Math 6530: K-Theory and Characteristic Classes

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1 Math 6530: K-Theory and Characteristic Classes Taught by Inna Zakharevich Notes by David Mehrle Cornell University Fall 2017 Last updated December 12, The latest version is online here.

2 Contents 1 Vector bundles Grassmannians Classification of Vector bundles Cohomology and Characteristic Classes Cohomology of Grassmannians Characteristic Classes Axioms for Stiefel-Whitney classes Some computations Cobordism Stiefel-Whitney Numbers Cobordism Groups Geometry of Thom Spaces L-equivalence and Transversality Characteristic Numbers and Boundaries K-Theory Bott Periodicity The K-theory spectrum Some properties of K-theory An example: K-theory of S Power Operations When is the Hopf Invariant one? The Splitting Principle Where do we go from here? The J-homomorphism The Chern Character and e invariant Student Presentations Yun Liu: Clifford Algebras Sujit Rao: Elementary Bott Periodicity Oliver Wang: Even periodic theories Shruthi Sridhar: Serre Swan Elise McMahon: Equivariant K-theory I Brandon Shapiro: Equivariant K-theory II David Mehrle: KR-theory

3 Contents by Lecture Lecture 01 on 23 August Lecture 02 on 25 August Lecture 03 on 28 August Lecture 04 on 30 August Lecture 05 on 1 September Lecture 06 on 6 September Lecture 07 on 8 September Lecture 08 on 11 September Lecture 09 on 13 September Lecture 10 on 18 September Lecture 11 on 20 September Lecture 12 on 22 September Lecture 13 on 25 September Lecture 14 on 27 September Lecture 15 on 29 September Lecture 16 on 04 October Lecture 17 on 06 October Lecture 18 on 11 October Lecture 19 on 13 October Lecture 20 on 16 October Lecture 21 on 20 October Lecture 22 on 23 October Lecture 23 on 25 October

4 Lecture 24 on 27 October Lecture 25 on 30 October Lecture 26 on 3 November Lecture 27 on 6 November Lecture 28 on 8 November Lecture 29 on 10 November Lecture 30 on 13 November Lecture 31 on 15 November Lecture 32 on 17 November Lecture 33 on 20 November Lecture 34 on 27 November Lecture 35 on 29 November Lecture 36 on 1 December

5 Lecture 01: Vector bundles 23 August 2017 Administrative There is a course webpage here. Office hours are Monday 1-2pm and Friday 2-3pm, but subject to change. There will be approximately four homework sets and a small final project for those who really want or need a grade. Homework must be typed. Inna s notes are on the class webpage, and are more complete than these. 1 Vector bundles What are we studying in this class? Mostly, we ll talk about vector bundles. These seem like geometric objects, but really they re topological objects. They come up most naturally when we talk about geometry of manifolds tangent lines to curves on a manifold don t ever intersect, and indeed they know nothing about one another. The fact that they might look like it is an illusion of the fact that we choose coordinates and go into R n. We should get away from coordinates then, and look at manifolds as intrinsic objects. Using this, we can define tangent spaces at a point. Definition 1.1. A vector bundle on a base space B is is a topological space E (the total space) together with a map p: E B such that for all b B, p 1 (b) has the structure of a vector space, and for all b B, there is a neighborhood U of b and an integer k (the rank) with a homeomorphism φ b : U R k p 1 (U) such that the following diagram commutes, U R k φ b p 1 (U) U and the restriction of φ b to each fiber is a linear homomorphism. Example 1.2. The trivial bundle pr 1 : B R k B is a vector bundle, called the trivial bundle of rank k over B. Remark 1.3. There are really two structures contained in the definition of a vector bundle. One is that of a fiber bundle, where the fibers are allowed to be anything: tori, spheres, or other things without vector space structure. The second thing is the linear structure on the fibers. 4

6 Lecture 01: Vector bundles 23 August 2017 Example 1.4. The tangent bundle to a manifold M embedded in R N is E = { (x, v) x M, v tangent to M at x } this is a subspace of M R N, and the projection onto the first factor is the map p: E M. But we want an intrinsic definition of tangent bundles that doesn t depend on the embedding. Example 1.5. Define the tangent bundle to an n-manifold M by TM = x M T x M as a set, with p: TM M defined in the obvious way. We can induce a topology on this set by the map x U T xm R n R n given by choosing coordinates around x in the first coordinate, and using the appropriate tangent vector in the second. Example 1.6. The normal bundle to an embedded n-manifold M R N is This has rank k = N n. ν = { (x, v) x M, v normal to M } M R N. Example 1.7. How do we construct vector bundles in general? If p: E B is a vector bundle, with homeomorphisms φ α : p 1 (U α ) U α R k φ β : p 1 (U β ) U β R k Then we have the composite homeomorphism φ 1 (U α U β ) R k α p 1 (U α U β ) (U α U β ) R k Restricting to the second coordinate, this gives an element GL k (R) above every point. This gives a smooth map φ β g αβ : U α U β GL k (R). These maps are called the transition functions, and they satisfy three things (a) g αα = id (b) g αβ (x) = g βα (x) 1 5

7 Lecture 01: Vector bundles 23 August 2017 (c) g αβ (x)g βγ (x)g γα (x) = id for x U α U β U γ. Proposition 1.8. Given an atlas {U α } of X and functions g αβ : U α U β GL k (R) satisfying (a) g αα = id (b) g αβ (x) = g βα (x) 1 (c) g αβ (x)g βγ (x)g γα (x) = id for x U α U β U γ, you can construct a vector bundle E = U α R k/ α (x, v) (x, g αβ (x) v) Remark 1.9. We ve been working with the assumption that the rank k is the same everywhere, but this is not required by the definition. We may have different rank in different connected components, but they are the same on the same connected components. We don t usually care about bundles with different ranks on different components though, so we ll almost always assume the rank is uniform. Remark We haven t used any properties of R, but we may use any other field (such as C). So far we haven t conclusively demonstrated that any vector bundles are nontrivial. Let s do that now. Definition An isomorphism of vector bundles over B is a map φ: E E such that (a) φ is an isomorphism, and (b) the diagram below commutes φ = E E p p and on each fiber it is a linear isomorphism. Definition Vect n (B) is the set of vector bundles of rank n over B. Question: What is Vect n (B)? Example If B =, then Vect n (B) = { R n }. B 6

