Section 70

Pencils of binary quartics

Part XI · Family 2.20: the K-polystable classificationPDF page 247

Put $V=\operatorname{Sym}^4\mathbb C^2$ and $\mathcal H=\operatorname{Gr}(2,V)$ . We regard an element $U\in \mathcal H$ as a pencil of binary quartics. Let $\mathcal S\subset\mathbb P(V)$ be the surface of squares $[q^2]$ , where $q$ is a nonzero binary quadratic form, and put $\mathcal H^\circ=\{U\in \mathcal H\mid\mathbb P(U)\cap\mathcal S=\varnothing\}.$ Multiplication $\operatorname{Sym}^2(\operatorname{Sym}^2\mathbb C^2)\longrightarrow\operatorname{Sym}^4\mathbb C^2$ is the projection onto the irreducible summand of dimension $5$ . Thus $\mathcal S$ is exactly the projected Veronese surface used in the description of twisted cubics on $V_5$ . If coordinates on $\mathbb P(V)$ are normalized as coefficients of $\binom4i x^{4-i}y^i$ , then the square of $ax^2+2bxy+cy^2$ has coordinates $[a^2:ab:(ac+2b^2)/3:bc:c^2]$ . In particular, the binary-form convention agrees with the parametrization in [3, Section 7.2]. The equivariant identification of the Hilbert scheme of twisted cubics on $V_5$ with $\mathcal H$ identifies $\mathcal H^\circ$ with the smooth twisted cubics. We denote the corresponding curve by $C_U$ and its blowup by $X_U$ .

For $p\in\mathbb P^1$ , choose a basis $f_0,f_1$ of $U$ with distinct orders of vanishing $a_0(p)=\operatorname{ord}_p(f_0)<a_1(p)=\operatorname{ord}_p(f_1).$ These integers do not depend on the adapted basis. They are called the vanishing sequence of $U$ at $p$ .

Lemma 70.1. For the Plücker linearization of the $\operatorname{SL}_2$ -action on $\mathcal H$ , a pencil $U$ is GIT-semistable if and only if $a_0(p)+a_1(p)\le4$ for every $p\in\mathbb P^1$ . It is GIT-stable if and only if all these inequalities are strict. Every $U\in \mathcal H^\circ$ is GIT-semistable. Such a pencil fails to be GIT-stable if and only if its vanishing sequence is $(1,3)$ at some point.

Proof. Every nontrivial one-parameter subgroup of $\operatorname{SL}_2$ is conjugate, up to a positive power, to $(x,y)\mapsto(t x,t^{-1}y)$ . At $p=\{x=0\}$ , the smallest weights of the two adapted basis vectors are $2a_0(p)-4$ and $2a_1(p)-4$ . The smallest weight of their exterior product is therefore $2(a_0(p)+a_1(p)-4)$ . The Hilbert–Mumford criterion gives the first two assertions, with the chosen sign convention.

Suppose now that $U\in \mathcal H^\circ$ . If $a_1(p)=4$ , then $U$ contains a fourth power, contrary to its definition. If $a_0(p)+a_1(p)>4$ and $a_1(p)\le3$ , the sequence is necessarily $(2,3)$ . In coordinates with $p=\{x=0\}$ this implies $U=x^2W$ for a two-dimensional space $W$ of binary quadratics. The projective line $\mathbb P(W)$ meets the conic of squares in $\mathbb P(\operatorname{Sym}^2\mathbb C^2)$ . Thus $U$ contains a square, again a contradiction. Consequently $a_0(p)+a_1(p)\le4$ . Equality can only have the form $(0,4)$ or $(1,3)$ , and the first possibility has already been excluded. ◻

Lemma 70.2. Every pencil in $\mathcal H^\circ$ which is not GIT-stable is $\operatorname{PGL}_2$ -equivalent to $U_{a,c}=\langle x^3y,\ xy^3+a x^2y^2+c x^4\rangle \qquad(a,c\in\mathbb C).$ Every displayed pencil belongs to $\mathcal H^\circ$ . Among them, precisely $U_{0,0}$ is GIT-polystable. Its stabilizer is $\mathbb C^*\rtimes\boldsymbol\mu_2$ . All the other displayed pencils have finite stabilizers, and they admit the degeneration $U_{a,c}(t)=\langle x^3y,\ xy^3+a t^2x^2y^2+c t^6x^4\rangle \longrightarrow U_{0,0}.$ Every member of this family lies in $\mathcal H^\circ$ .

Proof. Let $p$ have vanishing sequence $(1,3)$ . The quartic in $U$ with order $3$ at $p$ has another simple root $q\ne p$ . Choose coordinates with $p=\{x=0\}$ and $q=\{y=0\}$ . This quartic becomes $x^3y$ . After rescaling the other basis vector and subtracting a multiple of $x^3y$ , we obtain the displayed normal form.

Every quartic in $U_{a,c}$ is divisible by $x$ . A square divisible by $x$ is divisible by $x^2$ . But the only elements of $U_{a,c}$ divisible by $x^2$ are multiples of $x^3y$ , which are not squares. Hence $U_{a,c}\in \mathcal H^\circ$ , for all $a,c$ .

