N,M))/ns = 0,

for all kG-modules N , or as the least s ≥ 0 such that

limn→∞

(DimExt∗kG(M,M))/ns = 0.

It follows from the fact that Ext∗kG(M,M)) is a finitely generated module overH∗(G, k)/J(M), that the complexity of M is the dimension of the variety VG(M),which is the same as the Krull dimension of H∗(G, k)/J(M). Now if M is periodicthen its variety is a homogeneous subvariety of VG(k) of dimension one. It can onlybe a union of lines.

Exercise 3.2. Prove part (2) of the therorem.

Rank varieties. Assume now that G is an elementary abelian p-group. Let G =〈x1, x2, . . . , xn〉 and for α = (α1, α2, . . . , αn) ∈ kn, let uα = 1+

∑ni=1 αi(xi−1). Note

that uα is a unit of order p in kG, when α 6= 0. Associated to a kG-module M wecan define a rank variety

V rG(M) =

α ∈ kn |M↓〈uα〉 is not a free 〈uα〉−module

∪ 0

where uα is given as above and where M↓〈uα〉 denotes the restriction of M to thesubalgebra k〈uα〉 of kG.

Exercise 3.3. Prove that kG is free as a k〈uα〉-module, provided α 6= 0.

As an example, consider the situation in Exercise 1.2. Here let β = (b1, b2) anduβ = i1 + b1(x− 1) + b2(y − 1). Notice that uβ acts on M by the matrix

uβ ↔

(

1 0b1 + b2α 1

)

Now, uβ acts freely on M if and only if its matrix is conjugate to the matrix(

1 01 1

)

But this happens if and only if the matrix of uα−1 has rank 1. In turn, this happensif and only if b1 + b2α 6= 0. Hence the variety

V rG(M) = (b1, b2) ∈ k2|b1 + b2α = 0 = b(α, 1)|b ∈ k.

For another example let G = 〈z, y, x〉 be an elementary abelian group of order 8.Choose elements A,B,C ∈ k and define M to be a kG-module of dimension 4 such

14 JON F. CARLSON

that z, y, x act by the matrices

1 0 0 00 1 0 01 0 1 00 1 0 1

,

1 0 0 00 1 0 0A 0 1 00 B 0 1

,

1 0 0 00 1 0 00 C 1 01 0 0 1

.

Then the matrix of uα = 1 + α1(z − 1) + α2(y − 1) + α3(x− 1) is

1 0 0 00 1 0 0

α1 + Aα2 Cα3 1 0α3 α1 +Bα2 0 1

,

Again uα acts freely on M if and only if the rank of the matrix of uα − 1 is 2.But this happens if and only if the 2 × 2 minor in the lower left corner is not zero.Consequently, V r

G(M) = (α1, α2, α3) ∈ k3|(α1 + Aα2)(α1 +Bα2) + Cα23 = 0.

Exercise 3.4. Prove that V rG(M) is a homogeneous variety in the case that G is an

elementary abelian p-group.

Then we have the following result for any p.

Theorem 3.5. Let M be any kG-module. If p = 2 then, V rG(M) = VG(M) as subsets

of kn. If p > 2 then the map VG(M) −→ V rG(M) which sends α to αp = (αp1, . . . , α

pn)

is an inseparable isogeny (both injective and surjective). In particular, for α 6= 0,αp ∈ VG(M) (α ∈ VG(M) if p = 2) if and only if M↓〈uα〉 is not a free k〈uα〉-module.

We should emphasize that if v is a unit in kG such that v ≡ uα mod(Rad(kG)2)then M↓〈v〉 is a free k〈v〉-module if and only if αp 6∈ VG(M) (α 6∈ VG(M) ifp = 2). So for example the element x1x2x3 fails to act freely on M if and onlyif (1, 1, 1, 0, . . . , 0) ∈ VG(M).

Restrictions to shifted subgroups The proof of the theorem is rather compli-cated, but we can give a hint as to why it is true. It really has to do with restrictionsof cohomology elements to the cyclic subgroups of kG. We call these shifted cyclicsubgroups. With some work it is not difficult to prove the following. The proof isa matter of constructing a chain map from the projective resolution of the trivialmodule for the subgroup generated by uα to the projective resolution for the trivialmodule of G.

Proposition 3.6. Let α be the nonzero element α = (α1, . . . , αn) ∈ kn, and let

uα = 1 +n

∑

i=1

αi(xi − 1). Then U = 〈uα〉 is a cyclic group of order p and kU is a

subalgebra of kG. Hence the cohomology ring of U has the form given below. Weuse the same notation for the cohomology as in 2.15.

H∗(U, k) =

k[ζ ′, η′]/((η′)2) if p > 2

k[ζ ′], if p = 2.


For suitable choices of the generators, the restriction map reskG,kU : H∗(G, k) −→H∗(U, k) has the values reskG,kU(ηi) = αiη

′, resG,U(ζi) = αpi ζ′ for p > 2. Finally, for

p = 2, resG,U(ζi) = f(α)ζ ′.

In particular, we have the following.

Corollary 3.7. With the notation of the proposition, if ζ = f(ζ1, . . . , ζn) is a ho-mogeneous element of degree 2n in H∗(G, k) (degree n if p = 2), then

resG,U(ζ) = f(αp1, . . . , αpn)(ζ

′)n

(orresG,U(ζ) = f(α1, . . . , αn)(ζ

′)n

if p = 2).

The idea behind the proof of Theorem 3.5 is the following. Suppose that α ∈ kn

is a point in the rank variety V rG(M). Suppose that ζ ∈ J(M). Then we must have

thatresG,〈uα〉(ζ) · IdM = 0.

But by definition the restriction of M to 〈uα〉 is not a projective module. Becausemodules over the cyclic group 〈uα〉 are periodic and have finitely generated coho-mology, the only way that this can happen is if resG,〈uα〉(ζ) = 0. Consequently, α(or αp if p > 2) is in VG(M). This proves one of the inclusions. The other inclusionis considerably more complicated. It was proved by Avrunin and Scott [AS].

Dade’s Lemma and restrictions One of the reasons why this all works is con-tained in a result first proved by Dade [13], but which is a consequence of the equalityof the varieties 3.5.

Theorem 3.8. Suppose that G is an elementary abelian p-group and assume theabove notation. A kG-module M is projective if and only if its restriction to every〈uα〉 is projective.

More generally, maps on groups always induce maps on the maximal ideal spectraand on support varieties. In particular, suppose that H is a subgroup of G. Thenthe inclusion H −→ G induces a restriction map on cohomology rings resG,H :H∗(G, k) −→ H∗(H, k), and a corresponding map res∗G,H : VH(k) −→ VG(k) onmaximal ideal spectra. That is, the pullback of any maximal ideal in H∗(H, k) is amaximal ideal in H∗(G, k). Likewise, for M any kG-module, we have a map on thesupport varieties VH(M) −→ VG(M). A special case is the following, which can beseen very clearly for rank varieties.

Proposition 3.9. Suppose that H is a subgroup of G and that M is a kG-module.Then VH(M) = (res∗G,H)−1VG(M). In particular, MH is a free kH-module if andonly if resH,G(VH(k)) ∩ VG(M) = 0.

We should mention one further thing about modules over elementary abeliansubgroups. The result below also extends to module over any finite group. But weneed Quillen’s Theorem in order to make that extension.

16 JON F. CARLSON

Proposition 3.10. Suppose that G is an elementary abelian group of order pn andM is a kG-module. Let d = DimVG(M). Then the dimension of M is divisible bypn−d.

Proof. By Bezout’s Theorem there is a linear subspace V ⊆ kn = VG(k), such thatV ∩ VG(M) = 0 and Dim(V ) = r = n − d. Let α(1), . . . , α(r) be a basis for V .For each i let ui = uα(i). Then H = 〈u1, . . . , ur〉 is a shifted subgroup with theproperty that resG,H(VH(k)) ∩ VG(M) = 0. So by Proposition 3.9, M is free as akH-module. So the dimension of M is divisible by |H|.

