The point of type $A_4$

Proof. Use the global section graph of Lemma 43.1, with variables $a,b,e$ of divisor degrees $H-E,H-E,E$, respectively. For its candidate initial algebra replace $x_0,R$ in (51.1) by $ae,be$, and denote the resulting ring by $\mathcal R$. On the open set $ae\ne0$ its five equations are the polynomial graph (51.2). The locus $e=0$ is the affine cone over $V_5\cap\{x_0=R=0\}$, with the two free variables $a,b$; it has dimension four. The locus $a=0,e\ne0$ also has dimension four. The Pfaffian ideal therefore has codimension three. Its Pfaffian resolution makes the quotient Cohen–Macaulay, and the boundary has smaller dimension than the open graph. Thus $\mathcal R$ is a domain.

Give $a,e$ weight zero and $b$ weight five. The five Pfaffians have weights $4,8,6,5,7$. Replacing $x_0$ by $\ell$ and $R$ by $R+\gamma Q$ changes these equations only by terms of larger weight. The initial Pfaffian ring has the same bigraded Hilbert series as the original Cox presentation, by their codimension-three Pfaffian resolutions. Hence it is the entire associated graded ring. Localization at $ae$ is injective on it and on each divisor-degree component; it consequently gives the geometric point filtration, including all cancellations.

A Cox monomial of degree $mL$ with $h$ factors among the five $H$ variables and exponent $q$ in $b$ has exponents $2m-h-q$ in $a$ and $m-h$ in $e$. This gives exactly the two displayed inequalities. After setting $a=e=1$, the degree-one generators are $$\{1,u,P,Q,v,w,R\}\cup R\{1,u,P,Q,v,w,R\}.$$ The repetition of $R$ leaves thirteen generators. Every allowed monomial factors into $m$ such generators: put at most one $R$ alongside each of its $h$ $H$ factors and at most two $R$ factors in each remaining slot. Finally Riemann–Roch, or the same Pfaffian resolution, gives the displayed formula for $N_m$. ◻

Proof. Here is a separate Hilbert calculation. If $s_H$ is the sum of the weights of the five $H$ variables, $s_{ab}=w_a+w_b$, and $D$ is the last syzygy weight, the Pfaffian resolution has weight numerator $$1-\sum_{i=1}^5t^{2H}q^{d_i} +\sum_{i=1}^5t^{3H}q^{D-d_i}-t^{5H}q^D, \qquad \sum d_i=2D.$$ Differentiate with respect to $q$ and take the coefficient of degree $mL$. Expansion of the eight-variable geometric series gives $$\sum\text{weights} =\frac5{24}(5s_H+11s_{ab}+6w_e-3D)m^4+O(m^3).$$ For clarity, the monomials before taking the Pfaffian alternating sum are counted by $\sum_{h=m}^{s}(h+1)\binom{s-h+4}{4}$ in divisor degree $(s,-m)$; their weight sum is obtained by multiplying each summand by $s_H(s-h)/5+s_{ab}h/2+w_e(h-m)$. Taking the shifts $0,2H,3H,5H$, with the relation and syzygy weights above, gives the displayed leading coefficient. Since $N_m=(10/3)m^3+O(m^2)$, the normalized trace is $$\frac{5s_H+11s_{ab}+6w_e-3D}{16}.$$ In our case $(s_H,s_{ab},w_e,D)=(16,5,0,15)$. More explicitly, the finite weight sum is $$\frac{75}{4}m^4+\frac{55}{2}m^3+\frac{81}{4}m^2+\frac{11}{2}m,$$ whose quotient by $mN_m$ tends to $45/8$. Lemma 51.2 identifies this trace with the actual expected order. ◻

