1
$\begingroup$

OK, the heading was a bit tersely formulated...
If you have a quantum group and an irrep, you theoretically know the R matrix (mathematicians are a notoriously idle lot, they give the general formula and thus the problem is solved :-) - and the characteristic equation of the R matrix is a valid skein equation.

Now to the reverse process. Question 1: Are there really skein equations that can't be modeled with a R matrix? E.g. I heard that already the Kauffman 2-variable polynomial is unattainable this way, but I only heard it and never saw an actual proof. (I have no idea how the situation is for directed knots.)
Question 2. OK, assume we have a working R matrix, is there always a quantum group associated with that? ("Baxterization"??) Again, I think the answer is "no" for unoriented knots. (If no quantum group-based knot polynomial can distinguish mutants - also from the "so I heard" variety - the question is solved, since I have a R matrix doing this.)

$\endgroup$
3
  • 2
    $\begingroup$ Hauke -- the invariant that comes from the defining representation of $sl_4$ does distinguish Conway from Kinoshita-Teresaka. See liv.ac.uk/~su14/papers/mutantJKTR.pdf $\endgroup$
    – algori
    Commented Sep 21, 2011 at 15:42
  • $\begingroup$ Hauke -- I realize what I wrote was misleading, sorry. The invariant does come from the defining representation, but it has to be applied applied to the "triple copies" of the knots. If one wants an invariant of the knots themselves, one has to consider a different representation, namely the one with the Young diagram in the form of a triangle with three squares. See p. 3 of H. Morton's paper. $\endgroup$
    – algori
    Commented Sep 21, 2011 at 20:54
  • $\begingroup$ THX, the paper is interesting. (Eh, I'm not so good with all the equivalent notations - SL4, triangle irrep is also A_?, irrep (?,?,...,?) - could you enlighten me?) P.S. Of course I have my own opinion on the mutant matter, which is that an irrep with a pattern RxR->R1+...+Rn suffices if n>...eh, 5? But this is black magic, not math :-) $\endgroup$ Commented Sep 22, 2011 at 10:29

1 Answer 1

3
$\begingroup$

The Kauffman 2-variable knot polynomial probably can't be obtained from a quantum group if by this you mean the usual q-deformed universal enveloping algebras. If your two variables are $(r,q)$ and $r=\pm q^n$ then it can be obtained from quantum groups of type B,C or D (depending on $n$), see Wenzl's paper Comm. Math. Phys. 133 (1990) 383-432.

Generally a nice paper in which this kind of knot-invariant-to-quantum-group is discussed in Turaev and Wenzl's paper Math. Annalen 309 (1997), 411-461.

If you just start with an R-matrix and you want to define knot-invariants you need what Turaev calls an "enhanced Yang-Baxter operator." Given an R-matrix (enhanced or otherwise) it is possible to construct a braided Hopf algebra via the FRT (Faddeev-Reshetikhin-Takhtajan) construction but this will be unpleasant in general (infinite dimensional for sure).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .