# Comptes Rendus
Mathématique

Gloire Grâce Bockondas and Samuel Boissière

## Triple lines on a cubic threefold

Volume 361 (2023), p. 747-755

Published online: 11 May 2023

https://doi.org/10.5802/crmath.410

<figure><img src="image_1.png" /></figure>

This article is licensed under the

CREATIVE COMMONS ATTRIBUTION 4.0 INTERNATIONAL LICENSE.

http://creativecommons.org/licenses/by/4.0/
By using local computations explained in Remark 7, Murre claimed that the curve $M(X)$ of lines of the second type is smooth [7]. However, it is well-known to the experts that this curve can be singular and its singularities correspond to triple lines on $X$ (see [5, 6, 9]). Nevertheless, no complete proof does appear in the literature. The purpose of this paper is to fill this gap by giving a complete proof of Theorem 1 following the techniques of [7].

**Theorem 1.** The triple lines on a cubic threefold are exactly the singular points of the curve of lines of the second type.

Consider the Fermat cubic in $\mathbb{P}^4$ defined by

$$F_4 = \{x_0^3 + x_1^3 + x_2^3 + x_3^3 + x_4^3 = 0\}.$$

**Proposition 2.** The Fermat cubic $F_4 \subset \mathbb{P}^4$ contains exactly 135 triple lines.

## Acknowledgements

We would like to thank Xavier Roulleau, Daniel Huybrechts for helpful discussions, and Martí Lahoz, Juan Carlos Naranjo and Andrés Rojas for their comments.

## 2. Notation and preliminaries

For $X \subset \mathbb{P}^4$ a smooth complex cubic threefold, we denote by $F(X)$ the Fano surface of lines and $M(X)$ the curve of lines of the second type. Denote by $(x_0 : x_1 : x_2 : x_3 : x_4)$ the homogeneous coordinates on $\mathbb{P}^4$ and $p_{i,j}, 0 \le i < j \le 4$ the Plücker coordinates of the grassmannian of lines $\mathbb{G}(1,4) \subset \mathbb{P}^9$. Throughout this paper, we locally study the Fano surface of lines $F(X)$. For instance when we work in the affine chart $p_{0,1} = 1$ of $\mathbb{G}(1,4)$ isomorphic to $\mathbb{C}^6$, a point $[\ell] \in F(X) \subset \mathbb{G}(1,4)$ corresponds to a line $\ell$ on $X$ spanned by two points $v_0 = (1 : 0 : -p_{1,2} : -p_{1,3} : -p_{1,4})$ and $v_1 = (0 : 1 : p_{0,2} : p_{0,3} : p_{0,4})$. We take $(p_{0,2}, p_{0,3}, p_{0,4}, p_{1,2}, p_{1,3}, p_{1,4})$ as the local coordinates of $\mathbb{G}(1,4)$ on the affine chart $p_{0,1} = 1$ (see [2]).

Let $\ell$ be a line on $X$. There are two types of lines in $X$: lines with normal bundle $N_{\ell/X} \simeq \mathcal{O}_{\ell} \oplus \mathcal{O}_{\ell}$ called lines of the first type and those with $N_{\ell/X} \simeq \mathcal{O}_{\ell}(1) \oplus \mathcal{O}_{\ell}(-1)$ called lines of the second type [4]. Another description is as follows: consider the intersection $P \cap X$ of the cubic threefold $X$ with a plane $P \simeq \mathbb{P}^2$ containing $\ell$. We observe that $P \cap X = \ell \cup C$ where $C$ is a conic. It may happen that the conic degenerates, that is $P \cap X = \ell \cup \ell' \cup \ell''$. If $P \cap X = 2\ell \cup \ell'$ then we say the plane $P$ is tangent to $X$ at every point of $\ell$ [11]. The line $\ell$ is called a multiple line and $\ell'$ the residual line. We have in particular two cases: if $\ell \neq \ell'$ then $\ell$ is called a double line, and if $\ell = \ell'$, that is $P \cap X = 3\ell$, then $\ell$ is called a triple line.

**Proposition 3 ([4, Lemma 6.7]).** The line $\ell \subset X$ is of the second type if and only if there exists a unique 2-plane $P \supset \ell$ tangent to $X$ at every point of $\ell$. If $\ell \subset X$ is a line of the first type, then there is no 2-plane tangent to $X$ in all points of $\ell$.

It is worth noting that the multiple lines on $X$ are exactly the lines of the second type [7, Lemma 1.14].

## 3. Lines on a cubic threefold

Throughout this section we gather and extend some facts about lines on cubic threefolds.

