Two proofs of the off-shell charge identity

3. The off-shell charge identity

For arbitrary admissible geometric tangents define the asymptotic spatial charge

$$\delta E_B^{\mathrm{sp}}= \frac{dL^{d-3}}{16\pi G}\frac{\pi}{R} \int_B(R^2-r^2)\sum_i H^{(0)}_{ii}\,d^{d-1}y.$$

The difference between this charge and the holographic energy is the weighted boundary trace. The trace Ward identity suffices to make the energies equal. Conversely, equality for all balls on a slice implies pointwise vanishing of the trace there: divide by the integral of the weight and take the small-radius limit. Boundary conservation enters later.

Theorem 3.1. Defect as a weighted residual

For the stated geometric class, $D_B=\delta S_B-\delta E_B^{\mathrm{sp}}$ obeys

$$D_B=-\frac{1}{8\pi G}\int_{\Sigma_B}\xi_B^t E_{tt}(L/z)^{d-1}\,dz\,d^{d-1}y =-\frac{1}{16\pi G}\int_{\Sigma_B}V\,\delta R[q]\,dv_{\bar q}. \tag{3}$$

Here $q$ is the spatial metric variation on $\Sigma_B$ and $\delta R[q]$ is the scalar-curvature variation about $\bar q$ in direction $q$. In another frame the same formula uses the time-time contraction in that frame.

Covariant phase-space proof

The Einstein Lagrangian has presymplectic potential $\Theta$ and Noether charge $Q_\xi$:

$$\Theta=\frac{\epsilon_a}{16\pi G}\left(\bar\nabla_b h^{ab}-\bar\nabla^a h^b{}_b\right), \qquad Q_\xi=-\frac{\epsilon_{ab}}{16\pi G}\nabla^a\xi^b.$$

The contracted volume forms and index raising in the displayed background expressions use $\bar g$; the variation of $Q_\xi$ includes its full metric dependence. Define $\chi=\delta Q_\xi-\xi\cdot\Theta$ at fixed $\xi$. The linearized Noether identity about an on-shell background gives

$$d\chi=-\frac{1}{8\pi G}\xi^a E_{ab}\epsilon^b.$$

This is the Einstein specialization of the covariant phase-space identity [7], [1]. Vary the Noether current and subtract the derivative of $\xi\cdot\Theta$. The symplectic current drops out because $\mathcal L_\xi\bar g=0$.

At the bifurcation surface the pullback gives $\delta A/(4G)$: $\xi=0$ there and its normal boost has surface gravity $2\pi$. At the boundary the limiting pullback density is $\xi_B^t dL^{d-3}\operatorname{tr}_{\mathrm{sp}}H^{(0)}/(16\pi G)$, where $\operatorname{tr}_{\mathrm{sp}}H^{(0)}=\sum_i H^{(0)}_{ii}$. Both the radial derivative of $h$ and the variation of the volume form contribute. Stokes' theorem with charge orientation $\widetilde B-B$ gives the first equality in (3).

Independent slice proof

The spatial metric $\bar q=L^2z^{-2}(dz^2+dy^2)$ is time symmetric. Its Gauss constraint gives $E_{tt}=(L/z)^2\delta R/2$, so the slice constraint depends only on $q$. Let $D$ be the connection of $\bar q$ and define

$$U^A=V\left(D_Bq^{AB}-D^A\operatorname{tr}q\right) -q^{AB}D_BV+(\operatorname{tr}q)D^AV.$$

Integration by parts at the differential level yields

$$V\delta R-D_AU^A =q^{AB}\left(D_AD_BV-\bar q_{AB}\Delta V-V\operatorname{Ric}_{AB}\right).$$

Direct differentiation of $V=(\pi L/Rz)(R^2-r^2-z^2)$ gives $D_AD_BV=V\bar q_{AB}/L^2$. Since $\operatorname{Ric}_{AB}=-(d-1)\bar q_{AB}/L^2$, the right side vanishes. On the hemisphere $V=0$, and its outward derivative is $-2\pi$, so $U\cdot\nu=-2\pi\operatorname{tr}_{\widetilde B}q$.

On the face $z=\varepsilon$ the outward normal is $-z\partial_z/L$. The flux density tends to $dL^{d-3}\xi^t\sum_i H^{(0)}_{ii}$. To see the coefficient, write $\operatorname{tr}q=L^{-2}z^d\sum_i H_{ii}$. Radial gauge gives $D_Bq^{zB}=(z/L^2)\operatorname{tr}q$. The two leading $V$ terms in the flux contribute $d-1$ and $1$ times the same boundary density. The sum of the two surface fluxes is $16\pi G(\delta E_B^{\mathrm{sp}}-\delta S_B)$, proving the second equality in (3) independently of the spacetime form calculation.

The two routes give the same numerical normalization for the same tangent. The slice coefficient depends on the Einstein action normalization and the ball's surface gravity. Substituting the unit hyperbolic-time generator in only one of the charges changes the balance.