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The writhe-normalized ribbon trace is an unframed link invariant

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $k$ be a field and $C$ a $k$ -linear ribbon category, with $k$ -bilinear tensor product and ${End}_{C} (1) = k$ , let $X \in C$ be absolutely simple with $θ_{X} = λ {id}_{X}$ , $λ \in k^{\times}$ , and let $w$ be the exponent sum (Exponent sum and writhe of a braid). For $β \in B_{n}$ put

$J_{X} (\hat{β}) : = λ^{- w (β)} t_{n} (β) \in k .$

For the empty braid $e \in B_{0}$ , $w (e) = 0$ and $t_{0} (e) = 1$ , so $J_{X} (\emptyset) = 1$ . Its Markov class is isolated because stabilization requires $n \geq 1$ .

Then $J_{X}$ is invariant under conjugation in each $B_{n}$ and under both stabilizations $β \mapsto ι_{n} (β) σ_{n}^{\pm 1}$ (and hence under their inverses), so, by Markov's theorem (Markov's theorem for braid closures), $J_{X} (\hat{β})$ depends only on the oriented unframed link type of the closure $\hat{β}$ : if $\hat{β}$ and $\hat{β^{'}}$ are equivalent oriented links then $J_{X} (\hat{β}) = J_{X} (\hat{β^{'}})$ . If in addition the categorical dimension $d_{X} = {Tr}_{L} (j_{X})$ is a unit of $k$ , then $d_{X}^{- 1} J_{X}$ is the normalization with value $1$ on the unknot; no invertibility of $d_{X}$ is needed for invariance. The Axiom of Choice is used through Markov's theorem, and it supplies countable choice (AC implies DC implies countable choice) for the stabilization lemma; the trace computations themselves are finite.

Facts & Assumptions

Given: a $k$ -linear ribbon category $C$ with $k$ -bilinear tensor product and $End (1) = k$ , an absolutely simple object $X$ with $θ_{X} = λ {id}_{X}$ and $λ \in k^{\times}$ , the ribbon evaluation $t_{n}$ , and the exponent sum $w$ .

[L1]

The ribbon evaluation is $t_{n} (β) = {Tr}_{L} (j_{X^{\otimes n}} ρ_{n} (β))$ with $j = u θ$ a natural isomorphism, and the trace is cyclic: ${Tr}_{L} (a c) = {Tr}_{L} (c^{\lor \lor} a)$ for $a ⁣ : Y \to Y^{\lor \lor}$ and $c ⁣ : Y \to Y$ (The ribbon evaluation of an $X$ -colored closed braid, Basic properties of the categorical trace).

[L2]

For every braid $β$ one has $t_{n + 1} (ι_{n} (β) σ_{n}^{\pm 1}) = λ^{\pm 1} t_{n} (β)$ (The scalar twist controls the two Markov stabilizations); that lemma is stated under countable choice, which AC supplies here (AC implies DC implies countable choice).

[L3]

The exponent sum satisfies $w (ι_{n} (β) σ_{n}^{\pm 1}) = w (β) \pm 1$ and $w (γ β γ^{- 1}) = w (β)$ (Exponent sum and writhe of a braid).

[L4]

Two braids have equivalent oriented closures if and only if they are related by finitely many conjugations, stabilizations and destabilizations; this is Markov's theorem, proved under the Axiom of Choice (Markov's theorem for braid closures).

[F1]

The Axiom of Choice is assumed (The Axiom of Choice); it enters the argument through Markov's theorem [L4] and supplies countable choice for the stabilization lemma [L2] (AC implies DC implies countable choice). Absolute simplicity makes $θ_{X}$ act by the scalar $λ$ (The scalar twist controls the two Markov stabilizations).

