The exceptional-surface estimates

Proof. Take the second Veronese of the native ring of weights $(2,3,2)$. Lemma 5.2 gives the discrepancy and the boundary. Eliminating $u$ gives the equation of $B$. It misses the vertex of $G$; projection to $[V:W]$ is an isomorphism. The boundary meets it transversally in the three distinct stated points, proving the plt assertion and different. The alpha computation follows from Lemma 3.2. ◻

Proof. On $B$, $u=-VW$. Thus $u^iv^jw^k$ restricts, up to sign, to $V^{i+j}W^{i+k}$. The condition $jk=0$ says that $i$ is the smaller of these two exponents. The ordinary-degree bound is $n-i\le T$. Therefore all and only the binary monomials with both exponents at least $a$ occur. Their dimension is $n-2a+1$. The mean vanishing at either coordinate endpoint is $n/2$; at another point it is $(n-2a)/2$. Restoring $z^\eta$ adds $\eta/2$ at the three boundary points, a bounded correction.

For completeness put $M(q)=\min(1,2-q)$, $0\le q\le2$, after rescaling by $m$. With $a=\max(0,n-M)$ and $\lambda=n-2a$, the two intervals $[0,M]$, $[M,2M]$ give $$\int_0^{2M}\lambda\,dn=M^2,\qquad \int_0^{2M}\lambda\frac n2\,dn=\frac{M^3}2,\qquad \int_0^{2M}\frac{\lambda^2}2\,dn=\frac{M^3}3.$$ Since $\int_0^2M^2dq=4/3$ and $\int_0^2M^3dq=5/4$, division by $4/3$ gives the stated means. The two residue classes contribute equally to numerator and denominator, and give $N_m=(8/3)m^3+O(m^2)$. Endpoint and residue errors contribute $O(m^3)$ to order sums, so disappear after division by $mN_m$. ◻

Proof of Theorem 22.1. The residual-degree average after removing $B$ is $$R=\tfrac12S_X(G)-3S(W;B)=3-\frac{33}{16}=\frac{15}{16}.$$ Lemmas 3.3 and 22.2 give threshold at least $16/15$ for centres not contained in $B$. Along $B$, the curve term is $16/11$. The point terms are $16/15$ at $Q_V,Q_W$, $8/5$ at $Q_{V+W}$, and $16/5$ elsewhere. Thus $\delta(G,\Delta_G;W)\ge16/15$.

The positivity conditions are elementary here. The support is described by $0\le q\le2m$ and $0\le n\le2\min(m,2m-q)$ up to bounded residue corrections. Every interior integral ray has linearly growing curve dimension. For $W$, sufficiently small positive native weight relative to $m$ includes every weighted polynomial in $v,z,w$ of that weight, and hence a complete ample series on a Cartier Veronese. For $W^B$, the interior ray $(m,q,n)=(6k,k,3k)$ gives the complete series of degree $3k$ with $\eta=0$. Bounded support and eventual nonvanishing on the other interior rays also follow from the refinement construction in [4, Example 2.15], since Lemma 20.4 identifies the actual refinement. Both residue classes are retained as in Proposition 2.7. The plt assertions were checked above. Proposition 2.7 now gives $$\delta_p(X)\ge\min\{7/6,16/15\}=16/15.$$ ◻

Proof. Take the second Veronese of the native weights $(3,4,2)$ and use Lemma 5.2. On $V=1$ elimination gives $u=-W-U=-r^2$ and the displayed parameterization. The sole point outside this chart is $[0:0:1]$. Its index-three cover has coordinates $(U,V)$ and equation $U^2-UV^2-V=0$; it is smooth with $V=U^2+O(U^5)$. The quotient parameter is $U^3$, so the quotient curve is smooth and $r$ has a simple pole. The rational coordinate orders and the different follow from this chart and the two transverse boundary intersections in the affine chart. These charts also verify plt. ◻

