In a previous post I described a method of thinking about conformal compactifications, and I mentioned in passing that, in principle, the method should also apply to arbitrary signature pseudo-Euclidean space $\mathbb{R}^{p,q}$. A few days ago while visiting Oxford I had a conversation with Sergiu Klainerman where this came up, and we realised that we don't actually know what the conformal compactifications are! So let me write down here the computations in case I need to think about it again in the future.
Parametrise our pseudo-Euclidean space by coordinates $(t_1,\ldots,t_p,x_1,\ldots,x_q)$ where the $t$ directions are time-like and the $x$ directions are space-like. Using the method introduced in the previous article, we add to it two additional dimensions $(s,y)$ where $s$ is time-like and $y$ space-like to form $\mathbb{R}^{p+1,q+1}$. Let $C\subseteq \mathbb{R}^{p+1,q+1}$ be the null cone $\{|t|^2 + s^2 = |x|^2 + y^2\}$ and let $\Sigma$ be its intersection with $\{ s = y + 1\}$. This slice can be parametrised by \[ (t,x) \mapsto (s = \frac{x^2 - t^2 +1}{2}, t, y = \frac{x^2 - t^2 - 1}{2}, x)\] and one easily checks that the map above defines the obvious isometric diffeomorphism between $\mathbb{R}^{p,q}\to \Sigma$.
Now consider the intersection of $C$ with the $p + q + 1$ dimensional sphere given by $\{ |t|^2 + |x|^2 + s^2 + y^2 = 2\}$. For this set which we denote by $P$, the algebraic relation reduces to \[ s^2 + |t|^2 = 1\] and \[ y^2 + |x|^2 = 1\] giving the topology $\mathbb{S}^p \times \mathbb{S}^q$ for this set.
We see, however, this is again a double cover. For every $(s,t,x,y)\in C$ where $s\neq y$, there exists exactly one $\lambda \in \mathbb{R}\setminus \{0\}$ such that $\lambda(s,t,x,y) \in \Sigma$. In $P$ however, every point and its antipode correspond to the same line in $C$.
We also see that the set $\{s = y\}$ in $C$ (which, by the way, requires $|t|^2 = |x|^2$ to hold) is the conformal boundary. $\{s = y\}$ is a plane in $\mathbb{R}^{p+1,q+1}$ and separates it into two connected portions. It also splits $P$ by keeping only one copy of a point or its antipode. Let us look at the $s > y$ half.
Starting with $(t,x)$ we generate in $\Sigma$ the point with $s= \frac{x^2 - t^2 + 1}{2}$ and $y = \frac{x^2 - t^2 - 1}{2}$; note that this satisfies $s > y$. It thus suffices to find the positive scaling $\lambda$ to bring this point to the sphere of radius 2. We have that \[ t^2 + s^2 + x^2 + y^2 = t^2 + x^2 + \frac12( x^4 + t^4 - 2 x^2 t^2 + 1) = \frac{2}{\lambda^2}\] or \[\lambda = 2 \left( 2x^2 + 2 t^2 + x^4 + t^4 - 2 x^2 t^2 + 1\right)^{-1/2}.\] So let us write \[ (S,T,Y,X) = \lambda(s,t,y,x) \]
Let's see where some familiar sets map to: * $\{t = 0\} \implies S = 1, T = 0, Y = \frac{x^2 - 1}{x^2 + 1}, X = \frac{2x}{x^2 + 1}$ * For $t$ fixed, taking $|x| \to \infty$ brings us to $S\to 1, Y\to 1$. So there we have space-like infinity. * For $x$ fixed, taking $|t| \to \infty$ brings us to $S,Y \to -1$. That is time-like infinity. * Let $x = x_0 + x'\tau$ and $t = t_0 + t'\tau$ where $|t'|^2 = |x'|^2= 1$, and take $\tau\to \infty$ gives us that $S,Y \to \frac{x_0\cdot x' - t_0 \cdot t'}{\sqrt{ (x_0\cdot x' - t_0 \cdot t')^2 + 1}}$, while $(X,T) \to \frac{(x',t')}{\sqrt{(x_0\cdot x' - t_0 \cdot t')^2 + 1}}$.
From this we see that the Carter-Penrose diagram of $\mathbb{R}^{p,q}$ depends on the values of $p,q>0$. When $p = q = 1$ the diagram is exactly a square. When exactly one of $p,q$ is 1 we get the familiar diagram for the physical Minkowski space. And when $p,q > 1$ we have that the diagram looks like a isosceles right triangle: the two legs are the axes $|t| = 0$ and $|x| = 0$ respectively, and each point in the interior of the triangle corresponds to a manifold of type $\mathbb{S}^{p-1}\times\mathbb{S}^{q-1}$.
