So, due to the existence of isothermal coordinates, all 2-manifolds are conformally flat. The consequences of this are a bit confusing to me- this means one can conformally map, for instance, the sphere to some surface with identically zero curvature everywhere. What does this surface look like? Is it an infinitely large sphere? (Arbitrarily large radius would imply arbitrarily small curvature, yes?)

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    $\begingroup$ What isothermal coordinates are you referring to? $S^2$ can't have a flat metric by the Gauss-Bonnet theorem. Usually isothermal coordinates are a local thing. What do you mean by a global isothermal coordinates -- and how do you deduce they exist? There are certain flat metrics on 2-manifolds that have finitely many singularities, if that's what you're referring to. $\endgroup$ – Ryan Budney Aug 30 '11 at 0:11
  • $\begingroup$ Isothermal coordinates are any of the form $g_{ab} = e^\phi (\delta_{ab})$, for some smooth function $\phi$- that is, which are conformally flat. Their wikipedia page has a short outline of the proof of their existence for surfaces. $\endgroup$ – specterhunter Aug 30 '11 at 0:38
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    $\begingroup$ Well, as Ryan pointed out, you're talking about a local phenomenon. It is a very unfortunate standard abuse to speak of conformally flat manifolds when actually locally conformally flat manifolds are meant. One way to get those certain flat metrics Ryan mentions is by using quadratic differentials. Strebel's book mentioned on that Wikipedia page is the standard reference. $\endgroup$ – t.b. Aug 30 '11 at 0:54
  • $\begingroup$ I'm not clear on how you're going from a local condition (isothermal coordinates) to talking about a global map. Could you elaborate? $\endgroup$ – Ryan Budney Aug 30 '11 at 0:55
  • $\begingroup$ Isothermal coordinates just say that your manifold is locally isometric to a flat space. Actually your manifold is flat iff $e^{\phi} = 1$. $\endgroup$ – user40276 Apr 21 '14 at 23:19

one can conformally map, for instance, the sphere to some surface with identically zero curvature everywhere

Not the entire sphere (which the Gauss-Bonnet theorem would not let you, as Ryan Budney said). But you can conformally map the sphere minus a point onto a plane: this is precisely what the stereographic projection does. The sphere minus the North Pole is mapped onto the plane, as seen in the image below (source).

stereographic projection

A closed surface of genus $\ge 2$ can be built from a polygon in the hyperbolic plane by identifying certain sides. Since the hyperbolic polygon is conformally equivalent to a flat one (via the identity map), we again get a conformal map between a flat surface and almost all of the negatively curved surface.


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