# Hodge diamond of a Calabi-Yau fourfold

I am trying to compute the Hodge diamond of a Calabi-Yau fourfold which is a complete intersection inside of a projective bundle over a threefold base. I have computed the arithmetic genera $\chi_1$ and $\chi_2$ which give me two linear relations between the four independent Hodge numbers $h^{(1,1)}$, $h^{(1,2)}$, $h^{(1,3)}$ and $h^{(2,2)}$. So if I can compute two of them the arithmetic genera will give me the other two. Since my variety is a Calabi-Yau fourfold, $h^{(1,1)}$ is the second Betti number and $h^{(1,2)}$ is twice the third Betti number, so really all I need are the second and third Betti numbers to finish the calculation. As such, I wanted to use the Lefschetz hyperplane theorem to relate the cohomology of my fourfold to the cohomology of the ambient projective bundle, but it turns out that the divisors for which my fourfold is a complete intersection of are not ample (they are only relatively ample). So does anyone have any ideas on how to relate the cohomology of my variety to the cohomology of the projective bundle if the divisors for which my variety is a complete intersection of are only relatively ample?

• If you assume that $\pi$ and $\pi|_Y$ are smooth projective then you can use the Leray spectral sequence and the Lefschetz hyperplane theorem on the fibres to deduce that the maps $H^i(X) \to H^i(Y)$ are isomorphisms for $i < dim(Y) - dim(B)$. – naf Jun 23 '11 at 16:37
• Apparently we only need that $\pi$ be Kahler, i.e. that there exists a 2-class $\alpha$ on $X$ such that its restriction to each fiber of $\pi$ is a Kahler class. IIRC, this is a remark without proof in Voisin's book, in the chapter on the Leray spectral sequence. – Gunnar Þór Magnússon Jun 23 '11 at 19:46

## 1 Answer

I encountered a similar problem a few years ago, but then in dimension 3. In that case a master student wanted to calculate the hodge diamond of a threefold which was a hypersurface $W$ in a $P^2$-bundle over a del Pezzo surface. This threefold was birational to a singular hypersurface $Y$ in some weighted projective 4-space. Now you can apply Lefschetz' hyperplane theorem for $Y$, but it only works for the lower cohomology groups, since Poincar\'e duality might fail. There are methods to calculate the higher cohomology groups if the singularities of $Y$ are nice enough.

Then you consider a factorization of the birational map $W\dashrightarrow Y$, calculate exceptional divisors etc, in order to figure out the difference between the Hodge numbers of $W$ and of $Y$.

In the end this turned out to be too hard for the student in question, so I worked out some of the details. This particular example is worked out at the end of the first version (on arXiv) of my paper with Klaus Hulek on MW-groups of elliptic threefolds. The referee considered this example superfluous, therefore we omitted this example in the later versions and the published version of this paper.

• Hopefully the referee sees this question and your answer... – Mariano Suárez-Álvarez Jul 12 '11 at 2:24
• Is there a general procedure for finding the birationally equivalent $Y$? Also, is there any way to see this worked out example? – DZN Jul 12 '11 at 14:24
• If you want this particular example, go to arxiv look for a paper written by Klaus Hulek and me (there are only two such papers) and download the first version of the correct paper (rather than the most recent version). The method is a little ad hoc. I.e., in our example we started with $Y^2Z=X^3+aXZ+bZ^3$, where $a,b$ were in some $H^0(S,L^4)$ and $H^0(S,L^6)$ resp. and $X,Y,Z$ are some choice of vertical coordinates. Now you can consider $a,b$ also as elements from $K(S)=K(P^2)=k(s,t)$, homogenizing $a$ and $b$ yields a hypersurface $y^2=x^3+ax+b$ in some wps... – Remke Kloosterman Jul 12 '11 at 14:38
• ... Depending on your base type of equation etc. you should play around a little bit with the equations in order to find a birational model for your situation, where your threefold is a complete intersection of ample hypersurfaces (this should not be too hard, since you assumed that your ci is relatively ample) and where you can control the singularities (this is more annoying). – Remke Kloosterman Jul 12 '11 at 14:42