# Showing that the universal enveloping algebra of some $\mathfrak{g}$ is isomorphic to $\mathbb{C}[x_i,\partial/\partial x_i]$

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?

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Please have a look here and think of changing your title accordingly. –  draks ... Jul 21 '12 at 7:48
@draks Of course, I will edit the title shortly. –  math-visitor Jul 21 '12 at 7:49

$\mathfrak{g}$ 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 $\mathbb{C}[x_1, ... x_n]$.
Thanks Qiaochu! I thought the above elements are idempotents since $\left( x_i\frac{\partial}{\partial x_i}\right)\left( x_i\frac{\partial}{\partial x_i} \right) =x_i\frac{\partial}{\partial x_i}$. –  math-visitor Jul 21 '12 at 4:18
@math-visitor: no. $x_i$ is the operator describing multiplication by $x_i$, so in fact $\frac{\partial}{\partial x_i} x_i = x_i \frac{\partial}{\partial x_i} + 1$ by the product rule. –  Qiaochu Yuan Jul 21 '12 at 7:56