Proof. Fix $m,q,\eta$, put $T=\min(m,2m-q)-\eta$, and set $3i+2j+k=n$ in the standard monomial $u^iv^\eta z^jw^kb^q$. After removing $b^qv^\eta$, dividing by $V^n$ on $B$ gives the powers $r^{2i+j}$, with $ik=0$ and $i+j+k\le T$. Their span is exactly $$r^a\langle1,r,\ldots,r^{D-a}\rangle, \quad a=\max(0,n-T),\quad D=\lfloor2n/3\rfloor,\quad0\le n\le3T.$$ Indeed, the group $i=0$ supplies consecutive exponents from $\max(0,n-T)$ to $\lfloor n/2\rfloor$. In the group $k=0$, increasing $i$ by two and decreasing $j$ by three increases $2i+j$ by one; this supplies the remaining consecutive exponents up to $\lfloor2n/3\rfloor$, subject to the same lower bound.

The mean orders before the residue correction are $(a+D)/2$ at zero, $2n/3-(a+D)/2$ at infinity, and $(D-a)/2$ elsewhere. The factor $v^\eta$ adds $\eta/2$ at $Q_0,Q_1$ only. Set $M=\min(1,2-q)$ in the limit. The dimension function is $\lambda=2n/3$ for $0\le n\le M$ and $\lambda=M-n/3$ for $M\le n\le3M$. Direct integration gives $$\begin{array}{c|cccc} f&1&n&(a+2n/3)/2&(2n/3-a)/2\\\hline \int_0^{3M}\lambda f\,dn&M^2&4M^3/3&2M^3/3&2M^3/9. \end{array}$$ The last column is both the mean at infinity and the mean at an ordinary point in the limit. Integrating in $q$ and dividing by $4/3$ gives $5/8$ and $5/24$. The residue and rounding errors vanish as in the $D_4$ computation. ◻

Proof. The second relation loses $z^2$, whose secondary order is two. The two equations are a regular sequence. Modulo $e$ or $a$ they become $(v^2+zw,uw)$, which has height two. Hence $a,e$ are regular. Localization at $ae$ eliminates $u,b$ and gives a polynomial ring in $v,z,w$, proving that the ring is a domain. It remains injective after localizing at $w$. On the chart $a=e=w=1$, the equations are $u=-W-U$ and $b=W+U$. Their secondary orders are precisely those of the ordinary blowup of the origin in $(U,W)$. The fixed bidegree and native weight remove the common unit factors before restriction. This proves the required identification, including cancellations. The half order of $v$ means that its rational zero divisor is half of $(W=0)$; it causes only bounded residue corrections below. ◻

Proof. The leading monomials remain $v^2,uw$, so $ik=0$. Given $l$, the equations imply $$q=\min(l,n-l-2\kappa),\quad i=l-q,\quad j=\kappa-l, \quad k=n-l-2\kappa-q.$$ Nonnegativity and the two degree bounds are equivalent to the displayed interval and the bound on $\kappa$. On $V=1$, the restriction is, up to sign, $(W+U)^{i+q}U^jW^{\eta/2}$. After taking its leading form on the blowup, division by $W^{\kappa+\eta/2}$ gives $(r+1)^lr^{\kappa-l}$. Consecutive values of $l$ form a basis of the displayed complete polynomial space. Its fixed factors and moving degree give the means. The factor $W^{\eta/2}$ adds only the stated order at infinity. ◻

