Section 121

Reduction to exceptional rulings and joining lines

Part XVI · Family 3.8PDF page 405

We retain the notation of Section 117. In particular, $E$ is the strict transform of $\mathbb P^1\times Q$ , and $\theta\colon X_U\to\mathbb P^1$ is the del Pezzo fibration. Suppose that every ramification index of $\eta_U$ is at most three. For a ramification point $p\in C_U$ , the fibre $\pi^{-1}(p)$ is a curve isomorphic to $\mathbb P^1$ ; we call it its exceptional ruling.

Proposition 121.1. Let $\Sigma\subset X_U$ consist of the following curves:

the exceptional rulings over ramification points of index two or three;
in each fibre of type $2A_1$ , the proper transform of the line joining the two points of the plane centre;
in each fibre of type $A_2$ , the proper transform of the line joining the point of multiplicity three to the remaining point.

Then $\delta_p(X_U)\ge24/23$ for every $p\notin E\cup\Sigma$ . Moreover, $\delta_p(X_U)\ge12/11$ for every $p\in E$ .

We prove the assertion by surface flags and give their decompositions. The exceptional rulings in the statement contain the singular points of the fibres. Thus no singular point of a fibre is omitted from the remaining locus.

121.1 The resolved fibres

Let $S$ be a singular fibre and let $\mu\colon\widetilde S\to S$ be its minimal resolution. The morphism to $\mathbb P^2$ is the successive blow-up of four points, with infinitely near points following the smooth conic $Q$ . Write $h$ for the pullback of a line and $e_1,\ldots,e_4$ for the total transforms of the exceptional classes. Thus $h^2=1,\qquad e_i^2=-1,\qquad h e_i=e_i e_j=0\quad(i\ne j),$ and put $Q_0=2h-e_1-e_2-e_3-e_4,\qquad -K_{\widetilde S}=h+Q_0.$ The distinguished smooth curve in the class $Q_0$ is the proper transform of $E\cap S$ . We use $T_i$ for an irreducible terminal exceptional curve of class $e_i$ , write $R_{ij}=e_i-e_j$ for the indicated $(-2)$ -curve, and write $L_{ij}=h-e_i-e_j$ whenever this class is represented by an irreducible curve.

Lemma 121.2. The negative curves and reducible members of $|Q_0|$ are as follows. In the first column we record the multiplicities of the four-point plane centre. $\begin{array}{c|c|l} &(-2)\text{-curves}&(-1)\text{-curves}\\ \hline (2,1,1)&R_{12}&T_2,T_3,T_4,L_{12},L_{13},L_{14},L_{34}\\ (2,2)&R_{12},R_{34}&T_2,T_4,L_{12},L_{13},L_{34}\\ (3,1)&R_{12},R_{23}&T_3,T_4,L_{12},L_{14} \end{array}$ The complete list of reducible members is $\begin{array}{c|l} (2,1,1)&L_{12}+L_{34},\quad L_{13}+R_{12}+L_{14},\\ (2,2)&L_{12}+L_{34},\quad 2L_{13}+R_{12}+R_{34},\\ (3,1)&L_{12}+R_{23}+R_{12}+L_{14}. \end{array}$ The curves mapping to the remaining locus $\Sigma\cap S$ are, respectively, $T_2$ ; $T_2,T_4,L_{13}$ ; and $T_3,L_{14}$ .

Proof. For a nonexceptional $(-1)$ -curve of degree $d$ , adjunction gives $\sum m_i=3d-1$ and $\sum m_i^2=d^2+1$ . Cauchy–Schwarz forces $d\le1$ . Thus its class is $h-e_i-e_j$ ; an exceptional $(-1)$ -curve has class $e_i$ . Taking their nonnegative intersections with the effective $(-2)$ -curves gives precisely the displayed irreducible curves. They are terminal exceptional curves, chords of $Q$ , or its tangent lines. No line contains three of the four points, including their infinitely near conditions, since a line has intersection number two with the smooth conic. Thus the only irreducible $(-2)$ -curves are those displayed.

