Return to Answer

added 3 characters in body

Source Link

edited Sep 1 at 0:41

55.9k
7
151
240

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.

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 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.

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.

Source Link

created Aug 31 at 20:29

55.9k
7
151
240

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 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.