# Betti numbers of Proper nonprojective varieties

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.

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Sorry, I misunderstood the question, so my comment was stupid. – Sasha Feb 23 '13 at 7:55
No problem :) – LMN Feb 23 '13 at 7:59

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).

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Dimitri, regarding the evenness of the odd Betti numbers, this has indeed been known. It follows from Prop 5.3 of Deligne, Theoreme de Lefschetz... However, if you do find a proof using weak factorization, it would certainly be nice. – Donu Arapura Feb 23 '13 at 14:18
Thanks Dmitri, Donu. This is very helpful! – LMN Feb 23 '13 at 16:05
The evenness of odd Betti numbers is true for proper smooth varieties over any algebraically closed field, including positive characteristic ones (by a recent paper by Junecue Suh). – anon Feb 23 '13 at 17:10
anon, sorry to be silly - but since this isn't my area I just want to make sure. When you speaks of betti numbers in characteristic $p$, you referring to the algebraic de Rham complex? – LMN Feb 23 '13 at 18:19
@LMN: No, I speak of the $\ell$-adic Betti numbers, defined via $\ell$-adic \'etale cohomology for any prime $\ell \neq p$, with $p$ being the characteristic of the field. Some standard facts are: these numbers are independent of the choice of $\ell$ (for proper smooth varieties), and agree with the classical ones if the variety comes via reduction modulo $p$ from something in characteristic $0$. (They do not, however, always equal the de Rham Betti numbers.) In particular, Junecue's result proves evenness in characteristic $0$ as well. – anon Feb 23 '13 at 19:14