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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 $z=0$ (and $z=\infty$) 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?

Qmechanic
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user26866
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2 Answers2

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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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joshphysics
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TL;DR: The conformal boundary of (Euclidean) $AdS_{d+1}$ is easiest to analyse in stereographic coordinates (3), cf. Ref. 1. The bulk of $AdS_{d+1}$ is isomorphic to an open ball $B_{d+1}=\{y\in\mathbb{R}^{d+1}\mid y^2<1\}$ Hence the conformal boundary is isomorphic to a sphere $S^d=\{y\in\mathbb{R}^{d+1}\mid y^2=1\}$. Beware that the $AdS$ metric (4) becomes singular on the conformal boundary (5).

Embedding coordinates $(X^{-1},X^M)$ or $(X^{\pm},X^{\mu})$: $$\begin{align} -R^2~=~&-(X^{-1})^2+\sum_{M=0}^dX_M X^M~=~-X^+X^-+\sum_{\mu=0}^{d-1}X_{\mu} X^{\mu},\cr X^{\pm}~:=~&X^{-1}\pm X^d~>~0, \qquad X^{-1}~>~0, \end{align} \tag{1}$$ where $R$ is the $AdS$ radius.

Poincare coordinates $(x^{\mu},z)$: $$\begin{align} X^+ ~=~& \frac{x^2}{z} + z \quad\Rightarrow\quad dX^+ ~=~ \frac{2x_{\mu}dx^{\mu}}{z} - \frac{x^2 dz}{z^2} +dz, \cr X^- ~=~& \frac{R^2}{z} \quad\Rightarrow\quad dX^- ~=~ -\frac{R^2 dz}{z^2}, \cr X^{\mu} ~=~& \frac{Rx^{\mu}}{z} \quad\Rightarrow\quad dX^{\mu} ~=~ \frac{Rdx^{\mu}}{z} - \frac{Rx^{\mu}dz}{z^2},\cr x^2~:=~&\sum_{\mu=0}^{d-1}x_{\mu}x^{\mu}, \qquad z~>~0.\end{align}\tag{2}$$

Stereographic coordinates $(y^M)$: $$\begin{align} X^M ~=~& R\frac{2y^M}{1-y^2} \quad\Rightarrow\quad dX^M ~=~ 2R\frac{dy^M}{1-y^2} + 4Ry^M\frac{y_Ndy^N}{(1-y^2)^2}\cr X^{-1} ~=~& R\frac{1+y^2}{1-y^2}~>~R \quad\Rightarrow\quad dX^{-1} ~=~ 4R\frac{ y_Mdy^M}{(1-y^2)^2} \cr y^2~:=~&\sum_{M=0}^{d}y_My^M~<~1.\end{align}\tag{3}$$ Metric tensor: $$\begin{align} ds^2 ~=~& -dX^{-1}dX^{-1} + dX_M dX^M \cr ~=~& -dX^+ dX^- + dX_{\mu}dX^{\mu}\cr ~=~& R^2\frac{dz^2 + dx_{\mu}dx^{\mu}}{z^2} \cr ~=~& 4R^2 \frac{dy_M dy^M}{(1-y^2)^2}.\end{align}\tag{4}$$ The conformal boundary $S^d$ corresponds to $$y^2~=~1 \quad\Leftrightarrow\quad X^{-1}=\infty \quad\Leftrightarrow\quad z~=~0~\vee~z~=~\infty. \tag{5}$$ It is a conformal compactification of the $x$-spacetime $\mathbb{R}^d$.

References:

  1. E. Witten, Anti De Sitter Space And Holography, arXiv:hep-th/9802150; p.4-5.
  2. J.L. Petersen, Introduction to the Maldacena Conjecture on AdS/CFT, arXiv:hep-th/9902131; p.4.
  3. J. Kaplan, 2013 Lectures on AdS/CFT from the Bottom Up; section 5.2. (NB: Some eqs. in the PDF file are corrupted by some left parentheses.)
Qmechanic
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