The remaining lines of the second kind

Proof. The construction is the same weighted extraction as before; at $q$ one uses the pencil parameter $s$ in place of the vanishing normal coordinate. The formula in (47.1) gives its transition explicitly. On the coarse surface $[A:s]$ are homogeneous coordinates on a fixed projective line. The branch section is $$Q_h(w_0,w_1)A=w_1g_h(w_0,w_1)s,$$ where $Q_h=w_0^2+\beta w_0w_1-\varepsilon w_1^2$ and $g_h$ is either $w_0$ or $w_1$. The two coefficients on the right and left have no common zero: at the closed-orbit point this follows from $Q(0)=1$, and at $q$ it follows from smoothness of $C$. Thus the coarse surface is the trivial ruled surface, and $D$ is a smooth section of class $h+2c$. The weight-two chart is the quotient chart of weights $(1,1,0)$, so the extraction is plt, its discrepancy is $3$, and the different is $\frac12D$. Since $LZ=1$, adjunction gives the restrictions displayed in the statement.

For completeness, the primary ring here is the same prime Pfaffian ring used in Lemma 46.2; only the chosen pencil changes. The pair consisting of the initial forms of $f,g$ is a regular sequence. This can also be checked directly from its five equations. For $g=x_0$, the boundary component with $A\ne0$ has $u=w=0$ and $vJ+A^2=0$, while the part with $A=0$ is the remaining rank-one cone; the form $v+\beta A+\varepsilon J$ vanishes identically on neither three-dimensional component. For $g=w$, its open part $x_0\ne0$ has $J=0$, $A=u^2/x_0$, $T=u^3/x_0^2$, and the equation of $f$ solves for $v$. On its boundary $x_0=0$ one has $u=0$ and $vJ+A^2=0$, and again $f$ cuts the dimension by one. Hence in either case the simultaneous zero locus has dimension two. Lemma 43.1 therefore proves that the graph ring is integral and is the full associated algebra of global sections. Injection into its localized normal-coordinate model identifies it with the actual primary refinement, including its cancellations.

The Pfaffian weights and graph weights are unchanged. The Hilbert-series calculation therefore gives $$S_X(G)=\frac54\cdot\frac{29}{20}+\frac38\cdot2 =\frac{41}{16}.$$ A piece of anticanonical degree $m$ and primary degree $j$ has numerical class $mc+\frac j2h$ on $G$. Averaging gives $[D_W]$. ◻

Proof. Let $B=(A=0)$ and $B'=(s=0)$; they are disjoint horizontal sections. The principal filtrations by the original hyperplane sections $x_2,x_1$ and by the original pencil section induce the following lower bounds for fixed coefficients in the refined series: $$S(W;B)\ge\frac5{16},\qquad S(W;B')\ge\frac{11}{16},\qquad S(W;D)\ge\frac5{32}.$$ Indeed their primary initial sections are $A,u,s$. The last is the strict pencil section of class $F$; the hyperplane section $f$ instead has initial form $g(w)s$. The first has multiplicity one along $B$, the last has multiplicity one along $B'$, and the odd section $u$ has coarse divisor $D/2$. Choose a basis simultaneously compatible with the principal filtration and the primary filtration. Divisibility by a power of the original section implies divisibility of its initial section by the same power. Averaging the dimensions proves the asserted inequalities from $S_X(H)=5/16$ and $S_X(F)=11/16$. Additional vertical components have nonnegative coefficients and do not weaken them.

Choose the horizontal section $C$ through $r$. If $C\ne B$, the coefficient of $h$ in $[D_W]$ and the bound for $B$ give $S(W;C)\le41/32-5/16=31/32$. If $C=B$, use $B'$ instead and obtain the stronger bound $19/32$. Lemma 43.6, applied with the fixed curve $D$, gives $$S(W^C;r)\le \bigl([D_W]-\tfrac5{32}D\bigr)\cdot C =1-\frac5{16}=\frac{11}{16}.$$ Here $C^2=0$ and $r\notin D$, so the different on $C$ has coefficient zero at $r$. Local Abban–Zhuang adjunction now gives $\min\{32/31,16/11\}=32/31$. ◻

Proof. Points of $G$ outside $D$ are covered by Lemma 47.2. At a finite nonzero base point $w=c$, smoothness of the fibre says $Q(c)\ne0$: along $Z$ its ordinary normal derivative is $Q(c)\,dv$. Lemma 47.4 therefore handles the whole ruling over such a point, in both normalizations $g=1$ and $g=w$. The remaining points of $D$ are exactly the endpoint cases treated in Proposition 47.5: the closed-orbit point when $g=1$, the pencil base point at infinity when $g=1$, and the coincident closed-orbit and pencil base point when $g=w$. In the last normalization, every other specified base point may be chosen finite and nonzero as above. These cases exhaust the exceptional surface above smooth points of the fibre.

