Lines of the second kind: the ruled surface $\mathbb F_2$

Proof. At infinity put $t=x_0/x_5$, $a=x_1/x_5$, and $b=x_6/x_5$. These are regular parameters on $V_5$ near $q$. On their overlap with the first chart, $$a=tu,\qquad b=tuv-u^2-v/t.$$ On $X$, write $s=f/g$. The weighted initial equation at $q$ is $b=ts$; the variables $a,s,t$ are regular parameters there, and their normal weights are $(1,2,0)$. The two weighted ideals agree on the overlap. For the coarse exceptional surface put $U=u^2$ and $V=v$. Its transition to the infinity chart is $$U'=t^2U,\qquad S'=-t^{-1}U-t^{-2}V.$$ The associated rank-two bundle splits as $\mathcal O(1)\oplus \mathcal O(-1)$. Indeed its positive subbundle is generated on the finite chart by $(U,V)=(w,-1)$ and on the other chart by $(U',S')=(1,0)$. Consequently the coarse surface is $\mathbb F_2$, and its negative section has equation $U+wV=0$.

The weight-two chart of the extraction is the quotient of a smooth chart by an involution acting with weights $(1,1,0)$. It follows directly that the pair with its reduced exceptional divisor is plt, with different one half of the section $D=(u=0)$. The discrepancy is $1+2=3$. The displayed transition shows that $D^2=4$; hence $D\sim B+3c$. On the finite chart $D\cap B$ is given by $w=0$. Finally $LZ=1$ gives $L|_G=c$, and adjunction $K_G+\Delta_G=K_X|_G+3G|_G$ gives the stated restriction of $G$. ◻

Proof. Use homogeneous variables $(x_0,u,A,T,v,w,J)$ for the seven initial hyperplane sections. The candidate initial equations are $$vw-x_0A+u^2,\quad vJ-uT+A^2,\quad -x_0T+uA,\quad -x_0J-Aw,\quad -Tw-uJ.$$ They are the signed maximal Pfaffians, up to a common sign, of $$\begin{pmatrix} 0&0&J&-T&A\\ 0&0&-w&-u&x_0\\ -J&w&0&A&-u\\ T&u&-A&0&v\\ -A&-x_0&u&-v&0 \end{pmatrix}.$$ After inverting $x_0$, the equations give $A=(u^2+vw)/x_0$, $T=uA/x_0$, and $J=-wA/x_0$, so this open algebra is a polynomial algebra in $x_0^{\pm1},u,v,w$, of dimension four. On $x_0=0,A\ne0$, one has $u=w=0$ and $vJ+A^2=0$; $A,v,T$ give three independent parameters, with $A,v$ invertible. On $x_0=A=0,v\ne0$, the equations instead give $w=-u^2/v$ and $J=uT/v$, with parameters $v,u,T$. Finally, on $x_0=A=v=0$, one has $u=0$ and $Tw=0$, leaving a locus of dimension two in $T,w,J$. Thus the whole boundary has dimension three. The ideal has height three, so its Pfaffian resolution makes the quotient Cohen–Macaulay of pure dimension four. Since the boundary contains no component, $x_0$ is a nonzerodivisor. Injection into the polynomial localization then proves that the quotient is a domain. Its graded Pfaffian resolution has the same Hilbert series as the coordinate ring of $V$. The candidate equations belong to the initial ideal, and equality of Hilbert functions proves equality of the ideals. The localized graph in the normal parameters also identifies this weight filtration with the geometric normal filtration.

The pair $(J,x_0)$ is regular in this primary initial ring. Indeed, on $J=x_0=0$ its equations are $u^2+vw=0$, $uT=A^2$, $uA=Aw=Tw=0$. Their support has $A=0$. On $T\ne0$ one has $u=w=0$ and two parameters $T,v$; on $T=0$ it is the two-dimensional cone $u^2+vw=0$. Thus the common zero locus has dimension two. Lemma 43.1 applies to the graph equations $J=ea$, $x_0=eb$: their quotient is a domain and is the full initial section algebra. On $x_0=b=1$ it yields $e=1$, $a=-wA$, and the diagonal anticanonical algebra is exactly (46.1). Its fixed-degree components therefore include all cancellations among global sections.

