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The Cohomology Cup Product

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The Cohomology Cup Product The Cohomology Cup Product Dan Hathaway April 24, 2011 1 The Cup Product 1.1 Defining The Cup Product and Showing it is Well Defined Consider cohomology with coefficients in some fixed ring R. This is different from what we have been studying so far in class, because up till now our coefficients were in an arbitrary abelian group. We can L n take the direct sum n≥0 H (X) of all cohomology groups of a topological space X to obtain an abelian group. What we would like to do is add a multiplication operation on this group to get a ring. This multiplication will be the cup product. n n Given 2 Z (X), let [ ] represent the cohomology class of . Recall that C (X) = Hom(Cn(X);R), φ 2 Zn(X) means δ(φ) = 0, and φ 2 Bn(X) means φ = δ(φ0) for some (n − 1)-cochain φ0 2 Cn−1(X). L n n To define the cup product ^ on n≥0 H (X), it suffices to define [φ] ^ [ ] for [φ] 2 H (X) and [ ] 2 Hm(X) where n; m ≥ 0 are arbitrary (because then ^ can be defined component wise). To do this, we will use the symbol ^ to denote a product φ ^ of n and m cohains, and we will show that this operation is well defined on cohomology classes. Given φ 2 Cn(X) and 2 Cm(X), define φ ^ to be the (n + m)-cochain that satisfies the following formula for all σ : ∆n+m ! X: (φ ^ )(σ) = φ(σj[v0; :::; vn]) (σj[vn; :::; nn+m]): This formula makes sense because φ(σj[v0; :::; vn]) and (σj[vn; :::; vn+m]) are both elements of R, so they can be multiplied. What we want is an induced ^ operation Hn(X) × Hm(X) ! Hn+m(X): To get this, we must show that the ^ operation Cn(X) × Cm(X) ! Cn+m(X) maps Zn(X) × Zm(X) to Zn+m, and if [φ] = [φ0] and [ ] = [ 0], then [φ ^ ] = [φ0 ^ 0]. Both of these facts are implied by the following Lemma: Lemma: If φ 2 Cn(X) and 2 Cm(X), then δ(φ ^ ) = δφ ^ + (−1)nφ ^ δ . Proof: We will compute δ(φ ^ )(σ), (δφ ^ )(σ), and (φ ^ δ )(σ) separately for an arbitrary σ : ∆n+m+1 ! X. Computing δ(φ ^ ): n+m+1 X i δ(φ ^ )(σ) = (φ ^ )(@σ) = (φ ^ )( (−1) σj[v0; :::; v^i; :::; vn+m+1]) = i=0 n+m+1 X i (−1) (φ ^ )(σj[v0; :::; v^i; :::; vn+m+1]): i=0 The last equality follows because φ ^ is a homomorphism. Computing (δφ ^ )(σ): (δφ ^ )(σ) = (δφ)(σj[v0; :::; vn+1]) (σj[vn+1; :::; vn+m+1]) = φ(@(σj[v0; :::; vn+1])) (:::) = 1 n+1 n+1 X i X i φ( (−1) σj[v0; :::; v^i; :::; vn+1]) (:::) = (−1) φ(σj[v0; :::; v^i; :::; vn+1]) (:::): i=0 i=0 Computing (φ ^ δ )(σ) in the same way: (φ ^ δ )(σ) = φ(σj[v0; :::; vn])(δ )(σj[vn; :::; vn+m+1]) = φ(:::) (@(σj[vn; :::; vn+m+1])) = n+m+1 n+m+1 X i−n X i−n φ(:::) ( (−1) σj[vn; :::; v^i; :::; vn+m+1]) = (−1) φ(:::) (σj[vn; :::; v^i; :::; vn+m+1]): i=n i=n Thus, n+1 n X i (δφ ^ )(σ) + (−1) (φ ^ δ )(σ) = (−1) φ(σj[v0; :::; v^i; :::; vn+1]) (σj[vn+1; :::; vn+m+1])+ i=0 n+m+1 n X i−n (−1) (−1) φ(σj[v0; :::; vn]) (σj[vn; :::; v^i; :::; vn+m+1]): i=n Notice how the last term of the first sum cancels with the first term of the second sum. Comparing the resulting sum to what we calculated δ(φ ^ ) to be, we find that (δφ ^ )(σ) + (−1)n(φ ^ δ )(σ) = δ(φ ^ ). This completes the proof of the lemma. With this lemma in place, it is clear that if φ 2 Zn(X) and 2 Zm(X), then φ ^ 2 Zn+m(X): if δφ = 0 and δ = 0, then δ(φ ^ ) = δφ ^ + (−1)nφ ^ δ = 0 ^ + (−1)nφ ^ 0 = 0 + 0 = 0. Thus, ^ restricts to a map from Zn(X) × Zm(X) to Zn+m(X). Suppose now that φ, φ0 2 Zn(X), ; 0 2 Zm(X), φ − φ0 2 Bn(X), and − 0 2 Bm(X). We want to show that [φ ^ ] = [φ0 ^ 0]. Let φ − φ0 = δφ~ for some φ~ 2 Cn+1(X). The lemma gives us δ(φ~ ^ ) = δφ~ ^ + (−1)nφ ^ δ . Since 2 Zn(X), this means that δ = 0 so δ(φ~ ^ ) = δφ~ ^ . Thus, δφ~ ^ is a coboundary. That is, δφ~ ^ = (φ − φ0) ^ = (φ ^ ) − (φ0 ^ ) is a coboundary. That is, [φ ^ ] = [φ0 ^ ]. In the same way, we can show [φ0 ^ ] = [φ0 ^ 0]. This establishes [φ ^ ] = [φ0 ^ 0]. That is, ^ is a well defined map from Hn(X) × Hm(X) to Hn+m(X). 