Section 72

Reduced singular conic fibres: exact surface estimates

Part XI · Family 2.20: the K-polystable classificationPDF page 254

The following computations record an obstruction to extending the preceding smooth-fibre flag argument without an additional threefold estimate. They do not establish instability. Write $S=\operatorname{Bl}_{p_1,p_2,p_3,p_4}\mathbb P^2,\quad A=l-e_1-e_2,\quad D=l-e_3-e_4,\quad Z=A+D,\quad C=l.$ Assume that $S$ and $C$ are smooth and that $C$ avoids the four exceptional curves. Thus $A$ and $D$ meet transversally at one point $p_0$ , while $C$ meets each of $A,D$ once, possibly at $p_0$ . Put $(b,k)=\begin{cases}(2-u,1)&0\leq u\leq1,\\ (2-u,2-u)&1\leq u\leq2. \end{cases}$ Then the positive part restricted from the threefold is $M_u=(2b+k)l-b\sum e_i$ . In both intervals $k\leq b\leq2k$ .

Lemma 72.1. The pseudoeffective threshold of $A$ with respect to $M_u$ is $b+k$ . The negative part of $M_u-vA$ is $\begin{array}{c|c} v&N_{u,v}\\ \hline {[0,k]}&0\\ {[k,b]}&(v-k)D\\ {[b,b+k]}&(v-k)D+(v-b)(e_1+e_2). \end{array}$ The corresponding positive parts have volumes $4bk+k^2-2kv-v^2,\qquad 2k(2b+k-2v),\qquad 2(b+k-v)^2,$ and their intersections with $A$ are respectively $k+v$ , $2k$ , and $2(b+k-v)$ . Consequently, $S(W^S;A)=\frac{109}{104}.$ For $p\in A$ , the point estimate obtained from this flag is $S(W^{S,A};p)=\frac{79}{104} +\frac{15}{52}\boldsymbol1_{p=p_0} +\frac5{52}\boldsymbol1_{p\in e_1\cup e_2} +\frac5{104}\boldsymbol1_{p\in C}.$ In particular, at the node it is $109/104$ if $p_0\notin C$ , and $57/52$ if $p_0\in C$ .

Proof. The curves $D,e_1,e_2$ are mutually disjoint $(-1)$ -curves. Subtracting the displayed coefficients makes the positive part orthogonal to its negative part. Intersecting with the ten $(-1)$ -curves proves nefness. In the last chamber the positive part is $(b+k-v)(2l-e_3-e_4)$ . The nef divisor $2l-e_3-e_4$ has intersection $2(b+k-v)$ with $M_u-vA$ , so $v>b+k$ is not pseudoeffective. Squaring and intersecting with $A$ give all the displayed expressions.

For the curve moment integrate the three volume expressions with factor $3/26$ . For the point moment integrate $(P_{u,v}\cdot A)^2$ with factor $3/26$ . This gives $79/104$ . The negative component $D$ contributes $\frac6{26}\int_0^2\int_k^{b+k} (P_{u,v}\cdot A)(v-k)\,dv\,du=\frac{15}{52}.$ Each of $e_1,e_2$ contributes $\frac6{26}\int_0^2\int_b^{b+k} 2(b+k-v)(v-b)\,dv\,du=\frac5{52}.$ Finally the fixed part $(u-1)_+C$ contributes $5/104$ at its intersection with $A$ . All intersections are transverse. The node, the two exceptional endpoints, and any other point of $C\cap A$ are distinct; $C\cap A$ can coincide only with the node. ◻

Lemma 72.2. Let $\sigma:T\to S$ blow up $p_0$ , with exceptional curve $J$ . Then $S(W^S;J)=\frac{109}{52}+\frac5{104}\boldsymbol1_{p_0\in C}.$ The point moments on $J$ are bounded by $\frac{47}{104} +\frac{31}{52}\boldsymbol1_{q\in A'\cup D'} +\frac5{104}\boldsymbol1_{p_0\in C,\ q\in C'}.$ Here the three directions $A',D',C'$ are pairwise distinct whenever $C'$ meets $J$ . Thus the maximum is $109/104$ , attained at either branch direction.

