ALGEBRAIC TOPOLOGY: MATH 231BR NOTES

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1 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES AARON LANDESMAN CONTENTS 1. Introduction /25/ Overview Vector Bundles Tautological bundles on projective spaces and Grassmannians Operations on vector bundles /27/ Logistics Constructions with Vector Bundles Grassmannians and the universal bundle /29/ /1/ Characteristic Classes Leray-Hirsch Theorem /3/ Review /5/ /10/ Reviewing Leray-Hirsch Review of Chern and Stiefel-Whitney Classes Examples and Calculations /12/ Logistics Applications of Stiefel-Whitney classes Pontryagin Classes /17/ Theory on Pontryagin classes Calculations and Examples with Pontryagin Classes Euler Classes /19/ Review and Thom Classes Euler Classes Examples of Thom and Euler Classes /22/ More on Euler Classes K Theory /24/

2 2 AARON LANDESMAN Examples for K theory Reduced K theory Products /26/ Equivalent definitions of relative K theory Back to Smash Products /29/ Review Fredholm operators and index /2/ More on Fredholm operators The index of a family More examples of Fredholm operators /4/ Toeplitz Operators /7/ /9/ Review of infinite dimensional groups Real K theory Symplectic K theory /11/ Chern character /21/ Clifford Algebras and Clifford Modules /23/ How to obtain the Z/2-grading Bundles of Clifford modules /25/ Clifford Modules /28/ /30/ Review Calculating ^A Relating our computations to ^A /1/ /4/ Formalities Notations The exact sequence for K-theory in negative degrees /6/ Review Bordism and cobordism /8/ Oriented cobordism 107

3 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES /11/ Review /13/ Oriented bordism as a homology theory /15/ Review /18/ Review The relations between framed cobordism and homotopy groups /20/ Stabilization J-homomorphism /22/ Low stable homotopy groups More about Π Another integrality property for ^A /25/ Spectra More involved examples of spectra Fundamental groups of spectra /27/ Review Constructing the long exact sequence for cohomology Examples of cohomology theories for spectra Stable homotopy and stable homotopy 137

4 4 AARON LANDESMAN 1. INTRODUCTION Peter Kronheimer taught a course (Math 231br) on algebraic topology and algebraic K theory at Harvard in Spring 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 mathematics 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 suggestions to aaronlandesman@gmail.com. 1 This introduction has been adapted from Akhil Matthew s introduction to his notes, with his permission.

5 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES /25/ 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 zahariuc@math.harvard.edu (2) There will be slightly less than weekly homework. Homework 1 is up already 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) vector space structure on each fiber E x := φ 1 (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 U U with U X, the restriction of E to U, notated E U := φ 1 (U) E is trivial. Here, trivial means there exists a homeomorphism φ (2.1) E U U φ U φ id U R n U π or to U C n for the complex case, where is linear. φ Ex : E x {x} R n

6 6 AARON LANDESMAN Example 2.5. (1) The trivial vector bundle E = X R n with φ : X R n X the projection. (2) The Möbius vector bundle on S 1. Take Ẽ = I R and I = [0, 1]. Then, take E = Ẽ/ with the equivalence relation with (0, t) (1, t). (3) The tangent bundle of S n. Recall { } S n = x R n+1 : x = 1. Define TS n := { } (x, v) : x S n, v R n+1, x v = 0. We certainly have a projection map TS n S n where the fibers are vector spaces (TS n ) x = x R n+1. We shall now check local triviality, which we shall often not do in the future. Consider We have an inclusion Over U this is trivial with U := {x : x S n, x n+1 > 0}. TU TS n TU U R n (x, v) (x, π(v)) with π : R n+1 R n = { x R n+1 : x n+1 = 0 }. (4) If X R n is a smooth manifold, then TX = {(x, v) X R n : x X, v R n, 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 = id X. Lemma 2.7. The Möbius bundle is not isomorphic to the trivial vector bundle on the circle. We give two proofs. Proof 1. The Möbius 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 Möbius 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 (2.2) E p φ q F X

7 commutes, where φ is a homeomorphism and is a linear isomorphism. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 7 φ Ex : E x F x 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. Definition 2.9. Let CP n, or P n C { denote } x : x C n+1 : x is a 1 dimensional linear subspace Definition The tautological bundle over CP n be the bundle L := {(x, v) : x CP n, v x} where x is thought of simultaneously as a 1-dimensional space in C n+1 and a point of CP n. The projection is p : L CP n (x, v) x Exercise Define RP n and define the tautological bundle, showing it is a bundle. Exercise Show RP 1 = S 1. Show the tautological bundle π : L RP 1 = S 1 is isomorphic to the Möbius bundle. In particular, it is nontrivial. Definition A line bundle is a vector bundle with 1-dimensional fibers. Example The tangent bundle to S 2, TS 2 is nontrivial. One can see this by noting that if one removes the zero section from TS 2, one obtains a space which has the homotopy type of RP 3, which has fundamental group Z/2Z. In fact, by the Hairy ball theorem, TS 2 has no nowhere vanishing sections. Theorem The tangent bundle TS n 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 We define the Grassmannian { } G n (R N ) = x : x R N, x is a linear subspace with dim x = n. Exercise Show RP n = G 1 (R n+1 ). Definition We have a tautological bundle over the Grassmannian φ : E n G n (R N ) where E n := { } (x, v) G n (R N ) R N : v x

