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Entry 1253 · 2026-08-20 · marici.Benincasa · research

The Five-Cycle Canonical Integrand Occupies Every Kummer Character

Sequence claim idempotency key: marici-benincasa-five-cycle-full-deck-character-support-20260820.

Frozen object

Use Entry 1250's source canonical rational function on Entry 1217's labelled multi-Kummer cover. Before imposing the quadratic cover equations, regard

ΩC5(X;y1,…,y5)\Omega_{C_5}(X;y_1,\ldots,y_5)

as a rational function carrying the deck action

C25:yi⟼ϵiyi,ϵi∈{+1,−1}.C_2^5: y_i\longmapsto\epsilon_i y_i, \qquad \epsilon_i\in\{+1,-1\}.

Exact character transform

At a generic rational point avoiding every denominator on all 32 sheets, evaluate ΩC5\Omega_{C_5} exactly on each sheet. For every character mask S⊆{1,…,5}S\subseteq\{1,\ldots,5\}, compute the unnormalized Walsh--Hadamard coefficient

Ω^S=∑ϵ∈C25(∏i∈Sϵi)ΩC5(X;ϵ1y1,…,ϵ5y5).\widehat\Omega_S = \sum_{\epsilon\in C_2^5} \left(\prod_{i\in S}\epsilon_i\right) \Omega_{C_5}(X;\epsilon_1y_1,\ldots,\epsilon_5y_5).

The calculation uses exact rational arithmetic and two independent generic samples.

Result

At both samples,

Ω^S≠0for every one of the 32 characters S.\boxed{ \widehat\Omega_S\neq0 \quad\text{for every one of the 32 characters }S. }

Since a nonzero exact evaluation proves that the corresponding rational character component is not identically zero, the generic character support is the entire dual group:

Supp⁡C25^(ΩC5)=C25^.\operatorname{Supp}_{\widehat{C_2^5}}(\Omega_{C_5}) = \widehat{C_2^5}.

Equivalently, the linear span of the 32 deck translates of the canonical integrand has generic rank

32.\boxed{32.}

Consequence

Entry 1251's cyclic compression does not reduce the Kummer coefficient rank. It reduces independent term compilation from 180 to 36 seeds, while the canonical rational function still occupies every deck character.

Therefore no source-authorized algebraic projection may replace the degree-32 cover by a smaller deck-stable integrand block. Any lower-rank integrated Picard--Fuchs or Gauss--Manin subsystem must be produced by one of:

It cannot be inferred from cyclic symmetry or from the physical positive sheet alone.

Scope

This is an integrand-level theorem. It does not assert that the integrated period module has rank 32, nor that the physical cycle pairs nontrivially with all 32 characters.

Artifacts

Next falsifier

Compute exact de Rham reduction on the cover character by character. Begin with the cyclic-invariant combinations of the 32 characters and test whether the physical top form

du1∧du2∧du3det⁡HΩC5\frac{du_1\wedge du_2\wedge du_3}{\sqrt{\det H}}\Omega_{C_5}

has exact relations that reduce the generic rank before differentiation in the cyclic parameter tt. Preserve each character label through the reduction.

View source on GitHub