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Math 231br (Advanced Algebraic Topology) Lecture Notes Taught by: Professor Michael Hopkins Notetaker: Yuchen Fu Spring 2015

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Math 231br (Advanced Algebraic Topology) Lecture Notes

Taught by: Professor Michael HopkinsNotetaker: Yuchen Fu

Spring 2015

Contents

1 Introduction 3

2 Higher Homotopy Groups 42.1 Hurewicz’s Construction of Higher Homotopy Groups . . . . . . . . . . . . . . . . . . . . . . . 42.2 Relative Higher Homotopy Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3 Hurewicz Fibration, Serre Fibration and Fiber Bundle 53.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53.2 Relationship Among Them . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.3 Some Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4 Spectral Sequences 84.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84.2 Spectral Sequences: Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

5 More on Spectral Sequences 105.1 Serre Spectral Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

6 And More Spectral Sequences 116.1 The Extension Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116.2 Local Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

7 Snow Day 12

8 More on SSS, and the Hurewicz Theorem 138.1 Mappings on the E2 page; Transgression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138.2 The Hurewicz Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

9 Relative Hurewicz Theorem 149.1 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

10 Representability and Puppe Sequences (Guest Lecturer: Hiro Takana) 1510.1 Brown Representability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1510.2 Puppe Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

11 Cohomological SSS (Guest Lecturer: Xiaolin Shi) 1711.1 An Appetizer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1711.2 The Main Course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

12 A Glance at Model Categories, Part 1 (Guest Lecturer: Emily Riehl) 2012.1 Definition of Model Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

13 A Glance at Model Categories, Part 2 (Guest Lecturer: Emily Riehl) 2213.1 (Co)fibrant Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2213.2 Homotopy in Model Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

14 Serre Classes 2614.1 Serre Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

15 Mod C Hurewicz Theorem 28

16 Relatvie Mod C Hurewicz; Finiteness of Homotopy Groups of Spheres 3116.1 Relative mod C Hurewicz Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

17 Introduction to Steenrod Operations 3317.1 Motivating Steenrod Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

18 Construction of Steenrod Operations 36

1

19 More on Steenrod Operations 3819.1 Properties of the Steenrod Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

20 Mod 2 Cohomology of K(Z2, n) 40

21 Computing with Steenrod Squares (Guest Lecturer: Jeremy Hahn) 4221.1 Cohomology Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

22 Adem Relation and Bocksteins (Guest Lecturer: Xiaolin Shi) 4422.1 Adem Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4422.2 Bockstein Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

23 More on Adem Relation 4623.1 Some Tricks for Computation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4623.2 A New Treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

24 Even More on Adem Relation 4824.1 Higher Power Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4824.2 Application of Steenrod Algebra: Homotopy Groups of Spheres . . . . . . . . . . . . . . . . . . 49

25 Some Nontrivial Stable Homotopy Computation 5125.1 Killing Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5125.2 A Few Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5125.3 Back to Computation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

26 Introducing Cobordism 5426.1 Cobordism: An Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5426.2 Vector Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

27 Universal Vector Bundle 5627.1 Existence of universal vector bundle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

28 Stiefel-Whitney Classes (Guest Lecturer: Xiaolin Shi) 5828.1 Stiefel-Whitney Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

29 More on S-W Classes 6029.1 Examples of Stiefel-Whitney Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6029.2 Detour: Riemannian metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

30 Cohomology of the Grassmannian 62

31 Poincare-Thom Construction 64

32 Pontryagin-Thom and the MO Spectrum 66

33 Spectra 67

34 Thom Isomorphism 69

35 Cohomology of MO and Euler Classes 71

36 Describing MO 73

37 Finishing the Cobordism Story 75

2

1 Introduction

Did not attend as it was administrivia. Allow me to use this space to give some general information.These lecture notes are taken during Spring 2015 for Math 231br (Advanced Algebraic Topology) at Harvard.

The course was taught by Professor Michael Hopkins. The course is a continuation of Math 231a, which coversthe first three chapters of Allan Hatcher’s Algebraic Topology (henceforth referred to as simply “Hatcher”).The course is self-contained in that it does not assume background beyond that of Math 231a. However, due totime constraint, the last part regarding cobordism consisted largely of proof sketches, and a lot of details wereskipped. Thus for a deeper understanding of the material, reading one of the “core references” given belowalongside this note is highly recommended.

Regarding spectral sequences: in the middle of notetaking I discovered the preferable sseq package, soswitched to it. As a result you’ll see two ways of drawing spectral sequences in this document. I may unifythem in the future, but since drawing spectral sequence is a time-consuming task, I can’t make a promise.

Below I list the literature used/mentioned throughout this course:

• Background: Review of basic properties of fundamental groups, singular homologies and cohomologies.

Algebraic Topology by Allan Hatcher

• Core References: Contain proofs presented in this note, along with more details.

Vector Bundles and K Theory by Allan Hatcher

Cohomology Operations and Applications in Homotopy Theory by Robert E. Mosher

A Concise Course in Algebraic Topology by J. P. May

• Other References: Other materials mentioned throughout the course. Collected here for convenience.

Homotopy Theories and Model Categories by W.G. Dwyer and J. Spalinski

Cohomology Operations by Steenrod and Epstein

Notes on Cobordism Theory by Robert E. Stong

I’ve added some comments throughout this note, and they are all in the following format:(This is a comment — Charles).These notes may still contain numerous errors and typos, so please use at your own risk. I take responsibility

for all errors in the notes. There are some references here and there to the homework problems, which can alsobe found on my website.

3

2 Higher Homotopy Groups

Let (X, ∗) be a pointed space. πn(X) = f : Sn → X | f(∗) = ∗modulo pointed homotopy. Sometimes writtenas [Sn, X]∗. π0(X) is the set of path components. π1(X) is the fundamental group. So how do we describeπ2(X)? In particular how does it have a group structure? One straightforward way is S2 → S2∧S2 → X (gluef and g by f ∧ g), but there are other ways,

2.1 Hurewicz’s Construction of Higher Homotopy Groups

For instance, the following given by Hurewicz. Start with the set-theoretical part that πN (X) = f : In → X |f(∂In) = ∗ modulo homotopy. Then glue the Ins together as “adjacent cubes” gives the group structure.

Let A,X be topological spaces. Denote by XA the set of maps from A to X. One would like to topologizeXA such that continuous map Z → XA is exactly the same as continuous map Z × A → X. This can bedone by giving it the compact-open topology. Some point-set topology issues arise due to the fact that thespaces A and X are not necessarily compact Hausdorff, and is resolved by using compactly generated spaces(c.f. Steenrod, Convenient category of topological spaces).

Suppose A,Z,X have base points ∗. We only want to look at pointed maps i.e. to look at (X, ∗)(A,∗). Wemake this into a pointed space by assigning the base point ∗ := A → ∗. Now consider (Z, ∗) → (X, ∗)(A,∗).This then corresponds to a special map Z ×A→ X that factors through Z ×A/(∗×A∪Z ×∗), i.e. the smashproduct Z ∧A. Make Z ∧A into a pointed space by assigning ∗ := Z × ∗ ∪ ∗×A. Then (Z, ∗)→ (X, ∗)(A,∗)

corresponds to (Z ∧A, ∗)→ (X, ∗).Concretize this with (In, ∂In) = Sn = (I, ∂I) ∧ (In−1, ∂In−1) → (X, ∗), then we see it corresponds to

I/∂I → (X, ∗)(In−1,∂In−1). The latter is called Ωn(X), the n-fold loop space. More concretely, Ω(X) =Map((I, ∂I), (X, ∗)) = Map((S1, ∗), (X, ∗)), and Ωn(X) = Ω(n)(X). Now Hurewicz’s definition for higherhomotopy groups is simply πn(X) = π1(Ωn−1(X)). One can easily generalize this into πn(X) = πp(Ω

q(X)),where p+ q = n.

Uniqueness and Commutativity There are certainly two ways of “gluing”: f to the left of g (which wedenote by f ∗ g) and f on top of g (which we denote by f g). Then if we apply this to a 2-by-2 square, wesee that one have (a ∗ b) (c ∗ d) = (a c) ∗ (b d). Now the Eckmann-Hilton argument tells us that the twooperations are equal and both commutative. In particular we know that πn(X) is abelian for n > 1.

2.2 Relative Higher Homotopy Groups

For n > 0 and A a subspace of X, define πn(X,A) = Map((Dn, ∂Dn, ∗), (X,A, ∗)) modulo homotopy. Then weget a long exact sequence πn(A) → πn(X) → πn(X,A) → πn−1(A) → . . . → π0(A) → π0(X) is exact. (Somecautions need to be taken near the end of the exact sequence as to how to interpret it; in particular, startingfrom π1(X,A) we no longer have groups, but only pointed sets.)

Problem 1. Let ∂In = In−1∪Jn−1. Clearly Jn−1 is contractible and we have Jn−1 ⊆ ∂In ⊆ In. We can alsodefine πn(X,A) = Map((In, ∂In, Jn−1), (X,A, ∗)) modulo homotopy. Can we write πn(X,A) as a homotopygroup of some other space? (Hint: it is πn−1 of something else.)

Answer This is not hard—it is πn−1 of the path space P (X,A), which is a pointed space based on thefamily of paths I → X that begins at ∗X and ends in A, equipped with the compact-open topology andhas the base point ∗P (X,A) := I → ∗X . Let’s be more rigorous about this. Note that P (X,A) is the

same as Map((I1, ∂I1, J0), (X,A, ∗)). Let the base point be denoted as above. Then by directly expanding(In−1, ∂In−1)→ P (X,A) one sees that this corresponds to In → X such that ∂In−1 × I1 ∪ In−1 × J0 = Jn−1

goes to ∗X , and In−1 × I0 → A. Thus it corresponds to the definition above; the other way around is justformalism and we skip it.

4

3 Hurewicz Fibration, Serre Fibration and Fiber Bundle

Did not attend due to personal schedule conflict. I wrote the following notes to cover the material gap.Throughout this discussion, let E,B be basepointed topological spaces. For the sake of simplicity I’m notincluding all proofs; most of the proofs can be found in Hatcher, or this online note.

3.1 Definitions

Definition 1 (Homotopy Lifting Property). A map p : E → B is said to have the homotopy lifting propertywith respect to X if for any f, g in the following diagram, the diagonal map h : X×I → E in the diagram belowalways exists:

X E

X × I B

f

i p

g

h

where I is the unit [0, 1] and i is the inclusion x 7→ (x, 0).

Definition 2 (Relative Homotopy Lifting). p : E → B is said to have homotopy lifting property with respect toa pair (X,A) (where A ⊆ X) if the diagonal map h exists for any f, g and any given homotopy H : A× I → E:

X ∪A× I E

X × I B

f∪H

i p

g

h

where i is the obvious inclusion and we’re promised that H(a, 0) = f(a) for any a ∈ A ⊆ X.

The intuition is straightforward: to say p : E → B has HLP with respect to (X,A) is to say that anyhomotopy of X → B can be lifted to a homotopy X → E, provided that a homotopy already exists on A.

Definition 3 (Hurewicz and Serre Fibrations). We say p : E → B is a Hurewicz fibration if it has HLP forall space X, and say it is a Serre fibration if it has HLP for all CW complex X.

While we’re on this, let’s also define the dual notion for HLP and Hurewicz fibration, although we won’tmention them again until the end of this lecture. (The correct dual notion for Serre fibration, sometimes called“the Serre cofibration,” is the retract of relative cell complexes; see Lecture 13 for further discussions.)

Definition 4 (Homotopy Extension Property). Let i : A → X be a mapping of topological spaces. It satisfiesthe homotopy extension property with respect to some space Y if, given f and h0 in the following diagram:

Y X

Y I A

h0

hg 7→g(0) i

f

the diagonal map h always exist.

Definition 5 (Cofibration). A mapping i : A → X is called a cofibration if it satisfies HEP with respect toany space Y .

The concept of a fibration is closely related to that of a section, as the next fact shows.

Lemma 1. Suppose p : E → B is a Hurewicz (resp. Serre) fibration, and that we have a subspace pair (resp.CW pair) (X,A) where the inclusion is i : A→ X. Given the following diagram:

A E

X B

f

i ph

g

If either i or p is a homotopy equivalence (resp. weak equivalence), then h exists.

Proof. The Hurewicz case is trivial; for the Serre case one needs to inductively construct h one dimension at atime, then use the standard pasting trick (doubling the speed of homotopy at each higher dimension) to obtainh. See Theorem 5.1 of this note.

5

Proposition 1 (Existence of Sections). If p is a nonempty Hurewicz (resp. Serre) fibration, and B is acontractible space (resp. CW complex), then p admits a section, i.e. there is another map g such that p(g(x)) = xfor all x ∈ B.

Proof. Take X = B, g = 1B , A = ∗ ∈ B, f anything in the lemma above.

There are some other ways to define a Serre fibration which can be useful at times. Let us record themdown here.

Proposition 2. The following are equivalent:

1. p : E → B is a Serre fibration.

2. p : E → B has HLP for all disks.

3. p : E → B has HLP for all pairs (Dn, Sn−1).

4. p : E → B has HLP for all CW pairs (X,A).

Proof. For 2 to 3, note that (Dn × I,Dn × ∗ ∪ Sn−1 × I) is homeomorphic to (Dn × I,Dn × ∗). For 3 to 4, doinduction on the cell structure. The rest is trivial.

Example 1. The trivial fibration, E = B×F for some space F , and p being the projection onto the first factor,is obviously a Hurewicz fibration.

Definition 6 (Fiber Bundles). Given a Serre fibration p : E → B, if in addition there is a space F calleda fiber, such that we have the following local triviality condition: for each e ∈ E, there is a neighborhood Ueof p(e) in B, such that there is a homeomorphism ϕ : p−1(Ue) → Ue × F that makes the following diagramcommute:

p−1(Ue) Ue × F

Ue

ϕ

pπ1

where π1 is the projection onto the first factor, then we say that p is a fiber bundle.

Obviously, p−1(x) is homeomorphic to F for any x ∈ B, so we say F is the fiber of p, and sometimes writep as F → E → B. One can easily check the following:

Lemma 2. Any compositions and pullbacks of Hurewicz (resp. Serre) fibrations are again Hurewicz (resp.Serre) fibrations.

Remark 1. Though the pullback part is still true, it might be worth pointing out that the composition part is nolonger true for fiber bundles; however, under certain conditions (such as in the category of finite-dimensionalmanifolds, with smooth fiber bundles), we may still have the composition property. See here for a generaldiscussion.

3.2 Relationship Among Them

Some more words regarding the relationship among the three concepts.

Lemma 3. Being a Serre fibration is a local property; more concretely, suppose p : E → B is a map such thatevery point b has a neighborhood U such that p|p−1(U) is a Serre fibration. Then p is a Serre fibration.

Proof. Theorem 11 of this note.

Corollary 1. Every fiber bundle is a Serre fibration.

Proof. A trivial bundle is a Serre fibration; fiber bundles are locally trivial.

Example 2. It turns out if B is paracompact, a fiber bundle would also be a Hurewicz fibration. But here’s afiber bundle that is not a Hurewicz fibration. (source) Let R2 = R×0, 1, and let X = R2/(x, 0) ∼ (x, 1)∀x ∈R+, so it’s two copies of the real line glued together on the positive half. Let the (images of the) two copiesof R in X be called U and V . Define f((x, i)) = x be the “coordinate projection”, and note that it descendsdown to a map f : U ∩ V → R+, which cannot be extended to any continuous X → R+. (Note that g triviallydescends to X → R.) Define E to be the fiber bundle that is a trivial R+ bundle on both U and V , and theglueing function is one such that (x, a) = (x, f(x)a) where the LHS is over U and RHS over V . Now suppose

6

a section X → E exists; then in particular it becomes a section on subbundles, and by projecting to the secondfactor we obtain two functions gU : U → R+ and gV : V → R+ such that g = gV /gU is a function that agreeswith f on U ∩ V . However, it is possible to extend g to entire X: just let g(x) = gV (jV (x))/gU (jU (x)) wherejV is first map to the coordinate by g then send to the corresponding point on V , and jU is defined similarly.This yields a contradiction, so p cannot admit a section and is therefore not a Hurewicz fibration.

Example 3. And now an example of a Serre fibration that is not a fiber bundle. Let E = (x, y) ∈ R2 | 0 ≤y ≤ x ≤ 1 and B = I; let p be the projection onto the first factor.

So the inclusion order is Hurewicz⇒ Fiber Bundle⇒ Serre.

3.3 Some Properties

Theorem 3.1 (Long Exact Sequence of Serre Fibrations). Let p : E → B be a Serre fibration, and let F bethe basepoint fiber, then there is a (natural) long exact sequence:

. . .→ πn(F )i−→ πn(E)

p−→ πn(B)∂−→ πn−1(F )→ . . .→ π0(F )→ π0(E)→ π0(B)

Note that starting from π1(E) we only get exact sequence of pointed sets; before that it’s an exact sequence ofgroups. The long exact sequence follows from that of the pair (E,F ), using the fact that p : πn(E,F )→ πn(B)is an isomorphism.

The situation for (co)homology is not as simple as a long exact sequence, but instead a lot of them, puttogether in some systematic manner called the spectral sequence. This is the topic of the next lecture.

Now let us observe that it makes sense to talk about “the fiber” of a Hurewicz or Serre fibration.

Lemma 4. Any two fibers of a Hurewicz (resp. Serre) fibration are homotopy equivalent (resp. weakly equiva-lent), provided that the base space B is path connected.

Proof. See Problem 6 and 7 of Homework 1.

Because of this, we often also write a fibration as F → E → B when it is clear that B is path-connected.Now, if p : E → B and p′ : E′ → B′ are two Serre fibrations, and F, F ′ are the respective fibers over the

basepoints. A mapping between the two Serre fibrations is a pair of mapping f : E → E′, g : B → B′ such thatthe following diagram commute:

E E′

B B′

f

p p′

g

One can check that this induces a mapping h : F → F ′ between the fibers.

Lemma 5. If any two of the three maps in f, g, h are weak equivalences, then so is the third one.

Proof. Observe that a mapping between two Serre fibrations induce a mapping between the two associatedhomotopy LES. Now apply five lemma using the fact that two of the maps are weak equivalences (so theyinduce iso everywhere) to conclude that the third one also induces iso everywhere.

For a later discussion on model categories (in Lecture 13), we also include two results regarding the factor-ization of maps.

Lemma 6. Any map f : X → Y can be factored into Xg−→ X ′

h−→ Y , where g is a homotopy equivalence and his a Hurewicz fibration.

Proof. Let e0 be the map g 7→ g(0). Consider the pullback of f : X → Y along e0; this yields another space

Z = (x, γ) | x ∈ X, γ ∈ Y I , γ(0) = f(x). One can check that Zev1:(x,γ)7→γ(1)−−−−−−−−−−→ Y is a Hurewicz fibration, and

that Z(x,γ) 7→x−−−−−→ X is a homotopy equivalence, with the inverse s being attaching the constant path at x. Now

let X ′ = Z, g = s, h = ev1.

Lemma 7. Any map f : X → Y can be factored into Xg−→ X ′

h−→ Y , where g is a cofibration and h is ahomotopy equivalence.

Proof. Let X ′ be the mapping cylinder of f , g the inclusion into the cylinder, and h the retract onto the otherside of the cylinder.

7

4 Spectral Sequences

Introducing spectral sequences.

4.1 Motivation

Suppose we have a fiber bundle p : E → B of CW complexes. Let B(n) be the n-skeleton. Consider the cellular

chain complex CCelln (B) = Hn(B(n), B(n−1)) =⊕n cells

Hn(Dn, ∂Dn). Consider the pullback

E[n] E

B(n) B

where the function E[n] → B(n) is again a fiber bundle. Our goal is to understand the (co)homology ofE from that of B and F . Consider the long exact sequence, . . . → Hn(E[n−1], E[n−2]) → Hn(E[n], E[n−2]) →Hn(E[n], E[n−1])

d−→ Hn−1(E[n−1], E[n−2]) → . . .. If we understand the first and the third terms as well asthe mapping d, then we might understand the second term. Continuing this process, we might understandHn(E[∞], E[0]) = Hn(E).

Next, consider the pullback square

E′ E

Dn BΦ

where the Φ map is the characteristic map. For the pullback fiber bundle, since the base space is contractible,one can show that E′ is homotopy equivalent to Dn × F , and then from there observe that (E[n], E[n−1]) is

relatively homeomorphic to∐

n cells of B

(Dn×F, Sn−1×F ), where F is the fiber p−1(∗). Thus H∗(E[n], E[n−1]) =⊕

n cells

H∗(F × Dn, F × Sn−1) = Ccelln (B) ⊗ H∗(F ). If B has (trivial fundamental group?), the mapping

H∗(E[n], E[n−1]) → H∗(E

[n−1], E[n−2]) becomes Ccelln (B) ⊗ H∗(F ) → Ccelln−1(B) ⊗ H∗(F ), which is dcell ⊗ id(this is nontrivial). This indicates that we should look at H∗(B;H∗(F )) in order to obtain H∗(E).

(Note that Hatcher uses this way, but the argument is not as nice as what we’ll present below, e.g. thatdcell ⊗ id part is nontrivial to prove and invokes Hurewicz’s theorem, which we want to prove using spectralsequences.)

4.2 Spectral Sequences: Definition

Definition 7 (Spectral Sequence). A spectral sequence consists of a sequence (Er, dr : Er → Er) (each is achain complex) where d2

r = 0, Er abelian groups, together with an isomorphism Er+1 = ker dr/ im dr.

One place spectral sequences come from is an exact couple.

Definition 8 (Exact Couple). An exact couple is a long exact sequence . . .h−→ D

f−→ Dg−→ E−→ h−→ D

f−→ . . ., whichwe usually write as a triangle:

D D

E

f

gh

For instance, . . . → H∗(E[n−1]) → H∗(E

[n]) → H∗(E[n], E[n−1]) → . . .. We can let D =

⊕n

H∗(E[n]),

E =⊕n

H∗(E[n], E[n−1]), and with the obvious mappings they fit into an exact couple.

Suppose (D,E, f, g, h) is an exact couple. The derived exact couple is constructed as follows. Connect twoexact couple triangles side by side:

D D D

E E

f

g

f

gh h

8

Then we can get d = g h : E → E. One can check that d2 = 0. Let E′ = ker d/ im d, and D′ = im f .f ′ : D′ → D′ is now again given by f . g′ : D′ → E′ is given by g′(f(x)) = g(x). It’s in ker d and thus in E′.Now, for two choices x, x′ of x, their difference lies in h(E), and thus its image is in im d, i.e. the choice of xdoesn’t matter. Finally, h′ : E′ → D′ is given by h′(x) = h(x) where x ∈ ker d represents x in E′. One cancheck this yields another derived couple. Finally, we get a spectral sequence from a derived couple by lettingE1 = (E, d) and Er = (E′r−1, d

′r−1).

One place to get exact couples is from filtered chain complexes. Given a sequence of chain complexes

(C(1)∗ , C

(2)∗ , . . .). Let D = H∗(C

(n)), E = H∗(C(n)/C(n−1)), then this gives the long exact sequence of homology.

For instance, we can take C(n)r = Cr(E

[n]).Filtered chain complexes also come from double chain complexes. A double chain complex is a set Cij , i, j ≥

0, where we have vertical differentials dv : Cpq → Cp(q−1), and horizontal differentials dh : Cpq → C(p−1)q,such that the obvious diagram commutes (with an additional −1). For instance, C∗(X)⊗C∗(Y ) would yield adouble chain complex by letting dh and dv be the differentials on X and Y respectively. From a double chain

complex we get a total chain complex: Cn =⊕p+q=n

Cpq, and d : Cn → Cn−1 is dh+dv. One can check that with

the additional −1 this yields d2 = 0. There are two ways to filter this chain complex: one is C(n)∗ =

⊕p≤n

Cpq,

and another is⊕q≤n

Cpq.

9

5 More on Spectral Sequences

Again let’s consider a total chain complex Cn =⊕p+q=n

Cpq, and the associated filtered chain complexes Filtkn =⊕p+q=n,p≤k

Cpq. The exact couple is D = H∗(Filtk), E = H∗(Filtk/Filtk−1). Then we see that E = E1 is the

homology of the kth column (note that the quotient is pretty much just the corresponding entry on the kthcolumn). More precisely, E1

pq is the pth homology of the qth column, and the d : E → E differential turns outto be transition differential of the columns that sends Hp to Hp−1.

In a general exact couple, there is just one E-term. When it comes from a double chain complex, of course

E would have bi-grading: Er =⊕p,q

Erp,q. The mapping dr then sends Erp,q to Erp−r,q+r−1, as one can check.

Let’s get some examples. Consider the following double chain complex, where all the nonzero morphismsare isomorphism:

Z Z 0 0

0 Z Z 0

0 0 Z Z

We shall do the computation page by page. Since the E1 terms compute the column homologies, we seethat E1 page looks like this:

Z 0 0 0

0 0 0 0

0 0 0 Z

Now the E2 page stays the same because the arrow isn’t long enough yet. As for the d3, which reaches fromZ to Z, it is an interesting exercise to see that it’s actually an isomorphism.

Stable Terms Note that because everything is 0 outside the first quadrant, for every fixed (p, q), the termErpq eventually stabilizes as r →∞, and that stable term we call it E∞pq .

