I am trying to remember a result stating that under certain assumptions, given a transitive smooth action of a (compact?) Lie group on a smooth manifold, also the action of the semisimple factor is transitive. It could be due to Borel, but I cannot remember or find the reference.

Does anybody know?

Edit: the manifold must be compact and simply-connected. What else?


Yes, it's true: whenever $G$ is a compact Lie group and $X$ a simply connected (topological) manifold, and $G$ acts transitively continuously on $X$, then $G'=[G^0,G^0]$ acts transitively.

Indeed, one can write $X=G/H$. The $G^0$-orbits being open, $G^0$ acts transitively, so we can suppose that $G$ is connected. We have a fibre bundle $X=G/H\to G/HG'$, with connected fiber $HG'/H=G'/(H\cap G')$

Since fibers are connected, this induces an exact sequence $\pi_1(X)\to \pi_1(G/HG')\to 1$. Hence the torus $G/HG'$ is simply connected, which forces it to be a singleton. So $HG'=G$. This means that $G'$ acts transitively on $G/H=X$.

  • $\begingroup$ What is $G^0$? I thought identity component, but then you refer to $G/H G'$ being a torus, and connectedness of that group isn't obvious to me for $G$ disconnected …. $\endgroup$
    – LSpice
    Jun 23 '20 at 3:16
  • 1
    $\begingroup$ @LSpice Yes $G^0$ is the unit component. At the 1st sentence of the 2nd paragraph, I'm writing "so we can suppose that 𝐺 is connected". So when I refer to $G/HG'$ in the very next sentence I'm assuming $G$ connected... $\endgroup$
    – YCor
    Jun 23 '20 at 7:49

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