Quasi-simple finite groups of essential dimension 3

TL;DR

This paper classifies quasi-simple finite groups with an essential dimension of exactly 3, providing a complete understanding of their structure within this specific dimension.

Contribution

It offers the first comprehensive classification of quasi-simple finite groups of essential dimension 3, filling a gap in the understanding of their algebraic properties.

Findings

Identified all quasi-simple finite groups with essential dimension 3

Provided explicit descriptions of these groups

Enhanced understanding of the relationship between group structure and essential dimension

Abstract

We classify quasi-simple finite groups of essential dimension 3.

Equations

$${\mathrm{SL}}_2(7),{\mathrm{SL}}_2(11),{\mathrm{Sp}}_4(3),2.{\mathfrak{A}}_5,n.{\mathfrak{A}}_6,n.{\mathfrak{A}}_7\text{with n=2,3,6.}$$

$${\mathrm{PSL}}_2(7),{\mathrm{PSL}}_2(8),{\mathrm{PSL}}_2(11),{\mathrm{PSp}}_4(3),{\mathfrak{A}}_5,{\mathfrak{A}}_6,{\mathfrak{A}}_7.$$

$$E\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$${\mathrm{SL}}_2(7),n.{\mathfrak{A}}_6\text{with n=2,3,6.}$$

$${\mathrm{Tr}}_{H^2(S,\mathbb{C})}{\text{upalpha}}^{\ast }=2.$$

$$A=\left(\begin{array}{cc}1& 1\\ 5& -1\end{array}\right)B=\left(\begin{array}{cc}1& 5\\ 1& -1\end{array}\right)$$

$$A=\left(\begin{array}{cc}1& 2\\ -1& -1\end{array}\right)B=\left(\begin{array}{cc}4& 1\\ 1& -4\end{array}\right)$$

$$q(x):=\frac{1}{2}(x,x)\mathrm{mod}2.$$

$$q(x)=x_1^2+x_2^2+x_1{x}_2+x_3^2+x_4^2+x_3{x}_4+x_5^2+x_6^2+x_5{x}_6.$$

$$\lambda :\mathrm{Aut}(X)\hookrightarrow \mathrm{GL}(\mathrm{Pic}(X))={\mathrm{GL}}_8(\mathbb{Z})\stackrel{\det }{⟶}\{\pm 1\},$$

Full text

To the memory of Professor Alfred Lvovich Shmel’kin

**UDK ** 512.76

Keywords: essential dimension, group, algebraic variety, representation, Cremona group

1 Introduction

This paper is based on the author’s talk given at the Magadan conference.

Let $G$ be a finite group and let $V$ be a faithful representation of $G$ regarded as an algebraic variety. A compression is a $G$-equivariant dominant rational map $V\dashrightarrow X$ of faithful $G$-varieties. The essential dimension of $G$, denoted ${\operatorname{ed}}(G)$, is the minimal dimension of all faithful $G$-varieties $X$ appearing in compressions $V\dashrightarrow X$. This notion was introduced by J. Buhler and Z. Reichstein [BR97] in relation to some classical problems in the theory of polynomials. It turns out that the essential dimension depends only on the group $G$, i.e. it does not depend on the choice of linear representation $V$ [BR97, Theorem 3.1].

The computation of the essential dimension is a challenging problem of algebra and algebraic geometry. The finite groups of essential dimension $\leq 2$ have been classified (see [Dun13]). Simple finite groups of essential dimension $3$ have been recently determined by A. Beauville [Bea14] (see also [Ser10], [Dun10]):

1.1 Theorem**.**

The simple groups of essential dimension $3$ are ${\mathfrak{A}}_{6}$ and possibly ${\operatorname{PSL}}_{2}(11)$.

The essential dimension of $p$-groups was computed by Karpenko and Merkurjev [KM08]. In this short note we find all the finite quasi-simple groups of essential dimension $3$.

1.2 Definition**.**

A group $G$ is said to be quasi-simple if $G$ is perfect, that is, it equals its commutator subgroup, and the quotient of $G$ by its center is a simple non-abelian group.

The main result of this paper is the following.

1.3 Theorem**.**

Let $G$ be a finite quasi-simple non-simple group. If ${\operatorname{ed}}(G)=2$, then $G\simeq 2.{\mathfrak{A}}_{5}$. If ${\operatorname{ed}}(G)=3$, then $G\simeq 3.{\mathfrak{A}}_{6}$.

1.4 Notation**.**

Throughout this paper the ground field is supposed to be the field of complex numbers $\mathbb{C}$. We employ the following standard notations used in the group theory.

