Math 99R - Representations and Cohomology of Groups

Home , Bockstein homomorphism, Group cohomology, Tor functor

Math 99r - Representations and Cohomology of Groups

Taught by Meng Geuo Notes by Dongryul Kim Spring 2017

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

Contents

1 February 1, 2017 3 1.1 Chain complexes ...... 3 1.2 Projective resolution ...... 4

2 February 8, 2017 6 2.1 Left derived functors ...... 6 2.2 Injective resolutions and right derived functors ...... 8

3 February 14, 2017 10 3.1 Long exact sequence ...... 10

4 February 17, 2017 13 4.1 Ext as extensions ...... 13 4.2 Low-dimensional group cohomology ...... 14

5 February 21, 2017 15 5.1 Computing the ﬁrst cohomology and homology ...... 15

6 February 24, 2017 17 6.1 Bar resolution ...... 17 6.2 Computing cohomology using the bar resolution ...... 18

7 February 28, 2017 19 7.1 Computing the second cohomology ...... 19

1 Last Update: August 27, 2018 8 March 3, 2017 21 8.1 Group action on a CW-complex ...... 21

9 March 7, 2017 22 9.1 Induction and restriction ...... 22

10 March 21, 2017 24 10.1 Double coset formula ...... 24 10.2 Transfer maps ...... 25

11 March 24, 2017 27 11.1 Another way of deﬁning transfer maps ...... 27

12 March 28, 2017 29 12.1 Spectral sequence from a ﬁltration ...... 29

13 March 31, 2017 31 13.1 Spectral sequence from an exact couple ...... 32 13.2 Spectral sequence from a double complex ...... 32

14 April 4, 2017 34 14.1 Hochschild–Serre spectral sequence ...... 34

15 April 7, 2017 37 15.1 Homology and cohomology of cyclic groups ...... 37

16 April 11, 2017 39 16.1 Wreath product ...... 39

17 April 14, 2017 41 17.1 Cohomology of wreath products ...... 41

18 April 18, 2017 44 18.1 Atiyah completion theorem ...... 44 18.2 Skew-commutative graded diﬀerential algebras ...... 46

19 April 21, 2017 49 19.1 Cohomology of D2n ...... 49 19.2 The third cohomology ...... 50

20 April 25, 2017 52 20.1 Swan’s theorem ...... 52 20.2 Grothendieck spectral sequence ...... 53 20.3 Cechˇ 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 required universal 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 an epimorphism 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 {Cn}n∈Z is a sequence of objects Cn ∈ A with boundary maps dn : Cn → Cn−1 such that dn−1 ◦ dn = 0.

d2 d1 d0 d−1 d−2 ··· −→ C1 −→ C0 −→ C−1 −−→ C−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 dn Cn Cn−1

fn fn−1

δn Dn Dn−1 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 −n and cochain complexes given by Cn 7→ C .

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 (maybe not uniquely) lift f to f˜ making the diagram commute:

X f˜

f P Y

A projective R-module is a projective object in the category of R-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 says projectivity 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 ( 0 if n > 0 Hn(P•) = A if n = 0.

Deﬁnition 1.9. A positive complex is a complex (C•, d•) with Cn = 0 for all n < 0. Deﬁnition 1.10. An positive acyclic chain complex is a complex (C, d) with Cn = 0 for n < 0 and Hn(C) = 0 for n > 0. So a projective resolution is positive chain complex that is both acyclic and projective (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

ϕ H0(ϕ) (C• −→ D•) 7→ H0(C•) −−−−→ H0(D•) Math 99r Notes 5 induces a bijection

homotopy classes of ←→ {H0(C•) → H0(D•)}. chain maps C• → D•

Proof. So we have to prove two things: ﬁrst that we can get maps on chains from H0(C) → H0(D), and that if two maps give the same maps then they are chain homotopic. We are inductively going to lift maps.

Cn Cn−1 Cn−1

dn dn−1 Dn Dn−1 Dn−2

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 projective, we can lift this map. Now we show that two chain maps giving the same maps on homology is homotopic. We may assume that f• gives the zero map H0(C) → H0(D) and show 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 resolution of 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 , we saw that the set of morphisms A → B is the homotopy classes of a chain map from a projective complex to an acyclic complex.

Corollary 2.1. Any two projective resolutions P•,Q• of A are homotopic to each 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 ∈ A there 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 preserves the 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 compute its homology. We deﬁne the n-th left derived functor associated to P• as

P• Ln F (A) = Hn(F (P•)). Why is it a functor? If there is a map A → B then we get a map between projective resolutions, and then we get a map between homology.

Theorem 2.4. Suppose we have an additive functor F : A → B. Let P,Q be two assignment of projective resolutions to objects of A . There exists a P Q canonical natural isomorphism Ln F ' Ln F between the two functors.

The main idea is that there is a chain map PA → QA given by the identity idA and then this induces a Hn(PA) → Hn(QA). Lemma 2.5. An additive functor F : A → B induces an additive functor F : ChA → ChB, given by F (C•)n = F (Cn) and F (ψ•)n = F (ψn). This satisﬁes 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 can add vertical maps one by one. (i) This is clear, because F is additive and so the conditions translates exactly. (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 the sequences at each level after applying F .

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

PA A

ϕ• ψ• idA idA

QA A 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 ψ because 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 independent 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

n TorR(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 deﬁne the group ring as the ring

X RG = rigi : only ﬁnitely many gi are nonzero

gi∈G 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 M with an action of G, or an R-linear representation of G. We deﬁne the group homology

Hn(G, M) = TorRG(R,M), P P 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 a ∼ canonical 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

M N 0.

So we may deﬁne: Deﬁnition 2.10. An injective object I is an object such that there is always a 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

n n R F (A) = H (F (I•)). Math 99r Notes 9

If F is left exact, then R0F =∼ F . Also if I is injective then RnF (I) = 0 for n > 0. If F is a contravariant functor, then this can be thought of as a covariant functor F : A op → B. So you can do similar things. One thing to be careful is that if P ∈ A is projective, then P ∈ A op is injective. For instance, to define left derived functors, you first need to pick an injective resolution and apply F and 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

n n ExtR(M,N) = R (HomR(M, −))(N).

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

n n H (G, M) = ExtRG(R,M).

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

n n ExtR(M,N) = R (HomR(−,N))(M).

n What is the different between the two definitions of ExtR(M,N)? The first definition is first finding an injective resolution 0 → N → I• and compute the cohomology of

0 → Hom(M,I0) → Hom(M,I1) → · · · , and the second deﬁnition is ﬁnd a projective resolution P• → M → 0 and take the cohomology of

0 → Hom(P0,N) → Hom(P1,N) → · · · .

They turn out the be equivalent deﬁnitions. People use the second deﬁnition more 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 morphism δn : coker ∂n+1 → ker ∂n−1 such that Cn → Cn−1 factors as

δn Cn → coker ∂n+1 −→ ker ∂n−1 → Cn−1 and there exists a exact sequence

δn 0 → Hn(C) → coker ∂n+1 −→ 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 the map 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 → A0 → A → A00 → 0 is a short exact sequence of chain complexes in an abelian category, then there is a long exact sequence

0 00 ··· Hn(A ) Hn(A) Hn(A )

0 00 Hn−1(A ) Hn−1(A) Hn−1(A ) ··· .

Proof. This follows from the snake lemma on

0 0 0

0 00 Hn(A ) Hn(A) Hn(A )

0 00 coker ∂n coker ∂n coker ∂n 0

0 00 0 ker ∂n ker ∂n ker ∂n

0 00 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 → A0 → A → A00 → 0 be a short exact sequence in A and F : A → B be an additive functor. Suppose that A has enough projectives. 0 Then there exists a map wn : LnF (A) → Ln−1F (A ) such that

00 ··· LnF (A) LnF (A )

wn 0 Ln−1F (A ) Ln−1F (A) ···

00 ··· L0F (A) L0F (A ) 0 is a long exact sequence. Proof. We ﬁrst look at the projective resolutions

0 00 0 P• P• P• 0

0 A0 A A00 0.

But if we apply F , there is no reason for the sequence to remain exact. The key 0 00 idea is to make Pn = Pn ⊕ Pn . We look at the ﬁrst level.

