Reduction to finitely many points and a triangle example

Proof. Write $E=\mathbb P^1\times\mathbb P^1$, with $f$ a fibre of $E\to C_0$ and $s$ the other ruling class. The complete containing series for the $E$-refinement has class $(1+u)s+(4-2u)f$, for $0\le u\le1$, and no fixed part. Its moment along $s$ is $$\frac3{26}\int_0^1\int_0^{1+u} 2(1+u-v)(4-2u)\,dv\,du=\frac34.$$ For an irreducible curve $T\sim as+bf$ other than a fibre of $E\to C_0$, one has $a\ge1$, $b\ge0$. Monotonicity therefore gives $S(W^E;T)\le3/4$. Together with $S_X(E)=1/2$, adjunction at the generic point of $T$ gives $\delta_T(X)\ge4/3$.

A fibre $f$ of $E\to C_0$ is not contained in $\eta^{-1}(\Delta)$. Indeed $\eta|_f$ maps isomorphically onto a line. If this line were contained in $\Delta$, its general reduced conic fibre would have a component whose image is a line in $V_5$ through the fixed point $\pi(f)$. This would give infinitely many distinct lines through that point, which is impossible.

Here is an elementary justification of the last fact. Since $V_5$ is cut out by quadrics, the directions of lines through a point are cut out by quadrics in $\mathbb P(T_pV_5)=\mathbb P^2$. At least one is nonzero, for otherwise $V_5$ would contain its projective tangent three-space. A positive-dimensional component would thus have degree one or two. The union of the corresponding lines would be a surface of degree one or two in $V_5$. But $\operatorname{Pic}(V_5)=\mathbb Z[H]$ makes the degree of every effective surface a positive multiple of $H^3=5$. This contradiction proves finiteness. Now apply Lemma 71.8 to $f$. ◻

Proof. Choose regular parameters $(x,y,z)$ at $p$ with $A=(y=z=0)$. If a local section has order $k$ along $A$, its first nonzero normal term is $$\sum_{i+j=k}a_{ij}(x)y^iz^j.$$ Let $q=\min\operatorname{ord}_x a_{ij}$. Then $q$ is exactly the order of the leading section along $f_p$, whereas the order of the original section at $p$ is at most $k+q$. Terms of higher normal order may lower that total order; they cannot invalidate this upper bound.

In each $H^0(X,mL)$ choose a basis compatible with the point-order filtration and the single lexicographic refinement by $(k,q)$. Two filtrations of a finite-dimensional vector space admit a common adapted basis. The sum of the point orders is therefore at most the sum of the $k$-orders plus the sum of the $q$-orders. Divide by $m h^0(X,mL)$ and pass to the limit. These three sums give respectively the three moments in the assertion. ◻

Proof. For either component extraction, $S_X(F)=43/26$. The marked ruling has full moment $57/52$ for both normal-bundle types. The ordinary case was proved in Lemma 77.9. For the special case, use $q_0=[1:0:0]$, $q_D=[0:1:0]$, and the infinitely near point $q_E$ in direction $Z=0$. The residual monomials $X^iY^{b-i-k}Z^k$ satisfy $$0\le i\le a,\qquad \max(0,c_E-a+i,c_D-i)\le k\le b-i.$$ Integrating $k$ over this polygon with factor $3/13$ gives $11/104$, $59/104$, and $34/104$ on the intervals $[0,1]$, $[1,2]$, and $[2,3]$, respectively. Their sum is one. The fixed ruling in $R=s+f_0+T$ adds $5/52$. Thus its full moment is $57/52$ also. The preceding lemma now gives $$S_X(J)\le\frac{43}{26}+\frac{57}{52}=\frac{11}{4}.$$ The discrepancy of the ordinary point blowup is three. ◻

Proof. The earlier results treat divisor centres, smooth conic fibres, and every double fibre on the GIT-stable locus. Theorem 77.8, Proposition 78.3, and Corollary 78.6 treat closed point centres on every reduced fibre outside the set displayed above. In the possible equality case on a special-line surface, the retained term $9v(F)/26$ gives strict positivity.

Curve centres contained in $E$ are excluded by Lemma 79.1. For a curve not contained in $E$, choose a general point outside $E$. Over a reduced singular conic the new estimates there have the strict bound $104/99$; over a double fibre use Theorem 75.7; otherwise use Lemma 71.8. Thus all curve centres are excluded as well.

To identify the residual set, move a point of $Q$ to $[0:0:1]$ in the model of Lemma 73.1. The fibre equation is $$-y_1^2+z(f_0y_0+f_1y_1)+h_2z^2=0.$$ Its discriminant on $Q$ is $f_0^2$. At an intersection with $\Delta$, therefore $f_0=0$. If the discriminant is smooth there, this is a union of two distinct lines with vertex $[y_0:y_1:z]=[1:0:0]$, which lies on $E=(z=0)$. Conversely, a reduced conic whose node lies on $E$ has a double intersection with $E$, so its base point lies on $Q$. This proves the claimed bijection.

For a GIT-stable pencil $Q$ is not a component of $\Delta$, and every node of $\Delta$ has intersection multiplicity two with $Q$, by Proposition 76.1. The restriction $\Delta|_Q$ is the square of a nonzero binary cubic. Each remaining point of intersection also has multiplicity at least two. Since $\Delta\cdot Q=6$, there are at most $3-d$ of them. Finally apply Proposition 79.4. ◻

Proof. A triangle has three nodes. All lie on $Q$. Bezout’s theorem forces intersection multiplicity two at each and no further intersection. The pencil is GIT-stable by Proposition 76.1. Theorem 79.5 now leaves no possible centre of nonpositive beta invariant. The valuative criterion gives K-stability.

By Lemma 73.10, the three double fibres correspond to three distinct common roots of the pencil. Hence $U=g\langle x,y\rangle$ for a squarefree binary cubic $g$. Conversely such a pencil contains no square: multiplying $g$ by a linear form leaves at least two of its distinct roots with odd multiplicity. It therefore belongs to $\mathcal H^\circ$ and has three double fibres. A reduced nodal plane cubic with three nodes is a triangle. Finally $\operatorname{PGL}_2$ is transitive on unordered triples of distinct points, giving the displayed normal form and uniqueness. ◻