“Logarithmic” form of Kodaira Embedding

Question

Suppose you have a non-compact complex manifold $X$ with a Hodge metric, whose associated Kahler form has integral cohomology class. In the compact case, one would be able to conclude that $X$ is projective by the Kodaira embedding theorem. Are there some assumptions on the metric (e.g. describing the behavior near the boundary), short of fully assuming the existence of a compactification of $X$ and extension of the metric, that ensures $X$ is quasiprojective?

Answer 1

Sure, you should assume that there are sufficiently many functions which grow polynomially. Here is an example of such a result.

THEOREM: Let $M$ be a Stein variety equipped with a Kahler metric outside of singularities and an action of $Z$ by contracting homotheties. Consider the ring $R$ of $Z$-finite functions (a function is $Z$-finite if it generates a finite-dimensional $Z$-module). Then $Spec (R)$ is an affine variety, and $M$ is its analytization. Moreover, this algebraic structure is independent from the $Z$-action.

This theorem can be used to characterize algebraic cones, that is, cones of projective orbifolds which are smooth outside of origin.

More generally, if you manage to find sufficiently many holomorphic functions which grow to infinity near the boundary, the same conclusion can be obtained. This should lead to many theorems of form "let M be a Stein manifold with a certain type of a discrete group $\Gamma$ action, then $M$ is the analityzation of $Spec(R)$, where $R$ is the ring of $\Gamma$-finite functions".

In another direction, I suspect also that any complex ALE space (a complex manifold with flat structure at infinity and a Kahler, Calabi-Yau or a hyperkahler metric which is asymptotic to flat) is algebraic. This is known for hyperkahler ALE spaces in dimension=2 from the classification, but I could not find such a result for bigger dimension, even for hyperkahler ALE spaces (and I bet that for hyperkahler it's true in any dimension).