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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.

Unit normalization gives $n^{\nu }{\nabla }_{\mu }{n}_{\nu }=0$. Hence $K_{\mu \nu }=-P^{\alpha }{}_{\mu }{P}^{\beta }{}_{\nu }{\nabla }_{\alpha }{n}_{\beta }=-P^{\alpha }{}_{\mu }{\nabla }_{\alpha }{n}_{\nu }$. Set $F_{\mu \nu }={\nabla }_{\mu }{n}_{\nu }-{\nabla }_{\nu }{n}_{\mu }$. Frobenius theorem for a hypersurface normal gives $n_{\alpha }{F}_{\beta \gamma }+n_{\beta }{F}_{\gamma \alpha }+n_{\gamma }{F}_{\alpha \beta }=0$. Contracting with $n^{\alpha }{P}^{\beta }{}_{\mu }{P}^{\gamma }{}_{\nu }$ leaves $-P^{\beta }{}_{\mu }{P}^{\gamma }{}_{\nu }{F}_{\beta \gamma }=0$, precisely the vanishing antisymmetric part of the extrinsic curvature of a spatial hypersurface.

The traceless mixed extrinsic curvature of a spatial hypersurface can have leading solution ${\tilde{K}}^i{}_j=C^i{}_j(\mathbf{x})a_{\mathrm{loc}}^{-3}$. Cosmic inflation suppresses this anisotropic expansion. An anisotropic spatial shape can nevertheless freeze to a nonzero constant, so shear damping does not prove the absence of tensor cosmological perturbations.

For the normal flow with zero shift vector, $a_{\mathrm{loc}}=(\det h{)}^{1/6}$ obeys $H_{\mathrm{loc}}=N^{-1}{\partial }_t\log a_{\mathrm{loc}}=-K/3$ in the negative extrinsic curvature of a spatial hypersurface convention. The expansion scalar is $\theta =-K=3H_{\mathrm{loc}}$.

Write ${\gamma }_{ij}={}^{(3)}{g}_{ij}$, $\gamma =\det {\gamma }_{ij}$ and $D_i$ for the spatial covariant derivative. With the negative-shift convention, the extrinsic curvature of a spatial hypersurface is $K_{ij}=-({\dot{\gamma }}_{ij}+D_i{N}_j+D_j{N}_i)/(2N)$. Contracting with ${\gamma }^{ij}$ and using metric compatibility givesThe logarithmic determinant identity makes ${\gamma }^{ij}{\dot{\gamma }}_{ij}=\dot{\gamma }/\gamma$, henceThis is also the fractional volume expansion, with a minus sign from the extrinsic-curvature convention.

Define $a={\gamma }^{1/6}$ so that $\sqrt{\gamma }=a^3$. The useful unit-determinant conformal decomposition is ${\gamma }_{ij}=a^2{\tilde{\gamma }}_{ij}$ with $\det \tilde{\gamma }=1$, equivalently ${\tilde{\gamma }}_{ij}=a^{-2}{\gamma }_{ij}$. Multiplying by $a^2$ instead would be a different conformal rescaling, not this unit-determinant one; no conformal decomposition is needed for the trace identity itself.

With zero shift vector, $\dot{\gamma }/\gamma =6\dot{a}/a$ and the local volume Hubble parameter isAlong a normal trajectory proper time satisfies $ds=Ndt$, so $H=a^{-1}da/ds$ is the local physical volume-expansion rate. The printed equality with $\dot{a}/a$ requires the proper-time gauge $N=1$; zero shift alone does not set the lapse to unity. In a homogeneous Friedmann-Lemaître-Robertson-Walker metric it reduces to the ordinary Hubble parameter, while in an inhomogeneous spacetime it can vary from one normal trajectory to another.

Differentiate the spatial projection tensor before making any contractions:For the spatial projector derivative identity, its first term vanishes after projection on $\mu$. The definition of the extrinsic curvature of a spatial hypersurface then gives the stronger tensor identityHere $\lambda$ remains a free index. For a normal to a genuine foliation, the extrinsic curvature is symmetric: projecting ${\nabla }_{[\nu }{n}_{\mu ]}$ gives zero because locally $n_{\mu }$ is a scalar multiple of a time gradient. This is hypersurface orthogonality implies symmetric extrinsic curvature.

