#  $G$-Solid rational surfaces

Antoine Pinardin  _Antoine Pinardin_
School of Mathematics, The University of Edinburgh, Edinburgh EH9 3JZ, UK.
<EMAIL_ADDRESS>
###### Abstract.

We classify $G$-solid rational surfaces over the field of complex numbers.

##  1\. Introduction

We are interested in the equivariant birational geometry of rational surfaces
over the field of complex numbers for finite group actions. Let $S$ be a
rational surface, and $G$ be a finite group acting biregularly on $S$. Denote
by $\mathfrak{D}_{G}$ the family of smooth $G$-Del Pezzo surfaces with
invariant Picard group isomorphic to $\mathbb{Z}$, and by $\mathfrak{C}_{G}$
the family of smooth $G$-surfaces with a conic bundle structure and invariant
Picard rank $2$. The $G$-equivariant Minimal Model Program applied to a
resolution of singularities of $S$ implies that $S$ is $G$-birational to a
surface in $\mathfrak{D}_{G}\cup\mathfrak{C}_{G}$. We say that $S$ is
$G$-birationally solid, or simply $G$-solid if it is not $G$-birational to a
surface of $\mathfrak{C}_{G}$. The goal of this paper is to prove the
following result:

######  Theorem 1.1.

Let $S$ be a smooth Del Pezzo surface, and let $G$ a finite subgroup in
$\mathrm{Aut}(S)$ such that $\mathrm{Pic}^{G}(S)\cong\mathbb{Z}$. Then $S$ is
$G$-solid if and only if one of the following cases holds.

  * •

$K_{S}^{2}\leq 3$.

  * •

$K_{S}^{2}=4$ and $G$ does not fix a point on $S$.

  * •

$K_{S}^{2}=5$ and $G$ is not isomorphic to $\mathbb{Z}_{5}$ or $D_{5}$.

  * •

$K_{S}^{2}=6$ and $G$ is not isomorphic to $\mathbb{Z}_{6}$,
$\mathfrak{S}_{3}$, or $D_{6}$.

  * •

$K_{S}^{2}=8$, $S\cong\mathbb{P}^{1}\times\mathbb{P}^{1}$, and either

    * –

$G$ contains a subgroup isomorphic to $\mathfrak{A}_{4}$,

    * –

$G$ is conjugated in $\mathrm{Aut}(S)$ to a subgroup $H$ which contains
$(\frac{1}{y},x)$ and a subgroup of $\\{(ax,by),a,b\in\mathbb{C}^{*}\\}$ of
order greater than $2$, normal in $H$,

    * –

or $G$ is conjugated in $\mathrm{Aut}(S)$ to
$\left<(\frac{1}{y},x),(-x,-y),(-y,x)\right>\cong\mathbb{Z}_{2}^{2}\rtimes\mathbb{Z}_{4}$.

  * •

$K_{S}^{2}=9$, $S\cong\mathbb{P}^{2}$, $G$ does not fix a point on $S$, and
$G$ is not isomorphic to $\mathfrak{S}_{4}$ or $\mathfrak{A}_{4}$.

Notations for the cases $K_{S}^{2}=6$ and $K_{S}^{2}=8$ will be explained in
sections 5 and 4. A complete answer was already given for $K_{S}\leq 4$ by
Segre in [13], Manin in [10], Das Dores-Mauri in [8], and Dolgachev-Iskovskikh
in [7]. We will quickly recall these results in section 2. A partial answer is
given by Wolter in [15] for $K_{S}^{2}=5$, and Sakovics [11] gives a
classification of groups for which $\mathbb{P}^{2}$ is $G$-rigid, which we
will adapt for $G$-solidity. Remark that for $K_{S}^{2}=4$ or $K_{S}^{2}=9$,
we classify $G$-solid surfaces depending on whether or not $G$ fixes a point
on $S$. Dolgachev and Duncan give a classification of groups fixing a point on
Del Pezzo surfaces in [6].

For $S=\mathbb{P}^{2}$, we also prove the following result.

######  Theorem 1.2.

Let $S=\mathbb{P}^{2}$, and $G$ be a finite subgroup of $\mathrm{Aut}(S)$,
isomorphic to $\mathfrak{A}_{4}$ or $\mathfrak{S}_{4}$. Then the only $G$-Mori
fibre spaces $G$-birational to $S$ are $S$ itself with invariant Picard rank
$1$, and the Del Pezzo surface of degree $5$ with a $G$-conic bundle
structure.

In particular, for $G\cong\mathfrak{A}_{4}$ and $G\cong\mathfrak{S}_{4}$,
$\mathbb{P}^{2}$ is not $G$-birational to the Hirzebruch surface
$\mathbb{F}_{n}$, for any $n\in\mathbb{N}$.

A recent motivation for our work is given by Tschinkel, Yank, and Zhang in
[14]. Both $G$-solid surfaces and surfaces which are not $G$-birational to a
Hirzebruch surface are classes of divisors on threefolds which give rise to
Incompressible Divisorial Symbols, a modern tool used in the formalism of
Burnside groups to distinguish birational types of group actions. A recent
example of applications of these techniques by Cheltsov, Tshinkel, and Zhang
can be found in [5].

##  2\. Formalism and existing results

We recall here some results about the factorisation of $G$-equivariant
birational maps on surfaces into elementary links, and recall what is already
achieved on the matter.

######  Definition 2.1.

A $G$-equivariant Sarkisov link, or simply $G$-equivariant link, is a
$G$-equivariant birational map of one of the following types.

  * •

Links of type $\mathrm{I}$. These are commutative diagrams of the form

$\textstyle{X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{P}^{1}}$

where $X\in\mathfrak{D}_{G}$, $X^{\prime}\in\mathfrak{C}_{G}$, and $\sigma$ is
the blow up of an orbit $O$ of $G$. We say that the link is centered at $O$.

  * •

Links of type $\mathrm{II}$. These are commutative diagrams of the form

$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma}$$\scriptstyle{\tau}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi}$$\textstyle{X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi}$$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{S}$

where $X,X^{\prime}\in\mathfrak{D}_{G}$ and $S$ is a point, or
$X,X^{\prime}\in\mathfrak{C}_{G}$ and $\phi,\psi$ are underlying morphisms of
conic bundles. The map $\sigma$ is the blow up of an orbit $O$ of $G$ in $X$
and $\tau$ is a $G$-equivariant contraction of $(-1)$-curves on $Z$. Here too,
the link is said centered at $O$.

  * •

Links of type $\mathrm{III}$. These are inverses of the links of type $I$.

  * •

Links of type $\mathrm{IV}$. They only change the $G$-conic bundle structure
on $X\in\mathfrak{D}_{G}$.

We recall the main theorem about the Sarkisov program for surfaces. For
further details, the reader can refer to [9].

######  Theorem 2.2 ([9]).

Let $S$ be a $G$-rational surface in $\mathfrak{C}_{G}\cup\mathfrak{D}_{G}$,
and $\phi:S\dashrightarrow S^{\prime}$ be a $G$-equivariant birational map.
Then then $\phi$ admits a factorization into finitely many elementary links of
type $\mathrm{I}$-$\mathrm{IV}$.

Moreover, these links are fully classified by the author in the paper. We will
not recall here the full classification, as there is a lot of literature about
it. See for example [9] or [7].

The following result is often called the Segre-Manin theorem.

######  Proposition 2.3.

Let $S$ be a smooth Del Pezzo surface, and $G$ be a finite subgroup of
$\mathrm{Aut}(S)$.

  * •

If $K_{S}^{2}=1$, $S$ is $G$-superrigid.

  * •

IF $K_{S}^{2}=2$ or $K_{S}^{2}=3$, $S$ is $G$-rigid.

In [8], Das Daures and Mauri also classified finite groups $G$ for which a Del
Pezzo surface of degree $2$ or $3$ is $G$-superrigid.

Let $S$ be a Del Pezzo surface of degree $4$, and $G$ be a finite subgroup of
$\mathrm{Aut}(S)$. If $G$ fixes a point which is on the $(-1)$-curves, then
the $G$-invariant Picard rank $\rho^{G}(S)$ is at least $2$. This remark and a
straightforward application of the classification of Sarkisov links given in
[9] imply the following result.

######  Proposition 2.4.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ such that $\rho^{G}(S)=1$, where
$S$ is a Del Pezzo surface of degree $4$. The following conditions are
equivalent.

  * •

$S$ is $G$-birationally rigid,

  * •

$S$ is $G$-solid,

  * •

$G$ does not fix a point on $S$.

##  3\. Del Pezzo surfaces of degree $5$

Up to isomorphism, there is a unique Del Pezzo surface $S$ of degree $5$,
given by the blow-up of $\mathbb{P}^{2}$ in four points in general position.
For a nice description of its group of automorphisms, the reader can refer to
[1]. For what remains to be done about the $G$-solidity of $S$, we will only
need to recall that $\mathrm{Aut}(S)\cong\mathfrak{S}_{5}$, and use the
$G$-equivariant birational geometry of $\mathbb{P}^{2}$.

######  Lemma 3.1 ([7]).

Let $G\in\mathrm{Aut}(S)$. Then $\rho^{G}(S)=1$ if and only if $G$ is
isomorphic to $\mathfrak{S}_{5}$, $F_{5}$, $\mathfrak{S}_{4}$, $D_{5}$, or
$\mathbb{Z}_{5}$.

It is known that if $G$ is isomorphic to $\mathfrak{A}_{5}$ or
$\mathfrak{S}_{5}$, then $S$ is $G$-birationally superrigid. One can refer for
example to [2] or [3]. If $G$ is isomorphic to $F_{5}$, then Wolter proved
that $S$ is $G$-solid in [15] the only $G$-equivariant maps from $S$ are self
maps and maps to $\mathbb{P}^{1}\times\mathbb{P}^{1}$. What remains to study
is the $G$-solidity of $S$ for $G$ isomorphic to $\mathbb{Z}_{5}$ or $D_{5}$.

######  Proposition 3.2.

If $G$ is isomorphic to $\mathbb{Z}_{5}$ or $D_{5}$, then $S$ is not
$G$-solid.

###### Proof.

Consider the matrices $M=\begin{pmatrix}1&0&0\\\ 0&\mu_{5}&0\\\
0&0&\mu_{5}^{-1}\end{pmatrix}\in\mathrm{PGL}_{2}(\mathbb{C})$, where $\mu_{5}$
is a primitive $n$-th root of the unity, and $N=\begin{pmatrix}1&0&0\\\
0&0&1\\\ 0&1&0\end{pmatrix}$. We have $H:=\left<M\right>\cong\mathbb{Z}_{5}$,
and $H^{\prime}:=\left<M,N\right>\cong D_{5}$, under the action of both
groups, the point $(1:1:1)$ have an orbit of length $5$ in general position.
Blowing-up this orbit and contracting the proper transform of the conic
passing through the five points is an $G$-equivariant link from
$\mathbb{P}^{2}$ to a Del Pezzo surface of degree $5$, for
$G\in\\{H,H^{\prime}\\}$. On the other hand, the point
$(1:0:0)\in\mathbb{P}^{2}$ is fixed under the action of $G$ and $G^{\prime}$.
Hence, for $G\in\\{H,H^{\prime}\\}$, we can blow it up and get an
$G$-equivariant link to a $G$-surface of invariant Picard rank two with a
$G$-conic bundle structure. Since $\mathbb{Z}_{5}$ and $D_{5}$ are unique in
$\mathfrak{S}_{5}$ up to conjugacy, we conclude that $S$ is not
$\mathbb{Z}_{5}$-solid, and not $D_{5}$-solid either. ∎

##  4\. Del Pezzo surfaces of degree $8$

We choose to study here the case of Del Pezzo sufaces of degree $8$, as the
results will be useful for the degree $6$ case. There are two smooth Del Pezzo
surfaces of degree $8$. The projective plane blown-up in one point, whose
invariant Picard rank is $2$, and $\mathbb{P}^{1}\times\mathbb{P}^{1}$. The
focus of this section is the latter case, the surface
$S=\mathbb{P}^{1}\times\mathbb{P}^{1}$. For simplicity, a point of
$\mathbb{P}^{1}$ of the form $(1:a)$ will be simply denoted by $a$, and the
point $(0:1)$ by $\infty$. Recall that the automorphism group of
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ is isomorphic to
$(\mathrm{PGL}_{2}(\mathbb{C})\times\mathrm{PGL}_{2}(\mathbb{C}))\rtimes\mathbb{Z}_{2}$,
where $\mathbb{Z}_{2}$ acts on
$\mathrm{PGL}_{2}(\mathbb{C})\times\mathrm{PGL}_{2}(\mathbb{C})$ by permuting
the factors. Thus, we will often denote an element of $\mathrm{Aut}(S)$ by a
triple $(M,N,\sigma)$, where $M,N\in\mathrm{PGL}_{2}(\mathbb{C})$, and
$\sigma\in\mathbb{Z}_{2}$.

