LemmaStatement: AI-generatedProof: AI-generated

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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.

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Orthogonal projection away from the symmetry plane

Statement

At a point where the matrix $H$ of Gram determinant for the two symmetry directions is invertible, write its inverse as $H^{i j}$ and put $P (V) = V - X_{i} H^{i j} g (X_{j}, V)$ . Then $P (V)$ is orthogonal to both generators, $P (X_{i}) = 0$ , and $P$ is the identity on vectors orthogonal to both. The symmetric form $q (V, W) = g (V, W) - g (V, X_{i}) H^{i j} g (X_{j}, W)$ equals $g (P V, P W)$ and annihilates the orbit plane. All conventions are those of Stationary Lorentzian data and tensor conventions. No global quotient coordinate or positivity theorem is asserted.

Facts & Assumptions

[F1]

The metric is symmetric and the invertible Gram matrix has $H^{i j} H_{j k} = δ_{k}^{i}$ ; repeated indices $i, j$ are summed over $1, 2$ . Gram determinant for the two symmetry directions Stationary Lorentzian data and tensor conventions

Proof

**Given:**The metric, generators and invertible Gram matrix.

Goal: Check projection and the displayed form.

1.1givenF1

Direct substitution gives $g (X_{k}, P V) = g (X_{k}, V) - H_{k i} H^{i j} g (X_{j}, V) = 0$ . Substitution of $V = X_{k}$ gives $P (X_{k}) = 0$ . If $g (X_{j}, V) = 0$ for both $j$ , the defining formula gives $P V = V$ .

2.1F1step 1.1∎

Expand $g (P V, P W)$ . The two cross terms are each $- g (V, X_{i}) H^{i j} g (X_{j}, W)$ , and the quadratic term restores one copy by the inverse identity. The resulting expression is exactly $q (V, W)$ ; the projection property then shows it annihilates the orbit plane.

Depends on

Used by · 2

Dependency tree · 2 levels

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