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

I was recently trying to learn a little bit about group cohomology, but one point has been confusing me. According to wikipedia (http://en.wikipedia.org/wiki/Group_cohomology and some other sources on the internet), given a (topological) group $G$, we have that the group cohomology $H^n(G)$ is the same as the singular cohomology $H^n(BG)$ (with coefficients in a trivial $G$-module $M$). Moreover, it says that given any group $G$, if we don't care about its topology, we can always give it the discrete topology and look at the cohomology of $K(G,1)$. This seems to suggest that when $G$ has topology that we do care about, we can just look at $BG$ with whatever topology $G$ is supposed to have. The relevant citation in this section is a reference to a book called Cohomology of Finite Groups, but I was wondering if this result would work with groups such as $U(1)$ which are not finite? Moreover, it would seem then that there is some sort of natural way to define group cohomology to detect the topology of the group; for example maybe look at continuous $G$-modules and continuous cochains. However, I heard that when doing this, one has to be careful because in general, the category of continuous $G$-modules might not have enough injectives. Also, I found this article by Stasheff (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.bams/1183540920) which seems to suggest that for continuous cohomology, the we might not have the equality $H^n(BG)=H^n(G)$ between singular and group cohomology. I was wondering if someone could explain these connections to me (including what "continuous cohomology" is) and/or clarify what is happening? It would also be great if someone could tell me how one might compute something like $H^n(U(1);M)$ where $U(1)$ carries the discrete topology. Thanks.

share|improve this question
2  
I quote a comment from Bill Thurston: "For a discrete group, $BG=K(G,1)$ (as homotopy types), and the homology of the group is the same as the homology of this space. For a topological group, if $G_\delta$ denotes $G$ with discrete topology, then $K(G,1)=BG_\delta$, and often has quite different homology from $BG$. For instance, $B\mathbb{R}$ is trivial, verus $B\mathbb{R}_\delta$, with homology rank $2\omega$ in every dimension." –  Chris Gerig Aug 13 '11 at 6:31
add comment

3 Answers 3

up vote 11 down vote accepted

Short answer: you don't want to consider group cohomology as defined for finite groups for Lie groups like $U(1)$, or indeed topological groups in general. There are other cohomology theories (not Stasheff's) that are the 'right' cohomology groups, in that there are the right isomorphisms in low dimensions with various other things.

Long answer:

Group cohology, as one comes across it in e.g. Ken Brown's book (or see these notes), is all about discrete groups. The definition of group cocycles in $H^n(G,A)$, for $A$ and abelian group, can be seen to be the same as maps of simplicial sets $N\mathbf{B}G \to \mathbf{K}(A,n)$, where $\mathbf{B}G$ is the groupoid with one object and arrow set $G$, $N$ denotes its nerve and $\mathbf{K}(A,n)$ is the simplicial set corresponding (under the Dold-Kan correspondence) to the chain complex $\ldots \to 1 \to A \to 1 \to \ldots$, where $A$ is in position $n$, and all other groups are trivial. Coboundaries between cocycles are just homotopies between maps of simplicial spaces.

The relation between $N\mathbf{B}G$ and $K(G,1)$ is that the latter is the geometric realisation of the former, and the geometric realisation of $\mathbf{K}(A,n)$ is an Eilenberg-MacLane space $K(A,n)$, which represents ordinary cohomology ($H^n(X,A) \simeq [X,K(A,n)]$, where $[-,-]$ denotes homotopy classes of maps). This boils down to the fact that simplicial sets and topological spaces encode the same homotopical information. It helps that $N\mathbf{B}G$ is a Kan complex, and so the naive homotopy classes are the right homotopy classes, and so we have

$$sSet(N\mathbf{B}G,\mathbf{K}(A,n))/homotopy \simeq Top(BG,K(A,n))/homotopy = [BG,K(A,n)]$$

In fact this isomorphism is a homotopy equivalence of the full hom-spaces, not just up to homotopy.

If we write down the same definition of cocycles with a topological group $G$, then this gives the 'wrong' cohomology. In particular, we should have the interpretation of $H^2(G,A)$ as isomorphic to the set of (equivalence classes of) extensions of $G$ by $A$, as for discrete groups. However, we only get semi-direct products of topological groups this way, whereas there are extensions of topological groups which are not semi-direct products - they are non-trivial principal bundles as well as being group extensions. Consider for example $\mathbb{Z}/2 \to SU(2) \to SO(3)$.

The reason for this is that when dealing with maps between simplicial spaces, as $N\mathbf{B}G$ and $\mathbf{K}(A,n)$ become when dealing with topological groups, it is not enough to just consider maps of simplicial spaces; one must localise the category of simplicial spaces, that is add formal inverses of certain maps. This is because ordinary maps of simplicial spaces are not enough to calculate the space of maps as before. We still have $BG$ as the geometric realisation of the nerve of $\mathbf{B}G$, and so one definition of the cohomology of the topological group $G$ with values in the discrete group $A$ is to consider the ordinary cohomology $H^n(BG,A) = [BG,K(A,n)]$.

However, if $A$ is also a non-discrete topological group, this is not really enough, because to define cohomology of a space with values in a non-discrete group, you should be looking at sheaf cohomology, where the values are taken in the sheaf of groups associated to $A$. For discrete groups $A$ this gives the same result as cohomology defined in the 'usual way' (say by using Eilenberg-MacLane spaces).

So the story is a little more complicated than you supposed. The 'proper ' way to define cohomology for topological groups, with values in an abelian topological group (at least with some mild niceness assumptions on our groups) was given by Segal in

G Segal, Cohomology of topological groups, in: "Symposia Mathematica, Vol. IV (INDAM, Rome, 1968/69)", Academic Press (1970) 377–387

and later rediscovered by Brylinski (it is difficult to find a copy of Segal's article) in the context of Lie groups in this article.

share|improve this answer
3  
@ David: Re you short answer, one often does want to study $BG^d$ for $G$ a Lie group with the discrete topology, e.g. in the context of flat bundles and in algebraic K-theory via Quillen's definition $K(R)=H^*(BGL(R)^d_+)$. –  Paul Aug 14 '11 at 7:53
    
Flach's article "Cohomology of topological groups ..." contains a definition of cohomology of a topological group $G$ with values in a sheaf on top. spaces (including the case where the sheaf is represented by a topological G-module) in terms of the topos $BGG. –  Jakob Aug 24 '12 at 12:53
add comment

Continuous cohomology of a topological group G used to mean using the usual chain complex of multivaraible functions F:G^n \to A with the usual coboundary except that f is required to be continuous. in the smooth case, continuous or smooth gives the same cohomology. In the Lie case, van Est established very nice relations between this cohomology, the Lie algebra cohomology and the purely topological cohomology of the underlying space G

share|improve this answer
add comment

Most of the time, $H^n(G)$ means $H^n(BG)$ with the discrete topology, and topologists tend to write $K(G,1)$ rather than $BG$ to emphasize that $G$ is considered as a discrete group. This is because when $G$ comes with a natural topology then $BG$ is typically assumed to take the topology into account, specifically there is a fibration with base $BG$, contractible total space, and fiber $G$ with its given topology, and this specifies $BG$ up to homotopy type.

One way to study $K(U(1),1)$, is to observe that there is a short exact sequence of (discrete) groups $0\to Z\to R\to U(1)\to 0$.

Continuous cohomology is more involved, taking the topology into account, and is often defined by simplicial methods.

share|improve this answer
add comment

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.