• •

${\boldsymbol{\mu}}_{n}$ denotes the multiplicative group of order $n$ (in $\mathbb{C}^{*}$),

• •

${\mathfrak{A}}_{n}$ denotes the alternating group of degree $n$,

• •

${\operatorname{SL}}_{n}(q)$ (resp. ${\operatorname{PSL}}_{n}(q)$) denotes the special linear group (resp. projective special linear group) over the finite field $\mathbf{F}_{q}$,

• •

$n.G$ denotes a non-split central extension of $G$ by ${\boldsymbol{\mu}}_{n}$,

• •

${\operatorname{z}}(G)$ (resp. $[G,G]$) denotes the center (resp. the commutator subgroup) of a group $G$.

All simple groups are supposed to be non-cyclic.

2 Proof of Theorem 1.3

The following assertion is an immediate consequence of the corresponding fact for simple groups [Pro12].

2.1 Proposition**.**

Let $X$ be a three-dimensional rationally connected variety and let $G\subset\operatorname{Bir}(X)$ be a finite quasi-simple non-simple group. Then $G$ is isomorphic to one of the following:

[TABLE]

Proof.

We may assume that $G$ biregularly (and faithfully) acts on $X$. Let $Y:=X/{\operatorname{z}}(G)$ and $G_{1}:=G/{\operatorname{z}}(G)$. Then $Y$ is a three-dimensional rationally connected variety acted on by a finite simple group $G_{1}$. Then according to [Pro12] $G_{1}$ belongs to the following list:

[TABLE]

Since $G_{1}$ is perfect, there exists the* universal covering group* $\tilde{G}_{1}$, that is, a central extension of $G_{1}$ such that for any other extension $\hat{G}_{1}$ there is a unique homomorphism $\tilde{G}_{1}\to\hat{G}_{1}$ of central extensions (see e.g. [Kar93, §11.7, Theorem 7.4]). The kernel of $\tilde{G}_{1}\to G_{1}$ is the Schur multiplier ${\operatorname{M}}(G_{1})=H^{2}(G_{1},\mathbb{C}^{*})$ of $G_{1}$. Thus $G$ is uniquely (up to isomorphism) determined by $G/{\operatorname{z}}(G)$ and the homomorphism ${\operatorname{M}}(G_{1})\to{\operatorname{z}}(G)$. It is known that ${\operatorname{M}}(G_{1})\simeq{\boldsymbol{\mu}}_{2}$ in all the cases (2.1.2) except for ${\mathfrak{A}}_{6}$ and ${\mathfrak{A}}_{7}$ where the Schur multiplier is isomorphic to ${\boldsymbol{\mu}}_{6}$, and ${\operatorname{PSL}}_{2}(8)$ where the Schur multiplier is trivial (see [Kar93, §12.3, Theorem 3.2, §16.3, Theorem 3.2] and [CCN*+*85]). This gives us the list (2.1.1). ∎

2.1.3 Remark**.**

We do not assert that all the possibilities (2.1.1) occur. Using the technique developed in the works [Pro12], [PS16c], [PS16b], [PS16a] it should be possible to obtain a complete classification of actions of quasi-simple groups on rationally connected threefolds. However, this is much more difficult problem.

2.1.4 Corollary**.**

Let $G$ be a group satisfying the conditions of 2.1. Then $G$ has an irreducible faithful representation. The minimal dimension of such a representation $V$ is given by the following table [CCN*+*85]:

[TABLE]

2.2 Construction**.**

Let $G$ be a finite quasi-simple non-simple group having a faithful irreducible representation $V$. Let $\psi:V\dashrightarrow X$ be a compression with $\dim X={\operatorname{ed}}(G)$. Applying an equivariant resolution of singularities (see [AW97]) we may assume that $X$ is smooth (and projective). Furthermore, consider the compactification $\mathbb{P}:=\mathbb{P}(V\oplus\mathbb{C})$ and let $X\overset{\varphi}{\longleftarrow}Y\overset{f}{\longrightarrow}\mathbb{P}\supset V$ be an equivariant resolution of $\psi$, where $f$ is a birational morphism and $Y$ is smooth and projective. We also may assume that $f$ is passed through the blowup $\tilde{f}:\tilde{\mathbb{P}}\to\mathbb{P}$ of $0\in V\subset\mathbb{P}$. Let $\tilde{E}\subset\tilde{V}$ be the $\tilde{f}$-exceptional divisor and let $E\subset Y$ be its proper transform, and let $B:=\varphi(E)$. Thus we have the following $G$-equivariant diagram:

[TABLE]

The action of ${\operatorname{z}}(G)$ on $\tilde{E}\simeq\mathbb{P}(V)$ and on $E$ is trivial because $V$ is an irreducible representation. Hence, $G/{\operatorname{z}}(G)$ faithfully acts on $E$. By assumption the action of ${\operatorname{z}}(G)$ on $X$ is faithful. Hence $B\neq X$ and so $B$ is a rationally connected variety of dimension $<{\operatorname{ed}}(G)$.