0 0 00 00 0 P1 P1 ⊕ P1 P1 0

∂0 ∂ ∂00 0 0 00 00 0 P0 P0 ⊕ P0 P0 0

0 00 0 0 A0 A A00 0 f g

0 0 0

00 00 00 We ﬁrst lift the map P0 → A using the fact that P0 is projective and A → A 0 00 0 0 00 is surjective. Then take (a , a ) = f (a ) + 0(a ). Let us look at the next 00 00 0 step. Because g ◦ 0 ◦ ∂ = 0, we see that there is some ν : P1 → A such that 00 0 0 f ◦ ν = 0 ◦ ∂ . Then using surjectivity of P0 → A we can lift the map to 00 0 0 00 λ : P1 → P0 such that f λ = 0∂ . Deﬁne ∂(a0, a00) = (∂0(a0) − λ(a00), ∂00(a00)).

Then we check that

0 00 0 0 0 00 00 00 0 00 00 00 ∂(a , a ) = f (∂ (a ) − λ(a )) + 0∂ (a ) = −f λ(a ) − 0∂ (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 00 0 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

R 0 R R 00 · · · → Tor1 (A ,B) → Tor1 (A, B) → Tor1 (A ,B) → A0 ⊗ B → A ⊗ B → A00 ⊗ B → 0.

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

0 00 0 → HomR(B,A ) → HomR(B,A) → HomR(B,A ) 1 0 1 1 00 → ExtR(B,A ) → ExtR(B,A) → ExtR(B,A ) → · · · .

We also have

00 0 0 → HomR(A ,B) → HomR(A, B) → HomR(A ,B) 1 00 1 1 0 → ExtR(A ,B) → ExtR(A, B) → ExtR(A ,B) → · · · .

One property is that

n M M n RM M R ExtR Ai,B = ExtR(Ai,B), Torn Ai,B = Torn (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 interpretation n equivalence class of n-fold extensions ExtR(M,N) = 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 0 0 N Mn−1 ··· M0 M 0, where Pi are projective. Reﬂexivity and symmetry are clear. To check transi- 0 tivity, you look at the pullback of Pk → Mk ← Pk. Recall that

n n ExtR(M,N) = H (HomR(P•,N)), where P• → M is a projective resolution. Note that if you have a projective resolution, then you can construct maps and then the map Pn → N represents a 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 a n map {equivalence classes} → ExtR(M,N). n We now check that this map is bijective. Given an element [f] ∈ H (HomR(P•,N)), how do we construct the extension? First we may assume that Pn → N is surjective 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

n In the same way, if two extensions come from the same element in ExtR(M,N), then we can exhibit a equivalence. Math 99r Notes 14

Take a short exact sequence 0 → N → M0 → M → 0. The equivalence class of this can be seen to be the image of idM in the map 1 0 → Hom(N,N) → Hom(M,M0) → Hom(M,M) → ExtR(M,N) → · · · . 1 Then we get an element of ExtR(M,N). Let us see how this generalizes to the higher order. Given an exact sequence β 0 → N → M1 −→ M0 → M → 0, we get two short exact sequences 0 → N → M1 → E → 0 and 0 → E → M0 → M → 0, where E = im β. Then we get long exact sequences coming from Hom(−,N), 1 · · · → Hom(N,N) → ExtR(E,N) → · · · , 1 2 · · · → ExtR(E,N) → ExtR(M,N) → · · · . 2 The image of idN under the two maps gives an element of ExtR(M,N). This directly generalizes to higher n. There is an R-linear map n m n+m ExtR(M,N) × ExtR (N,L) → ExtR (M,L) which is called the Yoneda multiplication. How is this deﬁned? Given two extensions

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 ExtR(M,M). Then ExtR(M,N) has a • ExtR(M,M)-module structure. Using this multiplication, we can say that in the long exact sequence

0 → Hom(L, N) → Hom(L, M0) → Hom(L, M) → · · · → Extn(L, M) → Extn+1(L, N) → · · · , 1 the connecting maps are multiplication by x, where x ∈ ExtR(M,N) represents the sequence 0 → N → M0 → M → 0.

4.2 Low-dimensional group cohomology Recall that ∗ ∗ H (G, M) = ExtRG(R,M). Taking R = Z, we have 0 G H (G, M) = HomZG(Z,M) = HomG(Z,M) = M . Theorem 4.1. The ﬁrst cohomology is M-conjugate classes of splitting group extensions H1(G, M) = . 1 → M → M o G → G → 1 Math 99r Notes 15

5 February 21, 2017

n Last time we showed that ExtR(M,N) is the equivalence classes of n-fold extensions 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

M-conjugacy classes of group extensions H1(G, M) = . 1 → 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 a group homomorphism. In this case, we can write E = M o G as a semi-direct product. 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 o G 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 natural ZG-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) → Ext1 ( ,M) → Ext1 ( G, M) → · · · . ZG Z ZG Z

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

G 1 0 → M → M → HomZG(IG, M) → H (G, M) → 0. Proposition 5.3. A map of sets d : G → M deﬁnes a group homomorphism δ : IG → M given by δ(g − 1) = d(g). This δ is a ZG-linear if and only if d is a derivation. Proof. The map δ being G-equivariant means δ(h(g − 1)) = hδ(g − 1). This means 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 G action given by (gd)(h) = gd(h−1hg).) So we now have to study that map M → 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 conclusion is Der(G, M) H1(G, M) = . {(g 7→ gm − m): m ∈ M}

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

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

1 H (G, M) = {0 → M → E → Z → 0}.

ZG Recall that Hn(G, M) = Torn (Z,M). ∼ ab Proposition 5.5. H1(G, Z) = G/[G, G] = G . Proof. We again take the sequence 0 → IG → ZG → Z → 0. Then we get

ZG Tor1 (Z, ZG) → H1(G, Z) → Z ⊗ZG IG → Z ⊗ZG ZG → Z ⊗ZG Z → 0. Then

ZG id Tor1 (Z, ZG) → H1(G, Z) → Z ⊗ZG IG → Z −→ Z → 0.

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

Next time we are going to construct a space K(G, 1) such that π1(K) = G and πi(K) = 0 for i > 1. Then we get H1(G, Z) = H1(K(G, 1)). So H1 is the abelianization of π1. Math 99r Notes 17

6 February 24, 2017 6.1 Bar resolution We note that free R-modules are always projective. Let

n+1 Fn = free R-module on (n + 1)-tuples (g0, . . . , gn) ∈ G , and we deﬁne the maps

n X i ∂n(g0, . . . , gn) = (−1) (g0,..., gˆi, . . . , gn). i=1 Then we get a sequence

∂3 ∂2 ∂1 ··· −→ F2 −→ F1 −→ 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

−1 −1 −1 (g0, . . . , gn) = g0[g0 g1|g1 g2| · · · |gn−1gn], [g1|g2| · · · |gn] = (1, g1, g1g2, . . . , g1g2 ··· gn).

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

n−1 X i ∂n[g1| · · · |gn] = g1[g2| · · · |gn] + (−1) [g1| · · · |gigi+1| · · · |gn] i=1 n + (−1) [g1| · · · |gn−1].

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

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

Here, note that the cochain complex for cohomology is

d0 d1 0 → HomRG(F0,M) −→ HomRG(F1,M) −→· · · where the cochain complex are

i n C (G, M) = HomRG(Fi,M) = Fun(G ,M) Math 99r Notes 18 and the coboundary maps

n (d f)([g1| · · · |gn+1]) = f ◦ ∂n[g1| · · · |gn+1] n X i = g1f([g2| ... |gn+1]) + (−1) f([g1| ... |gigi+1| ... |gn+1]) i=1 n+1 + (−1) f([g1| ... |gn]).

