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For simplicity let's suppose $n=2$ and $\Bbb{P}^{n}_k = \operatorname{Proj}k[X,Y,Z]$ and $\Bbb{P}^{n\vee}_k = \operatorname{Proj}k[A,B,C]$. With the segre embeding

$$ \operatorname{Proj}k[X,Y,Z] \times \operatorname{Proj}k[A,B,C]\to\operatorname{Proj}k[AX,AY,\cdots, CZ] $$

and define "the incident variety" $\Sigma$ (or, "the incident correspondence", only find this terminology in Vakil's FOAG,No. 13.4.1) be the closed subscheme in $\operatorname{Proj}k[AX,AY,\cdots, CZ]$ vanishing $AX+BY+CZ$, i.e.,

$$ \Sigma := V(AX+BY+CZ) = \operatorname{Proj}\frac{k[AX,AY,\cdots, CZ]}{AX+BY+CZ} $$

Prove that $\Sigma \to \operatorname{Proj}k[A,B,C]$ is the projectivization of some locally free sheaf of rank 2 on $\operatorname{Proj}k[A,B,C]$, i.e., there is some locally free sheaf $\mathscr{E} $of rank 2 on $\operatorname{Proj}k[A,B,C]$, such that $\Sigma \to \operatorname{Proj}k[A,B,C]$ is the relative proj of $\operatorname{Sym}\mathscr{E}$:

$$ \Sigma = \mathcal{Proj}\operatorname{Sym}\mathscr{E} \to \operatorname{Proj}k[A,B,C] $$

I saw this question from David Eisenbud, Joe Harris, The geometry of schemes, Exercise III-25, Page104

The background for it is that I am continuing trying to gain some understanding the scheme theoretic dual projective space $\Bbb{P}_k^{n\vee}$, following a question I posted Two definitions of scheme theoretic dual projective space.

The following is what I have tried:

Looking at it in the affine open subscheme $\operatorname{Spec} k[U,V]$ where $U=B/A, V=C/A$, I got

$$ \operatorname{Proj}\frac{k[AX,AY,\cdots, CZ]}{AX+BY+CZ} \to \operatorname{Spec} k[U,V] $$ Then "since $A \neq 0$" (this makes no sense scheme theoretically, I just want to show what I have tried), I got (from no reason) $$ \operatorname{Proj}\frac{k[U,V][X,Y,Z]}{X+UY+VZ} \to \operatorname{Spec} k[U,V] $$

Then I failed to describe the left hand side of the above map as the symmetric sheaf of some locally free sheaf over $\operatorname{Spec} k[U,V]$, making less gluing them into a locally free sheaf on $\operatorname{Proj}k[A,B,C]$.

Is the above argument a right direction? If so, how to make it rigorous and conitue the argument? As in the comment of the answer here, the sheaf $\mathscr{E}$ seems to be some "equivalent class (maybe quotient out $\mathscr{O^*}$?) of the sheaf $\mathscr{O}(1)$", in such a way it corresponds to the inutition "A point of the dual projective space is an equivalence class of linear forms on the original vector space, which defines a hyperplane in the usual way." that Zhen Lin said. But I failed to make it concide with the above argument.

Thanks for reading, any help would be appreciated.

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    $\begingroup$ Your $n$s and your variables don't match up - $\operatorname{Proj} k[x,y,z]\cong\Bbb P^2$, not $\Bbb P^3$. Also, one tip for organization: it would be nice to put the problem before your work - it seems that in this post you start working on things first and then say what the problem is, and that's a little harder to follow. $\endgroup$
    – KReiser
    May 28, 2023 at 21:12
  • $\begingroup$ @KReiser Thank you. I fixed some typos and reorganized the text. $\endgroup$
    – onRiv
    May 29, 2023 at 0:14

1 Answer 1

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Still not totally accurate, but anyway I update the rest that I have tried as an answer, since it seems giving the desired sheaf.

By replacing $X$ to $-UY-VZ$, the scheme $\operatorname{Proj}\frac{k[U,V][X,Y,Z]}{X+UY+VZ}$ in fact is just $\operatorname{Proj}k[U,V][Y,Z]$. Hence on the affine open subscheme $\operatorname{Spec} k[U, V]$, the incident variety $\Sigma$ is the projectivization of the rank 2 free sheaf with generators $Y$ and $Z$.

Changing $U, V$ back to $B/A$ and $C/A$, the transition map from $\operatorname{Spec} k[B/A, C/A]$ to $\operatorname{Spec}k[A/B, C/B]$, by $Y = -A/B \cdot X -C/B \cdot Z$, is

\begin{aligned} \left(\begin{array}{l} Y \\ Z \end{array}\right) & =\left(\begin{array}{c} -\frac{A}{B} X-\frac{C}{B} Z \\ Z \end{array}\right) \\ & =\left(\begin{array}{cc} -\frac{A}{B} & -\frac{C}{B} \\ 0 & 1 \end{array}\right)\left(\begin{array}{l} X \\ Z \end{array}\right) \end{aligned}

and the other two transition maps have similar forms.

Hence "symbolically", it has the same form with the classic case treating dual projective space as linear forms.

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