braided tensor categories - Quantum group as (relative) Drinfeld double?

The most elementary construction I know of quantum groups associated to a finite dimensional simple Hopf algebra is to construct an algebra with generators $E_i$ and $F_i$ corresponding to the simple positive roots, and invertible $K_j$'s generating a copy of the weight lattice. Then one has a flurry of relations between them, and a coproduct defined on the generators by explicit formulas. These are not mortally complicated, but are still rather involved. Then come explicit checks of coassociativity, and compatibility between multiplication and comultiplication. Finally, one has the $R$-matrix which is an infinite sum with rather non-obvious normalizations. Enter more computations to verify $R$-matrix axioms.

I recall learning about a nice way to construct the quantum group, which in addition to requiring less formulas has the advantage of making it clear conceptually why it's braided.

I'm hoping someone can either point me to a reference for the complete picture, or perhaps fill in some of the details, since I only remember the rough outline. That, precisely, is my question.

I include the remarks below in hopes it will jog someone's memory.

You start with the tensor category $Vect_Lambda$ of $Lambda$-graded vector spaces, where $Lambda$ is the weight lattice. We have a pairing $langle,rangle:LambdatimesLambdato mathbb{Z}$, and we define a braiding $sigma_{mu,nu}:mu otimes nu to nuotimesmu$ to be $q^{langle mu,nu rangle}$. Here $q$ is either a complex number or a formal variable. We may need to pick some roots of $q$ if we regard it as a number; I don't remember (and am not too worried about that detail). Also, here we denoted by $mu$ and $nu$ the one dimensional vector space supported at $mu$ and $nu$ respectively, and we used the fact that both $muotimesnu$ and $nuotimesmu$ are as objects just $mu+nu$.

Okay, so now we're supposed to build an algebra in this category, generated by the $E_i$'s, which generators we regard as living in their respective gradings, corresponding to the simple roots. Here's where things start to get fuzzy. Do we take only the simples as I said, or do we take all the $E_alpha$'s, for all roots $alpha$? Also, what algebra do we build with the $E_i$'s? Of course it should be the positive nilpotent part of the quantum group, but since we build it as an algebra in this category, there may be a nicer interpretation of the relations? Anyways, let's call the algebra we are supposed to build here $U_q(mathfrak{n}^+)$. I definitely remember that it's now a bi-algebra in $Vect_Lambda$, and the coproduct is just $Delta(E_i)=E_iotimes 1 + 1otimes E_i$ (the pesky $K$ that appears there usually has been tucked into the braiding data). Now we take $U_q(mathfrak{n}^-)$ to be generated by $F_i$'s in negative degree, and we construct a pairing between $U_q(mathfrak{n}^+)$ and $U_q(mathfrak{n}^-)$. The pairing is degenerate, and along the lines of Lusztig's textbook, one finds that the kernel of the pairing is the q-Serre relations in each set of variables $E_i$ and $F_i$.

Finally, once we quotient out the kernel, we take a relative version of Drinfeld's double construction (the details here I also can't remember, but would very much like to), and we get a quasi-triangular Hopf algebra in $Vect_Lambda$. As an object in $Vect_Lambda$ it's just an algebra generated by the $E_i$'s and $F_i$'s, so no torus. But since we're working in this relative version, we can forget down to vector spaces, and along the way, we get back the torus action, because that was tucked into the data of $Vect_Lambda$ all along.

So, the construction (a) gives neater formulas for the products, coproducts, and relations (including the $q$-Serre relations), and (b) makes it clear why there's a braiding on $U_q(mathfrak{g})$ by building it as the double.

The only problem is that I learned it at a seminar where to my knowledge complete notes were never produced, and while I remember the gist, I don't remember complete details. Any help?

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