Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 54 / 1 / b / iii / Solution

Write $M=-a^2$ with $a>0$. On a constant-time slice, the proper distance from the axis to radius $r$ and the circumference at that radius areHence $C/s\to 2\pi a$. A smooth axis in a Riemannian metric requires the Euclidean limit $C/s\to 2\pi$, so regularity forces $a=1$, or $M=-1$. For $a\ne 1$ the conical singularity of negative-mass BTZ geometry has deficit angle $2\pi (1-a)$; for $a>1$ this is an angular excess. Even integer $a>1$ is singular with the stipulated angular period $2\pi$: a multiple covering does not make the tip a smooth point.

At $M=-1$, set $r=\ell \sinh \chi$. The metric tensor becomeswhich is regular at $\chi =0$ and describes Anti-de Sitter spacetime with nonperiodic time. For every $M<0$, $f=a^2+r^2/{\ell }^2$ stays positive. An outward radial null geodesic obeys $dt/dr=1/f$, and reaches the timelike conformal boundary in finite coordinate time, since ${\int }_0^{\infty }dr/f=\pi \ell /(2a)$. Thus no event horizon hides the conical axis. The geometry is nakedly singular except at $M=-1$. Its local curvature away from the axis is that of Anti-de Sitter spacetime; the argument is a global angular-identification test, not curvature blowup.