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The HOMFLYPT skein relation

Statement

Assume the Axiom of Choice. Let P be the oriented link invariant of The Hecke trace construction is an oriented link invariant, with coefficient ring R and variables l=us, m=s−s−1, v=s2 as in The HOMFLYPT coefficient ring. Let x,y be Artin words in the generators of Bn and 1≤i≤n−1, and let L+,L−,L0 be the oriented links represented by the closures of xσiy, xσi−1y and xy; these three braid words differ only at one crossing between the strands i,i+1, so the three link diagrams form a skein triple. Then l−1P(L+)−l P(L−)=m P(L0), and P(unknot)=1. Equivalently, in the normalisation of the trace tower, u−1P+−vu P−=(v−1)P0.

Facts & Assumptions

Given: AC (The Axiom of Choice), the invariant P of The Hecke trace construction is an oriented link invariant, a braid word xσiy in Bn and the corresponding skein triple (L+,L−,L0). The link-invariance assertion for P uses AC as recorded in its supplier; the skein computation itself is algebraic.

[F1]

P(L+)=ue(x)+e(y)+1αn−1tr⁡n(πn(xσiy)), P(L−)=ue(x)+e(y)−1αn−1tr⁡n(πn(xσi−1y)) and P(L0)=ue(x)+e(y)αn−1tr⁡n(πn(xy)), since all three words lie in Bn and α=(uz)−1 (The HOMFLYPT polynomial from the Hecke Markov trace, The exponent sum of a braid).

[F2]

Ti=vTi−1+(v−1) for every generator, and πn is multiplicative on words, so in H(n) πn(xσiy)=v πn(xσi−1y)+(v−1)πn(xy) (The Markov trace of an inverse Hecke generator, The Hecke generators satisfy the Artin relations and are units).

[F3]

tr⁡n is Λ-linear, so applying it to the identity of [F2] gives the corresponding relation between the three trace values (The Ocneanu Markov trace exists and is unique).

[F4]

l=us, m=s−s−1, s2=v, and (v−1)s−1=s−s−1=m in R (The HOMFLYPT coefficient ring).

[F5]

P(unknot)=1 (The Hecke trace construction is an oriented link invariant), and the closures of the three words represent the oriented links L+,L−,L0 of the statement (The closure of a geometric braid).

Proof

1.1F1F2F3

The trace identity. By [F2] and the Λ-linearity of the trace [F3], A:=vB+(v−1)C where A=tr⁡n(πn(xσiy)), B=tr⁡n(πn(xσi−1y)) and C=tr⁡n(πn(xy)) are the three trace values of [F1].

2.1F1step 1.1algebra

Normalisation. Multiply the identity of step 1.1 by ue(x)+e(y)αn−1 and use [F1]: u−1P(L+)=vu P(L−)+(v−1)P(L0), i.e. u−1P+−vuP−=(v−1)P0, the second displayed relation.

3.1F4F5step 2.1algebra∎

The (l,m) form. Divide the identity of step 2.1 by s and use [F4]: u−1s−1P+−vus−1P−=(v−1)s−1P0; here u−1s−1=(us)−1=l−1, vus−1=s2us−1=us=l because s2s−1=s, and (v−1)s−1=s−s−1=m. Hence l−1P+−lP−=mP0, which is the first displayed relation; the normalization P(unknot)=1 is [F5].

Remarks

  • The proof uses only the quadratic Hecke relation Ti=vTi−1+(v−1) and the linearity of the trace; no reduced or unreduced Burau matrix enters the skein relation, which is why the invariant is defined for all braids.
  • The variable dictionary is l=us, m=s−s−1, u2=z−/z; substituting z=z0=−1/(v+1), u=s turns the relation into the Jones skein relation of The Temperley-Lieb quotient and the Jones specialization.

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