# Arithmetic projective duality

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.

• Since $X^*$ is typically singular, what do you mean by $(X^*)^*$? – Jason Starr Jan 11 '17 at 14:37
• Take the closure of the variety obtained taking tangent hyperplanes on the smooth locus. – Bear Jan 11 '17 at 14:59

## 1 Answer

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.

• Not $a^\perp$ with multiplicity $2$? – Noam D. Elkies Jan 11 '17 at 23:26
• @NoamD.Elkies : you are right, $\mathcal{C}^*$ is $a^{\perp}$ with multiplicity $2$. Hence the dual of $\mathcal{C}^*$ is not even well-defined. – Libli Jan 12 '17 at 0:13
• 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. – Bear Jan 12 '17 at 12:02
• @Bear : how do you define the projective dual of a double line? – Libli Jan 12 '17 at 18:39