Global reduction and nodal auxiliary sections

Proof. This is [4, proof of Theorem 3.3, equation (3.6)], with $\lambda=1$. For completeness, choose an $m$-basis divisor compatible with the filtrations of both $T$ and $Q$, and write it as $a_mT+D'_m$, where $a_m\to S_X(T)$. The restriction $D'_m|_T$ is a basis divisor of $W^T$. The threshold assumption and convergence for refined series provide numbers $\lambda_m\to1$, with $\lambda_m>0$, such that $(T,\lambda_m D'_m|_T)$ is log canonical near $p$. Inversion of adjunction gives $$A_X(Q)\geq \lambda_m\mathop{\mathrm{ord}}_Q(D'_m)+\mathop{\mathrm{ord}}_Q(T).$$ Compatibility with $Q$ and passage to the limit give the assertion. A valuation has positive order on $T$ when its centre is contained in $T$. ◻

Proof. The intersection numbers are $H^3=H^2F=4$ and $HF^2=F^3=0$. For $0\leq t\leq2$, the positive and fixed parts of $L-tF$ are $$(P_t,N_t)= \begin{cases} (H+(1-t)F,0),&0\leq t\leq1,\\ ((2-t)H,(t-1)E),&1\leq t\leq2. \end{cases}$$ Indeed, $H$ and $F$ are nef. For $t>1$, restriction to an exceptional fibre of $\pi$ forces multiplicity at least $t-1$ along $E$, and the remaining divisor is $(2-t)H$. For $t>2$ the intersection with $H^2$ is negative. Consequently $$S_X(F)=\frac1{16}\left(\int_0^1(16-12t)\,dt+ \int_1^2 4(2-t)^3\,dt\right)=\frac{11}{16}.$$ The divisor $L-uH=(1-u)H+F$ is nef precisely for $0\leq u\leq1$ and is not pseudoeffective for $u>1$, since its intersection with $HF$ is then negative. Similarly $L-uE=(1-u)H+(1+u)F$ is nef for $0\leq u\leq1$ and is not pseudoeffective afterwards. Thus $$S_X(H)=\frac1{16}\int_0^1 4(1-u)^2(4-u)\,du=\frac5{16}, \qquad S_X(E)=\frac1{16}\int_0^1(16-24u+8u^3)\,du=\frac38.$$ ◻

Proof. The surface $F_0$ is isomorphic to its image on $V$: blowing up the Cartier curve $C$ on that hyperplane section is an isomorphism. The curve $C_0$ is smooth, lies in the smooth locus, and is anticanonical on $F_0$. The decomposition in Lemma 21.3 restricts to $-K_{F_0}$ for $0\leq t\leq1$ and to $(2-t)(-K_{F_0})$ for $1\leq t\leq2$, with fixed part $(t-1)C_0$ in the latter interval.

These restrictions give complete moving systems for the asymptotic calculation. For rational $0<t<1$, the divisor $L-tF$ is ample; the restriction sequence and vanishing give surjectivity onto $F_0$ for sufficiently divisible large multiples. For $1<t<2$, remove the fixed multiple of $E$. The remaining sections come from multiples of $H$ on $V$, and restriction to the hyperplane section is surjective since $V$ is arithmetically Cohen–Macaulay. The endpoints have measure zero. Thus the refined volume formula gives $$\begin{aligned} S(W^{F_0};Q) &=\frac3{16}\int_1^2 4(2-t)^2(t-1)\mathop{\mathrm{ord}}_Q(C_0)\,dt\\ &\quad+\frac3{16}\left(1+\int_1^2(2-t)^3\,dt\right) \int_0^\infty\mathop{\mathrm{vol}}(-K_{F_0}-vQ)\,dv\\ &=\frac1{16}\mathop{\mathrm{ord}}_Q(C_0)+\frac{15}{16}S_{F_0}(Q). \end{aligned}$$ This is the same calculation as [8, Lemma 2.1]; its proof uses $L^3=4d$ and works here with $d=4$. If the centre contains a point outside $C_0$, its order on $C_0$ is zero. Otherwise $\mathop{\mathrm{ord}}_Q(C_0)\leq A_{F_0}(Q)$, since $(F_0,C_0)$ is log canonical. Abban–Zhuang adjunction gives the displayed estimates. If $d_p\geq1$, the refined threshold is at least $1$, so Lemma 21.2, applied with $T=F_0$, gives $\beta_X(Q)\geq\frac5{16}\mathop{\mathrm{ord}}_Q(F_0)>0$ for the indicated centres. ◻

Proof. Blowing up $\pi(p)$ on the smooth quartic del Pezzo surface $\overline S$ gives a smooth cubic surface, since $\overline S$ contains no line through this point. Its exceptional curve is a line. Projection from this line gives a conic bundle with fibre class $h-2G$. Pullback after the other four blowups proves that $T$ is nef. The class $b$ is also nef, being the pullback of the elliptic fibration on $S$.

