Is a universal enveloping algebra a Poisson algebra?

Question

Let $g$ be a Lie algebra and $U(g)$ its enveloping algebra. I think that $U(g)$ is a Poisson algebra where the Poisson bracket is given by the Lie bracket of $g$ and Leibniz rule. Is this true? Thank you very much.

Answer 1

$T(g)$ is a Poisson algebra under the bracket given by $\{a,b\}=[a,b], a, b \in g$ and Leibniz rule: \begin{align} \{ab,c\} = a\{b,c\}+\{a,c\}b, \ a, b, c \in T(g). \end{align} To prove that $U(g)=T(g)/I$ is Poisson, we need to verify that $I$ is a Poisson ideal of $T(g)$. Let $I′=Span_{\mathbb{C}}\{ab−ba−[a,b]: a, b \in g\}$. Then $I=T(g)I′T(g)$. It suffices to verify that for all $x=ab-ba-[a,b]$, $a,b \in g$, $h,h′,c \in T(L)$, we have $\{hxh′,c\} \in I$.

We have \begin{align} \{hxh′,c\}=\{hxh′,c\}=hx\{h′,c\}+\{h,c\}xh′+h\{x,c\}h′. \end{align} This is in $I$ if $\{x,c\} \in I′$.

We have \begin{align} \{x,c\}&=\{ab−ba−[a,b],c\}\\ & = \{ab,c\}-\{ba,c\}-\{[a,b],c\} \\ &=a\{b,c\}+\{a,c\}b−b\{a,c\}−\{b,c\}a−\{[a,b],c\}. \end{align} Since $a,b,c,[a,b] \in g$, we have \begin{align} \{x,c\} &=a\{b,c\}+\{a,c\}b−b\{a,c\}−\{b,c\}a−\{[a,b],c\} \\ &=a[b,c]+[a,c]b−b[a,c]−[b,c]a−[[a,b],c] \\ &=([a,c]b−b[a,c]−[[a,c],b])+(a[b,c]−[b,c]a−[a,[b,c]])+[[a,c],b]+[a,[b,c]]−[[a,b],c]\\ &=([a,c]b−b[a,c]−[[a,c],b])+(a[b,c]−[b,c]a−[a,[b,c]]) \in I′. \end{align}