ALGEBRAIC TOPOLOGY: MATH 231BR NOTESweb.stanford.edu/...algebraic-topology-notes.pdf · Low stable...

ALGEBRAIC TOPOLOGY: MATH 231BR NOTES

AARON LANDESMAN

CONTENTS

1. Introduction 42. 1/25/16 52.1. Overview 52.2. Vector Bundles 52.3. Tautological bundles on projective spaces and Grassmannians 72.4. Operations on vector bundles 83. 1/27/16 83.1. Logistics 83.2. Constructions with Vector Bundles 83.3. Grassmannians and the universal bundle 104. 1/29/16 125. 2/1/16 155.1. Characteristic Classes 155.2. Leray-Hirsch Theorem 176. 2/3/16 186.1. Review 187. 2/5/16 218. 2/10/16 248.1. Reviewing Leray-Hirsch 248.2. Review of Chern and Stiefel-Whitney Classes 258.3. Examples and Calculations 269. 2/12/15 289.1. Logistics 289.2. Applications of Stiefel-Whitney classes 289.3. Pontryagin Classes 3110. 2/17/16 3110.1. Theory on Pontryagin classes 3110.2. Calculations and Examples with Pontryagin Classes 3310.3. Euler Classes 3411. 2/19/16 3511.1. Review and Thom Classes 3511.2. Euler Classes 3711.3. Examples of Thom and Euler Classes 3812. 2/22/16 3912.1. More on Euler Classes 3912.2. K Theory 4113. 2/24/16 44

1

2 AARON LANDESMAN

13.1. Examples for K theory 4413.2. Reduced K theory 4513.3. Products 4514. 2/26/16 4814.1. Equivalent definitions of relative K theory 4814.2. Back to Smash Products 4915. 2/29/16 5215.1. Review 5215.2. Fredholm operators and index 5316. 3/2/16 5516.1. More on Fredholm operators 5516.2. The index of a family 5616.3. More examples of Fredholm operators 5817. 3/4/2016 5817.1. Toeplitz Operators 6018. 3/7/2016 6119. 3/9/16 6519.1. Review of infinite dimensional groups 6519.2. Real K theory 6719.3. Symplectic K theory 6820. 3/11/2016 6920.1. Chern character 7020.2. 7321. 3/21/2016 7521.1. Clifford Algebras and Clifford Modules 7622. 3/23/2016 7822.1. How to obtain the Z/2-grading 7922.2. Bundles of Clifford modules 8023. 3/25/16 8123.1. Clifford Modules 8124. 3/28/16 8424.1. 8625. 3/30/16 8825.1. Review 8825.2. Calculating A 8925.3. Relating our computations to A 9126. 4/1/16 9327. 4/4/16 9727.1. Formalities 9727.2. Notations 9827.3. The exact sequence for K-theory in negative degrees 9928. 4/6/16 10128.1. Review 10128.2. Bordism and cobordism 10529. 4/8/2016 10629.1. Oriented cobordism 107

ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 3

30. 4/11/16 10830.1. Review 10831. 4/13/16 11231.1. Oriented bordism as a homology theory 11232. 4/15/16 11532.1. Review 11533. 4/18/16 11933.1. Review 11933.2. The relations between framed cobordism and homotopy groups 11934. 4/20/16 12234.1. Stabilization 12234.2. J-homomorphism 12335. 4/22/16 12635.1. Low stable homotopy groups 12635.2. More about Π3 12735.3. Another integrality property for A 12836. 4/25/16 13036.1. Spectra 13036.2. More involved examples of spectra 13136.3. Fundamental groups of spectra 13337. 4/27/16 13437.1. Review 13437.2. Constructing the long exact sequence for cohomology 13537.3. Examples of cohomology theories for spectra 13637.4. Stable homotopy and stable homotopy 137

4 AARON LANDESMAN

1. INTRODUCTION

Peter Kronheimer taught a course (Math 231br) on algebraic topology and algebraic Ktheory at Harvard in Spring 2016.

These are my “live-TEXed“ notes from the course. Conventions are as follows: Eachlecture gets its own “chapter,” and appears in the table of contents with the date.

Of course, these notes are not a faithful representation of the course, either in the math-ematics itself or in the quotes, jokes, and philosophical musings; in particular, the errorsare my fault. By the same token, any virtues in the notes are to be credited to the lecturerand not the scribe. 1

Thanks to James Tao for taking notes on the days I missed class.Please email suggestions to [email protected].

1This introduction has been adapted from Akhil Matthew’s introduction to his notes, with hispermission.

[email protected]


2. 1/25/16

2.1. Overview. This course will begin with(1) Vector bundles(2) characteristic classes(3) topological K-theory(4) Bott’s periodicity theorem (about the homotopy groups of the orthogonal and uni-

tary groups, or equivalently about classifying vector bundles of large rank onspheres)

Remark 2.1. There are many approaches to Bott periodicity. We give a proof in classfollowing an argument of Atiyah. This introduces index theory for Fredholm operatorsand related things.

Remark 2.2. K-theory is like ordinary homology (originally called an extraordinary ho-mology theory, but with the homology of a point not equal to Z).

The other archetype of a general homology theory like this is cobordism theory.If there’s time, we’ll also talk about generalized homology theories, using stable homo-

topy and spectra.

Some books, useful for this class include(1) Milnor and Stashelf’s Characteristic Classes(2) Atiyah’s K-theory(3) Atiyah’s collected works(4) Hatcher’s Vector bundles and K-theory (online and incomplete)

Remark 2.3. Atiyah’s K-theory addresses his book for someone who hasn’t taken 231a.For example, he proves Brouwer’s fixed point theorem.

logistical information:(1) Information for the CA: Name: Adrian Zahariuc email: [email protected](2) There will be slightly less than weekly homework. Homework 1 is up already.

2.2. Vector Bundles.

Definition 2.4. Let X be a topological space. A real (or complex) vector bundle on X (orover X) is a topological space Ewith a continuous map φ : E→ X and a real (or complex)vector space structure on each fiber Ex := φ−1(x). This must satisfy the additional condi-tion of being locally trivial, meaning that there is an open cover U of X so that for eachU ∈ U with U ⊂ X, the restriction of E to U, notated EU := φ−1(U) ⊂ E is trivial. Here,trivial means there exists a homeomorphism φ

(2.1)

EU U×Rn

U U

φ

φU π

id

or to U×Cn for the complex case, where

φ|Ex : Ex → x×Rn

is linear.

6 AARON LANDESMAN

Example 2.5. (1) The trivial vector bundle E = X × Rn with φ : X × Rn → X theprojection.

(2) The Mobius vector bundle on S1. Take E = I×R and I = [0, 1]. Then, take E = E/ ∼

with ∼ the equivalence relation with (0, t) ∼ (1,−t).(3) The tangent bundle of Sn. Recall

Sn =x ∈ Rn+1 : |x| = 1

.

Define

TSn :=(x, v) : x ∈ Sn, v ∈ Rn+1, x · v = 0

.

We certainly have a projection map TSn → Sn where the fibers are vector spaces(TSn)x ∼= x⊥ ⊂ Rn+1. We shall now check local triviality, which we shall often notdo in the future. Consider

U := x : x ∈ Sn, xn+1 > 0 .

We have an inclusion

TU→ TSn

Over U this is trivial with

TU→ U×Rn

(x, v) 7→ (x,π(v))

with π : Rn+1 → Rn =x ∈ Rn+1 : xn+1 = 0

.

(4) If X ⊂ Rn is a smooth manifold, then

TX = (x, v) ⊂ X×Rn : x ∈ X, v ∈ Rn, v is tangent to X at x .

(5) The normal bundle to X,

ν(X) = (x, v) : v is orthogonal to all vectors tangent to X at x .

Definition 2.6. Recall, a section of E→ X is a map s : X→ Ewhere φ s = idX.

Lemma 2.7. The Mobius bundle is not isomorphic to the trivial vector bundle on the circle.

We give two proofs.

Proof 1. The Mobius bundle is not orientable, but the trivial bundle is, as can be seen bydetermining whether the bundle remains connected or is disconnected, after removingthe image of the 0 section.

Proof 2. The Mobius bundle is not trivial because it has no nonvanishing section. as canbe seen by the intermediate value theorem.

Definition 2.8. Two vector bundles p : E→ X,q : F→ X are isomorphic over X if

(2.2)E F

X

φ

p

q


commutes, where φ is a homeomorphism and

φ|Ex : Ex → Fx

is a linear isomorphism.

2.3. Tautological bundles on projective spaces and Grassmannians. So far we’ve onlylooked at real vector bundles, but we will now consider complex ones.

Definition 2.9. Let CPn, or PnC denotex : x ⊂ Cn+1 : x is a 1 dimensional linear subspace

Definition 2.10. The tautological bundle over CPn be the bundle

L := (x, v) : x ∈ CPn, v ∈ x

where x is thought of simultaneously as a 1-dimensional space in Cn+1 and a point ofCPn. The projection is

p : L→ CPn

(x, v) 7→ x

Exercise 2.11. Define RPn and define the tautological bundle, showing it is a bundle.

Exercise 2.12. Show RP1 ∼= S1. Show the tautological bundle π : L → RP1 ∼= S1 isisomorphic to the Mobius bundle. In particular, it is nontrivial.

Definition 2.13. A line bundle is a vector bundle with 1-dimensional fibers.

Example 2.14. The tangent bundle to S2, TS2 is nontrivial. One can see this by noting thatif one removes the zero section from TS2, one obtains a space which has the homotopytype of RP3, which has fundamental group Z/2Z. In fact, by the Hairy ball theorem, TS2has no nowhere vanishing sections.

Theorem 2.15. The tangent bundle TSn is trivial if and only if n = 0, 1, 3, or 7.

Proof. Nontrivial, we may get to it later using Bott periodicity in the real vector bundlecase.

We next have an extension from projective spaces to Grassmannians.

Definition 2.16. We define the Grassmannian

Gn(RN) =

x : x ⊂ RN, x is a linear subspace with dim x = n

.

Exercise 2.17. Show RPn = G1(Rn+1).

Definition 2.18. We have a tautological bundle over the Grassmannian

φ : En → Gn(RN)

where

En :=(x, v) ∈ Gn(RN)×RN : v ∈ x

8 AARON LANDESMAN

2.4. Operations on vector bundles. We now discuss how to make new vector bundlesfrom old ones.

Definition 2.19. Given

p1 : E1 → X

p2 : E2 → X

we can form their direct sum E = E1 ⊕ E2 defined by a fiber product

(2.3)

E E1

E2 X.

We may note that for all x ∈ X, we have

Ex = (E1)x ⊕ (E2)x

is a family of vector spaces over X.

Definition 2.20. Given two bundles

p1 : E1 → X

p2 : E2 → X

we can form the tensor product bundle E1 ⊗ E2 defined as follows. If we have an openset U ⊂ Xwith E1|U and E2|U both trivial, then writing both as Ei|U ∼= U×Rni , we make

E|U ∼= U× (Rn1 ⊗Rn2)

and then make the topology on E compatible with these identifications.This has the property that the fiber Ex = (E1)x ⊗ (E2)x.

Remark 2.21. Soon, we’ll see a cleaner way to define the tensor product of two bundles.

3. 1/27/16

3.1. Logistics.(1) The first homework is actually up!(2) We can hand in completed problem sets in class.(3) Grading: 60% of the grade will be from the homework and 40% will be from a final

paper.(4) On February 3 and 5, Kronheimer will be away, and we’ll have a guest lecturer.

3.2. Constructions with Vector Bundles. We’ll next look at associated fiber bundles,starting with an example

Definition 3.1. Suppose V is a vector space, we can form its projectivization

PV = (V \ 0)/k×

where k = R or C. Given a vector bundle E → X we have fibers Ex and a projectivebundle P(Ex). As a set, we can take PE = ∪x∈XPEx. Better, we can define

PE = (E \ 0)/k×


where 0 is the image of the zero section in E.

Definition 3.2. Suppose we have a fiber bundle p : P → X with fiber F. We refer to X as abase. Then, p is a fiber bundle if there exists an open cover U so that for all U ∈ U, thereis a homeomorphism φU.

(3.1)

PU U× F

U U

φU

φ|PU π

Remark 3.3. A projective bundle is a fiber bundle with projective fibers.

Example 3.4. Say k = R. Take S(E) := (E \ 0)/R+. Then, F is a bundle with fiber a sphere.

Definition 3.5. An inner product, notated 〈, 〉x is an inner product on each Ex varyingcontinuously with x. In other words, in a local trivialization, where E is a productU×Rn,the matrix entries of the inner product are continuous functions of x.

Lemma 3.6. Inner products exist on E→ X if X admits partitions of unity.

Proof. Omitted, see, for example, math 230a notes.

Theorem 3.7. Partitions of unity exist on X if X is paracompact and Hausdorff. For example,these exist if X is a CW complex.

Proof. Omitted. See, for example, math 230a notes.

Example 3.8. We have the sphere bundle

S(E) ∼= v ∈ E : 〈v, v〉 = 1 .

Definition 3.9. Suppose we have a fiber bundle p : P → X and a topological group Hacting

P×H→ P

so that(1)

(3.2)

P×H P

X X

µ

p

That is, H acts on each fiber Px.(2) On each fiber Px, the group H acts simply transitively.

Then, p : P → X is a principal H bundle

Example 3.10. Suppose we have E→ X a real vector bundle. Let Px be the set of all linearisomorphisms ψ : Rn → Ex. This is is a principal GL(n, R) bundle. The action is given by

(ψ,h) 7→ ψ h

10 AARON LANDESMAN

where h : Rn → Rn and h ∈ GL(n, R). Then, take

P = ∪xPxso that

P = (x,ψ) : x ∈ X,ψ : Ex → Rn .

Remark 3.11. We know that on any connected set the dimension of the fiber is fixed. Ifthe space is disconnected, we will allow the dimensions to change. We will also almostalways assume the fiber is finite dimensional.

Remark 3.12. In local trivializations, the action map µ is

(U×H)×H→ U×H(x,g,h) 7→ (x,gh)

where in the local trivialization P ∼= U×H.

3.3. Grassmannians and the universal bundle. We’ll talk about R for illustration, in-stead of writing k = R or C. Last time we introduced the Grassmannian

Gk(Rn) = all k-dimensional subspace of Rn

= all orthonormal k-tuples /O(k)

The latter description gives the grassmannian the structure of this topological space.

Definition 3.13. Using the inclusions Rn ⊂ Rn+1, we obtain inclusions of Grassmannians

Gk(Rn) ⊂ Gk(Rn+1).

Then, we define Gk(R∞) is their union.

Definition 3.14. A map f : R∞ → Y is continuous if f|Rn is for all n. The same holds forGk(R

∞).

Definition 3.15. We have a tautological vector bundle Ek(Rn) → Gk(Rn) and the limit

forms a tautological vector bundle Ek(R∞)→ Gk(Rn). The fiber over x is xwith x ⊂ Rn

or R∞.

Definition 3.16. For simplification of notation, we use Gk = Gk(R∞). We let p : Ek → Gkdenote the tautological vector bundle.

Definition 3.17. Suppose we have q : F→ Y a vector bundle. Let f : X→ Y. Then, we canpull back F to X. We get E = f∗F, a vector bundle on X defined as the fiber product

(3.3)E F

X Y.

In more concrete terms, E ⊂ X× F is the set

E = (x,w) : f(x) = q(w) .

with a map p : E → X given by projection onto the first factor. We have Ex = p−1(x) ∼=Ff(x).


Lemma 3.18. The pullback constructed above is indeed a vector bundle.

Proof. Check local triviality by using local triviality on Y.

Example 3.19 (Restriction). Suppose we have f : X → Y. Then, f is an inclusion, andpulling back along f is called restriction.

Recall, paracompact means every cover has a locally finite refinement.

Lemma 3.20 (Important Lemma). If f0, f1 : X → Y are two homotopic maps, and if X isparacompact then f∗0(F) and f∗1(F) are isomorphic vector bundles.

Corollary 3.21. If X is contractible and paracompact (for example, if X is a ball) then every vectorbundle on X is trivial.

Proof. Apply Lemma 3.20 taking

f0 : X→ X

x 7→ 0

and

f1 : X→ X

x 7→ x

which are homotopic. But then, vector bundles over a point are all trivial.

Proof of Lemma 3.20. Suppose we have

f∗ : X× I→ Y

where I = [0, 1] be a homotopy from f0 to f1. We get E := (f∗)∗(F) which is a vector bundleon I× X. We now construct a fiber bundle over I× X

P :=(t, x,a) : a : E(0,x)

∼= E(t,x)

.

This is a fiber bundle with fiber GL(n, R).

Lemma 3.22. We claim P → I× X has a section.

Proof. Over 0 × X, there is an obvious section. There is an identity map from the fiberover (0, x) to itself. It is indeed possible to extend this section, using the homotopy liftingproperty (which crucially involves X being paracompact).

Since we have a section, we have a way of translating the identity section over 0 toanother section over 1, which gives the desired linear isomorphism.

That is, the definition of a section is precisely an isomorphism between the fibers over0 and 1 (and in fact for each t between 0 and 1).

Theorem 3.23. ForX paracompact and Hausdorff, a vector bundle onXwith fiber Rk classified upto isomorphism, are in bijection with homotopy classes of maps f : X → Gk. The correspondenceis realized by pulling back the universal bundle along f, sending f : X → Gk to E → X withE := f∗(Ek)

Proof. We’ll see this next class.

12 AARON LANDESMAN

4. 1/29/16

Theorem 4.1. Let X be paracompact.(1) If p : E→ X is a real or complex vector bundle or rank n, then there exists a map

X→ Gn(R∞)

so that E ∼= f∗(En), where En is the tautological bundle on Gn(R∞).(2) If we have f0, f1 : X→ Gn(R

∞) with f∗0(En) ∼= f∗1(En), then f0 ' f1. Further, conversely,if f0 ' f1 are homotopic, then the pullback of En by these two maps are isomorphic, as wesaw last class.

Remark 4.2. The statement is just saying that there is a bijection between

Vectn(X) := n− dimensional bundles on X

and homotopy classes of maps from X to Gn, notated [X,Gn].

Example 4.3. Whenn = 1 takeX = S1, and the real numbers. Then, Vect1(S1) is [S1, RP∞] =π1(RP

∞) ∼= Z/2Z. Indeed,

Vect1(S1) = trivial bundle, Mobius bundle .

Example 4.4. Taken = 1,X = S2 and the complex numbers. So, we’re looking at Vect1(S2).By the theorem, this is homotopy classes of maps

[S2, CP∞] = π2(CP

∞). Now, there is afiber bundle

(4.1)S1 S2n+1

CPn

and taking the limit, as n→∞, we get

(4.2)S1 S∞

CP∞and we can take the long exact sequence for a fibration. Since S∞ has trivial homotopygroups in all dimensions, we have πk(CP∞) = πk−1(S

1). In particular, πk(CP∞) = 0,except when k = 2, in which case π2(CP∞) = Z. So, the line bundles on S2 are isomorphicto Z. From the same argument, we the only complex line bundle on Sn for n > 2 is thetrivial bundle.

Proof of Theorem 4.1. (1) Given a vector bundle p : E→ X, there exists an open cover

U = Uα : α ∈ Iand trivializations

τα : E|Uα → Uα ×Rn.

Note, we can assume I is countable by paracompactness of X, as shown in the oldversion of Hatcher’s algebraic k theory book, Lemma 1.16. So, we can take I = N.


Now, by paracompactness, we can find a partition of unity subordinate to thiscover.

That is,

φα : X→ [0, 1]

where Suppφα ⊂ Uα and∑α∈Nφα = 1, with this sum locally finite.

We define σα by

(4.3)

E|Uα Uα ×Rn

Rn.

σn

We then consider φασα : E|Uα → Rn. Note that this extends continuously to all ofE.

Further, this map is linear on the fibers Ex → Rn, and varying continuously withx. Further, it is an isomorphism Ex → Rn for xwith φα(x) 6= 0. Now, define

Φ : E→ R∞ = ⊕α∈NRn

where

Φ = ⊕αφασα.

This mapΦ is(a) linear on the fibers of E(b) is injective on every fiber.

So, given thisΦ, we can now define a map to Gn(R∞) given by

f : X→ Gn(R∞)

x 7→ Φ(Ex)

Now, f has the property that there is a diagram

(4.4)

E En(R∞)

X Gn(R∞).

Ψ

p pn

f

so that E is the pullback of En(R∞). Now, Ψ is essentially the map Φ with somebookkeeping for the base Gn(R∞). More precisely, recall

En = (y, v) : y ⊂ R∞, y has dimension n, v ∈ yand define, for v ∈ Ex,

Ψ(v) := (Φ(Ex),Φ(v))

(2) We’ll now use the dictionary we established in the first part. Let f0, f1 satisfy

f∗0(En)∼= f∗1(En)

∼= E.

14 AARON LANDESMAN

Then, by the previous dictionary, we have

Φ0,Φ1 : E→ R∞which are linear on fibers and injective on fibers with

fi(x) = Φi(Ex) ⊂ R∞.

We want to find Φt : E → R∞ interpolating between Φ0,Φ1 which is linear onfibers and injective on each fiber. Then, take the homotopy

ft(x) = Φt(Ex).

Let’s start by trying linear interpolation. Suppose for the time being

that for every v ∈ E, we have that Φ0(v) is not a negative multiple ofΦ1(v), with v 6= 0.(4.5)

Then, we can simply take Φt(v) = (1− t)Φ0(v) + tΦ1(v). We can reduce to theabove assumption by the following trick:

Consider the linear map

A : R∞ → R∞(x1, x2, . . .) 7→ (x1, 0, x2, 0, x3, . . .) .

This is continuous because it is continuous on every Rn → R2n. Next, consider

B : R∞ → R∞(x1, . . .) 7→ (0, x1, 0, x2, . . .) .

Now, define

As(x) = (1− s)A(x) + sx.

Define Bs similarly. Then, As and Bs are both injective for all s ∈ [0, 1] .So, we can define a homotopyΦ0 ' Φ0 := A Φ0. These are maps E→ R∞, de-

fined using As, s ∈ [0, 1] . This homotopy is one through maps which are injectiveon the fibers. We can similarly apply this trick toΦ1 given by the homotopy to

Φ1 = B Φ1.Now assuming (4.5), we obtain homotopies

Φ0 ' Φ0 ' Φ1,' Φ1,where the middle equation uses the linear interpolation.

Remark 4.5. If the space was compact, we could actually realize every vector bundle asthe pullback of the tautological bundle on some finite grassmannian.

Remark 4.6. The trick of taking the maps A,B is similarly used in other places, such asshowing that S∞ is contractible.

Our next topic, which we’ll begin on Monday, is characteristic classes. We’ll talk aboutChern classes and Stiefel Whitney classes.


Definition 4.7. A characteristic class for real or complex vector bundles assigns to eachE → X (perhaps of a certain rank) a class c(E) ∈ H∗(X,G), for some group of coefficientsG.

This should be natural with respect to pullbacks, in a sense to be defined next time.

5. 2/1/16

5.1. Characteristic Classes.

Remark 5.1. The following is not a definition, it is just a description!A characteristic class c for real or complex rank r vector bundles assigns to each such

E→ X a class c(E) ∈ H∨(X) or H∗(X;G) satisfying naturality:If

(5.1)

E1 E2

X1 X2f

then

c(E1) = f∗ (c(E2)) .

The Chern classes for complex vector bundles

ci(E) := H2i(X)

for i = 0, 1, 2, . . .. The Stiefel-Whitney class

wi(E) ∈ Hi(X;Z/2)

for real vector bundles.

Remark 5.2. Suppose we have E → X be a rank k complex vector bundle. Then, E =

f∗(Ek) with X f−→ Gk(C∞) with Ek → Gk = Gk(C

∞) the tautological bundle.If c is a characteristic class, there is a universal characteristic class,

cU = c(Ek) ∈ H∗(Gk).

Then, c(E) = f∗(cU). So, a characteristic class corresponds to elements inH∗(Gk(C∞)),H∗(Gk(R∞)).

We’ll construct the Chern classes as characteristic classes of vector bundles, with thefollowing properties, in addition to naturality:

(1) (Whitney Sum Formula) If E = E ′ ⊕ E ′′, then

ci(E) =∑j+k=i

cj(E′)∪ ck(E ′′).

(2) We have c0(E) = 1 ∈ H0(X) for all E→ X. Here, 1means the element which is 1 oneach connected component.

(3) For L→ X a line bundle, c1(L) is defined by saying c1(E1) = −α where E1 → CP∞is the tautological bundle. (Recall H∗(G1(C∞)) = H∗(CP∞) = Z[α],α ∈ H2(CP∞),and we have a standard “preferred” choice of generator.)

16 AARON LANDESMAN

Remark 5.3. It turns out H∗(Gk(C∞)) ∼= Z [α1, . . . ,αk] with αi ∈ H2i(Gk(C∞)) equal to±ci(Ek). However, it does take quite some work to calculate this cohomology ring.

Remark 5.4. If we know X is paracompact, knowing c1(E1) = −α tells us what the Chernclass is. Even if X is more horrible, we can approximate X by a cell complex, meaning thatthere is a map from a cell complex inducing isomorphisms on cohomology and homology.To deal with such X, we usually talk about these approximations by cell complexes.

For Stiefel-Whitney classes we have analogous properties(1)

wi(E) =∑

wj(E′)∪wk(E ′′)

(2) w0(E) = 1(3) w1(E1) = −αR

1 , where we have E1 → RP∞ the tautological bundle (note the − signis extraneous as we’re working in Z/2 coefficients, and αR

1 ∈ H1(RP∞).

Definition 5.5. The total Chern class is

c(E) = c0(E) + c1(E) + · · · .

Remark 5.6. The Whitney sum formulas are usually combined as

c(E) = c0(E) + c1(E) + · · ·c(E) = c(E ′)∪ c(E ′′).

Remark 5.7. Let w1(L) ∈ H1(X;Z/2) for L a real line bundle, be

H1(X;Z/2) = Hom (H1(X), Z/2)= Hom (π1(X);Z/2) .

We can describe this geometrically as follows. Given [γ] ∈ π1(X), we have γ : S1 → X.Look at γ∗(L)→ S1. We assign

[γ] 7→ 0 if γ∗(L) is trivial1 if γ∗(L) is the Mobius bundle.

This ends up determining a map π1(X)→ Z/2, which curves out to be a homomorphismand gives w1(L).

Example 5.8. Suppose we have a line bundle L→M forM a 1 dimensional manifold. Forexample, suppose M = S1. Then, we get L→ S1 by w1(L)[S1] ∈ Z/2. Then, w1(L)[S61] isobtained by counting the zeros of a section s of L:

Choose a section s : S1 → Lwhere w has finitely many zeros. We will have

w1(L)[S1] =

∑x,s(x)=0

multx(s).

The total number of zeros will then always be even for the trivial bundle and odd for theMobius bundle.


Example 5.9. For X a compact, oriented, connected two manifold the fundamental class

[X] ∈ H2(X) ∼= Z,

for L→ X a complex line bundle, we have

c1(L)[X] ∈ Z

is an integer. It can be computed by taking a section

s : X→ L

with finitely many zeros and computing∑x,s(x)=0

multx(s).

5.2. Leray-Hirsch Theorem. We’ll now construct the Chern classes. To do so, we’ll needto delve into an aside, called the Leray-Hirsch Theorem

Let p : P → X be a fiber bundle with fiber F. Then,

H∗(X),H∗(P)

are both rings with a cup product. Then, H∗(P) is a module over H∗(X). The modulestructure of this map is

H∗(X)×H∗(P)→ H∗(P)

(a, c) 7→ p∗(a)∪ c.

Theorem 5.10 (Leray-Hirsch). Suppose Hk(F;R), where R is an arbitrary commutative ring, isa finitely generated, free Rmodule for all k. (For example, being free will automatically be satisfiedwhen R is a field.) Suppose, there exist classes

cj ∈ H∗(P;R)

indexed by j ∈N, whose restrictions to each fiber p−1(x) ∼= F form a basis form a basis for the freemodule H∗(F;R).

Then, H∗(P;R) is a free H∗(X;R) module with basis cj.Equivalently, if there exists cj, then for all c ∈ H∗(P;R), there exist unique aj ∈ H∗(X;R) so

that

c =∑j

p∗(aj)∪ cj.

In particular, p∗ is injective.

Proof.

i

Example 5.11 (Non-example, where the hypothesis fails). Taking S2 → S2 with fibers S1,thenH∗(S1) = Z in dimension 0, 1. The Leray-Hirsch theorem will not help because thereare not elements in the cohomology of S3 restricting to the generator of H1(S1).

18 AARON LANDESMAN

Example 5.12. Take E → X a complex rank n vector bundle over C and take P = PE, sothat O→ X has fiber CPn−1.

There’s a tautological line bundle L → P = PE so that the fiber at (x, `) ∈ P for x ∈X, ` ⊂ Ex is `. Let α = −c1(L) ∈ H2(P;Z).

Then, α|PEx is a generator of H2(CPn−1). The classes 1,α,α2, . . . ,αn−1 ∈ H∗(P) restrictto each fiber p−1(x) to give a basis of the cohomology of p−1(x). The Leray-Hirsch theoremimplies that H∗(P) is a free module over H∗(X).

6. 2/3/16

6.1. Review. Adrian Zahariuc is lecturing today.First, we recall key properties of Stiefel Whitney classes. Let E be a real vector bundle

over a paracompact space Xwith classes

wi(E) ∈ Hi(X;Z/2)

so that(1) wi(f∗E) = f∗wi(E) for f : X→ Y.(2) wi(E1 ⊕ E2) =

∑i+j=kwi(E1)∪wj(E2)

(3) wi(E) = 0 if rkE < 1.(4) w1(U) is a generator of H1(RP∞, Z/2) for U→ RP∞ the tautological bundle.

Next, recall the key properties of Chern classes, which are the complex analog of Stiefel-Whitney classes. Let E be a complex vector bundle over a paracompact X. We then haveci(E) ∈ H2i(X;Z). We then have

(1) ci(f∗E) = f∗ci(E) for f : X→ Y.(2) ci(E1 ⊕ E2) =

∑i+j=k ci(E1)∪ cj(E2)

(3) ci(E) = 0 if rkE < 1.(4) For U → CP∞, then c1(U) is a generator of H2(CP∞;Z) that pairs with the class

[CP1] ⊂ CP∞ to −1. (The pairing is just pull back to CP1.)Let X be paracompact and E a rank n R or C vector bundle over X. Last time, we

introduced PEπ−→ X and U over PE the tautological (line) bundle, over PE.

We already defined c1(L) and w1(L). Let

h = −c1(U) ∈ H2(PE;Z).

Then, 1,h,h2, . . . ,hn−1 restrict on all fibers to a basis of H∗(PE;Z). By the Leray-Hirschtheorem, we obtain the following fact.

Fact 6.1. Hp(PE;Z) is a free module over H∗(X;Z) generated by 1,h,h2, . . . ,hn−1.

Remark 6.2.

Definition 6.3. To know the algebra structure of Hp(PE;Z), we’d only need to know hn.There then exist unique ψi ∈ H2i(X;bz) so that

hn +

n∑i=1

π∗ψihn−i = 0.

We define the Chern classes are ci(E) := ψi.


Remark 6.4. These classes are in fact pullbacks of the classes from the Grassmannian,from naturality.

