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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)Reply

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)Reply

Ah. You mean "almost all" in this sense :-) Sigfpe 23:48, 24 March 2006 (UTC)Reply

See Teichmuller space for a sense of size.

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)Reply

in this article there are two different classifications:

under Uniformization theorem#Complex classification it is : the Riemann sphere, the complex plane, the unit disk in the complex plane.

and under Uniformization theorem#Geometric classification of surfaces it is the sphere (curvature +1), the Euclidean plane (curvature 0), the hyperbolic plane (curvature −1).

I am wondering how are these classifications related and secondly why is the " unit disk in the complex plane" special , you can just map it to the (complete) complex plane vv. WillemienH (talk) 08:47, 29 April 2015 (UTC)Reply

The French page to this article seems to be this one rather than the one already linked. Could any please change it? Thanks. — Preceding unsigned comment added by Nadya Inoubli (talk • contribs) 22:56, 8 June 2015 (UTC)Reply

We did compute the conformal mapping in the cases of this article in our recent preprint, which is available from here: https://arxiv.org/abs/2509.01978

I can provide the image files if there is interest to add them to this article. Arasila (talk) 08:51, 23 December 2025 (UTC)Reply