So, if you talk to the right people, they will tell you that the braiding of the category of representations of a quantum group are not unitary and that one can fix this by taking a different commutor (the cactus commutor). But I like the R-matrix better, and I think I know an inner product where it's unitary. I'm throwing this up here in hopes that it will look familiar to someone.

For those of you wondering why I would do such a thing, I should mention that I like categorification. Categorification is a funny thing, and one funny thing about it is that you can't really categorify just a vector space; you can only categorify a vector space with (maybe non-symmetric) inner product, and any map you categorify to an equivalence of categories must be unitary for this inner product. So if one is to categorify a linear operator, it had better be unitary with respect to something.

So, this inner product on V (x) W for any irreducibles V and W is the unique one (up to scalar) such that

$$(\Delta(F_i)v,w)=(v,\Delta(q^{-d}_iK_iE_i)w)$$ $$(\Delta(q^{-d}_iK_iE_i)v,w)= (v,\bar \Delta(F_i)w)$$

(Here, $\Delta(F)=F \otimes K^{-1} + 1\otimes F$ and $\bar \Delta(F)=F \otimes K + 1 \otimes F$).

Ring any bells with anyone?

**EDIT:** Anyone curious how this story actually works out should look at my papers Knot invariants and higher representation theory I and II. The upshot is that there is a bilinear form on a tensor product for which the R-matrix is an isometry.