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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. $\omega=e^z dx\wedge dy$).

My first question is: Do I have misunderstood something, or not?

Given a $\omega\in\mathcal{A}^p(M)$ 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 $d\omega=\phi\wedge\omega$.

My second question is : what is a necessary and sufficient condition for the complete integrability of $N$?

Any suggestion is welcome.

share|cite|improve this question
your counter example semms convincing. – Olivier Bégassat Aug 19 '11 at 22:57
@Olivier Bégassat: I have added a second issue related to the same remark in Abraham and Marsden. Do you have some suggestion? – Giuseppe Aug 22 '11 at 19:10
Also posted to MO:… – Willie Wong Aug 28 '11 at 21:30

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