= Solution
Identify the ambient real four-dimensional vector space with $2\times2$ real matrices by
$$
X(u,v,x,y)=\begin{pmatrix}u+x&y+v\\y-v&u-x\end{pmatrix},\qquad
-\det X=-u^2-v^2+x^2+y^2.
$$
For $A,B\in SL(2,\mathbb R)$, the linear action $X\mapsto AXB^{-1}$ preserves this quadratic form. It gives a group homomorphism to the <special orthogonal group> of signature $(2,2)$; its determinant on matrix space is $(\det A)^2(\det B)^{-2}=1$. The group $SL(2,\mathbb R)$ is connected: an oriented Gram–Schmidt factorization writes every element as a rotation times an upper triangular matrix with positive diagonal and determinant one, and each factor varies in a connected space. Thus the image lies in the identity component $SO_0(2,2)$.
If $AXB^{-1}=X$ for every $X$, taking $X=I$ gives $A=B$. A matrix commuting with every real $2\times2$ matrix is scalar, and its determinant-one condition then gives $A=B=\pm I$. The kernel is therefore exactly $\{(I,I),(-I,-I)\}$. Infinitesimally the action is $X\mapsto aX-Xb$, with $a,b$ traceless. If this vanishes for every $X$, then $a=b$ and it is scalar and traceless, hence zero. The differential is injective; source and target <Lie algebras> each have dimension six, so it is an isomorphism. The image is therefore an open subgroup of the connected target, which must be the entire target. Consequently
$$
\boxed{SO_0(2,2)\cong\frac{SL(2,\mathbb R)\times SL(2,\mathbb R)}{\{(I,I),(-I,-I)\}}.}
$$
The finite quotient is by the diagonal central subgroup, not by a sign in just one factor. The connected-component subscript is a genuine qualification of the source's displayed formula: the full $SO(2,2)$ is disconnected and cannot be isomorphic to a quotient of the connected product.
Let the anti-de Sitter radius be $\ell>0$. Its quadric is $-u^2-v^2+x^2+y^2=-\ell^2$, or $\det X=\ell^2$. Thus the map $X\mapsto g=X/\ell$ identifies this quadric smoothly and globally with $SL(2,\mathbb R)$. This is the <group model of three-dimensional anti-de Sitter spacetime>. The induced <Lorentzian metric> is bi-invariant, since the two matrix multiplication actions preserve the ambient quadratic form. At the identity, tangent matrices $a,b$ are traceless and
$$
g_\ell(a,b)=\frac{\ell^2}{2}\operatorname{Tr}(ab).
$$
Indeed, for a traceless matrix $a$, $\det a=-\frac12\operatorname{Tr}(a^2)$, and polarizing $-\det(\ell a)$ gives this metric. For $\mathfrak{sl}_2(\mathbb R)$ the <Killing form> is $K(a,b)=4\operatorname{Tr}(ab)$: in the basis $H=\operatorname{diag}(1,-1)$, $E=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $F=\begin{pmatrix}0&0\\1&0\end{pmatrix}$, the brackets are $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$, giving $K(H,H)=8$, $K(E,F)=4$ and the remaining independent pairings zero. Therefore $g_\ell=\ell^2K/8$.
The <canonical torsion-free connection on a Lie group> is the <Levi-Civita connection> of this <bi-invariant pseudo-Riemannian metric>. Its <Ricci tensor>, already calculated from the double bracket, is
$$
\boxed{\operatorname{Ric}=-\frac14K=-\frac2{\ell^2}g_\ell,\qquad R=-\frac6{\ell^2}.}
$$
Substitution into the vacuum <Einstein field equations> gives
$$
\operatorname{Ric}-\frac12Rg_\ell+\Lambda g_\ell
=\left(\frac1{\ell^2}+\Lambda\right)g_\ell=0,
\qquad \boxed{\Lambda=-\ell^{-2}.}
$$
Thus the group metric solves the three-dimensional vacuum equations with negative <cosmological constant>.
The antipodal map $X\mapsto-X$ is $g\mapsto-g$, multiplication by the central element $-I$. Hence \b[the antipodal quotient is $PSL(2,\mathbb R)=SL(2,\mathbb R)/\{\pm I\}$], with the same local curvature and cosmological constant. To see the global time issue, use
$$
u=\ell\cosh\rho\cos t,\quad v=\ell\cosh\rho\sin t,\quad
x=\ell\sinh\rho\cos\varphi,\quad y=\ell\sinh\rho\sin\varphi.
$$
The induced metric is
$$
ds^2=\ell^2\bigl(-\cosh^2\rho\,dt^2+d\rho^2+\sinh^2\rho\,d\varphi^2\bigr).
$$
On the quadric $t$ has period $2\pi$; curves with fixed $\rho,\varphi$ are closed timelike curves. The antipodal identification is $(t,\varphi)\sim(t+\pi,\varphi+\pi)$, and does not eliminate this problem. At $\rho=0$, even the half-period time shift closes after the antipodal quotient.
If global time is not identified, take the <universal cover>, with $t\in\mathbb R$. The topology of $SL(2,\mathbb R)$ is $S^1\times\mathbb R^2$, as follows from the same rotation–triangular decomposition, so the universal cover is diffeomorphic to $\mathbb R^3$ and carries the lifted group structure $\widetilde{SL(2,\mathbb R)}$. The metric and Einstein equation are unchanged locally. Here $t$ is a global time function with timelike gradient, excluding closed causal curves. \b[Unwrapped anti-de Sitter spacetime is the universal covering group, rather than the original embedded quadric or its antipodal quotient]; retaining the antipodal time-shift identification would wrap time again.
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