most recent 30 from http://mathoverflow.net 2013-05-22T20:58:11Z http://mathoverflow.net/feeds/user/14559 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24132/what-are-examples-of-mathematical-concepts-named-after-the-wrong-people-stigler/69830#69830 Yougeeaw 2011-07-08T21:08:57Z 2011-07-08T21:08:57Z

<p>The notion of Frobenius manifold is due to Dubrovin</p>

http://mathoverflow.net/questions/62544/homotopy-type-of-hilbert-schemes-of-points-of-mathbb-c2 Yougeeaw 2011-04-21T15:55:06Z 2011-04-22T16:24:52Z

<p>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...)</p>

http://mathoverflow.net/questions/62544/homotopy-type-of-hilbert-schemes-of-points-of-mathbb-c2/62634#62634 Yougeeaw 2011-09-01T18:10:10Z 2011-09-01T18:10:10Z

The idea is intuitively attractive, but I am unable to make it into a rigorous proof.

http://mathoverflow.net/questions/62544/homotopy-type-of-hilbert-schemes-of-points-of-mathbb-c2/62578#62578 Yougeeaw 2011-09-01T18:09:24Z 2011-09-01T18:09:24Z

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.

http://mathoverflow.net/questions/24132/what-are-examples-of-mathematical-concepts-named-after-the-wrong-people-stigler/69830#69830 Yougeeaw 2011-08-31T10:17:11Z 2011-08-31T10:17:11Z

Why -2 votes???

http://mathoverflow.net/questions/62544/homotopy-type-of-hilbert-schemes-of-points-of-mathbb-c2/62578#62578 Yougeeaw 2011-04-22T12:06:16Z 2011-04-22T12:06:16Z

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.