8 8 AARON LANDESMAN 2.4. Operations on vector bundles. We now discuss how to make new vector bundles from old ones. Definition Given p 1 : E 1 X p 2 : E 2 X we can form their direct sum E = E 1 E 2 defined by a fiber product (2.3) We may note that for all x X, we have is a family of vector spaces over X. Definition Given two bundles E E 1 E 2 X. E x = (E 1 ) x (E 2 ) x p 1 : E 1 X p 2 : E 2 X we can form the tensor product bundle E 1 E 2 defined as follows. If we have an open set U X with E 1 U and E 2 U both trivial, then writing both as E i U = U R n i, we make E U = U (R n 1 R n 2 ) and then make the topology on E compatible with these identifications. This has the property that the fiber E x = (E 1 ) x (E 2 ) x. Remark Soon, we ll see a cleaner way to define the tensor product of two bundles. 3. 1/27/ 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 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 E x and a projective bundle P(E x ). As a set, we can take PE = x X PE x. Better, we can define PE = (E \ 0) /k

9 where 0 is the image of the zero section in E. ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 9 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 φ U. (3.1) φ U P U U F φ PU π 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, x is an inner product on each E x varying continuously with x. In other words, in a local trivialization, where E is a product U R n, the matrix entries of the inner product are continuous functions of x. Lemma 3.6. Inner products exist on E X if X admits partitions of unity. Proof. Omitted, see, for example, math 230a notes. Theorem 3.7. Partitions of unity exist on X if X is paracompact and Hausdorff. For example, these exist if X is a CW complex. Proof. Omitted. See, for example, math 230a notes. Example 3.8. We have the sphere bundle S(E) = {v E : v, v = 1}. Definition 3.9. Suppose we have a fiber bundle p : P X and a topological group H acting so that (1) (3.2) P H P P H p µ P X X That is, H acts on each fiber P x. (2) On each fiber P x, the group H acts simply transitively. Then, p : P X is a principal H bundle Example Suppose we have E X a real vector bundle. Let P x be the set of all linear isomorphisms ψ : R n E x. This is is a principal GL(n, R) bundle. The action is given by (ψ, h) ψ h

10 10 AARON LANDESMAN where h : R n R n and h GL(n, R). Then, take so that P = x P x P = {(x, ψ) : x X, ψ : E x R n }. Remark 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 In local trivializations, the action map µ is where in the local trivialization P = U H. (U H) H U H (x, g, h) (x, gh) 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 G k (R n ) = { all k-dimensional subspace of R n } = { all orthonormal k-tuples } /O(k) The latter description gives the grassmannian the structure of this topological space. Definition Using the inclusions R n R n+1, we obtain inclusions of Grassmannians Then, we define G k (R ) is their union. G k (R n ) G k (R n+1 ). Definition A map f : R Y is continuous if f R n is for all n. The same holds for G k (R ). Definition We have a tautological vector bundle E k (R n ) G k (R n ) and the limit forms a tautological vector bundle E k (R ) G k (R n ). The fiber over x is x with x R n or R. Definition For simplification of notation, we use G k = G k (R ). We let p : E k G k denote the tautological vector bundle. Definition 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 (3.3) E F In more concrete terms, E X F is the set X Y. E = {(x, w) : f(x) = q(w)}. with a map p : E X given by projection onto the first factor. We have E x = p 1 (x) = F f(x).

11 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 11 Lemma 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 f 0, f 1 : X Y are two homotopic maps, and if X is paracompact then f0 (F) and f 1 (F) are isomorphic vector bundles. Corollary 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 and f 0 : X X x 0 f 1 : X X x x which are homotopic. But then, vector bundles over a point are all trivial. Proof of Lemma Suppose we have f : X I Y where I = [0, 1] be a homotopy from f 0 to f 1. We get E := (f ) (F) which is a vector bundle on I X. We now construct a fiber bundle over I X This is a fiber bundle with fiber GL(n, R). P := { (t, x, a) : a : E (0,x) = E(t,x) }. Lemma 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 For X paracompact and Hausdorff, a vector bundle on X with fiber R k classified up to isomorphism, are in bijection with homotopy classes of maps f : X G k. The correspondence is realized by pulling back the universal bundle along f, sending f : X G k to E X with E := f (E k ) Proof. We ll see this next class.

