(GJSZ) Homework II (due October 14), Math 602, Fall 2002.

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1 (GJSZ) Homework II (due October 14), Math 602, Fall BI (b). The objective of this section is to prove if G is a finite group acting trivially on Z and Q/Z, then Hom (G, Q/Z) = H 2 (G, Z). The first step if to define a homomorphism θ: Hom (G, Q/Z) H 2 (G, Z). We proceed as follows: Given any homomorphism χ: G Q/Z, let χ : G Q be any map representing χ, which means χ () = χ() (mod Z), for all G. Since χ is a homomorphism, χ(τ) = χ() + χ(τ). Writing χ () = χ() + k 1, χ (τ) = χ(τ) + k 2 and χ (τ) = χ(τ) + k 3, with k 1, k 2, k 3 Z, we have χ (τ) χ (τ) + χ () = χ(τ) + k 2 (χ(τ) + k 3 ) + χ() + k 1 = χ(τ) + k 2 χ() χ(τ) k 3 + χ() + k 1 = k 2 k 3 + k 1, which shows χ (τ) χ (τ) + χ () Z. Observe (δχ )(, τ) = χ (τ) χ (τ) + χ (), where χ is a 1-cochain in C 1 (G, Q), and so, (δχ )(, τ) B 2 (G, Q) and δ 2 (δχ ) 0, which means δχ is a 2-cocycle in Z 2 (G, Z). Furthermore, if χ is any other map representing χ, then χ χ is clearly integer valued, so δχ δχ is a 2-coboundary in B 2 (G, Z). If f is any 2-cocycle in Z 2 (G, Z), let [f] denote its image in H 2 (G, Z) (its cohomology class). As a consequence of the above, the map θ: Hom (G, Q/Z) H 2 (G, Z) given by θ(χ)(, τ) = [χ (τ) χ (τ) + χ ()] where χ is any cochain in C 1 (G, Q) representing χ, is indeed well-defined. Since the group operation in Hom (G, Q) is addition of functions, it is obvious θ is a homomorphism. We prove θ: Hom (G, Q/Z) H 2 (G, Z) is injective as follows. Assume θ(χ)(, τ) = 0 in H 2 (G, Z), which means χ (τ) χ (τ) + χ () is a 2-coboundary in B 2 (G, Z). So, there is some 1-cochain g C 1 (G, Z) so χ (τ) χ (τ) + χ () = g(τ) g(τ) + g() for all and τ in G. Since G is finite, if we let G = n, we can sum the above with respect to, and we get χ (τ) χ (τ) + χ () = g(τ) g(τ) + g(), 1

2 nχ (τ) χ () + χ () = ng(τ) g() + g(), so and since χ (τ) Q and g(τ) Z, nχ (τ) = ng(τ), for all τ G, χ (τ) = g(τ), for all τ G, which means χ 0 in Hom (G, Q/Z). Thus, θ is indeed injective. To prove surjectivity, we prove a general property of 2-cocycle in H 2 (G, A), when G is finite and A is any G-module. Recall if f is a 1-cochain, then and if f is a 2-cochain, then (δ 1 f)(, τ) = f(τ) f(τ) + f(), (δ 2 f)(, τ, ρ) = f(τ, ρ) f(τ, ρ) + f(, τρ) f(, τ). Assume G is a finite group, and let G = n. Then, if f is a 2-cochain, we can define the 1-cochain, g, by g() = f(, ρ). Let us compute Σ (δ 2 f)(, τ, ρ). We have (δ 2 f)(, τ, ρ) = f(τ, ρ) f(τ, ρ) + f(, τρ) If f is a 2-cocycle, (δ 2 f)(, τ, ρ) = 0, and we get f(, τ) = f(τ, ρ) f(τ, ρ) + f(, ρ) nf(, τ) = (δ 1 g)(, τ) nf(, τ). nf(, τ) = (δ 1 g)(, τ). Now, if f Z 2 (G, Z) is a 2-cocycle, by the above, there is a 1-cochain, g, in C 1 (G, Z) so nf(, τ) = (δ 1 g)(, τ), and thus, χ = g/n is a 1-cochain in C 1 (G, Q) so f(, τ) = (δχ )(, τ), 2

