Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 311 / 3

For any nonzero $g$-causal vector $v^a$,For timelike $v$, the first term is negative. For null $v$, a nonzero null vector cannot be orthogonal to a timelike vector, so $(t\cdot v{)}^2>0$. Thusin both cases. Every $g$-causal direction is strictly timelike for $\tilde{g}$, so its light cone is strictly wider.

Let $t^a$ exhibit the stable causality of $g$, so $g-t_a{t}_b$ has no closed timelike curves. SetThe vector $r^a$ remains timelike for $g^{\prime }$. Widening the cones of $g^{\prime }$ once more using $r$ giveswhich is causal by assumption. Hence

The relevant implications in the causal hierarchy areThe first follows from strict cone widening: failure of strong causality would give causal curves leaving and re-entering arbitrarily small neighborhoods, and the wider metric turns the limiting recurrence into a closed timelike curve. For the second, if distinct $p,q$ had $I^{+}(p)=I^{+}(q)$, arbitrarily short timelike segments near either point and this equality would produce a causal curve leaving and re-entering a sufficiently small causally convex neighborhood. ThereforeThe stated null geodesic completeness is stronger than needed: stable causality already gives the conclusion.

Write $I^{+}(U)=U=I^{-}(U)$. Since every chronological future and chronological past is open, $U$ is an open set. We show its complement is also open.

Take $p\notin U$ and choose $p_{-}\ll p\ll p_{+}$ in a convex normal neighborhood. The diamond $V=I^{+}(p_{-})\cap I^{-}(p_{+})$ is a neighborhood of $p$. If $x\in V\cap U$, then $p_{-}\ll x$ and $U=I^{-}(U)$ imply $p_{-}\in U$; then $p_{-}\ll p$ and $U=I^{+}(U)$ imply $p\in U$, a contradiction. Thus $V$ lies in the complement of $U$.

Hence $U$ is nonempty, open, and closed. Since $\mathcal{M}$ is a connected space,