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.


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.

  • $\begingroup$ Actually, they are happy to work over any field of characteristic zero. They just call it "$\rm C$" for convenience. $\endgroup$ – Theo Johnson-Freyd Feb 2 '10 at 5:09

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).


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