**Proposition 4 ([7, Corollary 1.9]).** The locus

$$M(X) = \{[\ell] \in F(X), \exists P \simeq \mathbb{P}^2 \mid P \cap X = 2\ell \cup \ell'\}$$

of lines of the second type on $X$ is a curve.
On the affine chart $p_{0,1} = 1$, the Fano surface $F(X)$ is defined by the vanishing locus of the terms $\phi^{i,j}(\ell)$ coming from the decomposition $F(p) = \sum_{i+j=3} t_0^i t_1^j \phi^{i,j}(\ell)$ for any point $p \in \ell \subset X$ with coordinates $t_0 v_0 + t_1 v_1$, where $\phi^{i,j}(\ell)$ are functions of the local Plücker coordinates of the line $\ell$. Any plane $P \simeq \mathbb{P}^2$ containing $\ell$ meets the plane $P_{0,1} = \{x_0 = 0, x_1 = 0\}$ at a unique point $v_2 = (0:0:\alpha_2:\alpha_3:\alpha_4)$ so that $P = \text{span}(\ell, v_2)$. Then the plane cubic $P \cap X$ has equation $F(t_0 v_0 + t_1 v_1 + t_2 v_2) = 0$ where $(v_0: v_1: v_2)$ are the projective coordinates of the plane $P$. Expanding in $t_2$ we have:

$$0 = F(t_0 v_0 + t_1 v_1) + t_2 \sum_{i=2}^{4} \frac{\partial F}{\partial x_i} (t_0 v_0 + t_1 v_1) \alpha_i + \frac{1}{2} t_2^2 \sum_{2 \le i,j \le 4} \frac{\partial^2 F}{\partial x_j \partial x_i} (t_0 v_0 + t_1 v_1) \alpha_i \alpha_j + t_2^3 F(v_2).$$

The line $\ell \subset P$ of equation $t_2 = 0$ is a line of the second type on $X$ if and only if $F(t_0 v_0 + t_1 v_1) = 0$ and the plane cubic equation is a multiple of $t_2^2$. For any point $p \in \ell \subset X$ with coordinates $t_0 v_0 + t_1 v_1$ on the affine chart $p_{0,1} = 1$, one can write

$$\frac{\partial F}{\partial x_i}(p) = \sum_{j+k=2} t_0^j t_1^k \phi_i^{j,k}(\ell) \quad (1)$$

where $\phi_i^{j,k}(\ell)$ are functions of the local Plücker coordinates. Then the line $\ell \subset X$ of equation $t_2 = 0$ is a second type line if and only if

$$\mathrm{m}(\ell) := \det \begin{pmatrix} \phi_2^{2,0}(\ell) & \phi_3^{2,0}(\ell) & \phi_4^{2,0}(\ell) \\ \phi_2^{1,1}(\ell) & \phi_3^{1,1}(\ell) & \phi_4^{1,1}(\ell) \\ \phi_2^{0,2}(\ell) & \phi_3^{0,2}(\ell) & \phi_4^{0,2}(\ell) \end{pmatrix} = 0. \quad (2)$$

The curve of lines of the second type on $X$ is thus locally given by

$$M(X) = \{\phi^{3,0}(\ell) = 0, \phi^{2,1}(\ell) = 0, \phi^{1,2}(\ell) = 0, \phi^{0,3}(\ell) = 0, \mathrm{m}(\ell) = 0\}.$$

For a generic cubic threefold $X \subset \mathbb{P}^4$ the locus $M(X)$ of lines of the second type is smooth [5, Proposition 2.13].

### 3.1. Triple lines

We recall that a triple line is a second type line $\ell \subset X$ such that $P \cap X = 3\ell$. To our knowledge, the locus of triple lines on a cubic threefold is shortly mentioned in the literature. We have the following lemma.

**Lemma 5 ([4, Lemma 10.15]).** The locus $\{[\ell] \in F(X), \exists P \simeq \mathbb{P}^2 | P \cap X = 3\ell\}$ of triple lines on $X$ defines a finite set.

The line of the second type $\ell \subset X$ of equation $t_2 = 0$ is a triple line if and only if the plane cubic equation is a multiple of $t_2^3$. For all point $p \in \ell \subset X$ with coordinates $t_0 v_0 + t_1 v_1$ one can write

$$\frac{\partial^2 F}{\partial x_i \partial x_j}(p) = \sum_{k+l=1} t_0^k t_1^l \phi_{i,j}^{k,l}(\ell) \quad (3)$$

where $\phi_{i,j}^{k,l}(\ell)$ are functions of the local Plücker coordinates on the affine chart $p_{0,1} = 1$. Then the line of the second type $\ell \subset X$ of equation $t_2 = 0$ is a triple line if and only if

$$\begin{pmatrix} \frac{1}{2} \phi_{2,2}^{1,0}(\ell) & \frac{1}{2} \phi_{3,3}^{1,0}(\ell) & \frac{1}{2} \phi_{4,4}^{1,0}(\ell) & \phi_{2,3}^{1,0}(\ell) & \phi_{2,4}^{1,0}(\ell) & \phi_{3,4}^{1,0}(\ell) \\ \frac{1}{2} \phi_{2,2}^{0,1}(\ell) & \frac{1}{2} \phi_{3,3}^{0,1}(\ell) & \frac{1}{2} \phi_{4,4}^{0,1}(\ell) & \phi_{2,3}^{0,1}(\ell) & \phi_{2,4}^{0,1}(\ell) & \phi_{3,4}^{0,1}(\ell) \end{pmatrix} \begin{pmatrix} \alpha_2^2 \\ \alpha_3^2 \\ \alpha_4^2 \\ \alpha_2 \alpha_3 \\ \alpha_2 \alpha_4 \\ \alpha_3 \alpha_4 \end{pmatrix} = 0 \quad (4)$$

holds.
4. Triple lines on a cubic threefold and the curve of lines of the second type

In this section we prove the following theorem by using the techniques of [7].

