# Complement of compact subset in Riemann surface

We have $S\subset S'$ Riemann surfaces with $S$ hyperbolic with boundary. For sufficiently large compact set $K\subset S$ does each component of $S'\setminus K$ have to be a hyperbolic Riemann surface? If so, how can this be shown? If not, can extra conditions provide such a result?

• What do you know about the compactness of $S$ and $S'$? – paf Aug 13 '16 at 13:06
• Is there any info for $S'$ like compact closed empty boundary or with/without finitely many points removed (finite type)? When you say hyperbolic do you mean a surface whose every connected component has the disc as a universal cover? Is $K$ arbitrary or are there some more specific restrictions? – Futurologist Aug 13 '16 at 13:15
• @Futurologist No assumptions on compactness. By hyperbolic I mean that the universal cover of $S$ is conformally isomorphic to the unit disk. No further restrictions on $K$. – Test123 Aug 13 '16 at 13:20
• But what do you mean by $K$ is large enough? It cannot be either one or two points in the case when $S'$ is the Riemann sphere... – Futurologist Aug 13 '16 at 13:31

## 1 Answer

I think this is more or less true if we assume $K$ is compact but contains at least three points. The set $S'\setminus K$ is a non-empty open subset of $S'$ and therefore a subsurface. Let $S_c$ be a connected component of $S'\setminus K$. It is a non-compact connected Riemann surface without boundary (since it's open) and let $p : \hat{S}_c \to S_c$ be its universal covering space. Then by the uniformization theorem $\hat{S}_c$ is conformally isomorphic to either the Riemann sphere, the plane or the unit disk. $\hat{S}_c$ cannot be the Riemann sphere as $S_c$ itself is not the Riemann sphere. Let us assume it is the plane $\mathbb{C}$ and thus the universal covering space of $S_c$ is $p : \mathbb{C} \to S_c$. Then the group of deck transformations of $p : \mathbb{C} \to S_c$ is the subgroup $\Gamma \subset \{z \mapsto az+b\}$ acting properly discontinuously on $\mathbb{C}$ without fixed points by conformal automorphisms of $\mathbb{C}$. But the restriction that there are no fixed points implies that $\Gamma \subset \{z \mapsto z+b\}$ , i.e. it's a discrete subgroup of the translation group. There are only two types of such groups -- either $\Gamma = \langle z \mapsto z+b \rangle$ or $\Gamma = \langle z \mapsto z+b, \, z \mapsto z+c\rangle$. Since by the uniformization theorem $S_c\cong\mathbb{C}/\Gamma$, there are two possibilities for $S_c$: In the first case it is conformally isomorphic to $\mathbb{C}\setminus \{0\}$, in the second case it is conformally isomorphic to a torus (i.e. аn elliptic curve). By construction, $S_c$ can be neither of those (it is not compact to be the torus and it is not obtained by the removal of two points from another Riemann surface because $K$ has at least three points). Therefore the universal covering space of $S_c$ can only be the unit disc, i.e. it is a hyperbolic Riemann surface.