2.3 Proposition**.**

Let $G$ be a finite quasi-simple non-simple group having a faithful irreducible representation. Then $G/{\operatorname{z}}(G)$ acts faithfully of a rationally connected variety of dimension $<{\operatorname{ed}}(G)$.

Proof.

Let $V$ be a faithful irreducible representation of $G$ and let $\psi:V\dashrightarrow X$ be a compression with $\dim X={\operatorname{ed}}(G)$. Apply the construction 2.2. Assume that $G$ has a fixed point $P\in X$. Then $G$ has faithful representation on the tangent space $T_{P,X}$. Let $T_{P,X}=\oplus T_{i}$ be the decomposition in irreducible components. At least one of them, say $T_{1}$ is non-trivial. Then $G/{\operatorname{z}}(G)$ faithfully acts on $\mathbb{P}(T_{1})$, where $\mathbb{P}(T_{1})<\dim X={\operatorname{ed}}(G)$. Thus we may assume that $G$ has no fixed points on $X$. By the construction 2.2 the variety $B$ is rationally connected and $\dim B<{\operatorname{ed}}(G)$. Since $G$ has no fixed points on $B$ and the group $G/{\operatorname{z}}(G)$ is simple, its action on $B$ must be effective. ∎

Comparing the list (2.1.1) with Theorem 3.1 we obtain the following.

2.4 Corollary**.**

Let $G$ be a finite quasi-simple non-simple group with ${\operatorname{ed}}(G)\leq 3$. Then for $G$ we have one of the following possibilities:

[TABLE]

Now we consider the possibilities of (2.4.1) case by case.

2.5 Lemma**.**

${\operatorname{ed}}(2.{\mathfrak{A}}_{5})=2$* and ${\operatorname{ed}}(3.{\mathfrak{A}}_{6})=3$.*

Proof.

Let us prove, for example, the second equality. Since $3.{\mathfrak{A}}_{6}$ has a faithful there-dimensional representation, ${\operatorname{ed}}(3.{\mathfrak{A}}_{6})\leq 3$. On the other hand, ${\mathfrak{A}}_{6}$ cannot effectively act on a rational curve. Hence, by Proposition 2.3  ${\operatorname{ed}}(3.{\mathfrak{A}}_{6})\geq 3$. ∎

2.5.1 Lemma**.**

Let $G$ be a quasi-simple non-simple group. Assume that $G\not\simeq 2.{\mathfrak{A}}_{5},\,3.{\mathfrak{A}}_{6}$. Assume also that $G$ contains a subgroup $\bar{H}$ such that

1. (i)

$\bar{H}$* is not abelian but its image $H\subset G/{\operatorname{z}}(G)$ is abelian,* 2. (ii)

for any action of $G/{\operatorname{z}}(G)$ on a rational projective surface the subgroup $H\subset G/{\operatorname{z}}(G)$ has a fixed point.

Then ${\operatorname{ed}}(G)\geq 4$.

Proof.

Since $\bar{H}/({\operatorname{z}}(G)\cap\bar{H})=H$, we have ${\operatorname{z}}(G)\cap\bar{H}\supset[\bar{H},\bar{H}]$ and $[\bar{H},\bar{H}]\neq\{1\}$ (because $\bar{H}$ is not abelian). Assume that ${\operatorname{ed}}(G)=3$. Apply construction 2.2. From the list (2.1.1) one can see that $G/{\operatorname{z}}(G)$ cannot faithfully act on a rational curve and by Corollary 2.1.4 $G$ has no fixed points on $X$. Hence, $B$ is a (rational) surface. By (ii) the group $\bar{H}$ has a fixed point, say $P$, on $B\subset X$. There is an invariant decomposition $T_{P,X}=T_{P,B}\oplus T_{1}$, where $\dim T_{1}=1$. The action of $[\bar{H},\bar{H}]$ on $T_{P,B}$ and $T_{1}$ is trivial. Hence it is trivial on $T_{P,X}$ and $X$, a contradiction. ∎

2.6 Proposition**.**

${\operatorname{ed}}({\operatorname{SL}}_{2}(7))=4$.

