Conics meeting 8 general lines

Question

I am trying to show that the number of plane conics in $\mathbb{P}^3$ meeting $8$ general lines is $92$, using what I know about intersection theory.

I started considering the tautological bundle $S$ of $G=G(2,3)$, and I took $P=P(\operatorname{Sym}^2S^{\vee})$. $P$ is a projective bundle of rank $5$ over $G$ whose fiber over the plane $H$ consists of the conics lying on $H$.

I also computed the Chow group of $P$, i.e. $$\mathbb{Z}[h,t]/(h^4,t^6+4ht^5+10h^2t^4+20h^3t^3)$$

What I cannot do is to find the cycle of the conics meeting a given line $L$ (I cannot even prove that it is a closed subscheme!). After that, I just need to elevate it to the eighth power and I should get $92$. Could anyone help?

Answer 1

Let $D$ be the divisor corresponding to conics in $P$ which meet a fixed line $L$, and I will show that $[D]=2h+t$. This is discussed in detail in section 9.7 of the book 3264 and all that, and actually I write nothing new.

There are two ways to compute this: a smart one and a stupid one.

The smart way: Let $U\subset P$ be the open set of pairs $(H,\xi)$ such that $H$ does not contain $L$. Note the complement of $U$ has codimension $2$, so it does not change the divisor classes. Consider the natural morphism $$\alpha\colon U \to L$$ which sends $(H,\xi)$ to $p=H\cap L$. Note that $\mathcal O_U(-1)$ is the tautological bundle, i.e. the fibre at $(H,\xi)$ is precisely the $1$-dimensional linear space of quadratic form $\mathcal Q \in \xi$. Therefore, the evaluation $$\mathcal Q \mapsto \mathcal Q(p)$$ defines a morphism of line bundles $$ \mathcal O_U(-1) \to \alpha^* \mathcal O_L(2)$$ and the desired divisor $D$ is precisely the zero locus of this morphism. This is equivalent to the zero locus of some section of the line bundle $\mathcal O_U(1) \otimes\alpha^* \mathcal O_L(2)$, which is exactly $t+2h$.

The stupid way: Since you have already computed the Chow ring, we can write $$[D]=a h + b t$$ for some (undetermined) integers. To solve $a$ and $b$, we test how it intersects with some special curves in $P$. Precisely, we can look at the following two curves:

• Fixed a general $H$ and let $\{C_\lambda \subset H\}$ be a general pencil of conics. This defines curve $\Gamma\subset P$.

• Fixed a general quadric surface $Q\subset \mathbb P^3$, and let $\{ H_\lambda\subset \mathbb P^3\}$ be a general pencil of (hyper)planes. Then the intersections $\{ H_\lambda \cap Q \}$ are conics, hence defines a curve $\Phi \subset P$.

We compute the intersection numbers between the divisors classes $h$, $t$, $[D]$ and the curves $[\Gamma]$, $[\Phi]$.

• $[\Phi]\cdot h$. On $\mathbb P^{3*}$, the image $\Phi$ meets $\mathcal O_{\mathbb P^{3*}}(1)$ at precisely one point, and hence easy to see $[\Phi]\cdot h=1$.
• $[\Phi]\cdot t$. This is the degree of the restriction of $\mathcal O_P(1)$ to the curve $\Phi$. We compute with its dual $\mathcal O_P(-1)$, as it has more clear geometric meaning: the fibre of $\mathcal O_P(-1)$ at $(H,\xi)$ is precisely the $1$-dimensional linear space $\xi$. Consider the $F\in H^0(\mathcal O_{\mathbb P^3}(2))$ which defines $Q$, then its restrictions to $H_\lambda$ give an nowhere-zero section of $\mathcal O_P(-1)|_\Phi$. Therefore, the restriction is a trivial bundle, and hence $[\Phi]\cdot t=0$
• $[\Phi] \cdot [D]$. This is equivalent to compute $[Q]\cdot[L]$, which is clearly to be $2$.

I will omit the computation with $[\Gamma]$ which is in the same sprit. After all this, you can get the intersetion numbers as follows:

	h	t	[D]
$[\Gamma]$	0	1	1
$[\Phi]$	1	0	2

Therefore, $$1= [\Gamma] \cdot [D] =[\Gamma] \cdot (a h+ b t)=b$$ and $$2=[\Phi] \cdot [D] = [\Phi] \cdot (ah+b t) =a.$$