6.2 Computing cohomology using the bar resolution In particular, we have 0 ∼ C (G, M) = HomRG(RGh[]i,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

1 (d f)([g1|g2]) = g1f([g2]) − f([g1g2]) + f([g1]). 1 Note that d f = 0 is equivalent to f([g1g2]) = g1f([g2])+f([g1]). This is exactly saying that f ∈ Der(G, M). This again shows Der(G, M) H1(G, M) = . {ga − a} Let us now compute H2(G, M). We have for f ∈ C2(G, M),

2 (d f)([g1|g2|g3]) = g1f([g2|g3]) − f([g1g2|g3]) + f([g1|g2g3]) − f([g1|g2]). So we can write {(f : G × G → M): d2f = 0} H2(G, M) = . {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

π 1 M i E G 1. s If g is a group homomorphism, then it splits. In general, we can ﬁnd a function f : 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 i E π G 1.

Since M is a normal subgroup of E, the group E acts on M by conjugation. If M is abelian, we further get an action on G on M, induced from E, by lifting g ∈ G tog ˜ ∈ E and letting i(g · m) =gi ˜ (m)˜g−1. If we are already given an ZG-module structure, we require the two actions to be the same. 1 We pick up from last time. The ﬁrst cohomology H (Z/2, F2) with the trivial 1 action is H (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 the group 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 completely recovered 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. We have

((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

π 1 M i E G 1 s1

1 M E G 1. 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 are equivalent 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). But this is ﬁne because you can use the normalized bar resolution deﬁned by

RGh[g1| · · · |gn]i F¯n = . RGh[g1| · · · |gn]: gi = 1 for some ii 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 have a cellular G-action on a CW-complex X˜. Assume that the action completely discontinuous 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) are ˜ quasi-isomorphic. This means that C∗(X; Z) is a free ZG-module resolution of Z. To compute H∗(G; Z), we need to take Z ⊗ZG (−) on the chain complex and look at the homology. We see that ˜ ˜ ˜ ∼ ˜ Z ⊗ZG Cn(X; Z) = Cn(X; Z)/hgx − x : x ∈ Cn(X; Z)i = Cn(X/G; Z). The second equality holds because simplices are simply connected and so there always 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 and πi(X) = 0 for i > 1. Whitehead’s theorem then tells us that K(G, 1) is unique up to homotopy. Let us now give a construction for X˜. For each g0, . . . , gn ∈ G, you make a n+1 simplicial complex so that the faces of the n-simplex [g0, . . . , gn] ∈ G are

n X i ∂[g0, . . . , gn] = (−1) [g0,..., gˆi, . . . , gn]. i=0 This is indeed homotopy equivalent to a point, because you can just shrink everything 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 to talk about induction and restriction.

9.1 Induction and restriction Consider an ring morphism Γ → Λ, so that Λ is an Γ-algebra. There are three things we can do: • Reduction: If M is a left Λ-module, then we can just consider M as a left Λ Γ-module. We write this as M ↓Γ or ResΓ M. • Induction: If N is a left Γ-module, then we can construct tensor product Γ Λ⊗Γ M as a left Λ-module and a right Γ-module. We write this as N ↑ N Λ or IndΓ N.

• Coinduction: If N is a left Γ-module, then we can construct HomΓ(Λ,N), Λ Λ which is a right Λ-module. We denote this as N ⇑ or CoindΓ N. Lemma 9.1. For Γ and Λ rings, M a left Λ-module, A a Λ-Γ bimodule, and N 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 a left Γ-module. Λ ∼ Λ (i) HomΓ(N, ResΓ M) = HomΛ(IndΓ ,M) Λ ∼ Λ (ii) HomΓ(ResΓ M,N) = HomΛ(M, CoindΓ N) Λ Λ Λ So ResΓ has a left adjoint IndΓ and a right adjoint CoindΓ . Proof. For (i) use A = Λ, and for (ii) use switch the roles of Λ and Γ and use A = Λ. Proposition 9.3. Suppose Γ ,→ Λ and Λ is a projective Γ-module. Let M and N be as above. n Λ ∼ n Λ (i) ExtΓ(N, ResΓ M) = ExtΛ(IndΓ N,M) Math 99r Notes 23

n Λ ∼ n Λ (ii) ExtΓ(ResΓ M,N) = ExtΛ(M, CoindΓ N)

Proof. (i) Suppose we have a projective resolution P• → N → 0 over Γ. Then n Λ n by deﬁnition, ExtΓ(N, ResΓ M) = H (Hom(P•,M)). Because Λ is projective, tensoring with Λ is exact, and we get a projective resolution Λ⊗ΓP• → Λ⊗ΓN → 0. Then

n Λ n n Λ ExtΛ(IndΓ N,M) = H (Hom(Λ ⊗Γ P•,M)) = H (Hom(P•, ResΓ 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 an inclusion 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

G ∗ G IndH N = RG ⊗RH N = (RG) ⊗RH N = HomRH (RG, N) = CoindH N.

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

G ∗ G ∗ G ∗ G ∗ ResH M = (ResH M) , IndH N = (IndH N) .

That is, the dual functor commutes with restriction and induction. Proposition 9.4. For ﬁnite groups H ≤ G, G ∼ G (i) HomRH (N, ResH M) = HomRG(IndH N,M), G ∼ G (ii) HomRH (ResH M,N) = HomRG(M, IndH N), n G ∼ n G (iii) ExtRH (N, ResH M) = ExtRG(IndH N,M), n G ∼ n G (iv) ExtRH (ResH M,N) = ExtRG(M, IndH M). If we apply (ii) to M = R, then we get

n R n G n G H (H,N) = ExtRH (Res R,N) = ExtRG(R, IndH N) = H (G, IndH N).

∗ Example 9.5. Let us H (Z/2; F2 ⊕ F2) with the action being switching the two components. Note that this is simply F2[Z/2] ⊗F2 F2. Then ( ∗ ∗ F2 ∗ = 0, H (Z/2, F2 ⊕ F2) = H ({1}, F2) = 0 ∗ > 0. Math 99r Notes 24

10 March 21, 2017

Last time we proved Λ ∼ Λ HomΓ(N, ResΓ M) = HomΛ(IndΓ N,M), Λ ∼ Λ HomΓ(ResΓ M,N) = Hom(M, CoindΓ N), and if Λ is projective over Γ, then the same thing is true for Ext. If H ⊆ G is a subgroup, then

∗ ∼ ∗ G ∼ G H (H,M) = H (G, CoindH M),H∗(H,M) = H∗(G, IndH M).

