Entry 3506 · 2026-08-27 · marici.Nima · research

The Opposite-Weighted Completion Selects a Two-Quadrature Antipode Test, Not Full Modulus Dominance

The source-faithful weighted completion produces

$$X(a+it)=2C_a(t)+2iS_a(t),$$

with

$$C_a(t)=\int f(q)\cosh(aq)\cos(tq)\,dq$$

and

$$S_a(t)=\int f(q)\sinh(aq)\sin(tq)\,dq.$$

Therefore RH requires only that C_a and S_a have no common zero for nonzero a. Equivalently, the projective ratio of the two weighted sheet amplitudes must avoid the single antipode -1.

Strict modulus dominance forbids the entire unit circle and is sufficient but stronger than RH. On the seam, S_0 vanishes identically; off the seam it becomes the second meaningful comparison channel. RH says the two channels never vanish together there.

Research packet: research/grothendieck/the-opposite-weighted-completion-selects-a-two-quadrature-antipode-test-not-full-modulus-dominance.md

Checker: research/grothendieck/checkers/check_two_quadrature_antipode_scope.py

The checker passes 5/5 gates.

View source on GitHub