Algebraic Topology: Math 231Br Notes

ALGEBRAIC TOPOLOGY: MATH 231BR NOTES

AARON LANDESMAN

CONTENTS 1. Introduction 4 2. 1/25/16 5 2.1. Overview 5 2.2. Vector Bundles 5 2.3. Tautological bundles on projective spaces and Grassmannians 7 2.4. Operations on vector bundles 8 3. 1/27/16 8 3.1. Logistics 8 3.2. Constructions with Vector Bundles 8 3.3. Grassmannians and the universal bundle 10 4. 1/29/16 12 5. 2/1/16 15 5.1. Characteristic Classes 15 5.2. Leray-Hirsch Theorem 17 6. 2/3/16 18 6.1. Review 18 7. 2/5/16 21 8. 2/10/16 24 8.1. Reviewing Leray-Hirsch 24 8.2. Review of Chern and Stiefel-Whitney Classes 25 8.3. Examples and Calculations 26 9. 2/12/15 28 9.1. Logistics 28 9.2. Applications of Stiefel-Whitney classes 28 9.3. Pontryagin Classes 31 10. 2/17/16 31 10.1. Theory on Pontryagin classes 31 10.2. Calculations and Examples with Pontryagin Classes 33 10.3. Euler Classes 34 11. 2/19/16 35 11.1. Review and Thom Classes 35 11.2. Euler Classes 37 11.3. Examples of Thom and Euler Classes 38 12. 2/22/16 39 12.1. More on Euler Classes 39 12.2. K Theory 41 13. 2/24/16 44 1 2 AARON LANDESMAN

13.1. Examples for K theory 44 13.2. Reduced K theory 45 13.3. Products 45 14. 2/26/16 48 14.1. Equivalent deﬁnitions of relative K theory 48 14.2. Back to Smash Products 49 15. 2/29/16 52 15.1. Review 52 15.2. Fredholm operators and index 53 16. 3/2/16 55 16.1. More on Fredholm operators 55 16.2. The index of a family 56 16.3. More examples of Fredholm operators 58 17. 3/4/2016 58 17.1. Toeplitz Operators 60 18. 3/7/2016 61 19. 3/9/16 65 19.1. Review of inﬁnite dimensional groups 65 19.2. Real K theory 67 19.3. Symplectic K theory 68 20. 3/11/2016 69 20.1. Chern character 70 20.2. 73 21. 3/21/2016 75 21.1. Clifford Algebras and Clifford Modules 76 22. 3/23/2016 78 22.1. How to obtain the Z/2-grading 79 22.2. Bundles of Clifford modules 80 23. 3/25/16 81 23.1. Clifford Modules 81 24. 3/28/16 84 24.1. 86 25. 3/30/16 88 25.1. Review 88 25.2. Calculating A^ 89 25.3. Relating our computations to A^ 91 26. 4/1/16 93 27. 4/4/16 97 27.1. Formalities 97 27.2. Notations 98 27.3. The exact sequence for K-theory in negative degrees 99 28. 4/6/16 101 28.1. Review 101 28.2. Bordism and cobordism 105 29. 4/8/2016 106 29.1. Oriented cobordism 107 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 3

30. 4/11/16 108 30.1. Review 108 31. 4/13/16 112 31.1. Oriented bordism as a homology theory 112 32. 4/15/16 115 32.1. Review 115 33. 4/18/16 119 33.1. Review 119 33.2. The relations between framed cobordism and homotopy groups 119 34. 4/20/16 122 34.1. Stabilization 122 34.2. J-homomorphism 123 35. 4/22/16 126 35.1. Low stable homotopy groups 126 35.2. More about Π3 127 35.3. Another integrality property for A^ 128 36. 4/25/16 130 36.1. Spectra 130 36.2. More involved examples of spectra 131 36.3. Fundamental groups of spectra 133 37. 4/27/16 134 37.1. Review 134 37.2. Constructing the long exact sequence for cohomology 135 37.3. Examples of cohomology theories for spectra 136 37.4. Stable homotopy and stable homotopy 137 4 AARON LANDESMAN

1. INTRODUCTION Peter Kronheimer taught a course (Math 231br) on algebraic topology and algebraic K theory at Harvard in Spring 2016. These are my “live-TEXed“ notes from the course. Conventions are as follows: Each lecture 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 errors are my fault. By the same token, any virtues in the notes are to be credited to the lecturer and 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 his permission. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 5

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 unitary groups, or equivalently about classifying vector bundles of large rank on spheres) Remark 2.1. There are many approaches to Bott periodicity. We give a proof in class following an argument of Atiyah. This introduces index theory for Fredholm operators and related things. Remark 2.2. K-theory is like ordinary homology (originally called an extraordinary homology 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 homotopy 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 (or over X) is a topological space E with a continuous map φ : E X and a real (or complex) −1 vector space structure on each fiber Ex := φ (x). This must satisfy the additional condition of being locally trivial, meaning that there is an open cover→ U of X so that for each −1 U ∈ U with U ⊂ X, the restriction of E to U, notated EU := φ (U) ⊂ E is trivial. Here, trivial means there exists a homeomorphism φ

φ n EU U × R

(2.1) φU π U id U or to U × Cn for the complex case, where n φ|Ex : Ex {x} × R is linear. → 6 AARON LANDESMAN

Example 2.5. (1) The trivial vector bundle E = X × Rn with φ : X × Rn X the projection. 1 (2) The Mobius¨ vector bundle on S . 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 n ∼ ⊥ n+1 (TS )x = x ⊂ R . We shall now check local triviality, which we shall often not do in the future. Consider → n U := {x : x ∈ S , xn+1 > 0} . We have an inclusion TU TSn Over U this is trivial with → TU U × Rn (x, v) 7 (x, π(v)) n+1 n n+1 → with π : R R = x ∈ R : 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 E where φ ◦ s = idX. Lemma 2.7. The Möbiusbundle 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 by determining whether the bundle remains connected or is disconnected, after removing the image of the 0 section. Proof 2. The Mobius¨ bundle is not trivial because it has no nonvanishing section. as can be seen by the intermediate value theorem. Definition 2.8. Two vector bundles p : E X, q : F X are isomorphic over X if φ E F → → (2.2) p q X ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 7 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 only looked at real vector bundles, but we will now consider complex ones. n n Deﬁnition 2.9. Let CP , or PC denote x : 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 of CPn. 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 is isomorphic 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 that if one removes the zero section from TS2, one obtains a space which has the homotopy type of RP3, which has fundamental group Z/2Z. In fact, by the Hairy ball theorem, TS2 has 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 bundle case. We next have an extension from projective spaces to Grassmannians. Definition 2.16. We define the Grassmannian N N Gn(R ) = x : x ⊂ R , x is a linear subspace with dim x = n .

n n+1 Exercise 2.17. Show RP = G1(R ). Deﬁnition 2.18. We have a tautological bundle over the Grassmannian N φ : En Gn(R ) where → N N En := (x, v) ∈ Gn(R ) × R : v ∈ x

8 AARON LANDESMAN

2.4. Operations on vector bundles. We now discuss how to make new vector bundles from 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 → E E1 (2.3)

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 open ∼ ni set U ⊂ X with E1|U and E2|U both trivial, then→ writing both as Ei|U = U × R , we make ∼ n1 n2 E|U = U × (R ⊗ R ) 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 projective bundle P(Ex). As a set, we can take PE = ∪x∈XPEx. Better, we can define × PE = (E \ 0)→/k ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 9

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 a base. Then, p is a fiber bundle if there exists an open cover U so that for all U ∈ U, there is a homeomorphism φU. →

φU PU U × F

(3.1) φ| π PU U U

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 h, ix is an inner product on each Ex varying continuously with x. In other words, in a local trivialization, where E is a product U × 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 : hv, vi = 1} . Definition 3.9. Suppose we have a fiber bundle p : P X and a topological group H acting P × H P → so that (1) → µ P × H P

(3.2) p

X X

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 linear n isomorphisms→ ψ : R 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 = ∪xPx → so that n P = {(x, ψ) : x ∈ X, ψ : Ex R } . Remark 3.11. We know that on any connected set the dimension of the fiber is fixed. If the space is disconnected, we will allow the dimensions→ to change. We will also almost always 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, instead of writing k = R or C. Last time we introduced the Grassmannian n n Gk(R ) = { all k-dimensional subspace of R } = { 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 n n+1 Gk(R ) ⊂ Gk(R ).

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 for G (R ). k ∞ n n Definition∞ 3.15. We have a tautological→ vector bundle Ek(R ) Gk(R ) and the limit n n forms a tautological vector bundle Ek(R ) Gk(R ). The fiber over x is x with x ⊂ R or R . ∞ → Definition∞ 3.16. For simplification of notation,→ we use Gk = Gk(R ). We let p : Ek Gk denote the tautological vector bundle. ∞ Definition 3.17. Suppose we have q : F Y a vector bundle. Let f : X Y. Then, we→ can pull back F to X. We get E = f∗F, a vector bundle on X defined as the fiber product → → E F (3.3)

X Y. In more concrete terms, E ⊂ X × F is the set E = {(x, w) : f(x) = q(w)} . −1 ∼ with a map p : E X given by projection onto the ﬁrst factor. We have Ex = p (x) = Ff(x). → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 11

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, and pulling 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 is ∗ ∗ paracompact then f0(F) and f1(F) are isomorphic vector bundles. Corollary 3.21. If X is contractible and paracompact→ (for example, if X is a ball) then every vector bundle 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 bundle on 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 fiber over (0, x) to itself. It is indeed→ possible to extend this section, using the homotopy lifting property (which crucially involves X being paracompact). Since we have a section, we have a way of translating the identity section over 0 to another section over 1, which gives the desired linear isomorphism. That is, the definition of a section is precisely an isomorphism between the fibers over 0 and 1 (and in fact for each t between 0 and 1). Theorem 3.23. For X paracompact and Hausdorff, a vector bundle on X with fiber Rk classified up to isomorphism, are in bijection with homotopy classes of maps f : X Gk. The correspondence is realized by pulling back the universal bundle along f, sending f : X Gk to E X with ∗ E := 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 f , f : X Gn(R ) with→f (En) = f (En), then f ' f . Further, conversely, 0 1 0 1 0 ∞1 if f ' f are homotopic, then the pullback of En by these two maps are isomorphic, as we 0 1 ∞ saw 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. When n = 1 take X = S1, and the real numbers. Then, Vect1(S1) is [S1, RP ] = π (RP ) =∼ Z/2Z. Indeed, 1 ∞ 1 1 ∞ Vect (S ) = { trivial bundle, Mobius¨ bundle } . Example 4.4. Take n = 1, X = S2 and the complex numbers. So, we’re looking at Vect1(S2). 2 By the theorem, this is homotopy classes of maps S , CP = π2(CP ). Now, there is a ﬁber bundle ∞ ∞ S1 S2n+1 (4.1)

CPn and taking the limit, as n , we get

1 → ∞ S S (4.2) ∞

CP and we can take the long exact sequence for a ﬁbration.∞ Since S has trivial homotopy π (CP ) = π (S1) π (CP ) = 0 groups in all dimensions, we have k k−1 . In particular,∞ k , k = 2 π (CP ) = Z S2 except when , in which case 2 ∞ . So, the line bundles on are isomorphic∞ to Z. From the same argument, we the only complex line bundle on Sn for n > 2 is the ∞ trivial bundle. Proof of Theorem 4.1. (1) Given a vector bundle p : E X, there exists an open cover

U = {Uα : α ∈ I} → and trivializations n τα : E|Uα Uα × R . Note, we can assume I is countable by paracompactness of X, as shown in the old version of Hatcher’s algebraic k theory→ book, Lemma 1.16. So, we can take I = N. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 13

Now, by paracompactness, we can ﬁnd a partition of unity subordinate to this cover. That is,

φα : X [0, 1]

where Supp φα ⊂ Uα and α∈N φα = 1, with this sum locally ﬁnite. We deﬁne σα by → P n E|Uα Uα × R (4.3) σn

Rn.

n We then consider φασα : E|Uα R . Note that this extends continuously to all of E. n Further, this map is linear on the fibers Ex R , and varying continuously with → n x. Further, it is an isomorphism Ex R for x with φα(x) 6= 0. Now, define n Φ : E R = ⊕α∈→NR → where ∞ → Φ = ⊕αφασα. This map Φ is (a) linear on the fibers of E (b) is injective on every fiber. So, given this Φ, 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 → Ψ E En(R )

(4.4) p p∞n

f X Gn(R ).

∞ so that E is the pullback of En(R ). Now, Ψ is essentially the map Φ with some bookkeeping for the base Gn(R ). More precisely, recall ∞ En = {(y, v) : y ⊂ R∞ , y has dimension n, v ∈ y}

and 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 ∗ ∼ ∗ ∼ f0(En) = f1(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 on fibers 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 (4.5) that for every v ∈ E, we have that Φ0(v) is not a negative multiple of Φ1(v), with v 6= 0.

Then, we can simply take Φt(v) = (1 − t)Φ0(v) + tΦ1(v). We can reduce to the above assumption by the following trick: Consider the linear map A : R R

(x1, x2, ...∞) 7 (x∞1, 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 homotopy Φ0 ' Φ˜ 0 := A ◦ Φ0. These are maps E R , defined using As, s ∈ [0, 1] . This homotopy is one through maps which are injective ∞ on the fibers. We can similarly apply this trick to Φ1 given by the homotopy→ to Φ˜ 1 = B ◦ Φ1. Now assuming (4.5), we obtain homotopies

Φ0 ' Φ˜ 0 ' Φ˜ 1, ' Φ1, where the middle equation uses the linear interpolation. Remark 4.5. If the space was compact, we could actually realize every vector bundle as the 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 as showing that S is contractible. Our next topic,∞ which we’ll begin on Monday, is characteristic classes. We’ll talk about Chern classes and Stiefel Whitney classes. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 15

Definition 4.7. A characteristic class for real or complex vector bundles assigns to each E X (perhaps of a certain rank) a class c(E) ∈ H∗(X, G), for some group of coefficients G. →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 → E1 E2 (5.1)

f X1 X2 then ∗ c(E1) = f (c(E2)) . The Chern classes for complex vector bundles 2i ci(E) := H (X) for i = 0, 1, 2, .... The Stiefel-Whitney class i wi(E) ∈ H (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 f (Ek) with X − Gk(C ) with Ek Gk = Gk(C ) the tautological bundle. If c is a characteristic class, there→ is a universal characteristic class, ∞ ∞ U ∗ → c→ = c(Ek) ∈ H (Gk). ∗ U ∗ ∗ Then, c(E) = f (c ). So, a characteristic class corresponds to elements in H (Gk(C )), H (Gk(R )). We’ll construct the Chern classes as characteristic classes of vector bundles, with∞ the ∞ following properties, in addition to naturality: (1) (Whitney Sum Formula) If E = E0 ⊕ E00, then 0 00 ci(E) = cj(E ) ∪ ck(E ). j+k=i X 0 (2) We have c0(E) = 1 ∈ H (X) for all E X. Here, 1 means the element which is 1 on each connected component. (3) For L X a line bundle, c1(L) is deﬁned→ by saying c1(E1) = −α where E1 CP H∗(G (C )) = H∗(CP ) = Z[α] α ∈ H2(CP ) is the tautological bundle. (Recall 1 , ∞, and we have a standard “preferred” choice of generator.) → ∞ ∞ → ∞ 16 AARON LANDESMAN ∗ ∼ 2i Remark 5.3. It turns out H (Gk(C )) = Z [α1, ... , αk] with αi ∈ H (Gk(C )) equal to ±c (E ). However, it does take quite some work to calculate this cohomology ring. i k ∞ ∞

Remark 5.4. If we know X is paracompact, knowing c1(E1) = −α tells us what the Chern class is. Even if X is more horrible, we can approximate X by a cell complex, meaning that there 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) 0 00 wi(E) = wj(E ) ∪ wk(E ) X (2) w0(E) = 1 R (3) w1(E1) = −α1 , where we have E1 RP the tautological bundle (note the − sign Z 2 αR H1(RP ) is extraneous as we’re working in / coefﬁcients,∞ and 1 ∈ . → 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(E0) ∪ c(E00).

1 Remark 5.7. Let w1(L) ∈ H (X; Z/2) for L a real line bundle, be 1 H (X; Z/2) = Hom (H1(X), Z/2)

= Hom (π1(X); Z/2) .

1 We can describe this geometrically as follows. Given [γ] ∈ π1(X), we have γ : S X. Look at γ∗(L) S1. We assign → 0 if γ∗(L) is trivial → [γ] 7 ∗ 1 if γ (L) is the Mobius¨ bundle.

This ends up determining a→ map π1(X) Z/2, which curves out to be a homomorphism and gives w1(L). Example 5.8. Suppose we have a line bundle→ L M for M a 1 dimensional manifold. For 1 1 1 example, suppose M = S . Then, we get L S by w1(L)[S ] ∈ Z/2. Then, w1(L)[S61] is obtained by counting the zeros of a section s of→L: 1 Choose a section s : S L where w has→ ﬁnitely many zeros. We will have w (L)[S1] = mult (s). → 1 x x,sX(x)=0 The total number of zeros will then always be even for the trivial bundle and odd for the Mobius¨ bundle. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 17

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 → multx(s). x,sX(x)=0 5.2. Leray-Hirsch Theorem. We’ll now construct the Chern classes. To do so, we’ll need to 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 module structure 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, is → a finitely generated, free R module for all k. (For example, being free will automatically be satisfied when 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 free module 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 = p (aj) ∪ cj. j X In particular, p∗ is injective.

Proof. i Example 5.11 (Non-example, where the hypothesis fails). Taking S2 S2 with ﬁbers S1, then H∗(S1) = Z in dimension 0, 1. The Leray-Hirsch theorem will not help because there 3 1 1 are not elements in the cohomology of S restricting to the generator→ of H (S ). 18 AARON LANDESMAN

Example 5.12. Take E X a complex rank n vector bundle over C and take P = PE, so that O X has fiber CPn−1. There’s a tautological→ line bundle L P = PE so that the fiber at (x, `) ∈ P for x ∈ 2 X, ` ⊂ E→x is `. Let α = −c1(L) ∈ H (P; Z). 2 n−1 2 n−1 ∗ Then, α|PEx is a generator of H (CP →). The classes 1, α, α , ... , α ∈ H (P) restrict to each fiber p−1(x) to give a basis of the cohomology of p−1(x). The Leray-Hirsch theorem implies 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 X with classes i wi(E) ∈ H (X; Z/2) so that ∗ ∗ (1) wi(f E) = f wi(E) for f : X Y. (2) wi(E1 ⊕ E2) = i+j=k wi(E1) ∪ wj(E2) (3) wi(E) = 0 if rk E < 1. → P 1 (4) w1(U) is a generator of H (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 have 2i ci(E) ∈ H (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 rk E < 1. → P 2 (4) For U CP , then c1(U) is a generator of H (CP ; Z) that pairs with the class [CP1] ⊂ CP −1 CP1 ∞to . (The pairing is just pull back to∞ .) Let X be paracompact→ ∞ and E a rank n R or C vector bundle over X. Last time, we π introduced PE − X and U over PE the tautological (line) bundle, over PE. We already defined c1(L) and w1(L). Let 2 → h = −c1(U) ∈ H (PE; Z). Then, 1, h, h2, ... , hn−1 restrict on all fibers to a basis of H∗(PE; Z). By the Leray-Hirsch theorem, 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. 2i There then exist unique ψi ∈ H (X; bz) so that n n ∗ n−i h + π ψih = 0. i=1 X We define the Chern classes are ci(E) := ψi. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 19

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 rk E < 1. → P 2 1 (4) For U CP , then c1(U) is a generator of H (CP ; Z) that pairs with the class [CP ] ⊂ CP −1 CP1 to .∞ (The pairing is just pull back to .)∞ 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 only need to check the Whitney sum formula. Before proving the Whitney sum formula, we will 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 or Hermitian metric.

