I've been reading the article "Relativistic Symmetry Groups" by Roger Penrose, which appeared in Group Theory in Non-Linear Problems: Lectures Presented at the NATO Advanced Study Institute on Mathematical Physics ed. A.O.Barut (1974). In the article he talked a bit about conformal compactification of space-time. The notion of a conformal infinity is, by now, a pretty standard topic in introductory courses in General Relativity. I mention this article of Penrose because it contains a few small facts which is often not discussed in said courses or textbooks.
Conformal compactification
Start with a pseudo-Riemannian manifold $(M,g)$, let $\tilde{g}$ be another pseudo-Riemannian metric on $M$, we say that $g$ and $\tilde{g}$ are conformal if there exists a positive scalar function $\phi$ on $M$ such that $\tilde{g} = \phi g$ (sufficient smoothness of the relevant quantities are always assumed). Observe that two conformal metrics measure angles the same way: recall that on a pseudo-Riemannian manifold $(M,g)$, given a point $p\in M$ and two non-null vectors $v,w\in T_pM$, the angle between the vectors can be defined by $g(v,w)^2 / [ g(v,v) g(w,w) ]$. (Notice that on Euclidean space, if $v,w$ form an angle $\theta$, then $v\cdot w = |v||w| \cos\theta$.) Thus if $\tilde{g}$ is conformal to $g$, they define the same angles: \[ \frac{\tilde{g}(v,w)^2}{\tilde{g}(v,v)\tilde{g}(w,w)} = \frac{\phi^2 g(v,w)^2}{\phi g(v,v) \phi g(w,w)} = \frac{g(v,w)^2}{g(v,v)g(w,w)}. \] In fact, this inference works the other way too. If $g,\tilde{g}$ are two pseudo-Riemannian metrics such that for any two vectors $v,w$ we have $g(v,w) = 0 \iff \tilde{g}(v,w) = 0$, then $g,\tilde{g}$ are conformal (up to a change of sign) by the above definition (see e.g. Exercise 14, Chapter 2 from B.O'Neill, Semi-Riemannian Geometry).
So, in plain English, two metrics are conformal if they measure angles the same way.
Now, let $(M,g)$ be a pseudo-Riemannian manifold that is non-compact. A conformal compactification of $(M,g)$ is a choice of a metric $\tilde{g}$ such that $(M,\tilde{g})$ can be isometrically embedded into a compact domain $\tilde{M}$ of a pseudo-Riemannian manifold $(M',g')$ (well, I am ignoring some regularity issues here). Let $\phi$ be the conformal factor as before. Then observe that any regular extension of $\phi$ to the conformal boundary $\partial\tilde{M} \subset M'$ must vanish on said boundary. This reflects the property of a conformal compactification that "brings infinity to a finite distance".
The simplest example of conformal compactification is the one-point compactification of Euclidean space via the stereographic projection. In this case, the target manifold $(M',g')$ is compact itself, taken to be standard sphere. The source manifold $(M,g)$ is Euclidean space with the standard metric, and the image set $\tilde{M}$ is taken to be the sphere minus the north pole.
Ambient space construction
Another way to think about conformal compactification is through the lens of the Fefferman-Graham ambient construction (or, more properly speaking, the reverse of that construction; also closely related to the foundations of twistor theory, which also appears in the article of Penrose).
We start by example. Consider the standard Minkowski space $\mathbb{R}^{1,d}$ with 1 time-dimension and $ d$ spatial dimensions. The collection of all null-lines through the origin forms the light cone $\mathcal{N}$, which except for the singularity at the origin is a null hyper-surface. This collection can be naturally associated to the celestial sphere: for each spatial direction, which we denote by $r = (x_1,\ldots,x_d)$ with $\sum x_i^2 = 1$, we can associate the null vector $(1,r)$.
Now take an arbitrary space-like hypersurface $\Sigma$ that does not pass through the origin, and consider the intersection $S = \Sigma \cap \mathcal{N}$. It is clear that $S$ must be a space-like submanifold of co-dimension 2, with the topology of the $d-1$ dimensional sphere. Furthermore, a bit of computation shows that the induced metric on $S$ will be $t^2$ times the standard metric of the sphere, where $t$ is the time coordinate of a point on $S$ under the embedding into $\mathbb{R}^{1,d}$.