2.6.1 Lemma**.**

Let $S$ be a smooth projective rational surface admitting the action of ${\operatorname{PSL}}_{2}(7)$. Let $H\subset{\operatorname{PSL}}_{2}(7)$ be a subgroup isomorphic to ${\boldsymbol{\mu}}_{2}\times{\boldsymbol{\mu}}_{2}$. Then $H$ has a fixed point on $S$.

Proof.

Since $H$ is abelian, according to [KS00] it is sufficient to show the existence of a fixed point on some birational model of $S$. By Theorem 3.1 we may assume that $S$ is either $\mathbb{P}^{2}$ or some special del Pezzo surface of degree $2$ (see 3.1(iii)). In the former case, $\mathbb{P}^{2}=\mathbb{P}(W)$, where $W$ is a three-dimensional irreducible representation of ${\operatorname{PSL}}_{2}(7)$. Then the abelian group $H\simeq{\boldsymbol{\mu}}_{2}\times{\boldsymbol{\mu}}_{2}$ has a fixed point on $\mathbb{P}^{2}=\mathbb{P}(W)$. Thus we assume that $S$ is a del Pezzo surface of degree $2$.

Let $\upalpha\in H$ be an element of order $2$. First assume that $\upalpha$ has a curve $C$ of fixed points. The image $\pi(C)$ under the anticanonical double cover $\pi:S\to\mathbb{P}^{2}$ must be a line (because the action on $\mathbb{P}^{2}$ is linear). Let $\upalpha^{\prime}\in H$, $\upalpha^{\prime}\neq\upalpha$ be another element of order $2$. Then $\upalpha^{\prime}(C)$ is also a curve $C$ of $\upalpha$-fixed points and $\pi(\upalpha^{\prime}(C))$ is also a line. Hence, $\pi(\upalpha^{\prime}(C))=\pi(C)$ and $\pi^{-1}(\pi(C))$ contains $\upalpha^{\prime}(C)$ and $C$. Since $\pi^{-1}(\pi(C))\sim-K_{S}$ and the fixed point locus of $\upalpha$ is smooth, we have $\upalpha^{\prime}(C)=C=\pi^{-1}(\pi(C))\sim-K_{S}$ and it is an ample divisor. Note that all the elements of order $2$ are conjugate in ${\operatorname{PSL}}_{2}(7)$. Hence $\upalpha^{\prime}$ also has a curve of fixed points, say $C^{\prime}$, and $C^{\prime}\sim-K_{S}$. Then the intersection points $C\cap C^{\prime}$ are fixed by $H=\langle\upalpha,\,\upalpha^{\prime}\rangle$.

Thus we may assume that any element $\upalpha\in H$ of order $2$ has only isolated fixed points. The holomorphic Lefschetz fixed point formula shows that the number of these fixed points equals $4\chi({\mathscr{O}}_{S})=4$. Then by the topological Lefschetz fixed point formula

[TABLE]

Since $\dim H^{2}(S,\mathbb{C})=8$ and all the eigenvalues of $\upalpha^{*}$ equal $\pm 1$, its determinant must be equal to $-1$ and so we have a non-trivial character of the group ${\operatorname{PSL}}_{2}(7)$. This contradicts the the fact that ${\operatorname{PSL}}_{2}(7)$ is simple. ∎

Proof of Proposition 2.6.

Consider the subgroup $\bar{H}\subset{\operatorname{SL}}_{2}(7)$ generated by the matrices

[TABLE]

Let $H$ be its image in ${\operatorname{PSL}}_{2}(7)$. It is easy to check that $A^{2}=B^{2}=-I$ and $[A,B]=-I$. Hence $\bar{H}$ is isomorphic to the quaternion group ${\operatorname{Q}}_{8}$ and $H\simeq{\operatorname{Q}}_{8}/{\operatorname{z}}({\operatorname{Q}}_{8})\simeq{\boldsymbol{\mu}}_{2}\times{\boldsymbol{\mu}}_{2}$. By Lemma 2.6.1 the group $\bar{H}$ has a fixed point on $B\subset X$. Hence we can apply Lemma 2.5.1. ∎

2.7 Proposition**.**

${\operatorname{ed}}(2.{\mathfrak{A}}_{6})=4$, ${\operatorname{ed}}(6.{\mathfrak{A}}_{6})\geq 4$.

Proof.

We are going to apply Lemma 2.5.1. Let $S$ be a projective rational surface acted by $G/{\operatorname{z}}(G)={\mathfrak{A}}_{6}$. By Theorem 3.1 we may assume that $S\simeq\mathbb{P}^{2}$. The group $2.{\mathfrak{A}}_{6}$ is isomorphic to ${\operatorname{SL}}_{2}(9)$ (see [CCN*+*85]). As in the proof of Proposition 2.6 take the subgroup $\bar{H}\subset{\operatorname{SL}}_{2}(9)$ generated by the matrices

[TABLE]

and apply Lemma 2.5.1.

