Basically, I want to know how one can see the $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ symmetry of AdS$_3$ explicitly.

AdS$_3$ can be defined as hyperboloid in $\mathbb{R}^{2,2}$ as
$$
X_{-1}^2+X_0^2-X_1^2-X_2^2=L^2
$$
where $L$ is the AdS radius. Since the metric of $\mathbb{R}^{2,2}$,
$$
ds^2=-dX_{-1}^2-dX_0^2+dX_1^2+dX_2^2,
$$
is invariant under $SO(2,2)$ transformations and also the hyperboloid defined above is invariant we can conclude that AdS$_3$ has an $SO(2,2)$ symmetry.

One can probably show with pure group theoretical arguments that the $SO(2,2)$ symmetry is isomorphic to an $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ symmetry. I would like to know however, if one can see this symmetry more explicitly in some representation of AdS$_3$?

I suppose a starting point might be, that one can write the hyperboloid constraint equation as
$$
\frac{1}{L^2}\text{det}\;\begin{pmatrix} X_{-1}-X_1 & -X_0+X_2 \\ X_0+X_2 & X_{-1}+X_1\end{pmatrix}=1
$$
i.e. there is some identification of the hyperboloid with the group manifold of $SL(2,\mathbb{R})$ itself. However, that does not tell us anything about the symmetries.

The only explanation that I have found (on page 12 of this pdf Master thesis) was that the group manifold of $SL(2,\mathbb{R})$ carries the Killing-Cartan metric
$$
g=\frac{1}{2}\text{tr}\,\left(g^{-1}dg\right)^2
$$
which is invariant under the actions
$$
g\rightarrow k_L\, g \qquad\text{and}\qquad g\rightarrow g\, k_R
$$
with $k_L,k_R\in SL(2,\mathbb{R})$. But how does one get from the metric on $\mathbb{R}^{2,2}$ to this Killing-Cartan metric? Also, I don't find this very explicit and was wondering if there is a more direct way.

This post imported from StackExchange Physics at 2014-10-27 19:55 (UTC), posted by SE-user physicus