The joining line in a fibre of type (3,1)

Proof. We first justify that these are the actual section spaces. Use the hypersurface realization $X\subset\mathbb F_1\times\mathbb P^2$ of class $h+2H$; its anticanonical class is $h+f+H$. The restriction map in every anticanonical degree $m\ge1$ is surjective. Indeed, the kernel has class $((m-1)h+mf,(m-2)H)$ and its first cohomology vanishes: for $m=1$ the factor $\mathcal O_{\mathbb P^2}(-1)$ is acyclic, and for $m\ge2$ the divisor $(m-1)h+mf$ on $\mathbb F_1$ is nef. Thus ambient sections give the whole anticanonical section space.

Write the ambient equation as $s_2q=s_1F$. The weights of $u,x_1,s_2$ for the ordinary curve extraction are one, and those of $v,x_0,x_2,s_1$ are zero. The initial ambient equation is $$s_2x_0x_2=s_1(uG_0+vx_1x_2).$$ This polynomial is irreducible: regarded as a polynomial in $s_2$, its two coefficients are coprime, since $AC\ne0$. Hence its initial ring is integral and the ambient weight filtration is the geometric valuation filtration. The homogeneous quotient therefore gives the actual initial section spaces, including all cancellations.

On the exceptional surface put $x_0=b_0$, $x_2=b_1$, $u=t$, $s_2=a$ and trivialize $v=s_1=1$. The initial relation gives the displayed equation for $x_1$. The ambient monomials of anticanonical degree $m$ show that the actual degree $(m,j)$ space is spanned by $$a^r t^i x_1^l b_0^b b_1^c,\qquad 0\le r\le m,\quad 0\le i\le2m-r,\quad l+b+c=m,\quad r+i+l=j.$$ All these sections vanish to order at least $\max(j-m,0)$ at $P_A,P_B,P_C$. If $j>2m$, the inequalities force both $l\ge j-2m$ and $i\ge j-2m$. Factoring $(x_1t)^k$ gives precisely the residual vanishing conditions in the statement. This proves containment in the asserted complete systems.

Here is a dimension check which also proves equality. Regard $S$ as the blow-up of $\mathbb P^2$ at four points, one infinitely near the first. Let $h,e_0,e_1,e_2,e_3$ be the usual orthogonal basis, with $\xi=h$, $f=h-e_0$, and $E_A=e_1,E_B=e_2,E_C=e_3$. The negative curves are $$e_0-e_1,\quad h-e_0-e_2-e_3,\quad e_1,\quad e_2,\quad e_3,\quad h-e_0-e_1.$$ The first two have square $-2$ and meet once; the others have square $-1$. This is a weak del Pezzo surface of degree five. Intersection with this list shows that every residual divisor in the statement is nef. Riemann–Roch and vanishing therefore give its dimension as $$d_{m,j}=\begin{cases} (j+1)(2m+j+2)/2,&0\le j\le m,\\ -j^2+4jm-3m^2/2+5m/2+1,&m\le j\le2m,\\ m(2n+1)+(n+1)(n+2)/2,&2m\le j\le3m. \end{cases}$$ Their sum is $(2m+1)(2m^2+2m+1)=h^0(X,-mK_X)$. Taking an associated graded preserves dimension, so every containment above is an equality. ◻

Proof. The preceding dimensions give $$\sum_{j=0}^{3m}j d_{m,j} =\frac{m(13m^3+18m^2+12m+3)}2.$$ Divide by $m(2m+1)(2m^2+2m+1)$ and pass to the limit. ◻

