The classification problem and the results proved here

Theorem 69.1. For $U\in\mathcal H^\circ$, the threefold $X_U$ is K-polystable if and only if $U$ is GIT-polystable. It is K-stable if and only if $U$ is GIT-stable. Every smooth member is K-semistable. More explicitly, the following assertions hold.

1. The pencil $U$ is GIT-semistable for the Plücker linearization. It fails to be GIT-stable exactly when it is equivalent under $\operatorname{PGL}_2$ to $$U_{a,c}=\langle x^3y,\ xy^3+a x^2y^2+c x^4\rangle.$$ The member $X_{U_{0,0}}$ is K-polystable. Every displayed member with $(a,c)\ne(0,0)$ is strictly K-semistable. The latter isomorphism classes are parametrized by $\mathbb P(1,3)$.

2. If $w_U$ is the Wronskian sextic of $U$, then all its roots have multiplicity at most three. Two triple roots characterize $X_{U_{0,0}}$. Exactly one triple root characterizes the nonclosed smooth GIT orbits and hence the strictly K-semistable members in (1). All the remaining pencils are GIT-stable.

3. Put $H=\pi^*\mathcal O_{V_5}(1)$, let $E$ be the exceptional divisor, and put $B=H-E$. Then $$S_X(E)=\frac12,\qquad S_X(B)=\frac{63}{104},\qquad S_X(H)=\frac{37}{104}.$$ Every prime divisor on $X$ has positive beta invariant. If $p$ lies on a smooth fibre of $\eta$, a suitable smooth $S\in|B|$ through that fibre satisfies $$A_X(v)-S_X(v)\ge\frac{41}{104}v(S)>0$$ for every divisorial valuation centred at $p$. Every valuation with curve centre not contained in the inverse image of the discriminant also has positive beta invariant.

4. At every double conic fibre $2R$, let $F$ be the reduced exceptional divisor of the blowup of the pulled-back maximal ideal of its base point. Then $$A_X(F)=3,\qquad S_X(F)=\frac{297}{104},\qquad A_X(F)-S_X(F)=\frac{15}{104}.$$ The different on $F\cong\mathbb P^1\times\mathbb P^1$ is $\Gamma/2$, where $\Gamma$ has bidegree $(2,1)$. The vanishing sequence $(1,3)$ occurs exactly when the point over $E\cap R$ is ramified for $\Gamma\to\mathbb P^1$ in the direction factor.

5. If a double fibre corresponds to a common root with vanishing sequence $(1,2)$, then $\delta_p(X)\ge104/99$ at every point of its support. In particular this holds along every double fibre of a GIT-stable member. The remaining possible centres of nonpositive beta invariant on such a member lie over reduced singular conics.

6. For the blowup $F$ of any component $A$ of a reduced singular fibre, $A_X(F)=2$ and $S_X(F)=43/26$. Its complete refined series is given in Lemma 77.5. If the fibre node is outside $E$ and $\pi(A)$ has trivial normal bundle in $V_5$, then $\delta_p(X)\ge104/99$ for every $p\in A$.

7. For a GIT-stable pencil, every valuation of nonpositive beta invariant must have centre at the node on $E$ of a reduced conic fibre. These points correspond to $\Delta_{\mathrm{sm}}\cap Q$, and there are at most $3-d$ of them, where $d$ is the number of nodes of the discriminant. At each of these points the ordinary point extraction satisfies $S_X(J)=11/4$ and $A_X(J)=3$. Its complete refinement gives $\delta_p(X)\ge104/103>1$ at every such node.

8. A triangular discriminant gives a K-stable member. There is one such member up to isomorphism, with pencil $U_\triangle=xy(x-y)\langle x,y\rangle$.

Consequently, the K-polystable smooth members are exactly those whose Wronskian has no triple root, together with the unique member $U_{0,0}$, whose Wronskian has two triple roots. The first collection consists of K-stable members; the latter member is K-polystable but not K-stable. Exactly one triple root gives strict K-semistability. An equivalent criterion using only the conic discriminant and the branch conic is given in Corollary 82.1.