(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.

  • $\begingroup$ @Timaeus Ah, I forgot the Wiki article uses $(+,-,\cdots,-)$. I'm using $(-,+,\cdots,+)$. The question does not change. $\endgroup$ – Ryan Unger Mar 21 '16 at 2:37

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.

  • $\begingroup$ Would you mind expanding a bit on "cutting a finite slit into a manifold doesn't change the topology". Would that slit change properties as being simple connected? $\endgroup$ – yess Mar 21 '16 at 12:20
  • $\begingroup$ @yess I think he's talking about cutting a notch into the end of a finite cylinder. This is certainly homeomorphic to the original cylinder. $\endgroup$ – Ryan Unger Mar 21 '16 at 12:56
  • $\begingroup$ @yess I edited the post to make it clear that I cut the slit from the edge. If you had a piece of paper and attached the top to the bottom to make a finite cylinder then you can cut the paper from the right and somewhere else cut it from the left and it had the same topology. Intuitively, the slit could open from a line to a V and then the center of the V can come upwards. At which point it's like you never cut it in the first place. $\endgroup$ – Timaeus Mar 21 '16 at 17:07
  • $\begingroup$ Can you come up with a counterexample that is simply connected and boundaryless? Right now the two answers have circles in them! $\endgroup$ – Ryan Unger Mar 21 '16 at 17:10
  • 1
    $\begingroup$ @0celo7 Neither spacetime is globally hyperbolic: I responded to your original post. As I always do for questions about mathematical relativity, otherwise the question goes through endless modifications of adding hypothesis, weakening conclusions, and restricting the equivalence to subclasses of spacetimes. So it becomes a discussion through edits. Whereas stack exchange is supposed to be about questions with definitive answers. If you want to ask whether globally hyperbolic simply connected boundryless spacetimes can be classified by topology and metric then it's a new question. $\endgroup$ – Timaeus Mar 21 '16 at 17:43

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.

  • $\begingroup$ How can the cylinder be globally hyperbolic? It's not even causal. $\endgroup$ – Ryan Unger Mar 21 '16 at 3:02
  • $\begingroup$ The warp coordinate is the space coordinate, not the time coordinate. Notice the difference with your example. $\endgroup$ – yess Mar 21 '16 at 3:04
  • $\begingroup$ Ah, yes, indeed. $\endgroup$ – Ryan Unger Mar 21 '16 at 3:07
  • $\begingroup$ Maybe there is a condition that's not on the causal ladder? Perhaps a condition on the topology of $M$ itself, like "no compact dimension if $M$ is a product" + global hyperbolicity (just a thought). $\endgroup$ – Ryan Unger Mar 21 '16 at 3:09
  • $\begingroup$ That sounds like a nice exercise. I will give it a try and if get something I will let you know. :) $\endgroup$ – yess Mar 21 '16 at 3:11

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 $\sigma$.

  • $\begingroup$ This should really be the accepted answer. $\endgroup$ – balu Aug 20 '18 at 13:28
  • $\begingroup$ @ Effore--I'm very unfamiliar with the notations being used, but a literal reading of the question would seem consistent with a literal reading of your answer as "Never". As this might invalidate "bouncing" cosmologies, could you clarify whether the spacelike curve might be of "vanishing" (in the sense of "asymptotically zero") length? (I was led into Ryan Unger's question while trying to understand how Pauli resolved his initial objection to particulate spin as violating GR's prohibition on relative motion at velocities greater than that of light in a vacuum.) $\endgroup$ – Edouard Apr 16 at 16:27

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