Contract $\rho$ with $\lambda$ in the stronger identity to obtain exactly the contraction displayed in the PDF:The last equality follows from the transversality of the extrinsic curvature. Thus the literal printed identity is valid, although both its sides vanish; the uncontracted identity explains its geometric origin.

First, the inverse metric tensor printed in the PDF has a missing factor: its lower-left entry must be $g^{i0}={\alpha }^{-2}{\beta }^i$, symmetric with $g^{0i}$. This correction follows either from symmetry or direct block multiplication and does not change the evolution equation used here.

Let $D_t={\partial }_t-{\beta }^k{\partial }_k$. Contract the spatial metric tensor evolution equation with ${\gamma }^{ij}$. Both shift-gradient terms contribute ${\partial }_k{\beta }^k$, givingThe Jacobi determinant derivative formula applies to every derivative of $\gamma =\det ({\gamma }_{ij})$ and gives $D_t\gamma =\gamma {\gamma }^{ij}{D}_t{\gamma }_{ij}$. Therefore the spatial determinant evolves asEquivalently, the spatial volume evolution identity is $D_t\sqrt{\gamma }=\sqrt{\gamma }({\partial }_k{\beta }^k-\alpha K)$, with $K$ the trace of the extrinsic curvature of a spatial hypersurface.

Use signature $(+---)$, the positive spatial metric ${\gamma }_{ij}$, and a future-directed unit normal $n_{\mu }=(N,0)$, so $n^{\mu }=N^{-1}(1,-N^i)$. This fixes the sign of the normal to agree with the specified $\Pi$. The negative normal covector printed in the source would instead give $-\Pi$; its orientation and the displayed derivative cannot both be retained. Use the operational negative-expansion convention for the extrinsic curvature of a spatial hypersurface, $K_{ij}=-({\dot{\gamma }}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$, which is the convention giving the requested expanding-universe sign. With this future normal and signature it equals $n_{i;j}$; the source's $-n_{i;j}$ identity instead uses its past-pointing normal.

For the scalar-field matter projections, put $s_i=D_i\varphi$ and $s^2={\gamma }^{ij}{s}_i{s}_j$. Decomposing the scalar gradient into normal and tangent parts gives ${\partial }_{\mu }\varphi {\partial }^{\mu }\varphi ={\Pi }^2-s^2$. Contracting the Klein-Gordon scalar stress-energy tensor with the normal and the spatial projection tensor therefore yieldsIn particular, the positive spatial metric raises the $i,j$ indices in $s^2$; raising them with the four-metric would reverse that spatial sign. These quantities are the energy density, momentum density and spatial stress in the normal frame.

For the homogeneous Spatially flat FLRW metric, $N_i=0$, ${\gamma }_{ij}=a^2{\delta }_{ij}$ and $s_i=0$. Consequently $\Pi =\dot{\bar{\varphi }}/\bar{N}$, $J_i=0$, $\bar{\rho }={\Pi }^2/2+V$, $\bar{P}={\Pi }^2/2-V$, andThe intrinsic Ricci scalar is zero and $K^2-K_{ij}{K}^{ij}=9H^2-3H^2=6H^2$. The Hamiltonian constraint thus gives the scalar-field Friedmann equation:The momentum constraint is identically satisfied, since its homogeneous spatial derivatives and $J_i$ vanish. The scalar equation reduces to ${\bar{N}}^{-1}\dot{\Pi }+3H\Pi +V_{,\varphi }=0$. Substituting $\Pi$ and multiplying by ${\bar{N}}^2$ gives the background inflaton equation of motion:No derivation of the supplied gravitational constraints is needed.

