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 $M$?

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.

A. Borel, Les bouts des espaces homogènes des groupes de Lie, Ann. Math. 58(3), 1953, 443-457, behind JSTOR's rapacious paywall here. $\endgroup$ – YCor Nov 13 '19 at 20:15