Proof. Record the successive orders $(a,q,j)$ in $u,b$, and the native degree $j$, taking the $b$ order first. The following degree-one sections give the vertices of a convex hull: $$\begin{array}{c|rrrrrr} \text{section}&1&w&Rw&R^2&Q&RQ\\\hline (a,q,j)&(0,0,0)&(0,0,1)&(0,1,6)&(0,2,10)&(5,0,6)&(5,1,11). \end{array}$$ Multiplying by the common unit $(u-1)^{2m}$ in degree $m$ verifies these orders without changing either filtration. The hull has halfspaces $$a,q\ge0,\quad -a-5q+j\le1,\quad -a-4q+j\le2, \quad 6a+25q-5j\le0.$$ Its volume is $10/3$ and its first moment is $(125/24,55/24,75/4)$. These numbers can be checked by coning to the origin the boundary triangles $$(1,4,0),(5,2,3),(2,1,0),(2,3,0), (5,3,0),(5,4,0),(5,2,1),(5,1,4),$$ where the six vertices are numbered in the order of the table. The sections $1,w,u,R$ show that, after adjoining the anticanonical degree, the generated lattice is $\mathbb Z^4$. The full-volume lemma, Lemma 34.2, applies: the hull volume equals the leading coefficient of $N_m$. Thus the centroid $(25/16,11/16,45/8)$ gives the actual moments, proving the two equalities.

The total expected degree of the residual zero divisors on $B$ is $$B\cdot M\left(S_X(G)-5S(W;B)\right) =\frac56\left(\frac{45}8-\frac{55}{16}\right)=\frac{175}{96}.$$ In each graded piece choose a basis compatible with the orders at $Q_0$ and $q$. The sum of these two orders is at most the degree of each zero divisor. Hence $S(W^B;q)\le175/96-25/16=25/96$. ◻

Proof. The curve is $K=\mathbb P(1,2)$, with different $\frac12[0:1]$. Put $y=v+u^2$ on the chart $w=1$, and let $P_0\in K$ be $y=0$. Order a monomial $u^ay^b$ by $(a+2b,b)$. Keeping native degree $j$ as the third record gives the actual flag $K\supset P_0$.

Here are all additional leading terms needed for a full-volume certificate. The first thirteen sections have anticanonical degree one, and the last two have degree two: $$\begin{array}{c|cc|c} \text{section}&(a,b,j)&\text{section}&(a,b,j)\\\hline 1&(0,0,0)&w&(0,0,1)\\ P&(0,1,4)&-R&(0,1,5)\\ -Rw&(0,1,6)&-PR&(0,2,9)\\ R^2&(0,2,10)&u&(1,0,2)\\ Q&(1,1,6)&-Ru&(1,1,7)\\ -RQ&(1,2,11)&-v&(2,0,3)\\ Rv&(2,1,8)&Q^2-PRv&(0,3,12)\\ R(PRv-Q^2)&(0,4,17)&& \end{array}$$ For example the $K$-initial forms of $P,Q,R$ are $y,uy,-y$, respectively; those of the last two sections are $y^3,y^4$. These identities follow by direct substitution of (51.2); thus the table includes genuine cancellation sections, not a guessed initial presentation.

The convex hull of the degree-normalized vectors has nine vertices $$\begin{array}{c|rrrrrrrrr} i&0&1&2&3&4&5&6&7&8\\\hline a&0&0&0&0&1&2&2&0&0\\ b&0&0&1&2&2&1&0&3/2&2\\ j&0&1&6&10&11&8&3&6&17/2. \end{array}$$ Its halfspaces are $$\begin{gathered} a,b\ge0,\quad b\le2,\quad -a-5b+j\le1,\quad -a-4b+j\le2,\\ 3a+8b-2j\le0,\quad5a+11b-2j\le5,\quad9a+20b-4j\le6. \end{gathered}$$ For an explicit integration certificate, cone to the origin its boundary triangles $$\begin{gathered} (6,7,0),(8,4,5),(8,4,3),(1,6,0),(2,4,5),(2,4,3),\\ (8,6,7),(8,6,5),(1,8,7),(1,7,0),(1,2,3),(1,8,3), (1,2,5),(1,6,5). \end{gathered}$$ The determinant formula gives volume $10/3$ and first moment $(103/48,307/96,75/4)$. The lattice is already generated by $1,w,u,P$, with anticanonical degree prepended. Lemma 34.2 therefore gives the exact centroid $$(\mu_a,\mu_b,\mu_j)=\left(\frac{103}{160},\frac{307}{320},\frac{45}8\right).$$ In particular $S(W;K)=\mu_a+2\mu_b=41/16$ and $S(W^K;P_0)=307/320$.

