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?

  • 1
    $\begingroup$ Nope. It's the same argument in dimension $1$, if you replace plane field by line field. $\endgroup$ Nov 10, 2014 at 23:56
  • $\begingroup$ Can you explain it carefully? $\endgroup$ Nov 11, 2014 at 11:14
  • $\begingroup$ 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. $\endgroup$ Nov 11, 2014 at 17:20
  • $\begingroup$ Ok, but if my 3-manifold (or my surface) is orientable? Your example is on a non orientable surface after all. $\endgroup$ Nov 11, 2014 at 18:51
  • $\begingroup$ 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. $\endgroup$ Nov 12, 2014 at 4:04


You must log in to answer this question.

Browse other questions tagged .