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This is well known that every point of a surface admits a neighbourhood with conformal coordinates (i.e. such that the metric is written $ds^2= e^u (dx^2 + dy^2)$). My first question is can we control the diffeomorphism which pass from original coordinates to conformal coordinates? In other word, let $u : B(0,1) \rightarrow R^3$ an immersion, then there exists a diffeomorphism $\phi$ of neighbourhood of $0$ such that $u\circ \phi$ is conformal. Can we control $\Vert \phi - Id\Vert_{C^1}$ w.r.t. $\Vert \vert u_x\vert^2 -\vert u_y \vert^2 +i \langle u_x, u_y\rangle \Vert_\infty$.

And what about the global case, i.e. considering maps from $S^2$ to $R^3$. Does there exists a proof of the existence of thiese coordinates which doesn't use the uniformization theorem but which consists in "gluing" all te local diffeomorphisms.

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Maybe this shows my age... but in my time such coordinates were called isothermal. Is conformal the accepted terminology now? (A more appropriate name might be conformally flat coordinates, by analogy with flat coordinates: those relative to which the connection one-form is identically zero.) –  José Figueroa-O'Farrill Oct 3 '10 at 15:32
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I don't think it's you, Jose (sorry for mangling the accent). In my time it was stilled called isothermal. But I disagree with the usage of (conformally) flat to refer to coordinates: I prefer flatness to be a condition on the intrinsic Riemannian structure, not an artefact of the coordinate system. –  Willie Wong Oct 3 '10 at 17:53
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Since the original coordinates can be arbitrarily bad, there is no reason to expect any control over the diffeomorphism. –  Deane Yang Oct 3 '10 at 18:28
    
Willie: I agree with you, but it is not unheard of (even in differential geometry circles) to refer to "flat coordinates" as any coordinate system on a flat riemannian manifold where flatness is manifest. It was in that spirit that I suggested tentatively "conformally flat coordinates". –  José Figueroa-O'Farrill Oct 3 '10 at 18:32
    
Concerning the case of the sphere, you can perform a stereographic projection (which is very explicit) to get your conformal coordinates. –  Benoît Kloeckner Oct 3 '10 at 19:47

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