# An Automorphism group of a rational surface: Not too big not too small

Kyounghee Kim  Department of Mathematics
Florida State University
Tallahassee, FL 32308<EMAIL_ADDRESS>
###### Abstract.

This article concerns the realization problem of subgroups of Coxeter groups.
We construct a subgroup $G$ of the Coxeter group $W_{15}$ such that $G$ is
realized as automorphism groups of a rational surface $X$. The automorphism
group $Aut(X)$ has six generators with positive entropy and $G\cong
Aut(X)^{*}\cong D_{3}\rtimes\mathbb{Z}$. We also show that there is an element
$\omega$ in $W_{14}$ is not realizable.

###### Key words and phrases:

an Automorphism Group of a Rational Surface, Coxeter Groups, Positive entropy

###### 2020 Mathematics Subject Classification:

14E07,14J26,37F99

##  1\. Introduction

Let $X$ be a rational surface over $\mathbb{C}$. If $f:X\to X$ is an
automorphism on $X$, due to Nagata [19] one can identify the induced linear
action $f^{*}:Pic(X)\to Pic(X)$ with an element of a Coxeter group $W_{X}$,
which is generated by special reflections in the group of isometry on a
lattice of signature $(1,n)$. With this identification, let us denote
$Aut(X)^{*}\subset W_{X}$ be the group of induced actions $f^{*}:Pic(X)\to
Pic(X)$ of $f\in Aut(X)$.

We say an element $\omega$ of a Coxeter group is realized by a rational
surface automorphism $f:X\to X$ if $\omega=f^{*}$ and also a subgroup $G$ of a
Coxeter group is realized by $Aut(X)$ if there is a rational surface $X$ such
that $G=Aut(X)^{*}$.

It is natural to ask which element $\omega\in W_{X}$ can be realized by an
automorphism on a rational surface. If $\omega$ is realized by a rational
automorphism $f:X\to X$, then the cyclic subgroup $\langle\omega\rangle$ is
realized by a group of automorphisms on a rational surface $X$. Thus, a more
interesting question asked in [14, 17, 7] must be identifying a subgroup of a
Coxeter group realized by an automorphism group of a rational surface. A
rational surface $X$ with an infinite $Aut(X)^{*}$ is quite special. To have
such a rational surface, one needs to carefully pick a finite set of points
$P=\\{p_{1},\dots,p_{n}\\},n\geq 9$ such that $X$ is obtained by successive
blowup of points in $P$. If $Aut(X)^{*}$ has a finite index in $W_{X}$, then
the automorphism group $Aut(X)$ is large, and we say the base locus $P$ of $X$
is Cremona Special. Cantat and Dolgachev [8] showed that there are only two
possible Cremona Special sets in $\mathbf{P}^{2}$ over $\mathbb{C}$: either
(1) $P$ is a Halphen set if $|P|=9$ or (2) $P$ is a Coble set if $|P|=10$.

If $f\in Aut(X)$ for some rational surface $X$ with infinite order, then the
cyclic group $\langle f^{*}\rangle=\\{(f^{*})^{n},n\in\mathbb{Z}\\}\subset
W_{X}$. Assuming that $f\in Aut(X)$ preserves an anticanonical cuspidal cubic,
McMullen [17] showed that the cyclic group $\langle f\rangle$ is a finite
index subgroup of the full group $Aut(X)$.

For a given element of the Coxeter group, finding a rational surface with an
automorphism with infinite order is quite delicate. Several methods (mostly
with an assumption of the existence of an invariant curve) to construct such
rational surfaces have been studied in [3, 4, 5, 10, 17, 22]. Yet there is
more to be discovered. Due to the rareness of such rational surfaces, it is
interesting to have an infinite non-cyclic automorphism group.

In this article, we present a rational surface whose automorphism group is
equivalent to the direct product of $D_{3}$ and $\mathbb{Z}$. Furthermore, we
show that the morphism from $Aut(X)$ to $Aut(X)^{*}\subset W_{X}$ is, in fact,
an isomorphism. Another interesting example of a realizable non-cyclic
subgroup of the Coxeter group was constructed by Blanc [5] using involutions
with a smooth cubic curve of fixed points. In Blanc’s construction, the
morphism from $Aut(X)$ to $W_{X}$ is not one-to-one.

######  Theorem A.

There is a rational surface $X$ such that its automorphism group
$\text{Aut}(X)$ is generated by $6$ distinct quadratic automoprhisms with
positive entropy and $\text{Aut}(X)\cong D_{3}\rtimes\mathbb{Z}$.

######  Theorem B.

There is a non-cyclic subgroup $G\subset W_{15}$ generated by six elements
such that $G$ is realized by automorphisms on a rational surface $X$ in
Theorem A, that is, $G\cong Aut(X)^{*}$. Furthermore $G$ is isomorphic to the
automorphism group $Aut(X)$

$G\cong\text{Aut}\,(X)$

The essentially same construction as in Theorem A and Theorem B would give
more infinite subgroups with multiple generators, which are realized by
automorphisms on rational surfaces.

######  Remark.

For each odd integer $n\geq 5$ such that $n\not\equiv 0$ (mod $3$), there is a
subgroup $G$ in the Coxeter group $W_{3n}$ such that $G$ is realized by
automorphisms on a rational surface $X_{3n}$ and
$G\cong\text{Aut}(X_{3n})\cong D_{3}\rtimes\mathbb{Z}$. The essentially same
construction as in Theorem 7.14 gives the corresponding rational surface
$X_{3n}$ and six quadratic automorphisms generating $\text{Aut}(X_{3n})$.

######  Remark.

For each odd integer $n\geq 5$ such that $n\equiv 0$ (mod $3$), there is a
subgroup $G$ in the Coxeter group $W_{3n}$ such that $G$ is realized by
automorphisms on a rational surface $X_{3n}$ and
$G\cong\text{Aut}(X_{3n})\cong(\mathbb{Z}/2\mathbb{Z})^{2}\rtimes\mathbb{Z}$.
In this case, the automorphism group is generated by $3$ quadratic
automorphisms. Because $n$ is divisible by $3$, an automorphism on $X_{3n}$
can not cyclically permute three irreducible components of the invariant
cubic.

In our construction, the base points are in the non-singular part of the anti-
canonical curve, which consists of three lines joining at a single point. The
non-cyclic normal subgroup in the Automorphism group corresponds to rotations
of three lines in the anti-canonical curve. If $n$ is even, we can not swap
two lines. (We will explain the reason in Lemma 5.3.) However, we can cyclicly
permute three lines.

######  Remark.

If $n$ is even and is not divisible by $3$, then there is a subgroup $G$ in
the Coxeter group $W_{3n}$ such that $G$ is realized by automorphisms on a
rational surface $X_{3n}$ and
$G\cong\text{Aut}(X_{3n})\cong\mathbb{Z}/3\mathbb{Z}\rtimes\mathbb{Z}$.

With iterated blowups, one can create any number of fixed elements under the
induced action on the Picard Group. Thus, the number of $p$-cycle is not an
obstruction for realization. If an automorphism $f:X\to X$ realizes an element
$\omega$ with a period $p$-cycle of $-1$ curves, then $f$ must be a lift of a
birational map with a periodic point with period $p$. Using this idea, the
following Theorem shows that not every element in $\cup_{n\geq 10}W_{n}$ is
realizable.

