Fix an algebraic integer $x\neq 0$. Does there exist a closed smooth manifold $M$ with a class $\rho\in H^{1}_{\mathrm {dR} }(M)$ and a smooth covering map $\phi:M\to M$ such that $\phi^*\rho=x\rho$?
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$\begingroup$ Did you really mean to ask about covering maps $M\to M$, or covering maps $\widetilde M\to M$? I strongly suspect the answer is negative in the former case; in the latter I'm not so sure. $\endgroup$– Mark GrantCommented Sep 8, 2020 at 15:15
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1$\begingroup$ @MarkGrant: In the latter case, the equation $\phi^*\rho = x\rho$ doesn't make sense. $\endgroup$– Michael AlbaneseCommented Sep 8, 2020 at 15:16
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$\begingroup$ @MarkGrant it's $M\to M$ $\endgroup$– user164740Commented Sep 8, 2020 at 15:17
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3$\begingroup$ This means that $x$ is an eigenvalue of a matrix with integer entries, hence an algebraic integer. $\endgroup$– Phil TostesonCommented Sep 8, 2020 at 15:18
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$\begingroup$ @PhilTosteson thank your for your remark $\endgroup$– user164740Commented Sep 8, 2020 at 15:19
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1 Answer
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Every nonzero algebraic integer $x$ in $\mathbb R$ is an eigenvalue of an $\mathbb R$-diagonalizable integer matrix $A \in M_n(\mathbb Z)$ with $\det(A) \neq 0$ for some $n$. So take the map of tori $A^t: \mathbb R^n /\mathbb Z^n \to \mathbb R^n/\mathbb Z^n$. This acts by $A^t$ on $\mathbb Z^n = H_1(\mathbb R^n /\mathbb Z^n)$ and hence by $A$ on $H^1(\mathbb R^n /\mathbb Z^n)$.
answered Sep 8, 2020 at 15:56
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$\begingroup$ Not quite! Only algebraic integers of norm $\pm 1$ occur this way. $\endgroup$ Commented Sep 8, 2020 at 16:06
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$\begingroup$ Thanks @Qiaochu-- I shouldn't have said $GL_n(\mathbb Z)$ (I was trying to be too brief). I think everything works the same when $det(A) \neq 0$. $\endgroup$ Commented Sep 8, 2020 at 16:12
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$\begingroup$ I guess the result is just a nontrivial covering map so yes. $\endgroup$ Commented Sep 8, 2020 at 16:13