The smooth corner of an $A_1$ fibre

Proof. Take the flag equation and the fibre equation to be $$x_0y_0+x_1y_1+x_2y_2=0,\qquad x_0y_1+x_2y_2=0.$$ The point under consideration is $([0:1:0],[0:0:1])$. On the chart $x_1=y_2=1$ put $u=x_0$, $a=x_2$, $b=y_0$. Then $y_1=-a-ub$, and the fibre equation is $$f=a(1-u)-u^2b.$$ The bad line has $u=a=0$, and its singular endpoint is $b=\infty$. Put $U=u(a+ub)$ and $V=a(a+ub)$. A basis of restricted bilinear forms is $$\mathcal H=\langle1,u,a,b,ub,ab,U,V\rangle.$$ For example, $U=-x_0y_1$ and $V=-x_2y_1$.

The other pencil generator $g$ is a unit at $p$. After rescaling, $$g=1+\alpha u+\beta b+\text{terms of weighted degree at least }2.$$ Moreover $\beta\ne0$: the restriction of $g$ to the bad line is $1+\beta b$, and its value at the singular endpoint is nonzero, since the base curve misses the singular points of the pencil. Multiplying the local anticanonical trivialization by the unit $g$, the section space becomes $g\mathcal H+f\mathcal H$.

In the order of (37.C1), choose the following fifteen sections: $$\begin{aligned} &g,gu,gb,f,gub,\\ &fu,fb,fa,R,fub,\\ &S,fab,fU,T,fV, \end{aligned}$$ where $$\begin{aligned} R&=gU-fu,\\ S&=gab-fb-(\alpha+1)fub,\\ T&=gV-fa-fub-(\alpha+1)fU-\beta fab. \end{aligned}$$ The weighted initial forms needed for the nontrivial reductions are: $$\begin{aligned} \operatorname{in}(R)&=u^2b,\\ \operatorname{in}(S)&=\beta ab^2+u^2b^2,\\ \operatorname{in}(fU)&=ua^2+u^2ab,\\ \operatorname{in}(T)&=\beta uab^2+u^2ab+u^3b^2,\\ \operatorname{in}(fV)&=a^3+uba^2. \end{aligned}$$ Taking least weights for $(h,1,0)$ gives (37.C1), since $h>1$ and $\beta\ne0$. Terms of higher weighted degree in $g$ cannot affect these identities. This proves the assertion for arbitrary $g$, including all special values of its remaining coefficients. ◻

Proof. Here is an integer certificate which proves the assertion in every degree. Number the exponent vectors in (37.C1) from $v_0$ to $v_{14}$ in the displayed order. The following twenty-four quadruples specify unimodular tetrahedra: (37.C2)$$\begin{array}{ccc} (0,1,4,5)&(1,4,5,8)&(0,2,3,4)\\ (0,3,4,5)&(7,11,12,14)&(11,12,13,14)\\ (3,7,11,12)&(2,6,10,13)&(6,10,11,13)\\ (2,4,6,13)&(2,3,4,6)&(3,4,5,6)\\ (3,5,6,12)&(3,6,11,12)&(4,5,6,9)\\ (4,6,9,13)&(4,8,9,13)&(4,5,8,9)\\ (5,6,9,12)&(5,8,9,12)&(8,9,12,13)\\ (9,11,12,13)&(6,9,11,12)&(6,9,11,13). \end{array} \tag{37.C2}$$ They form a simplicial complex. This can be checked without any geometric guess: lift the fifteen vectors to the respective heights $$\begin{aligned} (h_0,\ldots,h_7)&=(0,123,77,99,173,216,177,424),\\ (h_8,\ldots,h_{14})&=(483,308,516,495,521,582,906). \end{aligned}$$ For each quadruple, the affine function through its four lifted vertices is strictly below all eleven other lifted vertices. Consequently the tetrahedra are lower facets of one convex hull and meet along common faces. Each determinant $\det(v_j-v_i,v_k-v_i,v_l-v_i)$ has absolute value one. Their numbers of faces in dimensions $0,1,2,3$ are $15,50,60,24$.

Let $Q$ be their union. Every lattice point in $mQ$ is a sum of $m$ vertices of one tetrahedron, by unimodularity. It therefore gives a product of $m$ of the monomials (37.C1). Counting the relative interiors of all faces gives, for every $m\geqslant1$, $$\#(mQ\cap\mathbb Z^3) =15+50\binom{m-1}{1}+60\binom{m-1}{2} +24\binom{m-1}{3} =4m^3+6m^2+4m+1.$$ This is exactly $h^0(X,-mK_X)$. The inclusion supplied by Lemma 115.2 is therefore an equality in every degree. In particular, this argument does not infer generation of an initial algebra from computations in finitely many degrees.

Each tetrahedron has volume $1/6$. The sums of its four vertex coordinates, added over all tetrahedra in (37.C2), are $61,95,72$. It follows that $$\operatorname{vol}(Q)=4,\qquad \int_Q(u,a,b)=\left(\frac{61}{24},\frac{95}{24},3\right).$$ Division by $4$ gives the claimed moments. ◻

Proof of Proposition 115.1. The exceptional surface is $G=\mathbb P(1,2,1)$, the different is zero, and $A_X(G)=4$. By the preceding lemma, $$S_X(G)=\frac{61}{96}+2\cdot\frac{95}{96}+\frac34 =\frac{323}{96}.$$ We justify the bound on the refined series for arbitrary valuations. Fix a primary weighted-degree piece and a divisorial valuation over $G$. Choose $h$ sufficiently large that the weights $(h,1,0)$ separate all monomials in this piece. Choose a basis simultaneously compatible with the valuation filtration and this weight filtration. Only two filtrations on the already fixed primary piece are used. Lemma 115.3 identifies its limit with the corresponding monomial basis. Lower semicontinuity of log canonical thresholds applies to this one-parameter degeneration.

The limiting normalized basis divisor is supported on the three toric coordinate curves. Its coefficients tend to $61/96,95/96,3/4$. The toric boundary of $\mathbb P(1,2,1)$ is log canonical, including at its $\frac12(1,1)$ point. Hence the thresholds of these normalized basis divisors have lower limit at least $96/95$. The same bound holds for every valuation-adapted basis by the preceding degeneration. Thus $$\delta(G;W^G_{\bullet,\bullet})\geqslant\frac{96}{95}.$$ Abban–Zhuang adjunction now gives $$\delta_p(X)\geqslant \min\left\{\frac{384}{323},\frac{96}{95}\right\} =\frac{96}{95}.$$ The proof uses a weighted point extraction, actual section spaces, and an integer certificate for their refined initial algebra. It requires no Zariski decomposition on the extracted threefold. ◻