8 Lecture 02: Vector bundles 25 August 2017 Example If B = S 1, then there are several bundles just of rank 1: the tangent bundle TS 1, the trivial bundle S 1 R, and there s also the Möbius bundle. We can construct the Möbius bundle as follows. Take two open sets U and V on S 1 U V Then the Möbius bundle is (U R) (V R), glued on the left hand side by 1 and the right hand side by 1. Definition A section of p: E B is a map s: B E such that ps = id B. Example The zero section s 0 : B E, b (b, 0) sends a point in B to the zero vector in the corresponding fiber. Example If E = B R k is trivial, then there is an everywhere nonzero section: choose any nonzero x R k, and then the section is b (b, x). Lemma The tangent bundle on S 1 is trivial. Proof sketch. We can imagine both of these bundles as a circle with a line on each point, either tangent to the circle (TS 1 ) or normal to the circle (S 1 R). In either case, we can arrange the lines to make a cylinder by either rotating by π/2 in the plane of the circle or rotating by π/2 around a tangent vector. So these bundles look the same. Lemma The Möbius bundle is not trivial. Proof. Vector bundle isomorphisms preserve zero sections. So if E = E then so are E \ s 0 (B) = E \ s 0 (B). Now let E = TS1 = S 1 R and let E be the Möbius bundle. E \ s 0 (S 1 ) is S 1 R \ {0}, which is disconnected, but E \ s 0 (S 1 ) is connected (as we know from slicing a Möbius strip along the middle circle). Definition If a vector bundle has rank 1, we call it a line bundle. Example Let B = RP n. Let γ 1n RP n R n+1 be the bundle γ 1n = {(l, v) v l}. This is the tautological line bundle over RP n. 7

9 Lecture 02: Vector bundles 25 August 2017 Lemma γ 1n has no everywhere nonzero sections. Proof. Notice that RP n = S n /{±1}. A section s: RP n γ 1n is of the form s(±x) = (±x, t(x)x), where we write ±x for the image of the point x S n in RP n = S n /{±1}. We must have that t(x) = t( x), and t: S n R is an odd function, so t must hit zero somewhere. Hence, the section s cannot be everywhere nonzero. Lemma Let L be a line bundle over a base B. If L has an everywhere nonzero section, then it is trivial. Proof. Assume that s: B L is everywhere nonzero. Then define a map f: L B R by (b, v) (b, c), where v = c s(b) for a unique c. This is an isomorphism. Lemma Let f: E E be a map of vector bundles. Then f is an isomorphism if and only if it is a linear isomorphism on each fiber. Proof. Hatcher Lemma 1.1 Proposition p: E B is a trivial vector bundle if and only if there are n sections s 1,..., s n that are linearly independent at each point of b. Proof. First, if E = B R n, then set s i (b) = (b, e i ). Conversely, define f: E B R n by where we write v uniquely as in the basis s 1 (b),..., s n (b). (b, v) (b, (c 1,..., c n )), v = n c i s i (b) Example { ((cos TS 1 ) } = θ, sin θ), ( t sin θ, t cos θ) θ S 1, t R i=1 Define a map S 1 TS 1 by θ ( (cos θ, sin θ), ( sin θ, cos θ) ) This gives an everywhere nonzero section, which defines a basis of each fiber. Hence, TS 1 is trivial. 8

10 Lecture 02: Grassmannians 25 August 2017 Theorem If E B is a fiber bundle with fiber F, then there is a long exact sequence of homotopy groups π n (F) π n (E) π n (B) π n 1 (F) π 1 (F) π 1 (E) π 1 (B) Corollary For a vector bundle, π n (E) = π n (B) since the fibers are all contractible. 1.1 Grassmannians Definition The Grassmannian Gr n (R k ) of n-planes in R k is the space of n-dimensional linear subspaces of R k. Definition The Stiefel manifold V n (R k ) is the set of all orthogonal n- frames in R k. This is a subspace of (S k 1 ) n, and inherits its topology and manifold structure from that space. Fact V n (R k ) and Gr n (R k ) are compact. Proof. Notice that V n (R k ) is a closed subspace of a compact space, and therefore it is compact. There is an action of the orthogonal group O(n) on V n (R k ), and the quotient of V n (R k ) by this action is Gr n (R k ). Hence, Gr n (R k ) is compact. Lemma Gr n (R k ) is Hausdorff. Proof. It suffices to show that for any two n-planes ω 1, ω 2 in Gr n (R k ), there is a function to R which has different values on ω 1 and ω 2. For any point p R k, let f p (ω) be the Euclidean distance from ω to p. For any (v 1,..., v n ) V n (R k ) representing ω, f p (ω) = p p (p v 1 ) 2... (p v n ) 2. So f p is clearly continuous and well-defined as a function Gr n (R k ) R. If p ω 1 \ ω 2, then this gives the required function. Theorem Gr n (R k ) is a manifold. Proof. If ω is an n-plane, let ω be the orthogonal (k n)-plane in R k. Then let } U = {n-planes which do not meet ω This is homeomorphic to the set of graphs of linear maps ω ω, which is the space of n (k n) matrices. This is homemorphic to R n(k n). This gives an atlas on Gr n (R k ). 9

11 Lecture 03: Grassmannians 28 August 2017 The previous theorem also shows that dim Gr n (R k ) = n(k n). Since Gr n (R k ) is Hausdorff, we can try to construct a CW-structure on it. This is relatively simple once we figure out the correct cells to look at. Let p i : R k R i be the projection onto the first i coordinates, so p k is the identity and p 0 is constant. As i goes from k to 0 the dimension of an n-plane ω drops from n 0. Let σ i be the smallest integer j such that dim p j (ω) = j. The sequence σ = (σ 1,..., σ n ) is called a Schubert symbol. If we let e(σ) be the subset of Gr n (R k ) having σ as their Schubert symbol, we notice these are spaces whose n-planes have matrices with columns σ 1,..., σ n holding pivots after row reduction. These are the Schubert cells of Gr n (R k ). Remark Note that this doesn t rely on any properties of R other than that R m is homeomorphic to an open cell of dimension m. Thus we could have done the exact same analysis for Gr n (C k ). We want a Grassmannian of all n-planes, not just those in a particular dimension. We have inclusions Gr n (R k ) Gr n (R k+1 ) Gr n (R k+2 ) Gr n (R ), where R = i=1 R. Each of these inclusions respects the CW-structure on the Grassmannian Gr n (R k ), so we get a CW-structure on Gr n := Gr n (R ). Definition Gr n := Gr n (R ). Definition The universal bundle of n-planes is γ n := γ n, = { (ω, v) ω Gr n, v ω }. This may look like we just made things harder! Gr n and γ n are larger than their counterparts in R k. But for algebraic topologists, Gr n and γ n are much more natural. They have nice topological structure. Definition Let G be a topological group. Then EG is any weakly contractible space with a continuous free G-action and BG is the quotient of EG by this action. BG is called the classifying space of G. Example When G = Z, BZ = S 1 and EZ = R. When G = Z/2, B(Z/2) = RP and E(Z/2) = S. In general, for G discrete, BG is a K(G, 1) space, which means 1 i = 0 π i BG = G i = 1 0 i > 1 10

