Proof. We use labels for configurations of nodes and planes from [Reference Finkelnberg and Werner37].

• • $s=1, \ldots , 5$ , $p=0$ ; (J1–J5):

  $\mathrm {Cl}(X)={\mathbb Z}$ is freely generated by the hyperplane section, with trivial G-action, and $\mathrm {Pic}(\tilde X)$ is freely generated by the exceptional divisors of the blowup $\tilde X\to X$ and the pullback to $\tilde X$ of the basis of $\mathrm {Cl}(X)$ . The G-action permutes the exceptional divisors. So $\mathrm {Pic}(\tilde X)$ is a permutation module.

• • $s=4,5,6$ , $p=1$ ; (J6–J8):

  $\mathrm {Cl}(X)={\mathbb Z}^2$ is freely generated by the hyperplane section and the unique, necessarily G-stable, plane in X, with trivial G-action on their classes. Thus, $\mathrm {Pic}(\tilde X)$ is a permutation module.

• • $s=7$ , $p=2$ ; (J10):

  $\mathrm {Cl}(X)={\mathbb Z}^3$ is freely generated by the hyperplane section and the classes of the two planes in X, with G possibly permuting these two classes. Thus, $\mathrm {Pic}(\tilde X)$ is a permutation module.

• • $s=6,7$ , $p=3$ ; (J11–J12):

  $\mathrm {Cl}(X)={\mathbb Z}^3$ is freely generated by the classes of the three planes in X which form a tetrahedron (with one face missing), with G possibly permuting these classes. So $\mathrm {Pic}(\tilde X)$ is a permutation module.

Proof. Suppose first that there exists a $\langle \tau \rangle $ -equivariant blowup $\pi \colon X\to Y$ of a nonhyperelliptic curve $C\subset Y$ , where Y is a smooth threefold with trivial intermediate Jacobian. Since C is not hyperelliptic, it follows from Theorem 3 in [Reference Lauter49, Appendix] that

(1) $$ \begin{align} \mathrm{Aut}\big(\mathrm{IJ}(X)\big)\simeq\mathrm{Aut}(C)\times C_2, \end{align} $$

where the second factor corresponds to the action of multiplication by $(-1)$ . If C is pointwise fixed by $\tau $ , then $\tau $ acts trivially on $\mathrm {IJ}(X)$ . If $\tau $ acts faithfully on C, then its action on $\mathrm {IJ}(X)=\mathrm {J}(C)$ is induced by the action on C; thus, $\tau $ cannot project nontrivially to the second factor in (1).

The general case is treated similarly, using equivariant weak factorization.

Proof. Let $H=\langle \tau \rangle $ ; the action produces the symbol

$$ \begin{align*}(H, Y/H\curvearrowright k(S), (b)). \end{align*} $$

By [Reference Bogomolov and Prokhorov14], ${\mathrm H}^1(Y/H, \mathrm {Pic}(S))\neq 0$ , which implies that the symbol is incompressible; see [Reference Cheltsov, Tschinkel and Zhang24, Section 4]. Such symbols cannot appear for linear actions; see [Reference Tschinkel, Yang and Zhang53, Corollary 6.1].

Proof. Note that G leaves invariant a plane in $\mathbb {P}^4$ if and only if it leaves invariant a line. Thus, we may assume that there exists a G-stable plane in $\mathbb {P}^4$ . Linear projection from this plane induces a rational dominant map whose general fiber is a (possibly singular) cubic surface. Taking a G-equivariant resolution of indeterminacies of this map, a G-invariant resolution of singularities (if necessarily), and applying the G-equivariant Minimal Model program over $ {\mathbb P}^1$ , we obtain a G-birational map from X to a G-Mori fiber space with a positive-dimensional base.

Proof. Suppose that X is G-solid. If there exists a G-equivariant birational map , then $\mathbb {P}^3$ is G-solid, which contradicts [Reference Cheltsov and Sarikyan20]. Thus, the G-action on X is not linearizable. It follows from [Reference Avilov6, Reference Avilov4, Reference Avilov5] and the proofs of the main results in these papers that either X and G are as in (1), (2), (3), or X is the cubic threefold in (3) and $G\simeq C_4\rtimes C_5$ . Let us show that X is not G-solid in the latter case, contradicting the assumption.

Namely, suppose X is the threefold from (3), and $G\simeq C_4\rtimes C_5$ . By [Reference Avilov5, Reference Cheltsov, Sarikyan and Zhuang21], there exists the following $\mathfrak {S}_5$ -Sarkisov link:

where Q is the smooth quadric threefold

$$ \begin{align*}\{x_1x_2+x_2x_3 +\cdots +x_4x_5= 0\}\subset\mathbb{P}^4, \end{align*} $$

$\mathfrak {S}_5$ acts on Q by permuting the coordinates, $\chi $ is the birational map induced by the standard Cremona involution of $\mathbb {P}^4$ , $\alpha $ is the standard resolution of singularities, $\beta $ is a composition of $10$ Atiyah flops, and $\gamma $ is a blowup of an $\mathfrak {S}_5$ -orbit of length $5$ . Let $\eta \colon Q\to \mathbb {P}^3$ be the double cover induced by the projection from the $\mathfrak {S}_5$ -fixed point in $\mathbb {P}^4$ . Then $\eta $ is $\mathfrak {S}_5$ -equivariant, and $\mathbb {P}^3$ contains two skew lines $L_1$ and $L_2$ such that the curve $L_1+L_2$ is G-invariant. Let $C_1$ and $C_2$ be the preimages of these lines on Q. Then $C_1$ and $C_2$ are disjoint conics, and the curve $C_1+C_2$ is G-invariant. Blowing up these two conics, we obtain a G-equivariant birational map from Q to a conic bundle over $\mathbb {P}^1\times \mathbb {P}^1$ ; in particular, X is not G-solid, which contradicts our assumption. In fact, the G-action on Q is linearizable; see Section 6.