Proof. We give the integrals to make the normalization explicit. Rescale $(\kappa,n)$ by $m$ and write them as $(x,y)$. Put $$L=\max(0,(y-x-1)/3,(y-x-2)/2),\quad U=\min(x,y-2x,1+y-3x),\quad\lambda=U-L.$$ Integrate over $\Omega=\{0\le x\le2,\ L\le U\}$. The five integrands for dimension, order at $K$, and the three point types are $$\lambda,\quad\lambda x,\quad \lambda\left(x-\frac{L+U}{2}\right),\quad \lambda\frac{L+U}{2},\quad\frac{\lambda^2}{2}.$$ For an explicit piecewise evaluation, number the three entries of each maximum or minimum by $0,1,2$. The nonempty regions, with their five integrals in the order just listed, are $$\begin{array}{c|ccccc} (L,U)&I&S_K&S_0&S_{-1}&S_*\\\hline (0,0)&1/24&1/96&1/192&1/192&1/192\\ (0,1)&1/24&1/48&1/64&1/192&1/192\\ (1,0)&133/216&16/27&229/1728&265/576&229/1728\\ (1,1)&7/72&23/288&79/1728&59/1728&37/1728\\ (1,2)&7/27&13/36&4/27&23/108&1/18\\ (2,0)&23/108&461/1296&101/2592&821/2592&101/2592\\ (2,2)&7/108&157/1296&79/2592&235/2592&37/2592\\\hline \text{sum}&4/3&37/24&5/12&9/8&59/216 \end{array}$$ Each region is cut out by linear inequalities, so these are integrals of quadratic polynomials over rational polygons. For example, region $(0,0)$ is the triangle with vertices $(0,0),(1/2,3/2),(0,1)$, where $L=0,U=x$; it gives the first row directly. The defining max/min formulas specify every other region without a choice of subdivision.

Dividing the last four totals by $I=4/3$ gives the assertion. The two residue classes again give $N_m\sim2Im^3$ and cancel in the normalized means. All bounded rounding corrections vanish. At infinity the different has coefficient $1/2$, while it is zero elsewhere. The adjunction ratios are therefore $$\frac{3/2}{37/32}=\frac{48}{37},\quad \frac1{5/16}=\frac{16}5,\quad \frac1{27/32}=\frac{32}{27},\quad \frac{1/2}{59/288}=\frac{144}{59}.$$ They give the claimed local threshold. ◻

Proof. The quotient of $R_0$ by $w$ has relations $eau+v^2$ and $e^2ab+z^2$. It is a domain: $a,e$ are regular by the same height-two argument as before, and localization eliminates $u,b$. Thus powers of $w$ are saturated and compute the geometric filtration by $A$. Its associated graded ring retains $w$ as a free variable, and the two relations just displayed. The total $w$-order is the Hilbert-weight formula with $w_*=1$ and all other weights zero, hence $5/16$.

After localization, this ring is monomial with $u=-v^2$ and $b=-z^2$. Give $v$ order one, and hence $u$ order two, keeping $z,w,b$ at zero. Lemma 20.5, now applied to this multigraded ring, gives total average $v$-exponent $$\frac5{16}\cdot3-\frac5{32}\cdot2=\frac58.$$ Giving $z$ order one and $b$ order two similarly gives average $z$-exponent $$\frac5{16}+\frac{11}{16}\cdot2-\frac5{32}\cdot2=\frac{11}8.$$ On the coarse curve $A=\mathbb P(2,3)$, a section $z$ has rational order $1/3$ at $Q_\infty$, and $v$, whose square is $W$, has rational order $1/4$ at $Q_W$. These factors give $11/24$ and $5/32$.

Each fixed graded summand is a span of distinct monomials in a rational coordinate on $A$. At a nonzero finite point the vanishing sequence of $s$ distinct monomials is $0,1,\ldots,s-1$: the derivative determinant is a nonzero Vandermonde determinant in their exponents. Its mean is no larger than the mean order at either endpoint, since those endpoint orders form a set of distinct nonnegative numbers with integral differences. Summing gives the last bound.

At $Q_W$ the index-two chart has boundary coefficient $1/2$ on the transverse coordinate, so adjunction gives coefficient $1-(1-1/2)/2=3/4$. At $Q_\infty$ the index is three and the boundary misses the point, giving $2/3$. The smooth covers prove plt. The ratios away from $Q_\infty$ are at least $\min\{16/5,(1/4)/(5/32)\}=8/5$. ◻

Proof. All generator exponents in a fixed degree are $O(m)$, so support is bounded. The interval descriptions show nonvanishing on every integral interior ray for $W^B$ and $W^K$ after the bounded parity corrections; their lengths grow linearly there. The first series contains every native weighted polynomial in $v,z,w$ of weight $12k$ when $m=12k$, giving the complete ample series $\mathcal O_G(6k)$. For $W^B$, take $(m,q,n,\eta)=(6k,k,3k,0)$; the interval is $0\le\deg r\le2k$, with no fixed rational order at infinity, so it is a complete series of degree $2k$. For $W^K$ take $(m,n,\kappa,\eta)=(6k,3k,k,0)$; then $L_m=0,U_m=k$ and the complete series has degree $k$.