######  Theorem C.

There is an element $\omega\in W_{14}$ such that $\omega$ is not realizable
and thus the cyclic subgroup $\langle\omega\rangle\subset W_{14}$ is not
realizable as $Aut(X)^{*}$ for a rational surface $X$.

The unrealizable element $\omega\in W_{14}$ in Theorem C is not an essential
element in the sense that $\omega$ is conjugate into a subgroup $W_{10}\subset
W_{14}$. If a non-essential element is realized by a rational surface
automorphism $f:X\to X$, then the surface $X$ is non-minimal. Thus, the more
interesting/ correct question should be the following: “Is every essential
element in $\cup_{n\geq 10}W_{n}$ realizable?”

This article is organized as follows: Section 2 has basic background about
birational maps focusing on quadratic maps. Section 3 discusses the connection
to rational surface automorphisms and proves Theorem C. In Section 4, we
discuss properties of Salem numbers, which are useful to identify the
Automorphism group of the rational surface we construct. Sections 5 \- 6
provide the construction of the rational surface in Theorem A-B and describe
six quadratic automorphisms constructed using the Diller’s method in [10]. In
Section 7, we discuss properties of automorphisms on the rational surface
constructed in the previous Sections and prove Theorem A. The proof of Theorem
B is given in Section 8.

### Acknowledgement

The author thanks E. Hironaka for interesting discussions regarding Salem
numbers. The author also thanks the referee for a thorough reading and
thoughtful comments and suggestions.

### Statements and Declarations

This material is partly based upon work supported by the National Science
Foundation under DMS-1928930 while the author participated in a program hosted
by the Mathematical Sciences Research Institute in Berkeley, California,
during Spring 2022 semester.

##  2\. Quadratic birational maps

Due to Nagata [18, 19, 15], any automorphism $f:X\to X$ on a rational surface
$X$ with positive entropy is a lift of a birational map on $\mathbf{P}^{2}$.
There are three basic Cremona transformations:

$\displaystyle J_{3}:[x_{1}:x_{2}:x_{3}]\to[x_{2}x_{3}:x_{1}x_{3}:x_{1}x_{2}]$
$\displaystyle J_{2}:[x_{1}:x_{2}:x_{3}]\to[x_{1}x_{3}:x_{2}x_{3}:x_{1}^{2}]$
$\displaystyle
J_{1}:[x_{1}:x_{2}:x_{3}]\to[x_{1}^{2}:x_{1}x_{2}:x_{2}^{2}-x_{1}x_{3}].$

Each transformation $J_{i}$ is a quadratic involution with an exceptional
locus consisting of $i$ distinct irreducible components. It is well-known [9]
that a birational map on $\mathbf{P}^{2}$ can be written as a composition of
automorphisms on $\mathbf{P}^{2}$ and three involutions $J_{i},i=1,2,3$. In
fact, any quadratic birational map can be obtained from one of three
involutions, $J_{i}$ by pre- and post- composing automorphisms on
$\mathbf{P}^{2}$. For a birational map $f$ with the indeterminacy locus
$\mathcal{I}(f)$ and any curve $V$, let us use $f(V)$ for the set-theoretic
strict transformation $f(V\setminus\mathcal{I}(f))$.

###  2.1. The Cremona Involution

The Cremona involution $J_{3}$ has three distinct exceptional lines
$E_{i}:=\\{x_{i}=0\\}$ and three points of indeterminacy
$e_{1}:=[1:0:0],e_{2}:=[0:1:0]$ and $e_{3}:=[0:0:1]$. The map $J_{3}$ is
locally bi-holomorphic on the complement of the union of exceptional lines and
maps $E_{i}\setminus\\{e_{1},e_{2},e_{3}\\}$ to a point $e_{i}$ for each
$i=1,2,3$

$J_{3}:E_{i}\setminus\\{e_{1},e_{2},e_{3}\\}\mapsto e_{i}\qquad\text{ for all
}i=1,2,3.$

Since the map $J_{3}$ preserves a pencil of lines joining $e_{i}$ and a point
in $E_{i}$, the Cremona involution $J_{3}$ lifts to an automorphism on a
rational surface $X$ obtained by blowing up $\mathbf{P}^{2}$ along a set of
three points $\\{e_{1},e_{2},e_{3}\\}$. Let $\mathcal{E}_{i}$ be the
irreducible exceptional curve over $e_{i},i=1,2,3$ and $[\mathcal{E}_{0}]$ be
the class of the strict transformation of a generic line in $\mathbf{P}^{2}$.
The ordered set
$\\{[\mathcal{E}_{0}],[\mathcal{E}_{1}],[\mathcal{E}_{2}],[\mathcal{E}_{3}]\\}$
forms a geometric basis (See [13] for the definition.) for $Pic(X)$, in other
words $\\{[\mathcal{E}_{0}],[\mathcal{E}_{i}],i=1,\dots n\\}$ is a basis such
that

$[\mathcal{E}_{0}]\cdot[\mathcal{E}_{0}]=1,\ \
[\mathcal{E}_{i}]\cdot[\mathcal{E}_{i}]=-1\ \text{for }i=1,\dots n,\ \ \
\text{and}\ \ \ [\mathcal{E}_{i}]\cdot[\mathcal{E}_{j}]=0\ \text{for }i\neq
j.$

A geometric basis for a rational surface $X$ gives an isomorphism between
$Pic(X)$ and a lattice $\mathbb{Z}^{1,n}$ with signature $(1,n)$. (See Section
3.)

###  2.2. Two more Cremona Involutions

The involution $J_{2}$ has two distinct exceptional lines $E_{1},E_{3}$ and
two points of indeterminacy $e_{2},e_{3}$.

$J_{2}:E_{1}\setminus\\{e_{2},e_{3}\\}\mapsto e_{2},\qquad\text{and}\qquad
E_{3}\setminus\\{e_{2},e_{3}\\}\mapsto e_{3}.$

Unlike $J_{3}$, this involution $J_{2}$ requires iterated blowups to lift to
an automorphism: first blowing up two points $e_{2},e_{3}$, then a point
$e_{2}^{\prime}=\mathcal{E}_{2}\cap\\{x_{3}=0\\}$ where $\mathcal{E}_{2}$ is
the irreducible exceptional curve over $e_{2}$. Let $X$ be the resulting
rational surface then the ordered set
$\\{[\mathcal{E}_{0}],[\mathcal{E}^{\prime}_{2}],[\mathcal{E}_{2}+\mathcal{E}^{\prime}_{2}],[\mathcal{E}_{3}]\\}$
forms a geometric basis for $Pic(X)$ where $\mathcal{E}^{\prime}_{2}$ is the
exceptional curve over $e_{2}^{\prime}$.

The exceptional locus of the involution $J_{1}$ consists of one irreducible
line $E_{1}$ with multiplicity $3$, that is, one needs to blow up a point
three times to resolve the indeterminacy.

