A weighted extraction of the joining line

Proof. Choose coordinates so that the mixed fibre of the graph is defined by $f=x^3y^2$. Write $g=\sum_{i=0}^5 c_i x^i y^{5-i}$, where $c_0c_5\ne0$. On $Y_0\ne0$, put $a=Y_1/Y_0$, $b=Y_2/Y_0$, $q=b-a^2$, and let $t$ be the target coordinate vanishing on the mixed fibre. Set $$P=c_0+c_2b+c_4b^2,\qquad Q=c_1+c_3b+c_5b^2.$$ The two lifted graph sections are $$A=\frac{a^3+t(P+aQ)}{q},\qquad B=\frac{a^2b+t(aP+bQ)}{q}.$$ They satisfy $bA-aB=tP$ and $B=aA+a^2+tQ$. The anticanonical degree-one sections, in this trivialization, are $t^i\{1,a,b\}$ for $0\le i\le2$, together with $A,B,tA,tB$.

The morphism from the normal bundle of $\Gamma$ to the normal bundle $\mathcal O(1)\oplus\mathcal O$ of its image in $\mathbb P^1\times\mathbb P^2$ is an elementary transform at the two endpoints. At each endpoint its quotient is in the trivial summand: the graph is tangent to the fibre, whereas the joining line is transverse to the conic. Thus its kernel is $\mathcal O(1)\oplus\mathcal O(-2)$. Equivalently, in local coordinates near an endpoint the differential in the normal directions is $(a,w)\mapsto(a,bw)$. At the index-two endpoint the equation for $t$ contains an additional quadratic term, which has zero differential.

There is a coordinate-independent weighted extraction. If $I$ is the ideal of $\Gamma$, take the inverse image $J\subset I$ of the positive subbundle $\mathcal O(2)\subset I/I^2$. Locally $J=(w,a^2)$. The normalized blow-up of $J$ has weights $(1,2)$, discrepancy three, and different one half along the section corresponding to $a=0$. The coarse exceptional ruled surface need not be identified in this argument.

At the generic point of $\Gamma$, assign weights $1,2,0$ to $a,t,b$. The initial forms of $A,B$ are $tP/b$ and $a^2+tQ$. Subtracting linear combinations of $t,tb$, and rescaling, gives $t/b$ and $a^2+tb^2$. The same operation applies after multiplication by $t$. Putting $z=t/b$ yields the thirteen generators in the statement. They generate a subalgebra of the primary associated algebra. Lemma 99.2 below gives an exact dimension count after one further refinement. It proves equality of both associated algebras in every degree. ◻

Proof. The substitution $b\mapsto cb$, $a\mapsto a$, $z\mapsto c^{-3}z$ preserves $h$ and scales each of the thirteen sections. We may therefore suppose that $c=1$. Put $d=b-1$. In the primary degree-two piece use $$z,\quad dz,\quad d^2z,\quad h-3b^2z+3bz-z=a^2+d^3z.$$ In the primary degree-four piece use $$bz^2,\quad bdz^2,\quad bd^2z^2,\quad bhz-3b^3z^2+3b^2z^2-bz^2=ba^2z+bd^3z^2.$$ Together with $1,d,a,abz,ab^2z^2$, their leading forms are the thirteen monomials asserted in the statement. Their products give linearly independent leading forms whenever their exponent triples are distinct.

We count these products in every degree. The degree-one generators with first two coordinates $(2,0)$ allow third coordinate $0$ or $1$; those with first two coordinates $(0,2)$ allow $1$ or $2$. The generators with first two coordinates $(1,0)$, $(0,1)$, or $(0,0)$ allow every third coordinate between $0$ and $2$. Thus a triple $(a,k,z)$ is a product of $m$ generators if and only if there exist integers $r,s\ge0$ such that $$r\le\left\lfloor\frac a2\right\rfloor,\quad s\le\left\lfloor\frac k2\right\rfloor,\quad r+s\ge a+k-m,\quad s\le z\le2m-r.$$ Indeed, take $r$ generators of type $(2,0)$, $s$ of type $(0,2)$, $a-2r$ of type $(1,0)$, $k-2s$ of type $(0,1)$, and fill the remaining places with type $(0,0)$. Their possible third coordinates form the whole interval $[s,2m-r]$. Equivalently, $$\min\left\{\left\lfloor\frac a2\right\rfloor,2m-z\right\} +\min\left\{\left\lfloor\frac k2\right\rfloor,z\right\} \ge a+k-m.$$ Expanding the two minima gives precisely the inequalities and the parity exception in the statement.

