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With the convention $K_{ij}=({\dot{h}}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$ for extrinsic curvature, differentiation of the gravitational Lagrangian gives ${\pi }^{ij}=\sqrt{h}(K^{ij}-K{h}^{ij})$ in units $16\pi G=1$. It is a weight-one tensor density. In three spatial dimensions, $\pi =-2\sqrt{h}K$, soThe lapse function and shift vector have no time derivatives in this action and consequently have vanishing conjugate momenta. Changing the sign convention for extrinsic curvature changes the corresponding momentum relation; it must not be silently combined with this formula.

Using the lapse function $N$, shift vector $N^i$ and spatial metric $h_{ij}$, the Legendre transform in mechanics of the Einstein-Hilbert action gives, in units $16\pi G=1$,Here $\pi =h_{ij}{\pi }^{ij}$ and ${\pi }^{ij}$ is the canonical momentum of the spatial metric. Varying lapse and shift imposes the Hamiltonian constraint and momentum constraint. An asymptotic time translation requires an Arnowitt-Deser-Misner energy boundary term. The bulk constraints generate gauge evolution; the nonzero asymptotic energy comes from the boundary.

For a positive lapse function $\alpha$, shift vector ${\beta }^i$, and the convention $K_{ij}=-({\partial }_t{\gamma }_{ij}-{\mathcal{L}}_{\beta }{\gamma }_{ij})/(2\alpha )$ for extrinsic curvature of a spatial hypersurface, the time-coordinate harmonic slicing condition yields the displayed advective lapse equation. Combining ${\partial }_t(\sqrt{\gamma }/\alpha )={\partial }_i(\sqrt{\gamma }{\beta }^i/\alpha )$ with the spatial volume evolution identity proves it without setting the shift vector to zero.

For a positive lapse function $\alpha$, shift vector ${\beta }^i$ and spatial induced metric ${\gamma }_{ij}$, the block metric has $g_{00}=-{\alpha }^2+{\gamma }_{ij}{\beta }^i{\beta }^j$ and $g_{0i}={\gamma }_{ij}{\beta }^j$. Its block determinant is $g=-{\alpha }^2\det \gamma$. Therefore the metric volume tensor contracted with the future unit normal restricts to the spatial volume tensor: $n^{\mu }{ϵ}_{\mu ijk}=\sqrt{\det \gamma }[ijk]$.

With negligible shear, a minimally coupled scalar field serving as a clock, and $\Pi \ne 0$, the Hamiltonian constraint and momentum constraint imply $\dot{H}=-4\pi GN{\Pi }^2$ and ${\partial }_{i}H=-4\pi G\Pi {\partial }_i\varphi$. Differentiating the local Friedmann relation yields $\dot{\Pi }=-N(3H\Pi +V_{,\varphi })$ and ${\partial }_i\Pi =-(3H+V_{,\varphi }/\Pi ){\partial }_i\varphi$. Inserting these identities into ${\dot{\zeta }}_i$ cancels all terms, including gradients of the lapse function. The conserved covector becomes the gradient of a uniform-density curvature perturbation at linear adiabatic order. Dropping the shear constraint is essential; this is not a proof of conservation in an arbitrary non-attractor phase.

Retain the homogeneous lapse function $\bar{N}(t)$, and define the physical Hubble parameter $H=\dot{a}/(\bar{N}a)$. At first order the scalar metric perturbations give $\delta g_{00}=-2{\bar{N}}^2\Phi$, $\delta g_{0i}=a^2{B}_{,i}$, and $\delta g_{ij}=-2a^2(\Psi {\delta }_{ij}+E_{,ij})$. Applying the passive coordinate displacement component by component gives the scalar gauge transformations with a background lapseFor example the $0i$ transformation is $\delta {\tilde{g}}_{0i}=a^2{\partial }_{i}B-a^2{\partial }_i\dot{\lambda }+{\bar{N}}^2{\partial }_i{\xi }^0$. The spatial isotropic part acquires $-2a\dot{a}{\xi }^0{\delta }_{ij}$, explaining the positive sign in the transformation of $\Psi$.

Preserving synchronous gauge requires both transformed $\Phi$ and $B$ to remain zero. The first condition is ${\partial }_t(\bar{N}{\xi }^0)=0$, and the second is $\dot{\lambda }={\bar{N}}^2{\xi }^0/a^2$. Integrating them givesThus fixing the lapse function and shift vector perturbations does not exhaust the scalar coordinate freedom. The functions $C,D$ describe residual synchronous-gauge freedom; an integration-origin change in the integral is absorbed into $D$.

