# Comptes Rendus
Mathématique

Song Yang and Xun Yu

Rational cubic fourfolds in Hassett divisors

Volume 358, issue 2 (2020), p. 129-137

Published online: 15 June 2020

https://doi.org/10.5802/crmath.4

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CREATIVE COMMONS ATTRIBUTION 4.0 INTERNATIONAL LICENSE.

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Currently, there exist two popular point of views toward the rationality of smooth cubic fourfolds and both have associated $K3$ surfaces:

* Hassett's Hodge-theoretic result ([11, Theorem 5.1.3]): a smooth cubic fourfold $X$ has a Hodge-theoretically associated $K3$ surface if and only if its moduli point $[X] \in \mathscr{C}_d$ for some admissible value $d$ (i.e., $d > 6$, $d \equiv 0,2 \pmod 6$, $4 \nmid d$, $9 \nmid d$ and $p \nmid d$ for any odd prime $p \equiv 2 \pmod 3$);

* Kuznetsov's derived categorical conjecture ([16, Conjecture 1.1]): a smooth cubic fourfold $X$ is rational if and only if its Kuznetsov component $\mathrm{Ku}(X)$ is derived equivalent to a $K3$ surface (i.e., $\mathrm{Ku}(X)$ is called *geometric*), where $\mathrm{Ku}(X)$ is the right orthogonal to the exceptional collection $\{\mathcal{O}_X, \mathcal{O}_X(1), \mathcal{O}_X(2)\}$ in the bounded derived category of coherent sheaves on $X$.

It is important to notice that Kuznetsov's conjecture implies that a very general cubic fourfold is not rational, since for a very general cubic fourfold its Kuznetsov component can not be geometric. Addington-Thomas [2, Theorem 1.1] showed that for a smooth cubic fourfold $X$ if $\mathrm{Ku}(X)$ is geometric then $[X] \in \mathscr{C}_d$ for some admissible $d$, and conversely for any admissible value $d$, the set of cubic fourfolds $[X] \in \mathscr{C}_d$ for which $\mathrm{Ku}(X)$ is geometric is a Zariski open dense subset; see also Huybrechts [13] for the twisted version and a further study. Recently, based on Bridgeland stability conditions on $\mathrm{Ku}(X)$ constructed in [5, Theorem 1.2], Bayer-Lahoz-Macrì-Nuer-Perry-Stellari [4, Corollary 29.7] proved that for any admissible value $d$, $\mathrm{Ku}(X)$ is geometric for every $[X] \in \mathscr{C}_d$. So we now know that for a smooth cubic fourfold $X$ its Kuznetsov component $\mathrm{Ku}(X)$ is geometric if and only if $[X] \in \mathscr{C}_d$ for some admissible value $d$. Then one can restate Kuznetsov's conjecture as the following equivalent form.

**Conjecture 1.** A smooth cubic fourfold $X$ is rational if and only if $[X] \in \mathscr{C}_d$ for some admissible value $d$.

The first three admissible values are $14, 26, 38$. Every cubic fourfold in $\mathscr{C}_{14}$ is rational $[7, 9]$; see also [21, Theorem 2] for a different proof. Based on Kontsevich-Tschinkel [15, Theorem 1], Russo-Staglianò [21, Theorems 4, 7] finally showed that every cubic fourfold in $\mathscr{C}_{26}$ and $\mathscr{C}_{38}$ is rational; see also [20] for the construction of explicit birational maps. So far “if ” part of Conjecture 1 has been confirmed only for the three Hassett divisors $\mathscr{C}_{14}, \mathscr{C}_{26}, \mathscr{C}_{38}$. Thus finding rational cubic fourfolds in other Hassett divisors is of interest. The main result of this paper is the following.

**Theorem 2 (=Theorem 9).** Every Hassett divisor $\mathscr{C}_d$ contains a union of three subvarieties parametrizing rational cubic fourfolds, of codimension-two in $\mathscr{C}$.

