A simple branched covering is a branched covering with branching points of degree at most 2, in some context, it is also required to have at most one branching point in each fiber.

My question is one of the exercises from Donaldson's Riemann Surface: every Riemann surface can be realized as a simple branched covering over sphere.

Donaldson mentioned this result once again in chapter 14 when he talked about Hurwitz moves. And Donaldson said it was by Riemann-Roch theorem.

There is one proof in section 8 of http://www.jstor.org/stable/pdfplus/1970748.pdf?acceptTC=true&jpdConfirm=true, but it used more than Riemann-Roch.

  • $\begingroup$ Take a divisor of $\text{genus}+1$ points (crucial). Riemann-Roch says that the space of (nonconstant) meromorphic functions with (at worst) simple poles on the divisor is nonempty. $\endgroup$ – Chris Gerig Nov 6 '14 at 23:49
  • $\begingroup$ That only proves it's simple over infinity. It may have other branch points? $\endgroup$ – ZZY Nov 8 '14 at 0:08
  • $\begingroup$ Right, and now I don't see how you can deduce anything further. You can use Riemann-Hurwitz to see that it is possible for this branched cover to have degree 2 and 2g+2 simple branch points. And we can construct this explicitly, it's given on pg105 of Donaldson (Figure 7.3), generated by 180 degree rotation. $\endgroup$ – Chris Gerig Nov 8 '14 at 5:02
  • $\begingroup$ We tried that, but it only proves for g=2, it can be seen by dimension counting, real dimension of genus g surface is 6g-6, and for the rotation one, it's 2g+2 punctured sphere, so it's 4g-2, they are same iff g=2. And it's just the fact every genus 2 surface is hyperelliptic. One can use other special topological branched cover to prove for some other special genus, every such surface can be realized as branched cover of sphere with the fixed topological type. But it's hard for me to generalize to every genus. $\endgroup$ – ZZY Nov 9 '14 at 5:48
  • $\begingroup$ Huh? The 180 degree rotation quotient map is an example of a simple branched covering for any genus, where the surface has said symmetry. $\endgroup$ – Chris Gerig Nov 9 '14 at 8:20

Your Answer

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

Browse other questions tagged or ask your own question.