Untwisting the Cohomology with Twisted Coefficients - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:18:22Z http://mathoverflow.net/feeds/question/65724 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65724/untwisting-the-cohomology-with-twisted-coefficients Untwisting the Cohomology with Twisted Coefficients Chris Gerig 2011-05-22T18:22:58Z 2011-05-23T17:23:45Z <p>This question is set on a finite $2$-group $G$ and a subgroup $H$ of index $2$ (but perhaps the question could be answered for arbitrary orders/indexes).</p> <p>A while ago someone posted a question on this site, asking whether $|Ker(res^G_H)|\le 2\cdot|H_2(G,\mathbb{Z})|$, where $res^G_H:H_1(G,\mathbb{Z})\rightarrow H_1(H,\mathbb{Z})$ is the restriction map in dimension $1$ (also known as the Verlagerung transfer map $Ver$). Using the integral coefficients module $\tilde{\mathbb{Z}}$ with the twisted $G$-action $g\cdot z=-z$ for $g\notin H$, I was able to establish the relation $|Ker(res^G_H)|\le |H_2(G,\tilde{\mathbb{Z}})|$.</p> <p><strong>Question:</strong> Is there any feasible way to obtain $H_2(G,\tilde{\mathbb{Z}})$ from $H_2(G,\mathbb{Z})$, or even just relations of orders?</p> <p>There are the long exact sequences $\cdots\rightarrow H_2(G,\tilde{\mathbb{Z}})\stackrel{\delta}{\rightarrow} H_1(G)\stackrel{res}{\rightarrow}H_1(H)\rightarrow H_1(G,\tilde{\mathbb{Z}})$ and $\cdots\rightarrow H_2(G,\tilde{\mathbb{Z}})\rightarrow H_2(H)\stackrel{cor}{\rightarrow} H_2(G)\stackrel{\partial}{\rightarrow} H_1(G,\tilde{\mathbb{Z}})$ which arise from the short exact sequences of modules $\mathbb{Z}\hookrightarrow Ind^G_H\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ with the twisted-action on either the first or last $\mathbb{Z}$, respectively. But this ultimately leaves you with a relation based on images and kernels that doesn't give sufficient information. There is also the Serre spectral sequence $E_2^{p,q}=H_p(G/H,H_q(H,\tilde{\mathbb{Z}}))\Rightarrow H_{p+q}(G,\tilde{\mathbb{Z}})$, but it's too tough to crack open, and doesn't use the untwisted-homologies of $G$.</p> <p>This also leads to the general question: Is there a way to obtain $H_n(G,\tilde{M})$ from $H_n(G,M)$ in dimension $n\ge 1$ where $G$ is a $p$-group for arbitrary prime $p$? Here $\tilde{M}$ just refers to some twisting that you can put on some $M$.</p>