Math 99r - Representations and Cohomology of

Groups

Taught by Meng GeuoNotes by Dongryul Kim

Spring 2017

This course was taught by Meng Guo, a graduate student. The lectures weregiven at TF 4:30-6 in Science Center 232. There were no textbooks, and therewere 7 undergraduates enrolled in our section. The grading was solely based onthe final presentation, with no course assistants.

Contents

1 February 1, 2017 31.1 Chain complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Projective resolution . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 February 8, 2017 62.1 Left derived functors . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Injective resolutions and right derived functors . . . . . . . . . . 8

3 February 14, 2017 103.1 Long exact sequence . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 February 17, 2017 134.1 Ext as extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2 Low-dimensional group cohomology . . . . . . . . . . . . . . . . . 14

5 February 21, 2017 155.1 Computing the first cohomology and homology . . . . . . . . . . 15

6 February 24, 2017 176.1 Bar resolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176.2 Computing cohomology using the bar resolution . . . . . . . . . 18

7 February 28, 2017 197.1 Computing the second cohomology . . . . . . . . . . . . . . . . . 19

1 Last Update: August 27, 2018

8 March 3, 2017 218.1 Group action on a CW-complex . . . . . . . . . . . . . . . . . . . 21

9 March 7, 2017 229.1 Induction and restriction . . . . . . . . . . . . . . . . . . . . . . . 22

10 March 21, 2017 2410.1 Double coset formula . . . . . . . . . . . . . . . . . . . . . . . . . 2410.2 Transfer maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

11 March 24, 2017 2711.1 Another way of defining transfer maps . . . . . . . . . . . . . . . 27

12 March 28, 2017 2912.1 Spectral sequence from a filtration . . . . . . . . . . . . . . . . . 29

13 March 31, 2017 3113.1 Spectral sequence from an exact couple . . . . . . . . . . . . . . 3213.2 Spectral sequence from a double complex . . . . . . . . . . . . . 32

14 April 4, 2017 3414.1 Hochschild–Serre spectral sequence . . . . . . . . . . . . . . . . . 34

15 April 7, 2017 3715.1 Homology and cohomology of cyclic groups . . . . . . . . . . . . 37

16 April 11, 2017 3916.1 Wreath product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

17 April 14, 2017 4117.1 Cohomology of wreath products . . . . . . . . . . . . . . . . . . . 41

18 April 18, 2017 4418.1 Atiyah completion theorem . . . . . . . . . . . . . . . . . . . . . 4418.2 Skew-commutative graded differential algebras . . . . . . . . . . 46

19 April 21, 2017 4919.1 Cohomology of D2n . . . . . . . . . . . . . . . . . . . . . . . . . 4919.2 The third cohomology . . . . . . . . . . . . . . . . . . . . . . . . 50

20 April 25, 2017 5220.1 Swan’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5220.2 Grothendieck spectral sequence . . . . . . . . . . . . . . . . . . . 5320.3 Cech cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

2

Math 99r Notes 3

1 February 1, 2017

1.1 Chain complexes

In the context of and abelian category, we can define a chain complex.

Definition 1.1. An abelian category A is a category such that:

• HomA (A,B) is an abelian group,

• there exists a zero object 0 that is both initial and terminal, i.e., Hom(A, 0)and Hom(0, B) are always trivial,

• composition Hom(A,B)×Hom(B,C)→ Hom(A,C) is bilinear,

• it has finite products and coproducts (in this case they agree),

• for any φ : A → B there exists a kernel σ : K → A with the requireduniversal property,

• for any φ : A→ B there exists a cokernel σ : B → Q,

• every monomorphism is the kernel of its cokernel,

• every epimorphism is the cokernel of its kernel,

• any morphism φ : A → B factors like A → C → B so that A → C is anepimorphism and C → B is a monomorphism.

Example 1.2. The category of R-modules is an abelian category.

Definition 1.3. A functor F : A → B is called additive if Hom(A,B) →Hom(F (A), F (B)) is a group homomorphism.

Definition 1.4. A chain complex Cnn∈Z is a sequence of objects Cn ∈ Awith boundary maps dn : Cn → Cn−1 such that dn−1 dn = 0.

· · · d2−→ C1d1−→ C0

d0−→ C−1d−1−−→ C−2

d−2−−→ · · ·

Because we have the notion of a kernel, we can say that in a chain complex,there is a monomorphism im dn+1 → ker dn. We then define the homology as

Hn(C•, d•) = coker(im dn+1 → ker dn).

Definition 1.5. A cochain complex is (C•, d•) has maps dn : Cn → Cn+1

that satisfy dn+1 dn = 0. We likewise define the cohomology.

A morphism f• : (C•, d•)→ (D•, δ•) is a collection of maps fn : Cn → Dn

such that the diagram

Cn Cn−1

Dn Dn−1

fn

dn

fn−1

δn

Math 99r Notes 4

commute for all n. In this case, the chain map gives rise to maps

H•(f•) : H•(C•, d•)→ H•(D•, δ•)

on homology.Also there is an obvious equivalence between categories of chain complexes

and cochain complexes given by Cn 7→ C−n.

Definition 1.6. A chain homotopy between two maps f•, g• : (C, d)→ (D, δ)is a collection of maps hn : Cn → Dn+1 such that

hn−1d+ dhn = fn − gn.

1.2 Projective resolution

Definition 1.7. An object P ∈ C is called projective if you can always (maybenot uniquely) lift f to f making the diagram commute:

X

P Yf

f

A projective R-module is a projective object in the category ofR-modules.Note that this is equivalent to saying that HomR(P,−) is an exact functor.This is because HomR(P,−) is automatically a left exact functor and this saysprojectivity says that it preserves epimorphisms.

Definition 1.8. A projective resolution of A is a long exact sequence

· · · → Pn → Pn−1 → Pn−2 → · · · → P0 → A→ 0,

or equivalently a chain complex · · · → P1 → P0 with homology

Hn(P•) =

0 if n > 0

A if n = 0.

Definition 1.9. A positive complex is a complex (C•, d•) with Cn = 0 forall n < 0.

Definition 1.10. An positive acyclic chain complex is a complex (C, d) withCn = 0 for n < 0 and Hn(C) = 0 for n > 0.

So a projective resolution is positive chain complex that is both acyclic andprojective (i.e., consisting of projective objects).

Proposition 1.11. If (C, d) and (D, δ) are two positive chain complexes, (C, d)is projective, and (D, d0) is acyclic, then

(C•ϕ−→ D•) 7→ H0(C•)

H0(ϕ)−−−−→ H0(D•)

Math 99r Notes 5

induces a bijectionhomotopy classes of

chain maps C• → D•

←→ H0(C•)→ H0(D•).

Proof. So we have to prove two things: first that we can get maps on chainsfrom H0(C)→ H0(D), and that if two maps give the same maps then they arechain homotopic.

We are inductively going to lift maps.

Cn Cn−1 Cn−1

Dn Dn−1 Dn−2dn dn−1

Note that the composition Cn → Cn−1 → Dn−1 has image lying in the ker dn−1 =im dn. Then using that Dn → im dn is an epimorphism and that Cn is projec-tive, we can lift this map.

Now we show that two chain maps giving the same maps on homology ishomotopic. We may assume that f• gives the zero map H0(C) → H0(D) andshow that f• is null-homotopic. You can check this by lifting maps similarly.

Math 99r Notes 6

2 February 8, 2017

Last time we defined abelian categories, chain complexes and cochain complexes,associated homology and cohomology. Also we defined a projective resolutionof some object A ∈ A , which is basically a long exact sequence ending with· · · → A → 0 consisting of projective objects. For two objects A,B ∈ A , wesaw that the set of morphisms A → B is the homotopy classes of a chain mapfrom a projective complex to an acyclic complex.

Corollary 2.1. Any two projective resolutions P•, Q• of A are homotopic toeach other. In other words, there are chain maps φ : P• → Q• and ψ : Q• → P•such that φ ψ ∼ idQ and ψ φ ∼ idP .

Definition 2.2. We say that A has enough projectives if for any A ∈ Athere exists a projective P ∈ A and an epimorphism P → A.

This is equivalent to saying that any A ∈ A has a projective resolution,because we can inductively choose the projective that surjects onto the kernel.

Theorem 2.3. The category R−Mod has enough projectives.

2.1 Left derived functors

Suppose we have an additive functor F : A → B, i.e., functor that preservesthe additive group structure of Hom. If we apply F to a projective resolution

· · · → P2 → P1 → P0 → A→ 0,

we get a chain complex

· · · → F (P2)→ F (P1)→ F (P0)→ 0.

This is not in general exact, but it is a chain complex and so we can computeits homology. We define the n-th left derived functor associated to P• as

LP•n F (A) = Hn(F (P•)).

Why is it a functor? If there is a map A → B then we get a map betweenprojective resolutions, and then we get a map between homology.

Theorem 2.4. Suppose we have an additive functor F : A → B. Let P,Qbe two assignment of projective resolutions to objects of A . There exists acanonical natural isomorphism LPnF ' LQnF between the two functors.

The main idea is that there is a chain map PA → QA given by the identityidA and then this induces a Hn(PA)→ Hn(QA).

Lemma 2.5. An additive functor F : A → B induces an additive functorF : ChA → ChB, given by F (C•)n = F (Cn) and F (ψ•)n = F (ψn). Thissatisfies the following:

Math 99r Notes 7

(i) If Σ• : ψ• → ϕ• is a homotopy in two maps ψ•, ϕ• ∈ HomChA (C•, D•),then F (Σ•) : F (ψ•)→ F (ϕ•) is also a homotopy between F (ψ•), F (ϕ•) ∈HomChB(F (C•), F (D•)).

(ii) If C• ' D• are homotopic in ChA , then F (C•) ' F (D•) in ChB.

(iii) If F : A → B is an exact functor, then the induced functor F : ChA →ChB is also exact.

Proof. It is clear that F : ChA → ChB is an additive functor, because we canadd vertical maps one by one.

(i) This is clear, because F is additive and so the conditions translates ex-actly.

(ii) follows from (i).(iii) A sequence of chain complexes being exact is equivalent to the sequence

at each level being exact. So using exactness of F , we get the exactness of thesequences at each level after applying F .

Proof of Theorem 2.4. We already have maps idA between A. So we get twochain maps

PA A

QA A

ψ• idAϕ• idA

with ϕ ψ ' idPA and ψ ϕ ' idPA . Then applying F gives

F (ϕ) F (ψ) ' idF (PA), F (ψ) F (ϕ) ' idF (PA) .

This shows that H•(PA) ∼= H•(QA), and this does not depend on ϕ and ψbecause homotopic maps give exactly the same map on homology.

Naturality can be checked.

This theorem shows that the left derived functor LnF can be defined inde-pendent of the choice of the projective resolutions.

Example 2.6. For a fixed R-module B, take the functor

F : R−Mod→ R−Mod; A 7→ A⊗R B.

This gives left derived functors

TornR(A,B) = LnF (A),

and these are called the Tor functors.

Example 2.7. Let R be a ring and G be a (multiplicative) group. We definethe group ring as the ring

RG =

∑gi∈G

rigi : only finitely many gi are nonzero

Math 99r Notes 8

with the multiplication structure coming from the group relations (gi) · (gj) =(gigj).

Let M be an RG-module. This can also be thought of as an R-module Mwith an action of G, or an R-linear representation of G. We define the grouphomology

Hn(G,M) = TorRG(R,M),

where the RG-module structure on R is trivial, i.e., (∑rigi)r = (

∑ri)r.

Proposition 2.8. If F : A → B is a right exact functor, then there is acanonical natural isomorphisms L0F ∼= F .

Proof. The 0-th derived functor is defined as

L0F (A) = coker(F (P1)→ F (P0)).

If F is right exact, it preserves cokernels and so L0F (A) ∼= F (A).

Proposition 2.9. If P is projective, LnF (P ) = 0 for n > 0.

Proof. 0→ P → 0 is a resolution.

2.2 Injective resolutions and right derived functors

There is a dual of all the stuff we did so far. Recall that projective means

P

M N 0.

So we may define:

Definition 2.10. An injective object I is an object such that there is alwaysa lift

0 M N

P

This is same as saying that Hom(−, I) is exact.

Definition 2.11. An injective resolution of A is an exact sequence

0→ A→ I0 → I1 → I2 → · · · .

We can now define a right derived functor. For an additive functor F : A →B, its n-th right derived functor is defined as

RnF (A) = Hn(F (I•)).