12 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 G n (R ) so that E = f (E n ), where E n is the tautological bundle on G n (R ). (2) If we have f 0, f 1 : X G n (R ) with f 0 (E n) = f 1 (E n), then f 0 f 1. Further, conversely, if f 0 f 1 are homotopic, then the pullback of E n by these two maps are isomorphic, as we saw last class. Remark 4.2. The statement is just saying that there is a bijection between Vect n (X) := {n dimensional bundles on X} and homotopy classes of maps from X to G n, notated [X, G n ]. Example 4.3. When n = 1 take X = S 1, and the real numbers. Then, Vect 1 (S 1 ) is [S 1, RP ] = π 1 (RP ) = Z/2Z. Indeed, Vect 1 (S 1 ) = { trivial bundle, Möbius bundle }. Example 4.4. Take n = 1, X = S 2 and the complex numbers. So, we re looking at Vect 1 (S 2 ). By the theorem, this is homotopy classes of maps [ S 2, CP ] = π 2 (CP ). Now, there is a fiber bundle (4.1) S 1 S 2n+1 CP n and taking the limit, as n, we get (4.2) S 1 S CP and we can take the long exact sequence for a fibration. Since S has trivial homotopy groups in all dimensions, we have π k (CP ) = π k 1 (S 1 ). In particular, π k (CP ) = 0, except when k = 2, in which case π 2 (CP ) = Z. So, the line bundles on S 2 are isomorphic to Z. From the same argument, we the only complex line bundle on S n for n > 2 is the trivial bundle. Proof of Theorem 4.1. and trivializations (1) Given a vector bundle p : E X, there exists an open cover U = {U α : α I} τ α : E Uα U α R n. 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 So, we can take I = N.

13 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 13 Now, by paracompactness, we can find a partition of unity subordinate to this cover. That is, φ α : X [0, 1] where Supp φ α U α and α N φ α = 1, with this sum locally finite. We define σ α by (4.3) E Uα U α R n σ n R n. We then consider φ α σ α : E Uα R n. Note that this extends continuously to all of E. Further, this map is linear on the fibers E x R n, and varying continuously with x. Further, it is an isomorphism E x R n for x with φ α (x) = 0. Now, define where Φ : E R = α N R n Φ = α φ α σ α. 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 G n (R ) given by f : X G n (R ) x Φ(E x ) Now, f has the property that there is a diagram (4.4) Ψ E E n (R ) p p n f X G n (R ). so that E is the pullback of E n (R ). Now, Ψ is essentially the map Φ with some bookkeeping for the base G n (R ). More precisely, recall and define, for v E x, E n = {(y, v) : y R, y has dimension n, v y} Ψ(v) := (Φ(E x ), Φ(v)) (2) We ll now use the dictionary we established in the first part. Let f 0, f 1 satisfy f 0 (E n) = f 1 (E n) = E.

14 14 AARON LANDESMAN Then, by the previous dictionary, we have Φ 0, Φ 1 : E R which are linear on fibers and injective on fibers with f i (x) = Φ i (E x ) R. We want to find Φ t : E R interpolating between Φ 0, Φ 1 which is linear on fibers and injective on each fiber. Then, take the homotopy f t (x) = Φ t (E x ). Let s start by trying linear interpolation. Suppose for the time being (4.5) that for every v E, we have that Φ 0 (v) is not a negative multiple of Φ 1 (v), with v = 0. Then, we can simply take Φ t (v) = (1 t)φ 0 (v) + tφ 1 (v). We can reduce to the above assumption by the following trick: Consider the linear map A : R R (x 1, x 2,...) (x 1, 0, x 2, 0, x 3,...). This is continuous because it is continuous on every R n R 2n. Next, consider Now, define B : R R (x 1,...) (0, x 1, 0, x 2,...). A s (x) = (1 s)a(x) + sx. Define B s similarly. Then, A s and B s are both injective for all s [0, 1]. So, we can define a homotopy Φ 0 Φ 0 := A Φ 0. These are maps E R, defined using A s, s [0, 1]. This homotopy is one through maps which are injective on the fibers. We can similarly apply this trick to Φ 1 given by the homotopy to Φ 1 = B Φ 1. Now assuming (4.5), we obtain homotopies Φ 0 Φ 0 Φ 1, Φ 1, 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.

