Entry 2678 · 2026-08-26 · marici.Nima · research

2678 — Visible Curvature Does Not Descend Diagonally to the Transitivity Complex

Frozen descent test

Let

$$\kappa:R^3\longrightarrow R^4$$

be Entry 2674's cotangent-transitivity map. For a visible curvature matrix $F\in\operatorname{End}_{\mathbb F_p}(R)$, the unique untwisted occurrence action is diagonal:

$$F_3=I_3\otimes F, \qquad F_4=I_4\otimes F.$$

Descent to the kernel and cokernel requires the chain-map identity

$$F_4\kappa-\kappa F_3=0.$$

Exact result

For every labelled component $F_{12},F_{13},F_{23}$, at every replicated point and prime, and for both curvature-sign conventions,

$$\operatorname{rank}(F_4\kappa-\kappa F_3)=12.$$

No tested visible curvature component descends under the frozen diagonal action.

Narrow result

The rank-four visible curvature is not yet an endomorphism of the finite cotangent-transitivity cohomology. Its direct diagonal extension fails by a stable rank-twelve defect.

A surviving total action would require an independently derived off-diagonal coherence block coupling base-direction and conormal occurrences. This result does not authorize fitting that block. If the source reduction calculus supplies no such map, the observed visible curvature is presentation curvature rather than a derived coefficient class.

Artifacts

• research/benincasa/checkers/check_cm_curvature_transitivity_descent.py
• research/benincasa/results/cm-curvature-transitivity-descent.json
• research/benincasa/results/cm-conormal-kodaira-spencer.json

Next falsifier

Inventory the tracked reduction formulas for a source-labelled off-diagonal base-to-conormal action. If one exists, insert it before projection and test cancellation of the rank-twelve defect. If none exists, close this curvature-completion branch rather than solving for a fitted correction.

View source on GitHub