I don't know if this question is appropriate to this site, but I posted [here](https://math.stackexchange.com/questions/58568/on-a-remark-in-foundations-of-mechanics-2nd-edition-by-abraham-and-marsden) without an answer, so I tried this alternative.


Given a $2$-form $\omega$ on a manifold $M$, let us denote by $N$ the kernel of $\omega$, i.e. $N:=\{u\in TM : \omega(u,\cdot)=0\}$. Their Proposition 5.1.2 shows that if $\omega$ has constant rank (and is closed) then $N$ is a tangent distribution on $M$ (and completely integrable).


In the following remark they say that ``the reader can easily prove the converse of the previous conclusion''. While I understand that $N$ is a tangent distribution if and only if $\omega$ has constant rank. Instead I think that, for $\omega$ of constant rank, $N$ can be completely integrable even if $\omega$ is not closed, (e.g. $\omega=e^z dx\wedge dy$).


Starting from this consideration I have asked myself a question:  

Given a $\Omega\in\mathcal{A}^p(M)$, with $p>1$, whose rank is constant, let us define its kernel $N$ as above. Evidently $N$ is a tangent distribution on $M$, and I find it is completely integrable at least when there exists a $1$-form $\phi$ such that $d\Omega=\phi\wedge\Omega$. Clearly, if $\Omega$ is decomposable then the last condition is even necessary.


My question (**edited after the comment of Willie Wong**):



> 
> Is this last condition (the ``divisibility'' of $d\Omega$ by $\Omega$) necessary for the complete integrability of $N$ even when $\Omega$ is not decomposable? (Using Frobenius' Theorem I understand the case $p=1$, but what about the case $p>1$?.)
> 
> 
> 


Any suggestion and\or counterexample are welcome.



On a remark in Foundations of mechanics, 2nd Edition, by Abraham and Marsden
The answer is 'no'. To see why, just take any nondegenerate $2$-form $\omega$ on, say, $\mathbb{R}^6$, that has the property that $d\omega$ is not a multiple of $\omega$. (This will be true for a generic such $2$-form.) The kernel of this $\omega$ is trivial, but now, you can just regard it as being defined on $\mathbb{R}^8$, say, by pullback. Then the kernel has (constant) positive dimension and is obviously integrable, but $d\omega$ is still not a multiple of $\omega$. 


By the way, the theorem about integrability is true in much greater generality than for a single $p$-form: If $\mathcal{I}\subset\Omega^\*(M)$ is a graded ideal that is closed under exterior derivative and its `kernel' $N$, which is defined as the set of vectors $v\in T\_xM$ such that $\iota(v)\phi$ is in $\mathcal{I}\_x$ for all $\phi\in \mathcal{I}\_x$, has constant rank, then $N$ is integrable. This is a classic theorem of Cartan on 'Cauchy characteristics', and can, for example, be seen in *Exterior Differential Systems* by Bryant, et al.