The seven primary weights are $(0,1,2,3,2,0,2)$, the five Pfaffian weights are $(2,4,3,2,3)$, and the final shift is seven. Differentiating the Pfaffian Hilbert series gives $S_H(G)=\frac14\cdot10-\frac3{20}\cdot7=29/20$. The graph Hilbert series then gives $S_X(G)=\frac54(29/20)+\frac38\cdot2=41/16$. Equivalently this last number follows from the independent moment certificate below. A section of normal degree $j$ restricts to the class $m c+\frac j2(B+c)$; taking its average proves the class formula. ◻

Proof. Work at the generic point of $B$, invert $w$, and substitute $v=(A-u^2)/w$. The valuation is ordered first by normal degree $j=2a+b$, then by the exponent $a$ of $A$, and finally by the exponent $d$ of $w$. Its monomial coordinates are $(a,b,d)$ for $A^a u^b w^d$. The following nine points are valuations divided by section degree: $$\begin{array}{c|rrrrrrrrr} i&1&2&3&4&5&6&7&8&9\\\hline P_i&(0,0,0)&(0,0,1)&(0,2,-1)&(1,2,0)&(1,0,2)& (2,1,1)&(2,0,2)&(2,0,1/2)&(3/2,0,0). \end{array}$$ The first seven are supplied by $1,w,v,(wA)v,w(wA),(wA)(uA),(wA)^2$. The last two come from the degree-two sections $wA^4$ and $A^3$. They are genuine sections: writing $J=wA$ and $T=uA$, one has $$A^3=JAv+T^2,\qquad wA^4=J^2Av+JT^2.$$ Each product uses at most two hyperplane generators and at most two pencil generators. The signs of $J$ can harmlessly be reversed.

Their convex hull has volume $10/3$. For a directly checkable integral certificate, triangulate its facets from the origin with triples $$(8,4,6),(8,7,6),(4,2,5),(4,2,3), (7,4,5),(7,4,6),(8,4,3),(8,9,3).$$ The respective tetrahedral volumes are $5/12,1/2,1/3,1/3,2/3,1/3,1/2,1/4$. The sum of their first moments is $(307/96,103/48,5/3)$. Thus the centroid is $$\left(\frac{307}{320},\frac{103}{160},\frac12\right).$$ The valuation lattice is the full integer lattice, since $1,A,u,w$ are degree-one sections. After substituting $v=(A-u^2)/w$, the finite degree-one generating list in Lemma 46.2 consists of Laurent polynomials in $A,u,w$. Every exponent occurring in a product of $m$ generators has absolute value at most a fixed constant times $m$. Taking linear combinations and pivot monomials preserves this bound. Thus signed exponents satisfy the boundedness hypothesis of Lemma 34.2. Its volume agrees with $L^3/6=10/3$, so that lemma identifies these with the moments of the entire refinement. In particular it yields $S(W;B)=307/320$ and, independently, $S_X(G)=2(307/320)+103/160=41/16$.

On $B$, the relation is $u^2=-vw$. At $r_0$ the coordinate $v$ is nonzero, so the order of $A^{-a}A^a u^b w^d$ on the coarse curve is $b/2+d$. Its mean is $103/320+1/2=263/320$. The total degree of the refined divisors on $B$ is $$\left(\frac{41}{32}B+\frac{73}{32}c -\frac{307}{320}B\right)B=\frac{131}{80}.$$ For any other point, choose a basis compatible with that point and $r_0$ in each restricted graded piece. Subtracting the known mean at $r_0$ bounds the other mean by $131/80-263/320=261/320$. Finally a basis compatible with $B$ and a ruling $c_p$ restricts to a basis on $B$ whose order at $B\cap c_p$ is at least its order along $c_p$. The preceding two bounds therefore imply $S(W;c_p)\le263/320$ for every ruling. ◻

Proof. First consider $r\in G\setminus B$ and use its ruling $c_p$ as the surface flag. The residual-degree lemma, subtracting the mean along $B$ before restricting, bounds the point invariant by $$\left(\frac{41}{32}B+\frac{73}{32}c -\frac{307}{320}B\right)c_p=\frac{103}{320}.$$ The boundary $D$ meets $c_p$ transversally once. Hence the local discrepancy on $c_p$ is either one or one half, while $S(W;c_p)\le263/320$. Surface adjunction therefore gives a threshold at least $\min\{320/263,160/103\}$ at $r$.

