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Open questions in knot polynomials and surgery — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

The unknot defeats a universal coefficient-bound extension

Statement refuted

Assume full AC for the Jones invariant and ${A C}_{ω}$ for $c (K)$ . A classical knot is a smooth embedding $S^{1} ↪ S^{3}$ considered up to ambient isotopy. A regular diagram is a generic plane projection with only transverse double points, each marked with over/under information. The false universal extension is: for every classical knot $K$ , including the unknot, there is an integer $q$ with $2 \leq q \leq 3 c (K)$ and $f_{q} (K) \neq (0, 0)^{T}$ . Here $V_{K} (x)$ is the normalized library Jones polynomial with $V_{U} (x) = 1$ , $f_{q} (K) = (a_{q} (K), b_{q} (K))^{T}$ is defined by the unique expansion $V_{K} (x) = \sum_{q \geq 0} (a_{q} (K) + b_{q} (K) x) (x - x^{- 1})^{q}$ , and $c (K)$ is the least crossing number of a regular diagram of $K$ .

The boundary input is the unknot $U$ , represented by the crossing-free circle:

Statement references

Jones–Vassiliev coefficient vectors

Crossing number

Carmi and Cohen, On trivial Jones–Vassiliev polynomials, Theorem 6.8

Facts & Assumptions

Given: Full AC, ${A C}_{ω}$ , and the unknot $U$ with its one-component crossing-free diagram.

[F1]

The library Jones polynomial has $V_{U} (x) = 1$ , and the coefficient vector is obtained from the unique finite expansion in powers of $p = x - x^{- 1}$ (Jones–Vassiliev coefficient vectors).

[F2]

Under ${A C}_{ω}$ , $c (K)$ is the least element of the set of crossing counts of regular diagrams representing $K$ (Crossing number of a link).

[A1]

Full AC is the family-choice assertion (The Axiom of Choice).

[A2]

${A C}_{ω}$ selects an element from each countable family of nonempty sets (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Counterexample

1.1givenA1F1constructalgebra

Under [A1], $V_{U} (x) = 1$ by [F1]. Its unique coefficient expansion is $a_{0} (U) = 1$ , $b_{0} (U) = 0$ , and $a_{q} (U) = b_{q} (U) = 0$ for every $q \geq 1$ . Thus $f_{q} (U) = (0, 0)^{T}$ for every positive order.

1.2givenA2F2algebra

The displayed diagram has zero crossings, so under [A2] the set in [F2] contains $0$ ; all crossing counts are nonnegative. Hence $c (U) = 0$ , and there is no integer $q$ satisfying $2 \leq q \leq 3 c (U) = 0$ .

2.1givenstep 1.1step 1.2∎

The proposed universal statement includes $U$ , but steps 1.1 and 1.2 show that no required nonzero vector or admissible order exists for it. Therefore the statement is false.

Remarks

The source's actual coefficient claim assumes that $K$ is nontrivial, so this counterexample refutes only the broader universal extension. The presence or absence of DOI or journal metadata is bibliographic information; by itself it proves neither the source theorem nor its negation.

Remarks references

Carmi and Cohen, On trivial Jones–Vassiliev polynomials, Theorem 6.8

ExampleConstruction: AI-adaptedVerification: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

A trefoil Jones–Vassiliev coefficient calculation

Example

Assume full AC and ${A C}_{ω}$ . An oriented classical knot is an oriented smooth embedding $S^{1} ↪ S^{3}$ 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 $T = \hat{σ_{1}^{3}}$ 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 $V_{K} (x)$ be the library-normalized Jones polynomial, with $V_{U} (x) = 1$ for the unknot $U$ and $x = t^{1 / 2}$ in the convention $t = A^{- 4}$ . Put $p = x - x^{- 1}$ and write the unique finite expansion $V_{K} (x) = \sum_{q \geq 0} (a_{q} (K) + b_{q} (K) x) p^{q}$ ; define $f_{q} (K) = (a_{q} (K), b_{q} (K))^{T}$ .

In the library Jones convention $t = A^{- 4}$ , the state sum gives $V_{T} (t) = t + t^{3} - t^{4}$ . With $x = t^{1 / 2}$ and $p = x - x^{- 1}$ , the unique Jones–Vassiliev expansion is $V_{T} (x) = 1 - 3 p^{2} - 4 p^{4} - p^{6} + x (- 6 p^{3} - 5 p^{5} - p^{7}),$ so $f_{2} (T) = (- 3, 0)^{T} \neq 0$ . Carmi and Cohen's v1 preprint reports that every nontrivial knot has such a nonzero vector at some order $2 \leq q \leq 3 c (K)$ , and claims $V_{K} (x) = 1$ 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.

Example references

Jones–Vassiliev coefficient vectors

Kauffman bracket state sum

Bracket/Jones comparison

Carmi and Cohen, On trivial Jones–Vassiliev polynomials, Theorems 6.8–6.9

Kauffman, Combinatorial Knot Theory and the Jones Polynomial, §2

Roberts, Knots Knotes, §§4.3–4.4

Bar-Natan and Stoimenow, The Fundamental Theorem of Vassiliev Invariants, §1.1

Facts & Assumptions

Given: Full AC, ${A C}_{ω}$ , the two-strand braid $σ_{1}^{3}$ , and its closure $T$ .