**Theorem 6.** The triple lines on a cubic threefold are exactly the singular points of the curve of lines of the second type.

**Proof.** Let us start by fixing the notations. We can assume that $[\ell_0] \in \mathrm{M}(X)$ is a line of the second type given by

$$x_2 = 0, x_3 = 0, x_4 = 0.$$

On the affine chart $p_{0,1} = 1$ of $\mathbb{G}(1,4)$ the line $[\ell_0] \in \mathrm{M}(X)$ is thus given by

$$p_{0,2} = 0, p_{0,3} = 0, p_{0,4} = 0, p_{1,2} = 0, p_{1,3} = 0, p_{1,4} = 0.$$

We may assume after a linear change of variables $x_2, x_3$ and $x_4$ that the 2-plane $P = \mathrm{span}(\ell_0, v_2)$ tangent to $X$ at every point of $\ell_0$ is such that $\alpha_2 = 0, \alpha_3 = 0$ and $\alpha_4 = 1$. Then from the following equation

$$\begin{pmatrix} \phi_2^{2,0}(\ell_0) & \phi_3^{2,0}(\ell_0) & \phi_4^{2,0}(\ell_0) \\ \phi_2^{1,1}(\ell_0) & \phi_3^{1,1}(\ell_0) & \phi_4^{1,1}(\ell_0) \\ \phi_2^{0,2}(\ell_0) & \phi_3^{0,2}(\ell_0) & \phi_4^{0,2}(\ell_0) \end{pmatrix} \begin{pmatrix} \alpha_2 \\ \alpha_3 \\ \alpha_4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

we get

$$\phi_4^{2,0}(\ell_0) = 0, \phi_4^{1,1}(\ell_0) = 0, \phi_4^{0,2}(\ell_0) = 0. \qquad (5)$$

The proof goes as follows: we first compute the affine tangent space $T_{[\ell_0]}\mathrm{F}(X)$ of $\mathrm{F}(X)$ at the point $[\ell_0]$. Then we compute the affine tangent space $T_{[\ell_0]}\mathrm{M}(X)$ of $\mathrm{M}(X)$ at the point $[\ell_0]$. Afterwards we prove that the triple lines on $X$ are singular points of the curve $\mathrm{M}(X)$. Finally we prove that the singular points of the curve $\mathrm{M}(X)$ correspond to triple lines on $X$.

(1) Tangent space of $\mathrm{F}(X)$ at the point $[\ell_0]$. On the affine chart $p_{0,1} = 1$, a point in $\mathbb{C}^6$ with coordinates $(p_{i,j})_{i,j}$ belongs to the tangent space $T_{[\ell_0]}\mathrm{F}(X)$ if and only if

$$\sum_{i=2}^{4} (-p_{1,i} t_0 + p_{0,i} t_1) \left( t_0^2 \phi_i^{2,0}(\ell_0) + t_0 t_1 \phi_i^{1,1}(\ell_0) + t_1^2 \phi_i^{0,2}(\ell_0) \right) = 0$$

for all $(t_0 : t_1) \in \mathbb{P}^1$. Setting the coefficients of $t_0^3, t_0^2 t_1, t_0 t_1^2$ and $t_1^3$ each equal to zero one can see that for such a point the equality

$$\underbrace{\begin{pmatrix} 0 & 0 & \phi_2^{2,0}(\ell_0) & \phi_3^{2,0}(\ell_0) \\ \phi_2^{2,0}(\ell_0) & \phi_3^{2,0}(\ell_0) & \phi_2^{1,1}(\ell_0) & \phi_3^{1,1}(\ell_0) \\ \phi_2^{1,1}(\ell_0) & \phi_3^{1,1}(\ell_0) & \phi_2^{0,2}(\ell_0) & \phi_3^{0,2}(\ell_0) \\ \phi_2^{0,2}(\ell_0) & \phi_3^{0,2}(\ell_0) & 0 & 0 \end{pmatrix}}_{A} \begin{pmatrix} p_{0,2} \\ p_{0,3} \\ -p_{1,2} \\ -p_{1,3} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \qquad (6)$$

holds (see Equation (5)). The determinant of the matrix $A$ defined in Equation (6) can be expressed in terms of the resultant below

$$\det \begin{pmatrix} \phi_2^{2,0}(\ell_0) & 0 & \phi_3^{2,0}(\ell_0) & 0 \\ \phi_2^{1,1}(\ell_0) & \phi_2^{2,0}(\ell_0) & \phi_3^{1,1}(\ell_0) & \phi_3^{2,0}(\ell_0) \\ \phi_2^{0,2}(\ell_0) & \phi_2^{1,1}(\ell_0) & \phi_3^{0,2}(\ell_0) & \phi_3^{1,1}(\ell_0) \\ 0 & \phi_2^{0,2}(\ell_0) & 0 & \phi_3^{0,2}(\ell_0) \end{pmatrix} \qquad (7)$$

after permutation of the columns. For any point $p_0 \in \ell_0 \subset X$ with coordinates $(t_0 : t_1 : 0 : 0 : 0)$ in the affine chart $p_{0,1} = 1$ the polynomials

$$\frac{\partial F}{\partial x_i}(p_0) = \sum_{j+k=2} t_0^j t_1^k \phi_i^{j,k}(\ell_0)$$
vanish for $i \in \{0, 1\}$, and using Equation (5) so also does

$$ \frac{\partial F}{\partial x_4}(p_0) = \sum_{j+k=2} t_0^j t_1^k \phi_4^{j,k}(\ell_0). $$

