The singular point of an $A_1$ fibre

Proof. Write the two bilinear forms defining the flag threefold and this fibre as $$x_0y_0+x_1y_1+x_2y_2,\qquad x_0y_1+x_2y_2.$$ The singular point is $([0:1:0],[1:0:0])$. In the chart $x_1=y_0=1$ put $a=x_2$, $b=y_1$, $c=y_2$. The flag equation gives $x_0=-b-ac$, and the fibre equation becomes $$f=ac-b^2-abc.$$ A basis of the restrictions of bilinear forms can be chosen with the following eight least homogeneous terms: $$1,a,b,c,ab,b^2,bc,F.$$ Indeed one can use the actual functions $1,a,b,c,ab,ac-f,bc+ac^2,f$. The other generator $g$ of the pencil is a unit at $p$, since its base curve misses every singular point of a fibre. Rescale it so that $g(p)=1$. The base coordinate is $t=f/g$. The anticanonical bundle is $H+\mathcal O_{\mathbb P^1}(1)$ in the incidence presentation of $X$. Its sections are products of the eight bilinear sections with $1,t$, modulo the one relation $f=tg$. Removing $f$ from the first group gives precisely the fifteen least terms in the statement. Thus the displayed algebra is contained in the actual associated graded algebra.

We prove equality in every degree, rather than assuming generation of an associated graded ring in degree one. A monomial $a^r b^d c^q$ with $rq=0$ requires exactly $n$ factors among $a,b,c,ab,b^2,bc$, with $n$ as in (37.A1). For example, when $q=0$, pair as many $a$’s as possible with $b$’s, and use $b^2$ for the remaining pairs of $b$’s. Its $n$ factors may each carry one extra $F$, and the remaining $m-n$ slots may carry $1,F$, or $F^2$. This proves that every monomial in (37.A1) belongs to degree $m$. Conversely, reduce every factor $ac$ by $ac=F+b^2$. For the first summand $n$ decreases by at least one while $s$ increases by one; for the second $n$ does not increase. Hence both inequalities are preserved. Normal monomials are independent, since $ac-F-b^2$ has leading monomial $ac$ and they are its standard monomials.

For $n\ge1$ there are exactly $6n$ normal monomials of cost $n$. For $r=q=0$ there are two. For each of the two branches $r>0$ or $q>0$, there are $2(n-1)+(n+1)=3n-1$. Consequently the number in degree $m$ is $$(2m+1)+\sum_{n=1}^m6n(2m-n+1) =4m^3+6m^2+4m+1=h^0(X,-mK_X).$$ The inclusion of graded algebras is therefore equality in every degree. In particular these are the actual refined section spaces. ◻

Proof. In (37.A1) the prescribed gradings give $q=0$, $r=k$ when $k\ge0$ (and the symmetric formula when $k<0$), and $d=j-|k|-2s$. The two inequalities in (37.A1) say exactly $l\le s\le u$. All integers in this interval occur. Factoring its two endpoints proves (37.A2). This also proves completeness; the argument does not replace a section space by its Newton polygon. ◻