10.1 Double coset formula Consider two subgroups H,K ≤ G. Let us see what happens if we induce from K to G and then restrict from G to H. Theorem 10.1 (Mackey decomposition). Let M be an RK-module. Then

G G M H gKg−1 ResH IndK M = IndH∩gKg−1 ResH∩gKg−1 gM. HgK

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

G G M ResH IndH M = gM. g∈G/H

Proof. We have a decomposition M RG = R(HgK) HgK as a RH-RK bimodule. Then

G G M ResH IndK M = RG ⊗RK M = R(HgK) ⊗RK M HgK M = RH ⊗R(H∩gKg−1) R(gK) ⊗RK M HgK M H gKg−1 = IndH∩gKg−1 ResH∩gKg−1 gM. HgK Corollary 10.2. If M is an RK-module and N an RH-module, then

G G M G gKg−1 H (i) IndH M⊗RIndH N = IndH∩gKg−1 ((ResH∩gKg−1 gM)⊗R(ResH∩gKg−1 N)) HgK G G M G gKg−1 H (ii) HomR(IndK M, IndH N) = IndH∩gKg−1 HomR(ResH∩gKg−1 gM, ResH∩gKg−1 N) HgK Math 99r Notes 25

G G M gKg−1 H (iii) IndK M ⊗RG IndH N = ResH∩gKg−1 gM ⊗R(H∩gKg−1) ResH∩gKg−1 N HgK G G M H gKg−1 (iv) HomRG(IndH N, IndK M) = HomR(H∩gKg−1)(ResH∩gKg−1 N, ResH∩gKg−1 gM). HgK Lemma 10.3. For N an RH-module and M and RG-module, G G G (i) IndH (N ⊗R ResH M) = IndH N ⊗R M, G G G (ii) IndH HomR(N, ResH M) = HomR(IndH ,M). Proof of Corollary 10.2. (i) We have, by the decomposition,

G G G G G IndK M ⊗R IndH N = IndH (ResH IndK M ⊗R N) M G H gKg−1 = IndH (IndH∩gKg−1 ResH∩gKg−1 M ⊗R N) HgK M G gKg−1 H = IndH∩gKg−1 (ResH∩gKg−1 gM ⊗R ResH∩gKg−1 N). HgK 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,

RG G G M R(H∩gKg−1) gKg−1 H (v) Tor∗ (IndK M, IndH N) = Tor∗ (ResH∩gKg−1 gM, ResH∩gKg−1 N), HgK ∗ G G M ∗ H gKg−1 (vi) ExtRG(IndH N, IndK M) = ExtR(H∩gKg−1)(ResH∩gKg−1 N, ResH∩gKg−1 gM). HgK

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

∗ α∗ ∗ α ∗ H∗(H,M) −−→ H∗(G, M),H (G, M) −−→ H (H,M). If H ≤ G has ﬁnite index, there are canonical maps in the other direction:

G G ∗ ∗ TrH : H∗(G, M) → H∗(H,M), TrH : 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 injection ∗ M,→ HomRH (RG, M). Let us ﬁrst see how the maps α∗ and α are deﬁned.

∗ ∗ H∗(G, RG ⊗RH M) H∗(G, M) H (G, M) H (G, HomRH (RG, M))

α∗ α∗ ∗ H∗(H,M) H (H,M) Math 99r Notes 26

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

∗ ∗ H (G, RG ⊗RH M) H (G, M) H∗(G, M) H∗(G, HomRH (RG, M))

G G TrH TrH ∗ H (H,M) H∗(H,M) Math 99r Notes 27

11 March 24, 2017

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

11.1 Another way of deﬁning transfer maps For RG-modules M and M 0, there is a map

G 0 0 X −1 TrH : HomRH (M,M ) → HomRG(M,M ); φ 7→ (gφ : m 7→ gφ(g m)). g∈G/H

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) G G and β ∈ HomRG(M2,M3), then β ◦ TrH (α) = TrH (β ◦ α). G G (ii) Under a similar condition, TrH (β) ◦ α = TrH (β ◦ α). G K G (iii) If H ≤ K ≤ G are ﬁnite index subgroups, TrK TrH = TrH .

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

G X H TrK (α) = TrH∩gKg−1 (gα). HgK

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

0 0 HomRH (P•,M ) → HomRG(P•,M ).

n 0 n 0 Applying cohomology gives a map ExtRH (M,M ) → ExtRG(M,M ). In particular, 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 0 a left RG-module, there is a map

G 0 0 0 X −1 0 TrH : M ⊗RG M → M ⊗RH M ; m ⊗ m 7→ mg ⊗ g m . g∈G/H

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

G G (i) β ◦ TrH (α) = TrH (β ◦ α) G G (ii) TrH (β) ◦ α = TrH (β ◦ α) G K G (iii) TrK TrH = TrH

−1 G G X H gKg ∗ (iv) ResH TrK (α) = TrH∩gKg−1 ResH∩gKg−1 (gα) for α ∈ H (K,M). HgK

G G X H gKg−1 (v) TrH ResK (α) = ResH∩gKg−1 TrH∩gKg−1 (gα) for α ∈ H∗(K,M). HgK Math 99r Notes 28

G G ∗ (vi) TrH ResH (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

G X TrH :[m]G 7→ [gm]. g∈H\G

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

G X TrH : m 7→ gm. g∈G/H There is also a topological description. If H ≤ G is a finite index, there is a covering 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 that Hn(H,M) = 0 for some n. Then Hn(G, M) is annihilated by [G : H]. In particular, if [G : H] is invertible in M then Hn(G, M) = 0. G G Proof. For the first part, simply note that TrH ResH (α) = [G : H]α. For the second part, multiplication by [G : H] induces an isomorphism M → M and hence an isomorphism Hn(G, M) → Hn(G, M). Corollary 11.4. If |G| is finite, then Hn(G, M) is annihilated by |G| for all n > 0. If |G| is invertible in M, then Hn(G, M) = 0. Proposition 11.5. If [G : H] is invertible in R, then

G n 0 n 0 ResH : ExtRG(M,M ) ,→ ExtRH (M,M ) is injective. In particular, H∗(G, M) ,→ H∗(H,M).

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

n n H (G, M)(p) ,→ H (P,M), where P ≤ G is a p-Sylow subgroup. Deﬁnition 11.6. An element z ∈ Hn(H,M) is G-invariant if

H gHg−1 ResH∩gHg−1 z = ResH∩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.

n Proposition 11.7. H (G, M)(p) is isomorphic to the set of G-invariant elements in Hn(P,M). Math 99r Notes 29

12 March 28, 2017 12.1 Spectral sequence from a ﬁltration r r A spectral sequence consist of a collection the E -page Ep,q with diﬀerentials

r r r d : Ep,q → Ep−r,q+r−1. So the total degree decrease by 1 as p + q → p + q − 1. The ﬁltration degree p decreases by r. The next page can be computed as

r+1 r r Ep,q = H(E , d ). For cohomology, we will get

p,q p+r,q−r+1 p,q dr : Er → Er ,Er+1 = H(Er, dr). There are two ways to construct a spectral sequence. The first one is from a filtration 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, S where FpM are submodules of M and p FpM = M. To this ﬁltration is an associated graded module M M Gr M = Grp M = (FpM/Fp−1M). p p

Lemma 12.1. If f : M → M 0 preserves filtration, and it induces Gr M =∼ Gr M 0, then f is an isomorphism. Let us now assume that M is graded independent to the filtration. In this case, we require FpM to be a graded submodule. Then for fixed n, {FpMn} gives 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 complementary degree. Now consider a chain complex {C•, ∂} with FpC a filtration that is also a subcomplex of C. There is an induced filtration

FpC ∩ Z FpH(C) = im{H(FpC) → H(C•)} = . FpC ∩ B We can identify

FpH(C) FpC ∩ Z Grp H(C) = = . Fp−1H(C) (FpC ∩ B) + (Fp−1C ∩ Z) Math 99r Notes 30

Let us deﬁne r −1 ∞ Zp,q = FpCp+q ∩ ∂ Fp−rCp+q−1,Zp,q = FpCp+q ∩ Z and r r−1 ∞ Bp,q = FpCp+q ∩ ∂Fp+r−1Cp+q+1 = ∂Zp+r−1,Bp,q = FpCp+q ∩ B. Then we have 0 1 ∞ ∞ 1 0 FpC = Zp ⊇ Zp ⊇ · · · ⊇ Zp ⊇ Bp ⊇ · · · ⊇ Bp ⊇ Bp.