Consider the total chain complex C∗, and the associated homology Hn(C∗). It turns out that for each n, theE∞pqp+q=n terms yield a filtration of the homology group, more precisely we have 0 ⊆ F0 ⊆ . . . Fn ⊆ Fn+1 =Hn(C∗) where E∞pq = Fp+1/Fp.

5.1 Serre Spectral Sequence

Suppose we have a Serre fibration F → Ep−→ B, and suppose for the moment that B is simply connected. Then

there is a spectral sequence where E2pq = Hp(B;Hq(F )), and it converges to Hp+q(E), meaning that there is a

filtration Hn(E) whose associated graded group is⊕p+q=n

E∞pq .

Consider for example the fibration ΩSn+1 → PSn+1 p−→ Sn+1, where PX is the path space (and thusPSn+1 is contractible). We know the homology of PSn+1 and Sn+1, so this will give us the homology of ΩSn+1.(Charles: I believe SSAT has this example. Yep, Example 1.5 on Page 9 of Chapter 1.) The fact that (0, 0), whichis H0(ΩSn+1), stabilizes to Ho(PS

n+1) = Z, tells us that ΩSn+1 is connected, thus π1Sn+1 = π0(ΩSn+1) = 0

for n ≥ 1. Some more computations happens and we conclude that Hj(ΩSn+1) = Z if j is a multiple of n, and

0 otherwise.

Problem 2. Suppose X is simply connected and Hi(X) = 0 for 0 ≤ i < n+1, what can you say about H∗(ΩX)for ∗ < 2n?

Answer See Lecture 8.

10

6 And More Spectral Sequences

6.1 The Extension Problem

Note that Dr term is the direct sum of images in H∗(Filtr) → H∗(Filtk+r). For fixed n and sufficiently larger 0, Hn(Filtk+r) = Hn(C∗). At this point, Dr → Dr is the inclusion im(Hn(Filtk)) ⊆ im(Hn(Filkk+1)) ⊆. . . ⊆ Hn(C∗) and the Er term is the quotient of the images. Then for sufficiently large r, we have a filtrationF0 ⊆ . . . ⊆ Fn ⊆ Fn+1 = Hn(C∗) where Erpq = Fp/Fp+1, such that p + q = n. Then we have the filtration ofHn(C∗), but we still need to solve for the actual Hn(C∗). This is called solving the extension problem.

The chain complex we considered is the following:

0 Z 0 0 0 0

0 Z Z 0 0 0

0 0 Z Z 0 0

0 0 0 Z Z Z

2

2

2

2

2

0 2

We’ll do direct computation that looks kind of ugly. First do the computation with total chain complex:(Skip the long and boring computation.)Another example is given by considering both the spectral sequence and the total chain complex to inves-

tigate the H4 (filtration of three Z2s) and H5 (Z2) of the total space.Then we run the spectral sequence in a different filtration to solve the extension problem and observe that

H4 is actually Z8.

6.2 Local Systems

Definition 9. Let X be a space. A local system on X consists of:

1. A set Ax | x ∈ X of Abelian groups.

2. For every 1-simplex ∆[1]γ−→ X an isomorphism γ∗ : A∂1(γ) → A∂0(γ).

Such that for every 2-simplex c : ∆[2]→ X, which has edges c01, c02, c12, we have (c02)∗ = (c01)∗ (c12)∗, andsuch that if γ is constant, γ∗ = id.

Equivalently, consider the fundamental groupoid π≤1X of X, which is the category where objects are pointsin X, morphisms are homotopy classes of paths rel endpoints. Then a local system is a functor from π≤1X tothe Abelian groups.

Suppose X → B is a Hurewicz fibration, then b 7→ Hn(p−1(b)) gives a local system. In fact this is alsotrue for Serre fibration. (Use the fact that a weak equivalence X → Y induces isomorphism on all homologygroups.)

Now suppose X is a space with local system A. Then define Cn(X;A) =⊕

c:∆[n]→X

Ac0 where c0 is the image

of the first vertex under c. Define a differential d : Cn(X;A)→ Cn−1(X;A) =∑i

(−1)idi, where i ranges over

the index set for all the (n− 1) simplexes. Let dic to be the composition of ∆[n− 1]→ ∆[n] and c : ∆[n]→ X.(The rest is on the notes, we’ll figure it out later.) This then defines homologies with local coefficients.

11

7 Snow Day

Class cancelled due to heavy snow. The self-reading is on the construction of Serre Spectral Sequences, withnotes that I unfortunately cannot upload. This is a standard topic that requires some effort to build up correctly,so I’ll just refer readers to standard literature (e.g. Hatcher) for reference.

12

8 More on SSS, and the Hurewicz Theorem

8.1 Mappings on the E2 page; Transgression

Consider a spectral sequence. There are two regular chain complexes in there, the left column and the bottomrow. Thus there is a mapping from H∗(left column) → H∗(C), and there is another map H∗(C) → E2

∗,0 (thebottom row) by modding out everything else. These are called edge homomorphisms.

Suppose now we have a Serre filtration such that π≤1B is trivial. Then the edge homomorphism is inducedby the maps F → E and E → B.

Consider an element on the bottom row on the E2 page. Then the differentials are obstructions of suchelement ∈ H∗(B) from coming from H∗(E). Likewise, for an element on the left column, the kernel of F → Eare those that are “hit” by some differential coming from below.

Now suppose π≤1B is trivial, H∗(B) = 0 for ∗ < n, H∗(F ) = 0 for ∗ < m, then from the graph of the E2

page one sees that there is a canonical mapping defined as Hp(B)→ Hp−1(F ) for p < n+m+1. This mappingis usually denoted as τ and is called the transgression.

How do we understand the transgression geometrically? To have all of the dra = 0 for r < |a|−1 is equivalentto say that a is in the image of H∗(E,F ) (under the connecting homomorphism δ?). Define a ∈ Hp(B) to betransgressive if there is some a ∈ Hp(E,F ) such that it is the corresponding term in H∗(E,F ) by the mappingof pairs (E,F )→ (B, ∗). Then the transgression of a is defined as δ(a) where δ is the connecting homomorphismHn(E,F )→ Hn−1(F ).

Theorem 8.1. Consider the Serre Spectral Sequence ΩB → PB → B for π≤1(B) = ∗, Hp(B) = 0 for p < n.The the transgression τ : Hp(B)→ Hp−1(ΩB) is an isomorphism for p < 2n.

8.2 The Hurewicz Theorem

Theorem 8.2 (The Hurewicz Theorem). If π≤1(B) = ∗, and πi(B) = 0 for i < n for n ≥ 2, then the Hurewiczhomomorphism πn(B)→ Hn(B) is an isomorphism. In other words, the first nonvanishing homotopy group isisomorphic to the first nonvanishing homology.

Theorem 8.3 (Poincare). If B is connected, then the map π1(B)ab → H1(B) is an isomorphism.

Proof. Skipped.

Proof of the Hurewicz Theorem. Induction on n, starting with n = 1. (This is Poincare’s Theorem.) Supposen ≥ 2. Of course, πi(B) = 0 for i < n, which by induction implies Hi(B) = 0 for i < n. By the transgressiontheorem above, Hi(B) = Hi−1(ΩB) for i < 2n, which in particular tells us Hn(B) = Hn−1(ΩB) for n ≥ 2. Nowlook at the mapping Hn(PB,ΩB) → Hn−1(ΩB) and Hn(PB,ΩB) → Hn(B). But look at the correspondinghomotopy maps πn(PB,ΩB) → πn−1(ΩB) and πn(PB,ΩB) → πn(B). Hn(PB,ΩB) = Hn−1(ΩB) is iso bywhat we said above, Hn−1(ΩB) = πn−1(ΩB) by induction, πn(PB,ΩB) = πn−1(ΩB) by the contractibility ofPB (by the relative homotopy sequence), Hn(PB,ΩB) = Hn(PB/ΩB) = Hn(B) because and πn(PB,ΩB) =πn(B) by lemma 4.5 at here, and it follows that Hn(B) = πn(B).

As a corollary, we can get πi(Sn) = 0 for i < n, and πn(Sn) = Z. First realize that π1(Sn) = 0 by

Van-Kampen, then π2(Sn) = H2(Sn) = 0, etc. And then this pushes all the way to πn = Hn = Z.Now another example. Let A be any Abelian group, then we can construct M(A,n) by first taking a free

resolution 0 → F2 → F1 → A → 0 and the build the corresponding sequence of space∨Sn →

∨Sn →

M(A,n). Then by Hurewicz, we get a space that has πi = 0 for i < n and A for i = n. (See the end of lecture9 for the rest of this story.)

13

9 Relative Hurewicz Theorem

Theorem 9.1 (Relative Hurewicz Theorem). Suppose we have a pair of spaces (X,A), X is connected. Ifπi(X,A) = 0 for i < n, then Hi(X,A) = 0 for i < n, and πn(X,A) ∼= Hn(X,A) for n ≥ 3.

Recall that πn(X,A) = πn−1(F ) for some other space F , this is why we have n ≥ 3.

Proof. Now given f : A → X, then we can factor it into A∼=−→ A → X where the first map is a homotopy

equivalence, and the second map is a Hurewicz fibration. Recall that A = (γ, a) ∈ XI × A | γ(1) = a, or

equivalently, it is the pullback between f and XI γ(1)−−−→ A.Now let us look at the Serre Spectral Sequence of F → A → X. Then we see that Hp(X;Hq(F )) ⇒

Hp+q(A) = Hp+q(A).Now note that π1(A)→ π1(X)→ π1(X,A). The third one is 0 by assumption, thus the mapping π1(A)→

π1(X) is surjective. By the following fact, the entirity of π1(X) acts trivially on H∗(F ).

Exercise 1. If Y → B is a Serre fibration, and an α ∈ π1(B) is in the image of π1(Y ), then α acts triviallyon the homology of the fiber.

By assumption, πi(X,A) = πi−1(F ) = 0 for i < n. By the absolute Hurewicz Theorem, Hi(F ) = 0, soH∗(X;Hi(F )) = 0 for i < n−1. Then by the homology long exact sequence, for i < n−1, Hi(X) ∼= Hi(X) =⇒Hi(X,A) = 0 for i < n− 1, and Hi(A)→ Hi(X) is epi for n− 1, thus Hi(X,A) = 0 for i < n.

Since X is connected, H0(X;Hq(F )) = Hq(F ). By considering the transgression (?), we have the exactsequence:

Hn(A)→ Hn(X)→ Hn−1(F )→ Hn−1(A)→ Hn−1(X)→ 0.

On the other hand, we also have the following exact sequence:

Hn(A)→ Hn(X)→ Hn(X,A)→ Hn−1(A)→ Hn−1(A)→ . . .

If we can map the bottom sequence to the top sequence, then we can invoke the five lemma.There is a mapping A∪CF → A∪CF → X, from which (modulo some magic) we get a map (A∪CF,A)→

(X,A). Then we have Hn−1(F ) ∼= Hn(ΣF ) ∼= Hn(A ∪ CF,A) → Hn(X,A), which allows one to comparethe two exact sequences above and conclude that Hn(X,A) ∼= Hn−1(F ) ∼= πn−1(F ) (by absolute HurewiczTheorem), which is πn(X,A).

9.1 Application

Consider the fibration Snf−→ X → X ∪f Dn+1. Then Hn(X ∪Dn+1, X) = Hi(D

n+1, Sn+1) (by excision), whichis 0 for i < n+1 and Z for i = n+1. By the relative Hurewicz theorem, then we know that πi(X∪Dn+1, X) = 0for i < n + 1 and Z for i = n + 1 (n ≥ 2). In other words, πi(X) → πi(X ∪Dn+1) is an iso for i < n and epifor i = n. However, πn+1(X ∪Dn+1, X)→ πn(X) is induced by f , so we know that πi(X ∪Dn+1 ∼= πi(X) fori < n, and ∼= πi(X)/f for i < n.

This leads to the action of “killing homotopy groups.” More precisely, let∨Sn+1 → X → X ′, so the first

map is epi on πn+1, then πi(X) ∼= πi(X′) is iso for i < n+1, and πn+1(X ′) = 0. Now repeat with X ′ by adding

onto it the n+ 2 cells. Then we get a series of mappings X → PnX such that πi(X) ∼= πi(PnX) for i ≤ n, and

πi(PnX) = 0 for i > n, then we have a sequence PnX → Pn−1X → . . . which is called the Postnikov Tower

of X.Now recall from lecture 8 that we have, for any Abelian group A and any n ≥ 2, some MA such that

πiMA = A for i = n and 0 for i < n. Now PnMA has πn = A and πi = 0 for i 6= n. This is called theEilenberg-MacLane spaceK(A,n). This construction makes us lose some intuitive connection with geometry(the Eilenberg-Maclane space is highly non-geometrical), but it does yield information about geometry.

As an example, consider the fibration F → Sn → PnSn = K(Z, n). Then from the long exact sequence, wecan immediately conclude that πi(F ) = πiS

n for i > n, and 0 for i ≤ n. Now by Hurewicz theorem, HiF = 0for i < n + 1, thus πn+1S

n = πn+1(F ) ∼= Hn+1(F ). Keep going along this road, we can eventually computethings up to e.g. πn+14S

n. (See Lecture 11 for more details.)

14

10 Representability and Puppe Sequences (Guest Lecturer: HiroTakana)

10.1 Brown Representability

The big theorem to be deduced today is the following:

Theorem 10.1. Fix any abelian group A. Then for any pointed CW complex (X,x0), we have a naturalisomorphism [X,K(A,n)] ∼= Hn(X,A).

One says in this case that singular cohomology is represented by the Eilenberg-MacLane spaces. This is aspecial case of the more general Brown representation theorem. In particular, along with Whitehead’s theoremthis allows us to conclude that:

Corollary 2. Any two CW complexes that are both K(G,n) are homotopy equivalent.

Remark 2. If we take X to be a point, then we observe that [X,K(A,n)] = ∗ =⇒ H∗(X,A) = 0.

Remark 3. If we consider X = S0, then we see that [S0,K(A,n)] = π0K(A,n) = A if n = 0, and 0 if n > 0.

The strategy for proving this theorem is to show that the object on the left side actually satisfies theEilenberg-Steenrod axioms, then along with the fact that it behaves in the same way as singular cohomology onH0(∗), we can conclude the theorem from the uniqueness theorem1. More concretely, let us fix an abelian groupA. We shall show that there is a functor Kn : CWpairop → AbGrp given as (X,B)→ [(X,B),K(A,n)] thatsatisfies the Eilenberg-Steenrod axioms.

Of course, it would be comforting to know the following:

Remark 4. For any space X, [X,K(A,n)] is an abelian group.

Proof. Observe that [X,K(A,n)] = [X,Ω2K(A,n − 2)]. Now, a mapping f ∈ [X,ΩY ] is given by x 7→ fx :S1 → Y , so the composition of loops yield a group structure, and the second Ω guarantees commutativity, justas we observed for higher homotopy groups. Or one can observe that [X,Ω2K(A,n− 2)] = [Σ2X,K(A,n− 2)]and use the abelian group structure on [Σ2X,Y ] as explained in May’s Book2.

Now we need to check the E-S axioms. The additivity axiom is immediate from the natural isomorphism

[∨i

Xi,K(A,n)] ∼=∏i

[Xi,K(A,n)]. Homotopy and dimension axioms are trivial, and the excision axiom follows

from Mayer-Vietoris property, which is again trivial for the functor that we specified. So it remains to checkthe exactness axiom, which we’ll do by investigating the Puppe Sequence.

10.2 Puppe Sequences

Let us fix a CW pair A ⊆ X (but really any HEP pair works), so there is a monomorphism i : A → X. Nowconsider the following diagram:

A X

∗ X/A ∼= C(A) ∪i X

where on the bottom right (which is the pushout) we attach the cone C(A) along i to X, thereby collapsingA to a point and obtain X/A as the result. We can, however, continue this process by pushing out along thetrivial map again:

A X ∗

∗ X/A ∼= C(A) ∪i X (C(A) ∪i X)/X ∼= ΣA

The isomorphism on the bottom right is immediate if one draws a graph to convince oneself. Note that thisstep shows the difference between topological construction and the pure “algebraic” analogue, say for groups:if we do this for a group diagram, we would be gettin (X/A)/X = ∗ instead of ΣA. If we continue this processfor a few more steps we will get the following:

1This book, Theorem 1.31.2Here, page 58.

15

A X ∗

∗ X/A ∼= C(A) ∪i X (C(A) ∪i X)/X ∼= ΣA ∗

∗ ΣX Σ(X/A)

Now we need an easy observation:

Lemma 8. ΣX/A = ΣX/ΣA.

Proof. Omitted.

Then from the graph above we obtain the following exact sequence, known as the Puppe sequence:

A→ X → X/A→ ΣA→ ΣX → ΣX/ΣA→ Σ2A→ . . .

Moreover, any two consecutive maps of this sequence define a pushout.

Corollary 3. For any space Z, the sequence of pointed sets

[A,Z]← [X,Z]← [X/A,Z]← . . .

is exact.

Corollary 4. When Z ∼= K(A,n), for each CW pair (X,B), we have a long exact sequence of abelian groups.

This completes the proof of the theorem. As a final remark, we note that the same construction also yieldsan exact sequence for fibration, as can be read off from the following diagram:

. . . ΩE ΩB ∗

. . . ∗ F E

∗ B

16

11 Cohomological SSS (Guest Lecturer: Xiaolin Shi)

Today we will introduce the cohomological analogue of Serre Spectral Sequence and use it to do some compu-tations.

Definition 10 (Cohomological Serre Spectral Sequence). Given a Serre fibration F → E → B, we have aspectral sequence Ep,q2 = Hp(B;Hq(F )) ⇒ Hp+q(E), with differential dr : Ep,qr → Ep+r,q−r+1

r . The infinitypage again has a filtration: Hp+q(E) = F 0 ⊇ F 1 ⊇ . . . ⊇ F p+q, where Ep,q∞ = F p/F p+1.

The ring structure that cohomology carries extends naturally to a ring structure on the CSSS.

Proposition 3. There exists a multiplicative structure Ep,qr ×Es,tr → Ep+s,q+tr on each page, which we denoteby (x, y) 7→ xy. The differential dr acts as a derivation on this structure, i.e. dr(xy) = xdr(y) + (−1)|x|dr(x)y.

11.1 An Appetizer

Let us consider the Eilenberg-MacLane spaces. It is clear that K(Z, 1) = S1. (For a quick proof that the higherhomotopy groups vanish, consider Sn → S1 for n > 1; since Sn is simply connected, the map lifts to Sn → R,and since R is contractible this mapping must be nulhomotopic, and this property descends down to Sn → S1

via projection.) On the other hand, from the cohomology ring structure we know that K(Z, 2) = CP∞. Butsuppose we don’t already know this, can we deduce this directly from the definition of K(Z, 2)? Yes.

Let us consider the pathspace fibration ΩK(Z, 2)→ PK(Z, 2)→ K(Z, 2). As we noted before, ΩK(Z, 2) =K(Z, 1) = S1, and PK(Z, 2) is contractible. This allows us to write out the spectral sequence Ep,q2 =Hp(K(Z, 2);Hq(S1))⇒ Hp+q(∗), which is as follows:

2 0 0 0 0 0

1 H0 H1 H2 H3 H4

0 H0 H1 H2 H3 H4

0 1 2 3 4

Everything above the second line is zero, and the first two rows are also rather straightforward (Hk denotesthe kth cohomology of K(Z, 2), of course). The only nontrivial differential on the E2 page is then the arrowsfrom Hk to Hk+2 as indicated in the graph. Next, observe that E3 and later pages have no nontrivial differential,so the SSS collapses after the E2 page. Thus on the E∞ page, we have, on the left-bottom corner:

2 0 0 0 0 0

1 H0 ? ? ? ?

0 H0 H1 ? ? ?

0 1 2 3 4

Compare this with the diagram that we should get from the cohomology of a point:

17

2 0 0 0 0 0

1 0 0 0 0 0

0 Z 0 0 0 0

0 1 2 3 4

we immediate conclude that H0 = Z and H1 = 0. Furthermore, since the arrow indicated on the first graphkills the H0 at (0, 1), we know that that particular arrow must be injective; and since the H2 at (2, 0) also dies,we know it is also surjective and thus bijective. In this way, we immediately conclude that H2n(K(Z, 2)) = 0,and H2n+1(K(Z, 2)) = 0. It remains to identify the ring structure.

Now we have the following diagram as E2:

2 0 0 0 0 0

1 Z(a) 0 Z 0 Z

0 Z(1) 0 Z(x2) 0 Z(x4)

0 1 2 3 4

where some of the generators have been named and marked at the corner of the corresponding Z entries.Now, because a generates H0 at (0, 1), one can do direct computation and see that ax2 generates (2, 1), andsimilarly ax4 for (4, 1), etc. Choose the sign of a and x2 appropriately so that d(a) = x2, then since d(x2) = 0,we have d(ax2) = d(a)x2 + (−1)1ad(x2) = x2

2 = x4, and similarly x6 = x32, etc. This allows us to conclude that

H∗(K(Z, 2)) = Z[x2]. Finally, to conclude that this uniquely identifies K(Z, 2), we can either appeal to theresult from last lecture, or observe the fibration S1 → S∞ → CP∞, induced as the limit of S1 → S2n+1 → CPn,and investigate the homotopy long exact sequence associated with it (then apply Whitehead).

11.2 The Main Course

Theorem 11.1. π4(S3) = Z2.

This is the first stabilized term πS1 of the stable homotopy groups.

Proof. We start by considering the fibration X → S3 → K(Z, 3), where X is the homotopy fiber. As wementioned before, this effectively “kills π3(S3)” because, as one can observe from the homotopy long exactsequence, we have πi(X) = πi(S

3) for i ≥ 4 and πi(X) = 0 for i ≤ 3. However, this fibration itself is a bitawkward to use, so we “back up a step” by looking at the Puppe sequence: . . . → ΩS3 → ΩK(Z, 3) → X →S3 → K(Z, 3). We choose the fibration ΩK(Z, 3) → X → S3 to apply CSSS on. Then we get the E2 page asfollows:

18

6 Z(a3) 0 0 Z 0 0 0

5 0 0 0 0 0 0 0

4 Z(a2) 0 0 Z 0 0 0

3 0 0 0 0 0 0 0

2 Z(a) 0 0 Z 0 0 0

1 0 0 0 0 0 0 0

0 Z(1) 0 0 Z(x) 0 0 0

0 1 2 3 4 5 6

where the nontrivial differentials are hown. Let generators of the 0th column be named (1, a, a2, . . .) asshown in the graph, and the bottom one on the 3rd column be (x), then we know that the 3rd column’sgenerators are (x, ax, a2x, . . .).

Next note that the space X is 3-connected, so we can invoke the Hurewicz theorem to obtain that Hi(X) =0 =⇒ Hi(X) = 0 for i ≤ 3, and from which we immediately obtain that d3(a) = x, as the differential needsto kill both (0, 2) and (3, 0). In this manner, we can figure out the other differentials using the derivationproperty, e.g. d(an) = nan−1d(a) for |a| ≡ 0 (mod 2), and from this we can recover the E∞ page and see thatH2i+1(X) = Z/i for i ≥ 2. Then by the universal coefficient theorem, we have H2i(X) = Z/i for i ≥ 2, andthus H4(X) = Z/2 = π4(X) by Hurewicz. Since π4(X) = π4(S3), we have completed the proof.

19

12 A Glance at Model Categories, Part 1 (Guest Lecturer: EmilyRiehl)

Preliminary remarks: if A is a cell complex, X → Y is a weak homotopy equivalence, then [A,X] ∼= [A, Y ].Also we have Whitehead’s theorem: X ∼= Y between cell complex is a homotopy equivalence if and only if it’sa weak homotopy equivalence.

We’ll generalize this in an axiomatic framework for abstract homotopy theory, i.e. model category theory.This framework can also be used to do a few other things:

• One can use this to prove the equivalence of different homotopy theories (e.g. the case for simplicial sets);

• One can use this to get derived functor systematically (e.g. homotopy (co)limits).

• One can use this to present an (∞, 1)-category.

12.1 Definition of Model Categories

Definition 11 (DHKS3). A homotopical category is a category M with a class of maps W (the class ofequivalences, containing all isos) that satisfies either of the following:

• 2-of-3 property If Xf−→ Y

g−→ Z and two of f, g, g f are in W , then all of them are.

• 2-of-6 property If Xf−→ Y

g−→ Zh−→ N and h g, g f ∈W , then all of f, g, h, h g, g f, h g f are in

W .

(the second one is stronger than the first one, but in case for homotopical category they are equivalent.)

Definition 12 (Gabriel-Zisman). A homotopical category has a hopotopy category HoM = M [W−1] with a

localization functor Mγ−→ HoM (details omitted), given as follows:

• The objects of HoM are just objects of M .

• The morphism x→ y is a finite zig-zag x←− z1 −→ z2 . . . −→ y, up to equivalence described as the following:

xf←− z f−→ x = idx, and

xf←− z g←− y = x

gf←−− y.

If (M,W ) can be expanded to a model category then we can define HoM in a nicer way.

Definition 13 (Quillen / Joyal-Tierney). A model structure on a homotopical cateogry with all small limitsand colimits (M,W ) consists of a class C of cofibrations and a class F of fibrations, so that (C ∩W,F) and(C,F ∩W ) are weak factorization systems (defined below).