Proof. The blow-up $S\to G$ is an isomorphism near $F$. Normalize the primary degree by $u=j/m$. The nef residual divisors in Lemma 127.2 are $$D_u=\begin{cases} (u+1)h-e_0,&0\le u\le1,\\ (u+1)h-e_0-(u-1)(e_1+e_2+e_3),&1\le u\le2,\\ (7-2u)h-(3-u)(e_0+e_1+e_3)-e_2,&2\le u\le3. \end{cases}$$ For $u\ge2$ there is also the fixed divisor $(u-2)((x_1=0)+(t=0))$. Put $D_u-vF=P(u,v)+N(u,v)$ for its Zariski decomposition on this surface. Write $p=P(u,v)\cdot F$ and write $n$ for the coefficient of $E=e_0-e_1$ in $N(u,v)$. The following table gives all chambers. A polygon is specified by its vertices $(u,v)$; boundaries may be assigned to either adjacent chamber. $$\begin{array}{c|c|c} \text{vertices}&p&n\\\hline (0,0),(1,0),(1,1),(0,1)&u&0\\ (1,1),(1,2),(0,1)&u-v+1&v-1\\ (1,0),(2,0),(1,1)&u&0\\ (2,0),(3/2,3/2),(1,1)&u/2-v/2+1&(u+v-2)/2\\ (2,0),(2,3/2),(3/2,3/2)&2-2v/3&(3u+2v-6)/3\\ (1,2),(1,1),(3/2,3/2)&u-v+1&v-1\\ (2,3/2),(2,2),(1,2),(3/2,3/2)&4-2v&u+2v-4\\ (2,0),(3,0),(2,3/2)&4-u-2v/3&2v/3\\ (3,0),(2,2),(2,3/2)&10-3u-2v&2(u+v-3). \end{array}$$ These formulas follow by imposing $P\cdot C=0$ on the negative curves in the support and checking nonnegative intersections with the six curves listed above. The other components of $N$ are vertical and are disjoint from $F$. The volume is determined explicitly by $$P(u,v)^2=D_u^2-2\int_0^v p(u,s)\,ds,\qquad D_u^2=\begin{cases} u^2+2u,&0\le u\le1,\\ -2u^2+8u-3,&1\le u\le2,\\ u^2-10u+21,&2\le u\le3. \end{cases}$$

Set $k=\max(u-2,0)$. Integration over the displayed polygons gives $$\begin{array}{c|cccc} \text{range of }u&\iint P^2&\iint p^2&\iint 2p(n+k)&\iint 2pk\\\hline {[0,1]}&11/12&5/12&1/12&0\\ {[1,2]}&55/16&11/4&7/6&0\\ {[2,3]}&31/24&3/2&5/3&3/4. \end{array}$$ The normalization is $1/8$, since $h^0(X,-mK_X)=4m^3+O(m^2)$ and surface volumes are twice the leading coefficient of the dimension. Thus the fibre expected order is $271/384$, and the moving point contribution is $7/12$.

The only possible fixed contributions on $F$ are at its intersections with $E=(T=0)$ and with $(x_1=0)$. These points are distinct when $b_0b_1\ne0$. At the first point the additional contribution is $35/96$, and at the second it is $3/32$. Consequently the point expected orders are $$\frac{91}{96},\qquad\frac{65}{96},\qquad\frac7{12},$$ respectively at those two points and elsewhere. This proves the assertion. ◻

Proof of Theorem 127.1. Surface adjunction along the ruling fibre, followed by point adjunction, gives $$\delta(G;W^G)\big|_{b_0b_1\ne0} \ge\min\left\{\frac{384}{271},\frac{96}{91}\right\} =\frac{96}{91}.$$ Abban–Zhuang adjunction for the ordinary curve extraction now gives $$\delta_p(X)\ge \min\left\{\frac{16}{13},\frac{96}{91}\right\} =\frac{96}{91}.$$ No Zariski decomposition on a modification of the threefold is used. The all-degree calculation identifies the actual refined series, including its fixed factors. The displayed Zariski decompositions take place only on the blown-up exceptional surface. ◻

Proof. We calculate on the same surface $S$. For $u=j/m$, the full pullback of the refined series has nef moving class $D_u$ from Lemma 127.4, and fixed part $$(u-1)_+(e_1+e_2+e_3)+(u-2)_+(C+E+F_0),$$ up to bounded rounding terms. These exceptional fixed coefficients are essential for the calculation at $P_A$.

