Let $G$ be a Lie group and $H\subseteq G$ a closed subgroup. Then, $G/H$ is a smooth manifold and inherits a smooth action of $G$: $$G\times G/H\longrightarrow G/H,\quad g_1\cdot(g_2H)=g_1g_2H.$$

Question: Is this action proper?

If not, are there reasonable conditions so that it is proper? If the group $G$ is compact then of course it will be proper. But I am more interested in the case where $G$ is non-compact. For example, what happens if $G$ is the complexification of a compact Lie group $K$ and $H=T_{\Bbb C}$ is the complexification of a maximal torus $T$ in $K$?


It is not proper if $H$ is not compact since $H$ fixes the class of $1_G$ in $G/H$.

  • 1
    $\begingroup$ Why would that imply that the action is not proper? $\endgroup$ – Spenser Mar 23 '16 at 16:42
  • $\begingroup$ proper means that for every compact $K$, the adherence of the subset of elements $G_K=\{g\in G, g(K)\cap K\neq \phi\}$ is compact. Here, take $K =[1_G]$ the class of $1_G$, it is fixed by $H$, so $G_K$ contains $H$ and thus its adherence cannot be compact if $H$ is not compact. $\endgroup$ – Tsemo Aristide Mar 23 '16 at 16:45

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