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The AdS/CFT correspondence refers to the "boundary" of AdS space but I'm a little confused about what this means. Typically, one writes the AdS metric in the form $ds^2= \frac{L^2}{z^2}(-dt^2+d\vec x^2+dz^2)$ and then refers to the point $z=0$ as the boundary.

In what sense is this a boundary? AdS is maximally symmetric and so I'd think that there wouldn't be any special regions of it, such as a boundary. Relatedly, I could calculate the Ricci scalar and I'd of course find that it's constant everywhere (so that $z=0$ isn't a special point, in particular) and so the point $z=0$ seems like it's just a artifact of choosing poor coordinates. It appears reminiscent of the coordinate singularity that occurs in the Schwarzschild metric as one approaches the Schwarzschild radius (when using Schwarzschild coordiantes).

So is it truly a boundary? Does it matter?

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AdS is not a manifold with boundary in the standard sense (where neighborhoods of the boundary are diffeomorphic to neighborhoods of points on the boundary of some Euclidean half space). The boundary to which people often refer in this context is the so-called conformal boundary obtained through a conformal compactification of the spacetime.

In the conformal compactification construction, one maps the manifold $M$ being considered onto the interior of a compact manifold $\tilde M$ with boundary, and then one calls the boundary $\partial \tilde M$ of this manifold the conformal boundary of the original manifold.

More details here:

Conformal Compactification of spacetime

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  • $\begingroup$ @naturalnessquestions Sure thing. $\endgroup$ – joshphysics Oct 18 '13 at 3:16

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