K-stable smooth Fano threefolds of Picard rank two

Abstract

We prove that all smooth Fano threefolds in the families and are K-stable, and we also prove that smooth Fano threefolds in the family that satisfy one very explicit generality condition are K-stable.

1 Introduction

Recently, we witnessed huge progress on K-stability of Fano varieties [1, 15, 18, 22], which resulted in an immense breakthrough in the solution of the following problem:

Problem. Find all K-polystable smooth Fano threefolds.

Let us describe what is done in this direction. To do this, fix a smooth Fano threefold X. Then X belongs to one of the $105$ deformation families found by Iskovskikh, Mori, Mukai. These families are often labeled as , where the first digit stands for the rank of the Picard group of the threefolds in the family. For the detailed descriptions of the $105$ families, we refer the reader to [3].

The valuative criterion for K-stability [15, 18] allows us to explain when X is K-stable in relatively simple terms. To do this, let us remind what are $\beta $ -invariant and $\delta $ -invariant. Recall that E is a prime divisor over X if there is a birational morphism $f\colon \widetilde {X}\to X$ with normal $\widetilde {X}$ and a prime divisor $E\subset \widetilde {X}$ . We set $\beta (E)=A_X(E)-S_X(E)$ , where

$$ \begin{align*}S_X(E)=\frac{1}{(-K_X)^3}\int_{0}^{\infty}\mathrm{vol}\big(f^*(-K_X)-uE)du, \end{align*} $$

and $A_X(E)$ is the log discrepancy of the divisor E. Then, for every point $P\in X$ , we let

$$ \begin{align*}\delta_P(X)=\inf_{\substack{F/X\\ P\in C_X(F)}}\frac{A_{X}(F)}{S_X(F)}, \end{align*} $$

where the infimum is taken over all prime divisors over X whose centers on X contain P. Finally, we set

$$ \begin{align*}\delta(X)=\inf_{P\in X}\delta_P(X). \end{align*} $$

The valuative criterion says that X is K-stable (respectively, K-semistable) if $\beta (F)>0$ (respectively, $\beta (F)\geqslant 0$ ) for any prime divisor F over X. Hence, X is K-stable if $\delta (X)>1$ . We say that X is K-unstable if it is not K-semistable.

The simplest way to apply the valuative criterion is to check whether $\beta (S)<0$ or not for some irreducible surface S in the threefold X. This has been done in [15]. As a result, we know that X is K-unstable if it belongs to any of the following $26$ families:

To be precise, if X is contained in one of these $26$ families, then it contains an irreducible surface S such that $\beta (S)<0$ , so that X is K-unstable, which implies that X does not admit a Kähler–Einstein metric. Almost in every case, such surface S is not hard to find. For instance, if X is the unique smooth Fano threefold in the deformation family , then X is a blow up of $\mathbb {P}^3$ at a point, and S is the exceptional surface of the blow up.

We listed $26$ families of smooth Fano threefolds that contain no K-semistable members. There is another family that does not contain K-polystable members – the family . This family is quite special; see [3, § 5.10] for its detailed description, and it contains exactly two smooth Fano threefolds. One of them is K-semistable and not K-polystable, while another one is K-unstable. Moreover, it follows from [3, Main Theorem] that general members of the remaining $78$ families of smooth Fano threefolds are K-polystable.

Further, all K-polystable smooth Fano threefolds in $53$ families among these $78$ families are described in [2, 3, 4, 9, 13, 17, 19, 20, 23, 7, 8]. The remaining $25$ families are:

The deformation families contain non-K-polystable members, and [3, § 7] provides conjectures that describe K-polystable smooth Fano threefolds in these four families. On the other hand, all smooth Fano threefolds in the $21$ families

are conjectured to be K-stable [3]. The goal of this paper is to prove this for six families:

Main Theorem. Let X be a smooth Fano threefold contained in one of the following deformation families: . Then X is K-stable.

Therefore, to solve the problem posed above, it remains to find all K-polystable smooth Fano threefolds in the following $19$ families:

We hope that this will be done in a nearest future.

To prove Main Theorem, we use Abban–Zhuang theory [1] and its applications [3, 16]. A similar approach also works for almost all smooth Fano threefolds in the family . Namely, if X is a smooth Fano threefold in the family , we show that $\delta _P(X)>1$ for every point $P\in X$ that satisfy certain geometric conditions. As a result, we obtained

Auxiliary Theorem. Let X be a smooth Fano threefold in the deformation family . Recall that there exists the following Sarkisov link:

where V is a smooth cubic threefold in $\mathbb {P}^4$ , the morphism $\pi $ is a blow up of a smooth plane cubic curve, and $\phi $ is a morphism whose fibers are normal cubic surfaces. Suppose that

(★) $$ \begin{align} \text{no fiber of the morphism }\phi\text{ has a Du Val singular point of type }\mathbb{D}_5\text{ or }\mathbb{E}_6. \end{align} $$

Then X is K-stable.

Let us describe the structure of this paper. In Section 2, we prove auxiliary theorem, and we prove that all smooth Fano threefolds in the families and are K-stable. In Sections 3, 4, 5, 6, we prove that all smooth Fano threefolds in the families are K-stable, respectively. Note that Section 6 is very technical and long.

As we already mentioned, we use applications of Abban–Zhuang theory [1] which have been discovered in [3, 16]. For the background material, we refer the reader to [3, 16, 22].

2 Families

Fix $d\in \{1,2,3\}$ . Let V be one of the following smooth Fano threefolds:

• $\boxed{d=1}$ a smooth sextic hypersurface in $\mathbb {P}(1,1,1,2,3)$ ;

• $\boxed{d=2}$ a smooth quartic hypersurface in $\mathbb {P}(1,1,1,1,2)$ ;

• $\boxed{d=3}$ a smooth cubic threefold in $\mathbb {P}^4$ .

Then $-K_{V}\sim 2H$ for an ample divisor $H\in \mathrm {Pic}(V)$ such that $H^3=d$ and $\mathrm {Pic}(V)=\mathbb {Z}[H]$ . Let $S_1$ and $S_2$ be two distinct surfaces in the linear system $|H|$ , and let $\mathcal {C}=S_1\cap S_2$ . Suppose that the curve $\mathcal {C}$ is smooth. Then $\mathcal {C}$ is an elliptic curve by the adjunction formula. Let $\pi \colon X\to V$ be the blow up of the curve $\mathcal {C}$ , and let E be the $\pi $ -exceptional surface.

• • If $d=1$ , then X is a smooth Fano threefold in the deformation family .

• • If $d=2$ , then X is a smooth Fano threefold in the deformation family .

• • If $d=3$ , then X is a smooth Fano threefold in the deformation family .

Moreover, all smooth Fano threefolds in these families can be obtained in this way.

Note that $(-K_X)^3=4d$ . Moreover, we have the following commutative diagram:

where is the rational map given by the pencil that is generated by $S_1$ and $S_2$ , and $\phi $ is a morphism whose general fiber is a smooth del Pezzo surface of degree d.

The goal of this section is to show that X is K-stable in the case when $d=1$ or $d=2$ , and to show that X is K-stable in the case when $d=3$ and X satisfies the condition (★). To show that X is K-stable, it is enough to show that $\delta _O(X)>1$ for every point $O\in X$ . This follows from the valuative criterion for K-stability [15, 18].

Lemma 2.1. Let O be a point in X; let A be the fiber of the morphism $\phi $ such that $O\in A$ . Suppose that A has at most Du Val singularities at the point O. Then

$$ \begin{align*}\delta_O(X)\geqslant\left\{\begin{aligned} &\mathrm{min}\Big\{\frac{16}{11},\frac{16}{15}\delta_O(A)\Big\}\ \text{if }O\not\in E, \\ &\mathrm{min}\Big\{\frac{16}{11},\frac{16\delta_{O}(A)}{\delta_{O}(A)+15}\Big\}\ \text{if }O\in E. \end{aligned} \right. \end{align*} $$

Proof. Let u be a nonnegative real number. Then $-K_X-uA\sim _{\mathbb {R}} (2-u)A+E$ , which implies that divisor $-K_X-uA$ is pseudoeffective if and only if $u\leqslant 2$ . For every $u\in [0,2]$ , let us denote by $P(u)$ the positive part of Zariski decomposition of the divisor $-K_X-uA$ , and let us denote by $N(u)$ its negative part. Then

$$ \begin{align*}P(u)=\left\{\begin{aligned} &(2-u)A+E\ \text{if }0\leqslant u\leqslant 1, \\ &(2-u)H\ \text{if }1\leqslant u\leqslant 2, \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}N(u)=\left\{\begin{aligned} &0\ \text{if }0\leqslant u\leqslant 1, \\ &(u-1)E\ \text{if }1\leqslant u\leqslant 2. \end{aligned} \right. \end{align*} $$

Integrating, we get $S_X(A)=\frac {11}{16}$ . Using [1, Theorem 3.3] and [3, Corollary 1.102], we get

(2.1) $$ \begin{align} \delta_O(X)\geqslant \min\left\{\frac{1}{S_X(A)},\inf_{\substack{F/A\\ O\in C_A(F)}}\frac{A_A(F)}{S(W_{\bullet,\bullet}^A;F)}\right\}=\min\left\{\frac{16}{11},\inf_{\substack{F/A\\ O\in C_A(F)}}\frac{A_A(F)}{S(W_{\bullet,\bullet}^A;F)}\right\}, \end{align} $$

where the infimum is taken by all prime divisors F over the surface A with $O\in C_A(F)$ . The value $S(W_{\bullet ,\bullet }^A;F)$ can be computed using [3, Corollary 1.108] as follows:

$$ \begin{align*}S\big(W_{\bullet,\bullet}^A; F\big)= \frac{3}{4d}\int_1^{2}\big(P(u)\big\vert_{A}\big)^2(u-1)\mathrm{ord}_{F}\big(E\big\vert_{A}\big)du+ \frac{3}{4d}\int_{0}^{2}\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{A}-vF\big)dvdu. \end{align*} $$

Now, let F be any prime divisor over the surface A such that $O\in C_A(F)$ . Since

$$ \begin{align*}P(u)\big\vert_{A}=\left\{\begin{aligned} &-K_A\ \text{if }0\leqslant u\leqslant 1, \\ &(2-u)(-K_A)\ \text{if }1\leqslant u\leqslant 2, \end{aligned} \right. \end{align*} $$

we have

$$ \begin{align*} S\big(W_{\bullet,\bullet}^A; F\big)&= \frac{3}{4d}\int_1^{2}d(2-u)^2(u-1)\mathrm{ord}_{F}\big(E\big\vert_{A}\big)du+\\ &\quad+\frac{3}{4d}\int_{0}^{1} \int_{0}^{\infty}\mathrm{vol}\big(-K_A-vF\big)dvdu+\frac{3}{4d}\int_{1}^{2}\int_{0}^{\infty}\mathrm{vol}\big((2-u)(-K_A)-vF\big)dvdu=\\ &=\frac{\mathrm{ord}_{F}\big(E\big\vert_{A}\big)}{16}+\frac{3}{4d}\int_{0}^{\infty}\mathrm{vol}\big(-K_A-vF\big)dv+\frac{3}{4d}\int_{1}^{2}(2-u)^3\int_{0}^{\infty}\mathrm{vol}\big(-K_A-vF\big)dvdu=\\ &=\frac{\mathrm{ord}_{F}\big(E\big\vert_{A}\big)}{16}+\frac{3}{4d}\int_{0}^{\infty}\mathrm{vol}\big(-K_A-vF\big)dv+\frac{3}{16d}\int_{0}^{\infty}\mathrm{vol}\big(-K_A-vF\big)dv=\\ &=\frac{\mathrm{ord}_{F}\big(E\big\vert_{A}\big)}{16}+\frac{15}{16d}\int_{0}^{\infty}\mathrm{vol}\big(-K_A-vF\big)dv=\\ &=\frac{\mathrm{ord}_{F}\big(E\big\vert_{A}\big)}{16}+\frac{15}{16}S_A(F)\leqslant\frac{\mathrm{ord}_{F}\big(E\big\vert_{A}\big)}{16}+\frac{15A_{A}(F)}{16\delta_{O}(A)}.\end{align*} $$

Therefore, if $O\not \in E$ , then $\mathrm {ord}_{F}(E\vert _{A})=0$ , which implies that

$$ \begin{align*}S\big(W_{\bullet,\bullet}^A;F\big)\leqslant\frac{15A_{A}(F)}{16\delta_{O}(A)}. \end{align*} $$

Similarly, if $O\in E$ , then $\mathrm {ord}_{F}(E\vert _{A})\leqslant A_A(F)$ because $(A,E\vert _A)$ is log canonical, so that

$$ \begin{align*}S\big(W_{\bullet,\bullet}^A; F\big)=\frac{\mathrm{ord}_{F}\big(E\big\vert_{A}\big)}{16}+\frac{15}{16}S_A(F)\leqslant\frac{A_A(F)}{16}+\frac{15A_{A}(F)}{16\delta_{O}(A)}=\frac{\delta_{O}(A)+15}{16\delta_{O}(A)}A_{A}(F). \end{align*} $$

Now, using Equation (2.1), we obtain the required inequality.

Suppose X is not K-stable. Let us seek for a contradiction. Using the valuative criterion for K-stability [15, 18], we see that there exists a prime divisor $\mathbf {F}$ over X such that

$$ \begin{align*}\beta(\mathbf{F})=A_X(\mathbf{F})-S_X(\mathbf{F})\leqslant 0, \end{align*} $$

where $A_X(\mathbf {F})$ is a log discrepancy of the divisor $\mathbf {F}$ , and $S_X(\mathbf {F})$ is defined in [15] or [3, § 1.2]. Let Z be the center of the divisor $\mathbf {F}$ on X. Then Z is not a surface [3, Theorem 3.17]. We see that Z is an irreducible curve or a point. Let P be a point in Z. Then $\delta _P(X)\leqslant 1$ .

Lemma 2.2. One has $P\not \in E$ .

Proof. Let us compute $S_X(E)$ . Note that $S_X(E)<1$ by [3, Theorem 3.17]. Fix $u\in \mathbb {R}_{\geqslant 0}$ . Then $-K_X-uE$ is pseudoeffective $\iff -K_X-uE$ is nef $\iff u\leqslant 1$ . Thus, we have

$$ \begin{align*}S_X(E)=\frac{1}{4d}\int_{0}^{1}\big(-K_X-uE\big)^3du=\frac{1}{4d}\int_{0}^{1}d(2u^3-6u+4)du=\frac{3}{8}. \end{align*} $$

Suppose that $P\in E$ . Let us seek for a contradiction.

Note that $E\cong \mathcal {C}\times \mathbb {P}^1$ . Let $\mathbf {s}$ be a fiber of the projection $\phi \vert _{E}\colon E\to \mathbb {P}^1$ that contains P, and let $\mathbf {f}$ be a fiber of the projection $\pi \vert _{E}\colon E\to \mathcal {C}$ . Fix $u\in [0,1]$ , and take $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}-K_X-uE\big\vert_{E}-v\mathbf{s}\equiv (1+u-v)\mathbf{s}+d(1-u)\mathbf{f}. \end{align*} $$

This implies that

$$\begin{align*}(-K_X-uE)\big\vert_{E}-v\mathbf{s}\text{ is pseudoeffective }\iff (-K_X-uE)\big\vert_{E}-v\mathbf{s}\text{ is nef }\iff v\leqslant 1+v. \end{align*}$$

Therefore, using [3, Corollary 1.109], we get

$$ \begin{align*}S\left(W_{\bullet,\bullet}^{E};\mathbf{s}\right)=\frac{3}{4d} \int_{0}^{1} \int_{0}^{1+u} 2d(1-u)(1+u-v)dvdu=\frac{11}{16}. \end{align*} $$

Similarly, using [3, Theorem 1.112], we get

$$ \begin{align*}S\left(W_{\bullet,\bullet, \bullet}^{E,\mathbf{s}};P\right)=\frac{3}{4d}\int_{0}^{1} \int_{0}^{1+u}\big(d(1-u)\big)^{2}dvdu=\frac{5d}{16}. \end{align*} $$

Therefore, it follows from [3, Theorem 1.112] that

$$ \begin{align*}1\geqslant\frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant \min\left\{\frac{1}{S_{X}(E)}, \frac{1}{S\left(W_{\bullet,\bullet}^E,\mathbf{s}\right)}, \frac{1}{S\left(W_{\bullet,\bullet,\bullet}^{E,\mathbf{s}};P\right)}\right\}=\min \left\{\frac{8}{3}, \frac{16}{11}, \frac{16}{5d}\right\}\geqslant\frac{16}{15}>1, \end{align*} $$

which is a contradiction.

Let A be the fiber of the del Pezzo fibration $\phi $ such that A passes through the point P. Then A is a del Pezzo surface of degree $d\in \{1,2,3\}$ that has at most isolated singularities. In particular, we see that A is normal. Applying Lemmas 2.1 and 2.2, we obtain

Corollary 2.3. One has $\delta _P(A)\leqslant \frac {15}{16}$ .

Proof. Since $1\geqslant \frac {A_X(\mathbf {F})}{S_X(\mathbf {F})}\geqslant \delta _P(X)$ , we get $\delta _P(A)\leqslant \frac {15}{16}$ by Lemmas 2.1 and 2.2.

Corollary 2.4. The surface A is singular.

Proof. If A is smooth, then $\delta _P(A)\geqslant \delta (A)\geqslant \frac {3}{2}$ [3, § 2], which contradicts Corollary 2.3.

Let $\overline {S}$ be a general surface in $|H|$ that passes through $\pi (P)$ , and let S be the proper transform on X of the surface $\overline {S}$ . Then

• • the surface $\overline {S}$ is a smooth del Pezzo surface of degree d,

• • the surface $\overline {S}$ intersects the curve $\mathcal {C}$ transversally at d points,

• • the induced morphism $\pi \vert _{S}\colon S\to \overline {S}$ is a blow up of the points $\overline {S}\cap \mathcal {C}$ .

Observe that $\phi \vert _S\colon S\to \mathbb {P}^1$ is an elliptic fibration given by the pencil $|-K_S|$ . Set $C=A\big \vert _{S}$ . Then C is a reduced curve of arithmetic genus $1$ in $|-K_S|$ that has at most d components. In particular, if $d=1$ , then C is irreducible. Therefore, the following cases may happen:

1. (1) the curve C is irreducible, and C is smooth at P,

2. (2) the curve C is irreducible, and C has an ordinary node at P,

3. (3) the curve C is irreducible, and C has an ordinary cusp at P,

4. (4) the curve C is reducible.

Fix $u\in \mathbb {R}_{\geqslant 0}$ . Then $-K_X-uS$ is nef $\iff u\leqslant 1 \iff -K_X-uS$ is pseudoeffective. Using this, we see that

$$ \begin{align*}S_X(S)=\frac{1}{4d}\int_{0}^{1}(-K_X-uS)^3du=\frac{1}{4d}\int_{0}^{1}d(4-u)(1-u)^2du=\frac{5}{16}<1, \end{align*} $$

which also follows from [3, Theorem 3.17]. Moreover, if $u\in [0,1]$ , then

$$ \begin{align*}(-K_X-uS)|_S\sim_{\mathbb{R}}(1-u)(\pi\vert_{S})^*(-K_{\overline{S}})-K_S\sim_{\mathbb{R}}(1-u)\sum_{i=1}^{d}\mathbf{e}_i+(2-u)(-K_S), \end{align*} $$

where $\mathbf {e}_1,\ldots ,\mathbf {e}_d$ are exceptional curves of the blow up $\pi \vert _{S}\colon S\to \overline {S}$ .

Lemma 2.5. Suppose that C is irreducible. Then C is singular at the point P.

Proof. As in the proof of Lemma 2.2, it follows from [3, Theorem 1.112] that

$$ \begin{align*}1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant \min\left\{\frac{1}{S_X(S)},\frac{1}{S(W_{\bullet,\bullet}^S;C)}, \frac{1}{S(W_{\bullet, \bullet,\bullet}^{S,C};P)}\right\}, \end{align*} $$

where $S(W_{\bullet ,\bullet }^S;C)$ and $S(W_{\bullet , \bullet ,\bullet }^{S,C};P)$ are defined in [3, § 1.7]. Since we know that $S_X(S)<1$ , we see that $S(W_{\bullet ,\bullet }^S;C)\geqslant 1$ or $S(W_{\bullet , \bullet ,\bullet }^{S,C};P)\geqslant 1$ . Let us compute these numbers.

Let $P(u,v)$ be the positive part of the Zariski decomposition of $(-K_X-uS)\vert _{S}-vC$ , and let $N(u,v)$ be its negative part, where $u\in [0,1]$ and $v\in \mathbb {R}_{\geqslant 0}$ . Since

$$ \begin{align*}(-K_X-uS)\vert_{S}-vC\sim_{\mathbb{R}}(1-u)\sum_{i=1}^{d}\mathbf{e}_i+(2-u-v)C, \end{align*} $$

we see that $(-K_X-uS)\vert _{S}-vC$ is pseudoeffective $\iff v\leqslant 2-u$ . Moreover, we have

$$ \begin{align*}P(u,v)=\left\{\begin{aligned} &(1-u)\sum_{i=1}^{d}\mathbf{e}_i+(2-u-v)C\ \text{if }0\leqslant v\leqslant 1, \\ &(2-u-v)\Big(C+\sum_{i=1}^{d}\mathbf{e}_i\Big)\ \text{if }1\leqslant v\leqslant 2-u, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}N(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant 1, \\ &(v-1)\sum_{i=1}^{d}\mathbf{e}_i\ \text{if }1\leqslant v\leqslant 2-u. \\ \end{aligned} \right. \end{align*} $$

Thus, it follows from [3, Corollary 1.109] that

$$ \begin{align*} S\big(W^S_{\bullet,\bullet};C\big)& =\frac{3}{4d}\int_0^1\int_0^{2-u}P(u,v)^2dvdu=\\ &=\frac{3}{4d}\int_0^1\int_0^1d(1-u)(3-u-2v)dvdu+\frac{3}{4d}\int_0^1\int_1^{2-u}d(2-u-v)^2dvdu=\frac{11}{16}<1. \end{align*} $$

Thus, we conclude that $S(W_{\bullet , \bullet ,\bullet }^{S,C};P)\geqslant 1$ .

Since $P\not \in \mathbf {e}_1\cup \cdots \cup \mathbf {e}_d$ by Lemma 2.2, it follows from [3, Theorem 1.112] that

$$ \begin{align*} S\big(W_{\bullet, \bullet,\bullet}^{S,C};P\big)&=\frac{3}{4d}\int_0^1\int_0^{2-u}\big(P(u,v)\cdot C\big)^2dvdu=\\ &=\frac{3}{4d}\int_0^1\int_0^1d^2(u-1)^2dvdu+\frac{3}{4d}\int_0^1\int_1^{2-u}d^2(2-u-v)^2dvdu=\frac{5d}{16}<1, \end{align*} $$

which is a contradiction.

Now, let us show that C is reducible for $d\in \{1,2\}$ .

Lemma 2.6. Suppose that C is irreducible. Then $d=3$ and C has a cusp at P.

Proof. By Lemma 2.5, the curve C is singular at the point P.

Now, let $\sigma \colon \widetilde {S}\to S$ be the blow up of the point P; let $\mathbf {f}$ be the $\sigma $ -exceptional curve, and let $\widetilde {\mathbf {e}}_1,\ldots ,\widetilde {\mathbf {e}}_d,\widetilde {C}$ be the proper transforms on $\widetilde {S}$ of the curves $\mathbf {e}_1,\ldots ,\mathbf {e}_d,C$ , respectively. Then the curve $\widetilde {C}$ is smooth, and it follows from [3, Remark 1.113] that

$$ \begin{align*}1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\inf_{O\in\mathbf{f}}\frac{1}{S(W_{\bullet,\bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)}, \frac{2}{S(V_{\bullet,\bullet}^{S};\mathbf{f})},\frac{1}{S_X(S)}\right\}, \end{align*} $$

where $S(W_{\bullet , \bullet ,\bullet }^{\widetilde {S},\mathbf {f}};O)$ and $S(V_{\bullet ,\bullet }^{S};\mathbf {f})$ are defined in [3, § 1.7]. Since we know that $S_X(S)<1$ , we see that $S(V_{\bullet ,\bullet }^{S};\mathbf {f})\geqslant 2$ or there exists a point $O\in \mathbf {f}$ such that $S(W_{\bullet ,\bullet ,\bullet }^{\widetilde {S},\mathbf {f}};O)\geqslant 1$ .

Let us compute $S(V_{\bullet ,\bullet }^{S};\mathbf {f})$ . Fix $u\in [0,1]$ and $v\in \mathbb {R}_{\geqslant 0}$ . Since $\sigma ^*(C)\sim \widetilde {C}+2\mathbf {f}$ , we get

$$ \begin{align*}\sigma^*\big((-K_X-uS)\vert_{S}\big)-v\mathbf{f}\sim_{\mathbb{R}}(2-u)\widetilde{C}+(4-2u-v)\mathbf{f}+(1-u)\sum_{i=1}^{d}\widetilde{\mathbf{e}}_i. \end{align*} $$

Then the divisor $\sigma ^*((-K_X-uS)\vert _{S})-v\mathbf {f}$ is pseudoeffective $\iff v\leqslant 4-2u$ .

Let $P(u,v)$ be the positive part of the Zariski decomposition of $\sigma ^*((-K_X-uS)\vert _{S})-v\mathbf {f}$ , and let $N(u,v)$ be its negative part. Then

$$ \begin{align*}P(u,v)=\left\{\begin{aligned} &(2-u)\widetilde{C}+(4-2u-v)\mathbf{f}+(1-u)\sum_{i=1}^{d}\widetilde{\mathbf{e}}_i\ \text{if }0\leqslant v\leqslant \frac{d-du}{2}, \\ &\frac{8+d-4u-du-2v}{4}\widetilde{C}+(4-2u-v)\mathbf{f}+(1-u)\sum_{i=1}^{d}\widetilde{\mathbf{e}}_i\ \text{if }\frac{d-du}{2}\leqslant v\leqslant \frac{4+d-du}{2}, \\ &\frac{4-2u-v}{4-d}\Big(2\widetilde{C}+(4-d)\mathbf{f}+2\sum_{i=1}^{d}\widetilde{\mathbf{e}}_i\Big)\ \text{if }\frac{4+d-du}{2}\leqslant v\leqslant 4-2u, \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}N(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant \frac{d-du}{2}, \\ &\frac{2v+du-d}{4}\widetilde{C}\ \text{if }\frac{d-du}{2}\leqslant v\leqslant \frac{4+d-du}{2}, \\ &\frac{2v+du-2d}{4-d}\widetilde{C}+\frac{2v+du-4-d}{4-d}\sum_{i=1}^{d}\widetilde{\mathbf{e}}_i\ \text{if }\frac{4+d-du}{2}\leqslant v\leqslant 4-2u. \end{aligned} \right. \end{align*} $$

Thus, using [3, Corollary 1.109], we get

$$ \begin{align*} S\big(W_{\bullet,\bullet}^S;\mathbf{f}\big)&=\frac{3}{4d}\int_0^1\int_0^{\frac{d-du}{2}}\big(du^2-4du-v^2+3d\big)dvdu+\\ &\quad+\frac{3}{4d}\int_0^1\int_{\frac{d-du}{2}}^{\frac{4+d-du}{2}}\frac{d(1-u)(12-du+d-4u-4v)}{4}dvdu+\\ &\quad+\frac{3}{4d}\int_0^1\int_{\frac{4+d-du}{2}}^{4-2u}\frac{d(4-2u-v)^2}{4-d}dvdu=\frac{44+5d}{32}<2. \end{align*} $$

Therefore, there exists a point $O\in \mathbf {f}$ such that $S(W_{\bullet ,\bullet ,\bullet }^{\widetilde {S},\mathbf {f}};O)\geqslant 1$ .

Let us compute $S(W_{\bullet ,\bullet ,\bullet }^{\widetilde {S},\mathbf {f}};O)$ . Observe that

$$ \begin{align*}P(u,v)\cdot\mathbf{f}=\left\{\begin{aligned} &v\ \text{if }0\leqslant v\leqslant \frac{d-du}{2}, \\ &\frac{d-du}{2}\ \text{if }\frac{d-du}{2}\leqslant v\leqslant \frac{4+d-du}{2}, \\ &\frac{d(4-2u-v)}{4-d}\ \text{if }\frac{4+d-du}{2}\leqslant v\leqslant 4-2u. \end{aligned} \right. \end{align*} $$

Hence, it follows from [3, Remark 1.113] that

$$ \begin{align*} S(W_{\bullet,\bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)&=\frac{3}{4d}\int_0^1\int_0^{4-2u}\Big(\big(P(u,v)\cdot\mathbf{f}\big)\Big)^2dvdu+\\ &\quad+\frac{6}{4d}\int_0^1\int_0^{4-2u}\big(P(u,v)\cdot\mathbf{f}\big)\mathrm{ord}_O\big(N(u,v)\big|_{\mathbf{f}}\big)dvdu=\\ &=\frac{3}{4d}\int_0^1\int_0^{\frac{d-du}{2}}v^2dvdu+\\ &\quad+\frac{3}{4d}\int_{\frac{d-du}{2}}^{\frac{4+d-du}{2}}\Big(\frac{d-du}{2}\Big)^2dvdu+ \frac{3}{4d}\int_0^1\int_{\frac{4+d-du}{2}}^{4-2u}\Big(\frac{d(4-2u-v)}{4-d}\Big)^2dvdu+\\ &\quad+\frac{6}{4d}\int_0^1\int_0^{4-2u}\big(P(u,v)\cdot\mathbf{f}\big)\mathrm{ord}_O\big(N(u,v)\big|_{\mathbf{f}}\big)dvdu=\\ &=\frac{5d}{32}+\frac{3}{2}\int_0^1\int_0^{4-2u}\big(P(u,v)\cdot\mathbf{f}\big)\mathrm{ord}_O\big(N(u,v)\big|_{\mathbf{f}}\big)dvdu. \end{align*} $$

Therefore, if $O\not \in \widetilde {C}$ , we obtain $S(W_{\bullet ,\bullet ,\bullet }^{\widetilde {S},\mathbf {f}};O)=\frac {5d}{32}$ , which contradicts to $S(W_{\bullet ,\bullet ,\bullet }^{\widetilde {S},\mathbf {f}};O)\geqslant 1$ . Similarly, if $O\in \widetilde {C}$ and $\widetilde {C}$ intersects the curve $\mathbf {f}$ transversally at the point O, then

$$ \begin{align*}S(W_{\bullet,\bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)=\frac{5d}{32}+\frac{6}{4d}\int_0^1\int_0^{4-2u}\big(P(u,v)\cdot\mathbf{f}\big)\mathrm{ord}_O\big(N(u,v)\big|_{\mathbf{f}}\big)dvdu=\frac{44+5d}{64}<1, \end{align*} $$

which again contradicts $S(W_{\bullet ,\bullet ,\bullet }^{\widetilde {S},\mathbf {f}};O)\geqslant 1$ . Therefore, the curves $\widetilde {C}$ and $\mathbf {f}$ are tangent at O, which implies that C has a cusp at the point P.

Thus, to proceed, we may assume that $d=1$ or $d=2$ .

Now, let us consider the following commutative diagram:

where

• • $\rho $ is the blow up of the point $\widetilde {C}\cap \mathbf {f}$ ,

• • $\eta $ is the blow up of the intersection point of the $\rho $ -exceptional curve and the proper transform of the curve $\widetilde {C}$ ,

• • $\psi $ is the contraction of the proper transforms of both $(\sigma \circ \rho )$ -exceptional curves,

• • $\upsilon $ is the birational contraction of the proper transform of the $\eta $ -exceptional curve.