Therefore the equations

$$ \frac{\partial F}{\partial x_i}(p_0) = \sum_{j+k=2} t_0^j t_1^k \phi_i^{j,k}(\ell_0) = 0 $$

do not have a common root for $i \in \{2, 3\}$, otherwise the cubic threefold $X$ would be singular. The resultant (7) does not vanish and thus we have

$$ \det(A) \neq 0. \qquad (8) $$

The tangent space to $F(X)$ at the point $[\ell_0]$ is then given by

$$ T_{[\ell_0]} F(X) = \{p_{0,2} = 0, p_{0,3} = 0, p_{1,2} = 0, p_{1,3} = 0\} $$

in the affine chart $p_{0,1} = 1$.

(2) Tangent space of $M(X)$ at the point $[\ell_0]$. The tangent space $T_{[\ell_0]}M(X)$ is given by the intersection of the tangent space $T_{[\ell_0]}F(X)$ with the hypersurface defined by

$$ \sum_{0 \le i < 2 \le j \le 4} p_{i,j} \frac{\partial m}{\partial p_{i,j}}(\ell_0) = 0. $$

We differentiate the determinant in (2) with respect to $p_{i,j}$ and by using Equation (5) we get

$$ \det \begin{pmatrix} \phi_2^{2,0}(\ell_0) \phi_3^{2,0}(\ell_0) p_{0,4} \frac{\partial \phi_4^{2,0}}{\partial p_{0,4}}(\ell_0) + p_{1,4} \frac{\partial \phi_4^{2,0}}{\partial p_{1,4}}(\ell_0) & \phi_2^{1,0}(\ell_0) \phi_3^{1,0}(\ell_0) p_{0,4} \frac{\partial \phi_4^{1,0}}{\partial p_{0,4}}(\ell_0) + p_{1,4} \frac{\partial \phi_4^{1,0}}{\partial p_{1,4}}(\ell_0) & \phi_2^{0,0}(\ell_0) \phi_3^{0,0}(\ell_0) p_{0,4} \frac{\partial \phi_4^{0,0}}{\partial p_{0,4}}(\ell_0) + p_{1,4} \frac{\partial \phi_4^{0,0}}{\partial p_{1,4}}(\ell_0) \\ \phi_2^{1,1}(\ell_0) \phi_3^{1,1}(\ell_0) p_{0,4} \frac{\partial \phi_4^{1,1}}{\partial p_{0,4}}(\ell_0) + p_{1,4} \frac{\partial \phi_4^{1,1}}{\partial p_{1,4}}(\ell_0) & \phi_2^{0,1}(\ell_0) \phi_3^{0,1}(\ell_0) p_{0,4} \frac{\partial \phi_4^{0,1}}{\partial p_{0,4}}(\ell_0) + p_{1,4} \frac{\partial \phi_4^{0,1}}{\partial p_{1,4}}(\ell_0) & \phi_2^{1,2}(\ell_0) \phi_3^{1,2}(\ell_0) p_{0,4} \frac{\partial \phi_4^{1,2}}{\partial p_{0,4}}(\ell_0) + p_{1,4} \frac{\partial \phi_4^{1,2}}{\partial p_{1,4}}(\ell_0) \\ \phi_2^{0,2}(\ell_0) \phi_3^{0,2}(\ell_0) p_{0,4} \frac{\partial \phi_4^{0,2}}{\partial p_{0,4}}(\ell_0) + p_{1,4} \frac{\partial \phi_4^{0,2}}{\partial p_{1,4}}(\ell_0) & \phi_2^{0,1}(\ell_0) \phi_3^{0,1}(\ell_0) p_{0,4} \frac{\partial \phi_4^{0,1}}{\partial p_{0,4}}(\ell_0) + p_{1,4} \frac{\partial \phi_4^{0,1}}{\partial p_{1,4}}(\ell_0) & \phi_2^{1,2}(\ell_0) \phi_3^{1,2}(\ell_0) p_{0,4} \frac{\partial \phi_4^{1,2}}{\partial p_{0,4}}(\ell_0) + p_{1,4} \frac{\partial \phi_4^{1,2}}{\partial p_{1,4}}(\ell_0) \end{pmatrix} = 0. \quad (9) $$