Assume that each {FpCn} is finite, i.e., there exists a `(n) depending on n such that F`(n)Cn = Cn and F−`(n)Cn = 0. Then for each p and q, there exists a r (actually something like r = `(p + q − 1) + p does the job) such that r r+1 ∞ Zp,q = Zp,q = ··· = Zp,q. Likewise we have ∞ r+1 r Bp,q = ··· = Bp,q = Bp,q for some r. Let us now define Zr Er = p,q . p,q r r−1 Bp,q + Zp−1,q+1 r r r Because both Zp and Bp stabilizes, we can say that Ep,q converges (i.e., stabilizes) to ∞ Zp ∞ ∞ = Grp H(C) Bp + Zp−1 as r → ∞. On the 0-th page, we have 0 0 Zp,q FpCp+q Ep,q = 0 0 = = Grp,q C. Bp,q + Zp−1,q (sth in Fp−1Cp+q) + Fp−1Cp+q The 0-th differential d0 is induced from ∂ as

0 0 FpCp+q FpCp+q−1 0 d : Ep,q = → = Ep,q−1. Fp−1Cp+q Fp−1Cp+q−1 If you try hard, you will check that −1 1 FpC ∩ ∂ Fp−1C 0 0 Ep,q = = H(E , d ). ∂FpC + Fp−1C 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 i A

k j B. We have a differential d : B → B defined by jk, and it is a differential because jkjk = j0k = 0. Math 99r Notes 31

13 March 31, 2017

r r Last time we defined a spectral sequence {Ep,q, d } and discussed how it arises from an increasing filtration FpC• of a chain complex C•. We can define a r spectral sequence Ep,q that computes H•(C•). Actually the spectral sequence converges to ∞ ∼ Ep,q = Grp,q H∗(C•) in the case when the filtration is dimension-wise finite. The 0-th page is given by 0 Ep,q = FpCp+q/Fp−1Cp+q. There is a nice application. For X a finite-dimensional CW-complex, we sing ∼ cell would like to show that H∗ = H∗ . Consider the cellular chain complex sing C∗ and give a filtration given by sing sing (p) FpC∗ (X) = C∗ (X ). We have sing sing (p) FpC (X) C (X ) E0 = p+q = p+q = Csing(X(p),X(p−1)). p+q sing sing (p−1) p+q Fp−1Cp+q (X) Cp+q (X ) Then we see that ( 1 sing (p) (p−1) 0 if q > 0, Ep,q = Hp+q (X ,X ) = cell Cp (X) if q = 0. The spectral sequence then looks like Figure 1, where no dot means that there is a 0 there. q

d1 d1 d1 d1 p

sing sing (p) Figure 1: Spectral sequence coming from FpC∗ (X) = C∗ (X )

This sequence has 2 cell Ep,0 = Hp (X) and then all dr = 0 for r > 1. So the spectral sequence stays the same from the E2 page and we get sing ∼ cell H∗ (X) = H∗ (X), sing because the only nonzero graded associated in the ﬁltration of H∗ (X) is 2 cell E∗,0 = H∗ (X). Math 99r Notes 32

13.1 Spectral sequence from an exact couple Consider a series of maps A i A

k j B. 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 deﬁne A1 = i(A). Proposition 13.1. Under this construction,

i1 A1 A1

k1 j1 B1 is again an exact couple.

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

M M FpC• A = F C , i : F C ,→ F C ,B = . p • p−1 • p • F C p p p−1 • This induces a long exact sequence

i1 j1 k1 · · · → Hk(A) −→ Hk(A) −→ Hk(B) −→ Hk−1(A) → · · · . We then get a exact couple

i1 H∗(A) H∗(A)

k1 j1 H∗(B).

1 So B1 = E∗,∗, and if you run the construction iteratively, you are going to get r Br = E∗,∗ for r ≥ 0.

13.2 Spectral sequence from a double complex

As another example, let us consider a double complex {Cp,q} with diﬀerentials ∂0 and ∂00 such that ∂0 Cp,q Cp−1,q

∂00 ∂00 ∂0 Cp,q−1 Cp−1,q−1 Math 99r Notes 33 commutes and ∂02 = ∂002 = 0. This can also be considered as a chain complex in the category of chain complexes. Given this data, we can deﬁne a total complex

M 0 p 00 (TC)n = Cp,q,D = ∂ + (−1) ∂ . p+q=n

We can ﬁlter this total complex by M Fp(TC)n = Ci,n−i. i≤p

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 be r dimension-wise finite. Then {Ep,q} is going to compute H∗(TC). The zeroth page is L Fp(TCp+q) i≤p Ci,p+q−i Ep,q = = L = Cp,q Fp−1(TCp+q) i≤p−1 Ci,p+q−i

00 1 and the 0-th diﬀerentials is going to be simple ∂ . Then Ep,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

1 Ep,q = Hp(Cp(X) ⊗ C∗(Y )) = Hq(Y ; Cp(X)) = Hq(Y ) ⊗ Cp(X) and it follows that 2 Ep,q = Hp(X) ⊗ Hq(Y ). If you dig in, you will be able to check that dr = 0 for r ≥ 2. Then we get

M ∞ M 2 M Hn(X × Y ) = Hn(TC) = Ep,q = Ep,q = Hp(X) ⊗ Hq(Y ). p+q=n p+q=n p+q=n

This is the K¨unnethformula 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. Recall that Hn(G, M) = H∗(F• ⊗RG M), 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 double L complex (F ⊗ C)n = p+q=n Fp ⊗ Cq. Let us compute this homology using the spectral sequence. We have

1 Ep,q = Hq(Fp ⊗RG C•) = Fp ⊗RG Hq(C•) since Fq is projective. The next page will be

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

This spectral sequence, which is called the Hochschild–Serre spectral sequence is going to converges to Hp+q(G, C•). 0 ∼ Proposition 14.1. If f : C• → C• is a quasi-isomorphism, then H∗(G, C•) = 0 H∗(G, C•). There is another direction of running the spectral sequence. We can set

1 Ep,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

1 Ep,0 = H0(G, Cp) = (Cp)G = Cp/{gm − m}, and all other things in the ﬁrst page vanish. All dr for r > 2 should vanish because either the source or the target is zero. So we get

2 ∞ H∗(CG) = E = E = H∗(G, C•).

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

2 Ep,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 ⊗RG M = (F ⊗R M)G = ((F ⊗R M)H )Q = (F ⊗RH M)Q.

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

G G Q RG ⊗RH M = IndH ResH M = R[G/H] ⊗R M = Ind{1} M. So Q H∗(Q, Ind{1} M) = H∗({1},M) = 0 for ∗ > 0. Now we can ﬁnally apply the proposition and get the spectral sequence

2 Ep,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

−1 cg : h 7→ ghg , m 7→ gm gives an action −1 cg∗ : H∗(H,M) → H∗(gHg ,M).

Proposition 14.3. If g ∈ H, then cg∗ : H∗(M) → H∗(M) is the identity. Proof. Even on the chain level, we have

−1 cg∗ : F ⊗RH M → F ⊗R[gHg−1] M; x ⊗ m 7→ xh ⊗ 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 a spectral sequence in homology that gives

p,q p q p+q E2 = H (Q, H (H,M)) ⇒ H (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, M H ) → H1(G, M) → H1(H,M)Q → H2(Q, M H ) → H2(G, M). Proof. Look at the E2 page of the homology spectral sequence. r 2 2 Because the diﬀerentials d at E1,0 are all zero for r ≥ 2, we see that E1,0 = ∞ 2 E1,0. On the other hand, we have d that can possibly do something to E0,1, which is given by 2 2 d 2 0 → E2,0 −→ E0,1 → 0. Math 99r Notes 36

2 2 0 E0,1 E1,1

2 2 2 E0,0 E1,0 E2,0

0 0

Figure 2: E2 page of the Hochschild spectral sequence

Thus after taking the homology, we will get

3 2 2 3 0 → E2,0 → E2,0 → E0,1 → E0,1 → 0.

3 3 Furthermore, there are no more diﬀerential in and out of E0,1 and E2,0. Thus 3 ∞ 3 ∞ E0,1 = E0,1 and E2,0 = E2,0. That is,

∞ 2 2 ∞ 0 → E2,0 → E2,0 → E0,1 → E0,1 → 0.