The plane conics through the length-four centre form a pencil with no fixed component. Its base points are exactly those resolved by the four blow-ups, so $|Q_0|$ is base point free. A reducible plane member is a pair of lines, possibly equal. Since each line meets $Q$ in degree two, the possible pairs are exactly the pairs of chords or tangent lines in the displayed list. Adding the exceptional components gives the asserted equalities on $\widetilde S$ . The local equation from Lemma 117.3 identifies the terminal curves over multiple points with the exceptional rulings. ◻

121.2 The restricted divisors and the flag integrals

The threefold decomposition along $S$ is $\begin{array}{c|c|c} u&P(u)&N(u)\\ \hline {[0,1]}&H_1+H_2+(1-u)H_3&0\\ {[1,2]}&(2-u)H_1+H_2&(u-1)E. \end{array}$ In particular, $S_{X_U}(S)=\frac1{24} \left(\int_0^1(24-15u)\,du+ \int_1^2 3(2-u)(5-2u)\,du\right)=\frac56.$ These are the decompositions in [3, Lemma 5.77]. On $\widetilde S$ the positive part restricts to $h+Q_0$ for $0\le u\le1$ and to $h+(2-u)Q_0$ for $1\le u\le2$ . Thus we may use the parameter $c\in[0,1]$ and the divisor $A_c=h+cQ_0.$ The value $c=1$ is counted once more, with weight one, for the whole interval $0\le u\le1$ .

For a flag curve $L$ , write $A_c-vL=P_L(c,v)+N_L(c,v)$ for its surface Zariski decomposition, and denote its pseudo-effective threshold by $\tau_L(c)$ . For any integrand $F(c,v)$ , put $\mathcal I_L(F)= \int_0^{\tau_L(1)}F(1,v)\,dv+ \int_0^1\int_0^{\tau_L(c)}F(c,v)\,dv\,dc.$ If $p\in S\setminus(E\cup\operatorname{Sing}(S))$ and $L$ is a smooth flag curve through its inverse image $\widetilde p$ , then the surface flag quantities are $\begin{aligned} s_L&=\frac18\mathcal I_L(P_L^2),\\ t_{L,\widetilde p}&=\frac18\mathcal I_L((P_L\cdot L)^2) +\frac14\mathcal I_L\left((P_L\cdot L) \operatorname{ord}_{\widetilde p}(N_L|_L)\right). \end{aligned}$ Here $L$ is not the distinguished member $E\cap S$ . The fixed part $N(u)|_S=(u-1)(E\cap S)$ contributes nothing at $\widetilde p$ , since $p\notin E$ . Adjunction is crepant on $\widetilde S$ , and near $\widetilde p$ the resolution is an isomorphism. The Abban–Zhuang flag inequality therefore gives $\delta_p(X_U)\ge \min\left\{\frac65,\frac1{s_L},\frac1{t_{L,\widetilde p}}\right\}.$

Lemma 121.3. Suppose that $\widetilde p$ lies on a smooth member $L\in|Q_0|$ and is not on a terminal exceptional curve over a multiple point. Then $s_L=9/16$ and $t_{L,\widetilde p}\le23/24$ .

Proof. Put $B=e_1+e_2+e_3+e_4$ , interpreted as an effective sum of total transforms. The decomposition is $\begin{array}{c|c|c} v&P_L(c,v)&N_L(c,v)\\ \hline {[0,c]}&h+(c-v)Q_0&0\\ {[c,c+\tfrac12]}&(1+2c-2v)h&(v-c)B. \end{array}$ Both positive parts are nef, and the indicated exceptional support is negative definite. The threshold is $c+1/2$ . The volume and intersection on the first interval are $1+4c-4v$ and $2$ ; on the second they are $(1+2c-2v)^2$ and $2(1+2c-2v)$ . Substitution in the preceding integrals gives $s_L=\frac9{16},\qquad \frac18\mathcal I_L((P_L\cdot L)^2)=\frac{11}{12}.$ For a cluster of length $r$ , its contribution to $B$ is $R_{12}+2R_{23}+\cdots+(r-1)R_{r-1,r}+rT_r,$ with the indices shifted for other clusters. A smooth fibre of $|Q_0|$ is disjoint from all the $(-2)$ -curves and meets each terminal curve transversally once. At its intersection with $T_r$ the extra term is $\frac r4\left( \int_1^{3/2}2(3-2v)(v-1)\,dv+ \int_0^1\int_c^{c+1/2}2(1+2c-2v)(v-c)\,dv\,dc \right)=\frac r{24}.$ At all other points it is zero. In the case under consideration $r=1$ whenever such an intersection occurs. This gives the bound $11/12+1/24=23/24$ . ◻

121.3 Flags on the reducible conics

We give all the further decompositions needed for the proof. Throughout the following tables $0\le c\le1$ and $P_L=A_c-vL-N_L$ ; hence specifying $N_L$ specifies the positive part.

