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Donu's answer is correct but amounts to killing a fly with a gun shot: Greenberg proves a harder result than the one needed for the problem. Theorem. Let $G$ be a countable group. Then there exists a …

Your conjecture is false. Every nonorientable closed connected surface of negative Euler characteristic, admits a hyperbolic metric such that the surface is covered by 3 embedded disks. Hence, for eac …

Here is a translation of Strebel's definition. First of all, a quadratic differential on a complex manifold $M$ (holomorphic or not) is a smooth section of the symmetric square $S^2(T^{*(1,0)}M)$ of …

The following is not a real answer but an extensive comment on the OP. First, recall that an (open) Riemann surface is said to have type $P_{AB}$ (resp. $O_{AB}$) if it admits (resp., does not admit) …

In order to remove this question from the "unanswered list." Let $\epsilon>0$ be the Margulis constant for the hyperbolic plane (with curvature $-1$). Then for every complete hyperbolic surface $S$, i …

Expanding on my comments, here are some obstructions coming from Hausdorff dimension and self-similarity. One observation is that every nonelementary Kleinian group $\Gamma$ has positive critical exp …

Yes, this is true, but proving this is easier than finding a reference. Every finitely-generated matrix group (e.g. a lattice in $PSL(2, {\mathbb R})$ contains a torsion-free subgroup. The general re …

First of all, the constants $c_i$ will have to depend on the complex structure of $X$ since without prescribing a complex structure one cannot talks about dependence on a basis of the space of holomor …

The answer to your first question is positive. First of all, it suffices to consider the case when every point of your subset $C\subset S$ is a non-removable singularity of $f: S\to \mathbb C P^1$. (I …

I am not sure how well you know the Teichmuller theory, but the basic thing to understand is that $\mathcal T$ is not the space of hyperbolic metrics on $\Sigma$: It is the space of pairs $(\sigma, [\ …