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Given a (smooth) manifold $M$, are there any sufficient, intrinsic properties that would tell you there exists a (smooth) manifold $N$ such that $M$ is diffeomorphic to $TN?$ There are some obvious necessary conditions (like $M$ needs to be even-dimensional), but I'm more interested in whether any sufficient conditions are known.

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Can a manifold be a tangent bundle in two different ways? –  Mariano Suárez-Alvarez Aug 29 '12 at 17:05
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@ 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. –  Ian Agol Aug 29 '12 at 17:24
    
Ah! Of course. Thanks Agol. –  Mariano Suárez-Alvarez Aug 29 '12 at 17:26
    
It might be worth Googling "almost tangent structure" and looking at some of those papers, for a different perspective. –  Paul Reynolds Aug 29 '12 at 17:36
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@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/… ). –  Marco Golla Aug 31 '12 at 9:48
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2 Answers

Suppose $M$ is an open $2n$-manifold that is homotopy equivalent to a closed smooth $n$-manifold $N$, and suppose $n>2$. Then Haefliger's embedding theorem ensures that the homotopy equivalence $N\to M$ is homotopic to a smooth embedding. Moreover, by Siebenmann's open collar recognition theorem $M$ is diffeomorphic to the normal bundle to this embedding if and only if $M$ is the interior of a compact manifold with boundary such that the inclusion of the boundary induces an isomorphism on the fundamental group. Now it remains to check whether the normal bundle and tangent bundle to the embedding are isomorphic, which of course rarely happens.

A good example is when $N$ is an orientable $3$-manifold and $M=N\times \mathbb R^3$, which is precisely $TN$ because orientable $3$-manifolds are parallelizable. By above arguments, any two homotopy equivalent orientable $3$-manifolds have diffeomorphic tangent bundles. Specific examples can be found among lens spaces, such as $L(7,1)$ and $L(7,2)$.

The case $n=2$ seems more delicate.

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One necessary set of conditions:

Step 1: you need $M$ to have the homotopy-type of a submanifold $N$ half the dimension of $M$. This forces $M$ to be a vector bundle over $N$ with the fibre the same dimension as the base with some reasonable hypothesis (see texts on the h-cobordism theorem, like Kosinski, where these kinds of theorems are proven).

Step 2: Given a vector bundle, how do you know if it is the tangent bundle of the base space? For this you need an exponential map, or to show the vector bundle is diffeomorphic to a tubular neighbourhood of $\Delta N$ in $N \times N$. Or you could compare the classifying map of $TN$ with that of your vector bundle structure on $M$.

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Regarding Step 1: how does being the homotopy type of $N$ forces anything? Don't you need some assumptions like those in my answer? –  Igor Belegradek Aug 29 '12 at 18:13
    
Hi Igor, yes it really depends on what level of generality the question is asked in. Say $M$ is compact, and $N$ is compact without boundary -- so we will head for the conclusion that $M$ is a disc bundle over $N$. This side-steps the issue of ends and puts you into pre-Siebenmann mathematics. The idea is $N$ is embedded in $M$ (as you mention) so the complement of an open tubular neighbourhood of $N$ is $M$ becomes an h or s-cobordism depending on your assumptions, so it's a product. –  Ryan Budney Aug 29 '12 at 20:41
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