## Recovering the Alexander Polynomial from Ocneanu’s HOMFLYPT

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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?

-

## 1 Answer

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.

-

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 – Alexander Moll Oct 3 2011 at 16:47
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 $q=\pm 1$.  – Bruce Westbury Oct 3 2011 at 17:32
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 – Alexander Moll Oct 4 2011 at 3:15
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 – Alexander Moll Oct 4 2011 at 3:42
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 43, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9038128852844238, "perplexity_flag": "head"}