8
$\begingroup$

On page 9 of Kauffman's Formal Knot theory, Kauffman claims

The Alexander-Conway Polynomial is a true refinement of the Alexander Polynomial. Because it is defined absolutely (rather than up to sign and powers of variables) it is capable of distinguishing many links from their mirror images - a capability not available to the Alexander polynomial

Does anybody know of any examples of this happening? Or when two knots have the same Alexander polynomial but different Alexander-Conway polynomials?

| cite | improve this question | | | | |
$\endgroup$
  • 6
    $\begingroup$ I don't think it's possible for knots. But I suppose it's possible for 2-component links, but presumably for fairly simple reasons. For knots you can take the Alexander polynomial (ideal) and canonically make all the choices Kauffman is talking about, to essentially turn it into the Alexander-Conway polynomial. But for two or more component links it's less clear how to make the choices canonically. I haven't thought about this very carefully but that's my initial understanding of what Kaufmann is getting at. $\endgroup$ – Ryan Budney Oct 19 '18 at 22:11
  • 1
    $\begingroup$ I want to point out that although the Alexander-Conway polynomial cannot distinguish a knot from its mirror image, it can detect certain kinds of chirality. In fact, I made a conjecture many years ago that I wish someone would solve. Any amphicheiral knot has conway polynomial that splits as $f(z)f(-z)$ with coefficients taken modulo $4$. See arxiv.org/pdf/1608.04453.pdf $\endgroup$ – Jim Conant Dec 2 '18 at 1:00
7
$\begingroup$

Let $\overline{L}$ denote the mirror image of a link $L$. The (reduced) Alexander-Conway polynomial $D_L(t)$ of an $n$-component link $L$ satisfies $D_{\overline{L}}(t)=(-1)^{n+1}D_L(t)$. More generally, the multivariable potential function $\nabla_L(t_1,\ldots,t_n)$ also satisfies $\nabla_{\overline{L}}(t_1,\ldots,t_n)=(-1)^{n+1}\nabla_L(t_1,\ldots,t_n)$. So you can use these sign-refined versions to distinguish links from their mirror images for links with an even number of components.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.