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An unnormalized Hecke trace is not Markov invariant

Statement refuted

Assume AC for the link-invariance assertion about the corrected normalization. The unnormalized trace family ${tr}_{n}$ of The Ocneanu Markov trace exists and is unique, and more generally any normalization of the form $u^{e (β)} α^{n - 1} {tr}_{n} (π_{n} (β))$ whose parameters $u, α$ fail at least one of the two relations $u α z = 1$ and $u^{2} = z_{-} / z$ , is invariant under positive and negative Markov stabilizations of braids.

Facts & Assumptions

Given: AC (The Axiom of Choice), the Hecke tower over $Λ = Z [v^{\pm 1}, z]$ , the Ocneanu trace, a braid $β \in B_{n}$ , and a normalization $u^{e (β)} α^{n - 1} {tr}_{n} (π_{n} (β))$ with parameters in a commutative ring containing $Λ$ and in which $u, z$ are units. The counterexample calculation is algebraic; AC is included because [F4] cites the choice-qualified link-invariance result.

[F1]

${tr}_{n + 1} (x T_{n} y) = z {tr}_{n} (x y)$ and ${tr}_{n + 1} (x T_{n}^{- 1}) = z_{-} {tr}_{n} (x)$ for all $x, y \in H (n)$ , with $z_{-} = v^{- 1} (z + 1 - v)$ (The Ocneanu Markov trace exists and is unique, The Markov trace of an inverse Hecke generator).

[F2]

$z \neq z_{-}$ in the domain $Λ$ : indeed $z - z_{-} = (1 - v^{- 1}) (z + 1) \neq 0$ (The Markov trace of an inverse Hecke generator).

[F3]

$π_{n + 1} (β σ_{n}) = ι_{n} (π_{n} (β)) T_{n}$ and $π_{n + 1} (β σ_{n}^{- 1}) = ι_{n} (π_{n} (β)) T_{n}^{- 1}$ , and $e (β σ_{n}) = e (β) + 1$ , $e (β σ_{n}^{- 1}) = e (β) - 1$ (The Hecke generators satisfy the Artin relations and are units, The exponent sum of a braid).

[F4]

In the coefficient ring of The HOMFLYPT coefficient ring, $u^{2} = z_{-} / z$ ; with $α = (u z)^{- 1}$ from The HOMFLYPT polynomial from the Hecke Markov trace, this gives $u α z = 1$ and $u^{- 1} α z_{-} = 1$ . The link-invariance theorem The Hecke trace construction is an oriented link invariant proves these are the two stabilization factors for its normalization.

Counterexample

technique · counterexample: compute the two stabilization factors of the raw trace, then read off the cancellation conditions of the normalization

1.1F1F2F3algebra

The stabilization factors of the raw trace. By [F3] and [F1] ${tr}_{n + 1} (π_{n + 1} (β σ_{n})) = {tr}_{n + 1} (π_{n} (β) T_{n}) = z {tr}_{n} (π_{n} (β))$ , and similarly ${tr}_{n + 1} (π_{n + 1} (β σ_{n}^{- 1})) = z_{-} {tr}_{n} (π_{n} (β))$ . By [F2] the two factors $z$ and $z_{-}$ are distinct, so whenever ${tr}_{n} (π_{n} (β)) \neq 0$ the raw trace family takes different values on the two stabilizations and is therefore not Markov invariant.

2.1F1F3F4step 1.1algebra

The normalized family. For the normalization $F (β) : = u^{e (β)} α^{n - 1} {tr}_{n} (π_{n} (β))$ , [F3] gives $F (β σ_{n}) = u α z F (β)$ and $F (β σ_{n}^{- 1}) = u^{- 1} α z_{-} F (β)$ . Both stabilization factors equal $1$ precisely when $u α z = 1$ and $u^{- 1} α z_{-} = 1$ ; given the first relation $α = (u z)^{- 1}$ , and substituting this into the second gives $z_{-} / (u^{2} z) = 1$ , i.e. $u^{2} = z_{-} / z$ , since $u$ and $z$ are units. For the trivial braid $1 \in B_{1}$ one has $F (1) = 1$ , so $F (σ_{1})$ and $F (σ_{1}^{- 1})$ are the two factors themselves. If either factor differs from $1$ , that stabilization changes the value of this witness. Hence a normalization failing either relation is not invariant under both stabilizations; the two stabilized values need not differ from each other.

3.1F1F2F3F4step 2.1algebra∎

The explicit witness. Take $u = 1$ and $α = z^{- 1}$ in a ring containing $Λ$ with $z$ inverted. Then the positive factor is $u α z = 1$ , while the negative factor is $u^{- 1} α z_{-} = z_{-} / z = \frac{z + 1 - v}{v z}$ , which is not $1$ : the equality $\frac{z + 1 - v}{v z} = 1$ would give $z + 1 - v = v z$ , i.e. $(1 - v) (z + 1) = 0$ , and both factors $1 - v$ and $z + 1$ are nonzero in the domain $Λ$ . Concretely, for $β = 1 \in B_{1}$ one has ${tr}_{1} (1) = 1$ and hence $F (σ_{1}) = 1$ while $F (σ_{1}^{- 1}) = z_{-} / z \neq 1$ in $B_{2}$ ; the two closures are the same unknot, so $F$ is not an invariant of the closure, which refutes the displayed statement. The corrected normalization is the one of [F4], whose two factors are both $1$ .

Remarks

The two stabilization factors are the classical Markov parameters of the Ocneanu trace: the positive stabilization multiplies the trace by $z$ and the negative one by $z_{-}$ .
The counterexample is the reason the coefficient ring of The HOMFLYPT coefficient ring imposes both relations $u α z = 1$ and $u^{2} = z_{-} / z$ ; dropping either one destroys the invariance proved in The Hecke trace construction is an oriented link invariant.

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Sources

Joan S. Birman and Tara E. Brendle, Braids: A Survey, section 4.3 Theorem 12 and the two stabilization factors (printed pp. 47-49) (standard reference, not scraped)
Theo Johnson-Freyd, MATH 448 Reshetikhin-Turaev invariants, lecture notes 15 January 2016, Sections 1-3 (the two Markov stabilization factors z and z_-) (standard reference, not scraped)