Special lines and points away from the fibre node

Proof. For a special line, the exceptional surface over the line on $\operatorname{Bl}_{\pi(A)}V_5$ is $\mathbb F_2$. Write $s_2,f$ for its negative section and ruling. The projection divisor $M$ restricts to $s_2+2f$. It contracts $s_2$ to the vertex of a quadric cone. The smooth twisted cubic $\Gamma$ in the projection link is contained in this cone, and the restriction of $\Pi$ has class $2s_2+3f$. It is $$s_2+T_2,\qquad T_2\sim s_2+3f.$$ Indeed its nonexceptional image is the irreducible curve $\Gamma$; the remaining component is $s_2$, with the displayed multiplicity forced by the class. The curve $T_2$ is a section, hence smooth, and $T_2\cdot s_2=1$.

The elementary transformation taking this surface to $F$ is centred at a point $u\notin s_2$. Otherwise the resulting ruled surface would be $\mathbb F_3$, contrary to $F=\mathbb F_1$. Blow up $u$ and contract the strict transform of its ruling. The contracted ruling maps to $q_E$, and the new ruling through $q_E$ is the image of the exceptional curve over $u$. The condition $p_0\in E$ says precisely that $u\in T_2$: the second curve in the projection link then meets the original exceptional fibre at the intersection with $A$.

If $u\notin T_2$, both sections meet the contracted ruling once. Their images are $s$ and a section $T\sim s+3f$, both through $q_E$. The second marked point belongs to $R$ and avoids $s$, so it belongs to $T$. As a section, $T$ meets $f_D$ only at $q_D$, proving the assertion about $s\cap f_D$. Tangency of $T$ and $s$ at $q_E$ is allowed.

If $u\in T_2$, the strict transform of $T_2$ is disjoint from the contracted ruling. Its image therefore avoids $q_E$. The exceptional curve over $u$ supplies $f_0$ with coefficient one. Subtracting $s+f_0$ from $2s+3f$ gives $T\sim s+2f$. Its intersection with $s$ is one and is away from $q_E$.

For an ordinary line, perform the same transformation on $\mathbb F_0$ and its irreducible $(1,2)$-curve, as in Lemma 77.9. The residual image $T$ is irreducible, has class $2s+2f$, and has intersection zero with $s$. It is distinct from $s$, so it is disjoint from $s$. ◻

Proof. Contract $s$ to $q_0=[1:0:0]\in\mathbb P^2$, with homogeneous coordinates $[X:Y:Z]$. The point $q_E$ is infinitely near $q_0$ in direction $Z=0$, and take $q_D=[0:0:1]$. This normalization is possible because the two ruling fibres are different. For a normalized degree-$b$ monomial $X^iY^{b-i-k}Z^k$, the conditions of Lemma 77.5 are $$0\le i\le a,\qquad \max(0,c_E-a+i)\le k\le\min(b-i,b-c_D).$$ For the infinitely near condition, use $Y=w$, $Z=wz$ near $q_0$, and divide by $w^{b-a}$. The monomial becomes $w^{a-i}z^k$, whose order at $q_E$ is $a-i+k$. At $q_D$ its plane order is $b-k$.

The following bounds give the entire polygon explicitly: $$\begin{array}{c|c|c} t&i&k\\ \hline {[0,1]}&[0,t]&[0,1+t-i]\\ {[1,2]}&[0,t-1]&[0,2]\\ &[t-1,1]&[0,1+t-i]\\ &[1,t]&[i-1,1+t-i]\\ {[2,3]}&[0,1]&[0,6-2t]\\ &[1,4-t]&[i-1,7-2t-i]. \end{array}$$ Integrate over these polygons and then over $t$, with factor $3/13$. The integrands $1,i,a-i,k,b-i-k,g$ give respectively $$1,\quad63/104,\quad89/104,\quad84/104,\quad99/104,\quad10/104.$$ For another ruling put $$U(i)=\min(b-i,b-c_D)-\max(0,c_E-a+i).$$ After removing the two prescribed endpoint factors, the coefficient space is the complete binary series of degree $U(i)$. Its order sum at another base point has leading term $U(i)^2/2$. Integrating this expression gives $79/104$.

