# Question about $\rm AdS$ conformal boundary in Poincare coordinates

I've worked $$\rm AdS$$ using global coordinates and the ideas of a conformal boundary is plausible. Radial $$\rho$$ coordinate can be compactified and we can study its conformal boundary at $$\frac{\pi}{2}$$. However, when we go to the coordinate Patch:

$$ds^{2} = \frac{1}{z^{2}}(-dt^{2}+dz^{2}+d\vec{x}^{2})$$

I find very hard to find that $$\rho =\frac{\pi}{2}$$ corresponds to $$z=0$$ since that is the boundary of $$\rm AdS$$ in Poincare Patches. I do not know how to justify that $$z=0$$ is the boundary. I'm even more confused at finding the meaning of $$z=\infty$$. Is there a way to have the poincare patch and automaticaly find that $$z=0$$ is the boundary, without relating it to the global coordinates?

• At $z=0$ the metric blows up indicating clearly that it is an asymptotic boundary of the spacetime. $z=\infty$ is the boundary of the Poincar\'e patch. Remember that the Poincar\'e coordinates do not completely cover AdS so this boundary is fictitious'. Aug 27, 2020 at 18:06

Disclaimer: my notation here differs from yours: My variable $$z$$ is your variable $$1/z$$. So $$z=0$$ for me is $$z=\infty$$ for you and vice-versa.

If we think about the $$AdS_{p+1}$$ space as embedded on $$\mathbb{R}^{2,p}$$ as

$$X^{I}X_{I}=-R^{2}$$

The $$AdS_{p+1}$$ metric becomes the induced metric from $$\mathbb{R}^{2,p}$$, i.e.

$$ds^{2}=dX^{I}dX_{I}$$

The boundary is defined as the $$n_{I}X^{I}\rightarrow \infty$$ for all possible $$n_{I}$$ compatible with the constraint above. If we rescale the $$X^{I}$$ variables in order to maintain them finite as we approach the boundary we obtain a parameterization of the boundary in terms of projective coordinates $$\bar X^{I}$$

$$\bar X^{I}\cong \Lambda \bar X^{I},\qquad \bar X^{I} \bar X_{I} = 0$$

where the radius $$R$$ goes to zero because of the re-scaling.

Now, once we cover a patch of $$AdS_{p+1}$$ by coordinates, what we should look after is the intersection of boundary defined above with our patch. It is not guaranteed that our patch will cover the whole boundary.

The boundary have the topology $$\frac{S^{p-1}\times S^{1}}{\mathbb{Z}_2}$$ where $$S^{1}$$ is a closed time-like curve. Doing a universal cover opens this closed time-like curves into $$\mathbb{R}_t$$, which makes the boundary $$S^{p-1}\times\mathbb{R}_t$$

The Poincaré patch is the patch of $$AdS_{p+1}$$ covered by the Poncaré coordinates:

$$X^{+}=\left(\frac{1}{z}+z\,x^{\mu}x_{\mu}\right),\quad X^{-}=R^{2}z,\quad X^{\mu}=Rz\,x^{\mu}$$

where $$x^{0}$$ is the time coordinate and $$z>0$$. This does not cover the entire $$AdS_{p+1}$$ space but only the patch where $$X^{-}>0$$. At $$X^{-}\rightarrow 0$$, which in our coordinates is $$z\rightarrow 0$$, we have an horizon in which our time coordinate $$x^{0}$$ never cross. You can compare with the situation with the Rindler coordinates for flat space.

The part of the boundary of $$AdS_{p+1}$$ that is contained in our patch is given by $$X^{-}\rightarrow \infty$$, which in our coordinates is $$z\rightarrow \infty$$. This part of the boundary have the topology of $$\mathbb{R}^{p-1,1}$$ which is different than the topology of the whole boundary.

An interesting thing happens when we perform a Wick rotation $$x^{0}\rightarrow ix^{p}$$. The horizon $$z=0$$ closes to a point since the size of the hypersurface defined by holding $$z$$ fixed shrinks as $$z\rightarrow 0$$. The same phenomena happens when we do Wick rotation on Rindler coordinates. The horizon of the Rindler coordinates also shrinks to a point.

This means that the Poincaré coordinates cover almost all the Euclidean $$AdS_{p+1}$$ except singular point at $$z=0$$. This point turns out to be a point of the boundary of Euclidean $$AdS_{p+1}$$ and the topology of the boundary becomes

$$S^{p}$$

What happened is that the Wick rotation maps part $$\mathbb{R}^{p-1,1}$$ of the boundary to $$\mathbb{R}^{p}$$ and the horizon to mapped a point at infinity of $$\mathbb{R}^{p}$$, closing it to a $$S^{p}$$.

You could also make all this claims using the explicit form of the metric

$$ds^{2}=R^{2}\left(\frac{dz^{2}}{z^{2}} + z^{2}dx^{2}\right)$$

and define the boundary to be the place where the metric truly diverges, i.e. places where no change of coordinates can make the metric finite. This is usually what is done in the literature but I prefer to present a different look on that.