$J_{1}:E_{1}\setminus\\{e_{3}\\}\mapsto e_{3}.$

To resolve the indeterminacy, one needs to blow up the point $e_{3}$,
$e_{3}^{\prime}:=\mathcal{E}_{3}\cap\\{x_{1}=0\\}$, and
$e_{3}^{\prime\prime}:=\mathcal{E}_{3}^{\prime}\cap\\{x_{1}^{2}=x_{2}\\}$
where $\mathcal{E}_{3}$, $\mathcal{E}_{3}^{\prime}$ and
$\mathcal{E}_{3}^{\prime\prime}$ are irreducible exceptional curves over
$e_{3}$, $e_{3}^{\prime}$ and $e_{3}^{\prime\prime}$ (respectively). Again,
let $X$ be the resulting blowup space. In this case a geometric basis for
$Pic(X)$ is given by
$\\{[\mathcal{E}_{0}],[\mathcal{E}^{\prime\prime}_{3}],[\mathcal{E}^{\prime}_{3}+\mathcal{E}^{\prime\prime}_{3}],[\mathcal{E}_{3}+\mathcal{E}^{\prime}_{3}+\mathcal{E}^{\prime\prime}_{3}]\\}$.

######  Remark 2.1.

With the given geometric basis above, the lifts of all three involutions
$J_{i},i=1,2,3$ share the same linear action on $Pic(X)$.

###  2.3. Quadratic Birational Maps

It is known [9] that every quadratic birational map on $\mathbf{P}^{2}$ can be
written as $T^{-}\circ J_{i}\circ(T^{+})^{-1},i=1,2,3$ with
$T^{\pm}\in\text{Aut}\,(\mathbf{P}^{2})$. In fact, two Cremona involutions
$J_{1}$ and $J_{2}$ can be written as compositions of linear maps and $J_{3}$.
We say a birational map $\check{f}$ is basic if $\check{f}=T^{-}\circ
J_{3}\circ(T^{+})^{-1}$. A basic birational map $\check{f}=T^{-}\circ
J_{3}\circ(T^{+})^{-1}$ has three distinct exceptional lines $E^{+}_{i}$ and
the indeterminacy locus $\mathcal{I}=\\{p_{i}^{+},i=1,2,3\\}$ where

$E_{i}^{+}=T^{+}\\{x_{i}=0\\},\qquad\text{and}\qquad p_{i}^{+}=T^{-}e_{i}.$

Similarly, we have three exceptional lines $E_{i}^{-}$ and three points of
indeterminacy $p_{i}^{-}$ for $f^{-1}$. Let us denote $\check{f}(V)$ the
strict transform, $\check{f}(V\setminus\mathcal{I}(\check{f}))$, of $V$ under
$\check{f}$. We have

$\check{f}:E_{i}^{+}\mapsto
p_{i}^{-},\qquad\text{and}\qquad\check{f}^{-1}:E_{i}^{-}\mapsto
p_{i}^{+}\qquad\text{for }i=1,2,3.$

Note that any non-basic quadratic birational map is obtained from either
$J_{2}$ or $J_{1}$, and thus the number of irreducible components of its
exceptional locus is strictly less than $3$.

###  2.4. Dynamical Degree

Suppose $g:X\dasharrow X$ is a birational map on a compact complex manifold.
In this case, the Picard group $Pic(X)$ is identical to the Néron-Severi group
$NS(X)$. There is a natural induced linear map $g^{*}:Pic(X)\to Pic(X)$. The
dynamical degree $\lambda(g)$ is birational invariant and given by

$\lambda(g):=\lim_{n\to\infty}||(g^{n})^{*}||^{1/n}.$

We say $g$ is algebraically stable if the induced linear action
$g^{*}:Pic(X)\to Pic(X)$ satisfies the following condition $(g^{*})^{n}\ =\
(g^{n})^{*}$ for all $\ n\geq 1$. If $g$ is algebraically stable, the
dynamical degree is given by the spectral radius of the linear action $g^{*}$.

In general, a birational map
$\check{f}:\mathbf{P}^{2}\dasharrow\mathbf{P}^{2}$ does not satisfy the
algebraic stability condition. However, Diller and Favre [11] showed that one
can always construct an algebraically stable modification, that is, there is a
blowup $X=B\ell_{P}\mathbf{P}^{2}$ along a finite set $P$ of points such that
the lift $f:X\dasharrow X$ is algebraic stable and the dynamical degree of
$\check{f}$ is given by $\lambda(\check{f})=\lambda(f)$, the spectral radius
of the induced linear action $f^{*}$ on $Pic(X)$. Blanc and Cantat [6] showed
that a birational map $\check{f}$ lifts to an automorphism if and only if the
dynamical degree is a Salem number.

###  2.5. Orbit Data

Suppose $\check{f}$ is basic. For each exceptional line $E_{i}^{+}$, we define
$n_{i}\in\mathbb{N}\cup\\{\infty\\}$ by

$\left\\{\begin{aligned} &n_{i}\ =\
\min\\{n:\text{dim}\check{f}^{n}(E_{i}^{+})\lneq\text{dim}\check{f}^{n+1}(E_{i}^{+})\\},\\\
&n_{i}\ =\ \infty\qquad\text{ if }\ \ \text{dim}\check{f}^{n}(E_{i}^{+})=0\ \
\text{for all }n\geq 1\end{aligned}\right.$

If $n_{i}<\infty$, the last point of the orbit
$\check{f}^{n_{i}}E_{i}^{+}\in\mathcal{I}^{+}=\\{p_{1}^{+},p_{2}^{+},p_{3}^{+}\\}$.
Thus we can also define a permutation $\sigma\in\Sigma_{3}$ such that

$\check{f}^{n_{i}}E_{i}^{+}=p_{\sigma(i)}^{+}.$

We call this numerical information $n_{1},n_{2},n_{3}$ and a permutation
$\sigma$ the orbit data of the basic birational map $f$.

######  Remark 2.2.

For a non-basic quadratic birational map, we count the orbit length with
multiplicity with the identity permutation. For example, if
$\check{f}=T^{-}\circ J_{1}\circ(T^{+})^{-1}$, we let
$E_{1}^{+}=T^{+}\\{x_{1}=0\\}$ and the orbit data are $n_{1},n_{1},n_{1}$ and
the identity permutation.

######  Lemma 2.3.

If $\check{f}:\mathbf{P}^{2}\dasharrow\mathbf{P}^{2}$ is a basic quadratic
birational map, then so is the inverse $\check{f}^{-1}$. Furthermore If
$\check{f}$ has orbit data $n_{1},n_{2},n_{3}$ with a permutation
$\sigma\in\Sigma_{3}$, then the orbit data for $\check{f}^{-1}$ are given by
$n_{\sigma^{-1}(1)},n_{\sigma^{-1}(2)},n_{\sigma^{-1}(3)}$ with $\sigma^{-1}$.

###### Proof.

Since every quadratic map $\check{f}$ is given by a composition $T^{-}\circ
J_{i}\circ(T^{+})^{-1}$ for some $i=1,2,3$, it is obvious. ∎

###  2.6. Lifting to an Automorphism

Suppose the orbit data of a basic birational map $\check{f}$ consists of three
positive integers $n_{1},n_{2},n_{3}<\infty$ and let
$P_{i}=\\{p_{i,j}:=\check{f}^{j}E_{i}^{+},\ 1\leq j\leq n_{i}\\}$. Then
$\check{f}$ lifts to an automorphism on a blowup

$\pi:X=X_{3}\xrightarrow{\pi_{3}}X_{2}\xrightarrow{\pi_{2}}X_{1}\xrightarrow{\pi_{1}}X_{0}=\mathbf{P}^{2}$

where $\pi_{j}:X_{j}\to X_{j-1}$ is a blowup of a set of points
$P_{i_{j}}\subset X_{j-1}$ with $n_{i_{1}}\leq n_{i_{2}}\leq n_{i_{3}}$.

