Proof. This is well known [Reference Cheltsov and Prokhorov7], [Reference Coray and Tsfasman8].

Proof. Local computations.

Proof. Local computations.

Proof. The smoothness of the surface S easily follows from local computations. If $-K_S$ is ample, the remaining assertions are obvious. So, to complete the proof, we assume that $-K_S$ is not ample. Then the restriction $\omega \vert _{S}\colon S\to \mathbb {P}^1_{s,t}\times \mathbb {P}^1_{u,v}$ fits the commutative diagram

where $\alpha $ is a birational morphism that contracts all $(-2)$ -curves in S, and $\beta $ is a double cover branched over a singular curve of degree $(2,2)$ . Let $\ell $ be a $(-2)$ -curve in S. Then

$$ \begin{align*}(H_1+H_2)\cdot\ell=-K_S\cdot\ell=0, \end{align*} $$

so that $\omega (\ell )$ is a point in $\mathbb {P}^1_{s,t}\times \mathbb {P}^1_{u,v}$ . But $\pi _3(\ell )$ is a line in $\mathbb {P}^2_{x,y,z}$ that contains the point $\pi _3(C)$ . This shows that the curve $\ell $ is an irreducible component of a singular fiber of the conic bundle $\omega $ . Therefore, we see that $\omega (\ell )\in \Delta _{\mathbb {P}^1\times \mathbb {P}^1}$ . This implies that the conic bundle $\omega $ maps an irreducible component of the conic C to an irreducible component of the curve $\Delta _{\mathbb {P}^1\times \mathbb {P}^1}$ because S is a general surface in the linear system $|H_3|$ that contains the curve C.

If C is singular, an irreducible component of the curve $\Delta _{\mathbb {P}^1\times \mathbb {P}^1}$ is a curve of degree $(1,0)$ or $(0,1)$ , which is impossible [Reference Prokhorov15, §3.8]. Therefore, we see that the conic C is smooth and irreducible, and the curve $\omega (C)\cong C$ is an irreducible component of the discriminant curve $\Delta _{\mathbb {P}^1\times \mathbb {P}^1}$ . Since the conic bundle $\omega $ is standard [Reference Prokhorov15], the surface $\omega ^{-1}(\omega (C))$ is irreducible and nonnormal, which easily implies that the conic C is contained in its singular locus.

Choosing appropriate coordinates on $\mathbb {P}^1_{s,t}\times \mathbb {P}^1_{u,v}\times \mathbb {P}^2_{x,y,z}$ , we may assume that $\pi _3(C)=[0:0:1]$ , the conic C is given by $x=y=sv-tu=0$ , $([0:1],[0:1])$ is a smooth point of the curve $\Delta _{\mathbb {P}^1\times \mathbb {P}^1}$ , and the fiber $\omega ^{-1}([0:1],[0:1])$ is given by $s=u=xy=0$ . Then X is given by

$$ \begin{align*} &(a_1su+b_1sv+c_1 tu)x^2+(a_2su+b_2sv+c_2tu+tv)xy+\\ & \quad +b_4(sv-tu)xz+(a_3su+b_3sv+c_3tu)y^2+b_5(sv-tu)yz+(sv-tu)z^2=0 \end{align*} $$

for some numbers $a_1$ , $a_2$ , $a_3$ , $b_1$ , $b_2$ , $b_3$ , $b_4$ , $b_5$ , $c_1$ , $c_2$ , $c_3$ . One can check that $\Delta _{\mathbb {P}^1\times \mathbb {P}^1}$ indeed splits as a union of the curve $\omega (C)$ and the curve in $\mathbb {P}^1_{s,t}\times \mathbb {P}^1_{u,v}$ of degree $(2,2)$ that is given by

$$ \begin{align*} &a_1b_5^2stu^2-a_1b_5^2s^2uv+a_2b_4b_5s^2uv-a_2b_4b_5stu^2-a_3b_4^2s^2uv+a_3b_4^2stu^2-b_1b_5^2s^2v^2+\\& \quad +b_1b_5^2stuv+b_2b_4b_5s^2v^2-b_2b_4b_5stuv-b_3b_4^2s^2v^2+b_3b_4^2stuv-b_4^2c_3stuv+b_4^2c_3t^2u^2+\\& \quad +b_4b_5c_2stuv-b_4b_5c_2t^2u^2-b_5^2c_1stuv+b_5^2c_1t^2u^2+4a_1a_3s^2u^2+4a_1b_3s^2uv+4a_1c_3stu^2-\\& \quad -a_2^2s^2u^2-2a_2b_2s^2uv-2a_2c_2stu^2+4a_3b_1s^2uv+4a_3c_1stu^2+ 4b_1b_3s^2v^2+4b_1c_3stuv-\\& \quad -b_2^2s^2v^2-2b_2c_2stuv+4b_3c_1stuv+b_4b_5stv^2-b_4b_5t^2uv+4c_1c_3t^2u^2-c_2^2t^2u^2-2a_2stuv-\\& \quad -2b_2stv^2-2c_2t^2uv-t^2v^2=0. \end{align*} $$

The surface S is cut out on X by the equation $y=\lambda x$ , where $\lambda $ is a general complex number. Then the double cover $\beta \colon \overline {S}\to \mathbb {P}^1_{s,t}\times \mathbb {P}^1_{u,v}$ is branched over a singular curve of degree $(2,2)$ , which splits as a union of the curve $\omega (C)$ and the curve in $\mathbb {P}^1_{s,t}\times \mathbb {P}^1_{u,v}$ of degree $(1,1)$ that is given by

$$ \begin{align*} &\lambda^2 b_5^2tu-\lambda^2b_5^2sv+4\lambda^2a_3su+4\lambda^2b_3sv-2b_4\lambda b_5sv+2\lambda b_4b_5tu+\\& \quad +4\lambda^2c_3tu+4\lambda a_2su+4\lambda b_2sv-b_4^2sv+b_4^2tu+4\lambda c_2tu+4a_1su+4b_1sv+4c_1tu+4\lambda tv=0. \end{align*} $$

Since $\lambda $ is general and X is smooth, these two curves intersect transversally by two points, which implies the remaining assertions of the lemma.

Proof. Left to the reader.

Proof. One has $\tau =1$ . Moreover, for $u\in [0,1]$ , we have $N(u)=0$ and $P(u)|_S=-K_S+ (1-u)C$ . Let $L=-K_S+(1-u)C$ . Using Lemma 24 and arguing as in the proof of Lemma 27, we get

$$ \begin{align*} S\big(W^S_{\bullet,\bullet};F\big)&=\frac{1}{6}\int_0^1 4(1+(1-u))S_L(F)du\leqslant \\ & \quad \leqslant A_S(F)\int_0^1 \frac{4}{6}(1+(1-u)) \frac{19+8(1-u)+(1-u)^2}{24}du=\frac{143}{144}A_S(F) \end{align*} $$

for any prime divisor F over S such that $P\in C_S(F)$ . Then (3.1) gives $\delta _P(X)>1$ .

