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For constant sectional curvature $-1/a^2$, spin curvature is $[{\nabla }_{\mu },{\nabla }_{\nu }]=-{\gamma }_{\mu \nu }/(2a^2)$. The added Clifford term in $\mathcal{D}$ has the opposite commutator, making $\mathcal{D}$ flat. On a simply connected region, parallel transport of any initial spinor is path independent and constructs solutions of $\mathcal{D}ϵ=0$. Thus Anti-de Sitter spacetime and its universal cover admit the associated Killing spinors locally and, with appropriate global spin data, globally.

Embed radius-$L$ Anti-de Sitter spacetime as $X\cdot X=-L^2$ in signature with two timelike directions. Initial tangent $V_P$ satisfies $X_P\cdot V_P=0$ and $V_P\cdot V_P=-1$. The displayed curve remains on the hyperboloid and has unit timelike tangent; its ambient acceleration is normal, proving it is a geodesic. Every such curve reaches $-X_P$ after proper time $\pi L$. On the universal cover, this is a later antipodal event, not a closed causal loop. Global time advances by $2\pi$ over a complete ambient period.

The BTZ black hole is a quotient of three-dimensional Anti-de Sitter spacetime. In lapse-and-shift coordinates its outer and inner horizon radii are $r_{+}$ and $r_{-}$, and its outer-horizon surface gravity is

For the nonrotating BTZ black hole family with $M<0$ and angular period $2\pi$, the circumference of a small circle divided by its proper radius tends to $2\pi \sqrt{-M}$. A smooth axis requires this ratio to be $2\pi$, so only $M=-1$ gives regular Anti-de Sitter spacetime. Other values have a conical singularity, with deficit angle $\delta$ or an angular excess when $\delta <0$. The lapse $r^2/{\ell }^2-M$ is strictly positive, so outward radial null geodesics escape from arbitrarily near the axis: the defect is a naked singularity, despite finite local curvature away from it.

For ambient quadratic form $Q=-x_0^2-x_4^2+x_1^2+x_2^2$, associate $M=\left(\begin{array}{cc}{x}_4+x_1& x_2+x_0\\ x_2-x_0& x_4-x_1\end{array}\right)$. Its determinant is $-Q$, so the unit Anti-de Sitter spacetime hyperboloid is the special linear group of real two-by-two matrices. On its tangent space at the identity, the induced bilinear form is $\frac{1}{2}\mathrm{tr}(UV)$. The Killing form is $4\mathrm{tr}(UV)$, and left and right matrix multiplication preserve the ambient form. Consequently the induced bi-invariant pseudo-Riemannian metric is $B/8$. This model has periodic time; its causally unwrapped version is the universal covering Lie group.

A limit retaining the geometry near a horizon while discarding terms describing its distant asymptotic region. For a regular extremal brane throat, removing the constant from its harmonic function can yield an Anti-de Sitter spacetime factor. Local symmetries and supersymmetries of the limiting geometry need not extend to the full original spacetime.

An FLRW metric with $p=-{\varrho }_0$ and constant negative energy density ${\varrho }_0$ has ${\alpha }^2=-8\pi G{\varrho }_0/3>0$ and obeys ${\dot{a}}^2+k=-{\alpha }^2{a}^2$. Positive solutions exist only for $k=-1$. They are $a={\alpha }^{-1}\sin [\alpha (t-t_B)]$ on $0<\alpha (t-t_B)<\pi$. This is an open slicing of Anti-de Sitter spacetime, whose Ricci scalar is $-12{\alpha }^2$ and Kretschmann scalar is $24{\alpha }^4$. The zeros mark boundaries of this slicing rather than curvature singularities.

Represent the Anti-de Sitter spacetime quadric by $x\cdot x=-{\ell }^2$ in ${\mathbb{R}}^{3,2}$. Appending its timelike unit normal $x/\ell$ to an oriented tangent pseudo-orthonormal frame identifies the frame bundle with $SO(3,2)$. The stabilizer of that normal is $SO(3,1)$, so projection realizes $SO(3,2)/SO(3,1)$ as the base of a principal bundle. For the commonly used universal cover of the anti-de Sitter quadric, pull this bundle back along the covering map; the quadric itself has a periodic timelike coordinate.

