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:
- K. Holsheimer, Surface Charges, Holographic Renormalization,
and Lifshitz Spacetime, master thesis, Amsterdam. The PDF file is here.
Best Answer
I) First recall the fact that
This follows partly because:
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 $\eta_{\mu\nu}$.
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 $AdS_3$.