If a quadratic birational map $\check{f}=T^{-}\circ J_{2}\circ(T^{+})^{-1}$,
then the orbit lengths are $n_{1},n_{1},n_{3}$. Let
$P_{i}=\\{p_{i,j}:=\check{f}^{j}E_{i}^{+},\ 1\leq j\leq n_{i}\\}$ for $i=1,3$.
When both $n_{1},n_{3}<\infty$, $\check{f}$ lifts to an automorphism on a
rational surface

$\pi:X=X_{2}\xrightarrow{\pi_{2}}X_{1}\xrightarrow{\pi_{1}}X_{0}=\mathbf{P}^{2}$

where

  * •

$\pi_{1}$ is a successive blowup of two set of points $P_{1}$ and $P_{3}$ in
the increasing order of $n_{i}$, and

  * •

$\pi_{2}$ is a blowup of a set of points
$P_{2}:=\\{p_{2,j}:=\check{f}_{1}^{j}E_{1}^{+},\ 1\leq j\leq n_{1}\\}\subset
X_{1}$ where $f_{1}:X_{1}\to X_{1}$ is the induced biratioanl map.

Lastly, if a quadratic birational map $\check{f}=T^{-}\circ
J_{1}\circ(T^{+})^{-1}$ with $n_{1}<\infty$, then the rational surface
$\pi:X\to\mathbf{P}^{2}$ is constructed by successive blowups along a set of
poitns $P_{1}=\\{p_{1,j}:=\check{f}^{j}E_{1}^{+},\ 1\leq j\leq n_{1}\\}$,
$P_{2}:=\\{p_{2,j}:=\check{f}_{1}^{j}E_{1}^{+},\ 1\leq j\leq n_{1}\\}\subset
X_{1}$, and $P_{3}:=\\{p_{3,j}:=\check{f}_{2}^{j}E_{1}^{+},\ 1\leq j\leq
n_{1}\\}\subset X_{2}$.

In all three cases, a geometric basis of $Pic(X)$ is given by the classes of
total transformations of exceptional curves over a point in $P=\cup P_{i}$
together with the class of a generic line.

##  3\. Coxeter group $W_{n}$

Let $\mathbb{Z}^{1,n}$ be the unimodular lattice of signature $(1,n)$ with a
basis $(\mathbf{e}_{0},\mathbf{e}_{1},\dots,\mathbf{e}_{n})$ such that

$\mathbf{e}_{0}^{2}=1,\quad\mathbf{e}_{i}^{2}=-1\quad\text{ for }i\geq
1,\quad\text{ and }\quad\mathbf{e}_{i}\cdot\mathbf{e}_{j}=0\quad\text{for
}i\neq j.$

Suppose $n\geq 3$. Let

$\kappa_{n}=-3\mathbf{e}_{0}+\mathbf{e}_{1}+\mathbf{e}_{2}+\cdots+\mathbf{e}_{n},\qquad\text{and}$
$\alpha_{0}=\mathbf{e}_{0}-\mathbf{e}_{1}-\mathbf{e}_{2}-\mathbf{e}_{3},\quad\alpha_{i}=\mathbf{e}_{i}-\mathbf{e}_{i+1}\
\ \text{ for }i\geq 1.$

The sublattice $L_{n}$ with a basis
$(\alpha_{0},\alpha_{1},\dots,\alpha_{n-1})$ is the orthogonal complement of
the vector $\kappa_{n}$. Since $\alpha_{i}^{2}=-2$ for all $i\geq 0$, we can
define the reflection $\mathbf{s}_{i}$ through the vector $\alpha_{i}$ by

$\mathbf{s}_{i}:x\mapsto x+(x\cdot\alpha_{i})\alpha_{i}.$

These reflections generate the Coxeter group $W_{n}\subset O(L_{n})$ with
respect to the graph $E_{n}$ in Figure 1. Let $m_{i,i}=1$, and $m_{i,j}=3$ if
$\mathbf{s}_{i}$ and $\mathbf{s}_{j}$ are joined by an edge, then we have

$W_{n}=\langle\mathbf{s}_{0},\mathbf{s}_{1},\dots,\mathbf{s}_{n-1}\,|\,(\mathbf{s}_{i}\mathbf{s}_{j})^{m_{i,j}}=1\rangle.$

The subgroup $S_{n}$ generated by
$\\{\mathbf{s}_{1},\mathbf{s}_{2},\dots,\mathbf{s}_{n-1}\\}\subset W_{n}$ is
isomorphic to a permutation group on $n$ elements. Each element in $S_{n}$
permutes the basis elements $\mathbf{e}_{1},\dots,\mathbf{e}_{n}$. Let
$\mathbf{s}\in S_{n}$ such that $\mathbf{s}$ interchanges
$\mathbf{e}_{1}\leftrightarrow\mathbf{e}_{i}$,
$\mathbf{e}_{2}\leftrightarrow\mathbf{e}_{j}$,
$\mathbf{e}_{3}\leftrightarrow\mathbf{e}_{k}$ for some distinct indcies
$i,j,k$, and fixes everything else. Then we have

$\mathbf{s}_{ijk}=\mathbf{s}^{-1}\circ\mathbf{s}_{0}\circ\mathbf{s}\in W_{n}.$

This element $\mathbf{s}_{ijk}\in W_{n}$ acts on $\mathbb{Z}^{1,n}$ as a
reflection through the vector
$\mathbf{e}_{0}-\mathbf{e}_{i}-\mathbf{e}_{j}-\mathbf{e}_{k}$.

Figure 1. Coxeter Graph $E_{n}$

###  3.1. The smallest spectral radius in $W_{n}$.

Let $\rho(\omega)$ be the spectral radius of this linear map $\omega\in
W_{n}$. We say $\omega\in W_{n}$ is essential if $\omega$ is not conjugate
into a subgroup $W_{I}$ generated by a subset $\\{\mathbf{s}_{i},i\in
I\\}\subset\\{\mathbf{s}_{0},\mathbf{s}_{1},\dots,\mathbf{s}_{n-1}\\}$ for an
index set $I\subset\\{0,1,\dots,n-1\\}$:

$W_{I}\ =\ \langle s_{i},i\in I|(s_{j}s_{k})^{m_{j,k}}=1\rangle$

######  Theorem 3.1.

[16] If $\omega\in W_{n}$ is essential, then the spectral radius of $\omega$
is greater than or equal to the spectral radius of the Coxeter element
$\omega_{n}\in W_{n}$

$\rho(\omega)\geq\rho(\omega_{n}).$

The spectral radius of the Coxeter element of $W_{10}$ is given by Lehmer’s
number, $\rho(\omega_{n})$ is strictly increasing in $n$, and the spectral
radius of any element in $W_{n},n\leq 9$ is equal to $1$. Thus we have

######  Theorem 3.2.

[16] Suppose $\omega\in\bigcup_{n\geq 10}W_{n}$. If the spectral radius
$\rho(\omega)>1$ then $\rho(\omega)\geq\lambda_{\star}\approx 1.17628$ where
$\lambda_{\star}$ is the largest real root of the Lehmer’s polynomial

