Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\mathbb K$ be a field (of characteristic 0, say), $\mathfrak g$ a Lie bialgebra over $\mathbb K$, and $\mathcal U \mathfrak g$ its usual universal enveloping algebra. Then the coalgebra structure on $\mathfrak g$ is equivalent to a co-Poisson structure on $\mathcal U \mathfrak g$, i.e. a map $\hat\delta : \mathcal U \mathfrak g \to (\mathcal U \mathfrak g)^{\otimes 2}$ satisfying some axioms. A formal quantization of $g$ is a Hopf algebra $\mathcal U_\hbar \mathfrak g$ over $\mathbb K[[\hbar]]$ (topologically free as a $\mathbb K[[\hbar]]$-module) that deforms $\mathcal U \mathfrak g$, in the sense that it comes with an isomorphism $\mathcal U_\hbar \mathfrak g / \hbar \mathcal U_\hbar \mathfrak g \cong \mathcal U \mathfrak g$, and moreover that deforms the comultiplication in the direction of $\hat\delta$: $$\Delta = \Delta_0 + \hbar \hat\delta + O(\hbar^2),$$ where $\Delta$ is the comultiplication on $\mathcal U_\hbar \mathfrak g$ and $\Delta_0$ is the (trivial, i.e. which $\mathfrak g$ is primitive) comultiplication on $\mathcal U\mathfrak g$. This makes precise the "classical limit" criterion: "$\lim_{\hbar \to 0} \mathcal U_\hbar \mathfrak g = \mathcal U \mathfrak g$"

I am wondering about "the other" classical limit of $\mathcal U_\hbar \mathfrak g$. Recall that $\mathcal U\mathfrak g$ is filtered by declaring that $\mathbb K \hookrightarrow \mathcal U\mathfrak g$ has degree $0$ and that $\mathfrak g \hookrightarrow \mathcal U\mathfrak g$ has degree $\leq 1$ (this generates $\mathcal U\mathfrak g$, and so defines the filtration on everything). Then the associated graded algebra of $\mathcal U\mathfrak g$ is the symmetric (i.e. polynomial) algebra $\mathcal S\mathfrak g$. On the other hand, the Lie structure on $\mathfrak g$ induces a Poisson structure on $\mathcal S\mathfrak g$, one should understand $\mathcal U \mathfrak g$ as a "quantization" of $\mathcal S\mathfrak g$ in the direction of the Poisson structure. Alternately, let $k$ range over non-zero elements of $\mathbb K$, and consider the endomorphism of $\mathfrak g$ given by multiplication by $k$. Then for $x,y \in \mathfrak g$, we have $[kx,ky] = k(k[x,y])$. Let $\mathfrak g_k$ be $\mathfrak g$ with $[,]\_k = k[,]$. Then $\lim\_{k\to 0} \mathcal U\mathfrak g_k = \mathcal S\mathfrak g$ with the desired Poisson structure.

I know that there are functorial quantizations of Lie bialgebras, and these quantizations give rise to the Drinfeld-Jimbo quantum groups. So presumably I can just stick $\mathfrak g_k$ into one of these, and watch what happens, but these functors are hard to compute with, in the sense that I don't know any of them explicitly. So:

How should I understand the "other" classical limit of $\mathcal U_\hbar \mathfrak g$, the one that gives a commutative (but not cocommutative) algebra?

If there is any order to the world, in the finite-dimensional case it should give the dual to $\mathcal U(\mathfrak g^\*)$, where $\mathfrak g^\*$ is the Lie algebra with bracket given by the Lie cobracket on $\mathfrak g$. Indeed, B. Enriquez has a series of papers (which I'm in the process of reading) with abstracts like "functorial quantization that respects duals and doubles".

On answer that does not work: there is no non-trivial filtered $\hbar$-formal deformation of $\mathcal U\mathfrak g$. If you demand that the comultiplication $\Delta$ respect the filtration on $\mathcal U\mathfrak g \otimes \mathbb K[[\hbar]]$ and that $\Delta = \Delta_0 + O(\hbar)$, then the coassociativity constraints imply that $\Delta = \Delta_0$.

This makes it hard to do the $\mathfrak g \mapsto \mathfrak g_k$ trick, as well. The most naive thing gives terms of degree $k^{-1}$ in the description of the comultiplication.

share|improve this question
add comment

2 Answers

The answer is already in Drinfeld's "quantum groups" ICM report. To a QUE algebra $U =U_\hbar g$ with classical limit the Lie bialgebra $g$, he associates a QFSH algebra $U^\vee$ which turns out to be a formal deformation of the formal series Hopf algebra $\hat S(g)$; when $g$ is finite dimensional, this is the formal function ring over the formal group $G^\ast$ with Lie algebra $g^\ast$ and indeed the dual to $Ug^*$. All this was later developed in a paper by Gavarini (Ann Inst Fourier).

For example, if the deformation is trivial, so $U=Ug[[\hbar]]$, one finds $U^\vee$ to be the complete subalgebra generated by $\hbar g[[\hbar]]$ which is roughly speaking $U(\hbar g[[\hbar]])$ i.e. $U(g_\hbar)$ where $g_\hbar$ is $g[[\hbar]]$ but with Lie bracket multiplied by $\hbar$. So $U^\vee$ is a quasi-commutative algebra (a flat deformation of $\hat S(g)$ actually).

share|improve this answer
add comment
up vote 2 down vote accepted

The answer is essentially given in Kassel and Turaev, "Biquantization of Lie bialgebras", Pacific Journal of Mathematics, 2000 vol. 195 (2) pp. 297-369, MR1782170. They do the following: To a finite-dimensional Lie bialgebra $\mathfrak g$ over $\mathbb C$, they define a biassociative bialgebra $A_{u,v}(\mathfrak g)$, (topologically) free over $\mathbb C[u][[v]]$, such that:

  1. $A_{u,v}(\mathfrak g)$ is commutative module $u$ and cocommutative modulo $v$.
  2. $A_{u,v}(\mathfrak g) / (u,v) = \mathcal S\mathfrak g$, the symmetric algebra, with its induced Poisson and co-Poisson structures.
  3. $A_{u,v}(\mathfrak g) / (u)$ is a commutative Poisson bialgebra and its cobracket quantizes $\mathcal S(\mathfrak g)$ in the co-Poisson direction.
  4. $A_{u,v}(\mathfrak g) / (v)$ is a cocommutative co-Poisson bialgebra and its bracket quantizes $\mathcal S(\mathfrak g)$ in the Poisson direction. Indeed, $A_{u,v}(\mathfrak g) / (v,u-1) = \mathcal U\mathfrak g$.
  5. $A_{u,v}(\mathfrak g)$ is essentially dual to $A_{v,u}(\mathfrak g^*)$.

Thus the Etingof-Kazhdan quantization is $A_{u,v}(\mathfrak g) / (v-\hbar,u-1)$. More generally, we have $\hbar = uv$.

I still don't know if there's some way to look at this construction without variables $u,v$, and rather with (co-)filtrations. But, then again, I've only read the introduction to the paper. I also don't know if it works over other fields. Given that Kassel and Turaev rely on the Etingof-Kazhdan methods, which in turn rely on a Drinfeld associator, I assume that the method requires working over a $\mathbb Q$-algebra.

share|improve this answer
    
Actually, they are happy to work over any field of characteristic zero. They just call it "$\rm C$" for convenience. –  Theo Johnson-Freyd Feb 2 '10 at 5:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.