Timeline for Example of a manifold which is not a homogeneous space of any Lie group

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Feb 24, 2012 at 20:57	comment	added	Johannes Ebert		If $H$ has smaller rank, $S \subset H$ is a maximal torus of $H$, $S \subset T \subset G$ a maximal torus of $G$. The fibre bundle $T/S G/S \to G/T$ shows that $\chi(G/S)=0$. The fibre bundle $H/S \to G/S \to G/H$ shows $0=\chi(G/S) = \chi(H/S) \chi(G/H) = # W_H \chi(G/H)$ and so $\chi(G/H)\geq 0$.
Feb 24, 2012 at 20:54	comment	added	Johannes Ebert		@Dylan: I am pretty sure Hermann, as well as Mostow, does not assume compactness of $G$, only of $G/H$. If $G$ is compact, then in fact the result is much older and, nowadays, fairly easy. If $G$ is connected and compact and the rank of $H$ equals the rank of $G$, then there is a fibre bundle $H/T \to G/T \to G/H$, where $T$ is the maximal torus. By Hopf-Samelson, the Euler numbers of $G/T$ and $H/T$ are given by the order of the Weyl groups, hence both positive, so $\chi(G/H)>0$.
Feb 24, 2012 at 19:20	comment	added	Dylan Wilson		Are you sure they aren't assuming the group is compact? Mostow doesn't assume this. He also has some classification results.
Feb 24, 2012 at 17:08	comment	added	Johannes Ebert		This is also proven in R. Hermann "Compactification of homogeneous spaces. I." J. Math. Mech. 14 1965 655–678. Apparently Mostow did not know that paper.
Feb 24, 2012 at 6:59	comment	added	Dylan Wilson		Now I see this is also implied by Vitali's post... I wonder if Mostov knew this?
Feb 24, 2012 at 6:54	history	answered	Dylan Wilson	CC BY-SA 3.0