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Projective duality is a duality that associates to a (smooth) subvariety X of $\mathbb{P}^n$ the dual variety $X^*\subset\mathbb{P}^{n*}$ of tangent hyperplanes.

What makes the duality interesting is that if we work over an algebraically closed field of characteristic zero it is an actual duality, i.e. $(X^*)^*=X$.

Are there are references that analyse projective duality over not algebraically closed base fields or even dedekind schemes?

Thank you in advance for your help.

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  • $\begingroup$ Since $X^*$ is typically singular, what do you mean by $(X^*)^*$? $\endgroup$ Jan 11, 2017 at 14:37
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    $\begingroup$ Take the closure of the variety obtained taking tangent hyperplanes on the smooth locus. $\endgroup$
    – Bear
    Jan 11, 2017 at 14:59

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If I remember correctly, over any field of characteristic $2$, there is a standard example of a smooth conic $\mathcal{C} \subset \mathbb{P}^2$ and a point $a \in \mathbb{P}^2$ such that all lines going through $a$ are tangent to $\mathcal{C}$.

Hence, the projective dual to $\mathcal{C}$ is not integral. In the comments below, Noam notices that $\mathcal{C}^*$ is $a^{\perp}$ with multiplicity $2$. Hence the dual of $\mathcal{C}^*$ is not even well-defined.

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    $\begingroup$ Not $a^\perp$ with multiplicity $2$? $\endgroup$ Jan 11, 2017 at 23:26
  • $\begingroup$ @NoamD.Elkies : you are right, $\mathcal{C}^*$ is $a^{\perp}$ with multiplicity $2$. Hence the dual of $\mathcal{C}^*$ is not even well-defined. $\endgroup$
    – Libli
    Jan 12, 2017 at 0:13
  • $\begingroup$ Probably I'm missing something, but aren't you saying that that there is an example in char 2 of a conic whose projective dual is a double line? So the projective dual is well defined but the conic won't be the dual of its dual. I don't see a problem with that. $\endgroup$
    – Bear
    Jan 12, 2017 at 12:02
  • $\begingroup$ @Bear : how do you define the projective dual of a double line? $\endgroup$
    – Libli
    Jan 12, 2017 at 18:39

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