After restriction to a ruling and removal of its fixed order, the allowed exponents $i$ form an interval inside $[0,a]$. The endpoint point moments are the integrals of $i$ and $a-i$. Any other point has moment at most either endpoint moment, because for $l\le i\le h$ its order sum is $(h-l)^2/2$. This proves the ruling bound, also for $f_E,f_D$.

For the flag with first curve $s$, fix $i$ instead. The remaining binary exponents $k$ form the displayed interval. The two endpoint moments are the integrals of $k$ and $b-i-k$, and other points have no larger moment than either endpoint. This gives exactly the asserted point estimates on $s$. ◻

Proof. At a point of $F\setminus s$, use its ruling. The first moment is at most $99/104$. The fixed component $s$ contributes nothing at the chosen point, while the section $T$ contributes at most $5/52$. The further point moment is at most $(89+10)/104=99/104$.

At a point of $s$, use $s$ as first curve. Its full moment is $(89+10)/104=99/104$. At $s\cap f_D$, the point moment is $99/104$, since $T$ avoids this point. Elsewhere it is at most $$\frac{84}{104}+\frac{10}{104}(T\cdot s)_q\le1.$$ Except possibly at $q_E$, it is at most $99/104$: the section $T$ already meets $s$ at $q_E$, and has total intersection two. If the intersections are transverse, the same strict bound holds everywhere. These are surface flag estimates with all fixed components retained. They prove the surface assertions.

The first extraction has ratio $52/43>104/99$. The only possible surface point at which the estimate equals one is $q_E$, which maps to $r$. This proves the asserted strict local bounds away from $r$.

For completeness, the equality case on the surface still gives a strict threefold beta invariant. Choose a basis compatible with $v$ and the refinement, and write its pullback basis divisor as $a_mF+D_m'$, where $a_m\to43/26$ and $D_m'$ does not contain $F$. For coefficients $c_m\to1$ from below, the preceding estimates give log canonicity of $(F,c_mD_m'|_F)$. Inversion of adjunction gives, since $K_Y=f^*K_X+F$, $$A_X(v)-c_m v(D_m)\ge(2-c_ma_m)v(F).$$ Taking the limit proves the displayed inequality. Its right-hand side is positive for centres contained in $A$. ◻

Proof. We give the surface decompositions used in this computation. On $F^+$ use the total-transform plane basis $h,e_0,e_E,e_D$, with intersection matrix $\operatorname{diag}(1,-1,-1,-1)$. The moving divisors are $$D_t=\begin{cases} (1+t)h-e_0&0\le t\le1,\\ (1+t)h-e_0-(t-1)(e_E+e_D)&1\le t\le2,\\ (7-2t)h-(3-t)(e_0+e_E)-e_D&2\le t\le3. \end{cases}$$ The class of the unmarked ruling is $C=h-e_0$. In the ordinary case put $S=e_0$ and $Z=h-e_0-e_E-e_D$; in the special case put $S=e_0-e_E$ and use the same class $Z$. Thus $Z$ is the strict transform of $f_0$.

