8
$\begingroup$

Let $X$ be a smooth proper algebraic variety. Then $X$ has a dense affine open subset. In particular, any smooth proper algebraic variety has a dense Stein open subset as the complement of a divisor.

Does the same thing hold for compact complex manifolds? i.e.,

(1) Does any compact complex manifold have a dense Stein open subset?

(2) If (1) is true, can one take a dense Stein open subset as the complement of a divisor?

$\endgroup$
2
  • 5
    $\begingroup$ There are many compact complex manifolds which do not contain any divisors, e.g., a general complex torus of dimension at least two. $\endgroup$
    – naf
    May 31, 2019 at 8:05
  • 4
    $\begingroup$ @ulrich's remark is absolutely correct but does not address the question, because the complement of an open Stein subset of a compact complex manifold has no reason to be a divisor nor even to be analytic.For example the complement $U=\mathbb P^1(\mathbb C)\setminus K$ of the Cantor subset $K\subset [0,1]$ is a connected Stein dense open subset of $\mathbb P^1(\mathbb C)$ whose complement $K$ is certainly not analytic. By the way this is a great question because of its divine naïveté and (I think) diabolical difficulty... $\endgroup$ Jun 1, 2019 at 21:04

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.