The point extraction at an interior point

Proof. As at the index-two endpoint, write the graph as $ux^2y^3+vg=0$, with $g_0=1$, $g_5\ne0$, and use the cubic lifts $G_0,G_1$ and $R(c)$ of Lemma 98.2. Since $p$ lies on the open part of $\Gamma$, it has $t=x=0$ and $c\ne0,\infty$. A source scaling allows us to put $c(p)=1$. Put $z=c-1$. The conic equation $q=c-x^2=1+z-x^2$ is a unit near $p$, so the blow-up is an isomorphism there. Trivializing $L$ by this conic section gives the thirteen regular local functions $$t^i\{1,x,c\}\quad(0\le i\le2),\qquad A,B,tA,tB,$$ where $$A=\frac{c+tG_0}{q},\qquad B=xA+tR(c).$$ Put $$a=1+g_2+g_4,\qquad b=g_1+g_3+g_5,\qquad e=g_4-1,\qquad d=g_3+2g_5.$$ The first-order Taylor expansion of $G_0/q$ at $p$ is $a+bx+ez$. Thus $$U=A-1-at-btx-etz$$ has initial form $x^2$. The function $$V=B-x-bt-dtz-atx$$ satisfies the exact identity $$V=xU+btx^2+etxz+g_5tz^2.$$ Consequently its primary initial form is $$F'=x^3+t(bx^2+exz+g_5z^2).$$ The functions $tU$ and $tV$ are still linear combinations of the thirteen sections, since their definitions only require subtracting $t^2,t^2x,t^2z$ from $tA,tB$. Their initial forms are $tx^2$ and $tF'$. The nine other initial sections are $t^i\{1,x,z\}$.

Now take least auxiliary weights $(1,0,3)$ in $(x,z,t)$. The two terms $btx^2$ and $etxz$ have weights five and four, whereas $x^3$ and $g_5tz^2$ have weight three. Hence the new initial forms of $F',tF'$ are $x^3+g_5tz^2$ and $t(x^3+g_5tz^2)$. Since $g_5\ne0$, a scaling makes its coefficient one. This proves the claimed inclusion, directly from actual sections on $X$. ◻

Proof. Use the relation $z^2t=F-x^3$. If a product contains two $z$ factors and positive $t$-exponent, distribute at least one $t$ to these two factors. This redistribution is possible without changing the anticanonical degree: every generator type allows a consecutive interval of $t$-exponents starting at zero. For $1\le i+j\le4$, $$(zt^i)(zt^j)=Ft^{i+j-1}-x^3t^{i+j-1}.$$ Both terms are products of two degree-one generators. For the first, use $F,tF$ and $1,t,t^2$; for the second use $x^2,tx^2$ and $x,tx,t^2x$. Repetition decreases the number of $z$ factors, and ends with $s=0$ or $b\le1$. These normal monomials are linearly independent: $z^2t$ is the leading monomial of the single relation in $\mathbb C[x,z,t,F]/(z^2t-F+x^3)$.

For a normal monomial use $k$ factors of type $x^2t^i$, together with $a-2k$ factors of type $xt^i$, $b$ factors of type $zt^i$, and $c$ factors of type $Ft^i$. Fill unused places with $1,t,t^2$. This is possible exactly when $$\max(0,a+b+c-m)\le k\le\lfloor a/2\rfloor, \qquad s\le2m-k-c.$$ Taking the smallest $k$ gives the three inequalities in (I.1).

To count the basis, first take $s=0$. For fixed $b+c=n$, there are $n+1$ choices and $2(m-n)+1$ choices of $a$. If $s\ge1$, then $b=0$ or $1$. Put $n=m-b-c$; summing over $0\le a\le2n$ gives $$(2n+1)(m+b+n)-\frac{n(n+1)}2$$ choices of $(a,s)$. Thus the total is $$\begin{aligned} &\sum_{n=0}^m(n+1)(2m-2n+1)\\ &\quad+\sum_{n=0}^m\left((2n+1)(m+n)-\frac{n(n+1)}2\right)\\ &\quad+\sum_{n=0}^{m-1}\left((2n+1)(m+1+n)-\frac{n(n+1)}2\right) =\frac{10m^3+15m^2+11m+3}{3}. \end{aligned}$$ Lemma 94.2 therefore proves equality with the actual secondary initial algebra in every degree.

For the primary weight $a+b+3c+s$, the same three disjoint ranges give, respectively, $$\frac{m(m+1)(m+2)(3m+4)}6,\qquad \frac{m(m+1)(7m^2+7m+1)}2,\qquad \frac{m(21m^3+5m^2+3m+1)}6.$$ Here the first sum is $$\sum_{n=0}^m(n+1)(2m-2n+1)(m+n).$$ For each of the two other sums keep $b=0$ or $b=1$, put $c=m-b-n$, and sum over $0\le a\le2n$, $1\le s\le U$, where $U=\min(m+b+n,m+b+2n-a)$. The summand after summing in $s$ is $$U(a+3m-2b-3n)+\frac{U(U+1)}2.$$ Split the $a$-sum at $a=n$ and use $0\le n\le m-b$. This gives the two displayed expressions. Their total is $$\frac{m(15m^3+20m^2+15m+4)}2,$$ for every $m\ge0$. It follows directly that $S_X(G)=9/4$. ◻

Proof. For a monomial in (I.1), the primary degree and auxiliary weight are $$j=a+b+3c+s,\qquad q=a+3c+3s.$$ Thus $q-j=2s-b$. If $q-j\ge-1$, choose the unique $b\in\{0,1\}$ with $b\equiv j-q\pmod2$ and put $s=(q-j+b)/2$. If $q-j\le-2$, put $s=0$, $b=j-q$. The normal condition makes this choice unique. Set $k=q-3s=a+3c$. Substituting in (I.1) gives exactly $\ell\le c\le v$. Since $F=x^3+tz^2$, the component has the form (I.2).

