Two divisor refinements

Proof. The ideal of $C$ is generated, after twisting by $\mathcal O_M(1,3)$, by the equation of $S$ times sections of $\mathcal O_M(1,1)$ and by two lifts of $x(uf+vg)$ and $y(uf+vg)$ from $S$. This proves the assertion about $B$. All the proposed positive parts are nonnegative linear combinations of $L$, $B$, $T$ and $H$. In the first table, the first interval interpolates between $L$ and $2T+H$. For $u>1$, restriction to a fibre of $E\to C$ forces the component $(u-1)E$. After its removal, the remaining divisor is the pullback written in the table, and has no fixed divisorial component.

In the second table, $L-uT$ restricts to $(1-u)\ell_1+\ell_2$ on $D$. For $u>1$, its negative degree on the appropriate ruling forces multiplicity at least $u-1$ along $D$, because $D|_D=-\ell_1-\ell_2$. Removing this multiplicity gives $(2-u)B+(u-1)H$, which is globally generated when its coefficients are integral. The same conclusions for rational and real coefficients follow by taking multiples and continuity. Finally, beyond the stated thresholds the pushforward to $M$ has a negative coefficient of $H$, respectively $T$, and cannot be effective. ◻

Proof. For the first divisor, the intersection identities give $$P_D(u)^3= \begin{cases} 20-6u-6u^2-2u^3,&0\le u\le1,\\ 6(3-2u)^2,&1\le u\le3/2. \end{cases}$$ Consequently $$S_X(D)=\frac1{20} \left(\int_0^1(20-6u-6u^2-2u^3)\,du +\int_1^{3/2}6(3-2u)^2\,du\right) =\frac1{20}\left(\frac{29}{2}+1\right)=\frac{31}{40}.$$ For the fibre divisor, the two volume polynomials are $$P_T(u)^3= \begin{cases} 20-12u,&0\le u\le1,\\ (u-2)(u^2+2u-11),&1\le u\le2. \end{cases}$$ Thus $$S_X(T)=\frac1{20} \left(\int_0^1(20-12u)\,du +\int_1^2(u-2)(u^2+2u-11)\,du\right) =\frac1{20}\left(14+\frac{13}{4}\right)=\frac{69}{80}.$$ Both computations use volume on $X$, not on a weighted extraction. ◻

Proof. Take the ruling $Z\in|\ell_2|$ through $p$. On $D$, the moving parts in Lemma 90.1 have classes $$(1+u)(\ell_1+\ell_2)\quad(0\le u\le1),\qquad 2\ell_1+(6-4u)\ell_2\quad(1\le u\le3/2).$$ The fixed restriction in the second interval is $(u-1)C$. The volume of $a\ell_1+b\ell_2$ is $2ab$ when $a,b\ge0$. The first refinement therefore gives $$\begin{aligned} S(W^D;Z) &=\frac3{20}\int_0^1\int_0^{1+u} 2(1+u)(1+u-v)\,dv\,du\\ &\quad+\frac3{20}\int_1^{3/2}\int_0^{6-4u} 4(6-4u-v)\,dv\,du =\frac{61}{80}. \end{aligned}$$ For the flag $D\supset Z\supset\{p\}$ the moving contribution is $$\frac3{20}\left( \int_0^1\int_0^{1+u}(1+u)^2\,dv\,du +\int_1^{3/2}\int_0^{6-4u}4\,dv\,du\right)=\frac{69}{80}.$$ Since $C\to Q$ is an isomorphism, $C$ meets $Z$ transversely once. The fixed contribution is zero for $p\notin C$, and otherwise equals $$\frac6{20}\int_1^{3/2}\int_0^{6-4u}2(u-1)\,dv\,du =\frac1{20}.$$ All discrepancies in this smooth surface flag are one. The flag inequality, together with Lemma 90.2, now gives $$\delta_p(X)\ge \min\left\{\frac{40}{31},\frac{80}{61},\frac{80}{73}\right\} =\frac{80}{73}.$$ This reproduces [3, Lemma 5.68] with the two restriction integrals and the fixed contribution written separately. ◻

Proof. Only $p\notin D$ remains. Set $C_T=D|_T$. The moving restrictions are $-K_T$ for $0\le u\le1$ and $-K_T-(u-1)C_T$ for $1\le u\le2$; the fixed restrictions are supported on $C_T$. For any divisor $F$ over $T$ whose centre contains $p$, the fixed contribution is zero. Monotonicity of volume gives $$S(W^T;F)\le\frac3{10}\int_0^\infty \operatorname{vol}(-K_T-vF)\,dv =\frac65 S_T(F).$$ A smooth quartic del Pezzo surface satisfies $\delta(T)\ge4/3$ [3, Lemma 2.12]. Hence $S(W^T;F)\le(9/10)A_T(F)$. The first refinement has expected order $69/80$, so the Abban–Zhuang inequality yields $$\delta_p(X)\ge\min\{80/69,10/9\}=10/9.$$ ◻