Normal and Tangential Projections of Curvature

The four-dimensional Riemann tensor has three nontrivial projections relative to a spacelike slice. The fully spatial Gauss relation and the one-normal Codazzi relation contain only data intrinsic to a single slice; after contraction and use of Einstein’s equation, they become the Hamiltonian and momentum constraints. The two-normal Ricci relation contains a normal derivative of $K_{ij}$ and therefore belongs to evolution.

Required background. Hypersurface Normals and Tangential Projectors supplies the projected connection and induced tensors, while The Einstein Equation supplies the field equation used to interpret the curvature projections.

Helpful background. The Gibbons-Hawking-York Boundary Term supplies the boundary interpretation of the extrinsic-curvature data appearing in the projections.

The fully spatial Gauss relation

Fix

$$\gamma_{\mu\nu}=g_{\mu\nu}+n_\mu n_\nu, \qquad K_{\mu\nu} =-\gamma_\mu{}^\rho\gamma_\nu{}^\sigma\nabla_\rho n_\sigma.$$

Commuting projected covariant derivatives of a spatial vector gives the Gauss relation

$$\gamma_i{}^\mu\gamma_j{}^\nu \gamma_k{}^\rho\gamma_l{}^\sigma {}^{(4)}R_{\mu\nu\rho\sigma} ={}^{(3)}R_{ijkl} +K_{ik}K_{jl}-K_{il}K_{jk}.$$

No normal derivative occurs: every term is determined by $(\gamma_{ij},K_{ij})$ and spatial derivatives on one slice. Double contraction yields

$${}^{(3)}R+K^2-K_{ij}K^{ij} =2G_{\mu\nu}n^\mu n^\nu.$$

Using

$$G_{\mu\nu}+\Lambda g_{\mu\nu}=8\pi G T_{\mu\nu}, \qquad \rho=T_{\mu\nu}n^\mu n^\nu,$$

and $g(n,n)=-1$ gives the Hamiltonian constraint

$$\mathcal H \equiv{}^{(3)}R+K^2-K_{ij}K^{ij} -16\pi G\rho-2\Lambda=0.$$

This equation restricts data; it does not determine a second normal derivative of the metric.

The Codazzi relation and momentum constraint

Projecting three Riemann indices spatially and one along the normal gives, in an index order that fixes the sign,

$$n^\sigma\gamma_k{}^\rho \gamma_i{}^\mu\gamma_j{}^\nu {}^{(4)}R_{\rho\sigma\mu\nu} =D_jK_{ki}-D_iK_{kj}.$$

Contracting $i$ with $k$ produces

$$\gamma_i{}^\mu n^\nu{}^{(4)}R_{\mu\nu} =D_iK-D_jK^j{}_i.$$

The mixed metric projection vanishes, so neither the trace term in the Einstein tensor nor $\Lambda g_{\mu\nu}$ contributes. With the momentum-density convention

$$j_i=-\gamma_i{}^\mu n^\nu T_{\mu\nu},$$

one has $T(n,\partial_i)=-j_i$. The mixed Einstein equation is therefore equivalent to

$$\mathcal M_i \equiv D_jK^j{}_i-D_iK-8\pi G j_i=0.$$

The minus sign in the definition of $j_i$ is essential. Omitting it reverses the matter source while leaving the geometric Codazzi identity unchanged.

The two-normal Ricci relation

The remaining independent projection has two normal legs. For

$$(\partial_t)^\mu=\alpha n^\mu+\beta^\mu,$$

it is

$$\gamma_i{}^\mu n^\rho \gamma_j{}^\nu n^\sigma {}^{(4)}R_{\mu\rho\nu\sigma} =\frac1\alpha (\partial_t-\mathcal L_\beta)K_{ij} +\frac1\alpha D_iD_j\alpha +K_{ik}K^k{}_j.$$

Unlike Gauss and Codazzi, this expression contains $(\partial_t-\mathcal L_\beta)K_{ij}$. Combining it with the contracted Gauss relation gives the fully spatial projection of the four-dimensional Ricci tensor and hence the evolution equation for $K_{ij}$.

