Fibre refinements and the remaining centres

Proof. The identities $H^3=H^2F=6$ and $F^2=0$ give $L^3=24$. The positive and fixed parts of $L-uF$ are $$(P(u),N(u))= \begin{cases} (H+(1-u)F,0),&0\le u\le1,\\ ((2-u)H,(u-1)E),&1\le u\le2. \end{cases}$$ Indeed, $H$ and $F$ are nef. For $u>1$, the fibres of $E\to C$ force the fixed multiplicity $u-1$. Subtracting this multiple of $E$ leaves the nef divisor $(2-u)H$. The intersection with $HF$ shows that $u>2$ is not pseudoeffective. Therefore $$S_X(F)=\frac1{24}\left(\int_0^1(24-18u)\,du+ \int_1^2 6(2-u)^3\,du\right)=\frac{11}{16}.$$ Similarly $L-uE=(1-u)H+(1+u)F$ is nef for $0\le u\le1$, and its intersection with $HF$ is negative for $u>1$. Consequently $$S_X(E)=\frac1{24}\int_0^1(24-36u+12u^3)\,du=\frac38.$$ The restrictions of the positive parts to $S$ are $-K_S$ for $u\le1$ and $(2-u)(-K_S)$ for $u\ge1$. The fixed part restricts to $(u-1)C_S$ in the latter interval. The complete moving systems occur in the restriction calculation: for $0<u<1$ this follows from vanishing, and for $1<u<2$ it follows after removing the fixed part from restriction of multiples of $H$ to its member $S$ on $V$. For this last assertion one may use the restriction sequence and Kodaira vanishing on the smooth index-two Fano threefold $V$. The endpoints do not affect the integrals. The coefficients are $$\frac{3\cdot6}{24}\left(1+\int_1^2(2-u)^3\,du\right)=\frac{15}{16}, \qquad \frac{3\cdot6}{24}\int_1^2(2-u)^2(u-1)\,du=\frac1{16}.$$ This proves the asserted equality for the actual restricted series. All decompositions in this lemma take place on $X$ itself. ◻

Proof. The pair $(S,C_S)$ is log canonical, so $v(C_S)\le A_S(v)$. Lemma 110.1 gives $S(W^S;v)\le A_S(v)$. The basis-divisor proof of Abban–Zhuang adjunction, retaining the coefficient of $S$, gives $\beta_X(Q)\ge(1-S_X(S))\operatorname{ord}_Q(S)$. For clarity, choose a basis simultaneously adapted to $Q$ and $S$, write its basis divisor as $a_mS+D'_m$, and restrict $D'_m$ to $S$. The refined threshold bound gives numbers $c_m\to1$ for which $(S,c_mD'_m|_S)$ is log canonical near $p$. Inversion of adjunction then gives $$A_X(Q)\ge c_m S_m(Q)+(1-c_ma_m)\operatorname{ord}_Q(S).$$ Pass to the limit. If $p\notin E$, all relevant valuations of $S$ have zero order on $C_S$, and the refined threshold is at least $16/15$. The other Abban–Zhuang term is $16/11$. ◻

Proof. We give the surface calculations to fix the values needed here. On the minimal resolution, write the anticanonical class as $D$. At a smooth point off all lines, blow up the point, with exceptional curve $G$. The pencil classes $h-e_i$ in the plane model give the following Zariski decomposition. For $0\le v\le2$ its positive part is $D-vG$. For $2\le v\le3$ its negative part is $(v-2)Z$, where $Z$ has one of the following forms: $$Z=B_1+B_2+B_3,\qquad Z=2B_1+C_1+B_2,\qquad Z=3B_1+2C_1+C_2.$$ Here the $B_i$ are the proper transforms of the distinct moving pencil members through the point and have square $-1$; the $C_i$ are the intervening $(-2)$-curves. The three cases correspond to three distinct plane points, one infinitely near pair, and a chain of three infinitely near points. The curves $B_i$ meet $G$ in distinct points, and the $C_i$ do not meet $G$. In each case the proposed positive part is nef: its intersection with every negative curve is nonnegative, and the displayed divisor has zero intersection with the support of the negative part. These negative curves generate the effective cone of the resulting weak quintic del Pezzo surface. Equivalently, the same verification uses the plane classes $h-e_i$ and their residual exceptional chains. The squares and intersections are $$P(v)^2=\begin{cases}6-v^2,&0\le v\le2,\\ 2(3-v)^2,&2\le v\le3, \end{cases} \qquad P(v)G=\begin{cases}v,&0\le v\le2,\\ 2(3-v),&2\le v\le3. \end{cases}$$ They give $S_S(G)=5/3$. At any point $q\in G$, let $k$ be the coefficient of the component of $Z$ meeting $q$; put $k=0$ when there is no such component. Then $k\le3$ and $$S(W_S^G;q) \le\frac16\left(\int_0^2v^2\,dv+ \int_2^3 4(3-v)^2\,dv\right) +\frac{k}{3}\int_2^3 2(3-v)(v-2)\,dv =\frac23+\frac{k}{9}\le1.$$ On a smooth sextic surface $k\le1$, so this bound is $7/9$ and Abban–Zhuang gives $\delta_p(S)\ge6/5$.

