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Let $H_q(n)$ denote the Hecke algebra associated to the symmetric group $S(n)$: this is the $\mathbb{Z}[q^{\pm 1/2}]$ algebra generated by $T_1, \ldots, T_{n-1}$ satisfying the braid relations along with $T_i^2=(q-1)T_i + q$. Ocneanu's trace $\tau_z (T_{\omega_{\mu}}) = z^{l(\omega_{\mu})}$ defined on fundamental elements is the unique normalized trace on our Hecke algebra that can be jiggled to yield invariants of oriented links: this gives the two-parameter HOMFLYPT polynomials $P_L(q,z)$. These polynomials satisfy the skein relation

$$ \Big ( \frac{z}{z-q+1} \Big )^{1/2} P_{L_+} (q,z) - \Big ( \frac{z}{z-q+1} \Big )^{-1/2} P_{L_-}(q,z) = (q^{1/2} - q^{-1/2}) P_{L_0} (q,z).$$

which motivates the common change of variables $x = \sqrt{\frac{z}{z-q+1}}$, $y = q^{1/2} - q^{-1/2}$. Note that the target ring of this trace has to be at least $\mathbb{Z}[q^{\pm 1/2}, z^{\pm 1/2}, (z-q+1)^{ \pm 1/2}]$ if we want to write down the associated invariant $P_L(q,z)$.

Many people take the HOMFLYPT polynomials to be those obtained after the specialization $z=q^{N}/[N]$, which seems to be equivalent to $x=\sqrt{\frac{z}{z-q+1}} = q^{N/2}$. Setting $N=2$ recovers the Jones polynomial, and setting $N=0$ is supposed to recover the Alexander polynomial.

  • How am I supposed to correctly obtain the Alexander polynomial in terms of the original $q$ and $z$?

It seems that $x=q^{0/2}=1$ gives the correct specialization. Still, doesn't $\sqrt{\frac{z}{z-q+1}}=1$ force $q=1$, leaving $z$ free? How does the right-hand side of the skein relation survive then?

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This confusion arises because you have defined the HOMFLY polynomial over the ring $\mathbb{Z}(z,1/z,q,1/q)$. The correct ring is the ring generated by $z$,$1/z$, $q$, $1/q$, $\delta$ subject to the relation $z-1/z=\delta(q-1/q)$. This avoids denominators and can be specialised. This is a blow-up of the Laurent polynomial ring.

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  • $\begingroup$ I've edited the statement of the problem above. In this deduction, I'm not sure how to use this $\delta$ of yours. Thanks for your help! - Alex $\endgroup$
    – Alexander Moll
    Commented Oct 3, 2011 at 16:47
  • $\begingroup$ Your question is hard to follow. If you want to see the conventions I use see arxiv.org/abs/1007.2579. In particular $\delta$ is the value of a single closed loop. I am worried about your $[0]=1/1-q$ as $[0]=0$. I would normally put $z=q^n$. Putting $n=0$ then means $z-1/z=0$ so $\delta=0$ or $q=\pm 1$. $\endgroup$
    – Bruce Westbury
    Commented Oct 3, 2011 at 17:32
  • $\begingroup$ I've used the $q$ and $z$ from D. Goldschmidt's "Group Characters, Symmetric Functions, and Hecke Algebras" above. I'm hesitant to accept that $\delta$ is the value of a single closed loop, since HOMFLYPT returns 1 on the unknot katlas.org/wiki/Image:0_1.gif $\endgroup$
    – Alexander Moll
    Commented Oct 4, 2011 at 3:15
  • $\begingroup$ You're right that $[0] = \frac{1}{1-q}$ can't be the right (unbalanced) $q$-analog of $0$ - I've edited the question accordingly $\endgroup$
    – Alexander Moll
    Commented Oct 4, 2011 at 3:42

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