Let $M$ be a complex manifold or smooth algebraic variety, and let $X$ be a global vector field which is nonzero over “most” of $M$. Then I know that $X$ determines a foliation of $M$ by curves, possibly with singularities (wherever $X$ vanishes). However, if $X_1, \dots, X_k$ are global vectors fields which are linearly independent over “most” of $M$, then they need not induce a foliation of $M$ by $k$-surfaces, because the vector bundle spanned by $X_1, \dots, X_k$ might not be closed under the Lie bracket. (An obviously necessary condition, because the leaves are immersed submanifolds. It is also sufficient by Frobenius' theorem.)
In some cases that are of interest to me, rather than the vector fields $X_1, \dots, X_k$ themselves, what I actually have is their exterior product $X_1 \wedge \dots \wedge X_k$, i.e., is a single section of $\bigwedge^k TM$. Using this section, is it possible to determine whether the corresponding rank-$k$ subbundle of $TM$ is integrable?
EDIT: I guess another way to phrase my question (perhaps not 100% equivalently) is - does the Lie bracket on $TM$ usefully induce any additional structure on the exterior powers $\bigwedge^k TM$?