Section 95

The exceptional surface at an index-four point

Part XIII · Family 3.5: K-semistability and the complete smooth classificationPDF page 321

Let $p$ be the $A_3$ point of a fibre whose ramification index is four. We extract $p$ with weights $(1,2,2)$ . Its exceptional pair is $(G,\Delta)=(\mathbb P^2,\Lambda/2)$ , where $\Lambda$ is a line. Write $W=W^G$ for the actual refined series. The different is essential: the two points where the initial fibre conic meets $\Lambda$ have log discrepancy $1/2$ after adjunction to that conic.

Theorem 95.1. For the weight- $(1,2,2)$ extraction at an index-four point, $\delta(G,\Lambda/2;W)\ge\frac{80}{77},\qquad \delta_p(X)=\frac{400}{389}>1.$

Lemma 95.2 (The conic flag and its complete quotients). Choose native coordinates $x,z,w$ of weights $(1,2,2)$ , and put $c=x^2+z,\qquad t=zw-c^2.$ On the coarse exceptional plane use coordinates $U=z,\qquad V=x^2,\qquad W_0=w-z-2x^2.$ Then the initial fibre is the smooth conic $B=(UW_0-V^2=0),\qquad \Lambda=(V=0).$ Parametrize $B$ by $[U:V:W_0]=[p^2:pq:q^2]$ , and write $P=(p=0),\qquad Q=(q=0),\qquad R=(p+q=0).$ Thus $\Delta|_B=(P+Q)/2$ .

Fix $m\ge1$ . In the primary weight- $j$ piece $W_{m,j}$ , let $k$ be the order along $B$ . Put $n=j-4k$ and $\varepsilon\equiv n\pmod2$ , $0\le\varepsilon\le1$ . Set $q_k=\begin{cases}m,&0\le k\le m,\\2m-k,&m\le k\le2m,\end{cases}$ and $a=\max(\lfloor n/2\rfloor-q_k,0),\quad b=\max(\lfloor n/2\rfloor-m,0),\quad c_0=\max(n-m,0),\quad d=n-\varepsilon-a-b-c_0.$ For $0\le k\le2m$ and $n\ge0$ , the nonzero quotient is precisely $\xi^{\varepsilon}p^a q^b(p+q)^{c_0} H^0(\mathbb P^1,\mathcal O(d)), \tag{35.7}$ provided $d\ge0$ . Here $\xi^2=pq$ : this notation records the fractional fixed divisor $\varepsilon(P+Q)/2$ , using the index covers. It does not assert that $pq$ has a square root in the function field of $B$ . Each such quotient has dimension $d+1$ ; all other quotients are zero.

Proof. Use the homogeneous central model in Lemma 93.2, with centre $y(ux^4+vy^4)=0$ . Its affine blow-up chart has equation $t=zw-(x^2+z)^2$ . The change of coordinates in the statement gives $t=UW_0-V^2$ . The primary weight spaces on this model are the actual associated graded spaces of the original point filtration, as proved in that lemma. The defining section $t$ has weight four. Consequently the filtration by its order preserves the weight decomposition, and taking weight spaces commutes with taking the successive $t$ -quotients. No interchange of two arbitrary filtrations is being assumed.

The fibre $t=0$ is the plane blown up in a length-four cluster along a smooth conic and one further point of that conic. On its minimal resolution use $h,e_1,\ldots,e_5$ , where $e_1,\ldots,e_4$ are the total transforms for the successive cluster points and $e_5$ is the other point. The middle exceptional curve is $F=e_2-e_3$ . The restriction and vanishing argument used for Lemma 92.1 applies to the central model as well: it is a canonical Fano threefold, and the same nef divisors occur. After removing the fixed conic, which is disjoint from $F$ , the $k$ th fibre quotient is the complete space of plane forms of degree $d_k$ , with multiplicity $q_k$ at all five cluster points, where $d_k=\begin{cases}3m,&k\le m,\\5m-2k,&k\ge m.\end{cases}$ $d_k-q_k=3m-k$ when $k\ge m$ , whereas it is $2m$ when $k\le m$ ; also $d_k-2q_k=m$ in both intervals.

Here is a direct calculation of the $F$ -initial spaces. Use affine plane coordinates $x,c$ , with conic $z=c-x^2=0$ , and place the separate point at $[0:0:1]$ in coordinates $[1:x:c]$ . The length-four cluster ideal is $(z,x^4)$ ; its $q_k$ th power is complete. A plane monomial $x^i c^b$ must satisfy $i+b\le d_k$ and $b\le d_k-q_k$ . The first condition bounds its degree, and the second is exactly its multiplicity condition at $[0:0:1]$ . The cluster ideal is homogeneous for weights $(1,2)$ on $(x,c)$ , so these conditions split into homogeneous pieces.