12 Lecture 03: Grassmannians 28 August 2017 Remark B is one of the most mysterious functors in all of algebraic topology. It s evil insofar as it hides a lot of information, but at the same time it has lots of nice properties. The next theorem illustrates how in the case of O(n), BG has both nice homotopy-theoretic properties and a nice combinatorial description via Gr n. Theorem Gr n BO(n). Proof. First, we claim that EO(n) = V n (R ). To show this will suffice to prove the theorem, because we know that V n (R ) has a free action of O(n), and Gr n is the quotient of V n (R ) by this action. But what we don t know is that V n (R ) is weakly contractible. The map V n (R k ) S k 1 given by projecting an n-frame onto its last vector is a fiber bundle with fiber V n 1 (R k 1 ), considering R k 1 as the hyperplane orthogonal to the last vector. Thus there is a long exact sequence in homotopy π m+1 S k 1 π m V n 1 (R k 1 ) π m V n (R k ) π m S k 1 Since π m S k 1 = 0 for m < k 2, π m V n 1 (R k 1 ) = π m V n (R k ) for m < k 2. By iterating this and takin k large enough, we note that π m V n (R k ) = π m V 1 (R k n+1 ) = π m (S k n ). Thus for k large enough we can show that π m V n (R k ) = 0 for m < k n. Now consider π m V n (R ). An element in this group is a homotopy class of maps S m V n (R ) = k=n V n (R k ). Since S m is compact, this map factors through the inclusion V n (R k ) V n for some k. Assuming k is sufficiently large, we this map factors through V n (R k ). S m V n (R ) V n (R k ) And since the first map is nullhomotopic, then the composite must be as well. Thus, π m V n (R ) = 0. Remark A similar proof shows that Gr n (C ) BU(n). Remark In general, we cannot always factor a map from S m to a colimit through a finite stage, but it works if each map is given by a closed inclusion of Hausdorff spaces. 11

13 Lecture 04: Classification of Vector bundles 30 August Classification of Vector bundles How do we classify vector bundles? We will manage to classify them, but the result will be computationally useless for our purposes. We ll also spend a great deal of time figuring out how to make this result useful. Theorem There is a bijection of sets Vect n (B) = [B, Gr n ]. This result is magical! It gives a geometric classification through homotopical data! We can ignore geometric structure and instead use topological information. We need a few ingredients to prove Theorem Definition A space is paracompact if every open cover has a locally finite subcover. Lemma Given any open cover {U α } of a paracompact space X, there is a countable open cover {V i } such that: (a) for all i, V i is a disjoint union of spaces U α with U α U α. (b) there is a partition of unity {φ i } subordinate to V i. Definition If f: B B is a continuous map and p: E B is a vector bundle, then the pullback bundle of p: E B along f is the categorical pullback f (E) E p f B B. Explicitly, this is the set f (E) = {(b, e) b B, e E, f(b ) = p(e)}. What is the map [B, Gr n ] Vect n (B) in the theorem? It is given by sending the class of f: B Gr n to f γ n. [B, Gr n ] = Vect n (B) [f] [f γ n ] We should check that this map is well-defined. That is essentially the content of the next lemma. Lemma If f, g: X Y are homotopic and p: E toy is any vector bundle, then f E = g E. 12

14 Lecture 04: Classification of Vector bundles 30 August 2017 Proof. Let H: X I Y be a homotopy from f to g. In particular, H X {0} = f and H X {1} = g. Notice that H E is a bundle over X I, and the restriction of this bundle to X {0} is f E and the restriction to X {1} is g E. So it suffices to show that given any bundle E over X I, the restrictions over X {0} and X {1} are isomorphic. Case 1: First, assume that E is trivial. E = X I R k X I Let E 0 be the restriction of E to X {0}, and let E 1 be the restriction of E to X {1}. The required isomorphism between E 0 and E 1 is evident from the following diagram. E 0 = X {0} R k X {0} E 1 = X {1} R k X {1} Now for any map ρ: X [0, 1], let Γ ρ X I be the be the graph of ρ. This same proof as above shows that the restriction of E over Γ ρ is isomorphic to E 0, for any ρ. Let E ρ be the restriction of E to Γ ρ. Case 2: Dispose of the assumption that E is trivial, and instead assume that E is trivial over U I for U X open. Let ρ: X I be any function with support contained in U. Then E ρ = E 0 ; indeed, outside U, E ρ = E 0, and inside U, we can use the isomorphism from the previous case. Case 3: Assume only that E is a vector bundle over X I; no triviality assumptions. By Lemma 1.45, there is a countable open cover {U i } of X such that E is trivial over U i I. Let {φ i } be the subordinate partition of unity. Let with ψ 0 0, and ψ 1. ψ i = i φ j, j=1 13

15 Lecture 04: Classification of Vector bundles 30 August 2017 Claim that E ψ i = E ψj 1. This follows from case 2, because the support of ψ i ψ i 1 is contained within U i. Lemma For any vector bundle p: E B of rank n, the data of a map f: B Gr n such that E = f γ n is equivalent to the data of a map g: E R which is a linear injection on each fiber. Proof. Assume we are given a map f: B Gr n and an isomorphism E = f γ n. We have the data of this diagram: = E f γ n γ n R B f Gr n The map g is the composite along the top row of this diagram. Conversely, given a map g: E R that is a linear isomorphism of each fiber, define f(b) := g(p 1 (b)) Gr n. Note that p 1 (b) is an n-dimensional vector space, as a fiber of p: E B. Then applying g, we get an n-dimensional subspace of R. The vector bundle isomorphism E f γ n is as follows: E e f γ n (p(e), g(e)) Note that g(e) g(p 1 (p(e))) since e p 1 (p(e)). This is an isomorphism because g is a linear injection on the fibers; we can recover e uniquely from p(e) and g(e). Proof of Theorem We will prove both injectivity and surjectivity. Injectivity. Suppose that E = f γ n = (f ) γ n. We want to show that f f. By Lemma 1.48, take g, g : E R corresponding to these maps. Claim that it suffices to show that g g. Why? Suppose that G: E I R is a homotopy from g to g such that G E {t} is a linear injection on fibers for all t [0, 1]. Then we may define a homotopy F: B I Gr n between f and f by It s tempting to define F(b, t) = G(p 1 (b), t) Gr n. G(e, t) = g(e)t + g (e)(1 t), (1.1) 14

