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Take a complete smooth asymptotically flat spacelike initial-data hypersurface $\Sigma$, carrying a spin structure, without inner boundary. Let $n$ be its future unit normal, $h$ its induced metric, and use the extrinsic curvature convention $k_{ij}=-g({\nabla }_{i}n,e_j)$. Assume the Einstein constraints, the dominant energy condition, and the standard decay sufficient for finite ADM charges. In one asymptotically Cartesian end it is enough here to work with $h_{ij}={\delta }_{ij}+q_{ij}$, $q=O(r^{-1})$, $\partial q=O(r^{-2})$, $k=O(r^{-2})$, and the corresponding differentiated decay and integrability assumptions. The desired conclusion is positivity of the whole ADM four-momentum, not merely of the energy in one selected frame.

Introduce a commuting auxiliary spinor $ϵ$; this is not an anticommuting gravitino field. With $\overline{ϵ}=-{ϵ}^{†}{\gamma }^{\hat{0}}$ its causal Dirac spinor current is $K^{\mu }=\overline{ϵ}{\gamma }^{\mu }ϵ$, future causal. The Nester two-form isIt is real in these conventions and antisymmetric in $\mu ,\nu$. Differentiating, the two terms quadratic in first derivatives become equal after exchanging $\nu ,\rho$. The remaining terms are curvature commutators. The Einstein tensor contraction of spin curvature gives ${\gamma }^{\mu \nu \rho }{\nabla }_{\nu }{\nabla }_{\rho }=G^{\mu }{}_{\lambda }{\gamma }^{\lambda }/2$; the adjoint term supplies the other half. ThusThis is the central spinor-curvature identity, derived by differentiation rather than posited as a positivity formula.

Project onto $\Sigma$ and put $b^i=n_{\mu }{B}^{\mu i}$. In an adapted orthonormal frame $n_{\mu }=(-1,0,0,0)$, Clifford algebra givesIndeed the expansion of $|{\gamma }^i{\nabla }_iϵ{|}^2$ is ${\sum }_i|{\nabla }_iϵ{|}^2+({\nabla }_iϵ{)}^{†}{\gamma }^{ij}{\nabla }_jϵ$. With a Gaussian normal extension of the slice, the symmetric second fundamental form has zero contraction with the antisymmetric $B$, so the projected divergence is the intrinsic spatial divergence. This proves the Witten-Nester identityHere $D_i$ on $b$ denotes the spatial metric derivative. On spinors, ${\nabla }_i$ is the Sen spinor connection, the restriction of the spacetime connection, not just the intrinsic three-dimensional one:The extrinsic-curvature term is essential when the data have momentum.

Choose $ϵ$ to solve the Witten spinor equation, $\mathcal{D}ϵ=0$, approaching a prescribed constant ${ϵ}_{\infty }$ in the chosen end and zero constants in any other ends. The Einstein constraints identify $G_{\mu \nu }{n}^{\mu }{K}^{\nu }=8\pi G{T}_{\mu \nu }{n}^{\mu }{K}^{\nu }$. The dominant energy condition makes this nonnegative because both $n$ and $K$ are future causal. Consequently the right-hand side of the identity becomes a sum of nonnegative terms.

Existence of that spinor is an analytic part of the proof, not something established by the positivity identity alone. Extend ${ϵ}_{\infty }$ smoothly using a cutoff and solve $\mathcal{D}u=-\mathcal{D}{ϵ}_0$ for a decaying correction. The spatial principal symbol ${\gamma }^i{\xi }_i$ squares to $|\xi {|}_h^2$, so the equation is elliptic. For a compactly supported spinor, the integrated identity givesTogether with the weighted asymptotic estimates this gives the coercivity needed for elliptic existence of a Witten spinor. A decaying homogeneous solution would be parallel by the same identity and hence zero, excluding the kernel; the appropriate weighted elliptic solvability gives the correction. Under the stated smooth decay assumptions it has $ϵ-{ϵ}_{\infty }=O(r^{-1})$ and first derivative $O(r^{-2})$. The rigorous analytic step is supplied by the spinor existence proof. This also explains why completeness, asymptotic data and the absence of uncontrolled inner boundary terms are hypotheses of the argument.

