The complete refinement at a double fibre

Proof. The initial pairs and their graded normalizations were established in Proposition 74.3. In the $(1,2)$ model the centre is $(A,D-F,E)$ on $V_0$. Divide a quadratic in its ideal by $A$ at the generic point of $R$. The substitutions $$F=Bv,\qquad E=Av,\qquad D=Au+Bv,\qquad C^2=4ABu+(4B^2-A^2)v$$ give the leading sections. In the $(1,3)$ model the centre is $(B,D,F)$; divide by $B$ and use $D=Bu$, $F=Bv$, $E=Av$, $C^2=4B^2u-A^2v$. After choosing $a$ to vanish at the marked point, the six spaces in anticanonical degree one are $$\begin{array}{c|c|c} k&W_{1,k}&\dim W_{1,k}\\ \hline 0&H^0(\mathcal O_F(1,0))&2\\ 1&w\mathbb C&1\\ 2&H^0(\mathcal O_F(1,1))&4\\ 3&wH^0(\mathcal O_F(0,1))&2\\ 4&H^0(\mathcal O_F(1,2)\otimes\mathcal I_q)&5\\ 5&wvH^0(\mathcal O_F(0,1))&2. \end{array}$$ For example, in the first model the weight-four space is generated by the leading terms of $D(D-F)$, $F(D-F)$, $E(D-F)$, $E^2$, and $EF$, divided by $A$. These span $Au^2,Auv,Av^2,Buv,Bv^2$. The sixteen displayed sections exhaust $H^0(X,-K_X)$.

Multiplication gives subspaces of the actual associated graded ring. We first construct all the spaces in the statement using these degree-one sections. Suppose $j\le2m$. A monomial $a^i b^{m-\epsilon-i}u^{j-h}v^h$ belongs to the displayed ideal-power space exactly when $i+h\ge j-m$. Distribute the $j$ symbols $u,v$ among $m$ factors, with at most two in each factor. There are $i$ factors carrying $a$; these can carry two copies of $u$. Every other factor can carry at most one copy of $u$. The inequalities $j\le2m$ and $j-h\le m+i$ are precisely the capacity conditions. For $\epsilon=1$, one factor carries $w$ in place of an $a$ or $b$; its possibilities are the three odd rows of the table. The same capacity conditions apply. This constructs every required monomial.

If $j>2m$, put $r=2d+\epsilon$ and $a_0=m-r$. When $a_0<0$ the claimed space is zero. Otherwise take $r$ factors from the weight-five row and $a_0$ factors from the weight-four row. Their products span $$w^\epsilon Q^d v^r H^0\left(F,\mathcal O_F(a_0,2m-r)\otimes\mathcal I_q^{a_0}\right).$$ Indeed multiplication by all degree-$r$ binary forms in $u,v$ fills the indicated monomial space. A monomial of first degree $a_0=m-r$ vanishing to order $m$ at $q$ is necessarily divisible by $v^r$. Thus this is exactly the space asserted in the lemma.

It remains to prove that these subspaces exhaust the refinement. This is a dimension argument on the actual section ring, so it does not assume generation in anticanonical degree one. For $0\le j\le2m$ the constructed dimension is $$(m-\epsilon+1)(j+1) -\frac{n(n+1)}2,\qquad n=\max\{j-m,0\}.$$ For $j>2m$ it is $(a_0+1)(m+1+a_0/2)$. Summing the first formula over $j$ and $\epsilon$ gives $(m+1)(2m+1)^2-m(m+1)(m+2)/3$. The remaining nonzero terms have $a_0=0,\ldots,m-2$ and sum to $m(m-1)(4m+1)/6$. Their total is $$\frac{(2m+1)(13m^2+13m+6)}6=h^0(X,-mK_X),$$ by Riemann–Roch and vanishing. The constructed subspaces already have the full dimension. No extra graded sections can occur. ◻

Proof. The normalized leading dimensions $h(t)=\lim m^{-2}\dim W_{m,\lfloor mt\rfloor}$ are $t/2$, $t/2-(t-2)^2/8$, and $(5-t)(7-t)/2$ on the three intervals. Lemma 75.1 gives these limits directly and shows that no weight exceeds $5m$. Summing the leading sections of order at least $mt$ gives $\operatorname{vol}(f^*L-tT)=6\int_t^5h(s)\,ds$. Integration proves the formula, and a further integration gives $S_X(T)=297/104$, agreeing with the independent character computation. ◻

Proof. All these statements are integrals over the three displayed polygons. We give the bounds explicitly so that fixed-point corrections can be checked. For each row put $b_0=\max\{0,n-A\}$. Integrate first over $b_0\le y\le n$, with $n-y\le x\le A$, and then over $n\le y\le B$, with $0\le x\le A$. Integrating $x,y,A,B$ and $(\rho-2)_+$ over these regions, with factor $6/13$, gives the five averages. The unnormalized values are $13/12$, $83/48$, $33/16$, $73/24$, and $5/96$.

For the flag $\ell$, fix $y$. The allowed exponents in the other coordinate form the interval $[\max\{0,n-y\},A]$. A general point of $\ell$ therefore contributes one-half the square of the length of this interval. Its integral, before normalization, is $47/48$. At $q$, integrate $x$ instead, obtaining $13/12$. The fixed curve $\Gamma$ adds its average coefficient times its local intersection.