16 Lecture 05: Classification of Vector bundles 1 September 2017 but this doesn t necessarily work! This may pass through 0, but if g(p 1 (b)) and g (p 1 (b)) only intersect at 0 for all b, then we re fine. Since we re working in R, we have lots of space, so we can homotope things around. Define homotopies L o ((x 1, x 2,...), t) = t(x 1,...) + (1 t)(x 1, 0, x 2, 0) L e ((x 1, x 2,...), t) = t(x 1,...) + (1 t)(0, x 1, 0, x 2, 0,...) Then we may construct a homotopy from g to g via the following procedure: (1) Homotope g to be in all odd coordinates. (2) Homotope that to g living in even coordinates using (1.1). (3) Homotope from even coordinates back to all coordinates. This shows that the map [B, Gr n ] Vect n (B) is injective. Surjectivitiy. Suppose E is trivial, say E = B R k. Then take by Lemma 1.45 a countable cover {U i } with subordinate partition of unity {φ i }. Then define g i : E R n as follows: above U, take the composite E U U R n R n (b, v) φ i v outside of U, send everything to zero. Then we can define g: E R = (R n ) by e (g 1 (e), g 2 (e),...) By Lemma 1.48, this corresponds to the required f: B Gr n. Definition The classifying map f: B Gr n of an n-dimensional vector bundle p: E B is the preimage of [p] Vect n (B). Remark If instead we want to classify principal G-bundles for some group G over a base X, then the same proof gives a bijection Vect n (X) = [X, BG]. Definition The Whitney Sum of two vector bundles E B and E B is a new vector bundle E E B, which is the direct sum on fibers. Here are two descriptions of the Whitney sum: 15

17 Lecture 05: Cohomology and Characteristic Classes 1 September 2017 (1) as a pullback E B E E E B (2) Using Theorem 1.43, we have an isomorphism between vector bundles over B and homotopy classes of maps B Gr n. If E corresponds to f: B Gr m and E corresponds to Gr n, what does E E correspond to? f f : B B B f f Gr m Gr n Grm+n The map : Gr m Gr n Gr m+n comes from interleaving two copies of R and sending (ω, ξ) Gr m Gr n to the image under the interleaving. 2 Cohomology and Characteristic Classes Definition 2.1. The loop space of a pointed space X is the space ΩX consisting of all loops S 1 X with the weak topology. Definition 2.2. Let Z be a space. We write Z + for the space Z with a disjoint basepoint added, Z + := Z { }. Remark 2.3. Adding a basepoint is often a stupid operation when we re trying to look at maps into a space. For example, pointed maps S n Z + are stuck at the basepoint, so the homotopy groups are all trivial. On the other hand, maps out of Z + are perfectly fine to think about. Definition 2.4. Let CW denote the category of CW-complexes. Remark 2.5. We will not think about pairs of spaces (X, A); instead, we will look at the space X/A, and declare that the image of A under projection X X/A is the basepoint. Definition 2.6. The mapping cone of an inclusion α: A X is Cα = X CA / (x, 1) α(x) Definition 2.7. A generalized cohomology theory is a sequence of functors h n : CW op Ab together with natural suspension isomorphisms such that the following axioms hold. σ i : h i+1 (ΣX) h i (X) 16

18 Lecture 05: Cohomology and Characteristic Classes 1 September 2017 (a) homotopy invariance: If f 1, f 2 : X Y are homotopic, then h i (f 1 ) = h i (f 2 ). (b) exactness: If α: A X is an inclusion β: X Cα, then the sequence is exact. h i (Cα) h i (X) h i (A) (c) additivity: If {X j } is a collection of pointed spaces, then ( ) h i X j = h i (X j ) j Remark 2.8. There is an additional axiom, known as the dimension axiom. { (d) dimension: h i (S 0 ) = Z i = 0 0 otherwise But we don t include it because there is exactly one cohomology theory that satisfies all axioms (a)-(d): ordinary (singular) cohomology. If you ve seen cohomology before, you are probably wondering where the long exact sequence comes from. The axioms only have three-term sequences which are exact at the middle. But that s enough to reconstruct the long exact sequence. Consider the sequence of maps α β γ A X Cα Cβ Notice that Cα X/A, and Cβ ΣA, and Cγ ΣX. Then using the suspension isomorphisms, we have the long exact sequence of cohomology. h i (ΣX) h i (ΣA) h i (X/A) h i (X) h i (A) = σ i 1 = σ i 1 j h i 1 (X) h i 1 (A) Theorem 2.9. Suppose that we are given a sequence X 0, X 1,... of pointed spaces and weak equivalences X i ΩXi+1 for all i. Then the sequence of functors h n : CW op Ab defined by { [Y, h n Xn ] n 0 (Y) = [Y, Ω n X 0 ] n < 0 is a generalized cohomology theory. 17

19 Lecture 06: Cohomology and Characteristic Classes 6 September 2017 Remark The converse of this theorem actually holds as well; it s called the Brown Representability Theorem. Example Let X i = K(Z, i) be an Eilenberg-MacLane space. Then H i (Y) = [Y, K(Z, i)] { Z i = 0 H i (S 0 ) = [S 0, K(Z, i)] = π 0 K(Z, i) = 0 otherwise More generally, we can replace Z by any discrete group G to get singular cohomology with G-coefficients. Proof of Theorem 2.9. First, let s define the suspension isomorphism. We want to show that h n+1 (ΣY) = h n (Y). We have that h n+1 (ΣY) = [ΣY, X n+1 ] = [Y, ΩX n+1 ] = [Y, X n ] = h n (Y) We must also check the three axioms of a generalized cohomology theory. Homotopy invariance is clear, because we are only dealing with homotopy classes of maps. Additivity follows from the universal property of the product. It remains to check exactness. Consider where We want to show that α Y Z Cα, Cα := Z α CY = Z Y I/ (y, 0) (y, 0) (y, 1) α(y). [Cα, X n ] [Z, X n ] [Y, X n ] is exact at the middle. To show that the composite is zero, suppose that we are given f: Cα X n. Then f Y is null-homotopic, with a null-homotopy f CY : Y I X n, which is constant on Y {0}. Conversely, let f: Z X n be such that f Y is null-homotopic. Then there exists h: Y I X n such that h Y {0} is constant and h Y {1} = f Y. Define g: Cα X n by { g(z) = f(z) z Z g(y, t) = h(y, t) (y, t) Y I. Then g is a map whose image under [Cα, X n ] [Z, X n ] is f. 18