It remains to identify the boundary integral with the physical ADM charge. In the asymptotic symmetric coframe $e_i{}^a={\delta }_i{}^a+q_{ia}/2+O(r^{-2})$, the leading spatial spin connection isInsert this and the Sen correction into $b^i$. The real scalar part of ${\gamma }^{ij}{\gamma }^{kl}$ is ${\delta }^{il}{\delta }^{jk}-{\delta }^{ik}{\delta }^{jl}$; its two-gamma part has imaginary expectation and drops out. It follows that the intrinsic connection contributes $\frac{1}{2}({\partial }_j{q}_{ij}-{\partial }_i{q}_{jj})K_{\infty }^{\hat{0}}$. For the other term, symmetry of $k$ removes the three-spatial-gamma product, leaving $-(k_{ij}-k{h}_{ij})K_{\infty }^{\hat{j}}$. Thus the ADM boundary term of the Nester two-form has the expansionThe constant-spinor derivative contribution integrates to zero on a large coordinate sphere. Its coefficient is antisymmetric in $i,j$; the corresponding tangential vector on the sphere is divergence free, so integration by parts removes it. Equivalently, the flux of ${\gamma }^{ij}{\partial }_{j}u$ has identically zero divergence, with its asymptotic leading term having zero flux. The quadratic correction terms are $O(r^{-3})$ and their surface integrals vanish in the limit.

Define the ADM energy and ADM momentum with the same conventions:The boundary calculation then gives, including every normalization factor,This is the energy associated with the asymptotic spinor's causal translation. In the zero-momentum rest frame it is the ADM mass times $K_{\infty }^{\hat{0}}$; more generally it proves the full energy-momentum bound.

Finally apply the divergence theorem to the Witten identity. There is no inner boundary, and the spinors assigned zero at other ends give no charge there. ThereforeAsymptotic constant spinors can produce every future null direction. Normalize $K_{\infty }^{\hat{0}}=1$ and choose its spatial direction along $\mathbf{P}$; the inequality givesThis establishes the positive energy theorem and nonnegativity of the invariant mass when it is defined. For zero ADM energy, repeat the identity for a basis of asymptotic spinors: all become parallel and all matter contractions vanish. Their curvature integrability forces the spacetime curvature to vanish on the initial surface; the standard rigidity conclusion is that the data embed in Minkowski spacetime. Merely having a parallel null spinor in a local wave geometry would not by itself prove this global rigidity: completeness and asymptotic flatness remain essential.

A Killing spinor is a nonzero spinor parameter for which the fermionic supersymmetry variations vanish on a bosonic background. In minimal four-dimensional supergravity without a cosmological term this meansIt is an overdetermined equation and expresses an unbroken supersymmetry of the background. In the cosmological deformation the corresponding equation is ${\mathcal{D}}_{\mu }ϵ=0$; in theories with additional bosonic fields their contributions also enter the fermion variations.

For the undeformed theory the connection preserves both the spinor adjoint and gamma matrices. Consequently the Dirac currentobeys ${\nabla }_{\nu }{K}_{\mu }=0$. In particular ${\nabla }_{(\nu }{K}_{\mu )}=0$, so $K$ is a Killing vector, and in fact is parallel. For the real cosmological deformation, ${\nabla }_{\mu }ϵ=-(m/2){\gamma }_{\mu }ϵ$ and ${\nabla }_{\mu }\overline{ϵ}=(m/2)\overline{ϵ}{\gamma }_{\mu }$ instead give ${\nabla }_{\mu }{K}_{\nu }=(m/2)\overline{ϵ}[{\gamma }_{\mu },{\gamma }_{\nu }]ϵ$, whose symmetric part again vanishes.

