# Symmetry transformation in AdS space

In AdS/CFT papers the action of the SO(D,2) symmetry is usually given at the boundary where the transformations are just the conformal transformations (Poincare, scaling and special) for D+1 Minkowskian space.

I would like to know how the SO(1,2) transformations acts for arbitary point in AdS, lets say in coordinate system

$ds^2 = 1/z^2(dz^2+dx^2+dy^2)$.

Is is possible to write $z'=z(z,x,y)$, $x'=x(z,x,y)$ and $y'=y(z,x,y)$ so that the above metric stays invariant?

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You wrote a geometry with the Euclidean $(+++)$ signature which surely can't have that symmetry group. –  Luboš Motl Jun 13 '13 at 13:09
Your space is the hyperbolic space $H^3$ which has a global $SO(3,1)$ symmetry. $H^3$ could be written as $\frac{SO(3,1)}{SO(3)}$. $H^3$ could be considered as a "euclidean" version of $Ads3$ –  Trimok Jun 13 '13 at 13:22
OK! lets take $ds^2=1/(z^2)(-dt^2+dx^2+dy^2+dz^2)$ which surely is $AdS4$ ($SO(3,2)$). Can you write down transformations for that so that metric stays invariant? –  jancore Jun 13 '13 at 14:17

To see the whole symmetry, it is better to see $AdS4$ as a 4-surface in a 5-dimensional space.

We suppose that the 5-metrics is $ds^2 = dx^2 + dy^2 + dz^2 - dt^2 - du^2$

The 4-surface $AdS4$ is defined as :

$x^2 + y^2 + z^2 - t^2 - u^2 = -1$

With this definition, the whole $SO(3,2)$ symmetry, for transformations $(x,y,z,t,u) \rightarrow (x',y',z',t',u')$ is obvious.

Now, from the intrinsic 4-metric $ds^2=1/(z^2)(+dx^2+dy^2+dz^2-dt^2)$ some invariant-metrics transformations are easy to see:

• Three translations for the coordinates $t,x,y$
• Scale transformation : multiplying $t,x,y,z$ by a constant term $\lambda$
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Thanks Trimok! Where I can find transformations that are NOT SO easy to see? :) –  jancore Jun 13 '13 at 18:24
This is a good question.... If you accept infinitesimal transformations $dx,dy,dt \rightarrow dx',dy',dt'$, there is a $SO(2,1)$ symmetry. But probably someone else would have better ideas. –  Trimok Jun 13 '13 at 18:30