$t^{10}+t^{9}-t^{7}-t^{6}-t^{5}-t^{4}-t^{3}+t+1.$

###  3.2. Connection to Rational surface automorphisms

A rational surface $X$ is given by successive blowups $\pi:X\to\mathbf{P}^{2}$
with a factorization

$\pi:X=X_{n}\xrightarrow{\pi_{n}}X_{n-1}\xrightarrow{\pi_{n-1}}X_{n-2}\rightarrow\cdots\rightarrow
X_{1}\xrightarrow{\pi_{1}}X_{0}=\mathbf{P}^{2}$

where $\pi_{i}:X_{i}\to X_{i-1}$ is a blowup of a point $x_{i}\in X_{i-1}$.
Let $E_{i}\subset X_{i}$ denote an exceptional curve lying over $x_{i}\in
X_{i-1}$ and let $\mathcal{E}_{i}=(\pi_{i+1}\circ\dots\pi_{n})^{-1}E_{i}$ be
the total transform of $E_{i}$. If $x_{i}$ is an infinitely near point in
$\mathbf{P}^{2}$, i.e.
$(\pi_{1}\circ\dots\circ\pi_{i-1})\,x_{i}=(\pi_{1}\circ\dots\circ\pi_{j-1})\,x_{j}$
for some $j\lneq i$, then $\mathcal{E}_{i}$ is a reducible curve. Let
$e_{0}=\pi^{*}([\ell])$ where $\ell$ is a generic line in $\mathbf{P}^{2}$ and
$e_{i}=[\mathcal{E}_{i}]$ be the cohomology class of a curve $\mathcal{E}_{i}$
for $i=1,2,\dots,n$. These classes $e_{0},e_{1},\dots,e_{n}$ give a geometric
basis of the Picard lattice of $X$. With the intersection product, one can
check that

$e_{0}\cdot e_{0}=1,\quad e_{i}\cdot e_{i}=-1,\text{ for
}i=1,\dots,n,\quad\text{and }e_{i}\cdot e_{j}=0,\text{ for }i\neq j.$

Thus, there is a natural isomorphism

$\phi:\mathbb{Z}^{1,n}\to Pic(X)$

such that

$\phi(\mathbf{e}_{i})=e_{i},\quad i=0,\dots n$

Due to Nagata [19, 14], if $f:X\to X$ is an automorphism then there is an
element $\omega\in W_{n}$ such that for a linear action $f^{*}$ on $Pic(X)$

(3.1) $\phi\circ\omega=f^{*}\circ\phi.$

In this case, we say $\omega$ is realized by a rational surface automorphism
$f:X\to X$. Let us identify $f^{*}=\omega$ through the isomorphism $\phi$.

###  3.3. Not-realizable element

Let us consider $\omega\in W_{16}$ given by $\omega=\omega_{1}\cdot\omega_{2}$
where

(3.2) $\omega_{1}=\mathbf{s}_{0}\mathbf{s}_{1}\cdots\mathbf{s}_{9}\in W_{n\geq
10},\quad\text{and}\quad\omega_{2}=\mathbf{s}_{11}\mathbf{s}_{12}\mathbf{s}_{13}\in
W_{14}.$

The action $\omega$ on $\mathbb{Z}^{1,14}$ is given by

(3.3) $\displaystyle\omega\ :$ $\displaystyle\mathbf{e}_{0}\mapsto
2\mathbf{e}_{0}-\mathbf{e}_{1}-\mathbf{e}_{2}-\mathbf{e}_{3},$
$\displaystyle\mathbf{e}_{1}\mapsto\mathbf{e}_{0}-\mathbf{e}_{1}-\mathbf{e}_{3},\
\ \ \mathbf{e}_{2}\mapsto\mathbf{e}_{0}-\mathbf{e}_{1}-\mathbf{e}_{2},$
$\displaystyle\mathbf{e}_{3}\mapsto\mathbf{e}_{4}\mapsto\cdots\mapsto\mathbf{e}_{10}\mapsto\mathbf{e}_{0}-\mathbf{e}_{2}-\mathbf{e}_{3},$
$\displaystyle\mathbf{e}_{11}\mapsto\mathbf{e}_{12}\mapsto\mathbf{e}_{13}\mapsto\mathbf{e}_{14}\mapsto\mathbf{e}_{11}.$

######  Lemma 3.3.

If $\omega$ is realized by an automorphism $f:X\to X$ on a rational surface
$X$, then there is a rational surface $X_{1}$ obtained by blowing down four
$(-1)$-curves $\mathcal{E}_{i},i=11,\dots,14$ with $e_{i}=[\mathcal{E}_{i}]$
to $4$ distinct points in $X_{1}$ and the induced map $f_{1}:X_{1}\to X_{1}$
realizes $\omega_{1}\in W_{10}$ and there is a period $4$ cycle of points.

###### Proof.

The identification between $f^{*}$ and $\omega$ comes from a choice of a
geometric basis. Since $f$ is quadratic, due to [13, Theorem 5.2] we may
choose the geometric basis as defined in Section 2.6. Except for the class of
generic line, all other geometric basis elements correspond to $(-1)$ curves.
Thus, we can blowdown four $(-1)$ curves $\mathcal{E}_{i},i=11,\dots,14$ to
$4$ distinct points. ∎

######  Lemma 3.4.

If a rational surface automorphism $f_{1}:X_{1}\to X_{1}$ realizes
$\omega_{1}\in W_{10}$, then $f_{1}$ does not have a (non-fixed, no period 2)
period $4$ cycle.

###### Proof.

Suppose there is a curve $C$ of fixed points for $f^{n}_{1}$ for some $n\geq
2$. The characteristic polynomial of $\omega_{1}$ is the product of $(t-1)$
and the irreducible Lehmer’s Polynomial. Since no root of Lehmer’s Polynomial
is a root of unity and the $[C]$ an eigenvector of $\omega_{1}^{4}$ with the
eigenvalue $1$, it follows that the curve $C$ is, in fact, fixed. Thus all
periodic (non-fixed) points of $f_{1}$ are isolated. Since we have
$(f_{1}^{*})^{n}=(f_{1}^{n})^{*}$ and
$\text{Tr}\,\omega_{1}^{2}=\text{Tr}\,\omega_{1}^{4}$ , from Lefschetz’s fixed
point formula we see that there is no period $4$ cycle. ∎

###### Proof of Theorem C.

Theorem C is the immediate consequence of the previous two Lemmas. ∎

McMullen in [17] pointed out that $\mathbf{s}_{0}\in W_{8}$ can not be
realized with a rational surface automorphism $f:X\to X$ if $X$ is a blowup of
$\mathbf{P}^{2}$ along a set of $8$ distinct points in $\mathbf{P}^{2}$. On
the other hand, one can perform iterated blowups over the fixed points. Let
$X_{1}$ be the blowup of $\mathbf{P}^{2}$ along the set of four points
$x_{1}=[1:0:0],x_{2}=[0:1:0],x_{3}=[0:0:1]$ and $x_{4}=[1:1:1]$ and let
$e_{i}$ be the class of exceptional curves over $x_{i}$. Then the Cremona
involution $J_{3}$ lifts to an automorphism $f_{1}:X_{1}\to X_{1}$ with
$f_{1}^{*}=\mathbf{s}_{0}\in W_{4}$, that is $f_{1}^{*}e_{4}=e_{4}$. Since the
exceptional curve over $x_{4}$ is an invariant curve for $f_{1}$, there is a
fixed point $x_{5}$ in the exceptional curve. Thus $f_{1}$ lift to another
automorphism on a rational surface realizing $\mathbf{s}_{0}\in W_{5}$. By
repeating this procedure, we can realize $\mathbf{s}_{0}\in W_{n}$ for all
$n\geq 4$.

