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Gauss-Codazzi decomposition in Gaussian normal coordinates and the constraint equations in terms of the induced metric and second fundamental form

Statement

Let $(M, g, N)$ be a time-oriented smooth Lorentzian four-manifold, and let $Σ \subseteq M$ be a smooth spacelike embedded hypersurface with future unit normal $n$ , induced Riemannian metric $h$ and second fundamental form $K (X, Y) = g (\nabla_{X} n, Y)$ (Spacelike hypersurfaces, acausal sets, Cauchy sets, domains of dependence and Cauchy horizons, The future unit normal, induced metric and second fundamental form of a spacelike hypersurface). Write $D$ for the Levi-Civita connection of $h$ , $R^{h}$ for its curvature operator in the same operator convention as the ambient curvature, and $S_{h}$ for its scalar curvature. Explicitly distinguish the lowered tensor from the operator pairing by setting $R^{g} (A, B, C, D) : = g (R^{g} (C, D) B, A), R^{h} (A, B, C, D) : = h (R^{h} (C, D) B, A) .$ Thus $R_{a b c d}^{g} = R_{a b c d}$ in The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors; in particular $g (R^{g} (X, Y) Z, W) = R^{g} (W, Z, X, Y)$ . Let $G$ be the Einstein tensor defined there. At every $p \in Σ$ and for all $X, Y, Z, W \in T_{p} Σ$ :

(a) Gauss equation. $R^{g} (X, Y, Z, W) = R^{h} (X, Y, Z, W) + K (X, Z) K (Y, W) - K (X, W) K (Y, Z) .$ Equivalently, in operator order, $g (R^{g} (X, Y) Z, W) = h (R^{h} (X, Y) Z, W) + K (Y, Z) K (X, W) - K (X, Z) K (Y, W) .$

(b) Codazzi equation. $g (R^{g} (X, Y) Z, n) = (D_{Y} K) (X, Z) - (D_{X} K) (Y, Z) .$

(c) Normal projections. Define $({div}_{h} K) (Y) = h^{i j} (D_{e_{i}} K) (e_{j}, Y)$ , $τ = {tr}_{h} K = h^{i j} K_{i j}$ and $∣ K ∣_{h}^{2} = h^{i k} h^{j l} K_{i j} K_{k l}$ . Then $G (n, n) = \frac{1}{2} (S_{h} + τ^{2} - ∣ K ∣_{h}^{2}), G (n, Y) = ({div}_{h} K) (Y) - D_{Y} τ .$

(d) Constraint equations. If the tensor equation $G + Λ g = κ T$ holds at $p$ , for a smooth symmetric covariant $2$ -tensor $T$ and real constants $κ, Λ$ , then $S_{h} + τ^{2} - ∣ K ∣_{h}^{2} = 2 Λ + 2 κ T (n, n), ({div}_{h} K) (Y) - D_{Y} τ = κ T (n, Y) .$ With $ρ = T (n, n)$ and $j (Y) = - T (n, Y)$ , the second equation is $({div}_{h} K) (Y) - D_{Y} τ = - κ j (Y)$ .

Facts & Assumptions

Given: The smooth Lorentzian manifold, spacelike embedded hypersurface, normal and tensors in the Statement; in (d), the field equation at the selected point.

[L1]

The normal and second fundamental form are smooth, $g (n, n) = - 1$ , $g (n, X) = 0$ , $K$ is symmetric, $g (\nabla_{X} Y, n) = - K (X, Y)$ and $\nabla_{X} n$ is tangential with $h (\nabla_{X} n, Y) = K (X, Y)$ (The future unit normal, induced metric and second fundamental form of a spacelike hypersurface clauses (a)-(d)).

[L2]

The ambient connection is torsion-free and metric-compatible (Metric compatibility, torsion freedom and the metric-selected connection class, Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields); tangent vector fields have tangential bracket as proved in The future unit normal, induced metric and second fundamental form of a spacelike hypersurface proof 6.1.

