The special reduced fibre for second-kind lines

Proof. The morphism $f|_E$ is given by a base point free subsystem of $|s+2g|$ on $E\cong\mathbb F_2$; it contracts the negative section $s$ to $y_0$. Thus $f^{-1}(y_0)=s+Z$, where $Z$ is another line. Choose a general base line through $y_0$ and let $S$ be its inverse image. This surface is smooth: away from the fixed fibre this is Bertini, along its smooth locus it follows from smoothness of the conic bundle, and at its node a general base direction retains the nonzero derivative supplied by smoothness of $X$. The pencil induced on $E$ has fixed component $s$ and moving part a base point free pencil in $|2g|$. A general member of that pencil is $e_1+e_2$, where $e_1,e_2$ are distinct ruling fibres. Put $$A=H|_S=-K_S,\qquad F=B|_S=s+Z,\qquad R=e_1+e_2,\qquad c=E|_S=s+R.$$ The line $s$ is mapped isomorphically to $\ell$, and $Z$ is mapped to a line. Consequently $As=AZ=1$, while $Ae_i=0$. Adjunction on the smooth surface $S$ gives $$s^2=Z^2=-1,\quad e_i^2=-2,\quad sZ=se_i=1,\quad Ze_i=e_1e_2=0.$$ Here $sZ=1$ also follows directly from the reduced nodal conic fibre, and the remaining intersections follow from the geometry of $E$ and the equalities $Fe_i=1$. In particular, the three points $s\cap Z,s\cap e_1,s\cap e_2$ are distinct. The divisors $A,F$ are nef, and $A=c+F=2s+Z+R$.

The refined surface series still has moving and fixed classes given by (24.1). Its asymptotic completeness for $0<u<1$ follows from ampleness of $L-uB$. For $1<u<2$, remove the fixed multiple of $E$. Restriction of $H^0(X,kH)$ to $S$ is surjective for $k\gg0$: its obstruction is $H^1(X,(k-1)H+E)$, which is $H^1(V,\mathcal O_V(k-1))=0$, since $\pi_*\mathcal O_X(E)=\mathcal O_V$ and $R^i\pi_*\mathcal O_X(E)=0$ for $i>0$. The latter statements follow on the blow-up fibres from $\mathcal O_X(E)|_{\mathbb P^1}=\mathcal O_{\mathbb P^1}(-1)$. Hence the surface flag formulas apply to this weak del Pezzo surface.

First take the flag curve $Z$. The decomposition in Lemma 24.3 applies verbatim with its second component equal to $s$. Indeed, $A-Z=2s+R$ is nef: it is effective, has zero intersection with each component of its support, and has nonnegative intersection with every other irreducible curve. Moreover, $c|_Z$ is the single reduced point $s\cap Z$. Thus the constants are the same as in that lemma, and every point of $Z$ has local threshold at least $36/35$.

It remains to use the flag curve $s$. For $0\le u\le1$, put $a=1-u$, so the moving class is $D_u=A+aF$. For the parameter $v$ of this flag, its Zariski decompositions are $$\begin{array}{c|c|c} v&P_{u,v}&N_{u,v}\\\hline {[0,a]}&A+aF-vs&0\\ {[a,1]}&A+aF-vs-\frac{v-a}{2}R&\frac{v-a}{2}R\\ {[1,2+a]}&(2+a-v)\bigl(F+\frac12R\bigr)&\frac{v-a}{2}R+(v-1)Z. \end{array}$$ To verify nefness, $A+aZ$ is nef for $0\le a\le1$: the only possible negative intersection of the effective summand $aZ$ is with $Z$, where the full intersection is $1-a$. The first two intervals therefore interpolate between nef endpoint classes. For the other endpoint use $M=F+R/2$, an effective rational divisor with $Ms=1$, $MZ=Me_i=0$, hence nef. The negative parts are supported on mutually disjoint negative curves and are orthogonal to their positive parts. Finally $(A+aF-vs)M=2+a-v$, so $2+a$ is the pseudo-effective threshold.

Writing $d=P_{u,v}s$, the three intervals give $$\begin{array}{c|cc} v&P_{u,v}^2&d\\\hline {[0,a]}&3+4a-2v-v^2&1+v\\ {[a,1]}&(1+a)(3+a-2v)&1+a\\ {[1,2+a]}&(2+a-v)^2&2+a-v. \end{array}$$ The required inner integrals are $$\begin{split} I(a):=\int P_{u,v}^2\,dv&=2a^2+4a+\frac73,\\ J(a):=\int d^2\,dv&=-\frac{a^3}{3}+a^2+3a+\frac43,\\ K(a):=\int d\operatorname{coeff}_{e_i}(N_{u,v})\,dv &=\frac{(a+1)(a^2-4a+7)}{12},\\ M(a):=\int d\operatorname{coeff}_Z(N_{u,v})\,dv &=\frac{(a+1)^3}{6}. \end{split}$$ For $1\le u\le2$, scale the $a=0$ decomposition by $b=2-u$; the four inner integrals are respectively $b^3I(0)$, $b^3J(0)$, $b^3K(0)$, $b^3M(0)$. The original fixed part $(u-1)c=(u-1)(s+R)$ shifts the order along $s$ by $u-1$. After removing this coefficient of $s$, the original fixed divisor that remains on the flag curve is $(u-1)R|_s$, not $(u-1)c|_s$. These terms contribute $1/24$ to the expected order along $s$ and $1/24$ at each point $s\cap e_i$, respectively. Consequently $$\begin{split} S(W^S;s)&=\frac16\left(\int_0^1 I(a)\,da+\frac{I(0)}4\right)+\frac1{24} =\frac{35}{36},\\ S(W^{S,s};p)&=\frac{41}{72} +\frac{11}{36}\,\mathbf 1_{\{p\in s\cap R\}} +\frac29\,\mathbf 1_{\{p=s\cap Z\}}\le\frac78. \end{split}$$ For clarity, the first point coefficient is $\frac13(\int_0^1K(a)\,da+K(0)/4)+1/24=11/36$, and the second is $\frac13(\int_0^1M(a)\,da+M(0)/4)=2/9$. The point indicators cannot occur simultaneously. Abban–Zhuang adjunction now gives, for every $p\in s$, $$\delta_p(X)\ge\min\left\{\frac{72}{41},\frac{36}{35},\frac87\right\} =\frac{36}{35}.$$ This covers both components of the special fibre. ◻