Lemma 6.5. The Chern class satisfies the following properties.(1) ci(f∗E) = f∗ci(E) for f : X→ Y.(2) ci(E1 ⊕ E2) =

∑i+j=k ci(E1)∪ cj(E2)

(3) ci(E) = 0 if rkE < 1.(4) For U→ CP∞, then c1(U) is a generator ofH2(CP∞;Z) that pairs with the class [CP1] ⊂

CP∞ to −1. (The pairing is just pull back to CP1.)The Stiefel-Whitney class satisfies a similar class of properties as listed at the beginning.

Proof. The first and the last two follow immediately from the definition. We will onlyneed to check the Whitney sum formula. Before proving the Whitney sum formula, wewill need to develop some machinery. It will follow from the Splitting principle.

Lemma 6.6. Any real or complex vector bundle over a paracompact X admits a Riemannian orHermitian metric.

Proof. Use partitions of unity.

Corollary 6.7. Let E be a vector bundle over X and E ′ ⊂ E a subbundle. Then, there existsanother subbundle E ′′ ⊂ E such that E = E ′ ⊕ E ′′. That is, all short exact sequences of vectorbundles split.

Proof. Pick the orthogonal complement under the metric given by the above lemma.

Theorem 6.8. Let X be paracompact, and E a real or complex n bundle. Then, there exists a spaceΣ and a map q : Σ→ X so that

(1) q∗E is a direct sum of line bundles.(2) The map

q∗ : H∗(X;Z)→ H∗(Σ;Z)

is injective or

q∗ : H∗(X;Z/2)→ H∗(Σ;Z/2)

Proof. We will use induction on the rank. By induction, it’s enough to split off one linebundle. Take Σ = PE. Take q = π. Then, π∗E. So, we have an injective map

U → π∗E

where U is the tautological line bundle on PE. By Corollary 6.7, we have

π∗E = U⊕ V

where V is a bundle of lower rank. Then, Leray Hirsch implies q∗ is injective.

Remark 6.9. The difference between algebraic topology and algebraic geometry here isthat in algebraic topology, short exact sequences always split, while in algebraic geometrythey don’t, and so you can “pretend” they split, but really you’re just relating the first andlast terms of a short exact sequence to the middle term.

20 AARON LANDESMAN

To prove the Whitney sum formula, it suffices to show that if E = ⊕ni=1Li, for Li linebundles, then

c(E) =

n∏i=1

(1+ c1(Li)) .

Lemma 6.10. If L1 and L2 are line bundles over X, then c (L1 ⊗ L2) = c1(L1) + c1(L2).

Proof. There are maps fi : X→ CP∞ with f∗iU = Li. So, we can look at

(f1, f2) : X→ CP∞ ×CP∞.

Without loss of generality, we can assume X = CP∞ × CP∞. Then, Li = π∗iU. So, it’senough to prove this on the two skeleton. The two skeleton of CP∞×CP∞ is just twospheres glued at a point. The two line bundles are tautological one sphere and trivial onthe other sphere, and we can check the Chern class is the sum.

Exercise 6.11. Show the Chern class is indeed the sum of the two Chern classes of tauto-logical line bundles.

Lemma 6.12. Suppose L1, . . . ,Ln are line bundles over X. Suppose si is a section of Li vanishingat Zi ⊂ X.

Assume ∩iZi = ∅. Then,n∏i=1

c1(Li) = 0.

Proof. Observe si|X\Zi trivializes Li|X\Zi . That is,

c1(Li)|X\zi = 0.

We have a long exact sequences

(6.1) · · · H2(X,X \Zi) H2(X) H2(X \Zi) · · ·

We know c1(Li) maps to 0 under the second map. Therefore, there is some ξ ∈ H2(X,X \

Zi) mapping to c1(Li).

Remark 6.13. Recall that for X a topological space and A,B ⊂ X two subsets, we have amap

Hk(X,A;R)×Hl(X,B;R) ∪−→ Hk+l(X,A∪ B;R).

given by cup product.

Then, the product

ξi ∪ · · · ∪ ξn ∈ H2n(X,∪ni=1 (X \Zi)) = H2n(X,X) = 0.

Therefore, the product of these classes is 0, as claimed.


We now want to see

c (⊕ni=1Li) =n∏i=1

c(Li).

We have E = ⊕ni=1Li. We have the projectivization

π : PE→ X

and we have

U → π∗E = ⊕ni=1π∗Li.We now define the composition

πi : U→ ⊕ni=1π∗Li → π∗Li.

We can think of

πi ∈ Hom(U,π∗Li).

This section πi vanishes on L1⊕ · · · ⊕ Li⊕ · · · ⊕ Ln. These such loci intersect trivially. So,

0 = πc1

(U∨ ⊗ π∗Li

)= π (−c1(U) + π

∗c1(Li))

=

n∏i=1

(h+ π∗c1(Li)) .

When we expand the elementary functions in this product and get the Chern classes, sincethey are also just the elementary symmetric functions.

7. 2/5/16

Cliff Taubes is lecturing today. It may be a total non sequitur for the course.Today we’ll give some explicit examples of vector bundles and compute some Chern

classes using differential geometry techniques.

Example 7.1. Let E→ S1 be the Mobius bundle over the circle. The fibers are R. Think of

E =

(θ, v) ⊂ S1 ×R2 :

(cos θ sin θsin θ − cos θ

)v = v

.

This matrix has eigenvalues ±1. We’re picking out the +1 eigenspace.

Question 7.2. Why is this the Mobius bundle?

We can write this as a product bundle. We start with the circle S1.The condition above enforces

sin θv1 = (1+ cos θ)v2sin θv2 = (1− cos θ)v1.

We now have trivializations on the top and bottom halves of the circles, and local sectionsgiven by (

1sin θ1+cos θ

),( sin θ1−cos θ1

).

22 AARON LANDESMAN

Indeed, these two local sections agree on the overlaps. We can notate this combinedsection as (

1+ cos θsin θ

)which has a zero. This is nontrivial because it has no nonvanishing section.

Example 7.3. Now, we’ll look at a complex line bundle over S2, with fiber C. We’ll lookat E→ S2. Define

P :=

(iz ix− y

ix+ y −iz

).

Note that Pt = −P, so it is Hermitian. The trace is 0 and determinant is 1, so the eigenval-ues are ±i.

E =(~p, v) ⊂ S2 ×C2 : Pv = iv

.

where here ~p = (x,y, z) is a point on S2 ⊂ R3 with norm 1.

Example 7.4. Define

E∗ := ~p, v ⊂ S2 ×C3 : ~p× v = iv, where × refers to the cross product.

~p× (~p×~v) = −(|~p|2~v− ~p(~p ·~v)

).

We have

E∗ ⊕ E∗ = TS2 ⊗R C.

Example 7.5. We’ll now generalize our first example E. Start with a Riemann surface X(embedded in some real vector space Rn). Choose a fixed point q not on X. Consider thefunction

p 7→ p− q

p− q=Mp.

Of course,Mp implicitly depends on our fixed choice of q. We can construct

EX =(p, v) ⊂ X×C2 :Mpv = iv

.

Defining the function

fq : X→ S2

p 7→ p− q

|p− q|.

We now can realize EX as the pullback

(7.1)

EX E

X S2fq


That is, EX = f∗E.If q lies outside X, then the bundle will be trivial. If we vary q slightly, the bundle will

be isomorphic. In other words, the isomorphism class of the bundle is locally constant asa function of q. Then, as we move q very far away, we’re pulling the bundle back by amap to the two sphere where the image lies in an extremely small patch of S2, on whichthe bundle E on S2 is trivial.

Remark 7.6. If q lies inside the Riemann surface X, the bundle EX will not, in general, betrivial.

Example 7.7. Suppose we have a surface X ⊂ R3. Then, at each point on the surface, wecan define a normal direction ~np, written ~bn = (n1,n2,n3). We have a matrix satisfying(

in3 in1 −n2in1 +n2 −in3

)· v = iv.

We can define a bundle

E =(p, v) ⊂ X×C2 : ~np × v = iv

.

Fact 7.8. Suppose X is a closed 2-dimensional real oriented manifold and E → X is acomplex 1 dimensional vector bundle, then E is isomorphic to f∗F where F is the firstbundle on S2 we defined in this class.

Now, we’ll give some more examples of bundles over S2.

Example 7.9. Fix a positive integer n. Consider

H0(P1,OP1(n)),

defined to be the vector space of homogeneous complex polynomials in two variables ofdegree n.

Consider the operators

Lz := i

(u∂

∂u− v

∂

∂v

)Lx := i

(v∂

∂u+ u

∂

∂v

)Ly :=

(u∂

∂v− v

∂

∂u

),

which gives an action of the Lie algebra of SO(3) on this space of polynomials.Next, define

Lp := xLx + yLy + zLz,

and define the bundle En → S2 to be

En :=(p, v) ∈ S2 ×H0(P1,OP1(n) : Lpv = inv

.

24 AARON LANDESMAN

For example, let’s look at the case n = 1. Recall S2 ∼= CP1, and here a point (u, v) in CP1

corresponds to

x = 2re(uv)/(|u|2 + |v|2),

y = 2im(uv)/(|u|2 + |v|2),

z =|u|2 − |v|2

|u|2 + |v|2.

We have that the bundle E1 constructed above is the tautological bundle.

8. 2/10/16

8.1. Reviewing Leray-Hirsch. Let us start by recalling the Leray-Hirsch theorem:

Theorem 8.1. Suppose we have p : P → X a fiber bundle. Let cj ∈ Hδ(j)(P;R). Suppose that therestrictions of these classes to each fiber form a basis for H∗(Px;R), the cohomology of each fiber(meaning really that their pullbacks to the cohomology on each fiber for a basis for the cohomologyof that fiber). Here, Px = p−1(x) for all x.

Then, there’s an isomorphism

Dk(X)Ψ−→X E

k(X),

where for U ⊂ X,

Ek(U) := Hk(PU;R)

Dk(U) := ⊕jHk−δ(j)(u;R).

Here, PU = p−1(U). The map is given by

ΨU : Dk(U)→ Ek(U)

(a1,a2, . . .) 7→∑j

p∗(aj)∪ cj.

Proof. We’ll prove this for finite simplicial complexes. We have both E∗(U) and D∗(U)behave as follows:

(1) if U ι−→ V , there’s a map

ι∗ : E∗(V)→ E∗(U)

(2) the previous property is functorial(3) There is a Mayer Vietoris sequence as follows: For U,V in X, we have

(8.1)

0 Dk(U∪ V) Dk(U)⊕Dk(V) Dk(U∩ V Dk+1(U∪ V) 0

0 Ek(U∪ V) Ek(U)⊕ Ek(V) Ek(U∩ V) Ek+1(U∪ V) 0

Ψ Ψ Ψ Ψ


Now, if ΨU,ΨV and ΨU∩V are isomorphisms, the so is ΨU∪V . So, Ψ∆ is an isomorphism for∆ a simplex (isomorphic to a point). One can then add cells one at a time to obtain theproof for arbitrary simplicial complexes. One can then pass to arbitrary CW complexes,and finally arbitrary topological spaces.

To pass to infinite dimensional things, one can note that the nth homology only de-pends on the n+ 1 skeleton. It’s also true for cohomology by universal coefficients.

Then, to pass from cell complexes to general manifolds, we’ll need to perform anotherstep, which we omit.

8.2. Review of Chern and Stiefel-Whitney Classes. Adrian defined the Chern classes ina previous class. Let’s now discuss their real analog, Stiefel-Whitney Classes.

For p : E→ X, a real vector bundle, we’ll define

wi(E) ∈ Hi(X;Z/2).

Remark 8.2. Recall, we previously defined w1 as follows. Let E be a line bundle and let

f : X→ RP∞so that

f∗(Ltaut) = E

h ∈ H1(RP∞, Z/2) is the generator .

Then, we defined

wold1 (E) = f∗(h).

It won’t be immediate that our new definition is the same, although it isn’t too hard toshow they are the same.

Definition 8.3. Form

p : PE→ X

with fiber RPn−1, with n = rkE. There’s a tautological line bundle

Ltaut → PE.

Fiber by fiber, this is just OPEx(1) on the projective space fiber PEx. We have a class

h = woldi (Ltaut) ∈ Hi(PE;Z/2).

Then, we have classes

h0,h,h2, . . . ,hn−1,

which restrict to the fibers to give a basis of H∗ with Z/2 coefficients. Looking at hn ∈Hn(PE;Z/2), we know this is 0 on each fiber, although it need not be zero on the totalspace. By Leray-Hirsch, there exist unique classes

wi(E) ∈ Hi(X;Z/2)

so that

hn + p∗(w1(E))hn−1 + · · ·+ p∗(wn−1(E)h+ p∗(wn(E)) = 0.

Exercise 8.4. Check

26 AARON LANDESMAN

(1) w1 = wold1 for line bundles(2) Check naturality of wi.(3) Check the Whitney sum formula holds. That is, w(E1 ⊕ E2 = w(E1)w(E2) where

w = 1+w1 + · · · .Hint: The proof is the same as for Chern classes. The hardest part by far is the Whitneysum formula, which can be deduced from the splitting principle. This reduces to the casethat E1 and E2 are both sums of line bundles.

Lemma 8.5. Given X and E→ X there exists some

p : Z→ X

so that both(1) H∗(X;Z/2) → H∗(Z;Z/w)(2) p∗(E) decomposes as a direct sum of real line bundles.

Proof. Omitted. This was probably proven as the splitting principal by Adrian last week.

8.3. Examples and Calculations. Let’s start by examining how Chern classes behavewith tensor products. In fact, we saw this last week with Adrian.

Lemma 8.6. Let E1 → X and E2 → X be complex line bundles. Then,

c1(E1 ⊗ E2) = c1(E1) + c2(E2).

Proof. We have E1 = f∗1(Ltaut),E2 = f∗2(L

taut) where Ltaut → CP∞ is the tautologicalbundle. Then, take

f = (f1, f2) : X→ CP∞ ×CP∞.

Then, the universal case is L1 and L2 tautological on CP∞ × CP∞. One then must do acalculation just on the 2 skeleton, which is where the first cohomology appears to checkthis holds there. That is, we have two cells CP1 ×CP0 and CP0 ×CP1, which meet at thesingle points CP0 ×CP0. The second cohomology is Z⊕Z.

Corollary 8.7. We have

c1(E∨) = −c1(E)

for E a line bundle. Here, E∨ = Hom(E, C) where C is the trivial bundle.

Proof. Take E1 = E,E2 = E∨ in Lemma 8.6, using that E⊗ E∨ is trivial, as it has a nowherezero section corresponding to the identity in End(E) ∼= E⊗ E∨.

Corollary 8.8. For E of any rank,

ci(E∨) = (−1)ici(E).

Proof. Use the splitting principal. Let’s check when E = L1 ⊕ · · · ⊕ Ln. Then,

ci(E) = σi(c1(L1), . . . , c1(Ln)),

by the Whitney sum formula. That is,

c(E) =

n∏n=1

(1+ c1(Lj)

).


Then,

ci(E∨) = σi

(c1(L

∨1 ), . . . , c1(L

∨n ))

,

implying the result.

Question 8.9. Can we say anything about the Chern classes of E1⊗E2 for general complexvector bundles?

We can use the splitting principle with bare hands, although this gets very messy veryquickly.

Example 8.10. Let’s suppose E1 and E2 are rank 2 vector bundles. Suppose further thatc1(Ei) = 0.

Then, using the splitting principle, we may as well assume

E1 = L1 ⊕ L ′1` = c1(L1)

` ′ = c1(L′1).

Then, c1(E1) = `+ ` ′ = 0, so ` ′ = −`. Then,

c(E1) = (1+ `)(1+ ` ′) = 1− `2.

Similarly, we may write

c(E2) = 1−m2.

Then,

c2(E1) = −`2.

Then,

E1 ⊗ E2 ∼= L1 ⊗ L2 ⊕ L ′1 ⊗ L2 ⊕ L1 ⊗ L ′2 ⊕ L ′1 ⊗ L ′2.

Then,

c(E1 ⊗ E2) = (1+ `+m)(1+ `−m)(1− `+m)(1− `−m)

= 1− 2`2 − 2m2 + · · · .

So, for example,

c2(E1 ⊗ E2) = 2c2(E1) + 2c2(E2).

Example 8.11. Let’s compute Stiefel-Whitney classes of TRPn−1. For V ⊂ Rn of dimen-sion k, we have

TvGk(Rn) = Hom(V ,V⊥).

This, in fact, defines a chart for the Grassmannian of subspaces not intersecting V⊥ awayfrom 0.

In the case k = 1, we get

TRPn−1 = Hom(L,L⊥)

28 AARON LANDESMAN

where L = Ltaut is the tautological bundle and L⊥ ⊕ L ∼= Rn. We then have

Hom(L, Rn) ∼= Hom(L,L)⊕Hom(L,L⊥)∼= R⊕Hom(L,L⊥)∼= R⊕ TRPn−1

Therefore,

w(TRPn−1) = w(Hom(L, Rn))

= w(⊕nL∨)

=(w(L∨)

)n= (1+ h)n

with

h ∈ H1(RPn−1;Z/2).

Then,

wk(TRPn−1) =

(n

k

)hk.

9. 2/12/15

9.1. Logistics. Problem set 2 is available now. It is due on 2/19/2016. Future homeworkshould be submitted on canvas, uploaded.

9.2. Applications of Stiefel-Whitney classes. Last time, we sketched a proof thatwk(TRPn−1) =(nk

)hk ∈ Hk(RPn−1;Z/2) ∼= Z/2.

Example 9.1. The space RP4 does not immerse in R5. That is, there is no function f :RP4 → R5. Recall an immersion is a map with dfx is injective for all x ∈ RP4.

Recall we have NxM⊕ TxM ∼= Rn and so TM⊕NM = Rn

To see this, if f exists, there would exist a real line bundleN→ RP4 so that TRP4⊕N ∼=R5. But then,

w(TRP4)w(N) = w(R5) = 1.

We have w(TRP4) = 1+w1 +w2 +w3 +w4)(1+ u) = 1, where w1(N) = 1+ u. Then, wehave

1 = (1+ h+ 0+ 0+ h4)(1+ u)

since we are using Z/2 coefficients. Solving these equations, we have 1 · 1 = 1. In degree1, we have 1 · u+ h · 1 = 0, so u = −h = h. We next get hu = 0, and so h2 = 0. This is acontradiction because h2 6= 0 in the ring H∗(RP4;Z/2) and h is the generator of the firstcohomology, so h2 is nonzero.

The above has an analog over complex vector spaces and Chern classes.Take W to be a complex vector space. There is then an underlying real vector space by

restricting scalars, of dimension twice that of the complex vector space. That is,

dimRWR = 2dimCW.


Definition 9.2. In the other direction, if we are given V , a 2n dimensional real vectorspace, a complex structure on V is a complex vector spaceW withWR = V .

Remark 9.3. Equivalently, a complex structure is a map

J : V → V

is a real linear transformation with J2 = −1. Here,

(a+ ib)w = aw+ b(Jw)

where w ∈W.

Remark 9.4. We can similarly define a complex structure for real vector bundles.

Definition 9.5. A almost complex structure on a 2n dimensional manifoldM is a complexstructure J on TM.

Take [V] ∈ Gk(Cn). One can take a neighborhood of V in Gk(Cn), given by all vectorspaces intersecting V⊥ trivially. In other words, such a neighborhood is given by

Hom(V ,V⊥) → Gk(Cn)

f 7→ graph(f).

If [V] ∈ Gr, we can identify

T[V ](Grk(Cn)) ∼= Hom(V ,V⊥)R.

Then,

TGrk(Cn) = Hom(E,E⊥)R

where E is the tautological bundle, and E⊕ E⊥ = Cn.Suppose we are given an almost complex structure for TGrk(Cn) = Hom(V ,V⊥). Such

a bundle has is a complex vector bundle and has Chern classes.

Example 9.6. Let’s now compute ck(TCPn−1. We obtain

ck(TCPn−1) =

(n

k

)hk

with h ∈ H2(CPn−1;Z),hk ∈ H2k(CPn−1;Z) generators of the cohomology and k ≤ n− 1.The key difference is that here they appear with coefficients in Z instead of Z/2.

Example 9.7 (Complex surfaces in CP3). This example uses standard terminology fromalgebraic geometry. Feel free to skip it if you’re not familiar with the terminology.

TakeH→ CP3. The dual of the tautological bundle is c1(H) = h ∈ H2(CP3, Z). We takeH⊗H = H⊗2.

Here, we are confusing the divisor H with the line bundle corresponding to H. That is,the line bundle is OCP3(H).

Warning 9.8. In the following we will sometimes notate this just as H.

30 AARON LANDESMAN

Section of H are from linear maps a : Cn → C. Restricting a to L ⊂ Cn, we get a map

L→ C

where L is the tautological line bundle.That is, a ∈ Hom(L, C) and a is a section of L∨ = OCP3(H). If a is homogeneous of

degree d on Cn, we obtain that a ∈ Γ(OCP3(H⊗d)).

Given any section s ∈ Γ(OP3(H⊗d)), with the property that for every x ∈ X := V(s), we

have

ds|x : TxCP3 → OCP3(H

⊗d),

a complex linear nonzero map.This means every point x ∈ X is smooth in X, and that TxX is a complex C-linear sub-

space of TxCP3, because the tangent space at x is just the kernel of the linear map ds|x. Allthis amounts to is saying that this is a smooth almost complex manifold.

As an interesting special case, we could just take a smooth homogeneous polynomialof degree d. This would be a smooth complex surface in CP3.

We will now compute the Chern classes ci(X).

Remark 9.9. Recall that whenM is a manifold, we use the notation

wi(M) := wi(TM)

ci(M) := ci(TM),

the second case making sense only whenM is almost complex.

Now,

c(X)c(NX) = c(TCP3|X).

Let’s call c(X) = 1+ c1+ c2. There is no third Chern class as X is a 2 dimensional complexmanifold. We know c(NX) = c(H⊗d) because the derivative of s gives a map

TxCP3 ds−→ (H⊗d)s.

The kernel of this map is the tangent bundle, and so we have

TxCP3/TxX ∼= H⊗dx

and the left hand side is the normal bundle. Finally,

c(TCP3|X) = 1+ 4h+ 6h2 + 4h3).

Additionally, c(H⊗d) = (1+ dH). So, we have

(1+ c1 + c2)(1+ dh) = (1+ 4h+ 6h2 + 4h3)|X.

Writing down equations on the generators, we get

1 · 1 = 1c1 + dh = 4h

c2 + c1(dh|X) = 6h2|X.


Solving in the above, we get

c1(X) = (4− d)h|X

c2(X) = (6− (4− d)d)h|X.

Remark 9.10. The number (4− d) is quite interesting. In algebraic geometry, there is adistinction as to when this is positive, negative, or 0. For example, if we did this CP2, wewould have the first Chern class given by

c1 = (3− d)h.

This then tells us a curve of degree 1 and 2 is topologically a Riemann sphere if it issmooth. In degree 3, it will be a torus (of genus 1). In degree 4 it is a genus 3 surface, andin general, it has degree

(g−12

).

The curves of degree 1 and 2 are called Fano, the curves of degree 3 are called Calabi-Yau, and the curves of degree more than 3 are called general type.

9.3. Pontryagin Classes. For a real vector bundle V → X, we have a Pontryagin class

pi(V) ∈ H4i(X).Definition 9.11. We now describe the complexification construction.

Given V a real vector bundle, we have a complex vector bundle V ⊗R C. This is acomplex vector bundle satisfying

i(v⊗ λ) = v⊗ (iλ).

Define pi(V) = (−1)ic2i(V ⊗C).

Remark 9.12 (Reason for only considering even Pontryagin classes). As a complex vectorbundle

W := V ⊗C

has the property that W ∼= W, the complex conjugate, meaning that for j odd, we have(−1)jcj(W) = cj(W). In other words, 2cj(V ⊗C) = 0 for j odd. That is, it is 2 torsion. So,we usually only look at the even classes as the Pontryagin classes.

10. 2/17/16

10.1. Theory on Pontryagin classes.

Lemma 10.1. Recall that for V → X a real vector bundle we have V ⊗C a complex vector bundle.We then have V ⊗C ∼= V ⊗C, an isomorphism between a bundle and its conjugate.This implies that c2i+1(V ⊗C) = −c2i+1(V ⊗C).

Proof. This uses that L⊗ L ∼= C. Then, use the splitting principle, since W = ⊕Li, impliesck(w) = σk (c1(L1), . . .), where σk is the kth elementary symmetric function.

Recall that pk(V) := (−1)kc2k(V ⊗ C). are the Pontryagin classes for 1 ≤ k ≤ rkV/2,and p(V) = 1+ p1(V) + · · · .Lemma 10.2. For V1,V2 two real vector bundles, we have

p(V) = p(V1) · p(V2) +αwhere 2α = 0.

32 AARON LANDESMAN

Proof. We have

pk (V1 ⊕ V2) = (−1)k∑

n+m=2k

cn(V1 ⊗C)∪ cm(V2 ⊗C)

= α+ (−1)k∑

n,m≡0 mod 2

cn(V1 ⊗C)∪ cm(V2 ⊗C)

=∑i+j=k

(−1)ic2i (V1 ⊗C) (−1)jc2j(V2 ⊗C)

= α+∑i,j

pi(V1)pj(V2)

where α is some sum of elements of order 2.

Example 10.3. Take X to be(1) CP2

(2) CPn

(3) a hypersurface in CPn.Then, consider pk(TX), which are viewed as real vector bundles by restriction of scalarsof the complex vector bundleW = TX.

Question 10.4. Are the Pontryagin classes related to the Chern classes?

Lemma 10.5. Let W by a complex vector bundle and WR be the restriction of scalars of W to R.Then,

pk(WR) = (−1)k c2k(WR ⊗C).

Warning 10.6. Note that W 6∼= WR ⊗R C. In fact, the latter has rank twice as large as thatof the former.

Proof. Omitted.

Lemma 10.7. ForW a complex vector bundle, we have

WR ⊗R C ∼=W ⊕W.

Proof. We have multiplication by i corresponding to the linear transformation J

i :W →W

w 7→ iw

and

J :WR →WR

w 7→ iw.

We have a corresponding map

J⊗ 1 :WR ⊗R C→WR ⊗R C

w⊗ λ 7→ Jw⊗ λ.

This is complex linear with (J⊗ 1)2 = −1. We then have

WR ⊗C =W+ ⊕W−


are the ±i eigenspaces of J⊗ 1. Here, we’re doing linear algebra fiber by fiber. We thenobtain isomorphisms

W →W+

w 7→ w⊗ 1− (Jw)⊗ iand we similarly obtain

W →W−.

To see the maps are isomorphisms, we’ll just check the first one, defining the isomorphismtoW+. This indeed defines a mapW →WR ⊗C.

Exercise 10.8. Show this is a complex linear map. The former has a complex structurebecause we have multiplication by i. On the right hand side, we have tensored over C.

Exercise 10.9. Check that the image of this map lands on the +i eigenspace. Hint: Thesolution is essentially that

J⊗ 1 (w⊗ 1− (Jw)⊗ i) = Jw⊗ 1−(J2w

)⊗ i

= w⊗ i+ Jw⊗ 1= i(w⊗ 1− Jw⊗ i).

10.2. Calculations and Examples with Pontryagin Classes.

Example 10.10. Take X = CP2. What is p1(TX)? This is 4 real dimensional. We have

p1(TRX) = −c2(TRX⊗C)

= −c2(TX⊕ TX)= −c2(TX) − c2(TX) − c1(TX)c1(TX)

= −2c2(TX) + c21(TX).

where in the second line, this is the tangent bundle as a complex vector bundle.Now, we know the Chern classes of tangent bundles. For ci = ci(CP2),

c2 = 3h2

c1 = 3h

we have the resulting Pontryagin class is

p1(TX) = −2(3h2) + 9h2 = 3h2.

Definition 10.11. For X a closed oriented manifold of dimension n, with n = 4k, we havea pairing

pk(TX)[X] ∈ Z

with pk(TX) ∈ H4k, [X] ∈ H4k. We abbreviate this integer by pk[X].More generally, we denote

P(p1, . . . ,pr)[X]

34 AARON LANDESMAN

to be the pairing P(p1(TX), . . . ,pr(TX))[X] ∈ Z, where P is some polynomial of degree4k ∈ H∗.Example 10.12. We have, for example, p1[CP2] = 3.

Here is a fact, which is not too deep.

Fact 10.13. If X is a closed oriented smooth manifold of dimension 4, then p1[X] is alwaysdivisible by 3.

Here is a deeper fact.

Fact 10.14 (Rowin’s Theorem). If X is closed oriented smooth of dimension 4, and, inaddition, we have

w1(TX) = 0

w2(TX) = 0,

then p1[X] is divisible by 48.Note that w1(TX) = 0 follows from orientability, so w2(TX) = 0 is the main condition.

Example 10.15. Take X ⊂ CP3 a smooth quartic surface. Then, p1[X] = −48. In fact, wenow have the tools to compute this, using analogous methods to Example 10.10

10.3. Euler Classes. So far, we have wi, ci,pi. These have the properties

wi(V ⊗R) = wi(V)

pi(V ⊗R) = pi(V)

ci(W ⊗R) = ci(W).

We now define the Euler class. For an oriented real vector bundle V → X of rank r, wehave

c(V) ∈ Hr(X).

Remark 10.16. Note that

e(V ⊕R) = 0 ∈ Hr+1(X).For V a vector space isomorphic to Rr, we define

V0 = V \ 0.

Then,

Hr(V ,V0) = Z

and

Hs(V ,V0) = 0,

for s 6= r.Note

(D(V),S(V)) ∼=(V ,V0

)where D is the disk and S is the sphere. Take p : V → X be the vector bundle of rank rand let Vx = p−1(x).


Definition 10.17. An orientation for V → X is a choice of generator

(10.1) ux ∈ Hr(Vx,V0x) ∼= Z

satisfying coherence, meaning for all x ∈ X, there is some neighborhood U and a classuU ∈ Hr(VU,V0U) so that

uU|Vy,V0y= uy

for all y ∈ U.

Definition 10.18. A Thom class for V → X a vector bundle with an orientation is a class

u ∈ Hr(V ,V0)

such that u|Vx,V0x= ux holds for all x ∈ X.

Remark 10.19. Equivalently, we can write u ∈ Hr(D(V),S(V)), by some homotopiesrescaling the norm.

Remark 10.20 (Joke). The Thom class is essentially “one class to rule them all, and in thedarkness restrict them.”

Theorem 10.21. (1) Suppose V → X is an oriented vector bundle, for X paracompact. Then,a Thom class a exists.

(2) The Thom class is unique(3) When the Thom class u exists, there is an isomorphism

Hk(X)→ Hr+k(V ,V0)

a 7→ p∗(a)∪w.

for all k ∈ Z (even for negative k).

Proof. (1) Next time.(2) Next time.(3) Use the Leray-Hirsch theorem in the case of one generator. This isomorphism is

precisely that prescribed by Leray-Hirsch.

11. 2/19/16

11.1. Review and Thom Classes. Suppose we have p : V → X a rank n fiber bundle.Define V0 = V \ 0. If there exists ux ∈ Hn(Vx,V0x) for all x which is a generator and iscoherent, then this is an orientation for V . Then, a Thom class is an element u ∈ Hn(V ,V0)so that the restriction of u is ux.

The first main theorem about Thom classes is that they exist.

Theorem 11.1. (1) Suppose V → X is an oriented vector bundle, for X paracompact. Then, aThom class a exists.