Let us explicit Equation (9). For any point $p \in \ell \subset X$ with coordinates $t_0 v_0 + t_1 v_1$ in the affine chart $p_{0,1} = 1$ one can write $F(p) = F(t_0 v_0 + t_1 v_1)$. Differentiate Equation (1) with respect to $p_{u,4}$ for $u \in \{0, 1\}$ and using Equation (3) we have

$$ \sum_{j+k=2} t_0^j t_1^k \frac{\partial \phi_4^{j,k}}{\partial p_{u,4}}(\ell) = (-1)^u t_0^u t_1^{1-u} \sum_{k+l=1} t_0^k t_1^l \phi_{4,4}^{k,l}(\ell). $$

By identifying the homogeneous components in $t_0$ and $t_1$ we get

$$ \frac{\partial \phi_4^{j,k}}{\partial p_{u,4}}(\ell) = \phi_{4,4}^{j,k-1}(\ell), \quad \frac{\partial \phi_4^{j,k}}{\partial p_{u,4}}(\ell) = -\phi_{4,4}^{j-1,k}(\ell) $$

with $\phi_{4,4}^{2,-1}(\ell) = 0, \phi_{4,4}^{-1,2}(\ell) = 0$. Then Equation (9) becomes

$$ \phi(p_{0,4}, p_{1,4}) := \det \begin{pmatrix} \phi_2^{2,0}(\ell_0) \phi_3^{2,0}(\ell_0) & -p_{1,4}\phi_{4,4}^{1,0}(\ell_0) & -p_{1,4}\phi_{4,4}^{0,1}(\ell_0) \\ \phi_2^{1,1}(\ell_0) \phi_3^{1,1}(\ell_0) & p_{0,4}\phi_{4,4}^{1,0}(\ell_0) - p_{1,4}\phi_{4,4}^{0,1}(\ell_0) & p_{0,4}\phi_{4,4}^{0,2}(\ell_0) \\ \phi_2^{0,2}(\ell_0) \phi_3^{0,2}(\ell_0) & p_{0,4}\phi_{4,4}^{0,1}(\ell_0) & p_{0,4}\phi_{4,4}^{1,2}(\ell_0) \end{pmatrix} = 0. $$

Therefore the tangent space of the curve $M(X)$ at the point $[\ell_0]$ is given by

$$ T_{[\ell_0]} M(X) = \{p_{0,2} = 0, p_{0,3} = 0, p_{1,2} = 0, p_{1,3} = 0, \phi(p_{0,4}, p_{1,4}) = 0\} $$

in the affine chart $p_{0,1} = 1$. We now prove that the triple lines on $X$ are singular points of $M(X)$.
(3) *Triple lines and the curve M(X) of lines of the second type.* From Equation (4) the double line $[\ell_0] \in \mathrm{M}(X)$ is a triple line if and only if

$$ \begin{cases} \phi_{4,4}^{1,0}(\ell_0) = 0 \\ \phi_{4,4}^{0,1}(\ell_0) = 0 \end{cases} $$

which implies that $\phi(p_{0,4}, p_{1,4}) = 0$. Hence for $[\ell_0] \in \mathrm{M}(X)$ a triple line, the tangent space of the curve $\mathrm{M}(X)$ at the point $[\ell_0]$ is given by

$$ T_{[\ell_0]}\mathrm{M}(X) = \{p_{0,2} = 0, p_{0,3} = 0, p_{1,2} = 0, p_{1,3} = 0\} $$

in the affine chart $p_{0,1} = 1$. Its dimension is strictly greater than one. Therefore the triple lines on $X$ are singular points of $\mathrm{M}(X)$.

(4) *Singular points of the curve M(X) of lines of the second type.* We have seen that the curve of lines of the second type is given by $\mathrm{M}(X) = \{\phi^{3,0}(\ell) = 0, \phi^{2,1}(\ell) = 0, \phi^{1,2}(\ell) = 0, \phi^{0,3}(\ell) = 0, m(\ell) = 0\}$ in the affine chart $p_{0,1} = 1$. We are going to compute the Jacobian matrix of this curve at the point $[\ell_0]$. Keeping notation as above, we have for any point $p \in \ell \subset X$ with coordinates $t_0 v_0 + t_1 v_1$:

$$ F(p) = F(t_0 v_0 + t_1 v_1) = \sum_{i+j=3} t_0^i t_1^j \phi^{i,j}(\ell). \qquad (10) $$

Differentiating Equation (10) with respect to $p_{u,v}$ for $u \in \{0, 1\}$ and $v \in \{2, 3, 4\}$ we get

$$ \sum_{i+j=3} t_0^i t_1^j \frac{\partial \phi^{i,j}}{\partial p_{u,v}}(\ell) = (-1)^u t_0^u t_1^{1-u} \sum_{j+k=2} t_0^j t_1^k \phi_v^{j,k}(\ell). $$

