Reduced conic fibres

Proof. The first assertion is precisely [25, Appendix A, Lemma 24], applied to the smooth quartic del Pezzo surface $S$ and its smooth conic $F$. The second is the usual quartic del Pezzo estimate [3, Lemma 2.12]. These are the only external surface estimates used in the smooth-conic case. In particular, the first assertion does not require $p$ to avoid the lines on $S$. ◻

Proof. Let $R$ be any prime divisor over $S$ whose centre contains $p$. Lemma 40.1, (40.1) and (40.2) give $$\begin{aligned} \frac{S(W^S;R)}{A_S(R)} &\le\frac1{80} +\frac3{20}\int_0^1 4(1+t)\frac{19+8t+t^2}{24}\,dt +\frac3{20}\int_1^{3/2}4(3-2u)^3\frac34\,du\\ &=\frac1{80} +\frac1{40}\left(19+\frac{27}{2}+3+\frac14\right) +\frac9{20}\cdot\frac18\\ &=\frac2{160}+\frac{143}{160}+\frac9{160} =\frac{77}{80}. \end{aligned}$$ Thus the local threshold of $W^S$ is at least $80/77$. Abban–Zhuang adjunction along the smooth divisor $S$ gives $$\delta_p(X)\ge\min\left\{\frac{80}{41},\frac{80}{77}\right\} =\frac{80}{77}.$$ ◻

Proof. The divisor $M=H_S-Z$ is nef and $M^2=1$. Indeed, the linear forms vanishing on the line $Z$ generate its ideal in the anticanonical embedding, so $\mathcal O_S(H_S-Z)$ is globally generated. In the first interval $P_t(v)=(1-v)H_S+vM+tF$ is nef. In the second interval it is $M+(1+t-v)F$, hence is nef. Equation (40.3) gives $P_t(v)=(3+2t-2v)M$ in the last interval. In each case it is orthogonal to the components of $N_t(v)$. These components are disjoint $(-1)$-curves, so their intersection matrix is negative definite. This proves the asserted Zariski decompositions. Finally, $$(D_t-vZ)\cdot M=3+2t-2v,$$ which excludes pseudoeffectivity for $v>\tau_t$.

Intersection products give the table. In particular, $$\begin{aligned} D_t^2S_{D_t}(Z) &=\int_0^1(4+4t-2v-v^2)\,dv +\int_1^{1+t}(5+4t-4v)\,dv\\ &\quad +\int_{1+t}^{\tau_t}(3+2t-2v)^2\,dv =\frac{17+30t+12t^2}{6}. \end{aligned}$$ The common contribution to $D_t^2T_t(p)$ is $$\int_0^1(1+v)^2\,dv+\int_1^{1+t}4\,dv +\int_{1+t}^{\tau_t}(6+4t-4v)^2\,dv=3+4t.$$ At $p\in Z\cap T_i$ the extra contribution is $$2\int_{1+t}^{\tau_t}(6+4t-4v)(v-1-t)\,dv=\frac16.$$ At $o$ it is $$2\int_1^{1+t}2(v-1)\,dv +2\int_{1+t}^{\tau_t}(6+4t-4v)(v-1)\,dv =2t^2+t+\frac16.$$ The five intersection points on $Z$ are distinct. These calculations give (40.5). ◻