Math 99r Notes 9

If F is left exact, then R0F ∼= F . Also if I is injective then RnF (I) = 0 forn > 0.

If F is a contravariant functor, then this can be thought of as a covariantfunctor F : A op → B. So you can do similar things. One thing to be careful isthat if P ∈ A is projective, then P ∈ A op is injective. For instance, to defineleft derived functors, you first need to pick an injective resolution and apply Fand take the homology.

Example 2.12. Let M be a fixed R-module, and consider the functor

HomR(M,−) : R−Mod→ R−Mod; N 7→ HomR(M,N).

The Ext functor is defined as

ExtnR(M,N) = Rn(HomR(M,−))(N).

Example 2.13. Similarly, the group cohomology for an R-module M isdefined as

Hn(G,M) = ExtnRG(R,M).

Example 2.14. There is also a contravariant functor HomR(−, N). So we mayalso define

ExtnR(M,N) = Rn(HomR(−, N))(M).

What is the different between the two definitions of ExtnR(M,N)? The firstdefinition is first finding an injective resolution 0 → N → I• and compute thecohomology of

0→ Hom(M, I0)→ Hom(M, I1)→ · · · ,

and the second definition is find a projective resolution P• → M → 0 and takethe cohomology of

0→ Hom(P0, N)→ Hom(P1, N)→ · · · .

They turn out the be equivalent definitions. People use the second definitionmore often because they like projective resolutions.

Math 99r Notes 10

3 February 14, 2017

3.1 Long exact sequence

Lemma 3.1. Suppose C is a chain complex in A . There is a unique morphismδn : coker ∂n+1 → ker ∂n−1 such that Cn → Cn−1 factors as

Cn → coker ∂n+1δn−→ ker ∂n−1 → Cn−1

and there exists a exact sequence

0→ Hn(C)→ coker ∂n+1δn−→ ker ∂n−1 → Hn−1(C)→ 0.

Proof. Let us first construct this map. We have an epimorphism coker ∂n+1 →im ∂n, and a monomorphism im ∂n → ker ∂n−1. So composing these give themap δn. This construction also shows that

ker δn = ker(coker ∂n+1 → im ∂n) = Hn(C),

coker δn = coker(im ∂n → ker ∂n−1) = Hn−1(C).

Theorem 3.2. If 0 → A′ → A → A′′ → 0 is a short exact sequence of chaincomplexes in an abelian category, then there is a long exact sequence

· · · Hn(A′) Hn(A) Hn(A′′)

Hn−1(A′) Hn−1(A) Hn−1(A′′) · · · .

Proof. This follows from the snake lemma on

0 0 0

Hn(A′) Hn(A) Hn(A′′)

coker ∂′n coker ∂n coker ∂′′n 0

0 ker ∂′n ker ∂n ker ∂′′n

Hn−1(A′) Hn−1(A) Hn−1(A′′)

0 0 0

Then we immediately get the desired sequence.

Math 99r Notes 11

Theorem 3.3. Let 0 → A′ → A → A′′ → 0 be a short exact sequence in Aand F : A → B be an additive functor. Suppose that A has enough projectives.Then there exists a map wn : LnF (A)→ Ln−1F (A′) such that

· · · LnF (A) LnF (A′′)

Ln−1F (A′) Ln−1F (A) · · ·

· · · L0F (A) L0F (A′′) 0

wn

is a long exact sequence.

Proof. We first look at the projective resolutions

0 P ′• P• P ′′• 0

0 A′ A A′′ 0.

But if we apply F , there is no reason for the sequence to remain exact. The keyidea is to make Pn = P ′n ⊕ P ′′n . We look at the first level.

0 P ′1 P ′1 ⊕ P ′′1 P ′′1 0

0 P ′0 P ′0 ⊕ P ′′0 P ′′0 0

0 A′ A A′′ 0

0 0 0

∂′ ∂ ∂′′

ε′ ε ε′′ε0

f g

We first lift the map P ′′0 → A using the fact that P ′′0 is projective and A→ A′′

is surjective. Then take ε(a′, a′′) = fε′(a′) + ε0(a′′). Let us look at the nextstep. Because g ε0 ∂′′ = 0, we see that there is some ν : P ′′1 → A′ such thatf ν = ε0 ∂′′. Then using surjectivity of P ′0 → A′ we can lift the map toλ : P ′′1 → P ′0 such that fε′λ = ε0∂

′′. Define

∂(a′, a′′) = (∂′(a′)− λ(a′′), ∂′′(a′′)).

Then we check that

ε∂(a′, a′′) = fε′(∂′(a′)− λ(a′′)) + ε0∂′′(a′′) = −fε′λ(a′′)− ε0∂′′(a′′) = 0

and further that this is exact. Continue this.

Likewise we have the long exact sequence coming from right derived functors.

Math 99r Notes 12

Theorem 3.4. We have a long exact sequence

0→ R0F (A′)→ · · · → RnF (A)→ RnF (A′′)→ Rn+1F (A′)→ Rn+1F (A)→ · · · .

Corollary 3.5. L0F is right exact and R0F is left exact.

For example, we have the exact sequence

· · · → TorR1 (A′, B)→ TorR1 (A,B)→ TorR1 (A′′, B)

→ A′ ⊗B → A⊗B → A′′ ⊗B → 0.

So Tor measures the non-exactness of the (−⊗B) functor.Likewise we have

0→ HomR(B,A′)→ HomR(B,A)→ HomR(B,A′′)

→ Ext1R(B,A′)→ Ext1

R(B,A)→ Ext1R(B,A′′)→ · · · .

We also have

0→ HomR(A′′, B)→ HomR(A,B)→ HomR(A′, B)

→ Ext1R(A′′, B)→ Ext1

R(A,B)→ Ext1R(A′, B)→ · · · .

One property is that

ExtnR

(⊕Ai, B

)=⊕

ExtnR(Ai, B), TorRn

(⊕Ai, B

)=⊕

TorRn (Ai, B).

Math 99r Notes 13

4 February 17, 2017

Today we are going to talk more about the Ext functor. We will give a inter-pretation

ExtnR(M,N) =

equivalence class of n-fold extensions

0→ N →Mn−1 → · · · →M0 →M → 0.

4.1 Ext as extensions

We say that two extension are equivalent if there are maps

0 N Mn−1 · · · M0 M 0

0 N Pn−1 · · · P0 M 0

0 N M ′n−1 · · · M ′0 M 0,

where Pi are projective. Reflexivity and symmetry are clear. To check transi-tivity, you look at the pullback of Pk →Mk ← P ′k.

Recall that

ExtnR(M,N) = Hn(HomR(P•, N)),

where P• → M is a projective resolution. Note that if you have a projectiveresolution, then you can construct maps and then the map Pn → N representsa cycle in Hom(Pn, N).

Pn+1 Pn Pn−1 · · · P0 M 0

0 N Mn−1 · · · M0 M 0

Note that this element is independent of the resolution P• and so this gives amap equivalence classes → ExtnR(M,N).

We now check that this map is bijective. Given an element [f ] ∈ Hn(HomR(P•, N)),how do we construct the extension? First we may assume that Pn → N is sur-jective by direct summing Rk for sufficiently large k to the projective resolution.Then we can set

Pn+1 Pn Pn−1 · · · P0 M 0

0 N Pn−1/ ker f · · · P0 M 0

f

In the same way, if two extensions come from the same element in ExtnR(M,N),then we can exhibit a equivalence.

Math 99r Notes 14

Take a short exact sequence 0→ N →M0 →M → 0. The equivalence classof this can be seen to be the image of idM in the map

0→ Hom(N,N)→ Hom(M,M0)→ Hom(M,M)→ Ext1R(M,N)→ · · · .

Then we get an element of Ext1R(M,N).

Let us see how this generalizes to the higher order. Given an exact sequence

0 → N → M1β−→ M0 → M → 0, we get two short exact sequences 0 → N →

M1 → E → 0 and 0→ E →M0 →M → 0, where E = imβ. Then we get longexact sequences coming from Hom(−, N),

· · · → Hom(N,N)→ Ext1R(E,N)→ · · · ,

· · · → Ext1R(E,N)→ Ext2

R(M,N)→ · · · .

The image of idN under the two maps gives an element of Ext2R(M,N). This

directly generalizes to higher n.There is an R-linear map

ExtnR(M,N)× ExtmR (N,L)→ Extn+mR (M,L)

which is called the Yoneda multiplication. How is this defined? Given twoextensions

0→ N →Mn−1 → · · · →M0 →M → 0,

0→ L→ Qm−1 → · · · → Q0 → N → 0,

we define the product as

0→ L→ Qm−1 → · · · → Q0 →Mn−1 → · · · →M0 →M → 0.

This gives a graded ring structure on Ext•R(M,M). Then Ext•R(M,N) has aExt•R(M,M)-module structure. Using this multiplication, we can say that inthe long exact sequence

0→ Hom(L,N)→ Hom(L,M0)→ Hom(L,M)→· · · → Extn(L,M)→ Extn+1(L,N)→ · · · ,

the connecting maps are multiplication by x, where x ∈ Ext1R(M,N) represents

the sequence 0→ N →M0 →M → 0.

4.2 Low-dimensional group cohomology

Recall that

H∗(G,M) = Ext∗RG(R,M).

Taking R = Z, we have

H0(G,M) = HomZG(Z,M) = HomG(Z,M) = MG.

Theorem 4.1. The first cohomology is

H1(G,M) =

M -conjugate classes of splitting group extensions

1→M →M oG→ G→ 1

.

Math 99r Notes 15

5 February 21, 2017

Last time we showed that ExtnR(M,N) is the equivalence classes of n-fold ex-tensions 0→ N →Mn−1 → · · · →M0 → N → 0.

5.1 Computing the first cohomology and homology

Today we are going to compute the first cohomology of groups. We claim that

H1(G,M) =

M -conjugacy classes of group extensions1→M → E → G→ 1 that are splitting

.

A group extension means that we have a short exact sequence, i.e., E/M ∼=G. The extension is said to split if there is a section s : G → E that is also agroup homomorphism. In this case, we can write E = M oG as a semi-directproduct. The multiplicative structure is something like

(m1, g1)(m2, g2) = (m1 + g1m2, g1g2).

Definition 5.1. A map d : G→M is a derivation if d(gh) = gd(h) + d(g).

Proposition 5.2. For any section s : G→M oG of the form s(g) = (d(g), g),the map s is a group homomorphism if and only if d is a derivation.

Proof. Just compute s(g)s(h) = (d(g) + gd(h), gh).

Now we need to compute the first cohomology. Note that there is a naturalZG-linear map ZG→ Z, and we can look at its kernel:

0→ IG→ ZG→ Z→ 0.

This gives a long exact sequence

0→ HomZG(Z,M)→ HomZG(ZG,M)→ HomZG(IG,M)

→ Ext1ZG(Z,M)→ Ext1

ZG(ZG,M)→ · · · .

Since ZG is a free ZG-module, we see that

0→MG →M → HomZG(IG,M)→ H1(G,M)→ 0.

Proposition 5.3. A map of sets d : G → M defines a group homomorphismδ : IG→M given by δ(g − 1) = d(g). This δ is a ZG-linear if and only if d isa derivation.

Proof. The map δ being G-equivariant means δ(h(g − 1)) = hδ(g − 1). Thismeans d(hg)− d(h) = hd(g).

Math 99r Notes 16

Now this shows that the HomZG(IG,M) = Der(G,M). (Der(G,M) has Gaction given by (gd)(h) = gd(h−1hg).) So we now have to study that mapM → Der(G,M). By definition, any element m ∈M maps to (−)m : IG→M .So an element m ∈M maps to the derivation d(g) = gm−m. So the conclusionis

H1(G,M) =Der(G,M)

(g 7→ gm−m) : m ∈M.

Definition 5.4. We say that two section s1, s2;G→M oG are M-conjugateif there exists an m ∈M such that (m, 1)s1(g)(m, 1)−1 = s2(g).

This is equivalent to d1(g) = d2(g)+m−gm. So we get our theorem. Thinkabout how this relates to the interpretation

H1(G,M) = 0→M → E → Z→ 0.

Recall that Hn(G,M) = TorZGn (Z,M).

Proposition 5.5. H1(G,Z) ∼= G/[G,G] = Gab.

Proof. We again take the sequence 0→ IG→ ZG→ Z→ 0. Then we get

TorZG1 (Z,ZG)→ H1(G,Z)→ Z⊗ZG IG→ Z⊗ZG ZG→ Z⊗ZG Z→ 0.

