Section 119

The total ramification cases

Part XVI · Family 3.8PDF page 401

Proposition 119.1. Suppose that $\eta_U$ has a point of ramification index four. Then $X_U$ is K-polystable if and only if $U$ is equivalent to $\langle x^4,y^4\rangle$ . Every other such threefold is strictly K-semistable. Thus these strictly K-semistable threefolds are parametrized, up to isomorphism, by $\mathbb P(2,3)$ .

Proof. The threefold $X_{U_0}$ is K-polystable by [3, Proposition 5.74 and Remark 5.75]. For completeness, we give the degeneration proving the other assertion, which is [3, Lemma 7.7] in different coordinates. By Lemma 118.2, we may write the centre as $ux^4+v(y^4+ax^2y^2+bx^3y)=0, \qquad(a,b)\ne(0,0).$ Over $\mathbb A^1_t$ , replace its equation by $ux^4+v(y^4+t^2ax^2y^2+t^3bx^3y)=0.$ These centres form a smooth family of graphs. Blowing them up gives a smooth family of Fano threefolds, and hence its relative anticanonical divisor is relatively ample. The action $([u:v],[x:y],t)\longmapsto ([s^4u:v],[s^{-1}x:y],st)$ extends to the ambient product via the conic embedding and preserves this family. It provides its one-parameter equivariance, and all nonzero fibres are isomorphic to $X_{U_{a,b}}$ . The central fibre is $X_{U_0}$ , so this is a special test configuration after the anticanonical polarization is chosen.

On the central fibre the involution interchanging $u,v$ and $x,y$ conjugates the induced one-parameter subgroup to its inverse. Its Futaki invariant is therefore zero. The Donaldson–Futaki invariant of our test configuration is zero as well [3, Lemma 1.29]. It is not a product test configuration: the central pencil has two points of index four, while the general pencil has only one, so Lemma 117.2 shows that its central and general fibres are not isomorphic. Consequently the general fibre is not K-polystable. It is K-semistable by openness of K-semistability at the central fibre [3, Theorem 1.11], since all nonzero fibres are isomorphic. The last assertion follows from Lemmas 117.2 and 118.2. ◻