Timeline for multivalued holomorphic function on Riemann surfaces
Current License: CC BY-SA 4.0
7 events
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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 |
added 259 characters in body
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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 |
edited body; edited tags; edited title
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Jul 4, 2019 at 15:00 | history | asked | Yu Feng | CC BY-SA 4.0 |