The Alexander polynomial/Conway potential can be computed as the quantum invariant associated to (a certain quotient of) $\mathcal U_q(\mathfrak{sl}_2)$ for $q = i$. More generally this works for $q = \xi = e^{\pi i / r}$ a primitive $2r$th root of unity. The $r = 2$ case gives the Alexander polynomials, while $r > 2$ are known as ADO invariants [1] or colored Alexander invariants (with $r$ the color.) Another reference is [2].

EDIT: Based on Noah's comments, I'll be a little more specific about the construction. At $q = \xi$ a $2r$th root of unity, the elements $K^r, E^r, F^r$ become central. For the ADO invariants we want representations of $$\mathcal A = \mathcal{U}_\xi(\mathfrak{sl}_2)/(K^r - t, E^r, F^r)$$ where $t$ is an indeterminate (which will become the variable of the Alexander polynomial.) The $r$th ADO invariant of a link is the Reshetikhin-Turaev invariant assigning each strand to an $r$-dimensional representation of $\mathcal A$. (Actually, the quantum dimensions vanish, so you need a slight generalization of the RT construction. Also, there are multiple non-isomorphic $r$-dimensional irreps, but they are very closely related.)

There are few references computing examples of these invariants. In particular, I'm curious to know if anyone's computed a skein relation for them.

  1. Y. Akustu, T. Deguchi, T. Ohtsuki, Invariants of colored links, J. Knot Theory Ramifications1 (1992) 161184.
  2. Cristina Ana-Maria Anghel, A topological model for the coloured Alexander invariants arXiv:1906.04056
  • 1
    $\begingroup$ You won’t get a simple skein theory beyond the first and maybe second. But if you allow a lot of vertices then there’s Kauffman-Lins. $\endgroup$ – Noah Snyder Dec 4 '19 at 0:30
  • 1
    $\begingroup$ By "Kauffman-Lins" do you mean something like decomposing tensor products into direct sums of irreps and using these to get a graphical calculus? You wind up with a lot of sums over labeled vertices, but the braiding should at least act on each type of vertex in a relatively simple way. $\endgroup$ – Calvin McPhail-Snyder Dec 4 '19 at 2:51
  • 1
    $\begingroup$ Yes, that’s what I meant. There’s a book that does it in full detail. You’ll have to be careful about the usual Temperley-Lieb vs quantum sl_2 sign issue. press.princeton.edu/books/paperback/9780691036403/… $\endgroup$ – Noah Snyder Dec 4 '19 at 2:53
  • $\begingroup$ Does the recoupling theory there work directly? I'd think the fact that $K^r$ doesn't act by $1$ would mess things up, but I suppose it's not that hard to modify their theory to account for this. It's been a while since I've looked at that book. $\endgroup$ – Calvin McPhail-Snyder Dec 4 '19 at 2:56
  • $\begingroup$ On more thought I think I just don’t understand the construction you have in mind well enough. $\endgroup$ – Noah Snyder Dec 4 '19 at 16:15

Your Answer

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

Browse other questions tagged or ask your own question.