Math 231br (Advanced Algebraic Topology) Lecture Notes

Taught by: Professor Michael Hopkins Notetaker: Yuchen Fu

Spring 2015 Contents

1 Introduction 3

2 Higher Homotopy Groups 4 2.1 Hurewicz’s Construction of Higher Homotopy Groups ...... 4 2.2 Relative Higher Homotopy Groups ...... 4

3 Hurewicz Fibration, Serre Fibration and Fiber Bundle 5 3.1 Definitions ...... 5 3.2 Relationship Among Them ...... 6 3.3 Some Properties ...... 7

4 Spectral Sequences 8 4.1 Motivation ...... 8 4.2 Spectral Sequences: Definition ...... 8

5 More on Spectral Sequences 10 5.1 Serre Spectral Sequence ...... 10

6 And More Spectral Sequences 11 6.1 The Extension Problem ...... 11 6.2 Local Systems ...... 11

7 Snow Day 12

8 More on SSS, and the Hurewicz Theorem 13 8.1 Mappings on the E2 page; Transgression ...... 13 8.2 The Hurewicz Theorem ...... 13

9 Relative Hurewicz Theorem 14 9.1 Application ...... 14

10 Representability and Puppe Sequences (Guest Lecturer: Hiro Takana) 15 10.1 Brown Representability ...... 15 10.2 Puppe Sequences ...... 15

11 Cohomological SSS (Guest Lecturer: Xiaolin Shi) 17 11.1 An Appetizer ...... 17 11.2 The Main Course ...... 18

12 A Glance at Model Categories, Part 1 (Guest Lecturer: Emily Riehl) 20 12.1 Definition of Model Categories ...... 20

13 A Glance at Model Categories, Part 2 (Guest Lecturer: Emily Riehl) 22 13.1 (Co)fibrant Objects ...... 22 13.2 Homotopy in Model Categories ...... 22

14 Serre Classes 26 14.1 Serre Classes ...... 26

15 Mod C Hurewicz Theorem 28

16 Relatvie Mod C Hurewicz; Finiteness of Homotopy Groups of Spheres 31 16.1 Relative mod C Hurewicz Theorem ...... 31

17 Introduction to Steenrod Operations 33 17.1 Motivating Steenrod Squares ...... 34

18 Construction of Steenrod Operations 36

1 19 More on Steenrod Operations 38 19.1 Properties of the Steenrod Operations ...... 38

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

21 Computing with Steenrod Squares (Guest Lecturer: Jeremy Hahn) 42 21.1 Cohomology Operations ...... 42

22 Adem Relation and Bocksteins (Guest Lecturer: Xiaolin Shi) 44 22.1 Adem Relation ...... 44 22.2 Bockstein Homomorphisms ...... 44

23 More on Adem Relation 46 23.1 Some Tricks for Computation ...... 46 23.2 A New Treatment ...... 46

24 Even More on Adem Relation 48 24.1 Higher Power Operations ...... 48 24.2 Application of Steenrod Algebra: Homotopy Groups of Spheres ...... 49

25 Some Nontrivial Stable Homotopy Computation 51 25.1 Killing Cohomology ...... 51 25.2 A Few Notes ...... 51 25.3 Back to Computation ...... 52

26 Introducing Cobordism 54 26.1 Cobordism: An Overview ...... 54 26.2 Vector Bundles ...... 54

27 Universal Vector Bundle 56 27.1 Existence of universal vector bundle ...... 56

28 Stiefel-Whitney Classes (Guest Lecturer: Xiaolin Shi) 58 28.1 Stiefel-Whitney Classes ...... 58

29 More on S-W Classes 60 29.1 Examples of Stiefel-Whitney Classes ...... 60 29.2 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 covers the 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 to time constraint, the last part regarding cobordism consisted largely of proof sketches, and a lot of details were skipped. Thus for a deeper understanding of the material, reading one of the “core references” given below alongside this note is highly recommended. Regarding spectral sequences: in the middle of notetaking I discovered the preferable sseq package, so switched to it. As a result you’ll see two ways of drawing spectral sequences in this document. I may unify them 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 also be found on my website.

3 2 Higher Homotopy Groups

n Let (X, ∗) be a pointed space. πn(X) = {f : S → X | f(∗) = ∗} modulo pointed homotopy. Sometimes written n as [S ,X]∗. π0(X) is the set of path components. π1(X) is the fundamental group. So how do we describe 2 2 2 π2(X)? In particular how does it have a group structure? One straightforward way is S → S ∧ S → X (glue f and g by f ∧ g), but there are other ways,

2.1 Hurewicz’s Construction of Higher Homotopy Groups n For instance, the following given by Hurewicz. Start with the set-theoretical part that πN (X) = {f : I → 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 topologize XA such that continuous map Z → XA is exactly the same as continuous map Z × A → X. This can be done by giving it the compact-open topology. Some point-set topology issues arise due to the fact that the spaces 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,∗). We make 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 smash product 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 n−1 n−1 I/∂I → (X, ∗)(I ,∂I ). 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 higher n−1 q homotopy groups is simply πn(X) = π1(Ω (X)). One can easily generalize this into πn(X) = πp(Ω (X)), where p + q = n.

Uniqueness and Commutativity There are certainly two ways of “gluing”: f to the left of g (which we denote 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, we see that one have (a ∗ b) ◦ (c ∗ d) = (a ◦ c) ∗ (b ◦ d). Now the Eckmann-Hilton argument tells us that the two operations are equal and both commutative. In particular we know that πn(X) is abelian for n > 1.

2.2 Relative Higher Homotopy Groups n n For n > 0 and A a subspace of X, define πn(X,A) = Map((D , ∂D , ∗), (X, A, ∗)) modulo homotopy. Then we get a long exact sequence πn(A) → πn(X) → πn(X,A) → πn−1(A) → ... → π0(A) → π0(X) is exact. (Some cautions need to be taken near the end of the exact sequence as to how to interpret it; in particular, starting from π1(X,A) we no longer have groups, but only pointed sets.) Problem 1. Let ∂In = In−1 ∪ J n−1. Clearly J n−1 is contractible and we have J n−1 ⊆ ∂In ⊆ In. We can also n n n−1 define πn(X,A) = Map((I , ∂I ,J ), (X, A, ∗)) modulo homotopy. Can we write πn(X,A) as a homotopy group 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 the family of paths I → X that begins at ∗X and ends in A, equipped with the compact-open topology and has 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,J 0), (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 × J 0 = J n−1 n−1 0 goes to ∗X , and I × I → A. Thus it corresponds to the definition above; the other way around is just formalism 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 not including 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 property with respect to X if for any f, g in the following diagram, the diagonal map h : X × I → E in the diagram below always exists:

f X E h i p g X × I B

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 to a pair (X,A) (where A ⊆ X) if the diagonal map h exists for any f, g and any given homotopy H : A × I → E:

f∪H X ∪ A × I E h i p g X × I B

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 any homotopy 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 for all 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’t mention 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 satisfies the homotopy extension property with respect to some space Y if, given f and h0 in the following diagram: Y X h0 g7→g(0) i h Y I A 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 to any 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:

f A E h i p g X B 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 a time, then use the standard pasting trick (doubling the speed of homotopy at each higher dimension) to obtain h. 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 a contractible space (resp. CW complex), then p admits a section, i.e. there is another map g such that p(g(x)) = x for 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 them down 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, do induction 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 called a fiber, such that we have the following local triviality condition: for each e ∈ E, there is a neighborhood Ue −1 of p(e) in B, such that there is a homeomorphism ϕ : p (Ue) → Ue × F that makes the following diagram commute:

−1 ϕ p (Ue) Ue × F

p π1

Ue

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 write p 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 no longer true for fiber bundles; however, under certain conditions (such as in the category of finite-dimensional manifolds, with smooth fiber bundles), we may still have the composition property. See here for a general discussion.

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 that every 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 a fiber 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 copies of R in X be called U and V . Define f((x, i)) = x be the “coordinate projection”, and note that it descends + + down to a map f : U ∩ V → R , which cannot be extended to any continuous X → R . (Note that g trivially + descends to X → R.) Define E to be the fiber bundle that is a trivial R bundle on both U and V , and the glueing 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 second + + factor we obtain two functions gU : U → R and gV : V → R such that g = gV /gU is a function that agrees with f on U ∩ V . However, it is possible to extend g to entire X: just let g(x) = gV (jV (x))/gU (jU (x)) where jV 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.

2 Example 3. And now an example of a Serre fibration that is not a fiber bundle. Let E = {(x, y) ∈ R | 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 be the basepoint fiber, then there is a (natural) long exact sequence:

i p ∂ ... → πn(F ) −→ πn(E) −→ 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 of groups. 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, put together 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 p0 : E0 → B0 are two Serre fibrations, and F,F 0 are the respective fibers over the basepoints. A mapping between the two Serre fibrations is a pair of mapping f : E → E0, g : B → B0 such that the following diagram commute:

f E E0

p p0 g B B0

One can check that this induces a mapping h : F → F 0 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 associated homotopy LES. Now apply five lemma using the fact that two of the maps are weak equivalences (so they induce 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.

g Lemma 6. Any map f : X → Y can be factored into X −→ X0 −→h Y , where g is a homotopy equivalence and h is 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 ev :(x,γ)7→γ(1) Z = {(x, γ) | x ∈ X, γ ∈ Y I , γ(0) = f(x)}. One can check that Z −−−−−−−−−−→1 Y is a Hurewicz fibration, and (x,γ)7→x that Z −−−−−→ X is a homotopy equivalence, with the inverse s being attaching the constant path at x. Now 0 let X = Z, g = s, h = ev1.

g Lemma 7. Any map f : X → Y can be factored into X −→ X0 −→h Y , where g is a cofibration and h is a homotopy equivalence. Proof. Let X0 be the mapping cylinder of f, g the inclusion into the cylinder, and h the retract onto the other side 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 Cell (n) (n−1) M n n chain complex Cn (B) = Hn(B ,B ) = Hn(D , ∂D ). Consider the pullback n cells 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 of [n−1] [n−2] [n] [n−2] E from that of B and F . Consider the long exact sequence, ... → Hn(E ,E ) → Hn(E ,E ) → [n] [n−1] d [n−1] [n−2] Hn(E ,E ) −→ Hn−1(E ,E ) → .... If we understand the first and the third terms as well as the mapping d, then we might understand the second term. Continuing this process, we might understand [∞] [0] Hn(E ,E ) = Hn(E). Next, consider the pullback square

E0 E

Dn Φ B where the Φ map is the characteristic map. For the pullback fiber bundle, since the base space is contractible, one can show that E0 is homotopy equivalent to Dn × F , and then from there observe that (E[n],E[n−1]) is a n n−1 −1 [n] [n−1] relatively homeomorphic to (D ×F,S ×F ), where F is the fiber p (∗). Thus H∗(E ,E ) = n cells of B M n n−1 cell H∗(F × D ,F × S ) = Cn (B) ⊗ H∗(F ). If B has (trivial fundamental group?), the mapping n cells [n] [n−1] [n−1] [n−2] cell cell cell H∗(E ,E ) → H∗(E ,E ) becomes Cn (B) ⊗ H∗(F ) → Cn−1(B) ⊗ H∗(F ), which is d ⊗ 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. that dcell ⊗ id part is nontrivial to prove and invokes Hurewicz’s theorem, which we want to prove using spectral sequences.)

4.2 Spectral Sequences: Definition

Definition 7 (Spectral Sequence). A spectral sequence consists of a sequence (Er, dr : Er → Er) (each is a 2 chain complex) where dr = 0, Er abelian groups, together with an isomorphism Er+1 = ker dr/ im dr. One place spectral sequences come from is an exact couple. f g f Definition 8 (Exact Couple). An exact couple is a long exact sequence ... −→h D −→ D −→−→E −→h D −→ ..., which we usually write as a triangle:

f D D

h g E

[n−1] [n] [n] [n−1] M [n] For instance, ... → H∗(E ) → H∗(E ) → H∗(E ,E ) → .... We can let D = H∗(E ), n M [n] [n−1] E = H∗(E ,E ), and with the obvious mappings they fit into an exact couple. n Suppose (D, E, f, g, h) is an exact couple. The derived exact couple is constructed as follows. Connect two exact couple triangles side by side:

f f D D D

h g h g E E

8 Then we can get d = g ◦ h : E → E. One can check that d2 = 0. Let E0 = ker d/ im d, and D0 = im f. f 0 : D0 → D0 is now again given by f. g0 : D0 → E0 is given by g0(f(x)) = g(x). It’s in ker d and thus in E0. Now, for two choices x, x0 of x, their difference lies in h(E), and thus its image is in im d, i.e. the choice of x doesn’t matter. Finally, h0 : E0 → D0 is given by h0(x) = h(˜x) wherex ˜ ∈ ker d represents x in E0. One can check this yields another derived couple. Finally, we get a spectral sequence from a derived couple by letting 0 0 E1 = (E, d) and Er = (Er−1, dr−1). One place to get exact couples is from filtered chain complexes. Given a sequence of chain complexes (1) (2) (n) (n) (n−1) (C∗ ,C∗ ,...). Let D = H∗(C ), E = H∗(C /C ), then this gives the long exact sequence of homology. (n) [n] For instance, we can take Cr = Cr(E ). 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 a double chain complex by letting dh and dv be the differentials on X and Y respectively. From a double chain M complex we get a total chain complex: Cn = Cpq, and d : Cn → Cn−1 is dh +dv. One can check that with p+q=n 2 (n) M the additional −1 this yields d = 0. There are two ways to filter this chain complex: one is C∗ = Cpq, p≤n M and another is Cpq. q≤n

9 5 More on Spectral Sequences

M k Again let’s consider a total chain complex Cn = Cpq, and the associated filtered chain complexes Filtn = p+q=n M k k k−1 Cpq. The exact couple is D = H∗(Filt ),E = H∗(Filt /Filt ). Then we see that E = E1 is the p+q=n,p≤k homology of the kth column (note that the quotient is pretty much just the corresponding entry on the kth 1 column). More precisely, Epq is the pth homology of the qth column, and the d : E → E differential turns out to 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 r M r r r E would have bi-grading: E = Ep,q. The mapping dr then sends Ep,q to Ep−r,q+r−1, as one can check. p,q Let’s get some examples. Consider the following double chain complex, where all the nonzero morphisms are 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 see that 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 from Z 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 term r ∞ Epq eventually stabilizes as r → ∞, and that stable term we call it Epq . Consider the total chain complex C∗, and the associated homology Hn(C∗). It turns out that for each n, the ∞ {Epq }p+q=n terms yield a filtration of the homology group, more precisely we have 0 ⊆ F0 ⊆ ...Fn ⊆ Fn+1 = ∞ Hn(C∗) where Epq = Fp+1/Fp.

