# Inverse “Riemann mapping” [closed]

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?

-

## closed as off-topic by Benoît Kloeckner, Misha, Ryan Budney, David White, Carlo BeenakkerSep 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

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.
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