Proof. Use partitions of unity. Corollary 6.7. Let E be a vector bundle over X and E0 ⊂ E a subbundle. Then, there exists another subbundle E00 ⊂ E such that E = E0 ⊕ E00. That is, all short exact sequences of vector bundles 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 space Σ 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 line bundle. 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 is that in algebraic topology, short exact sequences always split, while in algebraic geometry they don’t, and so you can “pretend” they split, but really you’re just relating the ﬁrst and last terms of a short exact sequence to the middle term. 20 AARON LANDESMAN

n To prove the Whitney sum formula, it sufﬁces to show that if E = ⊕i=1Li, for Li line bundles, then n c(E) = (1 + c1(Li)) . i=1 Y 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 fi U = Li. So, we can look at

(f1∞, f2): X CP × CP . → ∞ ∞ ∗ Without loss of generality, we can assume X = CP × CP . Then, Li = πi U. So, it’s enough to prove this on the two skeleton.→ The two skeleton of CP × CP is just two ∞ ∞ spheres glued at a point. The two line bundles are tautological one sphere and trivial on ∞ ∞ the 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 tautological line bundles.

Lemma 6.12. Suppose L1, ... , Ln are line bundles over X. Suppose si is a section of Li vanishing at Zi ⊂ X. Assume ∩iZi = ∅. Then, n c1(Li) = 0. i=1 Y

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

c1(Li)|X\zi = 0. We have a long exact sequences

2 2 2 (6.1) ··· H (X, X \ Zi) H (X) H (X \ Zi) ···

2 We know c1(Li) maps to 0 under the second map. Therefore, there is some ξ ∈ H (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 a map ∪ Hk(X, A; R) × Hl(X, B; R) − Hk+l(X, A ∪ B; R). given by cup product. → Then, the product 2n n 2n ξi ∪ · · · ∪ ξn ∈ H (X, ∪i=1 (X \ Zi)) = H (X, X) = 0. Therefore, the product of these classes is 0, as claimed. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 21

We now want to see n n c (⊕i=1Li) = c(Li). i=1 n Y We have E = ⊕i=1Li. We have the projectivization π : PE X and we have ∗ →n ∗ U , π E = ⊕i=1π Li. We now deﬁne the composition → n ∗ ∗ πi : U ⊕i=1π Li π Li. We can think of → ∗→ πi ∈ Hom(U, π Li).

This section πi vanishes on L1 ⊕ · · · ⊕ L^i ⊕ · · · ⊕ Ln. These such loci intersect trivially. So, ∨ ∗ 0 = πc1 U ⊗ π Li ∗ = π (−c1(U) + π c1(Li)) n ∗ = (h + π c1(Li)) . i=1 Y When we expand the elementary functions in this product and get the Chern classes, since they 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 cos θ sin θ E = (θ, v) ⊂ S1 × R2 : v = v . → sin θ − cos θ

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)v2

sin θv2 = (1 − cos θ)v1. We now have trivializations on the top and bottom halves of the circles, and local sections given by 1 sin θ 1−cos θ sin θ , . 1+cos θ 1 22 AARON LANDESMAN

Indeed, these two local sections agree on the overlaps. We can notate this combined section as 1 + cos θ sin 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 look at E S2. Deﬁne iz ix − y P := . → ix + y −iz Note that Pt = −P, so it is Hermitian. The trace is 0 and determinant is 1, so the eigenvalues 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 2 3 E∗ := {~p, v} ⊂ S × C : ~p × v = iv , where × refers to the cross product. ~p × (~p × ~v) = − |~p|2~v − ~p(~p · ~v) . We have 2 E∗ ⊕ E∗ = TS ⊗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 the function p − q p 7 = M . p − q p Of course, M implicitly depends on our fixed choice of q. We can construct p → 2 EX = (p, v) ⊂ X × C : Mpv = iv . Defining the function 2 fq : X S p − q p 7 . → |p − q| E We now can realize X as the pullback →

EX E (7.1)

fq X S2 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 23

∗ 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 as a function of q. Then, as we move q very far away, we’re pulling the bundle back by a map to the two sphere where the image lies in an extremely small patch of S2, on which the bundle E on S2 is trivial.

Remark 7.6. If q lies inside the Riemann surface X, the bundle EX will not, in general, be trivial. Example 7.7. Suppose we have a surface X ⊂ R3. Then, at each point on the surface, we ~ can deﬁne a normal direction n~ p, written bn = (n1, n2, n3). We have a matrix satisfying in in − n 3 1 2 · v = iv. in1 + n2 −in3 We can deﬁne a bundle 2 E = (p, v) ⊂ X × C : n~ p × v = iv .

Fact 7.8. Suppose X is a closed 2-dimensional real oriented manifold and E X is a complex 1 dimensional vector bundle, then E is isomorphic to f∗F where F is the ﬁrst 2 bundle on S 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

0 1 H (P , OP1 (n)), deﬁned to be the vector space of homogeneous complex polynomials in two variables of degree n. Consider the operators ∂ ∂ L := i u − v z ∂u ∂v ∂ ∂ L := i v + u x ∂u ∂v ∂ ∂ L := u − v , y ∂v ∂u which gives an action of the Lie algebra of SO(3) on this space of polynomials. Next, deﬁne

Lp := xLx + yLy + zLz,

2 and deﬁne the bundle En S to be

2 0 1 E := (p, v) ∈ S × H (P , O 1 (n): L v = inv . n → P p

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), |u|2 − |v|2 z = . |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:

δ(j) Theorem 8.1. Suppose we have p : P X a fiber bundle. Let cj ∈ H (P; R). Suppose that the ∗ restrictions 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 cohomology −1 of that fiber). Here, Px = p (x) for all x. Then, there’s an isomorphism

k Ψ k D (X) − X E (X), where for U ⊂ X, → k Ek(U) := H (PU; R) k k−δ(j) D (U) := ⊕jH (u; R).

−1 Here, PU = p (U). The map is given by k k ΨU : D (U) E (U) ∗ (a1, a2, ...) 7 p (aj) ∪ cj. → j X Proof. We’ll prove this for ﬁnite simplicial→ complexes. We have both E∗(U) and D∗(U) behave as follows: ι (1) if U − 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

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

(8.1) Ψ Ψ Ψ Ψ

0 Ek(U ∪ V) Ek(U) ⊕ Ek(V) Ek(U ∩ V) Ek+1(U ∪ V) 0 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 25

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 the proof 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 depends 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 another step, which we omit. 8.2. Review of Chern and Stiefel-Whitney Classes. Adrian defined the Chern classes in a previous class. Let’s now discuss their real analog, Stiefel-Whitney Classes. For p : E X, a real vector bundle, we’ll define i wi(E) ∈ H (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 ∞ old ∗ w1 (E) = f (h). It won’t be immediate that our new definition is the same, although it isn’t too hard to show they are the same. Definition 8.3. Form p : PE X with fiber RPn−1, with n = rk E. 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 old taut → i h = wi (L ) ∈ H (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 total space. By Leray-Hirsch, there exist unique classes i wi(E) ∈ H (X; Z/2) so that n ∗ n−1 ∗ ∗ h + p (w1(E))h + ··· + p (wn−1(E)h + p (wn(E)) = 0. Exercise 8.4. Check 26 AARON LANDESMAN

old (1) w1 = w1 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 Whitney sum formula, which can be deduced from the splitting principle. This reduces to the case that 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 behave with 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). → ∗ taut → ∗ taut taut Proof. We have E1 = f1(L ), E2 = f2(L ) where L CP is the tautological bundle. Then, take ∞ f = (f1, f2): X CP × CP . →

Then, the universal case is L1 and L2 tautological∞ on CP∞ × CP . One then must do a calculation just on the 2 skeleton, which is where→ the ﬁrst cohomology appears to check ∞ ∞ this holds there. That is, we have two cells CP1 × CP0 and CP0 × CP1, which meet at the 0 0 single points CP × CP . 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 nowhere ∼ ∨ zero section corresponding to the identity in End(E) = E ⊗ E . Corollary 8.8. For E of any rank, ∨ i ci(E ) = (−1) ci(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, n c(E) = 1 + c1(Lj) . n=1 Y ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 27

Then, ∨ ∨ ∨ ci(E ) = σi c1(L1 ), ... , c1(Ln ) ,

implying the result.

Question 8.9. Can we say anything about the Chern classes of E1 ⊗ E2 for general complex vector bundles? We can use the splitting principle with bare hands, although this gets very messy very quickly.

Example 8.10. Let’s suppose E1 and E2 are rank 2 vector bundles. Suppose further that c1(Ei) = 0. Then, using the splitting principle, we may as well assume 0 E1 = L1 ⊕ L1` = c1(L1) 0 0 ` = c1(L1). 0 0 Then, c1(E1) = ` + ` = 0, so ` = −`. Then, 0 2 c(E1) = (1 + `)(1 + ` ) = 1 − ` . Similarly, we may write 2 c(E2) = 1 − m . Then, 2 c2(E1) = −` . Then, ∼ 0 0 0 0 E1 ⊗ E2 = L1 ⊗ L2 ⊕ L1 ⊗ L2 ⊕ L1 ⊗ L2 ⊕ L1 ⊗ L2. 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 dimension k, we have n ⊥ TvGk(R ) = Hom(V, V ). This, in fact, deﬁnes a chart for the Grassmannian of subspaces not intersecting V⊥ away from 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 ) n = w(L∨) = (1 + h)n with h ∈ H1(RPn−1; Z/2). Then, n w (TRPn−1) = hk. k k 9. 2/12/15 9.1. Logistics. Problem set 2 is available now. It is due on 2/19/2016. Future homework should be submitted on canvas, uploaded. n−1 9.2. Applications of Stiefel-Whitney classes. Last time, we sketched a proof that wk(TRP ) = n k k n−1 ∼ k h ∈ H (RP ; Z/2) = Z/2. Example 9.1. The space RP4 does not immerse in R5. That is, there is no function f : 4 5 4 RP R . Recall an immersion is a map with dfx is injective for all x ∈ RP . ∼ n n Recall we have NxM ⊕ TxM = R and so TM ⊕ NM = R 4 4 ∼ To→ see this, if f exists, there would exist a real line bundle N RP so that TRP ⊕ N = R5. But then, w(TRP4)w(N) = w(R5) = 1. → 4 We have w(TRP ) = 1 + w1 + w2 + w3 + w4)(1 + u) = 1, where w1(N) = 1 + u. Then, we have 1 = (1 + h + 0 + 0 + h4)(1 + u) since we are using Z/2 coefﬁcients. Solving these equations, we have 1 · 1 = 1. In degree 1, we have 1 · u + h · 1 = 0, so u = −h = h. We next get hu = 0, and so h2 = 0. This is a contradiction because h2 6= 0 in the ring H∗(RP4; Z/2) and h is the generator of the ﬁrst cohomology, 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,

dimR WR = 2 dimC W. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 29

Deﬁnition 9.2. In the other direction, if we are given V, a 2n dimensional real vector space, a complex structure on V is a complex vector space W with WR = 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 manifold M is a complex structure J on TM. n n Take [V] ∈ Gk(C ). One can take a neighborhood of V in Gk(C ), given by all vector spaces intersecting V⊥ trivially. In other words, such a neighborhood is given by ⊥ n Hom(V, V ) , Gk(C ) f 7 graph(f). → If [V] ∈ Gr, we can identify → n ∼ ⊥ T[V](Grk(C )) = Hom(V, V )R. Then, n ⊥ TGrk(C ) = Hom(E, E )R

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

n−1 Example 9.6. Let’s now compute ck(TCP . We obtain n c (TCPn−1) = hk k k

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 from algebraic geometry. Feel free to skip it if you’re not familiar with the terminology. 3 2 3 Take H CP . The dual of the tautological bundle is c1(H) = h ∈ H (CP , Z). We take H ⊗ 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 n ⊗d degree d on C , we obtain that a ∈ Γ(OCP3 (H )). ⊗d Given any section s ∈ Γ(OP3 (H )), with the property that for every x ∈ X := V(s), we have 3 ⊗d ds|x : TxCP OCP3 (H ), a complex linear nonzero map. This means every point x ∈ X is smooth→ in X, and that TxX is a complex C-linear sub- 3 space of TxCP , because the tangent space at x is just the kernel of the linear map ds|x. All this amounts to is saying that this is a smooth almost complex manifold. As an interesting special case, we could just take a smooth homogeneous polynomial of 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 when M is a manifold, we use the notation

wi(M) := wi(TM)

ci(M) := ci(TM), the second case making sense only when M is almost complex. Now, 3 c(X)c(NX) = c(TCP |X).

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

3 ds ⊗d TxCP − (H )s. The kernel of this map is the tangent bundle, and so we have → 3 ∼ ⊗d TxCP /TxX = Hx and the left hand side is the normal bundle. Finally, 3 2 3 c(TCP |X) = 1 + 4h + 6h + 4h ). Additionally, c(H⊗d) = (1 + dH). So, we have 2 3 (1 + c1 + c2)(1 + dh) = (1 + 4h + 6h + 4h )|X. Writing down equations on the generators, we get 1 · 1 = 1

c1 + dh = 4h 2 c2 + c1(dh|X) = 6h |X. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 31

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 a distinction as to when this is positive, negative, or 0. For example, if we did this CP2, we would 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 is smooth. In degree 3, it will be a torus (of genus 1). In degree 4 it is a genus 3 surface, and g−1 in general, it has degree 2 . 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 4i pi(V) ∈ H (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 a complex vector bundle satisfying i(v ⊗ λ) = v ⊗ (iλ). i Define pi(V) = (−1) c2i(V ⊗ C). Remark 9.12 (Reason for only considering even Pontryagin classes). As a complex vector bundle W := V ⊗ C has the property that W =∼ W, the complex conjugate, meaning that for j odd, we have j (−1) cj(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, implies ck(w) = σk (c1(L1), ...), where σk is the kth elementary symmetric function. k Recall that pk(V) := (−1) c2k(V ⊗ C). are the Pontryagin classes for 1 ≤ k ≤ rk V/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) + α where 2α = 0. 32 AARON LANDESMAN

Proof. We have k pk (V1 ⊕ V2) = (−1) cn(V1 ⊗ C) ∪ cm(V2 ⊗ C) n+m=2k X k = α + (−1) cn(V1 ⊗ C) ∪ cm(V2 ⊗ C) n,m≡X0 mod 2 i j = (−1) c2i (V1 ⊗ C) (−1) c2j(V2 ⊗ C) i+j=k X = α + pi(V1)pj(V2) i j X, 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 scalars of the complex vector bundle W = 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, k pk(WR) = (−1) c2k(WR ⊗ C). ∼ Warning 10.6. Note that W 6= WR ⊗R C. In fact, the latter has rank twice as large as that of the former. Proof. Omitted. Lemma 10.7. For W 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 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 33 are the ±i eigenspaces of J ⊗ 1. Here, we’re doing linear algebra fiber by fiber. We then obtain isomorphisms W W+ w 7 w ⊗ 1 − (Jw) ⊗ i → and we similarly obtain → W W−. To see the maps are isomorphisms, we’ll just check the first one, defining the isomorphism + to W . This indeed defines a map W W→R ⊗ C. Exercise 10.8. Show this is a complex linear map. The former has a complex structure because 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: The solution 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. 2 Example 10.10. Take X = CP . 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) 2 = −2c2(TX) + c1(TX). where in the second line, this is the tangent bundle as a complex vector bundle. 2 Now, we know the Chern classes of tangent bundles. For ci = ci(CP ), 2 c2 = 3h c1 = 3h we have the resulting Pontryagin class is 2 2 2 p1(TX) = −2(3h ) + 9h = 3h . Deﬁnition 10.11. For X a closed oriented manifold of dimension n, with n = 4k, we have a pairing

pk(TX)[X] ∈ Z 4k with pk(TX) ∈ H , [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 degree 4k ∈ H∗. 2 Example 10.12. We have, for example, p1[CP ] = 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 always divisible by 3. Here is a deeper fact. Fact 10.14 (Rowin’s Theorem). If X is closed oriented smooth of dimension 4, and, in addition, 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. 3 Example 10.15. Take X ⊂ CP a smooth quartic surface. Then, p1[X] = −48. In fact, we now 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, we have 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 r −1 and let Vx = p (x). → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 35

Deﬁnition 10.17. An orientation for V X is a choice of generator r 0 ∼ (10.1) ux ∈ H (Vx, Vx ) = Z → satisfying coherence, meaning for all x ∈ X, there is some neighborhood U and a class r 0 uU ∈ H (VU, VU) so that

u | 0 = u U Vy,Vy y 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) → u| 0 = u x ∈ X such that Vx,Vx x holds for all . Remark 10.19. Equivalently, we can write u ∈ Hr(D(V), S(V)), by some homotopies rescaling the norm. Remark 10.20 (Joke). The Thom class is essentially “one class to rule them all, and in the darkness 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. 0 n 0 Define V = V \ 0. If there exists ux ∈ H (Vx, Vx ) for all x which is a generator and is n 0 coherent, then this is an orientation for V. Then, a Thom class→ is an element u ∈ H (V, V ) 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, 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). → → 36 AARON LANDESMAN

(4) We have n−1 0 H (VU, VU) = 0. Proof. Recall from last time that the third part regarding the isomorphism follows from Leray-Hirsch. This also implies that the Thom class is unique. By the isomorphism, we have an isomorphism 0 n 0 H (U) H (VU, VU). If u , u0 ∈ Hn(V , V0 ) are two sections and u is a Thom class, then 0 = p∗(a)u ∈ U U U U → U U U H0(U). Here, a is a locally constant Z valued function on U. But, the assumption that 0 uU|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 n−1 0 H (VU, VU) = 0 because n−1 0 −1 H (VU, VU) = H (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 .

n−1 0 n 0 n 0 n 0 n 0 (11.1) H (V3, V3 ) H (VU, VU) H (V1, V1 ) ⊕ H (V2, V2 ) H (V3, V3 )

Note that u1|V = u2|V because both are Thom classes over U1 ∩ U2. By ex- U1∩U2 U1∩U2 actness of Mayer-Vietoris, there exists some uU mapping to (u1, u2) in the above exact sequence. To spell things out, the maps on the above long exact sequence for the latter three terms are 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 dimensional cell complex. One can do this by building this up skeleton by skeleton, inductively taking the 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 X which induces isomorphisms on homology and cohomology, and reduces the calculation on X to one on the cell complex. (This involves 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 = π1(u1) ∪ ∗ π2(u2) is a Thom class for V. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 37

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

n1 0 u1 ∈ H (V1, V1 ) n2 0 u2 ∈ H (V2, V2 ), both generators. Then we have

∗ ni 0 πi (ui) ∈ H (V1 × V2, V1 × V2) ∗ ∗ n 0 π1(u1 ∪ π2(u2) is a generator for H (V, V ). Hence, orientations of V1, V2 yield an orientation of V1 ⊕ V2 (called the Whitney sum) ∗ ∗ and u = π1(u1) ∪ π2(u2). 11.2. Euler Classes. n 0 Deﬁnition 11.3. Let u be the Thom class in H (V, V ). Let s0 : X V be the zero section of a vector bundle. The Euler class for an oriented rank n bundle V X, denoted e(V) is ∗ n s0(u) ∈ H (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, via the 0 section. In the case that the vector bundle→ were trivial, the Euler class is apparently trivial, 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 that D(V)/S(V) is the one point compactification of V if X is compact. This has a distinguished point which is the point corresponding to the sphere. When we throw out the boundary point, we get an open disc bundle, which is homeomorphic to the original V. Example 11.9. Take L CPn−1 to be the dual of the tautological line bundle with fiber C. 2 Look at LR, with fiber R . We have the Thom class 2 0 2 → u ∈ H (LR, LR) = H (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 as CPn \ ∗ So, we have 2 0 u ∈ H (LR, LR) 2 = H (D(LR), S(LR)) ∼ 2 = H (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. It follows that 2 n−1 e(L) = u|CPn−1 ∈ H (CP ) which is a generator h. By taking a limit, the same statement holds on CP . That is, the Euler class of the dual of the tautological bundle on CP . So, we get, the following corollary, ∞ Corollary 11.10. For L X a C line∞ bundle, we have 2 c(L) = e(LR) ∈ H (X). → Corollary 11.11. For W X, a complex vector bundle of rank n, we have 2n cn(W) = e(WR) ∈ H (X). → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 39

Proof. First, we’re using that the underlying real vector bundle of a complex vector bundle is orientable. If we write the complex coordinates as x + iy. Then, take the ordering 2n (x1, y1, x2, y2, ... , xn, yn). This deﬁnes an orientation of R . Then, use splitting principle, using the line bundle formula and multiplicativity of e and c. That is, if W = ⊕iLi, we have

cn(W) = c1(L1) ··· c1(Ln)

= e(L1) ··· e(Ln) = e(W). Deﬁnition 11.12. Suppose X is an oriented smooth closed n-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 i e(TX)[X] = (−1) bi i X where bi = rk H (X).

