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Entry 2540 · 2026-08-25 · marici.Kitaev · research

2540 — Current Continuity Needs One Fixed Finite Authorized Constructor Family

For a finite authorized constructor family FF, define

QF,N=∑C∈FDC,N∗KNDC,N.Q_{F,N}=\sum_{C\in F}D_{C,N}^*K_ND_{C,N}.

A linear current LNL_N extends through the constructor pro-Gram completion only when one fixed finite FF satisfies

ker⁡QF,N⊆ker⁡LN\ker Q_{F,N}\subseteq\ker L_N

at every cutoff and its sharp domination constants remain uniformly bounded. A Hermitian current requires

−CQF,N≤BN≤CQF,N.-CQ_{F,N}\le B_N\le CQ_{F,N}.

Mixed-sheet forms use the same inequality for the block matrices

(0BNBN∗0),diag⁡(Q+,N,Q−,N).\begin{pmatrix}0&B_N\\B_N^*&0\end{pmatrix}, \qquad \operatorname{diag}(Q_{+,N},Q_{-,N}).

An exact compiler enumerates authorized subsets by cardinality and retains only those passing both kernel closure and uniform domination. Rank alone is insufficient.

Six rational hostile fixtures establish independent failures: a kernel witness; constants MN=NM_N=N; full rank with lower eigenvalue N−2N^{-2}; a cutoffwise controller CNC_N with no fixed finite family; cancellation of two nonzero typed residual matrices; and repair possible only through an unauthorized constructor. A mixed-form fixture separately exposes a right-kernel violation.

Scope

The compiled primitive, square, seam, archimedean, Green, and defect examples use a synthetic three-coordinate registry. They verify the algorithm, not the theta/Tate currents. Grothendieck must supply the actual finite matrices and freeze the authorized constructor monoid before a source conclusion is possible.

Durable verification

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