######  Remark 4.1.

In the following cases, $S$ is not $G$-solid. We will call
$\mathbb{P}^{1}$-fibres a fibre of one of the the canonical projections from
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ to $\mathbb{P}^{1}$. If it is the
projection to the left (resp. right) factor, we will call it a vertical (resp.
vertical) $\mathbb{P}^{1}$-fibre.

  * •

If a point of $S$ has an orbit of length $2$ under the action of $G$, and if
the points of the orbit are not on the same $\mathbb{P}^{1}$-fibre, then we
can $G$-equivariantly blow-up these points and get a $G$-birational map to a
$G$-surface isomorphic to a Del Pezzo surface of degree $6$ with
$\rho^{G}(X)=2$, with a $G$-conic bundle structure.

  * •

If there is a pair of fixed points, then the blow-up of one of these points
followed by the contraction of the proper transforms of the two fibres of the
canonical projections to both $\mathbb{P}^{1}$ factors give a $G$-equivariant
birational map to $\mathbb{P}^{2}$, and subsequently blowing-up the image of
the other fixed point gives a $G$-equivariant birational map to the
$G$-surface $\mathbb{F}_{1}$ of invariant Picard rank $2$.

  * •

If there is a fixed point $P$ and an orbit of four points in general position,
we get a $G$-equivariant Sarkisov link to $\mathbb{P}^{2}$ as above, which can
be followed by the blow-up of the orbit of length $4$, leading to a
$G$-surface isomorphic to a Del Pezzo surface of degree $5$ with
$\rho^{G}(X)=2$, with a $G$-conic bundle structure.

Moreover, the classification of Sarkisov links given in [9] implies that, if
there exists a $G$-equivariant birational map $\phi:S\dashrightarrow
S^{\prime}$, where $\rho^{G}(S)=1$ and $\rho^{G}(S^{\prime})=2$, one of the
three possibilities above must appear in any decomposition of $\phi$ into
Sarkisov links. In particular, for $S$ to be $G$-solid, $G$ must be conjugated
in $\mathrm{Bir}(S)$ to a subgroup $G^{\prime}$ which has either a fixed point
or an orbit of length $2$.

###  4.1. Subgroups of $\mathrm{Aut}(S)$ in the normalizer of a maximal torus

We can embed $(\mathbb{C}^{*})^{2}$ as a maximal dense torus $\mathfrak{T}$ in
$S$ by the map $(a,b)\mapsto(1:a)\times(1:b)$. The action of
$(\mathbb{C}^{*})^{2}$ on itself by translation extends to a faithful action
on the whole variety $S$, by identifying $(\mathbb{C}^{*})^{2}$ to the
subgroup of $\mathrm{Aut}(S)$ whose elements are the maps
$(x,y)\mapsto(ax,by)$, for some $a,b\in\mathbb{C}^{*}$, or equivalently, with
the natural notation for elements of
$\mathrm{PGL}_{2}(\mathbb{C})^{2}\rtimes\mathbb{Z}_{2}$, the triples
$(M,N,0)$, with $M$ and $N$ diagonal matrices. We will denote this subgroup as
$\mathbb{T}$.

######  Remark 4.2.

We can simplify the notations for the elements of the normalizer
$N(\mathbb{T})$ of $\mathbb{T}$ in $\mathrm{Aut}(S)$. The normalizer of the
subgroup of diagonal matrices in $\mathrm{PGL}_{2}(\mathbb{C})$ is the
subgroup of matrices $M$ which are either diagonal or of the form
$\begin{pmatrix}0&1\\\ a&0\end{pmatrix}$. On the torus $\mathfrak{T}$, such
maps correspond to those of the form

$\displaystyle\phi:$
$\displaystyle(\mathbb{C}^{*})^{2}\rightarrow(\mathbb{C}^{*})^{2}$ (4.3)
$\displaystyle(x,y)$ $\displaystyle\mapsto\tau((\alpha f(x),\beta g(y)))$

where $\alpha,\beta\in\mathbb{C}^{*}$,
$f,g:\mathbb{C}^{*}\rightarrow\mathbb{C}^{*}$ are either the identity function
or $z\mapsto\frac{1}{z}$, and $\tau$ is either the identity of
$(\mathbb{C}^{*})^{2}$, or swaps the coordinates. When there is no ambiguity,
we will simply write $\phi=\tau((\alpha f(x),\beta g(y))$. Since the
automorphisms of the torus extend uniquely to the whole variety, we will often
use this notation for an automorphism of $\mathbb{P}^{1}\times\mathbb{P}^{1}$
which is in $N(\mathbb{T})$.

######  Example 4.4.

The automorphism $\left(\begin{pmatrix}0&1\\\
a&0\end{pmatrix},\begin{pmatrix}1&0\\\ 0&b\end{pmatrix},1\right)\in
N(\mathbb{T})$ will be denoted by $(ay,\frac{b}{x})$.

######  Lemma 4.5.

There is an exact sequence

(4.6) $\displaystyle 1\rightarrow\mathbb{T}\rightarrow
N(\mathbb{T})\stackrel{{\scriptstyle\omega}}{{\rightarrow}}D_{4}\rightarrow
1.$

###### Proof.

First, remark that $N(\mathbb{T})$ leaves the torus invariant. Indeed, let
$g\in N(\mathbb{T})$, and $P$ be a point of the torus $\mathbb{T}$.

$g(\mathbb{T}\cdot P)=g\mathbb{T}g^{-1}\cdot g(P)=\mathbb{T}\cdot g(P).$

In particular, the orbit of $g(P)$ is dense under the action of $\mathbb{T}$,
so that $g(P)$ is in the torus. The complement of the torus $\mathfrak{T}$ in
$S$ is the square
$C:=\pi_{1}^{-1}(0)\cup\pi_{1}^{-1}(\infty)\cup\pi_{2}^{-1}(0)\cup\pi_{2}^{-1}(\infty)$,
where $\pi_{1}$ and $\pi_{2}$ are the natural projections from
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ to the left and right factor,
respectively. Intersection number being preserved by automorphisms, any
element of $N(\mathbb{T})$ induces a symmetry of the square formed by $C$,
thus giving a natural surjective map $w:N(\mathbb{T})\rightarrow D_{4}.$ Its
kernel is $\mathbb{T}$, hence the exact sequence. ∎

We can naturally point out two elements which generate a group $H$ sent
isomorphically to $D_{4}$ by $\omega$. Let $r=(\frac{1}{y},x)$, and $s=(y,x)$.
These elements act the following way on the vertices of the square $C$.

$\displaystyle r:$
$\displaystyle(0,0)\mapsto(0,\infty)\mapsto(\infty,\infty)\mapsto(\infty,0)\mapsto(0,0),$
$\displaystyle s:$ $\displaystyle(0,\infty)\mapsto(\infty,0)\text{ and fixes
}(0,0)\text{ and }(\infty,\infty).$

######  Remark 4.7.

Any element $g$ of $N(\mathbb{T})$ mapped on the same image as $r$ by $\omega$
is actually equal to $r$ up to conjugation by an element of $\mathbb{T}$.
Indeed, let $g=(\frac{a}{y},bx)$, for some $a,b\in\mathbb{C}^{*}$, and
$t=(kx,ly)$, with $l^{2}=\frac{1}{ab}$ and $k=bl$. We get that $tgt^{-1}=r.$

######  Lemma 4.8.

A group $G\subset\mathrm{Aut}(S)$ leaves the torus $\mathfrak{T}$ invariant if
and only if $G$ is in the normalizer $N(\mathbb{T})$ of $\mathbb{T}$ in
$\mathrm{Aut}(S)$.

###### Proof.

We already proved that $N(\mathbb{T})$ left the torus invariant. Conversely,
assume that $g(\mathfrak{T})=\mathfrak{T}$, for some $g\in\mathrm{Aut}(S)$.
Then, there exists $h\in\left<r,s\right>\cong D_{4}$ such that
$\omega(gh)=\mathrm{id}$. Let $r=(\frac{1}{y},x)$ and remark that, for any
$t=(ax,by)\in\mathbb{T}$, $rtr^{-1}=(b^{-1}x,ay)\in\mathbb{T}$ and
$sts^{-1}=(bx,ay)\in\mathbb{T}$, so that $r,s\in N(\mathbb{T})$. Hence $h$,
and then $g$, are in the normalizer of $\mathbb{T}$. ∎

Let $G$ be a subgroup of $N(\mathbb{T})$. The restriction of $w$ to $G$
induces the exact sequence

$1\rightarrow T\rightarrow G\rightarrow W\rightarrow 1,$

where $W\subset D_{4}$ is the image of $G$ by $\omega$ in $\mathbb{W}$.

######  Remark 4.9.

We will often use the following facts.

  * •

The exact sequence $1\rightarrow T\rightarrow
G\stackrel{{\scriptstyle\omega}}{{\rightarrow}}W\rightarrow 1$ implies that
$G$ is generated by $T$ and elements mapped onto generators of $W$.

  * •

If $S$ is of $G$-invariant Picard rank $1$, $W$ must contain elements of the
form $tr$ or $ts$, for some $t\in\mathbb{T}$. Indeed, if neither $r$ nor $s$
belong to $W$, then $G$ leaves invariant the divisors $\\{(x,0)\\}$ and
$\\{(0,y)\\}$, up to linear equivalence. But together, they generate the whole
module $\mathrm{Pic}(S)$, isomorphic to $\mathbb{Z}^{2}$.

The following result from [4] is mentioned without proof by the authors, so we
will prove it here.

######  Lemma 4.10.

If $G$ is not conjugated in $N(\mathbb{T})$ to a group containing $r$, then
$S$ is not $G$-solid.

###### Proof.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ such that $\rho^{G}(S)=1$. Consider
the points $(0,0),(0,\infty),(\infty,\infty),(\infty,0)$, vertices of the
square $C$ formed by the complement of $\mathbb{T}$. These points being
permuted by the action of $G$, their orbits are of length at most $4$. But $W$
being a subgroup of $D_{4}$, they cannot be of length $3$. So if one of these
points is fixed, another one must be fixed as well. Hence, according to Remark
4.1, the only possibility for $S$ not to be $G$-solid is that these four
points lie on the same orbit. As remarked in 4.9, if $W$ does not contain
$\omega(s)$, then it contains $\omega(r)$, so that
$W=\left<\omega(r)\right>\cong\mathbb{Z}_{4}$. The maximal strict subgroup $H$
of $D_{4}$ containing $\omega(s)$ and not $\omega(r)$ is isomorphic to
$\mathbb{Z}_{2}^{2}$, generated by $\omega(s)$ and $\omega(r^{2})$. But under
the action of $H$, the vertices of $C$ only have orbits of length $2$. We
conclude that if the four vertices are on the same orbit, $W$ is either equal
to $\left<\omega(r)\right>\cong\mathbb{Z}_{4}$, or to
$\left<\omega(r),\omega(s)\right>\cong D_{4}$. Remark 4.7 allows us to
conclude that $r\in G$, up to conjugation by an element of $N(\mathbb{T})$. ∎

The fact below will allow us to take out many cases for studying the
$G$-solidity of $S$.