For $W^A$, use $m=24k$, order $k$ along $w$, and remaining native weight $12k$. The monomials $v^{4i}z^{3(k-i)}$, $0\le i\le k$, all satisfy the degree bounds, and restrict to the complete degree-$k$ series on the coarse curve. These displayed ample rays lie in the interiors of the respective support cones. The support and eventual nonvanishing for $W^A$ also follow directly from the two linear degree bounds and the bounded residues modulo the quadric relations. More generally, [4, Example 2.15] gives the interior-ray condition for the actual refinements identified above. The rational restrictions and their residue classes are treated as in Proposition 2.7; passing to a Cartier Veronese for the ample ray does not discard the other summands in the expected-order sums. All the extractions and curve adjunctions are plt by the quotient-chart computations above. ◻

Proof of Theorem 22.5. Consider a divisorial valuation on $(G,\Delta_G)$. If its centre is not contained in $A\cup B$, Lemma 3.2 gives local alpha at least $3/2$ and Lemma 3.3 gives threshold at least $(3/2)/(5/4)=6/5$.

For centres on $B$ away from $Q_0$, Lemma 22.7 gives ratios $12/5$ at $Q_1$, $8/5$ at $Q_\infty$, and $24/5$ elsewhere, together with the curve ratio $16/11$. At $Q_0$ use Lemma 22.12, obtaining $32/27$. For centres on $A$ away from $Q_\infty$ use Lemma 22.13, obtaining $8/5$; $Q_\infty$ was already treated using $B$. This covers every centre, including the curves themselves. Hence $$\delta(G,\Delta_G;W)\ge\min\{6/5,16/11,32/27,8/5\}=\frac{32}{27}.$$ The positivity conditions are supplied by Lemma 22.14. Consequently $$\delta_p(X)\ge\min\{9/8,32/27\}=\frac98.$$ The reverse inequality is supplied by $G$ itself, since $A_X(G)=9$ and $S_X(G)=8$. This proves the equality. ◻

Proof. Use coordinates $[a_0:z:x:y:w]$ on the fibre. The anticanonical normal forms are $$\begin{array}{c|cc} A_2+A_1&a_0z+x^2-xy&yw+xz\\ A_2+2A_1&a_0z+x^2&yw+xz\\ A_3+A_1&a_0z+x^2-xy&yw+z^2\\ A_3+2A_1&a_0z+x^2&yw+z^2. \end{array}$$ The first and third follow by linear changes from the equations in [13, §3.4]; the other two are the toric normal forms in the same classification. Their higher singular point is $[1:0:0:0:0]$. For example, on $a_0=1$, the second gives $yw-x^3$, and the fourth gives $yw+x^4$. In the two-node rows, the $y$- and $w$-coordinate points are both nodes. In each one-node row, only the $w$-coordinate point is a node; the $y$-coordinate point is smooth.

Write $\ell=a_0+$ a linear form in $z,x,y,w$, and extend the equations by $t$-multiples of linear forms. Smoothness of $V$ forces a nonzero $t\ell$ coefficient in the second equation. Set $\ell=ea$, $t=eb$ as before. The native weights are $$\begin{array}{c|rrrrr|cc|c|c} &z&x&y&w&b&d_1&d_2&A_X(G)&S_X(G)\\\hline A_2&4&2&3&3&6&4&6&8&101/16\\ A_3&2&1&2&2&4&2&4&5&4. \end{array}$$ Here $a,e$ have weight zero. The initial relations, after rescaling, are $$ea z+x^2,\qquad \begin{cases} e^2ab+yw+xz,&A_2,\\ e^2ab+yw+z^2,&A_3.\end{cases}$$ Modulo $e$ or $a$ the two polynomials have height two; after inverting $ae$ their quotient is a polynomial ring in $x,y,w$ and the two units. Thus the domain argument of Lemma 20.4 identifies the actual valuation filtration. The discrepancies are the sums of the three local parameter weights, and formula (20.1) gives the displayed expected orders.