Then

TorZG1 (Z,ZG)→ H1(G,Z)→ Z⊗ZG IG→ Z id−→ Z→ 0.

Now note that Z ⊗ZG IG = IG/(IG)2. This can be seen to be isomorphic tothe abelianization G/[G,G].

Next time we are going to construct a space K(G, 1) such that π1(K) = Gand πi(K) = 0 for i > 1. Then we get H1(G,Z) = H1(K(G, 1)). So H1 is theabelianization of π1.

Math 99r Notes 17

6 February 24, 2017

6.1 Bar resolution

We note that free R-modules are always projective. Let

Fn = free R-module on (n+ 1)-tuples (g0, . . . , gn) ∈ Gn+1,

and we define the maps

∂n(g0, . . . , gn) =

n∑i=1

(−1)i(g0, . . . , gi, . . . , gn).

Then we get a sequence

· · · ∂3−→ F2∂2−→ F1

∂1−→ F0 → R→ 0.

We first claim that this is acyclic. Define the homotopy hn : Fn → Fn+1 by

hn(g0, . . . , gn) = (1, g0, . . . , gn).

Then you can check that ∂hn + hn−1∂ = id for n ≥ 1. So this acyclic.We can further give a free RG-module structure on each Fn. We are going

to give the action g(g0, . . . , gn) = (gg0, . . . , ggn). We use the bar notation

(g0, . . . , gn) = g0[g−10 g1|g−1

1 g2| · · · |g−1n−1gn],

[g1|g2| · · · |gn] = (1, g1, g1g2, . . . , g1g2 · · · gn).

Then Fn is a free RG-module generated by [g1| · · · |gn]. Using this notation, wecan write

∂n[g1| · · · |gn] = g1[g2| · · · |gn] +

n−1∑i=1

(−1)i[g1| · · · |gigi+1| · · · |gn]

+ (−1)n[g1| · · · |gn−1].

Now we have a free (and hence projective) resolution F• → R → 0 and sowe have

Hi(G,M) = Hi(M ⊗RG F•),Hi(G,M) = Hi(HomRG(F•,M)).

Here, note that the cochain complex for cohomology is

0→ HomRG(F0,M)d0−→ HomRG(F1,M)

d1−→ · · ·

where the cochain complex are

Ci(G,M) = HomRG(Fi,M) = Fun(Gn,M)

Math 99r Notes 18

and the coboundary maps

(dnf)([g1| · · · |gn+1]) = f ∂n[g1| · · · |gn+1]

= g1f([g2| . . . |gn+1]) +

n∑i=1

(−1)if([g1| . . . |gigi+1| . . . |gn+1])

+ (−1)n+1f([g1| . . . |gn]).

6.2 Computing cohomology using the bar resolution

In particular, we have

C0(G,M) = HomRG(RG〈[]〉,M) ∼= M.

Then for the generator a ∈ C0(G,M), we have

(d0a)([g]) = ga− a.

Given a function f ∈ C1(G,M), we have

(d1f)([g1|g2]) = g1f([g2])− f([g1g2]) + f([g1]).

Note that d1f = 0 is equivalent to f([g1g2]) = g1f([g2])+f([g1]). This is exactlysaying that f ∈ Der(G,M). This again shows

H1(G,M) =Der(G,M)

ga− a.

Let us now compute H2(G,M). We have for f ∈ C2(G,M),

(d2f)([g1|g2|g3]) = g1f([g2|g3])− f([g1g2|g3]) + f([g1|g2g3])− f([g1|g2]).

So we can write

H2(G,M) =(f : G×G→M) : d2f = 0

f(g1, g2) = g1α(g2)− α(g1g2) + α(g1) : (α : G→M).

Theorem 6.1. H2(G,M) is the equivalence classes of group extensions 1 →M → E → G→ 1 with the given G-action.

Let us first describe the bijection. Suppose we are given a group extension

1 M E G 1.i π

s

If g is a group homomorphism, then it splits. In general, we can find a functionf : G×G→M such that

i(f(g, h)) = s(gh)s(g)−1s(h)−1.

This function f must satisfy

s(ghk) = i(f(gh, k))s(gh)s(k) = i(f(gh, k))i(f(g, h))s(g)s(h)s(k)

= i(f(g, k))s(g)s(hk) = i(f(g, hk))s(g)i(f(hk))s(h)s(k).

Math 99r Notes 19

7 February 28, 2017

A group extension of G by M is a short exact sequence

1 M E G 1.i π

Since M is a normal subgroup of E, the group E acts on M by conjugation. IfM is abelian, we further get an action on G on M , induced from E, by liftingg ∈ G to g ∈ E and letting i(g · m) = gi(m)g−1. If we are already given anZG-module structure, we require the two actions to be the same.

We pick up from last time. The first cohomology H1(Z/2,F2) with the trivialaction is H1(Z/2,F2) = F2. Indeed there are two splitting extensions

1→ Z/2→ Z/2⊕ Z/2→ Z/2→ 1,

with two sections s1 : 1 7→ (1, 1) and s2 : 1 7→ (0, 1).

7.1 Computing the second cohomology

We claim that H2(G,M) is the equivalence classes of 1 → M → E → G → 1.Let us choose any (set) section s. (If s is a group homomorphism, then thegroup splits.) Then we should be able to write

s(gh)i(f(g, h)) = s(g)s(h)

for some i : G×G→M . Then the group extension structure can be completelyrecovered by f and the ZG-module structure on M , because

i(m)s(g)i(n)s(h) = i(m+ gn)s(g)s(h) = i(m+ gn+ f(g, h))s(gh).

But an arbitrary f does not work, because it needs to satisfy associativity. Wehave

((a, g)(b, h))(c, k) = (a+ gb+ f(g, h), gh)(c, k)

= (a+ gb+ f(g, h) + ghc+ f(gh, k), ghk),

(a, g)((b, h)(c, k)) = (a, g)(b+ hc+ f(h, k), hk)

= (a+ gb+ ghc+ gf(h, k) + f(g, hk), ghk).

So the condition we get is

gf(h, k)− f(gh, k) + f(g, hk)− f(g, h) = 0.

This is precisely d2f = 0, that is, f is in the 2-cocycle.We now need to quotient out by equivalence. Suppose we have a diagram

1 M E G 1

1 M E G 1.

i π

s1

s2

Math 99r Notes 20

Since π s1 = π s2 = idG, there is a function c : G→M such that

s1(g) = i(c(g))s2(g).

Then we have

i(f1(g, h) + c(gh))s2(gh) = i(f1(g, h))s1(gh) = s1(g)s1(h)

= i(c(g))s2(g)i(c(h))s2(h) = i(c(g) + gc(h))s2(g)s2(h)

= i(c(g) + gc(h) + f2(g, h))s2(gh).

So we get (f1 − f2)(g, h) = gc(h) − c(gh) + c(g). This means that the two areequivalent if and only if they differ by a coboundary.

Theorem 7.1. The second group cohomology is

H2(G,M) = equivalence classes of 1→M → E → G→ 1.

Actually I lied, because there is also the condition f(1, g) = 0 = f(g, 1). Butthis is fine because you can use the normalized bar resolution defined by

Fn =RG〈[g1| · · · |gn]〉

RG〈[g1| · · · |gn] : gi = 1 for some i〉.

Math 99r Notes 21

8 March 3, 2017

8.1 Group action on a CW-complex

We are now going to talk about group homology via topology. Suppose we havea cellular G-action on a CW-complex X. Assume that the action completelydiscontinuous and has no fixed points. Associated to X is a (singular or cellular)chain complex

· · · → C2(X;Z)→ C1(X;Z)→ C0(X;Z)→ 0,

of ZG-modules. If X ' ∗, then the two chain complexes C∗(X;Z) and Z(0) arequasi-isomorphic. This means that C∗(X;Z) is a free ZG-module resolution ofZ.

To compute H∗(G;Z), we need to take Z⊗ZG (−) on the chain complex andlook at the homology. We see that

Z⊗ZG Cn(X;Z) = Cn(X;Z)/〈gx− x : x ∈ Cn(X;Z)〉 ∼= Cn(X/G;Z).

The second equality holds because simplices are simply connected and so therealways is a lifting. We give a name for this:

K(G, 1) = BGdiscrete = X/G.

We have then have an exact sequence G → X → X/G and so π1(X) = G andπi(X) = 0 for i > 1. Whitehead’s theorem then tells us that K(G, 1) is uniqueup to homotopy.

Let us now give a construction for X. For each g0, . . . , gn ∈ G, you make asimplicial complex so that the faces of the n-simplex [g0, . . . , gn] ∈ Gn+1 are

∂[g0, . . . , gn] =

n∑i=0

(−1)i[g0, . . . , gi, . . . , gn].

This is indeed homotopy equivalent to a point, because you can just shrinkeverything to [e] linearly. The G-action is trivially g[g0, . . . , gn] = [gg0, . . . , ggn].

Example 8.1. We have K(Z, 1) = S1 and K(Z/2, 1) = RP∞.

We have K(G×H, 1) = K(G, 1)×K(H, 1) and

H∗(G,R) = H∗(K(G, 1), R).

So for instance, Hn(Z/2,F2) = F2 for all n. There is also a cup product

H∗(G,M)×H∗(G,M)`−→ H∗(G,M).

Math 99r Notes 22

9 March 7, 2017

We might have a presentation during the last lectures. Today we are going totalk about induction and restriction.

9.1 Induction and restriction

Consider an ring morphism Γ → Λ, so that Λ is an Γ-algebra. There are threethings we can do:

• Reduction: If M is a left Λ-module, then we can just consider M as a leftΓ-module. We write this as M ↓Γ or ResΛ

Γ M .

• Induction: If N is a left Γ-module, then we can construct tensor productΛ⊗ΓM as a left Λ-module and a right Γ-module. We write this as N ↑Γ Nor IndΛ

Γ N .

• Coinduction: If N is a left Γ-module, then we can construct HomΓ(Λ, N),which is a right Λ-module. We denote this as N ⇑Λ or CoindΛ

Γ N .

Lemma 9.1. For Γ and Λ rings, M a left Λ-module, A a Λ-Γ bimodule, andN a left Γ-module, there is a natural isomorphism

HomΓ(N,HomΛ(A,M)) ∼= HomΛ(A⊗Γ N,M).

In other words, A⊗Γ (−) is left adjoint to HomΓ(A,−).

Proof. We construct the maps φ : HomΓ → HomΛ as

φ(α)(a⊗ n) = α(n)(a),

and ψ : HomΛ → HomΓ as

(ψ(β)(n))(a) = β(a⊗ n).

You can check that this is a bijection and isomorphism as left Γ-modules.

Proposition 9.2. Let Γ → Λ be a ring morphism, M be a left Λ-module, N aleft Γ-module.

(i) HomΓ(N,ResΛΓ M) ∼= HomΛ(IndΛ

Γ ,M)

(ii) HomΓ(ResΛΓ M,N) ∼= HomΛ(M,CoindΛ

Γ N)

So ResΛΓ has a left adjoint IndΛ

Γ and a right adjoint CoindΛΓ .

Proof. For (i) use A = Λ, and for (ii) use switch the roles of Λ and Γ and useA = Λ.

Proposition 9.3. Suppose Γ → Λ and Λ is a projective Γ-module. Let M andN be as above.

(i) ExtnΓ(N,ResΛΓ M) ∼= ExtnΛ(IndΛ

Γ N,M)

Math 99r Notes 23

(ii) ExtnΓ(ResΛΓ M,N) ∼= ExtnΛ(M,CoindΛ

Γ N)

Proof. (i) Suppose we have a projective resolution P• → N → 0 over Γ. Thenby definition, ExtnΓ(N,ResΛ

Γ M) = Hn(Hom(P•,M)). Because Λ is projective,tensoring with Λ is exact, and we get a projective resolution Λ⊗ΓP• → Λ⊗ΓN →0. Then

ExtnΛ(IndΛΓ N,M) = Hn(Hom(Λ⊗Γ P•,M)) = Hn(Hom(P•,ResΛ

Γ M)).

(ii) You do the same thing. Restrict the projection resolution of M to over Γ.This is easier.

Now suppose you have a group G and a subgroup H ≤ G. Then there is aninclusion RH → RG. We see that RG is a free RH-module.

If G is finite, then there is an isomorphism

RG ∼= HomR(RG,R) = (RG)∗

as RG-RG bimodules. Then

IndGH N = RG⊗RH N = (RG)∗ ⊗RH N = HomRH(RG,N) = CoindGH N.

