# Hyperbolic Riemann Surface

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

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If you look at the tags it looks like theOP is interested in the existence of positive Green's function which is certainly true for many values of the radius of the disc that is removed. –  Mohan Ramachandran May 3 '13 at 17:23
My goal is can say that there exists a Green's function on it. –  James May 6 '13 at 11:25

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.

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In the case where $X = \mathbb{C} / \Lambda$, nothing goes wrong when $r_x$ is too big, $X'$ will still be hyperbolic. It is possible as $r_x$ increases for the topological type of $X'$ to change (it could even become disconnected); but its hyperbolic nature will not change. –  Lee Mosher May 3 '13 at 13:51
A priori, in the non-compact setting, it is not enough to exclude the existence of entire curves in order to establish Kobayashi hyperbolicity (this is a necessary but not a sufficient condition). But for Riemann surfaces, of course, it is necessary and sufficient to be uniformized by the disc (so that in this case it suffices to check non-existence of entire curves). –  diverietti May 3 '13 at 15:46
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).