The exceptional ruling over an index-two point

Proof. The anticanonical Hilbert polynomial and its total weight are $$N_m=(2m+1)(2m^2+2m+1),\qquad W_m=\frac{m(2m+1)(16m^2+15m+8)}3.$$ Here the target affine coordinate has weight two and the three plane coordinates have weights $0,1,2$. Both generators of the centre have weight two. The resolution of its $m$-th ideal power has terms $$\bigoplus_{i=0}^{m-1}H^0\bigl(\mathcal O(2m-i-1,m-2)\bigr) \longrightarrow \bigoplus_{i=0}^{m}H^0\bigl(\mathcal O(2m-i,m)\bigr).$$ The average weights in these summands are respectively $3m-i-1$ and $3m-i$, including the weight-two shift in the first term. Their weighted dimension difference gives the displayed polynomial. The initial equations form a regular sequence, so the same resolution computes the associated graded spaces. Dividing $W_m$ by $mN_m$ and passing to the limit gives the assertion. ◻

Proof. Let $q$ be the conic equation and let $F$ be a quadratic lift of $uf+vg$. These generate the ideal of the centre. Twisting the standard resolution of its $m$-th power by $\mathcal O(2m,3m)$ shows that every section of $-mK_X$ is a sum of $$q^{m-i}F^i H^0\bigl(\mathcal O(2m-i,m)\bigr),\qquad 0\le i\le m.$$ There is no obstruction to taking global sections: the first cohomology of each term $\mathcal O(2m-i-1,m-2)$ is zero. Give the target affine coordinate weight two and the plane coordinates weights $0,1,2$. The initial forms of $q,F$ are $Y_0Y_2-Y_1^2$ and $uY_1^2+vY_0^2$, a regular sequence. The initial complex of the ideal-power resolution is therefore again exact. Consequently the filtered resolution is strict and the associated graded section space is generated by the sections of this initial ideal power. This argument accounts for all cancellations, not just the initial forms of degree-one sections. It also gives the geometric valuation filtration. On $u=Y_0=1$, after inverting $z=c-x^2$, the initial equation is $a(c-x^2)=x^2+t$; substituting $t=az-x^2$ identifies its ring with $\mathbb C[a,x,z,z^{-1}]$. This is injective on the initial section spaces, so a nonzero global initial section cannot acquire extra order at the generic point of the exceptional surface. After division by $q^m$, its sections have the form $$x^\epsilon a^iX^rC^sT^l,\qquad r+s+l=K,\quad 2r+s+\epsilon\le m,\quad 0\le i\le\min(m,2m-l).$$ These inequalities imply $l\ge h$ and weighted vanishing at least $2K-n$ at $P$. After removing $T^h$ this is exactly the condition in (124.1). At $Q$, homogenizing $a^i$ as a section of $mB$ gives vanishing at least $m-i+r+s\ge K-m$. This proves the inclusion.

We compute the asymptotic dimension. Resolve the weighted point $P$ by two ordinary blow-ups and blow up $Q$. Equivalently, start with $\mathbb P^2$ and blow up the point defining $\mathbb F_1$, then $P$, the point infinitely near $P$ in the direction $T$, and finally $Q$. Write $H,e_1,e_2,e_3,e_4$ for the standard classes. Thus $$\xi=H,\quad B=H-e_1,\quad T=H-e_2-e_3,\quad C=H-e_2-e_4, \quad V=H-e_1-e_4,\quad R=e_2-e_3.$$ Ignoring bounded rounding terms and writing $k=K/m$, the divisor after removing the fixed power of $T$ is $$D_k=\begin{cases} (k+1)H-e_1,&0\le k\le1/2,\\ (k+1)H-e_1-(k-1/2)(e_2+e_3),&1/2\le k\le1,\\ 2H-e_1-\frac12(e_2+e_3)-(k-1)e_4,&1\le k\le3. \end{cases}$$ Its negative part is zero up to $k=2$, is $(k-2)V$ for $2\le k\le5/2$, and is $$(k-2)V+(2k-5)C+(k-5/2)R$$ for $5/2\le k\le3$. The positive parts are nef: this follows by intersection with $$e_1,e_3,e_4,R, H-e_1-e_2, V, T, C,$$ which are precisely the negative curves of this weak del Pezzo surface. The volumes on the five successive intervals are $$k(k+2),\quad\frac{-2k^2+8k-1}{2},\quad \frac{-2k^2+4k+3}{2},\quad\frac{11-4k}{2},\quad 2(k-3)^2.$$ Their integrals are $7/24,23/24,13/6,1/2,1/12$, whose sum is four. The two parity choices cancel the factor $1/2$ in the surface Hilbert polynomial. This proves the dimension assertion. As a check on both the grading and the normalization, $$\frac14\int_0^3 2k\operatorname{vol}(D_k)\,dk=\frac83,$$ in agreement with Lemma 124.2.

