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.

• No, $\operatorname{Ad}$ (a group action) does not act through the Leibniz rule. That's what $\operatorname{ad}$ (a Lie algebra action) does. As for $\operatorname{Ad}$, it is given by $\operatorname{Ad}\left(g\right)\left(x_1 \otimes x_2 \otimes \cdots \otimes x_r\right) = \operatorname{Ad}\left(g\right)\left(x_1\right) \otimes \operatorname{Ad}\left(g\right)\left(x_2\right) \otimes \cdots \otimes \operatorname{Ad}\left(g\right)\left(x_r\right)$. – darij grinberg Feb 11 at 20:37
• @darijgrinberg Thanks, I was confusing between those two. – Seewoo Lee Feb 11 at 20:42
• Also, do you want $G$ to be connected or something like that? I don't see how to prove this otherwise. – darij grinberg Feb 11 at 20:43
• @darijgrinberg We may assume that $G = \mathrm{GL}(n, \mathbb{R})$ or $G = \mathrm{GL}(n, \mathbb{R})^{+}$, where the latter one is a connected component of the first one. I think Bump want to do only for this case. – Seewoo Lee Feb 11 at 20:45
• This kind of Zariski-density argument is highly elementary (it's just saying that a multivariate polynomial that vanishes on a nonzero ball must vanish identically). It's the use of the exponential map that is bothering me :) – darij grinberg Feb 11 at 21:36