The idea of the proof is simple: we first show any two Hassett divisors intersect by Theorem 7, which is of independent interest (for considerations of the intersections among Hassett divisors, see [2, 3, 7, 10] etc.), and finally we consider the intersections $\mathscr{C}_d \cap \mathscr{C}_{14}$, $\mathscr{C}_d \cap \mathscr{C}_{26}$ and $\mathscr{C}_d \cap \mathscr{C}_{38}$ for every Hassett divisor $\mathscr{C}_d$.

After completing this paper, Russo-Staglianò [22] announced the rationality of every cubic fourfold in $\mathscr{C}_{42}$. We remark that our method used for the proof of Theorem 2 also works in this case (in particular, it can be shown that the four intersections $\mathscr{C}_d \cap \mathscr{C}_{14}$, $\mathscr{C}_d \cap \mathscr{C}_{26}$, $\mathscr{C}_d \cap \mathscr{C}_{38}$, $\mathscr{C}_d \cap \mathscr{C}_{42}$ are mutually distinct).

Throughout this paper, we work over the complex number field $\mathbb{C}$.

### Acknowledgements

We would like to thank Professors Keiji Oguiso and Paolo Stellari for useful conversations. We also would like to thank the referee for helpful comments.
## 2. Lattice and Hodge theory for cubic fourfolds

In this section, we collect some known results on Hodge structures and lattices associated with smooth cubic fourfolds. We refer to [6, 11, 12, 14] for more detailed discussions, especially for the Hodge-theoretic aspect, and to [19, 23] for the basics of abstract lattice theory.

The cubic hypersurfaces in $\mathbb{P}^5$ are parametrized by $\mathbb{P}(H^0(\mathbb{P}^5, \mathcal{O}(3))) \cong \mathbb{P}^{55}$. Moreover, the smooth cubic hypersurfaces form a Zariski open dense subset $\mathcal{U} \subset \mathbb{P}^{55}$. Then the moduli space of smooth cubic fourfolds is the quotient space

$$ \mathcal{C} := \mathcal{U} / \mathrm{PGL}(6, \mathbb{C}) $$

which is a 20-dimensional quasi-projective variety.

Let $X$ be a smooth cubic fourfold. Then the cohomology $H^*(X, \mathbb{Z})$ is torsion-free and the Hodge numbers for the middle cohomology of $X$ are as follows:

$$ 0 \ 1 \ 21 \ 1 \ 0. $$

The Hodge-Riemann bilinear relations imply that $H^4(X, \mathbb{Z})$ is a unimodular lattice under the intersection form (.) of signature (21,2). Furthermore, as abstract lattices, [11, Proposition 2.1.2] implies the middle cohomology and the primitive cohomology

$$ L := E_8^{\oplus 2} \oplus U^{\oplus 2} \oplus I_{3,0} \cong H^4(X, \mathbb{Z}) $$

$$ L^0 := (h^2)^\perp \cong E_8^{\oplus 2} \oplus U^{\oplus 2} \oplus A_2 \cong H_{\mathrm{prim}}^4(X, \mathbb{Z}) $$

where the square of the hyperplane class $h$ is given as $h^2 = (1, 1, 1) \in I_{3,0}$ of which the intersection form is given by the identity matrix of rank 3, $A_2 = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, $U = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ the hyperbolic plane, $E_8$ is the unimodular positive definite even lattice of rank 8. Note that $L^0$ is an even lattice.

**Definition 3 (Hassett [11]).** A smooth cubic fourfold $X$ is called special if it contains an algebraic surface not homologous to a complete intersection.

The integral Hodge conjecture holds for smooth cubic fourfolds ([25, Theorem 18] or see also [4, Corollary 29.8] for a new proof). Thus, a smooth cubic fourfold $X$ is special if and only if the rank of the positive definite lattice

$$ A(X) := H^4(X, \mathbb{Z}) \cap H^{2,2}(X) $$

is at least 2.

**Definition 4 (Hassett [11]).** A labelling of a special cubic fourfold consists of a positive definite rank two saturated (i.e. the quotient group $A(X)/K$ is torsion free) sublattice

$$ K \subset A(X) \text{ such that } h^2 \in K, $$

and its discriminant $d$ is the determinant of the intersection form on $K$.