5.1 Serre Spectral Sequence

p Suppose we have a Serre fibration F → E −→ B, and suppose for the moment that B is simply connected. Then 2 there is a spectral sequence where Epq = Hp(B; Hq(F )), and it converges to Hp+q(E), meaning that there is a M ∞ filtration Hn(E) whose associated graded group is Epq . p+q=n p Consider for example the fibration ΩSn+1 → PSn+1 −→ Sn+1, where PX is the path space (and thus PSn+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), which n+1 n+1 n+1 n+1 n+1 is H0(ΩS ), stabilizes to Ho(PS ) = Z, tells us that ΩS is connected, thus π1S = π0(ΩS ) = 0 n+1 for n ≥ 1. Some more computations happens and we conclude that Hj(ΩS ) = 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 r Note that D term is the direct sum of images in H∗(Filtr) → H∗(Filtk+r). For fixed n and sufficiently large r r r  0, Hn(Filtk+r) = Hn(C∗). At this point, D → D 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 filtration r F0 ⊆ ... ⊆ Fn ⊆ Fn+1 = Hn(C∗) where Epq = Fp/Fp+1, such that p + q = n. Then we have the filtration of Hn(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 2 0 0 0 0 Z 2 Z 2 0 0 0 0 Z 2 Z 2 0 0 0 Z 0 Z 2 Z

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.

γ 2. For every 1-simplex ∆[1] −→ 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)∗, and such that if γ is constant, γ∗ = id.

Equivalently, consider the fundamental groupoid π≤1X of X, which is the category where objects are points in X, morphisms are homotopy classes of paths rel endpoints. Then a local system is a functor from π≤1X to the Abelian groups. −1 Suppose X → B is a Hurewicz fibration, then b 7→ Hn(p (b)) gives a local system. In fact this is also true for Serre fibration. (Use the fact that a weak equivalence X → Y induces isomorphism on all homology groups.) M Now suppose X is a space with local system A. Then define Cn(X; A) = Ac0 where c0 is the image c:∆[n]→X X i of the first vertex under c. Define a differential d : Cn(X; A) → Cn−1(X; A) = (−1) di, where i ranges over i 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, with notes 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 bottom 2 row. Thus there is a mapping from H∗(left column) → H∗(C), and there is another map H∗(C) → E∗,0 (the bottom 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 induced by the maps F → E and E → B. Consider an element on the bottom row on the E2 page. Then the differentials are obstructions of such element ∈ H∗(B) from coming from H∗(E). Likewise, for an element on the left column, the kernel of F → E are those that are “hit” by some differential coming from below. 2 Now suppose π≤1B is trivial, H∗(B) = 0 for ∗ < n, H∗(F ) = 0 for ∗ < m, then from the graph of the E page one sees that there is a canonical mapping defined as Hp(B) → Hp−1(F ) for p < n + m + 1. This mapping is 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 equivalent to say that a is in the image of H∗(E,F ) (under the connecting homomorphism δ?). Define a ∈ Hp(B) to be transgressive if there is somea ˜ ∈ Hp(E,F ) such that it is the corresponding term in H∗(E,F ) by the mapping of pairs (E,F ) → (B, ∗). Then the transgression of a is defined as δ(˜a) where δ is the connecting homomorphism Hn(E,F ) → Hn−1(F ).

Theorem 8.1. Consider the Serre Spectral Sequence ΩB → PB → B for π≤1(B) = ∗, H˜p(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 Hurewicz homomorphism πn(B) → Hn(B) is an isomorphism. In other words, the first nonvanishing homotopy group is isomorphic 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.) Suppose n ≥ 2. Of course, πi(B) = 0 for i < n, which by induction implies Hi(B) = 0 for i < n. By the transgression theorem above, Hi(B) = Hi−1(ΩB) for i < 2n, which in particular tells us Hn(B) = Hn−1(ΩB) for n ≥ 2. Now look at the mapping Hn(PB, ΩB) → Hn−1(ΩB) and Hn(PB, ΩB) → Hn(B). But look at the corresponding homotopy maps πn(PB, ΩB) → πn−1(ΩB) and πn(PB, ΩB) → πn(B). Hn(PB, ΩB) = Hn−1(ΩB) is iso by what we said above, Hn−1(ΩB) = πn−1(ΩB) by induction, πn(PB, ΩB) = πn−1(ΩB) by the contractibility of PB (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).

n n n As a corollary, we can get πi(S ) = 0 for i < n, and πn(S ) = Z. First realize that π1(S ) = 0 by n n Van-Kampen, then π2(S ) = H2(S ) = 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 _ n _ n resolution 0 → F2 → F1 → A → 0 and the build the corresponding sequence of space S → S →

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 lecture 9 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. If ∼ πi(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 γ(1) equivalently, it is the pullback between f and XI −−−→ 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 trivially on 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, so ∼ H∗(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 exact sequence:

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 compare ∼ ∼ the two exact sequences above and conclude that Hn(X,A) = Hn−1(F ) = πn−1(F ) (by absolute Hurewicz Theorem), which is πn(X,A).

9.1 Application

n f n+1 n+1 n+1 n+1 Consider the fibration S −→ X → X ∪f D . Then Hn(X ∪D ,X) = Hi(D ,S ) (by excision), which n+1 is 0 for i < n+1 and Z for i = n+1. By the relative Hurewicz theorem, then we know that πi(X ∪D ,X) = 0 n+1 for i < n + 1 and Z for i = n + 1 (n ≥ 2). In other words, πi(X) → πi(X ∪ D ) is an iso for i < n and epi n+1 n+1 ∼ for i = n. However, πn+1(X ∪ D ,X) → πn(X) is induced by f, so we know that πi(X ∪ D = πi(X) for ∼ i < n, and = πi(X)/f for i < n. _ This leads to the action of “killing homotopy groups.” More precisely, let Sn+1 → X → X0, so the first ∼ 0 0 0 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 n ∼ n onto it the n + 2 cells. Then we get a series of mappings X → P X such that πi(X) = πi(P X) for i ≤ n, and n n n−1 πi(P X) = 0 for i > n, then we have a sequence P X → P X → ... 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 n πiMA = A for i = n and 0 for i < n. Now P MA has πn = A and πi = 0 for i 6= n. This is called the Eilenberg-MacLane space K(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. n n n As an example, consider the fibration F → S → P S = K(Z, n). Then from the long exact sequence, we n can immediately conclude that πi(F ) = πiS for i > n, and 0 for i ≤ n. Now by Hurewicz theorem, HiF = 0 n ∼ for i < n + 1, thus πn+1S = πn+1(F ) = Hn+1(F ). Keep going along this road, we can eventually compute n things up to e.g. πn+14S . (See Lecture 11 for more details.)

14 10 Representability and Puppe Sequences (Guest Lecturer: Hiro Takana)

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 natural isomorphism [X,K(A, n)] =∼ H˜ n(X,A). One says in this case that singular cohomology is represented by the Eilenberg-MacLane spaces. This is a special case of the more general Brown representation theorem. In particular, along with Whitehead’s theorem this 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.

0 0 Remark 3. If we consider X = S , then we see that [S ,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 the Eilenberg-Steenrod axioms, then along with the fact that it behaves in the same way as singular cohomology on H0(∗), we can conclude the theorem from the uniqueness theorem1. More concretely, let us fix an abelian group A. We shall show that there is a functor Kn : CWpairop → AbGrp given as (X,B) → [(X,B),K(A, n)] that satisfies 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.

2 Proof. Observe that [X,K(A, n)] = [X, Ω K(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, just as 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 _ ∼ Y [ Xi,K(A, n)] = [Xi,K(A, n)]. Homotopy and dimension axioms are trivial, and the excision axiom follows i i from Mayer-Vietoris property, which is again trivial for the functor that we specified. So it remains to check the 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. Now consider 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 collapsing A to a point and obtain X/A as the result. We can, however, continue this process by pushing out along the trivial 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 this step 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 process for 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 yields an 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 a p,q p q p+q p,q p+r,q−r+1 spectral sequence E2 = H (B; H (F )) ⇒ H (E), with differential dr : Er → Er . The infinity p+q 0 1 p+q p,q p p+1 page again has a filtration: H (E) = F ⊇ F ⊇ ... ⊇ F , where E∞ = F /F . The ring structure that cohomology carries extends naturally to a ring structure on the CSSS.

p,q s,t p+s,q+t Proposition 3. There exists a multiplicative structure Er × Er → Er on each page, which we denote |x| by (x, y) 7→ xy. The differential dr acts as a derivation on this structure, i.e. dr(xy) = xdr(y) + (−1) dr(x)y.

11.1 An Appetizer

1 Let us consider the Eilenberg-MacLane spaces. It is clear that K(Z, 1) = S . (For a quick proof that the higher n 1 n n homotopy groups vanish, consider S → S for n > 1; since S is simply connected, the map lifts to S → R, n 1 and since R is contractible this mapping must be nulhomotopic, and this property descends down to S → S ∞ via projection.) On the other hand, from the cohomology ring structure we know that K(Z, 2) = CP . But suppose 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) = 1 p,q K(Z, 1) = S , and PK(Z, 2) is contractible. This allows us to write out the spectral sequence E2 = p q 1 p+q H (K(Z, 2); H (S )) ⇒ H (∗), 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 denotes the kth cohomology of K(Z, 2), of course). The only nontrivial differential on the E2 page is then the arrows k k+2 from H to H 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

0 1 we immediate conclude that H = Z and H = 0. Furthermore, since the arrow indicated on the first graph kills the H0 at (0, 1), we know that that particular arrow must be injective; and since the H2 at (2, 0) also dies, 2n we know it is also surjective and thus bijective. In this way, we immediately conclude that H (K(Z, 2)) = 0, 2n+1 and H (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. 0 Now, because a generates H at (0, 1), one can do direct computation and see that ax2 generates (2, 1), and similarly ax4 for (4, 1), etc. Choose the sign of a and x2 appropriately so that d(a) = x2, then since d(x2) = 0, 1 2 3 we have d(ax2) = d(a)x2 + (−1) ad(x2) = x2 = x4, and similarly x6 = x2, 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 the 1 ∞ ∞ 1 2n+1 n result from last lecture, or observe the fibration S → S → CP , induced as the limit of S → S → CP , and investigate the homotopy long exact sequence associated with it (then apply Whitehead).

11.2 The Main Course

3 Theorem 11.1. π4(S ) = Z2. S This is the first stabilized term π1 of the stable homotopy groups. 3 Proof. We start by considering the fibration X → S → K(Z, 3), where X is the homotopy fiber. As we 3 mentioned before, this effectively “kills π3(S )” because, as one can observe from the homotopy long exact 3 sequence, we have πi(X) = πi(S ) for i ≥ 4 and πi(X) = 0 for i ≤ 3. However, this fibration itself is a bit 3 awkward to use, so we “back up a step” by looking at the Puppe sequence: ... → ΩS → ΩK(Z, 3) → X → 3 3 S → K(Z, 3). We choose the fibration ΩK(Z, 3) → X → S to apply CSSS on. Then we get the E2 page as follows:

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,...) as shown in the graph, and the bottom one on the 3rd column be (x), then we know that the 3rd column’s generators are (x, ax, a2x, . . .). Next note that the space X is 3-connected, so we can invoke the Hurewicz theorem to obtain that H˜i(X) = i 0 =⇒ H˜ (X) = 0 for i ≤ 3, and from which we immediately obtain that d3(a) = x, as the differential needs to kill both (0, 2) and (3, 0). In this manner, we can figure out the other differentials using the derivation n n−1 property, e.g. d(a ) = na d(a) for |a| ≡ 0 (mod 2), and from this we can recover the E∞ page and see that 2i+1 H (X) = Z/i for i ≥ 2. Then by the universal coefficient theorem, we have H2i(X) = Z/i for i ≥ 2, and 3 thus H4(X) = Z/2 = π4(X) by Hurewicz. Since π4(X) = π4(S ), we have completed the proof.

19 12 A Glance at Model Categories, Part 1 (Guest Lecturer: Emily Riehl)

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’s a 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 of equivalences, containing all isos) that satisfies either of the following:

f g • 2-of-3 property If X −→ Y −→ Z and two of {f, g, g ◦ f} are in W , then all of them are.

f g • 2-of-6 property If X −→ Y −→ Z −→h 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 γ localization functor M −→ 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: f f x ←− z −→ x = idx, and f g g◦f x ←− z ←− y = x ←−− 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 limits and 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) such that: • 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 u Y ∃w l r X0 v Y 0

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 a 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 and contains 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 s1 X r1 A f g f B s2 Y r2 B

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. (This hints at the fact that homological algebra is a special case of homotopical algebra modulo the language of model structures.) 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 model categories 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: Emily Riehl)

Recall from last lecture the definition of model categories and weak factorization systems. As explained last time, the factorization axiom guarantees that any map in the cateogry can be factored into the composition of a trivial cofibration ∈ W ∩ C and a fibration ∈ F, and can be factored into a cofibration followed by a trivial fibration. Another quick remark: given the factorization and the lifting axioms of the weak factorization system, the closure 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 = retracts of 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 retracts of relative cell complexes built from Sn−1 → Dn. Dually, (C ∩ W, F) is the system where F are maps having the 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.

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

f Theorem 13.2 (Whitehead). If X, Y are cell complexes, then X −→ 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 category given above.

Definition 15. A is a cofibrant object if and only if the unique mapping ∅ → A from the initial object to A is in C. Dually, A is a fibrant object if and only if the unique mapping A → ∗ from A to the final object is in F. 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 complexes are 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 0, where the objects are bifibrant objects of M and the 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 id ` id A A A g f=(i0,i1)

cyl(A)

where g is a weak equivalence. The object is called good if f is a cofibration, and very good if additionally g is a trivial fibration. Note that in a model category we are always guaranteed to have a very good cylinder object. Also it’s worth noting 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 cofib ∅ A

cofib a A A A i1 id

i0 cofib cyl(A)

id A We have marked in the diagram the cofibrations that have been given by the assumptions. By the pushout a 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 an isomorphism, also is a weak equivalence, so from the 2-of-3 property we conclude that i0 and i1 are weak equivalences. Definition 17. Given two mappings f, g : A → X, a left homotopy between f and g is a mapping H such that the following diagram commutes:

f A i0

cyl(A) H X

i1

g A

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 dual notion 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). In addition, 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 A cyl(A) H X

cofib trivial cofib

cylgood(A)

Now suppose X is fibrant. Choose a good left homotopy H : cylgood(A) → X. Now consider the mapping g : cylgood(A) → A, which is guaranteed to be a weak equivalence by the cylinder object definition. By the factorization axiom, we can factor g into g2 ◦ g1, where g2 is a trivial fibration and g1 is a cofibration. By 2-of-3 g1 g2 property, this means that g1 is also trivial. We write this as cylgood(A) −→ cylvg(A) −→ A. Now consider the following diagram:

H cylgood(A) X H0 g1

cylvg(A) ∗

0 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.

