# 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 $\langle,\rangle:\Lambda\times\Lambda\to \mathbb{Z}$, and we define a braiding $\sigma_{\mu,\nu}:\mu \otimes \nu \to \nu\otimes\mu$ 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 $\mu\otimes\nu$ and $\nu\otimes\mu$ 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 $\Delta(E_i)=E_i\otimes 1 + 1\otimes E_i$ (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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This is a neat construction. I don't have a lot of intuition for presenting objects by generators and relations in a braided (nonsymmetric) category. You have to be a bit careful: global points, for example, do commute symmetrically. Anyway, my guess is that this is something close to a hands-on version of the Tannaka-Krein reconstruction for your braided category. –  Theo Johnson-Freyd Apr 8 '10 at 16:15
Hi David. Are you aware of the following paper of B. Enriquez: arxiv.org/abs/math/9904113 ? (I think Section 2 might be of some interest regarding your question) –  DamienC Apr 26 '11 at 23:03
Thanks Damien; I just got your message. I'll check it out. –  David Jordan May 11 '11 at 16:28

There is a detailed exposition of this in Majid's paper Double-bosonization of braided groups and the construction of $U_q(\mathfrak{g})$, Math Proc Cambridge Phil Soc 125(1). Especially appendix B where quantum group is obtained by a version of Tannaka-Krein duality for braided monoidal categories applied to the category of Yetter-Drinfeld modules of the positive part $U_q(\mathfrak{n})$ living in the category of comodules over a weakly quasitriangular group algebra $k\Lambda$ with respect to a braiding coming from the Cartan datum and a parameter $q$.

Here, the quantum groups are a special case of what Majid calls double bosonization which is the a briaded version of the Drinfeld double (the ordinary Drinfeld double arises by applying reconstruction to Yetter-Drinfeld modules in the symmetric monoidal category of $k$-vector spaces). It is not possible to define the Drinfeld double in a briaded monoidal category. This is why one needs to work with a fibre functor to vector spaces, and bosonizations (reflected by the commutator relations with the $K_i$ become necessary.

The nice thing about this construction is that the category of Yetter-Drinfeld modules is very special, it is the center of the category of $U_q(\mathfrak{n})$-modules. And $U_q(\mathfrak{n})$ is a Nichols algebra.

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I didn't realize this went all the way back to Majid. This is the perfect reference - thanks! –  David Jordan May 13 '13 at 0:46

Yes! It is indeed, to get the quantum group $\mathcal{U}_q(\mathfrak{g})$, you have to...

• start with a groupring $\mathcal{U}^0(\mathfrak{g})=\mathbb{k}[\mathbb{Z}^n]=\langle K_1,\ldots K_n\rangle_{Alg}$ representing the Cartan algebra $\mathcal{U}^0(\mathfrak{g})$, where $n$ is the rank of $\mathfrak{g}$

• consider the vectorspace $M=\langle E_1\ldots E_n\rangle_{Vect}$ for the simple roots $\alpha_1\ldots \alpha_n$ with

• an action of $\mathbb{k}[\Gamma]$, namely $K_i\otimes E_j\mapsto q^{(\alpha_i,\alpha_j)}E_j=:q_{ij}E_j$
• a graduation/coaction to $\mathbb{k}[\Gamma]$, namely $E_i\mapsto K_i\otimes E_i$
• therefore a braiding $E_i\otimes E_j\mapsto q_{ij}E_j\otimes E_i$
• form the tensor algebra $\mathcal{T}M$ modulo the Serre relations for the Cartan matrix a_{ij} $$ad_{E_i}^{1-a_{ij}}E_j=0$$ with the braided adjoint action resp. commutator $$ad_{E_i}(E_j)=[E_i,E_j]:=EiE_j-q_{ij}E_jE_i$$ This is the Borel part $\mathcal{U}_q(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^+)$, in this stage only a so-called braided Hopf algebra over $\mathbb{k}[\Gamma]$ with $\Delta(E_i)=1\otimes E_i + E_i\otimes 1$
• glue Borel part and Cartan algebra to a Radford biproduct $$\mathcal{U}^{\geq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})\ltimes\mathcal{U}^+(\mathfrak{g})$$ Especially the action and coaction of $\mathbb{k}[\Gamma]$ on $M$ automatically yield the relations $$K_iE_j=q_{ij}E_jK_j$$ $$\Delta(E_i)=K_i\otimes E_i\otimes 1$$
• Form the generalized Drinfel'd double, which means...
• Do the dual construction with $M^*=\langle F_1,\ldots F_n\rangle$ with action $K_i\otimes F_j\mapsto q^{-(\alpha_i,\alpha_j)}F_j$ to yield another Borel part $\mathcal{U}^-(\mathfrak{g})=\mathcal{U}(\mathfrak{n}^-)$ and another Radford biproduct $$\mathcal{U}^{\leq}(\mathfrak{g})=\mathcal{U}^0(\mathfrak{g})'\ltimes\mathcal{U}^-(\mathfrak{g})$$
• Quotient out an identification of both Cartan algebras $$U_q(\mathfrak{g})=(U_q^{\geq}(\mathfrak{g})\otimes U_q^{\leq}(\mathfrak{g}))^{\sigma}/(U^0_q(\mathfrak{g})= U^0_q(\mathfrak{g})')$$ the last step is called linking can nowadays (by A. Masuoka) be done via a Doi twist with a 2-cocycle, which causes the additional relations $$E_i F_i-F_i E_i=\frac{K_i-K_i^{-1}}{q_{ii}-q_{ii}^{-1}}$$

LITERATURE: e.g. Heckenberger: Nichols Algebras (Lecture Notes), 2008 (http://www.mi.uni-koeln.de/~iheckenb/na.pdf) page 35ff.

PS: This works equivalently for the truncated quantum groups. Here, the role of the Borel part $\mathcal{T}M/Serre$ is taken by a quotient Nichols algebra $\mathfrak{B}(M)$. There also appear some exotic example associated to Dynkin diagrams, that are impossible for semisimple Lie algebras (e.g. a certain triangle). See also the Wikipedia page "Nichols algebras" for links to more papers.

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${}^{free}U_q^\pm \twoheadrightarrow {}^{DK}U_q^\pm \twoheadrightarrow u_q^\pm \hookrightarrow {}^LU_q^\pm \hookrightarrow {}^{cofree}U_q^\pm$
${}^{free}U_q^+$ is the free associative algebra generated by the simple root operators $E_i$, which have the appropriate grading. It has the coproduct you described. Taking a quotient by Serre relations yields the first arrow. The operation of taking dual Hopf algebra with opposite comultiplication flips the direction of the diagram and changes the sign. Taking the double yields hybrid algebras, e.g., half Lusztig and half DeConcini-Kac. I don't understand why both $Vect^\Lambda_q$ and $Vect^\Lambda_{q^{-1}}$ play a role here.
The relative version of Drinfeld's double construction can be written on the category level as the E[2]-centralizer of the monoidal functor $Vect^\Lambda \to Rep ({}^{\ast}U_q^\pm)$, where $\ast$ indicates one of the choices above. I guess you can combine definition 2.5.1 with example 2.5.15 in Lurie's DAG VI: E[k]-algebras to get a description of this construction. Generically this seems to yield the category of representations of Lusztig's U-dot algebra instead of the original $U_q$.