2
$\begingroup$

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.

  • 0
    Yes, see [here](https://en.wikipedia.org/wiki/Universal_enveloping_algebra). The tensor algebra $T(L)$ of a Lie algebra $L$ is a Poisson algebra, and $U(L)=T(L)/I$, where $I$ is the two-sided ideal over $T(L)$ generated by elements of the form $ a ⊗ b − b ⊗ a − [ a , b ]$. – 2017-02-28
  • 0
    @DietrichBurde, thank you very much. Is $I$ a Poisson ideal of $T(L)$: $\{I, T(L)\} \subset I$? – 2017-02-28
  • 0
    May be an easier proof is given in Example $2.3$ [here](http://www.caressa.it/pdf/yas.pdf). – 2017-02-28
  • 0
    @DietrichBurde, thank you very much. – 2017-02-28
  • 0
    @DietrichBurde, I verified that $I$ is a Poisson ideal of $T(L)$. Let $I'=\{ab-ba-[a,b]\}$ (I omitted the tensor notation). Then $I=T(L)I'T(L)$. It suffices to verify that for all $x \in I'$, $h, h', c \in T(L)$, we have $\{hxh', c\} \in I$. According to [the article](https://en.wikipedia.org/wiki/Universal_enveloping_algebra), the Poisson bracket defined on $T(V)$ is given by \begin{align} \{d,e\}=[d,e], d, e \in L, \end{align} and Leibniz rule. 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 $[x,c] \in I'$. – 2017-02-28
  • 0
    We have \begin{align} x = ab-ba-[a,b] \end{align} for some $a,b \in L$. Therefore \begin{align} [x,c]&=[ab-ba-[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} $I'$ and $I$ can be defined as $\text{Span}_{\mathbb{C}}\{ab-ba-[a,b]: a,b \in L\}$, $I=T(L)I'T(L)$. In the above, we proved that for $a,b\in L$, $[h(ab-ba-[a,b])h', c] \in I$. This implies that $\{I, T(V)\} \subset I$. – 2017-02-28

1 Answers 1

2

$T(g)$ 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}