(2) The Thom class is unique(3) When the Thom class u exists, there is an isomorphism

Hk(X)→ Hr+k(V ,V0)

a 7→ p∗(a)∪w.

for all k ∈ Z (even for negative k).

36 AARON LANDESMAN

(4) We have

Hn−1(VU,V0U) = 0.

Proof. Recall from last time that the third part regarding the isomorphism follows fromLeray-Hirsch.

This also implies that the Thom class is unique. By the isomorphism, we have an iso-morphism

H0(U)→ Hn(VU,V0U).

If uU,u ′U ∈ Hn(VU,V0U) are two sections and uU is a Thom class, then ′U = p∗(a)uU ∈H0(U). Here, a is a locally constant Z valued function on U. But, the assumption thatu ′U|x = uU|x = ux, we obtain a(x) = 1 for all x. So, there is a unique Thom class.

In particular, if uU exists, then

Hn−1(VU,V0U) = 0

because

Hn−1(VU,V0U) = H−1(U) = 0.

So, to complete the proof, we only need show the Thom class exists.For this, we only need show that if U1,U2 are two open sets so that uU1 and uU2 exist,

then so does uU1∪U2 .We can see this by Mayer-Vietoris. Let V1 = VU1 ,V2 = VU2 ,V3 = VU1∩U2 = VU1 ∩ VU2 .

(11.1) Hn−1(V3,V03 ) Hn(VU,V0U) Hn(V1,V01 )⊕Hn(V2,V02 ) Hn(V3,V03 )

Note that u1|VU1∩U2 = u2|VU1∩U2because both are Thom classes over U1 ∩ U2. By ex-

actness of Mayer-Vietoris, there exists some uU mapping to (u1,u2) in the above exactsequence.

To spell things out, the maps on the above long exact sequence for the latter three termsare given by

(11.2) uU (u1,u2) u1|U1∩U2 − u2|U1∩U2 = 0.

Now, we can use this finitely many times to see that uX exists if X is a finite dimensionalcell complex. One can do this by building this up skeleton by skeleton, inductively takingthe union of the interiors of the k-cells.

To prove this for an arbitrary paracompact space, approximate it by a CW complex.That is, there exists a CW complex mapping to Xwhich induces isomorphisms on homol-ogy and cohomology, and reduces the calculation on X to one on the cell complex. (Thisinvolves going through a lot of machinery which is standard, but takes time to develop,so we omit it.)

Lemma 11.2. Say V = V1 ⊕ V2. Then, if u1,u2 are Thom classes for V1,V2, then u = π∗1(u1)∪π∗2(u2) is a Thom class for V .


Proof. Recall that for vector spaces, (V1 ⊕ V2)0 = (V1 × V2)0 ∪ (V1 × V2)0. If we have

u1 ∈ Hn1(V1,V01 )u2 ∈ Hn2(V2,V02 ),

both generators. Then we have

π∗i (ui) ∈ Hni(V1 × V2,V01 × V2)

π∗1(u1 ∪ π∗2(u2) is a generator for Hn(V ,V0).Hence, orientations of V1,V2 yield an orientation of V1 ⊕ V2 (called the Whitney sum)

and u = π∗1(u1)∪ π∗2(u2).

11.2. Euler Classes.

Definition 11.3. Let u be the Thom class in Hn(V ,V0). Let s0 : X → V be the zero sectionof a vector bundle. The Euler class for an oriented rank n bundle V → X, denoted e(V) iss∗0(u) ∈ Hn(X).

Remark 11.4. The section s0 : X → V becomes the map

(X, ∅) → (V ,V0) .

Then, the Thom class lives on the total space, which is fiber by fiber

(Rn, Rn \ 0) ,

which yields maps

u ∈ Hn(V ,V0)→ Hn(V) ∼= Hn(X)

where the last isomorphism is given by deformation retraction of the vector bundle, viathe 0 section. In the case that the vector bundle were trivial, the Euler class is apparentlytrivial, but if the Vector bundle is nontrivial, the Euler class may be nontrivial.

Lemma 11.5. We have multiplicativity

e(V1 ⊕ V2) = e(V1)∪ e(V2).

Proof. This follows immediately from Lemma 11.2.

Warning 11.6. Observe

c(C) = 1

but

e(R) = 0.

So,

c(W ⊕C) = c(W)

but

e(V ⊕R) = 0.

38 AARON LANDESMAN

11.3. Examples of Thom and Euler Classes. Let’s do a calculation from first principles.First, we need a definition

Definition 11.7. The Thom space of a bundle V is

D(V)/S(V),

by crushing everything on the boundary to a single point.

Lemma 11.8. The Thom space of V if X is compact is the one point compactification of V .

Proof. Note thatD(V)/S(V) is the one point compactification ofV ifX is compact. This hasa distinguished point which is the point corresponding to the sphere. When we throw outthe boundary point, we get an open disc bundle, which is homeomorphic to the originalV .

Example 11.9. Take L→ CPn−1 to be the dual of the tautological line bundle with fiber C.Look at LR, with fiber R2. We have the Thom class

u ∈ H2(LR,L0R) = H2(D(LR),S(LR)).

By an application of excision, we have

H2(D,S) = H2(D/S, ∗).Also, we know that the Thom space of LR is the one point compactification of V by

the above Lemma 11.8. We can then see that the tautological bundle lives inside CPn asCPn \ ∗

So, we have

u ∈ H2(LR,L0R)

= H2(D(LR),S(LR))

∼= H2(D(LR)/S(LR), ∗)∼= H2(CPn, ∗)

Then, u determines a class in

H2(CPn, ∗) = Z,

which is a generator in each fiber class. So, u = h ∈ H2(CPn) is the standard generator. Itfollows that

e(L) = u|CPn−1 ∈ H2(CPn−1)which is a generator h.

By taking a limit, the same statement holds on CP∞. That is, the Euler class of the dualof the tautological bundle on CP∞. So, we get, the following corollary,

Corollary 11.10. For L→ X a C line bundle, we have

c(L) = e(LR) ∈ H2(X).Corollary 11.11. ForW → X, a complex vector bundle of rank n, we have

cn(W) = e(WR) ∈ H2n(X).


Proof. First, we’re using that the underlying real vector bundle of a complex vector bundleis orientable. If we write the complex coordinates as x + iy. Then, take the ordering(x1,y1, x2,y2, . . . , xn,yn). This defines an orientation of R2n.

Then, use splitting principle, using the line bundle formula and multiplicativity of eand c.

That is, ifW = ⊕iLi, we have

cn(W) = c1(L1) · · · c1(Ln)= e(L1) · · · e(Ln)= e(W).

Definition 11.12. SupposeX is an oriented smooth closedn-manifold. We define e(TX)[X] ∈Z to be the Euler number or Euler characteristic.

Theorem 11.13. Suppose X is an oriented smooth closed n-manifold. Then,

e(TX) ∈ Hn

e(TX)[X] ∈ Z

is

e(TX)[X] =∑

(−1)ibi

where bi = rkHi(X).

Proof.

Example 11.14. We have e(TS6)[S6] = 1− 0+ 0− 0+ 0− 0+ 1 = 2.

Remark 11.15. Note, we have a fiber bundleW → S6 with fiber C3 so thatWR = TS6. Thisgives an almost complex structure on S6. It is a famous unsolved problem whether S6 isa complex manifold. But, we do have an almost complex structure, where S6 ⊂ R7 as theimaginary Cayley numbers. We have an almost complex structure given by J : TxS6 →TxS

6, satisfying J2 = −1, by v 7→ n · vwith n the normal vector.

Corollary 11.16. There exists a complex vector bundle W on S6 with c3(W)[S6]. Explicitly, thisis the the complex structure on the tangent bundle mentioned above.

12. 2/22/16

12.1. More on Euler Classes.

Remark 12.1. Let E→ X be a rank n oriented vector bundle over an n dimensional closedoriented manifold. We have e(E) ∈ Hn(X). We can consider the integer e(E)[X] ∈ Z.

If we have s : X → E a section transverse to 0, the zero section. That is, we will havefinitely many points where s is 0 and we also have that at all points with s(x) = 0,ds :TxX → Ex. We have a well defined sign ±1 corresponding to whether it preserves orreverses orientation.

Then the Euler class counts this sign. This is now made precise.

40 AARON LANDESMAN

Proposition 12.2. Let E → X be a rank n oriented vector bundle over an n dimensional closedoriented manifold. Suppose we have a section s : X→ E transverse to 0. Then,

e(E)[X] =∑

x:s(x)=0

sgn(ds|x)

where sgn counts the orientation. Here, the sign is whether the derivative of TxX ∼= Ex is orienta-tion preserving or reversing.

Example 12.3. If E = TX, s = ∇f, we have the Hessian Hess(f) is nondegenerate at thecritical points where s = 0. Here, the sign is (−1)i(f,x), where

i(f, x) = the number of negative entries in the diagonalization of Hess(f)x.

Here,

e(TX)[X] =∑x,∇f=0

(−1)i(f,x).

Theorem 12.4 (Poincare, Hopf). We have

e(TX)[X] =

n∑i=0

(−1)i rkHi(X).

Remark 12.5. One way to get intuition for this is to think of a triangulation of a manifold,and then you can try to construct the function fwhich is 0 at the vertices of each simplex,equals 1 at the barycenters of the edges, and equals n at the barycenters of the n simplices,and, further, the barycenters are the critical points of the function.

For instance, in 2 dimensions, the points are minima, the barycenters of triangles aremaxima, and the 1-barycenters are saddle points.

It will end up that

e(TX)[X] =

n∑i=0

(−1)i rkHi(X)

=∑

σ simplices

(−1)dimσ

=∑i

(−1)i rkC∆i

=∑i

(−1)iH∆i .

Remark 12.6. If E→ X is not oriented then, there’s a unique Thom class with Z/2 coeffi-cients, u ∈ Hn(E,E0;Z/2). This is characterized by the property that

u|Ex,E0x

is the generator for all x. The proof this exists is the same as the proof with Z coefficients,although there is no need to make an initial choice of orientation on the fibers, becausethere is a unique generator of Z/2.


Remark 12.7. We have a generator

e(2)(E) ∈ Hn(X;Z/2)

given by

e(2)(E) = s∗0(U).

If E is oriented, we have

e(2) = e mod 2,

where mod2means the restriction on cohomology induced by the map Z→ Z/2.

Proposition 12.8. We have

e(2)(E) = wn(E).

Proof. We first prove this for the case that rkE = n = 1. Then we use the splitting princi-ple.

12.2. K Theory. For now, we’ll Assam all vector bundles are complex and our base spaceX is a compact Hausdorff space.

Recall a couple quick observations about vector bundles on compact spaces.

Remark 12.9. (1) As we saw earlier in the course, given E → X, there exists a bundleF → X so that E⊕ F = CN. (Proof: Cover X with open sets U so that E|U is trivial

and obtain a map X f−→ Grn(CN). Identify E ∼= f∗(Etaut). Here, we take N to

be n · the number of open sets in the cover . Then, Etaut ⊕ E⊥taut = CN and setF = f∗(E⊥taut.)

(2) We don’t have any cancellation. That is, it is possible for

E⊕G ∼= E ′ ⊕Gbut not

E ∼= E ′.

Example 12.10. We can take E to be the trivial bundle on S2 of rank 2, E ′ to be the tangentbundle on S2, and G to be the trivial bundle of rank 1 on S2.

Definition 12.11. We define Vect(X) to be isomorphism classes of vector bundles on X. IfX is connected, we have a function

rk(X) : Vect(X)→ Z≥0

E 7→ rk(E).

Exercise 12.12 (Easy Exercise). Vector bundles form a semigroup under addition with 0the trivial element.

Remark 12.13. Given an abelian semigroup with 0, (A,+), we can form an abelian groupK(A) by adding in formal elements.

Example 12.14. We have K(N) = Z. We identify (n,m) ∼ (n+ k,m+ k) which we thinkof as n−m.

42 AARON LANDESMAN

Definition 12.15. Given a semigroup A, we define K(A) to be A× A/ ∼ where ∼ is thesmallest equivalence relation with

(a,b) ∼ (a+ z,b+ z)

for all a,b, z. We write [a,b] as the equivalence class.We notate

[a] := [a, 0] ∈ K(A)− [a] := [0,a] ∈ K(A) [a] − [b] = [a,b] ∈ K(A)

Warning 12.16. We can have [a] = [a ′] ∈ K(A) even when a 6= a ′ ∈ A, if cancellationdoes not hold (and crucially, we saw above it does not hold for vector bundles). To seethis, note

[a] =[a ′] ⇐⇒ (a, 0) ∼

(a ′, 0

)⇐⇒ (a+ z, z) =

(a ′ +w,w

)for some w, z⇐⇒ a+ z = a ′ + z for some z.

Definition 12.17. We write K(X) := K(Vect(X)). We write [E] − [F] ∈ K(X) for vectorbundles E and F. We then have

[E] =[E ′] ⇐⇒ E⊕G = E ′ ⊕G

for some G. In this case, we say E and E ′ are stably isomorphic.

Remark 12.18. Recall that for all F there exists G,Nwith

F⊕G ∼= CN.

Hence,

[E] − [F] = [E⊕G] − [F⊕G]

= [E⊕G] −[CN]

.

Define N := CN so that 1 = C, and taking E ′ = E⊕G, we can always write

[E] − [F] =[E ′]− [N] .

In particular, we can always assume what we are subtracting is a trivial bundle.

Lemma 12.19. We have

[E] =[E ′] ⇐⇒ there exists F with E⊕ F ∼= E ′ ⊕ F⇐⇒ there exists N so that E⊕CN = E ′ ⊕CN.

Proof. This follows from the fact that we can add G we have F⊕ G = CN. Here, we’reusing

E⊕ F⊕G ∼= E ′ ⊕ F⊕Gand so

E ′ ⊕CN ∼= E ′ ⊕CN.


Exercise 12.20. We have that K is a functor sending X 7→ K(X). That is, given f : X → Ygives

f∗ : Vect(Y)→ Vect(X)

f∗K(Y)→ K(X).

Exercise 12.21. If f0 ∼= f1 then f∗0 = f∗1 .

Remark 12.22. The group K(X) is a commutative ring under tensor product. That is, wehave multiplication on Vect(X) with 1 = C the multiplicative identity.

Then, one extends by linearity so that

([E1] − [F1]) · ([E2] − [F2]) = [E1 ⊗ E2] − [F1 ⊗ E2] − [E1 ⊗ F2] + [F1 ⊗ F2]= [E1 ⊗ E2 ⊕ F1 ⊗ F2] − [F1 ⊗ E2 ⊕ E1 ⊗ F2]

Example 12.23. Take X = S2. We have a map

E 7→ rk(E)

which induces a map

K(X)→ Z

[E] − [F] 7→ rk(E) − rk(F).

We have another map

K(X)→ Z

E 7→ c1(E)[S2]

.

Together, these induce a map

K(X)→ Z⊕Z

E 7→ (rk(E), c1(E)[S2])

Fact 12.24. If a line bundle has 0 Chern class on S2 it is trivial.

Using this fact, we see C 7→ (1, 0). We also have

b := [H] − 1 7→ (0, 1) .

Here, H is the tautological bundle on CP1. The ring structure is almost immediate, with

b · b = ([H⊗H⊕ 1⊗ 1] − [H⊗ 1⊕ 1⊗H])

We see this has rank 0 and c1 = 2− (1+ 1) = 0. This means b · b = 0 as an element ofK(X).

Hence, we conclude

K(S2) = Z[b]/(b2).

44 AARON LANDESMAN

13. 2/24/16

13.1. Examples for K theory.

Example 13.1. We have K(pt) = Z. We have rk([E] − [F]) = rk(E) − rk(F).

Example 13.2. Last time, we saw K(S2) is Z⊕Z as a group, and as a ring, it was Z[b]/b2.As a group, the copies of Z were from the rank and c1.

Example 13.3. We have K(S1) = Z, given by rank. A complex vector bundle over S1 istrivial.

Let’s see why these are all trivial. To make a vector bundle on Sn, note that this is thesame as a vector bundle on the top and bottom agreeing on the equator. Both are trivialas the disks are contractible. The identification is a way of identifying the fiber of eachvector bundle, which is a map

Sn−1 → GL(r, C).

If we have a family of such maps, we get a family of vector bundles. By homotopy in-variance, two such homotopic maps give isomorphic vector bundles, which correspondsto homotopy classes of maps [

Sn−1,GL(r, C]

.

This is sometimes called a clutching function. This turns out to be Vect(Sn).That said, there is an issue in the real case: If we start with a vector bundle E in the

complex case, we will have to choose an identification E|D1ψ−→ Cr. This is fine in the

complex case because these two maps are related by a change of basis, given by a mapD1 → GL(r, C). Here, D1 is contractible and GL(r, C) is path connected. So, up to homo-topy, this map is unique. However, GL(r, R) has two components, which means we havean orientation issue. Again, this is absent in the complex situation, as here, if we changeφ for φ ψwe get the same homotopy class of map.

In the case of S1, we have [S0,GL(r, C)

]= pt

which is trivial. Therefore, since GL(r, C) is path connected, there’s a unique map, and sothe vector bundle is determined by its rank.

Example 13.4. To understand vector bundles on S2 in a different way, we’re looking atmaps [

S1,GL(r, C)]∼=[S1,GL(1, C)

]=[S1, C \ 0

]= Z

where this Z is essentially determined by the first Chern class.


TABLE 1. The reduced K theory of Sn

n 0 1 2 3 4K(Sn) Z 0 Z 0 Z

Example 13.5. What is K(S3)? From the analysis for S1 above, we have[S2,GL(r, C)

]∼=[S2,U(r)

]pt.

We have

π2(G) = 0

for any Lie group G. For example, for U(2), we have

π2(U(2)) = π2(SU(2))

where

U(2)×U(1)× SU(2) ∼= S1 × S3.

Hence, once again, we have

K(S3) = Z,

given by the rank.

13.2. Reduced K theory.

Definition 13.6. For x0 ∈ X a point in a topological space, we have a map

rk : K(X)→ K(x0) ∼= Z

[E] − [F] 7→ [Ex0 ] − [Fx0 ] = rkx0 .

We define the reduced K theory of a space with basepoint (X, x0) to be K(X) := ker(rk)with rk as defined above.

Example 13.7. Let’s examine the reduced K theory of Sn.This table will follow from Bott periodicity. Here, the generator of Z above is b, with

b2 = 0.In dimension 2, the generator is the first Chern class. In dimension 4 has c2 = 1 and

rk = 2. But, in dimension 6 the generator has c3 = 2, which loosely comes from howtensor products behave with Chern classes.

13.3. Products. From E1 ∈ Vect(X1),E2 ∈ Vect(X2), we can form

E1 × E2 = p∗1(E1)⊕ p∗2(E2) ∈ Vect(X1 × X2).

From this, we obtain a ring map (i.e., a bilinear map)

K(X1)× K(X2)→ K(X1 × X2).

Question 13.8. How does K(X1 × X2) relate to K(X1)× K(X2)?

46 AARON LANDESMAN

In ordinary homology, if we start with (X1, x1), (X2, x2), we have

H∗(X1) := H∗(X1, x1)

and we get a map

p∗1 : H∗(X1)→ H∗(X1 × X2, x1× X2)

Similarly, we have a map

p∗2 : H∗(X2)→ H∗(X1 × X2,X1 × x2).

The product of these maps lies in

H∗(X1 × X2,X1 × x2∪ x1× X2) ∼= H∗(X1 × X2,X1 ∨X2)In good cases (e.g., when there is a neighborhood of the joined subset which retracts ontoit) the above is

H∗ ((X1 × X2)/ (X1 ∨X2))

Definition 13.9. We define the smash product of X1 and X2, notated X1 ∧X2 is

X1 ∧X2 := X1 × X2/X1 ∨X2.

We then obtain a map

H∗(X1)× H∗(X2)→ H∗(X1 ∧X2)

by the product map.

Example 13.10. As a set, the smash product of two spheres is Rn ×Rm, together with apoint. This ends up being the one point compactification of Rn+m.

Remark 13.11. Recall that if X is a locally compact space X has a one point compactifica-tion X+ which, as a set is X

∐∗ where an open neighborhood of the point ∗ at∞ is of

the form ∗∪ (X \K) where K ⊂ X is compact.For example, Sn = Rn ∪∞. Then, X+ = X

∐∗ is a homeomorphism if X is compact.

That is, the one point compactification of a compact space is just the disjoint union of apoint.

Lemma 13.12. If X, Y are two locally compact spaces

X+ ∧ Y+ = (X× Y)+

Example 13.13. For example, when X = Rn, Y = Rm, then X+ = Sn, Y+ = Sm.

Proof. This follows from the definition of the smash product. That is,

X+ ∧ Y+ =(X+ × Y+

)/(X+ ∨ Y+

).

Inside the compact space X+ ∧ Y+, we have a copy X × Y ⊂ X+ × Y+. The universalproperty of the one point compactification is that it is final among all compact spacescontaining X × Y. So, from the universal property of one point compactifications, wehave a continuous map

X+ × Y+ → (X× Y)+

sending X+ ∪ Y+ 7→ ∗. This is a continuous bijection map of compact Hausdorff spaces,hence a homeomorphism.


We’ll now define a similar product in K theory. We have

K(X+)× K(Y+)→ K(X+ ∧ Y+)

where ∧ denotes the smash product.

Example 13.14. We’ll construct a map

K(Sn)× K(Sm)→ K(Sn+m).

Example 13.15. The generator b ∈ K(S2), we get an element

bn ∈ S2n.

We will later see:

Theorem 13.16. The element bn is a generator for K(S2n) = Z.

Remark 13.17 (Detour). Let use consider K theory of pairs, K(X,A). Suppose A ⊂ Xcompact. Consider vector bundles E→ Xwith a trivialization on A given by

τ : E|A → Cn ×A.

We’ll now give at least two equivalent definitions of relative K theory.

Definition 13.18. We define

K(X,A) := K(X/A).

Lemma 13.19. Let A ⊂ X be a compact subset. The following are equivalent.(1) The relative K theory(2) Classes of triples (E, F, ) where

τ : E|A ∼= F|A

under the equivalence relation generated by(a)

(E, F, τ) ∼(E ′, F ′, τ ′

)if there exist isomorphisms

E→ E ′

F→ F ′

on X so that the diagram

(13.1)

E F

E ′ F ′

τ

τ ′

commutes.(b) We have

(E, F, τ) ∼(E⊕G, F⊕G, τ⊕ idG|A

).

Proof. We’ll see this next time.

48 AARON LANDESMAN

14. 2/26/16

14.1. Equivalent definitions of relative K theory. At the end of last time, we stated twodefinitions of the same object, and we have yet to complete the proof they are equivalent.

Proposition 14.1. The following two definitions of relative K theory are equivalent.(1) K(X,A) :=∼ K(X/A).(2) Equivalence classes of triples

(E, F, τ)/ ∼

where E, F are vector bundles over X and τ : E|A → F|A an isomorphism and the equiva-lence relation ∼ given by

(E, F, τ) ∼ (E⊕G, F⊕G, τ⊕ 1|G) .

Proof. First we give a map from the first definition above to the second definition.If we start with a vector bundle on X/A. A representative of an element of K(X/A)

is a difference[E]−[F]

so that rk E = rk F = n. Then, if p : X → X/A, we have p∗(E)is a vector bundle on X with a trivialization on A. Then, since p∗(E),p∗(F) both havetrivializations on A, they are, in particular isomorphic over A, and we get a preferredisomorphism τ : p∗(E)|A → p∗(F)|A.

Next, we give a map in the reverse direction. Start with [E, F, τ]. ChooseGwith F⊕G ∼=Cn. This enables us to assume our triples is of the form

[E⊕G, F, Cn, τ⊕ 1G] .

In other words, we can assume our representative is

[E, Cn, τ] ∈ K(X,A)

by performing a relabeling of the data above. We can now form the quotient space X/A.We can form a bundle E → X/A by identifying the fibers over points in A via the giventrivialization. Now, if we have a trivialization τ over A, we claim we can construct anextension of it to a trivialization on U for U ⊃ A a slightly larger open set. Once we knowthis, this guarantees E is a vector bundle.

This will follow from the following lemma.

Lemma 14.2. Given a vector bundle V → X where X is compact, a section s and A ⊂ X closed,then we can extend s to all of V → X.

Proof.

Exercise 14.3. Prove this Hint: Use a partition of unity. Essentially, on an open set, thisamounts to claiming that a continuous real valued function defined on an open set can beextended.

Now, apply the above lemma to V = E∨ ⊗Cn with E→ Cn. The map τ : E|ACn|A givesa section of V , which can be extended to all of X. This is an isomorphism when restrictedto A. The lemma allows us to extend this as a map τ+ : E→ Cn, on all of X, although thismay not be an isomorphism. However,

x : τ+(x) is an isomorphism Ex → Cnx


is open, as it is locally given by the vanishing of a determinant the locus where the deter-minant vanishes is closed.

Example 14.4. We saw last time K(S2) ∼= Z. This was generated by b = [H] − [C] . Wehave (S2, ∗) = D2/S1. So, K(D2,S1) = Z. This has generator

[C, C,×z] .

where we think of D2 ⊂ C ∼= R2, where we think of ×z : S1 → C× ⊂ Hom(C, C) as amap with winding number one.

Alternatively, if we just wanted a generator of K(D2,S1) we can write it as

[Cn, Cn, τ] .

It’s not hard to see we can take n = 1. Then, the map τ turns out to be τ : S1 → GL(1, C) ∼=C× to that corresponding to the generator of π1(C×) ∼= π1(S

1) ∼= Z.

14.2. Back to Smash Products. Recall we went through this detour on relative K theoryto understand smash products.

Definition 14.5. Recall if we have (X1,A1), (X2,A2) as two spaces, we can form (X,A) :=(X1 × X2,A1 × X2 ∪ X1 ×A2). For ai ∈ K(Xi,Ai), we define

a = a1 × a2 ∈ K(X,A).

Here,

ai := [Ei, Fi,σi]

define

a := [E, F, τ] ,

where

E := E1 ⊗ E2 ⊕ F1 ⊗ F2F := F1 ⊗ E2 ⊕ E1 ⊗ F2.

and

τ : E ∼= E1 ⊗ E2 ⊕ F1 ⊗ F2 → F1 ⊗ E2 ⊕ E1 ⊗ F2 ∼= F

when written as a 2× 2 matrix given as follows. First, extend σi over all of Xi. This maynot be an isomorphism outsideAi, but we now have σi : Ei → Fi. Then, the map given by(

σ1 ⊗ 1 −1⊗ σ∗21⊗ σ2 σ∗1 ⊗ 1

)where σ∗i denotes the adjoint of σi.

Remark 14.6. Note that the map τ defined above is unitary. Observe more generally that

w :=

(u −vv u

)

50 AARON LANDESMAN

with u, v ∈ C, we have

w∗w =

(u v−v u

)(u −vv u

)=

(uu+ vv 00 uu+ vv

)This has the property that it is invertible if either u over v is nonzero.

Lemma 14.7. The map τ from Definition 14.5 is invertible if either σ1 or σ2 is invertible.

Proof. Observe that τ∗τ is(σ∗1σ1 ⊗ 1+ 1⊗ σ∗2σ2 0

0 1⊗ σ∗2σ2 + σ∗1σ1 ⊗ 1

).

Note that this is a sum of two hermitian matrices with nonnegative eigenvalue. So, if oneof them is positive, their sum is positive.

Now, note that τ : E → F from Definition 14.5 is an isomorphism on A1 × X2 and onX1timesA1. We have a map

K(X1,A1)× K(X2,A2)→ K(X,A).

Exercise 14.8. Check this map is well defined and distributes over addition in either fac-tor. Hint: That is, check the following two parts:

(1) the case

E1 := E′1 ⊕ E ′′1

F1 := F′1 ⊕ F ′′1

and

σi :=

(σ ′i 00 σ ′′i

)and see the map distributes over this.

(2) To see well definedness, if E1 = G, F1 = G and σ1 = 1G : G|A → G|A then

a = a1 × a2 = 0 ∈ K(X,A).

For the latter case, when σ1 = id, we see τ becomes(1⊗ 1 −1⊗ σ∗21⊗ σ2 1⊗ 1

)This should element the element 0 in the product. To see this is the element 0, we onlyneed check that this map E → F is an isomorphism everywhere. This is an isomorphismeverywhere because the matrix above is always invertible, and we can in fact take a linearhomotopy for the upper right and lower left hand corners.


Remark 14.9. Suppose (X, x0), (Y,y0) are two spaces with basepoint, and their product is

(X× Y, x0 × Y ∪ X× y0) = (X× Y,X∨ Y) .

Then,

K(X)× K(Y)→ K(X× Y,X∨ Y)

∼= K(X∧ Y)

because

X∧ Y := X× Y/X∨ Y.

Example 14.10. In the case X = Sn, Y = S2, we have a map

K(Sn)× K(S2)→ K(Sn+2),

the map

K(Sn)→ K(Sn+2)

a 7→ ab

is an isomorphism, where b is the generator of K(S2).To unravel this, think of

K(Sn) ∼= K(Dn/Sn−1).

In this view, think of elements as a pair of vector bundles on the disk (which are alwaystrivial) with a trivialization on the boundary[

CN, CN, τ]

with

τ : Sn−1 → GL(N, C).

Equivalently, we can view τ as a map

τ : Rn \ 0→ GL(N, C).

Equivalently, we can view τ as a map

τ : Rn →MN×N(C)

with τ(v) invertible for v 6= 0. In the case n = 2, the generator is the Bott generator

τ(u) = u

for u ∈ R2 = C =M1×1(C).

Question 14.11. How do we get the generator for K(S4)?

This is given by b · b ∈ K(D4,S3). We just need to take the product of the one for S2with itself, which will be from the matrix given in Definition 14.5. More explicitly, wehave

τ : R4 →MN×N(C).

52 AARON LANDESMAN

Here, we take N = 2. We get

τ(u, v) =(u −v∗

v u∗

).

The Bott periodicity theorem says this is a generator for K(S4). Repeating this again for

σ1 = [u]

σ2 =

(u −u∗

v v∗

)will be the generator of (S6) = K(D6,S5).

15. 2/29/16

15.1. Review. Last time, we were discussing

K(Sn) = K(Dn,Sn−1) =[CN, CN, τ

],

with CN = Dn × CN and τ : Sn−1 → GL(N, C). We have τ : Rn → MN×N(C) with τ(z)invertible for z 6= 0. Bott periodicity implies that the generators of K(S2n) are given asfollows.

We now make a change of convention. In the base case n = 1, identify R2 ∼= C →M1×1(C). We take

C→M1×1(C)

z 7→ z.

For n = 2, take so we are dealing with S4,

R4 ∼= C2 →M2×2(C)

(z1, z2) 7→ (z1 −z2z2 z1.

)When n = 3, so we are dealing with S6, the map is

R6 ∼= C3 →M3×3(C)

(z1, z2, z3) 7→z1 −z2 −z3 0z2 z1 0 z3z3 0 z1 z20 z3 −z1 z2.

In S6, the Bott generator is given by something with c3(E)[S6] = ±2, which, as we saidpreviously, has something to do with the tangent bundle of S6.

In general this is given inductively by the matrix(σ1 ⊗ 1 −σ∗2 ⊗ 11⊗ σ2 1⊗ σ1

)Lemma 15.1 (Bott’s isomorphism). We have an isomorphism given by

K(X)→ K(X∧ S2)

where X is a pointed space.


