# Minkowski Space is not truly conformally compact

We say a manifold $(M,g)$ is conformally compact if it is the interior of some $(\overline M, \overline g)$, such that $$g = r^{-2}\overline g|_M,\quad \mathcal Z(r) = \partial M,\quad \text{and}\quad \mathrm{d}r_p \neq 0\ \text{for any}\ p\in \partial M.$$ Moreover, we say a conformally compact manifold is asymptotically flat if $|\mathrm{d} r|^2_{\overline g}\equiv 0$ on the boundary. (i.e. the first derivative of the defining function is null on the boundary.)

It is "well known" that Minkowski space is an asymptotically flat manifold - Penrose's famous compactification supposedly shows this. However, in his compactification, there are two points on the boundary for which $\mathrm{d}r_p = 0$: namely spacelike and timelike infinity. I've made my own coordinate free compactifications of Minkowski space, but they also have some singularities at conformal infinity.

How is this formalised? When people say "conformally compact" in literature, do they just mean "there is a dense subset of the boundary satisfying the above conditions"?

• Conformal compactification is e.g. discussed in my Phys.SE here. – Qmechanic Sep 7 '18 at 8:06