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

f g A A 0 i0 i0

0 cyl(A) H X cyl0(A) H X 0 i1 i1 g A A h

0 In order to construct the necessary cylinder object, consider the pushout of i1 and i0 as follows:

i0 A 0 cyl0(A)

i1 i˜1 0 i˜0 H cyl(A) 0 cyl00(A) H00

H X

0 0 Since H ◦ i1 = H ◦ i0 = g, we have the outer commutative square, and by the UMP of pushout we have the induced mapping H00 : cyl00(A) → X. Then we have the following diagram:

f A ˜0 i0◦i0

00 cyl00(A) H X ˜ 0 i1◦i1

A h

But we still need to verify that cyl00(A) is actually a cylinder object. This can be checked via the following diagram:

i0 A 0 cyl0(A)

i1 i˜1 i˜0 cyl(A) 0 cyl00(A)

A

˜ ˜0 Since pushout of trivial cofibrations are again trivial cofibrations, i1 and i0 are trivial (because A is cofibrant). On the other hand, cyl(A) → A and cyl0(A) → A are guaranteed to be trivial, so by 2-of-3 property we can conclude that cyl00(A) → A is trivial.

p p Theorem 13.3. If A is cofibrant, X −→ Y is trivial fibration, then [A, X] −→∗ [A, Y ] is an isomorphism. Note that this is the model categoric statement of Theorem 13.1.

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

f A i0

p cyl(A) H X Y

i1

g A

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

a f ` g ∅ X A A X f H0 p p k A Y cyl(A) H Y

As we mentioned above, ∅ → A is in C since A is cofibrant, and p ∈ F ∩W by assumption, then the mapping f exists by the lifting axiom, and thus we have surjectivity. Now for injectivity: suppose that we have pf 'l pg, a then consider the right diagram above, where A A → cyl(A) is a cofibration as we have shown above, so H0 exists, and thus f 'l g.

Proposition 6. If X is fibrant, A0 → h is a mapping, f, g : A → X, then f 'l g =⇒ fh 'l gh.

a k l Proof. As we mentioned before, we can choose a very good left homotopy A A → cylvg(A) −→ X for f ' g. On the other hand, consider the following diagram: ` 0 a 0 h h a A A A A cylvg(A) H0

0 0 h cylgood(A ) A A where the unmarked maps are canonical. It is clear that the right side map is a trivial fibration, and the left 0 0 0 k◦H side map is a cofibration, and thus H exists, and one sees that cylgood(A ) −−−→ X is a left homotopy between fh 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 then f '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 only 4 if f is a homotopy equivalence (i.e. there exists g : X → A such that f ◦ g ' 1X , g ◦ f ∼ 1A ). 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 (Lemma 4.24). Suppose f is a weak equivalence. Then by 2-of-3 property, we can factor it into a trivial cofibration q p followed by a trivial fibration. Suppose this is written as f : A −→ C −→ X. Now consider the following diagram:

∅ C s p X id X

∅ → X is cofibration because X is cofibrant, and C → X is a trivial fibration, so s exists, and clearly p◦s = idX . p Next observe that by the theorem we proved above, we know that [C,C] −→∗ [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 an inverse 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 lecture 3 3 11. We learned that H4(X) = Z2 = π4(S ) for X → S → 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 sequence K(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 that we have a bunch of 2-groups. The 0th row we have already calculated, on the other hand. Then we see that Z3 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 affect Z/3 at all. This suggests at the fact that we should be able to simplify these calculations significantly by only looking 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.

e1 n 3. Let S be the set of primes, the class of A such that ∀a ∈ A, ∃n = p1 . . . pne , 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.

f Definition 19. A map A −→ 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, which makes A → B an isomorphism if A/B ∈ C) for an abelian category A and a collection of objects C (meeting certain 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.

n n Proof. A ∈ C =⇒ A ⊕ A ∈ C because 0 → A → A ⊕ A → A → 0. Thus A ∈ C, i.e. Z ⊗ A ∈ C. Now m n m consider the exact sequence 0 → Z → Z → B → 0, then we have the sequence 0 → Tor(A, B) → A → 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 trivially on πnF . If HiF ∈ C for i > 0, HiE ∈ C for i > 0, then HiB ∈ C for i > 0, under some restrictive condition explained 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 finitely generated, 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 sequence 0 → (some quotient of H2E) → H2(B) → (a subgroup of H2F ) → 0. Since the two side terms are in C, so is the middle term. But we need some extra assumption to work on H3B, since there is a nontrivial mapping from (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 d2 prove the theorem. To be more detailed, let us look at the mapping HnB −→ 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 bottom ∞ row, which is then forced to be in C as well. Thus we can conclude that Er,0 ∈ C for arbitrary r. Suppose ∞ m m−1 Er,0 = Er,0 = ker dm/ im dm. Since im dm ∈ C, we know that ker dm ∈ C. Since Er,0 / ker dm = im dm ∈ C, it m−1 follows that Er,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

2 ∞ Recall the definition of a Serre class. Clearly, if Ep,q ∈ C, then so does Ep,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 that πnX → 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.

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

k In the next class, we’ll prove the even cooler result that πnS 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 the loopspace, then the inductive hypothesis settles the problem. However, note that given πiX = 0 mod C for i < n for some n does not implies π2X = π1ΩX = 0, so we cannot assume ΩX is simply connected as we did in 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, so let’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 us apply 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 these groups are in C. We know that the 0th row is 0 mod C, then from the universal coefficient theorem we see that the all the way to the row of n − 2, everything below is 0 mod C. Then since we know that the SSS converges to HnΩX, by SSS we have that HiΩX = 0C, i.e. HiΩF → HiΩX is iso mod C for i ≤ n − 1. This is the second half of the second point. Now look at the fibration ΩX → PX → X. By induction, HiX = 0 mod C for i < n − 1 because πiX = 0 mod C for i < n. Also by induction, we know that HiΩX = HiΩF = 0 mod C for i < n − 1. Now look at the SSS 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 mod C, which clears the (n − 1) column, which then yields that (n, 0) → (0, n − 1) is iso mod C. This yields the first half 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 a p-torsion abelian group, then Hn(K(A, 1); Z) is p-torsion for all n ≥ 1. 0 Proof. First, for any abelian group A, A is a filtered colimit A = colimA0⊆A,A’ finitely generatedA . If A is p-torsion, then each A0 is finitely generated and p-torsion. Since homology commutes with colimit, it suffices to prove the first part (and showing that the homologies are p-torsion). Now note that K(A × B, 1) = K(A, 1) × K(B, 1). 1 By Kunneth formula, we are reduced to A = Z and A = Z/p. Now K(Z, 1) = S , so no problem with A = Z. pn By the theorem that [K(A, m),K(B, m)] = Hom(A, B) (proof can be found here), we know that K(Z, 2) −→

29 n 5 1 n ∞ K(Z, 2) → K(Z/p , 2) is a fibration . If we back it up twice, we have S → K(Z/p , 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

n n p where all the differentials are multiplications by p (induced by K(Z, 2) −→ K(Z, 2)). Thus we have ∗ n n n H (K(Z/p , 1); Z) = Z[x]/(p x) for |x| = 2. By the universal coefficient theorem we have H2m−1(Z/p , 1); Z) = n Z/p 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 the fibration, but this is not very clear.

30 16 Relatvie Mod C Hurewicz; Finiteness of Homotopy Groups of Spheres

First, a warning: the assumption of simple connectivity is actually necessary. An example: take S2 ∨ S1, and certainly the homology groups are finitely generated, and π1 = Z, but π2 = ZZ by the universal cover of the 1 M graph (where S expands to Z, and H2 = Z = π2, which is not finitely generated) is not finitely generated n∈Z 3 3 Again consider the backing-up of S → K(Z, 3), which is K(Z, 2) → X → S . We used cohomology SSS on this to compute some homotopy groups of S3. Now fix some prime p and let C be the class of ` torsion groups such that (`, p) = 1. Then we know that H2p(X) = Z/p and HiX = 0 mod C for i < 2p. Then by mod C Hurewicz, we have πiX = 0 mod C for i < 2p and π2p(X) = Z/p mod C. Thus we know that the next mod C 3 3 3 nontrivial fundamental group for S after π3(S ) = Z is π2pS = Z/p ⊕ A for A being a torsion group of order smaller than p (because larger factors aren’t supposed to show up yet).

3 Corollary 7. πiS 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. Thus 3 we know that πiS = 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 generalizes to the fact that any simply connected space with finitely generated homology (i.e. finitely many cells at each level) yields finitely generated homotopy), we conclude:

3 Corollary 8. πiS 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 a root 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 C for 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 are simply 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. ∗ ∗ 1 2 Now recall that H (K(Z, 1); G) = H (S ; G) = G[x]/(x ) for |x| = 1. Now look at the CSSS of K(Z, 1) and K(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 that ∗ 2 H (K(2, Z); Q) = Qx2 for some |x2| = 2. Do this again, we see K(Z, 3) has Q-cohomology ring Q[x3]/x1. Keep ∗ ∗ 2 repeating this, we conclude that H (K(Z, 2n); Q) is Q[x2n], and H (K(Z, 2n + 1); Q) = Q[x2n+1]/x2n+1 is an exterior algebra. 2n−1 ∗ Note that S → K(Z, 2n − 1) is an iso in H (•; Q), and thus an iso in π∗ • (mod C) as the proposition 2n−1 2n−1 above indicates, and thus πi(S ) is finite for i > 2n − 1. (πi(S ) is trivial for i < 2n − 1.) n 2n What about the other half of the possibility, i.e. S for n even? We have S → K(Z, 2n), which is 2 2 2n x2n Q[e]/e → Q[x2n] explicitly given by e 7→ x2n where |e| = 2n. Now we have a fibration S → K(Z, 2n) −−→ 2n K(Z, 4n). Look at the Serre spectral sequence of the backup K(Z, 4n − 1) → S → K(Z, 2n) to get that ∗ 2n 2 2 2n H (S ; Q) = Q[x2n, x4n−1]/(x2n, x4n−1). We conclude that π∗S ⊗ Q is Q for ∗ = 2n, Q for ∗ = 4n − 1, and 2n 2n 2n 0 otherwise. As a corollary, π2nS = Z, π4n−1S = Z ⊕ a finite group, and πiS 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 an isomorphism for dim X < n and epimorphism for dim X = n. Proof. First let’s show the epimorphism. Consider the following diagram, where X is assumed to have dimension n, 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 map in [X(n−1),W ], so it suffices to deform all the mapping Dn → X → Z to Dn → W . Consider the following diagram where W 0 is the pullback of the two maps on the right:

Sn−1 W 0 W

Dn Dn Z

0 n n−1 0 Then as one can check, W and D has n-equivalence, and thus isomorphic for πn−1, thus S → W is nulhomotopic, so it extends to some Dn → W 0. Then the concatenation with W 0 → W gives us the required lifting (at least up to homotopy), and thus we can conclude that every X → Z can be deformed to some X → W for dim X = 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 lifting X × I → W exists, i.e. any homotopy in Z lifts to a homotopy in W . Thus maps that are zero in [X,Z] will be zero in the lifting, hence the monomorphism and isomorphism.

n Now let us address Steenrod’s problem. Suppose we have S → K(Z, n), then clearly it’s an iso on πi for n n i ≤ n, and epi for i = n + 1. This tells us that if dim X < n + 1, then [X,S ] = [X,K(Z, n)] = H (X; Z). In short, two maps of the same degree are homotopic (consider e.g. oriented n-manifolds). This result (or at least its earlier versions) was known in the 1930s. Also, this was part of the reason why cohomology was introduced. n On the other hand, we also know that the mapping S → 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 to figure 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 that 3 6 H (K(Z, 3)) = Z, H (K(Z, 3)) = Z/2, and further groups are 0. Using UCT, we have that H3(K(Z, 3)) = 3 Z,H5 = Z2. (See here for the details of this computation.) Or we can look at πi(K(Z, 3),S ) = 0 for i < 5 and Z/2 when i = 5 (which follows the homotopy LES). Then this yields that H5(K(Z, 3)) = Z2, from which the same 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) associated with the Eilenberg-Maclane space. By cellular homology, we can guess the following structure:

2 ... → Cn+3 = Z −→ 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)] = m n n H (W ; A). Consider S → K(Z, n) and extend it to S → K(Z, n) → K(Z, n + 3) and take the homotopy n fiber F → K(Z, n) → K(Z, n + 3), so that S → 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 see that 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). n+2 n+3 However, naively using K(Z2, ∗) doesn’t work either because H (K(Z2, n+2); Z) = 0 and H (K(Z2, n+ 2); Z) = Z2.

n n+2 Solution Recall that H (K(Z, n); Z) = Z and H (K(Z, n); Z2) = Z2. So instead we consider the fibration K(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 dim X = 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 one has a natural map onto Sn, so [X,Sn] has a group structure for dim X up to 2n. If we localize mod p, on the other 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)] → n−1 n+1 n [X,K(Z/2, n + 2)] → .... This yields H (X; Z) → H (X; Z/2) → [X,F ] → H (X; Z) → 0. Note that we “magically” now have a mapping (the first one) that increases the cohomology degree by 2 and reduces the 2 k k+2 coefficient mod 2. This is a new structure in the cohomology, denoted by Sq : H (X; Z) → H (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).

m 2m−1 2m−2 m 2m m Example 10. Consider [CP ,S ]. We have the long exact sequence H (CP ; Z) → H (CP ; Z2) → m 2m−1 2m−1 m m−1 [CP ,S ] → H (CP ; Z) = 0. The first one is Z generated by x (where x is a generator of 2 m m−1 2 m−1 H (CP ; Z)), the second one is Z2; by the Leibniz’s rule above, we see that the map sends x to Sq (x ) = m m 2m−1 (m − 1)x , so the mapping is multiplication by (m − 1). Thus we conclude that [CP ,S ] = 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 this external 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 a∪b

a×b

im∪in K(Z2, n) × K(Z2, m) K(Z2, n + m)

Note that [a ∪ b] = [b ∪ a] as cohomology classes, which is to say that (by abuse of notation) as maps, a ∪ b and b ∪ a are homotopic. The important observation that Steenrod made was that there is much information in this homotopy. Consider the following diagram, given a : X → K(Z2, n). This diagram is commutative up to homotopy:

a∪a X × X K(Z2, 2n) (x,y)7→(y,x) a∪a X × X

0 This means that we have a homotopy expressible as H : X × X × I → K(Z2, 2n) such that H(x, x , 0) = H(x0, x, 1), therefore in particular H factors through space X × X × I/{(x, y, 0) ∼ (y, x, 1)}. With a bit of 1 imagination, observe that this space is homeomorphic to S ×X×X/Z2, where the generator of Z2 sends (λ, x, y) 1 1 to (−λ, y, x). We write this space as S ×Z2 X × X, so we have the mapping h : S ×Z2 X × X → K(Z2, 2n). 1 h In general, let W be a space with an action of Z2, and let τ 6= 0 ∈ Z2. Then a mapping S ×Z2 W −→ Z is g the same thing as a map W −→ Z along with a homotopy g =∼ g ◦ τ. 1 2 Later we will prove that the map h : S ×Z2 X → X → K(Z2, 2n) extends to D × X × X → K(Z2, 2n), 2 ∞ which then extends to S ×Z2 X × X → K(Z2, 2n), etc. Eventually it extends to S ×Z2 X × X → K(Z2, 2n). ∞ p˜(a) a∪a This is the main idea: we look for a map S ×Z2 X × X −−→ K(Z2, 2n) extending X × X −−→ K(Z2, 2n). In 2n ∞ particular, note thatp ˜(a) ∈ H (S ×Z2 X × X; Z2).

n p˜ 2n ∞ Theorem 18.1. For any pair (X,A), there is a unique natural transformation H (X,A; Z2) −→ H (S ×Z2 2 6 (X,A) ; Z2), such that the following diagram commutes :

n p˜ 2n ∞ 2 H (X,A; Z2) H (S ×Z2 (X,A) ; Z2) a7→a∪a

2n 2 H ((X,A) ; Z2)

2 where (X,A) = (X × X,X × A ∪A×A A × X).

