Timeline for Hyperbolicity on Riemann Surfaces
Current License: CC BY-SA 3.0
17 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Aug 25, 2011 at 2:57 | vote | accept | ght | ||
Aug 24, 2011 at 19:48 | comment | added | Sam Nead | @All - Reading Val's question more carefully, I see that Val is not asking about hyperbolic surfaces (as he/she wrote) but rather asking about surfaces of hyperbolic type (as he/she also wrote!). So, Val and ght are asking the same question "Is every surface of hyperbolic type in fact Gromov hyperbolic?" The answer to this question is "No". This does not contradict anything I said about hyperbolic surfaces. I'll edit my answer to make this clear. | |
Aug 24, 2011 at 19:41 | history | edited | Sam Nead | CC BY-SA 3.0 |
hyperbolic versus hyperbolic type
|
Aug 24, 2011 at 16:27 | comment | added | j.c. | @ght I think there is some mismatch between your definitions of "surface" here. In his answer and comments, Sam Nead is working with hyperbolic surfaces in the sense of hyperbolic geometry, i.e. those with Riemannian metrics of constant negative Gaussian curvature. You are more concerned with Riemann surfaces of hyperbolic type, which only carry a conformal structure (a whole equivalence class of Riemannian metrics!). His last comment thus points out that a "Riemannian manifold of hyperbolic type" and a "hyperbolic surface" thus are not the same. | |
Aug 24, 2011 at 11:57 | comment | added | ght | @Sam - I think that you are contradicting yourself here. My question was: what is a sufficient condition for a surface of hyperbolic type to be Gromov's hyperbolic? Your answer: "...A simpler condition is that $\pi_{1}(S)$ is finitely generated..." Then Val and I asked you why this is true even when the fundamental group is trivial (simply connected) and you are saying that is not necessarily true in this case?? BTW, Val's question and my question are indeed the same. A simply connected surface is called of hyperbolic type if it is conformally equivalent to the unit disk. | |
Aug 24, 2011 at 9:17 | history | edited | Sam Nead | CC BY-SA 3.0 |
discussed the converse
|
Aug 24, 2011 at 9:09 | comment | added | Sam Nead | @ght - Your last question and Val's question are not the same. Val asked about a simply connected hyperbolic surface (ie the hyperbolic plane $H^2$). You are asking about a simply connected Riemann surface in the same conformal class as $H^2$. There are many Riemannian metrics on $H^2$, conformally equivalent to the usual one, that are nonetheless not Gromov hyperbolic. | |
Aug 24, 2011 at 1:04 | comment | added | ght | @Sam - Thanks. It is not entirely obvious to me at least, how do you prove that a simply connected 2-dimensional Riemannian manifold which is conformally equivalent to the unit disk is Gromov's hyperbolic (Val's question). Can you explain this in more detail? | |
Aug 22, 2011 at 20:44 | history | edited | Ian Agol | CC BY-SA 3.0 |
added 6 characters in body
|
Aug 22, 2011 at 13:06 | comment | added | Sam Nead | @ght - I've added more text to my answer. Let me know if you have further questions, or if this is enough. | |
Aug 22, 2011 at 13:05 | comment | added | Sam Nead | @Val - You are asking why the hyperbolic plane is Gromov hyperbolic. Here is a proof sketch: prove that any ideal triangle T in the hyperbolic plane is slim by showing that the legs of T are slim and giving an estimate for the diameter of the body. Now show that any triangle T' is contained in some ideal triangle. By the way, this proof uses the fundamental fact that all ideal triangles are related by isometries. | |
Aug 22, 2011 at 13:01 | history | edited | Sam Nead | CC BY-SA 3.0 |
Made correction as indicated by Agol.
|
Aug 22, 2011 at 11:45 | comment | added | Sam Nead | @Ian - right you are. I'll edit my answer. | |
Aug 21, 2011 at 18:37 | comment | added | Ian Agol | I think you also need the surface to have finite area to be q.i. to a tree, so that there are rank 1 cusps. Otherwise, the surface (with finitely generated fundamental group) will be geometrically finite, and have finitely many ends which are each q.i. to the hyperbolic plane or cusps. | |
Aug 20, 2011 at 0:56 | comment | added | Val | This might be a stupid question but how do you prove that a simply connected hyperbolic surface (i.e. conformally equivalent to the unit disk) is Gromov's hyperbolic? | |
Aug 20, 2011 at 0:17 | comment | added | ght | Thanks Sam! This is exactly what I meant. Can you please explain me a little bit more why this condition is sufficient? | |
Aug 19, 2011 at 21:22 | history | answered | Sam Nead | CC BY-SA 3.0 |