The point of a fibre with ramification index two

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 let $t=v/u$ be the target parameter. The centre is cut out by the conic and $F=tg(x,c)+f(x,c)$. Adding a multiple of the conic equation to $f$, and rescaling the target parameter, we may assume $$f=z+R(x,c),\qquad R(x,c)=A x^2+Bxc+C c^2,\qquad A\ne0,\qquad g(0,0)=1.$$ The inequality $A\ne0$ expresses ramification index two. On the blow-up chart $F=za$, put $w=a-1$. Then $x,z,w$ are regular parameters at $p$, and $$t g(x,c)=zw-R(x,c),\qquad c=z+x^2.$$ Thus the initial form of $t$ for the ordinary point valuation is $Q=zw-R(x,z)$.

The complete-intersection ideal sequence gives the fifteen anticanonical sections $$t^i\{1,x,c\}\quad(0\le i\le2),\qquad t^ia\{1,x,c\}\quad(0\le i\le1).$$ Subtracting the first group from the second gives their initial basis $$\{1,x,z,w,wx,wz\},\qquad Q\{1,x,z,w,wx,wz\},\qquad Q^2\{1,x,z\}.$$ These forms are independent; in particular, $wz$ and $Q$ are independent because $A\ne0$.

Now give $x,z,w$ the secondary weights $1,2,0$. The least secondary term of $Q$ is $zw-Ax^2$. Replacing $wz$ by $wz-Q=R(x,z)$ gives $Ax^2$, and replacing $Qwz$ by $Qwz-Q^2$ gives $Ax^2(zw-Ax^2)$. After rescaling $x$, these are precisely the generators in the statement. Products give an inclusion in every degree. We do not assume generation of the anticanonical ring in degree one. ◻

Proof. Straighten every factor $zw$ by $zw=x^2+t$. The only pairs of generators requiring this are $z,w$ and $z,wx$. Their products become $x^2+t$ and $x^3+xt$, respectively. The capacities for powers of $t$ show that each resulting term is a product of the same number of degree-one generators: $z$ has capacity two, whereas $w$, $wx$ and $x^2$ have capacity one. Each replacement reduces the number of $zw$ pairs. The resulting monomials are independent, since $zw$ is a leading monomial of the single relation.

If $c=0$, use $r$ factors $x^2$, $a-2r$ factors $x$ and $b$ factors $z$. The smallest possible $r$ is $\max(0,a+b-m)$. Such a distribution is possible exactly when $a+2b\le2m$, and its capacity for $t$ is $2m-r=\min(2m,3m-a-b)$. If $c\ge1$, use as many factors $wx$ as possible, namely $\min(a,c)$. The smallest necessary number of factors $x^2$ is $\max(0,a-m)$. This is possible exactly when $c\le m$ and $a+c\le2m$. The resulting capacity is $\min(2m-c,3m-a-c)$. Every smaller nonnegative power of $t$ can be distributed among these factors.

The two rows contain, respectively, $$\frac{(m+1)(11m^2+16m+6)}6,\qquad \frac{m(13m^2+9m+2)}6$$ monomials. Their sum is $(2m+1)(2m^2+2m+1)$, which equals $h^0(X,-mK_X)$ by Riemann–Roch and Kodaira vanishing. Thus the inclusion of Lemma 122.2 is an equality in every degree. ◻

Proof. The primary degree of $x^az^bw^ct^s$ is $a+b+c+2s$. In the first row of Lemma 122.3, put $b=m-n$. Then $0\le n\le m$, $0\le a\le2n$ and $0\le s\le U=\min(2m,2m+n-a)$. In the second row put $c=m-n$. Then $0\le n<m$, $0\le a\le m+n$ and $0\le s\le U=\min(m+n,2m+n-a)$. In either row the sum in $s$ is $$(U+1)(a+m-n)+U(U+1).$$ Split the first $a$-sum at $n$ and the second at $m$. The sums over the two rows are therefore $$\frac{m(m+1)(31m^2+43m+16)}6,\qquad \frac{m^2(23m^2+14m+3)}4.$$ Their sum is $$\frac{m(131m^3+190m^2+127m+32)}{12}.$$ The calculation holds for every nonnegative integer $m$. The all-degree equality in Lemma 122.3, or equivalently Lemma 94.2, identifies these primary degrees with the orders of the actual sections. Dividing by $m(2m+1)(2m^2+2m+1)$ and taking the limit proves the assertion. ◻

Proof. Let $j$ and $q$ be the primary and secondary degrees. Then $j=a+b+c+2s$, $q=a+2b+2s$, so $b=\max(q-j,0)$ and $c=\max(j-q,0)$. Put $k=j-b-c$. On the first row of the normal basis one has $$\ell=\max\left(0,\left\lceil\frac{k+2b-2m}{2}\right\rceil, k+b-3m\right),\qquad u=\min(2m,\lfloor k/2\rfloor),$$ and on the second row $$\ell=\max\left(0,\left\lceil\frac{k+c-2m}{2}\right\rceil, k+c-3m\right),\qquad u=\min(2m-c,\lfloor k/2\rfloor).$$ Set $A=k-2u$ and $d=u-\ell$. Whenever $d\ge0$, the corresponding component is precisely the binary space in the statement, since $\langle x^2,t\rangle=\langle x^2,zw\rangle$.

