### “The asymptotically hyperboloidal is not asymptotically null.”

By way of Roland Donninger, I learned today of the statement above which is apparently well-known in the numerical relativity community.

It may seem intuitively surprising: after all, the archetype of an asymptotically hyperboloidal surface is the hyperboloid as embedded in Minkowski space. Let $(t,r, \omega)\in \mathbb{R}\times\mathbb{R}_+ \times \mathbb{S}^{d-1}$ be the spherical coordinate system for the Minkowski space $\mathbb{R}^{1,d}$, the hyperboloid embeds in it as the surface $t^2 - r^2 = 1$. If you draw a picture we see clearly that the surface is asymptotic to the null cone $t = |r|$

The key, however, lies in the definition. For better or for worse, the definition under which the titular statement makes sense the following:

Definition
Let $(M,g)$ be an asymptotically simple space-time (or one for which one can define a Penrose compactification), and let $(\bar{M},\Omega^2 g)$ be the compactified space-time. We say that a hypersurface $\Sigma \subset M$ is asymptotically null if the $\bar{\Sigma}\cap \bar{M}$ transversely and the tangent space of $\bar{\Sigma}$ is null along $\partial\bar{M}$.

Now suppose near $\partial\bar{M}$ we can foliate via a double-null foliation $(u,v)$, with $\partial\bar{M} = \{ u = 0\}$. Let $x$ be a coordinate on $\partial\bar{M}$ so that $(u,v,x)$ form a coordinate system for a neighborhood of $\partial\bar{M}$. Assume that our surface $\Sigma$ can be written as a graph

$v = \phi(u,x)$

where $\phi$ is a $C^3$ function. Then the asymptotically null condition is just that $\partial_u \phi |_{u = 0} = 0$. Taking a Taylor expansion we have that this means

$v \approx \phi_{\infty}(x) + \phi^{(2)}_{\infty}(x) u^2$.

For the usual conformal compactification of Minkowski space, we have $u = \frac{\pi}{2} - \cot^{-1}\left( \frac{1}{r+t}\right)$. Hence we require that an asymptotically null surface to have convergence to the null surface at rate $O(1/(r+t)^2)$ (if $\phi$ is sufficiently differentiable; if we relax the differentiability at infinity we see that the above condition allows us to relax all the way to $O(1/(r+t)^{1+})$, but $O(1/(r+t))$ is not admissible).

On the other hand, the hyperboloid is given by $(r+t)(r-t) = -1 \implies r-t = v = O(1/(r+t))$ and so is not asymptotically null. And indeed, we can also check by direct computation that in the usual conformal compactification of Minkowski space, the limit of the hyperboloid at null infinity is space-like.

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