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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 $\deg(\mathrm{div}(f)) = 0$.

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 $X$ such that $D = \mathrm{div}(f)$?

Thanks!

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  • $\begingroup$ 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. $\endgroup$ – Zhen Lin Nov 4 '11 at 20:41
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    $\begingroup$ (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. en.wikipedia.org/wiki/Abel%E2%80%93Jacobi_map $\endgroup$ – user8268 Nov 4 '11 at 21:42
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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) = \Pi (z-z_i)^{n_i} $ is a rational function (hence meromorphic on $\mathbb{C}_{\infty}$) such that div$(f) = D$.

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

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