Cohomology and Poincar´eduality

Weiyi Zhang

Mathematics Institute, University of Warwick

December 18, 2013 2 Contents

1 Cohomology 5 1.1 Course description ...... 5 1.2 Cohomology ...... 5 1.3 Universal Coefficient Theorem ...... 7 1.3.1 Chain map and chain homotopy ...... 7 1.3.2 Hom functor ...... 8 1.3.3 Proof of Universal coefficient theorem, step 1 . . . . .9 1.4 Ext functor ...... 11 1.4.1 Universal coefficient theorem and K¨unnethformula for homology ...... 13 1.5 Singular cohomolgy ...... 14 1.6 The Eilenberg-Steenrod Axioms for cohomology ...... 15 1.7 Mayer-Vietoris Sequences ...... 16

2 Products 21 2.1 Cup Product for singular cohomology ...... 21 2.1.1 Cap product ...... 25 2.1.2 De Rham cohomology ...... 26 2.2 Cross Product and K¨unnethformula ...... 27 2.2.1 An alternative way to define cross and cup products . 29 2.3 Ljusternik-Schnirelmann category ...... 30 2.4 Higher products ...... 33

3 Poincar´eduality 37 3.1 Cohomology with compact supports ...... 37 3.2 Orientations for Manifolds ...... 40 3.3 Poincar´eduality theorem ...... 42

4 Applications of Poincar´eduality 47 4.1 Intersection form, Euler characteristic ...... 47 4.1.1 Intersection pairing ...... 47 4.1.2 Betti numbers and Euler characteristic ...... 49 4.2 Calculation of cohomology rings ...... 51

3 4 CONTENTS

n 4.2.1 Cohomology ring of CP ...... 51 n 4.2.2 Cohomology ring of RP and Borsum-Ulam ...... 51 4.2.3 Cohomology ring of lens spaces ...... 53 4.3 Degree and Hopf invariant ...... 55 4.3.1 Degree ...... 55 4.3.2 Hopf invariant ...... 56 4.4 Alexander duality ...... 58 4.5 Manifolds with boundary ...... 60 4.5.1 Linking number, Massey product ...... 63 4.6 Thom isomorphism ...... 64 Chapter 1

Cohomology and Universal Coefficient Theorem

1.1 Course description

Instructor: Weiyi Zhang Email: [email protected] Lecture time/room: Wednesday 1pm - 2pm MS.03 Thursday 5pm - 6pm CO D1.07 Monday 5pm - 6pm MA B1.01 Webpage: http://homepages.warwick.ac.uk/staff/Weiyi.Zhang/

Reference books:

• Allen Hatcher, “Algebraic Topology”, CUP 2002.

• Greenberg and Harper, “Algebraic Topology: a first course” : Addison- Wesley 1981

• William Fulton, “Algebraic Topology: a first course” : Springer 1995, GTM 153

• Bott and Tu, “Differential forms in Algebraic Topology”: Springer- Verlag 1999, GTM 82

Prerequisites: MA3H6 Algebraic Topology, MA3H5 Manifolds

1.2 Cohomology

Recall homology group is defined in the following lines. First, one forms a chain complex, which is a sequence of abelian groups and homomorphisms:

5 6 CHAPTER 1. COHOMOLOGY

∂n+1 ∂n ∂n−1 C• : ··· −→ Cn −→ Cn−1 −→ · · · The choices of the chain complex could be singular, simplicial, or cellular. Homomorphisms ∂n are called boundary map with the property ∂n ◦∂n+1 = 0 for all n. The homology groups (of the chain complex) are defined as

Zn(C•) Hn(C•) = , Bn(C•) where the elements of Zn(C•) := ker ∂n are called n-cycles, and elements of Bn(C•) := Im∂n+1 are called n-boundaries. The cohomology groups are defined in the similar lines as a dual object of homology groups. We first define the cochain group

n ∗ C = Hom(Cn,G) = Cn as the dual of the chain group Cn. And then define the coboundary map δn : n n+1 n C → C as the dual of boundary map: For f ∈ C , define (δnf)(x) := f(∂n+1x) for x ∈ Cn+1. It is easy to check δn ◦ δn−1 = 0. Then cohomology groups are define as

Zn(C•) Hn(C•) = , Bn(C•) n • where the elements of Z (C ) := ker δn are called n-cocycles, and elements n • of B (C ) := Imδn−1 are called n-coboundaries.

δ δ C• : ··· −→n−1 Cn −→δn Cn+1 −→n+1 · · · Let us calculate several examples. Take the coefficient group G = Z.

Example 1.2.1. Let C• be the chain complex · · · → 0 → Z → 0 → · · · th where Z is at k position. Then the homology groups are  n = k H (C ) = Z n • 0 n 6= k

• The corresponding cochain complex C is · · · ← 0 ← Z ← 0 ← · · · where th Z is still at k position. Then  n = k Hn(C•) = Z 0 n 6= k

m Example 1.2.2. Let C• be the chain complex · · · → 0 → Z → Z → 0 → · · · th th where Z are at k and (k − 1) positions. Then the homology groups are  n = k − 1 H (C ) = Zm n • 0 n 6= k − 1 1.3. UNIVERSAL COEFFICIENT THEOREM 7

• m The corresponding cochain complex C is · · · ← 0 ← Z ← Z ← 0 ← · · · th th where Z are still at k and (k − 1) position. But  n = k Hn(C•) = Zm 0 n 6= k

These examples show the difference of the free part Z and torsion part Zn. Actually if Cn is free abelian for all n and Hn(C) is finitely generated for all n. And Hn(C) = Fn ⊕ Tn where Fn is free abelian and Tn is torsion. n Then H (C) = Fn ⊕ Tn−1. This is the simplest form of Universal coefficient theorem which determines cohomology groups with arbitrary coefficients from homology with Z coefficients. Remark 1.2.3. For non-abelian group G, we could still define (co)homology, but the point is that usually Hn(C; G) do not have a group structure when n > 0, since Imδ need not to be a normal subgroup of kerδ.

1.3 Universal Coefficient Theorem

1.3.1 Chain map and chain homotopy Definition 1.3.1. A map f : C → D of chain complexes C and D is a sequence of homomorphisms f = {fn : Cn → Dn} such that ∂n ◦ fn = fn−1 ◦ ∂n. Proposition 1.3.2. The chain map f : C → D induces a homomorphism f∗ : Hn(C) → Hn(D) of homology groups.

Proof. By definition, it fn takes cycles in Zn(C) to cycles in Zn(D) and takes boundaries to boundaries. Hence it induces a homomorphism f∗ : Hn(C) → Hn(D). Definition 1.3.3. Two maps of chain complexes f, g : C → D are chain homotopic (denoted by f ' g) if there exists a sequence of maps T = {Tn : Cn → Dn+1} such that

∂n+1 ◦ Tn + Tn−1 ◦ ∂n = fn − gn.

We will save the subscripts for boundary maps later on. Proposition 1.3.4. If f and g are chain homotopic, then their induced homomorphisms of homology are equal.

Proof. Let T be a chain homotopy. For any zn ∈ Zn(C), we have f∗[zn] − g∗[zn] = [gn(zn) − fn(zn)] = [∂ ◦ T (zn) + T ◦ ∂(zn)] = [∂T (zn)] = 0. 8 CHAPTER 1. COHOMOLOGY

So chain homotopy is an equivalence relation. Definition 1.3.5. Two chain complexes C and D are called chain homotopy equivalent (C ' D), if there are chain maps f : C → D and g : D → C such that g ◦ f ' idC : C → C, f ◦ g ' idD : D → D. Each of them is called a chain homotopy equivalence. The next result follows from Propostion 1.3.4. Proposition 1.3.6. Chain homotopy equivalence induces isomorphism of homology groups. So if C ' D, then H∗(C) = H∗(D). The reverse is also true if C and D are complexes of abelian groups. All the above discussion works for cochain complexes with apparent mod- ifications.

1.3.2 Hom functor Notice that the functor Hom(·,G) is the key for cohomology. By defi- nition, Hom(H,G) is the set of all homomorphisms from H to G. It is an abelian group as well, and called homomorphism group. Let us look at it in a more abstract viewpoint. It is a countravariant functor, which means f : A → B induces f ∗ : Hom(B,G) → Hom(A, G) and if furthermore we have g : B → C then (g ◦ f)∗ = f ∗ ◦ g∗. The above discussion tells us Hom(·,G) is also a countravariant functor from Chain complexes to cochain complexes. It has the following properties:

• Splitting exactness: Hom(A1 ⊕ A2,G) = Hom(A1,G) ⊕ Hom(A2,G);

• Moreover for direct sum: Hom(⊕iAi,G) = ΠiHom(Ai,G);

f g • left exactness: If A → B → C → 0 is an exact sequence, then the induced sequence

f ∗ g∗ Hom(A, G) ←− Hom(B,G) ←− Hom(C,G) ←− 0 is exact. Proofs are left as exercises. The third item is the key for later applications.

Example 1.3.7. Hom(Zm,G) = {g ∈ G|mg = 0}. In fact, take Hom for m exact sequence 0 → Z → Z → Zm → 0, we obtain m 0 ← G ← G ← Hom(Zm,G) ← 0. By item three, the sequence is exact except possibly for the leftmost. So m Hom(Zm,G) = ker(G → G){g ∈ G|mg = 0}. And when mG 6= G, the sequence is not exact. In general, Hom functor does NOT keep exactness. 1.3. UNIVERSAL COEFFICIENT THEOREM 9

To state UCT, we need to introduce a functor Ext(H,G), which mea- sures the failure of Hom to be an exact functor. It is defined from a free resolution of the abelian group H: 0 → F1 → F0 → H → 0, with Fi = 0 for i > 1.This could be obtained in the following way. Choose a set of gen- erators for H and let F0 be a free abelian group with basis in one-to-one correspondence with these generators. Then we have a surjective homomor- phism f0 : F0 → H. The kernel of f0 is free as a subgroup of a free abelian group. We let F1 be the kernel and the inclusion to F0 as f1. It is an ex- act chain complex sequence. In summary, constructing a free resolution is equivalent to choosing a presentation for A. Take its dual cochain complex by Hom(F,G), we may lose its exactness, so could have its cohomology group, temporarily denoted by Hn(F ; G). For the above constructed resolution, Hn(F ; G) = 0 for n > 1. So the only interesting group is H1(F ; G). As we will show, it is independent of the resolution. There is a standard notation for that: Ext(H,G). The element in this group could also be interpreted as the isomorphism class of extensions of G by H, i.e. 0 → G → J → H → 0. Finally, the Universal coefficient theorem is ready to state.

Theorem 1.3.8. If a chain complex C of free abelian groups has homology groups Hn(C), then for each n, there is a natural short exact sequence:

n h 0 → Ext(Hn−1(C),G) → H (C; G) → Hom(Hn(C),G) → 0.

The sequence splits, so that

n H (C; G) = Ext(Hn−1(C),G) ⊕ Hom(Hn(C),G).

But the splitting is not natural.

1.3.3 Proof of Universal coefficient theorem, step 1 There is a natural choice of free resolution of the homology group

in q 0 → Bn(C) → Zn(C) → Hn(C) → 0.

1 ∗ So the H (F ; G) for this free resolution is exactly coker(in : Hom(Zn,G) → Hom(Bn,G)) by definition. Now, let us prove Universal coefficient theorem in two steps. Step 1: derive the split short exact sequence

∗ n h 0 → Cokerin−1 → H (C; G) → Hom(Hn(C),G) → 0.

Step 2: Prove H1(F ; G) only depends on H and G, but not the resolution. ∗ So cokerin−1 = Ext(Hn−1,G). We start with Step 1. 10 CHAPTER 1. COHOMOLOGY

Lemma 1.3.9. There is a natural homomorphism

n h : H (C; G) → Hom(Hn(C),G).

Proof. We first have the map in the cycle level: n For any cocycle α ∈ Z and any cycle z ∈ Zn, we let h(α)(z) = α(z). α ∈ Zn means δα = 0, i.e α∂ = 0. In other words, α vanishes on Bn. So h n descends to a map from Z to Hom(Hn,G). n Next if α ∈ B , then α = δβ = β∂. Hence α is zero on Zn. Thus there is a well defined quotient map h([α])([z]) = α(z) from Hn(C; G) to Hom(Hn(C),G). This is a homomorphism since:

h([α + β])([z]) = (α + β)(z) = α(z) + β(z) = h[α]([z]) + h[β]([z]).

Now there is a split short exact sequence

∂ 0 → Zn → Cn → Bn−1 → 0.

It splits since Bn−1 is free (for any generator of Bn−1, one could map it to a preimage of ∂. So it is not canonically chosen.). Thus we have p : Cn → Zn whose restriction to Zn is identity. We have the commutative diagram ∂ 0 Zn+1 Cn+1 Bn 0

0 ∂ 0 ∂ 0 Zn Cn Bn−1 0 Since the dual of a split short exact sequence is a split short exact se- quence (the splitting exactness of Hom), the following commutative diagram has exact rows, 0 Z∗ C∗ B∗ 0 n n δ n−1

0 δ 0 0 Z∗ C∗ B∗ 0 n+1 n+1 δ n This is a part of a short exact sequence of chain complexes. From this, we have a long exact sequence (because the differential of the complexes B and Z are trivial)

∗ i∗ ∗ in ∗ n ∗ n−1 ∗ Bn ←− Zn ←− H (C; G) ←− Bn−1 ←− Zn−1.

∗ The connection homomorphism is in because of definition: one takes an ∗ ∗ ∗ element of Zn, pulls back to Cn, applies δ to get an element in Cn+1, then 1.4. EXT FUNCTOR 11

∗ pull back to Bn. So we first extend a homomorphism f : Zn → G to 0 f : Cn → G, then composes it with ∂, finally view it as a map from Bn. So it is nothing but the restriction of f from Zn to Bn. We could also see from its dual operation: given b ∈ Bn, so b = ∂c, then the first step maps it to c, second takes ∂, thus get b back which is in Zn. The composition is the inclusion in. Hence we have

∗ n ∗ 0 ← Kerin ← H (C; G) ← Cokerin−1 ← 0.

The last step for Step 1 is

∗ Lemma 1.3.10. ker(in) = Hom(Hn(C),G)

∗ Proof. Because the elements of ker(in) are homomorphisms Zn → G that vanish on Bn, the same as homomorphisms Hn = Zn/Bn → G.

∗ Under this identification, the natural map h is the map 0 ← Kerin ← Hn(C; G). And the short exact sequence splits because of the induced map p∗.

1.4 Ext functor

Recall that to finish Step 2, we only need to prove that for any two (2-step) free resolutions of abelian group H, the homology groups are iso- morphic. So the notation Ext(H,G) is well defined. We have the following

Lemma 1.4.1. Given free resolutions F and F 0 of abelian groups H and H0, then every homomorphism α could be extended to a chain map from F to F 0:

f2 f1 f0 ··· F2 F1 F0 H 0

α2 α1 α0 α 0 0 0 0 f2 0 f1 0 f0 0 ··· F2 F1 F0 H 0

Furthermore, any two such chain maps extending α are chain homotopic.

Proof. Since Fi’s are free, it suffices to define αi on a basis of Fi. Given 0 0 0 x ∈ F0, then α(f0(x)) ∈ H . Since f0 is surjective, we have x such that 0 0 0 f0(x ) = α(f0(x)). We define α0(x) = x . 0 0 0 To define α1, for x ∈ F1, α0(f1(x)) lies in Imf1 = Kerf0 since f0α0f1 = 0 0 0 αf0f1 = 0. So define α1(x) = x such that α0(f1(x)) = f1(x ). Other αi could be constructed inductively by a similar way. 12 CHAPTER 1. COHOMOLOGY

To check any such chain maps are chain homotopic, we will only give a 0 proof for the case of 2-step free resolutions, i.e when Fi = Fi = 0 for n > 1. This is the case we need. The general case is similar and see page 194 of Hatcher for details. 0 If βi is another extension of α, then we want to find T0 : F0 → F1 and 0 0 T−1 : H → F0 such that αi − βi = fi+1Ti + Ti−1fi for i = 0, 1. We let 0 0 0 T−1 = 0. We let T0(x) = x such that f1(x ) = α0(x) − β0(x). This could 0 0 0 0 be done because f0β0(x) = f0α0(x) = αf0(x) and Imf1 = kerf0. Hence 0 α0 − β0 = f1T0. To check α1 − β1 = T0f1, we only need to check the relation 0 0 after composing f1 since it is injective. It is nothing but f1(α1 − β1) = (α0 − β0)f1. Corollary 1.4.2. For any two free resolutions F and F 0 of H, Hn(F ; G) = Hn(F 0; G).

Proof. It follows from above lemma and (cohomology version of) Proposition 1.3.6 by taking α = id : H → H and by looking at the composition of two chain maps, one from F to F 0 and the other from F 0 to F .

Hence we finished step 2 and thus the proof of Theorem 1.3.8. Now the calculation of cohomology groups is reduced to that of Ext. So next will be the properties of it.