To move from synchronous gauge to Newtonian gauge, choose $\lambda =-E_S$ to set $\tilde{E}=0$. Requiring $\tilde{B}=0$ then givesA scalar energy density transforms as $\delta {\rho }_N=\delta {\rho }_S-\dot{\bar{\rho }}{\xi }^0$. Dividing by the background energy density, which changes the contrast only at second order if perturbed in that denominator, yields the synchronous-to-Newtonian density transformation with a lapseThe last form uses background stress-energy conservation, $\dot{\bar{\rho }}=-3\bar{N}H(\bar{\rho }+\bar{P})$. With cosmic time $\bar{N}=1$ this is ${\delta }_N={\delta }_S-3(1+w)a^{2}H{\dot{E}}_S$; with conformal time $\bar{N}=a$ it is ${\delta }_N={\delta }_S-3(1+w)\mathcal{H}{E}_S^{\prime }$, where $\mathcal{H}=a^{\prime }/a$. Keeping these clock conventions distinct is essential to the factors of $a$.

Now impose the stipulated negligible shear and work where $\Pi \ne 0$. The Hamiltonian constraint, momentum constraint and local expansion evolution giveTime differentiation of the first equation, followed by $\dot{\varphi }=N\Pi$, gives $\dot{\Pi }=-N(3H\Pi +V_{,\varphi })$. Spatial differentiation of the same equation and use of the momentum constraint givesThese are the temporal and spatial identities needed for the nonlinear curvature covector. Differentiate its definition, keeping the spatially varying lapse function:The terms proportional to ${\partial }_{i}N$ cancel. The terms $-N{\partial }_{i}H$ and $-4\pi GN\Pi {\partial }_i\varphi$ cancel by the momentum constraint, and the remaining two terms cancel by the spatial momentum identity. Thereforeto the retained gradient order. No homogeneous-lapse assumption was needed.

For linear adiabatic cosmological perturbations, expansion of the local scale and scalar field gives ${\zeta }_i={\partial }_i[\Psi +\bar{H}\delta \varphi /\bar{\Pi }]$. For nonzero spatial modes, linearizing the Hamiltonian constraint and the momentum constraint givesThusThis uses the paper's overall sign for the uniform-density curvature perturbation; the opposite overall sign is also common. For an adiabatic single-field attractor, superhorizon conservation of single-field comoving curvature allows the primordial amplitude to be transported through later epochs without following every short-scale process. This proof specifically discarded the shear contribution to the momentum constraint and assumed a usable field clock. It is not a claim that every single-field background conserves every curvature mode: growing curvature perturbation in ultra-slow-roll inflation is a non-attractor counterexample, and $\Pi =0$ makes this particular covector undefined.

The harmonic equation, including critical points. All derivatives in this argument use the three-dimensional Levi-Civita connection of $\gamma$. Write $U_i={\nabla }_{i}U$, $|\nabla U{|}^2={\gamma }^{ij}{U}_i{U}_j$ and ${\Delta }_{\gamma }U={\gamma }^{ij}{\nabla }_i{\nabla }_{j}U$. The curvature relation for the static vacuum conformal spatial metric gives $R=2|\nabla U{|}^2$. Substituting it into the contracted Bianchi identity yieldsThe Hessian of a scalar is symmetric, so the last terms cancel:Where $\nabla U\ne 0$, this implies ${\Delta }_{\gamma }U=0$. On the interior of the critical set $\{\nabla U=0\}$, $U$ is locally constant and its Laplacian is also zero. Every other critical point is a limit of noncritical points, so continuity of the Laplacian givesThis avoids incorrectly dividing by a gradient at its zeros.

Integration by parts and rigidity. The usual whole-space energy identity isWith standard static asymptotic falloff $U=O(R^{-1})$ and $\nabla U=O(R^{-2})$, the last term is $O(R^{-1})$ and vanishes. Positivity of the spatial Riemannian metric then makes $U$ constant, and its limiting value fixes that constant to zero.

There is also a direct integration by parts proof requiring only the stated vanishing of $U$ at infinity, rather than an assumed flux decay rate. For a regular value $\epsilon >0$, the region ${\Omega }_{\epsilon }=\{U>\epsilon \}$ has compact closure. Completeness and the absence of an inner boundary ensure no missing boundary pieces; standard asymptotic flatness and $U\to 0$ on all ends keep this positive level set away from infinity. Its outward unit normal is $n=-\nabla U/|\nabla U|$. Multiplying the harmonic equation by $U$ and integrating givesBoth terms are nonpositive, so both vanish. A nonempty component with $U>\epsilon$ cannot have zero gradient throughout and boundary value $\epsilon$. Thus ${\Omega }_{\epsilon }$ is empty. Choose arbitrarily small regular values and apply the same argument to $-U$; it follows thatThis is vanishing harmonic function by level-set integration. It also makes explicit why an inner boundary would invalidate the conclusion.

