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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.

For $d{s}^2=-N^{2}d{t}^2+h_{ij}(d{x}^i-N^{i}dt)(d{x}^j-N^{j}dt)$ and the future unit normal, the convention $K_{ij}=-n_{i;j}$ gives $K_{ij}=-({\dot{h}}_{ij}+D_i{N}_j+D_j{N}_i)/(2N)$. Lower the shift vector with $h$. The sign of the shift terms changes when the opposite threading convention is used.

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 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 $\theta =-K=3H_{\mathrm{loc}}$.

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: $n^{\mu }{ϵ}_{\mu ijk}=\sqrt{\det \gamma }[ijk]$.

The momentum constraint is the equation obtained by varying the shift vector in the Hamiltonian formulation of general relativity. It expresses the tangential-normal projection of the Einstein field equations and generates spatial coordinate transformations. Writing $p^{ij}={\pi }^{ij}/\sqrt{h}$ avoids ambiguities about covariant derivatives of tensor densities.

For a minimally coupled scalar field with spatial induced metric $h_{ij}$, the field's derivative along the future unit normal is $\Pi$. If the shift vector enters as $d{x}^i-N^{i}dt$, then $\Pi =(\dot{\varphi }+N^i{\partial }_i\varphi )/N$. The canonical momentum of the density is ${\pi }_{\varphi }=\sqrt{h}\Pi$. Confusing this density with the normal derivative loses a volume factor.

Let ${\gamma }_{ij}$ denote the spatial induced metric and $\gamma =\det {\gamma }_{ij}$. Its conformally rescaled version ${\tilde{\gamma }}_{ij}={\gamma }^{-1/3}{\gamma }_{ij}$ has determinant one. The Jacobi determinant derivative formula consequently givesThus the determinant measures the volume-changing part of the spatial metric, while the unit-determinant part contains shape changes. Metric compatibility for the spatial covariant derivative also gives ${\gamma }^{ij}{D}_i{N}_j=D_i{N}^i$.

There is a sign inconsistency in the PDF. Contracting its defining expression for $K_{ij}$, with its threading convention $d{x}^i-N^{i}dt$ and its future normal $n^{\mu }=N^{-1}(1,N^i)$, givesnot the displayed trace with a minus sign before the divergence. This is the trace of extrinsic curvature with a negative shift. For a concrete check, take ${\gamma }_{ij}={\delta }_{ij}$, $N=1$ and $N^i=b{x}^i$. The defining formula gives $K_{ij}=-b{\delta }_{ij}$ and $K=-3b$, whereas the printed trace would give $+3b$. Reversing the shift convention can produce the other trace formula, but also changes the shift terms in $K_{ij}$ and the normal; the two conventions cannot be combined.

For the requested expansion interpretation, set the shift vector to zero. The sign discrepancy then disappears. A small coordinate volume has proper volume proportional to $\sqrt{\gamma }$, and proper time along its normal is $ds=Ndt$. Its expansion rate isThereforeThis is the local volume Hubble parameter, the mean of the three local directional expansion rates. In a homogeneous isotropic background $\gamma =a^6$, it reduces to $\dot{a}/(Na)$. The interpretation does not require the local expansion to be isotropic: the trace-free part of the extrinsic curvature can still describe shear.

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 $a$.

Write $h_{ij}={}^{(3)}{g}_{ij}$ for the induced metric and lower the shift vector with it, $N_i=h_{ij}{N}^j$. In the stated negative-shift convention the four-dimensional components and inverse components areMultiplying the given future unit normal by this metric tensor gives $n_{\mu }=(-N,0,0,0)$. In particular the tangential normal components vanish, but their covariant derivatives do not:The relevant Christoffel symbol isFor the second equality, the last parenthesis is $2h_{k\ell }{}^{(3)}{\Gamma }^{\ell }{}_{ij}$, so its contraction subtracts the two spatial connection terms in the symmetrized derivative of $N_i$. ThereforeThis is extrinsic curvature with a negative shift. Both the future-normal orientation and the definition $K=-\nabla n$ matter: replacing the threading by $d{x}^i+N^{i}dt$ reverses the shift terms. The spatial induced metric is generally a function of time as well as position; restricting it to time-independent data would simply make the displayed time derivative vanish.

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.

The unimodular spatial metric separates volume from shape: writing $h_{ij}=a^2{\tilde{h}}_{ij}$ and $a^6=h=\det h_{ij}$ gives $\det \tilde{h}=1$. Contract the extrinsic curvature with a negative shift with $h^{ij}$. Metric compatibility and the Jacobi determinant derivative formula giveConsequentlyFor zero shift vector, $\dot{h}/h=6\dot{a}/a$, soThe infinitesimal spatial volume is $\sqrt{h}{d}^{3}x=a^3{d}^{3}x$; its logarithmic change per unit normal proper time is $3H$. Thus this $H$ is the local volume Hubble parameter, reducing to the usual Hubble parameter in a homogeneous universe. The determinant-defined $a$ refers to fixed spatial coordinates, while this normal volume expansion has a geometric interpretation.

Use signature $(-,+,+,+)$ and write $h_{ij}$ for the spatial induced metric, $D_i$ for its spatial covariant derivative, and $q^2=h^{ij}{D}_i\varphi D_j\varphi$. The negative shift vector convention gives $n^{\mu }=N^{-1}(1,N^i)$ and $\Pi =n^{\mu }{\partial }_{\mu }\varphi$. Decomposing the scalar field derivative givesVarying the scalar field matter action with respect to the inverse metric tensor, including the variation of $\sqrt{-g}$, gives the stress-energy tensorRaising both indices gives the requested $T^{\mu \nu }$. Contracting with the unit normal twice, or once with the spatial projection tensor, givesProjecting both indices of the stress-energy tensor onto the spatial hypersurface givesTwo printed identifications require correction. The canonical momentum of the displayed density is ${\pi }_{\varphi }=\sqrt{h}\Pi$, not $\Pi$: explicitly ${\mathcal{L}}_m=N\sqrt{h}[{\Pi }^2/2-q^2/2-V]$ and differentiation with respect to $\dot{\varphi }$ supplies $\sqrt{h}$. Thus $\Pi$ is the normal scalar-field momentum. Also, the printed trace-free stress has an extra factor $1/2$ and uses ${\delta }_{ij}$ where the general spatial induced metric requires $h_{ij}$. In an orthonormal spatial frame the latter becomes ${\delta }_{ij}$, but the coefficient is still one. For a gradient $(q,0,0)$ in that frame, direct projection gives ${\tilde{S}}_{11}=2q^2/3$, rather than the printed $q^2/3$. The PDF's energy-density gradient is spatial ${\partial }_i\varphi {\partial }^i\varphi$; the TeX aid's spacetime index there is an OCR defect.

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.