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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...)

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  • $\begingroup$ Does dilation onto the origin look anything like a retract? $\endgroup$
    – JSE
    Commented Apr 21, 2011 at 18:56
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    $\begingroup$ The problem is that dilation doesn't fix $X^{[n]}_0$; it only fixes the homogeneous ideals. $\endgroup$
    – Ben Webster
    Commented Apr 21, 2011 at 19:17

2 Answers 2

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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)

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  • $\begingroup$ Very nice! I'm glad I now know that technique. $\endgroup$
    – Jim Bryan
    Commented Apr 21, 2011 at 22:30
  • $\begingroup$ 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. $\endgroup$
    – Yougeeaw
    Commented Apr 22, 2011 at 12:06
  • $\begingroup$ 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. $\endgroup$
    – Yougeeaw
    Commented Sep 1, 2011 at 18:09
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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.

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  • $\begingroup$ The idea is intuitively attractive, but I am unable to make it into a rigorous proof. $\endgroup$
    – Yougeeaw
    Commented Sep 1, 2011 at 18:10

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