This is a question about pathologies.


Let $X/\mathbb{C}$ be an irreducible projective variety smooth over $\mathbb{C}$. Then, the singular cohomology groups $H^i(X, \mathbb{C})$ have a hodge decompositon, and hodge theory tells us that the betti numbers (and hodge numbers) are not completely random. Eg, the odd betti numbers $b\_{2i+1}$ are even integers, Hard Lefschetz theorem, even betti numbers $b\_{2i}$ are nonzero, ... 


We can also see that the even Betti numbers of a variety $X$ as above are nonzero in "another" way: $X$ has a finite surjective map to $\mathbb{P}^n$ where $n= \dim X$. Then, because $X$ is Kahler, such a map induces an injective map on singular cohomology. Hence, the result follows from the corresponding result for projective space.


$\textbf{Question:}$


1.) Are there any nontrivial restrictions on the Betti numbers of smooth, irreducible proper (but non-projective) varieties?


For example, can I have such a 4-fold with Betti numbers $b\_0 = b\_8 = 1$, but all other $b\_i = 0$?


I find the situation a little disconcerting: for example, suppose I have a connected compact topological (or complex) manifold. If I know it's betti numbers are as directly above, then I would like to immediately say that it doesn't have the structure of an algebraic variety. However, the only thing I can say now is that a smooth complex variety with these Betti numbers can't be projective. If such a variety existed, then (by Ehresmann's thm) we can't put it in a proper smooth family over a curve with any projective variety. I'm no expert but that sounds pretty bad.



Betti numbers of Proper nonprojective varieties
I think, that many basic restrictions, that you have for complex projective varieties still hold for proper smooth complex varieties. Let me show that $b\_2>0$, and $b\_{2n-2}>0$.


Suppose that $X^{2n}$ is a proper smooth complex variety. Then there is a birational morphism $\phi: Y^{2n}\to X^{2n}$ from a *projective variety* $Y$ to $X$. Let $E\_1,...,E\_n$ be all the exceptional divisors of $\phi$. Since $Y$ is projective there is a divisor $D\subset Y$ that has a point $x$ in $Y^{2n}\setminus (E\_1,...E\_n)$ and there is a curve $C$ in $Y$ passing through $x$ and not contained in $D$. Consider finally the divisor $\phi(D)$ and the curve $\phi(C)$ in $X^{2n}$. Note that they intersect positively (since they have common point $\phi(x)$ and $\phi(C)$ does not belong to $ \phi(D)$).
So they represent non-zero cycles in $H\_2(X^{2n})$ and $H\_{2n-2}(X^{2n})$.


Also, it is very easy to see that $b\_1$ should be even. Indeed the fundamental group of a smooth complex variety is a birational invariant, and $b\_1$ depends only on $\pi\_1$.


I have an idea how using *weak factorization of birational maps* one can prove that all other $2k+1$ Betty numbers are even. But this requires some work, so I'll write this if I manage to work out details (I believe this should be a very well known fact).