Proof. We can choose the coordinates so that

$$ \begin{align*}\iota: (x_1,x_2,x_3,x_4,x_5)\mapsto(sx_2,tx_1,x_3,ux_4,vx_5), \end{align*} $$

for some $s,t,u,v\in k^\times $ , and X is given by

$$ \begin{align*}x_1x_2x_3+x_1q_1+x_2q_2+f_3=0, \end{align*} $$

for

$$ \begin{align*} q_1&=a_1x_4^2+a_2x_4x_5+a_3x_5^2 + a_4x_3^2+a_5x_3x_4+a_6x_3x_5, \\ q_2&=b_1x_4^2+b_2x_4x_5+b_3x_5^2 + b_4x_3^2+b_5x_3x_4+b_6x_3x_5, \\ f_3&=c_1x_3^3+x_3^2d(x_4,x_5)+x_3e(x_4,x_5) + r(x_4,x_5), \end{align*} $$

where

$$ \begin{align*} &\hskip -2.5cm d=d_1x_4+d_2x_5, \\ & \hskip -2.5cm e=e_1x_4^2+e_2x_4x_5+e_3x_5^2, \\ &\hskip -2.5cm r=r_1x_4^3+r_2x_4^2x_5+r_3x_4x_5^2+r_4x_5^3. \end{align*} $$

Since X is $\langle \iota \rangle $ -invariant, one has $\iota ^*(f)=stf$ , and thus, the zero loci of $q_1q_2$ and $f_3$ are preserved; and these polynomials cannot identically vanish, under our assumptions on singularities of X. Concretely,

$$ \begin{align*}\iota^* (f) = stx_1x_2x_3+sx_2\iota^*(q_2)+tx_1\iota^*(q_2)+\iota^*(f_3)=stf, \end{align*} $$

which implies that

(4) $$ \begin{align} \iota^*(q_2) = tq_1, \quad \iota^* (q_1) = sq_2, \end{align} $$

and

(5) $$ \begin{align} \iota^*(f_3)=stf_3. \end{align} $$

Expanding and substituting into (4), we obtain 12 equations:

$$ \begin{align*} & \hskip -2.8cm u^2a_1-tb_1=uva_2-tb_2=v^2a_3-tb_3=0, \\ & \hskip -2.8cm sa:1-u^2b_1=sa:2-uvb_2=sa:3-v^2b_3=0, \\ & \hskip -2.8cm a_4-tb_4=ua_5-tb_5=va_6-tb_6=0, \\ & \hskip -2.8cm sa:4-b_4=sa:5-ub_5=sa:6-vb_6=0, \end{align*} $$

and, writing down the (5) constraints on $f_3$ , additional equations

$$ \begin{align*} & c_1(1-st)=0, \\ & d_1(u-st)=d_2(v-st)=0,\\ & e_1(u^2-st)=e_2(uv-st)=e_3(v^2-st)=0, \\ & r_1(u^3-st)=r_2(uv-st)=r_3(uv^2-st)=r_4(v^3-st)=0, \end{align*} $$

in the variables

$$ \begin{align*}a_1,\ldots,a_6,b_1,\ldots,b_6,c_1,d_1,d_2,e_1,e_2,r_1,\ldots,r_4\in k. \end{align*} $$

Since X has nodes at $[1:0:0:0:0]$ and $[0:1:0:0:0]$ , we have

$$ \begin{align*}a_2^2-4a_1a_3\ne 0, \quad b_2^2-4b_1b_3\ne 0. \end{align*} $$

Thus, up to scaling $x_4,x_5$ and swapping them, we may further assume that one of the following holds:

• • $a_2=b_2=1$ , or

• • $a_2=b_1=b_3=1$ , $b_2=0$ , or

• • $a_1=b_1=b_3=1$ , $a_2=b_2=0$ .

The second option is impossible since $b_2=0$ forces $a_2=0$ . Solving the system of equations for the remaining two options using Magma, we obtain a complete set of solutions:

Here, we omitted solutions obtained by swapping coordinates $x_4$ and $x_5$ and scaling coordinates $x_1$ and $x_2$ . After an additional equivariant change of coordinates, we obtain the required assertion.

Proof. By the assumptions, G contains an element $\iota $ switching the nodes of X. It suffices to prove the required assertion for $G=\langle \iota \rangle $ . With the notation as above, we may assume that X is given by (2); that is,

$$ \begin{align*}f=x_1x_2x_3 + x_1q_2 + x_2q_2 + f_3=0, \end{align*} $$

and $\iota $ acts on the coordinates via

$$ \begin{align*}\iota: (x_1,x_2,x_3,x_4,x_5)\mapsto(sx_2,tx_1,x_3,ux_4,vx_5), \end{align*} $$

for roots of unity $s,t,u,v$ . Introducing new coordinates

$$ \begin{align*}w_1=\sqrt{t}x_1+\sqrt{s}x_2,\quad w_2=\sqrt{t}x_1-\sqrt{s}x_2, \end{align*} $$

we diagonalize $\iota $ , so that it acts via

(6) $$ \begin{align} \iota: (w_1,w_2,x_3,x_4,x_5)\mapsto(\lambda w_1,-\lambda w_2,x_3,ux_4,vx_5), \end{align} $$

where $\lambda =\sqrt {st}$ . The equation of X in the new coordinates becomes

$$ \begin{align*}f'=(w_1^2-w_2^2)x_3+(w_1+w_2)2\sqrt s q_1+(w_1-w_2)2\sqrt t q_2+4\sqrt{st}f_3=0. \end{align*} $$

Note that $\iota ^*(f')=\lambda ^2f'$ .

Recall that X is G-birational to the conic bundle (3). The conic bundle is not standard. In particular, the intermediate Jacobian of $\widetilde {X}$ and the Jacobian of the curve D are isomorphic, as principally polarized abelian varieties:

$$ \begin{align*}\mathrm{IJ}(\widetilde{X})\simeq \mathrm{IJ}(Y)\simeq \mathrm{J}(D), \end{align*} $$

where $\widetilde {X}$ is the standard desingularization of X, Y is the conic bundle in (3) and D is its discriminant curve in ${\mathbb P}^2_{x_3,x_4,x_5}$ given by the quartic form $h=q_1q_2-x_3f_3$ . Note that $\iota ^*(h)=\lambda ^2h$ .

Arguing as in the proof of [Reference Beauville9, Lemma 1], we see that $\iota $ acts faithfully on $\mathrm {IJ}(\widetilde {X})$ and preserves the principal polarization. However, the $\iota $ -action on coordinates $x_3,x_4,x_5$ induces an action on D and its Jacobian $\mathrm {J}(D)$ . We claim that the $\iota $ -actions on $\mathrm {IJ}(\widetilde X)$ and $\mathrm {J}(D)$ differ by multiplication by $-1$ . Since D is not hyperelliptic, this would imply that the G-action on X is not linearizable, by Remark 2.5.

