# quantum groups… not via presentations

Given a semisimple Lie algebra $\mathfrak g$ with Cartan matrix $a_{ij}$, the quantum group $U_q(\mathfrak g)$ is usually defined as the $\mathbb Q(q)$-algebra with generators $K_i$, $E_i$, $F_i$ (the $K_i$ are invertible and commute with each other) and relations $$\begin{split} K_iE_j &K_i^{-1}=q^{\langle\alpha_i,\alpha_j\rangle}E_j\qquad\qquad K_iF_j K_i^{-1}=q^{-\langle\alpha_i,\alpha_j\rangle}F_j\, \\\ &[E_i,F_j]=\delta_{ij}\frac{K_i-K_i^{-1}} {\quad q^{\langle\alpha_i,\alpha_i\rangle/2} -q^{-\langle\alpha_i,\alpha_i\rangle/2}\quad}\, \end{split}$$ along with two more complicated relations that I won't reproduce here.
One then defines the comultiplication, counit, and antipode by some more formulas.

Is there a way of defining $U_q(\mathfrak g)$ that doesn't involve writing down all those formulas?

In other words, is there a procedure that takes $\mathfrak g$ as input, produces $U_q(\mathfrak g)$ as output, and doesn't involve the choice of a Cartan subalgebra of $\mathfrak g$?

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not sure you will be satisfied, but there is FRT (Faddeev-Reshetikhin-Takhtadjan) approach, first define Fun_q(GL) via RTT=TTR relation, then extract dual Hopf algebra (i.e. Uq(GL)) analougsly to classical group as distributions in origin... –  Alexander Chervov Aug 22 '12 at 10:30
Would you consider a path through quantum geometric Satake and Tannakian reconstruction an admissible procedure? –  S. Carnahan Aug 22 '12 at 11:43
@Scott: If I use geometric Satake, Do I get U_q(g) out of that procedure, or do I just get its category of representations? @Alexander: Producing the dual Hopf algebra is just as good as producing U_q(g) itself. You're mentioning Fun_q(GL)... is the FRT approach something that only works for GL_n? –  André Henriques Aug 22 '12 at 12:40
@Andre As far as I remember in FRT paper they can work with ALL Lie algebras (well, may be classical ones). They do NOT choose Cartan subalgebra but RTT=TTR is explicit matrix relation so we choose basis in Lie algebra, but may be it does not much depend on basis choice... RTT=TTR is corollary of "universal" YangbBAxter R_{12}R_{23}R_{12} =R_{23}R_{12}R_{23}. –  Alexander Chervov Aug 22 '12 at 13:28

$\newcommand\g{\mathfrak{g}}$The answer to your question "is there a procedure that takes $\g$ as input, produces $U_q(\g)$ as output, and doesn't involve the choice of a Cartan subalgebra of $\g$?" is No. Not if you want it "canonical" in any sense. (Of course, if I wanted to cheat my way to a "yes," I could make choices that are equivalent to choosing a Cartan but not stated as such — I would construct for you a certain canonical homogeneous space, and then ask you to pick a point in it, ....)

The problem is that the automorphism group of $\g$ does not lift to $U_q(\g)$. Recall that the inner automorphism group is precisely the simplest group $G$ integrating $\g$ (take any connected group integrating $\g$ and mod out by its center). On the other hand, the inner automorphism "group" of $U_q(\g)$ is $U_q(\g)$ (or "$\operatorname{spec}(\operatorname{Fun}_q(G))$", depending on your point of view) itself. This certainly deforms the automorphisms. But there is not a procedure like you ask for, because you have to break some symmetry.

Here's a way to say this correctly: In a precise sense $U_q(\g)$ degenerates to $U(\g)$ as $q\to 1$, and this is part of the structure that I take you to mean when you write "$U_q(\g)$". In this degeneration, you can also study $\frac{\partial}{\partial q}(\dots)$ at $q=1$. In particular, looking at $\frac{\partial}{\partial q}\bigr|_{q=1}$ of the comultiplication on $U_q(\g)$ recovers the Lie cobracket on $\g$. But the Lie cobracket knows the Cartan subalgebra: it is precisely the kernel of the Lie cobracket. So $\operatorname{Aut}(\g)$ cannot lift to $U_q(\g)$ compatibly with all of this structure.

What does exist without choosing a Cartan subalgebra is the braided category of representations of $U_q(\g)$, although I would have to think a moment to recall how to write it down explicitly. (In the asymptotic limit $q = e^\hbar$ with formal $\hbar$, I do know how to write down $\operatorname{Rep}(U_{e^\hbar}\g)$ explicitly for any choice of Drinfel'd associator.) In particular, as a braided monoidal category, this category does have an action by $\operatorname{Aut}(\g)$.

