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.

  • $\begingroup$ 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? $\endgroup$ Nov 26, 2014 at 19:49
  • $\begingroup$ 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? $\endgroup$ Nov 26, 2014 at 20:08
  • $\begingroup$ 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 $U$. $\endgroup$ Nov 26, 2014 at 20:12
  • $\begingroup$ 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? $\endgroup$ Nov 26, 2014 at 20:32

1 Answer 1


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 $\Sigma$.

Moreover, any proof of Theorem 1 must use somehow that "complete hyperbolic metric" (otherwise what the word "bounded" can mean?). And your derivation does not use it.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .