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 [Reference Abban and Zhuang1, Theorem 3.3] and [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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.

Proof. Let us compute $S_X(E)$ . Note that $S_X(E)<1$ by [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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.

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.

Proof. If A is smooth, then $\delta _P(A)\geqslant \delta (A)\geqslant \frac {3}{2}$ [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, § 2], which contradicts Corollary 2.3.

Proof. As in the proof of Lemma 2.2, it follows from [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, § 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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.

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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, § 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 $S(W_{\bullet ,\bullet ,\bullet }^{\mathscr {S},\mathscr {F}};Q)\geqslant 1-\mathrm {ord}_{Q}(\Delta _{\mathscr {F}})=1$ .

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 [Reference Fujita16, 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 [Reference Fujita16]. Moreover, if Q is a point in G and Z is a smooth curve in G that passes through Q, then it follows from [Reference Fujita16, 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 [Reference Fujita16], 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 [Reference Fujita16, 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 [Reference Fujita16, 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.

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 [Reference Fujita16, Theorem 4.8], [Reference Fujita16, Corollary 4.17], [Reference Fujita16, Corollary 4.18] using notations introduced in [Reference Fujita16, § 4]. To start with, we apply [Reference Fujita16, 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 [Reference Fujita16, § 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 [Reference Fujita16, 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 [Reference Fujita16, § 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 [Reference Fujita16, 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 [Reference Fujita16, 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).

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.

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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Bruce and Wall6]. Now, arguing as in the proof of [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Lemma 2.13], we get $\delta _O(A)\geqslant \frac {3}{2}$ . So, applying Lemma 2.1 again, we get a contradiction as above.

Proof. Suppose A has singularity of type $\mathbb {D}_4$ . Then it follows from [Reference Bruce and Wall6] 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 [Reference Bruce and Wall6]. 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, § 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Corollary 1.109].

To do this, we use notations introduced in the proof of Lemma 2.1 applied to $O=P$ . Then using [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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.

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.

Proof. Local computations.

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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, § 1.7] for definitions of $S_X(S)$ , $S(W_{\bullet ,\bullet }^S;B)$ , $S(W_{\bullet , \bullet ,\bullet }^{S,B};P)$ .

Note that [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Theorem 1.112] requires S to have Du Val singularities, but S may have non-Du Val singularities. Nevertheless, we still can apply [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Theorem 1.112] here since the proof of [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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.

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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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.

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 [Reference Fujita16, § 4] to get a contradiction. Namely, applying [Reference Fujita16, 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 [Reference Fujita16, § 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 [Reference Fujita16, 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 [Reference Fujita16, 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 [Reference Fujita16, 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.

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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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.

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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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.

Proof. Local computations.

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.

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 $S(W_{\bullet ,\bullet ,\bullet }^{S,C};P)=1$ .

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.

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

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 [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, 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).

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 [Reference Brenton5, 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.