0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

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.