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Let $(M, g)$ be a smooth Lorenzian time-oriented manifold.

Is it possible for the Lorenzian metric induced topology to be different from that of the manifold topology, without CTCs?

We know that the two topologies are not necessarily the same and such matching can happen iff the spacetime is strongly causal.

So if the answer is yes and there exists such spacetime, they surely are not strongly causal.

I want to know how far this mismatch of topologies can go while avoiding CTCs. (avoiding also the case of "up to a metric fluctuation that can create a CTC")

Or does any topological mismatch/deviation result in CTCs(directly or up to metric fluctuation)?

Or in general:

Is there any relation at all between topological deviation, and the existence of CTCs(also up to a metric fluctuation) one way or another?

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  • $\begingroup$ Thanks. Before taking a closer look, I should ask, if in the final and initial states, the manifold topology and the Pseudo-Riemannian metric induced topology, each match separately or not? @A.V.S. $\endgroup$
    – Bastam Tajik
    Aug 15 at 16:59
  • $\begingroup$ Which topology would you induce with the metric, if it's pseudo Riemannian you don't really have a metric topology. $\endgroup$
    – Slereah
    Aug 15 at 19:22
  • $\begingroup$ @Slereah that's not true. mathoverflow.net/questions/266903/… $\endgroup$
    – Bastam Tajik
    Aug 15 at 20:10
  • $\begingroup$ @Slereah but I edit the question so that there's no ambiguity. $\endgroup$
    – Bastam Tajik
    Aug 15 at 20:51
  • $\begingroup$ @Slereah Nonetheless It's never mentioned in the question that the topology is metric, but that a topology can be defined, using a Lorentzian metric! $\endgroup$
    – Bastam Tajik
    Aug 15 at 21:06

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What do you mean by "how far this mismatch of topologies can go while avoiding CTCs"? You have already noticed that the two topologies coincide if and only if the spacetime is strongly causal. It is therefore sufficient to consider spacetimes without CTCs (causal) but not strongly causal to obtain that the two topologies do not coincide. Moreover in the case where spacetime is not strongly causal the Alexandrov topology ceases to be Hausdorff.

Or does any topological mismatch/deviation result in CTCs(directly or up to metric fluctuation)?

No, when the two topolgies do not coincide, there are not necessarily CTCs.

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  • $\begingroup$ The point of question is if such spacetime remains free of CTC even after fluctuations, something rather stable. $\endgroup$
    – Bastam Tajik
    Aug 18 at 21:24
  • $\begingroup$ For example in this paper by Witten arxiv.org/pdf/1901.03928.pdf that you shared somewhere else with me, on page 24 he gives a 2d example of such spacetimes. But I don't know if it's stable against fluctuations. $\endgroup$
    – Bastam Tajik
    Aug 18 at 21:27
  • $\begingroup$ Stable causality implies strong causality, but the viceversa is not true. So the two topologies could be the same even if the spacetime is not stable causal. $\endgroup$
    – Pipe
    Aug 18 at 21:30
  • $\begingroup$ stability and causality in general are distinct to me. Why do you think stable causal and stable strongly causal are the same? can one prove that any stable causal spacetime is at least strongly causal (regardless of dimensionality also) $\endgroup$
    – Bastam Tajik
    Aug 18 at 21:36
  • $\begingroup$ I know Hawking showed that strong causality and stability together are nontrivial and there are spacetimes that are strongly causal but not stable. what about stable but not strongly causal? @pipe $\endgroup$
    – Bastam Tajik
    Aug 18 at 21:39

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