Section 32

Geometry and the reduction to point centres

Part VI · Family 2.12: intersections of three bilinear divisorsPDF page 124

Family 2.12: intersections of three bilinear divisors

Let $X\subset\mathbb P^3\times\mathbb P^3$ be a smooth complete intersection of three divisors of bidegree $(1,1)$ . Write $H,H'$ for the two hyperplane classes restricted to $X$ . Both projections are blow-ups of smooth nonhyperelliptic curves of degree six and genus three. We denote their exceptional divisors by $E,E'$ . The standard determinantal description gives a smooth plane quartic $C=\{\det(z_0M_1+z_1M_2+z_2M_3)=0\}\subset\mathbb P^2, \qquad X=\{x^{\mathsf t}M_i y=0:1\le i\le3\}.$ In particular, every matrix on $C$ has rank three. These geometric facts hold for every smooth member; see [2, Proposition 2.1].

Lemma 32.1 (Intersection numbers and anticanonical sections). One has $\begin{gathered} L=-K_X=H+H',\quad E=3H-H',\quad E'=3H'-H,\\ H^3=(H')^3=1,\qquad H^2H'=H(H')^2=3. \end{gathered}$ so $L^3=20$ . For $m\ge0$ , the space $H^0(X,mL)$ is the bidegree $(m,m)$ part of $R=\mathbb C[x_0,\ldots,x_3,y_0,\ldots,y_3]/(F_1,F_2,F_3), \qquad F_i=x^{\mathsf t}M_i y.$ Its dimension is $\label{f212:eq:Nm} N_m=\frac{(2m+1)(5m^2+5m+3)}3 =\frac{10}3m^3+5m^2+\frac{11}3m+1.$

Proof. Adjunction gives $L=H+H'$ . The two blow-up canonical formulas give $4H-E=L=4H'-E'$ . In the ambient product, $[X]=(H+H')^3$ ; multiplication gives the four intersection numbers. The Koszul resolution of the three bilinear equations, twisted by $(m,m)$ , identifies the displayed quotient with global sections. Indeed, the terms have bidegrees $(m-r,m-r)$ , $0\le r\le3$ . A negative entry occurring here belongs to $\{-1,-2,-3\}$ , for which all cohomology on $\mathbb P^3$ vanishes. Nonnegative entries have no higher cohomology. Consequently $N_m=\sum_{r=0}^3(-1)^r\binom3r\binom{m-r+3}{3}^{\!2},$ where a summand with $m<r$ is zero. Expanding proves (32.1). ◻

We use the following published reduction explicitly as an external input.

Proposition 32.2 (Cheltsov–Li–Ma’u–Pinardin). If a prime divisor $F$ over $X$ satisfies $\beta_X(F)\le0$ , then its centre on $X$ is a closed point of $E\cap E'$ .

This is [1, Theorem A]. Thus it suffices to prove strict beta positivity for every divisor centred at every point of $E\cap E'$ . No group action is imposed. Nor do we assume that $E$ and $E'$ meet transversally.