By identifying the homogeneous components in $t_0$ and $t_1$ we get

$$ \frac{\partial \phi^{i,j}}{\partial p_{0,v}}(\ell) = \phi_v^{i,j-1}(\ell), \quad \frac{\partial \phi^{i,j}}{\partial p_{1,v}}(\ell) = -\phi_v^{i-1,j}(\ell) $$

with $\phi_v^{3,-1}(\ell) = 0, \phi_v^{-1,3}(\ell) = 0$. The Jacobian matrix of $\mathrm{M}(X)$ at the point $[\ell_0]$ is thus

$$ \begin{pmatrix} 0 & 0 & 0 & -\phi_2^{2,0}(\ell_0) & -\phi_3^{2,0}(\ell_0) & 0 \\ \phi_2^{2,0}(\ell_0) & \phi_3^{2,0}(\ell_0) & 0 & -\phi_2^{1,1}(\ell_0) & -\phi_3^{1,1}(\ell_0) & 0 \\ \phi_2^{1,1}(\ell_0) & \phi_3^{1,1}(\ell_0) & 0 & -\phi_2^{0,2}(\ell_0) & -\phi_3^{0,2}(\ell_0) & 0 \\ \phi_2^{0,2}(\ell_0) & \phi_3^{0,2}(\ell_0) & 0 & 0 & 0 & 0 \\ \frac{\partial m}{\partial p_{0,2}}(\ell_0) & \frac{\partial m}{\partial p_{0,3}}(\ell_0) & \frac{\partial m}{\partial p_{0,4}}(\ell_0) & \frac{\partial m}{\partial p_{1,2}}(\ell_0) & \frac{\partial m}{\partial p_{1,3}}(\ell_0) & \frac{\partial m}{\partial p_{1,4}}(\ell_0) \end{pmatrix} $$

using again Equation (5). Denote by $M_i$ its $5 \times 5$ minors where $i$ corresponds to the omitted column. The minors $M_1, M_2, M_4, M_5$ vanish and the singular points of the curve $\mathrm{M}(X)$ are the points $[\ell_0] \in \mathrm{M}(X)$ for which $M_3 = 0, M_6 = 0$ with

$$ M_3 = \frac{\partial m}{\partial p_{1,4}}(\ell_0) \det \begin{pmatrix} 0 & 0 & -\phi_2^{2,0}(\ell_0) & -\phi_3^{2,0}(\ell_0) \\ \phi_2^{2,0}(\ell_0) & \phi_3^{2,0}(\ell_0) & -\phi_2^{1,1}(\ell_0) & -\phi_3^{1,1}(\ell_0) \\ \phi_2^{1,1}(\ell_0) & \phi_3^{1,1}(\ell_0) & -\phi_2^{0,2}(\ell_0) & -\phi_3^{0,2}(\ell_0) \\ \phi_2^{0,2}(\ell_0) & \phi_3^{0,2}(\ell_0) & 0 & 0 \end{pmatrix} \\ M_6 = \frac{\partial m}{\partial p_{0,4}}(\ell_0) \det \begin{pmatrix} 0 & 0 & -\phi_2^{2,0}(\ell_0) & -\phi_3^{2,0}(\ell_0) \\ \phi_2^{2,0}(\ell_0) & \phi_3^{2,0}(\ell_0) & -\phi_2^{1,1}(\ell_0) & -\phi_3^{1,1}(\ell_0) \\ \phi_2^{1,1}(\ell_0) & \phi_3^{1,1}(\ell_0) & -\phi_2^{0,2}(\ell_0) & -\phi_3^{0,2}(\ell_0) \\ \phi_2^{0,2}(\ell_0) & \phi_3^{0,2}(\ell_0) & 0 & 0 \end{pmatrix}. $$
Since the determinant appearing in the expression of the minors $M_3$ and $M_6$ cannot vanish (see (8)) then one can see that the singular points of $M(X)$ are the points $[\ell_0]$ that satisfy

$$ \left\{ \begin{array}{l} \displaystyle \frac{\partial m}{\partial p_{1,4}}(\ell_0) = \det \begin{pmatrix} \phi_2^{2,0}(\ell_0) & \phi_3^{2,0}(\ell_0) & -\phi_{4,4}^{1,0}(\ell_0) \\ \phi_2^{1,1}(\ell_0) & \phi_3^{1,1}(\ell_0) & -\phi_{4,4}^{0,1}(\ell_0) \\ \phi_2^{0,2}(\ell_0) & \phi_3^{0,2}(\ell_0) & 0 \end{pmatrix} = 0 \\[1.5em] \displaystyle \frac{\partial m}{\partial p_{0,4}}(\ell_0) = \det \begin{pmatrix} \phi_2^{2,0}(\ell_0) & \phi_3^{2,0}(\ell_0) & 0 \\ \phi_2^{1,1}(\ell_0) & \phi_3^{1,1}(\ell_0) & \phi_{4,4}^{1,0}(\ell_0) \\ \phi_2^{0,2}(\ell_0) & \phi_3^{0,2}(\ell_0) & \phi_{4,4}^{0,1}(\ell_0) \end{pmatrix} = 0. \end{array} \right. $$

