Given that $G$ is a connected Lie Group and $\mathfrak{g}$ is its Lie Algebra. Denote $\tilde{G}$ as the universal covering Lie group of $G$. We know that $\mathrm{Aut}\left(\tilde{G}\right)=\mathrm{Aut}\left(\mathfrak{g}\right)$. We also know that $\mathrm{Aut}\left(G\right)$ is a closed subgroup of $\mathrm{Aut}\left(\tilde{G}\right)$, hence it's a Lie subgroup of $\mathrm{Aut}\left(\mathfrak{g}\right)$. Also We know that $\mathrm{Aut}\left(G\right)=\mathrm{Aut}\left(\mathfrak{g}\right)$ if $G$ is simply connected, as in such case we have $G=\tilde{G}$.
My question is, does the conclusion works from the other side? i.e.
For a connected (compact if needed) Lie group $G$, if $\mathrm{Aut}\left(G\right)=\mathrm{Aut}\left(\mathfrak{g}\right)$, do we know for sure that $G$ is simply connected?

  • $\begingroup$ What non-simply connected $G$ have you checked as examples? $\endgroup$ Commented May 27, 2022 at 15:05
  • $\begingroup$ $G=S^{2}$ for example, in which $\mathrm{Aut}\left(G\right)=\mathbb{Z}_{2}$ while $\mathrm{Aut}\left(\mathfrak{g}\right)=\mathbb{R}^{*}$. $\endgroup$
    – Zi Cui
    Commented May 27, 2022 at 15:36
  • $\begingroup$ Sorry, It's a typo here. It should be $S^{1}$ instead of $S^{2}$. Yet the rest is correct. $\endgroup$
    – Zi Cui
    Commented May 27, 2022 at 16:19
  • $\begingroup$ I thought that's a typo ... Good. What about some simple group like $SL_2(\mathbb R)$ or $SL_2(\mathbb C)$, or, for the compact case, $SU_2$? $\endgroup$ Commented May 27, 2022 at 16:48
  • 1
    $\begingroup$ You might want to check out the excellent answer to math.stackexchange.com/q/4419869/96384 (which contains far more info, but you should distill what you need in case $K= \mathbb R, \mathbb C$). As for $SU_2$, that is actually simply connected, so is irrelevenat for your question; instead you want to look at $SO_n$ and the answers (plus discussion in comments) here: math.stackexchange.com/q/3104437/96384 $\endgroup$ Commented May 27, 2022 at 17:18


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