Does the following "symmetric" 2nd cohomology group of a finite group with coefficients in $Z_2$ always vanish? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T05:53:50Z http://mathoverflow.net/feeds/question/108476 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108476/does-the-following-symmetric-2nd-cohomology-group-of-a-finite-group-with-coeffi Does the following "symmetric" 2nd cohomology group of a finite group with coefficients in $Z_2$ always vanish? Bruce Bartlett 2012-09-30T15:05:17Z 2012-10-13T08:32:46Z <p>Let $G$ be a finite group. Usually, a 2-cocycle on $G$ with values in $\mathbb{Z}_2 = \{+1, -1\}$ is a collection of signs $\epsilon_{g,h} \in \{+1, -1\}$, $g,h \in G$, satisfying the cocycle equation (written multiplicatively) $$\epsilon_{g,h} \epsilon_{gh, k} = \epsilon_{h,k} \epsilon_{g,hk}$$ And a 2-coboundary is a 2-cocycle with $\epsilon_{g,h} = \frac{t_g t_g}{t_{gh}}$ for all $g,h \in G$, with $t : G \rightarrow \{+1, -1\}$ an arbitrary map. Then the second cohomology group is $H^2(G, \mathbb{Z}_2)$ = {2-cocycles} / {2-coboundaries}. </p> <p>But suppose we demand that our 2-cocyles satisfy the 2-cocycle equation above together with the "conjugate-cyclic" symmetry $$\epsilon_{g,h} = \epsilon_{h^{-1} g^{-1}, g}$$ as well as the "conjugate symmetric" symmetry, $$\epsilon_{g,h} = \epsilon_{h^{-1}, g^{-1}}.$$ These symmetries make sense from a TQFT perspective if you draw the 2-cocycle as a bunch of trivalent vertices, when the first symmetry corresponds to counterclockwise rotation of the diagram and the second to a kind of vertical flip.</p> <p>And suppose now that the coboundaries given by $\{t_g\}$ satisfy $t_1 = 1$ and $t_g t_{g^{-1}} = 1$. This ensures that the "symmetric" 2nd cohomology group $H^2_{sym} (G) :=${ "symmetric" 2-cocycles} / {"symmetric" 2-coboundaries} makes sense. </p> <p>Question: Does this symmetric 2nd cohomology group always vanish?</p> <p>I've only checked one example, namely $G = \mathbb{Z}_2 \times \mathbb{Z}_2 = \langle a,b : a^2 = b^2 =(ab)^2 = 1 \rangle$. In normal cohomology, we have $H^2(G, \mathbb{Z}_2)$ = $(\mathbb{Z}_2)^3$ but according to my calculations, none of the non-trivial 2-cocycles respect the above symmetries, so $$H^2_{sym} (G, \mathbb{Z}_2) = 0.$$ Is this perhaps always true for arbitrary $G$? Or perhaps I have made a silly mistake. </p> http://mathoverflow.net/questions/108476/does-the-following-symmetric-2nd-cohomology-group-of-a-finite-group-with-coeffi/109519#109519 Answer by Bruce Bartlett for Does the following "symmetric" 2nd cohomology group of a finite group with coefficients in $Z_2$ always vanish? Bruce Bartlett 2012-10-13T08:32:46Z 2012-10-13T08:32:46Z <p>I'm pretty sure now that $H^2_{sym} (\mathbb{Z}_4 \times \mathbb{Z}_2, \mathbb{Z}_2) = \mathbb{Z}_2 \times \mathbb{Z}_2$, so it can be nonzero. For comparison, in ordinary group cohomology $H^2 (\mathbb{Z}_4 \times \mathbb{Z}_2, \mathbb{Z}_2) = (\mathbb{Z}_2)^3$.</p>