Let $\mathscr {F}$ be the $\upsilon $ -exceptional curve; let $\mathscr {C}$ be the proper transform on $\mathscr {S}$ of the curve C. Then $\mathscr {F}$ and $\mathscr {C}$ are smooth, $\mathscr {C}^2=-6$ , $\mathscr {F}^{2}=-\frac {1}{6}$ , $\mathscr {C}\cdot \mathscr {F}=1$ , and $\upsilon ^*(C)=\mathscr {C}+6\mathscr {F}$ .

Observe that $\mathscr {F}$ contains two singular points of the surfaces $\mathscr {S}$ , which are quotient singular points of type $\frac {1}{2}(1,1)$ and $\frac {1}{3}(1,1)$ . Denote these points by $Q_2$ and $Q_3$ , respectively. Note that $\mathscr {C}$ does not contain $Q_2$ and $Q_3$ . Write $\Delta _{\mathscr {F}}=\frac {1}{2}Q_2+\frac {2}{3}Q_3$ . Then, since $A_S(\mathscr {F})=5$ , it follows from [3, Remark 1.113] that

$$ \begin{align*}1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\inf_{Q\in\mathscr{F}}\frac{1-\mathrm{ord}_{Q}(\Delta_{\mathscr{F}})}{S(W_{\bullet,\bullet,\bullet}^{\mathscr{S},\mathscr{F}};Q)}, \frac{5}{S(V_{\bullet,\bullet}^{S};\mathscr{F})},\frac{1}{S_X(S)}\right\}. \end{align*} $$

But we already proved that $S_X(S)<1$ . Hence, we conclude that $S(V_{\bullet ,\bullet }^{S};\mathscr {F})\geqslant 5$ or there exists a point $Q\in \mathscr {F}$ such that $S(W_{\bullet ,\bullet ,\bullet }^{\mathscr {S},\mathscr {F}};Q)\geqslant 1-\mathrm {ord}_{Q}(\Delta _{\mathscr {F}})$ .

Let us compute $S(V_{\bullet ,\bullet }^{S};\mathscr {F})$ . Take $v\in \mathbb {R}_{\geqslant 0}$ . Set $P(u)=-K_X-u S$ . Then

$$ \begin{align*}\upsilon^*\big(P(u)|_S\big)-v\mathscr{F}\sim_{\mathbb{R}}(2-u)\mathscr{C}+(1-u)\sum_{i=1}^{d}\mathscr{E}_i+(12-6u-v)\mathscr{F}, \end{align*} $$

where $\mathscr {E}_i$ is the proper transform on $\mathscr {S}$ of the $(-1)$ -curve $\mathbf {e}_i$ . Using this, we conclude that the divisor $\upsilon ^*(P(u)|_S)-v\mathscr {F}$ is pseudoeffective $\iff v\leqslant 12-6u$ .

Let $\mathscr {P}(u,v)$ be the positive part of the Zariski decomposition of $\upsilon ^*(P(u)|_S)-v\mathscr {F}$ , and let $\mathscr {N}(u,v)$ be its negative part. Then

$$ \begin{align*}\mathscr{P}(u,v)=\left\{\begin{aligned} &(2-u)\mathscr{C}+(1-u)\sum_{i=1}^{d}\mathscr{E}_i+(12-6u-v)\mathscr{F}\ \text{if }0\leqslant v\leqslant d(1-u), \\ &\frac{12+d-(6+d)u-v}{6}\mathscr{C}+(1-u)\sum_{i=1}^{d}\mathscr{E}_i+(12-6u-v)\mathscr{F}\ \text{if }d(1-u)\leqslant v\leqslant 6+d-du, \\ &\frac{12-6u-v}{6-d}\Big(\mathscr{C}+\sum_{i=1}^{d}\mathscr{E}_i+(6-d)\mathscr{F}\Big)\ \text{if }6+d-du\leqslant v\leqslant 12-6u, \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}\mathscr{N}(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant d(1-u), \\ &\frac{v+d(u-1)}{6}\mathscr{C}\ \text{if }d(1-u)\leqslant v\leqslant 6+d-du, \\ &\frac{v+du-2d}{6-d}\mathscr{C}+\frac{v+du-6-d}{6-d}\sum_{i=1}^{d}\mathscr{E}_i\ \text{if }6+d-du\leqslant v\leqslant 12-6u. \end{aligned} \right. \end{align*} $$

This gives

$$ \begin{align*}\mathscr{P}(u,v)^2=\left\{\begin{aligned} &\frac{6du^2-24du-v^2+18d}{6}\ \text{if }0\leqslant v\leqslant d(1-u), \\ &\frac{d(1-u)(18-(d+6)u+d-2v)}{6}\ \text{if }d(1-u)\leqslant v\leqslant 6+d-du, \\ &\frac{d(12-6u-v)^2}{6(6-d)}\ \text{if }6+d-du\leqslant v\leqslant 12-6u. \end{aligned} \right. \end{align*} $$

Thus, using [3, Corollary 1.109] and integrating, we get

$$ \begin{align*}S(W_{\bullet,\bullet}^{\mathscr{S}};\mathscr{F})=\frac{3}{4d}\int_0^1\int_0^{12-6u}\mathscr{P}(u,v)^2dvdu=\frac{66+5d}{16}\leqslant\frac{19}{4}<5=A_S(\mathscr{F}) \end{align*} $$

since $d=1$ or $d=2$ . Thus, there is a point $Q\in \mathscr {F}$ such that $S(W_{\bullet ,\bullet ,\bullet }^{\mathscr {S},\mathscr {F}};Q)\geqslant 1-\mathrm {ord}_{Q}(\Delta _{\mathscr {F}})$ .

Now, using [3, Remark 1.103] again, we see that

$$ \begin{align*}S\big(W_{\bullet,\bullet,\bullet}^{\mathscr{S},\mathscr{F}};Q\big)&=\frac{3}{4d}\int_0^1\int_0^{12-6u}\Big(\big(\mathscr{P}(u,v)\cdot\mathscr{F}\big)\Big)^2dvdu+\\&\quad +\frac{6}{4d}\int_0^1\int_0^{12-6u}\big(P(u,v)\cdot\mathscr{F}\big)\mathrm{ord}_Q\big(\mathscr{N}(u,v)\big|_{\mathscr{F}}\big)dvdu.\end{align*} $$

On the other hand, we have

$$ \begin{align*}\mathscr{P}(u,v)\cdot\mathscr{F}=\left\{\begin{aligned} &\frac{v}{6}\ \text{if }0\leqslant v\leqslant d(1-u), \\ &\frac{d(1-u)}{6}\ \text{if }d(1-u)\leqslant v\leqslant 6+d-du, \\ &\frac{d(12-6u-v)}{6(6-d)}\ \text{if }6+d-du\leqslant v\leqslant 12-6u, \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}\mathscr{N}(u,v)\cdot\mathscr{F}=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant d(1-u), \\ &\frac{v-d(1-u)}{6}\ \text{if }d(1-u)\leqslant v\leqslant 6+d-du, \\ &\frac{v+du-2d}{6-d}\ \text{if }6+d-du\leqslant v\leqslant 12-6u. \end{aligned} \right. \end{align*} $$

In particular, we have

$$ \begin{align*}S(W_{\bullet,\bullet,\bullet}^{\mathscr{S},\mathscr{F}};Q)=\frac{5d}{96}+\frac{6}{4d}\int_0^1\int_0^{12-6u}\big(P(u,v)\cdot\mathscr{F}\big)\mathrm{ord}_Q\big(\mathscr{N}(u,v)\big|_{\mathscr{F}}\big)dvdu. \end{align*} $$

Hence, if $Q\not \in \mathscr {C}$ , then $\frac {1}{3}\leqslant 1-\mathrm {ord}_{Q}(\Delta _{\mathscr {F}})\leqslant S(W_{\bullet ,\bullet ,\bullet }^{\mathscr {S},\mathscr {F}};Q)=\frac {5d}{96}<\frac {1}{3}$ , which is absurd. Thus, we conclude that $Q=\mathscr {C}\cap \mathscr {F}$ . Then

$$ \begin{align*} S(W_{\bullet,\bullet,\bullet}^{\mathscr{S},\mathscr{F}};Q)&=\frac{5d}{96}+\frac{6}{4d}\int_0^1\int_0^{12-6u}\big(P(u,v)\cdot\mathscr{F}\big)\mathrm{ord}_Q\big(\mathscr{N}(u,v)\big|_{\mathscr{F}}\big)dvdu\leqslant\\ &\quad\leqslant\frac{5d}{96}+\frac{6}{4d}\int_0^1\int_0^{12-6u}\big(P(u,v)\cdot\mathscr{F}\big)\big(\mathscr{N}(u,v)\cdot\mathscr{F}\big)dvdu=\frac{11}{16}<1, \end{align*} $$

which is a contradiction, since $S(W_{\bullet ,\bullet ,\bullet }^{\mathscr {S},\mathscr {F}};Q)\geqslant 1-\mathrm {ord}_{Q}(\Delta _{\mathscr {F}})=1$ .

In particular, we conclude that either $d=2$ or $d=3$ .

Corollary 2.7. All smooth Fano threefolds in the family are K-stable.

Recall that A is the fiber of the del Pezzo fibration $\phi \colon X\to \mathbb {P}^1$ that passes through P. Note also that we have the following possibilities:

• • $d=2$ , and A is a double cover of $\mathbb {P}^2$ branched over a reduced quartic curve;

• • $d=3$ , and A is a normal cubic surface in $\mathbb {P}^3$ .

Observe that $C=S\cap A$ , where S is a general surface in $|\pi ^*(H)|$ that contains the point P. Since C is singular at P, the surface A must be singular at P, which confirms Corollary 2.4. Now, using classifications of reduced singular plane quartic curves and singular normal cubic surfaces [6], we see that $P=\mathrm {Sing}(A)$ , and one of the following three cases holds:

• • $d=2$ , and A is a double cover of $\mathbb {P}^2$ branched over four lines intersecting in a point;

• • $d=3$ , and A is a cone in $\mathbb {P}^3$ over a smooth plane cubic curve;

• • $d=3$ , and A has Du Val singular point of type $\mathbb {D}_4$ , $\mathbb {D}_5$ , or $\mathbb {E}_6$ .

Let us show that the first case is impossible.

Lemma 2.8. One has $d=3$ .

Proof. Suppose that $d=2$ . Then $P=\mathrm {Sing}(A)$ , and A is a double cover of $\mathbb {P}^2$ branched over a reduced reducible plane quartic curve that is a union of four distinct lines passing through one point. Let us seek for a contradiction.

Let $\alpha \colon \widetilde {X}\to X$ be the blow up of the point P, let $E_P$ be the $\alpha $ -exceptional divisor and let $\widetilde {A}$ be the proper transform on $\widetilde {X}$ of the surface A. Then $\widetilde {A}\cap E_P$ is a line $L\subset E_P\cong \mathbb {P}^2$ , and the surface $\widetilde {A}$ is singular along this line. Let $\beta \colon \overline {X}\to \widetilde {X}$ be the blow up of the line L, let $E_L$ be the $\beta $ -exceptional divisor, let $\overline {A}$ be the proper transform on $\overline {X}$ of the surface $\widetilde {A}$ and let $\overline {E}_P$ be the proper transforms on $\overline {X}$ of the surface $E_P$ . Then

• • $E_L\cong \mathbb {F}_2$ ,

• • the intersection $\overline {E}_P\cap E_L$ is the $(-2)$ -curve in $E_L$ ,

• • the surface $\overline {A}$ is smooth, and there exists a $\mathbb {P}^1$ -bundle $\overline {A}\to \mathcal {C}$ ,

• • $\overline {A}\cap E_L$ is a smooth elliptic curve that is a section of the $\mathbb {P}^1$ -bundle $\overline {A}\to \mathcal {C}$ ,

• • the surfaces $\overline {A}$ and $\overline {E}_P$ are disjoint,

• • $\overline {E}_P\cong \mathbb {P}^2$ and $\overline {E}_P\vert _{\overline {E}_P}\cong \mathcal {O}_{\mathbb {P}^2}(-2)$ .

There is a birational contraction $\gamma \colon \overline {X}\to \widehat {X}$ of the surface $\overline {E}_P$ such that $\widehat {X}$ is a projective threefold that has one singular point $O=\gamma (\overline {E}_P)$ , which is a terminal cyclic quotient singularities of type $\frac {1}{2}(1,1,1)$ . Thus, there exists the following commutative diagram

where $\sigma $ is a birational morphism that contracts the surface $\gamma (E_L)$ to the point P.

Let $G=\gamma (E_L)$ , let $\widehat {A}=\gamma (\overline {A})$ and let $\widehat {E}$ be the proper transform on $\widehat {X}$ of the surface E. Then $A_{X}(G)=4$ . Moreover, we have

$$ \begin{align*} \sigma^{*}(-K_X)&\sim 2\widehat{A}+\widehat{E}+8G, \\ \sigma^{*}(A)&\sim\widehat{A}+4G. \end{align*} $$

Note that $\widehat {A}\cong \overline {A}$ and $G\cong \mathbb {P}(1,1,2)$ , so we can identify G with a quadric cone in $\mathbb {P}^3$ . Note also that O is the vertex of the cone G. Moreover, by construction, we have $O\not \in \widehat {A}$ . Furthermore, the exists a $\mathbb {P}^1$ -bundle $\widehat {A}\to \mathcal {C}$ such that $G\vert _{\widehat {A}}$ is its section.

Let $\mathbf {g}$ be a ruling of the quadric cone G, let $\mathbf {l}$ be a fiber of the $\mathbb {P}^1$ -bundle $\widehat {A}\to \mathcal {C}$ and let $\mathbf {f}$ be a fiber of the $\mathbb {P}^1$ -bundle $\pi \circ \sigma \vert _{\widehat {E}}\colon \widehat {E}\to \mathcal {C}$ . Then $G\vert _{G}\sim _{\mathbb {Q}}-\mathbf {g}$ and $\widehat {A}\vert _{G}\sim _{\mathbb {Q}}4\mathbf {g}$ . Moreover, the intersections of the surfaces G, $\widehat {A}$ , $\widehat {E}$ with the curves $\mathbf {g}$ , $\mathbf {l}$ , $\mathbf {f}$ are given here:

Fix a nonnegative real number u. We have $\sigma ^{*}(-K_X)-uG\sim _{\mathbb {R}} 2\widehat {A}+\widehat {E}+(8-u)G$ , which implies that $\sigma ^{*}(-K_X)-uG$ is pseudoeffective $\iff u\in [0,8]$ . Furthermore, if $u\in [0,8]$ , then the Zariski decomposition of the divisor $\sigma ^{*}(-K_X)-uG$ can be described as follows:

$$ \begin{align*}P(u)=\left\{\begin{aligned} &2\widehat{A}+\widehat{E}+(8-u)G\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{9-u}{4}\widehat{A}+\widehat{E}+(8-u)G \ \text{if }1\leqslant u\leqslant 5, \\ &\frac{8-u}{3}\big(\widehat{A}+\widehat{E}+3G\big)\ \text{if }5\leqslant u\leqslant 8, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}N(u)=\left\{\begin{aligned} &0\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{u-1}{4}\widehat{A}\ \text{if }1\leqslant u\leqslant 5, \\ &\frac{u-2}{3}\widehat{A}+\frac{u-5}{3}\widehat{E}\ \text{if }5\leqslant u\leqslant 8, \\ \end{aligned} \right. \end{align*} $$

where $P(u)$ and $N(u)$ are the positive and the negative parts of the decomposition. Then

$$ \begin{align*}\mathrm{vol}\big(\sigma^{*}(-K_X)-uG\big)=P(u)^3=\left\{\begin{aligned} &8-\frac{u^3}{8}\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{18-3u}{2}\ \text{if }1\leqslant u\leqslant 5, \\ &\frac{(8-u)^{3}}{18}\ \text{if }5\leqslant u\leqslant 8. \\ \end{aligned} \right. \end{align*} $$

Integrating, we get $S_{X}(G)=\frac {27}{8}<4=A_{X}(G)$ . But [16, Corollary 4.18] gives

$$ \begin{align*}1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{A_{X}(G)}{S_{X}(G)}, \inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\}=\min\left\{\frac{32}{27},\inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\}, \end{align*} $$

where $\delta _{Q}(G,V^G_{\bullet ,\bullet })$ is defined in [16]. Moreover, if Q is a point in G and Z is a smooth curve in G that passes through Q, then it follows from [16, Corollary 4.18] that

$$ \begin{align*}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big) \geqslant\min\left\{\frac{1}{S\big(V^G_{\bullet,\bullet};Z\big)},\frac{1-\mathrm{ord}_Q(\Delta_Z)}{S\big(W^{G,Z}_{\bullet,\bullet,\bullet};Q\big)}\right\}, \end{align*} $$

where $S(V^G_{\bullet ,\bullet };Z)$ and $S(W^{G,Z}_{\bullet ,\bullet ,\bullet };Q)$ are defined in [16], and

$$ \begin{align*}\Delta_Z=\left\{\begin{aligned} &0\ \text{if }O\not\in Z, \\ &\frac{1}{2}O\ \text{if }O\in Z. \end{aligned} \right. \end{align*} $$

Let us show that $\delta _{Q}\big (G,V^G_{\bullet ,\bullet }\big )>1$ for every $Q\in G$ , which would imply a contradiction.

Let $\mathscr {C}=\widehat {A}\vert _{G}$ , and let $\ell $ be the curve in $|\mathbf {f}|$ that contains Q. Then $O\not \in \mathscr {C}$ and $O\in \ell $ so that $\Delta _{\mathscr {C}}=0$ and $\Delta _{\ell }=\frac {1}{2}O$ . Take $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}P(u)\big\vert_{G}-v\ell\sim_{\mathbb{R}}\left\{\begin{aligned} &(u-v)\mathbf{g}\ \text{if }0\leqslant u\leqslant 1, \\ &(1-v)\mathbf{g}\ \text{if }1\leqslant u\leqslant 5, \\ &\frac{8-u-3v}{3}\mathbf{g}\ \text{if }5\leqslant u\leqslant 8.\\ \end{aligned} \right. \end{align*} $$

Now, using [16, Theorem 4.8], we get

$$ \begin{align*} S\big(W^{G}_{\bullet,\bullet};\ell\big)&= \frac{3}{8}\int_0^8\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{G}-v\ell\big)dvdu=\frac{3}{8}\int_{0}^{1}\int_{0}^{u}\frac{(u-v)^{2}}{2}dvdu+\\ &\quad+\frac{3}{12}\int_{1}^{5}\int_{0}^{1}\frac{(1-v)^{2}}{4}dvdu+\frac{3}{12}\int_{5}^{8}\int_{0}^{\frac{8-u}{3}}\frac{(8-u-3v)^2}{18}dvdu=\frac{5}{16}. \end{align*} $$

Similarly, if $Q\not \in \mathscr {C}$ , then it follows from [16, Theorem 4.17] that

$$ \begin{align*} S\big(W_{\bullet,\bullet,\bullet}^{G,\ell};Q\big)&= \frac{3}{8}\int_{0}^{1}\int_{0}^{u}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+ \frac{3}{8}\int_{1}^{5}\int_{0}^{1}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+\\ &\quad+\frac{3}{8}\int_{5}^{8}\int_{0}^{\frac{8-u}{3}}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+F_Q= \frac{3}{8}\int_{0}^{1}\int_{0}^{u}\frac{(u-v)^{2}}{4}dvdu+\\ &\quad+\frac{3}{8}\int_{1}^{5}\int_{0}^{1}\frac{(1-v)^{2}}{4}dvdu+ \frac{3}{8}\int_{5}^{8}\int_{0}^{\frac{8-u}{3}}\frac{(8-u-3v)^2}{36}dvdu=\frac{5}{32} \end{align*} $$

so that $S(W_{\bullet ,\bullet ,\bullet }^{G,\ell };Q)=\frac {5}{32}$ , which implies that $\delta _{Q}(G,V^G_{\bullet ,\bullet })\geqslant \frac {16}{5}$ . Likewise, if $Q\in \mathscr {C}$ , then

$$ \begin{align*} S\big(W^{G}_{\bullet,\bullet};\mathscr{C}\big)&= \frac{3}{8}\int_{0}^{8}\big(P(u)^{2}\cdot G\big)\cdot\mathrm{ord}_{\mathscr{C}}\Big(N(u)\big\vert_{G}\Big)du+ \frac{3}{8}\int_0^8\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{G}-v\mathscr{C}\big)dvdu=\\ &=\frac{3}{8}\int_{1}^{5}\frac{u-1}{8}du+\frac{3}{8}\int_{5}^{8}\frac{(u-2)(8-u)^{2}}{54}du+\frac{3}{8}\int_0^1\int_{0}^{\frac{u}{4}}\frac{(u-4v)^{2}}{2}dvdu+\\ &\quad+\frac{3}{8}\int_{1}^{5}\int_{0}^{\frac{1}{4}} \frac{(1-4v)^{2}}{2}dvdu+\frac{3}{8}\int_{5}^{8}\int_{0}^{\frac{8-u}{12}}\frac{(8-u-12v)^{2}}{18}dvdu=\frac{11}{16} \end{align*} $$

and

$$ \begin{align*} S\big(W_{\bullet,\bullet,\bullet}^{G,\mathscr{C}};Q\big)&= \frac{3}{8}\int_{0}^{1}\int_{0}^{\frac{u}{4}}\Big(\big(P(u)\big\vert_{G}-v\mathscr{C}\big)\cdot\mathscr{C}\Big)^2dvdu+\\ &\quad+\frac{3}{8}\int_{1}^{5}\int_{0}^{\frac{1}{4}}\Big(\big(P(u)\big\vert_{G}-v\mathscr{C}\big)\cdot\mathscr{C}\Big)^2dvdu+\frac{3}{8}\int_{5}^{8}\int_{0}^{\frac{8-u}{12}}\Big(\big(P(u)\big\vert_{G}-v\mathscr{C}\big)\cdot\mathscr{C}\Big)^2dvdu=\\ &=\frac{3}{8}\int_{0}^{1} \int_{0}^{\frac{u}{4}}(2u-8v)^{2}dvdu+\frac{3}{8}\int_{1}^{5} \int_{0}^{\frac{1}{4}}(2-8v)^{2}dvdu+\\ &\quad+\frac{3}{8}\int_{5}^{8}\int_{0}^{\frac{8-u}{12}}(\frac{(16-2u-24v)^2}{9}dvdu=\frac{5}{8}. \end{align*} $$

This implies that $\delta _{Q}(G,V^G_{\bullet ,\bullet })\geqslant \min \left \{\frac {16}{11},\frac {8}{5}\right \}=\frac {16}{11}$ , which is a contradiction.

Corollary 2.9. All smooth Fano threefolds in the family are K-stable.

We see that $d=3$ so that A is a singular cubic surface in $\mathbb {P}^3$ such that $P=\mathrm {Sing}(A)$ . Let $\sigma \colon \widehat {X}\to X$ be the blow up of the point P, and let G be the $\sigma $ -exceptional surface. Denote by $\widehat {A}$ and $\widehat {E}$ the proper transforms on $\widehat {X}$ of the surfaces A and E, respectively.

Lemma 2.10. The surface A has Du Val singularities.

Proof. Suppose that A is a cone in $\mathbb {P}^3$ with vertex P. Take $u\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}\sigma^*(-K_X)-vG\sim_{\mathbb{R}}2\widehat{A}+\widehat{E}+(6-u)G. \end{align*} $$

Thus, the divisor $\sigma ^*(-K_X)-vG$ is pseudoeffective $\iff u\in [0,6]$ . Moreover, if $u\in [0,6]$ , then the Zariski decomposition of the divisor $\sigma ^*(-K_X)-vG$ can be described as follows:

$$ \begin{align*}P(u)=\left\{\begin{aligned} &2\widehat{A}+E+(6-u)G\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{7-u}{3}\widehat{A}+\widehat{E}+(6-u)G\ \text{if }1\leqslant u\leqslant 4, \\ &\frac{6-u}{2}(\widehat{A}+\widehat{E}+2G)\ \text{if }4\leqslant u\leqslant 6, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}N(u)=\left\{\begin{aligned} &0\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{u-1}{3}\widehat{A}\ \text{if }1\leqslant u\leqslant 4, \\ &\frac{u-2}{2}\widehat{A}+\frac{u-4}{2}\widehat{E}\ \text{if }4\leqslant u\leqslant 6, \\ \end{aligned} \right. \end{align*} $$

where $P(u)$ and $N(u)$ are the positive and the negative parts of the Zariski decomposition, respectively. Using this, we compute

$$ \begin{align*}S_{X}(G)=\frac{3}{12}\int_{0}^{1}u^{3}du+\frac{3}{12}\int_{1}^{4}udu+\frac{3}{12}\int_{4}^{6}\frac{u(6-u)^{2}}{4}du=\frac{43}{16}<3=A_X(G). \end{align*} $$

Let us apply [16, Theorem 4.8], [16, Corollary 4.17], [16, Corollary 4.18] using notations introduced in [16, § 4]. To start with, we apply [16, Corollary 4.18] to get

(2.2) $$ \begin{align} 1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{A_{X}(G)}{S_{X}(G)}, \inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\}=\min\left\{\frac{48}{43},\inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\}, \end{align} $$

where $\delta _{Q}(G,V^G_{\bullet ,\bullet })$ is defined in [16, § 4]. Let Q be an arbitrary point in the surface G, and let $\ell $ is a general line in $G\cong \mathbb {P}^2$ that contains Q. Then [16, Corollary 4.18] gives

$$ \begin{align*}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big) \geqslant\min\left\{\frac{1}{S\big(V^G_{\bullet,\bullet};\ell\big)},\frac{1}{S\big(W^{G,\ell}_{\bullet,\bullet,\bullet};Q\big)}\right\}, \end{align*} $$

where $S(V^F_{\bullet ,\bullet };\ell )$ and $S(W^{G,\ell }_{\bullet ,\bullet ,\bullet };Q)$ are defined in [16, § 4]. Take $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}P(u)\big\vert_{G}-v\ell\sim_{\mathbb{R}}\left\{\begin{aligned} &(u-v)\ell\ \text{if }0\leqslant u\leqslant 1, \\ &(1-v)\ell\ \text{if }1\leqslant u\leqslant 4, \\ &\frac{6-u-2v}{2}\ell\ \text{if }4\leqslant u\leqslant 6.\\ \end{aligned} \right. \end{align*} $$

Let $\mathscr {C}=\widehat {A}\vert _{G}$ . Then $\mathscr {C}$ is a smooth cubic curve in $G\cong \mathbb {P}^2$ . Let

$$ \begin{align*}N^\prime(u)=N(u)\big\vert_{G}=\left\{\begin{aligned} &0\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{u-1}{3}\mathscr{C}\ \text{if }1\leqslant u\leqslant 4, \\ &\frac{u-2}{2}\mathscr{C}\ \text{if }4\leqslant u\leqslant 6. \end{aligned} \right. \end{align*} $$

Now, using [16, Theorem 4.8], we get

$$ \begin{align*} S\big(W^{G}_{\bullet,\bullet};\ell\big)&=\frac{3}{12}\int_0^6\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{G}-v\ell\big)dvdu=\frac{3}{12}\int_{0}^{1}\int_{0}^{u}(u-v)^{2}dvdu+\\ &\quad+\frac{3}{12}\int_{1}^{4}\int_{0}^{1}(1-v)^{2}dvdu+\frac{3}{12}\int_{4}^{6}\int_{0}^{\frac{6-u}{2}}\left(\frac{6-u-2v}{2}\right)^{2}dvdu=\frac{5}{16}. \end{align*} $$

Similarly, it follows from [16, Theorem 4.17] that

$$ \begin{align*} S\big(W_{\bullet,\bullet,\bullet}^{G,\ell};Q\big)&= \frac{3}{12}\int_{0}^{1}\int_{0}^{u}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+ \frac{3}{12}\int_{1}^{4}\int_{0}^{1}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+\\ &\quad+\frac{3}{12}\int_{4}^{6}\int_{0}^{\frac{6-u}{2}}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+F_Q=\frac{3}{12}\int_{0}^{1}\int_{0}^{u}(u-v)^{2}dvdu+\\ &\quad+\frac{3}{12}\int_{1}^{4}\int_{0}^{1}(1-v)^{2}dvdu+\frac{3}{12}\int_{4}^{6}\int_{0}^{\frac{6-u}{2}}\left(\frac{6-u-2v}{2}\right)^{2}dvdu+F_Q=\frac{5}{16}+F_Q, \end{align*} $$

where

$$ \begin{align*} F_Q&=\frac{6}{12}\int_{1}^{4}\int_{0}^{1}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)\mathrm{ord}_Q\big(N^\prime(u)\big|_\ell\big)dvdu+\\ &\quad+\frac{6}{12}\int_{4}^{6} \int_{0}^{\frac{6-u}{2}}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)\mathrm{ord}_Q\big(N^\prime(u)\big|_\ell\big)dvdu\leqslant\\ &\quad\leqslant\frac{6}{12}\int_{1}^{4} \int_{0}^{1}\frac{(1-v)(u-1)}{3}dvdu+ \frac{6}{12}\int_{4}^{6} \int_{0}^{\frac{6-u}{2}}\frac{(6-u-2v)(u-2)}{4}dvdu=\frac{7}{12}. \end{align*} $$

So, we have $S(W_{\bullet ,\bullet ,\bullet }^{G,\ell };Q)\leqslant \frac {43}{48}$ . Then $\delta _{Q}(G,V^G_{\bullet ,\bullet })>1$ , which contradicts Equation (2.2).

Thus, we see that P is a Du Val singular point of the surface A of type $\mathbb {D}_4$ , $\mathbb {D}_5$ , $\mathbb {E}_6$ . Now, arguing as in the proof of [22, Lemma 9.11], we see that $\beta (\mathbf {F})>0$ if

1. (1) the inequality $\beta (G)>0$ holds,

2. (2) and for every prime divisor $\mathbf {E}$ over X such that $C_X(\mathbf {E})$ is a curve containing P, the following inequality holds:

   $$ \begin{align*}\frac{A_X(\mathbf{E})}{S_X(\mathbf{E})}\geqslant\frac{4}{3}. \end{align*} $$

Since $\beta (\mathbf {F})\leqslant 0$ by our assumption, we see that at least one of these conditions must fail.

Lemma 2.11. One has $\beta (G)\geqslant \frac {465}{2048}$ .

Proof. Let $\widehat {A}$ and $\widehat {E}$ be the proper transforms on $\widehat {X}$ of the surfaces A and E, respectively. Take $u\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}\sigma^*(-K_{X})-uG\sim\sigma^*(2H-E)-uG\sim\sigma^*(2A+E)-uG\sim 2\widehat{A}+\widehat{E}+(4-u)G, \end{align*} $$

which easily implies that the divisor $-K_{\widehat {X}}-G$ is pseudoeffective $\iff u\leqslant 4$ because we can contract the surfaces $\widehat {A}$ and $\widehat {E}$ simultaneously after flops. Then

$$ \begin{align*}\beta(G)=A_X(G)-S_X(G)=3-\frac{1}{12}\int_0^4\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)du. \end{align*} $$

Note that $\sigma ^*(-K_{X})-uG$ is nef for $u\in [0,1]$ because the divisor $-K_X$ is very ample. Thus, if $u\in [0,1]$ , then

$$ \begin{align*}\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)=\big(\sigma^*(-K_{X})-uG\big)^3=12-u^3. \end{align*} $$

Similarly, if $1\leqslant u\leqslant \frac {3}{2}$ , then

$$ \begin{align*}\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)\leqslant\mathrm{vol}\big(\sigma^*(-K_{X})-G\big)=\big(\sigma^*(-K_{X})-G\big)^3=11. \end{align*} $$

Finally, let us estimate $\mathrm {vol}(\sigma ^*(-K_{X})-uG)$ in the case when $4\geqslant u>\frac {3}{2}$ .