In the case $G=6.{\mathfrak{A}}_{6}$, let $Z\subset{\operatorname{z}}(G)$ be the subgroup of order $3$. Then $6.{\mathfrak{A}}_{6}/Z\simeq 2.{\mathfrak{A}}_{6}\simeq{\operatorname{SL}}_{2}(9)$. Let $\hat{H}$ be the inverse image of the subgroup of ${\operatorname{SL}}_{2}(9)$ generated by $A$ and $B$ from (2.7.1) and let $\bar{H}\subset\hat{H}$ be the Sylow $2$-subgroup. Then, as above, $\bar{H}\simeq{\operatorname{Q}}_{8}$ and we can apply Lemma 2.5.1. ∎

2.7.2 Remark**.**

Since $6.{\mathfrak{A}}_{6}$ has a six-dimensional faithful representation, one has ${\operatorname{ed}}(6.{\mathfrak{A}}_{6})\leq 6$. However, we are not able to compute the precise value.

3 Appendix: simple subgroups in the plane Cremona group

The following theorem can be easily extracted from the classification [DI09]. For convenience of the reader we provide a relatively short and self-contained proof.

3.1 Theorem** ([DI09]).**

Let $G\subset\operatorname{Cr}_{2}(\mathbb{C})$ be a finite simple subgroup. Then the embedding $G\subset\operatorname{Cr}_{2}(\mathbb{C})$ is induced by one of the following actions:

1. (i)

${\mathfrak{A}}_{5}$, ${\operatorname{PSL}}_{2}(7)$, or ${\mathfrak{A}}_{6}$ acting on $\mathbb{P}^{2}$, 2. (ii)

${\mathfrak{A}}_{5}$* acting on the del Pezzo surface of degree $5$;* 3. (iii)

${\operatorname{PSL}}_{2}(7)$* acting on some special del Pezzo surface of degree $2$ which can be realized as a double cover of $\mathbb{P}^{2}$ branched in the Klein quartic curve;* 4. (iv)

${\mathfrak{A}}_{5}$* acting on $\mathbb{P}^{1}\times\mathbb{P}^{1}$ through the first factor.*

3.2 Remark**.**

It is known (see [DI09, §8] and [Che14, §B]) that the above listed actions are not conjugate in $\operatorname{Cr}_{2}(\mathbb{C})$.

Proof.

Applying the standard arguments (see e.g. [DI09, §3]) we may assume that $G$ faithfully acts on a smooth projective rational surface $X$ which is either a del Pezzo surface or an equivariant conic bundle. Moreover, one has $\operatorname{rk}\operatorname{Pic}(X)^{G}=1$ (resp. $\operatorname{rk}\operatorname{Pic}(X)^{G}=1$) in the del Pezzo (resp. conic bundle) case.

First, consider the case where $X$ has an equivariant conic bundle structure $\pi:X\to B\simeq\mathbb{P}^{1}$. Since the group $G$ is simple, it acts faithfully either on the base $B$ or on the general fiber. Hence $G$ is embeddable to ${\operatorname{PGL}}_{2}(\mathbb{C})$. This is possible only if $G\simeq{\mathfrak{A}}_{5}$. We claim that $\pi$ is a $\mathbb{P}^{1}$-bundle. Assume that $\pi$ has a degenerate fiber $F$. Then its components $F^{\prime},F^{\prime\prime}\subset F$ must be switched by an element $\upalpha\in G$ of order $2$. The intersection point $P=F^{\prime}\cap F^{\prime\prime}$ is fixed by $\upalpha$ and the action of $\upalpha$ on the tangent space $T_{P,X}$ is diagonalizable. Since the tangent directions to $F^{\prime}$ and $F^{\prime\prime}$ are interchanged, the action of $\upalpha$ on $T_{P,X}$ has the form ${\operatorname{diag}}(1,-1)$. This means that $\upalpha$ has a curve $C$ of fixed points passing through $P$. Clearly, $C$ dominates $B$ and so the action of $\upalpha$ on $B$ is trivial. All elements of order $2$ in $G\simeq{\mathfrak{A}}_{5}$ are conjugate and generate $G$. Hence $G$ acts on $B$ trivially. But then the point $P$ must be fixed by $G$ and $G$ has a faithful two-dimensional representation on $T_{P,X}$, a contradiction.