On the other hand, one of the basic theorem we had is that the homology of the total chain complex is given by the ﬁltration. Then we have a short exact sequence ∞ ∞ 0 → E0,1 → H1(G, M) → E1,0 → 0. We can then concatenate the two exact sequences to get

∞ 2 2 ∞ 0 → E2,0 → E2,0 → E0,1 → H1(G, M) → E1,0 → 0.

∞ ∞ ∞ ∞ Finally note that E0,2, E1,1, and E2,0 give a ﬁltration on H2(G, M). So E2,0 appear as a quotient of H2(G, M). Hence we get

2 2 ∞ H2(G, M) → E2,0 → E0,1 → H1(G, M) → E1,0 → 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 that Z[G] = Z[x, x−1] as a ring. We have a resolution

·(1−x) x7→1 0 → Z[G] −−−−→ Z[G] −−−→ Z → 0.

This is related to the fact that BZ = S1 which has ﬁnite homology or cohomology. Anyways, we can compute homology H∗(Z,A) for a Z[G]-module A as f 0 → A −→ A → 0. Then G H0(Z,A) = coker f = AG,H1(Z,A) = ker f = A . In the other direction, we are going to have

f ∗ 0 ← HomZ[G](Z[G],A) = A ←− HomZ[G](Z[G],A) = A ← 0 and so 0 G 1 H (Z,A) = A ,H (Z,A) = AG. We can make a similar computation for G = Z/p. Let x be the generator of G = Z/p. The resolution is given by

1−x 1+x+···+xn−1 1−x x7→1 · · · → Z[Z/n] −−−→ Z[Z/n] −−−−−−−−−→ Z[Z/n] −−−→ Z[Z/n] −−−→ Z.

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

1−x 1+x+···+xn−1 1−x x7→1 · · · → A −−−→ A −−−−−−−−−→ A −−−→ A −−−→ Z. This immediately shows

G H0 = A/ha − axi = AG,H2i+1 = H1 = A /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 = ker N/ im(1 − x),H2i = H2.

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

i dimFp H (Z/p; Fp) = 1 Math 99r Notes 38 for all i. The cup product structure (which comes from the comultiplication map B• → B• ⊗ B•) can be computed as ( ∗ F2[x] p = 2, H (Z/p; Fp) = 2 2 Fp[x, y ]/(x ) p > 2.

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

1 1 1 α 2 H (Z/p, Z) → H (Z/p, Z) → H (Z/p, Fp) −→ H (Z/p, Z) → · · · and y can be taken to be the image of α(x) under the map H2(Z/p, Z) → 2 H (Z/p, Fp) coming from the reduction map Z → Fp. This procedure has a name. There is always a map

∗ ∗+1 H (X; Fp) → H (X; Fp) and this is called a Bockstein homomorphism. In the case p = 2 we have y = 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 permutation group. We have an exact sequence

1 → Z/3 → Σ3 → Z/2 → 1.

Take k = F2. Then the pages are given by

( p 2 p q H (Z/2, F2) q = 0 Ep,q = H (Z/2,H (Z/3, F2)) = 0 q > 0.

This shows that p ∼ p H (Σ3, F2) = H (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

×p G o Z/p = G o Z/p, where the action is given by

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

Then there is a short exact sequence

p p 1 → G → G o Z/p → Z/p → 1.

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

s,t s t p s+t E2 = H (Z/p, H (G , Fp)) ⇒ H (G o Z/p, Fp).

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

∗ p O ∗ H (G , Fp) = H (G, Fp). p

Because the action of Z/p on Gp is given by shifting the elements, the induced ∗ p action of Z/p on H (G , Fp) is also going to be just shifting the tensors. L Lemma 16.1. Let Λ = i Λi be a graded Fp-vector space. Give a Z/p-action ⊗p ⊗p on Λ by cyclic shifting. Then as a Fp[Z/p]-module, Λ is a direct sum of a free 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

∗ p M M H (G , Fp) = Fp[Z/p] ⊕ Fp.

Then we see that

∗ ∗ p M ∗ M ∗ E2 = H (Z/p, H (G , 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 see that λi ⊗ · · · ⊗ λi lives in the pi-th graded pieces and so will come out in the pi-th cohomology. This shows that,

 t p /p H (G , Fp)Z s = 0 s,t s t p  E2 = H (Z/p, H (G , Fp)) = 0 s > 0, p - t  s ip p H (Z/p, H (G , Fp)) t = ip. Math 99r Notes 40

s t p Figure 3: Spectral sequence from H (Z/p, H (G , Fp))

In particular, if t = ip then

s,ip M s E2 = H (Z/p, Fp). x a basis of n H (G,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 o Z/p = Gp o Z/p. Then there is an exact sequence and so Gp → G o Z/p → Z/p. So we have a spectral sequence s,t s s p s+t E2 = H (Z/p, H (G , Fp)) ⇒ H (G o Z/p, Fp). t p Because H (G , Fp) is a direct sum of Fp[Z/p] and Fp, we get some description s,t of E2 .

17.1 Cohomology of wreath products For s = 0, we have

0,t s p Z/p ⊗p i E2 = H (G , Fp) = {λ : λ ∈ H (G, Fp)} ⊕ (free Fp[Z/p]). Recall that if A ≤ B is of ﬁnite index, there is a transfer map tr : H∗(A, M) → H∗(B,M), and we have X res ◦ tr = g. g∈B/A So applying this to Gp ,→ G o Z/p gives

X p−1 ∗ p ∗ p res ◦ tr = g = (1 + T + ··· + T ): H (G , Fp) → H (G , Fp). g∈Z/p Then we see that actually

0,t ⊗p i p E2 = {λ : λ ∈ H (G , Fp)} ⊕ (image of res ◦ tr).

0,∗ ∗ res ∗ Lemma 17.1. If x ∈ E2 comes from x ∈ im(H (Q, k) −−→ H (P, k)), then x is a permanent cycle, i.e., d∗x = 0. Proof. Consider the map

P i Q Q/P

i P P {1}.

Then there is a map of spectral sequences

H∗(Q/P, H∗(P, k)) H∗(Q, k)

res H∗({1},H∗(P, k)) H∗(P, k).

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 permanent cycle. Math 99r Notes 42

We note that

s ⊗p s ⊗p H (Z/p, Fphλ i) = H (Z/p, Fp) ⊗ hλ i. ∗ ∼ 2 j ⊗p j But recall that H (Z/p, Fp) = Fp[x, y]/(x ). So the basis are y ⊗λ and xy ⊗ 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 find out all the differentials. Let us go back into the case of the wreath product. The image of the restriction map is a permanent cycle, and so we may as well just ignore that and look at the other parts.

⊗p 0,ip Lemma 17.2. The diﬀerential on λi ∈ Er has to be zero. Proof. If we have a short exact sequence H → G → G/H, this gives a ﬁbration BH → 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. n Given an element λn ∈ H (BG, Z/p), it corresponds to a map fn : BG → K(Z/p, n). Then this induces a map

∗ ∗ ∗ fn : H (K(Z/p), n) → H (BG),

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

BGp BG o Z/p BZ/p p fi

p p K(Z/p, i) K(Z/p, i) ×Z/p EZ/p BZ/p

This gives a map of spectral sequence

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

s t p s+t H (BZ/p; H (BG ; Fp)) H (BG o Z/p; Fp)

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

t p t p H (K(Z/p, n) ; Fp) → H (BG ; Fp).