Proof. We have $\tau =\frac {3}{2}$ . Moreover, we have

$$ \begin{align*}P(u)=\left\{\begin{aligned} &(1-u)H_1+H_2+H_3,\ \text{if }0\leqslant u\leqslant 1, \\ &(2-u)H_2+(3-2u)H_3,\ \text{if }1\leqslant u\leqslant \frac{3}{2}, \\ \end{aligned} \right. \end{align*} $$

and

$$ \begin{align*}N(u)=\left\{\begin{aligned} &0,\ \text{if }0\leqslant u\leqslant 1, \\ &(u-1)E_2,\ \text{if }1\leqslant u\leqslant \frac{3}{2}.\\ \end{aligned} \right. \end{align*} $$

Note also that $E_2\vert _{S}$ is a smooth genus $3$ curve contained in the smooth locus of the surface S.

Recall that S is a quintic del Pezzo surface with at most Du Val singularities and that the restriction morphism $\pi _2\vert _{S}\colon S\to \mathbb {P}^1_{u,v}$ is a conic bundle. Note that the morphism $\pi _3\vert _{S}\colon S\to \mathbb {P}^2_{x,y,z}$ is birational. Let C be a fiber of the conic bundle $\pi _2\vert _{S}$ , and let L be the preimage in S of a general line in $\mathbb {P}^2_{x,y,z}$ . Then $-K_S\sim C+L$ and

$$ \begin{align*}P(u)\big\vert_{S}\sim_{\mathbb{R}}\left\{\begin{aligned} &C+L,\ \text{if }0\leqslant u\leqslant 1, \\ &(2-u)C+(3-2u)L,\ \text{if }1\leqslant u\leqslant \frac{3}{2}. \\ \end{aligned} \right. \end{align*} $$

Since $2L-C$ is pseudoeffective, the divisor $\frac {7-4u}{3}(-K_S)-(2-u)C-(3-2u)L$ is also pseudoeffective.

Let F be a divisor over S such that $P\in C_S(F)$ . Then it follows from Lemma 27 that

$$ \begin{align*} S\big(W^S_{\bullet,\bullet};F\big)&\leqslant\frac{1}{6}A_S(F)\int_1^{\frac{3}{2}}(u-1)\big(P(u)\big\vert_{S}\big)^2du+\frac{1}{6}\int_{0}^{\frac{3}{2}}\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{S}-vF\big)dvdu=\\&=\frac{7}{288}A_S(F)+\frac{1}{6}\int_{0}^{1}\int_0^\infty \mathrm{vol}\big(-K_S-vF\big)dvdu+\\&\quad+\frac{1}{6}\int_{1}^{\frac{3}{2}}\int_0^\infty\mathrm{vol}\big((2-u)C+(3-2u)L-vF\big)dvdu\leqslant\\&\leqslant\frac{7}{288}A_S(F)+\frac{1}{6}\int_{0}^{1}5\frac{A_S(F)}{\delta_P(S)}du+\frac{1}{6}\int_{1}^{\frac{3}{2}}\int_0^\infty\mathrm{vol}\Bigg(\frac{7-4u}{3}\big(-K_S\big)-vF\Bigg)dvdu=\\&=\frac{7}{288}A_S(F)+\frac{5}{6\delta_P(S)}A_S(F)+\frac{1}{6}\int_{1}^{\frac{3}{2}}\Bigg(\frac{7-4u}{3}\Bigg)^3\int_0^\infty\mathrm{vol}\big(-K_S-vF\big)dvdu\leqslant\\&\leqslant\frac{7}{288}A_S(F)+\frac{5}{6\delta_P(S)}A_S(F)+\frac{1}{6}\int_{1}^{\frac{3}{2}}\Bigg(\frac{7-4u}{3}\Bigg)^35\frac{A_S(F)}{\delta_P(S)}du=\\&=\frac{7}{288}A_S(F)+\frac{5}{6\delta_P(S)}A_S(F)+\frac{25}{162\delta_P(S)}A_S(F)=\Bigg(\frac{80}{81\delta_P(S)}+\frac{7}{288}\Bigg)A_S(F). \end{align*} $$

Then $\delta _{P}(S;W^S_{\bullet ,\bullet })\geqslant \frac {1}{\frac {80}{81\delta _P(S)}+\frac {7}{288}}=\frac {2,592\delta _P(S)}{2,560+63\delta _P(S)}$ and the required assertion follows from (3.1).

Proof. We have $\tau =1$ . Moreover, for $u\in [0,1]$ , we have $N(u)=0$ and $P(u)|_S=-K_S+2(1-u)C$ . On the other hand, it follows from Lemma 7 that S is a smooth del Pezzo surface of degree $2$ , and the restriction map $\pi _3\vert _{S}\colon S\to \mathbb {P}^2_{x,y,z}$ is a double cover that is ramified over a smooth quartic curve. Therefore, applying the Galois involution of this double cover to C, we obtain another smooth irreducible curve $Z\subset S$ such that $C+Z\sim -2K_S$ , $C^2=Z^2=0$ and $C\cdot Z=4$ , which gives

$$ \begin{align*}P(u)|_S-vC\sim_{\mathbb{R}}\Big(\frac{5}{2}-2u-v\Big)C+\frac{1}{2}Z. \end{align*} $$

Then $P(u)\vert _{S}-vC$ is pseudoeffective $\iff P(u)\vert _{S}-vC$ is nef $\iff v\leqslant \frac {5}{2}-2u$ . Thus, we have

$$ \begin{align*}\mathrm{vol}\big(P(u)\vert_{S}-vC\big)=\big(-K_S+2(1-u)C\big)^2=10-8u-4v \end{align*} $$

and $P(u,v)\cdot C=2$ . Now, integrating, we obtain $S(W_{\bullet ,\bullet }^S;C)=\frac {31}{36}$ and $S(W_{\bullet ,\bullet ,\bullet }^{S,C};P)=1$ .

Proof. We have $\tau =1$ . Moreover, for $u\in [0,1]$ , we have $N(u)=0$ and $P(u)|_S=-K_S+(1-u)C$ . Since S is a smooth del Pezzo surface, the restriction map $\omega \vert _{S}\colon S\to \mathbb {P}^1_{s,t}\times \mathbb {P}^1_{u,v}$ is a double cover ramified over a smooth elliptic curve. Therefore, using the Galois involution of this double cover, we get an irreducible curve $Z\subset S$ such that $C+Z\sim -K_S$ , $C^2=Z^2=0$ , and $C\cdot Z=2$ , which gives

$$ \begin{align*}P(u)|_S-vC\sim_{\mathbb{R}}(2-u-v)C+Z. \end{align*} $$

Then $P(u)\vert _{S}-vC$ is pseudoeffective $\iff P(u)\vert _{S}-vC$ is nef $\iff v\leqslant 2-u$ . Thus, we have

$$ \begin{align*}\mathrm{vol}\big(P(u)\vert_{S}-vC\big)=\big(-K_S+(1-u)C\big)^2=8-4u-4v \end{align*} $$

and $P(u,v)\cdot C=2$ . Now, integrating, we obtain $S(W_{\bullet ,\bullet }^S;C)=\frac {7}{9}$ and $S(W_{\bullet ,\bullet ,\bullet }^{S,C};P)=1$ .

Proof. We have $\tau =1$ . Moreover, for $u\in [0,1]$ , we have $N(u)=0$ and $P(u)|_S=-K_S+(1-u)C$ . It follows from Lemma 6 that S contains two $(-2)$ -curves $\mathbf {e}_1$ and $\mathbf {e}_2$ such that $-K_S\sim 2C+\mathbf {e}_1+\mathbf {e}_2$ . On the surface S, we have $C^2=0$ , $C\cdot \mathbf {e}_1=C\cdot \mathbf {e}_2=1$ , $\mathbf {e}_1^2=\mathbf {e}_2^2=-2$ , and

$$ \begin{align*}P(u)|_S-vC\sim_{\mathbb{R}}(3-u-v)C+\mathbf{e}_1+\mathbf{e}_2. \end{align*} $$

Then $P(u)\vert _{S}-vC$ is pseudoeffective $\iff v\leqslant 3-u$ . Moreover, we have

$$ \begin{align*}P(u,v)&=\left\{\begin{aligned} &(3-u-v)C+\mathbf{e}_1+\mathbf{e}_2,\ \text{if }0\leqslant v\leqslant 1-u, \\ &\frac{3-u-v}{2}\big(2C+\mathbf{e}_1+\mathbf{e}_2\big),\ \text{if }1-u\leqslant v\leqslant 3-u, \\ \end{aligned} \right. \\N(u,v)&=\left\{\begin{aligned} &0,\ \text{if }0\leqslant v\leqslant 1-u, \\ &\frac{u+v-1}{2}(\mathbf{e}_1+\mathbf{e}_2),\ \text{if }1-u\leqslant v\leqslant 3-u, \\ \end{aligned} \right. \\\mathrm{vol}\big(P(u)\vert_{S}-vC\big)&= \left\{\begin{aligned} &8-4u-4v,\ \text{if }0\leqslant v\leqslant 1-u, \\ &(u+v-3)^2,\ \text{if }1-u\leqslant v\leqslant 3-u. \\ \end{aligned} \right.\end{align*} $$

Now, integrating $\mathrm {vol}(P(u)\vert _{S}-vC)$ , we obtain $S(W_{\bullet ,\bullet }^S;C)=\frac {8}{9}$ .

To compute $S(W_{\bullet ,\bullet ,\bullet }^{S,C};P)$ , observe that $F_P(W_{\bullet ,\bullet ,\bullet }^{S,C})=0$ , because $P\not \in \mathbf {e}_1\cup \mathbf {e}_2$ , since S is a general surface in $|H_3|$ that contains C. On the other hand, we have

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

Hence, integrating $(P(u,v)\cdot C)^2$ , we get $S(W_{\bullet ,\bullet ,\bullet }^{S,C};P)=\frac {7}{9}$ as required.

Proof. We have $\tau =1$ . For $u\in [0,1]$ , we have $N(u)=0$ and $P(u)|_S\sim _{\mathbb {R}}-K_S+(1-u) (C+C^\prime )$ , where $C^\prime $ is the irreducible curve in S such that $C+C^\prime $ is the fiber of the conic bundle $\pi _3$ that passes through the point P. Since $P\not \in E_1\cap E_2$ , we see that $P\not \in C^\prime $ .

By Lemma 6, the surface S is a smooth del Pezzo surface of degree $4$ , so we can identify it with a complete intersection of two quadrics in $\mathbb {P}^4$ . Then C and $C^\prime $ are lines in S, and S contains four additional lines that intersect C. Denote them by $L_1$ , $L_2$ , $L_3$ , and $L_4$ , and let $Z=L_1+L_2+L_3+L_4$ . Then the intersections of the curves C, $C^\prime $ , and Z on the surface S are given in the table below.

Observe that $-K_S\sim _{\mathbb {Q}}\frac {3}{2}C+\frac {1}{2}C^\prime +\frac {1}{2}Z$ . This gives $P(u)\vert _{S}-vC\sim _{\mathbb {R}}(\frac {5}{2}-u-v)C+ (\frac {3}{2}-u)C^\prime +\frac {1}{2}Z$ , which implies that $P(u)\vert _{S}-vC$ is pseudoeffective $\iff v\leqslant \frac {5}{2}-u$ .

Moreover, we have

$$ \begin{align*}P(u,v)&=\left\{\begin{aligned} &\Big(\frac{5}{2}-u-v\Big)C+\Big(\frac{3}{2}-u\Big)C^\prime+\frac{1}{2}Z,\ \text{if }0\leqslant v\leqslant 1, \\ &\Big(\frac{5}{2}-u-v\Big)(C+C^\prime)+\frac{1}{2}Z,\ \text{if }1\leqslant v\leqslant 2-u, \\ &\Big(\frac{5}{2}-u-v\Big)(C+C^\prime+Z),\ \text{if }2-u\leqslant v\leqslant \frac{5}{2}-u, \\ \end{aligned} \right.\\[4pt]N(u,v)&=\left\{\begin{aligned} &0,\ \text{if }0\leqslant v\leqslant 1, \\ &(v-1)C^\prime,\ \text{if }1\leqslant v\leqslant 2-u, \\ &(v-1)C^\prime+(v+u-2)Z,\ \text{if }2-u\leqslant v\leqslant \frac{5}{2}-u, \\ \end{aligned} \right.\\[4pt]P(u,v)\cdot C&=\left\{\begin{aligned} &1+v,\ \text{if }0\leqslant v\leqslant 1, \\ &2,\ \text{if }1\leqslant v\leqslant 2-u, \\ &10-4u-4v,\ \text{if }2-u\leqslant v\leqslant \frac{5}{2}-u, \end{aligned} \right.\\[4pt]\mathrm{vol}\big(P(u)\vert_{S}-vC\big)&= \left\{\begin{aligned} &8-v^2-4u-2v,\ \text{if }0\leqslant v\leqslant 1, \\ &9-4u-4v,\ \text{if }1\leqslant v\leqslant 2-u, \\ &(5-2u-2v)^2,\ \text{if }2-u\leqslant v\leqslant \frac{5}{2}-u. \\ \end{aligned} \right. \end{align*} $$

Now, integrating $\mathrm {vol}(P(u)\vert _{S}-vC)$ and $(P(u,v)\cdot C)^2$ , we get $S(W_{\bullet ,\bullet }^S;C)=1$ and

$$ \begin{align*} S\big(W_{\bullet, \bullet,\bullet}^{S,C};P\big)=\frac{5}{6}+F_P\big(W_{\bullet,\bullet,\bullet}^{S,C}\big)&=\frac{5}{6}+\frac{1}{3}\int_0^1\int_0^{\frac{5}{2}-u}\big(P(u,v)\cdot C\big)\cdot \mathrm{ord}_P\big(N(u,v)\big|_C\big)dvdu\leqslant\\ &\leqslant\frac{5}{6}+\frac{1}{3}\int_0^1\int_2^{\frac{5}{2}-u}(10-4u-4v)(v+u-2)dvdu=\frac{31}{36}, \end{align*} $$

because $P\not \in C^\prime $ , and the curves Z and C intersect each other transversally.

Proof. Suppose that $\beta (\mathbf {F})\leqslant 0$ . Then $\delta _P(X)\leqslant 1$ for every point $P\in \mathfrak {C}$ . Let us seek for a contradiction.

Let $S_1$ be a general surface in the linear system $|H_1|$ . Then $S_1$ is smooth. Hence, if $S_1\cap \mathfrak {C}\ne \varnothing $ , then $\delta _P(X)\leqslant 1$ for every point $P\in S_1\cap \mathfrak {C}$ , which contradicts Corollary 11. We see that $S_1\cdot \mathfrak {C}=0$ . Similarly, we see that $S_2\cdot \mathfrak {C}=0$ for a general surface $S_2\in |H_2|$ . So, we see that $\omega (\mathfrak {C})$ is a point.

Let C be the scheme fiber of the conic bundle $\omega $ over the point $\omega (\mathfrak {C})$ . Then $\mathfrak {C}$ is an irreducible component of the curve C. If the fiber C is smooth, then we $\mathfrak {C}=C$ .

Suppose that C is smooth. If S is a general surface in the linear system $|H_1+H_2|$ that contains $\mathfrak {C}$ , then $S(W_{\bullet ,\bullet }^S;\mathfrak {C})=\frac {31}{36}<1$ by Lemma 14, which contradicts Corollary 8. So, the curve C is singular.

Note that $\pi _3(\mathfrak {C})$ is a line in $\mathbb {P}^2_{x,y,z}$ . On the other hand, the discriminant curve $\Delta _{\mathbb {P}^2}$ is an irreducible smooth quartic curve in $\mathbb {P}^2_{x,y,z}$ . Therefore, in particular, the line $\pi _3(\mathfrak {C})$ is not contained in $\Delta _{\mathbb {P}^2}$ . Now, let P be a general point in $\mathfrak {C}$ , let Z be the fiber of the conic bundle $\pi _3$ that passes through P, and let S be a general surface in $|H_3|$ that contains the curve Z. Then Z and S are both smooth, and it follows from Lemma 6 that S is a del Pezzo of degree $4$ , so that $\delta _P(X)>1$ by Lemma 9.

Proof. Apply Lemmas 15–17 and Corollary 13.

Proof. Apply Lemma 14 and Corollary 13.

Proof. The proof of this lemma is similar to the proof of [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Lem. 2.12]. Namely, as in that proof, we will apply [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Th. 1.7.1], [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Cor. 1.7.12], and [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Cor. 1.7.25] to get (♣) and (♠). Let us use notations introduced in [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Sect. 1] applied to S polarized by the ample divisor L.

First, we suppose that P is not contained in any line in S. In particular, the conic C is smooth. Let $\sigma \colon \widetilde {S}\to S$ be the blowup of the point P, let E be the exceptional curve of the blowup $\sigma $ , and let $\widetilde {C}$ be the proper transform on $\widetilde {S}$ of the conic C. Then $\widetilde {S}$ is a smooth cubic surface in $\mathbb {P}^3$ , and there exists a unique line $\mathbf {l}\subset \widetilde {S}$ such that $-K_{\widetilde {S}}\sim \widetilde {C}+E+\mathbf {l}$ . Take $u\in \mathbb {R}_{\geqslant 0}$ . Then

$$ \begin{align*}\sigma^*(L)-uE\sim_{\mathbb{R}}(1+t)\widetilde{C}+(2+t-u)E+\mathbf{l}, \end{align*} $$

which implies that $\sigma ^*(L)-uE$ is pseudoeffective $\iff u\leqslant 2+t$ . Similarly, we see that

$$ \begin{align*}\mathscr{P}(u)&\sim_{\mathbb{R}}\left\{\begin{aligned} &(1+t)\widetilde{C}+(2+t-u)E+\mathbf{l},\ \text{if }0\leqslant u\leqslant 2, \\ &(3+t-u)\widetilde{C}+(2+t-u)E+\mathbf{l},\ \text{if }2\leqslant u\leqslant2+t, \\ \end{aligned} \right.\\[8pt]\mathscr{N}(u)&=\left\{\begin{aligned} &0,\ \text{if }0\leqslant u\leqslant 2, \\ &(u-2)\widetilde{C},\ \text{if }2\leqslant u\leqslant 2+t,\\ \end{aligned} \right.\\[8pt]\mathscr{P}(u)\cdot E&=\left\{\begin{aligned} &u,\ \text{if }0\leqslant u\leqslant 2, \\ &2,\ \text{if }2\leqslant u\leqslant2+t, \\ \end{aligned} \right.\\[8pt]\mathrm{vol}\big(\sigma^*(L)-uE\big)&=\left\{\begin{aligned} &4+4t-u^2,\ \text{if }0\leqslant u\leqslant 2, \\ &4(2+t-u),\ \text{if }2\leqslant u\leqslant2+t, \\ \end{aligned} \right. \end{align*} $$

where we denote by $\mathscr {P}(u)$ the positive part of the Zariski decomposition of the divisor $\sigma ^*(L)-uE$ , and we denote by $\mathscr {N}(u)$ its negative part. This gives

$$ \begin{align*}S_L(E)=\frac{8+12t+3t^2}{6(1+t)}. \end{align*} $$

Moreover, applying [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Cor. 1.7.25], we obtain

$$ \begin{align*}S(W^E_{\bullet,\bullet};Q)\leqslant\frac{4+6t+3t^2}{6(1+t)} \end{align*} $$

for every point $Q\in E$ . Note that $A_S(E)=2$ . Thus, it follows from [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Cor. 1.7.12] that

$$ \begin{align*}\delta_P(S,L)\geqslant\frac{6(1+t)}{4+6t+3t^2}>\frac{24}{19+8t+t^2}. \end{align*} $$

To complete the proof of the lemma, we may assume that S contains a line $\ell $ such that $P\in \ell $ . Then $\ell \cdot C=0$ or $\ell \cdot C=1$ . If $\ell \cdot C=0$ , then $\ell $ must be an irreducible component of the conic C. Let us apply [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Th. 1.7.1] and [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Cor. 1.7.25] to the flag $P\in \ell $ to estimate $\delta _P(S,L)$ . Take $u\in \mathbb {R}_{\geqslant 0}$ . Let $P(u)$ be the positive part of the Zariski decomposition of the divisor $L-u\ell $ , and let $N(u)$ be its negative part. We must compute $P(u)$ , $N(u)$ , $P(u)\cdot \ell $ , and $\mathrm {vol}(L-u\ell )$ .

