Section 56

Completion for Family 2.16

Part IX · Family 2.16: conics on intersections of two quadricsPDF page 205

Proposition 56.1 (Every point of a double fibre). If $p\in\Lambda$ and $g^{-1}(g(p))=2\Lambda$ , then $\delta_p(X)\ge88/85$ .

Proof. Apply Abban–Zhuang to the plt curve extraction of Lemma 54.2. The refined threshold over any point of $\Lambda$ is bounded below by the global surface threshold. Lemmas 55.1 and 55.4 give $\delta_p(X)\ge \min\left\{\frac{3}{57/22},\frac{88}{85}\right\} =\frac{88}{85}.$ To recall the local argument, choose a threefold basis compatible with $F$ and a given prime divisor over $X$ , and write its pull-back as $S_m(F)F+\Gamma_m$ . The finite-level bounds in the proof of Lemma 55.4 make $(F,D/2+\lambda\Gamma_m|_F)$ log canonical for every fixed $\lambda<88/85$ and large $m$ . Inversion of adjunction gives log canonicity of $(X^\sharp,F+\lambda\Gamma_m)$ near $F$ . The pull-back boundary for the original basis divisor has coefficient $1-A_X(F)+\lambda S_m(F)\le1$ on $F$ , so it too is log canonical there. By properness, the centre on $X^\sharp$ of any divisor whose centre on $X$ contains $p$ meets $\omega^{-1}(p)$ . This proves the required inequality for all such divisors. Passing to the limit gives the stated bound. Bounded support, the ample-series condition, and the different were verified above. The estimate includes $p=q=\Lambda\cap E$ . ◻

Proof of Theorem 53.1 and the Family 2.16 assertion of Theorem 1.1. Every plane conic fibre is smooth, reduced with two components, or a double line. Proposition 53.3 and Proposition 56.1 therefore give $\delta_p(X)\ge\min\{176/171,88/85\}=176/171$ for every point $p$ . For any prime divisor $V$ over $X$ , choose a point of its centre. The definition of $\delta_p$ gives $A_X(V)/S_X(V)\ge176/171$ . Taking the infimum proves the global bound, and hence K-stability. ◻

Example 56.2 (A smooth member with a double fibre). In coordinates $[s:x:z:t_0:t_1:t_2]$ on $\mathbb P^5$ , take $Q_1=z t_0-s t_1-x t_2, \qquad Q_2=x^2+s z+t_0t_2+t_1^2+t_2^2, \qquad V=(Q_1=Q_2=0).$ The determinant of the Hessian matrix of $Q_1+\lambda Q_2$ is $4\lambda^6+4\lambda^2-1$ . It has six distinct roots and its leading coefficient is nonzero. Simultaneous diagonalization therefore gives six distinct pencil eigenvalues, which proves that the intersection of the quadrics is smooth. The plane $t_0=t_1=t_2=0$ cuts out the smooth conic $x^2+s z=0$ . Blow it up. Over $[y_0:u:v]=[1:0:0]$ , the strict transform equations are $z=0$ and $x^2=0$ , giving a double line. For its leading model, $R=su,\qquad T=tv,\qquad q=s^2u+t^2v, \qquad P=([1:0],[0:1]).$ Here $P$ is a ramification point of $D\to\mathbb P^1_{[u:v]}$ . Thus the limiting monomial calculation already occurs on the actual exceptional surface. The branch valuation has expected order $85/44$ and discrepancy two for $(F,D/2)$ , so the refined surface threshold is exactly $88/85$ in this example. In particular, smoothness of $X$ does not rule out the fibres treated in this part.

Remark 56.3 (Avoiding a Zariski decomposition on the extracted threefold). We have not computed a Zariski decomposition of $\omega^*L-tF$ on the weighted threefold extraction. The volume integrals in Lemma 53.2 take place on $X$ itself. For the new extraction, the prime initial algebra in Lemma 54.3 identifies the actual section filtration, and Lemma 54.4 proves completeness of its point-ideal spaces. Their dimensions compute $S_X(F)$ exactly. The surface estimate then uses a degeneration preserving the different and explicit order sums. These statements provide the required vanishing information; no unproved decomposition on $X^\sharp$ is substituted for them.