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?



Does the vanishing of the Poisson bracket on $S(\mathfrak{g})^{\mathfrak{g}}$ inspire the disover of Duflo's isomorphism theorem?

> 
> 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. 


