# How to make sense of $\mathcal{I}^-$ as a Cauchy surface rigorously?

In some references, like Hawking's derivation of black hole radiation, one considers that $$\mathcal{I}^-$$ is a Cauchy surface. One recent reference with such a claim is the paper "Soft Hair on Black Holes" (page 7 after equation 2.8), in verbis:

In the absence of stable massive particle or black holes, $$\mathcal{I}^+$$ ($$\mathcal{I}^-$$) is a Cauchy surface

Well by definition:

A Cauchy Surface for the spacetime $$(M,g)$$ is a surface $$\Sigma\subset M$$ such that every inextendible causal curve on the structure $$(M,g)$$ hits $$\Sigma$$ exactly once.

So part of the definition requires $$\Sigma$$ to be a subset of spacetime points. But strictly speaking in a rigorous framework $$\mathcal{I}^-$$ is not a place of the physical spacetime $$(M,g)$$. It is actually defined just on the conformal completion to one unphysical spacetime $$(\hat{M},\hat{g})$$ as the boundary $$\mathcal{I}=\partial \hat{M}$$.

So when one says that $$\mathcal{I}^{-}$$ is a Cauchy surface, how can we actually understand that statement from a rigorous standpoint?

Does it mean that when $$(M,g)$$ is globally hyperbolic so is any unphysical conformal completion $$(\hat{M},\hat{g})$$ and that any inextendible causal curve in $$(M,g)$$ when extended with respect to the $$(\hat{M},\hat{g})$$ structure (on which it is not inextendible) will hit $$\mathcal{I}^-$$ exactly once?

This seems like a problem because timelike geodesics doesn't hit $$\mathcal{I}^-$$. So what is going on here in the end?

• $\mathcal I^-$ is null, so it is also strange to call it a Cauchy surface. – MBN Jan 22 at 16:27
• @MBN, is it not possible for a Cauchy surface to be null? Perhaps that is something very basic that I'm missing. But if that's the case there is even more reason to ask what people are doing when they claim that it is a Cauchy surface. I've added one modern reference with such a claim. – user1620696 Jan 22 at 16:40

## 1 Answer

I think you have to look at the motivation for adjoining $$\mathcal{I}^-$$ and the motivation for the definition of a Cauchy surface.

The reason we adjoin idealized boundary points to a spacetime is to make life more convenient, so we can make definitions and calculations simpler by avoiding special-casing and explicit limits. So it's perfectly natural to want to allow Cauchy surfaces to include boundary points, and mean-spirited to want to forbid it purely because someone stated a definition that doesn't literally allow that.

The reason we define a Cauchy surface the way we do is so that if we describe all the conditions on the Cauchy surface, we can predict the future evolution of the spacetime and the matter fields.

Your source says:

In the absence of stable massive particle or black holes, $$\mathcal{I}^+$$ ($$\mathcal{I}^-$$) is a Cauchy surface

If there are no stable massive particles, then a timelike world-line extending back to $$i^-$$ can never be the world-line of a particle. Therefore we don't need information about $$i^-$$ to give us the initial conditions.

MBN says in a comment:

$$\mathcal{I}^-$$ is null, so it is also strange to call it a Cauchy surface.

Normally we don't want one point on a Cauchy surface to be causally related to another point on it, because then we might have a constraint that would have to be satisfied by the state of the matter fields on the surface. But this doesn't seem like a concern for $$\mathcal{I}^-$$. I guess it is imaginable for a photon to start on $$\mathcal{I}^-$$ and stay on $$\mathcal{I}^-$$, but then it seems like we wouldn't care about that photon, since it would never make an appearance on stage. Maybe there is some way that this is formalized, I don't know: --

Are there null geodesics inside null infinity?

• Thanks. Your point is that $\mathcal{I}^-$ behaves as a Cauchy surface in the sense that one can also get one well-posed initial value problem for massless fields with initial data defined on $\mathcal{I}^-$? – user1620696 Jan 23 at 1:10
• @user1620696: Yes. – Ben Crowell Jan 24 at 16:20