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Let $X\subset\mathbb{P}^n$ be e normal variety, $L\subset\mathbb{P}^n$ a linear subspace, and $Y = X\cap L$ a linear section. Assume that $Y$ is also normal. In particular, we have that $Sing(X)$ has codimension at least two in $X$, and $Sing(Y)$ has codimension at least two in $Y$.

Does there exist any generalization of the Lefschetz hyperplane theorem to higher codimension linear sections of singular varieties ensuring that the restriction map

$$Cl(X)\rightarrow Cl(Y)$$

is an isomorphism or at least surjective?

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  • 1
    $\begingroup$ You would need some assumptions. For instance, a rank 5 quadric $X$ in $\mathbb{P}^n$ has $\operatorname{Cl}(X)= \mathbb{Z}$, but it admits a hyperplane section $Y$ of rank 4, hence $\operatorname{Cl}(Y)=\mathbb{Z}^2 $. $\endgroup$ – abx May 16 at 8:02
  • $\begingroup$ In your example we have $Sing(X) = Sing(Y)$ if I got it right. In the case I am interested in I know that $Sing(X) = Sing(Y)\cap L$ but $Sing(X)\neq Sing(Y)$. $\endgroup$ – Jessica_90 May 16 at 8:16
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    $\begingroup$ Look for Grothendieck-Lefschetz theorey, for instance this paper: arxiv.org/pdf/1601.05846.pdf $\endgroup$ – Hailong Dao May 16 at 16:28
  • $\begingroup$ In Theorem 1 (v) here dima.unige.it/~badescu/attivita%20scientifica/… they say that if $Y\subset \mathbb{P}^n$ is normal of dimension at least three and $Y$ can be set-theoretically defined by at most $n-3$ equations then $Pic(\mathbb{P}^n)\rightarrow Pic(Y)$ is an isomorphism. Is not this in contradiction with abx example? We could take for instace a quadric cone with vertex a point in $\mathbb{P}^4$. $\endgroup$ – Jessica_90 May 17 at 6:27
  • $\begingroup$ No contradiction. $\operatorname{Pic}(Y) $ is not the same thing as $\operatorname{Cl}(Y) $. $\endgroup$ – abx May 17 at 14:28

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