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.

  • $\begingroup$ 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 $\operatorname{Ad}\left(g\right)\left(x_1 \otimes x_2 \otimes \cdots \otimes x_r\right) = \operatorname{Ad}\left(g\right)\left(x_1\right) \otimes \operatorname{Ad}\left(g\right)\left(x_2\right) \otimes \cdots \otimes \operatorname{Ad}\left(g\right)\left(x_r\right)$. $\endgroup$ – darij grinberg Feb 11 at 20:37
  • $\begingroup$ @darijgrinberg Thanks, I was confusing between those two. $\endgroup$ – Seewoo Lee Feb 11 at 20:42
  • $\begingroup$ Also, do you want $G$ to be connected or something like that? I don't see how to prove this otherwise. $\endgroup$ – darij grinberg Feb 11 at 20:43
  • $\begingroup$ @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. $\endgroup$ – Seewoo Lee Feb 11 at 20:45
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    $\begingroup$ 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 :) $\endgroup$ – darij grinberg Feb 11 at 21:36

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