The valuation of $F$ has weights $(1,2)$ on $(x,z)$ . A form contributing order $n$ after division by the cluster exceptional factor therefore has weight $N=n+2q_k$ . Put $l=\lfloor N/2\rfloor$ . Write such a form as $x^\varepsilon (x^2)^l f(c/x^2)$ . The bounds just given say that the exponents of $f$ lie between $b_-=\max(0,N-d_k),\qquad b_+=\min(l,d_k-q_k).$ Membership in $(z,x^4)^{q_k}$ additionally says that $f(h)$ is divisible by $(h-1)^M$ , where $M=\max(0,2q_k-l).$ Indeed a term $x^{N-2i}z^i$ belongs to this ideal precisely when $N+2i\ge4q_k$ .

Since $h$ and $h-1$ have distinct zeros, the resulting polynomial space is exactly $h^{b_-}(h-1)^M\mathbb C[h]_{\le b_+-b_--M}.$ On $B$ one has $z=p^2$ , $x^2=pq$ , $c=p(p+q)$ . Dividing by the cluster factor $z^{q_k}=p^{2q_k}$ gives $\xi^\varepsilon p^{l-2q_k+M}q^{l-b_+}(p+q)^{b_-} H^0(\mathbb P^1,\mathcal O(b_+-b_--M)).$ Substituting $d_k-2q_k=m$ reduces its exponents to (35.7). This proves completeness and the dimension statement in every degree. ◻

Example 95.3. Take $m=2$ , $k=0$ , $n=4$ . Then $a=b=0$ , $c_0=2$ , $d=2$ , so the quotient is $(p+q)^2H^0(\mathbb P^1,\mathcal O(2))$ . The maximal sums of basis orders at $P,Q,R$ are respectively $3,3,9$ : the first two are $0+1+2$ , while the last is $2+3+4$ . For $n=3$ the quotient has the additional fixed divisor $(P+Q)/2$ . This is why the parity term appears in the finite-level formulas, even though its contribution tends to zero after dividing by $mN_m$ .

Lemma 95.4 (The conic and point moments, computed separately). For refinement along $B$ one has $S(W;B)=\frac{69}{80}.$ Writing $W^B$ for the resulting series on $B$ , its point moments are $\begin{gathered} S(W^B;P)=\frac{77}{160},\qquad S(W^B;Q)=\frac{27}{64},\qquad S(W^B;R)=\frac9{10},\\ S(W^B;O)=\frac{25}{64}\quad(O\notin\{P,Q,R\}). \end{gathered} \tag{35.8}$ The average degree of the residual plane divisors after removing their $B$ -components is $113/160$ .

Proof. Use the quotients (35.7). Their contributions to the order sums at $B,P,Q,R,O$ are respectively $k(d+1),\quad (a+\varepsilon/2+d/2)(d+1),\quad (b+\varepsilon/2+d/2)(d+1),\quad (c_0+d/2)(d+1),\quad d(d+1)/2.$ These are the maximal sums for complete binary spaces, with the fixed factors included. Sum them and divide by $mN_m$ .

We spell out the integrals. Put $u=k/m$ , $v=n/m$ and $q(u)=\begin{cases}1,&0\le u\le1,\\2-u,&1\le u\le2,\end{cases}$ $A=(v/2-q(u))_+,\quad C=(v-1)_+,\quad B_0=(v/2-1)_+,\quad D=v-A-B_0-C,$ where $r_+=\max(r,0)$ . Integrate over $D\ge0$ , $v\ge0$ , $0\le u\le2$ . Floors and parity change the sums by $O(m^3)$ . The following are the unnormalized integrals, with measure $dv\,du$ : $\begin{array}{c|rrrrrrr} &D&uD&(A+D/2)D&(B_0+D/2)D&(C+D/2)D&D^2/2&(4u+v)D\\ \hline 0\le u\le1&2&1&11/12&11/12&2&5/6&7\\ 1\le u\le2&4/3&15/8&11/16&47/96&1&15/32&221/24. \end{array} \tag{35.9}$ For the first row, split the $v$ -interval at $1,2,3$ ; the values of $D$ are $v,1,3-v$ . For the second row, use $q=2-u$ and the following five regions: $\begin{array}{c|c|c} q&v&D\\ \hline [0,1]&[0,\min(1,2q)]&v\\ {}[1/2,1]&[1,2q]&1\\ {}[0,1/2]&[2q,1]&v/2+q\\ {}[0,1]&[\max(1,2q),2]&1+q-v/2\\ {}[0,1]&[2,2+q]&2+q-v. \end{array}$ These regions give each entry by integrating a quadratic polynomial. For example the last column of (35.9) sums to $389/24$ , while the first sums to $10/3$ . Multiplication by $3/10$ , the leading normalization of $1/(mN_m)$ , gives $S_X(G)=389/80$ again. The second column gives $69/80$ . Columns three through six give (35.8). Thus the computation also checks the normalization independently of the threefold character sum.

