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?


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.

| cite | improve this answer | |
  • $\begingroup$ What evidence is there that the universe is globally hyperbolic? $\endgroup$ – Doc May 21 at 1:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.