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Suppose that G is a finite group, then we have the following map f which takes an element z in the center of G and a 3-cohomology class w and returns a 2-cohomology class f(z,w) (for concreteness let's take coefficients to be C* everywhere).

$f(z,w)(x,y) = \frac{w(z, x, y) w(x, y, z)}{w(x, z, y)}.$

Is this map ever nontrivial? That is can you find a group G, a central element z, and an element of H^3 such that f(z,w) is not the trivial element of H^2?

The motivation for this question is that it should give an example where Z(G) did not lift to a subcategory of the Drinfel'd center Z(Vec(G,w)).

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I assume you tried $Z_p^3$? –  Ian Agol Sep 19 '10 at 15:58
This looks to me like a special case of the construction: take an element $z$ in the center of $G$, get a homomorphism of groups $\mathbb{Z}\times G$ sending $(n,g)\mapsto z^ng$, consider the induced map $H^*(G,M)\to H^*(\mathbb{Z}\times G,M)\approx H^*(\mathbb{Z},\mathbb{Z})\otimes H^*(G,M)$, then get a map $H^q(G,M)\to H^{q-1}(G,M)$ by evaluating on the generator of $H_1(\mathbb{Z},\mathbb{Z})$. Is that right? –  Charles Rezk Sep 19 '10 at 16:07
@Charles. Suppose this is a special case, does that buy us anything? –  Chris Schommer-Pries Oct 6 '10 at 21:36

2 Answers 2

up vote 7 down vote accepted

To answer Chris' (and maybe Ian's) question: The map that Charles describes is nontrivial for q=3 in the cases $G=Z^3, M=Z$ and $G=(Z/2)^3, M=Z/2$, the latter answering the original question (if Charles is right). The proof is easy since the cohomology rings are polynomial, respectively exterior, algebras.

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Following up on what has already been said, this is an explanation taken from "Group cohomology and gauge equivalence of some twisted quantum doubles", by Geoffrey Mason and Siu-Hung Ng.

Let $G$ be a finite abelian group. We denote by $Z^3(G,\mathbb C^{\ast})$ the group of normalized $3$-cocycles. For any $\omega\in Z^3(G,\mathbb C^{\ast})$ and $g\in G$ we have the map $$\omega_g (x,y)=\frac{\omega(g,x,y)\omega(x,y,g)}{\omega(x,g,y)}$$ Now let $Z^3(G,\mathbb C^{\ast})_{ab}$ denote the set of all normalized $3$-cocycles $\omega$ for which $\omega_g$ is a $2$-coboundary for all $g\in G$, and let $H^3(G,\mathbb C^{\ast}) _{ab}$ be the corresponding set of cohomology classes. It is not hard to check that $H^3(G,\mathbb C^{\ast})_{ab}$ is a subgroup of $H^3(G,\mathbb C^{\ast})$, and your question is asking for an example when it is a proper subgroup.

An easier way to do this is to look for a different description of $H^3(G,\mathbb C^{\ast})_{ab}$. It is not hard to check that $\omega_{g}(x,y)=\omega_g(y,x)$ for all $(x,y)\in G\times G$ iff $\omega_g$ is a $2$-coboundary. So if you define the map $\psi^{\ast}: H^3(G,\mathbb C^{\ast})\to Hom(\bigwedge^3 G,\mathbb C^{\ast})$ $$\psi^{\ast}([\omega])(x,y,z)=\frac{\omega(x,y,z)\omega(y,z,x)\omega(z,x,y)}{\omega(y,x,z)\omega(z,y,x)\omega(x,z,y)}=\frac{\omega_z(x,y)}{\omega_z(y,x)}$$ then $H^3(G,\mathbb C^{\ast})_{ab}$ is precisely the kernel of $\psi^{\ast}$ (Lemma 7.4 in the paper above). Now $\psi^{\ast}$ is surjective so the question becomes: when is $Hom(\bigwedge^3 G,\mathbb C^{\ast})$ non-trivial? This is the case whenever $G$ is the direct sum of at least $3$ cyclic factors, in particular any $(\mathbb Z/n\mathbb Z)^3$ works.

An explicit example in this case is $\omega(x,y,z)=\mu^{x_1y_2z_3}$ where $x=(x_1,x_2,x_3)$ etc. and $\mu$ is a primitive $n$th root of unity. Then we have $\psi^{\ast}([\omega])(x,y,z)=\mu^{\det(x,y,z)}$ which is non-trivial. In particular $\omega_x$ is non-trivial in $H^2$.

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If I'm not mistaken, $H^*(G,A)_{ab}$ is the notation MacLane used for $H^*(K(G,2),A)$, before the discovery of Eilenberg-MacLane spaces. –  S. Carnahan Aug 25 '11 at 3:13
@S.C. Yes, I think so, and his definition was similar to the second definition above (as the kernel of $\psi^{\ast}$). –  Gjergji Zaimi Aug 25 '11 at 11:16

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