TITLE: Example of a manifold which is not a homogeneous space of any Lie group
QUESTION [74 upvotes]: Every manifold that I ever met in a differential geometry class was a homogeneous space: spheres, tori, Grassmannians, flag manifolds, Stiefel manifolds, etc.  What is an example of a connected smooth manifold which is not a homogeneous space of any Lie group?
The only candidates for examples I can come up with are two-dimensional compact surfaces of genus at least two.  They don't seem to be obviously homogeneous, but I don't know how to prove that they are not.  And if there are two-dimensional examples then there should be tons of higher-dimensional ones.
The question can be trivially rephrased by asking for a manifold which does not carry a transitive action of a Lie group.  Of course, the diffeomorphism group of a connected manifold acts transitively, but this is an infinite-dimensional group and so doesn't count as a Lie group for my purposes.
Orientable examples would be nice, but nonorientable would be ok too.

REPLY [10 votes]: I would think that many examples from 3-manifold theory would work. Take any compact, oriented, irreducible 3-manifold $M$ whose torus decomposition is nontrivial and has at least one hyperbolic piece. Such examples at least have no locally homogeneous Riemannian metric, as a consequence of Thurston's analysis of the 8 geometries of 3-manifold theory. A specific example of this sort can be obtained from a hyperbolic knot complement in $S^3$ by deleting an open solid torus neighborhood of the knot and doubling across the resulting 2-torus boundary; the doubling torus produces a characteristic $Z^2$ subgroup of $\pi_1(M)$. These examples have universal cover homeomorphic to $R^3$, and so they have trivial $\pi_2$. By the homotopy exact sequence there would be a quotient group $\pi_1(G) / \pi_1(H)$ identified with a subgroup of $\pi_1(G/H)$ whose quotient set is $\pi_0(H)$. Perhaps, in order to get a proof, one can analyze this situation by considering the intersection of the $Z^2$ subgroup with the $\pi_1(G) / \pi_1(H)$ subgroup.