The refined estimate on the exceptional surface

Proof. Use the second weights $(r,s,a,b,z)=(1,0,2,0,1)$. Put $A=as^2$, $B=br^2$ and write $j=2h+\varepsilon$, where $\varepsilon\in\{0,1\}$. The primary weight is $2N+\varepsilon$, where $N=\alpha+\beta+h$; the second weight is $Q=i+j+2\alpha$. Set $$R=Q-\varepsilon-2N,\qquad S=m-Q,\qquad d=N-K-L,$$ $$K=\max\left\{0,\left\lceil\frac{Q-m}{2}\right\rceil,Q-2m\right\},\quad L=\max\left\{0,\left\lceil\frac{2N+\varepsilon-Q}{2}\right\rceil\right\}.$$ The corresponding space is zero unless $L\le m$ and $d\ge0$. Otherwise it is exactly $$z^\varepsilon r^R s^S A^K B^L\operatorname{Sym}^d\langle A,B\rangle.$$ Indeed, the original inequalities become $\alpha\ge K$, $\beta\ge L$, $\alpha+\beta\le N$, and $\beta\le m$. Fixing $\beta=L$ and varying $\alpha$ gives the basis $A^i(A+B)^{d-i}$ of the remaining binary space. The integers $R,S$ may be negative, but the factors $A^KB^L$ make every displayed section regular. In particular, this computes the entire space, including all cancellations.

For the asymptotic sums, put $q=Q/m$, $n=N/m$, $$k(q)=\max\{0,(q-1)/2,q-2\},\quad l(n,q)=\max\{0,n-q/2\},\quad d(n,q)=\min\{n,q/2\}-k(q).$$ The domain is $0\le q\le4$, $k(q)\le n\le q/2+1$. Splitting at $q=1,3$ and $n=q/2$ gives six elementary integrals. Their total mass is $\int d\,dn\,dq=11/6$. Since the two parities have the same leading term, normalized moments are $\frac6{11}\int d(\cdots)\,dn\,dq$. The bounds for the coefficients of the invariant curves are as follows: $$\begin{array}{c|c|c} \text{curve}&\text{integrand after the factor }d&\text{moment}\\\hline r=0&q-2n+2l+d&21/44\\ s=0&1-q+2k+d&23/44\\ a=0&k+d/2&67/88\\ b=0&l+d/2&109/176\\ A+\kappa B=0\ (\kappa\ne0)&d/2&37/176 \end{array}$$ The last row is the maximal average order of a basis of binary forms of degree $d$ at one prescribed point. At $D$ there is also the odd factor $z$, whose contribution is $O(1/m)$.

The curves $A+\kappa B=0$ are smooth. They meet only at $P=(r=a=0)$ and $P'=(s=b=0)$, with contact order two. At $P$, blow up with weights $(1,2)$ in $(r,a)$; at $P'$, use weights $(1,2)$ in $(s,b)$. The total moments of these valuations are $$\frac6{11}\int qd\,dn\,dq=2,\qquad \frac6{11}\int(1+2n-q)d\,dn\,dq=\frac{155}{88}.$$ These are moments of the whole basis divisor; one must not add the separate maxima in the table. The boundary $\frac12D$ has value $1$ for either weighted valuation, whose log discrepancy on $G$ is $3$. Thus the two required weighted inequalities hold. All curve coefficients are strictly below one, even after adding $\frac12D$. The ordinary blow-up inequality follows from the weighted inequality and the bound one for the coefficient of the transverse coordinate line. The displayed blow-ups resolve the arrangement, so the resulting pairs are log canonical.

For precision at the equality at $P$, the sum of the second weights is exactly $2mh_m$ for every $m$. Indeed, the mean weights of the hyperplane, line-pencil, and quadric-pencil coordinates are respectively $1,1,0$. Both equations have weight $2$, equal to the mean weights of their divisor classes. Every shifted term of the Koszul character therefore has mean $2m$, proving the identity. Thus there is no positive error at this corner. At the other corner and along curves all inequalities have a fixed positive margin for large $m$. Applying this to bases compatible with an arbitrary divisorial valuation and the second weight filtration, separately in each primary graded piece, and using semicontinuity of log canonical thresholds, proves the assertion. ◻

Proof. Suppose first that $f$ is a square. After rescaling, $f=s^2$. The coefficient of $r^2$ in $g$ is nonzero by coprimality. Taking the maximum initial form for the weights $(r,s,a,b,z)=(1,0,2,0,1)$ gives $z^2+as^2+br^2$, after rescaling $b$. Suppose instead that $g$ is a square. Take $g=r^2$. Coprimality gives $d\ne0$. The minimum initial form for the same weights is $z^2+ad s^2+br^2$; the omitted term $acr s$ has weight three. Rescale $a$.

