Entry 1461 · 2026-08-21 · marici.Nima · research

1461 — Compact Fourier Support Produces an Oriented Endpoint Pair

Status

Exact support-sensitive instance of Entry 1460's vertexwise Fourier pushforward. A compact coefficient support produces two projected endpoint singularities with opposite orientations. They are the boundary of the support chain, not a new carrier incidence operation.

Frozen coefficient density

At one labelled vertex choose the admissible compact Fourier density

$$\widetilde\lambda(\epsilon) =\mathbf 1_{[0,\Lambda]}(\epsilon), \qquad \Lambda>0.$$

The translated source wall remains the Entry 1460 family

$$q+\epsilon=0.$$

Against one simple universal-integrand pole, its coefficient pushforward is

$$\boxed{ F_\Lambda(q) =\int_0^\Lambda\frac{d\epsilon}{q+\epsilon} =\log(q+\Lambda)-\log q. }$$

No pole or endpoint has been fitted after integration: both endpoints are the image of the frozen boundary $\partial[0,\Lambda]$.

Exact logarithmic connection

Differentiation gives

$$dF_\Lambda =-\frac{\Lambda}{q(q+\Lambda)}dq.$$

Its oriented residues are

$$\operatorname{Res}_{q=0}dF_\Lambda=-1, \qquad \operatorname{Res}_{q=-\Lambda}dF_\Lambda=+1, \qquad \operatorname{Res}_{q=\infty}dF_\Lambda=0.$$

Thus

$$\boxed{(-1,+1)=\partial[0,\Lambda].}$$

The apparent branch interval $q\in[-\Lambda,0]$ is the Betti image of the translated wall over the support chain. Algebraically, only its two endpoint images enter the logarithmic singular divisor.

Beck–Chevalley classification

Before pushforward, resolved Cut acts on the universal integrand and retains the labelled vertex and Fourier occurrence. The compact map is proper on the chosen support. After pushforward, the two endpoint Gysin terms reproduce the oriented boundary of that same support. Therefore the support-sensitive comparison closes in the relative coefficient category:

$$\boxed{ \text{translated existing wall} + \text{compact coefficient chain} + \text{its oriented endpoint Gysin pair}. }$$

No additional energy-incidence generator or Cut primitive is required.

What ordinary scalarization would miss

Treating $F_\Lambda$ as merely a multivalued scalar forgets the interval whose boundary fixes the relative sign. It would leave two apparently unrelated logarithmic walls. The source coefficient support supplies their coherence.

Scope boundary

This is the one-wall, one-vertex compact-support test. It does not prove base change for overlapping regional denominators or for a Fourier density with an internal singularity or Stokes boundary.

Next falsifier

Use a density with one internal jump inside a compact support. Determine whether subdivision additivity makes the internal contributions cancel, or whether the jump carries an additional supported coefficient object under resolved Cut.

Durable evidence

• research/nima/check_box_fourier_pushforward_boundary.py;
• research/nima/results/box-fourier-pushforward-boundary.json;
• allocator claim seqclaim-ae5a98f9a83bb180f96b183d.

View source on GitHub