To compute the action of $\iota $ on $\mathrm {IJ}(\widetilde {X})$ , recall that its tangent space at zero ${\mathcal T}_0 \,\mathrm {IJ}(\widetilde {X})$ is isomorphic to ${\mathrm H}^2(\widetilde {X},\Omega ^1_{\widetilde {X}})$ . We show that ${\mathrm H}^2(\widetilde {X},\Omega ^1_{\widetilde {X}})^\vee $ is canonically isomorphic to the linear subspace in

$$ \begin{align*}{\mathrm H}^0\big(\mathbb{P}^4,\Omega^4_{\mathbb{P}^4}\otimes\mathcal{O}_{\mathbb{P}^4}(2X)\big)\cong {\mathrm H}^0\big(\mathbb{P}^4,\mathcal{O}_{\mathbb{P}^4}(K_{\mathbb{P}^4}+2X)\big)\cong {\mathrm H}^0\big(\mathbb{P}^4,\mathcal{O}_{\mathbb{P}^4}(1)\big) \end{align*} $$

consisting of all sections that vanish at the nodes of X. The proof is essentially contained in [Reference Cynk33]. We follow the proof of [Reference Beauville9, Lemma 1]. Let $\pi : \widetilde {\mathbb P}^4\to {\mathbb P}^4$ be the blowup of ${\mathbb P}^4$ centered at two nodes $p_1$ and $p_2$ of X, and identify $\widetilde X$ with the strict transform of X in $\widetilde {\mathbb P}^4.$ The exact sequence

$$ \begin{align*}0\to ({\mathcal N}_{\widetilde{X}/\widetilde{{\mathbb P}^4}})^\vee\to\Omega_{\widetilde{{\mathbb P}^4}\vert\widetilde{X}}^1\to\Omega^1_{\widetilde{X}}\to 0 \end{align*} $$

gives rise to a $\langle\iota\rangle$ –equivariant exact sequence

$$ \begin{align*}0\to {\mathrm H}^2(\widetilde{X},\Omega_{\widetilde{X}}^1)\to{\mathrm H}^3(\widetilde{X},({\mathcal N}_{\widetilde{X}/\widetilde{{\mathbb P}^4}})^\vee)\to {\mathrm H}^3(\widetilde X,\Omega^1_{\widetilde{{\mathbb P}^4}\vert\widetilde{X}})\to 0. \end{align*} $$

By [Reference Cynk33], the dimension of ${\mathrm H}^3(\widetilde X,\Omega ^1_{\widetilde {{\mathbb P}^4}\vert \widetilde {X}})$ equals the defect of X, which is $0$ in our case. It follows that

$$ \begin{align*}{\mathrm H}^2(\widetilde{X},\Omega_{\widetilde{X}}^1)\cong{\mathrm H}^3(\widetilde{X},({\mathcal N}_{\widetilde{X}/\widetilde{{\mathbb P}^4}})^\vee). \end{align*} $$

Similarly, the $\langle\iota\rangle$ –equivariant exact sequence

$$ \begin{align*}0\to{\mathcal O}_{\widetilde{{\mathbb P}^4}}(-2\widetilde X)\to{\mathcal O}_{\widetilde{{\mathbb P}^4}}(-\widetilde{X})\to ({\mathcal N}_{\widetilde{X}/\widetilde{{\mathbb P}^4}})^\vee\to0 \end{align*} $$

and the vanishing of ${\mathrm H}^i(\widetilde {{\mathbb P}^4},{\mathcal O}_{\widetilde {{\mathbb P}^4}}(-\widetilde X))$ ([Reference Cynk33, Corollary 2]) provide an $C_2$ -isomorphism

$$ \begin{align*}{\mathrm H}^3(\widetilde{X},({\mathcal N}_{\widetilde{X}/\widetilde{{\mathbb P}^4}})^\vee)\cong{\mathrm H}^4(\widetilde{\mathbb P}^4,{\mathcal O}_{\widetilde{{\mathbb P}^4}}(-2\widetilde{X})). \end{align*} $$

By Serre duality, we have canonical isomorphisms between

$$ \begin{align*}{\mathrm H}^4(\widetilde{\mathbb P}^4,{\mathcal O}_{\widetilde{{\mathbb P}^4}}(-2\widetilde{X}))^\vee\cong{\mathrm H}^0(\widetilde{{\mathbb P}^4},K_{\widetilde{{\mathbb P}^4}}\otimes{\mathcal O}_{\widetilde{{\mathbb P}^4}}(2\widetilde{X}))\cong{\mathrm H}^0(\widetilde{\mathbb{P}^4},\Omega^4_{\widetilde{\mathbb{P}^4}}\otimes\mathcal{O}_{\widetilde{\mathbb{P}^4}}(2\widetilde X)). \end{align*} $$

So we have a $\langle\iota\rangle$ –equivariant isomorphism

$$ \begin{align*}{\mathrm H}^2(\widetilde{X},\Omega^1_{\widetilde{X}})^\vee\cong{\mathrm H}^0(\widetilde{{\mathbb P}^4},K_{\widetilde{{\mathbb P}^4}}\otimes{\mathcal O}_{\widetilde{{\mathbb P}^4}}(2\widetilde{X})). \end{align*} $$

Let $E_1$ and $E_2$ be the exceptional divisors of $\pi $ over $p_1$ and $p_2$ , respectively. By adjunction,

$$ \begin{align*}K_{\widetilde{{\mathbb P}^4}}=\pi^*({\mathcal O}_{{\mathbb P}^4}(-5))\otimes{\mathcal O}_{\widetilde{{\mathbb P}^4}}(3E_1+3E_2), \end{align*} $$

and

$$ \begin{align*}{\mathcal O}_{\widetilde{{\mathbb P}^4}}(2\widetilde{X})=\pi^*({\mathcal O}_{{\mathbb P}^4}(6))\otimes{\mathcal O}_{\widetilde{{\mathbb P}^4}}(-4E_1-4E_2). \end{align*} $$

Then we know

$$ \begin{align*}K_{\widetilde{{\mathbb P}^4}}\otimes{\mathcal O}_{\widetilde{{\mathbb P}^4}}(2\widetilde{X})=\pi^*({\mathcal O}_{{\mathbb P}^4}(1))\otimes{\mathcal O}_{\widetilde{{\mathbb P}^4}}(-E_1-E_2). \end{align*} $$

