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An orthonormal coframe is a collection of independent differential forms $E^{\hat{a}}=e^{\hat{a}}{}_{\mu }d{x}^{\mu }$ satisfying $g_{\mu \nu }={\eta }_{\hat{a}\hat{b}}{e}^{\hat{a}}{}_{\mu }{e}^{\hat{b}}{}_{\nu }$, with $\eta =\mathrm{diag}(-1,1,1,1)$. Locally one can diagonalize the metric tensor and rescale the resulting covectors, or use Gram-Schmidt orthogonalization for a symmetric bilinear form, choosing a timelike direction first. The coframe is not unique: a local Lorentz transformation preserves the metric. For the Levi-Civita connection, solve Cartan's first structure equation and metric compatibility together:These determine the connection 1-forms. The antisymmetry applies to the lowered pair of indices; a mixed time-space pair instead has ${\omega }^0{}_i={\omega }^i{}_0$. Next Cartan's second structure equation gives the curvature 2-forms, and reading off their coefficients determines the Riemann curvature tensor:This also states the curvature-sign convention. Contracting gives the Ricci tensor, and its coordinate components are $R_{\mu \nu }=e^{\hat{a}}{}_{\mu }{e}^{\hat{b}}{}_{\nu }{R}_{\hat{a}\hat{b}}$.

On a static patch with $V>0$, chooseWrite $h=\sqrt{V}$, $a=h^{\prime }$, $b=h/r$, and $c=\cot \theta /r$. The exterior derivative givesThe nonzero independent connection 1-forms are consequentlywith ${\omega }^1{}_0={\omega }^0{}_1$, spatial mixed pairs antisymmetric, and the remaining forms zero. For example,Similarly ${\Omega }^0{}_2={\omega }^0{}_1\wedge {\omega }^1{}_2=-ab{E}^0\wedge E^2$, and the angular derivatives in $d{\omega }^1{}_3$ cancel against ${\omega }^1{}_2\wedge {\omega }^2{}_3$. The resulting Cartan curvature forms for a static spherical metric areFor the last entry, $d(-c{E}^3)=E^2\wedge E^3/r^2$ and ${\omega }^2{}_1\wedge {\omega }^1{}_3=-V{E}^2\wedge E^3/r^2$. The remaining curvature 2-forms follow from ${\Omega }_{ab}=-{\Omega }_{ba}$.

Put $A=V^{\prime \prime }/2$, $B=V^{\prime }/(2r)$, and $C=(1-V)/r^2$. Reading the Riemann curvature tensor from the forms and contracting gives the diagonal orthonormal Ricci tensorFor example $R_{00}=R^1{}_{010}+R^2{}_{020}+R^3{}_{030}=A+2B$, while $R_{22}=-B-B+C$. All off-diagonal entries vanish because the curvature 2-forms only involve their own index planes. In the coordinate basis the nonzero components areThe angular Einstein field equations with the negative cosmological constant give $(rV{)}^{\prime }=1+3r^2/l^2$. Integrating introduces one real constant, conventionally written $-2m$:Substitution gives $V^{\prime \prime }/2+V^{\prime }/r=3/l^2$, so the time and radial Einstein field equations hold as well. The solution is the Schwarzschild--anti-de Sitter black hole family, including its zero and negative mass limits.

For $m>0$, $V$ increases strictly from $-\infty$ at $r=0$ to $+\infty$ at infinity. Its unique positive simple zero $r_h$ satisfies $r_h^3+l^2{r}_h=2m{l}^2$. This zero is a coordinate singularity, not a curvature singularity: with the ingoing coordinate $v=t+\int dr/V$, the radial metric is $-Vd{v}^2+2dvdr$ and is regular there. Inside the future black-hole region, $\nabla r={\partial }_v+V{\partial }_r$ is future timelike, so every future-directed causal trajectory decreases $r$ and cannot return to the asymptotic anti-de Sitter boundary. Conversely, every exterior radius admits an outward radial null geodesic reaching that boundary, since ${\int }_r^{\infty }dr/V$ converges. Thus $m>0$ gives an event horizon at the unique positive zero of $V$. Here the event horizon is defined relative to the asymptotic timelike boundary of Anti-de Sitter spacetime. For $m=0$, $V=1+r^2/l^2>0$ and the spacetime is pure Anti-de Sitter spacetime, with no black-hole event horizon. For $m<0$, $V=1+2|m|/r+r^2/l^2>0$, so there is also no event horizon; the mass term produces a naked singularity at $r=0$. The field equation itself does not impose a positive mass integration constant.