Timeline for Family with a fixed special fiber over finite fields
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 15, 2013 at 23:11 | comment | added | Matt | Alright, then I think you mean that $X'$ should be over $\mathbb{Z}_p$ otherwise the dimensions won't make any sense. Every smooth projective curve always lifts to characteristic $0$ because the obstructions to deforming both the curve and an ample line bundle lie in $H^2$ which vanishes since it is a curve. So by Grothendieck's Existence Theorem the formal lift algebraizes. Surfaces are more delicate. There are known results like every K3 surface lifts to characteristic $0$, but there are also known surfaces that do not lift. | |
| Mar 15, 2013 at 19:40 | comment | added | Naga Venkata | @Matt: I am interested in curves and surfaces. We can assume that $\dim X'= \dim X+1$. | |
| Mar 15, 2013 at 15:52 | comment | added | Matt | This might be standard, but a variety over $\mathbb{Q}_p$ with "special fiber ..." just means a scheme $\mathfrak{X}/\mathbb{Z}_p$ whose generic fiber is $X'$ and special fiber is $X$ right? There are lots of examples of when this can't happen. Sometimes they formally lift, but aren't projective. Sometimes the deformation theory is highly obstructed. I'm not sure there can ever be some general conditions on $X$ that would guarantee a lift. Are you interested in a more specific type of variety or at least of a particular dimension. | |
| Mar 15, 2013 at 7:40 | comment | added | JRN | I suspect it is related to this: tea.mathoverflow.net/discussion/1552/string-of-down-votes | |
| Mar 15, 2013 at 7:10 | comment | added | Yemon Choi | Why the downvote? | |
| Mar 15, 2013 at 1:40 | history | asked | Naga Venkata | CC BY-SA 3.0 |