This identification can also be taken the opposite way: let $(\mathbb{S}^{d-1},h)$ be the standard sphere with the standard metric, then any conformal metric to $h$ can be realized as a section of the "future half" of the light cone $\mathcal{N}^+$, by taking the point $r = (x_1,\ldots,x_d)$ with $\sum x_i^2 = 1$ (using the coordinates of the standard embedding of the standard sphere into Euclidean space) on $\mathbb{S}^{d-1}$ to the point $(\sqrt{\phi}, \sqrt{\phi}r) \in \mathbb{R}^{1,d}$ where $\phi$ is the conformal factor. In other words, the future light cone $\mathcal{N}^+$, seen as a line bundle over $\mathbb{S}^{d-1}$, is a realization of the conformal structure of the standard sphere.
The rather neat thing now is that if we take, instead of a space-like slice $\Sigma$, the null plane $t = x_1 + 1$, the cut $S$ becomes a paraboloid, which under a small change of coordinates is isometric to the flat Euclidean space of $(d-1)$ dimensions. This change of coordinates is precisely the stereographic projection!
So far the material presented is more or less standard.
Ambient space construction for Minkowski space
In the previous section we compactified Euclidean space. Now we ask if something similar can be done to Minkowski space. The answer, as it turns out, is yes. The first fact to observe is the following: given an arbitrary pseudo-Riemannian manifold $(M,g)$, we can trivially embed it in a manifold of two higher dimensions by adjoining a time-like and a space-like dimension as follows: let $\hat{M} = M \times \mathbb{R}^{1,1}$ and let the metric $\hat{g}$ on $\hat{M}$ be the product metric $g\times \eta^{1,1}$ where the latter component is the standard (1+1) dimensional Minkowski metric. Now take the null hypersurface $\Sigma \subset \hat{M}$ given by $\Sigma = M \times \{ s = u + 1\}$ where $(s,u)$ are the standard time-space coordinates on $\mathbb{R}^{1,1}$. Observe the following: under this construction, any horizontal slice of $\Sigma$ will be isometric to $M$.
Now one may be tempted to try and compactify arbitrary non-compact Riemannian manifolds. Comparing the situation in the paragraph above with the compactification of the Euclidean space, we see that in order to exploit the null cone structure, it is necessary that the set of all null-rays in the ambient space $\hat{M}$ emanating from a fixed point $p$ must not refocus at a later time $s$ (remember it is by traveling along the null geodesics that we identify points on the compact manifold with the non-compact manifold). Using the fact that a geodesic on a product manifold projects down to geodesics on each multiplicand, we see that this condition is equivalent to asking that the manifold $M$ has infinite injectivity radius at the point $p$. Furthermore, in order for the compactification to be defined for all points on $M$, it is necessary that for any point $q\in M$, $q$ can be joined to $p$ with a geodesic. The two conditions together imply that $M$ should have a point $p$ at which the exponential map defines a global diffeomorphism between $M$ and $T_pM$. This occurs, for example, when $M$ has non-positive curvature and is simply connected. But this by itself is not enough. To use the construction of the previous section, $M$ itself has to admit a scaling symmetry about the point $p$. More precisely, we need that the radial vector field $r \partial_r$ be a conformal isometry of $M$. This is to guarantee that the various sections of the null cone $\mathcal{N}^+$ in $\hat{M}$ are conformal to each other. So together we know the following must be true of $M$: it is topologically $\mathbb{R}^{d-1}$, and we can pick $p$ to be the origin. The metric can be written in radial coordinates as $ds^2 = dr^2 + r^2 \gamma d\omega^2$, where $\gamma d\omega^2$ is a fixed metric on the sphere $\mathbb{S}^{d-2}$. But regularity at the origin will impose the condition that $\gamma d\omega^2$ is in fact the standard metric, which means that the method given only works for Euclidean space.
Getting back to Minkowski space though: observe that Minkowski space also has the conformal symmetry: the vector field $t\partial_t + r\partial_r$ generates a "radial" conformal isometry, where now $r$ is the spatial radial direction. So the construction will go through. (We can actually do this for arbitrary signature flat space $\mathbb{R}^{m,n}$. Here we'll just work with Minkowski space.) To recap: we've embedding $\mathbb{R}^{d,1}$ into $\mathbb{R}^{d+1,2}$, which we parametrize by $(t,s,u,x_1,\ldots,x_d)$, where $t,s$ are time-like and the rest space-like, with $(t,x_1,\ldots,x_d)$ parametrizing the original Minkowski space. The null "cone" now becomes the set $\{t^2 + s^2 = u^2 + |x|^2\}$. We see that the intersection of the null cone with the set $s = u+1$ is a horizontal slice, and is intrinsically isometric to the standard Minkowski space.
