Explicit Coquasi-Triangular Quantised Coordinate Algebra of a Complex Semi-Simple Lie Group?

Question

Let $SL_q(N)$ be usual quantised coordinate algebra of the special linear group. As is well-known, this is co-quasi-triangular algebra with coquasi-triangular structure given by $$ R(u^i_j \otimes u^k_l) = q^{-\frac{1}{2}}.(q^{\delta_{ij}}\delta_{im}\delta_{jn} + (q-q^{-1})\theta (i-j)\delta_{in}\delta_{jm}), $$

Now consider the much more general case of $G_q$ the quantised coordinate algebra of a complex semi-simple Lie group. This is defined dually in terms of a Drinfield-Jimbo quantised enveloping Lie algebra $\mathfrak{g}$. As is also well-known, these algebras are also co-quasi-triangular. My question is does there exist a general formula for the co-quasi-triangular structure $R(u^i_j \otimes u^k_l)$ in terms of the Cartan data of $\mathfrak{g}$?

Answer 1

The coefficients of $R$ are essentially the coefficients of the braiding of the vector representation of $U_q(\mathfrak{g})$. So, more or less, you are asking for a general formula in terms of Cartan data for the braiding on the vector representation.

I've never seen a general formula for these, although the book by Klimyk and Schmudgen does give a general formula for the vector representations of the 4 infinite families of complex simple Lie algebras. The formulas for these are in Chapter 8. The stuff on coquasitriangularity of the corresponding quantized function algebras is in Chapter 10.

There is a pretty nice way to compute these things by hand if necessary. Remark 2.1 in the paper De Rham complex for quantized irreducible flag manifolds says:

Note that the braiding $\hat{R}$ is uniquely determined if one demands that [the braiding] is a $U_q(\mathfrak{g})$-module homomorphism satisfying \begin{equation} \label{braid} \hat{R}_{V,W} (v \otimes w) = q^{(wt(v), wt(w))} w \otimes v + \sum_i w_i \otimes v_i, \end{equation} where $wt(w_i) \prec wt(w)$ and $wt(v_i) \succ wt(v)$.

Here $V$ and $W$ are any finite-dimensional $U_q(\mathfrak{g})$-modules. The point is that this immediately determines the braiding on $v \otimes w$ whenever $v$ is a highest weight vector. These vectors generate $V \otimes W$ as a $U_q(\mathfrak{g})$-module.

Now you can go ahead and act on $v \otimes w$ with the $F_i$'s (lowering the weights of the vectors), and use the fact that $\hat{R}$ has to commute with the action of the $F_i$'s to compute $\hat{R} (v \otimes w)$ for all weight vectors $v$ and $w$.

This is not an explicit formula, but it does determine the braiding in terms of the pairings between the weights of the vector representation, which is not exactly the Cartan data, but close.