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Timeline for determinant of normal bundle ample

Current License: CC BY-SA 3.0

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Feb 17, 2014 at 21:30 comment added Damian Rössler @Jason Starr. A simple abelian variety has that property.
Feb 17, 2014 at 17:27 history edited user45766 CC BY-SA 3.0
deleted 148 characters in body; edited title
Feb 17, 2014 at 16:22 comment added Georges Elencwajg @user45766: It is extremely unpleasant that you have completely deleted the original question which just asked for a formula for the determinant of the normal bundle. I have given you precisely such a formula , but now my answer looks like a complete non sequitur because of your modifications. Please modify your post in order that the original question is re-established and add your question on ampleness below.
Feb 17, 2014 at 13:35 answer added Jason Starr timeline score: 2
Feb 17, 2014 at 13:28 review Close votes
Feb 17, 2014 at 15:48
Feb 17, 2014 at 13:22 comment added Jason Starr @abx: I think there is a precise question here, "Is there any smooth projective variety $X$ other than $\mathbb{P}^n$ such that for every smooth subvariety $Y$, $\text{det}\ \mathcal{N}_{Y/X}$ is ample?"
Feb 17, 2014 at 13:09 comment added abx What is the question? Do you want $X$ to be a surface, and $Y$ a curve? Then of course $\mathcal{N}_{Y/X}$ is ample iff $Y^2>0$. Please edit your question so that it makes reasonable sense (or delete it).
Feb 17, 2014 at 12:29 answer added Georges Elencwajg timeline score: 3
Feb 17, 2014 at 12:22 history edited user45766 CC BY-SA 3.0
added 160 characters in body; edited title
Feb 17, 2014 at 12:16 comment added user5117 Dear user45766, ok. That seems like quite a different (and much more focused) question. You might consider editing your question to clarify this.
Feb 17, 2014 at 12:13 comment added user45766 I am interested in the problem under what kind of assumptions $\mathrm{det}\mathcal{N}_{Y/X}$ is ample?
Feb 17, 2014 at 11:59 comment added user5117 Do you know the adjunction formula?
Feb 17, 2014 at 11:55 history asked user45766 CC BY-SA 3.0