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Let $\pi:E\to M$ be a vector bundle over a closed smooth manifold and supose $\Pi:F\to E$ is a fibre bundle over the total space of $\pi$. I'd like to know if, restricted to $E_p$, the second bundle $Pi$ is trivializable (moreover, homogeneous) as a fibre bundle over the vector space $E_p$. I belive this is true if $\Pi$ is a vector bundle over $E$, but I haven't written down a proof yet. I'll appreciate any help. Thanks.

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Since $E_p$ is a vector space, it is contractible. This should be enough to guarantee that any fibre bundle over $E_p$ is trivializable. – Somnath Basu Nov 18 '10 at 0:07
Are you supposing $\Pi$ is a principal bundle? If not, then it can't be homogeneous, at least by how I understand homogeneous bundles (=quotients of Lie groups by subgroups). And As Somnath says, locally trivial bundles over a fin. dim. vector space (with the usual topology) are globally trivial. – David Roberts Nov 18 '10 at 0:16
up vote 2 down vote accepted

If $E \to M$ is a vector bundle then it is a weak equivalence. (being a vector bundle means it is a fibration and then look at the LES in homotopy and as Somnath points out the fiber is contractible.) So having a bundle on E is the same as having a bundle on M up to homotopy.

But as for your question, up to isomorphism there is only one vector bundle over a contractible space. It is definitely trivial over each fiber of the vector bundle you started with.

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@Sean: you are killing flies with a sledgehammer. For any vector bundle, the zero section is a homotopy inverse to the bundle projection. – Johannes Ebert Nov 18 '10 at 8:52
You are right, that is even better. My first thoughts are typically "is there an LES we can use?" Not to mention you then need to apply Whiteheads Thm. – Sean Tilson Nov 18 '10 at 12:44
Thanks everyone for your comments, thanks Sean for your answer. @David Roberts: by homogeneous I mean that the trivialization of the bundle is given by the following diffeomorphisms of the base: $x\mapsto x+y$, that is, homogeneous under the action of the vector space as an additive group on the total space given by translations. Apparently what it takes for such a "meta-bundle" to be homogeneous in this sense is that it is diffeomorphic to the fibred product of the two bundles over $M$. – Óscar Guajardo Dec 1 '10 at 21:43

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