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It is well known that a necessary and sufficient condition for a compact Kähler manifold $\mathcal{X}$ to be a projective algebraic variety is that it admit a positive holomorphic line bundle $L \rightarrow \mathcal{X}$. The positivity of $L$ yields a proper holomorphic embedding in projective space and Chow's theorem tells us that the image is cut out by homogeneous polynomials. Here I regard the existence of a positive holomorphic line bundle (meaning a line bundle whose Chern connection has positive curvature) as a differential-geometric condition.

Now suppose that $X$ is a smooth, complete variety over $\mathbb{C}$ in the algebraic category, where $X$ is not necessarily projective. We have an analytification functor (à la Serre's GAGA) which when applied to $X$ yields a compact complex manifold $X^{an}$. In this way, we can think of $X^{an}$ as algebraic in the complex category. Note of course that $X^{an}$ need not be projective, nor even Kähler. Hironaka's example yields non-Kähler examples. My question is:

Is there some differential-geometric condition that tests when a compact complex manifold $\mathcal{X}$ is algebraic (i.e. $\mathcal{X} = X^{an}$ is the analytification of a smooth complete variety $X$ over $\mathbb{C}$)?

We know in particular that such an $\mathcal{X}$ will be a Moishezon manifold (its algebraic dimension $a(\mathcal{X})$ equals $\mbox{dim}_{\mathbb{C}}$($\mathcal{X}$)) but it is known that there are Moishezon manifolds that are not algebraic (i.e. not the analytification of a variety), so the condition of being Moishezon is not sufficient. That said, it is known (due to Artin) that every Moishezon manifold is the analytification of a proper algebraic space, which is defined here by Wikipedia and is a generalization of a scheme. So my question asks for a differential-geometric condition which is almost 'parallel' to asking when an algebraic space is in fact a scheme.

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  • $\begingroup$ Related: MO/4573. I doubt there will be a differential geometric criterion because analysis hardly detects whether there is a Zariski topology. With this respect, the question which manifolds are analytified analytic spaces is more natural and as you noted, the answer is Moishezon manifolds. By the way, there are differential geometric criteria to detect this, e.g., existence of semi-positive line bundles (Grauert-Riemenschneider conjecture) or rational Kähler currents (Ji-Shiffman). $\endgroup$ – Ben Apr 18 '16 at 21:33
  • $\begingroup$ Don't take me wrong, I'm not saying it's not a good question. I asked myself and others the very same question quite often in the past, but I gave up; saying that the other (easier) one is more natural is kind of the easy way out, I have to admit. $\endgroup$ – Ben Apr 18 '16 at 21:40
  • $\begingroup$ @Ben: not at all - I understand what you mean, regarding naturality of the question. I'm still just curious enough to ask though in case someone knows, and keen to learn more about this circle of ideas in general. Thanks for pointing out this interesting relation to the Grauert-Riemenschneider conjecture. $\endgroup$ – john Apr 18 '16 at 22:14
  • $\begingroup$ I should also say that "differential-geometric" condition may overstate the strength of what I would be satisfied with. There's no single word, but I would be interested to hear of any condition that just doesn't mention the analytification functor and which could be phrased only in the language/category of complex manifolds. For example, I'm content with "full algebraic dimension" as a condition to be Moishezon (though, as I say, the Grauert-Riemenschneider relation is very interesting). $\endgroup$ – john Apr 18 '16 at 22:15
  • $\begingroup$ Just a comment since some things you said seemed to imply that you may not have heard this fact. But, if $X$ is a complex Kahler manifold, then it's projective if if and only if it's Moishezon. So, any algebraic compact complex manifold which is non-projective is not Kahler. There is also some fact that algebraicity implies non-trivial $H^2_\text{sing}$. Another necessary condition for something to be the analytification of a proper smooth variety is the degeneration of the Hodge spectral sequence, this shouldn't hold for an arbitrary (non-Kahler) complex manifold, but still holds $\endgroup$ – Alex Youcis Apr 20 '16 at 14:25
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A theorem of Demailly and Pǎun (https://arxiv.org/abs/math/0105176) gives a condition for a compact complex manifold to be of Fujiki class $\mathcal{C}$ (that is bimeromoprhic to a Kähler manifold). A manifold is Kähler if and only if it supports a Kähler current. A Kähler current is a real positive $(1,1)$-current $T$, such that $T - \varepsilon \omega > 0$ for a constant $\varepsilon > 0$ and positive hermitian $(1,1)$-form $\omega$.

The homology class of such current is integral if and only if the manifold is projective (this is the standard Kodaira embedding theorem).

A similar theorem for Moishezon manifolds was proved by Popovici (https://arxiv.org/abs/math/0603738). It states, roughly speaking, that a complex manifold is Moishezon if and only if it supports a semi-positive and strictly positive almost everywhere real $(1,1)$-current with integral homology class.

The integrality condition is hardly a differential-geometric one, but you cannot escape it (at least, as far, is we know by now), and I don't really think one might give a more differential-geometric answer.

In a general situation it is, of course, rather hard to verify the Popovici condition. Another "geometric" approach to studying algebraic dimensions of complex manifolds is to study the algebraic reduction. For any complex manifold $X$ there exists a normal projective variety $\overline{X}$ and a meromorphic map $\alpha \colon X \to \overline{X}$, such that any meromorphic function on $X$ can be lifted from $\overline{X}$. The variety $\overline{X}$ is unique up to birational equivalence.

Being Moishezon is equivalent to $\alpha$ being a birational equivalence. More generally, $a(X) = \dim_{\mathbb C}(\overline{X})$.

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