##  4\. Salem Numbers

A Salem number $\tau$ is an algebraic integer such that the minimal polynomial
is a reciprocal polynomial of degree $\geq 4$ such that there are exactly two
real roots outside the unit circle and all other roots are (non-real) complex
with modulus $1$. If an element $\omega\in W_{n}$ of a Coxeter group has the
spectral radius $>1$, then the spectral radius is a Salem number. Since the
induced action $f^{*}|_{Pic(X)}$ of a rational surface automorphism $f:X\to X$
can be identified with an element in a Coxeter group, Salem numbers play an
essential role. Blanc and Cantat [6] showed that a birational map $\check{f}$
on $\mathbf{P}^{2}$ lifts to a rational surface automorphism with positive
entropy if and only if the dynamical degree $\delta(\check{f})$ is a Salem
number. In this section, we present the useful results on Salem numbers. The
following results of Salem are discussed and proved in the survey paper by
Smyth [21].

######  Proposition 4.1.

[20] If $\tau$ is a Salem number of degree $d$, then so is $\tau^{n}$ for all
$n\in\mathbb{N}$.

The following proposition by Salem gives a specific idea about the number
fields containing Salem numbers. The last two statements are specifically
useful for our results.

######  Proposition 4.2.

[20]

  1. (1)

A number field $K$ is of the form $\mathbb{Q}(\tau)$ for some Salem number
$\tau$ if and only if $K$ has a totally real subfield $\mathbb{Q}(\alpha)$ of
index $2$, and $K=\mathbb{Q}(\tau)$ with $\tau+\tau^{-1}=\alpha$, where
$\alpha>2$ is an irrational algebraic integer, all of whose conjugates not
equal to $\alpha$ lie in $(-2,2)$.

  2. (2)

Suppose that $\tau$ and $\tau^{\prime}$ are Salem numbers with
$\tau^{\prime}\in\mathbb{Q}(\tau)$. Then
$\mathbb{Q}(\tau^{\prime})=\mathbb{Q}(\tau)$ and, if $\tau^{\prime}>\tau$,
then $\tau^{\prime}/\tau$ is also a Salem number in $\mathbb{Q}(\tau)$.

  3. (3)

If $K=\mathbb{Q}(\tau)$ for some Salem number $\tau$, then there is a Salem
number $\tau_{1}\in K$ such that the set of Salem numbers in $K$ consists of
the power of $\tau_{1}$.

I would like to thank Eko Hironaka for helping me with the following
proposition.

######  Proposition 4.3.

Suppose $\alpha\neq\beta$ are two distinct Salem numbers. Then one and only
one of the the followings occurs

  1. (1)

there is a Salem number $\tau$ such that $\alpha=\tau^{m}$ and
$\beta=\tau^{n}$ for some $m,n\geq 1$

  2. (2)

the products of $\alpha,\beta$ and their reciprocals,
($\alpha\beta,\alpha/\beta,\beta/\alpha,$ and $1/(\alpha\beta)$) are not Salem
numbers.

###### Proof.

Suppose there is no Salem number $\tau$ such that $\alpha=\tau^{m}$ and
$\beta=\tau^{n}$ for some $m,n\geq 1$. Let $K$ be the splitting field of
$\alpha$ over $\mathbb{Q}$ and let $H=\text{Gal}(K/\mathbb{Q})$ be the Galois
group of $K$ over $\mathbb{Q}$. Then $\beta\not\in K$. Now consider the
smallest Galois extension $L$ over $\mathbb{Q}$ which contains both Salem
numbers $\alpha$ and $\beta$ and let $G=\text{Gal}(L/\mathbb{Q}$). Since
$K\subset L$ is a splitting field, by the fundamental Theorem of Galois
theory, we have a normal subgroup $N\lhd G$ such that

$N=\\{g\in G:g(a)=a,\forall a\in K\\}.$

Thus every element in $N$ fixes all Galois conjugate of $\alpha$ including
$\alpha$ itself. Since $\beta\not\in K$, there is $g\in N\lhd G$ such that
$g(\beta)\neq\beta$, in other words, $g(\beta)$ is a Galois conjugate of
$\beta$. Thus we have either $g(\beta)=1/\beta$ or $|g(\beta)|=1$. Since $g\in
N\lhd G$, we have $g(\alpha\beta)=\alpha g(\beta)$. If $|g(\beta)|=1$, then
$g(\alpha\beta)$ is a non-real complex number outside the unit circle and thus
$\alpha\beta$ has a Galois conjugate outside the unit circle. If
$g(\beta)=1/\beta$, then $\alpha\beta$ is conjugate to a real number
$\alpha/\beta\neq 1/(\alpha\beta)$. In either case, we see that $\alpha\beta$
is not Salem. Similarly we can conclude that $\alpha/\beta,\beta/\alpha,$ and
$1/(\alpha\beta)$ are not Salem numbers. ∎

##  5\. Quadratic rational surface automorphisms

Searching for basic birational maps on $\mathbf{P}^{2}$ with given orbit data
is a challenging problem since it involves solving systems of high-degree
algebraic equations with a large number of variables. With the extra condition
of the existence of an invariant cubic, McMullen [17] and Diller [10] provide
recipes to construct birational maps which lift to rational surface
automorphisms. For a given orbit data with a specified invariant cubic,
Diller’s Method gives an explicit formula for the corresponding birational
map. This method also works for the higher dimension [1]. In this article, we
are considering basic quadratic birational maps fixing three lines joining at
a single point with orbit data

(5.1) $n_{1}=n_{2}=n_{3}=n\geq 4,\qquad\text{and }\ \ \sigma\in S_{3}$

###  5.1. Quadratic birational map fixing three concurrent lines

Let
$L_{j}=\\{\gamma_{i}(t):t\in\mathbb{C}\cup\\{\infty\\}\\}\subset\mathbf{P}^{2}$
be the line given by the parametrizations

(5.2)
$\gamma_{1}(t)=[-t:1:1],\quad\gamma_{2}(t)=[t:1:0],\quad\text{and}\quad\gamma_{3}(t)=[t:0:1]$

Let $C=L_{1}\cup L_{2}\cup L_{3}$ be a reducible cubic consisting of three
lines joining at a single point $[1:0:0]$. Let us fix affine coordinates
$(t,i)=\gamma_{i}(t)$ for a point in the cubic $C$.

