Let us consider Minkowski spacetime. Let $(u,r,x^A)$ be retarded coordinates with $x^A$ coordinates on the sphere. Future null infinity is described here as the $r\to \infty$ limit with $(u,x^A)$ fixed.

One usually introduces $\mathcal{I}^+_-$, the past boundary of $\mathcal{I}^+$, as the $u\to -\infty$ limit with $x^A$ fixed.

On the other hand to describe spatial infinity $i^0$ one introduces a set of hyperbolic coordinates $(\rho,\tau,x^A)$ outside of the lightcone of the origin $r\geq t$. Spatial infinity is the hyperboloid described as the $\rho\to \infty$ with $(\tau,x^A)$ fixed and is denoted $\mathbb{H}_3^+$.

This hyperboloid has a future boundary which can be considered to be the $\tau \to \infty$ limit with $x^A$ fixed.

This hyperbolic foliation is shown in the figure from "Advanced Lectures on GR":

enter image description here

Now in these same lecture notes, the author says that the future boundary of $\mathbb{H}_3^+$ is the same as the past boundary of $\mathcal{I}^+$ (end of page 71 and begining of page 72):

The boundary hyperbolic metric is now a smooth codimension 1 manifold which resolves $i^0$. It intersects null infinity at two spheres denoted by $\mathcal{I}^+_-$ and $\mathcal{I}^-_+$ which are respectively the past limit of future null infinity and the future limit of past null infinity. In the hyperbolic description, $\mathcal{I}^+_-$ coincides with the sphere at the future time $\tau\to \infty$ of the boundary hyperboloid, and $\mathcal{I}^-_+$ is the sphere at the past $\tau\to-\infty$ of the boundary hyperboloid.

This is confusing me a lot for a variety of reasons:

  1. Looking at the picture, all hyperboloids of the foliation seem to intersect $\mathcal{I}^+$ at $u = 0$. Therefore I fail to see how the intersection of the hyperboloid with $\mathcal{I}^+$ corresponds to $\mathcal{I}^+_-$ which lies at $u\to -\infty$.

  2. Using $u = -\rho e^{-\tau}$, taking $\tau \to \infty$ first I get $u\to 0$ regardless of $\rho$. This seems to confirm what we see in the picture that "the boundary of all hyperboloids intersect $\mathcal{I}^+$ at $u = 0$".

  3. On the other hand, if I take $\rho\to \infty$ we have $u\to -\infty$ regardless of $\tau$ which seems to imply that the whole hyperboloid intersects $\mathcal{I}^+$ at $u\to -\infty$ which looks very weird. Moreover it really seems that $u$ is even ill-defined at the boundary of the hyperboloid, since its value depends on the path taken to get there.

In summary: why the past boundary $\mathcal{I}^+_-$ of $\mathcal{I}^+$, which is located at $u\to -\infty$ is the same thing as the intersection of the hyperboloid resolving $i^0$ and $\mathcal{I}^+$, which seems to be at $u =0$?


1 Answer 1


The picture is very misleading. Firstly, since we are trying to look at the spatial infinity and metric around it, only one of the paths you've described above is correct.

The question is how to reach spatial infinity? If you take $\tau \rightarrow \infty$ first, then you're implicitly assuming that $\rho<< \tau$. In which case, you'll never reach a spacelike infinity. In fact a surface where $\rho \rightarrow \tau-$(I'm taking this to imply $\rho$ approaches $\tau$ but is less than $\tau$ at all points), is timelike everywhere and asymptotically becomes null. So, this path has to be discarded.

Thus the only way to reach $i_0$ in this co-ordinate system is to keep $\tau$ fixed and take $\rho \rightarrow \infty$. That way, as you've already observed, the limit of the surface is at $u \rightarrow -\infty$.

So, how does this foliation actually look? It looks something like this (pardon the extremely bad quality of the picture): Blown up Spatial infinity

  • $\begingroup$ This picture is wrong. You cannot have a usual (codimension 1) description of $\mathscr{I}^{+}$ if you are blowing up $\mathcal{i}_0$ to be a codimension 1 boundary (instead of codimension 2 “corner” in the usual asymptotic null infinity picture). $\endgroup$
    – A.V.S.
    Mar 28, 2020 at 4:10
  • $\begingroup$ The picture is different co-ordinate system for null and spatial infinity. That's implied. $\endgroup$
    – Ari
    Mar 28, 2020 at 4:14
  • $\begingroup$ But that is my point. The “misleading picture” in the OP covers entirety of Minkowski spacetime. There is no room for different coordinate systems. $\endgroup$
    – A.V.S.
    Mar 28, 2020 at 4:21

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