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Existence of meromorphic functions

  1. Jan 13, 2012 #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.
     
  2. jcsd
  3. Jan 13, 2012 #2

    morphism

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    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.
     
    Last edited: Jan 13, 2012
  4. Jan 13, 2012 #3

    mathwonk

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    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/
     
    Last edited: Jan 13, 2012
  5. Jan 13, 2012 #4

    lavinia

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    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.
     
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