All of the refined bounds just cited are at least $80/79$. The primary ratio is $A_X(G)/S_X(G)=3/(41/16)=48/41>1$. Applying the local Abban–Zhuang inequality to $G$ gives the lower bound $\min\{48/41,80/79\}=80/79$. Points where the fibre is singular are handled separately by the point-extraction lemmas and are not assumed to satisfy $Q(c)\ne0$. ◻

Proof. Put $z=w-c$, $r=\alpha+\gamma c$, and $U_c=A-cv$. Rescale the pencil section by the nonzero constant $g(c)$ and denote the resulting section again by $F$. Thus $F|_{w=c}=f|_{w=c}$. Record exponents of $(u,v,z)$, taking the normal degree $u+2v$ first, then the $z$-order, and then the $u$-order. The following leading vectors are supplied by actual degree-one anticanonical sections: $$(0,0,0),(0,0,1),(2,0,1),(3,0,0), (0,2,0),(1,2,0),(0,2,1).$$ Indeed use $1,z$, then $qzA-czF$, $qT-cFu$, $Fv$, $FT$, and $FzA$. All these are actual anticanonical sections. In particular $zF=(w-c)F$ is a difference of two products of a hyperplane section and the pencil section. In the case $g=w$, it can equally be written $cf-c^2(f/w)$; no multiplication by an additional hyperplane factor is used. On $z=0$ we have $A=u^2+cv$ and $F=ru^2+qv$, so the asserted initial terms are nonzero because $c,q\ne0$.

The degree-two section $$R=q^2FT^2-F^2\bigl(c(q+1)U_c^2+qc^2U_cv\bigr)$$ restricts on $z=0$ to $Fu^6$. This follows from the polynomial identity $$U^3=q^2U(U+cV)^2-(rU+qV) \bigl(c(q+1)U^2+qc^2UV\bigr),\qquad q=1+rc.$$ Thus $R$ supplies the normalized vector $(3,1/2,0)$. All its summands use two hyperplane factors and at most two pencil factors, so they are genuine sections of $2L$.

The hull of these eight vectors has volume $10/3$ and first moment $(79/24,21/8,35/24)$. Its centroid is $(79/80,63/80,7/16)$. The lattice is the full lattice after adjoining the remaining degree-one sections $u$ and $v$. The full-volume lemma therefore gives $$S(W;C_c)=7/16,\qquad S(W^{C_c};D\cap C_c)=79/160.$$ The factor $1/2$ in the second equality comes from the coarse uniformizer $u^2/v$ at the point of the different. The total expected degree on the weighted ruling is $S_X(G)/2=41/32$. At any other point, simultaneous bases give an upper bound $41/32-79/160=63/80$.

The discrepancy of $C_c$ is one. On the ruling the different has coefficient $1/2$ at $D\cap C_c$ and no other support. Hence Abban–Zhuang adjunction gives the lower bound $$\min\{16/7,\ (1/2)/(79/160),\ 1/(63/80)\}=80/79.$$ ◻

Proof. We give the three local computations together, retaining the primary degree in every calculation. Denote the primary degree by $j$. On the chart $s=1$ of the coarse exceptional surface, the seven $H$-sections and their primary degrees are (47.2)$$\label{f214:eq:F0-endpointH} (1,w,u,A,uA,wA,\chi s),\qquad (0,0,1,2,3,2,2),$$ and the two $F$-sections are $1,s$, of primary degrees zero and two. The branch cover is smooth, with coordinate $u$, and the local relation is $$u^2=Q(w)A-r(w)s,\qquad Q(0)=1.$$ The three cases are $$\begin{array}{c|cc} &r(w)&\chi\\\hline \mathrm{I}:\ \text{closed-orbit point not on }C&w&1\\ \mathrm{II}:\ C\cap Z\text{ not closed-orbit}&w&w\\ \mathrm{III}:\ C\cap Z\text{ closed-orbit}&w^2&w. \end{array}$$ In Case II the coordinate $w$ is the reciprocal of the original affine coordinate on the line: all $H$-sections are divided by the original coordinate. In particular the last $H$-section is $ws$, rather than $s$. This distinction changes the expected order. Multiplying each fixed-degree section by a common power of the unit $Q$ clears its denominator, and changes none of the orders used below.

In Cases I and II, blow up the coarse surface at the point. Its exceptional curve $K\cong\mathbb P^1$ has discrepancy $3/2$ for the pair with boundary $\frac12D$. On the cover its valuation has weights $(u,w)=(1,2)$ and ramification index two. Let $P_D\in K$ be the point of the strict transform of $D$. After taking this weighted initial form, take the lowest $u$ exponent. Record the resulting exponents and the primary degree as $(a,b,j)$. The coarse order at $P_D$ is $a/2$.

