+ 3 like - 0 dislike
485 views

1. Is the only necessary ingredient that the isometry group of AdS matches the conformal group in one dimension less or are there other prerequisites to build a holographic connection?

2. How does one demonstrate that the isometry group of $AdS_{d+1}$ is $SO(d,2)$? I cannot find any references that do this explicitly which is what I need. Is this because it takes a long time to show this?

This post imported from StackExchange Physics at 2016-01-06 09:18 (UTC), posted by SE-user user11128

+ 3 like - 0 dislike

1) That's not the only ingredient -- it's a prerequisite for holography. In reality, with holographic duality one always means a precise mapping from observables in a gravity theory in AdS to observables in a CFT that lives on the boundary. So holography is much richer: it prescribes for example how you can compute a Wilson loop on the boundary CFT in terms of gravity.

2) No, it's in fact almost trivial. $AdS_{d+1}$ with radius $R$ can be defined as the solution to

$$\eta_{\mu \nu} X^\mu X^\nu = R^2$$

with $\eta_{\mu \nu} = (1,1,-1,\ldots,-1)$ and where $X^\mu$ lives in $\mathbb{R}^{d+2}$. But $SO(2,d)$ is precisely the group that leaves the quadratic form $\eta_{\mu \nu} X^\mu X^\nu$ invariant.

If this is too abstract, think of the sphere $S^2$. It can be defined as the set of points $X^\mu \in \mathbb{R}^3$ that obey

$$\delta_{\mu \nu} X^\mu X^\nu = R^2.$$

Its isometry group is $SO(3)$ because this leaves $X^2$ invariant.

This post imported from StackExchange Physics at 2016-01-06 09:18 (UTC), posted by SE-user Hans Moleman
answered Jan 6, 2016 by (30 points)
+ 0 like - 0 dislike

1. From the requirement of matching physical observables on both sides of the correspondence one can deduce a more general principle behind matching symmetries: global symmetries corresponding to Noether currents that are in principle observable have to match. This does not only include spacetime symmetries like the conformal group, but for instance extance applies to supersymmetric R-symmetry. The latter happens to correspond to the isometry group of the compact manifold ($S^5$ in the case of $AdS_5\times S^5$).
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.