Let $ M $ be a compact 3-manifold which is the total space of a fiber bundle (with connected nontrivial fiber and base) where neither the base nor the fiber is a hyperbolic surface. Then $ M $ always admits a Thurston geometry.
Here is a proof:
For circle bundles over a surface $ \Sigma $ $$ S^1 \to M \to \Sigma $$ A closed 3-manifold $ M $ is a Seifert fiber space if and only if it admits one of $ S^3,S^2 \times E^1,E^3,Nil,\tilde{SL_2},H^2 \times E^1 $ geometry (this is Theorem 1 of https://personal.math.ubc.ca/~mihmar/SeifertGeometry.pdf). A circle bundle is a trivial type of Seifert fiber space (it is exactly the case when the base 2 dimensional orbifold is actually a 2 dimensional manifold). So every circle bundle over a closed surface admits one of the six fibered geometries. Since our base space is non hyperbolic then the total space of the bundle has solvable $ \pi_1 $ so the geometry must in particular be one of $ S^3,S^2 \times E^1,E^3,Nil $.
For $ T^2 $ bundles over $ S^1 $ $$ T^2 \to M \to S^1 $$ we have by this question https://mathoverflow.net/questions/416611/torus-bundles-and-compact-solvmanifolds that $ M $ must be homeomorphic to a solvmanifold (thus diffeomorphic since we are in dimension 3). Then by this question https://mathoverflow.net/questions/414531/3-dimensional-solvmanifolds-and-thurston-geometries/416509#416509 we can conclude that $ M $ admits a geometry (specifically one of $ E^3, Nil,Solv $).
The two $ S^2 $ bundles over the circle $$ S^2 \to M \to S^1 $$ correspond to the trivial bundle and the mapping torus of the antipodal map of $ S^2 $. Both manifolds admit $ S^2 \times E^1 $ geometry.
The only $ \mathbb{R}P^2 $ bundle over the circle $$ \mathbb{R}P^2 \to M \to S^1 $$ is the trivial bundle and it admits $ S^2 \times E^1 $ geometry. It is Riemannian homogeneous.
The four Klein bottle bundles over the circle $$ K \to M \to S^1 $$ are exactly the four compact flat non-orientable 3-manifolds. They admit $ E^3 $ geometry. Exactly two of them are solvmanifolds see, the other two are not Riemannian homogeneous
So my question is whether this same sort of result holds for non-compact three manifolds. Basically the idea is what if the 2d part of the bundle admits a flat or round metric then can we still conclude that the total space of the bundle admits a Thurston geometry.
-If the base is a line or plane then the bundle is trivial so the total space of the bundle admits either $ E^1 \times S^2 $ or $ E^3 $ geometry depending on the geometry of the fiber.
So one of my questions is Let $ M $ be the total space of a bundle with fiber a non-hyperbolic surface ( so only interesting ones left are plane or Moebius strip) and base a circle. Then must $ M $ admit geometry?
And also Let $ M $ be the total space of a circle bundle over a non hyperbolic surface (so only interesting one left is Moebius strip), must $ M $ admit a geometry?
Finally, Let $ M $ be the total space of a line bundle over a non hyperbolic surface (so interesting bases are sphere, projective plane, torus, Klein bottle, Moebius strip). Must $ M $ admit a geometry?