Until further notice, $G$ will denote a subgroup of $N(\mathbb{T})$ such that
$\rho^{G}(S)=1$, containing $r=(\frac{1}{y},x)$.

######  Remark 4.11.

If $T$ is of order greater or equal to $5$, then $S$ is $G$-solid. Indeed,
points of the torus have an orbit of length at least $|T|$, and links centered
at an orbit of length greater than five are links of type $\mathrm{II}$
leading back to $S$ and sending $G$ to a conjugated group in
$\mathrm{Aut}(S)$. Orbits of the points in the complement of the torus are not
in general position. Moreover, the toric part $T$ of $G$ cannot be isomorphic
to $\mathbb{Z}_{3}$ or $\mathbb{Z}_{4}$. Indeed, let $(ax,by)\in T$. Since $T$
is normal in $G$, $gtg^{-1}=(\frac{1}{b}x,ay)$ must be in $T$. But one can
immediately check that this is not the case for any of the possible generators
of $T$, when $T$ is isomorphic to $\mathbb{Z}_{3}$, $\mathbb{Z}_{4}$. Hence,
we are only left with the cases where $T$ is isomorphic to the trivial group,
$\mathbb{Z}_{2}$, $\mathbb{Z}_{2}^{2}$, or $\mathbb{Z}_{5}$.

######  Proposition 4.12.

If $T$ is trivial, then $S$ is not $G$-solid. The options for $G$ are the
following.

  * •

A group isomorphic to $\mathbb{Z}_{4}$, generated by $r$.

  * •

A group isomorphic to $D_{4}$, generated by $r$ and $(y,x)$.

  * •

A group isomorphic to $D_{4}$, generated by $r$ and $(-y,-x)$.

###### Proof.

Recall that we have the exact sequence

$1\rightarrow T\rightarrow
G\stackrel{{\scriptstyle\omega}}{{\rightarrow}}W\rightarrow 1,$

where $W$ is a subgroup of $D_{4}$. Since $T$ is trivial, $G$ is isomorphic to
$W$. Since $G$ contains $r=(\frac{1}{y},x)$, generating a subgroup isomorphic
to $\mathbb{Z}_{4}$, either $G\cong\mathbb{Z}_{4}$, or $G\cong D_{4}$. In the
first case, $G$ is generated by $r$, and has two fixed points in general
position: $(1,1)$ and $(-1,-1)$.

In the second case, $G$ is isomorphic to $D_{4}$. Hence, $G$ is generated by
$r$ and an element $h\in G$ which fixes the points $(0,0)$ and
$(\infty,\infty)$, and swaps the points $(0,\infty)$ and $(\infty,0)$. In
other words, $\omega(h)=\omega(s)$, where $s=(x,y)$, so that $h=ts=(ay,bx)$,
for some $a,b\in\mathbb{C}^{*}$. But the structure of $D_{4}$ implies that
$(y,\frac{1}{x})=r^{-1}=hrh=(a^{2}y,\frac{1}{x})$. Hence, $a^{2}=1$. Finally,
since $h$ is of order $2$, we get $ab=1$. So the only two options for $h$ are
$h=(y,x)$, or $h=(-y,-x)$.

If $G$ is generated by $(\frac{1}{y},x)$ and $(y,x)$, the points $(1,1)$ and
$(-1,-1)$ are fixed by the action. If $G$ is generated by $(\frac{1}{y},x)$
and $(-y,-x)$, the points $(1,1)$ and $(-1,-1)$ form an orbit of length $2$.
By Remark 4.1, $S$ is not $G$-solid in any of these two cases. ∎

######  Proposition 4.13.

If $T\cong\mathbb{Z}_{2}$, there are the following options and only them:

  * •

$G=\left<t,r\right>\cong\mathbb{Z}_{4}\times\mathbb{Z}_{2}$, with $t=(-x,-y)$.
In this case, $S$ is not $G$-solid.

  * •

$G=\left<t,r,s\right>\cong\mathbb{Z}_{2}\times D_{4}$, with $t$ as above. The
surface $S$ is not $G$-solid in this case either.

  * •

$G=\left<t,r,h\right>\cong\mathbb{Z}_{2}^{2}\rtimes\mathbb{Z}_{4}$, with $t$
and $g$ as above, and $h=(-y,x)$. In this case, $S$ is $G$-solid.

###### Proof.

Here, the toric part $T$ of $G$ is generated by an element $t$ of order $2$,
that is to say, either $t=(-x,-y)$, $t=(-x,y)$, or $t=(y,-x)$. But since $T$
is normal in $G$, $t$ must commute with $r$, so that $t=(-x,-y)$. If
$W\cong\mathbb{Z}_{4}$, $G$ is generated by $r$ and $t$, and isomorphic to
$\mathbb{Z}_{4}\times\mathbb{Z}_{2}$. Then $\\{(1,1),(-1,-1)\\}$ is an orbit
of length $2$ in general position, and $S$ is not $G$-solid.

Now assume that $W$ is isomorphic to $D_{4}$. Then there is an element $h\in
G$ such that $\omega(h)=\omega(s).$ We deduce that $h=(ay,bx)$, for some
$a,b\in\mathbb{C}^{*}$. If $\mathrm{o}(h)=2$, then $a=b^{-1}$. Using classical
relations in the group $D_{4}$, we get $\omega(hrh^{-1})=\omega(r)^{-1}$. So
$\tau hrh^{-1}=r^{-1}$, for some $\tau\in T=G\cap\mathbb{T}$. But $\tau
hrh^{-1}=\tau\cdot(a^{2}y,\frac{1}{x})$, and $r^{-1}=(y,\frac{1}{x})$, so
$\tau=\mathrm{Id}$, and $a=\pm 1$. Hence, $h=(y,x)$ or $h=(-y,-x)$. But these
two elements only differ by a composition with $t$, so we may assume that
$h=s=(y,x)$. We have $G=\left<t,r,s\right>\cong\mathbb{Z}_{2}\times D_{4}$. In
this case, there is an orbit of length $2$ in general position :
$\\{(1,1),(-1,-1)\\}$. Hence, $S$ is not $G$-solid.

The only other possibility is $\mathrm{o}(h)=4$. Since $h^{2}=(abx,aby)$, this
implies that $a^{2}b^{2}=1$ and $ab\neq 1$, so that $ab=-1$ and
$h=(ax,-\frac{y}{a})$. Using again the fact that $r^{-1}=\tau
hrh^{-1}=\tau\cdot(a^{2}y,\frac{1}{x})$, for some $\tau\in T$, we deduce that
$h=(-y,x)$ or $h=(y,-x)$. Since these two possibilities are inverses of each
other, we may assume that $h=(-y,x)$. The group $G$ is generated by $r$ and
$h$, and is isomorphic to $G\cong\mathbb{Z}_{2}^{2}\rtimes\mathbb{Z}_{4}$. It
has no fixed points, no orbit of length $2$ or $4$ in general position. Hence,
$S$ is $G$-solid. ∎

######  Proposition 4.14.

If $T\cong\mathbb{Z}_{2}^{2}$, then $S$ is $G$-solid, and $G$ is conjugated in
$N(\mathbb{T})$ to one of the following groups :

  * •

$G\cong\mathbb{Z}_{2}^{2}\rtimes\mathbb{Z}_{4}$, generated by $r$ and
$t=(-x,y)$,

  * •

$G\cong\mathbb{Z}_{2}^{2}\wr\mathbb{Z}_{2}$, generated by $r$, $t=(-x,y)$, and
$s=(y,x)$,

  * •

$G\cong\mathbb{Z}_{2}^{3}\rtimes\mathbb{Z}_{4}$, generated by $r$, $t=(-x,y)$,
and $h=(iy,ix)$.111This semidirect product is the group with GAP ID (32,6).

###### Proof.

Assume $W\cong\mathbb{Z}_{4}$. Since $T\subset(\mathbb{C}^{*})^{2}$ is
isomorphic to $\mathbb{Z}_{2}^{2}$, it is generated by $t_{1}=(-x,y)$ and
$t_{2}=(x,-y)$. We have $rt_{1}r^{-1}=t_{2}$, so
$G=\left<r,t_{1}\right>\cong\mathbb{Z}_{2}^{2}\rtimes\mathbb{Z}_{4}$.

Now assume that $W\cong D_{4}$, so that there exists an element $h\in G$ such
that $\omega(h)=\omega(s)$. If $\mathrm{o}(h)=2$, we get $h=(ay,\frac{x}{a})$.
But $r^{-1}=\tau hrh^{-1}=\tau\cdot(a^{2}y,\frac{1}{x})$, so
$\tau=\mathrm{id}$ and $a=\pm 1$, or $\tau=t_{1}=(-x,y)$ and $a=\pm i$. Up to
composition by an element of $T$, we get $h=s=(y,x)$ or $h=(-ix,iy)$. In the
first case, we have that $G=\left<t_{1},t,h\right>$ is isomorphic to
$\mathrm{SmallGroup}(32,27)$, and in the second case to
$\mathrm{SmallGroup}(32,6)$.

If $\mathrm{o}(h)=4$, we get $h=(ay,-\frac{x}{a})$, which only differs from
the previous case by a composition with an element of $T$, so $G$ is still in
one of the two above options. There is no other possibility for $h$.

In all those cases, there is no fixed point, and no orbit of length two or
four in general position, so that $S$ is $G$-solid, by Remark 4.1. ∎

The only remaining possibility is $T\cong\mathbb{Z}_{5}$.

######  Proposition 4.15.

If $T\cong\mathbb{Z}_{5}$, $G$ is isomorphic to the Frobenius group
$F_{5}\cong\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4}$.

###### Proof.

Assume $T\cong\mathbb{Z}_{5}$. It is generated by an element of the form
$t=(\mu,\mu^{l})$ or $t=(\mu^{k},\mu)$. Since
$r(\mu^{k},\mu)^{-1}r^{-1}=(\mu,\mu^{-k})$, we may assume that
$t=(\mu,\mu^{l})$. But then $rtr^{-1}=(\mu^{-l},\mu)$, and this element is in
$\left<t\right>$ if and only if $l=2$ or $l=3$. Assume that there is an
element $h\in G$ such that $\omega(h)=s\in D_{4}$. Such element is of the form
$(ay,bx)$. But $hth^{-1}=(\mu^{l},\mu)$, which is not in $T$. Since this
subgroup is normal in $G$, we get a contradiction. Hence, $G$ is generated by
$(\mu,\mu^{l})$ and $r$, with $l=2$ or $l=3$. In both cases, $G$ is isomorphic
to the Frobenius group $F_{5}\cong\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4}$. ∎

The $G$-solidity of $S$ in this case is due to [15]. The focus of this paper
being the Del Pezzo surface of degree $5$, we will simply restate it for a
focus on $\mathbb{P}^{1}\times\mathbb{P}^{1}$.

######  Proposition 4.16.

If $T\cong\mathbb{Z}_{5}$, so that $G\cong F_{5}$, then $S$ is $G$-solid.

