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Let $G$ be a Lie group and $\mathfrak{g}$ its Lie algebra. Then $\mathfrak{g}$ acts on $G$ by \begin{align} \xi.f(X) = \frac{d}{dt}|_{t=0} f(e^{t\xi}X), \ \xi \in \mathfrak{g}, X \in G, f \in C^{\infty}(G). \end{align} Do we have $\xi.(fg)=(\xi.f) g + f \xi.g$, $f, g \in C^{\infty}(G)$?

Let $U(\mathfrak{g})$ be the enveloping algebra of $\mathfrak{g}$. Let $\xi, \eta \in U(\mathfrak{g})$. How to compute $(\xi \eta).f$, where $f \in C^{\infty}(G)$? Let $1$ be the identity element in $U(\mathfrak{g})$. What is the result of $1.f$, $f \in C^{\infty}(G)$?

Thank you very much.

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I am not completely sure what you mean by "acting on $G$". What you write out in the beginning of the question is just the natural identification of $\mathfrak g$ with the space of right invariant vector fields on $G$. Of course, any such vector field acts as a derivation on $C^\infty(G)$ (as any vector field does), so you do get $\xi\cdot(fg)=(\xi\cdot f)g+f\xi\cdot g$. This means that you interpret any element of $\mathfrak g$ as a first order differential operator acting on $C^\infty(G)$.

Correspondingly, you get an identification of $\mathcal U(\mathfrak g)$ as the space of all right invariant differential operators acting on $C^\infty(G)$. In this identification $1\cdot f =f$ and $(\xi\eta)\cdot f=\xi\cdot(\eta\cdot f)$, so you simply compose the two first order operators determined by $\xi$ and $\eta$ to get a second order operator. This is well defined by definition of the Lie bracket.

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  • $\begingroup$ thank you very much. Sorry, I made a typo. I need to say "acting on $C^{\infty}(G)$" ( not "acting on $G$" ). $\endgroup$ – LJR Mar 1 '18 at 16:25

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