In [11], Hassett defined $\mathcal{C}_d$ as the set of special cubic fourfolds $X$ with labelling of discriminant $d$. Moreover, Hassett [11, Theorem 1.0.1] showed that $\mathcal{C}_d \subset \mathcal{C}$ is an irreducible divisor and is nonempty if and only if

$$ d > 6 \text{ and } d \equiv 0,2 \pmod{6}. \qquad (*) $$

The following proposition is a generalization of [11, Theorems 1.0.1].

**Proposition 5 ([12, Proposition 12 and p. 43]).** Fix a positive definite lattice $M$ of rank $r \ge 2$ admitting a saturated embedding

$$ M \subset L \text{ such that } h^2 \in M. $$
We denote by $\mathscr{C}_M \subset \mathscr{C}$ the smooth cubic fourfolds $X$ admitting algebraic classes with this lattice structure

$$h^2 \in M \subset A(X) \subset L.$$

Then $\mathscr{C}_M$ has codimension $r-1$ and there exists a cubic fourfold $[X] \in \mathscr{C}_M$ with $A(X) = M$, provided $\mathscr{C}_M$ is nonempty. Moreover, $\mathscr{C}_M$ is nonempty if and only if there exists no sublattice $K \subset M$, $h^2 \in K$, with $K = K_2$ or $K_6$, where $K_2 = \begin{pmatrix} 3 & 1 \\ 1 & 1 \end{pmatrix}$ and $K_6 = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$.

This proposition is crucial for our purpose, so we sketch a proof for the convenience of readers.

**Sketch of proof.** Suppose $\mathscr{C}_M$ is nonempty. If $K_6 \subset M$ is a sublattice with $h^2 \in K_6$, then there is a smooth cubic fourfold $X$ such that $A(X) \cap \langle h^2 \rangle^\perp$ contains an element $r$ with $(r.r) = 2$ and this contradicts Voisin [24, Section 4, Proposition 1]; furthermore, Hassett [11, Theorem 4.4.1] excludes the case when $K_2 \subset M$ is a sublattice with $h^2 \in K_2$.

Conversely, suppose that there exists no rank two sublattice $K \subset M$, $h^2 \in K$, with $K = K_2$ or $K_6$. Since the signature of $L$ is $(21, 2)$ and $M \subset L$ is positive definite, by a standard argument, one can always find $\omega \in L \otimes_Z \mathbb{C}$ such that

$$(\omega, \omega) = 0, \ (\omega, \bar{\omega}) < 0 \text{ and } L \cap \omega^\perp = M.$$

According to the description of the image of the period map for cubic fourfolds (Laza [17, Theorem 1.1] and Looijenga [18, Theorem 4.1]), one has a smooth cubic fourfold $X$ and an isometry $\phi: H^4(X, \mathbb{Z}) \xrightarrow{\cong} L$ mapping the square of the hyperplane class to $h^2 \in L$ and a generator of $H^{3,1}(X)$ to $\omega$. Thus $M = A(X)$ and hence $\mathscr{C}_M$ contains $[X]$ and nonempty. $\square$

In the rest of the text, we will frequently use the following lemma to check the nonemptiness condition in the Proposition 5.

**Lemma 6.** Let $M \subset L$ be a positive definite saturated sublattice and $h^2 \in M$. Then the following three conditions are equivalent:

(i) there exists no sublattice $K \subset M$, $h^2 \in K$, with $K = K_2$ or $K_6$;

(ii) there exists no $r \in M$ such that $(r.r) = 2$ (i.e., $M$ does not represent 2);

(iii) for any $0 \neq x \in M$, $(x.x) \geq 3$.

In particular, if $M$ satisfies one of the three equivalent conditions, then $\phi \neq \mathscr{C}_M \subset \mathscr{C}_{M'}$ for any saturated sublattice $M' \subset M \subset L$ such that $h^2 \in M'$.