Proposition 1.4.3. Computation properties:

1. Ext(H ⊕ H0,G) = Ext(H,G) ⊕ Ext(H0,G)

2. Ext(H,G) = 0 if H is free abelian

3. Ext(Zn,G) = G/nG. Proof. For the first, we take direct sum of the free resolution. For the second, we use 0 → H → H → 0 as the resolution to calculate. For the last, we use

n 0 −→ Z −→ Z −→ Zn −→ 0. and the calculation in Example 1.3.7.

Corollary 1.4.4. If the homology groups of chain complex C of free abelian groups are finitely generated and Hn(C) = Fn ⊕ Tn where Fn is free and Tn n is torsion, then H (C; Z) = Fn ⊕ Tn−1. Proof. It follows from Theorem 1.3.8 and Proposition 1.4.3, and the fact Hom(Zm, Z) = 0, Hom(Z, Z) = Z. Next property shows how Ext fucntor remedies the left exactness of Hom functor. 1.4. EXT FUNCTOR 13

Proposition 1.4.5. Let 0 → A → B → C → 0 be a short exact sequence of abelian groups. Then there is a six-term exact sequence 0 → Hom(C,G) → Hom(B,G) → Hom(A, G) → Ext(C,G) → Ext(B,G) → Ext(A, G) → 0.

1.4.1 Universal coefficient theorem and K¨unnethformula for homology

Instead of Hom, we apply ⊗ to a free resolution 0 → F1 → F0 → H → 0. This operation is right exact, so by similar idea as Ext, we use Tor to measure its non-exactness, i.e. Tor is the first (and the only non-trivial) homology of the new complex. Then we have Theorem 1.4.6. If C is a chain complex of free complex of free abelian groups, then there are natural short exact sequences

0 → Hn(C) ⊗ G → Hn(C; G) → T or(Hn−1(C),G) → 0 for all n. These sequences split, though not naturally. The proof is almost identical to that of Theorem 1.3.8 after establishing corresponding properties for Tor. See Hatcher’s Appendix A to Chapter 3 for details. Finally, we want to remark that the universal coefficient theorem in homology is a special case of the K¨unneththeorem. We first introduce teh tensor products of chain complexes. Let (C, ∂) and (D, ∂) be chain complexes, where Ci and Di are zero for i < 0. The tensor product of chain complexes is M (C ⊗ D)n = Cp ⊗ Dq, p+q=n with differential p ∂(cp ⊗ dq) = (∂cp) ⊗ dq + (−1) cp ⊗ (∂dq). It indeed defines a chain complex since p p−1 p ∂∂(cp⊗dq) = ∂((∂cp)⊗dq+(−1) cp⊗(∂dq)) = (−1) ∂cp⊗∂dq+(−1) ∂cp⊗∂dq = 0.

Tensor product of chain maps is defined as (f ⊗g)(cp ⊗dq) = (fpcp)⊗(gqdq). Please check it commutes with ∂. We also know that chain homotopy is kept by tensor product. Theorem 1.4.7 (K¨unnethformula). For a free chain complex C and an arbitrary chain complex D, there is a natural short exact sequence M M 0 → Hp(C)⊗Hq(D) → Hn(C⊗D) → T or(Hp(C),Hq(D)) → 0. p+q=n p+q=n−1 It splits but not canonically.

To get Theorem 1.4.6, we take D as the complex D0 = G and Di = 0 for i 6= 0. 14 CHAPTER 1. COHOMOLOGY

1.5 Singular cohomolgy

This is an explicit example of choosing the chain complex and cochain complex. n n+1 Recall that a standard singular n-simplex is the convex set ∆ ⊂ R consisting of all (n + 1)-tuples of real numbers (v0, ··· , vn) with vi ≥ 0, v0 + n ··· + vn = 1. A singular n-simplex in X is a continuous map σ : ∆ → X. The singular chain group Sn(X) is the free abelian group generated by the singular n-simplices. The boundary homomorphism ∂ : Sn(X) → Sn−1(X) P i is defined as ∂(σ) = i(−1) σ|[v0, ··· , vˆi, ··· , vn]. Then we could define singular homology and denoted by Hn(X). n Fix coefficient group G, we set S (X) = Hom(Sn(X),G) be the cochain group. The coboundary map δ : Sn(X; G) → Sn+1(X; G) is the dual of ∂. n ∂ φ So any φ ∈ S (X; G) δφ is the composition Sn+1(X) −→ Sn(X) −→ G. Explicitly, for σ : ∆n+1 → X,

X i δφ(σ) = (−1) φ(σ|[v0, ··· , vˆi, ··· , vn+1]). i

For a pair (X,A), we could also talk about relative groups. Let Sn(X,A) = Sn(X) and Sn(X,A; G) = Hom(S (X,A),G). The elements of Sn(X,A; G) Sn(A) n are n-cochains taking the value 0 on singular n-simplices in A. Hence we have relative groups H∗(X,A; G) := H∗(S∗(X,A; G)). A map of pairs f : (X,A) → (Y,B) induces homomorphisms f ∗ : Sn(Y,B; G) → Sn(X,A; G) and f ∗ : Hn(Y,B; G) → Hn(X,A; G). We could describe relative cohomology groups in terms of exact se- quences. Recall that the short exact sequence

i j 0 → Sn(A) −→ Sn(X) −→ Sn(X,A) → 0 splits. By splitting exactness of Hom functor, the dual

∗ j∗ 0 ← Sn(A; G) ←−i Sn(X; G) ←− Sn(X,A; G) ← 0 is a split exact sequence as well. Since i∗ and j∗ commute with δ, so it induces a long exact sequence of cohomolgy groups

j∗ ∗ · · · −→ Hn(X,A; G) −→ Hn(X; G) −→i Hn(A; G) −→δ Hn+1(X,A; G) −→ · · · .

Let us describe the connecting homomorphism Hn(A; G) −→δ Hn+1(X,A; G). For φ ∈ Zn(A; G), we first extend it to a cochain φ¯ ∈ Sn(X; G), by taking value 0 on singular simplices not in A. Then δX (φ¯) = φ∂¯ ∈ Sn+1(X; G). It is in Sn+1(X,A; G) because original φ is a cocycle in A, i.e taking value X 0 in Bn(A), which means δ (φ¯) = φ∂¯ takes value 0 on Sn+1(A). Fi- nally it is in Zn+1(X,A; G) because δX,A(δX φ¯) = δX (δX φ¯) = 0. Its class [δX φ¯] ∈ Hn+1(X,A; G) is δ[φ]. 1.6. THE EILENBERG-STEENROD AXIOMS FOR COHOMOLOGY 15

A more general long exact sequence is for a triple (X, A, B), induced by

∗ j∗ 0 ← Sn(A, B; G) ←−i Sn(X,B; G) ←− Sn(X,A; G) ← 0.

When B is a point, it induces the long exact sequence for reduced cohomol- ogy.

1.6 The Eilenberg-Steenrod Axioms for cohomol- ogy

For simplicity, let us omit the coefficient G in out notation. A cohomology theory consists of 3 functions: 1. For any integer n and any pair of spaces (X,A), we have an abelian group Hn(X,A).

2. For any integer n and any map of pairs f :(X,A) → (Y,B)(f maps A to B), we obtain homomorphism f ∗ : Hn(Y,B) → Hn(X,A).

3. For any integer n and any pair (X,A), we have a connecting homo- morphism δ : Hn(A) → Hn+1(X,A). These functions satisfy the following 7 axioms:

1. Unit: If id :(X,A) → (X,A) is the identity, id∗ is the identity.

2. Composition: (g ◦ f)∗ = f ∗ ◦ g∗.

3. Naturality: Given f :(X,A) → (Y,B), the following diagram com- mutes:

n n H (A) ∗ H (B) f|A δ δ

Hn+1(X,A) Hn+1(Y,B) f ∗

4. Exactness: The following sequence is exact:

j∗ ∗ · · · −→ Hn(X,A) −→ Hn(X) −→i Hn(A) −→δ Hn+1(X,A) −→ · · · .

Here i : A → X and j :(X, ∅) → (X,A) are inclusions.

5. Homotopy: If f ' g are homotopic maps of pairs, then f ∗ = g∗.

6. Excision: Given (X,A) and U ⊂ X such that U¯ ⊂ int(A). Then the inclusion i :(X\U, A\U) → (X,A) induces isomorphisms in cohomol- ogy. 16 CHAPTER 1. COHOMOLOGY

7. Dimension: Let P be a one-point space, then

 G n = 0 Hn(P ) = 0 n 6= 0

There are many generalized cohomology theories which satisfy all the ax- ioms except probably the dimension axiom. The Eilenberg-Steenrod unique- ness theorem says that there is a unique cohomology theory satisfying all the axioms in the category of finite CW complexes. Unfortunately the proof is beyond the scope of current course. But we could prove a weaker version. Suppose Hn(X,A) and Kn(X,A) are cohomology theories, and φ : Hn(X,A) → Kn(X,A) is a natural transformation of cohomology theo- ries, i.e. it commutes with induced homomorphisms and with coboundary homomorphisms in long exact sequence of pairs.

Theorem 1.6.1 (Weak form of Eilenberg-Steenrod Uniqueness). Suppose φ : Hn(X) → Kn(X) is an isomorphism when X = {pt}. Then φ is an isomorphism for any finite CW complex.

We left the proof to the end of next section. A few more words about Eilenberg-Steenrod axioms. If we want to prove uniqueness for a larger category, more conditions are needed for the uniqueness. For the category of CW complexes, an eighth axiom is added to guarantee its uniqueness. This is included in the discussion of Hatcher’s book.

Milnor Addition: Let X = tXα be disjoint union. Then the induced Q ∗ n Q n homomorphism α iα : H (X) → α H (Xα) is an isomorphism.

This axiom has force only if there are infinitely many Xα. The finite sum case is a corollary of the following Mayer-Vietoris Sequence. There are also examples of cohomology theories which are not additive (by James and Whitehead).

1.7 Mayer-Vietoris Sequences

By using Eilenberg-Steenrod Axioms, we can form the Mayer-Vietoris Sequence. The model in our mind is X = A ∪ B with A and B open in X. But for the ease of application, we use the following setting.

Definition 1.7.1. For U ⊂ A ⊂ X, the map (X\U, A\U) → (X,A) is an excision if the induced homomorhism Hn(X,A) → Hn(X\U, A\U) is an isomorphism for all n. 1.7. MAYER-VIETORIS SEQUENCES 17

Example 1.7.2. The inclusion (D+,Sn−1) → (Sn,D−) is an excision. We cannot apply the Excision Axiom directly. But look at the exact sequences from exactness axiom corresponding to the two pairs, we have Hi+1(Sn,D−) = Hi+1(Sn) = Hi(Sn−1) = Hi+1(D+,Sn−1) when i > 0 by noticing Hn(D) = Hn(pt). The rest of two identities also follow from the exact sequence by knowing H0(D) = H0(Sn) = G. Definition 1.7.3. Suppose A, B ⊂ X, such that • A ∪ B = X

• Both (A, A ∩ B) → (X,B) and (B,A ∩ B) → (X,A) are excisions. Then (X; A, B) is an excisive triad. Example 1.7.4. Let X = A∪B with A and B open in X. Then (A, A∩B) = (X\U, B\U) where U = B\(A ∩ B). Since X\U = A, U is closed and U ⊂ B = int(B). Then the excision axiom shows this is an excision. Example 1.7.5. (Sn; D+,D−) is an excisive triad. Theorem 1.7.6. Suppose (X; A, B) is an excisive triad. Then there is a long exact sequence

(j∗ ,j∗ ) i∗ −i∗ · · · −→ Hn(X) −→A B Hn(A)⊕Hn(B) A−→B Hn(A∩B) −→δ Hn+1(X) → · · ·

Recall that a similar MV sequence is derived for homology as the long exact sequence associated to the short exact sequence

0 → C∗(A ∩ B) → C∗(A) ⊕ C∗(B) → C∗(A + B) → 0.

We could prove it similarly for singular cohomology say, but bot for a general cohomology theory. We need the following algebraic lemma: Lemma 1.7.7 (Barratt-Whitehead). Suppose we have the following com- mutative diagram with exact rows

fn gn hn ··· An Bn Cn An−1 ···

αn βn γn αn−1 0 0 0 0 fn 0 gn 0 hn 0 ··· An Bn Cn An−1 ···

0 in which every third map γi : Ci → Ci is an isomorphism. Then there is a long exact sequence

0 −1 0 (fn,αn) 0 βn−fn 0 hnγn gn · · · −→ An −→ Bn ⊕ An −→ Bn −→ An−1 −→ · · · . 18 CHAPTER 1. COHOMOLOGY

This is a standard diagram chasing argument. We would like to leave this as an exercise. So we only sketch the proof.

Proof. This is a chain complex is easy to check in terms of commutativity of the diagram. To check the exactness, we have 3 parts: 0 0 1. ker(βn −fn) ⊂ im(fn, αn). Assume βn(b) = fn(d). By commutativity and exactness at Bn, we have a ∈ An such that fn(a) = b. We have αn(a) − 0 0 0 0 0 −1 0 d ∈ kerfn = imhn+1. So hn+1(c ) = αn(a) − d. Let a = a − hn+1γn+1(c ). 0 0 Then (fn(a ), αn(a )) = (b, d). −1 0 0 −1 0 2. ker(hnγn gn) ⊂ im(βn − fn). Let e ∈ ker(hnγn gn). By exactness −1 0 0 at Cn, we have b ∈ Bn such that gn(b) = γn gn(e). So βn(b) − e ∈ kergn = 0 0 imfn. Choose any element d in it, we have e = βn(b) − fn(d). −1 0 3. ker(fn−1, αn−1) ⊂ im(hnγn gn). First find an element in Cn. Then 0 an element in Bn. Proof. (of the theorem) We have

j∗ ··· Hn−1(X) B Hn−1(B) δ Hn(X,B) hn Hn(X) ···

∗ ∗ ∗ jA iB γn jA i∗ h0 ··· Hn−1(A) A Hn−1(A ∩ B) δ Hn(A, A ∩ B) n Hn(A) ···

γn is an isomorphism since excisive triad. Hence we have the MV sequence by Barratt-Whitehead.

Example 1.7.8 (Cones and suspensions). Given a space X, the cone of X is X × [0, 1] C(X) := (x, 1) ∼ (y, 1), ∀x, y ∈ X and suspension as X × [0, 1] Σ(X) := . (x, 0) ∼ (y, 0), (x, 1) ∼ (y, 1), ∀x, y ∈ X

Then (Σ(X); C+(X),C−(X)) is an excisive triad. Use MV sequence and k k+1 C+(X) ' C−(X) ' pt, we have H (X) = H (ΣX). If X is connected, 1 0 H (ΣX) = 0 and H (ΣX) = Z. Especially ΣSn = Sn+1, so it could be used to calculate the cohomology of Sn inductively. Complete the calculation. Given a continuous map f : Sn−1 → A for n ≥ 1. We have A t Dn X = C(f) := A ∪ Dn = f f(x) ∼ x, ∀x ∈ Sn−1 Notice that if f ' g : Sn−1 → A, then C(f) ' C(g). Also f extends to a map f : Dn → C(f). 1.7. MAYER-VIETORIS SEQUENCES 19

Proposition 1.7.9. 1. The inclusion A → X induces isomorphisms Hq(A) = Hq(X) for q 6= n, n − 1. 2. There is an exact sequence f ∗ 0 → Hn−1(X) → Hn−1(A) → Hn−1(Sn−1) → Hn(X) → Hn(A) → 0 Proof. Use sequence for pairs (or MV sequence). For both items, use Hi(X,A) = Hi(Dn,Sn−1) = Hi−1(Sn−1) and previous calculation for H∗(Sn).

n+1 n C \0 Example 1.7.10. Recall CP = ∼ where x ∼ y if there exists λ 6= 0 such that λx = y. n n−1 2n 2n−1 n−1 We write CP = CP ∪f D , where f : S → CP is the natural quotient map. The above sequence tells us for 0 < m < 2n, 2n−1 2n−1 2n n 0 → H (S ) → H (CP ) → 0, m n m n−1 0 → H (CP ) → H (CP ) → 0. 0 n By induction and H (CP ) = Z since it is path connected, we have  m = 0, 2, ··· , 2n Hm( P n) = Z C 0 otherwise Now we can finish the proof of Theorem 1.6.1 Proof. We use induction. The dimension axiom gives us the dimension 0 case. Assume it is done for all complexes with dimension less that or equal to n−1, and X be a dimension n cell complex. So X is obtained by attaching n-cells to an (n − 1) cell complex. So we could do these attachment one by one (but A will possibly have dimension n then). Hence the statement is reduced to the following: Suppose the result is n true for A, prove the statement is true for X = C(f) := A∪f D . Proposition 1.7.9 item 1 tells us φ : Hq(X) → Kq(X) is an isomorphism for q 6= n − 1, n. For the rest, look at

0 Hn−1(X) Hn−1(A) Hn−1(Sn−1) Hn(X) Hn(A) 0

' α '

0 Kn−1(X) Kn−1(A) Kn−1(Sn−1) Kn(X) Kn(A) 0

As we have shown in the suspension calculation, α is also an isomor- phism. So five lemma completes the proof. The five lemma also shows us H∗(X,A) = K∗(X,A). Notice that for infinite CW complexes, we need Milnor additivity in the above argument. A telescope argument could reduce infinite dimensional case to finite dimensional case. 20 CHAPTER 1. COHOMOLOGY Chapter 2

Products

In this chapter, we take the coefficients in a commutative ring R with a unit.The most common choices are Z, Zm or Q.

