Question

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 (x) K^-1 + 1(x)F and \bar Delta(F)=F (x) K + 1 (x) F).
Ring any bells with anyone?

Answer 1

You're being thrown off by a poor choice of names that Drinfel'd made. (Hey, he had to name a lot of stuff, so I can't really blame him for getting one wrong.) By "unitarization" he means "modifying it so that it's square is 1" which has nothing whatsoever to do with being unitary. The usual R-matrix should be unitary wrt to the usual *-structures on quantum groups if q is a number of the right form (either size 1 or real depending on the *-structure).