Reduction to the non-reduced conics

Proof. The effective cone is generated by $D,F_1,F_2$. The intersection numbers are $$D^3=2,\qquad D^2F_1=-2,\qquad D^2F_2=-1,\qquad DF_1F_2=2,\qquad F_1^2=F_2^2=0.$$ The divisor $-K_X-uD$ is nef for $0\leq u\leq1$, and is not pseudoeffective for $u>1$. Its cube is $14-6u-6u^2-2u^3$. Hence $$S_X(D)=\frac1{14}\int_0^1(14-6u-6u^2-2u^3)\,du =\frac{17}{28}.$$ For the two fibres the positive parts are $$\begin{aligned} P_1(u)&=\begin{cases} D+(1-u)F_1+2F_2,&0\leq u\leq\frac12,\\ 2(1-u)D+(1-u)F_1+2F_2,&\frac12\leq u\leq1, \end{cases}\\ P_2(u)&=\begin{cases} D+F_1+(2-u)F_2,&0\leq u\leq1,\\ (2-u)D+F_1+(2-u)F_2,&1\leq u\leq2. \end{cases} \end{aligned}$$ The corresponding negative parts are $0,(2u-1)D$ and $0,(u-1)D$, respectively. These decompositions follow by intersecting with the two rulings of $D$ and with a conic fibre; equivalently, they are [43, Propositions 5.1 and 5.2]. Therefore $$\begin{aligned} S_X(F_1)&=\frac1{14}\left( \int_0^{1/2}(14-18u)\,du+ \int_{1/2}^1\bigl(24(1-u)^2-8(1-u)^3\bigr)\,du\right) =\frac{45}{112},\\ S_X(F_2)&=\frac1{14}\left( \int_0^1(14-9u)\,du+ \int_1^2\bigl(6(2-u)^2-(2-u)^3\bigr)\,du\right) =\frac{45}{56}. \end{aligned}$$ For completeness, the divisorial-stability reduction in [43, Section 5] uses the effective cone to reduce the verification to $F_2$, by the criterion of Fujita quoted there. The last computation gives $\beta_X(F_2)=11/56>0$, and thus proves the stated divisorial stability. This argument does not require any condition on the singular fibres. ◻

Proof. Write $D\simeq\mathbb P^1\times\mathbb P^1$, and denote the ruling classes by $l_1,l_2$. Since $D|_D=-l_1-l_2$ and $(-K_X)|_D=l_1+l_2$, the restriction of $-K_X-uD$ is $(1+u)(l_1+l_2)$. Choose the ruling $L\in|l_1|$ through $p$. For $0\leq v\leq1+u$, the divisor $(1+u-v)l_1+(1+u)l_2$ is nef, and beyond this interval it is not pseudoeffective. There are no fixed components in these restrictions. Thus the two successive expected vanishing orders are $$\begin{aligned} S(W^D_{\bullet,\bullet};L) &=\frac3{14}\int_0^1\int_0^{1+u} 2(1+u-v)(1+u)\,dv\,du=\frac{45}{56},\\ S(W^{D,L}_{\bullet,\bullet,\bullet};p) &=\frac3{14}\int_0^1\int_0^{1+u} (1+u)^2\,dv\,du=\frac{45}{56}. \end{aligned}$$ Abban–Zhuang adjunction for the flag $X\supset D\supset L\supset p$ and Lemma 85.1 now give $$\delta_p(X)\geq \min\left\{\frac{28}{17},\frac{56}{45},\frac{56}{45}\right\} =\frac{56}{45}.$$ This also reproduces [43, Lemma 6.1]. ◻

Proof. Let $p$ be a general point of the centre of $F$. If $p\in D$, apply Lemma 85.2. Hence we may assume that $p\notin D$. Let $S_i$ be the fibre of $\pi_i$ containing $C_0$; thus $S_i\sim F_i$. If $S_2$ is smooth along $C_0$, then [43, Lemma 6.2] gives $\delta_p(X)>1$.