Why quaternion groups have exotic endotrivial modules. The followingexample is somewhat premature in our course, because a full proof requires the factthat a kG-module M is endotrivial if and only if its restriction to every elementaryabelian p-subgroup is endotrivial. This fact will be proved in the next section.Assuming that fact, the example is a very good illustration of what we want to dowith the support variety technology.

Suppose that G = 〈x, y|x2 = y2 = (xy)2, x4 = 1〉 is a quaternion group of order8. Let z = x2 be the central element of order 2 in G. Let G = G/〈z〉. Then Gis a fours group, an elementary abelian group of order 4. So we know that for anysubgroup of order 4 such as C = 〈x〉 in G, Ω2(k)↓C ∼= k ⊕ (kC)2, since Ω2(kC) ∼= k.Let M = (z − 1)Ω2(k). Then (M)C is a free kC-module, where C = C/〈z〉. Theimplication is that VG(M) is not all of VG(k) = k2. Hence it must have dimensionone and must be a union of lines.

The cohomology ring of H∗(G, k) is well known and it is easy enough to computethat the dimension of Ω2(k) is 9 (see Exercise 2.4). This means that the dimensionof M is 4. Now if the support variety VG(M) is a union of r lines, then by Theorem3.1, M is a direct sum of r modules each having dimension at least 2. Consequently,it must be that the support variety is a union of at most two lines.

Next we notice that the three F2-rational lines in VG(k) correspond to the actualsubgroups of order 2 of G which act freely on M . So they are not in the supportvariety. On the other hand, Ω2(k) is certainly defined over F2, and so its variety isF2-rational. That is, it is the zero set in k2 of a polynomial with coefficients in F2.Hence the only choice is that VG(M) consists of the two lines through the points(1, α) and (1, α2) where α is a cube root of unity in F4.

Therefore M ∼= N1 ⊕ N2 where VG(N1) is the line through (1, α) and VG(N2) isthe line through (1, α2). The module

M = M/m ∈M |(z − 1)m = 0


is isomorphic to M with the isomorphism give by multiplication by z − 1. SoM ∼= L1 ⊕ L2 where (z − 1)L1 = N1 and (z − 1)L2 = N2. So M has the form

L1 L2

k

N1 N2

where multiplication by z− 1 takes the top to the bottom. Now let U be the kernelof the quotient map M −→ L2. Then U has the form

L1

k

N1 N2

Finally, let W = U/N2. The thing to note is that as a module over Z = 〈z〉, W hasthe form WZ

∼= k ⊕ kZ2, which is an endotrivial module. Because Z is the uniquenontrivial elementary abelian subgroup of G, this information is sufficient to provethat W is an endotrival module (see 5.5 in a later section).

We should note that the modules constructed above, were discovered by Dade[14] using direct computations some thirty years ago. These are the only endotrivialmodules for any p-group that can not be defined over the prime field Fp.

Extending Splittings The following is another illustration of how the supportvarieties can be used. Assume thatG is a p-group which has a central cyclic subgroupZ = 〈z〉 of order p. Let G = G/〈z〉.

Exercise 3.11. Suppose that N is a kG-module and that N has the property that(z − 1)p−1N = 0. Hence, N is a module over the algebra R = k〈z〉/((z − 1)p−1).Assume that N is a free R-module. So NZ is a direct sum of copies of R. Supposethat VG((z − 1)p−2N) = V1 ∪ V2 where V1 ∩ V2 = 0. There exist L1 and L2 suchthat (z − 1)p−2N = L1 ⊕ L2 where VG(L1) = V1 and VG(L2) = V2. Show that thereexist submodules N1 and N2 such that N = N1 ⊕N2 and for i = 1, 2, we have thatLi = (z − 1)p−2Ni.

Hint: Note that VG((z − 1)iN/(z − 1)i+1N) = VG((z − 1)p−2N). The statementconcerning the existence of L1 and L2 is a direct consequence of Theorem 3.1. In thesame way, we must have that (z−1)p−3N/(z−1)p−2N ∼= L′

1⊕L′2 where VG(L′

i) = Vi.So there is another exact sequence

0 // W // (z − 1)p−3N // L2// 0

18 JON F. CARLSON

where W contains (z − 1)p−2N and W/(z − 1)p−2N ∼= L′1. So we have a sequence

0 // L2// W/L1

// L′1

// 0

where L2 is the quotient (z − 1)p−2N/L1. Because VG(L2) ∩ VG(L1) = 0 the lastsequence splits. Now continue from here.

4. The cohomology ring of a dihedral groups.

In this section, we briefly describe the mod-2 cohomology ring H∗(G, k) in thecase that G is a dihedral 2-group. This example will be used extensively in whatfollows. Through assume that G = 〈x, y|x2 = y2 = (xy)4 = 1〉 is a dihedral groupof order 8. Let k be a field of characteristic 2. Let X = x+ 1 and Y = y + 1. Thenwe can see that X2 = Y 2 = 0. We can also show the following.

Exercise 4.1. Show that XYXY = (XY )2 = NG = Y XYX = (Y X)2.

In fact, we can show further that

Lemma 4.2. The group algebra kG has the form

kG ∼= k〈X, Y 〉/(X2, Y 2, XY XY − Y XYX).

That is, kG is the quotient of a polynomial ring in noncommuting variables X andY by the ideal generated by X2, Y 2 and XYXY − Y XYX.

The proof is reasonably straightforward, and mainly comes from noting that theclasses of the elements 1, X, Y,XY,XYX, Y XY and XYXY form a k-basis for thequotient. As there are exactly eight of these elements and because we have seenthat X and Y satisfy the relations that we gave, the quotient must be isomorphicto the group algebra.

The projective resolution. Then we can form a kG-projective resolution of thetrivial kG-module k as follows.

Proposition 4.3. A minimal projective resolution of k has the form

. . . // P2∂2

// P1∂1

// P0ε

// k // 0

where for all n, Pn ∼= (kG)n+1 and the boundary maps are given by

∂1 =

(

XY

)

, ∂2 =

X 0Y XY XYX

0 Y

, ∂3 =

X 0 0Y XY X 0

0 Y XYX0 0 Y

∂4 =

X 0 0 0Y XY X 0 0

0 Y XY XYX 00 0 Y XYX0 0 0 Y

,


∂5 =

X 0 0 0 0Y XY X 0 0 0

0 Y XY X 0 00 0 Y XYX 00 0 0 Y XYX0 0 0 0 Y

, · · ·

Note that because the variables do not commute, the matrices should be multipliedon the right. So for example if (a, b, c) ∈ P2

∼= kG3 then consider (a, b, c) as a rowvector and multiply

∂2(a, b, c) =(

a b c)

X 0Y XY XYX

0 Y

= (aX + bY XY, bXYX + cY ) ∈ P1.

Cocycles and chain maps. Now we can compute the generators for the cohomol-ogy ring H∗(G, k). We are going to write our cohomology products as compositionsof chain maps, so it is necessary to get a chain map for each of the generators. Notethat P1

∼= (kG)2 so there are two cohomology generators in degree one, representedby the cocycles η1, η2 : P1 −→ k, given by

η1(a, b) = ε(a) and η2(a, b) = ε(b).

Then we lift the cocycle η1 to a chain map η1∗

. . . // P2∂2

//

η1,2

P1∂1

//

η1,1

η1

A

A

A

A

A

A

A

A

P0ε

// k // 0

. . . // P1∂1

// P0ε

// k

It can be seen that the chain map for η1 has the form

η1,1(a, b) = a, η1,2(a, b, c) = (a, bY X), η1,3(a, b, c, d) = (a, b, 0)

η1,4(a, b, c, d, e) = (a, b, cY X, 0), η1,5(a, b, c, d, e, f) = (a, b, c, 0, 0), · · ·

It can be checked that this is a chain map by verifying that η1,i−1∂i = ∂i−1η1,i foreach i.

