Section 105

The weighted extraction and its actual section ring

Part XIV · Family 3.6: double conics and weighted curve extractionsPDF page 356

Write a double conic as $2\ell$ . We use the normalized blow-up of its fibre ideal, rather than the ordinary blow-up of its support. The latter has exceptional surface $\mathbb F_3$ , since $N_{\ell/X}=\mathcal O_{\mathbb P^1}(1)\oplus \mathcal O_{\mathbb P^1}(-2)$ . Indeed, the derivative of the conic equation along its double line gives a surjection $\mathcal O(1)\oplus\mathcal O^{\oplus2}\to\mathcal O(2)$ . It vanishes on $\mathcal O(1)$ , and the other two entries are relatively prime binary quadratics, so the kernel is as stated. In the weighted construction below the exceptional pair is a product of two projective lines with a smooth boundary. Put $E_C=E$ and $E_\Lambda=R$ in this section.

Lemma 105.1. Let $X$ be the blowup of $\mathbb P^3$ along a smooth complete intersection $C=\{q_0=q_1=0\}$ of two quadrics and a line $\Lambda$ disjoint from $C$ . Put $A=H-E_\Lambda$ and $B=2H-E_C$ . Suppose that a fibre of the conic bundle $\eta=(|A|,|B|):X\to\mathbb P^1\times\mathbb P^1$ is $2\ell$ . There are homogeneous coordinates $[r:s:z:w]$ on $\mathbb P^3$ such that $\Lambda=\{s=w=0\},\qquad \pi(\ell)=\{z=w=0\},\qquad q_0=z^2+w(c_0r+c_1s+c_2z+c_3w),$ where $(c_0,c_1)\ne(0,0)$ . If $f(r,s)=s(c_0r+c_1s),\qquad g(r,s)=q_1(r,s,0,0),$ then $f$ and $g$ have no common zero on $\mathbb P^1$ . In particular, $g(1,0)\ne0$ . On a neighbourhood of the whole curve $\ell$ , the threefold is given in $\mathbb P^2_{[r:s:z]}\times\mathbb A^2_{a,b}$ by an equation whose terms of smallest weight, for $\operatorname{wt}(z)=1$ and $\operatorname{wt}(a)=\operatorname{wt}(b)=2$ , are $z^2+af(r,s)+bg(r,s).$

Proof. The plane corresponding to the first coordinate of $\eta(\ell)$ contains $\Lambda$ , and its intersection with the corresponding quadric is the double line $2\pi(\ell)$ . The quadric cannot be smooth, since a tangent plane section of a smooth quadric is a pair of distinct lines. Every singular member of the pencil through a smooth elliptic quartic is a rank-three quadric. Choose the plane $w=0$ and its double line $z=w=0$ ; the displayed form follows. The rank-three condition is precisely $(c_0,c_1)\ne(0,0)$ . The zeroes of $f$ are the point $\Lambda\cap\pi(\ell)=[1:0:0:0]$ and the vertex of $q_0=0$ , counted with multiplicity. The first point does not belong to $C$ . The vertex does not belong to $C$ either, since otherwise the complete intersection $\{q_0=q_1=0\}$ would be singular. Thus $f$ and $g$ are relatively prime.

In the chart of the pencil of planes through $\Lambda$ with $w=as$ , the blowup of $\Lambda$ is a trivial $\mathbb P^2$ -bundle with fibre coordinates $[r:s:z]$ . The second pencil has local equation $q_0+bq_1=0$ . Substitution gives $z^2+as(c_0r+c_1s+c_2z+c_3as)+bq_1(r,s,z,as)=0.$ This equation contains the whole strict transform of $\pi(\ell)$ and has the stated initial form. The use of the blowup chart is essential at $\Lambda\cap\pi(\ell)$ : the coordinate $a$ remains regular there. ◻

Lemma 105.2. In the notation of Lemma 105.1, let $\sigma:Y\to X$ be the normalized blowup of the ideal $(a,b)$ , and let $G$ be its reduced exceptional divisor. Then $\sigma$ is the weighted blowup of $\ell$ with transverse weights $(1,2)$ , and $A_X(G)=3,\qquad G\cong\mathbb P^1\times\mathbb P^1,\qquad (K_Y+G)|_G=K_G+\frac12D,$ where $D=\{Uf(r,s)+Vg(r,s)=0\}\subset \mathbb P^1_{[r:s]}\times\mathbb P^1_{[U:V]}$ is a smooth curve of bidegree $(2,1)$ . The pair $(Y,G)$ is purely log terminal.