[L3]

The operator is $R (X, Y) Z = \nabla_{X} \nabla_{Y} Z - \nabla_{Y} \nabla_{X} Z - \nabla_{[X, Y]} Z$ , and $R_{a b c d} = g (R (e_{c}, e_{d}) e_{b}, e_{a})$ , ${Ric}_{b d} = g^{a c} R_{a b c d}$ , $S = g^{b d} {Ric}_{b d}$ and $G = Ric - \frac{1}{2} S g$ (The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors). Curvature is tensorial and is antisymmetric in its first pair and last pair of lowered slots (Tensoriality of the curvature, its coordinate formula and its point-normal metric formula clause (a), The algebraic symmetries of the curvature tensor and the symmetry of the Ricci and Einstein tensors).

[L4]

Tangential trace, inverse metric and contractions use the usual tensor operations (Tensor fields of type (r,s) on a Lorentzian four-manifold and their components, The inverse metric, musical isomorphisms and index raising). The pointwise field equation, when assumed in (d), may be evaluated on $(n, n)$ and $(n, Y)$ directly; this uses no global field-equation premise.

Proof

technique · direct

1.1L1L2L3algebra

(Induced connection.) On tangent fields define $D_{X} Y = (\nabla_{X} Y)^{⊤}$ , where $V^{⊤} = V + g (V, n) n$ . This defines a smooth connection on $Σ$ because tangential differentiation is independent of ambient extensions. By [L1], $\nabla_{X} Y = D_{X} Y + K (X, Y) n$ and $\nabla_{X} n = A X$ with $h (A X, Y) = K (X, Y)$ . Projecting the torsion identity gives $D_{X} Y - D_{Y} X = [X, Y]$ , and restricting metric compatibility gives $X h (Y, Z) = h (D_{X} Y, Z) + h (Y, D_{X} Z)$ . Thus $D$ is torsion-free and $h$ -compatible. Uniqueness follows directly by solving these two identities for $2 h (D_{X} Y, Z)$ : it equals $X h (Y, Z) + Y h (X, Z) - Z h (X, Y) - h (X, [Y, Z]) + h (Y, [Z, X]) + h (Z, [X, Y])$ . Hence $D$ is the Levi-Civita connection of $h$ , with its curvature and Ricci/scalar contractions defined in the same slot order as [L3].

2.1step 1.1L1L2L3algebra

(Full tangential and normal curvature decomposition.) Expand $\nabla_{X} \nabla_{Y} Z$ using step 1.1: its tangential part is $D_{X} D_{Y} Z + K (Y, Z) A X$ , and its coefficient of $n$ is $K (X, D_{Y} Z) + X (K (Y, Z))$ . Subtract the analogous expression with $X, Y$ exchanged and subtract $\nabla_{[X, Y]} Z$ . The tangential part is $R^{h} (X, Y) Z + K (Y, Z) A X - K (X, Z) A Y$ . The coefficient of $n$ is $X (K (Y, Z)) - Y (K (X, Z)) + K (X, D_{Y} Z) - K (Y, D_{X} Z) - K ([X, Y], Z)$ . Writing $(D_{X} K) (Y, Z) = X (K (Y, Z)) - K (D_{X} Y, Z) - K (Y, D_{X} Z)$ , and using $D_{X} Y - D_{Y} X = [X, Y]$ , that coefficient is $(D_{X} K) (Y, Z) - (D_{Y} K) (X, Z)$ . Pairing the tangential part with $W$ yields the operator Gauss display, and replacing $(X, Y, Z, W)$ by $(Z, W, Y, X)$ yields its lowered-slot display. Pairing the normal part with $n$ multiplies its coefficient by $g (n, n) = - 1$ , yielding exactly (b).

3.1step 2.1L3L4algebra

(Double tangential contraction.) Choose any tangent basis $e_{i}$ at $p$ and complete it by $e_{0} = n$ . Then $g^{00} = - 1$ , $g^{0 i} = 0$ and $g^{i j} = h^{i j}$ . Let $U_{i j} = R_{0 i 0 j}$ and $Q = h^{i k} h^{j l} R_{i j k l}$ in the lowered-slot convention of [L3]. Contracting the Gauss identity of step 2.1 gives $Q = S_{h} + τ^{2} - ∣ K ∣_{h}^{2}$ : its first contraction is the intrinsic scalar by definition, its $K_{i k} K_{j l}$ contraction is $τ^{2}$ , and its $K_{i l} K_{j k}$ contraction is $∣ K ∣_{h}^{2}$ . No normal derivative enters $Q$ .

