# Misplaced complex analysis intuition on Riemann Surfaces

Next week I will be giving a lecture, based on Chapter 2.6 from Jost's book Compact Riemann Surfaces. He states the following theorem:

Theorem 1 (Jost Theorem 2.6.2) Let $S$ and $\Sigma$ be Riemann Surfaces, assume that $\Sigma$ is hyperbolic and complete with respect to the hyperbolic metric. Then any bounded holomorphic map $f: S\backslash\{p\}\to \Sigma$ extends to a holomorphic map $\overline{f}: S \to \Sigma$.

Now, the definition of hyperbolic that Jost is using is not the standard definition, he defines the hyperbolic metric $d_H$ on $\Sigma$ to be

$d_H(p,q) := \inf \{ \sum\limits_{i=1}^n d(z_i,w_i): n \in \mathbb{N},\ p_0,\ p_1,\dots,\ p_n \in \Sigma,\ p_0 = p,\ p_n = q,$ $\phantom{spacespacespace} f_i: D \to \Sigma \ \text{holomorphic}, \ f_i(z_i) = p_{i-1},\ f_i(w_i) = p_i \}$.

For reference purposes this is apparently a special case of the "Kobayashi metric".

Let $D$ be the unit disk in the complex plane. From complex analysis, we have the following theorem (I paraphrase from Rudin's Real and Complex Analysis):

Theorem 2 Suppose $f: D\backslash \{0\} \to \mathbb{C}$ is holomorphic and bounded. Then $f$ has a removable singularity at $0$.

Now, if we consider a coordinate chart $\psi_1$ on $S$ in a neighborhood of $\{p\}$, with $\psi_1(p) = 0$, and an atlas $\{(\phi_n,U_n)\}$ on $\Sigma$, we can write $\phi_i \circ f \circ \psi_1^{-1}: D\backslash \{0\} \to D$.

On immediately seeing the statement of the theorem, I want to claim that this map is bounded. Is this true? And, if so, can we not simply apply Theorem 2 to obtain Theorem 1? My complex analysis intuition says this should be trivial, but Jost produces a much more complicated proof, which lacks intuition.

• Without assuming that $p$ is a removable singularity, how do you deduce that there is a neighbourhood $V$ of $p$ such that $f(V\setminus\{p\})$ is contained in a coordinate neighbourhood? – Daniel Fischer Nov 26 '14 at 19:49
• I don't want to claim that $f(D\{p\}$ is contained in a single coordinate neighborhood. But I think locally you should be able to write it as such? – Michael Pinkard Nov 26 '14 at 20:08
• To use Riemann's removable singularity theorem, you need $\psi_1(p)$ to be an isolated singularity of some chart representation. It could be (well, by the theorem, it couldn't) that $\psi_1(f^{-1}(U))$ lies entirely in one half-plane with $\psi_1(p)$ on the boundary for every coordinate domain $U$. – Daniel Fischer Nov 26 '14 at 20:12
• So you are saying that, because we don't know a priori what kind of singularity $f$ has at $p$, that we can't assume that the image of any neighborhood of $p$ is contained in a single coordinate neighborhood on $\Sigma$, and therefore we can't apply the Riemann removable singularity theorem? – Michael Pinkard Nov 26 '14 at 20:32

The mistake in your derivation is that $\phi_i\circ f\circ\psi_1^{-1}$ is not defined in the whole punctured neighborhood of $0$, therefore Theorem 2 is not appicable. It is not defined because $\phi_i$ is defined only on some part of $\Sigma$.