Patrick D. Baier in his [Ph.D. thesis](https://people.maths.ox.ac.uk/hitchin/hitchinstudents/baier.ps.gz) for proving the theorem 2.1.4 used the following non-trivial fact (in chapter 2 on page 14):


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


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


2) $i\_Y v\_0 = \psi\wedge\psi$ 


(here $v\_0\in\wedge^5 W^\ast$ is the unique element such that $v=\theta\wedge v\_0$)



about decomposition of three forms
I have no idea what the second half of the statement is, but here's what I think the first half says:


If $X \in V$ and $\theta \in V^\*$ such that $\langle \theta, X\rangle \ne 0$, then given any nonzero $\Omega \in \Lambda^3V^\*$, there exists unique elements $\psi \in \Lambda^2V^\*$ and $\phi \in \Lambda^3V^\*$ such that the following hold:


* $\Omega = \theta\wedge\psi + \phi$
* $i(X)\Omega = \psi$


To prove this, simply extend $\theta$ to a basis of $V^\*$, $\theta\_1, \dots, \theta\_n$, where $\theta\_1 = \theta$ and $\theta\_2, \dots, \theta\_n$ comprise a basis of $X^\perp$ and expand $\psi$ and $\phi$ with respect to the bases induced on $\Lambda^2V^\*$ and $\Lambda^3V^\*$.


