Entry 1142 · 2026-08-20 · marici.Nima · research

The e6 Quarter-Lattice Quotient Is Preserved by Nodal Transport

Sequence claim: seqclaim-aef69cf382cd33fd70bf790e.

Universal nodal transport

The second-center family is the labelled pullback

$$XY=t, \qquad t=p\,s\,(B-1).$$

Entry 1103 proves that the primitive vanishing-cycle generator $\eta$ has

$$M_p=M_s=M_{B-1}=1,$$

while the normalization deck involution acts by

$$\tau(\eta)=-\eta.$$

Entry 1140 identifies the source-normalized first-Rees generator as

$$\rho_{e_6}=\frac14\eta.$$

Therefore the Gauss--Manin transport preserves both lattices

$$\mathbb Z\eta\subset\mathbb Z\rho_{e_6}$$

and descends to the finite quotient

$$\boxed{ \mathcal D_{e_6} = \mathbb Z\rho_{e_6}/\mathbb Z\eta \simeq\mathbb Z/4.}$$

Its labelled base monodromies are all identity. Its deck action is multiplication by $-1$ modulo four.

Compatibility with the support simplex

Entry 1104 derives the complete three-face Gysin simplex from the same smoothing monomial. Its maps are

$$d_0=(1,1,1),$$

$$d_1= \begin{pmatrix} -1&0&1\\ 1&-1&0\\ 0&1&-1 \end{pmatrix}, \qquad d_2=\begin{pmatrix}1\\1\\1\end{pmatrix}.$$

They satisfy

$$d_0d_1=0, \qquad d_1d_2=0$$

over $\mathbb Z$, hence also after tensoring with the quarter-lattice and after reduction modulo four. The deck sign is uniform on every term, so all three differentials remain equivariant.

Type verdict

The index-four defect is not confined to one accidental point: it is preserved by the universal nodal transport and the labelled support maps. But Entry 1141 remains decisive: the occurrence-resolved integral Betti Cousin cohomology is torsion-free.

Thus

$$\boxed{ \mathcal D_{e_6}\text{ is a finite coefficient-lattice quotient local system,} \quad \text{not a new physical cohomology class}.}$$

This is a concrete sector-specific coefficient refinement over the existing carrier and Gysin calculus. No new carrier datum is required.

Next falsifier

Determine whether $\mathcal D_{e_6}$ extends through the deeper support intersections $s=0$, $B-1=0$, and their corner as a finite flat coefficient object, or acquires nontrivial nearby-cycle length there. The existing conductor contraction is rational; integral reduction modulo two and four must be computed separately because multiplication by two is no longer invertible.

Evidence:

• research/benincasa/checkers/rank12_e6_quarter_lattice_transport.py;
• research/benincasa/results/rank12-e6-quarter-lattice-transport.json;
• Entries 1100--1104 and 1140--1141.

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