###### Proof.

The orbits of the points in the complement of the torus $\mathbb{T}$ are not
in general position. On the other hand, points of the torus have orbits of
length at least $5$, so that the only possible $G$-Sarkisov link which is not
a birational involution leads to a Del Pezzo surface of degree $5$ with
invariant Picard rank $1$. This surface is $G$-solid, according to [15]. ∎

###  4.2. Subgroups which do not normalize any maximal torus

Let us now study the $G$-solidity of $S=\mathbb{P}^{1}\times\mathbb{P}^{1}$
for groups which are not in the normalizer of any maximal torus. Let $G$ be a
subgroup of $\mathrm{Aut}(S)$ such that $\rho^{G}(S)=1$. There is an exact
sequence of the form

(4.17) $\displaystyle 1\rightarrow H\times_{D}H\rightarrow
G\rightarrow\mathbb{Z}_{2}\rightarrow 1,$

where $H\times_{D}H$ stands for the fibre product of a subgroup $H$ of
$\mathrm{PGL}_{2}(\mathbb{C})$ with itself, over a subgroup $D$ which is
normal in $H$. Moreover, $H$ is isomorphic to
$\mathbb{Z}_{n},D_{n},\mathfrak{A}_{4},\mathfrak{S}_{4},$ or
$\mathfrak{A}_{5}$. We will briefly recall the formalism of fibre products and
justify the existence of the above exact sequence. Although already described
for example in [12], we will need for our problem a deeper study of the fibre
product $H\times_{D}H$.

######  Definition 4.18.

Let $\phi:G\rightarrow D$ and $\psi:H\rightarrow D$ be surjective group
morphisms. The fibre product of $G$ and $H$ over $D$, according to $\phi$ and
$\psi$ is the subgroup of $G\times H$ consisting of pairs $(g,h)$ such that
$\phi(g)=\psi(h)$. It is denoted by $G\times_{D,\phi,\psi}H$.

We will often omit the morphisms in the notation and simply denote the fibre
product as $G\times_{D}H$.

######  Lemma 4.19 (Goursat’s Lemma).

Let $G$ and $H$ be groups. The subgroups of $G\times H$ are exactly the fibre
products of the form $G^{\prime}\times_{D}H^{\prime}$.

Recall that
$\mathrm{Aut}(\mathbb{P}^{1}\times\mathbb{P}^{1})\simeq(\mathrm{PGL}_{2}(\mathbb{C})\times\mathrm{PGL}_{2}(\mathbb{C}))\rtimes\mathbb{Z}_{2}$.
Consider the group homomorphism

$\displaystyle\delta:\mathrm{Aut}(S)$ $\displaystyle\rightarrow\mathbb{Z}_{2}$
$\displaystyle(M,N,i)$ $\displaystyle\mapsto i.$

Geometrically, an element of $\mathrm{Aut}(S)$ is mapped to $1$ if and only if
it swaps the rulings of $\mathbb{P}^{1}\times\mathbb{P}^{1}$.

######  Lemma 4.20.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ such that $\rho^{G}(S)=1$, and
$\delta_{G}$ be the restriction of $\delta$ to $G$. Up to conjugation of $G$
in $\mathrm{Aut}(S)$,
$\ker(\delta_{G})=(H\times_{D}H)\times\\{e\\}\subset\mathrm{PGL}_{2}^{2}(\mathbb{C})\rtimes\mathbb{Z}_{2}$,
for some subgroup $H$ of $\mathrm{PGL}_{2}(\mathbb{C})$.

###### Proof.

Call $\pi_{1}$ and $\pi_{2}$ the canonical projections of
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ onto the first and second factor,
respectively. Since $\mathrm{Pic}(S)\cong\mathbb{Z}^{2}$ is generated by a
fibre of $\pi_{1}$ and a fibre of $\pi_{2}$, $\rho^{G}(S)=1$ implies that
there is an element of $G$ which swaps the fibres of $\pi_{1}$ and $\pi_{2}$.
Hence $\delta_{G}:G\rightarrow\mathbb{Z}_{2}$ is surjective. Its kernel is a
subgroup of
$\mathrm{PGL}_{2}(\mathbb{C})\times\mathrm{PGL}_{2}(\mathbb{C})\times\\{e\\}\subset(\mathrm{PGL}_{2}(\mathbb{C})\times\mathrm{PGL}_{2}(\mathbb{C}))\rtimes\mathbb{Z}_{2}$.
Applying Goursat’s Lemma, it is of the form $H\times_{D,\phi,\psi}H^{\prime}$,
where $H$ and $H^{\prime}$ are subgroups of $\mathrm{PGL}_{2}(\mathbb{C})$.
Let $(h,h^{\prime},0)\in H\times_{D,\phi,\psi}H^{\prime}$, with the natural
notation for elements of
$(\mathrm{PGL}_{2}(\mathbb{C})\times\mathrm{PGL}_{2}(\mathbb{C}))\rtimes\mathbb{Z}_{2}$,
and denote $g=(N,N^{\prime},1)$, with some elements
$N,N^{\prime}\in\mathrm{PGL}_{2}(\mathbb{C})$. Then
$g(h,h^{\prime},0)g^{-1}=(Nh^{\prime}N^{-1},N^{\prime}hN^{\prime-1},0)$, so
that, $H\times_{D,\phi,\psi}H^{\prime}$ being normal in $G$,
$H\times_{D,\phi,\psi}H^{\prime}=gH\times_{D,\phi,\psi}H^{\prime}g^{-1}=(N,N^{\prime},0)H\times_{D,\phi,\psi}H^{\prime}(N,N^{\prime},0)^{-1}$.
Denoting $\alpha=(I,N^{\prime},0)$, we deduce that $\alpha
H\times_{D,\phi,\psi}H^{\prime}\alpha^{-1}=H\times_{D,\phi,\xi}H$, where

$\displaystyle\xi:H$ $\displaystyle\rightarrow D$ $\displaystyle h$
$\displaystyle\mapsto\psi(N^{\prime-1}hN^{\prime}),$

and we have $\ker(\delta_{\alpha
G\alpha^{-1}})=H\times_{D,\phi,\xi}H\times\\{e\\}$. ∎

The exact sequence (4.17) immediately follows from Lemma 4.20. The
classification of subgroups of $\mathrm{PGL}_{2}(\mathbb{C})$ implies that $H$
is isomorphic to $\mathbb{Z}_{n}$, $D_{n}$, $\mathfrak{A}_{4}$,
$\mathfrak{S}_{4}$, or $\mathfrak{A}_{5}$. First, we can take off the first
two possibilities.

######  Lemma 4.21.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ which is not in the normalizer of a
maximal torus, and such that $\rho^{G}(S)=1$. With the notations of (4.17),
$H$ cannot be isomorphic to $\mathbb{Z}_{n}$ or $D_{n}$.

###### Proof.

Assume that $H$ is isomorphic to $\mathbb{Z}_{n}$ or $D_{n}$. Then there are
points $P_{1}$ and $P_{2}$ of $\mathbb{P}^{1}$ which are either fixed by $H$
or forming an orbit of length $2$. Consider the divisors
$L_{1}=\\{(P_{1},y),y\in\mathbb{P}^{1}\\},$ and
$L_{2}=\\{(P_{2},y),y\in\mathbb{P}^{1}\\}$. The divisor $L_{1}+L_{2}$ is
invariant by $H\times_{D}H$. Let $g\in G$ such that $G$ is generated by
$H\times_{D}H$ and $g$. This elements swaps the rulings of
$\mathbb{P}^{1}\times\mathbb{P}^{1}$, so that the divisor
$D=L_{1}+L_{2}+g(L_{1})+g(L_{2})$ forms a square. To prove that $G$ is in the
normalizer of a maximal torus, we are going to show that it leaves $D$
invariant. Let $\alpha\in G$. Since $H\times_{D}H$ is normal in $G$ and
$g^{2}\in H\times_{D}H$, we can write $\alpha=hg^{i}=g^{i}h^{\prime}$, for
some $h,h^{\prime}\in H\times_{D}H$ and $i\in\\{0,1\\}$. We get:

$\displaystyle\alpha(D)$
$\displaystyle=g^{i}h(L_{1}+L_{2})+g^{i}hg(L_{1}+L_{2})$
$\displaystyle=g^{i}(L_{1})+g^{i}(L_{2})+g^{i+1}h^{\prime}(L_{1}+L_{2})$
$\displaystyle=g^{i}(L_{1})+g^{i}(L_{2})+g^{i+1}(L_{1})+g^{i+1}(L_{2})$
$\displaystyle=D.$

We conclude using Corollary 4.8. ∎

The following group theoretic result is about the fibre product of a group
with itself. Roughly speaking, if a group $G$ splits in a nice way, then a
fibre product of the form $G\times_{D,\phi,\psi}G$ always admits a subgroup
isomorphic to $G$. It will allow us to give a simple characterization of
subgroups of $N(\mathbb{T})$.

######  Lemma 4.22.

Let a group $D$ act on a group $N$, and $G=N\rtimes D$. Let
$\phi,\psi:G\rightarrow D$ be morphisms such that
$\ker(\phi)=\ker(\psi)=N\times\\{e\\}$. Then $G\times_{D,\phi,\psi}G$ has a
subgroup isomorphic to $G$.

###### Proof.

The subgroups $N\times\\{e\\}=\ker(\psi)$ and $\\{e\\}\times D$ having trivial
intersection in $G$ means that the restriction of $\psi$ to $\\{e\\}\times D$
is an isomorphism. So we can define the set
$\tilde{D}=\\{(d,\psi^{-1}\circ\phi(d)),d\in\\{e\\}\times D\\}$. It is a
subgroup of $G\times_{D,\phi,\psi}G$, isomorphic to $D$. Denote $\tilde{N}$
the subgroup $\ker(\phi)\times\\{\mathrm{id}\\}$ of $G\times G$. It is a
subgroup of $G\times_{D,\phi,\psi}G$, isomorphic to $\ker(\phi)$. Consider the
projection $\pi$ from $G\times G$ to the first factor $G$. Its restriction
$\pi^{\prime}$ to $\tilde{N}\tilde{D}$ is surjective by construction of
$\tilde{N}$ and $\tilde{D}$. If $\pi^{\prime}(g,g^{\prime})=e$, then $g=e$.
But since $\tilde{N}$ and $\tilde{D}$ have trivial intersection, $e=e.e$ is
the only decomposition of $e$ into a product of an element of $\ker(\psi)$ and
an element of $\\{e\\}\times D$. Hence,
$(g,g^{\prime})=(e.e,e.\psi^{-1}\circ\phi(e))=(e.e)$, so that $\pi^{\prime}$
is an isomorphism between $\tilde{N}\tilde{D}\subset G\times_{D,\phi,\psi}G$
and $G$. ∎

We get the immediate consequence.

######  Corollary 4.23.

In the following cases, a fibre product of the form $G\times_{D,\phi,\psi}G$
contains a subgroup isomorphic to $G$.

  * •

$G$ is simple,

  * •

$G$ is abelian,

  * •

$G$ is isomorphic to $\mathfrak{A}_{n}$ or $\mathfrak{S}_{n}$ for some $n$.

###### Proof.

Let $G$ be in one of the above cases, and $N$ be a normal subgroup of $G$.
There is no other subgroup isomorphic to $N$ normal in $G$, and we always have
$G=ND$, for some group $D$ such that $N\cap D=\\{\mathrm{id}\\}$. The
conclusion follows from Lemma 4.22. ∎

Although we do not need it for our study, we wondered if the above result was
true in general, that is to say if a fibre product of the form $G\times_{D}G$
always admitted a subgroup isomorphic to $G$. It may not be immediate to see a
counterexample. We found the smallest one, which is interesting for our study,
since it can be chosen as the fibre product $H\times_{D}H$ in the exact
sequence (4.17).