Constructing Steenrod Operations First, from now on, as long as we’re talking about Steenrod algebra, n ∆ one should always assume the coefficient group to be Z2. Start with the diagonal cup product H (X) −→ n n 2n ∞ ∞ ∆ H (X) ⊗ H (X) → H (X × X). Observe that we have the canonical map h : RP × X → S ×Z2 X −→ ∞ 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:

n p˜ 2n ∞ 2n H (X) H (S ×Z2 X × X) H (X × X) P h∗ ∆∗

2n ∞ 2n H (RP × X) H (X)

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 Steenrod operations:

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 thisp ˜. Given any (X,A) and a class a ∈ Hn(X,A), there is a map n (X,A) → K(Z2, n), ∗) such that a is pulled back from the canonical mapping in ∈ H (K(Z2, n), ∗), therefore n 2n ∞ 2 it suffices to show that there is a unique choicep ˜(i ) ∈ H (S ×Z2 (K(Z2, n), ∗) ), and the rest of the proof 2n 2 n ⊗2 follows from naturality. By Kunneth formula, we know that H ((K(Z2, n), ∗) ) = H (K(Z2, n)) = Z2, so 2n ∞ 2 2n 2 it suffices to show that the mapping H (S ×Z2 (K(Z2, n), ∗) ) → H ((K(Z2, n), ∗) ) is an isomorphism. This follows from the following proposition immediately. Proposition 15. Suppose (X,A) is (n − 1)-connected, then there is a SES:

2n ∞ 2 2m 2 n n 1−flip n n 0 → H (S ×Z2 (X,A) ) → H ((X,A) ) = H (X,A) ⊗ H (X,A) −−−−→ H (X,A) ⊗ H (X,A) → 0.

k 2 k+1 2 Proof. We first show that for every k ≥ 1, the map S ×Z2 (X,A) → S ×Z2 (X,A) induces an isomorphism 2n 2n k+1,Sk 2 2n+1 k+1,Sk 2 on H (•; Z2). In particular, we need H (S ×Z2 (X,A) ; Z2) = H (S ×Z2 (X,A) ; Z2) = 0. The k+1 k 27 pair in consideration is relative homeomorphic to (D ×Z2,S ×Z2)×Z2 (X,A) ; since both pairs are excisive, k+1 k 2 ∼ k+1 k 2 this induces isomorphism on cohomology. But note that (D ×Z2,S ×Z2)×Z2 (X,A) = (D ,S )×(X,A) , and the cohomology for the latter is H∗(Dk+1,Sk) ⊗ H∗(X,A)⊗2, which is 0 for degree less than 2n + 2 (by relative Hurewicz theorem). Now that we have confirmed that the degree of the sphere doesn’t matter, we verify the original claim with S∞ replaced with S1. Apply the same argument as above, but this time with the pair (S1,S0). We have the following maps: 0 2 1 2 1 0 2 ... ← S ×Z2 (X,A) ← S ×Z2 (X,A) ← (S ,S ) ×Z2 (X,A) ← ... 2n 1 0 2 2n 1 2 where, by the same argument above, H ((S ,S ) ×Z2 (X,A) ) = H (S × (X,A) ) = 0, so we have the following map 2n 1 2 2n 0 2 0 → H (S ×Z2 (X,A) ) → H (S ×Z2 (X,A) ) → ..., 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

i ∞ 2 Last time we showed that if (X,A) is (n − 1)-connected, then H (S ×Z2 (X,A) ) = 0 for i < 2n, and 2n ∞ 2 n ⊗2 H (S ×Z2 (X,A) ) are the elements in H (X,A) invariant under the tensor factor transposition. 2n ∞ In particular, let (X,A) = Kn, then we have a corollary that there is a unique element β(in) ∈ H (S ×Z2 2 i n Kn) restricting to in ⊗ in. Thus, the P operator that we mentioned last time uniquely defines Sq : H (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 X 4. Cartan formula: Sqn(xy) = Sqi(x)Sqj(y). i+j=n

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 Steenrod algebra (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.

n i n+i Claim 5 Suppose x ∈ H (X; Z2) and i < 0, then Sq (x) ∈ H (x) for n + i < n. By naturality, this should hold 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( , n)) ιn Z2

Sqi Sqi

n+i n+i H (X) H (K(Z2, n)) But when n + i < n, the right bottom object is zero, as Eilenberg-Maclane spaces have vanishing lower cohomologies, 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 from n m comparing coefficients. Note that x pulls back from some in ∈ H Kn, y pulls back from some im ∈ H Km. Now consider the following diagram in the universal case (which we don’t know commutes yet), in which we use the shorthand Kj = (K(Z2, j), ∗):

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

38 n m H (Kn) × H (Km) p˜×p˜ ∪

n+m 2n ∞ 2 2m ∞ 2 H (Kn × Km) H (S ×Z2 Kn) × H (S ×Z2 Km) p˜ h∗×h∗ ∪ 2n+2m ∞ 2 2n 2 2m 2 H (S ×Z2 (Kn × Km) ) = Z2 H (Kn) × H (Km)

h∗ ∪ 2n+2m 2 H ((Kn × Km) )

k 2k ∞ 2 As a general observation, if (X,A) is (n−1)-connected and H (X,A) = Z2, then H (S ×Z2 (X,A) ) = Z2 k ⊗2 because it is a subgroup of H (X,A) = 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 image in 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 either they are equal (and both nonzero), or at least one of them is the zero map. Since the general case pulls back from the universal case, we conclude that either P (xy) = P (x)P (y) for all x, y, or one of them is constantly zero. 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) are nonzero at some point. ∗ ∞ n+m n m n+m n+m 2(n+m) To do this, consider H (RP ) = Z2[t]. We know that Sq (t t ) = Sq (t ) = t . On the other hand, Sqn(tn) = Sqm(tm) = t2(n+m), so neither is zero, and we have thus obtained the Cartan formula.

0 n Claim 3 Again let’s work with the universal case. Observe that Sq (in) ∈ H (K(Z2, n), Z2) = Z2. Moreover, 0 n n ∼ any homomorphism Sq : H (X) → H (X) is given, in the universal case, by a map in [K(Z2, n),K(Z2, n)] = n H (K(Z2, n), Z2) = Z2, so there are only two natural maps: the identity map, and the zero map. Thus it suffices to show that Sq0 is nonzero somewhere. n n ∞ n n 2 ∞ The example that we consider is (R , R − {0}). Again look at h : S ×Z2 (R , R − {0}) ← RP × n n 2n 2n (R , R −{0})). So let’s figure out what happens to H of the first space. Note that H of the second space is n ∞ n n n n n n n H (RP )⊗H (R , R −{0}) = Z2, which is generated by t x for some nonzero generator of H (R , R −{0}). We would like to show thatp ˜(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 = 2n 2n n 2n (R ,R − {0}). Now consider how R s embed into R : they embed as the “axes”, and the Z2 factor will act by flipping these two axes (i.e. flipping around the “diagonal”). However, we can choose an alternative basis by choosing the diagonals as the “axes” (i.e. a different embedding), then we see that Z2 acts as −1 ∞ n n on one of them, and 1 on the other. This allows us to rewirte the mapping as (S ×Z2 (R+, R+ − {0})) × n n ∞ n n (R−, R− − {0}) ← RP × (R , R − {0})), in which case Z2 now does not act on the first parenthesis, so we have factored this map into product of two maps, where the second factor is the identity, and the firt factor is ∞ ∞ ∞ n n RP = S ×Z2 ({0}, ∅) → S ×Z2 (R , R − {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 ∞ 1 1 ∞ n = 1. Now note S ×Z2 (R , R − 0) has the same cohomology as S ×Z2 ((−1, 1), −1, 1) by a direct ∗ ∞ ∗ ∞ transformation. We want to show that H (S ×Z2 ((−1, 1), −1, 1)) → H (S ×Z2 (−1, 1)) is an isomorphism. 1 ∞ 1 ∞ By the long exact sequence of a pair, it suffices to note that H (S ×Z2 −1, 1) = H (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 case ∗ is rather tricky, but today we can establish the H (K(Z2, n); Z2) case. (From now on, omit the Z2 coefficient.) ∗ ∗ ∞ 1 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 ∈ H . ∗ Now we want to consider H (K(Z2, 2)). Consider the fibration K(Z2, 1) → ∗ → K(Z2, 2) and its SSS (we’ll only write generators now).

5 x5 0 ? ? ? ? ?

4 x4 0 ? ? ? ? ?

3 x3 0 ? ? ? ? ?

2 2 2 x 0 x y2 ? ? ? ?

1 x 0 xy2 ? ? ? ?

2 0 1 0 y2 y3 y2 ? ?

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) is 2 simply connected. By transgression, we must have d2(x) = y2 for some y2. We know that d2(x ) = 2xd2(x) = 0, 2 2 3 so x does not hit (2, 1) and thus the mapping d2(xy2) = y2 must be nonzero. Now, we know that d2(x ) targets 2 2 2 2 x d2(x) = x y2. On the other hand, x must be transgressive, so d3(x ) = y3 for some y3 ∈ (3, 0), which might be zero or might not. Technically, you can do this to figure out the rest of the E2 page and thus the entire spectral sequence (see your homework). k The important thing to figure out is the differentials on x2 . 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).

n 1. If x ∈ H∗(F ) is transgressive, then so is x2 for all n.

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

n n We shall let the target of x2 be called y2 +1. The proof of Borel’s theorem is done in the homework using Zeeman comparison theorem,

Koszul Resolution Consider the abstract chain complex C∗ = F2[x][y2, y3, . . . , y2n+1,...]. Define the dif- 2n n 14 ( 8 4 2 ferential d(x ) = y2n+1. Then this extends to all x in the obvious manner: e.g. d(x ) = d x · x · x ) = k 1 10 6 O 2n (Leibniz Rule) = x 2y3+x y5+x y9. More abstractly, we want a map from vector space F2{1, x } to F2[x] n=0 k O O 2n by first sending it to F2[x] then to F2[x]. In particular, note that the colimit of the F2{1, x }, w.r.t. k n=0 2n k, is isomorphic to F2[x]. Let en = x 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 = F2{1, en} ⊗ F2[y2n+1]. q q q where den = y2n+1, dy2n+1 = 0, dy2n+1en = y2n+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 and 0 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 and j k suppose π1B acts trivially. If x ∈ H (x) is transgressive, then Sq (x) is transgressive and the transgression of Sqk(x) is the Sqk of the transgression of x.

40 Proof. Start by considering the definition of transgression:

Hj+1(E,F ) Hj(F ) δ p∗

Hj+1(B)

to say that γ is the tresgression of x is to say that p∗(y) = δ(x). Now apply Sqk to everything, and it suffices to show that the diagram still commutes. We need Sqk(δx) = δ(Sqkx). Consider the definition of the connecting homomorphism. The following are all isomorphisms in cohomology:

(E ∪ CF,CF ) (E ∪ CF, ∗)

(E,F )

And we see there’s a mapping E∪CF → ΣF , which then induces a mapping Hk+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 check that the diagram above still works after Sqk.

2 1 1 Now using the Kudo regression theorem, we know that x = Sq (x) transgresses to Sq (y2) = y3. Similarly, 2 2 2 ( x = Sq (x ) transgresses to Sq y3) = y5. Therefore we have:

∗ 1 2 1 4 2 1 Corollary 9. H (K(Z2, 2)) = Z2[y, Sq y, Sq Sq y, Sq Sq Sq y, . . .].

Now consider the same settting but applied to K(Z2, 2) → ∗ → K(Z2, 3). By Borel’s theorem, we can conclude that K(Z2, 3) again has a polynomial algebra cohomology. In particular, this allows us to compute the 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: Jeremy Hahn)

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

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

k Last time we used Serre Spectral Sequence to compute H (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 defined as (im − 2im−1) + (im−1 − 2im−2) + ... + (i2 − 2i1) + i1.

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

21.1 Cohomology Operations

l Definition 23. Fix an integer l, a cohomology operation of degree i is a natural transformation H (•, F2) → l+i H (•, F2). Examples: in degree 5, we have Sq5, Sq2Sq3 = Sq(2,3), Sq5 + Sq(2,3), etc.

∗ ∗+i Definition 24. A stable cohomology operations of degree i is a natural transformation H (•, F2) → H (•, 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 current setting, so strictly speaking we don’t have to worry about the term “stable” that much.

l Remark 7. H (•, F2) is represented by K(F2, l). By Yoneda’s lemma, cohomology operations are in bijection l+i I with H (K(F2, l)), and these are sums of Sq 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 on products 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. Otherwise suppose 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 before since Sq(1,1) = 0.

n ∞ ∞ Proof of the Theorem. We first need a lemma. Consider x1 . . . xn ∈ H (RP × ... × RP ) = F2[x1, . . . , xn]; | {z } n copies ∗ ∞ ∞ by naturality, it corresponds to a map (x1 . . . xn) : RP × ... × RP → K(F2, n). | {z } n copies

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

42 ∗ l Lemma 11. The natural map (x1 . . . xn) produces a monomorphism in H (•, F2) for all l ≤ 2n. l l ∞×n Proof of the Lemma. We’d like to show that H (K(F2, n)) ,→ H (RP ) for l ≤ 2n. For this, we want to I show that the images of Sq (ιn) are linear independent for I admissible of total degree ≤ 2n. The images are I Sq (x1 . . . xn). Each of these will be symmetric polynomials in the xi, so we can express them in terms of I elementary symmetric polynomials (σ1 = x1 + ... + xn, σ2,...). We want to calculate Sq (σn). The rest is just calculation, and follow the claims below.

i Claim 1 Sq (σn) = σnσi.

I Claim 2 Sq (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 and n+i b define the same element in H (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 of 1-dimensional classes. Some side notes: Steenrod squares are examples of primary cohomology operations; it turns out that there are also higher cohomology operations, which one discovers from the study of Adam spectral sequences. It turns out 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 on all 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 arbitrary Steenrod product to an admissible one.