The total expected degree on $K$ is $\frac12S(W;K)=41/32$. Simultaneous bases at $P_0$ and any other point give expected order at most $41/32-307/320=103/320$ at the latter. The discrepancy there is at least $1/2$, while $P_0$ has discrepancy one. Thus the refined threshold is at least $\min\{320/307,160/103\}=320/307$. ◻

Proof. In each piece of the $T$-associated graded series put the filtration induced by the subspaces divisible by $R^q$. Taking associated graded for the $T$ filtration preserves the dimensions of each of these subspaces. Consequently the sum of their induced $R$ weights is precisely the original $B$ weight. Choose a basis of this graded piece compatible with both the induced $R$ filtration and the point-order filtration at $p$, and lift it with the indicated $T$ and $R$ orders. Since $T\ne B$, the $T$-initial form of a section divisible by $R^q$ is divisible by $(R|_T)^q$. All these zeros lie away from $p$.

The expected degree of the zero divisors on $T$ before removing these zeros is $M\cdot T\,(S_X(G)-dS(W;T))$. Their expected degree on $B\cap T$ is at least $B\cdot T\,S(W;B)$. Subtract and use $M^2=1/6$, $B\sim5M$ and Lemmas 51.3 and 51.4. This gives the assertion. The argument uses two filtrations in each already restricted graded piece; it makes no assertion about a basis compatible with three unrelated filtrations on the original section space. ◻

Proof. For $c\ne0$ adjoin the homogeneous cancellation section $r=aeu-cw^2$ to $\mathcal R$. It has divisor degree $2H$ and native weight two. The quotient $\mathcal R/(r)$ is a domain: on $ae\ne0$ eliminate $u$ to obtain a polynomial graph, while on $ae=0$ its support forces $w=u=v=P=0$ and has dimension three. The quotient is Cohen–Macaulay of dimension four, so this boundary contributes neither a component nor an embedded component. Its open graph is reduced. The prime principal ideal $(r)$ and its powers are saturated with respect to $ae$. Thus localization identifies the actual $T_c$ filtration with the $r$-adic filtration, whose associated graded is $\mathcal R/(r)[r_\mathrm{new}]$. Its normalized trace is $S_X(2H)=5/32$.

The same argument applies to the prime elements $u$ and $w$, both of divisor degree $H$. For $u=0$, the support on $ae=0$ additionally satisfies $v=P=0$ and $Qw=0$, and again has dimension three; for $w=0$ it satisfies $u=v=P=0$ and has dimension three. This proves the required saturation and gives trace $S_X(H)=5/16$. These are computations with actual principal filtrations of the Cox ring.

Every point with $w\ne0$ belongs to one of the $T_c$. This curve is smooth there and has zero different. Lemma 51.7 bounds its point term away from $B$ by $5/8$ if $c\ne0$, and by $25/48$ if $c=0$. Its first term is $32/5$ or $16/5$. The remaining points outside $B$ lie on $T_\infty$. Its two orbifold points belong to $B$, so the different vanishes at the points under consideration. The same lemma gives point term at most $5/16$, while its first term is $16/5$. Abban–Zhuang adjunction proves the last assertion. ◻

Proof of Theorem 51.1. The curve $B$ has first ratio $16/11$. Away from $Q_0$, Lemma 51.4 bounds its point terms by $25/96$, whereas the discrepancies are at least $1/3$. At $Q_0$, Lemma 51.5 gives ratio at least $\min\{48/41,320/307\}=320/307$. Lemma 51.8 covers the complement of $B$, with stronger bounds. Hence $\delta(G;W)\ge320/307$.

All series have bounded support, as is explicit in Lemma 51.2. The first series contains an ample series: every weighted polynomial of native degree $6k$ occurs when $m=12k$, giving the complete series $H^0(G,\mathcal O_G(6k))$. The subsequent series are the actual purely log terminal refinements, so the bounded-support and ample-series assertions follow as in [4, Example 2.15]; the displayed full-dimensional certificates also give nonvanishing on interior rays. The different and purely log terminal conditions were checked above. Proposition 2.7 therefore applies and gives $$\delta_p(X)\ge\min\left\{\frac6{45/8},\frac{320}{307}\right\} =\frac{320}{307}.$$ No Zariski decomposition on the weighted blowup of the threefold has been used. The first filtration is identified by a prime Pfaffian initial ring; the further moments come from actual sections, and the estimates on the complementary locus retain their zeros on a second curve. ◻