######  Example 4.24.

The smallest occurrence of a group $H$ such that a fibre product of the form
$H\times_{D,\phi,\psi}H$ does not contain a subgroup of isomorphic to $H$ is
$H\simeq D_{4}$, $D=\mathbb{Z}_{2}^{2}$, and

$\displaystyle\phi:r$ $\displaystyle\mapsto(0,1)$ $\displaystyle s$
$\displaystyle\mapsto(1,0)$ $\displaystyle\psi:r$ $\displaystyle\mapsto(1,0)$
$\displaystyle s$ $\displaystyle\mapsto(1,1).$

We can even use this example in our context: there exists a subgroup
$G\subset\mathbb{P}^{1}\times\mathbb{P}^{1}$ of invariant Picard rank equal to
$1$, and such that $\ker\delta_{G}=H\times_{\mathbb{Z}_{2}^{2},\phi,\psi}H$,
with $H\cong D_{4}$. We generate $H\times H$ by
$(r,I),(s,I),(I,r),(I,s)\in\mathrm{PGL}_{2}(\mathbb{C})^{2}$, where:

$\displaystyle r$ $\displaystyle=\begin{pmatrix}1&0\\\ 0&i\end{pmatrix},\text{
and}$ $\displaystyle s$ $\displaystyle=\begin{pmatrix}0&1\\\
1&0\end{pmatrix},$

and define the morphisms $\phi$ and $\psi$ as above. Let $G$ be generated by
$H\times_{D,\phi,\psi}H$ and $g=(M,N,1)$, where

$\displaystyle M$ $\displaystyle=\begin{pmatrix}1&0\\\
0&\mu_{8}\end{pmatrix},\text{ and}$ $\displaystyle N$
$\displaystyle=\begin{pmatrix}1&0\\\ 0&\mu_{8}^{-1}\end{pmatrix},$

$\mu_{8}$ being a square root of $i$. One can check that
$H\times_{\mathbb{Z}_{2}^{2},\phi,\psi}H$ is normal in $G$, and since $g$ is
of order $2$, we have
$\ker(\delta_{G})=H\times_{\mathbb{Z}_{2}^{2},\phi,\psi}H$.

######  Lemma 4.25.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ with invariant Picard rank $1$,
which is not in the normalizer of any maximal torus. Then $G$ has a subgroup
isomorphic to $\mathfrak{A}_{4}$.

###### Proof.

The group $G$ satisfying an exact sequence of the form (4.17), it has a
subgroup isomorphic to $H\times_{D}H$, where $H$ is $\mathbb{Z}_{n}$, $D_{n}$,
$\mathfrak{A}_{4}$, $\mathfrak{S}_{4}$, or $\mathfrak{A}_{5}$. Then by Lemma
4.21, $H$ is isomorphic to $\mathfrak{A}_{4}$, $\mathfrak{S}_{4}$, or
$\mathfrak{A}_{5}$. Hence, by Lemma 4.22, $G$ has a subgroup isomorphic $H$,
so it has a subgroup isomorphic to $\mathfrak{A}_{4}$. ∎

The converse statement is true, regardless of the Picard rank.

######  Lemma 4.26.

The normalizer $N(\mathbb{T})$ of the torus $\mathbb{T}$ does not have any
subgroup isomorphic to $\mathfrak{A}_{4}$.

###### Proof.

Recall $N(\mathbb{T})$ satisfies the exact sequence (4.6):

$\displaystyle 1\rightarrow\mathbb{T}\rightarrow
N(\mathbb{T})\stackrel{{\scriptstyle\omega}}{{\rightarrow}}D_{4}\rightarrow
1.$

There is no surjective map with abelian kernel from $\mathfrak{A}_{4}$ to a
subgroup of $D_{4}$. Hence the result. ∎

We chose to separate the two above lemmas since the latter does not require
the Picard group to be of rank one. In particular, we proved the following.

######  Proposition 4.27.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ with invariant Picard rank $1$.
Then $G$ is not isomorphic to a subgroup of $N(\mathbb{T})$ if and only if $G$
contains a subgroup isomorphic to $\mathfrak{A}_{4}$.

Finally, we will use the above group theoretic results for our geometric
problems, as obstructions for the orbits of $G$ to be small enough to generate
Sarkisov links leading to $G$-conic bundles.

######  Proposition 4.28.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ with invariant Picard rank $1$,
which is not conjugated to a subgroup of $N(\mathbb{T})$. There is no fixed
point under the action of $G$, and no orbit of length $2$.

###### Proof.

Let $P$ be a point of $\mathbb{P}^{1}$, and $p$ be the length of its orbit
under the action of $H\subset\mathrm{PGL}_{2}(\mathbb{C})$, with the notations
of the exact sequence (4.17). Then a point of the form
$(P,Q)\in\mathbb{P}^{1}\times\mathbb{P}^{1}$ will have an orbit of length at
least $p$ under the action of $H\times_{D}H\subset G$. Moreover, summing up
Lemmas 4.21 and 4.22, we know that $H$ has a subgroup $A$ isomorphic to
$\mathfrak{A}_{4}$. Up to conjugacy in $\mathrm{PGL}_{2}(\mathbb{C})$, $A$ is
generated by the matrices $\begin{pmatrix}1&0\\\ 0&\epsilon\end{pmatrix}$ and
$\begin{pmatrix}1&2\\\ 1&-1\end{pmatrix}$. Under this action, there is no
fixed point on $\mathbb{P}^{1}$. Since $\mathfrak{A}_{4}$ does not have a
subgroup of index $2$, there is no orbit of length $2$ either. Hence, all
orbits in $\mathbb{P}^{1}\times\mathbb{P}^{1}$ under the action of $G$ are of
length strictly bigger than $2$. ∎

######  Proposition 4.29.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ with invariant Picard rank $1$,
which is not in the normalizer of a maximal torus. Then $S$ is $G$-solid.

###### Proof.

By Remark 4.1, for $S$ not to be $G$-solid, $G$ must be conjugated in
$\mathrm{Bir}(S)$ to a subgroup $G^{\prime}$ of $\mathrm{Aut}(S)$ with
invariant Picard rank $1$ which either has a fixed point or an orbit of length
$2$. By Proposition, 4.28, such group $G^{\prime}$ must be conjugated in
$\mathrm{Aut}(S)$ to a subgroup of $N(\mathbb{T})$. But by Proposition 4.27,
there is no subgroup of $N(\mathbb{T})$ isomorphic to $G^{\prime}$. ∎

Summing up the results of this section, we get the following classification of
groups of automorphisms for which $\mathbb{P}^{1}\times\mathbb{P}^{1}$ is
$G$-solid.

######  Theorem 4.30.

Let $G$ be a subgroup of automorphisms of $\mathbb{P}^{1}\times\mathbb{P}^{1}$
with invariant Picard rank $1$. Then $(G,\mathbb{P}^{1}\times\mathbb{P}^{1})$
is not $G$-solid if and only if it is in the normalizer a maximal torus, and
in one of the following cases.

  * •

The image of $G$ in $D_{4}$ by $\omega$ does not contain the subgroup
isomorphic to $\mathbb{Z}_{4}$.

  * •

$G$ is conjugated in $\mathrm{Aut}(\mathbb{P}^{1}\times\mathbb{P}^{1})$ to one
of the following groups.

    * –

$\mathbb{Z}_{4}$, generated by $g=(\frac{1}{y},x)$,

    * –

$D_{4}$, generated by $g$, and $(y,x)$,

    * –

$D_{4}$, generated by $g$ and $(-y,-x)$,

    * –

$\mathbb{Z}_{4}\times\mathbb{Z}_{2}$, generated by $g$ and $t=(-x,-y)$,

    * –

$\mathbb{Z}_{2}\times D_{4}$, generated by $t$ and $g$ as above, and
$h=(y,x)$.

For conciseness and a presentation which does not require any background about
the toric structure of $\mathbb{P}^{1}\times\mathbb{P}^{1}$, applying
Proposition 4.27, we can reformulate the this theorem in the following way.

######  Theorem 4.31.

Let $G$ be a subgroup of automorphisms of $\mathbb{P}^{1}\times\mathbb{P}^{1}$
with invariant Picard rank $1$. Then $\mathbb{P}^{1}\times\mathbb{P}^{1}$ is
$G$-solid if one of the following cases holds:

  * •

$G$ contains a subgroup isomorphic to $\mathfrak{A}_{4}$.

  * •

$G$ is conjugated in $\mathrm{Aut}(S)$ to a group $H$ which contains
$(\frac{1}{y},x)$ and a subgroup of $\\{(ax,by),a,b\in\mathbb{C}^{*}\\}$ of
order greater than $2$, normal in $H$.

  * •

$G$ is conjugated in $\mathrm{Aut}(S)$ to
$\left<(\frac{1}{y},x),(-x,-y),(-y,x)\right>\cong\mathbb{Z}_{2}^{2}\rtimes\mathbb{Z}_{4}$

##  5\. Del Pezzo surfaces of degree $6$

Up to isomorphism, there is a unique smooth Del Pezzo surface of degree $6$.
It is obtained by blowing-up $\mathbb{P}^{2}$ in three points $P_{1},P_{2}$
and $P_{3}$ in general position. We will denote by $E_{i}$ the exceptional
curve contracted to the point $P_{i}$, and by $D_{ij}$ the proper transform of
the line passing through $P_{i}$ and $P_{j}$. Recall that there is a split
exact sequence

$1\rightarrow\mathbb{T}\rightarrow\mathrm{Aut}(S)\stackrel{{\scriptstyle\omega}}{{\rightarrow}}D_{6}\rightarrow
1,$

where $\omega$ is given by the action of $G$ on the hexagon formed by the
$(-1)$-curves of $S$. The group $\mathbb{T}\cong(\mathbb{C}^{*})^{2}$ is the
lift on $S$ of the diagonal torus of $\mathbb{P}^{2}$. This subgroup will be
denoted by $\mathbb{T}$.

We will use the well known embedding of $S$ in
$\mathbb{P}^{2}\times\mathbb{P}^{2}$ given by $xu=yv=zw$, where
$(x:y:z)\times(u:v:w)$ stands for the coordinates in
$\mathbb{P}^{2}\times\mathbb{P}^{2}$. This model is presented for instance in
[1]. Explicitly, an element $(a,b)\in(\mathbb{C}^{*})^{2}$ corresponds to the
map $(x:y:z)\times(u:v:w)\rightarrow(x:ay:bz)\times(u:a^{-1}v:b^{-1}w)$. The
maps

$\displaystyle r$
$\displaystyle:=(x:y:z)\times(u:v:w)\mapsto(w:u:v)\times(z:x:y),\text{ and}$
$\displaystyle s$
$\displaystyle:=(x:y:z)\times(u:v:w)\mapsto(x:z:y)\times(u:w:v)$

generate a subgroup of $\mathrm{Aut}(S)$ isomorphic to $D_{6}$, and the
quotient $\mathrm{Aut}(S)/\mathbb{T}\cong D_{6}$ is generated by the images of
$r$ and $s$. Geometrically, $r$ acts on the hexagon formed by the
$(-1)$-curves as an elementary rotation, and $s$ acts as a reflection of the
hexagon which does not fix any vertex. We have the relations
$r^{6}=s^{2}=(rs)^{2}=\mathrm{id}$, giving the classical presentation of
$D_{6}$.

######  Lemma 5.1.

Up to conjugation by an element of $\mathbb{T}$, any element $r^{\prime}$ such
that $\omega(r^{\prime})=\omega(r)$ is actually equal to $r$.

