# Why spatial infinity is a point and not an $S^2$?

First a disclaimer, this question already has been asked here, but as pointed out in comments, more detail was required.

So this is a more detailed version. Let $(\mathbb{R}^4,\eta)$ be Minkowski spacetime. Introduced advanced/retarded null coordinates $(u,v,\theta,\phi)$. Intuitively speaking, we would like to define $\mathscr{I}^+$ as $(u,\infty,\theta,\phi)$ and $\mathscr{I}^-$ as $(-\infty,v,\theta,\phi)$ since these would be the endpoints of the null geodesics of interest.

This is of course nonsense, but we get around this compactifying along the null lines. We introduce $U = \arctan u$ and $V = \arctan v$. This yields a coordinate system $(U,V,\theta,\phi)$ for Minkowski spacetime.

Now we can consider the spacetime obtained by letting $U$ and $V$ take on all possible values. Let's call this $N$.

The points $(u,\infty,\theta,\phi)$ thus would correspond to $(u,\pi/2,\theta,\phi)$ which are not in $\mathbb{R}^4$ but are in $N$. So we can actualy define $\mathscr{I}^+$ inside $N$. The same happens to $(-\infty,v,\theta,\phi)$ which is just $(-\pi/2,v,\theta,\phi)$ and albeit not being in $\mathbb{R}^4$ are in $N$ and allows us to define $\mathscr{I}^-$.

Now spatial infinity to me would mean "take $r\to \infty$ with $t$ fixed". Since $u = t-r$ and $v = t+r$ this would be $u\to-\infty$ and $v\to \infty$. This would be $U = -\pi/2$ and $V = \pi/2$. So it seems we should define spatial infinity as the set of points of $N$ with coordinates $(-\pi/2,\pi/2,\theta,\phi)$.

But this is an $S^2$, while everyone says that $i^0$ is a point. My point is that all the transformations of coordinates did not involve $\theta,\phi$ so even for $U=-\pi/2,V=\pi/2$ it seems $\theta,\phi$ can take on any values. So it seems $i^0$ is a set of points, and $S^2$.

What am I missing here? Why $i^0$ is a point and not an $S^2$?

Edit: this claim is also made on a text by Penrose from 1964. He quotes that in coordinates $(U,V,\theta,\phi)$ the metric is:

$$ds^2=\dfrac{1}{\cos^2 U\cos^2 V}\left(dUdV-\frac{1}{4}\sin^2(U-V)(d\theta^2-\sin^2\theta d\phi^2)\right)$$

With the initial range $-\pi/2 < U,V < \pi/2$ this is just an awkward parametrization of Minkowski spacetime.

Now Penrose quotes that $i^0$ given by $U = -\pi/2$ and $V = \pi/2$ in the "bigger spacetime" are points because in the metric $\sin^2 (U-V)=0$ for these coordinates.

But the bigger manifold" with coordinates $U,V$ isn't given. We are trying to find what it is. It is this $N$. Now why the fact that $\sin^2 (U-V)=0$ at $i^0$ implies that it is a point and not an $S^2$?

• $i^0=\mathscr I^+\cap\mathscr I^-$, right? If you take $\mathbb R^3$ with polar coordinates, the point $r=0$ is a point, although in principle $\theta,\phi$ are arbitrary there. The same thing happens here: $i^0$ is a point, even if $\theta,\phi$ can take any value there (they cannot, because they are not well-defined coordinates there; spatial infinity is a singularity of your coordinate system). Or did I misunderstand something? – AccidentalFourierTransform Aug 9 '18 at 3:05
• Maybe it's much easier to picture it on a plane. Take $\mathbb R^2$, with polar coordinates $r,\theta$. "Infinity" is at "$r=\infty$", but this is not an $S^1$, in spite of what it may seem. Indeed, once you compactify to a Riemann sphere, infinity is the north pole, which is clearly a point. The $S^1$ that foliate the plane become singular there -- $r=\infty$ is equivalent to $r=0$ -- and you get a point instead of a circle. – AccidentalFourierTransform Aug 9 '18 at 3:12
• @AccidentalFourierTransform your second argument is circular: you're saying if you compactift by adding a single point, infinity is a single point. The question is, why do you compactify to a sphere and not to a disk (the latter has a circle at infinity) – doetoe Aug 10 '18 at 2:01
• @AccidentalFourierTransform I get your intuition. You say that for all $\theta,\phi$ the coordinates $(-\pi/2,\pi/2,\theta,\phi)$ represent the same point. The issue is how this comes out of the formalism in the question. Up to the point we define $U,V$ we are just changing coordinates in Minkowski spacetime. We then want to extend the ranges of $U,V$ to add in the missing points. My naive view was to extend $U,V$ from $-\infty$ to $\infty$. But I believe somehow that that is the problem. – user1620696 Aug 10 '18 at 2:36
• If we just extend like that (allow $U,V$ to go freely from $-\infty$ to $\infty$ to describe the bigger manifold containing the original spacetime), I see no reason to have all these $(\theta,\phi)$ as a single point. Is the extension the problem? If so how it should be done? – user1620696 Aug 10 '18 at 2:36