Proof. Example 11.14. We have e(TS6)[S6] = 1 − 0 + 0 − 0 + 0 − 0 + 1 = 2. 6 3 6 Remark 11.15. Note, we have a ﬁber bundle W S with ﬁber C so that WR = TS . This gives an almost complex structure on S6. It is a famous unsolved problem whether S6 is 6 7 a complex manifold. But, we do have an almost→ complex structure, where S ⊂ R as the 6 imaginary Cayley numbers. We have an almost complex structure given by J : TxS 6 2 TxS , satisfying J = −1, by v 7 n · v with n the normal vector. 6 6 → Corollary 11.16. There exists a complex vector bundle W on S with c3(W)[S ]. Explicitly, this is 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 closed oriented 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 have ﬁnitely 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 or reverses 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 closed oriented manifold. Suppose we have a section s : X E transverse to 0. Then, → e(E)[X] = sgn(ds| ) → x x:sX(x)=0 ∼ where sgn counts the orientation. Here, the sign is whether the derivative of TxX = Ex is orientation preserving or reversing. Example 12.3. If E = TX, s = ∇f, we have the Hessian Hess(f) is nondegenerate at the critical 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] = (−1)i(f,x). x,X∇f=0 Theorem 12.4 (Poincare, Hopf). We have n i e(TX)[X] = (−1) rk Hi(X). i=0 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 f which 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 are maxima, and the 1-barycenters are saddle points. It will end up that n i e(TX)[X] = (−1) rk Hi(X) i=0 X = (−1)dim σ σ simplicesX i ∆ = (−1) rk Ci i X i ∆ = (−1) Hi . i X 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| 0 Ex,Ex 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, because there is a unique generator of Z/2. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 41

Remark 12.7. We have a generator n e(2)(E) ∈ H (X; Z/2) given by ∗ e(2)(E) = s0(U). If E is oriented, we have

e(2) = e mod 2, where mod2 means 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 rk E = n = 1. Then we use the splitting principle. 12.2. K Theory. For now, we’ll Assam all vector bundles are complex and our base space X 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 bundle N F X so that E ⊕ F = C . (Proof: Cover X with open sets U so that E|U is trivial f N ∼ ∗ and obtain a map X − Grn(C ). Identify E = f (Etaut→ ). Here, we take N to ⊥ N be→n · the number of open sets in the cover . Then, Etaut ⊕ Etaut = C and set ∗ ⊥ F = f (Etaut.) → (2) We don’t have any cancellation. That is, it is possible for E ⊕ G =∼ E0 ⊕ G but not E =∼ E0.

Example 12.10. We can take E to be the trivial bundle on S2 of rank 2, E0 to be the tangent bundle 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. If X 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 0 the trivial element. → Remark 12.13. Given an abelian semigroup with 0, (A, +), we can form an abelian group K(A) by adding in formal elements. Example 12.14. We have K(N) = Z. We identify (n, m) ∼ (n + k, m + k) which we think of as n − m. 42 AARON LANDESMAN

Definition 12.15. Given a semigroup A, we define K(A) to be A × A/ ∼ where ∼ is the smallest 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] = [a0] ∈ K(A) even when a 6= a0 ∈ A, if cancellation does not hold (and crucially, we saw above it does not hold for vector bundles). To see this, note [a] = a0 (a, 0) ∼ a0, 0 (a + z, z) = a0 + w, w for some w, z ⇐⇒ a + z = a0 + z for some z. ⇐⇒ Definition 12.17. We write K(X) := K(Vect(X)). We write [E] − [F] ∈ K(X) for vector bundles E and F. We then have⇐⇒ [E] = E0 E ⊕ G = E0 ⊕ G for some G. In this case, we say E and E0 are stably isomorphic. ⇐⇒ Remark 12.18. Recall that for all F there exists G, N with F ⊕ G =∼ CN. Hence, [E] − [F] = [E ⊕ G] − [F ⊕ G] h i = [E ⊕ G] − CN .

Deﬁne N := CN so that 1 = C, and taking E0 = E ⊕ G, we can always write [E] − [F] = E0 − [N] . In particular, we can always assume what we are subtracting is a trivial bundle. Lemma 12.19. We have [E] = E0 there exists F with E ⊕ F =∼ E0 ⊕ F there exists N so that E ⊕ CN = E0 ⊕ CN. ⇐⇒ Proof. This follows from the fact that we can add G we have F ⊕ G = CN. Here, we’re ⇐⇒ using E ⊕ F ⊕ G =∼ E0 ⊕ F ⊕ G and so E0 ⊕ CN =∼ E0 ⊕ CN. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 43

Exercise 12.20. We have that K is a functor sending X 7 K(X). That is, given f : X Y gives f∗ : Vect(Y) Vect(X)→ → f∗K(Y) K(X). → Exercise 12.21. If f =∼ f then f∗ = f∗. 0 1 0 1 → Remark 12.22. The group K(X) is a commutative ring under tensor product. That is, we have 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 h 2i E 7 c1(E) S . → Together, these induce a map → K(X) Z ⊕ Z 2 E 7 (rk(E), c1(E)[S ]) → Fact 12.24. 0 S2 If a line bundle has Chern→ class on it is trivial. Using this fact, we see C 7 (1, 0). We also have b := [H] − 1 7 (0, 1) . → H CP1 Here, is the tautological bundle on . 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 of K(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 is trivial. Let’s see why these are all trivial. To make a vector bundle on Sn, note that this is the same as a vector bundle on the top and bottom agreeing on the equator. Both are trivial as the disks are contractible. The identiﬁcation is a way of identifying the ﬁber of each vector 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 corresponds to homotopy classes of maps h i 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 ψ r complex case, we will have to choose an identiﬁcation E|D1 − C . This is ﬁne in the complex case because these two maps are related by a change of basis, given by a map 1 1 D GL(r, C). Here, D is contractible and GL(r, C) is path connected.→ So, up to homotopy, this map is unique. However, GL(r, R) has two components, which means we have an orientation→ issue. Again, this is absent in the complex situation, as here, if we change φ for φ ◦ ψ we get the same homotopy class of map. In the case of S1, we have h i S0, GL(r, C) = pt

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

Example 13.4. To understand vector bundles on S2 in a different way, we’re looking at maps h i h i S1, GL(r, C) =∼ S1, GL(1, C) h i = S1, C \ {0} = Z where this Z is essentially determined by the ﬁrst Chern class. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 45 n TABLE 1. The reduced K theory of S

n 0 1 2 3 4 K˜ (Sn) Z 0 Z 0 Z

Example 13.5. What is K(S3)? From the analysis for S1 above, we have h i h i 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 how tensor products behave with Chern classes.

13.3. Products. From E1 ∈ Vect(X1), E2 ∈ Vect(X2), we can form ∗ ∗ E1 × E2 = p1(E1) ⊕ p2(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(X × X ) relate to K˜ (X ) × K˜ (X )? 1 2 → 1 2 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 ∗ ∗ ∗ p1 : H˜ (X1) H (X1 × X2, {x1} × X2) Similarly, we have a map ∗ ∗ → ∗ p2 : 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 onto it) 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 a point. 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 compactification 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. n n ` + For example, S = R ∪ . 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 a ` point. ∞ 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 universal property of the one point compactification is that it is final among all compact spaces containing X × Y. So, from the universal property of one point compactifications, we have a continuous map X+ × Y+ (X × Y)+ sending X+ ∪ Y+ 7 ∗. This is a continuous bijection map of compact Hausdorff spaces, hence a homeomorphism. → → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 47

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 ∈ S˜2n. 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 ⊂ X compact. Consider vector bundles E X with a trivialization on A given by n τ : E|A C × 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) ∼ E0, F0, τ0 if there exist isomorphisms E E0 F F0 → on X so that the diagram → E τ F (13.1)

0 E0 τ F0 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 two definitions 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 equivalence 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 have trivializations on A, they are, in particular isomorphic over A,→ and we get a preferred ∗ ∗ isomorphism τ : p (E^)|A p (^F)|A. Next, we give a map in the reverse direction. Start with [E, F, τ]. Choose G with F ⊕ G =∼ n C . This enables us to assume→ our triples is of the form n [E ⊕ G, F, C , 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 Eˇ X/A by identifying the fibers over points in A via the given trivialization. Now, if we have a trivialization τ over A, we claim we can construct an extension of it to a trivialization→ on U for U ⊃ A a slightly larger open set. Once we know this, this guarantees Eˇ 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, this amounts to claiming that a continuous real valued function defined on an open set can be extended. ∨ n n n Now, apply the above lemma to V = E ⊗ C with E C . The map τ : E|AC |A gives a section of V, which can be extended to all of X. This is an isomorphism when restricted + n to A. The lemma allows us to extend this as a map τ : E→ C , on all of X, although this may not be an isomorphism. However, + n x : τ (x) is an isomorphism Ex → Cx → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 49 is open, as it is locally given by the vanishing of a determinant the locus where the determinant vanishes is closed. Example 14.4. We saw last time K˜ (S2) =∼ Z. This was generated by b = [H] − [C] . We have (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 a map with winding number one. 2 1 Alternatively, if we just wanted a generator of K(D , S ) 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) =∼ × × ∼ 1 ∼ C to that corresponding to the generator of π1(C ) = π1(S ) = Z. 14.2. Back to Smash Products. Recall we went through this detour on relative→ K theory to 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 ⊗ F2 F := 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 may → not be an isomorphism outside Ai, but we now have σi : Ei Fi. Then, the map given by ∗ σ1 ⊗ 1 −1 ⊗ σ2 ∗ → 1 ⊗ σ2 σ1 ⊗ 1 ∗ where σi denotes the adjoint of σi. Remark 14.6. Note that the map τ deﬁned above is unitary. Observe more generally that u −v w := v u 50 AARON LANDESMAN

with u, v ∈ C, we have u v u −v w∗w = −v u v u uu + vv 0 = 0 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 ∗ ∗ σ1σ1 ⊗ 1 + 1 ⊗ σ2σ2 0 ∗ ∗ . 0 1 ⊗ σ2σ2 + σ1σ1 ⊗ 1 Note that this is a sum of two hermitian matrices with nonnegative eigenvalue. So, if one of them is positive, their sum is positive.

Now, note that τ : E F from Deﬁnition 14.5 is an isomorphism on A1 × X2 and on X1timesA1. 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 factor. Hint: That is, check the following two parts: (1) the case 0 00 E1 := E1 ⊕ E1 0 00 F1 := F1 ⊕ F1 and 0 σi 0 σi := 00 0 σi 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 ⊗ σ2 1 ⊗ σ2 1 ⊗ 1 This should element the element 0 in the product. To see this is the element 0, we only need check that this map E F is an isomorphism everywhere. This is an isomorphism everywhere because the matrix above is always invertible, and we can in fact take a linear homotopy for the upper right→ and lower left hand corners. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 51

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 always trivial) with a trivialization on the boundary h i 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 n → τ : R 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 2 for u ∈ R = 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 S2 with itself, which will be from the matrix given in Deﬁnition 14.5. More explicitly, we have 4 τ : R MN×N(C).

→ 52 AARON LANDESMAN

Here, we take N = 2. We get u −v∗ τ(u v) = , v u∗ . The Bott periodicity theorem says this is a generator for K˜ (S4). Repeating this again for

σ1 = [u] u −u∗ σ = 2 v v∗ will be the generator of (˜S6) = K˜ (D6, S5).

15. 2/29/16 15.1. Review. Last time, we were discussing h i K˜ (Sn) = K(Dn, Sn−1) = CN, CN, τ ,

N n N n−1 n with C = D × C and τ : S GL(N, C). We have τ : R MN×N(C) with τ(z) invertible for z 6= 0. Bott periodicity implies that the generators of K˜ (S2n) are given as follows. → → 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, 4 ∼ →2 R = C M2×2(C) z1 −z2 (z1, z2) 7 → z2 z1. 6 When n = 3, so we are dealing with S ,→ the map is 6 ∼ 3 R = C M3×3(C)   z1 −z2 −z3 0 → z2 z1 0 z3  (z1, z2, z3) 7   z3 0 z1 z2  0 z3 −z1 z2. 6 → 6 In S , the Bott generator is given by something with c3(E)[S ] = ±2, which, as we said previously, has something to do with the tangent bundle of S6. In general this is given inductively by the matrix ∗ σ1 ⊗ 1 −σ2 ⊗ 1 1 ⊗ 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. → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 53

Proof. Applying Bott’s isomorphism to X+ for X, with X 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 Hc(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 in class. So, we only need define α and verify it is a one sided inverse. To do this, we’ll need to define Fredholm operators.

15.2. Fredholm operators and index. Let H be a Hilbert spaces. That is, a separable space with a countable complete orthonormal system C0, C1, ... . Take i aiei ∈ H with 2 |ai| < . i P PDefinition 15.7. Define L(H1, H2) to be bounded linear operators ∞ A : H1 H2 so that there exists a C with ||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) ker A 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 ker A − 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 ker B , ker AB. So, we have an exact sequence

0 ker B ker AB ker A → (15.1)

cokerB cokerAB cokerA 0. This should follow from a suitable operation of the snake lemma. Using this exact sequence, we deduce AB is Fredholm. We then have dim ker B − dim ker AB + dim ker A − 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 ker B ⊂ ker AB. 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 ker SL 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 also has index 0. This is another special case of Lemma 15.9. Lemma 15.12. Invertible operators are an open set in A. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 55

Proof. First, we reduce to showing it for a neighborhood of 1 ∈ A, and then translate. Consider 1 + B with ||B|| < 1. Here, ||Bh|| ||B||op := sup . h6=0 ||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. P P 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 another lemma 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 manifest the 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 of

ﬁnite codimension such that the composite πH2 ◦ A ◦ iH1 is an isomorphism, where → H A H

i π (16.1) H1 H2

H1 H2.

To see this equivalence, if we had a Fredholm operator A, we take H2 = im A, H1 = ⊥ (ker A) . For the other direction, if we have such an H1, H2 with this property, we obtain that dim ker A ≤ codimH1, dim cokerA ≤ codimH2.

For ﬁner H1, H2 invertibility of πH2 ◦ A ◦ iH1 is an open condition for A, since the invertible operators are an open set of all operators. Then, we have

F = ∪H1,H2 A : 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 closed span 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 stabilize at being equal to H. It could not stabilize at anything smaller than H, because as n0 approaches , 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)) = Hn for 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 of this set over all n is an open cover. Remark 16.4. Recall that if A ∈ F we have index A = dim ker A − dim cokerA ∈ Z. Now, for A ∈ F, we can think of [ker A] − [cokerA] ∈ K(∗) =∼ Z. So, the index of a Fredholm operator can be thought of as a formal difference of spaces in the 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 [ker A(x)] − [cokerA(x)] ∈ K(X). → We’d like to call this index A ∈ K(X). However, the problem is that the set of ker A(x) do not form a vector bundle. The dimension 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. The kernel 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, then Sx := ker A(x), as x ∈ X forms a vector bundle over X.

Proof. By the Fredholm operator property, this dimension is ﬁnite. Pick x0 ∈ X, and work near x0. Let S0 = ker A(x0). Consider

B(x): H H ⊕ S0

A(x) 7 (A(x), πS0 ) → → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 57

Note that B(x0) is invertible. The property of being invertible is open, so B(x) is invertible for x near x0. Further, B(x) is invertible and −1 B(x) ({0} ⊕ S0) = ker A(x) = Sx.

Thus, for x ∈ U, there is some open neighborhood x0 ∈ U with a trivialization ∪x∈USx = −1 U × S0 using B(x) . 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 → ker pn ◦ A(x) forms a vector bundle. Further, ∼ n cokerpn ◦ A(x) = H/Hn = C for all n. Now, we define the index of A to be n index A = [ker(pn ◦ A)] − [C ] . 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 we replaced n by n + 1. Note that the index is still defined for n + 1 if it is defined for n. We have 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 hpn(A)h, eni. This implies that as vector bundles, we have→ → ker pn+1A = ker pnA ⊕ C. Hence, replacing n by n + 1 yields n [ker(pn ◦ A) ⊕ C] − [C ⊕ C] ∈ K(X) which is the same element as n [ker(pn ◦ A)] − [C ] . 58 AARON LANDESMAN

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

index A1 = index A2 → 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 → index A ∈ K ([1, 2] × X) . Restricting to n = 1, 2 we get the two indices. Remark 16.11. In fact, the map index : [X, F] K(X) is bijective, although this will take some work, and we won’t show this now. To prove this, one would need Kuiper’s theorem,→ which says that the group of invertible linear maps H H preserving the inner product, notated U(H), is contractible when H 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 in operator 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 + K0, with K, K0 are compact operators. Proof. This follows from the proposition→ above together with one of the lemmas from the end 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. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 59

Lemma 17.2. If K ∈ K and {hi}i≥1 is a bounded sequence in H, then the sequence {Khi}i≥1 has 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 kKn − Kkop 0. For every n ≥ 1, the sequence ∼ N(n) {Knhi}i≥1 ⊂ im (Kn) = R → has a convergent subsequence, since it lives in a bounded subset of RN(n). Diagonalizing shows that there exists a index subset I ⊂ N so that {Knhi}i∈I converges for all n. Do a “3-argument” to see that {Khi}i≥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 suffices to show that every bounded sequence in ker(1 + K) is convergent. Now {hi}i≥1 ⊂ ker(1 + K) bounded implies that {Khi}i≥1 has a convergent subsequence by the Lemma. But {Khi}i≥1 is {−hi}i≥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. Since ker(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 {hi}i≥1 is a bounded sequence, then passing to a subsequence I ⊂ N, we know that {Khi}i∈I is convergent, so the ì are convergent, implying that {hi} are convergent. So hi h in H, and

(1 + K)h = lim(1 + K)hi → = lim `i = `, so ` ∈ im (1 + K). (b) If {hi}i≥1 is not a bounded subsequence, then, passing to a subsequence, we may assume that khik . Consider these vectors of unit length: h ¯ i → ∞ hi = . khik

Applying case (a) to h¯ i in place of hi, we ﬁnd that {h¯ i} admits a convergent subsequence h¯ i h¯ . But kh¯ k = 1, so h¯ 6= 0. A contradiction results from the fact that

→ (1 + K)h¯ = lim(1 + K)h¯ i `i = lim khik = 0.

