When you start studying algebraic geometry, you are told that we study algebraic varieties because they are the complex manifolds which we can write down expliclity, and that all of the interesting behavior of a complex manifold can be seen in the algebraic structure. There are even theorems which make this precise in various ways, such as Chow’s Theorem or GAGA. You are also (I hope!) cautioned that this is not actually true. There are many complex manifolds, even compact ones, which are not algebraic, and you may be shown one or two examples.

However, you probably aren’t told why these manifolds are not algebraic. In this post, I want to take two of these standard examples and sketch the proofs. These examples are: the quotient of by powers of the matrix , where is a real number greater than 1, and the quotient of under translations by a generic lattice of rank 4.

In the course of the argument, I’ll get you to the point where you are ready to understand, and be interested in, the Hodge Conjecture and the classification of surfaces. I might write some posts on those topics, if I can think of anything to say which is more interesting or more informative than the sources I’ve linked.

In both examples I want to discuss, the obstacle is that the complex manifold in question does not have enough hypersurfaces. More specifically, we are going to use the following lemma:

**Lemma:** If is a compact smooth algebraic variety, then has a complex subvariety of (complex) codimension 1 whose homology class is nontrivial.

Sketch of proof: The best way to prove a homology class is nontrivial is to exhibit another homology class with which it has nontrivial intersection product. (Intersection product is the product on homology obtained by Poincare dualizing cup product.) The great fact is this: if is a smooth compact complex manifold of dimension and and are complex subvarieties of dimensions and , such that is a nonempty finite set of points, then the homology classes of and have nonzero intersection product. (In fact, their product is always at least .)

So our strategy is to write down a hypersurface and a curve in , so that the hypersurface meets but does not intersect the curve. If is algebraic, then it contains an affine open , which can be written as a closed subvariety of . Pick a point of , and pick a generic hyperplane through and a generic plane through . Then and are a hypersruface and a curve in respectively. Take their closures in to get a codimension one subvariety and a dimension one subvariety which meet in a nonempty, zero dimensional set. (Note that these closures may not be smooth, so we need to have intersection theory for arbitrary complex subvarieties in a smooth variety.)

It is also true, by a very similar proof, that any irreducible algebraic subvariety of a compact smooth algebraic variety has nontrivial cohomology class.Now, let’s look at our first example: the quotient of by powers of the matrix . Topologically, this is the product of the three-sphere and the one-sphere. So by Künneth’s theorem, is trivial. Immediately, we see that no hypersurface can have nontrivial cohomology class.

The second example, for a rank four lattice , is more interesting. Here is topologically , so . However, it is not easy for a homology class to be the class of a complex subvariety. Write the complex coordinates on as . Then is a complex-valued one form on . This notation scared me when I first saw it, so let me assure you that it does not mean anything complicated. The function is a smooth, complex valued function on , so we can take its differential and get the one-form . Apart from taking complex values, it is no worse than any ordinary one form. The one form is invariant under translation by , so it descends to the quotient . Sinilarly, descends to a one form on , so is a two form. Now, suppose that is a complex curve in . The restriction of to is identically zero (exercise!). So . Thus, if is the homology class of , then .

Now, here comes the computational bit. Let , , and be a basis of the lattice . Take a basis of and pair it against . If you chose the same basis that I did, you should get the six maximal minors of the matrix . If is chosen generically, there is no reason for any integral combination of these minors to be zero. So there is no integral homology class, and hence no homology class at all, which can be the class of a holomorphic currve in .

This has some odd consequences for trying to create a moduli space of algebraic complex structures on . We can specify a complex structre on by just specifying the 2 by 4 matrix of complex numbers . We have to impose that the columns this matrix span a discrete lattice, which removes a set of measure 0. Also, changing bases in doesn’t effect the complex structure that we get, so we only want to consider 2 by 4 matrices up to the action of . In other words, complex structures on (up to diffeomorphisms homotopic to the identity) are parameterized by the grassmannian , minus a set of measure zero. (If we want to just look at isomorphism classes of complex manifolds, and not specify a homotopy class of isomorphism to , we need to take a further quotient by the discrete group , but we will not consider this issue here.)

However, if we then which to consider which of these structures are algebraic, life becomes interesting. Every single element of might be the homology class of a holomorphic curve. The set of points in where that class is holomorphic form a complex subvariety of dimension 3. Inside this three-dimensional variety, the actual algebraic locus is an open subset. (There is a positivity condition, which I ~~find a little hard to justify from the perspective I’ve taken here~~ just figured out how to explain nicely. See the comments below.) So, in order to describe all of the algebraic complex structures on , we need an infinite (though countable) list of complex varieties.

For those who are curious about that positivity condition: Consider any complex linear projection of to . Let be the pullback of the area form. Then descends to a (real-valued) two form on the quotient . We must have for every such projection. Exercise: why?