Section 82

The complete classification

Part XI · Family 2.20: the K-polystable classificationPDF page 292

Proof of Theorem 69.1. The first assertion follows from Lemmas 70.1 and 70.2 and Proposition 70.4. Corollary 70.8 proves the second assertion. The moments and the valuative reductions are Lemmas 71.1, 71.5, 71.7 and 71.8. Finally, Lemmas 73.4 and 73.11, together with Proposition 74.3, prove the fourth assertion. The fifth assertion is Theorem 75.7, together with the preceding reductions. The sixth follows from Proposition 77.3 and Theorem 77.8. The seventh combines Theorem 79.5 with Theorem 80.1, and the eighth is Corollary 79.6. If $U$ is GIT-stable, the finite-node reduction and the strict point estimate leave no valuation of nonpositive beta invariant. The valuative criterion gives K-stability. The nonstable pencils are exactly the displayed $U_{a,c}$ . Their K-polystability and strict K-semistability are Proposition 70.4; $U_{0,0}$ is not K-stable because its automorphism group is positive-dimensional. These facts prove every direction of the classification, and also prove K-semistability of all smooth members. ◻

The classification can be read directly from the Wronskian sextic:

Wronskian condition	K-stability of $X_U$
Every root has multiplicity at most two	K-stable
Exactly two triple roots	K-polystable, not K-stable
Exactly one triple root	Strictly K-semistable

There are no other cases on $\mathcal H^\circ$ . The second row is the single orbit $\langle x^3y,xy^3\rangle$ . The third consists of $\langle x^3y,xy^3+a x^2y^2+c x^4\rangle$ with $(a,c)\ne(0,0)$ , parametrized up to isomorphism by $\mathbb P(1,3)$ . This explicitly describes all exceptions to K-stability, including which exception is K-polystable.

82.1 An intrinsic classification by the conic bundle

The criterion does not require finding a binary quartic pencil. Let $\pi:X\to V_5$ be the birational contraction, with exceptional divisor $E$ , and let $\eta:X\to\mathbb P^2$ be the other extremal contraction. Let $\Delta$ be the reduced nodal cubic discriminant of $\eta$ , and let $Q$ be the smooth branch conic of the finite double cover $\eta|_E:E\to\mathbb P^2$ . If $Q$ is not a component of $\Delta$ , put $I_b(\Delta,Q)=\dim_{\mathbb C} \mathcal O_{\mathbb P^2,b}/(\Delta,Q).$ Here the equations in the denominator are local equations of the two curves. This finite integer is their local intersection multiplicity.

Corollary 82.1. Every smooth member $X$ of Family 2.20 is K-semistable. Exactly one of the following alternatives holds.

If $Q\subset\Delta$ , then $\Delta=Q\cup M$ , where $M$ is a line transverse to $Q$ . The threefold is K-polystable but not K-stable, and is isomorphic to $X_{\langle x^3y,xy^3\rangle}$ .
If $Q\not\subset\Delta$ and $I_b(\Delta,Q)=2$ at every node $b$ of $\Delta$ , then $X$ is K-stable. In particular every member with smooth discriminant is K-stable.
If $Q\not\subset\Delta$ and $I_b(\Delta,Q)>2$ at some node $b$ of $\Delta$ , then $X$ is strictly K-semistable. Such a node is unique, and its intersection multiplicity is either four or six.

Thus $X$ is K-polystable precisely in the first two alternatives.

Proof. Proposition 76.1 identifies the three alternatives with, respectively, the closed nonstable orbit, the GIT-stable locus, and the nonclosed semistable locus. Theorem 69.1 now supplies every claimed K-stability conclusion; the sufficiency in the second case uses Theorem 80.1.