So we have

$$ \underbrace{\begin{pmatrix} -M_{13} & M_{23} \\ -M_{23} & M_{33} \end{pmatrix}}_{B} \begin{pmatrix} \phi_{4,4}^{1,0}(\ell_0) \\ \phi_{4,4}^{0,1}(\ell_0) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \qquad (11) $$

where $M_{13}$, $M_{23}$ and $M_{33}$ are the $2 \times 2$ minors of the matrix

$$ \begin{pmatrix} \phi_2^{2,0}(\ell_0) & \phi_3^{2,0}(\ell_0) & \phi_4^{2,0}(\ell_0) \\ \phi_2^{1,1}(\ell_0) & \phi_3^{1,1}(\ell_0) & \phi_4^{1,1}(\ell_0) \\ \phi_2^{0,2}(\ell_0) & \phi_3^{0,2}(\ell_0) & \phi_4^{0,2}(\ell_0) \end{pmatrix}. $$

This matrix is of rank two for if it was of rank one then one could find a common root of the equations $\frac{\partial F}{\partial x_i}(p_0) = \sum_{j+k=2} t_0^j t_1^k \phi_i^{j,k}(\ell_0) = 0$ for $i \in \{2, 3, 4\}$ with $p_0 \in \ell_0$ a point with coordinates $(t_0 : t_1 : 0 : 0 : 0)$ in the affine chart $p_{0,1} = 1$. Let us consider the matrix $B$ defined in Equation (11). If $\det(B) = 0$ we can assume that a column is a multiple of the other one. Then one could find a common root of $\frac{\partial F}{\partial x_i}(p_0) = \sum_{j+k=2} t_0^j t_1^k \phi_i^{j,k}(\ell_0) = 0$ for $i \in \{2, 3\}$ and $X$ would be singular using Equation (5). Hence $\det(B) \neq 0$ and Equation (11) holds if and only if $(\phi_{4,4}^{1,0}(\ell_0), \phi_{4,4}^{0,1}(\ell_0))$ vanishes, which are the necessary and sufficient conditions for the line $[\ell_0] \in M(X)$ to be a triple line. Therefore the singular points of $M(X)$ are the triple lines on $X$. $\square$

As a consequence of Theorem 1 one can deduce that the curve $M(X)$ of lines of the second type is reduced.

**Remark 7.** Let $\ell_0$ be a line on $X$ given by $x_2 = 0, x_3 = 0, x_4 = 0$. Then the equation of $X$ may be written

$$ F(x_0, \dots, x_4) = x_2 q_2(x_0, \dots, x_4) + x_3 q_3(x_0, \dots, x_4) + x_4 q_4(x_0, \dots, x_4) = 0 $$

where $q_i(x_0, \dots, x_4)$ are homogeneous polynomials of degree two. If moreover $\ell_0$ is a line of the second type so that the plane $P = \operatorname{span}(\ell_0, v_2)$ tangent to $X$ in all points of $\ell_0$ is such that $\alpha_2 = 0, \alpha_3 = 0$ and $\alpha_4 = 1$ then the equation of $X$ may take the form

$$ F(x_0, \dots, x_4) = x_2 q_2(x_0, \dots, x_4) + x_3 q_3(x_0, \dots, x_4) + x_4^2 l(x_0, \dots, x_4) = 0 $$

where $l(x_0, \dots, x_4)$ is a linear polynomial. Set $l(x_0, \dots, x_4) = a_0 x_0 + a_1 x_1 + a_2 x_2 + a_3 x_3 + a_4 x_4$. Using the techniques of [7] we get

$$ a_0 = \frac{1}{2} \phi_{4,4}^{1,0}(\ell_0) \quad \text{and} \quad a_1 = \frac{1}{2} \phi_{4,4}^{0,1}(\ell_0). $$

So for $\ell_0$ a triple line on $X$ we have $a_0 = 0, a_1 = 0$. As corrected in [6, p. 17], Murre erroneously wrote $l(x_0, x_1) = a_0 x_0 + a_1 x_1$ for the linear polynomial instead in equation (13) of [7, p. 167], implying that for $\ell_0 \subset X$ a triple line this polynomial vanishes and $X$ contains the plane $\{x_2 =$
0, $x_3 = 0$]: the cubic threefold X would therefore be singular. Whereas for $\ell_0$ a triple line on X the equation of X may be written

$$F(x_0, \dots, x_4) = x_2 q_2(x_0, \dots, x_4) + x_3 q_3(x_0, \dots, x_4) + k x_4^3 = 0$$

with $k \neq 0$ and the cubic threefold X remains smooth.

## 5. Counting triple lines on the Fermat cubic

In this section we prove the following proposition.

**Proposition 8.** *The Fermat cubic $F_4 \subset \mathbb{P}^4$ contains exactly 135 triple lines.*

*Proof.* In the affine chart $p_{0,1} = 1$ the set $M(F_4)$ is a non smooth curve given by the following equations:

$$\begin{align*}
p_{1,2}^3 + p_{1,3}^3 + p_{1,4}^3 - 1 &= 0 \\
p_{0,2} p_{1,2}^2 + p_{0,3} p_{1,3}^2 + p_{0,4} p_{1,4}^2 &= 0 \\
p_{0,2}^2 p_{1,2} + p_{0,3}^2 p_{1,3} + p_{0,4}^2 p_{1,4} &= 0 \\
p_{0,2}^3 + p_{0,3}^3 + p_{0,4}^3 + 1 &= 0 \\
(p_{0,4} p_{1,3} - p_{0,3} p_{1,4})(p_{0,4} p_{1,2} - p_{0,2} p_{1,4})(p_{0,3} p_{1,2} - p_{0,2} p_{1,3}) &= 0
\end{align*}$$

