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Edit: After helpful comments, I now know that I am concerned with flat fiber bundles up to fiber preserving homotopy.

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

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

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

Now, how nice can we assume this $\pi_1$-action to be? I would like to use additional structure in my problem without giving up too much generality. In particular, is it very restricting to only look at flat fiber bundles of the form

(ii) $E \cong \widetilde{B} \times_{\pi_{1}} |L|$,

where $L$ is a simplicial $\pi_{1}$-complex? What, if we further assume the action on $L$ to be regular?

I can't think of a flat bundle that is not of form (ii). Do you know counterexamples that clarify what (ii) cannot describe? (possibly with relaxed conditions on $E$, $F$, $B$.) And are there theorems that specify exact conditions on a flat bundle to be of form (ii)?

I am particularly interested in the case $B$, $F$ compact, $\pi_{1}$ infinite.

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  • $\begingroup$ I don't understand what is being asked, exactly. You've defined flat fiber bundles as twisted products, and then you ask if it's reasonable to assume something about the action of $\pi_1$. Adding hypotheses is quite often reasonable, if you can deduce something interesting from them. Without more information about your motivation, it seems impossible to say anything mathematical here. $\endgroup$
    – Dan Ramras
    Commented Sep 15, 2017 at 3:22
  • $\begingroup$ You are right, 'reasonable' was very vague. (edited the question a bit) I am trying to understand how much smaller the category of fiber bundles of form (ii) is versus those of form (i). Is there even a difference? Does that difference vanish if we look at those categories modulo homotopy equivalences? $\endgroup$
    – ort96
    Commented Sep 15, 2017 at 22:30
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    $\begingroup$ If you consider flat fibre bundles "as is", the action can be as ugly as you want. For example, take $B=S^1$, then the action is described by a single homeomorphism $\Phi$ of $F$. Try $F=S^1$ and let $\Phi$ be a rotation by an irrational multiple of $\pi$. Then $\Phi$ cannot be simplicial. Or are you allowed to consider actions up to isotopy or spaces as fibres up to some equivalence in your specific problem? $\endgroup$
    – Sebastian Goette
    Commented Sep 17, 2017 at 18:25
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    $\begingroup$ I think one would need to have specific goals in mind in order to really give an answer here. For instance, are you interested in fiber-wise homotopy equivalences? Are there some types of invariants you want to compute, for which you are hoping to replace a given bundle by something simplicial that will have the same invariants? $\endgroup$
    – Dan Ramras
    Commented Sep 18, 2017 at 18:38
  • $\begingroup$ Thanks. Yes, fiberwise homotopy equivalence is the correct notion for my problem. $\endgroup$
    – ort96
    Commented Sep 19, 2017 at 19:35

1 Answer 1

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Up to homotopy, everything works. For simplicity, assume that $F$ is connected. Let $G=\pi_1$ with discrete topology, $EG$ the classifying space for $G$; then first replace $F$ with $EF= F\times EG$, the action $G\times F\to F$ with $G\times EF\to EF$. The latter is now properly discontinuous, $EF/G$ has homotopy type of a CW complex $Y$, hence, you get a $G$-equivariant homotopy-equivalence $F\to EF\to \tilde{Y}$, where $\tilde{Y}$ is the universal cover of $Y$. Now, take $\tilde{B}\times_G \tilde{Y}$.

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  • $\begingroup$ Thanks! I'll have to read more on classifying spaces to fully understand this. Doesn't $F \cong \widetilde {Y} $ imply $F $ simply connected? So perhaps $\widetilde {Y}$ is not universal? $\endgroup$
    – ort96
    Commented Sep 21, 2017 at 18:33
  • $\begingroup$ @ort96: Yes, you are right, I was thinking about simply-connected $F$. The cover $\tilde{Y}\to Y$ is the one corresponding to the kernel of the natural homomorphism $\pi_1(Y)\to G$, so its fundamental group will be isomorphic to $\pi_1(F)$. The point is that $EG$ is contractible. $\endgroup$
    – Misha
    Commented Sep 22, 2017 at 1:08

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