1.2 Cohomology Ring L n With the cup product defined, it is worth checking that this operation on n≥0 H (X) indeed gives us a ring structure. We have that ^ is associative because given [φ] 2 Hn(X), [ ] 2 Hm(X), and [γ] 2 Hk(X), we have ((φ ^ ) ^ γ)(σ) = φ(σj[v0; :::; vn]) (σj[vn; :::; vn+m])γ(σj[vn+m; :::; vn+m+k]) = (φ ^ ( ^ γ))(σ) so ([φ] ^ [ ]) ^ [γ] = [φ] ^ ([ ] ^ [γ]). We also have that ^ distributes with addition, because if φ, φ0 2 Zn(X) and 2 Zm(X), then 0 0 0 0 ((φ+φ ) ^ )(σ) = (φ+φ )(σj[v0; :::; vn]) (σj[vn; :::; vn+m]) = φ(:::) (:::)+φ (:::) (:::) = (φ ^ +φ ^ )(σ): The same is true for the second argument. Thus, we do indeed have a ring structure on cohomology. In L n fact, we have more than just a ring structure: n≥0 H (X) forms an R-algebra (which can be easily verified). 1.3 Taking the Cohomology Ring is Functorial We have shown that every topological space has associated with it a cohomology ring. The question arises as to if there is a ring homomorphism induced by a continuous function between topological spaces. This turns out to be the case. 2 That is, let X and Y be topological spaces and let f : X ! Y be a continuous function. For each n, ∗ n n we have an induced group homomorphism fn : H (Y ) ! H (X). This induces a group homomorphism ∗ L n L n ∗ f : n≥0 H (Y ) ! n≥0 H (X). We claim that this map f is also a ring homomorphism. This is because if φ is an i-cochain on Y and is a j-cochain on Y , then ∗ ∗ ∗ ∗ (f (φ) ^ f ( ))(σ) = (f (φ))(σj[v0; :::; vi])(f ( ))(σj[vi; :::; vi+j]) = ∗ φ((f ◦ σ)j[v0; :::; vi]) ((f ◦ σ)j[vi; :::; vi+j]) = (φ ^ )(f ◦ σ) = f (φ ^ )(σ): Thus, on cohomology classes we have [f ∗(φ)] ^ [f ∗( )] = [f ∗(φ ^ )]. Thus, for elements a; b 2 L n ∗ ∗ ∗ ∗ n≥0 H (Y ) we have f (a) ^ f (b) = f (a ^ b). That is, f is a ring homomorphism. We have that if X, Y , and Z are topological spaces with the continuous maps f : X ! Y and ∗ ∗ ∗ ∗ ∗ ∗ g : Y ! Z, then fn ◦gn = (f ◦g)n, for all n. This means that f ◦g = (f ◦g) . The fact that these maps L n L n commute is not affected by the additional ring structure that is put on n≥0 H (X), n≥0 H (Y ), and L n n≥0 H (Z). Thus, the operation F of taking the cohomology ring is a (contravariant) functor from the category of topological spaces to the category of rings. 1.4 Cup Product Is Supercommutative L n i j Consider the cohomology ring n≥0 H (X). Let a 2 H and b 2 H . As long as R is commutative (a fact which we are assuming), it turns out that a ^ b = (−1)ijb ^ a. We mention this fact here because we will need this for having a well defined multiplication operation on the tensor product of two cohomology rings. We omit the proof of supercommutivity for brevity. 2 Computing the Cohomology Ring of Simplicial Complexes The definition we gave for ^ can be modified to apply to simplicial complexes. Using this definition, we will compute the cohomology ring of various simplicial complexes. 2.1 Using Z as a Coefficient Ring From now on in this section, we will use Z instead of an arbitrary ring R. Recall an important tool from a past homework assignment for calculating the cohomology of a simplicial complex given that we are using Z for our coefficient group: if all the homology groups Hn(X) are free abelian, then the canonical n homomorphism from the abelian group H (X) to the abelian group Hom(Hn(X); Z) is an isomorphism. We will use this result to compute the cohomology of a space as a first step before we investigate the ring structure induced by the cup product. 2.2 Cohomology Ring of a Point First things first, we will compute the cohomology ring of X where X is a space with a single point. Once this is done, we will know the cohomology ring of any contractible space. In class we showed H0(X) = Z and Hi(X) = 0 for i 6= 0. Since every homology group of X is free abelian, we have i ∼ 0 ∼ H (X) = Hom(Hi(X); Z) for all i. Since H0(X) = Z, we have H (X) = Hom(Z; Z). Since Hi(X) = 0 for i ≥ 0, we have Hi(X) =∼ Hom(0; Z) = 0.
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