Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $Y\subset \mathbb{P}^n$ be a smooth projective variety and let $H$ be a smooth hypersurface in $\mathbb{P}^n$ such that $Z=Y\cap H$ is smooth. How are the normal bundles of the various embeddings of $Z$ related?

More precisely, does either of the following hold?

(1) $N_{Z|H}=N_{Y|\mathbb{P}^n}|_Z$?

(2) $N_{Z|\mathbb{P}^n}=N_{Y|\mathbb{P}^n}|_Z\oplus N_{H|\mathbb{P}^n}|_Z$?

Any hints or references would be greatly appreciated!

share|improve this question
1  
I assume you intend that $Y$ and $H$ intersect transversely? Then (1) is correct (as is $N_{Z/Y}\cong N_{H/\Bbb P^n}$). (2) seems to be true, as (1) should split the exact sequence $$0\to N_{Z/Y}\to N_{Z/\Bbb P^n}\to N_{Y/\Bbb P^n}\big|_Z\to 0.$$ –  Ted Shifrin Jan 2 at 15:10
    
Thanks for the comment! Could you add some details? Why is $N_{Z|Y}=N_{H|P^n}$? And how does this imply (1)? –  Bonanza Jan 2 at 17:57
1  
When I make it home to a regular computer, I'll try to give you a more complete answer if no one else has done so. I hope to get to it tomorrow, weather permitting! –  Ted Shifrin Jan 2 at 18:14
    
That would be great, thank you! –  Bonanza Jan 2 at 19:27

1 Answer 1

up vote 4 down vote accepted

$\require{AMScd} \newcommand{\P}{\Bbb P^n}$Preliminary comment: Your title "hyperplane section" is more specific than you wrote in the body of the question. It holds for any smooth hypersurface $H$ (or one that is smooth where it intersects $Y$). In fact, the result holds more generally for $Z=Y\cap V$ for any smooth subvarieties $Y,V$.

OK, so here we go. First, by definition (in the algebraic category), the normal bundle of a submanifold (smooth subvariety) $Y\subset X$ is given by $N_{Y/X} = TX\big|_Y/TY$.

Next, a crucial observation is that transversality of two submanifolds (smooth subvarieties) $H,Y\subset X$ is the condition that at each point $x\in H\cap Y$ we have $T_x H + T_x Y = T_x X$. This is equivalent to saying that $T_x H \to T_x X/T_x Y = N_x(Y/X)$ is surjective.

So, now we make a ridiculous commutative diagram that puts $Z=H\cap Y$ in its proper place. $$\begin{CD} @. 0 @. 0 @. 0 @. \\ @. @VVV @VVV @VVV @. \\ 0 @>>> TZ @>>> TY\big|_Z @>>> N_{Z/Y} @>>> 0 \\ @. @VVV @VVV @V{\phi}VV @. \\ 0 @>>> TH\big|_Z @>{\alpha}>> T\P\big|_Z @>>> N_{H/\P}\big|_Z @>>> 0 \\ @. @VVV @V{\beta}VV @VVV @. \\ 0 @>>> N_{Z/H} @>{\iota}>> N_{Y/\P}\big|_Z @>>> ? @>>> 0 \\ @. @VVV @VVV @VVV @.\\ @. 0 @. 0 @. 0 @. \end{CD}$$ Now, by our transversality hypothesis, $\beta\circ\alpha$ is surjective, and so $\iota$ is surjective; i.e., $\iota$ is an isomorphism. This means that we must have $?=0$. We're done, by symmetry, but, having come this far, we apply the so-called Nine Lemma and conclude that $\phi$ is an isomorphism as well. This completes the proof of (1).

Item (2) follows from another application of the Nine Lemma. $$\begin{CD} @. 0 @. 0 @. 0 @. \\ @. @VVV @VVV @VVV @. \\ 0 @>>> TZ @>>> TY\big|_Z @>>> N_{Z/Y} @>>> 0 \\ @. @| @VVV @V{\phi}VV @. \\ 0 @>>> TZ @>>> T\P\big|_Z @>>> N_{Z/\P}\big|_Z @>>> 0 \\ @. @VVV @VVV @VVV @. \\ 0 @>>> 0 @>>> N_{Y/\P}\big|_Z @>>> K @>>> 0 \\ @. @VVV @VVV @VVV @.\\ @. 0 @. 0 @. 0 @. \end{CD}$$ Letting $K=\text{coker}(\phi)$, the Nine Lemma tells us that $K\cong N_{Y/\P}\big|_Z$. But we already know that $N_{Y/\P}\big|_Z\cong N_{Z/H}$, and so we have the short exact sequence $$0\to N_{Z/Y} \to N_{Z/\P} \to N_{Z/H} \to 0\,.$$ But this exact sequence splits, since we could swap the rôles of $Y$ and $H$. Thus, as desired, $N_{Z/\P}\cong N_{Z/Y}\oplus N_{Z/H}$. Nowhere in this argument did we use anything about $H$'s being a hypersurface (i.e., $N_{H/\P}$'s being a line bundle).

share|improve this answer
    
Excellent. Thank you. –  Bonanza Jan 4 at 20:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.