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The scalar twist controls the two Markov stabilizations

Statement

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). 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 (Absolutely simple objects) so that the twist acts as $θ_{X} = λ {id}_{X}$ for a unique $λ \in k^{\times}$ (Twist and ribbon structure), and let $t$ be the ribbon evaluation of The ribbon evaluation of an $X$ -colored closed braid. Fix the convention of The ribbon trace equals the framed-closure evaluation that a positive stabilization closes to the positive curl with $F_{X} (φ_{X}^{+}) = θ_{X}$ . Then for every $n \geq 1$ and $β \in B_{n}$ , with $ι_{n}$ the standard inclusion of Markov conjugation and stabilization moves,

$t_{n + 1} (ι_{n} (β) σ_{n}) = λ t_{n} (β), t_{n + 1} (ι_{n} (β) σ_{n}^{- 1}) = λ^{- 1} t_{n} (β) .$

With the opposite drawing convention, in which the positive stabilization closes to the inverse curl, the two scalars are exchanged; the pair of formulas must always be fixed by the local curl picture. No semisimplicity or dimension hypothesis is used beyond absolute simplicity of $X$ and $End (1) = k$ .

Facts & Assumptions

Given: ${A C}_{ω}$ ; a $k$ -linear ribbon category $C$ with $k$ -bilinear tensor product and $End (1) = k$ , an absolutely simple object $X$ with $θ_{X} = λ {id}_{X}$ , $λ \in k^{\times}$ , an integer $n \geq 1$ and a braid $β \in B_{n}$ .

[L1]

Under ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) the ribbon evaluation satisfies $t_{n} (β) = F_{X} ({\hat{β}}^{f r})$ for the blackboard-framed closure, and the closure of $ι_{n} (β) σ_{n}^{\pm 1}$ is the closure of $β$ with one full twist $φ_{X}^{\pm 1}$ inserted on a band, evaluated to $θ_{X}^{\pm 1}$ by the functor (The ribbon trace equals the framed-closure evaluation, A ribbon object defines a unique framed-tangle evaluation functor).

[L2]

The positive stabilization of Markov conjugation and stabilization moves is $β \mapsto ι_{n} (β) σ_{n}$ and the negative stabilization is $β \mapsto ι_{n} (β) σ_{n}^{- 1}$ .

[L3]

An absolutely simple object $X$ has $End (X) = k {id}_{X}$ , so every automorphism of $X$ , in particular $θ_{X}$ , is a scalar $λ {id}_{X}$ with $λ \in k^{\times}$ (Absolutely simple objects).

[L4]

The twist is a natural automorphism of the identity and the ribbon structure satisfies the dual-compatibility (Twist and ribbon structure).

Proof

technique · direct

1.1L1L2givenconstruct

Inserting the curl. By [L1] the value $t_{n + 1} (ι_{n} (β) σ_{n}^{\pm 1})$ equals $F_{X}$ of the blackboard-framed closure of $β$ with one full twist generator $φ_{X}^{\pm 1}$ inserted in a band, the sign being fixed by the declared convention that positive stabilization corresponds to the positive curl.

1.2L1L4construct

Sliding the curl to the seam. In the framed tangle calculus the inserted full twist can be slid along its band without changing the morphism of the framed oriented tangle category: the curl-slide relations move a small curl past crossings and past the cup and cap ends of a band, and the twist is natural [L4], so the framed closure of $β$ with the curl inserted anywhere on a band is the same framed tangle as the closure of $β$ with the twist inserted at the closure seam of that band. Moving the curl to the seam and evaluating, the twist acts on the last tensor factor $X$ of $X^{\otimes n}$ before the closure pairing is taken, so $t_{n + 1} (ι_{n} (β) σ_{n}^{\pm 1}) = {Tr}_{L} ⁣ (j_{X^{\otimes n}} (ρ_{n} (β) \circ (1^{\otimes (n - 1)} \otimes θ_{X}^{\pm 1}))),$ where the insertion is the twist on the last tensor factor; this is the same formula obtained by applying the closure-comparison lemma to the modified diagram.

2.1L3step 1.2algebra

Evaluating the scalar. By [L3] the twist is $θ_{X}^{\pm 1} = λ^{\pm 1} {id}_{X}$ , so the insertion in step 1.2 is multiplication by the scalar $λ^{\pm 1}$ and can be taken out of the trace: ${Tr}_{L} (j ρ_{n} (β) λ^{\pm 1}) = λ^{\pm 1} {Tr}_{L} (j ρ_{n} (β)) = λ^{\pm 1} t_{n} (β)$ , because tensor product and composition are $k$ -bilinear: in the defining evaluation--coevaluation composite a scalar multiple of the input becomes the same scalar multiple of the composite. The identification $End (1) = k$ then identifies that composite with a scalar. This gives the two displayed formulas.

2.2L1step 1.1given

Convention warning. The identification of positive stabilization with the positive curl is a drawing convention: with the opposite convention the inserted curl in step 1.1 is $φ_{X}^{\mp 1}$ , so the two scalars in the display are exchanged. The pair of formulas is therefore always fixed against the local curl picture, as stated.

3.1step 1.1step 1.2step 2.1step 2.2∎

Conclusion. Steps 1.1--2.1 prove $t_{n + 1} (ι_{n} (β) σ_{n}^{\pm 1}) = λ^{\pm 1} t_{n} (β)$ , and step 2.2 records the convention dependence. No semisimplicity or dimension hypothesis is used beyond $End (1) = k$ and absolute simplicity of $X$ ; the only choice principle used is ${A C}_{ω}$ , consumed exactly through the closure comparison of [L1], which constructs the functor $F_{X}$ .

Depends on

Used by

The unnormalized ribbon trace is not an unframed Markov invariant Counterexample
Writhe normalization cancels a ribbon kink Example
The writhe-normalized ribbon trace is an unframed link invariant Theorem

Dependency tree · two levels

32 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

V. G. Turaev, Quantum Invariants of Knots and 3-Manifolds (de Gruyter Studies in Mathematics 18, 1994) (standard reference, not scraped)
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)