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Aug 31, 2012 at 10:19 comment added johndoe for what it's worth, another necessary condition is orientability.
Aug 31, 2012 at 9:48 comment added Marco Golla @OP: as for some necessary condition, you also need a symplectic structure with lots of Lagrangian sections. @Mariano: among other examples, exotic spheres qualify (see mathoverflow.net/questions/31690/… ).
Aug 31, 2012 at 9:09 history edited Benoît Kloeckner
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Aug 30, 2012 at 4:22 comment added Chris Gerig Haha sorry I read it so quickly that I saw "connected".
Aug 30, 2012 at 4:10 comment added Ian Agol @ Chris: contractible manifolds have trivial tangent bundles (that's basically why I assumed contractibility).
Aug 30, 2012 at 3:37 comment added Chris Gerig Wait, "the tangent bundle is trivializable" isn't a general statement, right? We're implicitly assuming our $M_i$ are 3-manifolds (as the explicit example is)?
Aug 29, 2012 at 17:40 answer added Igor Belegradek timeline score: 9
Aug 29, 2012 at 17:36 comment added Paul Reynolds It might be worth Googling "almost tangent structure" and looking at some of those papers, for a different perspective.
Aug 29, 2012 at 17:29 answer added Ryan Budney timeline score: 5
Aug 29, 2012 at 17:24 comment added Ian Agol @ Mariano: Yes, for example there are contractible manifolds $M_1, M_2$ which are not homeomorphic, but $M_1 \times \mathbb{R} \cong M_2\times \mathbb{R}$. So $TM_1\cong TM_2$ since the tangent bundle is trivializable. E.g. $\mathbb{R}^3$ and the Whitehead manifold.
Aug 29, 2012 at 17:05 comment added Mariano Suárez-Álvarez Can a manifold be a tangent bundle in two different ways?
Aug 29, 2012 at 16:50 history asked Jon Cohen CC BY-SA 3.0