15 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 Characteristic Classes. 5. 2/1/16 Remark 5.1. The following is not a definition, it is just a description! A characteristic class c for real or complex rank r vector bundles assigns to each such E X a class c(e) H (X) or H (X; G) satisfying naturality: If (5.1) E 1 E 2 then f X 1 X 2 c(e 1 ) = f (c(e 2 )). The Chern classes for complex vector bundles for i = 0, 1, 2,.... The Stiefel-Whitney class for real vector bundles. c i (E) := H 2i (X) w i (E) H i (X; Z/2) Remark 5.2. Suppose we have E X be a rank k complex vector bundle. Then, E = f (E k ) with X f G k (C ) with E k G k = G k (C ) the tautological bundle. If c is a characteristic class, there is a universal characteristic class, c U = c(e k ) H (G k ). Then, c(e) = f (c U ). So, a characteristic class corresponds to elements in H (G k (C )), H (G k (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 = E E, then c i (E) = c j (E ) c k (E ). j+k=i (2) We have c 0 (E) = 1 H 0 (X) for all E X. Here, 1 means the element which is 1 on each connected component. (3) For L X a line bundle, c 1 (L) is defined by saying c 1 (E 1 ) = α where E 1 CP is the tautological bundle. (Recall H (G 1 (C )) = H (CP ) = Z[α], α H 2 (CP ), and we have a standard preferred choice of generator.)

16 16 AARON LANDESMAN Remark 5.3. It turns out H (G k (C )) = Z [α 1,..., α k ] with α i H 2i (G k (C )) equal to ±c i (E k ). However, it does take quite some work to calculate this cohomology ring. Remark 5.4. If we know X is paracompact, knowing c 1 (E 1 ) = 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) w i (E) = w j (E ) w k (E ) (2) w 0 (E) = 1 (3) w 1 (E 1 ) = α R 1, where we have E 1 RP the tautological bundle (note the sign is extraneous as we re working in Z/2 coefficients, and α R 1 H1 (RP ). Definition 5.5. The total Chern class is c(e) = c 0 (E) + c 1 (E) +. Remark 5.6. The Whitney sum formulas are usually combined as c(e) = c 0 (E) + c 1 (E) + c(e) = c(e ) c(e ). Remark 5.7. Let w 1 (L) H 1 (X; Z/2) for L a real line bundle, be H 1 (X; Z/2) = Hom (H 1 (X), Z/2) = Hom (π 1 (X); Z/2). We can describe this geometrically as follows. Given [γ] π 1 (X), we have γ : S 1 X. Look at γ (L) S 1. We assign { 0 if γ (L) is trivial [γ] 1 if γ (L) is the Möbius bundle. This ends up determining a map π 1 (X) Z/2, which curves out to be a homomorphism and gives w 1 (L). Example 5.8. Suppose we have a line bundle L M for M a 1 dimensional manifold. For example, suppose M = S 1. Then, we get L S 1 by w 1 (L)[S 1 ] Z/2. Then, w 1 (L)[S61] is obtained by counting the zeros of a section s of L: Choose a section s : S 1 L where w has finitely many zeros. We will have w 1 (L)[S 1 ] = mult x (s). x,s(x)=0 The total number of zeros will then always be even for the trivial bundle and odd for the Möbius bundle.

17 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 17 Example 5.9. For X a compact, oriented, connected two manifold the fundamental class for L X a complex line bundle, we have [X] H 2 (X) = Z, c 1 (L)[X] Z is an integer. It can be computed by taking a section with finitely many zeros and computing x,s(x)=0 s : X L mult x (s) 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) p (a) c. Theorem 5.10 (Leray-Hirsch). Suppose H k (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 c j 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 c j. Equivalently, if there exists c j, then for all c H (P; R), there exist unique a j H (X; R) so that c = j p (a j ) c j. In particular, p is injective. Proof. i Example 5.11 (Non-example, where the hypothesis fails). Taking S 2 S 2 with fibers S 1, then H (S 1 ) = Z in dimension 0, 1. The Leray-Hirsch theorem will not help because there are not elements in the cohomology of S 3 restricting to the generator of H 1 (S 1 ).

18 18 AARON LANDESMAN Example Take E X a complex rank n vector bundle over C and take P = PE, so that O X has fiber CP n 1. There s a tautological line bundle L P = PE so that the fiber at (x, l) P for x X, l E x is l. Let α = c 1 (L) H 2 (P; Z). Then, α PEx is a generator of H 2 (CP n 1 ). The classes 1, α, α 2,..., α n 1 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/ 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 w i (E) H i (X; Z/2) so that (1) w i (f E) = f w i (E) for f : X Y. (2) w i (E 1 E 2 ) = i+j=k w i(e 1 ) w j (E 2 ) (3) w i (E) = 0 if rk E < 1. (4) w 1 (U) is a generator of H 1 (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 c i (E) H 2i (X; Z). We then have (1) c i (f E) = f c i (E) for f : X Y. (2) c i (E 1 E 2 ) = i+j=k c i(e 1 ) c j (E 2 ) (3) c i (E) = 0 if rk E < 1. (4) For U CP, then c 1 (U) is a generator of H 2 (CP ; Z) that pairs with the class [CP 1 ] CP to 1. (The pairing is just pull back to CP 1.) Let X be paracompact and E a rank n R or C vector bundle over X. Last time, we introduced PE π X and U over PE the tautological (line) bundle, over PE. We already defined c 1 (L) and w 1 (L). Let h = c 1 (U) H 2 (PE; Z). Then, 1, h, h 2,..., h n 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. H p (PE; Z) is a free module over H (X; Z) generated by 1, h, h 2,..., h n 1. Remark 6.2. Definition 6.3. To know the algebra structure of H p (PE; Z), we d only need to know h n. There then exist unique ψ i H 2i (X; bz) so that n h n + π ψ i h n i = 0. i=1 We define the Chern classes are c i (E) := ψ i.

