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

$\begingroup$ 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 nonLie group ones. $\endgroup$ – Hauke Reddmann Apr 7 '12 at 14:07

$\begingroup$ 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. $\endgroup$ – Bruce Westbury Apr 7 '12 at 15:23