Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A$ be an abelian surface. And consider the Hilbert scheme $A^{[n]}$ of $n$ points on $A$, which is a desingularisation of the $n$-th symmetric product $S^nA$. We have a composed map $$ \varphi:A^{[n]}\to S^nA\to A\,, $$ where the second map is the sum.

We now define $K^{[[n]]}:=\varphi^{-1}(0)$.

Question: Has anybody computed the Hodge numbers of $K^{[[n]]}$?

I know that Lothar Göttsche has worked on closely related things. But I was unable to find precisely what I am looking for.

EDIT: 1) any information about low-dimension examples is very welcome!

2) a formula exists (by Göttsche-Soergel, see comments below) but in order to get actual numbers one should probably use a computer.

share|improve this question
2  
In the paper by Göttsche and Soergel, Math. Ann. 1993, I believe they compute the Hodge numbers of $S^{[n]}$ for any surface, but you probably knew this, and it isn't quite what you want. If someone doesn't answer in a day or two, I'll think about it when I have more time. –  Donu Arapura Dec 2 '11 at 14:46
1  
Here are some more things you probably already know. First, since these varieties are hyperk\"ahler, they have $h^{1,0}=h^{0,1}=0$ and $h^{2,0}=h^{0,2}=1$. Second, the cohomology group H^2 carries the structure of a lattice, and in fact (see arxiv:1012.4155, Prop 3.2) it is isomorphic to the lattice 3U+<-2(n+1)> (U=hyperbolic plane), so has rank 7; therefore $h^{1,1}=5$. –  Artie Prendergast-Smith Dec 2 '11 at 15:48
    
@Artie Prendergast-Smith: I am a beginner in that subject, and I actually didn't know about $h^{1,1}=5$. Thank you for that. –  DamienC Dec 2 '11 at 16:33
    
@Donu Arapura: yes, I was told (very) recently about that paper. This is true that there is a formula (this is Corollary 1 -of Theorems 6 and 7- of the paper) but I must admit that at the moment I find it diffucult to make computations with it (but this only tells you about my incompetence - the formula is really nice). –  DamienC Dec 2 '11 at 16:40
    
Dear Damien, you're welcome. I'm a little confused by your edit 2): Donu seems to be saying that Goettsche--Soergel give a formula for the Hodge numbers of the Hilbert scheme of points on a surface, rather than Kummer varieties as in your question. Am I missing something? –  Artie Prendergast-Smith Dec 2 '11 at 16:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.