Hilbert scheme of points on a complex surface - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:45:14Z http://mathoverflow.net/feeds/question/13305 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13305/hilbert-scheme-of-points-on-a-complex-surface Hilbert scheme of points on a complex surface Vamsi 2010-01-29T00:14:17Z 2010-02-23T13:56:35Z <p>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)</p> http://mathoverflow.net/questions/13305/hilbert-scheme-of-points-on-a-complex-surface/13311#13311 Answer by Andrea Ferretti for Hilbert scheme of points on a complex surface Andrea Ferretti 2010-01-29T00:46:32Z 2010-01-29T11:51:51Z <p>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.</p> http://mathoverflow.net/questions/13305/hilbert-scheme-of-points-on-a-complex-surface/13313#13313 Answer by Ben Webster for Hilbert scheme of points on a complex surface Ben Webster 2010-01-29T01:04:40Z 2010-01-29T01:04:40Z <p>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.</p> http://mathoverflow.net/questions/13305/hilbert-scheme-of-points-on-a-complex-surface/13331#13331 Answer by Allen Knutson for Hilbert scheme of points on a complex surface Allen Knutson 2010-01-29T03:35:52Z 2010-01-29T03:35:52Z <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/13305/hilbert-scheme-of-points-on-a-complex-surface/13335#13335 Answer by David Speyer for Hilbert scheme of points on a complex surface David Speyer 2010-01-29T04:30:45Z 2010-02-23T13:56:35Z <p>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 "<a href="http://math.berkeley.edu/~mhaiman/ftp/hilb-qtcat/discrete-math.pdf" rel="nofollow">(t,q)-Catalan numbers and the Hilbert scheme</a>", Discrete Math. 193 (1998), 201-224.</p> <p>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 <code>$\mathbb{Z}_{\geq 0}^2$</code>, 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 $S_0$.</p> <p>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 $S \times \mathbb{CP}^{\infty}$. </p> <p>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.</p>