2.1 Cup Product for singular cohomology

There is a product structure in cohomolgy.We start with singular coho- mology. Let σ : ∆m+n → X be a singular simplex. By the front m-face or mσ, we mean σ|[v0, ··· , vm]. Similarly, for the back n-face or σn, we mean σ|[vm, ··· , vm+n]. Then we define the cup product at chain level: Definition 2.1.1. Given φ ∈ Sm(X) and ψ ∈ Sn(X), the cup product φ ∪ ψ ∈ Sm+n(X) is defined by

(φ ∪ ψ)(σ) = φ(σ|[v0, ··· , vm]) · ψ(σ|[vm, ··· , vm+n]). To get intuition for this definition, think about S1 ∨ S1 ∨ S2 and T 2. Details in class. Lemma 2.1.2. For φ ∈ Sm(X) and ψ ∈ Sn(X), we have

δ(φ ∪ ψ) = δφ ∪ ψ + (−1)mφ ∪ δψ.

Proof. For σ : ∆m+n+1 → X, we have

m+1 X i δφ ∪ ψ(σ) = (−1) φ(σ|[v0, ··· , vˆi, ··· , vm+1])ψ(σ|[vm+1, ··· , vm+n+1]) i=0

m+n+1 m X i (−1) (φ∪δψ)(σ) = (−1) φ(σ|[v0, ··· , vm])ψ(σ|[vm, ··· , vˆi, ··· , vm+n+1]) i=m Add and cancel the last of the first sum and the first of the second sum, we have δ(φ ∪ ψ)(σ).

21 22 CHAPTER 2. PRODUCTS

Corollary 2.1.3. There is a well defined cup product

Hm(X) × Hn(X) −→∪ Hm+n(X).

Proof. Zm(X) × Zn(X) −→∪ Zm+n(X). If φ ∈ Zm(X), φ ∪ δψ = (−1)kδ(φ ∪ ψ). Similarly Bm(X) × Zn(X) −→∪ Bm+n(X).

If our φ ∈ Sm(X,A) ⊂ Sm(X), then φ ∪ ψ ∈ Sm+n(X,A). Thus we have

Hm(X,A) × Hn(X) −→∪ Hm+n(X,A)

Hm(X) × Hn(X,A) −→∪ Hm+n(X,A) If A, B ⊂ X are relatively open subsets of A ∪ B. Then one can define

Hm(X,A) × Hn(X,B) −→∪ Hm+n(X,A ∪ B) as follows. At cochain level, take φ ∈ Sm(X,A) and ψ ∈ Sn(X,B), we have clearly φ ∪ ψ ∈ Sm+n(X,A) ∩ Sm+n(X,B). However, it is not Sm+n(X,A ∪ B). So we need the fact that the inclusion S∗(A) + S∗(B) ⊂ S∗(A ∪ B) induces isomorphism in homology, which is proved in Algebraic Topology module (Proposition 2.21 of Hatcher). This tells us cochain complex S∗(A∪ B,A) ∩ S∗(A ∪ B,B) has trivial cohomology. Look at the short exact sequence

0 → S∗(X,A∪B) → S∗(X,A)∩S∗(X,B) → S∗(A∪B,A)∩S∗(A∪B,B) → 0.

We know the inclusion S∗(X,A ∪ B) → S∗(X,A) ∩ S∗(X,B) induces iso- morphism on cohomology.

Lemma 2.1.4. For a map f : X → Y , the induced maps f ∗ : Hn(Y ; R) → Hn(X; R) satisfy f ∗(α ∪ β) = f ∗(α) ∪ f ∗(β), and similarly in the relative case.

Proof. Let α and β represented by φ ∈ Sm and ψ ∈ Sn respectively. Let σ : ∆m+n → X. So f(σ) is a (m + n)-simplex in Y . Then

f ∗(φ ∪ ψ)(σ) =(φ ∪ ψ)(f(σ))

=φ(f(σ)|[v0, ··· , vm])ψ(f(σ)|[vm, ··· , vm+n]) ∗ ∗ =f (φ)(σ|[[v0, ··· , vm])f ψ(σ|[vm, ··· , vm+n) =f ∗(φ) ∪ f ∗(ψ)(σ)

It is easy to check that the cup product is associative (even at chain level): (φ∪ψ)∪θ = φ∪(ψ ∪θ). We also know there is a unit: Let 1 ∈ S0(X) be the function takes value 1 on any point of X, then 1 ∪ a = a. 2.1. CUP PRODUCT FOR SINGULAR COHOMOLOGY 23

Proposition 2.1.5. If α ∈ Hm(X; R) and β ∈ Hn(X; R) then

α ∪ β = (−1)mnβ ∪ α.

Proof. For a singular n-simplex σ :[v0, ··· , vn] → X, letσ ¯ = σ ◦ r where r is the linear map determined by r(vi) = vn−i. Then we define ρ : Sn(X) → n(n+1) Sn(X) by ρ(σ) = nσ¯ where n = (−1) 2 . We want to show ρ is a chain map which is chain homotopic to the identity. Once we have that, the theorem follows:

∗ ∗ (ρ φ ∪ ρ ψ)(σ) = φ(mσ|[vm, ··· , v0])ψ(nσ|[vm+n, ··· , vm])

∗ ρ (ψ ∪ φ)(σ) = m+nψ(σ|[vm+n, ··· , vm])φ(σ|[vm, ··· , v0]) mn with m+n = (−1) mn, show that

ρ∗φ ∪ ρ∗ψ = (−1)mnρ∗(ψ ∪ φ) at chain level. Since ρ induces identity in cohomology, we have α ∪ β = (−1)mnβ ∪ α when passing to the cohomology level. Now the proof reduces to the following two lemmas.

Lemma 2.1.6. ρ is a chain map. Proof. For an n-simplex σ,

X i ∂ρ(σ) = n (−1) σ|[vn, ··· , vˆn−i, ··· , v0] i

X i ρ∂(σ) =ρ( (−1) σ|[v0, ··· , vˆi, ··· , vn]) i X n−i =n−1 (−1) σ|[vn, ··· , vˆn−i, ··· , v0] i n They are equal since n = (−1) n−1. Lemma 2.1.7. This chain map ρ is chain homotopic to the identity and so it induces the identity homomorphism in cohomology.

Proof. We need the fact that there is a natural division of [v0, ··· , vn] × I into n + 1 simplices. If we denote (vi, 0) by vi and (vi, 1) by wi, then these simplices are σi = [v0, ··· , vi, wi, ··· , wn]. I.e. trace out along the bottom until position i, then jump to top, trace out the rest start with position i. We define P : Sn(X) → Sn+1(X) by

X i P (σ) = (−1) n−i(σ ◦ π)|[v0, ··· , vi, wn, ··· , wi], i 24 CHAPTER 2. PRODUCTS where π : ∆ × I → ∆ is a projection. We will leave out σ ◦ π for notational simplicity.

X i j ∂P (σ) = (−1) (−1) n−i[v0, ··· , vˆj, ··· , vi, wn, ··· , wi] j≤i X i i+1+n−j + (−1) (−1) n−i[v0, ··· , vi, wn, ··· , wˆj, ··· , wi] j≥i

X i j P ∂(σ) = (−1) (−1) n−i−1[v0, ··· , vi, wn, ··· , wˆj, ··· , wi] ij n−i Since n−i = (−1) n−i−1, the terms with i 6= j cancel in the two sums. The terms with j = i give X n[wn, ··· , w0]+ n−i[v0, ··· , vi−1, wn, ··· , wi] i>0 X n+i+1 + (−1) n−i[v0, ··· , vi, wn, ··· , wi+1] − [v0, ··· , vn] i

∂P (σ) + P ∂(σ) = n[wn, ··· , w0] − [v0, ··· , vn] = ρ(σ) − σ. Hence P is the chain homotopy relates ρ to identity. Remark 2.1.8. As we mentioned in the class, this proof should be read using picture. P is the map from a simplex to get an oriented cylinder with simplicial division. Then ∂P is the oriented boundary of the cylinder. P ∂ is the part of the boundary without top and bottom and with opposite sign. Cancellations could also be interpreted: the cancellation on i 6= j is the cancellation on the boundary without top and bottom. Then second cancella- tion above is the intersection face of different simplices in the division with different orientation. To summarize, we have Theorem 2.1.9 (The cohomology ring). Let X be a topological space and R a commutative ring with identity. Then M H∗(X; R) = Hi(X; R) i is a graded commutative ring with identity under the cup product, i.e. if α ∈ Hm(X; R) and β ∈ Hn(X; R) then α ∪ β = (−1)mnβ ∪ α. Moreover H∗(X; R) is an R-algebra. 2.1. CUP PRODUCT FOR SINGULAR COHOMOLOGY 25

∗ n Z[an] 0 Example 2.1.10. H (S ; ) = 2 , where the generator of H corre- Z an n sponds to 1 and the generator of H is an.

Actually, we are ready to calculate the cohomology ring of projective n n spaces RP and CP . But I would like to leave these after Poincar´eduality. We remark that all the above results extend to relative case, i.e replacing H∗(X; R) with H∗(X,A; R).

2.1.1 Cap product

For any space X, there is a bilinear pairing operation between cochains and chains.

q Definition 2.1.11. Let a ∈ S (X) and σ ∈ Sp+q(X). Then the cap product a ∩ σ ∈ Sp(X) is

a ∩ σ =< a, σ|[vp, ··· , vp+q] > ·σ|[v0, ··· , vp], or a ∩ σ =< a, σq > ·pσ.

The cap product at chain level has the following properties.

∗ Proposition 2.1.12. 1. Duality: For a, b ∈ S (X), c ∈ S∗(X), we have

< a ∪ b, c >=< a, b ∩ c > .

∗ 2. Associativity: For a, b ∈ S (X), c ∈ S∗(X), we have

(a ∪ b) ∩ c = a ∩ (b ∩ c).

3. Unit: 1 ∩ c = c.

∗ 4. Naturality: Let f : X → Y be a map. For b ∈ S (Y ) and c ∈ S∗(X), we have ∗ b ∩ (f∗c) = f∗(f b ∩ c).

All these are easy to derive from the definition and the properties of cup product. To define cap product on (co)homology, we need

q Proposition 2.1.13. For a ∈ S (X) and σ ∈ Sp+q(X), we have

∂(a ∩ σ) = (−1)p(δa) ∩ σ + a ∩ (∂σ). 26 CHAPTER 2. PRODUCTS

Proof. For simplicity, we omit the notation σ in the calculation.

p+q X i a ∩ ∂[v0, ··· , vp+q] = (−1) a ∩ [v0, ··· , vˆi, ··· , vp+q] i=0 p X i = (−1) < a, [vp, ··· , vp+q] > [v0, ··· , vˆi, ··· , vp] i=0 p+q p−1 X i−p+1 + (−1) (−1) < a, [vp−1 ··· , vˆi, ··· , vp+q] > [v0, ··· , vp−1] i=p−1 p−1 =∂(a ∩ σ) + (−1) < a, ∂[vp−1 ··· , vp+q] > [v0, ··· , vp−1] =∂(a ∩ σ) + (−1)p−1(δa) ∩ σ

Hence it induces the cap product between homology and cohomology:

q ∩ H (X) × Hp+q(X) → Hp(X).

2.1.2 De Rham cohomology Let M be a smooth manifold of dimension n now. Let Ωq(M) be the (real linear) space of q-forms, let d :Ωq(M) → Ωq+1(M) be the exterior differential. Then we have the de Rham cochain complex

0 −→d Ω0(M) −→d Ω1(M) −→d · · · −→d Ωn(M) −→d 0.

We denote he space of closed (exact) k-forms, i.e k-forms ω with dω = 0 k k (ω = dη respectively), by ZdR(M) (and BdR(M) respectively). We denote ∗ its cohomology by HdR(M), which is called the de Rham cohomology of M. The cup product for de Rham cohomology is just the wedge product ω∧η. Since for k-form ω

d(ω ∧ η) = dω ∧ η + (−1)kω ∧ dη we know it descends to a product on cohomology by the same argument as Corollary 2.1.3. A notable fact for de Rham cohomology is its cup product is graded commutative even at chain level: x ∧ y = (−1)|x||y|y ∧ x. There are several reasons prevent us from exploiting the Eilenberg-Steenrod ∗ uniqueness theorem to claim it is isomorphic to singular cohomology H (X; R). The main reason is de Rham cohomology is only defined for smooth mani- folds. The lack of relative version of de Rham theory could be compensated by Thom isomorphism. However, de Rham theorem does ensure this isomorphism. 2.2. CROSS PRODUCT AND KUNNETH¨ FORMULA 27

∗ ∗ Theorem 2.1.14 (de Rham theorem). HdR(M) = H (M; R) as cohomology rings. More precisely, we need two facts for this theorem: sm 1. Let Sq (M; R) be the real space spanned by smooth singular q-simplices σ : ∆q → M, then the inclusion

sm S∗ (M; R) → S∗(M; R) is a chain homotopic equivalence. Then its dual

∗ ∗ S (M; R) → Ssm(M; R) is a cochain homotopic equivalence. 2. We could take integration of a q-form on a singular chain of dimension q, this is a bilinear function Z q sm Ω (M) × Sq (M; R) → R, (ω, σ) 7→ ω. σ

Stokes theorem tells us Z Z ω = dω. ∂σ σ In other words, exterior differential is dual to boundary map. This pro- vides us a cochain map

∗ ∗ Ω (M) → Ssm(M; R). If we show this is a cochain homotopic equivalence, we finish the proof of de Rham theorem. The proof proceeds as the same pattern as the proof of Poincar´eduality which we will provide in next chapter.

2.2 Cross Product and K¨unnethformula

We want to understand the cohomology ring of a product space. Let us first define the cross product in cohomology.

Definition 2.2.1. If p1 : X × Y → X and p2 : X × Y → Y are the projections, then define the cross product by

∗ ∗ m+n x × y = p1(x) ∪ p2(y) ∈ H (X × Y ; R) for x ∈ Hm(X; R) and y ∈ Hn(Y ; R).

m+n m+n To be more precise, for σ : ∆ → X × Y , let σ1 = p1 ◦ σ : ∆ → X m+n m n and σ2 = p2 ◦ σ : ∆ → Y . Then for φ ∈ S (X) and ψ ∈ S (Y ), we have

(φ × ψ)(σ) = φ(σ1|[v0, ··· , vm])ψ(σ2|[vm, ··· , vm+n]) 28 CHAPTER 2. PRODUCTS

We could also extend this to a relative version

Hm(X,A) × Hn(Y,B) → Hm+n(X × Y,X × B ∪ A × Y ).

Since × is bilinear, it factors through the tensor product to give a linear map (also denoted ×)

∗ ∗ ∗ H (X; R) ⊗R H (Y ; R) → H (X × Y ; R). Proposition 2.2.2. For any a, c ∈ H∗(X; R) and b, d ∈ H∗(Y ; R) with c ∈ Hm(X; R) and b ∈ Hn(Y ; R) we have

(a × b) ∪ (c × d) = (−1)mn(a ∪ c) × (b ∪ d).

∗ ∗ ∗ ∗ mn ∗ ∗ Proof. Because (p1a ∪ p2b) ∪ (p1c ∪ p2d) = (−1) p1(a ∪ c) ∪ p2(b ∪ d). Exercise: By above proposition and induction for dimension n, prove the cohomology ring H∗(T n) for a n-torus is isomorphic to exterior algebra

ΛZ[x1, ··· , xn]. Any generator has dimension 1. Recall that the exterior algebra ΛR[x1, x2 ··· ] over a commutative ring R with identity is the free R-module with basis the finite product xi1 ··· xik , i1 < ··· < ik, with the multiplication defined by the rules xixj = −xjxi and 2 xi = 0. The following K¨unnethformula generalizes the previous example.

