2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.