# Past boundary of $\mathcal{I}^+$ and future boundary of the hyperboloid resolving $i^0$

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":

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 $$u =0$$?

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 $$u \rightarrow -\infty$$.
• 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). – A.V.S. Mar 28 '20 at 4:10