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

Let $X=\mathbb C^2$, let $X^{[n]}$ be the Hilbert scheme of length $n$ 0-cycles in $X$, and let $X^{[n]}_0$ be the closed subscheme formed by the 0-cycles supported at 0. As far as I know $X^{[n]}_0$ and $X^{[n]}$ have the same homotopy type. Can anybody suggest a proof? (according to Nakajima this can be proved by adapting an argument of Slodowy (Four lectures on simple groups and singularities, Section 4.3) but I am unable to do it...)

share|improve this question
    
Does dilation onto the origin look anything like a retract? –  JSE Apr 21 '11 at 18:56
1  
The problem is that dilation doesn't fix $X^{[n]}_0$; it only fixes the homogeneous ideals. –  Ben Webster Apr 21 '11 at 19:17

2 Answers 2

up vote 18 down vote accepted

It is a general folklore result, that if ${\mathbb C}^\times$ acts on a smooth complex variety $X$ so that the fixed point set $X^{{\mathbb C}^\times}$ is proper and the limit ${{\rm lim}_{\lambda \to 0} \lambda z } $ exists for every $z\in X$, then the downward flow $$D:=\{ z\in X | {\rm lim}_{\lambda \to \infty} \lambda z \mbox{ exists}{\}}$$ is a retract of $X$. This can be proved by Morse theory arguments as in Kirwan's book or you can prove it by first showing that the imbedding induces

$$H_*(D;{\mathbb Z})\to H_{*}(X;{\mathbb Z})$$ an isomorphism (by induction with respect to an appropriate ordering of the set of components of $X^{{\mathbb C}^\times}$). Similarly you can show that the fundamental groups are also isomorphic. Then the relative Hurewitz theorem will tell you that they are weakly homotopy equivalent, and as they are varieties so CW-complexes, Whitehead's theorem imply that they are homotopy equivalent.

For the Hilbert scheme you can use the ${\mathbb C}^\times$ action induced from the dilation on ${\mathbb C}^2$ mentioned above, which will have the required property, due to the fact that the Hilbert-Chow morphism is proper.

(EDIT: more detailed discussion of this argument can now be found in Corollary 1.3.6 of the paper http://arxiv.org/abs/1309.4914)

share|improve this answer
    
Very nice! I'm glad I now know that technique. –  Jim Bryan Apr 21 '11 at 22:30
    
Thank you. I can follow this until proving the isomorphism between the homologies. I have some problems with the homotopy. But I will review the argument again. –  Yougeeaw Apr 22 '11 at 12:06
    
Yep, you get to proving the isomorphism of the homotopy groups by using relative Hurewicz. And then you prove that the two spaces are homotopy equivalent. –  Yougeeaw Sep 1 '11 at 18:09

Here's how I understand the argument in Slodowy's book. Sticking with Tamas' notation, let $X$ denote the Hilbert scheme and $D$ the punctual Hilbert scheme. You'd like to define a deformation retraction from $X$ to $D$ by sending every point to its limit under the dilation action. Unfortunately, as Ben points out, $D$ is not fixed by the dilation action, and this function isn't even continuous! So that doesn't work.

Instead, you choose a closed tubular neighborhood $A$ of $D$ such that $D$ is a deformation retract of $A$; this is possible by a basic result in algebraic topology (Slodowy cites Spanier's book). You then use the dilation action to define a deformation retraction from $X$ to $A$. More precisely, send every element $x\in X$ to the first element of $A$ that it hits when you dilate it inward. This is a perfectly well-behaved map, and it does the trick.

share|improve this answer
    
The idea is intuitively attractive, but I am unable to make it into a rigorous proof. –  Yougeeaw Sep 1 '11 at 18:10

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.