ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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Unknot, trivial color, and a three-crossing braid value

Example

For every finite braid word $β$ whose closure is a knot in the declared generic matrix convention, the $N = 1$ generic value is $1$ ; the one-strand identity braid has generic value $1$ for every $N \geq 1$ . Let $T$ be the closure of the positive two-strand braid $σ_{1}^{3}$ in the declared crossing convention. With $t = q^{- 1} = A^{- 4}$ , its $N = 2$ generic value is $J_{2}^{g e n} (T; q) = t + t^{3} - t^{4}$ . In addition, assuming the Axiom of Choice and Countable Choice, for every oriented knot $K \subset R^{3} \subset S^{3}$ the fundamental-color generic value satisfies $J_{2}^{g e n} (K; q) = V_{K} (q^{- 1})$ , where $V_{K}$ is the library Jones polynomial normalized by $V_{◯} = 1$ .

The braid diagram uses braid height increasing downward. At each positive generator, the strand entering on the right passes over the strand entering on the left; the closure joins each bottom endpoint to the top endpoint in the same position. The resulting closed braid has one component. For the diagram convention, closure arcs are routed outside the rectangular braid box without adding crossings. The diagram is explanatory; the verification uses the displayed matrices.

Example references

Murakami, An Introduction to the Volume Conjecture, §2.4 and Example 2.5

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

Fundamental ribbon trace and bracket formula

Normalized bracket and library Jones polynomial

Facts & Assumptions

Given: The admissible generic color matrices, their normalization from the preceding definition, and the positive two-strand braid $σ_{1}^{3}$ ; for the general fundamental-color identity, assume the Axiom of Choice and Countable Choice.

[F1]

The generic matrix value is $J_{N}^{g e n} (β; q) = λ_{N}^{- w (β)} Tr (G_{N}^{\otimes n} ρ_{N} (β)) / d_{N}$ , with $d_{N} \neq 0$ in $Q (A)$ (Normalized colored Jones sequence).

[F2]

For the fundamental ribbon color and any blackboard-framed braid closure, the ribbon trace satisfies $t_{n} (β) = δ ⟨ \hat{β} ⟩$ , with $δ = - A^{2} - A^{- 2}$ and twist scalar $- A^{3}$ ; this statement assumes Countable Choice (The fundamental quantum sl2 ribbon trace matches the Hecke and bracket evaluations).

[F3]

Under the Axiom of Choice, the library Jones polynomial satisfies $V_{K} (t) = f_{K} (t^{- 1 / 4})$ , where $f_{K}$ is the writhe-normalized Kauffman bracket and $t = A^{- 4}$ (The bracket state sum matches the library Jones specialization).

[F4]

Every braid group element is represented by a finite word in the Artin generators and their inverses, and under AC the Artin presentation identifies these words with geometric braid classes (The braid group by Artin presentation, The Artin presentation is complete for geometric braids).

[F5]

A geometric braid closure joins each strand endpoint to the endpoint prescribed by its permutation (The closure of a geometric braid).

[F6]

The positive Artin half twist has determinant crossing sign $+ 1$ in the declared oriented projection (Oriented crossing signs, diagram writhe, and blackboard framing).

[F7]

The opposite geometric half twist is the group inverse of the positive half twist (The elementary geometric half twist, its support disc, and its opposite).

[F8]

The exponent sum of a word is the sum of its signed generator exponents (The exponent sum of a braid).

[F9]

Under AC, every nonempty oriented link in $R^{3}$ is equivalent to a closed geometric braid with at least one strand (Alexander's theorem: every link is a closed braid).

[A1]

The Axiom of Choice (The Axiom of Choice).

[A2]

Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Verification

technique · direct matrix evaluation

Given: [F1] and the matrix data in the preceding definition.

1.1F1givenalgebra

For $N = 1$ , the basis has one vector, $E = F = 0$ , $K_{L} = 1$ , $c_{1} = G_{1} = 1$ , $d_{1} = λ_{1} = 1$ . Every braid generator therefore acts as $1$ , and the normalized value of every such knot closure is $1$ . For the unknot at any $N$ , the one-strand identity braid has $w = 0$ and trace $Tr (G_{N}) = d_{N}$ , so its normalized value is also $1$ .

