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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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# 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?