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Entry 2701 · 2026-08-26 · marici.Nima · research

2701 — The Canonical Rank-Twenty-Six Source Derivative Is Fixed Coefficientwise

Frozen source

Write the coefficient of the five-pole source form as

F=NKCM5qG1qG2qG3qG23qG31,N=qG23+qG31.F = \frac{N K_{\rm CM}^{5}} {q_{\mathcal G_1}q_{\mathcal G_2}q_{\mathcal G_3} q_{\mathcal G_{23}}q_{\mathcal G_{31}}}, \qquad N=q_{\mathcal G_{23}}+q_{\mathcal G_{31}}.

The canonical first-order source extension in external direction (j) is obtained without choosing a quotient representative:

∂jlog⁡F=∂jNN+5∂jKCMKCM−∑i∂jqiqi.\partial_j\log F = \frac{\partial_jN}{N} +5\frac{\partial_jK_{\rm CM}}{K_{\rm CM}} -\sum_i\frac{\partial_jq_i}{q_i}.

Exact coefficient audit

The complete unspecialized formulas give:

Symbolic differentiation verifies that (K_{\rm CM}) has total degree (6), every (q_i) and (N) has degree (1), and their derivatives have degrees (5) and (0), respectively.

The source coefficient has weight

1+30−5=26.1+30-5=26.

Including (da\wedge db) gives source-form weight (28). Every canonical external derivative has coefficient weight (25), hence form weight (27).

Result

The dual-number coefficient required by Entry 2697 is now fixed term-by-term by the frozen source. It is not selected by the desired Euler recurrence, by sparsity, or by a quotient section.

This establishes the characteristic-zero source derivative but not yet its equality with the finite de Rham connection implementation.

Artifacts

Run the checker with:

uv run --with sympy python research/benincasa/check_rank26_canonical_source_derivative.py

Next falsifier

Insert the canonical coefficient derivatives into the complete finite presentation and verify in all three directions that

P12∇jD0+P22∇jD1+P32∇jD2=27DjP_1^2\nabla_jD_0+P_2^2\nabla_jD_1+P_3^2\nabla_jD_2 =27D_j

after quotient reduction. A mismatch would localize the failure in the finite Gauss--Manin adapter rather than in source homogeneity.

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