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ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)
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Fixed-vector transverse angular coefficient 8π/3

Example

Let e∈R3 be a unit vector and PT(n)=I3−nn⊤ for n∈S2. Then the angular coefficient in the coherent infrared family with b(n)=λPT(n)e, λ∈R∖{0}, is ∫S2∣b(n)∣2 dΩ=8πλ23.

Facts & Assumptions

Given: The unit vector e, the scalar λ≠0, and the transverse projection and angular integral in Logarithmic coherent infrared tail: finite energy, no Fock limit.

[F1]

Logarithmic coherent infrared tail: finite energy, no Fock limit uses PT(n)=I3−nn⊤ and Cb=∫S2∣b(n)∣2dΩ for a measurable transverse b∈L2(S2;C3).

[A1]

e∈R3 is a unit vector and λ∈R∖{0}, as stated in the Example.

[F2]

Agreement with the existing polar sphere measure: the Euclidean chart measure on S2 equals polar surface measure and is invariant under orthogonal transformations.

[F3]

Surface integration on compact C1 hypersurfaces: surface integration is the Euclidean chart integral with density det⁡(DXTDX).

[F4]

A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in Rn: coordinate hyperplanes and singletons have Lebesgue measure zero, hence chart seams and poles have zero area when their coordinate parameter sets are null.

[A2]

The Axiom of Countable Choice (ACω): assume the same Countable Choice carried by the coherent-tail and surface-measure suppliers; no stronger choice principle is used.

Verification

Given: [A1]–[A2], [F1]–[F4] and the specified b(n)=λPT(n)e.

1.1A1A2F1F2F3F4algebra

Since [A1] gives ∣e∣=1 and ∣n∣=1, ∣PT(n)e∣2=e⊤(I3−nn⊤)e=1−(n⋅e)2. Rotational invariance in [F2] permits taking e=(0,0,1). Use the chart X(u,ϕ)=(1−u2cos⁡ϕ,1−u2sin⁡ϕ,u) for −1<u<1 and 0<ϕ<2π. Its tangent vectors are orthogonal with squared lengths (1−u2)−1 and 1−u2, so [F3] gives area density 1. The omitted azimuthal seam is a coordinate line in sphere graph charts, and each pole is a singleton in its graph chart; these parameter sets are null by [F4], and their chart integrals vanish. Thus with u=n3 the angular integral is 2π∫−11(1−u2) du.

2.1A1A2F1step 1.1algebra∎

The integral equals 2π[u−u3/3]−11=8π/3. Multiplying by λ2 yields Cb=8πλ2/3>0, which is the coefficient required by [F1].

Depends on

Dependency tree · 2 levels

Sources