Let ${\frak g}$ be a complex semisimple Lie algebra and $G$ a connected Lie group with Lie algebra ${\frak g}$. Let $\tilde{G}$ be a universal covering group of $G$.

Take $X\in{\frak g}$ and consider the two adjoint orbits $$\begin{align*} {\cal O}_X &= \{{\rm Ad}_gX:g\in G\}\subseteq{\frak g} \\ \tilde{{\cal O}}_X &= \{{\rm Ad}_gX:g\in \tilde{G}\}\subseteq{\frak g}. \end{align*} $$

Question: Is ${\cal O}_X=\tilde{{\cal O}}_X$?

I know that if $H$ is any other Lie group with Lie algebra ${\frak g}$ then its adjoint orbit need not coincide with those of $G$, but the only counterexample I know is when $H$ is not connected and $G$ is its identity component. Here the situation is different as both groups are connected.


Let $\pi\colon \tilde G\rightarrow G$ be the covering map. Let $g\in\tilde G$. Since $\pi$ is a morphism of Lie groups, one has $$ \mathrm{Ad}_{\pi( g)}\circ D_{\tilde e}\pi=D_{\tilde e}\pi\circ \mathrm{Ad}_{g}, $$ where $e$ is the neutral element of $G$, and $\tilde e$ that of $\tilde G$. Identifying $$ \tilde{\frak g}=T_{\tilde e}\tilde G=T_eG={\frak g} $$ through the differential $D_{\tilde e}\pi$ of $\pi$, one gets $$ \mathrm{Ad}_{\pi(g)}=\mathrm{Ad}_{g}. $$ It follows that $\tilde{\mathcal{O}}_X=\mathcal O_X$, since $\pi$ is surjective.


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