A principal bundle with structural Lie group $G$ is locally a product $U\times G$, with changes of trivialization of the form $(x,g)\mapsto (x,t(x)g)$. Define a right action locally by $(x,g)\cdot h=(x,gh)$. It is independent of trivialization because $t(x)(gh)=(t(x)g)h$. It is smooth and obeys the group-action law. Right multiplication in $G$ is free and transitive, so the global action is free and each fiber is a single orbit. Equivalently these properties and equivariant local triviality can be included in the definition of a principal bundle.

For a representation $\rho :G\to GL(V)$, form the associated vector bundleProjection sends $[p,v]$ to the base point of $p$. A local section $s(x)$ gives coordinates $[s(x),v]$ and hence a local product $U\times V$; on overlaps the principal transition functions act through $\rho$. Because these maps are linear in $v$, the fibers have well-defined vector-space structures.

For an $n$-dimensional manifold, its frame bundle has points $u:{\mathbb{R}}^n\to T_{x}M$ that are linear isomorphisms. Right action is composition $u\cdot A=u\circ A$ for $A\in GL(n,\mathbb{R})$. Coordinate frames provide local trivializations. Its associated bundle for the defining representation is the tangent bundle, by the well-defined map $[u,v]\mapsto u(v)$. A metric restricts the frames to linear isometries and reduces the group to $O(r,s)$; choosing orientation gives the corresponding special-orthogonal reduction.

In Minkowski spacetime choose a reference oriented pseudo-orthonormal frame at the origin. Every such frame is specified by its base point $a\in {\mathbb{R}}^{3,1}$ and a Lorentz transformation $L\in SO(3,1)$. The Poincaré group has multiplicationand its right Lorentz action is $(a,L)\cdot h=(a,Lh)$. Henceis the oriented orthonormal frame bundle of Minkowski spacetime. The same argument with $O(3,1)$ gives the full unoriented bundle. If time orientation is fixed, use the orthochronous components throughout; the use of $SO$ in the printed quotient is an orientation reduction.

For four-dimensional de Sitter spacetime, use the spacelike hyperboloid $x\cdot x={\ell }^2$ in ${\mathbb{R}}^{4,1}$. An element of $SO(4,1)$ maps a fixed unit spacelike vector to $x/\ell$, and its other columns give an oriented pseudo-orthonormal tangent frame. The stabilizer of that unit vector is $SO(3,1)$. This identifies the oriented frame bundle of de Sitter spacetime and its projection:The right stabilizer action changes the tangent frame without changing $x$.

For Anti-de Sitter spacetime, take $x\cdot x=-{\ell }^2$ in ${\mathbb{R}}^{3,2}$. A unit timelike normal has orthogonal complement of signature $(3,1)$, so its stabilizer is again $SO(3,1)$. Appending this normal to a tangent frame realizes the oriented orthonormal frame bundle of anti-de Sitter spacetime asConnected components or orientation/time-orientation restrictions must be chosen consistently. This quotient is the anti-de Sitter quadric, which has a periodic timelike coordinate. If “anti-de Sitter spacetime” denotes its universal cover, pull the frame bundle back to that cover, rather than identify the covered spacetime with the unmodified quotient.

Take the causally unwrapped universal cover of Anti-de Sitter spacetime, with curvature radius $L$. In global coordinates,The ordinary radial Penrose diagram is a half-strip: the left edge $\rho =0$ is a regular center and the right edge $\rho =\pi /2$ is a timelike conformal boundary. For displaying signed spatial motion, the figure doubles the radial diameter to $-\pi /2<\rho <\pi /2$. Its two ends represent antipodal spatial directions of the higher-dimensional boundary; they are not a claim that the full AdS boundary has two disconnected components.

The vertical edges are spatial infinity at every finite global time, forming the timelike boundary $\mathbb{R}\times S^{d-2}$. The arrows mark the future and past timelike ends $T\to \pm \infty$, conventionally $i^{+}$ and $i^{-}$. There are no separate null-infinity edges like Minkowski ${\mathcal{I}}^{\pm }$: radial null rays satisfy $dT=\pm d\rho$ and reach the timelike boundary in finite global time. Their physical affine parameter is nevertheless infinite there. The drawing is a finite excerpt of an infinitely tall strip, not an identification of its top and bottom. On the original hyperboloid global time is periodic and there are closed timelike curves; unwrapping it is essential to the usual physical maximally extended AdS spacetime.

