Rigidity of isometries of finite covers of Riemann surfaces

Let $$\Theta$$ and $$\Sigma$$ be compact Riemann surfaces with $$\widetilde\Theta=\widetilde\Sigma=\mathbb{H}^2$$ (so $$\theta$$ and $$\Sigma$$ are quotients of $$\mathbb{H}^2$$ by Fuchsian groups) such that $$\rho:\Theta\rightarrow\Sigma$$ is a finite cover (locally isometric not branched).

What can we say about $$\textrm{Isom}(\Theta)$$ if we have a description of $$\textrm{Isom}(\Sigma)$$? More specifically, is every element of $$\textrm{Isom}(\Theta)$$ a lift of an element of $$\textrm{Isom}(\Sigma)$$?

If this is too general, then assume that $$\Sigma$$ is constructed from a Fricke canonical polygon with $$4g$$ sides. Note that $$\textrm{Isom}(\Theta)$$ and $$\textrm{Isom}(\Sigma)$$ are necessarily finite.

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As an additional restriction assume that the Fuchsian group $$\Gamma$$ used to construct $$\Sigma$$ is torsion-free and arithmetic (i.e. $$\Gamma$$ is commensurable with a Fuchsian group derived from a quaternion.

• This holds for generic surfaces $\Sigma$ but not in general, e.g. for arithmetic hyperbolic surfaces $\Sigma$ since the commensurator of $\pi_1(\Sigma)$ is dense in $PSL(2,{\mathbb R})$. I think, every Riemann surface admits a "Fricke canonical polygon", so maybe you have in mind some further restrictions. – Moishe Kohan Mar 23 at 21:50
• Thank you and yes I think you are right about the polygons. Are there any situations when it does hold? – Sam Hughes Mar 23 at 22:24
• As I said, "generically" this is true: There exists an open and dense subset of the moduli spaces of $\Sigma$ for which all automorphisms of $\Theta$ come from the covering map and from $\Sigma$ itself. In turn, generically $\Sigma$ has no nontrivial automorphisms except for genus 2 surfaces which are all hyperelliptic. – Moishe Kohan Mar 23 at 22:45
• Sorry I'm not too familiar with moduli spaces so I might be being a bit slow here. Going the other way, you say it may not hold for an arithmetic hyperbolic surface. If you could explain in more detail why as answer (or give an example) that would be great. – Sam Hughes Mar 23 at 23:13

1 Answer

Let $$g$$ denote the genus of $$\Sigma$$. I will say that a Riemann surface $$\Sigma$$ is maximal if for every holomorphic covering map $$p: \Sigma'\to \Sigma$$ the group of conformal automorphisms of $$\Sigma'$$ consists only of elements which are lifts of automorphisms of $$\Sigma$$ (this includes automorphisms of the covering map $$p$$: these are lifts of the identity map $$\Sigma\to \Sigma$$).

Suppose that $$g\ge 3$$. Then there exists an open and dense subset $$U$$ of the moduli space $${\mathcal M}_g$$ of genus $$g$$ surfaces such that every $$\Sigma\in U$$ is uniformized by a (torsion-free) maximal Fuchsian subgroup of $$PSL(2, {\mathbb R})$$. Here a Fuchsian subgroup is called maximal if it is not contained in any strictly larger Fuchsian subgroup.

See

L. Greenberg, Maximal Fuchsian groups. Bull. Amer. Math. Soc. 69 (1963) 569–573.

and also (for further results in this direction)

L. Greenberg, Maximal groups and signatures. Discontinuous groups and Riemann surfaces (Proc. Conf., Univ. Maryland, College Park, Md., 1973), pp. 207–226. Ann. of Math. Studies, No. 79, Princeton Univ. Press, Princeton, N.J., 1974.