**Proof.** First of all, (ii) $\Rightarrow$ (i) is clear since both $K_2$ and $K_6$ represent 2.

Secondly, (i) $\Rightarrow$ (ii). Suppose that there exists $r \in M$ such that $(r.r) = 2$. We denote by $K \subset M$ the sublattice generated by $h^2$ and $r$. Hence, the Gram matrix of $K$ with respect to the basis $(h^2, r)$ is

$$\begin{pmatrix} (h^2.h^2) & (h^2.r) \\ (r.h^2) & (r.r) \end{pmatrix} = \begin{pmatrix} 3 & a \\ a & 2 \end{pmatrix}.$$

Replacing $r$ by $-r$ if necessary, we may and will assume $a \geq 0$. Since $K$ is positive definite, we have $a^2 < 6$ and thus $a = 0, 1, 2$. If $a = 0$ (resp. 2), then $K$ is isometric to $K_6$ (resp. $K_2$), contradiction. If $a = 1$, then $h^2 - 3r \in (h^2)^\perp = L^0$ and $((h^2 - 3r).(h^2 - 3r)) = 15$, an odd number, contradicting to the fact $L^0$ is even.

Finally, clearly (iii) implies (ii). Conversely, we show (ii) implies (iii). By hypothesis, we may assume that there is $r \in M$ with $(r.r) = 1$. Then let $K \subset M$ be the sublattice generated by $h^2$ and $r$. Hence, the Gram matrix of $K$ with respect to the basis $(h^2, r)$ is

$$\begin{pmatrix} 3 & a \\ a & 1 \end{pmatrix}.$$

where $a = (h^2.r)$. Replacing $r$ by $-r$ if necessary, we may and will assume $a \geq 0$. Since $K$ is positive definite, we have $a^2 < 3$ and thus $a = 0, 1$. If $a = 0$, then $r \in (h^2)^\perp = L^0$ and $(r.r) = 1$, an odd
number, contradicting to the fact $L^0$ is even. If $a = 1$, then $K$ is isometric to $K_2$ and $K$ represents 2, contradiction. □

**3. Intersections of Hassett divisors**

In this section, we prove Theorem 2 (=Theorem 9) and discuss some related results (Theorem 7 and Theorem 13).

Firstly, we setup some notations for latter use. Let

$$L = E_8^{\oplus 2} \oplus U_1 \oplus U_2 \oplus I_{3,0},$$

where $U_1$ and $U_2$ are two copies of $U$. The standard basis of $U$ consists of isotropic elements $e, f$ with $(e.f) = 1$. We denote the standard basis of $U_i$ by $e_i, f_i$, $i = 1, 2$, and denote by $h^2$ the element $(1, 1, 1) \in I_{3,0} \subset L$.

We will use the following theorem, an interesting result for itself, to prove Theorem 9.

**Theorem 7.** Any two Hassett divisors intersect, i.e., $\mathscr{C}_{d_1} \cap \mathscr{C}_{d_2} \neq \emptyset$ for any two integers $d_1$ and $d_2$ satisfying $(\star)$. Moreover, there exists a smooth cubic fourfold X and a codimension-two subvariety $\mathscr{C}_{A(X)} \subset \mathscr{C}$ such that $[X] \in \mathscr{C}_{A(X)} \subset \mathscr{C}_{d_1} \cap \mathscr{C}_{d_2}$ and $A(X)$ is a rank 3 lattice with discriminant $d_1 d_2/3$, except if both $d_1$ and $d_2$ are $\equiv 2 \pmod 6$, in which case the discriminant is $(d_1 d_2 - 1)/3$.

**Proof.** By definition, an integer $d$ satisfies $(\star)$ if $d > 6$ and $d \equiv 0, 2 \pmod 6$. Therefore, the proof is divided into three cases:

**Case 1:** $d_1 \equiv 0 \pmod 6$ and $d_2 \equiv 0 \pmod 6$. Suppose $d_1 = 6n_1$, $d_2 = 6n_2$ and $n_1, n_2 \ge 2$. We consider the rank 3 lattice

$$M := \langle h^2, \alpha_1, \alpha_2 \rangle \subset L$$

generated by $h^2$, $\alpha_1 := e_1 + n_1 f_1$ and $\alpha_2 := e_2 + n_2 f_2$. Then the Gram matrix of $M$ with respect to the basis $(h^2, \alpha_1, \alpha_2)$ is