Proof. At every point of $\ell$ at least one of $f$ and $g$ is nonzero. If $g\ne0$ , the equation of $X$ allows us to eliminate $b$ , and $(a,b)=(a,z^2)$ in the local ring. If $f\ne0$ , the same argument, with $a$ eliminated, gives $(a,b)=(b,z^2)$ ; this equality follows as well from the implicit function theorem, or from the completed local ring. The normalized blowup of $(a,z^2)$ is the weighted blowup with $\operatorname{ord}_G(z)=1$ and $\operatorname{ord}_G(a)=2$ . Its log discrepancy is $1+2=3$ .

The exceptional divisor is the relative weighted hypersurface $Z^2+Uf(r,s)+Vg(r,s)=0$ with weights $(1,2,2)$ in $[Z:U:V]$ . The projection to $[U:V]$ identifies its coarse space with $\mathbb P^1_{[r:s]}\times\mathbb P^1_{[U:V]}$ : indeed, the second Veronese replaces $Z$ by $T=Z^2$ , and the equation eliminates $T$ . Here $Z$ has degree one along the first factor, so that $T$ , $Uf$ and $Vg$ have the same degree. Along $Z=0$ the transverse weighted chart has stabilizer of order two. It follows either by adjunction on the index cover, or directly from the cyclic quotient chart of the $(1,2)$ -weighted blowup, that the different is $D/2$ . There is no other divisorial stabilizer. Since $f$ and $g$ are relatively prime, $D$ is the graph of the morphism $[r:s]\mapsto[-g(r,s):f(r,s)]$ ; it is smooth and has bidegree $(2,1)$ . The local weighted blowup of a smooth codimension-two centre with coprime transverse weights is purely log terminal. This also follows here from the quotient charts and the smoothness of $D$ . ◻

Remark 105.3. If $c_0=0$ , the line $\Lambda$ passes through the vertex of the singular quadric. The quadric fibre of $|B|$ is then nonnormal. For example, blowing up $\{y=z=0\}$ in the quadric $x^2-yz=0$ gives, in the chart $y=tz$ , the surface $x^2-tz^2=0$ , singular along $x=z=0$ . Thus a proof using only sextic del Pezzo surfaces with Du Val singularities would omit these smooth threefolds. The weighted extraction above applies without alteration: in this case $f$ is a square, while $f$ and $g$ remain relatively prime and $D$ remains smooth.

Lemma 105.4. Let $T=\operatorname{Bl}_\Lambda(\mathbb P^3)\times\mathbb P^1$ , so that $X\subset T$ has class $2H+B$ . Write the Cox coordinates of $T$ as $r,z,s,a_0,a_1,b_0,b_1$ , of respective divisor classes $H,H,H-A,A,A,B,B.$ Assign these coordinates weights $0,1,0,0,2,0,2$ . Then the initial equation of $X$ is $F_0=b_0z^2+b_0a_1s(c_0r+c_1a_0s)+b_1g(r,a_0s).$ This equation is irreducible. The induced weight filtration on $H^0(X,-mK_X)$ is the filtration defined by $G$ .

Proof. The initial equation follows from the normal form above, after homogenizing $s$ by $a_0$ . Regard $F_0$ as a polynomial of degree one in $a_1$ . Its coefficient is $b_0s(c_0r+c_1a_0s)$ , and its constant term is $b_0z^2+b_1g(r,a_0s)$ . These polynomials are relatively prime: no factor of $s(c_0r+c_1a_0s)$ divides the term $b_0z^2$ , and $b_0$ does not divide $b_1g$ . In the case $c_0=0$ the extra factor $a_0$ also does not divide the constant term. Gauss’s lemma proves irreducibility.

For $m\ge1$ the restriction sequence for $X\subset T$ is exact on sections in degree $m(H+A+B)$ . Indeed, after subtracting $2H+B$ the relative degree is $m-2$ . For $m=1$ the relative line bundle $\mathcal O_{\mathbb P^2}(-1)$ is acyclic; for $m\ge2$ the pushforward is a direct sum of line bundles with nonnegative degrees on both base factors. Thus the relevant section spaces are the homogeneous pieces of the Cox ring modulo the equation of $X$ .