Every matrix in $SU(2)$ has the unique formThe four real coordinates of $(\alpha ,\beta )$ give a smooth bijection with smooth inverse onto the unit three-sphere. Matrix multiplication identifies this model with the unit quaternions. Its round metric is invariant under left and right multiplication by unit quaternions. The actionhas kernel $\{(1,1),(-1,-1)\}$: a kernel pair must have $a=b$ from $q=1$ and must commute with all quaternions. Its derivative is injective, and both dimensions are six, givingThe full isometry group is $O(4)$. Quaternion conjugation reverses orientation and supplies the additional component. Thus an arbitrary isometry has either the displayed left-right form or that form composed with conjugation.

For the Anti-de Sitter spacetime model, writeThe hyperboloid with this quadratic form equal to $-1$ is exactly $SL(2,\mathbb{R})$. Its induced Lorentzian metric is preserved by $M\mapsto AM{B}^{-1}$ for $A,B\in SL(2,\mathbb{R})$. At the identity the metric is $g(U,V)=\frac{1}{2}\mathrm{tr}(UV)$ on traceless matrices, so it is the Killing metric divided by eight. This proves the geometric identification, including the metric rather than only the underlying manifold.

A trivial left-right action again forces $A=B$ to be a scalar commuting with all determinant-one matrices, hence $A=B=\pm I$. The differential is injective and both dimensions are six, soThis is the existing left-right double cover of SO0(2,2). The full hyperboloid isometry group is $O(2,2)$. Its other components can be obtained by also using $M\mapsto M^T$ and $M\mapsto CM{C}^{-1}$, $C=\mathrm{diag}(1,-1)$. Transposition reverses one timelike coordinate; conjugation by $C$ reverses one timelike and one spacelike coordinate, so these generate the four component classes. All isometries of these constant-curvature quadrics come from ambient orthogonal transformations: a value and an orthonormal derivative at one point determine an isometry, and the ambient group realizes every such choice.

Here ${\mathrm{AdS}}_3$ means the hyperboloid with periodic time. The physically common causally unwrapped spacetime is its universal cover, modeled by the universal covering Lie group of $SL(2,\mathbb{R})$, and is not itself $SL(2,\mathbb{R})$. Its isometries lift the above local symmetries, with the global covering and central identifications changed accordingly.

Use cosmic time $t$ here; derivatives with respect to conformal time will be used from Question 2 onward. Conservation of pressureless matter and radiation in cosmology makes ${\rho }_m{a}^3$ and ${\rho }_r{a}^4$ constant. DefineThe Friedmann equation is a mechanical energy equation,For nonnegative dust and radiation, at least one nonzero, $V$ tends to $-\infty$ as $a\to 0$ and to $+\infty$ as $a\to \infty$, whileThus every energy line meets the potential once. An expanding solution reaches this turning point and then contracts: $a_{tt}=-V^{\prime }(a)/2<0$. The Friedmann effective potential for matter, radiation and vacuum is sketched below, alongside the positive-vacuum case.

To bound the full cosmic time from the initial zero of $a$ to its final zero, use the Friedmann acceleration equation:Put maximum expansion at $t=0$, with $a(0)=a_{\max }$ and $a_t(0)=0$. The forced-oscillator identity isUntil the first zero and for $0\le t\le \pi /(2\omega )$, the kernel is nonnegative and the bracket is nonpositive. Hence $a(t)\le a_{\max }\cos (\omega t)$, so contraction cannot remain at positive $a$ beyond $\pi /(2\omega )$. Applying the same argument backward from maximum expansion bounds the expanding half. Therefore the negative-vacuum Friedmann lifetime bound isAt either endpoint, nonzero dust or radiation has divergent density. If both are absent, the same estimate instead bounds the open Anti-de Sitter spacetime coordinate patch; its coordinate zeros are not a finite lifetime of the maximally extended vacuum spacetime.

Use ambient quadratic form $Q=-(X^0{)}^2-(X^4{)}^2+(X^1{)}^2+(X^2{)}^2$. The printed sine coordinate has a duplicated superscript: it must be $X^2$, not another $X^1$. With that correction the embedding lies on $Q=-1$. Put $s=\sqrt{1+r^2}$. Differentiating the two timelike coordinates giveswhereas the two spatial coordinates give $(d{X}^1{)}^2+(d{X}^2{)}^2=d{r}^2+r^{2}d{\theta }^2$. Their difference is the induced Lorentzian metricThe polar coordinate $\theta$ collapses at $r=0$; this is a coordinate degeneracy, not a degeneracy of the metric.