For a fixed valuation, extend a compatible basis of each actual piece to a compatible basis of the containing space in (124.1). All the vanishing orders are nonnegative. Thus the sum for the smaller space is no greater than the sum for the larger space. Since their total dimensions have ratio tending to one, the same comparison holds for expected vanishing orders. Notice that finite-degree equality of the two spaces is not required.

We also record why the containing spaces are a multigraded series. Before factoring $T^h$, their class is $mB+(j/2)\xi$ and their conditions are the homogeneous inequalities $$v_T\ge j/2-m,\qquad v_{P,(1,2)}\ge j-m, \qquad v_Q\ge j/2-m.$$ Here $v_T(x)=v_{P,(1,2)}(x)=0$ and $v_Q(x)=1/2$. The factor $x^\epsilon$ and the rounding of these inequalities give exactly $h,A,b$ above. These conditions are preserved under multiplication. The support is bounded by $0\le j\le6m$; on any rational ray $0<j/(2m)<1/2$ the even pieces are the complete series of the ample class $mB+(j/2)\xi$. One may clear the denominator by a Veronese grading; the odd pieces have the same leading moments. Thus the multigraded Abban–Zhuang estimates apply to the containing series as well. ◻

Proof. The point $P$ does not lie on $D_x$. Extract it with weights $(1,2)$ on $(C,T)$. Its exceptional curve is the image of $e_3$ on the resolution above. Its log discrepancy is three. The exceptional curve meets the contracted curve $R$ at a point $q_R$; the different there is $1/2$. Let $q_T$ be its intersection with the strict transform of $T$.

For clarity we give the complete numerical computation. Write $P(k,v)+N(k,v)$ for the Zariski decomposition of $D_k-ve_3$. The negative parts on its ten chambers are listed below; each chamber is the convex hull of the indicated vertices in the $(k,v)$-plane. Put $U=H-e_1-e_2$.

The support matrices are negative definite, and subtracting the listed parts gives a nef divisor on each chamber. Thus the table also verifies the decomposition directly.

The normalization is $1/4$. The fixed order along $e_3$ is $(2k-1)_+$. Integration gives $$\frac14\int\!\!\int P(k,v)^2\,dv\,dk=\frac{295}{384},\qquad \frac14\int_0^3(2k-1)_+\operatorname{vol}(D_k)\,dk =\frac{649}{384}.$$ Their sum is $S(W;e_3)=59/24$. The moving point moment is $$\frac14\int\!\!\int(P(k,v)\cdot e_3)^2\,dv\,dk =\frac{43}{128}.$$ At $q_R$ the additional fixed coefficient before contracting $R$ is $N_R(k,v)+(k-1/2)_+$; at $q_T$ it is $N_T(k,v)+(k-1)_+$. The corresponding additional moments are $165/128$ and $191/384$. Hence the uncontracted point moments are $13/8$ and $5/6$.

When $R$ is contracted, the pullback of the exceptional curve is $e_3+\frac12R$. Thus its refinement at $q_R$ subtracts one half of the first expected order. The actual point moment there is $$\frac{13}{8}-\frac12\frac{59}{24}=\frac{19}{48}.$$ Taking account of the different, the flag bound is $$\min\left\{\frac{3}{59/24},\frac{1/2}{19/48}, \frac1{5/6},\frac1{43/128}\right\}=\frac65.$$ This proves the lemma. ◻

Proof. We give both fibre calculations, since the fibre through $P$ has a different strict transform. On the smooth surface $R$, write $$B=h-e_1,\quad T=h-e_2-e_3,\quad C=h-e_2-e_4,\quad V=h-e_1-e_4,\quad R_0=e_2-e_3.$$ Its negative curves are $$e_1,e_3,e_4,h-e_1-e_2,V,T,C,R_0.$$ The first seven have self-intersection $-1$, and $R_0^2=-2$. They generate the effective cone. The different is half the strict transform of the curve $D_x=h-e_4$.

For the limiting weight $u=K/m$, put $$D_u=\begin{cases} (u+1)h-e_1,&0\le u\le1/2,\\ (u+1)h-e_1-(u-1/2)(e_2+e_3),&1/2\le u\le1,\\ 2h-e_1-\tfrac12(e_2+e_3)-(u-1)e_4,&1\le u\le3. \end{cases}$$ There is also the fixed divisor $(u-1)_+T$, where $a_+=\max(a,0)$.

