Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 58 / 5 / a / Solution

Introduce conformal timeIt satisfies $d\eta /dt=\mathrm{sech}t$ and $\cos \eta =\mathrm{sech}t$, so the metric becomes $d{s}^2={\sec }^2\eta (-d{\eta }^2+d{\xi }^2)$. Choose null coordinatesThus $f(u,v)=\sec [(v-u)/2]$ is positive on $-\pi <v-u<\pi$. Choosing the two null coordinates with these orientations gives the requested positive coefficient of $dudv$.

There is a global-topology distinction in the source: unrestricted $\xi \in \mathbb{R}$ gives the universal cover of two-dimensional de Sitter spacetime. The usual two-dimensional de Sitter spacetime has $\xi$ periodic modulo $2\pi$. Indeed its standard embedding is $X^0=\sinh t$, $X^1=\cosh t\cos \xi$, $X^2=\cosh t\sin \xi$, and is unchanged by $\xi \mapsto \xi +2\pi$. The local metric and the null-coordinate calculation apply to either choice.