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CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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The next Jones–Wenzl recursion denominator vanishes at p=6

Statement refuted

For every even p≥4 and every primitive 2p-th root A, the standard Jones–Wenzl recursion for the BHMV Kauffman bracket can be continued to every strand number by cancelling any zero quantum-integer denominator against a zero numerator.

Facts & Assumptions

Given: The BHMV conventions at p=6, a primitive twelfth root A, and the standard recursive formula for Jones–Wenzl projectors.

[F1]

The BHMV root-of-unity skein parameter: r=p/2, δ=−A2−A−2, and [1]=1, [2]=δ, [3]=δ[2]−[1].

[F2]

Jones–Wenzl recursion over unit quantum integers: the next projector in the standard recursion uses the inverse of the next quantum integer.

Counterexample

Given: Take p=6 and any primitive twelfth root A.

1.1givenF1algebra

Here r=3, q=−A2 has q2 of order three, and q+q−1=−1. Thus δ=−1, [2]=δ=−1, and [3]=δ[2]−[1]=(−1)(−1)−1=0.

2.1F2step 1.1algebra∎

The recursion for p3 requires division by [3]. Since [3]=0 in C, no multiplicative inverse exists, so this recursive expression is undefined. In particular, cancellation cannot assign it a value. The calculation refutes the asserted universal continuation; it makes no claim that a projector at rank three cannot be constructed by a different method.

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