19 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) c i (f E) = f c i (E) for f : X Y. (2) c i (E 1 E 2 ) = i+j=k c i(e 1 ) c j (E 2 ) (3) c i (E) = 0 if rk E < 1. (4) For U CP, then c 1 (U) is a generator of H 2 (CP ; Z) that pairs with the class [CP 1 ] CP to 1. (The pairing is just pull back to CP 1.) 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 E E a subbundle. Then, there exists another subbundle E E such that E = E E. 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 is injective or q : H (X; Z) H (Σ; Z) 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 20 AARON LANDESMAN To prove the Whitney sum formula, it suffices to show that if E = n i=1 L i, for L i line bundles, then n c(e) = (1 + c 1 (L i )). Lemma If L 1 and L 2 are line bundles over X, then c (L 1 L 2 ) = c 1 (L 1 ) + c 1 (L 2 ). Proof. There are maps f i : X CP with f i U = L i. So, we can look at i=1 (f 1, f 2 ) : X CP CP. Without loss of generality, we can assume X = CP CP. Then, L i = 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 Show the Chern class is indeed the sum of the two Chern classes of tautological line bundles. Lemma Suppose L 1,..., L n are line bundles over X. Suppose s i is a section of L i vanishing at Z i X. Assume i Z i =. Then, n c 1 (L i ) = 0. i=1 Proof. Observe s i X\Zi trivializes L i X\Zi. That is, We have a long exact sequences c 1 (L i ) X\zi = 0. (6.1) H 2 (X, X \ Z i ) H 2 (X) H 2 (X \ Z i ) We know c 1 (L i ) maps to 0 under the second map. Therefore, there is some ξ H 2 (X, X \ Z i ) mapping to c 1 (L i ). Remark Recall that for X a topological space and A, B X two subsets, we have a map given by cup product. Then, the product H k (X, A; R) H l (X, B; R) H k+l (X, A B; R). ξ i ξ n H 2n (X, n i=1 (X \ Z i)) = H 2n (X, X) = 0. Therefore, the product of these classes is 0, as claimed.

21 We now want to see ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 21 c ( n i=1 L i) = n c(l i ). i=1 We have E = n i=1 L i. We have the projectivization and we have We now define the composition We can think of π : PE X U π E = n i=1 π L i. π i : U n i=1 π L i π L i. π i Hom(U, π L i ). This section π i vanishes on L 1 ^L i L n. These such loci intersect trivially. So, ( ) 0 = πc 1 U π L i = π ( c 1 (U) + π c 1 (L i )) n = (h + π c 1 (L i )). When we expand the elementary functions in this product and get the Chern classes, since they are also just the elementary symmetric functions. i=1 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 S 1 be the Möbius bundle over the circle. The fibers are R. Think of { ( ) } E = (θ, v) S 1 R 2 cos θ sin θ : 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 Möbius bundle? We can write this as a product bundle. We start with the circle S 1. The condition above enforces sin θv 1 = (1 + cos θ)v 2 sin θv 2 = (1 cos θ)v 1. We now have trivializations on the top and bottom halves of the circles, and local sections given by ( ) ( 1 sin θ ) sin θ, 1 cos θ. 1+cos θ 1

22 22 AARON LANDESMAN Indeed, these two local sections agree on the overlaps. section as ( ) 1 + cos θ sin θ We can notate this combined 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 S 2, with fiber C. We ll look at E S 2. Define ( ) iz ix y P :=. ix + y iz Note that P t = P, so it is Hermitian. The trace is 0 and determinant is 1, so the eigenvalues are ±i. { } E = ( p, v) S 2 C 2 : Pv = iv. where here p = (x, y, z) is a point on S 2 R 3 with norm 1. Example 7.4. Define, where refers to the cross product. We have E := { p, v} S 2 C 3 : p v = iv p ( p v) = ( ) p 2 v p( p v). E E = TS 2 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 R n ). Choose a fixed point q not on X. Consider the function p p q p q = M p. Of course, M p implicitly depends on our fixed choice of q. We can construct { } E X = (p, v) X C 2 : M p v = iv. Defining the function We now can realize E X as the pullback f q : X S 2 p p q p q. (7.1) E X E f q X S 2