For the additional numerical assertion, when $Q$ is not a component the restriction $\Delta|_Q$ is the square of a nonzero binary cubic. All intersection multiplicities are therefore even and their sum is six. Every discriminant node lies on $Q$ and contributes at least two. An intersection multiplicity greater than two is four or six, and there cannot be two such points. The second alternative includes the case with no discriminant nodes: its condition is then vacuous. ◻

Remark 82.2. The third alternative tests only the nodes of $\Delta$ . A smooth point of $\Delta$ with $I_b(\Delta,Q)=4$ or $6$ does not satisfy its hypothesis. Such a point corresponds to a reduced conic fibre, and the point-refinement theorem treats its node on $E$ . Confusing these two types of nodes would give an incorrect classification.

82.2 Explicit examples of all alternatives

Use the projective bundle and coordinates of Lemma 73.1. Thus $Ay_0+By_1+Cy_2=Ay_1+By_2+Cy_3=0,$ and $q_0$ is specified by $y_0y_2-y_1^2=Cq_0,\quad y_0y_3-y_1y_2=-Bq_0,\quad y_1y_3-y_2^2=Aq_0.$ Here $E=(z=0)$ and $Q=(B^2=4AC)$ .

Equation of $X$	Equation of $\Delta$	Stability
$q_0+zy_0+Az^2=0$	$4A^2C-AB^2+C^3=0$	K-stable
$q_0+Bz^2=0$	$B(4AC-B^2)=0$	K-polystable, not K-stable
$q_0+zy_2+Bz^2=0$	$A^2C+4ABC-B^3=0$	Strictly K-semistable
$q_0+zy_3+Bz^2=0$	$A^3+4ABC-B^3=0$	Strictly K-semistable

Lemma 82.3. All assertions in the table hold. In its first row, $\Delta$ is smooth and has contact order six with $Q$ at $[1:0:0]$ . In its third and fourth rows the only discriminant node is $[0:0:1]$ , with intersection multiplicity four and six, respectively.

Proof. The determinants of the conic matrices give the displayed cubics. For the first cubic the partial derivatives are $8AC-B^2,-2AB,4A^2+3C^2$ , which have no common projective zero. For each of the last two cubics their only common zero is $[0:0:1]$ . On $C=1$ the quadratic terms there are $A(A+4B)$ and $4AB$ , respectively, so these singularities are ordinary nodes. The second cubic is the union of $Q$ and the transverse line $B=0$ .

Parametrize $Q$ by $[A:B:C]=[s^2:2st:t^2]$ . The restrictions of the four discriminant equations are, in their displayed order, $t^6,\qquad0,\qquad s^4t^2,\qquad s^6.$ This proves the contact assertions, including the distinction between a smooth discriminant point in the first row and a singular discriminant point in the last row.

We check smoothness of the threefolds as well as of their discriminants. Over a smooth discriminant point the conic matrix has rank two, and its determinant has nonzero differential; this makes the total space smooth at the fibre node. At every discriminant node in the table the conic matrix has rank one. Near such a base point, completing a square writes the equation as $u^2+s\,q_1(v,w)+t\,q_2(v,w)+\text{terms of base order at least two}=0.$ The quadratic initial term of the determinant is a nonzero multiple of $\det(sq_1+tq_2)$ . Since the discriminant has an ordinary node, this is a product of distinct linear forms. The quadratics $q_1,q_2$ have no common zero in $\mathbb P^1$ : otherwise, choosing that zero as $[1:0]$ would make this determinant a square. Thus at every point of the double line at least one base derivative of the equation is nonzero. The total space is smooth. Points on smooth fibres and the other points of reduced fibres are automatically smooth.

Finally consider the flat seven-parameter family $q_0+z\sum f_i y_i+z^2(h_0A+h_1B+h_2C)=0$ in the fixed projective bundle. Its smooth parameter locus is a nonempty open subset of $\mathbb A^7$ , hence connected, and contains the model of a member of Family 2.20. Adjunction gives $-K_X=(H+B)|_X$ . The evaluation map of $\mathcal O(1)\oplus G$ embeds the projective bundle in $\mathbb P^6\times\mathbb P^2$ ; thus $H+B$ is very ample. All four equations are therefore in the same Fano deformation family. Their stability follows from Corollary 82.1. ◻