Adjoint action on universal enveloping algebra Let $G$ be a Lie group, $\mathfrak{g} = \operatorname{Lie}(G)$ be a Lie algebra and $U\mathfrak{g}$ be the universal enveloping algebra of $\mathfrak{g}$. I want to show that if $D\in Z(U\mathfrak{g})$ is in center, then the extended adjoint action $\operatorname{Ad}:G\to \operatorname{End}(U\mathfrak{g})$ satisfies $\operatorname{Ad}(g)D = D$. 
This is an exercise 2.2.5 in Bump's automorphic form. The adjoint action $\operatorname{Ad}$ of $G$ on $U\mathfrak{g}$ is defined by
$$
\operatorname{Ad}(g)(x_{1}\otimes \cdots \otimes x_{r}) = \operatorname{Ad}(g)x_{1}\otimes \operatorname{Ad}(g)x_{2} \otimes \cdots \otimes \operatorname{Ad}(g)x_{r} 
$$
where $x_1, x_2, \ldots, x_r \in \mathfrak{g}$; but I'm not sure how to proceed. 
 A: I do not agree with the argument in the comments; it is not obvious why the adjoint action of $\mathfrak g$ on $U(\mathfrak g)$ vanishes on the center; in general $\operatorname{Ad}_x(D) \neq xD - Dx$.
Identify $U(\mathfrak g)$ with the algebra of left invariant differential operators on $G$ (connected Lie group) in the standard way. Then you want to show that the operator corresponding to $D \in Z(U(\mathfrak g))$ is invariant under conjugation. Equivalently, right invariant. Now take $X \in \mathfrak g$, $t \in \mathbb R$, $b$ a real analytic function in a neighborhood of $1$ and consider
$$f(g, t) = R_{e^{tX}} D R_{e^{-tX}} b(g) \,,$$
where $R_g$ denotes right translation, for $g, t$ in a small neighborhood $U \times I$ of $(1, 0)$. For fixed $g$ this is real analytic as a function of $t$. We have:

*

*$f(g, 0) = D b(g)$;

*$\frac d{dt} f(g, 0) =(XD - DX) b(g) = 0$, because $D$ is central;

*$\frac {d^2}{dt^2} f(g, 0) =(X^2D - 2XDX + DX^2) b(g) = 0$;

*$\frac {d^n}{dt^n} f(g, 0) = \sum_{i=0}^n \binom ni X^i D (-X)^{n-i} = DX^n \sum_{i=0}^n \binom ni (-1)^i = 0$ for all $n > 0$.

Therefore $f(g, t) = D b(g)$ for all $g \in U$ and $t \in I$.
For $t \in I$, the differential operators $D$ and $R_{e^{tX}} D R_{e^{-tX}}$ coincide in a neighborhood of $1$ and therefore on all of $G$ by left-invariance, so they determine the same element of $U(\mathfrak g)$.
Because $G$ is connected and $I$ can be chosen to be the same for all
bounded $X \in \mathfrak g$, it follows that $R_{g} D R_{g^{-1}} = D$ for all $g \in G$.