There exists a birational morphism $\pi \colon S\to \mathbb {P}^2$ that blows up five points $O_1,\dots ,O_5\in \mathbb {P}^2$ such that no three of them are collinear. For every $i\in \{1,\ldots ,5\}$ , let $\mathbf {e}_i$ be the $\pi $ -exceptional curve such that $\pi (\mathbf {e}_i)=O_i$ . Similarly, let $\mathbf {l}_{ij}$ be the strict transform of the line in $\mathbb {P}^2$ that contains $O_i$ and $O_j$ , where $1\leqslant i<j\leqslant 5$ . Finally, let B be the strict transform of the conic on $\mathbb {P}^2$ that passes through the points $O_1,\dots ,O_5$ . Then $\mathbf {e}_1,\ldots ,\mathbf {e}_5,\mathbf {l}_{12},\ldots ,\mathbf {l}_{45},B$ are all lines in S, and each extremal ray of the Mori cone $\overline {\mathrm {NE}}(S)$ is generated by a class of one of these $16$ lines.

Suppose that the conic C is irreducible. Then $C\cdot \ell =1$ . In this case, without loss of generality, we may assume that $\ell =\mathbf {e}_1$ and $C\sim \mathbf {l}_{12}+\mathbf {e}_2$ . If $0\leqslant t\leqslant 1$ , then

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &L-u\ell,\ \text{if }0\leqslant u\leqslant 1, \\ &L-u\ell-(u-1)(\mathbf{l}_{12}+\mathbf{l}_{13}+\mathbf{l}_{14}+\mathbf{l}_{15}),\ \text{if }1\leqslant u\leqslant 1+t, \\ &L-u\ell-(u-1)(\mathbf{l}_{12}+\mathbf{l}_{13}+\mathbf{l}_{14}+\mathbf{l}_{15})-(u-t-1)B, \ \text{if }1+t\leqslant u\leqslant\frac{3+t}{2}, \\ \end{aligned} \right. \\[5pt]N(u)&=\left\{\begin{aligned} &0, \ \text{if }0\leqslant u\leqslant 1, \\ &(u-1)(\mathbf{l}_{12}+\mathbf{l}_{13}+\mathbf{l}_{14}+\mathbf{l}_{15}),\ \text{if }1\leqslant u\leqslant 1+t, \\ &(u-1)(\mathbf{l}_{12}+\mathbf{l}_{13}+\mathbf{l}_{14}+\mathbf{l}_{15})+(u-t-1)B, \ \text{if }1+t\leqslant u\leqslant\frac{3+t}{2}, \\ \end{aligned} \right.\\[5pt]P(u)\cdot\ell&=\left\{\begin{aligned} &1+t+u,\ \text{if }0\leqslant u\leqslant 1, \\ &5+t-3u,\ \text{if }1\leqslant u\leqslant 1+t, \\ &6+2t-4u,\ \text{if }1+t\leqslant u\leqslant\frac{3+t}{2}, \\ \end{aligned} \right.\\[5pt]\mathrm{vol}\big(L-u\ell\big)&=\left\{\begin{aligned} &4(1+t)-2u(1+t)-u^2, \ \text{if }0\leqslant u\leqslant 1, \\ &(2-u)(4+2t-3u),\ \text{if }1\leqslant u\leqslant 1+t, \\ &(3+t-2u)^2,\ \text{if }1+t\leqslant u\leqslant\frac{3+t}{2}, \\ \end{aligned} \right.\end{align*} $$

and $L-u\ell $ is not pseudoeffective for $u>\frac {3+t}{2}$ . Similarly, if $t\geqslant 1$ , then

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &L-u\ell,\ \text{if }0\leqslant u\leqslant 1, \\ &L-u\ell-(u-1)(\mathbf{l}_{12}+\mathbf{l}_{13}+\mathbf{l}_{14}+\mathbf{l}_{15}),\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \\[6pt]N(u)&=\left\{\begin{aligned} &0, \ \text{if }0\leqslant u\leqslant 1, \\ &(u-1)(\mathbf{l}_{12}+\mathbf{l}_{13}+\mathbf{l}_{14}+\mathbf{l}_{15}), \ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\\[6pt]P(u)\cdot\ell&=\left\{\begin{aligned} &1+t+u,\ \text{if }0\leqslant u\leqslant 1, \\ &5+t-3u,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\\[6pt]\mathrm{vol}\big(L-u\ell\big)&=\left\{\begin{aligned} &4(1+t)-2u(1+t)-u^2, \ \text{if }0\leqslant u\leqslant 1, \\ &(2-u)(4+2t-3u),\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \end{align*} $$

and $L-u\ell $ is not pseudoeffective for $u>2$ . Then

$$ \begin{align*}S_L\big(\ell\big)= \left\{\begin{aligned} &\frac{17+4t-t^2}{24},\ \text{if }0\leqslant t\leqslant 1, \\ &\frac{2+3t}{3(1+t)},\ \text{if }t\geqslant 1. \\ \end{aligned} \right. \end{align*} $$

Observe that $P\not \in \mathbf {l}_{ij}$ for every $1\leqslant i<j\leqslant 5$ . Thus, if $t\leqslant 1$ , then [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Cor. 1.7.25] gives

$$ \begin{align*}S(W^{\ell}_{\bullet,\bullet};P)= \left\{\begin{aligned} &\frac{19+8t+t^2}{24},\ \text{if }P\in B, \\ &\frac{9+15t+3t^2+t^3}{12(1+t)},\ \text{if }P\not\in B. \\ \end{aligned} \right. \end{align*} $$

Similarly, if $t\geqslant 1$ , then [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Cor. 1.7.25] gives

$$ \begin{align*}S\big(W^{\ell}_{\bullet,\bullet};P\big)=\frac{5+6t+3t^2}{6(1+t)}. \end{align*} $$

Now, using [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Th. 1.7.1], we get (♣).

To complete the proof of the lemma, we may assume that the conic C is reducible. In this case, we let $\ell $ be an irreducible component of the conic C that contains P. Without loss of generality, we may assume that $\ell =\mathbf {e}_1$ and $C=\mathbf {e}_1+B$ . Then

