Let $X$ be a smooth, projective variety over $k$ and $Y$ a smooth closed subvariety of codimension r. Let $\mathcal{N}_{Y/X}$ be the rank r normal bundle. Is it possible to determine when $\mathrm{det}\mathcal{N}_{Y/X}$ ?
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$\begingroup$ Do you know the adjunction formula? $\endgroup$– user5117Commented Feb 17, 2014 at 11:59
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$\begingroup$ I am interested in the problem under what kind of assumptions $\mathrm{det}\mathcal{N}_{Y/X}$ is ample? $\endgroup$– user45766Commented Feb 17, 2014 at 12:13
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1$\begingroup$ @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?" $\endgroup$– Jason StarrCommented Feb 17, 2014 at 13:22
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1$\begingroup$ @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. $\endgroup$– Georges ElencwajgCommented Feb 17, 2014 at 16:22
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1$\begingroup$ @Jason Starr. A simple abelian variety has that property. $\endgroup$– Damian RösslerCommented Feb 17, 2014 at 21:30
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2 Answers
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Given an exact sequence of locally free sheaves $0\to E'\to E\to E''\to 0$ on, say, a ringed space, there is a canonical isomorphism $\text {det} E=\text {det} E'\otimes \text {det} E''$: this is pure multilinear algebra.
Now in your situation apply this to the exact sequence defining the normal bundle $$ 0\to T_Y\to T_X|Y \to N_{Y/X}\to 0$$ and get $\text {det} (T_X|Y)=\text {det} T_Y\otimes \text {det} N_{Y/X}$.
From this you obtain your required formula $$\text {det} N_{Y/X}=\text {det} T_X|Y \otimes \text {det} T_Y^*$$.
Classically geometers express this in terms of canonical bundles $K_X=\text {det}\: T_X^*, K_Y=\text {det}\: T_Y^*$ as $$K_Y=K_X|Y\otimes \text {det} N_{Y/X}$$ This is the adjunction formula mentioned by Artie: I wrote this answer to show that it is a triviality!
answered Feb 17, 2014 at 12:29
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$\begingroup$ This is an answer to the original question which just asked for a formula for the determinant, and I gave such a formula here . The original question did not even mention the word "ample" ! $\endgroup$ Commented Feb 17, 2014 at 16:28
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$\begingroup$ The more interesting question is under what kind of assumptions $\mathrm{det}\mathcal{N}_{Y/X}$ is ample ? As far as I know for $X$ a projective space this is true, but what is for example for Hirzebruch surfaces or $\mathbb{P}^1\times\mathbb{P}^1$? $\endgroup$ Commented Feb 17, 2014 at 17:29
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2$\begingroup$ Well that new question may be more interesting but it is not what you asked originally. $\endgroup$ Commented Feb 17, 2014 at 17:37
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As others have pointed out, the question is somewhat imprecise. Here is one precise formulation (of course I do not know if this is what the OP has in mind): "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?" The OP asks about $\mathbb{P}^1 \times \mathbb{P}^1$ or other Hirzebruch surfaces. Of course these varieties fail the condition. In general, if $f:X\to Z$ is a fiber-type contraction and if $Y\subset X$ is a general fiber, then $\mathcal{N}_{Y/X}$ is $\mathcal{O}_Y^{\oplus r}$ where $r$ equals $\text{dim}(Z)$. So for any $X$ as above, $X$ admits no fiber-type contraction. If one allows $Y$ to be somewhat singular, probably one can rule out all contractions.
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2$\begingroup$ Good question, but I am not sure there is a reasonable answer. For instance, an abelian variety will have this property if and only if it is simple. $\endgroup$– abxCommented Feb 17, 2014 at 13:47