Points on smooth fibres of the conic bundle

Proof. The intersection numbers follow from $H\cdot C_0=3$, $g(C_0)=0$ and $\deg N_{C_0/V_5}=4$. The two nef divisors $H$ and $B$ vanish on the fibres of the two extremal contractions, respectively, and hence generate the nef cone. If $D\ne E$ is a prime divisor, write $D\sim aH-bE$. Its pushforward gives $a\geq1$, while intersection with a general conic fibre gives $2(a-b)\geq0$. Thus the effective cone is generated by $E$ and $B$. Therefore $-K_X-uE$ is nef for $0\leq u\leq1$ and is not pseudoeffective for $u>1$. Its volume is $40-18(1+u)^2+4(1+u)^3$. Similarly, $-K_X-uH$ is nef precisely up to its pseudoeffective threshold $1$, and its volume is $5(2-u)^3-9(2-u)+4$. For $B$, the positive and negative parts are $$\begin{array}{c|c|c} u&P(u)&N(u)\\ \hline {[0,1]}&(2-u)H-(1-u)E&0\\ {[1,2]}&(2-u)H&(u-1)E. \end{array}$$ Consequently, $$S_X(B)=\frac1{26}\left( \int_0^1\left(5(2-u)^3-9(2-u)(1-u)^2+4(1-u)^3\right)du +\int_1^2 5(2-u)^3du\right)=\frac{63}{104}.$$ The other two expected orders are separately $$\begin{aligned} S_X(E)&=\frac1{26}\int_0^1\bigl(40-18(1+u)^2+4(1+u)^3\bigr)du=\frac12,\\ S_X(H)&=\frac1{26}\int_0^1\bigl(5(2-u)^3-9(2-u)+4\bigr)du=\frac{37}{104}. \end{aligned}$$

Write $E\cong\mathbb F_n$, with negative section $s$ and fibre $f$ of $E\to C_0$. Since the normal bundle has degree $4$, $n$ is even, and $-E|_E=s+(n-4)f/2$. Hence $B|_E=s+(n+2)f/2$. Its nefness implies $n\leq2$. If $n=2$, then $B\cdot s=0$ but $-K_X\cdot s=3$. This is impossible: $s$ would be a component of a conic-bundle fibre, whose total anticanonical degree is $2$, and $-K_X$ is ample. Therefore $n=0$. The morphism $\eta|_E$ is projection of the quadric $E\subset\mathbb P^3$ embedded by $|\mathcal O(1,1)|$ from a point outside it. It is a finite double cover with smooth branch conic. ◻

Proof. Take the inverse image of a general line through $q$. Since $\eta$ is smooth along $Z$, Bertini’s theorem gives a surface smooth along $Z$ and away from $Z$. We can also arrange that its intersection with $E$ is a smooth member of $|B|_E|=|\mathcal O(1,1)|$: the lines tangent to the branch conic of the finite double cover $\eta|_E$ form a proper subset of the pencil. Let $S_0=\pi(S)$. Along $C_0$, choose local coordinates $(t,x,y)$ with $C_0=(x=y=0)$ and write the equation of $S_0$ as $a(t)x+b(t)y+O((x,y)^2)=0$. If $a(t)=b(t)=0$ at some point, then $S\cap E$ contains the whole exceptional fibre over that point. This is impossible because $S\cap E$ is a smooth $(1,1)$-curve. Thus $S_0$ is smooth along $C_0$, and it is smooth elsewhere because $S$ is. Since $C_0$ is a Cartier divisor on $S_0$, its blowup is the identity. Hence $\pi|_S:S\to S_0$ is an isomorphism, and $S$ is a quintic del Pezzo surface. On it $C^2=1$ and $(-K_S)\cdot C=3$. The plane model associated to $|C|$ contracts four disjoint $(-1)$-curves and identifies $C$ with the strict transform of a line avoiding the four points. Finally $Z\sim B|_S=-K_S-C$, giving the stated class. Its intersections with the $e_i$ equal $1$; since $Z$ is smooth, they are transverse. ◻

Proof. Put $T=e_1+e_2+e_3+e_4$. On $S$, $C=l$, $Z=2l-T$, and $C\cdot Z=2$. The positive part $P(u,v)$ of $P(u)|_S-vZ$ and its negative part $N(u,v)$ are given by $$\begin{array}{c|c|c|c} u&v&P(u,v)&N(u,v)\\ \hline {[0,1]}&[0,2-u]&(2-u-v)Z+C&0\\ {[0,1]}&[2-u,\frac52-u]&(\frac52-u-v)(Z+T)&(v+u-2)T\\ {[1,2]}&[0,2-u]&(2-u-v)Z+(2-u)C&0\\ {[1,2]}&[2-u,\frac32(2-u)]&(\frac32(2-u)-v)(Z+T)&(v+u-2)T. \end{array}$$ Indeed, in the second and fourth rows the displayed positive part is orthogonal to every $e_i$, while the $e_i$ are disjoint negative curves. The other displayed divisors are nef, as follows by intersecting the ten $(-1)$-curves of the quintic del Pezzo surface. The volume becomes zero at the right endpoint in each row. These observations prove every claimed decomposition and threshold.

We justify that these volume calculations compute the actual refined series. For an integral nef moving part $M$ on $X$, one has $M-B=K_X+(M+H)$. The divisor $M+H$ is nef and big, so Kawamata–Viehweg vanishing gives $H^1(X,\mathcal O_X(M-B))=0$. Consequently restriction to $S$ is surjective, after removing the fixed multiple of $E$. Likewise, for each integral nef part $P$ in the table, $P-Z=K_S+(P+C)$, and $P+C$ is nef and big because $C$ is nef and $C^2=1$. Thus restriction from $S$ to $Z$ is surjective after removing the displayed fixed part. The resulting series on $Z\cong\mathbb P^1$ is complete. The formulas below follow either by summing dimensions of these complete series and taking limits, or by the multigraded Abban–Zhuang volume formulas. Endpoint rounding contributes $o(1)$ to the normalized moments.

