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.


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.


1 Answer 1


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).

  • 3
    $\begingroup$ 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. $\endgroup$ Feb 23, 2013 at 14:18
  • $\begingroup$ 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). $\endgroup$
    – anon
    Feb 23, 2013 at 17:10
  • $\begingroup$ 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? $\endgroup$
    – LMN
    Feb 23, 2013 at 18:19
  • $\begingroup$ Donu, thanks for the reference, I'll have a look (and will try to see if indeed I have an alternative proof :) ). LMN, you are welcome :) $\endgroup$ Feb 23, 2013 at 18:34
  • 2
    $\begingroup$ @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. $\endgroup$
    – anon
    Feb 23, 2013 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.