Let $X$ be a complex manifold endowed with a holomorphic closed 2-form $\omega$ whose associated map $\omega : TX \to T^\*X$ is invertible. Can we always embed $X$ as an open subset of a compact complex manifold $Y$ endowed with a holomorphic Poisson structure $\pi : T^\*Y \to TY$ such that $\pi|\_X = \omega^{-1}$? We may assume that $X$ is quasi-projective, or other reasonable assumptions, if that helps.


Note that this holds, for example, if $X$ is the cotangent bundle of a compact manifold.



When does a holomorphic symplectic manifold compactify to a Poisson manifold?
No, not even under the nicest possible algebraicity assumptions like quasiprojective.


Let $Z$ be the product of two curves of genus $\geq 2$. Choose a nonzero $2$-form $\omega$ on $Z$, the wedge of a nonzero one-form on each of the two curves. Let $X$ be obtained from $Z$ by removing the locus where $\omega$ vanishes. Then $X$ clearly has a nowhere vanishing (and thus invertible as $\dim X=2$) two-form.


Now any compactification $Y$ of $X$ is a compact complex manifold birational to the algebraic variety $Z$, thus is Moishezon, and because it's a surface, [must be projective algebraic](https://mathoverflow.net/a/51752/18060). Since $Z$ is a minimal surface, $Y$ is the blow-up of $Z$ at finitely many points. So if the Poisson structure on $X$ extends to $Y$, then it extends to $Z$ less finitely many points, hence to $Z$.


But $Z$ does not have a nontrivial Poisson structure since $T Z \otimes TZ $ is a sum of line bundles of negative degree and thus has no global sections.