Definition 14. A weak factorization system of a category M consists of two classes of maps (L,R) suchthat:

• Factorization Any f ∈M can be factored into r l, where l ∈ L and r ∈ R.

• Lifting If l ∈ L, r ∈ R, then the following diagram lifts:

X Y

X ′ Y ′

l

u

r∃w

v

Note that this generalizes both HEP and HLP.

• Closure L = l | l r ∀r ∈ R, R = r | l r ∀l ∈ L, where refers to the lifting diagram above(note this yields a Galois connection).

(Reference: Homotopy theories and model categories. W. G. Dwyer and J. Spalinski.)

Example 4. On the category Set, (mono, epi) is a weak factorization system. The factorization is the cograph

factorization X → X∐

Y → Y , and the lifting property and the closure property can be checked manually.

3Homotopy Limit Functors on Model Categories and Homotopical Categories.

20

Lemma 9. If Z is the class of maps with the left lifting property (i.e. being the L in the definition above)with respect to some class R, then Z is closed under coproduct, pushout, retract, transfinite composition andcontains the isos. (The dual version for right lifting property also holds.)

Proof. We prove the case for retract. Recall that f is a retract of g if there exists a diagram of the following:

A X A

B Y B

s1

f

r1

g f

s2 r2

such that r1 s1 = idA, r2 s2 = idB . Then the rest of the proof is trivial. (...)

Example 5. Some examples of model structures are as follows.

• (Top, homotopy equivalences, closed Hurewicz cofibrations, fibrations)

• (Top, weak homotopy equivalences, retracts of relative cell complexes, Serre fibrations)

• (Chain complexes over R-module, homotopy equivalences, Hurewicz cofibrations, Hurewicz fibrations)

• (Chain complexes over R-module, quasi-isomorphisms, retracts of relative cell complexes, Serre fibrations)

Note that the third and the fourth ones are the respective analogies of the first and the second ones. (Thishints at the fact that homological algebra is a special case of homotopical algebra modulo the language of modelstructures.)

Remark 5. A few remarks to be made before the end of lecture.

1. The model axioms are self-dual.

2. If M is a model category, and X is an object in M , then X/M and M/X (the “slice categories”) are modelcategories with a forgetful functor X/M → X that creates weak equivalences, cofibrations and fibrations.

21

13 A Glance at Model Categories, Part 2 (Guest Lecturer: EmilyRiehl)

Recall from last lecture the definition of model categories and weak factorization systems. As explained lasttime, the factorization axiom guarantees that any map in the cateogry can be factored into the composition ofa trivial cofibration ∈ W ∩ C and a fibration ∈ F , and can be factored into a cofibration followed by a trivialfibration.

Another quick remark: given the factorization and the lifting axioms of the weak factorization system, theclosure property is equivalent to the “retract closure” property that both L and R are closed under all retracts.This can be proved using a “retract argument” c.f. this note.

Today we consider the second example listed in last lecture, (Top, weak homotopy equivalences, C = retractsof relative cell complexes, F = Serre fibrations). This construction is due to Quillen. More explicitly, (C,F∩W )is the system where F ∩W are maps having the right lifting property against Sn−1 → Dn, and C are retractsof relative cell complexes built from Sn−1 → Dn. Dually, (C ∩W,F) is the system where F are maps havingthe RLP against Dn → Dn × I and C ∩W are retracts of rel. cell complexes built from Dn → Dn × I.

Today’s object is to prove the following two theorems at the full generality of model categories.

Theorem 13.1. If A is a cell complex, f is a weak equivalence X → Y , then [A,X]f∗−→ [A, Y ] is an isomor-

phism.

Theorem 13.2 (Whitehead). If X, Y are cell complexes, then Xf−→ Y is a weak homotopy equivalence if and

only if it is a homotopy equivalence.

13.1 (Co)fibrant Objects

Cell complexes are known as the fibrant-cofibrant (sometimes known as bifibrant) objects in the model categorygiven above.

Definition 15. A is a cofibrant object if and only if the unique mapping ∅ → A from the initial object to Ais in C. Dually, A is a fibrant object if and only if the unique mapping A→ ∗ from A to the final object is inF . A is called fibrant-cofibrant (or bifibrant) if both conditions are met.

Example 6. In the model category above, all objects are fibrant (this is an easy exercise), and the cell complexesare cofibrant.

13.2 Homotopy in Model Categories

Any model category is equipped with a natural notion of homotopy such that

1. The whitehead theorem is true;

2. There exists well-defined homotopy classes of maps;

3. The category HoM is equivalent to the category M ′, where the objects are bifibrant objects of M andthe morphisms are homotopy classes of maps.

Definition 16. A cylinder object cyl(A) for A ∈M is an object such that the following factorization exists:

A∐

A A

cyl(A)

f=(i0,i1)

id∐id

g

where g is a weak equivalence. The object is called good if f is a cofibration, and very good if additionally gis a trivial fibration.

Note that in a model category we are always guaranteed to have a very good cylinder object. Also it’s worthnoting that cylinder objects are not uniquely defined.

Proposition 4. Given a good cylinder object cyl(A) with A a cofibrant, i0 and i1 are trivial cofibrations.

Proof. Consider the following diagram, where the square is the defining pushout for coproduct:

22

∅ A

A A∐

A

cyl(A)

A

cofib

cofib

i1 id

i0

id

cofib

We have marked in the diagram the cofibrations that have been given by the assumptions. By the pushout

property of cofibrations, the two maps into A∐

A are cofibrations, then since cofibrations are closed under

composition, i0 and i1 are cofibrations. By definition, cyl(A) → A is a weak equivalence, and id, being anisomorphism, also is a weak equivalence, so from the 2-of-3 property we conclude that i0 and i1 are weakequivalences.

Definition 17. Given two mappings f, g : A → X, a left homotopy between f and g is a mapping H suchthat the following diagram commutes:

A

cyl(A) X

A

i0

f

H

i1

g

We denote this by f 'l g.

Note that the “left” comes from the fact that model categories are self-dual. In particular, there is a dualnotion of path objects and the associated notion of right homotopy.

Remark 6. If f 'l g : A → X, then there exists a good left homotopy (where the cylinder object is good). Inaddition, if X is fibrant, then there exists a very good left homotopy.

Proof. Just consider the following diagram, where the factorization through cylgood(A) is granted by the fac-torization axiom.

A∐

A cyl(A) X

cylgood(A)

cofib

H

trivial cofib

Now suppose X is fibrant. Choose a good left homotopy H : cylgood(A) → X. Now consider the mappingg : cylgood(A) → A, which is guaranteed to be a weak equivalence by the cylinder object definition. By thefactorization axiom, we can factor g into g2 g1, where g2 is a trivial fibration and g1 is a cofibration. By 2-of-3

property, this means that g1 is also trivial. We write this as cylgood(A)g1−→ cylvg(A)

g2−→ A. Now consider thefollowing diagram:

cylgood(A) X

cylvg(A) ∗

g1

H

H′

By assumption, X → ∗ is fibration and g1 is trivial cofibration, so by the lifting axiom, the mapping H ′

exists, and one sees that cylvg(A) yields a very good homotopy.

Proposition 5. If A is cofibrant, then 'l defines an equivalence relation on Hom(A,X).

Note that regardless of whether A is cofibrant or not (?), we can always define [A,X] = Hom(A,X)/ 'l.

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Proof. We skip the proof for reflexivity and symmetry. (Just choose the cylinder object judiciously.) Let usprove transitivity. Suppose we have two homotopies, given by the following diagrams:

A

cyl(A) X

A

i0

f

H

i1

g

A

cyl′(A) X

A

i′0

g

H′

i′1

h

In order to construct the necessary cylinder object, consider the pushout of i1 and i′0 as follows:

A cyl′(A)

cyl(A) cyl′′(A)

X

i1

i′0

H′i1

H

i′0

H′′

Since H i1 = H ′ i′0 = g, we have the outer commutative square, and by the UMP of pushout we have theinduced mapping H ′′ : cyl′′(A)→ X. Then we have the following diagram:

A

cyl′′(A) X

A

i′0i0

f

H′′

i1i′1

h

But we still need to verify that cyl′′(A) is actually a cylinder object. This can be checked via the followingdiagram:

A cyl′(A)

cyl(A) cyl′′(A)

A

i1

i′0

i1

i′0

Since pushout of trivial cofibrations are again trivial cofibrations, i1 and i′0 are trivial (because A is cofibrant).On the other hand, cyl(A) → A and cyl′(A) → A are guaranteed to be trivial, so by 2-of-3 property we canconclude that cyl′′(A)→ A is trivial.

Theorem 13.3. If A is cofibrant, Xp−→ Y is trivial fibration, then [A,X]

p∗−→ [A, Y ] is an isomorphism.

Note that this is the model categoric statement of Theorem 13.1.

Proof. First we would like to check that p∗ is well defined. Suppose we have f 'l g, then by the followinggraph, we directly have pf 'l pg.

A

cyl(A) X Y

A

i0

f

H p

i1

g

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Now we want to show surjectivity. Consider the following commutative square on the left, where k : A→ Yis some chosen element of [A, Y ]:

∅ X

A Y

pf

k

A∐

A X

cyl(A) Y

f∐g

pH′

H

As we mentioned above, ∅ → A is in C since A is cofibrant, and p ∈ F ∩W by assumption, then the mappingf exists by the lifting axiom, and thus we have surjectivity. Now for injectivity: suppose that we have pf 'l pg,

then consider the right diagram above, where A∐

A→ cyl(A) is a cofibration as we have shown above, so H ′

exists, and thus f 'l g.

Proposition 6. If X is fibrant, A′ → h is a mapping, f, g : A→ X, then f 'l g =⇒ fh 'l gh.

Proof. As we mentioned before, we can choose a very good left homotopy A∐

A→ cylvg(A)k−→ X for f 'l g.

On the other hand, consider the following diagram:

A′∐

A′ A∐

A cylvg(A)

cylgood(A′) A′ A

h∐h

H′

h

where the unmarked maps are canonical. It is clear that the right side map is a trivial fibration, and the left

side map is a cofibration, and thus H ′ exists, and one sees that cylgood(A′)

kH′−−−→ X is a left homotopy betweenfh and gh.

Proposition 7. Suppose f, g : A → X. If A is cofibrant then f 'l g =⇒ f 'r g, and if X is fibrant thenf 'r g =⇒ f 'l g.

Proof. Omitted. Some clever lifting argument.

Theorem 13.4. Let f : A→ X where A and X are bifibrant objects. Then f is a weak equivalence if and onlyif f is a homotopy equivalence (i.e. there exists g : X → A such that f g ' 1X , g f ∼ 1A

4).

Of course, this is the model categoric version of Whitehead’s theorem, i.e. Theorem 13.2.

Proof. We will only prove the only if direction; for the other one, see Dwyer and Spalinski’s notes (Lemma4.24). Suppose f is a weak equivalence. Then by 2-of-3 property, we can factor it into a trivial cofibration

followed by a trivial fibration. Suppose this is written as f : Aq−→ C

p−→ X. Now consider the following diagram:

∅ C

X X

ps

id

∅ → X is cofibration because X is cofibrant, and C → X is a trivial fibration, so s exists, and clearly ps = idX .

Next observe that by the theorem we proved above, we know that [C,C]p∗−→ [C,X] is an isomorphism, so in

particular p∗([p s]) = [p s p] = [p] = p∗([1C ]), so we know that p s ∼ 1C . Dually, this also gives us aninverse r to q such that r q = 1A and q r = 1C . Together this yields a two-sided homotopy inverse for f .

4Since 'l and 'r are the same in this case, we are justified to just write '. [A,X] in this case is often written as π(A,X).

25

14 Serre Classes

Welcome back, Mike. Okay now recall the Hurewicz theorem and the computation that we had in lecture11. We learned that H4(X) = Z2 = π4(S3) for X → S3 → K(Z, 3). Suppose we then use the Eilenberg-Maclane fibration X5 → X → K(Z/2, 4), back up a step along the Puppe sequence and consider the sequenceK(Z/2, 3)→ X5 → X (i.e. its Spectral Sequence):

3 Z/2

2 0

1 0

0 Z 0 0 0 Z/2 0 Z/3 0 Z/4 0 Z/5

0 1 2 3 4 5 6 7 8 9 10

We don’t know much about the 0th column (except the part revealed by Hurewicz), but it looks like thatwe have a bunch of 2-groups. The 0th row we have already calculated, on the other hand. Then we see thatZ3 can’t be effectively killed when it reaches these groups, nor do the following Z/5,Z/7, etc on the 0th row.Moreover, until the 6th page for Z/3 (and 8th page for Z/4, etc.), the calculation on earlier pages will not affectZ/3 at all. This suggests at the fact that we should be able to simplify these calculations significantly by onlylooking at some particular groups “locally”, and this is where the idea of Serre classes kick in.

14.1 Serre Classes

Definition 18. A Serre Class of abelian groups is a class C of abelian groups such that if 0 → A → B →C → 0 is an exact sequence then B ∈ C iff A,C ∈ C.

Example 7. Some Serre classes are listed below:

1. The class of torsion abelian groups.

2. For some fixed p, the class of p-torsion abelian groups.

3. Let S be the set of primes, the class of A such that ∀a ∈ A, ∃n = pe11 . . . pnen, pi ∈ S, na = 0.

4. The class of finitely generated abelian groups.

Intuitively, the Serre classes are the elements that we hope to ignore in our computation.

Definition 19. A map Af−→ B is an epi mod C if coker f ∈ C, is mono mod C if ker f ∈ C, and is iso mod C

if both hold.

Homological algebra extends naturally to the mod C version.

Exercise 2. Prove that the mod C version of the five lemma holds.

More systematically speaking, there is a universal functor F : A → AC (the localization functor, whichmakes A → B an isomorphism if A/B ∈ C) for an abelian category A and a collection of objects C (meetingcertain closure conditions), such that (epi, mono, iso) mod C translates to (epi, mono, iso) in AC .

Example 8. Let A be abelian groups, and C be the torsion abelian groups. Then A → B is (epi, mono, iso)mod C if and only if A⊗Q→ B ⊗Q is (epi, mono, iso). In this case, ⊗Q is the universal functor F .

Proposition 8. If C is a Serre class, A ∈ C, B is finitely generated, then A⊗B,Tor(A,B) ∈ C.

Proof. A ∈ C =⇒ A ⊕ A ∈ C because 0 → A → A ⊕ A → A → 0. Thus An ∈ C, i.e. Zn ⊗ A ∈ C. Nowconsider the exact sequence 0 → Zm → Zn → B → 0, then we have the sequence 0 → Tor(A,B) → Am →An → A⊗B → 0, from which we can conclude that Tor(A,B), A⊗B ∈ C.

26

Corollary 5. If HiX is finitely generated for each i, and A ∈ C, then for each i, Hi(X;A) ∈ C.

Proposition 9. For A → X, if Hi(X,A) ∈ C, then HiA→ HiX is iso mod C.

Proof. Break up the long exact sequence of homology.

Proposition 10. Suppose F → E → B is a Serre fibration such that π0B = 0, π1B is abelian acting triviallyon πnF . If HiF ∈ C for i > 0, HiE ∈ C for i > 0, then HiB ∈ C for i > 0, under some restrictive conditionexplained below.

Proof. Consider the E2 page mod C.

2 0 ? ? ? ?

1 0 ? ? ? ?

0 Z ? ? ? ?

0 1 2 3 4

From the assumption we know that the 0th column is all 0C (x ∈ C, x = 0 mod C, x ∈ 0C are the same thing.).Naturally the next step is to try to clear the rest of the page. Let’s first see a general attack strategy that,though unfortunately won’t work in this case, gives us some interesting insights.

First consider the tail of the homological LHE: . . . → H1E → H1B → 0. This directly yields H1B ∈ C.Now look at (1, 1): It’s H1(B;H1F ). For the sake of argument, suppose we also know that Hi(B) are finitelygenerated, then we can conclude that Hi(B;HiF ) ∈ C by an universal coefficient argument. Now look at (2, 0);There is a possible differential H2E → H2B → H1F by transgression. Then we have a short exact sequence0 → (some quotient of H2E) → H2(B) → (a subgroup of H2F ) → 0. Since the two side terms are in C, so isthe middle term. But we need some extra assumption to work on H3B, since there is a nontrivial mappingfrom (3, 0)→ (1, 1) on E2. So unfortunately this strategy gets stuck here.

Now to make things work, let us add the extra assumption that C is either

1. closed under arbitrary sums, or

2. consisting entirely of finitely generated groups.

Consider case 1. if C has all infinite sums, then for A ∈ C, B arbitrary, we would still have A⊗B,Tor(A,B) ∈C, so Hi(X;A) ∈ C. So we have the following diagram mod C:

2 0 0 0 0 0

1 0 0 0 0 0

0 Z ? ? ? ?

0 1 2 3 4

If one is familiar with SSS, one can directly read off the SSS that HiE → HiB is iso mod C and thus

prove the theorem. To be more detailed, let us look at the mapping HnBd2−→ Hn−2(B;H1F ), and get the

corresponding SES 0→ ker d2 → HnB → im d2 → 0. Since im d2 is a subgroup of something 0C , so is itself 0C ;so it remains to prove that ker d2 is 0C .

To do this, let us look at the E∞ page. We see that HnE, which is in C, is the filtration of n groups(everything on the diagonal but the bottom one) that are in C already, along with the last one on the bottomrow, which is then forced to be in C as well. Thus we can conclude that E∞r,0 ∈ C for arbitrary r. Suppose

E∞r,0 = Emr,0 = ker dm/ im dm. Since im dm ∈ C, we know that ker dm ∈ C. Since Em−1r,0 / ker dm = im dm ∈ C, it

follows that Em−1r,0 ∈ C, and continue this argument we eventually get that ker d2 ∈ C and thus HiB ∈ C.

We’ll do case 2, and the applications that require it, in the next class. (We didn’t; just assume it.)

27

15 Mod C Hurewicz Theorem

Recall the definition of a Serre class. Clearly, if E2p,q ∈ C, then so does E∞p,q. Similarly, if everything along a

diagonal n on the E∞ page is in C, then so is HnE. (This is how Serre class interacts well with SSS.)

Theorem 15.1 (Mod C Hurewicz Theorem). Suppose C is a Serre class satisfying

1. A,B ∈ C =⇒ A⊗B,Tor(A,B) ∈ C (for instance, when C consists of fin. gen. abelian groups), and

2. A ∈ C =⇒ Hn(K(A,n);Z) ∈ C for each n.

Suppose X is simply connected, then if πi = 0 mod C for i < n, then HiX = 0 mod C for i < n and thatπnX → HnX is an isomorphism mod C.

Example 9. If C is a collection of fin. gen. abelian groups staisfying the condition above, then if X is fin.gen. and HnX are fin. gen. for each n, then πnX are fin. gen.

Corollary 6. πnSk is finitely generated for k ≥ 2.

In the next class, we’ll prove the even cooler result that πnSk is finite for k > n.

Proof. By induction on n. When n = 2 we have the usual Hurewicz theorem. So we may assume n > 2.

First Attempt Recall that in the regular proof, the inductive step was given by the following diagram:

πiX HiX

πi−1ΩX Hi−1ΩX

∼= ∼=

where the right equivalence holds in a range of i using SSS. We reduce dimension by 1 by going to theloopspace, then the inductive hypothesis settles the problem. However, note that given πiX = 0 mod C fori < n for some n does not implies π2X = π1ΩX = 0, so we cannot assume ΩX is simply connected as we didin the regular proof, so we can’t “bootstrap” this proof naively.

Hotfix To resolve this, consider the following fibration:

F → X → K(π2X, 2).

Then we have π≤2F = 0, πiF = πiX for i > 2. Then look at the following:

ΩF → ΩX → ΩK(π2X, 2) = K(π2X, 1).

Then ΩF is simply connected, and πiΩF = πiΩX for i ≥ 2.We will establish the following equivalences:

• πiX = πi−1ΩX = πi−1ΩF mod C for all i, and

• HiX = Hi−1ΩX = Hi−1ΩF mod C for i ≤ n.

Then the inductive hypothesis applied to ΩF gives the result. The first equality is trivial by construction, solet’s look at the second one.

Note By universal coefficient theorem, the two hypotheses on the Serre class together imply that for each n,A,B ∈ C =⇒ Hn(K(A, 1);B) ∈ C.

28

Second Equality Consider again the SES ΩF → ΩX → K(π2X, 1). One can check that in a looped fibration,π1B acts trivially on HiF . So we can apply Serre spectral sequence without using local coefficients. Let usapply SSS.

n ? ? ? ? ? ? ?

n− 1 0 ? ? ? ? ? ?

n− 2 0 0 0 0 0 0 0

. . . . . . . . . . . . . . . . . . . . . . . .

0 Z 0 0 0 0 0 0

0 1 2 3 4 5 6

Consider the mod C SSS above. The 0th column now consists of homologies of ΩF , so until Hn−1ΩF thesegroups are in C. We know that the 0th row is 0 mod C, then from the universal coefficient theorem we see thatthe all the way to the row of n− 2, everything below is 0 mod C. Then since we know that the SSS convergesto HnΩX, by SSS we have that HiΩX = 0C , i.e. HiΩF → HiΩX is iso mod C for i ≤ n− 1. This is the secondhalf of the second point.

Now look at the fibration ΩX → PX → X. By induction, HiX = 0 mod C for i < n− 1 because πiX = 0mod C for i < n. Also by induction, we know that HiΩX = HiΩF = 0 mod C for i < n− 1. Now look at theSSS of the fibration above.

n ? ? ? ? ? ? ?

n− 1 ? ? ? ? ? ? ?

n− 2 0 0 0 0 0 ? ?

. . . . . . . . . . . . . . . . . . . . . . . .

0 Z 0 0 0 0 ? ?

0 1 2 . . . n− 2 n− 1 n

By induction and universal coefficient, everything to the left of (n− 1) column and below (n− 1) row is 0(except at (0, 0)). Then looking at action of d2 on (n− 1, 0) (it injects into a 0C) we see that (n− 1, 0) is 0 modC, which clears the (n− 1) column, which then yields that (n, 0)→ (0, n− 1) is iso mod C. This yields the firsthalf of the second point.

Proposition 11. If A is finitely generated, then Hn(K(A, 1);Z) is finitely generated for all n ≥ 0. If A is ap-torsion abelian group, then Hn(K(A, 1);Z) is p-torsion for all n ≥ 1.

Proof. First, for any abelian group A, A is a filtered colimit A = colimA′⊆A,A’ finitely generatedA′. If A is p-torsion,

then each A′ is finitely generated and p-torsion. Since homology commutes with colimit, it suffices to prove thefirst part (and showing that the homologies are p-torsion). Now note that K(A × B, 1) = K(A, 1) ×K(B, 1).By Kunneth formula, we are reduced to A = Z and A = Z/p. Now K(Z, 1) = S1, so no problem with A = Z.

By the theorem that [K(A,m),K(B,m)] = Hom(A,B) (proof can be found here), we know that K(Z, 2)pn−→

29

K(Z, 2) → K(Z/pn, 2) is a fibration5. If we back it up twice, we have S1 → K(Z/pn, 1) → K(Z, 2) = CP∞.Now we look at what happens in cohomology SSS:

4 0 0 0 0 0 0 0

3 0 0 0 0 0 0 0

2 0 0 0 0 0 0 0

1 Ze 0 Zex 0 Zex2 0 Zex3

0 Z1 0 Zx 0 Zx2 0 Zx3

0 1 2 3 4 5 6

where all the differentials are multiplications by pn (induced by K(Z, 2)pn−→ K(Z, 2)). Thus we have

H∗(K(Z/pn, 1);Z) = Z[x]/(pnx) for |x| = 2. By the universal coefficient theorem we have H2m−1(Z/pn, 1);Z) =Z/pn for m ≥ 1. In particular, this means H0 = Z, H2m = 0 for m ≥ 1.

5I’m still not sure how to go about this, I think there might be some functor K(−, n) or a delooping functor that preserves thefibration, but this is not very clear.

30

16 Relatvie Mod C Hurewicz; Finiteness of Homotopy Groups ofSpheres

First, a warning: the assumption of simple connectivity is actually necessary. An example: take S2 ∨ S1, andcertainly the homology groups are finitely generated, and π1 = Z, but π2 = ZZ by the universal cover of the

graph (where S1 expands to Z, and H2 =⊕n∈Z

Z = π2, which is not finitely generated) is not finitely generated

Again consider the backing-up of S3 → K(Z, 3), which is K(Z, 2) → X → S3. We used cohomology SSSon this to compute some homotopy groups of S3. Now fix some prime p and let C be the class of ` torsiongroups such that (`, p) = 1. Then we know that H2p(X) = Z/p and HiX = 0 mod C for i < 2p. Then by modC Hurewicz, we have πiX = 0 mod C for i < 2p and π2p(X) = Z/p mod C. Thus we know that the next mod Cnontrivial fundamental group for S3 after π3(S3) = Z is π2pS

3 = Z/p⊕A for A being a torsion group of ordersmaller than p (because larger factors aren’t supposed to show up yet).

Corollary 7. πiS3 6= 0 for infinitely many i values (one for each p).