On a smooth sextic surface, a line $T$ has adjacent lines $U,V$. For $0\le v\le1$, $D-vT$ is nef. For $1\le v\le2$, the negative part is $(v-1)(U+V)$. Thus $$P(v)^2=\begin{cases}6-2v-v^2,&0\le v\le1,\\ (2-v)(4-v),&1\le v\le2, \end{cases} \qquad P(v)T=\begin{cases}1+v,&0\le v\le1,\\ 3-v,&1\le v\le2. \end{cases}$$ Consequently $S_S(T)=1$, and the secondary invariant is $7/9$ away from $T\cap(U\cup V)$ and $1$ at these two intersections. This proves the assertion on all smooth fibres.

For an $A_1$ fibre, let $T$ be one of the two lines not through the node. On the resolution, $T$ meets a $(-1)$-curve $U$ meeting the exceptional $(-2)$-curve $R$. For $1\le v\le2$ the negative part is $(v-1)(2U+R)$; for $v\le1$ it is zero. The same displayed formula for $P(v)^2$ holds, because $(2U+R)^2=-2$. Thus $S_S(T)=1$, and the secondary invariant is $7/9$ away from $T\cap U$. The latter point lies on a line through the node and is excluded from the assertion. This finishes the proof. ◻

Proof. First consider a point $p\in C_S$ on a smooth fibre. If $p$ is off the six lines, Lemmas 110.1 and 110.3 give $$\delta_p(X)\ge\min\left\{\frac{16}{11}, \frac{16(6/5)}{15+6/5}\right\}=\frac{32}{27}>1.$$ If $p$ lies on just one line $T$, refine once more by $T$. The first invariant is $15/16$. Since $C_ST=1$, the intersection at $p$ is transverse; the secondary invariant is at most $$\frac{15}{16}\cdot\frac79+\frac1{16}=\frac{19}{24}.$$ Hence $\delta_p(X)\ge16/15$.

Only the corners of the hexagon of lines remain. There are finitely many pairs $(S,p)$ with $S$ smooth and $p\in C_S$ such a corner. To see this, view $V$ as the incidence variety of points and lines in $\mathbb P^2$. The contact plane at $(x,\ell)$ is the span of the tangent directions of the two projection fibres. At a corner, these are the two line directions in $S$, so $T_pC$ lies in the contact plane. The induced morphism $$T_C\longrightarrow\mathcal O_V(1,1)|_C$$ is not identically zero. Otherwise the second projection would send every point of the smooth plane cubic given by the first projection to its tangent line. This tangent-line map has degree six with respect to $\mathcal O_{(\mathbb P^2)^*}(1)$: the three quadratic partial derivatives of its smooth cubic equation have no common zero on the cubic. But the second projection of $C$ has degree three, a contradiction. The displayed morphism therefore has only finitely many zeros. For each zero there is at most one smooth pencil member whose tangent plane is the contact plane, since the two pencil sections have independent differentials along the smooth complete intersection $C$.

A curve in $E$ dominating the pencil base meets a general fibre at points avoiding this finite set. At any such point the preceding strict local bound excludes the curve as a nonpositive-beta centre. ◻

Proof. A centre meeting a smooth fibre outside $E$ is excluded by Lemma 110.2. A horizontal curve in $E$ is excluded by Lemma 110.5. The divisor $E$ itself has $S_X(E)=3/8<1$. These observations exclude all horizontal centres. A divisorial centre in a fibre is the fibre itself and has $S_X(F)=11/16<1$. Any remaining centre lies in a fibre and has dimension at most one. At every point outside the stated lines, Lemma 110.3 and the strict inequality in Lemma 110.2 exclude it. The singular point of either singular fibre lies on the stated lines. ◻