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Let $M$ be a strongly causal Lorentzian manifold. If $M$ has dimension 4, a theorem of Hawking, King, and McCarthy (see Thm 5) says that $M$ is determined up to conformal isomorphism by its class of null geodesic curves (where the parameterization of the null geodesic is forgotten) and thence by the causal relation $J^+$ on $M$.

Question: Does this theorem also hold in dimensions other than 4?

My hunch would be that it continues to hold in higher dimensions, but in lower dimensions I'm not so sure. At any rate, I don't see how to adapt Hawking, King, and McCarthy's argument (which involves choosing coordinates using a certain configuration of null geodesics) to dimension 2.

EDIT: As came up in the comments, I should clarify that a priori I don't want to assume anything about the smooth structure on $M$. In fact, I would prefer not to assume anything about the topology either -- just take $M$ as a set of points, equipped with the relation $J^+$. From this data, when can one recover the topological, smooth, and conformal structures on $M$?

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    $\begingroup$ Yes, it holds in all dimensions. The conformal structure at every point is uniquely defined by the set of all null vectors (O'Neill, Semi-Riemannian Geometry, Chapter 2, exercise 14). Any mapping that preserves null geodesics preserve the set of null vectors, and hence determine the metric pointwise up to a conformal factor. $\endgroup$
    – Willie Wong
    Commented Dec 6, 2021 at 20:32
  • $\begingroup$ It fails for Riemannian, or for negative of Riemannian, but works fine for all other signatures. $\endgroup$
    – Ben McKay
    Commented Dec 6, 2021 at 20:33
  • $\begingroup$ @WillieWong I'm confused -- surely determining the conformal structure is stronger than determining the smooth structure? $\endgroup$
    – Tim Campion
    Commented Dec 9, 2021 at 3:27
  • $\begingroup$ Ah I see -- yes, part of the point of my question is to not assume anything about the smooth structure from the start. So one can't even talk about tangent vectors a priori, and so I don't think the exercise you cite applies. $\endgroup$
    – Tim Campion
    Commented Dec 9, 2021 at 3:29
  • $\begingroup$ @TimCampion but what do you mean by a "strongly causal Lorentzian manifold" if you don't have a smooth structure, or a topological structure? None of the phrases "manifold", "Lorentzian", and "strongly causal" have to my knowledge established meaning without at least a topological structure. $\endgroup$
    – Willie Wong
    Commented Dec 9, 2021 at 13:06

2 Answers 2

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Trying to recover as much of the topology/geometry from the causal order as possible has been studied quit a bit since the early paper of Hawking et al that you cite. A quick summary of my understanding of the situation is that there are some purely order-theoretic topologies (like Alexander or Scott topologies) that reproduce the topology of the original Lorentzian manifold when the causal relation is sufficiently non-pathological. However, it is very difficult to find a set of conditions on partially ordered sets (posets) that single out precisely those posets that come from the causal order of smooth (or even topological) Lorentzian manifolds. AFAIK, that is an open problem . (Please correct me if I'm wrong!)

You can find a relatively recent collection of results in that direction here:

Martin, Keye; Panangaden, Prakash, A domain of spacetime intervals in general relativity, Commun. Math. Phys. 267, No. 3, 563-586 (2006). ZBL1188.83071 arXiv:gr-qc/0407094

One can start walking up and down the citation tree from here to find more recent progress.

Added: If one restricts attention to generalizing just Thm 5 of Hawking-King-McCarthy to other dimensions, as requested by Tim, then this was done in Thm 1.2 here

Peleska, Jan, A characterization for isometries and conformal mappings of pseudo- Riemannian manifolds, Aequationes Math. 27, 20-31 (1984). ZBL0539.53017.

All dimension higher than 2 are covered, which agrees with Willie's counter example.

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  • $\begingroup$ Right, but I'm not asking about characterizing which partial orders are the causal order on a Lorentzian manifold. I'm asking, given a partial order, and granted that there exists a Lorentzian manifold $M$ whose causal order is isomorphic to that partial order, whether is $M$ unique up to conformal isomorphism. I'd think that might be a less wide-open question, especially since the answer is known in dimension 4. $\endgroup$
    – Tim Campion
    Commented Dec 9, 2021 at 17:04
  • $\begingroup$ @TimCampion OK, see the added reference. $\endgroup$
    – Igor Khavkine
    Commented Dec 9, 2021 at 22:22
  • $\begingroup$ Great, thanks! The answer is that the causal relation determines the conformal structure in dimensions $\geq 3$. In combination with Willie Wong's negative answer below for dimension 2, this seems to be a complete answer. $\endgroup$
    – Tim Campion
    Commented Dec 9, 2021 at 22:27
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    $\begingroup$ Incidentally, the difference is due, in part, to the fact that for two dimensional manifolds the conformal group is huge. You can approximate the non-differentiable mapping in my example with smooth ones. That the results hold in higher dimensions can be interpreted as saying that this particular type of non-compactness is forbidden. $\endgroup$
    – Willie Wong
    Commented Dec 9, 2021 at 22:32
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In 2 dimensions the question posted in this comment has a negative answer.

Consider the standard Minkowski space with double null coordinates $(u,v)$ in which the metric is $ds^2 = - du~dv$.

Any strictly increasing bijection $\phi: \mathbb{R}\to \mathbb{R}$ induces a mapping $(u,v) \mapsto (\phi(u),v)$ that is an causal isomorphism.

Choosing $\phi$ to be continuous and not differentiable you get an example of an $\mathscr{M}$-homeomorphism that preserves null geodesics and is not a diffeomorphism (in the terminology of Hawking-King-McCarthy; compare to Theorem 5).

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