Here is a full chamber list. A domain specified by vertices means their convex hull in the $(t,v)$-plane. On each domain $P=D_t-vC-N$ is the positive part. For the last table take $C=S$. The coefficients not displayed in $N$ are zero. $$\begin{gathered}\text{ordinary line},\quad C=h-e_0\\ \begin{array}{c|l|l} t&\text{vertices}&N\\ \hline {[0,1]}&(0,0),(1,0),(1,1),(0,1)&0\\ &(0,1),(1,1),(1,2)&(v-1)S\\ {[1,2]}&(1,0),(2,0),(2,1),(1,1)&0\\ &(1,1),(2,1),(1,2)&(v-1)S\\ &(2,1),(2,2),(1,2)&(t+2v-4)S+(t+v-3)Z\\ {[2,3]}&(2,0),(3,0),(2,1)&0\\ &(3,0),(2,2),(2,1)&(2t+2v-6)S+(t+v-3)Z. \end{array}\end{gathered}$$ $$\begin{gathered}\text{special line},\quad C=h-e_0\\ \begin{array}{c|l|l} t&\text{vertices}&N\\ \hline {[0,1]}&(0,0),(1,0),(1,1),(0,1)&0\\ &(0,1),(1,1),(1,2)&(v-1)(S+e_E)\\ {[1,2]}&(1,0),(2,0),(1,1)&0\\ &(2,0),(2,2),(1,1)&(t+v-2)S/2\\ &(1,1),(2,2),(1,2)&(v-1)S+(v-t)e_E\\ {[2,3]}&(2,0),(3,0),(2,2)&vS/2. \end{array}\end{gathered}$$ $$\begin{gathered}\text{special line},\quad C=S=e_0-e_E\\ \begin{array}{c|l|l} t&\text{vertices}&N\\ \hline {[0,1]}&(0,0),(1,0),(1,1)&ve_E\\ {[1,2]}&(1,0),(2,0),(2,1)&0\\ &(1,0),(2,1),(1,1)&(v-t+1)e_E\\ &(1,1),(2,1),(2,2)&(2v-t)e_E+(v-1)Z\\ {[2,3]}&(2,0),(3,0),(2,1)&0\\ &(3,0),(3,1),(2,2),(2,1)&(2t+2v-6)e_E+(t+v-3)Z. \end{array}\end{gathered}$$ To verify the decompositions, intersect $P$ with $S,e_E,e_D,Z$. These generate the negative curves in the respective weak del Pezzo configurations; $P$ has nonnegative intersection with each and is orthogonal to every component of $N$. The upper edges give the pseudoeffective thresholds.

The curve moment is $(3/26)\iint P^2\,dv\,dt$; the free point moment is $(3/26)\iint(P\cdot C)^2\,dv\,dt$. A negative component with coefficient $n$ contributes $(3/13)\iint(P\cdot C)n\,dv\,dt$ at its intersection with $C$. The displayed domains give $$\begin{array}{c|cc|l} \text{flag}&\text{curve}&\text{free point}& \text{negative-part corrections relevant away from }f_0\\ \hline \text{ordinary},\ f_b&37/52&17/26&2/13\text{ at }s\cap f_b\\ \text{special},\ f_b&79/104&63/104&1/4\text{ at }s\cap f_b\\ \text{special},\ s&89/104&79/104&0\text{ away from }q_E. \end{array}$$ The exceptional curves over the marked points and the curve $Z$ are disjoint from an unmarked ruling. In the last row their intersections with $s$ lie over $q_E$, which is excluded here. Thus no marked-point correction is omitted in this table.

In the ordinary case, Lemma 78.1 gives $R=f_0+T$, with $T\cdot f_b=2$ and $T\cap s=\varnothing$. At $s\cap f_b$ the total moment is $(68+16)/104=84/104$. Elsewhere it is at most $(68+20)/104=88/104=11/13$.

In the special case, $R=s+f_0+T$, with $T$ a section. At points off $s$ the moment is at most $(63+10)/104$. For the flag $s$, its fixed component adds $10/104$ to the first moment. The residual section $T$ meets $s$ transversally once, away from $f_0$, so the further point moment is at most $(79+10)/104$. This proves all assertions. ◻

Proof. The inverse image of such a point on the component extraction is an unmarked ruling. The preceding flags cover this entire ruling. Apply Abban–Zhuang refinement, including the first ratio $52/43$. For an ordinary component the minimum is $13/11$; for a special component the stated common bound is $104/99$. ◻