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Today I showed, using some ad hoc algebraic topology, that if $\Sigma$ is a Riemann surface and $\mathfrak{p} \subset \Sigma$ is a finite set of points, then there is another Riemann surface $S$ and a branched covering $\phi : S \to \Sigma$ which has non-trivial branch points EXACTLY on the inverse image of $\mathfrak{p}$.

I'm suspicious that this was known or used elsewhere, and that it possibly follows trivially from some more sophisticated algebraic geometric mechanism. Does this ring a bell to anybody?

flag	asked Apr 2 2010 at 0:41 Jesse Gell-Redman 71●2

Except, of course, if Sigma has genus 0 and p is a single point. This is pretty easy from an algebraic topology perspective, so I don't see the point in looking for a deep proof. – David Speyer Apr 2 2010 at 0:57

The deeper fact is that the Riemmann surface so defined will be algebraic; this is Riemann's existence theorem. – David Speyer Apr 2 2010 at 0:59

right, genus not zero. – Jesse Gell-Redman Apr 2 2010 at 1:06

Doesn't Riemann's existence theorem only hold in the compact case? – Qiaochu Yuan Apr 2 2010 at 2:00

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Jesse, choose eff. divisor $\mathfrak{m}$ with supp. $\mathfrak{p}$, degree $d$. Can make proj. alg. curve $\Sigma_ {\mathfrak{m}}$ with normalization $\Sigma \rightarrow \Sigma_ {\mathfrak{m}}$ crushing $\mathfrak{m}$ to pt (if $\mathfrak{m} \ne 0$). Pic variety $J_ {\mathfrak{m}}$ of $\Sigma_ {\mathfrak{m}}$ yields a ton of such abelian covers if $J_ {\mathfrak{m}} \ne 0$ (dim = $g+d-1$!). This "geometric class field theory" is analytically concrete, and works algebraically over any perfect field; very useful! See Serre's book "Algebraic groups and class fields", esp. Ch. 1. – BCnrd Apr 2 2010 at 7:09

H. Hasson · Answer 1 · 2010-04-02 00:56:18Z UTC

Yes. The fundamental group of this Riemann surface minus those branch points is $< a_1, b_1, ..., a_g, b_g, c_1, ..., c_r| [a_1,b_1]...[a_g,b_g]c_1...c_r=1>$ (where $g$ is the genus, and $r$ is the number of the soon-to-be branch points). We have to guarantee that those will be branch points. Take any group generated by $r-1$ non trivial elements (such that their product isn't $1$). Map $\pi_1$ to that group such that each $a_i$ and $b_i$ go to 1, and each $c_i$ ($i$ going from $1$ to $r-1$) will go to said generators of the chosen group. Let $c_r$ go the the inverse of what $c_1...c_{r-1}$ goes to. Then this will correspond to some topological cover of $\Sigma$. Riemann's existence theorem says that we can make any (finite) topological cover into an algebraic cover. So this topological cover corresponds to a Riemann surface dominating it. It's an easy exercise to show that the ramification index at each of the preimages of your branch points (let's say branch point number $i$) is the order of the image of $c_i$ in the chosen group.

philip314 · Answer 2 · 2010-04-02 09:20:15Z UTC

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Yes. (Is this already a sufficient answer to the stated question?)

ps.: As a lot sensible keywords like Riemanns existence thm have been written I would only like to give you a great (the best?!) reference for this: Tamas Szamuely's Fundamental Groups and Galois Groups, in your case chapter 3.

answered Apr 2 2010 at 9:20

philip314
757●5●9

You might also want to read mathoverflow.net/questions/2704/… – philip314 Apr 2 2010 at 9:59

Branched coverings of Riemann surfaces with specified branch points.

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