# Cell decomposition of punctual Hilbert scheme of points on $A^n$?

Hello Everyone,

I am thinking on calculating the motive of Hilbert scheme of points on a smooth variety, of any dimension. There is a famous paper by Gottsche that calculates the Hilbert scheme of points on a smooth surface, which uses a cell decomposition of punctual Hilbert scheme of points on the plane. I am wondering if a similar results hold for punctual Hilbert scheme of any dimension, or at least is it known that such cell decomposition exists?

I apologize if this is a stupid question.

Best, minimax

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I'm not exactly sure what properties you're looking for in your cell decomposition, but Hilbert schemes of points in $\mathbb A^n$ are, in general, neither smooth nor irreducible. For a given number of points in $\mathbb A^3$, even the number of irreducible components of the Hilbert scheme unknown except in some examples. – Dustin Cartwright Nov 13 '12 at 19:46
@Dustin: I am wondering if the punctual Hilbert scheme has a cell decomposition into linear pieces, I.e. cells of the form A^n. If such a decomposition exists, then I suspect one might use techniques in Gottsche's paper to show the motive of Hilbert scheme of points can be written as a polynomial of symmetric power and the Lefschetz motive L=[A^1]... I am interested to know some examples or references that may be related :) – minimax Nov 13 '12 at 20:11
If you pick a generic monomial ordering, do the Groebner basins give a cell decomposition? Alternatively (and this is just a rephrasing of the same idea), does the Bialynicki-Birula decomposition give cells despite the scheme not being smooth? – Alexander Woo Nov 14 '12 at 3:13
In dimension 3 there is a nice formula for the virtual motive arXiv:0909.5088, link.springer.com/article/10.1007/s00222-012-0408-1. In that paper we also speculate what the formula for the virtual motive should be in any dimension. It is a speculation more than a conjecture since when the Hilbert scheme is singular and not dimension three, it isn't clear what the correct definition of virtual motive should be. I expect that the ordinary motive is poorly behaved compared to the virtual motive (when they are smooth the two definitions coincide up to a shift of the Lefschetz motive.) – Jim Bryan Dec 31 '12 at 22:21