Existence of meromorphic functions

In summary, a proof that every Riemann surface has a non-constant meromorphic function is difficult, but it is possible if you have access to the Riemann-Roch theorem.
  • #1
lavinia
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What is a proof that every Riemann surface has a non-constant meromorphic function? Is it even true?

I was wondering this because if it is true then for compact Riemann surfaces without boundary
one can use the meromorphic function to produce a meromorphic 1 form whose degree - the sum of the orders of its zeros and poles - is twice its genus minus 2.

The cool thing is that a meromorphic 1 form can be combined with the metric scaling factor in isothermal coordinates to show that the total Gauss curvature is the sum of the indices of the vector field associated to the meromorphic function. this gives the Gauss Bonnet theorem without proving anything general about the sum of the indices of a vector field with isolated zeros
and without resorting to parallel translation arguments on geodesic triangles.
 
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  • #2
If the Riemann surface is noncompact, then in fact it admits a nonconstant holomorphic function. This result I believe is due to Stein.

The result that there exist nonconstant meromorphic functions on a compact Riemann surface is part of Riemann-Roch theory. In fact, in common treatments of the Riemann-Roch theorem, this result is used in the proof. However, if you have access to the Riemann-Roch theorem by other means, then you can deduce from it the result as follows. Let X be a compact Riemann surface of genus g. If D is a divisor on X, then Riemann-Roch (or rather, Riemann's inequality) tells us that
[tex] \dim H^0(X, \mathcal O_D) \geq 1 - g + \deg D. [/tex]
Here [itex]H^0(X, \mathcal O_D)[/itex] is the space of meromorphic functions f on X such that the coeffs of (f)+D are nonnegative. If we let x be any point of X, and if we let D be the divisor (g+1)x, then the RHS of the above inequality will be equal to 2. This not only shows that X admits nonconstant meromorphic functions, but in fact it shows that X admits a meromorphic function with a single pole.
 
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  • #3
it is very difficult to prove the existence of a non constant meromorphic function on a riemann surface, or at least it uses a lot of analysis (of course maybe you think this is an easy subject, hilbert space, closed operators,...).

see pages 441-442 of hARTSHORNES ALGEBRAIC GEOMETRY.

http://books.google.com/books?id=3r...&q=existence of meromorphic functions&f=false

robert gunning has a freely downloadable book on the topic on his webpage.

http://www.math.princeton.edu/~gunning/
 
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  • #4
thanks to you, Mathwonk and Morphism. I had no idea. It's funny how you think about one thing and you end up in places you never expected.
 

Related to Existence of meromorphic functions

What is a meromorphic function?

A meromorphic function is a complex-valued function that is defined and holomorphic (single-valued and differentiable) everywhere except for a set of isolated points called poles. At these poles, the function is not defined or is infinite.

How is a meromorphic function different from a holomorphic function?

A holomorphic function is defined and differentiable everywhere in the complex plane, while a meromorphic function is only defined and differentiable everywhere except at its poles. In other words, a meromorphic function is a generalization of a holomorphic function.

Can a meromorphic function have an infinite number of poles?

Yes, a meromorphic function can have an infinite number of poles, but they must be isolated points. This means that there must be a non-zero distance between each pole. For example, the function f(z) = cot(z) has an infinite number of poles at z = nπ, where n is any integer.

What is the relationship between meromorphic functions and rational functions?

Every rational function can be expressed as a meromorphic function, but not every meromorphic function is a rational function. This is because rational functions can have both poles and zeros, while meromorphic functions only have poles.

Why are meromorphic functions important in complex analysis?

Meromorphic functions are important in complex analysis because they provide a way to extend the concept of differentiability to functions with poles. They also play a crucial role in the study of complex integration and the residue theorem, which are important tools in complex analysis and its applications.

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