Thus, $\pi$ is a $\mathbb{P}^{1}$-bundle and $X$ is a Hirzebruch surface $\mathbb{F}_{n}$. Let $F_{1}$ be a fiber and let $F_{1},\dots,F_{r}$ be its orbit. By making elementary transformations in $F_{1},\dots,F_{r}$ one obtains an equivariant birational maps $\mathbb{F}_{n}\dashrightarrow\mathbb{F}_{n+r}$ and (only if $n\geq r$) $\mathbb{F}_{n}\dashrightarrow\mathbb{F}_{n-r}$. There are such orbits for $r=12$, $20$, $30$, and $60$. Using this trick we may replace $\mathbb{F}_{n}$ with $\mathbb{F}_{n^{\prime}}$, where $n^{\prime}=0$ or $1$. If $n^{\prime}=1$, then contracting the negative section we obtain the action on $\mathbb{P}^{2}$ with a fixed point. This is impossible for $G\simeq{\mathfrak{A}}_{5}$. Hence, we may assume that $X\simeq\mathbb{P}^{1}\times\mathbb{P}^{1}$. Additional elementary transformations allow to trivialize the action on the second factor. This is the case (iv). See [Che14, §B] for details.

From now on we assume that $X$ is a del Pezzo surface***More generally, actions of simple groups on del Pezzo surfaces with log terminal singularities were studied in [Bel15]. with $\operatorname{rk}\operatorname{Pic}(X)^{G}=1$. We consider the possibilities according to the degree $d=K_{X}^{2}$.

Case $d=1$. This case cannot occur, as $|-K_{X}|$ has one base point $P$, and $G$ has to act on $T_{P,X}$ effectively. Hence $G\subset{\operatorname{GL}}(T_{P,X})$. However, there are no simple finite subgroups in ${\operatorname{GL}}_{2}(\mathbb{C})$, a contradiction.

Case $d=2$. Then the anticanonical map $X\to\mathbb{P}^{2}$ is a double cover whose branch divisor $B\subset\mathbb{P}^{2}$ is a smooth quartic. The action of $G$ in $X$ descends to $\mathbb{P}^{2}$ so that $B$ is $G$-stable. Therefore, $G\subset\operatorname{Aut}(B)$. According to the Hurwitz bound $|G|\leq 168$. Moreover, $\operatorname{Aut}(B)$ contains no elements of order $5$. Then the only possibility is $G\simeq{\operatorname{PSL}}_{2}(7)$ and $B=\{x_{1}^{3}x_{2}+x_{2}^{3}x_{3}+x_{3}^{3}x_{1}=0\}$. We get the case (iii).

Case $d=3$. Then $X$ is a cubic surface in $\mathbb{P}^{3}$. The action of $G$ on the lattice $\Lambda:=K_{X}^{\perp}\subset\operatorname{Pic}(X)$ is faithful. Hence our group $G$ has a representation on the vector space $\Lambda/2\Lambda=(\mathbf{F}_{2})^{6}$ over the field $\mathbf{F}_{2}$. The intersection form induces an even quadratic form on $\Lambda$ and, therefore, it induces a quadratic form

[TABLE]

Take a standard basis $\mathbf{h},\mathbf{e}_{1},\dots,\mathbf{e}_{6}$ in $\operatorname{Pic}(X)$ with $(\mathbf{h},\mathbf{h})=1$, $(\mathbf{h},\mathbf{e}_{i})=0$, $(\mathbf{e}_{i},\mathbf{e}_{j})=-\delta_{j}^{i}$. Then using the basis $\mathbf{e}_{1}-\mathbf{e}_{2}$, $\mathbf{e}_{2}-\mathbf{e}_{3}$, $\mathbf{e}_{4}-\mathbf{e}_{5}$, $\mathbf{e}_{5}-\mathbf{e}_{6}$, $\mathbf{h}-\mathbf{e}_{1}-\mathbf{e}_{2}-\mathbf{e}_{3}$, $\mathbf{h}-\mathbf{e}_{4}-\mathbf{e}_{5}-\mathbf{e}_{6}$ of $\Lambda$ we can write $q(x)$ in the following form:

[TABLE]