Now let C be the Serre class of all torsion abelian groups. Then HiX = 0 mod C, so πiX = 0 mod C. Thuswe know that πiS

3 = Z for i = 3 and torsion when i > 3. But since πi for i > 3 is also finitely generated(because the homology of S3 is finitely generated, then we invoke mod C Hurewicz; note that this generalizesto the fact that any simply connected space with finitely generated homology (i.e. finitely many cells at eachlevel) yields finitely generated homotopy), we conclude:

Corollary 8. πiS3 is finite for i > 3.

This idea of “looking at homotopy groups locally” turned out to be quite far-reaching and can be seen as aroot for the development of chromatic homotopy theory.

16.1 Relative mod C Hurewicz Theorem

Theorem 16.1. Let A → X be a pair of spaces, where X and A are both simply connected. Additionally,assume the assumption that we had in lecture 14. If πi(X,A) = 0 mod C for i < n, then Hi(X,A) = 0 mod Cfor i < n, and πn(X,A)→ Hn(X,A) is iso mod C.

Proof. Let F → A→ X be the fibration. Invoke the Serre Spectral Sequence.

n ? ? ? ? ? ? ?

n− 1 0 0 0 0 0 0 0

n− 2 0 0 0 0 0 0 0

. . . . . . . . . . . . . . . . . . . . . . . .

0 Z 0 0 0 0 0 0

0 1 2 . . . n− 2 n− 1 n

Then we see that Hi(A) → Hi(X) is iso for i < n − 1 and epi for i = n − 1 modC, and thus Hi(X,A) =0 mod C for i < n. The rest of the proof is similar as before and we skip.

Suppose C is the class of torsion abelian groups.

Lemma 10. A map A→ B is iso mod C iff A⊗Q→ B ⊗Q is iso.

Proof. Easy.

Observing this, it is easy to prove the following:

Proposition 12. Suppose we have f : X → Y and that H∗(X), H∗(Y ) are finitely generated, X and Y aresimply connected, then the following are equivalent:

31

1. H∗(X;Q) ∼= H∗(Y ;Q);

2. H∗(X;Q) ∼= H∗(Y ;Q);

3. π∗X → π∗Y is iso mod C; and

4. π∗X ⊗Q→ π∗Y ⊗Q is iso mod C.

Now recall that H∗(K(Z, 1);G) = H∗(S1;G) = G[x]/(x2) for |x| = 1. Now look at the CSSS of K(Z, 1) andK(Z, 2) (i.e. K(Z, 1) → PK(Z, 2) → K(Z, 2) with rational coefficients: cohomology of K(Z, 2) (Q, 0,Q, 0, . . .)is the 0th row, that of K(Z, 1) (Q,Q, 0, 0, . . .) is 0th column. Then by staring at the CSSS we get thatH∗(K(2,Z);Q) = Qx2 for some |x2| = 2. Do this again, we see K(Z, 3) has Q-cohomology ring Q[x3]/x2

1. Keeprepeating this, we conclude that H∗(K(Z, 2n);Q) is Q[x2n], and H∗(K(Z, 2n+ 1);Q) = Q[x2n+1]/x2

2n+1 is anexterior algebra.

Note that S2n−1 → K(Z, 2n− 1) is an iso in H∗(•;Q), and thus an iso in π∗ • (mod C) as the propositionabove indicates, and thus πi(S

2n−1) is finite for i > 2n− 1. (πi(S2n−1) is trivial for i < 2n− 1.)

What about the other half of the possibility, i.e. Sn for n even? We have S2n → K(Z, 2n), which is

Q[e]/e2 → Q[x2n] explicitly given by e 7→ x2n where |e| = 2n. Now we have a fibration S2n → K(Z, 2n)x22n−−→

K(Z, 4n). Look at the Serre spectral sequence of the backup K(Z, 4n − 1) → S2n → K(Z, 2n) to get thatH∗(S2n;Q) = Q[x2n, x4n−1]/(x2

2n, x24n−1). We conclude that π∗S

2n ⊗Q is Q for ∗ = 2n, Q for ∗ = 4n− 1, and0 otherwise. As a corollary, π2nS

2n = Z, π4n−1S2n = Z⊕ a finite group, and πiS

2n is finite for other is.Note that there is a systematic treatment of rational homotopy theory via model category theory, due to

Quillen.

32

17 Introduction to Steenrod Operations

Problem 3 (Steenrod’s Problem). If X is a CW complex of dimension ≤ n+ 1, then what is [X,Sn]?

Definition 20. W → Z is an n-equivalence between 1-connected spaces if

1. πi(Z,W ) = 0 for i ≤ n,

2. Hi(Z,W ) = 0 for i ≤ n,

3. πiW → πiZ is isomorphism for i < n and epimorphism for i = n, and

4. HiW → HiZ is the same.

Proposition 13. If W → Z is an n-equivalence and X is a CW-complex, then [X,W ] → [X,Z] is anisomorphism for dimX < n and epimorphism for dimX = n.

Proof. First let’s show the epimorphism. Consider the following diagram, where X is assumed to have dimensionn, and let X(n−1) be its (n− 1)-skeleton.

Sn−1 X(n−1) W

Dn X Z

Since W → Z is (n − 1)-equivalent, it follows induction that every map in [X(n−1), Z] comes from some mapin [X(n−1),W ], so it suffices to deform all the mapping Dn → X → Z to Dn → W . Consider the followingdiagram where W ′ is the pullback of the two maps on the right:

Sn−1 W ′ W

Dn Dn Z

Then as one can check, W ′ and Dn has n-equivalence, and thus isomorphic for πn−1, thus Sn−1 → W ′ isnulhomotopic, so it extends to some Dn → W ′. Then the concatenation with W ′ → W gives us the requiredlifting (at least up to homotopy), and thus we can conclude that every X → Z can be deformed to someX →W for dimX = n.

Now we look at the isomorphism claim. Consider the following graph:

X × I W

X × I Z

Suppose dim(X) < n, then all cells of X × I have dimension ≤ n, so as we have argued above, the liftingX × I → W exists, i.e. any homotopy in Z lifts to a homotopy in W . Thus maps that are zero in [X,Z] willbe zero in the lifting, hence the monomorphism and isomorphism.

Now let us address Steenrod’s problem. Suppose we have Sn → K(Z, n), then clearly it’s an iso on πi fori ≤ n, and epi for i = n+ 1. This tells us that if dimX < n+ 1, then [X,Sn] = [X,K(Z, n)] = Hn(X;Z). Inshort, two maps of the same degree are homotopic (consider e.g. oriented n-manifolds). This result (or at leastits earlier versions) was known in the 1930s. Also, this was part of the reason why cohomology was introduced.

On the other hand, we also know that the mapping Sn → K(Z, n) is iso on Hi, i ≤ n and epi on Hn+1.Thus we know that Hn(K(Z, n);Z) = Z and Hn+1(K(Z, n);Z) = 0. But if we want to go further, we need tofigure out Hn+2. This motivates our further study on the (co)homology of Eilenberg-Maclane spaces.

Let’s try to figure out the cohomology of K(Z, 3) with SSS on its path fibration:

K(Z, 2)→ ∗ → K(Z, 3).

33

4 Zx2 0 0 Zx2i 0 0 ?

3 0 0 0 0 0 0 ?

2 Zx 0 0 Zxi 0 0 ?

1 0 0 0 0 0 0 ?

0 Z1 0 0 Zi 0 0 α

0 1 2 3 4 5 6

In the CSSS above, α is some element such that α2 = 0 and is killed by xi. Then we can conclude thatH3(K(Z, 3)) = Z, H6(K(Z, 3)) = Z/2, and further groups are 0. Using UCT, we have that H3(K(Z, 3)) =Z, H5 = Z2. (See here for the details of this computation.) Or we can look at πi(K(Z, 3), S3) = 0 for i < 5 andZ/2 when i = 5 (which follows the homotopy LES). Then this yields that H5(K(Z, 3)) = Z2, from which thesame result can be deduced.

Now consider the SSS for the next fibration in line:

K(Z, 3)→ ∗ → K(Z, 4).

5 Z2 0 0 0 0 0 0

4 0 0 0 0 0 0 0

3 Z 0 0 0 0 0 0

2 0 0 0 0 0 0 0

1 0 0 0 0 0 0 0

0 Z 0 0 0 Z 0 Z2

0 1 2 3 4 5 6

where (0, 6) comes from transgression. Continuing this process, we conclude that Hn+2(K(Z, n);Z) = Z2.

17.1 Motivating Steenrod Squares

Now we take a more geometric point of view and start considering the chain complex C∗K(Z, n) associatedwith the Eilenberg-Maclane space. By cellular homology, we can guess the following structure:

. . .→ Cn+3 = Z 2−→ Cn+2 = Z→ Cn+1 = 0→ Cn = Z→ . . .

We want to map K(Z, n) into something else so that we can get rid of the Z2 homology at n+ 2.

Naive Approach For the rest of this discussion, keep in mind of the general fact that [W,K(A,m)] =Hm(W ;A). Consider Sn → K(Z, n) and extend it to Sn → K(Z, n) → K(Z, n + 3) and take the homotopyfiber F → K(Z, n)→ K(Z, n+ 3), so that Sn → K(Z, n) factors through F as follows:

F K(Z, n) K(Z, n+ 3)

Sn

34

Take it back a step to K(Z, n + 2) → F → K(Z, n), and consider the SSS. If we look at the SSS, we seethat Z at (0, n+ 2) maps to Z2 at (n+ 3, 0), so it actually does not kill the Z2. We want a Z2 at (0, n+ 2).

However, naively using K(Z2, ∗) doesn’t work either because Hn+2(K(Z2, n+2);Z) = 0 and Hn+3(K(Z2, n+2);Z) = Z2.

Solution Recall that Hn(K(Z, n);Z) = Z and Hn+2(K(Z, n);Z2) = Z2. So instead we consider the fibrationK(Z2, n+ 1)→ F → K(Z, n) obtained from backing up the fibration F → K(Z, n)→ K(Z2, n+ 2).

Before we continue, a few words about [X,F ]. First, observe that for dimX = n+1 we have [X,Sn]→ [X,F ]being an isomorphism. Additionally, Sn ∨ Sn and Sn × Sn have cohomology agree up to 2n, but the first onehas a natural map onto Sn, so [X,Sn] has a group structure for dimX up to 2n. If we localize mod p, on theother hand, then we actually have a group structure all the way up.

Now consider the long exact sequence [X,K(Z, n − 1)] → [X,K(Z/2, n + 1)] → [X,F ] → [X,K(Z, n)] →[X,K(Z/2, n + 2)] → . . .. This yields Hn−1(X;Z) → Hn+1(X;Z/2) → [X,F ] → Hn(X;Z) → 0. Note thatwe “magically” now have a mapping (the first one) that increases the cohomology degree by 2 and reduces thecoefficient mod 2. This is a new structure in the cohomology, denoted by Sq2 : Hk(X;Z)→ Hk+2(X;Z/2).

Proposition 14. The Steenrod square Sq2 has the following properties:

1. additivity,

2. x ∈ H2(X) =⇒ Sq2(x) = x2 (mod 2), and

3. Sq2(xy) = Sq2(x)y + xSq2(y).

Example 10. Consider [CPm, S2m−1]. We have the long exact sequence H2m−2(CPm;Z)→ H2m(CPm;Z2)→[CPm, S2m−1] → H2m−1(CPm;Z) = 0. The first one is Z generated by xm−1 (where x is a generator ofH2(CPm;Z)), the second one is Z2; by the Leibniz’s rule above, we see that the map sends xm−1 to Sq2(xm−1) =(m− 1)xm, so the mapping is multiplication by (m− 1). Thus we conclude that [CPm, S2m−1] = Z2 for m odd,and 0 for m even.

Eventually we shall be able to calculate the cohomology of Eilenberg-Maclane spaces using this machinery(this was done by Cartan and Serre).

35

18 Construction of Steenrod Operations

Let’s start with Brown’s representability theorem. We know that there is an natural isomorphism [X,K(A,n)] ≡Hn(X,A), so we shall not differentiate between the two structures and abuse the notation a bit in what follows.

Recall the external cup product used in Kunneth formula: Hi(X)×Hj(Y )→ Hi(X)⊗Hj(Y )→ Hi+j(X×Y ). Given two maps f : [X,K(A,n)] and g : [X,K(A,m)], we shall write f∪g to represent their image under thisexternal cup product. In particular, let a, b be two mappings X → K(Z2, n) and X → K(Z2,m) respectively,and let im be the canonical mapping K(A,n)→ K(A,n+m). The the following diagram commutes:

X

X ×X

K(Z2, n)×K(Z2,m) K(Z2, n+m)

a∪b∆

a×b

im∪in

Note that [a ∪ b] = [b ∪ a] as cohomology classes, which is to say that (by abuse of notation) as maps, a ∪ band b ∪ a are homotopic. The important observation that Steenrod made was that there is much informationin this homotopy.

Consider the following diagram, given a : X → K(Z2, n). This diagram is commutative up to homotopy:

X ×X K(Z2, 2n)

X ×X

a∪a

(x,y)7→(y,x) a∪a

This means that we have a homotopy expressible as H : X ×X × I → K(Z2, 2n) such that H(x, x′, 0) =H(x′, x, 1), therefore in particular H factors through space X × X × I/(x, y, 0) ∼ (y, x, 1). With a bit ofimagination, observe that this space is homeomorphic to S1×X×X/Z2, where the generator of Z2 sends (λ, x, y)to (−λ, y, x). We write this space as S1 ×Z2

X ×X, so we have the mapping h : S1 ×Z2X ×X → K(Z2, 2n).

In general, let W be a space with an action of Z2, and let τ 6= 0 ∈ Z2. Then a mapping S1 ×Z2 Wh−→ Z is

the same thing as a map Wg−→ Z along with a homotopy g ∼= g τ .

Later we will prove that the map h : S1 ×Z2 X → X → K(Z2, 2n) extends to D2 ×X ×X → K(Z2, 2n),which then extends to S2 ×Z2

X ×X → K(Z2, 2n), etc. Eventually it extends to S∞ ×Z2X ×X → K(Z2, 2n).

This is the main idea: we look for a map S∞ ×Z2X ×X p(a)−−→ K(Z2, 2n) extending X ×X a∪a−−→ K(Z2, 2n). In

particular, note that p(a) ∈ H2n(S∞ ×Z2X ×X;Z2).

Theorem 18.1. For any pair (X,A), there is a unique natural transformation Hn(X,A;Z2)p−→ H2n(S∞ ×Z2

(X,A)2;Z2), such that the following diagram commutes6:

Hn(X,A;Z2) H2n(S∞ ×Z2(X,A)2;Z2)

H2n((X,A)2;Z2)

p

a7→a∪a

where (X,A)2 = (X ×X,X ×A ∪A×A A×X).

Constructing Steenrod Operations First, from now on, as long as we’re talking about Steenrod algebra,

one should always assume the coefficient group to be Z2. Start with the diagonal cup product Hn(X)∆−→

Hn(X) ⊗ Hn(X) → H2n(X × X). Observe that we have the canonical map h : RP∞ × X → S∞ ×Z2X

∆−→S∞ ×Z2 X × X (note that this is well-defined since the image of the diagonal map is invariant under the Z2

action). Assuming the theorem above, we get the following factorization of the diagonal cup product:

Hn(X) H2n(S∞ ×Z2 X ×X) H2n(X ×X)

H2n(RP∞ ×X) H2n(X)

P

p

h∗ ∆∗

6For the vertical map, see next lecture.

36

The diagonal map P is often called the total power operation. Observe that its codomain is H∗(RP∞ ×X) = Z2[t] ⊗H∗(X) = H∗(X)[t] for |t| = 1. It’s important because it yields a construction of the Steenrodoperations:

P (x) = x2 + Sqn−1(x)t+ . . .+ Sq0(x)tn,

where we define Sqj(0 ≤ j ≤ n) as the coefficient of tn−j in the expression above. In particular, Sqn(x) = x2.

Proof of Theorem 18.1. Now we describe this p. Given any (X,A) and a class a ∈ Hn(X,A), there is a map(X,A) → K(Z2, n), ∗) such that a is pulled back from the canonical mapping in ∈ Hn(K(Z2, n), ∗), thereforeit suffices to show that there is a unique choice p(in) ∈ H2n(S∞ ×Z2

(K(Z2, n), ∗)2), and the rest of the prooffollows from naturality. By Kunneth formula, we know that H2n((K(Z2, n), ∗)2) = Hn(K(Z2, n))⊗2 = Z2, soit suffices to show that the mapping H2n(S∞ ×Z2 (K(Z2, n), ∗)2) → H2n((K(Z2, n), ∗)2) is an isomorphism.This follows from the following proposition immediately.

Proposition 15. Suppose (X,A) is (n− 1)-connected, then there is a SES:

0→ H2n(S∞ ×Z2 (X,A)2)→ H2m((X,A)2) = Hn(X,A)⊗Hn(X,A)1−flip−−−−→ Hn(X,A)⊗Hn(X,A)→ 0.

Proof. We first show that for every k ≥ 1, the map Sk×Z2 (X,A)2 → Sk+1×Z2 (X,A)2 induces an isomorphismon H2n(•;Z2). In particular, we need H2n(Sk+1,Sk×Z2 (X,A)2;Z2) = H2n+1(Sk+1,Sk×Z2 (X,A)2;Z2) = 0. Thepair in consideration is relative homeomorphic to (Dk+1×Z2, S

k×Z2)×Z2(X,A)27; since both pairs are excisive,

this induces isomorphism on cohomology. But note that (Dk+1×Z2, Sk×Z2)×Z2

(X,A)2 ∼= (Dk+1, Sk)×(X,A)2,and the cohomology for the latter is H∗(Dk+1, Sk) ⊗H∗(X,A)⊗2, which is 0 for degree less than 2n + 2 (byrelative Hurewicz theorem).

Now that we have confirmed that the degree of the sphere doesn’t matter, we verify the original claim withS∞ replaced with S1. Apply the same argument as above, but this time with the pair (S1, S0). We have thefollowing maps:

. . .← S0 ×Z2(X,A)2 ← S1 ×Z2

(X,A)2 ← (S1, S0)×Z2(X,A)2 ← . . .

where, by the same argument above, H2n((S1, S0) ×Z2(X,A)2) = H2n(S1 × (X,A)2) = 0, so we have the

following map0→ H2n(S1 ×Z2

(X,A)2)→ H2n(S0 ×Z2(X,A)2)→ . . . ,

so it remains to check that the next map on the first sequence is 1−flip, which is a boilerplate definition check.

7details of this homeomorphism can be found here.

37

19 More on Steenrod Operations

Last time we showed that if (X,A) is (n − 1)-connected, then Hi(S∞ ×Z2(X,A)2) = 0 for i < 2n, and

H2n(S∞ ×Z2 (X,A)2) are the elements in Hn(X,A)⊗2 invariant under the tensor factor transposition.In particular, let (X,A) = Kn, then we have a corollary that there is a unique element β(in) ∈ H2n(S∞×Z2

K2n) restricting to in ⊗ in. Thus, the P operator that we mentioned last time uniquely defines Sqi : Hn(X)→

Hn+i(X) for i ≤ n.

19.1 Properties of the Steenrod Operations

Steenrod operations have many nice properties (all products are cup products, and x ∈ Hn(X)):

1. Sqi are homomorphisms.

2. Sqn(x) = x2.

3. Sq0(x) = x.8

4. Cartan formula: Sqn(xy) =∑i+j=n

Sqi(x)Sqj(y).

5. Sqi(x) = 0 for all i < 0.

6. Stability, i.e. the following square commutes:

Hm(X) SqnHm+n(X)

Hm(ΣX) SqnHm+n(ΣX)

7. Sqn(x) = 0 for n > dim(X).

In addition, all of these properties are natural. On Friday, these properties uniquely identify the Steenrodalgebra (Turned out we didn’t.)

Proof. Now let’s prove each of these properties.

Claim 1 and 2 We have shown those above in the construction.

Claim 5 Suppose x ∈ Hn(X;Z2) and i < 0, then Sqi(x) ∈ Hn+i(x) for n+ i < n. By naturality, this shouldhold for the case of K(Z2, n) as well. However, consider the following square with respect to ιn : X → K(Z2, n):

x ∈ Hn(X) Hn(K(Z2, n))

Hn+i(X) Hn+i(K(Z2, n))

Sqi Sqi

ιn

But when n + i < n, the right bottom object is zero, as Eilenberg-Maclane spaces have vanishing lowercohomologies, and thus we must have Sqi(x) = 0.

Claim 4 Equivalently, it suffices to show P (xy) = P (x)P (y), as the Cartan formula then follows fromcomparing coefficients. Note that x pulls back from some in ∈ HnKn, y pulls back from some im ∈ HmKm.Now consider the following diagram in the universal case (which we don’t know commutes yet), in which weuse the shorthand Kj = (K(Z2, j), ∗):

8This one requires explicit computation on cohomology, and is not derivable from the algebraic structure of Steenrod algebraalone. For instance, in the algebraic construction of Steenrod algebra over a Galois field, the Sq0 in fact corresponds to theFrobenius map.

38

Hn(Kn)×Hm(Km)

Hn+m(Kn ×Km) H2n(S∞ ×Z2 K2n)×H2m(S∞ ×Z2 K

2m)

H2n+2m(S∞ ×Z2(Kn ×Km)2) = Z2 H2n(K2

n)×H2m(K2m)

H2n+2m((Kn ×Km)2)

∪p×p

p h∗×h∗∪

h∗ ∪

As a general observation, if (X,A) is (n−1)-connected and Hk(X,A) = Z2, then H2k(S∞×Z2(X,A)2) = Z2

because it is a subgroup of Hk(X,A)⊗2 = Z2 × Z2 invariant under the flip isomorphism.Thus we see the object second-to-bottom on the left column above is nothing other than Z2. On the other

hand, since the lower rectangle commutes, we see that both maps factor through Z2 followed by h∗, so the imagein the last row is Z2 (it’s not 0 otherwise P = 0, which we refute later). Now in the universal case, the mapping(im, in) 7→ P (im ∪ in) and (im, in) 7→ P (im) ∪ P (in) both are mappings in Hom(Z2 × Z2,Z2) = Z2, so eitherthey are equal (and both nonzero), or at least one of them is the zero map. Since the general case pulls backfrom the universal case, we conclude that either P (xy) = P (x)P (y) for all x, y, or one of them is constantlyzero. So we can prove the theorem as long as we can see that both (x, y) 7→ P (xy) and (x, y) 7→ P (x)P (y) arenonzero at some point.

To do this, consider H∗(RP∞) = Z2[t]. We know that Sqn+m(tntm) = Sqn+m(tn+m) = t2(n+m). On theother hand, Sqn(tn) = Sqm(tm) = t2(n+m), so neither is zero, and we have thus obtained the Cartan formula.

Claim 3 Again let’s work with the universal case. Observe that Sq0(in) ∈ Hn(K(Z2, n),Z2) = Z2. Moreover,any homomorphism Sq0 : Hn(X)→ Hn(X) is given, in the universal case, by a map in [K(Z2, n),K(Z2, n)] ∼=Hn(K(Z2, n),Z2) = Z2, so there are only two natural maps: the identity map, and the zero map. Thus itsuffices to show that Sq0 is nonzero somewhere.

The example that we consider is (Rn,Rn − 0). Again look at h : S∞ ×Z2(Rn,Rn − 0)2 ← RP∞ ×

(Rn,Rn−0)). So let’s figure out what happens to H2n of the first space. Note that H2n of the second space isHn(RP∞)⊗Hn(Rn,Rn−0) = Z2, which is generated by tnx for some nonzero generator of Hn(Rn,Rn−0).We would like to show that p(x) gets map to tnx.

In general, note that (V, V −0)× (W,W −0) = (V ×W,V ×W −0). In particular, (Rn, Rn−0)2 =(R2n, R2n − 0). Now consider how Rns embed into R2n: they embed as the “axes”, and the Z2 factor willact by flipping these two axes (i.e. flipping around the “diagonal”). However, we can choose an alternativebasis by choosing the diagonals as the “axes” (i.e. a different embedding), then we see that Z2 acts as −1on one of them, and 1 on the other. This allows us to rewirte the mapping as (S∞ ×Z2 (Rn+,Rn+ − 0)) ×(Rn−,Rn− − 0) ← RP∞ × (Rn,Rn − 0)), in which case Z2 now does not act on the first parenthesis, so wehave factored this map into product of two maps, where the second factor is the identity, and the firt factor isRP∞ = S∞ ×Z2

(0, ∅)→ S∞ ×Z2(Rn,Rn − 0). By Kunneth formula, we are reduced to showing that this

factor map is nonzero in Hn.Before we proceed further, however, observe that Sq0(xy) = Sq0(x)Sq0(y), so it suffices to assume that

n = 1. Now note S∞ ×Z2(R1,R1 − 0) has the same cohomology as S∞ ×Z2

((−1, 1),−1, 1) by a directtransformation. We want to show that H∗(S∞ ×Z2

((−1, 1),−1, 1))→ H∗(S∞ ×Z2(−1, 1)) is an isomorphism.

By the long exact sequence of a pair, it suffices to note that H1(S∞ ×Z2−1, 1) = H1(S∞) = 0.

Claim 6 and 7 are formal consequences of the ones above.