Similarly the chain map for η2 has the form

η2,1(a, b) = b, η2,2(a, b, c) = (bXY, c), η2,3(a, b, c, d) = (0, c, d)

η2,4(a, b, c, d, e) = (0, cXY, d, e), η1,5(a, b, c, d, e, f) = (0, 0, d, e, f), · · ·

Exercise 4.4. Verify that the chain maps are as stated.

Products. Now for the products notice that ε η1,1 η1,2 : P2 −→ k is given by ε η1,1η1,2(a, b, c) = ε(a). Likewise, εη2,1η2,2(a, b, c) = ε(c) and εη1,1η2,2(a, b, c) =ε(bXY ) = 0. From the last calculation we conclude that η1η2 = 0. From the firsttwo calculations we deduce that there is a new generator of cohomology in degree

20 JON F. CARLSON

2 represented by the cocycle ζ : P2 −→ k by ζ(a, b, c) = ε(b). As before we cancalculate the chain map that lifts ζ. It has the form

ζ2(a, b, c) = b, ζ3(a, b, c, d) = (b, c)

ζ4(a, b, c, d, e) = (b, c, d), ζ5(a, b, c, d, e, f) = (b, c, d, e), · · ·

We should check at least informally, that this is everything. For example in degree6, the cocycle that takes (a, b, c, d, e, f, g) to ε(c) is the composition η1.1η1,2ζ4ζ6 andso this represents the class η2

1ζ2. In this way see that

(a, b, c, d, e, f, g) 7→ ε(a) represents η61

(a, b, c, d, e, f, g) 7→ ε(b) represents η41ζ

(a, b, c, d, e, f, g) 7→ ε(c) represents η21ζ

2

(a, b, c, d, e, f, g) 7→ ε(d) represents ζ3

(a, b, c, d, e, f, g) 7→ ε(e) represents η22ζ

2

(a, b, c, d, e, f, g) 7→ ε(f) represents η42ζ

(a, b, c, d, e, f, g) 7→ ε(g) represents η62

So we see that we have the complete cohomology in degree 6. In general we have

Proposition 4.5. Let G = D8 the dihedral group of order 8. The cohomology ringof G has the structure

H∗(G, k) ∼= k[η1, η2, ζ]/I

where I = (η1η2) is the ideal generated by η1η2.

We have not attempted to give a rigorous proof of the proposition, but it shouldbe clear that a proof could be constructed from out calculations.

5. Elementary abelian subgroups in cohomology and representations

In this section, we investigate the role that the elementary abelian subgroups ofa group play in the representation theory and the cohomology of the group. Themain result is a theorem of Quillen [21], written in the early 1970’s, showed that theelements of the cohomology ring of a group could be detectected up to nilpotenceby their restrictions to the elementary abelian subgroups. The first application tomodule theory appeared in the paper of Chouinard (see [Ch]) where he proved,among other things, that a kG-module is projective if and only its restriction toevery elementary abelian p-subgroup is projective. Other application by Alperinand Evens [AE1, AE2], Avrunin [Av] and others [AS, C1] quickly followed.

The actual theorem that we prove is stated in terms of the existence of a modulewith a filtration. The interpretation as a result about cohomology takes some effort.The proof uses a theorem of Serre on the vanishing of a certain product of Bocksteinelements. There are several ways of deriving Serre’s Theorem, all of which are toocomplicated to treat in this short course. The Bockstein elements will be explainedin terms of certain exact sequences of induced modules which are easy to understand.The approach to the theory was developed in [C2].

The Main Theorem. First we state our main theorem, and we can see a fewexamples.


Theorem 5.1. There exists an integer τ , depending only on G, and a finitely gen-erated kG-module V such that the direct sum k ⊕ V has a filtration

0 = L0 ⊆ L1 ⊆ · · · ⊆ Lτ = k ⊕ V

with the property that for each i = 1, . . . , τ, there is an elementary abelian p-subgroupEi ⊆ G and a kEi-module Wi such that

Li/Li−1∼= W ↑G

i .

The modules V,W1, . . . ,Wτ are finitely generated.

So the theorem says that the trivial module k is a direct summand of a modulethat is filtered by modules induced from elementary abelian p-subgroups. The firstthing we should prove is that the same holds for any module.

Corollary 5.2. Let M be any kG-module. Then M is a direct summand of a modulethat is filtered by modules induced from elementary abelian p-subgroups. Moreoverif M is finite dimensional, then such a filtration can be found where every modulein the filtration is of finite dimension.

Proof. Assume that τ , V , Li, etc. are exactly as in the statement of the theorem.Now tensor everything with M . We get that M ⊗ (k ⊕ V ) has a filtration

0 = M ⊗ L0 ⊆M ⊗ L1 ⊆ · · · ⊆M ⊗ Lτ = M ⊗ (k ⊕ V ).

Now it is a matter of verifying that

M ⊗ Li/(M ⊗ Li−1) ∼= M ⊗W ↑Gi

∼= (M ⊗Wi)↑G,

using Frobenius Reciprocity.

Example 5.3. Suppose that G = 〈x|x9 = 1〉 and p = 3. Then let X = x −1 and observe that kG ∼= k[X]/(X9). Let V = Ω(k) ∼= kG/(X8). Let a be agenerator of k and let b be a generator of V . The k ⊕ V has a basis consistingof a, b,Xb,X62b,X3b,X4b,X5b,X6b,X7b. Let L1 be the submodule generated bya + X5b. Then L1 has a basis a + X5b,X6b,X7b and L1congk

↑GH where H = 〈x3〉.

Likewise, if we let L2 be the submodule generated by X3b and a, then (k⊕Ω(k))/L2

and L2/L1 are both isomorphic to k↑GH . Hence, the theorem holds in this case.

Some consequences of the Main Theorem. Now as an application we canprove the theorem of Chouinard.

Theorem 5.4. Suppose that M is a kG-module. Then M is projective if and onlyif ME is projective for every elementary abelian p-subgroup E of G.

Proof. If M is a projective kG-module, then its restriction to any subgroup H is aprojective kH-module. Consequently, the challenge is to prove the other direction.So assume that ME is a free kE-module for every elementary abelian p-subgroup Eof G. Let τ , V , Li, etc. be exactly as in the theorem. As in the last corollary, thestrategy is to tensor everything with M . This time, for each i we have that

M ⊗ Li/(M ⊗ Li−1) ∼= M ⊗W ↑Gi

∼= (MEi⊗Wi)

↑G.

22 JON F. CARLSON

Because MEiis a free kEi-module, we have that M ⊗ Li/(M ⊗ Li−1) is a free kG-

module. This implies that M ⊕ (M ⊗ V ) is a free module. It follows that M isprojective.

The next is really a corollary of the corollary.

Corollary 5.5. A kG-module M is an endotrivial module if and only if its restric-tion to every elementary abelian p-subgroup of G is an endotrivial module.

Proof. It should be clear from the definition that if M is an endotrivial module, thenits restriction to every subgroup of G is an endotrivial module. Therefore we canassume that ME is an endotrivial module for every elementary abelian p-subgroupof G. In particular, for any elementary abelian subgroup E 6= 1, this says thatHomk(M,M)E ∼= (M ⊗ M∗)E ∼= kE ⊕ (proj), and the dimension of M must becongruent to 1 or −1 modulo p. So, DimM is not divisible by p. Now, we have atrace map

Homk(M,M)Tr

// k

which simply sends any matrix α to its trace. In addition, the identity homomor-phism on M is fixed by elements of G. Consequently, the map

γ : k // Homk(M,M)

such that γ(a) = (a/DimM) IdM , is a homomorphism. In addition Tr(γ(a)) = afor all a ∈ k. Let K denote the kernel of the trace map. We have an exact sequence

0 // K // Homk(M,M)Tr

// k // 0

which is split and is also split on restriction to any elementary abelian p-subgroupof G. Hence, Homk(M,M)E ∼= kE ⊕ KE It follows that KE is a projective modulefor every elementary abelian p-subgroup E of G. By Chouinard’s Theorem, K is aprojective, and hence also injective kG-module. Therefore the sequence splits andM is an endotrivial module.

