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.

The normal binormal is invariant under oriented boosts

Statement

For the normal pair in Normalized Komar data on a nonrotating horizon section, replace $n, s$ by $n^{'} = \cosh β n + \sinh β s$ and $s^{'} = \sinh β n + \cosh β s$ , where $β$ is any supplied smooth real function on $C$ . Then $n^{'}, s^{'}$ have the same normalizations and $b_{a b}^{'} = b_{a b}$ . Thus $Q_{C} [T]$ does not depend on this change of oriented normal frame.

Facts & Assumptions

[F1]

The normal pair has Gram matrix $diag (- 1, 1)$ and the functional uses only its binormal and the section area measure. Normalized Komar data on a nonrotating horizon section

[F2]

The real functions $\sinh$ and $\cosh$ are defined on all of $R$ ; for every real $x$ , $\cosh^{2} x - \sinh^{2} x = 1$ and $\cosh x > 0$ . Apply these facts pointwise at $x = β (p)$ for $p \in C$ . The six hyperbolic functions and their natural domains Addition formulas, identities, parity, and derivatives of the hyperbolic functions

Proof

**Given:**The normal pair and boost in the Statement.

Goal: Check preservation of the binormal.

1.1givenF1F2

Bilinearity and $\cosh^{2} β - \sinh^{2} β = 1$ give $g (n^{'}, n^{'}) = - 1$ , $g (s^{'}, s^{'}) = 1$ and $g (n^{'}, s^{'}) = 0$ . Both remain normal to $C$ , and the boost preserves the future and normal-plane orientation choices.

2.1F1F2step 1.1∎

Expanding $s_{a}^{'} n_{b}^{'} - n_{a}^{'} s_{b}^{'}$ cancels the pure $n_{a} n_{b}$ and $s_{a} s_{b}$ terms; the remaining multiplier of $s_{a} n_{b} - n_{a} s_{b}$ is $\cosh^{2} β - \sinh^{2} β = 1$ . The functional therefore has exactly the same integrand.

Depends on

Used by · 2

Dependency tree · 2 levels

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