Alphabeta Math
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.

Zero boundary expansions without maximality

Statement

Let C‾=C∪S be the smooth exterior section with boundary supplied under the regular stationary-vacuum hypotheses. Let n be its future unit normal and ν the unit normal to S pointing into C‾ (toward the end), and put k(X,Y)=−g(∇Xn,Y). With γ=g∣TS, define the unaveraged boundary mean curvature by HS=tr⁡γ[(X,Y)↦g(∇Xν,Y)] for X,Y∈TS, using covariant derivatives along S. If the two horizon null expansions vanish at S, then HS=0 and tr⁡Sk=0, where tr⁡Sk=tr⁡γ(k∣TS×TS). In particular both expressions θ±=±HS+tr⁡Sk vanish, independently of the sign convention for k. No maximality is assumed.

Facts & Assumptions

Given: The data and hypotheses of the Statement.

Goal: The two zero boundary traces and their stated sign independence.

[F1]

The regular-spacetime hypotheses are formulated in ZFC. The exterior section is a smooth spacelike manifold with smooth compact boundary S⊂Σ0. The exterior Cauchy component has a complete asymptotically Euclidean closure Regular non-degenerate stationary vacuum spacetime

[F2]

The two smooth null horizons H+ and H− contain S, intersect transversely there, and have zero screen expansion. Their screen expansion is the trace of ([X],[Y])↦g(∇XL,Y) for a local nonzero radical field L; zero expansion is unchanged by nonzero smooth rescaling. Regular non-degenerate stationary vacuum spacetime

[F3]

The induced data have a smooth future unit normal n, induced metric h=g∣TC‾, and second fundamental form k(X,Y)=−g(∇Xn,Y). The surface expansion of a null normal ℓ is tr⁡γ[(X,Y)↦g(∇Xℓ,Y)]. Spacelike sections, maximality and lapse–shift data Null expansions, screen quotients and focal data

[F4]

For a spacelike two-surface in a time-oriented Lorentzian four-manifold, the normal plane is Lorentzian with two future null rays. A null hypersurface with radical RL has tangent space L⊥. On a spacelike section its quotient screen is isometric to the section tangent plane, and surface null second fundamental forms scale by χefL=efχL. These assertions do not require an affine generator. Local null normal pairs and the positive screen quotient

Proof

Given: The data of the Statement and [F1]–[F4].

Goal: Identify the two horizon traces with the boundary trace formulas and solve them.

1.1F1F2F4

At p∈S, the inclusion TpS⊂TpC‾ makes γ positive definite. For either horizon let RL be its radical line. Since TpS⊂TpH±=L⊥, L is normal to S. The line RL cannot intersect the positive-definite plane TpS nontrivially. Their dimensions are one and two, so TpH±=TpS⊕RL. Thus S is a spacelike section of each horizon, and [F4] identifies its tangent metric with each screen metric.

2.1F1F2F3F4step 1.1algebra

The fields n,ν are orthogonal unit normals to S, with squared lengths −1,1. They span its normal plane, whose null lines are exactly R(n+ν) and R(n−ν), because g(an+bν,an+bν)=−a2+b2. Put ℓ±=n±ν. They are future null, since g(ℓ±,n)=−1, and satisfy g(ℓ+,ℓ−)=−2. The two horizon radical lines are distinct: if they agreed, [F4] would give equal tangent hyperplanes, contrary to their transverse intersection. Hence, after a local interchange of horizon labels, each ℓ± is a smooth positive multiple a±L±∣S of a smooth future horizon radical field.

3.1givenF2F3F4step 1.1step 2.1algebra

Under the screen isometries of step 1.1, the horizon form restricts to χL±∣S(X,Y)=g(∇X(L±∣S),Y) on TS; differentiation in a tangent direction depends only on the restriction to S. Its γ-trace is therefore zero by [F2]. Write a±=ef± locally and apply the surface rescaling law in [F4] to obtain tr⁡γχℓ±=0. Linearity of the connection and the definitions in the Statement and [F3] give χℓ±(X,Y)=−k(X,Y)±g(∇Xν,Y) and consequently 0=−tr⁡Sk±HS. No affine parametrization is used.

4.1givenstep 3.1algebra∎

Adding the two equalities of step 3.1 gives −2tr⁡Sk=0; subtracting them gives 2HS=0. Thus both θ±=±HS+tr⁡Sk vanish for the normal pointing into the exterior. Replacing k by −k changes the sign of its trace and preserves both zero values. The argument has imposed no maximality condition.

Depends on

Used by · 2

Dependency tree · 2 levels

Each result is drawn at its shortest distance from this item. Arrows run from prerequisites to the items that use them; this item has a heavier outline. External source nodes link directly to the cited text. The view shows up to two dependency steps; open a prerequisite to continue.