It follows that we can canonically identify ${\mathrm H}^2(\widetilde {X},\Omega ^1_{\widetilde {X}})^\vee $ with linear subspace in ${\mathrm H}^0(\mathbb {P}^4,\Omega ^4_{\mathbb {P}^4}\otimes \mathcal {O}_{\mathbb {P}^4}(2X))$ , (or equivalently, in ${\mathrm H}^0({\mathbb P}^4,{\mathcal O}_{{\mathbb P}^4}(1))$ ), which consists of all sections that vanish at $p_1$ and $p_2.$ Now we can compute the induced G-action on ${\mathcal T}_0\mathrm {IJ}(\widetilde X)^\vee $ explicitly. Set

$$ \begin{align*}z_2=\frac{w_2}{w_1}, \, z_3=\frac{x_3}{w_1}, \, z_4=\frac{x_4}{w_1}, \, z_5=\frac{x_5}{w_1}, \end{align*} $$

and consider the rational $4$ -forms

$$ \begin{align*}z_3\omega, \, z_4\omega,\, z_5\omega, \qquad \text{where}\quad\omega=\frac{dz_2\wedge dz_3\wedge dz_4\wedge dz_5}{\left(f'(1,z_2,z_3,z_4,z_5)\right)^2}. \end{align*} $$

These give sections of ${\mathrm H}^0(\mathbb {P}^4,\Omega ^4_{\mathbb {P}^4}\otimes \mathcal {O}_{\mathbb {P}^4}(2X))$ , forming a basis of the subspace consisting of sections that vanish at the nodes of X. One computes

$$ \begin{align*}\iota^*(z_2)=-z_2,\quad\iota^*(z_3)=\frac1\lambda z_3,\quad\iota^*(z_4)=\frac{u}{\lambda}z_4,\quad\iota^*(z_5)=\frac{v}{\lambda}z_5 \end{align*} $$

and

$$ \begin{align*}\iota^*(f'(1,z_2,\ldots,z_5)^2)=\iota^*\left(\frac{f'(w_1,w_2,x_3,x_4,x_5)^2}{w_1^6}\right)=\frac{f'(1,z_2,\ldots,z_5)^2}{\lambda^2}. \end{align*} $$

Using these, we see that $\iota $ acts on ${\mathcal T}_0 \,\mathrm {IJ}(\widetilde {X})^\vee $ with eigenvalues

$$ \begin{align*}-\frac{uv}{\lambda^2},\quad -\frac{u^2v}{\lambda^2},\quad-\frac{uv^2}{\lambda^2}. \end{align*} $$

Similarly, to compute the action of $\iota $ on $\mathrm {J}(D)$ , we note that ${\mathcal T}_0 \,\mathrm {J}(D)^\vee $ is canonically isomorphic to

$$ \begin{align*}{\mathrm H}^0\big(\mathbb{P}^2,\Omega^2_{\mathbb{P}^2}\otimes\mathcal{O}_{\mathbb{P}^2}(D)\big). \end{align*} $$

Set $y_4=\frac {x_4}{x_3}$ and $y_5=\frac {x_5}{x_3}$ . The rational $2$ -forms

$$ \begin{align*}\frac{dy_4\wedge dy_5}{h(1,y_4,y_5)},\quad y_4\frac{dy_4\wedge dy_5}{h(1,y_4,y_5)},\quad y_5\frac{dy_4\wedge dy_5}{h(1,y_4,y_5)} \end{align*} $$

define sections of ${\mathrm H}^0(\mathbb {P}^2,\Omega ^2_{\mathbb {P}^2}\otimes \mathcal {O}_{\mathbb {P}^2}(D))$ , forming its basis. One has

$$ \begin{align*}\iota^*(y_4)=uy_4,\quad \iota^*(y_5)=vy_5,\quad \iota^*(h(1,y_4,y_5))=\lambda^2h(1,y_4,y_5), \end{align*} $$

and $\iota $ acts on ${\mathcal T}_0 \,\mathrm {J}(D)^\vee $ with eigenvalues

$$ \begin{align*}\frac{uv}{\lambda^2},\quad \frac{u^2v}{\lambda^2},\quad \frac{uv^2}{\lambda^2}. \end{align*} $$

This shows that the $\iota $ -action on $\mathrm {IJ}(\widetilde {X})\simeq \mathrm {J}(D)$ differs from the action on $\mathrm {J}(D)$ induced by the action on D by multiplication by $-1$ , as claimed. Therefore, the G-action on X is not linearizable.

Proof. Let $X\subset {\mathbb P}^4$ be a 3-nodal cubic threefold given by (7), with $\mathrm {Aut}(X)$ not fixing any node. There exists an exact sequence

(8) $$ \begin{align} 0\to H\to\mathrm{Aut}(X)\stackrel{\rho}{\to}{\mathfrak S}_3\to 0 \end{align} $$

and a $\sigma _{123}\in \mathrm {Aut}(X)$ acting transitively on the nodes, so that $\rho (\sigma _{123})=(1,2,3)$ . The zeroes of $q_1,q_2,q_3$ define at most six points on ${\mathbb P}^1_{x_4,x_5}$ ; thus,

(9) $$ \begin{align} \sigma_{123}=\begin{pmatrix} 0&0&s_3&0&0\\ s_1&0&0&0&0\\ 0&s_2&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&\zeta_6^r \end{pmatrix} \end{align} $$

for some $s_1,s_2,s_3\in k^\times $ , where $\zeta _6=e^{\frac {2\pi i}{6}}$ . We have the following cases:

1. (a) $\mathrm {gcd}(q_1,q_2,q_3)=1$ . We may assume that

   $$ \begin{align*}q_1=x_4^2+bx_4x_5+x_5^2, \qquad b\in k,\quad b\ne\pm 2. \end{align*} $$
   The cyclic action on $x_1,x_2$ and $x_3$ implies that $q_2$ and $q_3$ are multiples of $\sigma _{123}^*(q_1)$ and $\sigma _{123}^*(q_2)$ , respectively, and $\sigma _{123}^*(f_3)=s_1s_2s_3f_3.$ The torus action on the coordinates $x_1,x_2$ and $x_3$ allows us to assume that
   $$ \begin{align*} q_2&=\sigma_{123}^*(q_1),\\ q_3&=\sigma_{123}^*(q_2). \end{align*} $$
   Since $q_1$ , $q_2$ , $q_3$ are coprime, we have $r\ne 0, 3$ . Thus, $r=1$ or $r=2$ .