Here are the integrations for the moments. The first region has $s=0$ and $y=b/m$, $0\le y\le1$. The second consists of the two rows $b=0,1$ and has $y=s/m$, $0\le y\le2$; it is counted twice. Writing $k/m$ again as $k$, the lower and upper bounds for $c/m$ are $$\begin{array}{c|c|c} &L&U\\ \hline z&\max(0,k+2y-2)&k/3\\ t&\max(0,k-2,y+k-3)&\min(2-y,k/3). \end{array}$$ Integrate where $L\le U$ and put $D=U-L$. The first row is bounded by $k\le3(1-y)$, so $U=k/3$ throughout. The exact integrals, including the multiplicity two in the second row, are (I.4)$$\begin{array}{c|cccccc} &\int D&\int(k-3U+3D/2)D&\int(b/m)D&\int(s/m)D &\int LD&\int D^2/2\\ \hline z&1/3&1/6&1/12&0&1/36&1/18\\ t&3&23/12&0&7/3&7/24&5/8. \end{array} \tag{I.4}$$ For example, the first region has two chambers, separated by $k=2(1-y)$; direct integration of $D=k/3$ on the lower chamber and $D=2-2y-2k/3$ on the upper gives its first row. For the second region the five chambers are obtained by choosing which affine functions give $L,U$. The polygons in $(y,k)$ are $$\begin{array}{c|c|l} L&U&\text{vertices}\\ \hline 0&2-y&(2,0),(2,1),(3/2,3/2)\\ 0&k/3&(0,0),(2,0),(3/2,3/2),(1,2),(0,2)\\ k-2&k/3&(0,3),(0,2),(1,2),(1,3)\\ y+k-3&2-y&(2,1),(1,3),(3/2,3/2)\\ y+k-3&k/3&(1,3),(1,2),(3/2,3/2). \end{array}$$ Integration of the displayed polynomials on these polygons gives the second row of (I.4). Multiplying the summed columns by $3/10$ gives (I.3); lattice boundaries, floors and ceilings give only lower-order terms.

The fixed $x$-coefficient is $\alpha-3h=1/80$. The average ordinary degree is consequently $$S_X(G)=(\alpha-3h)+b_0+t_0+3f_0+6h=\frac94.$$ ◻

Proof. Away from the coordinate vertices the pencil members are disjoint and smooth, so the component bounds suffice. At $[1:0:0]$, only the two transverse coordinate lines occur.

At $[0:1:0]$, put $z=1$. The branches have equations $t=\lambda x^3$. Three successive ordinary blow-ups separate them. The final exceptional valuation has weights $(1,3)$ and log discrepancy four. Its order on $B$ is $a+3c+3f+6h$, independently of $\nu_0,\nu_\infty$. The preceding two exceptional inequalities follow by interpolation with the coefficient bound on $x=0$.

At $[0:0:1]$, put $t=1$. The nonzero finite members are cusps $z^2=\lambda x^3$. Write $$u=a+3\nu_\infty,\qquad v=b+2\nu_0,\qquad e=f+2h-\nu_\infty-\nu_0.$$ The regular toric subdivision with exceptional primitive vectors $(1,1),(2,3),(1,2)$ resolves all these cusps simultaneously. Their strict transforms meet the $(2,3)$ exceptional curve transversely at distinct points, away from the coordinate corners. The corresponding orders and log discrepancies are $$\begin{array}{c|c|c} (1,1)&u+v+2e&2\\ (2,3)&2u+3v+6e&5\\ (1,2)&u+2v+3e&3. \end{array}$$ The middle order is $2a+3b+6f+12h$. The first equals one third of this order plus $u/3$; the last equals half this order plus $v/2$. Since $u,v\le1$, the asserted bounds imply all three exceptional inequalities. After this subdivision the total transform has simple normal crossings and all coefficients are at most one. This verifies log canonicity at every point and for every divisorial valuation. ◻

Proof of the theorem. An arbitrary basis of each binary component (I.2) produces a divisor of the form in the preceding lemma, with $$a=\alpha_m-3h_m,\quad b=b_m,\quad c=t_m,\quad f=f_m.$$ Indeed the sum of orders of a basis of degree-$D$ binary forms at any point is at most $D(D+1)/2$, while its total degree is $D(D+1)$. The limiting component bounds are $$\frac58,\quad\frac{13}{30},\quad\frac{217}{240}, \quad\frac3{10},\quad\frac{49}{240}.$$ The smooth tangency corner has order $29/8$ and discrepancy four; the cusp corner has order $25/8$ and discrepancy five. The largest of all these ratios is $29/32$. Thus all these basis divisors have log canonical threshold at least any fixed $\tau<32/29$ for all sufficiently large $m$.

For a divisor over $G$, choose a basis of each primary piece compatible with its valuation filtration and the auxiliary weight filtration. The least-weight degeneration has the binary components (I.2), because the normal-basis count proved equality of algebras. Lower semicontinuity of log canonical thresholds therefore gives the same bound for the original valuation-adapted basis divisor. It follows that $\delta(G;W^G)\ge32/29$. Finally, Abban–Zhuang adjunction gives $$\delta_p(X)\ge\min\left\{\frac{3}{9/4},\frac{32}{29}\right\} =\frac{32}{29}.$$ Again the proof needs no Zariski decomposition on the threefold extraction: the actual refined series is identified by sections and an exact Hilbert count, and its thresholds are controlled by two explicit surface resolutions of a pencil. ◻