Lemma. If $f$ is a nonzero meromorphic function on a compact Riemann surface, then $\deg\left ( \operatorname{div}\left ( f \right ) \right )=0.$ Proven in Rick Miranda's book, page 130.
Question. Is the inverse statement true ?
Let $D$ be a divisor on the compact Riemann surface $X$ with $\deg\left ( D \right )=0,$ then there exists a meromorphic $f$ on $X$ such that $\operatorname{div}\left ( f \right )=D$ ?
Divisor of meromorphic functions on compact Riemann surfaces
complex-analysismeromorphic-functionsriemann-surfaces
Best Answer
This is true if and only if genus is zero (i.e. $X=\mathbb{P}^1$).
The set of all equivalent classes of degree zero divisors forms an abelian variety of dimension $g$, called "Jacobian".
When $g(X)=0$, you get what you want. The geometry of $\mathbb{P}^1$ is simple.
When $g(X)=1$, the Jacobian of $X$ is in fact isomorphic to $X$. Pick $P_0 \in X$, you have non-canonical isomorphism $X \longrightarrow \operatorname{Jac}(X)$, $Q \longmapsto Q-P_0$. (It is an exercise that $Q-P_0$ is not principal.) So you see that even in the $g=1$ case, you already have many many degree $0$ divisor classes that are NOT principal.