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.

Null second fundamental form descends to the screen quotient

Statement

For a branch in Regular bifurcate null horizon data before rigidity, suppose its smooth generator $L$ is geodesic up to reparametrization, $\nabla_{L} L = κ L$ for a smooth scalar $κ$ . Then $B (X, Y) = g (\nabla_{X} L, Y)$ on $T H$ defines a symmetric bilinear form on $T H / R L$ . Its trace in the positive quotient metric agrees with the expansion computed on every spacelike section transverse to $L$ . Thus the expansion uses no preferred section. The quotient and metric conventions are those of Null expansions, screen quotients and focal data, and the connection is as in Stationary Lorentzian data and tensor conventions.

Facts & Assumptions

[F1]

The null generator spans the radical of the hypersurface tangent metric. Its quotient metric is positive definite, and a transverse spacelike section identifies isometrically with the quotient. The connection is torsion-free and metric-compatible. Regular bifurcate null horizon data before rigidity Null expansions, screen quotients and focal data Stationary Lorentzian data and tensor conventions

Proof

**Given:**The null hypersurface and pregeodesic generator in the Statement.

Goal: Establish descent, symmetry and section independence.

1.1givenF1

Metric compatibility gives $B (X, L) = \frac{1}{2} X g (L, L) = 0$ . The pregeodesic equation gives $B (L, Y) = κ g (L, Y) = 0$ . For tangent extensions $X, Y$ , metric compatibility and $g (L, X) = g (L, Y) = 0$ give $B (X, Y) - B (Y, X) = - g (L, \nabla_{X} Y - \nabla_{Y} X) = - g (L, [X, Y]) = 0$ , since $[X, Y]$ is tangent to the hypersurface. Thus $B$ annihilates the kernel in both slots and is symmetric.

2.1F1step 1.1∎

Changing a tangent representative by a multiple of $L$ therefore leaves $B$ unchanged. By the quotient metric and section identification in F1, its trace on the quotient equals its trace in any section representative. The definition of expansion is exactly that trace.

Depends on

Used by · 2

Dependency tree · 2 levels

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