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It is known that on a hyperelliptic surface the set of Weierstrass points and the set of ramification points of the extension of the projection map $(x,y)\mapsto x$ to $\mathbb{P}^1$ coincide.

However, I am not sure if this is the case for a general Riemann surface. Maybe $p$ a Weierstrass point iff there is a ramified covering of $\mathbb{P}^1$ with $p$ as a ramification point.

It seems like I can't do much in either direction to prove this, but it also seems like something similar to this proposition would be nice to have since ramification provides a nice geometric intuition.

AnonymousCoward
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  • For future reference, this overflow post contains what I was looking for http://mathoverflow.net/questions/9204/what-do-weierstrass-points-look-like – AnonymousCoward Jun 02 '11 at 16:26

1 Answers1

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There is no relationship like the one you proposed:

  1. If $p$ is an arbitrary point on a regular projective curve $C$, then by Riemann-Roch there is a function $f$ having a zero of order $>1$ at $p$. The covering induced by $f$ is ramified at $p$.

Of course one could require $p$ to be the only ramification point. But then:

  1. By the Riemann-Hurwitz formula a covering $C\rightarrow\mathbb{P}^1$ alwways has more than $1$ ramification point provided that $C$ has genus $>0$ (maybe plus some additional requirements in the case of characteristic $>0$ and in the case $K$ is not algebraically closed).

H

Hagen Knaf
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