Indeed, we chose hi so that hi ⊥ ker(1 + K), and the same is true for h¯ i, so their limit h¯ is also orthogonal to ker(1 + K). This contradicts kh¯ k 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 we have dim(im (1 + K))⊥ = dim ker(1 + K∗), and the latter is finite dimensional. In fact, K is compact, so K∗ is compact, and we ∗ apply 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 unit circle (although any two separable Hilbert spaces are isomorphic as Hilbert spaces), and the inner product is defined as 1 2π hh, gi = h(eiθ) g(eiθ) dθ. 2π Z0 n With this convention, a complete orthonormal system is given by en = z , where n ∈ Z. We have a decomposition

H = H+ ⊕ H−

H+ = closed span of {en}n≥0

H− = closed span of {en}n<0. 1 In other words, H+ consists of those functions on S which extend as analytic functions on the (open?) unit disc. If f : S1 C is continuous (or L ), then the following operator is in A(H): ∞ → Mf : h 7 fh. This is because kMfhk ≤ sup(f) khk. In fact, if f is continuous,then kMfk = sup(f), where kMfk 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, −1 −1 and hence index(Mf) = 0. We have Mf = Mg where g(z) = z , 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-shift operator with respect to the basis {en}n≥0. In the same vein,→

en−1 if n ≥ 1 Ag(en) = 0 otherwise.

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

The operators Af and Ag exemplify the following result: ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 61

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 other words, MfMg = 1. We claim that →

AgAf = 1 + K where K is compact, and similarly 0 AgAf = 1 + K 0 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, the operator 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 because

P− ◦ Mf ◦ ιH+ : H+ H− is compact if f is continuous, and B ◦ K is compact whenever B is bounded and K is compact. To see that P− ◦ Mf ◦ ιH+ is compact, use the→ following argument: −r (1) If f(z) = z for r ∈ N, then P− ◦ Mf ◦ ιH+ is finite rank because it sends i=0 aiei r to i=0 aiei−r. ∞ −1 P (2) If f is a polynomial in z, z , i.e. a finite Laurent series, then P− ◦ Mf ◦ ιH+ is a finite linearP combination of finite rank operators, hence finite rank. (3) If f is any continuous function, we use approximation by polynomials: there exists a sequence {fk}k≥1 of functions realized as finite Laurent series such that fk f 1 uniformly on S . Then Mfk 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→+ is compact. → → 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

2 1 and try to construct an analogous Fredholm operator Af. Let H = L (S ), and consider N > an element h ∈ H(N) = H ⊗ C , which is a column vector (h1, ... , hN) of functions in 2 1 1 L (S ). Viewing f as a matrix of functions in C(S ), we deﬁne Mfh to be the product of N this matrix with the vector h. Let H+(N) = H+ ⊗ C , and deﬁne P+ : H(N) H+(N) as before. Let

Af = P+ ◦ Mf ◦ ιH+(N). → As before, we have Afg = AfAg + K where K is a compact operator. 1 Now let E be any N-dimensional vector space. Given f : S GL(E), we get Af on H+ ⊗ E. More generally, given vector spaces E and F, we let GL(E, F) be the space of 1 invertible linear maps E F. Given f : S GL(E, F), we get→ a Fredholm operator Af : H+ ⊗ E H+ ⊗ F. 2 Consider an element of→K(X × R ), where X→is a compact topological space. By our discussion of→ relative K-theory, and our deﬁnition of K-theory of non compact spaces, we have 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 bundles 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, we can 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 the first 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 → 1 fx : S GL(Ex, Fx).

By the above recipe, each fx gives us a Fredholm operator Afx . Hence we have a family of → Fredholm operators Afx parameterized by x ∈ X. Given such a family A : X F, we have seen how to obtain an element of K(X) given by index(A), i.e. n → [ker(Pn ◦ A)] − [C ] for n sufficiently large. There is one subtlety: this definition takes F to be the space of Fredholm operators on a fixed Hilbert space H, but the above paragraph gives a family of Fredholm 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 by using the map Pn : H+ ⊗ Ex H+ ⊗ Ex in place of the previous Pn. → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 63

Definition 18.1. The map α : 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 homomorphisms, we can assume that a = [E]. Here E X is a vector bundle, β(E) = a · b where 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, 1 fx :S 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 of

Afx 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 Ke(X) in place of K(X), where the map 2 β : Ke(X) Ke(X ∧ S ) is given by multiplication by b ∈ K(R2) = Ke(S2). To spell things out, the codomain of β is → 2 2 1 2 Ke(X ∧ S ) = K(X × D , X × S ∪ {x0} × D ) where x0 ∈ X is the basepoint. Temporarily ignoring the fact that the pair of vector 2 bundles are isomorphic on {x0} × D , we still have a map 2 2 1 2 α : Ke(X ∧ S ) = K(X × D , X × S ∪ {x0} × D ) K(X), and we still have a family of maps 1 → fx : S GL(Ex, Fx)

→ 64 AARON LANDESMAN for all x ∈ X. In this case, for x = x0, this map extends: 2 fx0 : D GL(Ex0 , Fx0 ), 2 i.e. fx0 has a given null-homotopy. The isomorphism on {x0} × D gives an identification → Ex0 = Fx0 for which fx0 = 1 after the given homotopy. This means that ker(A ) = coker(A ) = 0. fx0 fx0 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) := Ke(X ), where X+ = X ∪ { } is the one-point compactification. More precisely, for locally compact X, 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) 2 i23(c · v) ∈ K(X × Y × R ) 2 2 where i23 : X × R × Y X × Y × R . Now we can apply αX×Y:

αX×Y(i23(c · v)) ∈ K(X × Y). →

Lemma 18.3. We have αX×Y(i23(c · v)) = αX(c) · v Proof. This is because they are the same family of vector spaces in x ∈ X. As before, we may assume that c = [π∗E, π∗F, f] as before, where π : X × R2 X, and v = [V] for some vector bundle V Y. Then n αX(c) = [ker(Pn ◦ Afx )] − [C ], → → ∗ ∗ 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 n αX(c) · v = [ker(Pn ◦ Afx ) ⊗ Vy] − [C ⊗ Vy] where (x, y) varies over X × Y. Furthermore, αX×Y(i23(c · v)) equals exactly the same m thing. To see this, write Vy = C for ﬁxed y ∈ Y, so that fx ⊗ IdVy maps (π∗E)⊕m = π∗E ⊗ Cm π∗F ⊗ Cm = (π∗F)⊕m,

→ ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 65

f z ∈ S1 M acting by x on each ﬁber (depending on ). Therefore fx⊗IdVy is a block-diagonal 2 1 matrix of functions in L (S ), whose blocks are m copies of Mfx . The projection that makes A P fx⊗IdVy can be taken to be the previous n on each summand. Combining this with the previous observation, we conclude that it applies also to the case where X and Y are only locally compact. We wish to apply this to the case Y = R2. In this case, 2 2 2 2 i23 : X × R × R X × R × R still interchanges the last two factors, but it gives the identity on K(X × R2 × R2), because → 2 2 + 4 i23 is homotopic to IdX×R2×R2 . Explicitly, on each ﬁber (R × R ) = S , we are doing a higher-dimensional rotation. Now for c ∈ K(X × R2), we have

βXαX(c) = αX(c) · b by 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 this equals

αX×R2 (c · b) = αX×R2 βX×R2 (c) = c (by the ﬁrst part). “I can follow the formal steps, but I don’t claim to understand exactly where you are swindled.”

19. 3/9/16 β 19.1. Review of inﬁnite dimensional groups. Recall we have K˜ (X) − K˜ (X ∧ S2) is an isomorphism, so K˜ (Sn) =∼ K˜ (Sn+2). We have h i [E] − CN ∈ K˜ (Sn) →

with N = rk E and h i h i [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 h n−1 → i S , GL(N, C) . We have a function φ : Sn−1 GL(N, C) with φ ∼ φ ⊕ 1 with → φ 0 φ ⊕ 1 = 0 1 66 AARON LANDESMAN

TABLE 2. Groups of Spheres

n n K˜ (S ) πn−1(U) 0 Z 1 0 0 2 Z Z 3 0 0 4 Z Z

an N + 1 × N + 1 matrix and φ ⊕ 1 : Sn−1 GL(N + 1, C). We have →h i 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 h i K˜ (Sn) = Sn−1, U . Recall h n−1 i ∼ S , GL = πn−1(GL). So, here is a table The 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)). 2n−1 2N−1 This is an isomorphism if πn−2(S ) = 0, πn−1(S ) = 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 a conjugate linear map J : W W with j2 = 1. → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 67

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 = h1, I, J, Ki Lemma 19.4. Left H vector spaces (quaternion vector spaces) are in bijection with complex vector spaces W 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) 2 we can form W = W1 ⊗C W2, J = J1 ⊗ J2 so that J = ±1. So, if

(W1, J1) , (W2, J2) are quaternion vector spaces then (W, J is a real vector space. Similarly, if we took two real vector spaces, we would get a real vector space. If we took one quaternion vector space 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 product 4 H ⊗C H, we get (C , 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 semigroup 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 KOg(X) similarly to the complex case. Definition 19.7. We can define the Symplectic K theory KSp(X) to be the semigroup of Quaternionic vector bundles, with representatives [E] − [F]. We have n KOg(S ) as equivalence classes represented by h i [E] − RN ,

we have n KOg(S ) = { real vector bundles E / ∼} h i = 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 n KOg(S ) πn−1(O) 0 Z 1 Z/2Z Z/2Z 2 Z/2Z Z/2Z 3 0 0

TABLE 4. Symplectic K groups of spheres

n n KSpg (S ) πn−1(Sp) 0 Z 1 0 0 2 0 0 3 0 0 4 Z Z

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

(19.2) O(N − 1) O(N) SN−1.

∼ 3 ∼ This gives an exact sequence on homotopy groups. We have π1(SO(3)) = π1(RP ) = Z. 19.3. Symplectic K theory. Deﬁnition 19.8. Let’s write GL(N, H) for H linear invertible maps HN HN. Inside this, 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) 3 for k < 7. So, πn−1(Sp) = πn−1(S ) for n < 7. 3 3 3 Remark 19.9. We have π3(S ) 3 [f], with f := id : S S . We have f(q) = q with 3 4 2 S ⊂ H. This gives an element b4 ∈ KSpg (S . This element is a “close friend” of b ∈ K˜ (S ). Remark 19.10. We have maps → KOg(X) × KOg(Y) KOg(X ∧ Y). KOg(X) × KSpg (Y) KSpg (X ∧ Y).KSpg (X) × KSpg (Y) KOg(X ∧ Y). → → → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 69

TABLE 5. Symplectic K groups of spheres

n n KSpg (S ) πn−1(Sp) 0 Z 1 0 0 2 0 0 3 0 0 4 Z Z 5 Z/2Z Z/2Z 6 Z/2Z Z/2Z

TABLE 6. Real K groups of spheres

n n KOg(S ) πn−1(O) 0 Z 1 Z/2Z Z/2Z 2 Z/2Z Z/2Z 3 0 0 4 Z Z 5 0 0 6 0 0 7 0 0 8 Z Z

We also have β∗, which is multiplication by b∗ giving maps 4 KOg(X) KSpg (X ∧ S ), 4 KSpg (X) KOg(X ∧ S ). → Hence, we have a map → 8 KOg(X) KOg(X ∧ S ). Corollary 19.11. We have n→+4 n KOg(S ) = KSpg (S ) n+4 n KSpg (S ) = KOg(S ). Proof. Omitted. Example 19.12. We can now expand the table of KOg, KSpg . 20. 3/11/2016 Let’s relate K-theory with characteristic classes. Recall that, to each complex vector bundle E X, we have associated Chern classes ci(E). We can furthermore deﬁne ci(a) for a ∈ K(X), as follows. Consider the map → M × Vect(X) H2i(X)

→ 70 AARON LANDESMAN where M × H2i(X) = units 2i = {1 + t1 + t2 + ··· + tN, ti ∈ H (X)}. L 2i This is because, for any t ∈ i>0 H , 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 because an inﬁnite sum of terms is not a valid element of a direct sum. In such a case, we should instead consider × Vect(X) H2i(X)

This is the map c : E 7 c(E), which is a semigroupQ homomorphism by the Whitney sum formula → c(E ⊕ F) = c(E) ^ c(F). The universal property of groupiﬁcation gives us a corresponding map

× c :(K(X), +) H2i(X) Q where c([E] − [F]) = c(E)c(F)−1.

For example, if H CP is the dual of the tautological bundle, so that c1 = h is the 2 standard generator of H (∞CP ), then our deﬁnition says → c∞(−[H]) = 1 − h + h2 − h3 + ··· .

20.1. Chern character. Given variables x1, ... , xn, we have the elementary symmetric polynomials, 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 elementary symmetric polynomials σk. For example, we have

xi = σ1 2 2 Xxi = σ1 − 2σ2 3 3 X xi = σ1 − 3σ1σ + 3σ3 k Xxi = sk(σ1, ... , σn). X ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 71

2i Deﬁnition 20.2. The k-th Chern character chk(E) ∈ H (X; Q) of a rank-n complex vector bundle E X is 1 ch (E) := s (c (E), ... , c (E)). k k! k 1 n → 0 By convention, let ch0(E) = rank(E) ∈ H (X), where the rank is a locally constant function on X. Further, deﬁne 2i ch(E) = chi(E) ∈ H (X; Q). ∞ i=0 i≥0 X Y For example, the above formulae show that

ch1(E) = c1(E) 1 ch (E) = (c2 − 2c ) 2 2 1 2 1 ch (E) = (c3 − 3c c + 3c ). 3 6 1 1 2 3 This deﬁnition is motivated by the splitting principle: if

E = L1 ⊕ · · · ⊕ Ln, then ci(E) = σi(c1(L1), ... , c1(Ln)), so that n k sk(c1(E), ... , cn(E)) = c1(Li) i=1 X by the deﬁnition of sk above. Then n k c1(Li) ch (E) = ∈ H2k(X; Q) k k! i=1 Xn ch(E) = ec1(Li). i=1 X Here, for x ∈ H2(X; Q), we have deﬁned 1 ex = 1 + x + x2 + · · · ∈ H2i(X; Q). 2 i≥0 Y For vector bundles E and F, we use the splitting principle to write

E = L1 ⊕ · · · ⊕ Ln F = M1 ⊕ · · · ⊕ Mm. Then ch(E ⊕ F) = ch(E) + ch(F), since both sides equal n m ec1(Li) + ec1(Mj). i=1 j=1 X X 72 AARON LANDESMAN

This would have happened for any power series deﬁnition of ch. What is special about our particular deﬁnition is that the Chern class of the tensor product takes on a simple form. In fact, let xi = c1(Li) and yj = c1(Mj) for brevity, and note that M E ⊗ F = Li ⊗ Mj i,j ch(E ⊗ F) = ec1(Li⊗Mj) i j X X = exi+yj i j X X !   x y = e i  e j  i j X X = ch(E) ch(F).

The upshot of this is

Theorem 20.3. The map ch : K(X) H2i(X; Q) i≥0 Y is a ring homomorphism. →

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

n n Proof. The n-th roots of unity {ζi}i=1, i.e. the solutions to x − 1 = 0, are such that σi = 0 n−1 for all 1 ≤ i ≤ n − 1, and σn = (−1) . On the other hand,

n n sn = ζi i=1 Xn = 1 i=1 X = n, so the previous reasoning implies that

n−1 sn(0, ... , 0, (−1) ) = n.

Corollary 20.5. If E X has c1 = ··· = cn−1 = 0, then 1 → ch (E) = (−1)n−1n · c (E) n n! n (−1)n−1 = c (E). (n − 1)! n ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 73

20.2. Let N 2n [E] − [C ] ∈ Ke(S ) =∼ Z 2n be the generator. What is cN(E)[S ]? On the problem sets, we found bundles E such that

2 c1(E)[S ] = 1 4 c2(E)[S ] = 1 6 c3(E)[S ] = 2.

2 2 2 (Here c1(E)[S ] means c1(E) evaluated on the fundamental class [S ] of S .) Bott’s theorem n 2 tells us that the generator is b where b ∈ Ke(S ). We had b = [H] − [C] where c1(b) = 2 2 c1(H) = h was a generator of H (S ; Z). Therefore

ch(b) = 0 + h = h,

since rank(b) = 0. Now

b × · · · × b ∈ K(S2 × · · · × S2, T),

where n T = [ S2 × · · · × {∗} × · · · × S2 j=1

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

b × · · · × b ∈ K(S2 × · · · × S2, T) 2n =∼ Ke(S ),

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

π : S2 × · · · × S2 S2n

is the quotient map used to construct the smash product.→ We see that ∗ n 2n ∗ n 2 2 π cn(b )[S ] = cn(π (b ))[S × · · · × S ] 2 2 = cn(b × · · · × b)[S × · · · × S ] n = (n − 1)!(−1) 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- 2n mediate chi(•)’s are zero because cohomology of S vanishes in those dimensions. Now this equals n (n − 1)!(−1) (h1 × · · · × hn) 74 AARON LANDESMAN

2 2 2 2 where h1 ∈ H (S × · · · × S ) is the pullback of h from the i-th copy of S . Now multiplicativity of ch gives

n ch(bn)[S2n] = h[S2] i=1 Yn = 1 i=1 Y = 1.

Therefore

Theorem 20.6.

n 2n n cn(b )[S ] = (−1) (n − 1)!

2n 2n Corollary 20.7. For E S , chn(E)[S ] 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 with J2 = 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 tensors by C, the lower map takes a quaternion vector space and only remembers the complex multiplication. For instance, we have a commutative diagram where all maps are isomorphisms:

KOg(S0) Ke(S0)

rk rk Z id Z

Similarly, we have a commutative diagram

KSpg (S0) Ke(S0)

rkH rkC ×2 Z Z ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 75

4 4 From before, b4 ∈ KSpg (S ) has c2(b4) = 1, so – as C-vector bundles – or in Ke(S ), we have 2 2 4 4 b4 = b , i.e. b4 = b , since complex bundles on S are classiﬁed by their Chern class.

