The singular point of an $A_2$ fibre

Proof. Choose the flag and fibre equations $$x_0y_0+x_1y_1+x_2y_2=0,\qquad x_0y_1+x_1y_2=0.$$ The singular point is $([0:0:1],[1:0:0])$. On $x_2=y_0=1$ set $a=x_1$, $b=y_1$, $c=y_2$. Then $x_0=-ab-c$ and the fibre equation is $(a-b)c-ab^2$. Put $A=a-b$ and $C=c-b^2$. These are regular parameters and the fibre equation becomes $F=AC-b^3$. The following eight actual bilinear sections form a basis: $$\begin{gathered} 1,\ A,\ b,\ C+b^2,\ Ab+b^2,\ bC+Ab^2+2b^3,\\ (C+Ab+2b^2)(C+b^2),\ F. \end{gathered}$$ Their least terms for weights $(3,2,3)$ are $1,A,b,C,b^2,bC,C^2,F$. For example the sixth and seventh sections are $-x_0y_1$ and $-x_0y_2$. The other pencil generator $g$ is a unit at this point; normalize $g(p)=1$. In the incidence model the base coordinate is $t=F/g$. Anticanonical sections are products of the eight bilinear sections with $1,t$, modulo $F=tg$. Removing $F$ from the first group gives seven sections and $t$ times all eight. Their initial forms are exactly those displayed below. This proves their inclusion in the actual associated graded section algebra. The next lemma proves equality by an exact dimension count in every degree. ◻

Proof. First consider the seven forms $1,A,b,C,b^2,bC,C^2$. The least number of their nonconstant factors needed to write $A^ab^dC^c$ is $a+\lceil(d+c)/2\rceil$. Thus a monomial satisfying the displayed conditions is a product of $m$ of the fifteen initial forms: use $n$ nonconstant factors, multiply any of these by $F$ when necessary, and use $1,F,F^2$ for the remaining $m-n$ factors. The possible total powers of $F$ are precisely $0,\ldots,2m-n$.

Conversely, reduce every product by $AC=F+b^3$. Replacing $AC$ by $b^3$ preserves $a+\lceil(d+c)/2\rceil$. Replacing it by $F$ increases $s$ by one and decreases this integer by at least one. Hence both inequalities are preserved. The resulting monomials have $ac=0$. They are linearly independent: they are the normal monomials in $\mathbb C[A,b,C,F]/(AC-b^3-F)$ for a monomial order with leading term $AC$.

For fixed $n\geqslant1$, there are $4n+1$ possibilities with $a=0$, and $2n-1$ with $a>0$ and $c=0$. There are therefore $6n$ possibilities. The total number is $$2m+1+\sum_{n=1}^m6n(2m-n+1) =(2m+1)(2m^2+2m+1).$$ This equals $h^0(X,-mK_X)$, by Riemann–Roch and vanishing, since $(-K_X)^3=24$. Thus products of the displayed initial forms already have the dimension of the entire section space. In particular, the calculation identifies the actual initial algebra in every degree. ◻

Proof. The two limiting regions are $$\begin{aligned} P_A&=\{(a,d,s):a,d,s\geqslant0,\ a+d/2\leqslant1, \ s+a+d/2\leqslant2\},\\ P_C&=\{(c,d,s):c,d,s\geqslant0,\ (c+d)/2\leqslant1, \ s+(c+d)/2\leqslant2\}. \end{aligned}$$ Their overlap contributes only $O(m^2)$ lattice points. Their volumes sum to $4$. Integrating first in $s$ gives $$\begin{gathered} \int_{P_A}a=\frac5{12},\qquad \int_{P_C}c=\frac53,\qquad \int_{P_A}d+\int_{P_C}d=\frac52,\\ \int_{P_A}s+\int_{P_C}s=\frac{11}{4}. \end{gathered}$$ Division by $4$ gives the asserted moments. The order along $G$ of a basis monomial is $3a+2d+3c+6s$. Its mean is therefore $$3\cdot\frac5{48}+2\cdot\frac58+ 3\cdot\frac5{12}+6\cdot\frac{11}{16}=\frac{111}{16}.$$ The normal basis is compatible with the weighted order, so this is $S_X(G)$. There is also an exact all-degree verification. Among the $6n$ monomials $A^ab^dC^c$ of cost $n\ge1$, the branch $a=0$ has $d+c=2n-1$ or $2n$ and total weight $20n^2$. The branch $a>0$, $c=0$ has total weight $$\sum_{a=1}^{n-1}(8n-2a-2)+3n=7n^2-6n+2.$$ Thus their combined weight is $27n^2-6n+2$. Including powers of $F$, whose weight is six, gives $$\begin{aligned} &6m(2m+1)+\sum_{n=1}^m\bigl( (2m-n+1)(27n^2-6n+2)\\ &\quad +18n(2m-n)(2m-n+1)\bigr) =\frac{m(111m^3+158m^2+105m+26)}4. \end{aligned}$$ Dividing by $m(4m^3+6m^2+4m+1)$ gives the same limit. ◻

Proof. The coarse description follows by taking the third Veronese of the weighted coordinate ring. The generic stabilizer on $b=0$ has order three, giving the stated different.

