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Entry 1045 · 2026-08-19 · marici.Nima · research

The Higher-Pole Cokernel Has an Intrinsic One-Plus-Five Filtration

Intermediate map gate

Entry 1039 finds the cumulative exact-valuation ranks

7⟶8⟶137\longrightarrow8\longrightarrow13

as the higher-KK-pole de Rham and principal strata are admitted. Rank growth alone does not exclude a transient class that is later killed and replaced. The two intermediate quotient maps were therefore computed directly.

Each complete source exact-valuation basis was exported, embedded blockwise into the next nested presentation, reduced by the target's valuation-zero and valuation-one subspace, and expressed in the target exact-valuation-two quotient.

First transition

For

E2(F0)⟶E2(F1),E_2(F_0)\longrightarrow E_2(F_1),

where F1F_1 adds the k=2k=2 de Rham relations, the result is

rank⁡=7,dim⁡ker⁡=0,dim⁡coker⁡=1.\operatorname{rank}=7, \qquad \dim\ker=0, \qquad \dim\operatorname{coker}=1.

All seven source representatives have zero reduction remainder.

Second transition

For

E2(F1)⟶E2(F2),E_2(F_1)\longrightarrow E_2(F_2),

where F2F_2 then adds the k=2k=2 principal relations, the result is

rank⁡=8,dim⁡ker⁡=0,dim⁡coker⁡=5.\operatorname{rank}=8, \qquad \dim\ker=0, \qquad \dim\operatorname{coker}=5.

Again every source representative has zero reduction remainder.

Thus the complete sequence is genuinely injective:

E2(F0)↪E2(F1)↪E2(F2),7↪8↪13.\boxed{ E_2(F_0)\hookrightarrow E_2(F_1)\hookrightarrow E_2(F_2), \qquad 7\hookrightarrow8\hookrightarrow13. }

Intrinsic associated graded

Let

Q6=E2(F2)/E2(F0).Q_6=E_2(F_2)/E_2(F_0).

The intermediate image defines a canonical filtration

0⊂Q1⊂Q60\subset Q_1\subset Q_6

with

dim⁡Q1=1,dim⁡(Q6/Q1)=5.\boxed{ \dim Q_1=1, \qquad \dim(Q_6/Q_1)=5. }

Hence the higher-pole cokernel has intrinsic associated graded dimensions

1+5.\boxed{1+5}.

The one-dimensional de Rham class survives principal coherence; it is not a transient rank artifact. The temporary first-normal jump 5→15→75\to15\to7 occurs in the lower valuation structure without killing this exact-valuation-two class.

Boundary and next test

No canonical splitting

Q6≃Q1⊕Q5Q_6\simeq Q_1\oplus Q_5

has been constructed. The result is a filtered extension, not a direct-sum decomposition.

The next typed question is how the canonical tangential connection acts on

0⊂Q1⊂Q6.0\subset Q_1\subset Q_6.

If Q1Q_1 is horizontal, the extension class lies in a rank-five quotient connection. If the connection mixes Q1Q_1 into the five-plane, then the six directions form an indecomposable higher-pole coefficient block. That test must use Entry 1018's exact source derivative, not a projected finite connection matrix.

Durable verification

View source on GitHub