1.2F1givenconstructalgebra

The strand permutation of $σ_{1}^{3}$ is $(12)^{3} = (12)$ , so its two endpoints are interchanged; pairing each bottom endpoint with the top endpoint in the same position therefore produces one component. At $N = 2$ , the only nonzero terms of $R_{2}$ are $r = 0, 1$ . On the ordered basis $(v_{+} v_{+}, v_{+} v_{-}, v_{-} v_{+}, v_{-} v_{-})$ they give $c_{2} = (\begin{array}{cccc} A & 0 & 0 & 0 \\ 0 & 0 & A^{- 1} & 0 \\ 0 & A^{- 1} & A - A^{- 3} & 0 \\ 0 & 0 & 0 & A \end{array})$ and $G_{2} \otimes G_{2} = diag (A^{4}, 1, 1, A^{- 4})$ . The middle block $B = (\begin{array}{cc} 0 & A^{- 1} \\ A^{- 1} & A - A^{- 3} \end{array})$ satisfies $tr (B^{3}) = A^{3} - A^{- 9}$ by direct multiplication; the other two diagonal entries of $c_{2}^{3}$ are $A^{3}$ , hence $Tr ((G_{2} \otimes G_{2}) c_{2}^{3}) = A^{7} + A^{3} + A^{- 1} - A^{- 9}$ . Here $w (σ_{1}^{3}) = 3$ , $λ_{2} = - A^{3}$ , and $d_{2} = - A^{2} - A^{- 2}$ .

1.3A1F4F5F6F7F8algebra

Write a braid word as $β = \prod_{j = 1}^{m} σ_{i_{j}}^{ε_{j}}$ with $ε_{j} \in {1, - 1}$ , using [F4] under [A1]. Each letter gives one crossing in the braid box. The closure pairs endpoints as prescribed by [F5]; in the diagram convention fixed in the Example, the joining arcs are routed outside the rectangular braid box without added crossings. A positive letter contributes $+ 1$ to writhe by [F6]. For a negative letter the opposite half twist [F7] exchanges the over- and under-strands; the determinant in the crossing-sign definition changes sign when its two vectors are exchanged, so that crossing contributes $- 1$ . Thus $wr (\hat{β}) = \sum_{j} ε_{j} = w (β)$ by [F8].

2.1F1step 1.2algebra

Substitution into [F1] using step 1.2 yields $J_{2}^{g e n} (T; q) = \frac{A^{- 9} (A^{7} + A^{3} + A^{- 1} - A^{- 9})}{A^{2} + A^{- 2}} = A^{- 4} + A^{- 12} - A^{- 16} = t + t^{3} - t^{4}$ , where $t = A^{- 4} = q^{- 1}$ .

2.2A1A2F1F2F9step 1.3givenalgebra

Given an oriented knot $K \subset R^{3} \subset S^{3}$ , choose a braid closure $\hat{β}$ representing it by [F9]. By [F2], the fundamental-color ribbon trace is $δ$ times the raw bracket and the twist is $- A^{3}$ . Since $d_{2} = δ$ , formula [F1] gives $J_{2}^{g e n} (K; q) = (- A^{3})^{- w (β)} ⟨ \hat{β} ⟩$ . By step 1.3 this is the writhe-normalized bracket $f_{K} (A)$ . The use of [F9] requires [A1], and the framed trace in [F2] requires [A2].

3.1A1A2F2F3step 2.2given∎

The bracket/Jones theorem [F3] now gives $f_{K} (A) = V_{K} (t)$ at $t = A^{- 4} = q^{- 1}$ . Its hypothesis uses AC [A1], while the ribbon-trace identification in [F2] uses ACω [A2], already supplied in step 2.2; the proof keeps those axiom branches separate. Therefore $J_{2}^{g e n} (K; q) = V_{K} (q^{- 1})$ .

Remarks

The all-color ribbon and Laurent-polynomial prerequisites are now supplied locally. The general fundamental-color identification uses the actual matrix-to-bracket and bracket-to-Jones suppliers linked above, with their fixed A/B smoothing chart, positive Artin crossing convention, and separate AC and countable-choice hypotheses. The displayed finite matrix calculation fixes the variable transport $t = q^{- 1} = A^{- 4}$ .

Depends on

Used by

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Sources

Hitoshi Murakami, An Introduction to the Volume Conjecture, §2.4 and Example 2.5 (standard reference, not scraped)
Louis H. Kauffman, Combinatorial Knot Theory and the Jones Polynomial, §2 (standard reference, not scraped)