Timeline for Lorentzian metrics on a disk up to conformal equivalence
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 23, 2013 at 8:24 | comment | added | pmnev | @ Ben McKay: Thanks a lot, I beleive Luo-Stong's theorem answers my question. | |
| Aug 22, 2013 at 17:33 | comment | added | Ben McKay | Tilla Weinstein's book discusses the Luo-Stong theorem, which apparently gives both necessary conditions and sufficient conditions, and she gives an explicit counterexample. | |
| Aug 22, 2013 at 17:27 | comment | added | Ben McKay | Maybe the answer is in Tilla Weinstein's book An Introduction to Lorentz Surfaces. | |
| Aug 22, 2013 at 17:25 | comment | added | pmnev | @Misha On a general Lorentzian disk one also cannot have a closed null-curve, if there were one, we would have a simply connected domain (the interior of the null-curve) with a non-vanishing vector field tangent to the boundary, which would contradict the hairy ball theorem. Concerning null-curves 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. | |
| Aug 22, 2013 at 17:24 | comment | added | Ben McKay | Sorry, I didn't read the question carefully enough. | |
| Aug 22, 2013 at 16:30 | comment | added | Misha | @pmnev: Null-curves on the discs is Minkowski plane is never recurrent, for instance; also, one null-curve cannot be asymptotic to another one. However, I would like to see an example of nontrivial dynamics on symply-connected Lorentzian surfaces as there might be some topological obstructions. | |
| Aug 22, 2013 at 14:46 | comment | added | pmnev | 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. | |
| Aug 22, 2013 at 14:05 | history | answered | Ben McKay | CC BY-SA 3.0 |