LemmaStatement: AI-generatedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Extrinsic curvature decay on the original stationary end

Statement

On the $t = 0$ end of the original regular stationary geometry, let $h = g_{i j} d x^{i} d x^{j}$ and $T = \partial_{t} = N n + X$ , with $n$ the future normal chosen so that $N > 0$ there. From $g_{00} = - 1 + 2 M_{0} / r + O_{6} (r^{- 2})$ , $g_{0 i} = O_{6} (r^{- 2})$ , and $g_{i j} = δ_{i j} + O_{6} (r^{- 1})$ , one obtains $N = 1 + O_{6} (r^{- 1})$ , $X = O_{6} (r^{- 2})$ and $k = O_{5} (r^{- 3})$ . Consequently ${tr}_{h} k = O_{5} (r^{- 3})$ . This conclusion does not assume the section maximal or $M_{0}$ positive.

Facts & Assumptions

Given: The data and hypotheses of the Statement.

Goal: The asserted conclusion with its stated scope.

[F1]

The end coefficients are stationary and satisfy spatial differentiated Cartesian bounds through order six; the original section agrees with $t = 0$ sufficiently far out. Regular non-degenerate stationary vacuum spacetime Differentiated asymptotic bounds and parameter uniformity The exterior Cauchy component has a complete asymptotically Euclidean closure

[F2]

The lapse-shift decomposition and sign convention are $T = N n + X$ and $k = - g (\nabla n, \cdot)$ . Spacelike sections, maximality and lapse–shift data

Proof

**Given:**The data and hypotheses of the Statement.

Goal: The asserted conclusion with its stated scope.

[F1] The end coefficients are stationary and satisfy spatial differentiated Cartesian bounds through order six; the original section agrees with $t = 0$ sufficiently far out. Regular non-degenerate stationary vacuum spacetime Differentiated asymptotic bounds and parameter uniformity The exterior Cauchy component has a complete asymptotically Euclidean closure

[F2] The lapse-shift decomposition and sign convention are $T = N n + X$ and $k = - g (\nabla n, \cdot)$ . Spacelike sections, maximality and lapse–shift data

1.1F1F2algebra

The end metric $h$ is uniformly positive for large $r$ , its inverse is $δ + O_{6} (r^{- 1})$ , and $X^{i} = h^{i j} g_{0 j} = O_{6} (r^{- 2})$ . The identity $N^{2} = - g_{00} + h (X, X)$ gives $N = 1 + O_{6} (r^{- 1})$ , using the positive root. These inverse and square-root bounds follow by differentiating the identities and Leibniz rule up to order six.

2.1F1F2algebra∎

Stationarity gives $0 = \partial_{t} h = - 2 N k + L_{X} h$ . The coordinate formula $(L_{X} h)_{i j} = X^{a} \partial_{a} h_{i j} + h_{a j} \partial_{i} X^{a} + h_{i a} \partial_{j} X^{a}$ is $O_{5} (r^{- 3})$ . Divide by $2 N$ and contract with $h^{- 1}$ to obtain both claimed curvature and trace bounds.

Depends on

Used by · 5

Dependency tree · 2 levels

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