In both cases the action preserves the Cox grading. The initial equation is prime, so localization introduces no extra initial relations. Its different remains the smooth curve of bidegree $(2,1)$ with coefficient $1/2$. The preceding lemma applies to the complete initial series. Semicontinuity of log canonical thresholds, with bases chosen separately in each primary graded piece, proves the assertion. ◻

Proof. A spanning monomial has the form $$r^i z^{2h+\varepsilon}s^k a^\alpha b^\beta, \qquad i+2h+\varepsilon+k=m,\quad 0\le\alpha\le m+k,\quad0\le\beta\le m,$$ where $\varepsilon=0$ or $1$. Put $n=\alpha+\beta+h$ and replace $z^{2h}$ by $(-af-bg)^h$. The remaining polynomial has bidegree $(m-\varepsilon,n)$. At either $p_1$ or $p_2$ its order is at least $\alpha+h=n-\beta\ge n-m$. At $p_3$ its order is at least $k+\beta+h=n-\alpha+k\ge n-m$. At $p_4$ its order is at least $\beta+h=n-\alpha\ge n-2m$. These inequalities are preserved by taking linear combinations, proving the inclusions.

Here the inclusions are equalities, as can be seen without any asymptotic argument. For $\varepsilon\in\{0,1\}$, the actual piece is contained in the direct sum $$\bigoplus_{k=0}^n a^{n-k}b^k g^{(k-m)_+}s^{(n-m-k)_+} \operatorname{Sym}^{m-\varepsilon-2(k-m)_+-(n-m-k)_+}\langle r,s\rangle,$$ where a summand with negative exponent is zero. Indeed, in expanding $(af+bg)^h$, suppose that $j$ factors contribute $af$. The final exponent of $b$ is $k=\beta+h-j$. The coefficient contains $g^{h-j}$ and $s^{k_0+j}$, where $k_0$ is the original exponent of $s$. Since $\beta\le m$ and $\alpha\le m+k_0$, these powers are at least $(k-m)_+$ and $(n-m-k)_+$ respectively.

Conversely, the displayed direct sum is the whole fat-point space. The two points on $a=0$ impose the power of $g$, and the point $b=s=0$ imposes the power of $s$. The fourth condition is automatic: if $k<n-2m$, the required power of $s$ already exceeds the horizontal degree. This description also makes sense when $g$ has a double root, with its full scheme-theoretic power retained.

Put $a'=(k-m)_+$ and $b'=(n-m-k)_+$. For $a'=0$ there are $m+1$ choices of $k$, and for $a'>0$ there is one; the same statement holds for the other exponent. Summing dimensions over all $n$ and both parities gives $$\begin{aligned} &(m+1)^2(2m+1) +(m+1)\frac{m(m-1)}2+(m+1)m^2+{m\choose3}\\ &\quad =\frac{(2m+1)(11m^2+11m+6)}6=h^0(X,-mK_X). \end{aligned}$$ For example, the last term is $\sum_{a'=1}^{\lfloor m/2\rfloor}(m-2a')^2={m\choose3}$. Since every actual piece is contained in its displayed space and the sums of their dimensions agree, all inclusions are equalities. In particular, this accounts for every possible cancellation among the original generators.

Set $$D_t=A+tB-(t-1)_+(e_1+e_2+e_3)-(t-2)_+e_4,$$ and put $F_1=B-e_1-e_2$, $F_2=B-e_3-e_4$, and $R=A-e_3$. The negative part $N_t$ of $D_t=P_t+N_t$ and the square of its positive part are $$\begin{array}{c|c|c} t&N_t&P_t^2\\\hline {[0,1]}&0&2t\\ {[1,3/2]}&0&2t-3(t-1)^2\\ {[3/2,2]}&(t-3/2)F_1&5/2-(t-1)^2\\ {[2,5/2]}&(t-3/2)F_1+(t-2)F_2&11/2-2t\\ {[5/2,3]}&(2t-4)F_1+(t-2)F_2+(2t-5)R&2(3-t)^2. \end{array}$$ Here the negative curves in each indicated support have negative definite intersection matrix and $P_t$ is orthogonal to that support. To check nefness it is enough, by linearity in each chamber, to check the endpoints. The only endpoint requiring explanation is $$P_{5/2}=A+B-\frac12(e_1+e_2)-e_3.$$ For an irreducible curve $\Gamma=uA+vB-\sum m_i e_i$ different from the exceptional curves and the indicated fibres, intersection with $F_1$ gives $m_1+m_2\le u$, and intersection with the horizontal fibre through $p_3$ gives $m_3\le v$. Hence $$P_{5/2}\cdot\Gamma=u+v-\frac12(m_1+m_2)-m_3\ge\frac u2\ge0.$$ The excluded fibres and exceptional curves are checked directly. The other endpoints are $A$, $A+B$, $P_2=P_{5/2}+\frac12B$, and $P_{3/2}=P_2+\frac12e_3$; for the last one its intersection with $e_3$ is $1/2$, and all other intersections increase. Finally, $$P_t=(3-t)(2A+2B-e_1-e_2-2e_3)\quad\text{for }5/2\le t\le3.$$ The divisor $M=2P_{5/2}$ is nef and $D_t\cdot M=6-2t$ for $t\ge2$. Thus no $D_t$ with $t>3$ is pseudoeffective.