39

20 Mod 2 Cohomology of K(Z2, n)

We still want to compute the cohomology of Eilenberg-Maclance spaces. As we said before, the Z-coefficient caseis rather tricky, but today we can establish the H∗(K(Z2, n);Z2) case. (From now on, omit the Z2 coefficient.)

Let’s start with the only one that we’re sure of. H∗(K(Z2, 1);Z2) = H∗(RP∞;Z2) = F2[x] for some x ∈ H1.Now we want to consider H∗(K(Z2, 2)). Consider the fibration K(Z2, 1) → ∗ → K(Z2, 2) and its SSS (we’llonly write generators now).

5 x5 0 ? ? ? ? ?

4 x4 0 ? ? ? ? ?

3 x3 0 ? ? ? ? ?

2 x2 0 x2y2 ? ? ? ?

1 x 0 xy2 ? ? ? ?

0 1 0 y2 y3 y22 ? ?

0 1 2 3 4 5 6

The first column is given above. We know that (1, 0) is zero and in fact 1st column is zero since K(Z2, 2) issimply connected. By transgression, we must have d2(x) = y2 for some y2. We know that d2(x2) = 2xd2(x) = 0,so x2 does not hit (2, 1) and thus the mapping d2(xy2) = y2

2 must be nonzero. Now, we know that d2(x3) targetsx2d2(x) = x2y2. On the other hand, x2 must be transgressive, so d3(x2) = y3 for some y3 ∈ (3, 0), which mightbe zero or might not. Technically, you can do this to figure out the rest of the E2 page and thus the entirespectral sequence (see your homework).

The important thing to figure out is the differentials on x2k . We (i.e. the students) shall show the following:

Proposition 16. Let F → E → B be a fibration and let π1B acts trivially on the fiber, and H∗(E) = H∗(pt).

1. If x ∈ H∗(F ) is transgressive, then so is x2n for all n.

2. Borel’s Theorem (see homework) Assume cohomology works on Z2 coefficients. The cohomology of Bis given by the polynomial algebra generated by the images of all the transgressions.

We shall let the target of x2n be called y2n+1. The proof of Borel’s theorem is done in the homework usingZeeman comparison theorem,

Koszul Resolution Consider the abstract chain complex C∗ = F2[x][y2, y3, . . . , y2n+1, . . .]. Define the dif-ferential d(x2n) = y2n+1. Then this extends to all xn in the obvious manner: e.g. d(x14) = d(x8 · x4 · x2) =

(Leibniz Rule) = x12y3+x10y5+x6y9. More abstractly, we want a map from vector space

k⊗n=0

F21, x2n to F2[x]

by first sending it to⊗k

F2[x] then to F2[x]. In particular, note that the colimit of the

k⊗n=0

F21, x2n, w.r.t.

k, is isomorphic to F2[x]. Let en = x2n and let eI = ei1 . . . eik for I = (i1, . . . , ik), then eI form a basis of F2[x].Now, notice that the complex C∗ is the tensor product of the Dn(n ≥ 0), where Dn = F21, en ⊗ F2[y2n+1].where den = y2n+1, dyq2n+1 = 0, dyq2n+1en = yq2n+1den. Then by explicit computation, we find that the homol-ogy of Dn is Z2 at 0 and 0 at higher levels. And as we tensor these Dn together, we see that H0(C∗) = Z2 and0 at higher levels. This construction is often known as the koszul resolution.

So back to our old discussion, why is x transgressive? Steenrod Algebra! Note that x ∈ Hn =⇒ Sqn(x) =x2. We have the Kudo transgression theorem which works for any SSS: suppose we have F → E → B andsuppose π1B acts trivially. If x ∈ Hj(x) is transgressive, then Sqk(x) is transgressive and the transgression ofSqk(x) is the Sqk of the transgression of x.

40

Proof. Start by considering the definition of transgression:

Hj+1(E,F ) Hj(F )

Hj+1(B)

δ

p∗

to say that γ is the tresgression of x is to say that p∗(y) = δ(x). Now apply Sqk to everything, and it sufficesto show that the diagram still commutes.

We need Sqk(δx) = δ(Sqkx). Consider the definition of the connecting homomorphism. The following areall isomorphisms in cohomology:

(E ∪ CF,CF ) (E ∪ CF, ∗)

(E,F )

And we see there’s a mapping E∪CF → ΣF , which then induces a mappingHk+1(E∪CF ) = Hk+1(E,F )←Hk+1(ΣF ) ∼= Hk(F ). So the connecting homomorphism is just the suspension isomorphism followed by a map.Then it is clear that this connecting homomorphism commutes with Sqk. Finally, it is a straightforward checkthat the diagram above still works after Sqk.

Now using the Kudo regression theorem, we know that x2 = Sq1(x) transgresses to Sq1(y2) = y3. Similarly,x2 = Sq2(x2) transgresses to Sq(y3) = y5. Therefore we have:

Corollary 9. H∗(K(Z2, 2)) = Z2[y, Sq1y, Sq2Sq1y, Sq4Sq2Sq1y, . . .].

Now consider the same settting but applied to K(Z2, 2) → ∗ → K(Z2, 3). By Borel’s theorem, we canconclude that K(Z2, 3) again has a polynomial algebra cohomology. In particular, this allows us to computethe cohomology of all K(Z2, n). Next time we’ll introduce a nice way to clean things up.

41

21 Computing with Steenrod Squares (Guest Lecturer: JeremyHahn)

Recall from the last lecture that we have the natural operations Sqk : Hk(•,F2) → H∗+k(•,F2) satisfying theSteenrod axioms.

Definition 21. For any sequence I = (im, im−1, . . . , i1), we use SqI to denote the composition Sqim . . .Sqi1 .

Last time we used Serre Spectral Sequence to compute Hk(K(F2, n)).

Definition 22. A sequence I = (im, . . . , i1) is admissible if ik ≥ 2ik−1 for all k. The excess of I is then definedas (im − 2im−1) + (im−1 − 2im−2) + . . .+ (i2 − 2i1) + i1.

Theorem 21.1. The cohomology H∗(K(F2, n)) = F2[SqIιn] where I runs over all admissible sequences ofexcess < n9.

21.1 Cohomology Operations

Definition 23. Fix an integer l, a cohomology operation of degree i is a natural transformation H l(•,F2) →H l+i(•,F2).

Examples: in degree 5, we have Sq5, Sq2Sq3 = Sq(2,3), Sq5 + Sq(2,3), etc.

Definition 24. A stable cohomology operations of degree i is a natural transformation H∗(•,F2)→ H∗+i(•,F2)commuting with the suspension isomorphism, i.e. H∗(X) ∼= H∗+1(ΣX).

Note that every cohomology operation can be extended to a stable cohomology operation in our currentsetting, so strictly speaking we don’t have to worry about the term “stable” that much.

Remark 7. H l(•,F2) is represented by K(F2, l). By Yoneda’s lemma, cohomology operations are in bijectionwith H l+i(K(F2, l)), and these are sums of SqI by what we said above.

Remark 8. All apparently “non-admissible” operations are in fact admissible, e.g. Sq(2,3) = Sq5 + Sq(4,1).

So this brings in the question: how can we prove that two stable cohomology operations are the same?

Theorem 21.2. Suppose a, b are both degree i stable cohomology operations, and suppose that they agree onproducts of 1-dimensional classes. Then a and b are the same stable cohomology operation.

Corollary 10. Sqi is the unique degree i stable cohomology operations satisfying:

1. The Cartan formula, and

2. Sq0(x) = x, Sq1(x) = x2, Sqi(x) = 0 for i ≥ 2 when x is a 1-dimensional class.

Let’s see a few examples on how to use this theorem first.

Example 11. Sq1Sq1 = Sq(1,1) = 0.

Proof. If x is one dimensional, then Sq1(Sq1(x)) = Sq1(x2) = xSq1(x) + Sq1(x)x = 2xSq1(x) = 0. Otherwisesuppose x = yz, where y, z are smaller products of 1-dimensional classes. Then Sq1(Sq1(yz)) = Sq1(Sq1(y)z+ySq1(z)) = Sq(1,1)(y)z + Sq1(y)Sq1(z) + Sq1(y)Sq1(z) + ySq(1,1)z = 0 by induction.

Example 12. Sq1Sq2 = Sq3. (Note that the RHS is admissible.)

Proof. If x is a 1-dimensional class then both sides are 0. Now suppose x = yz, then Sq3(yz) = Sq3(y)z +Sq2(y)Sq1(z)+Sq1(y)Sq2(z)+ySq3(z) = Sq(1,2)(y)z+Sq2(y)Sq1(z)+Sq1(y)Sq2(z)+ySq(1,2)z (by induction).On the other hand,Sq(1,2)(yz) = Sq1(Sq2(yz)) = Sq1(Sq2(y)z + Sq1(y)Sq1(z) + ySq2(z)) = Sq(1,2)(y)z +Sq2(y)Sq1(z) + Sq(1,1)(y)Sq1(z) + Sq1(y)Sq(1,1)(z) + Sq1(y)Sq2(z) + ySq(1,2)(z), which is what we had beforesince Sq(1,1) = 0.

Proof of the Theorem. We first need a lemma. Consider x1 . . . xn ∈ Hn(RP∞ × . . .× RP∞︸ ︷︷ ︸n copies

) = F2[x1, . . . , xn];

by naturality, it corresponds to a map (x1 . . . xn)∗ : RP∞ × . . .× RP∞︸ ︷︷ ︸n copies

→ K(F2, n).

9Mike’s note says ≤ n, but that might have been a mistake.

42

Lemma 11. The natural map (x1 . . . xn)∗ produces a monomorphism in H l(•,F2) for all l ≤ 2n.

Proof of the Lemma. We’d like to show that H l(K(F2, n)) → H l(RP∞×n) for l ≤ 2n. For this, we want toshow that the images of SqI(ιn) are linear independent for I admissible of total degree ≤ 2n. The images areSqI(x1 . . . xn). Each of these will be symmetric polynomials in the xi, so we can express them in terms ofelementary symmetric polynomials (σ1 = x1 + . . .+xn, σ2, . . .). We want to calculate SqI(σn). The rest is justcalculation, and follow the claims below.

Claim 1 Sqi(σn) = σnσi.

Claim 2 SqI(σn) = σnσim . . . σi1 +(sum of monomials of smaller lexigraphical order) where I = (im, . . . , i1).

Now we want to show that a, b : Hn(•)→ Hn+i(•) are the same for all n. Of course, this is saying that a andb define the same element in Hn+i(K(F2, n)), which is the same to say that a(ιn) = b(ιn) for all n. By stability,it suffices to prove the statement for n > i. By the lemma, it suffices to show that a(x1 . . . xn) = b(x1 . . . xn)(of course we’re abusing the notations a bit). But we have already assumed that a and b agree on products of1-dimensional classes.

Some side notes: Steenrod squares are examples of primary cohomology operations; it turns out that thereare also higher cohomology operations, which one discovers from the study of Adam spectral sequences. It turnsout that though the primary cohomology operations do not completely capture the structure of the cohomology,along with the higher cohomology operations they do; in other words, two cohomology theories that agree onall cohomology operations would be equal to each other.

43

22 Adem Relation and Bocksteins (Guest Lecturer: Xiaolin Shi)

Recall that we have proven the axiomatized description of the Steenrod operations and the uniqueness theorem.Let’s just restate it here:

Theorem 22.1 (Uniqueness). The Steenrod squares are the unique cohomological operations such that:

• Sq0(x) = 0.

• Sq|x|(x) = x2.

• Sqk(x) = 0 for k > |x|.

• Sqk commutes with the suspension isomorphism.

• Cartan formula holds.

We also briefly touched on–though did not prove–the Adem relation, which allows us to write an arbitrarySteenrod product to an admissible one.

22.1 Adem Relation

Proposition 17 (Adem Relation). SqaSqb =

ba/2c∑j=0

(b− j − 1

a− 2j

)Sqa+b−jSqj.

Keep in mind that everything is mod 2.

Example 13. Sq3 = Sq1Sq2 (we have shown this last time).

Example 14. Sq2Sq3 =

(3− 0− 1

2− 2 · 0

)Sq5Sq0 +

(3− 1− 1

2− 2 · 1

)Sq4Sq1 = Sq5 + Sq4Sq1 =⇒ Sq5 = Sq2Sq3 +

Sq4Sq1 (since we work mod 2).

Today we talk about the Bockstein homomorphisms.

22.2 Bockstein Homomorphisms

We already know that H∗(K(Z2, n);Z2) = Z2[SqIιn] for admissible I with excess less than n. We computedthis via the Serre spectral sequence. By the same method, one can compute H∗(K(Z, n);Z2) = Z2[SqJ ιn] withadmissible J with excess less than n, and that if J = (jm, . . . , j1), then j1 > 1. Today we’d like to computeH∗(K(Z/2k, n);Z2) for k > 1 based on the facts above. (From now on we drop the Z2 coefficient.)

First step, let’s consider n = 1. Consider the SES 0 → Z 2k−→ Z → Z/2k → 0. We get a fibration

K(Z, 1)2k−→ K(Z, 1)→ K(Z/2k, 1)10. We know that K(Z, 1) = S1. Now we run the Serre spectral sequence on

this fibration: Ep,q2 = Hp(K(Z/2k, 1);Hq(S1))⇒ Hp+q(S1). This means we have to only consider two rows.11

0 2 4 6

0

2

4

Z2(1) Z2(ι1) Z2(βk) Z2(ι1βk) Z2(β2k) Z2(ι1β

2k) Z2(β3

k)

Z2(a) Z2(aι1) Z2(aβk)Z2(aι1βk)Z2(aβ2k)Z2(aι1β

2k)Z2(aβ3

k)

10How do we know this is a simple fibration?11I’m using the sseq package for spectral sequences from now on.

44

Both rows are (H0, H1, . . .), and we know that H0 = H1 = Z2 since we’re with Z2 coefficients. On E∞we should have Z2 for H1(E), so (0, 1) must have been killed, and thus the first nontrivial differential isa monomorphism, and because (2, 0) must be killed we know that that particular differential is in fact anisomorphism, and thus H2 = Z2. Continue this process to the right, we see that Hj = Z2 for every j.

Now let us assign names to the generators as indicated in the graph above (note that ι1 is the funda-mental class of K(Z2k , 1)), and compute the rest using the Leibniz rule. Then we see that H∗(K(Z/2k, 1) =Z2[ι1, βk]/(ι21 = 0).12 This yields that H∗(K(Z/2k, 1) = Z2[ι1, βk]/(ι21 = 0) = Z2[ι1]/(ι21 = 0) ⊗ Z2[βk] =H∗(K(Z, 1))⊗H∗(K(Z, 2)).

In general, this calculation yields that H∗(K(Z/2k, n);Z2) = H∗(K(Z, n)) ⊗ H∗(K(Z, n + 1)), which one

can write as Z2[SqIιn, SqJβk]. One can then consider the pullback fibration K(Z/2k, n) → K(Z, n + 1)

2k−→K(Z, n+1), then the βk ∈ Hn+1(K(Z/2k, n)) is the restriction from ιn+1 ∈ Hn+1(K(Z, n+1)), so we have βk ∈[K(Z/2k, n),K(Z/2, n+ 1)]. This yields a stable natural transformation βk : Hn(X;Z/2k)→ Hn+1(X;Z2).

But where do we get this βk, really? Consider the fibration Z2 → Z/2k+1 → Z/2k, and consider itsassociated fiber sequence:

. . .→ Hn(X;Z2)→ Hn(X;Z/2k+1)→ Hn(X;Z/2k)βk−→ Hn+1(X;Z2)→ . . .

In the special case of k = 1, we have β1 : Hn(X;Z2)→ Hn+1(X;Z2), which coincides with Sq1.

12How do we know that βk 6= ι21? We can compare its action as a cohomological operation with ι21.

45

23 More on Adem Relation

Again, welcome back Mike. Today we would like to prove the Adem relation.

23.1 Some Tricks for Computation

First, a fast trick to compute binomial coefficients mod 2. The validity of the following method can be easilyproven.

Proposition 18. Suppose p is a prime and m = (a1 . . . ar)p, n = (b1 . . . br)p, then

(m

n

)=

r∏i=1

(arbr

)(mod p).

Example 15.

(5

4

)=

((101)2

(100)2

)=

(1

1

)(0

0

)(1

0

)= 1 (mod 2).

Second, recall that for x ∈ Hd(X), we the total power operation P (x) ∈ H2d(X × RP∞) = Sqd(x) +Sqd−1(x)t+ . . .+ Sq0(x)td ∈ H∗(X)[t], where the Steenrod operations arise as coefficients of this polynomial.In order to make bookkeeping easier, another way of writing this is by the lower index: Sq0(x) + Sq1(x)t +. . .+ Sqd(x)td, where Sqd−i(x) = Sqi(x).

23.2 A New Treatment

We can also write the Adem relation in the lower-index form. Suppose we apply P again. For the sake of

clarity, let’s write the P (x) above as Pt(x). So we have Pt(x) =∑i

Sqi(x)ti. Now consider the formal operator

L(t) =∑i

Sqiti, then L(s) · L(t) =

∑i,j

SqjSqisitj , which we can consider as a generating function for SqjSqi.

We have proven before that Pt is a ring homomorphism, which yields additivity and the Cartan formula. Onemore detail for calculation: suppose X = RP∞, and 0 6= x ∈ H1(RP∞), then Pt(x) = x2 + xt.

Now suppose x ∈ Hd(X). Let us calculate Pt(Ps(x)). It equals Pt(∑i

Sqi(x)si) =∑i

Pt(Sqi(x))Pt(si) =∑

i

∑j

SqjSqi(x)tjPt(s)i =

∑i

∑j

SqjSqi(x)tj(s2 + st)i = L(t) · L(s2 + st). Now we shall prove that that

L(s) · L(s2 + st) is symmetric in s and t, and thus we have the Adem Relation:

L(t) · L(s2 + st) = L(s) · L(t2 + st).

Symmetric Property. We need to show that Pt Ps(x) = Ps Pt(x) for x being a product of 1-dimensionalclasses (then invoke the uniqueness theorem of stable cohomology operations). Since both sides are ring ho-momorphisms, it suffices to do the case in which x is a 1-dimensional class. Now, Pt Ps(x) = Pt(x

2 + sx) =(x2 + xt)2 + (s2 + st)(x2 + xt) = x4 + (t2 + s2 + st)x2 + (s2t+ st2)x, which is symmetric.

(We’ll show later that this symmetric property actually stems from the very definition of things, and don’treally rely on the lucky calculation above.)

Now we show that the form above of the Adem relation coincides with the usual one. We want∑i,j

(s2 + st)itjSqiSqj =∑i,j

(t2 + st)isjSqiSqj .

Let u = s2 + st, then the left side becomes∑i,j

uitjSqiSqj . How do we handle the right side? Use residues.

Suppose we have a formal Laurent series f(x) =∑i>−∞

aixi, then Resx=0f(x)dx = a−1, which is the coefficient

ofdx

x. Also, suppose x = g(y), g(0) = 0, g′(0) is nonzero, so g−1 exists by the inverse function theorem.

Theorem 23.1. Resx=0f(x)dx = Resy=0f(g(y))g′(y)dy.

Proof. See any complex analysis textbook.

46

Now look at∑i,j

uitjSqiSqj , so∑n

tnSqnSqm is the coefficient of um in it.13 Now u = s2 + st, du = tds

(where s is the variable), thus we know that

∑n

tnSqnSqm = coefficient of um in∑i,j

uitjSqiSqj = Resu=0uisjSqjSqium+1

du

which further simplifies to

Ress=0ti+1(s+ t)isj

sm+1(s+ t)m+1SqiSqjds = Ress=0

ti+1(s+ t)i−m−1sj

sm+1dsSqiSqj

which is the coefficient of sm in∑i,j

ti+1(s+ t)i−m−1sjSqiSqj , thus tnSqnSqm =

(i−m− 1

m− j

)t2i+j−2mSqjSqi,

from which it is immediate by a substitution that

SqnSqm =

(i−m− 1

2i−m− n

)Sq2m+n−2iSqi.

13Sorry for the notation mess.

47

24 Even More on Adem Relation

Let’s first investigate a more elementary way to prove the relation Ps(Pt(x)) = Pt(Ps(x)). Recall the diagram:

RP∞ ×X S∞ ×Z2 (X ×X)

K(Z2, 2n)

Pt(x)Pt(x)

If we apply this action again, we get

RP∞ × RP∞ ×X S∞ ×Z2(S∞ ×Z2

(X ×X))2

K(Z2, 4n)

Ps(Pt(x))Ps(Pt(x))

Consider the space S∞× (S∞×S∞×X2×X2) that “covers” the space on top right. There’s one Z2 actingon the outside, and there’s Z2 × Z2 acting on the inside, and the outer Z2 acts on the Z2 × Z2 by exchangingthe factors. So we can write the space S∞ ×Z2 (S∞ ×Z2 (X ×X))2 as (S∞)3 ×Z2oZ2 X

4 → K(Z2, 4n), where odenotes the wreath product.

As a motivating example, suppose X is a point. The fundamental group of RP∞ ×RP∞ ×X in this case isZ2×Z2, and the fundamental group of (S∞)3/(Z2 oZ2) is Z2 oZ2, so the trivial mapping Z2×Z2 → ∗, inducedby Ps Pt, actually factors as Z2 × Z2 → Z2 o Z2 → ∗. So what is this embedding? Note that by definition,Σn o Σk ⊆ Σnk, so Z2 o Z2 ⊆ Σ4 acts on four elements (namely two copies of the former Z2). In particular, ifthe two copies are 1, 2 and 3, 4 respectively, then the generator of the first Z2 (the “outer” Z2) embeds as(13)(24), and the generator of the second Z2 (the “inner” Z2) embeds as (12)(34). One should note that thesetwo actions are conjugate in Σ4 by the element (23), as this conjugation relation is really why Adem relationwould hold.

24.1 Higher Power Operations

It is easy to observe that S∞ ∼= (x, y) ∈ R∞ | x 6= y14. To generalize this, let Ck(R∞) denote the kthconfiguration spaces, i.e. (x1, . . . , xk) ∈ R∞ | xi 6= xj. The configuration spaces are contractible spaces

with free Σk actions. It turns out that BΣk = Ck(R∞)/Σk (modding out the natural action of Σk, i.e. taking

unordered size-k sets). Define the kth extended power of Xx−→ K(Z2, n) as the unique extension to Pk(x)

(which we omit the definition) such that the following graph commutes and that Pk(x) i = xk:

X BΣk ×X Ck(R∞)×Σk Xk

K(Z2, nk)

i

Pk(x)Pk(x)

The fact giving the Adem relation is that P2 P2 factors through a total 4th power operation i.e. the mapin fact factors as follows:

BΣ2×BΣ2×X = B(Σ2×Σ2)×X → C2(R∞)3×Z2oZ2X4 → C4(R∞)×ZoZ2

X4 → C4(R∞)×Σ4X4 → K(Z2, 4n)

where the only not-so-canonical map C2(R∞)3 ×Z2oZ2X4 → C4(R∞) ×Z2oZ2

X4 comes from the following

construction of C23 → C4: the elements in the two “inner” C2s are two pairs of points (a1, a2), (b1, b2), such

that a1 6= a2, b1 6= b2, and that each pair is indexed by by a pair of points in X; the outer C2 then has twopoints ca 6= cb, indexed by the two inner pairs repsectively. We would like to consider only the two innerpairs, but of course they might collide; so what we need to do is use ca and cb to send them to two differentdirections so that they are apart. More precisely, consider the continuous mapping: let the “ball” containinga1 and a2 be of radius ra, and the other “ball” containing b1 and b2 be of radius rb, then send the two “balls”along the directions c1 and c2 by some large value depending on max(ra, rb) and cos(θ) where θ is the anglebetween ca and cb, such that the center of the two balls are 3 max(ra, rb) apart, meaning that they are disjoint,so (a1, a2, b1, b2) ∈ C4 now.

14The latter deformation retracts to the former, as one can visually see from the finite cases.

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Then, due to the following fact and the obvious fact that C4(R∞)×Σ4X4 = BΣ4×X, the two maps under

consideration are homotopic in their BΣ2 ×BΣ2 ×X → BΣ4 ×X part, since we mentioned in the last sectionthat the induced actions are conjugate. Since P4 is uniquely defined, the second part BΣ4×X → K(Z2, 4n) isunique, so we have shown that the two maps are homotopic.

Proposition 19. Suppose we have f1, f2 : H → G are conjugate, i.e. ∃g ∈ G.gf1g−1 = f2, then Bf1, Bf2 :

BH → BG are homotopic.

Proof. Skipped.

It remains to show that the map factors. In fact, we will show the stronger statement that H4n(C4(R∞)×ZoZ2

X4) = H4n(C4(R∞) ×Σ4X4). Without loss of generality we consider the universal case w.r.t. cohomology

pullback, i.e. when X = K(Z2, n) and A = ∗, so that H∗(X,A) = 0 for ∗ < n by Hurewicz. Suppose G isa group acting on (X,A)k for some k, and that S is a contractible space with free G action. Let B = S/G,then B is a classifying space of G, i.e. π1B = G, higher homotopy groups vanish, and S is the universal cover.Suppose E is an arbitrary space with G action. Then we have the following theorem:

Theorem 24.1. If H∗(E) = 0 for ∗ < r, then Hr(S ×G E)→ Hr(S ×E)→ Hr(E) is a monomorphism withimage those elements invariant under G.

