Draft paper – On the uncoloured quantum link invariants from U_{q}(osp(1|2n)) and U_{-q}(so(2n+1))

Last week I finished a draft of a small paper the idea for which came to me at ICM in Madrid in August, possibly while talking to Nathan Geer after my talk. Pretty soon afterwards, I wrote down the beginnings of the idea at this post. It’s taken a fair bit of work to get it to this draft stage. I’m happy to e-mail a pdf of the draft paper to whomever requests it.

Essentially, the paper says that the quantum link invariants from the algebras mentioned in the title, obtained by colouring each component of the link with the respective irreducible (2n+1)-dimensional representations, are related by a change of variables. The thing is, it may not necessarily be possible to directly convert an invariant from one quantum algebra to the invariant from the other quantum algebra, as you have to actually look at the parent link invariant from the Birman-Wenzl-Murakami algebra in two indeterminates (r and s).

Let’s look at it more closely. Let L be a link and let F(L, r, s) be the Kauffman link invariant obtained from the Birman-Wenzl-Murakami algebra as outlined in Wenzl’s paper Quantum groups and subfactors of type B, C, and D. Here, r and s are the indeterminates in the BMW algebra.

Let q be non-zero and a generic complex number, ie, not a root of unity. F(L, r, s) is a sum of some sort: let us write it as F(L, r, s) = X + Y. Now, it transpires that
F(L, -q2n, q) looks as if it is a quantum link invariant obtained when you colour each component of your link with the irreducible (2n+1)-dimensional representation of
Uq(osp(1|2n)) and that F(L, q2n, –q) looks as if it is a corresponding quantum link invariant for U-q(so(2n + 1)).

Now I wrote F(L, r, s) = X + Y before without specifying what X and Y were. Let me say that X is that component of the sum which does not vanish when you make the substitution (r, s) = (-q2n, q) and Y does vanish when you make this substitution. It transpires that X similarly is nonvanishing when you make the substitution
(r, s) = (q2n, –q) and that Y also does vanish under this substitution. So, we have
F(L, -q2n, q) = X(-q2n, q) and F(L, q2n, –q) = X(q2n, –q) where the meanings are obvious. So, if you can write down F(L, -q2n, q) and make the substitutions
(-q2n, q) -> (q2n, –q), you get F(L, q2n, –q). The problem is, that you don’t a priori know whether some power of q in F(L, -q2n, q) is the image of r (-q2n) or of s (q) in
F(L, r, s), and so you don’t know what substitution to make.

Nevertheless, in principle, one can make such substitutions and obtain F(L, q2n, –q). Similarly, one can start from F(L, q2n, –q) and obtain F(L, -q2n, q) by making the correct substitutions. In some sense, then, these quantum link invariants are interchangeable. It seems to me that it could be quite difficult to actually make these substitutions, the easiest way would be to know what F(L, r, s) is, but this might be difficult to do!

You obtain the Kauffman invariant F(L, r, s) by using the trace function tr in BMW; the result I discuss here, assuming that it’s true (and I think that it is), uses tr at its core. The details are in Wenzl’s paper.