The one-parameter subgroup $(x,y)\mapsto(t x,t^{-1}y)$ gives exactly the displayed family after rescaling its basis vectors. The pencil $U_{0,0}$ has vanishing sequence $(1,3)$ at each of its two common roots, and sequence $(0,1)$ elsewhere. For every one-parameter subgroup with zero Hilbert–Mumford weight its limit is a pencil spanned by the two monomials of orders $1$ and $3$ , and is therefore $\operatorname{PGL}_2$ -equivalent to $U_{0,0}$ . The orbit-closure form of the Hilbert–Mumford criterion shows that this orbit is closed in the semistable locus.

If $(a,c)\ne(0,0)$ , the point $p$ is the unique point with vanishing sequence $(1,3)$ . Indeed, when $c\ne0$ , it is the unique common root of the pencil. When $c=0$ and $a\ne0$ , the other common root is $q$ , where the sequence is $(1,2)$ . Every stabilizing transformation fixes $p$ , the unique quartic with order $3$ there, and hence its other root $q$ . It is therefore diagonal. The diagonal action sends $(a,c)$ to $(t^2a,t^6c)$ , so the stabilizer is finite. On the other hand, $U_{0,0}$ is preserved by the diagonal torus and by interchanging $x$ and $y$ ; preservation of its two common roots shows that these are all its automorphisms. Consequently no $U_{a,c}$ with $(a,c)\ne(0,0)$ is equivalent to $U_{0,0}$ , and its orbit is not closed. ◻

Remark 70.3. The nonclosed smooth orbits have a particularly simple parametrization. Two nonzero pairs $(a,c)$ and $(a',c')$ give equivalent pencils if and only if $(a',c')=(t^2a,t^6c)$ for some $t\in\mathbb C^*$ . Thus these orbits are parametrized by $\mathbb P(1,3)$ , or equivalently by $c/a^3\in\mathbb P^1$ . This is a parametrization of isomorphism classes of the strictly K-semistable examples below; it is not their GIT quotient. Their entire collection is contracted to the single closed orbit $U_{0,0}$ by the GIT quotient.

Proposition 70.4. The threefold $X_{U_{0,0}}$ is K-polystable. For every $(a,c)\ne(0,0)$ , the threefold $X_{U_{a,c}}$ is strictly K-semistable. Consequently, if $X_U$ is K-polystable and $U\in \mathcal H^\circ$ , then either $U$ is GIT-stable or $U$ is equivalent to $U_{0,0}$ .

Proof. The pencil $U_{0,0}$ gives the unique smooth member of Family $2.20$ with infinite automorphism group, and its K-polystability is proved in [3, Proposition 5.43].

Pull back the universal curve on $V_5\times \mathcal H$ by the map $t\mapsto U_{a,c}(t)$ of Lemma 70.2. Every fibre is a smooth twisted cubic. Therefore this is a smooth family of regularly embedded curves in $V_5\times\mathbb A^1$ . Blowing up the family gives a smooth projective morphism $\mathcal X\to\mathbb A^1$ , all of whose fibres are Fano threefolds in Family $2.20$ . The relative anticanonical divisor is relatively ample. The construction is equivariant for the displayed one-parameter subgroup, and over $\mathbb A^1\setminus\{0\}$ it is equivariantly isotrivial with fibre $X_{U_{a,c}}$ . Hence it is a special test configuration of $X_{U_{a,c}}$ with central fibre $X_{U_{0,0}}$ .

K-semistability is open in a smooth family of Fano varieties; see [3, Theorem 1.11]. Since the central fibre is K-polystable, some nonzero fibre is K-semistable, and all nonzero fibres are isomorphic. Thus $X_{U_{a,c}}$ is K-semistable. The Donaldson–Futaki invariant of this test configuration is the Futaki invariant of the induced action on its central fibre. It is zero: on a K-polystable variety both a product test configuration and the one induced by the inverse action have nonnegative Donaldson–Futaki invariants, which are negatives of one another.

Finally, $X_{U_{a,c}}$ is not isomorphic to $X_{U_{0,0}}$ . An isomorphism of two such blowups preserves the unique birational extremal contraction, the other extremal contraction being a conic bundle. It descends to an automorphism of $V_5$ carrying one centre to the other. Since $\operatorname{Aut}(V_5)=\operatorname{PGL}_2$ , equivariance of the Hilbert-scheme description would then identify the corresponding pencils. Lemma 70.2 rules this out. Our test configuration is therefore nonproduct, proving strict K-semistability. ◻

Example 70.5. The pencil $\langle x^4+y^4,x^3y\rangle$ belongs to $\mathcal H^\circ$ and is GIT-stable. To check the first assertion, a square with nonzero coefficient of $x^4+y^4$ would have the form $(A x^2+Bxy+C y^2)^2$ , with $AC\ne0$ . The vanishing of its $xy^3$ coefficient forces $B=0$ , but then its $x^2y^2$ coefficient is $2AC\ne0$ , a contradiction. The remaining pencil member $x^3y$ is not a square. The pencil has no common root, so its first vanishing order is always $0$ , and its second order is at most $3$ , because a fourth-order zero would be a square. Thus it is GIT-stable by Lemma 70.1.