Proof. Applying Bott’s isomorphism toX+ forX, withX locally compact, we have K(X+) ∼=K(X+ ∧ S2). Hence, we have

K(X) ∼= K(X×R2).

Remark 15.2. So far, we’ve only defined K theory for compact spaces. Recall, that (X×Y)+ = X+ ∧ Y+ where X, Y are locally compact.

Definition 15.3. If X is locally compact, we define the K group of X to be

K(X) := K(X+),

that is, the reduced K group of the one point compactification of X.

Remark 15.4. This makes sense because if X is already compact, since in this case, X+ =X∐

∞, and so K(X∐∞) ∼= K(X).

Remark 15.5. K theory of locally compact spaces is like a cohomology theory with com-pact support H∗c(X).

Theorem 15.6. For X locally compact, we have

β : K(X)→ K(X×R2)

a 7→ a · b

where b ∈ K(R2) is the Bott generator, coming from the generator for S2.

We’ll prove this much later in Theorem 18.2. To prove this, we’ll define an inverse map

α : K(X×R2)→ K(X).

We’ll show αβ(a) = a,a ∈ K(X) and then we’ll show βα(c) = c, c ∈ K(X×R2.Note that the second composition follows formally for the following reason. If αβ = 1

then βα = 1 which is probably described in Hatcher’s book, although it was omitted inclass.

So, we only need define α and verify it is a one sided inverse. To do this, we’ll need todefine Fredholm operators.

15.2. Fredholm operators and index. Let H be a Hilbert spaces. That is, a separablespace with a countable complete orthonormal system C0,C1, . . . . Take

∑i aiei ∈ H with∑

i |ai|2 <∞.

Definition 15.7. Define L(H1,H2) to be bounded linear operators

A : H1 → H2

so that there exists a Cwith ||Ah|| ≤ C||h|| for all h. We define

A(H) := L(H,H).

Definition 15.8. A bounded linear operator A : H1 → H2 ∈ L(H1,H2) is Fredholm if(1) kerA is finite dimensional,(2) cokerA is finite dimensional.

54 AARON LANDESMAN

We write F(H1,H2), or just F for Fredholm operators.The index of a Fredholm operator is

index(A) = dim kerA− dim cokerA

for A ∈ F(H1,H2).

Lemma 15.9. If A,B ∈ F(H1,H2), then so are AB,BA and

index(AB) = index(A) + index(B).

Proof. We always have an inclusion kerB → kerAB. So, we have an exact sequence

(15.1)0 kerB kerAB kerA

cokerB cokerAB cokerA 0.

This should follow from a suitable operation of the snake lemma. Using this exact se-quence, we deduce AB is Fredholm. We then have

dim kerB− dim kerAB+ dim kerA− dim cokerB = 0.

The proof for BA holds with A and B reversed.

Lemma 15.10. If AB and BA are Fredholm, with A,B ∈ A(H), then so are A,B.

Proof. We have

kerB ⊂ kerAB.

We have im B ⊃ im BA. We then have that B is Fredholm if AB and BA are.

Example 15.11. (1) If A ∈ A(H) is invertible, then it is Fredholm, and its index is 0.(2) Consider left and shifts. By a left shift, we mean the map

SL : H→ H

(a0,a1, . . .) 7→ (a1,a2, . . .) .

We have kerSL is spanned by (1, 0, . . .) and the cokernel is 0, so the index is 1.(3) The right shift is defined analogously, by

SR : H→ H

(a0,a1, . . .) 7→ (0,a0,a1, . . .) .

This has index −1.Consider the projection operator

pr : H→ H

(a0,a1, . . . ,an,an+1, . . .) 7→ (0, . . . , 0,an,an+1, . . .)

This has kernel and cokernel of dimension n, so the index is 0.We have SR SL = id, which has index 0. This is a special case of Lemma 15.9.SR SL = pr, where pr is the projection forgetting the first coordinate. Indeed, this alsohas index 0. This is another special case of Lemma 15.9.

Lemma 15.12. Invertible operators are an open set in A.


Proof. First, we reduce to showing it for a neighborhood of 1 ∈ A, and then translate.Consider 1+Bwith ||B|| < 1. Here,

||B||op := suph 6=0

||Bh||

||h||.

The series 1−B+B@ −B3+ · · · converges in A because∑

||Bk|| <∞. The sum∑

(−1)kBk

is an inverse for 1+B in A. Let this open set be U.Then, if A is invertible, there is an open set U around 1, with

A · u : u ∈ U .

is an open neighborhood A consisting of invertible elements.

Next time, we’ll show

Lemma 15.13. Fredholm operators F ⊂ A forms an open set.

16. 3/2/16

16.1. More on Fredholm operators. Picking up where we left off last time, here is anotherlemma on Fredholm operators.

Lemma 16.1. The set of Fredholm operators F ⊂ A is open.

Proof. Let us now rewrite the condition of being Fredholm in a way that makes manifestthe openness of the condition to be Fredholm.

An operator A : H → H is Fredholm if and only if there exists H1 ⊂ H,H2 ⊂ H both offinite codimension such that the composite πH2 A iH1 is an isomorphism, where

(16.1)

H H

H1 H2.

A

πH2iH1

To see this equivalence, if we had a Fredholm operator A, we take H2 = im A,H1 =

(kerA)⊥. For the other direction, if we have such an H1,H2 with this property, we obtainthat dim kerA ≤ codimH1, dim cokerA ≤ codimH2.

For finer H1,H2 invertibility of πH2 A iH1 is an open condition for A, since the in-vertible operators are an open set of all operators.

Then, we have

F = ∪H1,H2A : πH2 A iH1 is invertible.

This is a union of open sets, hence open.

Let H be the complete orthonormal system e0, e1, . . .. Then, define Hn to be the closedspan of

en, en+1, . . . .

Here, codimHn = n. Let

pn : H→ H

(a0,a1, . . . an−1,an, . . .) 7→ (0, . . . , 0,an, . . .)

56 AARON LANDESMAN

be the orthogonal projection onto Hn.

Lemma 16.2. For A ∈ F there exists and n0 such that

im pn A = Hn.

for all n ≥ n0.Proof. We know im A has finite codimension. Choose n0 so that

im A+ span(e0, . . . , en0−1

)= H.

This is possible because this sequence depending on n eventually stabilizes. It must sta-bilize at being equal to H. It could not stabilize at anything smaller than H, because as n0approaches∞, we approach a complete orthonormal system for H.

Lemma 16.3. If X is compact and if A : X→ F is continuous, there exists some n so that

im (pn A(x)) = Hnfor all x ∈ X.

Proof. For fixed n,

A : im pn A = Hn

is an open set. This holds since every open cover has a finite subcover, and the union ofthis set over all n is an open cover.

Remark 16.4. Recall that if A ∈ F we have indexA = dim kerA−dim cokerA ∈ Z. Now,for A ∈ F, we can think of

[kerA] − [cokerA] ∈ K(∗) ∼= Z.

So, the index of a Fredholm operator can be thought of as a formal difference of spaces inthe K theory of a point.

Remark 16.5. For compact X, this doesn’t yet work. For A : X→ F, we’d like to say

[kerA(x)] − [cokerA(x)] ∈ K(X).We’d like to call this indexA ∈ K(X).

However, the problem is that the set of kerA(x) do not form a vector bundle. Thedimension of this kernel may jump as x varies in X.

Example 16.6. For example, take H = CN, take X = R, and take A(t) = t · id, t ∈ R. Thekernel of this jumps at t = 0.

But, this does work in some cases, as we’ll now see.

16.2. The index of a family.

Lemma 16.7. If A(x) : x ∈ X is a continuous family of surjective Fredholm operators, thenSx := kerA(x), as x ∈ X forms a vector bundle over X.

Proof. By the Fredholm operator property, this dimension is finite. Pick x0 ∈ X, and worknear x0. Let S0 = kerA(x0). Consider

B(x) : H→ H⊕ S0A(x) 7→ (A(x),πS0)


Note that B(x0) is invertible. The property of being invertible is open, so B(x) is invertiblefor x near x0.

Further, B(x) is invertible and

B(x)−1 (0⊕ S0) = kerA(x)= Sx.

Thus, for x ∈ U, there is some open neighborhood x0 ∈ Uwith a trivialization∪x∈USx =U× S0 using B(x)−1.

Definition 16.8 (Index of a family). Given a compact space X and

A : X→ F,

choose n so that

im (pnA(x)) = Hn

for all x ∈ X. We have

pn A : H→ H.

From Lemma 16.7, we have that

kerpn A(x)forms a vector bundle. Further,

cokerpn A(x) = H/Hn ∼= Cn

for all n.Now, we define the index of A to be

indexA = [ker(pn A)] − [Cn] .

Lemma 16.9. The definition of the index of a family Definition 16.8 is well defined.

Proof. We need to check the definition is independent of our choice of n. Suppose wereplaced n by n+ 1. Note that the index is still defined for n+ 1 if it is defined for n. Wehave an exact sequence

(16.2) 0 ker(pn A) ker(pn+1 A) C 0

where the last map is

ker(pn A)→ C

h 7→ 〈pn(A)h, en〉.This implies that as vector bundles, we have

kerpn+1A = kerpnA⊕C.

Hence, replacing n by n+ 1 yields

[ker(pn A)⊕C] − [Cn ⊕C] ∈ K(X)which is the same element as

[ker(pn A)] − [Cn] .

58 AARON LANDESMAN

Lemma 16.10. If A1,A2 : X→ F are homotopic, then

indexA1 = indexA2

in K(X). In particular, the index defines a map

index : [X,F]→ K(X).

Proof. We have a homotopy

A : [1, 2]× X→ F

and we obtain an element

indexA ∈ K ([1, 2]× X) .

Restricting to n = 1, 2we get the two indices.

Remark 16.11. In fact, the map index : [X,F] → K(X) is bijective, although this will takesome work, and we won’t show this now.

To prove this, one would need Kuiper’s theorem, which says that the group of invert-ible linear maps H→ H preserving the inner product, notated U(H), is contractible whenH is a Hilbert space.

16.3. More examples of Fredholm operators.

Definition 16.12. A map K : H→ H is of finite rank if im (K) is finite dimensional.

Definition 16.13. The set of compact operators

K ⊂ A(H)

is the closure of the set of finite rank operators.

We have that K is compact if and only if there exists Ki of finite rank n with Ki → K inoperator norm.

Next, time, we’ll see:

Proposition 16.14. If K : H→ H is a compact open, then 1+K ∈ F.

Corollary 16.15. A map A : H → H is in F if it is “invertible modulo compact operators,”meaning there exists B so that AB = 1+K,BA = 1+K ′, with K,K ′ are compact operators.

Proof. This follows from the proposition above together with one of the lemmas from theend of last class.

17. 3/4/2016

From last time, we defined

K = compact operators

= closure(finite-rank operators

)F = Fredholm operators.

We want to prove

Proposition 17.1. If K ∈ K, then 1+K ∈ F.


Lemma 17.2. If K ∈ K and hii≥1 is a bounded sequence in H, then the sequence Khii≥1has a convergent subsequence.

In other words, the K-image of the closed unit ball in H is sequentially compact.

Proof. Choose a sequence Kn of finite rank with ‖Kn − K‖op → 0. For every n ≥ 1, thesequence

Knhii≥1 ⊂ im (Kn) ∼= RN(n)

has a convergent subsequence, since it lives in a bounded subset of RN(n). Diagonalizingshows that there exists a index subset I ⊂ N so that Knhii∈I converges for all n. Do a“3ε-argument” to see that Khii≥1 is also Cauchy.

Proof of Proposition 17.1. We need to prove three things:(1) ker(1+ K) is finite-dimensional. In fact, since any Hilbert space satisfying (every

bounded sequence has a convergent subsequence) is finite-dimensional, it sufficesto show that every bounded sequence in ker(1+K) is convergent. Now hii≥1 ⊂ker(1 + K) bounded implies that Khii≥1 has a convergent subsequence by theLemma. But Khii≥1 is −hii≥1, so the latter is convergent.

(2) im (1+ K) is closed. In fact, consider a sequence ì ∈ im (1+ K) such that ì →` ∈ H. By definition, for each i, we have ì = (1+ K)hi for some hi ∈ H. Sinceker(1+K) can be nonempty, these hi are not yet unique. To make a unique choice,we shall require that hi ⊥ ker(1+K).

Now there are two possibilities: (a) If hii≥1 is a bounded sequence, then pass-ing to a subsequence I ⊂ N, we know that Khii∈I is convergent, so the ì areconvergent, implying that hi are convergent. So hi → h in H, and

(1+K)h = lim(1+K)hi

= lim ì

= `,

so ` ∈ im (1+K).(b) If hii≥1 is not a bounded subsequence, then, passing to a subsequence, we

may assume that ‖hi‖→∞. Consider these vectors of unit length:

hi =hi‖hi‖

.

Applying case (a) to hi in place of hi, we find that hi admits a convergent sub-sequence hi → h. But ‖h‖ = 1, so h 6= 0. A contradiction results from the factthat

(1+K)h = lim(1+K)hi

= limì‖hi‖

= 0.

Indeed, we chose hi so that hi ⊥ ker(1+ K), and the same is true for hi, so theirlimit h is also orthogonal to ker(1+K). This contradicts ‖h‖ 6= 0 in ker(1+K).

60 AARON LANDESMAN

(3) We now know that ker(1+K) is closed, so that

codimim (1+K) = dim(im (1+K))⊥.

But a bounded operator on a Hilbert space has an adjoint operator K∗, and wehave

dim(im (1+K))⊥ = dim ker(1+K∗),and the latter is finite dimensional. In fact, K is compact, so K∗ is compact, and weapply the first step with K∗ in place of K.

Corollary 17.3. If A is invertible modulo K, then A ∈ F.

17.1. Toeplitz Operators. Consider the Hilbert space H = L2(S1) with S1 ⊂ C the unitcircle (although any two separable Hilbert spaces are isomorphic as Hilbert spaces), andthe inner product is defined as

〈h,g〉 = 1

2π

∫2π0h(eiθ)g(eiθ)dθ.

With this convention, a complete orthonormal system is given by en = zn, where n ∈ Z.We have a decomposition

H = H+ ⊕H−

H+ = closed span of enn≥0H− = closed span of enn<0.

In other words, H+ consists of those functions on S1 which extend as analytic functionson the (open?) unit disc. If f : S1 → C is continuous (or L∞), then the following operatoris in A(H):

Mf : h 7→ fh.This is because ‖Mfh‖ ≤ sup(f) ‖h‖. In fact, if f is continuous,then ‖Mf‖ = sup(f),where ‖Mf‖ denotes operator norm.

Definition 17.4. Define an operator Af : H+ → H+ by

Afh = P+(fh),

where P+ : H→ H+ is the projection that kills H−. Symbolically,

Af = P+ Mf ιH+ .

Example 17.5. If f(z) = z, then Mf : en 7→ en+1 for all n, so it is an isomorphism H → H,and hence index(Mf) = 0. We haveM−1

f =Mg where g(z) = z−1, a.k.a. z.Now pass to H+, in the following sense. Af : H+ → H+ is an operator for which

Af(en) = en+1 for all n ≥ 0. This time, index(Af) = −1, since Af is just a right-shiftoperator with respect to the basis enn≥0. In the same vein,

Ag(en) =

en−1 if n ≥ 10 otherwise.

Therefore index(Ag) = 1, since Ag is a left-shift operator with respect to the basis enn≥0.

The operators Af and Ag exemplify the following result:


Proposition 17.6. If f is continuous and nowhere zero, then Af : H+ → H+ is Fredholm.

Proof. Since f is nowhere zero, we may define g by g(z) = 1/f(z), so that fg = 1. In otherwords,MfMg = 1. We claim that

AgAf = 1+K

where K is compact, and similarly

AgAf = 1+K′

where K ′ is compact. By Corollary 17.3, this implies that Af is Fredholm. More generally,we claim that, for any continuous functions f and g, not necessarily with g = 1/f, theoperator

Agf −AgAf

is compact. Indeed,

Agf = P+ Mgf ιH+

= P+ Mg Mf ιH+

AgAf = P+ Mg P+ Mf ιH+

Agf −AgAf = P+ Mg (1− P+) Mf ιH+

= P+ Mg P− Mf ιH+ .

This is compact becauseP− Mf ιH+ : H+ → H−

is compact if f is continuous, and B K is compact whenever B is bounded and K iscompact. To see that P− Mf ιH+ is compact, use the following argument:

(1) If f(z) = z−r for r ∈N, then P− Mf ιH+ is finite rank because it sends∑∞i=0 aiei

to∑ri=0 aiei−r.

(2) If f is a polynomial in z, z−1, i.e. a finite Laurent series, then P− Mf ιH+ is a finitelinear combination of finite rank operators, hence finite rank.

(3) If f is any continuous function, we use approximation by polynomials: there existsa sequence fkk≥1 of functions realized as finite Laurent series such that fk → f

uniformly on S1. ThenMfk →Mf in operator norm, so P− Mfk ιH+ → P− Mf ιH+ . Since the former have finite rank by part 2, we conclude that P− Mf ιH+ iscompact.

18. 3/7/2016

From last time, we had a continuous map

f : S1 → GL(1, C) = C \ 0

and a Fredholm map

Af : H+ → H+

Af = P+ Mf ιH+→H.

To generalize this, consider a continuous map

f : S1 → GL(N, C),

62 AARON LANDESMAN

and try to construct an analogous Fredholm operator Af. Let H = L2(S1), and consideran element h ∈ H(N) = H⊗CN, which is a column vector (h1, . . . ,hN)> of functions inL2(S1). Viewing f as a matrix of functions in C(S1), we define Mfh to be the product ofthis matrix with the vector h. Let H+(N) = H+ ⊗CN, and define P+ : H(N) → H+(N) asbefore. Let

Af = P+ Mf ιH+(N).

As before, we haveAfg = AfAg +K

where K is a compact operator.Now let E be any N-dimensional vector space. Given f : S1 → GL(E), we get Af on

H+ ⊗ E. More generally, given vector spaces E and F, we let GL(E, F) be the space ofinvertible linear maps E → F. Given f : S1 → GL(E, F), we get a Fredholm operatorAf : H+ ⊗ E→ H+ ⊗ F.

Consider an element of K(X × R2), where X is a compact topological space. By ourdiscussion of relative K-theory, and our definition of K-theory of non compact spaces, wehave

K(X×R2) = K(X×D2,X× S1),and an element of the right hand side looks like [V ,W, f] where V and W are vector bun-dles on X×D2, and f is a map such that

f : V |X×S1 →W|X×S1

is an isomorphism. Since X × D2 deformation retracts to X, (isomorphism classes of)vector bundles on the former are equivalent to vector bundles on the latter. Therefore, wecan write

[V ,W, f] = [π∗E,π∗F, f]

where E, F → X are vector bundles on X, and π : X×D2 → X is the projection onto thefirst factor. Viewed fiber-by-fiber over x ∈ X, this consists of a family of vector spaces Ex,a family of vector spaces Fx, and a family of maps

fx : S1 → GL(Ex, Fx).

By the above recipe, each fx gives us a Fredholm operator Afx . Hence we have a family ofFredholm operators Afx parameterized by x ∈ X.

Given such a family A : X → F, we have seen how to obtain an element of K(X) givenby index(A), i.e.

[ker(Pn A)] − [Cn]

for n sufficiently large. There is one subtlety: this definition takes F to be the space ofFredholm operators on a fixed Hilbert spaceH, but the above paragraph gives a family ofFredholm operators on different Hilbert spaces, i.e.

Afx ∈ F(H+ ⊗ Ex,H+ ⊗ Fx).The spaces H+ ⊗ Ex and H+ ⊗ Fx vary as x ∈ X vary, but we can still form the index byusing the map

Pn : H+ ⊗ Ex → H+ ⊗ Exin place of the previous Pn.


Definition 18.1. The mapα : K(X×R2)→ K(X)

is defined as sending[V ,W, f] = [π∗E,π∗F, f] 7→ index(Af)

as described above.

Theorem 18.2. This is an inverse to the Bott map β, i.e.

αβ(a) = a, a ∈ K(X)βα(c) = c, c ∈ K(X×R2).

Proof. The first part is proved as follows. Since α and β are clearly group homomor-phisms, we can assume that a = [E]. Here E→ X is a vector bundle, β(E) = a · bwhere

b = Bott class ∈ K(D2,S1) = K(R2)

given by[C, C, f] f : S1 → C− 0 f(z) = z.

Unwinding this definition, we have

β(E) = a · b= [π∗E,π∗E, f]

where, for all x ∈ X,

fx :S1 → GL(Ex,Ex)z 7→ z⊗ 1Ex .

Therefore αβ(a) = index(Af) for f as above. For each x ∈ X, fx(z) = z gives

Afx : H+ ⊗ Ex → H+ ⊗ Ex,which is (left-shift)⊗ 1Ex . The kernel of (left-shift) is one-dimensional, so the kernel ofAfx is canonically Ex, and surjectivity of (left-shift) implies that coker(Afx) = 0. Hence

index(Af) = [E] − [0]

= a,

as desired.The second part is proved by a formal argument. The first part can be carried out with

K(X) in place of K(X), where the map

β : K(X)→ K(X∧ S2)

is given by multiplication by b ∈ K(R2) = K(S2). To spell things out, the codomain of βis

K(X∧ S2) = K(X×D2,X× S1 ∪ x0×D2)where x0 ∈ X is the basepoint. Temporarily ignoring the fact that the pair of vectorbundles are isomorphic on x0×D2, we still have a map

α : K(X∧ S2) = K(X×D2,X× S1 ∪ x0×D2)→ K(X),

and we still have a family of maps

fx : S1 → GL(Ex, Fx)

64 AARON LANDESMAN

for all x ∈ X. In this case, for x = x0, this map extends:

fx0 : D2 → GL(Ex0 , Fx0),

i.e. fx0 has a given null-homotopy. The isomorphism on x0×D2 gives an identificationEx0 = Fx0 for which fx0 = 1 after the given homotopy. This means that

ker(Afx0 ) = coker(Afx0 ) = 0.

This shows that the above argument carries through for reduced K-theory as well.This observation allows us to extend to locally compact X, because

K(X) := K(X+),

where X+ = X∪ ∞ is the one-point compactification. More precisely, for locally compactX, we have

β : K(X)→ K(X×R2)

α : K(X×R2)→ K(X)

and αβ = 1K(X).Now consider compact spaces X and Y, which have maps αX,βX,αY ,βY defined as

above. For two elements

c ∈ K(X×R2) and v ∈ K(Y),we can form

αX(c) · v ∈ K(X× Y)c · v ∈ K(X×R2 × Y)i23(c · v) ∈ K(X× Y ×R2)

where i23 : X×R2 × Y → X× Y ×R2. Now we can apply αX×Y :

αX×Y(i23(c · v)) ∈ K(X× Y).

Lemma 18.3. We haveαX×Y(i23(c · v)) = αX(c) · v

Proof. This is because they are the same family of vector spaces in x ∈ X. As before, wemay assume that c = [π∗E,π∗F, f] as before, where π : X×R2 → X, and v = [V] for somevector bundle V → Y. Then

αX(c) = [ker(Pn Afx)] − [Cn],

c · v = [π∗E⊗ V ,π∗F⊗ V , f⊗ IdV ],

where we are pulling back π∗E and V to X×R2 × Y before tensoring them. Now

αX(c) · v = [ker(Pn Afx)⊗ Vy] − [Cn ⊗ Vy]where (x,y) varies over X × Y. Furthermore, αX×Y(i23(c · v)) equals exactly the samething. To see this, write Vy = Cm for fixed y ∈ Y, so that fx ⊗ IdVy maps

(π∗E)⊕m = π∗E⊗Cm → π∗F⊗Cm = (π∗F)⊕m,


acting by fx on each fiber (depending on z ∈ S1). Therefore Mfx⊗IdVy is a block-diagonalmatrix of functions in L2(S1), whose blocks arem copies ofMfx . The projection that makesAfx⊗IdVy can be taken to be the previous Pn on each summand.

Combining this with the previous observation, we conclude that it applies also to thecase where X and Y are only locally compact.

We wish to apply this to the case Y = R2. In this case,

i23 : X×R2 ×R2 → X×R2 ×R2

still interchanges the last two factors, but it gives the identity on K(X×R2×R2), becausei23 is homotopic to IdX×R2×R2 . Explicitly, on each fiber (R2 ×R2)+ = S4, we are doing ahigher-dimensional rotation.

Now for c ∈ K(X×R2), we have

βXαX(c) = αX(c) · bby definition. Applying the Lemma with Y = R2 and v = b, this equals

αX×R2(i23(c · b)),

where c · b ∈ K(X×R2×R2). But the homotopy in the previous paragraph says that thisequals

αX×R2(c · b) = αX×R2βX×R2(c)

= c (by the first part).

“I can follow the formal steps, but I don’t claim to understand exactly where you areswindled.”

19. 3/9/16

19.1. Review of infinite dimensional groups. Recall we have K(X)β−→ K(X∧ S2) is an

isomorphism, so K(Sn) ∼= K(Sn+2). We have

[E] −[CN]∈ K(Sn)

with N = rkE and

[E] −[CN]∼= [E⊕C] −

[CN+1

].

Then, K(Sn) are the equivalence classes of E → Sn up to isomorphism with E ∼ E⊕ C.Rank N vector bundles on Sn are given by the clutching function[

Sn−1,GL(N, C)]

.

We have a function

φ : Sn−1 → GL(N, C)

with φ ∼ φ⊕ 1with

φ⊕ 1 =(φ 00 1

)

66 AARON LANDESMAN

TABLE 2. Groups of Spheres

n K(Sn) πn−1(U)0 Z

1 0 02 Z Z

3 0 04 Z Z

an N+ 1×N+ 1matrix and

φ⊕ 1 : Sn−1 → GL(N+ 1, C).

We have

K(Sn) =[Sn−1,GL

]with

GL = ∪N≥1GL(N)

given by the limit

GL(N) → GL(N+ 1) → · · ·φ 7→ φ⊕ 1 · · · .

Then, GL(N) deformation retracts to U(N) with

U = ∪NU(N)

and

K(Sn) =[Sn−1,U

].

Recall [Sn−1,GL

]∼= πn−1(GL).

So, here is a tableThe following is Bott periodicity, which we have seen last time.

Theorem 19.1. We have a fiber sequence

(19.1) U(N− 1) U(N) S2N−1

giving maps

πn−1(U(N− 1))→ πn−1(U(N)).

This is an isomorphism if πn−2(S2n−1) = 0,πn−1(S2N−1) = 0. That is, if n− 1 < 2N− 1,n ≤2N− 1.

Lemma 19.2. We have that real vector space V is the same as a complex vector space W with aconjugate linear map

J :W →W

with j2 = 1.


Proof. Staring with Rn, we can send it to (Cn, J) where J is the conjugation map z 7→ z.In general, we send V 7→ (V ⊗R C, J) with J(v⊗ λ) = v⊗ λ. In the reverse direction, we

send (W, J) 7→ w ∈W : Jw = w .

Definition 19.3. The quaternion algebra is

H = 〈1, I, J,K〉

Lemma 19.4. Left H vector spaces (quaternion vector spaces) are in bijection withcomplex vector spacesW with a conjugate linear map J :W →W with J2 = −1.

Proof. Similar to the real case.

If we start with

(W1, J1) , (W2, J2)

we can formW =W1 ⊗C W2, J = J1 ⊗ J2 so that J2 = ±1. So, if

(W1, J1) , (W2, J2)

are quaternion vector spaces then (W, J is a real vector space. Similarly, if we took tworeal vector spaces, we would get a real vector space. If we took one quaternion vectorspace and one real vector space, their tensor product would be quaternion.

Example 19.5. If we take two quaternion vector spaces and take their tensor productH⊗C H, we get (C4, J), which is a four dimensional real vector space.

19.2. Real K theory. Recall that we defined K(X) to be the abelian group from the semi-group of complex vector bundles on X.

In the real case, we have the following definition.

Definition 19.6. We define KO(X) to be the semigroup group of real vector bundles on X.Representatives can be written as [E] − [F]. We can define KO(X) similarly to the complexcase.

Definition 19.7. We can define the Symplectic K theory KSp(X) to be the semigroup ofQuaternionic vector bundles, with representatives [E] − [F].

We have

KO(Sn)

as equivalence classes represented by

[E] −[RN]

,

we have

KO(Sn) = real vector bundles E / ∼

=[Sn−1, 0

]= πn−1(O).

where E ∼ E⊕R. Here O = ∪NO(N).

68 AARON LANDESMAN

TABLE 3. Real K groups of spheres

n KO(Sn) πn−1(O)0 Z

1 Z/2Z Z/2Z2 Z/2Z Z/2Z3 0 0

TABLE 4. Symplectic K groups of spheres

n KSp(Sn) πn−1(Sp)0 Z

1 0 02 0 03 0 04 Z Z

To compute some of the above entries, we have a sequence

(19.2) O(N− 1) O(N) SN−1.

This gives an exact sequence on homotopy groups. We have π1(SO(3)) ∼= π1(RP3) ∼= Z.

19.3. Symplectic K theory.

Definition 19.8. Let’s write GL(N, H) for H linear invertible maps HN → HN. Insidethis, there is the group Sp(N) ⊂ GL(N, H) with

Sp(N) := GL(N, H)∩O(4N).

Write

Sp := ∪nSp(N).

Note that Sp(1) is isometries R4 → R4 respecting I, J,K. That is, it is of the form q 7→ qλ

with |λ| = 1. So, Sp(1) ∼= S3 ∼= SU(2). In the fibration picture, we have

(19.3) Sp(1) Sp(2) S7.

The exact sequence of homotopy groups satisfies

πk(Sp(1)) ∼= πk(Sp)

for k < 7. So, πn−1(Sp) = πn−1(S3) for n < 7.

Remark 19.9. We have π3(S3) 3 [f], with f := id : S3 → S3. We have f(q) = q withS3 ⊂H. This gives an element b4 ∈ KSp(S4. This element is a “close friend” of b ∈ K(S2).Remark 19.10. We have maps

KO(X)× KO(Y)→ KO(X∧ Y).

KO(X)× KSp(Y)→ KSp(X∧ Y).KSp(X)× KSp(Y)→ KO(X∧ Y).


TABLE 5. Symplectic K groups of spheres

n KSp(Sn) πn−1(Sp)0 Z

1 0 02 0 03 0 04 Z Z

5 Z/2Z Z/2Z6 Z/2Z Z/2Z

TABLE 6. Real K groups of spheres

n KO(Sn) πn−1(O)0 Z

1 Z/2Z Z/2Z2 Z/2Z Z/2Z3 0 04 Z Z

5 0 06 0 07 0 08 Z Z

We also have β∗, which is multiplication by b∗ giving maps

KO(X)→ KSp(X∧ S4),

KSp(X)→ KO(X∧ S4).

Hence, we have a map

KO(X)→ KO(X∧ S8).

Corollary 19.11. We have

KO(Sn+4) = KSp(Sn)

KSp(Sn+4) = KO(Sn).

Proof. Omitted.

Example 19.12. We can now expand the table of KO, KSp.

20. 3/11/2016

Let’s relate K-theory with characteristic classes. Recall that, to each complex vectorbundle E → X, we have associated Chern classes ci(E). We can furthermore define ci(a)for a ∈ K(X), as follows. Consider the map

Vect(X)→ (⊕H2i(X)

)×

70 AARON LANDESMAN

where (⊕H2i(X)

)×= units

= 1+ t1 + t2 + · · ·+ tN, ti ∈ H2i(X).