∗ ∗ ∗ Theorem 2.2.3. The cross product H (X; R)⊗RH (Y ; R) → H (X×Y ; R) is an isomorphism of rings if X and Y are CW complexes and Hk(Y ; R) is a finitely generated free R-module for all k. We want to use Theorem 1.6.1. We show the theorem for finite CW complexes. For general CW complexes, as we mentioned before, we need Milnor additivity axiom. Consider the following functors:

n i j h (X,A) = ⊕i+j(H (X,A; R) ⊗R H (Y ; R), kn(X,A) = Hn(X × Y,A × Y ; R). We have φ : hn(X,A) → kn(X,A) given by the cross product. So we need to check 1. h∗ and k∗ are cohomology theories. 2. φ is a natural transformation. Proof. First we check h∗ and k∗ are cohomology theories. All axioms are easy to verify. A few words for exactness axiom. The exactness for k∗ is trivial. For h∗, it is where we use the freeness of Hk(Y ; R). Naturality of φ with respect to maps between spaces is from the natu- rality of cup products. Natuarality with respect to the coboundary maps is to check the following diagram commutes. We save R in our notation. 2.2. CROSS PRODUCT AND KUNNETH¨ FORMULA 29

Hk(A) × Hl(Y ) δ×id Hk+1(X,A) × Hl(Y )

× ×

Hk+l(A × Y ) δ Hk+l+1(X × Y,A × Y )

To check this, start with an element (a, b) represented by cocycle φ ∈ Sk(A) and ψ ∈ Sl(Y ). Extend φ to a cochain φ¯ ∈ Sk(X; R). Then (φ, ψ) ¯ ∗ ¯ ∗ maps rightward to (δφ, ψ) and then downward to πX (δφ) ∪ πY (ψ). On the ∗ ∗ other direction, (φ, ψ) maps downward to πX (φ)∪πY (ψ) and then rightward ∗ ¯ ∗ ∗ ¯ ∗ to δ(πX (φ) ∪ πY (ψ)) = πX (δφ) ∪ πY (ψ). I hope the chain level δ and connecting homomorphism δ do not confuse you!

Example 2.2.4. Now it is more straightforward to show

∗ n H (T ) = ΛZ[x1, ··· , xn]. Similarly, one could also show

∗ n m [an, am] H (S × S ) = 2 2 . an = 0, am = 0

2.2.1 An alternative way to define cross and cup products We could construct cup product for CW complexes. By virtue of Eilenberg- Steenrod uniqueness, this cup product is the same as that of singular coho- mology. The cross product in this setting is quite natural. Start with that of chain level. Take cells ei ∈ X and ej ∈ Y , then we could send it to product cell ei × ej in X × Y . One could extend this map by tensor product from C∗(X) ⊗ C∗(Y ) to C∗(X × Y ). Then for a pair of cocycles z1, z2 of X and Y , it thus yields a cocycle z1 × z2. This defines Hi(X) × Hj(Y ) → Hi+j(X × Y ).

One could check it is the same as our previous defined cross product. Then by using the diagonal map ∆ : X → X × X, x 7→ (x, x), we can define cup product as the composition

∗ Hk(X) × Hl(X) →× Hk+l(X × X) ∆→ Hk+l(X).

This is our previous cup product since

∗ ∗ ∗ ∗ ∆ (a × b) = ∆ (p1(a) ∪ p2(b)) = a ∪ b. But unfortunately, this cup product is not defined at the level of cellular cochains, which prevents us to prove the properties. To resolve this issue, 30 CHAPTER 2. PRODUCTS one need to find cellular map which is homotopic to ∆. This is in general true for maps between CW complexes. And for our case X → X × X, the map is actually a slight modification of P (called Alexander-Whitney chain approximation) used in the proof of graded commutativity of cup product. We do not go through the detail again.

2.3 Ljusternik-Schnirelmann category

Definition 2.3.1. The Ljusternik-Schnirelmann category cat(X) of a topo- logical space X is defined to be the smallest integer k such that there is an open covering {Ui}1≤i≤k of X such that each inclusion Ui ,−→ X is null- homotopic (we call Ui is contractible in X), i.e homotopic to a constant map.

Example 2.3.2. Notice the subtly of the definition. Sn−1 is contractible in Dn, although Sn−1 itself is not contractible space.

Example 2.3.3. cat(S1) = 2. Actually, for any suspension Σ(X), cat(Σ(X)) ≤ 2 since it could be covered by two contractible sets C+(X) and C−(X).

Notice cat(M) < ∞ if M is a compact manifold since it could be covered with finitely many sets homeomorphic to open discs. And in fact cat(M) ≤ dim M + 1. In general, there is no reason for cat(X) to be finite.

Definition 2.3.4. The cup length cl(X) of a topological space X is defined to be

mi cl(X) := max{n|there exist αi ∈ H (X) with mi > 0 such that α1∪· · ·∪αn 6= 0}.

Proposition 2.3.5. For any space X, we have cl(X) < cat(X).

Sn Proof. Suppose cat(X) = n, so X = j=1 Uj with each Uk contractible in X. We denote the inclusion by ik. Since ik is nullhomotopic, its induced ∗ ∗ ∗ ∗ homomorphism ik could be decomposed as H (X) → H (pt) → H (Uk). So ∗ q q when q > 0, ik = 0 : H (X) → H (Uk). By the cohomology exact sequence ∗ q q for the pair (X,Uk), we know jk : H (X,Uk) → H (X) is surjective. So for ∗ ∗ ∗ any ξk ∈ H (X), we have ηk ∈ H (X,Uk) such that ξk = jk(ηk). Look at the commute diagram

∗ ∗ ∪ ∗ n H (X,U1) × ··· × H (X,Un) H (X, ∪k=1Uk)

∗ ∗ ∗ j1 jn j H∗(X) × ··· × H∗(X) ∪ H∗(X) 2.3. LJUSTERNIK-SCHNIRELMANN CATEGORY 31

∗ ∗ ∗ Hence ξ1 ∪ · · · ∪ ξn = j1 (η1) ∪ · · · ∪ jn(ηn) = j (η1 ∪ · · · ∪ ηn) = 0. The last ∗ n ∗ equality is because of H (X, ∪k=1Uk) = H (X,X) = 0.

Example 2.3.6. An n-torus T n has cat(T n) = n + 1.

Example 2.3.7. So for any suspension, cl(Σ(X)) = 1 if X is not weakly contractible. This tells us cup product does not commute with suspension and hence not a stable property.

Ljusternik-Schnirelmann use the notion of cat to study critical points. Their main theorem is the following

Theorem 2.3.8. Let M be a smooth connected compact manifold and f : M → R is a smooth function. Then f has at least cat(X) critical points. Example 2.3.9. Any smooth function on T 2 has at least 3 critical points. Can you construct a smooth function on T 2 with exactly 3 critical points? (Hint: use the viewpoint in the proof of the following theorem, construct a vector field on torus with 3 singular points.)

Actually, Ljusternik-Schnirelmann category is an example of general cat- egory (It has nothing to do with “category theory”). We assume X be a locally contractible path connected space.

X Definition 2.3.10. A category is an assignment ν : 2 → N ∪ {0} (where 2X denotes the set of all subsets in X, i.e. the power set) satisfying the following axioms:

• Continuity: for every A ∈ 2X there exists an open set U ⊃ A such that ν(A) = ν(U).

• Monotonicity: if A, B ∈ 2X with A ⊂ B then ν(A) ≤ ν(B).

• Subadditivity: for any A, B ∈ 2X we have ν(A ∪ B) ≤ ν(A) + ν(B).

• Naturality: if φ : X → Y is a homeomorphism then for any A ∈ 2X , νY (φ(A)) = νX (A).

• Normalization: ν(∅) = 0, and if A = {x0, ··· , xn} is a finite set then ν(A) = 1.

To prove Theorem 2.3.8, we prove the following more general proposition.

Proposition 2.3.11. Let X be a locally contractible path connected compact metric space, and φt a global flow on X. Suppose there exists a Lyapunov function Φ: X → R such that Φ strictly decreases along non-constant orbits of φt. Then Φ has at least ν(X) critical points where ν is any category. 32 CHAPTER 2. PRODUCTS

Proof. Let Xc := Φ−1(−∞, c]. A critical value for Φ is a value such that Φ−1(c) contains a constant orbit. If c is not critical, then for sufficiently small δ > 0, we could find t > 0 such that

c+δ c−δ φt(X ) ⊂ X , since Φ strictly decreases away from the constant orbits. If c is a critical level and U is a neighorhood of Φ−1(c) then for small δ > 0, we have t > 0

c+δ c−δ φt(X \U) ⊂ X .

By naturality and monotonicity, ν(Xc+δ\U) ≤ ν(Xc−δ). For j = 1, ··· ,N = ν(X), let

c cj := sup{c|ν(X ) < j}.

Then c1 = min{Φ} and cN = max{Φ}. Note that cj is a critical value of Φ for each j. −1 Now we want to prove either cj < cj+1 or Φ (cj) contains infinitely many critical points. If the latter happens, the theorem follows immediately. −1 So we assume the latter does not happen, say Φ (cj) = {x0, ··· , xn}. Then by continuity axiom, there exists a neighborhood U of {x0, ··· , xn} such that ν(U) = 1. Then by subadditivity, we have

ν(Xcj +δ) ≤ν(Xcj +δ\U) + 1 ≤ν(Xcj −δ) + 1 ≤j

Hence cj+1 ≥ cj + δ > cj. Thus c1 < ··· < cN are N = ν(X) different critical points. This completes the proof.

Then let us finish the proof of Theorem 2.3.8: Give M a Riemannian metric and let ∇f denote the gradient of f with respect to this metric, i.e. it is the unique vector field determined by

< (∇f)p,Vp >= dfp(Vp) for every vector field V . The critical points of f are precisely the zeros of ∇f. Let φt be the associated flow of ∇f, i.e.

dφ (p) t = −(∇f) . dt φt(p)

We claim that f is a Lyapunov function for φt: 2.4. HIGHER PRODUCTS 33

d dφ (p) (f ◦ φ (p)) =df( t ) dt t dt dφ (p) = < (∇f) , t > φt(p) dt dφ (p) dφ (p) =− < t , t > dt dt ≤0

dφt(p) with equality holds if and only if dt = 0, i.e. p is a critical point of f, and φt(p) is a constant orbit p. This completes the proof.

2.4 Higher products

Later we will see that the linking number of two spheres Sp and Sq in Rp+q+1 will be understood as cup product of the cohomolgy ring of the complement H∗(Rp+q+1\(Sp ∪Sq)). There are links with 3 components with each two of them are unlinked, but nonetheless all three are link. The most famous example is the Borromean rings. This more complicated linking phenomenon for three or more spheres suggest the existence of higher cup product: the Massey product. We start with Massey triple product. Assume [u], [v], [w] are cohomology classes of dimension p, q and r re- spectively, represented by u ∈ Zp(X), v ∈ Zq(X) and w ∈ Zr(X). If [u] ∪ [v] = 0 = [v] ∪ [w], then we introduce a (set of) new cohomology classes. For the notation, we introduceu ¯ = (−1)1+deg uu. Since [u]∪[v] = 0, we have s ∈ Cp+q−1(X) such that δs =u ¯∪v. Similarly, we have t ∈ Cq+r−1(X) such that δt =v ¯ ∪ w. The elements ¯ ∪ w +u ¯ ∪ t determines a cocycle in Zp+q+r−1(X):

δ(¯s ∪ w +u ¯ ∪ t) =(−1)p+qδs ∪ w + (−1)pu¯ ∪ δt =(−1)p+qu¯ ∪ v ∪ w + (−1)p+q+1u¯ ∪ v ∪ w = 0

We define the Massey triple product as the set of all such cohomology classes

< [u], [v], [w] >= {[¯s ∪ w +u ¯ ∪ t]|δs =u ¯ ∪ v, δt =v ¯ ∪ w}.

There are indeterminacy from the different choices of representatives. But we could identify them as following. Proposition 2.4.1. The Massey triple product < [u], [v], [w] > is an ele- ment of quotient group Hp+q+r−1(X)/([u]∪Hq+r−1(X)+Hp+q−1(X)∪[w]). Proof. We need to show different choices of u, v, w, s, t do not affect the coset in Hp+q+r−1(X) given above. We only check that of s. Others are left as exercise. 34 CHAPTER 2. PRODUCTS

If s and s0 are chosen such that δs =u ¯ ∪ v = δs0, then

(¯s ∪ w +u ¯ ∪ t) − (¯s0 ∪ w +u ¯ ∪ t) = (¯s − s¯0) ∪ w, which is in Hp+q−1(X) ∪ [w] as cohomology classes.

The Massey product was used to prove the Jacobi identity for Whitehead product in homotopy group. We could define higher order Massey products. When two triple products < [u], [v], [w] > and < [v], [w], [x] > are defined, and if 0 ∈< [u], [v], [w] > and 0 ∈< [v], [w], [x] >, then we could find

δY1 = t¯0 ∪ w +u ¯ ∪ t1, δY2 = t¯1 ∪ x +v ¯ ∪ t2 where δt0 =u ¯ ∪ v, δt1 =v ¯ ∪ w, δt2 =w ¯ ∪ x. Then we form a subset < [u], [v], [w], [x] > in H|u|+|v|+|w|+|x|−2(X) whose elements are

u¯ ∪ Y2 + t¯0 ∪ t2 + Y¯1 ∪ x.

This is called a fourfold product. As we mentioned in the class, it is better to understand the whole picture by matrix u s   v t  w   u t0 Y1  v t1 Y2    w t2  x

We can do inductively to define n-fold Massey product. < [a1,1], [a2,2], ··· , [an,n] > to be the set of elements of the forms

a¯1,1a2,n +a ¯1,2a3,n + ··· +a ¯1,n−1an,n for all solutions of the equations

δai,j =a ¯i,iai+1,j +a ¯i,i+1ai+2,j + ··· +a ¯i,j−1aj,j, 1 ≤ i ≤ j ≤ n, (i, j) 6= (1, n).

Hence, to ensure the set is non-empty, we need the vanishing of many lower order Massey product. We remark that the Massey products are defined for (homology) of a k differential graded algebra (DGA) A. It is a graded algebra A = ⊕k≤0A with a differential d : A → A of degree +1, such that 1. A is graded commutative, i.e

x · y = (−1)kly · x, x ∈ Ak, y ∈ Al 2.4. HIGHER PRODUCTS 35

2. d is a derivation, i.e.

d(x · y) = dx · y + (−1)kx · dy, x ∈ Ak

3. d2 = 0

Examples include cohomology ring with 0 as its differential and de Rham complex (Ω∗, d) on a manifold. Rational homotopy theory of Quillen and Sullivan is built to understand (real) homotopy group by the structure of DGA. For example, a manifold on which all Massey products vanish is a formal manifold: its real homotopy type follows (“formally”) from its real cohomology ring. Deligne-Griffiths- Morgan-Sullivan proved that all K¨ahlermanifolds are formal. 36 CHAPTER 2. PRODUCTS Chapter 3

Poincar´eduality

We want to prove the Poincar´eduality in this chapter. For a compact n- p n n manifold M, this asserts that H (M ) is isomorphic to Hn−p(M ). It is the most important result in this course, and has lots of important applications. Poincar´e’soriginal proof used the idea of dual cell structures (explained in the class by pictures). It is rather intuitive and geometric proof. However, we will lose some generality if we use this method. Hence, we use the proof of Milnor. The basic idea of Milnor’s proof is very natural and explained as follows. We know that any n-manifold n is a union of open subsets, each of which is homeomorphic to R . It is n natural to first prove (certain version of) the theorem for R , and then use Mayer-Vietoris sequences to prove the case of a finite union of open subsets. Finally, it passes to the case of an infinite union by a direct limit argument. We will then state and prove a more general version which is applicable to n noncompact manifolds since we have to first deal with R . For this reason, we need to introduce cohomology with compact supports.

3.1 Cohomology with compact supports

Suppose M is a topological space. The singular cochains with compact support is defined as α ∈ Sp(M) such that there is a compact set K ⊂ M p p p such that α ∈ S (M,M\K) ⊂ S (M), i.e. α|M\K = 0. Write Sc (M) for ∗ the set of all these cochains. Note that δ preserves Sc (M). Hence

p p ∗ Definition 3.1.1. Hc (M) := H (Sc (M)) is the cohomology with compact support of M.

∗ ∗ Observe that if M is compact Hc (M) = H (M). We will need to cal- ∗ culate Hc (M) in general for the proof of Poincar´eduality. We want to understand it by relative cohomology groups. So we introduce the defini- tion of a direct limit.

37 38 CHAPTER 3. POINCARE´ DUALITY

Definition 3.1.2. A directed system of abelian groups {Gα|a ∈ A} is a collection of abelian groups indexed by a partially ordered set A satisfying

1. For all a, b ∈ A there exists c ∈ A such that a ≤ c and b ≤ c.

2. For all a ≤ b there exists a homomorphism fab : Ga → Gb such that faa = id and if a ≤ b ≤ c, fac = fbc ◦ fab.

Recall that a partially ordered set (or poset) is a set A along with a binary relation ≤ which is reflexive, antisymmetric and transitive:

1. Reflexive: a ≤ a;

2. Antisymmetric: if a ≤ b and b ≤ a, then a = b;

3. Transitive: if a ≤ b and b ≤ c, then a ≤ c.

A partially ordered set with property 1 in Definition 3.1.2 is called a directed set. The main example of directed sets in our mind is the set of compact subsets with partial ordering ⊂ and f is the inclusion.

Definition 3.1.3. Given a direct system {Gα|a ∈ A}, the Direct Limit (colimit) is defined to be

lim G := (⊕ G )\N −→ a a∈A a a∈A where N ⊂ ⊕a∈AGa is the subgroup generated by x − fab(x) where x ∈ Ga and a ≤ b.

Direct limit has the following universal property. Actually, this property characterizes direct limit.