• • If $r=1$ , then $b=0$ ; the entries in (9) are $s_1=s_2=s_3=\pm 1$ , and

  $$ \begin{align*} q_1&=x_4^2+x_5^2, \\ q_2&=x_4^2+\zeta^2_6x_5^2, \\ q_3&=x_4^2+\zeta^4_6x_5^2, \end{align*} $$
  There are subcases:

  • – $f_3(x_4,x_5)\not \equiv 0.$ Then $\sigma _{123}$ fixes the points defined by $f_3$ in ${\mathbb P}^1$ . And up to isomorphism, $f_3=dx_4^3$ or $dx_4x_5^2$ , for some $d\ne 0$ . Since $\sigma _{123}^*(f_3)=s_1s_2s_3f_3$ , the latter is impossible. So $f_3=dx_4^3$ and $s_1=1$ . This gives $\rho (\mathrm {Aut}(X))=C_3$ . However, any $\gamma \in H$ takes the form

    $$ \begin{align*}(x_1,x_2,x_3,x_4,x_5)\mapsto(t_1x_1,t_2x_2,t_3x_3,t_4x_4+t_5x_5,t_6x_4+t_7x_5), \end{align*} $$
    for some $t_j\in k^\times $ . Since $\gamma $ leaves (7) invariant, one finds $H=C_2=\langle \sigma _{2}^3\rangle $ and $\mathrm {Aut}(X)\simeq C_6=\langle \sigma _{2}\rangle $ , where
    $$ \begin{align*}\sigma_2\colon (x_1,x_2,x_3,x_4,x_5)\mapsto (x_2,x_3,x_1,x_4,\zeta_6 x_5). \end{align*} $$

  • – $f_3(x_4,x_5)\equiv 0$ . Then

    $$ \begin{align*}H\simeq C_2^2=\langle\iota,\sigma_{2}^3\rangle, \quad \mathrm{Aut}(X)\simeq C_3\rtimes\mathfrak{D}_4 = \langle \sigma_{2}, \sigma_{7}\rangle, \end{align*} $$
    where
    (10) $$ \begin{align} \iota\colon (x_1,x_2,x_3,x_4,x_5) & \mapsto (x_1,x_2,x_3,-x_4,-x_5),\qquad \end{align} $$
    (11) $$ \begin{align} \sigma_{7} \colon (x_1,x_2,x_3,x_4,x_5)&\mapsto (\zeta^4_6x_2,\zeta^2_6x_1,x_3,\zeta^2_6x_5,\zeta_6 x_4). \end{align} $$

• • If $r=2$ , then

  $$ \begin{align*} q_1&=x_4^2+bx_4x_5+x_5^2, \\ q_2&=x_4^2+\zeta^2_6 bx_4x_5 +\zeta^4_6 x_5^2, \\ q_3&=x_4^2+\zeta^4_6bx_4x_5+\zeta^2_6x_5^2, \end{align*} $$
  $b\ne 0$ and $s_1=s_2=s_3=\pm 1$ . When $b\ne 1$ , $q_1,q_2,q_3$ define six points in ${\mathbb P}^1$ , but when $b=1$ , they define three points. There are subcases:

  • – $f_3(x_4,x_5)=el_1l_2l_3$ , $e\in k^\times $ . Then $\sigma _{123}$ permutes the points defined by $f_3$ in ${\mathbb P}^1$ ; that is,

    $$ \begin{align*}l_1=x_4+dx_5, \quad l_2=x_4+\zeta^2_6dx_5, \quad l_3=x_4+\zeta^4_6 dx_5, \quad d\in k^\times. \end{align*} $$
    In this case, $\sigma _{123}$ takes the form
    $$ \begin{align*}\sigma_{1}\colon (x_1,x_2,x_3,x_4,x_5)\mapsto (x_2,x_3,x_1,x_4,\zeta^2_6 x_5). \end{align*} $$
    One finds that $H=0$ and
    $$ \begin{align*}\mathrm{Aut}(X)\simeq \begin{cases} C_3=\langle\sigma_{1}\rangle & d^6\ne 1,\\ {\mathfrak S}_3=\langle\sigma_{1},\sigma_{3}\rangle, & d^3=1,\\ {\mathfrak S}_3=\langle\sigma_{1},\sigma_{3}'\rangle, & d^3=-1. \end{cases} \end{align*} $$
    $$ \begin{align*}\sigma_{3}\colon (x_1,x_2,x_3,x_4,x_5)\mapsto (\zeta^2_6x_2,\zeta^4_6x_1,x_3,\zeta^4_6x_5,\zeta^2_6 x_4), \end{align*} $$
    $$ \begin{align*}\sigma_{3}'\colon (x_1,x_2,x_3,x_4,x_5)\mapsto (\zeta^2_6x_2,\zeta^4_6x_1,x_3,\zeta_6x_5,\zeta^5_6 x_4). \end{align*} $$

  • – $f_3=l_1l_2^2$ . Then $\sigma _{123}$ fixes two points defined by $f_3$ in ${\mathbb P}^1$ , and $f_3=dx_4^2x_5$ or $dx_4x_5^2$ , for some $d\in k^\times $ . But then (7) cannot be $\sigma _{123}$ -invariant. So this case does not exist.

  • – $f_3=l^3$ . Then $f_3=dx_4^3$ or $dx_5^3$ , for some $d\ne 0$ . One finds

    $$ \begin{align*}H\simeq 0, \quad \mathrm{Aut}(X)\simeq C_3=\langle\sigma_{1}\rangle. \end{align*} $$

  • – $f_3\equiv 0$ . Then $\mathrm {Aut}(X)$ contains the involution $\iota $ from (10). Up to a twist by $\iota $ , we may assume $\sigma _{123}=\sigma _1$ . Note that $\mathrm {Aut}(X)$ also contains $\sigma _3.$ Using the same argument to find H as above, one gets that when $b^2\ne -2$ ,

    $$ \begin{align*}H\simeq C_2=\langle\iota\rangle, \quad \mathrm{Aut}(X) \simeq C_2\times\mathfrak{S}_3 =\langle\iota,\sigma_{1},\sigma_{3}\rangle; \end{align*} $$
    when $b^2=-2$ ,
    $$ \begin{align*}H\simeq Q_8, \quad \mathrm{Aut}(X)\cong\mathrm{GL}_2(\mathbb{F}_3)=\langle \iota,\sigma_{1},\sigma_{3}, \sigma_{9}\rangle, \end{align*} $$
    $$ \begin{align*}\sigma_{9}: (x_1,x_2,x_3,x_4,x_5)\mapsto\left(x_2,\zeta^5_6x_1,\zeta_6 x_3, \frac{\zeta_6 bx_4+x_5}{1-\zeta^2_6},\frac{\zeta_6 x_4+bx_5}{1-\zeta^2_6}\right). \end{align*} $$