The causal property uses an ordinary commuting spinor. Take an orthonormal frame with $({\gamma }^0{)}^{†}=-{\gamma }^0$, $({\gamma }^i{)}^{†}={\gamma }^i$, and $\overline{ϵ}={ϵ}^{†}{\gamma }^0$. Then $K^0=-{ϵ}^{†}ϵ$. Reverse its time orientation and write $V=-K$, soThe matrices ${\alpha }^i$ are Hermitian and obey $\{{\alpha }^i,{\alpha }^j\}=2{\delta }^{ij}$. For every spatial unit vector $u$, $(u_i{\alpha }^i{)}^2=1$, whence $|u_i{V}^i|\le V^0$. Choose $u$ along the spatial part of $V$ to obtainA nonzero Killing spinor cannot vanish at an isolated point: its defining connection equation transports its value along every curve. Thus on a connected background this is a nonzero, nowhere-spacelike causal Dirac spinor current. The estimate does not need the Majorana condition, but it does need commuting components; Grassmann bilinears have no such ordinary positivity order.

For the positive energy theorem, take a complete asymptotically flat spacelike hypersurface with a spin structure $\Sigma$, with suitable decay, Einstein constraint equations, and matter satisfying the dominant energy condition. Assume there are no inner boundaries, or that any inner-boundary contribution has the required nonnegative sign. The auxiliary commuting spinor approaches an arbitrary constant ${ϵ}_{\infty }$ at spatial infinity. Define the real Nester two-formAntisymmetry removes second derivatives except through the Riemann curvature tensor. Differentiation and the three-gamma spinor curvature identity used above giveThe first term comes from differentiating the two spinor factors. For the second term, replace ${\nabla }_{\nu }{\nabla }_{\rho }$ by half their commutator; its spin-curvature contraction is $G^{\mu \nu }{\gamma }_{\nu }/2$ from each conjugate term. This is the reason the Einstein tensor, and hence the matter energy density, enters the boundary-charge identity.

In an orthonormal frame adapted to the future normal of $\Sigma$, put ${\chi }_i={\nabla }_iϵ$. Since ${\gamma }^0{\gamma }^{0ij}=-{\gamma }^{ij}$,Here ${\nabla }_i$ is the spacetime spin connection pulled back to $\Sigma$, including its extrinsic curvature. Choose the auxiliary spinor to solve the elliptic Witten spinor equationThe standard asymptotic-flatness and completeness hypotheses permit this boundary-value problem; the positivity identity also excludes a decaying homogeneous zero mode. It is this hypersurface equation that removes the negative square. A spacetime Killing spinor is not assumed for a general initial-data set.

Stokes' theorem and $G_{\mu \nu }={\kappa }^2{T}_{\mu \nu }$ now giveThe dominant energy condition makes the second integrand nonnegative because $n$ and $V$ are future causal. Evaluating the two-form with the asymptotic spin connection identifies its surface integral with the ADM energy and momentum:For example its energy term uses $B^{0i}=\frac{1}{2}({\partial }_j{h}_{ij}-{\partial }_i{h}_{jj})|{ϵ}_{\infty }{|}^2$ to first order in the asymptotic spatial metric tensor perturbation. With ${\kappa }^2=8\pi G$ this is exactly the ADM normalization. Since $P_i{\alpha }^i$ has eigenvalues $\pm |\mathbf{P}|$, positivity for every boundary spinor provesKilling spinors describe saturation: their spatial derivative term vanishes, and the contracted matter term must also vanish for the corresponding charge to be zero. They characterize preserved supersymmetry rather than being prerequisites for positivity. Under the usual rigidity hypotheses, zero total four-momentum makes the nonnegative integral vanish for a full basis of asymptotic spinors; their parallel extensions force a vanishing Riemann curvature tensor and give the Minkowski vacuum. A single Killing spinor with null current alone does not justify asserting that every component of the total momentum vanishes.

