Poincar´e Duality Section 3.3 239 The Duality Theorem The form of Poincar´e duality we will prove asserts that for an R orientable closed k → n manifold, a certain naturally defined map H (M; R) Hn−k(M; R) is an isomor- phism. The definition of this map will be in terms of a more general construction called cap product, which has close connections with cup product. Cap Product Let us see how cup product leads naturally to cap product. For an arbitrary space X and coefficient ring R let us for simplicity write Ck(X; R) as Ck and its k = k ∈ k ∈ ` dual C (X; R) HomR(Ck,R) as C . For cochains ϕ C and ψ C we have their + cup product ϕ ` ψ ∈ Ck ` . When we regard ϕ ` ψ as a function of ψ for fixed ϕ, + + we get a map C`→Ck ` , ψ , ϕ ` ψ. One might ask whether this map C`→Ck ` is → the dual of a map Ck+` C` . The answer turns out to be yes, and this map on chains will be the cap product map α , α a ϕ. In order to define α a ϕ it suffices by linearity to take α to be a singular simplex + σ : ∆k `→X , and then the formula for σ a ϕ is rather like the formula defining cup product: a = | ··· | ··· σ ϕ ϕ σ [v0, ,vk] σ [vk, ,vk+`] To say that for fixed ϕ the map α , α a ϕ has as its dual the map ψ , ϕ ` ψ is --------------a→ϕ -----→ψ ` to say that the composition Ck+` C ` R is the map α , (ϕ ψ)(α),orin other words, (∗)ψ(αaϕ) = (ϕ ` ψ)(α) + This holds since for σ : ∆k `→X we have | | ψ(σ a ϕ) = ψ ϕ σ|[v , ···,v ] σ|[v , ···,v + ] 0 k k k ` = | ··· | ··· = ` ϕ σ [v0, ,vk] ψ σ [vk, ,vk+`] (ϕ ψ)(σ ) Another way to write the formula (∗) that may be more suggestive is in the form hα a ϕ, ψi=hα, ϕ ` ψi h− −i × i→ where , is the map Ci C R evaluating a cochain on a chain. × k→ a In order to show that the cap product Ck+` C C` , (α, ϕ) , α ϕ, induces a × k → corresponding cap product Hk+`(X; R) H (X; R) H`(X; R) we need a formula for ∂(α a ϕ). This could be derived by a direct calculation similar to the one for cup 240 Chapter 3 Cohomology products, but it is easier to use the duality relation hα a ϕ, ψi=hα, ϕ ` ψi to reduce to the cup product formula. We have h∂(α a ϕ), ψi=hαaϕ, δψi=hα, ϕ ` δψi = (−1)k hα, δ(ϕ ` ψ)i−hα, δϕ ` ψi = (−1)k h∂α,ϕ ` ψi−hα, δϕ ` ψi = (−1)k h∂α a ϕ, ψi−hαaδϕ, ψi ∈ ` Since this holds for all ψ C and C` is a free R module, we must have ∂(α a ϕ) = (−1)k(∂α a ϕ − α a δϕ) From this it follows that the cap product of a cycle α and a cocycle ϕ is a cycle. Further, if ∂α = 0 then ∂(α a ϕ) =(α a δϕ), so the cap product of a cycle and a coboundary is a boundary. And if δϕ = 0 then ∂(α a ϕ) =(∂α a ϕ), so the cap product of a boundary and a cocycle is a boundary. These facts imply that there is an induced cap product a × k ----------------------- → Hk+`(X; R) H (X; R) H ` (X; R) This is R linear in each variable. ` `→ k+` a → We have seen that the map ϕ : C C is dual to ϕ : Ck+` C` , but when we pass to homology and cohomology this may no longer be true since cohomology is not always just the dual of homology. What we have instead is a slightly weaker relation h between cap and cup product in the form H`(XR)→ Hom ( H(XR)R) ; ; , ` R → of the commutative diagram at the right. → ϕ ()ϕ When the maps h are isomorphisms, for h k + ` (XR)→Hom ( ()XR R) example when R is a field or when R = Z H ; R Hk +` ; , and the homology groups of X are free, then the map ϕ ` on cohomology is the dual of a ϕ on homology. In these cases cup and cap product determine each other, at least if one assumes finite generation so that cohomology determines homology as well as vice versa. However, there are examples where cap and cup products are not equivalent when R = Z and there is torsion in homology. We shall also need relative forms of the cap product. Using the same formulas as before, one checks that there are relative cap product maps a × k ----------------------- → Hk+`(X, A; R) H (X; R) H ` (X, A; R) a × k ----------------------- → Hk+`(X, A; R) H (X, A; R) H ` (X; R) × k → For example, in the second case the cap product Ck+`(X; R) C (X; R) C`(X; R) × k restricts to zero on the submodule Ck+`(A; R) C (X, A; R), so there is an