Automorphisms of Fano threefolds of rank 2 and degree 28

Abstract

We describe the automorphism groups of smooth Fano threefolds of rank 2 and degree 28 in the cases where they are finite.

1 Introduction

A smooth Fano threefold of Picard rank 2 and degree 28 is the blow-up of a smooth quadric threefold $Q\subset \mathbb {P}^4$ in a smooth rational quartic curve $C_4\subset Q$. Isomorphism classes of such threefolds form an irreducible two-dimensional family, which according to the Mori-Mukai classification corresponds to family 2.21. Let $\pi :X\rightarrow Q$ be such a threefold. Then the action of $\text {Aut}(Q,C_4)$ on Q lifts to an action on X, so that we may identify it with a subgroup of $\text {Aut}(X)$.

By a result of Cheltsov-Przyjalkowski-Shramov ([3, Lemma 9.2]), we have that either $\text {Aut}(X)$ is finite, or $\text {Aut}(X)\cong \text {Aut}(Q,C_4)\times \mathbb {Z}_2$, where upto isomorphism $\text {Aut}(Q,C_4)$ is described as follows:

• (1) There is a unique smooth threefold in family 2.21, unique upto isomorphism, such that $\text {Aut}(Q,C_4)\cong \textrm{PGL}_2(\mathbb {C})$,

• (2) There is a one-dimensional family of non-isomorphic smooth threefolds in family 2.21 such that $\text {Aut}(Q,C_4)\cong \mathbb {G}_m\rtimes \mathbb {Z}_2$,

• (3) There is a unique smooth threefold in family 2.21 such that $\text {Aut}(Q,C_4)\cong \mathbb {G}_a\rtimes \mathbb {Z}_2$.

The goal of this paper is to describe $\text {Aut}(X)$ when it is finite, where X is a smooth threefold in family 2.21. Our main result is the following:

Theorem 1.1

Let X be a smooth Fano threefold of rank 2 and degree 28. Then $\textrm{Aut}(X)\cong \textrm{Aut}(Q,C_4)\times \mathbb {Z}_2$. Furthermore, if $\textrm{Aut}(Q,C_4)$ is finite then it is isomorphic to $\mathbb {Z}_2\times \mathbb {Z}_2,\mathbb {Z}_2$ or 0.

We prove this theorem in two parts: Theorem 2.1 and Theorem 3.1.

Remark 1.2

The factor of $\mathbb {Z}_2$ appearing in the factorisation $\textrm{Aut}(X)\cong \textrm{Aut}(Q,C_4)\times \mathbb {Z}_2$ is generated by an involution g, which may be described as follows:

Let $\mathfrak {d}$ denote the restriction to Q of the linear system of quadric hypersurfaces in $\mathbb {P}^4$ which contain $C_4$, and let $\phi :Q\dashrightarrow \mathbb {P}^4$ be the corresponding rational map. The image of $\phi$ is a smooth quadric threefold, and $\phi$ contracts the intersection of the secant variety of $C_4$ with Q, V, onto a smooth rational curve $C_4'\subset Q'$. The base locus of $\phi$ is equal to $C_4$, and there is a birational morphism $\pi ':X\rightarrow Q'$, where X is the blow-up of Q along $C_4$. This morphism contracts the strict transform, $E'$, of V onto the curve $C_4'$. Thus, there is a commutative diagram

In [3], it is shown that in cases (1) and (2) of the above classification, there exists a basis of $\mathfrak {d}$ such that $Q'=Q$ and $C_4'=C_4$, so that $\phi$ lifts to an involution $g\in \textrm{Aut}(X)$. We will show in Theorem 3.1 that this is always the case.

