A Riemann surface is said to be:

-Potential-theoretically hyperbolic if it has a non-constant bounded subharmonic function.

-Poincaré hyperbolic if it is covered by the unid disk.

Are this definitions equivalents? And why??


2 Answers 2


Potential theoretic hyperbolic implies poincare hyperbolic. This is just part of proof of uniformisation theorem. The other direction is false since any compact Riemann surface of genus at least two is poincare hyperbolic.


I would add to Mohan's response, which is correct. Even if the surface is non-compact, it might be Poincare hyperbolic but not potential theoretically hyperbolic. For example, if you take any compact Riemann surface and remove a finite number of points, then it is Stein, so has subharmonic functions, but no non-constant bounded subharmonic functions.

There are many parabolic (i.e. open but not potential theoretically hyperbolic) Riemann surfaces. See the book of Nakai and Sario for a lot more on the classification problem.


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