20 Lecture 06: Cohomology of Grassmannians 6 September 2017 Remark Our goal is to use this to understand [B, Gr n ]. As remarked before, however, this is hopeless. But, we understand [B, K(Z, i)] and [Gr n, K(Z, i)]. Then given [f] [B, Gr n ], we get a map f : [Gr n, K(Z, i)] [B, Gr n ]. [Gr n, K(Z, i)] f [B, K(Z, i)] = = H i (Gr n ) H i (B) Although we have no hope of computing homotopy classes of maps [B, Gr n ] even for spheres, we do know the cohomology of spheres! The moral is that these maps give invariants of vector bundles, which we can compute. These are called characteristic classes. The strategy is (a) compute H (Gr n, Z/2), (b) compute im f as an invariant of E B, (c) hope that this retains useful information. 2.1 Cohomology of Grassmannians Example Let s first consider the case n = 1, where Gr 1 = RP. This has another useful description of RP as a quotient of S by a Z/2 action. S has a cells structure with two zero-cells, two one-cells, two two-cells, etc. The action of Z/2 = {±1} switches the cells in each dimension. Depending on whether or not the dimension is even, the action of Z/2 switches the orientation when it swaps the cells. Hence, the dimension of the boundaries is 2 in even dimensions and 0 in odd dimensions. But with Z/2 coefficients, the dimension of the boundaries is always zero. Hence, in each dimension, and as a ring, with x in degree 1. H (RP, Z / 2 ) = Z / 2 H (RP, Z / 2 ) = Z / 2 [x], We could theoretically do a similar thing using the Schubert cell structure on Grassmannians, but that uses a lot of (really cool) combinatorics that we don t have time for. So we ll do something harder. Remark The next theorem is one of those theorems whose proof is less useful than its consequences. 19

21 Lecture 07: Cohomology of Grassmannians 8 September 2017 I used to think that proofs were important and theorems were just made up, but now I think that theorems are important and proofs are just made up. Mike Hopkins (paraphrased) Definition Let B be a paracompact space, and p: E B a vector bundle over B. Let D(E) be the unit disk bundle, and S(E) the sphere bundle (boundary of the disk bundle). Then define the Thom space of E Th(E) := D(E)/ S(E) For general B, we define Th(E) as the one-point compactification of E. Example Let E = B R n. Then Th(E) = B + S n. Theorem 2.17 (Thom Isomorphism Theorem). Let p: E B be an n-dimensional fiber bundle. There exists a natural class c H n (Th(E), Z/2) such that the restriction of c to any fiber F is a generator of H n (S n ) and the map H i (B +, Z/2) b Φ H i+n (Th(E), Z/2) p (b) c is an isomorphism for all i. (This is not a ring map!) Definition The class c in the Thom Isomorphism Theorem is called the Thom class. Remark The Thom class c of a bundle p: E B is natural in the following sense: given f: B B, the Thom class of f (E) is f (c), where f : H n (Th(E); Z/2) H n (Th(E); Z/2). Theorem As a ring, with w i in degree i. H (Gr n+, Z/2) = Z / 2 [w 1,..., w n ] Remark If we have an oriented bundle, then the Thom isomorphism theorem holds with Z coefficients. The problem with unoriented bundles is that we struggle choose generators. Remark In algebra, we often add together elements of different degrees. But in topology, the different degrees in a cohomology ring come from different dimension. So adding elements of different degrees is weird and doesn t quite make sense. Yet we do it anyway when we define total Chern classes and total Whitney classes. Allen Knutson calls these abominations. If we re topologists, we try to avoid working with elements of mixed degree. So we won t talk about total Chern classes or total Whitney classes here. 20

22 Lecture 07: Cohomology of Grassmannians 8 September 2017 Definition Let p: E B be a vector bundle and let c be the Thom class of E. Let j: D(E) Th(E) be the projection from the disk bundle onto the Thom bundle. Define e := (p ) 1 (j c) H n (B; Z/2). We will call this the Z/2-Euler class (this is not standard terminology!). Remark Like the Thom class, the Z/2-Euler class is natural in the sense that given a vector bundle p: E B with Z/2-Euler class e and a function f: B B, the Z/2-Euler class of f (E) is f (e). Remark Here is a geometric description of the Euler class. Let p: E B be a vector bundle where B is a manifold. Let s: B E be a generic section. Let ζ be the zero section. Then s(b) ζ(b) is the Poincaré dual of e. Therefore, a nonzero Euler class implies that there do not exist everywhere-nonzero sections, and hence the bundle is nontrivial. Lemma For γ n Gr n, the Z/2-Euler class is nonzero. Proof. By Remark 2.24, it suffices to find any bundle with nonzero Z/2-Euler class. If we find such a bundle, it will be a pullback of the universal bundle by the classification of vector bundles (Theorem 1.43). Therefore, a bundle with a nonzero Z/2-Euler class shows that the Z/2-Euler class of the universal bundle is nonzero. We will show that the universal bundle γ n,n+1 Gr n (R n+1 ) is nontrivial. Notice that Gr n (R n+1 ) = Gr 1 (R n+1 ) = RP n by taking the orthogonal compliment of any n-plane. Now consider the isomorphism RP n = S n /{±1}. We may think of γ n,n+1 as the quotient of a bundle over S n by ±1. In particular, the identification is (x, v) ( x, v). We must show that this bundle over RP n has a nontrivial Z/2-Euler class. By Remark 2.25, it is enough to find a section of this bundle that intersects the zero section transversely. Then the Poincaré dual of that will be the Z/2-Euler class. The section we choose is s: RP n γ n,n+1 that takes a point x to the projection of (1, 0,..., 0) onto x. This is zero only at (1, 0,..., 0) and at ( 1, 0,..., 0) in RP n, so intersects the zero section transversely. Hence, it defines a nonzero element of H 0 (RP n ; Z/2) = H 0 (Gr n (R n+1 ); Z/2), and its Poincaré dual is a nonzero element of H n (Gr n (R n+1 )), and therefore the Z/2-Euler class of γ n,n+1 is nonzero. 21

