quantum groups - An inner product that makes the R-matrix unitary

I'm pretty late to the party here, and Ben, it seems that you already have a satisfactory answer to your question, but I thought for the sake of completeness I would just post this in case anybody stumbles across this question and is still curious.

The key observation is that the eigenvalues of the braiding induced by the $R$-matrix are of the form $pm q^{t}$, where $2t$ is a rational number in the image of the bilinear pairing of the weight lattice with itself. This follows from a result of Drinfeld which says that the square of the braiding acts as $Delta(C^{-1})C otimes C$, where $C$ is the quantum Casimir element. $C$ acts as the scalar $q^{(lambda, lambda + 2rho)}$ in the irrep $V_{lambda}$, and the statement about the eigenvalues of the braiding follows from this.

If you want the braiding to be unitary, and you want that statement to have some meaning in the category of representations of $U_q(mathfrak{g})$, then you need a $*$-structure on the quantum group. There are three cases to consider: $q in mathbb{R}, q in i mathbb{R}$ and $mathfrak{g}=mathfrak{sp}(2n,mathbb{C})$, and $|q|=1$.

The first two cases are ruled out because the eigenvalues of the braiding will not have modulus one, and hence the braiding cannot be unitary. This leaves only the case $|q|=1$.

If you want this inner product to be natural/compatible with the quantum group infrastructure, then it should probably be invariant under the action of $U_q(mathfrak{g})$ in the sense that $langle av, w rangle = langle v, a^*wrangle$ for $a in U_q(mathfrak{g})$ and $v,w in V otimes V$. Of course a $*$-structure on $U_q(mathfrak{g})$ needs to be specified in order to make sense of this. For $|q|=1$ the standard choice is $K_i^* = K_i, E_i^*=-E_i, F_i^*=-F_i$, although this can also be modified by a diagram automorphism of $mathfrak{g}$.

To find an invariant inner product on $V otimes V$, one way is to take the tensor product of an invariant inner product on $V$ with itself (if $V$ is an irrep then this is unique up to a scalar multiple if it exists). However, after a brief glance in Chari-Pressley, I'm not sure if the finite-dimensional irreps have an invariant inner product for the case $|q|=1$. I didn't do any further checking, but I would be interested to know if there are invariant inner products for irreps in that case.

Also, this doesn't rule out the possibility of inner products on $V otimes V$ that do not come from inner products on $V$.

Summary of answer: the $R$-matrix cannot be made unitary in any inner product for $q in mathbb{R}$ or $q in i mathbb{R}$. For $|q|=1$ the situation is unclear to me.