# Why do we need 3 variables to parametrize $\scr{I}^\pm$ in a Penrose diagram?

In the figure we can see the Penrose diagram for Minkowski space If I understand correctly, $$i^-$$ and $$\scr{I}^-$$ have coordinates $$r=\infty$$ and $$t=-\infty$$ while $$i^+$$ and $$\scr{I}^+$$ have coordinates $$r=\infty$$ and $$t=+\infty$$. I would think, then that they can be parametrized with two variales, namely, the angles in a two-sphere. However, in many textbooks (like this review by Strominger https://arxiv.org/abs/1703.05448, on page 13) they claim that, for example. $$\scr{I}^-$$ is a three dimensional surface that can be thought of as the product of a two-sphere and a null line. Where is this extra degree of freedom to parametrize the null infinity coming from?

My attempt at an answer: The only thing I can think of is that, although $$r$$ and $$t$$ are infinite at the boundaries, there's still the degree of freedom to choose the ratio between them. I'm thinking that the worldline of a massive particle can end up at $$r=\infty$$ and $$t=\infty$$ as well as the worldline of a massless particle, but the relation between $$r$$ and $$t$$ will be different in each case since the massless particle followed a null geodesic all the way to infinite while the massive particle followed a timelike geodesic. Is this correct?

• The point is that $v = t + r$ is constant along ingoing null geodesics. Changing coordinates to $(v,r,x^A)$ where $x^A$ are coordinates on a sphere, one reaches $\mathcal{I}^-$ by fixing $(v,x^A)$ and taking the radial coordinate to infinity. In this sense, past null infinity is formally the endpoints of ingoing null geodesics infinitely far away. More rigorous discussions involve a conformal compactification to bring null infinity to finite coordinate values on a bigger manifold. Details of the rigorous approach can be found in Wald's GR book. – user1620696 Nov 29 '19 at 23:05
• The only difference between your self-answer and @user1620696's comment is that you talk about the ratio of r to t, while they talk about t+r, which seems more accurate. It seems to me that we also get the right answer by simply adding two to the dimensionality of every feature on the diagram, which makes sense because a 2-sphere's worth of symmetry is hidden in the diagram. – user4552 Nov 30 '19 at 1:06
• I think it is relevent also to ask why it is different for spacelike and timelike infinites. – MBN Dec 2 '19 at 10:13

the regions $$\scr{I}^\pm$$ are reached by travelling on a light ray (i.e. a null geodisic) which satisfies $$r=\pm ct + r_0$$ where $$c$$ is the speed of light and $$r_0$$ is just the initial position at $$t=0$$. This means that geodisics going towards $$\scr{I}^+$$ will have a constant $$u=r-ct$$ while geodisics going towards $$\scr{I}^-$$ will have constant $$v=r+ct$$. This null variables basically label all the different null geodesics and are a good coordinate to parametrize the null infinity.
In conclusion, in order to fully parametrize $$\scr{I}^\pm$$ you need to give the angular variables in the 2-sphere plus $$u$$ (if it's $$\scr{I}^+$$) or $$v$$ (if it's $$\scr{I}^-$$ to label which null geodesic did you use to get there.
Here's another way to think about it. It's true that both $$r$$ and $$t$$ go to infinity, but the surface is still three dimensional for the following reason. If $$u = t - r$$, and $$r \to \infty$$ while $$u$$ remains fixed, then $$t = u + r$$ should be thought of as $$t$$ "shifted by an infinite constant." While $$r$$ is "set" to infinity, bringing the dimension down from $$4$$ to $$3$$, $$t$$ is merely "shifted" by infinity, keeping the dimension fixed at $$3$$. A similar thing happens at the black hole horizon.