Proof of the Main Theorem: Reduction to p-groups. We have stated themain theorem for an arbitrary finite group, but it is really only necessary to proveit for p-groups. That is we have the following.

Proposition 5.6. Suppose that Theorem 5.1 is true for the Sylow p-subgroup P ofa finite group G. Then the theorem holds for G.

Proof. By our assumption, there exists a number τ , a kP -module V , elementaryabelian subgroups Ei of P and kEi modules Wi such that kP ⊕ V has a filtration

0 = L0 ⊆ L1 ⊆ · · · ⊆ Lτ = kP ⊕ V,

where Li/Li−1∼= W ↑P

i . Now we induce the entire structure to G obtaining thesequence

0 = L↑G0 ⊆ L↑G

1 ⊆ · · · ⊆ L↑Gτ = k↑GP ⊕ V ↑G.


Now the first point is that k is a direct summand of k↑GP . That is we have a map

φ : k↑GP∼= kG ⊗kP k −→ k given by φ(g ⊗ a) = ga = a for g ∈ G and a ∈ k. and

a map θ : k −→ k↑GP by θ(a) = (1/|G : P |)∑

(g ⊗ a) , where the sum is over acomplete set of representatives of the left cosets of P in G. It can be checked thatφθ is the identity on k. Next we check that

(L↑Gi )/(L↑G

i−1)∼= (Li/Li−1)

↑G ∼= (W ↑Pi )↑G ∼= W ↑G

i

by the transitivity of the induction. Hence the theorem holds for G.

Proof of the Main Theorem: Serre’s Theorem. Serre’s Theorem, which westate below, is the result that tells us that the cohomology and representation theoryof an elementary abelian p-group is different from that of an arbitrary p-group. Itis phrased in terms of the vanishing of a certain product of cohomology elements,but we will see that it has deep implications for the structure of modules over groupalgebras.

In our development we will not give a proof of Serre’ Theorem. The original proofused the action of the Steenrod algebra on the group cohomology. Other proofs havebeen given by Kroll [Kr] using Chern classes and by Pakianathan and Yalcin [PY]using LS-categories. Here is the statement of the theorem.

Theorem 5.7. [Se1]. Let Fp = Z/pZ be the field with p elements. Suppose that G isa p-group which is not elementary abelian. Then there is a sequence γ1, γ2, . . . , γn ∈H1(G, k) of nonzero elements such that if p = 2 then

γ1 · · ·γn = 0,

while if p > 2, then

β(γ1) · · ·β(γn) = 0,

where β is the Bockstein map.

In the case that G is a dihedral group of order 8, we know that η1η2 = 0 where η1

and η2 are the two degree one elements as in the last section. So Serre’s Theorem istrue in this case.

Suppose first that p = 2 and that ζ ∈ H1(G, k) where k = F2. Then ζ asan element in Ext1

kG(k, k) is represented by an exact sequence of the form in thebottom row of the diagram

0 // Ω(k) //

ζ′

kGε

//

θ

k //

0

0 // k // U // k // 0,

where ζ ′ represents ζ and U is the pushout. Now note that U has dimension 2.Hence (Rad kG)2 is in the kernel of θ. Recall also that

xy − 1 ≡ (x− 1) + (y − 1) mod(Rad(kG))2.

24 JON F. CARLSON

So the set x : x ∈ G | θ(x − 1) = 0 is actually a maximal subgroup H of G.

Moreover, U ∼= k↑GH is the induced module from H. As a consequence, we see thatζ is represented by the sequence

0 // k // k↑GH// k // 0.

Now suppose that p > 2. For our purposes it is not necessary to understandthe Bockstein map. Rather we only want to know the extensions in Ext2

kG(k, k) =H2(G, k) that represents the Bockstein of an element of degree one. First noticethat H1(G, k) ∼= Hom(G, k) by an argument that is almost precisely the same asthat in the case that p = 2. In the same way as before, if ζ ∈ H1(G, k) then ζ isrepresented by a sequence

0 // k // k↑GH /((Rad kG)2 · k↑GH ) // k // 0,

for the corresponding maximal subgroup H. The Bockstein of γ is represented bythe sequence

0 // k // k↑GH(x−1)

// k↑GH// k // 0

Exercise 5.8. Suppose that G = 〈x | x9 = 1〉 and H = 〈x3〉 is the unique maximalsubgroup of G. Show that the sequence

0 // k // k↑GH(x−1)

// k↑GH// k // 0

represents the zero element in H2(G, k).

With the above analysis, we see that Serre’s Theorem is equivalent to the follow-ing. The point is that the product of the cohomology elements is represented by thesplice of the sequences.

Theorem 5.9. Suppose that G is a p-group which is not elementary abelian. Thenthere is a sequence of maximal subgroups H1, . . . , Hn of G and an exact sequence

E : 0 // k // Cn−1// · · · // C1

// C0// k // 0

such that

(1) Ci ∼= k↑GHifor i = 1, . . . , n, and

(2) the class of E in ExtnkG(k, k) is zero.

Proof of the Main Theorem: Construction of the module U ∼= k ⊕ V . Wedo this part only for the special case that G = D8, the dihedral group of order 8.The reader should be able to see how the general case would proceed. The notationis the same as in the last section. In particular, let G = 〈x, y|x2 = y2 = (xy)4 = 1〉.Let E = 〈x, (xy)2〉 and F = 〈y, (xy)2〉 be the two elementary abelian subgroups ofG. We have an exact sequence

E : 0 // k // k↑GE// k↑GF

// k // 0


which represents the element η1η2 = 0 in H2(G, k) = Ext2kG(k, k). This is the splice

of the two sequence representing η1 and η2. Next we let C be the complex obtainedby truncating the two ends (the copies of k) off of the ends of the sequence E . Thusthe complex C has the form

0 // C1// C2

// 0

where C1∼= k↑GE and C2

∼= k↑GF . Then the homology of C is given by

Hn C =

k if n = 0, 1,

0 otherwise.

Now let (P∗, ε) be a projective resolution of k:

(P∗, ε) : . . . // P2∂2

// P1∂1

// P0ε

// k // 0.

We consider the complex (P ⊗ C)∗

. . . //

P2 ⊗ C1

⊕P3 ⊗ C0

//

P1 ⊗ C1

⊕P2 ⊗ C0

//

P0 ⊗ C1

⊕P1 ⊗ C0

// P0 ⊗ C0// 0

Note that the complex

C(0)∗ : 0 // 0 // C0

// 0

is a subcomplex of C∗. Also the complex

C(1)∗ : 0 // C1

// 0 // 0

is a quotient complex of C∗. That is, we have a exact sequences

0 // C(0)∗

// C∗ // C(1)∗

// 0

0 // (C(0) ⊗ P )∗ // (C ⊗ P )∗ // (C(1) ⊗ P )∗ // 0

of complexes. Now we notice that

H∗(C(0) ⊗ P ) =

H0(C(0)∗ ) ⊗ H0(P∗) ∼= k↑GE if n = 0

0 otherwise

and

H∗(C(1) ⊗ P ) =

H1(C(1)∗ ) ⊗ H0(P∗) ∼= k↑GE if n = 1

0 otherwise

In particular, we see that (C(0)⊗P )∗ is a projective resolution of k↑GE , and (C(1)⊗P )∗is projective resolution of k↑GF except that it is shifted by one degree.

26 JON F. CARLSON

Now let Γ∗ = Γ∗(C ⊗ P ) be the truncation of the complex (C ⊗ P )∗ at degree 2.That is, Γ∗ has the form

. . . //

P3 ⊗ C1

⊕P4 ⊗ C0

//

P2 ⊗ C1

⊕P3 ⊗ C0

//

P1 ⊗ C1

⊕P2 ⊗ C0

// 0.

Notice that for n > 2, Hn(Γ) = Hn(C ⊗ P ) = 0. So Γ∗ is a shift by two degrees of aprojective resolution of H2(Γ∗). Let U = Ω−2(k)⊗H2(Γ∗). It remains to prove thatU has the properties that we want.