Suppose that $S_2$ is singular at some point of $C_0$. The restriction $\pi_2|_D$ is a projection $\mathbb P^1\times\mathbb P^1\to\mathbb P^1$, so $S_2$ is smooth along $D$. Thus its singular point on $C_0$ is outside $D$. By [43, Lemma 3.5], $S_1$ is a smooth sextic del Pezzo surface. The conic is $C_0=L_3+L_4$, where $L_3,L_4$ are $(-1)$-curves meeting transversely. Moreover, $D|_{S_1}=L_1+L_2$ is a disjoint union of two $(-1)$-curves. We may number them so that $L_1L_3=L_2L_4=1$ and $L_1L_4=L_2L_3=0$.

Let $W^{S_1}_{\bullet,\bullet}$ be the series obtained by refining $H^0(X,-mK_X)$ with respect to $S_1$. The calculation in the corrected [43, Lemma 6.3] proves $$\delta_p(S_1;W^{S_1}_{\bullet,\bullet})\geq1.$$ We recall the relevant numbers, so that the non-strict inequality is explicit. For $p\in L_3\setminus(D\cup L_4)$, the flag beginning with $L_3$ has first expected vanishing order $1$ and point expected vanishing order $39/56$. The same holds with $L_3,L_4$ interchanged. At $p=L_3\cap L_4$, blow up $p$ on $S_1$ and denote the exceptional curve by $e$. Its log discrepancy is $2$ and its first expected vanishing order is $2$. The next expected vanishing order is $11/28$ away from the two points corresponding to $L_3,L_4$, and is $1$ at either of these two points. These are all the terms in the surface adjunction estimate. In particular, its lower bound is $1$, including at $L_3\cap L_4$.

It is essential to retain the positive remainder in the first adjunction step. Applied to a valuation centred in $S_1$, the last surface estimate gives $$S_X(F)\leq A_X(F)-\bigl(1-S_X(S_1)\bigr) \operatorname{ord}_F(S_1).$$ Indeed, choose a basis simultaneously compatible with the $S_1$-filtration and $\operatorname{ord}_F$, and write its basis divisor as $a_mS_1+\Omega_m$. Then $a_m\to S_X(S_1)$ and $\Omega_m|_{S_1}$ is the corresponding refined basis divisor. The refined threshold estimate says that $(S_1,(1-\epsilon)\Omega_m|_{S_1})$ is log canonical near $p$ for all sufficiently divisible $m$, with an error tending to zero. Inversion of adjunction gives $$A_X(F)\geq\operatorname{ord}_F(S_1) +(1-\epsilon)\operatorname{ord}_F(\Omega_m).$$ Letting $m$ tend to infinity and then $\epsilon$ tend to zero proves the inequality. This is the strictness argument underlying the equality discussion in [43, Section 8].

Here the centre of $F$ is contained in $C_0\subset S_1$, so $\operatorname{ord}_F(S_1)>0$. Lemma 85.1 therefore gives the strict inequality $$\beta_X(F)\geq\frac{67}{112}\operatorname{ord}_F(S_1)>0.$$ Thus the non-strict surface estimate is sufficient. ◻

Proof. The centre is neither a divisor, by Lemma 85.1, nor a subvariety meeting $D$, by Lemma 85.2. Suppose first that the centre $Z$ is a curve not contracted by $\pi_2$. The general cubic fibre is smooth, and only finitely many fibres of $\pi_2$ are singular. A general point $p\in Z$ therefore lies on a smooth cubic fibre. Applying [43, Lemma 6.2] at $p$ gives $\beta_X(F)>0$.

Suppose next that $Z$ is contained in a cubic fibre $S_2$ but is not contracted by $\pi$. The cubic fibre $S_2$ is normal by [43, Lemma 3.2]. The rationality argument in the proof of [43, Lemma 3.4] then shows that its singularities are Du Val. In particular, its singular locus is finite. There are consequently only finitely many conic fibres in $S_2$ meeting this singular locus. A general point $p\in Z$ avoids their union. The cubic surface $S_2$ is smooth along the conic through this point, so [43, Lemma 6.2] again gives a contradiction.

Thus a curve centre is contained in a conic fibre. A point centre is of course contained in its conic fibre as well. If that fibre were reduced, Lemma 85.3 would give the same contradiction. This proves the assertion. ◻