$$\begin{pmatrix} 3 & 0 & 0 \\ 0 & 2n_1 & 0 \\ 0 & 0 & 2n_2 \end{pmatrix}.$$

Therefore, $M \subset L$ is positive definite saturated sublattice such that $h^2 \in M$. In addition, for any nonzero $v = xh^2 + y\alpha_1 + z\alpha_2 \in M$, where $x, y, z$ are integers, we have

$$(v.v) = 3x^2 + 2n_1y^2 + 2n_2z^2 \ge 3$$

since $n_1, n_2 \ge 2$ and at least one of the integers $x, y, z$ is nonzero. Hence, the embedding $M \subset L$ satisfies Lemma 6(iii). Thus, by Lemma 6 and Proposition 5, $\mathscr{C}_M \subset \mathscr{C}$ is nonempty and has codimension 2, and there exists a cubic fourfold $[X] \in \mathscr{C}_M$ with $A(X) = M$. Thus $A(X)$ is a rank 3 lattice of discriminant $\mathrm{disc}(A(X)) = d_1 d_2/3$. Moreover, we consider the sublattices

$$K_{d_1} := \langle h^2, \alpha_1 \rangle \subset M$$

with discriminant $d_1$, and

$$K_{d_2} := \langle h^2, \alpha_2 \rangle \subset M$$

with discriminant $d_2$. Clearly, both $K_{d_1}$ and $K_{d_2}$ are saturated sublattices of $M$. Applying Lemma 6 and Proposition 5 again, we obtain $[X] \in \mathscr{C}_M \subset \mathscr{C}_{K_{d_1}} = \mathscr{C}_{d_1}$ and $[X] \in \mathscr{C}_M \subset \mathscr{C}_{K_{d_2}} = \mathscr{C}_{d_2}$. Consequently, $[X] \in \mathscr{C}_M \subset \mathscr{C}_{d_1} \cap \mathscr{C}_{d_2}$ is what we want.

**Case 2:** $d_1 \equiv 0 \pmod 6$ and $d_2 \equiv 2 \pmod 6$. Given $d_1 = 6n_1$ and $d_2 = 6n_2 + 2$ with $n_1 \ge 2, n_2 \ge 1$. We consider the rank 3 sublattice

$$M := \langle h^2, \alpha_1, \alpha_2 + (0,0,1) \rangle \subset L$$
where $(0, 0, 1) \in I_{3,0}$. Then the Gram matrix of $M$ with respect to the basis $(h^2, \alpha_1, \alpha_2 + (0, 0, 1))$ is

$$ \begin{pmatrix} 3 & 0 & 1 \\ 0 & 2n_1 & 0 \\ 1 & 0 & 2n_2 + 1 \end{pmatrix} $$

Thus, $M \subset L$ is positive definite saturated sublattice with $h^2 \in M$. Furthermore, for any nonzero $v = xh^2 + y\alpha_1 + z(\alpha_2 + (0, 0, 1)) \in M$, we get

$$ (v.v) = 2x^2 + 2n_1 y^2 + 2n_2 z^2 + (x+z)^2 \geq 3 $$

since $n_1 \geq 2$, $n_2 \geq 1$ and at least one of the integers $x, y, z$ is nonzero. Hence, by Lemma 6 and Proposition 5, we conclude that $\mathscr{C}_M \subset \mathscr{C}$ is nonempty and has codimension 2, and there exists a cubic fourfold $[X] \in \mathscr{C}_M$ with $A(X) = M$. Thus $A(X)$ is a rank 3 lattice of discriminant disc($A(X)$) = $d_1d_2/3$. Similarly, we consider the sublattices:

$$ K_{d_1} := \langle h^2, \alpha_1 \rangle \subset M $$

of discriminant $d_1$, and

$$ K_{d_2} := \langle h^2, \alpha_2 + (0, 0, 1) \rangle \subset M $$

of discriminant $d_2$. Again Lemma 6 and Proposition 5 imply $[X] \in \mathscr{C}_M \subset \mathscr{C}_{K_{d_1}} = \mathscr{C}_{d_1}$ and $[X] \in \mathscr{C}_M \subset \mathscr{C}_{K_{d_2}} = \mathscr{C}_{d_2}$. Consequently, $[X] \in \mathscr{C}_M \subset \mathscr{C}_{d_1} \cap \mathscr{C}_{d_2}$ is what we wanted.