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &L-u\ell,\ \text{if }0\leqslant u\leqslant 1, \\ &L-u\ell-(u-1)B,\ \text{if }1\leqslant u\leqslant 1+t, \\ &L-u\ell-(u-t-1)(\mathbf{l}_{12}+\mathbf{l}_{13}+\mathbf{l}_{14}+\mathbf{l}_{15})-(u-1)B, \ \text{if }1+t\leqslant u\leqslant\frac{3+2t}{2}, \\ \end{aligned} \right. \\[4pt]N(u)&=\left\{\begin{aligned} &0, \ \text{if }0\leqslant u\leqslant 1, \\ &(u-1)B,\ \text{if }1\leqslant u\leqslant 1+t, \\ &(u-t-1)(\mathbf{l}_{12}+\mathbf{l}_{13}+\mathbf{l}_{14}+\mathbf{l}_{15})+(u-1)B, \ \text{if }1+t\leqslant u\leqslant\frac{3+2t}{2}, \\ \end{aligned} \right.\\[4pt]P(u)\cdot\ell&=\left\{\begin{aligned} &1+u,\ \text{if }0\leqslant u\leqslant 1, \\ &2,\ \text{if }1\leqslant u\leqslant 1+t, \\ &6+4t-4u,\ \text{if }1+t\leqslant u\leqslant\frac{3+2t}{2}, \\ \end{aligned} \right. \\[4pt]\mathrm{vol}\big(L-u\ell\big)&=\left\{\begin{aligned} &4(1+t)-2u-u^2,\ \text{if }0\leqslant u\leqslant 1, \\ &5+4t-4u,\ \text{if }1\leqslant u\leqslant 1+t, \\ &(3+2t-2u)^2,\ \text{if }1+t\leqslant u\leqslant\frac{3+2t}{2}, \\ \end{aligned} \right. \end{align*} $$

and the divisor $L-u\ell $ is not pseudoeffective for $u>\frac {3+2t}{2}$ . This gives

$$ \begin{align*}S_L\big(\ell\big)=\frac{17+30t+12t^2}{24(1+t)}. \end{align*} $$

Moreover, using [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Cor. 1.7.25], we compute

$$ \begin{align*}S\big(W^{\ell}_{\bullet,\bullet};P\big)= \left\{\begin{aligned} &\frac{19+30t+12t^2}{24(1+t)},\ \text{if }P\in B, \\ &\frac{19+24t}{24(1+t)},\ \text{if }P\in\mathbf{l}_{12}\cup\mathbf{l}_{13}\cup\mathbf{l}_{14}\cup\mathbf{l}_{15}, \\ &\frac{3+4t}{4(1+t)},\ \text{otherwise}. \\ \end{aligned} \right. \end{align*} $$

Now, using [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Th. 1.7.1], we get (♠) as claimed.

Proof. We let $P_0$ be the singular point of the surface S, and let $\ell _0$ be the $\pi $ -exceptional curve. Then it follows from [Reference Coray and Tsfasman8] that there exists a birational morphism $\pi \colon \widetilde {S}\to \mathbb {P}^2$ such that $\pi (\ell _0)$ is a line, the map $\pi $ blows up three points $Q_1$ , $Q_2$ , and $Q_3$ contained in $\pi (\ell _0)$ and another point $Q_0\in \mathbb {P}^2\setminus \pi (\ell _0)$ .

For $i\in \{0,1,2,3\}$ , let $\mathbf {e}_i$ be the $\pi $ -exceptional curve such that $\pi (\mathbf {e}_i)=Q_i$ . For every $i\in \{1,2,3\}$ , let $\ell _i$ be the strict transform of the line in $\mathbb {P}^2$ that passes through $Q_0$ and $Q_i$ . Then $\ell _0$ , $\ell _1$ , $\ell _2$ , $\ell _3$ , $\mathbf {e}_0$ , $\mathbf {e}_1$ , $\mathbf {e}_2$ , and $\mathbf {e}_3$ are the only irreducible curves in the surface $\widetilde {S}$ that have negative self-intersections. Moreover, the intersections of these curves are given in the following table:

Note that $\eta (\ell _1)$ , $\eta (\ell _2)$ , $\eta (\ell _3)$ , $\eta (\mathbf {e}_0)$ , $\eta (\mathbf {e}_1)$ , $\eta (\mathbf {e}_2)$ , and $\eta (\mathbf {e}_3)$ are all lines contained in the surface S. Among them, only the lines $\eta (\mathbf {e}_1)$ , $\eta (\mathbf {e}_2)$ , and $\eta (\mathbf {e}_3)$ pass through the singular point $P_0$ .

For $(a_0,a_1,a_2,a_3,b_0,b_1,b_2,b_3)\in \mathbb {R}^8$ , we write

$$ \begin{align*}[a_0,a_1,a_2,a_3,b_0,b_1,b_2,b_3] := \sum_{i=0}^3 a_i \ell_i + \sum_{i=0}^3 b_i \mathbf{e}_i \in \mathrm{Pic} (\widetilde{S}) \otimes \mathbb{R}. \end{align*} $$

If $P=P_0$ , then $C=\ell _0$ , which implies that $\tau =2$ and

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &[-u, 1, 1, 1, 2, 0, 0, 0],\ \text{if }0\leqslant u\leqslant 1, \\ &[-u, 1, 1, 1, 2, 1-u, 1-u, 1-u],\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \\[6pt]N(u)&=\left\{\begin{aligned} &0,\ \text{if }0\leqslant u\leqslant 1, \\ &(u-1)(\mathbf{e}_1+\mathbf{e}_2+\mathbf{e}_3),\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\\[6pt]P(u)\cdot C&=\left\{\begin{aligned} &2,\ \text{if }0\leqslant u\leqslant 1, \\ &3 -u,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \quad P(u)^2=\left\{\begin{aligned} &5 - 2 u^2,\ \text{if }0\leqslant u\leqslant 1, \\ &(4-u)(2-u),\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \end{align*} $$

which implies that $S_S(C)=\frac {17}{15}$ and $S(W^{C}_{\bullet , \bullet };O)=1$ . Therefore, using (♢), we obtain $\delta _{P_0}(S)=\frac {15}{17}$ .

To proceed, we may assume that $P\ne P_0$ . If $O\in \mathbf {e}_0$ , we let $C=\mathbf {e}_0$ . Then $\tau =2$ , and

$$ \begin{align*}P(u)&=\left\{\begin{aligned} & [0, 1, 1, 1, 2-u, 0, 0, 0],\ \text{if }0\leqslant u\leqslant 1, \\ &[0, 2-u, 2-u, 2-u, 2-u, 0, 0, 0],\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \\[6pt]N(u)&=\left\{\begin{aligned} &0,\ \text{if }0\leqslant u\leqslant 1, \\ &(u-1) (\ell_1 + \ell_2 + \ell_3),\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \\[6pt]P(u)\cdot C&=\left\{\begin{aligned} &1+u,\ \text{if }0\leqslant u\leqslant 1, \\ &4-2u,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \quad P(u)^2=\left\{\begin{aligned} &5-2u-u^2,\ \text{if }0\leqslant u\leqslant 1, \\ &2(2-u)^2,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \end{align*} $$

which implies that $S_S(C)=\frac {13}{15}$ and $S(W^{C}_{\bullet , \bullet };O)\leqslant \frac {13}{15}$ , so that $\delta _P(S)=\frac {15}{13}$ by (⧫).