Next degenerate by the secondary weights $w_x=1$, $w_z=2$, all other weights zero. The initial relations become (22.1)$$\label{f210:ln:Antoric} ea z+x^2,\qquad e^2ab+yw.$$ Again they are a regular sequence, are regular with respect to $a,e$, and give a polynomial ring after localization. Thus this is the actual torus limit of every refined summand, including cancellations. It is monomial in $x,y,w$, with $z=-x^2$ and $b=-yw$ on the unit chart.

For $A_2$, the coarse exceptional surface is $\mathbb P(2,1,1)_{U,Y,W}$ with $U=x^3$, $Y=y$, $W=w$, and different $\frac23(U=0)$. For $A_3$, it is $\mathbb P^2_{U,Y,W}$ with $U=x^2$ and different $\frac12(U=0)$. In (22.1), the mean exponent of $x$ is $$\frac5{16}(1+2)-\frac5{32}\,2=\frac58.$$ The mean exponent of each of $y,w$ is $$\frac5{16}+\frac{11}{16}-\frac5{32}=\frac{27}{32}.$$ Consequently the coordinate divisor coefficients of the monomial basis limit are $$\begin{array}{c|ccc|ccc} &c_U&c_Y&c_W&A(U)&A(Y)&A(W)\\\hline A_2&5/24&27/32&27/32&1/3&1&1\\ A_3&5/16&27/32&27/32&1/2&1&1. \end{array}$$ Lemma 5.3 gives $\delta(G,\Delta_G;W)\ge32/27$. The first discrepancy ratios are $128/101$ and $5/4$, respectively, both larger than $32/27$. Apply Proposition 2.7.

All these series have bounded support. To check positivity, note that on the unit chart the section space contains every monomial of total degree at most $m$ in the three local parameters. For a range of small positive native degrees this contains the complete weighted homogeneous series on $G$, giving an interior ample ray after a fixed Veronese. Eventual nonvanishing on every other integral interior ray follows from [4, Example 2.15], applied to the actual point refinement. This verifies all the positivity hypotheses used in adjunction. ◻

Proof. These are linear changes of the anticanonical normal forms and the toric models in [13, §3.4]. The smooth-locus tables in [12, §6.2, Tables 27, 28, 31–33, 35] identify the lines in the statement. To specify the identification in the $A_4$ case, take $(x_0,x_1,x_2,x_3,x_4)=(u,w,z,v,a_0)$ in [13, §3.4]. Its curve $E_5$ is $(u=z=v=0)$ and meets the second exceptional component of the $A_4$ chain. This is the line with smooth-locus threshold $24/29$. In the $D_5$ case it is the unique line. In the mixed cases the indicated lines join the higher singularity to a node.

The point $o=[1:0:0:0:0]$ is the higher singularity. The differential of $q_2$ vanishes at $o$. Along the displayed line, the differentials at a smooth point are independent precisely under the nonvanishing conditions stated. This also follows directly from the two equations. ◻

Proof. Write the equations of $V$ as $Q_i=q_i+tL_i$, with $q_i$ from Lemma 22.17. Since $dq_2(o)=0$ and $V$ is smooth, the coefficient of $ta_0$ in $Q_2$ is nonzero. The curve $C$ avoids $o$, so the coefficient of $a_0$ in its other linear equation $\ell$ is nonzero as well.

First replace $u,z$ by themselves plus constant multiples of $t$, to remove the linear $t$-terms of the two equations at $p$. The second equation becomes $$wu+\varepsilon vz+tc\quad\text{or}\quad wu+z^2+tc,$$ where $\varepsilon\ne0$ and $c$ is a linear form vanishing at $p$. Its restriction to $\ell_0$ has a nonzero $a_0$ coefficient. All new $tv,tz,tu,t^2$ terms are absorbed into $tc$. Next replace $z$ by a nonzero multiple of $z+ku$, with $k$ constant, to remove the linear $u$-term of $Q_1$ at $p$. In the first form of $Q_2$ this only adds an $uv$-term; in the second it adds $uz,u^2$ terms.

