Definition: Let $(V,\Omega)$ be a symplectic vector space, we define
$\perp:\Lambda ^k(V^\*)\to\Lambda ^{k-2}(V^{\ast})$
by $\perp(\omega)=i\_{X\_{\Omega}}(\omega)$ 


here if $(e\_1,e\_2,...e\_n,f\_1,f\_2,...f\_n)$ is basis of symplectic vector space then $\Omega=e\_{1}^{\*}\wedge f\_{1}^{\*}+...+e\_{n}^{\*}\wedge f\_{n}^{\*} $
and $\ X\_\Omega =e\_{1}\wedge f\_{1}+...+e\_{n}\wedge f\_{n}$
and $X\in V $ and $\omega\_X:=i\_X(\omega)$ .


**Question**: Let $Sp(V,\Omega)$ be symplectic group, then I want to see the operator 
$q\_{\omega}(X)=\perp^2(\omega\_X\wedge\omega\_X)$ is invariant under $Sp(V,\Omega)$-action
In fact if $S\in Sp(V,\Omega) $ 


then have we $q\_{S^\*\omega}(X)=S^{\ast}q\_{\omega}(X)$ ?



$q_{S^*\omega}(X)=S^{\ast}q_{\omega}(X)$ ?
[Edit: I repeat the question.
$X\in V$. 


$X\_{\Omega} = \Omega^{-1}$, the dual symplectic structure on $V^\*$.


$\Lambda^2S(X\_\Omega)=X\_{\Omega}$ for all $S\in Sp(V)$.


$q\_\omega(X) = i\_{X\_{\Omega}}i\_{X\_{\Omega}}(i\_X\omega\wedge i\_X\omega).$]


Then
$S^\star(q\_\omega(X)) = q\_{S^\star\omega}(S^{-1}X)$ 
since $S^\star(i\_X\omega) = i\_{S^{-1}X}(S^\star\omega)$.