Proof. The strict transforms $A',D'$ are disjoint $(-2)$ -curves. For $0\leq v\leq k$ , the divisor $\sigma^*M_u-vJ$ is nef. For $k\leq v\leq2b+k$ , its negative part is $(v-k)(A'+D')/2$ . The positive part has square $\begin{cases} 4bk+k^2-v^2&0\leq v\leq k,\\ 2k(2b+k-v)&k\leq v\leq2b+k, \end{cases}$ and intersection with $J$ equal to $v$ and $k$ , respectively. For nefness, one may write the positive part in the second interval as $(b-(v-k)/2)\sigma^*Z+k(\sigma^*l-J),$ a nonnegative combination of the two nef pencils $\sigma^*Z$ and $\sigma^*l-J$ . At the threshold it equals $k(\sigma^*l-J)$ . The nef divisor $\sigma^*l-J$ has intersection $2b+k-v$ with $\sigma^*M_u-vJ$ , proving the threshold.

Integrating the two squares gives $109/52$ . The fixed curve $C$ adds $\frac3{26}\int_1^2 5(2-u)^2(u-1)\,du=\frac5{104}$ exactly when it passes through $p_0$ . For the second refinement, integrating $(P_{u,v}\cdot J)^2$ gives $47/104$ . The coefficient $(v-k)/2$ of either branch contributes $31/52$ at its direction; the strict transform of the fixed curve contributes $5/104$ at its direction. These directions are distinct because $C,A,D$ are three distinct lines through the same point in the plane model. ◻

Remark 72.3. Neither lemma is a destabilizing calculation for the threefold. The first curve moment exceeds its surface log discrepancy $1$ , while the node moment exceeds its surface log discrepancy $2$ . This says precisely that these unmodified surface Abban–Zhuang estimates fail to prove the desired threefold assertion. A successful proof must exploit more information, for example the positive first-stage margin $41v(S)/104$ and a comparison with the conic-bundle coordinates, or a different extraction.

Lemma 72.4. Keep the preceding assumptions, and let $v$ be a divisorial valuation centred at $p\in A$ on $X$ . If $p\neq p_0$ or $p_0\notin C$ , then $A_X(v)-S_X(v)\geq\frac{46v(S)-5A_X(v)}{104}.$ If $p=p_0\in C$ , then $A_X(v)-S_X(v)\geq\frac{51v(S)-10A_X(v)}{104}.$ In particular these inequalities are strict whenever $A_X(v)/v(S)<46/5$ in the first case, or $A_X(v)/v(S)<51/10$ in the second case.

Proof. Choose a basis adapted simultaneously to $v$ and the single lexicographic flag $X\supset S\supset A\supset\{p\}$ . Write its normalized basis divisor as $D_m=a_mS+D'_m$ , and write $D'_m|_S=b_mA+R_m$ , where $R_m$ does not contain $A$ . Then $a_m\to63/104$ , $b_m\to109/104$ , and $(R_m\cdot A)_p$ tends to the point moment in Lemma 72.1. Away from the node this moment is at most $89/104$ . At the node it is $109/104$ off $C$ and $114/104$ on $C$ . Consequently, one can choose positive coefficients $c_m$ with $c_mb_m\leq1$ and $c_m(R_m\cdot A)_p\leq1$ , tending to $c=104/109$ in the first case and $c=104/114$ in the second case.

Both $A$ and $S$ are smooth at $p$ . Applying inversion of adjunction first to $(S,A+c_mR_m)$ and then to $(X,S+c_mD'_m)$ gives $A_X(v)-c_m v(D_m)\geq(1-c_ma_m)v(S).$ All boundaries used here are effective. The basis is adapted to $v$ , so $v(D_m)\to S_X(v)$ . Passing to the limit and dividing by $c$ yields $A_X(v)-S_X(v)\geq(1/c-63/104)v(S)-(1/c-1)A_X(v),$ which is exactly the assertion. ◻