## about decomposition of three forms

patrick D Baier in his PhD thesis in chapter2 in page 14 for proving the theorem 2.1.4 used of following non-trivial fact

Let $0\neq X\in V$(here $V$ is of dimension 6) , $W^*=Ann(X)$ and $\Omega\in\wedge ^3V^{\star}$.Then we can find unique elements $\psi\in\wedge ^2V^{\star}$ and $\phi\in\wedge ^3V^{\star}$ such that $\Omega=\theta\wedge\psi+\phi$ (I think this decomposition is from hitchin in his paper about geometry of three forms in six dimension) so that $i_X\Omega=\psi$ (why?) and moreover there are unique elements $r\in\mathbb{R}$ and $Y\in Ker(\theta)$ (here $\theta\in V^*$ with $\theta (X)=1$) such that (why?)we have

1) $i_{(rX)}v=\psi\wedge \phi$

2) $i_Yv_0=\psi\wedge\psi$

(here $v_0\in\wedge^5W^*$ is the unique element such that $v=\theta\wedge v_0)$

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From the thesis, I see that one is supposed to make some choices of $v \in \Lambda^6 V^*$ and $\theta \in V^*$. $i_X v$ and $\psi \wedge \phi$ are both in $\ker\; i_X \cong \Lambda^5 W^* \subset \Lambda^5 V^*$, so must be proportional. $\ker \theta \to \Lambda^4 W^*, \; Y \mapsto i_Y v_0$ is an isomorphism. – Johannes Nordström Feb 3 at 10:12
Dear @Johannes, why we have $keri_X≅Λ^5W^∗$ – Hassan Jolany Feb 3 at 14:12