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Corollary 4.1: area and pointwise limiting flux rigidity

PDF p. 1: Abstract; PDF p. 7: §1.4; PDF p. 27: Corollary 4.1 and proof

Statement

Adopt the regular non-degenerate stationary vacuum construction, with its original exterior section spacelike and maximal, ADM mass one and stationary Killing field T not identically zero on the bifurcation sphere S. Use the single primitive and regular Harrison family gq,Fq, 0≤q≤1, supplied in this construction, with the mass-one end normalization required by the Harrison flux theorem. Work in ZFC. Orient S by the outward end-sphere convention and set γq=gq∣TS and dμq=dμγq. Define eq,bq by (⋆gqFq)∣TS=eq dμq and Fq∣TS=bq dμq. Then

∣S∣g0=4π,e0=1,b0=0

pointwise on S. The area radius, defined as ∣S∣g0/(4π), is therefore one. Moreover ∫S(e02+b02) dμ0≤4π and ∫Se0 dμ0=4π.

Facts & Assumptions

[F1]

The upper area theorem gives ∣S∣g0≤4π; the lower area theorem gives ∣S∣gq≥4π(1−q2) for sufficiently small q>0. Theorem 4.1: limiting horizon area upper bound Theorem 4.2: charged horizon area lower bound

[F2]

The regular Harrison result gives a jointly smooth metric family on a neighborhood of the fixed sphere and a two-form family continuous in q and smooth in spacetime. Proposition 2.1: Einstein–Maxwell equations and regularity For q∈(0,1], the spacelike section supplied by the stable-bifurcation result makes γq Riemannian; at q=0, the upper-area interface treats S as spacelike, so γ0 is Riemannian as well. Nonnegative principal eigenvalue of the horizon stability operator Theorem 4.1: limiting horizon area upper bound The horizon and surface conventions define the positive area form and the two density ratios in the Statement. Horizon null frame, flux densities and stability operator notation Surface measure, flux densities and null normal gauge

[F3]

For each q∈(0,1], the stable-bifurcation result supplies an individually admissible maximal section and a smooth spacelike extension across S, in which the compact sphere has an outward normal-graph tube. These are precisely the section, extension and tube hypotheses of the stable-horizon result, which gives ∫S(eq2+bq2) dμq≤4π for this q. Nonnegative principal eigenvalue of the horizon stability operator Stable-horizon flux-energy and area-charge inequalities

[F4]

With the stated primitive, end normalization and orientation, the Harrison flux theorem gives ∫Seq dμq=4π1−q2 and ∫Sbq dμq=0. Proposition 2.1: electric and magnetic fluxes Horizon null frame, flux densities and stability operator notation

[F5]

On a fixed compact oriented surface, the compact-flux lemma gives convergence of area and of the flux and squared-density integrals under uniform convergence of Riemannian metrics and their associated restricted two-forms. It also proves pointwise e=1, b=0 when area and electric flux are both 4π and the squared-density integral is at most 4π. Compact flux saturation and pointwise density rigidity

[F6]

The Lorentzian Hodge-calculus lemma gives (⋆gF)ab=12(volg)abcdFcd in a positively oriented four-dimensional chart, with both indices of F raised by g−1. The volume form uses the metric and the fixed spacetime orientation. Lorentzian exterior calculus and Hodge star

Proof

Given: The construction and hypotheses in the Statement.

Goal: Prove the limiting area, flux identities, and pointwise rigidity.

Proof technique: direct.

1.1givenF2F5F6

Let qj∈(0,1] decrease to zero. On the compact fixed sphere, the metric regularity in [F2] gives uniform convergence γqj→γ0. In local positively oriented coordinates the component formula in [F6] expresses the Hodge star algebraically in the metric, its inverse and the oriented volume density. Since the limiting Lorentzian metric is nondegenerate, the same regularity and continuity give uniform convergence of the restricted forms Fqj∣TS and (⋆gqjFqj)∣TS to their q=0 counterparts. The convergence assertion of [F5] therefore applies to area, flux and squared-density integrals.

2.1F1F3F4step 1.1algebra

Passing to the limit in the lower area bound of [F1] gives ∣S∣g0≥4π; its upper bound gives the reverse inequality. Thus ∣S∣g0=4π. For each qj∈(0,1], use its section, extension and tube from [F3] to obtain the stable-horizon estimate. Passing to the limit with step 1.1 gives ∫S(e02+b02) dμ0≤4π. Likewise [F4] and step 1.1 give ∫Se0 dμ0=4π and ∫Sb0 dμ0=0.

3.1F2F5step 2.1algebra∎

The sphere is compact, connected, oriented and smooth; dμ0 is a smooth positive measure and e0,b0 are continuous. The equality assertion of [F5], with the three integral conditions from step 2.1, gives e0=1 and b0=0 at every point. Equivalently, expansion of the nonnegative integrand gives 0≤∫S((e0−1)2+b02) dμ0≤4π−8π+4π=0. The area-radius definition and ∣S∣g0=4π give radius one.

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