A $2$- form $\alpha$ on a n- manifold $M$ is called anti symplectic if for every $x\in M$, $\{ v\in T\_{x} M \mid i\_{v} \alpha=0 \}$ is a $n-2$ dimensional subspace of $T\_{x}M$. So we obtain a $n-2$ dimensional subbundle of $T M$. We denote it by $\tilde{\ker} \;\alpha$


*The first question:*



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> Under what condition on $\alpha$, the distribution $\tilde{\ker} \;\alpha$ is integrable?
>  (I am motivated by Frobenious condition $\alpha \wedge d\alpha=0$, so I search for an algebraic condition)
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*The second question:*



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> Assume that $\omega$ is a symplectic 2-form on a 2n- manifold. Can we write $\omega$ in the global form $\omega=\sum\_{i=1}^{n} \alpha\_{i}$ where each $\alpha\_{i}$ is anti symplectic form.(I think that the local argument and then using partition of unity, does not work)
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Anti_symplectic 2-forms
For your first question: The condition for integrability of $\mathrm{ker}\alpha$ when $\alpha$ is decomposable is that $\mathrm{d}\alpha = \lambda\wedge \alpha$ for some $1$-form $\lambda$.


For your second question: The answer is 'no' in general. If this can be done, then the tangent bundle of $M$ can be written as the direct sum of $n$ oriented $2$-plane bundles (i.e., complex line bundles), and it is easy to give examples of manifolds for which such a decomposition cannot hold. For example, just take $\mathbb{CP}^2$ with its standard symplectic structure. Its tangent bundle is not the sum of two complex line bundles of this kind for reasons having to do with its Chern and Euler classes.