23 Lecture 07: Cohomology of Grassmannians 8 September 2017 Definition We have Th(E) = D(E)/S(E). This yields a long exact sequence in cohomology H i j (Th(E); Z/2) H i (D(E); Z/2) H i (S(E); Z/2) H i+1 (Th(E); Z/2) Now apply the Thom Isomorphism Theorem (Theorem 2.17), and notice moreover that H i (B) = H i (D(E)) since B D(E). j H i (Th(E); Z/2) H i (D(E); Z/2) H i (S(E); Z/2) H i+1 (Th(E); Z/2) Φ = = e H i n (B + ; Z/2) H i (B; Z/2) H i (S(E); Z/2) H i n+1 (B; Z/2) The bottom row here is called the Gysin sequence. Proof of Theorem Proof by induction on n, using the Gysin sequence for the universal bundle γ n Gr n. If n = 0, Gr 0 is a point, and H (Gr 0 ; Z/2) is a polynomial ring on zero generators. If n > 0, assume that H (Gr n 1 ; Z/2) = Z/2[w 1,..., w n 1 ]. The sphere bundle on the universal bundle γ n is S(γ n ) = {(ω, v) v ω, v = 1} There is a natural projection p : S(γ n ) Gr n 1 given by S(γ n ) (ω, v) p Gr n 1 ω v This defines a fiber bundle with fiber S consisting of all unit vectors orthogonal to ω v. Since S is contractible, p induces isomorphisms on all homotopy groups by Theorem 1.27 and therefore also on cohomology rings: The diagram Gives a ring homomorphism H (S(γ n ); Z/2) = H (Gr n 1 ; Z/2). p Gr n S(γ n ) Gr n 1 η: H (Gr n ; Z/2) H (S(γ n ); Z/2) = H (Gr n 1 ; Z/2) (2.1) So we may replace the terms H i (S(γ n ); Z/2) in the Gysin sequence for the universal bundle γ n Gr n, to get the following sequence. H i n e (Gr n+ ; Z/2) H i η (Gr n ; Z/2) H i (Gr n 1 ; Z/2) H i n+1 (Gr n+ ; Z/2) 22

24 Lecture 08: Characteristic Classes 11 September 2017 For i < n 1, the ring map η is an isomorphism because the groups H i n+1 (Gr n+ ; Z/2) and H i n (Gr n+ ; Z/2) vanish. This moreover means that for each generator w j H (Gr n 1 ; Z/2), there is a unique w j H (Gr n ; Z/2) such that η(w j ) = w j for j < n 1. For i = n 1, the map H 0 (Gr n ; Z/2) e H n (Gr n ; Z/2) is injective by Lemma Therefore, we have the following diagram. 0 e 0 H n 1 (Gr n ) H n 1 (Gr n 1 ) H 0 (Gr n+ ) H n (Gr n ) = Z/2 1 e = 0 This shows that η is an isomorphism in degree n 1. This means that there must be some w n 1 H (Gr n ; Z/2) such that η(w n 1 ) = w n 1 H (Gr n 1 ; Z/2). Now because H (Gr n 1 ; Z/2) is generated by w 1,..., w n 1 as a ring and η is a ring homomorphism, it must be surjective in each degree. Hence, the Gysin sequence splits into short exact sequences for all i: 0 H i n e (Gr n+ ) H i η (Gr n ) H i (Gr n 1 ) 0. So define w n = e H n (Gr n ). Claim that w 1,..., w n are generators for H (Gr n ; Z/2) as a polynomial ring. To show this, it suffices to show that for all i, every element of H i (Gr n ; Z/2) can be uniquely written as a polynomial in w 1,..., w n. For i < n, this follows because η is an isomorphism H i (Gr n 1 ; Z/2) = H i (Gr n ; Z/2). For i n, we proceed by induction. Let x H i (Gr n ; Z/2). Then η(x) H i (Gr n 1 ; Z/2) is polynomial in w 1,..., w n 1. Since ker(η) = im( e), we may write x = p(w 1,..., w n 1 ) + w n y for y H i n (Gr n ; Z/2). By induction, y is polynomial in w 1,..., w n, and therefore x is as well. So any element of H (Gr n ; Z/2) may be written as a polynomial in w 1,..., w n. 2.2 Characteristic Classes Definition A characteristic class for n-dimensional real vector bundles is a function ξ assigning to each vector bundle E p B an element ξ(e) H i (B; Z/2) for some i such that: (a) ξ(e) depends only on the isomorphism class of E; 23

25 Lecture 08: Characteristic Classes 11 September 2017 (b) for any f: B B, ξ(f (E)) = f (ξ(e)). Lemma Characteristic classes for n-dimensional real vector bundles correspond to (homogeneous) elements of H i (Gr n ; Z/2). Proof. First, given a characteristic class ξ, we get an element ξ(γ n ) of H i (Gr n ; Z/2). On the other hand, given a cohomology class c H i (Gr n ; Z/2), we get a characteristic class ξ defined by ξ(e) := f (c), where f: B Gr n is the classifying map for E B. One can check that this is a bijection. Here is another proof of Lemma Proof. If we identify n-dimensional real vector bundles with their classifying maps, then we may think of such bundles as the collection of morphisms represented by the functor [, Gr n ]. In this situation, a characteristic class is a natural transformation [ ; Gr n ] = H ( ; Z/2), which corresponds by the Yoneda lemma to an element of H (Gr n ; Z/2). Recall that H (Gr n ; Z/2) = Z/2[w 1,..., w n ] with w i in degree i. In the course of the proof, we constructed in equation (2.1) a map η: H (Gr n ; Z/2) H (Gr n 1 ; Z/2) that is an isomorphism in degrees less than n. This is the map of polynomial rings that evaluates the last generator w n at 0. η Z/2[w 1,..., w n ] Z/2[w 1,..., w n 1 ] w n 0 We also canonically defined w n = e when demonstrating Theorem 2.20 by induction. Definition The Stiefel-Whitney classes are the ones associated via Lemma 2.29 to the generators w i of H (Gr n ; Z/2). They are written w i (E) for a vector bundle E B. Remark Since all elements of H (Gr n ; Z/2) are polynomial in the w i, it follows from Lemma 2.29 that we can learn everything about characteristic classes by studying the Stiefel-Whitney classes. Usually, when characteristic classes are introduced, they are given with four axioms. Here, we will prove these axioms. 24

