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Apr 13, 2017 at 12:58 history edited CommunityBot
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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
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