most recent 30 from http://mathoverflow.net 2013-05-18T23:15:24Z http://mathoverflow.net/feeds/question/39273 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39273/is-the-following-map-from-zg-x-h3g-c-h2g-c-ever-nontrivial Noah Snyder 2010-09-19T04:18:17Z 2011-08-25T11:14:18Z

<p>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).</p> <p>$f(z,w)(x,y) = \frac{w(z, x, y) w(x, y, z)}{w(x, z, y)}.$</p> <p>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?</p> <p>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)).</p>

http://mathoverflow.net/questions/39273/is-the-following-map-from-zg-x-h3g-c-h2g-c-ever-nontrivial/73253#73253 Peter Teichner 2011-08-20T00:21:11Z 2011-08-20T00:21:11Z

<p>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. </p>

http://mathoverflow.net/questions/39273/is-the-following-map-from-zg-x-h3g-c-h2g-c-ever-nontrivial/73622#73622 Gjergji Zaimi 2011-08-25T02:26:50Z 2011-08-25T11:14:18Z

<p>Following up on what has already been said, this is an explanation taken from <a href="http://arxiv.org/abs/math/0002246" rel="nofollow">"Group cohomology and gauge equivalence of some twisted quantum doubles"</a>, by Geoffrey Mason and Siu-Hung Ng.</p> <p>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 <code>$H^3(G,\mathbb C^{\ast}) _{ab}$</code> 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.</p> <p>An easier way to do this is to look for a different description of <code>$H^3(G,\mathbb C^{\ast})_{ab}$</code>. It is not hard to check that <code>$\omega_{g}(x,y)=\omega_g(y,x)$</code> 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 <code>$H^3(G,\mathbb C^{\ast})_{ab}$</code> 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. </p> <p>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 $H^2$.</p>