We further have, where we denote M∗ = HomR(M,R),

ResGHM∗ = (ResGHM)∗, IndGH N

∗ = (IndGH N)∗.

That is, the dual functor commutes with restriction and induction.

Proposition 9.4. For finite groups H ≤ G,

(i) HomRH(N,ResGHM) ∼= HomRG(IndGH N,M),

(ii) HomRH(ResGHM,N) ∼= HomRG(M, IndGH N),

(iii) ExtnRH(N,ResGHM) ∼= ExtnRG(IndGH N,M),

(iv) ExtnRH(ResGHM,N) ∼= ExtnRG(M, IndGHM).

If we apply (ii) to M = R, then we get

Hn(H,N) = ExtRRH(ResR,N) = ExtnRG(R, IndGH N) = Hn(G, IndGH N).

Example 9.5. Let us H∗(Z/2;F2 ⊕ F2) with the action being switching thetwo components. Note that this is simply F2[Z/2]⊗F2

F2. Then

H∗(Z/2,F2 ⊕ F2) = H∗(1,F2) =

F2 ∗ = 0,

0 ∗ > 0.

Math 99r Notes 24

10 March 21, 2017

Last time we proved

HomΓ(N,ResΛΓ M) ∼= HomΛ(IndΛ

Γ N,M),

HomΓ(ResΛΓ M,N) ∼= Hom(M,CoindΛ

Γ N),

and if Λ is projective over Γ, then the same thing is true for Ext. If H ⊆ G isa subgroup, then

H∗(H,M) ∼= H∗(G,CoindGHM), H∗(H,M) ∼= H∗(G, IndGHM).

10.1 Double coset formula

Consider two subgroups H,K ≤ G. Let us see what happens if we induce fromK to G and then restrict from G to H.

Theorem 10.1 (Mackey decomposition). Let M be an RK-module. Then

ResGH IndGKM =⊕HgK

IndHH∩gKg−1 ResgKg−1

H∩gKg−1 gM.

Here gM is same as M as abelian groups, but has the gKg−1-action (gkg−1)(gm) =gkm. Furthermore, if H is normal in G, then

ResGH IndGHM =⊕

g∈G/H

gM.

Proof. We have a decomposition

RG =⊕HgK

R(HgK)

as a RH-RK bimodule. Then

ResGH IndGKM = RG⊗RK M =⊕HgK

R(HgK)⊗RK M

=⊕HgK

RH ⊗R(H∩gKg−1) R(gK)⊗RK M

=⊕HgK

IndHH∩gKg−1 ResgKg−1

H∩gKg−1 gM.

Corollary 10.2. If M is an RK-module and N an RH-module, then

(i) IndGHM⊗RIndGH N =⊕HgK

IndGH∩gKg−1((ResgKg−1

H∩gKg−1 gM)⊗R(ResHH∩gKg−1 N))

(ii) HomR(IndGKM, IndGH N) =⊕HgK

IndGH∩gKg−1 HomR(ResgKg−1

H∩gKg−1 gM,ResHH∩gKg−1 N)

Math 99r Notes 25

(iii) IndGKM⊗RG IndGH N =⊕HgK

ResgKg−1

H∩gKg−1 gM⊗R(H∩gKg−1) ResHH∩gKg−1 N

(iv) HomRG(IndGH N, IndGKM) =⊕HgK

HomR(H∩gKg−1)(ResHH∩gKg−1 N,ResgKg−1

H∩gKg−1 gM).

Lemma 10.3. For N an RH-module and M and RG-module,

(i) IndGH(N ⊗R ResGHM) = IndGH N ⊗RM ,

(ii) IndGH HomR(N,ResGHM) = HomR(IndGH ,M).

Proof of Corollary 10.2. (i) We have, by the decomposition,

IndGKM ⊗R IndGH N = IndGH(ResGH IndGKM ⊗R N)

=⊕HgK

IndGH(IndHH∩gKg−1 ResgKg−1

H∩gKg−1 M ⊗R N)

=⊕HgK

IndGH∩gKg−1(ResgKg−1

H∩gKg−1 gM ⊗R ResHH∩gKg−1 N).

The other ones can be done similarly.

The third and fourth formulas work for Ext and Tor.

Corollary 10.4. For M and RK-module and N and RH-module,

(v) TorRG∗ (IndGKM, IndGH N) =⊕HgK

TorR(H∩gKg−1)∗ (ResgKg

−1

H∩gKg−1 gM,ResHH∩gKg−1 N),

(vi) Ext∗RG(IndGH N, IndGKM) =⊕HgK

Ext∗R(H∩gKg−1)(ResHH∩gKg−1 N,ResgKg−1

H∩gKg−1 gM).

10.2 Transfer maps

Given any group homomorphism α : H → G, and M a both RG-module andan RG-module, there are induced maps

H∗(H,M)α∗−−→ H∗(G,M), H∗(G,M)

α∗−−→ H∗(H,M).

If H ≤ G has finite index, there are canonical maps in the other direction:

TrGH : H∗(G,M)→ H∗(H,M), TrGH : H∗(H,M)→ H∗(G,M).

These are called transfer maps.Let us see how to construct this. Let us name our inclusion map α : H → G.

There is a canonical surjection RG ⊗RH M → M and a canonical injectionM → HomRH(RG,M). Let us first see how the maps α∗ and α∗ are defined.

H∗(G,RG⊗RH M) H∗(G,M)

H∗(H,M)

α∗

H∗(G,M) H∗(G,HomRH(RG,M))

H∗(H,M)α∗

Math 99r Notes 26

Now in the case when the index H ≤ G is finite, coinduction is the same asinduction. So we can apply the homology or cohomology in the other way.

H∗(G,RG⊗RH M) H∗(G,M)

H∗(H,M)TrGH

H∗(G,M) H∗(G,HomRH(RG,M))

H∗(H,M)TrGH

Math 99r Notes 27

11 March 24, 2017

Last time we defined the transfer map for [G : H] <∞ using induction. Thesemaps also can be defined directly.

11.1 Another way of defining transfer maps

For RG-modules M and M ′, there is a map

TrGH : HomRH(M,M ′)→ HomRG(M,M ′); φ 7→∑

g∈G/H

(gφ : m 7→ gφ(g−1m)).

Proposition 11.1. Let H ≤ G be a finite index subgroups.

(i) Let M1 be a RH-module and M2,M3 be RG-modules. If α ∈ HomRH(M1,M2)and β ∈ HomRG(M2,M3), then β TrGH(α) = TrGH(β α).

(ii) Under a similar condition, TrGH(β) α = TrGH(β α).

(iii) If H ≤ K ≤ G are finite index subgroups, TrGK TrKH = TrGH .

(iv) If H,K ≤ G and α ∈ HomRH(M1,M2), then

TrGK(α) =∑HgK

TrHH∩gKg−1(gα).

Given any projective resolution P• → M as an RG-module, it is also aprojective resolution as an RH-module. This gives a transfer map

HomRH(P•,M′)→ HomRG(P•,M

′).

Applying cohomology gives a map ExtnRH(M,M ′)→ ExtnRG(M,M ′). In partic-ular, letting M = R gives a map H∗(H,M)→ H∗(G,M).

You can do a similar thing for homology. For M a right RG-module and M ′

a left RG-module, there is a map

TrGH : M ⊗RGM ′ →M ⊗RH M ′; m⊗m′ 7→∑

g∈G/H

mg ⊗ g−1m′.

Proposition 11.2. Let H ≤ G be a subgroup of finite index.

(i) β TrGH(α) = TrGH(β α)

(ii) TrGH(β) α = TrGH(β α)

(iii) TrGK TrKH = TrGH

(iv) ResGH TrGK(α) =∑HgK

TrHH∩gKg−1 ResgKg−1

H∩gKg−1(gα) for α ∈ H∗(K,M).

(v) TrGH ResGK(α) =∑HgK

ResHH∩gKg−1 TrgKg−1

H∩gKg−1(gα) for α ∈ H∗(K,M).

Math 99r Notes 28

(vi) TrGH ResGH(α) = [G : H]α for α ∈ H∗(G,M).

On the zeroth homology and cohomology, there is a very explicit description.Recall that H0(G,M) = M/(gm−m). Then the transfer map is given by

TrGH : [m]G 7→∑

g∈H\G

[gm].

In the other direction, H0(G,M) = MH . We have

TrGH : m 7→∑

g∈G/H

gm.

There is also a topological description. If H ≤ G is a finite index, there is acovering map BH → BG of index [G : H]. Using the theory of covering spaces,you can construct the transfer maps.

Corollary 11.3. Let M be an RG-module with [G : H] <∞, and assume thatHn(H,M) = 0 for some n. Then Hn(G,M) is annihilated by [G : H]. Inparticular, if [G : H] is invertible in M then Hn(G,M) = 0.

Proof. For the first part, simply note that TrGH ResGH(α) = [G : H]α. For thesecond part, multiplication by [G : H] induces an isomorphism M → M andhence an isomorphism Hn(G,M)→ Hn(G,M).

Corollary 11.4. If |G| is finite, then Hn(G,M) is annihilated by |G| for alln > 0. If |G| is invertible in M , then Hn(G,M) = 0.

Proposition 11.5. If [G : H] is invertible in R, then

ResGH : ExtnRG(M,M ′) → ExtnRH(M,M ′)

is injective. In particular, H∗(G,M) → H∗(H,M).

Denote by Hn(G,M)(p) the p-primary component that vanishes under mul-tiplication by some power of p. If G, is finite, we get an embedding

Hn(G,M)(p) → Hn(P,M),

where P ≤ G is a p-Sylow subgroup.

Definition 11.6. An element z ∈ Hn(H,M) is G-invariant if

ResHH∩gHg−1 z = ResgHg−1

H∩gHg−1 gz

for all g ∈ G.

In the case when H is normal in G, there is an action of G/H on H∗(H,M).Then being G-invariant is exactly being invariant under this action.

Proposition 11.7. Hn(G,M)(p) is isomorphic to the set of G-invariant ele-ments in Hn(P,M).

Math 99r Notes 29

12 March 28, 2017

12.1 Spectral sequence from a filtration

A spectral sequence consist of a collection the Er-page Erp,q with differentials

dr : Erp,q → Erp−r,q+r−1.

So the total degree decrease by 1 as p+ q → p+ q − 1. The filtration degree pdecreases by r. The next page can be computed as

Er+1p,q = H(Er, dr).

For cohomology, we will get

dr : Ep,qr → Ep+r,q−r+1r , Ep,qr+1 = H(Er, dr).

There are two ways to construct a spectral sequence. The first one is from afiltration and the second is from a exact couple.

For an R-module M , consider an increasing filtration on M ,

· · · ⊆ Fp−1M ⊆ FpM ⊆ Fp+1M ⊆ · · · ⊆M,

where FpM are submodules of M and⋃p FpM = M . To this filtration is an

associated graded module

GrM =⊕p

GrpM =⊕p

(FpM/Fp−1M).

Lemma 12.1. If f : M → M ′ preserves filtration, and it induces GrM ∼=GrM ′, then f is an isomorphism.

Let us now assume that M is graded independent to the filtration. In thiscase, we require FpM to be a graded submodule. Then for fixed n, FpMngives a filtration on Mn. Then we define

Grp,q = FpMp+q/Fp−1Mp+q.

We call p the filtration degree, p+ q the total degree, and q the complementarydegree.

Now consider a chain complex C•, ∂ with FpC a filtration that is also asubcomplex of C. There is an induced filtration

FpH(C) = imH(FpC)→ H(C•) =FpC ∩ ZFpC ∩B

.

We can identify

GrpH(C) =FpH(C)

Fp−1H(C)=

FpC ∩ Z(FpC ∩B) + (Fp−1C ∩ Z)

.

Math 99r Notes 30

Let us define

Zrp,q = FpCp+q ∩ ∂−1Fp−rCp+q−1, Z∞p,q = FpCp+q ∩ Z

and

Brp,q = FpCp+q ∩ ∂Fp+r−1Cp+q+1 = ∂Zr−1p+r−1, B∞p,q = FpCp+q ∩B.

Then we have

FpC = Z0p ⊇ Z1

p ⊇ · · · ⊇ Z∞p ⊇ B∞p ⊇ · · · ⊇ B1p ⊇ B0

p .