For the matrix model of anti-de Sitter three-space, defineThen $\det M=-Q$. This linear identification of ambient space with real two-by-two matrices identifies $Q=-1$ with $SL(2,\mathbb{R})$. Its inverse recovers each $X$ by sums and differences of matrix entries, so the identification is global. The identity corresponds to $X^4=1$ and the other coordinates zero. For a traceless tangent matrix $U$, the induced squared norm isPolarization gives $g_I(U,V)=\frac{1}{2}\mathrm{tr}(UV)$.

The Killing form of ${\mathfrak{sl}}_2(\mathbb{R})$ is $B(U,V)=4\mathrm{tr}(UV)$. This normalization can be checked without assuming it: for $H=\mathrm{diag}(1,-1)$, $E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $F=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, the brackets are $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$. Taking traces of their three-dimensional adjoint matrices gives $B(H,H)=8$, $B(E,F)=B(F,E)=4$, and all other basis pairings zero. These agree with $4\mathrm{tr}(UV)$.

Both left and right multiplication by matrices in $SL(2,\mathbb{R})$ preserve $-\det M$, hence its polarized ambient bilinear form and the induced metric. Equality at the identity therefore propagates throughout the group:In particular the Killing-form Einstein metric calculation gives $\mathrm{Ric}=-2g$, consistent with unit-radius three-dimensional Anti-de Sitter spacetime.

Every element of $SO(2,2)$ preserves the ambient quadratic form, hence the hyperboloid and its induced metric, and so acts by isometries. Define the left-right double cover of SO0(2,2) byThis is a group homomorphism, since successive actions multiply the two factors in the same order. It preserves $\det M$; as a transformation of the four-dimensional matrix space it has determinant $(\det A{)}^2(\det B^{-1}{)}^2=1$. Thus it takes values in $SO(2,2)$.

If its action is the identity, applying it to $M=I$ gives $A=B$. Commuting with every matrix then forces $A$ to be scalar. The condition $\det A=1$ leaves only $A=I$ or $A=-I$. HenceThe differential is $(a,b):M\mapsto aM-Mb$. A zero differential implies $a=b$ and that $a$ is scalar; tracelessness makes $a=b=0$. It is therefore an isomorphism onto the target Lie algebra, since both dimensions are six. The image is an open subgroup. Since $SL(2,\mathbb{R})$ is connected, the image is connected and is exactly the identity component $S{O}_0(2,2)$: an open subgroup of a connected group is also closed and must be the whole group. Thus the precise covering statement isIt is not onto the disconnected full $SO(2,2)$. For example, conjugation $M\mapsto DM{D}^{-1}$, $D=\mathrm{diag}(1,-1)$, sends $(X^0,X^4,X^1,X^2)$ to $(-X^0,X^4,X^1,-X^2)$ and belongs to the other component of $SO(2,2)$. It reverses the orientation of the negative two-plane and is not in the image of the connected source.

The Adjoint representation of a Lie group of $SL(2,\mathbb{R})$ preserves its Killing form of metric signature $(2,1)$ and gives a second double cover,Its kernel is the scalar centre; its differential is injective and both dimensions are three, so the same connected-image argument proves surjectivity onto the identity component. The full $SO(2,1)$ again has an additional component. Thus $S{O}_0(2,1)$ is locally the same Lie group geometry as $SL(2,\mathbb{R})$, with its Killing metric, and globally is its central quotient.

With the printed periodic time, the group model has topology $S^1\times {\mathbb{R}}^2$ and closed timelike curves, since varying $t$ at fixed spatial position gives a timelike circle. The causally unwrapped version of Anti-de Sitter spacetime instead has $t\in \mathbb{R}$ and is the universal covering Lie group. It must not be substituted silently for the periodic spacetime specified here.

For bosonic embedding coordinates $X^M(\sigma )$, the supermembrane action truncates toThe first term is its induced worldvolume volume and the second is its Wess-Zumino brane coupling. A membrane of the opposite orientation reverses $q$.