Proposition 3.1.4 (Universal property of direct limit). Any homomor- phisms φa : Ga → H such that φa = φb ◦fab for any pair a ≤ b factor through lim G . That is, there exists a unique homomorphism φ : lim G → H −→a∈A a −→a∈A a such that φ ◦ h = φ for all a ∈ A. Here h : G → lim G are the in- a a a a −→a∈A a clusion maps.

fab Ga Gb ha hb

lim G a∈A a φa −→ φb φ H 3.1. COHOMOLOGY WITH COMPACT SUPPORTS 39

Proof. For any g ∈ G , we denote its equivalence class in lim G as [g]. a −→a∈A a So ha(g) = [g]. Let us first construct a φ: φ([g]) = φa(g). It is easy to check that it makes the diagram commutes. Let us check it is well defined. Suppose there are g1 ∈ Ga and g2 ∈ Gb such that [g1] = [g2]. Then we know there is a c such that a ≤ c and b ≤ c, and fac(g1) = fbc(g2). Then we see that

φa(g1) = φc ◦ fac(g1) = φc ◦ fbc(g2) = φb(g2).

Then we prove the uniqueness. Otherwise there is another φ0 : lim G → −→a∈A a 0 H such that φ ◦ ha = φa for all a ∈ A. Then

0 0 φ ([g]) = φ ◦ ha(g) = φa(g) = φ([g]).

In other words, φ0 = φ.

∗ Now we could give an alternative definition of Hc (M) in terms of direct limit. The compact subsets K ⊂ M form a directed set under inclusion. For K ⊂ L, we have the inclusion (M,M\L) → (M,M\K), and thus the homomorphism Hp(M,M\K) → Hp(M,M\L). Hence we have the direct limit lim Hp(M,M\K). −→ K⊂M p p This is equal to Hc (M) we defined at beginning. It is easy to see Hc (M) ⊂ lim Hp(M,M\K) by definition. For the other inclusion, each element −→K⊂M of lim Hp(M,M\K) is represented by a cocycle in Sp(M,M\K) for −→K⊂M some compact K, hence the inclusion at cochain level. And such a cocycle is zero in lim Hp(M,M\K) if and only if it is in Bp(M,M\L) for some −→K⊂M compact L ⊃ K, hence the inclusion passes to the cohomology level.

∗ n n Example 3.1.5. We compute Hc (R ). Since every compact set of R is contained in the closed ball DR(0) of some radius R ∈ N, we have lim Hp( n, n\D (0)) = lim Hp( n, n\K). −→ R R R −→ R R R∈N K⊂Rn n n−1 Now for any R > 0, R \DR(0) is homotopy equivalent to S . Hence n n the long exact sequence for pairs (R , R \DR(0)) gives us  m = n Hm( n, n\D (0)) = Z R R R 0 m 6= n n n n n n n Since the map H (R , R \DR(0)) → H (R , R \DR+1(0)) corresponds to the inclusions are isomorphism, we conclude

 m = n Hm( n) = Z c R 0 m 6= n 40 CHAPTER 3. POINCARE´ DUALITY

∗ This example tells us Hc (M) is NOT a cohomology theory in the sense of Eilenberg-Steenrod. Since it is not a homotopy invariant: A point is ∗ ∗ ∗ n ∗ n compact, so Hc (pt) = H (pt), but Hc (R ) is not the same as H (R ). n Actually, it is even not a functor: the constant map R → pt does not induce a map on cohomology with compact support.

3.2 Orientations for Manifolds

Suppose M is an n-manifold. For each x ∈ M, choose an open ball U with x ∈ U. Then by excision,

Hn(M,M\x) = Hn(U, U\x) = Z.

Definition 3.2.1. A local orientation µx for M at x is a choice of one of the two possible generators for Hn(M,M\x).

If x, y ∈ U then we have homomorphisms ρx and ρy induced by the inclusion of pairs

ρx ρy Hn(M,M\x) ← Hn(M,M\U) → Hn(M,M\y).

So a generator for Hn(M,M\U) gives a local orientation at any point in U.

Definition 3.2.2. An orientation of M is a function x 7→ µx subject to the following continuity condition: Given any point x ∈ M, there exists a neighborhood U of x and an element µU ∈ Hn(M,M\U) such that ρy(µU ) = µy for each y ∈ U. We say M is orientable if there exists an orientation. And if the orien- tation is fixed, M is oriented. One could translate this into the connectedness of the double cover

M˜ = {µx|x ∈ M, µx is a local orientation of M at x}. We will not go through this construction, see Hatcher for details. But we summarize two useful criterion Proposition 3.2.3. 1. If M is simply connected, then M is orientable.

1 2. Suppose H (M; Z2) = 0, then M is orientable. n n n Example 3.2.4. 1. R , S , CP are orientable. 2. If M and N are orientable, then M × N is orientable. To determine which manifolds are not orientable, we have the following lemma. For the proof, we need relative Mayer-Vietoris sequence which is assigned in Example sheet 1:

· · · → Hn(X,A∩B) → Hn(X,A)⊕Hn(X,B) → Hn(X,A∪B) → Hn−1(X,A∩B) → · · · 3.2. ORIENTATIONS FOR MANIFOLDS 41

Lemma 3.2.5. Let M be a n-manifold, and K ⊂ M be a compact subset. Then

1. Hi(M,M\K) = 0 for i > n;

2. Suppose x 7→ ax is an orientation of M. Then there is a unique class aK ∈ Hn(M,M\K) whose image in Hn(M,M\x) is ax for all x ∈ K.

Proof. We want to use the relative Mayer-Vietoris sequence for a triple (X, A, B):

· · · → Hp(X,A ∩ B) → Hp(X,A) ⊕ Hp(X,B) → Hp(X,A ∪ B) → · · · We write our proof in 4 steps: 1. Suppose the lemma is true for K1, K2 and K1 ∩ K2, we want to prove it is true for K1 ∪ K2 as well. Then we take X = M, A = M\K1, B = M\K2, so

· · · → Hp+1(M,M\(K1∩K2)) → Hp(M,M\K) → Hp(M,M\K1)⊕Hp(M,M\K2) → · · ·

By assumption, if p > n the left and right term are both zero, so Hp(M,M\K) = 0. For the second statement, we know the map Hn(M,M\K1)⊕Hn(M,M\K2) →

Hn(M,M\(K1 ∩K2)) is the difference map aK1 −aK2 . By uniqueness, it has to be zero. So we have aK from the exact sequence. This is unique because, Hn+1(M,M\(K1 ∩ K2)) = 0. n 2. We reduce the problem to the case M = R . Any compact set K ⊂ M can be written as K1 ∪ · · · ∪ Km where each Ki is contained in a n neighborhood which is homeomorphic to a ball in R . Then apply step 1 and induction on K1 ∪ · · · ∪ Km−1, Km and their intersection. n n 3. Suppose M = R and K ⊂ R is a compact convex subset. For any point x ∈ K, and S is a large (n−1)-sphere with centre x. Then n n S is a deformation retract of both R \x and R \K. Hence the map n n n n Hi(R , R \K) → Hi(R , R \x) is an isomorphism for each i. Induction also shows this is true when K is a finite union of compact convex sets. n 4. Now suppose K ⊂ R is an arbitrary compact subset and β ∈ n n n Hi(R , R \K). We choose a relative cycle z with [z] = β. Let C ∈ R \K be the union of the images of the boundary of singular simplices in z. C is compact (could be ∅ if z is closed in the absolute sense), the distance from K to C is some real number δ > 0. Cover K by finitely many balls with centres in K and radii < δ. Let N be the union of these balls and so K ⊂ N and z defines a class βN ∈ n n Hi(R , R \N) so that the restriction ρK (βN ) = β. 42 CHAPTER 3. POINCARE´ DUALITY

If i > n then by step 3, βN = 0 so β = 0. This with step 2 finishes the first part of the lemma. If i = n, step 1 and step 3 also construct aN and then aK = ρK (aN ) 0 such that ρx(aN ) = ax and ρx(aK ) = ax. We prove the uniqueness: if aK 0 is another choice, let β = aK − aK . So ρx(β) = 0 for any x ∈ K, especially when x is one of the centres of the balls to define N. By step 3 again, β is zero 0 n n on these balls and thus on N. Hence aK −aK = β = 0 ∈ Hn(R , R \K). When M is closed (i.e. compact without boundary), take K = M, we have Corollary 3.2.6. Suppose M is a connected closed n-manifold. Then

1. Hi(M) = 0 if i > n;

2. M is orientable if and only if Hn(M) = Z. If M is not orientable, Hn(M) = 0.

Proof. Need a bit words for non-orientable case. If Hn(M) 6= 0, take a cycle z 6= 0. We take a cell decomposition of M. Then at two sides of any n−1-dimensional cell, the coefficient of z is the same. Since M is connected, the coefficient on all n-cells are the same. This gives us an orientation on M.

2n 2n 2n+1 Example 3.2.7. RP is not orientable, since H2n(RP ) = 0. RP is 2n+1 orientable since H2n+1(RP ) = Z. In particular, if M itself is compact, then there is one and only one µM ∈ Hn(M) with the required property. This class µ = µM is called the fundamental homology class of M.

3.3 Poincar´eduality theorem

The Poincar´eduality for compact manifolds could be stated now. Theorem 3.3.1. Let M be a compact and oriented n-manifold, then the homomorphism

p D : H (M) → Hn−p(M), α 7→ α ∩ µM is an isomorphism. It actually follows from a more general theorem (which we will prove), for any oriented manifolds. Before stating the result, we need to explain the notations. First observe that for any pair (X,A), the cap product gives rise to a paring i S (X,A) ⊗ Sn(X,A) → Sn−i(X) 3.3. POINCARE´ DUALITY THEOREM 43 and hence to pairing

i H (X,A) ⊗ Hn(X,A) → Hn−i(X).

For oriented M, we define the duality map

p D : Hc (M) → Hn−p(M) as follows. For any a ∈ Hp(M) = lim Hp(M,M\K), choose a representative c −→ a0 ∈ Hp(M,M\K) and set

0 D(a) = a ∩ µK .

This is well defined since for K ⊂ L, we have the restriction

ρK : Hn(M,M\L) → Hn(M,M\K) with ρK (µL) = µK . Then the naturality of the cap product tells us the following diagram commutes:

Hi(M,M\K) ∩µK

Hn−i(M)

∩µL Hi(M,M\L)

Theorem 3.3.2. Let M be an oriented n-manifold, then the homomorphism

p D : Hc (M) → Hn−p(M) is an isomorphism.

n Proof. 1. We first prove it for M = R . Given a closed ball B, we know that n n n n n Hn(R , R \B) = Z with generator µB. Hence H (R , R \B) = Z and by n n n universal coefficient theorem, the homomorphism h : H (R , R \B; Z) → n n Hom(Hn(R , R \B); Z) is an isomorphism. Then there exists a generator a such that < a, µB >= 1. Now the identity

< 1 ∪ a, µB >=< 1, a ∩ µB >

n shows that a ∩ µB is a generator of H0(R ) = Z. Thus ∩µB gives an iso- n n n n morphism H (R , R \B) → H0(R ) for all B. Hence by universal property of direct limit, the map D is an isomorphism in the case i = n. The cases i 6= n is obvious since it maps 0 to 0. 2. Suppose M = U ∪ V and that the theorem is true for U, V, U ∩ V . ∗ We first construct Mayer-Vietoris for Hc :

p−1 p p p p · · · → Hc (M) → Hc (U ∩ V ) → Hc (U) ⊕ Hc (V ) → Hc (M) → · · · 44 CHAPTER 3. POINCARE´ DUALITY

This is obtained from relative Mayer-Vietoris

Hp(M,M\(K ∩L)) → Hp(M,M\K)⊕Hp(M,M\L) → Hp(M,M\(K ∪L)) and excisions

Hp(M,M\(K ∩ L)) = Hp(U ∩ V,U ∩ V \(K ∩ L))

Hp(M,M\K) = Hp(U, U\K) Hp(M,M\L) = Hp(V,V \L) Now if we know the following diagram of exact sequence is commutative (up to a sign)

p p p p δ p+1 Hc (U ∩ V ) Hc (U) ⊕ Hc (V ) Hc (M) Hc (U ∩ V )

D D D D ∂ Hn−p(U ∩ V ) Hn−p(U) ⊕ Hn−p(V ) Hn−p(M) Hn−p−1(U ∩ V )

then five lemma will finish this step. The first two squares are easily seen to commute at the chain level. Much less simple is the third square, which we will show commutes up to a sign. Notice we only need to show the following is commutative

Hp(M,M\(K ∪ L)) δ Hp+1(U ∩ V,U ∩ V \(K ∩ L))

∩µK∪L ∩µK∩L ∂ Hn−p(M) Hn−p−1(U ∩ V )

Let A = M\K and B = M\L. Then the δ map is obtained from the short exact sequence

0 → S∗(M,A) ∩ S∗(M,B) → S∗(M,A) ⊕ S∗(M,B) → S∗(M,A ∩ B) → 0

Recall that we use the fact S∗(M,A ∪ B) → S∗(M,A) ∩ S∗(M,B) induces isomorphism on cohomology. For a cocycle φ ∈ S∗(M,A ∩ B), we write φ = ∗ ∗ φA − φB for φA ∈ S (M,A) and φB ∈ S (M,B). Then δ[φ] is represented ∗ ∗ by the cocycle δφA = δφB ∈ S (M,A) ∩ S (M,B). Similarly if z ∈ S∗(M) represents a homology class then ∂[z] = [∂zU ], where z = zU − zV with zU ∈ S∗(U) and zV ∈ S∗(V ). Via barycentric subdivision, the class µK∪L can be represented by a chain α that is a sum α = αU\L + αU∩V + αV \K of chains in three open sets U\L, U ∩ V , and V \K respectively. By uniqueness of µK∩L the chain αU∩V represents µK∩L, since the other two chains lie in the complement of K ∩ L. Similarly the chain αU\L + αU∩V represents µK . 3.3. POINCARE´ DUALITY THEOREM 45

Now let φ be a cocycle representing an element in Hp(M,M\(K ∪ L)). By δ, it maps to δφA. Continuing downward to the bottom, we obtain δφA ∩ n−p−1 αU∩V , which represents the same homology class as (−1) φA ∩ ∂αU∩V , since n−p ∂(φA ∩ αU∩V ) = (−1) δφA ∩ αU∩V + φA ∩ ∂αU∩V .

For the other way, φ is first mapped to φ ∩ α ∈ Hn−p(M). Write is as a sum of a chain in U and a chain in V :

φ ∩ α = φ ∩ αU\L + φ ∩ (αU∩V + αV \K ). and by definition ∂[φ ∩ α] = [∂(φ ∩ αU\L)] ∈ Hn−k−1(U ∩ V ).

∂(φ ∩ αU\L) = φ ∩ ∂αU\L = φA ∩ ∂αU\L = −φA ∩ ∂αU∩V .

The second equality is because φB is zero on M\L. The last equality follows from αU\L + αU∩V = µK which is a chain in U\K. This completes step 2. 3. Suppose M is the union of a direct system of open subsets {Ui}i∈I with the property that if K is a compact subset of M then K is contained in some Ui. Then if we have

Hp(M) = lim Hp(U ),H (M) = lim H (U ), c −→ c i n−p −→ n−p i i∈I i∈I and know the theorem is true for each Ui, then the theorem follows for M since the direct limit of a system of isomorphisms is an isomorphism. We prove H (M) = lim H (U ), the other follows by the same argu- p −→i∈I p i ment. By the universal property of directed limit, we have homomorphism

lim H (U ) → H (M). −→ p i p i∈I

To show it is surjective: if z ∈ Sp(M) is a cycle then there exists a compact set such that [z] ∈ im(Hp(K) → Hp(X)). Assume K ⊂ Ui. Then [z] ∈ im(H (U ) → H (M) and so [z] ∈ im(lim H (U ) → H (M)). p i p −→i∈I p i p For the injectivity, take a cycle z in Ui and assume it is a boundary of K in X. Then take j so that K ⊂ U hence its inclusion into lim H (U ) j −→i∈I p i is 0. n 4. Suppose M is an open subset of R . If M is convex, it follows from n step 1 since then M is homeomorphic to R . We can find convex open ∞ sets V1,V2, ··· such that M = ∪i=1Vi (for example, those open discs whose centres have rational coordinates). Then by step 2, this is true for V1∪· · ·∪Vr r ∞ for each r. And by step 3, it is true for ∪r(∪i=1Vi) = ∪i=1Vi = M. 5. M is arbitrary. Consider the family of all open subsets U of M such that Poincar´eduality holds for U. This family is nonempty. In view of step 3, we could apply Zorn’s lemma to this family to choose a maximal open set V belonging to it. If V 6= M, then there is an open subset B ⊂ M such that 46 CHAPTER 3. POINCARE´ DUALITY

n B is homeomorphic to R , and B is not contained in V . We apply step 2 and step 4 (for the intersection) to conclude Poincar´eduality also holds for V ∪ B, contradicting the maximality of V . Thus V = M.

