Timeline for Can the Causal Structure recover the manifold topology for non-chronological spacetimes?
Current License: CC BY-SA 4.0
18 events
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| May 17 at 13:00 | comment | added | Bastam Tajik | Later the question would be how much of the topology is fixed in such case by the causal structure, to me topological dimension can be also a degree of freedom while other Algebraic topological structures are fixed(e.g. the fundamental group etc.) | |
| May 17 at 12:38 | comment | added | Bastam Tajik | In case it is so, which I really hope so, finding an $\ll$-isomorphism that is not conformally locally everywhere isometric to the spinning Cosmic String, is enough to prove that the topology cannot be recovered for SCS. @WillieWong | |
| May 16 at 11:26 | comment | added | Bastam Tajik | I guess by order preserving you mean, $\ll$-isomorphism. @WillieWong | |
| May 16 at 1:42 | comment | added | Willie Wong | The reverse implication doesn't hold, which is my problem with how you quoted the theorem. A manifold homeomorphism is not necessarily a $\mathcal{P}$-homeomorphism (precisely because the $\mathcal{P}$ topology is in general strictly finer). While I don't have a counterexample available, it is not clear to me that order preserving + manifold homeomorphic implies $\mathcal{P}$-homeomorphic. | |
| May 15 at 14:01 | history | edited | Bastam Tajik | CC BY-SA 4.0 |
Nothing changed but PS added and an article attached
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| May 15 at 11:33 | comment | added | Bastam Tajik | @WillieWong Nonetheless the question is robust in that it asks the topology restrictions imposed by (at least) the causal structure in the particular case of the Spinning Cosmic String. | |
| May 15 at 11:08 | comment | added | Bastam Tajik | But Unfortunetly, the contraposition of the HKM becomes useless, doesn't it? Since $\sim \rho$-homeomorphism doesn't imply $\sim \mathcal{M}$-homeomorphism. @WillieWong | |
| May 15 at 10:52 | comment | added | Bastam Tajik | And in general the $\rho$ topology is strictly finer than the manifold topology $\mathcal{M}$, which means that a $\rho$-homeomorphism is also an $\mathcal{M}$-homeomorphism. @WillieWong | |
| May 15 at 10:49 | comment | added | Bastam Tajik | @WillieWong Take a look at the corollary 5.7: projecteuclid.org/journals/… | |
| May 15 at 10:30 | comment | added | Bastam Tajik | Moreover I used an indirect reference: philarchive.org/archive/WTHOON page 7. @WillieWong | |
| May 15 at 1:22 | comment | added | Willie Wong | I have doubts about how you quoted Hawking--King--McCarthy. The closest statement I finding in that paper is Theorem 6, but this assumes the mapping is a homeomorphism with respect to the path topology that they define, which is not the same as the manifold topology. Can you clarify with a more precise reference? | |
| May 14 at 22:00 | history | edited | Bastam Tajik | CC BY-SA 4.0 |
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| May 14 at 11:25 | history | edited | Bastam Tajik | CC BY-SA 4.0 |
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| May 13 at 19:45 | answer | added | Tim Campion | timeline score: 5 | |
| May 13 at 19:28 | comment | added | Bastam Tajik | Upvote without comment=flattering | |
| May 13 at 19:26 | history | edited | Bastam Tajik | CC BY-SA 4.0 |
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| May 13 at 15:59 | comment | added | Bastam Tajik | Downvote without comment=unprofessional and irresponsible | |
| May 13 at 15:24 | history | asked | Bastam Tajik | CC BY-SA 4.0 |