26 Lecture 08: Characteristic Classes 11 September 2017 Lemma Given f: B B, w i (f E) = f (w i (E)). Lemma For any vector bundle E B, w i (E ε k ) = w i (E), where ε k is a trivial bundle of rank k. Proof. It suffices to show this for k = 1. Let f: B Gr n be the classifying map for E B. Consider E ε 1 γ n ε 1 B f Gr n By Lemma 2.29 and the pullback property of characteristic classes, we have that On the other hand, w i (E ε 1 ) = w i (f (γ n ε 1 )) = f (w i (γ n ε 1 )) w i (E) = w i (f (γ n )) = f (w i (γ n )). So it suffices to show that w i (γ n ε 1 ) = w i (γ n ). To compute w i (γ n ε 1 ), we first need to find its classifying map g: Gr n Gr n+1. Then w i (γ n ε 1 ) := g (w i ) H (Gr n ; Z/2) w i (γ n ) := w i H (Gr n ; Z/2) Note that in the first line, w i is the polynomial generator of H (Gr n+1 ; Z/2), and on the second line, w i is the polynomial generator of H (Gr n ; Z/2). If g = η: H (Gr n+1 ; Z/2) H (Gr n ; Z/2), then we re done. This is what we claim. To that end, recall that η constructed as follows. The map S(γ n+1 ) (ω, v) Gr n ω v defines a fiber bundle with fiber S ; since S is contractible, the long exact sequence of homotopy gives an isomorphism H (S(γ n+1 ); Z/2) = H (Gr n ; Z/2). Then η is the composite of this map with the map induced on cohomology by S(γ n+1 ) D(γ n+1 ) Gr n+1. 25

27 Lecture 08: Characteristic Classes 11 September 2017 Altogether in one diagram, the map on cohomology induced by the following determines η. i Gr n+1 D(γ n+1 ) S(γ n+1 ) Gr n (ω, v) ω v To see that g = η, let s verify that the pullback of γ n+1 Gr n+1 along g gives γ n ε 1. This involves two pullbacks we first pull back γ n+1 Gr n+1 along the map S(γ n+1 ) Gr n+1 and show that it splits as t t with t trivial, and then show that t is isomorphic to the pullback of γ n Gr n along the weak equivalence S(γ n+1 ) Gr n. First, pull the universal bundle γ n+1 Gr n+1 back to a bundle Ẽ over S(γ n+1 ). An element of Ẽ looks like (ω, v, u) with v, u ω and v a unit vector. γ n+1 Ẽ i Gr n+1 D(γ n+1 ) S(γ n+1 ) This has a section t: S(γ n+1 ) Ẽ given by (ω, v) (ω, v, v) Because v is a unit vector, this is an everywhere nonzero section. Therefore, Ẽ contains a trivial bundle given by the image of t. So we decompose Ẽ = t t. Note that elements of t look like (ω, v, u) with u v and v a unit vector. Now remains to see what the pullback of the universal bundle γ n Gr n along S(γ n+1 ) Gr n looks like. We hope it looks like t. t t γ n S(γ n+1 ) Gr n Given (λ, w) γ n, the pullback along the map S(γ n+1 ) Gr n looks like (λ, v, w) with w v λ. This is exactly t. Now, the pullback of γ n all the way along g: Gr n Gr n+1 is γ n ε 1. This shows that g = η. We now know that Therefore, w i (g (γ n+1 )) = ηw i (γ n+1 ). η w i (γ n+1 ) = { 0 if i n + 1 w i = w i (γ n ) if i n 26

28 Lecture 09: Axioms for Stiefel-Whitney classes 13 September 2017 We will write the Künneth Theorem down here, because we will need it. Theorem 2.34 (Künneth). For cohomology with coefficients in a field k, H (X Y; k) = H (X; k) k H (Y; k). 2.3 Axioms for Stiefel-Whitney classes Theorem 2.35 (Whitney Sum Formula). w i (E E ) = j+k=i w j (E) w k (E ) Proof. We will first prove this for Grassmannians, as usual. Consider the bundle γ m γ n Gr m Gr n. The classifying map of this bundle is : Gr m Gr n Gr m+n We want to compute (w i ), which is by Lemma 2.29 the i-th Whitney class of the sum γ m γ n. Proof by induction on m + n. The base case is trivial. For m + n > 0, let g n : Gr n 1 Gr n be the map that induces η: H (Gr n ; Z/2) H (Gr n 1; Z/2), where η(w i ) = w i for i < n and η(w n ) = 0. We know that H (Gr n Gr m ; Z/2) = Z/2[w n1,..., w nn ] Z/2[w m1,..., w mm ] Moreover, w i is some polynomial in the w ij, say (w i ) = q i (w m1,..., w nn ). Now consider g m 1: Gr m 1 Gr n Gr m Gr n. Evaluating on w i, we have (g m 1) w i = q i (w m 1,1,... w m 1,m 1, 0, w n1,..., w nn ) (2.2) On the other hand, g m is the classifying map of the bundle γ m 1 ε 1. Therefore, by Lemma 2.29, we have (g m 1) w i = w i (γ m 1 ε 1 γ n ) 27

29 Lecture 09: Axioms for Stiefel-Whitney classes 13 September 2017 Now by Lemma 2.33, this is And then by induction, we have w i (γ m 1 γ n ) = w i (γ m 1 γ n ) j+k=i Equating (2.2) and (2.3), we have q i (w m1,..., w nn ) Analogously, q i (w m1,..., w nn ) j+k=i j+k=i w j (γ m 1 ) w k (γ n ) (2.3) w mj w nk (mod w mm ). w mj w nk (mod w nn ). So by the Chinese Remainder Theorem, q i (w m1,..., w nn ) = w mj w nk (mod w mm w nn ). j+k=i If i < m + n, this congruence must be equality because w mm w nn has grading m + n. If i > m + n, w i = 0, so its pullback must also be zero, and the formula on the right is zero as well. The only case that remains to check is when i = m + n. That is, we must check that (w m+n ) = w mm w nn. Notice that w m+n is the Z/2-Euler class of γ m+n, w mm is the Z/2-Euler class of γ m, w nn is the Z/2-Euler class of γ n. This equality is true for Thom classes by the Künneth theorem, since H ( D(E E ) ) / S(E E ); Z / 2 = H ( ) ( D(E) / S(E) ; Z / 2 H D(E ) ) / S(E ); Z / 2 This implies that it is also true for Z/2-Euler classes, since the Z/2-Euler class is a pullback of the Thom class. Hence, we have shown that w i (γ m γ n ) = w mj w nk. j+k=i Now let E, E be any two bundles over B, of dimensions m and n. Assume that E and E are classified by maps f, f. Now consider the pullback diagram E E γ m+n f f B B B Gr m Gr n Gr m+n 28