induced × k → a cap product Ck+`(X, A; R) C (X, A; R) C`(X; R). The formula for ∂(σ ϕ) still holds, so we can pass to homology and cohomology groups. There is also a more general relative cap product a ∪ × k ----------------------- → Hk+`(X, A B; R) H (X, A; R) H ` (X, B; R), Poincar´e Duality Section 3.3 241 ∪ defined when A and B are open sets in X , using the fact that Hk+`(X, A B; R) can + = + be computed using the chain groups Cn(X, A B; R) Cn(X; R)/Cn(A B; R),asin the derivation of relative Mayer–Vietoris sequences in §2.2. Cap product satisfies a naturality property that is a little more awkward to state than the corresponding result for cup product since both covariant and contravariant functors are involved. Given a map f : X→Y , the relevant induced maps on homology and cohomology fit into the diagram shown below. It does not quite make sense to say this diagram commutes, but the spirit of ` → Hk()X H()X Hk- ()X ` commutativity is contained in the formula → → f → f f ∗ ` → f∗(α) a ϕ = f∗ α a f (ϕ) Hk()Y H()Y Hk-`()Y a = which is obtained by substituting fσ for σ in the definition of cap product: fσ ϕ | ··· | ··· ϕ fσ [v0, ,v`] fσ [v`, ,vk]. There are evident relative versions as well. Now we can state Poincar´e duality for closed manifolds: Theorem 3.30 (Poincar´e Duality). If M is a closed R orientable n manifold with ∈ k →- fundamental class [M] Hn(M; R), then the map D : H (M; R) Hn−k(M; R) de- fined by D(α) = [M] a α is an isomorphism for all k. Recall that a fundamental class for M is an element of Hn(M; R) whose image in | ∈ Hn(M x; R) is a generator for each x M . The existence of such a class was shown in Theorem 3.26. Example 3.31: Surfaces. Let M be the closed orientable surface of genus g .In ∗ Example 3.7 we computed the cup product structure in H (M; Z) directly from the definitions using simplicial cohomology, and the same procedure could be used to compute cap products. However, let us instead deduce the cap product structure from the cup product structure. This is possible since we are in the favorable situation where the cohomology groups are exactly the Hom duals of the homology groups. ` Using the notation from Example 3.7, the homomorphism αi sends βi to γ and all a other αj ’s and βj ’s to 0, so the dual homomorphism αi sends [M] to the homology a = class [bi] Hom dual to βi . Thus [M] αi [bi] and so the Poincar´e dual of αi is = =− D(αi) [bi]. Similarly we have D(βi) [ai] where the minus sign comes from the ` =− relation βi αi γ. It is tempting to try to interpret Poincar´e duality geometrically, replacing coho- mology by homology which is more geometric, and taking explicit cycles representing = homology classes. Thus the geometric interpretation of the relation D(αi) [bi] might be the statement that the cycle bi is the Poincar´e dual of the cycle ai . The special feature of these two cycles is that they intersect transversely in a single point, while all other aj ’s and bj ’s are disjoint from ai and bi , or can be made disjoint after a homotopy. There are some difficulties with this geometric interpretation of Poincar´e duality in terms of geometric intersections of cycles, however. For example, take the 242 Chapter 3 Cohomology genus 1 case when the surface is a torus. Here there are other homology classes be- sides [b1] that are represented by cycles that are embedded loops intersecting a1 + transversely in one point, namely each homology class [b1] m[a1] is represented by such a loop, a loop going once around the torus in the b1 direction and m times in the a1 direction. Thus there is not a unique homology class Poincar´e dual to [a1],in contrast with the situation for cohomology where the inverse of the Poincar´e duality a isomorphism [M] gives a unique cohomology class β1 dual to [a1]. The reason for this difference between homology and cohomology is that the isomorphism between 1 H1 and its Hom dual H is not canonical, but depends on a choice of basis, just as in linear algebra there is no canonical isomorphism between a vector space and its dual vector space.