Timeline for Nontrivial tangent bundle that is diffeomorphic to the trivial bundle
Current License: CC BY-SA 2.5
8 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 19, 2017 at 2:43 | comment | added | user21574 | $T^*G\cong G\times \mathfrak g^*\cong G\times \mathbb R^{n}$ | |
| Apr 2, 2012 at 0:54 | comment | added | Misha | @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. | |
| Mar 18, 2011 at 15:36 | comment | added | Somnath Basu | @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! | |
| Mar 18, 2011 at 6:30 | comment | added | Dylan Wilson | 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. | |
| Mar 18, 2011 at 1:22 | comment | added | Igor Belegradek | @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 mathoverflow.net/questions/58685/trivial-fiber-bundle. | |
| Mar 18, 2011 at 0:12 | comment | added | David Roberts♦ | 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. | |
| Mar 17, 2011 at 23:25 | history | asked | Faisal | CC BY-SA 2.5 |