1 Z = KOg(S0) Ke(S0) = Z

2 Z = KOg(S4) Ke(S4) = Z

1 Z = KOg(S8) Ke(S8) = Z

2 Z = KOg(S12) Ke(S12) = Z

The second row follows from the preceding commutative diagram, since multiplication 4 0 by b4 relates KOg(S ) and KSpg (S ). 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| ≤ 1} S(V) := {v : |v| = 1}. The key property of ν is that, for each x ∈ X, r ν|(D(Vx),S(Vx)) is ‘the’ generator of H (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, then K(D(Vx), S(Vx)) = Z, from our computation of Ke(Sn). We ask: does there exist ν ∈ K(D(V), S(V)) such that, for each 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- 2n lying the Bott generator of K(D(Vx), S(Vx)) = K(R ), which came from taking tensor powers 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 inner product h−, −iV . 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) = −2hu, viV 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 the general Clifford relation, which→ follows by expanding general u and v with respect to an orthonormal basis. Deﬁnition 21.1. A Z/2-graded vector space (over C) is a vector space with an operator : S S such that 2 = 1. In this case, we have the decomposition S = S+ ⊕ S− into the ±1 eigenspaces of . This structure is equivalent to a decomposition of S into two vector spaces,→ but it will be convenient to encode this structure into an operator . If S1 and S2 are Z/2-graded, with grading operators 1 and 2, then S = S1 ⊗ S2 is Z/2-graded, with operator = 1 ⊗ 2. Equivalently, we have + + + − − S = S1 ⊗ S2 ⊕ S1 ⊗ S2 − + − − + S = S1 ⊗ S2 ⊕ S1 ⊗ S2 . When S has an inner product, we shall require that be unitary. Deﬁnition 21.2. A (complex) Clifford module for V (a 2k-dimensional real inner product space) is a Z/2-graded complex inner product space (S, ) 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, ◦ γ(v) = −γ(v) ◦ . (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) = −2hu, viV IdS holds for all u, v ∈ V. Remark 21.3. The requirement that dim V be even is imposed because the correct deﬁnition of ‘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 0 −σ(v)∗ γ(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.

→ ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 77

Proposition 21.4. Let S1 and S2 be Clifford modules for V1 and V2, respectively. Then S := S1 ⊗ S2 is a Clifford module for V1 ⊕ V2, where

γ : V1 ⊕ V2 End(S) is deﬁned by → γ(v1, v2) := γ1(v1) ⊗ IdS2 + 1 ⊗ γ2(v2). Proof. We check the Clifford relation: γ(v1, v2)γ(u1, u2) + γ(u1, u2)γ(v1, v2) = γ1(v1) ⊗ IdS2 + 1 ⊗ γ2(v2) γ1(u1) ⊗ IdS2 + 1 ⊗ γ2(u2) + γ1(u1) ⊗ IdS2 + 1 ⊗ γ2(u2) γ1(v1) ⊗ IdS2 + 1 ⊗ γ2(v2) = γ1(v1)γ1(u1) + γ1(u1)γ1(v1) ⊗ IdS2 2 + 1 ⊗ γ2(v2)γ2(u2) + γ2(u2)γ2(v2) + (terms that cancel because 1 ◦ γ1(•) = −γ1(•) ◦ 1) = −2 IdS1 ⊗hv1, u1i + hv2, u2i ⊗ 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 + 1 ⊗ γ2 ⊗ 1 + 1 ⊗ 2 ⊗ γ3 no 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 module is S = S+ ⊕ S− = C ⊕ C with 1 0 = 0 −1 and x 0 −x − iy γ := y x − iy 0 , which we can abbreviate as 0 −z γ (z) = 1 z 0 2k 2 2 For R = R ⊕ · · · ⊕ R , the standard Clifford module is Sk := S1 ⊗ · · · ⊗ S1 with k terms. ∼ 2k (*) Any Clifford module S for V = R 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 by 2 mapping v to γ(v). This map sends S1 to the Bott generator of K(R ), and similarly Sk is 2k sent to the Bott generator of K(R ). → Passing from vector spaces to vector bundles, let X be a compact topological space, and let V X be a real vector bundle with inner product, so that the fibers Vx are isomorphic to R2k. → 78 AARON LANDESMAN

Definition 21.6. A complex Clifford module S X is a Z/2-graded complex vector bundle with 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 what we have done above, we can construct such a ν if we have: • A Clifford module S X for V X, such that ∼ 2k • for each x ∈ X, Sx is the standard Clifford module for Vx = R . Given such an S, applying our→ previous→ construction (*) fiber by fiber gives the desired ν ∈ K(V).

22. 3/23/2016 From last time, we reduced the problem of finding a class ν to finding a bundle of Clifford modules, each fiber of which is the standard Clifford module. This time, we classify 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) is complex-linear, such that γ(v)∗ = −γ(v), and the Clifford relation holds. Proposition 22.1. If S is a Clifford module, and S0 ⊂ S a Clifford submodule, meaning→ that S0 is preserved by all γ(v), then (S0)⊥ 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 ‘irreducible’ means there is no nontrivial Clifford submodule. Lemma 22.3. (Schur) If S is irreducible and φ : S S is a Clifford-homomorphism, meaning 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 Clifford submodule of S, hence 0, so φ equals its eigenvalue. Proposition 22.4. The standard Clifford module (of dimension 2k) for V = R2k is irreducible. Moreover, every irreducible Clifford module has dimension 2k, and any Clifford module of dimension 2k is irreducible. Any such Clifford module (irreducible of dimension 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 is irreducible by induction on k, the base case k = 0 being trivial. Consider η = γ(e1)γ(e2) ∈ End(S). ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 79

Because γ(e1) and γ(e2) anti-commute by the Clifford relations, 2 2 2 η = −γ(e1) γ(e2) = −1, so S = S0 ⊕ S00, where S0 = ker(η − i) and S00 = ker(η + i) are the eigenspaces of η. The purpose of having two γ(ei) terms in the definition of η is that now η commutes with 0 00 γ(ei) for all i > 2. Therefore S and S are Clifford modules for the smaller vector space 2k−2 R := span(e3, ... , e2k). 0 00 0 Furthermore, we have dim S = dim S because γ(e1) restricts to isomorphisms from S to S00 and vice versa, because it anti-commutes with η. Since we assumed that dim S = 2k, it follows that dim S0 = dim S00 = 2k−1. By the inductive hypothesis, they are irreducible Clifford modules for R2k−2. Now, considering S again as a Clifford module for R2k = V, the previous paragraph shows that: either S is irreducible, or S decomposes into irreducibles S0 ⊕ S00. But S0 is not γ(e1)-invariant, so the former possibility holds. To see that S is isomorphic to the standard Clifford module of R2k = V, we start with 0 00 the decomposition S = S ⊕ S , with γ(e1) acting as the matrix 0 −1 1 0 , and start to (recursively) pick out the tensor product structure that defines the standard Clifford module. 22.1. How to obtain the Z/2-grading. Pick an orthonormal basis V = R2k, and consider the operator k := (−i) γ(e1)γ(e2) ··· γ(e2k) n/2 = (−i) γ(e1)γ(e2) ··· γ(en). Note that γ(ej) = −γ(ej) for all j = 1, ... , 2k. Also, 2 n/2 (n) 2 2 = (−1) (−1) 2 γ(e1) ··· γ(en) = 1. But an with these properties was our definition of a Z/2-grading. Is this the only way to define a grading? Suppose 1 and 2 are two potentially different Z/2-gradings on an irreducible Clifford module S. Schur’s lemma then implies that 1 = ±2. This sign comes as no surprise: changing the order of the ei in our definition of above, we can in fact change by a sign. In other words, the we chose depends on the orientation class of the basis {ei}, and our uniqueness result shows that it is well-defined given a choice of orientation class. To summarize: Proposition 22.5. For V = R2k, there exists a unique irreducible Clifford module S with dim S = 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 topological space, and V X a real vector bundle of rank 2k. We seek a complex vector bundle S X, which is made into a Clifford module bundle by a ﬁber-wise action γ : 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, then c G 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 0 0 c 0 (S , γ ) are two spin structures for V→|RaX. For all x ∈ X, both Sx and Sx are isomorphic to the standard Clifford module on Vx, so 0 Lx := HomCliff(Vx)(Sx, Sx) is a one-dimensional complex vector space, but not canonically isomorphic to C since 0 Sx and Sx 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 ∼ 0 0 (S, γ) = (S , γ ). In general, → Definition 22.7. Call L X the difference line bundle of (S, γ) and (S0, γ0). We have this transitivity result: → Proposition 22.8. Consider three Clifford modules (S1, γ1), (S2, γ2), (S3, γ3), and let L12, L23, L13 ∼ be the corresponding difference line bundles. Then canonically L13 = L12 ⊗ L23.

Proof. The map is given by composing the homomorphisms S1 S2 and S2 S3 given by 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 (S0, γ0) 7 L, which is the difference from (S, γ). This map is also surjective: given L, we obtain → → S00 = S ⊗ L 00 γ (v) = γ(v) ⊗ IdL . Since (S00, γ00) and (S0, γ0) have the same difference L from S, the previous Proposition shows that they are isomorphic. This does not quite address existence. To ﬁx this, suppose that X is a CW-complex, and try and build up S X for V X. Suppose we have already deﬁned a spin-c structure m S(m) X = (m-skeleton of X), → → and consider an attaching map → φ : ∂Dm+1 = Sm Xm,

→ ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 81

which also gives a map ψ : Dm+1 Xm+1 which is injective on em+1 := int(Dm+1). Even m ∗ though ψ may not be an inclusion on S , we can consider φ (S(m)), which is a spin-c ∗ m+1 structure for ψ (V)|Sn . The problem→ of extending the spin-c structure onto the cell D ∗ m+1 becomes: Can I extend φ (S(m)) on the boundary over all of D , as a spin-c structure for ψ∗(V)? Since Dm+1 is contractible, it has a unique spin-c structure, corresponding to the standard Clifford module for one fiber. The question is: does this trivial spin-c structure, m ∗ restricted to S , coincide with the spin-c structure φ (S(m)) obtained by pullback? If yes, then we can extend over Dm+1. Proposition 22.10. If m ≥ 3, then any complex line bundle L Sm is trivial. m m−1 Proof. This is because complex line bundles on S are classified by [S , U(1)]. → 3 Corollary 22.11. If X 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 trivial on 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. 2 Corollary 22.14. If w1(V) = w2(V) = 0, then V X is trivial on X , so a spin-c structure exists. → → 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 2k with 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 = S with 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- 2 1 2 2 tion. So, Schur’s lemma implies J = λ · 1S for λ ∈ S . Also, J ◦ J = J ◦ J so λ = λ, and so λ = ±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 x 0 −z γ = 1 y z 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 2 J1(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 define Jk in terms of J1 and Jk−1. Definition 23.8. Use

Sk = Sk−1 ⊗ S1 so that

Jk−1 ⊗ J1 if Jk−1 is even. Jk = Jk−1 ⊗ ε1J1 if Jk−1 is odd. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 83

k − 1 even and real even and Quaternionic odd and real odd and Quaternionic k odd and Quaternionic odd and real even and real even and Quaternionic

TABLE 7. summary of oddness, evenness, realness, and Quaternionicness via induction

Lemma 23.9. Jk satisﬁes Jk ◦ γk = γk ◦ Jk. Proof. We want to check

Jk ◦ γk(vk−1, v1) = γk(vk−1, v1) ◦ Jk 2k−1 2 2k where (vk−1, v1) ∈ R ⊕ R = R . There are two cases to check, depending on whether 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, 2 2 2 2 we get Jk = (Jk−1 ⊗ J1) = Jk−1 ⊗ J1. Similarly, if Jk−1 is odd, we have 2 2 2 Jk = Jk−1 ⊗ (ε1J1) 2 = −Jk−1 ⊗ (ε1ε1J1J1) 2 = Jk−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 k εk = (−i) γ(e1)γ(e2) ··· γ(e2k) k and so εkJk = (−1) Jkεk. This tells us oddness and evenness. So, in a table, We have

(Sk, γk, Jk) 2k 2k the standard real Clifford module for R . The εk form an orientation for R . For V X an oriented real rank 2k vector bundle, a spin structure is a vector bundle ∼ S X with 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 odd 1 Quaternionic odd 2 Quaternionic even 3 real odd 4 real even 5 Quaternionic odd

TABLE 8. Table of realness, Quaternionicness, oddness, and evenness of Clifford operators, this is four periodic.

Remark 23.12. If V = V 0 ⊕ R2, a spin structure for V 0 gives one for V and visa versa. 0 Given (S, γ) for V, set S to be the i or −i eigenspace of γ(e1)γ(e2). Similarly, for V = V 0 ⊕ V 00 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 being one for V ⊕ R. 4 + − 3 4 Example 23.14. For V = R , take S2 = S2 ⊕ S2 . For v ∈ S ⊂ R , we have an isomorphism γ(v): S+ =∼ S−, + − 4 3 + − and we get (S , S , γ) ∈ KSp(D , S ). Here, S2 , S2 are quaternion vector spaces and γ is a quaternion linear map between them. In fact, this is the Bott generator 4 b4 ∈ KSp(R ). 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 rk V = 8m or rk V = 8m + 4, then there exists a class → KO(D(V), S(V)) if rk V = 8m bV ∈ KSp(D(V), S(V)) if rk V = 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), where S = S+ ⊕ S−, 0 −σ∗ → γ = σ 0 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 85

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. If 2k ∈ 8Z, the same statement holds for the map ψ : KO(X) KO(V). If 2k ∈ 8Z + 4, the same 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 actually prove that this map is an isomorphism, but→ our proof of Bott periodicity contains all the ingredients. The latter concerns the trivial vector bundle X × R2k, while the former concerns a nontrivial vector bundle p : V X. Just as before, the definition of ψ−1 begins by writing down Fredholm operators. Just as in Bott periodicity, we can extend this result to locally compact X, for suitable V. → If f : X Y is a proper map of locally compact topological spaces, we obtain a pullback map f∗ : K(Y) K(X), as follows. We consider the induced map f+ : X+ Y+ of + ∗ + ∗ the compactifications,→ and recall that K(X) := Ke(X ), so f comes from (f ) on reduced + + K-theory of X and→ Y . → One important special case of this is the pushforward map: If X is locally compact, and ι : U X is an inclusion of an open subset, then there is a map ι+ : X+ U+, which is a bijection on U, and collapses all of (X \ U) ∪ { } to { } ⊂ U+. In this way, we get a map ι∗ : 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 k−r + k + r He (X ) = He (X ∧ S ). Motivated by this, we defined, for r ≥ 0, K−r(X) := K(X × Rr) −r + + r Ke (X ) := Ke(X ∧ S ). Note that these higher K-groups are periodic in r: by Bott periodicity, we have an isomorphism 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 definition for negative r. But→ we can use Bott periodicity to extend the definition, i.e. for n ≥ 0, pick k with 8k − n ≥ 0, and define KOn(X) := KO(X × R8k−n). Because the periodicity isomorphisms are not just any isomorphism, but the preferred Bott isomorphism, everything is canonical so far. Now, given a proper map f : X Y between locally compact spaces, we obtain a pullback map f∗ : KOn(Y) KOn(X), coming from the map X × Rr Y × Rr which is the identity on Rr. Similarly, given ι : U X with U an open subset of X, we obtain a r r → pushforward map ι∗ : K (U→) K (X). → → → 86 AARON LANDESMAN

Similarly, if V X is spin, and rank V = 8k, the ‘Thom isomorphism’ of the preceding theorem 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 rank V→= 8k restriction. If rank V = r, and V X has a spin structure, we can consider V ⊕ R8k−r, whose total space is just V × R8k−r. Since this 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 = rank V. These ‘dimensions’ n and n + r agree with those of the original Thom map ψ : Hm(X) H→m+r(D(V), S(V)), r = rank V, 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 the normal bundle of X in Y, so that s = rank V is the codimension of X in Y. Further, suppose that V is oriented and given a spin structure. By the Tubular Neighborhood Theorem,→ we have a containment X ⊂ U ⊂ Y where U is an open tubular neighborhood of Y, and U is homeomorphic to V. If we have a metric, we might write U = D(V), where we think of U as a bundle of -radius disks over X. In this situation, we have maps

ψ ι KOn(X) KOn+s(V) = KOn+s(U) ∗ KOn+s(Y).

n n+s We call the composition ι! : KO (X) KO (Y). On the other hand, given a map f : X Y which is not necessarily an embedding, we 8k 8k can embed X ⊂ R for sufﬁciently→ large k, and thereby obtain a map fe : X Y × R . This modiﬁed fe is an embedding; we have→ ‘lifted X into certain small free dimensions.’ Furthermore, this feis essentially uniquely determined by f, in the sense that→ any two fe’s with the same 8k are ambient isotopic to one another, and two fe’s with different 8k’s can 0 be related by including R8k ⊂ R8k . In this way, we again obtain a map m m+s f! : KO (X) KO (Y)

depending only on f. Remark: as an element of K(X), the normal bundle NX/Y is given by → R8k + f∗(TY) − TX.

(what was the relation to Aaron’s question, which asked about a spin structure on NX/Y?) t If Y itself is R , 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 two uniquely determines a spin structure on the third. Now consider the special case f : X R0. The above construction gives an embedding fe: X R8k, and then a map n → n−dim X f! : KO (X) KO ({pt}), → → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 87

dim X e.g. f! : KO (X) KO({pt}) = Z. To appreciate the relation to ordinary cohomology, recall that an oriented manifold X comes equipped with a homomorphism Hdim X(X) X called ‘evaluation on→ the orientation class [X].’ In the de Rham world, this is the integra- tion map → [ω] 7 ω, ZX ω ∈ Hdim X(X) where is a top-form. To compute→ this integral, we could integrate chart by chart. Alternatively, we could embed X , RN, consider ω ∈ Ωn(X), and construct N N an extension ωe ∈ Ω (R ) by ﬁrst pulling back to U, then multiplying by a form that represents the Thom class. This map ι! : ω →7 ωe is analogous to the ι! that we deﬁned above. Another interesting case of the above map →f! is 0 − dim X dim X f! : KO (X) KO ({pt}) = KOg(S ). 0 dim X If, for example, dim X = 8k, what does 1 ∈ KO (X) = KO (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 power series k √ yi/2 a^(y1, ... , yk) = √ sinh( y /2) i=1 i Y yi = 1 − i + ··· . 24 P Note that t/ sin h is an even power series, so only integer powers of yi appear. Suppose we 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’

p1 7 A = 1 − + (p2 − 2p ) + ··· . b 24 5760 1 2

Theorem 24.2. If X is compact and spin, then f!(1) ∈ Z equals Ae([X]), where [X] is the orientation class. For example, if dim X = 8, then 1 7p2 − 4p [X] 27 · 45 1 2 2 In particular, in this example, we find that p1(X) − 2p2(X) is divisible by 5760 when dim X = 8. The reason the formula is so complicated is the fact that the definition of + − these Thom maps relied on bV , which relied on S and S , whose definitions are quite elaborate. 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.

2 ∼ Let V1 be an oriented bundle with ﬁber R = C. So, we have V1 = (W1)R with W1 a complex line bundle. We have

+ − 0 S1 := S1 ⊕ S1 =: L1 ⊕ L1 0 + 0 That is, L1, L1 are just renamings of S1 , S1, and are just complex line bundles. We have a map ∼ 0 J : L1 = L1, 0 mapping with complex conjugation. So, if x = c1(L1) then −x = c1(L1). 2 The ﬁbers of V1 are R . Recall we have a Clifford multiplication map 0 σ∗ γ := σ 0 .

