Homogeneous Manifold Deformation Retracts Onto Compact Submanifold

Question

Let G be a connected Lie group. Then by 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 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?

Answer 1

Consider the following example. Let $G=SL(2, {\mathbb R})$ and $\Gamma< G$ a discrete subgroup which is free of infinite rank. Form the quotient manifold $M=G/\Gamma$. Then $G$ acts on $M$ via left multiplication: The action is smooth and transitive, thus, $M$ is homogeneous. But $M$ has non-finitely generated fundamental group, hence, cannot be homotopy-equivalent to a compact manifold.

Of course, the answer is different for Riemannian homogeneous manifolds, i.e. for Riemannian manifolds $M$ admitting transitive isometric Lie group actions $G\times M\to M$. Such an action has compact point-stabilizer $H< G$ and, thus, $M$ is homeomorphic to $G/H$. Taking a maximal compact subgroup $K< G$ containing $H$, we see that $G/H$ is homotopy-equivalent to $K/H$, which is a compact manifold.

Edit. Here is a semi-explicit construction. Start with countably many round circles $C_n, n\in {\mathbb Z}- \{0\}$, in the complex plane ${\mathbb C}$, whose centers lie on the x-axis and which bound pairwise disjoint open disks. For instance, take centers which are even integers and unit radii. For each circle $C=C(a,r)$ in the complex plane define the inversion $J_C$ in this circle by the formula: $$ J_C(z)= \frac{r^2}{\bar{z} +a} -a. $$ Now, for each $n$ let $g_n$ denote the composition of the inversions $$ g_n=J_{C_n}\circ J_{C_{-n}}. $$ These will be linear-fractional transformations of the extended complex plane preserving the upper half-plane: $$ g_n(z)= \frac{a_n z+b_n}{c_n z+ d_n}, a_nd_n -b_n c_n=1, a_n, b_n, c_n, d_n\in {\mathbb R}. $$ I leave it to you to compute the coefficients in terms of the centers and the radii.

It is a standard fact that the transformations $g_n$ freely generate a discrete subgroup of $PSL(2, {\mathbb R})$. My favorite reference for this staff is Beardon's book "The Geometry of Discrete Groups."