We also have Poncar´eduality for non-orientable manifolds, but only for Z2 coefficient. Let M be an arbitrary n-manifold. For each point x ∈ M, µx denotes the unique non-zero element of the local homology group Hn(M,M\x; Z2). And for each compact subset K, the same argument of Lemma 3.2.5 gives us the unique element µK of Hn(M,M\K; Z2) such that ρx(µK ) = µx for all x ∈ K. Now we define homomorphism

p H (M,M\K; Z2) → Hn−p(M; Z2), x 7→ x ∩ µK .

This induces the homomorphism

p D2 : Hc (M; Z2) → Hn−p(M; Z2).

Theorem 3.3.3. For any n-manifold M, then the homomorphism

p D2 : Hc (M; Z2) → Hn−p(M; Z2) is an isomorphism. Chapter 4

Applications of Poincar´e duality

4.1 Intersection form, Euler characteristic

4.1.1 Intersection pairing

Let M be a closed connected orientable n-manifold. We let µ ∈ Hn(M) be the orientation, i.e. the unique element such that the image of µ in Hn(M,M\x) is a generator. Now we have a pairing on cohomology ring induced by cup product:

k n−k ∪ n ∩µ <, >: H (M) × H (M) −→ H (M) −→ Z In other words, < a, b >= (a ∪ b)(µ). This map is called the intersection forms. One could also define the pairing on homology by taking Poincar´e duality. Poincar´eduality simply tells us that the intersection form is non-singular when we take the free part. Corollary 4.1.1. Suppose we take coefficients in a field F . then the inter- section form <, >: Hk(M; F ) × Hn−k(M; F ) → F is non-singular. The same conclusion if we look at the the pairing Hk(M; ) Hn−k(M; ) Z × Z → . T ors T ors Z Here the intersection form is nonsingular means both < α, · > and < ·, α > are isomorphisms if α is non-zero.

Proof. Consider the composition

n−k h D∗ k H (M; R) −→ HomR(Hn−k(M; R),R) −→ HomR(H (M; R),R)

47 48 CHAPTER 4. APPLICATIONS OF POINCARE´ DUALITY

Recall (from Example sheet 1) that h is an isomorphism for the above two cases. And here D∗ is the Hom-dual of the Poincar´eduality map D : Hk → Hn−k. Notice D∗(h(α))(β) = α ∩ (β ∩ µ) = (α ∪ β)(µ) =< α, β > .

Since both D∗ and h are isomorphisms, their composition is an isomorphism and hence the intersection form is non-singular.

Later we will just write Hm(M) for the free part if there is no confusion. Corollary 4.1.2. Let M be even dimensional (n = 2m) orientable manifold. Then the pairing m m H (M) × H (M) → Z m is unimodular: choose a basis u1, ··· , uk ∈ H (M), then the matrix A = (aij) with aij =< ui, uj > has det A = ±1. And when m is even, it is symmetric; m is odd, it is anti-symmetric. Proof. The second conclusion is easy, only prove the first one. Take a basis m u1, ··· , uk ∈ H (M), then we know there is a dual basis v1, ··· , vk such that < ui, vj >= δij. Let A = (aij) where aij =< ui, uj >. Let B is matrix P of base change: vj = k bkjuk. Then X X δij =< ui, vj >= bkj < ui, uk >= aikbkj, k k i.e. AB = I. Since both of A, B are of Z coefficients, then det A = ±1. Now, if an orientable M has dimension 4m, the intersection pairing is a symmetric unimodular bilinear form. So the eigenvalues are all real numbers. + We will denote the numbers of its positive and negative eigenvalues by b2m − and b2m respectively. Their sum is Betti number b2m. Their difference is an important invariant called signature, denote by

+ − σ(M) = b2m − b2m. For orientable manifolds of dimensions other than 4m, let σ(M) = 0. There is another viewpoint of the intersection pairing, from homology. ∗ If x, y ∈ H∗(M) and ξ, η ∈ H (M) and x = Dξ, y = Dη. Then we define

x · y =< ξ, η >= (ξ ∪ η)(µM ) which is also a non singular pairing by above corollary. Assume X,Y are closed oriented submanifold of dimensions i and j re- spectively with i + j = n. We also assume they intersect transversally, i.e. at each point x ∈ X ∩ Y ,

TxX + TxY = TxM. 4.1. INTERSECTION FORM, EULER CHARACTERISTIC 49

Then the intersection is also a submanifold of dimension 0, thus finite num- ber of points. Then each x ∈ X ∩Y has a sign (x) determined by comparing the orientations of TxX + TxY and TxM. Let a = i∗(µX ) and b = i∗(µY ) where i is the inclusion. Then the intersection number could be calculated as X a · b = (x). (4.1) x∈X∩Y The intuition is very clear. We choose a singular decomposition of manifold M such that it also induces singular decomposition of X and Y . So each intersection point will be a vertex of singular simplices. Since oriented means P we have µM = si where si are all simplices. And we also have similar formula for µX and µY constitutes of sub-simplices of si, but might have −1 −1 P a sign. Then PD (a) ∪ PD (b)( si) is nonzero only when si contains intersection point. And at each intersection point, and any si containing it, the evaluation is just a check of whether the orientations are matched. To make this argument rigorous, we need to make the definition of the cohomology class PD−1(a) clearer such that it satisfies the property we want: restricting to each normal direction is just 1. This needs the Thom isomorphism theorem.

4.1.2 Betti numbers and Euler characteristic Let us introduce Betti numbers and Euler characteristic. For a CW complex M with finitely many cells in each dimension, let

i i bi = rank(Hi(M; Z)) = dim Hi(M; Q) = rank(H (M; Z)) = dim H (M; Q)). bi(M) is called the i-th Betti number. The Euler characteristic of M is

n X i χ(M) = (−1) bi(M). i=0

P∞ i More generally, we have the Poincar´eseries PM (t) = i=0 bi(M)t , and χ(M) = PM (−1). Give M a finite cell structure, let αi(M) be the number of i-cells, then P i P i we have χ = i(−1) αi(M). If we let QM (t) = i αit (M), this actually follows from the following more general result

Theorem 4.1.3. Let M be a finite CW complex. then

QM (t) − PM (t) = (1 + t)R(t).

Proof. First if we have a short exact sequence for Abelian groups

0 → A → B → C → 0 50 CHAPTER 4. APPLICATIONS OF POINCARE´ DUALITY and A, B, C are finitely generated, then rankB = rankA + rankC. Then we look at short exact sequences

0 → Zq → Cq → Bq−1 → 0

0 → Bq → Zq → Hq → 0 We have X q X q X q (rankCq)t = (rankZq)t + t (rankBq)t , q q q

X q X q X q (rankZq)t = (rankBq)t + (rankHq)t . q q q Add them we have X q QM (t) = PM (t) + (1 + t) (rankBq)t . q

From this alternative definition of Euler characteristic, it is easy to see χ(M) has counting property: Proposition 4.1.4. 1. If A and B are subcomplex of a finite CW com- plex M, χ(A ∪ B) + χ(A ∩ B) = χ(A) + χ(B). 2. If M˜ is a k-sheeted covering of M, then χ(M˜ ) = kχ(M). Then we have the following restrictions on Euler characteristic of mani- folds. Proposition 4.1.5. 1. If M is an odd dimensional closed manifold then χ(M) = 0. 2. If M is an orientable 4k +2-dimensional manifold, then χ(M) is even. Proof. If M is orientable and odd dimension, then by Poincar´eduality, χ(M) = 0. If not orientable, we know its double covering M˜ we constructed 1 ˜ before is orientable and χ(M) = 2 χ(M) = 0. For the second statement, we know bi = bn−i by Poincar´eduality. So

χ(M) ≡ b2k+1 mod 2. 2k+1 We want to prove b2k+1 is even. Choose a basis of H (M), and let A be the matrix of intersection form under this basis. Since α ∪ β = −β ∪ α when both are 2k + 1 dimensional. So A is antisymmetric: A = −AT . So

det A = det AT = det(−A) = (−1)b2k+1 det A.

Since the matrix is non-degenerate, so b2k+1 is even. A non-orientable manifold could violate the second statement. For ex- 2 ample, χ(RP ) = 1. 4.2. CALCULATION OF COHOMOLOGY RINGS 51

4.2 Calculation of cohomology rings

4.2.1 Cohomology ring of CP n We have calculated that  m = 0, 2, ··· , 2n Hm( P n) = Z C 0 otherwise

We claim the cohomology ring [α] H∗( P n; ) = Z C Z αn+1 2 n n−1 where α has degree 2, i.e α ∈ H (CP ; Z). Since the inclusion CP → n i CP induces an isomorphism on H for i ≤ 2n − 2, by induction on n, 2i n i H (CP ; Z) is generated by α for i < n. By the corollary, there is an integer m such that the product α ∪ mαn−1 = mαn = 1. Hence m = ±1, and our conclusion follows. Notice S2 ×S4 has the same cohomology groups 3 as CP . But by K¨unnethformula, [α, β] H∗(S2 × S4; ) = Z . Z α2, β2

∗ 3 Z[α] This is not isomorphic to H (CP ; Z) = α4 . n The same calculation works for Quaternionic projective space HP = n+1 H \0 ∼ where x ∼ y if there exists λ ∈ H\0 such that λx = y. We conclude that [α] H∗( P n; ) = Z H Z αn+1 where α has degree 4. Let O be the Cayley numbers. They are non-associative. We can form 1 8 2 OP = S and OP and [α] H∗( P 2; ) = Z . O Z α3

4.2.2 Cohomology ring of RP n and Borsum-Ulam Recall that  Z m = 0 or m = n = 2k + 1 n  Hm(RP ; Z) = Z2 m odd , 0 < m < n  0 otherwise

Hence by universal coefficient theorem, we know

 0 ≤ m ≤ n Hm( P n; ) = Z2 R Z2 0 otherwise 52 CHAPTER 4. APPLICATIONS OF POINCARE´ DUALITY

n Using Theorem 3.3.3, and the same argument above for CP , we con- clude that [x] H∗( P n; ) = Z2 . R Z2 xn+1 Now as an application of this calculation, we have

m n Lemma 4.2.1. Suppose we have a continuous map f : RP → RP such m n that f∗ 6= 0 : H1(RP ; Z2) → H1(RP ; Z2), then m ≤ n. 1 ∗ 1 n Proof. Since H (X; G) = Hom(H1(X),G), we know f 6= 0 : H (RP ; Z2) → 1 m H (RP ; Z2). 1 n ∗ 1 m Take ξ 6= 0 ∈ H (RP ; Z2), then η = f (ξ) 6= 0 ∈ H (RP ; Z2). By the calculation of cohomology ring, we know ηm = f ∗(ξm) 6= 0. So ξm 6= 0 ∈ m n H (RP ; Z2), which means m ≤ n. Lemma 4.2.2. Let σ be a path connecting an antipodal on Sn. Then un- n n der the quotient map π : S → RP , it becomes a singular cycle π∗(σ) n representing a nonzero element in H1(RP ; Z2). Proof. We use the cellular decomposition induced from the natural one of 0 1 n 0 S ⊂ S ⊂ · · · ⊂ S . Let σ be the one connecting S . When n = 1, π∗(σ) 1 1 1 rotates along RP = S odd number turns. It is nontrivial in H1(RP ; Z2). When n > 1, take path τ with the same end points of σ in S1 ⊂ Sn. By 1 induction, π∗(τ) represents nonzero element in H1(RP ; Z2). Since the in- 1 n clusion H1(RP ; Z2) → H1(RP ; Z2) is an isomorphism. So π∗(τ) is nonzero n n in H1(RP ; Z2) as well. On the other hand, σ − τ is singular cycle in S n with n > 1, so it is a boundary. Hence π∗(σ) is nonzero in H1(RP ; Z2). Theorem 4.2.3. There is no continuous map f : Sn+1 → Sn such that f(−x) = −f(x).

n+1 n Proof. If there is such a map, then it gives a map g : RP → RP . Take n+1 σ connecting antipodal of S , it is mapped to a path f∗(σ) connecting n n+1 the antipodal of S . Hence by Lemma 4.2.2, g∗ 6= 0 : H1(RP ; Z2) → n H1(RP ; Z2). Then Lemma 4.2.1 finishes the proof. Corollary 4.2.4 (Borsuk-Ulam). Let f : Sn → Rn be a continuous map. Then there exists x ∈ Sn such that f(x) = f(−x).

Proof. If not, construct f(x) − f(−x) g : Sn → Sn−1, g(x) = . ||f(x) − f(−x)||

We have g(−x) = −g(x), contradicting Theorem 4.2.3.

Corollary 4.2.5 (Ham sandwich). Let A1, ··· ,Am be m measurable sets in m R . Then we have a hyperplane P which bisects each Ai. 4.2. CALCULATION OF COHOMOLOGY RINGS 53

m+1 m m Proof. Consider in R . Fix x0 ∈/ R . For any vector v ∈ S , construct a m+1 hyperplane orthogonal to v and pass through x0. It divides R and thus m m R into two parts. We record the volume of the Ai ⊂ R in the half space determined by the direction of v, by fi(v). Hence we have a continuous map

m m f : S → R , v 7→ (f1(v), ··· , fm(v)).

By corollary 4.2.4, there exists v wuch that f(v) = f(−v). This hyperplane bisects each Ai.

n Another application is the Ljusternik-Schnirelmann category cat(RP ) = n n n + 1. This is because cl(RP ) = n, so cat(RP ) ≥ n + 1. On the other n hand, it is not hard to construct a smooth function on RP with n + 1 critical points (exercise).

4.2.3 Cohomology ring of lens spaces

Given an integer m > 1 and integers l1, ··· , ln relatively prime to m, 2n−1 define the lens space L = Lm(l1, ··· , ln) to be the orbit space S /Zm of 2n−1 n the unit sphere S ⊂ C with the action of Zm generated by

2πil1/m 2πiln/m ρ(z1, ··· , zn) = (e z1, ··· , e zn).

The condition li is coprime to m means Zm acts freely. Thus the projec- tion S2n−1 → L is a covering space. When m = 2, ρ is the antipodal map 2n−1 and L2 = RP . L has a CW structure with one cell ek for each k ≤ 2n − 1 and the resulting cellular chain complex is

0 m 0 0 m 0 0 → Z −→ Z −→ Z −→ · · · −→ Z −→ Z −→ Z → 0

Therefore   Z k = 0, 2n − 1 Hk(Lm(l1, ··· , ln)) = Zm k odd , 0 < k < 2n − 1  0 otherwise

By universal coefficient theorem,

 0 ≤ k ≤ 2n − 1 Hk(L (l , ··· , l ); ) = Zm m 1 n Zm 0 otherwise

Since the cohomology group only depends on m and n, later on we 2n−1 2n−1 will denote the lens space by Lm or simply L . To calculate the cup 1 2n−1 2 2n−1 product, we let α ∈ H (L ; Zm) and β ∈ H (L ; Zm) be generators. 2i 2n−1 i 2i+1 2n−1 We claim that H (L ; Zm) is generated by β and H (L ; Zm) is generated by αβi. 54 CHAPTER 4. APPLICATIONS OF POINCARE´ DUALITY

w e w

T3 e3 e0

e T0 v T2 e

e2 e1 T1

w e w

Use induction, we assume it is true for L2n−1 and want to show it for L2n+1. Using the inclusion L2n−1 → L2n+1 which induces isomorphism in cohomology for 0 ≤ k ≤ 2n − 1 by comparing the cellular chain complexes, k 2n+1 we may assume the claim holds for H (L ; Zm) with k ≤ 2n − 1. By n−1 n corollary 4.1.1, there exists λ ∈ Zm such that β ∪ λαβ = λαβ generates 2n+1 2n+1 n H (L ; Zm). So λ has to a generator of Zm and therefore αβ is a 2n+1 2n+1 n generator of H (L ; Zm). It also implies that β is a generator of 2n 2n n H (L ; Zm), otherwise αβ would have order less than m. ∗ 2n−1 To complete the calculation of the ring H (L ; Zm), we need to com- pute α2. By graded commutativity, we have α ∪ α = −α ∪ α. So if m is odd, α2 = 0. When m = 2k, we claim α2 = kβ. We use the fact that the 2-skeleton 1 2 2n−1 1 S ∪fm e of L is the circle S attached by a 2-cell with a map of degree m. We first get the 2-skeleton a ∆-complex structure by subdivide an m-gon into m triangles Ti around a central vertex v, and identify all the outer edges by rotations of the m-gon. We call the faces in a counterclockwise order T0, ··· ,Tm−1 and the the rays from v which bound Ti by ei and ei+1. Then we choose a representative φ for α which assigns value 1 to the boundary edge. The condition φ is a cocycle means φ(ei) + φ(e) = φ(ei+1), which means we could take φ(ei) = i in Zm. Then by definition of the cup product, (φ ∪ φ)(Ti) = φ(ei)φ(e) = i. Since the sum 0 + 1 + ··· + (m − 1) is k in Zm, P 2 we know φ ∪ φ evaluates as k on Ti. This means α = kβ. Hence

[α, β] H∗(L2n−1; ) = Z2k+1 Z2k+1 α2 = 0, βn = 0

[α, β] H∗(L2n−1; ) = Z2k Z2k α2 = kβ, βn = 0 4.3. DEGREE AND HOPF INVARIANT 55

4.3 Degree and Hopf invariant

4.3.1 Degree We can define degree of a map f : M n → N n between closed oriented connected manifolds. Indeed, by orientable manifolds Hn(M) = Hn(N) = Z and thus f∗ : Hn(M) → Hn(N) maps the generator µM to an integer multiple k of µN . We call this k := deg(f) the degree of the map f. The degree has natural composition property: deg(f)deg(g) = deg(f ◦ g). Since n n H (M) = H (N) = Z as well, we can define degree as the corresponding integer for the cohomology f ∗ : Hn(N) → Hn(M). Apparently, these two definitions result the same number.

