Group cohomology vs. topological cohomology in the case of non-trivial action - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:47:31Z http://mathoverflow.net/feeds/question/15727 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15727/group-cohomology-vs-topological-cohomology-in-the-case-of-non-trivial-action Group cohomology vs. topological cohomology in the case of non-trivial action Randy Brown 2010-02-18T16:42:35Z 2010-02-19T02:45:33Z <p>When A is an abelian group with trivial G-action (G being a discrete group) we get that H<sup>n</sup>(G,A)&#8773;H<sup>n</sup>(BG,A). Is there a similar connection between group cohomology and topological cohomology if A is a non-trivial &#8484;G-module? What <em>can</em> we say in that case?</p> http://mathoverflow.net/questions/15727/group-cohomology-vs-topological-cohomology-in-the-case-of-non-trivial-action/15732#15732 Answer by Sammy Black for Group cohomology vs. topological cohomology in the case of non-trivial action Sammy Black 2010-02-18T17:26:54Z 2010-02-18T17:26:54Z <p>You have to introduce the language of sheaves. Then, the coefficients can be described succinctly as a locally constant sheaf.</p> <p>Alternatively, you can view the coefficient system as a bundle over your space $BG$ with $A$ fibers. Locally, the coefficients behave as if the $G$-action were trivial, but if one considers a loop that represents a nontrivial element of $\pi_1(BG) = G$, monodromy comes into play.</p> http://mathoverflow.net/questions/15727/group-cohomology-vs-topological-cohomology-in-the-case-of-non-trivial-action/15735#15735 Answer by Chris Schommer-Pries for Group cohomology vs. topological cohomology in the case of non-trivial action Chris Schommer-Pries 2010-02-18T18:10:56Z 2010-02-18T18:10:56Z <p>This is an example of twisted cohomology. In general, for a generalized cohomology theory (spectrum) E and a space X you can talk about E-twists over X. This is a certain structure on X. Given a particular twist $\tau$, you can then form the $\tau$-twisted cohomology $E^\tau(X)$. This was the subject of a <a href="http://arxiv.org/abs/1002.3004" rel="nofollow">recent ArXiv paper</a> by Ando-Blumberg-Gepner. One way to think of this is that E-cohomology of X is naturally graded by the E-twists. (This viewpoint is especially when E is an $E_\infty$-ring spectrum, in which case twists can be added and there is a cup product-type formula). </p> <p>In the case of ordinary cohomology, a twist reduces to a local system on X. A <em>local system</em> is the same thing as a functor from the fundamental groupoid of X to the category of abelian groups. This can equivalently be thought of as a locally constant sheaf of abelian groups on X, connecting up with Sammy Black's answer. For more general types of cohomology, these don't correspond to sheaves in the usual sense. If you allow "sheaves of spectra" you can get this work, but that is a difficult and long story. </p> <p>In the case that $X = BG$ (with G discrete), then the fundamental groupoid of X is equivalent to the category G, i.e. the category with a single object with automorphisms the group G. Then a local system, i.e. a functor $G \to Ab$ is exactly the same as a G-module. Then the twisted cohomology of $BG$ with this twist is exactly the group cohomology of G with values in the module. There is also a similar homology story. </p> http://mathoverflow.net/questions/15727/group-cohomology-vs-topological-cohomology-in-the-case-of-non-trivial-action/15777#15777 Answer by Paul for Group cohomology vs. topological cohomology in the case of non-trivial action Paul 2010-02-19T02:45:33Z 2010-02-19T02:45:33Z <p>For any space $X$ with $\pi_1X=G$ and any $ZG$ module $A$, define $C_*(X;A)$ to be $C_*(\tilde{X})\otimes_{ZG}A$ and define $C^* (X;A)$ to be $Hom_{ZG}(C_*(\tilde{X}),A)$, where $C_*(\tilde{X})$ is the<br /> (cellular, or singular) chain complex of the universal cover of $X$ viewed as a free $ZG$ complex. (supressing notation involving left v.right action). Then take co/homology. It makes no difference whether the action is trivial or not. If $A$ has a trivial action, then $C_*(\tilde{X})\otimes_{ZG}A=C_*(X)\otimes A$ and $Hom_{ZG}(C_*(\tilde{X}),A)=Hom(C_*(X),A)$. Taking $X=BG$ answers your question, assuming you like topology and define group homology in terms of $BG$</p> <p>If you prefer to start with an algebraic definition of group cohomology (eg bar construction) then you need some argument that relates the algebra to a cell structure for $BG$: but you need this argument whether or not your coefficients are twisted.</p>