Extrinsic curvature of a spatial hypersurface ($K_{ij}$)

With future unit normal $n^{\mu }$, the convention $K_{\mu \nu }=-{\perp }^{\rho }{}_{\mu }{\nabla }_{\rho }{n}_{\nu }$ measures how the hypersurface is embedded in spacetime.

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.

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 $d{s}^2=-N^{2}d{t}^2+h_{ij}(d{x}^i-N^{i}dt)(d{x}^j-N^{j}dt)$ 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.

For the threading convention $d{x}^i-N^{i}dt$ and $K_{ij}=-({\dot{\gamma }}_{ij}+D_i{N}_j+D_j{N}_i)/(2N)$, contract with ${\gamma }^{ij}$. Metric compatibility gives ${\gamma }^{ij}{D}_i{N}_j=D_i{N}^i$, and the Jacobi determinant derivative formula gives ${\gamma }^{ij}{\dot{\gamma }}_{ij}={\partial }_t\log \det \gamma$. With zero shift, $-K/3=(3N{)}^{-1}{\partial }_t\log \sqrt{\gamma }$ is the local volume Hubble parameter. Reversing the shift convention reverses both the normal's spatial component and the shift terms; the two conventions must not be mixed.

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 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.

For a Ricci-flat spacetime with the stated Gauss–Codazzi equations for a spatial hypersurface and induced orientation ${\widetilde{ϵ}}_{abc}=n^d{ϵ}_{dabc}$, the electric part of the Weyl tensor and magnetic part of the Weyl tensor satisfyThe electric relation comes from the vanishing spatial projection of the Ricci tensor. The magnetic relation combines the normal curvature projection with the sign ${ϵ}_{acde}{n}^c=-{\widetilde{ϵ}}_{ade}$.

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.

For the convention $K_{\mu \nu }=-{\perp }^{\rho }{}_{\mu }{\nabla }_{\rho }{n}_{\nu }$, contraction of the Gauss equation gives

The Hamiltonian constraint is the normal-normal projection of the Einstein field equations onto a spatial hypersurface. With the extrinsic-curvature convention used here,

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.

For a spatially flat Friedmann-Lemaître-Robertson-Walker metric, negative extrinsic curvature, and $K=-3H+\kappa$, the Hamiltonian constraint gives the displayed scalar perturbation equation. Indeed $K^2=9H^2-6H\kappa$ and $K^i{}_j{K}^j{}_i=3H^2-2H\kappa$ to first order. The spatial Ricci scalar is $4\Delta \Phi$; subtract the background equation $6H^2=16\pi G\bar{\rho }$ and divide by four.