Matching between $i^+$ and ${\cal I}^+$ I have a confusion about the paper "Asymptotic symmetries of QED and Weinberg’s soft photon theorem" by Campiglia & Laddha. They use the hyperbolic slicing of the region inside the ligthcone of the origin to resolve $i^+$. We have $t \geq r$ and we define $$\tau=\sqrt{t^2-r^2},\quad \rho=\dfrac{r}{\sqrt{t^2-r^2}}\tag{1}.$$
Now let $\mathbb{H}_3^\tau$ be the constant $\tau$ surfaces. One wants to identify $i^+$ with the $\tau \to \infty$ limit which we denote $\mathbb{H}_3^+$. Now I get the impression that $\partial \mathbb{H}_3^+$ is often matched with ${\cal I}^+_+$ but I can't see where this comes from.
For finite $\tau$, the authors claim $\mathbb{H}_3^\tau$ intersect ${\cal I}^+$ at $u = t-r = 0$. In that case to have a Cauchy surface one matches $\partial \mathbb{H}_3^\tau$ with the sphere at $u=0$ on ${\cal I}^+$ and adjoins the $u < 0$ portion of ${\cal I}^+$.
So for all $\tau$ the boundary $\partial\mathbb{H}_3^\tau$ matches with the sphere at $u=0$ of ${\cal I}^+$. Then suddenly when $\tau \to \infty$ the boundary $\partial \mathbb{H}_3^+$ matches with ${\cal I}^+_+$.
Indeed I'm able to argue that a point in $\mathbb{H}_3^\tau$ with coordinates $(\tau,\rho,z,\bar{z})$ if we take $\rho\to \infty$ and translate to retarded coordinates goes to a point on $\mathscr{I}^+$ with $u =0$ and the same $(z,\bar{z})$. But for infinite $\tau$ I really don't understand this matching.
Why does $\partial \mathbb{H}_3^+$ is matched with ${\cal I}^+_+$ when for all finite $\tau$ the boundary $\partial \mathbb{H}_3^\tau$ is matched with the cut $u=0$ of ${\cal I}^+$?
 A: As I recall, the idea is to take the hyperboloid ${\mathbb H}_\tau$ to be the Cauchy slice for massive particles and the portion of ${\cal I}^+$ described by $u<0$ as the Cauchy slice for massless particles. This can always be done by translating the origin of the lightcone sufficiently far into the future (essentially sending $u \to u + u_0$) so that all massless particles exit the spacetime at $u< 0$.
In this case, since there is no flux through the $u>0$ portion of ${\cal I}^+$, the fields are frozen (i.e. the field strengths vanish) in the region $ u> 0$. It then follows that for any field
$$
\phi|_{\partial {\mathbb H}_\tau} = \phi|_{u=0} = \phi|_{{\cal I}^+_+}.
$$
They use this property to match the fields accordingly. Note that we are NOT asserting that $\partial {\mathbb H}_\tau = {\cal I}^+_+$, only that the field values on these boundaries are equal.

ASIDE - The same idea is utilized when spatial infinity is blown up into de Sitter slices. The boundary of the de Sitter slices is at $u=0$ and in that case one assumes that all massless particles exit the system at $u>0$. Fields are therefore frozen in the region $u<0$ so $\phi|_{u=0} = \phi|_{{\cal I}^+_-}$.