Example 4.3.1. A reflection of Sn along a great circle has degree −1, since it changes the orientation. Hence the antipodal map a sending x 7→ −x has degree (−1)n+1 since it is a composition of n + 1 reflections.

This example has lots of corollaries. We only show a few.

Corollary 4.3.2. 1. If f, g : Sn → Sn are maps such that f(x) 6= g(x) for all x then f is homotopic to a ◦ g.

2. If f : Sn → Sn has no fixed points then it is homotopic to the antipodal map, and thus has degree (−1)n+1.

Proof. By assumption

(1 − t)f(x) − tg(x) x 7→ ||(1 − t)f(x) − tg(x)|| is a well defined homotopy from f to a ◦ g. The second follows from the first by taking g = id.

We can also get some information for group actions on Sn. First note 2n−1 n that S could be viewed as the unit sphere in C , thus it admits a free 1 iθ 2n−1 action of S , i.e. z 7→ e z. Especially, it tells us that Zm could act on S freely. But for S2n

2n Corollary 4.3.3. Suppose a group G acts freely on S . Then G ≤ Z2. Proof. By assumption, each non-trivial element g ∈ G has no fixed point, thus has degree −1 by above corollary. Hence there is at most one such element, otherwise the composition would give a map of degree 1 which has to be trivial.

n n Proposition 4.3.4. Given f : CP → CP , there exists an integer k such that deg f = kn. 56 CHAPTER 4. APPLICATIONS OF POINCARE´ DUALITY

2 n ∗ Proof. Let u be a generator of H (CP ), then f (u) = ku for some constant k. Hence f ∗(un) = f ∗(u)n = knun. By definition, deg f = kn.

2n n Proposition 4.3.5. If f : S → CP with n > 1 then deg(f) = 0. Proof. f ∗(u) = 0 since H2(S2n) = 0. So f ∗(un) = f ∗(u)n = 0.

Usually, the philosophy behind the last statement is: if there is a map f : M → N with deg(f) 6= 0, then M should have more complicated topology then N. Exercise: 1. Construct maps of any integer degrees for f : Sn → Sn. (Hint: start with S1 and then use suspension.) 2. Prove that any n-fold covering map has degree n.

4.3.2 Hopf invariant Hopf invariant is a sort of degree when studying the maps S2n−1 → Sn. In general, given a map f : Sm → Sn with m ≥ n, we can form the CW complex Sn t Dm+1 C(f) := Sn ∪ Dm+1 = . f f(x) ∼ x, ∀x ∈ Sm The homotopy type of C(f) depends only on the homotopy class of f. We could use Proposition 1.7.9 to calculate the (co)homology group of it. For n example, if m = n and f has degree d. Then H (C(f)) = Z|d|, which detects degree up to sign. When m > n, we calculate that the cohomology of C(f) has Z in dimen- sions 0, n and m + 1. Especially when m = 2n − 1, we have chance to use cup product to detect something nontrivial. In this case, choose generators α ∈ Hn(C(f)) and β ∈ H2n(C(f)), then the ring structure of H∗(C(f)) is determined by α2 = H(f)β for an integer H(f) which is called the Hopf invariant of f. If f is a constant map then C(f) = Sn ∨ S2n and H(f) = 0. Also, H(f) is always zero for odd n since α2 = −α2 in this case.

2 1 3 Example 4.3.6. 1. n = 2. We use S = CP and view S as the unit 2 3 2 sphere in C . The map S → S is defined as

(z0, z1) 7→ [z0 : z1].

From the definition, it is a bundle S1 → S3 → S2, which is called the 2 Hopf bundle. It is easy to see that C(f) = CP . Thus H(f) = 1 since ∗ 2 Z[α] H (CP ) = α3 . 2. n = 4. Replacing the field C by H, some construction yields the fiber 3 7 4 2 bundle S → S → S . And C(f) = HP and H(f) = 1. 4.3. DEGREE AND HOPF INVARIANT 57

3. n = 8. Use Caylay octonion. We have Hopf bundle S7 → S15 → S8. 2 And C(f) = OP and H(f) = 1.

0 1 1 4. n = 1. It is the covering map, viewed as bundle S → S → RP . It is measured by its degree, which is 2.

It is a fundamental theorem of Adams says that maps S2n−1 → Sn of Hopf invariant 1 only exists when n = 2, 4, 8. It has many interesting corollaries:

n 1. R is a division algebra only for n = 1, 2, 4, 8.

2. Sn has n linearly independent tangent vector fields only for n = 0, 1, 3, 7.

3. The only fiber bundle Sp → Sq → Sr occur only when (p, q, r) = (0, 1, 1), (1, 3, 2), (3, 7, 4), (7, 15, 8).

One could also define the Hopf invariant in terms of degree. Let y, z be two different regular values for a map f : S2n−1 → Sn, then the manifolds f −1(y) and f −1(z) could be oriented and the linking number is defined as the degree of the function: Let M and N be two manifolds of dimension n − 1 in S2n−1. Choose a point p ∈ S2n−1 which is not in M or N, and 2n−1 2n−1 think S \p as R . Then the linking number link(M,N) of M and N is defined as the degree of the map

x − y g : M × N → S2n−2, (x, y) 7→ . ||x − y||

Let us understand this definition in terms of low dimensional examples. First 0 1 1 is the toy example: the linking of two S in S or R . Let the coordinate of S0 be {a, b} and {c, d} respectively. Then the map g is determined by the order of these numbers. For example, if a < c < b < d, then two S0 are linked both from our common sense and from the formula since g maps to 1 once and −1 thrice, so degree one. If a < b < c < d, then degree is zero since maps to −1 four times. And if a < c < d < b, it maps to 1 and −1 both twice. But the degree is 0 as well since g(a, c) and g(a, d) is considered as opposite orientation because that of c and d. 1 3 3 2 2 A more realistic example is for two S in S or R . So g : T → S . For a point v in the unit sphere, the orthogonal projection of the link to the plane perpendicular to v gives a link diagram on plane. A point in T 2 sends to v corresponds to a crossing in the link diagram where γ1 is over γ2. A neighborhood of it is mapped to a neighborhood of v preserving or reversing orientation depending on the sign of the crossing. Thus the it is just a signed counting of the number of times g covers v. 58 CHAPTER 4. APPLICATIONS OF POINCARE´ DUALITY

There is a more concrete formula (Gauss formula) which could be gen- eralized to higher dimension

Z T1 Z T2 1 (γ ˙1, γ˙2, γ1 − γ2) link(γ1, γ2) = 3 dt1dt2. 4π 0 0 ||γ1 − γ2|| It is nothing but an integration interpretation of degree. Now, the Hopf invariant H(f) = link(f −1(y), f −1(z)) for any two regu- lar values y and z. To understand the equivalence of these definitions, we understand the cup product as the intersection of the Poincar´edual of co- cycles. For the toy model above, S1 bounds a D2. Pairs of two points in S1 are linked if and only if two semi circle in D2 with the pairs as end points intersect. When glue the boundary S1 by a double covering to another S1, 2 1 we get RP , and two semi-circle above become two S . For two S1’s, which are considered as inverse image of a regular value of map S3 → S2, we have similar story. But now S3 bounds D4 where one 2 1 could consider the picture in C , and S bounds an immersed disk. Then the intersection number of these two surfaces in D4 is exactly the same as the linking number of two S1 in S3. One could prove this fact by pulling two circles until they touch. So the intersection is at the boundary and easy to look at. Finally, since S1 are fibers of the map, after gluing they will become a point and the original surfaces will become a closed one. For an explicit example bearing in mind, one could consider the Hopf 1 3 2 bundle S → S → S where the fiber is (z1, z2) with a fixed ratio. For 1 1 3 example, S × 0, 0 × S ⊂ S ⊂ C × C are two of them.

4.4 Alexander duality

Let us first introduce notations. The reduced (co)homology H˜i(M) = i i Hi(M, x) and H˜ (M) = H (M, x). So it only differs from original (co)homology at i = 0.

Theorem 4.4.1. If K is a compact, locally contractible, nonempty, proper n ˜ n ˜ n−i−1 subspace of S , then Hi(S \K; Z) = H (K; Z) for all i. n Proof. We first argue it for i 6= 0. By Poincar´eduality, Hi(S \K; Z) = n−i n Hc (S \K). By definition of cohomology with compact supports,

Hn−i(Sn\K) = lim Hn−i(Sn\K,U\K) c −→ where U is taken as open neighborhoods of K. By excision, Hn−i(Sn\K,U\K) = Hn−i(Sn,U). And by long exact sequence for pairs, Hn−i(Sn,U) = H˜ n−i−1(U) when i 6= 0. Now if we can show

lim H˜ n−i−1(U) = H˜ n−i−1(K), −→ 4.4. ALEXANDER DUALITY 59 the argument for i 6= 0 is complete. To show this, we use the fact that K is a retract of some neighborhood n U0 in S since it is locally contractible. Thus in the directed limit we can only choose these open neighborhood U ⊂ U0 which could be retracted to K. This implies the surjectivity of the map lim H∗(U) → H∗(K) since we −→ can pull back the cohomology of K to that of U. To prove the injectivity, n n any U ⊂ U0 is regarded as a subspace of R ⊂ S . The linear homotopy n U × I → R from the identity to the retraction U → K takes K × I to K, hence takes V × I to U for some (small) neighborhood V of K by compactness of I. Hence the inclusion V,−→ U is homotopic to the retraction V → K ⊂ U. Thus the restriction H∗(U) → H∗(V ) factors through H∗(K). Therefore if an element of H∗(U) restrict to 0 in H∗(K), it will be zero in H∗(V ) and thus in lim H∗(U). −→ The only difference for i = 0 case is we do not have Hn(Sn,U) = H˜ n−1(U). Instead we have the short exact sequence

0 → H˜ n−1(U) → Hn(Sn,U) → Hn(Sn) → 0

Take directed limit, we have seen the first term becomes lim H˜ n−1(U) = −→ n−1 n H˜ (K). By Poincar´eduality, the middle term is H0(S \K) and the last n ˜ n ˜ n−1 is H0(S ) = Z. So this sequence tells us H0(S \K) = H (K).

In the proof, we know that the locally contractible condition is used to guarantee lim H˜ n−i−1(U) = H˜ n−i−1(K). This is not always true. Look at −→ the the following example.

Example 4.4.2. Let K denote the subset of the graph of the function y = 1 sin( x ) for x 6= 0 and y-axis with |x|, |y| ≤ 1. Since there are three path 0 components, H (K; Z) is free abelian of rank 3. However, for the directed limit lim H0(U), we only need calculate it for −→ open path connected neighborhoods of K, and thus lim H0(U) = . −→ Z Notice that K is not locally contractible at origin. Hence, it is also not a CW complex. This is called topologist’s sine curve which is a typical example of a connected but not path connected space. We also notice that this fails the Alexander duality.

This theorem has many interesting applications. Let us start with the lowest nontrivial dimension n = 2.

Corollary 4.4.3. Let K ⊂ S2 be a simple closed curve, then S2\K has two components.

˜ 2 1 1 2 Proof. Alexander duality says that H0(S \K) = H (S ) = Z. So H0(S \K) = 2 Z . 60 CHAPTER 4. APPLICATIONS OF POINCARE´ DUALITY

For n = 3, if we take K as knots. The Alexander duality simply tells us that we cannot distinguish different knots from their homology groups. A result of Gordon-Luecke tells us that the fundamental group of the knot complement determines the knot. Actually, if we choose K ⊂ Sn as a space homeomorphic to Sm, we will have the Alexander duality as well and the proof is a delightful use of Mayer-Vietoris sequence. Especially, this works for a Alexander horned sphere. It is an example homeomorphic to D3 with its boundary homeomorphic to S2, but its complement is not simply connected. However, by Alexander duality, its complement has trivial first homology.

Exercise: A non-orientable closed surface cannot be embedded in S3 as a submanifold.

Finally, we want to add that if we work on Cech cohomology instead of singular cohomology, then the locally contractible condition could be re- moved. This is because Cech cohomology Hˇ q(K) = lim Hq(U) for all neigh- −→ borhoods of K. Especially, for Cech cohomology H0 detects the connected components instead of path connected components. Notice that it does not contradict to the Eilenberg-Steenrod uniqueness since all CW complexes are locally contractible, thus path connected if it is connected. A final remark on the definition of Cech cohomology. It is defined first to each open cover U = {Uα} of a given space X and associate a simpli- cial complex N(U) called its nerve. This associates a vertex to each Uα and a set of k + 1 vertices spans a k-simplex if the corresponding Uα have nonempty intersection. For a refinement V = {Vβ} of U, which means each Vβ is contained in some Uα, then these inclusion would induce a simpli- cial map N(V) → N(U). Then the Cech cohomology Hˇ q(X) is defined as lim Hq(N(U)). −→

4.5 Manifolds with boundary

An n-manifold with boundary is a Hausdorff space M in which each point n has an open neighborhood homeomorphic either to R (called interior point) n n or to the half space R+ = {(x1, ··· , xn) ∈ R |xn ≥ 0} (called boundary point). An interior point x ∈ M has Hn(M,M\x) = Z. A boundary point n x corresponds to a point (x1, ··· , xn) ∈ R+ with xn = 0. By excision, we + + have Hn(M,M\x) = Hn(Rn , Rn \0) = 0. If M is a compact manifold with boundary, then ∂M has a collar neigh- borhood, i.e. an open neighoborhood homeomorphic to ∂M × [0, 1) by a homeomorphism sending ∂M to ∂M × 0. A compact manifold M with boundary is called orientable if M˚ := M\∂M is orientable. If ∂M × [0, 1) is a collar neighborhood of ∂M, then 1 Hi(M, ∂M) = Hi(M\∂M, ∂M × (0, 2 )). So Lemma 3.2.5 gives a relative 4.5. MANIFOLDS WITH BOUNDARY 61 fundamental class, denoted as [M, ∂M] restricting to a given orientation at each point of M\∂M. The following tells how to relate relative fundamental class to µ∂M . Later, for simplicity, we will write it as [∂M]. Proposition 4.5.1. An orientation of M determines an orientation of ∂M.

Proof. Consider an open neighborhood U of a point x ∈ ∂M which is home- n omorphic to an open half disk in R+. Let V = ∂U = U ∩ ∂M and let y ∈ U˚ = U\V . We have the following isomorphisms

Hn(M,˚ M˚\U˚) =Hn(M,˚ M˚\y)

=Hn(M,M\y)

=Hn(M,M\U˚) ∂ →Hn−1(M\U,M˚ \U)

=Hn−1(M\U,˚ (M\U˚)\x)

=Hn−1(∂M, ∂M\x)

=Hn−1(∂M, ∂M\V ) The connecting homomorphism is that of the triple (M,M\U,M˚ \U) that is an isomorphism since H∗(M,M\U) = H∗(M,M) = 0. The isomorphism that follows comes from the observation that the inclusion (M\U˚)\x → M\U is a homotopy equivalence. The next to last isomorphism is given by excision of M˚\U˚.

Especially, ∂[M, ∂M] restricts to a generator of Hn−1(∂M, ∂M\x) for all x ∈ ∂M and thus is the fundamental class [∂M] determined by the orientation of ∂M which is induced from that of M. We have the following

Theorem 4.5.2 (Lefschetz duality). Suppose M is a compact oriented n- manifold with boundary, then the homomorphisms

p D : H (M) → Hn−p(M, ∂M), α 7→ α ∩ [M, ∂M]

p D : H (M, ∂M) → Hn−p(M), α 7→ α ∩ [M, ∂M] are isomorphisms. And the following diagram is commutative.