Proof. The curve $C_S$ does not contain $Z$, since it is a smooth irreducible curve of genus two. Thus it contributes nothing to $\mathcal S_Z$. Equation (40.4) gives $$\begin{aligned} \mathcal S_Z &=\frac3{20}\int_0^1 \frac{17+30t+12t^2}{6}\,dt +\frac3{20}\int_1^{3/2}4(3-2u)^3\frac{17}{24}\,du\\ &=\frac9{10}+\frac{17}{320}=\frac{61}{64}. \end{aligned}$$ For $p\ne o$, equation (40.5) reads $$T_t(p)=\frac{3+4t}{4(1+t)}+\frac{r_p}{24(1+t)}.$$ The fixed part restricts to the divisor $(u-1)C_S|_Z$ in the tail. Its contribution at $p$ is $\mu_p/80$. Therefore $$\begin{aligned} \mathcal T_p &=\frac{\mu_p}{80} +\frac3{20}\int_0^1\left(3+4t+\frac{r_p}{6}\right)\,dt +\frac3{20}\int_1^{3/2}4(3-2u)^3 \left(\frac34+\frac{r_p}{24}\right)\,du\\ &=\frac{\mu_p}{80}+\frac34+\frac{r_p}{40} +\frac9{160}+\frac{r_p}{320} =\frac{129}{160}+\frac9{320}r_p+\frac{\mu_p}{80}. \end{aligned}$$ Here the fixed term has the same coefficient $1/80$ as in (40.1): summing all orders along $Z$ restores the leading dimension $P(u)|_S^2$ of the surface series.

Since $C_S\sim2H_S-F$, one has $C_S\cdot Z=2$. Hence $0\le\mu_p\le2$; also $r_p\le1$, because the $T_i$ are disjoint. This proves $$\mathcal T_p\le\frac{258+9+8}{320}=\frac{55}{64}.$$ Abban–Zhuang adjunction for the flag $p\in Z\subset S\subset X$ now gives $$\delta_p(X)\ge \min\left\{\frac{80}{41},\frac{64}{61},\frac{64}{55}\right\} =\frac{64}{61}.$$ ◻

Proof. The complementary class $H_S-F$ is a basepoint-free conic pencil. This standard fact can also be checked in the plane presentation above: $H_S-F=h-e_1$. If a line $D$ distinct from $Z,Z^{\dagger}$ passed through $o$, then $F\cdot D\ge2$, whereas nefness of $H_S-F$ would give $F\cdot D\le H_S\cdot D=1$. Thus the only lines through $o$ are the two components of $F$. Neither can be a component of a fibre of $|H_S-F|$, since $(H_S-F)\cdot Z=(H_S-F)\cdot Z^{\dagger}=1$. Every singular fibre of a conic pencil on a smooth del Pezzo surface is a union of two lines. It follows that $G$ is smooth. The equalities $G\cdot Z=G\cdot Z^{\dagger}=1$ prove transversality and the assertions about their strict transforms.

We have $M=\sigma^*(G+F/2)-R$. If $D\subset S$ is an irreducible curve different from $G$, then $$M\cdot\widetilde D =G\cdot D+\frac12F\cdot D-\operatorname{mult}_oD\ge0:$$ the local intersection with the smooth curve $G$ is at least $\operatorname{mult}_oD$, and $F$ is nef. Direct calculation gives $M\cdot\widetilde G=0$ and $M\cdot R=1$. Hence $M$ is nef, and $M^2=(H_S-F/2)^2-1=1$. ◻

Proof. Put $c=3+2t$ and write $\sigma^*D_t-vR=P_t(v)+N_t(v)$. The Zariski decomposition is $$N_t(v)= \begin{cases} 0,&0\le v\le1,\\ \dfrac{v-1}{2}(\widetilde Z+\widetilde Z^{\dagger}), &1\le v\le2+2t,\\[4pt] \dfrac{v-1}{2}(\widetilde Z+\widetilde Z^{\dagger}) +(v-2-2t)\widetilde G,&2+2t\le v\le c. \end{cases}$$ Indeed, $\sigma^*H_S-R$ is globally generated: it is the system of hyperplanes through $o$ in the anticanonical embedding after blowing up their base point. Thus $\sigma^*D_t-vR$ is nef for $0\le v\le1$. In the second interval $P_t(v)$ interpolates between $\sigma^*H_S-R+t\sigma^*F$ and $M$, so it is nef. In the last interval $P_t(v)=(c-v)M$. The negative components are disjoint negative curves and are orthogonal to $P_t(v)$ in each interval. This proves the decompositions. Finally, $$(\sigma^*D_t-vR)\cdot M=c-v,$$ so the pseudoeffective threshold is exactly $c$.