∗ ⊗n ⊗p Since f (ιp ) = λn , we can use functoriality to obtain data of the diﬀerential ⊗p on λn . Math 99r Notes 43

⊗p We want to show that all differentials on λn are zero. It suffices to check ⊗p that all differentials on ιn are zero, by the above argument. But by the t Hurewicz theorem and the universal coefficients theorem, we have H (K(Z/p, n), Fp) = s,t 0 for t ≤ n−1. So we have E2 = 0 for 0 < t < np. So the only possible nonzero 0,np np+1,0 differential is dnp+1 : E → E . Here is a general fact for Serre spectra sequences (or Hochschild–Serre spectral sequences). The bottom row of the infinity page is

∗,0 ∗ ∗ p E∞ = im H (BZ/p; Fp) → H (K(Z/p, i) ×Z/p EZ/p) . If show that this map is injective, then nothing on the bottom row can be 0,np np+1,0 killed and so dnp+1 : E → E has to be zero. This can be shown by exhibiting 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 deﬁne W1 = Z/p, Wr = Wr−1 o Z/p. P i If n = aip and 0 ≤ ai < p, then the p-Sylow subgroups of Sn can be written as a1 a2 ak W1 × W2 × · · · × Wk . So knowing about the cohomology of the wreath products is going to tell us something 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 {(f : G → ): f(xy) = f(x) + f(y)} H1(G, ) = Z = 0. Z 0 The next cohomology can be computed as 2 H (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 of G, or equivalently 1-dimensional characters of G. So there seems to be some connection 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 established easier if we use topology. If you think about it a bit, complex representations of G are somewhat like complex vector bundles over BG. Rank k complex vector bundles over BG are [BG, BU(n)], and this is quite similar to group homomorphisms G → U(n). So it seems reasonable to expect representations of G to be related to K∗(BG). Now K∗ is a cohomology theory. There is something called an Atiyah–Hirzebruch spectral sequence that computes a cohomology theory in terms of its coefficient ring. For a space X, the set of complex vector bundles over X has a additive structure, 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 ring 0 morphism K (X) → Z given by mapping to the dimension of the fiber over x0. We further define the ideal 0 0 Ke (X) = ker(K (X) → Z). Theorem 18.2 (Bott periodicity). For any finite CW-complex X,

Ke 0(Σ2X) = Ke 0(X). Ke 0(−) is a representable functor, and this allows us to deﬁne a cohomology theory Ke −n(X) = Ke 0(ΣnX), which will further extend to the positive side using Bott periodicity as

( 0 n Ke (X) 2 | n, ∗ ∗ Ke (X) = K (X) = Ke (X+). Ke 0(ΣX) 2 - n, Math 99r Notes 45

Definition 18.3. For a CW-complex X such that each skeleton Xn is finite, define ∗(X) = lim K∗(Xn). K ←− Just like complex vector bundles, there is an additive structure, coming from ⊕, and a multiplicative structure, coming from ⊗, on the set of complex representations 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 × n} C → BG, (v, x) ∼ (gv, gx) 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 its dimension. 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 α extends to αˆ : R[(G) → K ∗(BG). Theorem 18.4. For every ﬁnite G, αˆ 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 know all the representations: they are simply the 1-dimensional ones 1, ρ, ρ2, . . . , ρn−1. So the ring R(G) is given by

n n R(G) =∼ Z[ρ]/(ρ − 1) =∼ Z[σ]/((σ + 1) − 1), and I(G) = (σ). So the completion is going to look like

n R[(G) = Z[[σ]]/((σ + 1) − 1)

But how do you compute K ∗(BG)? If you recall, there is a Serre spectral sequence s t s+t H (B; H (F ; Z)) ⇒ H (E; Z) Math 99r Notes 46 and for group cohomology this corresponds to the Hochschild–Serre spectral sequence Hs(G/H, Ht(H,M)) ⇒ Hs+t(G, M). For a general cohomology theory, there is the Atiyah–Hirzebruch spectral sequence 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 immediately give what the ring exactly is, but you at least see what the associated graded is. This can be checked to be exactly the associated graded for R[(G). . Actually, the mapα ˆ will induce a map on the associated graded GR(\Z/nZ) → GK ∗(B(Z/nZ)) and by following the generators, it can be checked that this map is an isomorphism. It follows from this thatα ˆ is a homeomorphism. Let me make a remark about how Theorem 18.4 is proved. We checked that this is true for ﬁnite groups. Using representation theory, you can check that if H/G has quotient G/H =∼ Z/qZ for some prime q, then the statement for H implies the statement for G. This proves the theorem for solvable G. Next use the Brauer’s theorem on elementary groups to show that this proves the general case.

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 paper published by John Tate, Homology of Noetherian rings and local rings. First thing 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 write these as just R-algebras) if these satisfy: Lλ (1) X is graded by X = λ=−∞ Xλ such that XiXj ⊆ Xi+j,

(2) X<0 = 0, 1 ∈ X0, Xλ is ﬁnitely generated, Math 99r Notes 47

λµ 2 (3) xy = (−1) yx where x ∈ Xλ and y ∈ Xµ, x = 0 if λ is odd, 2 (4) dXλ ⊆ Xλ−1, d = 0, λ (5) d(xy) = (dx)y + (−1) x(dy) for x ∈ Xλ and y ∈ Xµ. From this you produce a chain complex

d d d ··· −→ Xλ −→ Xλ−1 −→· · · .

Then H•(X) is going to have a natural structure of an algebra. If X is free and acyclic, 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 of an 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 is supposed 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 . Then λ T x = (−1) xT 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

(2) Yλ = Xλ ⊕ Xλ−pT ⊕ Xλ−2pT ⊕ · · · with the relations (i + j)! T (i)T (j) = T (i+j),T (i)x = xT (i), dT (i) = tT (i−1). i!j! This does to trick.

In general, for τi ∈ Hp−1(X), you can construct Y = XhT1,...,Tni such that

(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-algebra X such that H0(X) = R/M. Math 99r Notes 48

Proof. Let X be the union X0 ⊆ X1 ⊆ · · · where we construct them in the 0 1 0 following way. Write M = (t1, . . . , tn). We let X and X = X hT1,...,Tni. 1 Now H0(X ) = R/M. 1 1 Now let s1, . . . , sm ∈ Z1(X ) be such that si generate H1(X ). Then we can 2 1 2 2 set X = X hS1,...,Smi so that H1(X ) = 0 and H0(X ) = R/M. You can go on. Proposition 18.12. Let X,Y be free acyclic algebras such that H(X) = R/M and H(Y ) = R/N. Let j : X → R/M and k : Y → R/N be the induced maps. Then j⊗1 (R/M) ⊗ Y ←−− 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)hui 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) = Khσi where σ is the class of s. We can make Xhsi with dS = s be the free resolution of K. Now we can actually compute Tor.

TorR(K,L) = H((X ⊗ L)hsi) = H(P(X ⊗ L)S(i)) ∞ X = H(X ⊗ L)U i = H(X ⊗ L)hui. i=1 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 cohomology n 2 −1 of D2n = hσ, τ|σ = τ = 1, στσ = σ i. The basic method is the Hochschild– Serre spectral sequence. Example 19.1. We are going to have

Exti ( [τ],A) = Exti ( ,A) = Hi( /n, A). ZD2n Z Z[Z/n] Z Z Lemma 19.2. Let 1 → H → G → G/H → 1 be a short exact sequence and consider 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 checked because this tensoring is exact.

Corollary 19.3. Let A be a ZG-module, regarded as a ZH-module. Then Exti ( [G/H],A) = Exti ( ,A) = Hi(H,A). ZG Z ZH Z We ﬁrst take a projective resolution of Z as a Z[G/H]-module, and the take the projective resolution of of them as ZG-modules. Then we get

...... ∂ ∂ ∂

C01 C11 C21 ··· d10 d11 d21 ∂ ∂ ∂

C00 C10 C20 ··· d00 d10 d20 ∂ ∂ ∂

B0 B1 B2 ··· Z d0 d1 d2

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

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

Ext∗ (B ,A) = Ext∗ (B ⊗ [G/H],A) ZG ∗ ZG ∗ Z[G/H] Z = Ext∗ ( [G/H],A) ⊗ B = Hi(H,A) ⊗ B . ZG Z Z[G/H] ∗ Z[G/H] i So we get the Hochschild–Serre spectral sequence H∗(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 explicitly construct the chain complex. You can track all the diﬀerentials, because we can make 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 map α : N → E and an action en such that (1) α(en) = eα(n)e−1, (2) α(n)n0 = nn0n−1. Theorem 19.6. For a group G and an RG-module A, there is a bijection between 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 if there are maps ϕN and ϕE such that

A N E G

ϕN ϕE A N 0 E0 G.