Now the question is how to get a compact section of this null "cone". If we intersect with the plane $\{s = 1\}$, we see that the set is now parametrized by $t^2 + 1 = u^2 + |x|^2$, which is a hyperboloid! The coordinates remains unbounded, so it is not a good choice. Similarly if we intersect with the plane $\{u = 1\}$. In fact, these two slices corresponds to the de Sitter and anti-de Sitter space respectively. The trick, as noted by Penrose in his article, is to intersect the null "cone" against the 5 dimensional sphere given by $t^2 + s^2 + u^2 + |x|^2 = 2$. The intersection now, after simple algebraic manipulations, satisfies \[ t^2 + s^2 = 1 \] and \[u^2 + |x|^2 = 1\] which implies that it is topologically $\mathbb{S}^1\times\mathbb{S}^d$.
Unlike the case of the compactification of Euclidean space, this object $M'$ is not the closure of the image $\tilde{M}$ of the Minkowski space. In fact, it is a double cover! One way to see it is that each null line is twice-represented in $M'$: the points $(t,s,u,x)$ and $(-t,-s,-u,-x)$ are two distinct points in $M'$ that corresponds to the same null line. Another way to see this is that the 5 dimensional sphere we use for the intersection not only has a piece in the part of the ambient space $\hat{M}$ where $s >0$, it also has a piece where $s < 0$. Therefore an accurate model for this in the Euclidean case is also to intersect with a sphere: but since the forward and backward light cones are essentially disconnected, we'll pick up a "double cover" in the sense that we end up with two copies of the standard sphere.
Connection to the Carter-Penrose diagram
Now you ask: "Wait a second here! What happened to the conformal infinities that we learned about in schools? The future and past null infinities and such? How do they fit into the picture?" Well, as is clear from the Carter-Penrose diagram, the usual conformal compactification of the Minkowski space-time does not have topology $\mathbb{S}^1 \times \mathbb{S}^d$. So what gives?
Remember the bit about the double cover? This turns out to be a very special feature of conformally flat space-times. The picture in the previous section is obtained by taking two copies of the standard Carter-Penrose diagram of Minkowski space and gluing them together, so that one is folded over the other. More precisely, the gluing identifies the past null infinity of one copy with the future null infinity of the other copy. A side effect is that the space-like infinity of one copy now becomes simultaneously the future and past-time-like infinities of the other copy.
Another way to think about the picture is the following: recall that the future time-infinity compactifies to a single point such that Minkowski space sits inside its past light cone. And the boundaries (the past light cone itself) corresponds to future null infinity. Swapping the word future with past we have an analogous picture for the past time-infinity. For the spatial infinity, however, we have that the Minkowski space lies completely to the outside of the light cone emanating from that point. So locally around the "point at infinity", if we fit a copy of the spatial infinity and a copy of the future time infinity and a copy of the past time infinity, we end up having exact the right amount to piece together a local copy of Minkowski space!
Now why is this particularly nice picture usually omitted from textbooks? The reason was alluded to earlier: it only works for conformally flat space-times. (A space-time is said to be conformally flat if it admits a metric conformal to it with zero curvature.) And restricting to the Einstein-vacuum equations, which asks the metric to also be Ricci-flat, means that the only solution for which such a picture can be constructed is Minkowski space. The problem is one of the mass. Consider the simplest case where the metric is the Schwarzschild metric with mass $m$. The Schwarzschild solution is Ricci-flat, but not conformally flat. This means that it has a non-vanishing Weyl curvature. Now, the Weyl curvature is a conformal invariant, and the mass contributes asymptotically to the Weyl curvature on the order of $m/r^3$, where $r$ is the "radial" coordinate of the metric in familiar form. The requirement that the conformal transformation be non-degenerate at the "boundaries at infinity" means that around the boundary at infinity, we have a "regular" coordinate $R'$ which corresponds to $1/r$, therefore in an orthonormal frame adapted to the conformal coordinates, we see that the Weyl curvature decays like $m R'$ in a neighborhood of the region $\{R' = 0\}$.
Now why is this significant? Looking at the compactification procedure for Minkowski space, it is clear that we glued the two copies such that the space-time is reflectively symmetric across the surface $\{R'=0\}$. But such an extension is only possible if the function to be extended has no "odd" parts in its Taylor expansion. To look at it another way, to extend the function (corresponding to Weyl curvature) smoothly to the other side of $\{R' = 0\}$, we need to match the first derivative, which in the case that the first derivative is non-vanishing requires an odd (not even) extension across the surface. This is incompatible with the desire to glue two of the same space-time together along the boundary.
One may object to the desire to glue two of the same space-time in a reflectively symmetric manner. After all, we are gluing the past boundary to a future boundary. I claim that in order to have a good way of even identifying points on the two boundaries, one needs some sort of reflection or time-translation symmetry, and this implies that the gluing will be symmetrical across that surface.
