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The Markov trace of an inverse Hecke generator

Statement

In the Hecke tower $H (1) \subset H (2) \subset \dots$ over $Λ = Z [v^{\pm 1}, z]$ : (a) each generator is invertible with $T_{i}^{- 1} = v^{- 1} T_{i} + (v^{- 1} - 1)$ and $T_{i} - v T_{i}^{- 1} = v - 1$ ; (b) for the Ocneanu trace of The Ocneanu Markov trace exists and is unique put $z_{-} : = v^{- 1} (z + 1 - v) \in Λ$ ; then for every $n \geq 1$ , every $x \in H (n)$ ${tr}_{n + 1} (x T_{n}^{- 1}) = z_{-} {tr}_{n} (x)$ ; (c) $z - z_{-} = (1 - v^{- 1}) (z + 1) \neq 0$ as an element of the domain $Λ$ , so the two formal generic stabilisation factors differ. Under specialization they can agree; for example $z = - 1$ gives $z_{-} = z$ .

Facts & Assumptions

Given: The Hecke tower over $Λ = Z [v^{\pm 1}, z]$ , an integer $n \geq 1$ , an element $x \in H (n)$ and the Ocneanu trace. No choice principle is used.

[F1]

$H (n)$ is the $Λ$ -algebra with generators $T_{1}, \dots, T_{n - 1}$ , quadratic relations $T_{i}^{2} = (v - 1) T_{i} + v$ , braid relations and distant commutations (The generic type-A Hecke algebra).

[F2]

The Ocneanu trace satisfies (M1)--(M4), and the two-sided form ${tr}_{n + 1} (u T_{n} v) = z {tr}_{n} (u v)$ for $u, v \in H (n)$ (The Ocneanu Markov trace exists and is unique).

[F3]

$Λ = Z [v^{\pm 1}, z]$ is a polynomial ring over the Laurent ring $Z [v^{\pm 1}]$ , hence a domain, and $v$ is a unit with inverse $v^{- 1}$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, The Laurent polynomial ring as the principal localisation of Z[t] at t, Units, powers and the domain property of the Laurent polynomial ring).

Proof

1.1F1F3algebra

Inverses. From $T_{i}^{2} = (v - 1) T_{i} + v$ of [F1] multiply by $v^{- 1}$ : $v^{- 1} T_{i}^{2} = (1 - v^{- 1}) T_{i} + 1$ , so $T_{i} (v^{- 1} T_{i} + (v^{- 1} - 1)) = 1$ ; the same computation with the order reversed gives $(v^{- 1} T_{i} + (v^{- 1} - 1)) T_{i} = 1$ , so $T_{i}$ is a unit with $T_{i}^{- 1} = v^{- 1} T_{i} + (v^{- 1} - 1)$ ; then $T_{i} - v T_{i}^{- 1} = T_{i} - T_{i} - (1 - v) = v - 1$ .

2.1F2step 1.1algebra

Traces of inverses. By (M2) and step 1.1, $T_{n}^{- 1} = v^{- 1} T_{n} + (v^{- 1} - 1)$ in $H (n + 1)$ , so ${tr}_{n + 1} (x T_{n}^{- 1}) = v^{- 1} {tr}_{n + 1} (x T_{n}) + (v^{- 1} - 1) {tr}_{n + 1} (x) = v^{- 1} z {tr}_{n} (x) + (v^{- 1} - 1) {tr}_{n} (x) = z_{-} {tr}_{n} (x)$ , where the middle equality uses (M4) in its form $x \in H (n)$ and the two-sided form [F2]; this proves the displayed negative-stabilization identity.

3.1F3algebra∎

Distinctness of the generic factors. Direct expansion in the domain $Λ$ gives $z - z_{-} = z - v^{- 1} (z + 1 - v) = z (1 - v^{- 1}) - (v^{- 1} - 1) = (1 - v^{- 1}) (z + 1)$ ; since $1 - v^{- 1} \neq 0$ and $z + 1 \neq 0$ in the domain $Λ$ of [F3], the product is nonzero. Hence the positive and negative stabilisations multiply the trace by distinct formal generic factors $z$ and $z_{-}$ . They may coincide after specialization, as at $z = - 1$ .

Depends on

Used by

An unnormalized Hecke trace is not Markov invariant Counterexample
The Hecke trace skein calculation for a three-crossing braid Example
The Hecke generators satisfy the Artin relations and are units Lemma
The Hecke trace construction is an oriented link invariant Theorem
The HOMFLYPT skein relation Theorem

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Sources

Joan S. Birman and Tara E. Brendle, Braids: A Survey, section 4.3 printed p. 48 (the inverse relation x_i^{-1} = t^{-1}x_i + (t^{-1}-1)) and printed p. 49 (the two stabilisation factors) (standard reference, not scraped)