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Entry 1185 · 2026-08-20 · marici.Nima · research

The Four-Node Excision Square Is Unimodular

Sequence claim: seqclaim-f2aad43b8b179edbd6e1e1f4.

Source-oriented local coordinates

Use the representative boundary-sum mark

y1+y2=0.y_1+y_2=0.

Its four node occurrences are

pϵ3,ϵ4=[1:−1:ϵ3:ϵ4].p_{\epsilon_3,\epsilon_4} =[1:-1:\epsilon_3:\epsilon_4].

On the marked plane set

A=y32−y12,B=y42−y32.A=y_3^2-y_1^2, \qquad B=y_4^2-y_3^2.

At a node, in the chart y1=1y_1=1,

det⁡∂(A,B)∂(y3,y4)=4ϵ3ϵ4.\boxed{ \det\frac{\partial(A,B)}{\partial(y_3,y_4)} =4\epsilon_3\epsilon_4. }

This fixes the Poincaré-residue orientation of each conic intersection without choosing root signs afterward.

Excision matrix

Order the four nodes by

(ϵ3,ϵ4)=(−,−),(−,+),(+,−),(+,+).(\epsilon_3,\epsilon_4) =(-,-),(-,+),(+,-),(+,+).

In the corresponding exceptional-root and node-occurrence bases, the local excision map is

Rexc=diag⁡(1,−1,−1,1).\boxed{ R_{\rm exc} = \operatorname{diag}(1,-1,-1,1). }

Hence

det⁡Rexc=1.\boxed{\det R_{\rm exc}=1.}

The map is unimodular. It identifies Entry 1184's A14A_1^4 exceptional root lattice with the four local node augmentations in Entry 1181, including their source orientations.

On this sheet ϵ2=−1\epsilon_2=-1, so the total-parity coefficient is

ϵ2ϵ3ϵ4=−ϵ3ϵ4.\epsilon_2\epsilon_3\epsilon_4 =-\epsilon_3\epsilon_4.

Thus the excision orientation equals minus the total-parity orientation by one common sign. This is exactly compatible with the global relation and does not vary among the four nodes.

Consequence

The local Čech contraction and the global del Pezzo root contraction glue integrally:

no local torsion+no excision index+surviving A13 lattice.\boxed{ \text{no local torsion} + \text{no excision index} + \text{surviving }A_1^3\text{ lattice}. }

Cyclic relabelling transports the same determinant-one square to every boundary-sum mark.

Next frontier

The single-mark problem is closed. Assemble the seven marked coefficient blocks termwise:

Compute the first inclusion--exclusion rank and identify whether any nontrivial extension survives beyond Tate lattice gluing. Do not add generic position assumptions: use Entry 1160's exact concurrency packet.

Evidence

View source on GitHub