Here, 0 σ : V1 Hom(L1, L1) x 7 x − iy = z y → .

Then, → 0 σ : W1 Hom(L1, L1). So, → ∨ 0 −c1(W1) = c1 L1 ⊗ L1

= c1(L1 ⊗ L1)

= −2c1(L1). ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 89

We also get

x = c1(L1) 1 = c (W ) 2 1 1 1 = e(V ) 2 1 1 =: t 2 1 In the last line, t1 is just deﬁned to be the Euler class of V1.

25.2. Calculating A^ . Now, the Pontryagin class is 4 p1(V1) ∈ H (X; Q).

Lemma 25.1. If we set

y1 := p1(V1)

t1 := e(V1). We have the relation that 2 y1 = t1. Proof.

p1(V1) = −c2(V1 ⊗R C)

= −c2(W1 ⊗R C)

= −c2(W1 ⊕ W1)

= − [(1 + c1(W1))(1 − c1(W1))]H4 2 = c1(W1) 2 = t1. Lemma 25.2. We have + − ch(S1 − S1 ) = 2 sinh t1/2 Proof. Indeed, + − 0 ch(S1 − S1 ) = ch(L1 − L1) 0 = ch(L1) − ch(L1) c (L ) c (L0 ) = e 1 1 − e 1 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 n + − + − ch(S − S ) = ch(Si − Si ) i=1 Yn = 2 sinh ti/2 i=1 Y 2 where ti = 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. Let V be the normal bundle. Then, V gets a spin structure. We have V or 0 ∼ N D (V) = U ⊂ R , 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: 2k Theorem 25.5. Let uV ∈ H (D(V), S(V)) be the Thom class of V and let p : V X be the projection. Retain the notation given above. Then, there exists a unique a(V) ∈ H∗(X) so that we ∗ can 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 k k e(V) = e(Vi) = ti. i=1 i=1 Y Y So, we have n n ! 2 sinh ti/2 = ti ∪ a(V) i=1 i=1 Y Y These are odd power series in ti, so we can consider ! ! sinh ti/2 ti a(V) = ti · . t /2 i i i i Y Y Y 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 Q ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 91

that X is a manifold embedded into RN. We’re skipping the intricacies here, but using the universal example, you can justify cancellation to deduce n sinh t /2 a(V) = i t /2 i=1 i Y √ n sinh y /2 = √ i , y /2 i=1 i Y 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 last time, which was √ y /2 A^ (V) = √i . sinh yi/2 Y 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 a degree 1 map SN D(V)/S(V) given by crushing the outside. This is the same map we got from pushing forwards elements of K theory. → So, from last time,→ we have a map N i∗ : U R N bV 7 i!(1) ∈ K(R ). → We have computed → ∗ N ch(i!(1)) = A^ [X] ∈ H (S , Q), as all power series here have rational coefﬁcients. So, we have A^ [X] is some rational number. Using Bott periodicity, we concluded that for all x ∈ Ke(SN), ∗ N h Ni ch(x) ∈ He S ; Q S ∈ Q is actually an integer, because we computed the value on the generator of the cohomology of 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 have p 1 [X] 24 in dimension 4. We have 1 7p2 − 4p [X] 27 · 45 1 2 in dimension 8. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 93

Lemma 25.11. As an extra bonus, A^ [X] ∈ 2Z if dim X =∼ 4 mod 8. ∼ Proof. In the case that dim X = 4 mod 8. Then, bV ∈ im KSp(V). The map KSp(RN) K(RN) ×2 Z −− 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 bundle TM = Rn, where n = dim M. → Remark 26.2. Note that M is parallelizable if and only if there exist vector ﬁelds v1, ... , vn on M which 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 + 1 with e(V)[Sn+1] = 1. Proof. Think of Sn ⊂ Rn+1. A vector v ∈ Sn, with |v| = 1. n If S is parallelizable, there exist u1, ... , un which are vector ﬁelds and everywhere linearly independent. These can be thought of as maps n n+1 ui : S R .

Further, we may assume that ui(v) ⊥ v and → hui(v), uj(v)i = 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 s is 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 n+1 n+1 ∼ wn+1(V) 6= 0 ∈ H (S , 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, 4 m or 8, we will show that every V S has wm(V) = 0. Since m = n + 1, this is what we wanted to show. The idea is to use Bott periodicity. We have → w(V) = w(V ⊕ Rr). m So, we are really saying that wm(c) = 0 for c ∈ KO(S ) for m 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) ⊗ pY(F). → ∗ ∗ Take ai := wi(E). We won’t distinguish between ai and pX(ai) ∈ H (X × Y, Z/2). Similarly, notate→ bj := w→j(F). We’re used to writing w(E) = 1 + a1 + ··· + ar. Deﬁne r r−1 ∗ wT (E) = T + a1T + ··· + ar ∈ H (X; Z/2)[T]. If

E = ⊕iLi F = ⊕jMj, then r wT (E) = (T + xi) i=1 Yr wT (F) = (T + yi) . i=1 Y Here,

xi := w1(Li)

yi := w1(Mi). Then,

wT (E ⊗ F) = wT (⊕i,jLi ⊗ Mj) = 1 + xi + yj . i j Y, Then, when one expands this out, one is supposed to replace σk(xi) with ak and replace σk(yj) with bk. Then, one should see what one gets, and note that it is independent of whether E, F actually split, using the splitting principle. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 95

26.0.4. Step 4. This step wanders a bit. (26.1) m Assume r = rk E = 8, (or, we could also take 2 ), and assume that a1 = ··· a7 = 0 and a8 = a. Therefore, 8 wT (E) = T + a.

Remark 26.6. If X were S8, this would be inevitable, as there is no cohomology in dimensions between 1 and 7. So, 8 8 (1 + xi) = T + a. i=1 Y Since we’re working over a field of characteristic 2, if we also take the splitting field, so that we work over a field with an eighth root of a, x8 = a, we have

T 8 + a = (T + x)8. So, formally, we can consider the expansion T + x + yj , j i Y Y as we can assume all the x roots are the same from the above assumptions. Therefore, s 8 8 8 T + x + yj = T + x + yj j i j=1 Y Y Y 8 8 = T + a + yj j Y 8 8 8 = P(T , a, σ(y1, ... , ys)) 8 8 = P(T , a, σk(y1, ... , ys) ) 8 8 8 = P(T , a, b1, ... , bs). (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, 8 wT (E ⊗ F) = P(T , a, 0) s = T 8 + a i=1 Y = (T 8 + 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

→ ∗ 8 s → wT (E F) ∈ H (S × S , Z/2) 8+s 8 s Proof. Note that w8+s(E F) ∈ H (S × S , Z/2) is independent of F, using the previous s step. So, w8+s(E (F − R )) = 0. 8 Or, this also holds because all wk(E F) are pulled back from S .

26.0.6. Step 6. Apply Corollary 26.7 with the Bott generator on S8,

h 8i b8 := E − R , where E is some rank 8 bundle. This says

0 wT (b8 · f) = T , = 1. since all wk vanish for b8 · f (you only needed to check the top class vanished because the only nonvanishing cohomologies of Sn are 0 and n). Hence, this shows that all spheres of dimension more than 8 cannot be parallelizable, since we have shown the desired statement 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/2 2 Z/2 3 0 4 Z 5 0 6 0 7 0 8 Z

TABLE 9. Bott periodicity these groups are 0 for numbers which are not powers of 2, and hence they are not parallelizable. It is elementary to see that S0, S1, S3, S7 are parallelizable, as follows from existence of algebras R, C, H, O. (What we needed here was the existence of a bilinear map, so that multiplication by v is invertible and there is an identity element.) In each case, the vector desired vector ﬁeld v 7 ui(v) is given by multiplication

v 7 Iiv → → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 97

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, Ke to be a cohomology theory. Let (X, A) be compact pairs and X a pointed space. Suppose for all X, he(X) is a pointed set or group. Suppose he : { pointed compact spaces } { pointed sets } is a contravariant functor. The category of pointed compact spaces is that whose objects are pointed spaces and whose morphisms are homotopy→ classes of pointed maps. Example 27.1. One example of such a functor is reduced K-theory, he = Ke. Remark 27.2. We’ll require the following axioms: (1) he(pt) = {0} . (2) If i j (27.1) A X X/A

where i is an inclusion is exact at he(X) in

j∗ i∗ (27.2) he(X/A) he(X) he(A)

(but the ﬁrst and last maps need not be injections or surjections) (3) If A is contractible then

j∗ : he(X/A) he(X) is injective. → Remark 27.3. Surjectivity follows from the first two conditions, since the first implies he(A) = 0, and the exactness from the second condition is surjectivity. So being injective is equivalent to an isomorphism. Lemma 27.4. Reduced K theory, Ke, has these properties. Proof. The first and one direction of the second condition of Remark 27.2 are clear. In the second 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] ∈ Ke(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) = Ke(X/A). To check the third condition, we have ([E] , [F] , τ) ∈ K(X, A) = Ke(X/A) and E =∼ F on X after stabilizing (direct summing with some CN). (Following from the assumption 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 [0, 1] × X CX := . {0} × 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 CX ΣX := . {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 the smash product 1 S ∧ 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 he(ΣX) =∼ he(SX). Proof. Use the third axiom from Remark 27.2. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 99

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 he satisfying the three axioms from Remark 27.2, deﬁne for q ≥ 0, −q q he (X) := he(S X) to be Ke−q. Theorem 27.12. For i j (27.3) A X X/A, we have the exact sequence.

j∗ i∗ he−1(X/A) he−1(X) he−1(A) (27.4) δ j∗ i∗ he(X/A) he(X) he(A).

Proof. We have

he(S(X/A)) he(SX) he(SA) (27.5)

he(SX/SA) he(SX) he(SA) where the vertical maps are equality. The bottom sequence is exact by the second axiom of 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 X ∪ CA h(X ∪ CA) =∼ h e e CA =∼ he(X/A). using property three of Remark 27.2, since CA is contractible. 100 AARON LANDESMAN

(2) We have X ∪ CA = SA X Using the above two observations, we have that δ is the map

−1 he (A) = he(SA) X ∪ CA = h e X he(X ∩ CA) X ∪ CA =∼ h →e CA = he(X/A).

Observe that the map on the third line above is the only line which is not necessarily an isomorphism. 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 latter two 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 having positive second coordinate and the cone over X having negative second coordinate. That is, inside

X I1 × X I2 × X a a we have the equivalence relations

a ∼ (11, a)

x ∼ (12, x) .

The second axiom of Remark 27.2 implies that he(X) is exact in

(27.6) he(X/A) he(X) he(A).

So, the second axiom of Remark 27.2, applied to (X ∪ CA, X) at X ∪ CA in

X∪CA (27.7) he CA he(X ∪ CA) he(X) is exact in the middle. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 101

We further have vertical isomorphisms

X∪CA he CA he(X ∪ CA) he(X) (27.8)

X∪CA he(SA) he CA he(X).

Unravelling the bottom objects, we can identify the maps with

he (SA) he(X/A) he(X) (27.9)

he−1(A) he0 (X/A) he0(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∪C2X h (X ∪ C A ∪ C X) h(X ∪ C A). e X∪C1A e 1 2 e 1

We can identify the above with vertical isomorphisms h X∪C1A∪C2X h (X ∪ C A ∪ C X) h(X ∪ C A) e X∪C1A e 1 2 e 1

(27.11) X∪C1A he (SX) he X he(X/A)

he−1 (X) he−1 (A) he(X/A)

This is exact at the middle Exercise 27.13. Check this. These maps are not actually the same maps (the left map is the one with the problem. The issue is that we use a copy I2 and I1 of the interval, but it terms out to have the same image, and gives the desired isomorphisms. But, the differ by automorphisms of the respective sets, so the images and kernels are the same.

28. 4/6/16 28.1. Review. Last class, we deﬁned he(X), where X is a pointed space. We deﬁned −1 q he (X) := he(S X), 102 AARON LANDESMAN

and we got the exact sequence

he−q(X/A) he−q(X) he−q(A) (28.1)

he−q+1(X/A) he−q+1(X) ··· .

This continues down until

he−1(X/A) he−1(X) he−q(A) (28.2)

he0(X/A) he0(X) he0(A).

For he = Ke or he = KOg, we get periodicity maps −q ∼ −q−8 β = β8 : Ke = Ke and β commutes with the maps in the long exact sequence. Recall we have K(X, A) = Ke(X/A) for compact pairs (X, A). Deﬁne n K (X, A) := Ke(X/A). Example 28.1. We have K0(S2n) = Z ⊕ Z. This has generators C 2 of rank 1 and c1(C) = 0, and on S , the other generators is H − C of rank 0 and c1(H − C) = 1. We have K1(S2n) given by 1 2n 2n K (S ) = Ke(S {+}) 1 2n = Ke (S a) = 0 To see the penultimate equality, take the union of S2n {+}, and we take a basepoint 2n 2n x0 ∈ S , we have a pair of points {x0, +} ⊂ S {+}. Looking at the sequence of this pair in reduced K-theory, we have ` ` (28.3) Ke1(S2n Ke1(S2n {+}) Ke1 (x +) ` ` where we know Ke1(x +) = 0. We also have Ke1(S2n) = Ke0(S2n+1). Therefore, Ke1(S2n +). ` ` ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 103

Example 28.2. We have K0(S2n+1) = Z. By a similar computation to the above example, we have K1(S2n+1) = Ke1(S2n+1) = Z. Example 28.3. We have Ke−1(X) = Ke0(SX). We know the suspension is a union of two cones over X, meeting along X. A vector bundle on SX is two vector bundles on the cone over X (necessarily trivial on each cone separately) glued together. Therefore, −1 0 Ke (X) = Ke (SX) = [X, U] where U = lim U(N). Example 28.4. 0 1 0 1 Ke (RP ) = Ke (S ) = 0. We also have 1 1 1 1 Ke (RP ) = Ke (S ) = Z. Example 28.5. Let’s look at the K theory of RP1. We have a sequence

i j (28.4) RP1 RP2 RP1/RP1.

Note that RP2/RP1 =∼ S2. So, we get an exact sequence

∂ j∗ i∗ (28.5) Ke−1(RP1) Ke0(S2) Ke0(RP2) Ke0(RP1 0.

This sequence is

(28.6) Z Z Ke0(RP2) 0

So, we have Ke0(RP2) =∼ Z/d for some d, where d comes from the map ∂ corresponding to 1 2 2 multiplication by d : Z Z. We also have a sequence corresponding to S ∪φ1 e = D . This is contractible, so Ke∗(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 map between this exact sequence and the one above for RP2. On the level of groups it maps this S1 to RP1 given by multiplication by 2. From this, one can deduce d = 2, by chasing the diagram. Exercise 28.6. Make the above description precise. So, 0 2 Ke (RP ) = 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 map K0(S2) K0(S2). So, one just needs to know that pullbacks of bundles on S2 by a map of degree d is multiplication by degree d. Example→ 28.7. What is Ke1(RP2)? Similarly to the previous example, we have a sequence

∂ (28.8) Ke0(RP1) Ke1(S2) Ke1(RP2) Ke1(RP1) Kf2(S2).

This sequence is

×2 (28.9) 0 0 K1(RP2) Z Z

using for the last term that Ke2 = Ke0. and so K1(RP2) = 0. Example 28.8. Let’s compute the K theory of RP3. We have a sequence

∂ (28.10) 0 Ke0(RP3) Ke0(RP2) Ke1(S3)

This is

(28.11) 0 Ke0(RP3) Z/2 Z 0.

So, Ke0(RP3) = Z/2. Example 28.9. Question: What is Ke1(RP3). Using a similar calculation to the above, we have

∂ i∗ (28.12) Ke0(RP2) Ke1(S3) Ke1(RP3) Ke1(RP2).

This yields an exact sequence

(28.13) Z/2 Z Ke1(RP3) 0

and so Ke1(RP3) =∼ Z. Example 28.10. When we try to compute K1(RP3), and look at the corresponding sequence, something new gets in our way. We have the analogous sequence

∂ f (28.14) Ke−1(RP3) Ke0(S4) Ke0(RP4) Ke0(RP3) r Ke1(S4)

This sequence is → f (28.15) Z Z Ke0(RP4) Z/2 0.

We now have to compute the multiplication for the ∂ map. Here, this map is multiplication by 2. From this, we know the image of f is Z/2. Therefore, Ke0(RP4) is an extension of Z/2 by Z/2. Therefore, it is either Z/2 ⊕ Z/2 or Z/4. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 105

Exercise 28.11. Show that the answer is Z/4. Hint: One way to prove this, which works in dimension 4, but will fail in higher dimension is to ﬁnd a bundle F on RP4 so that F ⊕ F has nonzero characteristic classes. From the above sequence, the map

0 4 0 3 Ke (RP ) Ke (RP )

RP3 K0(RP4) is surjective. So, any bundle on extends→ to one on e . In fact, the map 0 4 0 3 Ke (RP ) Ke (RP )

is also surjective. From this, deduce there exists a complex line bundle L RP4 with 2 4 → 2 w2(LR) 6= 0 ∈ H (RP , Z/2). This is because there exists such a line bundle on RP , and so the line bundle extends. Therefore, we have w4(L ⊕ L) 6= 0. Then, x :=→ L − [C] ∈ Ke0(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 theory which we can find. Example 28.12. In general, for real projective spaces, one gets K-theory Z/2k for increas- ing k. For Ke given A, X, X/A, one has a periodic sequence of length 6. For KOg, one has a 24 term exact sequence for i ranging from 0 to 7. One can probably use all the 0’s for KOg(Sn) to calculate KOg(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 n manifolds } / ∼

where M0 ∼ M1 if there is a smooth compact (n + 1) manifold M with boundary N := ∂M so that N = M0 M1. Example 28.14. `We have

Λ0 = {∅, ∗}

Note that any even number of points is cobordant to the empty set, by drawing a bunch of intervals. Example 28.15. For 1 manifolds, we have

Λ1 = {∅} .

Note that S1 is cobordant to ∅ via D2. Example 28.16. We have

2 Λ2 = ∅, RP , although we didn’t show this today. 106 AARON LANDESMAN

29. 4/8/2016 2 Picking up from last time, we wish to show that Λ2 = {∅, RP }. One way to see that RP2 is nontrivial (under cobordism) is via the Stiefel-Whitney numbers: Deﬁnition 29.1. Let M be an n-dimensional manifold. Let I be a multi-index, i.e. n n w = w 1 ··· w k , I i1 ik n where r nrir = n, so that wI ∈ H (M; Z/2). We obtain a number by evaluating on the fundamental class [M] ∈ Hn(M; Z/2), to obtain wI[M] ∈ Z/2. P Lemma 29.2. If M0 and M1 are cobordant, in the sense that M0 ∼ 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 to describing which possible sets of Stiefel-Whitney numbers arise among n-dimensional manifolds.

Proof. If N is an (n + 1)-dimensional manifold realizing the cobordism M0 ∼ M1, in the sense that ∂N = M0 t M1, then we have

TN|M0 = TM0 ⊕ R

TN|M1 = TM1 ⊕ R, where the trivial bundles R corresponds to the outward normal direction. The Whitney sum formula then tells us that n wi(TN)|wi = wI(Mi) ∈ H (Mi; Z/2) for i ∈ {1, 2}. We also know that, in the long exact sequence of the pair (N, ∂N), we have this 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]. 2 2 2 2 The fact that RP 6∼ ∅ now follows from either w1[RP ] = 1 6= 0 or w2[RP ] = 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 T 2#T 2# ··· #T 2 where T 2 = S2 is the torus, or RP2#RP2# ··· #RP2.

