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Let $p: E \rightarrow B$ be a flat fiber bundle with fiber $F$ where $E$, $B$, $F$ are compact, smooth manifolds.

I am looking for a counterexample for the following statement:

$E \cong \widetilde{B} \times_G F'$, where $G$ is a finite quotient of $\pi_1(B)$ acting on a compact smooth manifold $F'$ and $\widetilde{B}$ is the cover of $B$ associated to that quotient. "$\cong$" stands for a) fiberwise diffeomorphic, b) fiberwise homotopy equivalent.

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This is one aspect of this question Compute cohomology of flat fiber bundles - does this always work?, more precisely formulated.

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    $\begingroup$ If your monodromy representation has values in $\operatorname{GL}(n,\mathbf C)$ then your statement is true, according to Deligne and Sullivan. $\endgroup$
    – Jarek Kędra
    Commented Sep 28, 2017 at 22:18

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Any finitely presented group occurs as the fundamental group of a smooth compact manifold, so the question reduces to whether we can find a finitely presented group $\pi$ acting on a smooth compact manifold $F$ which does not factor through a finite quotient up to homotopy. In turn, it's enough to find an action whose induced action on the homology groups of $F$ doesn't factor through a finite quotient.

An easy example that comes to mind is $\pi = GL_2(\mathbb{Z}), F = T^2$, since here the action on homology $H_1(T^2)$ is even faithful. Any infinite subgroup of $GL_2(\mathbb{Z})$ will do as well, including the copy of $\mathbb{Z}$ generated by any element of infinite order, and so we can get $3$-manifold examples which are $T^2$-bundles over $S^1$ by taking mapping tori.

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    $\begingroup$ Concretely, take $\left(\begin{smallmatrix}1 & 1\\0 & 1\end{smallmatrix}\right)$ and form the mapping torus, as Qiaochu suggested, and compute its fundamental group. It is isomorphic to $N(3,\mathbf Z)$ -- the uppertriangular matrices with integer coefficients. Then show that this group does not contain a subgroup isomorphic to $\mathbf Z^3$. $\endgroup$
    – Jarek Kędra
    Commented Sep 28, 2017 at 22:09
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    $\begingroup$ Right, this 3-manifold is concretely the Heisenberg manifold $H_3(\mathbb{R})/H_3(\mathbb{Z})$. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 28, 2017 at 22:43

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