All free indices below are spatial, and $K={\gamma }^{\mu \nu }{K}_{\mu \nu }$. Contract the vacuum Ricci tensor using $g^{\mu \nu }={\gamma }^{\mu \nu }-n^{\mu }{n}^{\nu }$:Reversing both antisymmetric curvature pairs identifies the last term as $-E_{\alpha \beta }$. The spatial Gauss–Codazzi equations for a spatial hypersurface then giveFor the magnetic part of the Weyl tensor, moving the normal to the first slot of the metric volume tensor introduces a minus sign:The remaining curvature indices are spatial, so the normal projection in the Gauss–Codazzi equations for a spatial hypersurface givesThe interchange $\mu \leftrightarrow \nu$ makes the two spatial covariant derivative terms equal. Thus the constants, with the paper's orientation and extrinsic-curvature convention, areIn particular, the sign of the magnetic expression must include the minus from moving $n$ past the first volume-tensor index. These formulas reconstruct the spatial electric part of the Weyl tensor and magnetic part of the Weyl tensor from the induced metric and the extrinsic curvature of a spatial hypersurface.

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.

Differentiate the normalization of the timelike unit normal with the metric-compatible covariant derivative:Thus the derivative of the normal is already orthogonal to the normal in its second index. Expanding the second spatial projection tensor in the definition of the extrinsic curvature of a spatial hypersurface givesTo prove symmetry using Frobenius theorem, set $F_{\alpha \beta }={\nabla }_{\alpha }{n}_{\beta }-{\nabla }_{\beta }{n}_{\alpha }$. Hypersurface orthogonality is equivalent toContract this with $n^{\alpha }{P}^{\beta }{}_{\mu }{P}^{\lambda }{}_{\nu }$. The last two terms vanish because a projected normal vanishes, while $n^{\alpha }{n}_{\alpha }=-1$ in the first term. HenceThe antisymmetric part of the doubly projected covariant derivative of $n$ is therefore zero. This proves that hypersurface orthogonality implies symmetric extrinsic curvature:No geodesic assumption for the normal congruence is needed; its normal acceleration may be nonzero.

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.

The definition of the extrinsic curvature of a spatial hypersurface and ${\perp }^{\rho }{}_{\mu }={\delta }^{\rho }{}_{\mu }+n^{\rho }{n}_{\mu }$ givewhere $a_{\nu }=n^{\rho }{\nabla }_{\rho }{n}_{\nu }$ is the normal acceleration. HenceDifferentiating $n^{\nu }{n}_{\nu }=-1$ shows $n^{\nu }{a}_{\nu }=0$, so the acceleration is spatial as required.

Insert $Z_{\mu }={\Theta }_{\mu }+n_{\mu }\Theta$ from part (i) into the projected divergence:The term containing ${\nabla }_{\mu }\Theta$ vanishes because ${\perp }^{\mu \nu }{n}_{\nu }=0$. The remaining contraction isby the definition of the extrinsic curvature of a spatial hypersurface. Therefore

Let $\mathcal{D}={\bar{N}}^{-1}{\partial }_t$ and $\bar{\Pi }=\mathcal{D}\bar{\varphi }$. For negative-expansion extrinsic curvature of a spatial hypersurface, its trace perturbation is $\kappa$. With $\delta \Pi =\mathcal{D}\delta \varphi -\bar{\Pi }\Psi$, scalar evolution is $\mathcal{D}\delta \Pi -\Psi \mathcal{D}\bar{\Pi }+3H\delta \Pi -\kappa \bar{\Pi }-\Delta \delta \varphi +V_{,\varphi \varphi }\delta \varphi =0$. Combining background acceleration terms produces the standard lapse source $-\bar{\Pi }(\kappa +\mathcal{D}\Psi -3H\Psi )+2\Psi V_{,\varphi }$.

Differentiating the spatial projection tensor and projecting its derivative and covariant indices leaves $-n^{\lambda }{K}_{\sigma \rho }$, where $K$ is the extrinsic curvature of a spatial hypersurface. Contracting $\lambda$ with $\rho$ gives zero because $K$ is transverse. This distinction separates the informative tensor identity, with a free normal index, from its vanishing trace. Hypersurface orthogonality implies symmetric extrinsic curvature.