22.1 Adem Relation

ba/2c X b − j − 1 Proposition 17 (Adem Relation). SqaSqb = Sqa+b−jSqj. a − 2j j=0 Keep in mind that everything is mod 2. Example 13. Sq3 = Sq1Sq2 (we have shown this last time). 3 − 0 − 1 3 − 1 − 1 Example 14. Sq2Sq3 = Sq5Sq0 + Sq4Sq1 = Sq5 + Sq4Sq1 =⇒ Sq5 = Sq2Sq3 + 2 − 2 · 0 2 − 2 · 1 Sq4Sq1 (since we work mod 2). Today we talk about the Bockstein homomorphisms.

22.2 Bockstein Homomorphisms

∗ I We already know that H (K(Z2, n); Z2) = Z2[Sq ιn] for admissible I with excess less than n. We computed ∗ J this via the Serre spectral sequence. By the same method, one can compute H (K(Z, n); Z2) = Z2[Sq ιn] with admissible J with excess less than n, and that if J = (jm, . . . , j1), then j1 > 1. Today we’d like to compute ∗ H (K(Z/2k, n); Z2) for k > 1 based on the facts above. (From now on we drop the Z2 coefficient.) k 2 k First step, let’s consider n = 1. Consider the SES 0 → Z −→ Z → Z/2 → 0. We get a fibration k 2 k 10 1 K(Z, 1) −→ K(Z, 1) → K(Z/2 , 1) . We know that K(Z, 1) = S . Now we run the Serre spectral sequence on p,q p k q 1 p+q 1 11 this fibration: E2 = H (K(Z/2 , 1); H (S )) ⇒ H (S ). This means we have to only consider two rows.

4

2

2 2 3 Z2(a) Z2(aι1) Z2(aβk)Z2(aι1βk)Z2(aβk)Z2(aι1βk)Z2(aβk)

2 2 3 0 Z2(1) Z2(ι1) Z2(βk) Z2(ι1βk) Z2(βk) Z2(ι1βk) Z2(βk)

0 2 4 6

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

44 0 1 0 1 Both rows are (H ,H ,...), and we know that H = H = Z2 since we’re with Z2 coefficients. On E∞ 1 we should have Z2 for H (E), so (0, 1) must have been killed, and thus the first nontrivial differential is a monomorphism, and because (2, 0) must be killed we know that that particular differential is in fact an 2 j isomorphism, and thus H = Z2. Continue this process to the right, we see that H = Z2 for every j. Now let us assign names to the generators as indicated in the graph above (note that ι1 is the funda- ∗ k mental class of K(Z2k , 1)), and compute the rest using the Leibniz rule. Then we see that H (K(Z/2 , 1) = 2 12 ∗ k 2 2 Z2[ι1, βk]/(ι1 = 0). This yields that H (K(Z/2 , 1) = Z2[ι1, βk]/(ι1 = 0) = Z2[ι1]/(ι1 = 0) ⊗ Z2[βk] = ∗ ∗ H (K(Z, 1)) ⊗ H (K(Z, 2)). ∗ k ∗ ∗ In general, this calculation yields that H (K(Z/2 , n); Z2) = H (K(Z, n)) ⊗ H (K(Z, n + 1)), which one k I J k 2 can write as Z2[Sq ιn, Sq βk]. One can then consider the pullback fibration K(Z/2 , n) → K(Z, n + 1) −→ n+1 k n+1 K(Z, n+1), then the βk ∈ H (K(Z/2 , n)) is the restriction from ιn+1 ∈ H (K(Z, n+1)), so we have βk ∈ k n k n+1 [K(Z/2 , n),K(Z/2, n + 1)]. This yields a stable natural transformation βk : H (X; Z/2 ) → H (X; Z2). k+1 k But where do we get this βk, really? Consider the fibration Z2 → Z/2 → Z/2 , and consider its associated fiber sequence:

n n k+1 n k βk n+1 ... → H (X; Z2) → H (X; Z/2 ) → H (X; Z/2 ) −→ H (X; Z2) → ... n n+1 1 In the special case of k = 1, we have β1 : H (X; Z2) → H (X; Z2), which coincides with Sq .

12 2 2 How do we know that βk 6= ι1? We can compare its action as a cohomological operation with ι1.

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 easily proven.   r   m Y ar Proposition 18. Suppose p is a prime and m = (a . . . a ) , n = (b . . . b ) , then = (mod p). 1 r p 1 r p n b i=1 r 5 (101)  101 Example 15. = 2 = = 1 (mod 2). 4 (100)2 1 0 0

d 2d ∞ d Second, recall that for x ∈ H (X), we the total power operation P (x) ∈ H (X × RP ) = Sq (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 + d i ... + Sqd(x)t , where Sqd−i(x) = Sq (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 X i clarity, let’s write the P (x) above as Pt(x). So we have Pt(x) = Sqi(x)t . Now consider the formal operator i X i X i j L(t) = Sqit , then L(s) · L(t) = SqjSqis t , which we can consider as a generating function for SqjSqi. i i,j We have proven before that Pt is a ring homomorphism, which yields additivity and the Cartan formula. One ∞ 1 ∞ 2 more detail for calculation: suppose X = RP , and 0 6= x ∈ H (RP ), then Pt(x) = x + xt. d X i X i Now suppose x ∈ H (X). Let us calculate Pt(Ps(x)). It equals Pt( Sqi(x)s ) = Pt(Sqi(x))Pt(s ) = i i X X j i X X j 2 i 2 SqjSqi(x)t Pt(s) = SqjSqi(x)t (s + st) = L(t) · L(s + st). Now we shall prove that that i j i j 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-dimensional classes (then invoke the uniqueness theorem of stable cohomology operations). Since both sides are ring ho- 2 momorphisms, it suffices to do the case in which x is a 1-dimensional class. Now, Pt ◦ Ps(x) = Pt(x + 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’t really rely on the lucky calculation above.) Now we show that the form above of the Adem relation coincides with the usual one. We want

X 2 i j X 2 i j (s + st) t SqiSqj = (t + st) s SqiSqj. i,j i,j

2 X i j Let u = s + st, then the left side becomes u t SqiSqj. How do we handle the right side? Use residues. i,j X i Suppose we have a formal Laurent series f(x) = aix , then Resx=0f(x)dx = a−1, which is the coefficient i>−∞ dx of . Also, suppose x = g(y), g(0) = 0, g0(0) is nonzero, so g−1 exists by the inverse function theorem. x 0 Theorem 23.1. Resx=0f(x)dx = Resy=0f(g(y))g (y)dy. Proof. See any complex analysis textbook.

46 X i j X n m 13 2 Now look at u t SqiSqj, so t SqnSqm is the coefficient of u in it. Now u = s + st, du = tds i,j n (where s is the variable), thus we know that

i j X X u s SqjSqi tnSq Sq = coefficient of um in uitjSq Sq = Res du n m i j u=0 um+1 n i,j

which further simplifies to

ti+1(s + t)isj ti+1(s + t)i−m−1sj Res Sq Sq ds = Res dsSq Sq s=0 sm+1(s + t)m+1 i j s=0 sm+1 i j

X i − m − 1 which is the coefficient of sm in ti+1(s + t)i−m−1sjSq Sq , thus tnSq Sq = t2i+j−2mSq Sq , i j n m m − j j i i,j from which it is immediate by a substitution that

 i − m − 1  Sq Sq = Sq Sq . n m 2i − m − n 2m+n−2i i

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) Pt(x) P˜t(x)

K(Z2, 2n)

If we apply this action again, we get

∞ ∞ ∞ ∞ 2 RP × RP × X S ×Z2 (S ×Z2 (X × X)) Ps(Pt(x)) P˜s(P˜t(x))

K(Z2, 4n)

∞ ∞ ∞ 2 2 Consider the space S × (S × S × X × X ) that “covers” the space on top right. There’s one Z2 acting on the outside, and there’s Z2 × Z2 acting on the inside, and the outer Z2 acts on the Z2 × Z2 by exchanging ∞ ∞ 2 ∞ 3 4 the factors. So we can write the space S ×Z2 (S ×Z2 (X × X)) as (S ) ×Z2oZ2 X → K(Z2, 4n), where o denotes the wreath product. ∞ ∞ As a motivating example, suppose X is a point. The fundamental group of RP × RP × X in this case is ∞ 3 Z2 × Z2, and the fundamental group of (S ) /(Z2 o Z2) is Z2 o Z2, so the trivial mapping Z2 × Z2 → ∗, induced by 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, if the 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 these two actions are conjugate in Σ4 by the element (23), as this conjugation relation is really why Adem relation would hold.

24.1 Higher Power Operations ∞ ∼ ∞ 14 ˜ ∞ It is easy to observe that S = {(x, y) ∈ R | x 6= y} . To generalize this, let Ck(R ) denote the kth ∞ configuration 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 x ˜ unordered size-k sets). Define the kth extended power of X −→ K(Z2, n) as the unique extension to Pk(x) k (which we omit the definition) such that the following graph commutes and that Pk(x) ◦ i = x :

i ˜ ∞ k X BΣk × X Ck(R ) ×Σk X Pk(x) P˜k(x)

K(Z2, nk)

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

˜ ∞ 3 4 ˜ ∞ 4 ˜ ∞ 4 BΣ2 ×BΣ2 ×X = B(Σ2 ×Σ2)×X → C2(R ) ×Z2oZ2 X → C4(R )×ZoZ2 X → C4(R )×Σ4 X → K(Z2, 4n) ˜ ∞ 3 4 ˜ ∞ 4 where the only not-so-canonical map C2(R ) ×Z2oZ2 X → C4(R ) ×Z2oZ2 X comes from the following 3 construction of C˜2 → C˜4: the elements in the two “inner” C˜2s 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 C˜2 then has two points ca 6= cb, indexed by the two inner pairs repsectively. We would like to consider only the two inner pairs, but of course they might collide; so what we need to do is use ca and cb to send them to two different directions so that they are apart. More precisely, consider the continuous mapping: let the “ball” containing a1 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 angle between 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) ∈ C˜4 now.

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

48 ˜ ∞ 4 Then, due to the following fact and the obvious fact that C4(R ) ×Σ4 X = BΣ4 × X, the two maps under consideration are homotopic in their BΣ2 × BΣ2 × X → BΣ4 × X part, since we mentioned in the last section ˜ that the induced actions are conjugate. Since P4 is uniquely defined, the second part BΣ4 × X → K(Z2, 4n) is unique, so we have shown that the two maps are homotopic.

−1 Proposition 19. Suppose we have f1, f2 : H → G are conjugate, i.e. ∃g ∈ G.gf1g = f2, then Bf1, Bf2 : BH → BG are homotopic. Proof. Skipped. 4n ˜ ∞ It remains to show that the map factors. In fact, we will show the stronger statement that H (C4(R )×ZoZ2 4 4n ˜ ∞ 4 X ) = H (C4(R ) ×Σ4 X ). 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 is a 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:

∗ r r r Theorem 24.1. If H (E) = 0 for ∗ < r, then H (S ×G E) → H (S × E) → H (E) is a monomorphism with image 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- 0 n tical columns on the E2 page. Everything is zero until row n, where the first element becomes H (B,H (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 just the invariants of Hn(E) under G. Since nothing can ever get cancelled, this fully determines the SSS. So n n H (S ×G E) → H (F ) is injective by edge homomorphism. 4 ˜ ∞ But back to our story. Specializing with E = (K(Z2, n), ∗) , r = 4n, S = C4(R ) and G = Σ4 and 4n ˜ ∞ 4 4n ˜ ∞ 4 G = Σ2 o Σ2 respectively, we see that H (C4(R ) ×ZoZ2 X ) and H (C4(R ) ×Σ4 X ) (or really, their images r ⊗4 under monomorphism) are respectively just elements of H (K(Z2, n), ∗) invariant under Σ2 o Σ2 and Σ4 k 4 ∧4 respectively. However, note that in our case, (X,A) = (K(Z2, n), ∗) = K(Z2, n) as one can check from 4n ∧4 ∧4 definition, so the target space is H (K(Z2, n), ∗) ) = Z2 = Z2. It is clear that the invariant of this Z2 under both 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 an ending note, there is yet another pure homological-theoretical explanation to this story that involves explicitly computing out all cohomologies involved. The takeaway from that story is that the only possible obstruction to 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 is Steenrod 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 is n n equivalent to ignoring all odd torsions.) We shall calculate π∗(S ) localized at 2 by replacing S by an easier space 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) for i < 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.

i i Remark 9. An equivalent condition is that H (X; Z2) ←− H (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, n if we can build spaces that cohomologically approximate S mod C2, then we can us them to compute the n homotopy groups of S mod C2. n ∗ I n Take S → K(Z, n). We know H (K(Z, n)) = Z2[Sq ιn], and more explicitly for the latter space, H = n+1 Z2(ιn) (single Z2 summand generated by ιn), H = 0 (because I must have first term greater than 1), so ∗ n 2 up till here it agrees with H (S ); but the next one, generated by Sq ιn, is probably different. So let’s modify K(Z, n) so it has closer comohological behavior. Sq2 In particular, consider the following diagram, where X1 → K(Z, n) −−→ 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:

49 K(Z2, n + 1) X1

2 n Sq S K(Z, n) K(Z2, n + 2)

n Note that since πn(K(Z2, n + 2)) = 0, the mapping from πn(X1) → πn(K(Z, n)) is surjective, so S → n 2 K(Z, n) lifts to S → X1. Run Serre spectral sequence (diagram omitted), we see that ιn+1 hits Sq ιn by 1 3 3 2 2 3 1 transgression, and Sq ιn+1 = Sq ιn by transgression theorem, and thus Sq ιn+1 = Sq Sq ιn = Sq Sq ιn = 0 2 1 5 by Adem relation. Similarly Sq Sq ιn+1 hits Sq ιn. Consider what’s left in the cohomology of X1, one can check that it’s Z2 in dimension n, one in dimension (n + 3), 2 classes in (n + 4), etc. We’ll continue this computation in the next lecture.

50 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 style to 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 sketch the overall strategy. n Start with the mapping S → K(Z, n) = X0 → K(Z2, n + 2), and choose X1 to be the homotopy fiber of n X0 → K(Z2, n + 2). As we said in last class, there is a lifting f : S → X1. Now choose the smallest m such m m m n m that H f : H X1 → H S is not a monomorphism, and let A be the kernel of H f. Then we have the n mapping S → X1 → K(A, m), and we can continue the construction by letting X2 be the homotopy fiber of X1 → K(A, m), etc. (That m actually increases is not completely trivial, but is not hard to observe from the SSS 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 Y of E-M spaces that is an approximation of homotopy groups on low dimensions: f : W → K(πi(W ), i) = X0, i ∗ such that f is surjective in H (•; Z2). Let m be the smallest integer such that there is a nontrivial kernel in m m H (X0) → H (W ), and let A be this kernel. Then we have W → X0 → K(A, m), and we can continue this ∗ process. 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 thus n homotopy approximations, of S mod C2. The homotopy of each Xi, on the other hand, is easy to compute inductively thanks to the homotopy LES of fibrations. Remark 10. This technique is called “killing cohomology one at a time”. Frank Adams later observed that if instead 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

a Consider the fibration Xk+1 → Xk −→ 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 is 1 the first nonzero element on the fiber column, on dimension n. Let (1) = Sq ιn, then (1) may cause trouble for ∗ ∗ H (Xk+1) approximating H (Xk), because it may not get killed. Of course, if it does, things work out rather nicely.