We can explicitly describe the threefolds appearing in [3, Lemma 9.2]. Let us fix some notation. Observe that after a projective transformation $C_4$ is the image of the Veronese embedding of $\mathbb {P}^1$ in $\mathbb {P}^4$:

$$\begin{aligned} \mathbb {P}^1&\rightarrow \mathbb {P}^4 \\ [u:v]&\mapsto [u^4:u^3v:u^2v^2:uv^3:v^4]. \end{aligned}$$

The space of global sections of $\mathcal {I}_{C_4}(2)$ is generated by the following quadratic forms:

$$\begin{aligned} f_0&= x_3^2 - x_2x_4, \\ f_1&= x_2x_3 - x_1x_4,\\ f_2&= x_2^2 - x_0x_4,\\ f_3&= x_1x_2 - x_0x_3,\\ f_4&= x_1^2 - x_0x_2,\\ f_5&= 3x_2^2 - 4x_1x_3 + x_0x_4, \end{aligned}$$

where $\mathcal {I}_{C_4}$ is the ideal sheaf of $C_4$ in $\mathbb {P}^4$, and $x_0$, $x_1$, $x_2$, $x_3$, $x_4$ are homogeneous coordinates on $\mathbb {P}^4$. Observe that the standard $\textrm{PGL}_2(\mathbb {C})$-action on $C_4$ lifts to an action on $\mathbb {P}^4$ such that $C_4$ is invariant. We fix the following subgroups of $\textrm{PGL}_2(\mathbb {C})$:

$$\begin{aligned}&\mathbb {Z}_2,\text { generated by } \begin{pmatrix}0 &{} 1 \\ 1 &{} 0\end{pmatrix},\\&\mathbb {G}_m,\text { consisting of matrices } \begin{pmatrix}1 &{} 0 \\ 0 &{} t\end{pmatrix} \text { for every } t\in \mathbb {G}_m,\\&\mathbb {G}_a,\text { consisting of matrices } \begin{pmatrix}1 &{} t \\ 0 &{} 1\end{pmatrix} \text { for every } t\in \mathbb {G}_a. \end{aligned}$$

Now we can describe $\textrm{Aut}(X)$ for the threefolds listed before:

Example 1.3

([1, Section 5.9]). Let Q be the quadric given by the equation

$$\begin{aligned}(1-4s^2)f_2+f_5=0,\end{aligned}$$

for some $s\in \mathbb {C}\setminus \{-1,0,1\}$. Then Q is $\mathbb {G}_m$-invariant and $\mathbb {Z}_2$-invariant, and conversely any smooth quadric admitting a faithful $\mathbb {G}_m$-action is isomorphic, via an element of $\textrm{PGL}_2(\mathbb {C})$, to a quadric given by an equation of this form. Moreover, we have the following:

$$\begin{aligned}\textrm{Aut}(Q,C_4)\cong {\left\{ \begin{array}{ll} \mathbb {G}_m\rtimes \mathbb {Z}_2, &{} s\ne \pm \frac{1}{2}, \\ \textrm{PGL}_2(\mathbb {C}), &{} s=\pm \frac{1}{2}. \end{array}\right. }\end{aligned}$$

The involution g described before is given by:

$$\begin{aligned}\tau :[x_0:x_1:x_2:x_3:x_4]\mapsto [f_4:sf_3:s^2f_2:sf_1:f_0].\end{aligned}$$

See [1, Remark 5.52] for an explanation of why $\tau \circ \tau :Q\dashrightarrow Q$ is the identity map on $Q\setminus C_4$.

Example 1.4

Suppose that the quadric Q is given by the equation

$$\begin{aligned}f_0+f_5=0.\end{aligned}$$

Then Q is $\mathbb {G}_a$-invariant and $\mathbb {Z}_2$-invariant, and $\textrm{Aut}(Q,C_4)\cong \mathbb {G}_a\rtimes \mathbb {Z}_2$. We will prove in case (2) of Theorem 3.1 that the blow-up of Q in $C_4$ admits an action of the involution g.

Remark 1.5

Recall that for a finite subgroup $G\subset \text {Aut}(Y)$, a variety Y is called G-Fano if it has terminal singularities, $-K_Y$ is ample and $\text {Cl}(Y)^G$ is rank 1. It is proven in [4] that the Hilbert scheme of conics on a smooth threefold X from the family No2.21 is isomorphic to $\mathbb {P}^1\times \mathbb {P}^1$, with the degenerate conics being parameterised by a smooth curve $C\subset \mathbb {P}^1\times \mathbb {P}^1$ of bidegree (2, 2). If X is G-Fano for the group $G=\langle g\rangle \cong \mathbb {Z}_2$, then this curve must be invariant upon swapping the two factors of $\mathbb {P}^1$.

