Suppose we have a spacetime $(M,g)$, and denote by $J^+(p)$ the set of points that lie in the causal future of $p$, i.e. $x \in J^+(p)$ iff there is a future-directed timelike curve $\gamma: [0,1]\rightarrow M$ such that $\gamma(0) = p$ and $\gamma(1) = x$.

Question: For any pair of elements $p,q$ in a globally-hyperbolic spacetime, is it always the case that $J^+(p)$ and $J^+(q)$ intersect?

A standard result (see e.g. here, Thm 3.78, page 49) is that every globally hyperbolic spacetime admits some Cauchy surface $\mathcal{S}$, and moreover $M$ is isometric to the product $\mathcal{S} \times \mathbb{R}$.

Without loss of generality we can assume that $p$ and $q$ lie on the same Cauchy surface since the result is trivially true if $p,q$ are timelike/lightlike events, and if $p$ and $q$ lie on different slices, say $p \in \mathcal{S} \times \{t\}$ and $q \in \mathcal{S}\times \{t'\}$ where $t <t'$, then we can just look at some point $p'\in J^+(p)$ that lies on $\mathcal{S}\times\{t'\}$ (such a point always exists since we can pick any inextendible future timelike curve $\gamma$ passing through $p$ and use the definition of a Cauchy surface to conclude that $\gamma$ will pass through $\mathcal{S}\times \{t'\}$) and use that $p<p'$ implies $J^+(p')\subset J^+(p)$.

How can I proceed from here? I was hoping to define some sort of spacelike geodesic connecting $p$ and $q$, and then pick an appropriate element $r \in \mathcal{S} \times \{t+l\}$, where $l$ is the length of the spacelike geodesic connecting $p$ and $q$. Is this the right approach?


1 Answer 1


In a globally-hyperbolic spacetime, does every pair of elements have overlapping light cones?

No. Standard cosmological models of our own universe are a counterexample. There are cosmological horizons, so future light cones do not all overlap, but the universe is globally hyperbolic.

  • $\begingroup$ What evidence is there that the universe is globally hyperbolic? $\endgroup$
    – Doc
    May 21, 2020 at 1:35

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