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.

1 Example 17. Consider K(Z2, n) → K(Z4, n) → K(Z2, n) and its SSS. We have ιn and Sq ιn at n and n + 1 0 1 0 respectively on the base, and ιn and Sq ιn at n and n + 1 on the fiber. The problem is that we have a mapping 0 1 from ιn to Sq ιn, and it’s not clear immediately how this differential will work.

p In general, suppose F −→i E −→ B is the fibration, a nice graphical way to represent SSS with Bockstein 0 0 homomorphisms is to connect u at (n, 0) and u at (n + 1, 0) with a r-fold line to indicate βr(u) = u (which n 2r n+1 0 is a H (X, Z ) → H (X, Z), and connect v at (0, m) and v at (0, m + 1) with a s-fold line to indicate a 0 0 βs(v) = v . Then, if there is a transgression τ(v) = u , then in the total homology of E there is a nontrivial ∗ r+s ∗ 0 Bockstein homomorphism βr+s such that i β p (u) = v . This is known as the Bockstein Lemma and detailed discussion can be found on page 106 of this book.

51 In general, this tells us that the problem of using the technique we have so far to compute all homotopy groups of spheres is that eventually we’ll hit problem at dimension 15, where a class just “pops up” and we can’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

2 n Sq S K(Z, n) K(Z2, n + 2)

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

b5

b4,1

b4

b3,1

b3

b2,1

b2

b1

b0

a4,2 a5,2 1 a0 a2 a3 a4 a5 a6 a7

We know that b0 goes to a2, thus b1 goes to a3; further computations now call for Adem relations. b2 and 2 2 2 3 1 thus b3 goes to zero because db2 = Sq ιn+1 = Sq Sq ιn = Sq Sq ιn = 0. Some more computation shows that b2,1 goes to a5. But we have a pair (a4, b3,1) that is probably nontrivial; in fact, a4 → b3,1 is a Z4 bockstein 1 1 homomorphism on the kernel of Sq (since Sq a4 = a5 gets killed) by the Bockstein lemma above. This allows us to write down the cohomology of X1 in a small range.

52 c2

c1

c0

b3 1 a0 b2 a4 b3,1

The next class we have to kill is b2 (the package I’m using is unable to draw double arrow, so I’m using blue color instead). So we map X1 → K(Z2, n + 3) and take the fiber X2, and back up to get K(Z2, n + 2) → X2. 2 Draw SSS again as in the graph above. c0 hits b2, c1 hits b3. c2 hits Sq b2 = b3,1, so we have the trouble with (a4, c3), the same situation again, which this time yields a Z8 Bockstein. Repeat this process again with X2 → 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)

2 n Sq S K(Z, n) K(Z2, n + 2)

Thus we have the conclusion:

n n n n Theorem 25.1. After localizing at 2, we have πnS = Z, πn+1S = Z2, πn+2S = Z2, πn+3S = Z8. More detailed calculation for up to the 7th stable homotopy group is carried out in this book. But as we have seen above, this method has limitations (at dimension 15). As a matter of fact, when you run this on other 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.

53 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 equivalence relation 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 a 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: a Lemma 13. Ωn(X) is a group. In particular, M M is null-cobordant. a a 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 the proof. In particular, let f : M → X = U ∪V , and then f −1(U), f −1(V ) is a covering of M. By smooth partition of unity, one can choose a function that’s 0 on one of them, 1 on the other of them, and in the intersection it’s only increasing. Choose a submanifold L in the intersection and map it to a regular value, then we have f|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 of cobordism classes of maps from pairs of smooth n-manifolds (M, ∂M) to (X,A). We’ll show this in a completely different approach from what we said above, namely the approach by Pontryagin and Thom, by identifying it with the homotopy group of something else (Thom spaces), and then compute them. (Don’t worry, unlike spheres, 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 ring k under cartesian product that is isomorphic to Z2[x2, x4, x5,... | i 6= 2 − 1], and then Ωn(X) is just the regular homology 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 some obvious 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 vector bundle then is a continuous map S → Grn(W ). The second approach is consider the maps p : V → S, and take −1 Vx = p (x), and somehow we need this to be a vector space. We mainly need some + : V ×S V → V that is commutative, associative, 0 being the unit, and for every λ ∈ R, all operations are R-homomorphisms. Unfortunately, this na¨ıve approach won’t quite work because we can’t guarantee local triviality. This condition says that for each x ∈ S has a neighborhood U ⊇ S, such that VU , being the pullback of U → S

54 n n and V → S, is isomorphic to U × R and such that VU → U is pullback of projection U × R → U by the isomorphism. (This notion was actually not clear for a while in history.) Once we add this condition in, things start to work out nicely. We’ll show in next lecture that over finite CW complex, every vector bundle is pullback over some universal n vector bundle over the Grassmannian. For example, take the case of R . There is a tautological vector bundle n given by Vk → Grk(R ) such that the fiber over a subspace H of dimension k is just H. In other words, n n Vk ⊆ Grk(R ) × R = {(H, x) | x ∈ H}. n ⊥ We need to check local triviality. Suppose H is a point in Grk(R ), then we have its complement H in n 0 n 0 ⊥ R . Then the needed U = {H ⊆ R | H ∩ H = 0}, i.e. spaces that projects to the entirity of H when n ∼ 0 projected by the projection π : R → H. On U, we can use π to identify V |U = U × H:(H , x) ∈ V |U ⊆ n n 0 Grk(R ) × R 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 ⊕ V W, V ⊗W , V 7→ S (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 fiber bundles. (How? we apply the operations fiber-wise, and check local nontriviality by checking that its result on the 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 (the point at infinity), and it’s a good idea to collapse all of these points to one. To do this, we instead use the Thom complex: Thom(X,V ) = XV = SV /X (collapsing the infinity section). It turns out this construction has 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...

55 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 n ∗ [X, Grk(R )] → 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 p f X Y

where we define the f ∗(V ) = {(x, v) ∈ X × V | f(x) = p(v)}, and then a fiber bundle on the right induces n a fiber bundle on the left, where the fibers are the same on both bundles. Also recall that Vk → Grk(R ) is n the tautological k-plane bundle defined in the last class, where Vk = {(H, v) | H ⊆ R , 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:

g V W

X

V ×X V V

g×g g

W ×X W W

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, then

the 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 obvious 0 0 k 0 k thing holds. Choose a local trivial neighborhood U ⊆ X of x such that V |U 0 = U × R , W |U 0 = U × R . 0 0 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 X 15 for each x, and ρα(x) = 1 for all x. α Proposition 21. Suppose V → X is a vector bundle, Z ⊆ X is closed, and s : Z → V is a section, then there exists a section s0 : X → V extending s. k k Proof. First suppose V = X × R , then a section is a map s(x) = (x, t(x)) for some t : X → R . Then the case k holds by Tietze’s theorem. Now let X be covered by local trivial neighborhoods {Uα}, where V |Uα = Uα × R , 0 and let Zα = Z ∩ Uα. Then by case 1, we have some sα extending sα for each sα defined by restricting s to Zα. 0 X 0 Now, choose a partition of unity ρ, then we claim that ραsα define a section on X, and ραsα is a section α 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 the pullback 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 vector bundles. 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 things are the same, so they are isomorphic in a neighborhod of X × {t}. Since X is compact, there is an isomorphism on X × (t − , t + ) for some  > 0, which shows that ∀s ∈ (t − , t + ), Vs ∼ Vt.

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

56 Corollary 11. Vectk(X) is a homotopy functor of X. In other words, if f, g : X → Y are homotopic, and V → 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 → X to be the restriction of V to X × {0}. Write π : X × I → X to be the projection. Then there is an isomorphism ∼ ∗ V = 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 algebraic vector bundles. Proposition 23. Suppose X is compact Hausdorff, V → X is a vector bundle, then for n  0 there is a surjective map

n X × R V

X

k Proof. Let U1,...,Um be a finite open cover such that V |Ui is trivial. Let ei : Ui → R → V |Ui be the P M k i ρiei isomorphism, and choose a partition of unity ρi, then ρiei is surjective on the support of ρi, so X× R −−−−−→ i V is surjective. Note that if X is locally contractible, we can choose U1,...,Um to work for all vector bundles.

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

57 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 all vector spaces are pullbacks of universal bundles. Today we talk about characteristic classes, which are data that we attach to vector bundles, which can be used to identify whether the bundle is trivial or not. They are to 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 some i particular elements of H (X; Z2). There are two ways of constructing S-W classes. One relies on geometry (using BO and BU); a cleaner way is 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 0 V

f X0 X

i ∗ i 0 The S-W classes wi(V ) ∈ H (X) pulls back to f wi(v) ∈ H (X ), and naturality says they are precisely 0 wi(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 > dim V .

∞ ∞ ∞ 4. Non-triviality: Consider Gr1(R ) = RP , there is a (canonical) line bundle V1 → RP , which only 1 ∞ has dimension 1 and thus only a single S-W class w1(V1) ∈ H (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 have n a fiber bundle RP → P (V ) → X. There is a canonical line bundle L over P (V ), such that it pulls back to ∞ ∞ V1 → Gr1(R ) = RP . Thus we have the following graph:

L V1

n−1 ∞ ∞ RP P (V ) Gr1(R ) = RP

X

∗ ∗ ∞ ∗ ∗ n−1 Then take the induced map on H on the middle line to get H (RP ) → H (P (V )) → H (RP ) that sends generators to generators (nontrivial; needs a check). Thus there exists a x ∈ H1(P (V )) so that the 1 n−1 ∗ ∗ 2 n−1 restriction of x to H (RP ) is a generator. Then by Leray-Hirsch, H (P (V )) = H (X)[1, x, x , . . . , x ], n n−1 which means that x must be linearly expressible in 1, x, . . . , x . Then define wi to be the coefficients such n n−1 n−2 that x + w1(v)x + w2(v)x + ... + 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 0 V

f X0 X

58 ∗ Let xV and xV 0 be the canonical generators defined above. We know that f xV = xV 0 , then from the uniqueness 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’s ∞ ∞ really 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 open ∼ sets 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 dim V = m, dim V = n. Then define w ∈ H∗(P (V ⊕ V )), where w = xm + w (V )xm−1 + 1 2 Vi 1 2 V1 V1⊕V2 1 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⊕V2 pulls 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

0 V = L1 ⊕ ...Ln V

F (X) X

such that V 0 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 V −→π X. Pull it back along P (π) P (V ) −−−→ 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 then we’re done. Finally, note that the cohomology map on the bottom is H∗(X) → H∗(X)[1, x, . . . , xn−1] so it’s injective. Now suppose V → X is the canonical line bundle. Then triviality and non-triviality specifies all S-W classes on the canonical line bundle. By naturality, this specifies S-W classes for all line bundles (since they are all pullbacks of the canonical bundle), then by the splitting principle and the Cartan formula, along with the injectivity, this allows us to specify S-W classes for all vector bundles.

59 29 More on S-W Classes

29.1 Examples of Stiefel-Whitney Classes n n Consider the class RP , over which there is the totalogical line bundle L → RP . We saw that the total X i S-W class wt(L) = 1 + tx (this is a new notation: wt(L) = wi(V )t where t is a formal variable), thus i M X k w (L) = x, where 0 6= x ∈ H1( n; ). Let V = L, then w (V ) = w (L)k = (1 + tx)k = tixi, then 1 RP Z2 t t i k k w = xi. Now note that L lies in the trivial bundle n+1, and we have a short exact sequence of vector i i R n+1 bundles 0 → L → R → H → 0, where H is called the hyperplane bundle, and the SES is called the Euler Sequence. 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. 1 X Thus we know that L⊕H is the trivial bundle, thus w (L)·w (H) = 1, and thus w (H) = = tixi. t t t 1 + tx n+1 We know that dim H = n, so wn+1(H) = 0, and thus wn+1 = x = 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 of n linear maps l → H, and thus we see that the tangent bundle T RP = Hom(L, H). Note that Hom(L, R) = L. Now we hom into the exact sequence above to get

n+1 ∗ 0 → Hom(L, L) = ∗ → Hom(L, R ) = ⊕n+1L = ⊕n+1L → Hom(L, H) = T → 0 n + 1 Thus w (T ) · 1 = (1 + tx)n+1, hence the S-W class of T n is w = xk. t RP k k

29.2 Detour: Riemannian metric Let V be a vector space.

h·i Definition 28. A Riemannian metric on V is a symmetric bilinear map V × V −→ R such that it is positive definite: hv, vi > 0 for v 6= 0.

h·i 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 X a metric h·iα for each local neighborhood, let sα be a partition of unity, then sαh·iα is a Riemannian metric α over V . Example 18. If 0 → U → V → W → 0 is an exact sequence of vector bundles, then given a metric h·i over V , definte W 0 = U ⊥ ⊆ V w.r.t. the metric, then W 0 → V → W is an isomorpism, so V = U ⊕ W .

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

• ν has dimension n − dim M.

n • ν ⊕ T =∼ M × R .

• wt(ν) · wt(T ) = 1.

60 (1 + tx)2 Now, note that T 13 ⊕ 1 = 14L, so w (T ) = (1 + tx)14, and w (ν) = (1 + tx)−14 = = RP t t (1 + tx)16 1 + t2x2 = 1 + t2x2. Thus w (ν) = 0, w (ν) 6= 0. Conclusion: 13 does not immerse into 14, for if it did, 1 + t16x16 1 2 RP RP 9 14 w2(ν) would be 0 since dim ν = 1. Along the same way, we can conclude that RP is not immersive into RP , 15 but it might immerse into RP . Of course, this does not allow you to construct the immersions. Well, we do have some results:

2n−1 Theorem 29.1 (Whitney). Every smooth manifold of dimension n immerses in R . In fact, we have something better:

2n−α(n) Theorem 29.2 (P. Cohen). Every smooth compact manifold of dimension n immerses in R , where α is the Hamming weight.