Here is the nonvanishing that will be essential. On $\ell_0$, write $$\ell=Aa_0+Bw,\qquad c=Da_0+E_0w, \qquad A\ne0,\ D\ne0.$$ The restrictions of $\ell$ and $c$ are independent because $\ell(p)\ne0$ and $c$ has a simple zero at $p$. Thus $DB-AE_0\ne0$, and (22.5)$$\label{f210:pt:wline} w=\frac{D}{DB-AE_0}\ell-\frac{A}{DB-AE_0}c \quad\text{on }\ell_0.$$ Both coefficients are nonzero. Globally we can therefore write $$w=\alpha\ell+\lambda c+\text{a linear combination of }v,z,u,t, \qquad \alpha\lambda\ne0.$$ The linear changes already made do not change this assertion on the line.

The linear parts of the two equations eliminate $z,u$. Hence $v,c,t$ are regular parameters on the chart $\ell=1$. The lowest ordinary parts of $Q_1,Q_2$ are $$\ell z+v^2+t(\gamma_1c+\gamma_2v+\gamma_3t), \qquad \alpha\ell u+tc,$$ respectively. Terms in $u,z$ multiplied by a function vanishing at $p$ have order at least three. In particular $u,z$ have order two. After substituting $\ell=ea$, $t=eb$, set $r=\alpha au+bc$. The second equation now reads $$er+\kappa vz+\lambda uc+\mu uv+\nu ebu+ \text{terms of ordinary order at least four}=0.$$ Its cubic part is (22.3). The term $\lambda uc$, together with $u=-bc/\alpha$ in degree two on the unit chart, contributes a nonzero multiple of $bc^2$. None of the other displayed terms contributes $bc^2$. Thus $r$ has order exactly three. The defining relation $r-\alpha au-bc=0$ has initial part (22.4). This proves the assertions about orders and initial equations. Equality with the full geometric filtration is established next. ◻

Proof. The original presentation with the additional variable $r$ has two quadric relations and the relation $r-\alpha au-bc$. We show that the three initial forms are a regular sequence and that $a,e$ are non-zero-divisors in their quotient.

Modulo $e$, their ideal is $$(v^2,\ \kappa vz+\lambda uc+\mu uv,\ \alpha au+bc).$$ Its height is three. Indeed, setting $v=0$ leaves $uc=0$ and $\alpha au+bc=0$; every component has codimension two in the remaining variables because $\alpha\lambda\ne0$.

Modulo $a$, the last equation is $bc=0$. On the component $b=0$, the first equation is $v^2=0$, and the second reduces, after $v=0$, to $er+\lambda uc=0$, which is a nonzero further equation. On the component $c=0$, the first equation is $$v^2+\gamma_2ebv+\gamma_3e^2b^2=0.$$ It is not divisible by $e$. The second equation has coefficient $e$ in $r$, so it shares no factor with this quadratic. Thus the ideal modulo $a$ also has height three.

After inverting $ae$, one eliminates $u,z,r$, in that order, and obtains a polynomial ring in the three parameters $v,c,b$, with $a^{\pm1},e^{\pm1}$ adjoined. The original three-generator initial ideal therefore has height three; the preceding codimension checks rule out components on $a=0$ or $e=0$. It is a complete intersection, hence Cohen–Macaulay, and $a,e$ are non-zero-divisors. Since its localization is a domain, it is a domain itself.