An informal conjecture of Y. Prokhorov is that the invariance of this curve is a sufficient condition for X to be G-Fano. It is proven in [2] that every smooth curve in $\mathbb {P}^1\times \mathbb {P}^1$ of bidegree (2, 2) is invariant. As a corollary to Theorem 1.1, we have that every smooth threefold X in the family No2.21 is G-Fano, so that Prokhorov’s informal conjecture is true. For a detailed discussion of G-Fano threefolds, see [6].

Remark 1.6

Smooth threefolds in the family No2.21 are parametrised by $\mathbb {P}^5\setminus \Delta$, where $\Delta \subset \mathbb {P}^5$ is the discriminant locus of singular quadrics. The group $\textrm{PGL}_2(\mathbb {C})$ acts on this space, and it follows from Theorem 1.1 that any two threefolds in family 2.21 are isomorphic if and only if their corresponding points in the parameter space $\mathbb {P}^5\setminus \Delta$ lie in the same $\textrm{PGL}_2(\mathbb {C})$-orbit. Moreover, the moduli space of smooth GIT-polystable threefolds in family 2.21 is given by the GIT quotient

$$\begin{aligned}(\mathbb {P}^5\setminus \Delta )//\textrm{PGL}_2(\mathbb {C}).\end{aligned}$$

2 Computation of $\text {Aut}(Q,C_4)$

The first half of proving Theorem 1.1 is the computation of $\textrm{Aut}(Q,C_4)$, which we will do in this section. The result we will prove is:

Theorem 2.1

Let Q be a smooth quadric threefold containing the quartic curve $C_4$. If $\textrm{Aut}(Q,C_4)$ is finite, then it is isomorphic to either $\mathbb {Z}_2\times \mathbb {Z}_2$, $\mathbb {Z}_2$ or 0.

The following lemma will be useful:

Lemma 2.2

Let $Q\subset \mathbb {P}^4$ be any quadric hypersurface containing the curve $C_4$. Suppose that $\textrm{Aut}(Q,C_4)$ is finite, and contains an element of finite order $n>2$. Then Q is singular.

Proof

Since $\textrm{Aut}(Q,C_4)\subseteq \textrm{Aut}(\mathbb {P}^4,C_4)\cong \textrm{PGL}_2(\mathbb {C})$, we may identify $\textrm{Aut}(Q,C_4)$ with a subgroup of $\textrm{PGL}_2(\mathbb {C})$. Moreover, by considering the action of $\textrm{PGL}_2(\mathbb {C})$ on the parameter space $\mathbb {P}^5$, we identify $\textrm{Aut}(Q,C_4)$ with the stabiliser of the point of $\mathbb {P}^5$ corresponding to Q. Fix a finite cyclic subgroup $G\subset \text {PGL}_2(\mathbb {C})$ of order n, and let $g_1\in G$ be a generator. Then $g_1$ fixes precisely two distinct points of $\mathbb {P}^1$, which upto projective transformation are [0 : 1] and [1 : 0]. Thus

$$\begin{aligned}g_1=\begin{pmatrix} 1 &{} 0 \\ 0 &{} \zeta \end{pmatrix},\end{aligned}$$

for some primitive $n^\text {th}$ root of unity $\zeta$. Then $g_1$ acts on $\mathbb {P}^5$ by:

$$\begin{aligned}\mapsto [\zeta ^6a_0:\zeta ^5a_1:\zeta ^4a_2:\zeta ^3a_3:\zeta ^2a_4:\zeta ^4a_5],\end{aligned}$$

and we can read off the points of $\mathbb {P}^5$ whose stabiliser contains G:

• $n=2$: $(\mathbb {P}^5)^G = \big \{[a_0:0:a_2:0:a_4:a_5],[0:a_1:0:a_3:0:0]\big \}$,