Proof. Here are explicit finite sums for the calculation. Sum over $-m\le k\le m$, $|k|\le j\le4m$, retaining only $u\ge l$ as in Lemma 112.3. Every summand has weight $D+1$. The quantities to sum are $$\max(k,0),\quad\max(-k,0),\quad j-|k|-2u,\quad l,\quad D.$$ For even $m$ the resulting sums, in this order, are $$\begin{aligned} A_m=C_m&=\frac{m(m+1)(5m^2+5m+2)}6,\\ B_m&=\frac{m(m^3+18m^2+18m+8)}8,\\ Q_m&=\frac{m(65m^3+130m^2+100m+32)}{48},\\ R_m&=\frac{m(67m^3+38m^2+8m-8)}{24}. \end{aligned}$$ These identities follow by substituting the displayed lower and upper endpoints and summing the arithmetic progressions. They also provide a direct calculation of all the constants: divide the first four sums by $mN_m$ and the last by $2mN_m$. Floors and ceilings change only lower-order terms, so the same limits hold without the parity restriction. Finally the average ordinary degree is $$2\alpha+\beta+2\varphi+4h=\frac{121}{48}.$$ By definition this is $S_X(G)$. The sum of the primary degrees is in fact $$\frac{m(121m^3+170m^2+113m+28)}{12}$$ for every $m\ge0$. To check this without a parity argument, group the normal monomials by the cost $n$ in (37.A1). For $n\ge1$, the $6n$ monomials $a^rb^dc^q$ have total ordinary degree $$(4n-1)+2\left((n-1)(4n-1)+\frac{3n(n+1)}2\right) =11n^2-3n+1.$$ The first term comes from $r=q=0$; in either remaining branch, $1\le r<n$ contributes degrees $2n-1,2n$, while $r=n$ contributes degrees $n,n+1,\ldots,2n$ (and similarly for $q$). Writing $M=2m-n$, summation over $0\le s\le M$ consequently gives $$2m(2m+1)+\sum_{n=1}^m \left((2m-n+1)(11n^2-3n+1) +6n(2m-n)(2m-n+1)\right),$$ which evaluates to the stated polynomial. Equality of the normal algebra with the actual refinement was proved in Lemma 112.2, so these are the valuation weights. ◻

Proof. The first four quantities bound the coefficients of all components. Away from $[0:0:1]$ and $[1:0:0]$ the support has simple normal crossings. At each of these two points the pencil is $a=\lambda b^2$ in local coordinates. Two successive ordinary blow-ups resolve all its members simultaneously. The exceptional valuations have weights $(1,1)$ and $(2,1)$, and log discrepancies two and three. Their orders on $B$ are at most $\alpha+\beta+\varphi+3h$ and exactly $2\alpha+\beta+2\varphi+4h$, respectively. After the second blow-up all distinct branches meet the last exceptional curve at different points. Thus the stated inequalities give a log resolution with all boundary coefficients at most one. This proves the claim for every valuation, including valuations above the exceptional curves. ◻

Proof of Proposition 112.1. For an arbitrary basis of a complete degree-$D$ binary space, the sum of the vanishing orders at any prescribed point of $\mathbb P^1$ is at most $D(D+1)/2$, whereas its total degree is $D(D+1)$. Consequently arbitrary bases in the components (37.A2) give divisors of the form in Lemma 112.5. The limiting component bounds are $$\frac{107}{192},\quad\frac{35}{48},\quad\frac{11}{16}, \quad\frac{67}{192}.$$ The limiting exceptional orders are $13/8$ and $121/48$, whose ratios to the discrepancies are $13/16$ and $121/144$. The largest of all six ratios is $121/144$. Hence for every $\tau<144/121$ all sufficiently high-degree basis divisors in these components give log canonical pairs after multiplication by $\tau$, uniformly over the choices of bases.

Now fix a divisorial valuation over $G$. In each already fixed primary-degree piece choose a basis compatible with its valuation filtration and the displayed torus-weight filtration; this uses only the two-filtration lemma. Its least-weight terms are bases in (37.A2). Degeneration under this one-parameter subgroup and lower semicontinuity of log canonical thresholds give the same bound for the original valuation-adapted basis. Passing to the limit gives $$\delta(G;W^G)\ge\frac{144}{121}.$$ The ordinary point blow-up is a plt extraction and has zero different. Abban–Zhuang adjunction therefore gives $$\delta_p(X)\ge\min\left\{\frac{3}{121/48}, \frac{144}{121}\right\} =\frac{144}{121}.$$ No Zariski decomposition on $Y$ has been used. The calculation used actual sections, an exact Hilbert count in every degree, and a resolution of a conic pencil on its exceptional plane. ◻