For completeness, the following integrations give all the coefficients. Divide the indices by $m$, write $y=b/m$ or $y=c/m$, and let $k$ denote $k/m$. In each row $0\le y\le1$. The limiting lower and upper bounds for $s/m$ are $$\begin{array}{c|c|c|c} &\text{range of }k&L&U\\\hline z &[0,2-2y]&0&k/2\\ z &[2-2y,4]&(k+2y-2)/2&k/2\\ z &[4,5-y]&k+y-3&2\\\hline w &[0,2-y]&0&k/2\\ w &[2-y,4-2y]&(k+y-2)/2&k/2\\ w &[4-2y,4-y]&(k+y-2)/2&2-y\\ w &[4-y,5-2y]&k+y-3&2-y. \end{array}$$ Put $d=U-L$. Integration with respect to $dk\,dy$ gives $$\begin{array}{c|ccccc} &\int d&\int(k-2U+d)d&\int yd&\int Ld&\int d^2/2\\\hline z&11/6&9/8&5/8&7/6&13/24\\ w&13/6&13/8&7/8&85/96&71/96. \end{array}$$ Divide the summed columns by four, since $m h^0(X,-mK_X)=4m^4+O(m^3)$. This gives $\alpha=11/16$, $b_0=5/32$, $c_0=7/32$, $s_0=197/384$ and $h_0=41/128$, where $a_0=\alpha-2h_0=3/64$. The rounding and boundary terms contribute $O(m^3)$ to the sums. As a check, $$a_0+b_0+c_0+2s_0+4h_0=\frac{131}{48}.$$

Finally, a basis of a complete binary space of degree $d$ has total zero multiplicity $d(d+1)$ on the pencil parameter line. Its total order at any prescribed point is at most $d(d+1)/2$. This proves the last assertion, including the two reducible members of the pencil. ◻

Proof of Theorem 122.1. Consider the pencil $T_\lambda=(zw-\lambda x^2=0)$ on $G$. Its special members are $T_0=(z=0)+(w=0)$ and $T_\infty=2(x=0)$. Lemma 122.5 bounds the coefficients of every basis divisor of the secondary initial series. The bounds on $(x=0)$, $(z=0)$, $(w=0)$, $T_1$ and all other irreducible members are $$\frac{11}{16},\qquad\frac{61}{128},\qquad \frac{69}{128},\qquad\frac56,\qquad\frac{41}{128}.$$

The pencil has base points $[0:0:1]$ and $[0:1:0]$. Blow up each twice, the second time at the common point of the strict transforms of the finite pencil members. The total boundary then has simple normal crossings. At the first base point the orders on the first and second exceptional curves are $$a_0+b_0+s_0+2h_0+\nu_\infty,\qquad a_0+2b_0+2s_0+4h_0=\frac83,$$ where $0\le\nu_\infty\le h_0$ is the moving coefficient of $T_\infty$. At the second base point the orders are obtained by replacing $b_0$ with $c_0$, and the second order is $67/24$. Their log discrepancies are two and three, respectively. The first order is half the sum of the second order and $a_0+2\nu_\infty$, the coefficient of $(x=0)$; hence its ratio is bounded by the larger of the second ratio and $11/16$. These blow-ups and the component bounds check all divisorial valuations. The largest ratio of order to discrepancy is $67/72$. It follows that for each $\tau<72/67$, every such basis divisor has log-canonical threshold at least $\tau$ for all sufficiently large $m$.

For any divisorial valuation over $G$, choose in each primary piece a basis compatible with both that valuation and the secondary weight filtration. Two filtrations admit a common adapted basis. Diagonal degeneration gives a basis of the secondary initial series. Lower semicontinuity of log-canonical thresholds gives the same lower bound for the original basis divisor. Passing to the limit yields $\delta(G;W^G)\ge72/67$. Abban–Zhuang adjunction and Lemma 122.4 give $$\delta_p(X)\ge \min\left\{\frac{144}{131},\frac{72}{67}\right\} =\frac{72}{67}.$$ We have not computed a Zariski decomposition on the blow-up of the threefold. The exact normal basis determines the actual refined series and its expected orders. All subsequent blow-ups occur on its exceptional plane, to resolve the conic pencil. ◻