• $n=3$: $(\mathbb {P}^5)^G = \big \{[a_0:0:0:a_3:0:0], [0:a_1:0:0:a_4:0]\big \}$,

• $n=4$: $(\mathbb {P}^5)^G = \big \{[a_0:0:0:0:a_4:0]\big \},$

• $n>4$: $(\mathbb {P}^5)^G=\varnothing$,

where the numbers $a_0,a_1,a_2,a_3,a_4,a_5$ are all arbitrary complex numbers. One checks that for $n>2$, the corresponding threefolds which have finite $\textrm{Aut}(Q,C_4)$ are all singular.

Now let us recall the following classification theorem for quadric threefolds which contain the curve $C_4$:

Theorem 2.3

([5]) Let $Q\subset \mathbb {P}^4$ be a smooth quadric containing $C_4$. Then there exists an automorphism $\phi \in \textrm{PGL}_2(\mathbb {C})$ such that $\phi (Q)$ is given by one of the following equations:

• (1) $\mu (f_0+f_4)+\lambda f_2+f_5=0$, for some $\lambda \in \mathbb {C}\setminus \{1,-3\}$ and $\mu \in \mathbb {C}\setminus \{2,-2\}$ such that $\mu ^2\ne -\lambda ^2-2\lambda +3$,

• (2) $f_0+\lambda f_2+ f_5=0$, for some $\lambda \in \mathbb {C}\setminus \{1,-3\}$,

• (3) $f_1+f_5=0$.

Let us find $\text {Aut}(Q,C_4)$ in each of these cases.

Proof of Theorem 2.1

We may assume that $\mu \ne 0$ in case (1), and $\lambda \ne 0$ in case (2), as otherwise the threefolds are isomorphic to those which are described in Example 1.3 and Example 1.4. Then $\textrm{Aut}(Q,C_4)$ is finite, and since $\textrm{Aut}(Q,C_4)$ is isomorphic to a subgroup of $\textrm{PGL}_2(\mathbb {C})$, it must be isomorphic to one of the following groups:

$$\begin{aligned}0,\mathbb {Z}_n,\mathbb {Z}_2\times \mathbb {Z}_2,D_{2n},\mathfrak {A}_4,\mathfrak {S}_4,\mathfrak {A}_5,\end{aligned}$$

where $\mathfrak {S}_n$ (resp. $\mathfrak {A}_n$) is the symmetric (resp. alternating) group on n letters. Then by Lemma 2.2 the only possibilities are that $\textrm{Aut}(Q,C_4)$ is isomorphic to $\mathbb {Z}_2\times \mathbb {Z}_2, \mathbb {Z}_2$ or 0.

Suppose Q is in case (1). Then Q admits an action of $\mathbb {Z}_2\times \mathbb {Z}_2$, generated by $g_1,g_2\in \textrm{PGL}_2(\mathbb {C})$, which are given by:

$$\begin{aligned}g_1=\begin{pmatrix}1 &{} 0 \\ 0 &{} -1\end{pmatrix},g_2=\begin{pmatrix}0 &{} 1 \\ 1 &{} 0 \end{pmatrix}.\end{aligned}$$

Hence, $\textrm{Aut}(Q,C_4)\cong \mathbb {Z}_2\times \mathbb {Z}_2$.

Suppose that Q is in case (2). Then Q admits an action of the group $\mathbb {Z}_2$, generated by the element $g_1$. Suppose that $\textrm{Aut}(Q,C_4)\cong \mathbb {Z}_2\times \mathbb {Z}_2$, and let $g\in \textrm{Aut}(Q,C_4)$ be a non-trivial element distinct from $g_1$. Considering the standard action of $\textrm{PGL}_2(\mathbb {C})$ on $\mathbb {P}^1$, observe that $g_1$ fixes the points [0 : 1] and [1 : 0], and since $gg_1=g_1g$, we see that g must swap these points. Since g has order 2, it must be equal to either $g_2$ or $g_1g_2$. The threefold Q is not invariant under either of these.

