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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:

if D is a divisor on X with deg(D) = 0, then exists a meromorphic function f on X s.t. D = div(f)


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I'm quite sure the answer is no... for example, there is no function on the complex torus with only one simple zero and one simple pole. – Zhen Lin Nov 4 '11 at 20:41
(degree-$0$ divisors)/(principal divisors) is the Jacobian variety of your Riemann surface. It is a complex torus of (complex) dimension $g$ (= the genus of the surface). See e.g. – user8268 Nov 4 '11 at 21:42
up vote 2 down vote accepted

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.

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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$(f) = D$.

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Shouldn't the sum in the definition of f(z) be a product? – jmc May 8 '12 at 15:44

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