There seems to be a lot of theorems allowing to prove restricted cases of this (eg. uniformization, classification theorem for compact surfaces) . Intuitively, it seems true, but I've never seen a proof of the general case.
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2$\begingroup$ How is what you are asking for different from uniformization? $\endgroup$– Igor RivinCommented Feb 21, 2015 at 14:26
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$\begingroup$ Uniformization applies to Riemann surfaces .I'm asking for 2d manifolds in general. As far as I know, you can assign a complex structure to a manifold if it's orientable. It would need an extension to non-orientable manifolds. $\endgroup$– ZeroCommented Feb 21, 2015 at 14:32
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1$\begingroup$ A non-orientable manifold admits a double covering by an orientable manifold. $\endgroup$– abxCommented Feb 21, 2015 at 14:49
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$\begingroup$ How does one prove that given D. a double cover of a manifold S, and a constant curvature metric for D, you can 'push' the metric trough the covering map to obtain a constant-curvature metric for S? $\endgroup$– ZeroCommented Feb 21, 2015 at 15:02
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This is a standard consequence of the uniformization, but I agree that locating a reference may be a challenge. You can find details in my survey http://arxiv.org/abs/1306.1256, see Theorem 2.2 where it is shown that any open (smooth connected) 2-manifold admits a complete metric of constant negative curvature. The same proof works for compact manifolds but in this case you need to allow for two other possibilities of constant curvature 0 and 1, and it is clear from the proof why this is the case.
answered Feb 21, 2015 at 16:54
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$\begingroup$ 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). $\endgroup$– ZeroCommented Feb 21, 2015 at 18:12
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$\begingroup$ 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) . $\endgroup$– ZeroCommented Feb 21, 2015 at 18:27
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1$\begingroup$ 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. $\endgroup$ Commented Feb 21, 2015 at 19:48
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$\begingroup$ I see your point now (unique path lifting property) . I was confused. $\endgroup$– ZeroCommented Feb 21, 2015 at 20:44
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$\begingroup$ I did not mean to say anything about lifting property. $\endgroup$ Commented Feb 21, 2015 at 21:31