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The Hecke trace construction is an oriented link invariant

Statement

Assume the Axiom of Choice. Let $P$ be the Hecke-trace polynomial of The HOMFLYPT polynomial from the Hecke Markov trace with coefficient ring $R$ of The HOMFLYPT coefficient ring. Then:

(1) conjugation: $P (\hat{γ β γ^{- 1}}) = P (\hat{β})$ for all $n \geq 1$ and all $β, γ \in B_{n}$ ;

(2) positive stabilization: $P (\hat{β σ_{n}}) = P (\hat{β})$ for all $β \in B_{n}$ ;

(3) negative stabilization: $P (\hat{β σ_{n}^{- 1}}) = P (\hat{β})$ for all $β \in B_{n}$ ;

(4) consequently, if $β \in B_{n}$ and $β^{'} \in B_{m}$ are braids whose closures are ambient-isotopic oriented links, then $P (\hat{β}) = P (\hat{β^{'}})$ . The assignment $\hat{β} \mapsto P (\hat{β})$ is therefore a well-defined invariant of oriented links in $S^{3}$ , taking values in $R$ and normalized by $P (unknot) = 1$ and $P (\emptyset) = u z$ ; it is denoted $P (L)$ for an oriented link $L$ .

Facts & Assumptions

Given: AC (The Axiom of Choice), the coefficient ring $R$ , the Hecke tower $H (1) \subset H (2) \subset \dots$ over $Λ$ , the trace family ${tr}_{n}$ , the homomorphism $π_{n} : B_{n} \to H (n)^{\times}$ , the exponent sum $e$ , and the polynomial $P$ of The HOMFLYPT polynomial from the Hecke Markov trace. AC is used through [F7], whose closed-braid equivalence theorem assumes AC.

[F1]

$P (\hat{β}) = u^{e (β)} α^{n - 1} {tr}_{n} (π_{n} (β))$ for $β \in B_{n}$ , and this is well defined from the braid element (The HOMFLYPT polynomial from the Hecke Markov trace).

[F2]

The Ocneanu trace satisfies (M1)-(M4): ${tr}_{n} (1) = 1$ , ${tr}_{n + 1} \circ ι_{n} = {tr}_{n}$ , ${tr}_{n} (x y) = {tr}_{n} (y x)$ and ${tr}_{n + 1} (x T_{n}) = z {tr}_{n} (x)$ for $x \in H (n)$ ; also ${tr}_{n + 1} (x T_{n} y) = z {tr}_{n} (x y)$ for $x, y \in H (n)$ (The Ocneanu Markov trace exists and is unique).

[F3]

Each generator is a unit with $T_{i}^{- 1} = v^{- 1} T_{i} + (v^{- 1} - 1)$ and $T_{i} = v T_{i}^{- 1} + (v - 1)$ , and with $z_{-} : = v^{- 1} (z + 1 - v)$ one has ${tr}_{n + 1} (x T_{n}^{- 1}) = z_{-} {tr}_{n} (x)$ for $x \in H (n)$ (The Markov trace of an inverse Hecke generator).

[F4]

$e : B_{n} \to Z$ is a homomorphism with $e (σ_{i}) = 1$ , so $e (γ β γ^{- 1}) = e (β)$ , $e (β σ_{n}) = e (β) + 1$ and $e (β σ_{n}^{- 1}) = e (β) - 1$ (The exponent sum of a braid).

[F5]

$π_{n}$ is a group homomorphism with $π_{n} (σ_{i}) = T_{i}$ , so $π_{n + 1} (β σ_{n}) = ι_{n} (π_{n} (β)) T_{n}$ and $π_{n + 1} (β σ_{n}^{- 1}) = ι_{n} (π_{n} (β)) T_{n}^{- 1}$ under the inclusion $B_{n} \to B_{n + 1}$ and $ι_{n} : H (n) \to H (n + 1)$ , and $π_{n} (γ β γ^{- 1}) = π_{n} (γ) π_{n} (β) π_{n} (γ)^{- 1}$ (The Hecke generators satisfy the Artin relations and are units).

[F6]

$R$ is commutative, $u, z$ are units, and $u α z = 1$ and $u^{- 1} α z_{-} = 1$ with $α = (u z)^{- 1}$ , $z_{-} = v^{- 1} (z + 1 - v)$ (The HOMFLYPT coefficient ring).

[F7]

Markov's theorem together with the moves of Markov conjugation and stabilization moves: two braids have ambient-isotopic oriented closures if and only if they are related by conjugation, by positive and negative stabilizations and by the inverse destabilizations (Markov's theorem for braid closures); each move changes the closure by an ambient isotopy preserving orientation (Markov moves preserve the oriented closure up to isotopy, The closure of a geometric braid).

[F8]

Under AC, every nonempty oriented link is equivalent to the closure of a braid with $n \geq 1$ ; the empty link is the closure of the unique braid in $B_{0}$ (Alexander's theorem: every link is a closed braid). The geometric braid has an Artin-word representative by The Artin presentation surjects onto the geometric braid group, so the algebraic trace formula applies. A link in $S^{3}$ can first be moved off infinity as in the conventions of Oriented links in the three-sphere and ambient isotopy (also used by the closure supplier).

