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The stationary maximal vacuum divergence identity with the article sign

Statement

On a smooth maximal spacelike hypersurface in a vacuum spacetime (here meaning $Ric (g) = 0$ ), decompose the stationary Killing field as $T = N n + X$ , use $k (U, V) = - g (\nabla_{U} n, V)$ , and write $D$ for the Levi–Civita connection of the induced Riemannian metric $h$ . Spatial indices range from $1$ to $3$ , and are raised and lowered with $h$ . Then $D_{i} X_{j} + D_{j} X_{i} = 2 N k_{i j}, D_{i} k^{i j} = 0, D_{i} (k^{i j} X_{j}) = N ∣ k ∣_{h}^{2} .$ Here $∣ k ∣_{h}^{2} = h^{i a} h^{j b} k_{i j} k_{a b}$ . These are conditional local identities on an already supplied maximal hypersurface; they do not construct it.

Statement references

Spacelike sections, maximality and lapse–shift data

Stationary Lorentzian data and tensor conventions

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

Gradient hessian and divergence connection formulas

Killing fields and conserved currents, and on-shell conservation of the Hilbert stress-energy tensor

Product connection on tensor and hom bundles

Induced connections commute with contraction and permutation

The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors

Facts & Assumptions

[F1]

The hypersurface data fix $n, h, k, N, X$ , maximality ${tr}_{h} k = 0$ , and the stationary data fix the ambient metric-compatible connection and the local-flow Killing condition. Spacelike sections, maximality and lapse–shift data Stationary Lorentzian data and tensor conventions

[F2]

For this Lorentzian ambient metric, Proof 1.1 of the constraint supplier proves $D_{U} V = (\nabla_{U} V)^{⊤}$ , hence $h (D_{U} V, W) = g (\nabla_{U} V, W)$ for tangent fields. Its Statement (c) gives $G (n, Y) = ({div}_{h} K) (Y) - D_{Y} {tr}_{h} K$ , where $K (U, V) = g (\nabla_{U} n, V) = - k (U, V)$ is symmetric. Gauss-Codazzi decomposition in Gaussian normal coordinates and the constraint equations in terms of the induced metric and second fundamental form

[F3]

Clause (a) characterizes $L_{T} g = 0$ by the covariant Killing equation $g (\nabla_{U} T, V) + g (\nabla_{V} T, U) = 0$ . Its smooth oriented time-oriented Lorentzian four-dimensional hypotheses are supplied by the stationary data. Killing fields and conserved currents, and on-shell conservation of the Hilbert stress-energy tensor Stationary Lorentzian data and tensor conventions

[F4]

On $T Σ$ and its dual use the product tensor connection induced by $D$ . It differentiates tensor products by the Leibniz rule and commutes with contractions and permutations. Since $D$ is $h$ -compatible, $D h = 0$ ; differentiating $h^{i a} h_{a j} = δ^{i}_{j}$ with these rules also gives $D h^{- 1} = 0$ . Product connection on tensor and hom bundles Induced connections commute with contraction and permutation Spacelike sections, maximality and lapse–shift data

[F5]

For the Riemannian metric $h$ , the divergence of a smooth vector field $Y$ is the trace $D_{i} Y^{i}$ . Gradient hessian and divergence connection formulas

[F6]

The scalar curvature is ${Scal}_{g} = g^{a b} {Ric}_{a b}$ and the Einstein tensor is $G = Ric - \frac{1}{2} {Scal}_{g} g$ . These are geometric tensor definitions, independent of a physical field-equation postulate. The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors

Proof

Given: The smooth stationary Lorentzian data, the supplied maximal spacelike hypersurface, and the geometric vacuum hypothesis $Ric (g) = 0$ , with the conventions and interfaces in [F1]–[F6].

Goal: Derive the three local identities with the displayed sign.

1.1givenF1F2F4

Apply the tangential projection identity in [F2] to the tangent shift $X$ . It gives $g (\nabla_{U} X, V) = h (D_{U} X, V)$ for tangent $U, V$ . Metric compatibility and the dual/product connection in [F4] identify the right side in components with $D_{i} X_{j}$ . The projection identity is valid for a spacelike hypersurface in a Lorentzian manifold; no Riemannian ambient-metric hypothesis is used.

1.2givenF1F2F4F6

The hypothesis $Ric (g) = 0$ implies ${Scal}_{g} = 0$ and $G = 0$ by [F6]. In [F2], $K = - k$ and maximality give ${tr}_{h} K = 0$ , so its mixed normal projection yields ${div}_{h} K = 0$ . Hence $h^{i a} D_{i} k_{a j} = 0$ . Raising $j$ and commuting the inverse metric with $D$ using [F4] gives exactly $D_{i} k^{i j} = 0$ .

2.1givenF1F2F3step 1.1

By [F3], $0 = g (\nabla_{U} T, V) + g (\nabla_{V} T, U)$ . Substitute $T = N n + X$ . The terms $U (N) g (n, V)$ and $V (N) g (n, U)$ vanish by normal orthogonality, and the two normal-derivative terms sum to $- 2 N k (U, V)$ , using the symmetry in [F2]. Step 1.1 converts the remaining terms to $h (D_{U} X, V) + h (D_{V} X, U)$ . Thus $D_{i} X_{j} + D_{j} X_{i} = 2 N k_{i j}$ .

3.1F2F4F5step 2.1step 1.2algebra∎

Set $Y^{i} = k^{i j} X_{j}$ . By [F5] its divergence is $D_{i} Y^{i}$ . The tensor product and contraction rules of [F4] give $D_{i} Y^{i} = (D_{i} k^{i j}) X_{j} + k^{i j} D_{i} X_{j}$ . Step 1.2 eliminates the first term. Symmetry of $k$ and step 2.1 give $k^{i j} D_{i} X_{j} = \frac{1}{2} k^{i j} (D_{i} X_{j} + D_{j} X_{i}) = N k^{i j} k_{i j} = N ∣ k ∣_{h}^{2}$ . This proves the final identity. Reversing the second fundamental form convention reverses the displayed divergence sign but leaves its zero-integral implication unchanged.

Depends on

Used by · 1

A supplied complete maximal slice has zero extrinsic curvature in the zero-shift case

Dependency tree · 2 levels

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