Frobenius condition in terms of Lie brackets Let $\alpha$ be a $1$-form and $\xi = \ker \alpha$.
Frobenius theorem tells us that $\xi$ is integrable iff $\alpha\wedge{\rm d}\alpha = 0
.$
In the book "Introduction to Contact Topology" from Hansjorg Geiges, he claims (page $3$) that in terms of Lie bracket this is equivalent to $[X,Y] \in \xi$ $\forall ~X,Y \in \xi$, where $X \in \xi$ means that $X$ is a smooth vector field and $X_p \in \xi_p\ \forall p \in M$
I don't understand why that claim is true. A proof of this fact would be most welcomed.
Thank in advance.
 A: We use that $${\rm d}\alpha(X,Y) = X(\alpha(Y)) - Y(\alpha(X)) -\alpha([X,Y]). \qquad (\ast)$$
Suppose that $\alpha \wedge{\rm d}\alpha = 0$ and let's check that $\xi$ is closed by the Lie bracket. Let $X,Y \in \xi$. Then $(\ast)$ becomes ${\rm d}\alpha(X,Y) = -\alpha([X,Y])$. Evaluating $\alpha \wedge {\rm d}\alpha$ at the triple $([X,Y],X,Y)$ and using again that $Y\in \xi$ gives $$0 = -\alpha([X,Y])^2,$$whence $[X,Y] \in \xi$.
Now suppose that $\xi$ is closed by the Lie bracket and let's check that $\alpha\wedge {\rm d}\alpha = 0$. Take $p \in M$. If $\alpha_p = 0$, done. Otherwise take a local frame $X_1,\cdots,X_n$ around $p$ with the $n-1$ first ones in $\xi$. It suffices to check that $\alpha \wedge {\rm d}\alpha $ evaluated at $(X_i,X_j,X_k)$ is zero. If the index $n$ repeats, done. If the index $n$ does not appear, done. If the index $n$ appears once, we can suppose that $k = n$ and $i,j < n$. Then $$(\alpha \wedge {\rm d}\alpha)(X_i,X_j,X_n) = \alpha(X_i){\rm d}\alpha(X_j,X_n) - \alpha(X_n) {\rm d}\alpha(X_i,X_j) = 0,$$since $\alpha(X_i) = 0$ because $X_i \in \xi$ and ${\rm d}\alpha(X_i,X_j) = 0$, since $X_i,X_j \in \xi$, $\xi$ is closed under Lie bracket, and we use $(\ast)$ again.
