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)