show/hide this revision's text 2 cured tex problems

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 $G^$ G^\ast$ with Lie algebra $g^$ 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).

show/hide this revision's text 1

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 $G^$ with Lie algebra $g^$ 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).