4
$\begingroup$

Consider a Poisson-Lie group $G$, with whatever additional requirements (quasi-triangular, compact, simply connected).

We can consider $G$ as a Poisson Manifold and apply Kontsevich formality to obtain a star-product on $C^\infty(G)$, or equivalently a Maurer-Cartan element in $\star\in\Gamma (D_{\mathrm{poly}}^\bullet(G))$, the dg-LA of (formal) polydifferential operators.

Conversely we can consider the Lie algebra $\mathfrak g$, and then we can deform $U\mathfrak g$ using Etinghof-Kazhdan deformation quantization to a non-cocommutative Hopf algebra. Let us assume $\mathfrak g$ is quasi-triangular so that we only deform the coproduct $\Delta_\hbar \colon U_\hbar\mathfrak g\to U_\hbar\mathfrak g\otimes U_\hbar \mathfrak g$.

These two types of deformation quantization have to be related. The first thing to note is that $U{\mathfrak g}^{\otimes n}$ can be interpreted as invariant polydifferential operators on $C^\infty(G)$. Furthermore, since $\Delta_\hbar$ is an algebra morphism, it is determined by its restriction $\Delta_\hbar|_{\mathfrak g}\colon \mathfrak g\to U_\hbar\mathfrak g\otimes U_\hbar \mathfrak g$. The fact that $\Delta_\hbar$ is an algebra morphism means that this is a cocycle, and hence should integrate to a map $G\to U_\hbar\mathfrak g\otimes U_\hbar \mathfrak g$. Coassociativity of $\Delta_\hbar$ should translate to this being star product, modulo some details. We note that this star product is invariant (in the same sense of invariance as the Poisson structure on $G$.) Differentiating invariant star products should then also produce deformed coproducts on the universal enveloping algebra.

Does this give an equivalence between invariant star products and deformations of $U\mathfrak g$? Invariant star products are the Maurer-Cartan elements of some dgLA. Can we interpret deformations of $U\mathfrak g$ as Maurer-Cartan elements of some appropriate dgLA, and are the dgLA's quasi-isomorphic?

This seems like a fairly basic question, but I have been unable to find any literature discussing this.

$\endgroup$
2
  • $\begingroup$ Can you please say what kind of invariance you are talking about? The identification of $U\mathfrak g$ with invariant differential operators uses the left( or right) multiplication, whereas the Kontsevich star product of a Poisson Lie structure is never inariant with respect to the left multiplication unless the Poisson tensor is zero. $\endgroup$ Commented Aug 9, 2019 at 16:17
  • $\begingroup$ I left this vague, because it is clear there should be some sort of invariance property, but I haven't figured out exactly what it should be. Note that the Poisson structure on a Poisson-Lie group is always zero at the origin (and linear in a neighborhood of the identity). However, the Poisson bivector is still determined by its germ at the identity. I had a similar kind of invariance in mind for the star product. But perhaps one can not extend it beyond a (formal) neighborhood of the identity. $\endgroup$ Commented Aug 11, 2019 at 10:06

0

You must log in to answer this question.