1. (b) When $q_1=l_1l, q_2=l_2l$ and $q_3=l_3l$ and $l_1, l_2, l_3$ are coprime. Then $\mathrm {Aut}(X)$ fixes the point in ${\mathbb P}^1_{x_4,x_5}$ defined by l and acts as $C_3$ on the three points defined by $l_1, l_2$ and $l_3$ . This implies that $r=2$ in (9), and that

   $$ \begin{align*}l=x_4, \quad l_1=x_4+x_5,\quad l_2=x_4+\zeta_3 x_5, \quad l_3=x_4+\zeta^{2}_3x_5. \end{align*} $$

   Then either $f_3=l^{\prime 3}$ defines one point and $\sigma _{123}$ fixes the point, or $f_3=l_1'l_2'l_3'$ , defining three distinct points, with $\sigma _{123}$ permuting them (i.e., $f_3=ax_4^3+bx_5^3$ for some $a,b\in k$ ). Since X is 3-nodal, one has $b\ne 0$ and $(a,b)\ne (0,1)$ . From the form of $f_3$ , one sees $\sigma _{123}=\sigma _1$ . Here, $H=0$ since any element in H fixes four points on ${\mathbb P}^1$ , defined by $l,l_1,l_2,l_3$ , and acts trivially on $x_4, x_5$ . Moreover, one can show that no action on ${\mathbb P}^1$ fixes two points defined by l and $l_1$ and swaps those defined by $l_2$ and $l_3$ at the same time. Therefore, $\rho (\mathrm {Aut}(X))=C_3$ and

   $$ \begin{align*}H\simeq 0, \quad \mathrm{Aut}(X)\simeq C_3=\langle\sigma_{1}\rangle. \end{align*} $$

1. (c) $q_1=q_2=q_3=q$ : We may assume that $q=x_4x_5$ . In this case, the exact sequence (8) splits and

   $$ \begin{align*}\quad \mathrm{Aut}(X)\simeq H\times{\mathfrak S}_3, \end{align*} $$
   with the factor ${\mathfrak S}_3$ acting via permutations of $x_1, x_2$ and $x_3$ and trivially on $x_4, x_5$ . Moreover, it is easy to see that H must act faithfully on ${\mathbb P}^1_{x_4,x_5}$ . Since H preserves the pair of points defined by q in ${\mathbb P}^1_{x_4,x_5}$ , it is either cyclic or dihedral. Assume that $H\ne 1$ . Then the structure of $\mathrm {Aut}(X)$ depends on $f_3$ as follows:

• • $f_3=dl^3$ , for some $d\ne 0$ and linear form l in $x_4$ and $x_5$ . Then H has a fixed point (i.e., H is a cyclic group). Moreover, H swaps two points, and thus, $H\simeq C_2$ with $l=x_4+x_5$ and H is generated by swapping coordinates $x_4$ and $x_5.$

• • $f_3=dl_1^2l_2$ , for some $d\ne 0$ and linear forms $l_1$ and $l_2$ defining two distinct points in ${\mathbb P}^1$ . Then H fixes two points defined by $l_1$ and $l_2$ , and swaps two points defined by q. Similarly, we have $H\simeq C_2$ with

  $$ \begin{align*}l_1=x_4-x_5,\quad l_2=x_4+x_5, \end{align*} $$
  where H is generated by swapping $x_4$ and $x_5.$

• • $f_3=dl_1l_2l_3$ , defining three distinct points. There are subcases:

  • – H permutes three points defined by $f_3$ and swaps two points defined by q. Then $H\simeq {\mathfrak S}_3$ , generated by

    $$ \begin{align*} \sigma_8:(x_1,x_2,x_3,x_4,x_5)&\mapsto (x_1,x_2,x_3,\zeta_3 x_4, \zeta^2_3 x_5)\\ \sigma_6:(x_1,x_2,x_3,x_4,x_5)&\mapsto (x_1,x_2,x_3,x_5, x_4), \end{align*} $$
    and
    (12) $$ \begin{align} l_1=x_4+x_5, \quad l_2=\zeta_3 x_4+\zeta_3^2x_5, \quad l_3=\zeta^2_3x_4+\zeta_3 x_5. \end{align} $$

  • – H permutes three points defined by $f_3$ and fixes two points defined by q; thus, $H\simeq C_3$ , and $l_1,l_2,l_3$ are as in (12). However, we know that X admits an additional symmetry swapping two points in q as in the case above.

  • – H fixes the point defined by $l_1$ . Then H swaps two points defined by q and two points defined by $l_2$ and $l_3$ because otherwise, H is trivial. In this case $H\simeq C_2$ , with

    $$ \begin{align*}l_1=x_4+x_5, \quad l_2=ax_4+bx_5, \quad l_3=bx_5+ax_4, \end{align*} $$
    for some $a,b\ne 0$ , and $\left (\frac ab\right )^3\ne 1$ ; H is generated by the involution swapping $x_4$ and $x_5$ .

Proof. Arguing as in Example 2.10, let ${\mathcal X}\to {\mathbb A}^1_k$ be the family given by $f_b$ . For a very general b, ${\mathcal X}_b$ is a 3-nodal cubic described as Case (5) in Proposition 4.1. The special fiber ${\mathcal X}_0$ is a 9-nodal cubic, and the G-action fixes a smooth genus 1 curve on ${\mathcal X}_0$ . From computations in Section 9, there exists an (H1)-obstruction to stable linearizability of the G-action on ${\mathcal X}_0$ . The six additional nodes form two G-orbits with trivial stabilizer. By Proposition 2.9, a very general element in the family ${\mathcal X}$ is not G-stably linearizable.

Proof. Let $\phi \colon \mathrm {Aut}(X)\to \mathfrak {S}_4$ be the homomorphism given by the action on the nodes of X. Since $\mathrm {Aut}(X)$ does not fix nodes, we may assume that there is a $\iota \in \mathrm {Aut}(X)$ such that $\phi (\iota )=(12)(34)$ or $\phi (\iota )=(1234)$ .

Suppose that $\phi (\iota )=(12)(34)$ . Then $\iota $ is given by

$$ \begin{align*}(x_1,x_2,x_3,x_4,x_5)\mapsto(x_2+sx_5,x_1+sx_5,x_3+sx_5,x_4+sx_5,tx_5), \end{align*} $$

for some $s,t\in k$ such that $t\ne 0$ . Considering how $\iota $ acts on (13), we see that $s=0$ or $a_1=a_2=a_3$ . In the former case, we have $b_1=b_2$ and $b_3=b_4$ , which implies $t=1$ , because otherwise, $t=-1$ and $a=a_1=a_2=a_3=0$ , which implies that X is not 4-nodal. Thus, if $(s,t)=(0,1)$ and $\mathrm {im}(\phi )\simeq C_2$ , then we are in the case ( $C_2$ ).

