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Kauffman bracket state sum and the Jones polynomial — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A type-I kink changes the raw bracket

Statement refuted

The false statement is: let $A$ be an indeterminate and, for every nonempty regular oriented diagram $D$ of an unframed knot with finite crossing set $C (D)$ , let $⟨ D ⟩ = \sum_{S : C (D) \to {A, B}} A^{a (S) - b (S)} (- A^{2} - A^{- 2})^{∣ S ∣ - 1},$ where $a (S)$ and $b (S)$ count A- and B-smoothings, the over-strand runs southwest to northeast, A joins the left and right endpoint pairs with weight $A$ , B joins the top and bottom endpoint pairs with weight $A^{- 1}$ , and $∣ S ∣$ is the number of state circles. Then $⟨ D ⟩$ depends only on the ambient-isotopy class of the represented oriented knot.

Statement references

Kauffman bracket state sum

Type-I curl factors

Oriented type-I move

Kauffman, An Invariant of Regular Isotopy (1990), §§III and VI

Kauffman, Combinatorial Knot Theory and the Jones Polynomial (2022), §2

Roberts, Knots Knotes (1999), §§4.1 and 4.3

Funar, A Crash Course in Quantum Invariants of Knots and 3-Manifolds, Chapter 3, §1.1

Facts & Assumptions

Given: The false statement above and the bracket state sum on nonempty regular oriented diagrams after their strand arrows are forgotten.

[F1]

A crossing-free one-component diagram has one state with one circle and bracket $(- A^{2} - A^{- 2})^{1 - 1} = 1$ (Kauffman bracket smoothing and loop value).

[F2]

Adding a positive type-I kink multiplies the raw bracket by $- A^{3}$ (Kauffman bracket curl and writhe correction).

[F3]

An oriented type-I kink is an allowed local Reidemeister replacement and each such local replacement is realized by an ambient isotopy of the oriented link (Oriented Reidemeister moves, Each oriented Reidemeister move is realized by an ambient isotopy).

Counterexample

1.1F1construct

Let $D_{0}$ be a crossing-free diagram of the oriented unknot. By [F1], $⟨ D_{0} ⟩ = 1$ .

2.1F2F3algebra∎

Obtain $D_{1}$ by adding one positive type-I kink to $D_{0}$ . By [F3] the two diagrams represent ambient-isotopic unframed oriented knots, while [F2] gives $⟨ D_{1} ⟩ = - A^{3}$ . Since $A$ is an indeterminate, $- A^{3} \neq 1$ in $Z [A^{\pm 1}]$ . Thus the proposed value differs on two diagrams of the same unframed knot, refuting the statement.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Trefoil, mirror, and unknot bracket calculations

Example

Assume the Axiom of Choice. Let $A$ be an indeterminate. Let $D_{+}$ be the positive trefoil diagram, defined as the closure of the two-strand braid $σ_{1}^{3}$ , where $σ_{1}$ is the anticlockwise geometric half twist with the left label passing below and the right label above the midpoint, and both strands are oriented in the braid direction. Let $D_{-}$ be its mirror image, and let $U$ be the crossing-free one-component unknot diagram. Fix a unit viewing normal $n$ and orient the projection plane $P$ so $(u, v)$ is positive exactly when $(n, u, v)$ is positive in the oriented ambient $3$ -space. At each crossing use the local source chart whose over-strand runs southwest to northeast; the A-smoothing joins the left endpoints and right endpoints with weight $A$ , and the B-smoothing joins the top endpoints and bottom endpoints with weight $A^{- 1}$ . Transport these labels by orientation-preserving changes of local coordinates. For a diagram $D$ , let $C (D)$ be its finite crossing set. Each state is a map $S : C (D) \to {A, B}$ ; let $a (S)$ and $b (S)$ count the smoothing types and $∣ S ∣$ count its circles. Define $⟨ D ⟩ = \sum_{S : C (D) \to {A, B}} A^{a (S) - b (S)} (- A^{2} - A^{- 2})^{∣ S ∣ - 1}$ and $f_{D} (A) = (- A^{3})^{- wr (D)} ⟨ D ⟩$ , where writhe is the sum of determinant crossing signs, computed as $sgn \det_{P} (v_{o v e r}, v_{u n d e r})$ after fixing a unit viewing normal $n$ and using the induced orientation of $P$ above. With $t = A^{- 4}$ and Jones polynomial $V_{D} (t) = f_{D} (t^{- 1 / 4})$ , the values are $V_{D_{+}} (t) = t + t^{3} - t^{4}, V_{D_{-}} (t) = t^{- 1} + t^{- 3} - t^{- 4}, V_{U} (t) = 1.$

Example references

Kauffman bracket state sum

Jones specialization comparison

Kauffman, An Invariant of Regular Isotopy (1990), §§III and VI

Kauffman, Combinatorial Knot Theory and the Jones Polynomial (2022), §2

Roberts, Knots Knotes (1999), §§4.1–4.4

Funar, A Crash Course in Quantum Invariants of Knots and 3-Manifolds, Chapter 3, §§1.1–2.3

Verification

Given: AC and the closure $D_{+} = \hat{σ_{1}^{3}}$ with its three positive crossings.