###### Proof.

Let $r^{\prime}$ be such automorphism. Since the kernel of $\omega$ is the
normal subgroup $\mathbb{T}$, we have $r^{\prime}=tr,$ for some
$t\in\mathbb{T}$. Explicitly, there exist $(a,b)\in(\mathbb{C}^{*})^{2}$ such
that $r^{\prime}=(w:au:bv)\times(z:a^{-1}x:b^{-1}y)$. Let
$t=(c,d)\in\mathbb{T}\cong(\mathbb{C}^{*})^{2}$. We have

$tr^{\prime}t^{-1}=(x:y:z)\times(u:v:w)\mapsto(w:acd^{-1}u:bcv)\times(z:a^{-1}c^{-1}dx:b^{-1}c^{-1}y).$

Setting $c=b^{-1}$ and $d=ab^{-1}$, we get $tr^{\prime}t^{-1}=r$. ∎

Let $G$ be a subgroup of $\mathrm{Aut}(S)$, such that $\rho^{G}(S)=1$. We will
start our investigation of the $G$-equivariant birational geometry of $S$ by
the following remark, which can be immediately deduced from the classification
of Sarkisov links given in [9].

######  Remark 5.2.

The only $G$-equivariant Sarkisov link $S\dashrightarrow X$ which leads to a
surface $X$ which is not isomorphic to $S$ is the blow-up of a point $P$ which
is not in the exceptional locus, followed by the contraction of the proper
transforms of the lines passing through $P$ and $P_{i}$, for
$i\in\\{1,2,3\\}$. In particular, $X$ is isomorphic to
$\mathbb{P}^{1}\times\mathbb{P}^{1}$. Hence, any $G$-equivariant birational
map from $S$ to a rational surface $S^{\prime}$ of $G$-invariant Picard rank
$2$ must be decomposed in the following way.

$X\stackrel{{\scriptstyle\phi}}{{\rightarrow}}\mathbb{P}^{1}\times\mathbb{P}^{1}\stackrel{{\scriptstyle\psi}}{{\rightarrow}}S^{\prime},$

where $\phi$ and $\psi$ are two $G$-equivariant birational maps. In Remark 4.1
of Section 4, we point out some birational maps from
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ leading to a rational surface with
invariant Picard rank $2$, to what we will refer when proving that the Del
Pezzo surface of degree $6$ is not $G$-solid, for some given group $G$.

In particular, this remark implies the following lemma:

######  Lemma 5.3.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ of invariant Picard rank $1$, not
isomorphic to a subgroup of $D_{6}$. Then $S$ is $G$-solid.

###### Proof.

Assume that $S$ is not $G$-solid. Then, by Remark 5.2, there exists a subgroup
$G^{\prime}$ of $\mathrm{Aut}(S)$ birationally conjugated to $G$, which fixes
a point $P$ in general position. But in this case,
$G^{\prime}\cap\mathbb{T}=\mathrm{id}$, so that $G^{\prime}$ is mapped
isomorphically by $\omega$ to a subgroup of $D_{6}$. ∎

######  Lemma 5.4.

If the invariant Picard group of $S$ is isomorphic to $\mathbb{Z}$, the image
of $G$ by $\omega$ in $D_{6}$ must contain the subgroup of $D_{6}$ isomorphic
to $\mathbb{Z}_{6}$, or the subgroup isomorphic to $\mathfrak{S}_{3}$ acting
transitively on the $(-1)$-curves of $S$.

###### Proof.

Assume that $G$ does not act transitively on the $(-1)$-curves of $S$. Just
straightforwardly checking all the possible subgroups of $D_{6}$, one can
check that one of the divisors $E_{1}+E_{2}+E_{3}$, or $D_{12}+D_{23}+D_{13}$,
or $E_{i}+E_{jk}$, with $j,k\neq i$, is invariant by $G$. In all those cases,
by Castelnuovo’s contractibility criterion, there exists a $G$-equivariant
morphism to either $\mathbb{P}^{2}$ or $\mathbb{P}^{1}\times\mathbb{P}^{1}$. ∎

######  Corollary 5.5.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ such that $\rho^{G}(S)=1$. Then $G$
is not isomorphic to any of the groups
$\mathbb{Z}_{2}^{2},\mathbb{Z}_{3},\mathbb{Z}_{2},\\{\mathrm{id}\\}$.

The only remaining groups of interest are $G\cong D_{6}$,
$G\cong\mathbb{Z}_{6}$, and $G\cong\mathfrak{S}_{3}$. Let us study the case of
$D_{6}$.

######  Proposition 5.6.

If $G$ is a subgroup of $\mathrm{Aut}(S)$ isomorphic to $D_{6}$ such that
$\rho^{G}(S)=1$, then $S$ is not $G$-solid.

###### Proof.

Assume that $G\cong D_{6}$. There is no quotient of $D_{6}$ isomorphic to
$\mathbb{Z}_{6}$, so we must have $\omega(D_{6})=D_{6}$, by Lemma 5.4. In this
case, the group $G$ is generated by an element $g$ such that
$\omega(g)=\omega(r)$, and an element $h$ such that $\omega(h)=\omega(s)$. By
Lemma 5.1, $g=r$ up to conjugation by an element of $\mathbb{T}$, and its
unique fixed point is $P_{0}=(1:1:1)\times(1:1:1)$. The isomorphism $h$ is of
the form $(x:y:z)\times(u:v:w)\mapsto(x:az:bz)\times(u:a^{-1}w:b^{-1}v)$.
Since $\mathrm{o}(h)=2$, be get $b=a^{-1}$. Moreover,
$hr^{2}h=(x:y:z)\times(u:v:w)\mapsto(az:x:a^{-2}y)\times(a^{-1}w:u:a^{2}v)$.
But the structure of $D_{6}$ implies that
$hr^{2}h=r^{-2}=(x:y:z)\times(u:v:w)\mapsto(z:x:y)\times(w:u:v)$, so that
$a=1$. Hence, $h=s=(x:y:z)\times(u:v:w)\mapsto(x:z:y)\times(u:w:v)$. This
automorphism also fixes $P_{0}$, so that there exists a $G$-equivariant link
from $S$ to $\mathbb{P}^{1}\times\mathbb{P}^{1}$ centered at $P_{0}$.

We conclude that $S$ is not $G$-solid by using Theorem 4.31 to show that
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ is not $H$-solid, for any subgroup $H$ of
$\mathrm{Aut}(\mathbb{P}^{1}\times\mathbb{P}^{1})$ of invariant Picard group
$\mathbb{Z}$. Indeed, $D_{6}$ does not have any subgroup isomorphic to
$\mathfrak{A}_{4}$, or any subgroup isomorphic to $\mathbb{Z}_{4}$. ∎

######  Proposition 5.7.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ isomorphic to $\mathbb{Z}_{6}$ and
such that $\rho^{G}(S)=1$. Then $S$ not $G$-solid.

###### Proof.

First, remark that the restriction $\omega|_{G}$ of $\omega$ to $G$ is
injective. Indeed, if not, then the image of $G$ by $\omega$ is either
isomorphic to $\mathbb{Z}_{3}$, isomorphic to $\mathbb{Z}_{2}$, or trivial. By
Lemma 5.4, $\rho^{G}(S)>1$, contradicting our assumption. The group $G$ is
then generated by an element $g$ such that $\omega(g)=\omega(r)$. By Lemma
5.1, up to conjugation by an element of $\mathbb{T}$,
$g=r=(x:y:z)\times(u:v:w)\mapsto(w:u:v)\times(z:x:y)$. The only fixed point of
$H$ on $S$ is $(1:1:1)\times(1:1:1)$.

The only subgroup of index $2$ is $\mathbb{Z}_{3}$, generated by
$g^{2}=(x:y:z)\times(u:v:w)\mapsto(y,z,x)\times(v,w,u)$. Let $\mu$ be a
primitive third root of the unity. The fixed points of $g^{2}$ are the points
of the form $M_{k}=(1:\mu^{k}:\mu^{2k})\times(1:\mu^{2k}:\mu^{k})$, for
$k=0,1,2$. The point $M_{0}$ is fixed by $G$, while $\\{M_{1},M_{2}\\}$ form
an orbit of length $2$ not lying on the $(-1)$-locus. We can blow-up $M_{0}$,
and contract three $(-1)$-curves to get a link $\phi$ to
$\mathbb{P}^{1}\times\mathbb{P}^{1}$. By the following Lemma, $M_{1}$ and
$M_{2}$ cannot be sent by $\phi$ to points on the same $\mathbb{P}^{1}$-fibre.
Hence, we can blow-up the image by $\phi$ of the orbit $\\{M_{1},M_{2}\\},$
and get a $G$-equivariant map to a surface of $\mathfrak{C}_{G}$. ∎

The following Lemma deals with subgroups of
$\mathrm{Aut}(\mathbb{P}^{1}\times\mathbb{P}^{1})$, but we choose to state it
here as it is only used in the proof of Proposition 5.7.

######  Lemma 5.8.

Let $G\cong\mathbb{Z}_{6}$ be a subgroup of
$\mathrm{Aut}(\mathbb{P}^{1}\times\mathbb{P}^{1})$ such that $\rho^{G}(S)=1$.
Assume that $G$ has an orbit of length $2$ formed by points $P$ and $Q$. Then
these two points cannot lie on the same $\mathbb{P}^{1}$-fibre.

###### Proof.

As detailed in Section 4, there is an exact sequence $1\rightarrow
G^{\prime}\rightarrow
G\stackrel{{\scriptstyle\omega}}{{\rightarrow}}\mathbb{Z}_{2}\rightarrow 1$.
An element $g\in G$ such that $\omega(g)=1$ must swap the two rulings of
$\mathbb{P}^{1}\times\mathbb{P}^{1}$. This fact and the structure of
$\mathbb{Z}_{6}$ imply that $G^{\prime}$ is isomorphic to $\mathbb{Z}_{3}$,
and that $G$ is generated by $G^{\prime}$ and an element $g$ of order $2$
which sends vertical $\mathbb{P}^{1}$-fibres to horizontal ones, and vice-
versa. Since $\mathbb{Z}_{3}$ cannot have an orbit of length $2$, we must have
$Q=g(P)$. Suppose that $P$ and $Q$ are on the same $\mathbb{P}^{1}$-fibre $F$.
Without loss of generality, we may assume that $F$ is horizontal. But then
$F^{\prime}=g(F)$ is vertical, and contains $g(P)=Q$, as well as $g(Q)=P$.
Hence, the points $P$ and $Q$ are also on the same vertical fibre. This yields
$P=Q$, which contradicts our assumption. ∎

######  Proposition 5.9.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ isomorphic to $\mathfrak{S}_{3}$
and such that $\rho^{G}(S)=1$. Then $S$ is not $G$-solid.

###### Proof.

If the restriction $\omega|_{G}$ of $\omega$ to $G$ is not injective, then the
image of $G$ by $\omega$ is either isomorphic to $\mathbb{Z}_{3}$, isomorphic
to $\mathbb{Z}_{2}$, or trivial. By Lemma 5.4, $S$ is not $G$-solid. Assume
that $\omega|_{G}$ is injective. We have two possibilities. Either $\omega(G)$
does not act transitively on the $(-1)$-curves, in which case $\rho^{G}(S)\neq
1$ by Lemma 5.4, or it does. Assume we are in the latter case. The group $G$
is generated by an element $g$ such that $\omega(g)=\omega(r^{2})$, and an
element $h$ such that $\omega(h)=\omega(rs)$. Geometrically, $rs$ acts on the
hexagon formed by the $(-1)$-curves as a reflection which fixes two opposite
vertices. By Lemma 5.1, we have
$g=(x:y:z)\times(u:v:w)\mapsto(y:z:x)\times(v,w,u)$ up to conjugation by an
element of the torus. Since $\ker(\omega)=\mathbb{T}$, $h=trs$ for some
$t\in\mathbb{T}$. Explicitly, $h$ is of the form
$(x:y:z)\times(u:v:w)\mapsto(w:av:bu)\times(z:a^{-1}y:b^{-1}x)$, for some
$(a,b)\in(\mathbb{C}^{*})^{2}$. We get
$h^{2}=(x:y:z)\times(u:v:w)\mapsto(b^{-1}x:y:bz)\times(bu:v:b^{-1}w)$, and
knowing that $\mathrm{o}(h)=2$, we deduce $b=1$. Moreover,
$hgh=(x:y:z)\times(u:v:w)\mapsto(a^{-1}y:az:x)\times(av:a^{-1}w:u)$, but
relations in $D_{6}$ imply that
$hgh=g^{-1}=(x:y:z)\times(u:v:w)\mapsto(y:z:x)\times(v:w:u)$. Hence, $a=1$, so
that $h=(x:y:z)\times(u:v:w)\mapsto(w:v:u)\times(z:y:x)$. The three points of
the form $(1:\mu^{2k}:\mu^{k})\times(1:\mu^{k}:\mu^{2k})$ are fixed by the
action of $G$, so that $S$ is $G$-birational to $\mathbb{F}_{1}$ of invariant
Picard rank $2$, by Remark 4.1 of the Section 4. ∎

Summing up all the above results, we get the following.

