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Let $M$ be a compact complex manifold of dimension three.

Let us say that a divisor $D$ on $M$ is big if there is a constant $C>0$ such that $$ h^0(M, \mathcal O_M(nD)) > C n^3 $$ for sufficiently large $n \in \mathbb N$.

Assume that $M$ has a big divisor. My question is

Is $M$ projective?

If the answer for the above question is no, can adding some topological conditions to $M$ change the answer?

What if $M$ is simply-connected and $h^{2,0}=h^{0,2}=0$ with trivial canonical class?

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    $\begingroup$ You probably mean $n^3$ instead of $3^n$. $\endgroup$
    – abx
    Commented Feb 15, 2023 at 6:44
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    $\begingroup$ @abx, you're right. It is corrected now. $\endgroup$
    – Basics
    Commented Feb 15, 2023 at 6:46

1 Answer 1

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The existence of a big divisor implies that $M$ is Moichizon so the "only obstruction" to projectivity is that $M$ may not be Kahler. In fact, this can happen even for nonprojective varieties.

Hironaka's construction gives for any projective $3$-fold $X$ equipped with curves $C, C' \subset X$ intersecting transversally at two points $P, Q$ a birational modification $f : \tilde{X} \to X$ where $\tilde{X}$ is not projective (see Hartshorne appendix B).

However, we can choose $X$ such that it is,

(1) simply connected,

(2) has $h^{2,0} = h^{0,2}$ and,

(3) has a big divisor $D$ disjoint from $C, C'$ (which consequently cannot be ample).

Then,

$$ H^0(\tilde{X}, \mathcal{O}_X(n f^* D)) = H^0(\tilde{X}, f^* \mathcal{O}_X(n D)) = H^0(X, \mathcal{O}_X(n D) \otimes f_* \mathcal{O}_{\tilde{X}}) \\ \supset H^0(X, \mathcal{O}_X(n D)) $$

and since $D$ is big, this grows as a degree $3$ polynomial. Since the topological properties are birational invariants we see that $\tilde{X}$ satisfies the required properties but is not projective.

Concretely, choose $X$ to be the blowup of $\mathbb{P}^3$ at a point $P \in \mathbb{P}^3$. The exceptional fiber is $\mathbb{P}^2$ which indeed contains two suitable curves $C, C' \subset X$. Choose $D$ to be the pullback of any ample hyperplane $\mathbb{P}^2 \subset \mathbb{P}^3$ which does not contain the special point $P$. Furthermore, $X$ is rational so $h^{2,0} = h^{0,2} = 0$ and $\pi_1(X) = 0$.

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