Timeline for Euler characteristic of Cauchy surface in Lorentz manifold
Current License: CC BY-SA 3.0
22 events
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| Feb 16, 2014 at 6:31 | comment | added | David Hillman | @Misha: yes, my polyhedra are intersections of closed half-spaces (the sides are on null hyperplanes). Convex. Thanks for all your help; you've given me some things to investigate. | |
| Feb 16, 2014 at 5:57 | comment | added | Misha | of $R^{2,1}$, then interesting examples abound (Margulis space-times). On the other hand, I am sure that in this setting existence of tiles does not imply completeness. You really have to do some reading, say, read the survey that I linked. | |
| Feb 16, 2014 at 5:56 | comment | added | Misha | @DavidHillman: I see what you mean. It depends on your notion of polyhedron. Do you mean an intersection of closed half-spaces in $R^{2,1}$? Then, if I remember correctly, there are no interesting examples (meaning, examples whose fundamental group contains a free subgroup). Here I am assuming, in addition, that you have completeness since it is unclear to me why completeness follows from existence of such tiles (in a similar setting of tiling of hyperbolic plane, there are examples of incomplete metrics which admit such tiling). But if you allow nonconvex tiles, still closed as subsets... | |
| Feb 16, 2014 at 5:05 | comment | added | David Hillman | Oops: Misha's reply to my comment about open balls showed up for some reason after I was re-editing it, so things are out of order. Okay, right, no metric balls. Silly me. So I'll try saying it this way. Each polyhedron looks like a polyhedron in Minkowski space. If you take any point on the boundary of a polyhedron and consider all the polyhedrons hitting that point, the complex of these polyhedra looks like it's in Minkowski space. So geodesics extend, because they extend in Minkowski space. (By "looks like it's in Minkowski space" I mean there is the appropriate isometry.) | |
| Feb 16, 2014 at 5:02 | comment | added | Misha | Do you really want to use r-balls with respect to the Lorentzian metric? Maybe in presence of a compact Cauchy hypersurface this is a good concept, I am not sure. With this assumption, I do not have any examples apart from the Minkowski space-time itself. | |
| Feb 16, 2014 at 4:48 | comment | added | David Hillman | @Misha: Okay, more brain failure; ignore what I said about compact neighborhoods. The key in my situation is that there exists a number r>0 such that for every point there is an open ball centered at that point of radius r that looks like Minkowski space. Now we get completeness, right? (I will look at your example, but it will take some time; have to understand things like torsion free discrete cocompact subgroup of SO(2,1).) | |
| Feb 16, 2014 at 3:52 | comment | added | Misha | David: You have to be careful with completeness in Lorentzian setting. The right definition is that all geodesics extend indefinitely (more formally, the exponential map is defined on the entire tangent bundle). The Cauchy criterion, in contrast, is no longer useful since you do not have a metric space, so thinking in terms of R-balls is not a good idea. | |
| Feb 15, 2014 at 20:59 | comment | added | Misha | @DavidHillman: This condition is insufficient for completeness. Look at the example I gave. If flat Lorentzian manifold is also compact then it is complete ( this is not obvious). | |
| Feb 15, 2014 at 20:56 | comment | added | David Hillman | @Misha: on the topic of completeness: the reason I say the manifolds would be geodesically complete is that the rules for assembling polyhedra are such that each point has a compact neighborhood that looks like Minkowski space. I previously discussed the setup a bit in <mathoverflow.net/questions/153970> and learned that I was working with $(G,X)$ structures. But I guess now that the word compact above makes them more restrictive. Is completeness the main extra property that arises? | |
| Feb 14, 2014 at 20:35 | history | edited | Misha | CC BY-SA 3.0 |
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| Feb 14, 2014 at 19:36 | comment | added | David Hillman | @Misha: you're right, these locally flat Lorentz manifolds produced by tilings would be complete. | |
| Feb 14, 2014 at 6:59 | history | edited | Misha | CC BY-SA 3.0 |
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| Feb 14, 2014 at 6:38 | history | edited | Misha | CC BY-SA 3.0 |
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| Feb 14, 2014 at 1:28 | comment | added | David Hillman | Yes, of course, sorry, brain failure. Compact spacetimes: not interesting to me. (And that result about Euler characteristic of compact spacetimes is probably irrelevant.) I was thinking about compact Cauchy surfaces, closed surfaces that you make from finitely many tiles (in my tiling example), the classic case being in 2d where your surfaces are circles and spacetime is a cylinder. And also interested in the case where the surfaces aren't compact, such as Minkowski space. And, compact or no, the flat case is of particular interest. | |
| Feb 13, 2014 at 20:47 | comment | added | Misha | @DavidHillman: Before we get into your new questions (I have something to say about them but will do it later), you should clarify the question which you asked in the post. Namely, What is supposed to be compact: Cauchy hypersurface or the entire space-time? I hope, you understand the difference. | |
| Feb 13, 2014 at 20:43 | comment | added | David Hillman | I think in general with tiles and local rules for assembling them you can get arbitrary flat Lorentz manifolds including compact ones (definitely seems true in 2d) so I'm looking for any results, compact or not, that might suggest something about my situation. I have a feeling that the finiteness of the tiling system might make compact results relevant, but who knows. | |
| Feb 13, 2014 at 20:40 | comment | added | David Hillman | Thanks for your replies...my DG is not up to speed so it will take me some time to digest. Here is where the question came from. I'm trying to find a certain kind of tiling system of Minkowski space, where there are finitely many polyhedral tile shapes (up to Poincare equivalence). There are local rules for assembling them into a Cauchy surface, and rules for evolving the surface. In 3d I haven't found an interesting example and am wondering if there is something I should know about Euler characteristic that would restrict my search. (to be continued...) | |
| Feb 13, 2014 at 12:14 | comment | added | Robert Bryant | I think your understanding of what physicists would want (and what is physically permissible) is correct, but the OP wrote 'compact $n$-dimensional Lorentzian spacetime', implying that the spacetime itself is to be compact. As you say, perhaps the OP did not intend this and will so inform us. | |
| Feb 13, 2014 at 12:05 | comment | added | Misha | @RobertBryant: Maybe I misinterpreted the question, but my understanding of Cauchy hypersurfaces is that physicists want to have a manifold of the form $M\times R$, foliated by Cauchy hypersurfaces $M\times t$ ("time machines" are not allowed). Hopefully, OP will clarify if he wanted a compact space-time of a compact Cauchy hypersurface, that was unclear from the question. | |
| Feb 13, 2014 at 9:47 | comment | added | Robert Bryant | However, your locally flat ambient Lorentzian manifold $C/\Gamma$ is not compact, which the OP wanted (though, of course, such an example would not be of physical interest). | |
| Feb 13, 2014 at 5:29 | history | edited | Misha | CC BY-SA 3.0 |
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| Feb 13, 2014 at 4:49 | history | answered | Misha | CC BY-SA 3.0 |