Vakil’s FOAG gives the definition of the dual peojective space via introducing new indeterminates: (sorry that I have to quote it as a screenshot)
And the answer in Scheme theoretic dual of $\mathbb P^n_k$ gives a coordinate free definition, as $(\Bbb{P}V)^\vee = \Bbb{P}(V^\vee )$. How is this coordinate free definition connect to the relation $a_0 x_0 + a_1 x_1 + \cdots a_n x_n = 0$? Given $a_i$ the dual base for $x_i$, shouldn’t it be $a_i x_j =\delta_{ij}$?
Thank you in advance.
Best Answer
The incidence variety is $I\subset\mathbb P(V)\times\mathbb P(V^\vee)$ consisting of $(v,v^\vee)\in\mathbb P(V)\times\mathbb P(V^\vee)$ such that $v^\vee(v)=0$, which is well-defined since the condition is invariant under constant multiplication. Now writing the equation in terms of bases recovers Vakil's description.
In terms of the Proj construction, $\mathbb P(V)=\mathrm{Proj}(\mathrm{Sym}^\bullet V^\vee)$ and $\mathbb P(V^\vee)=\mathrm{Proj}(\mathrm{Sym}^\bullet V)$, so $$\mathbb P(V)\times \mathbb P(V^\vee)=\mathrm{Proj}(\bigoplus_{n\ge0}\mathrm{Sym}^nV^\vee\otimes \mathrm{Sym}^n V).$$ Now the incidence variety is defined by the ideal generated by the element $\mathrm{id}_{V}\in\mathrm{hom}(V,V)\cong V\otimes V^\vee$, which gives an element of degree $1$ in $\bigoplus_{n\ge0}\mathrm{Sym}^nV^\vee\otimes \mathrm{Sym}^n V$.