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Let $X$ be a(n even dimensional) smooth complex projective variety in $\mathbb{P}^N$, and let $X^{\vee}$ be its dual variety; up to an higher degree Veronese embedding of $X$, I assume that $X^{\vee}$ is a hypersurface (that is, $X$ has no defect).

By construction, for any $H\in X^{\vee}$, the corresponding hyperplane cuts $X$ in a singular section $X_H=X\cap H$.

By Theorem of Reflexivity, for any $(P,H)\in X\times X^{\vee}_{sm}$ such that $H$ is tangent to $X$ at $P$ is equivalent to say the hyperplane corresponding to $P$ in $\overset{\vee}{\mathbb{P}^N}$ is tangent to $X^{\vee}$ at $H$.

Question: assuming that the singular points of $X_H$ are only double ordinary (for simplicity, all singularities are nodes, so $X_H$ is a nodal section), what can I say about $H\in X^{\vee}$?

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  • $\begingroup$ What kind of property are you asking for? $H$ is a smooth point if and only if $X_H$ has exactly one ordinary double point. What else? $\endgroup$ – abx May 16 at 10:26
  • $\begingroup$ 1) I am not assuming that $X_H$ has exactly only one node. 2) How can I see the equivalence between the smoothness of $H$ and the "nodality" of $X_H$? $\endgroup$ – Armando j18eos May 16 at 10:33
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    $\begingroup$ One reference is SGA 7, Exp. XVII, Proposition 3.5. If you have $n$ nodes, $X^{\vee}$ has $n$ branches locally around $H$, each branch being smooth. $\endgroup$ – abx May 16 at 12:06
  • $\begingroup$ Perfect! This is very enlightening. $\endgroup$ – Armando j18eos May 16 at 12:54
  • $\begingroup$ Also see Tevelev's Projectively Dual Varieties. $\endgroup$ – AG learner May 16 at 15:58

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