# $q_{S^*\omega}(X)=S^{\ast}q_{\omega}(X)$ ?

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 $q_{S^*\omega}(X)=S^{\ast}q_{\omega}(X)$ ?

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[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 $S^\star(i_X\omega) = i_{S^{-1}X}(S^\star\omega)$.