Timeline for one "big" Hilbert scheme?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 15, 2013 at 13:59 | comment | added | IMeasy | welll yes.... it is actually what I meant! | |
| Mar 10, 2013 at 5:17 | comment | added | user30379 | Dear IMeasy: My above comment can stand on its own merits without needing "validation". :) | |
| Mar 9, 2013 at 20:44 | comment | added | IMeasy | Hey put your comment in an answer so that I validate it! | |
| Mar 9, 2013 at 18:25 | comment | added | user30379 | Yes, the Hilbert scheme classifying flat closed (finitely presented) subschemes of a projective space is a disjoint union indexed by the various Hilbert polynomials. The functor makes sense without reference to ample line bundles or Hilbert polynomials (and as such is an algebraic space when one goes beyond the setting of projective geometry). The Hilbert polynomial is a means of cutting it into (not obviously) finite-type pieces in the projective case; there is nothing like that available for Hilbert functors of more general proper finitely presented schemes in place of projective spaces. | |
| S Mar 9, 2013 at 17:48 | vote | accept | IMeasy | ||
| Mar 9, 2013 at 17:48 | vote | accept | IMeasy | ||
| S Mar 9, 2013 at 17:48 | |||||
| Mar 9, 2013 at 17:47 | vote | accept | IMeasy | ||
| Mar 9, 2013 at 17:48 | |||||
| Mar 9, 2013 at 17:47 | comment | added | IMeasy | don't take me bad... but, to obtain a Hilbert scheme, doesn't one need to fix the Hilbert polynomial? The Hilb-scheme parametrizing all subchemes is just the disjoint union of the different Hilb-schemes? | |
| Mar 9, 2013 at 10:30 | answer | added | Damien L | timeline score: 0 | |
| Mar 9, 2013 at 9:48 | answer | added | Marc Palm | timeline score: 0 | |
| Mar 9, 2013 at 9:35 | history | asked | IMeasy | CC BY-SA 3.0 |