If $a_1=a_2=a_3$ , then after a coordinate change, we may assume that $a_1=a_2=a_3=0$ . In this case, we get

$$ \begin{align*}0=s=a(1-t^3)=b_3-b_4t^2=b_4-b_3t^2=b_2-b_1t^2=b_1-b_2t^2. \end{align*} $$

Since X is 4-nodal, this gives $a=0$ , $b_1=-b_2$ , $b_3=-b_4$ and $t=\pm i$ . Hence, if $\mathrm {im}(\phi )\simeq C_2$ , then we are in the case ( $C_4$ ).

Now, we suppose that $\phi (\iota )=(1234)$ . Then $\iota $ is given by

$$ \begin{align*}(x_1,x_2,x_3,x_4,x_5)\mapsto(x_4+sx_5,x_1+sx_5,x_2+sx_5,x_3+sx_5,tx_5) \end{align*} $$

for some $s,t\in k$ such that $t\ne 0$ . Then

• • $a=2a_2b_4-a_2^3$ , $a_1=2a_2-a_3$ , $b_1=b_2=b_3=b_4$ , or

• • $a_1=a_3$ , $b_1=b_2=b_3=b_4$ , or

• • $a_1=a_2=a_3$ .

In the former case, X is not 4-nodal. If $a_1=a_2=a_3$ , then after a coordinate change, we may assume that $a_1=a_2=a_3=0$ , which gives

$$ \begin{align*}0=s=a(1-t^3)=b_2-b_3t^2=b_3-b_4t^2=b_4-b_1t^2=b_1-b_2t^2, \end{align*} $$

so, after an appropriate scaling of $x_5$ , we see that

• • $a\ne 1$ , $b_1=b_2=b_3=b_4=1$ , or

• • $a=0$ , $b_1=1$ , $b_2=-1$ , $b_3=1$ , $b_4 =-1$ , $t=i$ , or

• • $a=0$ , $b_1=1$ , $b_2=-\zeta _8^2$ , $b_3=-1$ , $b_4=\zeta _8^2$ , $t=\zeta _8$ ,

which implies that we are in cases ( $\mathfrak {S}_4$ ), ( $C_4$ ), ( $C_8$ ), respectively.

If $a_1=a_3$ and $b_1=b_2=b_3=b_4$ , then after a coordinate change, we may assume that $a_1=a_3=0$ . If $a_2=0$ , then we are in the case ( $\mathfrak {S}_4$ ). Finally, if $a_2\ne 0$ , then, scaling $x_5$ , we may further assume that $a_1=1$ , so X is given by

$$ \begin{align*} x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4+ax_5^3+b_1x_5^2\big(x_1+x_2+x_3+x_4\big)+x_5\big(x_1x_3+x_2x_4\big)=0, \end{align*} $$

which gives $\mathrm {Aut}(X)=\langle \sigma _1,\sigma _2,\sigma _3\rangle \cong \mathfrak {D}_4$ , so we are in the case ( $\mathfrak {D}_4$ ).

To proceed, we may assume that $\mathrm {im}(\phi )\not \simeq C_2$ and $\mathrm {im}(\phi )\not \simeq C_4$ . Then, up to a coordinate change, one of the following four cases holds:

• • $\mathrm {im}(\phi )=\langle (12)(34),(14)(23)\rangle \simeq C_2^2$ ,

• • $\mathrm {im}(\phi )=\langle (12)(34),(1234)\rangle \simeq \mathfrak {D}_4$ ,

• • $\mathrm {im}(\phi )=\langle (12)(34),(14)(23),(123)\rangle \simeq \mathfrak {A}_4$ ,

• • $\mathrm {im}(\phi )=\langle (12)(34),(14)(23),(1234),(123)\rangle \simeq \mathfrak {S}_4$ .

Since $\mathrm {im}(\phi )$ contains $(12)(34)$ or $(1234)$ , the cubic X must be given by one of the equations explicitly described above. Using additional symmetries of X, we conclude that we are in one of the cases ( $C_2^2$ ), ( $\mathfrak {D}_4$ ), ( $\mathfrak {S}_4$ ), or the cubic X is given by (13) with

$$ \begin{align*}a\ne 0,\quad a_1=a_2=a_3=0,\quad b_1=b_3=1,\quad b_2=b_4=-1, \end{align*} $$

or

$$ \begin{align*}a_1=1, \quad a_2=\zeta_3,\quad a_3=\zeta_3^2,\quad b_1=b_2=b_3=b_4=0. \end{align*} $$

In the first of the latter two cases, X has eight nodes, and in the last case, the singularities of X are not nodes. This completes the proof of the theorem.

Proof. Let ${\mathcal X}\to {\mathbb A}^1_k$ be the family given by $f_b$ . The generic fiber ${\mathcal X}_b$ is a 4-nodal cubic of the type $(C_2)$ in Theorem 5.1. The special fiber ${\mathcal X}_{9/4}$ is an 8-nodal cubic, with an (H1)-obstruction to stable linearizability of the G-action by Corollary 8.3. The additional four nodes have trivial stabilizer and thus are $BG$ -rational singularities. Applying Proposition 2.9 and Example 2.8, one concludes that a very general member in the family ${\mathcal X}$ is not G-stably linearizable.

Proof. Let ${\mathcal X}\to {\mathbb A}^1_k$ be the family consisting of $X_a$ . The special fiber $cX_0$ is a 10-nodal cubic, with an (H1)-obstruction to stable linearizability of the G-action, from computations in [Reference Cheltsov, Tschinkel and Zhang24]. The additional six nodes have $C_2$ -stabilizers. They are $BG$ -rational singularities, by Example 2.8. Applying Proposition 2.9 and Example 2.8, one concludes that a very general fiber is not G-stably linearizable.

Proof. We may assume that no singular point of X is G-fixed. If X contains a G-stable line that is disjoint from $\Pi $ , then the G-action on $X_{2,2}$ is linearizable, by Lemma 1.1. Conversely, if the G-action on $X_{2,2}$ is linearizable, then it follows from [Reference Hassett and Tschinkel42, Theorem 24] that $X_{2,2}$ contains a G-stable line $\ell $ . And $P\not \in \ell $ because otherwise, the preimage of $\ell $ on X would be a G-fixed singular point. Similarly, we see that $\ell $ must be disjoint from the four lines in $X_{2,2}$ containing P. Then $\ell $ is mapped by $\phi ^{-1}$ to a G-stable line in X that is disjoint from the plane $\Pi $ .