If $O\in \ell _1$ , we let $C=\ell _1$ . In this case, we have $\tau =2$ , and

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &[0, 1-u, 1, 1, 2, 0, 0, 0],\ \text{if }0\leqslant u\leqslant 1, \\ &[1-u, 1-u, 1, 1, 3-u, 2-2u, 0, 0],\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \\[6pt]N(u)&=\left\{\begin{aligned} &0,\ \text{if }0\leqslant u\leqslant 1, \\ &(u-1)(\ell_0+\mathbf{e}_0+2\mathbf{e}_1),\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\\ [6pt]P(u)\cdot C&=\left\{\begin{aligned} &1+u,\ \text{if }0\leqslant u\leqslant 1, \\ &4-2u,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \quad P(u)^2=\left\{\begin{aligned} &5-2u-u^2,\ \text{if }0\leqslant u\leqslant 1, \\ &2(2-u)^2,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\end{align*} $$

so that $S_S(C)=\frac {13}{15}$ . If $O\in \ell _1\setminus (\mathbf {e}_0\cup \mathbf {e}_1)$ , then $S(W^{C}_{\bullet , \bullet };O)=\frac {11}{15}$ . If $O=\ell _1\cap \mathbf {e}_1$ , then $S(W^{C}_{\bullet , \bullet };O)=1$ . Thus, using (⧫), we see that $\delta _P(S)=\frac {15}{13}$ if $O\in \ell _1\setminus \mathbf {e}_1$ , and $\delta _P(S)\geqslant 1$ if $O=\ell _1\cap \mathbf {e}_1$ .

Similarly, $\delta _P(S)=\frac {15}{13}$ if $O\in \ell _2\setminus \mathbf {e}_2$ or $O\in \ell _3\setminus \mathbf {e}_3$ , and $\delta _P(S)\geqslant 1$ if $O=\ell _2\cap \mathbf {e}_2$ or $O=\ell _3\cap \mathbf {e}_3$ .

If $O\in \mathbf {e}_1$ , we let $C=\mathbf {e}_1$ . In this case, we have $\tau =2$ , and

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &\Big[-\frac{u}{2}, 1, 1, 1, 2, -u, 0, 0\Big],\ \text{if }0\leqslant u\leqslant 1, \\ &\Big[-\frac{u}{2}, 2-u, 1, 1, 2, -u, 0, 0\Big],\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \\[6pt]N(u)&=\left\{\begin{aligned} &\frac{u}{2} \ell_0,\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{u}{2} \ell_0 + (u-1) \ell_1,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\\[6pt]P(u)\cdot C&=\left\{\begin{aligned} &\frac{2+u}{2},\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{4 - u}{2},\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \quad P(u)^2=\left\{\begin{aligned} &5-2u-\frac{u^2}{2},\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{(6-u)(2-u)}{2}, \ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \end{align*} $$

which implies that $S_S(C)=1$ and $S(W^{C}_{\bullet , \bullet };O)\leqslant \frac {13}{15}$ if $O\in \mathbf {e}_1\setminus \ell _0$ , so that $\delta _P(S)=1$ by (⧫).

Likewise, we see that $\delta _P(S)=1$ in the case when $O\in \mathbf {e}_2$ or $O\in \mathbf {e}_3$ . Thus, to complete the proof, we may assume that P is not contained in any line in S.

Now, we let C be the unique curve in the pencil $|\ell _1+\mathbf {e}_1|$ that contains P. By our assumption, the curve C is smooth and irreducible. Then $\tau =2$ , and

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &\Big[-\frac{u}{2}, 1-u, 1, 1, 2, -u, 0, 0\Big],\ \text{if }0\leqslant u\leqslant 1, \\ &\Big[-\frac{u}{2}, 1-u, 1, 1, 3-u, -u, 0, 0\Big],\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \\[6pt]N(u)&=\left\{\begin{aligned} &\frac{u}{2} \ell_0,\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{1}{2} u \ell_0 + (u-1)\mathbf{e}_0,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\\[6pt]P(u)\cdot C&=\left\{\begin{aligned} &\frac{4-u}{2},\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{3(2-u)}{2},\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \quad P(u)^2=\left\{\begin{aligned} &5-4u+\frac{u^2}{2},\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{3(2-u)^2}{2},\ \text{if }1\leqslant u\leqslant 2. \\ \end{aligned} \right.\end{align*} $$

Then $S_S(C)=\frac {11}{15}$ and $S(W^{C}_{\bullet , \bullet };O)=\frac {23}{30}$ . Thus, it follows from (⧫) that $\delta _P (S)\geqslant \frac {30}{23}>\frac {15}{13}$ .

Proof. Let $\mathbf {e}_1$ and $\mathbf {e}_2$ be $\eta $ -exceptional curves. Then $\widetilde {S}$ contains $(-1)$ -curves $\ell _1$ , $\ell _2$ , $\ell _3$ , $\ell _4$ , and $\ell _5$ such that the intersections of the curves $\ell _1$ , $\ell _2$ , $\ell _3$ , $\ell _4$ , $\ell _5$ , $\mathbf {e}_1$ , and $\mathbf {e}_2$ on $\widetilde {S}$ are given in the following table.

The curves $\eta (\ell _1)$ , $\eta (\ell _2)$ , $\eta (\ell _3)$ , $\eta (\ell _4)$ , and $\eta (\ell _5)$ are the only lines in S. Moreover, we have $\ell =\eta (\ell _1)$ , and $\eta (\ell _1)$ , $\eta (\ell _2)$ , an $\eta (\ell _5)$ are the only lines in S that contain a singular point of the surface S.

As in the proof of Lemma 25, for $(a_1,a_2,a_3,a_4,a_5,b_1, b_2)\in \mathbb {R}^7$ , we write

$$ \begin{align*}[a_1,a_2,a_3,a_4,a_5,b_1, b_2] := \sum_{i=1}^5 a_i \ell_i + \sum_{i=1}^2 b_i \mathbf{e}_i \in \mathrm{Pic} (\widetilde{S}) \otimes \mathbb{R}. \end{align*} $$

If $O\in \ell _1\setminus (\mathbf {e}_1\cup \mathbf {e}_2)$ , we let $C=\ell _1$ . In this case, we have $\tau =3$ , and

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &\Big[1-u, 1, 1, 1, 1, \frac{2-u}{2}, \frac{2-u}{2}\Big], \ \text{if }0\leqslant u\leqslant 2, \\ &[1-u, 3-u, 3-u, 0, 0, 0],\ \text{if }2\leqslant u\leqslant 3, \\ \end{aligned} \right. \\[6pt]N(u)&=\left\{\begin{aligned} &\frac{u}{2} (\mathbf{e}_1 + \mathbf{e}_2),\ \text{if }0\leqslant u\leqslant 2, \\ &(u-2) (\ell_2 + \ell_5) + (u-1)(\mathbf{e}_1 + \mathbf{e}_2), \ \text{if }2\leqslant u\leqslant 3, \\ \end{aligned} \right. \\[6pt]P(u)\cdot C&=\left\{\begin{aligned} &1, \ \text{if }0\leqslant u\leqslant 2, \\ &3-u, \ \text{if }2\leqslant u\leqslant 3, \\ \end{aligned} \right. \quad P(u)^2=\left\{\begin{aligned} &5 - 2u, \ \text{if }0\leqslant u\leqslant 2, \\ &(3-u)^2, \ \text{if }2\leqslant u\leqslant 3, \\ \end{aligned} \right. \end{align*} $$

which implies that $S_S(C)=\frac {19}{15}$ and $S(W^{C}_{\bullet , \bullet };O)\leqslant \frac {17}{15}$ , so that $\delta _P(S)=\frac {15}{19}$ by (⧫).

