What is the Conformal Boundary of AdS Space?

[kau]

Somehow I can't relate two things and confused over this.
What I understand when someone say that some spacetime has conformal boundary it means that I can write the metric conformally to some other metric where the coordinates are finite ..So it has boundary.
Now I just read something on Ads Conformal boundary which i can't understand much.
Consider (d+2) dim spacetime with two negative eigenvalue of the metric and imposde the following condition
$-x0^{2}+ \Sigma{ x^{i 2} }- x^{{d+1} ^{2}} = -L^{2}$ doing this give you the AdS space.
Now to understand the conformal boundary of this spacetime the logic that is put forward is the following:
For large $X^{M}$ this $-x0^{2}+ \Sigma{ x^{i 2} }- x^{{d+1} ^{2}} = -L^{2}$ The reason behind that I think since we have positive and negative sign. So in large value limit that contributes very small quantity which we can assume to be zero. (please correct me if I am wrong in this statement) . But the condition is it has to become -L^2 to be a part of AdS.. So in some sense it has to have some end somewhere.
And then they defined the boundary as the set of points which is on null geodesic originating from the centre of (d+2) dim spacetime and then end at null cone at infinity. Can someone explain this part ?

 

[samalkhaiat]

Somehow I can't relate two things and confused over this.
What I understand when someone say that some spacetime has conformal boundary it means that I can write the metric conformally to some other metric where the coordinates are finite ..So it has boundary.
Now I just read something on Ads Conformal boundary which i can't understand much.
Consider (d+2) dim spacetime with two negative eigenvalue of the metric and imposde the following condition
$-x0^{2}+ \Sigma{ x^{i 2} }- x^{{d+1} ^{2}} = -L^{2}$ doing this give you the AdS space.
Now to understand the conformal boundary of this spacetime the logic that is put forward is the following:
For large $X^{M}$ this $-x0^{2}+ \Sigma{ x^{i 2} }- x^{{d+1} ^{2}} = -L^{2}$ The reason behind that I think since we have positive and negative sign. So in large value limit that contributes very small quantity which we can assume to be zero. (please correct me if I am wrong in this statement) . But the condition is it has to become -L^2 to be a part of AdS.. So in some sense it has to have some end somewhere.
And then they defined the boundary as the set of points which is on null geodesic originating from the centre of (d+2) dim spacetime and then end at null cone at infinity. Can someone explain this part ?

Reading the above, it is not at all clear to me how much you know about the conformal group $C(1,n-1)$ and its global action.

1) Globally, the conformal group $C(1,n-1)$ acts not on the Minkowski space $\mbox{M}^{(1,n-1)}$ but on its conformal compactification $\mbox{M}_{c}^{(1,n-1)}$. This is an n-dimensional compact manifold isomorphic to $S^{n-1} \times S^{1} / \mathbb{Z}_{2}$.

2) The basic idea behind Ads/CFT is the fact that the conformal boundary of $\mbox{Ads}_{n+1}$ is a 2-fold covering of $\mbox{M}_{c}^{(1,n-1)}$, i.e. $\partial(\mbox{Ads}_{n+1}) = S^{n-1} \times S^{1}$.

If you understand where the above two points come from, then it is easy to understand the relation $\mbox{M}^{(1,n-1)} \cong \mbox{M}_{c}^{(1,n-1)} - \{ \mathcal{K}_{\infty} \}$, where $\{ \mathcal{K}_{\infty} \} \subset \mathbb{R}^{(2,n)}$ is the set of points at infinity (projective cone).