1
$\begingroup$

NOTICE: This is not a question about research.

Hi guys. I'm studying Chow forms from the book Discriminants, Resultants, and Multidimensional Determinants of Andrei Zelevinsky and Izrail' Moiseevič Gel'fand. In order to introduce Chow forms the authors prove this proposition.

enter image description here enter image description here

There is something I don't understand and that of course some algebraic geometers can explain me.

(i) The authors use the term generic pencil when they define the degree. What is the precise meaning of the word generic? Where can I find more about the definition of degree of hypersurfaces in grasmannians?

(ii) To prove that the restriction of the sheaf $\mathcal{O}_{G(k,n)}(1)$ to $P_{NM}$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(1)$ the author claim that it suffices to prove that any section of $\mathcal{O}_{G(k,n)}(1)$ has a exactly a zero on $P_{NM}$. Why?

enter image description here

$\endgroup$
2
  • 3
    $\begingroup$ i) : the term generic pencil in $G(k,n)$ means the following. Take two generic $\mathbb{C}^{k-1}$ (say M) and a generic $\mathbb{C}^{k+1}$, say $N$ in the fixed $\mathbb{C}^n$. <<Generic>> means that there are Zariski open subsets of $G(k-1,n)$ and $G(k+1,n)$ where the choices of $M$ and $N$ can be made in. Then the set of $\mathbb{C}^k$ containing $M$ and contained in $N$ is easily seen to be a $\mathbb{P}^1$. $\endgroup$
    – Libli
    Mar 25, 2018 at 0:37
  • 3
    $\begingroup$ ii) : We know that $\mathrm{Pic}(\mathbb{P}^1) = \mathbb{Z}$, hence to determine $\mathcal{O}_{G(k,n)}(1)|_{P_{MN}}$, we only have to find the degree of an effective divisor representing $\mathcal{O}_{G(k,n)}(1)|_{P_{MN}}$. A section of $\mathcal{O}_{G(k,n)}(1)$ naturally restricts to a section of $\mathcal{O}_{G(k,n)}(1)|_{P_{MN}}$. If the zero locus of the restricted section vanishes exactly in one point, then the corresponding effective divisor is a point, that is a divisor of degree $1$. Means that $\mathcal{O}_{G(k,n)}(1)|_{P_{MN}} = \mathcal{O}_{\mathbb{P}^1}(1)$. $\endgroup$
    – Libli
    Mar 25, 2018 at 0:43

0