6
$\begingroup$

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!

$\endgroup$
6
+50
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.