Timeline for From abstract to concrete complex projective varieties
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 22, 2012 at 22:38 | comment | added | David Feldman | @Will Yes, please. Posting I didn't know whether this was a mere exercise for an expert or an open problem. Evidently a little of both. And please say something more about the not a complete intersection story. | |
| Aug 22, 2012 at 21:05 | comment | added | Damian Rössler | Sorry, I just realized that I misread the definition (I thought the hypersurfaces were elements of the multisets). | |
| Aug 22, 2012 at 20:49 | comment | added | David Feldman | I doubt that my ${\cal H}_X$ determines $X$ up to isomorphism: suspect that ${\cal H}_X$ functions as a de facto discrete invariant insensitive to moduli. And I never asked about determining $X$ up to isomorphism, only ${\cal H}_X$. | |
| Aug 22, 2012 at 20:36 | comment | added | Will Sawin | This will not be so well-behaved for embeddings where $X$ is not a complete intersection. When it is, you can extract that multiset from the Hilbert polynomial of $X$. The set of Hilbert polynomials of line bundles of $X$ can be described using Hirzebruch-Riemann-Roch and the intersection theory on $X$. Is that what you're looking for? | |
| Aug 22, 2012 at 20:17 | comment | added | Damian Rössler | If I understand well, the invariant $H_X$ determines $X$ up to isomorphism (as an abstract variety). I don't understand what it would mean to determine $X$ up to isomorphy from familiar invariants. | |
| Aug 22, 2012 at 17:57 | comment | added | Donu Arapura | OK, I misunderstood. | |
| Aug 22, 2012 at 17:56 | comment | added | David Feldman | @Donu I don't understand. I intended my ${\cal H}_X$ to vary over all embeddings into projective space. | |
| Aug 22, 2012 at 17:51 | comment | added | Donu Arapura | A few quick comments: (a) This depends on the choice of embedding into projective space. So it's not an invariant of $X$ alone. (b) It is more natural to consider the ideal of polynomials vanishing along $X$. | |
| Aug 22, 2012 at 17:38 | history | asked | David Feldman | CC BY-SA 3.0 |