integrability of ker $\omega$ in symplectic case

Question

How can we prove that if $(M,\omega)$, is pre-symplectic and d$\omega=0$ then ker$\omega$ is integrable?.

Answer 1

In more invariant language, you can see this as follows: let $X,Y$ be vector fields contained in $\ker\omega$. Then $$ \mathcal{L}_X\,\omega = d(i_X\omega)+i_X(d\omega) = 0 $$ using Cartan's Magic Formula. Then, using the identity (see e.g. here) $$ i_{[X,Y]}\omega = [\mathcal{L}_X,i_Y]\,\omega$$ we see that $i_{[X,Y]}\omega = 0$. It follows that $\ker \omega$ is involutive, and hence by Frobenius' Theorem, integrable. Application of Frobenius' Theorem requires that $\ker\omega$ defines a subbundle of $TM$, which is equivalent to $\omega$ having constant rank.