Existence of meromorphic functions

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Discussion Overview

The discussion revolves around the existence of non-constant meromorphic functions on Riemann surfaces, exploring both compact and non-compact cases. Participants examine theoretical implications, connections to the Riemann-Roch theorem, and the complexity of proving such existence.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions whether every Riemann surface has a non-constant meromorphic function and discusses implications for compact Riemann surfaces.
  • Another participant asserts that non-compact Riemann surfaces admit non-constant holomorphic functions, referencing a result attributed to Stein.
  • A participant mentions that the existence of non-constant meromorphic functions on compact Riemann surfaces is part of Riemann-Roch theory and provides a reasoning based on Riemann's inequality.
  • Concerns are raised about the difficulty of proving the existence of non-constant meromorphic functions, suggesting that it involves complex analysis and referencing specific literature for further reading.
  • One participant expresses gratitude for the insights shared by others, indicating a sense of discovery in the discussion.

Areas of Agreement / Disagreement

Participants present differing views on the existence of non-constant meromorphic functions, particularly distinguishing between compact and non-compact Riemann surfaces. The discussion remains unresolved regarding the general proof of existence for all Riemann surfaces.

Contextual Notes

The discussion highlights the dependence on the Riemann-Roch theorem for compact surfaces and the complexity involved in proving existence, with references to specific texts that may contain relevant information.

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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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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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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