Choose the outward normal vector $n=(\cos \phi ,\sin \phi ,0)$ and the positive conventionThe ambient connection vanishes in Cartesian coordinates. Since ${\partial }_{\phi }n=e_{\phi }/\rho$ and ${\partial }_{\hat{z}}n=0$, the extrinsic curvature components are $K_{\phi \phi }=\rho$, $K_{\phi \hat{z}}=K_{\hat{z}\hat{z}}=0$. Taking the trace using the induced metric,Reversing the normal or using the negative extrinsic-curvature convention gives $-1/\rho$. The two principal curvatures are $1/\rho$ and zero in the chosen convention; the averaged mean curvature would be $K/2$, so it must not be confused with the requested trace.

The two coordinate tangent vectors of the embedding areTheir Euclidean inner products give the pullback of a Riemannian metric, equivalently the induced metric:Locally set $u=\rho \phi$. The line element becomes $d{u}^2+d{\hat{z}}^2$, so the induced Riemann curvature tensor vanishes identically. This is the intrinsic flatness of a circular cylinder. Its bending in the ambient space is extrinsic curvature, not intrinsic curvature; periodicity of the angular coordinate does not change the local flatness.

With vanishing extrinsic curvature, the momentum constraint is identically satisfied and the vacuum Hamiltonian constraint reduces to $\mathcal{R}=0$. To evaluate that Ricci scalar, write ${\psi }_i={\partial }_i\psi$, raise these intermediate indices with ${\delta }^{ij}$, and use ${\gamma }^{ij}={\psi }^{-4}{\delta }^{ij}$. The Levi-Civita connection isSubstitution into the stated curvature convention givesTracing with ${\gamma }^{ij}$ cancels the gradient-square terms:Here the Laplacian and norm on the right are those of the flat Euclidean metric. Thus the time-symmetric conformally flat vacuum initial data constraints becomeThe equivalence uses $\psi \ne 0$. This identity is also the three-dimensional specialization of scalar curvature under conformal rescaling; locally the conformal exponent is $2\log |\psi |$, so either fixed nonzero sign of $\psi$ gives the same metric.

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.

Write $\mathcal{D}={\bar{N}}^{-1}{\partial }_t$, $\bar{\Pi }=\mathcal{D}\bar{\varphi }$ and $\Delta =a^{-2}{\nabla }^2$. Since the background scalar has no spatial gradient,Also $\delta P=\bar{\Pi }\mathcal{D}\delta \varphi -{\bar{\Pi }}^2\Psi -V_{,\varphi }\delta \varphi$. These are the linearized scalar-field matter projections; gradient energy first contributes quadratically.

There is a source convention defect in the shear formula. For the stated positive shift $N_i=a^2{B}_{,i}$ and spatial metric with $+2E_{,ij}$, direct substitution into the extrinsic curvature gives the positive-shift scalar shear conventionThere is no extra Laplacian in the definition of ${\chi }_{+}$. Equivalently use $\chi =-{\chi }_{+}$: then the trace-free term has the printed minus sign and $\kappa =3(\mathcal{D}\Phi +H\Psi )-\Delta \chi$. This latter convention retains the printed momentum-constraint form. The source mixes these two choices.

At first order $\delta (K^2)=-6H\kappa$ and $\delta (K^i{}_j{K}^j{}_i)=-2H\kappa$. Combining their difference with $\delta {}^{(3)}R=4\Delta \Phi$ givesFor the momentum constraint, differentiating the tensor above givesFor nonzero Fourier wavenumber, and absorbing the homogeneous integration mode into the background, the consistent momentum constraint is

Linearizing the scalar equation, the shift-advection term and lapse-gradient times scalar-gradient term vanish at this order because the background is spatially homogeneous. The remaining equation isInsert $\delta \Pi$, differentiate its lapse term, and use $\mathcal{D}\bar{\Pi }+3H\bar{\Pi }=-V_{,\varphi }$. The two background-acceleration terms combine into $2\Psi V_{,\varphi }+3H\bar{\Pi }\Psi$. Hence the perturbed Klein-Gordon equation with background lapse isThis scalar evolution equation and the Hamiltonian constraint agree with the displayed targets after the shear conventions are repaired.

