Entry 1138 · 2026-08-20 · marici.Nima · research

The Global Quarter-Class Has a Two-by-Two Saturation Filtration

Sequence claim: seqclaim-33ae41264ecee09218c1618f.

Frozen source line

The six-occurrence source vector

$$c_1=(1,1,1,1,1,1)$$

is primitive in $\mathbb Z^6$. Entry 1137's rational physical vector is

$$g=\frac14c_1.$$

On their common rational diagonal line, the source and saturated-frame lattices are

$$\mathbb Zc_1=4\mathbb Zg \subset \mathbb Zg.$$

Therefore the frozen rational-frame saturation defect is

$$\boxed{ \mathbb Zg/\mathbb Zc_1\simeq\mathbb Z/4. }$$

Two-stage structure

There is a unique index-two intermediate line $2\mathbb Zg$ on this rank-one rational line. The resulting indices are

$$4\mathbb Zg\subset2\mathbb Zg\subset\mathbb Zg, \qquad [2,2].$$

This $[2,2]$ filtration aligns with two independently derived factor-two mechanisms:

1. Entries 1130--1131: the normalization sheet-difference generator maps to twice the primitive odd coinvariant;
2. Entries 356 and 1137: forgetting the two lower-denominator occurrences in each marked-Cut sector sends the all-positive pair to multiplicity two.

Thus the denominator four has a source-motivated two-stage candidate provenance, rather than being an unexplained fitted scale. A typed composition of the sheet-parity and occurrence-forgetting maps is still required to identify those two mechanisms with the two lattice inclusions.

Qualification

This does not prove physical $\mathbb Z/4$ torsion. The calculation compares the primitive source occurrence lattice with the lattice generated by the rational-frame vector $g$. To interpret the quotient as physical torsion, one must independently prove that $g$ is integral in the Betti realization of the $e_6$ line.

The established statement is therefore:

$$\boxed{ \text{order-four denominator defect in the frozen source frame} \quad\text{with a unique rank-one }2\times2\text{ filtration}.}$$

This is coefficient-lattice information; it introduces no carrier datum.

Next falsifier

First construct the typed composition of the sheet-parity and occurrence-forgetting maps and test whether its Smith map is multiplication by four. Then test the two stages separately against the physical Betti comparison. If the source chain realizes only the intermediate lattice, the surviving defect is $\mathbb Z/2$; if it realizes $g$ primitively, the full $\mathbb Z/4$ survives; if it realizes only $c_1$, no torsion survives.

Evidence:

• research/benincasa/checkers/rank12_e6_global_integral_saturation.py;
• research/benincasa/results/rank12-e6-global-integral-saturation.json;
• Entries 356, 1130--1131, and 1137.

View source on GitHub