TITLE: Showing that the universal enveloping algebra of some $\mathfrak{g}$ is isomorphic to $\mathbb{C}[x_i,\partial/\partial x_i]$
QUESTION [5 upvotes]: The universal enveloping algebra $U(\mathfrak{g})$ of a Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ is defined to be 
$$
\dfrac{\mathbb{C}\oplus\mathfrak{g}\oplus ( \mathfrak{g}\otimes \mathfrak{g})\oplus (\mathfrak{g}\otimes \mathfrak{g}\otimes \mathfrak{g})\oplus\ldots}{\langle a\otimes b-b\otimes a -[a,b]\rangle}, 
$$which means it is an associative tensor algebra generated by the vector space $\mathfrak{g}$ mod out all elements of the form $ a\otimes b-b\otimes a -[a,b]$, where $a,b\in \mathfrak{g}$. 
Suppose $\mathfrak{g}=\mathbb{C}[e,p,q]/\langle [p,q]=e,[p,e]=0,[q,e]=0
\rangle$. Then we have a (ring) isomorphism
$$
U(\mathfrak{g})/\langle e-1\rangle  \cong \mathbb{C}[x, \partial/\partial x]  
$$ where we can define the map to be $U(\mathfrak{g})\stackrel{\phi}{\rightarrow}\mathbb{C}[x,\partial/\partial x]$ 
by sending 
$$
p \mapsto \partial/\partial x 
\mbox{ and }
q \mapsto x. 
$$
One can check that since 
$$
\phi([p,q])=[\partial/\partial x,x]= 1-x\dfrac{\partial}{\partial x}, 
$$ we take $e=x\dfrac{\partial}{\partial x}$, which gives us the isomorphism $U(\mathfrak{g})/\langle 1-e\rangle \cong \mathbb{C}[x,\partial/\partial x]$. 

Now what should $\mathfrak{g}$ be so that $U(\mathfrak{g})/I\cong \mathbb{C}[x_1,x_2,\ldots, x_n,\partial /\partial x_1,\partial /\partial x_2, \ldots, \partial /\partial x_n]$?  

I could be wrong but I'm guessing that if we take $\mathfrak{g}$ to be the algebra $\mathbb{C}[p_i,q_i,e_i]/I$ where $I$ is generated by brackets of the form 
$$
[p_i,q_j]= \left\{ \begin{aligned}
e_i &\mbox{ if } i=j\\
-q_j p_i &\mbox{ if } i\not= j \\
\end{aligned}
\right.
$$
$$
[p_i,e_j]= \left\{ \begin{aligned}
0 &\mbox{ if } i=j\\
- p_i &\mbox{ if } i\not= j \\
\end{aligned}
\right.
$$
$$
[q_i,e_j]= \left\{ \begin{aligned}
0 &\mbox{ if } i=j\\
q_i &\mbox{ if } i\not= j, \\
\end{aligned}
\right.
$$
then $U(\mathfrak{g})/\langle e_i-1\rangle\cong \mathbb{C}[x_1,x_2,\ldots, x_n,\partial /\partial x_1,\partial /\partial x_2, \ldots, \partial /\partial x_n]$ where we take the map to be 
$$
p+\langle e_i-1\rangle \mapsto \partial/\partial x_i
$$
and 
$$
q+\langle e_i-1\rangle \mapsto  x_i. 
$$

Also, since the order of multiplication matters, i.e.,
  $$
(\overbrace{\partial/\partial x_1 + \partial/\partial x_2}^{\deg -1?})(\overbrace{x_1}^{\deg 1}) = 1+0=\overbrace{1}^{\deg 0}
$$ while 
  $$
x_1(\partial/\partial x_1 + \partial/\partial x_2) = x_1 \partial /\partial x_1 + x_1\partial/\partial x_2, 
$$ how should one think about multiplication in the skew polynomial algebra? 

Finally, I just noticed that $x_i\dfrac{\partial}{\partial x_i}$ form mutually orthogonal idempotent elements in $\mathbb{C}[x_i,\partial/\partial x_i]$. That is,
$$
\left(\sum_i x_i \dfrac{\partial}{\partial x_i}\right)^2 = \left(\sum_i x_i \dfrac{\partial}{\partial x_i}\right).$$ I am guessing that these are the only non-scalar idempotent elements in the algebra. 

Do these idempotent elements have a geometric interpretation?

REPLY [4 votes]: $\mathfrak{g}$ should be a Heisenberg Lie algebra. A nice coordinate-invariant way to describe these is as $V \oplus \mathbb{C}e$ where $V$ is a symplectic vector space and the symplectic form gives the Lie bracket, which takes values in $\mathbb{C}e$ (which is central). This is precisely the Lie algebra generated by the operators $x_1, x_2, ... x_n, \frac{\partial}{\partial x_1}, ... \frac{\partial}{\partial x_n}$ acting on $\mathbb{C}[x_1, ... x_n]$. 
I do not believe that the elements you wrote down are idempotent.