From zero Ricci curvature to Minkowski spacetime. In dimension three the Schouten tensor is $S_{ij}=R_{ij}-R{\gamma }_{ij}/4$, so it vanishes here. The three-dimensional curvature reconstruction then gives $R_{ijpq}=0$: $\gamma$ is a flat Riemannian manifold. This use of three-dimensional curvature from the Ricci tensor is essential; zero Ricci tensor would not by itself imply flatness in four dimensions.

A complete connected flat spatial manifold has Euclidean universal cover. Standard asymptotic flatness, with an ordinary Euclidean end, excludes a nontrivial free Euclidean quotient: a nontrivial fixed-point-free Euclidean isometry contains a translational or screw component, and its cyclic quotient has at most quadratic volume growth, incompatible with a three-dimensional Euclidean end. Extra identifications cannot restore cubic growth. Thus the complete spatial metric is globally Euclidean, not just locally flat. With $U=0$ and the standard global time coordinate the four-metric becomesTherefore horizonless static vacuum rigidity gives Minkowski spacetime as the sole solution under these hypotheses.

Allowing horizons. A black hole exterior can be static and asymptotically flat without being Minkowski. The Schwarzschild spacetime is the basic example. Its lapse function vanishes at the event horizon, so $U$ is unbounded below there; the horizon introduces an inner end or boundary in the spatial reduction, and the previous energy argument no longer has its hypotheses. In fact for Schwarzschild,so this conformal quotient terminates at $r=2M$ and is not the complete nonsingular quotient used above.

With the usual regularity, connected-horizon and global hypotheses of the static vacuum black-hole uniqueness theorem, the nontrivial black-hole exterior is Schwarzschild. Rotating Kerr black holes are stationary but not static and therefore are not alternatives in this question. This statement concerns a regular domain outside the horizon; the Schwarzschild interior contains a curvature singularity. If “globally static” and “nonsingular everywhere” are retained literally, Schwarzschild does not meet them. Allowing horizons means relaxing the horizonless global hypotheses to the corresponding static exterior problem.

Use the equivalent density form of harmonic slicing from part (ii). With $\sqrt{-g}=\alpha \sqrt{\gamma }$, positive lapse function $\alpha$, and the corrected inverse metric tensor, its time component becomesThus $D_t(\sqrt{\gamma }/\alpha )=(\sqrt{\gamma }/\alpha ){\partial }_i{\beta }^i$, orThe spatial volume evolution identity gives $D_t\log \sqrt{\gamma }={\partial }_i{\beta }^i-\alpha K$. Subtraction yields $D_t\log \alpha =-\alpha K$, hence the harmonic lapse evolution isThe shift vector need not vanish; its transport term is essential in a general 3+1 decomposition of spacetime.

Take the orientation induced by the future-directed unit normal and positive lapse function $\alpha$. In coordinates adapted to the 3+1 decomposition of spacetime, any component ${ϵ}_{\ell ijk}$ with all four indices spatial vanishes: only three distinct spatial indices are available. The shift vector therefore contributes nothing, andHere $[ijk]$ is the Levi-Civita symbol, and the last equality uses the metric determinant in a 3+1 decomposition, $|g|={\alpha }^2\gamma$. The right side is precisely the spatial metric volume tensor, with component $\sqrt{\gamma }$ at $ijk=123$, the appropriate permutation sign, and zero for repeated indices. HenceMore invariantly, the spatial volume form is the restriction to the hypersurface of the contraction of the spacetime volume form with $n$.

The same invariant interval can be written as $g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }={\tilde{g}}_{\alpha \beta }d{\tilde{x}}^{\alpha }d{\tilde{x}}^{\beta }$. Applying the chain rule to the differentials proves the metric transformation lawFor first-order scalar cosmological perturbations and ${\xi }^{\mu }=(T,{\partial }^{i}L)$, compare perturbations at the same background coordinate label. Expanding the Jacobians and the shifted background gives the linear metric gauge-transformation lawFor the background ${\bar{g}}_{00}=-a^2$, ${\bar{g}}_{ij}=a^2{\delta }_{ij}$, the Lie derivative components areHere primes denote conformal time and $\mathcal{H}$ is the conformal Hubble parameter. These expressions establish all four alternatives, including any chosen pair.