If $O\in \mathbf {e}_1$ , then $C=\mathbf {e}_1$ . In this case, we have $\tau =2$ , and

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &[1, 1, 1, 1, 1, 1-u, 1],\ \text{if }0\leqslant u\leqslant 1, \\ &[3-2u, 2-u, 1, 1, 1, 1-u, 2-u],\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\\[6pt]N(u)&=\left\{\begin{aligned} &0,\ \text{if }0\leqslant u\leqslant 1, \\ &2(u-1)\ell_1 + (u-1) \ell_2 + (u-1) \mathbf{e}_2,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\\[6pt]P(u)\cdot C&=\left\{\begin{aligned} &2u,\ \text{if }0\leqslant u\leqslant 1, \\ &3-u,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \quad P(u)^2=\left\{\begin{aligned} & 5-2u^2,\ \text{if }0\leqslant u\leqslant 1, \\ &(2-u)(4-u), \ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \end{align*} $$

which implies that $S_S(C)=\frac {17}{15}$ and $S(W^{C}_{\bullet , \bullet };O)\leqslant \frac {19}{15}$ , so that $\delta _P(S)\geqslant \frac {19}{15}$ by (♢).

On the other hand, we already know that $S_S(\ell )=\frac {19}{15}$ , which implies that $\delta _P(S)=\frac {19}{15}$ if $P=\eta (\mathbf {e}_1)$ . Similarly, we see that $\delta _P(S)=\frac {19}{15}$ if $P=\eta (\mathbf {e}_2)$ . Hence, we may assume that $O\not \in \mathbf {e}_1\cup \mathbf {e}_2\cup \ell _1$ .

If $O\in \ell _2$ , we let $C=\ell _2$ . In this case, we have $\tau =2$ , and

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &\Big[1, 1-u, 1, 1, 1, \frac{2-u}{2}, 1\Big],\ \text{if }0\leqslant u\leqslant 1, \\ &\Big[1, 1-u, 2-u, 1, 1, \frac{2-u}{2}, 1\Big],\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \\[6pt]N(u)&=\left\{\begin{aligned} &\frac{u}{2} \mathbf{e}_1,\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{u}{2} \mathbf{e}_1 + (u-1) \ell_3,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\\[6pt]P(u)\cdot C&=\left\{\begin{aligned} &\frac{2+u}{2},\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{4-u}{2},\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \quad P(u)^2=\left\{\begin{aligned} & 5-2u - \frac{u^2}{2},\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{(6-u)(2-u)}{2},\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \end{align*} $$

which implies that $S_S(C)=1$ and $S(W^{C}_{\bullet , \bullet };O)\leqslant \frac {13}{15}$ , so that $\delta _P(S)=1$ by (⧫).

Similarly, we see that $\delta _P(S)=1$ if $O\in \ell _5$ . Hence, if P is contained in a line in S that passes through a singular point of the surface S, then $\delta _P(S)=1$ . Thus, we may assume that $O\not \in \ell _2\cup \ell _2$ .

If $P\in \ell _3$ , we let $C=\ell _3$ . In this case, we have $\tau =2$ , and

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &[1, 1, 1-u, 1, 1, 1, 1],\ \text{if }0\leqslant u\leqslant 1, \\ & [1, 3-2u, 1-u, 2-u, 1, 2-u, 1],\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \\[6pt]N(u)&=\left\{\begin{aligned} &0,\ \text{if }0\leqslant u\leqslant 1, \\ & (u-1) (\ell_4 + 2 \ell_2 + \mathbf{e}_1),\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\\[6pt]P(u)\cdot C&=\left\{\begin{aligned} &1+u,\ \text{if }0\leqslant u\leqslant 1, \\ &4-2u,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \quad P(u)^2=\left\{\begin{aligned} &5-2u - u^2,\ \text{if }0\leqslant u\leqslant 1, \\ &2(2-u)^2,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \end{align*} $$

which implies that $S_S(C)=\frac {13}{15}$ and $S(W^{C}_{\bullet , \bullet };O)\leqslant \frac {13}{15}$ , so that $\delta _P(S)=\frac {15}{13}$ by (⧫).

Similarly, we see that $\delta _P(S)=\frac {15}{13}$ if $O\in \ell _4$ . Therefore, we may also assume that $O\not \in \ell _3\cup \ell _4$ .

Let C be the curve in the pencil $|\ell _2 + \ell _3|$ that contains O. Then C is smooth and irreducible, since O is not contained in the curves $\ell _1$ , $\ell _2$ , $\ell _3$ , $\ell _4$ , $\ell _5$ , $\mathbf {e}_1$ , and $\mathbf {e}_2$ by assumption. Then $\tau =2$ , and

$$ \begin{align*}P(u)&=\left\{\begin{aligned} &\Big[1, 1-u, 1-u, 1, 1, \frac{2-u}{2}, 1\Big],\ \text{if }0\leqslant u\leqslant 1, \\ &\Big[1, 1-u, 1-u, 2-u, 1, \frac{2-u}{2}, 1\Big],\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \\[6pt]N(u)&=\left\{\begin{aligned} &\frac{u}{2} \mathbf{e}_1,\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{u}{2} \mathbf{e}_1 + (u-1) \ell_4,\ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right.\\[6pt]P(u)\cdot C&=\left\{\begin{aligned} &\frac{4-u}{2},\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{3(2-u)}{2}, \ \text{if }1\leqslant u\leqslant 2, \\ \end{aligned} \right. \quad P(u)^2=\left\{\begin{aligned} &5-4u+\frac{u^2}{2},\ \text{if }0\leqslant u\leqslant 1, \\ &\frac{3(2-u)^2}{2},\ \text{if }1\leqslant u\leqslant 2. \\ \end{aligned} \right. \end{align*} $$

This implies that $S_S(C)=\frac {11}{15}$ and $S(W^{C}_{\bullet , \bullet };O)=\frac {23}{30}$ , so that $\delta _P(S)\geqslant \frac {30}{23}>\frac {15}{13}$ by (⧫).

Proof. Since the log pair $(S, c_j E_j|_S)$ is log canonical at P, we conclude that $\mathrm {ord}_F(E_j|_S)\leqslant \frac {A_S(F)}{c_j}$ . Thus, we get the first inequality in (♢). Moreover, since $P(u)|_S=-K_S-N(u)|_S$ , we have

$$ \begin{align*}\int_0^\tau\int_0^\infty\mathrm{vol}(P(u)|_S-vF\big)dvdu\leqslant\int_0^\tau (-K_S)^2 S_S(F)du =\tau (-K_S)^2 S_S(F)\leqslant A_S(F) \frac{\tau (-K_S)^2}{\delta_P(S)}. \end{align*} $$

Hence, the assertion follows.