First, let $L=L_{12}$ in type $(2,1,1)$ or $(2,2)$ , and put $B_L=R_{12}+2T_2$ and $M=L_{34}$ . The same table applies to $L=L_{34}$ in type $(2,1,1)$ with $B_L=T_3+T_4$ , $M=L_{12}$ , and to $L=L_{34}$ in type $(2,2)$ with $B_L=R_{34}+2T_4$ , $M=L_{12}$ : $\begin{array}{c|l|l} v&N_L&P_L\cdot L\\ \hline {[0,c]}&0&1+v\\ {[c,1]}&(v-c)B_L&1+2c-v\\ {[1,1+c]}&(v-c)B_L+(v-1)M&2(1+c-v). \end{array}$ The respective values of $P_L^2$ are $1+4c-2v-v^2,\quad 1+4c+2c^2-2v-4cv+v^2,\quad 2(1+c-v)^2.$

For $L=L_{13}$ in type $(2,1,1)$ the decomposition is $\begin{array}{c|l|l} v&N_L&P_L\cdot L\\ \hline {[0,c]}&\tfrac v2R_{12}&1+\tfrac v2\\ {[c,2c]}&\tfrac v2R_{12}+(v-c)T_3&1+c-\tfrac v2\\ {[2c,1+c]}&(v-c)(R_{12}+T_3)+(v-2c)T_2&1+2c-v. \end{array}$ The respective volumes are $1+4c-2v-\frac{v^2}{2},\quad 1+4c+c^2-2v-2cv+\frac{v^2}{2},\quad 1+4c+3c^2-2v-4cv+v^2.$ For $L_{14}$ interchange $3$ and $4$ .

Finally, for $L=L_{12}$ in type $(3,1)$ we have $\begin{array}{c|l|l} v&N_L&P_L\cdot L\\ \hline {[0,\tfrac32c]}&\tfrac v3R_{12}+\tfrac{2v}3R_{23}&1+\tfrac v3\\ {[\tfrac32c,1+c]}&(v-c)(R_{12}+2R_{23})+(2v-3c)T_3&1+2c-v. \end{array}$ The two volumes are $1+4c-2v-\frac{v^2}{3},\qquad 1+4c+3c^2-2v-4cv+v^2.$

These are Zariski decompositions: every coefficient of the displayed negative part is nonnegative in its stated range, its support is negative definite, and its components are orthogonal to the positive part. Intersecting with the complete list in Lemma 121.2 shows that the positive part is nef. The cone of curves of these weak del Pezzo surfaces is generated by those negative curves. At the right endpoint the positive part has square zero; the same intersection inequalities show that the class ceases to be pseudo-effective beyond it.

Lemma 121.4. For the flag curves just listed, at a point outside $E$ , the contracted $(-2)$ -curves, and the curves mapping to $\Sigma$ in Lemma 121.2, the following bounds hold: $\begin{array}{c|c|c|c|c} \text{type}&L&s_L&\frac18\mathcal I_L((P_L\cdot L)^2) &t_{L,\widetilde p}\text{ at most}\\ \hline (2,1,1)&L_{12}&19/24&11/16&19/24\\ (2,1,1)&L_{13},L_{14}&29/32&115/192&27/32\\ (2,1,1)&L_{34}&19/24&11/16&27/32\\ (2,2)&L_{12},L_{34}&19/24&11/16&19/24\\ (3,1)&L_{12}&179/192&115/192&115/192. \end{array}$

Proof. Integrating the displayed volume and intersection polynomials gives the third and fourth columns. We describe every allowed correction for the final column. For the first decomposition table, the correction at $L\cap M$ is $5/48$ . At a point of $L\cap T_i$ where $B_L$ contains $T_i$ with coefficient one, it is $5/32$ . Coefficient two gives $5/16$ , but this occurs only on a remaining terminal curve and is excluded. For the second decomposition table, the only allowed correction is at $L_{13}\cap T_3$ (or $L_{14}\cap T_4$ ), and it is $47/192$ . The intersection with $R_{12}$ is excluded. In the final table, the only negative component meeting $L_{12}$ is $R_{23}$ , so there is no allowed correction. These intersections are distinct and transverse, as follows either from the blow-up description or from the chords and tangents to the smooth conic. Thus no two of the listed corrections occur at the same allowed point. Adding them to the fourth column proves the result. ◻

Proof of Proposition 121.1. The assertion on $E$ is [3, Lemma 5.76]. If $p$ lies outside $E$ on a smooth fibre of $\theta$ , use [3, Lemma 5.77]. Otherwise resolve its singular fibre as above. Since $p\notin\Sigma$ , the surface is smooth at $p$ . If the member of $|Q_0|$ through $\widetilde p$ is smooth, Lemma 121.3 applies. If it is reducible, Lemma 121.2 and the exclusions defining $\Sigma$ show that $\widetilde p$ lies on one of the curves of Lemma 121.4. In every case both $s_L$ and $t_{L,\widetilde p}$ are at most $23/24$ , so the flag inequality gives $\delta_p(X_U)\ge24/23$ . This proves the assertion and covers all points outside the stated remaining locus. ◻