with

$$(p_{0,4} p_{1,3} - p_{0,3} p_{1,4})(p_{0,4} p_{1,2} - p_{0,2} p_{1,4})(p_{0,3} p_{1,2} - p_{0,2} p_{1,3}) = m(\ell)$$

its local equation in the Fano surface $F(F_4)$. Consider the quadrics

$$Q_1 = p_{0,4} p_{1,3} - p_{0,3} p_{1,4}, \quad Q_2 = p_{0,4} p_{1,2} - p_{0,2} p_{1,4}, \quad Q_3 = p_{0,3} p_{1,2} - p_{0,2} p_{1,3}.$$

We note that the intersection of the Fano surface $F(F_4)$ with $Q_1, Q_2$ and $Q_3$, denoted by $M_1(F_4), M_2(F_4)$ and $M_3(F_4)$ respectively, are smooth curves that correspond to the irreducible components of $M(F_4)$.

The intersection points $[\ell]$ of the curves $M_1(F_4)$ and $M_2(F_4)$ are given by the equations:

$$\begin{align*}
p_{0,2} p_{1,3}^3 - p_{0,2} p_{1,3}^4 - p_{1,3} p_{0,2} + p_{0,3} p_{1,3}^3 - p_{0,3} p_{1,3}^4 + 1, \quad p_{1,2}^3 + p_{1,3}^3 - 1, \\
p_{0,2} p_{0,3}, p_{0,2} p_{1,2}, p_{0,3} p_{1,3}, p_{1,2} p_{1,3}, p_{0,4}, p_{1,4}.
\end{align*}$$

If $p_{1,3} = 0$ then we have 9 intersection points given by the following equations:

$$p_{0,3}^3 = -1, \quad p_{1,2}^3 = 1, \quad p_{0,2} = 0, \quad p_{0,4} = 0, \quad p_{1,4} = 0.$$

We get $v_2 = (0:0:0:0:1)$. We have thus $f(t_0 v_0 + t_1 v_1 + t_2 v_2) = t_2^3$ implying that $\ell$ is a triple line on the Fermat cubic. Also the Jacobian matrix of the curve $M(F_4)$ at the point $[\ell]$ is not of rank 5; $[\ell]$ is a singular point of $M(F_4)$. We get 9 triple lines with coordinates $(0, p_{0,3}, 0, p_{1,2}, 0, 0)$ such that $p_{0,3}^3 = -1$ and $p_{1,2}^3 = 1$.

If $p_{1,3} \neq 0$ then we have 9 intersection points given by the equations:

$$p_{0,2}^3 = -1, \quad p_{1,3}^3 = 1, \quad p_{0,3} = 0, \quad p_{0,4} = 0, \quad p_{1,2} = 0, \quad p_{1,4} = 0.$$

They correspond to triple lines with coordinates $(p_{0,2}, 0, 0, 0, p_{1,3}, 0)$ with $p_{0,2}^3 = -1$ and $p_{1,3}^3 = 1$.

Similarly we find the intersection points $[\ell]$ of the curves $M_1(F_4)$ and $M_3(F_4)$, $M_2(F_4)$ and $M_3(F_4)$. We get 18 triple lines at each intersection. We note that the intersection of the three curves $M_1(F_4), M_2(F_4)$ and $M_3(F_4)$ is empty. Therefore there are 54 triple lines in the affine chart $p_{0,1} = 1$.

We do the same computations to count the triple lines on the Fermat cubic in the other affine charts of $\mathbb{G}(1, 4)$ by using SageMath [10]. However a triple line on $F_4$ can be viewed in many affine charts and then counted several times. In order to avoid such a situation we use the Plücker stratification (see [3]). In the stratum $p_{0,1} = 0, p_{0,2} = 1$ we get 36 triple lines, we get 18 triple lines
in the stratum $p_{0,1} = p_{0,2} = 0, p_{0,3} = 1$ and in the stratum $p_{0,1} = \cdots = p_{0,4} = 0, p_{1,2} = 1$, we get 9 triple lines in the stratum $p_{0,1} = \cdots = p_{1,2} = 0, p_{1,3} = 1$ whereas the other strata contain no triple line. So there are exactly 135 triple lines on the Fermat cubic $F_4 \subset \mathbb{P}^4$. $\square$

**Remark 9.** It is clear that the curve $M(F_4)$ of lines of the second type of the Fermat cubic $F_4 \subset \mathbb{P}^4$ is reducible. The Fano surface $F(F_4)$ of the Fermat cubic contains 30 elliptic curves that intersect in 135 points [9]. These intersection points are exactly the triple lines on the Fermat cubic. We wish to point out that the elliptic curves of the Fano surface $F(F_4)$ are exactly the irreducible components of the curve $M(F_4)$ of lines of the second type. However, whether the singularities of $M(X)$ are always nodal is still an open question.

## References

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