about decomposition of three forms

Question

Patrick D. Baier in his Ph.D. thesis 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 $v=\theta\wedge v_0$)

Answer 1

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 $\Lambda^3V^*$.

Answer 2

Dear @Hassan, about the first question: why $\iota_X \Omega=\psi$ please note that $\psi\in\Lambda^2 W^* $ and $\phi\in\Lambda^3 W^* $. So the identity comes from $\theta(X)=1$. About the second question me too I'd like to know why.

Cheers