3 Now, since f(, τ) = (δχ )(, τ) and f Z 2 (G, Z) we also have which means (δχ )(, τ) = χ (τ) χ (τ) + χ () Z, χ (τ) χ (τ) + χ () = 0 in Q/Z, i.e., χ represents a homomorphism χ Hom (G, Q/Z). Therefore, we have θ(χ) = [f], where χ Hom (G, Q/Z) is the homomorphism represented by χ. This proves the surjectivity of θ. Finally, we proved θ: Hom (G, Q/Z) H 2 (G, Z) is an isomorphism. BI (c) We have the G-pairing Z A A, given by (n, a) na. By part (a) with p = 2 and q = 0, we obtain a pairing H 2 (G, Z) H 0 (G, A) H 2 (G, A) given by the cup-product. We know from class H 0 (G, A) = A G = {a A a = a} for all G. Thus, in view of part (b), we have a pairing Hom (G, Q/Z) A G H 2 (G, A). For any χ Hom (G, Q/Z) and any a A G, using the definition of part (a), the isomorphism Hom (G, Q/Z) = H 2 (G, Z), and the cup-product we have H 2 (G, Z) H 0 (G, A) H 2 (G, A), (δχ a)(, τ) = [δχ (, τ)(τ a)] where χ is any map representing χ. Since a A G, we have τ a = a, and so, (δχ a)(, τ) = [δχ (, τ)a] = [(χ (τ) χ (τ) + χ ())a], for any χ representing χ. For any α A, if ξ = N α = G α, we get [ (δχ ξ)(τ, ρ) = δχ (τ, ρ) ] [ ] α = δχ (τ, ρ)α. G Observe δχ (τ, ρ)α is a 2-cocycle in Z 2 (G, A). Thus, if we prove for every cocycle f Z 2 (G, A), the 2-cocycle N f(τ, ρ) = f(τ, ρ) is a 2-coboundary, we will have shown (δχ N α)(, τ) = 0. Let u f (ρ) = f(, ρ). Since f is a 2-cocycle, we have G 0 = f(τ, ρ) f(τ, ρ) + f(, τρ) f(, τ). 3

4 If we sum the above over, we get Therefore, 0 = G f(τ, ρ) G f(τ, ρ) + G f(, τρ) G f(, τ) = N f(τ, ρ) f(, ρ) + f(, τρ) f(, τ) G G G = N f(τ, ρ) u f (ρ) + u f (τρ) u f (τ). N f(τ, ρ) = u f (τ) + u f (ρ) u f (τρ), which proves N f(τ, ρ) is a 2-coboundary. Thus, for any α A. (δχ N α)(, τ) = 0. However, observe N A A G, since if ξ = G α, then for any τ G, ( ) τ ξ = τ α = τ α = α = ξ. G G G Consequently, we have the pairing Hom (G, Q/Z) (A G /N A) H 2 (G, A). We still need to prove u f (τ) A G, since this is needed in part (d). For this, we sum the cocycle condition over τ. We get 0 = f(τ, ρ) f(τ, ρ) + f(, τρ) f(, τ) 0 = f(τ, ρ) f(τ, ρ) + f(, τρ) f(, τ) ( ) = f(τ, ρ) f(τ, ρ) + f(, τ) f(, τ) = u f (ρ) u f (ρ), and so, u f (ρ) = u f (ρ), for all G, as desired. BI (d) Now, we assume G = {1, 0, 2 0,..., n 1 0 } is a finite cyclic group. Let χ 0 : G Q/Z be the homomorphism defined by χ 0 ( 0 ) = 1/n (mod Z) and let χ 0 be the 1-cocycle in C 1 (G, Q) representing χ 0 defined by χ 0( k 0) = k/n, for k = 0,..., n 1. We have the homomorphism θ: A G /N A H 2 (G, A) defined by θ(α) = δχ 0 α. First, we prove injectivity. If θ(a) = 0, this means there is some 1-cochain g C 1 (G, A) so (δχ 0 α)(, τ) = (χ 0(τ) χ 0(τ) + χ 0(τ))α = g(τ) g(τ) + g() 4

5 for all, τ G. If we sum the above over G, we get ( χ 0(τ) χ 0(τ) + ) χ 0() α = g(τ) g(τ) + g(), ( ) nχ 0(τ) χ 0() + χ 0() α = g(τ) g() + g(), which yields nχ 0(τ)α = g(τ). If we set τ = 0, since χ 0( 0 ) = 1/n, we get α = g( 0 ), and this shows α N A, as desired. Now, we need to prove the surjectivity of θ. Unfortunately, this has eluded all our attemps! All we can do is this: For any 2-cocycle f Z 2 (G, A), if there is some a A G so (δχ 0 α)(, τ) = [(χ 0(τ) χ 0(τ) + χ 0(τ))a] = [f(, τ)], then by summing over G as above, we get [ ] [χ 0(τ)a] = f(, τ), Since χ 0( k 0) = k/n, this yields [χ 0(τ)a] = [u f (τ)]. u f ( k 0) = ka, k = 0,..., n 1, and in particular, we must have a = u f ( 0 ). We also observed { δχ 0 (0 h, 0) k = χ 0(0 h ) χ 0(0 h+k ) + χ 0(0) k 0 if 0 h + k n 1 = 1 if n h + k 2n 2. 5