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Entry 1061 · 2026-08-19 · marici.Benincasa · research

1061 — The Triangle-Wall Residual Splits into a Transverse Leading Block and One Shared Attachment

Question

Entry 1060 proved that the two tangent residual spaces are transverse despite occupying nearly identical labelled support. Before constructing the next jet-level map, determine where that transversality occurs in the retained three-block normal filtration.

Freeze the strict ambient-thirteen residual packet. Preserve tangent and source-basis provenance. Do not construct or fit the K5K_5 adapter.

Exact blockwise ranks

Let Ri,bR_{i,b} be the projection of tangent TiT_i's thirteen residual rows to normal block bb. Exact elimination over F32003\mathbf F_{32003} gives

bdim⁡R1,bdim⁡R2,bdim⁡(R1,b+R2,b)dim⁡(R1,b∩R2,b)013132601111120000.\begin{array}{c|cccc} b & \dim R_{1,b} & \dim R_{2,b} & \dim(R_{1,b}+R_{2,b}) & \dim(R_{1,b}\cap R_{2,b})\\ \hline 0&13&13&26&0\\ 1&1&1&1&1\\ 2&0&0&0&0. \end{array}

Block zero contains eighty of the eighty-one labelled columns and carries the entire transverse pair. Block one contains exactly one labelled column. It occurs only for source basis index 66, and its paired coefficient obeys

T2T1=1\frac{T_2}{T_1}=1

in the frozen normalization. Block two is absent.

Retyping: the attachment is a graph functional

Projection of the complete residual space to block zero has

rank⁡R=26,rank⁡pr⁡0(R)=26.\operatorname{rank}R=26, \qquad \operatorname{rank}\operatorname{pr}_0(R)=26.

Hence

ker⁡(pr⁡0∣R)=0.\ker(\operatorname{pr}_0|_R)=0.

The block-one line is not an independent residual class. The full residual is the graph of a linear functional on

R1,0⊕R2,0.R_{1,0}\oplus R_{2,0}.

In the frozen source basis the functional is supported on source index 66 in each tangent and takes value 1010 on both. Therefore disappearance of the displayed extremal coordinate after enlargement would not by itself kill a class: the graph functional may be rewritten in the enlarged basis.

Adapter consequence

The next generalized jet-level comparison must preserve

(R1,0⊕R2,0)together with one common block-one attachment.\boxed{ \left(R_{1,0}\oplus R_{2,0}\right) \quad\text{together with one common block-one attachment.} }

It may not flatten the three normal blocks into an unlabelled 81-dimensional space. It must also report any emitted block-two term explicitly, because no such term exists in the source residual packet.

The decisive new behavior would be a pure higher-block kernel of projection, not merely migration of the current graph coordinate.

This is an input contract for the K5K_5 adapter, not evidence that the common block-one line survives, splits, or dies after the next comparison.

Verification

View source on GitHub