Finally suppose that Q is in case (3), and suppose that $\textrm{Aut}(Q,C_4)$ is non-trivial. Then it contains an element, g, of order 2. Since g fixes two distinct points of $\mathbb {P}^1$, it must be equal to $g_2$, $g_1g_2$, or be given by a matrix of the form

$$\begin{aligned} \begin{pmatrix} 1 &{} a \\ b &{} -1 \end{pmatrix}, \text { for some } a,b\in \mathbb {C}\text { such that } \quad ab\ne -1. \end{aligned}$$

One checks that $g_2$ nor $g_1g_2$ leave Q invariant, and if g is given by a matrix of the above form then g(Q) is given by the equation:

$$\begin{aligned}{} & {} 4bf_0+2(1-3bc)f_1-3c(1-bc)f_2+2c^2(3-bc)f_3-4c^3f_4\\{} & {} \quad +(bc^2-2b^2c^2-4bc-c-2)f_5=0\end{aligned}$$

Clearly $g(Q)\ne Q$, so that $\textrm{Aut}(Q,C_4)$ has to be trivial. $\square$

3 Existence of the additional involution

The second half of proving Theorem 1.1 is the assertion that $\textrm{Aut}(X)\cong \textrm{Aut}(Q,C_4)\times \mathbb {Z}_2$, which we will do in this section. The result is:

Theorem 3.1

Let X be a smooth Fano threefold in family No2.21. Then there exists an involution $g\in \textrm{Aut}(X)$ such that $\textrm{Aut}(X)\cong \textrm{Aut}(Q,C_4)\times \langle g\rangle$.

Proof

We proceed case-by-case, according to the classification in Theorem 2.3.

3.1 Case (1): Q is given by $\mu (f_0+f_4)+\lambda f_2+f_5=0$

Observe that the linear system of quadrics which contain $C_4$ is 5-dimensional, so it is more natural to express members of family No2.21 in terms of fourfolds. Let us show how to do this. Fix the Veronese surface $S_4\subset \mathbb {P}^5$ given by the embedding:

$$\begin{aligned} \upsilon :\mathbb {P}^2&\rightarrow \mathbb {P}^5 \\ {[}x:y:z]&\mapsto [x^2:xy:y^2:yz:z^2:xz]. \end{aligned}$$

The space of global sections of $\mathcal {I}_{S_4}(2)$ is generated by the quadratic forms:

$$\begin{aligned} g_0&= x_3^2 - x_2 x_4,\\ g_1&= x_3 x_5 - x_1 x_4,\\ g_2&= x_5^2 - x_0 x_4, \\ g_3&= x_1 x_5-x_0 x_3, \\ g_4&= x_1^2 - x_0 x_2, \\ g_5&= x_1 x_3 - x_2 x_5, \end{aligned}$$

where $x_0,x_1,x_2,x_3,x_4,x_5$ are homogeneous coordinates on $\mathbb {P}^5$.

Consider the following rational map:

$$\begin{aligned} \phi :\mathbb {P}^5&\dashrightarrow \mathbb {P}^5 \\ {[}x_0:x_1:x_2:x_3:x_4:x_5]&\mapsto [g_0:g_1:g_2:g_3:g_4:g_5]. \end{aligned}$$

I claim that $\phi$ is a birational involution. The following observation is due to I. Dolgachev: we can identify $\mathbb {P}^5$ with the space of symmetric $3\times 3$ matrices, upto scaling. Then under this identification, the rational map above is:

$$\begin{aligned} \phi :\mathbb {P}^5&\dashrightarrow \mathbb {P}^5 \\ \begin{pmatrix} x_0 &{} x_1 &{} x_5 \\ x_1 &{} x_2 &{} x_3 \\ x_5 &{} x_3 &{} x_4\end{pmatrix}&\mapsto \begin{pmatrix} x_3^2 - x_2 x_4 &{} x_3 x_5 - x_1 x_4 &{} x_1 x_3 - x_2 x_5 \\ x_3 x_5 - x_1 x_4 &{} x_5^2 - x_0 x_4 &{} x_1 x_5-x_0 x_3 \\ x_1 x_3 - x_2 x_5 &{} x_1 x_5-x_0 x_3 &{} x_1^2 - x_0 x_2\end{pmatrix} \end{aligned}$$

But this is the same map as taking a matrix M to its adjoint $\textrm{adj}(M)$. Thus it follows from the relation $\textrm{adj}(\textrm{adj}(A))=\textrm{det}(A)^{n-2}A$ for any $n\times n$ matrix A that $\phi$ is a birational involution.