Then it is easy to see that the Arf invariant of $q(x)$ equals $1$. The group preserves the intersection form and the quadratic form $q(x)$. Therefore, there is a natural embedding $G\hookrightarrow{\operatorname{O}}_{6}(\mathbf{F}_{2})^{-}$. Since $G$ is simple, $G\subset[{\operatorname{O}}_{6}(\mathbf{F}_{2})^{-},{\operatorname{O}}_{6}(\mathbf{F}_{2})^{-}]$. It is known that $[{\operatorname{O}}_{6}(\mathbf{F}_{2})^{-},{\operatorname{O}}_{6}(\mathbf{F}_{2})^{-}]\simeq{\operatorname{PSp}}_{4}(3)$ (see e.g. [CCN*+*85]). Moreover, $G$ is isomorphic to one of the following groups: ${\operatorname{PSp}}_{4}(3)$, ${\mathfrak{A}}_{6}$ or ${\mathfrak{A}}_{5}$. On the other hand, $G$ faithfully acts on $H^{0}(X,-K_{X})\simeq\mathbb{C}^{4}$. Then the only possibility is $G\simeq{\mathfrak{A}}_{5}$. The defining equation $\psi(z)=0$ of $X\subset\mathbb{P}^{3}$ is a cubic invariant on $H^{0}(X,-K_{X})$. Hence the representation on $H^{0}(X,-K_{X})$ is the standard irreducible representation of ${\mathfrak{A}}_{5}$. Then in a suitable basis we have $\psi=z_{1}^{3}+\cdots+z_{4}^{3}-(z_{1}+\dots+z_{4})^{3}$. The fixed point locus of an element $\upalpha\in G$ of order two is a union of a line and three isolated points. By the topological Lefschetz fixed point formula the action of $\upalpha$ on $\Lambda$ is diagonalizable as follows: $\upalpha={\operatorname{diag}}(1,1,1,1,-1,-1)$. This implies that the representation of $G$ on $\Lambda$ is the sum of the irreducible four-dimensional representation and the trivial one. This contradicts the minimality assumption $\operatorname{rk}\operatorname{Pic}(X)^{G}=1$.

Case $d=4$. Then $X=X_{2\cdot 2}=Q^{\prime}\cap Q^{\prime\prime}\subset\mathbb{P}^{4}$ is an intersection of two quadrics. The group $G$ acts on the pencil of quadrics $\langle Q^{\prime},Q^{\prime\prime}\rangle\simeq\mathbb{P}^{1}$ leaving invariant the subset of five singular elements. It is easy to see that in this case the action on $\langle Q^{\prime},Q^{\prime\prime}\rangle$ must be trivial. Hence, $G$ fixes vertices $P_{1},\dots,P_{5}$ of five $G$-stable quadratic cones $Q_{i}\in\langle Q^{\prime},Q^{\prime\prime}\rangle$. Since these points $P_{1},\dots,P_{5}$ generate $\mathbb{P}^{5}$, $G$ must be abelian, a contradiction.

Case $d=5$. A del Pezzo surface of degree $5$ is unique up to isomorphism. Consider the (faithful) action of $G$ on the space $\operatorname{Pic}(X)\otimes\mathbb{C}$ and on the orthogonal complement $K_{X}^{\perp}\subset\operatorname{Pic}(X)\otimes\mathbb{C}$. The intersection form induces a non-degenerate quadratic form on $K_{X}^{\perp}$. Hence $G$ faithfully acts on a two-dimensional quadric in $\mathbb{P}^{3}$. Then the only possibility is $G\simeq{\mathfrak{A}}_{5}$. One can see that $\operatorname{Aut}(X)$ isomorphic to the symmetric group $\mathfrak{S}_{5}$ and so a del Pezzo surface of degree $5$ admits an action of ${\mathfrak{A}}_{5}$. We get the case (ii).

Case $6\leq d\leq 8$. Then the action of $G$ on $\operatorname{Pic}(X)\simeq\mathbb{Z}^{10-d}$ must be trivial. This contradicts the minimality assumption $\operatorname{rk}\operatorname{Pic}(X)^{G}=1$.

Case $d=9$. Then $X=\mathbb{P}^{2}$. So $G\subset{\operatorname{PGL}}_{3}(\mathbb{C})$, and by the classification of finite subgroups in ${\operatorname{PGL}}_{2}(\mathbb{C})$ we get the case (i).

Theorem 3.1 is proved. ∎

3.3 Theorem** ([DI09], [Tsy13]).**

Let $G\subset\operatorname{Cr}_{2}(\mathbb{C})$ be a finite quasi-simple non-simple subgroup. Then $G\simeq 2.{\mathfrak{A}}_{5}$.

Proof.

