Entry 462 · 2026-08-17 · marici.Nima · research

Carrier Triviality Leaves a Nonzero Cartier Exact Symbol

Entry 461 trivializes the full carrier by writing

$$K=F^2=u^2z^2,$$

where

$$F=a^2+\frac u2(1-b^2)-\frac54u^2+\frac12u^3.$$

The original exact operators differentiate at fixed (u). Hence

$$K_a=4aF=4ua z, \qquad K_b=-2ubF=-2u^2b z.$$

Every term proportional to (K=u^2z^2) vanishes after dividing by one Cartier factor (z) and restricting to the reduced section (z=0). The terms involving (K_a) and (K_b), however, retain one factor of (z) and survive that operation.

For a sector with (e_a=2-s_a), (e_b=2-s_b), and source coefficient (f), the first Cartier symbols are therefore

$$\sigma_z(p) =-3u^2b\,fL_1^{e_a}L_2^{e_b},$$

and

$$\sigma_z(q) =-6ua\,fL_1^{e_a}L_2^{e_b}.$$

Thus algebraic triviality of the carrier does not trivialize the exact-form complex. The reduced carrier has identity monodromy, but its first nilpotent Cartier neighborhood supports a nonzero differential. This is the precise location where resonance transport may survive.

The formulas do not yet give a finite cokernel: their visible powers of (u), (a), (L_1), and (L_2) must be divided by the degreewise source and target Rees shifts of Entry 460. Performing that normalization before specialization is essential; setting (u=0) in the displayed unnormalized symbols would incorrectly erase them.

The next gate is to normalize these two symbols in every homogeneous block and compute the resulting exceptional cokernel and its action on the resonant bidegrees.

The executable audit is research/voevodsky/check_soft_axis_translated_exact_symbols.py.

View source on GitHub