This is because, for any t ∈ ⊕i>0H2i, we have

(1+ t)−1 = 1− t+ t2 − · · · .

This is all well and good when H2i(X) = 0 for large i. If not, there is a subtlety becausean infinite sum of terms is not a valid element of a direct sum. In such a case, we shouldinstead consider

Vect(X)(∏

H2i(X))×

This is the map c : E 7→ c(E), which is a semigroup homomorphism by the Whitney sumformula

c(E⊕ F) = c(E) ^ c(F).

The universal property of groupification gives us a corresponding map

c : (K(X),+)(∏

H2i(X))×

wherec([E] − [F]) = c(E)c(F)−1.

For example, if H → CP∞ is the dual of the tautological bundle, so that c1 = h is thestandard generator of H2(CP∞), then our definition says

c(−[H]) = 1− h+ h2 − h3 + · · · .

20.1. Chern character. Given variables x1, . . . , xn, we have the elementary symmetricpolynomials, with k = 1, . . . ,n:

σk(x1, . . . , xn) = x1 · · · xk + · · ·+ xn−k+1 · · · xn.

Theorem 20.1. Every symmetric polynomial in the xi’s is a polynomial in the elementarysymmetric polynomials σk.

For example, we have ∑xi = σ1∑x2i = σ

21 − 2σ2∑

x3i = σ31 − 3σ1σ+ 3σ3∑

xki = sk(σ1, . . . ,σn).


Definition 20.2. The k-th Chern character chk(E) ∈ H2i(X;Q) of a rank-n complex vectorbundle E→ X is

chk(E) :=1

k!sk(c1(E), . . . , cn(E)).

By convention, let ch0(E) = rank(E) ∈ H0(X), where the rank is a locally constant functionon X.

Further, define

ch(E) =∞∑i=0

chi(E) ∈∏i≥0H2i(X;Q).

For example, the above formulae show that

ch1(E) = c1(E)

ch2(E) =1

2(c21 − 2c2)

ch3(E) =1

6(c31 − 3c1c2 + 3c3).

This definition is motivated by the splitting principle: if

E = L1 ⊕ · · · ⊕ Ln,

thenci(E) = σi(c1(L1), . . . , c1(Ln)),

so that

sk(c1(E), . . . , cn(E)) =n∑i=1

c1(Li)k

by the definition of sk above. Then

chk(E) =n∑i=1

c1(Li)k

k!∈ H2k(X;Q)

ch(E) =n∑i=1

ec1(Li).

Here, for x ∈ H2(X;Q), we have defined

ex = 1+ x+1

2x2 + · · · ∈

∏i≥0H2i(X;Q).

For vector bundles E and F, we use the splitting principle to write

E = L1 ⊕ · · · ⊕ LnF =M1 ⊕ · · · ⊕Mm.

Thench(E⊕ F) = ch(E) + ch(F),

since both sides equaln∑i=1

ec1(Li) +

m∑j=1

ec1(Mj).

72 AARON LANDESMAN

This would have happened for any power series definition of ch. What is special aboutour particular definition is that the Chern class of the tensor product takes on a simpleform. In fact, let xi = c1(Li) and yj = c1(Mj) for brevity, and note that

E⊗ F =⊕i,j

Li ⊗Mj

ch(E⊗ F) =∑i

∑j

ec1(Li⊗Mj)

=∑i

∑j

exi+yj

=

(∑i

exi

)∑j

eyj

= ch(E) ch(F).

The upshot of this is

Theorem 20.3. The map

ch : K(X)→∏i≥0H2i(X;Q)

is a ring homomorphism.

Lemma 20.4. The coefficient of σn in sn is (−1)n−1n.

Proof. The n-th roots of unity ζini=1, i.e. the solutions to xn − 1 = 0, are such that σi = 0

for all 1 ≤ i ≤ n− 1, and σn = (−1)n−1. On the other hand,

sn =

n∑i=1

ζni

=

n∑i=1

1

= n,

so the previous reasoning implies that

sn(0, . . . , 0, (−1)n−1) = n.

Corollary 20.5. If E→ X has c1 = · · · = cn−1 = 0, then

chn(E) =1

n!(−1)n−1n · cn(E)

=(−1)n−1

(n− 1)!cn(E).


20.2. Let

[E] − [CN] ∈ K(S2n) ∼= Z

be the generator. What is cN(E)[S2n]? On the problem sets, we found bundles E such that

c1(E)[S2] = 1

c2(E)[S4] = 1

c3(E)[S6] = 2.

(Here c1(E)[S2] means c1(E) evaluated on the fundamental class [S2] of S2.) Bott’s theoremtells us that the generator is bn where b ∈ K(S2). We had b = [H] − [C] where c1(b) =c1(H) = hwas a generator of H2(S2;Z). Therefore

ch(b) = 0+ h = h,

since rank(b) = 0. Now

b× · · · × b ∈ K(S2 × · · · × S2, T),

where

T =n⋃j=1

S2 × · · · × ∗× · · · × S2

is the locus which is quotiented out when taking the smash product. Therefore

b× · · · × b ∈ K(S2 × · · · × S2, T)∼= K(S2n),

so that b× · · · × b = π∗(bn) where bn ∈ K(S2n) is the canonical generator, where

π : S2 × · · · × S2 → S2n

is the quotient map used to construct the smash product. We see that

π∗(cn(b

n)[S2n])= cn(π

∗(bn))[S2 × · · · × S2]= cn(b× · · · × b)[S2 × · · · × S2]= (n− 1)!(−1)n chn(b× · · · × b)= (n− 1)!(−1)n ch(b× · · · × b),

where the last line follows because ch0(b× · · · × b) = 0 since rank(b) = 0, and the inter-mediate chi(•)’s are zero because cohomology of S2n vanishes in those dimensions. Nowthis equals

(n− 1)!(−1)n(h1 × · · · × hn)

74 AARON LANDESMAN

where h1 ∈ H2(S2 × · · · × S2) is the pullback of h from the i-th copy of S2. Now multi-plicativity of ch gives

ch(bn)[S2n] =n∏i=1

h[S2]

=

n∏i=1

1

= 1.

Therefore

Theorem 20.6.

cn(bn)[S2n] = (−1)n(n− 1)!

Corollary 20.7. For E→ S2n, chn(E)[S2n] is an integer.

Recall that KO and KSp were defined as formal differences of ‘vector bundles with J,’i.e. pairs (W, J) where W ∈ VectC(X) and J : W → W is a conjugate-linear operator withJ2 = 1 for the real case, and J2 = −1 for the quaternion case. Forgetting J gives maps

KO(X)→ K(X)

KSp(X)→ K(X),

or similarly for reduced K-theory. The upper map takes a real vector space and tensorsby C, the lower map takes a quaternion vector space and only remembers the complexmultiplication. For instance, we have a commutative diagram where all maps are isomor-phisms:

KO(S0) K(S0)

Z Z

rk rk

id

Similarly, we have a commutative diagram

KSp(S0) K(S0)

Z Z

rkH rkC

×2


From before, b4 ∈ KSp(S4) has c2(b4) = 1, so – as C-vector bundles – or in K(S4), we haveb4 = b

2, i.e. b24 = b4, since complex bundles on S4 are classified by their Chern class.

Z = KO(S0) K(S0) = Z

Z = KO(S4) K(S4) = Z

Z = KO(S8) K(S8) = Z

Z = KO(S12) K(S12) = Z

1

2

1

2

The second row follows from the preceding commutative diagram, since multiplicationby b4 relates KO(S4) and KSp(S0). The isomorphisms between the entries of the right-hand column are given by b2.

21. 3/21/2016

Recall how we defined the Euler class of a real rank-r oriented vector bundle V → X.We proved the existence of the Thom class

ν ∈ Hr(D(V),S(V)),

where

D(V) := v : |v| ≤ 1S(V) := v : |v| = 1.

The key property of ν is that, for each x ∈ X,

ν|(D(Vx),S(Vx)) is ‘the’ generator of Hr(D(Vx),S(Vx);Z) = Z,

where ‘the’ refers to the generator singled out by the orientation on V .Now we consider a K-theoretic analogue. Let X be a compact space. If we assume that

r is even, thenK(D(Vx),S(Vx)) = Z,

from our computation of K(Sn). We ask: does there exist ν ∈ K(D(V),S(V)) such that, foreach x ∈ X,

ν|D(Vx),S(Vx) is a generator of K(D(Vx),S(Vx))?

To answer this, it helps to have a better understanding of the algebraic structure under-lying the Bott generator of K(D(Vx),S(Vx)) = K(R2n), which came from taking tensorpowers of the Bott generator b ∈ K(R2).

76 AARON LANDESMAN

21.1. Clifford Algebras and Clifford Modules. Let V be a real vector space with an innerproduct 〈−,−〉V . Look at bilinear maps

γ : V × S→ S,

where S is a complex inner product space, satisfying the Clifford relation, i.e.

γ(v)γ(u) + γ(u)γ(v) = −2〈u, v〉V IdS

for all u, v ∈ V , as operators S → S. For example, if |v| = 1, taking u = v gives γ(v)2 =− IdS. If u ⊥ v, then γ(u)γ(v) = −γ(u)γ(v). In fact, these two special cases imply thegeneral Clifford relation, which follows by expanding general u and v with respect to anorthonormal basis.

Definition 21.1. A Z/2-graded vector space (over C) is a vector space with an operatorε : S → S such that ε2 = 1. In this case, we have the decomposition S = S+ ⊕ S− into the±1 eigenspaces of ε. This structure is equivalent to a decomposition of S into two vectorspaces, but it will be convenient to encode this structure into an operator ε.

If S1 and S2 are Z/2-graded, with grading operators ε1 and ε2, then S = S1 ⊗ S2 isZ/2-graded, with operator ε = ε1 ⊗ ε2. Equivalently, we have

S+ = S+1 ⊗ S+2 ⊕ S

−1 ⊗ S

−2

S− = S+1 ⊗ S−2 ⊕ S

−1 ⊗ S

+2 .

When S has an inner product, we shall require that ε be unitary.

Definition 21.2. A (complex) Clifford module for V (a 2k-dimensional real inner productspace) is a Z/2-graded complex inner product space (S, ε) and a linear map

γ : V → End(S)

satisfying the following conditions:(1) γ(v) is odd, for all v ∈ V . This means that γ(v) sends S+ to S−, and S− to S+. In

other words, ε γ(v) = −γ(v) ε.(2) γ(v) is skew-adjoint for all v, i.e. γ(v)∗ = −γ(v).(3) The Clifford relation

γ(v)γ(u) + γ(u)γ(v) = −2〈u, v〉V IdSholds for all u, v ∈ V .

Remark 21.3. The requirement that dimV be even is imposed because the correct definitionof ‘Clifford algebra’ for odd-dimensional vector spaces V is not given by the above.

Taking (i) and (ii) implies that γ(v) is block-diagonal, of form

γ(v) =

(0 −σ(v)∗

σ(v) 0

),

with respect to the decomposition S = S+ ⊕ S−. With this notation, (iii) implies that,whenever |v| = 1, we have γ(v)2 = −1, i.e.

σ(v)∗σ(v) = IdS+ ,

or in other words σ(v) : S+ → S− is unitary.


Proposition 21.4. Let S1 and S2 be Clifford modules for V1 and V2, respectively. ThenS := S1 ⊗ S2 is a Clifford module for V1 ⊕ V2, where

γ : V1 ⊕ V2 → End(S)

is defined byγ(v1, v2) := γ1(v1)⊗ IdS2 + ε1 ⊗ γ2(v2).

Proof. We check the Clifford relation:

γ(v1, v2)γ(u1,u2) + γ(u1,u2)γ(v1, v2) =(γ1(v1)⊗ IdS2 + ε1 ⊗ γ2(v2)

)(γ1(u1)⊗ IdS2 + ε1 ⊗ γ2(u2)

)+(γ1(u1)⊗ IdS2 + ε1 ⊗ γ2(u2)

)(γ1(v1)⊗ IdS2 + ε1 ⊗ γ2(v2)

)=(γ1(v1)γ1(u1) + γ1(u1)γ1(v1)

)⊗ IdS2

+ ε21 ⊗(γ2(v2)γ2(u2) + γ2(u2)γ2(v2)

)+ (terms that cancel because ε1 γ1(•) = −γ1(•) ε1)

= −2(

IdS1 ⊗〈v1,u1〉+ 〈v2,u2〉 ⊗ IdS2)

.

Note that tensor product of Clifford modules is associative: for S1,S2,S3, the γ of S1 ⊗S2 ⊗ S3 is

γ := γ1 ⊗ 1⊗ 1+ ε1 ⊗ γ2 ⊗ 1+ ε1 ⊗ ε2 ⊗ γ3no matter how the tensor product is parenthesized.

With this next definition, we begin our approach to the Bott generator of K(R2):

Definition 21.5. For V = R2 with the standard inner product, the standard Clifford moduleis S = S+ ⊕ S− = C⊕C with

ε =

(1 00 −1

)and

γ

(xy

):=

(0 −x− iy

x− iy 0

),

which we can abbreviate as

γ1(z) =

(0 −zz 0

)For R2k = R2 ⊕ · · · ⊕R2, the standard Clifford module is Sk := S1 ⊗ · · · ⊗ S1 with k terms.

(*) Any Clifford module S for V ∼= R2k gives an element uS ∈ K(V) = K(D(V),S(V)) =Z, in the following way: uS = [S+,S−, τ] where τ : v : |v| = 1 → GL(S+,S−) is given bymapping v to γ(v). This map sends S1 to the Bott generator of K(R2), and similarly Sk issent to the Bott generator of K(R2k).

Passing from vector spaces to vector bundles, let X be a compact topological space, andlet V → X be a real vector bundle with inner product, so that the fibers Vx are isomorphicto R2k.

78 AARON LANDESMAN

Definition 21.6. A complex Clifford module S → X is a Z/2-graded complex vector bundlewith inner product, and a bundle map

γ : V → End(S),

i.e. a continuous family of maps

γx : Vx → End(Sx)

for all x ∈ X, satisfying (i)–(ii) on each fiber.

We seek ν ∈ K(V) = K(D(V),S(V)) equal to a generator on each fiber K(Vx). With whatwe have done above, we can construct such a ν if we have:

• A Clifford module S→ X for V → X, such that• for each x ∈ X, Sx is the standard Clifford module for Vx ∼= R2k.

Given such an S, applying our previous construction (*) fiber by fiber gives the desiredν ∈ K(V).

22. 3/23/2016

From last time, we reduced the problem of finding a class ν to finding a bundle ofClifford modules, each fiber of which is the standard Clifford module. This time, weclassify ungraded Clifford modules for R2k = V , with the standard inner product, i.e.pairs (S,γ) where S is a complex vector space with inner product, and γ : V → End(S) iscomplex-linear, such that γ(v)∗ = −γ(v), and the Clifford relation holds.

Proposition 22.1. If S is a Clifford module, and S ′ ⊂ S a Clifford submodule, meaning thatS ′ is preserved by all γ(v), then (S ′)⊥ is also a Clifford submodule.

Proof. This is because each γ(v) is a skew-adjoint operator.

Corollary 22.2. Every Clifford module S is a direct sum of irreducible ones, where ‘irre-ducible’ means there is no nontrivial Clifford submodule.

Lemma 22.3. (Schur) If S is irreducible and φ : S→ S is a Clifford-homomorphism, mean-ing that it respects the action of the algebra, i.e.

φ γ(v) = γ(v) φ for all v ∈ V ,

then φ is multiplication by some λ ∈ C.

Proof. As with other variants of Schur’s lemma, note that φ must have some eigenvalue(since we are working with complex vector spaces), which must be a proper Cliffordsubmodule of S, hence 0, so φ equals its eigenvalue.

Proposition 22.4. The standard Clifford module (of dimension 2k) for V = R2k is irre-ducible. Moreover, every irreducible Clifford module has dimension 2k, and any Cliffordmodule of dimension 2k is irreducible. Any such Clifford module (irreducible of dimen-sion 2k) is isomorphic to the standard Clifford module.

Proof. Let (S,γ) be a Clifford module of dimension 2k for V = R2k. We prove that S isirreducible by induction on k, the base case k = 0 being trivial.

Considerη = γ(e1)γ(e2) ∈ End(S).


Because γ(e1) and γ(e2) anti-commute by the Clifford relations,

η2 = −γ(e1)2γ(e2)

2 = −1,

so S = S ′ ⊕ S ′′, where S ′ = ker(η− i) and S ′′ = ker(η+ i) are the eigenspaces of η. Thepurpose of having two γ(ei) terms in the definition of η is that now η commutes withγ(ei) for all i > 2. Therefore S ′ and S ′′ are Clifford modules for the smaller vector space

R2k−2 := span(e3, . . . , e2k).

Furthermore, we have dimS ′ = dimS ′′ because γ(e1) restricts to isomorphisms from S ′

to S ′′ and vice versa, because it anti-commutes with η. Since we assumed that dimS = 2k,it follows that dimS ′ = dimS ′′ = 2k−1. By the inductive hypothesis, they are irreducibleClifford modules for R2k−2.

Now, considering S again as a Clifford module for R2k = V , the previous paragraphshows that: either S is irreducible, or S decomposes into irreducibles S ′ ⊕ S ′′. But S ′ isnot γ(e1)-invariant, so the former possibility holds.

To see that S is isomorphic to the standard Clifford module of R2k = V , we start withthe decomposition S = S ′ ⊕ S ′′, with γ(e1) acting as the matrix(

0 −11 0

),

and start to (recursively) pick out the tensor product structure that defines the standardClifford module.

22.1. How to obtain the Z/2-grading. Pick an orthonormal basis V = R2k, and considerthe operator

ε := (−i)kγ(e1)γ(e2) · · ·γ(e2k)= (−i)n/2γ(e1)γ(e2) · · ·γ(en).

Note thatεγ(ej) = −γ(ej)ε for all j = 1, . . . , 2k.

Also,

ε2 = (−1)n/2(−1)(n2)γ(e1)

2 · · ·γ(en)2

= 1.

But an εwith these properties was our definition of a Z/2-grading.Is this the only way to define a grading? Suppose ε1 and ε2 are two potentially different

Z/2-gradings on an irreducible Clifford module S. Schur’s lemma then implies that ε1 =±ε2. This sign comes as no surprise: changing the order of the ei in our definition of εabove, we can in fact change ε by a sign. In other words, the ε we chose depends on theorientation class of the basis ei, and our uniqueness result shows that it is well-definedgiven a choice of orientation class.

To summarize:

Proposition 22.5. For V = R2k, there exists a unique irreducible Clifford module S withdimS = 2k. If furthermore V is oriented, then S has a canonical Z/2-grading.

80 AARON LANDESMAN

22.2. Bundles of Clifford modules. Return to the situation where X is a compact topo-logical space, and V → X a real vector bundle of rank 2k. We seek a complex vectorbundle S → X, which is made into a Clifford module bundle by a fiber-wise actionγ : V → End(S), such that, for each x ∈ X,

(Sx,γx) ∼= (standard Clifford module for Vx).

Definition 22.6. This S → X is called a spinc structure for the bundle V → X. (The c-superscript stands for C as opposed to R. For example, if G is a compact Lie group, thenGc denotes its complexification.)

This definition is most commonly applied when X is a smooth manifold and V = TX.In this case, a spinc structure for TX→ X is said to be a spinc structure for X.

We investigate the existence and uniqueness of such an (S,γ). First, suppose (S,γ) and(S ′,γ ′) are two spinc structures for V |RaX. For all x ∈ X, both Sx and S ′x are isomorphicto the standard Clifford module on Vx, so

Lx := HomCliff(Vx)(Sx,S′x)

is a one-dimensional complex vector space, but not canonically isomorphic to C sinceSx and S ′x are not actually equal. In this way, we obtain a complex line bundle L → X.If L happens to be a trivial vector bundle, choosing a nowhere zero section shows that(S,γ) ∼= (S ′,γ ′). In general,

Definition 22.7. Call L→ X the difference line bundle of (S,γ) and (S ′,γ ′).

We have this transitivity result:

Proposition 22.8. Consider three Clifford modules (S1,γ1), (S2,γ2), (S3,γ3), and let L12,L23,L13be the corresponding difference line bundles. Then canonically L13 ∼= L12 ⊗ L23.Proof. The map is given by composing the homomorphisms S1 → S2 and S2 → S3 givenby sections of L12 and L23.

Corollary 22.9. Given (S,γ), we get an injective map

isomorphism classes of spin-c structures→ line bundles

sending (S ′,γ ′) 7→ L, which is the difference from (S,γ). This map is also surjective: givenL, we obtain

S ′′ = S⊗ Lγ ′′(v) = γ(v)⊗ IdL .

Since (S ′′,γ ′′) and (S ′,γ ′) have the same difference L from S, the previous Propositionshows that they are isomorphic.

This does not quite address existence. To fix this, suppose that X is a CW-complex, andtry and build up S→ X for V → X. Suppose we have already defined a spin-c structure

S(m) → Xm = (m-skeleton of X),

and consider an attaching map

φ : ∂Dm+1 = Sm → Xm,


which also gives a map ψ : Dm+1 → Xm+1 which is injective on em+1 := int(Dm+1). Eventhough ψ may not be an inclusion on Sm, we can consider φ∗(S(m)), which is a spin-cstructure for ψ∗(V)|Sn . The problem of extending the spin-c structure onto the cell Dm+1

becomes: Can I extend φ∗(S(m)) on the boundary over all of Dm+1, as a spin-c structurefor ψ∗(V)?

Since Dm+1 is contractible, it has a unique spin-c structure, corresponding to the stan-dard Clifford module for one fiber. The question is: does this trivial spin-c structure,restricted to Sm, coincide with the spin-c structure φ∗(S(m)) obtained by pullback? If yes,then we can extend over Dm+1.

Proposition 22.10. Ifm ≥ 3, then any complex line bundle L→ Sm is trivial.

Proof. This is because complex line bundles on Sm are classified by [Sm−1,U(1)].

Corollary 22.11. If X3 admits a spin-c structure S(3), then X admits a spin-c structure S, i.e.the obstruction only exists on the lower-dimensional cells. In particular, if V → X is trivialon X3, then a spin-c structure for V → X exists.

Lemma 22.12. IF a real vector bundle V → X is trivial on X2, then it is trivial on X3.

Proof. This is because π2(O(N)) = 0. In fact, all finite-dimensional Lie groups have π2 =0.

Corollary 22.13. If V → X is trivial on X2, then a spin-c structure exists.

Corollary 22.14. If w1(V) = w2(V) = 0, then V → X is trivial on X2, so a spin-c structureexists.

Proof. Show that w1(V) = w2(V) = 0 implies that V |X2 is trivial, by extending cell-by-cell.

In fact, the necessary and sufficient condition is that w1(V) = 0, and w2(V) is the Z/2-reduction of a class in H2(X;Z).

23. 3/25/16

23.1. Clifford Modules. Let V be a real vector space of dimension 2kwith inner product.Let S be an irreducible Clifford module with γ Clifford multiplication. Orient V . This

is canonical over Z/2. Let ε be the grading.

Definition 23.1. A real Clifford module (S,γ, J) is a Clifford module (S,γ) with J : S→ S

and isometry where S = Swith C-conjugate multiplication so that

γ(v)J = Jγ(v)

for all v ∈ V . Briefly, Jγ = γJ.

Lemma 23.2. If, in addition, S is irreducible, then J2 = ±1.

Proof. Note that J2 : S → S is a complex linear map commuting with Clifford multiplica-tion. So, Schur’s lemma implies J2 = λ · 1S for λ ∈ S1. Also, J2 J = J J2 so λ = λ, and soλ = ±1.

Definition 23.3. So, J is either

82 AARON LANDESMAN

(1) real, if J = 1(2) Quaternionic, if J2 = −1

Also, J is either(1) even if εJ = Jε, so

J : S+ → S+

S− → S−.

(2) odd if εJ = −Jε.

In this way, there are four sorts of J’s. For V = R2n, we have the standard

(Sk,γk, εk) .

We’ll see there is also a standard Jk. In fact, there’s only one Jk up to sign.Now, take S1 = C⊕C Take γ1 to be

γ1

(xy

)=

(0 −zz 0

)where z = x+ iy. Set

J1(a,b) = (−b,a).

Lemma 23.4. Here, J1γ = γJ1.

Proof.

Exercise 23.5. Prove this.

Lemma 23.6. J1 is Quaternionic.

Proof. We have

J21(a,b) = J1(−b,a)

= (−a,−b)= (−a,−b).

Lemma 23.7. We have J1 is odd.

Proof. We can easily check J1ε1 = −ε1J1.

For the standard Clifford modules, we know define the Jk inductively. We will defineJk in terms of J1 and Jk−1.

Definition 23.8. Use

Sk = Sk−1 ⊗ S1so that

Jk =

Jk−1 ⊗ J1 if Jk−1 is even.Jk−1 ⊗ ε1J1 if Jk−1 is odd.


k− 1 even and real even and Quaternionic odd and real odd and Quaternionick odd and Quaternionic odd and real even and real even and Quaternionic

TABLE 7. summary of oddness, evenness, realness, and Quaternionicnessvia induction

Lemma 23.9. Jk satisfies Jk γk = γk Jk.

Proof. We want to check

Jk γk(vk−1, v1) = γk(vk−1, v1) Jkwhere (vk−1, v1) ∈ R2k−1 ⊕ R2 = R2k. There are two cases to check, depending onwhether Jk−1 is odd or even. Let’s check the odd case.

Exercise 23.10. Check that this holds when Jk−1 is even.

When Jk−1 is odd, we have

(Jk−1 ⊗ ε1J1) (γk−1 ⊗ 1+ 1⊗ εk−1 ⊗ γ1) = Jk−1γk−1 ⊗ ε1J1 + Jk−1εk−1 ⊗ ε1J1γ1= γk−1Jk−1 ⊗ ε1J1 + εk−1Jk−1 ⊗ γ1ε1J1= (γk−1 ⊗ 1+ εk−1 ⊗ γ1) (Jk−1 ⊗ ε1J1) .

Question 23.11. Is Jk real or Quaternionic, even or odd?

To see this, just take the definition of J and square it, and use induction. If Jk−1 is even,we get J2k = (Jk−1 ⊗ J1)2 = J2k−1 ⊗ J21. Similarly, if Jk−1 is odd, we have

J2k =(J2k−1

)⊗ (ε1J1)

2

= −J2k−1 ⊗ (ε1ε1J1J1)

= J2k−1 ⊗ 1.Similarly, if Jk−1 is Quaternionic, then Jk is real.(1) If Jk−1 even and real, then Jk is of quaternion type.(2) If Jk−1 is even and Quaternionic, then Jk is real.(3) If Jk−1 is odd and real, Jk is real.(4) If Jk−1 is odd and Quaternionic, Jk is Quaternionic.

For evenness and oddness, we have

εk = (−i)kγ(e1)γ(e2) · · ·γ(e2k)and so εkJk = (−1)kJkεk. This tells us oddness and evenness.

So, in a table,We have

(Sk,γk, Jk)

the standard real Clifford module for R2k. The εk form an orientation for R2k.For V → X an oriented real rank 2k vector bundle, a spin structure is a vector bundle

S→ Xwith γ, J so that for all x ∈ X, on the fiber, each Sx ∼= Sk, with γk, Jk.

84 AARON LANDESMAN

k real or Quaternionic even or odd1 Quaternionic odd2 Quaternionic even3 real odd4 real even5 Quaternionic odd

TABLE 8. Table of realness, Quaternionicness, oddness, and evenness ofClifford operators, this is four periodic.

Remark 23.12. If V = V ′ ⊕R2, a spin structure for V ′ gives one for V and visa versa.Given (S,γ) for V , set S ′ to be the i or −i eigenspace of γ(e1)γ(e2). Similarly, for V =V ′ ⊕ V ′′ a spin structure for any two determines one for the third.

Remark 23.13. We can define a spin structure for an odd rank vector bundle V by beingone for V ⊕R.

Example 23.14. For V = R4, take S2 = S+2 ⊕ S−2 . For v ∈ S3 ⊂ R4, we have an isomor-

phism

γ(v) : S+ ∼= S−,

and we get (S+,S−,γ) ∈ KSp(D4,S3). Here, S+2 ,S−2 are quaternion vector spaces and γ isa quaternion linear map between them. In fact, this is the Bott generator

b4 ∈ KSp(R4).

Example 23.15. Similarly, for R8, we get(S+,S−,γ

)∈ KO(R8)

the generator.

Corollary 23.16. If V → X admits a spin structure, X is compact, and rkV = 8m or rkV =8m+ 4, then there exists a class

bV ∈KO(D(V),S(V)) if rkV = 8m

KSp(D(V),S(V)) if rkV = 8m+ 4

with the property that

bV |(D(V),S(V))

is the generator for each x ∈ X. Then, bV =(S+,S−,γ

), where we regard S+,S− as bundles

pulled back on D(V) pulled back from X.

24. 3/28/16

Let p : V → X be an oriented R2k-bundle which admits a spin structure (S,γ,σ), whereS = S+ ⊕ S−,

γ =

(0 −σ∗

σ 0

)


and σ : V → Hom(S+,S−). We obtained a class in K(V) = K(D(V),S(V)) defined by

bV = [p∗S+,p∗S−,σ].

If 2k ∈ 8Z, we have a real structure bV ∈ KO(V). If 2k ∈ 8Z + 4, we have a Quater-nionic structure bV ∈ KSp(V). We think of bV as being analogous to the Thom class.Furthermore, there is an analogue of the Thom isomorphism, as follows:

Theorem 24.1. The map ψ : K(X) → K(V) given by a 7→ p∗(a) · bV is an isomorphism. If2k ∈ 8Z, the same statement holds for the map ψ : KO(X) → KO(V). If 2k ∈ 8Z + 4, thesame statement holds for the map ψ : KSp(X)→ KSp(V).

For example, ψ sends the trivial class 1 ∈ K(X) to bV ∈ K(V). We won’t actuallyprove that this map is an isomorphism, but our proof of Bott periodicity contains allthe ingredients. The latter concerns the trivial vector bundle X×R2k, while the formerconcerns a nontrivial vector bundle p : V → X. Just as before, the definition of ψ−1 beginsby writing down Fredholm operators. Just as in Bott periodicity, we can extend this resultto locally compact X, for suitable V .

If f : X→ Y is a proper map of locally compact topological spaces, we obtain a pullbackmap f∗ : K(Y) → K(X), as follows. We consider the induced map f+ : X+ → Y+ ofthe compactifications, and recall that K(X) := K(X+), so f∗ comes from (f+)∗ on reducedK-theory of X+ and Y+.

One important special case of this is the pushforward map: If X is locally compact, andι : U → X is an inclusion of an open subset, then there is a map ι+ : X+ → U+, which is abijection on U, and collapses all of (X \U) ∪ ∞ to ∞ ⊂ U+. In this way, we get a mapι∗ : K(U)→ K(X) by the previous paragraph.

We wish to view K(X) as the degree 0 functor of an extraordinary cohomology theory.For ordinary cohomology, we have

Hk−r(X+) = Hk(X+ ∧ Sr).

Motivated by this, we defined, for r ≥ 0,

K−r(X) := K(X×Rr)

K−r(X+) := K(X+ ∧ Sr).