Using K-theory, one can obtain some better results using more sophisticated cohomology theories. In the n n+k end, however, the answer to the question “What is the smallest k such that RP → R ” is not known up till this 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:

n n ∞ RP × RP RP

n+k RP

∞ and take the generator x for the cohomology of RP , which pulls back to (x + y) on the left for generators x, 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:

n+k Theorem 29.3 (Hirsch). A n-dimensional smooth manifold M immerses into R if and only if there exists n+k ν → M, with dim ν = k and that ν ⊕ T = M × R . 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-dimensional manifold. Then if either m < n, or m = n and M open in N, then the space Imm(M,N) of smooth immersions of M into N is weakly homotopy equivalent to Mono(TM,TN), the space of vector bundle monomorphisms from TM to TN. This idea leads to a more generalized principle known as the h-principle, which has applications in geometry and PDEs. (c.f. here)

61 30 Cohomology of the Grassmannian

n+k ∞ Let G = lim Grk( ), also written as Grk( ), which is also BO(k) (the classifying space of O(k)). We n→∞ R R ∼ showed that for paracompact Hausdorff X,[X,G] = Vectk(X), given by pulling back the universal bundle Vk. i ∞ Let wi = wi(Vk) ∈ H (Grk(R )) be the S-W classes of the universal bundle, then S-W classes of all vector ∗ ∗ bundles are obtained as pullbacks, i.e. ui = wi(f Vk) = f wi. ∗ Theorem 30.1. Z2[w1, . . . , wn] → H (G) is an isomorphism. n+k ⊥ Theorem 30.2. On Grk(R ), we have Vk and its orthogonal complement Vk of dimension n, having S-W 0 0 0 0 0 ∗ n+k classes w1, . . . , wn. Then the map Z2[w1, . . . , wk, w1, . . . , wn]/{wi · wi = 1 for all i} → H (Grk(R )) is an isomorphism. We shall prove the first theorem only.

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

 ∗ ... 1 0 0 0 0 0 0 0   ∗ ... 0 ∗ ... 1 0 0 0 0    ......  ∗ ... 0 ∗ ... 0 ∗ ... 1 0 n+k 1 2 n+k Given some k-plane V in R , associate with it the sequence dim(V ∩R ), dim(V ∩R ),..., dim(V ∩R ) 1 2 n+k where R ⊆ R ⊆ ... ⊆ R is some chosen sequence of subspaces. This sequence eventually goes to k, but we need to keep track where it bumps up (we say that it bumps up at the “jump” dimensions ji). After scaling we can put each basis vector as ending in 1 and thus having ji − 1 free dimensions; by subtracting previous 5 basis vectors we can get the form as above. For instance, in the following element of Gr3(R ): 1 1 0 0 0 1 3 2 0 0 0 1 1 1 3

i we have the sequence (dim(V ∩ R )) = (0, 1, 2, 2, 3), thus the jump dimensions are (2, 3, 5). By counting the number 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 ∞ n L1⊕...⊕Ln ∞ 16 (RP ) −−−−−−−→ Grn(R ) of nth product of the tautological line bundle One can check the composition ∗ ∗ ∞ n Z2[w1, . . . , wn] → H (G) → H ((RP ) ) = Z2[z1, . . . , zn] sends wi → σi, the ith elementary symmetric Σn function in z1, . . . , zn. We know from classical invariant theory that Z2[z1, . . . , zn] = Z2[σ1, . . . , σn]; in particular, 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, dim Z2[w1, . . . , wk]m ≥ dim Hm(G), since we already showed that it’s monomorphic. Consider the Poincare Series of the Grassman- X m m X m nian PG = dim H (G)t , and that of the polynomial ring PW = dim Z2[w1, . . . , wk]mt . m m

X m About Poincare Series Let A∗ be a graded ring over Z2, then P (A∗) = dim Amt . Some properties: 1. P (A ⊗ B) = P (A)P (B). (Easy to check.) 1 2. Suppose A = [x] for |x| = l, then P (A) = . Z2 1 − tl k Y 1 X Thus we know that P = (this is also known as the partition generating function) = ρ ti, so W 1 − ti i i=1 i ρi is the number of ways to write i in the form s1 + 2s2 + ... + ksk. This in turn one-to-one corresponds to a sequence 0 ≤ a1 ≤ ... ≤ ak ≤ i, where ai = s1 + ... + si; this corresponds, as we discussed above, to a cell in X G of dimension ai = i, so ρi equals the number of i-cells in G. Finally, recall that the number of i-cells of i 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 map X → BG such that γ is the pullback of π along this map.

62 0 X 0 A few words on the second theorem. If we write wi · wi = 1 + ri, then wi · wi = 1 is equivalent to i ri = 0 ∀i. In fact, the sequence (ri) form a regular sequence (as defined in commutative algebra). 5 0 0 0 For instance, consider Gr2(R ), then the cohomology is given by Z2[w1, w2, w1, w2, w3]/{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 m (1 − qm) ... (1 − qm−r+1) the number of cells of Grassmannians of finite dimensions. (Write = r if r ≤ m, r q (1 − q) ... (1 − q ) 5 n and 0 otherwise. Then the expression above is simply . The general statement is that counts 2 t k q k-dimensional subspaces of an n-dimensional vector space over Fq. Also, I guess I should mention that this is also 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 . We say a smooth map f : M → W is transverse in Z if for all z ∈ Z and x ∈ M such that f(x) = z, we have TxM ⊕ 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 a closed K ⊆ M, then can choose the homotopy to be constant on a neighborhood of K. Also, there is an extended version 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 necessary to 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 transverse map, which is constant in a neighborhood. Thus we get a new homotopy H0 of f, g that is transverse to x. Let N = (H0)−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, then we can say it is B(V )/S(V ), where B(V ) = {v ∈ V | kvk ≤ 1}, S(V ) = {v ∈ V | kvk = 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 check properties: • (X × Y )V ⊕W = XV ∧ Y W ;

• 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 the point 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 to X. Let M = f −1(X) ⊆ Sn+k, then it has dimension k. Then we can interpret the homotopy of XV in terms of M. To do so, we need some properties about M. It of course comes in with an embedding in Sn+k, and also a map f : M → X. Moreover, it also comes from an isomorphism ν ∼ f ∗V , where ν is the normal bundle to the 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 is −1 −1 −1 transverse in X. If f (X) = M0, g (X) = M1, then N = H (X) is a cobordism of M0 and M1, which comes with all the associated data that is compatible with the data from the two submanifolds. This gives a V well-defined map from πn+kX to the set {The three maps mentioned above}. Theorem 31.1 (Pontryagin-Thom). This map is a bijection. Note that: n+k 1. We can let the embedding be M → R instead, because clearly the image and thus the preimage are away from the infinity points respectively. 2. What’s the additive structure on the RHS set? Move things by a translation we can take the disjoint union of the embeddings, which doesn’t change the elements up to cobordism. n+k Proof. What we describe below is known as the Pontryagin-Thom construction. Suppose we have M ⊆ R ⊆ Sn+k, f : M → X and ν → f ∗V , where V → X is a vector bundle over a manifold. Recall the tubular neighbor- n+k hood theorem from baby topology: suppose a manifold M embeds in R , then there exists a neighborhood n+k U of M in R , and a diffeomorphism U → ν, such that the zero section M → ν is the same as the inclusion M → U posted composed by the diffeomorphism. Thus we can choose a tubular neighborhood U of M in n+k n+k n+k n+k ν R , and get a continuous map S → S /(S − U) → M such that x maps to x if x ∈ U, and ∗ if x∈ / U. This map is called the Pontryagin-Thom collapse. But recall that the following diagram:

ν V

f M X

induces a map M ν → 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 itself might 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 smooth manifold, then we have the following isomorphism regarding the Thom spectrum (defined in the next lecture – Charles):

V n+k ∗ πn+kX = {M → X,M → R , 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 g N if X is not a smooth manifold. Consider the map X −→ Grn(R ) 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

g N X Grn(R )

V N Vn We observe that X → Grn(R ) 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 g V N Vn N N Vn L → X −→ Grn(R ) is transverse to Grn(R ) ⊆ Grn(R ) . N V When X → Grn(R ) is a Serre fibration, every map L → X is homotopic to a map that is transverse to the zero section. (My raw note says here “we can move a bit such that we don’t move in a neighborhood of the base point.” I’m not sure what’s going on here. – Charles) f ι n+k ∗ Now suppose we have the data (M −→ X,M −→ R , f V ∼ v), then we get an element of the homotopy n+k h n+k group. Suppose ι1, ι2 : M → R are two isotopic embeddings, i.e. there is an embedding M × I −→ R × I ∗ ∗ that 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 of n ι. However, for N  0, by a general position argument we see that any two embeddings M → R are isotopic, (so the space of embeddings lim Emb(M, n) is contractible, because we can take M to be Sn × M), so if n→∞ R we can embed into even larger spaces we can probably get rid of ι. Well, if we admit to a larger embedding n+k n+k+1 ∗ M → R → R , then the normal bundle data becomes f V ⊕R ∼ v⊕R. Then the new data corresponds to the homotopy Sn+k+1 → XV ⊕R = ΣXV . If r : Sn+k → XV , then this new map is simply Σr. Thus if we t V f ∗ consider lim πn+k+tΣ X , then this can be specified by the data M −→ X and f V =s ν, where =s is a stable t→∞ isomorphism (see below). s t At this moment it is fitting to introduce some more notation. Let us write πnZ = lim πn+tΣ Z and call t→∞ it the nth stable homotopy group. For instance, by Freudenthal suspension theorem, stable homotopy groups s 0 πkS are well-defined. N N Definition 31. Two vector bundles V,W on X are stably isomorphic if V ⊕ R ∼ W ⊕ R for N  0. n n n n n Example 19. TS → S and S × R → S are not isomorphic unless n = 1, 3, 7, in which cases the sphere n n n has trivial tangent bundle. However, these two bundles are stably isomorphic, for TS ⊕ R =∼ S × R × R. Remark 14. In terms of commutative algebra, any vector bundle is a finitely generated projective module over n n n the space of continuous functions over the base space; then S × R → S is a free module, so we have an example 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 have ∗ to mention that data? Well, suppose M is of dimension k, then the stable isomorphism f V =s ν is given by ∗ j ∼ n+k+j ∗ n+k+j that TM ⊕ f V ⊕ R = R . Note that M embeds into f V , so there is a mapping M → Grk(R ). Thus the tangent bundle / normal bundle is classified by a map into the Grassmannian. Thus if we take n+j+k X = Grn+j(R ) 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 map ∞ ∞ ∞ Grn(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 πn+kMO(n) = k-manifolds/cobordism. Here n→∞ 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 lim πn+kMO(n) where MO(n) n→∞ ∞ is the Thom complex Thom(Grk(R )). Note that we have the following diagram:

f ∞ M Grk(R )

n+k ∗ R f Vk

We should be able to multiply with another space X, and add the trivial bundle for X, then because V ⊕W V W (X × Y ) = X ∧ Y , we also obtain πn+k(MO(n) ∧ X). So we also observe that the cobordism classes of k-manifolds M × X is lim πn+k(MO(n) ∧ X). n→∞

Definition 32 (Spectrum). A spectrum E consists of a sequence of spaces En and connecting maps tn :ΣEn → En+1.

n n n n+1 0 Example 20. E = {S }, where En = S and ΣS → S is the usual map. This is the sphere spectrum S .

∞ n In general, if X is any pointed space, there is a suspension spectrum Σ X given by just EnX = Σ X. 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}. For instance, 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 weak equivalence.

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 πn+kEn. n→∞ There is a lot of details in the construction of spectra, but we’ll not go deep into the theory here. Note that of course the definition πkE makes sense for k ∈ Z since eventually (n + k) is positive. For example, consider −m −m −m j−m m > 0 and let S be the spectrum given by Sj = ∗ for j < m, and Sj = S for j ≥ m. Then −m −m n−m π−mS = lim πn−mSn = lim πn−mS = . n→∞ n→∞ Z

Definition 34. A map E → F of spectra consists of a collection of maps En → Fn such that it commutes with tn. 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.

m Proof. Given E = {En}, define a new spectrum E given by En = lim Ω En+m. (We ignore some technical m→∞ 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 involves some careful detail checking.)

67 Proposition 33. Suppose A → X is a map, E is a spectrum. Then there is a LES

δ ... → πk(E ∧ A) → πk(E ∧ X) → πk(E ∧ (X ∪ CA)) −→ 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)) is exactness of the middle term in the LES associated with X → (X ∪ CA) → (X ∪ CA) ∪ CX = ΣA. Note that πk(E ∧ A) → πk(E ∧ X) → πk(E ∧ (X ∪ CA)) is zero because A → X → X ∪ CA is zero. We have the following diagram:

Sn+k Dn+k+1 Sn+k+1 x y

En ∧ X En ∧ (X ∪ CA) En ∧ ΣA

The y in the diagram is defined by the diagram and is an element of πk+1E ∧ ΣA = πkE ∧ A. And it turns out that y gets sent to x in the first map. To see this, extend the map to the right again to have the bottom be En ∧ 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 πn+kK(A, n)∧X. When X = ∗, πk(HA) = 0 n→∞ if k 6= 0 and A if k = 0, so HAk(•) also satisfies the dimension axiom. So in fact HAk(X) = Hk(X; A). In other words, HkX = lim πn+k(K(A, n) ∧ X). n→∞ Theorem 33.1. Every homology theory comes from a spectrum.

k n Definition 35 (Cohomology Theory of Spectra). Define E (X,A) = [Σ (X ∪ CA), En+k] for some large n.

68 34 Thom Isomorphism

There is a relative version of the Serre spectral sequence that goes as follows: given a Serre spectral sequence p p F → E −→ B, and subspaces F0 ⊆ F , E0 ⊆ E such that F0 → E0 −→ B is again a Serre fibration, then for each p q p+q −0 subspace B0 ⊆ B, we have a spectral sequence H (B,B0; H (F,F0)) ⇒ H (E,E0 ∪ p (B0)). In particular, p q p+q taking B0 to be a point, we get H (B; H (F,F0)) ⇒ H (E,E0). Note that if π1B does not act trivially, we’ll need to use local coefficients. n Now consider a vector bundle V of dimension n over a space X. Then the pointwise fiber is R . Consider the sphere bundle SV whose pointwise fiber is the one-point compactification, i.e. Sn. Then as we defined before, Thom(X,V ) = SV /X. Now the trivial bundle ∗ → X → X is a subbundle of Sn → SV → X in the sense above, so applying the spectral sequence we obtain Hp(X; Hq(Sn, ∗)) ⇒ Hp+q(Thom(X,V )) (we use local coefficients if applicable). Corollary 13. Hn+p(Thom(X,V )) =∼ Hp(X; Hn(Sn, ∗)). n n Example 24 (An example of Thom isomorphism). Consider mod 2 cohomology. We have H (S ) = Z2, and ∗+n ∼ ∗ Aut(Z2) is trivial, so π1X acts trivially, and thus H (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 element n n V V u ∈ H (Thom(X,V )) having the property that, for each x ∈ X, the image of u in H (Sx ) (where Sx is the fiber over x of SV → X) is a generator. Observation: He ∗(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) g p∗⊗1

H∗(X) ⊗ H∗(SV ,X)

where p∗ is the map induced by p : SV → X, and ∪ is the relative cup product. Note that the presentation of Thom(X,V ) is different from the standard one, where it is considered as B(V )/S(V ). Another way to see the module 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) → He ∗+n(Thom(X,V )). It is possible to prove this using Serre spectral sequences. Here’s another proof sketch:

n 1. First prove it for V = X × R , 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, since ˜ ∗ ∼ ˜ ∗ V H (Thom(X,V )) = H (Sx ) 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 the choice 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, so d 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 isomorphism Hk(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.

m m m • Same setting as above. Consider the embedding M,→ R × N → Thom(N, R ) (where R × N is the m ν trivial bundle of rank m), then Pontryagin-Thom gives a map Thom(N, R ) → M , where ν is of rank m + d. Then we have the following diagram:

k+d+m m k+d+m ν H (Thom(N, R )) He (M )

Hk+d(N) Hk(M)

69 Note that we can obtain this even when f : M → N is not an embedding. This map has a lot of interpretations, 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 Thom isomorphism with some additional constructions.