The volume and the degree on $R$ are $$\begin{array}{c|c|c} v&P_t(v)^2&P_t(v)\cdot R\\ \hline {[0,1]}&4+4t-v^2&v\\ {[1,2+2t]}&5+4t-2v&1\\ {[2+2t,3+2t]}&(3+2t-v)^2&3+2t-v. \end{array}$$ Consequently $$\begin{aligned} S_{D_t}(R) &=\frac1{4(1+t)}\left( \int_0^1(4+4t-v^2)\,dv +\int_1^{2+2t}(5+4t-2v)\,dv +\int_{2+2t}^{3+2t}(3+2t-v)^2\,dv\right)\\ &=\frac{3+2t}{2}. \end{aligned}$$ The log discrepancy is $A_S(R)=2$ and the different on $R$ is zero. For $q\in R$, the next refined expected order is $$U_t(q)=\frac1{4(1+t)}\left( \int_0^c(P_t(v)\cdot R)^2\,dv +2\int_0^c(P_t(v)\cdot R)\mathop{\mathrm{ord}}_q(N_t(v)|_R)\,dv\right).$$ The common first integral is $$\int_0^1v^2\,dv+\int_1^{2+2t}1\,dv +\int_{2+2t}^{3+2t}(3+2t-v)^2\,dv=\frac53+2t.$$ At either $\widetilde Z\cap R$ or $\widetilde Z^{\dagger}\cap R$, the additional integral is $$\int_1^{2+2t}(v-1)\,dv +\int_{2+2t}^{3+2t}(3+2t-v)(v-1)\,dv =2t^2+3t+\frac76.$$ At $\widetilde G\cap R$ it is $$2\int_{2+2t}^{3+2t}(3+2t-v)(v-2-2t)\,dv=\frac13.$$ Thus $$U_t(q)= \begin{cases} \dfrac{12t^2+30t+17}{24(1+t)}, &q\in(\widetilde Z\cup\widetilde Z^{\dagger})\cap R,\\[4pt] \dfrac12,&q=\widetilde G\cap R,\\[4pt] \dfrac{6t+5}{12(1+t)},&\text{otherwise}. \end{cases}$$ The first value is the largest, and $$\frac{3+2t}{4}-\frac{12t^2+30t+17}{24(1+t)} =\frac1{24(1+t)}>0.$$ Abban–Zhuang adjunction on the blown-up surface therefore gives $$\delta_o(S,D_t)\ge \min\left\{\frac{2}{S_{D_t}(R)}, \inf_{q\in R}\frac1{U_t(q)}\right\} =\frac4{3+2t}.$$ This is a bound for all divisors whose centres on $S$ contain $o$. Indeed, the log canonicity statement near the whole exceptional curve also controls strict transforms of curve centres passing through $o$. ◻

Proof. Choose $S$ as above. For any prime divisor $R'$ over $S$ whose centre contains $o$, Lemma 40.7 gives $$S_{D_t}(R')\le\frac{3+2t}{4}A_S(R'),\qquad S_{H_S}(R')\le\frac34A_S(R').$$ The second inequality is the same lemma at $t=0$. Hence (40.1) gives $$\begin{aligned} \frac{S(W^S;R')}{A_S(R')} &\le\frac1{80} +\frac3{20}\int_0^1(1+t)(3+2t)\,dt +\frac9{20}\int_1^{3/2}(3-2u)^3\,du\\ &=\frac1{80} +\frac3{20}\left(3+\frac52+\frac23\right) +\frac9{160}\\ &=\frac2{160}+\frac{148}{160}+\frac9{160} =\frac{159}{160}. \end{aligned}$$ Thus $\delta_o(S;W^S)\ge160/159$. Adjunction along $S$ now yields $$\delta_o(X)\ge\min\left\{\frac{80}{41},\frac{160}{159}\right\} =\frac{160}{159}.$$ ◻

Proof. Use Lemma 40.2, Lemma 40.4 or Lemma 40.9, according to the position of $p$. Their lower bounds are all at least $160/159$. By the definition of $\delta_p(X)$, every $V$ whose centre contains $p$ satisfies $S_X(V)\le(159/160)A_X(V)$, giving the stated beta inequality. ◻