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The Riemann mapping theorem states, that any simply connected domain $U \subset \mathbb C$ can be conformally mapped to the open unit disk $D$. I.e. there is a Diffeomorphism $\Psi: D \to U$ such that $\Psi^* g_{U}=e^{u}g_{D}$ where $u\in C^\infty(D)$ and $g$ denotes the Euklidean metric on $U$ respectively $D$.

My question is now whether the "inverse" is true, i.e.: given a function $u\in C^\infty(D)$, is there a domaine $U$ and a diffeomorphism $\Psi: D \to U$ such that $\Psi^* g_{U}=e^{u}g_{D}$ holds?

If it doesn't hold in general: Under what conditions on $u$ does it hold?

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closed as off-topic by Benoît Kloeckner, Misha, Ryan Budney, David White, Carlo Beenakker Sep 26 '13 at 12:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Benoît Kloeckner, Misha, Ryan Budney, David White, Carlo Beenakker
If this question can be reworded to fit the rules in the help center, please edit the question.

$g_U$ is a flat metric, thus so is $\Psi^* g_U$. On the other hand, $e^u g_D$ usually is not flat. Without context, this looks closer to homework than to a research-level question. – Benoît Kloeckner Sep 25 '13 at 10:39
Thanks for pointing this out! This is of course an obvious obstruction which I should have noticed myself – twch Sep 25 '13 at 14:48
up vote 1 down vote accepted

It does not hold in general, as Benoit Kloeckner explained. More precisely, there are two obstacles, one local and one global. The local obstacle is the Gaussian curvature, $-e^{-2u}\Delta u$. It must be zero for the pull-back of the Euclidean metric (which means that $u$ must be harmonic). But there is also a global obstacle. For a given metric of zero curvature on the disc, there exists a conformal local homeomorphism to the plane such that your metric is the pull-back of the Euclidean metric. However this local homeomorphism is not necessary a global homeomorphism. There are Euclidean surfaces conformally equivalent to the unit disc but not isometric to any region in the plane.

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Thanks, I think I the global obstructions you mention can be understood like this: For the isometry we need a biholomorphic function $\Phi:D\to \Phi(D)$ with $|\Phi'|=e^{1/2 u}$. If $\Delta u=0$ the Poincaré lemma gives us a function $v$ s.t. $u+iv$ is holomorphic (study differential form $\omega=-u_ydx +u_x dy$). Taking a holomorphic function $\Phi:D\to \mathbb C$ with $\Phi'(z)=e^{\frac{1}{2}(u(z)+iv(z))}$ is the unique holomorphic solution (up to an additive constant and a constant phase), that fullfills the condition on the derivative. It is of course locally bijective, but not globally. – twch Sep 25 '13 at 14:45

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