**Case 3:** $d_1 \equiv 2 \pmod 6$ and $d_2 \equiv 2 \pmod 6$. Assume $d_1 = 6n_1+2$ and $d_2 = 6n_2+2$ with $n_1, n_2 \geq 1$. We consider the rank 3 sublattice

$$ M := \langle h^2, \alpha_1 + (0, 1, 0), \alpha_2 + (0, 0, 1) \rangle \subset L $$

here $(0, 1, 0) \in I_{3,0}$. Then the Gram matrix of $M$ with respect to the basis $(h^2, \alpha_1 + (0, 1, 0), \alpha_2 + (0, 0, 1))$ is

$$ \begin{pmatrix} 3 & 1 & 1 \\ 1 & 2n_1+1 & 0 \\ 1 & 0 & 2n_2+1 \end{pmatrix} $$

Thus, $M \subset L$ is positive definite saturated sublattice such that $h^2 \in M$. For any nonzero $v = xh^2 + y(\alpha_1 + (0, 1, 0)) + z(\alpha_2 + (0, 0, 1)) \in M$, we obtain

$$ (v.v) = x^2 + 2n_1 y^2 + 2n_2 z^2 + (x+y)^2 + (x+z)^2 \geq 3 $$

since $n_1, n_2 \geq 1$ and at least one of the integers $x, y, z$ is nonzero. Hence, Lemma 6 and Proposition 5 concludes that $\mathscr{C}_M \subset \mathscr{C}$ is nonempty and has codimension 2, and there exists a cubic fourfold $[X] \in \mathscr{C}_M$ with $A(X) = M$. Thus $A(X)$ is a rank 3 lattice of discriminant disc($A(X)$) = $(d_1d_2 - 1)/3$. Moreover, we consider

$$ K_{d_1} := \langle h^2, \alpha_1 + (0, 1, 0) \rangle \subset M $$

with discriminant $d_1$ and

$$ K_{d_2} := \langle h^2, \alpha_2 + (0, 0, 1) \rangle \subset M $$

with discriminant $d_1$. By Lemma 6 and Proposition 5, we obtain $[X] \in \mathscr{C}_M \subset \mathscr{C}_{K_{d_1}} = \mathscr{C}_{d_1}$ and $[X] \in \mathscr{C}_M \subset \mathscr{C}_{K_{d_2}} = \mathscr{C}_{d_2}$. As a consequence, $[X] \in \mathscr{C}_M \subset \mathscr{C}_{d_1} \cap \mathscr{C}_{d_2}$ is what we wanted. This finishes the proof of Theorem 7. □

**Remark 8.** Note that it has been known for 20 years that $\mathscr{C}_8 \cap \mathscr{C}_{14} \neq \emptyset$ (Hassett [10]) and proved more recently that $\mathscr{C}_8$ intersects every Hassett divisor (Addington–Thomas [2, Theorem 4.1]). It is also shown that $\mathscr{C}_8 \cap \mathscr{C}_{14}$ has five irreducible components ($[3, 7]$). Moreover, [7, p. 166] has mentioned that $\mathscr{C}_{14}$ intersects many other divisors $\mathscr{C}_d$, however, it is not obvious to see which Hassett divisors intersect with $\mathscr{C}_{14}$.
Consequently, Theorem 7 not only generalizes [2, Theorem 4.1] but also implies that $\mathscr{C}_{14}$ intersects all Hassett divisors. Because of the same reason, we may conclude the main result of the current paper.

**Theorem 9 (=Theorem 2).** Every Hassett divisor $\mathscr{C}_d$ contains a union of three subvarieties parametrizing rational cubic fourfolds, of codimension-two in $\mathscr{C}$.

