|
Let X be a compact Riemann surface, and f a meromorphic function on X. There's a theorem telling us that deg(div(f)) = 0. But is also true the inverse statement? I mean is it true that:
Thanks! |
|||||||
|
|
That is not true. Here a counterexample: consider a Riemann surface $X$ of genus $g \geq 1$. Fix $p, q \in X$ distinct points and consider the divisor $D = p - q$. This divisor has degree $0$, but it is not principal, because on the contrary there would be a holomorphic map $f: X \rightarrow \overline{\mathbb{C}}$ of degree equal to $1$ (for it has single simple zero/infinity value), and it is well known that a such map with this property is an isomorphism. That is a contradiction, since $g(\overline{\mathbb{C}}) = 0$. Look for Abel-Jacobi Theorem for necessary and sufficient conditions for a divisor be principal. |
|||
|
|
|
Rafael answer shows that if every divisor on $X$ is principal then $X$ is isomorphic to the Riemann sphere $\mathbb{C}_{\infty}$. The converse also holds true, i.e., every degree $0$ divisor $D$ on $\mathbb{C}_{\infty}$ is principal. To see this just note that if $D = \sum n_i \cdot z_i + n_{\infty}\cdot \infty$ then $f(z) = \sum (z-z_i)^{n_i} + z^{-n_{\infty}}$ is a rational function (hence meromorphic on $\mathbb{C}_{\infty}$) such that div$(f) = D$. |
|||
|
|
