Timeline for Vector field on a K3 surface with 24 zeroes
Current License: CC BY-SA 3.0
12 events
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Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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Oct 17, 2015 at 21:56 | comment | added | mkemeny | K3s have no (nontrivial and global) algebraic (or even analytic) vector fields, see page 15 of math.uni-bonn.de/people/huybrech/K3Global.pdf .The best I imagine you are going to get is something like the answer given by Danny Ruberman. | |
Oct 15, 2015 at 20:56 | comment | added | David Roberts♦ | @AllenKnutson that's what I was hoping. Otherwise some nice analytic vector fields, given by equations, would be perfectly good... | |
Oct 15, 2015 at 20:49 | comment | added | David Roberts♦ | @potentiallydense well, your name is describing me well, so I don't mind. Note that I'm not by any stretch of the imagination an algebraic geometer! | |
Oct 15, 2015 at 19:28 | comment | added | Allen Knutson | Are there $K3$s with enough algebraic vector fields to make this happen? | |
Oct 15, 2015 at 13:12 | comment | added | Lazzaro Campeotti | Dear David, a smooth projective surface minus finitely many points cannot be affine, because functions which are regular outside a subset of codimension 2 are regular everywhere. (Sorry to keep being annoying.) | |
Oct 15, 2015 at 12:07 | answer | added | Danny Ruberman | timeline score: 8 | |
Oct 15, 2015 at 9:48 | comment | added | David Roberts♦ | Even better would be a K3 minus 24 points that is affine, ie zero locus of vector field is the intersection with plane at infinity of a projective K3. | |
Oct 15, 2015 at 9:45 | history | edited | David Roberts♦ | CC BY-SA 3.0 |
added 23 characters in body
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Oct 15, 2015 at 9:45 | comment | added | David Roberts♦ | Erg, I meant projective variety, not affine! | |
Oct 15, 2015 at 8:40 | comment | added | Lazzaro Campeotti | Nice question. Sorry to pick a nit, but what does "affine K3" mean? | |
Oct 14, 2015 at 22:38 | history | asked | David Roberts♦ | CC BY-SA 3.0 |