23 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 23 That is, E X = 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 S 2, on which the bundle E on S 2 is trivial. Remark 7.6. If q lies inside the Riemann surface X, the bundle E X will not, in general, be trivial. Example 7.7. Suppose we have a surface X R 3. Then, at each point on the surface, we can define a normal direction n p, written bn = (n 1, n 2, n 3 ). We have a matrix satisfying ( ) in3 in 1 n 2 v = iv. in 1 + n 2 in 3 We can define a bundle E = { } (p, v) X C 2 : 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 bundle on S 2 we defined in this class. Now, we ll give some more examples of bundles over S 2. Example 7.9. Fix a positive integer n. Consider H 0 (P 1, O P 1(n)), defined to be the vector space of homogeneous complex polynomials in two variables of degree n. Consider the operators ( L z := i u u v ) v ( L x := i v ) L y := u + u v ( u v v u which gives an action of the Lie algebra of SO(3) on this space of polynomials. Next, define ), L p := xl x + yl y + zl z, and define the bundle E n S 2 to be { } E n := (p, v) S 2 H 0 (P 1, O P 1(n) : L p v = inv.

24 24 AARON LANDESMAN For example, let s look at the case n = 1. Recall S 2 = CP 1, and here a point (u, v) in CP 1 corresponds to x = 2re(uv)/( u 2 + v 2 ), y = 2im(uv)/( u 2 + v 2 ), z = u 2 v 2 u 2 + v 2. We have that the bundle E 1 constructed above is the tautological bundle. 8. 2/10/ Reviewing Leray-Hirsch. Let us start by recalling the Leray-Hirsch theorem: Theorem 8.1. Suppose we have p : P X a fiber bundle. Let c j H δ(j) (P; R). Suppose that the restrictions of these classes to each fiber form a basis for H (P x ; R), the cohomology of each fiber (meaning really that their pullbacks to the cohomology on each fiber for a basis for the cohomology of that fiber). Here, P x = p 1 (x) for all x. Then, there s an isomorphism where for U X, Here, P U = p 1 (U). The map is given by D k (X) Ψ X E k (X), E k (U) := H k (P U ; R) D k (U) := j H k δ(j) (u; R). Ψ U : D k (U) E k (U) (a 1, a 2,...) j p (a j ) c j. Proof. We ll prove this for finite simplicial complexes. We have both E (U) and D (U) behave as follows: (1) if U ι V, there s a map ι : E (V) E (U) (2) the previous property is functorial (3) There is a Mayer Vietoris sequence as follows: For U, V in X, we have 0 D k (U V) D k (U) D k (V) D k (U V D k+1 (U V) 0 (8.1) Ψ Ψ Ψ Ψ 0 E k (U V) E k (U) E k (V) E k (U V) E k+1 (U V) 0

25 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 25 Now, if Ψ U, Ψ V and Ψ U V are isomorphisms, the so is Ψ U 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 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 w i (E) H i (X; Z/2). Remark 8.2. Recall, we previously defined w 1 as follows. Let E be a line bundle and let f : X RP so that Then, we defined f (L taut ) = E h H 1 (RP, Z/2) is the generator. w old 1 (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 RP n 1, with n = rk E. There s a tautological line bundle L taut PE. Fiber by fiber, this is just O PEx (1) on the projective space fiber PE x. We have a class Then, we have classes h = w old i (L taut ) H i (PE; Z/2). h 0, h, h 2,..., h n 1, which restrict to the fibers to give a basis of H with Z/2 coefficients. Looking at h n H n (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 so that Exercise 8.4. Check w i (E) H i (X; Z/2) h n + p (w 1 (E))h n p (w n 1 (E)h + p (w n (E)) = 0.

26 26 AARON LANDESMAN (1) w 1 = w old 1 for line bundles (2) Check naturality of w i. (3) Check the Whitney sum formula holds. That is, w(e 1 E 2 = w(e 1 )w(e 2 ) where w = 1 + w 1 +. 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 E 1 and E 2 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 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 E 1 X and E 2 X be complex line bundles. Then, c 1 (E 1 E 2 ) = c 1 (E 1 ) + c 2 (E 2 ). Proof. We have E 1 = f 1 (Ltaut ), E 2 = f 2 (Ltaut ) where L taut CP is the tautological bundle. Then, take f = (f 1, f 2 ) : X CP CP. Then, the universal case is L 1 and L 2 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 CP 1 CP 0 and CP 0 CP 1, which meet at the single points CP 0 CP 0. The second cohomology is Z Z. Corollary 8.7. We have c 1 (E ) = c 1 (E) for E a line bundle. Here, E = Hom(E, C) where C is the trivial bundle. Proof. Take E 1 = E, E 2 = 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, c i (E ) = ( 1) i c i (E). Proof. Use the splitting principal. Let s check when E = L 1 L n. Then, by the Whitney sum formula. That is, c i (E) = σ i (c 1 (L 1 ),..., c 1 (L n )), c(e) = n ( 1 + c1 (L j ) ). n=1