######  Theorem 5.10.

Let $G$ be a subgroup of $\mathrm{Aut}(S)$ such that $\rho^{G}(S)=1$. Then $S$
is $G$-solid if and only if $G$ is not isomorphic to $\mathbb{Z}_{6}$,
$\mathfrak{S}_{3}$, or $D_{6}$.

##  6\. The projective plane

The only remaining smooth Del Pezzo surface is $\mathbb{P}^{2}$. Most of the
work has been done by D.Sakovics in [11]. A subgroup of
$\mathrm{PGL}_{2}(\mathbb{C})$ is called intransitive if it fixes a point of
$\mathbb{P}^{2}$, and is called transitive otherwise.

######  Theorem 6.1 ([11]).

The projective plane is $G$-rigid if and only if $G$ is transitive and not
isomorphic to $\mathfrak{S}_{4}$ or $\mathfrak{A}_{4}$.

Moreover, if $G$ is intransitive, then we can $G$-equivariently blow-up a
fixed point and get a surface of $\mathfrak{C}_{G}$, so that $S$ is not
$G$-solid. Hence, the only remaining cases to study are those of
$\mathfrak{S}_{4}$ and $\mathfrak{A}_{4}$. In this section, we are going to
give the full birational geometry of the projective plane for groups of
automorphisms isomorphic to $\mathfrak{S}_{4}$ or $\mathfrak{A}_{4}$. In other
words, we are going to list all the $G$-Mori fibre spaces $X$ such that there
exists a $G$-equivariant birational map $\mathbb{P}^{2}\dashrightarrow X$.

######  Lemma 6.2.

Subgroups of $\mathrm{PGL}_{2}(\mathbb{C})$ isomorphic to $\mathfrak{S}_{4}$
and subgroups of $\mathrm{PGL}_{2}(\mathbb{C})$ isomorphic to
$\mathfrak{A}_{4}$ are unique up to conjugacy in
$\mathrm{PGL}_{2}(\mathbb{C})$.

###### Proof.

The canonical projection
$\pi:\mathrm{GL}_{3}(\mathbb{C})\rightarrow\mathrm{PGL}_{2}(\mathbb{C})$
induces a surjection
$\mathrm{SL}_{3}(\mathbb{C})\rightarrow\mathrm{PGL}_{2}(\mathbb{C})$. Let
$G\subset\mathrm{PGL}_{2}(\mathbb{C})$ be a subgroup of
$\mathrm{PGL}_{2}(\mathbb{C})$ isomorphic to $\mathfrak{S}_{4}$, and consider
its lift $G^{\prime}\subset\mathrm{SL}_{3}(\mathbb{C})$ by the above
projection. The kernel of $\pi$ is $\\{I_{3},\mu I_{3},\mu^{2}I_{3}\\}$, where
$\mu$ is a primitive cube root of the unity. Thus, we have an extension

$\mathbb{Z}_{3}\rightarrow G^{\prime}\rightarrow G.$

But $\\{I_{3},\mu I_{3},\mu^{2}I_{3}\\}$ lies in the center of
$\mathrm{GL}_{3}(\mathbb{C})$, and there is no non-trivial central extension
of $\mathfrak{S}_{4}$ in $\mathrm{GL}_{3}(\mathbb{C})$.

Thus, $G^{\prime}\cong\mathbb{Z}_{3}\times\mathfrak{S}_{4}$, and its subgroup
$\\{\mathrm{id}\\}\times\mathfrak{S}_{4}$ is sent isomorphically to $G$, so
there exists a subgroup of $\mathrm{GL}_{3}(\mathbb{C})$ isomorphic to
$\mathfrak{S}_{4}$, whose projection in $\mathrm{PGL}_{2}(\mathbb{C})$ is $G$.

Hence, we only have to check that faithful representations of degree $3$ of
$\mathfrak{S}_{4}$ map to the same subgroup of $\mathrm{PGL}_{2}(\mathbb{C})$
once we compose them with the projection $\pi$. But the only such
representations are the standard one and its product with sign representation.
Both induce the same subgroup of $\mathrm{PGL}_{2}(\mathbb{C})$.

Finally, the group $\mathfrak{S}_{4}$ has a unique conjugacy class of
subgroups isomorphic to $\mathfrak{A}_{4}$. Hence, the above lemma implies
that there is a unique conjugacy class of subgroups isomorphic to
$\mathfrak{A}_{4}$ in $\mathrm{PGL}_{2}(\mathbb{C})$. ∎

######  Proposition 6.3.

Let $G\cong\mathfrak{S}_{4}$ be a subgroup of $\mathrm{Aut}(\mathbb{P}^{2})$.
The only $G$-equivariant Sarkisov links starting from $\mathbb{P}^{2}$ are:

  * •

A link of type $II$ of the form

$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma}$$\scriptstyle{\tau}$$\textstyle{\mathbb{P}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau}$$\textstyle{\mathbb{P}^{2}}$

where $Z$ is a Del Pezzo surface of degree $6$, and $\tau$ is the standard
Cremona involution.

  * •

A link of type $I$ of the form

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma}$$\scriptstyle{\pi}$$\textstyle{\mathbb{P}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\textstyle{\mathbb{P}^{1}}$

where $\pi:X\rightarrow\mathbb{P}^{1}$ is a $G$-conic bundle on a Del Pezzo
surface of degree $5$ with invariant Picard rank $2$.

The only $G$-equivariant Sarkisov link starting from $X$ is the inverse of • ‣
6.3, leading back to $\mathbb{P}^{2}$.

###### Proof.

Up to conjugacy in $\mathrm{PGL}_{2}(\mathbb{C})$, the subgroup isomorphic to
$\mathfrak{S}_{4}$ is generated by the matrices $a:=\begin{pmatrix}0&0&1\\\
1&0&0\\\ 0&1&0\end{pmatrix}$, and $b:=\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
0&0&1\end{pmatrix}$. Let $G$ be the subgroup of $\mathrm{PGL}_{2}(\mathbb{C})$
generated by $A=\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 0&0&1\end{pmatrix}$,
$B=\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 0&0&1\end{pmatrix}$,
$C=\begin{pmatrix}0&0&1\\\ 1&0&0\\\ 0&1&0\end{pmatrix}$, and
$D=\begin{pmatrix}1&0&0\\\ 0&0&1\\\ 0&1&0\end{pmatrix}$. This group is
isomorphic to $\mathfrak{S}_{4}$. Let us see the possible orbits on which a
Sarkisov links can be centered. Recall that two isomorphic subgroups of
$\mathfrak{S}_{4}$ are always conjugated, so that it is enough to find an
occurrence of each subgroup up to isomorphism in the study of the possible
stabilizers.

  * •

There is no fixed point under the above action.

  * •

Remark that the subgroup of $G$ generated by $a:=\begin{pmatrix}0&0&1\\\
1&0&0\\\ 0&1&0\end{pmatrix}$, and $b:=\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
0&0&1\end{pmatrix}$ is isomorphic to $\mathfrak{A}_{4}$. It is the only
subgroup of $G$ of index two and has no fixed point, so that $G$ does not have
any orbit of length $2$.

  * •

The only subgroup of index three up to conjugation is $D_{4}$, generated by
$\begin{pmatrix}-1&0&0\\\ 0&0&-1\\\ 0&1&0\end{pmatrix}$, and
$\begin{pmatrix}1&0&0\\\ 0&0&1\\\ 0&1&0\end{pmatrix}$. Its fixed points are
$O_{3}:=\\{(1:0:0),(0:1:0),(0:0:1)\\}$, and $O_{3}$ is the only orbit of
length $3$ under the action of $G$. It is the center of the link of type $II$:

$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma}$$\scriptstyle{\tau}$$\textstyle{\mathbb{P}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau}$$\textstyle{\mathbb{P}^{2}}$

where $Z$ is a Del Pezzo surface of degree $6$, and $\tau$ is the standard
Cremona involution.

  * •

The group $G$ does have an orbit of length $4$ in general position. Indeed,
$G$ has a unique subgroup of index $4$. It is isomorphic to
$\mathfrak{S}_{3}$, and generated by $\begin{pmatrix}0&0&1\\\ 1&0&0\\\
0&1&0\end{pmatrix}$, and $\begin{pmatrix}1&0&0\\\ 0&0&1\\\
0&1&0\end{pmatrix}$. Their only common fixed point is $(1:1:1)$, and its orbit
under the action of $G$ is $O_{4}:=\\{(1:1:1),(-1:1:1),(1:-1:1),(1:1:-1)\\}$.
Hence, there is a $G$-equivariant link of type $I$ of the form:

(6.8)

where $\pi:X\rightarrow\mathbb{P}^{1}$ is a $G$-conic bundle on a Del Pezzo
surface of degree $5$, with invariant Picard rank $2$.

  * •

There is an orbit of length $6$, but not in general position. There are two
subgroups of index $6$ of $G$. The first one is $\mathbb{Z}_{2}^{2}$,
generated by $A=\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 0&0&1\end{pmatrix}$, and
$B=\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 0&0&1\end{pmatrix}$. Its only fixed point
is $(0:0:1)$, and its orbit is $O_{3}$. The other subgroup of index $6$ of $G$
is $Z_{4}$, generated by $\begin{pmatrix}1&0&0\\\ 0&0&1\\\
0&-1&0\end{pmatrix}$. Its fixed points are $(0:-i:1)$, $(0,i,1)$, and
$(1:0:0)$. The orbit of the last one is $O_{3}$, and the two others have the
orbit $O_{6}:=\\{(0:i:1),(0:-i:1),(1:0:i),(1:0:-i),(i:1:0),(-i:1:0)$. These
six points are not in general position, as they all lie on the Fermat conic
$x^{2}+y^{2}+z^{2}=0$.

  * •

There is no orbit of length $8$. The only subgroup of index $8$ is
$\mathbb{Z}_{3}$, generated by $\begin{pmatrix}0&0&1\\\ 1&0&0\\\
0&1&0\end{pmatrix}$. Its fixed points are $(1:1:1)$,
$(1:\mu_{3}:\mu_{3}^{2})$, and $(1:\mu_{3}^{2}:\mu_{3})$, where $\mu_{3}$ is a
primitive cube root of the unity. The orbit of $(1:1:1)$ is of length $4$, and
$(1:\mu_{3}:\mu_{3}^{2})$, and $(1:\mu_{3}^{2}:\mu_{3})$ lie on the same orbit
of length $8$:
$O_{8}:=\\{(1:\mu_{3}:\mu_{3}^{2}),(1:\mu_{3}:\mu_{3}^{2}),(1:\mu_{3}:\mu_{3}^{2}),(1:\mu_{3}:\mu_{3}^{2}),(1:\mu_{3}^{2}:\mu_{3}),(1:\mu_{3}^{2}:\mu_{3}),(1:\mu_{3}^{2}:\mu_{3}),(1:\mu_{3}^{2}:\mu_{3})\\}$.
But these eight points are on the Fermat conic $x^{2}+y^{2}+z^{2}=0$, hence
are not in general position.

