# Are there skein relations for the colored Alexander/ADO invariants?

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
• 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. – Noah Snyder Dec 4 '19 at 0:30
• 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. – Calvin McPhail-Snyder Dec 4 '19 at 2:51
• 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/… – Noah Snyder Dec 4 '19 at 2:53
• 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. – Calvin McPhail-Snyder Dec 4 '19 at 2:56
• On more thought I think I just don’t understand the construction you have in mind well enough. – Noah Snyder Dec 4 '19 at 16:15