Timeline for Do all surfaces (2d riemanian manifolds) admit constant curvature?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 21, 2015 at 22:20 | comment | added | Zero | If a deck-transformation has a fixed-point, then it is the identity (trivial) transformation and all points are fixed points . So you consider only transformations without fixed points. Sorry for misunderstandings. | |
| Feb 21, 2015 at 21:56 | comment | added | Igor Belegradek | You give a wrong reason. The group of deck-transformations never has a fixed point. | |
| Feb 21, 2015 at 21:37 | comment | added | Zero | That's what was implied by the no-fixed point rule, I thought (if a deck has a fixed point, and we have path-connectedness, then it's the identity transformation) . | |
| Feb 21, 2015 at 21:31 | comment | added | Igor Belegradek | I did not mean to say anything about lifting property. | |
| Feb 21, 2015 at 20:44 | comment | added | Zero | I see your point now (unique path lifting property) . I was confused. | |
| Feb 21, 2015 at 19:48 | comment | added | Igor Belegradek | I am not sure why you say the $S^2$ case is harder. The conformal automorphisms of $S^2$ then are $z\to \frac{az+b}{cz+d}$ or their compositions with $z\to \bar z$, and there is a fixed point unless it is a very special kind of isometry. I do not know why total curvature is relevant. | |
| Feb 21, 2015 at 18:27 | comment | added | Zero | Given that the total curvature of the sphere is 2, and the total curvature of a manifold (that might have the sphere as universal map) is an integer, this argument (if made rigorous) simplifies things to considering only the involutory transformations (ie f(f(x)) = x) . | |
| Feb 21, 2015 at 18:12 | comment | added | Zero | Thanks :) .I sketched something similar to 2.2 upon understanding the uniformization theorem (however I wasn't confident enough it's correct).I modeled H^2 as the Poncaire disc, then showed that the Poincare metric commutes with all possible transformations on the unit disk that preserve the conformal class (the complex automorphisms of the unit disk, and their reflections).I tried to figure out a similar trick with the metric induced by stereographicly by the Riemann sphere, but that turns out to be more difficult (the extended complex plane has the most possible mobius transformations). | |
| Feb 21, 2015 at 17:44 | vote | accept | Zero | ||
| Feb 21, 2015 at 16:54 | history | answered | Igor Belegradek | CC BY-SA 3.0 |