Consider a short exact sequence of Abelian groups -- I'm happy to assume they're finite as a toy example: $$ 0 \to H \to G \to G/H \to 0\ . $$

I want to understand the classifying space of $G$. Since $BH \cong EG/H$, $G/H$ acts on $BH$ and we can write $BG \cong E(G/H) \times_{G/H} EG/H$. Thus, we have a fiber bundle (which I'll write horizontally) $$ BH \to BG \to B(G/H) $$

On the other hand, the central extension is classified by an element of the group cohomology $H^2(G/H,H)$ which is the same as $H^2(B(G/H),H)$. The latter is an element in the homotopy class of maps $[B(G/H),K(H,2)]$ and $K(H,2)\cong BBH$. This map looks like it classifies a principal $BH$ bundle over $B(G/H)$. I find it hard to imagine that this 'principal' $BH$ bundle is not 'the same' as the bundle above, so the question is, how do you see that? From this construction, it's not even obvious to me that the above bundle is a principal bundle.

I would guess (and being a poor physicist, I'm not so up on my homotopy theory), there's a sense that the classifying space of an Abelian group is an 'Abelian group', and taking classifying spaces of an exact sequence gives you back an 'exact sequence'. That gets you a 'principal bundle' (aren't quotation marks fun?), but even then I'm not sure how to see that the classifying map of this bundle is the same as the class in group cohomology.

Any references to the needed background would also be greatly appreciated.

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    $\begingroup$ I don't know the story well enough to write a good answer, and I'm sure someone else does, but the short answer is "Yes." $\endgroup$ – Ben Webster Jan 17 '10 at 21:30

Yes. The principal bundles are the same and your guess that $BA$ is an abelian group is exactly right. A good reference for this story, and of Segal's result that David Roberts quotes, is Segal's paper:

G. Segal. Cohomology of topological groups, Symposia Mathematica IV (1970), 377- 387.

The functors $E$ and $B$ can be described in two steps. First you form a simplicial topological space, and then you realize this space. It is easy to see directly that $EG$ is always a group and that there is an inclusion $G \rightarrow EG$, which induces the action. The quotient is $BG$. Under suitable conditions, for example if $G$ is locally contractible (which includes the discrete case), the map $EG\rightarrow BG$ will admit local sections and so $EG$ will be a $G$-principal bundle over $BG$. This is proven in the appendix of Segal's paper, above. There are other conditions (well pointedness) which will do a similar thing.

The inclusion of $G$ into $EG$ is a normal subgroup precisely when $G$ is abelian, and so in this case $BG$ is again an abelian group.

I believe your question was implicitly in the discrete setting, but the non-discrete setting is relevant and is the subject of Segal's paper. Roughly here is the answer: Given an abelian (topological) group $H$, the $BH$-princical bundles over a space $X$ are classified by the homotopy classes of maps $[X, BBH]$. When $H$ is discrete, $BBH = K(H,2)$. If $X = K(G,1)$ for a discrete group $G$, these correspond to (central) group extensions:

$$H \rightarrow E \rightarrow G$$

If $G$ has topology, then the group extensions can be more interesting. For example there can be non-trivial group extensions which are trivial as principal bundles. Easy example exist when H is a contractible group. However Segal developed a cohomology theory which classifies all these extensions. That is the subject of his paper.

  • $\begingroup$ As Scott notes in his answer, it's homotopy classes of pointed maps [X, BBH]... $\endgroup$ – Chris Schommer-Pries Jan 18 '10 at 2:53
  • $\begingroup$ Actually, it is the non-discrete case that I'm interested in -- I just figured that the discrete case would be an easier thing to think about. Now to see if I can get my hands on that Segal paper. Thanx! $\endgroup$ – Aaron Bergman Jan 18 '10 at 15:23

Your picture is essentially correct, except you need to specify that the maps take basepoints to basepoints (i.e., they are pointed maps).

BG is given as the homotopy fiber of a pointed map from B(G/H) to BBH. Applying the based loop space functor $\Omega$ to the pointed maps yields the sequence of group homomorphisms. This does not require G to be abelian. There is a subclass of abelian central extensions, which are those that can be delooped again to maps from BB(G/H) to BBBH.

In general, group homomorphisms can be delooped once to maps of pointed spaces, while abelian group homomorphisms can be delooped arbitrarily many times, to infinite loop maps of infinite loop spaces. There is an equivalence between grouplike homotopy commutative spaces and double loop spaces, but since G is discrete, the double loop space property is strictified for free.

Edit: A standard reference for this material is the second chapter in Adams, Infinite Loop Spaces.


@There is an equivalence between grouplike homotopy commutative spaces and double loop spaces

No, grouplike homotopy commutative is not enough. Double loop spaces have much more in thee way of higher homotopies, even if the space were to have a strictly associative homotopy commutative structure. That was one of the motivations for operads. Also see JF Adams: 10 types of H-spaces.


A quick and dirty answer is that if we regard the groups as one-object groupoids, they again form a short exact sequence of groupoids, and the geometric realisation of this sequence is the $BH$ bundle you mention. $BH$ is a (normal) closed subgroup of $BG$ by a result of Segal, so all that needs to be checked for this to be a principal bundle is that $BG \to B(G/H)$ admits local sections. I believe this is true because the groups involved are discrete hence well-pointed - the inclusion of the identity is a closed cofibration (a paper by Baez and Stevenson has relevant calculations for topological groups in a different context).


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