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  Conformal compatification of Minkowski and AdS

+ 4 like - 0 dislike

How do I show that the compactification of Minkowski is given by the quadric $$uv-\eta_{ij}x^{i}x^{j}=0$$ with an overall scale equivalence in the coordinates.I get that for $v \neq 0$, the surface can be parametrized with the Minkowski coordinates. Now for $v=0$, I can have arbitrary values of $u$, which means basically two values, $ u=0 $ and $u\neq 0$. So are the infinities mapped to these points ? After that is it obvious that the conformal group acts on the space time defined by the quadric ?

From, $$uv-\eta_{ij}x^{i}x^{j}=1$$ if I have to show that the boundary of $AdS_{d+1}$ is Minkowski in $d$ dimensions, how do I take the limit ?

This post imported from StackExchange Physics at 2014-07-01 10:35 (UCT), posted by SE-user Sourav
asked Jul 1, 2014 in Theoretical Physics by Sourav (20 points) [ no revision ]

1 Answer

+ 2 like - 0 dislike

The conformal compactification is supposed to belong to the projective space so we still identify points along the rays (equivalence classes under scaling) $$ (u,v,x_i)\sim \lambda (u,v,x_i), \quad \lambda \neq 0$$ Then there is the quadric equation you wrote down – an equation that respects the identification above – so both added variables $u,v$ are pretty much removed.

For $v\neq 0$, you may scale $v$ using the equivalence above to $v=1$, and $u$ is determined by the quadric. So the $v\neq 0$ part of the conformal compactification may be parameterized by $x_i$, just like you said.

For $v=0$, we clearly get "new points" that are added to the Minkowski space. So the resulting space isn't quite the same as the Minkowski space. It has new points. If it were exactly the same, we wouldn't call it "the conformal compactification of the Minkowski space" but just "the Minkowski space" (in other coordinates).

The points you get for $v=0$ may have an arbitrary $u$ but you still have the equation which reduces to $$ \eta_{ij} x^i x^j = 0$$ and the scaling equivalence. The latter allows us to set $u=1$, for example. So the points added from $v=0$ are in one-to-one correspondence with the null vectors $x^i$ in the Minkowski space. You may also visualize these new $v=0$ points differently. If you scale the vector $(u,v,x^i)$ with $v=0$ so that you get $v=1$, both $u$ and $x^i$ will be scaled to infinite values. More precisely, imagine $v=\epsilon$, $u=c_u/\epsilon$, $x^i= c_x^i / \epsilon$. Here $c_u$ is calculated from the quadric but the point is that we are adding classes of points $c^x_i/\epsilon$ which are at infinity – both in the spacelike, timelike, and null directions. Some special discussion would be needed to describe the topology near the null transition between the two regions etc.

These special subtleties make the interpretation of the conformal compactification a bit complicated. The result is simple for the 1+1-dimensional Minkowski space. The conformal compactification is actually $S^1\times S^1$. The Penrose causal diagram looks like $I^1\times I^1$, the product of two line intervals (the diamond) but the conformal compactification completes it and connects the endpoints of both line intervals to produce a circle. Because of the Penrose causal diagram discussion, we may see that we're not really adding points at infinity from "generic directions" but only those that are close to the light cone. Generic points at infinity would have $\eta_{ij}x^ix^j$ scaling like $1/\epsilon^2$ but the conformal compactification only picks those where this scales as $1/\epsilon$.

If you add $1$ on the right hand side of the quadratic equation, you don't get just a "conformal compactification of the AdS space". You get the AdS space itself! The global AdS space may be defined as the hyperboloid given exactly by the equation you wrote down, without any scaling identifications in this case (the equation isn't scale invariant due to the $1$ term on the right hand side, so it would be impossible). And indeed, one may see that $AdS_{d+1}$ has the symmetry $SO(d,2)$ from the (by one) higher-dimensional space where the hyperboloid is embedded.

answered Jul 1, 2014 by Luboš Motl (10,268 points) [ no revision ]
Yeah, that was a mistake, that is the AdS space itself, not compact. But how do I show from that equation, that the boundary of $AdS_{d+1}$ is Minkowski in $d$ dimensions ?

@Sourav / The $AdS_{d+1}$ could be considered as defined by the equation $uv-\eta_{ij}x^{i}x^{j}=1$ in a $(d+2)$ flat space defined by the metrics $ds^2= du \,dv - \eta_{ij}dx^{i}dx^{j}$. Now, to obtain the metrics of the $AdS_{d+1}$ space, you may express $du$ as a function of $v, x^i, dv, dx^i$. When $v \to +\infty$, you will find that $du \to 0$, and by consequence $du \,dv \to 0$. So, when $v \to +\infty$, the metrics of the $AdS_{d+1}$ space goes like $ds^2_{AdS}\approx \eta_{ij}dx^{i}dx^{j}$, that is a $d$-dimensional Minkowski space.

Dear Sourav, the boundary of the global AdS, the hyperboloid above, is actually $S^{d-1}\times R$, the spatial dimensions are compact. The signature is Minkowskian, it is locally Minkowskian, and it is exactly Minkowskian if one only takes the Poincare patch of the AdS, but for the whole AdS, the conformal boundary includes the sphere. This is well-explained at the beginnings of lectures on AdS/CFT e.g. arxiv.org/abs/hep-th/9905111


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