From the bar resolution, H3(G, A) is the set of some maps f : G×G×G → A quotiented out by some subgroup. Now given an exact sequence

0 → A −→i N −→α E −→π G → 1, with a section s0 : G → E, we can deﬁne a function

−1 f(g, h) = s0(g)s0(h)s0(gh) . 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 factor k : 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 → A such 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 cohomology class. 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 z0 = z + ∂c. A similar computation shows that z is independent on the choice of the ﬁrst section s0. You can further check that equivalent 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 function z : G × G × G → A. Let F be the free group generated by G. There is a natural embedding s0 : G → F , and this association gives an epimorphism π : F → G. Then we get an exact sequence

0 → A −→i A × ker π −→α F −→π 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 × ker π 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 projective modules. We want to associate to each module a rank, but this is hard because we are trying to do this over an arbitrary ring, in particular, the group ring ZΓ. We can deﬁne the rank as

rk P = 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

(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 for finite Γ. Theorem 20.1 (Swan). (P ) = ρ(P ) for all projective P . This was proved by Swan in 1960 and by Bass in 1976 and 1979. One motivation 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 are the same. If F is a free Z-module, then trZ(idF ) = n. But because our ring is noncommutative, and so trace can’t be defined in an invariant way. Even in the 1-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. We can lift F i π P id P. Then we have tr(α) = tr(απi) = tr(iαπ), Math 99r Notes 53 where iαπ : F → F . We can take this as a definition. We still have to show that this well-defined. Consider

0 π β j F P P 0 F 0. i α π0

Then the identity tr(iαπ0i0βπ) = tr(i0βπiαπ0) implies

tr(αβ) = tr(iαβπ) = tr(i0βαπ0) = tr(βα).

This also implies well-deﬁnedness. Now let ϕ : α → β be a ring morphism. This induces a map T ϕ : 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 there exists 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 multiplication map µb : B → B. This gives a mapϕ ˜ : B → T (A), and we have

0 0 ϕ˜(bb − b b) = trA(µbµb0 ) − trA(µb0 µβ) = 0.

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

Γ(P ) = T (ϕ)(RZΓ(P )). Likewise, we have

ρΓ(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 of the 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 resolution 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 complex admits 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 and get an injective resolution of C•. Definition 20.6. A fully injective resolution of C• of this form is called a Cartan–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 ij ij (i+1)j ij ij i(j+1) 2 2 ∂1 : C → C and ∂2 : C → C satisfying ∂1 = ∂2 = 0 and ∂1∂2 = ∂2∂1. Definition 20.8. The totalization of the bicomplex is the complex M Tot(Cij)n = Cij. i+j=n There are two canonical filtrations of the Tot(C). There is

p n M r,n−r p n M n−r,r F(1)(Tot(C)) = C ,F(2)(Tot(C)) = C . r≥p r≥p

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:

ij ij i(j−1) ij ij (i−1)j H(1) = ker(∂1 )/ im(∂1 ),H(2) = ker(∂2 )/ im(∂2 ).

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

(1) p,q p •,q (2) p,q p q,• E2 = H (H(1) (C)), E2 = H (H(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 be additive. 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, and assume A and B have enough injectives. Assume that F (Q) is G-acyclic for p,q all Q injective. Then for all A ∈ A , there exists a spectral sequence Er such that p,q p q p+q E2 = R G(R F (A)) =⇒ R (FG)(A). Proof. Let A → C• be an injective resolution. Then take a fully injective resolution 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 spectral sequences (1)E and (2)E. WE have ( 0 q > 0 (1)Ep,q = Hp(H•,q(GI)) = Hp(RqG(FC•)) = 2 (1) 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, p,q ∼ p,q that splits, and so you can show G(H(2) (I)) = H(2) (GI). Then we see that

(2) p q,• p q,• ∼ p q E2 = H (H(2) (GI)) = H (G(H(2) (I))) = R G(R F (A)).

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

p q p+q H (Y,R f∗(F )) =⇒ H (X, F ). Math 99r Notes 56

20.3 Cechˇ cohomology This was a talk by Julian Salazar.

Deﬁnition 20.13. A presheaf on a category C with S -values is a functor C 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 {Ui}i∈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 a sheaf of R-algebras. Moreover, Ω∗(−) is a sheaf of skew-commutative diﬀerential algebras.

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

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

Deﬁnition 20.18. Γ(X, −): Sh(X) → Γ(X, OX )−Mod is left exact, and we deﬁne 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 n n Hsing(X; A) = H (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 Cechˇ cohomology. Let U = {Uα}α∈I be an open cover. Let P be a presheaf on X. Definition 20.21. The Cechˇ complex is defined with objects

ˇp Y C (U, P) = P(Ui0 ∩ · · · ∩ Uip )

i0<···

p+1 X (dα) = (−1)kα . i0,...,ip+1 i0,...,ˆik,...,ip+1 k=0

The Cechˇ cohomology Hˇ (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 cohomology of the nerve of the category coming from the open cover. Example 20.22. With respect to the standard open cover of S1 with two ˇ ˇ 0 1 ˇ 1 1 contractible sets, the Cech cohomology is H (S , ZS1 ) = Z and H (S , ZS1 ) = Z. Now let us apply the Grothendieck spectral sequence. This says that if F : A → B and G : B → C are additive functors with F taking injectives to G-acyclics, then there is a spectral sequence

p,q p q p+q E2 = R G(R F (−)) =⇒ R (G ◦ F )(−). Note that the Hochschild–Serre spectral sequence is an instance of this, with the functors (−)N and (−)G/N . Consider the diagram

Sh(X) i PSh(X)

Hˇ 0(X,−) Γ(X,−) OX −Mod.

This gives a spectral sequence

p,q ˇ p q p+q E2 = H (U , H (F )) =⇒ H (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. Then you recover the Mayer–Vietoris sequence with sheaves as coeﬃcients. If F is the 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 Hˇ p(U, F ) =∼ Hp(X, F ). This is used in computing things like cohomology of line bundles over projective space. There is the canonical aﬃne open cover U = {U0,U1,U2,...} and n this allows us to to compute H(Pk , O(n)). Index abelian category, 3 Hattori–Stallings rank, 53 acyclic, 4 Hochschild–Serre spectral associated graded module, 29 sequence, 34 Atiyah completion theorem, 45 homology, 3 bar notation, 17 induction, 22 bar resolution, 17 injective, 8 bicomplex, 54 injective resolution, 8 Bockstein homomorphism, 38 invariant element, 28

Cartan–Eilenberg resolution, 54 left derived functor, 6 Cechˇ cohomology, 56 chain complex, 3 M-conjugate, 16 chain homotopy, 4 Mackey decomposition, 24 cochain complex, 3 morphism cohomology, 3 of chain complexes, 3 coinduction, 22 positive complex, 4 cross module, 50 projective, 4 projective resolution, 4 derivation, 15 diﬀerential graded algebra, 46 rank, 52 double complex, 32 reduction, 22 right derived functor, 8 Eilenberg–MacLane space, 21 enough projectives, 6 sheaf, 56 exact couple, 32 sheaf cohomology, 56 Ext functor, 9 split, 15

ﬁltration, 29 Tor functor, 7 transfer maps, 25 Grothendieck spectral sequence, 55 group cohomology, 9 wreath product, 39 group homology, 8 group ring, 7 Yoneda multiplication, 14