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Jan 30, 2020 at 21:22 comment added Yu Feng The unit disk is a hyperbolic Riemann surface and the unit disk removes a finite number of points is a hyperbolic Riemann surface. We can construct a negative subharmonic function on it: $\log\mid z\mid$, where $z$ is the complex coordinate on the disk. The complex plane $\mathbb{C}$ and $\mathbb{C}$ removes a finite number of points are parabolic Riemann surfaces. The surface does not need to have finite genus and finite Euler characteristic.
Jan 30, 2020 at 21:22 comment added Yu Feng Thank you for your consideration. There are some equivalent definitions for hyperbolic Riemann surface. Let $M$ be an open Riemann surface, then the following are equivalent: (1) There exist a Green's function on $M$ (with singularity at some point $P\in M$). (2) There exist a non-constant negative subharmonic function on $M$ (= there exists a non-constant bounded subharmonic function on $M$). (3) Brownian motion on $M$ is transient. (4) The maximum principle does not hold (for every compact set $K$).
Jan 30, 2020 at 21:17 history edited Yu Feng CC BY-SA 4.0
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Jan 30, 2020 at 15:45 comment added Dmitri Panov Dear Yu, is there some other characterisation of hyperbolic surfaces that doesn't use sub-harmonic functions? It seems to me that according to your definition a surface is hyperbolic if an only if it is not a punctured surface? If I am wrong, could you please give me the simplest example that contradict my statement? Also, do you assume that your surface has finite genus and has finite Euler characteristic?
Jan 25, 2020 at 18:03 answer added Dmitri Panov timeline score: 2
Jul 5, 2019 at 8:27 history edited Yu Feng CC BY-SA 4.0
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Jul 4, 2019 at 15:00 history asked Yu Feng CC BY-SA 4.0