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The unnormalized ribbon trace is not an unframed Markov invariant

Statement refuted

The claim refuted is: the unnormalized ribbon evaluation $t_{n} (β)$ is invariant under Markov stabilization, hence defines an invariant of oriented unframed links. Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Take $k$ of characteristic $\neq 2$ and let $C$ be the category of $Z / 2$ -graded finite-dimensional vector spaces with the sign braiding $c (v \otimes w) = (- 1)^{∣ v ∣ ∣ w ∣} w \otimes v$ and the twist $θ_{V} = (- 1)^{∣ v ∣} v$ on homogeneous vectors, which makes $C$ a ribbon category. Let $X$ be the odd one-dimensional object, so $End (X) = k$ and $θ_{X} = - {id}_{X}$ , giving $λ = - 1$ . Then The scalar twist controls the two Markov stabilizations gives

$t_{2} (σ_{1}) = λ t_{1} (e) = - t_{1} (e),$

while the closures of $e \in B_{1}$ and of $σ_{1} \in B_{2}$ are both the unknot, the second being the positive stabilization of the first. Since $t_{1} (e) = {Tr}_{L} (j_{X}) = d_{X} = 1 \neq 0$ , the two values $1$ and $- 1$ differ, so the unnormalized trace is not invariant under positive stabilization and is not an unframed link invariant; writhe normalization is required, as in the companion example.

Facts & Assumptions

Given: ${A C}_{ω}$ ; a field $k$ with $char k \neq 2$ , the category $C$ of $Z / 2$ -graded finite-dimensional vector spaces with the sign braiding and the parity twist, and the odd one-dimensional object $X$ .

[L1]

A twist on a braided rigid monoidal category is a natural automorphism $θ$ of the identity with $θ_{V \otimes W} = (θ_{V} \otimes θ_{W}) c_{W, V} c_{V, W}$ and dual-compatibility $(θ_{V})^{\lor} = θ_{V^{\lor}}$ ; a ribbon category is a braided rigid monoidal category with a twist (Twist and ribbon structure).

[L2]

Assume $End (1) = k$ and $X$ absolutely simple, so $θ_{X} = λ {id}_{X}$ with $λ \in k^{\times}$ ; then $t_{n + 1} (ι_{n} (β) σ_{n}^{\pm 1}) = λ^{\pm 1} t_{n} (β)$ (The scalar twist controls the two Markov stabilizations, Absolutely simple objects); that lemma is stated under countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), which is assumed here.

[L3]

The positive stabilization of $e \in B_{1}$ is $σ_{1} \in B_{2}$ , and a stabilization preserves the oriented closure up to ambient isotopy: the closures of a braid and of its stabilizations are equivalent oriented links (Markov conjugation and stabilization moves, Markov moves preserve the oriented closure up to isotopy, The closure of a geometric braid).

[F1]

The ribbon evaluation is $t_{n} (β) = {Tr}_{L} (j_{X^{\otimes n}} ρ_{n} (β))$ with $j = u θ$ ; for $n = 1$ and the trivial braid $e$ , $t_{1} (e) = {Tr}_{L} (j_{X}) = d_{X}$ , the categorical dimension of $X$ (The scalar twist controls the two Markov stabilizations and the defining formula of the ribbon evaluation).

Counterexample

1.1L1givenalgebra

The model is a ribbon category. The usual graded duals of finite-dimensional spaces give rigidity. The sign braiding is a symmetric braiding, so it is in particular braided; the parity twist is a natural automorphism of the identity, and on homogeneous vectors $θ_{V \otimes W} = (- 1)^{∣ v ∣ + ∣ w ∣}$ equals $(θ_{V} \otimes θ_{W}) c_{W, V} c_{V, W}$ because the double braiding is the identity; the dual of a homogeneous space has the same parity, so $θ_{V^{\lor}} = (θ_{V})^{\lor}$ . Hence [L1] makes $C$ a ribbon category.

1.2L1F1L2givenalgebra

The odd line and its twist eigenvalue. The odd one-dimensional space $X$ has $End (X) = k$ , so it is absolutely simple, and $θ_{X} = - {id}_{X}$ , so $λ = - 1 \in k^{\times}$ because $char k \neq 2$ . Its categorical dimension is $d_{X} = t_{1} (e) = 1$ : the defining composite of [F1] is ${Tr}_{L} (j_{X})$ with $j_{X} = u_{X} θ_{X}$ , and on the odd line the Drinfeld morphism is $u_{X} = - {id}_{X}$ while $θ_{X} = - {id}_{X}$ , so $j_{X} = {id}_{X}$ and $d_{X} = {Tr}_{L} ({id}_{X}) = 1$ .

2.1L2step 1.2algebra

The two values differ. By [L2] applied to $β = e \in B_{1}$ and its positive stabilization $σ_{1} \in B_{2}$ , $t_{2} (σ_{1}) = λ t_{1} (e) = (- 1) \cdot 1 = - 1$ , while $t_{1} (e) = 1$ by step 1.2. The two values differ.

3.1L3step 2.1given

The closures are the same oriented link. By [L3] the closures of $e$ and of its stabilization $σ_{1}$ are equivalent oriented links; both are the unknot, the closure of $σ_{1}$ being a one-component diagram with a single positive kink. If the unnormalized evaluation were an invariant of oriented unframed links, it would take equal values on the closures of $e$ and $σ_{1}$ ; step 2.1 shows that it does not.

4.1step 1.1step 1.2step 2.1step 3.1∎

Conclusion. The unnormalized ribbon trace takes the values $- 1$ and $1$ on two braids whose closures are equivalent oriented links, so it is not invariant under positive stabilization and not an unframed link invariant; the writhe factor is necessary, as the companion example shows. All computations are finite; the only choice principle used is ${A C}_{ω}$ , consumed through the stabilization lemma [L2].

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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)