Skip to main content
added 170 characters in body
Source Link

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...)

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?

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...)

Source Link

Homotopy type of Hilbert schemes of points of $\mathbb C^2$

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?