Assume that each FpCn is finite, i.e., there exists a `(n) depending on nsuch that F`(n)Cn = Cn and F−`(n)Cn = 0. Then for each p and q, there existsa r (actually something like r = `(p+ q − 1) + p does the job) such that

Zrp,q = Zr+1p,q = · · · = Z∞p,q.

Likewise we haveB∞p,q = · · · = Br+1

p,q = Brp,q

for some r.Let us now define

Erp,q =Zrp,q

Brp,q + Zr−1p−1,q+1

.

Because both Zrp and Brp stabilizes, we can say that Erp,q converges (i.e., stabi-lizes) to

Z∞pB∞p + Z∞p−1

= GrpH(C)

as r →∞.On the 0-th page, we have

E0p,q =

Z0p,q

B0p,q + Z0

p−1,q

=FpCp+q

(sth in Fp−1Cp+q) + Fp−1Cp+q= Grp,q C.

The 0-th differential d0 is induced from ∂ as

d0 : E0p,q =

FpCp+qFp−1Cp+q

→ FpCp+q−1

Fp−1Cp+q−1= E0

p,q−1.

If you try hard, you will check that

E1p,q =

FpC ∩ ∂−1Fp−1C

∂FpC + Fp−1C= H(E0, d0).

Later we will define dr : Er → Er and see that Er+1 = H(Er, dr).There is another way of getting a spectral sequence, from an exact couple

A A

B.

i

jk

We have a differential d : B → B defined by jk, and it is a differential becausejkjk = j0k = 0.

Math 99r Notes 31

13 March 31, 2017

Last time we defined a spectral sequence Erp,q, dr and discussed how it arisesfrom an increasing filtration FpC• of a chain complex C•. We can define aspectral sequence Erp,q that computes H•(C•). Actually the spectral sequenceconverges to

E∞p,q∼= Grp,qH∗(C•)

in the case when the filtration is dimension-wise finite. The 0-th page is givenby

E0p,q = FpCp+q/Fp−1Cp+q.

There is a nice application. For X a finite-dimensional CW-complex, wewould like to show that Hsing

∗ ∼= Hcell∗ . Consider the cellular chain complex

Csing∗ and give a filtration given by

FpCsing∗ (X) = Csing

∗ (X(p)).

We have

E0p+q =

FpCsingp+q (X)

Fp−1Csingp+q (X)

=Csingp+q (X(p))

Csingp+q (X(p−1))

= Csingp+q (X(p), X(p−1)).

Then we see that

E1p,q = Hsing

p+q (X(p), X(p−1)) =

0 if q > 0,

Ccellp (X) if q = 0.

The spectral sequence then looks like Figure 1, where no dot means thatthere is a 0 there.

d1 d1 d1 d1

p

q

Figure 1: Spectral sequence coming from FpCsing∗ (X) = Csing

∗ (X(p))

This sequence hasE2p,0 = Hcell

p (X)

and then all dr = 0 for r > 1. So the spectral sequence stays the same from theE2 page and we get

Hsing∗ (X) ∼= Hcell

∗ (X),

because the only nonzero graded associated in the filtration of Hsing∗ (X) is

E2∗,0 = Hcell

∗ (X).

Math 99r Notes 32

13.1 Spectral sequence from an exact couple

Consider a series of maps

A A

B.

i

jk

We want this to be exact, i.e.,

ker j = im i, ker k = im j, ker i = im k.

Such a thing is called an exact couple. If we define d = jk, then d2 = jkjk = 0.So we may take the homology B1 = H(B) of B with respect to d. Let us defineA1 = i(A).

Proposition 13.1. Under this construction,

A1 A1

B1

i1

j1k1

is again an exact couple.

For example, for a filtration FpC• of a chain complex C•, you can take

A =⊕p

FpC•, i : Fp−1C• → FpC•, B =⊕p

FpC•Fp−1C•

.

This induces a long exact sequence

· · · → Hk(A)i1−→ Hk(A)

j1−→ Hk(B)k1−→ Hk−1(A)→ · · · .

We then get a exact couple

H∗(A) H∗(A)

H∗(B).

i1

j1k1

So B1 = E1∗,∗, and if you run the construction iteratively, you are going to get

Br = Er∗,∗ for r ≥ 0.

13.2 Spectral sequence from a double complex

As another example, let us consider a double complex Cp,q with differentials∂′ and ∂′′ such that

Cp,q Cp−1,q

Cp,q−1 Cp−1,q−1

∂′

∂′′ ∂′′

∂′

Math 99r Notes 33

commutes and ∂′2 = ∂′′2 = 0. This can also be considered as a chain complexin the category of chain complexes.

Given this data, we can define a total complex

(TC)n =⊕p+q=n

Cp,q, D = ∂′ + (−1)p∂′′.

We can filter this total complex by

Fp(TC)n =⊕i≤p

Ci,n−i.

If we assume that Cp,q has finitely many nonzero entries for every fixed p + q,(e.g., when Cp,q 6= 0 only when p, q > 0) then our filtration is going to bedimension-wise finite. Then Erp,q is going to compute H∗(TC).

The zeroth page is

Ep,q =Fp(TCp+q)

Fp−1(TCp+q)=

⊕i≤p Ci,p+q−i⊕i≤p−1 Ci,p+q−i

= Cp,q

and the 0-th differentials is going to be simple ∂′′. Then E1p,q = Hq(Cp,∗) and

so on.There is an application. Let k be a field and consider the double complex

Cp,q = Cp(X)⊗k Cq(Y ), C∗(−) = C∗(−; k).

Then

E1p,q = Hp(Cp(X)⊗ C∗(Y )) = Hq(Y ;Cp(X)) = Hq(Y )⊗ Cp(X)

and it follows thatE2p,q = Hp(X)⊗Hq(Y ).

If you dig in, you will be able to check that dr = 0 for r ≥ 2. Then we get

Hn(X × Y ) = Hn(TC) =⊕p+q=n

E∞p,q =⊕p+q=n

E2p,q =

⊕p+q=n

Hp(X)⊗Hq(Y ).

This is the Kunneth formula for over a field.

Math 99r Notes 34

14 April 4, 2017

14.1 Hochschild–Serre spectral sequence

This is about homology of a group with coefficient in a chain complex. Recallthat

Hn(G,M) = H∗(F• ⊗RGM),

where F• is a projective resolution of R. We have computed H0(G,M) = MG =M/gm−m.

Given a chain complex C•, we can define

H∗(G,C•) = H∗(F• ⊗ C•)

where the homology on the right side is considered as a homology as a doublecomplex (F ⊗ C)n =

⊕p+q=n Fp ⊗ Cq.

Let us compute this homology using the spectral sequence. We have

E1p,q = Hq(Fp ⊗RG C•) = Fp ⊗RG Hq(C•)

since Fq is projective. The next page will be

E2p,q = Hp(F• ⊗Hq(C•)) = Hp(G,Hq(C•)).

This spectral sequence, which is called the Hochschild–Serre spectral se-quence is going to converges to Hp+q(G,C•).

Proposition 14.1. If f : C• → C ′• is a quasi-isomorphism, then H∗(G,C•) ∼=H∗(G,C

′•).

There is another direction of running the spectral sequence. We can set

E1p,q = Hq(F• ⊗RG Cp) = Hq(G,Cp).

This spectral sequence should also compute the same homology H∗(G,C•).Let us put in an assumption (∗) that H∗(G,Cp) = 0 for ∗ > 0. (For instance,

set each Cp to be a free RG-module.) Then

E1p,0 = H0(G,Cp) = (Cp)G = Cp/gm−m,

and all other things in the first page vanish. All dr for r > 2 should vanishbecause either the source or the target is zero. So we get

H∗(CG) = E2 = E∞ = H∗(G,C•).

Proposition 14.2. Under the assumption (∗) on C•, the spectral sequence

E2p,q = Hp(G,Hq(C•))

computes Hp+q(CG).

Math 99r Notes 35

Suppose we have a short exact sequence

1→ H → G→ Q = G/H → 1.

Let F• be a projective resolution of R as an RG-module. Then

F ⊗RGM = (F ⊗RM)G = ((F ⊗RM)H)Q = (F ⊗RH M)Q.

In the spectral sequence, take C• = F• ⊗RH M . To apply the proposition, weneed to check that Cp = Fp ⊗M satisfies the condition (∗). It suffices to checkthat RG⊗RH M satisfies this condition. But recall that

RG⊗RH M = IndGH ResGHM = R[G/H]⊗RM = IndQ1M.

SoH∗(Q, IndQ1M) = H∗(1,M) = 0 for ∗ > 0.

Now we can finally apply the proposition and get the spectral sequence

E2p,q = Hp(Q,Hq(H,M)) ⇒ Hp+q(G,M).

To make sense of this, we need to specify the Q-action on Hq(H,M). The action

cg : h 7→ ghg−1, m 7→ gm

gives an actioncg∗ : H∗(H,M)→ H∗(gHg

−1,M).

Proposition 14.3. If g ∈ H, then cg∗ : H∗(M)→ H∗(M) is the identity.

Proof. Even on the chain level, we have

cg∗ : F ⊗RH M → F ⊗R[gHg−1] M ; x⊗m 7→ xh−1 ⊗ hm = x⊗m.

So it is the identity on homology.

So this indeed gives an RQ-module structure on H∗(H,M). There is also aspectral sequence in homology that gives

Ep,q2 = Hp(Q,Hq(H,M)) ⇒ Hp+q(G,M).

Corollary 14.4. There are exact sequences

H2(G,M)→ H2(G,MH)→ H1(H,M)Q → H1(G,M)→ H1(Q,MH)→ 0,

0→ H1(Q,MH)→ H1(G,M)→ H1(H,M)Q → H2(Q,MH)→ H2(G,M).

Proof. Look at the E2 page of the homology spectral sequence.Because the differentials dr at E2

1,0 are all zero for r ≥ 2, we see that E21,0 =

E∞1,0. On the other hand, we have d2 that can possibly do something to E0,1,which is given by

0→ E22,0

d2−→ E20,1 → 0.

Math 99r Notes 36

E20,0 E2

1,0 E22,0

E20,1 E2

1,10

0

0

0

Figure 2: E2 page of the Hochschild spectral sequence

Thus after taking the homology, we will get

0→ E32,0 → E2

2,0 → E20,1 → E3

0,1 → 0.

Furthermore, there are no more differential in and out of E30,1 and E3

2,0. ThusE3

0,1 = E∞0,1 and E32,0 = E∞2,0. That is,

0→ E∞2,0 → E22,0 → E2

0,1 → E∞0,1 → 0.

On the other hand, one of the basic theorem we had is that the homology ofthe total chain complex is given by the filtration. Then we have a short exactsequence

0→ E∞0,1 → H1(G,M)→ E∞1,0 → 0.

We can then concatenate the two exact sequences to get

0→ E∞2,0 → E22,0 → E2

0,1 → H1(G,M)→ E∞1,0 → 0.

Finally note that E∞0,2, E∞1,1, and E∞2,0 give a filtration on H2(G,M). So E∞2,0appear as a quotient of H2(G,M). Hence we get

H2(G,M)→ E22,0 → E2

0,1 → H1(G,M)→ E∞1,0 → 0.

This is precisely the exact sequence we want.

Math 99r Notes 37

15 April 7, 2017

15.1 Homology and cohomology of cyclic groups

Let us compute the group cohomology or homology with G = Z. We know thatZ[G] = Z[x, x−1] as a ring. We have a resolution

0→ Z[G]·(1−x)−−−−→ Z[G]

x 7→1−−−→ Z→ 0.

This is related to the fact that BZ = S1 which has finite homology or coho-mology. Anyways, we can compute homology H∗(Z, A) for a Z[G]-module Aas

0→ Af−→ A→ 0.

ThenH0(Z, A) = coker f = AG, H1(Z, A) = ker f = AG.

In the other direction, we are going to have

0← HomZ[G](Z[G], A) = Af∗←− HomZ[G](Z[G], A) = A← 0

and soH0(Z, A) = AG, H1(Z, A) = AG.

We can make a similar computation for G = Z/p. Let x be the generator ofG = Z/p. The resolution is given by

· · · → Z[Z/n]1−x−−−→ Z[Z/n]

1+x+···+xn−1

−−−−−−−−−→ Z[Z/n]1−x−−−→ Z[Z/n]

x7→1−−−→ Z.

Indeed, RP∞ = B(Z/2) has cohomology ring Z/2[x] with coefficients Z/2.Algebraically, we need to compute the homology of

· · · → A1−x−−−→ A

1+x+···+xn−1

−−−−−−−−−→ A1−x−−−→ A

x 7→1−−−→ Z.

