My (hopefully) simple question is this: Given an involutive ($C^1$) plane field $P$ on a manifold $M$, say an open subset of $R^n$, generated by a linearly independent set of vector fields $(X^1,\ldots,X^k)$, does there always exist a set of vector fields $(Y^1,\ldots,Y^k)$ generating $P$ for which the Lie bracket $[ Y^i,Y^j]$ of any two is zero; i.e. the vector fields commute?
I've seen this question but that is for a specific pair of vectorfields. I've also seen a proof of the Frobenius theorem in which the author, at each point of $M$, finds a local chart in which the spanning vector fields are the first $k$ coordinates, but it is not clear (to me, at least) whether or not the construction is local or global.
I would welcome any responses that help me figure this out.
Edit: A related question is, if this is not possible in general, suppose I knew the leaves of the foliation were proper and each was the graph of a function on some domain in $R^n$. Would that make a difference?
Best Answer
The construction given by the proof of Frobenius is indeed local. There will be topological restrictions that do not allow what you want globally. For example, if the leaves of the foliation are even-dimensional spheres, then they admit no nowhere-zero vector field, let alone a global basis of vector fields.