But the category is strictly less data than the Hopf algebra. Namely, "Tannakian reconstruction" is the statement that $U_q(\g)$ is the Hopf algebra of endomorphisms of a certain braided coalgebra in $\operatorname{Rep}(U_{q}\g)$ (or it is the Hopf dual of the Hopf algebra of co-endomorphisms of a certain braided algebra in the category of Ind-objects in $\operatorname{Rep}(U_{q}\g)$, if for you representations are finite-dimensional), and you cannot choose this coalgebra canonically. This coalgebra is unique up to isomorphism, but certainly not up to unique isomorphisms (or else $U_q(\g)$ would be trivial). The failure of this coalgebra to exists up to canonical isomorphism is essentially the same problem as above.

In a precise way, this is failure of there to exist a canonical isomorphisms between different choices of the coalgebra is analogous of the failure of the "fundamental group" of a topological space to be an honest group. Recall that a pointed topological space has a fundamental group, which is a group determined up to canonical isomorphism. For comparison, a non-pointed but path-connected topological space has a group assigned to each point, and these are non-canonically isomorphic. Thus a non-pointed path-connected topological space has a "fundamental group up to conjugation," also known as a connected groupoid.

What this should all mean, although I'm not going to try to write down the correct definition, is that there does exist canonically associated to $\g$ a "Hopf algebroid" which is noncanonically equivalent as a Hopf algebroid to any choice of $U_q(\g)$. Or, at least, I'm confident of everything in my answer in the $\hbar\to 0$ asymptotics, and have thought less about the finite-$q$ case, and so I'm generalizing intuition from that setting, but I think it's all correct.

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To be clear, are you saying there is a Tannakian reconstruction for Uq(g) up to non-unique isomorphism? –  B. Bischof Aug 22 '12 at 22:57
Thank you Theo for your insightful answer. I'm quite curious about the Hopf algebroid you mentioned in your last paragraph... can you say a bit more about it? –  André Henriques Aug 23 '12 at 9:48
@B. Bischof: I believe so, but I could be mistaken. Certainly to have a chance of any two fibers being isomorphic I had better word over an algebraically closed field — otherwise you should expect that the fibers correspond to "Galois actions" of the quantum group on the field. But I'm trusting intuition from the case of actual algebraic groups. –  Theo Johnson-Freyd Aug 23 '12 at 12:13
@B. Bischof: That said, most likely even if there are other fiber functors, I can ask for those that degenerate to "the" fiber functor as $q\to 1$, and these should all be isomorphic, I'd expect. –  Theo Johnson-Freyd Aug 23 '12 at 12:14
@Andre: Not really — again I'm going based on intuition from the group case. What I'd expect is that there is a category whose objects are faithful braided monoidal functors from "$\mathrm{Rep}(U_q(\g)$" to some version of VECT; equivalently, the objects of the category are braided coalgebras which are generators of the category. Morphisms are all linear maps, but this category should be some version of "Hopf algebroid" where the comultiplication encodes the various coalgebra structures, e.g. the grouplike elements are the coalgebra homomorphisms. –  Theo Johnson-Freyd Aug 23 '12 at 12:27

Some possible partial answers might be:

• one could follow Lusztig and do away with the Lie algebra completely, just starting from a root datum. Then do some geometry...
• Majid's reinterpretation of Lusztig's construction, as exposited in his "A Quantum Groups Primer", is a (good, IMHO) attempt to explain where the formulae come from. The definition of the positive part of $U_{q}(\mathfrak{g})$ as natural (braided) object acted on by "the Cartan part" explains most of the formulae. Taking the Drinfeld double "explains" the cross-relations.
• another explanation for the formulae, especially the quantized Serre relations, is given via the Ringel-Hall categorical approach (Hall algebras); one should also mention Green at this point. This has been extended of late to double Hall algebras, trying to construct the whole quantum group and not just the positive part, but this is still essentially done via the Drinfeld double construction, just at a categorical level.
• a quite non-standard route would be to go from $\mathfrak{g}$ to the algebraic group $G$, construct the quantized coordinate ring - where you might find the formulae more to your taste and/or better motivated along Grothendieck/Manin lines (see the quantum groups book of Brown-Goodearl for example) - then dualize to get $U_{q}(\mathfrak{g})$. I say non-standard because most people want to go the other way, to figure out what the quantized coordinate ring should be.

Apologies for the dearth of precise references: I can try to provide some if any of the above is helpful. Kassel's book specifically covers the FRT construction, by the way.

Edit: The full Hopf structure, as opposed to just the algebra structure, is - again IMHO - essentially canonically determined. The coproduct is pretty canonical, by comparison with the natural one for $U(\mathfrak{g})$. There's almost no choice for the counit and the antipode formula is (if I remember correctly) forced from the requirement that various maps are algebra morphisms. Alternatively, the Hopf structure can be seen to drop out of the Drinfeld double construction; that might not be a helpful thing to say, of course.

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The double construction of $\mathfrak g$ picks out a particular Cartan subalgebra. Indeed, the coproduct picks out the Cartan subalgebra, unless you're going to play games with "gauge equivalence" or something. I agree that counit and antipode maps are forced: this is equivalent to saying that an associative algebra cannot have more than one unit, and that an element in a unital associative algebra cannot have more than one inverse. This is in contrast with the coproduct, which is determined up to isomorphism but any particular choice of coproduct is actual data. –  Theo Johnson-Freyd Aug 22 '12 at 14:42