# Homogeneous manifold deformation retracts onto compact submanifold

Let $$G$$ be a connected Lie group. Then by a theorem of Cartan there is a diffeomorphism $$G \cong K \times \mathbb{R}^n$$ where $$K$$ is a maximal compact subgroup of $$G$$. Now, let $$M$$ be a homogeneous manifold. In other words, there exists a Lie group $$G$$ acting transitively on $$M$$. Is it true that $$M$$ deformation retracts onto a compact submanifold? Slightly stronger, is it true that there is a diffeomorphism $$M \cong K \times \mathbb{R}^n$$ where $$K$$ is a compact submanifold of $$M$$?

• You can find information about the topology of homogeneous spaces of Lie groups in A. Borel, Les bouts des espaces homogènes des groupes de Lie, Ann. Math. 58(3), 1953, 443-457, behind JSTOR's rapacious paywall here. – YCor Nov 13 '19 at 20:15
• Crossposted from MathSE: math.stackexchange.com/questions/3433385 – YCor Nov 13 '19 at 23:01

Mostow-Karpelevich theorem says that if $$G/G^\prime$$ is a homogeneous space where $$G$$, $$G^\prime$$ are Lie groups with finitely many connected components, and maximal compact subgroups $$K\supset K^\prime$$, respectively, then $$G/G^\prime$$ is a vector bundle over $$K/K^\prime$$. In fact, it is a homogeneous $$K$$-vector bundle. A more precise statement is in [Mostow, G. D., Covariant fiberings of Klein spaces II, Amer. J. Math. 84 (1962), 466–474]. There are examples where the bundle is nontrivial.
Another result in this direction addresses the situation when $$G$$ is solvable, and $$G^\prime$$ is any closed subgroup. Then $$G/G^\prime$$ is a vector bundle over a compact of the form $$H/K^\prime$$ where $$H$$ is solvable Lie group and $$H^\prime$$ is a closed subgroup. See [Auslander, L. and Tolimieri, R., Splitting theorems and the structure of solvmanifolds, Ann. of Math. (2) 92 (1970), 164–173].
In order for the quotient to be a vector bundle some assumptions are clearly necessary. For example consider a nonabelian free discrete subgroup of $$SL(2,\mathbb R)$$, and lift it to a nonabelian free discrete subgroup $$F$$ of the universal cover $$\widetilde{SL}(2,\mathbb R)$$. The latter is diffeomorphic to $$\mathbb R^3$$, so the quotient $$\widetilde{SL}(2,\mathbb R)/F$$ is homotopy equivalent to a wedge of circles, and in particular it is not a vector bundle over a manifold.