Lemma 29.4. These cobordism classes are equal: [M0#M1] = [M0 t M1]. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 107

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 n ”top” of the cylinders by a cylinder D × [2, 3].

Lemma 29.5. In Λn, we have M t M ' ∅.

Proof. Consider the manifold N = M × [0, 1]. 2 The result Λ2 = {∅, RP } now follows from the above Lemmas, although with the observation that T 2 ∼ ∅ via the solid torus N = S1 × D2.

Proposition 29.6. The set Λn is an abelian group under disjoint union, with identity element [∅] and inverses −[M] = [M]. L Proposition 29.7. The set n=0 Λn is a commutative ring, with identity element [pt] and [M ] × [M ] := [M × M ] multiplication 0 1 ∞ 0 1 . 29.1. Oriented cobordism.

Deﬁnition 29.8. Let Ωn be the set of equivalence classes of oriented n-manifolds, under the equivalence relation M0 ∼ M1 whenever there exists an oriented (n + 1)-dimensional manifold N such that ∂N = (−M0) t M1. Here −M0 denotes M0 with the opposite orientation. 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 normal vector e0 ∈ TpN, as well as an ordered basis (e1, ... , en) for Tp(∂N) ⊂ TpN. The ordered basis (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 direct sum decomposition

T(x,y)(A × B) = TxA ⊕ TyB, whereby an oriented basis for the left hand side is given by taking an oriented basis for TxA, and then an oriented basis for TyB. Example 29.9. For an oriented n-manifold M, orient N = [0, 1] × M by the product orientation. Then ∂N = {1} × M − {0} × M according to the ‘outward normal ﬁrst’ boundary orientation.

Proposition 29.10. The set Ωn is an abelian group under [M1] + [M2] := [M1 t M2]. In fact, [∅] is the identity element, and inverse are given by −[M] := [−M], i.e. opposite orientation. L Proposition 29.11. The set n=0 Ωn is a ring under [M1] × [M2] = [M1 × M2] with the [pt] pt product orientation. The identity∞ element is , where is a ‘positive point.’ (An orientation of a point is a sign ±1.) This ring is not commutative but graded commutative:

(dim M1)(dim M2) [M1] × [M2] = (−1) [M2] × [M1]. 108 AARON LANDESMAN

Remark 29.12. The two orientations of a point deserves clariﬁcation. For us, an orientation of a vector space V means a choice of basis vector of the one-dimensional vector space Λdim V V. Since λ0{0} = R canonically, we see that a point has two orientations, ±1, and furthermore the +1 ∈ R can be singled out as the canonical orientation of the point. In this way, every orientable connected (possibly zero-dimensional) manifold has exactly two orientations.

Proposition 29.13. We have Ω0 = Z as abelian groups. Proof. Positive points can be canceled with negative points by an arc joining them. The result follows because, the boundary of each one-dimensional manifold (with boundary) has total index zero.

Proposition 29.14. We have Ω1 = {0} as abelian groups. Proof. The only orientable 1-manifolds are disjoint unions of circles, and a circle is trivial- 2 ized via a disk D .

Proposition 29.15. We have Ω2 = {0} as abelian groups. Proof. All connected 2-manifolds are connected sums of tori, and tori are oriented cobordant to ∅. n1 nr Lemma 29.16. If [M ] = [M ] in Ωn, then p [M ] = p [M ] where p = p ··· p where 0 1 I 0 I 1 I i1 ir r 4nrir = dim M are the Pontryagin classes. PProof. Similar to the analogous result for Stiefel-Whitney number.

Theorem 29.17. We have Ω3 = {0}, Ω4 = Z, and Ω5 = 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.) 4 4 To see that Ω4 = Z, we note that two 4-manifolds satisfy M0 ∼ M1 in Ω4 if and only if 2 p1[M0] = p1[M1], where this inequality takes place in 3Z. A generator is given by CP , 2 for which p1[CP ] = −3. Instead of carrying on with the futile task of directly classifying n-dimensional manifolds, we use more algebraic topology machinery: we shall realize Ωn as the homotopy group of a space, to be described below. This will help us compute Ωn. For N ≥ n, let Ωn,N be the set of equivalence classes of oriented n-manifolds M embedded in RN, modulo the equivalence relation of ‘cobordism within RN,’ in the sense that M0 ∼ M1 whenever there exists an (n + 1)-dimensional manifold N embedded in N [0, 1] × R , such that ∂N = −M0 t M1. Remark 29.18. Consider n-manifolds M embedded in RN. For N sufﬁciently large (depending on n), any two embeddings will be ambient isotopic to one another. For this reason, Ωn,N should be thought of as an approximation to Ωn, especially for large N. 30. 4/11/16

30.1. Review. Recall Ωk from last time, which was cobordism classes of oriented k-manifolds. n+k We deﬁne Ωk,n to be classes of oriented k-manifolds M ⊂ R . The equivalence relation is given by smoothly embedded manifolds N ⊂ Rn+k × I, where I = [0, 1]. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 109

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 embedding theorem. Theorem 30.2 (Whitney embedding theorem). Let M be a manifold. There exists an embedding M , RN−1 if N − 1 ≥ 2k + 1. Proof. Given M compact and smooth, we can embed M , RN, with N ∈ Z. Given → N N N−1 an embedding M , R , we can project pu : R R , projecting orthogonally to a unit vector u, can we choose u so that pu(M) is an embedding.→ You do a dimension count, which you then→ need to back up with Sard’s→ theorem. We have to avoid u which is parallel to x1 − x2 for xi ∈ M. You also have to avoid u parallel to tangent vectors to M. The elements parallel is to pairs of points is 2k dimensional and the tangent vectors are 2k − 1 dimensional. Here, u ∈ SN−1. So, by Sard’s theorem this is an embedding if N − 1 > 2k. Surjectivity follows from the Whitney embedding theorem, Theorem 30.2. For the isomorphism, we will need to embed the cobordism N similarly. N N Let Gn(R ( be a grassmannian. Let Ge(R ) be the grassmannian of oriented n-dimensional subspaces. We have a two fold cover N N Gen(R ) Gn(R ). There is a tautological oriented vector bundle → N En,N Gen(R ) so that the ﬁber over a point is the corresponding oriented n-dimensional plane. As N , we get a universal oriented bundle→

En Gen(R ). → ∞ Deﬁnition 30.3. The Thom space ∞ → B(En) τ(En) := . S(En) ◦ As a set, this is the union of the open unit balls B (En) ∪ { }. We similarly have a Thom space for ﬁnite Grassmannians,

B(En,N) ∞ τ(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 construction

πk+n(τ(En)) Ωk,n. Take → [f] ∈ πk+n (τ (En,N)) which is a representation of k+n f : S τ(En,N) ⊂ τ (En) .

Here, τ(En) is an infinite cell complex which is the union of the finite cell complexes → τ(En,N). Every compact subset of τ(En) is contained in τ(En,N) for some N ∈ Z. n+k n+k −1 ◦ ◦ Think of S = R ∪ { } . Set U = f (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∞ by a∞ smooth map, so we may assume f|U is smooth and transverse to Z, where Z is the zero section 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 normal bundle to Z is oriented, since the tautological bundle is oriented. Then, we can pull back the 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 get a 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 was f−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]

−1 → ◦ where M = f (Z), and Z was 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 much about the particular bundle En. But, to make the inverse map, we will need to use that this is the universal bundle to the grassmannian.

What is the inverse map Ωk,n πk+n(τ(En))? Start with M ⊂ Rk+n with an oriented normal bundle V M. The universal property of the grassmannian we get a corresponding→ map N → M Gen(R ).

→ ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 111

Here, we can take N = k + n. That is, our map is k+n f1 : M Gen(R ) 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 of M in Rk+n. Hence, combining the two maps above, we have a map → ◦ f2 : U B (En,N). −1 k+n k+n c Then, f2 (Z) = M. In S = R ∪ , let U denote the complement of U. Extend f2 k+n → to f : S τ(En,N) given by sending the points where f2 is not yet defined to the point . That is extend f2 to f by ∞ c → f(U ) = { } ∈ 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 and k+n k+n+1 M1 are cobordant via a manifold W ⊂ R × [0, 1] ⊂ R . To show well definedness, we’ll construct a map k+n S × [0, 1] τ En,N0 with N0 = k + n + 1. We then take a tubular neighborhood and construct a similar map → k+n to the map above. Both M0, M1 give maps from S to τ(En,N), we will show the maps become homotopic after the inclusion i : τ (En,N) ⊂ τ En,N0 . For M ⊂ Rk+n, we get a map k+n f : S τ(En,N)

→ 112 AARON LANDESMAN which then gives the map k+n i ◦ f : S τ(En,N0 ), which gives a well deﬁned map → Ωk,n πk+n(τ(En,N0 )) [M] 7 [i ◦ f] . → with N0 = 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 ring will become homology groups of a point.

31. 4/13/16 31.1. Oriented bordism as a homology theory. Today, we’ll discuss oriented bordism as a 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, it follows 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 special orthogonal 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. If X 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 the manifold, since the maps are determined. If X has a basepoint x0 ∈ X, we obtain a map

i∗ : MSOk({x0}) MSOk(X)

→ ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 113 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 another map →

p∗ : MSOk(X) Ωk. →

Note that i∗ is injective because p∗i∗ = 1Ωk . → Deﬁnition 31.4. We deﬁne the corresponding reduced group

MSOk(X) MSO] k(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, h i M −M, σ σ is a boundary of a a [N, τ] with N = [0, 1] × M and τ = σ ◦ pM. Example 31.7. We have

MSO^2 ({x1, x0}) = 0. To see this, we have ∼ ∼ MSO^2 ({x1, x0}) = MSO2 ({x1}) = Ω2 = 0, since every 2-dimensional oriented manifold is a boundary. E.g., a torus is the boundary of a solid torus. 114 AARON LANDESMAN

Definition 31.8. We define ∼ MSOk(X, A) := MSO] k(X/A) = MSO] k(X ∪ CA) The first is the definition, the latter is an isomorphism. Lemma 31.9. If we have maps

i j (31.1) A X X/A,

the sequence

i∗ j∗ (31.2) MSO] k(A) MSO] k(X) MSO] k(X/A).

is exact at MSO] k(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 straightforward 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 MSO] k factors through MSO] k(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

i j (31.4) A X X ∪ CA.

It sufﬁces to prove the lemma with X/A replaced by X ∪ CA, as they are homotopic, using a previous lemma that MSOk groups are preserved under homotopy. Suppose

j∗ [M, σ] = 0

in MSO] k(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 → A × [0, 1] CA = A × {0} We have a coordinate function t on [0, 1]. We can think of t : X ∪ CA [0, 1] −1 with t|X = 1. Picture t as the height function. In N we have U ⊂ N be U = τ (CA). In fact, 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 can 1 further assume t = 2 is a regular value (by Sard’s theorem),→ meaning that the derivative 1 −1 is surjective when t = 2 . So M1/2 = (t ◦ τ) (1/2) ⊂ U, is a smooth submanifold. 0 −1 0 Let N = (t ◦ τ) [1/2, 1]. We have a restriction τ := τ|N0 . We have σ1/2 : M1/2 A × {1/2}. → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 115

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 ◦ τ0 : N0 X. → This shows that [M, σ] ∈ ker j∗ satisﬁes → [M, σ] = [M1/2, r ◦ σ1/2] Corollary 31.10. (See the discussion of abstract he from before.) There exists a longer exact sequence (31.5)

MSO] k(A) MSO] k(X) MSO] k(X/A) MSO] k(SA) MSO] k(SX) ···

Lemma 31.11. We have ∼ MSO] k(SA) = MSO] k−1(A). Proof. The idea is that if we look at the suspension of A, inside we have the level set A × {1/2}. We have a coordinate function t in the vertical direction. Say we have a k- dimensional M mapping to SX, so that

[M, σ] ∈ MSO] k(SA) Inside this M we have M0 := σ−1(1/2). That is, M0 = (t ◦ σ)−1 (1/2). We have that M0 ⊂ M is a smooth k − 1 dimensional manifold. We then have 0 0 σ|M : M A × {1/2} = A, and construct 0 → M , σ|M0 ∈ MSO] k−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 S×[0,1] 32.1. Review. Recall that SA := ∼ and we were showing that MSO] k(SA) = MSO] k−1(A). The idea was that inside SA we have A × {1/2}, and we construct the map by sending 0 −1 0 σ : M SA to M = σ (1/2) and the map σ|M0 : M A. Remark 32.1. Recall that we set up → → MSOk(SA) MSO] k(SA) := . MSOk(pt) Lemma 32.2. But, we could have equivalently deﬁned

MSO] k−1(A) ⊂ MSOk−1(A) as the subset 0 0 0 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 σ00 : M00 A, where M00 is not necessarily a boundary, we can construct a map σ00 pt : M00 −M00→ A and a a → pt : M00 A and taking M0 := M00 −M00, we see M0 is the boundary of a cylinder. Thus, every map can 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 η MSOk(SA) − 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 + 0 [0, 1/2] is M ). Here, M = ∂M and M is a boundary. So, im η ⊂ MSO] k−1, where we regard MSO] k−1 ⊂ MSOk−1 as the subset which comes from boundaries. 0 0 In fact, it’s not hard to see that im η = MSO] k−1. If we have a map σ : M A, we can take a collar neighborhood of M0. So, we can construct a map σ : M SA extending σ0, showing the map η is surjective. → What is the kernel of η? We have that MSOk( basepoint ) ⊂ ker η. → To see that this is the full kernel, we have [M, σ] giving a map σ : M SA. We have 0 −1 → M := σ (A × {1/2}). and a map σ0 : M0 A × {1/2} =∼ A. Suppose M0 = ∂N0 and σ0 extends to τ0 : N0 A. The existence of this extension τ0 is precisely what it means for [M0, σ0] = 0.→ Given this, we want to show that h→ i [M, σ] = Mf, σe where σe 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 M0. 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+ and M− and take a small collar around it. More precisely, we have + 0 − 0 Mf = M ∪ N × {ε} M ∪ N × {−ε} . a ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 117

and there is a cobordism between M0 and M. So far, we have an h i [M, σ] = Mf, σe , with im σe ∩ (A × {1/2}) = ∅ So, we can construct a map

σe : Mf SA where the image lies in the two base points of SA, using that the image of σe does not 1 → intersect A × 2 .

Remark 32.4. There are some clear variants of Lemma 32.3. We can similarly deﬁne MOk as unoriented bordism and MUk as complex bordism. We want this notion to be stable with respect to adding trivial summands. More precisely, a stable complex structure on V Z a vector bundle over a manifold Z means 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 → k+2 k k+2 ∼ k J1, R ∼ J2, V ⊕ R ⊕ C if there is an isomorphism V ⊕ R = V ⊕ R ⊕ C where k the J1, J2 agree on V ⊕ R , 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 collection of r pointwise independent sections where r = rk V. Definition 32.6. A stable framing of a vector bundle→ V Z is a framing of V ⊕ Rk for some 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 large n. 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 first k+n define Πk,n(pt), to be equivalence classes of k-dimensional manifolds M ⊂ R together with a stable framing of ν(M), the normal bundle to M. Here, in more detail, the equivalence relation is given by M0 ∼ M1 if there exists a manifold N and a framing for ν(N), the normal bundle to N in Rk+n+1 which extends the framing of M0, M1. We have Πk,n(pt) eventually stabilizes, by Lemma 32.10 and define Πk(pt) to be this stabilization. 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 bundle.

Exercise 32.11. Check this by constructing Πk as manifolds together with a framing for its tangent 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 directions for each point. If we have a left and right directions, the points cancel by a framed ∼ arc. Therefore, Π0 = Z.

Lemma 32.14. We have Π1. Proof. We will have M = S1 or a union of S1’s. We can embed a circle in R2. There are two trivializations of its normal bundle, as either the outward one or inward one. (Here, we mean trivializations up to homotopy.) If we put S1 in R3, these two trivializations become the 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 get all others as φ0(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 done for 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 map S1 O(2). → So, we want to answer the question: Question→ 32.16. What are the maps S1, O(2) =∼ S1, S1 S1? 1 1 Of course, they are Z × Z because S , S =∼ Z Z. ` If we instead embedded S1 ⊂ R4, the normal bundle would be S1 × R3. Then, we’d 1 1 ∼` 1 3 ∼ want to understand S , O(3) . First, S , SO(3) = S , PC = Z/2. Therefore, h i S1, O(3) =∼ Z/2 Z/2. This is what we get before thinking about cobordisms.a ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 119 ∼ 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] k+n k+n with M ⊂ R is a smooth, compact k-manifold embedded in R and φ = (φ1, ... , φn) is a normal framing. That is, each φi is a normal vector field. k+n The equivalence relation is defined by M1 ∼ M0 if there exists a cobordism N ⊂ R × k+n [0, 1] with a normal frame restricting to those of M0 and M1 on the boundaries R × {0} , Rk+n × {1}. Remark 33.1. The point of introducing this definition is that, in disguise, this is the problem 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 result of this subsection is to relate framed cobordism to homotopy groups of spheres. Proposition 33.2. There is an isomorphism n ∼ η : πk+n(S ) = Πk,n. n Proof. We will first describe the map η. Given [f] ∈ πk+n(S ), where f : Rk+n ∪ Rn ∪ . This can be taken to be smooth by a suitable approximation. Let y ∈ Rn be a regular value, meaning that M = 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 of n tangent vectors for R , call them e1, ... , en. Then, the map f is given so that the inverse image of y ∈ Rn is some submanifold of Rn+k. Further, we have an isomorphism df : ∼ n νx(M) = TyR . 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 different representatives f0 and f1 with regular values y0 and y1. Let’s check the independence of this choice corresponding∞ to k-manifolds M0 and M1. 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 k+n n eF : R ∪ × [0, 1] (R ∪ ) × [0, 1] (x, t) 7 (F(x), t) . ∞ → ∞ −1 Choose a path γ from y0 to y1 so that it is transverse to eF, meaning that N := eF (γ([0, 1])) → −1 −1 is a smooth k + 1 manifold with boundary. Here, eF (γ(1)) = M1, eF (γ(0)) = M0 . Then, extends e1, ... , en to be a framing of ν(γ). Use deF to obtain a framing for ν(N) so that ∼ dF : νx(n) = ν (γ). ej Fg(x) 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 and surjective. We’ll check surjectivity, as injectivity is similar. Given some (M, φ) with M ⊂ 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 n k+n α : D1(R ) × M R

(v, x) 7 x + ε viφi(x). → i X where → v1  . v =  .  vn and D1 is the open disk of radius 1. The tubular neighborhood theorem tells us there is a bundle over M whose fibers are disks. The framing gives us an identification of the fiber n bundle as the product bundle D1(R ) × M. For ε > 0 small, the above map α restricts to a diffeomorphism n D1(R ) × M U(M) for U(M) ⊂ Rk+n some open neighborhood of M. So, we have U(M) =∼ (D◦)n × M. So, we have→ a map p U (M) − D◦(Rn) =∼ Rn. Call this composition f0. Further, (f0)−1(0) = M and the framings agree. Then, we can extend the map f0 to all of Rn+k ∪ , which→ initially is a map

