Points of type $A_3$

Proof. By Lemma 43.4, the image of $p$ belongs to the two-dimensional orbit. In the translated chart used there the fibre has equation $$u^2+(b+1)uv+bv^2+bu^3+buvw-u^2w-vw^2, \qquad b\ne1.$$ With weights $(2,2,1)$ its initial form is $B_b=u^2+(b+1)uv+bv^2-vw^2$. The other two normal hyperplane coordinates have initial forms $vw$ and $v(u+v)$. Put $c=(1-b)^{-1}$ and make the weighted linear change $u'=cu+(c-1)v$, $v'=v$, $w'=cw$. It gives $P'=cP$, $Q'=cQ$ and $B'_0=c^2B_b$. Thus it identifies the complete space of leading hyperplane sections, together with its distinguished fibre section, with the displayed model.

The equations (50.1) are the five maximal Pfaffians of $$\begin{pmatrix} 0&h&u+v&-w&0\\ -h&0&-P&u&-v\\ -(u+v)&P&0&B&-Q\\ w&-u&-B&0&P\\ 0&v&Q&-P&0 \end{pmatrix}.$$ After inverting $h$ they give the polynomial algebra in $h^{\pm1},u,v,w$. On $h=0$, the branch $v\ne0$ has $w=0$, $u=-v$ and $vB+vQ+P^2=0$, so its closure has dimension three. On the branch $v=0$, the remaining equations in $u,w,P,Q$ are $u^2=Pw$, $uQ=P^2$, $uP=wQ$. They define an affine cone over a twisted cubic; $B$ is free, and this branch also has dimension three. The Pfaffian ideal has height three and its resolution is Cohen–Macaulay. Thus $h$ is regular and the algebra is a domain. Its Hilbert series is $(1+3t+t^2)/(1-t)^4$. The algebra generated by the displayed leading hyperplane sections embeds in the associated graded section algebra. Equality of their Hilbert functions proves equality of the algebras. In particular, this argument accounts for every cancellation among sections.

The substituted Cox algebra is also a domain. On $e\ne0$ this follows from the preceding paragraph. The boundary $h=B=0$ in (50.1) has dimension two; hence after substitution $e=0$ has dimension four, whereas the whole algebra has dimension five. In particular $h,B$ are a regular sequence in the primary initial algebra. Lemma 43.1 proves that $e$ is regular, that the graph is a domain, and that its divisor-degree components give the actual full initial global section algebra.

Here is the weight calculation separately. Let $t$ record $H$-degree, $r$ record $F$-degree, and $s$ record extraction weight. The weighted Hilbert series is $$\frac{1-t^2\sum_{d\in\mathcal D}s^d +t^3\sum_{d\in\mathcal D}s^{11-d}-t^5s^{11}} {(1-ts)^{}(1-ts^2)^2(1-ts^3)(1-ts^4) (1-r)(1-rs^4)(1-t/r)}, \qquad\mathcal D=(3,4,4,5,6).$$ The coefficient of $t^mr^m$ at $s=1$ is $N_m=(2m+1)(5m^2+5m+3)/3$. Differentiating in $s$ and then setting $s=1$ gives the total weight $$T_m=\frac{m(355m^3+510m^2+377m+102)}{24}.$$ For example, this coefficient can be checked by summing over the exponent $i$ of $e$: a numerator term $\varepsilon t^as^d$ contributes $$\varepsilon\sum_{i=0}^{m-a} {m-a-i+4\choose4}(m+i+1) \left(\frac{12}{5}(m-a-i)+2(m+i)+d\right).$$ It follows that $S_X(G)=\lim T_m/(mN_m)=71/16$. The discrepancy is $2+2+1=5$. ◻

Proof. On $w=1$, write $b=B$. Then $v=(u^2-b)/(1-u)$. Multiplication by the common unit $(1-u)^{2m}$ clears the denominators of all degree-$m$ sections. The flag $D\supset q$ therefore records the lowest $b$-power and then the lowest $u$-power. With coordinates $(\alpha,\beta,j)$ recording the $u$-order, the $b$-order, and the native degree, respectively, the degree-one pivot vectors are $$\begin{gathered} (0,0,0),(0,0,1),(0,1,4),(0,1,5),(0,2,8),\\ (1,0,2),(1,1,6),(2,0,2),(2,0,3),\\ (2,1,6),(2,1,7),(3,0,4),(3,1,8). \end{gathered}$$ For example $Q|_D=u^3/(1-u)^2$, which gives $(3,0,4)$, and $BQ$ gives $(3,1,8)$. These thirteen vectors are distinct, so these leading terms give a basis in degree one. Their convex hull has vertices $$(0,0,0),(0,0,1),(0,1,5),(0,2,8), (2,0,2),(2,1,6),(3,0,4),(3,1,8).$$ Its volume is $10/3$, its first moment is $(95/24,55/24,355/24)$, and its centroid is $$(19/16,11/16,71/16).$$ The degree-prefixed vectors generate the full lattice: use $1,w,u,B$. The semigroup-volume lemma identifies these with the actual asymptotic moments, since $N_m=(10/3)m^3+O(m^2)$. This proves the first two claims.