Proof of the Main Theorem: The filtration on U . A slight variation on ourprevious observation reveals that Γ′

∗ = Γ∗(C(0) ⊗ P ) is a subcomplex of Γ∗, and

moreover the quotient complex (Γ/Γ′)∗ is isomorphic to Γ′′∗ = Γ∗(C

(1) ⊗ P ). So wehave an exact sequence of complexes

0 // Γ′∗

// Γ∗// Γ′′

∗// 0

and a corresponding long exact sequence on homology

0 // H2(Γ′) // H2(Γ) // H2(Γ

′′) // 0

Next thing that we should observe is that H2(Γ′) ∼= Ω2(k↑GF ) ⊕ (proj) and that

H2(Γ′′) ∼= Ω1(k↑GE ) ⊕ (proj). Hence we have an exact sequence

0 // Ω2(k↑GF ) ⊕ (proj) // H2(Γ) // Ω1(k↑GE ) ⊕ (proj) // 0

Then tensoring with Ω−2(k) we get

0 // k↑GF ⊕ (proj) // U // Ω−1(k↑GE ) ⊕ (proj) // 0

Finally, we observe that Ω−1(k↑GE ) ∼= (Ω−1(kE))↑G. Thus we have the filtration thatwe want.

Proof of the Main Theorem: Verifying that U ∼= k⊕V . The fact that η1η2 = 0means that the sequence E is equivalent to the split sequence

0 // kId

// k0

// kId

// k // 0

That in turn means that the complex is C∗ is equivalent to a split complex D∗ ofthe form

0 // k0

// k // 0

by a sequence of chain maps that induce isomorphisms on homology. For the D∗ wecan prove the following.

Lemma 5.10. Suppose that the complex Γ∗(D ⊗ P ) is as above. Then

H2(Γ(D ⊗ P )) ∼= Ω2(k) ⊕ Ω1(k) ⊕ (proj)

.


Proof. The point is that (D⊗P )∗ is the direct sum of (D(0) ⊗P )∗ ∼= P∗ and (D(1) ⊗P )∗ ∼= P∗[1], where by P∗[1] we mean the resolution P∗ shifted one degree. Hencethe homology, after applying the trunction, Γ is precisely as stated.

It remains to prove that equivalences of complexes do not subtantially change theoutcome of our process. To this end, suppose that we have a complex U∗ and a chain

map φ : U∗// C∗ of the form

U∗

φ

: 0 //

U1//

φ1

U0//

φ0

0

C∗ : 0 // C1// C0

// 0.

which induces an isomorphism on homology. Then we obtain a chain map

φ⊗ 1 : (U ⊗ P )∗ // (C ⊗ P )∗

which also induces an isomorphism on homology. Unfortunately, it is not necessarythat the induced chain map

Γ(φ⊗ 1) : Γ∗(U ⊗ P ) // Γ∗(C ⊗ P )

induce an isomorphism on homology. However, it does induce an isomorphism inthe stable category, and this is the way we should view these objects.

From a practical standpoint we need to alter to (U ⊗P )∗ by adding a totally split

complex of projective modules so that the resulting map φ : (U ⊗ P )∗ // (C ⊗ P )∗is a surjection. This is accomplished as follows.

Suppose we have a chain map between two complexes θ∗ : I∗// J∗ such that

in degree n, the module map θn : In // Jn fails to be surjective. Then we find a

projective module Q with the property that there is a map α : Q // Jn such that

(θn, α) : In ⊕Q // Jn is surjective. It sufficient to choose Q to be a projective

cover of the cokernel of θn. Now alter the complex I∗ be taking direct sum with the

exact complex Q // Q , and alter θ as in the diagram

. . . // In+1// In ⊕Q

(∂,Id)//

(θn,α)

In−1 ⊕Q //

(θn−1,∂′α)

In−2// . . .

. . . // Jn∂′

// Jn−1// . . .

Assuming that we have followed the above proceedure whenever necessary, wehave now a surjective chain map

φ′ : V∗// (C ⊕ P )∗

where V∗∼= (U ⊗P )∗⊕Q∗ where Q∗ is a totally split complex of projective modules.

The important thing is that Hn(Γ(V)) ∼= Hn(Γ(U ⊗P ))⊕Hn(Γ(Q)), and the homol-ogy Hn(Γ(Q)) is zero except possibly in degree 2 where it is a projective module. In

28 JON F. CARLSON

particular, we have that

Hn(Γ(V)) ∼= Hn(Γ(U ⊗ P )) ⊕ (proj)

In addition, the chain map φ′ : V∗// (C ⊗ P )∗ induces an isomophism on ho-

mology. Let K∗ denote the kernel of φ′. We see that the terms in K∗ are projectivemodules. There is a long exact sequence on homology

. . . // H2(K) // H2(V)φ′∗// H2(C ⊗ P ) // H1(K) // . . .

Because φ′ induces an isomorphism on homology, we must have that Hn(K) = 0 ofall n. Thus, K∗ is an exact complex of projective modules. It follows that Hn(Γ(K))is zero except possibly in degree 2 where it might be a projective module. Hence bythe same arguments as before, we have that

H2(V) ∼= H2(Γ(C ⊗ P )) ⊕ (proj) ∼= H2(Γ(U ⊗ P )) ∼= Ω2(k) ⊕ Ω(k) ⊕ (proj) .

So finally, we can see that

U = Ω−2(H2(Γ(U ⊗ P ))) ∼= k ⊕ Ω−1(k) ⊕ (proj),

as was to be proved.

Remarks on the general proof. In a full proof of the theorem we would followthe lines of what we have done for the example of the dihedral group. Usually, thefiltration would be more complicated because the complex C∗ would have more thantwo terms. So the filtration would have to be demonstrated by a sequence of exactsequences of complexes.

In addition, the argument of the example would be only the induction step inthe argument of a general proof. The point is that the arugment only shows thatthe trivial module k is a direct summand of a module which can be filtered bymodules induced from maximal subgroups of G. In the case that G ∼= D8, themaximal subgroups are elementary abelian and the proof is complete. In a moregeneral situation, the maximal subgroup H might not be elementary abelian. Thenwe must assume by induction that the theorem is true for the maximal subgroupand replace that section of the complex by a module which is filtered by modulesinduced from the elementary abelian subgroups of H. Each time this is done themodule V in the theorem, aquires new direct summands. For a complete version ofthe proof see the original paper [C2] or the book [10].

6. Quillen’s Dimension Theorem

The object of this section is to prove a form of Quillen’s Dimension Theorem andsee some of its consequences. The statement of the theorem concerns varieties ormaximal ideal spectra of rings. But the proof is really about the structure of thering. Indeed, in the proof we see something of the connection between the ringstructure and the structure of its variety.

To state and prove the main theorem of this section, we adopt the notation of thelast section. From the statement of Theorem 5.1 we know that there is a natural


number τ , a finite dimensional kG-module U and a sequence of elementary abeliansubgroups E1, . . . , Eτ such that there is a filtration

0 = L0 ⊆ L1 ⊆ . . . Lτ = k ⊕ V

where for each i, Li/Li−1∼= W ↑G

i for some kEi-module Wi. The proof of some of thecorollaries involved tensoring everything in this system with another module M anddeducing thatM is a direct summand of a module that is filtered by modules that arethe inductions of restrictions ofM to elementary abelian subgroups. It is noteworthythat the sequence of subgroups does not change from the sequence E1, . . . , En ofelementary abelian p-subgroups in the theorem. Hence, we can consider these itemsas invariants of the group G, although their selection is certainly not unique. Whatfollows is the theorem that we need.

Theorem 6.1. Suppose that we have homogeneous elements ζ1, . . . , ζτ ∈ H∗(G, k)with the property that resG,Ei

(ζi) = 0. Then ζτ · · · ζ2ζ1 = 0.