Asymptotic Riemann–Roch in these finitely many rational chambers gives $$h^0(T,mD_{n/m})=\frac{m^2}{2}P_{n/m}^2+O(m),$$ uniformly for $0\le n\le3m$. Replacing $mA$ by $(m-1)A$ changes the total over $n$ by $O(m^2)$. Direct integration yields $$\int_0^3P_t^2\,dt=\frac{11}{3},\qquad \frac3{11}\int_0^3 2tP_t^2\,dt=\frac{243}{88}.$$ There are two parities. Their sum therefore has leading term $\frac{11}{3}m^3$, as claimed. ◻

Proof. The restriction to $D$ of the first projection is an isomorphism. Thus no two of the four points lie on a fibre of the first projection. The displayed intersection numbers follow from $a^2=b^2=0$, $ab=1$, $e_i^2=-1$. To check the list, consider an irreducible curve other than the ten listed curves, of class $da+eb-\sum m_i e_i$. Intersection with $h_1$ and $h_2$ gives $m_1+m_2\le d$ and $m_3+m_4\le d$. Consequently $$-K_S\cdot\Gamma=2d+2e-\sum m_i\ge2e.$$ The pencil generated by the two products defining $D$ becomes base-point-free on $S$, so $\widetilde D$ is nef. Hence $-K_S=\widetilde D+b$ is nef, and $(-K_S)^2=4$. Thus every negative curve is a $(-1)$-curve or a $(-2)$-curve. The inequality forces $e=0$ for any additional negative curve. Its image would then be a fibre of the first projection, which gives one of the already listed $v_i$. This proves the assertion. We also record a direct nefness test used below. If $P=xa+yb-\sum z_i e_i$ has nonnegative intersection with these ten curves and with $a,b$, then $$z_i\ge0,\qquad y\ge z_i,\qquad x\ge z_1+z_2,\qquad x\ge z_3+z_4.$$ Thus $\sum z_i^2\le y\sum z_i\le2xy$, and $P$ lies in the closure of the positive cone. The same argument shows that every irreducible curve other than the ten negative curves has nonnegative square and lies in that closure. The Hodge index theorem therefore gives $P\Gamma\ge0$ for every such curve. This proves that $P$ is nef without any further assertion about the Mori cone. Finally, each fibre of the second projection meets $D$ in two distinct points and hence transversally. The first projection restricts isomorphically to $D$, so its fibres are transverse to $D$ as well. Thus the three tangent directions at each $p_i$ are pairwise distinct. ◻

Proof. For $i=1,3,4$ write $$L_t-we_i=P_i(t,w)+R_i(t,w)$$ for its Zariski decomposition, and put $d_i(t,w)=P_i(t,w)e_i$. Let $r_i(t,w)$ and $s_i(t,w)$ be the coefficients of $v_i$ and $h_j$ in $R_i(t,w)$, respectively. No other component of the negative part meets $e_i$. The formulae for the moments are $$\begin{aligned} S(e_i)&=\frac6{11}\iint (w+c_i(t))d_i(t,w)\,dw\,dt,\\ S_\mathrm{free}(e_i)&=\frac6{11}\iint\frac{d_i(t,w)^2}{2}\,dw\,dt,\\ S_{v_i}(e_i)&=\frac6{11}\iint\left(\frac{d_i(t,w)^2}{2} +r_i(t,w)d_i(t,w)\right)\,dw\,dt,\\ S_{h_j}(e_i)&=\frac6{11}\iint\left(\frac{d_i(t,w)^2}{2} +s_i(t,w)d_i(t,w)\right)\,dw\,dt, \end{aligned}$$ where $c_1=c_3=M$ and $c_4=N$. The integration is over the effective range $w\ge0$, $0\le t\le3$; pieces with $d_i=0$ make no contribution. All these integrals are integrals of quadratic polynomials over rational polygons. Their unnormalized values are $$\begin{array}{c|c|c|c|c|c} i&\iint d_i&\iint(w+c_i)d_i&\iint d_i^2/2& \iint r_i d_i&\iint s_i d_i\\ \hline 1&11/6&71/32&59/96&5/24&25/32\\ 3&11/6&67/32&2/3&7/24&15/32\\ 4&11/6&61/32&17/32&23/96&29/48 \end{array}$$ The following tables give the chamber polygons and the three linear functions $d_i,r_i,s_i$. Each polygon is the convex hull of the listed vertices $(t,w)$; its boundary has measure zero and may belong to either adjacent chamber.

