Alphabeta Math
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Compactly supported background with a nonzero heavy-fermion anomaly variation

Physical scope

A fixed smooth compactly supported background-field calculation inside the perturbative local U(1)_V threshold model; it is illustrative and makes no empirical or global-anomaly claim.

Example

In the trivial-bundle H=SU(3)×U(1)V threshold model on natural-unit Minkowski space M=R1,3 with η=diag⁡(−1,1,1,1), orientation dx0∧dx1∧dx2∧dx3, and zero SU(3) background, use a smooth real compactly supported U(1)V potential B and parameter α, with G=dB and B↦B+dα, θ↦θ−α. Choose a smooth compactly supported cutoff χ that equals 1 on a neighborhood of supp⁡α, with α≥0. Let λ>0 have mass dimension two and set B=λχ(x)(x0dx1+x2dx3), normalizing α so that λ2∫Mα d4x=1. For the charge-one positive-chirality field, neutral mass partner, and MH>0 of the stated formal model, this evaluates the exact local identity from the preceding theorem. It makes no assertion about a finite-momentum vertex or the Fourier support of the chosen compactly supported backgrounds. The consistent heavy-sector and Wess–Zumino variations are equal, nonzero, and each equal +1/(12π2).

Facts & Assumptions

Given: The theorem's charge, transformation, and consistent-anomaly convention, together with the smooth compactly supported cutoff and normalized nonnegative test parameter.

[F1]

Heavy-chiral threshold Wess–Zumino anomaly match: the heavy-sector consistent variation and Wess–Zumino variation both equal +124π2∫MαG∧G.

[A1]

The support of α lies where χ=1, α≥0, and λ2∫Mα d4x=1.

Verification

Given: [F1], [A1], and the orientation dx0∧dx1∧dx2∧dx3.

1.1givenA1constructalgebra

On a neighborhood of supp⁡α, B=λ(x0dx1+x2dx3), so G=dB=λ(dx0∧dx1+dx2∧dx3). The cutoff makes B globally smooth and compactly supported.

2.1step 1.1algebra

The two summands of G have degree two, so they commute under the wedge product; each self-wedge is zero. Hence G∧G=2λ2dx0∧dx1∧dx2∧dx3 on supp⁡α.

3.1F1A1step 2.1algebra∎

Integrating against α and using [A1] gives ∫MαG∧G=2λ2∫Mα d4x=2. Therefore both variations in [F1] equal +2/(24π2)=+1/(12π2).

Depends on

Dependency tree · 2 levels

Sources