CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

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At loop value zero, a local generator is nilpotent but nonzero

Statement refuted

The assertion “if the loop value is $0$ , then every local generator is $0$ ” is false for $R = Z$ , loop value $δ = 0$ , and rank $n = 2$ .

Facts & Assumptions

Given: $R = Z$ , $δ = 0$ , and $n = 2$ .

[F1]

The rank-two diagram module is free on its loop-free noncrossing pairings; its identity and cap-cup diagram are distinct basis elements. (Temperley–Lieb planar diagram data and endomorphism modules, Stacking is associative and horizontal juxtaposition is strict monoidal)

[F2]

At loop value $0$ , the local relation is $d_{1}^{2} = 0$ , and $u_{1} \mapsto d_{1}$ under the presentation map. (The elementary diagram generators satisfy the local Temperley–Lieb relations)

[F3]

The presentation map $P_{2} (Z, 0) \to {T L}_{2} (Z, 0)$ is an isomorphism. (The generator presentation is the planar diagram algebra over every commutative ring)

Counterexample

technique · direct

1.1F1algebra

In cyclic boundary order the four points have exactly three pairings: $(1, 2) (3, 4)$ , $(1, 3) (2, 4)$ , and $(1, 4) (2, 3)$ . The middle pairing crosses, while the other two are the identity and $d_{1}$ . They are distinct members of the diagram indexing set, so $d_{1}$ is a nonzero standard basis vector in the free abelian module [F1].

2.1F2F3step 1.1∎

By [F2], $d_{1}^{2} = δ d_{1} = 0$ . Since the isomorphism [F3] sends $u_{1}$ to this nonzero $d_{1}$ , injectivity gives $u_{1} \neq 0$ and $u_{1}^{2} = 0$ . This is the claimed counterexample.

Remarks

This boundary case does not affect the diagram basis: the loop parameter changes multiplication coefficients, not which pairings index the free module.

Depends on

Used by

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