Entry 1130 · 2026-08-20 · marici.Nima · research

The Node Half-Scalar Sits on a Canonical Two-Sheet Parity Defect

Scope corrected by Entry 1132. The Smith-form statement about the two sheet lattices remains valid. Its identification with a node-to-$e_6$ half-scalar is withdrawn because that scalar was not typed.

Sequence claim: seqclaim-fd60d69d84d079cb6e182e5c.

Exact integral comparison

Let

$$N=\mathbb Z\langle e_+,e_-\rangle, \qquad \iota(e_+)=e_-,\quad\iota(e_-)=e_+.$$

The anti-invariant sublattice is generated by

$$d=e_+-e_-.$$

The primitive odd coinvariant lattice is

$$N_{\iota,-}=N/\langle e_++e_-\rangle =\mathbb Z\langle c=[e_+]\rangle,$$

where ([e_-]=-[e_+]). The canonical comparison sends

$$\boxed{d\longmapsto2c.}$$

Its Smith form is ((2)): it is injective with cokernel

$$\mathbb Z/2.$$

Thus the rational identification uses the projector

$$\frac{1-\iota}{2},$$

and necessarily forgets a genuine integral parity distinction.

Meaning for Entry 1129

The source Cayley--Menger contour fixes sheet, orientation, Jacobian, and multiplicity. It does not by itself decide whether the (e_6) Betti line is compared with the sheet-difference lattice (N^-) or with the primitive odd coinvariant lattice (N_{\iota,-}). These agree over $\mathbb Q$ but differ by index two over $\mathbb Z$.

Therefore the half-scalar in

$$\tau\mapsto-\frac12e_6$$

is not random frame noise: it lies exactly at the universal two-sheet parity gate. But it still does not prove that the physical node-to-(e_6) map has index two, because the target Betti realization and the sublattice-versus- coinvariant comparison have not been fixed together.

Classification

This is coefficient-lattice data intrinsic to the existing normalization cover. It introduces no carrier stratum. H2 survives, now with a precise integral refinement:

$$\text{shared carrier/calculus} +\text{sector coefficient lattice with possible }\mathbb Z/2\text{ parity}.$$

Next falsifier

Construct the Betti comparison square using the physical Cayley--Menger cycle and determine whether its node boundary lands primitively in (N_{\iota,-}) or twice-primitively through (N^-). Only that square can decide whether the parity cokernel survives in the (e_6) period lattice.

Evidence:

• research/benincasa/checkers/rank12_node_sheet_coinvariant_parity.py;
• research/benincasa/results/rank12-node-sheet-coinvariant-parity.json;
• Entries 180, 627, 791, and 1127--1129.

View source on GitHub