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.

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.

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 projection $X\mapsto X_{\parallel }$ is tensorial. Although a covariant derivative is not $C^{\infty }$-linear in its second argument, the extra term in ${\nabla }_{X_{\parallel }}(f{Y}_{\parallel })$ is proportional to $Y_{\parallel }$, whose contraction with the normal vanishes. Thus $K(X,Y)$ is $C^{\infty }$-linear in both arguments and defines a $(0,2)$ tensor on the hypersurface.

Since $n_a{Y}_{\parallel }^a=0$,Therefore the extrinsic curvature is

With a future unit timelike normal in signature $(-,+,+,+)$, this convention makes the extrinsic curvature trace equal to the normal congruence's expansion, $K={\nabla }_{\mu }{n}^{\mu }$. An expanding flat Friedmann-Lemaître-Robertson-Walker metric has $K_{ij}=a^{\prime }{\delta }_{ij}$ and $K=3a^{\prime }/a^2$. This is the opposite sign to the negative-expansion convention used in many 3+1 decomposition of spacetime formulas. Projection identities and momentum constraints must use the same convention throughout.

Riemannian geometry studies smooth manifolds equipped with a Riemannian metric: an inner product at each tangent space, varying smoothly with position. This local measuring rule determines angles, arc length, Riemannian distance and geodesics. An inhabitant of a surface can discover its geometry by measuring within it, without knowing how it sits in a larger Euclidean space.

A circular cylinder is a particularly useful example. Its surface bends in space, yet unrolling it preserves the induced metric: a small patch has the geometry of a flat sheet. By contrast, a round sphere has positive Gaussian curvature. The distinction between intrinsic flatness of a circular cylinder and extrinsic curvature separates the geometry an inhabitant measures from the way an outside observer sees the surface bend.

Geodesics supply the analogue of straight lines. On a round sphere they follow great circles; on a circular cylinder they become straight lines when lifted to an unrolled sheet. Their arc length lets us compare candidate routes, as in great-circle distance and Riemannian distance on a circular cylinder. The same measuring rule also defines differential operators such as the Laplace-Beltrami operator, connecting geometry with diffusion and spectral geometry.

Bernhard Riemann's 1854 inaugural lecture proposed studying spaces of arbitrary dimension through their infinitesimal measuring rules, including a positive quadratic expression for the squared line element. This moved the subject beyond a particular surrounding space: the measuring rule itself became part of the geometry. ^[ref].

With negative extrinsic curvature convention, the background trace is $\bar{K}=-3H$, where $H=\dot{a}/(a\bar{N})$. The scalar expansion perturbation is $\kappa =K+3H$. Use the explicit ADM conventionwith ${\gamma }_{ij}=a^2[(1-2\Phi ){\delta }_{ij}+2{\partial }_i{\partial }_{j}E]$. Substituting into $K_{ij}=-({\dot{\gamma }}_{ij}-D_i{\beta }_j-D_j{\beta }_i)/(2N)$ giveswhere $\chi =a^2(\dot{E}-B)/\bar{N}$ is the scalar shear potential and $\Delta =a^{-2}{\nabla }^2$.

Thus the positive metric cross term and ${\beta }_i=+a^2{\partial }_{i}B$ accompany this choice of $\chi$; changing the shift sign requires changing its definition too. For the future unit-normal congruence, the expansion is $-K=3H-\kappa$, so positive $\kappa$ means reduced expansion. A homogeneous perturbation with $\Psi =0$, $B=E=0$ checks the sign: increasing $\Phi$ decreases the local spatial scale and gives $\kappa =3\dot{\Phi }/\bar{N}$.

