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I don't know about schemes and every definition of a Hilbert scheme (quite naturally!) involves schemes. But, the Hilbert scheme of points on a complex surface is known to be smooth (Fogarty). So is there a concrete description of it as a complex manifold? (For instance in the case of n=2 it is a blowup of XxX along the diagonal)

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Given a codimension $d$ ideal $I$ in $R = {\mathbb C}[x,y]$, the quotient ring $R/I$ can be thought of as a $d$-dimensional vector space with actions of two commuting operators $x,y$ and a "cyclic" vector $1$ that generates it as an $R$-module.

Consequently, if your surface is the plane you can think of the Hilbert scheme as the space of pairs of commuting matrices on $d$-space, but then take the (open) set in there of pairs $(x,y)$ that admit a cyclic vector, and then divide that variety by the conjugation action of $GL(d)$. To see smoothness, you might first show this open set is smooth, then that the $GL(d)$ action is free and proper.

All this is spelled out in Nakajima's book (which I'm guessing is the one Andrea Ferretti meant to reference), except I think he shows smoothness by analyzing the tangent spaces.

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Thanks. The description in terms of matrices was concrete indeed But, how does one describe the hilbert scheme of X where X is a general complex 2-manifold? I mean, in Gottsche's slides (for a talk) it is written that atleast as a set it is the collection {(x1,Q1),(x2,Q2),....(xk,Qk)} where where xk is a point on X and Qk is "the quotient ring of holomorphic functions at xk" (any idea as to what this means?). –  Vamsi Jan 29 '10 at 5:36
    
I'm guessing that the n points are sitting at k spots, so one can think of it as a disjoint union of k fat points, and Q_i is the coordinate ring of the ith fat point. So yeah, that's as a set; it would be some work from that definition even to understand the topology of what happens when points collide. –  Allen Knutson Jan 29 '10 at 12:23
    
Is there a natural map from Hilb_n(C^2) to $(\mathbb{C}^2)^n/S_n$ (or maybe to $((\mathbb{C}^\times)^2)^n/S_n$)? Something like (pair of matrices) goes to (their eigenvalues)? I've heard there's a map like this that's supposed to be a resolution of singularities, but I don't know the details, or what this means exactly... –  Peter Samuelson Feb 23 '10 at 19:17
    
@Peter: It's called the "Hilbert-Chow morphism", and more generally goes from Hilbert schemes to Chow varieties. This particular instance is described in detail in e.g. Brion & Kumar's book on Frobenius splitting, where they use the fact that it's a crepant resolution of singularities to show that the Hilbert scheme is Frobenius split. –  Allen Knutson Feb 24 '10 at 3:30

Here is a geometric description in the case of $H_n(\mathbb{C}^2)$. This is meant to be a geometric rewrite of Proposition 2.6 in Mark Haiman's "(t,q)-Catalan numbers and the Hilbert scheme", Discrete Math. 193 (1998), 201-224.

Let $S= (\mathbb{C}^2)^n/S_n$; notice that this is an orbifold. Let $S_0$ be the open dense set where the $n$ points are distinct. For $D$ an $n$-element subset of $\mathbb{Z}_{\geq 0}^2$, let $A_{D}$ be the polynomial $\det( x_i^{a} y_i^{b})$, where $(a, b)$ ranges over the elements of $D$ and $i$ runs from $1$ to $n$. For any $D$ and $D'$, the ratio $A_D/A_{D'}$ is a meromorphic function on $S$, and is well defined on $S_0$.

Map $S_0$ into $S_0 \times \mathbb{CP}^{\infty}$ where the homogenous coordinates on $ \mathbb{CP}^{\infty}$ are the $A_{D}$'s. (Only finitely many of the $A_D$'s are needed, but it would be a little time consuming to say which ones.) The Hilbert scheme is the closure of $S_0$ in $S \times \mathbb{CP}^{\infty}$.

Algebraically, we can describe this as the blow up of $S$ along the ideal generated by all products $A_D A_{D'}$. Haiman points out that the reduction of this ideal is the locus where two of the points collide and speculates that this ideal may be reduced. If his speculation is correct, then we can describe $H_n(\mathbb{C}^2)$ geometrically as the blow up of $(\mathbb{C}^2)^n/S_n$ along the reduced locus where at least two of the points are equal.

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I just made some edits to this answer for clarity. In particular, I added the last sentence. –  David Speyer Feb 23 '10 at 13:57
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Haiman proves, in his mammoth paper arxiv.org/abs/math.AG/0010246 , that this speculation is correct. –  David Speyer Mar 14 '10 at 20:43

Actually, the construction of the Hilbert scheme doesn't have to involve schemes at all. For $\mathbb{C}^2$, it's just the space of ideals of codimension $d$ in $\mathbb{C}[x,y]$. That's enough to say what it is as a topological space as a subspace of all codimension $d$ subspaces.

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There is a "concrete" description of the Hilbert scheme of points on surfaces in the book of Nakajima. As far as I remember it doesn't really use the general machinery of Hilbert schemes.

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