Let $\sigma :\widetilde{\mathbb {P}}^5\rightarrow \mathbb {P}^5$ be the blow-up of $\mathbb {P}^5$ in $S_4$, and let E be the exceptional divisor. Observe that for general divisors $\widetilde{H}\in |\sigma ^*\mathcal {O}_{\mathbb {P}^5}(1)|$ and $\widetilde{Q}\in |\sigma ^*\mathcal {O}_{\mathbb {P}^5}(2)-E|$, we have that $\widetilde{H}\cap \widetilde{Q}$ is a smooth element of family 2.21.

Since $\phi$ has base locus equal to $S_4$, it lifts to a biregular involution $g\in \textrm{Aut}(\widetilde{\mathbb {P}}^5)$ which swaps the linear systems $|\widetilde{H}|$ and $|\widetilde{Q}|$. Thus, the intersection $\widetilde{H}\cap g(\widetilde{H})$ is $\langle g\rangle$-invariant, for any $\widetilde{H}$. We will now show that every smooth element X of family 2.21 which is in case (1) of Theorem 2.3 is isomorphic to a subvariety of $\widetilde{\mathbb {P}}^5$ of the form $\widetilde{H}\cap g(\widetilde{H})$, for some hyperplane $H\subset \mathbb {P}^5$, and therefore possesses an involution not coming from $\textrm{Aut}(Q,C_4)$.

So fix such a threefold X. Then the quadric Q is given by the equation

$$\begin{aligned}\mu (f_0+f_4)+\lambda f_2+f_5=0,\end{aligned}$$

for some $\lambda \in \mathbb {C}\setminus \{1,-3\}$ and $\mu \in \mathbb {C}\setminus \{2,-2\}$ such that $\mu ^2\ne -\lambda ^2-2\lambda +3$. Let us choose roots a, b of the equations

$$\begin{aligned} (\mu +2)x^4+2\lambda -2&=0, \\ (\mu + 2)x^4+\mu -2&=0, \end{aligned}$$

respectively, so that the equation of Q becomes:

$$\begin{aligned} \frac{2 - 2b^4}{1 + b^4}(f_0+f_4)+ \frac{1-2a^4 + b^4}{1 + b^4}f_2 + f_5=0. \end{aligned}$$

(3.2)

Now consider the following hypersurfaces in $\mathbb {P}^5$:

$$\begin{aligned} H&=\{x_0=a^2x_2+b^2x_4\},\\ Q_2&=\{g_0=a^2g_2+b^2g_4\}. \end{aligned}$$

We have that the intersections $H\cap S_4$ and $Q_2\cap S_4$ are smooth, so that the intersection of their strict transforms, $\widetilde{H}\cap \widetilde{Q}_2\subset \widetilde{\mathbb {P}}^5$, is a smooth member of family 2.21. Moreover, $Q_2=\phi (H)$, so that $\widetilde{H}\cap \widetilde{Q}_2$ is $\langle g\rangle$-invariant. Consider the projective transformation $\psi :\mathbb {P}^5\rightarrow \mathbb {P}^5$ given by the matrix¹

$$\begin{aligned} \begin{pmatrix} 1 &{} 0 &{} 0 &{} 0 &{} b^2 &{} -2b \\ 0 &{} -a &{} 0 &{} ab &{} 0 &{} 0 \\ 0 &{} 0 &{} a^2 &{} 0 &{} 0 &{} 0 \\ 0 &{} -a &{} 0 &{} -ab &{} 0 &{} 0 \\ 1 &{} 0 &{} 0 &{} 0 &{} b^2 &{} 2b \\ 1 &{} 0 &{} 0 &{} 0 &{} -b^2 &{} 0 \end{pmatrix}. \end{aligned}$$