For scalar metric potentials $d{s}^2=a^2[-(1+2A)d{\tau }^2+2B_{,i}d\tau d{x}^i+((1-2\psi ){\delta }_{ij}+2E_{,ij})d{x}^{i}d{x}^j]$ and positive extrinsic curvature, $\delta K^i{}_j=a^{-1}[-({\psi }^{\prime }+\mathcal{H}A){\delta }^i{}_j+{\partial }^i{\partial }_j(E^{\prime }-B)]$. The scalar shear terms cancel between its divergence and trace gradient. With $j_i=-P^{\mu }{}_i{T}_{\mu \nu }{n}^{\nu }$, the momentum constraint becomes $2{\partial }_i({\psi }^{\prime }+\mathcal{H}A)/a=-8\pi G{j}_i$. Reversing the convention for $K$ reverses the geometric left side.

Choose the ADM shift vector ${\beta }^i$ throughwith $N=\bar{N}(1+\Psi )$ and ${\gamma }_{ij}=a^2[(1-2\Phi ){\delta }_{ij}+2{\partial }_i{\partial }_{j}E]$. For negative extrinsic curvature, $K_{ij}=-({\dot{\gamma }}_{ij}-D_i{\beta }_j-D_j{\beta }_i)/(2N)$. The scalar shear potential isRaised spatial derivatives use $a^{-2}{\delta }^{ij}$, so $\Delta =a^{-2}{\nabla }^2$. The expression follows by keeping the trace-free part of ${\dot{\gamma }}_{ij}$ and the symmetrized shift gradient in the extrinsic-curvature formula. The sign of $B$ is therefore tied to the metric cross term; it is not determined by the name “shift” alone.

The positive-shift scalar shear convention instead writes ${\chi }_{+}=-\chi$ for this same explicit metric. Its trace-free and trace shear terms both reverse their displayed sign. Stating the line element prevents a mistaken combination of a minus-cross-term shift definition with a plus-cross-term extrinsic-curvature formula.

Restrict the spacetime spin connection to a spacelike hypersurface. With future unit normal $n$ and convention $k_{ij}=-g({\nabla }_{i}n,e_j)$, it differs from the intrinsic spin connection by the displayed extrinsic curvature term. Its spatial Dirac operator is the operator in the Witten spinor equation; replacing it by the intrinsic operator would omit the ADM momentum contribution.

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.

For a spatial initial slice with vanishing extrinsic curvature, the vacuum momentum constraint is automatic and the Hamiltonian constraint requires zero Ricci scalar. In three spatial dimensions, the indicated conformally flat metric has $\mathcal{R}=-8{\psi }^{-5}\Delta \psi$, with a nonzero conformal factor. Thus a harmonic function supplies the data on its domain. A positive example is $\psi =1+a/r$ on a punctured or exterior region. There is no nonconstant globally smooth example on all of Euclidean three-space tending to one at infinity, by the maximum principle for harmonic functions. Time symmetry of an initial slice does not imply that its entire evolved spacetime is static.

A hypersurface is totally geodesic when its intrinsic geodesics are also ambient geodesics. Equivalently, for a non-null hypersurface with the induced metric, its extrinsic curvature vanishes. The normal part of the ambient covariant derivative along a curve of the tangent is the second fundamental form, so its vanishing identifies the two geodesic equations.

This is a local statement: an ambient geodesic starting tangent to the hypersurface stays in it for a sufficiently short interval. Global containment needs the relevant completeness or closedness hypothesis. An open line segment inside the Euclidean plane has zero extrinsic curvature, but an ambient straight line leaves the segment in finite time. The whole line is totally geodesic and complete. A great circle on a round sphere is another complete example.