Talk:Uniformization theorem
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Add a reference to the Gauss-Bonnet theorem, which determines the sign of the curvature when the surface is of finite type.
--Mosher 14:34, 21 September 2005 (UTC)
Why does it say "almost all" surfaces are hyperbolic? This only makes sense if you have a measure on the space of "all" surfaces. We haven't talked about such a measure. So unless someone objects some time soon I'm going to delete that.
--sigfpe 8 Feb 2006
There are infinitely many (homeomorphism classes of) surfaces. Only finitely many of these do not admit a hyperbolic structure. So deleting "almost all" would be a mistake. Replacing "almost all" by "all but finitely many" might be more accurate, I suppose...
Sam nead 16:32, 1 March 2006 (UTC)
Ah. You mean "almost all" in this sense :-) Sigfpe 23:48, 24 March 2006 (UTC)
I think there's a lot that needs to be clarified here. "The uniformization theorem" means a few different things, depending on context. It's not at all clear that the statements, "Every surface has a geometric structure," and "Every simply connected Riemann surface is either S2, the complex plane, or the upper half plane," are equivalent, and the article kind of talks about all these facts at once. For one, the fact that the conformal automorphism group of H is exactly the same as the hyperbolic isometry group is nontrivial. The fact that the universal cover of a constant curvature manifold is the same, with isometric covering translations, is worth mentioning. I think the relationship between complex structures, conformal classes, and constant curvature metrics should really be explained well for this article.67.169.47.149 (talk) 08:05, 31 January 2009 (UTC)
