The joining line of a fibre of type (2,2)

Proof. Write the conic and graph as $$q=UV-x^2=0,\qquad F=vx^2+uG(U,x,V)=0,$$ with $[u:v]$ the target coordinates. Put $G_0(U,V)=G(U,0,V)$. Its coefficients of $U^2$ and $V^2$ are nonzero, since the graph has no base point at either ramification point. The joining line is the strict transform of $u=x=0$.

Let $e$ define the exceptional divisor of the original blow-up, and let $s_1=q/e$, $s_2=F/e$. These are global sections of classes $E$, $2H-E$, $T+2H-E$, respectively, satisfying $$es_1=UV-x^2,\qquad es_2=vx^2+uG(U,x,V).$$ Give $u,x,s_2$ weights $2,1,2$, and all other displayed sections weight zero. These weights agree with the transverse valuation at the generic point of $\Gamma$. Their initial identities are $$es_1=UV,\qquad es_2=vx^2+uG_0(U,V).$$ We shall identify the entire refined series by an exact dimension count below; no assertion about generators of the whole Cox ring is needed.

Along $\Gamma$, the sections $v$ and $s_1$ are units. Set both equal to one, so $e=UV$, and write $t=u$, $a=s_2$. The exceptional weighted directions satisfy $$x^2=UVa-tG_0(U,V),$$ with weights $1,2,2$ on $x,a,t$. Their coarse surface is $G=\mathbb P^1_{U,V}\times\mathbb P^1_{a,t}$. The generic index two along $x=0$ gives the different $D/2$, where $$D=(UVa-tG_0(U,V)=0).$$ The polynomials $UV$ and $G_0$ have no common zero, so $D$ is the graph of a morphism of degree two and is smooth. On the open part of $\Gamma$, the normal coordinates are $x,u$; at the endpoints one uses $x,s_2$. Both descriptions give weights $(1,2)$ and log discrepancy three. ◻

Proof. Products of the global sections in the preceding proof of class $m(2T+3H-E)$, after setting $v=s_1=1$, have initial forms $$(UV)^\alpha x^\ell t^i a^c H^0(\mathbb P^1,\mathcal O(m-2\alpha-\ell)),$$ where $c\le m+\alpha$, $i+c\le2m$ and $\ell\le m-2\alpha$. These statements follow directly by comparing the three divisor classes of the monomial. Write $\ell=\varepsilon+2r$ and replace $x^{2r}$ by $(UVa-tG_0)^r$. Then $K=i+c+r$.

First suppose $K\le2m$. Terms with $r=0$ already span the space in the statement. To see this, expand a section of $\mathcal O(n,K)$ as $$\sum_i f_i(U,V)a^{K-i}t^i.$$ Vanishing to order $k$ at both $P,Q$ is equivalent to divisibility of $f_i$ by $(UV)^{\max(k-i,0)}$. This is precisely the inequality $c\le m+\alpha$ above. Every term with $r>0$ satisfies the same vanishing condition and adds nothing outside this space.

If $K>2m$, the inequality $i+c\le2m$ forces $r\ge K-2m=h$. Remove the common factor $(UVa-tG_0)^h$. The remaining fibre degree is $2m$, the base degree is $n-2h$, and the required vanishing at both points is $m$. The preceding argument applies again. Thus the displayed space is contained in the actual refined piece. We check equality by counting dimensions in all degrees.

For $d,e_0\ge0$ and $e_0\ge k$, the binary expansion used above gives $$h^0(\mathcal O(d,e_0)\otimes\mathcal I_P^k\mathcal I_Q^k) =(d+1)(e_0-k+1)+r(d-r),\qquad r=\min(k,\lfloor d/2\rfloor).$$ Combine the two parity rows, and put $n=\lfloor m/2\rfloor$. Their dimensions at fixed $K$ are $$\begin{array}{c|c} 0\le K\le m &(2m+1)(K+1)\\ K=m+k,\ 1\le k\le m &(2m+1)(m+1)+r(2m-1-2r),\quad r=\min(k,n)\\ K=2m+h,\ 1\le h\le n &(2d+1)(m+1)+d(d-1)/2,\quad d=m-2h. \end{array}$$ Summing the three rows gives $(2m+1)(2m^2+2m+1)$, as one checks separately for $m=2n$ and $m=2n+1$. Riemann–Roch and Kodaira vanishing give the same value for $h^0(X,-mK_X)$. Since associated graded spaces preserve dimension, all the inclusions are equalities. ◻

Proof. Put $u=K/m$. The limiting fixed divisor is $h\widetilde D$, and the remaining class is $L_u=dB+eT-k(E_1+E_2)$, with $$\begin{array}{c|cccc} u&d&e&k&h\\ \hline {[0,1]}&1&u&0&0\\ {[1,2]}&1&u&u-1&0\\ {[2,5/2]}&5-2u&2&1&u-2. \end{array}$$ Its negative part is $rC$, where $r=\max(0,k-d/2)$, and its positive part has square $$V(u)=2de-2k^2+2r^2.$$ Indeed its intersections with $E_i$, $B-E_i$ and $C$ are $k-r$, $e-k$ and $d-2k+2r$, respectively, and are nonnegative.

