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Let $\Sigma$ be a compact Riemann surface of genus $g$, and $f: \Sigma \to \mathbb{C}$ a meromorphic function. Take $U \subset \Sigma$ an open disk in $\Sigma$ biholomorphic to a disk in $\mathbb{C}$, such that $f|_{U}$ is a holomorphic function. By the principal of analytic continuation, $f|_{U}$ (or its germ) should determine $f$ and $\Sigma$. Is there any formula or construction for this determination?

More generally, suppose $f$ is a holomorphic global section of some holomorphic vector bundle $E$ over $\Sigma$. Construct $f_{1}|_{{U}}$ similarly in a local trivialization (projected to the first element in the chosen basis). Can we reconstruct the relevant sub-bundle of $E$?

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  • $\begingroup$ If $\Sigma$ is compact then $f$ is constant $\endgroup$
    – Henri
    Commented Aug 31 at 17:08
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    $\begingroup$ I assume you meant for $f$ to be meromorphic. What you want is probably here: en.wikipedia.org/wiki/…. $\endgroup$
    – Vik78
    Commented Aug 31 at 17:19
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    $\begingroup$ Principal is not the same as principle. And you cannot reconstruct a Riemann surface from a germ of a holomorphic map. $\endgroup$
    – Moishe Kohan
    Commented Aug 31 at 18:13
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    $\begingroup$ The germ of a meromorphic function cannot distinguish between whether its domain is a Riemann surface or one of its covering spaces. $\endgroup$
    – Daniel Asimov
    Commented Aug 31 at 19:08
  • $\begingroup$ @DanielAsimov While all meromorphic functions on $\Sigma$ naturally extends to the covering spaces of $\Sigma$, such extension should admit symmetry, and should descend back to $\Sigma$. In principle shouldn't we able to mod out all symmetries to get the ultimately covered surface? $\endgroup$
    – Student
    Commented Sep 1 at 21:52

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Elaborating on some comments:

Let $S$ be the Riemann sphere. Consider triples $(a,g,b)$ where $a\in S$, $b\in S$, and $g$ is the germ at $a$ of an analytic map from a neighborhood of $a$ to $S$ that takes $a$ to $b$. Make the set of all such triples into a one-dimensional complex manifold as follows.

(It is not a manifold in the strictest sense, because it has uncountably many connected components, but we will be interested in just one of its components.)

A chart is given by any pair $(U,f)$ where $U\subset S$ is open and $f:U\to S$ is analytic. Use the obvious bijection between $U$ and set of all points $(a,g,f(a))$ where $a\in U$ and $g$ is the germ of $f$ at $a$.

So given the germ, at some point $a\in S$, of a meromorphic function, you get a connected one-dimensional complex manifold, whose points are all the germs that can be obtained by analytic continuation from that one.

Usually this is not compact, but if what you start with is a pair $(U,f)$ that corresponds to an open subset of $\Sigma$ and the restriction of a global map $\Sigma\to S$ then it will be, if not $\Sigma$, then some compact Riemann surface that has $\Sigma $ as a branched cover.

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