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  Conformal compatification of Minkowski and AdS

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How do I show that the compactification of Minkowski is given by the quadric uvηijxixj=0

with an overall scale equivalence in the coordinates.I get that for v0, 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 u0. 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, uvηijxixj=1

if I have to show that the boundary of AdSd+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

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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,xi)λ(u,v,xi),λ0

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 v0, you may scale v using the equivalence above to v=1, and u is determined by the quadric. So the v0 part of the conformal compactification may be parameterized by xi, 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 ηijxixj=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 xi in the Minkowski space. You may also visualize these new v=0 points differently. If you scale the vector (u,v,xi) with v=0 so that you get v=1, both u and xi will be scaled to infinite values. More precisely, imagine v=ϵ, u=cu/ϵ, xi=cix/ϵ. Here cu is calculated from the quadric but the point is that we are adding classes of points cxi/ϵ 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 S1×S1. The Penrose causal diagram looks like I1×I1, 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 ηijxixj scaling like 1/ϵ2 but the conformal compactification only picks those where this scales as 1/ϵ.

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 AdSd+1 has the symmetry SO(d,2) from the (by one) higher-dimensional space where the hyperboloid is embedded.

This post imported from StackExchange Physics at 2014-07-01 10:35 (UCT), posted by SE-user Luboš Motl
answered Jul 1, 2014 by Luboš Motl (10,278 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 AdSd+1 is Minkowski in d dimensions ?

This post imported from StackExchange Physics at 2014-07-01 10:35 (UCT), posted by SE-user Sourav
@Sourav / The AdSd+1 could be considered as defined by the equation uvηijxixj=1 in a (d+2) flat space defined by the metrics ds2=dudvηijdxidxj. Now, to obtain the metrics of the AdSd+1 space, you may express du as a function of v,xi,dv,dxi. When v+, you will find that du0, and by consequence dudv0. So, when v+, the metrics of the AdSd+1 space goes like ds2AdSηijdxidxj, that is a d-dimensional Minkowski space.

This post imported from StackExchange Physics at 2014-07-01 10:35 (UCT), posted by SE-user Trimok
Dear Sourav, the boundary of the global AdS, the hyperboloid above, is actually Sd1×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

This post imported from StackExchange Physics at 2014-07-01 10:35 (UCT), posted by SE-user Luboš Motl

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