First consider a point of $b_0=0$ different from $P_A$. For $D_u-vF_A=P+N$, the negative parts are as follows. If $0\le u\le1$, put $s=(v-1)_+$; then $$N=sE+(v+s)e_1.$$ If $1\le u\le2$, put $r=(v+1-u)_+$, $s=(v-1)_+$ and $t=(u+v-3)_+$; then $$N=(s+t)E+tF_0+(r+s+t)e_1.$$ If $2\le u\le3$, put $t=(u+v-3)_+$; then $$N=2tE+tF_0+3te_1.$$ The thresholds in these three ranges are $u+1$, $2$ and $6-2u$, respectively. The nef classes $h$ for $u\le1$ and $h-e_2$ for $u\ge1$ exclude larger values of $v$. As before, these formulas are verified by zero intersection with the support, nonnegative intersection with the six negative curves, and negative definiteness of the support. Integration gives $$\frac18\int\!\int P^2\,dv\,du=\frac34, \qquad \frac18\int\!\int(P\cdot F_A)^2\,dv\,du=\frac7{12}.$$ The only other negative curve meeting $F_A$ is $e_1$, and their intersection maps to $P_A$. None of $C,E,F_0$ meets $F_A$. Thus at every other point of the original fibre the flag bound is at least $\min(4/3,12/7)=4/3$.

It remains to consider $P_A$. Extract this point by the ordinary blow-up; its exceptional curve is $e_1$ and its log discrepancy over $G$ is two. For $D_u-ve_1=P+N$ we give the complete negative parts in a compact form. For $0\le u\le1$, $$N=\frac{(v-1)_+}{2}E+(v-u)_+F_A.$$ For $1\le u\le2$, set $r=(u+v-2)_+$ and $s=(3u+v-6)_+$; then $$N=\left(\frac r2+\frac s6\right)E+\frac{s}{3}F_0 +(v-1)_+F_A.$$ For $2\le u\le3$, set $s=(3u+v-9)_+$; then $$N=\frac{2v+s}{3}E+\frac{v+2s}{3}F_0+s e_3 +(v-1)_+F_A.$$ The thresholds are $2u+1$, $3$ and $7-2u$, respectively. The nef classes $2h-e_0-e_1$ for $u\le1$ and $2h-e_0-e_1-e_2$ for $u\ge1$ exclude larger values. These formulas comprise twelve chambers and pass the same intersection checks. Write $l=P\cdot e_1$ and put $k=(u-2)_+$. The integrals are $$\begin{aligned} \frac18\int\!\int P^2&=\frac{107}{96},\qquad \frac18\int_1^3(u-1)D_u^2\,du=\frac{65}{96},\\ \frac18\int\!\int l^2&=\frac{115}{288},\qquad \frac14\int\!\int l(N_E+k)=\frac{79}{144},\\ \frac14\int\!\int lN_{F_A}&=\frac{101}{288},\qquad \frac14\int\!\int lk=\frac3{32}. \end{aligned}$$ All double integrals use the stated thresholds and measure $dv\,du$. For example, they are evaluated by splitting at the zeros of the affine functions inside $(\ )_+$. The first expected order, including the fixed coefficient of $e_1$, is therefore $43/24$.

The curves $E,F_A,C$ meet $e_1$ at three distinct points. In local coordinates at $P_A$ their tangent directions are $T=0$, $b_0=0$ and $b_0-\gamma T=0$, with $\gamma\ne0$. The point moments at these three points and elsewhere are, respectively, $$\frac{91}{96},\qquad\frac34,\qquad \frac{71}{144},\qquad\frac{115}{288}.$$ There is no different for this ordinary point extraction. Consequently $$\delta_{P_A}(G;W^G)\ge \min\left\{\frac{48}{43},\frac{96}{91},\frac43, \frac{144}{71},\frac{288}{115}\right\} =\frac{96}{91}.$$ This covers the whole exceptional fibre over the simple endpoint. The threefold adjunction bound is then $$\delta_p(X)\ge \min\left\{\frac{16}{13},\frac{96}{91}\right\} =\frac{96}{91}.$$ ◻