Recovering the Alexander Polynomial from Ocneanu's HOMFLYPT - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:40:58Z http://mathoverflow.net/feeds/question/77014 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77014/recovering-the-alexander-polynomial-from-ocneanus-homflypt Recovering the Alexander Polynomial from Ocneanu's HOMFLYPT Alexander Moll 2011-10-03T06:14:50Z 2012-05-23T01:09:57Z <p>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</p> <p>$$\Big ( \frac{z}{z-q+1} \Big )^{1/2} P_{L_+} (q,z) - \Big ( \frac{z}{z-q+1} \Big )^{-1/2} P_{L_-}(q,z) = (q^{1/2} - q^{-1/2}) P_{L_0} (q,z).$$</p> <p>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 $P_L(q,z)$.</p> <p>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. </p> <ul> <li>How am I supposed to correctly obtain the Alexander polynomial in terms of the original $q$ and $z$?</li> </ul> <p>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?</p> http://mathoverflow.net/questions/77014/recovering-the-alexander-polynomial-from-ocneanus-homflypt/77023#77023 Answer by Bruce Westbury for Recovering the Alexander Polynomial from Ocneanu's HOMFLYPT Bruce Westbury 2011-10-03T08:50:16Z 2011-10-03T08:50:16Z <p>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.</p>