27 ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 27 Then, ) c i (E ) = σ i (c 1 (L 1 ),..., c 1(L n ), implying the result. Question 8.9. Can we say anything about the Chern classes of E 1 E 2 for general complex vector bundles? We can use the splitting principle with bare hands, although this gets very messy very quickly. Example Let s suppose E 1 and E 2 are rank 2 vector bundles. Suppose further that c 1 (E i ) = 0. Then, using the splitting principle, we may as well assume E 1 = L 1 L 1 l = c 1(L 1 ) l = c 1 (L 1 ). Then, c 1 (E 1 ) = l + l = 0, so l = l. Then, Similarly, we may write Then, Then, Then, So, for example, c(e 1 ) = (1 + l)(1 + l ) = 1 l 2. c(e 2 ) = 1 m 2. c 2 (E 1 ) = l 2. E 1 E 2 = L1 L 2 L 1 L 2 L 1 L 2 L 1 L 2. c(e 1 E 2 ) = (1 + l + m)(1 + l m)(1 l + m)(1 l m) = 1 2l 2 2m 2 +. c 2 (E 1 E 2 ) = 2c 2 (E 1 ) + 2c 2 (E 2 ). Example Let s compute Stiefel-Whitney classes of TRP n 1. For V R n of dimension k, we have T v G k (R n ) = 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 TRP n 1 = Hom(L, L )

28 28 AARON LANDESMAN where L = L taut is the tautological bundle and L L = R n. We then have Therefore, with Then, Hom(L, R n ) = Hom(L, L) Hom(L, L ) = R Hom(L, L ) = R TRP n 1 w(trp n 1 ) = w(hom(l, R n )) = w( n L ) ( = w(l ) = (1 + h) n ) n h H 1 (RP n 1 ; Z/2). w k (TRP n 1 ) = 9. 2/12/15 ( ) n h k. k 9.1. Logistics. Problem set 2 is available now. It is due on 2/19/2016. Future homework should be submitted on canvas, uploaded ( Applications of Stiefel-Whitney classes. Last time, we sketched a proof that w k (TRP n 1 ) = n ) k h k H k (RP n 1 ; Z/2) = Z/2. Example 9.1. The space RP 4 does not immerse in R 5. That is, there is no function f : RP 4 R 5. Recall an immersion is a map with df x is injective for all x RP 4. Recall we have N x M T x M = R n and so TM NM = R n To see this, if f exists, there would exist a real line bundle N RP 4 so that TRP 4 N = R 5. But then, w(trp 4 )w(n) = w(r 5 ) = 1. We have w(trp 4 ) = 1 + w 1 + w 2 + w 3 + w 4 )(1 + u) = 1, where w 1 (N) = 1 + u. Then, we have 1 = (1 + h h 4 )(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 h 2 = 0. This is a contradiction because h 2 = 0 in the ring H (RP 4 ; Z/2) and h is the generator of the first cohomology, so h 2 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, dim R W R = 2 dim C W.

29 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 W R = V. Remark 9.3. Equivalently, a complex structure is a map J : V V is a real linear transformation with J 2 = 1. Here, where w W. (a + ib)w = aw + b(jw) 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. Take [V] G k (C n ). One can take a neighborhood of V in G k (C n ), given by all vector spaces intersecting V trivially. In other words, such a neighborhood is given by If [V] Gr, we can identify Then, Hom(V, V ) G k (C n ) f graph(f). T [V] (Gr k (C n )) = Hom(V, V ) R. TGr k (C n ) = Hom(E, E ) R where E is the tautological bundle, and E E = C n. Suppose we are given an almost complex structure for TGr k (C n ) = Hom(V, V ). Such a bundle has is a complex vector bundle and has Chern classes. Example 9.6. Let s now compute c k (TCP n 1. We obtain ( ) n c k (TCP n 1 ) = h k k with h H 2 (CP n 1 ; Z), h k H 2k (CP n 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 CP 3 ). This example uses standard terminology from algebraic geometry. Feel free to skip it if you re not familiar with the terminology. Take H CP 3. The dual of the tautological bundle is c 1 (H) = h H 2 (CP 3, 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 O CP 3(H). Warning 9.8. In the following we will sometimes notate this just as H.