These tables are obtained directly from the intersection matrix in Lemma 106.6: for a proposed negative support $C_1,\ldots,C_k$, solve $$\sum_j n_j(C_jC_l)=(L_t-we_i)C_l,$$ then impose $n_j\ge0$ and $(L_t-we_i-\sum n_jC_j)C\ge0$ for each of the ten negative curves. These inequalities give exactly the polygons in the tables. For example, the chamber $0\le w\le t\le1$ has no negative part and $d_i=w$; it contributes $\int_0^1\int_0^t w^2/2\,dw\,dt=1/24$ to the free-point integral.

The log discrepancy of $e_i$ over $(G,D/2)$ is $3/2$. At the point $\widetilde D\cap e_i$, adjunction gives boundary coefficient $1/2$, and the relevant refined moment is the free-point entry of the table, which is strictly smaller than $1/2$. At every other point of $e_i$ the adjunction boundary is zero, and all refined moments are strictly smaller than one. The first moments are strictly smaller than $3/2$. The Abban–Zhuang inequality therefore proves the last assertion. ◻

Proof. Write $L_t-w\widetilde D=P(t,w)+R(t,w)$ and $d=P(t,w)\widetilde D$. The seven chamber polygons and their linear functions are as follows.

The only components of $R$ meeting $\widetilde D$ are the exceptional curves $e_i$, so they produce no fixed part at an unmarked point. Direct integration gives $$\iint d=\frac{11}{6},\qquad \iint wd=\frac{37}{96},\qquad \iint\frac{d^2}{2}=\frac{27}{16}.$$ Multiplying by $6/11$ gives the asserted moments. Since $A_{G,D/2}(D)=1/2$, both terms in the resulting Abban–Zhuang minimum are strictly greater than one. ◻

Proof. Use the horizontal fibre through $p$. If it is different from the two marked fibres, its strict transform has class $b$. For the two marked fibres use $\alpha$ and $\beta$, respectively. Write $P(t,v)+N(t,v)$ for the Zariski decomposition of $T_t-vC$, where $C$ is this flag curve. The following table gives the integrated curve and point moments. All entries already include the factor $3/11$. $$\begin{array}{c|c|c|c} C&\text{curve moment}&\text{point}&\text{point moment}\\ \hline b&115/264&\text{general}&31/66\\ &&J_1,J_2&125/264\\ &&J_3&23/44\\ &&J_4&31/66\\ \hline \alpha&67/88&\text{general}&37/88\\ &&J_3&23/44\\ &&J_4&37/88\\ \hline \beta&109/176&\text{general}&37/88\\ &&J_1,J_2&79/176 \end{array}$$ Here an entry $J_i$ means the point where the flag meets $J_i$. Intersections with exceptional curves are omitted from the last two blocks because they lie over the marked points, all of which belong to $D$ and are excluded from the statement.

For completeness, the chamber data for this calculation are given below. A polygon is specified by its vertices $(t,v)$; its listed negative part determines the positive part as $P=T_t-vC-N$. On every listed polygon the coefficients of $N$ are nonnegative, its support has negative definite intersection matrix, $P$ is nef, and $P\cdot N=0$. These assertions can be checked against the ten negative curves listed above. The polygons cover the big region; outside them the volume is zero. Integrate $$P^2\quad\text{for the curve moment},\qquad (P\cdot C)^2+2(P\cdot C)\operatorname{ord}_p(N|_C) \quad\text{for the point moment}.$$ In the second expression a component equal to $C$ is omitted from $N$: its chamber has $P\cdot C=0$ and makes no point contribution.

The point moment bounds the coefficient at $p$ of the normalized restricted basis divisor on $C$. Choose the finite-level bases simultaneously adapted to the tested valuation and to $C$, and apply inversion of adjunction as in Proposition 104.4. This bounds the quotient of expected vanishing by discrepancy by the larger of the curve and point moments. The branch curve is absent from a neighbourhood of $p$, so the discrepancy of $(G,\tfrac12D)$ there is the ordinary surface discrepancy. The largest relevant entry is $67/88$. ◻