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Is there an example of a smooth $n$-manifold $M$ whose tangent bundle is nontrivial as a bundle but is nonetheless (abstractly) diffeomorphic to the trivial bundle $M \times \mathbb{R}^n$?

(This question was inspired by Trivial fiber bundle.)

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A trivial observation: such an isomorphism would cover an automorphism of $M$. A possibly unhelpful observation: One could try to put a Riemannian metric on $M$ and see what constraints this gives in the light of my first remark. – David Roberts Mar 18 '11 at 0:12
@David, is no reason for the diffeomorphism $TM\to M\times \mathbb R^n$ to cover a diffemorphism of $M$. @Faisal, the answer to your question is almost certainly yes, but I do not see an explicit example at the moment, cf. my answer to – Igor Belegradek Mar 18 '11 at 1:22
I think I can show that the unit tangent bundle over $S^5$ is diffeomorphic to the trivial one... but I'm having trouble otherwise, since most of the diffeomorphisms I can think of rely on theorems about compact spaces. – Dylan Wilson Mar 18 '11 at 6:30
@Dylan - That was the example I was thinking of too! It is clear that $TS^5$ is not the trivial rank $5$ bundle over $S^5$. It only has $3$ linearly independent nowhere vanishing vector fields (for lack of anything elementary one can use Adams' vector fields on spheres result). Although I don't know how to do it myself, if you can prove (extend your sphere bundle result to vector bundles) your claim, we're done! – Somnath Basu Mar 18 '11 at 15:36
@Dylan and @Somnath: Tangent bundles over spheres never give such examples. Moreover, in general, if the homotopy-equivalence $M\to M$ induced by a diffeomorphism $TM\to M\times {\mathbb R}^n$ is homotopic to a diffeomorphism $M\to M$, then $TM$ is trivial. See "Diffeomorphism of total spaces and equivalence of bundles" by De Sapio and Waldschap. This still leaves open the case when $M$ is a homotopy-sphere though. – Misha Apr 2 '12 at 0:54

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