30 30 AARON LANDESMAN Section of H are from linear maps a : C n C. Restricting a to L C n, 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 = O CP 3(H). If a is homogeneous of degree d on C n, we obtain that a Γ(O CP 3(H d )). Given any section s Γ(O P 3(H d )), with the property that for every x X := V(s), we have ds x : T x CP 3 O CP 3(H d ), a complex linear nonzero map. This means every point x X is smooth in X, and that T x X is a complex C-linear subspace of T x CP 3, 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 CP 3. We will now compute the Chern classes c i (X). Remark 9.9. Recall that when M is a manifold, we use the notation w i (M) := w i (TM) c i (M) := c i (TM), the second case making sense only when M is almost complex. Now, c(x)c(nx) = c(tcp 3 X ). Let s call c(x) = 1 + c 1 + c 2. 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 T x CP 3 ds (H d ) s. The kernel of this map is the tangent bundle, and so we have T x CP 3 /T x X = H d x and the left hand side is the normal bundle. Finally, c(tcp 3 X ) = 1 + 4h + 6h 2 + 4h 3 ). Additionally, c(h d ) = (1 + dh). So, we have (1 + c 1 + c 2 )(1 + dh) = (1 + 4h + 6h 2 + 4h 3 ) X. Writing down equations on the generators, we get 1 1 = 1 c 1 + dh = 4h c 2 + c 1 (dh X ) = 6h 2 X.

31 Solving in the above, we get ALGEBRAIC TOPOLOGY: MATH 231BR NOTES 31 c 1 (X) = (4 d)h X c 2 (X) = (6 (4 d)d)h X. Remark 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 CP 2, we would have the first Chern class given by c 1 = (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 in general, it has degree ( g 1) 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 Pontryagin Classes. For a real vector bundle V X, we have a Pontryagin class p i (V) H 4i (X). Definition 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 Define p i (V) = ( 1) i c 2i (V C). i(v λ) = v (iλ). 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 ( 1) j c j (W) = c j (W). In other words, 2c j (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 Theory on Pontryagin classes /17/16 Lemma 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 c 2i+1 (V C) = c 2i+1 (V C). Proof. This uses that L L = C. Then, use the splitting principle, since W = L i, implies c k (w) = σ k (c 1 (L 1 ),...), where σ k is the kth elementary symmetric function. Recall that p k (V) := ( 1) k c 2k (V C). are the Pontryagin classes for 1 k rk V/2, and p(v) = 1 + p 1 (V) +. Lemma For V 1, V 2 two real vector bundles, we have where 2α = 0. p(v) = p(v 1 ) p(v 2 ) + α

32 32 AARON LANDESMAN Proof. We have p k (V 1 V 2 ) = ( 1) k n+m=2k = α + ( 1) k c n (V 1 C) c m (V 2 C) n,m 0 mod 2 c n (V 1 C) c m (V 2 C) = ( 1) i c 2i (V 1 C) ( 1) j c 2j (V 2 C) i+j=k = α + i,j p i (V 1 )p j (V 2 ) where α is some sum of elements of order 2. Example Take X to be (1) CP 2 (2) CP n (3) a hypersurface in CP n. Then, consider p k (TX), which are viewed as real vector bundles by restriction of scalars of the complex vector bundle W = TX. Question Are the Pontryagin classes related to the Chern classes? Lemma Let W by a complex vector bundle and W R be the restriction of scalars of W to R. Then, p k (W R ) = ( 1) k c 2k (W R C). Warning Note that W = WR R C. In fact, the latter has rank twice as large as that of the former. Proof. Omitted. Lemma For W a complex vector bundle, we have W R R C = W W. Proof. We have multiplication by i corresponding to the linear transformation J and We have a corresponding map i : W W w iw J : W R W R w iw. J 1 : W R R C W R R C w λ Jw λ. This is complex linear with (J 1) 2 = 1. We then have W R C = W + W

33 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 and we similarly obtain W W + w w 1 (Jw) i 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 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 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 J 2 w i = w i + Jw 1 = i(w 1 Jw i) Calculations and Examples with Pontryagin Classes. Example Take X = CP 2. What is p 1 (TX)? This is 4 real dimensional. We have p 1 (T R X) = c 2 (T R X C) = c 2 (TX TX) = c 2 (TX) c 2 (TX) c 1 (TX)c 1 (TX) = 2c 2 (TX) + c 2 1 (TX). where in the second line, this is the tangent bundle as a complex vector bundle. Now, we know the Chern classes of tangent bundles. For c i = c i (CP 2 ), we have the resulting Pontryagin class is c 2 = 3h 2 c 1 = 3h p 1 (TX) = 2(3h 2 ) + 9h 2 = 3h 2. Definition For X a closed oriented manifold of dimension n, with n = 4k, we have a pairing p k (TX)[X] Z with p k (TX) H 4k, [X] H 4k. We abbreviate this integer by p k [X]. More generally, we denote P(p 1,..., p r )[X]