Let Z be a general hyperplane section of the cubic surface A that passes through P, and let $\widehat {Z}$ be its proper transform on the threefold $\widehat {X}$ . Then Z is an irreducible cuspidal cubic curve, and $\widehat {Z}\subset \widehat {A}$ . Observe that $(\sigma ^*(-K_{X})-uG)\cdot \widehat {Z}=3-2u$ and $\widehat {A}\cdot \widehat {Z}=-4$ , so $\widehat {A}$ is contained in the asymptotic base locus of the divisor $\sigma ^*(-K_{X})-uG$ for $u>\frac {3}{2}$ . Moreover, if $\sigma ^*(-K_{X})-uG\sim _{\mathbb {R}}\widehat {D}+\lambda \widehat {A}$ for $\lambda \in \mathbb {R}_{\geqslant 0}$ and an effective $\mathbb {R}$ -divisor $\widehat {D}$ whose support does not contain $\widehat {A}$ , then $\widehat {Z}\not \subset \widehat {D}$ , which implies that

$$ \begin{align*}0\leqslant \widehat{D}\cdot\widehat{Z}=\Big(\sigma^*(-K_{X})-uG-\lambda \widehat{A}\Big)\cdot\widehat{Z}=3-2u-4\lambda \end{align*} $$

so that $\lambda \geqslant \frac {3-2u}{4}$ . Thus, if $4\geqslant u>\frac {3}{2}$ , then

$$ \begin{align*}\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)\leqslant\mathrm{vol}\Big(\sigma^*(2H-E)-uG-\frac{2u-3}{4}\widehat{A}\Big). \end{align*} $$

Moreover, if $4\geqslant u>\frac {3}{2}$ , then

$$ \begin{align*}\sigma^*(2H-E)-uG-\frac{2u-3}{4}\widehat{A}\sim_{\mathbb{R}}\frac{11-2u}{4}\sigma^{*}(H)-\frac{7-2u}{2}\sigma^*(E)-\frac{3}{2}G. \end{align*} $$

Therefore, if $4\geqslant u>\frac {3}{2}$ , then

$$ \begin{align*}\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)\leqslant\mathrm{vol}\Big(\frac{11-2u}{4}\sigma^{*}(H)-\frac{7-2u}{2}\sigma^*(E)\Big)=\mathrm{vol}\Big(\frac{11-2u}{4}H-\frac{7-2u}{2}E\Big). \end{align*} $$

Furthermore, if $\frac {7}{2}\geqslant u>\frac {3}{2}$ , then $\frac {11-2u}{4}H-\frac {7-2u}{2}E$ is nef so that

$$ \begin{align*}\mathrm{vol}\Big(\frac{11-2u}{4}H-\frac{7-2u}{2}E\Big)=\Big(\frac{11-2u}{4}H-\frac{7-2u}{2}E\Big)^3=\frac{25-6u}{16}. \end{align*} $$

Similarly, if $4\geqslant u>\frac {7}{2}$ , then

$$ \begin{align*}\mathrm{vol}\Big(\frac{11-2u}{4}H-\frac{7-2u}{2}E\Big)=\Big(\frac{11-2u}{4}H\Big)^3=\frac{(11-2u)^3}{256}. \end{align*} $$

Now, we can estimate $\beta (G)$ as follows

$$ \begin{align*} \beta(G)&=3-\frac{1}{12}\int_0^4\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)du\geqslant 3-\frac{1}{12}\int_0^{1}(12-u^3)du-\frac{1}{12}\int_1^{\frac{3}{2}}11du-\\&\quad-\frac{1}{12}\int_{\frac{3}{2}}^{\frac{7}{2}}\frac{25-6u}{16}du-\frac{1}{12}\int_{\frac{7}{2}}^4\frac{(11-2u)^3}{256}du=3-\frac{5679}{2048}=\frac{465}{2048} \end{align*} $$

as claimed.

Therefore, there exists a prime divisor $\mathbf {E}$ over X such that $C_X(\mathbf {E})$ is a curve, $P\in C_X(\mathbf {E})$ , and $A_X(\mathbf {E})<\frac {4}{3}S_X(\mathbf {E})$ . Set $Z=C_X(\mathbf {E})$ . Then $\delta _O(X)<\frac {4}{3}$ for every point $O\in Z$ .

Lemma 2.12. One has $Z\subset A$ , and Z is a line in the cubic surface A.

Proof. Let O be a general point in Z, and let $A_O$ be the fiber of $\phi $ that passes through O. If $Z\not \subset A$ , then $A_O$ is smooth, so that $\delta _O(A_O)\geqslant \frac {3}{2}$ by [3, Lemma 2.13], which gives

$$ \begin{align*}\frac{4}{3}>\frac{A_X(\mathbf{E})}{S_X(\mathbf{E})}\geqslant\delta_O(X)\geqslant\mathrm{min}\Bigg\{\frac{16}{11},\frac{16\delta_O(A_O)}{\delta_O(A_O)+15}\Bigg\}\geqslant\mathrm{min}\Bigg\{\frac{16}{11},\frac{16\times\frac{3}{2}}{\frac{3}{2}+15}\Bigg\}=\frac{16}{11}>\frac{4}{3} \end{align*} $$

by Lemma 2.1. This shows that $Z\subset A$ and $A_O=A$ .

To complete the proof of the lemma, we have to show that Z is a line in the surface A. Suppose that Z is not a line. Then the point O is not contained in a line in the surface A because A contains finitely many lines [6]. Now, arguing as in the proof of [3, Lemma 2.13], we get $\delta _O(A)\geqslant \frac {3}{2}$ . So, applying Lemma 2.1 again, we get a contradiction as above.

Now, our auxiliary theorem follows from the following lemma:

Lemma 2.13. The surface A does not have a singular point of type $\mathbb {D}_4$ .

Proof. Suppose A has singularity of type $\mathbb {D}_4$ . Then it follows from [6] that, for a suitable choice of coordinates x, y, z, t on the projective space $\mathbb {P}^3$ , one of the following cases hold:

1. (A) $A=\{tx^2=y^3-z^3\}\subset \mathbb {P}^3$ ,

2. (B) $A=\{tx^2=y^3-z^3+xyz\}\subset \mathbb {P}^3$ .

Note that $P=[0:0:0:1]$ , and A contains six lines [6]. In case (A), these lines are

$$ \begin{align*} L_1&=\{x=y-z=0\},\\[3pt] L_2&=\{x=y-\omega_3z=0\},\\[3pt] L_3&=\{x=y+\omega_3^2z=0\},\\[3pt]L_4&=\{t=y-z=0\},\\[3pt] L_5&=\{t=y+\omega_3z=0\},\\[3pt] L_6&=\{t=y+\omega_3^2z=0\}, \end{align*} $$

where $\omega _3$ is a primitive cube root of unity. In case (B), these lines are

$$ \begin{align*} L_1&=\{x=y-z=0\},\\ L_2&=\{x=y-\omega_3z=0\},\\ L_3&=\{x=y-\omega_3^2z=0\},\\ L_4&=\{x+3(y-z)=y-z-9t=0\},\\ L_5&=\{x+3\omega_3(y-\omega_3z)=\omega_3y-\omega_3^2z-9t=0\},\\ L_6&=\{x+3\omega_3^2(y-\omega_3^2z)=\omega_3^2y-\omega_3z-9t=0\}. \end{align*} $$

Note that $P=L_1\cap L_3\cap L_3$ , $P\not \in L_4\cup L_5\cup L_6$ and $-K_A\sim 2L_1+L_4\sim 2L_2+L_5\sim 2L_3+L_6$ .

By Lemma 2.12, we may assume that $Z=L_1$ .

Recall that $S_X(A)=\frac {11}{16}$ ; see the proof of Lemma 2.1. Using [3, Theorem 1.112], we get

$$ \begin{align*}\frac{4}{3}>\frac{A_X(\mathbf{E})}{S_X(\mathbf{E})}\geqslant \min\left\{\frac{1}{S_X(A)},\frac{1}{S(W_{\bullet,\bullet}^A;L_1)}\right\}=\min\left\{\frac{16}{11},\frac{1}{S(W_{\bullet,\bullet}^A;L_1)}\right\}, \end{align*} $$

where $S(W_{\bullet ,\bullet }^A;L_1)$ is defined in [3, § 1.7]. Therefore, we conclude that $S(W_{\bullet ,\bullet }^A;L_1)<\frac {4}{3}$ . Let us compute $S(W_{\bullet ,\bullet }^A;L_1)$ using [3, Corollary 1.109].

To do this, we use notations introduced in the proof of Lemma 2.1 applied to $O=P$ . Then using [3, Corollary 1.109] and computations from the proof of Lemma 2.1, we get

$$ \begin{align*}S\big(W_{\bullet,\bullet}^A; L_1\big)=\frac{1}{4}\int_{0}^{1}\int_0^\infty \mathrm{vol}\big(-K_A-vL_1\big)dvdu+\frac{1}{4}\int_{1}^{2}\int_0^\infty \mathrm{vol}\big((2-u)(-K_A)-vL_1\big)dvdu \end{align*} $$

since $L_1\not \subset \mathrm {Supp}(N(u))$ since $L_1\not \subset E$ . Let us compute $S(W_{\bullet ,\bullet }^A; L_1)$ . Take $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}-K_A-vL_1\sim_{\mathbb{R}}(2-v)L_1+L_4. \end{align*} $$

Thus, the divisor $-K_A-vL_1$ is pseudoeffective $\iff v\leqslant 2$ since $L_4^2=-1$ . Fix $v\in [0,2]$ . Let $P(u,v)$ be the positive part of the Zariski decomposition of the divisor $-K_A-vL_1$ , and let $N(u,v)$ be its negative part. Then

$$ \begin{align*}P(u,v)=\left\{\begin{aligned} &(2-v)L_1+L_4\ \text{if }0\leqslant v\leqslant 1, \\ &(2-v)(L_1+L_4)\ \text{if }1\leqslant v\leqslant 2, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}N(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant 1, \\ &(v-1)L_4\ \text{if }1\leqslant v\leqslant 2. \end{aligned} \right. \end{align*} $$

Thus, if $0\leqslant v\leqslant 1$ , then $\mathrm {vol}(-K_A-vL_1)=3-2v$ because $L_1^2=0$ and $L_1\cdot L_4=0$ . Similarly, if $1\leqslant v\leqslant 2$ , then $\mathrm {vol}(-K_A-vL_1)=(v-2)^2$ . This gives

$$ \begin{align*}\frac{1}{4}\int_{0}^{1}\int_0^\infty \mathrm{vol}\big(-K_A-vL_1\big)dvdu= \frac{1}{4}\int_{0}^{1}\int_0^1(3-2v)dvdu+\frac{1}{4}\int_{0}^{1}\int_1^2(v-2)^2dvdu=\frac{7}{12} \end{align*} $$

and

$$ \begin{align*} &\frac{1}{4}\int_{1}^{2}\int_0^\infty \mathrm{vol}\big((2-u)(-K_A)-vL_1\big)dvdu= \frac{1}{4}\int_{1}^{2}\int_0^\infty(2-u)^3\mathrm{vol}\big((-K_A)-vL_1\big)dvdu=\\ &\quad=\frac{1}{4}\int_{1}^{2}\int_0^1(2-u)^3(3-2v)dvdu+\frac{1}{4}\int_{1}^{2}\int_1^2(2-u)^3(v-2)^2dvdu=\frac{7}{48}. \end{align*} $$

Combining, we get $S(W_{\bullet ,\bullet }^A;L_1)=\frac {35}{48}<\frac {3}{4}$ . This is a contradiction.

3 Family .

Let R be a smooth surface of degree $(2,4)$ in $\mathbb {P}^1\times \mathbb {P}^2$ ; let $\pi \colon X\to \mathbb {P}^1\times \mathbb {P}^2$ be a double cover ramified over the surface R. Then X is a smooth Fano threefold in the family . Moreover, all smooth Fano threefolds in this family can be obtained this way.

Let $\mathrm {pr}_1\colon \mathbb {P}^1\times \mathbb {P}^2\to \mathbb {P}^1$ and $\mathrm {pr}_2\colon \mathbb {P}^1\times \mathbb {P}^2\to \mathbb {P}^2$ be the projections to the first and the second factors, respectively. Set $p_1=\mathrm {pr}_1\circ \pi $ and $p_2=\mathrm {pr}_2\circ \pi $ . We have the following commutative diagram:

where $p_1$ is a fibration into del Pezzo surfaces of degree $2$ , and $p_2$ is a conic bundle.

Lemma 3.1. Let S be a fiber of the morphism $p_1$ . Then S is irreducible and normal.

Proof. Since $p_1\colon X\to \mathbb {P}^1$ is a Mori fiber space, any fiber of $p_1$ is irreducible and reduced. Moreover, any fiber of $\mathrm {pr}_1|_R$ is reduced by local computations. Thus, the assertion follows.

Lemma 3.2. Let S be a fiber of the morphism $p_1$ , let C be a fiber of the morphism $p_2$ and let P be a point in $S\cap C$ . Then S or C is smooth at P.

Proof. Local computations.

Now, we are ready to prove that X is K-stable. Recall from [10] that $\mathrm {Aut}(X)$ is finite. Thus, the threefold X is K-stable if and only if it is K-polystable [22].

Let $\tau $ be the Galois involution of the double cover $\pi \colon X\to \mathbb {P}^1\times \mathbb {P}^2$ , and let $G=\langle \tau \rangle $ . Suppose that X is not K-polystable. Then it follows from [24, Corollary 4.14] that there exists a G-invariant prime divisor $\mathbf {F}$ over X such that

$$ \begin{align*}\beta(\mathbf{F})=A_X(\mathbf{F})-S_X(\mathbf{F})\leqslant 0. \end{align*} $$

Let Z be the center of this divisor on X. Then Z is not a surface by [3, Theorem 3.17]. Hence, we see that either Z is a G-invariant irreducible curve, or Z is a G-fixed point. Let us seek for a contradiction.

Let P be a general point in Z, and let S be the fiber of $p_1$ that passes through P.

Lemma 3.3. The surface S is singular at P.

Proof. Suppose that S is smooth at P. Let B be a general curve in $|-K_S|$ that contains P. Then B is a smooth curve in S. Applying [3, Theorem 1.112], we get

$$ \begin{align*}1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant \min\left\{\frac{1}{S_X(S)},\frac{1}{S(W_{\bullet,\bullet}^S;B)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{S,B};P)}\right\}. \end{align*} $$

Since $S_X(S)<1$ by [3, Theorem 3.17], we see that $S(W_{\bullet ,\bullet }^S;B)\geqslant 1$ or $S(W_{\bullet , \bullet ,\bullet }^{S,B};P)\geqslant 1$ . We refer the reader to [3, § 1.7] for definitions of $S_X(S)$ , $S(W_{\bullet ,\bullet }^S;B)$ , $S(W_{\bullet , \bullet ,\bullet }^{S,B};P)$ .

Note that [3, Theorem 1.112] requires S to have Du Val singularities, but S may have non-Du Val singularities. Nevertheless, we still can apply [3, Theorem 1.112] here since the proof of [3, Theorem 1.112] remains valid in our case because S is smooth along B.

Let us compute $S(W_{\bullet ,\bullet }^S;B)$ and $S(W_{\bullet , \bullet ,\bullet }^{S,B};P)$ . Take $u\in \mathbb {R}_{\geqslant 0}$ and $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$\begin{align*}-K_X-uS\text{ is nef }\iff -K_X-uS\text{ is pseudoeffective }\iff u\leqslant 1, \end{align*}$$

Similarly, if $u\in [0,1]$ , then $(-K_X-uS)\vert _{S}-vB\sim _{\mathbb {R}}(1-v)(-K_S)$ , so

$$\begin{align*}(-K_X-uS)\vert_{S}-vB\text{ is nef }\iff (-K_X-uS)\vert_{S}-vB\text{ is pseudoeffective }\iff v\leqslant 1. \end{align*}$$

Now, applying [3, Corollary 1.109], we get

$$ \begin{align*}S(W_{\bullet,\bullet}^S;B)=\frac{3}{6}\int_0^1\int_0^1\big((1-v)(-K_S)\big)^2dvdu=\frac{1}{2}\int_0^1\int_0^12(1-v)^2dv=\frac{1}{3}<1. \end{align*} $$

Similarly, using [3, Theorem 1.112], we get

$$ \begin{align*}S(W_{\bullet,\bullet,\bullet}^{S,B},P)=\frac{3}{6}\int_0^1\int_0^1\big((1-v)(-K_S)\cdot B\big)^2dvdu=\frac{2}{3}<1. \end{align*} $$

But we already know that $S(W_{\bullet ,\bullet }^S;B)\geqslant 1$ or $S(W_{\bullet , \bullet ,\bullet }^{S,B};P)\geqslant 1$ . This is a contradiction.

If Z is a curve, then S is smooth at P by Lemma 3.1 because P is a general point in Z. Hence, we conclude that $Z=P$ because S is singular at the point P by Lemma 3.3. Recall that Z is G-invariant. This implies that $\tau (P)\in R$ .

Let C be the fiber of $p_2$ that passes through P. Then C is smooth at P by Lemma 3.2 because S is singular at P. Since $\tau (P)\in R$ , we see that C is irreducible and smooth.

Let T be a sufficiently general surface in linear system $|p_2^*(\mathcal {O}_{\mathbb {P}^2}(1))|$ that contains C. Since C is smooth, it follows from Bertini’s theorem that the surface T is smooth.

As in the proof of Lemma 3.3, it follows from [3, Theorem 1.112] that

$$ \begin{align*}1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{1}{S_X(T)},\frac{1}{S(W_{\bullet,\bullet}^T;C)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{T,C};P)}\right\}. \end{align*} $$

Moreover, it follows from [3, Theorem 3.17] that $S_X(T)<1$ . Thus, we conclude that

$$ \begin{align*}\max\big\{S(W_{\bullet,\bullet}^T;C),S(W_{\bullet, \bullet,\bullet}^{T,C};P)\big\}\geqslant 1. \end{align*} $$

In fact, since P is the center of the divisor $\mathbf {F}$ on X, [3, Theorem 3.17] gives

(3.1) $$ \begin{align} \max\big\{S(W_{\bullet,\bullet}^T;C),S(W_{\bullet, \bullet,\bullet}^{T,C};P)\big\}>1. \end{align} $$

Now, let us compute $S(W_{\bullet ,\bullet }^T;C)$ and $S(W_{\bullet , \bullet ,\bullet }^{T,C};P)$ using the results obtained in [3, § 1.7].

Take $u\in \mathbb {R}_{\geqslant 0}$ and $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$\begin{align*}-K_X-uT\text{ is nef }\iff -K_X-uT\text{ is pseudoeffective }\iff u\leqslant 1, \end{align*}$$

Similarly, if $u\in [0,1]$ , then

$$\begin{align*}(-K_X-uT)\vert_{T}-vC\text{ is nef }\iff (-K_X-uT)\vert_{T}-vC\text{ is pseudoeffective }\iff v\leqslant 1-u \end{align*}$$

because $(-K_X-uT)\vert _{T}-vC\sim _{\mathbb {R}}S|_T+(1-u-v)C$ . So, using [3, Corollary 1.109], we get

$$ \begin{align*}S(W_{\bullet,\bullet}^T;C)=\frac{3}{6}\int_0^1\int_0^{1-u}\big(S|_T+(1-u-v)C\big)^2dvdu=\frac{1}{2}\int_0^1\int_0^{1-u}4(1-u-v)dvdu=\frac{1}{3}<1. \end{align*} $$

Hence, it follows from Equation (3.1) that $S(W_{\bullet ,\bullet ,\bullet }^{T,C},P)>1$ . Now, using [3, Theorem 1.112], we get

$$ \begin{align*}S(W_{\bullet,\bullet,\bullet}^{T,C},P)=\frac{3}{6}\int_0^1\int_0^{1-u}\Big(\big(S|_T+(1-u-v)C\big)\cdot C\Big)^2dvdu=\frac{3}{6}\int_0^1\int_0^{1-u}4\,dvdu=1, \end{align*} $$

which is a contradiction. This shows that X is K-stable.

Corollary 3.4. All smooth Fano threefolds in the family are K-stable.

4 Family .

Let $\mathscr {S}$ and $\mathscr {S}^\prime $ be smooth cubic surfaces in $\mathbb {P}^3$ such that their intersection is a smooth curve of genus $10$ . Set $\mathscr {C}=\mathscr {S}\cap \mathscr {S}^\prime $ , and let $\pi \colon X\to \mathbb {P}^3$ be the blow up of the curve $\mathscr {C}$ . Then X is a smooth Fano threefold in the family , and every smooth Fano threefold in this family can be obtained in this way. Moreover, there exists a commutative diagram

where is a map that is given by the pencil generated by the surfaces $\mathscr {S}$ and $\mathscr {S}^\prime $ , and $\phi $ is a fibration into cubic surfaces. Note that $-K_X^3=10$ and $\mathrm {Aut}(X)$ is finite [10].

Let $H=\pi ^*(\mathcal {O}_{\mathbb {P}^3}(1))$ , and let E be the $\pi $ -exceptional surface. Then $-K_X\sim 4H-E$ , the morphism $\phi $ is given by the linear system $|3H-E|$ , and $E\cong \mathscr {S}\times \mathbb {P}^1$ .

The goal of this section is to prove that X is K-stable. Suppose that X is not K-stable. Let us seek for a contradiction. First, using the valuative criterion for K-stability [15, 18], we see that there exists a prime divisor $\mathbf {F}$ over X such that

$$ \begin{align*}\beta(\mathbf{F})=A_X(\mathbf{F})-S_X(\mathbf{F})\leqslant 0. \end{align*} $$

Let Z be the center of the divisor $\mathbf {F}$ on X. Then Z is not a surface by [3, Theorem 3.17]. Therefore, either Z is an irreducible curve or Z is a point. Fix a point $P\in Z$ .

Let A be the surface in $|3H-E|$ that contains P. Fix $u\in \mathbb {R}_{\geqslant 0}$ . Let $\mathscr {P}(u)$ be the positive part of the Zariski decomposition of $-K_X-uA$ , and let $\mathscr {N}(u)$ be its negative part. Then

$$ \begin{align*}-K_X-uA\sim_{\mathbb{R}}(4-3u)H-(1-u)E\sim_{\mathbb{R}}\Big(\frac{4}{3}-u\Big)A+\frac{1}{3}E. \end{align*} $$

This implies that $-K_X-uA$ is pseudoeffective $\iff u\leqslant \frac {4}{3}$ . Moreover, we have

$$ \begin{align*}\mathscr{P}(u)=\left\{\begin{aligned} &(4-3u)H-(1-u)E\ \text{if }0\leqslant u\leqslant 1, \\ &(4-3u)H\ \text{if }1\leqslant v\leqslant \frac{4}{3}, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}\mathscr{N}(u)=\left\{\begin{aligned} &0\ \text{if }0\leqslant u\leqslant 1, \\ &(u-1)E\ \text{if }1\leqslant u\leqslant \frac{4}{3}. \\ \end{aligned} \right. \end{align*} $$

Integrating, we obtain $S_X(A)=\frac {67}{120}<1$ , which also follows from [3, Theorem 3.17].

Note that $\pi (A)$ is a normal cubic surface in $\mathbb {P}^3$ , and $\pi (A)$ is smooth along the curve $\mathscr {C}$ . In particular, we see that $A\cong \pi (A)$ , and A is smooth along the intersection $E\cap A$ .

Lemma 4.1. The surface A is singular at the point P.

Proof. Suppose that A is smooth at P. Let C be a general curve in $|-K_A|$ that passes through the point P. Then C is a smooth irreducible elliptic curve. Take $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}\mathscr{P}(u)\vert_{S}-vC\sim_{\mathbb{R}}\left\{\begin{aligned} &(1-v)C\ \text{if }0\leqslant u\leqslant 1, \\ &(4-3u-v)C\ \text{if }1\leqslant u\leqslant \frac{4}{3}. \end{aligned} \right. \end{align*} $$

Therefore, using [3, Corollary 1.109], we obtain

$$ \begin{align*}S\big(W_{\bullet,\bullet}^A;C\big)=\frac{3}{10}\int_0^1\int_0^13(1-v)^2dvdu+\frac{3}{10}\int_1^{\frac{4}{3}}\int_0^{4-3u}3(4-3u-v)^2dvdu=\frac{13}{40}. \end{align*} $$

Similarly, using [3, Theorem 1.112], we obtain

$$ \begin{align*} S\big(W_{\bullet,\bullet,\bullet}^{A,C};P\big)\leqslant \frac{3}{10}\int_0^1\int_0^1\big(3(1-v)\big)^2dvdu+\frac{3}{10}\int_1^{\frac{4}{3}}\int_0^{4-3u}\big(3(4-3u-v)\big)^2dvdu+\\ +\underbrace{\frac{6}{10}\int_1^{\frac{4}{3}}\int_0^{4-3u}3(4-3u-v)(u-1)dvdu}_{\text{if } P\in E}=\frac{39}{40}+\frac{1}{120}=\frac{59}{60}. \end{align*} $$

Therefore, it follows from [3, Theorem 1.112] that

$$ \begin{align*}1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{1}{S_X(A)},\frac{1}{S(W_{\bullet,\bullet}^A;C)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{A,C};P)}\right\} \geqslant\min\left\{\frac{120}{67},\frac{40}{13},\frac{60}{59}\right\}>1, \end{align*} $$

which is absurd.

Corollary 4.2. The point P is not contained in the surface E.

Since $A\cong \pi (A)$ , we may consider A as a cubic surface in $\mathbb {P}^3$ . Then

• • either $\mathrm {mult}_P(A)=2$ and A has Du Val singularities.

• • or $\mathrm {mult}_P(A)=3$ and A is a cone over a plane smooth cubic curve with vertex P.

Lemma 4.3. One has $\mathrm {mult}_P(A)\ne 3$ .

Proof. Let $\sigma \colon \widehat {X}\to X$ be a blow up of the point P, and let G be the $\sigma $ -exceptional surface. Denote by $\widehat {A}$ and $\widehat {E}$ the proper transforms on $\widehat {X}$ of the surfaces A and E, respectively. Suppose that $\mathrm {mult}_P(A)=3$ . Take $u\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}\sigma^*(-K_X)-vG\sim_{\mathbb{R}}\frac{1}{3}\widehat{E}+\frac{4}{3}\widehat{A}+(4-u)G. \end{align*} $$

Thus, the divisor $\sigma ^*(-K_X)-vG$ is pseudoeffective $\iff u\in [0,4]$ . Moreover, if $u\in [0,4]$ , then the Zariski decomposition of the divisor $\sigma ^*(-K_X)-vG$ can be described as follows:

$$ \begin{align*}P(u)=\left\{\begin{aligned} &\frac{1}{3}\widehat{E}+\frac{4}{3}\widehat{A}+(4-u)G\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{1}{3}\widehat{E}+\frac{5-u}{3}\widehat{A}+(4-u)G\ \text{if }1\leqslant u\leqslant 4, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}N(u)=\left\{\begin{aligned} &0\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{u-1}{3}\widehat{A}\ \text{if }1\leqslant u\leqslant 4, \\ \end{aligned} \right. \end{align*} $$

where $P(u)$ and $N(u)$ are the positive and the negative parts of the Zariski decomposition, respectively. Using this, we compute

$$ \begin{align*}S_{X}(G)=\frac{3}{10}\int_{0}^{1}u^{3},du+\frac{3}{10}\int_{1}^{4}udu=\frac{93}{40}<3=A_{X}(G). \end{align*} $$

As in the proof of Lemma 2.10, let us use results from [16, § 4] to get a contradiction. Namely, applying [16, Corollary 4.18], we get

$$ \begin{align*}1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{A_{X}(G)}{S_{X}(G)}, \inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\}= \min\left\{\frac{40}{31},\inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\}, \end{align*} $$

where $\delta _{Q}(G,V^G_{\bullet ,\bullet })$ is defined in [16, § 4]. So, there is $Q\in G$ such that $\delta _{Q}(G,V^G_{\bullet ,\bullet })<\frac {40}{31}$ .

Let $\ell $ is a general line in $G\cong \mathbb {P}^2$ that contains Q. Then [16, Corollary 4.18] gives

$$ \begin{align*}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big) \geqslant\min\left\{\frac{1}{S\big(V^F_{\bullet,\bullet};\ell\big)},\frac{1}{S\big(W^{G,\ell}_{\bullet,\bullet,\bullet};Q\big)}\right\}. \end{align*} $$

Let us compute $S(V^G_{\bullet ,\bullet };\ell )$ and $S(W^{G,\ell }_{\bullet ,\bullet ,\bullet };Q)$ . Take $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}P(u)\big\vert_{G}-v\ell\sim_{\mathbb{R}}\left\{\begin{aligned} &(u-v)\ell\ \text{if }0\leqslant u\leqslant 1, \\ &(1-v)\ell\ \text{if }1\leqslant u\leqslant 4. \end{aligned} \right. \end{align*} $$

Let $\widehat {\mathscr {C}}=\widehat {A}\vert _{G}$ . Then $\widehat {\mathscr {C}}$ is a smooth cubic curve in $G\cong \mathbb {P}^2$ . Let

$$ \begin{align*}N^\prime(u)=N(u)\big\vert_{G}=\left\{\begin{aligned} &0\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{u-1}{3}\widehat{\mathscr{C}}\ \text{if }1\leqslant u\leqslant 4. \end{aligned} \right. \end{align*} $$

Now, using [16, Theorem 4.8], we get

$$ \begin{align*} S\big(W^{G}_{\bullet,\bullet};\ell\big)&=\frac{3}{10}\int_0^4\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{G}-v\ell\big)dvdu=\\ &=\frac{3}{10}\int_{0}^{1}\int_{0}^{u}(u-v)^{2}dvdu+\frac{3}{10}\int_{1}^{4}\int_{0}^{1}(1-v)^{2}dvdu=\frac{13}{40}. \end{align*} $$

Similarly, it follows from [16, Theorem 4.17] that $S(W_{\bullet ,\bullet ,\bullet }^{G,\ell };Q)$ can be computes as follows:

$$ \begin{align*} &\frac{3}{10}\int_{0}^{1}\int_{0}^{u}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+\frac{3}{10}\int_{1}^{4}\int_{0}^{1}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+F_Q=\\ &\quad=\frac{3}{12}\int_{0}^{1}\int_{0}^{u}(u-v)^{2}dvdu+\frac{3}{10}\int_{1}^{4}\int_{0}^{1}(1-v)^{2}dvdu+F_Q=\frac{13}{40}+F_Q, \end{align*} $$

where $F_Q=0$ if $Q\not \in \widehat {A}\vert _{G}$ , and

$$ \begin{align*}F_Q=\frac{6}{10} \int_{1}^{4} \int_{0}^{1}\frac{(1-v)(u-1)}{3}dvdu=\frac{9}{20} \end{align*} $$

otherwise. This gives $S(W_{\bullet ,\bullet ,\bullet }^{G,\ell };Q)\leqslant \frac {31}{40}$ . Combining the estimates, we get $\delta _{Q}(G,V^G_{\bullet ,\bullet })\geqslant \frac {40}{31}$ , which is a contradiction. This completes the proof of the lemma.