Note that these higher K-groups are periodic in r: by Bott periodicity, we have an isomor-phism β : K−r(X) → K−r−2(X). Similarly, we have an isomorphism KO−r−8(X) ∼= K−r(X).Because R−r is not well-defined for r < 0, we cannot directly apply the above definitionfor negative r. But we can use Bott periodicity to extend the definition, i.e. for n ≥ 0, pickkwith 8k−n ≥ 0, and define

KOn(X) := KO(X×R8k−n).

Because the periodicity isomorphisms are not just any isomorphism, but the preferred Bottisomorphism, everything is canonical so far.

Now, given a proper map f : X → Y between locally compact spaces, we obtain apullback map f∗ : KOn(Y) → KOn(X), coming from the map X×Rr → Y ×Rr which isthe identity on Rr. Similarly, given ι : U → X with U an open subset of X, we obtain apushforward map ι∗ : Kr(U)→ Kr(X).

86 AARON LANDESMAN

Similarly, if V → X is spin, and rankV = 8k, the ‘Thom isomorphism’ of the precedingtheorem extends to the higher K-groups as well, i.e.

ψ : KOn(X)→ KOn(V)

is an isomorphism.We can use ‘stability’ to remove the rankV = 8k restriction. If rankV = r, and V → X

has a spin structure, we can consider V ⊕R8k−r, whose total space is just V ×R8k−r. Sincethis latter vector bundle does have rank = 8k, we can write down the map

ψ : KO(X)→ KO(V ×R8k−r) =: KOr(V),

or, more generally,ψ : KOn(X)→ KOn+r(V), r = rankV .

These ‘dimensions’ n and n+ r agree with those of the original Thom map

ψ : Hm(X)→ Hm+r(D(V),S(V)), r = rankV ,

where instead of requiring that V have a spin structure, we require that V is oriented.

24.1. Now suppose X ⊂ Y is an embedding of smooth manifolds, and let V → X be thenormal bundle of X in Y, so that s = rankV is the codimension of X in Y. Further, supposethat V is oriented and given a spin structure. By the Tubular Neighborhood Theorem, wehave a containment X ⊂ U ⊂ Y where U is an open tubular neighborhood of Y, and U ishomeomorphic to V . If we have a metric, we might write U = Dε(V), where we think ofU as a bundle of ε-radius disks over X. In this situation, we have maps

KOn(X) KOn+s(V) = KOn+s(U) KOn+s(Y).ψ ι∗

We call the composition ι! : KOn(X)→ KOn+s(Y).On the other hand, given a map f : X → Y which is not necessarily an embedding, we

can embed X ⊂ R8k for sufficiently large k, and thereby obtain a map f : X → Y ×R8k.This modified f is an embedding; we have ‘lifted X into certain small free dimensions.’Furthermore, this f is essentially uniquely determined by f, in the sense that any two f’swith the same 8k are ambient isotopic to one another, and two f’s with different 8k’s canbe related by including R8k ⊂ R8k ′ . In this way, we again obtain a map

f! : KOm(X)→ KOm+s(Y)

depending only on f. Remark: as an element of K(X), the normal bundleNX/Y is given by

R8k + f∗(TY) − TX.

(what was the relation to Aaron’s question, which asked about a spin structure onNX/Y?)If Y itself is Rt, then to given a spin structure on V = NY/X is equivalent to giving

a spin structure on TX, i.e. a ‘spin structure on X.’ This follows from a previous result,which said that if V1 = V2 ⊕ V3 are vector bundles, then a spin structure on any twouniquely determines a spin structure on the third.

Now consider the special case f : X→ R0. The above construction gives an embeddingf : X→ R8k, and then a map

f! : KOn(X)→ KOn−dimX(pt),


e.g. f! : KOdimX(X) → KO(pt) = Z. To appreciate the relation to ordinary cohomology,recall that an oriented manifold X comes equipped with a homomorphismHdimX(X)→ Xcalled ‘evaluation on the orientation class [X].’ In the de Rham world, this is the integra-tion map

[ω] 7→ ∫Xω,

where ω ∈ HdimX(X) is a top-form. To compute this integral, we could integrate chartby chart. Alternatively, we could embed X → RN, consider ω ∈ Ωn(X), and constructan extension ω ∈ ΩN(RN) by first pulling back to U, then multiplying by a form thatrepresents the Thom class. This map ι! : ω 7→ ω is analogous to the ι! that we definedabove.

Another interesting case of the above map f! is

f! : KO0(X)→ KO−dimX(pt) = KO(SdimX).

If, for example, dimX = 8k, what does 1 ∈ KO0(X) = KOdimX(X), what is f!(1) ∈KO(pt) = Z? The answer is surprisingly complicated: given the total Pontryagin class

p(X) = 1+ p1(X) + · · ·+ pk(X),we introduce formal variables y1, . . . ,yk, such that

pj(X) = σj(y1, . . . ,yk).

where the σj are the elementary symmetric polynomials. Consider the symmetric powerseries

a(y1, . . . ,yk) =k∏i=1

√yi/2

sinh(√yi/2)

= 1−

∑i yi24

+ · · · .

Note that t/ sinh is an even power series, so only integer powers of yi appear. Supposewe write a as a power series in

σ1(y1, . . . ,yk),σ2(y1, . . . ,yk), . . . ,σk(y1, . . . ,yk)

and substitute pj for σj. Then we get a class ‘A-roof’

A = 1−p124

+7

5760(p21 − 2p2) + · · · .

Theorem 24.2. If X is compact and spin, then f!(1) ∈ Z equals A([X]), where [X] is theorientation class. For example, if dimX = 8, then

1

27 · 45

(7p21 − 4p2

)[X]

In particular, in this example, we find that p21(X) − 2p2(X) is divisible by 5760 whendimX = 8. The reason the formula is so complicated is the fact that the definition ofthese Thom maps relied on bV , which relied on S+ and S−, whose definitions are quiteelaborate.

88 AARON LANDESMAN

25. 3/30/16

25.1. Review. Today, we’ll do a calculation to explain where the characteristic class A,defined last time, comes from.

Let V be an oriented vector space of dimension 2k with a spin structure. Say S =S+ ⊕ S−. We can form the formal difference [S+] − [S−] ∈ K(X).

This has a Chern character. Let’s calculate

ch([S+]−[S−])

.

Write

V = V1 ⊕ · · · ⊕ Vk.

We have a special case that Vi has rank 2 and each Vi has a spin structure Si. Define

S := ⊗iSi.

Let V1 be an oriented bundle with fiber R2 ∼= C. So, we have V1 = (W1)R with W1 acomplex line bundle. We have

S1 := S+1 ⊕ S

−1 =: L1 ⊕ L ′1

That is, L1,L ′1 are just renamings of S+1 ,S ′1, and are just complex line bundles. We have amap

J : L1 ∼= L′1,

mapping with complex conjugation. So, if x = c1(L1) then −x = c1(L′1).

The fibers of V1 are R2. Recall we have a Clifford multiplication map

γ :=

(0 σ∗

σ 0

).

Here,

σ : V1 → Hom(L1,L ′1)(xy

)7→ x− iy = z.

Then,

σ :W1 → Hom(L1,L ′1).

So,

−c1(W1) = c1

(L∨1 ⊗ L ′1

)= c1(L1 ⊗ L1)= −2c1(L1).


We also get

x = c1(L1)

=1

2c1(W1)

=1

2e(V1)

=:1

2t1

In the last line, t1 is just defined to be the Euler class of V1.

25.2. Calculating A. Now, the Pontryagin class is

p1(V1) ∈ H4(X;Q).

Lemma 25.1. If we set

y1 := p1(V1)

t1 := e(V1).

We have the relation that

y1 = t21.

Proof.

p1(V1) = −c2(V1 ⊗R C)

= −c2(W1 ⊗R C)

= −c2(W1 ⊕W1)

= − [(1+ c1(W1)) (1− c1(W1))]H4

= c1(W1)2

= t21.

Lemma 25.2. We have

ch(S+1 − S−1 ) = 2 sinh t1/2

Proof. Indeed,

ch(S+1 − S−1 ) = ch(L1 − L′1)

= ch(L1) − ch(L′1)

= ec1(L1) − ec1(L′1)

= et1/2 − e−t1/2

= 2 sinh t1/2.

90 AARON LANDESMAN

Corollary 25.3. For V = V1 ⊕ · · ·Vk, S = S1 ⊗ · · · ⊗ Sk, we have

ch(S+ − S−) =

n∏i=1

ch(S+i − S−i )

=

n∏i=1

2 sinh ti/2

where t2i = yi = p1(Vi).

Proof. Use the preceding lemmas.

As in the previous class, let X be a compact manifold with spin structure. Let X → RN

be an embedding. LetV be the normal bundle. Then, V gets a spin structure. We haveV orD0(V) ∼= U ⊂ RN, with U a tubular neighborhood. Then, let bV ∈ K(V) = K(D(V),S(V).We have

bV =(p∗S

+,p∗S−,σ)

.

Question 25.4. Can we compute ch(bV) ∈ H∗(D(V),S(V))?

Recall the Thom isomorphism theorem:

Theorem 25.5. Let uV ∈ H2k(D(V),S(V)) be the Thom class of V and let p : V → X be theprojection. Retain the notation given above. Then, there exists a unique a(V) ∈ H∗(X) so that wecan write ch(bV) = uV ∪ p∗a(V).

Restricting the Thom isomorphism theorem to X can look at

ch(bV)|X = e(V)∪ a(V).So,

ch(S+ − S−) = e(V)∪ a(V),because the restriction of bV to X is S+ − S−.

Now, recall

e(V) =

k∏i=1

e(Vi) =

k∏i=1

ti.

So, we haven∏i=1

2 sinh ti/2 =

(n∏i=1

ti

)∪ a(V)

These are odd power series in ti, so we can consider(∏i

ti

)a(V) =

(∏i

ti

)·∏i

sinh ti/2ti/2

.

The second term is even in each ti separately. We’d like to cancel the∏i ti from the left

and right. We can’t cancel in general. To deduce we can cancel them, you have to use


that X is a manifold embedded into RN. We’re skipping the intricacies here, but using theuniversal example, you can justify cancellation to deduce

a(V) =

n∏i=1

sinh ti/2ti/2

=

n∏i=1

sinh√yi/2

√yi/2

,

where the last inequality is only as formal power series.

25.3. Relating our computations to A. Now, let V be This looks similar to A defined lasttime, which was

A(V) =∏ √

yi/2

sinh√yi/2

.

This is a symmetric power series in yi. Hence, also a power series in

σj(y1, . . . ,yk),

hence a power series in pj(V).Since,

A(V ⊕W) = A(V)A(W).

We get a map

(Vect(X),⊕)→ (H∗(X))× .

Elements of the latter are power series starting with 1. So, we get a map

K(X)→ (H∗(X, Q))×

Therefore, for −V ∈ K(X), we got

a(V) = A(−V)

=1

A(V).

Now, going back to the situation that X is embedded in RN, we have

V ⊕ TX = Rr

on X. So,

a(V) = A(TX) := A(X).

So, we can conclude the following:

Theorem 25.6. Let uV be the Thom class. Then,

ch(bV) = uV ∪ p∗A(X) ∈ H∗(D(V),S(V)).

92 AARON LANDESMAN

Remark 25.7. Observe that we have a fundamental class

[D(V),S(V)] ∈ HN(D(V),S(V)).

So, we can calculate

ch(bV) [D(V),S(V)] .

By the Thom isomorphism,

ch(bV) [D(V),S(V)] = A(X)[X]

=: A[X]

where the last equality is just notation.

Remark 25.8. In fact, using excision, where U is the tubular neighborhood of X ⊂ V ,

H∗(D(V),S(V)) = H∗(D(V),D(V) \D0(V))

= H∗(RN, RN \U)

This has a map to H∗(SN). We have a map H∗(RN, RN \U) → H∗(SN). because there is adegree 1 map SN → D(V)/S(V) given by crushing the outside. This is the same map wegot from pushing forwards elements of K theory.

So, from last time, we have a map

i∗ : U→ RN

bV 7→ i!(1) ∈ K(RN).

We have computed

ch(i!(1)) = A[X] ∈ H∗(SN, Q),

as all power series here have rational coefficients. So, we have A[X] is some rationalnumber.

Using Bott periodicity, we concluded that for all x ∈ K(SN),

ch(x) ∈ H∗(SN;Q

) [SN]∈ Q

is actually an integer, because we computed the value on the generator of the cohomologyof SN.

Hence, we obtain the following corollary.

Corollary 25.9. For a compact smooth manifold, A[X] ∈ Z.

Remark 25.10. (At the end of last time this was written incorrectly.) We havep124

[X]

in dimension 4. We have1

27 · 45

(7p21 − 4p2

)[X]

in dimension 8.


Lemma 25.11. As an extra bonus, A[X] ∈ 2Z if dimX ∼= 4 mod 8.

Proof. In the case that dimX ∼= 4 mod 8. Then, bV ∈ im KSp(V). The map

KSp(RN)→ K(RN)

Z×2−−→ Z.

So, in fact, p1[X] ∈ 48Z.

26. 4/1/16

Today, we’ll talk about an application of periodicity for KO.

Definition 26.1. A smooth manifold M is parallelizable if TM →M is the trivial bundleTM = Rn, where n = dimM.

Remark 26.2. Note thatM is parallelizable if and only if there exist vector fields v1, . . . , vnonMwhich are linearly independent at each point.

Theorem 26.3. The sphere Sn is parallelizable if and only if n = 0, 1, 3, 7.

Proof. We will prove this in several steps.

26.0.1. Step 1. In this step, we prove the following lemma.

Lemma 26.4. If Sn is parallelizable, there exists a real vector bundle V on Sn+1 of rank n+ 1with e(V)[Sn+1] = 1.

Proof. Think of Sn ⊂ Rn+1. A vector v ∈ Sn, with |v| = 1.If Sn is parallelizable, there exist u1, . . . ,un which are vector fields and everywhere

linearly independent. These can be thought of as maps

ui : Sn → Rn+1.

Further, we may assume that ui(v) ⊥ v and

〈ui(v),uj(v)〉 = δij.

In other words, we get a map

ψ : Sn → SO(n+ 1)

v 7→ [v,u1(v), . . . ,un(v)] .

Then, we will use ψ as a clutching function, or transition function for a rank (n + 1)bundle V on Sn+1.

On an open set V = Rn+1 on the boundary, we have s(v) = v. If s is a section, where sis the first vector, we can extend this as a section with one transverse 0 at the north pole.Therefore, e(V), which is the number of 0’s of s, counted with multiplicity, is equal to ±1,which defined up to orientation.

94 AARON LANDESMAN

26.0.2. Step 2.

Corollary 26.5. If Sn is parallelizable, then there exists a vector bundle V → Sn+1 with

wn+1(V) 6= 0 ∈ Hn+1(Sn+1, Z/2) ∼= Z/2.

Proof. This follows from Lemma 26.4.

26.0.3. Step 3. In order to prove our main theorem, we will show that unless m = 1, 2, 4or 8, we will show that every V → Sm has wm(V) = 0. Since m = n+ 1, this is what wewanted to show. The idea is to use Bott periodicity.

We have

w(V) = w(V ⊕Rr).

So, we are really saying that wm(c) = 0 for c ∈ KO(Sm) form 6= 1, 2, 4, 8.Look at wj for bundles of the form

E F→ X× Y,

given pX : E→ X,pY : F→ Y vector bundles, where E F := p∗(E)⊗ p∗Y(F).Take ai := wi(E). We won’t distinguish between ai and p∗X(ai) ∈ H∗(X × Y, Z/2).

Similarly, notate bj := wj(F). We’re used to writing w(E) = 1+ a1 + · · ·+ ar. Define

wT (E) = Tr + a1T

r−1 + · · ·+ ar ∈ H∗(X;Z/2)[T ].

If

E = ⊕iLiF = ⊕jMj,

then

wT (E) =

r∏i=1

(T + xi)

wT (F) =

r∏i=1

(T + yi) .

Here,

xi := w1(Li)

yi := w1(Mi).

Then,

wT (E⊗ F) = wT (⊕i,jLi ⊗Mj)

=∏i,j

(1+ xi + yj

).

Then, when one expands this out, one is supposed to replace σk(xi) with ak and replaceσk(yj) with bk. Then, one should see what one gets, and note that it is independent ofwhether E, F actually split, using the splitting principle.


26.0.4. Step 4. This step wanders a bit.(26.1)Assume r = rkE = 8, (or, we could also take 2m), and assume that a1 = · · ·a7 = 0 and a8 = a.

Therefore,

wT (E) = T8 + a.

Remark 26.6. If Xwere S8, this would be inevitable, as there is no cohomology in dimen-sions between 1 and 7.

So,

8∏i=1

(1+ xi) = T8 + a.

Since we’re working over a field of characteristic 2, if we also take the splitting field, sothat we work over a field with an eighth root of a, x8 = a, we have

T8 + a = (T + x)8.

So, formally, we can consider the expansion∏j

∏i

(T + x+ yj

),

as we can assume all the x roots are the same from the above assumptions. Therefore,∏j

∏i

(T + x+ yj

)=

s∏j=1

(T8 + x8 + y8j

)=∏j

(T8 + a+ y8j

)= P(T8,a,σ(y81, . . . ,y

8s))

= P(T8,a,σk(y1, . . . ,ys)8)

= P(T8,a,b81, . . . ,b8s).

(26.2)Now, assume on Y that all 8th powers of classes in Hk(Y, Z/2) are equal to 0 for k > 0.

Hence,

wT (E⊗ F) = P(T8,a, 0)

=

s∏i=1

(T8 + a

)= (T8 + a)s.

96 AARON LANDESMAN

26.0.5. Step 5.

Corollary 26.7. If E → S8 has rank 8 (playing the role of Y from the previous step) and F → Ss

has rank s (playing the role of X from the previous step), then

wT (E F) ∈ H∗(S8 × Ss, Z/2)

Proof. Note thatw8+s(E F) ∈ H8+s(S8× Ss, Z/2) is independent of F, using the previousstep. So, w8+s(E (F− Rs)) = 0.

Or, this also holds because all wk(E F) are pulled back from S8.

26.0.6. Step 6. Apply Corollary 26.7 with the Bott generator on S8,

b8 := E−[R8]

,

where E is some rank 8 bundle. This says

wT (b8 · f) = T0,= 1.

since allwk vanish for b8 · f (you only needed to check the top class vanished because theonly nonvanishing cohomologies of Sn are 0 and n). Hence, this shows that all spheres ofdimension more than 8 cannot be parallelizable, since we have shown the desired state-ment for KO(S8+s), when s > 0.

26.0.7. Step 7. Recall the following table from Bott periodicity. Using the above table,

n KO(Sn)1 Z/22 Z/23 04 Z

5 06 07 08 Z

TABLE 9. Bott periodicity

these groups are 0 for numbers which are not powers of 2, and hence they are not par-allelizable. It is elementary to see that S0,S1,S3,S7 are parallelizable, as follows fromexistence of algebras R, C, H, O. (What we needed here was the existence of a bilinearmap, so that multiplication by v is invertible and there is an identity element.) In eachcase, the vector desired vector field v 7→ ui(v) is given by multiplication

v 7→ Iiv


where 1, . . . , I1, . . . , I2r form an orthonormal basis for the corresponding real algebra.Equivalently, there is a map

Sn → SO(n+ 1)

v 7→ (1 · v, I1 · v, I2, ·v, . . . , I2r · v) .

Remark 26.8. This was first proved as a consequence of the Hopf invariant 1 problem,using a harder k theory.

27. 4/4/16

27.1. Formalities. First, we give some formalities for K, K to be a cohomology theory.Let (X,A) be compact pairs and X a pointed space. Suppose for all X, h(X) is a pointed

set or group. Suppose

h : pointed compact spaces → pointed sets

is a contravariant functor. The category of pointed compact spaces is that whose objectsare pointed spaces and whose morphisms are homotopy classes of pointed maps.

Example 27.1. One example of such a functor is reduced K-theory, h = K.

Remark 27.2. We’ll require the following axioms:(1) h(pt) = 0 .(2) If

(27.1) A X X/Ai j

where i is an inclusion is exact at h(X) in

(27.2) h(X/A) h(X) h(A)j∗ i∗

(but the first and last maps need not be injections or surjections)(3) If A is contractible then

j∗ : h(X/A)→ h(X)

is injective.

Remark 27.3. Surjectivity follows from the first two conditions, since the first im-plies h(A) = 0, and the exactness from the second condition is surjectivity. Sobeing injective is equivalent to an isomorphism.

Lemma 27.4. Reduced K theory, K, has these properties.

Proof. The first and one direction of the second condition of Remark 27.2 are clear. In thesecond condition of the only nontrivial point is to show

ker i∗ ⊂ im j∗

To see this, note that an element of ker i∗ can be written as

[E] − [F] ∈ K(X)

98 AARON LANDESMAN

such that

E|A ∼= F|A

after stabilizing (i.e., after direct summing both with a high power of the trivial bundle⊕CN).

So, we have a triple

([E] , [F] , τ) ∈ K(X,A).

From this isomorphism τ, we can crush A to a point, since

K(X,A) = K(X/A).

To check the third condition, we have

([E] , [F] , τ) ∈ K(X,A) = K(X/A)

and E ∼= F on X after stabilizing (direct summing with some CN). (Following from theassumption that their image agrees under j∗.) Now, we have(

CM, CM, τ)

.

We have a map

τ : A→ GL(M, C)→ isoms(CM, CM).

To be 0, we need to extend τ to all of X. That is, we need this τ to be the constant map,which is invertible if A is contractible.

27.2. Notations.

Definition 27.5. Write CX for the cone on X, defined by

CX :=[0, 1]× X0× X .

Note that X has a preferred basepoint, which is the image of 0× X.

Definition 27.6. From the cone, we can form the unreduced suspension, defined by

ΣX :=CX

1× X .

This has two distinguished points which are the images of

1× X, 0× X.

Remark 27.7. If X itself has a basepoint x0 ∈ X, we have the reduced suspension is thesmash product

S1 ∧X := (I× X)(.∂I× X)∪ I× x0.

Definition 27.8. We’ll write SX for the reduced suspension S1 ∧X.

Lemma 27.9. We have

h(ΣX) ∼= h(SX).

Proof. Use the third axiom from Remark 27.2.


Definition 27.10. For q ∈ Z≥1, define

SqX = Sq ∧X

= S1 ∧ · · ·∧ S1 ∧X= S · · · SX,

where S is written q times.

27.3. The exact sequence for K-theory in negative degrees.

Definition 27.11. Given h satisfying the three axioms from Remark 27.2, define for q ≥ 0,

h−q(X) := h(SqX)

to be K−q.

Theorem 27.12. For

(27.3) A X X/A,i j

we have the exact sequence.

(27.4)

h−1(X/A) h−1(X) h−1(A)

h(X/A) h(X) h(A).

j∗ i∗

δ

j∗ i∗

Proof. We have

(27.5)

h(S(X/A)) h(SX) h(SA)

h(SX/SA) h(SX) h(SA)

where the vertical maps are equality. The bottom sequence is exact by the second axiomof Remark 27.2.

We have a pair (X,A), and can form X∪CA glued by

X∐CA

a ∼ (1,a).

Make the following observations.(1) Observe that

h(X∪CA) ∼= h

(X∪CACA

)∼= h(X/A).

using property three of Remark 27.2, since CA is contractible.

100 AARON LANDESMAN

(2) We have

X∪CAX

= SA

Using the above two observations, we have that δ is the map

h−1(A) = h(SA)

= h

(X∪CAX

)→ h(X∩CA)

∼= h

(X∪CACA

)= h(X/A).

Observe that the map on the third line above is the only line which is not necessarily anisomorphism.

Consider the following three pairs.

(1) (X,A)(2) (X∪CA,X)(3) (X∪C1A∪C2X,X∪C1A).

Here, C1,C2 are not iterated cones, but just a gluing of X,C1A, and CX where the lattertwo cones are identified with A and X along the end opposite their cone point.

If we want, we can think of this as a subset of X× [−1, 1] with the cone over A havingpositive second coordinate and the cone over X having negative second coordinate.

That is, inside

X∐

I1 × X∐

I2 × X

we have the equivalence relations

a ∼ (11,a)x ∼ (12, x) .

The second axiom of Remark 27.2 implies that h(X) is exact in

(27.6) h(X/A) h(X) h(A).

So, the second axiom of Remark 27.2, applied to (X∪CA,X) at X∪CA in

(27.7) h(X∪CACA

)h(X∪CA) h(X)

is exact in the middle.


We further have vertical isomorphisms

(27.8)

h(X∪CACA

)h(X∪CA) h(X)

h(SA) h(X∪CACA

)h(X).

Unravelling the bottom objects, we can identify the maps with

(27.9)

h (SA) h(X/A) h(X)

h−1(A) h0 (X/A) h0(X).

Checking the commutativity of the maps is a little more subtle.The third property of Remark 27.2 implies exactness in the middle of

(27.10) h(X∪C1A∪C2XX∪C1A

)h (X∪C1A∪C2X) h(X∪C1A).

We can identify the above with vertical isomorphisms

(27.11)

h(X∪C1A∪C2XX∪C1A

)h (X∪C1A∪C2X) h(X∪C1A)

h (SX) h(X∪C1AX

)h(X/A)

h−1 (X) h−1 (A) h(X/A)

This is exact at the middle

Exercise 27.13. Check this. These maps are not actually the same maps (the left map isthe one with the problem. The issue is that we use a copy I2 and I1 of the interval, but itterms out to have the same image, and gives the desired isomorphisms. But, the differ byautomorphisms of the respective sets, so the images and kernels are the same.

28. 4/6/16

28.1. Review. Last class, we defined h(X), where X is a pointed space. We defined

h−1(X) := h(SqX),

102 AARON LANDESMAN

and we got the exact sequence

(28.1)

h−q(X/A) h−q(X) h−q(A)

h−q+1(X/A) h−q+1(X) · · · .

This continues down until

(28.2)

h−1(X/A) h−1(X) h−q(A)

h0(X/A) h0(X) h0(A).

For h = K or h = KO, we get periodicity maps

β = β8 : K−q ∼= K−q−8

and β commutes with the maps in the long exact sequence. Recall we have

K(X,A) = K(X/A)

for compact pairs (X,A). Define

Kn(X,A) := K(X/A).

Example 28.1. We have

K0(S2n) = Z⊕Z.

This has generators

C

of rank 1 and c1(C) = 0, and on S2, the other generators isH−C of rank 0 and c1(H−C) =1.

We have K1(S2n) given by

K1(S2n) = K(S2n∐

+)

= K1(S2n)

= 0

To see the penultimate equality, take the union of S2n∐

+, and we take a basepointx0 ∈ S2n, we have a pair of points x0,+ ⊂ S2n

∐+. Looking at the sequence of this

pair in reduced K-theory, we have

(28.3) K1(S2n K1(S2n∐

+) K1 (x∐

+)

where we know K1(x∐

+) = 0. We also have K1(S2n) = K0(S2n+1). Therefore, K1(S2n∐

+).


Example 28.2. We have K0(S2n+1) = Z. By a similar computation to the above example,we have K1(S2n+1) = K1(S2n+1) = Z.

Example 28.3. We have K−1(X) = K0(SX). We know the suspension is a union of twocones over X, meeting along X. A vector bundle on SX is two vector bundles on the coneover X (necessarily trivial on each cone separately) glued together. Therefore,

K−1(X) = K0(SX)

= [X,U]

where U = limU(N).

Example 28.4.

K0(RP1) = K0(S1) = 0.

We also have

K1(RP1) = K1(S1) = Z.

Example 28.5. Let’s look at the K theory of RP1. We have a sequence

(28.4) RP1 RP2 RP1/RP1.i j

Note that RP2/RP1 ∼= S2. So, we get an exact sequence

(28.5) K−1(RP1) K0(S2) K0(RP2) K0(RP1 0.∂ j∗ i∗

This sequence is

(28.6) Z Z K0(RP2) 0

So, we have K0(RP2) ∼= Z/d for some d, where d comes from the map ∂ corresponding tomultiplication by d : Z → Z. We also have a sequence corresponding to S1 ∪φ1 e2 = D2.This is contractible, so K∗(D2) = 0. We have a sequence

(28.7) S1 D2 S2.

The corresponding map for this sequence is multiplication by 1. We then have a mapbetween this exact sequence and the one above for RP2. On the level of groups it mapsthis S1 to RP1 given by multiplication by 2. From this, one can deduce d = 2, by chasingthe diagram.

Exercise 28.6. Make the above description precise.

So,

K0(RP2) = Z/2.

For this, we needed to know that given a map S1 → S1 of degree d yields a map

f∗ : K−1(S1)→ K−1(S1)

104 AARON LANDESMAN

is also multiplication by 1. One can show this by reducing it to a question about a mapK0(S2)→ K0(S2). So, one just needs to know that pullbacks of bundles on S2 by a map ofdegree d is multiplication by degree d.

Example 28.7. What is K1(RP2)?Similarly to the previous example, we have a sequence

(28.8) K0(RP1) K1(S2) K1(RP2) K1(RP1) K2(S2).∂

This sequence is

(28.9) 0 0 K1(RP2) Z Z×2

using for the last term that K2 = K0. and so K1(RP2) = 0.

Example 28.8. Let’s compute the K theory of RP3. We have a sequence

(28.10) 0 K0(RP3) K0(RP2) K1(S3)∂

This is

(28.11) 0 K0(RP3) Z/2 Z 0.

So, K0(RP3) = Z/2.

Example 28.9. Question: What is K1(RP3). Using a similar calculation to the above, wehave

(28.12) K0(RP2) K1(S3) K1(RP3) K1(RP2).∂ i∗

This yields an exact sequence

(28.13) Z/2 Z K1(RP3) 0

and so K1(RP3) ∼= Z.

Example 28.10. When we try to compute K1(RP3), and look at the corresponding se-quence, something new gets in our way. We have the analogous sequence

(28.14) K−1(RP3) K0(S4) K0(RP4) K0(RP3)→ r K1(S4)∂ f

This sequence is

(28.15) Z Z K0(RP4) Z/2 0.f

We now have to compute the multiplication for the ∂ map. Here, this map is multiplica-tion by 2. From this, we know the image of f is Z/2. Therefore, K0(RP4) is an extensionof Z/2 by Z/2. Therefore, it is either Z/2⊕Z/2 or Z/4.


Exercise 28.11. Show that the answer is Z/4. Hint: One way to prove this, which worksin dimension 4, but will fail in higher dimension is to find a bundle F on RP4 so that F⊕ Fhas nonzero characteristic classes. From the above sequence, the map

K0(RP4)→ K0(RP3)

is surjective. So, any bundle on RP3 extends to one on K0(RP4). In fact, the map

K0(RP4)→ K0(RP3)

is also surjective. From this, deduce there exists a complex line bundle L → RP4 withw2(LR) 6= 0 ∈ H2(RP4, Z/2). This is because there exists such a line bundle on RP2, andso the line bundle extends. Therefore, we have w4(L ⊕ L) 6= 0. Then, x := L − [C] ∈K0(RP4) satisfies 2x 6= 0. In particular, the group cannot be Z/2⊕Z/2.

Carrying this on, you obtain a similar uncertainty in RP6, although RP5 has K theorywhich we can find.

Example 28.12. In general, for real projective spaces, one gets K-theory Z/2k for increas-ing k. For K given A,X,X/A, one has a periodic sequence of length 6. For KO, one has a24 term exact sequence for i ranging from 0 to 7.

One can probably use all the 0’s for KO(Sn) to calculate KO(RPn) up to n = 4 or so.