Proof. We know that there exists a kG-module V such that U = k⊕V has a filtration

0 ⊆ L0 ⊆ L1 ⊆ . . . ⊆ Lτ = U

where Li/Li−1∼= W ↑G

i for some kEi-module Wi. Now for any m, Hm(G, k) =ExtmkG(k, k) is embedded in ExtmkG(U, U) as a direct summand. So we may identifyζi with its inclusion into Extmi

kG(U, U), wheremi is the degree of ζi. Moreover, becausethe sums are direct, if we show that the product ζn · · · ζ1 is zero as an element ofExt∗kG(U, U), then we are done.

For each i we have an exact sequence

0 // Li−1ji

// Liqi

// W ↑Gi

// 0

The long exact sequence in cohomology, is natural with respect to multiplication bycohomology elements. So we have a diagram

. . . // ExtrkG(U, Li−1)(ji)∗

//

ζi

ExtrkG(U, Li)(qi)∗

//

ζi

ExtrkG(U,W ↑Gi ) //

ζi

. . .

. . . // Extr+mi

kG (U, Li−1)(ji)∗

// Extr+mi

kG (U, Li)(qi)∗

// Extr+mi

kG (U,W ↑Gi ) // . . .

where the vertical maps are right multiplication by ζi. In addition, there is a com-mutative diagram

ExtrkG(U,W ↑Gi )

∼=//

ζi

ExtrkEi(U,Wi)

resG,Ei(ζi)

Extr+mi

kG (U,W ↑Gi )

∼=// Extr+mi

kEi(U,Wi)

where the horizontal maps are isomorphisms. The horizontal isomorphisms arecertified by the Eckmann-Shapiro Lemma (see below). By the hypothesis of thetheorem, we know that resG,Ei

(ζi) = 0. Therefore we must have that multiplication

by ζi is zero on ExtrkG(U,W ↑Gi ).

30 JON F. CARLSON

Now we can complete the proof of the theorem. Consider first the case i = τ (soLi = U) and r = 0. From the two diagrams above, we conclude that (qτ )∗(IdU ·ζτ) =0 and hence IdU ·ζτ = ζτ = (jτ )∗(φτ) for some φτ ∈ Extmτ

kG(U, Lτ−1). Now assumeby induction that ζτ · · · ζi+1 = (jτ )∗ · · · (ji+1)∗(φi+1) for some φi+1 in ExtrkG(U, Li),where r = mi+1 + . . .+mτ . Then we have that

ζτ · · · ζi = ((jτ )∗ · · · (ji+1)∗φi+1)ζi = (jτ )∗ · · · (ji+1)∗(φi+1ζi).

It is also true that φi+1ζi = (ji)∗φi for some element φi ∈ Extr+mi

kG (U, Li−1). If wepersist along this path, we eventually prove that

ζτ · · · ζ1 = (jτ )∗ · · · (j1)∗(φ1)

for some φ1 ∈ ExtskG(U, L0). As L0 = 0, the product is 0. This finishes theproof.

Exercise 6.2. Suppose that M is a kG-module and N is a kH-module where H isa subgroup of G. Prove that

HomkG(M,N↑G) ∼= HomkH(MH , N)

and also that

ExtnkG(M,N↑G) ∼= ExtnkH(MH , N)

for any n. Hint: For f : MH// N a kH-homomorphism, let ψ(f) : M // N↑G

be defined by ψ(f)(m) =∑

gH g ⊗ f(g−1m). Likewise, for f : M // N↑G define

θ(f) : MH// N by θ(f)(

∑

gH g ⊗ mg) = f(m1). The sums should be over a

complete set of representative of the left cosets of H in G. And we always assumethat the representative of the identity coset H is the element 1 ∈ G.

Corollary 6.3. Suppose that I is the ideal in H∗(G, k) consisting of all elementsζ having the property that the restriction resG,E(ζ) = 0 for all elementary abelianp-subgroups E of G. Then I τ = 0.

That is the set of all elements in H∗(G, k) which vanish on restriction to everyelementary abelian subgroup of G is a nilpotent ideal. Because, H∗(G, k) is finitelygenerated as a k-algebra, it is noetherian as a ring and its Jacobson radical isnilpotent. The ideal I in the corollary is clearly in the Jacobson radical. If E is anelementary abelian p-subgroup of G, then H∗(E, k)/Rad(H∗(E, k)) is a polynomialring. Consequently we have the following.

Corollary 6.4. The Jacobson radical of H∗(G, k) is the ideal consisting of all ζ suchthat resG,E(ζ) ∈ Rad(H∗(E, k)) for all elementary abelian p-subgroups E of G.

Actually we can say more. Let E1, . . . , Er be a complete set of representatives ofthe conjugacy classes of maximal elementary abelian p-subgroups of G. Now if E isany elementary abelian p-subgroup then E is conjugate to a subgroup of some Ei

and the restriction from G to E factors through the restriction map from G to Ei.Consequently, the Jacobson radical can be characterized as the intersection of the


kernels of the composition of the restriction map to Ei with the quotient map ontoH∗(Ei, k)/Rad(H∗(Ei, k)). So let pi be the kernel of the composition

H∗(G, k) // H∗(Ei, k) // H∗(Ei, k)/Rad(H∗(Ei, k)).

Then we can prove the following.

Corollary 6.5. The ideals pi are the minimal primes in H∗(G, k). Moreover

Rad(H∗(G, k)) = ∪ri=1pi.

The ideals pi are prime because the image of H∗(G, k) in H∗(Ei, k)/Rad(H∗(Ei, k))is an integral domain. It takes a bit more work to show that these are the minimalprimes. We leave this to the reader to ponder.

Next we want to restate all of this in terms of the geometry. For this we needsome commutative algebra. In particular, the following is essentially Theorem 9.3and Exercise 9.3 of Matsumora’s book [Mats].

Theorem 6.6. Let A and B be finitely generated graded commutative algebras overa field k, with A ⊆ B. We assume that B is finitely generated as a module over A.Then for any prime ideal p in A we have that

(1) There exists a prime ideal of B lying over p and(2) B has only a finite number of prime ideals that lie over p.

Quillen’s Theorem is the following.

Theorem 6.7. Suppose that E1, . . . , Er are a complete set of representatives of theconjugacy classes of maximal elementary abelian p-subgroups of the finite group G.Then

VG(k) =t

⋃

i=1

res∗G,Ei(VEi

(k))

In particular, each of the maps res∗G,Eiis finite-to-one on varieties and the images

res∗G,Ei(VEi

(k)) are the irreducible components of VG(k).

Proof. If α ∈ VG(k) is a maximal ideal of the ring H∗(G, k), then one of the minimalprimes pi must be contained in α for some i. Therefore, res∗G,Ei

(α) is a maximalideal within the image resG,Ei

(H∗(G, k)) in H∗(Ei, k). Because H∗(Ei, k) is a finitelygenerated module over resG,Ei

(H∗(G, k)), the previous theorem applies. Hence, thereare finitely many prime ideals γ1, . . . , γr in VEi

(k) that lie over α. Finally, sinceH∗(Ei, k)/γj is finitely generated as a module over the field H∗(G, k)/α, then eachγj must be a maximal ideal.

7. Properties of support varieties

In this section, we explore and justify a couple of the properties of support varietiesthat we want for the next section. In view of the fact that we have omitted theproof of so many other things, the proofs of these properties can be considered tobe complete. Our aim is not a complete verification of the results, but rather wewant to give some flavor of the type of proofs that are possible in the area.

32 JON F. CARLSON

Recall that the variety of kG-module M is the variety VG(M) = VG(J(M)) ⊆VG(k) of the ideal J(M) which is the annihilator in H∗(G, k) of the cohomology ringExt∗kG(M,M) of M . That is, it is the closed set of all maximal ideals in H∗(G, k)that contain J(M). Using methods that are very similar to those of the last section,we can prove the follow theorem. This theorem was first proved by Alperin andEvens [AE2] and independently by Avrunin [Av].

