# Cohomology of a fiber bundle with fiber $H$ and base space $BG$

Are there any general results on the (integral) cohomology of fiber bundle, where the fiber is a compact group $H$ (continuous or discrete) and the base space is the classifying space $BG$ of another compact group $G$ (continuous or discrete). Any literature references is much appreciated.

Since we have two groups $G$ and $H$. I wonder if the result can be expressed as group cohomology of the two groups.

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Serre's spectral sequence? –  Fernando Muro Nov 12 '12 at 21:35
I wonder if the Serre's spectral sequence can be expressed in terms of group cohomology of $H$ and $G$. –  Xiao-Gang Wen Nov 12 '12 at 21:50
Do you have in mind a definition of group cohomology that does not involve the cohomology of the classifying space? –  S. Carnahan Nov 12 '12 at 23:00
There is an elementary definition of group cohomology without involving the topological cohomology of the classifying space. (See Wiki en.wikipedia.org/wiki/Group_cohomology ) I stress group cohomology since we may need G-module where the group has non-trivial action. Using classifying space to define group cohomology with a non-trivial G-module, we may need the "local coefficient system" which I do not understand. This is why I prefer to state the results in terms of group cohomology. –  Xiao-Gang Wen Nov 12 '12 at 23:50

A typical example would be the case when $G$ is a subgroup of $H$. Then $(EH\times H)/G$ (diagonal action) is 1. homotopy equivalent to $H/G$, and 2. fibered over $BG=EH/G$ with fiber $H$. Note that this works both in the Lie case and the discrete case but in the latter case what we get is not very interesting since the fiber of our fibration is a potentially infinite discrete space.

[upd: There is one thing one can extract from this though: the $i$-th cohomology group of $G$ with coefficients in the infinite product $\Pi_{h\in H}\mathbb{Z}_{(h)}$ is $\Pi_{h G\in H/G} \mathbb{Z}_{(hG)}$ when $i=0$ and is 0 otherwise; this may be of some use when $G$, or its index in $H$, is finite.]

On the other hand, if $G$ is normal in $H$ one can go a bit further: $BH=EH/H$ is the quotient of $BG=EH/G$ by a free action of $H/G$. So, as above, we construct a fibration over $B(H/G)$ with fiber $BG$ and total space $BH$. If we now take an $H$-module $M$ (i.e., a local system on $BH$) we get the Hochschild-Serre spectral sequence

$$E_2^{pq}=H^p(H/G,H^q(G,M))\Rightarrow H^{p+q}(H,M).$$

There are lots of references where this is discussed. One could take a look e.g. at the original paper by Hochschild and Serre (Cohomology of group extensions, Transactions AMS 1953).

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Let $X$ be the fiber bundle with fiber $H$ and base space $BG$. I wonder do we have: $E_2^{pq}=H_g^p(H,H_g^q(G,M))\Rightarrow H^{p+q}(X,M)$ ? –  Xiao-Gang Wen Nov 13 '12 at 0:36
Xiao-Gang -- there are potentially lots of fiber bundles with fiber $H$ and base $BG$. If we have one that comes from an action of $G$ on $H$, then we do have a Serre spectral sequence converging to the cohomology of the total space, which is $\sim H/G$, but its $E_2$ term looks quite different from what you describe. The Hochschild-Serre sequence has a similar (but still different) $E_2$ but converges to the cohomology of $H$. –  algori Nov 13 '12 at 0:53
Indeed, we are considering fiber bundles with fiber $H$ and base $BG$ that comes from an action of $G$ on $H$. But $G$ may not be a subgroup of $H$. So we do not have $H/G$. –  Xiao-Gang Wen Nov 13 '12 at 2:39
Xiao-Gang -- re "We do not have..": yes we do, assuming the action comes from a group homomorphism $f:G\to H$, except that the action will be not free in general (and the base of the fibration will be $BIm(f)=BG/\ker(f)$). –  algori Nov 13 '12 at 2:49
algori: Thank you for explaining, but as a physicist, I am still confused (maybe over something simple). Do you mean $H^*(X,M)=H^*(H/G,M)$? But if $G$ is not a subgroup of $H$, we cannot compute the quotient group $H/G$. But do you mean $H/Im(f)$ instead of $H/G$? If the action is trivial, do you have $H^*(X,M)=H^*(H,M)$? (which does not look right, since $X=H\times BG$ in this case). –  Xiao-Gang Wen Nov 13 '12 at 3:17