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ort96
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compute cohomology of flat fiber bundles - does this alwas work?

Setting

Let $p: E \rightarrow B$ be a flat fiber bundle with fiber $F$ where $E$, $B$, $F$ are compact, smooth manifolds. Then $E$ has the form of a twisted product

$E \cong \widetilde{B} \times_{\pi_{1}} F$,

where $\widetilde{B}$ is the universal cover of $B$, $\pi_{1}$ is the fundamental group of $B$ and $F$ carries a $\pi_{1}$-action.

We consider cohomology with field coefficients.

Background

$\pi_{1}$ finite would be very convenient, for instance then we'd have $$H^*(E) = H^*(\widetilde{B} \times F)^{\pi_{1}} = (H^*(\widetilde{B}) \times H^*(F))^{\pi_{1}}. $$ (Since then $\widetilde{B} \times F$ is compact and the diagonal action is free.) This product structure on the cohomology of $E$ can be very useful.

Sadly, $\pi_{1}$ is often infinite. However, sometimes we can still reduce the structure group to a finite one. If, for instance, there is a triangulation on $F$ such that the $\pi_{1}$-action on $F$ is simplicial, we can do the following:

The group action is in this case a group homomorphism $$ \phi: \pi_1 \rightarrow \{\text{Permutations of Vertices} \}, $$ where the right hand side is finite because $F$ is compact. In particular $G:= \pi_1/ker(\phi) $ is finite. We have $$ E \cong (\widetilde{B}/ker(\phi)) \times_{G} F $$ and thus reduced our structure group to a finite one. The diagonal action is still free and $\widetilde{B}/ker(\phi) \times F$ is compact, so we get $$H^*(E) = H^*(\widetilde{B}/ker(\phi) \times F)^{G} = (H^*(\widetilde{B}/ker(\phi)) \times H^*(F))^{G}. $$

Question

Are there compact flat fiber bundles where we cannot do this trick to compute cohomology? I am looking for an explicit example. (Since we want to compute cohomology and perhaps characteristic classes, we are only concerned with fiber bundles up to fiber preserving homotopy equivalences.)

Related

In this post simplicial structure on a flat fiber bundle I learned, that we can always replace $F$ by a simplicial complex (roughly $F \times E\pi_1$) but loose compactness. (So the above trick does not work with this.)

Here Equivalence of Flat Fiber Bundles vs Equivalence of Group Actions on the Fiber we are looking into what $\pi_{1}$-actions on $F$ result in equivalent fiber bundles. (No answers as of yet though.)

ort96
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