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In Miranda's book on algebraic curves and Riemann surfaces, Miranda writes:

It is a basic and highly nontrivial result that a compact Riemann surface has nonconstant meromorphic functions on it [...] The theory involved in producing meromorphic functions for an unknown compact Riemann surface is rather technical analysis and functional analysis. After one has access to meromorphic functions, however, the theory is completely algebraic, or at least can be made to be so.

I've seen this claim a number of other places as well. It seems unnatural to use real analysis to prove a theorem about Riemann surfaces, which are geometric/algebraic objects. Is there genuinely no purely geometric/algebraic way to realize an abstract Riemann surface as a branched cover of the Riemann sphere?

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In what sense is a Riemann surface an algebraic object before you know that they're secretly algebraic curves (which requires knowing that they have enough meromorphic functions)? –  Qiaochu Yuan Mar 26 '13 at 21:01
    
They're not, but you can still consider the moduli space of algebraic curves of given genus and the moduli space of Riemann surfaces of a given genus and to prove that they're the same. The spaces have the same number of dimensions and the former lies in the latter. Maybe you could use some sort of continuity argument to prove that the former can't be a proper subset of the latter. –  Jonah Sinick Mar 26 '13 at 21:18
    
Jonah -- I can't see how you can possibly make this argument rigorous without using just the sorts of facts that you're trying to avoid. The only map accessible to you will be one from the analytification of the algebraic moduli space to the analytic moduli space, and of course just because they have the same dimension tells you nothing; you can't use algebraic arguments becuase you've been forced to leave the algebraic world. –  user30035 Mar 26 '13 at 22:43
    
wwcanard — Ok, so I agree that algebraic arguments won't be relevant. Still, one can hope for a more geometric argument. –  Jonah Sinick Mar 26 '13 at 23:36
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(1) If you don't want to use analysis, why do you care about a theorem that says that you get to use analysis? (2) How do you compute the analytic deformation space without analytic input? (uniformization or coherent sheaves or...) (3) Yes, if you have deformation theory I think you can prove that the algebraic moduli space is open in the analytic moduli space, but what then? Maybe use a compactification? –  Ben Wieland Mar 27 '13 at 22:45

2 Answers 2

This deep fact is essentially the same as the uniformization theorem. The problem is how to construct at least one holomorphic or meromorphic form with prescribed singularity. All known proofs use some Analysis, and none of them is simple. Once you have Uniformization, it is easy to construct holomorphic forms.

A good modern proof (in full generality) is contained in the book of J. Hubbard, Teichmuller Theory. The complexity of the proof depends on how exactly you define a Riemann surface. A Riemann surface is a 1-dimensional complex manifold. A 1-dimensional complex manifold is separable (has a countable base). This fact is a part of the modern statement of the Uniformization theorem. However, if you include this separability to the DEFINITION of a Riemann surface, the proof substantially simplifies. Especially simple proof (assuming separabiity) can be found in Goluzin's book Geometric theory of functions of a complex variable.

All Riemann surfaces arising in real life are easily seen to be separable (give me an example if I am wrong), so there is no real harm if we include this in the definition:-)

But with or without this separability condition, all proofs of existence of holomorphic forms or of the uniformization use Analysis. Complex or real, I don't see a sharp distinction between these two.

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Dear Alexandre, you write All Riemann surfaces arising in real life are easily seen to be separable. This is actually always true, also in unreal life, and is a theorem of Radó –  Georges Elencwajg Mar 27 '13 at 9:41
    
Just a nitpick: All connected Riemann surfaces are separable. –  Misha Mar 27 '13 at 16:48
    
George, you misunderstood what I said. That all Riemann surfaces are separable is a theorem. This theorem is non-trivial. I said that I do not know any APPLICATIONS of this theorem, because most Riemann surfaces which arise in real life are EASILY seen to be separable (without any need to appeal to this theorem), just from the way they arise. In other words, I said that this theorem is not useful. –  Alexandre Eremenko Mar 28 '13 at 1:06
    
I think that it's much easier than uniformization. The proof I'm thinking of is the one in Gunning's book and Narasimhan's book (and many others). The heart of it is showing that for any holomorphic line bundle $L$, the vector spaces $H^0(S,L)$ and $H^1(S,L)$ are finite-dimensional. This requires some functional analysis, but nothing all that deep. –  Andy Putman Mar 30 '13 at 4:07
    
Which book of Gunning? –  Alexandre Eremenko Mar 30 '13 at 12:17

(Deleted incorrect suggestion).

I think one can use uniformization and the construction of automorphic functions on the universal cover to produce meromorphic functions. A google search for these terms found these notes.

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Thanks Ian. If I understand correctly, the fact that 1-forms have holomorphic representatives requires the Hodge Theorem, which in turn is usually proved via the theory of elliptic operators, though I found this question mathoverflow.net/questions/28265/… (which will take some time to digest). The only proofs of the uniformization theorem that I've seen use hard analysis. Maybe the use of hard analysis is inevitable. –  Jonah Sinick Mar 26 '13 at 21:44
    
I guess one could argue that even the proof in the genus 1 case requires hard analysis (in that one has to verify the convergence of infinite series), but somehow it's still quite conceptual. –  Jonah Sinick Mar 26 '13 at 21:47
    
@Jonah: I'm not sure what you have in mind. If you consider the definition of a Riemann surface, there are charts which map to the complex plane, such that the overlaps of charts are realized by germs of holomorphic maps. If you start with this definition, I suppose one could try to construct a map of the surface to the sphere in a purely local way, by developing maps of the charts. The issue is that this is likely to only produce a map on the universal cover. To get things to match up with respect to covering translations, one would have to deform these holomorphic maps, which seems analytic. –  Ian Agol Mar 26 '13 at 22:26
    
@Ian: I guess what I want to do is (i) show that the collection of genus g Riemann surfaces with a meromorphic map f to the sphere is dense in the moduli space of genus g Riemann surfaces and then (ii) show that an infinitesimal deformation X' of a Riemann surface X with a meromorphic map to the sphere itself has a meromorphic map to the sphere by infinitesimally deforming f. Presumably this is what the analytic proofs implicitly do: I'll have to think about how they correspond to this mental model. –  Jonah Sinick Mar 27 '13 at 0:18
    
Dear Ian, A higher genus curve may not admit a non-constant holomorphic map to an elliptic curve, if I'm not getting myself confused, so I'm not sure the strategy of your first paragraph will always work. I'm pretty sure the strategy of your second paragraph does, though (and maybe it goes back to Poincare, who might have invented Poincare series in part for this purpose?). Best wishes, –  Emerton Mar 27 '13 at 2:04

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