For completeness, the count is also explicit. Write $a=2r+\epsilon$, $k=2s+\eta$, where $\epsilon,\eta\in\{0,1\}$. The condition is $r+s+\epsilon+\eta\le m$. Put $N=m-r-s-\epsilon-\eta$. The number of possible third coordinates is $$2m+1-\max\{0,s-N\}-\max\{0,r-N\}.$$ If $$Q(n)=\sum_{r+s\le n}\max\{0,2r+s-n\},$$ then summing first over $s$ gives $$Q(n)=\begin{cases} n(n+2)(2n+5)/24&\text{if $n$ is even},\\ (n+1)(n+3)(2n+1)/24&\text{if $n$ is odd}. \end{cases}$$ Set $Q(-1)=Q(-2)=0$. Consequently the number of distinct products is $$\begin{aligned} &(2m+1)\left(\binom{m+2}{2}+2\binom{m+1}{2}+\binom m2\right)\\ &\quad -2\bigl(Q(m)+2Q(m-1)+Q(m-2)\bigr) =\frac{10m^3+15m^2+11m+3}{3}. \end{aligned}$$ This is $h^0(X,-mK_X)$. Thus the products already fill the entire filtered vector space, and there are no further leading forms in any degree. Notice that the semigroup is not normal: $(1,1,1)$ is the first missing lattice point. Its missing boundary points must not be silently inserted in the finite-degree basis. ◻

Proof. Let $P$ be the polytope in the preceding lemma with $m=1$. The missing lattice points belong to a two-dimensional face. There are $O(m^2)$ of them, so they do not change the limits of normalized first moments. Over the triangle $T=\{a,k\ge0\mid a+k\le2\}$ the interval in the third coordinate is $$\max\left\{0,\frac{a+2k-2}{2}\right\}\le z\le 2-\max\left\{0,\frac{2a+k-2}{2}\right\}.$$ Each of the two cut-off triangular caps has volume $1/3$; hence $\operatorname{vol}(P)=4-2/3=10/3$. The integral of $a$ over $P$ is $$\int_P a=\frac83-\frac16-\frac5{12}=\frac{25}{12}.$$ Here the first term is its integral over $T\times[0,2]$. For the lower cap the base triangle has vertices $(0,1),(0,2),(2,0)$, and the integral of $a(a+2k-2)/2$ is $1/6$. For the upper cap the base triangle has vertices $(1,0),(2,0),(0,2)$, and the integral of $a(2a+k-2)/2$ is $5/12$. The involution $(a,k,z)\mapsto(k,a,2-z)$ preserves $P$. Therefore $$\frac1{\operatorname{vol}(P)}\int_P(a,k,z) =\left(\frac58,\frac58,1\right).$$ The primary order is $a+2z$, while the secondary order is $k$. This proves the first two assertions.

Fix a primary degree $j$ and a secondary degree $k$. In the basis of Lemma 99.2 the remaining exponents satisfy $a=j-2z$. Their allowed values of $z$ form an interval of consecutive integers. The parity exception can remove only its lower endpoint. Thus this piece, on the weighted fibre, is exactly $$a^{a_0}z^{z_0} \langle (a^2)^d,(a^2)^{d-1}z,\ldots,z^d\rangle$$ for some nonnegative integers $a_0,z_0,d$. At $z=0$ its vanishing sequence is $z_0,z_0+1,\ldots,z_0+d$. At the coarse point $a=0$ it is $a_0/2,a_0/2+1,\ldots,a_0/2+d$. The rational offset is the order in the induced rational line bundle; it is also obtained by taking an even multiple before passing to the coarse fibre. At any other point the vanishing sequence is $0,1,\ldots,d$. Hence the last sequence has average no larger than the sequence at $a=0$. Summing over $j,k$ gives the asserted moments. These are complete restricted series, so the estimates hold for bases adapted to any chosen point, rather than only for the displayed monomial basis. ◻

Proof. The extraction with transverse weights $(1,2)$ has $A_X(G)=3$. Over the indicated open set its different is one half of the section $a=0$. The fibre $F_c$ is not contained in this different. The different induced on its coarse curve is therefore $\frac12\{a=0\}$. Its point discrepancies are $1/2$ at $a=0$ and $1$ everywhere else. By Lemma 99.3, the ratios on the fibre are $8/5$ at $a=0$, exactly $1$ at $z=0$, and at least $16/5$ elsewhere. The Abban–Zhuang refinement along $F_c$ gives a bound $\min\{8/5,1\}=1$ on the exceptional surface above $p$. Applying the same refinement to $G$ gives $$\delta_p(X)\ge\min\left\{\frac3{21/8},\frac{1}{5/8},1\right\}=1.$$ The equality in the last point estimate only proves semistability. ◻