Integrable subbundle

Question

Let $D\subset TM$ is a integrable smooth regular subbundle and $f_{1},...f_{k}$ is smooth local frame for $Ann(D)$. Why $\omega=f_{1}\wedge ....\wedge f_{k}$ is closed form?

Answer 1

It is not necessarily closed. For example, take $M=\mathbb R^3$ with coordinates $(x,y,z)$, and let $D$ be the subbundle spanned everywhere by $\partial/\partial z$. We could take $f_1 = dx$ and $f_2 = (z^2 +1) dy$, and then $f_1\wedge f_2$ is not closed.

What is true is that in a neighborhood of each point, it is always possible to find $1$-forms $f_1,\dots,f_k$ whose wedge product is closed. In fact, it's always possible to have each $f_i$ individually closed. This follows from the Frobenius theorem -- in a neighborhood of each point, there are coordinates $(x^1,\dots,x^n)$ such that $D$ is annihilated by $dx^1,\dots,dx^k$.