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(See here for notation.) In Minkowski space, if $p\prec q$, then there is no spacelike curve $c:[0,1]\to \mathbb{R}^{n-1,1}$ with $c(0)=p$ and $c(1)=q$. This is obvious from a spacetime diagram. Here a "spacelike curve" is a $C^1$ mapping from a nondegenerate inverval into a Lorentzian manifold with everywhere spacelike tangent vector.

Take the cylinder $M=S^1\times\mathbb{R}$ with the Lorentzian metric $\mathrm{d}s^2=-\mathrm{d}\theta^2+\mathrm{d}t^2$. This spacetime is not causal, the circles $t=\mathrm{const.}$ are closed timelike curves. Let $\Sigma$ be a "surface" of constant $\theta$. It is clear that if $p,q\in\Sigma$, then there exists a spacelike curve $c$ connecting the two. But if they are close enough, then there exists a timelike curve $\lambda$ connecting the two as well. So $p\prec q$ (or the other way around). Thus the above result for Minkowski space does not apply to all spacetimes.

Thus my question is: what are the necessary and sufficient conditions on the spacetime topology and metric so that $p\prec q$ $\implies$ there exists no spacelike curve connecting $p$ and $q$?

Some observations: $J^+(p)\neq M$ for any $p$ is necessary. (Otherwise $p\prec q$ for all $q\in M$, in particular those connected to $p$ by a spacelike curve.) So $M$ should be causal, and probably even strongly causal. Perhaps one can reverse the question: if $p,q$ can be connected by a spacelike curve, what are the conditions for which $p\not\prec q$? Then, suppose $p$ and $q$ are contained in a Cauchy surface. There is a spacelike curve connecting them, but they cannot be connected by a causal curve because a causal curve intersects a Cauchy surface once.

Ryan Unger
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3 Answers3

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There exist no necessary and sufficient conditions on the topology and the metric that guarantee that when $p$ is in the causal past of $q$ then there exists no spacelike curve between the two events.

Why? Because there exists two Lorentzian manifolds, both with a Minkowski metric, and both with the same topology, one which has the property you desire and one which does not.

For the topology of the manifold, each manifold can have the topology of the cylinder: $\mathbb S\times \mathbb R.$

One manifold can literally be $$\{(a,b,c):a^2+b^2=1,c\in[0,10]\},$$ with time going around in the circular direction and space going in the direction of length ten.

The other manifold can be the same manifold but excluding the points $$\{(1,0,c):c\geq 0.01\},$$ and the points $$\{(0,1,c):c\leq 9.99\},$$ with time going around in the circular direction and space going in the direction of length ten.

The topology of the second manifold equals the topology of the first (cutting a finite slit out of a manifold, starting at an edge, doesn't change the topology). And the topology of the first is the same as the topology of the cylinder ($\mathbb S\times \mathbb R$).

So if someone told you your topology was $\mathbb S\times \mathbb R$ and told you that your metric was Minkowski, then you might have the first one, or you might have the second (or something else). And the first one does not have the property you want (the causal future of any event is the whole manifold), and the second one does.

So knowing the metric and knowing the topology simply isn't enough information.

Timaeus
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In any spacetime (even Minkowski) of dimension $n\ge 3$ any two causally related events are connected by a spacelike curve. The idea is as follows. Connect $p$ and $q$ with a causal curve $\sigma$. Take a metric mith slightly larger cones $g'>g$, so that $\sigma$ becomes $g'$-timelike. Now there is a $g'$-lightlike curve (hence $g$-spacelike) which connects $p$ to $q$, it is obtained as an helix winding over a cylinder with axis $\sigma$.

Ettore Minguzzi
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The canonical Lorentzian cylinder $(R × S^1 , g = −dt^2 + dθ^2 )$ is globally hyperbolic and you can find points that can be joined causally and also with a spacelike curve. For example, take a null geodesic that join $p,q$ and notice that you can find always a spacelike curve by taking the helix that join these points.

See Fig. 13

Therefore any causal condition from the hierarchy can not be a sufficient condition such that $p\prec q$ implies that $p,q$ can not be join by a spacelike curve.

On the other hand if $p,q$ belong to a Cauchy surface, by definition a Cauchy surface is acausal, $p\not\prec q$

There is an interesting paper about how to determine a cauchy surface from intrinsic properties.

yess
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