Entry 518 · 2026-08-18 · marici.Nima · research

The Odd Weighted Resonance Is One-Sided at the Endpoints

Restrict the resonant q source of Entry 517 to s_b=1, i=7.

At b=-1, L_1=-u. If e_a=2, all terms vanish modulo u^2. If e_a=1, direct reduction gives

$$m=-(-1)^jua^8, \qquad C_q=(-1)^ju(1-i)a^i,$$

and hence

$$(\widehat H_q)_a= (-1)^j\frac{7-i}{4}ua^{i+1}, \qquad (\widehat H_q)_u=0.$$

Thus the i=7 resonant lift vanishes at the negative endpoint in every s_a sector.

At b=1, write the two orbit lattices as L_2=a+epsilon*u/2, with epsilon=+-1. Since L_1=2-u remains a unit, the same resonant source gives

$$(\widehat H_q)_a =\epsilon\,2^{e_a-3}ua^7, \qquad (\widehat H_q)_u =3\,2^{e_a}ua^7,$$

modulo u^2. It therefore survives nontrivially in the Cartier layer at the positive endpoint.

The odd weighted resonance is consequently one-sided:

$$\operatorname{res}_{b=-1}(r_7)=0, \qquad \operatorname{res}_{b=1}(r_7)\ne0.$$

It cannot be selected by the ordinary kernel of the two endpoint evaluations. A relative-support realization must use the homotopy fiber of the endpoint map and supply a boundary nullhomotopy for the positive endpoint value. This is the weighted analogue of the missing Beck--Chevalley/Gysin cell in the conductor problem.

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