Characterization of involutiveness

In terms of the ideal

Let $\mathcal{I}(D)$ be the ideal of ${\mathrm{\Omega }}^{\ast }(M)$ given by the dual description of the distribution.

A distribution $D$ is involutive if and only if $\mathcal{I}(D)$ is a differential ideal.

(See [Warner_1983] proposition 2.30)

Proof
Suppose $D$ is involutive, and let $\omega \in \mathcal{I}(D$ be a 1-form. Given $X,Y\in \mathrm{\Gamma }(M,D)$, $d\omega (X,Y)=0$, by the infinitesimal Stokes' theorem. Therefore, by using proposition lemma of belongingness to ideal, $d\omega \in \mathcal{I}(D$. Since $\mathcal{I}(D)$ is generated by 1-forms, it is true for any of its elements.

Conversely, suppose $\mathcal{I}(D)$ is a differential ideal, and take $X,Y\in \mathrm{\Gamma }(M,D)$. To see that $[X,Y]\in \mathrm{\Gamma }(M,D)$ observe that $d\omega (X,Y)=0$ and so $\omega ([X,Y])=0$ for any $\omega \in \mathrm{\Gamma }(M,D^{\ast })$. Therefore $[X,Y]\in \mathrm{\Gamma }(M,D^{\ast }$.$◼$

In terms of the Pfaffian system

It can also be characterized in terms of $D^{\ast }$ (see dual description of the distribution): for every $\omega \in D^{\ast }$ it must be

for every $X,Y\in D$. See [Lychagin_2021] page 5/77.