For the monomials with $a=r\geqslant0,c=0$, write $d=j+3k$, where $0\leqslant j\leqslant2$, and put $\ell=k+s$. Fixing the primary weighted degree and the torus weight fixes $r,j,\ell$. The corresponding component is $$A^r b^jF^{\ell-K}\, \operatorname{Sym}^{K}\langle z,F\rangle, \quad K=\min\left\{\ell, \left\lfloor\frac{2m-2r-j}{3}\right\rfloor, 4m-2\ell-2r-j\right\}.$$ Indeed, its normal monomials have exactly $0\leqslant k\leqslant K$. For $a=0,c=r>0$ the same statement holds with $C^r$ in place of $A^r$ and $$K=\min\left\{\ell, \left\lfloor\frac{2m-r-j}{3}\right\rfloor, 4m-2\ell-r-j\right\}.$$ Thus these are complete binary spaces, including at the ends of the ranges.

In a complete binary space of degree $K$, the sum of the orders of any basis at a fixed point of $\mathbb P^1$ is at most $0+1+\cdots+K$. The normal basis has mean $k=K/2$ in each block. Since $d=j+3k$ and $j\leqslant2$, Lemma 113.3 gives mean $K/2=5/24$. The mean fixed exponent of $F$ is consequently $11/16-5/24=23/48$. The other two fixed coefficients are the moments of $a,c$. The factor $b^j$ has normalized coefficient tending to zero. ◻

Proof. We explain the estimate for divisors obtained from torus-compatible bases. This suffices: fix one primary weighted-degree piece of the refined series. For any divisorial valuation choose a basis of this piece simultaneously compatible with its valuation filtration and the torus filtration. These are two filtrations on one vector space; the primary weighted degree has already been fixed. Repeat for each primary piece and take the torus limit. The initial bases remain bases of the displayed complete binary spaces. Lower semicontinuity of log canonical thresholds reduces the assertion to these limits.

Resolve the vertex of $G$ and blow up the two smooth base points of the pencil $AC=\lambda z$. On this surface the strict transforms of all pencil members, the two coordinate lines, the different, and the exceptional curves have simple normal crossings. Distinct pencil members meet each exceptional curve over a base point at distinct points. The reducible member $AC=0$ has its two components separated by the resolution of the vertex.

The limiting expected orders and log discrepancies of the relevant prime divisors are as follows. The boundary is always $(2/3)\{z=0\}$. $$\begin{array}{c|c|c} \text{divisor}&\text{expected order}&\text{log discrepancy}\\ \hline A=0&5/16&1\\ C=0&5/8&1\\ z=0&5/24&1/3\\ F=0&11/16&1\\ AC=\lambda z,\quad\lambda\ne0,1,\infty&\leqslant5/24&1\\ \text{exceptional curve over }[0:1:0]&1&4/3\\ \text{exceptional curve over }[1:0:0]&21/16&4/3\\ \text{exceptional curve over }[0:0:1]&15/32&1 \end{array}$$ The first five entries follow from Lemma 113.4; the member $F=0$ has its additional fixed coefficient $23/48$. At $[0:1:0]$ the orders of $A,z,F$ are all one, while $C$ is a unit. Hence the expected order is $5/48+(5/8)/3+11/16=1$. At $[1:0:0]$ it is $5/12+(5/8)/3+11/16=21/16$. The resolution of the $\frac12(1,1)$ vertex has order $1/2$ on each of $A,C$ and zero on $z,F$. The reducible pencil member contributes at most $5/24$, so its expected order is $(5/48+5/12)/2+5/24=15/32$.

Every ratio of the middle entry to the last is at most $63/64$. The simple normal crossings criterion therefore proves the claim. More precisely, for each real $t<64/63$, all the relevant coefficients satisfy the log canonical inequalities for all sufficiently large $m$. The preceding binary bound is uniform in the pencil member, so the same conclusion holds for every valuation-adapted basis. Passing to the asymptotic threshold and then letting $t$ tend to $64/63$ gives the assertion. ◻

Proof. The weighted blow-up is plt and $A_X(G)=3+2+3=8$. Apply the Abban–Zhuang inequality using Lemmas 113.3 and 113.5. ◻