Can the Causal Structure recover the manifold topology for non-chronological spacetimes?

Question

Given a time-oriented spacetime $(M,g)$, a binary relation $\ll$ can be defined on this spacetime where $p \ll q$ for $p, q \in M$ if and only if there exists a time-like path connecting $p$ and $q$.

A $\ll$-isomorphism between the two time-oriented spacetimes $(M_1, g_1)$ and $(M_2, g_2)$ is a bijective map $\phi :M_1 \rightarrow M_2$ such that $\forall p, q \in M_1 $: $p\ll q$ if and only if $\phi(p) \ll \phi(q)$.

A spacetime is called chronological if and only if there exists no $p \in M$ where $p \ll p$.

A theorem due to Hawking, King, McCarthy and improved by Malament 1977 states that:

Let $\phi$ be a $\ll$-isomorphism between the time-oriented spacetimes $(M_1, g_1)$ and $(M_2, g_2)$. If $\phi$ is a homeomorphism, then it's a conformal isometry.

On the other hand, a theorem due to Malament states that:

Given the time-oriented $\ll$-isomorphic spacetimes $(M_1, g_1)$ and $(M_2, g_2)$, if the spacetimes are distinguishing(both past and future), then $\phi$ is a smooth conformal isometry.

This theorem means literally: the causal structure for distinguishing spacetimes alone, can determine the spacetime [always the pair $(M, g)$] up to a smooth conformal isometry!

Now given this context, take the spacetime of a spinning Cosmic String $(M, g)$, where $M$ is homeomorphic to $\mathbb{R}^4 - \mathbb{R}^2$ with its Euclidean subspace topology and $g$ is everywhere locally:

$$g= -(dt + a \, d \phi)^2 + d\rho^2 + \kappa^2 \rho^2 + d\phi^2 + dz^2 \ \text{where} \ \ a > 0$$

Obviousely this spacetime is not even chronological, let alone distinguishing!

Suppose the time-oriented spacetime $(N, h)$ is $\ll$-isomorphic with $(M, g)$, meaning that they have the same causal structure,

My question is:

Does there exist any non-trivial conformal isometry $(M, g)$ and $(N,h)$? Or in general what are the common topological features of $M$ and $N$?

In case such nontrivial conformal isometry does not exists, based on HKM theorem, one can safely argue that the manifold topology cannot be fixed by the causal structure.

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PS: The topology assumed in HKM theorem is the path topology $\mathcal{P}$, which is strictly finer than the manifold topology $\mathcal{M}$: $\mathcal{M} \subsetneqq \mathcal{P}$.

Moreover the theorem 6 in the HKM paper assumes strong causality, which is unnecessary as shown by Malament 1977

Also trivially a $\mathcal{P}$-homeomorphism is trivially an $\mathcal{M}$-homeomorphism.

Answer 1

Regarding the title question: No, the topology cannot be recovered from the chronology relation for non-chronological spacetimes. For example, there are plenty of distinct Lorentzian manifolds for which every point chronologically precedes every other point.

For an example of the example, let $\mathbb R^{d+1}$ be Minkowski space and quotient by $\mathbb Z^{d+1}$ to get a Lorentzian torus $T^{d+1}$. In this manifold, every point chronologically precedes every other point. Moreover, for $d \geq 0$ this manfold has continuum-many points. So there is a bijection $T^{d+1} \to T^{d’+1}$ for any $d,d’$, and this bijection preserves the relation $\ll$. Nevertheless for $d \neq d’$ we have that $T^{d+1}$ is not conformally equivalent to $T^{d’+1}$.

For a more general example of the example, let $M^d$ be any Riemannian manifold whatsoever. Then the manifold $N^{d+1} := \mathbb R \times M^d$ has a natural Lorentzian metric which is the product metric, but with a minus sign inserted on the $\mathbb R$ (time) coordinate. In this metric, the point $(t,x)$ chronologically precedes the point $(t+1,y)$ whenever the Riemannian distance from $x$ to $y$ in $M$ is $<1$. You can quotient this by $\mathbb Z$ to get a Lorentzian metric on $S^1 \times M$. In this metric every point chronologically precedes every point in the same path component.