The point of a fibre with ramification index three

Proof. Write the conic as $YC-X^2=0$. On the chart $Y=u=1$, put $x=X/Y$, $c=C/Y$, $z=c-x^2$ and $t=v/u$. The curve blown up is the complete intersection of the conic equation and a section of $\mathcal O_{\mathbb P^1\times\mathbb P^2}(1,2)$. After changing this second equation by a multiple of the conic equation, it has the form $$F=tg(x,c)+xc+\gamma c^2,\qquad g(0,0)=1.$$ Indeed the corresponding quartic on the conic has a root of multiplicity three at $x=0$, and its leading coefficient can be normalized to one.

On the blow-up chart $F=za$, the functions $x,z,a$ are regular parameters at $p$. Put $w=a-x$. With weights $(1,2,1)$ on $x,z,w$, the initial form of the original target parameter is $zw-x^3$; the initial forms of $c$ and $a$ are $x^2+z$ and $w+x$. This also exhibits the $A_2$ singularity of the fibre. Weighted adjunction gives the exceptional pair and discrepancy asserted.

The complete-intersection ideal sequence identifies $H^0(X,-K_X)$ with the direct sum of the conic equation times $H^0(\mathcal O(2,1))$ and $F$ times $H^0(\mathcal O(1,1))$. The kernel sheaf $\mathcal O(1,-1)$ has no cohomology, so this describes the whole section space and the sum is direct. Divide these sections by the exceptional equation $z$. Their initial forms are precisely the fifteen sections in the statement.

The polynomial $t=zw-x^3$ is homogeneous of secondary weight three. Replacing $a$ by $a-x$ gives $w$, and similarly gives $tw$. The least secondary terms of $c$, $xa$ and $ca$ are $z$, $x^2$ and $xz$, respectively. Products of these actual initial sections prove the asserted inclusion. We have not assumed that the anticanonical ring is generated in degree one. ◻

Proof. Straighten factors containing both $z$ and $w$ using $zw=x^3+t$. For $(zt^i)(wt^j)$, one has $i\le2$, $j\le1$; distribute the $t$-exponent of $x^3t^{i+j}$ between $x$ and $x^2$, with capacities two and one. The remaining term $t^{i+j+1}$ is a product of two listed generators. For the additional generator $xz$, use $$(xzt^i)(wt^j)=(x^2t^i)(x^2t^j)+xt^{i+j+1},\qquad i,j\le1.$$ Again both terms are products of two generators. Each replacement reduces the number of $zw$ pairs. The resulting monomials with $bc=0$ are independent, since $zw$ is the leading monomial of the single relation for a suitable monomial order.

For $c=0$, use $r$ factors $x^2$, $q$ factors $xz$, $a-2r-q$ factors $x$ and $b-q$ factors $z$, and fill the remaining places with $1$. The smallest possible $r+q$ is $\max(0,a+b-m)$. Such a distribution exists exactly when $b\le m$ and $a+b\le2m$. The total capacity for $t$ is then $2m-r-q=\min(2m,3m-a-b)$, and every smaller nonnegative integer can be distributed. For $b=0$, use $r$ factors $x^2$, $a-2r$ factors $x$, and $c$ factors $w$. Taking $r=\max(0,a+c-m)$ gives the second row.

The first row contributes $(m+1)(5m^2+6m+2)/2$ monomials. The second contributes $m^2(3m+1)/2$. For example, the latter count is obtained by setting $n=m-c$ and summing $(2n+1)(m+n+1)-n(n+1)/2$ for $0\le n<m$. Their sum is $N_m$. Since $(-K_X)^3=24$, Riemann–Roch and Kodaira vanishing give the same value for $h^0(X,-mK_X)$. Taking an associated graded space preserves dimension. The inclusion of Lemma 123.2 is therefore an equality in every degree. ◻

Proof. The primary degree of a normal monomial is $a+2b+c+3s$. For the first row of Lemma 123.3, put $b=m-n$. The ranges are $0\le n\le m$, $0\le a\le m+n$ and $0\le s\le U=\min(2m,2m+n-a)$. Summation in $s$ gives $$(U+1)(a+2m-2n)+\frac32U(U+1).$$ For the second row put $c=m-n$. The ranges are $0\le n<m$, $0\le a\le2n$ and $0\le s\le U=\min(m+n,m+2n-a)$, and the sum in $s$ is $$(U+1)(a+m-n)+\frac32U(U+1).$$ Splitting both $a$-sums at $a=n$ gives the respective totals $$\frac{m(m+1)(125m^2+141m+46)}{12},\qquad \frac{m^2(59m^2+6m-5)}{12}.$$ Consequently, for every $m\ge0$, $$\sum(a+2b+c+3s)=\frac{m(2m+1)(46m^2+45m+23)}6.$$ The equality of algebras in Lemma 123.3 preserves primary degrees, as in Lemma 94.2. This is therefore the exact weight sum of the actual valuation filtration. Divide by $mN_m$ and let $m$ tend to infinity to obtain $23/6$. ◻

Proof. The equations for the degrees give $j=a+2b+c+3s$, $q=a+3c+3s$. Thus $a=k-3s$, and substituting in the normal-basis inequalities gives $\ell\le s\le u$. Replacing $t$ by $zw-x^3$ identifies their span with the stated complete binary space.