Next, observe that if $$F< PSL(2,{\mathbb R})$$ is a maximal (torsion-free) Fuchsian subgroup then the Riemann surface $$H^2/F$$ is maximal. Indeed, a finite holomorphic covering map $$p: \Sigma'\to \Sigma$$ comes from a finite-index subgroup $$F'< F$$: $$\Sigma'=H^2/F'$$. The preimage of the group of conformal automorphisms of $$\Sigma'$$ in $$PSL(2, {\mathbb R})$$ is a Fuchsian subgroup $$F''< PSL(2,{\mathbb R})$$ containing $$F'$$ as a finite index subgroup. Then the subgroup of P$$SL(2,{\mathbb R})$$ generated by $$F, F''$$ would be still Fuchsian, hence, equal to $$F$$ by the maximality assumption.

Thus, "generic" Riemann surfaces $$\Sigma$$ of genus $$\ge 3$$ are maximal. This argument fails in the case of $$g=2$$ but only because every genus 2 Riemann surface is hyperelliptic, it admits a hyperelliptic involution. By digging a bit deeper in Greenberg's papers, you can see that every generic genus 2 Riemann surface is again maximal.

I will write more on arithmetic surfaces later.

Edit. Here are more details on the arithmetic case. I will need a definition and a theorem:

Definition. Two subgroups $$H_1, H_2$$ of a group $$G$$ are called commensurable if $$H_1\cap H_2$$ has finite index in both $$H_1, H_2$$. Given a subgroup $$H< G$$ the commensurator of $$H$$ in $$G$$, denoted $$Comm_G(H)$$, is the subset consisting of elements $$g\in G$$ such that the subgroups $$H^g:= gHg^{-1}< G$$ and $$H$$ are commensurable.

It is easy to see that $$Comm_G(H)$$ is a subgroup of $$G$$.

Theorem (Borel?). Let $$G$$ be a connected semisimple Lie group. Then for every arithmetic subgroup $$\Gamma< G$$, the commensurator $$Comm_G(\Gamma)$$ is dense in $$G$$.

I will use this in the case $$G=PSL(2, {\mathbb R})$$. You can find a proof for instance in the book by Machlachlan and Reid "The arithmetic of hyperbolic 3-manifolds". You can easily convince yourself that in the case $$\Gamma=PSL(2, {\mathbb Z})$$ the commensurator $$Comm_G(\Gamma)$$ contains $$PSL(2, {\mathbb Q})$$.

This theorem has a much harder converse due to Margulis (that I will not need): If the commensurator of a lattice is nondiscrete then the lattice is arithmetic.

Proposition. Every arithmetic Riemann surface $$\Sigma$$ is non-maximal.

Proof. Let $$F< PSL(2, {\mathbb R})$$ be a Fuchsian subgroup. Then for every $$g\in Comm_G(F)$$ the subgroup $$F^g$$ normalizes a finite index subgroup $$F'$$ of $$F$$ and, hence, acts on the Riemann surface $$H^2/F'$$ by automorphisms. If the surface $$\Sigma=H^2/F$$ were maximal, all the Fuchsian groups of the form $$F^g$$ would be contained in a fixed Fuchsian group $$\hat{F}$$ (obtained by lifting all automorphisms of $$\Sigma$$). However, the union $$\bigcup_{g\in Comm_G(F)} F^g$$ is stable under conjugation via elements of $$Comm_G(F)$$. Hence, the subgroup $$M$$ generated by this union is also normalized by $$Comm_G(F)$$. In view of the density of $$Comm_G(F)$$ in $$G$$, the subgroup $$M$$ has to be dense as well. This means that $$M$$ cannot be contained in a discrete subgroup $$\hat{F}$$ as above, proving non-maximality of the surface $$\Sigma$$. qed

This proof is nonconstructive. You will obtain explicit examples by considering a non-uniform lattice in $$G$$ such as $$F=PSL(2, {\mathbb Z})$$ and looking at its conjugates $$F^g$$ by, say, linear-fractional maps of the form $$z\mapsto az$$, $$a\in {\mathbb Q}^2$$.

• This is a fantastic answer, thank you very much! – Sam Hughes Mar 25 at 20:47