**Proof.** Applying Theorem 7 to the pairs of integers $(d_1, d_2) = (d, 14)$, $(d, 26)$, $(d, 38)$. Then there exist three smooth cubic fourfolds $X_1, X_2$ and $X_3$ such that

$$[X_1] \in \mathscr{C}_{A(X_1)} \subset \mathscr{C}_d \cap \mathscr{C}_{14} \subset \mathscr{C}_d,$$

$$[X_2] \in \mathscr{C}_{A(X_2)} \subset \mathscr{C}_d \cap \mathscr{C}_{26} \subset \mathscr{C}_d,$$

$$[X_3] \in \mathscr{C}_{A(X_3)} \subset \mathscr{C}_d \cap \mathscr{C}_{38} \subset \mathscr{C}_d,$$

where $\mathscr{C}_{A(X_1)}, \mathscr{C}_{A(X_2)}$, and $\mathscr{C}_{A(X_3)}$ are subvarieties of codimension-two in $\mathscr{C}$. Here $A(X_1), A(X_2)$ and $A(X_3)$ are three different rank 3 lattices of discriminants:

* if $d \equiv 0 \pmod{6}$, then $\mathrm{disc}(A(X_1)) = 14d/3$, $\mathrm{disc}(A(X_2)) = 26d/3$ and $\mathrm{disc}(A(X_3)) = 38d/3$;

* if $d \equiv 2 \pmod{6}$, then $\mathrm{disc}(A(X_1)) = (14d - 1)/3$, $\mathrm{disc}(A(X_2)) = (26d - 1)/3$ and $\mathrm{disc}(A(X_3)) = (38d - 1)/3$.

By definition of $\mathscr{C}_{A(X_i)}$ (see Proposition 5), a smooth cubic fourfold $[X] \in \mathscr{C}_{A(X_i)}$ only if there exists a saturated embedding $A(X_i) \subset A(X)$. Since $A(X_1), A(X_2)$ and $A(X_3)$ are rank 3 lattices of different discriminants, it follows that there is no saturated embedding $A(X_i) \subset A(X_j)$ if $i \neq j$. Therefore, $[X_i] \notin \mathscr{C}_{A(X_j)}$ if $i \neq j$ and $\mathscr{C}_{A(X_1)}, \mathscr{C}_{A(X_2)}$, and $\mathscr{C}_{A(X_3)}$ are three different subvarieties of codimension-two in $\mathscr{C}$.

Moreover, since every smooth cubic fourfold in $\mathscr{C}_{14}, \mathscr{C}_{26}$ and $\mathscr{C}_{38}$ is rational ([7, 21]), so every smooth cubic fourfold in $\mathscr{C}_{A(X_1)}, \mathscr{C}_{A(X_2)}$ and $\mathscr{C}_{A(X_3)}$ is rational. Therefore, $\mathscr{C}_{A(X_1)}, \mathscr{C}_{A(X_2)}$ and $\mathscr{C}_{A(X_3)}$ are three different codimension-two subvarieties which parametrize rational cubic fourfolds. This completes the proof of Theorem 9. $\square$

Our main result also motivates the following natural question:

**Question 10.** Suppose that $d$ satisfies $(\star)$ and $d$ is not an admissible value. Does the Hassett divisor $\mathscr{C}_d$ contain a union of countably infinite codimension-two subvarieties in $\mathscr{C}$ parametrizing rational cubic fourfolds?

The answer to Question 10 has already been known for $\mathscr{C}_8$ and $\mathscr{C}_{18}$ ([1, 10]).

**Corollary 11.** The answer to Question 10 is yes if the "if" part of Conjecture 1 holds.

Returning to Conjecture 1, as a by-product of Theorem 9 (=Theorem 2), we have the following.

**Corollary 12.** For every admissible value $d$, the Hassett divisor $\mathscr{C}_d$ contains a union of three subvarieties parametrizing rational cubic fourfolds, of codimension-two in $\mathscr{C}$.

To obtain more information about the Hassett divisors, it is of importance to notice that Addington-Thomas [2, Theorem 4.1] showed that for any $d$ satisfying $(\star)$ there exists a cubic fourfold $[X] \in \mathscr{C}_8 \cap \mathscr{C}_d$ such that $[X] \in \mathscr{C}_d$ for some admissible value $d'$. Even if it is conjectured to be rational, however, it is still unknown whether such a $X$ is rational or not. Using the idea of the proof of Theorem 7 and Theorem 9, we obtain a generalization of [2, Theorem 4.1].