Lemma 5.1 identifies the polynomial-weight associated graded ring with this quotient. The orders in Lemma 22.18 give a map to the local point-associated graded ring. On inverting $ae$ it is the polynomial identification just described. Since the quotient injects into this localization, no nonzero global initial form can vanish locally. Finally, the degrees of $a,e$ form a lattice basis, so setting them equal to one is injective on any fixed bidegree after removing its common Laurent factor. This verifies the second assertion of Lemma 5.1, including all cancellations. ◻

Proof. The discrepancy is three because $p$ is a smooth point of a threefold. Apply (20.2) to the actual filtration established in Lemma 22.19. The four $H$-weights sum to $2+2+1+1=6$, the weight of $b$ is one, and the relation weights are $2,3,2$. The new $L$-generator has weight three. Therefore $$S_X(G)=\frac5{16}\,6+\frac{11}{16} -\frac5{32}(2+3)+\frac14(3-2) =\frac{65}{32}.$$ The last term is required: using the seven old generators would not compute the actual expected order. ◻

Proof. Apply the one-parameter subgroup with weights $(2,0,5)$ on $(v,c,b)$. The corresponding secondary weights on the presentation of $R_p$ are $$\begin{array}{c|rrrrrrrr} &u&z&v&c&b&a&e&r\\\hline \text{secondary weight}&5&4&2&0&5&0&0&5. \end{array}$$ The initial relations are (22.6)$$\label{f210:pt:toricring} eaz+v^2=0,\qquad er+\lambda uc=0,\qquad \alpha au+bc=0.$$ In fact the two terms of the first equation have weight four, while the omitted $ebc,ebv,e^2b^2$ terms have weights $5,7,10$. In the second equation $er,uc$ have weight five, whereas $vz,uv,ebu$ have weights $6,7,10$. Thus no coefficient specialisation has been assumed.

Modulo $e$, the ideal in (22.6) is $(v^2,uc,\alpha au+bc)$, of height three. Modulo $a$ it is $(v^2,er+\lambda uc,bc)$, again of height three. After inverting $ae$ it is a polynomial ring in $v,c,b$. The same regularity and localization argument proves that this is the actual torus limit of each refined space. On the unit chart its generators are monomials: $$u=-\frac1\alpha bc,\qquad z=-v^2,\qquad r=\frac\lambda\alpha bc^2.$$

We compute the three coordinate coefficients separately. For the exponent of $v$, give $v,z$ weights $1,2$ and all other variables weight zero. The relation weights are $(2,0,0)$, and (20.2) gives $$c_v=\frac5{16}(1+2)-\frac5{32}\,2=\frac58.$$ For the exponent of $c$, give $u,c,r$ weights $1,1,2$ and all other variables weight zero. The relation weights are $(0,2,1)$, so $$c_c=\frac5{16}(1+1)-\frac5{32}\,2+\frac14(2-1)=\frac9{16}.$$ For the exponent of $b$, give $u,b,r$ weight one and the other variables weight zero. The relation weights are $(0,1,1)$, giving $$c_b=\frac5{16}+\frac{11}{16}-\frac5{32}+\frac14(1-1)=\frac{27}{32}.$$ As a check, $c_v+c_c+c_b=65/32=S_X(G)$, since $-G|_G=\mathcal O_{\mathbb P^2}(1)$. The monomial basis divisor therefore tends to $$\frac58(v=0)+\frac9{16}(c=0)+\frac{27}{32}(b=0).$$ Its largest coefficient is $27/32$. Lemma 5.3 proves the assertion.

For bounded support, let $h$ be the total exponent in the four $H$-generators, $q$ the exponent in $b$, and $k$ that in $r$. The remaining exponents are $2m-h-q-2k$ in $a$ and $m-h-k$ in $e$. Their nonnegativity gives $h+k\le m$ and $q\le2m-h-2k$, so every exponent and every filtration order is $O(m)$. To exhibit an interior ample ray, observe that all monomials in the regular parameters $v,c,b$ of total degree at most $m$ lift to sections of $mL$: for exponents $i,j,k$, the lift is $v^ic^jb^ka^{2m-i-j-k}e^{m-i-j}$. Thus $W_{m,J}$ contains the complete degree-$J$ series on $\mathbb P^2$ for $0\le J\le m$. Eventual nonvanishing on every integral interior ray follows from [4, Example 2.15], because Lemma 22.19 identifies the actual point refinement. This verifies the positivity conditions in Abban–Zhuang adjunction. ◻