Section 37

Strict adjunction and completion of the proof

Part VI · Family 2.12: intersections of three bilinear divisorsPDF page 141

Lemma 37.1 (Keeping the positive extraction term). Let $\sigma:Y\to X$ be the ordinary blow-up of $p$ , with exceptional divisor $G$ . Suppose that the pivot means of its refined series have maximum one. Then every prime divisor $F$ over $X$ with centre $p$ satisfies $\beta_X(F)\ge\bigl(3-S_X(G)\bigr)\mathop{\mathrm{ord}}_F(G) =\frac34\mathop{\mathrm{ord}}_F(G)>0.$

Proof. Choose an $m$ -basis of $H^0(X,mL)$ compatible with $\mathop{\mathrm{ord}}_G$ and $\mathop{\mathrm{ord}}_F$ . Write its basis divisor as $\sigma^*D_m=S_m(L;G)G+D'_m.$ Its restriction $D'_m|_G$ is a basis divisor of the actual refinement. Lemma 34.1 bounds the thresholds of all such basis divisors, so there are $\lambda_m\to1$ , $0<\lambda_m\le1$ , for which $(G,\lambda_mD'_m|_G)$ is log canonical. This step does not require a basis simultaneously adapted to three filtrations. Inversion of adjunction gives that $(Y,G+\lambda_mD'_m)$ is log canonical near $G$ . Therefore $\begin{aligned} A_X(F)&=A_Y(F)+2\mathop{\mathrm{ord}}_F(G)\\ &\ge\lambda_m S_m(L;F) +\bigl(3-\lambda_m S_m(L;G)\bigr)\mathop{\mathrm{ord}}_F(G). \end{aligned}$ Taking limits proves the claimed inequality. Finally, $\mathcal O_Y(-G)=\mathfrak m_p\mathcal O_Y$ , so $\mathop{\mathrm{ord}}_F(G)=\mathop{\mathrm{ord}}_F(\mathfrak m_p)>0$ when the centre is $p$ . ◻

Proof of Theorem 1.1 for Family 2.12. Let $F$ be a prime divisor over $X$ . If $\beta_X(F)\le0$ , Proposition 32.2 says its centre is a point $p\in E\cap E'$ . Use the normal form of Lemma 33.1, and write $Q_i=A_it^2+B_ita+C_itb+D_iab.$ There are four exhaustive possibilities.

If $D_2D_3\ne0$ , divide by the $ab$ coefficients and translate $a,b$ by multiples of $t$ . This gives the normal form of Corollary 35.2. Its refined threshold is at least $320/253$ , and the ordinary blow-up has $A_X(G)/S_X(G)=4/3$ . Proposition 2.7 gives $\delta_p(X)\ge320/253>1$ , a contradiction.

If exactly one of $D_2,D_3$ vanishes, exchange the two factors if necessary so that $D_2\ne0,D_3=0$ . Normalize $Q_2=ab+\kappa t^2$ . The one-sided Jacobian condition gives $B_3\ne0$ , so rescale $Q_3$ to $t(a+Cb+\gamma t)$ . The alternative $C=\gamma=\kappa=0$ would give a common factor other than $t$ , excluded by Lemma 33.3. Lemma 35.3 and ordinary extraction give $\delta_p(X)\ge160/137>1$ , again a contradiction.

Suppose $D_2=D_3=0$ and $B_2C_3-B_3C_2\ne0$ . Solve $B_2u+C_2v=A_2,\qquad B_3u+C_3v=A_3,$ and replace $a,b$ by $a+ut,b+vt$ . This removes both $t^2$ terms, without changing the two factor spans. Since $C_2B_3\ne0$ , rescaling the quadrics now gives $Q_2=t(\beta a+b),\qquad Q_3=t(a+\gamma b),\qquad\beta\gamma\ne1.$ Lemma 35.4 gives refined threshold at least one and maximum pivot mean exactly one. Lemma 37.1 then gives $\beta_X(F)>0$ , a contradiction. The minimum form of adjunction alone would yield only $\delta_p(X)\ge1$ here; retaining the exceptional coefficient is essential.

Finally, suppose $D_2=D_3=0$ and $B_2C_3-B_3C_2=0$ . The smooth Jacobian conditions force the two residual linear forms to be nonzero and proportional. Lemma 36.1 identifies the exact weighted initial ring, and Lemma 36.2 supplies the plt extraction of weights $(2,1,1)$ with zero different. Its expected order is $13/4$ and discrepancy four. The refined threshold is at least $40/37$ by Lemma 36.3. Thus $\delta_p(X)\ge\min\left\{\frac{16}{13},\frac{40}{37}\right\} =\frac{40}{37}>1,$ which gives the final contradiction. Hence every prime divisor over $X$ has positive beta invariant, and $X$ is K-stable. ◻

Example 37.2 (How the different certificates are used). These are illustrations of the local polynomial computations; the global hypotheses are verified separately by the normal-form lemmas. For $Q_2=ab+t^2$ and $Q_3=ab+ta+tb+2t^2$ , the first ordinary certificate gives the means $(7/10,253/320,243/320)$ . The extra degree-two sections are needed to account for the full volume $10/3$ .

For $Q_2=ab+t^2$ and $Q_3=ta$ , the last row of the one-vanishing-coefficient certificate applies, giving $(137/160,253/320,193/320)$ . This illustrates why the special coefficients cannot simply be discarded by a generality assumption.

For $Q_2=tb$ and $Q_3=ta$ , the pivot algebra is already monomial. Its means are $(1,5/8,5/8)$ . The coordinate line $t=0$ has discrepancy one and expected order one on the refined plane. Nevertheless, the threefold divisors centred at $p$ have strictly positive beta by Lemma 37.1.

In the repeated-residual case, the weighted model uses $Y=-t-ab$ and $Q=t(b-a)-a^2b$ . The section $P_1=b^2Q+abQ-a^2bY$ has leading monomial $tb^3$ : its lower lexicographic terms are determined by cancellation. The further sections $P_2,P_3$ in Lemma 36.3 complete the asymptotic volume certificate. Ignoring these cancellations would give an incomplete calculation of the refined series.

Remark 37.3 (Why no threefold Zariski decomposition is required). The proof never computes a Zariski decomposition on either point extraction. The complete-intersection and localization arguments identify the actual valuation filtrations, including every cancellation. Weighted Hilbert functions then compute $S_X(G)=9/4$ or $S_X(T)=13/4$ directly. On the exceptional surface, explicit section identities and full-volume moment certificates control all basis divisors by degeneration. The only place where the surface bound equals one is handled by the strict adjunction inequality. These algebraic steps supply the information for which one would otherwise seek positive and fixed parts on a threefold modification.