[F1] The algorithm enumerates every smoothing state once and adds $A^{a (S) - b (S)} δ^{∣ S ∣ - 1}$ , with $δ = - A^{2} - A^{- 2}$ (Kauffman bracket state sum).

[F2] In the declared chart, a positive braid crossing has operator $C_{1} = A I + A^{- 1} e_{1}$ , with $I$ the two through strands and $e_{1}$ the top-cap, bottom-cup diagram (Hecke–bracket trace dictionary and local descent).

[F3] The Temperley–Lieb closure trace assigns $δ^{c (D)}$ to a basis diagram, where $c (D)$ counts circles after joining corresponding top and bottom endpoints. The closure of $I$ has two circles, and the closure of $e_{1}$ has one (Unnormalized diagram closure and partial trace).

[F4] Stacking satisfies $e_{1}^{2} = δ e_{1}$ (The elementary diagram generators satisfy the local Temperley–Lieb relations).

[F5] Each positive Artin half twist has crossing sign $+ 1$ ; mirroring swaps over and under and negates the determinant sign (Oriented crossing signs, diagram writhe, and blackboard framing, The elementary geometric half twist, its support disc, and its opposite).

[F6] The normalized bracket of every nonempty oriented link equals the library Jones specialization after $t = A^{- 4}$ (The bracket state sum matches the library Jones specialization).

[F7] The closure of the identity one-strand braid is the crossing-free one-component unknot (The closure of a geometric braid).

1.1F1F2F3F4construct

For a state with $k$ B-smoothings, the smoothed braid is the closure of a stack with $k$ copies of $e_{1}$ and $3 - k$ identity layers, because each A-smoothing is $I$ by [F2]. If $k = 0$ , the identity closure has two circles by [F3]. If $k \geq 1$ , the first $e_{1}$ connects the two closing arcs into one circle; each additional $e_{1}$ contributes one removed loop by $e_{1}^{2} = δ e_{1}$ in [F4]. Thus the state has $k$ circles. There are $(\binom{3}{k})$ states with $k$ B-smoothings, each with weight $A^{3 - 2 k} δ^{∣ S ∣ - 1}$ . The complete state table is $\begin{array}{cccl} k & states & circles & total contribution \\ 0 & 1 & 2 & A^{3} δ \\ 1 & 3 & 1 & 3 A \\ 2 & 3 & 2 & 3 A^{- 1} δ \\ 3 & 1 & 3 & A^{- 3} δ^{2} \end{array}$ Each term follows from [F1] and the circle counts in [F3–F4].

2.1F1F5F6step 1.1algebra

Substituting $δ = - A^{2} - A^{- 2}$ into the table gives $\begin{aligned} ⟨ D_{+} ⟩ & = A^{3} δ + 3 A + 3 A^{- 1} δ + A^{- 3} δ^{2} \\ = (- A^{5} - A) + 3 A + (- 3 A - 3 A^{- 3}) + (A + 2 A^{- 3} + A^{- 7}) \\ = A^{- 7} - A^{- 3} - A^{5} . \end{aligned}$ The braid has three positive crossings, so [F5] gives $wr (D_{+}) = 3$ . Therefore $f_{D_{+}} (A) = (- A^{3})^{- 3} (A^{- 7} - A^{- 3} - A^{5}) = A^{- 4} + A^{- 12} - A^{- 16} = t + t^{3} - t^{4} .$ By [F6], this is the library Jones value.

3.1F1F5F6F7step 2.1algebra∎

Mirroring keeps the projection and swaps over and under. To return the over-strand to the source chart's southwest-to-northeast position, rotate the local chart through a positive quarter-turn; this exchanges the top/bottom and left/right pairings. Thus the A- and B-smoothings interchange, so state-by-state $⟨ D_{-} ⟩ (A) = ⟨ D_{+} ⟩ (A^{- 1})$ . The determinant sign changes sign when the over and under tangent vectors are exchanged, hence $wr (D_{-}) = - 3$ by [F5]. It follows directly that $f_{D_{-}} (A) = f_{D_{+}} (A^{- 1}) = t^{- 1} + t^{- 3} - t^{- 4}$ , and [F6] identifies this with $V_{D_{-}} (t)$ . By [F7], the crossing-free one-circle diagram has one state with one circle and writhe $0$ , so its raw bracket is $1$ and [F6] gives $V_{U} (t) = 1$ .