Hence, we see that the surface A has Du Val singularities. Let S be a general surface in the linear system $|H|$ that contains P. Then S is smooth, and $-K_X-uS\sim _{\mathbb {R}} (4-u)H-E$ . Hence, the divisor $-K_X-uS$ is pseudoeffective $\iff $ it is nef $\iff u\leqslant 1$ . Then

$$ \begin{align*}S_X(S)=\frac{3}{10}\int_{0}^{1}(-K_X-uS)^3du=\frac{3}{10}\int_{0}^{1}u(1-u)(7-u)du=\frac{13}{40}<1. \end{align*} $$

Let $C=A\vert _{S}$ . Then C is a reduced curve in $|-K_S|$ that is singular at P, and $C\cong \pi (C)$ . Moreover, the curve $\pi (C)$ is a general hyperplane section of the cubic surface $\pi (A)\subset \mathbb {P}^3$ that passes through the point $\pi (P)$ . Therefore, since $\pi (A)$ is not a cone by Lemma 4.3, we conclude that the curve C is irreducible. Hence, one of the following two cases holds:

1. (1) the curve C has an ordinary node at P,

2. (2) the curve C has an ordinary cusp at P.

Let $\Pi =\pi (S)$ . Then $\Pi $ is a plane in $\mathbb {P}^3$ such that $\pi (P)\in \Pi $ and $\Pi \cap \pi (A)=\pi (C)$ , and the morphism $\pi \vert _{S}\colon S\to \Pi $ is a composition of blow ups of $9$ intersection points $\Pi \cap \mathscr {C}$ , which we denote by $O_1,\ldots ,O_9$ . Note that $\pi (C)$ is a reduced plane cubic curve that passes through these nine points, and $\pi (C)$ is smooth away from $\pi (P)$ .

Lemma 4.4. The curve C cannot have an ordinary double point at the point P.

Proof. For each $i\in \{1,\ldots ,9\}$ , let $L_i$ be the proper transform on S of the line in $\Pi $ that passes through P and $O_i$ . Then $L_i\ne L_j$ for $i\ne j$ since $\pi (C)$ is irreducible. We have

$$ \begin{align*}(-K_X-uS)\big\vert_{S}\sim_{\mathbb{R}}\frac{1-u}{6}\sum_{i=1}^9L_i. \end{align*} $$

Let $\sigma \colon \widehat {S}\to S$ be the blow up of S at the point P, let $\mathbf {f}$ be the $\sigma $ -exceptional curve and let $\widehat {C},\widehat {L}_1,\ldots ,\widehat {L}_{9}$ be the proper transforms on $\widehat {S}$ of the curves $C,L_1,\ldots ,L_9$ , respectively. Then $\widehat {L}_1,\ldots ,\widehat {L}_{9}$ are disjoint. On the surface $\widehat {S}$ , we have

$$ \begin{align*}\widehat{C}^2=-4, \mathbf{f}^2=\widehat{L}_1^2=\cdots=\widehat{L}_9^2=-1, \widehat{L}_1\cdot\mathbf{f}=\cdots=\widehat{L}_9\cdot\mathbf{f}=1, \widehat{C}\cdot\mathbf{f}=2, \end{align*} $$

and $\widehat {C}\cdot \widehat {L}_1=\cdots =\widehat {C}\cdot \widehat {L}_9=0$ .

Fix $u\in [0,1]$ . Let v be a nonnegative real number. Then

(4.1) $$ \begin{align} \sigma^*\big((-K_X-uS)\big\vert_S\big)-v\mathbf{f}\sim_{\mathbb{R}}\frac{5+u}{6}\widehat{C}+\frac{1-u}{6}\sum_i^9\widehat{L}_i+\frac{19-7u-6v}{6}\mathbf{f}. \end{align} $$

So, the divisor $\sigma ^*((-K_X-uS)|_S)-v\mathbf {f}$ is pseudoeffective $\iff $ it is nef $\iff v\leqslant \frac {19-7u}{6}$ . Let $P(u,v)$ and $N(u,v)$ be the positive and the negative parts of its Zariski decomposition. Then, using Equation (4.1), we compute

$$ \begin{align*}P(u,v)=\left\{\begin{aligned} &\frac{5+u}{6}\widehat{C}+\frac{1-u}{6}\sum_i^9\widehat{L}_i+\frac{19-7u-6v}{6}\mathbf{f}\ \text{if }0\leqslant v\leqslant \frac{3-3u}{2}, \\ &\frac{19-7u-6v}{12}\widehat{C}+\frac{1-u}{6}\sum_i^9\widehat{L}_i+\frac{19-7u-6v}{6}\mathbf{f}\ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 3-u, \\ &\frac{19-7u-6v}{12}\big(\widehat{C}+2\sum_i^9\widehat{L}_i+2\mathbf{f}\big)\ \text{if }3-u\leqslant v\leqslant \frac{19-7u}{6}, \end{aligned} \right. \end{align*} $$

$$ \begin{align*}N(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant \frac{3-3u}{2}, \\ &\frac{-3+3u+2v}{4}\widehat{C}\ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 3-u, \\ &\frac{2v+3u-3}{4}\widehat{C}+(v+u-3)\sum_i^9\widehat{L}_i\ \text{if }3-u\leqslant v\leqslant \frac{19-7u}{6}, \end{aligned} \right. \end{align*} $$

$$ \begin{align*}\mathrm{vol}\Big(\sigma^*\big((-K_X-uS)\big|_S\big)-v\mathbf{f}\Big)=\left\{\begin{aligned} &(1-u)(7-u)-v^2\ \text{if }0\leqslant v\leqslant \frac{3-3u}{2}, \\ &\frac{(1-u)(37-13u-12v)}{4}\ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 3-u, \\ &\frac{(19-7u-6v)^2}{4}\ \text{if }3-u\leqslant v\leqslant \frac{19-7u}{6}, \end{aligned} \right. \end{align*} $$

$$ \begin{align*}P(u,v)\cdot\mathbf{f}=\left\{\begin{aligned} &v \ \text{if }0\leqslant v\leqslant \frac{3-3u}{2}, \\ &\frac{3(1-u)}{2} \ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 3-u, \\ &\frac{3(19-7u-6v)}{2}\ \text{if }3-u\leqslant v\leqslant \frac{19-7u}{6}. \end{aligned} \right. \end{align*} $$

Indeed, the divisor $P(u,v)$ above is effective, and it has a nonnegative intersection with every irreducible component of $\operatorname {Supp}\left (P(u,v)\right )$ , which implies that it is nef in every case. Now, using [3, Corollary 1.109], we get

$$ \begin{align*} S\big(W_{\bullet,\bullet}^{\widehat{S}};\mathbf{f}\big)&= \frac{3}{10}\int_0^1\int_{0}^{\frac{19-7u}{6}}\mathrm{vol}\Big(\sigma^*\big((-K_X-uS)\big|_S\big)-v\mathbf{f}\Big)dudv=\\ &=\frac{3}{10}\int_0^1\int_0^{\frac{3-3u}{2}}\big((1-u)(7-u)-v^2\big)dv+\frac{3}{10}\int_0^1\int_{\frac{3-3u}{2}}^{3-u}\frac{(1-u)(37-13u-12v)}{4}dv+\\ &\quad+\frac{3}{10}\int_0^1\int_{3-u}^{\frac{19-7u}{6}}\frac{(19-7u-6v)^2}{4}dvdu=\frac{767}{480}<2=A_S(\mathbf{f}). \end{align*} $$

Moreover, if Q is a point in $\mathbf {f}$ , then [3, Remark 1.113] gives

$$ \begin{align*} &S\big(W_{\bullet,\bullet,\bullet}^{\widehat{S},\mathbf{f}};Q\big)= F_{Q}+\frac{3}{10}\int_{0}^{1}\int_{0}^{\frac{19-7u}{6}}\big(P(u,v)\cdot\mathbf{f}\big)^2dvdu=F_Q+\frac{3}{10}\int_0^1\int_0^{\frac{3-3u}{2}}v^2dv+\\ &\quad+\frac{3}{10}\int_0^1\int_{\frac{3-3u}{2}}^{3-u}\Bigg(\frac{3(1-u)}{2}\Bigg)^2dv+\frac{3}{10}\int_0^1\int_{3-u}^{\frac{19-7u}{6}}\Bigg(\frac{3(19-7u-6v)}{2}\Bigg)^2dvdu=F_Q+\frac{147}{320}, \end{align*} $$

where

$$ \begin{align*}F_Q=\frac{6}{10}\int_{0}^{1}\int_{0}^{\frac{19-7u}{6}}\big(P(u,v)\cdot\mathbf{f}\big)\mathrm{ord}_Q\Big(N(u,v)\big\vert_{\mathbf{f}}\Big)dvdu, \end{align*} $$

which implies the following assertions:

• • if $Q\not \in \widehat {C}\cup \widehat {L}_1\cup \cdots \cup \widehat {L}_{9}$ , then $F_Q=0$ ;

• • if $Q\in \widehat {L}_1\cup \cdots \cup \widehat {L}_{9}$ , then

  $$ \begin{align*}F_Q=\frac{6}{10}\int_0^1\int_{3-u}^{\frac{19-7u}{6}}\frac{3(19-7u-6v)(v+u-3)}{2}dvdu=\frac{1}{960}; \end{align*} $$

• • if $Q\in \widehat {C}$ and $\widehat {C}$ intersects $\mathbf {f}$ transversally at P, then

  $$ \begin{align*} F_Q=&\frac{6}{10}\int_0^1\int_{\frac{3-3u}{2}}^{3-u}\frac{3(1-u)(2v+3u-3)}{8}dvdu+\\ &\frac{6}{10}\int_0^1\int_{3-u}^{\frac{19-7u}{6}}\frac{3(19-7u-6v)(2v+3u-3)}{8}dvdu=\frac{643}{1920};\quad \quad \quad \quad \quad \end{align*} $$

• • if $Q\in \widehat {C}$ and $\widehat {C}$ is tangent to $\mathbf {f}$ at the point P, then $F_Q=\frac {643}{960}$ .

Thus, if C has a node at P, then $S(W_{\bullet ,\bullet ,\bullet }^{\widehat {S},\mathbf {f}};Q)\leqslant \frac {305}{384}$ , so [3, Remark 1.113] gives

$$ \begin{align*}1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\inf_{Q\in\mathbf{f}}\frac{1}{S(W_{\bullet,\bullet,\bullet}^{\widehat{S},\mathbf{f}};Q)}, \frac{2}{S(V_{\bullet,\bullet}^{S};\mathbf{f})},\frac{1}{S_X(S)}\right\} \geqslant\min\left\{\frac{384}{305},\frac{960}{767},\frac{40}{13}\right\}=\frac{960}{767}>1, \end{align*} $$

which is a contradiction.

Lemma 4.5. The curve C cannot have an ordinary cusp at the point P.

Proof. Suppose C has a cusp. Let L be an irreducible curve in S such that $\pi (L)$ is a line and $\pi (L)\cap \pi (C)=\pi (P)$ . Then $(-K_X-uS)\vert _{S}\sim _{\mathbb {R}}(1-u)L+C$ .

Now, we consider the following commutative diagram:

where $\sigma _1$ is the blow up of P, $\sigma _2$ is the blow up of the point in the $\sigma _1$ -exceptional curve contained in the proper transform of C, $\sigma _3$ is the blow up of the point in the $\sigma _2$ -exceptional curve contained in the proper transform of C, $\upsilon $ is the birational contraction of the proper transforms of $\sigma _1\circ \sigma _2$ -exceptional curves and $\sigma $ is the birational contraction of the proper transform of the $\sigma _3$ -exceptional curve. Then $\widehat {S}$ has two singular points:

1. (1) a cyclic quotient singularity of type $\frac {1}{2}(1,1)$ , which we denote by $Q_2$ ;

2. (2) a cyclic quotient singularity of type $\frac {1}{3}(1,1)$ , which we denote by $Q_3$ .

Let $\mathbf {f}$ be the $\sigma $ -exceptional curve, let $\widehat {C}$ be the proper transform on $\widehat {S}$ of the curve C and let $\widehat {L}$ be the proper transform of the curve L. Then the curves $\mathbf {f}$ , $\widehat {C}$ , $\widehat {L}$ are smooth. Moreover, it is not very difficult to check that $Q_2\in \mathbf {f}\ni Q_3$ , $Q_2\not \in \widehat {C}\not \ni Q_3$ , $Q_2\in \widehat {L}\not \ni Q_3$ . Further, we have $A_S(\mathbf {f})=5$ , $\sigma ^*(C)\sim \widehat {C}+6\mathbf {f}$ , $\sigma ^*(L)\sim \widehat {L}+3\mathbf {f}$ . On the surface $\widehat {S}$ , we have

$$ \begin{align*}\widehat{L}^2=-\frac{1}{2}, \widehat{L}\cdot\widehat{C}=0, \widehat{C}\cdot \mathbf{f}=\frac{1}{2}, \widehat{C}^2=-6, \widehat{C}\cdot \mathbf{f}=1, \mathbf{f}^2=-\frac{1}{6}. \end{align*} $$

Note that $Q_2=\mathbf {f}\cap \widehat {L}$ , and $\widehat {C}$ intersects $\mathbf {f}$ transversally by one point.

Fix $u\in [0,1]$ . Let v be a nonnegative real number. Then

(4.2) $$ \begin{align} \sigma^*\big((-K_X-uS)\big\vert_{S}\big)-v\mathbf{f}\sim_{\mathbb{R}}(1-u)\widehat{L}+\widehat{C}+(9-3u-v)\mathbf{f}. \end{align} $$

Thus, the divisor $\sigma ^*((-K_X-uS)|_S)-v\mathbf {f}$ is pseudoeffective $\iff $ it is nef $\iff v\leqslant 9-3u$ . Let $P(u,v)$ be the positive part of the Zariski decomposition of $\sigma ^*((-K_X-uS)|_S)-v\mathbf {f}$ , and let $N(u,v)$ be the negative part. Then, using Equation (4.2), we compute

$$ \begin{align*}P(u,v)=\left\{\begin{aligned} &(1-u)\widehat{L}+\widehat{C}+(9-3u-v)\mathbf{f}\ \text{if }0\leqslant v\leqslant 3-3u, \\ &(1-u)\widehat{L}+\frac{9-3u-v}{6}\widehat{C}+(9-3u-v)\mathbf{f}\ \text{if }3-3u\leqslant v\leqslant 8-2u, \\ &\frac{9-3u-v}{6}\big(6\widehat{L}+\widehat{C}+6\mathbf{f}\big)\ \text{if }8-2u\leqslant v\leqslant 9-3u, \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}N(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant 3-3u, \\ &\frac{v+3u-3}{6}\widehat{C}\ \text{if }3-3u\leqslant v\leqslant 8-2u, \\ &\frac{v+3u-3}{6}\widehat{C}+(v+2u-8)\widehat{L}\ \text{if }8-2u\leqslant v\leqslant 9-3u. \end{aligned} \right. \end{align*} $$

This gives

$$ \begin{align*}\mathrm{vol}\Big(\sigma^*\big((-K_X-uS)\big|_S\big)-v\mathbf{f}\Big)=\left\{\begin{aligned} &(1-u)(7-u)-\frac{v^2}{6}\ \text{if }0\leqslant v\leqslant 3-3u, \\ &\frac{(1-u)(17-5u-2v)}{2}\ \text{if }3-3u\leqslant v\leqslant 8-2u, \\ &\frac{(9-3u-v)^2}{2}\ \text{if }8-2u\leqslant v\leqslant 9-3u, \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}P(u,v)\cdot\mathbf{f}=\left\{\begin{aligned} &\frac{v}{6}\ \text{if }0\leqslant v\leqslant 3-3u, \\ &\frac{1-u}{2}\ \text{if }3-3u\leqslant v\leqslant 8-2u, \\ &\frac{9-3u-v}{2}\ \text{if }8-2u\leqslant v\leqslant 9-3u. \end{aligned} \right. \end{align*} $$

Now, we use [3, Corollary 1.109] to get

$$ \begin{align*} S\big(W_{\bullet,\bullet}^{\widehat{S}};\mathbf{f}\big)&=\frac{3}{10}\int_0^1\int_0^{3-3u}\Big((1-u)(7-u)-\frac{v^2}{6}\Big)dv+\\ &\quad+\frac{3}{10}\int_0^1\int_{3-3u}^{8-2u}\frac{(1-u)(17-5u-2v)}{2}dv+\frac{3}{10}\int_0^1\int_{8-2u}^{9-3u}\frac{(9-3u-v)^2}{2}dvdu=\frac{173}{40}. \end{align*} $$

Similarly, if Q is a point in $\mathbf {f}$ , then [3, Remark 1.113] gives

$$ \begin{align*} S\big(W_{\bullet,\bullet,\bullet}^{\widehat{S},\mathbf{f}};Q\big)&=F_{Q}+\frac{3}{10}\int_{0}^{1}\int_{0}^{9-3u}\big(P(u,v)\cdot\mathbf{f}\big)^2dvdu=\\ &=F_Q+\frac{3}{10}\int_0^1\int_0^{3-3u}\Big(\frac{v}{6}\Big)^2dvdu+\frac{3}{10}\int_0^1\int_{3-3u}^{8-2u}\Big(\frac{1-u}{2}\Big)^2dvdu+\\ &\quad+\frac{3}{10}\int_0^1\int_{8-2u}^{9-3u}\Big(\frac{9-3u-v}{2}\Big)^2dvdu=F_Q+\frac{5}{32}, \end{align*} $$

where $F_Q$ can be computes as follows:

• • if $Q\ne \widehat {C}\cap \mathbf {f}$ and $Q\ne \widehat {L}\cap \mathbf {f}$ , then $F_Q=0$ ;

• • if $Q=\widehat {L}\cap \mathbf {f}$ , then

  $$ \begin{align*}F_Q=\frac{6}{10}\int_0^1\int_{8-2u}^{9-3u}\frac{(9-3u-v)(v+2u-8)}{2}dvdu=\frac{1}{80}; \end{align*} $$

• • if $Q=\widehat {C}\cap \mathbf {f}$ , then

  $$ \begin{align*} F_Q=&\frac{6}{10}\int_0^1\int_{3-3u}^{8-2u}\frac{(1-u)(v+3u-3)}{12}dv+\\ &+\frac{6}{10}\int_0^1\int_{8-2u}^{9-3u}\frac{(9-3u-v)(v+3u-3)}{12}dudv=\frac{193}{480}. \end{align*} $$

Therefore, we conclude that

$$ \begin{align*}S\big(W_{\bullet,\bullet,\bullet}^{\widehat{S},\mathbf{f}};Q\big)= \left\{\begin{aligned} &\frac{5}{32}\ \text{if }Q\not\in\widehat{C}\cup\widehat{L}, \\[5pt] &\frac{27}{80} \ \text{if }Q=\widehat{L}\cap\mathbf{f}, \\[5pt] &\frac{67}{120}\ \text{if }Q=\widehat{C}\cap\mathbf{f}. \end{aligned} \right. \end{align*} $$

Let $\Delta _{\mathbf {f}}=\frac {1}{2}Q_2+\frac {2}{3}Q_3$ . Then, using [3, Remark 1.113] and our computations, we get

$$ \begin{align*}1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\inf_{Q\in\mathbf{f}}\frac{1-\mathrm{ord}_Q\big(\Delta_{\mathbf{f}}\big)}{S(W_{\bullet,\bullet,\bullet}^{\widehat{S},\mathbf{f}};Q)}, \frac{5}{S(V_{\bullet,\bullet}^{S};\mathbf{f})},\frac{1}{S_X(S)}\right\} \geqslant \min \left\{\frac{40}{13}, \frac{200}{173}, \frac{120}{67}\right\}=\frac{200}{173}>1, \end{align*} $$

which is a contradiction.

Combining Lemmas 4.4 and 4.5, we obtain a contradiction.

Corollary 4.6. All smooth Fano threefolds in the family are K-stable.

5 Family (Verra threefolds)

Smooth Fano threefolds in the family can be described as follows:

1. (a) smooth divisors of degree $(2,2)$ in $\mathbb {P}^2\times \mathbb {P}^2$ , which are known as Verra threefolds,

2. (b) double covers of the (unique) smooth divisor in $\mathbb {P}^2\times \mathbb {P}^2$ of degree $(1,1)$ branched over smooth anticanonical K3 surfaces.

Note that every double cover of a smooth divisor in $\mathbb {P}^2\times \mathbb {P}^2$ of degree $(1,1)$ branched over a smooth anticanonical surface is K-stable [14, Example 4.4]. In fact, this also implies that general Verra threefold is K-stable [3, Example 3.14].

The goal of this section is to prove that all smooth Verra threefolds are K-stable.

Let X be a smooth divisor of degree $(2,2)$ in $\mathbb {P}^2\times \mathbb {P}^2$ , let $\pi _1\colon X\to \mathbb {P}^2$ be the projection to the first factor of $\mathbb {P}^2\times \mathbb {P}^2$ , and let $\pi _2\colon X\to \mathbb {P}^2$ be the projection to the second factor. Then $\pi _1$ and $\pi _2$ are conic bundles [21]. Set $H_1=\pi _1^*(\mathcal {O}_{\mathbb {P}^2}(1))$ and $H_2=\pi _2^*(\mathcal {O}_{\mathbb {P}^2}(1))$ . Then

$$ \begin{align*}-K_X\sim H_1+H_2, \end{align*} $$

and the group $\mathrm {Pic}(X)$ is generated by $H_1$ and $H_2$ . Note that $\mathrm {Aut}(X)$ is finite [10]. Thus, the threefold X is K-stable $\iff $ it is K-polystable. See [22] for details.

Lemma 5.1. Fix a point $P\in X$ . Let $C_1$ be the fiber of the conic bundle $\pi _1$ that contains P, and let $C_2$ be the fiber of $\pi _2$ that contains P. Then $C_1$ or $C_2$ is smooth at P.

Proof. Local computations.

Let $\Delta _1$ and $\Delta _2$ be the discriminant curves of the conic bundles $\pi _1$ and $\pi _2$ , respectively. Then $\Delta _1$ and $\Delta _2$ are reduced curves of degree $6$ that have at most ordinary double points as singularities. For basic properties of the discriminant curves $\Delta _1$ and $\Delta _2$ , see [21, § 3.8]. In particular, we know that no line in $\mathbb {P}^2$ can be an irreducible component of these curves.

Lemma 5.2. Fix a point $P\in X$ . Let $C_2$ be the fiber of the conic bundle $\pi _2$ that contains P, let S be a general surface in $|H_2|$ that contains $C_2$ . Then one of the following cases holds:

1. (1) $C_2$ is smooth, S is smooth, the divisor $-K_S$ is ample;

2. (2) $C_2$ is smooth, S is smooth, the divisor $-K_S$ is nef, the surface S has exactly four irreducible curves that have trivial intersection with the divisor $-K_S$ , these curves are disjoint and none of them passes through P, and $C_2\subset \mathrm {Sing}(\pi _1^{-1}(\Delta _1))$ ;

3. (3) $C_2$ is singular and reduced, S is smooth, the divisor $-K_S$ is ample;

4. (4) $C_2$ is not reduced, $P\not \in \mathrm {Sing}(S)\subset \mathrm {Supp}(C_2)$ , and $\mathrm {Sing}(S)$ consists of two points, which are ordinary double points of the surface S, and $-K_S$ is ample.

Proof. The assertion about the singularities of the surface S are local and well known.

We have $-K_S\sim H_1\vert _S$ and $K_S^2=2$ , so S is a weak del Pezzo surface of degree $2$ , and the restriction $\pi _1\vert _{S}\colon S\to \mathbb {P}^2$ is the anticanonical morphism. Let $\ell $ be an irreducible curve in the surface S such that

$$ \begin{align*}-K_S\cdot \ell=0. \end{align*} $$

Then $\ell $ is an irreducible component of the fiber $\pi _1^{-1}(\pi _1(\ell ))$ . But $\pi _2(\ell )$ is the line $\pi _2(S)$ , which implies that the scheme fiber $\pi _1^{-1}(\pi _1(\ell ))$ is singular, which implies that $\pi _1(\ell )\in \Delta _1$ . Since $\ell \cap C_2\ne \varnothing $ and $\pi _2(S)$ is a general line in $\mathbb {P}^2$ that passes through the point $\pi _2(P)$ , we see that $\pi _1(C_2)\subset \Delta _1$ . This implies that $C_2$ is irreducible and reduced.

Let $R=\pi _1^{-1}(\pi _1(C_2))$ . Then the surface R is singular along a curve that is isomorphic to the conic $\pi _1(C_2)\cong C_2$ . Let $f\colon \widetilde {R}\to R$ be the blow up of this curve. Then $\widetilde {R}$ is smooth, and the composition morphism $\pi _1\circ f$ induces a $\mathbb {P}^1$ -bundle $\widetilde {R}\to \widetilde {C}_2$ , where $\widetilde {C}_2$ is double cover of the conic $\pi _1(C_2)$ that is branched over the eight points $\pi _1(C_2)\cap (\Delta _1-\pi _1(C_2))$ . In particular, we see that $\widetilde {C}_2$ is an irrational curve, which implies that $C_2=\mathrm {Sing}(R)$ .

Vice versa, if the fiber $C_2$ is a smooth conic, and the conic $\pi _1(C_2)$ is an irreducible component of the discriminant curve $\Delta _1$ , then it follows from the Bertini theorem that

$$ \begin{align*}S\cdot\pi_1^{-1}(\pi_1(C_2))=2C_2+\ell_1+\ell_2+\cdots+\ell_k, \end{align*} $$

where $\ell _1,\ell _2,\ldots ,\ell _k$ are k distinct irreducible reduced curves in X that are irreducible components of the fibers of the natural projection $\pi _1^{-1}(\pi _1(C_2))\to \pi _1(C_2)$ . Since

$$ \begin{align*}4=2H_2^2\cdot H_1=H_2\cdot S\cdot\pi_1^{-1}(\pi_1(C_2))=H_2\cdot\Big(2C_2+\sum_{i=1}^k\ell_i\Big)=k, \end{align*} $$

we see that S contains exactly $4$ irreducible curves that intersects trivially with $-K_S$ . Now, the generality in the choice of S implies that none of these curves contains P.

Example 5.3. Actually, the case (2) in Lemma 5.2 can happen. Indeed, in the assumption and notations of Lemma 5.2, let $P=([0:0:1],[0:0:1])$ , and suppose that X is given by

$$ \begin{align*}(u^2+2uw+v^2+2w^2)x^2+(uv-w^2)xy+(uw-2uv+3v^2+w^2)y^2+(uw+v^2)z^2=0, \end{align*} $$

where $([u:v:w],[x:y:z])$ are coordinates on $\mathbb {P}^2\times \mathbb {P}^2$ . Then the threefold X is smooth. For instance, this can be checked using the following Magma script:

Observe that the fiber $C_1$ is a singular reduced curve given by $u=v=2x^2-xy+y^2=0$ , the fiber $C_2$ is a smooth curve that is given by $x=y=uw+v^2=0$ , and the discriminant curve $\Delta _1$ is a union of the conic $\pi _1(C_2)$ and the irreducible quartic plane curve given by

$$ \begin{align*} 8u^3v-4u^3w-11u^2v^2+16u^2vw-12u^2w^2+8uv^3-\\ -28uv^2w+14uvw^2-16uw^3-12v^4-28v^2w^2-7w^4=0. \end{align*} $$

As in case (2) in Lemma 5.2, we have $C_2=\mathrm {Sing}(\pi _1^{-1}(\pi _1(C_2)))$ .

Let us prove that X is K-stable. Suppose it is not. Using the valuative criterion [15, 18], we see that there exists a prime divisor $\mathbf {F}$ over X such that

$$ \begin{align*}\beta(\mathbf{F})=A_X(\mathbf{F})-S_X(\mathbf{F})\leqslant 0. \end{align*} $$

Let Z be the center of the divisor $\mathbf {F}$ on X. Then Z is not a surface by [3, Theorem 3.17].

Let P be any point in Z, let $C_1$ be the fiber of the conic bundle $\pi _1$ that contains P and let $C_2$ be the fiber of the conic bundle $\pi _2$ that contains P. By Lemma 5.1, at least one curve among $C_1$ or $C_2$ is smooth at P. We may assume that $C_2$ is smooth at P.

Let S be a general surface in $|H_2|$ that contains $C_2$ . Then S is smooth by Lemma 5.2. Moreover, one of the following three cases holds:

1. (1) $C_2$ is smooth, the divisor $-K_S$ is ample;

2. (2) $C_2$ is smooth, $\pi _1(C_2)\subset \Delta _1$ , the divisor $-K_S$ is nef, the surface S has exactly four irreducible curves that have trivial intersection with the divisor $-K_S$ , these curves are disjoint and none of them passes through P;

3. (3) $C_2$ is singular and reduced, the divisor $-K_S$ is ample.

Let C be the curve in X that is defined as follows:

• • if $C_2$ is smooth and irreducible, we let $C=C_2$ .

• • if $C_2$ is reducible, we let C be its irreducible component that contains P.

Note that $-K_S\sim H_1\vert _{S}$ and $K_S^2=2$ . Let $\eta \colon S\to \mathbb {P}^2$ be the restriction morphism $\pi _1\vert _{S}$ . Then $\eta $ is an anticanonical morphism of the surface S, which is generically two-to-one. Hence, the morphism $\eta $ induces an involution $\tau \in \mathrm {Aut}(S)$ . We let $C^\prime =\tau (C)$ .

Now, let u be a nonnegative real number. Then we have $-K_X-uS\sim _{\mathbb {R}} H_1+(1-u)H_2$ , so $-K_X-uS$ is nef $\iff -K_X-uS$ is pseudoeffective $\iff u\leqslant 1$ . Then $S_X(S)=\frac {5}{12}$ . Now, let us use notations introduced in [3, § 1.7]. Using [3, Theorem 1.112], we get

$$ \begin{align*}1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant \min\left\{\frac{1}{S_X(S)},\frac{1}{S(W_{\bullet,\bullet}^S;C)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{S,C};P)}\right\}, \end{align*} $$

where $S(W_{\bullet ,\bullet }^S;C)$ and $S(W_{\bullet , \bullet ,\bullet }^{S,C};P)$ are defined in [3, § 1.7]. Hence, since $S_X(S)<1$ , we get

(5.1) $$ \begin{align} \max\left\{S(W_{\bullet,\bullet}^S;C),S(W_{\bullet, \bullet,\bullet}^{S,C};P)\right\}\geqslant 1. \end{align} $$

Moreover, if $Z=P$ , then it also follows from [3, Theorem 1.112] that

(5.2) $$ \begin{align} \max\left\{S(W_{\bullet,\bullet}^S;C),S(W_{\bullet, \bullet,\bullet}^{S,C};P)\right\}>1. \end{align} $$

Let us estimate $S(W_{\bullet ,\bullet }^S;C)$ and $S(W_{\bullet , \bullet ,\bullet }^{S,C};P)$ using results obtained in [3, § 1.7].