This immediately shows

H0 = A/〈a− ax〉 = AG, H2i+1 = H1 = AG/NA, H2i = H2.

where N = 1 + x+ x2 + · · ·+ xn−1. This doesn’t have a very good description.For cohomology, we can do the same thing and get

H0 = AG, H2i+1 = H1 = kerN/ im(1− x), H2i = H2.

We can compute Hi(Z/p;Fp) with trivial action on Fp using this. Becausethe N map is going to be zero, all maps are zero and thus we get

dimFp Hi(Z/p;Fp) = 1

Math 99r Notes 38

for all i. The cup product structure (which comes from the comultiplicationmap B• → B• ⊗B•) can be computed as

H∗(Z/p;Fp) =

F2[x] p = 2,

Fp[x, y2]/(x2) p > 2.

There is another relation between x and y. There is a connected homomorphism

H1(Z/p,Z)→ H1(Z/p,Z)→ H1(Z/p,Fp)α−→ H2(Z/p,Z)→ · · ·

and y can be taken to be the image of α(x) under the map H2(Z/p,Z) →H2(Z/p,Fp) coming from the reduction map Z→ Fp.

This procedure has a name. There is always a map

H∗(X;Fp)→ H∗+1(X;Fp)

and this is called a Bockstein homomorphism. In the case p = 2 we havey = Sq1(x), but Sq1(x) = x2 if x ∈ H1. This is why we can take y = x2.

Finally, let us try to compute the cohomology with G = Σ3 the permutationgroup. We have an exact sequence

1→ Z/3→ Σ3 → Z/2→ 1.

Take k = F2. Then the pages are given by

E2p,q = Hp(Z/2, Hq(Z/3,F2)) =

Hp(Z/2,F2) q = 0

0 q > 0.

This shows thatHp(Σ3,F2) ∼= Hp(Z/2,F2).

This can also been seen from the fact that Σ3 = GL2(F2).

Math 99r Notes 39

16 April 11, 2017

16.1 Wreath product

If you have a group G, we define the wreath product as

G o Z/p = G×p o Z/p,

where the action is given by

x · (g1, . . . , gp) = (g2, . . . , gp, g1).

Then there is a short exact sequence

1→ Gp → Gp o Z/p→ Z/p→ 1.

We want to compute H∗(G o Z/p,Fp) over Fp. There is a spectral sequence

Es,t2 = Hs(Z/p,Ht(Gp,Fp)) ⇒ Hs+t(G o Z/p,Fp).

To compute this, we need to know the Fp[Z/p]-module structure on H∗(Gp,Fp).First note that by the Kunneth formula,

H∗(Gp,Fp) =⊗p

H∗(G,Fp).

Because the action of Z/p on Gp is given by shifting the elements, the inducedaction of Z/p on H∗(Gp,Fp) is also going to be just shifting the tensors.

Lemma 16.1. Let Λ =⊕

i Λi be a graded Fp-vector space. Give a Z/p-actionon Λ⊗p by cyclic shifting. Then as a Fp[Z/p]-module, Λ⊗p is a direct sum of afree Fp[Z/p]-module and the trivial module (Fp-vector space with trivial action)generated by some of λi ⊗ · · · ⊗ λi.

Using this lemma, we can decompose

H∗(Gp,Fp) =⊕

Fp[Z/p]⊕⊕

Fp.

Then we see that

E2 = H∗(Z/p,H∗(Gp,Fp)) =⊕

H∗(Z/p,Fp[Z/p])⊕⊕

H∗(Z/p,Fp).

We know hat H∗(Z/p,Fp[Z/p]) = 0 for ∗ > 0. For the second factor, we seethat λi ⊗ · · · ⊗ λi lives in the pi-th graded pieces and so will come out in thepi-th cohomology. This shows that,

Es,t2 = Hs(Z/p,Ht(Gp,Fp)) =

Ht(Gp,Fp)Z/p s = 0

0 s > 0, p - tHs(Z/p,Hip(Gp,Fp)) t = ip.

Math 99r Notes 40

s

t

Figure 3: Spectral sequence from Hs(Z/p,Ht(Gp,Fp))

In particular, if t = ip then

Es,ip2 =⊕

x a basis ofHn(G,Fp)

Hs(Z/p,Fp).

So our spectral sequences looks something like Figure 3.

Proposition 16.2. The E2 page collapses, i.e., E2 = E∞.

Note that if there is a map between exact sequence of groups

1 N G Q 1

1 M H P 1

you get a map between the Hochschild–Serre spectral sequences.

E2 = H∗(Q,H∗(N, k)) H∗(G, k)

F2 = H∗(P,H∗(M,k)) H∗(H, k)

We remark that this is also true for fibrations and the Serre spectral sequence.

Math 99r Notes 41

17 April 14, 2017

Recall that the wreath product is defined as G oZ/p = GpoZ/p. Then there isan exact sequence and so Gp → G oZ/p→ Z/p. So we have a spectral sequence

Es,t2 = Hs(Z/p,Hs(Gp,Fp)) ⇒ Hs+t(G o Z/p,Fp).

Because Ht(Gp,Fp) is a direct sum of Fp[Z/p] and Fp, we get some descriptionof Es,t2 .

17.1 Cohomology of wreath products

For s = 0, we have

E0,t2 = Hs(Gp,Fp)Z/p = λ⊗p : λ ∈ Hi(G,Fp) ⊕ (free Fp[Z/p]).

Recall that ifA ≤ B is of finite index, there is a transfer map tr : H∗(A,M)→H∗(B,M), and we have

res tr =∑

g∈B/A

g.

So applying this to Gp → G o Z/p gives

res tr =∑g∈Z/p

g = (1 + T + · · ·+ T p−1) : H∗(Gp,Fp)→ H∗(Gp,Fp).

Then we see that actually

E0,t2 = λ⊗p : λ ∈ Hi(Gp,Fp) ⊕ (image of res tr).

Lemma 17.1. If x ∈ E0,∗2 comes from x ∈ im(H∗(Q, k)

res−−→ H∗(P, k)), then xis a permanent cycle, i.e., d∗x = 0.

Proof. Consider the map

P Q Q/P

P P 1.

i

i

Then there is a map of spectral sequences

H∗(Q/P,H∗(P, k)) H∗(Q, k)

H∗(1, H∗(P, k)) H∗(P, k).

res

Since the bottom spectral sequence collapses at zero, all differentials are zero.So in order for anything in the image of res to live, it has to be a permanentcycle.

Math 99r Notes 42

We note that

Hs(Z/p,Fp〈λ⊗p〉) = Hs(Z/p,Fp)⊗ 〈λ⊗p〉.

But recall that H∗(Z/p,Fp) ∼= Fp[x, y]/(x2). So the basis are yj⊗λ⊗p and xyj⊗λ⊗p. Furthermore, this multiplicative structure is preserved on the differential:

d(x⊗ λ⊗p) = d(x)⊗ λ⊗p ± x⊗ d(λ⊗p).

So in most cases, knowing the differential on the t = 0 column is enough to findout all the differentials.

Let us go back into the case of the wreath product. The image of therestriction map is a permanent cycle, and so we may as well just ignore thatand look at the other parts.

Lemma 17.2. The differential on λ⊗pi ∈ E0,ipr has to be zero.

Proof. If we have a short exact sequence H → G→ G/H, this gives a fibrationBH → BG→ B(G/H). This gives a Serre spectral sequence

Hs(B(G/H),Ht(BH, k)) ⇒ Hs+t(BG, k),

where the calligraphic H means the local coefficients.Given an element λn ∈ Hn(BG,Z/p), it corresponds to a map fn : BG →

K(Z/p, n). Then this induces a map

f∗n : H∗(K(Z/p), n)→ H∗(BG),

and the fundamental class ιn ∈ Hn(K(Z/p, n)) corresponding to id is going tomap to λn. Then we have a sequence of fibrations.

BGp BG o Z/p BZ/p

K(Z/p, i)p K(Z/p, i)p ×Z/p EZ/p BZ/p

fpi

This gives a map of spectral sequence

Hs(BZ/p;Ht(K(Z/p, n)p;Fp) Hs+t(K(Z/p, i)p ×Z/p EZ/p;Fp)

Hs(BZ/p;Ht(BGp;Fp)) Hs+t(BG o Z/p;Fp)

Here, note that for s = 0 we are going to get

Ht(K(Z/p, n)p;Fp)→ Ht(BGp;Fp).

Since f∗(ι⊗np ) = λ⊗pn , we can use functoriality to obtain data of the differentialon λ⊗pn .

Math 99r Notes 43

We want to show that all differentials on λ⊗pn are zero. It suffices to checkthat all differentials on ι⊗pn are zero, by the above argument. But by theHurewicz theorem and the universal coefficients theorem, we haveHt(K(Z/p, n),Fp) =0 for t ≤ n−1. So we have Es,t2 = 0 for 0 < t < np. So the only possible nonzerodifferential is dnp+1 : E0,np → Enp+1,0.

Here is a general fact for Serre spectra sequences (or Hochschild–Serre spec-tral sequences). The bottom row of the infinity page is

E∗,0∞ = im(H∗(BZ/p;Fp)→ H∗(K(Z/p, i)p ×Z/p EZ/p)

).

If show that this map is injective, then nothing on the bottom row can bekilled and so dnp+1 : E0,np → Enp+1,0 has to be zero. This can be shown byexhibiting a section of the fibration, and this can be done.

Example 17.3. We have D8 = Z/2 o Z/2 and so

H∗(D8,F2) = F2[x, y, e]/(xy).

Let us defineW1 = Z/p, Wr = Wr−1 o Z/p.

If n =∑aip

i and 0 ≤ ai < p, then the p-Sylow subgroups of Sn can be writtenas

W a11 ×W

a22 × · · · ×W

akk .

So knowing about the cohomology of the wreath products is going to tell ussomething about the cohomology of symmetric groups.

Math 99r Notes 44

18 April 18, 2017

18.1 Atiyah completion theorem

This was a presentation of mine. This is based on Atiyah’s 1961 paper Charac-ters and cohomology of finite groups. For a finite group G, what is H∗(G,Z)?We have

H1(G,Z) =(f : G→ Z) : f(xy) = f(x) + f(y)

0= 0.

The next cohomology can be computed as

H2(G,Z) = 1→ Z→ E → G→ 1 = HomGrp(G,Q/Z) = HomGrp(G,U(1)).

But if you think about it, these are 1-dimensional complex representations ofG, or equivalently 1-dimensional characters of G. So there seems to be someconnection between the finite complex representations of G and H∗(G,Z).

Here is the big idea. We want to find a relation between

H∗(G,Z) = H∗(BG;Z) ←→ representations of G.

This is an algebraic problem, but it turns out the connection can be establishedeasier if we use topology. If you think about it a bit, complex representationsof G are somewhat like complex vector bundles over BG. Rank k complexvector bundles over BG are [BG,BU(n)], and this is quite similar to grouphomomorphisms G→ U(n). So it seems reasonable to expect representations ofG to be related toK∗(BG). NowK∗ is a cohomology theory. There is somethingcalled an Atiyah–Hirzebruch spectral sequence that computes a cohomologytheory in terms of its coefficient ring.

For a space X, the set of complex vector bundles over X has a additivestructure, coming from ⊕, and a multiplicative structure, coming from ⊗.

Definition 18.1. For a finite CW-complex X, we define

K0(X) = groupification of complex vector bundles /X.

This has a natural ring structure. If X has a base point x0, there is a ringmorphism K0(X)→ Z given by mapping to the dimension of the fiber over x0.We further define the ideal

K0(X) = ker(K0(X)→ Z).

Theorem 18.2 (Bott periodicity). For any finite CW-complex X,

K0(Σ2X) = K0(X).

K0(−) is a representable functor, and this allows us to define a cohomology

theory K−n(X) = K0(ΣnX), which will further extend to the positive sideusing Bott periodicity as

Kn(X) =

K0(X) 2 | n,K0(ΣX) 2 - n,

K∗(X) = K∗(X+).

Math 99r Notes 45

Definition 18.3. For a CW-complex X such that each skeleton Xn is finite,define

K ∗(X) = lim←−K∗(Xn).

