Skip to main content
8 events
when toggle format what by license comment
Mar 7, 2013 at 3:32 vote accept Fei YE
Mar 1, 2013 at 8:21 answer added Sándor Kovács timeline score: 7
Feb 26, 2013 at 11:25 comment added Chris Brav As for classification, in general no. Sometimes if you fix some additional geometric conditions, say smooth Fano threefold, then there is a classification of those with Picard number 1. But for other geometric conditions, say Calabi-Yau threefolds, then (last time I checked) there is not yet a classification of those with Picard number 1.
Feb 26, 2013 at 10:31 answer added Yuri Zarhin timeline score: 11
Feb 26, 2013 at 8:54 comment added Serge Lvovski There are many. Not only most Grassmannians, but most flag varieties (of semisimple groups) with Picard number one are not complete intersections. A still bigger series of examples is as follows. Consider an arbitrary smooth projective variety $X\subset\mathbb P^n$ with Picard number one. Even if it is a complete intersection in $\mathbb P^n$, its $m$-fold embedding, for $m\gg0$, is not a complete intersection anymore (proof: canonical class of a c.i. must be a multiple of the class of hyperplane section).
Feb 26, 2013 at 8:29 comment added Fei YE Thanks a lot for the comment. Are there other types?
Feb 26, 2013 at 6:24 comment added Atsushi Kanazawa For example, Grassmannian $Gr(k,n)$ is in general not complete intersection of any projective spaces, but is smooth projective and has Picard number $1$.
Feb 26, 2013 at 5:44 history asked Fei YE CC BY-SA 3.0