0 U(M) f Rn ∞ (33.1)

Rn+k ∪ Rn ∪

∞ ∞ ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 121

by deﬁning f(z) = for z ∈/ U(M). This is continuous, since, as we approach the boundary of U(M), we go to in Rn ∪ . n This shows that [∞M, φ] is in the image of πk+n(S ). 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 a framed N ⊂ Rk+n × [0, 1], and want a homotopy. One dimension lower, we just did this for 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 homotopy groups.∞ 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, we ∼ saw Π0,n = Z. When n = 2, we had a framing which is either left handed or right handed n ∼ n ∼ at any point. Indeed, we have π0+n(S ) = πn(S ) = Z. Example 33.5. In the case k = 1, we saw that things get fairly confusing, fairly quickly. start with a 1-dimensional manifold M. 2 Start with k = n = 1. Take M ⊂ R to be the unit circle. Take φ = {φ1} to be the ∂ 2 outward normal ∂r . This class is [M, φ] = 0. This is 0 because this is a boundary of of D , 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, in 1 this case, Π1+1 = π2(S ) = 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 φ is the outward normal and φ = ∂ . By the same construction using a hemi- 1 2 ∂x3 sphere, we get that in this case [M, φ] = 0. However, in the case of R3, there are more options 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} . k+1 Here, we can take φ1 to be the outward normal in R and φ2, ... , φn the remaining coordinate directions. But, already in the R3 case, given a map g : Sk O(n), k consider the new framing φg deﬁned by, for x ∈ S , → ∗ φg(x) = g(X)φ (x). 122 AARON LANDESMAN

We get a map h k i ∼ n S , O(n) Πk,n = πk+n(S ). Restricting this to SO(n), we get a map → n J : πk(SO(n)) πk+n(S ), 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 stabilization. 34.1. Stabilization. Last semester, in 231a, a proof of the stabilization of n n+1 πk+n(S ) πk+n+1(S ). This map was induced by the suspension map, sending → f : Sk+n Sn 7 Sf : SSk+n SSn. The map → n → n+→1 πk+n(S ) πk+n+1(S ) [f] 7 [Sf] which is onto if n ≥ k + 1 and an isomorphism→ if n ≥ k + 2. s This is a special case of the Freudenthal→ suspension theorem. Defining πk as the stabi- i lization of the kth homotopy group of spheres, the limit of πk+i(S ), we have s n πk = πk+n(S ) for any n ≥ k + 2. Last time, we saw n ∼ πk+n(S = Πk,n), k+n where Πk,n is normal framed k-manifolds M ⊂ R , notated as pairs [M, φ]), where φ is a framing. We have a natural map

Πk,n Πk,n+1 sending M ⊂ Rk+n , Rk+n+1, and sending a framing → φ = {φ1, ... , φn} 7 {φ0, ... , φn} → with φ0 = en+k+1. We constructed vertical maps →

n n+1 πk+n(S ) πk+n+1(S ) (34.1)

Πk,n Πk,n+1 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 123

Exercise 34.1. Show that up to possibly annoying orientations and signs, this diagram commutes. Deﬁnition 34.2. Deﬁne

Πk := lim Πk n. n , Exercise 34.3. For vector bundles U, V on a space X with W := U ⊕ V, a stable framing →∞ of any two of U, V, and W determines a stable framing for the third. Hint: If we have a stable framing for U and W, 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 ⊕ k j R is trivialized by φ then there exists a j0 trivialization π of V ⊕ R , 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, by Exercise 34.3.

Remark 34.6. Hence, Πk is equivalence classes of [M, ψ] with M a compact k-manifold and 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 O where O is the orthogonal group for large M. In fact, this plays well with the associated group structure of Πk, which comes down to O being a group, and the group structures→ being 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 ∼ s in Πk = πk. Given any map g : M O, we get an associated map φg := g→· φ∗ g s and a corresponding element [M, φ ] ∈ πk. 124 AARON LANDESMAN

We have a group homomorphism s n J : πk(O) πk+n(S ) s = πk → for sufﬁciently large n. We’d like to understand what the J homomorphism.

s k πk(O) πk =∼ 1 Z/2 − Z/2 0 2 0 − ? 3 Z →? 4 0 → 5 0 → 6 0 7 Z 8 Z/2

TABLE 10. A table of homotopy groups of the orthogonal group

3 ∼ In the case when k = 1, we have π4(S ) = Z/2. The order of im J for k = 4m − 1 was settled by Mahowald around 1970. Looking at the case when k = 4m − 1, the image of s JZ π4m−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 to determine→ J(d · g) = 0, for d ∈ Z. We’ll get a necessary condition on d, which will give a lower bound for the size of the image of J. Lemma 34.8. If J(d · g) = 0, then

d(2m − 1)!Qm ∈ Z, where A^ m = Qm · pm, where Qm is some coefﬁcient, and A^ m is the top dimensional class of A^ 4m (introduced several classes ago) in H , 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 the 0 element. Take φ1 := g · ( the bounding framing ), where g : Sk O is the generator of πk(O). We have ∼ s [M, φ1] = J(g) ∈ Πk = πk. → Then, taking d times this generator corresponds to a disjoint union of spheres. If J(d · g) = 0, there is some manifold N which is a boundary for d copies of the sphere. That is, N has a stable framing ψ extending the copies of φ1, with φ1 a framing on each copy of M. Out of this picture, we can make a closed manifold N^ := N ∪ D4m ∪ · · · D4m, ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 125

so that the resulting manifold is closed. The disks are attached to the bottom of the spheres. We have that TN is stably trivial but TN^ is not stably trivial, since on the boundary of disks, we’ll have the framing φ∗, if we trivialize the disks (since φ∗ is the unique stable framing extending to these disks. At these boundaries of the disks, the framings differ by g, which is the Bott generator as 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 top and 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 → 4m (2m − 1)! if m ≡ 0 mod 2 pm(g)[S ] = 2(2m − 1)! if m ≡ 1 mod 2. Then, the answer for N^ , we have d(2m − 1)! if m ≡ 0 mod 2 pm[N^ ] = 2d(2m − 1)! if m ≡ 1 mod 2. as it is a gluing of d classes corresponding to the Bott generator, so it is d times the corresponding 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 Z if m ≡= 0 mod 2 A^ N^ ∈ 2Z if m ≡ 1 mod 2

When p1, ... , pm−1 = 0, we have that A^ m = Qm · pm, where Qm is some coefﬁcient, and 4m A^ m is the top dimensional class in H . Then, A^ N^ = Qm · pm N^ −691 = p N^ . 2615348736000 m 126 AARON LANDESMAN

Remark 34.10. We have B Q = (−1)m 2m m 2 (2m) ! Corollary 34.11. d is divisible by the denominator of

B2m . 4m Corollary 34.12. The image of J is a cyclic group of order divisible by n, as given in the following 5 table In particular, π8(S ) has order at least 24.

m n 1 24 2 240 3 504 4 480 5 264 6 65520 . . . .

TABLE 11. Table of scaled Bernoulli numbers.

5 Proof. The statement about π8(S ) 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 order 24. 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 manifold M = 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 the ∼ tangent 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 h i S1, SO =∼ Z/2. So,

Π1 = Z/2 = {0, 1} , where 1 corresponds to S1 with the Lie group framing. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 127

Note that we have TS1 ⊕ R =∼ R2 on S1. If we identify the trivial summand of R2, we can take one of the summands to be the outward normal. If we try to extend the Lie group framing 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 same s as the group operation on πk. We have the following stable groups in low dimensions

Stable homotopy group where it stabilizes framed manifold generators group image of J? s 2 π0 π2(S ) • Z No s 3 1 π1 π4(S ) (S , Lie ) Z/2 Yes s 4 1 1 π2 π6(S ) (S × S , Lie ) Z/2 No s 5 3 ∼ π3 π8(S ) (S = SU(2), Lie ) Z/24 Yes s 6 π4 π10(S ) 0 No s 7 π5 π12(S ) 0 No s 8 3 3 π6 π14(S ) (S × S , Lie ) Z/2 No s 9 7 π7 π16(S ) (S , Octonian ) Z/240 Yes s 10 8 1 π8 π18(S ) SU(3), Lie ), (S , φ ) 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 the s generator is the closest analog. In the π8 case, the second Z/2 is the image of the J homomorphism, but the full group is not generated by the image J homomorphism.

35.2. More about Π3. s Remark 35.4. Why is π3 = Z/24. Last time, we showed there was at least a Z/24, but why is that precisely the group? s Here, we’re looking at π3 = Π3. Fact 35.5. Note the following facts (the first two are on par with the difficulty with a problem set problem): (1) A three dimensional manifold has trivial tangent bundle if and only if M is oriented, if and only if TM is stably trivial. (2) A connected four manifold N with nonempty boundary has trivial TN = R4 if and only 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. 3 (4) Π3 is generated by im J. That is, given a framed three manifold, (M , φ), we can find (N, φ) so that the boundary of N is the union of S3, with the trivial framing φ1, and (M3, φ). This fact is closely related to the previous fact, and will follow from the previous one. 128 AARON LANDESMAN ∼ To show Π3 = Z/24, we know the generator maps to something of order which is a multiple of 24. We want to show it maps to an element of order precisely 24. To show it is order precisely 24, we start with 24 copies of (S3, φ1) and find a manifold (N, φ) with framing whose boundary is 24 such copies. We can exhibit such a 4-manifold N as a quartic surface N^ ⊂ P3, which is a K3 quartic surface, which has Euler number 24. This has w2 = 0. To see the Euler number is 24, one would want to show there is some vector field with 24 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 restrict correctly. 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 are products of the normal direction with I, J, and K. 35.3. Another integrality property for A^ . We’ll discuss the Hirzebruch signature theorem and the L-genus now.

Example 35.6. Let X be a smooth 8-manifold, suppose p1[X] = 0 and take the A^ genus 7p2 − 4p A^ [X] = 1 2 [X] 5760 4 = − p [X] 5760 2 1 = − p [X]. 1440 2

Corollary 35.7. Suppose X is spin and p1 = 0. Then, since A^ [X] is an integer, we have that 1440 | p2[X]. This holds if X \ pt has trivial tangent bundle. Proof. It’s immediate from the above, since 7p2 − 4p A^ [X] = 1 2 [X] 5760 4 = − p [X] 5760 2 1 = − p [X]. 1440 2 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. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 129

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 have n √ yi 1 1 2 √ = 1 + p1 + (7p2 − 24p ) + ··· tanh y 3 45 1 i=1 i Y =: 1 + L1 + L2 + ··· .

where the yi are formal variables, and pi are the elementary symmetric functions of the yi In other words, the L-genera are the terms of the expansion of the above power series. Theorem 35.9. We have 4k σ(X ) = Lk[X] 7 σ(X8) = p [X] 45 2 if p1 = 0, where σ(X4): H4(X) × H4(X) Z (α, β) 7 (α ∪ β) [X], → Proof. Omitted. → 8 Corollary 35.10. If X is spin and p1 = 0, then 45 1440 | σ, 7 that is, 1440 7 · = 7 · 32 | σ(X). 45 8 Proof. This follows from the previous results stating that σ(X ) = 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 there −1 were such a homeomorphism, we would have σ(X) = 8 = 1440 p2[X], but we know 7 · 32 divides 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 diffeomorphic. These are called exotic 7 spheres. Proof. It sufﬁces to construct the manifold described above. One way to do it is to construct it explicitly using plumbing. A more mysterious way to construct it is to exhibit it from the hypersurface 2 3 5 2 2 E := x1 + x2 + x3 + +x4 + x5 in C5 and construct Z as E ∩ B10, where B10 is the unit ball in C5. Then, Z is a smooth hypersurface in B10 of dimension 8. The boundary of Z is Z ∩ S9, which turns out to be 7 homeomorphic but not diffeomorphic to S .

36. 4/25/16 36.1. Spectra. Recall that we were talking about stabilization. Today, we’ll discuss spectra, 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, 1 En are pointed spaces, εn are pointed spaces, and SEn := S ∧ 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 ask for other similar variants. n n ∼ Example 36.3. (1) Consider the sphere spectrum given by En = S and εn : SS = Sn+1 the natural isomorphism. (2) If we are given E and a space X we can form a new spectrum E ∧ X with

E ∧ X := {(En ∧ X, εn ∧ 1)}n∈Z . (3) If S is the sphere spectrum, we can form n S ∧ X = {(S X, 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 is indexed by n ∈ Z. Spectra were lated studied and reﬁned by Spanier, G. Whitehead, and Adams. Remark 36.5. One might note that we have often been indexing examples by positive integers, but it turns out that we will only really care about spectra indexed by n for all n bigger than some chosen n0. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 131

36.2. More involved examples of spectra. Example 36.6 (Thom spectrum). Recall we have the universal vector bundle (which has unfortunate 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 , Gr . We have ι E = R × En. This gives rise to a map ∞ n+1 n+1 ∞ τ(R × En) τ(En+1 → or, taking one point compactifications, a map ε → S(τ(En)) − n τ(En+1). This enables us to define the Thom spectrum, → MO = {(τ(En), εn)}n One 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 G if m = n πm(K(G, n)) = 0 otherwise. There are maps ε SK(G, n) − 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 a tautology in the category of sets, So, giving a map ε SK(G, n) − n K(G, n + 1) is the same as giving a map → K(G, N) Ω (K(G, n + 1)) , where by definition ΩX is the loop space of X, defined as → 1 ΩX := XS , where AB is defined to be maps from B to A. 132 AARON LANDESMAN

Because h m S1 i πm(ΩK(G, n + 1)) = S , K(G, n + 1) h i = S1 ∧ Sm, K(G, n + 1)

= πm+1 (K(G, n + 1)) G if m + 1 = n + 1 = 0 otherwise. Fact 36.8. Eilenberg MacLane spaces not only exist, but but are unique in the sense that for any two spaces satisfying the property that their nth fundamental group is G, and all other fundamental groups are 0, then we have a uniquely determined isomorphism on homotopy 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 We have reduced K-theory, defined by Ke(X) := [X, F] where F is by definition the Fredholm operators on a given Hilbert space H. We have h i 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 F if n ≡ 0 mod 2 Bn := ΩF 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. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 133

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 h k+n i πk+n(En) = S , En

S h k+n i − SS , SEn h i = Sk+n+1, SE → n (εn)∗ h k+n+1 i −−− S , En+1 . We get a map → πk+n(En) πk+n+1(En+1). Deﬁne → π (E) := π (E ) k −lim k+n n Example 36.12. We have ∼ → n πk(S) = πk+n(S ) for n large is the nth stable homotopy group of a sphere. Example 36.13. For a space X, we have π (S ∧ X) = π (SnX) k −lim k+n s =: πk(X), the kth stable homotopy group of X. → Example 36.14. Let’s compute π (G) := π (K(G n)) k −lim k+n , 0 if k 6= 0 = →G if k = 0. Example 36.15. We have Z if k ≡ 0 mod 2 πk(B) = 0 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 above observation for π0, and Bott periodicity.

Fix E = {(En, εn)} and we’ll use E to construct generalized, reduced, cohomology and homology groups, which we’ll notate as k h (X), hk(X). 134 AARON LANDESMAN

Remark 36.17. For example, we can define these homology and cohomology groups for finite cell complexes, or other reasonable classes of spaces. Remark 36.18. We’ll now give the definitions, which are straightforward, but checking long exact sequence for cohomology is much easier than checking the long exact sequences hold for homology. Definition 36.19. We define the generalized, reduced cohomology groups as hk(X) := [Sn E ] −lim , k+n , where the limit is taken over maps n S → n [S X, Ek+n] − [SS X, SEk+n] h n+1 i S X, Ek+n+1 . → The generalized, reduced homology groups of X, notated hk(X), are h → i h (X) := Sk+n E ∧ X k −lim , n = πk(E ∧ X) Remark 36.20. As we see in the definition,→ a stable homotopy group is an example of a generalized homology theory. k k Remark 36.21. Really, we should notate hk(X) as hk(X, E), and similarly h (X) as h (X, E), since the groups depend on a spectrum E. Each E gives a generalized homology and cohomology theory, both defined in homotopy. Remark 36.22. One could equivalently define h i hk(X) = Sn EX −lim , k+n h i = Sn−k EX −lim , n → X = π−k(E ), where → X X E := En. n Next time, we’ll prove these give the expected long exact sequence and illustrate what these 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) := [SnX E ] −lim , n+k , h n+k i hk(X) := S , X ∧ En=k . → ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 135

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 property 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 sequence on hk is a limit of exact sequences, hence exact. We now deduce an exact sequence, we get

··· hk(SX) hk(SA) (37.4)

hk(X/A) hk(X) hk(A).

Remark 37.3. Recall that this boundary map came from a map X ∪ CA, attached along A, to CX ∪ X ∪ CA, where the cones were glued along X and A. This inclusion induces a map 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) = [SnSA E ] −lim , n+k h i = Sn+1A E −lim , n+k = → [SmA E ] −lim , m+k−1 = h→k−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 generalized homology series. The hk case was done first for spectra that are convergent, meaning there 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. n One can conclude this exact sequence from stabilization of πn+k(S ). Then, one can approximate 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 h (X) = π (SnX) k −lim k+n s = πk(X). For the sphere spectrum, → hk(X) = stable cohomotopy groups h i = SnX Sk+n −lim , k = πs (X). → Example 37.9. The Thom spectra have the following homology theory. We have, h (X) = π (X ∧ τ(E )) k −lim k+n n+k = MSO] k(X) → where En+k is not the spectra but the universal bundle over the grassmannian. Here, we were using the Thom space for the oriented grassmannian, but if we took the Thom space for the unoriented grassmannian, we would have gotten MOg k(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 MacLane spectrum. We have Here, [X, K(k, G)] = Hk(X; G). ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 137 where K(k, G) is the Eilenberg MacLane space. In this way, Eilenberg-MacLane spaces play the role of the classifying spaces for cohomology. We have [SnX, K(n + k, G)] = Hn+k(SnX; G) = Hk(X; G). Then, [SnX K(n + k G)] = Hn(X; G) −lim , , , because the maps between the terms in this spectra are all the expected isomorphisms. s k 37.4. Stable homotopy and stable→ homotopy. We’ll now discuss πk and πs , and see that familiar things like fundamental classes and Poincare duality carry over. Definition 37.12. We have nonreduced versions s Πk(X) = πk X {∗} and Πk(X) is defined as the limit as the limit n a 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 now defined Πk(X) in the context of our generalized homology theories. We have h n+k n i Πk(X) = S , S (X {∗} h i = Rn+k ∪ ,a(Rn × X)+ . If X is compact, this is equal to homotopy classes of proper maps ∞ f Rn+k − Rn × X. For nice spaces X, we can approximate f by a map → f = (f1, f2) with smooth first component n+k n f1 : R R . Taking −1 → n+k M := f1 (p) ⊂ R , M is a stably framed k manifold. In the rest of the class, we my use framed to mean stably framed. Now, restricting f2 to M, 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) where M 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 the fundamental class in Πn(X) be? Here, we need a stably framed manifold, but don’t the need manifold to be compact or oriented. 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 of M = X, → We also have Poincare duality. Recall that by deﬁnition Πk(X) is the limit as n of homotopy 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 k ∼ Π (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 regular value of f. Take M := f−1(p), which has dimension n − k and a trivialized normal bundle ν(M). For M ⊂ 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 product M ⊂ Rm × X with 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.