Just like complex vector bundles, there is an additive structure, comingfrom ⊕, and a multiplicative structure, coming from ⊗, on the set of complexrepresentations of G. So we can again make this into a ring R(G). In this case,we have a pretty interpretation

R(G) ⊆ (ring of class functions G→ C)

by looking at the characters.Given a complex representation ρ : G→ GL(n,C), we can glue this into the

principal G-bundle EG→ BG by

(v, x) ∈ EG× Cn(v, x) ∼ (gv, gx)

→ BG,

which is going to be a complex vector bundle over BG. This gives a ring map

α : R(G)→ K 0(BG) → K ∗(BG).

There is an augmentation map R(G)→ Z that sends a representation to itsdimension. Let I(G) be its kernel. This gives a filtration

R(G) ⊇ I(G) ⊇ I(G)2 ⊇ · · · ,

and we can can complete R(G) with respect to the filtration. Then the map αextends to

α : R(G)→ K ∗(BG).

Theorem 18.4. For every finite G, α is an homeomorphism.

Let us just check this for cyclic groups G = Z/nZ.

Example 18.5. Let us look at this map for cyclic groups G = Z/nZ. We knowall the representations: they are simply the 1-dimensional ones 1, ρ, ρ2, . . . , ρn−1.So the ring R(G) is given by

R(G) ∼= Z[ρ]/(ρn − 1) ∼= Z[σ]/((σ + 1)n − 1),

and I(G) = (σ). So the completion is going to look like

R(G) = Z[[σ]]/((σ + 1)n − 1)

But how do you compute K ∗(BG)? If you recall, there is a Serre spectralsequence

Hs(B;Ht(F ;Z)) ⇒ Hs+t(E;Z)

Math 99r Notes 46

and for group cohomology this corresponds to the Hochschild–Serre spectralsequence

Hs(G/H,Ht(H,M)) ⇒ Hs+t(G,M).

For a general cohomology theory, there is the Atiyah–Hirzebruch spectral se-quence

Hs(B;Kt(F )) ⇒ Ks+t(E)

and if you have some faith, you will believe that there is a spectral sequence

Hs(G/H,K t(BH)) ⇒ K s+t(BG).

In particular, Hs(G,Kt(∗))⇒ K s+t(BG).

Example 18.6. We can try to compute K ∗(B(Z/nZ)). This doesn’t imme-diately give what the ring exactly is, but you at least see what the associ-ated graded is. This can be checked to be exactly the associated graded for

R(G). . Actually, the map α will induce a map on the associated graded

G R(Z/nZ) → GK ∗(B(Z/nZ)) and by following the generators, it can bechecked that this map is an isomorphism. It follows from this that α is ahomeomorphism.

Let me make a remark about how Theorem 18.4 is proved. We checked thatthis is true for finite groups. Using representation theory, you can check that ifH / G has quotient G/H ∼= Z/qZ for some prime q, then the statement for Himplies the statement for G. This proves the theorem for solvable G. Next usethe Brauer’s theorem on elementary groups to show that this proves the generalcase.

Corollary 18.7. For any finite G, K 2k+1(BG) = 0.

Corollary 18.8. There is a spectral sequence

Hs(G/H, R(H)) ⇒ R(G).

In particular, Hs(G,Z)⇒ R(G).

18.2 Skew-commutative graded differential algebras

This was a presentation by Carlos Albors-Riera. This is based on a paperpublished by John Tate, Homology of Noetherian rings and local rings. Firstthing I should talk about is what this thing is.

Definition 18.9. An associative algebra X and a R-linear map d : X → X con-stitute an skew-commutative graded differential R-algebra (I will writethese as just R-algebras) if these satisfy:

(1) X is graded by X =⊕λ

λ=−∞Xλ such that XiXj ⊆ Xi+j ,

(2) X<0 = 0, 1 ∈ X0, Xλ is finitely generated,

Math 99r Notes 47

(3) xy = (−1)λµyx where x ∈ Xλ and y ∈ Xµ, x2 = 0 if λ is odd,

(4) dXλ ⊆ Xλ−1, d2 = 0,

(5) d(xy) = (dx)y + (−1)λx(dy) for x ∈ Xλ and y ∈ Xµ.

From this you produce a chain complex

· · · d−→ Xλd−→ Xλ−1

d−→ · · · .

Then H•(X) is going to have a natural structure of an algebra. If X is free andacyclic, and R = X0, we have a free resolution

· · · → X2 → X1 → R→ B/M,

where M = B0 is the boundaries.

Proposition 18.10. Let p > 0. There exists a canonical extension Y ⊇ X ofan R-algebra such that

(0) t has degree p− 1,

(1) Yλ = Xλ for λ < p,

(2) Bp−1(Y ) = Bp−1(X)⊕Rt.

Proof. When p is odd, consider the free X-module XT with basis 1, T where T issupposed to be in degree p. Let us define Y = X⊕XT . Then Yλ = Xλ+Xλ−pT .Because deg T is odd, we are right to set T 2 = 0. Define t = dT . ThenTx = (−1)λxT for x ∈ Xλ and

d(xT ) = (dx)T + (−1)λx(dT ).

You can check that this satisfies all the conditions.For p even, let Y be the free module on 1, T, T (2), . . . where T (k) is supposed

to be in degree pk. Then you can consider

Yλ = Xλ ⊕Xλ−pT ⊕Xλ−2pT(2) ⊕ · · ·

with the relations

T (i)T (j) =(i+ j)!

i!j!T (i+j), T (i)x = xT (i), dT (i) = tT (i−1).

This does to trick.

In general, for τi ∈ Hp−1(X), you can construct Y = X〈T1, . . . , Tn〉 suchthat

(1) Y ⊇ X and Yλ = Xλ for λ < p,

(2) Hp−1(Y ) = Hp−1(X)/(Rτ1 + · · ·+Rτn).

Theorem 18.11. For M ⊆ R and ideal, there exists a free acyclic R-algebraX such that H0(X) = R/M .

Math 99r Notes 48

Proof. Let X be the union X0 ⊆ X1 ⊆ · · · where we construct them in thefollowing way. Write M = (t1, . . . , tn). We let X0 and X1 = X0〈T1, . . . , Tn〉.Now H0(X1) = R/M .

Now let s1, . . . , sm ∈ Z1(X1) be such that si generate H1(X1). Then we canset X2 = X1〈S1, . . . , Sm〉 so that H1(X2) = 0 and H0(X2) = R/M . You cango on.

Proposition 18.12. Let X,Y be free acyclic algebras such that H(X) = R/Mand H(Y ) = R/N . Let j : X → R/M and k : Y → R/N be the induced maps.Then

(R/M)⊗ Y j⊗1←−− X ⊗ Y 1⊗k−−→ X ⊗R/N

induce isomorphisms

H(R/M ⊗ Y ) ∼= H(X ⊗ Y ) ∼= H(X ⊗R/N).

Definition 18.13. We define TorR(R/M,R/N) = H(X ⊗ Y ).

Theorem 18.14. Let M,N be ideals and let a ∈ MN be a non-zero divisor.Let R = aR and set K = R/M , L = R/N . Then

TorR(K,L) = TorR(K,L)〈u〉

with u in degree 2.

Proof. Let X be a free, acyclic R-algebra such that H(X) = Kλ = (R/M)λ.Then

d(NX1) = NdX1 = NM

and so we can choose s ∈ NX1 such that ds = a. Let X = X/aX, with s ∈ X1.You can show that H(X) = K〈σ〉 where σ is the class of s. We can make X〈s〉with dS = s be the free resolution of K.

Now we can actually compute Tor.

TorR(K,L) = H((X ⊗ L)〈s〉) = H(∑

(X ⊗ L)S(i))

=

∞∑i=1

H(X ⊗ L)U i = H(X ⊗ L)〈u〉.

Math 99r Notes 49

19 April 21, 2017

19.1 Cohomology of D2n

This was a talk by Max Hopkins. We are going to talk about the cohomologyof D2n = 〈σ, τ |σn = τ2 = 1, στσ = σ−1〉. The basic method is the Hochschild–Serre spectral sequence.

Example 19.1. We are going to have

ExtiZD2n(Z[τ ], A) = ExtiZ[Z/n](Z, A) = Hi(Z/n,A).

Lemma 19.2. Let 1 → H → G → G/H → 1 be a short exact sequence andconsider a resolution · · · → C1 → C0 → Z→ 0 over ZH. Then

(1) ZG⊗ZH Z = ZG/H,

(2) · · · → ZG⊗ZH C1 → ZG⊗ZH C0 → ZG/H → 0 is a resolution over ZG.

Proof. (1) can be checked by the coset decomposition. (2) can also be checkedbecause this tensoring is exact.

Corollary 19.3. Let A be a ZG-module, regarded as a ZH-module. Then

ExtiZG(Z[G/H], A) = ExtiZH(Z, A) = Hi(H,A).

We first take a projective resolution of Z as a Z[G/H]-module, and the takethe projective resolution of of them as ZG-modules. Then we get

......

...

C01 C11 C21 · · ·

C00 C10 C20 · · ·

Z B0 B1 B2 · · ·

∂ ∂ ∂

∂ ∂d10

∂d11 d21

∂ ∂d00

∂d10 d20

ε d0 d1 d2

Here we can lift the maps to get d00, d10, . . ., but we don’t know if the liftedmaps are going to be chain complexes. But this can be fixed.

Anyways, we can pass the complex through HomZG(−,M) and take thevertical cohomology. Then we get

Ext∗ZG(B∗, A) = Ext∗ZG(B∗ ⊗Z[G/H] Z[G/H], A)

= Ext∗ZG(Z[G/H], A)⊗Z[G/H] B∗ = Hi(H,A)⊗Z[G/H] Bi.

So we get the Hochschild–Serre spectral sequenceH∗(G/H,H∗(H,M))⇒ H∗(G).Anyways, we are going to apply it to

1→ Z/n→ D2n → Z/2→ 1.

Math 99r Notes 50

We already know a nice resolution of Z over Z[Z/n], and so we can explicitlyconstruct the chain complex. You can track all the differentials, because we canmake our maps explicit.

Theorem 19.4. The cohomology ring of D2n with coefficients F2 is

H∗(D2n ,F2) = F2[x, y, w]/(xy)

where x, y are in dimension 1 and w is in dimension 2.

19.2 The third cohomology

This was a presentation by Stefan Spataru.

Definition 19.5. A cross module is a pair of groups (N,E) with a mapα : N → E and an action en such that

(1) α(en) = eα(n)e−1,

(2) α(n)n′ = nn′n−1.

Theorem 19.6. For a group G and an RG-module A, there is a bijectionbetween H3(G,A) and the equivalence classes of short exact sequences

0→ A→ N → G→ E → 1.

Here, the equivalence relation is the one generated by to sequence equivalent ifthere are maps ϕN and ϕE such that

A N E G

A N ′ E′ G.

ϕN ϕE

From the bar resolution, H3(G,A) is the set of some maps f : G×G×G→ Aquotiented out by some subgroup.

Now given an exact sequence

0→ Ai−→ N

α−→ Eπ−→ G→ 1,

with a section s0 : G→ E, we can define a function

f(g, h) = s0(g)s0(h)s0(gh)−1.

This f is going to satisfy

f(g, h)f(gh, k) = s(g)f(h, k)f(g, hk).

We can lift this to a map F : G ×G → N , again put in a modifying factork : G×G×G→ N to make

F (g, h)F (gh, k) = k(g, h, k) · s(g)F (h, k) · F (g, hk).

Math 99r Notes 51

The image of k has to be in the image of i. So there is a map z : G×G×G→ Asuch that k = i z. You can check that d3z = 0.

Now let us show that z is independent of the choice of F , in as a cohomologyclass. If we choose another

F2(g, h) = F1(g, h)c(g, h),

then we would get

c(g, h)c(gh, k) = s(g)c(h, k)c(g, hk).

Then d2c = 0 and so z′ = z + ∂c. A similar computation shows that z isindependent on the choice of the first section s0. You can further check thatequivalent sequences give the same element in H3(G,A). Then we get a well-defined map from equivalence classes to H3(G,A).

We now want to show that this map is bijective. Consider any functionz : G×G×G→ A. Let F be the free group generated by G. There is a naturalembedding s0 : G → F , and this association gives an epimorphism π : F → G.Then we get an exact sequence

0→ Ai−→ A× kerπ

α−→ Fπ−→ G→ 1.

Here α is given by (a, x) 7→ x.

Proposition 19.7. (1) The kernel of π is generated by the image of T .(2) A× kerπ is generated by the image of i and the image of T .

Here, T is the section that was named F before.