It remains to show that there is no $G$-equivariant link starting from $X$,
except the inverse of the link (6.8). For this, we will show that all the
orbits under the action of $G$ lifted on $X$ have several points on the same
fibre of the conic bundle. The birational map $\gamma$ is given by the linear
system $|C|$ of conics passing through all the points of the orbit $O_{4}$.
The curves $x^{2}-y^{2}=0$ and $x^{2}-z^{2}=0$ form a basis of this linear
system. Hence, up to a change of basis, the map $\gamma$ is of the form
$(x:y:z)\mapsto(x^{2}-y^{2}:x^{2}-z^{2})$. The image of $G$ by $\gamma$ is
isomorphic to $\mathfrak{S}_{3}$. Hence the kernel $N$ of the induced morphism
$G\rightarrow\mathrm{Aut}(\mathbb{P}^{1})$ is isomorphic to
$\mathbb{Z}_{2}^{2}$, and generated by the matrices
$A=\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 0&0&1\end{pmatrix}$, and
$B=\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 0&0&1\end{pmatrix}$. The action of $N$ on
the smooth conics of this system is faithful, hence $N$ does not fix any point
on the regular fibres of the conic bundle. ∎

We get the following immediate consequences.

######  Corollary 6.9.

Let $G$ be a subgroup of $\mathrm{Aut}(\mathbb{P}^{2})$ isomorphic to
$\mathfrak{S}_{4}$. The projective plane is not $G$-solid, but not
$G$-birational to any Hirzebruch surface.

######  Corollary 6.10.

Let $G$ be a subgroup of $\mathrm{Aut}(\mathbb{P}^{2})$ isomorphic to
$\mathfrak{S}_{4}$. Then
$\mathrm{Bir}^{G}(\mathbb{P}^{2})=\left<G,\tau\right>\cong\mathfrak{S}_{4}\times\mathbb{Z}_{2}$,
where $\tau$ is the standard Cremona involution.

###### Proof.

The elements of $G$ are the automorphisms of the form
$(x:y:z)\mapsto\sigma(\alpha x:\beta y:\gamma z)$, where $\sigma$ is a
permutation of the coordinates, and $\alpha,\beta,\gamma\in\\{-1,1\\}$. The
involution $\tau:(x:y:z)\dashrightarrow(yz:xz:xy)$ commutes with all these
elements. ∎

The remaining case to study is that of $G\cong\mathfrak{A}_{4}$. The result
will be the same as for $G\cong\mathfrak{S}_{4}$.

######  Proposition 6.11.

Let $G\cong\mathfrak{A}_{4}$ be a subgroup of $\mathrm{Aut}(\mathbb{P}^{2})$.
The only $G$-equivariant Sarkisov links starting from $\mathbb{P}^{2}$ are:

  * •

A link of type $II$ of the form

$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma}$$\scriptstyle{\tau}$$\textstyle{\mathbb{P}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau}$$\textstyle{\mathbb{P}^{2}}$

where $Z$ is a Del Pezzo surface of degree $6$, and $\tau$ is the standard
Cremona involution.

  * •

Three links of type $I$ of the form

(6.16)

where $\pi:X\rightarrow\mathbb{P}^{1}$ is a $G$-conic bundle on a Del Pezzo
surface of degree $5$ with invariant Picard rank $2$.

  * •

A one parameter family of links of type $II$ of the form

$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma}$$\scriptstyle{\tau}$$\textstyle{\mathbb{P}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i_{a}}$$\textstyle{\mathbb{P}^{2}}$

where $\sigma$ is the blow-up of an orbit of six points, $\tau$ is the
$G$-equivariant contraction of eight $(-1)$-curves, and $i_{a}$ is a
birational involution.

The only $G$-equivariant Sarkisov link starting from $X$ is the inverse of
6.16, leading back to $\mathbb{P}^{2}$.

###### Proof.

We use again Lemma 6.2 to affirm that $\mathfrak{A}_{4}$ is unique up to
conjugacy in $\mathrm{PGL}_{2}(\mathbb{C})$. Up to conjugacy in
$\mathrm{PGL}_{2}(\mathbb{C})$, the subgroup isomorphic to $\mathfrak{A}_{4}$
is generated by the matrices $a:=\begin{pmatrix}0&0&1\\\ 1&0&0\\\
0&1&0\end{pmatrix}$, and $b:=\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
0&0&1\end{pmatrix}$. Isomorphic subgroups of $\mathfrak{A}_{4}$ are in the
same conjugacy class.

  * •

There is no fixed point under the above action.

  * •

There is no subgroup of index $2$.

  * •

The only subgroup of index $3$ is $N=\mathbb{Z}_{2}^{2}$, generated by
$A=\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 0&0&1\end{pmatrix}$, and
$B=\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 0&0&1\end{pmatrix}$. Its only fixed point
is $(0:0:1)$, and its orbit is $O_{3}=\\{(1:0:0),(0:1:0),(0:0:1)\\}$. The only
link centered at this orbit is teh link of type $II$

$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma}$$\scriptstyle{\tau}$$\textstyle{\mathbb{P}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau}$$\textstyle{\mathbb{P}^{2}}$

where $\tau$ is the standard Cremona involution.

  * •

The unique subgroup of index $4$ is $\mathbb{Z}_{3}$, generated by
$\begin{pmatrix}0&0&1\\\ 1&0&0\\\ 0&1&0\end{pmatrix}$, and have three
independant fixed points : $(1,1,1)$, $(1,\mu_{3},\mu_{3}^{2})$, and
$(1,\mu_{3}^{2},\mu_{3})$. They give rise to three distinct orbits of length
$4$ in general position: $O_{4}:=\\{(1:1:1),(-1:1:1),(1:-1:1),(1:1:-1)\\}$,
$O^{\prime}_{4}:=\\{(1:\mu_{3}:\mu_{3}^{2}),(-1:\mu_{3}:\mu_{3}^{2}),(1:-\mu_{3}:\mu_{3}^{2}),(1:\mu_{3}:-\mu_{3}^{2})\\}$,
and
$O^{\prime\prime}_{4}:=\\{(1:\mu_{3}^{2}:\mu_{3}),(-1:\mu_{3}^{2}:\mu_{3}),(1:-\mu_{3}^{2}:\mu_{3}),(-1:-\mu_{3}^{2}:\mu_{3})\\}$.
In each case, the points are in general position, and blowing them up we get a
$G$-equivariant link of type $1$

(6.21)

where $\sigma$ is the blow-up of one of the above orbits of length $4$. We
will denote the three possibilities for $X$ respectively by $X_{1}$, $X_{2}$,
and $X_{3}$.

  * •

The unique subgroup of index $6$ of $G$ is isomorphic to $\mathbb{Z}_{2}$,
generated by $\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 0&0&1\end{pmatrix}$. Its fixed
points are the points of $O_{3}$, and those of the form $(0:1:a)$, with $a\neq
0$. They form orbits of length $6$ of the form
$O_{a}:=\\{(0:1:a),(a:0:1),(1:a:0),(0:-1:a),(a:0:-1),(-1:a:0)\\}$. The only
Sarkisov link centered at such orbit is the link of type $II$

$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma}$$\scriptstyle{\tau}$$\textstyle{\mathbb{P}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i_{a}}$$\textstyle{\mathbb{P}^{2}}$

where $\sigma$ is the blow-up of $O_{a}$, $\tau$ is the $G$-equivariant
contraction of eight $(-1)$-curves, and $i_{a}$ is the birational involution
$(x:y:z)\mapsto(f_{1}(x,y,z):f_{2}(x,y,z):f_{3}(x,y,z))$, where

$\displaystyle f_{1}(x,y,z)$
$\displaystyle=\left(a^{12}+1\right)x^{2}y^{2}z+a^{10}\left(-y^{4}\right)z+2a^{8}y^{2}z^{3}-a^{6}z^{5}+2a^{4}x^{2}z^{3}-a^{2}x^{4}z,$
$\displaystyle f_{2}(x,y,z)$
$\displaystyle=\left(a^{12}+1\right)x^{2}yz^{2}-a^{10}x^{4}y+2a^{8}x^{2}y^{3}-a^{6}y^{5}+2a^{4}y^{3}z^{2}-a^{2}yz^{4},$
$\displaystyle f_{3}(x,y,z)$
$\displaystyle=\left(a^{12}+1\right)xy^{2}z^{2}+a^{10}(-x)z^{4}+2a^{8}x^{3}z^{2}-a^{6}x^{5}+2a^{4}x^{3}y^{2}-a^{2}xy^{4}.$

We will now show that there is no $G$-equivariant Sarkisov link starting from
any of the $G$-surfaces $X_{1}$, $X_{2}$, and $X_{3}$ of degree $5$, except
the inverse of the link (6.21). The $G$-surface $X_{1}$ is the same as $X$ in
the proof of Proposition 6.3, and the proof is the same. We proceed in the
same way for the two other cases. The $G$-surface $X_{2}$ is the blow-up of
$\mathbb{P}^{2}$ in the points of
$O^{\prime}_{4}:=\\{(1:\mu_{3}:\mu_{3}^{2}),(-1:\mu_{3}:\mu_{3}^{2}),(1:-\mu_{3}:\mu_{3}^{2}),(1:\mu_{3}:-\mu_{3}^{2})\\}$.
The linear system of conics passing through these points is generated by $\mu
x^{2}-z^{2}=0$ and $(\mu+1)x^{2}+y^{2}=0$. The $G$-surface $X_{3}$ is the
blow-up of $\mathbb{P}^{2}$ in the points of
$O^{\prime}_{4}:=\\{(1:\mu_{3}^{2}:\mu_{3}),(-1:\mu_{3}^{2}:\mu_{3}),(1:-\mu_{3}^{2}:\mu_{3}),(1:\mu_{3}^{2}:-\mu_{3})\\}$.
The linear system of conics passing through these points is generated by $\mu
x^{2}-y^{2}=0$ and $(\mu+1)x^{2}+z^{2}=0$. In both of these cases, and the
subgroup $N\cong\mathbb{Z}_{2}^{2}$ of $G$ acts faithfully on each smooth
conic of the system, hence does not fix any point in the fibres of the conic
bundle. ∎

Here as well, we immediately get the following consequences.

######  Corollary 6.22.

Let $G$ be a subgroup of $\mathrm{Aut}(\mathbb{P}^{2})$ isomorphic to
$\mathfrak{A}_{4}$. The projective plane is not $G$-solid, but not
$G$-birational to any Hirzebruch surface.

######  Corollary 6.23.

Let $G$ be a subgroup of $\mathrm{Aut}(\mathbb{P}^{2})$ isomorphic to
$\mathfrak{A}_{4}$. Then
$\mathrm{Bir}^{G}(\mathbb{P}^{2})=\left<G,\tau,i_{a}|a\in\mathbb{C}^{*}\right>$,
where $\tau$ is the standard Cremona involution.

Summing up the results of this section, we can conclude on the $G$-birational
solidity of the projective plane.

######  Theorem 6.24.

Let $G$ be a finite subgroup of $\mathrm{Aut}(\mathbb{P}^{2})$. The following
assertions are equivalent.

  * •

$S$ is $G$\- birationally rigid,

  * •

$S$ is $G$-birationally solid,

  * •

$G$ is transitive and not isomorphic to $\mathfrak{S}_{4}$ or
$\mathfrak{A}_{4}$.

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