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.

Algebraic reconstruction from orbit and horizontal metric data

Statement

For the projection data of Orthogonal projection away from the symmetry plane, define one-forms $σ^{i} (V) = H^{i j} g (X_{j}, V)$ . Then the full metric is $g = H_{i j} σ^{i} \otimes σ^{j} + q$ . Thus agreement of $H$ , the one-forms $σ^{i}$ and $q$ on a common domain implies agreement of the full metric there. The Gram-matrix and tensor conventions are those of Gram determinant for the two symmetry directions and Stationary Lorentzian data and tensor conventions. This is an algebraic reconstruction identity, not an assertion that an Ernst-map uniqueness theorem alone supplies the global one-forms or identifies the domains.

Facts & Assumptions

[F1]

The inverse Gram matrix defines $P$ and $q$ as in the projection lemma, and every vector decomposes as $V = X_{i} σ^{i} (V) + P V$ . Orthogonal projection away from the symmetry plane Gram determinant for the two symmetry directions Stationary Lorentzian data and tensor conventions

[F2]

A one-form is a smooth covector field, of type $(0, 1)$ , and a covariant two-tensor field assigns a bilinear form to each tangent space. Equality of such fields is pointwise equality of these multilinear maps. Smoothness of a covector or covariant two-tensor field is tested by smooth component functions in each chart. Tensor fields of type (r,s) on a Lorentzian four-manifold and their components

[F3]

For covectors $α, β$ on the same tangent space, the tensor-product definition gives $(α \otimes β) (V, W) = α (V) β (W)$ . We apply this definition pointwise to the one-forms in the Statement. The tensor product of multilinear tensors

[F4]

For each finite $r \in N$ , if a real matrix-valued function on an open Euclidean domain has $C^{r}$ entries and nonzero determinant everywhere, its inverse has $C^{r}$ entries. We use this with chart dimension four and matrix size two. Matrix inversion preserves Cᵏ regularity where the determinant is nonzero

[F5]

For every finite $r \in N$ , finite componentwise sums and products of $C^{r}$ Euclidean maps are $C^{r}$ ; hence the same finite operations preserve smoothness. Cᵏ Euclidean maps are closed under componentwise algebra and composition

Proof

**Given:**The pointwise tensors and decomposition in the Statement.

Goal: Recover the full metric from these data.

1.1givenF1

Inserting the vector decompositions into $g (V, W)$ , both mixed terms vanish because $P V$ and $P W$ are orthogonal to the generators. The orbit term is $H_{i j} σ^{i} (V) σ^{j} (W)$ and the remaining term is $g (P V, P W) = q (V, W)$ .

1.2F1F2F4F5algebra

In a smooth coordinate chart, $H_{i j} = g_{a b} X_{i}^{a} X_{j}^{b}$ is a finite sum of products of smooth component functions of the metric and generators. Thus $H_{i j}$ and $\det H = H_{11} H_{22} - H_{12} H_{21}$ are smooth. The nonzero determinant locus is open by continuity. On the chart portion of that locus, apply the matrix-inversion result for every finite $r$ to obtain smooth $H^{i j}$ . Each $σ^{i}$ is pointwise linear in its vector argument and has components $σ_{a}^{i} = H^{i j} g_{b a} X_{j}^{b}$ , smooth by the finite-operation result. The component criterion makes these covectors smooth one-forms. The formula for $q$ in the projection input also has smooth components by the same finite-operation argument.

2.1F1F2F3step 1.1step 1.2∎

The tensor-product evaluation rule gives $(H_{i j} σ^{i} \otimes σ^{j} + q) (V, W) = H_{i j} σ^{i} (V) σ^{j} (W) + q (V, W)$ . The equality in step 1.1 holds at every point for every pair of tangent vectors, so pointwise equality of bilinear maps gives the asserted covariant-tensor identity. Equal tensors on its right side therefore give equal metrics on a common domain. It gives no mechanism for constructing that domain or controlling boundary conditions, which remain part of the external global classification input.

Depends on

Used by · 1

Connected smooth axisymmetric vacuum Kerr classification

Dependency tree · 2 levels

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