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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?

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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 '10 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 '10 at 0:59
right, genus not zero. – Jesse Gell-Redman Apr 2 '10 at 1:06
Doesn't Riemann's existence theorem only hold in the compact case? – Qiaochu Yuan Apr 2 '10 at 2:00
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 '10 at 7:09

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.

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Aren't you missing the case $r=1$? – Dylan Thurston Apr 2 '10 at 3:50
Yes... So for that case let's take the group to be non-abelian. Then it has a non-trivial commutator [x,y]. Let a_1 go to x, b_1 go to y, c_1 go to [x,y]^(-1), and the rest to 1. This is of course only possible for g>=1; but for g=0, as David mentioned, this is impossible. – H. Hasson Apr 2 '10 at 4:06
This seems interesting, but I'm also confused by this answer. What exactly is the cover? I'm even confused by what is covered: is the mentioned $\Sigma$ the Riemann surface with points removed, or just the surface itself? I guess the reason I'm confused is that I'm used to covers corresponding to subgroups of $\pi_1$, but here you seem to say that the map from $\pi_1$ to somewhere correspond to covers. Also, at what point do you fill back in the branch points you removed? How do we know the resulting cover is finite? Thank you (if you have time to clarify), sorry if I'm just stupidly confused. – Ilya Grigoriev Apr 6 '10 at 2:04
No problem, that's how we learn. The point is not that maps from pi_1 correspond to covers, but that finite quotients (by not nec. normal subgroups) of pi_1 correspond to finite covers. I'm mapping surjectively to a finite group, G; so I know this corresponds to a finite normal cover with deck transformations G (the quotient I'm thinking of is pi_1/kernel of this map). The pi_1 is of Sigma without the branch points. Once we get this cover, and algebraize it, there's only one way to fill in the points (this is true for curves; see 1.6 in Hartshorne). – H. Hasson Apr 7 '10 at 16:35

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.

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You might also want to read… – user2146 Apr 2 '10 at 9:59

As mentioned by David, the answer is "no": If $\Sigma$ is the Riemann sphere and $\mathfrak{p}$ consists of a single point, then there is no such covering. (Indeed, in this case $\Sigma\setminus\mathfrak{p}$ is simply-connected, so there is only one ramified cover, and this cover has degree one ...

If you have more than one point, or $\Sigma$ is not the sphere, then the answer is "yes". In the case of the sphere with two points, you can use $z\mapsto z^d$ (up to conformal change of coordinate), and this is the only choice.

Otherwise, your cover $S$ can be chosen to be the unit disc. Indeed, consider the surface as an "orbifold" with given some ramification indices at your given points. It is known that this orbifold has a universal covering, which means that there is a holomorphic function, on the sphere, plane or disc, which is ramified in each preimage of the ramification points, with multiplicity a multiple of the given ramification index. In most cases, in particular if the ramification indices are taken sufficiently large, the universal cover will be the unit disc, as claimed.

This argument also works for infinite but discrete subsets of $\Sigma$.

In fact, it is not hard to see that the result will be true for any countable collection of branched values. (Here of course the map will not be an orbifold covering. However, using the orbifold covering argument we first see that we prove the result for the disc and a discrete set of points, and then apply the orbifold argument again ...)

(Of course, even if the surface $\Sigma$ is compact, this argument gives a non-compact cover - which may or may not be what you had in mind.)

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