The division is thus structural rather than a component-counting convention:

• zero normal derivatives of $K_{ij}$ give intrinsic restrictions on each slice;
• one normal derivative of $K_{ij}$ supplies dynamical evolution;
• the contracted Bianchi identity propagates the constraints when the evolution equations and matter conservation hold.

The Gauss and Codazzi derivations and the last two-normal projection appear in Gourgoulhon 2007, §2.5, pp. 34–37 and §§3.4.1–3.4.2, pp. 47–50, PDF; their Einstein-equation contractions are given in Gourgoulhon 2007, §4.1.3, pp. 53–54, PDF.

Minkowski and Milne checks

On inertial Minkowski slices,

$${}^{(3)}R_{ijkl}=0, \qquad K_{ij}=0, \qquad \alpha=1, \qquad \beta^i=0.$$

Every curvature projection vanishes term by term, and the vacuum constraints hold for $\Lambda=0$.

Milne slices give a sharper sign test because neither intrinsic nor extrinsic curvature vanishes even though the ambient Riemann tensor does. With $\gamma_{ij}=\tau^2q_{ij}$,

$${}^{(3)}R_{ijkl} =-\frac1{\tau^2} (\gamma_{ik}\gamma_{jl}-\gamma_{il}\gamma_{jk}), \qquad K_{ij}=-\frac1\tau\gamma_{ij}.$$

The quadratic $K$ terms in the Gauss relation are

$$K_{ik}K_{jl}-K_{il}K_{jk} =\frac1{\tau^2} (\gamma_{ik}\gamma_{jl}-\gamma_{il}\gamma_{jk}),$$

and cancel the intrinsic curvature exactly. Since $D_iK_{jk}=0$, Codazzi also vanishes. Finally,

$$\partial_\tau K_{ij} =-\frac1{\tau^2}\gamma_{ij}, \qquad K_{ik}K^k{}_j =\frac1{\tau^2}\gamma_{ij},$$

so the two terms on the right of the Ricci relation cancel. The scalar check is

$${}^{(3)}R+K^2-K_{ij}K^{ij} =-\frac6{\tau^2}+\frac9{\tau^2}-\frac3{\tau^2}=0.$$

All three projections therefore recover flat four-dimensional spacetime.

Common pitfalls

Gauss and Codazzi are geometric identities, not field equations. They become Einstein constraints only after the curvature contractions are related to $T_{\mu\nu}$.

A constraint contains no second time derivative by construction. Its failure indicates inadmissible data or departure from the Einstein solution set, not a different gauge choice.

Index order fixes the Codazzi sign. State the projected Riemann indices and the definition of $j_i$ before quoting the momentum constraint.

Exercise: recover both Einstein constraints

Starting from the scalar Gauss relation

$${}^{(3)}R+K^2-K_{ij}K^{ij}=2G(n,n)$$

and the contracted Codazzi relation above, derive $\mathcal H=0$ and $\mathcal M_i=0$ including $\Lambda$ and matter.

Solution

The normal-normal Einstein equation gives

$$G(n,n)+\Lambda g(n,n)=8\pi G\rho.$$

Since $g(n,n)=-1$, $G(n,n)=8\pi G\rho+\Lambda$. Multiplying by two and substituting into scalar Gauss yields

$${}^{(3)}R+K^2-K_{ij}K^{ij} =16\pi G\rho+2\Lambda,$$

which is $\mathcal H=0$.

For the mixed projection, $g(n,\partial_i)=0$, so

$$R(n,\partial_i)=G(n,\partial_i) =8\pi G T(n,\partial_i)=-8\pi G j_i.$$

Codazzi gives $R(n,\partial_i)=D_iK-D_jK^j{}_i$. Multiplying by $-1$ therefore gives

$$D_jK^j{}_i-D_iK=8\pi G j_i,$$

which is $\mathcal M_i=0$. The cosmological term has no mixed projection.

References

• Gourgoulhon, Éric. 3+1 Formalism and Bases of Numerical Relativity. Lecture notes, arXiv:gr-qc/0703035v1 (2007). arXiv.