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ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A trefoil Jones–Vassiliev coefficient calculation

Example

Assume full AC and ACω. An oriented classical knot is an oriented smooth embedding S1↪S3 considered up to ambient isotopy. A regular diagram is a generic plane projection with only transverse double points, each marked with over/under information; c(K) is the least crossing count among regular diagrams representing K. Let T=σ13^ be the oriented closure of three positive two-strand geometric half twists, with each strand oriented in the braid direction. The figure shows the closure; its three crossings are positive under the library determinant convention. For any classical knot K, let VK(x) be the library-normalized Jones polynomial, with VU(x)=1 for the unknot U and x=t1/2 in the convention t=A−4. Put p=x−x−1 and write the unique finite expansion VK(x)=∑q≥0(aq(K)+bq(K)x)pq; define fq(K)=(aq(K),bq(K))T.

c¾31positivetwo-strandbraidclosure

In the library Jones convention t=A−4, the state sum gives VT(t)=t+t3−t4. With x=t1/2 and p=x−x−1, the unique Jones–Vassiliev expansion is VT(x)=1−3p2−4p4−p6+x(−6p3−5p5−p7), so f2(T)=(−3,0)T≠0. Carmi and Cohen's v1 preprint reports that every nontrivial knot has such a nonzero vector at some order 2≤q≤3c(K), and claims VK(x)=1 implies that K is the unknot. This finite calculation checks one example only; it does not verify either general claim or the preprint's machine-geometry proof.

Facts & Assumptions

Given: Full AC, ACω, the two-strand braid σ13, and its closure T.

[F1]

A state of a nonempty diagram chooses A or B at each crossing, contributes Aa−bδ∣S∣−1, and has δ=−A2−A−2 (Kauffman bracket smoothing and loop value).

[F2]

In the stated chart a positive braid crossing is the operator AI+A−1e1, where I is the through pairing and e1 is the cap-cup pairing (Bracket smoothings of positive and negative braid crossings).

[F3]

Closing a basis diagram identifies each top endpoint with the corresponding bottom endpoint and assigns one factor δ to each resulting circle; stacking removes each closed cycle with the same factor (Unnormalized diagram closure and partial trace, Temperley–Lieb planar diagram data and endomorphism modules).

[F4]

The cap-cup diagram named e1 in [F2] is the elementary diagram generator d1, and it satisfies d12=δd1 (Temperley–Lieb planar diagram data and endomorphism modules, The elementary diagram generators satisfy the local Temperley–Lieb relations).

[F5]

The positive elementary half twist has endpoint permutation (1 2), and a braid closure has one component for each cycle of its endpoint permutation (The elementary geometric half twist, its support disc, and its opposite, The closure of a geometric braid).

[F6]

The positive geometric Artin half twist has determinant crossing sign +1 (Oriented crossing signs, diagram writhe, and blackboard framing).

[F7]

The writhe-normalized bracket of a diagram D representing K is fK(A):=(−A3)−wr⁡(D)⟨D⟩ (The bracket state sum matches the library Jones specialization).

[F8]

In the library convention t=A−4, the exact Jones value of the positive trefoil closure T=σ13^ is VT(t)=t+t3−t4 (The Temperley-Lieb quotient and the Jones specialization).

[F9]

The library Jones polynomial is an ambient-isotopy invariant and VU(t)=1 (The Temperley-Lieb quotient and the Jones specialization).

[F10]

For p=x−x−1, every Laurent Jones polynomial has a unique expansion ∑q(aq+bqx)pq, and fq=(aq,bq)T (Jones–Vassiliev coefficient vectors, Unique Laurent decomposition for the Jones–Vassiliev basis).

[F11]

Under ACω, c(K) is the least number of crossings among regular diagrams of K (Crossing number of a link).

[A1]

Full AC supplies the library Jones-invariance hypothesis (The Axiom of Choice).

[A2]

ACω is the countable-choice hypothesis in the crossing-number definition (The Axiom of Countable Choice (ACω)).

Verification

1.1givenF1F2F3F4construct

For a state with k B-smoothings, [F2] gives a stack of k copies of e1 and 3−k copies of I. If k=0, the closure has two circles by [F3]. If k≥1, repeated use of e12=δe1 from [F4] and the one-circle closure of e1 in [F3] gives exactly k circles. There are (3k) choices, so [F1] gives the complete grouped state sum kstates∣S∣contribution012A3δ1313A2323A−1δ313A−3δ2.

2.1F5F6F7F8step 1.1algebra

Substituting δ=−A2−A−2 into the grouped sum gives ⟨T⟩=A3δ+3A+3A−1δ+A−3δ2=A−7−A−3−A5. By [F5] the closure is one knot component, and [F6] gives writhe 3. Applying [F7] therefore gives fT(A)=(−A3)−3⟨T⟩=A−4+A−12−A−16. On the other hand, [F8] gives the exact Jones value t+t3−t4, which at t=A−4 is the same Laurent polynomial. Thus the direct normalized state-sum calculation agrees with the library Jones value for this trefoil.

3.1givenF10step 2.1algebra

Write xn=An(p)+Bn(p)x. From x2=px+1, multiplication by x gives An+1=Bn and Bn+1=An+pBn. Starting with x0=1 and x1=x yields nAnBn21p3p1+p241+p22p+p352p+p31+3p2+p461+3p2+p43p+4p3+p573p+4p3+p51+6p2+5p4+p681+6p2+5p4+p64p+10p3+6p5+p7. This lists every recurrence step needed for the powers in VT(x)=x2+x6−x8.

4.1F10step 3.1algebra

Combining the rows for n=2,6,8 gives a(p)=1−3p2−4p4−p6 and b(p)=−6p3−5p5−p7. Uniqueness in [F10] therefore gives f2(T)=(−3,0)T≠0.

5.1givenA1A2F9F11step 2.1step 4.1∎

The computed Jones polynomial is not 1. Under [A1], [F9] gives VU=1 and invariance under ambient isotopy, so T is nontrivial. The displayed braid has three crossings, so under [A2] and [F11] its crossing-number set contains 3 and c(T)≤3. This example verifies a nonzero coefficient at order 2 for this single knot; the general bound and unknot-detection claim in the Example section are only reported from the source and are not used as premises.

Depends on

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