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Let $X$ be a variety contained in $\mathbb{P}^N$ and let $Y$ be a smooth hyperplane section of $X$. I have read in page 54 of Voisin's book "Hodge theory and complex algebraic geometry II" the following:

there exists a Lefschetz pencil of hyperplane sections of $X$ of which $Y$ is one member.

I would like to know why this is true. Any reference for it will be also very helpful. Thank you very much :)!

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A pencil of hyperplane sections of $X$ corresponds to a line in the dual projective space $\check{\mathbb{P}}^N$. It is a Lefschetz pencil if and only if the line is transverse to the projectively dual variety $$ \check{X} \subset \check{\mathbb{P}}^N. $$ So, a restatement of your claim is that for a point $y \in \check{\mathbb{P}}^N$ not on $\check{X}$ there is a line through $y$ transverse to $\check{X}$, which is obvious --- just consider the linear projection map $$ \pi_y \colon \check{X} \to \check{\mathbb{P}}^{N-1} $$ from $y$, choose a point in the target not on the critical locus of $\pi_y$, and take the corresponding line.

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  • $\begingroup$ Dear Sasha thank you very much for your answer!... could you tell me how $\pi_y$ is defined and what you mean for the critical locus of $\pi_y$? $\endgroup$
    – Roxana
    Commented Apr 2, 2023 at 10:30
  • $\begingroup$ $\pi_y$ is the linear projection $\mathbb{P}^N \dashrightarrow \mathbb{P}^{N-1}$ with center at $y$, restricted to $\check{X}$. The critical locus, or rather the set of critical values, is the image of the union of the singular locus of $\check{X}$ and the set of points of $\check{X}$ at which the differential of $\pi_y$ is not surjective. $\endgroup$
    – Sasha
    Commented Apr 2, 2023 at 11:52
  • $\begingroup$ Dear Sasha thank you very much! for the linear projection $\pi_y$ with center in $y$, you mean a notion similar to the Stereographic projection? could you suggest a reference where I can read what exactly you mean :)... $\endgroup$
    – Roxana
    Commented Apr 2, 2023 at 14:07
  • $\begingroup$ Choose coordinates such that $y = (1:0:\dots:0)$. Then $$\pi_y(x_0:x_1:\dots:x_n) = (x_1:\dots:x_n).$$ $\endgroup$
    – Sasha
    Commented Apr 2, 2023 at 14:56
  • $\begingroup$ Dear Sasha thank you very much for the clarifications :)!... I have the following questions: 1). Why I need to choose a point in the target not on the critical locus of $\pi_y$? $\endgroup$
    – Roxana
    Commented Apr 4, 2023 at 21:59

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