Taking the initial form of a principal ideal gives the ideal $(F_0)$ . As its quotient is a domain, localization in a coordinate chart meeting $\ell$ is injective. In that chart the same filtration is the $(1,2)$ -weight filtration just computed on the transverse parameters. Consequently a nonzero initial section cannot disappear after restriction to the weighted normal cone. This identifies the global weight filtration with $\operatorname{ord}_G$ , including all possible cancellations between sections. ◻

Lemma 105.5 (The weighted curve moment). For the extraction of Lemma 105.2, the weights on the product-graph coordinates are $(H_0,H_1,H_2,H_3)=(0,0,1,2),\qquad (U_0,U_1)=(V_0,V_1)=(0,2).$ Its two equations have degrees $(2,1,0)$ and $(1,0,1)$ and both have weight $2$ . One has $S_X(G)=\frac{243}{88}.$

Proof. The initial product graph is the same variety as the initial Cox hypersurface in Lemma 105.4. Indeed, its plane-graph equation is homogeneous for these weights. Its two base centres are $\{z^2+w(c_0r+c_1s)=g(r,s)=0\},\qquad \{s=w=0\}.$ They are disjoint because $g(1,0)\ne0$ , and the first is a complete intersection. Thus the initial product graph is their integral blow-up graph, with the displayed regular sequence of two equations. This also verifies directly the integrality needed in the Koszul calculation. Write $h_m=h^0(X,-mK_X)$ and let $T_m$ be the sum of the weights. The Koszul resolution gives $\begin{aligned} h_m&={m+3\choose3}(m+1)^2- \left({m+1\choose3}+{m+2\choose3}\right)m(m+1) +{m\choose3}m^2\\ &=\frac{(2m+1)(11m^2+11m+6)}6. \end{aligned}$ The average weight in $H^0(\mathbb P^3\times\mathbb P^1\times\mathbb P^1, \mathcal O(a,b,c))$ is $3a/4+b+c$ . Put $N_f={m+1\choose3}m(m+1),\qquad N_g={m+2\choose3}m(m+1),\qquad N_{fg}={m\choose3}m^2.$ The shifts by the two equations therefore give $T_m=\frac{11m}{4}h_m+\frac12N_f-\frac14N_g-\frac14N_{fg} =\frac{m(81m^3+118m^2+83m+22)}8.$ Taking the limit of $T_m/(mh_m)$ proves the assertion. Integrality of the initial graph ensures that these are the actual valuation weights. ◻

Definition 105.6. For the exceptional divisor $G$ , let $V_m^{\ge j}=\{s\in H^0(X,-mK_X):\operatorname{ord}_G(s)\ge j\}, \qquad W_{m,j}=V_m^{\ge j}/V_m^{\ge j+1}.$ These are the actual refined spaces on $G$ . Put $h_m=h^0(X,-mK_X)$ . Their level- $m$ basis boundaries are normalized by $mh_m$ , summing over all $j$ and a basis in each $W_{m,j}$ . The different is $D/2$ , so throughout the surface calculation $\delta(G,D/2;W)=\inf_F\frac{A_{G,D/2}(F)}{S(W;F)}.$ For an odd weight the factor $z$ denotes a rational section whose square cuts out $D$ ; its contribution to a basis boundary is $D/2$ . This convention retains the different, which is an additional fixed boundary independent of $m$ . More precisely, if $a_G,b_G$ denote the two ruling classes, the weight- $j$ space has rational divisor class $ma_G+(j/2)b_G$ . Writing $j=2n+\varepsilon$ , its polynomial part has class $(m-\varepsilon)a_G+nb_G$ , and its fixed half-divisor is $\varepsilon D/2$ .

Lemma 105.7. On the chart $a_0=b_0=1$ , the direct sum of the actual spaces at level $m$ is spanned, modulo $z^2+af+bg=0$ , by $r^i z^j s^k a^\alpha b^\beta, \qquad i+j+k=m,\quad0\le\alpha\le m+k,\quad0\le\beta\le m.$ The primary weight is $j+2\alpha+2\beta$ . All primary weights between zero and $6m$ occur, and $W_{m,2m}$ contains the complete ample series $H^0(G,\mathcal O_G(m,m))$ . Thus the bounded-support and ample-series hypotheses of Abban–Zhuang theory hold.

Proof. The Cox degrees in Lemma 105.4 give exactly these inequalities after setting $a_0=b_0=1$ . The same lemma proves that their span is the actual associated graded space. Its largest primary weight is $6m$ . Every intermediate integral weight occurs, by taking products of the degree-one generators of weights $0,1,2,3,4,5,6$ . Taking $j=0$ , $i+k=m$ , and $\alpha+\beta=m$ gives all monomials of bidegree $(m,m)$ on $G$ . The ray $(m,2m)$ is interior to the support, and $\mathcal O_G(1,1)$ is ample. ◻

Example 105.8. For $m=1$ , the even pieces, after the square relation is used, have dimensions $2,4,3,1$ in weights $0,2,4,6$ . The odd pieces have dimensions $1,2,1$ in weights $1,3,5$ . Their total is $14$ , and their weighted sum is $38$ . Thus the first-level expected order is $19/7$ ; its limit, computed above, is $243/88$ .