Proof. Use Serre spectral sequence on the fiber bundle E → S×GE → B. Consider the SSS, which has two iden-tical columns on the E2 page. Everything is zero until row n, where the first element becomes H0(B,Hn(E)).The action of the fundamental group is nontrivial so we have to consider local coefficients; but, as one ob-serves from inspecting the chain complex, this action is just the action of G, so the 0th cohomology is justthe invariants of Hn(E) under G. Since nothing can ever get cancelled, this fully determines the SSS. SoHn(S ×G E)→ Hn(F ) is injective by edge homomorphism.

But back to our story. Specializing with E = (K(Z2, n), ∗)4, r = 4n, S = C4(R∞) and G = Σ4 andG = Σ2 oΣ2 respectively, we see that H4n(C4(R∞)×ZoZ2

X4) and H4n(C4(R∞)×Σ4X4) (or really, their images

under monomorphism) are respectively just elements of Hr(K(Z2, n), ∗)⊗4 invariant under Σ2 o Σ2 and Σ4

respectively. However, note that in our case, (X,A)k = (K(Z2, n), ∗)4 = K(Z2, n)∧4 as one can check fromdefinition, so the target space is H4n(K(Z2, n), ∗)∧4) = Z∧4

2 = Z2. It is clear that the invariant of this Z2 underboth Z2 o Z2 and Σ4 is Z2 itself. Hence the isomorphism.

Moral of the story: Adem relation holds because something is invariant under Σ4, not just Z2 o Z2. As anending note, there is yet another pure homological-theoretical explanation to this story that involves explicitlycomputing out all cohomologies involved. The takeaway from that story is that the only possible obstructionto the Eilenberg-Maclane structure from having a E∞ spectra is the higher homotopy groups of E-M spaces,which of course don’t exist. (Charles: I might sketch out that construction in my blog. The reference here isSteenrod and Epstein, Cohomology Operations.)

24.2 Application of Steenrod Algebra: Homotopy Groups of Spheres

Let C2 be the Serre class consisting of group A such that nA = 0 for n 0 and n odd. (modding out C2 isequivalent to ignoring all odd torsions.) We shall calculate π∗(S

n) localized at 2 by replacing Sn by an easierspace which is an equivalence mod C2 through a range.

Lemma 12. Suppose X,Y have finitely generated homology and X → Y is an isomorphism in Hi(X;Z2) fori < n and epi in i = n. Then Hi(X;Z)→ Hi(Y ;Z) is iso mod C2 for i < n and epi mod C2 for i = n.

Proof. Easy by the UCT.

Remark 9. An equivalent condition is that Hi(X;Z2)←− Hi(Y ;Z2) is iso for i < n and epi for i = n.

What we can say from the conditions above is that πi(X) = πi(Y ) mod C2 for i < n. Thus in particular,if we can build spaces that cohomologically approximate Sn mod C2, then we can us them to compute thehomotopy groups of Sn mod C2.

Take Sn → K(Z, n). We know H∗(K(Z, n)) = Z2[SqIιn], and more explicitly for the latter space, Hn =Z2(ιn) (single Z2 summand generated by ιn), Hn+1 = 0 (because I must have first term greater than 1), soup till here it agrees with H∗(Sn); but the next one, generated by Sq2ιn, is probably different. So let’s modifyK(Z, n) so it has closer comohological behavior.

In particular, consider the following diagram, where X1 → K(Z, n)Sq2−−→ K(Z2, n + 2) is a fibration (X1 is

the homotopy fiber) and K(Z2, n+ 1)→ X1 → K(Z, n) is its one-step backup:

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K(Z2, n+ 1) X1

Sn K(Z, n) K(Z2, n+ 2)Sq2

Note that since πn(K(Z2, n + 2)) = 0, the mapping from πn(X1) → πn(K(Z, n)) is surjective, so Sn →K(Z, n) lifts to Sn → X1. Run Serre spectral sequence (diagram omitted), we see that ιn+1 hits Sq2ιn bytransgression, and Sq1ιn+1 = Sq3ιn by transgression theorem, and thus Sq3ιn+1 = Sq2Sq2ιn = Sq3Sq1ιn = 0by Adem relation. Similarly Sq2Sq1ιn+1 hits Sq5ιn. Consider what’s left in the cohomology of X1, one cancheck that it’s Z2 in dimension n, one in dimension (n + 3), 2 classes in (n + 4), etc. We’ll continue thiscomputation in the next lecture.

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25 Some Nontrivial Stable Homotopy Computation

Professor Hopkins recently discovered a new puzzle website made by A. Watanuki, a graduate student at RIMS,that generates spectral sequence computation problems called Subedoku, which are presented in a similar styleto the famous Sudoku puzzles. The website can be found here.

25.1 Killing Cohomology

Keep in mind that the overall goal of ours is to compute homotopy groups of spheres. Below we shall sketchthe overall strategy.

Start with the mapping Sn → K(Z, n) = X0 → K(Z2, n + 2), and choose X1 to be the homotopy fiber ofX0 → K(Z2, n + 2). As we said in last class, there is a lifting f : Sn → X1. Now choose the smallest m suchthat Hmf : HmX1 → HmSn is not a monomorphism, and let A be the kernel of Hmf . Then we have themapping Sn → X1 → K(A,m), and we can continue the construction by letting X2 be the homotopy fiber ofX1 → K(A,m), etc. (That m actually increases is not completely trivial, but is not hard to observe from theSSS on the backup fibration, as we have seen in the last class.)

More generally, in order to compute the cohomology of some space W , start with a mapping into a product

of E-M spaces that is an approximation of homotopy groups on low dimensions: f : W →∏i

K(πi(W ), i) = X0,

such that f is surjective in H∗(•;Z2). Let m be the smallest integer such that there is a nontrivial kernel inHm(X0) → Hm(W ), and let A be this kernel. Then we have W → X0 → K(A,m), and we can continue thisprocess. In the end W → X∞ will be an isomorphism in H∗(•;Z2).

The last piece of the story, of course, is that X1, X2, . . . are now cohomological approximations, and thushomotopy approximations, of Sn mod C2. The homotopy of each Xi, on the other hand, is easy to computeinductively thanks to the homotopy LES of fibrations.

Remark 10. This technique is called “killing cohomology one at a time”. Frank Adams later observed that ifinstead of doing this one at a time, we kill all cohomologies simultaneously, we can obtain even stronger results;this leads to the Adams spectral sequence.

25.2 A Few Notes

Consider the fibration Xk+1 → Xka−→ K(Z2, n + 1) and the backup K(Z2, n) → Xk+1 → Xk. Consider the

spectral sequence of the latter. On the base, a lies on dimension n + 1 as the first nonzero element, and ιn isthe first nonzero element on the fiber column, on dimension n. Let (1) = Sq1ιn, then (1) may cause trouble forH∗(Xk+1) approximating H∗(Xk), because it may not get killed. Of course, if it does, things work out rathernicely.

Example 16. Start with ∗ → K(Z4, 2)→ K(Z2, 2). Take the fiber X1, do the SSS calculation on K(Z2, 1)→X1 → K(Z4, 2) to observe that X1 = K(Z2, 2); then do the same construction again, and observe that X2 = ∗,so we have the following diagram:

K(Z1) K(Z2, 2) K(Z2, 2)

∗ K(Z4, 2) K(Z2, 2)

which tells us that H∗(∗) = H∗(∗). Yeah.

Example 17. Consider K(Z2, n)→ K(Z4, n)→ K(Z2, n) and its SSS. We have ιn and Sq1ιn at n and n+ 1respectively on the base, and ι′n and Sq1ι′n at n and n+ 1 on the fiber. The problem is that we have a mappingfrom ι′n to Sq1ιn, and it’s not clear immediately how this differential will work.

In general, suppose Fi−→ E

p−→ B is the fibration, a nice graphical way to represent SSS with Bocksteinhomomorphisms is to connect u at (n, 0) and u′ at (n + 1, 0) with a r-fold line to indicate βr(u) = u′ (whichis a Hn(X,Z2r ) → Hn+1(X,Z), and connect v at (0,m) and v′ at (0,m + 1) with a s-fold line to indicate aβs(v) = v′. Then, if there is a transgression τ(v) = u′, then in the total homology of E there is a nontrivialBockstein homomorphism βr+s such that i∗βr+sp∗(u) = v′. This is known as the Bockstein Lemma anddetailed discussion can be found on page 106 of this book.

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In general, this tells us that the problem of using the technique we have so far to compute all homotopygroups of spheres is that eventually we’ll hit problem at dimension 15, where a class just “pops up” and wecan’t kill it with our current technique.

25.3 Back to Computation

Recall that last time we went up to the second level of the tower:

K(Z2, n+ 1) X1

Sn K(Z, n) K(Z2, n+ 2)Sq2

Now let aI = SqIιn in K(Z, n), and let bI = SqIιn+1 in K(Z2, n+ 1). Consider the SSS:

1 a0 a2 a3 a4 a5

a4,2

a6

a5,2

a7

b0

b1

b2

b3

b2,1

b4

b3,1

b5

b4,1

We know that b0 goes to a2, thus b1 goes to a3; further computations now call for Adem relations. b2 andthus b3 goes to zero because db2 = Sq2ιn+1 = Sq2Sq2ιn = Sq3Sq1ιn = 0. Some more computation shows thatb2,1 goes to a5. But we have a pair (a4, b3,1) that is probably nontrivial; in fact, a4 → b3,1 is a Z4 bocksteinhomomorphism on the kernel of Sq1 (since Sq1a4 = a5 gets killed) by the Bockstein lemma above. This allowsus to write down the cohomology of X1 in a small range.

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1 a0 b2b3

a4 b3,1

c0

c1

c2

The next class we have to kill is b2 (the package I’m using is unable to draw double arrow, so I’m using bluecolor instead). So we map X1 → K(Z2, n+ 3) and take the fiber X2, and back up to get K(Z2, n+ 2)→ X2.Draw SSS again as in the graph above. c0 hits b2, c1 hits b3. c2 hits Sq2b2 = b3,1, so we have the troublewith (a4, c3), the same situation again, which this time yields a Z8 Bockstein. Repeat this process again withX2 → K(Z8, 4), we finally kill the pair, as one can check. Thus we have the following diagram so far:

K(Z8, n+ 3) X3

K(Z2, n+ 2) X2 K(Z2, n+ 4)

K(Z2, n+ 1) X1 K(Z2, n+ 3)

Sn K(Z, n) K(Z2, n+ 2)Sq2

Thus we have the conclusion:

Theorem 25.1. After localizing at 2, we have πnSn = Z, πn+1S

n = Z2, πn+2Sn = Z2, πn+3S

n = Z8.

More detailed calculation for up to the 7th stable homotopy group is carried out in this book. But as wehave seen above, this method has limitations (at dimension 15). As a matter of fact, when you run this onother localizations, e.g. localized at 3, you hit the same problem much more quickly.

Remark 11. Next topic is spectra and cobordism. Review your smooth manifolds and transversality.

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26 Introducing Cobordism

26.1 Cobordism: An Overview

Recall the days of singular homology (back when AT was easy)? There’s something closely related to that,which doesn’t have an official name, but people call it (co)bordism homology,

Definition 25. For X, let Ωn(X) be the set of all maps from smooth n-manifolds M to X up to an equivalencerelation known as cobordism.

Definition 26. Two maps fi : Mi → X, i ∈ 1, 2, is said to be cobordant if there is some h : N → X where

N is a smooth (n+ 1)-manifold, such that ∂N = M1

∐M2, and the restriction onto Mi of h is fi.

Figure 1: Illustrations from Wikipedia.

Of course, Ωn(X) is a commutative monoid under disjoint union. But in fact it has an identity:

Lemma 13. Ωn(X) is a group. In particular, M∐

M is null-cobordant.

Proof. Let X = M × I, then X is a (n+ 1)-manifold with boundary (M∐

M)∐∅.

For every X → Y , we can define Ω(X)→ Ω(Y ), and we can in fact prove Mayer-Vietoris, but we’ll skip theproof. In particular, let f : M → X = U∪V , and then f−1(U), f−1(V ) is a covering of M . By smooth partitionof unity, one can choose a function that’s 0 on one of them, 1 on the other of them, and in the intersectionit’s only increasing. Choose a submanifold L in the intersection and map it to a regular value, then we havef |L : L→ U ∪ V is the connecting homomorphism in this homology.

In general, we know that Ωn(X) is a generalized homology theory, where Ωn(X,A) is defined as the set ofcobordism classes of maps from pairs of smooth n-manifolds (M,∂M) to (X,A). We’ll show this in a completelydifferent approach from what we said above, namely the approach by Pontryagin and Thom, by identifyingit with the homotopy group of something else (Thom spaces), and then compute them. (Don’t worry, unlikespheres, they are easy to compute.)

The big question, of course, is what is Ωn(X). But before that, what is Ωn(∗)? We’ll show that it is a ringunder cartesian product that is isomorphic to Z2[x2, x4, x5, . . . | i 6= 2k − 1], and then Ωn(X) is just the regularhomology under the coefficient group of this ring. From here on, many more stories unfold.

26.2 Vector Bundles

Informal Definition A vector bundle over S is (roughly) a family of vector spaces parametrized by S, i.e.for each x ∈ S, there should be a corresponding Vx vector space, such that the map is continuous in someobvious way.

There are two approaches to realize this idea. Suppose all of the Vx live in a huge fixed vector space W .On one hand, we can look at Grn(W ), the Grassmannian. (We computed its cohomology last term.) A vectorbundle then is a continuous map S → Grn(W ). The second approach is consider the maps p : V → S, and takeVx = p−1(x), and somehow we need this to be a vector space. We mainly need some + : V ×S V → V that iscommutative, associative, 0 being the unit, and for every λ ∈ R, all operations are R-homomorphisms.

Unfortunately, this naıve approach won’t quite work because we can’t guarantee local triviality. Thiscondition says that for each x ∈ S has a neighborhood U ⊇ S, such that VU , being the pullback of U → S

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and V → S, is isomorphic to U × Rn and such that VU → U is pullback of projection U × Rn → U by theisomorphism. (This notion was actually not clear for a while in history.) Once we add this condition in, thingsstart to work out nicely.

We’ll show in next lecture that over finite CW complex, every vector bundle is pullback over some universalvector bundle over the Grassmannian. For example, take the case of Rn. There is a tautological vector bundlegiven by Vk → Grk(Rn) such that the fiber over a subspace H of dimension k is just H. In other words,Vk ⊆ Grk(Rn)× Rn = (H,x) | x ∈ H.

We need to check local triviality. Suppose H is a point in Grk(Rn), then we have its complement H⊥ inRn. Then the needed U = H ′ ⊆ Rn | H ′ ∩ H⊥ = 0, i.e. spaces that projects to the entirity of H whenprojected by the projection π : Rn → H. On U , we can use π to identify V |U ∼= U × H: (H ′, x) ∈ V |U ⊆Grk(Rn)× Rn 7→ (H ′, π(x)) ∈ U ×H.

Any operation on vector spaces should also translate to vector bundles. For example, V 7→ V ∗, V,W 7→ V ⊕W,V ⊗W , V 7→ SV (one-point compactification), V 7→ P (V ) (projectivization), V 7→ Grk(V ) (Grassmannian),etc. Notice that some of these produce topological spaces, not vector spaces, in which case we product fiberbundles. (How? we apply the operations fiber-wise, and check local nontriviality by checking that its result onthe trivial (local) product bundle remains trivial.)

A remark: V 7→ SV produces a fiber bundle over X. But note that now every fiber has a special point (thepoint at infinity), and it’s a good idea to collapse all of these points to one. To do this, we instead use theThom complex: Thom(X,V ) = XV = SV /X (collapsing the infinity section). It turns out this constructionhas a lot of beautiful properties.

So, for the record, how would SSS of this fiber bundle look like? We’ll have H∗(X,H∗(Sn, ∗))⇒ H∗(SV , X).But if we draw it out, we just have a row representing the shifted cohomology of the base. Disappointing? Not,because remember the Steenrod operations...

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27 Universal Vector Bundle

27.1 Existence of universal vector bundle

Let Vectk(X) be the set of vector bundles over X of dimension k, modulo the isomorphism relation.

Theorem 27.1 (Existence of Universal Bundle). If X is a compact Hausdorff space, then for n 0 the map[X,Grk(Rn)]→ Vectk(X) that sends f to the pullback bundle f∗Vk for some universal Vk, is an isomorphism.

Recall that by the pullback we mean the following diagram:

f∗V V

X Y

p

f

where we define the f∗(V ) = (x, v) ∈ X × V | f(x) = p(v), and then a fiber bundle on the right inducesa fiber bundle on the left, where the fibers are the same on both bundles. Also recall that Vk → Grk(Rn) isthe tautological k-plane bundle defined in the last class, where Vk = (H, v) | H ⊆ Rn,dim(H) = k, v ∈ H.Finally recall the definitions of maps of vector bundles. Suppose V → X,W → X are vector bundles on X,then a map betwen them is a map V →W such that the obvious maps commute:

V W

X

g

V ×X V V

W ×X W W

g×g g

and the restriction to each fiber g|Vx : Vx →Wx is a linear map.

Proposition 20. Suppose V,W are vector bundles over X, g : V →W is a map between vector bundles, thenthe set x ∈ X | g|Vx is an isomorphism is open.

Proof. Suppose x ∈ X is a point such that g|Vx is iso, then we need a neighborhood U such that the obviousthing holds. Choose a local trivial neighborhood U ′ ⊆ X of x such that V |U ′ = U ′ × Rk, W |U ′ = U ′ × Rk.Then g(x, v) = (x, ρ(x)v) where ρ : U ′ →Mn(R). Now GLn(R) ⊆Mn(R) is open, so we can let U = y ∈ U ′ |det(ρ(y)) 6= 0.

Now recall the partition of unity: suppose Uα is an open cover of a paracompact Hausdorff space X.Then there exists a set of functions ρα : X → I such that suppαρα ⊆ Uα, ρα(x) 6= 0 for only finitely many α

for each x, and∑α

ρα(x) = 1 for all x. 15

Proposition 21. Suppose V → X is a vector bundle, Z ⊆ X is closed, and s : Z → V is a section, then thereexists a section s′ : X → V extending s.

Proof. First suppose V = X ×Rk, then a section is a map s(x) = (x, t(x)) for some t : X → Rk. Then the caseholds by Tietze’s theorem. Now let X be covered by local trivial neighborhoods Uα, where V |Uα = Uα×Rk,and let Zα = Z ∩Uα. Then by case 1, we have some s′α extending sα for each sα defined by restricting s to Zα.

Now, choose a partition of unity ρ, then we claim that ραs′α define a section on X, and

∑α

ραs′α is a section

extending s on X.

Theorem 27.2. If V is a vector bundle over X × I, X compact Hausdorff, then V0 ∼ V1, where Vt is thepullback of V along x 7→ (x, t).

Proof. Note that we get a function I → Vectk(X) where the right side is the isomorphism classes of vectorbundles. We will show that this map is locally constant, and thus constant since I is connected. More precisely,fix some t ∈ I. Compare V and π∗Vt where π : X × I → X is the projection. Over X × t, these two thingsare the same, so they are isomorphic in a neighborhod of X×t. Since X is compact, there is an isomorphismon X × (t− ε, t+ ε) for some ε > 0, which shows that ∀s ∈ (t− ε, t+ ε), Vs ∼ Vt.

15Note that Dn is paracompact Hausdorff, so in fact all CW complexes are paracompact Hausdorff.

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Corollary 11. Vectk(X) is a homotopy functor of X. In other words, if f, g : X → Y are homotopic, andV → Y is a vector bundle, then f∗V ∼ g∗V .

Here’s a generalization to the theorem above, whose proof we skip:

Proposition 22. Suppose X is a paracompact Hausdorff, and V → X × I is a vector bundle, Write V0 → Xto be the restriction of V to X ×0. Write π : X × I → X to be the projection. Then there is an isomorphismV ∼= π∗V0 extending the identity map over X × 0.

Note that this is not true for categories in which one doesn’t have partitions of unity, e.g. for algebraicvector bundles.

Proposition 23. Suppose X is compact Hausdorff, V → X is a vector bundle, then for n 0 there is asurjective map

X × Rn V

X

Proof. Let U1, . . . , Um be a finite open cover such that V |Ui is trivial. Let ei : Ui → Rk → V |Ui be the

isomorphism, and choose a partition of unity ρi, then ρiei is surjective on the support of ρi, soX×⊕i

Rk∑i ρiei−−−−−→

V is surjective. Note that if X is locally contractible, we can choose U1, . . . , Um to work for all vector bundles.

Finally, let t be the surjective map above, such that Rn t−→ Vx. For each x ∈ X, define Hx ⊆ Rn to bethe orthogonal complement of the kernel of t. Define f : X → Grk(Rn) that sends x to Hx, then t gives anisomorphism V → f∗Vk.

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28 Stiefel-Whitney Classes (Guest Lecturer: Xiaolin Shi)

Recall that we showed in the last lecture that there is a correspondance Vectn(X) 7→ [X,Grn(R∞)], and allvector spaces are pullbacks of universal bundles. Today we talk about characteristic classes, which are datathat we attach to vector bundles, which can be used to identify whether the bundle is trivial or not. They areto vector bundles as (co)homology are to spaces. (The following construction can be found in [VBKT].)

28.1 Stiefel-Whitney Classes

Definition 27. Let V → X be a n-dimensional vector bundle. The Stiefel-Whitney classes wi(V ) are someparticular elements of Hi(X;Z2).

There are two ways of constructing S-W classes. One relies on geometry (using BO and BU); a cleaner wayis to uniquely identify the S-W classes with some properties, then construct them algebraically from cohomology.We’ll go with the second approach.

Proposition 24. The following properties hold for S-W classes:

1. Naturality. In the following pullback of vector bundles:

V ′ V

X ′ Xf

The S-W classes wi(V ) ∈ Hi(X) pulls back to f∗wi(v) ∈ Hi(X ′), and naturality says they are preciselywi(V

′).

2. Cartan formula: Given V1 → X,V2 → X, one can canonically construct the direct sum bundle V1⊕V2 →X. Define the total S-W class w(V ) = 1 + w1(V ) + w2(V ) + . . ., then w(V1 ⊕ V2) = w(V1) · w(V2).

3. Triviality: wi(V ) = 0 for i > dimV .

4. Non-triviality: Consider Gr1(R∞) = RP∞, there is a (canonical) line bundle V1 → RP∞, which onlyhas dimension 1 and thus only a single S-W class w1(V1) ∈ H1(RP∞). The condition is that w1(V1) 6= 0.

The Construction Take a vector bundle V → X, and projectivize it to get P (V )→ X, where we then havea fiber bundle RPn → P (V ) → X. There is a canonical line bundle L over P (V ), such that it pulls back toV1 → Gr1(R∞) = RP∞. Thus we have the following graph:

L V1

RPn−1 P (V ) Gr1(R∞) = RP∞

X

Then take the induced map on H∗ on the middle line to get H∗(RP∞) → H∗(P (V )) → H∗(RPn−1) thatsends generators to generators (nontrivial; needs a check). Thus there exists a x ∈ H1(P (V )) so that therestriction of x to H1(RPn−1) is a generator. Then by Leray-Hirsch, H∗(P (V )) = H∗(X)[1, x, x2, . . . , xn−1],which means that xn must be linearly expressible in 1, x, . . . , xn−1. Then define wi to be the coefficients suchthat xn + w1(v)xn−1 + w2(v)xn−2 + . . .+ wn(v) = 0. To be complete, define higher S-W classes to be trivial.

The Properties Let’s check naturality. Take the following pullback:

V ′ V

X ′ Xf

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Let xV and xV ′ be the canonical generators defined above. We know that f∗xV = xV ′ , then from theuniqueness of the linear relation above we immediately have the naturality.

Triviality is trivial. Let’s now see non-triviality. Take the projectivization P (V1) → RP∞, what is it? It’sreally just the identity map RP∞ → RP∞. Now let the canonical generator be x, then we have x+w1(V1)·1 = 0,thus x = w1(V1) 6= 0 since we work on Z2.

Finally, let’s check the Cartan formula. Take the projectivization P (V1 ⊕ V2)→ X, and there are two opensets U1, U2 covering the total space P (V1⊕V2), given as Ui = P (V1⊕V2)−P (Vi) ∼= P (V3−i), as one can check.Now let dimV1 = m, dimV2 = n. Then define wVi ∈ H∗(P (V1 ⊕ V2)), where wV1

= xmV1⊕V2+ w1(V1)xm−1

V1⊕V2+

. . .+ wm(V1) · 1 = 0, and the similar equality holds for wV2. More concretely, take Ui → P (V1 ⊕ V2), and take

the cohomology map H∗(P (V1 ⊕ V2)) → H∗(Ui), and consider the image wVi : since xV1⊕V2pulls back to xVi ,

both images wV1 and wV2 would vanish as we pull back, by the defining relations. So the classes are nontrivial,but trivial after being pulled back. In other words, wVi ∈ H∗(P (V1 ⊕ V2), Ui).