The M2-brane supergravity solution admits the harmonic functionwith arbitrary centre positions $y_a$ in eight transverse dimensions. Away from the sources, the Laplacian of each term is zero. The unrestricted positions describe static separated membranes carrying aligned charges, so their separations have no potential. Physically, their gravitational attraction is cancelled by their four-form charge repulsion. This is the M2-brane probe no-force identity, which can also be checked directly without relying only on the existence of the multicentre solution.

Keep the field-strength ordering printed in the paper and use $F=dA$. If ${\mathrm{vol}}_{012}=d{x}^0\wedge d{x}^1\wedge d{x}^2$, a gauge vanishing at infinity isThe last equality uses the sign acquired by moving a one-form past a three-form. For this convention the parallel probe has $q=-1$. In static gauge, at a fixed transverse position, its induced worldvolume metric is $h_{ij}=H^{-2/3}{\eta }_{ij}$, hence $\sqrt{-\det h}=H^{-1}$ andUsing the opposite convention for $F$, $A$ and orientation together yields the same cancellation. Changing only the probe charge produces an anti-membrane, with nonconstant potential $T_2(2H^{-1}-1)$. Thus the orientation cannot be dropped from the argument.

The cancellation also survives an expansion for slow transverse motion. With $v^2=|\dot{y}{|}^2$,There is no static potential and no position dependence in the leading kinetic coefficient.

Near an isolated positive membrane charge, write $H\sim R^6/r^6$ with $r=|y-y_a|$ and $R^6=Q_a$. The regular membrane near-horizon limit removes the asymptotically flat constant and givesSet $z=R^3/(2r^2)$. ThenThus the Anti-de Sitter spacetime radius is half the sphere radius. In orthonormal units the four-form has magnitude $|F|=6/R$ along the AdS volume form, the Freund-Rubin compactification flux.

The full asymptotically flat solution has spinors $ϵ=H^{-1/6}{ϵ}_0$ subject to one oriented membrane projector, leaving sixteen real parameters. In the throat, the Killing spinor equation separates into the standard AdS and sphere equations with correlated signs fixed by the flux. There are four real AdS${}_4$ solutions and eight real $S^7$ solutions for the required signs, soThis is maximal supersymmetry of AdS4 times S7: the additional sixteen spinors need not extend through the asymptotically flat region. A multicentre geometry therefore remains globally sixteen-supersymmetric even though each regular local throat has this enhancement. The classical supergravity description is controlled when $R$ is large compared with the eleven-dimensional Planck length; singular harmonic multipoles not representing regular positive membrane centres do not automatically have the same throat.

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.

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.

Assume the level set is regular and non-null. The latter hypothesis is necessary: a null hypersurface has a degenerate induced metric and hence no ordinary intrinsic Levi-Civita connection determined by that metric. Normalize the normal vector and define the non-null hypersurface projection byHere $h$ denotes the induced metric restricted to tangent vectors, while its ambient extension annihilates $n$. The intrinsic covariant derivative is $D_{X}Y=P{\nabla }_{X}Y$ for tangent vector fields. Projection of the ambient torsion-free connection and metric compatibility shows that $D$ is the Levi-Civita connection of $h$. Define the extrinsic curvature by $K(X,Y)=g({\nabla }_{X}n,Y)$, or $K_{ab}=P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$. It is symmetric: differentiating the defining function twice and projecting removes the derivative of its normalization, leaving a multiple of its symmetric Hessian. This sign convention makes the scalar second fundamental form $g(n,{\nabla }_{X}Y)=-K(X,Y)$.

Let $S$ be the tangent linear map characterized by $h(SX,Y)=K(X,Y)$. Differentiating $g(n,n)=\epsilon$ and $g(n,Y)=0$ gives the Gauss formulaFix the Riemann curvature tensor conventionApplying the Gauss formula twice givesSubtracting the expression with $X,Y$ interchanged, and then subtracting the bracket derivative, provesThus the Gauss–Codazzi equations for a non-null hypersurface give the intrinsic Riemann curvature tensor asAll free indices here are tangent to the hypersurface. The normal component of the same calculation isAfter antisymmetrizing and using $D_{X}Y-D_{Y}X=[X,Y]$, this yields the curved-ambient Codazzi equationThe extrinsic curvature measures how tangent planes turn in the ambient manifold; the Gauss equation says that this turning contributes to intrinsic curvature even when the ambient Riemann curvature tensor is zero.