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Take a general spacetime that is not strongly causal.

Call this spacetime $(M, g) $ where $M$ is a connected time-oriented manifold and $g$ is the Lorentzian metric that satisfies the Einstein's Field Equation.

The topology considered is the manifold topology throughout the whole text which is strictly finer than the Alexandrov topology iff the spacetime is not strongly causal.

Let $U$, $V$ be open subsets (manifold topology) of a spacetime $(M,g)$, with $V\subset U$. $V$ is called causally convex in $U$ if any causal curve contained in $U$ with endpoints in $V$, is entirely contained in $V$.

A point $p$ is called strongly causal iff given any neighborhood $U$ (manifold topology) of $p$ there exists a neighborhood $V\subset U$, $p\in V$, such that $V$ is causally convex in $U$.

Call the set of all points that are not strongly causal in $(M,g)$ spacetime: $N$.

My question:

What are the general features of such subset $N$? Is it a submanifold of $N$? Is it immersed or embedded or anything at all? Is it Lorentzian in case of being a manifold at all?

Even some hints about what goes on around one such point is very useful.

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    $\begingroup$ While I don't know enough about mathematical pyhsics, I can at least comment that the easiest way to show that a subset $S\subset M$ is a submanifold, is to use the implicit function theorem. Thus we have to write the subset the preimage of a submanifold $N\subset N'$ under some smooth map $f:M\to N'$. Often the submanifold can be written as the solution of a system of equations and thus these equations define $f$ and we are looking at the preimage of $0$. Then one still has to verify the condition on the differential. $\endgroup$
    – HenrikRüping
    Commented Sep 11, 2023 at 13:43
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    $\begingroup$ You are more likely to get an answer if you define the interval topology, that would make your question self-contained. $\endgroup$
    – Ryan Budney
    Commented Sep 11, 2023 at 14:31
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    $\begingroup$ At the very least, we know that $N$ is closed since strong causality at a given point is an open condition. We also know that $N$, if nonvoid, cannot consist of a single point: for each $p\in N$ there are $q_\pm\in J^{\pm}(p)\smallsetminus\{p\}$ such that $I^\mp(r)\supset I^\mp(q_\pm)$ for all $r\in J^\pm(p)$, which on their turn imply that $q_\pm\in N$ (or course, we could still have $q_+=q_-$). This is argued e.g. in Lemma 4.16 and Remark 4.17, pp. 31-32 of R. Penrose, Tecniques of Differential Topology in Relativity (SIAM, 1972). $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Sep 14, 2023 at 20:48
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    $\begingroup$ (continued) In the previous comment, $I^{+/-}(p)$ and $J^{+/-}(p)$ are respectively the chronological future / past of $p$ and the causal future / past of $p$ in $(M,g)$. Moreover, the interior of $N$ contains all points of $(M,g)$ contained in some closed timelike curve = vicious points (the latter forming an open subset) of $(M,g)$, and the boundary of $N$ is ruled by past- or future-inextendible null geodesics. Paragraphs 4.26 though 4.32, pp. 34-38 of R. Penrose, ibid. discuss this and other properties of $N$, see also B. Carter, Gen. Rel. Grav. 1 (1971) 349-391. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Sep 14, 2023 at 21:30
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    $\begingroup$ No, $N$ cannot have isolated points. This is made clear by the characterization of $N$ provided in my last comment above. This is discussed in more detail in the aforementioned references by Penrose and Carter. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Sep 14, 2023 at 21:33

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