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Set $z=e^y$. The metric becomesso these are Poincare coordinates on anti-de Sitter spacetime with unit curvature radius. They cover one patch, not the whole global extension.

On the two-dimensional section with the three transverse $x_i$ suppressed, choose embedding coordinates $X_{-1}=\sec \rho \cos T$, $X_0=\sec \rho \sin T$, $X_d=\tan \rho$. The Poincaré transformation isChoose the connected component containing $T=\rho =0$. Its denominator is positive, givingIts two sloping edges are Poincaré horizons, null coordinate boundaries at $z\to \infty$. The right vertical edge is the $z\to 0$ timelike conformal boundary. Substituting the transformation into $z^{-2}(d{z}^2-d{t}^2)$ gives ${\sec }^2\rho (d{\rho }^2-d{T}^2)$, verifying the chart and its null edges.

The full coordinates in this part have $y\in \mathbb{R}$, hence the whole shaded $z>0$ patch. If one additionally retains only the Randall–Sundrum bulk $y\ge 0$ as in Question4, keep $z\ge 1$. The brane $y=0$, or $z=1$, has $\cos T=\cos \rho -\sin \rho$ and is drawn in orange; the retained side contains $T=0$, $\rho <0$. A patch horizon is not a curvature singularity or the timelike conformal boundary. A geodesic can cross it in finite proper time while the patch coordinate $t$ diverges.

An orthonormal coframe writes a metric as $g={\eta }_{ab}{\theta }^a\otimes {\theta }^b$ with constant diagonal signature matrix $\eta$. In Cartan's first structure equation, solveThese are the torsion-free and metric-compatibility conditions and determine the Levi-Civita connection. Then Cartan's second structure equation givesContract $R^a{}_{bad}$ to obtain the Ricci tensor and then contract again for the scalar curvature. This fixes the curvature convention; reversing the definition of the Riemann curvature tensor reverses the resulting curvature signs.

On $z>0$, choose ${\theta }^0=dt/z$, ${\theta }^1=dx/z$, ${\theta }^2=dy/z$, ${\theta }^3=dz/z$, with $\eta =\mathrm{diag}(-1,1,1,1)$. For $a=0,1,2$,A torsion-free metric connection therefore hasFor example, ${\Omega }^a{}_3=-d{\theta }^a=-{\theta }^a\wedge {\theta }^3$, while for $a,b<3$,Metric antisymmetry supplies the remaining components, giving uniformlyThus the metric has constant sectional curvature $-1$. In four dimensions,It is an Einstein manifold, with Einstein constant $-3$ in the stated curvature convention.

These are Poincare coordinates on anti-de Sitter spacetime of unit radius. Constant curvature gives the maximal ten-dimensional local isometry algebra $\mathfrak{so}(3,2)$. This can also be checked directly: set $u=(t,x,y)$, ${\eta }_{ab}=\mathrm{diag}(-1,1,1)$ and $u_a={\eta }_{ab}{u}^b$. The ten independent Killing vector fields areTranslations and Lorentz transformations leave the numerator and $z$ unchanged. A dilation rescales numerator and denominator equally. For $K_a$, direct differentiation gives ${\mathcal{L}}_{K_a}{\eta }^{(4)}=4u_a{\eta }^{(4)}$ and $K_a(z)=2u_{a}z$, so the conformal factor cancels and ${\mathcal{L}}_{K_a}g=0$.

Hence the maximally extended Anti-de Sitter spacetime has connected isometry group locally $S{O}_0(3,2)$, with the appropriate covering group if one unwraps its time coordinate. The displayed coordinates cover only a patch. The ten local generators do not all give globally complete flows preserving that patch; for example special conformal flows can cross its horizon. The manifest complete patch symmetries include the boundary Poincare transformations and positive dilations. This distinguishes local maximal symmetry from the global isometries of a chosen coordinate domain.

The boundary of a chosen Poincare coordinates on anti-de Sitter spacetime patch inside the full spacetime is a null coordinate horizon. It is not the timelike conformal boundary, and not a curvature singularity. Timelike geodesics can cross it in finite proper time while the patch time diverges. In a signed-diameter conformal strip, the patch containing $T=\rho =0$ satisfies $|T|<\pi /2+\rho$.