Let $M$ be a connected closed conformal oriented manifold.

Assume there exist conformal covering maps $\phi_k:M\to M$ of all degrees $k\geq 1$. Is $M\cong S^1$ then?

Can we at least rule out $\mathrm{dim}(M)=3$?

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    $\begingroup$ A finite product of copies of $S^1$ also works $\endgroup$ Sep 7, 2020 at 10:50
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    $\begingroup$ @FrancescoPolizzi does it? Wouldn't you only get squares, or third powers and so on? $\endgroup$
    – user164740
    Sep 7, 2020 at 10:51
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    $\begingroup$ @JoeT: Not at all. If $M^2$ is the square torus, then there is a conformal covering map of degree $k = a^2+b^2$, where $a$ and $b$ are integers. $\endgroup$ Sep 7, 2020 at 11:26
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    $\begingroup$ @RobertBryant my bad. But still, that's not all integers. $\endgroup$
    – user164740
    Sep 7, 2020 at 11:27
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    $\begingroup$ @JoeT: I didn't say it was. $\endgroup$ Sep 7, 2020 at 11:28

2 Answers 2


Here is a partial answer: If there is such a conformal manifold $M$ of dimension $n\ge 2$, then $M$ admits a flat metric. The reason is that the sequence of conformal covering maps $\phi_k: M\to M$ cannot contain a subsequence converging to a conformal map. Hence, the universal conformal covering $\tilde{M}$ cannot admit a compatible Riemannian metric for which the lifts $\tilde\phi_k: \tilde{M}\to \tilde{M}$ are isometric. Thus, by Ferrand's solution of Lichnerowicz conjecture

Ferrand, Jacqueline, The action of conformal transformations on a Riemannian manifold, Math. Ann. 304, No. 2, 277-291 (1996). ZBL0866.53027.

the manifold $\tilde M$ is either conformal to $S^n$ (which is, of course, impossible) or to $E^n$.

Thus, the problem essentially reduces to the one of flat tori and there should be an algebraic argument proving that $n=1$ in this setting:

Suppose that $\Gamma< Isom(E^n)$ is a discrete cocompact torsion-free subgroup such that the manifold $M=E^n/\Gamma$ admits a covering $\phi: M\to M$ of degree $d$. Then $\phi$ lifts to an affine conformal map $\Phi: E^n\to E^n$. Let $\Lambda< \Gamma$ be the the translation lattice in $\Gamma$. Then $\Phi \Lambda \Phi^{-1}= \Lambda'$ is index $d$ sublattice. In other words, $\Phi$ projects to a degree $d$ conformal self-map $\psi: A\to A$, where $A= E^n/\Lambda$ is a flat torus.

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    $\begingroup$ I'm not quite sure what your last sentence means. For flat tori, the space of covering maps of a given degree always has positive dimension, so you can't literally mean 'finitely many'. Also, you can't mean that only a finite number of primes can be the degree of a conformal covering map. For the square torus in dimension 2, for example, the degree can be any number of the form $a^2+b^2\not=0$ where $a$ and $b$ are integers, and there are infinitely many primes of this form: $2$ and all the primes of the form $4n{+}1$. $\endgroup$ Sep 8, 2020 at 9:07
  • $\begingroup$ @RobertBryant: I removed that sentence, I did not think enough about the problem when I wrote it. $\endgroup$ Sep 8, 2020 at 14:24

This is the second part of the answer. Suppose $E^n$ is a flat torus admitting a conformal self-map $\varphi_d$ of degree $d$ for every $d=1,2,3,\ldots$. We prove that this is only possible when $n=1$.

Algebraic reformulation: Fix a positive definite symmetric bilinear form $Q$ on $\mathbb{R}^n$, $n\geq 2$. Call an integer $n\times n$ matrix $M$ conformal if $M^t Q M$ is a positive real multiple of $Q$. Degree of such a matrix is $\det M$. We prove that it is not possible to have a conformal matrix of degree $d$ for each $d=1,2,3,\ldots$. Suppose the contrary, i.e. there is such an integer matrix $M_d$ for every $d$.

First, by taking determinants for every $d$ we find the coefficient of proportionality $$ M_d^t Q M_d= d^{2/n} Q. $$ Normalize $Q$ so that $Q_{11}=1$. For any vectors $u,v$ denote $(u,v)=u^t Q v$. Let $v_d$ be the first column of $M_d$. Then we have $$ (v_d, v_d) = d^{2/n}\qquad (d=1,2,3,\ldots) $$ We claim that this is impossible. Consider the case $n=2$ first. Note that no two among $v_1, v_2, v_3$ can be collinear. Hence $v_3=\alpha v_1 + \beta v_2$ for some $\alpha, \beta\in\mathbb{Q}$. This allows to compute all the entries of $Q$ out of $\alpha, \beta$ and deduce that they are rational. So we have $a,b,c\in\mathbb{Q}$ so that the equation $a x^2 + b xy + c y^2=d$ has solutions in integers for every $d$, but $b^2-4a c<0$. This is impossible: by Chebotarev density theorem one can choose a prime $p$ such that $p$ doesn't divide the numerators and the denominators of $a,b,c$ and the equation $a x^2 + b x + c=0$ has no roots mod $p$. Setting $d=p$ leads to a contradiction.

Now consider the case $n\geq 3$. Consider the numbers of the form $p^{2/n}$ for prime numbers $p>n$. They are linearly independent over $\mathbb{Q}$ because the field extension generated by $p^{2/n}$ is ramified at $p$, and can only be further ramified at the divisors of $n$. On the other hand, consider the sequence of integer $n\times n$ matrices $v_p v_p^t$. We have an infinite sequence of elements of a finite dimensional vector space, so there must be a linear relation $$ \sum_{i=1}^N c_i v_{p_i} v_{p_i}^t = 0 \qquad ((c_1,c_2,\ldots,c_N)\in\mathbb{Q}^N\setminus \{0\}) $$ for prime numbers $p_1,p_2,\ldots,p_N>n$. This implies $$ \sum_{i=1}^N c_i p_i^{2/n} = \sum_{i=1}^N c_i (v_{p_i}, v_{p_i}) = \sum_{i=1}^N c_i \operatorname{trace}(Q v_{p_i} v_{p_i}^t)=0, $$ a contradiction.


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