28.2. Bordism and cobordism. Next time, we’ll do this in more detail, but we start now.

Definition 28.13. First, we have a notion of unoriented bordism. Consider the set

Λn := smooth compact nmanifolds / ∼

whereM0 ∼M1 if there is a smooth compact (n+ 1) manifoldMwith boundaryN := ∂Mso that N =M0

∐M1.


Λ0 = ∅, ∗

Note that any even number of points is cobordant to the empty set, by drawing a bunchof intervals.

Example 28.15. For 1manifolds, we have

Λ1 = ∅ .

Note that S1 is cobordant to ∅ via D2.


Λ2 =

∅, RP2

,

although we didn’t show this today.

106 AARON LANDESMAN

29. 4/8/2016

Picking up from last time, we wish to show that Λ2 = ∅, RP2. One way to see thatRP2 is nontrivial (under cobordism) is via the Stiefel-Whitney numbers:

Definition 29.1. LetM be an n-dimensional manifold. Let I be a multi-index, i.e.

wI = wn1i1· · ·wnkik ,

where∑r nrir = n, so that wI ∈ Hn(M;Z/2). We obtain a number by evaluating on the

fundamental class [M] ∈ Hn(M;Z/2), to obtain wI[M] ∈ Z/2.

Lemma 29.2. IfM0 andM1 are cobordant, in the sense thatM0 ∼M1 in Λn, then

wI[M0] = wI[M1],

for all I as in the above Definition.

Remark 29.3. In fact, the converse holds, so the problem of determining Λn reduces todescribing which possible sets of Stiefel-Whitney numbers arise among n-dimensionalmanifolds.

Proof. If N is an (n+ 1)-dimensional manifold realizing the cobordism M0 ∼ M1, in thesense that ∂N =M0 tM1, then we have

TN|M0 = TM0 ⊕R

TN|M1 = TM1 ⊕R,

where the trivial bundles R corresponds to the outward normal direction. The Whitneysum formula then tells us that

wi(TN)|wi = wI(Mi) ∈ Hn(Mi;Z/2)

for i ∈ 1, 2. We also know that, in the long exact sequence of the pair (N,∂N), we havethis map:

Hn+1(N,∂N;Z/2) Hn(∂N;Z/2) Hn(N;Z/2)

[N] [∂N] = [M0] + [M1] 0

Therefore [M0] = [M1] considered as elements of Hn(N;Z/2), implying that

wI(TN)[M0] = wI(TN)[M1].

The fact that RP2 6∼ ∅ now follows from either w21[RP2] = 1 6= 0 or w2[RP2] = 1 6= 0.To see that these are all the elements of Λ2, we discuss the classification of 2-manifolds.

Any connected 2-manifold is either a connected sum

T2#T2# · · · #T2

where T2 = S2 is the torus, orRP2#RP2# · · · #RP2.

Lemma 29.4. These cobordism classes are equal: [M0#M1] = [M0 tM1].


Proof. The cobordism is realized by a manifold N built from M0 × [0, 1] and M1 × [0, 1].Consider n-disks on M0 × 1 and M1 × 1, and attach these n-dimensional disks at the”top” of the cylinders by a cylinder Dn × [2, 3].

Lemma 29.5. In Λn, we haveMtM ' ∅.

Proof. Consider the manifold N =M× [0, 1].

The result Λ2 = ∅, RP2 now follows from the above Lemmas, although with theobservation that T2 ∼ ∅ via the solid torus N = S1 ×D2.

Proposition 29.6. The set Λn is an abelian group under disjoint union, with identity ele-ment [∅] and inverses −[M] = [M].

Proposition 29.7. The set⊕∞n=0Λn is a commutative ring, with identity element [pt] and

multiplication [M0]× [M1] := [M0 ×M1].

29.1. Oriented cobordism.

Definition 29.8. Let Ωn be the set of equivalence classes of oriented n-manifolds, underthe equivalence relation M0 ∼ M1 whenever there exists an oriented (n+ 1)-dimensionalmanifoldN such that ∂N = (−M0)tM1. Here −M0 denotesM0 with the opposite orien-tation.

Here, we need to specify the boundary orientation on ∂N. The usual convention is‘outward normal first.’ This means that, at each p ∈ ∂N, we take an outward normalvector e0 ∈ TpN, as well as an ordered basis (e1, . . . , en) for Tp(∂N) ⊂ TpN. The orderedbasis (e1, . . . , en) is declared to be positively oriented for Tp(∂N) precisely when the basis(e0, e1, . . . , en) is positively oriented for TpN.

If A and B are oriented manifolds, then A×B is oriented at (x,y) ∈ A×B by the directsum decomposition

T(x,y)(A× B) = TxA⊕ TyB,

whereby an oriented basis for the left hand side is given by taking an oriented basis forTxA, and then an oriented basis for TyB.

Example 29.9. For an oriented n-manifold M, orient N = [0, 1]×M by the product orien-tation. Then

∂N = 1×M− 0×Maccording to the ‘outward normal first’ boundary orientation.

Proposition 29.10. The set Ωn is an abelian group under [M1] + [M2] := [M1 tM2]. Infact, [∅] is the identity element, and inverse are given by −[M] := [−M], i.e. oppositeorientation.

Proposition 29.11. The set⊕∞n=0Ωn is a ring under [M1] × [M2] = [M1 ×M2] with the

product orientation. The identity element is [pt], where pt is a ‘positive point.’ (An orien-tation of a point is a sign ±1.) This ring is not commutative but graded commutative:

[M1]× [M2] = (−1)(dimM1)(dimM2)[M2]× [M1].

108 AARON LANDESMAN

Remark 29.12. The two orientations of a point deserves clarification. For us, an orientationof a vector space V means a choice of basis vector of the one-dimensional vector spaceΛdimVV . Since λ00 = R canonically, we see that a point has two orientations, ±1, andfurthermore the +1 ∈ R can be singled out as the canonical orientation of the point. In thisway, every orientable connected (possibly zero-dimensional) manifold has exactly twoorientations.

Proposition 29.13. We haveΩ0 = Z as abelian groups.

Proof. Positive points can be canceled with negative points by an arc joining them. Theresult follows because, the boundary of each one-dimensional manifold (with boundary)has total index zero.

Proposition 29.14. We haveΩ1 = 0 as abelian groups.

Proof. The only orientable 1-manifolds are disjoint unions of circles, and a circle is trivial-ized via a disk D2.

Proposition 29.15. We haveΩ2 = 0 as abelian groups.

Proof. All connected 2-manifolds are connected sums of tori, and tori are oriented cobor-dant to ∅.

Lemma 29.16. If [M0] = [M1] in Ωn, then pI[M0] = pI[M1] where pI = pn1i1· · · pnrir where∑

r 4nrir = dimM are the Pontryagin classes.

Proof. Similar to the analogous result for Stiefel-Whitney number.

Theorem 29.17. We haveΩ3 = 0,Ω4 = Z, andΩ5 = Z/2.

Proof. The statement Ω3 = 0 says that every three-manifold is the boundary of a four-manifold with boundary. (Didn’t mention how to prove this.)

To see that Ω4 = Z, we note that two 4-manifolds satisfy M40 ∼M

41 in Ω4 if and only if

p1[M0] = p1[M1], where this inequality takes place in 3Z. A generator is given by CP2,for which p1[CP2] = −3.

Instead of carrying on with the futile task of directly classifying n-dimensional mani-folds, we use more algebraic topology machinery: we shall realize Ωn as the homotopygroup of a space, to be described below. This will help us computeΩn.

For N ≥ n, let Ωn,N be the set of equivalence classes of oriented n-manifolds M em-bedded in RN, modulo the equivalence relation of ‘cobordism within RN,’ in the sensethat M0 ∼ M1 whenever there exists an (n + 1)-dimensional manifold N embedded in[0, 1]×RN, such that ∂N = −M0 tM1.

Remark 29.18. Consider n-manifolds M embedded in RN. For N sufficiently large (de-pending on n), any two embeddings will be ambient isotopic to one another. For thisreason,Ωn,N should be thought of as an approximation toΩn, especially for large N.

30. 4/11/16

30.1. Review. RecallΩk from last time, which was cobordism classes of oriented k-manifolds.We defineΩk,n to be classes of oriented k-manifoldsM ⊂ Rn+k. The equivalence relationis given by smoothly embedded manifolds N ⊂ Rn+k × I, where I = [0, 1].


Proposition 30.1. The inclusion map

Ωk,n → Ωk

is surjective if n ≥ k+ 1 and is an isomorphism if n ≥ k+ 2.

Proof. We will not prove this whole result, but it follows from the Whitney embeddingtheorem.

Theorem 30.2 (Whitney embedding theorem). Let M be a manifold. There exists an embed-dingM → RN−1 if N− 1 ≥ 2k+ 1.

Proof. Given M compact and smooth, we can embed M → RN, with N ∈ Z. Givenan embedding M → RN, we can project pu : RN → RN−1, projecting orthogonally toa unit vector u, can we choose u so that pu(M) is an embedding. You do a dimensioncount, which you then need to back up with Sard’s theorem. We have to avoid u whichis parallel to x1 − x2 for xi ∈ M. You also have to avoid u parallel to tangent vectors toM. The elements parallel is to pairs of points is 2k dimensional and the tangent vectorsare 2k− 1 dimensional. Here, u ∈ SN−1. So, by Sard’s theorem this is an embedding ifN− 1 > 2k.

Surjectivity follows from the Whitney embedding theorem, Theorem 30.2. For the iso-morphism, we will need to embed the cobordism N similarly.

LetGn(RN( be a grassmannian. Let G(RN) be the grassmannian of orientedn-dimensional

subspaces. We have a two fold cover

Gn(RN)→ Gn(R

N).

There is a tautological oriented vector bundle

En,N → Gn(RN)

so that the fiber over a point is the corresponding oriented n-dimensional plane. AsN→∞, we get a universal oriented bundle

En → Gn(R∞).

Definition 30.3. The Thom space

τ(En) :=B(En)

S(En).

As a set, this is the union of the open unit balls B(En) ∪ ∞. We similarly have a Thomspace for finite Grassmannians,

τ(En,N) :=B(En,N)

S(En,N),

which is a finite dimensional cell complex.

Proposition 30.4 (Thom). There is an isomorphism

Ψk,N : πk+n (τ (En))→ Ωk,n.

There is an analogous statement for the unoriented grassmannian Gn(R∞) and Λk,n.

110 AARON LANDESMAN

Proof. Here are the maps in both directions (i.e., Ψk,N and its inverse). We give a construc-tion

πk+n(τ(En))→ Ωk,n.

Take

[f] ∈ πk+n (τ (En,N))

which is a representation of

f : Sk+n → τ(En,N) ⊂ τ (En) .

Here, τ(En) is an infinite cell complex which is the union of the finite cell complexesτ(En,N). Every compact subset of τ(En) is contained in τ(En,N) for some N ∈ Z.

Think of Sn+k = Rn+k ∪ ∞ . Set U = f−1 (B(En,N)) . Here, B(En,N) is τ− ∞, where∞ corresponds to the collapsed boundary of the balls over each fiber. So, U ⊂ Rk+n.

Then, f is a continuous map between smooth manifolds. This can be approximated bya smooth map, so we may assume f|U is smooth and transverse to Z, where Z is the zerosection in the target. That is, we may assume that for x ∈ f−1(Z), we have

im dfx ⊕ Tf(x)Z = Tf(x)B(En,N).

Hence, f−1(Z) is a submanifold of U ⊂ Rk+n of codimension n.So, M = f−1(Z) ⊂ Rk+n. This is a k-dimensional manifold. Further, this is oriented

because we only need to orient the normal bundle. We can do this because the normalbundle to Z is oriented, since the tautological bundle is oriented. Then, we can pull backthe orientation.

Note, there were some choices along the way. For example, we chose a representative[f].

Exercise 30.5. Show that this was independent of the choice of f. That is, if f0 ∼ f1, we geta cobordism N in Rk+n × [0, 1] .

Altogether, we got a map πk+n(τ(En))→ Ωk,n sending f to the manifold M which wasf−1(Z) ⊂ Rk+n, and we saw above it was an oriented k-manifold.

That is, we have

πk+n (τ(En))→ Ωk,n

[f] 7→ [M]

whereM = f−1(Z), and Zwas the zero section of B(En,N).

Remark 30.6. We only used that we had a zero section Z in the Thom space, and not muchabout the particular bundle En. But, to make the inverse map, we will need to use thatthis is the universal bundle to the grassmannian.

What is the inverse mapΩk,n → πk+n(τ(En))?

Start with M ⊂ Rk+n with an oriented normal bundle V →M. The universal propertyof the grassmannian we get a corresponding map

M→ Gn(RN).


Here, we can take N = k+n. That is, our map is

f1 :M→ Gn(Rk+n)

x 7→ Vx.

This f1 extends to a map of bundles

V → En,N

(x, v) 7→ (Vx, v) .

(with N = k+n).Hence, we get a map

B(V)→ B(En,N).

By the tubular neighborhood theorem, for ε small, we have a homeomorphism

B(V)∼=−→ U

(x, v) 7→ x+ ε · v

where U is an open neighborhood ofM in Rk+n. Hence, combining the two maps above,we have a map

f2 : U→ B(En,N).

Then, f−12 (Z) = M. In Sk+n = Rk+n ∪∞, let Uc denote the complement of U. Extend f2to f : Sk+n → τ(En,N) given by sending the points where f2 is not yet defined to the point∞. That is extend f2 to f by

f(Uc) = ∞ ∈ τ (En,N) .

Hence, we get

[f] ∈ πn+k (τ (En,N))

with N = k+n.Now, we have to ask whether this map is well defined, and whether this is indeed an

inverse to our first map.First, let’s understand well definedness. That is, we are asking whether if we have

another M0 cobordant to our M1, will the map agree for the two manifolds. Say M0 andM1 are cobordant via a manifoldW ⊂ Rk+n × [0, 1] ⊂ Rk+n+1.

To show well definedness, we’ll construct a map

Sk+n × [0, 1]→ τ(En,N ′

)with N ′ = k+ n+ 1. We then take a tubular neighborhood and construct a similar mapto the map above. Both M0,M1 give maps from Sk+n to τ(En,N), we will show the mapsbecome homotopic after the inclusion

i : τ (En,N) ⊂ τ(En,N ′

).

ForM ⊂ Rk+n, we get a map

f : Sk+n → τ(En,N)

112 AARON LANDESMAN

which then gives the map

i f : Sk+n → τ(En,N ′),

which gives a well defined map

Ωk,n → πk+n(τ(En,N ′))

[M] 7→ [i f] .

with N ′ = 1+ k+n.

Exercise 30.7. Show these maps are mutually inverse. (This is fairly routine, but it’s fas-cinating that this construction can be done.)

Next, we’ll augment Ω and construct a generalized homology theory so that this ringwill become homology groups of a point.

31. 4/13/16

31.1. Oriented bordism as a homology theory. Today, we’ll discuss oriented bordism asa homology theory. We’ll discuss CW-pairs (X,A).

Remark 31.1. One advantage of dealing with such nice pairs is that if A is contractible, itfollows that X ∼= X/A, because the homotopy extension property holds.

For example, it is useful that X∩CA ∼= X/A.

We want to define oriented bordism groups MSOk(X), where SO stands for specialorthogonal groups.

Definition 31.2. We define MSOk(X) to be bordism classes of pairs

(M,σ)

where M is a smooth, compact, oriented, and k-dimensional manifold and σ :M→ X. IfX is a point, then σ is uniquely determined. We introduce the equivalence relations

(M0,σ0) ∼ (M1,σ1)

if(1) there exists an oriented cobordism N so that ∂N = (−M0)

∐M1.

(2) There exists

τ : N→ X

satisfying

τ|Mi = σi

for i ∈ 0, 1 .

Example 31.3. We have MSOk(pt) is simply Ωk, since the only interesting datum is themanifold, since the maps are determined. If X has a basepoint x0 ∈ X, we obtain a map

i∗ : MSOk(x0)→MSOk(X)


by functoriality: If we have a map f : Y → X, we obtain a corresponding map

f∗ : MSOk(Y)→MSOk(X)

[M,σ] 7→ [M, f σ] .

We also have a map in the other direction, sending p : X→ x0, which induces anothermap

p∗ : MSOk(X)→ Ωk.

Note that i∗ is injective because p∗i∗ = 1Ωk .

Definition 31.4. We define the corresponding reduced group

MSOk(X) :=MSOk(X)

i∗Ωk

Lemma 31.5. If f0 ∼ f1 : X→ Y then

(f0)∗ = (f1)∗ : MSOk(X)→MSOk(Y).

Proof. Indeed, the maps are given by (f0)∗ sending

[M,σ] 7→ [M, f0 σ]and (f1)∗ is given by sending

[M,σ] 7→ [M, f1 σ] .

Indeed, the homotopy F between f0 and f1 determines the cobordism by taking N :=M× [0, 1] and the corresponding map to X given by F σ. That is, the cobordism is

[M× [0, 1] , F σ] ,

with F the homotopy between f0 and f1.

Lemma 31.6. Addition in cobordism is disjoint union.

Proof. Indeed, we have

− [M,σ] = [−M,σ] .

So, [M∐

−M,σ∐

σ]

is a boundary of

[N, τ]

with N = [0, 1]×M and τ = σ pM.


MSO2 (x1, x0) = 0.

To see this, we have

MSO2 (x1, x0) =MSO2 (x1) ∼= Ω2 ∼= 0,

since every 2-dimensional oriented manifold is a boundary. E.g., a torus is the boundaryof a solid torus.

114 AARON LANDESMAN

Definition 31.8. We define

MSOk(X,A) := MSOk(X/A) ∼= MSOk(X∪CA)The first is the definition, the latter is an isomorphism.

Lemma 31.9. If we have maps

(31.1) A X X/A,i j

the sequence

(31.2) MSOk(A) MSOk(X) MSOk(X/A).i∗ j∗

is exact at MSOk(X). As a slight variant, we have an exact sequence

(31.3) MSOk(A) MSOk(X) MSOk(X,A)

Proof. We will prove the first exact sequence (not the variant). As usual, it’s quite straight-forward to show j∗ i∗ = 0 because j i : A → X/A is constant, mapping A to a point.Hence, it factors through a point, and so the corresponding map on MSOk factors throughMSOk(pt) = 0. Hence, im i∗ ⊂ ker j∗. We only need check the reverse inclusion.

Next, replace X/A by the homotopy equivalent space X∪CA. We have maps

(31.4) A X X∪CA.i j

It suffices to prove the lemma with X/A replaced by X∪CA, as they are homotopic, usinga previous lemma that MSOk groups are preserved under homotopy. Suppose

j∗ [M,σ] = 0

in MSOk(X ∪ CA). Then, we have a cobordism [N, τ] with τ : N → X ∪ CA, with ∂N =M∪M∗ and τ(M∗) ⊂ p, where p is the cone point of CA.

We have

CA =A× [0, 1]A× 0

We have a coordinate function t on [0, 1].We can think of

t : X∪CA→ [0, 1]

with t|X = 1. Picture t as the height function. In N we have U ⊂ N be U = τ−1(CA). Infact, we can take U := (t τ)−1(0, 1), with t τ : N → [0, 1]. Without loss of generality,by approximating t by a smooth function, we may assume t is smooth on U. We canfurther assume t = 1

2 is a regular value (by Sard’s theorem), meaning that the derivativeis surjective when t = 1

2 . SoM1/2 = (t τ)−1 (1/2) ⊂ U, is a smooth submanifold.Let N ′ = (t τ)−1 [1/2, 1]. We have a restriction τ ′ := τ|N ′ . We have σ1/2 : M1/2 →

A× 1/2.


Then, the union of X and A × [1/2, 1] deformation retracts to X via the map r : X ∪(A× [1/2, 1])→ X. Then,

σ τ ′ : N ′ → X.

This shows that [M,σ] ∈ ker j∗ satisfies

[M,σ] = [M1/2, r σ1/2]

Corollary 31.10. (See the discussion of abstract h from before.) There exists a longer exact se-quence(31.5)

MSOk(A) MSOk(X) MSOk(X/A) MSOk(SA) MSOk(SX) · · ·

Lemma 31.11. We have

MSOk(SA) ∼= MSOk−1(A).

Proof. The idea is that if we look at the suspension of A, inside we have the level setA× 1/2. We have a coordinate function t in the vertical direction. Say we have a k-dimensionalMmapping to SX, so that

[M,σ] ∈ MSOk(SA)

Inside thisMwe haveM ′ := σ−1(1/2). That is,M ′ = (t σ)−1 (1/2).We have thatM ′ ⊂M is a smooth k− 1 dimensional manifold. We then have

σ| ′M :M ′ → A× 1/2 = A,

and construct [M ′,σ|M ′

]∈ MSOk−1(A).

This determines a map. We have to check this is well defined and is an isomorphism,which we omit for today, but may come back to next time.

32. 4/15/16

32.1. Review. Recall that SA := S×[0,1]∼ and we were showing that MSOk(SA) = MSOk−1(A).

The idea was that inside SA we have A× 1/2, and we construct the map by sendingσ :M→ SA toM ′ = σ−1(1/2) and the map σ|M ′ :M ′ → A.

Remark 32.1. Recall that we set up

MSOk(SA) :=MSOk(SA)

MSOk(pt).

Lemma 32.2. But, we could have equivalently defined

MSOk−1(A) ⊂ MSOk−1(A)

as the subset

Sk−1(A) :=[M ′,σ

]:M ′ is an oriented boundary ,σ :M ′ → A

.

116 AARON LANDESMAN

Proof. The point is that

MSOk−1(A) = MSOk−1(pt)⊕ Sk−1(A)

To see this, in general, if we have a map σ ′′ : M ′′ → A, where M ′′ is not necessarily aboundary, we can construct a map

σ ′′∐

pt :M ′′∐

−M ′′ → A

and

pt :M ′′ → A

and takingM ′ :=M ′′∐

−M ′′, we seeM ′ is the boundary of a cylinder. Thus, every mapcan be written as the sum of a map from a boundary and a constant map.

Lemma 32.3. Observe that ⊕ is a well defined map

MSOk(SA)η−→MSOk−1(A).

The map η is well defined and surjective. Further, the kernel is precisely MSOk( basepoint ).

Proof. We can write M = M+ ∪M− (the preimage of [1/2, 1] is M+ and the preimage of[0, 1/2] is M−). Here, M ′ = ∂M+ and M ′ is a boundary. So, im η ⊂ MSOk−1, where weregard MSOk−1 ⊂ MSOk−1 as the subset which comes from boundaries.

In fact, it’s not hard to see that im η = MSOk−1. If we have a map σ ′ :M ′ → A, we cantake a collar neighborhood of M ′. So, we can construct a map σ :M → SA extending σ ′,showing the map η is surjective.

What is the kernel of η? We have that MSOk( basepoint ) ⊂ kerη.To see that this is the full kernel, we have

[M,σ]

giving a map

σ :M→ SA.

We have

M ′ := σ−1(A× 1/2).

and a map

σ ′ :M ′ → A× 1/2 ∼= A.

Suppose M ′ = ∂N ′ and σ ′ extends to τ ′ : N ′ → A. The existence of this extension τ ′ isprecisely what it means for [M ′,σ ′] = 0. Given this, we want to show that

[M,σ] =[M, σ

]where σ is constant with no base point. That is, we want to show it comes from MSOk( basepoint ).For this, we want to construct a cobordism between M and M ′. For this, we take M×[a,b] and we cut M× a into two pieces M+ and M−. We add in a manifold joining M+

andM− and take a small collar around it. More precisely, we have

M =(M+ ∪N ′ × ε

)∐ (M− ∪N ′ × −ε

).


and there is a cobordism betweenM ′ andM.So far, we have an

[M,σ] =[M, σ

],

with

im σ∩ (A× 1/2) = ∅

So, we can construct a map ˜σ : M→ SA

where the image lies in the two base points of SA, using that the image of σ does notintersect A×

12

.

Remark 32.4. There are some clear variants of Lemma 32.3. We can similarly defineMOkas unoriented bordism and MUk as complex bordism. We want this notion to be stablewith respect to adding trivial summands.

More precisely, a stable complex structure on V → Z a vector bundle over a manifold Zmeans that we have a map

J :(V ⊕Rk

)→ (V ⊕Rk

)for some large enough k with J2 = −1. We quotient by the equivalence relation that(J1, Rk+2

)∼(J2,V ⊕Rk ⊕C

)if there is an isomorphism V ⊕Rk+2 ∼= V ⊕Rk ⊕C where

the J1, J2 agree on V ⊕Rk, and J2 is the standard multiplication by i operator on C.

We can similarly define spin bordism and framed bordism.

Definition 32.5. A framing of a vector bundle V → Z is a trivialization. That is, a collec-tion of r pointwise independent sections where r = rkV .

Definition 32.6. A stable framing of a vector bundle V → Z is a framing of V ⊕Rk forsome k, modulo the equivalence relations described above.

(For example, a stable framing for V gives a stable framing for V ⊕R.)

We now want to define framed bordism more precisely

Remark 32.7. Recall that Ωk = MSOk(pt). We have an isomorphism Ωk ∼= Ωk,n for largen. This is bordism classes of maps

M → Rk+n

with N ⊂ Rk+n × [0, 1] .

Definition 32.8. Framed bordism of a space X, notated Πk, is defined as follows. We firstdefineΠk,n(pt), to be equivalence classes of k-dimensional manifoldsM ⊂ Rk+n togetherwith a stable framing of ν(M), the normal bundle toM.

Here, in more detail, the equivalence relation is given by M0 ∼ M1 if there exists amanifoldN and a framing for ν(N), the normal bundle toN in Rk+n+1 which extends theframing ofM0,M1.

We have Πk,n(pt) eventually stabilizes, by Lemma 32.10 and define Πk(pt) to be thisstabilization.

118 AARON LANDESMAN

Warning 32.9. We will often omit pt and denote Πk(pt) by Πk.

Lemma 32.10. We have that Πk,n(pt) is independent of n for n sufficiently large.

Proof. We have equivalence classes of M ⊂ Rk+n embedded with a framed normal bun-dle.

Exercise 32.11. Check this by constructing Πk as manifolds together with a framing for itstangent bundle, and use this to show the Πk,n eventually stabilize.

Question 32.12. What is Π0?

Lemma 32.13. We have Π0 ∼= Z.

Proof. If we start with a manifold in R, then we have that framings correspond to direc-tions for each point. If we have a left and right directions, the points cancel by a framedarc. Therefore, Π0 ∼= Z.

Lemma 32.14. We have Π1.

Proof. We will haveM = S1 or a union of S1’s. We can embed a circle in R2. There are twotrivializations of its normal bundle, as either the outward one or inward one. (Here, wemean trivializations up to homotopy.) If we put S1 in R3, these two trivializations becomethe same, as we can flip the circle around in R3.

If we have R3 ⊃ S1, we have a normal bundle S1 ×R2. Given one framing φ, we getall others as

φ ′(θ) = a(θ) ·φ(θ)with

a : S1 → GL(2, R).

Remark 32.15. Of course, there was nothing special about the circle, this could be donefor any manifold.

Or, we could have taken

a : S1 → O(2).

So, the interesting thing to do is to take a framing and “rotate it” by an interesting mapS1 → O(2).

So, we want to answer the question:

Question 32.16. What are the maps[S1,O(2)

]∼=[S1,S1

∐S1]?

Of course, they are Z×Z because[S1,S1

]∼= Z∐

Z.If we instead embedded S1 ⊂ R4, the normal bundle would be S1 ×R3. Then, we’d

want to understand[S1,O(3)

]. First,

[S1,SO(3)

]∼=[S1, P3C

]∼= Z/2. Therefore,[

S1,O(3)]∼= Z/2

∐Z/2.

This is what we get before thinking about cobordisms.


After following this process some more, we find Π1,3 ∼= Z, after quotienting by cobor-dism. We find Π1,4 ∼= Z/2. In general, we find Π1,n ∼= Z/2 for n ≥ 4.

We’ll sketch this result in more detail next time.

33. 4/18/16

33.1. Review. Recall last time we defined Πk,n, which is equivalence classes of pairs

[M,φ]

withM ⊂ Rk+n is a smooth, compact k-manifold embedded in Rk+n andφ = (φ1, . . . ,φn)is a normal framing. That is, each φi is a normal vector field.

The equivalence relation is defined byM1 ∼M0 if there exists a cobordismN ⊂ Rk+n×[0, 1] with a normal frame restricting to those of M0 and M1 on the boundaries Rk+n ×0 , Rk+n × 1.

Remark 33.1. The point of introducing this definition is that, in disguise, this is the prob-lem of understanding the homotopy groups of spheres. Recall that this is similar to Ωk,which was isomorphic to the homotopy group of some Thom space.

33.2. The relations between framed cobordism and homotopy groups. The main resultof this subsection is to relate framed cobordism to homotopy groups of spheres.

Proposition 33.2. There is an isomorphism

η : πk+n(Sn) ∼= Πk,n.

Proof. We will first describe the map η. Given [f] ∈ πk+n(Sn), where

f : Rk+n ∪∞→ Rn ∪∞.

This can be taken to be smooth by a suitable approximation. Let y ∈ Rn be a regularvalue, meaning thatM = f−1(y) ⊂ Rk+n is a smooth k-dimensional manifold.

Further,M has a normal framing φ1, . . . ,φn as follows: If we take the standard basis oftangent vectors for Rn, call them e1, . . . , en. Then, the map f is given so that the inverseimage of y ∈ Rn is some submanifold of Rn+k. Further, we have an isomorphism df :νx(M) ∼= TyRn. There exist unique

φi(x) ∈ νx(M)

so that

df(φi(x)) = ei.

These define elements φ1, . . . ,φn, and so we get (M,φ).Along the way of this construction we chose a regular value y, and we chose an iden-

tification for Si ∼= Ri ∪∞ with i = n,n + k. For example, we might choose differentrepresentatives f0 and f1 with regular values y0 and y1. Let’s check the independence ofthis choice corresponding to k-manifoldsM0 andM1.

Construct a homotopy

F :(

Rk+n ∪∞)× [0, 1]→ Rn

120 AARON LANDESMAN

which can be taken to be smooth as follows. We get a map

F :(

Rk+n ∪∞)× [0, 1]→ (Rn ∪∞)× [0, 1]

(x, t) 7→ (F(x), t) .

Choose a path γ from y0 to y1 so that it is transverse to F, meaning thatN := F−1(γ([0, 1]))is a smooth k+ 1manifold with boundary. Here, F−1(γ(1)) =M1, F−1(γ(0)) =M0 . Then,extends e1, . . . , en to be a framing of ν(γ). Use dF to obtain a framing for ν(N) so that

dFj : νx(n) ∼= νF(x)

(γ).

Therefore,

[M0,φ0] = [M1,φ1] ∈ πk,n.

Now, we want to check that η is a bijection. We want to show the map is injective andsurjective. We’ll check surjectivity, as injectivity is similar.

Given some (M,φ) withM ⊂ Rk+n, we’d like to construct a map

f : Rk+n ∪∞→ Rn ∪∞so that η ([f]) = [M,φ].

