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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!

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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 '14 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 '14 at 17:57
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 '14 at 18:14
That would be great, thank you! – Bonanza Jan 2 '14 at 19:27
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).

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Excellent. Thank you. – Bonanza Jan 4 '14 at 20:38

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