There is no fixed component along $Z$ in $N(u)|_S$. The surface volumes in the four rows are, respectively, $$9-4u-4v,\quad4(\tfrac52-u-v)^2,\quad (2-u)(10-5u-4v),\quad(6-3u-2v)^2.$$ The surface formula is therefore $$\begin{aligned} S(W^S;Z)=\frac3{26}\bigg(&\int_0^1\int_0^{2-u}(9-4u-4v)\,dv\,du\\ &+\int_0^1\int_{2-u}^{5/2-u}4(\tfrac52-u-v)^2\,dv\,du\\ &+\int_1^2\int_0^{2-u}(2-u)(10-5u-4v)\,dv\,du\\ &+\int_1^2\int_{2-u}^{3(2-u)/2}(6-3u-2v)^2\,dv\,du\bigg) =\frac{171}{208}. \end{aligned}$$ Write $\Omega_i$ for the four integration regions and $a(u,v)=P(u,v)\cdot Z$. Its four expressions are $2$, $4(5/2-u-v)$, $2(2-u)$, and $4(3(2-u)/2-v)$. The threefold fixed part restricts to $(u-1)_+C$. Thus the point formula gives $$S(W^{S,Z};p)=\frac3{26}\sum_i\iint_{\Omega_i}a(u,v)^2\,dv\,du +\frac6{26}\sum_i\iint_{\Omega_i}a(u,v) \left((u-1)_+d_p+n(u,v)r_p\right)\,dv\,du,$$ where $n(u,v)=v+u-2$ in the second and fourth rows, and is zero otherwise. The three integrals, including their normalization factors, are $47/52$, $5/104$, and $5/208$. This proves the formula. Since $C\cap T=\varnothing$, one cannot have both $d_p>0$ and $r_p>0$. Moreover $d_p\leq2$ and $r_p\leq1$, because the $e_i$ are disjoint. The assertions follow. ◻

Proof. For each sufficiently divisible $m$, choose a basis of $H^0(X,-mK_X)$ adapted simultaneously to $v$ and the single lexicographic flag $X\supset S\supset Z\supset\{p\}$. This requires compatibility with two filtrations. Let $D_m$ be its normalized basis divisor, and write $D_m=a_mS+D'_m$. On $S$, write $D'_m|_S=b_mZ+R_m$, with $R_m$ effective and not containing $Z$. Let $c_m=(R_m\cdot Z)_p$. The flag moment formulas give $$a_m\longrightarrow\frac{63}{104},\qquad b_m\longrightarrow\frac{171}{208},\qquad c_m\longrightarrow S(W^{S,Z};p)\leq1.$$ Choose positive $\lambda_m\to1$ such that $\lambda_mb_m\leq1$ and $\lambda_mc_m\leq1$. For example, take $$\lambda_m=\min\{1,1/\max(b_m,c_m)\},$$ omitting reciprocals of zero. Since $Z$ is smooth at $p$, inversion of adjunction gives that $(S,Z+\lambda_mR_m)$ is log canonical near $p$. Therefore $(S,\lambda_mD'_m|_S)$ is log canonical there. A second application gives that $(X,S+\lambda_mD'_m)$ is log canonical near $p$. Both boundaries are effective. Evaluating at $v$ yields $$A_X(v)-\lambda_mv(D_m)\geq(1-\lambda_ma_m)v(S).$$ The basis is adapted to $v$, so $v(D_m)=S_m(v)\to S_X(v)$. Taking the limit proves the assertion. For a valuation with centre $Z$, apply the same argument near a general point of $Z$. ◻

Proof. For $D\ne E$, its pushforward is a nonzero effective divisor of class $a\mathcal O_{V_5}(1)$ for an integer $a\geq1$. Write $D\sim aH-bE=aB+(a-b)E$. The description of the effective cone gives $a-b\geq0$. Thus $D-B$ is effective. Monotonicity of volume implies $S_X(D)\leq S_X(B)=63/104$. ◻

Proof. At a point outside $R$, the intersection of the smooth conic fibre with $E$ is either absent or transverse at that point. Lemma 71.3 and the iterated Abban–Zhuang estimate give $$\delta_p(X)\geq\min\left\{\frac{104}{63},\frac{208}{171}, \frac{104}{99}\right\}=\frac{104}{99}.$$ For the ramification curve, write $E=\mathbb P^1\times\mathbb P^1$, with rulings $s,f$ chosen so that $f$ is a fibre of $\pi|_E$. The ramification formula gives $R\sim s+f$. Since $-E|_E=s-2f$ and $H|_E=3f$, one has $$(-K_X-uE)|_E=(1+u)s+(4-2u)f\quad(0\leq u\leq1).$$ There is no fixed part in this interval. Thus $$S(W^E;R)=\frac3{26}\int_0^1\int_0^{1+u} 2(1+u-v)(4-2u-v)\,dv\,du=\frac{63}{104}.$$ The first-stage moment is $S_X(E)=1/2$, so $\delta_R(X)\geq\min\{2,104/63\}=104/63$. Finally, let $T$ be a curve not contained in $\eta^{-1}(\Delta)$. If $T\ne R$, a general point of $T$ lies outside both $R$ and $\eta^{-1}(\Delta)$, so the first estimate excludes every valuation with centre $T$. If $T=R$, use the second estimate. ◻