Symmetries of AdS$_3$, $SO(2,2)$ and $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$

Question

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 Ref. 1) 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.

References:

1. K. Holsheimer, Surface Charges, Holographic Renormalization, and Lifshitz Spacetime, master thesis, Amsterdam. The PDF file is here.

Answer 1

I) First recall the fact that

$SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ is (the double cover of) the identity component $SO^{+}(2,2;\mathbb{R})$ of the split orthogonal group $O(2,2;\mathbb{R})$.

This follows partly because:

1. There is a bijective isometry from the split real space $(\mathbb{R}^{2,2},||\cdot||^2)$ to the space of $2\times2 $ real matrices $({\rm Mat}_{2\times 2}(\mathbb{R}),\det(\cdot))$, $$\mathbb{R}^{2,2} ~\ni~\tilde{x}~=~(x^0,x^1,x^2,x^3)~\mapsto~ m~=~\begin{pmatrix}x^0+x^3 & x^1 +x^2\\ x^1 -x^2 & x^0-x^3\end{pmatrix}~\in~ {\rm Mat}_{2\times 2}(\mathbb{R}) , $$ $$ ||\tilde{x}||^2 ~=~x^{\mu} \eta_{\mu\nu}x^{\nu} ~=~\det(m).$$ We use here the sign convention $(+,-,+,-)$ for the split metric $\eta_{\mu\nu}$.

2. There is a group action $\rho: SL(2,\mathbb{R})\times SL(2,\mathbb{R})\times {\rm Mat}_{2\times 2}(\mathbb{R}) \to {\rm Mat}_{2\times 2}(\mathbb{R})$ given by $$(g_L,g_R)\quad \mapsto\quad\rho(g_L,g_R)m~:= ~g_L m g_R^t,$$ $$ g_L,g_R\in SL(2,\mathbb{R}),\qquad m\in {\rm Mat}_{2\times 2}(\mathbb{R}), $$ which is length preserving, i.e. $(g_L,g_R)$ is a split-orthogonal transformation. In other words, there is a Lie group homomorphism
   $$\rho: SL(2,\mathbb{R})\times SL(2,\mathbb{R}) \quad\to\quad O({\rm Mat}_{2\times 2}(\mathbb{R}),\mathbb{R})~\cong~ O(2,2;\mathbb{R}) , \qquad \rho(\pm {\bf 1}_{2 \times 2}, \pm {\bf 1}_{2 \times 2})~=~{\bf 1}_{{\rm Mat}_{2\times 2}(\mathbb{R})}.$$

[It is interesting to compare with similar constructions for other signatures, cf. e.g. this Phys.SE post.]

II) Next define the anti-de-Sitter space $AdS_3$ with negative cosmological constant $\Lambda<0$ as the hypersurface

$$ AdS_3 ~:=~ \det{}^{-1}(\{\Lambda^{-1}\})~:=~\{ m\in {\rm Mat}_{2\times 2}(\mathbb{R}) \mid \det(m)~=~\Lambda^{-1}\} $$

endowed with the induced metric. Note that the above group action $\rho$ preserves the anti-de-Sitter space:

$$\rho: SL(2,\mathbb{R})\times SL(2,\mathbb{R})\times AdS_3\quad \to\quad AdS_3.$$

In fact, the Lie group $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ induces in this way global isometries on $AdS_3$.