# Trivialization of holomorphic symplectic 2-form

Let $X$ be a holomorphic symplectic compact manifold with a fixed holomorphic 2-form $\omega$. $\omega$ yields an isomorphism $\phi:T_{X} \rightarrow \Omega_{X}$ via $$v \mapsto \phi(v)=\omega(v,-).$$ Given a holomorphic two form $\alpha \cup \beta$, where $\alpha,\beta$ are holomorphic 1-form, we have $\alpha \cup \beta=C\omega$ for some constant $C$. I initially thought $$C=\omega(\phi^{-1}(\alpha),\phi^{-1}(\beta))$$ but I cannot prove this. It there an explicit way to write $C$ down?

Edit Sorry, I implicitly assumed that $\dim_{\mathbb{C}}X=2$ above. A similar formula should hold in general, as Robert says below.

In case $\dim_{\mathbb{C}}X=2$ this is a simple linear algebra and it certainly holds. So please ignore my question.

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@Koopa: Are you assuming that $X$ has (complex) dimension $2$? If not, then the equation $a\cup b = C \omega$ can only hold if $C=0$, since $\omega^2$ is not zero. If the dimension is $2$, then your question is really a linear algebra question. If the dimension of $X$ is $2n>2$, then you might want to define $C$ by something like the equation $$a\cup b \cup \omega^{n-1} = (C/n)\ \omega^n.$$ – Robert Bryant Aug 20 '12 at 22:28
You are right, Robert. I was a bit confused; I initially had K3 surfaces in my mind and thought that a similar formula should hold in general. My question doesn't really make sense. In higher dimensional case, the equation should be something like yours. Thank you very much for pointing this out. – Koopa Aug 21 '12 at 0:29