For $0\leq v\leq2a$, the proposed positive part lies on the segment between the nef classes $ah+b$ and $aT+b$. On the second interval it is a nonnegative combination of $T$ and $b$. The negative part is supported on the curve $\widetilde B$ of square $-4$, and $P_{a,v}\widetilde B=0$. These conditions prove that this is the Zariski decomposition. Finally $$T\cdot(ah+b-vG)=4a+4-2v,$$ which is negative for $v>2+2a$. This proves the claimed threshold. ◻

Proof. The surface-volume and curve-restriction formulas of Abban–Zhuang theory, with the original normalization $L^3=16$, have factor $3/16$. Complete positive-part systems give upper bounds for the actual restricted systems, so the following upper bounds suffice without an assertion of completeness at boundary parameters. Lemma 21.6 gives $$P_{a,v}^2=\begin{cases} 4a^2+8a-v^2,&0\leq v\leq2a,\\ 8a+8a^2-4av,&2a\leq v\leq2+2a, \end{cases} \quad P_{a,v}\cdot G=\begin{cases}v,&v\leq2a,\\2a,&v\geq2a.\end{cases}$$ The expected order of $G$ is therefore at most $$\begin{aligned} &\frac3{16}\int_0^1\left(\int_0^{2a}(4a^2+8a-v^2)\,dv +\int_{2a}^{2+2a}(8a+8a^2-4av)\,dv\right)da\\ &\quad =\frac3{16}\int_0^1\left(\frac{16}3a^3+16a^2+8a\right)da=2. \end{aligned}$$ The moving contribution to a point term on $G$ is at most $$\frac3{16}\int_0^1\left(\int_0^{2a}v^2\,dv+ \int_{2a}^{2+2a}4a^2\,dv\right)da =\frac3{16}\int_0^1\left(\frac83a^3+8a^2\right)da=\frac58.$$ At a transverse intersection with $\widetilde B$, the fixed-part contribution is $$\frac3{16}\int_0^1\int_{2a}^{2+2a} 2(2a)(v/2-a)\,dv\,da =\frac3{16}\int_0^1 4a\,da=\frac38.$$ Thus the point term is at most $5/8+3/8=1$. Since $A_S(G)=2$ and the different on $G$ is zero, surface adjunction gives a local threshold at least $\min\{2/2,1\}=1$. ◻

Proof of Theorem 21.1. Apply Lemma 21.2 with $T=S$, using Lemma 21.7 and $S_X(S)=5/16$. For a divisor centred at $p$ one has $\mathop{\mathrm{ord}}_Q(S)>0$, which proves the assertion. ◻

Proof. This is the smooth-locus content of the complete quartic tables in [12, §6.2, Tables 21–35]. A smooth quartic del Pezzo surface has threshold $4/3$. For the singular surfaces, the values below one in their smooth loci are exactly those in the table. In particular, the smooth-locus bound for a $D_4$ surface is one, while that for a surface with a single $A_2$ or $A_3$ singularity is at least one. The equations and incidences of the special lines are given in Lemma 22.17. ◻

Proof. Every fibre is smooth along $C$, since the two hyperplanes cutting out $C$ have independent differentials on $V$ there. Thus $p\notin E$. A general hyperplane section through $\pi(p)$ is smooth and meets $C$ transversally at four distinct points. There are only finitely many lines on $V$ through $\pi(p)$. Indeed, their directions are cut out by two quadrics in the projectivized tangent space $\mathbb P^2$; a positive-dimensional intersection would give a plane or a quadric cone surface in $V$, or the two quadrics would coincide and give a quadric cone surface. Such a surface has degree $1$ or $2$, whereas $\operatorname{Pic}(V)=\mathbb Z[H]$ makes every effective divisor have degree divisible by $4$. Hence $\overline S$ can be chosen to contain none of these lines.

The general hyperplane section of the integral fibre through its singular point is irreducible, by Bertini irreducibility. Here projection from that point has two-dimensional image: otherwise the fibre would be a cone with that vertex. Since the fibre is nondegenerate in $\mathbb P^4$, such a vertex has embedding dimension $4$, whereas a hypersurface in the smooth threefold $V$ has embedding dimension at most $3$. Locally an $A_n$ singularity has equation $xy+z^{n+1}=0$. Restriction to a general smooth surface through the point has a nondegenerate quadratic term in two variables, and so gives an ordinary node. The section has arithmetic genus $1$ and is irreducible; the node accounts for its entire arithmetic genus, so it has no other singularities. Its strict transform on $S$ therefore has the required properties. ◻

Proof of Theorem 1.1 for Family 2.10. Let $Q$ be a prime divisor over $X$, and put $Z=c_X(Q)$. If $Z$ dominates $\mathbb P^1$, then $Z$ meets a smooth fibre. Example 21.5 gives $\beta_X(Q)>0$.

Otherwise $Z$ is contained in a fibre $F_0$. Suppose first that $Z$ meets its smooth locus, and choose a point $p$ of this intersection. If $p$ is on a special line, Theorem 19.1(3) gives $\delta_p(X)\ge32/27>1$, so $\beta_X(Q)>0$. If $p$ is not on a special line, Lemma 21.9 gives $\delta_p(F_0)\ge1$. The strict fibre estimate in Lemma 21.4 then gives $$\beta_X(Q)\ge\frac5{16}\mathop{\mathrm{ord}}_Q(F_0)>0,$$ because $Z\subset F_0$.

The remaining possibility is that $Z$ is a singular point of $F_0$. Quartic Du Val del Pezzo surfaces have only $A$-type, $D_4$, or $D_5$ singularities. For an $A$-type point, Lemma 21.10 and Theorem 21.1 give $\beta_X(Q)>0$. For a $D_4$ or $D_5$ point, apply Theorem 19.1(1)–(2). Thus every prime divisor over $X$ has positive $\beta_X$, and the valuative criterion proves K-stability. ◻