Edit: This has been answered in this thread Are there compact flat fiber bundles with "truly" infinite structure group?.


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 of characteristic zero or not dividing the order of $\pi_1$. (as pointed out by Neil Strickland)

Since we want to compute cohomology and perhaps characteristic classes, we are only concerned with fiber bundles up to fiber preserving homotopy equivalences.


$\pi_{1}$ 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 $$H^*(E) = H^*(\widetilde{B}/ker(\phi) \times F)^{G} = (H^*(\widetilde{B}/ker(\phi)) \times H^*(F))^{G}. $$


Are there compact flat fiber bundles where we cannot do this trick to compute cohomology? I am looking for an explicit example.


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.)

  • $\begingroup$ could you fix the typo in the title? $\endgroup$
    – YCor
    Sep 28, 2017 at 10:19
  • 5
    $\begingroup$ For $H^*(E)=H^*(\widetilde{B}\times F)^{\pi_1}$ you need the the field of coefficients to have characteristic zero, or characteristic not dividing the order of $\pi_1$. $\endgroup$ Sep 28, 2017 at 10:20

1 Answer 1


Let $G=\mathbb{Z}$ act on $S^1$ by an irrational rotation: this defines a flat fibre bundle $S^1 \to E \to S^1$. As $\mathbb{Z}$ is a free group this action can be deformed to the trivial action, and so this fibre bundle is isomorphic to the trivial bundle (as a non-flat fibre bundle).

There is clearly no $G$-invariant triangulation of $S^1$. If you try to interpret your formula nonetheless you get $(H^*(\mathbb{R}) \otimes H^*(S^1))^G \cong H^*(S^1)$ which is too small.

  • $\begingroup$ But I can replace this first by the isomoprhic trivial bundle. My formula then just replaces the universal covering $\mathbb{R}$ by the trivial covering $S^1$ and everything works. I meant a counterexample that is not homotopy equivalent to something where it works. $\endgroup$
    – ort96
    Sep 28, 2017 at 11:29

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