# Minimal degree of holomorphic function in a Riemann Surface

Fixed a genus $g$; what is the minimal integer $d$ such that for any Riemann Surface $X$ of genus $g$ there exist a positive divisor $D$ such that $l(D)\ge 2$ and $deg(D)=d$?

A Riemann-Roch argument gives that we can construct a divisor of degree $g$ with $l(D)>1$, but i don't know how to show that there exist curves of genus $g$ such that they don't have (nontrivial) meromorphic function in $l(D)$ if $deg(D)<g$,or something like that.

• $l(D)\ge 2$ because i don't want to consider the constant functions; in low genus i think that is different (for example every riemann surface of genus 2 is hyperelleptic, then his map is minimal degree is 2), but i ask in general. – lecheconmilo Nov 22 '16 at 21:17
• Sorry I messed up :) Yes $l(D) \ge 2$ means there is a non-constant meromorphic function $f$ such that $(f)+D$ is a positive divisor, and $l(D) \ge 2$ since the vector space contains $C_1 f(z)+C_2$. In the case of $g = 1$, the minimum is $deg(D)=2$ and every $D,deg(D)=2$ works, while in the case $g=2$ it is $deg(D)=2$ but not every $D$ works. – reuns Nov 22 '16 at 21:19
• I think you can find the answer in chapter 4 and 5 – reuns Nov 22 '16 at 22:01