As far as I know, interesting results for open Riemann surfaces are quite rare. One of them is the theorem of Gunning and Narasimhan, which asserts that every connected open Riemann surface admits a holomorphic immersion into the complex plane. Another example is given by the theorem of Behnke and Stein, which says that every connected open Riemann surface is a Stein manifold. Is there any more interesting results for open Riemann surfaces?
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4 Answers
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The results on open Riemann surfaces are not "rare". They are just well forgotten. I only list a few books which deal with open Riemann surfaces: MR0114911 (Zbl 0196.33801), MR0228671 (Zbl 0152.27401), MR0159935 (Zbl 0112.30801), MR0264064 (Zbl 0199.40603), MR1973182 (Zbl 1162.14310). It is true that there are "too many" Riemann surfaces, and not too much can be said about "all of them". However, there is a highly non-trivial classification, and some subclasses are very important. For example, the class of "hyperelliptic" surfaces of infinite genus plays a very important role in the study of Schrodinger operators and their finite difference analog, see for example the works of Sodin and Yuditskii, MR1288838 (Zbl 1041.47502), and Kean, Moerbeke, MR0397076 (Zbl 0319.34024). Abelian covers of compact surfaces is another class of open Riemann surfaces that was studied a lot: MR0740581 (Zbl 0541.60076).
Added: here is a recent non-trivial result about arbitrary open Riemann surfaces: https://arxiv.org/abs/2103.16702.
answered Oct 29, 2012 at 14:08
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2$\begingroup$ +1 for "just well forgotten". $\endgroup$ Commented Oct 30, 2012 at 9:45
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$\begingroup$ Is it possible to decode those MR references without an MR account? $\endgroup$– BettinaCommented Feb 24 at 19:05
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1$\begingroup$ @Bettina: I added references to Zentralblatt zbmath.org which is free. $\endgroup$ Commented Feb 25 at 12:55
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$\begingroup$ @AlexandreEremenko Thank you very much, I appreciate it! $\endgroup$– BettinaCommented Feb 25 at 20:04
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For example, a theorem of Grauert and Röhrl asserts that every holomorphic vector bundle on a non-compact Riemann surface is trivial.
You can find this result (and its proof) in the book of O. Forster, Lectures on Riemann Surfaces.
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The Riemann uniformization theorem.
answered Oct 30, 2012 at 0:32
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I hope that the following result by Bishop and myself might be "interesting": Every open Riemann surface is a cover over the sphere, branched over only three points. Equivalently, every open Riemann surface is equilaterally triangulable.
See Belyi functions on non-compact surfaces; or: Building Riemann surfaces from equilateral triangles and https://arxiv.org/abs/2103.16702 .
answered Jun 21, 2021 at 12:40