For the boundary, take the lowest $w$-power and then the highest $u$-power within each native-degree piece. The degree-one leading terms, together with the degree-two sections $Q^2-Bv^2$ and $B(Q^2-Bv^2)$, give a convex hull with vertices $$\begin{gathered} (0,0,0),(0,0,1),(0,1,0),(0,1,1),(2,0,1),\\ (2,1,1),(3,1,0),(4,0,0), (\tfrac12,\tfrac32,0),(\tfrac32,\tfrac32,0). \end{gathered}$$ Here the coordinates are the normalized exponents of $(u,v,w)$. The two new vertices follow from $Q^2-Bv^2=v^3(u+v+w^2)$. The hull has volume $10/3$ and centroid $(453/320,187/320,7/16)$. Its $w$-moment is therefore $7/16$. On the coarse surface, the divisor of the weighted section $w$ is $\frac12\{z=0\}$, giving $S(W;\{z=0\})=7/32$. All the volume and moment assertions are obtained by triangulating these explicitly listed rational polyhedra and applying the determinant formula for the volume and centroid of a tetrahedron. ◻

Proof. The point $q$ is smooth and lies outside $\Delta_G$. Blow it up with weights $(1,2)$ in the coordinates $(u,v)$, and denote the exceptional curve by $K=\mathbb P(1,2)$. Its log discrepancy is $3$, and its different has coefficient $1/2$ at the point $u=0$. Put $y=v-u^2$ and let $p\in K$ be the ordinary point $y=0$. Take the lowest $(1,2)$-weight and then the lowest $y$-power. The degree-one pivot vectors $(\alpha,\beta,j)$ are exactly the first thirteen vectors in the preceding proof. Indeed the initial terms of $v,P,Q,B$ are respectively $u^2+y,u^2+y,u(u^2+y),-y$. The same full-volume certificate gives $$S(W;K)=\frac{19}{16}+2\frac{11}{16}=\frac{41}{16}, \qquad S(W^K;p)=\frac{11}{16}.$$ The expected degree of the residual divisors on $K$ is $41/32$. For any ordinary point $r\ne p$, choose a basis simultaneously adapted to $p$ and $r$ in every graded piece. The sum of their multiplicities is at most the degree; hence $S(W^K;r)\le19/32$.

At the orbifold point use instead the lowest $u$-power after taking the $(1,2)$-initial form. The degree-one pivot vectors are $$\begin{gathered} (0,0,0),(0,0,1),(0,1,2),(0,1,3),(0,1,4),(0,1,5),\\ (0,2,6),(0,2,7),(0,2,8),(1,0,2), (1,1,4),(1,1,6),(1,2,8). \end{gathered}$$ Their hull again has volume $10/3$, with centroid $$(7/16,17/16,71/16).$$ Since $u^2/v$ is a coarse uniformizer at $u=0$, this gives $S(W^K;\{u=0\})=7/32<1/2$. Consequently $$\delta(K,\operatorname{Diff}_K;W^K) \ge\min\{16/11,32/19,16/7\}=16/11.$$ Applying Abban–Zhuang gives the required bound $\min\{3/(41/16),16/11\}=48/41$. ◻

Proof. Let $H$ denote the line class on the coarse surface $G=\mathbb P^2$. The expected class of a basis divisor is $(71/32)H$, and $D\sim2H$. On $D$, the expected degree after restriction is $2(71/32-2\cdot11/16)=27/16$. At every point other than $q$, simultaneous bases with the $q$-filtration therefore give a refined $S$-invariant at most $27/16-19/16=1/2$. The different of $D$ has coefficient $1/2$ at its two intersections with $\{z=0\}$, and is zero elsewhere. Thus the refined threshold is at least one on $D\setminus\{q\}$; the preceding lemma treats $q$. Since $S(W;D)=11/16<1$, Abban–Zhuang proves the assertion along $D$.

For completeness, consider a point outside $D$. Choose a line $C$ through it; if the point lies on $\{z=0\}$, take this boundary line itself. In each $C$-graded piece choose a basis simultaneously adapted to the induced $D$-filtration and the chosen point. These bases lift to bases adapted to both original divisor filtrations. In particular the sum of their induced $D$-weights is the original $D$-weight sum. Every residual divisor contains its corresponding multiple of $D|_C$, whose support avoids the point under consideration. If $s=S(W;C)$, its point invariant is consequently at most $$\frac{71}{32}-s-2\frac{11}{16} =\frac{27}{32}-s\le\frac{27}{32}.$$ Before restriction, a common basis for $C$ and $D$ gives an effective averaged residual class $(27/32-s)H$. Hence $s\le27/32$. For $C\ne\{z=0\}$ this is below its discrepancy $1$. For the boundary line its discrepancy is $1/2$ and its invariant is $7/32$, by Lemma 50.2; the restriction bound is then $5/8<1$. The induced different on this boundary line is zero. Abban–Zhuang proves the assertion off $D$ as well. ◻