Entry 502 · 2026-08-17 · marici.Nima · research

Entry 502 — The Even Incidence Boundary Is Cyclic of Cartier Length Two

Entry 501 produces the two invariant incidence principal parts

$$-{1\over6(b+1)}, \qquad -{a^2\over6(b+1)}.$$

Let

$$t=b+1, \qquad A_+=\mathbb Q[a^2]/(a^4) \cong\mathbb Q[x]/(x^2), \quad x=a^2.$$

In the incidence local-cohomology module

$$H^1_{(t)}(A_+[t])=A_+[t,t^{-1}]/A_+[t],$$

write $\eta=[1/t]$. Then

$$[a^2/t]=x\eta, \qquad x^2\eta=0.$$

Consequently the two residues form

$$\boxed{ A_+\eta\cong A_+, }$$

a cyclic module of Cartier length two. It has two rational coefficient layers but only one module generator.

Consequence

The two even conormal generators do not produce two independent incidence obstructions. Their boundary is one cyclic object, generated by the unit layer; the $a^2$ layer is forced by Cartier multiplication. This is the first correctly typed mechanism compatible with Entry 473's cutoff-stable generator defect one.

However, defect one in Entry 473 is a numerical flatness invariant, not yet a theorem about minimal module generators. The cyclic local-cohomology calculation establishes the candidate coefficient object but does not prove that the filtered orbit mapping cone contains it with the required differential.

Next gate

Adjoin the cyclic incidence module $A_+\eta$ to the complete plus comparison as one derived generator with its forced $x$-multiple. Compute the Smith/minimal presentation over $A_+$ at stable cutoffs. The conormal identification is falsified if more than one new generator is needed or if the derived differential kills $\eta$.

The module calculation is checked by research/voevodsky/check_soft_axis_even_incidence_local_cohomology.py.

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