Then the intersection of the fourfolds

$$\begin{aligned} \psi (H)&=\{x_2-x_5=0\}\\ \psi (Q_2)&=\{(1-b^4)(g_0+g_4)-a^4g_2+2(1+b^4)g_5=0\} \end{aligned}$$

is given as a subvariety of $\mathbb {P}^4$ by Eq. 3.2, which defines X. Thus $X\cong \widetilde{H}\cap \widetilde{Q}_2$.

It remains to show that the birational involution g commutes with the action of $\text {Aut}(Q,C_4)$.

Consider the subgroup $G\subset \textrm{PGL}_6(\mathbb {C})$ generated by the commuting involutions

$$\begin{aligned} \alpha :[x_0:x_1:x_2:x_3:x_4:x_5]&\mapsto [x_0:x_1:x_2:-x_3:x_4:-x_5]\\ \beta :[x_0:x_1:x_2:x_3:x_4:x_5]&\mapsto [x_0:-x_1:x_2:-x_3:x_4:x_5]. \end{aligned}$$

Then $\alpha$ and $\beta$ commute with the birational involution described previously:

$$\begin{aligned} \phi :[x_0:x_1:x_2:x_3:x_4:x_5]\mapsto [x_3^2 - x_2 x_4&: x_3 x_5 - x_1 x_4:x_5^2 - x_0 x_4: \\&:x_1 x_5-x_0 x_3: x_1^2 - x_0 x_2:x_1 x_3 - x_2 x_5], \end{aligned}$$

The hypersurfaces H and $Q_2$ are G-invariant. Moreover $S_4$ is G-invariant, so that G is isomorphic to a subgroup of $\textrm{Aut}(Q,C_4)$. Thus since $\textrm{Aut}(Q,C_4)\cong \mathbb {Z}_2\times \mathbb {Z}_2$ by Theorem 2.1, we have that $\textrm{Aut}(Q,C_4)\cong G$. So we see that $\textrm{Aut}(Q,C_4)$ commutes with the involution g.

For the remaining cases, we will compute bases for the linear system $\mathfrak {d}$ of quadrics sections of $Q\subset \mathbb {P}^4$ containing the curve $C_4$ such that the corresponding rational map is an involution, and commutes with the action of $\textrm{Aut}(Q,C_4)$.

3.2 Case (2): Q is given by $f_0+\lambda f_2+f_5=0$

Let us make the substitution $\lambda =1-4s^2$, for some $s\in \mathbb {C}\setminus \{-1,0,1\}$. Consider the rational map:

$$\begin{aligned} \iota :Q&\dashrightarrow \mathbb {P}^4 \\ [x_0:x_1:x_2:x_3:x_4]&\mapsto \left[ f_4+\frac{s^2}{2}f_2-\frac{1}{16}f_0: \frac{s}{4}f_1+sf_3: s^2f_2: sf_1: f_0\right] . \end{aligned}$$

Observe that it has base locus equal to $C_4$, so indeed corresponds to the linear system $\mathfrak {d}$. To see that the map $\iota$ is a birational involution, consider the following rational parametrisation of Q,

$$\begin{aligned} p:\mathbb {P}^3&\dashrightarrow Q \\ [x_0:x_2:x_3:x_4]&\mapsto \left[ x_0x_3 : s^2x_0x_4 - s^2x_2^2 + x_2^2 - \frac{1}{4}x_2x_4 + \frac{1}{4}x_3^2 : x_2x_3 : x_3^2 : x_3x_4\right] . \end{aligned}$$

This is a rational inverse to the projection $Q\dashrightarrow \mathbb {P}^3$ from the point [0 : 1 : 0 : 0 : 0]. Moreover, it is an isomorphism between the open subsets $\mathbb {P}^3\setminus \Pi$ and $Q\setminus V$, where $\Pi \subset \mathbb {P}^3$ is the plane given by $x_3=0$, and $V\subset Q$ is the singular quadric surface given by the intersection of Q with the plane $x_3=0$, this latter variety being the closure of the union of lines through [0 : 1 : 0 : 0 : 0]. Let Z be the curve $p^{-1}(C_4)$, which is a quartic rational curve in $\mathbb {P}^3$.

