noncompact Riemannian homogeneous is trivial vector bundle over compact homogeneous

Question

Is it true that a manifold $ E $ admits a metric with respect to which the isometry group is transitive ($ E $ is Riemannian homogeneous) if and only if $ E $ is the total space of a $ K $ equivariant vector bundle where $ K $ is a compact group acting transitively on the base $ B $ of the bundle?

If the vector bundle $ E $ is not assumed equivariant then this fails. The Moebius strip is the total space of a vector bundle but it is not Riemannian homogeneous.

This question has a slightly similar flavor to:

Homogeneity of a projective vector bundle

Answer 1

The only if direction fails. That is, there are $K$-equivariant vector bundles which are not homogeneous. For example, the Mobius band has the form $O(2)\times_{O(1)} \mathbb{R}$, and is not Riemannian homogeneous as you mention.

On the other hand, it seems the if direction is true. In fact, I think I can prove that $M$ must be diffeomorphic to the trivial bundle over a compact homogeneous space, at least if $M$ is connected.

Let $G$ denote the identity component of the isometry group (which still acts transitively). Fix a point $p\in M$ and let $G_p$ denote the isotropy group. Then $G_p$ is compact. This follows because the action is proper (see, e.g, this paper), and then $G_p\times \{p\}\subseteq G\times M$ is the inverse image of the compact set $\{(p,p)\}\subseteq M\times M$ under the map $G\times M\rightarrow M\times M$ given by $(g,m)\mapsto(gm,m)$. Moreover, because the action is proper, we have a diffeomorphism $M\cong G/G_p$.

Writing $H$ for the maximal compact subgroup of $G$, we therefore have (up to conjugacy) inclusions $G_p\subseteq H\subseteq G$. From this, one can form the fiber bundle $H/G_p\rightarrow G/G_p\rightarrow G/H$.

Now, $G/H$ is diffeomorphic to Euclidean space, which is contractible. Thus, this bundle is trivial, so $G/G_p$ is diffeomorphic to $(H/G_p)\times (G/H)$.