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I'm turning here (a variation of) a question asked by a friend of mine. For the purposes of this question I will say that a compact complex manifold is projective if it is isomorphic to a subvariety of $\mathbb{P}^n$ and algebraic if it is isomorphic to the complex analytic space associated to a scheme.

One often finds many examples of compact complex manifolds which are not projective. For instance Hopf and Inoue surfaces, some K3, some tori... Usually the proof shows that either this manifold is not Kahler, or the Kahler cone does not intersect the lattice of integral cohomology. But of course this does not tell us anything about their being algebraic in the sense outlined above.

So there are two questions. First, what is an example of a compact complex manifold which is not algebraic? Probably this is standard, but I don't have any reference in mind. I think that the complex analytic space associated to a smooth algebraic space which is not a scheme will do, but I'm not expert of algebraic spaces, so I don't have even such an example in mind.

The second, subtler, question is: are the examples above of nonprojective complex manifold also non algebraic? How one can prove such a statement?

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Most complex tori $C^g/Lambda$ aren't algebraic. An algebraic complex torus would be an abelian variety and hence projective, but for a generic Lambda the variety will have no non-constant meromorphic functions on it at all other than the constants. –  Kevin Buzzard Jan 29 '10 at 20:50
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Can you sketch better why an algebraic complex torus is necessarily projective? This is exactly what I am missing: as I told in the question I know many classes of nonprojective varieties, but I cannot exclude that these are the analytification of some nonprojective complete scheme. –  Andrea Ferretti Jan 30 '10 at 1:44
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4 Answers

up vote 8 down vote accepted

I wrote a blog post about some of the standard examples of nonalgebraic compact complex manifolds.

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Thank you. As I told in the questions I know the standard examples, but I was just missing the arguments which go from non-projectivity to non-algebraicity. It turns out the question was more trivial then I thought. :-/ –  Andrea Ferretti Jan 30 '10 at 2:35
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The simplest example of a complex analytic non-algebraic manifold (and hence, non-projective) is probably the Hopf surface. Indeed, any smooth complete complex algebraic variety, projective or not, is bimeromorphically Kaehler,so its cohomology admits a Hodge decomposition, which can't exist for the Hopf surface, since its first Betti number is odd.

Any smooth complex algebraic variety is Moishezon, i.e. the transcendence degree of the field of meromorphic functions equals the dimension. All Moishezon surfaces are algebraic and even projective (Kodaira), but starting from dimension 3 there are Moishezon non-algebraic varieties. Here is an example (given in Hironaka's thesis). Let $C$ be a nodal plane cubic (or any other curve with one node and no other singularities) in $\mathbf{P}^3(\mathbf{C})$ and let $P$ be the singular point of $C$. Take a Euclidean neighborhood $U$ of $P$ such that $U\cap C$ is analytically two branches $C_1$ and $C_2$ intersecting transversally. Let $X$ be $\mathbf{P}^3(\mathbf{C})\setminus\{P\}$ blown up along $C\setminus{\{}P{\}}$ and let $Y$ be the result of blowing up $U$ along $C_1$ and then blowing up the result along the proper preimage of $C_2$. Note that $Y$ exists only in the analytic category.

Both $X$ any $Y$ map to $\mathbf{P}^3(\mathbf{C})$ and the parts of both $X$ and $Y$ over $U\setminus P$ can be naturally identified. So we glue them together to get an analytic manifold $Z$. It is Moishezon, since it is bimeromorphic to $\mathbf{P}^3(\mathbf{C})$. Let us show that it is not algebraic. Let $L$ be the preimage of a point in $C\setminus P$ and let $L_i,i=1,2$ be the preimage of a point of $C_i\setminus P$. The preimage of $P$ itself is two transversal lines, $L'$ and $L''$, the first of which appears after the first blow-up and the second one after the second. We have $[L]=[L_1]=[L_2], [L_2]=[L''],[L_1]=[L']+[L'']$. (Here I really wish I could draw you a picture!) So $[L']=0$ i.e. we have a $\mathbf{P}^1$ inside $Z$ which is homologous to zero. This is impossible for an algebraic variety (as David writes in his blog posting).

Hironaka's thesis also contains examples of complete, algebraic but not projective manifolds constructed in a similar fashion.

upd: woops, wrote this answer in a hurry just before going out for drinks; missed a couple of things as a result. These have now been fixed. The curve $C$ is a nodal plane cubic, not conic. Also, David doesn't actually show in his posting that the class of an irreducible curve in a smooth complete variety is non-trivial, but this is easy anyway: let $Z$ be the ambient smooth complete variety and let $W$ be an irreducible curve. Take a smooth point $Q$ of $W$ and let $U$ be an affine neighborhood containing $Q$. There is an irreducible hypersurface $S$ through $Q$ in $U$ that does not contain $W\cap U$ and intersects $W$ transversally at $Q$. So the closure $\bar S$ of $S$ in $Z$ does not contain $W$ and intersects $W$ transversally at at least one point. So the Poincar\'e dual class of $\bar S$ takes a positive value on the class of $W$.

Note that the class of a reducible curve may well be zero.

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Why bimeromorphically Kahler varieties admit a Hodge decomposition? –  Andrea Ferretti Jan 30 '10 at 2:36
    
Andrea -- see e.g. Peters-Steenbrink, Mixed Hodge structure, theorem 2.29. –  algori Jan 31 '10 at 0:05
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These issues are discussed to some extent at this previous MO question.

For instance, if a compact complex manifold is algebraic then its field of meromorphic functions has transcendence degree over $\mathbb C$ equal to its dimension. So any compact complex manifold without as many meromorphic functions as one usually expects has no chance of being algebraic.

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The two simplest examples I know are Abelian varieties (the only algebraic complex Tori, dimension g(g-1)/2 inside the space of complex Tori which is dimension g^2 - David writes about them in his post), and K3 surfaces. Rough sketch for K3s: a 20 dimensional complex moduli space, but the algebraic tangent space to any moduli point is 19 dimensional.

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Abelian varieties are complex projective. –  Anweshi Jan 29 '10 at 22:38
    
Can you expand a bit further on K3? The argument I know is like: a very general point in the period space will have no integral points of type (1,1), hence the surface will not be projective. But I fail to see how this implies that the surface is not algebraic. Your argument seems a bit different, so i'd be glad to know. –  Andrea Ferretti Jan 30 '10 at 1:52
    
@Anweshi: Abelian varieties are, complex Tori in dimension > 1 are generally not. I admit my phrasing is extremly poor –  David Lehavi Jan 30 '10 at 7:20
    
@Andrea: I will somtime today –  David Lehavi Jan 30 '10 at 7:23
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