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For the positive-expansion convention for extrinsic curvature, write ${\nabla }_{\mu }{n}_{\nu }=K_{\mu \nu }-n_{\mu }{a}_{\nu }$ with normal acceleration $a_{\nu }=n^{\rho }{\nabla }_{\rho }{n}_{\nu }$. The Ricci commutator applied to $n$ gives the difference of its two second-derivative contractions. Projecting tangentially, the acceleration term in that commutator cancels exactly the acceleration term converting a spacetime divergence of $K$ into its fully projected spatial covariant derivative. Einstein's equations then identify the displayed equation as the momentum constraint.

With signature $(-,+,+,+)$, future unit normal $n$ and $K_{ab}=-P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$, the Gauss–Codazzi equations become $(\perp R{)}_{abcd}={\mathcal{R}}_{abcd}+K_{ac}{K}_{db}-K_{ad}{K}_{cb}$ and $(\perp R{)}_{abcd}{n}^d=-D_a{K}_{bc}+D_b{K}_{ac}$, where in the second expression only the first three indices are projected. The intrinsic curvature and spatial covariant derivative use the spatial induced metric. The normal is timelike, so the normal-sign convention differs from a spacelike normal in a positive-definite ambient metric.

For a unit timelike normal $n$, the induced metric is $h_{\mu \nu }=g_{\mu \nu }+n_{\mu }{n}_{\nu }$. The spatial covariant derivative projects the derivative and all free indices with the spatial projection tensor $P^{\mu }{}_{\nu }={\delta }^{\mu }{}_{\nu }+n^{\mu }{n}_{\nu }$. Metric compatibility gives ${\nabla }_{\lambda }{h}_{\mu \nu }=({\nabla }_{\lambda }{n}_{\mu })n_{\nu }+n_{\mu }{\nabla }_{\lambda }{n}_{\nu }$; projecting kills the normal factor in each term. Thus the spatial covariant derivative is compatible with the induced metric.

Write $q_{\alpha \beta }=g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ for the induced metric. By definition of the spatial covariant derivative,The spacetime Levi-Civita connection is metric compatible, so $\nabla g=0$. The remaining terms contain either $P_{\alpha }{}^{\sigma }{n}_{\sigma }$ or $P_{\beta }{}^{\lambda }{n}_{\lambda }$, each zero because $n^{\mu }{n}_{\mu }=-1$. ThereforeThis verifies metric compatibility of the spatial covariant derivative directly, without assuming that the unprojected derivative of $q$ vanishes.

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.

Write $q_{\alpha \beta }=g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ for the induced metric. The spatial covariant derivative projects its derivative index and both tensor indices:Four-dimensional metric compatibility removes the derivative of $g$. Each remaining term contains an undifferentiated normal on a projected index and hence vanishes. ThereforeThis establishes metric compatibility of the spatial covariant derivative directly from its projected definition.

Differentiating the normal's normalization givesso $n^{\mu }{\nabla }_{\nu }{n}_{\mu }=0$. Consequently the second projection in the extrinsic curvature definition is redundant:Finally differentiate $P^{\lambda }{}_{\nu }={\delta }^{\lambda }{}_{\nu }+n^{\lambda }{n}_{\nu }$ and project:The first term vanishes by $P^{\nu }{}_{\beta }{n}_{\nu }=0$, and the second is precisely the defining curvature. The positive sign follows from the specified positive-expansion convention, rather than from an independent projector convention.

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.

With the $+---$ signature and $n^2=1$, direct expansion of the two spatial projection tensors givesThis restriction is negative definite on spatial vectors. The positive spatial induced metric is instead $h_{\alpha \beta }=n_{\alpha }{n}_{\beta }-g_{\alpha \beta }=-q_{\alpha \beta }$, whose pullback to a time slice is the $h_{ij}$ used in the line element. The spatial metric sign for a unit timelike normal is important here: the plus sign in the PDF's claimed equality is incompatible with its normal normalization: $g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ is not even transverse to $n^{\alpha }$.

The spatial covariant derivative of a spatial tensor projects every index, including its derivative index. Projection only on the derivative index is sufficient for a scalar but not for a general tensor. Using metric compatibility of the spacetime Levi-Civita connection, we obtainEach term contains a normal contracted with its spatial projection tensor. Consequently $\boxed{D_i{h}_{jk}=0}$, and the negative spatial restriction also satisfies $D_i{q}_{jk}=0$. This proves the requested metric compatibility of the spatial covariant derivative after correcting the source's metric sign. Contracting indices gives the particular expression written in the question.

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.

Using the normalization of the timelike unit normal, the spatial projection tensor satisfiesLikewise $P^{\mu }{}_{\alpha }{n}^{\alpha }=0$ and $P^{\mu }{}_{\alpha }{P}^{\alpha }{}_{\nu }=P^{\mu }{}_{\nu }$. The spacetime form of the induced metric is thereforeIts restriction to the slice is the given three-metric. Metric compatibility of the four-dimensional Levi-Civita connection givesApply the three projectors defining the spatial covariant derivative. In the first summand $P^{\beta }{}_j{n}_{\beta }=0$, and in the second $P^{\alpha }{}_i{n}_{\alpha }=0$. Both summands vanish, soThis proves the metric compatibility of the spatial covariant derivative without treating the spatial projection tensor itself as covariantly constant.

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 inverse spacetime metric decomposes asContracting the covariant derivative of $Z_{\nu }$ and recognizing the fully projected contraction as its spatial covariant derivative gives

For a minimally coupled scalar field, put $q^2=h^{ij}{D}_i\varphi D_j\varphi$ using the spatial covariant derivative. Spatial and normal projections of its stress-energy tensor give $\rho ={\Pi }^2/2+q^2/2+V$, ${\mathcal{J}}_i=-\Pi D_i\varphi$, and $S_{ij}=D_i\varphi D_j\varphi +h_{ij}({\Pi }^2/2-q^2/2-V)$. The trace-free stress is $D_i\varphi D_j\varphi -h_{ij}{q}^2/3$, with coefficient one. This makes the scalar anisotropic stress second order in a spatial gradient expansion.