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Let $X$ be a smooth, projective variety over $\mathbb{C}$ of dimension $n$ satisfying the property that for every $i \ge 0$, $H^{i,i}(X,\mathbb{C}) \cap H^{2i}(X,\mathbb{Q})=\mathbb{Q}c_1(\mathcal{O}_X(1))^i$. Is it then true for a very general smooth hyperplane section $Y \subset X$, we have $H^{i,i}(Y,\mathbb{C}) \cap H^{2i}(Y,\mathbb{Q})=\mathbb{Q}c_1(\mathcal{O}_Y(1))^i$? In particular, is it true that for a fixed integer $n$ and a very general complete intersection subvariety $Y$ of $\mathbb{P}^n$, we have $H^{i,i}(Y,\mathbb{C}) \cap H^{2i}(Y,\mathbb{Q})=\mathbb{Q}c_1(\mathcal{O}_Y(1))^i$? We know that this holds true for very general hypersurfaces in $\mathbb{P}^n$.

Any hint/reference will be most welcome.

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  • $\begingroup$ The question would make more sense with $\Bbb{Z}$ replaced by $\Bbb{Q}$; then the answer (for the last question) would be positive, with some exceptions. See Deligne's Exposé 19 in SGA 7. $\endgroup$
    – abx
    Commented Mar 11, 2020 at 5:28
  • $\begingroup$ @abx Thank you, I have edited the question by replacing $\mathbb{Z}$ by $\mathbb{Q}$. $\endgroup$
    – Jana
    Commented Mar 11, 2020 at 5:38
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    $\begingroup$ Now Theorem 1.4 in Deligne's Exposé 19 gives you the best general result you can hope for. $\endgroup$
    – abx
    Commented Mar 11, 2020 at 5:49

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I guess that this is not true even for hypersurfaces. Take a smooth quadric in $Y=Q \subset \mathbb{P}^{3} = X$. Then $h^{1,1}(Q) = b_{2}(Q) = 2$, so the second cohomology cannot be generated by $1$ element.

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  • $\begingroup$ But, very general hypersurfaces in $\mathbb{P}^3$ of degree at least 4 has Picard number $1$. This is the Noether-Lefschetz theorem. $\endgroup$
    – Jana
    Commented Mar 11, 2020 at 5:38

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