A property of quantum group R matrices?

Question

Assume Q is a quantum Lie group which allows a R matrix (with the usual quantum Yang-Baxter equation). Take the Jordan normal form J of R. Does a Q exist with offdiagonal J elements (i.e. R has defective eigenvalues)?

score 2 · Answer 1 · answered Apr 06 '12 at 15:36

I think a much stronger statement is true although I am missing some of the details.

The starting data is: $K$ is a commutative ring, $M$ is a $K$ -linear braided monoidal category, and $V$ is an object of $M$ .

Then for $n>0$ we have an algebra homomorphism from the group algebra of the $n$ -string braid group $KB_n$ to $End(\otimes^n V)$ . Then we can ask the general question: is the image of this homomorphism a semisimple algebra? There is a weaker version. For every object $W\in M$ , $Hom(W,\otimes^n V)$ is a representation of $B_n$ ; then the question is: are all these representations completely reducible?

Now let $U$ be a Drinfeld-Jimbo quantised enveloping algebra over the field $K=\mathbb{Q}(q)$ and let $M$ be the category of finite dimensional representations. Then I propose that, for any $V$ all the above questions are answered affirmatively.

The proof is an application of the following:

Define a *-algebra to be an algebra with an antiinvolution. Then I believe that a sub *-algebra of a semisimple *-algebra is semisimple.

Then to apply this to the problem in hand it remains to show that $End(\otimes^n V)$ is a semisimple *-algebra and that the image is a sub *-algebra.

Have I overlooked something?

Only to formulate it in terms that stupid me is familiar with :-) But in any case, close enough: then the obvious converse is, a) there are R matrices (even invertible one) which are Jordan nondiagonal, b) such matrices can't be the R matrix of a quantum Lie group, in fact c) your answer to my previous question (application of Schurs Lemma) throws out most of them anyway, i.e. d) a R matrix list obtained by merely solving the CYBE will contain non-Lie group ones. — Hauke Reddmann, Apr 07 '12 at 14:07
I am not sure what you are asking. In any case, you seem to be trying to start a discusson and MO is (intentionally) not suited for discussions. — Bruce Westbury, Apr 07 '12 at 15:23

score 2 · Accepted Answer · answered Apr 06 '12 at 18:51

What is meant by a Quantum Lie Group here?

If we are talking about the quantized enveloping algebras of complex semisimple lie algebras $U_q(\mathfrak{g})$ then it is a fact that the action of the universal $R$ -matrix on the tensor product of finite dimensional representations is semisimple i.e. its Jordan form is diagonal

A property of quantum group R matrices?

2 Answers2