0
$\begingroup$

Take the time-oriented Lorentzian spacetime $(M, g)$ that is not strongly causal.

In such case it is shown that the Alexandrov topology and the Manifolds topology deviate such that the manifold topology is strictly finer than the Alexandrov topology.

My question is that:

The necessary and sufficient conditions under which two points that are distinguishable under the Manifold topology, are indistinguishable under the Alexandrov topology for a generic such spacetime?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

I am not sure that is what you are really asking but if the Alexandrov topology is Hausdorff, it is the manifold topology and the spacetime is strongly causal. See for example Thm. 4.75 in

Minguzzi, E., Lorentzian causality theory, Living Rev. Relativ. 22, Paper No. 3, 202 p. (2019). ZBL1442.83021.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Well, true that if the two topologies do not match then the Alexandrov topology is not Hausdorff. But "not being Hausdorff", doesn't mean that the two points are indistinguishable. There are weaker versions of distinguishability. I want the two points to share the same neighbourhood filter under Alexandrov topology and have non-equal neighbourhood filters under the manifold topology. Is this possible at all? $\endgroup$
    – Bastam Tajik
    Commented Oct 20, 2023 at 15:47
1
$\begingroup$

What you seem (to me) to be asking is under which conditions on a Lorentzian manifold its Alexandrov topology not even $T_0$. If that is the case, then it is easy to see that if $(M,g)$ is not chronological then the Alexandrov topology cannot be $T_0$. More precisely, one can show that pairs of points lying in some closed timelike curve cannot be separated by the Alexandrov topology.

To wit, let $p,q\in M$ such that $p\ll q$ and $q\ll p$, where $\ll$ is the chronology relation (a strict partial order) on $(M,g)$. Let now $r,s\in M$ such that $r\ll p\ll s$ so that $p\in(r,s)\doteq\{t\in M\ |\ r\ll t\ll s\}$, recalling that subsets of the form $(r,s)$ for $r\ll s$ form a basis for the Alexandrov topology of $(M,g)$. Since $p\ll q$ (resp. $q\ll p$), we have that $r\ll q$ (resp. $q\ll s$), therefore $q\in(r,s)$ as well. This already shows that the Alexandrov topology is not $T_1$, but since the roles of $p,q$ above are symmetric, we also conclude that the Alexandrov topology is not even $T_0$, as claimed.

The above argument shows that the $T_0$ condition for the Alexandrov topology is a sufficient condition for the chronology condition on $(M,g)$ to hold. R. Penrose provides examples (see Techniques of Differential Topology in Relativity, SIAM, 1972, particularly Remark 4.25, pp. 34) of weaker failures of separation conditions for the Alexandrov topology - more precisely:

  • A (future- and past-)distinguishing space-time $(M,g)$ in which the Alexandrov topology is $T_1$ but not Hausdorff (see Figure 23, pp. 29 of R. Penrose, ibid.);
  • A distinguishing space-time $(M,g)$ in which the Alexandrov topology is not $T_1$ (see Figure 26, pp. 35 of R. Penrose, ibid.). This example does seem to be $T_0$ with respect to the Alexandrov topology, though.

Both these examples, being distinguishing, are causal and hence chronological. It is not clear to me whether (let alone how) separation conditions below the Hausdorff condition for the Alexandrov topology on a Lorentzian manifold $(M,g)$ fit into the causal ladder of space-time conditions between chronology and strong causality, though.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .