# Conformal Compactification of spacetime

I have been reading Penrose's paper titled "Relativistic Symmetry Groups" where the concept of conformal compactification of a space-time is discussed. My other references have been this and this. In case you cannot see the paper above, let me describe what I understand so far:

The idea of conformal compactification is to bring points at "infinity" on a non-compact pseudo-Riemannian manifold $M$ (equipped with metric $g$) to a finite distance (in a new metric) by a conformal rescaling of the metric ${\tilde g} = \Omega^2 g$. This is done so that $(M,{\tilde g})$ can be isometrically embedded into a compact domain ${\tilde M}$ of another (possibly non-compact) pseudo-Riemannian manifold $M'$. This allows us to discuss the asymptotic behaviour of the manifold under consideration.

In the references above, they mention the following without explanation:

Then observe that any regular extension of $\phi$[$=\Omega^2$] 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”.

This I do not understand. Firstly, what is meant by conformal boundary? Secondly, why should $\Omega = 0$ on the conformal boundary?

Is there any good reference for this material?

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Often (like in the case of the standard compactifications of $\mathbb R^{d,1}$ or $\mathrm{AdS}_{d+1}$), the non-compact manifold $M$ that we start with is mapped to the interior of a compact manifold $\tilde M$ that also happens to be a manifold with boundary. In these cases, we define the conformal boundary as $\partial\tilde M$ as indicated by Ben Crowell.

Note, however, that this does not always happen. Consider, for example, the standard stereographic projection which maps $M=\mathbb R^2$ onto $\tilde M=S^2\setminus \{(0,0,1)\}$ where in my notation I'm treating the sphere as an embedded submanifold of $\mathbb R^3$ with north pole $(0,0,1)$. Notice, in this case, that $\tilde M$ is not the interior of a compact manifold with boundary; when we include the north pole, we get $S^2$ which is a compact manifold without boundary.

In contrast, consider $\mathbb R^{d,1}$ with metric $$ds^2 = -dt^2 + dr^2 + r^2 d\Omega_{d-1}^2$$ where $d\Omega_{d-1}^2$ is the metric on $S^{d-1}$. Let $\vec\theta$ be coordinates on the sphere, then the diffeomorphism $$f(t,r,\vec\theta) = (T(t,r,\vec\theta), R(t,r,\vec\theta), \vec\theta)$$ where \begin{align} T(r,t,\vec\theta) &= \tan^{-1}(t+r) + \tan^{-1}(t-r)\\ R(r,t,\vec\theta) &= \tan^{-1}(t+r) -\tan^{-1}(t-r) \end{align} leaves the sphere factor unchanged but maps all of the $(r,t)$ plane to the interior of the triangular region in the $(R,T)$ plane satisfying $$0\leq R \leq \pi, \qquad |T|\leq \pi-R$$ This region does have a boundary (the edges of the triangle) which allows us to define the conformal boundary of $\mathbb R^{d,1}$.

As for the $\Omega^2 = 0$ constraint, this is how I think of it intuitively (and pretty imprecisely). The new metric $\tilde g$ on the compact manifold is related to the original metric $g$ by the conformal factor: $$\tilde g = \Omega^2 g$$ Now in the original manifold, as you go out to infinity, $g$ allows for distances between points to be arbitrarily large. But after the compactification, all points will be some finite distance away from one another. In order for this to happen, distances between points need to be multiplied by a smaller and smaller number as you go further and further out so that the product remains finite. The factor $\Omega^2$ multiplying $g$ is precisely what does this for you. This roughly is what the quote means when it says

This reflects the property of a conformal compactification that “brings infinity to a finite distance."

By the way, I found it useful to explicitly go through the $\mathrm{AdS}_{d+1}$ example. In particular, you can, for example, verify for yourself that for the explicit mapping written above, the conformal factor is $$\Omega(t,r,\vec\theta)^2 = \frac{1}{\frac{1}{4}(1+(r-t)^2)(1+(r+t)^2)}$$ which vanishes as $r,t\to\infty$

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Thank you. That explained a lot. –  Prahar Jun 14 '13 at 18:53
@Prahar Sure thing; I know the literature on this stuff can definitely be improved. I really can't emphasize enough how useful it is in this context to get your hands dirty by explicitly computationally going through some standard conformal compactifications. Anyway cheers! –  joshphysics Jun 14 '13 at 19:02
Yes! I agree with you. That's what I've been doing. –  Prahar Jun 14 '13 at 20:47

Firstly, what is meant by conformal boundary?

I think the conformal boundary is being defined in the quoted material, as $\partial \tilde{M}$. This powerpoint seems to confirm that this is the definition. In more lowbrow terms, I think it refers to idealized surfaces such as $\mathscr{I}^+$, $i^0$, etc.

Secondly, why should Ω=0 on the conformal boundary?

I think pp. 2-3 (pp. 11-12 in the pdf) of the master's thesis explain this. It's in order to satisfy #2 on the list of requirements. The general idea is that according to the correspondence principle, no point in spacetime is supposed to differ in its local properties from any other point. If we didn't satisfy requirement #2, then a point P on the boundary would have special properties, e.g., the space of points lying at a finite affine distance from P would have a lower dimension than the dimension of the manifold.

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Minor comment to the answer (v2): It is best to supply title, author, etc, of link, so we can reconstruct the link in case of future link rot. –  Qmechanic Jun 18 '13 at 20:05