Theorem 7.1. If M is a finitely generated kG-module, then

VG(M) =⋃

E∈A

res∗G,E(VE(M))

where A is the set of elementary abelian p-subgroups of G.

The proof follows the same ideas as that of Quillen’s Theorem 6.7. We need toshow that a homogeneous element ζ in H∗(G, k) annihilates Ext∗kG(M,M) if andonly if its restriction every elementary abelian subgroup E of G annihilates thecohomology of ME.

The theorem shows that we can measure the variety of a module M by lookingat its restrictions. An example of a place where we might want to do this is inthe computation of the varieties of the modules Lζ . The modules Lζ are defined as

follows. Let ζ be a nonzero element of Hm(G,K) and suppose that ζ : Ωm(k) −→ k

is a cocycle representing ζ. The module Lζ is defined to be the kernel of ζ. Tobe consistent, we let Lζ = Ωm(k) ⊕ Ω(k) in the case that ζ is the zero element ofHm(G, k). In any event, we have an exact sequence

Eζ : 0 // Lζ // Ωm(k) ⊕ (proj)ζ

// k // 0.

If ζ = 0, then the projective direct summand (proj) can be assumed to be theprojective cover of k. Otherwise, (proj) is the zero module. The support variety ofLζ is given in the following.

Proposition 7.2. Suppose that ζ ∈ Hm(G, k). Assume that m is even if p > 2.Then VG(Lζ) = VG(ζ) is the set of all maximal ideals that contain ζ.

Proof. Let α be any element of VG(k). Then there exists an elementary abelian p-subgroup E = 〈x1, . . . , xr〉, such that α contains the kernel of the restriction to kE.

That is, the homomorphism α : H∗(G, k) // k factors through the restriction to

H∗(E, k). Now we use the rank variety of the module (see 3.5). We know that thereis a cyclic shifted subgroup U = 〈uα〉 where

uα =r

∑

i=1

αi(xi − 1)

such that α factors through restriction to U . If α /∈ VG(ζ), then resG,U(ζ) is notnilpotent, and hence it is not zero. By assumption, if p > 2, then m must be even.Therefore, whether p is even or odd, Ωm(kU) ∼= k, and Ωm(k)↓U ∼= kU⊕(proj). Thus,the sequence Eζ must split on restriction to U because resG,U(ζ) 6= 0. Therefore Lζis free as a kU -module and α /∈ VG(Lζ). If, on the other hand, α /∈ VG(Lζ), then(Lζ)↓U is free, resG,U(ζ) 6= 0 and α /∈ VG(ζ).


Next we turn our attention to the connectedness theorem. It says that we canunder some circumstances tell if a module is indecomposable by knowing its variety.For the proof of this theorem, we need the Tensor Product Theorem for supportvarieties. We assume that it is true without proof.

Theorem 7.3. Suppose that M and N are kG-modules. Then

VG(M ⊗N) = VG(M) ∩ VG(N).

We also need a theorem on the annihilation of cohomology.

Proposition 7.4. Let ζ ∈ Hn(G, k) and let M be a finitely generated kG-module.Then ζ annihilates Ext∗kG(M,M) if and only if

M ⊗ Lζ ∼= Ω(M) ⊕ Ωn(M) ⊕ (proj) .

Proof. The element ζ as an element of ExtnkG(k, k) ∼= Ext1kG(Ωn−1(k), k) is repre-

sented by the sequence

Eζ : 0 // k // Ω−1(Lζ) // Ωn−1(k) // 0.

This can be seen from the diagram

0

0

Lζ

Lζ

0 // Ωn(k) //

ζ′

Pn−1//

Ωn−1(k) // 0

0 // k //

Ω−1(Lζ) //

Ωn−1(k) // 0

0 0

where Pn−1 is the n − 1 term in the projective resolution of k and ζ ′ is a cocyclerepresentating ζ. Then we have that ζ · IdM ∈ ExtnkG(M,M) ∼= Ext1

kG(Ωn−1(M),M)is represented by the sequence

Eζ ⊗M : 0 // M // Ω−1(Lζ) ⊗M // Ωn−1(k) ⊗M // 0.

If ζ annihilates the cohomology of M , then ζ · IdM = 0 and the sequence Eζ ⊗Msplits. Hence the middle term

Ω−1(Lζ) ⊗M ∼= Ω−1(Lζ ⊗M) ⊕ (proj)

is the direct sum of the two end terms. Now we need only translate everything byΩ to complete the necessary condition of the proposition. Conversely if M ⊗ Lζ ∼=Ω(M) ⊕ Ωn(M) ⊕ (proj), then the sequence splits and ζ · IdM = 0.

34 JON F. CARLSON

Actually in the above analysis we also need to know if the middle term of an exactsequence is the direct sum of the two end terms, then the sequence splits. That is,we need the following.

Exercise 7.5. Suppose that R is a finite dimensional algebra over a field k and that

0 // A // B // C // 0

is an exact sequence of finitely generated R-modules such thatB ∼= A⊕C. Prove thatthe sequence splits. Hint: Show by a dimension argument that the connectinghomomorphism δ in the long exact sequence

0 // HomR(C,A) // HomR(C,B) // HomR(C,C)δ

// Ext1R(C,A) // . . .

is the zero map. Use this fact to get a splitting of the sequence.

Now we turn to the connectedness theorem. One implication of the theorem is thatthe variety of an indecomplable module is connected as a projective variety. Thatis, if we have two subvarieties V1 and V2 in VG(k), the statement that V1 ∩ V2 = 0is the same as saying that the projective varieties V 1 and V 2 intersect in the emptyset.

Theorem 7.6. Suppose that M is a finitely generated kG-module with the propertythat VG(M) = V1 ∪ V2 where V1 and V2 are nonzero closed subvarieties such thatV1 ∩ V2 = 0. Then M ∼= M1 ⊕M2 where VG(M1) = V1 and VG(M2) = V2.

Proof. First, notice that we can assume that neither V1 nor V2 is zero, and proceedby induction on the sum of the dimensions of V1 and V2. We should observe in thecourse of the proof that the minimal case in which both V1 and V2 have dimension1 is covered in the argument that follows.

For some n there exists an element ζ ∈ Hn(G, k) with the properties that V1 ⊆VG(ζ) and dim(V2 ∩ VG(ζ)) ≤ dimV2. That is, ζ can be chosen to be in the idealthat defines V1 but not in the ideal of any component of V2. Likewise chooseγ ∈ Hm(G, k) such that V2 ⊆ VG(γ) and dim(V1 ∩ VG(γ)) ≤ dim V1. We use

the symbol ζ to denote a cocycle ζ : Ωn(k) // k that represents the cohomol-

ogy class ζ. Let Ωn(γ) : Ωm+n(k) // Ωn(k) be a representative of the class of γ

in HomkG(Ωm+n(k),Ωn(k)) ∼= Hm(G, k). Then the composition ζ Ωn(γ) representsthe cup product ζγ.


Hence we have the diagram

0

0

E : 0 // Ωn(Lγ) // Lζγ ⊕Q //

Lζ //

0

0 // Ωn(Lγ) // Ωm+n(k) ⊕QΩn(γ)

//

ζγ

Ωn(k) //

ζ

0

k

k

0 0

with exact rows and columns. Here Q is a projective module which is added toΩm+n(k) to be certain that the map Ωn(γ) is surjective. Often it will be the zeromodule.

Now notice that VG(M) ⊆ VG(Lζγ) = VG(ζγ) = VG(ζ) ∪ VG(γ). Consequently,some power ζ tγt of ζγ annihilates the cohomology of M . Without loss of generality,we replace ζ by ζ t and γ by γt so that ζγ annihilates Ext∗kG(M,M).

Now consider the tensor product sequence E⊗M . We should observe that VG(Lζ⊗M) = V1∪ (V2∩VG(ζ)). Then by induction, Lζ⊗M ∼= X1⊕X2, where VG(X1) = V1

and VG(X2) = V2∩VG(ζ). Likewise, Ωn(Lγ)⊗M ∼= Y1⊕Y2 where VG(Y1) = V1∩VG(γ)and VG(Y2) = V2. Hence the sequence E ⊗M has the form

0 // Y1 ⊕ Y2// Ωm+n(M) ⊕ Ω(M) ⊕ (proj) // X1 ⊕X2

// 0.

