LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Kauffman bracket evaluation of framed links in the three-sphere

Statement

Assume the Axiom of Countable Choice and let $R = Z [A^{\pm 1}]$ and $δ = - A^{2} - A^{- 2}$ . For a nonempty framed unoriented link $L$ in $S^{3}$ , choose a regular planar diagram $D$ of its core. Let $m_{j} \in Z$ be the full-twist correction on the $j$ th component from the blackboard framing of $D$ to the specified framing of $L$ , counting an A-type twist positively and its inverse negatively as in the framed diagram calculus, and put $m = \sum_{j} m_{j}$ . For a state $S$ of $D$ , let $a (S)$ and $b (S)$ be its numbers of A- and B-smoothings and $∣ S ∣$ its number of resulting circles. The assignment $E (\emptyset) = 1, E (L) = δ (- A^{3})^{m} \sum_{S} A^{a (S) - b (S)} δ^{∣ S ∣ - 1}$ is independent of the diagram and twist corrections and extends $R$ -linearly to a map $E : K_{A} (S^{3}) \to R$ satisfying $E ([\emptyset]) = 1$ .

Statement references

Kauffman bracket skein module

Kauffman bracket smoothing and loop value

Framed diagram calculus

Józef H. Przytycki, Fundamentals of Kauffman Bracket Skein Modules (1998), §2, Theorem 2.3(a)

Sean Pohorence, 3-manifold Invariants and TQFTs (2016), §3.1

Facts & Assumptions

Given: Countable choice, a framed unoriented link in $S^{3}$ , and its regular diagram and framing corrections.

[A1]

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

[F1]

For a nonempty regular diagram $D$ , the Kauffman bracket is the finite sum $⟨ D ⟩ = \sum_{S} A^{a (S) - b (S)} δ^{∣ S ∣ - 1}$ with $δ = - A^{2} - A^{- 2}$ (Kauffman bracket smoothing and loop value).

[F2]

The raw bracket is unchanged by every local Reidemeister II and III replacement in the fixed source-chart convention (Kauffman bracket R2 and R3 identities).

[F3]

Framed isotopies in $S^{3}$ admit finite planar-isotopy/Reidemeister sequences; a type-I move changes the blackboard framing by one signed twist (Framed link diagrams and local Reidemeister moves in plane products).

[F4]

The skein module imposes $[L_{\times}] = A [L_{A}] + A^{- 1} [L_{B}]$ and $[L ⊔ O] = δ [L]$ for a zero-framed trivial circle (Kauffman bracket skein module of a three-manifold).

Proof

1.1A1F1F2F3givenalgebra

The state sum in [F1] is invariant under planar isotopy because that isotopy bijects crossings, states, and state circles. By [F3], diagrams of the same framed link are connected by a finite move sequence. For RII and RIII, [F2] leaves the raw bracket unchanged and the framing correction is unchanged. For an A-type R1 kink, the local state contribution is $A δ + A^{- 1} = - A^{3}$ ; for the opposite B-type kink it is $A + A^{- 1} δ = - A^{- 3}$ . For the A-type kink [F3] decreases the correction by one and for the B-type kink it increases the correction by one, so the product $(- A^{3})^{m} ⟨ D ⟩$ is unchanged. Hence $E (L)$ is independent of diagram and framing correction.

2.1F1F3F4step 1.1algebra

At a crossing, partition the finite state set by whether that crossing is smoothed A or B. This gives $⟨ D_{\times} ⟩ = A ⟨ D_{A} ⟩ + A^{- 1} ⟨ D_{B} ⟩$ . In a framed skein triple the exterior framed strips agree; place the same twist marks on those common arcs and none on the blackboard-framed local tangles. Their total correction $m$ is the same for the crossing and both smoothings. Multiplying the state identity by $δ (- A^{3})^{m}$ proves that $E$ sends the crossing relation in [F4] to zero.

3.1F1F4step 1.1step 2.1algebra∎

If $L$ is nonempty, adjoining a disjoint zero-framed trivial circle adds one circle to every state and multiplies the bracket by $δ$ . Thus $E (L ⊔ O) = δ E (L)$ . For the empty link, its one-circle diagram has bracket $1$ , so the defining assignment gives $E (O) = δ E (\emptyset)$ . The evaluation therefore annihilates every trivial-circle relation in [F4]. Its $R$ -linear extension to the free module factors through the quotient $K_{A} (S^{3})$ and sends the empty class to $1$ .

Depends on

Used by

The Kauffman bracket skein module of S³ is rank one Theorem

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Józef H. Przytycki, Fundamentals of Kauffman Bracket Skein Modules, arXiv:math/9809113 (1998), §2, Theorem 2.3(a) (standard reference, not scraped)
Sean Pohorence, 3-manifold Invariants and TQFTs (January 2016), §3.1, pp. 11–12 (standard reference, not scraped)
Thang T. Q. Lê, The Colored Jones Polynomial and the AJ Conjecture, §6.2, Example 6.4, printed p. 30 (standard reference, not scraped)