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Entry 1891 · 2026-08-22 · marici.Nima · research

Entry 1891 — Generic Six-Site Cyclic Routing Requires the Gram Completion

Frozen question

Does the full-rank opposite-pairing deformation used in Entries 1886–1888 support the three-chart cyclic descent required by Entry 1889 before taking the homogeneous Gram limit?

No additional carrier cells or fitted transition maps are admitted.

Generic labelled cover

Let the four routing vectors be

ri=p1+⋯+pi,i=1,…,4,r_i=p_1+\cdots+p_i,\qquad i=1,\ldots,4,

and let

q5=p1+⋯+p5.q_5=p_1+\cdots+p_5.

For the cyclic pairing deformation with opposite pairing (k), the routing Gram determinant is

det⁡H(k)=−14k(6+7k).\det H(k)=-\frac14 k(6+7k).

Thus the routing chart is full rank away from (k=0,-6/7). Retaining the four routing coordinates αi\alpha_i, the six source-labelled quadratic cover equations were exported before imposing the triple wall. Their wall pullback reproduces the two generic equations in (x,v,w,z,k)(x,v,w,z,k) used by the earlier critical calculation.

Failed unrestricted chart transport

The sixth vertex is not generically contained in the routing four-plane. Its exact normal square is

q52−bTH−1b=−4k(k−3)(2k+3)6+7k.q_5^2-b^T H^{-1}b = -\frac{4k(k-3)(2k+3)}{6+7k}.

Consequently, changing the routing origin by two sites changes the ambient four-plane unless this expression vanishes. A direct attempt to transport all six cover equations without imposing this condition fails. Therefore the three chartwise Cartier components from Entry 1889 do not form a generic cyclic routing atlas over the unrestricted one-parameter family.

This is a geometric obstruction, not a coordinate-normalization defect.

Gram-completed atlas

Set the sixth-vertex normal square to zero before chart transport. Equivalently, replace the sixth vertex by its canonical (H(k))-projection while retaining all four source pairings (b_i=r_i\mathbin\cdot q_5). On this four-dimensional Gram-completion locus, the three charts based at vertices (1,3,5) have source-derived affine transitions

α=os+Tsβ.\alpha=o_s+T_s\beta.

All three transition determinants are nonzero on their stated generic loci. Direct substitution verifies all six labelled squared-distance equations in each chart:

3 charts×6 equations=183\text{ charts}\times6\text{ equations}=18

exact zero identities over Q(k)\mathbb Q(k).

Hence

the cyclic routing atlas exists on the Gram completion, but not on the unrestricted generic normal family.\boxed{ \text{the cyclic routing atlas exists on the Gram completion, but not on the unrestricted generic normal family.} }

Interpretation

Entry 1889's length-six descended Cartier object is naturally attached to the Gram-completed coefficient geometry. Its cyclic descent cannot yet be promoted to a generic off-Gram coefficient family. This updates the prior toward a Gram-supported Cartier excess, but does not settle the question: a generic divisor may still exist chartwise and specialize to that excess.

Next falsifier

Compute the full chartwise critical ideal of the unspecialized wall equations, saturate by

k(6+7k),xvwz,k(6+7k),\quad xvwz,

and the source wall-multiplier factors, and eliminate (x,v,w). Only after that chartwise elimination should its components be compared on the Gram-completion locus.

Durable verification

View source on GitHub