This relies on the tubular neighborhood theorem. We have a map

α : D1(Rn)×M→ Rk+n

(v, x) 7→ x+ ε∑i

viφi(x).

where

v =

v1...vn

and D1 is the open disk of radius 1. The tubular neighborhood theorem tells us there is abundle over M whose fibers are disks. The framing gives us an identification of the fiberbundle as the product bundle D1(Rn)×M. For ε > 0 small, the above map α restricts toa diffeomorphism

D1(Rn)×M→ U(M)

for U(M) ⊂ Rk+n some open neighborhood ofM.So, we have U(M) ∼= (D)n ×M. So, we have a map

U (M)p−→ D(Rn) ∼= Rn.

Call this composition f ′. Further, (f ′)−1(0) = M and the framings agree. Then, we canextend the map f ′ to all of Rn+k ∪∞, which initially is a map

(33.1)

U(M) Rn

Rn+k ∪∞ Rn ∪∞f ′


by defining f(z) =∞ for z /∈ U(M). This is continuous, since, as we approach the bound-ary of U(M), we go to∞ in Rn ∪∞.

This shows that [M,φ] is in the image of πk+n(Sn).We now sketch injectivity, which will conclude the proof. Suppose f0 and f1 and regular

values y0 and y1. Suppose these map under η to framed fibers (M0,φ0) and (M1,φ1)which agree in Πk,n. We’d like to show f0 and f1 are homotopic.

That is, we can to construct a homotopy between f0 and f1. That is, we’re given aframed N ⊂ Rk+n × [0, 1], and want a homotopy. One dimension lower, we just did thisfor surjectivity, using the tubular neighborhood theorem, and mapping the other stuff to∞. The remainder of this proof uses the same idea to construct this homotopy.

Remark 33.3. The above proposition gives another proof of the stabilization of homotopygroups.

Example 33.4. Last time, we looked at

Π0,n = framed 0-manifold / ∼

where ∼ identifies two manifolds which are framed cobordant. In dimension n = 1, wesaw Π0,n ∼= Z. When n = 2, we had a framing which is either left handed or right handedat any point. Indeed, we have π0+n(Sn) ∼= πn(S

n) ∼= Z.

Example 33.5. In the case k = 1, we saw that things get fairly confusing, fairly quickly.start with a 1-dimensional manifoldM.

Start with k = n = 1. Take M ⊂ R2 to be the unit circle. Take φ = φ1 to be theoutward normal ∂

∂r . This class is [M,φ] = 0. This is 0 because this is a boundary of of D2,the upper hemisphere of S2. Further, this framing extends to an outward normal vector.That is,

[M,φ] = (∂N,ψ)∂N .

The only other possibility is the inward pointing normal, which is also a boundary. So, inthis case, Π1+1 = π2(S1) = 0.

Next, let’s look at n = 2. So, M ⊂ R3. Take M = S1, the unit circle. Say M ⊂ R2 ⊂ R3.Here, we can take our frame to be

φ = (φ1,φ2)

where φ1 is the outward normal and φ2 = ∂∂x3

. By the same construction using a hemi-sphere, we get that in this case [M,φ] = 0. However, in the case of R3, there are moreoptions for vector fields. We’ll now examine this in more generality.

Say we have Sk ⊂ Rk+1 ⊂ Rk+n. Then, we have a framing

φ∗ = φ1, . . . ,φn .

Here, we can take φ1 to be the outward normal in Rk+1 and φ2, . . . ,φn the remainingcoordinate directions. But, already in the R3 case, given a map

g : Sk → O(n),

consider the new framing φg defined by, for x ∈ Sk,

φg(x) = g(X)φ∗(x).

122 AARON LANDESMAN

We get a map [Sk,O(n)

]→ Πk,n∼= πk+n(S

n).

Restricting this to SO(n), we get a map

J : πk(SO(n))→ πk+n(Sn),

which is a homomorphism known as the J-homomorphism.The first interesting case of this is k = 3. We’ll see more of this next time.

34. 4/20/16

Before returning to the J-homomorphism discussed last time, we first discuss stabiliza-tion.

34.1. Stabilization. Last semester, in 231a, a proof of the stabilization of

πk+n(Sn)→ πk+n+1(S

n+1).

This map was induced by the suspension map, sending

f : Sk+n → Sn 7→ Sf : SSk+n → SSn.

The map

πk+n(Sn)→ πk+n+1(S

n+1)

[f] 7→ [Sf]

which is onto if n ≥ k+ 1 and an isomorphism if n ≥ k+ 2.This is a special case of the Freudenthal suspension theorem. Defining πsk as the stabi-

lization of the kth homotopy group of spheres, the limit of πk+i(Si), we have

πsk = πk+n(Sn)

for any n ≥ k+ 2.Last time, we saw

πk+n(Sn ∼= Πk,n),

where Πk,n is normal framed k-manifolds M ⊂ Rk+n, notated as pairs [M,φ]), where φis a framing.

We have a natural map

Πk,n → Πk,n+1

sendingM ⊂ Rk+n → Rk+n+1, and sending a framing

φ = φ1, . . . ,φn 7→ φ0, . . . ,φn

with φ0 = en+k+1.We constructed vertical maps

(34.1)

πk+n(Sn) πk+n+1(S

n+1)

Πk,n Πk,n+1


Exercise 34.1. Show that up to possibly annoying orientations and signs, this diagramcommutes.

Definition 34.2. Define

Πk := limn→∞Πk,n.

Exercise 34.3. For vector bundles U,V on a space X with W := U⊕ V , a stable framingof any two of U,V , and W determines a stable framing for the third. Hint: If we have astable framing for U andW, it means(

U⊕Rj)⊕ V ∼=W

is isomorphic to

RN1 ⊕ V ∼= RM,

so V is also be stably trivial, by definition.

Remark 34.4. Concretely, the above exercise means that if W is trivialized by ψ and U⊕Rk is trivialized by φ then there exists a j0 trivialization π of V ⊕Rj, for all j ≥ j0 so that

φ⊕ π = ψ⊕(

standard basis for Rj+k)

.

Example 34.5. If we think of

TM⊕ ν(M) = Rk+n

for M ⊂ Rk+n, then the tangent bundle is trivial if and only if the normal bundle is, byExercise 34.3.

Remark 34.6. Hence, Πk is equivalence classes of [M,ψ] with M a compact k-manifoldand ψ a trivialization of TM⊕Rj for some j.

More generally, given stable frames ψ for M, others are given by g,ψ with g : M → Owhere O is the orthogonal group for large M. In fact, this plays well with the associatedgroup structure of Πk, which comes down to O being a group, and the group structuresbeing compatible.

34.2. J-homomorphism. Recall last time we defined a J homomorphism.

Definition 34.7. Fix a dimension k, take M = Sk. Take φ∗ to be a stable framing of TSk,which bounds

[M,φ∗] = 0

in Πk ∼= πsk.Given any map

g :M→ O,

we get an associated map

φg := g ·φ∗

and a corresponding element [M,φg] ∈ πsk.

124 AARON LANDESMAN

We have a group homomorphism

J : πk(O)→ πsk+n(Sn)

= πsk

for sufficiently large n.We’d like to understand what the J homomorphism.

k πk(O) πsk

1 Z/2∼=−→ Z/2

2 0 0−→ ?3 Z → ?4 05 06 07 Z

8 Z/2

TABLE 10. A table of homotopy groups of the orthogonal group

In the case when k = 1, we have π4(S3) ∼= Z/2. The order of im J for k = 4m− 1 wassettled by Mahowald around 1970. Looking at the case when k = 4m− 1, the image ofJZ→ πs4m−1 is some cyclic finite group.

Take the generator g of πk(O) ∼= Z, given by Bott’s theorem, and ask if it is possible todetermine J(d · g) = 0, for d ∈ Z. We’ll get a necessary condition on d, which will give alower bound for the size of the image of J.

Lemma 34.8. If J(d · g) = 0, then

d(2m− 1)!Qm ∈ Z,

where Am = Qm · pm, where Qm is some coefficient, and Am is the top dimensional class of A(introduced several classes ago) in H4m, and pm is the Pontryagin class.

Proof. Since J is a group homomorphism, we have

J(d · g) = d · J(g).

Taking M = Sk = S4m−1, we can take φ∗ to be the bounding framing of TM, to give the0 element. Take φ1 := g · ( the bounding framing ), where g : Sk → O is the generator ofπk(O). We have

[M,φ1] = J(g) ∈ Πk ∼= πsk.

Then, taking d times this generator corresponds to a disjoint union of spheres. If J(d ·g) =0, there is some manifoldNwhich is a boundary for d copies of the sphere. That is,N hasa stable framing ψ extending the copies of φ1, with φ1 a framing on each copy ofM.

Out of this picture, we can make a closed manifold

N := N∪D4m ∪ · · ·D4m,


so that the resulting manifold is closed. The disks are attached to the bottom of thespheres.

We have that TN is stably trivial but TN is not stably trivial, since on the boundary ofdisks, we’ll have the framing φ∗, if we trivialize the disks (since φ∗ is the unique stableframing extending to these disks.

At these boundaries of the disks, the framings differ by g, which is the Bott generatoras a map S4m−1 → O.

Using this, we can calculate all Pontryagin classes of TN, call them

p1, . . . ,pm.

where dim N = 4m.First, let’s find p1. The inclusion N → N induced an isomorphism

Hj(N)→ Hj(N)

when j < 4m− 1 because N = N∪ ∪iCi, where Ci are 4m dimensional cells.So, πi(TN) = 0 for i < m because the tangent bundle of N is stably trivial, so all

Pontryagin classes of TN are 0.The only possibly nonzero Pontryagin class is pm(N).

Question 34.9. What is pm[N] ∈ Z?

As a warm up, start with S4m, and note that if we start with a stable framing for the topand bottom of S4m, with boundaries glued by the Bott generator.

This gives a map V → S4m, which is a generator of KO(S4m).Letting g be the Bott generator, on S4m we previously found

pm(g)[S4m] =

(2m− 1)! ifm ≡ 0 mod 22(2m− 1)! ifm ≡ 1 mod 2.

Then, the answer for N, we have

pm[N] =

d(2m− 1)! ifm ≡ 0 mod 22d(2m− 1)! ifm ≡ 1 mod 2.

as it is a gluing of d classes corresponding to the Bott generator, so it is d times the corre-sponding value on the sphere above.

Note that N is spin because w1 = w2 = 0, as TN is trivial on the 2-skeleton.Also, recall that for spin manifolds, we have the result that

A[N]∈

Z ifm ≡= 0 mod 22Z ifm ≡ 1 mod 2

When p1, . . . ,pm−1 = 0, we have that Am = Qm · pm, whereQm is some coefficient, andAm is the top dimensional class in H4m. Then,

A[N]= Qm · pm

[N]

=−691

2615348736000pm[N]

.

126 AARON LANDESMAN

Remark 34.10. We have

Qm = (−1)mB2m

2 (2m) !

Corollary 34.11. d is divisible by the denominator ofB2m4m

.

Corollary 34.12. The image of J is a cyclic group of order divisible by n, as given in the followingtable In particular, π8(S5) has order at least 24.

m n1 242 2403 5044 4805 2646 65520...

...

TABLE 11. Table of scaled Bernoulli numbers.

Proof. The statement about π8(S5) follows from the preceding statements because 8− 5 =3, which is 4m− 1 for m = 1, and when m = 1, the corresponding cyclic group has order24.

35. 4/22/16

35.1. Low stable homotopy groups.

Example 35.1. Let G be a lie group, such as S1 or S3 = SU(2). Given a framed manifoldM = G, we can take φ to be a basis of left invariant vector fields.

To trivialize the tangent bundle to G, look at multiplication by g

×g : G→ G

h 7→ g · h,

which will trivialize TG by sending a tangent vector at the identity to the section of thetangent bundle which are translates of vectors. In fact, this translation identifies TG ∼= TeG.

This is called a Lie group framing.

Example 35.2. A stable framing, for a fixed orientation is a class[S1,SO

]∼= Z/2.

So,

Π1 = Z/2 = 0, 1 ,

where 1 corresponds to S1 with the Lie group framing.


Note that we have TS1 ⊕R ∼= R2 on S1. If we identify the trivial summand of R2, wecan take one of the summands to be the outward normal. If we try to extend the Lie groupframing to the center of the circle, we cannot extend it to a vector field on a disk.

Note that Πk is a group, with group operation given by disjoint union. This is the sameas the group operation on πsk. We have the following stable groups in low dimensions

Stable homotopy group where it stabilizes framed manifold generators group image of J?πs0 π2(S

2) • Z Noπs1 π4(S

3) (S1, Lie ) Z/2 Yesπs2 π6(S

4) (S1 × S1, Lie ) Z/2 Noπs3 π8(S

5) (S3 ∼= SU(2), Lie ) Z/24 Yesπs4 π10(S

6) 0 Noπs5 π12(S

7) 0 Noπs6 π14(S

8) (S3 × S3, Lie ) Z/2 Noπs7 π16(S

9) (S7, Octonian ) Z/240 Yesπs8 π18(S

10) SU(3), Lie ), (S8,φ1) Z/2⊕Z/2 No

TABLE 12. A table of stable homotopy groups and stable framing groups.

Remark 35.3. It turns out this forms a graded ring, and all elements of this ring are nilpo-tent. The last column marks whether the generator is the image of the J homomorphism.If S7 were an associative algebra, the generator would be the Lie group framing, but thegenerator is the closest analog. In the πs8 case, the second Z/2 is the image of the J homo-morphism, but the full group is not generated by the image J homomorphism.

35.2. More about Π3.

Remark 35.4. Why is πs3 = Z/24. Last time, we showed there was at least a Z/24, butwhy is that precisely the group?

Here, we’re looking at πs3 = Π3.

Fact 35.5. Note the following facts (the first two are on par with the difficulty with aproblem set problem):

(1) A three dimensional manifold has trivial tangent bundle if and only if M is ori-ented, if and only if TM is stably trivial.

(2) A connected four manifoldNwith nonempty boundary has trivial TN = R4 if andonly if it has Stiefel Whitney classes w1 = w2 = 0 (that is, it is a Spin manifold).

(3) (This fact is harder and more interesting) Every three manifold bounds a spin 4-manifold. That is, given any three manifold M3 there is some 4-manifold N4 with∂N4 =M3.

(4) Π3 is generated by im J. That is, given a framed three manifold, (M3,φ), we canfind (N,φ) so that the boundary of N is the union of S3, with the trivial framingφ1, and (M3,φ). This fact is closely related to the previous fact, and will followfrom the previous one.

128 AARON LANDESMAN

To show Π3 ∼= Z/24, we know the generator maps to something of order which is amultiple of 24. We want to show it maps to an element of order precisely 24. To show itis order precisely 24, we start with 24 copies of (S3,φ1) and find a manifold (N,φ) withframing whose boundary is 24 such copies. We can exhibit such a 4-manifold N as aquartic surface N ⊂ P3, which is a K3 quartic surface, which has Euler number 24. Thishas w2 = 0.

To see the Euler number is 24, one would want to show there is some vector field with24 transverse 0’s.

When we remove a neighborhood of the 24 zeros, we’ll be left with 24 copies of S3.We can further get the framing data for the S3, and it can be checked that these restrictcorrectly. We start with a nonzero vector field V which has 24 transverse 0’s and define.on N := N \ neighborhoods of zeros of V . Then, TN has a Quaternionic structure

(1, I, J,K)

acting on TN. Then, we can take the framing

Ψ = V , IV , JV ,KV ,

on N. The corresponding framing on S3 is given by the three vector fields which areproducts of the normal direction with I, J, and K.

35.3. Another integrality property for A. We’ll discuss the Hirzebruch signature theo-rem and the L-genus now.

Example 35.6. Let X be a smooth 8-manifold, suppose p1[X] = 0 and take the A genus

A[X] =7p21 − 4p25760

[X]

= −4

5760p2[X]

= −1

1440p2[X].

Corollary 35.7. Suppose X is spin and p1 = 0. Then, since A[X] is an integer, we have that1440 | p2[X].

This holds if X \ pt has trivial tangent bundle.

Proof. It’s immediate from the above, since

A[X] =7p21 − 4p25760

[X]

= −4

5760p2[X]

= −1

1440p2[X].

and A[X] is an integer.To see that X \ pt having trivial tangent bundle implies X is spin and p1 = 0, then

removing a point is the same as removing an 8 cell will preserve values of p1 = w1 =w2 = 0.


We can now get a handle on p2[X] using the Hirzebruch signature theorem.

Definition 35.8. Recall the L-genus relates to√y

tanh√y

,

and we haven∏i=1

√yi

tanh√yi

= 1+1

3p1 +

1

45(7p2 − 24p

21) + · · ·

=: 1+ L1 + L2 + · · · .

where the yi are formal variables, and pi are the elementary symmetric functions of theyi In other words, the L-genera are the terms of the expansion of the above power series.

Theorem 35.9. We have

σ(X4k) = Lk[X]

σ(X8) =7

45p2[X]

if p1 = 0, where

σ(X4) : H4(X)×H4(X)→ Z

(α,β) 7→ (α∪ β) [X],

Proof. Omitted.

Corollary 35.10. If X8 is spin and p1 = 0, then

1440 |45

7σ,

that is,

7 · 144045

= 7 · 32 | σ(X).

Proof. This follows from the previous results stating that σ(X8) = 7/45 · p2[X] and 1440 |

p2.

Remark 35.11. Observe that we can construct an 8-manifold with boundary Z, which has(1) w1 = w2 = 0(2) p1 = 0(3) The boundary of Z, ∂Z has the same homology as S7(4) π1 = 0(5) σ(Z) = 8 (which is, in particular, not divisible by 7 · 32 from the Corollary 35.10.

So, from Corollary 35.10, we have

∂Z 6∼= S7,where ∼= above means diffeomorphism (although there is a homeomorphism. If therewere such a homeomorphism, we would have σ(X) = 8 = −1

1440p2[X], but we know 7 · 32divides σ(X), whereas here it does not, and it fails by a factor of 28.

130 AARON LANDESMAN

Corollary 35.12. There exist 7-manifolds which are homotopy equivalent to S7 but not diffeomor-phic. These are called exotic 7 spheres.

Proof. It suffices to construct the manifold described above. One way to do it is to con-struct it explicitly using plumbing.

A more mysterious way to construct it is to exhibit it from the hypersurface

E := x21 + x32 + x

53 ++x24 + x

25

in C5 and construct Z as E ∩ B10, where B10 is the unit ball in C5. Then, Z is a smoothhypersurface in B10 of dimension 8. The boundary of Z is Z ∩ S9, which turns out to behomeomorphic but not diffeomorphic to S7.

36. 4/25/16

36.1. Spectra. Recall that we were talking about stabilization. Today, we’ll discuss spec-tra, which is, loosely speaking, homotopy theory with stabilization baked in.

Definition 36.1. A spectrum is a collection

E = (En, εn)n∈Z

where

εn : SEn → En+1,

En are pointed spaces, εn are pointed spaces, and SEn := S1 ∧ En.

From now on, everything will be done with base points.

Remark 36.2. One might ask that En are cell complexes and εn are cellular maps, or askfor other similar variants.

Example 36.3. (1) Consider the sphere spectrum given by En = Sn and εn : SSn ∼=Sn+1 the natural isomorphism.

(2) If we are given E and a space Xwe can form a new spectrum E∧Xwith

E∧X := (En ∧X, εn ∧ 1)n∈Z .

(3) If S is the sphere spectrum, we can form

S∧X = (SnX, εn)n ,

the suspension spectrum.

Remark 36.4. The name spectrum is not related to the spectrum of a linear operator. His-torically this was introduced by Lima, which can be thought of as a sequence of numbers(as in the spectrum of a hydrogen atom). Likely, this is called a spectrum just because it isindexed by n ∈ Z. Spectra were lated studied and refined by Spanier, G. Whitehead, andAdams.

Remark 36.5. One might note that we have often been indexing examples by positiveintegers, but it turns out that we will only really care about spectra indexed by n for all nbigger than some chosen n0.


36.2. More involved examples of spectra.

Example 36.6 (Thom spectrum). Recall we have the universal vector bundle (which hasunfortunate notation as it has nothing to do with a spectra)

En → Grn(R∞),

with corresponding Thom space τ(En) := D(En)/S(En).Given V ⊂ R∞, an n-dimensional vector space, we can form R⊕V ⊂ R1+∞, and hence

a map ι : Grn → Grn+1. We have ι∗En+1 ∼= R× En. This gives rise to a map

τ(R× En)→ τ(En+1

or, taking one point compactifications, a map

S(τ(En))ε−→n τ(En+1).

This enables us to define the Thom spectrum,

MO = (τ(En), εn)nOne can define the oriented Thom spectrum,MSO similarly.

Example 36.7 (Eilenberg-MacLane spaces). Recall that for an abelian group G, for n > 1,we have the Eilenberg-MacLane space

K(G,n)

with

πm(K(G,n)) =

G ifm = n

0 otherwise.

There are maps

SK(G,n) ε−→n K(G,n+ 1)

for all n, which we next describe.For reasonable pointed spaces, we have

[A∧B,C] ∼= [B,CA],

with CA is the space of pointed maps (A,C) with the compact open topology. This is atautology in the category of sets,

So, giving a map

SK(G,n) ε−→n K(G,n+ 1)

is the same as giving a map

K(G,N)→ Ω (K(G,n+ 1)) ,

where by definitionΩX is the loop space of X, defined as

ΩX := XS1,

where AB is defined to be maps from B to A.

132 AARON LANDESMAN

Because

πm(ΩK(G,n+ 1)) =[Sm,K(G,n+ 1)S

1]

=[S1 ∧ Sm,K(G,n+ 1)

]= πm+1 (K(G,n+ 1))

=

G ifm+ 1 = n+ 1

0 otherwise.

Fact 36.8. Eilenberg MacLane spaces not only exist, but but are unique in the sense thatfor any two spaces satisfying the property that their nth fundamental group is G, and allother fundamental groups are 0, then we have a uniquely determined isomorphism onhomotopy groups between two such spaces.

So, there exists a unique map

δn : K(G,n)→ ΩK(G,n+ 1)

giving the identity on homotopy groups, which is n-dimensional.

Example 36.9 (Bott Spectrum). We define the Bott spectrum B. Suppose X is compact Wehave reduced K-theory, defined by

K(X) := [X,F]

where F is by definition the Fredholm operators on a given Hilbert space H. We have[S2 ∧X,F

]∼= [X,F]

since K-theory of the second suspension of X is isomorphic to that of X. Equivalently,

[X,ΩΩF] ∼= [X,F] .

If we define

Bn :=

F if n ≡ 0 mod 2ΩF if n ≡ 1 mod 2

then we have a map

SBn → Bn+1,

or equivalently a map

Bn → ΩBn+1 .

when n is odd (since there is tautologically a map from

Bn ∼= ΩF → ΩF ∼= ΩBn+1.

So, the isomorphism

[X,ΩΩF] ∼= [X,F] .

corresponds to a homotopy equivalence

F ∼= Ω2F,

and this gives us and εn for n even.


Remark 36.10. There’s similarly an 8-step periodic spectrum from periodicity of KO.

36.3. Fundamental groups of spectra.

Definition 36.11. For E a spectrum, consider

πk+n(En) =[Sk+n,En

]S−→ [

SSk+n,SEn]

=[Sk+n+1,SEn

](εn)∗−−−→ [

Sk+n+1,En+1]

.

We get a map

πk+n(En)→ πk+n+1(En+1).

Define

πk(E) := lim−→πk+n(En)Example 36.12. We have

πk(S) ∼= πk+n(Sn)

for n large is the nth stable homotopy group of a sphere.

Example 36.13. For a space X, we have

πk(S∧X) = lim−→πk+n(SnX)=: πsk(X),

the kth stable homotopy group of X.

Example 36.14. Let’s compute

πk(G) := lim−→πk+n(K(G,n))

=

0 if k 6= 0G if k = 0.


πk(B) =

Z if k ≡ 0 mod 20 if k ≡ 1 mod 2.

For example, we have π0(F) = Z, as classified by the index.

Exercise 36.16. Show this computes the stable fundamental groups of B using the aboveobservation for π0, and Bott periodicity.

Fix E = (En, εn) and we’ll use E to construct generalized, reduced, cohomology andhomology groups, which we’ll notate as

hk(X),hk(X).

134 AARON LANDESMAN

Remark 36.17. For example, we can define these homology and cohomology groups forfinite cell complexes, or other reasonable classes of spaces.

Remark 36.18. We’ll now give the definitions, which are straightforward, but checkinglong exact sequence for cohomology is much easier than checking the long exact se-quences hold for homology.

Definition 36.19. We define the generalized, reduced cohomology groups as

hk(X) := lim−→ [Sn,Ek+n] ,

where the limit is taken over maps

[SnX,Ek+n]S−→ [SSnX,SEk+n]→ [

Sn+1X,Ek+n+1]

.

The generalized, reduced homology groups of X, notated hk(X), are

hk(X) := lim−→ [Sk+n,En ∧X]= πk(E∧X)

Remark 36.20. As we see in the definition, a stable homotopy group is an example of ageneralized homology theory.

Remark 36.21. Really, we should notate hk(X) as hk(X,E), and similarly hk(X) as hk(X,E),since the groups depend on a spectrum E. Each E gives a generalized homology andcohomology theory, both defined in homotopy.

Remark 36.22. One could equivalently define

hk(X) = lim−→ [Sn,EXk+n]

= lim−→ [Sn−k,EXn]

= π−k(EX),

where (EX)n:= EXn.

Next time, we’ll prove these give the expected long exact sequence and illustrate whatthese cohomology and homology groups are.

37. 4/27/16

37.1. Review. Last time, we saw how a spectra

(En, εn)n∈Z

gives rise to a general cohomology and homology theory.

hn(X) := lim−→ [SnX,En+k] ,

hk(X) :=[Sn+k,X∧ En=k

].


Question 37.1. Why are these generalized (co)homology theories?

We’ll deal with the cohomology theory, since it’s easier.

37.2. Constructing the long exact sequence for cohomology.

Proposition 37.2. Cohomology as defined above is a generalized cohomology theory.

Proof. Suppose we have a sequence

(37.1) A X X/A

Note that we get a corresponding exact sequence

(37.2) hk(X/A) hk(X) hk(A),

which is exact at hk(X), which is straightforward. Indeed, the homotopy extension prop-erty for (X,A) implies that

(37.3) [X/A,B] [X,B] [A,B]

is always exact in the middle, for sufficiently nice pairs of spaces (X,A). Therefore, the se-quence on hk is a limit of exact sequences, hence exact. We now deduce an exact sequence,we get

(37.4)

· · · hk(SX) hk(SA)

hk(X/A) hk(X) hk(A).

Remark 37.3. Recall that this boundary map came from a map X ∪ CA, attached alongA, to CX ∪ X ∪CA, where the cones were glued along X and A. This inclusion induces amap from X/A to SA, and gave a map on cohomology.

So, we need that

hk(SA) = hk−1(A),

which follows from the definition, because

hk(SA) = lim−→ [SnSA,En+k]

= lim−→ [Sn+1A,En+k]

= lim−→ [SmA,Em+k−1]

= hk−1(A).

This implies the usual long exact sequence.

136 AARON LANDESMAN

Remark 37.4. In general, we would have

h∗(S0) 6= ordinary cohomology H∗(pt).

and but if the two were equal, it would coincide with ordinary cohomology.

Remark 37.5. It takes more work to check our definition of homology forms a generalizedhomology series. The hk case was done first for spectra that are convergent, meaningthere exists n0 so that En is (n−n0)-connected.

Definition 37.6. Here, k-connected means

πr(X) = 0

for r ≤ k.

Example 37.7. The sphere Sn is n− 1 connected.

One can conclude this exact sequence from stabilization of πn+k(Sn). Then, one canapproximate a general spectrum by a convergent one.

37.3. Examples of cohomology theories for spectra.

Example 37.8. For the sphere spectrum, we have the homology theory

hk(X) = lim−→πk+n(SnX)= πsk(X).

For the sphere spectrum,

hk(X) = stable cohomotopy groups

= lim−→ [SnX,Sk+n]

= πks (X).

Example 37.9. The Thom spectra have the following homology theory. We have,

hk(X) = lim−→πk+n(X∧ τ(En+k))

= MSOk(X)

where En+k is not the spectra but the universal bundle over the grassmannian. Here, wewere using the Thom space for the oriented grassmannian, but if we took the Thom spacefor the unoriented grassmannian, we would have gotten MOk(X).

Example 37.10. For the Bott spectrum, we recover K-theory. We have

hk(X) = Kk(X)

hk(X) k-homology

Example 37.11. Ordinary homology or cohomology is recovered from Eilenberg MacLanespectrum. We have Here,

[X,K(k,G)] = Hk(X;G).


where K(k,G) is the Eilenberg MacLane space. In this way, Eilenberg-MacLane spacesplay the role of the classifying spaces for cohomology. We have

[SnX,K(n+ k,G)] = Hn+k(SnX;G) = Hk(X;G).

Then,

lim−→ [SnX,K(n+ k,G)] = Hn(X;G),

because the maps between the terms in this spectra are all the expected isomorphisms.

37.4. Stable homotopy and stable homotopy. We’ll now discuss πsk and πks , and see thatfamiliar things like fundamental classes and Poincare duality carry over.

Definition 37.12. We have nonreduced versions

Πk(X) = πsk

(X∐

∗)

and Πk(X) is defined as the limit as the limit n→∞ of homotopy classes of

Rm × X f−→ Rm+k

for f proper.

Remark 37.13. This agrees with the notion of Πk = Πk(pt) defined long ago. We’ve nowdefined Πk(X) in the context of our generalized homology theories.

We have

Πk(X) =[Sn+k,Sn(X

∐∗]

=[Rn+k ∪∞, (Rn × X)+

].

If X is compact, this is equal to homotopy classes of proper maps

Rn+k f−→ Rn × X.

For nice spaces X, we can approximate f by a map

f = (f1, f2)

with smooth first component

f1 : Rn+k → Rn.

Taking

M := f−11 (p) ⊂ Rn+k,

M is a stably framed kmanifold. In the rest of the class, we my use framed to mean stablyframed.

Now, restricting f2 toM, we get

f2|M :M→ X,

As in the case of MSO, we have the following proposition.

Proposition 37.14. We can identify

Πk(X) = cobordism classes of pairs (M,φ, f)

whereM is a k-manifold, φ is a stable framing, and f :M→ X.

138 AARON LANDESMAN

Proof. Omitted.

Question 37.15. Suppose we have an n-manifold which is compact. What should thefundamental class in Πn(X) be?

Here, we need a stably framed manifold, but don’t the need manifold to be compact ororiented.

The definition is elementary.

Definition 37.16. Take the fundamental class to be

[M,φ, f] ∈ Πk(M)

to be

M = X

f := id :M→ X

and φ to be the given stable framing ofM = X,

We also have Poincare duality. Recall that by definition Πk(X) is the limit as n → ∞ ofhomotopy classes of

Rm × X f−→ Rm+k

for f proper.

Theorem 37.17 (Poincare duality). If X is a stably framed n-manifold, we have

Πk(X) ∼= Πn−k(X).

Proof. Omitted.

Remark 37.18. When X is a manifold, we can take f to be smooth and p ∈ Rm a regularvalue of f. TakeM := f−1(p), which has dimension n− k and a trivialized normal bundleν(M). ForM ⊂ Rm × X, if TX is trivialized then TM is too. Say φ is the trivialization.

We then have a manifold X, a Euclidean space Rm, and inside the productM ⊂ Rm×Xwith a stable framing φ. We have a projection

pX :M→ X.

We have a triple

(M,φ,pX) ∈ Πn−k(X).So, this gives a geometric description of Πk(X) and proves Poincare duality without too

much difficulty.

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