Hyperbolic Riemann Surface

Question

Let $X$ be a compact Riemann surface and $x\in X$. Is $X - \overline{D(x,r_x)}$ hyperbolic?

Answer 1

If a domain on a Riemann surface has a component of the frontier with at least two distinct points then it is potential theoretic hyperbolic. This result goes back to Osgood. Existence of green 's function is connected to existence of barrier at some frontier point. The existence of barriers is local. The solution of the local problem can be found in the book of Ahlfors-Sario titled Riemann Surfaces .

Answer 2

Here's an answer from a different point of view then Henri's. It may happen that $X'$ is disconnected but in that case I'll just argue one component at a time. The Riemann surface $X'$ is noncompact, and if it is nonhyperbolic then by the Riemann mapping theorem it must be conformally equivalent (biholomorphic) to either $\mathbb{C}$ or $\mathbb{C}-$(point). Each of its one or two ends must therefore be a removable singularity. But that's impossible if $r_x > 0$ (justifying this impossibility probably involves a case analysis not unlike what Henri proposes).

Answer 3

Let $X'=X\setminus \overline {D(x,r_x)}$.

If $X$ is already hyperbolic, then the answer is yes (there is no entire curves in $X'$).

If $X=\mathbb C, \mathbb C^*$, then $X'$ has no entire curves by Picard's little theorem, so it is hyperbolic.

If $X=\mathbb C/\Lambda$, then $\pi_1(X')$ is non trivial, and different from $\mathbb Z$ or $\mathbb Z^2$, so $X'$ is hyperbolic. This works if $r_x$ is not too big. Else you could imagine that $X'$ becomes $\mathbb C^*$.

If $X$ is $\mathbb P^1$, then your surface is biholomorphic to $\mathbb D$ (it is simply connected and strictly realized inside the complex plane).

In conclusion, $X'$ is always hyperbolic.