Proof. Since G does not fix any of the nodes, either $G\simeq C_2\times {\mathfrak S}_3$ or $G\simeq C_2\times C_3$ . In both cases, we may assume that G preserves the subset $\{p_1,p_2\}$ and $\{p_3,p_4,p_5\}$ . Then G pointwise fixes the line $l\subset \mathbb {P}^4$ that passes through the points $[1:1:0:0:0]$ and $[0:0:1:1:1]$ . Observe that $\iota (l)=l$ , so that the intersection $l\cap Q$ contains G-fixed points, which implies the assertion.

Proof. We may assume the G action is generated by

$$ \begin{align*} (x_1,\ldots,x_5)& \mapsto(x_4,x_3,x_2,x_1,x_5), \\ (x_1,\ldots,x_5)& \mapsto(\zeta x_1,\zeta^2x_2,\zeta^3x_3,\zeta^4x_4,x_5), \end{align*} $$

where $\zeta =e^{\frac {2\pi i}{5}}$ . Smooth G-invariant quadrics $Q_{a,b}$ are given by

$$ \begin{align*}ax_1x_4-bx_2x_3+x_5^2=0 \end{align*} $$

for $a,b\ne 0$ . Notice that each $Q_{a,b}$ is ${\mathfrak D}_5$ -isomorphic to $Q_{1,1}$ with the same ${\mathfrak D}_5$ -action under a change of variables

$$ \begin{align*}x_1'={\sqrt{a}}x_1,x_2'={\sqrt{b}}x_2,x_3'={\sqrt{b}}x_3,x_4'={\sqrt{a}}x_4,x_5'= x_5. \end{align*} $$

Consider a G-invariant conic

$$ \begin{align*}C=\{x_2=x_3=0\}\cap Q_{1,1} \end{align*} $$

and a G-invariant twisted cubic curve

$$ \begin{align*}R=\{x_5=x_1x_3-x_2^2=x_2x_4-x_3^2=0\}\cap Q_{1,1}. \end{align*} $$

The system of quadric hypersurfaces on ${\mathbb P}^4$ containing both C and R induces a G-equivariant birational map $Q_{1,1}\dashrightarrow {\mathbb P}^3$ ; see, for example, [Reference Araujo, Castravet, Cheltsov, Fujita, Kaloghiros, Martinez-Garcia, Shramov, Süß and Viswanathan3, Section 5.10].

Proof. It is easy to see that $\mathrm {Aut}(Q)$ contains a unique subgroup isomorphic to $C_4\rtimes C_5$ . Thus, we may change coordinates on $\mathbb {P}^4$ as we need and, in particular, assume that Q is given by

$$ \begin{align*}\sum_{i=1}^5x_i^2+i\sum_{1\leqslant i<j\leqslant 5}x_ix_j=0, \end{align*} $$

and that the action of G on Q is given by

$$ \begin{align*} (x_1,x_2,x_3,x_4,x_5)&\mapsto(x_2,x_3,x_4,x_5,x_1),\\ (x_1,x_2,x_3,x_4,x_5)&\mapsto(x_1,x_3,x_5,x_2,x_4). \end{align*} $$

Then Q contains the following smooth quintic elliptic curve:

$$ \begin{align*}\left\{\begin{aligned} &x_1^2+i(x_3x_4+x_2x_5)=0,\\ &x_2^2+i(x_4x_5+x_3x_1)=0,\\ &x_3^2+i(x_5x_1+x_4x_2)=0,\\ &x_4^2+i(x_1x_2+x_5x_3)=0,\\ &x_5^2+i(x_2x_3+x_1x_4)=0. \end{aligned} \right. \end{align*} $$

Blowing up Q along this curve, we obtain the claim; cf. [Reference Cheltsov and Pokora19].

Proof. Enumerating all (conjugacy classes of) subgroups G of ${\mathfrak S}_6$ which do not fix any point among $p_1,\ldots , p_6$ , and computing all G-invariant cubics singular at $p_1,\ldots ,p_6$ , we obtain the list of (families of) 6-nodal cubics whose automorphism groups do not fix any of the nodes. These are the eight families of cubics listed above. Since $\mathrm {Aut}(X)\subset {\mathfrak S}_6$ , one can find the full automorphism groups $\mathrm {Aut}(X)$ .

Proof. Since $\sigma $ does not fix any node, we may assume that

$$ \begin{align*}\sigma=(1,3)(2,5)(4,6) \end{align*} $$

and X is one of the cases in Proposition 7.3. From the diagram (17), we know that

$$ \begin{align*}\mathrm{rk}\,\mathrm{Cl}^{\langle\sigma\rangle}(X)+3=\mathrm{rk}\,\mathrm{Cl}^{\langle\sigma\rangle}(\widetilde X)=\mathrm{rk}\,\mathrm{Cl}^{\langle\sigma\rangle}(\widehat X)=\mathrm{rk}\,\mathrm{Cl}^{\langle\sigma\rangle}(S)+1, \end{align*} $$

where S is a smooth cubic surface contained in a hyperplane $H\subset {\mathbb P}^4$ . By Lefschetz fixed-point theorem, one has [Reference Dolgachev and Iskovskikh36, Section 6]

$$ \begin{align*}\mathrm{rk}\, \mathrm{Cl}^{\langle\sigma\rangle}(S)=\frac12\left(7+\mathrm{Tr}_2\left(\sigma^*\right)\right), \end{align*} $$

$$ \begin{align*}\mathrm{Tr}_2(\sigma^*)=s-2+\sum_i(2-2g_i), \end{align*} $$

where $\mathrm {Tr}_2(\sigma ^*)$ is the trace of $\sigma ^*$ -action on ${\mathrm H}^2(S,{\mathbb C})$ , s is the number of isolated $\sigma $ -fixed points on S, and $g_i$ are the genera of fixed curves. In our case, we compute that the induced $\sigma $ -action on $H\simeq {\mathbb P}^3$ has weights $(1,1,1,-1)$ . The fixed locus $S^\sigma $ consists of one point and a smooth cubic curve. Substituting into the formulae above we obtain

$$ \begin{align*}\mathrm{rk} \,\mathrm{Cl}^{\langle\sigma\rangle}(S)=3, \end{align*} $$

which implies $\mathrm {rk}\, \mathrm {Cl}^{\langle \sigma \rangle }(X)=1. $