On the coarse plane a primary-degree- $j$ section has divisor degree $j/2$ . Consequently, after removing the conic components, the average degree is $\frac12 S_X(G)-2S(W;B)=\frac{389}{160}-\frac{138}{80} =\frac{113}{160}.$ ◻

Lemma 95.5 (Controlling points off the conic). Let $D_m=e_mB+R_m$ be a basis divisor of $W$ compatible with order along $B$ . Thus $R_m\ge0$ and $B\not\subset\operatorname{Supp}(R_m)$ . Then $e_m\longrightarrow69/80,\qquad \deg R_m\longrightarrow113/160, \qquad\operatorname{coeff}_{\Lambda}(R_m)\le27/64+o(1),$ where the error bound is independent of the chosen compatible basis. For every $\tau<80/77$ , the pair $(G,\Lambda/2+\tau D_m)$ is log canonical off $B$ for all large $m$ .

Proof. The first two assertions follow from the preceding lemma. Restriction of the residual sections to $B$ gives bases of the quotients (35.7). Since $\Lambda$ and $B$ meet transversely at $Q$ , $\operatorname{coeff}_{\Lambda}(R_m) \le\operatorname{mult}_{Q}(R_m|_B) \le S_m(W^B;Q).$ This proves the uniform third assertion; it does not require a basis compatible with a third filtration.

Write $R_m=b_m\Lambda+R'_m$ , where $R'_m$ has no $\Lambda$ -component. For all large $m$ , $\frac12+\tau b_m<1,\qquad \tau\deg R_m<1.$ Off $B\cup\Lambda$ the boundary is $\tau R'_m$ . Its degree is less than one, so it is log canonical: at any point its multiplicity is at most its degree, and the standard multiplicity bound gives $\operatorname{lct}_q(G;R'_m)\ge1/\operatorname{mult}_q(R'_m)$ . Along $\Lambda\setminus B$ , the divisor $\tau R'_m|_\Lambda$ has degree less than one and hence all coefficients less than one. Inversion of adjunction gives log canonicity of $(G,\Lambda+\tau R'_m)$ there. Decreasing the $\Lambda$ -coefficient to $1/2+\tau b_m$ proves the assertion. ◻

Proof of Theorem 95.1. Fix $\tau<80/77$ . For every basis divisor $D_m=e_mB+R_m$ compatible with $B$ , the coefficients of $(\Lambda/2)|_B+\tau R_m|_B$ are at most one for all large $m$ , by (35.8). The bounds at $P,Q,R$ and the other points are uniform over the choice of basis. The restrictive inequality is the one at $P$ : $\frac12+\tau\frac{77}{160}<1.$ Also $\tau e_m<1$ . Inversion of adjunction along the smooth conic therefore gives log canonicity of $(G,\Lambda/2+\tau D_m)$ near $B$ . Lemma 95.5 proves it on the complement.

For any divisorial valuation over $G$ , choose in each primary piece a basis compatible with its filtration and the $B$ -filtration. The resulting order is the maximal basis order for that valuation. The preceding global log-canonicity statement, followed by $m\to\infty$ and $\tau\to80/77$ , proves $\delta(G,\Lambda/2;W)\ge80/77$ . Abban–Zhuang adjunction for the point extraction gives $\delta_p(X)\ge\min\left\{\frac{5}{389/80},\frac{80}{77}\right\} =\frac{400}{389}.$ The exceptional divisor $G$ itself gives the opposite inequality, so equality holds.

No Zariski decomposition on the weight- $(1,2,2)$ blow-up of $X$ has been used. The commuting homogeneous fibre filtration and the plane ideal $(c-x^2,x^4)^{q_k}$ give the exact section spaces. Their complete binary quotients control the conic; the residual plane degree controls its complement. Keeping the different $\Lambda/2$ is necessary at both stages. ◻