When reading the "Lectures on Riemann Surfaces" by Otto Forster on page 37, he claimed that

Suppose $X$ is a Riemann surface and $f:X\to D^{*}$( $D^*$ is the punctured unit disk $\{z\in\mathbb{C}:0<|z|<1\}$) is an unbranched holomorphic covering map. Then one of the following holds:

(i) If the covering has an infinite number of sheets, then there exists a biholomorphic mapping $\varphi:X\to H$ of $X$ onto the left half plane such that the following diagram commutes $$ \require{AMScd} \begin{CD} X @>{\varphi}>> H\\ @V{f}VV @VV{\exp}V \\ D^* @>{id}>> D^* \end{CD} $$ (ii) If the covering is $k$-sheeted( $k<\infty$), then there exists a biholomorphic mapping $\varphi:X\to D^*$ such that following diagram commutes, where $p_k:D^*\to D^*$ is the mapping $z\to z^k$ . $$ \require{AMScd} \begin{CD} X @>{\varphi}>> D^*\\ @V{f}VV @VV{p_k}V \\ D^* @>{id}>> D^* \end{CD} $$

My question is, are there any deep explanations for this theorem? Why $D^*$ has such a good property that infinite and finite sheet can both be turned into some function we are familiar with? Can other Riemann surface other than $D^*$ have the similar property? Thank you for your help!


The universal covering space of the punctured disk $D^*$ in the category of Riemann surfaces is the upper-half plane, with the exponential map $H\to D^*$. The coverings $X \to D^*$ are classified by the subgroups of $\pi_1(D^*) = \mathbf Z$. The covering corresponding to $(0)$ is the full covering space $H \to D^*$ with the exponential map, whereas the the other coverings are $D^* \xrightarrow{z^k} D^*$. These, together, account for all of the subgroups of $\mathbf Z$, so all coverings of $D^*$ must fall in either category.

  • $\begingroup$ Thanks! So I wonder if there is other surface that has similar property? Or a covering of other surfaces that cannot be turned into logarithm or power root? $\endgroup$ – Golbez Dec 4 '13 at 2:27
  • $\begingroup$ @Golbez You are welcome! Which property precisely? $\endgroup$ – Bruno Joyal Dec 5 '13 at 6:45
  • $\begingroup$ That covering has an infinite number of sheets, then exists $\phi$ to make the diagram commutes. If there is finite number of sheets, also exists $\phi$ to make the second diagram commutes. Also,are there any "counterexample"s that this kind of $\phi$ does not exists? Thanks! If these two questions are answered, I will give a $50$ reputation bonus. $\endgroup$ – Golbez Dec 5 '13 at 7:59
  • $\begingroup$ @Golbez I still don't really understand what you are looking for. Are you just looking for some other space (other than $D^*$) whose coverings we can completely classify? What do you mean by "there exists $\phi$ which makes the diagram commutes"? What do you want the target of $\phi$ to be? $\endgroup$ – Bruno Joyal Dec 6 '13 at 22:40
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    $\begingroup$ @Golbez Well, the classification of coverings can get quite complicated. For instance if you take the plane minus 2 points, then the fundamental group is free on 2 generators, and this is a huge, non-abelian group. Surprisingly, however, the universal covering is just an open disc. (I don't know if this is the kind of stuff you wanted...) $\endgroup$ – Bruno Joyal Dec 10 '13 at 17:07

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