To complete the proof, notice that Ext1kG(X1, Y2) = 0 and also that Ext1

kG(X2, Y1) =0. That is, Y2 can not extend X1 and Y1 can not extend X2. The only conclusionpossible is that E ⊗M is the direct sum of two exact sequences, an extension N1 ofX1 by Y1 and another extension N2 of X2 by Y2. So in the middle term we have

Ωm+n(M) ⊕ Ω(M) ⊕ (proj) ∼= N1 ⊕N2.

where VG(N1) ⊆ W1 and VG(N2) ⊆ W2. The Krull-Schmidt Theorem guaranteethat Ω(M) and Ωm+n(M) decompose as desired, and so does M .

8. The rank of the group of endotrivial modules

Recall that a kG-module is an endotrivial module if M ∗ ⊗M ∼= Homk(M,M) ∼=k ⊕ (proj). It is not difficult to see that if M is an endotrivial module, then M ∼=M0 ⊕ (proj) where M0 is an indecomposable endotrivial module. Hence we canput an equivalence relation on the set of endotrivial modules, by saying that M isequivalent to N if and only if there exist projective modules P and Q such thatM ⊕P ∼= N ⊕Q. Let [M ] denote the equivalence class of an endotrivial module M .The group of endotrivial modules consists of the set T (G) of equivalence classes of

36 JON F. CARLSON

endotrivial modules, with the operation

[M ] + [N ] = [M ⊗N ].

This is an abelian group and the inverse of a class [M ] is the class of the dual [M ∗].There are two important theorems which we need to assume here. The first result

which got the subject was proved by Dade [13] more than 25 years ago.

Theorem 8.1. Suppose that G is an abelian p-group and that M is an endotrivialkG-module. Then M ∼= Ωn(k) ⊕ (proj) for some n.

We know that if G is a cyclic p-group then Ω2(k) ∼= k. Also, if |G| = 2 thenΩ(k) ∼= k. So we get immediately that

Corollary 8.2. Let G be an abelian p-group then

T (G) =

0 if |G| = 2,

Z/2Z if G is cyclic and |G| > 2

Z otherwise.

The other theorem that we want was originally due to Puig [20], but it also followsfrom the complete classification [9] of endotrivial modules.

Theorem 8.3. Let G be any group. Let A be a complete set of representatives ofthe conjugacy classes of maximal elementary abelian p-subgroups of G. Then thekernel of the product of the restriction maps

∏

E∈A res∗G,E : T (G) //∏

E∈A T (E)

is a finite group.

For the proof of our theorem, we also want Alperin’s result [1] on the fusion ofelementary abelian subgroups.

Theorem 8.4. Suppose that E and F are two elementary abelian subgroups of Gboth having rank at least 3. Let M be an endotrivial kG-module and assume thatME

∼= Ωa(k) while MF∼= Ωb(k). Then a = b.

From now on we assume that the rank of G is at least 2. Let E1, . . . , En bemaximal elementary abelians subgroups of G with the following properties.

• If the rank of G is 2 then E1, . . . , En is a complete set of representatives ofthe conjugacy classes of maximal elementary abelian p-subgroups of G.

• If the rank of G is greater than 2, then E1, . . . , En−1 is a complete set ofrepresentatives of the conjugacy classes of maximal elementary abelian p-subgroups rank 2 of G and En has rank larger than 2.

Define the type of an endotrivial module M to be the n-tuple a1, . . . , an whereMEi

∼= Ωai(k)⊕(proj). From the theroems of Alperin and Puig we get the following.

Proposition 8.5. The map T (G) // Zn which takes [M ] to Type(M) has finite

kernel.


Consequently, the image of the Type map is a torsion free subgroup. of∏

T (Ei).Our objective is to prove that the cokernel of the Type map is finite. In other wordswe want the following. This theorem was first proved for p-groups by Alperin [1],where he showed that we could use relative syzygies to construct endotrivial moduleshaving sufficiently diverse types. The proof we give here is very different. For onething, we do not have to assume that G is a p-group. It shows that set of generatorsfor a torsion free subgroup of T (G) having the right rank can be constructed bycarving up the modules Ωn(k) for certain n.

Theorem 8.6. Assume that the p-rank of G is at least 2. Then the torsion-freerank of the group T (G) is the number n.

Proof. Notice that if n = 1, then there is nothing more to prove. So for the remainderof the proof assume that n > 1.

By Quillen’s Dimension Theorem, we know that

VG(k) =⋃

E∈C

res∗G,E(VE(k)).

If E has p-rank two, then VE(k) is a plane, and the intersection of res∗G,E(VE(k))with res∗G,F (VF (k)), for F not conjugate to E, is a subvariety W of dimension one.In fact, the intersection W is precisely the subvariety res∗G,Z(VZ(k)) where Z is asubgroup of order p in the center of some Sylow p-subgroup of G. Thus we canget that for some m > 0, there exists an element ζ ∈ Hm(G, k) with the propertythat VG(ζ), the set of all maximal ideals containing ζ, intersects W transversely.It is sufficient here to choose the element ζ so that resG,Z(ζ) is not nilpotent, orequivalently, so that resG,Z(ζ) 6= 0 and that m is even if p > 2. Assume that suchan element has been chosen.

Let ζ ′ : Ωm(k) → k be a cocycle that represents ζ. The kernel of ζ ′, Lζ , is a modulehaving support variety VG(Lζ) = VG(ζ). Then, the support variety is disconnected.That is, we have that

VG(Lζ) = V1 ∪ V2 ∪ · · · ∪ Vn

where

Vi =⋃

E∈Ci

res∗G,E(VE(Lζ)).

Notice here that when i < n and in any case that Ei is a maximal elementary abeliangroup of rank 2, we have that Vi = res∗G,Ei

(VEi(Lζ)) = res∗G,Ei

(VEi(resG,Ei

(ζ))).Moreover, because VG(Lζ) is transverse to W , it is necessary that Vi ∩ Vj = 0.Consequently, Lζ decomposes as

Lζ = L1 ⊕ L2 ⊕ · · · ⊕ Ln

where VG(Li) = Vi.

38 JON F. CARLSON

For any i < n, let V ′i = ∪j 6=iVj and let L′

i = ⊕j 6=iLj. Then we have a diagram

0

0

L′i

L′i

0 // Lζ //

Ωm(k)ζ′

//

k // 0

0 // Li //

Ni//

k // 0

0 0

where Ni is the pushout.We claim that Ni is an endotrivial module. For the proof it is only necessary to

show that the restriction of Ni to any maximal elementary abelian p-subgroup E ofG is an endotrivial module. Consider first the restriction to Ei. Because V ′

i , whichis the support variety of L′

i, intersects res∗G,Ei(VE(k)) in the set 0 we conclude

that (L′i)Ei

is a projective kE-module. Hence on restriction to the subgroup Ei, themiddle column splits, and (Ni)Ei

∼= Ωm(k) ⊕ (proj). Thus, (Ni)Eiis an endotrivial

module. On the other hand, suppose we consider Ej for j 6= i. Then by a similarargument, Li is projective on restriction to Ej. So this time the bottom row in thediagram splits on restriction to Ej, and we have that (Ni)Ej

∼= k ⊕ (proj).Hence, we have constructed a collection N1, . . . , Nn−1 of endotrivial modules.

Moreover, we know the restriction of any Ni to any maximal elementary abelian p-subgroup of G. In particular, we know that the Type of Ni is (0, . . . , 0, m, 0, . . . , 0).Therefore, we have that the classes of the modules Ω(k), N1, . . . , Nn−1 generate atorsion-free subgroup of T (G) that has rank n. This proves the theorem.

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Department of Mathematics, University of Georgia, Athens, Georgia 30602, USA

E-mail address : [email protected]