Let $P(u)=-K_X-uS$ . Then [3, Corollary 1.109] gives

$$ \begin{align*}S\big(W^S_{\bullet,\bullet};C\big)=\frac{1}{4}\int_0^1\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{S}-vC\big)dvdu. \end{align*} $$

Let $P(u,v)$ be the positive part of the Zariski decomposition of the divisor $P(u)\vert _{S}-vC$ , and let $N(u,v)$ be its negative part, where $u\in [0,1]$ and $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}S\big(W_{\bullet,\bullet,\bullet}^{S,C};P\big)=F_{P}+\frac{1}{4}\int_{0}^{1}\int_{0}^{\infty}\big(P(u,v)\cdot C\big)^2dvdu \end{align*} $$

by [3, Theorem 1.112], where

$$ \begin{align*}F_P=\frac{1}{2}\int_{0}^{1}\int_{0}^{\infty}\big(P(u,v)\cdot C\big)\mathrm{ord}_P\Big(N(u,v)\big\vert_{C}\Big)dvdu. \end{align*} $$

Lemma 5.4. Suppose that $C_2$ is smooth, and $-K_S$ is ample. Then

$$ \begin{align*} S(W_{\bullet,\bullet}^S;C)&=\frac{13}{24},\\ S(W_{\bullet, \bullet,\bullet}^{S,C};P)&=1. \end{align*} $$

Proof. We have $C\cdot C^\prime =4$ and $(C^\prime )^2=C^2=0$ . Fix $u\in [0,1]$ and $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}P(u)\vert_{S}-vC\sim_{\mathbb{Q}}\Big(\frac{3}{2}-u-v\Big)C+\frac{1}{2}C^\prime, \end{align*} $$

which implies that $P(u)\vert _{S}-vC$ is pseudoeffective $\iff P(u)\vert _{S}-vC$ is nef $\iff v\leqslant \frac {3}{2}-u$ . If $0\leqslant u\leqslant 1$ and $0\leqslant v\leqslant \frac {3}{2}-u$ , then $P(u,v)=(\frac {3}{2}-u-v)C+\frac {1}{2}C^\prime $ and $N(u,v)=0$ , so

$$ \begin{align*}S\big(W^S_{\bullet,\bullet};C\big)=\frac{1}{4}\int_0^1\int_0^{\frac{3}{2}-u}\Bigg(\Big(\frac{3}{2}-u-v)C+\frac{1}{2}C^\prime\Bigg)^2dvdu= \frac{1}{4}\int_0^1\int_0^{\frac{3}{2}-u}6-4u-4v\,dvdu=\frac{13}{24}. \end{align*} $$

Similarly, we see that $F_P=0$ and $P(u,v)\cdot C=2$ , which gives $S(W_{\bullet ,\bullet ,\bullet }^{S,C};P)=1$ .

Lemma 5.5. Suppose that $C_2$ is smooth, $-K_S$ is not ample. Then

$$ \begin{align*} S(W_{\bullet,\bullet}^S;C)&=\frac{7}{12},\\ S(W_{\bullet, \bullet,\bullet}^{S,C};P)&=\frac{5}{6}. \end{align*} $$

Proof. In this case, we have the following commutative diagram:

where $\phi $ is a birational map that contracts four disjoint $(-2)$ -curves, and $\overline {S}$ is a del Pezzo surface of degree $2$ that has $4$ isolated ordinary double points, and $\pi $ is a double cover that is ramified in a reducible quartic curve that is a union of two irreducible conics such that $\eta (C)$ is one of these two conics. In particular, we have $C=\tau (C)$ .

Let $E_1$ , $E_2$ , $E_3$ , $E_4$ be the $\phi $ -exceptional curves. Fix $u\in [0,1]$ and $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}P(u)\vert_{S}-vC\sim_{\mathbb{Q}}(2-u-v)C+\frac{1}{2}(E_1+E_2+E_3+E_4) \end{align*} $$

so the divisor $P(u)\vert _{S}-vC$ is pseudoeffective $\iff v\leqslant 2-u$ . Moreover, we have

$$ \begin{align*}P(u,v)=\left\{\begin{aligned} &(2-u-v)C+\frac{1}{2}(E_1+E_2+E_3+E_4)\ \text{if }0\leqslant v\leqslant 1-u, \\ &(2-u-v)\Big(C+\frac{1}{2}(E_1+E_2+E_3+E_4)\Big)\ \text{if }1-u\leqslant v\leqslant 2-u, \\ \end{aligned} \right. \end{align*} $$

$$ \begin{align*}N(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant 1-u, \\ &\frac{u+v-1}{2}(E_1+E_2+E_3+E_4)\ \text{if }1-u\leqslant v\leqslant 2-u, \\ \end{aligned} \right. \end{align*} $$

$$ \begin{align*}\mathrm{vol}\big(P(u)\vert_{S}-vC\big)= \left\{\begin{aligned} &(6-4u)-4\ \text{if }0\leqslant v\leqslant 1-u, \\ &2(2-u-v)^2\ \text{if }1-u\leqslant v\leqslant 2-u. \\ \end{aligned} \right. \end{align*} $$

Now, integrating $\mathrm {vol}(P(u)\vert _{S}-vC)$ , we obtain $S(W_{\bullet ,\bullet }^S;C)=\frac {7}{12}$ .

To compute $S(W_{\bullet ,\bullet ,\bullet }^{S,C};P)$ , we first observe that $F_{P}=0$ because $P\not \in E_1\cup E_2\cup E_3\cup E_4$ since S is a general surface in $|H_2|$ that contains $C_2$ . On the other hand, we have

$$ \begin{align*}P(u,v)\cdot C=\left\{\begin{aligned} &2\ \text{if }0\leqslant v\leqslant 1-u, \\ &4-2u-2v\ \text{if }1-u\leqslant v\leqslant 2-u. \\ \end{aligned} \right. \end{align*} $$

Hence, integrating $(P(u,v)\cdot C)^2$ , we get $S(W_{\bullet ,\bullet ,\bullet }^{S,C};P)=\frac {5}{6}$ as required.

Lemma 5.6. Suppose that $C_2$ is singular. Then

$$ \begin{align*} S(W_{\bullet,\bullet}^S;C)&=\frac{3}{4},\\ S(W_{\bullet, \bullet,\bullet}^{S,C};P)&\leqslant \frac{11}{12}. \end{align*} $$

Proof. The curve $C_2$ consists of two irreducible components: the curve C and another curve, which we denote by L. The curves C and L are smooth and intersects transversally at one point. Note that $P\ne C\cap L$ since $C_2$ is smooth at the point P by assumption.

The intersections of the curves C, L and $C^\prime =\tau (C)$ on S are given in the table below.

Fix $u\in [0,1]$ and $v\in \mathbb {R}_{\geqslant 0}$ . Since $C+C^\prime \sim -K_S$ , we have

$$ \begin{align*}P(u)\vert_{S}-vC\sim_{\mathbb{Q}}(2-u-v)C+(1-u)L+C^\prime, \end{align*} $$

so $P(u)\vert _{S}-vC$ is pseudoeffective $\iff v\leqslant 2-u$ . Moreover, if $0\leqslant u\leqslant \frac {1}{2}$ , then

$$ \begin{align*}P(u,v)=\left\{\begin{aligned} &(2-u-v)C+(1-u)L+C^\prime\ \text{if }0\leqslant v\leqslant 1, \\ &(2-u-v)(C+L)+C^\prime\ \text{if }1\leqslant v\leqslant \frac{3}{2}-u, \\ &(2-u-v)(C+L+C^\prime)\ \text{if }\frac{3}{2}-u\leqslant v\leqslant 2-u, \end{aligned} \right. \end{align*} $$

$$ \begin{align*}N(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant 1, \\ &(v-1)L\ \text{if }1\leqslant v\leqslant \frac{3}{2}-u, \\ &(v-1)L+(2v+2u-3)C^\prime\ \text{if }\frac{3}{2}-u\leqslant v\leqslant 2-u, \end{aligned} \right. \end{align*} $$

$$ \begin{align*}\mathrm{vol}\big(P(u)\vert_{S}-vC\big)= \left\{\begin{aligned} &6-v^2-4u-2v\ \text{if }0\leqslant v\leqslant 1, \\ &7-4u-4v \ \text{if }1\leqslant v\leqslant \frac{3}{2}-u, \\ &4(u+v-2)^2\ \text{if }\frac{3}{2}-u\leqslant v\leqslant 2-u. \end{aligned} \right. \end{align*} $$

Similarly, if $\frac {1}{2}\leqslant u\leqslant 1$ , then

$$ \begin{align*}P(u,v)=\left\{\begin{aligned} &(2-u-v)C+(1-u)L+C^\prime\ \text{if }0\leqslant v\leqslant \frac{3}{2}-u, \\ &(2-u-v)(C+2C^\prime)+(1-u)L\ \text{if }\frac{3}{2}-u\leqslant v\leqslant 1, \\ &(2-u-v)(C+L+C^\prime)\ \text{if }1\leqslant v\leqslant 2-u, \end{aligned} \right. \end{align*} $$

$$ \begin{align*}N(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant \frac{3}{2}-u, \\ &(2v+2u-3)C^\prime\ \text{if }\frac{3}{2}-u\leqslant v\leqslant 1, \\ &(v-1)L+(2v+2u-3)C^\prime\ \text{if }1\leqslant v\leqslant 2-u, \end{aligned} \right. \end{align*} $$

$$ \begin{align*}\mathrm{vol}\big(P(u)\vert_{S}-vC\big)= \left\{\begin{aligned} &6-v^2-4u-2v\ \text{if }0\leqslant v\leqslant \frac{3}{2}-u, \\ &(5-2u-3v)(3-2u-v)\ \text{if }\frac{3}{2}-u\leqslant v\leqslant 1, \\ &4(u+v-2)^2\ \text{if }1\leqslant v\leqslant 2-u. \end{aligned} \right. \end{align*} $$

Hence, integrating $\mathrm {vol}(P(u)\vert _{S}-vC)$ , we get $S(W_{\bullet ,\bullet }^S;C)=\frac {3}{4}$ .

Now, let us compute $S(W_{\bullet , \bullet ,\bullet }^{S,C};P)$ . If $0\leqslant u\leqslant \frac {1}{2}$ , then

$$ \begin{align*}P(u,v)\cdot C=\left\{\begin{aligned} &1+v\ \text{if }0\leqslant v\leqslant 1, \\ &2\ \text{if }1\leqslant v\leqslant \frac{3}{2}-u, \\ &8-4u-4v\ \text{if }\frac{3}{2}-u\leqslant v\leqslant 2-u. \end{aligned} \right. \end{align*} $$

Similarly, if $\frac {1}{2}\leqslant u\leqslant 1$ , then

$$ \begin{align*}P(u,v)\cdot C=\left\{\begin{aligned} &1+v\ \text{if }0\leqslant v\leqslant \frac{3}{2}-u, \\ &7-4u-3v\ \text{if }\frac{3}{2}-u\leqslant v\leqslant 1, \\ &8-4u-4v\ \text{if }1\leqslant v\leqslant 2-u. \end{aligned} \right. \end{align*} $$

Then integrating, we get $S(W_{\bullet ,\bullet ,\bullet }^{S,C};P)=\frac {145}{192}+F_{P}$ . To compute $F_{P}$ , let us recall that $P\not \in L$ . Hence, if $P\not \in C^\prime $ , then $F_{P}=0$ , which implies that $S(W_{\bullet ,\bullet ,\bullet }^{S,C};P)=\frac {145}{192}<\frac {11}{12}$ as required.

We may assume that $P\in C\cap C^\prime $ . If $C^\prime $ intersects C transversally at P, then

$$ \begin{align*}\mathrm{ord}_P\Big(N(u,v)\big\vert_{C}\Big)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant\frac{3}{2}-u, \\ &2u+2v-3\ \text{if }\frac{3}{2}-u\leqslant v\leqslant 2-u, \end{aligned} \right. \end{align*} $$

which gives

$$ \begin{align*} F_P&=\frac{1}{2}\int_{0}^{\frac{1}{2}}\int_{\frac{3}{2}-u}^{2-u}(8-4u-4v)(2u+2v-3)dvdu+\\ &\quad+\frac{1}{2}\int_{\frac{1}{2}}^{1}\int_{1}^{\frac{3}{2}-u}(7-4u-3v)(2u+2v-3)dvdu\\&\quad+\frac{1}{2}\int_{\frac{1}{2}}^{1}\int_{\frac{3}{2}-u}^{2-u}(8-4u-4v)(2u+2v-3)dvdu=\frac{31}{384}, \end{align*} $$

so $S(W_{\bullet ,\bullet ,\bullet }^{S,C};P)=\frac {145}{192}+\frac {31}{384}=\frac {107}{128}<\frac {11}{12}$ . If $C^\prime $ is tangent to C at the point P, then

$$ \begin{align*}\mathrm{ord}_P\Big(N(u,v)\big\vert_{C}\Big)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant\frac{3}{2}-u, \\ &2(2u+2v-3)\ \text{if }\frac{3}{2}-u\leqslant v\leqslant 2-u,\\ \end{aligned} \right. \end{align*} $$

which gives $F_P=\frac {31}{192}$ , so $S(W_{\bullet ,\bullet }^{S,C};P)=\frac {145}{192}+\frac {31}{192}=\frac {11}{12}$ .

Now, using Equation (5.2) and Lemmas 5.4, 5.5, 5.6, we see that Z is a curve.

Lemma 5.7. One has $H_1\cdot Z\geqslant 1$ and $H_2\cdot Z\geqslant 1$ .

Proof. If $H_2\cdot Z=0$ , then $Z=C$ , which is impossible by Equation (5.1) and Lemmas 5.4, 5.5, 5.6. Hence, we see that $H_2\cdot Z\geqslant 1$ and $\pi _2(Z)$ is a curve. Let us show that $H_1\cdot Z\geqslant 1$ .

Suppose that $H_1\cdot Z=0$ . Then Z must be an irreducible component of the curve $C_1$ . If $C_1$ is reduced, then arguing exactly as in the proofs of Lemmas 5.4, 5.5, 5.6, we obtain a contradiction with [3, Corollary 1.109]. Thus, we see that $C_1$ is not reduced.

So far, the point P was a point in Z. Let us choose $P\in Z$ such that $\pi _2(P)\in \pi _2(Z)\cap \Delta _2$ . Then $C_2$ is singular. But it is smooth at P by Lemma 5.1, which fits our assumption above. Then $S(W_{\bullet ,\bullet }^S;C)=\frac {3}{4}$ and $S(W_{\bullet , \bullet ,\bullet }^{S,C};P)\leqslant \frac {11}{12}$ by Lemma 5.6, which contradicts Equation (5.1).

Both $\pi _1(Z)$ and $\pi _2(Z)$ are curves. Similar to what we did in the proof of Lemma 5.7, let us choose the point $P\in Z$ such that $\pi _1(P)\in \Delta _1$ . Then $C_1$ is singular at P, which implies that $C_2$ is smooth at P by Lemma 5.1. Now, using Equation (5.1) and Lemmas 5.5 and 5.6, we see that $C=C_2$ , the curve $C_2$ is smooth, the divisor $-K_S$ is ample.

We see that S is a del Pezzo surface, and $\eta \colon S\to \mathbb {P}^2$ is a double cover ramified in a smooth quartic curve, so we are almost in the same position as in the proof of Lemma 5.4. But now, we have one small advantage: the point $\eta (P)$ is contained in the ramification divisor of the double cover $\eta $ because $C_1$ is singular at P. Then $|-K_S|$ contains a unique curve that is singular at P. Denote this curve by R. We have the following possibilities:

1. (a) R is an irreducible curve that has a nodal singularity at P;

2. (b) R is an irreducible curve that has a cuspidal singularity at P;

3. (c) $R=R_1+R_2$ for two $(-1)$ -curves in S that intersect transversally at P;

4. (d) $R=R_1+R_2$ for two $(-1)$ -curves in S that are tangent at P.

Let $f\colon \widetilde {S}\to S$ be the blow up of the point P. Denote by $\widetilde {R}$ and $\widetilde {C}$ the proper transforms on the surface $\widetilde {S}$ of the curves R and C, respectively. Fix $u\in [0,1]$ and $v\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}f^*\big(P(u)\vert_{S}\big)-vE\sim_{\mathbb{Q}} \widetilde{R}+(1-u)\widetilde{C}+(3-u-v)E. \end{align*} $$

Let $\widetilde {P}(u,v)$ be the positive part of the Zariski decomposition of $\widetilde {R}+(1-u)\widetilde {C}+(3-u-v)E$ , and let $\widetilde {N}(u,v)$ its negative part. Then it follows from [3, Remark 1.113] that

(5.3) $$ \begin{align} 1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{1}{S_X(S)}, \frac{2}{S(W_{\bullet,\bullet}^{\widetilde{S}};E)},\inf_{O\in E}\frac{1}{S(W_{\bullet, \bullet,\bullet}^{\widetilde{S},E};O)}\right\}, \end{align} $$

where $S(W_{\bullet ,\bullet }^{\widetilde {S}};E)$ and $S(W_{\bullet , \bullet ,\bullet }^{\widetilde {S},E};O)$ are defined in [3] similar to $S(W_{\bullet ,\bullet }^S;C)$ and $S(W_{\bullet , \bullet ,\bullet }^{S,C};P)$ . These two numbers can be computed using [3, Remark 1.113]. Namely, we have

$$ \begin{align*}S\big(W_{\bullet,\bullet}^{\widetilde{S}};E\big)=\frac{1}{4}\int_0^1\int_0^{\infty}\big(\widetilde{P}(u,v)\big)^2dvdu \end{align*} $$

and

$$ \begin{align*}S\big(W_{\bullet, \bullet,\bullet}^{\widetilde{S},E};O\big)=\frac{1}{4}\int_0^1\int_0^{\infty}\Big(\big(\widetilde{P}(u,v)\cdot E\big)\Big)^2dvdu+ F_O, \end{align*} $$

where O is a point in E and

$$ \begin{align*}F_O=\frac{1}{2}\int_0^1\int_0^{\infty}\big(\widetilde{P}(u,v)\cdot E\big)\mathrm{ord}_O\Big(\widetilde{N}(u,v)\big|_{E}\Big)dvdu. \end{align*} $$

Let us use these formulas to estimate $S(W_{\bullet ,\bullet }^{\widetilde {S}};E)$ and $S(W_{\bullet , \bullet ,\bullet }^{\widetilde {S},E};O)$ .

If the curve R is irreducible, the intersections of the curves $\widetilde {C}$ , $\widetilde {R}$ , E can be computed as follows: $\widetilde {C}^2=-1$ , $\widetilde {C}\cdot \widetilde {R}=0$ , $\widetilde {C}\cdot E=1$ , $\widetilde {R}^2=-2$ , $\widetilde {R}\cdot E=2$ , $E^2=-1$ . If R is reducible, then $\widetilde {R}=\widetilde {R}_1+\widetilde {R}_2$ for two smooth irreducible curves $\widetilde {R}_1+\widetilde {R}_2$ such that the intersection form of the curves $\widetilde {C}$ , $\widetilde {R}_1$ , $\widetilde {R}_1$ and E is given in the following table:

Then $\widetilde {R}+(1-u)\widetilde {C}+(3-u-v)E$ is pseudoeffective $\iff v\leqslant 3-u$ . Moreover, we have

$$ \begin{align*}\widetilde{P}(u,v)=\left\{\begin{aligned} &\widetilde{R}+(1-u)\widetilde{C}+(3-u-v)E\ \text{if }0\leqslant v\leqslant 2-u, \\ &(1-u)\widetilde{C}+(3-u-v)(E+\widetilde{R})\ \text{if }2-u\leqslant v\leqslant 2, \\ &(3-u-v)(E+\widetilde{R}+\widetilde{C})\ \text{if }2\leqslant v\leqslant 3-u, \\ \end{aligned} \right. \end{align*} $$

$$ \begin{align*}\widetilde{N}(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant 2-u, \\ &(v+u-2)\widetilde{R}\ \text{if }2-u\leqslant v\leqslant 2, \\ &(v+u-2)\widetilde{R}+(v-2)\widetilde{C}\ \text{if }2\leqslant v\leqslant 3-u,\\ \end{aligned} \right. \end{align*} $$

$$ \begin{align*}\big(\widetilde{P}(u,v)\big)^2= \left\{\begin{aligned} &6-v^2-4u\ \text{if }0\leqslant v\leqslant 2-u, \\ &14+2u^2+4uv+v^2-12u-8v\ \text{if }2-u\leqslant v\leqslant 2, \\ &2(3-u-v)^2\ \text{if }2\leqslant v\leqslant 3-u, \\ \end{aligned} \right. \end{align*} $$

Now, integrating, we obtain $S(W_{\bullet ,\bullet }^{\widetilde {S}};E)=\frac {17}{12}$ .

Fix a point $O\in E$ . To estimate $S(W_{\bullet , \bullet ,\bullet }^{\widetilde {S},E};O)$ , first we observe that

$$ \begin{align*}\widetilde{P}(u,v)\cdot E= \left\{\begin{aligned} &v\ \text{if }0\leqslant v\leqslant 2-u, \\ &4-2u-v\ \text{if }2-u\leqslant v\leqslant 2, \\ &6-2u-2v\ \text{if }2\leqslant v\leqslant 3-u.\\ \end{aligned} \right. \end{align*} $$

Therefore, integrating $(\widetilde {P}(u,v)\cdot E)^2$ , we obtain $S(W_{\bullet , \bullet ,\bullet }^{Y,\widetilde {S}};O)=\frac {13}{24}+F_O$ .

If $O\not \in \widetilde {C}\cup \widetilde {R}$ , then $F_O=0$ . Similarly, if $O\in \widetilde {C}$ , then $O\not \in \widetilde {R}$ , which gives

$$ \begin{align*}F_O=\frac{1}{2}\int_0^1\int_2^{3-u}(6-2u-2v)(v-2)dvdu=\frac{1}{24}. \end{align*} $$

Finally, if $O\in \widetilde {R}$ , then $O\not \in \widetilde {C}$ , which gives

$$ \begin{align*}F_O\leqslant\frac{1}{2}\int_0^1\int_{2-u}^{2}2(4-2u-v)(v+u-2)dvdu+\frac{1}{2}\int_0^1\int_{2}^{3-u}2(6-2u-2v)(v+u-2)dvdu=\frac{7}{24}. \end{align*} $$

Summarizing, we get $S(W_{\bullet , \bullet ,\bullet }^{\widetilde {S},E};O)\leqslant \frac {5}{6}$ for every point $O\in E$ , which contradicts Equation (5.3) because $S_X(S)=\frac {5}{12}$ and $S(W_{\bullet ,\bullet }^{\widetilde {S}};E)=\frac {17}{12}<2$ . This shows that X is K-stable.

Corollary 5.8. All smooth Fano threefolds in the family are K-stable.

6 Family

Now, let us fix three smooth quadric hypersurfaces $\mathscr {Q}$ , $\mathscr {Q}^\prime $ and $\mathscr {Q}^{\prime \prime }$ in $\mathbb {P}^4$ such that their intersection is a smooth curve of degree $8$ and genus $5$ . We set $\mathscr {C}=\mathscr {Q}\cap \mathscr {Q}^\prime \cap \mathscr {Q}^{\prime \prime }$ . Let $\pi \colon X\to \mathscr {Q}$ be the blow up of the smooth curve $\mathscr {C}$ . Then X is a smooth Fano threefold, which is contained in the family . Moreover, all smooth Fano threefolds in this family can be obtained in this way. Note that $-K_X^3=14$ and $\mathrm {Aut}(X)$ is finite [10].

The pencil generated by the surfaces $\mathscr {Q}^\prime \vert _{\mathscr {Q}}$ and $\mathscr {Q}^{\prime \prime }\vert _{\mathscr {Q}}$ gives a rational map , which fits the following commutative diagram:

where $\phi $ is a fibration into quartic del Pezzo surfaces. Let E be the $\pi $ -exceptional surface, and let $H=\pi ^*(\mathcal {O}_{\mathbb {P}^4}(1)\vert _{\mathscr {Q}})$ . Then $-K_X\sim 3H-E$ , the morphism $\phi $ is given by the linear system $|2H-E|$ , and $E\cong \mathscr {C}\times \mathbb {P}^1$ .

Lemma 6.1. Let S be a fiber of the morphism $\phi $ . Then S has at most Du Val singularities.

Proof. Since $E\vert _{S}$ is a smooth curve, the surface S is smooth along $E\vert _{S}$ , which implies that it has at most isolated singularities. Hence, we conclude that S is normal and irreducible.

Note that $S\cong \pi (S)$ . Suppose that the singularities of the surface $\pi (S)$ are not Du Val. Then it follows from [5, Theorem 1] that $\pi (S)$ is a cone in $\mathbb {P}^4$ over a quartic elliptic curve. Let P be the vertex of the cone $\pi (S)$ , and let $T_P$ be the hyperplane section of the quadric hypersurface $\mathscr {Q}$ that is singular at P. Then $T_P$ contains all lines in $\mathscr {Q}$ that pass through P, which implies that $T_P$ contains $\pi (S)$ . This is impossible since $T_P$ is a quadric cone.

The goal of this section is to prove that X is K-stable. To do this, we fix a point $P\in X$ . By [15, 18], to prove that X is K-stable, it is enough to show that $\delta _P(X)>1$ .

Let S be the fiber of the morphism $\phi $ containing P. Then S is a quartic del Pezzo surface, and S has at most Du Val singularity at P by Lemma 6.1. Moreover, if S is singular at P, then P is a singular point of the surface S of type $\mathbb {A}_1$ , $\mathbb {A}_2$ , $\mathbb {A}_3$ , $\mathbb {A}_4$ , $\mathbb {D}_4$ or $\mathbb {D}_5$ , see [12].

Lemma 6.2. If $\delta _P(S)>\frac {54}{55}$ or $P\in \mathrm {Sing}(S)$ and $\delta _P(S)>\frac {27}{28}$ , then $\delta _P(X)>1$ .

Proof. Take $u\in \mathbb {R}_{\geqslant 0}$ . Since $S\sim 2H-E$ , we have

$$ \begin{align*}-K_X-uS\sim_{\mathbb{R}} (3-2u)H-(u-1)E\sim_{\mathbb{R}}\Big(\frac{3}{2}-u\Big)S+\frac{1}{2}E. \end{align*} $$

Using this, we conclude that the divisor $-K_X-uS$ is pseudoeffective if and only if $u\leqslant \frac {3}{2}$ . For $u\leqslant \frac {3}{2}$ , let $P(u)$ be the positive part of Zariski decomposition of the divisor $-K_X-uS$ , and let $N(u)$ be the negative part of Zariski decomposition of the divisor $-K_X-uS$ . Then

$$ \begin{align*}P(u)=\left\{\begin{aligned} &-K_X-uS\ \text{if }0\leqslant u\leqslant 1, \\ &(3-2u)H\ \text{if }1\leqslant u\leqslant \frac{3}{2}, \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}N(u)=\left\{\begin{aligned} &0\ \text{if }0\leqslant u\leqslant 1, \\ &(u-1)E\ \text{if }1\leqslant u\leqslant \frac{3}{2}. \end{aligned} \right. \end{align*} $$

This gives

$$ \begin{align*}S_{X}(S)=\frac{1}{14}\int_{0}^{\frac{3}{2}}\big(P(u)\big)^3du=\frac{1}{14}\int_{0}^{1}(14-12u)du+\frac{1}{14}\int_{1}^{\frac{3}{2}} 2(3-2u)^{3}du=\frac{33}{56}<1. \end{align*} $$

Now, using [1, Theorem 3.3] and [3, Corollary 1.102], we get

(6.1) $$ \begin{align} \delta_P(X)\geqslant \min\left\{\frac{1}{S_X(S)},\inf_{\substack{F/S\\ P\in C_S(F)}}\frac{A_S(F)}{S(W_{\bullet,\bullet}^S;F)}\right\}, \end{align} $$

where the infimum is taken by all prime divisors F over the surface S such that $P\in C_S(F)$ , and $S(W_{\bullet ,\bullet }^S;F)$ can be computed using [3, Corollary 1.108] as follows:

$$ \begin{align*}S\big(W_{\bullet,\bullet}^S; F\big)=\frac{3}{14}\int_0^{\frac{3}{2}}\big(P(u)\big\vert_{S}\big)^2\mathrm{ord}_{F}\big(N(u)\big\vert_{S}\big)du+\frac{3}{14}\int_{0}^{\frac{3}{2}}\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{S}-vF\big)dvdu. \end{align*} $$

Let $E_S=E|_S$ . Then $E_S\in |-2K_S|$ and $E_S\cong \mathscr {C}$ . Moreover, the surface S is smooth in a neighborhood of the curve $E_S$ . Furthermore, we have

$$ \begin{align*}P(u)\big\vert_{S}=\left\{\begin{aligned} &-K_S\ \text{if }0\leqslant u\leqslant 1, \\ &(3-2u)(-K_S)\ \text{if }1\leqslant u\leqslant \frac{3}{2}, \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}N(u)\big\vert_{S}=\left\{\begin{aligned} &0\ \text{if }0\leqslant u\leqslant 1, \\ &(u-1)E_S\ \text{if }1\leqslant u\leqslant \frac{3}{2}. \end{aligned} \right. \end{align*} $$

Let F be any prime divisor over S such that $P\in C_S(F)$ . Then

$$ \begin{align*} S\big(W_{\bullet,\bullet}^S; F\big)&=\frac{3}{14}\int_{1}^{\frac{3}{2}}4(u-1)(3-2u)^{2}\mathrm{ord}_{F}\big(E_{S}\big)du+\\ &\quad+\frac{3}{14}\int_{0}^{1} \int_{0}^{\infty}\mathrm{vol}\big(-K_S-vF\big)dvdu+\\ &\quad+\frac{3}{14}\int_{1}^{\frac{3}{2}}\int_{0}^{\infty}\mathrm{vol}\big((3-2u)(-K_S)-vF\big)dvdu=\\ &=\frac{\mathrm{ord}_{F}\big(E_{S}\big)}{56}+\frac{3}{14}\int_{0}^{\infty}\!\mathrm{vol}\big(-K_S-vF\big)dv+\frac{3}{14}\int_{1}^{\frac{3}{2}}\!(3-2u)^3\int_{0}^{\infty}\!\mathrm{vol}\big(-K_S-vF\big)dvdu=\\ &=\frac{\mathrm{ord}_{F}\big(E_{S}\big)}{56}+\frac{3}{14}\int_{0}^{\infty}\mathrm{vol}\big(-K_S-vF\big)dv+\frac{3}{112}\int_{0}^{\infty}\mathrm{vol}\big(-K_S-vF\big)dv=\\ &=\frac{\mathrm{ord}_{F}\big(E_{S}\big)}{56}+\frac{27}{112}\kern-1pt\int_{0}^{\infty}\!\kern-1pt\mathrm{vol}\big(-K_S-vF\big)dv=\frac{\mathrm{ord}_{F}\big(E_{S}\big)}{56}+\frac{27}{28}S_S(F)\kern1.3pt{\leqslant}\kern1.3pt\frac{A_{S}(F)}{56}+\frac{27A_{S}(F)}{28\delta_{P}(S)} \end{align*} $$

because log pair $(S,E_S)$ is log canonical. Therefore, if $\delta _P(S)>\frac {54}{55}$ , Equation (6.1) gives $\delta _P(X)>1$ . Similarly, if $P\in \mathrm {Sing}(S)$ , $P\not \in E_S$ , so $\mathrm {ord}_{F}(E_{S})=0$ , which implies that

$$ \begin{align*}S\big(W_{\bullet,\bullet}^S; F\big)=\frac{27}{28}S_S(F)\leqslant\frac{27A_{S}(F)}{28\delta_{P}(S)}. \end{align*} $$

Hence, in this case, it follows from Equation (6.1) that $\delta _P(X)>1$ provided that $\delta _P(S)>\frac {27}{28}$ .

Corollary 6.3. If S is smooth, then $\delta _P(X)>1$ .

Proof. If S is smooth, then $\delta (S)=\frac {4}{3}$ by [3, Lemma 2.12]. Now, apply Lemma 6.2.

Corollary 6.4. If P is not contained in a line in S, then $\delta _P(X)>1$ .

Proof. If P is not contained in a line in S, then P is a smooth point of the surface S, and the blow up of the surface S at this point is a (possibly singular) cubic surface in $\mathbb {P}^3$ . Thus, arguing exactly as in the end of the proof of [3, Lemma 2.12], we obtain $\delta _P(S)\geqslant \frac {3}{2}$ , which implies that $\delta _P(X)>1$ by Lemma 6.2.