Math 99r Notes 52

20 April 25, 2017

20.1 Swan’s theorem

This was a talk by Aaron Slipper. This version is related to ranks of projectivemodules. We want to associate to each module a rank, but this is hard becausewe are trying to do this over an arbitrary ring, in particular, the group ring ZΓ.

We can define the rank as

rkP = rkZ(P ⊗ZΓ Z).

Now rank of Z-modules are well-defined, and so we get a good notion of rank.Then rk(ZΓ)n = n. Brown calls this

ε(P ) = rkZ(Z⊗ZΓ P ).

There is also more simple definition

ρ(P ) = rkZ(P )/|Γ|.

It is not a priori clear that this is even an integer. Also, this is defined only forfinite Γ.

Theorem 20.1 (Swan). ε(P ) = ρ(P ) for all projective P .

This was proved by Swan in 1960 and by Bass in 1976 and 1979. Onemotivation for looking at this is to define Euler characteristics of groups.

The idea is to find another definition of rank and show that all three arethe same. If F is a free Z-module, then trZ(idF ) = n. But because our ring isnoncommutative, and so trace can’t be defined in an invariant way. Even in the1-dimensional case, x 6= axa−1.

The idea of Bass is to look at the abelianization T (A) = A/[A,A], where[A,A] is the ideal generated by ab− ba. Then we can define the trace map

tr : Matn(A)→ A/[A,A].

Then for α an n×m matrix and β an m× n matrix, we have

tr(αβ) = tr(βα).

So we have a well-defined trace for endomorphisms of free modules.Now we need to define it for general projective modules. Let α : P → P be

a map, and let π : F P be a surjection from a free module of finite rank. Wecan lift

F

P P.

πi

id

Then we havetr(α) = tr(απi) = tr(iαπ),

Math 99r Notes 53

where iαπ : F → F . We can take this as a definition. We still have to showthat this well-defined. Consider

F P P ′ F ′.π

i

β

α

j′

π′

Then the identity tr(iαπ′i′βπ) = tr(i′βπiαπ′) implies

tr(αβ) = tr(iαβπ) = tr(i′βαπ′) = tr(βα).

This also implies well-definedness.Now let ϕ : α → β be a ring morphism. This induces a map Tϕ : T (A) →

T (B). Let α : P → P be a map. This induces a map B⊗Aα : B⊗AP → B⊗AP .Then we get the identity

trB(B ⊗A α) = T (ϕ)(trA(α)).

This follows from the free case.We can do the analogous thing for restriction. Let ϕ : A → B be a ring

homomorphism such that B is a finitely generated left A-module. Then thereexists a map trB/A : T (B)→ T (A) such that for any α : P → P ,

trA(α) = trB/A(trB(α)).

Let me try to define this map. For any b ∈ B, there is a right multiplicationmap µb : B → B. This gives a map ϕ : B → T (A), and we have

ϕ(bb′ − b′b) = trA(µbµb′)− trA(µb′µβ) = 0.

So the map factors through B → T (B) → T (A). This is going to be the maptrB/A.

Definition 20.2. The Hattori–Stallings rank is defined as R(P ) = tr(idP ).

If Γ is a finite group, then there is an augmentation map ϕ : ZΓ→ Z and

εΓ(P ) = T (ϕ)(RZΓ(P )).

Likewise, we haveρΓ(P ) = τ(RZΓ(P ))

for the map τ : T (ZΓ)→ Z with τ : 1 7→ 1 and γ 7→ 0.Note that if P 6= 0, then ρΓ(P ) > 0. In particular, ρΓ(ZΓ) = 1, and this

shows that ZΓ is not decomposable into two projective ZΓ-modules.

20.2 Grothendieck spectral sequence

This was a talk by George Torres. This computes the right derived functors ofthe composition of two functors. Let A be an abelian category.

Math 99r Notes 54

Definition 20.3. Given a (co)chain complex C• we define an injective reso-lution of C• as Ip,• → Ip+1,• with commuting maps.

Any injective resolution induces exact sequences

0→ Zp(C•)→ Zp(I•,0)→ Zp(I•,1)→ · · · ,0→ Bp(C•)→ Bp(I•,0)→ Bp(I•,1)→ · · · .

This further gives an exact sequence

0→ Hp(C•)→ Hp(I•,0)→ Hp(I•,1)→ · · · .

Definition 20.4. An injective resolution is called fully injective if the previ-ous exacts sequences are injective resolutions.

Proposition 20.5. If A has enough injectives, then every cochain complexadmits a fully injective resolution.

Proof. We can first find an injective resolution of 0 → Bn(C•) → Zn(C•) →Hn(C•)→ 0. Then using the exact sequence, 0→ Zn(C•)→ Cn → Bn+1(C•)→0, you get an injective resolution of Cn. Then you can put them together andget an injective resolution of C•.

Definition 20.6. A fully injective resolution of C• of this form is called aCartan–Eilenberg resolution.

These are used in defining hyper-derived categories, etc.

Definition 20.7. A bicomplex is a lattice of objectives Cij ∈ A with maps∂ij1 : Cij → C(i+1)j and ∂ij2 : Cij → Ci(j+1) satisfying ∂2

1 = ∂22 = 0 and

∂1∂2 = ∂2∂1.

Definition 20.8. The totalization of the bicomplex is the complex

Tot(Cij)n =⊕i+j=n

Cij .

There are two canonical filtrations of the Tot(C). There is

F p(1)(Tot(C))n =⊕r≥p

Cr,n−r, F p(2)(Tot(C))n =⊕r≥p

Cn−r,r.

These two induce two spectral sequences, which we will call (1)E and (2)E.There are two differentials, and so there are two homologies:

Hij(1) = ker(∂ij1 )/ im(∂

i(j−1)1 ), Hij

(2) = ker(∂ij2 )/ im(∂(i−1)j2 ).

Proposition 20.9. Let A be cocomplete and let Cij be a bicomplex. Then

(1)Ep,q2 = Hp(H•,q(1) (C)), (2)Ep,q2 = Hp(Hq,•(2) (C)),

and they both converges to Hp+q(Tot(C)).

Math 99r Notes 55

Definition 20.10. Let A have enough injectives and let T : A → B beadditive. Then Q ∈ A is right T -acyclic if RiT (Q) = 0 for i > 0.

Theorem 20.11. Let F : A → B and G : B → C be additive functors, andassume A and B have enough injectives. Assume that F (Q) is G-acyclic forall Q injective. Then for all A ∈ A , there exists a spectral sequence Ep,qr suchthat

Ep,q2 = RpG(RqF (A)) =⇒ Rp+q(FG)(A).

Proof. Let A → C• be an injective resolution. Then take a fully injectiveresolution of F (C•).

......

...

0 I01 I11 I21 · · ·

0 I00 I10 I20 · · ·

0 C0 C1 C2 · · ·

Now I is a bicomplex and so GI is a bicomplex. Then we have two spectralsequences (1)E and (2)E. WE have

(1)Ep,q2 = Hp(H•,q(1) (GI)) = Hp(RqG(FC•)) =

0 q > 0

Hp(GF (C•)) q = 0.

So this collapses at this point, and so

Hn(Tot(GI)) = Rn(GF )(A).

On the other hand, let us look at (2)E. There is a short exact sequence

0→ Zp(I•,q)→ Ip,q → Bp+1(I•,q)→ 0,

that splits, and so you can show G(Hp,q(2) (I)) ∼= Hp,q

(2) (GI). Then we see that

(2)E2 = Hp(Hq,•(2) (GI)) = Hp(G(Hq,•

(2) (I))) ∼= RpG(RqF (A)).

This finishes the proof.

Example 20.12 (Leray spectral sequence). Consider X a topological space andF an abelian sheaf. Consider a continuous map f : X → Y and take F = f∗(−)and G = Γ(Y,−). Then we have a spectral sequence

Hp(Y,Rqf∗(F )) =⇒ Hp+q(X,F ).

Math 99r Notes 56

20.3 Cech cohomology

This was a talk by Julian Salazar.

Definition 20.13. A presheaf on a category C with S -values is a functorC op → S . For a topological space, a presheaf on it is one for C = Op(X),the open sets of X with inclusion maps.

Definition 20.14. A sheaf is a presheaf F such that the following hold:given any open cover Uii∈I and sections σi ∈ F (Ui) satisfying σα|Uα∩Uβ =σβ |Uα∩Uβ for all α, β ∈ I, there exists a unique σ ∈ F (X) such that σ|Uα = σα.

Example 20.15. For X a smooth manifold, the space of p-form Ωp(−) is asheaf of R-algebras. Moreover, Ω∗(−) is a sheaf of skew-commutative differentialalgebras.

Definition 20.16. A ringed space is a pair (X,OX) of a topological space Xand a sheaf of rings OX on X.

Proposition 20.17. The category Sh(X) = Sh(X,OX−Mod) has enough in-jectives.

Definition 20.18. Γ(X,−) : Sh(X) → Γ(X,OX)−Mod is left exact, and wedefine sheaf cohomology as

Hi(F ) = RiΓ(F ).

It’s nice conceptually, but it is really hard to compute.

Proposition 20.19. Let X be compactly-generated weakly-Hausdorff space.Then

Hnsing(X;A) = Hn(X,AX),

where AX is the constant sheaf.

Proposition 20.20. Hp,q(X) ∼= Hq(X,Ωp) for complex analytic spaces X.

Now let us defined Cech cohomology. Let U = Uαα∈I be an open cover.Let P be a presheaf on X.

Definition 20.21. The Cech complex is defined with objects

Cp(U,P) =∏

i0<···<ip

P(Ui0 ∩ · · · ∩ Uip)

and differentials

(dα)i0,...,ip+1=

p+1∑k=0

(−1)kαi0,...,ik,...,ip+1.

The Cech cohomology H(U,P) is the cohomology coming from this complex.

Math 99r Notes 57

If you think about this, this is actually looking at the simplicial cohomologyof the nerve of the category coming from the open cover.

Example 20.22. With respect to the standard open cover of S1 with twocontractible sets, the Cech cohomology is H0(S1,ZS1) = Z and H1(S1,ZS1) =Z.

Now let us apply the Grothendieck spectral sequence. This says that ifF : A → B and G : B → C are additive functors with F taking injectives toG-acyclics, then there is a spectral sequence

Ep,q2 = RpG(RqF (−)) =⇒ Rp+q(G F )(−).

Note that the Hochschild–Serre spectral sequence is an instance of this, withthe functors (−)N and (−)G/N .

Consider the diagram

Sh(X) PSh(X)

OX−Mod.

i

Γ(X,−)H0(X,−)

This gives a spectral sequence

Ep,q2 = Hp(U ,Hq(F )) =⇒ Hp+q(X,F ).

Here Hq(F ) is the sheaf U 7→ Hq(U,F ).

Example 20.23 (Mayer–Vietoris). Consider U = U, V . Let’s draw the E2

page. There are going to be two columns, because there are two open sets. Thenyou recover the Mayer–Vietoris sequence with sheaves as coefficients. If F isthe constant sheaf, then you recover Mayer–Vietoris for ordinary cohomology.

Theorem 20.24 (Leray). If F ∈ Sh(X) and F is acyclic on every open in U,we have

Hp(U,F ) ∼= Hp(X,F ).

This is used in computing things like cohomology of line bundles over pro-jective space. There is the canonical affine open cover U = U0, U1, U2, . . . andthis allows us to to compute H(Pnk ,O(n)).

Index

abelian category, 3acyclic, 4associated graded module, 29Atiyah completion theorem, 45

bar notation, 17bar resolution, 17bicomplex, 54Bockstein homomorphism, 38

Cartan–Eilenberg resolution, 54Cech cohomology, 56chain complex, 3chain homotopy, 4cochain complex, 3cohomology, 3coinduction, 22cross module, 50

derivation, 15differential graded algebra, 46double complex, 32

Eilenberg–MacLane space, 21enough projectives, 6exact couple, 32Ext functor, 9

filtration, 29

Grothendieck spectral sequence, 55group cohomology, 9group homology, 8group ring, 7

Hattori–Stallings rank, 53Hochschild–Serre spectral

sequence, 34homology, 3

induction, 22injective, 8injective resolution, 8invariant element, 28

left derived functor, 6

M -conjugate, 16Mackey decomposition, 24morphism

of chain complexes, 3

positive complex, 4projective, 4projective resolution, 4

rank, 52reduction, 22right derived functor, 8

sheaf, 56sheaf cohomology, 56split, 15

Tor functor, 7transfer maps, 25

wreath product, 39

Yoneda multiplication, 14

58