There are two parity rows in Lemma 126.3. Their leading terms agree, and $$\int_0^{5/2}V(u)\,du=4,$$ which agrees with $h^0(X,-mK_X)=4m^3+O(m^2)$. The weight divided by $m$ is $2u$. Consequently $$S_X(G)=\frac14\int_0^{5/2}2uV(u)\,du=\frac{67}{24}.$$ All contributions of the parity factor are of lower order. ◻

Proof. Use the fibre $F=B$ through $q$. It avoids $P,Q$. Subtract $vF$ from $L_u$, and write $b=d-v$, with $0\le b\le d$. Its negative part is $r_bC$, where $r_b=\max(0,k-b/2)$. The positive square and its degree on $F$ are $$V_F=2be-2k^2+2r_b^2,\qquad l=e-r_b.$$ The fixed divisor restricted to $F$ is $r_b(F\cap T_0)+h(F\cap D)$. These two points are distinct. Therefore $$S(W^G;F)=\frac14\int\!\int_0^d V_F\,db\,du=\frac{11}{24}.$$ At a point outside $D$, the subsequent point moment is bounded by $$\frac12\int\!\int_0^d l\left(n_q+\frac l2\right)db\,du,$$ where $n_q=r_b$ at $F\cap T_0$ and is zero otherwise. These integrals are $5/6$ and $69/128$, respectively. For clarity, the chamber contributions, with the normalizing factors included, are $$\begin{array}{c|c|ccc} u&b&S(W^G;F)&n_q=0&n_q=r_b\\ \hline {[0,1]}&[0,1]&1/8&1/12&1/12\\ {[1,3/2]}&[0,2u-2]&3/64&11/128&7/64\\ {[1,3/2]}&[2u-2,1]&3/32&11/128&11/128\\ {[3/2,2]}&[0,1]&7/48&19/96&67/192\\ {[2,5/2]}&[0,5-2u]&3/64&11/128&79/384. \end{array}$$ The point bounds follow by replacing the restricted series by the complete series of its degree; this gives an upper bound for its maximal total vanishing. Boundary and rounding terms have lower order. Since $F$ has log discrepancy one and $\Delta|_F$ has no support at $q$, the surface flag inequality gives $$\delta_q(G,\Delta;W^G)\ge \min\{24/11,6/5,128/69\}=6/5.$$ ◻

Proof. Use $D$ as the first member of the flag. Its log discrepancy is $1/2$, and its adjoint boundary is zero. Set $\tau=\min(d/2,e)$. For $0\le v\le\min(k,\tau)$, the negative part of $L_u-v\widetilde D$ remains $rC$, since $\widetilde D\cdot C=0$. Its positive square and degree on $\widetilde D$ are $$V_D=2(d-2v)(e-v)-2(k-v)^2+2r^2,\qquad l_D=d+2e-2k-4v.$$ If $k\le v\le\tau$, which is possible only when $r=0$, the negative part is $(v-k)(E_1+E_2)$, and $$V_D=2(d-2v)(e-v),\qquad l_D=d+2e-4v.$$ The positive parts are nef by the intersections with the five negative curves listed above. The threshold is $\tau$; at its endpoint the positive part has square zero, and beyond it a nef ruling class has negative intersection.

The negative curves meet $\widetilde D$ only at $P,Q$, with $C$ disjoint from $\widetilde D$. Thus there is no fixed point contribution at $q$. The first moment and the following point moment satisfy $$S(W^G;D)=\frac14\int\left(hV(u)+\int_0^\tau V_D\,dv\right)du =\frac{11}{48},$$ $$S(W^{G,D};q)\le\frac14\int\!\int_0^\tau l_D^2\,dv\,du =\frac{175}{192}.$$ Here are the integrations, including the factor $1/4$: $$\begin{array}{c|c|cc} u&v&\int V_D/4&\int l_D^2/4\\ \hline {[0,1/2]}&[0,u]&1/128&7/192\\ {[1/2,1]}&[0,1/2]&7/192&1/6\\ {[1,3/2]}&[0,u-1]&3/64&17/96\\ {[1,3/2]}&[u-1,1/2]&7/384&25/192\\ {[3/2,2]}&[0,1/2]&7/96&13/48\\ {[2,5/2]}&[0,(5-2u)/2]&3/128&25/192. \end{array}$$ The fixed contribution $\frac14\int hV(u)\,du$ is $3/128$. The complete restricted series again bounds the point moment from above. Applying the flag inequality gives $$\delta_q(G,\Delta;W^G)\ge \min\left\{\frac{1/2}{11/48},\frac{192}{175}\right\} =\frac{192}{175}.$$ ◻

Proof of Theorem 126.1. Every point of $G$ above $p$ is covered by one of the preceding two lemmas. Abban–Zhuang adjunction for the weighted curve extraction therefore gives $$\delta_p(X)\ge \min\left\{\frac{3}{67/24},\frac{192}{175}\right\} =\frac{72}{67}>1.$$ We have not computed a Zariski decomposition on the weighted extracted threefold. The actual refined spaces were identified from its graded section algebra. All divisor decompositions used afterwards are on the explicitly described surface $S$. ◻