Homogeneous universal cover implies existence of a geometric structure

Question

Let $ M $ be a compact connected manifold with universal cover $ \widetilde{M} $. Suppose that $ \widetilde{M} $ is homogeneous. That is, there exists a Lie group acting transitively on $ \widetilde{M} $. Then does there exist a geometric structure on $ M $ in the sense of the wikipedia page on the Thurston geometries (https://en.wikipedia.org/wiki/Geometrization_conjecture#The_eight_Thurston_geometries)?

That is, does there exists a Lie group $ G $ with subgroups $ K $ and $ \Gamma $ such that $ K $ is compact, $ G/K $ is simply connected, $ \Gamma $ is discrete, and $$ \Gamma \backslash G/K \cong M $$

This is certainly true for dimensions 1 (the circle) and 2 (all surfaces are constant curvature see for example this question https://mathoverflow.net/questions/198100/do-all-surfaces-2d-riemanian-manifolds-admit-constant-curvature and thus are locally symmetric and indeed space forms see my question Is every surface locally symmetric?). Is it true for dimension 3 because of Thurston's Geometrization conjecture? Is it true for $ n \geq 4 $ ?

Answer 1

Let $S$ be the compact connected orientable genus 1 surface with two boundary circles $c_1, c_2$. Consider the product manifold $S\times S^1$. Its boundary is the disjoint union of two tori $T_i=c_i\times S^1, i=1, 2$. Pick points $p_i\in c_i$ and $q\in S^1$, $i=1,2$. Consider a homeomorphism $f: T_1\to T_2$ which sends $c_1\times \{q\}$ to $\{p_2\}\times S^1$ and $\{p_1\}\times S^1$ to $c_2\times \{q\}$. The most important thing about $f$ is that the unoriented loop $f(\{p_1\}\times S^1)$ is not isotopic to the unoriented loop $\{p_2\}\times S^1$ in the torus $T_2=c_2\times S^1$. It follows that the manifold $M$ obtained from $S\times S^1$ by gluing its boundary components via $f$ is not a Seifert manifold. Its JSJ decomposition is given by splitting it open along the image $T$ of $T_1$ (and $T_2$) in $M$. Since $S\times S^1$ is irreducible and has incompressible boundary, the manifold $M$ is also irreducible. Moreover, $M$ is Haken: It contains incompressible torus $T$. Hence, its universal cover is homeomorphic to $R^3$.

Lastly, one proves that $M$ does not admit any of Thurston's geometries. Since $M$ is not Seifert, one only needs to observe that $M$ is not hyperbolic (because it contains an incompressible torus) and does not admit Sol-geometry. The latter is, e.g. because $\pi_1(M)$ contains $\pi_1(S)$ which is a nonabelian free group, while fundamental groups of Sol-manifolds are solvable.