Trivialization of holomorphic symplectic 2-form - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T02:30:33Zhttp://mathoverflow.net/feeds/question/105123http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/105123/trivialization-of-holomorphic-symplectic-2-formTrivialization of holomorphic symplectic 2-formKoopa2012-08-20T21:59:25Z2012-08-21T00:25:38Z
<p>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? </p>
<p><strong>Edit</strong>
Sorry, I implicitly assumed that $\dim_{\mathbb{C}}X=2$ above. A similar formula should hold in general, as Robert says below. </p>
<p>In case $\dim_{\mathbb{C}}X=2$ this is a simple linear algebra and it certainly holds. So please ignore my question. </p>