Hq−1(M) Hq−1(∂M) δ Hq(M, ∂M) Hq(M)

D D D D ∂ Hn−q−1(M, ∂M) Hn−q(∂M) Hn−q(M) Hn−q(M, ∂M) 62 CHAPTER 4. APPLICATIONS OF POINCARE´ DUALITY

Proof. We just apply Theorem 3.3.2 to M\∂M. Via a collar neighbor- p p hood, we have H (M, ∂M) = Hc (M\∂M). And obviously, Hn−p(M) = p Hn−p(M\∂M). Hence D : H (M, ∂M) → Hn−p(M) is an isomorphism. The commutativity could be checked by inspecting the definition and using the boundary formula for cap product. Leave the details as an exercise. p Finally by five lemma, D : H (M) → Hn−p(M, ∂M) is an isomorphism as well. p For general manifold with boundary, we also similarly have Hc (M) iso- p morphic to Hn−p(M, ∂M), and Hc (M, ∂M) isomorphic to Hn−p(M) if we define Hp(M, ∂M) = lim Hi(M, (M − K) ∪ ∂M). c −→ Now, we want to know what kind of n-manifolds could be the boundary of an n + 1-dimensional manifold with boundary? From the classification of surfaces, we now each orientable closed surface is the boundary of certain 3-manifold. What about the non-orientable ones? Theorem 4.5.3. Let an n-manifold M n = ∂W n+1. Then χ(M) is even. Proof. When n is odd, it follows from Corollary 4.1.5. When n is even, take the double of W , which is obtained by take two copies W + and W − of W and glue them along the boundary. We denote it by 2W . So W + ∪ W − = 2W and W + ∩ W − = M. Hence, we have χ(2W ) + χ(M) = χ(W +) + χ(W −) = 2χ(W ). Since 2W is an odd dimensional manifold, χ(2W ) = 0. Hence χ(M) = 2χ(W ), which is an even number. Example 4.5.4. The double of a M¨obiusband is a Klein bottle. The double of an annulus of dimension 2 is a torus. The double of a disk is a sphere. Whose double is a torus? 2 For non-orientable surface, RP is not a boundary. But one can check that the Klein bottle is a boundary. More generally, one can check that 2k 2k RP and CP are not boundaries. The next result relates the signature with the boundaries. Theorem 4.5.5. Let M 4k = ∂W 4k+1 where W is a compact oriented 4k+1- manifold, then σ(M) = 0.

Proof. We use R as coefficient for (co)homology. We denote [M] = µM . The inclusion i : M → W induces homomorphism i∗ : H2k(W ) → H2k(M). Let U = Im i∗. For u = i∗(w) ∈ U, we have ∗ ∗ ∗ < u, u >=< i (w) ∪ i (w), [M] >=< i (w ∪ w), [M] >=< w ∪ w, i∗[M] >= 0

The last equality is because i∗∂ = 0 in the long exact sequence of the pair (W, M) and [M] = ∂[W, M]. We have seen that the diagram 4.5. MANIFOLDS WITH BOUNDARY 63

∗ H2k(W ) i H2k(M) δ H2k+1(W, M)

D D D

∂ i∗ H2k+1(W, M) H2k(M) H2k(W )

∗ ∗ is commutative. So rkImi = rk ker δ = rk ker i∗. Since i and i∗ are ∗ dual homomorphisms (that is where we use R coefficient), so rk cokeri = rk ker i∗. Hence

rkH2k(M) = rkImi∗ + rk cokeri∗ = 2rkU.

Let V = H2k(M) and the positive/negative eigenspace of the intersection form < v, v > would decompose it as V ±. The intersection form is 0 on linear subspace U, so V + ∩ U = 0. Hence rkV + + rkU ≤ rkV . Similarly rkV − + rkU ≤ rkV . However, the intersection is nonsingular, so rkV + + rkV − = rkV .Thus rkV ± = rkU and σ(M) = 0.

This is proved by Thom, as part of his cobordism theory.

4.5.1 Linking number, Massey product We could interpret the linking number of two cycles in Euclidean space by cup product in the complementary space. For example, suppose that Sp and Sq are disjoint spheres in Sn where n = p+q +1, and 1 ≤ p < q ≤ n−2. By Alexander duality theorem, the complementary space Sn\(Sp ∪ Sq) has cohomology group Z in dimensions p, q and p+q. Therefore the cup product of the generators in dimensions p and q will be a certain multiple of the generator of cohomology class in dimension p + q. It can be shown that this multiple is just the linking number of Sp and Sq. Another slightly different but easier to generalize definition is the following: we consider open neighoborhood of the linked spheres and let them be U1,U2, and M n is the complement of U1,U2 in S and let the boundary be B = B1 ∪ B2. Let w and v be the generators of Hp and Hq respectively. And let the n n−1 generators µ ∈ H (M,B), µi ∈ H (Bi). These could be chosen to the ones compatible with the orientations induced from Sn. Then there is a inclusion g : B → M. We have the exact sequence

g∗ Hn−1(M) → Hn−1(B) →δ Hn(M,B) → 0

Since µi are mapped to µ, the linking number is just the number m such ∗ that g (w ∪ v) = m(µ1 − µ2). To understand it by intersections for two S1 in S3. Then the w, v cor- 1 responds to singular discs D1,D2 bounds the two S . And the generator of 2 3 1 1 H (S \(S ∪ S )) corresponds to a path connecting B1 and B2. Hence the 64 CHAPTER 4. APPLICATIONS OF POINCARE´ DUALITY intersection, if in general position, is a signed count of these paths. Can you see the second viewpoint from this interpretation? Similar idea could be applied to understand the Massey product. Now let us have three spheres in Sn, any two of them have linking number 0. The generators of cohomology in dimensions between 0 and n − 1 are denoted by w1, w2, w3. We could similarly define µi and the same exact sequence. Notice g∗ is injective since the 3 cohomology groups are FREE abelian groups of dimensions 2, 3, 1. Another view to see this is by the naturality of Alexander duality, i.e. the following diagram commutes:

Hq(X) Hq(Y )

D D

n n Hn−q−1(S \X) Hn−q−1(S \Y )

Now triple product could be understood as higher linking number of links, like the ones for Borromean ring.

Theorem 4.5.6. There exists an integer m13 such that

∗ g < w1, w2, w3 >= m13(µ1 − µ3).

For Borromean ring, this integer is 1. Here is a sketch of the proof. If n−1 ∗ x ∈ H (M) and g (x) = a1µ1 + a2µ2 + a3µ3, it follows from the exactness that a1 + a2 + a3 = 0. Hence we only need to prove the coefficient µ2 ∗ ∗ in g < w1, w2, w3 > is 0, or g2 < w1, w2, w3 >= 0 for g2 : B2 → M. ∗ ∗ Actually, we could show g2w1 = g2w3 = 0, which follows from the naturality of Alexander duality and the fact that pairwise linking numbers are 0.I.e. ∗ understand g2 on q1th cohomology as n n ¯ Hp1 (U1) ⊕ Hp1 (U2) ⊕ Hp1 (U3) → Hp1 (S \B2) = Hp1 (U2) ⊕ Hp1 (S \U2)

n ¯ Then w1 corresponds to the generator of Hp1 (U1) where U1 ⊂ S \U2 and the degree of n ¯ Hp1 (U1) → Hp1 (S \U2) is just the linking number of S1 and S2. Exercise: Think about how to use the intersection viewpoint to under- stand the m13 of Borromean ring.

4.6 Thom isomorphism

Let B be a manifold. A vector bundle π : E → B of rank n is a family of n-dimensional real vector spaces {Ex}x∈B, with E := tx∈M Ex and π : 4.6. THOM ISOMORPHISM 65

E → B mapping Ex to x, equipped with a topology for E such that π is continuous and the following local triviality condition holds: For each x ∈ B there exists a neighborhood U of x and a homeomorphism

−1 n t : E|U := π (U) → U × R which is fibre preserving in the sense that for all x ∈ B the restriction of t n on Ey is a vector space isomorphism onto R . The space E is called the total space, B is called the base space. The map π is called the projection. A continuous map s : B → E such that πs = 1 is called a section of E. We could view B as a subset of E via the zero section x 7→ 0 ∈ Ex. Then denote E0 = E\B. k Typical examples include the trivial bundle B × R ; the tangent bundle TM = tx∈M TxM. Let {(Ui, φi)} be an atlas for M. The diffeomorphism n φi : Ui → R induces a map n n n (φi)∗ : TUi → TR = R × R and the normal bundle N = NS/M of S in M which is determined by the exact sequence 0 → TS → TM |S → N → 0. In the previous statements, we make use of a fact which is a good exercise.

Exercise: A vector bundle of rank n is trivial if and only if it has n sections which are linearly independent on π−1(x) for ∀x ∈ B.

n A section of TM is a vector field on M. Especially, it implies that T R is trivial but TS2 is not trivial. Similar to that of a manifold, a local orientation at x ∈ B is a preferred n generator µx ∈ H (Ex,Ex\0). A vector bundle is called orientable if for every point x ∈ B, there is a neighborhood x ∈ U ⊂ B such that the there n −1 −1 is a cohomology class µU ∈ H (π (U), π (U)0) such that µU |Ex = µx. Theorem 4.6.1. Let π : E → B be an oriented vector bundle of rank n. Then

m 1. H (E,E0) = 0 for m < n.

n 2. There exists a unique cohomology class u ∈ H (E,E0), called the n Thom class such that for all x ∈ B, the restriction of u to H (Ex,Ex\0) = Z is the prefered generator determined by the orientation. 3. The map m m+n T : H (B) → H (E,E0) α 7→ π∗α ∪ u is an isomorphism. 66 CHAPTER 4. APPLICATIONS OF POINCARE´ DUALITY

Proof. Let us only prove the case when B is compact. Then we can choose finite covering {Ui} such that on each Ui, E is a trivial bundle. n 1. First consider the case of trivial bundle E = B × R . Then by K¨unnethformula. We have

∗ ∗ n n ∗ H (B) ⊗ H (R , R \0) = H (E,E0), and hence m m−n m−n H (E,E0) = H (B) ⊗ Z = H (B). n So H (E,E0) = Z, and we choose u as the generator corresponding to the orientation. Then the theorem is verified in this case. 2. We construct u by induction. Suppose B = V ∪ W where the where the assertions of the theorem hold for E|V ,E|W ,E|V ∩W . Considering the long exact sequence of the pair (E,E0) we have

m−1 m m m H (E|V ∩W ,E0|V ∩W ) → H (E,E0) → H (E|V ,E0|V )⊕H (E|W ,E0|W )

The the assertion 1 follows from assumption on V,W . For m = n, we have

n n n n 0 → H (E,E0) → H (E|V ,E0|V )⊕H (E|w,E0|W ) → H (E|V ∩W ,E0|V ∩W ) → · · ·

By assumption, the Thom classes uV and uW exist and are unique. By uniqueness, they have the same image in EV ∩W , namely uV ∩W . Thus they n form a cohomology class u ∈ H (E,E0). This class is uniquely defined since n−1 H (E,E0) = 0. To show the last assertion, we consider the following diagram

Hm(V ) ⊕ Hm(W ) Hm(V ∩ W ) δ Hm+1(B)

T T T

m+n m+n m+n δ m+n+1 H (E|V ,E0|V ) ⊕ H (E|W ,E0|W ) H (E|V ∩W ,E0|V ∩W ) H (E,E0)

If we could show it commutes then the 5-lemma would give the right T is also isomorphism. Again, the point is to show the second square commutes. m+n Let choose a representative φ ∈ S (E,E0) of u. Then the restrictions φV , φW and φV ∩W represent the Thom classes uV , uW and uV ∩W respectively. Now take a ∈ Hk(V ∩ W ) and a representative ψ. Suppose δa = b and k k if we write ψ = ψV − ψW where ψV ∈ S (V ) and ψW ∈ S (W ), we have [δψV ] = b. Hence

∗ ∗ T δ(a) = π (b) ∪ u = π [δψV ] ∪ u.

Next ∗ ∗ δT (a) = δ(π (a) ∪ uV ∩W ) = [δπ (ψV ) ∪ φV ] = T δ. 4.6. THOM ISOMORPHISM 67

∗ Second equality is because φV is closed, the last is because π commutes with δ since it is a cochain map. 3. Suppose B is covered by finitely many open sets B1, ··· ,Bk such that the bundle EBi is trivial for each Bi. We will prove by induction on k that the theorem is true for E. It is trivial for k = 1. For k > 1, we can assume the assertions are true for E|B1∪···∪Bk−1 and E|(B1∪···∪Bk−1)∩Bk . Hence by step 2, the theorem follows.

Now let M n be a closed smooth manifold and S is a codimension k closed submanifold which is cooriented, i.e. the normal bundle NS is an oriented vector bundle. So a natural coorientation would be induced from orienta- tions of TS and TM. The tubular neighborhood theorem states that every submanifold S in M has a tubular neighborhood which is diffeomorphic to the normal bundle. Then we could indentify such a tubular neighborhood with our normal bundle NS/M . And the Thom isomorphism applies to the normal bundle gives

∗ T ∗+k ∗+k ∗+k H (S) → H (NS,NS\S) → H (M,M\S) → H (M).

Without confusion, we denote the image of 1 ∈ H0(S) in this sequence by Φ as the image of u ∈ Hk(M,M\S) in Hk(M), and called the the Thom class of S. Actually, this is the inverse of Poincar´ewith respect to S, i.e

Proposition 4.6.2. i∗[S] = Φ ∩ [M].

Proof. First, if we denote the inclusion κ : NS → M, i : S → M and the retraction r : NS → S. Then i∗ ◦ r∗ = k∗ on homology. We have

∗ Φ ∩ [M] = u ∩ [M,M\S] = u ∩ k∗[N,N\S] = i∗ ◦ r∗(k u ∩ [N,N\S]).

Thus Φ ∩ [M] is i∗ image of some element of Hn−k(S). In other words, Φ ∩ [M] = t · i∗[S]. We must show t = 1. We only need to prove it in N. For any point p ∈ S, we choose a neighborhood U ⊂ S, such that (V,U) = n n−k −1 (R , R ), where V = π (U) and π : N → S is the normal bundle. Then by definition of fundamental class, [M] and [S] restrict to fundamental n n−k classes µ0 and ν0 of R and R respectively. And u is pulled back to the n−k n Thom class u0 of R in R . Let g : V → N be the inclusion. Then if we read the relation in Hn(N,N\Q) where Q ⊂ V is the set corresponding to k R , we have

∗ i∗[S] = g∗ν0 = g∗(g u ∩ µ0) = u ∩ g∗(µ0) = u ∩ [M] = t · i∗[S].

The second inequality is because of the elementary relation u0 ∩ µ0 = ν0 in trivial bundle case in Step 1 of the Thom isomorphism. Hence t = 1. 68 CHAPTER 4. APPLICATIONS OF POINCARE´ DUALITY

Now we could finish the proof of Equation (4.1) by virtue of Proposition 4.6.2. Φx ∩ [M] is just (x) for any x ∈ X ∩ Y where the coorientation is induced from that of X and Y . (It is an exercise to check it. Notice we use normal bundle for the first and use tangent bundle for the second.) Hence the proof is reduced to prove

Proposition 4.6.3. ΦX∩Y = ΦX ∪ ΦY . Equivalently, i∗[X ∩ Y ] = i∗[X] · i∗[Y ].

Proof. It is helpful to introduce some self-explained super and sub scripts to our Thom classes and inclusions.

M ΦX∩Y ∩ [M] =(iX∩Y )∗[X ∩ Y ] M Y =(iY )∗(iX∩Y )∗[X ∩ Y ] M Y =(iY )∗(uX∩Y ∩ [Y ]) M M ∗ M =(iY )∗((iY ) uX ∩ [Y ]) M =ΦX ∩ (iY )∗[Y ] =ΦX ∩ (ΦY ∩ [M])

=(ΦX ∪ ΦY ) ∩ [M]

Thom isomorphism has lots of interesting applications, e.g. Lefschetz fixed point theorem. Let us mention another: Euler class.

Definition 4.6.4. Let E → M be a vector bundle and o : M → E be the zero section. Then the Euler class e(E) is the image of Thom class uE under the composition n n o∗ n H (E,E0) → H (E) → H (M).

For the special case when S is a submanifold of M and E is taken as its normal bundle. The Euler class e(S) is the pull back of the Thom class through i : S → (M,M\S). In this case, Euler class could be viewed as the obstruction of deforming S to M\S.

Proposition 4.6.5. If the inclusion S ⊂ M is homotopic to a map f : S → M whose image is contained in M\S, then e(S) = 0.

Proof. Hence the inclusion i : S → (M,M\S) is homotopic to a map φ : S → (M,M\S) which is factored through (M\S, M\S). Hence in cohomology i∗ = φ∗ factors through H∗(M\S, M\S) = 0. So i∗ = 0 and hence e(N) = 0. 4.6. THOM ISOMORPHISM 69

Proposition 4.6.6. T (e(S)) = u ∪ u where u ∈ Hk(M,M\S) is the Thom class.

Proof. T (e) = T (i∗(u)) = π∗i∗(u) ∪ u = u ∪ u.

Corollary 4.6.7. If the codimension k of S in M is odd, then 2e(S) = 0.

The name of Euler class is from the following

Theorem 4.6.8. Let M be an oriented n-manifold, ∆ : M → M × M be the diagonal map. Then

1. < e(M), [∆M ] >= χ(M) where χ is the Euler characteristic.

2. The normal bundle of the diagonal (∆M ) is isomorphic to TM. The Euler class is a characteristic class in the following sense

Definition 4.6.9. A cohomology class c(E) ∈ H∗(M) associated to any vector bundle E → M is called a characteristic class if it is natural with respect to pullbacks in the sense that c(f ∗E) = f ∗(c(E)).