∞ 0 ∞ Take MO(k) = Thom(Grk(R )). The kth cohomology of this object is Z2 = H (Grk(R )). Now consider u MO(k) −→ 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)?

j Proposition 34. Sq (u) = g(wj, u), which we’ll just write as wju from now on.

j 0 0 Proof Sketch. By Thom isomorphism, Sq (u) = wju for some wj which are the S-W classes of some bundle V 0 ∗ ∗ 0 over X. One then check that they satisfy 1) naturality: for every f : X → Y , we have wj(f V ) = f (wj(V )); 0 ∞ 2. the Cartan formula; and 3. wj(L) 6= 0 where L is the tautological line bundle over R . But these axioms 0 characterize S-W classes, so wj = wj.

70 35 Cohomology of MO and Euler Classes

The big theorem today is the following: u Theorem 35.1. MO(k) −→ 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 do i I on the Thom class: Sq u 7→ wiu, but it is harder to compute general Sq u because we have to analyze the S-W classes. Well, let’s look at the proof. For this kind of statement, here’s a tried-and-true technique: find something that maps into the first space, and prove that the composition mapping is a monomorphism. (We’ve seen this a few times before in this note. – Charles) Definition 36 (Euler Class). Suppose V → X is a vector bundle of dimension n. Consider the Thom complex Thom(X,V ), and there is a zero section X → Thom(X,V ). Then u ∈ Hn(Thom) pulls back to the Euler class n e(V ) ∈ H (X, Z2). Why is it called the Euler class? After all, granted that Euler was a genius, his time was a bit too early for things such as vector bundles and cohomologies. Well, suppose M is a manifold, then e(TM) ∈ Hn and he(TM), [∗]i = 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 the rest 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.

0 0 Corollary 14. If e(V ) 6= 0, then V 6= V ⊕ R for any V .

Proposition 37. e(V ) = wn(V ).

n Proof. Under the Thom isomorphism, e(V ) = u · e(V ) = u · u = Sq (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 ) → X −−−−−−−→zero section Thom(X,V ) is exact.

∞ ∞ ∞ f Example 25. Let L be the tautological bundle over RP , then S(L) = S = ∗. Thus in S(V ) → RP −→ ∞ Thom(RP ,L), the map f is a homotopy equivalence, so it induces isomorphism on all cohomologies, in particular u pulls back to a nontrivial class.

∞ k L1⊕...⊕Lk ∞ Example 26. Consider (RP ) −−−−−−−→ Grk(R ). The second term has Thom complex MO(k), and by ∞ ∧k the example above, the first term has Thom complex (RP ) .

∞ k via BO(k) Now let’s go back to MO(k) → K(Z2, k). Consider the composite map (RP ) −−−−−−−→ MO(k) → K(Z2, k), then ι pulls back to u and then to e(L1 ⊕ ... ⊕ Lk) = x1 . . . xk. Recall the proof of stable cohomology operations being determined by its product-of-one-dimensional behaviors, and by that same argument this induces a monomorphism through a range of dimensions. At this moment let’s revisit something left out a few lectures ago. Recall that πnMO = lim πn+kMO(k). 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

71 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 standard Hurewicz argument. Now a spectra perspective. Recall that we have uk : MO(k) → K(Z2, k), and it is compatible with suspension of the MO, and thus we have a mapping of spectra MO → HZ2. j n+j Definition 37 (Cohomology of Spectra). If E = {En} is a spectrum, then H (E) = lim H (En). n→∞

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 for n > 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 compactible with suspension, it does have Steenrod operations, yet there is no relation between Sq∗(X) and the dimension of X. ∗ I Recall that the cohomology of H (K(Z2, n)) is Z2[Sq ] for admissible I and excess ≤ n. So the cohomology ∗+k I of the HZ2 spectrum, given by lim H (K(Z2, k)), is a Z2 vector space with basis Sq where I is admissible. k→∞ In addition, since we can compose Steenrod operations, it has the structure of an algebra; it is called the Steenrod algebra. As a formal definition, the Steenrod algebra is the algebra over Z2 generated by formal symbols Sq1, Sq2,... modulo the Adem relation. Remark 17. How do we know that Adem relations eventually move any monomials to the reduced form? we n X can assign moments to I. In particular, if I = (i1, . . . , in) then moment(I) = sis. One can check that when s=1 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.

72 36 Describing MO

∞ ∞ Recall the setup: we have Grk(R ) = BO(k), and we write lim Grk(R ) = BO. We also have MO = k→∞ {MO(k)} where MO(k) = Thom(BO(k); Vk). And we are interested in πn(MO) = lim πn+kMO(k). k→∞ ∗ By Thom isomorphism, we know H (MO(k)) = uZ2[w1, . . . , wk]. Now recall the map induced by the Thom u 1 class MO(k) −→ K(Z2, k). Let’s see what is not hit by pulling back along this map: ι goes to u, Sq ι hits 2 3 2 uw1, Sq ι hits uw2, Sq ι hits uw2, but note that we haven’t hit uw1. So there is in fact another thing, i.e. 2 u,uw1 what we really have is MO(k) −−−−→ K(Z2, k) × K(Z2, k + 2). More generally, let HZ2 = {K(Z2, k)} and 2 write X = {K(Z2, k + 2)}. Note that πiX = Z2 where i = 2, and 0 otherwise, so we can write X = Σ HZ2, using the fact that in spectra, πiX = πi+1ΣX. Using this language, we can write the map identified earlier as 2 MO → HZ2 ∨ Σ HZ2. This is again surjective, but we can obtain an isomorphism if we keep going: ∞ ∼ _ s(i) ∼ Theorem 36.1. MO = Σ HZ2 for some {s(i)}i≥0, where by = we mean (mod 2) weak equivalence. i=0 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 generated |ei| A-module through dimension n < 2k. Let {ei} be a (finite) basis, i.e. ei ∈ H (X; Z2). Now consider the map Y ∗ X → K(Z2, |ei|). In cohomology, we know that H (K(Z2, |ei|)) is the module over admissible Is. In the I range ∗ < n there is no product, so it is a vector space with basis Sq ι|ei|, which is just Aι|ei|. Now since the ∗ M 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 algebraic structure that we can play with. Here’s a rough analogy: recall that if H is a subgroup of G, then one can consider the action of H on G (as a set), and then note that this action is free. What we have here, the freeness of H∗(MO) over A, is something in the similar spirit. (The technically accurate statement is that “Steenrod algebra is the endomorphism ring of the Eilenberg-Maclane spectrum HZ2.” Compare this with the statement of 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 we pass to the Thom complex, then we have the mapping MO(k) ∧ MO(l) → MK(k + l). Passing to homology gives H∗MO ⊗ H∗MO → H∗MO, making H∗MO into a commutative ring that is isomorphic to H∗BO via the Thom isomorphism. (The proof is a matter of bookkeeping.) Then we have an induced map, via the UCT, ψ H∗(MO) ⊗ H∗(MO) ←− H∗(MO), making it a coalgebra with a counit.

ψ Definition 38. A coalgebra over k is a vector space M equipped with a coproduct map M −→ M ⊗ M and the counit M −→ k, such that it satisfies the coassociativity: 1 ⊗ ψ ◦ ψ = ψ ⊗ 1 ◦ ψ, and the counital law: 1 ⊗  ◦ ψ =  ⊗ 1 ◦ ψ = id. X Note that Steenrod is also a coalgebra. Recall that we have Sqn = Sqi ⊗ Sqj, so A is in fact a i+j=n co-commutative coalgebra.

ψ Claim 1. The coproduct A −→ A ⊗ A is a ring homomorphism.

This makes the Steenrod algebra a Hopf algebra, which has both an algebra structure and a coalgebra structure. In general, a Hopf algebra can have an action · on a coalgebra, which is an action that is compactible with the coproducts, i.e. the following diagram commutes (where σ2,3 swaps the 2nd and the 3rd factors):

A ⊗ M · M

ψ⊗ψ ψ A ⊗ A ⊗ M ⊗ M M ⊗ M

σ2,3 ·⊗· A ⊗ M ⊗ A ⊗ M

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.

73 Theorem 36.2 (Milnor-Moore). If A is a connected graded Hopf-algebra, acting on a connected graded coalgebra M, 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 . (In the case of Steenrod algebra, ker  is M M M 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 ⊗A M. 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. The claim 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 ψ to check it’s mono, look at the following: A ⊗ M → M −→ 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 induces A ⊗ Fi → M ⊗ Fi. By induction on i, we assume mono for degree up to i − 1. Now consider the LES 0 → 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 is mono; and the first one is mono by assumption.

74 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 for ∗ HZ2. In fact, π∗(MO) = HomA(H (MO), Z2).

X k So how big is πn(MO)? Consider the Poincare series P (MO, t) = dim πk(MO)t ; we also have X k X k k _ |S| P (H∗MO, t) = dim Hk(MO)t = dim H (MO)t . Since MO = Σ HZ2, we know that π∗(MO) has S ∗ ∗ 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 1 P (H∗MO) = . (1 − t)(1 − t2)(1 − t3) ...

On the other hand, a basis of admissible monomoials gives that 1 P (A) = . (1 − t)(1 − t3)(1 − t7) ... (1 − t2n−1)

Thus we know that Y 1 P (π MO) = . ∗ 1 − tn n6=2j −1

Our goal today is the following theorem, which says the structure of π∗(MO) is exactly the one predicted by the Poincare series.

j Theorem 37.1. πn(MO) is Z2[xn | n > 0, n 6= 2 − 1].

Observe that we have π∗(MO) ,→ H∗(MO) = H∗(BO). So what is H∗(BO)? Recall the mapping ∗ ∗ ∞ k Z2[w1, . . . , wk] = H (BO(k)) → H ((RP ) ) = Z[x1, . . . , xk], where xi = wi(Li). In this map, wn 7→ Y X n ∗ σ(x1, . . . , xk) where (1 + txi) = σnt (σn are the symmetric functions). In fact, H (BO(k)) is the i ring of symmetric invariants in Z[x1,...,Zk]. ˜ ∞ ∞ ⊗k ˜ ∞ Corollary 18. Write Hk(RP ) = H∗(RP ) , then Sym(HkCP ) → H∗(BO) is surjective. However, as we look at the Poincare series, we can conclude that it is in fact an isomorphism.

∞ j Proof. Choose bi ∈ H∗RP such that hx , bii = δij (the Kronecker pairing on cohomology, where x is the ∞ generator of H∗RP ), then the former is Z2[b1, b2,...]. Now compare the Poincare series. 0 0 2 Lemma 14. Suppose bi ∈ HiBO are elements satisfying bi = bi mod I where I = (b1, b2,...), then the map 0 0 from Z2[b1, b2,...] to H∗BO is an isomorphism.

Let sn = sn(w1, w2,...) be the symmetric function given by the Newton polynomials. In other words, they n p+1 X p are functions such that sp(σ1, . . . , σn) = (−1) xk, where σi are the elementary symmetric functions. k=1 n Then one can see that sn ∈ H (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.

∞ n Example 28. If L is the tautological bundle over RP , then sn(L) = x . In cohomology, this says that ∗ n n ∞ ∞ sn ∈ H (BO) pulls back to x in H (RP ). Let i : RP → BO, then bn 7→ i∗bn, the latter of which we ∗ n sometimes also write as bn. Then hsn, bn) = hsn, i∗bni = hi sn, bni = hx , bni = 1.

Corollary 19. hsn, bmi = δmn.

Proposition 40. Suppose a ∈ Hk(BO) and b ∈ Hl(MO) such that k, l > 0, k + l = n, then hsn, abi = 0.

75 Proof. hψ(a ⊗ b)i = hψ(sn), a ⊗ bi = hsn ⊗ 1 + 1 ⊗ sn, a ⊗ bi = 0 (consider degree).

0 0 2 0 Theorem 37.2. bn ∈ Hn(BO) satisfies bn = bn mod I iff hsn, bni = 1. 0 0 Corollary 20. {bn} are polynomial generators iff hsn, bni = 1.

Now consider the Hurewicz homomorphism πk(MO) → Hk(MO). Consider a manifold M of dimension k, 17 k and consider [M] representing M in πk(MO) , then it maps to H([M ]) ∈ Hk(MO) = Hk(BO). Theorem 37.3. Given w ∈ H∗(BO), then hw, H([M k])i = hw(ν), [M]i, where ν is the normal bundle of an N embedding M ⊆ R .

We are interested in hsk,H([M])i = hsk(ν), [M]i. But note that ν + TM is the trivial bundle of dimension N, thus sk(ν) + sk(TM) = 0, thus sk(v) = sk(TM) (mod 2); hence we conclude that

n Theorem 37.4. If Mi for i 6= 2 − 1, i > 0 is a sequence of manifolds satisfying hsi(TM), [M]i = 1, then π∗(MO) = Z2[[Mi]]. 0 Proof. We just need to prove that we have a polynomial algebra of the right size. The elements bi = H([Mi]), i 6= n 2 − 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.

n n n So what are the generators, explicitly? Recall that T RP = (n + 1)L, thus sn(T RP ) = (n + 1)x , thus n n hsnT (RP ), [RP ]i = n + 1, so we have half of the generators (if n is even). For a while it was a mystery what the odd generators are. 1 m −1 First construction was given by Dold by considering the Z2 action on S ⊗ CP /{(λ, t) ∼ (λ , t)}. The second one given by Milnor, by considering certain hypersurfaces (called Milnor hypersurfaces) of degree 2, a b X a b denoted by M ⊆ RP × RP . For instance, we can let M be the zero set xiyi. Let i : M,→ RP × RP i≤min(a,b) be the embedding, and ν the normal bundle. The RHS has a line bundle L1 ⊗ L2, let x = w1(L1), y = w1(L2), ∗ ∗ a b then [M] = x + y. Consider TM + i (L1 ⊗ L2) = i T (RP × RP ). Suppose a, b > 1 and a + b = n − 1. Then ∗ n ∗ n n ∗ n n−1 a sn(TM) = i ((x+y) )−i ((a+1)x +(b+1)y ), and hsn(TM), [M]i = hi (x+y) , [M]i = h(x+y) , [RP × n + 1 b]i = (mod 2). If n odd and n 6= 2k − 1, then n + 1 = 2k(2a + 1) = 2k+1q + 2k for some q > 0; RP a a + b write this as a + b, then = 1. Thus we know that we have obtained all the generators. This completes a 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 complex conjugation of the variety, then up to cobordism, the complex variety is the square of the fixed points. For n ∼ n 2 example, CP =cb (RP ) .

17 Careful: [M] is also sometimes used to refer to the fundamental class in HkM.

76