Let $V$ be a complex vector space of dimension $2n$, with a nondegenerate $2-$form $\omega$. Then the differential form $\alpha_v(-)=\omega(v,-)$ at $v\in\mathbb{P}(V)$ is a (local) contact form on $\mathbb{P}(V)$.
I want to show that $d\imath_v\omega=\omega$ then I can get the result, but I cannot show this. Suppose I take a vector space of dimension $2$, and I take the form to be $dz_1\wedge dz_2$, then I take $v=(1,0)$, so I get $d\imath_v\omega(v)=0$. I think I make a mistake somewhere, can someone figure it out?
Updated: I find out the above one is not a problem as the space is one dimensional. So I should say this: take $V$ to be dimension $2n$, and $\omega=\sum dx_i\wedge dy_i$, where $dx_i,dy_i$ are holomorphic. Then WLOG suppose $v=(1,0,...,0)$, which is dual to $dx_1$. Then $d\imath_v\omega(v)=0$.