Let $\xi \subset \text{T}M$ be a integrable plane field on a smooth 3-manifold (i.e. the tangent field of a foliation). Is it true that $\mathcal{F}$ is coorientable?
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1Nope. It's the same argument in dimension $1$, if you replace plane field by line field. – Ryan Budney Nov 10 '14 at 23:56
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Can you explain it carefully? – Antonio Alfieri Nov 11 '14 at 11:14
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The tangent bundle of the Klein bottle splits as a direct sum of a trivial bundle and a non-trivial bundle. It's that non-trivial line bundle you're interested in, so your problem is equivalent to finding a non-zero vector field on the Klein bottle -- just take the complementary line bundle once you've found that. The vectors field comes from a fixed point free action of $S^1$ on the Klein bottle. – Ryan Budney Nov 11 '14 at 17:20
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Ok, but if my 3-manifold (or my surface) is orientable? Your example is on a non orientable surface after all. – Antonio Alfieri Nov 11 '14 at 18:51
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Orientability isn't enough in general. For 2-manifolds it is, but not for 3-manifolds. For example, $S^1 \times \mathbb R^2$ has an integrable plane field where the foliation consists of a Moebius band and a 1-parameter family of cylinders. You should think of this as a regular neighbourhood of a Moebius band. – Ryan Budney Nov 12 '14 at 04:04