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For any finite dimensional Lie algebra $\mathfrak{g}$, we know that the universal enveloping algebra $U(\mathfrak{g})$ is a deformation of the symmetric algebra $S(\mathfrak{g})$. In fact let's define $$ U_t(\mathfrak{g}):=\text{T}(\mathfrak{g})/(X\otimes Y-Y\otimes X-t[X,Y]). $$ Then $S(\mathfrak{g})=U_0(\mathfrak{g})$ and $U(\mathfrak{g})=U_1(\mathfrak{g})$. Moreover we have the symmetrization map $$ I_{PBW}:S(\mathfrak{g})\longrightarrow U_t(\mathfrak{g}) $$ which pulls back the product on $U_t(\mathfrak{g})$ to a product on $S(\mathfrak{g})$. We call it the star product and denote it by $*_t$.

Obviously $*_t$ is different from the original product on $S(\mathfrak{g})$. In fact we can prove that the first order deformation of the product is governed by the $\textit{Poisson bracket}$ on $S(\mathfrak{g})$. More precisely the Poisson bracket is defined to be $ \text{{a,b}} := c^k _ {ij} X_k \cdot \partial^i a \cdot \partial^j b$ ( $c^k _ {ij}$ is the structure constant of $\mathfrak{g}$ ) and we can prove that $$ a *_t b= ab+\frac{t}{2}\text{{a,b}}+O(t^2). $$

Furthermore, we have the following result

  1. The Poisson bracket vanishes on the invariant subalgebra $S(\mathfrak{g})^{\mathfrak{g}}$. This is almost the definition.

  2. The symmetrization map $I_{PBW}$ maps $S(\mathfrak{g})^{\mathfrak{g}}$ isomorphically (as vector spaces, not as algebras) onto the center $Z(U(\mathfrak{g}))$.

  3. (Duflo's Isomorphism Theorem) We can precompose a map $D: S(\mathfrak{g})\rightarrow S(\mathfrak{g})$ such that the composition restrict to $S(\mathfrak{g})^{\mathfrak{g}}$ is an $\textit{algebraic isomorphism}:~S(\mathfrak{g})^{\mathfrak{g}}\rightarrow Z(U(\mathfrak{g}))$.

The Duflo's Isomorphism Theorem is of course highly non-trivial and we can refer to Calaque and Rossi's book http://math.univ-lyon1.fr/~calaque/LectureNotes/LectETH.pdf, as well as well as many other resources, for further discussions.

I usually wonder that (maybe historically, maybe not) how people could expect that there is an algebraic isomorphism between $S(\mathfrak{g})^{\mathfrak{g}}$ and $Z(U(\mathfrak{g}))$. The thing we can notice is that the first order deformation, which is the Poisson bracket, vanishes. We know that it is a necessary condition (at least it should vanish in the second Hochschild cohomology) to find an algebraic isormorphism.

My question is: Does the vanishing of the Poisson bracket plays an important role in finding and proving Duflo's isomorphism theorem? Or it is just an literally first step?

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My question is: Does the vanishing of the Poisson bracket plays an important role in finding and proving Duflo's isomorphism theorem? Or it is just an literally first step?

Let $A_0$ be a Poisson algebra and $A$ a deformation quantization of $A_0$ (assume we are in a context when it exists).

Assume you have a quantization map $Q:A_0\to A$, by which I mean a section of the classical limit map $A\to A/(\hbar)=A_0$.

Then for any two elements $a,b\in A_0$, $[Q(a),Q(b)]=\hbar\{a,b\}+O(\hbar^2)$.

Hence if you want to have $Q(ab)=Q(a)Q(b)$ you must at least assume that $\{a,b\}=0$.

My (non-)answer to your question is then:

the vanishing of the Poisson bracket is a necessary requirement if you want a statement of Duflo-type. It is just a first step.

The actual history comes from the Harish-Chandra isomomorphism. Duflo noticed that the original formula could be written for any Lie algebra, without any use of roots and similar stuff specific to the semi-simple case.

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