(5.3) $C=\\{(t,i):=\gamma_{i}(t)|t\in\mathbb{C}\cup\\{\infty\\},i=1,2,3\\}$

The given parametrization in (5.5) ensures that if three points on the cubic,
$p_{1}=(t_{1},1),p_{2}=(t_{2},2),$ and $p_{3}=(t_{3},3)$ are co-linear than
$t_{1}+t_{2}+t_{3}=0$. We say a birational map $\check{f}$ on $\mathbf{P}^{2}$
properly fixes $C$ in (5.3) if $\check{f}(C)=C$ and none of the points of
indeterminacy $p_{i}^{\pm}$ are the common intersection point $C_{sing}$ of
$C$.

Suppose $\check{f}:\mathbf{P}^{2}\dasharrow\mathbf{P}^{2}$ be a quadratic map
that properly fixes $C$. For a given irreducible curve
$V\subset\mathbf{P}^{2}$ of degree $k$, the degree of the image $\check{f}(V)$
is $2k-(m_{1}+m_{2}+m_{3})$ where $m_{i}$ is the vanishing order of the point
of indeterminacy $p_{i}^{+}$ in $V$. Since the invariant cubic $C$ consists of
three lines, each irreducible component $L_{i}$ should contain exactly one
point of indeterminacy. It follows that there are three distinct points of
indeterminacy, and thus we have

######  Lemma 5.1.

If a quadratic birational map
$\check{f}:\mathbf{P}^{2}\dasharrow\mathbf{P}^{2}$ properly fixes a concurrent
three lines $C$, then $\check{f}$ is basic.

We may assume that $p_{i}^{+}\in L_{i}\setminus\\{C_{sing}\\}$ for all
$i=1,2,3$. It follows that $\check{f}$ permutes three lines. Suppose
$\check{f}(L_{i})=L_{\tau(i)}$ for some permutation $\tau\in S_{3}$. Since the
exceptional line $E_{1}^{+}$ intersects $L_{j}$ at one point and since for
$j=2,3$, the indeterminant point $p_{j}^{+}$ is contained in both $E_{1}^{+}$
and $L_{j}$, we have non-indeterminate point $q_{1}\in L_{1}\cap E_{1}^{+}$.
Similarly we have $L_{i}\cap
E_{i}^{+}=\\{q_{i}\\}\not\in\mathcal{I}(\check{f})$ for all $i=1,2,3$, and
$p_{i}^{-}=\check{f}(E_{i}^{+})\in L_{\tau(i)}$ for all $i=1,2,3.$

######  Lemma 5.2.

Suppose $\check{f}$ is a basic quadratic map properly fixing $C=L_{1}\cup
L_{2}\cup L_{3}$. And suppose $\check{f}(L_{i})=L_{\tau(i)}$ for some $\tau\in
S_{3}$. Then after renaming it, if necessary, we have

$p_{i}^{+},q_{i}\in L_{i},\qquad\text{and }\quad p_{i}^{-}\in
L_{\tau(i)}\qquad\text{ for all }i=1,2,3$

where $q_{i}$ is the unique non-indeterminate intersection point $q_{i}\in
E^{+}_{i}\cap(C\setminus\mathcal{I})$.

Because of the previous Lemma 5.2, we have few restrictions on the orbit data
for a basic quadratic map $\check{f}$ properly fixing $C$.

######  Lemma 5.3.

Suppose $\check{f}$ is a basic quadratic map properly fixing $C=L_{1}\cup
L_{2}\cup L_{3}$ with orbit data $n_{1},n_{2},n_{3},\sigma\in S_{3}$. Suppose
$\check{f}(L_{i})=L_{\tau(i)}$ for some $\tau\in S_{3}$. If $n_{i}<\infty$ for
all $i$, then we have

  * •

if $\tau=Id$ is identity, then so is $\sigma$.

  * •

if $\tau$ permutes $i$ and $j$, then either(1) both $n_{i}$ and $n_{j}$ are
odd and $\sigma=\tau$, or (2) both $n_{i}$ and $n_{j}$ are even and
$\sigma=Id$.

  * •

if $\tau$ is a cyclic permutation, then either (1) $n_{i}\equiv 0$ (mod $3$)
for all $i$ and $\sigma=Id$, (2) $n_{i}\equiv 1$ (mod $3$) for all $i$ and
$\sigma=\tau$, or (3) $n_{i}\equiv 2$ (mod $3$) for all $i$ and
$\sigma=\tau^{2}$.

###### Proof.

Notice that for each $i=1,2,3$, we have $q_{i},p_{i}^{+}\in L_{i}$ and
$\check{f}^{j}(q_{i})=f^{j}(E_{i}^{+})$ for all $j\leq n_{i}$. Since
$\check{f}^{j}(q_{i})\in L_{\tau^{j}(i)}$ and
$\check{f}^{n_{i}}(q_{i})=p_{\sigma(i)}^{+}\in L_{\sigma(i)}$, we have

$\sigma(i)=\tau^{n_{i}}(i)\qquad\text{for all }i=1,2,3,$

and thus the statement of this Lemma follows. ∎

Suppose all orbit lengths of a basic quadratic map are equal to five,
$n_{1}=n_{2}=n_{3}=5$. Then all six permutations in $S_{3}$ are possible
candidates for basic quadratic birational maps properly fixing three
concurrent lines. The explicit formulas for the minimal polynomials (up to
cyclotomic factors) of the dynamical degree using orbit data are given in [2]:
With orbit lengths $n_{1},n_{2},n_{3}\in\mathbb{N}$, we have

  * •

a cyclic permutation :

$\chi(t)\ =\
(t-1)((t^{n_{1}}+1)(t^{n_{2}}+1)(t^{n_{3}}+1)+1)-(t^{n_{1}+n_{2}+n_{3}}-1),$

  * •

a transposition $(i\,j)$ with $\\{i,j,k\\}=\\{1,2,3\\}$:

$\chi(t)\ =\
(t-1)(t^{n_{k}}(t^{n_{i}}+1)(t^{n_{j}}+1)-t^{n_{i}}-t^{n_{j}}-2)-(t^{n_{i}+n_{j}}-1)(t^{n_{k}}-1),$

  * •

an identity permutation :

$\chi(t)\
=(t-2)t^{n_{1}+n_{2}+n_{3}}+(t^{n_{1}+n_{2}}+t^{n_{2}+n_{3}}+t^{n_{1}+n_{3}})-t(t^{n_{1}}+t^{n_{2}}+t^{n_{3}})+2t-1$

With these formulas, we see that all six cases share the same dynamical degree
given by the largest real root $\delta$ of a Salem polynomial $\chi(t)$ of
degree $4$

(5.4) $\chi_{5}(t)=t^{4}-2t^{3}+t^{2}-2t+1,\qquad\text{and }\quad\delta\approx
1.8832$

Furthermore, It is known that the dynamical degree is the largest real root of
a Salem polynomial, which is determined by orbit data. Also, it is known [2]
that if the dynamical degree is bigger than $1$ then the dynamical degree is
strictly increasing in orbit lengths. It follows that there are at most six
distinct orbit data for quadratic birational maps fixing concurrent lines with
the dynamical degree $\delta\approx 1.8832$ in (5.4).

###  5.2. Diller’s Method

To construct a basic map with the orbit data in (5.1) which properly fixes
three concurrent lines defined in (5.3), we apply the construction by Diller
in [10, 1] in which all possible invariant cubics were discussed. The original
Theorem in [10] assumes that a basic birational map properly fixes a cusp
cubic. The discussion for a basic birational map properly fixing three
concurrent lines is in Section $4$ in [10]. In [1], the same method is applied
to the basic maps in the higher dimensions. This Section only gives the
necessary components for constructing a basic birational map fixing three
concurrent lines. The following Theorem is slightly modified from the original
versions [1, Theorem 2.6] for the mappings properly fixing pre-determined
three concurrent lines.