Now, let T be a surface in the linear system $|H|$ such that $P\in T$ , and let $\mathcal {Q}=\pi (T)$ . Then $\mathcal {Q}$ is a hyperplane section of the hypersurface $\mathscr {Q}$ , so both $\mathcal {Q}$ and T are irreducible. In the following, we will choose T such that the surface $\mathcal {Q}$ is smooth so that $\mathcal {Q}\cong \mathbb {P}^1\times \mathbb {P}^1$ .

Lemma 6.5. Suppose that T is a general surface in the linear system $|H|$ such that $P\in T$ . Then the (scheme) intersection $S\cap T$ is an irreducible reduced curve.

Proof. Let $\rho \colon \widetilde {S}\to S$ be a blow up of the quartic del Pezzo surface S at the point P, and let Z be the proper transform of the curve $T\vert _{S}$ on the surface $\widetilde {S}$ . Then $|Z|$ has no base points and gives the morphism $\eta \colon \widetilde {S}\to \mathbb {P}^3$ that fits the following commutative diagram:

where is a projection from P. Moreover, if P is a smooth point of the surface S, then $Z^2=3$ , and the image of the morphism $\eta $ is an irreducible cubic surface in $\mathbb {P}^3$ . Similarly, if P is a singular point of the surface S, then we have $Z^2=4-\mathrm {mult}_P(S)=2$ , and the image of the morphism $\eta $ is an irreducible quadric surface. Therefore, we conclude that the curve Z must be irreducible and reduced (by Bertini theorem), which implies that the intersection $S\cap T$ is also irreducible and reduced.

Remark 6.6. Suppose that S is singular at P, and T is a general surface in $|H|$ that passes through the point P. Then, choosing appropriate coordinates $[x:y:z:t:w]$ on $\mathbb {P}^4$ , we may assume that $\pi (P)=[0:0:0:0:1]$ , and the surface $\pi (S)$ is given in $\mathbb {P}^4$ by

$$ \begin{align*}\left\{\begin{aligned} &at^2+btx+f_2(x,y,z)=0, \\ &wt=g_2(x,y,z), \\ \end{aligned} \right. \end{align*} $$

where a and b are complex numbers, $f_2(x,y,z)$ and $g_2(x,y,z)$ are nonzero quadratic homogeneous polynomials. In the chart $w\ne 0$ , the surface $\pi (S)$ is given by

$$ \begin{align*}\left\{\begin{aligned} &at^2+btx+f_2(x,y,z)=0, \\ &t=g_2(x,y,z), \\ \end{aligned} \right. \end{align*} $$

where now we consider x, y, z, t as affine coordinates on $\mathbb {C}^4$ . Then $\pi (S)\cap \mathcal {Q}$ is cut out on the surface $\pi (S)$ by $c_1x+c_2y+c_3z+c_4t=0$ , where $c_1$ , $c_2$ , $c_3$ , $c_4$ are general numbers. The affine part of the surface $\pi (S)$ is isomorphic to the hypersurface in $\mathbb {C}^3$ given by

$$ \begin{align*}ag_2^2(x,y,z)+bxg_2(x,y,z)+f_2(x,y,z)=0, \end{align*} $$

and the affine part of the curve $\pi (S)\cap \mathcal {Q}$ is cut out by $c_1x+c_2y+c_3z+c_4g_2(x,y,z)=0$ . If P is a singular point of the surface S of type $\mathbb {D}_4$ or $\mathbb {D}_5$ , then $S\cap T$ has an ordinary cusp at the point P, which easily implies that the intersection $S\cap T$ is reduced and irreducible. Similarly, if P is a Du Val singular point of the surface S of type $\mathbb {A}_1$ , $\mathbb {A}_2$ , $\mathbb {A}_3$ or $\mathbb {A}_4$ , then the intersection $S\cap T$ has an isolated ordinary double singularity at P.

Observe that the morphism $\pi \colon X\to \mathscr {Q}$ induces a birational morphism $\varpi \colon T\to \mathcal {Q}$ , and the morphism $\phi \colon X\to \mathbb {P}^1$ induces a fibration $\varphi \colon T\to \mathbb {P}^1$ that both fit the following commutative diagram:

where is a map given by the pencil generated by the curves $\mathscr {Q}^\prime \vert _{\mathcal {Q}}$ and $\mathscr {Q}^{\prime \prime }\vert _{\mathcal {Q}}$ . In the following, we will always choose $T\in |H|$ such that the quadric surface $\mathcal {Q}$ is smooth, and T is either smooth or has one isolated ordinary singularity, which would imply that a general fiber of the induced fibration $\varphi \colon T\to \mathbb {P}^1$ is a smooth elliptic curve. Let $\mathcal {C}=S\vert _{T}$ . Then $\mathcal {C}$ is the fiber of the (elliptic) fibration $\varphi $ that contains the point P.

Let u be a nonnegative real number. Then $-K_X-uT\sim _{\mathbb {R}}(3-u)H-E\sim _{\mathbb {R}}(1-u)H+S$ , which implies that $-K_X-uT$ is nef $\iff -K_X-uT$ is pseudoeffective $\iff u\in [0,1]$ . Integrating, we get $S_{X}(T)=\frac {9}{28}<1$ . For simplicity, we let $P(u)=-K_X-uT$ .

Lemma 6.7. Suppose that S is singular at P. Then $\delta _P(X)>1$ .

Proof. Now, let us choose $T\in |H|$ such that T is a general surface in $|H|$ that contains P. Then T and $\mathcal {Q}$ are smooth, and $\varpi $ is a blow up of the eight intersection points $\mathscr {C}\cap \mathcal {Q}$ . Moreover, by Lemma 6.5, the curve $\mathcal {C}$ is an irreducible singular curve of arithmetic genus $1$ . Thus, we have $P=\mathrm {Sing}(\mathcal {C})$ . Furthermore, using Remark 6.6, we see that

• • either $\mathcal {C}$ has an isolated ordinary double singularity at P,

• • or the curve $\mathcal {C}$ has an ordinary cusp at the point P.

Recall that $\mathcal {Q}$ is a smooth quadric surface, so that it contains exactly two lines that pass though $\pi (P)$ . Since T is chosen to be general, these lines are disjoint from $\mathscr {C}\cap \mathcal {Q}$ . Denote by $L_1$ and $L_2$ the proper transforms of these lines on T. Then

$$ \begin{align*}P(u)\big\vert_{T}\sim_{\mathbb{R}}\big((1-u)H+S\big)\big\vert_{T}\sim_{\mathbb{R}}(1-u)(L_1+L_2)+\mathcal{C}. \end{align*} $$

Now, we let $\sigma \colon \widetilde {T}\to T$ be the blow up of the point P, we let $\mathbf {e}$ be the $\sigma $ -exceptional curve, and we denote by $\widetilde {\mathcal {C}}$ , $\widetilde {L}_1$ , $\widetilde {L}_2$ the proper transforms on $\widetilde {T}$ of the curves $\mathcal {C}$ , $L_1$ , $L_2$ , respectively. Take a nonnegative real number v. Then

(6.2) $$ \begin{align} \sigma^*\big(P(u)\big\vert_{T}\big)-v\mathbf{e}\sim_{\mathbb{R}}\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(4-2u-v)\mathbf{e}. \end{align} $$

Note that the curves $\widetilde {\mathcal {C}}$ , $\widetilde {L}_1$ , $\widetilde {L}_2$ are disjoint. Moreover, we have $\widetilde {L}_{1}^2=\widetilde {L}_{2}^2=-1$ and $\widetilde {\mathcal {C}}^2=-4$ . Thus, using Equation (6.2), we see that $\sigma ^*(P(u)\vert _{T})-v\mathbf {e}$ is pseudoeffective $\iff v\leqslant 4-2u$ .

Let $\widetilde {P}(u,v)$ be the positive part of Zariski decomposition of the divisor $\sigma ^*(P(u)\vert _{T})-v\mathbf {e}$ , and let $\widetilde {N}(u,v)$ be its negative part. Then it follows from [3, Remark 1.113] that

(6.3) $$ \begin{align} \delta_P(X)\geqslant\min\left\{\frac{1}{S_X(T)}, \frac{2}{S(W_{\bullet,\bullet}^{\widetilde{T}};\mathbf{e})},\inf_{O\in \mathbf{e}}\frac{1}{S(W_{\bullet, \bullet,\bullet}^{\widetilde{T},\mathbf{e}};O)}\right\}, \end{align} $$

where $S(W_{\bullet ,\bullet }^{\widetilde {T}};\mathbf {e})$ and $S(W_{\bullet , \bullet ,\bullet }^{\widetilde {T},\mathbf {e}};O)$ are defined in [3, § 1.7], and these two numbers can be computed using formulas described in [3, Remark 1.113]. Namely, we have

$$ \begin{align*}S\big(W_{\bullet,\bullet}^{\widetilde{T}};\mathbf{e}\big)=\frac{3}{14}\int_0^1\int_0^{4-2u}\big(\widetilde{P}(u,v)\big)^2dvdu \end{align*} $$

and

$$ \begin{align*}S\big(W_{\bullet, \bullet,\bullet}^{\widetilde{T},\mathbf{e}};O\big)=\frac{3}{14}\int_0^1\!\int_0^{4-2u}\Big(\big(\widetilde{P}(u,v)\cdot\mathbf{e}\big)\Big)^2dvdu+\frac{3}{7}\int_0^1\!\int_0^{4-2u}\kern-1.5pt\big(\widetilde{P}(u,v)\cdot\mathbf{e}\big)\mathrm{ord}_O\Big(\widetilde{N}(u,v)\big|_{\mathbf{e}}\Big)dvdu, \end{align*} $$

where O is a point in $\mathbf {e}$ . Moreover, using Equation (6.2), we compute $\widetilde {P}(u,v)$ and $\widetilde {N}(u,v)$ as follows:

$$ \begin{align*}\widetilde{P}(u,v)=\left\{\begin{aligned} &\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(4-2u-v)\mathbf{e}\ \text{if }0\leqslant v\leqslant 2-2u, \\ &\frac{4-2u-v}{2}\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(4-2u-v)\mathbf{e}\ \text{if }2-2u\leqslant v\leqslant 3-u, \\ &\frac{4-2u-v}{2}\big(\widetilde{\mathcal{C}}+2(\widetilde{L}_{1}+\widetilde{L}_{2})+2\mathbf{e}\big)\ \text{if }3-u\leqslant v\leqslant 4-2u, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}\widetilde{N}(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant 2-2u, \\ &\frac{v-2+2u}{2}\widetilde{\mathcal{C}}\ \text{if }2-2u\leqslant v\leqslant 3-u, \\ &\frac{v-2+2u}{2}\widetilde{\mathcal{C}}+(v-3+u)(\widetilde{L}_{1}+\widetilde{L}_{2}) \ \text{if }3-u\leqslant v\leqslant 4-2u.\\ \end{aligned} \right. \end{align*} $$

Thus, we have

$$ \begin{align*}\big(\widetilde{P}(u,v)\big)^2=\left\{\begin{aligned} &2(1-u)(5-u)-v^{2}\ \text{if }0\leqslant v\leqslant 2-2u, \\ &2(1-u)(7-3u-2v)\ \text{if }2-2u\leqslant v\leqslant 3-u, \\ &2(4-2 u-v)^{2}\ \text{if }3-u\leqslant v\leqslant 4-2u, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}\widetilde{P}(u,v)\cdot\mathbf{e}=\left\{\begin{aligned} &v\ \text{if }0\leqslant v\leqslant 2-2u, \\ &2(1-u)\ \text{if }2-2u\leqslant v\leqslant 3-u, \\ &2(4-2u-v)\ \text{if }3-u\leqslant v\leqslant 4-2u. \\ \end{aligned} \right. \end{align*} $$

Now, integrating, we get $S(W_{\bullet ,\bullet }^{\widetilde {T}};\mathbf {e})=\frac {51}{28}<2$ .

Let O be an arbitrary point in $\mathbf {e}$ . If $O\not \in \widetilde {L}_1\cup \widetilde {L}_2\cup \widetilde {\mathcal {C}}$ , then we compute $S(W_{\bullet , \bullet ,\bullet }^{\widetilde {T},\mathbf {e}};O)=\frac {4}{7}$ . Similarly, if $O\in \widetilde {L}_1\cup \widetilde {L}_2$ , then $S(W_{\bullet , \bullet ,\bullet }^{\widetilde {T},\mathbf {e}};O)=\frac {17}{28}$ . Finally, if $O\in \widetilde {\mathcal {C}}$ , then

$$ \begin{align*} S\big(W_{\bullet, \bullet,\bullet}^{\widetilde{T},\mathbf{e}};O\big)&=\frac{4}{7}+\frac{3}{7}\int_{0}^{1}\int_{2-2u}^{3-u} 2(1-u)\frac{v-2+2 u}{2}\mathrm{ord}_O\big(\widetilde{\mathcal{C}}\big\vert_{\mathbf{e}}\big)dvdu+\\ &\quad+\frac{3}{7}\int_{0}^{1}\int_{3-u}^{4-2u} 2(4-2u-v)\frac{v-2+2 u}{2}\mathrm{ord}_O\big(\widetilde{\mathcal{C}}\big\vert_{\mathbf{e}}\big)dvdu=\frac{4}{7}+\frac{17}{56}\mathrm{ord}_O\big(\widetilde{\mathcal{C}}\big\vert_{\mathbf{e}}\big). \end{align*} $$

Hence, if $\widetilde {\mathcal {C}}$ intersects $\mathbf {e}$ transversally, then $S(W_{\bullet , \bullet ,\bullet }^{\widetilde {T},\mathbf {e}};O)<1$ so that $\delta _P(X)>1$ by Equation (6.3).

Therefore, to complete the proof of the lemma, we may assume that $\widetilde {\mathcal {C}}$ is tangent to $\mathbf {e}$ . This means that $\mathcal {C}$ has a cusp at P, and the intersection $\widetilde {\mathcal {C}}\cap \mathbf {e}$ consists of a single point.

Now, as in the proof of Lemma 4.5, we consider the following commutative diagram:

where $\gamma $ is a composition of the blow up of the point $\widetilde {\mathcal {C}}\cap \mathbf {e}$ with the blow up of the unique intersection point of the proper transforms of the curves $\widetilde {\mathcal {C}}$ and $\mathbf {e}$ , $\upsilon $ is the birational contraction of all $(\sigma \circ \gamma )$ -exceptional curves that are not $(-1)$ -curve, and $\varsigma $ is the birational contraction of the proper transform of the unique $\gamma $ -exceptional curve that is $(-1)$ -curve. Then $\widehat {T}$ has two singular points:

1. (1) a cyclic quotient singularity of type $\frac {1}{2}(1,1)$ , which we denote by $O_2$ ;

2. (2) a cyclic quotient singularity of type $\frac {1}{3}(1,1)$ , which we denote by $O_3$ .

Let $\mathbf {f}$ be the $\varsigma $ -exceptional curve, let $\widehat {\mathcal {C}}$ be the proper transform on $\widehat {T}$ of the curve $\mathcal {C}$ and let $\widehat {L}_1$ and $\widehat {L}_2$ be the proper transforms on $\widehat {T}$ of the curves $L_1$ and $L_2$ , respectively. Then the curves $\mathbf {f}$ , $\widehat {\mathcal {C}}$ , $\widehat {L}_1$ , $\widehat {L}_2$ are smooth, and the curve $\mathbf {f}$ contains both points $O_2$ and $O_3$ , which are not contained in $\widehat {\mathcal {C}}$ . Moreover, we have

$$ \begin{align*}\widehat{L}_1\cap\widehat{L}_2=\widehat{L}_1\cap\mathbf{f}=\widehat{L}_2\cap\mathbf{f}=O_3. \end{align*} $$

Furthermore, we have $A_T(\mathbf {f})=5$ , $\varsigma ^*(\mathcal {C})\sim \widehat {\mathcal {C}}+6\mathbf {f}$ , $\varsigma ^*(L_1)\sim \widehat {L}_1+2\mathbf {f}$ , $\varsigma ^*(L_2)\sim \widehat {L}_2+2\mathbf {f}$ , and the intersection form of the curves $\mathbf {f}$ , $\widehat {\mathcal {C}}$ , $\widehat {L}_1$ and $\widehat {L}_2$ is given in the following table:

For a nonnegative real number v, we have

$$ \begin{align*}\varsigma^*\big(P(u)\big\vert_{T}\big)-v\mathbf{f}\sim_{\mathbb{R}}\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(10-4u-v)\mathbf{f}, \end{align*} $$

which implies that the divisor $\varsigma ^*(P(u)\vert _{T})-v\mathbf {f}$ is pseudoeffective if and only if $v\leqslant 10-4u$ because the intersection form of the curves $\widehat {\mathcal {C}}$ , $\widehat {L}_1$ , $\widehat {L}_2$ is negative definite.

Let $\widehat {P}(u,v)$ be the positive part of Zariski decomposition of the divisor $\varsigma ^*(P(u)\vert _{T})-v\mathbf {f}$ , and let $\widehat {N}(u,v)$ be its negative part. Set $\Delta _{\mathbf {f}}=\frac {1}{2}O_2+\frac {2}{3}O_3$ . By [3, Remark 1.113], we get

(6.4) $$ \begin{align} \delta_P(X)\geqslant\min\left\{\frac{1}{S_X(T)}, \frac{5}{S(W_{\bullet,\bullet}^{\widehat{T}};\mathbf{f})},\inf_{O\in \mathbf{f}}\frac{1-\mathrm{ord}_O\big(\Delta_{\mathbf{f}}\big)}{S(W_{\bullet, \bullet,\bullet}^{\widehat{T},\mathbf{f}};O)}\right\}, \end{align} $$

where $S(W_{\bullet ,\bullet }^{\widehat {T}};\mathbf {f})$ and $S(W_{\bullet , \bullet ,\bullet }^{\widehat {T},\mathbf {f}};O)$ are defined as $S(W_{\bullet ,\bullet }^{\widetilde {T}};\mathbf {e})$ and $S(W_{\bullet , \bullet ,\bullet }^{\widetilde {T},\mathbf {e}};O)$ used earlier. Moreover, it follows from [3, Remark 1.113] that

$$ \begin{align*}S\big(W_{\bullet,\bullet}^{\widehat{T}};\mathbf{f}\big)=\frac{3}{14}\int_0^1\int_0^{10-4u}\big(\widehat{P}(u,v)\big)^2dvdu \end{align*} $$

and

$$ \begin{align*}S\big(W_{\bullet, \bullet,\bullet}^{\widehat{T},\mathbf{f}};O\big)=\frac{3}{14}\int_0^1\!\kern-1pt\int_0^{10-4u}\!\Big(\big(\widehat{P}(u,v)\cdot\mathbf{f}\big)\Big)^2dvdu+\frac{3}{7}\int_0^1\!\kern-1pt\int_0^{10-4u}\!\big(\widehat{P}(u,v)\cdot\mathbf{f}\big)\mathrm{ord}_O\Big(\widehat{N}(u,v)\big|_{\mathbf{f}}\Big)dvdu. \end{align*} $$

Moreover, we compute $\widehat {P}(u,v)$ and $\widehat {N}(u,v)$ as follows:

$$ \begin{align*}\widehat{P}(u,v)=\left\{\begin{aligned} &\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(10-4u-v)\mathbf{f}\ \text{if }0\leqslant v\leqslant 4-4u, \\ &\frac{10-4u-v}{6}\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(10-4u-v)\mathbf{f}\ \text{if }4-4u\leqslant v\leqslant 9-3u, \\ &\frac{10-4u-v}{6}\big(\widehat{\mathcal{C}}+6(\widehat{L}_{1}+\widehat{L}_{2})+6\mathbf{f}\big)\ \text{if }9-3u\leqslant v\leqslant 10-4u, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}\widehat{N}(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant 4-4u, \\ &\frac{v-4+4u}{6}\widehat{\mathcal{C}}\ \text{if }4-4u\leqslant v\leqslant 9-3u, \\ &\frac{v-4+4u}{6}\widehat{\mathcal{C}}+(v-9+3u)(\widetilde{L}_{1}+\widetilde{L}_{2})\ \text{if }9-3u\leqslant v\leqslant 10-4u. \\ \end{aligned} \right. \end{align*} $$

This gives

$$ \begin{align*}\big(\widehat{P}(u,v)\big)^2=\left\{\begin{aligned} &2(1-u)(5-u)-\frac{v^{2}}{6}\ \text{if }0\leqslant v\leqslant 4-4u, \\ &\frac{2(1-u)(19-7u-2v)}{3}\ \text{if }4-4u\leqslant v\leqslant 9-3u, \\ &\frac{2(10-4u-v)^{2}}{3}\ \text{if }9-3u\leqslant v\leqslant 10-4u, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}\widehat{P}(u,v)\cdot\mathbf{f}=\left\{\begin{aligned} &\frac{v}{6}\ \text{if }0\leqslant v\leqslant 4-4u, \\ &\frac{2(1-u)}{3}\ \text{if }4-4u\leqslant v\leqslant 9-3u, \\ &\frac{2(10-4 u-v)}{3}\ \text{if }9-3u\leqslant v\leqslant 10-4u. \\ \end{aligned} \right. \end{align*} $$

Now, integrating, we get $S(W_{\bullet ,\bullet }^{\widehat {T}};\mathbf {f})=\frac {135}{28}<A_{T}(\mathbf {f})=5$ .

Let O be a point in $\mathbf {f}$ . If $O\not \in \widehat {L}_1\cup \widehat {L}_2\cup \widehat {\mathcal {C}}$ , then

$$ \begin{align*} S\big(W_{\bullet, \bullet,\bullet}^{\widehat{T},\mathbf{f}};O\big)&=\frac{3}{14}\int_{0}^{1}\int_{0}^{4-4u}\left(\frac{v}{6}\right)^{2}dvdu+\frac{3}{14}\int_{0}^{1}\int_{4-4u}^{9-3u}\left(\frac{2(1-u)}{3}\right)^{2}dvdu+\\ &\quad+\frac{3}{14}\int_{0}^{1}\int_{9-3u}^{10-4 u}\left(\frac{2(10-4 u-v)}{3}\right)^{2}dvdu=\frac{13}{63}.\quad\quad\quad \end{align*} $$

Similarly, if $O=\mathbf {f}\cap \widetilde {\mathcal {C}}$ , then

$$ \begin{align*} S\big(W_{\bullet, \bullet,\bullet}^{\widehat{T},\mathbf{f}};O\big)&=\frac{13}{63}+ \frac{3}{7}\int_{0}^{1}\int_{4-4u}^{9-3u}\frac{2(1-u)}{3}\times\frac{v-4+4 u}{6}dvdu+\\[5pt] &\quad+\frac{3}{7}\int_{0}^{1}\int_{9-3u}^{10-4u}\frac{2(10-4 u-v)}{3}\times\frac{v-4+4u}{6}dvdu=\frac{13}{63}+\frac{193}{504}=\frac{33}{56}. \end{align*} $$

Likewise, if $O=O_3$ , we compute $S(W_{\bullet , \bullet ,\bullet }^{\widehat {T},\mathbf {f}};O)=\frac {3}{14}$ . So, using Equation (6.4), we get $\delta _P(X)>1$ .

Thus, to prove that $\delta _P(X)>1$ , we may assume that S is singular, but $P\not \in \mathrm {Sing}(S)$ .

Lemma 6.8. Suppose that $P\not \in E$ . Then $\delta _P(X)>1$ .

Proof. Recall that E is the $\pi $ -exceptional surface. Using Corollary 6.4, we may assume that S contains a line L that passes through P. Then $\pi (L)$ is a line in $\mathscr {Q}$ , $\pi (L)\cap \mathscr {C}\ne \varnothing $ , and one of the following cases holds:

1. Case 1: the line $\pi (L)$ intersects the curve $\mathscr {C}$ transversally at two points,

2. Case 2: the line $\pi (L)$ is tangent to the curve $\mathscr {C}$ at their single intersection point.

Now, let us choose T to be a sufficiently general surface in $|H|$ that passes through L. If the intersection $\pi (L)\cap \mathscr {C}$ consists of two points, then $\varpi \colon T\to \mathcal {Q}$ is a blow up of eight distinct points of the transversal intersection $\mathcal {Q}\cap \mathscr {C}$ , which implies that T is smooth. On the other hand, if $L\cap \mathscr {C}$ consists of one point, then T has one ordinary double point, which is not contained in the curve L. We have $\mathcal {C}=S\vert _{T}=L+Z$ , where Z is a smooth rational irreducible curve such that $\pi (Z)$ is a smooth twisted cubic in $\mathscr {Q}$ . The twisted cubic curve $\pi (Z)$ in $\mathscr {Q}$ intersects the curve $\mathscr {C}$ transversally by six distinct points, which we denote by $Q_3,Q_4,Q_5,Q_6,Q_7,Q_8$ . Moreover, if $\pi (L)\cap \mathscr {C}$ consists of two distinct points, we denote these points by $Q_1$ and $Q_2$ . Likewise, if $\pi (L)\cap \mathscr {C}$ consists of one point, we let $Q_1=Q_2=\pi (Z)\cap \mathscr {C}$ . Then

1. Case 1: the morphism $\varpi \colon T\to \mathcal {Q}$ is the blow up of the points $Q_1,Q_2,\ldots ,Q_8$ ,

2. Case 2: the morphism $\varpi \colon T\to \mathcal {Q}$ is a composition of the blow up of the points $Q_3,\ldots ,Q_8$ with a weighted blow up with weights $(1,2)$ of the point $Q_1=Q_2$ .

Since T is a general surface in $|H|$ that contains the line L, we may assume that $P\not \in Z$ . Likewise, we may assume further that Z is contained in the smooth locus of the surface T. Moreover, we may also assume that the quadric surface $\mathcal {Q}$ does not contain lines that pass through one point in the set $\{Q_1,Q_2,\pi (P)\}$ and one point in $\{Q_3,Q_4,Q_5,Q_6,Q_7,Q_8\}$ . Indeed, let $\mathcal {Q}^\prime $ , $\mathcal {Q}^{\prime \prime }$ , $\mathcal {Q}^{\prime \prime \prime }$ be the hyperplane sections of the quadric $\mathscr {Q}$ that are singular at the points $Q_1$ , $Q_2$ , $\pi (P)$ , respectively. Then $\mathcal {Q}^\prime $ , $\mathcal {Q}^{\prime \prime }$ , $\mathcal {Q}^{\prime \prime \prime }$ are cones, $\pi (L)\subset \mathcal {Q}^\prime \cap \mathcal {Q}^{\prime \prime }\cap \mathcal {Q}^{\prime \prime \prime }$ , and every line in $\mathscr {Q}$ containing a point in $\{Q_1,Q_2,\pi (P)\}$ is a ruling of one of these cones. On the other hand, we have $\mathscr {C}\not \subset \mathcal {Q}^\prime \cup \mathcal {Q}^{\prime \prime }\cup \mathcal {Q}^{\prime \prime \prime }$ because $\mathscr {C}$ is not contained in a hyperplane. This implies that the quadric threefold $\mathscr {Q}$ contains at most finitely many lines that pass through a point in $\{Q_1,Q_2,\pi (P)\}$ and a point in $\mathscr {C}\setminus \{Q_1,Q_2\}$ . Therefore, we can choose the surface $T\in |H|$ such that $L\subset T$ , but $\mathcal {Q}=\pi (T)$ does not contain any of these lines.

Let us identify $\mathcal {Q}=\mathbb {P}^1\times \mathbb {P}^1$ such that the line $\pi (L)$ is a curve in $\mathcal {Q}$ of degree $(0,1)$ . Denote by $\mathbf {e}_1,\ldots ,\mathbf {e}_8$ the $\varpi $ -exceptional curves such that $\varpi (\mathbf {e}_1)=Q_1,\ldots ,\varpi (\mathbf {e}_8)=Q_8$ . Let $\mathbf {g}_3,\ldots ,\mathbf {g}_8$ be the strict transforms on T of the curves in $\mathcal {Q}$ of degree $(1,0)$ that pass through the points $Q_3,\ldots ,Q_8$ , respectively. Then $L,Z,\mathbf {e}_1,\ldots ,\mathbf {e}_8,\mathbf {g}_3,\ldots ,\mathbf {g}_8$ are smooth irreducible rational curves. In Case 1, their intersections are given in the following table:

In Case 2, we have $\mathbf {e}_1=\mathbf {e}_2$ , and $\mathbf {e}_1$ contains the singular point of T so that $\mathbf {e}_1^2=-\frac {1}{2}$ . The remaining intersection numbers are exactly the same as in Case 1.

Observe that $P\not \in Z\cup \mathbf {g}_3\cup \mathbf {g}_4\cup \mathbf {g}_5\cup \mathbf {g}_6\cup \mathbf {g}_7\cup \mathbf {g}_8\cup \mathbf {e}_1\cup \mathbf {e}_2$ since $P\not \in E$ by assumption.

Recall that $P(u)=-K_X-uT$ is nef $\iff P(u)$ is pseudoeffective $\iff u\in [0,1]$ . Let v be a nonnegative real number. Then, in both Cases 1 and 2, we have

(6.5) $$ \begin{align} P(u)\big\vert_{T}-vL\sim_{\mathbb{R}}\frac{9-5u-4v}{4}L+\frac{3+u}{4}Z+\frac{5-5u}{4}\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i, \end{align} $$

which implies that the divisor $P(u)\vert _{T}-vL$ is pseudoeffective $\iff v\leqslant \frac {9-5u}{4}$ .

