Timeline for Reference Request for Hilbert Schemes
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 18, 2016 at 3:40 | comment | added | AHusain | I thought Haiman's lectures were more understandable than Nakajima. | |
| Feb 17, 2016 at 6:45 | answer | added | Joey | timeline score: 4 | |
| Feb 16, 2016 at 2:03 | comment | added | Qfwfq | Ok, I see! BTW, I just opened a related question :) | |
| Feb 16, 2016 at 1:58 | comment | added | Hamed | @Qfwfq Of course, you're right. For the purpose of this question translation invariance is redundant, and I should've excluded it. I included it honestly as a force of habit. The reason we keep saying translation invariant in our work is, another avenue of the study is finding a basis for for these polynomials, in other words: what subset of Jack polynomials is the basis of this space (in this part of the work keeping the exact form of the polynomials is important so we do not do this coordinate change). | |
| Feb 16, 2016 at 0:40 | comment | added | Qfwfq | @Hamed: alright but, up to a linear change of variables, you have a polynomial $f(z_1,\cdots, z_n)$ which is periodic of period $1$ in the variable $z_n$, hence constant in $z_n$, so essentially a polynomial in $n-1$ variables... | |
| Feb 14, 2016 at 21:29 | comment | added | Hamed | @JasonStarr I actually tried that book, it is a little bit above my pay-grade. Another way I could've asked this question is: A step-by-step program to get ready to understand that book. | |
| Feb 14, 2016 at 21:26 | comment | added | Hamed | @Qfwfq Oh, a polynomial $f(z_1, \cdots, z_n)$ is translation invariant if for any constant $c\in \mathbb{C}$, you have $f(z_1+c, \cdots, z_n+c)=f(z_1, \cdots, z_n)$. Another way to define it is through the operator $D=\sum \partial_i$, then $f$ is translation invariant if $Df=0$. Edit: We are interested in multi-variable polynomials (these $z_i$ are in physics the coordinates of a particle). | |
| Feb 14, 2016 at 19:24 | comment | added | Qfwfq | Probably I'm misunderstanding your terminology, but how can a (non constant) polynomial $P(x)$ be translation invariant? | |
| Feb 14, 2016 at 14:56 | comment | added | Jason Starr | Nakajima has a textbook about Hilbert schemes: "Lectures on Hilbert Schemes of Points on Surfaces". Here is a link to the webpage for the book: bookstore.ams.org/ulect-18 | |
| Feb 14, 2016 at 7:04 | comment | added | roy smith | read mumford's lectures on curves on an algebraic surface. | |
| Feb 14, 2016 at 2:19 | comment | added | Lev Borisov | Step one: talk to Davesh Maulik :) | |
| Feb 14, 2016 at 2:01 | history | asked | Hamed | CC BY-SA 3.0 |