3.2step 1.1step 2.1L3L4algebra

(Exact mixed contraction.) From step 2.1 in the same basis, $R_{0 i j k} = g (R (e_{j}, e_{k}) e_{i}, n) = (D_{e_{k}} K) (e_{j}, e_{i}) - (D_{e_{j}} K) (e_{k}, e_{i})$ . First-pair antisymmetry gives $R_{i 0 j k} = (D_{e_{j}} K) (e_{k}, e_{i}) - (D_{e_{k}} K) (e_{j}, e_{i})$ . Therefore ${Ric}_{0 k} = h^{i j} R_{i 0 j k} = h^{i j} (D_{e_{j}} K) (e_{k}, e_{i}) - h^{i j} (D_{e_{k}} K) (e_{j}, e_{i}) = ({div}_{h} K)_{k} - D_{e_{k}} τ$ , using symmetry of $K$ and $D h^{- 1} = 0$ . Since $g_{0 k} = 0$ , $G_{0 k} = {Ric}_{0 k}$ . This proves the second formula in (c).

4.1step 3.1L3algebra

(Exact scalar and Einstein contractions.) By [L3] and the two pair antisymmetries, ${Ric}_{00} = h^{i j} R_{i 0 j 0} = h^{i j} U_{i j}$ and ${Ric}_{i j} = - R_{0 i 0 j} + h^{k l} R_{k i l j} = - U_{i j} + h^{k l} R_{k i l j}$ . Hence $h^{i j} {Ric}_{i j} = Q - h^{i j} U_{i j} = Q - {Ric}_{00}$ , and $S = - {Ric}_{00} + h^{i j} {Ric}_{i j} = Q - 2 {Ric}_{00}$ . Since $g_{00} = - 1$ , $G_{00} = {Ric}_{00} + \frac{1}{2} S = \frac{1}{2} Q$ . This proves the first formula in (c) without needing any value or normal derivative of $U$ .

5.1step 3.1step 4.1step 3.2L4algebra∎

(Constraints.) Evaluate $G + Λ g = κ T$ on $(n, n)$ and $(n, Y)$ . As $g (n, n) = - 1$ and $g (n, Y) = 0$ , the results are $G (n, n) = Λ + κ T (n, n)$ and $G (n, Y) = κ T (n, Y)$ ; insert steps 3.1, 3.2 and 4.1 and multiply the first equation by $2$ . The stated $j$ form follows from $j (Y) = - T (n, Y)$ . No global, compactness, completeness or field-equation hypothesis is used in (a)-(c), and in (d) only the pointwise field equation is used.

Remarks

In any Gaussian normal chart in which $n = \partial_{0}$ , $g = - (d x^{0})^{2} + h_{i j} (x^{0}, x) d x^{i} d x^{j}$ , one has $K_{i j} = \frac{1}{2} \partial_{0} h_{i j}$ , $Γ^{0}_{i j} = K_{i j}$ , $Γ^{k}_{0 i} = h^{k l} K_{l i}$ , $Γ^{a}_{00} = Γ^{0}_{0 i} = 0$ and $Γ^{k}_{i j} =^{h} Γ^{k}_{i j}$ . Direct substitution into Tensoriality of the curvature, its coordinate formula and its point-normal metric formula clause (b) gives $R^{0}_{i 0 j} = \partial_{0} K_{i j} - K_{j k} K^{k}_{i}$ , so $R_{0 i 0 j} = K_{i k} K^{k}_{j} - \partial_{0} K_{i j}$ . The all-spatial formula is $R_{i j k l} = R_{i j k l}^{h} + K_{i k} K_{j l} - K_{i l} K_{j k}$ , and the one-normal formula is $R_{0 i j k} = (D_{e_{k}} K) (e_{i}, e_{j}) - (D_{e_{j}} K) (e_{i}, e_{k})$ . These formulas use the lowered-slot order explicitly defined above. The proof does not require the construction of a Gaussian chart.

Reversing $n$ reverses $K$ and both mixed projections; the Gauss quadratic terms and $τ^{2} - ∣ K ∣_{h}^{2}$ remain unchanged. In particular replacing $K$ by $- K$ reverses the divergence expression, rather than leaving it invariant.

Gauss-Codazzi decomposition in Gaussian normal coordinates and the constraint equations in terms of the induced metric and second fundamental form

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

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