Let $P(u,v)$ be the positive part of Zariski decomposition of the divisor $P(u)\vert _{T}-vL$ , and let $N(u,v)$ be its negative part. Then it follows from [3, Theorem 1.112] that

(6.6) $$ \begin{align} \delta_P(X)\geqslant\min\left\{\frac{1}{S_X(T)},\frac{1}{S(W_{\bullet,\bullet}^{T};L)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{T,L};P)}\right\}, \end{align} $$

where

$$ \begin{align*}S\big(W_{\bullet,\bullet}^{T};L\big)=\frac{3}{14}\int_0^1\int_0^{\frac{9-5u}{4}}\big(P(u,v)\big)^2dvdu \end{align*} $$

and

$$ \begin{align*}S\big(W_{\bullet, \bullet,\bullet}^{T,L};P\big)=\frac{3}{14}\int_0^1\!\int_0^{\frac{9-5u}{4}}\Big(\big(P(u,v)\cdot L\big)\Big)^2dvdu+\frac{3}{7}\int_0^1\!\int_0^{\frac{9-5u}{4}}\big(P(u,v)\cdot L\big)\mathrm{ord}_P\Big(N(u,v)\big|_{L}\Big)dvdu. \end{align*} $$

Let us compute $S(W_{\bullet ,\bullet }^{T};L)$ and $S(W_{\bullet , \bullet ,\bullet }^{T,L};P)$ . If $0\leqslant u\leqslant \frac {1}{3}$ , then, using Equation (6.5), we get

$$ \begin{align*}P(u,v)=\left\{\begin{aligned} &\frac{9-5u-4v}{4}L+\frac{3+u}{4}Z+\frac{5-5u}{4}\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if }0\leqslant v\leqslant 1, \\[-1pt] &\frac{9-5u-4v}{4}\big(L+\mathbf{e}_1+\mathbf{e}_2\big)+\frac{3+u}{4}Z+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if }1\leqslant v\leqslant \frac{3-3u}{2}, \\[-1pt] &\frac{9-5u-4v}{4}\big(L+Z+\mathbf{e}_1+\mathbf{e}_2\big)+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 2-u, \\[-1pt] &\frac{9-5 u-4 v}{4}\Big(L+Z+\mathbf{e}_1+\mathbf{e}_2+\sum_{i=3}^8\mathbf{g}_i\Big)\ \text{if }2-u\leqslant v\leqslant \frac{9-5u}{4}, \\ \end{aligned} \right. \end{align*} $$

$$ \begin{align*}N(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant 1, \\[-1pt] &(v-1)\big(\mathbf{e}_1+\mathbf{e}_2\big) \ \text{if }1\leqslant v\leqslant \frac{3-3u}{2}, \\[-1pt] &(v-1)\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{2v+3u-3}{2}Z\ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 2-u, \\[-1pt] &(v-1)\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{2v+3u-3}{2}Z+(v-2+u)\sum_{i=3}^{8}\mathbf{g}_{i}\ \text{if }2-u\leqslant v\leqslant \frac{9-5u}{4}, \\ \end{aligned} \right. \end{align*} $$

$$ \begin{align*}\big(P(u,v)\big)^2=\left\{\begin{aligned} &2u^2+(2v-12)u-2v^2-2v+10 \ \text{if }0\leqslant v\leqslant 1, \\[-1pt] &2u^2+(2v-12)u-6v+12\ \text{if }1\leqslant v\leqslant \frac{3-3u}{2}, \\[-1pt] &\frac{13u^2+16uv+4v^2-42u-24v+33}{2}\ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 2-u, \\[-1pt] &\frac{(9-5u-4v)^{2}}{2}\ \text{if }2-u\leqslant v\leqslant \frac{9-5u}{4}, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}P(u,v)\cdot L=\left\{\begin{aligned} &1-u+2v\ \text{if }0\leqslant v\leqslant 1, \\[-1pt] &3-u\ \text{if }1\leqslant v\leqslant \frac{3-3u}{2}, \\[-1pt] &6-4u-2v\ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 2-u, \\[-1pt] &2(9-5u-4v)\ \text{if }2-u\leqslant v\leqslant \frac{9-5u}{4}. \\ \end{aligned} \right. \end{align*} $$

Similarly, if $\frac {1}{3}\leqslant u\leqslant 1$ , then, using Equation (6.5), we get

$$ \begin{align*}P(u,v)=\left\{\begin{aligned} &\frac{9-5u-4v}{4}L+\frac{3+u}{4}Z+\frac{5-5u}{4}\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if }0\leqslant v\leqslant \frac{3-3u}{2}, \\[-1pt] &\frac{9-5u-4v}{4}(L+Z)+\frac{5-5u}{4}\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 1, \\[-1pt] &\frac{9-5u-4v}{4}\big(L+Z+\mathbf{e}_1+\mathbf{e}_2\big)+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if }1\leqslant v\leqslant 2-u, \\[-1pt] &\frac{9-5 u-4 v}{4}\Big(L+Z+\mathbf{e}_1+\mathbf{e}_2+\sum_{i=3}^8\mathbf{g}_i\Big)\ \text{if }2-u\leqslant v\leqslant \frac{9-5u}{4}, \\ \end{aligned} \right. \end{align*} $$

$$ \begin{align*}N(u,v)=\left\{\begin{aligned} &0\ \text{if }0\leqslant v\leqslant \frac{3-3u}{2}, \\ &\frac{2v+3u-3}{2}Z\ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 1, \\ &(v-1)\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{2v+3u-3}{2}Z\ \text{if }1\leqslant v\leqslant 2-u, \\ &(v-1)\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{2v+3u-3}{2}Z+(v-2+u)\sum_{i=3}^{8}\mathbf{g}_{i}\ \text{if }2-u\leqslant v\leqslant \frac{9-5u}{4}, \\ \end{aligned} \right. \end{align*} $$

$$ \begin{align*}\big(P(u,v)\big)^2=\left\{\begin{aligned} &2u^2+(2v-12)u-2v^2-2v+10\ \text{if }0\leqslant v\leqslant \frac{3-3u}{2}, \\ &\frac{(1-u)(29-13u-16v)}{2}\ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 1, \\ &\frac{13u^2+16uv+4v^2-42u-24v+33}{2}\ \text{if }1\leqslant v\leqslant 2-u, \\ &\frac{(9-5u-4v)^{2}}{2}\ \text{if }2-u\leqslant v\leqslant \frac{9-5u}{4}, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}P(u,v)\cdot L=\left\{\begin{aligned} &1-u+2v\ \text{if }0\leqslant v\leqslant \frac{3-3u}{2}, \\ &4-4u\ \text{if }\frac{3-3u}{2}\leqslant v\leqslant 1, \\ &6-4u-2v\ \text{if }1\leqslant v\leqslant 2-u, \\ &2(9-5u-4v)\ \text{if }2-u\leqslant v\leqslant \frac{9-5u}{4}. \\ \end{aligned} \right. \end{align*} $$

Therefore, we have $P\not \in \mathrm {Supp}(N(u,v))$ because $P\not \in Z\cup \mathbf {g}_3\cup \mathbf {g}_4\cup \mathbf {g}_5\cup \mathbf {g}_6\cup \mathbf {g}_7\cup \mathbf {g}_8\cup \mathbf {e}_1\cup \mathbf {e}_2$ . So, integrating $(P(u,v))^2$ and $(P(u,v)\cdot L)^2$ , we get $S(W_{\bullet ,\bullet }^{T};L)=\frac {423}{448}$ and $S(W_{\bullet , \bullet ,\bullet }^{T,L};P)=\frac {79}{84}$ , which implies that $\delta _P(X)>1$ by Equation (6.6).

By Lemma 6.8, to show that $\delta _P(X)>1$ , we may assume that $P\in E$ . Then $\pi (P)\in \mathscr {C}$ . Now, let us choose T to be a sufficiently general surface in $|H|$ that contains the point P so that $\mathcal {Q}$ is a sufficiently general hyperplane section of the quadric $\mathscr {Q}$ that contains $\pi (P)$ . Then T is smooth, and $\varpi \colon T\to \mathcal {Q}$ is a blow up of eight points of the intersection $\mathcal {Q}\cap \mathscr {C}$ .

Let $Q_1=\pi (P)$ , let $Q_2,\ldots ,Q_8$ be the remaining seven points of the intersection $\mathcal {Q}\cap \mathscr {C}$ and let $\mathbf {e}_1,\ldots ,\mathbf {e}_8$ be the $\varpi $ -exceptional curves such that $\varpi (\mathbf {e}_1)=Q_1,\ldots ,\varpi (\mathbf {e}_8)=Q_8$ . For every $u\in [0,1]$ , set

$$ \begin{align*}t(u)=\inf\Big\{v\in \mathbb R_{\geqslant 0} \ \big|\ \text{the divisor }P(u)\big|_T-v\mathbf{e}_1\text{ is pseudoeffective}\Big\}, \end{align*} $$

and fix a real number $v\in [0, t(u)]$ . Let $P(u,v)$ and $N(u,v)$ be the positive and the negative parts of the Zariski decomposition of the $\mathbb {R}$ -divisor $P(u)|_T-v\mathbf {e}_1$ , respectively. Then

(6.7) $$ \begin{align} \delta_P(X)\geqslant\min\left\{\frac{1}{S_X(T)},\frac{1}{S(W_{\bullet,\bullet}^{T};\mathbf{e}_1)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{T,\mathbf{e}_1};P)}\right\} \end{align} $$

by [3, Theorem 1.112], where

$$ \begin{align*}S\big(W_{\bullet,\bullet}^{T};\mathbf{e}_1\big)=\frac{3}{14}\int_0^1\int_0^{t(u)}\big(P(u,v)\big)^2dvdu \end{align*} $$

and

$$ \begin{align*} S\big(W_{\bullet, \bullet,\bullet}^{T,\mathbf{e}_1};P\big)&=\frac{3}{14}\int_0^1\int_0^{t(u)}\Big(\big(P(u,v)\cdot\mathbf{e}_1\big)\Big)^2dvdu+\\[5pt] &\quad+\frac{3}{7}\int_0^1\int_0^{t(u)}\big(P(u,v)\cdot \mathbf{e}_1\big)\mathrm{ord}_P\Big(N(u,v)\big|_{\mathbf{e}_1}\Big)dvdu. \end{align*} $$

Let us compute $S(W_{\bullet ,\bullet }^{T};\mathbf {e}_1)$ and $S(W_{\bullet , \bullet ,\bullet }^{T,\mathbf {e}_1};P)$ .

Recall that $\varphi \colon T\to \mathbb {P}^1$ is an elliptic fibration, which is given by the linear system $|-K_T|$ . As in the proof of Lemma 6.8, let us identify $\mathcal {Q}=\mathbb {P}^1\times \mathbb {P}^1$ .

Lemma 6.9. Let F be a curve in $|-K_T|$ . Then F is irreducible and reduced.

Proof. Suppose that F is reducible or nonreduced. Then the curve $\pi (F)$ is also reducible or nonreduced, and every irreducible component of the curve F is a smooth $(-2)$ -curve. But $\pi (F)$ is a curve in $\mathcal {Q}$ of degree $(2,2)$ that passes through the points $Q_1,Q_2,\ldots ,Q_8$ . Therefore, we have one of the following possibilities:

1. (1) $\mathcal {Q}$ contains a line that passes through $Q_1$ and one point among $Q_2,\ldots ,Q_8$ ,

2. (2) $\mathcal {Q}$ contains a line that passes through two points among $Q_2,\ldots ,Q_8$ ,

3. (3) $\mathcal {Q}$ contains a conic that passes through $Q_1$ and three points among $Q_2,\ldots ,Q_8$ .

Recall that $\mathcal {Q}$ is a general hyperplane section of the quadric $\mathscr {Q}$ that contains $Q_1=\pi (P)$ . As we already mentioned in the proof of Lemma 6.8, the quadric $\mathscr {Q}$ contains finitely many lines that pass through $Q_1$ and a point in $\mathscr {C}\setminus Q_1$ . Thus, since $\mathcal {Q}$ is assumed to be general, the quadric $\mathcal {Q}$ does not contain any of these lines so that $\mathcal {Q}$ does not contain a line that passes through $Q_1$ and a point among $Q_2,\ldots ,Q_8$ .

Similarly, a parameter count implies that $\mathcal {Q}$ does not contain secant lines of the curve $\mathscr {C}$ so that $\mathcal {Q}$ does not contain a line that passes through two points among $Q_2,\ldots ,Q_8$ ,

Finally, we suppose that $\mathcal {Q}$ contains an irreducible conic C that passes through $Q_1$ and three points among $Q_2,\ldots ,Q_8$ . Let be a linear projection from the point $Q_1$ . Then $\rho (\mathscr {C})$ is a curve of degree $7$ , and the induced map is an isomorphism because $\mathscr {C}$ is an intersection of quadrics. Similarly, all points $\rho (Q_2),\ldots ,\rho (Q_8)$ are distinct. Then $\rho (C)$ is a three-secant line of the curve $\rho (\mathscr {C})$ . Note that $\rho (\mathscr {C})$ contains one-parameter family of three-secants [11, Appendix A]. But $\rho (\mathcal {Q})$ is a general plane in $\mathbb {P}^3$ , which implies that $\rho (\mathcal {Q})$ does not contain three-secant lines of the curve $\rho (\mathscr {C})$ – a contradiction.

Corollary 6.10. Let $\gamma $ be a class in the group $\mathrm {Pic}(T)$ such that $-K_T\cdot \gamma =1$ and $\gamma ^2=-1$ . Then the linear system $|\gamma |$ consists of a unique $(-1)$ -curve.

Proof. Apply the Riemann–Roch theorem, Serre duality and Lemma 6.9.

Let us use Corollary 6.10 to describe infinitely many $(-1)$ -curves in the surface T. Namely, let $\ell _1$ and $\ell _2$ be any curves in $\mathcal {Q}=\mathbb {P}^1\times \mathbb {P}^1$ of degrees $(1,0)$ and $(0,1)$ , respectively. For $n\in \mathbb {Z}_{\geqslant 0}$ and $i\in \{2,3,4,5,6,7,8\}$ , let $B_{n,1,1}$ , $B_{n,1,2}$ , $B_{n,2,i}$ , $B_{n,3}$ , $B_{n,4,i}$ be the classes

$$ \begin{align*}\varpi^*\big(a_1\ell_1+a_2\ell_2\big)-\sum_{i=1}^8b_i\mathbf{e}_i\in\mathrm{Pic}(T), \end{align*} $$

where $a_1,a_2,b_1,\ldots ,b_8$ are nonnegative integers given in the following table:

By Corollary 6.10, each linear system $|B_{n,1,1}|$ , $|B_{n,1,2}|$ , $|B_{n,2,i}|$ , $|B_{n,3}|$ , $|B_{n,4,i}|$ contains a unique $(-1)$ -curve. Hence, we can identify the classes $B_{n,1,1}$ , $B_{n,1,2}$ , $B_{n,2,i}$ , $B_{n,3}$ , $B_{n,4,i}$ with $(-1)$ -curves in $|B_{n,1,1}|$ , $|B_{n,1,2}|$ , $|B_{n,2,i}|$ , $|B_{n,3}|$ , $|B_{n,4,i}|$ , respectively. Set

$$ \begin{align*} B_{n, 1}&=B_{n, 1, 1}+B_{n, 1, 2},\\ B_{n, 2}&=B_{n,2,2}+B_{n,2,3}+B_{n,2,4}+B_{n,2,5}+B_{n,2,6}+B_{n,2,7}+B_{n,2,8},\\ B_{n, 4}&=B_{n,4,2}+B_{n,4,3}+B_{n,4,4}+B_{n,4,5}+B_{n,4,6}+B_{n,4,7}+B_{n,4,8}. \end{align*} $$

Note that irreducible components of each curve $B_{n,1}$ , $B_{n,2}$ , $B_{n,4}$ are disjoint $(-1)$ -curves, and $B_{n,1}\cap B_{n,2}=\varnothing $ , $B_{n,2}\cap B_{n,3}=\varnothing $ , $B_{n,3}\cap B_{n,4}=\varnothing $ , $B_{n,4} \cap B_{n+1,1}=\varnothing $ for each $n\geqslant 0$ .

Now, we let $I^\prime _{0,1}=[0,\frac {1}{3}]$ and $I^{\prime \prime }_{0,1}=[\frac {1}{3},\frac {3}{8}]$ . For every $n\in \mathbb {Z}_{> 0}$ , we also let

$$ \begin{align*} I_{n, 1}^{\prime}&=\left[\frac{-1+4 n+14 n^{2}}{6 n+14 n^{2}}, \frac{1+13 n+21 n^{2}}{3+16 n+21 n^{2}}\right],\\ I_{n, 1}^{\prime \prime}&=\left[\frac{1+13n+21n^{2}}{3+16n+21n^{2}}, \frac{3+35n+49 n^{2}}{8+42 n+49 n^{2}}\right]. \end{align*} $$

For every $n\in \mathbb {Z}_{\geqslant 0}$ , we let

$$ \begin{align*} I_{n,2}^{\prime}&=\left[\frac{3+35 n+49n^{2}}{8+42 n+49 n^{2}}, \frac{3+22 n+28 n^{2}}{6+26 n+28 n^{2}}\right],\\ I_{n,2}^{\prime\prime}&=\left[\frac{3+22n+28 n^{2}}{6+26 n+28 n^{2}}, \frac{2+7 n}{3+7 n}\right],\\ I_{n,3}^{\prime}&=\left[\frac{2+7 n}{3+7 n}, \frac{21+50 n+28 n^{2}}{26+54 n+28 n^{2}}\right],\\ I_{n, 3}^{\prime\prime}&=\left[\frac{21+50 n+28 n^{2}}{26+54 n+28 n^{2}}, \frac{39+91 n+49 n^{2}}{48+98 n+49 n^{2}}\right],\\ I_{n,4}^{\prime}&=\left[\frac{39+91 n+49 n^{2}}{48+98 n+49 n^{2}}, \frac{19+41 n+21 n^{2}}{23+44 n+21 n^{2}}\right],\\ I_{n,4}^{\prime\prime}&=\left[\frac{19+41 n+21 n^{2}}{23+44 n+21 n^{2}}, \frac{17+32 n+14 n^{2}}{20+34 n+14 n^{2}}\right]. \end{align*} $$

Set $I_{n,1}=I_{n,1}^\prime \cup I_{n,1}^{\prime \prime }$ , $I_{n,2}=I_{n,2}^\prime \cup I_{n,2}^{\prime \prime }$ , $I_{n,3}=I_{n,3}^\prime \cup I_{n,3}^{\prime \prime }$ , $I_{n,4}=I_{n,4}^\prime \cup I_{n,4}^{\prime \prime }$ . Then

$$ \begin{align*}[0,1)=\bigcup_{n \in \mathbb{Z}_{\geqslant 0}} \Big(I_{n, 1} \cup I_{n, 2} \cup I_{n, 3} \cup I_{n, 4}\Big), \end{align*} $$

the intervals $I_{n,1}$ , $I_{n,2}$ , $I_{n,3}$ , $I_{n,4}$ have positive volumes, and all their interiors are disjoint. Let us analyze $P(u,v)$ and $N(u,v)$ when u is contained in one of these intervals.

First, we deal with $u\in I_{n,1}$ . If $u\in I_{n,1}^{\prime }$ and $v\in \left [0,\frac {2+14 n+28 n^{2}-u(1+14 n+28 n^{2})}{1+7 n+7 n^{2}}\right ]$ , then

$$ \begin{align*} P(u, v)&=\frac{19+70 n+84 n^{2}-u\left(16+70 n+84 n^{2}\right)-v\left(8+28 n+21 n^{2}\right)}{8+28 n+21 n^{2}}\mathbf{e}_{1}+\\ &\quad+\frac{3+35n+49 n^{2}-u\left(8+42 n+49 n^{2}\right)}{8+28 n+21 n^{2}}B_{n, 1}+ \frac{1-4n-14n^{2}+u\left(6 n+14 n^{2}\right)}{8+28 n+21 n^{2}}B_{n, 2} \end{align*} $$

and $N(u,v)=0$ . The same holds if $u\in I_{n,1}^{\prime \prime }$ and $v\in \left [0,\frac {7+26 n+28n^{2}-u(6+26 n+28 n^{2})}{3+10 n+7 n^{2}}\right ]$ . Then

$$ \begin{align*} \big(P(u,v)\big)^{2}&=10-12u+2u^{2}-2v-v^{2},\\ P(u,v)\cdot\mathbf{e}_1&=1+v. \end{align*} $$

Similarly, if $u\in I_{n,1}^\prime $ and $v\in \left [\frac {2+14 n+28 n^{2}-u\left (1+14 n+28 n^{2}\right )}{1+7 n+7 n^{2}}, \frac {7+26 n+28 n^{2}-u\left (6+26 n+28 n^{2}\right )}{3+10 n+7 n^{2}}\right ]$ , then

$$ \begin{align*} P(u, v)&=\frac{19+70n+84n^{2}-u(16+70 n+84 n^{2})-v(8+28 n+21 n^{2})}{8+28 n+21 n^{2}}\mathbf{e}_{1}+\\ &\quad+\frac{\big(19+70n+84n^{2}-u(16+70 n+84 n^{2})-v(8+28 n+21 n^{2})\big)(1+7n+7n^{2})}{8+28n+21n^{2}}B_{n,1}+\\ &\quad+\frac{1-4n-14n^{2}+u(6 n+14 n^{2})}{8+28 n+21 n^{2}}B_{n,2} \end{align*} $$

and

$$ \begin{align*}N(u,v)=\left(u\left(1+14 n+28 n^{2}\right)+v\left(1+7n+7n^{2}\right)-2-14n-28n^{2}\right)B_{n,1}. \end{align*} $$

Then

$$ \begin{align*} \big(P(u,v)\big)^{2}&=10-12 u+2 u^{2}-2 v-v^{2}+\\ &\quad+2\left(u\left(1+14 n+28 n^{2}\right)+v\left(1+7 n+7 n^{2}\right)-2-14n-28n^{2}\right)^{2}\quad\quad\quad\quad \end{align*} $$

and

$$ \begin{align*} P(u,v)\cdot\mathbf{e}_1&=5+56n+280n^2+588n^3+392n^4-\\ &\quad-2u(1+7n+7n^2)(1+14n+28n^2)-v(1+28n+126n^2+196n^3+98n^4). \end{align*} $$

Likewise, if $u\in I_{n,1}^{\prime \prime }$ and $v \in \left [\frac {7+26 n+28 n^{2}-u\left (6+26 n+28 n^{2}\right )}{3+10 n+7 n^{2}}, \frac {2+14 n+28 n^{2}-u\left (1+14 n+28 n^{2}\right )}{1+7 n+7 n^{2}}\right ]$ , then

$$ \begin{align*} P(u, v)&=\frac{(19+70 n+84 n^{2}-u(16+70 n+84 n^{2})-v(8+28 n+21 n^{2})}{8+28 n+21 n^{2}}\mathbf{e}_{1}+\\ &\quad+\frac{\big(19+70 n+84 n^{2}-u(16+70 n+84 n^{2})-v(8+28 n+21 n^{2})\big)(1+n)(3+7n)}{8+28 n+21 n^{2}}B_{n,2}+\\ &\quad+\frac{3+35 n+49 n^{2}-u\left(8+42 n+49 n^{2}\right)}{8+28 n+21 n^{2}}B_{n,1} \end{align*} $$

and

$$ \begin{align*}N(u,v)=\left(u\left(6+26 n+28 n^{2}\right)+v\left(3+10 n+7 n^{2}\right)-7-26n-28n^{2}\right)B_{n,2}. \end{align*} $$

Then

$$ \begin{align*} &\big(P(u,v)\big)^{2}=10-12 u+2 u^{2}-2 v-v^{2}+\\ &\quad+7\left(u\left(6+26 n+28 n^{2}\right)+v\left(3+10 n+7 n^{2}\right)-7-26n-28n^{2}\right)^{2}\end{align*} $$

and

$$ \begin{align*} P(u,v)\cdot\mathbf{e}_1&=148+1036 n+2751 n^{2}+3234 n^{3}+1372 n^{4}-\\ &\quad-14u(1+n)(1+2 n)(3+7 n)^{2}-v\left(62+420 n+994 n^{2}+980 n^{3}+343 n^{4}\right). \end{align*} $$

If $u \in I_{n, 1}^{\prime }$ and $v\in \left [\frac {7+26 n+28 n^{2}-u\left (6+26 n+28 n^{2}\right )}{3+10 n+7 n^{2}}, \frac {19+70 n+84 n^{2}-u\left (16+70 n+84 n^{2}\right )}{8+28 n+21 n^{2}}\right ]$ , then

$$ \begin{align*} P(u,v)&=\frac{19+70 n+84 n^{2}-u\left(16+70 n+84 n^{2}\right)-v\left(8+28 n+21 n^{2}\right)}{8+28 n+21 n^{2}}\mathbf{e}_1+\\ &\quad+\frac{\big(19+70 n+84 n^{2}-u\left(16+70 n+84 n^{2}\right)-v\left(8+28 n+21 n^{2}\right)\big)\left(1+7n+7 n^{2}\right)}{8+28 n+21 n^{2}}B_{n,1}+\\ &\quad+\frac{\big(19+70 n+84 n^{2}-u\left(16+70 n+84 n^{2}\right)-v\left(8+28 n+21 n^{2}\right)\big)(1+n)(3+7n)}{8+28 n+21 n^{2}}B_{n,2} \end{align*} $$

and

$$ \begin{align*} N(u,v)&=\left(u\left(1+14 n+28 n^{2}\right)+v\left(1+7 n+7 n^{2}\right)-2-14n-28n^{2}\right)B_{n,1}+ \\ &\quad+\left(u\left(6+26 n+28 n^{2}\right)+v\left(3+10 n+7 n^{2}\right)-7-26n-28n^{2}\right)B_{n,2}. \end{align*} $$

The same holds if $u\in I_{n,1}^{\prime \prime }$ and $v\in \left [\frac {2+14 n+28 n^{2}-u\left (1+14 n+28 n^{2}\right )}{1+7 n+7 n^{2}}, \frac {19+70 n+84 n^{2}-u\left (16+70 n+84 n^{2}\right )}{8+28 n+21 n^{2}}\right ]$ . In both cases, we have

$$ \begin{align*} \big(P(u,v)\big)^{2}&=\left(19+70 n+84 n^{2}-u\left(16+70 n+84 n^{2}\right)-v\left(8+28 n+21 n^{2}\right)\right)^{2},\\ P(u,v)\cdot\mathbf{e}_1&=\left(8+28 n+21 n^{2}\right)\left(19+70 n+84 n^{2}-u\left(16+70 n+84 n^{2}\right)-v\left(8+28 n+21 n^{2}\right)\right). \end{align*} $$

Hence, if $u\in I_{n,1}$ , then

$$ \begin{align*}t(u)=\frac{19+70 n+84 n^{2}-u\left(16+70 n+84 n^{2}\right)}{8+28 n+21 n^{2}}. \end{align*} $$

Now, we deal with $u\in I_{n,2}$ . If $u\in I_{n,2}^{\prime }$ and $v\in \left [0,\frac {7+26 n+28 n^{2}-u\left (6+26 n+28 n^{2}\right )}{3+10 n+7 n^{2}}\right ]$ , then

$$ \begin{align*} P(u,v)&=\frac{17+56 n+56 n^{2}-u\left(15+56 n+56 n^{2}\right)-7v(1+n)(1+2 n)}{7(1+n)(1+2n)}\mathbf{e}_{1}+ \\ &\quad+\frac{(1+n)(2+7 n)-u(1+n)(3+7n)}{7(1+n)(1+2n)} B_{n, 2}+\frac{u\left(8+42 n+49 n^{2}\right)-3-35n-49n^{2}}{7(1+n)(1+2n)}B_{n, 3} \end{align*} $$

and $N(u,v)=0$ . The same holds if $u\in I_{n,2}^{\prime \prime }$ and $v\in \left [0,\frac {15+42n+28n^{2}-u\left (14+42 n+28 n^{2}\right )}{6+14 n+7 n^{2}}\right ]$ . Then

$$ \begin{align*} \big(P(u,v)\big)^{2}&=10-12u+2u^{2}-2v-v^{2},\\ P(u,v)\cdot\mathbf{e}_1&=v+1. \end{align*} $$

If $u\in I_{n,2}^\prime $ and $v \in \left [\frac {7+26 n+28 n^{2}-u\left (6+26 n+28 n^{2}\right )}{3+10 n+7 n^{2}}, \frac {15+42 n+28 n^{2}-u\left (14+42 n+28 n^{2}\right )}{6+14 n+7 n^{2}}\right ]$ , then

$$ \begin{align*} P(u,v)&=\frac{17+56n+56 n^{2}-u(15+56 n+56 n^{2})-7v(1+n)(1+2n)}{7(1+n)(1+2n)}\mathbf{e}_{1}+\\&\quad +\frac{(1+n)(3+7n)\big(17+56n+56 n^{2}-u(15+56 n+56 n^{2})-7v(1+n)(1+2n)\big)}{7(1+n)(1+2n)}B_{n, 2}+\\ &\quad+\frac{u\left(8+42 n+49 n^{2}\right)-3-35n-49n^{2}}{7(1+n)(1+2 n)}B_{n,3}, \end{align*} $$

$$ \begin{align*}N(u,v)=\big(u\left(6+26 n+28 n^{2}\right)+v\left(3+10 n+7n^{2}\right)-7-26n-28n^{2}\big)B_{n, 2}, \end{align*} $$

$$ \begin{align*} \big(P(u,v)\big)^{2}&=10-12 u+2 u^{2}-2 v-v^{2}+\\ &\quad+7\left(u\left(6+26 n+28 n^{2}\right)+v\left(3+10 n+7 n^{2}\right)-7-26n-28n^{2}\right)^{2}, \end{align*} $$

$$ \begin{align*} P(u,v)\cdot\mathbf{e}_1&=148+1036 n+2751 n^{2}+3234 n^{3}+1372 n^{4}-\\ &\quad-14u(1+n)(1+2 n)(3+7 n)^{2}-\\ &\quad-v\left(62+420 n+994 n^{2}+980 n^{3}+343 n^{4}\right). \end{align*} $$

Similarly, if $u\in I_{n,2}^{\prime \prime }$ and $v\in \left [\frac {15+42 n+28 n^{2}-u\left (14+42 n+28 n^{2}\right )}{6+14 n+7 n^{2}}, \frac {7+26 n+28 n^{3}-u\left (6+26 n+28 n^{2}\right )}{3+10 n+7 n^{2}}\right ]$ , then

$$ \begin{align*} P(u,v)&=\frac{17+56+56 n^{2}-u\left(15+56 n+56 n^{2}\right)-7v(1+n)(1+2 n)}{7(1+n)(1+2n)}\mathbf{e}_{1}+\\ &\quad+\frac{(6+14 n+7 n^{2})\big(17+56+56 n^{2}-u\left(15+56 n+56 n^{2}\right)-7v(1+n)(1+2 n)\big)}{7(1+n)(1+2 n)}B_{n, 3}+\\ &\quad+\frac{(1+n)(2+7n)-u(1+n)(3+7n)}{7(1+n)(1+2n)}B_{n,2}, \end{align*} $$

$$ \begin{align*}N(u,v)=\big(u\left(14+42n+28n^{2}\right)+v\left(6+14n+7n^{2}\right)-15-42n-28n^{2}\big)B_{n,3}, \end{align*} $$

$$ \begin{align*} \big(P(u, v)\big)^{2}=10-12 u+2 u^{2}-2 v-v^{2}+\\ +\left(u\left(14+42 n+28 n^{2}\right)+v\left(6+14 n+7 n^{2}\right)-15-42n-28n^{2}\right)^{2}, \end{align*} $$

$$ \begin{align*} P(u,v)\cdot\mathbf{e}_1&=7(1+n)(13+53 n+70n^{2}+28 n^{3})-\\&\quad -14u(1+n)(1+2n)\left(6+14 n+7 n^{2}\right)-7v(1+n)^2\left(5+14 n+7 n^{2}\right). \end{align*} $$

Likewise, if $u\in I_{n,2}^{\prime }$ and $v\in \left [\frac {15+42 n+28 n^{2}-u\left (14+42 n+28 n^{2}\right )}{6+14 n+7 n^{2}}, \frac {17+56 n+56 n^{2}-u\left (15+56 n+56 n^{2}\right )}{7(1+n)(1+2 n)}\right ]$ , then

$$ \begin{align*} P(u,v)&=\frac{17+56n+56 n^{2}-u\left(15+56 n+56 n^{2}\right)-v(7(1+n)(1+2 n))}{7(1+n)(1+2n)}\mathbf{e}_{1}+\\ &\quad+\frac{(1+n)(3+7n)\big(17+56n+56 n^{2}-u\left(15+56 n+56 n^{2}\right)-7v(1+n)(1+2 n)\big)}{7(1+n)(1+2n)}B_{n, 2}+\\ &\quad+\frac{\left(6+14 n+7 n^{2}\right)\big(17+56n+56 n^{2}-u\left(15+56 n+56 n^{2}\right)-7v(1+n)(1+2n)\big)}{7(1+n)(1+2n)}B_{n,3} \end{align*} $$

and

$$ \begin{align*} N(u,v)&=\big(u(6+26 n+28 n^{2})+v(3+10 n+7 n^{2})-7-26n-28n^{2}\big) B_{n, 2}+\\ &\quad+\big(u\left(14+42 n+28 n^{2}\right)+v\left(6+14 n+7 n^{2}\right)-15-42n-28 n^{2}\big)B_{n,3}. \end{align*} $$

