Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

When A is an abelian group with trivial G-action (G being a discrete group) we get that Hn(G,A)≅Hn(BG,A). Is there a similar connection between group cohomology and topological cohomology if A is a non-trivial ℤG-module? What can we say in that case?

share|improve this question
5  
See Hatcher's "Algebraic Topology", the section on Local Coefficients. –  Steve D Feb 18 '10 at 16:56
    
Helpful stuff on this question in Mike Shulman's answer here: mathoverflow.net/questions/2390/… –  Hunter Brooks Feb 18 '10 at 19:36

3 Answers 3

up vote 8 down vote accepted

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 recent ArXiv paper 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).

In the case of ordinary cohomology, a twist reduces to a local system on X. A local system 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.

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.

share|improve this answer

You have to introduce the language of sheaves. Then, the coefficients can be described succinctly as a locally constant sheaf.

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.

share|improve this answer

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
(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 $BG$

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.