For $\ell_0\ne\ell$, subtract a multiple $z\ell_0$ from the series. Restriction to $\ell_0$ has fixed order $h=\max\{0,n-B+z\}$ at $a=0$ and moving degree $A-h$. Here $0\le z\le B-n+\min\{n,A\}$. The point moment away from $a=0$ is the integral of $(A-h)^2/2$. The correction at $a=0$ is the integral of $h(A-h)$. Their unnormalized values are $47/48$ and $5/48$. The curve moment is obtained by integrating $(B-\max\{0,n-x\})^2/2$ over $0\le x\le A$ and then over $\rho$; it equals $21/16$, giving $63/104$ after normalization. These computations also follow directly by counting the monomials in Lemma 75.1. ◻

Proof. Blow up $q$ and denote its exceptional curve by $J$. Its log discrepancy for $(F,\Gamma/2)$ is $3/2$. The polygon moment of $x+y$, with the fixed-curve contribution included, gives $$S(W;J)=\frac12+\frac{83}{104}+\frac5{208}=\frac{275}{208}.$$ The different on $J$ is one-half of the point determined by the tangent of $\Gamma$. This point differs from both coordinate directions: transversality excludes the direction of $\ell$, and $\Gamma\to R$ is an isomorphism, excluding the other ruling.

To compute the second refinement, slice $\mathcal P_\rho$ by $x+y=z$. The interval of possible $y$ is $[\max\{0,z-A\},\min\{B,z\}]$, for $n\le z\le A+B$. The integral of one-half the square of its length is $37/48$, or $37/104$ after normalization. The two coordinate directions have moments $1/2$ and $83/104$. At the tangent direction of $\Gamma$ the moment is $37/104+5/208=79/208$. Thus the Abban–Zhuang bound is $$\min\left\{\frac{312}{275},\ 2,\ \frac{104}{83},\ \frac{104}{79},\ \frac{104}{37}\right\}=\frac{312}{275}.$$ All restricted spaces used here are complete binary spaces with the displayed fixed endpoint orders, by the exact monomial description. ◻

Proof. The points $p,q$ lie on different fibres of both rulings. Choose coordinates putting them at opposite corners. If $x,y$ are the exponents at $q$, the exponents at $p$ are $x'=A-x$, $y'=B-y$. Here $y'=0$ denotes the tangent ruling through $p$. The curve $\Gamma$ has local equation $y'=c(x')^2+\text{higher terms}$, with $c\ne0$. Use the weighted blowup of weights $(1,2)$ in these two local coordinates, and denote its exceptional curve by $J$. Then $A_{F,\Gamma/2}(J)=3-2/2=2$. The moment is $$S(W;J)=\left(\frac{99}{104}-\frac12\right) +2\left(\frac{73}{52}-\frac{83}{104}\right) +2\frac5{208}=\frac{89}{52}.$$

The coarse curve $J$ is $\mathbb P^1$. Its different consists of two distinct points of coefficient $1/2$: the quotient point and the point cut out by the initial equation of $\Gamma$. The other coordinate direction has coefficient zero. For the second refinement slice by $z=x'+2y'$. With $K=A+B-n$, the interval of allowed $y'$ is $$\left[\max\left\{0,\frac{z-A}{2},z-K\right\}, \min\left\{B,\frac z2\right\}\right],$$ when its length is nonnegative. The integrals of one-half the square of its length over the three $\rho$-intervals are $17/192$, $19/64$, and $5/96$. Their sum is $7/16$. The general point moment on $J$ is consequently $21/104$. The point belonging to $\Gamma$ has moment $21/104+5/208=47/208$. The quotient point has moment $(99/104-1/2)/2=47/208$, and the other coordinate point has moment $73/52-83/104=63/104$. These give $$\delta_p(F,\Gamma/2;W)\ge \min\left\{\frac{104}{89},\frac{104}{47}, \frac{104}{63},\frac{104}{21}\right\} =\frac{104}{89}.$$ The division by two at the quotient point is necessary: it is the order on the coarse weighted exceptional curve. ◻

Proof. We first prove $\delta(F,\Gamma/2;W)\ge104/99$. The marked point is treated by Lemma 75.5. Every ramification point of the direction projection is treated by Lemma 75.6. For another point, use its horizontal ruling. On $\ell$, the only point of $\Gamma$ other than $q$ is $q'$, and its refined point moment is $99/208$, with discrepancy $1/2$. All other point moments there are $47/104$, with discrepancy one. The first moment is $83/104$.

On a different ruling, a transverse intersection with $\Gamma$ has point moment $99/208$, again with discrepancy $1/2$. It cannot also have $a=0$, since $\Gamma$ meets that vertical ruling only at $q$. Points outside $\Gamma$ have moment at most $1/2$ and discrepancy one. The first moment is $63/104$. Thus every remaining flag gives a bound at least $104/99$.

Apply Abban–Zhuang refinement on the threefold extraction. Its first ratio is $$\frac{A_X(T)}{S_X(T)}=\frac{3}{297/104}=\frac{104}{99}.$$ The pair on its exceptional surface is $(F,\Gamma/2)$, whose refined threshold has just been bounded below by the same number. Inversion of adjunction gives the bound in a neighbourhood of the whole exceptional surface, and hence along $R$ on $X$. This includes all valuations whose centres contain a point of $R$. One can equivalently apply the finite-basis proof: choose coefficients tending to $104/99$ from below, use the exact refined spaces above, and apply effective adjunction on the extraction before taking limits. ◻