The same holds if $u\in I_{n,2}^{\prime \prime }$ and $v\in \left [\frac {7+26 n+28 n^{2}-u\left (6+26 n+28 n^{2}\right )}{3+10 n+7 n^{2}}, \frac {17+56 n+56 n^{2}-u\left (15+56 n+56 n^{2}\right )}{7(1+n)(1+2 n)}\right ]$ . Moreover, in both cases, we have

$$ \begin{align*}P(u,v)\cdot\mathbf{e}_1=14(1+n)(1+2 n)\left(17+56 n+56 n^{2}-u\left(15+56 n+56 n^{2}\right)-7v(1+n)(1+2 n)\right) \end{align*} $$

and

$$ \begin{align*}\big(P(u,v)\big)^{2}=2 \left(17+56 n+56 n^{2}-u\left(15+56 n+56 n^{2}\right)-7v(1+n)(1+2n)\right)^{2}. \end{align*} $$

Thus, if $u\in I_{n,2}$ , then

$$ \begin{align*}t(u)=\frac{17+56n+56n^{2}-u\left(15+56n+56n^{2}\right)}{7(1+n)(1+2n)}. \end{align*} $$

Now, we deal with $u\in I_{n,3}$ . If $u\in I_{n,3}^{\prime }$ and $v\in \left [0,\frac {15+42 n+28 n^{2}-u\left (14+42 n+28 n^{2}\right )}{6+14 n+7 n^{2}}\right ]$ , then

$$ \begin{align*} P(u,v)&=\frac{59+112 n+56 n^{2}-u\left(57+112 n+56n^{2}\right)-7v(1+n)(3+2n)}{7(1+n)(3+2n)}\mathbf{e}_{1}+ \\ &\quad+\frac{39+91 n+49 n^{2}-u\left(48+98 n+49 n^{2}\right)}{7(1+n)(3+2n)}B_{n, 3}+\frac{u(1+n)(3+7n)-(1+n)(2+7n)}{7(1+n)(3+2n)}B_{n,4} \end{align*} $$

and $N(u,v)=0$ . The same holds if $u\in I_{n,3}^{\prime \prime }$ and $v\in \left [0,\frac {31+58 n+28 n^{2}-u\left (30+58 n+28 n^{2}\right )}{11+18 n+7 n^{2}}\right ]$ . Then

$$ \begin{align*} \big(P(u,v)\big)^{2}&=10-12u+2u^{2}-2v-v^{2},\\ P(u,v)\cdot\mathbf{e}_1&=1+v. \end{align*} $$

If $u\in I_{n,3}^\prime $ and $v \in \left [\frac {15+42 n+28 n^{2}-u\left (14+42 n+28 n^{2}\right )}{6+14 n+7 n^{2}}, \frac {31+58 n+28 n^{2}-u\left (30+58 n+28 n^{2}\right )}{11+18 n+7 n^{2}}\right ]$ , then

$$ \begin{align*} P(u,v)&=\frac{59+112n+56 n^{2}-u\left(57+112 n+56 n^{2}\right)-7v(1+n)(3+2n)}{7(1+n)(3+2n)}\mathbf{e}_{1}+\\ &\quad+\frac{\left(6+14n+7 n^{2}\right)\big(59+112n+56 n^{2}-u\left(57+112 n+56 n^{2}\right)-7v(1+n)(3+2n)\big)}{7(1+n)(3+2n)}B_{n, 3}+\\&\quad +\frac{u(1+n)(3+7n)-(1+n)(2+7n)}{7(1+n)(3+2n)}B_{n,4} \end{align*} $$

$$ \begin{align*}N(u,v)=\big(u\left(14+42 n+28 n^{2}\right)+v\left(6+14 n+7 n^{2}\right)-15-42n-28n^{2}\big)B_{n,3}, \end{align*} $$

$$ \begin{align*} \big(P(u, v)\big)^{2}&=10-12 u+2 u^{2}-2 v-v^{2}+\\ &\quad+\left(u\left(14+42 n+28 n^{2}\right)+v\left(6+14 n+7 n^{2}\right)-15-42n-28 n^{2}\right)^{2}, \end{align*} $$

$$ \begin{align*} P(u,v)\cdot\mathbf{e}_1&=7(1+n)\big(13+53 n+70 n^{2}+28 n^{3}\big)-\\&\quad -14u(1+n)(1+2n)\left(6+14 n+7 n^{2}\right)-7v(1+n)^2\left(5+14n+7n^{2}\right). \end{align*} $$

Similarly, if $u\in I_{n,3}^{\prime \prime }$ and $v \in \left [\frac {31+58 n+28 n^{2}-u\left (30+58 n+28 n^{2}\right )}{11+18 n+7 n^{2}}, \frac {15+42 n+28 n^{2}-u\left (14+42 n+28 n^{2}\right )}{6+14 n+7 n^{2}}\right ]$ , then

$$ \begin{align*} P(u,v)&=\frac{59+112n+56n^{2}-u(57+112n+56n^{2})-7v(1+n)(3+2n)}{7(1+n)(3+2n)}\mathbf{e}_{1}+\\ &\quad+\frac{(1+n)(11+7n)\big(59+112n+56n^{2}-u(57+112n+56n^{2})-7v(1+n)(3+2n)\big)}{7(1+n)(3+2n)}B_{n,4}+\\ &\quad+\frac{39+91n+49n^{2}-u\left(48+98n+49n^{2}\right)}{7(1+n)(3+2n)}B_{n,3},\end{align*} $$

$$ \begin{align*}N(u,v)=\big(u\left(30+58n+28n^{2}\right)+v\left(11+18n+7 n^{2}\right)-31-58n-28n^{2}\big)B_{n,4}, \end{align*} $$

$$ \begin{align*} \big(P(u, v)\big)^{2}&=10-12 u+2 u^{2}-2v-v^{2}+\\ &\quad+7\left(u\left(30+58n+28n^{2}\right)+v\left(11+18n+7n^{2}\right)-31-58n-28n^{2}\right)^{2}, \end{align*} $$

$$ \begin{align*} P(u,v)\cdot\mathbf{e}_1&=2388+8372 n+10983 n^{2}+6370 n^{3}+1372 n^{4}-\\ &\quad-14u(1+n)^{2}(11+7n)(15+14n)-v\left(846+2772 n+3346 n^{2}+1764 n^{3}+343 n^{4}\right). \end{align*} $$

If $u\in I_{n,3}^{\prime }$ and $v\in \left [\frac {31+58 n+28 n^{2}-u\left (30+58 n+28 n^{2}\right )}{11+18 n+7 n^{2}}, \frac {59+112 n+56 n^{2}-u\left (57+112 n+56 n^{2}\right )}{7(1+n)(3+2 n)}\right ]$ , then

$$ \begin{align*} P(u, v)&=\frac{59+112 n+56 n^{2}-u\left(57+112 n+56 n^{2}\right)-7v(1+n)(3+2n)}{7(1+n)(3+2n)}\mathbf{e}_1+\\&\quad +\frac{\left(6+14 n+7 n^{2}\right)\big(59+112 n+56 n^{2}-u\left(57+112 n+56 n^{2}\right)-7v(1+n)(3+2n)\big)}{7(1+n)(3+2n)}B_{n,3 }+\\ &\quad+\frac{(1+n)\left(11+7n\right)\big(59+112 n+56 n^{2}-u\left(57+112 n+56 n^{2}\right)-7v(1+n)(3+2n)\big)}{7(1+n)(3+2n)}B_{n, 4} \end{align*} $$

and

$$ \begin{align*} N(u,v)&=\left(u\left(14+42 n+28 n^{2}\right)+v\left(6+14 n+7 n^{2}\right)-15-42n-28n^{2}\right)B_{n,3}+\\ &\quad+\left(u\left(30+58n+28n^{2}\right)+v\left(11+18n+7n^{2}\right)-31-58n-28n^{2}\right)B_{n, 4}. \end{align*} $$

The same holds if $u\in I_{u,3}^{\prime \prime }$ and $v\in \left [\frac {15+42 n+28 n^{2}-u\left (14+42 n+28 n^{2}\right )}{6+14 n+7 n^{2}}, \frac {59+112 n+56 n^{2}-u\left (57+112 n+56 n^{2}\right )}{7(1+n)(3+2n)}\right ]$ . In both cases, we have

$$ \begin{align*}P(u,v)\cdot\mathbf{e}_1=14(1+n)(3+2n)\left(59+112n+56 n^{2}-u\left(57+112 n+56 n^{2}\right)-7v(1+n)(3+2 n)\right) \end{align*} $$

and

$$ \begin{align*}\big(P(u,v)\big)^{2}=2\left(59+112n+56n^{2}-u\left(57+112n+56n^{2}\right)-7v(1+n)(3+2n)\right)^{2}. \end{align*} $$

Therefore, if $u\in I_{n,3}$ , then

$$ \begin{align*}t(u)=\frac{59+112n+56n^{2}-u\left(57+112n+56n^{2}\right)}{7(1+n)(3+2n)}. \end{align*} $$

Finally, we deal with $u\in I_{n,4}$ . If $u\in I_{n,4}^{\prime }$ and $v\in \left [0,\frac {31+58n+28n^{2}-u\left (30+58n+28n^{2}\right )}{11+18n+7n^{2}}\right ]$ , then

$$ \begin{align*} P(u, v)&=\frac{103+182 n+84 n^{2}-u\left(100+182 n+84 n^{2}\right)-v\left(36+56 n+21 n^{2}\right)}{36+56n+21 n^{2}}\mathbf{e}_{1}+\\ &\quad+\frac{17+32 n+14 n^{2}-u\left(20+34 n+14 n^{2}\right)}{36+56n+21n^{2}}B_{n,4}+\frac{u\left(48+98 n+49 n^{2}\right)-39-91n-49n^{2}}{36+56n+21 n^{2}}B_{n+1,1} \end{align*} $$

and $N(u,v)=0$ . The same holds when $u\in I_{n,4}^{\prime \prime }$ and $v\in \left [0,\frac {44+70 n+28 n^{2}-u\left (43+70 n+28 n^{2}\right )}{15+21 n+7 n^{2}}\right ]$ . In both cases, we compute

$$ \begin{align*} \big(P(u,v)\big)^{2}&=10-12u+2u^{2}-2 v-v^{2},\\ P(u,v)\cdot\mathbf{e}_1&=1+v. \end{align*} $$

If $u\in I_{n,4}^\prime $ and $v \in \left [\frac {31+58 n+28 n^{2}-u\left (30+58 n+28 n^{2}\right )}{11+18 n+7 n^{2}}, \frac {44+70 n+28 n^{2}-u\left (43+70 n+28 n^{2}\right )}{15+21 n+7 n^{2}}\right ]$ , then

$$ \begin{align*} P(u, v)&=\frac{103+182n+84n^{2}-u(100+182n+84n^{2})-v(36+56n+21n^{2})}{36+56n+21n^{2}}\mathbf{e}_{1}+\\ &\quad+\frac{(1+n)(11+7n)\big(103+182n+84n^{2}-u(100+182n+84n^{2})-v(36+56n+21n^{2})\big)}{36+56n+21n^{2}}B_{n,4}+\\ &\quad+\frac{u(48+98n+49n^{2})-39-91n-49n^{2}}{36+56n+21n^{2}}B_{n+1,1}, \end{align*} $$

$$ \begin{align*}N(u,v)=\left(u\left(30+58 n+28 n^{2}\right)+v\left(11+18 n+7 n^{2}\right)-31-58n-28n^{2}\right)B_{n,4}, \end{align*} $$

$$ \begin{align*} \big(P(u,v)\big)^{2}&=10-12 u+2 u^{2}-2 v-v^{2}+\\ &\quad+7\left(u\left(30+58 n+28 n^{2}\right)+v\left(11+18 n+7 n^{2}\right)-31-58n-28n^{2}\right)^{2}, \end{align*} $$

$$ \begin{align*} P(u,v)\cdot\mathbf{e}_1&=2388+8372 n+10983 n^{2}+6370 n^{2}+1372 n^{4}-\\ &\quad-14u(1+n)^{2}(11+7n)(15+14 n)-\\ &\quad-v\left(846+2772 n+3346 n^{2}+1764 n^{3}+343 n^{4}\right). \end{align*} $$

If $u\in I_{n,4}^{\prime \prime }$ and $v \in \left [\frac {44+70 n+28 n^{2}-u\left (43+70 n+28 n^{2}\right )}{15+21 n+7 n^{2}}, \frac {31+58 n+28 n^{2}-u\left (30+58 n+28 n^{2}\right )}{11+18 n+7 n^{2}}\right ]$ , then

$$ \begin{align*} P(u,v)&=\frac{103+182n+84n^{2}-u(100+182n+84n^{2})-v(36+56n+21n^{2})}{36+56 n+21n^{2}}\mathbf{e}_{1}+\\ &\quad+\frac{(15+21n+7n^{2})\big(103+182n+84n^{2}-u(100+182n+84n^{2})-v(36+56n+21n^{2})\big)}{36+56 n+21n^{2}}B_{n+1,1}+\\ &\quad+\frac{17+32n+14n^{2}-u\left(20+34n+14n^{2}\right)}{36+56 n+21n^{2}}B_{n,4} \end{align*} $$

and

$$ \begin{align*}N(u,v)=\left(u\left(43+70 n+28 n^{2}\right)+v\left(15+21 n+7 n^{2}\right)-44-70n-28 n^{2}\right)B_{n+1,1}. \end{align*} $$

Moreover, in this case, we have

$$ \begin{align*} \big(P(u,v)\big)^{2}&=10-12 u+2 u^{2}-2 v-v^{2}+\\ &\quad+2\left(u\left(43+70 n+28 n^{2}\right)+v\left(15+21 n+7 n^{2}\right)-44-70n-28n^{2}\right)^{2}\end{align*} $$

and

$$ \begin{align*} P(u,v)\cdot\mathbf{e}_1&=1321+3948 n+4396 n^{2}+2156 n^{3}+392 n^{4}-\\ &\quad-2u\left(15+21 n+7 n^{2}\right)\left(43+70 n+28 n^{2}\right)-\\&\quad -v\left(449+1260 n+1302 n^{2}+588 n^{3}+98 n^{4}\right). \end{align*} $$

If $u\in I_{n,4}^{\prime }$ and $v\in \left [\frac {44+70 n+28 n^{2}-u\left (43+70 n+28 n^{2}\right )}{15+21 n+7 n^{2}},\frac {103+182 n+84 n^{2}-u\left (100+182 n+84 n^{2}\right )}{36+56 n+21 n^{2}}\right ]$ , then

$$ \begin{align*} P(u,v)&=\frac{103+182n+84n^{2}-u(100+182n+84n^{2})-v(36+56n+21n^{2})}{36+56n+21n^{2}}\mathbf{e}_1+\\ &\quad+\frac{(1+n)(11+7n)\big(103+182n+84n^{2}-u(100+182n+84n^{2})-v(36+56n+21n^{2})\big)}{36+56n+21n^{2}}B_{n,4}+\\ &\quad+\frac{(15+21 n+7n^{2})\big(103+182n+84n^{2}-u(100+182n+84n^{2})-v(36+56n+21n^{2})\big)}{36+56n+21n^{2}}B_{n+1,1} \end{align*} $$

and

$$ \begin{align*} N(u,v) &=\big(u(30+58 n+28 n^{2})+v(11+18 n+7 n^{2})-31-58n-28n^{2}\big)B_{n,4} +\\&\quad +\big(u(43+70 n+28 n^{2})+v(15+21 n+7 n^{2})-44-70n-28n^{2}\big)B_{n+1,1}. \end{align*} $$

The same holds when $u\in I_{n,4}^{\prime \prime }$ and $v \in \left [\frac {31+58n+28 n^{2}-u\left (30+58n+28 n^{2}\right )}{11+18n+7 n^{2}},\frac {103+182n+84n^{2}-u\left (100+182 n+84n^{2}\right )}{36+56n+21n^{2}}\right ]$ . Moreover, in both cases, we have

$$ \begin{align*}P(u,v)\cdot\mathbf{e}_1=(36+56n+21n^{2})\big(103+182n+84n^{2}-u(100+182n+84n^{2})-v(36+56n+21n^{2})\big) \end{align*} $$

and

$$ \begin{align*}\big(P(u,v)\big)^{2}=\big(103+182n+84n^{2}-u(100+182n+84 n^{2})-v(36+56 n+21 n^{2})\big)^{2} \end{align*} $$

Thus, if $u\in I_{n,4}$ , then

$$ \begin{align*}t(u)=\frac{103+182n+84n^{2}-u\left(100+182n+84n^{2}\right)}{36+56n+21n^{2}}. \end{align*} $$

Now, we are ready to compute $S(W_{\bullet ,\bullet }^{T};\mathbf {e}_1)$ . Namely, for every $i\in \{1,2,3,4\}$ , we set

$$ \begin{align*}S_{n,i}=\frac{3}{14}\int_{I_{n,i}}\int_0^{t(u)}\big(P(u,v)\big)^2dvdu. \end{align*} $$

Then

$$ \begin{align*}S\big(W_{\bullet,\bullet}^{T};\mathbf{e}_1\big)=\sum_{n=0}^{\infty}\Big(S_{n,1}+S_{n,2}+S_{n,3}+S_{n,4}\Big). \end{align*} $$

On the other hand, integrating, we get

$$ \begin{align*}S_{n,1}= \left\{\begin{aligned} &\frac{84365}{114688}\ \text{if }n=0, \\ &\frac{(8+28n+21n^{2})A_{n,1}}{448 n^{4}(1+n)(2+7n)^{4}(3+7n)^{4}(4+7n)^{4} (1+7n+7n^{2})}\ \text{if }n\geqslant 1, \\ \end{aligned} \right. \end{align*} $$

where

$$ \begin{align*} A_{n,1}&=1536+109312 n+2935552 n^{2}+42681728 n^{3}+386407488 n^{4}+2335296292 n^{5}+\\ &\quad+9789648099 n^{6}+29038364761 n^{7}+61312905318 n^{8}+91454579804 n^{9}+\\ &\quad+94035837280 n^{10}+63317750608 n^{11}+25088413952 n^{12}+4427367168 n^{13}.\quad\quad \end{align*} $$

Similarly, we get

$$ \begin{align*}S_{n,2}=\frac{(1+2n)A_{n,2}}{4(1+n)(2+7n)^{4}(3+7n)^{4}(4+7n)^{4} (6+14n+7n^{2})}, \end{align*} $$

where

$$ \begin{align*} A_{n,2}&=1618654+31459234 n+271069253 n^{2}+1362423916 n^{3}+4419070194 n^{4}+9654348284n^{5}+\\ &\quad+14368501182 n^{6}+14362052096 n^{7}+9209328422 n^{8}+3412762192n^9+553420896n^{10}. \end{align*} $$

Likewise, we get

$$ \begin{align*}S_{n,3}=\frac{(3+2n)A_{n,3}}{4(1+n)(3+7n)^{4}(6+7n)^{4}(8+7n)^{4}(11+7n)(6+14 n+7 n^{2})}, \end{align*} $$

where

$$ \begin{align*} A_{n,3}&=1167997914+15454923336 n+91492878645 n^{2}+319934133575 n^{3}+\\ &\quad+734395997090 n^{4}+1162203105378 n^{5}+1294197714054 n^{6}+1014406754242 n^{7}+\\ &\quad+548632346402 n^{8}+195059453722 n^{9}+41045383120 n^{10}+3873946272n^{11}. \end{align*} $$

Finally, we get

$$ \begin{align*}S_{n,4}=\frac{(36+56 n+21 n^{2})A_{n,4}}{448(1+n)^{4}(6+7n)^{4}(8+7n)^{4}(10+7n)^{4}(11+7n)(15+21n+7 n^{2})}, \end{align*} $$

where

$$ \begin{align*} A_{n,4}&=365613573312+4021500121920 n+20341847967024 n^{2}+\\ &\quad+62650071283024 n^{3}+131072047236004 n^{4}+196698030664492 n^{5}+217823761840153 n^{6}+\\ &\quad+180219167765455 n^{7}+111395400841326 n^{8}+50802960251820 n^{9}+16615457209344 n^{10}+\\ &\quad+3690223711216 n^{11}+498816700928 n^{12}+30991570176 n^{13}. \end{align*} $$

Then, adding, we get

$$ \begin{align*}S\big(W_{\bullet,\bullet}^{T};\mathbf{e}_1\big)=\sum_{n=0}^{\infty}\Big(S_{n,1}+S_{n,2}+S_{n,3}+S_{n,4}\Big)\approx 0.976712233\ldots<1. \end{align*} $$

Finally, let us compute $S(W_{\bullet , \bullet ,\bullet }^{T,\mathbf {e}_1};P)$ . For every $i\in \{1,2,3,4\}$ , we set

$$ \begin{align*} M_{n,i}^\prime&=\frac{3}{14}\int_{I_{n,i}^\prime}\int_0^{t(u)}\Big(\big(P(u,v)\cdot\mathbf{e}_1\big)\Big)^2dvdu,\\ M_{n,i}^{\prime\prime}&=\frac{3}{14}\int_{I_{n,i}^{\prime\prime}}\int_0^{t(u)}\Big(\big(P(u,v)\cdot\mathbf{e}_1\big)\Big)^2dvdu.\quad\quad\quad\quad\quad\quad\quad \end{align*} $$

Then

$$ \begin{align*}S\big(W_{\bullet, \bullet,\bullet}^{T,\mathbf{e}_1};P\big)= \sum_{n=0}^{\infty}\sum_{i=1}^{4}\Big(M_{n,i}^\prime+M_{n,i}^{\prime\prime}\Big)+\frac{3}{7}\int_{0}^1\int_0^{t(u)}\big(P(u,v)\cdot \mathbf{e}_1\big)\mathrm{ord}_P\Big(N(u,v)\big|_{\mathbf{e}_1}\Big)dvdu. \end{align*} $$

On the other hand, integrating, we get

$$ \begin{align*}M_{n,1}^\prime= \left\{\begin{aligned} &\frac{1403}{22268}\ \text{if }n=0, \\ &\frac{(1+n)A_{n,1}^\prime}{448n^{4}(1+3n)^{4}(3+7n)^{4}(1+7n+7 n^{2})}\ \text{if }n\geqslant 1, \\ \end{aligned} \right. \end{align*} $$

where

$$ \begin{align*} A_{n,1}^\prime&=1+81 n+2535 n^{2}+37209 n^{3}+301046 n^{4}+1459736 n^{5}+\\ &\quad+4420190 n^{6}+8425410 n^{7}+9821448 n^{8}+6392736 n^{9}+1778112 n^{10}. \end{align*} $$

Similarly, we get

$$ \begin{align*}M_{n, 1}^{\prime\prime}=\frac{(1+7 n+7 n^{2})A_{n,1}^{\prime\prime}}{28(1+n)(1+3 n)^{4}(2+7 n)^{4}(3+7 n)^{4}(4+7 n)^{4}}, \end{align*} $$

where

$$ \begin{align*} A_{n,1}^{\prime\prime}&=480574+12906866 n+157271760 n^{2}+1149521334 n^{3}+5612285145 n^{4}+\\&\quad +19278934535 n^{5}+47770884833 n^{6}+86016481159 n^{7} +111679016743 n^{8}+\\ &\quad+101939513907 n^{9}+62077730148 n^{10}+22635902898 n^{11}+3735591048 n^{12}. \end{align*} $$

Likewise, we have

$$ \begin{align*}M_{n,2}^{\prime}=\frac{(6+14n+7n^{2})A_{n,2}^{\prime}}{224(1+n)(1+2 n)^{3}(2+7 n)^{4}(3+7 n)^{4}(4+7 n)^{4}}, \end{align*} $$

and

$$ \begin{align*}M_{n,2}^{\prime\prime}=\frac{11780+111142 n+430951 n^{2}+875637 n^{3}+978656 n^{4}+566832 n^{5}+131712 n^{6}}{224(1+2 n)^{3}(3+7 n)^{4}( 6+14 n+7 n^{2})}, \end{align*} $$

where

$$ \begin{align*} A_{n,2}^{\prime}&=1561176+35176776 n+356105548 n^{2}+2137950448 n^{3}+\\ &\quad+8458603286 n^{4}+23158717414 n^{5}+44778314889 n^{6}+61151030584 n^{7}+\\ &\quad+57807289939 n^{8}+36026947376 n^{9}+13321631568 n^{10}+2213683584 n^{11}. \end{align*} $$

Similarly, we have

$$ \begin{align*}M_{n,3}^{\prime}=\frac{(11+7n)A_{n,3}^{\prime}}{224(1+n)^{3}(3+7 n)^{4}(13+14 n)^{4} (6+14 n+7 n^{2})}, \end{align*} $$

where

$$ \begin{align*} A_{n,3}^{\prime}&=13726028+164541190 n+859036123 n^{2}+2564002455 n^{3}+4823323519 n^{4}+\\&\quad +5933644367 n^{5}+4776917782 n^{6}+2428774768 n^{7}+708314208 n^{8}+90354432 n^{9}. \end{align*} $$

Likewise, we have $M_{n,3}^{\prime \prime }=\frac {(6+14 n+7 n^{2})A_{n,3}^{\prime \prime }}{224(1+n)^{3}(6+7 n)^{4}(8+7 n)^{4}(11+7 n)(13+14n)^{4}}$ , where

$$ \begin{align*} A_{n,3}^{\prime\prime}&=67760261208+703706084640 n+3313300067388 n^{2}+9335574166156 n^{3}+\\ &\quad+17489294547578 n^{4}+22873117200584 n^{5}+21308562209725 n^{6}+14139587568253 n^{7}+\\ &\quad+6548997703738 n^{8}+2016283621072 n^{9}+371345421216 n^{10}+30991570176 n^{11}. \end{align*} $$

Similarly, we see that

$$ \begin{align*}M_{n,4}^{\prime}=\frac{(15+21n+7n^{2})A_{n,4}^{\prime}}{28(1+n)^{4}(6+7 n)^{4}(8+7 n)^{4}(11+7 n)(23+21 n)^{4}}, \end{align*} $$

where

$$ \begin{align*} A_{n,4}^{\prime}&=88135013250+967134809574 n+4853884596732 n^{2}+\\ &\quad+14732868828434 n^{3}+30120687035243 n^{4}+43697011451345 n^{5}+46124583653603 n^{6}+\\ &\quad+35692827118809 n^{7}+20096052100397 n^{8}+8028312817917 n^{9}+\\ &\quad+2160120347280 n^{10}+351456857766 n^{11}+26149137336 n^{12}. \end{align*} $$

Finally, we have

$$ \begin{align*}M_{n,4}^{\prime\prime}=\frac{(11+7n)A_{n,4}^{\prime\prime}}{448(1+n)^{4}(10+7 n)^{4}(23+21 n)^{4} (15+21 n+7 n^{2})}, \end{align*} $$

where

$$ \begin{align*} A_{n,4}^{\prime\prime}&=7582266167+59702225967 n+210973884925 n^{2}+440580768679 n^{3}+\\ &\quad+602090743422 n^{4}+562572998512 n^{5}+363945674554 n^{6}+160955181870 n^{7}+\\ &\quad+46566357768 n^{8}+7957643904 n^{9}+609892416 n^{10}. \end{align*} $$

Now, adding terms together, we see that

(6.8) $$ \begin{align} S\big(W_{\bullet, \bullet,\bullet}^{T,\mathbf{e}_1};P\big)\leqslant 0.974+\frac{3}{7}\int_{0}^1\int_0^{t(u)}\big(P(u,v)\cdot \mathbf{e}_1\big)\mathrm{ord}_P\Big(N(u,v)\big|_{\mathbf{e}_1}\Big)dvdu. \end{align} $$

Now, for every $i\in \{1,2,3,4\}$ and any irreducible component $\ell $ of the curve $B_{n,i}$ , we let

$$ \begin{align*} F_{n,i}&=\frac{3}{7}\int_{0}^1\int_0^{t(u)}\big(P(u,v)\cdot \mathbf{e}_1\big)\mathrm{ord}_\ell\big(N(u,v)\big)\big(\ell\cdot\mathbf{e}_1\big)dvdu=\\ &=\frac{3}{7}\int_{I_{n,i-1}}\int_0^{t(u)}\big(P(u,v)\cdot \mathbf{e}_1\big)\mathrm{ord}_\ell\big(N(u,v)\big)\big(\ell\cdot\mathbf{e}_1\big)dvdu+\\ &\quad+\frac{3}{7}\int_{I_{n,i}}\int_0^{t(u)}\big(P(u,v)\cdot \mathbf{e}_1\big)\mathrm{ord}_\ell\big(N(u,v)\big)\big(\ell\cdot\mathbf{e}_1\big)dvdu+\\ &\quad+\frac{3}{7}\int_{I_{n,i+1}}\int_0^{t(u)}\big(P(u,v)\cdot \mathbf{e}_1\big)\mathrm{ord}_\ell\big(N(u,v)\big)\big(\ell\cdot\mathbf{e}_1\big)dvdu, \end{align*} $$

where we set $I_{0,0}=\varnothing $ , $I_{n,5}=I_{n+1,1}$ for $n\geqslant 0$ and $I_{n,0}=I_{n-1,4}$ for $n\geqslant 1$ . Since irreducible components of the curve $B_{n,i}$ are disjoint, we get

$$ \begin{align*}\frac{3}{7}\int_{0}^1\int_0^{t(u)}\big(P(u,v)\cdot \mathbf{e}_1\big)\mathrm{ord}_P\Big(N(u,v)\big|_{\mathbf{e}_1}\Big)dvdu\leqslant \sum_{n=0}^{\infty}\sum_{i=1}^{4}F_{n,i}. \end{align*} $$

On the other hand, each $F_{n,i}$ is not difficult to compute. For instance, we have

$$ \begin{align*} F_{0,1}&=\frac{3}{7}\int_{0}^{\frac{1}{3}}\int_{2-u}^{\frac{7-6u}{3}}(v+u-2)(5-2u-v)dvdu+\\ &\quad+\frac{3}{7}\int_{0}^{\frac{1}{3}}\int_{\frac{7-6 u}{3}}^{\frac{19-16u}{8}}8(u+v-2)(19-16u-8v)dvdu+\\ &\quad+\frac{3}{7}\int_{\frac{1}{3}}^{\frac{3}{8}}\int_{2-u}^{\frac{19-16 u}{8}}8(u+v-2)(19-16u-8v)dvdu=\frac{281}{32256}. \end{align*} $$

Similarly, we see that

$$ \begin{align*}F_{n,1}=\frac{3(1+7 n+7 n^{2} )^{2}}{2 n^{2}(1+3 n)(-1+7 n)(1+7 n)(2+7 n)(3+7 n)^{2}(4+7 n)(2+21 n)} \end{align*} $$

for $n\geqslant 1$ . Likewise, we get

$$ \begin{align*}F_{n,2}= \left\{\begin{aligned} &\frac{5}{3584}\ \text{if }n=0, \\ &\frac{1+n}{112 n(1+2 n)(1+3 n)(2+7 n)(3+7 n)^{2}(4+7 n)}\ \text{if }n\geqslant 1. \\ \end{aligned} \right. \end{align*} $$

Likewise, for every $n\geqslant 0$ , we have

$$ \begin{align*}F_{n,3}=\frac{15 (6+14n+7n^{2})^{2}}{4(1+n)^{2}(1+2 n)(2+7 n)(3+7 n)^{2}(4+7 n)(6+7 n)(8+7 n)(13+14 n)} \end{align*} $$

and

$$ \begin{align*}F_{n,4}=\frac{(11+7 n)^{2}}{112(1+n)^{2}(3+7 n)(6+7 n)(8+7 n)(10+7 n)(13+14 n)(23+21 n)}. \end{align*} $$

Now, one can easily check that the total sum of all $F_{n,1}$ , $F_{n,2}$ , $F_{n,3}$ , $F_{n,4}$ is at most $0.014$ . This and Equation (6.8) give $S(W_{\bullet , \bullet ,\bullet }^{T,\mathbf {e}_1};P)\leqslant 0.974+0.014=0.988$ . Using Equation (6.7), we get $\delta _P(X)>1$ .

Corollary 6.11. All smooth Fano threefolds in the family are K-stable.