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A Moishezon manifold $M$ is a compact connected complex manifold such that the field of meromorphic functions on $M$ has transcendence degree equal to the complex dimension of $M$. There exists a Moishezon manifold which is not projective for example and I believe that Moishezon manifold form a good class of manifolds.

The existence of the non-projective Moishezon manifold surprised me because there are lots of meromorphic functions on it; in my understanding (it turns out to be incorrect), a non-projective compact complex manifold has very few meromorphic functions and very few complex submanifolds.

My question is how should one think of the difference among projective manifolds, non-projective manifolds, and Moishezon manifold? Are there any good idea about how these three classes of manifolds are characterized heuristically?

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I think your question is too vague. Like how should one think of the difference among rational numbers, real numbers and complex numbers? –  Makoto Kato Sep 19 '12 at 1:26
    
I'm not going to elaborate on this in an answer, because I don't know if it is the same as what you are asking for. To me, non-projective complex manifolds come up in the following way. Take the space of K3 surfaces. It is this large 20-dimensional space. If you pick a projective one, this is the same as choosing an ample line bundle on it. Moving to nearby ones can be done by deforming in one of the "20 directions." It turns out that you can only deform in 19 of the directions if you want to bring that line bundle along, so you get non-projective ones by deforming in a "bad" direction. –  Matt Sep 19 '12 at 13:13
    
But those deformed $K3$'s aren't Moishezon (at least, most of them aren't). Most of them contain no holomorphic curves at all, while Moishezon manifolds contain complex subvarieties of every smaller dimension. –  David Speyer Sep 19 '12 at 13:57
    
@MakotoKato, I agree that my question is vague mainly because I don't really know what I should ask for; any idea is welcome. –  M. K. Sep 19 '12 at 16:52
    
@Matt Yes, a generic K3 surface is non-projective and generic deformation of an algebraic surface gives you non-algebraic one. Algebraic K3 surfaces are just like rational numbers in real numbers. –  M. K. Sep 19 '12 at 16:58

1 Answer 1

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From the point of view of complex geometry there is a rough partial hierarchy of compact complex manifold; these are the projective, Moishezon, Kähler and Fujiki manifolds.

A compact complex manifold $X$ is projective if it can be embedded into a projective space $\mathbb P^N$. By Serre's GAGA theorem this means that it is algebraic, i.e., it can be defined as the zero locus of a finite number of polynomials. In diffeo-geometric terms, a manifold is projective if and only if it admits a holomorphic line bundle $L \to X$ with a positively curved hermitian metric. This is Kodaira's embedding theorem.

A larger class of manifolds are the Kähler manifolds. These are $X$ that admit a hermitian metric $h$ (on the tangent bundle), whose Kähler form is closed. This means that the smooth $(1,1)$-form $\omega = - Im h$ is $d$-closed, or $d \omega = 0$. Any projective manifold is Kähler, since the curvature form of a positively curved metric on a line bundle defines a Kähler metric. The converse is false; a Kähler manifold is projective if and only if it admits a Kähler metric $\omega$ whose cohomology class is integral (in which case one can find a line bundle on $X$ such that the metric is the curvature form of a positive hermitian metric by a lemma of André Weil).

A Moishezon manifold can now be characterized as a modification of a projective manifold, so there exists a generically 1-1 meromorphic map $X \to Y$, where $Y$ is projective. This is equivalent to $X$ admitting an integral Kähler current, which is roughly a closed $(1,1)$-form $T$ that has distribution coefficients (in local coordinates) whose cohomology class is integral. We also demand a positivity property: if $\omega$ is hermitian, then for every small $\epsilon > 0$ the current $T - \epsilon \omega$ should be positive.

A Fujiki manifold is then a modification of a Kähler manifold. Similarly, this is equivalent to $X$ admitting a Kähler current $T$, without conditions on its cohomology class. Any Moishezon manifold is thus a Fujiki manifold.

These classes do absolutely not cover all compact complex manifolds. To see this, we remark that the Hodge decomposition holds for all of these classes (this is slightly nontrivial). The Hodge decomposition fails for the Hopf surface, so that general nonprojective (non-Kähler) manifold does not fit into any of these classes. Little is known about general non-Kähler manifolds, but a conjecture of Bogomolov states that they can all be obtained via algebraic constructions on foliations of projective manifolds.

For details on these differential geometric characterizations you could look at Jean-Pierre Demailly's book on complex geometry and his lecture notes on transcendental methods in algebraic geometry (both available for free on his website).

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@ Gunnar Thank you very much for the detailed answer. So there is a hierarchy "projective $\subset$ Kaehler $\subset$ Moishezon $\subset$ Fijiki". Do you know any easy example which is not projective but Kaehler? –  M. K. Oct 4 '12 at 7:25
    
Dear M.K., beware that the hierarchy is a "box"; so a projective manifold is both Kahler and Moishezon, but a Moishezon manifold is in general not Kahler. A general complex torus is Kahler but non-projective. You can play with the parameters in my answer to tinyurl.com/cwgmkla and get an explicit example of a torus that admits no subvarieties (and it thus in particular non-projective). –  Gunnar Þór Magnússon Oct 4 '12 at 8:23

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