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Decompose the mixed extrinsic curvature of a spatial hypersurface into $K^i{}_j=\frac{1}{3}K{\delta }^i{}_j+{\tilde{K}}^i{}_j$. In the long-wavelength approximation in cosmology, the inflationary shear damping law makes ${\tilde{K}}^i{}_j$ fall as the inverse local volume. During cosmic inflation, many e-folds therefore suppress the anisotropic expansion, while $K$ still measures the local isotropic expansion through the local volume Hubble parameter.

Neglecting spatial gradients at leading order gives approximately independent expanding patches. Restrict also to scalar cosmological perturbations, and neglect the independent frozen tensor cosmological perturbation. The remaining spatial scale can be written as a background scale times a local fluctuation:This is the nonlinear scalar metric in a cosmological gradient expansion. The exponential keeps the metric positive and permits finite $\zeta$; no expansion in its amplitude is needed. Scalar lapse and shift perturbations describe the remaining freedom in time slicing and spatial threading. Choose $N=1+\Psi$ and $N_i=-a_{\mathrm{bg}}^2{\partial }_{i}B$. The negative shift vector convention then produces the positive mixed term $2a_{\mathrm{bg}}^2{\partial }_{i}Bdtd{x}^i$.

The exact completed-square form hasThe quadratic shift expression must be understood as this norm. If the raised derivative in the displayed ansatz is intended as a flat derivative, the corresponding factors of the spatial scale are needed. In either convention the shift-square term is of second gradient order, and the leading long-wavelength approximation in cosmology is unaffected.

Damping of shear motivates locally isotropic scalar expansion, not the disappearance of every tensor perturbation. A time-independent anisotropic shape can survive; the conformally flat scalar ansatz additionally neglects that tensor sector.

Set $h=\det h_{ij}$. Contract the extrinsic curvature with a negative shift with $h^{ij}$. The Jacobi determinant derivative formula and metric compatibility giveConsequentlyThe local volume element is $\sqrt{h}{d}^{3}x$, so for zero shift vector the proper-time derivative along the normal is $d/ds=N^{-1}{\partial }_t$ andHere $\theta$ is the expansion scalar. Writing $a_{\mathrm{loc}}=h^{1/6}$ defines a local linear scale from this volume, whence the local volume Hubble parameter isThe factor of three converts volume expansion into linear expansion, and $N$ converts coordinate time into proper time. This agrees with the usual Hubble parameter in a homogeneous FLRW metric.

For a volume-shape decomposition the unimodular spatial metric is ${\gamma }_{ij}=a_{\mathrm{loc}}^{-2}{h}_{ij}$, with determinant one. If instead the printed positive power $a_{\mathrm{loc}}^2{h}_{ij}$ is taken literally, its determinant is $a_{\mathrm{loc}}^{12}$; it is a conformal rescaling but not the unimodular shape metric. The trace calculation does not need that rescaling.

Write $h_{ij}$ for the spatial metric tensor and lower the shift vector with $h_{ij}$. For the negative-shift convention, the four-dimensional metric and its inverse have componentsThe normal has covariant components $n_{\mu }=(-N,0,0,0)$, so $n_{\mu }{n}^{\mu }=-1$ and it annihilates every tangent vector to the spatial slice. Since $n_i=0$, its covariant derivative gives $n_{i;j}=N{\Gamma }^0{}_{ij}$ and therefore $K_{ij}=-N{\Gamma }^0{}_{ij}$.

Compute the required Christoffel symbol using the symmetric connection formula. Its time-index term and spatial-index term combine to giveThe spatial expression in parentheses is ${\dot{h}}_{ij}+D_i{N}_j+D_j{N}_i$, where $D$ is the covariant derivative of $h$. Thus the extrinsic curvature with a negative shift isThis derivation fixes both signs from the actual 3+1 decomposition of spacetime, rather than transferring a formula using the opposite shift vector convention. The connection formula used here has $g_{\lambda \kappa ,\nu }$ as its second differentiated term; the printed repeated-index version is a typographical error.

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.

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

The spatial metric is stationary and only ${\beta }^r=b(r)=Mr/(r+M{)}^2$ is nonzero. The evolution equation therefore saysthe Lie derivative of the spatial metric along the shift vector. For the radial component,soFor the angular components,and spherical symmetry suppliesAll off-diagonal components vanish.

Contracting with the inverse spatial metric gives the mean curvatureThus

With negative extrinsic curvature convention, the background trace is $\bar{K}=-3H$. The scalar expansion perturbation is $\kappa =K+3H$. For lapse function $N=\bar{N}(1+\Psi )$, induced metric $h_{ij}=a^2[(1-2\Phi ){\delta }_{ij}+2E_{,ij}]$ and shift vector $N_i=-a^2{B}_{,i}$, it is $\kappa =3(\dot{\Phi }/\bar{N}+H\Psi )-\Delta \chi$, where $\chi$ is the scalar shear potential. The expansion of the future normal congruence is $-K$, so its perturbation has the opposite sign.

For the negative extrinsic curvature convention and a shift vector entering the line element as $d{x}^i-N^{i}dt$, the trace-free scalar curvature perturbation is $\delta K^i{}_j{}_{\mathrm{TF}}=-({\partial }^i{\partial }_j-{\delta }_j^i\Delta /3)\chi$. Here spatial raised derivatives use $a^{-2}{\delta }^{ij}$, so $\Delta =a^{-2}{\nabla }^2$. Changing the sign of the metric shift vector changes the corresponding definition of $\chi$.

For zero shift vector, negative-expansion extrinsic curvature $K=-3H$ and $S^i{}_j={\tilde{K}}^i{}_j$, retain the leading spatial-gradient equations of a minimally coupled scalar. They give $\dot{\varphi }=N\Pi$, $\dot{\Pi }=-N(3H\Pi +V_{,\varphi })$, $\dot{H}=-4\pi GN{\Pi }^2-N{S}^i{}_j{S}^j{}_i/2$, and ${\dot{S}}^i{}_j=-3NH{S}^i{}_j$. The first-gradient momentum constraint is ${\partial }_{i}H=-4\pi G\Pi {\partial }_i\varphi -{\nabla }_j{S}^j{}_i/2$. Shear decays as $a^{-3}$ during expansion. Its neglect after inflationary damping yields the equations underlying conservation of the nonlinear curvature covector.

In a 3+1 decomposition of spacetime with $K_{ij}=-({\partial }_t{\gamma }_{ij}-{\mathcal{L}}_{\beta }{\gamma }_{ij})/(2\alpha )$, contract the metric evolution with ${\gamma }^{ij}$ and use the Jacobi determinant derivative formula. Here $\gamma =\det {\gamma }_{ij}$, $K={\gamma }^{ij}{K}_{ij}$, $\alpha$ is the lapse function, and $\beta$ is the shift vector. The identity separates spatial transport and coordinate compression from the geometric volume change caused by extrinsic curvature.