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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    $\begingroup$ I assume you tried $Z_p^3$? $\endgroup$ – Ian Agol Sep 19 '10 at 15:58
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    $\begingroup$ 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? $\endgroup$ – Charles Rezk Sep 19 '10 at 16:07
  • $\begingroup$ @Charles. Suppose this is a special case, does that buy us anything? $\endgroup$ – Chris Schommer-Pries Oct 6 '10 at 21:36

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.


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 $H^2$.

  • $\begingroup$ 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. $\endgroup$ – S. Carnahan Aug 25 '11 at 3:13
  • $\begingroup$ @S.C. Yes, I think so, and his definition was similar to the second definition above (as the kernel of $\psi^{\ast}$). $\endgroup$ – Gjergji Zaimi Aug 25 '11 at 11:16

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