30 Lecture 09: Axioms for Stiefel-Whitney classes 13 September 2017 Then w i (E E ) = (f f ) w i = (f f ) w mj w nk = j+k=i j+k=i w j (E) w k (E ) H (B B; Z/2) = w i (E) w k (E ) H (B; Z/2) j+k=i Remark Another way to see the that the product of Euler classes is again an Euler class, at least for manifolds, is to use the fact that the Euler class is Poincaré dual to the intersection of a generic section with the zero section. Given sections ψ: Gr m γ m of γ m, and φ: Gr n γ n of γ n, this gives a section ψ φ of γ m γ n. The product of the intersection of ψ with the zero section and the intersection of φ with the zero section is equal to the intersection of ψ φ with the zero section of γ m γ n. Remark There is another easier proof of the Whitney sum formula that uses the splitting principle to reduce the proof to the case of line bundles. From there, the only ingredient is the Thom isomorphism theorem. Let s summarize what we know about Stiefel-Whitney classes. Theorem 2.38 ( Axioms for Stiefel-Whitney Classes). (1) For every j 0, there is a Stiefel-Whitney class w j (E) H j (B; Z/2), with w 0 (E) = 1 and w j (E) = 0 if j is larger than the rank of E. (2) Given any map f: B B, w j (f (E)) = f (w j (E)) (3) w i (E E ) = w j (E) w k (E ) j+k=i (4) For γ n Gr n, w n (γ n ) = 0. These are often taken as the axioms for Stiefel-Whitney classes, and in fact characterize them uniquely. We will prove this later once we ve discussed the splitting principle. To see the utility of these axioms, let s prove Lemma 2.33 using them. Lemma 2.39 (Lemma 2.33, repeated). For any vector bundle E B, w i (E) = w i (E ε k ), where ε k is a trivial bundle of rank k. 29

31 Lecture 10: Some computations 18 September 2017 Proof. Claim that w i (ε k ) = 0 for i > 0 or w i (ε k ) = 1 for i = 0 over any base. Given a trivial bundle ε k B, the classifying map factors through a point ε k γ k r B Gr k Therefore, w i (ε k ) = r (w i (ε k 0)) = { 1 i = 0 0 else. Now by the Whitney Sum Formula, w i (E ε n ) = w j (E) w k (ε n ) = w i (E). j+k=i 2.4 Some computations In this section, we will only use the four axioms of Stiefel-Whitney classes that we proved previously in Theorem Proposition If E = E, then w i (E) = w i (E ) for all i. Proof. If E = E, then their classifying maps are homotopic. Proposition For all i > 0 and any base B, w i (ε k ) = 0. Proof. Consider the pullback diagram. ε k ε k B pr { } This shows that w i (ε k B) = pr w i (ε k { }). And w i (ε k { }) H i ({ }) = 0. Proposition If E is a rank n bundle over a paracompact base B with an everywhere nonzero section, then w n (E) = 0. Moreover, if E has k everywhere independent sections, then w n (E) = w n 1 (E) = = w n k+1 (E) = 0. 30

32 Lecture 10: Some computations 18 September 2017 Proof. Let s 1,..., s k be everywhere independent sections of E. Let E be the span of s 1,..., s k ; it is a trivial bundle of rank k: E = ε k. Let E be the orthogonal compliment of E, so E = (E ) (this is where we use paracompactness). Then E = E E = ε k E. Now we may use the Whitney sum formula w i (E) = w j (ε k ) w k (E ) = w i (E ). j+k=i Note that E has rank n k. Therefore, w i (E) = 0 if i > n k. This gives a bound on the number of possible linearly independent sections of E. Proposition For every k, there exists a unique polynomial q i such that whenever E E = ε n, w i (E ) = q i (w 1 (E), w 2 (E),..., w i (E)). Proof. By induction on i. When i = 0, w 0 (E ) = 1. When i = 1, w 1 (ε n ) = w 0 (E) w 1 (E ) + w 1 (E) w 0 (E ) But w 1 (ε n ) = 0 by Proposition 2.41, and w 0 (E) = w 0 (E ) = 1. Hence, we have w 1 (E ) = w 1 (E) = w 1 (E), where the last equality holds because we work mod 2. Now suppose that q 0,..., q i 1 exist. Then 0 = w i (ε n ) = w k (E) w j (E ) k+j=i = w i (E ) + k+j=i j<i = w i (E ) + Then by rearranging, we have k+j=k j<i w i (E ) = q i (w 1 (E),..., w k (E)) := w k (E) w j (E ) w k (E) q j (w 1 (E),..., w j (E)) k+j=i j<i w k (E) q j (w 1 (E),..., w j (E)) Definition We write w i (E) for q i (w 1 (E),..., w i (E)). These are called many things, among them dual/orthogonal/normal Stiefel-Whitney classes. 31

33 Lecture 10: Some computations 18 September 2017 Definition The total Stiefel-Whitney class of a bundle is w(e) := w 0 (E) + w 1 (E) + w 2 (E) +... H (Gr n ; Z/2). This is well-defined because for j > rank(e), w j (E) = 0. Remark We call total Stiefel-Whitney classes an abomination because they sum elements of mixed degree, and therefore doesn t have a clear geometric interpretation. Using the total Stiefel-Whitney class, we can rewrite the Whitney sum formula as w(e)w(e ) = w(e E ). In the case of Proposition 2.43, the dual Stiefel-Whitney classes of E are the coefficients of the inverse power series of w(e), so w(e)w(e) = 1. Sometimes, it is convenient to use an abomination. The following is a consequence of Proposition Lemma 2.47 (Whitney Duality Theorem). Let TM be the tangent bundle to M R N, and let ν be the normal bundle. Then w i (ν) = w i (TM). Remark Note that w i (TM) is independent of the embedding! Hence, the class of a normal bundle is independent of the embedding of M into R N for some N. In fact, we need an embedding TM R N, but instead only an immersion; the tangent bundle doesn t notice if two places far apart map to the same place in R N, only that the tangent space is locally nicely included in R N. This can give us bounds on the dimension N into which we can immerse a manifold. Proposition Let E B be a bundle, and assume that B is compact. Then there is some E such that E E = ε N for some N. Proof. As in Lemma 1.48, it suffices to construct g: E R N that is a linear injection on fibers. For each x B there is some U x such that p 1 (U x ) = U x R n. By Urysohn s Lemma there is a map φ x : B [0, 1] which is 0 outside of U x and nonzero at x. Then {φ 1 x (0, 1]} is an open cover of B. Since B is compact, there is a finite subcover U 1,..., U m defined by φ 1,..., φ m. These define maps g 1,..., g m : U i R n defined by g i : (x, v) φ i (x)v. 32

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