# Conformal covers of all degrees

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 $$\mathrm{dim}(M)=3$$?

• A finite product of copies of $S^1$ also works Sep 7, 2020 at 10:50
• @FrancescoPolizzi does it? Wouldn't you only get squares, or third powers and so on?
– user164740
Sep 7, 2020 at 10:51
• @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. Sep 7, 2020 at 11:26
• @RobertBryant my bad. But still, that's not all integers.
– user164740
Sep 7, 2020 at 11:27
• @JoeT: I didn't say it was. Sep 7, 2020 at 11:28

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.

• 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 $4n{+}1$. Sep 8, 2020 at 9:07
• @RobertBryant: I removed that sentence, I did not think enough about the problem when I wrote it. 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.