As in the proof of Theorem 3.1 we may assume that $G$ faithfully acts on a smooth projective rational surface $X$ which is either a del Pezzo surface with $\operatorname{rk}\operatorname{Pic}(X)^{G}=1$ or an equivariant conic bundle with $\operatorname{rk}\operatorname{Pic}(X)^{G}=2$. Let $\bar{G}:=G/{\operatorname{z}}(G)$. Then $\bar{G}$ is a simple group acting on a rational surface $X/{\operatorname{z}}(G)$. Hence $\bar{G}$ is embeddable to $\operatorname{Cr}_{2}(\mathbb{C})$ and by Theorem 3.1 we have $\bar{G}\simeq{\mathfrak{A}}_{5}$, ${\mathfrak{A}}_{6}$, or ${\operatorname{PSL}}_{2}(7)$. Therefore, as in the proof of Proposition 2.1 we have one of the following possibilities: $G\simeq 2.{\mathfrak{A}}_{5}$, ${\operatorname{SL}}_{2}(7)$, or $n.{\mathfrak{A}}_{6}$ for $n=2$, $3$, or $6$. If $X$ has an equivariant conic bundle structure $\pi:X\to B\simeq\mathbb{P}^{1}$, then $G$ non-trivially acts either on the base $B$ or on the general fiber. This is possible only if $G\simeq 2.{\mathfrak{A}}_{5}$.

Assume that $X$ is a del Pezzo surface with $\operatorname{rk}\operatorname{Pic}(X)^{G}=1$. Let $Z\subset{\operatorname{z}}(G)$ be a cyclic subgroup of prime order $p$ and let $\pi:X\to Y:=X/Z$ be the quotient. The surface $Y$ is rational and $\bar{G}:=G/Z$ faithfully acts on $Y$.

First, consider the case where $Z$ has only isolated fixed points. If $p=2$, then by the holomorphic Lefschetz formula the number of fixed points equals $4$. These points cannot be permuted by $G$, so they are fixed by $G$. Similarly, in the case $p=3$ denote by $n_{0}$ (resp., $n_{1}$, $n_{2}$) the number of fixed points with action of type $\frac{1}{3}(1,-1)$ (resp., $\frac{1}{3}(1,1)$, $\frac{1}{3}(-1,-1)$). Then again by the holomorphic Lefschetz formula $n_{1}=n_{2}$, $n_{0}+n_{1}=3$. Hence there are at most three points of each type and so, as above, all these points are fixed by $G$. Since the groups ${\operatorname{SL}}_{2}(7)$ and $n.{\mathfrak{A}}_{6}$ cannot act faithfully on the tangent space to a fixed point, the only possibility is $G\simeq 2.{\mathfrak{A}}_{5}$.

Now consider the case where the fixed point locus $X^{Z}$ of $Z$ is one-dimensional. Let $C$ be the union of all the curves in $X^{Z}$. Note that $C$ is smooth because it is a part of the fixed point locus. Clearly, $C$ is $G$-invariant. Then the class of $C$ must be proportional to $-K_{X}$ in $\operatorname{Pic}(X)$. Hence $C$ is ample and connected. Since $C$ is smooth, it is irreducible. The group $G/Z$ non-trivially acts on $C$. Hence $C$ cannot be an elliptic curve. Moreover, if $C$ is rational, then $G/Z\simeq{\mathfrak{A}}_{5}$ and we are done. Thus we can write $C\sim-aK_{X}$ with $a>1$. If $-K_{X}$ is very ample, then $C$ is contained in a hyperplane section and $a=1$, a contradiction. Thus it remains to consider only two possibilities: $K_{X}^{2}=1$ and $2$. If $K_{X}^{2}=1$, then the anticanonical linear system has a unique base point, say $O$. Since, the representation of $G$ in the tangent space $T_{O,X}$ is faithful, the only possibility is $G\simeq 2.{\mathfrak{A}}_{5}$. Finally assume that $K_{X}^{2}=2$. Then the anticanonical map is a double cover $\Phi_{|-K_{X}|}:X\to\mathbb{P}^{2}$. The action of $Z$ of $\mathbb{P}^{2}$ must be trivial (otherwise $a=1$). Hence $p=2$, $Z$ is generated by the Geiser involution $\gamma$, and $C$ is the ramification curve of $\Phi_{|-K_{X}|}$. On the other hand, there exists the following homomorphism

[TABLE]

where $\lambda(\gamma)=-1$. Since our group $G$ is perfect, $\gamma\notin G$, a contradiction. ∎

3.4 Remark**.**

In the same manner one can describe the actions of $G=2.{\mathfrak{A}}_{5}$ on rational surfaces, i.e. the embeddings $2.{\mathfrak{A}}_{5}\hookrightarrow\operatorname{Cr}_{2}(\mathbb{C})$), see [Tsy13] for details.

Acknowledgements. I would like to thank the referee for careful reading and numerous helpful comments and suggestions.