The $00$ component is $\delta g_{00}=-2a^{2}A$, immediately giving the lapse function perturbation transformation. The $0i$ component is $a^2{\partial }_{i}B$, giving the shift vector scalar-potential transformation. The trace of the spatial perturbation is $6a^{2}C$, so tracing the spatial Lie derivative gives the transformation of $C$. Its trace-free scalar part is $2a^2({\partial }_i{\partial }_j-{\delta }_{ij}{\nabla }^2/3)E$, giving the transformation of $E$. ThusAs usual, identifying scalar potentials from their derivatives uses the standard boundary conditions, or nonzero Fourier modes, to remove homogeneous ambiguities.

At first order, tensor cosmological perturbations are spatial transverse-traceless tensors. The coordinate-generated spatial perturbation consists of a trace term and symmetrized derivatives of the displacement. In Fourier space, the latter terms carry a factor $k_i$ or $k_j$. The transverse-traceless projector removes these longitudinal terms and the trace, including the derivative of a transverse vector displacement. Therefore the tensor perturbation is gauge invariant at linear order around the homogeneous background. This is a first-order statement, not a claim of automatic invariance at arbitrary perturbative order.

Perform the Legendre transform in mechanics, using ${\dot{h}}_{ij}=2N{K}_{ij}+D_i{N}_j+D_j{N}_i$. The trace contraction ${\pi }^{ij}{K}_{ij}=\sqrt{h}(K_{ij}{K}^{ij}-K^2)$ givesTo avoid differentiating a tensor density as though it were an ordinary tensor, set $p^{ij}={\pi }^{ij}/\sqrt{h}$. Integration by parts with the Levi-Civita connection of $h$ givesafter dropping the surface term. AlsoTherefore the requested Hamiltonian isVarying the lapse function and shift vector imposes the Hamiltonian constraint $\mathcal{H}=0$ and momentum constraint ${\mathcal{H}}_i=0$. The total canonical generator also contains multipliers for the vanishing lapse and shift canonical momenta. With an asymptotic boundary, surface terms must be restored to define the Arnowitt-Deser-Misner energy.

There are no ${\partial }_{t}N$ or ${\partial }_t{N}^i$ terms in the displayed Einstein-Hilbert action, so the lapse function and shift vector have zero canonical momenta. Their equations impose constraint equations in field theory rather than independent propagation; they are Lagrange multipliers in the Hamiltonian formulation of general relativity.

With the question's positive sign for $K_{ij}=({\dot{h}}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$, vary ${\dot{h}}_{ij}$ at fixed $h,N,N^i$. Since $\delta K_{ij}=\delta {\dot{h}}_{ij}/(2N)$,Thus the canonical momentum of the spatial metric is a weight-one tensor density. Write $\pi =h_{ij}{\pi }^{ij}$; in three spatial dimensions $\pi =-2\sqrt{h}K$, soThe action has suppressed its overall $1/(16\pi G)$, and this normalization is retained here. Reversing the sign convention for extrinsic curvature would reverse the momentum relation; the calculation uses exactly the convention supplied in this question.

Use a dot for the coordinate-time derivative and let $\bar{N}$ be the homogeneous background lapse function. The passive linear transformation is $\delta {\tilde{g}}_{\mu \nu }=\delta g_{\mu \nu }-{\mathcal{L}}_{\xi }{\bar{g}}_{\mu \nu }$. Its time-time and time-space components yield the scalar gauge transformations with a background lapse:For example, ${\bar{g}}_{00}=-{\bar{N}}^2$ gives $-({\mathcal{L}}_{\xi }\bar{g}{)}_{00}=2\bar{N}\dot{\bar{N}}{\xi }^0+2{\bar{N}}^2{\dot{\xi }}^0$; ${\bar{g}}_{0i}=0$ gives $-({\mathcal{L}}_{\xi }\bar{g}{)}_{0i}={\bar{N}}^2{\partial }_i{\xi }^0-a^2{\partial }_i\dot{\lambda }$. These establish the signs directly.

To preserve synchronous gauge in cosmology, both transformed quantities must remain zero. HenceIntegrating gives the residual synchronous-gauge freedomThe integration lower limit can be absorbed into $D$. Spatially homogeneous additions to $\lambda$ generate no spatial displacement, so they are physically irrelevant to this scalar parametrization.

The spatial transformation also gives $\tilde{\Phi }=\Phi +(\dot{a}/a){\xi }^0$ and $\tilde{E}=E+\lambda$. Thus fixing the lapse and shift perturbations does not fix the entire coordinate system; the arbitrary functions $C,D$ can still change the remaining scalar cosmological perturbations.

