Is it correct that a disk endowed with a metric of Lorentzian signature (smooth up to the boundary) is always conformally equivalent to some simply connected domain on Minkowski plane (with the metric induced from Minkowski)?
It seems the answer is no. See the paper "An Analogue of the Riemann Mapping Theorem for Lorentz Metrics" by R. S. Kulkarni, Proc. R. Soc. Lond. A 9 September 1985 vol. 401 no. 1820 117130.

$\begingroup$ Thanks for the reference, I am trying to understand whether the answer to the question can be inferred from this paper. $\endgroup$ – pmnev Aug 22 '13 at 14:47
Here is the main idea. Suppose that the metric extends smoothly to the boundary. Such a metric has two null directions at each point. Follow one until you reach the boundary, and then switch to the other one. Watch each boundary point move, according the first null direction, and then watch it move according to the second one. These are two continuous maps on the boundary. You can pretty easily come up with examples for which the resulting dynamical system is quite messy. I think that Vladimir Arnol'd once said he studied these dynamical systems in his doctoral thesis. Global invariants of these dynamical systems prove that there is an infinite dimensional moduli space of Lorentzian metrics on the closed disk. The ones on the open disk can get even worse.

$\begingroup$ Right, I understand that, but I did not ask whether all Lorentzian metrics on a disk are conformal to Minkowski metric on a standard disk. In terms of this dynamical system, my question is whether one arising from given Lorentzian metric on a disk can be realized on some closed curve (boundary of a domain) in Minkowski plane. $\endgroup$ – pmnev Aug 22 '13 at 14:46

$\begingroup$ @pmnev: Nullcurves on the discs is Minkowski plane is never recurrent, for instance; also, one nullcurve cannot be asymptotic to another one. However, I would like to see an example of nontrivial dynamics on symplyconnected Lorentzian surfaces as there might be some topological obstructions. $\endgroup$ – Misha Aug 22 '13 at 16:30

$\begingroup$ Sorry, I didn't read the question carefully enough. $\endgroup$ – Ben McKay Aug 22 '13 at 17:24

$\begingroup$ @Misha On a general Lorentzian disk one also cannot have a closed nullcurve, if there were one, we would have a simply connected domain (the interior of the nullcurve) with a nonvanishing vector field tangent to the boundary, which would contradict the hairy ball theorem. Concerning nullcurves asymptotically approaching one another, I would expect that this is impossible on a disk too; that was a part of the question. On a cylinder, as opposed to the disk, one can have these effects, as e.g. in Misner space. $\endgroup$ – pmnev Aug 22 '13 at 17:25

$\begingroup$ Maybe the answer is in Tilla Weinstein's book An Introduction to Lorentz Surfaces. $\endgroup$ – Ben McKay Aug 22 '13 at 17:27