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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

Use the convention ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ and $R_{ab}=R^c{}_{acb}$. Each independent two-form has only one coordinate-plane wedge, so the Ricci tensor is diagonal. Contracting the six coefficients givesBut $2B+F=2(1+q/2)+(1-q)=3$, soIn coordinate-independent form, $R_{\mu \nu }=-3g_{\mu \nu }$. Thus the vacuum Einstein field equations $G_{\mu \nu }+\Lambda g_{\mu \nu }=0$ hold withThe curvature radius is one in the metric's normalization. The parameter $\alpha$ changes the individual curvature 2-forms but cancels from their Ricci contraction. For $\alpha =0$ this is Anti-de Sitter spacetime; for nonzero $\alpha$ it is the planar Einstein metric with cubic radial function, with a nontrivial Weyl tensor rather than universally constant sectional curvature. The result is local on regular coordinate patches; a zero of $F$ invalidates this particular static coframe, not the tensor equation in a suitable regular extension.

In the Polonyi model, the Kähler metric is $K_{z\bar{z}}=1$. The relevant Kähler covariant derivative of a superpotential isThe supergravity auxiliary field is $F^z=-e^{|z{|}^2/2}\overline{D_{z}W}$, up to an irrelevant common phase convention. Thus a constant vacuum preserves supersymmetry exactly when this auxiliary field vanishes. For $m=0$, the superpotential and scalar potential vanish identically, and every constant scalar value is a supersymmetric vacuum.

For $m\ne 0$, put $⟨z⟩=x+iy$. Its supersymmetry condition becomesSince $\beta >0$, it requires $y=0$ and $x^2+\beta x+1=0$. Hence the Polonyi supersymmetry branches areAt these points $W\ne 0$, so the supergravity F-term potential is negative, $V=-3e^K|W{|}^2$: these are supersymmetric Anti-de Sitter spacetime vacua, not zero-energy ones. The condition $D_{z}W=0$ also makes them stationary, as follows by differentiating the supergravity F-term potential.

For $0<\beta <2$ and $m\ne 0$, the auxiliary field cannot vanish anywhere, so any vacuum has supersymmetry breaking. For $\beta \ge 2$, a stationary vacuum at any other scalar value still breaks supersymmetry; the parameter condition alone does not determine which vacuum is selected. In particular, a nontrivial zero-energy vacuum cannot preserve supersymmetry: $D_{z}W=0$ and $V=0$ would also imply $W=0$, whereas $W=0$ gives $z=-\beta$ and $D_{z}W=m^2\ne 0$.

For an affine parameter $\lambda$, the geodesic Lagrangian isVarying $S=\int \mathcal{L}d\lambda$ gives the geodesic equations

The rotational Killing vector fields of the spherical symmetry conserve the angular-momentum vector. Its direction is fixed, so every nonradial orbit lies in the plane through the symmetry centre orthogonal to that vector. A spatial rotation can make this the equatorial plane. Equivalently, the initial conditions $\theta =\pi /2$ and $\dot{\theta }=0$ solve the $\theta$ equation uniquely. This is the planarity of geodesics in spherical symmetry; a radial geodesic has zero angular momentum and may be assigned any plane.

On $\theta =\pi /2$, the cyclic coordinates $t$ and $\varphi$ produce geodesic conserved quantities from Killing vectors,For a timelike geodesic parametrized by proper time, the proper-time normalization is $g_{ab}{\dot{x}}^a{\dot{x}}^b=-1$. Substitution of $E$ and $L$ gives the first-order radial equationThis is the timelike geodesic effective potential for static Anti-de Sitter spacetime with curvature radius one.

A geodesic passing through $r=0$ must have $L=0$. If it moves away from the origin, $E>1$, andTaking proper time $s=0$ at departure givesIt turns around at $s=\pi /2$ and first returns to the origin atThis energy-independent refocusing is the characteristic radial timelike geodesic in anti-de Sitter spacetime result; for curvature radius $a$, the answer is $\pi a$.

The state–operator correspondence maps $\mathcal{O}$ to the lowest one-particle state on $S^4\times \mathbb{R}$. On a sphere of radius $r$, cylinder energy equals scaling dimension divided by $r$. The primary has $\Delta =5$.

A translation generator raises the dimension by one and transforms as a vector of $SO(5)$. States with no angular momentum arise from scalar descendant pairs $P^2$, so the allowed descendants areTheir dimensions are $5+2n$, and hence the one-quantum, zero-angular-momentum energies areThese are the $l=0$ normal-mode energies of a massless scalar in global Anti-de Sitter spacetime.