Now we claim that wV1· wV2

= 0, which follows the following diagram:

H∗(P (V1 ⊕ V2), U1)×H∗(P (V1 ⊕ V2), U2) H∗(P (V1 ⊕ V2), U1 ∪ U2)) = 0

H∗(P (V1 ⊕ V2))×H∗(P (V1 ⊕ V2)) H∗(P (V1 ⊕ V2))

Now the Cartan formula comes out by multiplying the coefficients in 0 = wV1⊕V2= wV1

· wV2.

To prove uniqueness we need an important principle:

Proposition 25 (Splitting Principle). Let V → X be a dimensional n vector bundle, then there is a pullback

V ′ = L1 ⊕ . . . Ln V

F (X) X

such that V ′ is the direct sum of n line bundles, and f∗ : H∗(X)→ H∗(F (X)) is an injection.

Proof. The strategy is to split off a line bundle once at a time. Start with Vπ−→ X. Pull it back along

P (V )P (π)−−−→ X to obtain a new bundle P (π)∗V → P (V ). We know that there’s a canonical ling bundle L lying

over P (V ), so P (π)∗V = L⊕L⊥ (fiberwise orthogonal complement). Do this construction repeatedly and thenwe’re done. Finally, note that the cohomology map on the bottom is H∗(X) → H∗(X)[1, x, . . . , xn−1] so it’sinjective.

Now suppose V → X is the canonical line bundle. Then triviality and non-triviality specifies all S-W classeson the canonical line bundle. By naturality, this specifies S-W classes for all line bundles (since they are allpullbacks of the canonical bundle), then by the splitting principle and the Cartan formula, along with theinjectivity, this allows us to specify S-W classes for all vector bundles.

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29 More on S-W Classes

29.1 Examples of Stiefel-Whitney Classes

Consider the class RPn, over which there is the totalogical line bundle L → RPn. We saw that the total

S-W class wt(L) = 1 + tx (this is a new notation: wt(L) =∑i

wi(V )ti where t is a formal variable), thus

w1(L) = x, where 0 6= x ∈ H1(RPn;Z2). Let V =⊕k

L, then wt(V ) = wt(L)k = (1 + tx)k =∑(

k

i

)tixi, then

wi =

(k

i

)xi. Now note that L lies in the trivial bundle Rn+1, and we have a short exact sequence of vector

bundles 0→ L→ Rn+1 → H → 0, where H is called the hyperplane bundle, and the SES is called the EulerSequence. It is a fundamental object in topology and geometry.

In algebraic geometry, this sequence does not split, but in topology the situation is different, as

Proposition 26. Every short exact sequence of vector bundles split.

Thus we know that L⊕H is the trivial bundle, thus wt(L) ·wt(H) = 1, and thus wt(H) =1

1 + tx=∑

tixi.

We know that dimH = n, so wn+1(H) = 0, and thus wn+1 = xn+1 = 0. It is also straightforward to show that

Proposition 27. Every vector bundler over R is isomorphic to its dual.

Next, for every l ∈ L, one can consider a neighborhood of l (and thus the tangent space at l) as the graph oflinear maps l → H, and thus we see that the tangent bundle TRPn = Hom(L,H). Note that Hom(L,R) = L.Now we hom into the exact sequence above to get

0→ Hom(L,L) = ∗ → Hom(L,Rn+1) = ⊕n+1L∗ = ⊕n+1L→ Hom(L,H) = T → 0

Thus wt(T ) · 1 = (1 + tx)n+1, hence the S-W class of TRPn is wk =

(n+ 1

k

)xk.

29.2 Detour: Riemannian metric

Let V be a vector space.

Definition 28. A Riemannian metric on V is a symmetric bilinear map V × V 〈·〉−→ R such that it is positivedefinite: 〈v, v〉 > 0 for v 6= 0.

Definition 29. A Riemannian metric on a vector bundle is a symmetric bilinear map V ×X V〈·〉−→ X ×R that

is positive definite.

Proposition 28. Every vector bundle over a paracompact space has a Riemannian metric.

Proof. Obvious when the bundle is trivial. For general V → X, choose a locally trivial covering Uα. Choose

a metric 〈·〉α for each local neighborhood, let sα be a partition of unity, then∑α

sα〈·〉α is a Riemannian metric

over V .

Example 18. If 0 → U → V → W → 0 is an exact sequence of vector bundles, then given a metric 〈·〉 overV , definte W ′ = U⊥ ⊆ V w.r.t. the metric, then W ′ → V →W is an isomorpism, so V = U ⊕W .

Question “13” What is the smallest k such that RP13 immerses into RP13+k?Recall that an immersion is a map such that the induced map of tangent spaces is injective everywhere.For f : M → Rn an immersion, define ν →M such that νx is the orthogonal complemenet of DfTx ⊆ Rn.

Then we know that

• ν has dimension n− dimM .

• ν ⊕ T ∼= M × Rn.

• wt(ν) · wt(T ) = 1.

60

Now, note that TRP13 ⊕ 1 = 14L, so wt(T ) = (1 + tx)14, and wt(ν) = (1 + tx)−14 =(1 + tx)2

(1 + tx)16=

1 + t2x2

1 + t16x16= 1 + t2x2. Thus w1(ν) = 0, w2(ν) 6= 0. Conclusion: RP13 does not immerse into RP14, for if it did,

w2(ν) would be 0 since dim ν = 1. Along the same way, we can conclude that RP9 is not immersive into RP14,but it might immerse into RP15.

Of course, this does not allow you to construct the immersions. Well, we do have some results:

Theorem 29.1 (Whitney). Every smooth manifold of dimension n immerses in R2n−1.

In fact, we have something better:

Theorem 29.2 (P. Cohen). Every smooth compact manifold of dimension n immerses in R2n−α(n), where αis the Hamming weight.

Using K-theory, one can obtain some better results using more sophisticated cohomology theories. In theend, however, the answer to the question “What is the smallest k such that RPn → Rn+k” is not known up tillthis day. The best results we have are obtained by considering this problem in terms of the axial map problem,which is to consider the following diagram, where k is the same answer as above:

RPn × RPn RP∞

RPn+k

and take the generator x for the cohomology of RP∞, which pulls back to (x+ y) on the left for generatorsx, y, and try to argue how the graph works when we raise to the (n+ k)th power.

The best result regarding immersion that we have is the following theorem by Hirsch:

Theorem 29.3 (Hirsch). A n-dimensional smooth manifold M immerses into Rn+k if and only if there existsν →M , with dim ν = k and that ν ⊕ T = M × Rn+k.

The result, known as Smale-Hirsch Theory, applies to greater generality:

Theorem 29.4 (Smale-Hirsch). Let M be a smmooth m-dimensional manifold, and N a smooth n-dimensionalmanifold. Then if either m < n, or m = n and M open in N , then the space Imm(M,N) of smooth immersionsof M into N is weakly homotopy equivalent to Mono(TM, TN), the space of vector bundle monomorphismsfrom TM to TN .

This idea leads to a more generalized principle known as the h-principle, which has applications in geometryand PDEs. (c.f. here)

61

30 Cohomology of the Grassmannian

Let G = limn→∞

Grk(Rn+k), also written as Grk(R∞), which is also BO(k) (the classifying space of O(k)). We

showed that for paracompact Hausdorff X, [X,G] ∼= Vectk(X), given by pulling back the universal bundle Vk.Let wi = wi(Vk) ∈ Hi(Grk(R∞)) be the S-W classes of the universal bundle, then S-W classes of all vectorbundles are obtained as pullbacks, i.e. ui = wi(f

∗Vk) = f∗wi.

Theorem 30.1. Z2[w1, . . . , wn]→ H∗(G) is an isomorphism.

Theorem 30.2. On Grk(Rn+k), we have Vk and its orthogonal complement V ⊥k of dimension n, having S-Wclasses w′1, . . . , w

′n. Then the map Z2[w1, . . . , wk, w

′1, . . . , w

′n]/wi · w′i = 1 for all i → H∗(Grk(Rn+k)) is an

isomorphism.

We shall prove the first theorem only.

Preparation: Cellular Structure of Grassmannians Recall the cell structure on Grk(Rn+k): there is acell corresponding to each sequence 0 ≤ a1 ≤ . . . ≤ ak ≤ n, with dimension a1 + . . . + ak. The interior is thek-plane whose basis has form

∗ . . . 1 0 0 0 0 0 0 0∗ . . . 0 ∗ . . . 1 0 0 0 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .∗ . . . 0 ∗ . . . 0 ∗ . . . 1 0

Given some k-plane V in Rn+k, associate with it the sequence dim(V ∩R1),dim(V ∩R2), . . . ,dim(V ∩Rn+k)

where R1 ⊆ R2 ⊆ . . . ⊆ Rn+k is some chosen sequence of subspaces. This sequence eventually goes to k, butwe need to keep track where it bumps up (we say that it bumps up at the “jump” dimensions ji). After scalingwe can put each basis vector as ending in 1 and thus having ji − 1 free dimensions; by subtracting previousbasis vectors we can get the form as above. For instance, in the following element of Gr3(R5):1 1 0 0 0

1 3 2 0 00 1 1 1 3

we have the sequence (dim(V ∩ Ri)) = (0, 1, 2, 2, 3), thus the jump dimensions are (2, 3, 5). By counting thenumber of asterisks in the matrix above, it is clear that for each jump dimension sequence 0 < x1 < . . . < xk ≤n+k, the dimension of the associated cell is (x1−1) + . . .+ (xk−k); taking ai = xi− i we get the result above.

Proof of the First Theorem. First let us show that the map is a monomorphism. Consider the classifying map

(RP∞)nL1⊕...⊕Ln−−−−−−−→ Grn(R∞) of nth product of the tautological line bundle16 One can check the composition

Z2[w1, . . . , wn] → H∗(G) → H∗((RP∞)n) = Z2[z1, . . . , zn] sends wi → σi, the ith elementary symmetricfunction in z1, . . . , zn. We know from classical invariant theory that Z2[z1, . . . , zn]Σn = Z2[σ1, . . . , σn]; inparticular, σ1, . . . , σn are algebraically independent, so the composite map is injective.

Now we need to show the map is an epimorphism. It suffices to show that for each m, dimZ2[w1, . . . , wk]m ≥dimHm(G), since we already showed that it’s monomorphic. Consider the Poincare Series of the Grassman-

nian PG =∑m

dimHm(G)tm, and that of the polynomial ring PW =∑m

dimZ2[w1, . . . , wk]mtm.

About Poincare Series Let A∗ be a graded ring over Z2, then P (A∗) =∑

dimAmtm. Some properties:

1. P (A⊗B) = P (A)P (B). (Easy to check.)

2. Suppose A = Z2[x] for |x| = l, then P (A) =1

1− tl.

Thus we know that PW =

k∏i=1

1

1− ti(this is also known as the partition generating function) =

∑i

ρiti, so

ρi is the number of ways to write i in the form s1 + 2s2 + . . .+ ksk. This in turn one-to-one corresponds to asequence 0 ≤ a1 ≤ . . . ≤ ak ≤ i, where ai = s1 + . . . + si; this corresponds, as we discussed above, to a cell in

G of dimension∑i

ai = i, so ρi equals the number of i-cells in G. Finally, recall that the number of i-cells of

G upper bounds the dimension of Hi(G), so we’re done.

16Recall that given any principal bundle π : EG → BG and any G-principal bundle γ : Y → X, there is a classifying mapX → BG such that γ is the pullback of π along this map.

62

A few words on the second theorem. If we write wi · w′i = 1 +∑i

ri, then wi · w′i = 1 is equivalent to

ri = 0 ∀i. In fact, the sequence (ri) form a regular sequence (as defined in commutative algebra).For instance, considerGr2(R5), then the cohomology is given by Z2[w1, w2, w

′1, w

′2, w

′3]/r1, r2, . . . , r5. Then

we have(1− t)(1− t2) . . . (1− t5)

(1− t)(1− t2)(1− t)(1− t2)(1− t3)

this is the Gaussian binomial coefficients, and they are always polynomials, and one can check that they count

the number of cells of Grassmannians of finite dimensions. (Write

(m

r

)q

=(1− qm) . . . (1− qm−r+1)

(1− q) . . . (1− qr)if r ≤ m,

and 0 otherwise. Then the expression above is simply

(5

2

)t

. The general statement is that

(n

k

)q

counts

k-dimensional subspaces of an n-dimensional vector space over Fq. Also, I guess I should mention that this isalso used in the study of quantum groups – Charles.)

63

31 Poincare-Thom Construction

Here we’ll use some facts about transversality. Let W,M be manifolds, Z ⊆ W be a submanifold of W . Wesay a smooth map f : M → W is transverse in Z if for all z ∈ Z and x ∈ M such that f(x) = z, we haveTxM ⊕ TzZ → Tf(x)W surjective.

Proposition 29. Every M → W is homotopy to one that is transverse in Z. If f is already transverse on aclosed K ⊆M , then can choose the homotopy to be constant on a neighborhood of K. Also, there is an extendedversion for M with boundary.

Suppose we have a map f : Sn+k → Sn, and pick a point x ∈ Sn. We can move f by a homotopy if necessaryto make it transverse in x. Suppose f, g : Sn+k → Sn, then we can choose a homotopy H : Sn+k × I → Sn

such that it’s transverse to x on the two ends of the homotopy. Then there is a homotopy of H to a transversemap, which is constant in a neighborhood. Thus we get a new homotopy H ′ of f, g that is transverse to x. LetN = (H ′)−1(x), then N is a cobordism of f−1(x) and g−1(x).

The Goal We would like to have a tool to link homotopy and cobordism. This is given by the Pontryagin-Thom construction, as we describe below.

Suppose V → X is a vector bundle of dimension n, where X is a smooth manifold. Define Thom(X,V ) =XV , where we one-point compactify every fiber of V , then mod out X. If we give V a Riemannian metric, thenwe can say it is B(V )/S(V ), where B(V ) = v ∈ V | ‖v‖ ≤ 1, S(V ) = v ∈ V | ‖v‖ = 1. In what follows,we shall also use R to denote the trivial bundle X × R→ X. The Thom space has the following easy to checkproperties:

• (X × Y )V⊕W = XV ∧ YW ;

• In particular, XV⊕R = ΣXV .

Now let f : Sn+k → XV , and note that XV is no longer a smooth manifold, but it is smooth away from thepoint of infinity. So to handle this we extend the transversality theorem, noting that it is local property:

Remark 12. In the transversality theorem above, W doesn’t have to be smooth in a neighborhood of Z.

Of course, X ⊆ XV , so let’s move f by a homotopy if necessary, then we can assume that f is transverse toX. Let M = f−1(X) ⊆ Sn+k, then it has dimension k. Then we can interpret the homotopy of XV in termsof M . To do so, we need some properties about M . It of course comes in with an embedding in Sn+k, and alsoa map f : M → X. Moreover, it also comes from an isomorphism ν ∼ f∗V , where ν is the normal bundle tothe embedding M → Sn+k, given by Df .

Now suppose we have a homotopy H : Sn+k × I → XV between f = H(−, 0) and g = H(0,−) that istransverse in X. If f−1(X) = M0, g−1(X) = M1, then N = H−1(X) is a cobordism of M0 and M1, whichcomes with all the associated data that is compatible with the data from the two submanifolds. This gives awell-defined map from πn+kX

V to the set The three maps mentioned above.

Theorem 31.1 (Pontryagin-Thom). This map is a bijection.

Note that:

1. We can let the embedding be M → Rn+k instead, because clearly the image and thus the preimage areaway from the infinity points respectively.

2. What’s the additive structure on the RHS set? Move things by a translation we can take the disjointunion of the embeddings, which doesn’t change the elements up to cobordism.

Proof. What we describe below is known as the Pontryagin-Thom construction. Suppose we have M ⊆ Rn+k ⊆Sn+k, f : M → X and ν → f∗V , where V → X is a vector bundle over a manifold. Recall the tubular neighbor-hood theorem from baby topology: suppose a manifold M embeds in Rn+k, then there exists a neighborhoodU of M in Rn+k, and a diffeomorphism U → ν, such that the zero section M → ν is the same as the inclusionM → U posted composed by the diffeomorphism. Thus we can choose a tubular neighborhood U of M inRn+k, and get a continuous map Sn+k → Sn+k/(Sn+k − U) → Mν such that x maps to x if x ∈ U , and ∗ ifx /∈ U . This map is called the Pontryagin-Thom collapse. But recall that the following diagram:

ν V

M Xf

induces a map Mν → XV , thus the concatenation of this map is the inverse direction that we need.

64

A general reference for cobordism is Robert E. Stong’s Notes on Cobordism Theory, though the book itselfmight be somewhat hard to read.

65

32 Pontryagin-Thom and the MO Spectrum

Last class we proved, using transversality, that if V → X is a vector bundle of dimension n, X is a smoothmanifold, then we have the following isomorphism regarding the Thom spectrum (defined in the next lecture –Charles):

πn+kXV = M → X,M → Rn+k, f∗V ∼ v/cobordism

Remark 13. X doesn’t have to be a smooth manifold.

Let’s observe that it still makes sense to talk about transversality of Sn+k → XV over some x ∈ XV , even

if X is not a smooth manifold. Consider the map Xg−→ Grn(RN ) given in the universal bundle construction,

explicitly by V 7→ g∗Vn. Since we are only interested in homotopy group, we can change space up to a homotopy,so might as well suppose that g is a Serre fibration. Then so is

V Vk

X Grn(RN )g

We observe that XV → Grn(RN )Vn is a Serre fibration away from ∗.

Definition 30. Let L be a manifold with boundary, L → XV is said to be transverse to the zero section if

L→ XV g−→ Grn(RN )Vn is transverse to Grn(RN ) ⊆ Grn(RN )Vn .

When X → Grn(RN ) is a Serre fibration, every map L→ XV is homotopic to a map that is transverse tothe zero section. (My raw note says here “we can move a bit such that we don’t move in a neighborhood of thebase point.” I’m not sure what’s going on here. – Charles)

Now suppose we have the data (Mf−→ X,M

ι−→ Rn+k, f∗V ∼ v), then we get an element of the homotopy

group. Suppose ι1, ι2 : M → Rn+k are two isotopic embeddings, i.e. there is an embedding M × I h−→ Rn+k × Ithat restricts to ιi on two ends, then this h gives a cobordism between (M,f∗V ∼ v1, ι1) and (M,f∗V ∼ v, ι2),so they represent the same homotopy class. Thus the homotopy element depends only on the isotopy type ofι. However, for N 0, by a general position argument we see that any two embeddings M → Rn are isotopic,(so the space of embeddings lim

n→∞Emb(M,Rn) is contractible, because we can take M to be Sn ×M), so if

we can embed into even larger spaces we can probably get rid of ι. Well, if we admit to a larger embeddingM → Rn+k → Rn+k+1, then the normal bundle data becomes f∗V ⊕R ∼ v⊕R. Then the new data correspondsto the homotopy Sn+k+1 → XV⊕R = ΣXV . If r : Sn+k → XV , then this new map is simply Σr. Thus if we

consider limt→∞

πn+k+tΣtXV , then this can be specified by the data M

f−→ X and f∗V =s ν, where =s is a stable

isomorphism (see below).At this moment it is fitting to introduce some more notation. Let us write πsnZ = lim

t→∞πn+tΣ

tZ and call

it the nth stable homotopy group. For instance, by Freudenthal suspension theorem, stable homotopy groupsπskS

0 are well-defined.

Definition 31. Two vector bundles V,W on X are stably isomorphic if V ⊕ RN ∼W ⊕ RN for N 0.

Example 19. TSn → Sn and Sn × Rn → Sn are not isomorphic unless n = 1, 3, 7, in which cases the spherehas trivial tangent bundle. However, these two bundles are stably isomorphic, for TSn ⊕ R ∼= Sn × Rn × R.

Remark 14. In terms of commutative algebra, any vector bundle is a finitely generated projective module overthe space of continuous functions over the base space; then Sn × Rn → Sn is a free module, so we have anexample of a projective module that, when summed with a free module, becomes free.

Now we’d like to remove the normal bundle data as well. Is there a universal X such that we don’t haveto mention that data? Well, suppose M is of dimension k, then the stable isomorphism f∗V =s ν is given bythat TM ⊕ f∗V ⊕ Rj ∼= Rn+k+j . Note that M embeds into f∗V , so there is a mapping M → Grk(Rn+k+j).Thus the tangent bundle / normal bundle is classified by a map into the Grassmannian. Thus if we takeX = Grn+j(Rn+j+k) and V = Vn+j , the data M → X already gives the normal bundle data as well.

The structure that we eventually arrive at is the sequence of spaces Grn(R∞) = BO(n). It has a uni-versal n dimensional bundle Vn, and the Thom complex Thom(Grn, Vn) = MO(n). Note that there is a mapGrn(R∞)→ Grn+1(R∞⊕R) = Grn+1(R∞), and the Vn+1 pulls back to Vn⊕R, then we see that πn+kMO(n)→πn+k+1ΣMO(n)→ πn+k+1MO(n+1), and one observes that lim

n→∞πn+kMO(n) = k-manifolds/cobordism. Here

we are looking at the homotopy group of a spectrum.

66

33 Spectra

We have proven that the set of cobordism classes of k manifolds is the same as limn→∞

πn+kMO(n) where MO(n)

is the Thom complex Thom(Grk(R∞)). Note that we have the following diagram:

M Grk(R∞)

Rn+k f∗Vk

f

We should be able to multiply with another space X, and add the trivial bundle for X, then because(X × Y )V⊕W = XV ∧ YW , we also obtain πn+k(MO(n) ∧X). So we also observe that the cobordism classesof k-manifolds M ×X is lim

n→∞πn+k(MO(n) ∧X).

Definition 32 (Spectrum). A spectrum E consists of a sequence of spaces En and connecting maps tn : ΣEn →En+1.

Example 20. E = Sn, where En = Sn and ΣSn → Sn+1 is the usual map. This is the sphere spectrum S0.

In general, if X is any pointed space, there is a suspension spectrum Σ∞X given by just EnX = ΣnX.

Example 21. The spectrum MO = MO(n).

Example 22. Suppose E is a spectrum and X a pointed space, then we can form E ∧ X = En ∧ X. Forinstance, Σ∞X = S0 ∧X.

Definition 33. E is an Ω-spectrum if each of the maps En → ΩEn+1, which are the adjoints of tn, is a weakequivalence.

In some discussions, what we call “spectrum” is called a “prespectrum,” and “Ω-spectra” are called “spec-tra”. (This is used primarily in homological algebra.) Now if E is a spectrum, we set πkE = lim

n→∞πn+kEn.

There is a lot of details in the construction of spectra, but we’ll not go deep into the theory here. Note thatof course the definition πkE makes sense for k ∈ Z since eventually (n + k) is positive. For example, considerm > 0 and let S−m be the spectrum given by S−mj = ∗ for j < m, and S−mj = Sj−m for j ≥ m. Then

π−mS−m = lim

n→∞πn−mS

−mn = lim

n→∞πn−mS

n−m = Z.

Definition 34. A map E → F of spectra consists of a collection of maps En → Fn such that it commutes withtn. It is a weak equivalence if it induces an isomorphism of πk for all k ∈ Z.

Proposition 30. Every spectrum is weakly equivalent to a Ω-spectrum.

Proof. Given E = En, define a new spectrum E given by En = limm→∞

ΩmEn+m. (We ignore some technical

details.) Then there is a canonical mapping E → E, the latter of which is an Ω-spectrum.

Proposition 31. If E is an Ω-spectrum, then for n + k ≥ 0, the map πn+kEn → πn+k+1En+1 is an isomor-phism.

Proposition 32. Suppose E is a spectrum, X is a space. Then the obvious map πk(E ∧X)→ πk+1(E ∧ΣX)is an isomorphism.

The proof below is sometimes called a roller-coaster proof, due to its associated diagram.

Proof Sketch. We have the following diagram:

πn+k(En ∧X) πn+k+1(En+1 ∧X) πn+k+2(En+2 ∧X) . . .

πn+k+1(En ∧ ΣX) πn+k+2(En+1 ∧ ΣX) πn+k+3(En+2 ∧ ΣX) . . .

Now note that the colimits of the two rows must be the same. (The actual proof is not this easy and involvessome careful detail checking.)

67

Proposition 33. Suppose A→ X is a map, E is a spectrum. Then there is a LES

. . .→ πk(E ∧A)→ πk(E ∧X)→ πk(E ∧ (X ∪ CA))δ−→ πk−1(E ∧A)→ . . .

where δ is given by πk(E ∧ (X ∪ CA))→ πk(E ∧ ΣA)∼=←− πk−1(E ∧A).

Proof. We just need to check exactness at πk(En ∧ X) since, for instance, exactness at πk(E ∧ (X ∪ CA)) isexactness of the middle term in the LES associated with X → (X ∪CA)→ (X ∪CA) ∪CX = ΣA. Note thatπk(E ∧A)→ πk(E ∧X)→ πk(E ∧ (X ∪CA)) is zero because A→ X → X ∪CA is zero. We have the followingdiagram:

Sn+k Dn+k+1 Sn+k+1

En ∧X En ∧ (X ∪ CA) En ∧ ΣA

x y

The y in the diagram is defined by the diagram and is an element of πk+1E ∧ΣA = πkE ∧A. And it turns outthat y gets sent to x in the first map. To see this, extend the map to the right again to have the bottom beEn ∧ ΣX, then check that the vertical arrow is again x.

