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



Conformal covers of all degrees
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**](http://dx.doi.org/10.1007/BF01446294), Math. Ann. 304, No. 2, 277-291 (1996). [ZBL0866.53027](https://zbmath.org/?q=an:0866.53027).


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.