Here are the integrations giving the five limits. Write $y=b/m$ on the first branch and $y=c/m$ on the second, with $0\le y\le1$, and write $k$ for $k/m$. The lower and upper bounds for $s/m$ are as follows: $$\begin{array}{c|c|c|c} &\text{range of }k&L&U\\ \hline z &[0,2-y]&0&k/3\\ z &[2-y,5-y]&(k+y-2)/3&k/3\\ z &[5-y,6]&(k+y-3)/2&k/3\\ z &[6,7-y]&(k+y-3)/2&2\\ \hline w &[0,2-2y]&0&k/3\\ w &[2-2y,5-2y]&(k+2y-2)/3&k/3\\ w &[5-2y,6-3y]&(k+2y-3)/2&k/3\\ w &[6-3y,7-4y]&(k+2y-3)/2&2-y. \end{array}$$ Put $d=U-L$. Integration with respect to $dk\,dy$ gives $$\begin{array}{c|ccccc} &\int d&\int(k-3U+3d/2)d&\int yd&\int Ld&\int d^2/2\\ \hline z&5/2&7/4&13/12&229/144&83/144\\ w&3/2&23/24&11/24&41/48&5/16. \end{array}$$ Divide the summed columns by four, with the third column contributing separately to $b_0$ and $c_0$. This proves the assertion, since $mN_m=4m^4+O(m^3)$ and the boundary and rounding terms contribute $O(m^3)$ to the weighted sums. The fixed $x$ coefficient is $\alpha-3h=1/96$. As a check, the average weighted degree is $$(\alpha-3h)+2b_0+c_0+3\sigma+6h=\frac{23}{6},$$ in agreement with Lemma 123.4. ◻

Proof. The first five inequalities bound the coefficients of all components. Away from the coordinate vertices the pencil members are smooth and disjoint. At $[1:0:0]$, only $Z_0$ and $W_0$ occur, and they cross transversely.

At $[0:1:0]$, take the index-two cover of the chart $z=1$. Its coordinates are $x,w$, with the involution $(x,w)\mapsto(-x,-w)$. This cover is quasi-étale, so log canonicity can be checked upstairs. The branches are $x=0$ and $w=\lambda x^3$. Put $$u=a+3\nu_\infty,\qquad v=c+s+2h-\nu_\infty.$$ Three successive ordinary blow-ups separate the branches. Their exceptional valuations have weights $(1,i)$, for $i=1,2,3$, and have orders $u+iv$ on $B$ and log discrepancies $1+i$. The component inequality gives $u\le1$, whereas $$u+3v=a+3c+3s+6h=\alpha+3(c+s+h)\le4.$$ Linear interpolation gives $u+iv\le1+i$ for $i=1,2$. After the third blow-up the distinct branches meet the last exceptional curve at distinct points. The total transform has simple normal crossings, with every coefficient at most one. This proves log canonicity at the index-two vertex.

The chart $w=1$ at $[0:0:1]$ is smooth. Its branches are $x=0$ and $z=\lambda x^3$. The same three blow-ups, with $b$ in place of $c$, use the second corner inequality. This checks every point on a log resolution, and hence every divisorial valuation over $G$. ◻

Proof of Theorem 123.1. Use the pencil $T_\lambda=(zw-\lambda x^3=0)$ on $G$, with $T_0=(z=0)+(w=0)$ and $T_\infty=3(x=0)$. In a complete binary space of degree $d$, the sum of the orders of any basis at a prescribed point is at most $d(d+1)/2$. Consequently every basis divisor of the secondary initial series has the form in Lemma 123.7, with limiting coefficients $$a=\frac1{96},\quad b=\frac{13}{48},\quad c=\frac{11}{96},\quad s=\frac{11}{18},\quad h=\frac29.$$ The five component bounds in that lemma are $$\frac{65}{96},\quad\frac{71}{144},\quad\frac{97}{288}, \quad\frac56,\quad\frac29.$$ The two corner orders are $169/48$ and $383/96$, and their log discrepancies are four. Thus the largest ratio of order to log discrepancy is $383/384$. For every $\tau<384/383$, all these basis divisors have log-canonical threshold at least $\tau$ for sufficiently large $m$. The three ordinary surface blow-ups at each base point of the pencil, described in Lemma 123.7, check all valuations over $G$, including those above its index-two point.

Fix any divisorial valuation over $G$. In each primary piece choose a basis compatible with this valuation and the secondary weight filtration. Two filtrations admit such a common adapted basis. Its diagonal degeneration gives a basis divisor of the initial series just considered. Lower semicontinuity of log-canonical thresholds gives the same bound for the original basis divisor. Taking limits proves $\delta(G;W^G)\ge384/383$. Abban–Zhuang adjunction now gives $$\delta_p(X)\ge \min\left\{\frac{A_X(G)}{S_X(G)},\delta(G;W^G)\right\} \ge\min\left\{\frac{24}{23},\frac{384}{383}\right\} =\frac{384}{383}.$$ No Zariski decomposition on the weighted blow-up of the threefold has been used. The actual sections, the exact normal basis in every degree and the complete binary spaces replace this calculation. The subsequent blow-ups occur only on the exceptional surface to resolve its pencil. ◻