Proof

1.1F1F2F4F5F7

Conjugation. For $β, γ \in B_{n}$ the closures of $γ β γ^{- 1}$ and $β$ are ambient-isotopic by [F7]; to see the invariance algebraically, [F4] gives $e (γ β γ^{- 1}) = e (β)$ , and [F5] gives $π_{n} (γ β γ^{- 1}) = π_{n} (γ) π_{n} (β) π_{n} (γ)^{- 1}$ , so the trace property (M3) of [F2] gives ${tr}_{n} (π_{n} (γ β γ^{- 1})) = {tr}_{n} (π_{n} (β))$ ; the normalising factors are therefore equal and $P (\hat{γ β γ^{- 1}}) = P (\hat{β})$ .

1.2F1F2F4F5F6algebra

Positive stabilization. Let $β \in B_{n}$ ; then $π_{n + 1} (β σ_{n}) = ι_{n} (π_{n} (β)) T_{n}$ by [F5], so by (M4) of [F2] and (M2), ${tr}_{n + 1} (π_{n + 1} (β σ_{n})) = {tr}_{n + 1} (ι_{n} (π_{n} (β)) T_{n}) = z {tr}_{n} (π_{n} (β))$ . Including this trace multiplier, the stabilized value is $P (\hat{β σ_{n}}) = u^{e (β) + 1} α^{n} z {tr}_{n} (π_{n} (β)) = (u α z) P (\hat{β})$ by [F1] and [F4]. Since $u α z = 1$ by [F6], $P (\hat{β σ_{n}}) = P (\hat{β})$ .

1.3F1F3F4F5F6algebra

Negative stabilization. Similarly $π_{n + 1} (β σ_{n}^{- 1}) = ι_{n} (π_{n} (β)) T_{n}^{- 1}$ by [F5], and [F3] gives ${tr}_{n + 1} (ι_{n} (π_{n} (β)) T_{n}^{- 1}) = z_{-} {tr}_{n} (π_{n} (β))$ . Including the trace multiplier, $P (\hat{β σ_{n}^{- 1}}) = u^{e (β) - 1} α^{n} z_{-} {tr}_{n} (π_{n} (β)) = (u^{- 1} α z_{-}) P (\hat{β})$ by [F1] and [F4]. Since $u^{- 1} α z_{-} = 1$ by [F6], $P (\hat{β σ_{n}^{- 1}}) = P (\hat{β})$ .

2.1F1F2F7F8step 1.1step 1.2step 1.3∎

Invariance on link types. By [F7] two braids with ambient-isotopic oriented closures are connected by a finite chain of braid relations, conjugations, stabilizations and destabilizations; braid relations do not change the element of $B_{n}$ , hence do not change $P$ by [F1]; conjugation is step 1.1 and the two stabilizations are steps 1.2 and 1.3, while a destabilization is the reverse of one of these equalities; each move preserves the isotopy class of the closure by [F7]. Hence $P$ is constant along the chain. By [F8] every nonempty oriented link has a braid representative, so these values define an invariant on every nonempty link type. The empty link has only its zero-strand representative by the page-count property of [F7], and its separate value $P (\emptyset) = u z$ from [F1] is invariant. The unknot is the closure of $1 \in B_{1}$ , and $P = u^{0} α^{0} {tr}_{1} (1) = 1$ by (M1) of [F2].

Remarks

The two normalising constants are exactly the ones forced by the stabilizations: the positive move scales the trace by $z$ , the negative by $z_{-}$ , and the relations of the coefficient ring make the corresponding factors $u α z$ and $u^{- 1} α z_{-}$ equal to $1$ .
The invariant is the HOMFLYPT polynomial in the normalisation of The HOMFLYPT coefficient ring; its skein relation is The HOMFLYPT skein relation and its Jones specialization is The Temperley-Lieb quotient and the Jones specialization.

Depends on

Used by

An unnormalized Hecke trace is not Markov invariant Counterexample
The Temperley-Lieb quotient and the Jones specialization Definition
The Hecke trace skein calculation for a three-crossing braid Example
Khovanov-Rozansky homology categorifies the HOMFLYPT polynomial Theorem
The HOMFLYPT skein relation Theorem

Cited to discharge well-definedness by The HOMFLYPT polynomial from the Hecke Markov trace.

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Sources

Joan S. Birman and Tara E. Brendle, Braids: A Survey, section 4.3 Theorem 12 and its proof outline (printed pp. 47-49): Ocneanu's trace and Markov's theorem give the HOMFLYPT invariant (standard reference, not scraped)
Theo Johnson-Freyd, MATH 448 Reshetikhin-Turaev invariants, lecture notes 15 January 2016, Sections 1-3 (the Ocneanu trace and the Markov property) (standard reference, not scraped)