Corollary 12. A spectrum gives a generalized homology theory (X,A) 7→ Ek(X,A) = πkE ∧ (X ∪ CA).

Example 23 (The singular homology spectrum). Let HA = K(A,n), where ΣK(A,n)→ K(A,n+1) is given

by the adjoint of K(A,n)∼=−→ ΩK(A,n+1). Then HAk(X) = lim

n→∞πn+kK(A,n)∧X. When X = ∗, πk(HA) = 0

if k 6= 0 and A if k = 0, so HAk(•) also satisfies the dimension axiom. So in fact HAk(X) = Hk(X;A). Inother words, HkX = lim

n→∞πn+k(K(A,n) ∧X).

Theorem 33.1. Every homology theory comes from a spectrum.

Definition 35 (Cohomology Theory of Spectra). Define Ek(X,A) = [Σn(X ∪ CA), En+k] for some large n.

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34 Thom Isomorphism

There is a relative version of the Serre spectral sequence that goes as follows: given a Serre spectral sequence

F → Ep−→ B, and subspaces F0 ⊆ F , E0 ⊆ E such that F0 → E0

p−→ B is again a Serre fibration, then for eachsubspace B0 ⊆ B, we have a spectral sequence Hp(B,B0;Hq(F, F0))⇒ Hp+q(E,E0 ∪ p−0(B0)). In particular,taking B0 to be a point, we get Hp(B;Hq(F, F0)) ⇒ Hp+q(E,E0). Note that if π1B does not act trivially,we’ll need to use local coefficients.

Now consider a vector bundle V of dimension n over a space X. Then the pointwise fiber is Rn. Considerthe sphere bundle SV whose pointwise fiber is the one-point compactification, i.e. Sn. Then as we definedbefore, Thom(X,V ) = SV /X. Now the trivial bundle ∗ → X → X is a subbundle of Sn → SV → X in thesense above, so applying the spectral sequence we obtain Hp(X;Hq(Sn, ∗)) ⇒ Hp+q(Thom(X,V )) (we uselocal coefficients if applicable).

Corollary 13. Hn+p(Thom(X,V )) ∼= Hp(X;Hn(Sn, ∗)).

Example 24 (An example of Thom isomorphism). Consider mod 2 cohomology. We have Hn(Sn) = Z2, andAut(Z2) is trivial, so π1X acts trivially, and thus H∗+n(Thom(X,V )) ∼= H∗(X).

Thom Isomorphism Suppose V is a vector bundle over X of dimension n. A Thom class for V is an elementu ∈ Hn(Thom(X,V )) having the property that, for each x ∈ X, the image of u in Hn(SVx ) (where SVx is thefiber over x of SV → X) is a generator.

Observation: H∗(Thom(X,V )) is a module over H∗(X). To see this, observe that H∗(Thom(X,V )) =H∗(SV , X), and the action is given by the g map in the following diagram:

H∗(SV )⊗H∗(SV , X) H∗(SV , X)

H∗(X)⊗H∗(SV , X)

p∗⊗1g

where p∗ is the map induced by p : SV → X, and ∪ is the relative cup product. Note that the presentation ofThom(X,V ) is different from the standard one, where it is considered as B(V )/S(V ). Another way to see themodule structure is by considering H∗(Thom(X,V )) = H∗(V, V −X).

Theorem 34.1 (Thom isomorphism). Multiplication by a Thom class u (i.e. b 7→ g(b, u)) is an isomorphism

from H∗(X)→ H∗+n(Thom(X,V )).

It is possible to prove this using Serre spectral sequences. Here’s another proof sketch:

1. First prove it for V = X × Rn, the trivial bundle, using suspension isomorphism;

2. Then by Meyer-Vietoris, we can see it holds when X is covered by finitely many sets on which V is trivial;

3. Pass to the limit.

Some words on Thom classes. A unique Thom class exists for every V in mod 2 cohomology, sinceH∗(Thom(X,V )) ∼= H∗(SVx ) for each x. When we use Z coefficients it is no longer unique; it is, however,unique for simply connected space. For orientable manifolds, the choice of a Thom class is equivalent to thechoice of an orientation.

Let’s see some Thom isomorphisms in action.

• Suppose we have M of dimension n, N of dimension n+ d being smooth manifolds, and f : M → N , sod is dimension of the normal bundle ν. There is the Pontryagin-Thom collapse map N → Thom(M,ν),which induces Hk+d(Thom(M,ν)) → Hk+d(N). If we precompose this with the Thom isomorphismHk(M) → Hk+d(Thom(M,ν)), we obtain a mapping Hk(M) → Hk+d(N), which at a first glance is“going the wrong way.” This map is called the pushforward map.

• Same setting as above. Consider the embedding M → Rm ×N → Thom(N,Rm) (where Rm ×N is thetrivial bundle of rank m), then Pontryagin-Thom gives a map Thom(N,Rm) → Mν , where ν is of rankm+ d. Then we have the following diagram:

Hk+d+m(Thom(N,Rm)) Hk+d+m(Mν)

Hk+d(N) Hk(M)

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Note that we can obtain this even when f : M → N is not an embedding. This map has a lot ofinterpretations, but is often referred to as “integration over the fibers.”

We can get this map from another perspective:

Hk(M) Hk+d(N)

Hn−k(M) Hn−k(N)

where the two vertical maps are given by Poincare duality. In fact, Poincare duality is just Thomisomorphism with some additional constructions.

Take MO(k) = Thom(Grk(R∞)). The kth cohomology of this object is Z2 = H0(Grk(R∞)). Now consider

MO(k)u−→ K(Z2, k); as we’ll see, it induces monomorphism on cohomology through a certain range. The

question is, what would be the Steenrod operations of u? i.e. what is Sqj(u)?

Proposition 34. Sqj(u) = g(wj , u), which we’ll just write as wju from now on.

Proof Sketch. By Thom isomorphism, Sqj(u) = w′ju for some w′j which are the S-W classes of some bundle Vover X. One then check that they satisfy 1) naturality: for every f : X → Y , we have w′j(f

∗V ) = f∗(w′j(V ));2. the Cartan formula; and 3. w′j(L) 6= 0 where L is the tautological line bundle over R∞. But these axiomscharacterize S-W classes, so w′j = wj .

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35 Cohomology of MO and Euler Classes

The big theorem today is the following:

Theorem 35.1. MO(k)u−→ K(Z2, k) induces a monomorphism on cohomology through dimension 2k.

We know the cohomology of the space on the right, and we know what individual Steenrod operations doon the Thom class: Sqiu 7→ wiu, but it is harder to compute general SqIu because we have to analyze the S-Wclasses.

Well, let’s look at the proof. For this kind of statement, here’s a tried-and-true technique: find somethingthat maps into the first space, and prove that the composition mapping is a monomorphism. (We’ve seen thisa few times before in this note. – Charles)

Definition 36 (Euler Class). Suppose V → X is a vector bundle of dimension n. Consider the Thom complexThom(X,V ), and there is a zero section X → Thom(X,V ). Then u ∈ Hn(Thom) pulls back to the Euler classe(V ) ∈ Hn(X,Z2).

Why is it called the Euler class? After all, granted that Euler was a genius, his time was a bit too earlyfor things such as vector bundles and cohomologies. Well, suppose M is a manifold, then e(TM) ∈ Hn and〈e(TM), [∗]〉 = χ(M) by Poincare duality, where χ is the Euler characteristic of manifolds.

Now is a good time to list some properties of Euler class:

Remark 15. In the next few propositions, tensors, cup product and direct sums are external. In particular,the external cup product is what Hatcher calls the cross product of cohomology.

Proposition 35. e(V ⊕W ) = e(V )e(W ).

Proof. Recall that Thom(X × Y, V ⊕W ) = Thom(X,V ) ∧ Thom(Y,W ). Thus uV⊕W = uV ⊗ uW . Then therest follows from naturality, by looking at the following diagram:

Thom(V ⊗W ) Thom(V ⊗W ) Thom(V ) ∧ Thom(W )

X X ×X

∼=

Proposition 36. If V is the trivial bundle, e(V ) = 0.

Proof. The zero section from X to the Thom space of the trivial bundle is null.

Corollary 14. If e(V ) 6= 0, then V 6= V ′ ⊕ R for any V ′.

Proposition 37. e(V ) = wn(V ).

Proof. Under the Thom isomorphism, e(V ) = u · e(V ) = u · u = Sqn(u) = wn(V ).

Corollary 15. If L is a line bundle, then e(L) = w1(L).

Remark 16. Give V a metric, let S(V ) be the unit sphere, then Thom(X,V ) is the mapping cone of S(V )→ X.

In particular, S(V )→ Xzero section−−−−−−−→ Thom(X,V ) is exact.

Example 25. Let L be the tautological bundle over RP∞, then S(L) = S∞ = ∗. Thus in S(V ) → RP∞ f−→Thom(RP∞, L), the map f is a homotopy equivalence, so it induces isomorphism on all cohomologies, inparticular u pulls back to a nontrivial class.

Example 26. Consider (RP∞)kL1⊕...⊕Lk−−−−−−−→ Grk(R∞). The second term has Thom complex MO(k), and by

the example above, the first term has Thom complex (RP∞)∧k.

Now let’s go back to MO(k) → K(Z2, k). Consider the composite map (RP∞)kvia BO(k)−−−−−−−→ MO(k) →

K(Z2, k), then ι pulls back to u and then to e(L1⊕ . . .⊕Lk) = x1 . . . xk. Recall the proof of stable cohomologyoperations being determined by its product-of-one-dimensional behaviors, and by that same argument thisinduces a monomorphism through a range of dimensions.

At this moment let’s revisit something left out a few lectures ago. Recall that πnMO = limk→∞

πn+kMO(k).

How do we know this actually stabilizes? Well, we have the suspension homomorphism ΣMO(k)→MO(k+1).Recall, on the other hand, that BO(k)→ BK(k + 1) is an iso in cohomology through dimension 2k. Then as

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we shift to MO via multiplication by Thom class, by what we just showed above, we know that ΣMO(k) →MO(k+1) is iso in cohomology through 2k+1. Thus πn+kMO(k) is independent of k for k > n via a standardHurewicz argument.

Now a spectra perspective. Recall that we have uk : MO(k) → K(Z2, k), and it is compatible withsuspension of the MO, and thus we have a mapping of spectra MO → HZ2.

Definition 37 (Cohomology of Spectra). If E = En is a spectrum, then Hj(E) = limn→∞

Hn+j(En).

Corollary 16. The map MO → HZ2 is a monomorphism on cohomology.

In general, cohomology of a space is a ring along with Steenrod operations, such that Sqn(x) = 0 forn > dim(X). On the contrary, cohomology of a spectrum is not a ring (note that in suspension isomorphism,all the cup products go to zero, so there is no ring structure), but since Steenrod operations are compactiblewith suspension, it does have Steenrod operations, yet there is no relation between Sq∗(X) and the dimensionof X.

Recall that the cohomology of H∗(K(Z2, n)) is Z2[SqI ] for admissible I and excess ≤ n. So the cohomologyof the HZ2 spectrum, given by lim

k→∞H∗+k(K(Z2, k)), is a Z2 vector space with basis SqI where I is admissible.

In addition, since we can compose Steenrod operations, it has the structure of an algebra; it is called theSteenrod algebra. As a formal definition, the Steenrod algebra is the algebra over Z2 generated by formalsymbols Sq1, Sq2, . . . modulo the Adem relation.

Remark 17. How do we know that Adem relations eventually move any monomials to the reduced form? we

can assign moments to I. In particular, if I = (i1, . . . , in) then moment(I) =

n∑s=1

sis. One can check that when

you apply an Adem relation, the moment decreases.

We shall eventually prove that two vector bundles are cobordant if and only if they have the same Stiefel-Whitney classes.

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36 Describing MO

Recall the setup: we have Grk(R∞) = BO(k), and we write limk→∞

Grk(R∞) = BO. We also have MO =

MO(k) where MO(k) = Thom(BO(k);Vk). And we are interested in πn(MO) = limk→∞

πn+kMO(k).

By Thom isomorphism, we know H∗(MO(k)) = uZ2[w1, . . . , wk]. Now recall the map induced by the Thom

class MO(k)u−→ K(Z2, k). Let’s see what is not hit by pulling back along this map: ι goes to u, Sq1ι hits

uw1, Sq2ι hits uw2, Sq3ι hits uw2, but note that we haven’t hit uw21. So there is in fact another thing, i.e.

what we really have is MO(k)u,uw2

1−−−−→ K(Z2, k) × K(Z2, k + 2). More generally, let HZ2 = K(Z2, k) andwrite X = K(Z2, k + 2). Note that πiX = Z2 where i = 2, and 0 otherwise, so we can write X = Σ2HZ2,using the fact that in spectra, πiX = πi+1ΣX. Using this language, we can write the map identified earlier asMO → HZ2 ∨ Σ2HZ2. This is again surjective, but we can obtain an isomorphism if we keep going:

Theorem 36.1. MO ∼=∞∨i=0

Σs(i)HZ2 for some s(i)i≥0, where by ∼= we mean (mod 2) weak equivalence.

We will prove this by showing that H∗(MO) is a free module over A, the Steenrod algebra.

Example 27. Suppose X is a (k − 1)-connected space, and suppose that H∗(X;Z2) is free, finitely generatedA-module through dimension n < 2k. Let ei be a (finite) basis, i.e. ei ∈ H |ei|(X;Z2). Now consider the map

X →∏

K(Z2, |ei|). In cohomology, we know that H∗(K(Z2, |ei|)) is the module over admissible Is. In the

range ∗ < n there is no product, so it is a vector space with basis SqIι|ei|, which is just Aι|ei|. Now since the

range also guarantees that there is no nontrivial contribution by the Kunneth formula, so H∗(X) ←⊕

Aι|ei|is an isomorphism for ∗ < n, thus by mod C Hurewicz the map between associated spaces is a weak equivalence.

We know that A → H∗(MO) given by a 7→ au is a monomorphism, but in fact there is some algebraicstructure that we can play with. Here’s a rough analogy: recall that if H is a subgroup of G, then one canconsider the action of H on G (as a set), and then note that this action is free. What we have here, the freenessof H∗(MO) over A, is something in the similar spirit. (The technically accurate statement is that “Steenrodalgebra is the endomorphism ring of the Eilenberg-Maclane spectrum HZ2.” Compare this with the statementof Theorem 36.1.)

Consider the canonical map BO(k)×BO(l)→ BO(k+ l). This allows us to pull Vk+l back to Vk⊕Vl. If wepass to the Thom complex, then we have the mapping MO(k) ∧MO(l) → MK(k + l). Passing to homologygives H∗MO ⊗ H∗MO → H∗MO, making H∗MO into a commutative ring that is isomorphic to H∗BO viathe Thom isomorphism. (The proof is a matter of bookkeeping.) Then we have an induced map, via the UCT,

H∗(MO)⊗H∗(MO)ψ←− H∗(MO), making it a coalgebra with a counit.

Definition 38. A coalgebra over k is a vector space M equipped with a coproduct map Mψ−→ M ⊗M and

the counit Mε−→ k, such that it satisfies the coassociativity: 1 ⊗ ψ ψ = ψ ⊗ 1 ψ, and the counital law:

1⊗ ε ψ = ε⊗ 1 ψ = id.

Note that Steenrod is also a coalgebra. Recall that we have Sqn =∑i+j=n

Sqi ⊗ Sqj , so A is in fact a

co-commutative coalgebra.

Claim 1. The coproduct Aψ−→ A⊗A is a ring homomorphism.

This makes the Steenrod algebra a Hopf algebra, which has both an algebra structure and a coalgebrastructure. In general, a Hopf algebra can have an action · on a coalgebra, which is an action that is compactiblewith the coproducts, i.e. the following diagram commutes (where σ2,3 swaps the 2nd and the 3rd factors):

A⊗M M

A⊗A⊗M ⊗M M ⊗M

A⊗M ⊗A⊗M

·

ψ⊗ψ ψ

σ2,3·⊗·

For example, A is a Hopf algebra, and H∗(MO) is a coalgebra, and A acts on H∗(MO) by the Cartan formula.Also note that everything we look at is graded and connected: A = A∗, A∗ = 0 when ∗ < 0, A0 = k; M = M∗,M∗ = 0 for ∗ < 0, M0 = k, ψ1 = 1⊗ 1.

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Theorem 36.2 (Milnor-Moore). If A is a connected graded Hopf-algebra, acting on a connected graded coalgebraM , where 1 ∈M0 is sent to 1⊗ 1 by ψ, and if a 7→ a · 1 is a monomorphism, then M is free over A.

Proof. First some side observations. A maps to k by the counit ε. (In the case of Steenrod algebra, ker ε is

generated by all the squares.) Now suppose M =⊕

A, then k ⊗A M =⊕

k ⊗A A =⊕

k. This tells us

where the generators of M in terms of A are. This motivates looking at k ⊗AM .Let M = k ⊗A M . Choose a vector space section σ : M → M , such that its concatenation with M → M

is id, then there is a mapping A ⊗M → M given by a ⊗ x 7→ a · x which is a map of left A-modules. Theclaim is that this map is an isomorphism with our assumptions. First we need to check it’s an epimorphism.This is fairly straightforward via a proof by induction on the grading. (Kind of like Nakayama’s lemma.) Now

to check it’s mono, look at the following: A ⊗M → Mψ−→ M ⊗M → M ⊗M (all tensored over k), and

show this entire thing is a mono. The key step is to define Fi = M∗≤i ⊆ M . The big map above inducesA ⊗ Fi → M ⊗ Fi. By induction on i, we assume mono for degree up to i − 1. Now consider the LES0→ A⊗ Fi−1 →M ⊗ Fi →M ⊗ Fi/Fi−1 → 0, and note that the last one is given by a⊗ x 7→ a · 1⊗ x, so it ismono; and the first one is mono by assumption.

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37 Finishing the Cobordism Story

We showed that the cobordism ring is the same as the homotopy groups of the MO spectrum, and that

the cohomology of MO is free over the Steenrod algebra, i.e. MO ∼=∨

Σ∗HZ2 (or, equivalently, MO(k) is

equivalent to a product of K(Z2, j) for some js through dimension 2k).

Corollary 17. The Hurewicz homomorphism πn(MO) → Hn(MO) is a monomorphism since it is true forHZ2. In fact, π∗(MO) = HomA(H∗(MO),Z2).

So how big is πn(MO)? Consider the Poincare series P (MO, t) =∑

dimπk(MO)tk; we also have

P (H∗MO, t) =∑

dimHk(MO)tk =∑

dimHk(MO)tk. Since MO =∨S

Σ|S|HZ2, we know that π∗(MO) has

basis S over Z2, and H∗(MO) = Z2[S]⊗H∗(HZ2) = Z2[S]⊗A. In other words, P (H∗MO) = P (A)P (π∗MO).We know that H∗(MO) = H∗(BO) by Thom isomorphism, which is just Z2[w1, w2, . . .], thus

P (H∗MO) =1

(1− t)(1− t2)(1− t3) . . ..

On the other hand, a basis of admissible monomoials gives that

P (A) =1

(1− t)(1− t3)(1− t7) . . . (1− t2n−1).

Thus we know that

P (π∗MO) =∏

n 6=2j−1

1

1− tn.

Our goal today is the following theorem, which says the structure of π∗(MO) is exactly the one predictedby the Poincare series.

Theorem 37.1. πn(MO) is Z2[xn | n > 0, n 6= 2j − 1].

Observe that we have π∗(MO) → H∗(MO) = H∗(BO). So what is H∗(BO)? Recall the mappingZ2[w1, . . . , wk] = H∗(BO(k)) → H∗((RP∞)k) = Z[x1, . . . , xk], where xi = wi(Li). In this map, wn 7→σ(x1, . . . , xk) where

∏i

(1 + txi) =∑

σntn (σn are the symmetric functions). In fact, H∗(BO(k)) is the

ring of symmetric invariants in Z[x1, . . . , Zk].

Corollary 18. Write Hk(RP∞) = H∗(RP∞)⊗k, then Sym(HkCP∞)→ H∗(BO) is surjective. However, as welook at the Poincare series, we can conclude that it is in fact an isomorphism.

Proof. Choose bi ∈ H∗RP∞ such that 〈xj , bi〉 = δij (the Kronecker pairing on cohomology, where x is thegenerator of H∗RP∞), then the former is Z2[b1, b2, . . .]. Now compare the Poincare series.

Lemma 14. Suppose b′i ∈ HiBO are elements satisfying b′i = bi mod I2 where I = (b1, b2, . . .), then the mapfrom Z2[b′1, b

′2, . . .] to H∗BO is an isomorphism.

Let sn = sn(w1, w2, . . .) be the symmetric function given by the Newton polynomials. In other words, they

are functions such that sp(σ1, . . . , σn) = (−1)p+1n∑k=1

xpk, where σi are the elementary symmetric functions.

Then one can see that sn ∈ Hn(BO). (One should think of them as characteristic classes of vector bundles.)

Proposition 38. sn(V ⊕W ) = sn(V ) + sn(W ).

Proof. Follows from the splitting principle.

Proposition 39. Equivalently, under the coproduct map H∗(BO) → H∗(BO) ⊗ H∗(BO), we have sn 7→sn ⊗ 1 + 1⊗ sn. An element having such property is called a primitive element.

Example 28. If L is the tautological bundle over RP∞, then sn(L) = xn. In cohomology, this says thatsn ∈ H∗(BO) pulls back to xn in Hn(RP∞). Let i : RP∞ → BO, then bn 7→ i∗bn, the latter of which wesometimes also write as bn. Then 〈sn, bn) = 〈sn, i∗bn〉 = 〈i∗sn, bn〉 = 〈xn, bn〉 = 1.

Corollary 19. 〈sn, bm〉 = δmn.

Proposition 40. Suppose a ∈ Hk(BO) and b ∈ Hl(MO) such that k, l > 0, k + l = n, then 〈sn, ab〉 = 0.

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Proof. 〈ψ(a⊗ b)〉 = 〈ψ(sn), a⊗ b〉 = 〈sn ⊗ 1 + 1⊗ sn, a⊗ b〉 = 0 (consider degree).

Theorem 37.2. b′n ∈ Hn(BO) satisfies b′n = bn mod I2 iff 〈sn, b′n〉 = 1.

Corollary 20. b′n are polynomial generators iff 〈sn, b′n〉 = 1.

Now consider the Hurewicz homomorphism πk(MO)→ Hk(MO). Consider a manifold M of dimension k,and consider [M ] representing M in πk(MO)17, then it maps to H([Mk]) ∈ Hk(MO) = Hk(BO).

Theorem 37.3. Given w ∈ H∗(BO), then 〈w,H([Mk])〉 = 〈w(ν), [M ]〉, where ν is the normal bundle of anembedding M ⊆ RN .

We are interested in 〈sk, H([M ])〉 = 〈sk(ν), [M ]〉. But note that ν + TM is the trivial bundle of dimensionN , thus sk(ν) + sk(TM) = 0, thus sk(v) = sk(TM) (mod 2); hence we conclude that

Theorem 37.4. If Mi for i 6= 2n − 1, i > 0 is a sequence of manifolds satisfying 〈si(TM), [M ]〉 = 1, thenπ∗(MO) = Z2[ [Mi] ].

Proof. We just need to prove that we have a polynomial algebra of the right size. The elements b′i = H([Mi]), i 6=2n − 1, and the elements b2n−1 form the polynomial algebra generators for H∗(MO), thus Z2[ [Mi] ] →π∗(MO)→ H∗(MO) is a monomorphism with the same Poincare series.

So what are the generators, explicitly? Recall that TRPn = (n + 1)L, thus sn(TRPn) = (n + 1)xn, thus〈snT (RPn), [RPn]〉 = n+ 1, so we have half of the generators (if n is even). For a while it was a mystery whatthe odd generators are.

First construction was given by Dold by considering the Z2 action on S1 ⊗ CPm/(λ, t) ∼ (λ−1, t). Thesecond one given by Milnor, by considering certain hypersurfaces (called Milnor hypersurfaces) of degree 2,

denoted by M ⊆ RPa×RPb. For instance, we can let M be the zero set∑

i≤min(a,b)

xiyi. Let i : M → RPa×RPb

be the embedding, and ν the normal bundle. The RHS has a line bundle L1 ⊗L2, let x = w1(L1), y = w1(L2),then [M ] = x+ y. Consider TM + i∗(L1 ⊗ L2) = i∗T (RPa × RPb). Suppose a, b > 1 and a+ b = n− 1. Thensn(TM) = i∗((x+y)n)− i∗((a+1)xn+(b+1)yn), and 〈sn(TM), [M ]〉 = 〈i∗(x+y)n, [M ]〉 = 〈(x+y)n−1, [RPa×

RPb]〉 =

(n+ 1

a

)(mod 2). If n odd and n 6= 2k − 1, then n + 1 = 2k(2a + 1) = 2k+1q + 2k for some q > 0;

write this as a+ b, then

(a+ b

a

)= 1. Thus we know that we have obtained all the generators. This completes

the classification of manifolds up to cobordism.

Example 29. If one has a complex variety defined over the reals, and consider the fixed points under complexconjugation of the variety, then up to cobordism, the complex variety is the square of the fixed points. Forexample, CPn ∼=cb (RPn)2.

17Careful: [M ] is also sometimes used to refer to the fundamental class in HkM .

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