For $r\in B\setminus\{r_0\}$, use $B$ as the flag. The different has no support at $r$, and the preceding lemma gives threshold at least $\min\{320/307,320/261\}=320/307$. Thus the refined threshold is at least $320/307$ at every point of $G$ except possibly $r_0$, whose image on $X$ is $p_0$. Finally $A_X(G)/S_X(G)=48/41>320/307$. Abban–Zhuang adjunction for the weighted curve extraction proves the assertion. ◻

Proof. The point $p_0$ is a singular point of the fibre, so it is not on $C$. Normalize $q=C\cap Z$ to $w=\infty$ as above. Smoothness of $C$ at $q$ then gives $\varepsilon\ne0$. Rescaling $w$ and the fibre equation makes its normal initial form $(w-1)A$. The weighted exceptional surface, boundary, and numerical class of the refined series are the same as in Lemma 46.1. The exact generators are now $H_i,(w-1)A H_i$, with the same seven $H_i$ as before.

In the coordinates $(A,u,w)$, the following degree-one leading monomials are obtained by elementary row operations: $$\begin{split} &(0,0,1),(2,1,0),(0,0,0),(2,0,1),(2,0,0),\\ &(0,2,-1),(1,0,2),(1,2,-1),(1,1,1). \end{split}$$ Number these points consecutively from one to nine. For example the last three are obtained from $wA$, $uA$, and $v$ by multiplication by $(w-1)A$ and subtraction of the lower terms. Their convex hull has volume $10/3$ and first moment $(25/8,55/24,7/8)$, giving centroid $(15/16,11/16,21/80)$. A tetrahedral certificate from the origin is $$(8,9,6),(8,9,2),(8,5,2),(4,5,2), (9,7,1),(9,6,1),(4,9,7),(4,9,2),$$ with volumes $1/2,2/3,1/3,1/3,1/6,1/3,1/2,1/2$. Thus $S(W;B)=15/16$.

We also need the point $r_1\in B$ over $w=1$. Put $z=w-1$ and multiply every degree-$m$ section by the common unit $w^m$. Degree-one generators are then $$w,w^2,uw,Aw,uAw,A-u^2,w^2A,$$ and their products with $zA$. In the order $(j,A,z)$ their leading exponents $(a,b,d)$ have the ten vertices $$(1,2,1),(2,1,1),(2,0,2),(1,1,0),(0,2,0), (0,0,0),(1,0,0),(2,0,1),(1,0,2),(0,0,1).$$ This hull again has volume $10/3$; its first moment is $(25/8,55/24,65/24)$. One can use the triples $$(1,3,2),(9,1,3),(8,3,2),(4,1,5),(4,1,2), (10,1,5),(10,9,1),(8,4,2),(8,7,4),$$ whose volumes are $2/3,2/3,1/3,1/3,1/3,1/3,1/3,1/6,1/6$. The full-volume lemma gives $S(W^B;r_1)=13/16$. The total degree on $B$ is now $51/32$, so at every other point its mean is at most $51/32-13/16=25/32$. In particular every ruling has mean at most $13/16$.

Away from $B$, subtracting its mean before restricting to a ruling leaves point invariant at most $41/32-15/16=11/32$. The different on that ruling has coefficient at most one half, giving refined threshold at least $\min\{16/13,16/11\}$. On $B\setminus\{r_0\}$ the different vanishes and the threshold is at least $16/15$. Abban–Zhuang adjunction and $A_X(G)/S_X(G)=48/41$ give the assertion. ◻