Besides the sections in (47.2) and their products with $s$, we use the degree-two section $$T=(uA)^2+(sw)(swA)-(wA)(sA).$$ Every factor on the right is a degree-one anticanonical section. On the chart $s=1$, put $N=u^2+w$, so $A=N/Q$. If $Q=1+\beta w+\gamma w^2$, then $$Q^2T=N\bigl(u^4+\beta w^3+\gamma w^4\bigr).$$ Its weighted initial form is $u^4(u^2+w)$, and its recorded vector is $(4,1,6)$. In Case II we also have $$T'=(uA)(suA)+(s\chi s)(swA)-(swA)(sA).$$ Since $\chi=w$, this is $sT$ as a local rational function, and has vector $(4,1,8)$. The notation $s\chi s$ denotes the product of the $F$-section $s$ with the $H$-section $\chi s$; thus every factor is again a genuine degree-one anticanonical section.

The convex hulls of the degree-normalized certified vectors have the following vertices: $$\begin{array}{c|l} \mathrm{I}&(0,0,0),(0,0,4),(0,1,0),(0,2,2),(0,2,4),\\ &(1,1,5),(2,0,2),(2,1/2,3)\\[2pt] \mathrm{II}&(0,0,0),(0,0,2),(0,1,0),(0,1,4),(0,2,2),(0,2,4),\\ &(1,1,5),(2,0,2),(2,0,4),(2,1/2,3),(2,1/2,4). \end{array}$$ Both have Euclidean volume $20/3$. The value lattice is $\{(a,b,j):a\equiv j\pmod2\}$, of index two; after including the anticanonical degree, the displayed actual sections generate this full lattice. Thus the normalized volume is $10/3$, the leading coefficient of $N_m$. The full-volume lemma gives the exact centroids $$\mathrm{I}:\left(\frac{49}{80},\frac{29}{40},\frac{41}{16}\right), \qquad \mathrm{II}:\left(\frac{117}{160},\frac{233}{320},\frac{41}{16}\right).$$ These figures follow directly by coning the listed rational faces to the origin and summing determinants; in particular their native-degree component provides an independent check against $S_X(G)=41/16$.

Dividing the cover weight $a+2b$ by two gives $$\begin{array}{c|cc} &S(W;K)&S(W^K;P_D)\\\hline \mathrm{I}&33/32&49/160\\ \mathrm{II}&35/32&117/320. \end{array}$$ The different on $K$ is $\frac12P_D$. Its residual divisors have total expected degree $S(W;K)$. Simultaneous bases at $P_D$ and any other point show that the latter point terms are at most $29/40$ and $233/320$, respectively. Hence Abban–Zhuang gives $$\mathrm{I}:\quad \min\left\{\frac{16}{11},\frac{80}{49},\frac{40}{29}\right\} =\frac{40}{29}, \qquad \mathrm{II}:\quad \min\left\{\frac{48}{35},\frac{160}{117},\frac{320}{233}\right\} =\frac{160}{117}.$$

Consider Case III. After rescaling the cover coordinate and $A$, its relation can be written $u^2=w^2s-Q(w)A$, with $Q(0)=1$. Blow up the coarse point with weights $(w,A)=(1,2)$. The exceptional curve $K\cong\mathbb P(1,2)$ has discrepancy $3-\frac12\cdot2=2$. On the cover this is the ordinary blowup of $(w,u)$; its divisorial ramification index is one. Put $u=w+Y$, and refine first by total degree in $w,Y$, and then by the $Y$ order. This is the point $P_A$ of $A=0$ on $K$.

The degree-one sections give the thirteen vectors, in the order $(Y,w,j)$, $$\begin{gathered} (0,0,0),(0,1,0),(0,1,1),(1,1,2),(1,2,3),(1,2,2),(0,1,2),\\ (0,0,2),(0,1,3),(1,1,4),(1,2,5),(1,2,4),(0,1,4). \end{gathered}$$ There are two more certified degree-two sections: $$U=(uA)^2-(wA)(swA)=-Q A^3, \qquad U'=(uA)(suA)-(swA)^2=-sQ A^3.$$ Since the initial form of $A$ is $-Y(2w+Y)$, their vectors are $(3,3,6)$ and $(3,3,8)$. The resulting hull has vertices $$(0,0,0),(0,0,2),(0,1,0),(0,1,4),(1,1,4), (1,2,2),(1,2,5),(3/2,3/2,3),(3/2,3/2,4).$$ It has volume $10/3$, generates the full value lattice, and has centroid $$\left(\frac{187}{320},\frac{353}{320},\frac{41}{16}\right).$$ Thus $S(W;K)=27/16$ and $S(W^K;P_A)=187/320$. The total expected degree on the coarse weighted curve is $\frac12S(W;K)=27/32$. At any other point it is therefore at most $83/320$.

The different of $K$ consists of two distinct points of coefficient $1/2$: its quotient point and its intersection with the strict transform of $D$. Neither is $P_A$. The two cover points $u=w$ and $u=-w$ map to $P_A$; they have the same vanishing order because each fixed primary-degree piece has fixed parity in $u$. Consequently the curve threshold is at least $\min\{320/187,160/83\}$, while the first ratio is $2/(27/16)=32/27$. This proves the last asserted bound. ◻