Write $h_{ij}={}^{(3)}{g}_{ij}$, and retain only first-order scalar cosmological perturbations. The inverse induced metric and inverse lapse function areThe flat background spatial connection vanishes, and $N_i=-a^2{B}_{,i}$ is already first order. ConsequentlyThe comma on $B_{,i}$ matters: this is the gradient of the scalar shift vector potential, not an unrelated vector field. Substituting the time derivative of $h_{ij}$ into the negative extrinsic curvature convention givesRaising the first index with the perturbed inverse induced metric is essential. Its $\Phi$ and $E$ terms cancel the corresponding terms multiplying the background Hubble parameter, leavingFor the scalar shear potential $\chi =a^2(\dot{E}-B)/\bar{N}$, raised spatial derivatives mean ${\partial }^i=a^{-2}{\delta }^{ij}{\partial }_j$ to the required order. Thus the final term is $-{\partial }^i{\partial }_j\chi$. Taking the trace identifies the scalar expansion perturbation:Separating the trace and trace-free parts now yieldsAll signs follow from the source's $d{x}^i-N^{i}dt$ shift vector convention and negative extrinsic curvature definition.

Choose the asymptotic time translation: $N=1+O(r^{-1})$, $N^i=O(r^{-1})$, with their derivatives $O(r^{-2})$, in asymptotically Cartesian coordinates. Use the usual differentiable falloff $h_{ij}={\delta }_{ij}+O(r^{-1})$, ${\partial }_k{h}_{ij}=O(r^{-2})$, ${\pi }^{ij}=O(r^{-2})$, and analogous falloff for allowed variations; impose the standard additional conditions needed for finite bulk integrals. These variations need not fix the Arnowitt-Deser-Misner energy.

Only the spatial curvature term produces the leading nonvanishing surface variation. Applying the variation of the Ricci scalar and applying integration by parts to its two derivative terms yieldsThe terms involving derivatives of the asymptotic lapse function, and the shift boundary terms, vanish under these falloff conditions. The displayed surface integral need not vanish: its integrand is $O(r^{-2})$ and the area is $O(r^2)$.

Hamilton's equations require a Hamiltonian with a variation expressed by the bulk functional derivatives alone for the allowed variations. The unwanted term is exactly $-\delta E_{\mathrm{ADM}}$, so differentiability of the gravitational Hamiltonian at spatial infinity requiresIts variation cancels the surface term. On the constraint surface, $H_{\mathrm{total}}=E_{\mathrm{ADM}}$. Restoring units inserts $1/(16\pi G)$ in the surface integral; the paper has set this factor to one. The original PDF has the standard equation ${\partial }_t{h}_{ij}=\delta H/\delta {\pi }^{ij}$; the extra $t$ in the extracted TeX is a transcription error.

In the paper's convention, ${\dot{h}}_{ij}=2N{K}_{ij}+D_i{N}_j+D_j{N}_i$. The Legendre transform in mechanics therefore givesIndeed ${\pi }^{ij}{K}_{ij}=\sqrt{h}(K_{ij}{K}^{ij}-K^2)$, and inversion of the canonical momentum of the spatial metric givesTo integrate the shift term without ambiguity about tensor densities, put $p^{ij}={\pi }^{ij}/\sqrt{h}$. Metric compatibility of the spatial covariant derivative gives, up to a boundary term,Thus the Hamiltonian formulation of general relativity hasVariation of the lapse function imposes the Hamiltonian constraint $\mathcal{H}=0$, while variation of the shift vector imposes the momentum constraint ${\mathcal{H}}_i=0$.

The action contains neither $\dot{N}$ nor ${\dot{N}}^i$, so their conjugate momentum vanishes. They impose constraints rather than independent evolution equations: after the Legendre transform in mechanics, the lapse function and shift vector are Lagrange multipliers in the Hamiltonian formulation of general relativity.

Use precisely the paper's sign convention for extrinsic curvature, $K_{ij}=({\dot{h}}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$. At fixed $h_{ij},N,N^i$,Consequently the canonical momentum of the spatial metric isThis is a weight-one tensor density, with indices raised and lowered using $h_{ij}$. Its trace and inverse relation areThe sign of $K_{ij}$ used here is opposite to the convention in extrinsic curvature of a spatial hypersurface; mixing the two conventions would reverse the momentum formula.

Comparison with the 3+1 decomposition of spacetime gives the diagonal spatial metricThe mixed metric coefficient is ${\beta }_r=g_{tr}=M/r$. Raising its index with ${\gamma }^{rr}=r^2/(r+M{)}^2$ yields the shift vectorNowand $g_{tt}=-{\alpha }^2+{\beta }_i{\beta }^i=-(r-M)/(r+M)$. The positive lapse function is consequently