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Is there a classification of branched coverings of the closed unit disk $\mathbb{D} =\{z\in \mathbb{C} \ | \ |z| \leq 1 \}$? Here we consider only branched covering projections which restrict to unbranched covering projections on the boundary. I am especially interested in the following question:

Which Riemann surfaces $X$ can/cannot be obtained as branched coverings of $\mathbb{D}$?

What I can prove:

  • since $\mathbb{D}$ is not closed, no closed $X$ can be realized as a branched covering of it
  • some computations with Riemann-Hurwitz formula show that if there is only one branching point $p \in \mathbb{D}$ then $X$ must be homeomorphic to a disk.
  • for a covering $f:X\to \mathbb{D}$ of degree $N\geq 2$ with $r$ branching points on $\mathbb{D}$ we have $$ \chi (X) = (1-r)N + \sum_{i=1}^r s_i $$ where $s_i$ is the cardinality of the fiber over the $i$-th branching point. But this formula does not give any restriction, since it allows for any value of $\chi$(X).

Thanks in advance for any reference/suggestion.

Lor
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  • The Riemann-Hurwitz formula is valid for compact Riemann surfaces.. – Malik Younsi Sep 23 '14 at 15:03
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    Regarding your question, it might be relevant to say that any $n$-connected planar domain can be mapped onto the unit disk by an holomorphic branched covering fo degree $n$, see http://math.stackexchange.com/questions/95570/does-constant-modulus-on-boundary-of-annulus-imply-constant-function/95581#95581 – Malik Younsi Sep 23 '14 at 15:09
  • Sorry for the improper use of Riemann-Hurwitz. I probably want to consider covering of the closed unit disk which give unbranched covers of the boundary. I will edit the question accordingly. – Lor Oct 06 '14 at 14:26

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