# Can every non-compact Kähler manifold be realized as the analytification of a smooth variety?

I was just thinking about how we have nice theorems relating compact K?hler manifolds to the algebraic setting, but I was wondering if anything interesting holds in the non-compact case?

i.e. Given a non-compact K?hler manifold,$$M$$, does there exist a smooth variety $$X$$ such that $$M \cong X^{an}$$.

• No, for example $\mathbb{C}-\mathbb{Z}$ wouldn't be, because analytic spaces associated to algebraic varieties would have finitely generated homology. Even if you imposed finiteness conditions, it would fail, e.g. take the disk.... – Donu Arapura Feb 21 at 3:04
• Also, by Hartog's theorem, for the open complement $M$ in a compact, Kaehler manifold $\overline{M}$ of a proper closed analytic subvariety $\partial \overline{M}$ of codimension $\geq 2$, the field of meromorphic functions on $M$ equals the field of meromorphic functions on $\overline{M}$. Thus, if $\overline{M}$ is not Moishezon, i.e., if it is not projective, then the open complement $M$ cannot be Moishezon. Thus, $M$ is not biholomorphic to the underlying complex analytic space of a finite type, separated, complex algebraic space. – Jason Starr Feb 21 at 3:11