[F1]

A state of a nonempty diagram chooses A or B at each crossing, contributes $A^{a - b} δ^{∣ S ∣ - 1}$ , and has $δ = - A^{2} - A^{- 2}$ (Kauffman bracket smoothing and loop value).

[F2]

In the stated chart a positive braid crossing is the operator $A I + A^{- 1} e_{1}$ , where $I$ is the through pairing and $e_{1}$ 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 $e_{1}$ in [F2] is the elementary diagram generator $d_{1}$ , and it satisfies $d_{1}^{2} = δ d_{1}$ (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 $f_{K} (A) : = (- A^{3})^{- wr (D)} ⟨ D ⟩$ (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 $T = \hat{σ_{1}^{3}}$ is $V_{T} (t) = t + t^{3} - t^{4}$ (The Temperley-Lieb quotient and the Jones specialization).

[F9]

The library Jones polynomial is an ambient-isotopy invariant and $V_{U} (t) = 1$ (The Temperley-Lieb quotient and the Jones specialization).

[F10]

For $p = x - x^{- 1}$ , every Laurent Jones polynomial has a unique expansion $\sum_{q} (a_{q} + b_{q} x) p^{q}$ , and $f_{q} = (a_{q}, b_{q})^{T}$ (Jones–Vassiliev coefficient vectors, Unique Laurent decomposition for the Jones–Vassiliev basis).

[F11]

Under ${A C}_{ω}$ , $c (K)$ 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]

${A C}_{ω}$ is the countable-choice hypothesis in the crossing-number definition (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Verification

1.1givenF1F2F3F4construct

For a state with $k$ B-smoothings, [F2] gives a stack of $k$ copies of $e_{1}$ and $3 - k$ copies of $I$ . If $k = 0$ , the closure has two circles by [F3]. If $k \geq 1$ , repeated use of $e_{1}^{2} = δ e_{1}$ from [F4] and the one-circle closure of $e_{1}$ in [F3] gives exactly $k$ circles. There are $(\binom{3}{k})$ choices, so [F1] gives the complete grouped state sum $\begin{array}{cccc} k & states & ∣ S ∣ & contribution \\ 0 & 1 & 2 & A^{3} δ \\ 1 & 3 & 1 & 3 A \\ 2 & 3 & 2 & 3 A^{- 1} δ \\ 3 & 1 & 3 & A^{- 3} δ^{2} \end{array}$ .

2.1F5F6F7F8step 1.1algebra

Substituting $δ = - A^{2} - A^{- 2}$ into the grouped sum gives $⟨ T ⟩ = A^{3} δ + 3 A + 3 A^{- 1} δ + A^{- 3} δ^{2} = A^{- 7} - A^{- 3} - A^{5}$ . By [F5] the closure is one knot component, and [F6] gives writhe $3$ . Applying [F7] therefore gives $f_{T} (A) = (- A^{3})^{- 3} ⟨ T ⟩ = A^{- 4} + A^{- 12} - A^{- 16}$ . On the other hand, [F8] gives the exact Jones value $t + t^{3} - t^{4}$ , 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 $x^{n} = A_{n} (p) + B_{n} (p) x$ . From $x^{2} = p x + 1$ , multiplication by $x$ gives $A_{n + 1} = B_{n}$ and $B_{n + 1} = A_{n} + p B_{n}$ . Starting with $x^{0} = 1$ and $x^{1} = x$ yields $\begin{array}{cll} n & A_{n} & B_{n} \\ 2 & 1 & p \\ 3 & p & 1 + p^{2} \\ 4 & 1 + p^{2} & 2 p + p^{3} \\ 5 & 2 p + p^{3} & 1 + 3 p^{2} + p^{4} \\ 6 & 1 + 3 p^{2} + p^{4} & 3 p + 4 p^{3} + p^{5} \\ 7 & 3 p + 4 p^{3} + p^{5} & 1 + 6 p^{2} + 5 p^{4} + p^{6} \\ 8 & 1 + 6 p^{2} + 5 p^{4} + p^{6} & 4 p + 10 p^{3} + 6 p^{5} + p^{7} \end{array}$ . This lists every recurrence step needed for the powers in $V_{T} (x) = x^{2} + x^{6} - x^{8}$ .

4.1F10step 3.1algebra

Combining the rows for $n = 2, 6, 8$ gives $a (p) = 1 - 3 p^{2} - 4 p^{4} - p^{6}$ and $b (p) = - 6 p^{3} - 5 p^{5} - p^{7}$ . Uniqueness in [F10] therefore gives $f_{2} (T) = (- 3, 0)^{T} \neq 0$ .

5.1givenA1A2F9F11step 2.1step 4.1∎

The computed Jones polynomial is not $1$ . Under [A1], [F9] gives $V_{U} = 1$ 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 $c (T) \leq 3$ . 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.

Open questions in knot polynomials and surgery — Examples and Counterexamples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

The unknot defeats a universal coefficient-bound extension

Statement refuted

Statement references

Facts & Assumptions

Counterexample

Remarks

Remarks references

A trefoil Jones–Vassiliev coefficient calculation

Example

Example references

Facts & Assumptions

Verification

Sources