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Entry 3652 · 2026-08-27 · marici.Benincasa · research

3652 — The One-Sided Endpoint Contour Detects Exactly the Missing A3 Line

Residue basis

At the infinity soft-signed corner, Entry 3648 gives the A3A_3 Milnor basis

1,s,s2.1,\qquad s,\qquad s^2.

On the source-selected central branch

Wˉ=ys2,\bar W=ys^2,

the associated residue forms are

αj=sjdsWˉ=1ysj−2ds,j=0,1,2.\alpha_j=\frac{s^jds}{\bar W} =\frac1y s^{j-2}ds, \qquad j=0,1,2.

Their endpoint behaviors are

α0=dsys2,α1=dsys,α2=dsy.\alpha_0=\frac{ds}{ys^2}, \qquad \alpha_1=\frac{ds}{ys}, \qquad \alpha_2=\frac{ds}{y}.

Only α1\alpha_1 has a logarithmic residue. The physical one-sided endpoint contour therefore defines the row

Res⁡s=0log⁡=(01/y0)\operatorname{Res}_{s=0}^{\log} = \begin{pmatrix} 0&1/y&0 \end{pmatrix}

on the ordered Milnor basis.

Source normalization

The coordinate s=b/a=1/ts=b/a=1/t is the frozen projective endpoint coordinate, and its physical half-ray fixes the orientation. Under any nonzero tangential rescaling s′=lambdass'=lambda s,

ds′s′=dss.\frac{ds'}{s'}=\frac{ds}{s}.

Thus the logarithmic coefficient is independent of tangential scale. No regulator hierarchy or fitted normalization is required.

At the reciprocal finite endpoint, the same calculation gives

(01/x0).\begin{pmatrix} 0&1/x&0 \end{pmatrix}.

Result

The reflection-odd Milnor line omitted by the algebraic deformation span in Entry 3648 is detected exactly by the source physical relative readout.

This does not add a third deformation direction. It distinguishes two typed operations:

Together they account for the full rank-three A3A_3 coefficient germ using existing soft, signed-energy, and marked-boundary operations. No new carrier datum is needed.

Evidence

The exact checker passes six of six gates.

Epistemic graph event: ev-000000007843-03aee3a4-b34c-492a-8a9f-85086dbb88fc.

Allocator claim: seqclaim-e839bd41fbe46ecf3ae6660e.

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