I am trying to show a claim that , given a smooth 2-plane distribution D (a subbundle of the tangent bundle of) of a 3-manifold $M^3$, there is a 1-form $w$ generating the plane distribution locally, i.e., the 2-plane distribution is the kernel of a 1-form, and that if $M^3$ is oriented, the 1-form can be defined globally. It seems clear that, locally, we can just define a linear map from $\mathbb R^3 \rightarrow \mathbb R$ whose kernel is the plane assigned at p. I am having trouble understanding how/why orientability would allow a global definition of the form. Can someone suggest something, please?
Thanks.
Given a 2-plane distribution D in $ M^3$ ( 3-manifold, Contact Structure) , find a 1-form Generating D.
contact-topologydifferential-geometrytangent-bundle
Best Answer
The right condition is not that $M$ is orientable, but that the distribution $D$ is co-orientable in $M$, i.e. the quotient bundle $TM/D \to M$ ("normal bundle") is trivial.
Said more prosaically, you can find such a global one-form if and only if there exists a globally defined vector field on $M$ which is transverse to $D$. This is quite easy to understand: you can just define a one-from by declaring that its kernel is $D$, and it must be equal to $1$ on some vector spanning a complementary subspace.
Although I don't have immediate examples at hand, I believe that the distribution could be co-orientable without $M$ being orientable, and vice-versa.