In contrast, $\langle x^3y,xy^3+x^4\rangle$ gives a strictly K-semistable smooth threefold, while $\langle x^3y,xy^3\rangle$ gives a K-polystable smooth threefold. These examples show why one must distinguish K-polystability from K-stability in this family.

Remark 70.6. The preceding argument proves a necessary GIT criterion for K-polystability and completely identifies the smooth nonclosed GIT orbits. It does not prove that every GIT-stable pencil gives a K-stable threefold. The required implication is proved below by the valuative estimates and Theorem 80.1; neither reductivity of the stabilizer nor positivity of the CM line bundle proves it by itself.

70.1 A sextic test

Let $f,g$ be a basis of a pencil $U\subset\operatorname{Sym}^4\mathbb C^2$ . Its Wronskian is the binary sextic $\operatorname{Wr}(U)=\left[\frac{\partial f}{\partial x}\frac{\partial g}{\partial y}-\frac{\partial f}{\partial y}\frac{\partial g}{\partial x}\right]\in\mathbb P(\operatorname{Sym}^6\mathbb C^2).$ Changing the basis multiplies the displayed form by a nonzero constant, so its projective class is well defined.

Lemma 70.7. The Wronskian defines a finite surjective $\operatorname{SL}_2$ -equivariant morphism $\operatorname{Wr}\colon\operatorname{Gr}(2,\operatorname{Sym}^4\mathbb C^2)\longrightarrow\mathbb P(\operatorname{Sym}^6\mathbb C^2).$ If $a_0(p)<a_1(p)$ is the vanishing sequence of $U$ at $p$ , then $\operatorname{mult}_p\operatorname{Wr}(U)=a_0(p)+a_1(p)-1.$

Proof. The homogeneous Jacobian is linear in $f\wedge g$ and is equivariant for $\operatorname{SL}_2$ . It never vanishes identically for independent quartics. Indeed, on an affine chart, vanishing would give $(f/g)'=0$ and hence $f/g$ constant in characteristic zero. Thus the corresponding linear projection of the Plücker embedding has no base point on the Grassmannian, and defines a morphism. Its pullback of $\mathcal O_{\mathbb P^6}(1)$ is the Plücker line bundle. Since this is ample, a positive-dimensional fibre would contain a projective curve on which an ample line bundle is trivial, which is impossible. Hence the morphism is finite. Both source and target have dimension $6$ , so it is surjective.

For the local calculation, take $p=\{x=0\}$ and write $f(z,1)=z^{a_0}u(z)$ and $g(z,1)=z^{a_1}v(z)$ , where $u(0)v(0)\ne0$ . Euler’s identity gives the homogeneous Jacobian on this chart as $4(f'g-fg')$ . Its first nonzero coefficient is $4(a_0-a_1)u(0)v(0)$ at order $a_0+a_1-1$ . This coefficient is nonzero. The asserted multiplicity follows. ◻

Corollary 70.8. Let $U\in \mathcal H^\circ$ . Its Wronskian has no root of multiplicity greater than $3$ . Moreover:

$U$ is GIT-stable if and only if every root of its Wronskian has multiplicity at most $2$ ;
$U$ is GIT-polystable and not GIT-stable if and only if its Wronskian consists of two distinct triple roots;
$U$ has a nonclosed GIT-semistable orbit if and only if its Wronskian has exactly one triple root.

In the last case $X_U$ is strictly K-semistable. Thus a necessary condition for $X_U$ to be K-polystable is that its Wronskian have no triple root, or have precisely two triple roots. The sufficiency of the first alternative is proved in Section 82.

Proof. The multiplicity formula and Lemma 70.1 prove the first assertion and (1). In the nonstable case use Lemma 70.2. Direct differentiation gives $\operatorname{Wr}(U_{a,c})=[x^3(2y^3+a xy^2-c x^3)].$ The second factor has a nonzero $y^3$ coefficient, so the displayed root $x=0$ has multiplicity exactly $3$ . Another triple root occurs if and only if $2y^3+a xy^2-c x^3$ is the cube of a linear form. Its $x^2y$ coefficient is zero; comparing coefficients in $2(y-dx)^3$ forces $d=0$ , and then $a=c=0$ . The claim follows from Proposition 70.4. ◻

Example 70.9. For the smooth GIT-stable pencil $U=\langle x^4+y^4,x^3y\rangle$ used above, the Wronskian is $\operatorname{Wr}(U)=[x^2(x^4-3y^4)].$ It has one double root and four simple roots, in agreement with the criterion. For $U_{0,1}$ the Wronskian is $[x^3(2y^3-x^3)]$ , with one triple root and three simple roots. For $U_{0,0}$ it is $[x^3y^3]$ .