Fix an algebraic integer $x\neq 0$. Is there a closed smooth 4-manifold $M$ with a class $\rho\in H^{1}_{\mathrm {dR} }(M)$ and a smooth covering map $\phi:M\to M$ such that $\phi^*\rho=x\rho$?
Is there such a manifold of the form $P\times S^1$?
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2$\begingroup$ We need $x$ an algebraic integer (and not just an algebraic number). If $x$ is an algebraic unit (i.e. $x^{-1}$ is an algebraic integer) then $x$ is an eigenvalue of a matrix in $Sp_{2g}(\mathbb Z)$ and so $x$ is an eigenvalue of a map from a surface to itself (or a $3$-manifold by taking the product with $S^1$.) So the difficulty is for algebraic integers that are not units. $\endgroup$– Will SawinCommented Sep 8, 2020 at 17:17
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2$\begingroup$ Very few closed 3-manifolds admit nontrivial self coverings: I think you only get Seifert manifolds (and not all of them). $\endgroup$– Moishe KohanCommented Sep 8, 2020 at 17:59
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$\begingroup$ @MoisheKohan $H^1$ of a Seifert manifold is either $H^1$ of the underlying surface or $H^1$ of the surface + $H^1$ of the fiber. So maybe you only get units again for Seifert manifolds ( algebraic integers of degree at most 3 from the 3-torus). Do you know a reference? $\endgroup$– Will SawinCommented Sep 9, 2020 at 15:00
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1$\begingroup$ @WillSawin Seifert manifolds are not fibrations, but rather singular fibrations. If you like, you can think of them as fiber bundles over an orbifold. But there is no Serre SS available here. $\endgroup$– mmeCommented Sep 12, 2020 at 12:05
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1$\begingroup$ @MikeMiller It seems Moishe gave a simpler argument, but there is a Leray spectral sequence associated to any morphism, and not just a fibration, and it also works for orbifolds, and even for nastier spaces like algebraic stacks - why wouldn't it? $\endgroup$– Will SawinCommented Sep 12, 2020 at 14:04
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