Then $\iota (p(\mathbb {P}^3\setminus (\Pi \cup Z))$ lies in Q, and since $p(\mathbb {P}^3\setminus (\Pi \cup Z))=Q\setminus (V\cup C_4)$ is dense in $Q\setminus C_4$, it follows that $\iota$ is a rational self-map of Q. To see that $\iota$ is an involution on $Q\setminus C_4$, observe that $\iota \circ \iota \circ p$ is equal to the map

$$\begin{aligned} \mathbb {P}^3&\dashrightarrow Q \\ [x_0:x_1:x_2:x_3]&\mapsto \left[ x_0: \frac{(-4s^2 + 4)x_2^2 - x_2x_4 + 4s^2x_0x_4 + x_3^2}{4x_3}: x_2: x_3: x_4\right] , \end{aligned}$$

which is equal to the identity morphism on $\mathbb {P}^3\setminus \Pi$. Thus $\iota \circ \iota$ is equal to the identity morphism on $Q\setminus (V\cup C_4)$, so that it is equal to the identity morphism on $Q\setminus C_4.$

Let us prove that $\iota$ commutes with the action of $\textrm{Aut}(Q,C_4)$. If $s\ne \pm \frac{1}{2}$ then by Theorem 2.1, $\textrm{Aut}(Q,C_4)=\langle g_1\rangle$, where $g_1$ is the linear transformation

$$\begin{aligned}\mapsto [x_0:-x_1:x_2:-x_3:x_4].\end{aligned}$$

Then it is plain that $\iota$ commutes with $g_1$. If $s=\pm \frac{1}{2}$, then Q is the quadric described in Example 1.4, and $\textrm{Aut}(Q,C_4)$ contains subgroup isomorphic to $\mathbb {G}_a$ consisting of automorphisms of the form

$$\begin{aligned}{}[x_0:x_1:x_2:x_3:x_4]\mapsto [x_0+4tx_1+6t^2x_2+4t^3x_3+t^4x_4&:x_1+3tx_2+3t^2x_3+x_4: \\&:x_2+2tx_3+t^2x_4:x_3+tx_4:x_4], \end{aligned}$$

for every $t\in \mathbb {C}$. One can easily see that $\iota$ commutes with each of these automorphisms.

3.3 Case (3): Q is given by $f_1+f_5=0$

Let $\sigma$ be the rational map

$$\begin{aligned} \sigma :Q&\dashrightarrow \mathbb {P}^4 \\ [x_0:x_1:x_2:x_3:x_4]&\mapsto \left[ \frac{1}{64}f_0 - \frac{1}{4}f_1 + \frac{3}{2}f_2 + 16f_3 + 64f_4 :\right. \\&\left. :-\frac{1}{8}f_0 + \frac{3}{2}f_1 + 2f_2 + 32f_3 :f_0 - 8f_1 + 16f_2:-8f_0 + 32f_1:64f_0\right] . \end{aligned}$$

By replacing the map $\iota$ with $\sigma$, the map p by the rational parametrisation

$$\begin{aligned} \mathbb {P}^3&\dashrightarrow Q \\ [x_1:x_2:x_3:x_4]&\mapsto \left[ \frac{4x_1x_3 + x_1x_4 - 3x_2^2 - x_2x_3}{x_4}:x_1:x_2:x_3:x_4\right] , \end{aligned}$$

and the varieties $\Pi$ and V with $\{x_4=0\}$, it follows verbatim from the proof of case (2) that $\sigma$ is a birational involution of Q with base locus equal to $C_4$. By Theorem 2.1, $\textrm{Aut}(Q,C_4)$ is trivial in this case, so there is nothing left to prove. $\square$