Is a universal enveloping algebra a Poisson algebra? 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.
 A: $T(g)$ 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} 