Proof

technique · direct

1.1L1L3givenalgebra

Conjugation invariance. Fix $γ \in B_{n}$ ; write $ρ = ρ_{n}$ and $j = j_{X^{\otimes n}}$ . Then $t_{n} (γ β γ^{- 1}) = {Tr}_{L} (j ρ (γ) ρ (β) ρ (γ)^{- 1})$ . Apply cyclicity [L1] with $a = j ρ (γ)$ and $c = ρ (β) ρ (γ)^{- 1}$ : $t_{n} (γ β γ^{- 1}) = {Tr}_{L} ((ρ (β) ρ (γ)^{- 1})^{\lor \lor} j ρ (γ)) = {Tr}_{L} (ρ (β)^{\lor \lor} (ρ (γ)^{- 1})^{\lor \lor} j ρ (γ)) .$ Naturality of $j$ at the morphism $ρ (γ)^{- 1}$ gives $(ρ (γ)^{- 1})^{\lor \lor} j = j ρ (γ)^{- 1}$ , so the argument equals $ρ (β)^{\lor \lor} j$ . Applying cyclicity again, ${Tr}_{L} (ρ (β)^{\lor \lor} j) = {Tr}_{L} (j ρ (β)) = t_{n} (β)$ . Since $w (γ β γ^{- 1}) = w (β)$ by [L3], this gives $J_{X} (γ β γ^{- 1}) = J_{X} (β)$ .

2.1L2L3step 1.1algebra

Invariance under stabilizations. By [L2], $t_{n + 1} (ι_{n} (β) σ_{n}^{\pm 1}) = λ^{\pm 1} t_{n} (β)$ , while $w (ι_{n} (β) σ_{n}^{\pm 1}) = w (β) \pm 1$ by [L3]. Hence $λ^{- w (ι_{n} β σ_{n}^{\pm 1})} t_{n + 1} (ι_{n} (β) σ_{n}^{\pm 1}) = λ^{- w (β) \mp 1} λ^{\pm 1} t_{n} (β) = λ^{- w (β)} t_{n} (β) = J_{X} (\hat{β}) .$ Destabilizations are the inverses of stabilizations, so $J_{X}$ is also invariant under them.

3.1L4F1step 1.1step 2.1

Markov's theorem. Steps 1.1 and 2.1 show that $J_{X}$ is unchanged by each of the Markov moves and their inverses. By Markov's theorem [L4], two braids are related by such moves exactly when their closures are equivalent oriented links; therefore $J_{X} (\hat{β})$ depends only on the oriented unframed link type of $\hat{β}$ . This uses AC through Markov's theorem, and the countable-choice input to [L2] used in step 2.1 is also supplied by AC, as recorded in [F1] and [L4].

4.1L1step 3.1algebra

The normalization clause. The value $J_{X}$ is an invariant of oriented links, so multiplying it by a unit $d_{X}^{- 1}$ leaves an invariant. On the closure of the identity $e \in B_{1}$ , which is the unknot, one has $w (e) = 0$ and $t_{1} (e) = {Tr}_{L} (j_{X}) = d_{X}$ , so $d_{X}^{- 1} J_{X} (\hat{e}) = d_{X}^{- 1} d_{X} = 1$ . If $d_{X}$ is not a unit of $k$ , the normalization by $d_{X}^{- 1}$ is not available, but the invariance statement of step 3.1 does not use it.

5.1step 1.1step 2.1step 3.1step 4.1∎

Conclusion. Steps 1.1--2.1 prove that the writhe-normalized ribbon trace is invariant under Markov equivalence and hence an invariant of oriented unframed link types of closures, and step 4.1 gives the further unknot normalization when $d_{X}$ is invertible. Every categorical computation is finite; the Axiom of Choice is consumed by Markov's theorem and by the countable-choice input to the stabilization lemma.

Depends on

Used by

Writhe normalization cancels a ribbon kink Example

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Sources

P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories (AMS Mathematical Surveys and Monographs 205), author's final version (standard reference, not scraped)
V. G. Turaev, Quantum Invariants of Knots and 3-Manifolds (de Gruyter Studies in Mathematics 18, 1994) (standard reference, not scraped)