Welcome to MathOverflow

Question

5

1

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.

edited Dec 2 2011 at 16:42

asked Dec 2 2011 at 14:08

DamienC
3,576●15●23

1

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 2011 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 2011 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 2011 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 2011 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 2011 at 16:55

show 3 more comments

Hodge numbers of Kummer varieties

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.