One perturbation, two ball readouts

One bulk perturbation supplies both a first area variation and an asymptotic charge for each boundary ball. Their difference integrates the linearized Einstein residual against the ball's Killing weight. Two independent calculations give that identity: one from covariant phase space, one from scalar-curvature variation on the bulk slice. All centers and radii on one slice recover its Hamiltonian constraint. A finite determining family of Lorentz frames, with all slices in each frame, recovers the boundary-index contractions. The Bianchi and Ward identities then complete the local equation. The explicit non-gauge examples below show why recovery of that equation and reconstruction of the metric are different questions.

Read this edition with Second-Order Detection by Boundary Balls. Here the fixed-frame kernel exhibits what first variations can miss. The companion's all-radii canonical-energy result shows how the same family of balls can detect a transverse AdS perturbation at second order.

1. What the ball first law can recover

Faulkner, Guica, Hartman, Myers and Van Raamsdonk showed that the complete family of ball first laws recovers the linearized gravitational equations about pure AdS for first variations around the vacuum [1]. The key identity turns the difference between the two ball readouts into an integral of a local bulk residual. The converse depends on which balls are available and on the conditions at the conformal boundary.

This edition reconstructs the Einstein branch within the source–readout architecture [2]. The vacuum ball and its bulk Killing wedge use the normalization from the spherical modular-transport construction [3], [4]. The family of balls plays a role analogous to the determining null directions of the local horizon construction [5], with spatial inversion as an additional step. The nonlinear spin-two bootstrap through transport and projective reduction [6] starts from different input laws and reaches a different endpoint.

The calculation starts with one common off-shell metric tangent. An elementary local inversion shows what the defects recover, and metric-level examples show what restricted ball families miss. The holographic dictionary, the Einstein area prescription and the boundary first law are inputs. We first calculate the charge defect as a geometric observable and then use the dictionary to identify it with the first-law defect. The reconstruction is at first order about pure AdS and at classical bulk order.

2. Common source and the ball projections

Use boundary dimension $d\geq 2$, $c=\hbar=1$, and Poincaré coordinates:

$$\bar g=\frac{L^2}{z^2}(dz^2+\eta_{\mu\nu}\,dx^\mu dx^\nu), \qquad \Lambda=-\frac{d(d-1)}{2L^2}.$$

Write $E_{ab}=\delta(G_{ab}+\Lambda g_{ab})$ for the unscaled Einstein residual. The metric tangent $h$ has radial gauge $h_{za}=0$, with $h_{\mu\nu}=z^{d-2}H_{\mu\nu}$ and $H$ locally $C^2$ up to $z=0$. Its boundary value is $H^{(0)}$. The balls are bounded, so this geometric class imposes no global spatial decay. The geometric tests accept arbitrary tangents in that class. To compare a tangent with a CFT state, add the state–geometry dictionary and the flat-space Ward conditions:

$$\eta^{\mu\nu}H^{(0)}_{\mu\nu}=0, \qquad \partial^\mu H^{(0)}_{\mu\nu}=0. \tag{1}$$

Those conditions alone do not supply a CFT state for every off-shell test.

One tangent, many apertures

The source history contains the reference state and geometry and their first variations. An aperture is a ball $B$ in a chosen boundary frame together with its background bulk wedge. Each aperture evaluates the same $h$. Set $s=t-t_0$, $y=x-x_0$, and $r=|y|$ in the rest coordinates of $B$. The bulk Killing field is

$$\xi_B=\frac{\pi}{R}\left[(R^2-s^2-r^2-z^2)\partial_t-2s(y^i\partial_i+z\partial_z)\right].$$

Its boundary limit is the ball conformal Killing field. It vanishes at the bifurcation hemisphere $\widetilde B$, given by $s=0$ and $r^2+z^2=R^2$. The upper half-ball $\Sigma_B$ is bounded by $B$ and $\widetilde B$. The factor $2\pi R$ relative to hyperbolic time gives surface gravity $2\pi$, the normalization appropriate to the dimensionless modular energy. On the slice the positive lapse is $V=(L/z)\xi_B^t$.

The one-sided energy projection

For a CFT perturbation the energy projection is

$$\delta E_B^{\mathrm{hol}}=\frac{\pi}{R}\int_B(R^2-r^2)\,\delta\langle T_{tt}\rangle\,d^{d-1}y, \qquad \delta\langle T_{\mu\nu}\rangle=\frac{dL^{d-3}}{16\pi G}H^{(0)}_{\mu\nu}.$$

The weight comes from vacuum ball modular flow. The full vacuum modular implementer includes a complementary action, whereas the first law uses the one-sided variation. With a regulator it is the variation in the expectation of the reference modular Hamiltonian. In the continuum the first law applies to perturbations admitting the relevant first variations, with the relative-entropy derivative vanishing at the reference state. A density matrix for the diamond algebra is unnecessary.

The area projection and its cutoff endpoint

At Einstein order the entropy projection is $\delta S_B=\delta A(\widetilde B)/(4G)$. Varying the metric at fixed embedding gives

$$\delta A(\widetilde B)=\frac{RL^{d-3}}{2}\int_B z^{2-d} \left(\delta^{ij}-\frac{y^iy^j}{R^2}\right)h_{ij}\,d^{d-1}y, \qquad z=\sqrt{R^2-r^2}. \tag{2}$$

In the $y$ chart the induced metric is $L^2z^{-2}(\delta_{ij}+y_i y_j/z^2)$. Its inverse is $z^2L^{-2}(\delta^{ij}-y^iy^j/R^2)$ and the volume density is $RL^{d-1}z^{-d}$. The hemisphere is extremal and the time-slice extrinsic curvature vanishes. To apply the same formula to the perturbed extremal surface, the cutoff endpoint term must also vanish. In the $z$ chart the background radius is $r(z)=\sqrt{R^2-z^2}$, giving radial canonical momentum

$$L^{d-1}z^{1-d}r^{d-2}\frac{r'}{\sqrt{1+(r')^2}} =-\frac{L^{d-1}}{R}r^{d-2}z^{2-d}.$$

The time momentum vanishes on the background. At fixed $z$ cutoff with regular angular parametrization, a uniform radial coordinate displacement $o(z^{d-2})$ suffices for the endpoint term to vanish. Assume that falloff for the anchored first variation. It includes the stronger $O(z^{d+1})$ condition used in the construction. For $d=3$ the sufficient criterion is $o(z)$. It does not exclude other ways for the integrated endpoint to vanish, such as cancellations.