First take a fibre $F$ whose base point differs from $0,\infty,-1$. Its strict transform has class $B$. For $D_u-vB$, put $$r=(v-1)_+,\quad s=(v-\max(1,2-u))_+,\quad c=(2u+2v-5)_+,\quad b=(u-2)_+.$$ The negative and positive parts are $$N=r e_1+sT+bV+cC+\frac c2R_0,\qquad P=D_u-vB-N.$$ The threshold is $$\tau(u)=\begin{cases} 1+u,&0\le u\le1/2,\\ 3/2,&1/2\le u\le1,\\ 2-u/2,&1\le u\le2,\\ 3-u,&2\le u\le3. \end{cases}$$ These formulas follow by imposing zero intersection with the support of $N$. On each resulting chamber the support matrix is negative definite and $P$ has nonnegative intersection with all eight negative curves. At $v=\tau(u)$ its square is zero; a nef ruling class excludes larger values.

Put $l=P\cdot B$. Integration gives $$\frac14\int_0^3\int_0^{\tau(u)}P^2\,dv\,du=\frac{69}{128}, \qquad \frac14\int_0^3\int_0^{\tau(u)}l^2\,dv\,du=\frac{79}{192}.$$ The first number is $S(W^G;F)$ and the second bounds the moving point moment. The extra point contributions, obtained by integrating $l$ times the fixed coefficient with factor $1/2$, are $$\begin{array}{c|c|c} \text{point on }F&\text{fixed coefficient}&\text{extra contribution}\\ \hline F\cap e_1&r&7/384\\ F\cap T&s+(u-1)_+&57/128\\ F\cap C&c&3/64. \end{array}$$ The other negative curves do not meet this fibre. Thus the point bounds are $55/128$, $329/384$, $11/24$ at these three points, and $79/192$ elsewhere. These points are distinct: any coincidence would lie over $0$, $\infty$ or $-1$. The point $F\cap D_x$ is also distinct from them and has log discrepancy $1/2$ on the adjoint curve, whereas its moment is $79/192$. Therefore the flag inequality gives $$\delta_q(G,D_x/2;W^G)\ge \min\left\{\frac{128}{69},\frac{384}{329},\frac{96}{79}\right\} =\frac{384}{329}.$$

Now let the base point be $-1$. The strict transform of its fibre is $F=h-e_1-e_2$. Set $k_0=\min((u-1/2)_+,1/2)$ and $$r=(v-1)_+,\quad s=(v-2k_0)_+,\quad c=(2u+v-5)_+,\quad b=(u-2)_+.$$ This time the Zariski decomposition is $$N=r e_1+s e_3+bV+cC+\frac{v+s+c}{2}R_0, \qquad P=D_u-vF-N,$$ with threshold $\tau(u)=1+u$ for $0\le u\le1$, and $\tau(u)=3-u$ for $1\le u\le3$. The same intersection checks verify this decomposition. Writing $l=P\cdot F$, one obtains $$\frac14\int\!\int P^2\,dv\,du=\frac{67}{96},\qquad \frac14\int\!\int l^2\,dv\,du=\frac{11}{32},\qquad \frac12\int\!\int lr\,dv\,du=\frac1{16}.$$ Thus the point moment is at most $13/32$ at $F\cap e_1$ and $11/32$ elsewhere, except at $F\cap R_0$, which maps to the excluded point $P$. The fixed curve $T$ is disjoint from this strict transform. At $F\cap D_x$ the moment $11/32$ is below the adjoint log discrepancy $1/2$. Consequently $$\delta_q(G,D_x/2;W^G)\ge \min\left\{\frac{96}{67},\frac{32}{13},\frac{16}{11}\right\} =\frac{96}{67}>\frac{384}{329}.$$ All displayed integrals are over $0\le u\le3$ and $0\le v\le\tau(u)$. Splitting at the zeros of the affine functions inside $(\ )_+$ gives ordinary integrals of quadratic polynomials. The complete series of degree $ml+O(1)$ on the flag curve bounds the maximal point vanishing from above. Hence the computation applies also if a restriction map is not surjective; the rounding terms have lower order. ◻

Proof of Theorem 124.1. The excluded points on $\Gamma$ have coordinates $a=\infty$ and $a=0$, respectively. For every other base point, the fibre flags of Lemma 124.5 cover all points of the extracted surface except possibly $P$. At $P$ use Lemma 124.4. The domination in Lemma 124.3 applies to every divisorial valuation, so these bounds hold for the actual refined series. Consequently its local thresholds over $p$ are at least $384/329$. Lemma 120.3 and Lemma 124.2 now give $$\delta_p(X)\ge \min\left\{\frac{3}{8/3},\frac{384}{329}\right\} =\frac98.$$ The only decompositions in this argument occur on the explicitly blown-up exceptional surface. ◻