**Theorem 13.** If $d_1$ and $d_2$ satisfy $(\star)$, then $\mathscr{C}_{14} \cap \mathscr{C}_{d_1} \cap \mathscr{C}_{d_2}$ contains a codimension-three subvariety in $\mathscr{C}$ parametrizing rational cubic fourfolds.
**Proof.** Analogously to the proof of Theorem 7, we only need to consider three cases:

**Case 1.** Given $d_1 = 6n_1$ and $d_2 = 6n_2$ with $n_1, n_2 \ge 2$. We consider the rank 4 sublattice

$$M := \langle h^2, v, \alpha_1, \alpha_2 \rangle \subset L$$

where $v = (3, 1, 0) \in I_{3,0} \subset L$, $\alpha_1 := e_1 + n_1 f_1$ and $\alpha_2 := e_2 + n_2 f_2$. Then the Gram matrix of $M$ with respect to the basis $(h^2, v, \alpha_1, \alpha_2)$ is

$$\begin{pmatrix} 3 & 4 & 0 & 0 \\ 4 & 10 & 0 & 0 \\ 0 & 0 & 2n_1 & 0 \\ 0 & 0 & 0 & 2n_2 \end{pmatrix}$$

Thus, $M \subset L$ is positive definite saturated sublattice with $h^2 \in M$. For any nonzero $v = x_1 h^2 + x_2 v + x_3 \alpha_1 + x_4 \alpha_2 \in M$, we have

$$(v \cdot v) = 2(x_1 + 2x_2)^2 + x_1^2 + 2x_2^2 + 2n_1 x_3^2 + 2n_2 x_4^2 \ge 3$$

since $n_1, n_2 \ge 2$ and at least one of the integers $x_i$ is nonzero ($i = 1, 2, 3, 4$). Hence, Lemma 6 and Proposition 5 conclude that $\mathcal{C}_M$ is nonempty and has codimension 3. In addition, we consider the lattices $K_{14} = \langle h^2, v \rangle$ and $K_{d_i} := \langle h^2, \alpha_i \rangle \subset M$ with discriminant $d_i$. By Lemma 6 and Proposition 5, we obtain $\mathcal{C}_M \subset \mathcal{C}_{K_{d_1}} = \mathcal{C}_{d_1}$ and also $\mathcal{C}_M \subset \mathcal{C}_{K_{d_2}} = \mathcal{C}_{d_2}$. Consequently, $\emptyset \ne \mathcal{C}_M \subset \mathcal{C}_{14} \cap \mathcal{C}_{d_1} \cap \mathcal{C}_{d_2}$ is what we wanted, since every cubic fourfold in $\mathcal{C}_{14}$ is rational.

Since Case 2 and Case 3 are the same as Case 1, we just give the main ingredients and left the details to the interested readers.

**Case 2.** Given $d_1 = 6n_1$ and $d_2 = 6n_2 + 2$ with $n_1 \ge 2, n_2 \ge 1$. We consider the rank 4 sublattice

$$M := \langle h^2, v, \alpha_1, \alpha_2 + (0, 0, 1) \rangle \subset L$$

and its sublattices $K_{14} = \langle h^2, v \rangle$, $K_{d_1} := \langle h^2, \alpha_1 \rangle \subset M$ and $K_{d_2} := \langle h^2, \alpha_2 + (0, 0, 1) \rangle \subset M$.

**Case 2.** Given $d_1 = 6n_1 + 2$ and $d_2 = 6n_2 + 2$ with $n_1, n_2 \ge 1$. We consider the rank 4 sublattice

$$M := \langle h^2, v, \alpha_1 + (0, 1, 0), \alpha_2 + (0, 0, 1) \rangle \subset L$$

and its sublattices $K_{14} = \langle h^2, v \rangle$, $K_{d_1} := \langle h^2, \alpha_1 + (0, 1, 0) \rangle \subset M$ and $K_{d_2} := \langle h^2, \alpha_2 + (0, 0, 1) \rangle \subset M$. $\square$

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