######  Theorem 5.4.

[10, 1] Let $C$ be the union of three lines joining at the same point defined
in (5.3). Suppose $a\in\mathbf{C}^{*}$ and $t_{j}^{+}\in\mathbf{C}$,
$j=1,2,3$, are distinct paramenters satisfying
$t_{1}^{+}+t_{2}^{+}+t_{3}^{+}\neq 0$. Then there is a unique basic birational
map $\check{f}=T^{-}\circ J_{3}\circ(T^{+})^{-1}$, $b\in\mathbf{C}$, and a
permutation $\tau\in S_{3}$ such that

  * •

$\check{f}$ properly fixes $C$ with $\check{f}|_{C}$ is given by
$\check{f}|_{C}((t,i)=(at+b,\tau(i))$

  * •

$(t_{i}^{+},i)=T^{-}(e_{i})$

  * •

$b=\frac{1}{3}a\sum t_{i}^{+}$

Suppose $\check{f}=T^{-}\circ J_{3}\circ(T^{+})^{-1}$ is a basic birational
map properly fixing $C=L_{1}\cup L_{2}\cup L_{3}$ in (5.3) and
$\check{f}(L_{i})=L_{\tau(i)}$ for some permutation $\tau\in S_{3}$. With the
affine coordinates $(t,i)=\gamma_{i}(t)$ for $C$ and the parametrizations
$\gamma_{i}$ defined in (5.5), we have the restriction map on the cubic $C$
given by $f|_{C}:(t,i)\mapsto(at+b,\tau(i)),$ for some $a,b\in\mathbb{C}.$ By
conjugating with an automorphism on $\mathbf{P}^{2}$, we may assume that
$f|_{C}(1,i)=(1,\tau(i))$, that is,

$f|_{C}:(t,i)\mapsto(a(t-1)+1,\tau(i))\qquad\text{for some
}a\in\mathbb{C}\setminus\\{0\\}.$

Based on Lemma 5.2, let us suppose that $p_{i}^{+},q_{i}\in L_{i}$ for
$i=1,2,3$ and thus set

(5.5) $p_{i}^{+}=\gamma(t_{i}^{+},i),\qquad
p_{i}^{-}=\gamma(t_{i}^{-},\tau(i))\qquad\text{and }\qquad
q_{i}=\gamma(s_{i},i)=\gamma(f^{-1}|_{C}(t_{i}^{-},\tau(i))).$

Let us also suppose that the orbit data of $\check{f}$ is given by (5.1) with
appropriate permutation $\sigma$ with respect to the Lemma 5.3. Since each
$n_{i}=n$, we have

(5.6)
$(t_{\sigma(i)}^{+},\sigma(i))=\check{f}_{C}^{n-1}(t_{i}^{-},\tau(i))=(a^{n-1}(t_{i}^{-}-1)+1,\sigma(i))$

Since the exceptional line $E_{1}^{+}$ passes through $p_{2}^{+},p_{3}^{+}$
and $q_{1}$, we have $s_{1}+t_{2}^{+}+t_{3}^{+}=0$. Thus using (5.5) and
(5.6), we get

(5.7)
$a^{n+1}(t^{-}_{\sigma^{-1}(2)}+t^{-}_{\sigma^{-1}(3)}-2)=1-t_{1}^{-}-3a$

Similarly we have $s_{2}+t_{1}^{+}+t_{3}^{+}=0$ and
$s_{3}+t_{1}^{+}+t_{2}^{+}=0$. Thus we get

(5.8) $\displaystyle
a^{n+1}(t^{-}_{\sigma^{-1}(1)}+t^{-}_{\sigma^{-1}(3)}-2)=1-t_{2}^{-}-3a$
$\displaystyle
a^{n+1}(t^{-}_{\sigma^{-1}(1)}+t^{-}_{\sigma^{-1}(2)}-2)=1-t_{3}^{-}-3a$

On the other hand, a generic line $h$ will intersect each line $L_{i}$ at one
point, say $x_{i}=\gamma(r_{i},i)$ and $x_{1}+x_{2}+x_{3}=0$. Furthermore, the
degree of $\check{f}=2$ and thus $\\#|\check{f}(h)\cap C|=6$. Since $h$ also
intersects each exceptional line at a single point, we have
$p_{i}^{-}\in\check{f}(h)$ for all $i=1,2,3$. This gives another condition on
the parameters for critical images $\sum a(r_{i}-1)+1+\sum_{i}t_{i}^{-}=a(\sum
r_{i})+3(1-a)+\sum t_{i}^{-}=0$, i.e.

(5.9) $\sum_{i}t_{i}^{-}=3(a-1)$

And similarly we have $\sum_{i}t_{i}^{+}=3\,\frac{1-a}{a}$. Solving $4$
equations (5.7\- 5.9), we get

$a^{n+1}-2a^{n}+2a-1=0\quad\text{and}\quad t_{i}^{-}=\frac{-\,3a}{1+2a^{n}}\ \
\text{for all }i=1,2,3.$

######  Proposition 5.5.

Suppose $n\geq 4$. There exists a basic birational map $\check{f}$ properly
fixing three concurrent lines with orbit lengths $n,n,n$. This birational map
$\check{f}$ lifts to an automorphism and is given by the largest root of the
polynomial

(5.10) $\chi_{n}(t)=t^{n+1}-2t^{n}+2t-1$

###### Proof.

Let $\alpha$ is a non-cyclotomic root of $\chi_{n}(t)$ and let

(5.11) $t_{1}\ =\ \frac{-\,3\alpha}{1+2\alpha^{n}},\quad\text{and}\quad
t_{j}=\frac{-\,3\alpha^{j}}{1+2\alpha^{n}},j=2,3,\dots,n$

Then $3t_{n}\neq 0$ and thus by Theorem 5.4 we know there exists a basic
birational map $\check{f}$ such that $\check{f}|_{C}:(t,i)=(\alpha
t+(1-\alpha),\tau(i))$ for some $\tau\in S_{3}$ with $p_{j}^{+}=(t_{n}+1,j)\in
C$ for $j=1,2,3$. Since all three orbit lengths $n_{i}$ are finite, it follows
that $\check{f}$ lifts to an automorphism on a rational surface obtained by
blowing up the points in the orbits of exceptional lines
$\\{(t_{j}+1,\tau^{j}(i)),i=1,2,3,j=1,\dots,n\\}$. ∎

######  Theorem 5.6.

Let $C$ be a cubic curve with three lines joining at a single point. Suppose
$n\geq 4$ such that $n$ is not divisible by both $2$ and $3$. There are
exactly six quadratic birational maps $\check{f}_{\sigma},\sigma\in S_{3}$
properly fixing $C$ with orbit data $n,n,n$ and a permutation $\sigma$ such
that all six maps share the same base locus and the same dynamical degree
$\delta_{n}$ where $\delta_{n}$ is the largest real root of $\chi_{n}(t)$
given in (5.10).

###### Proof.

If a quadratic birational map $\check{f}$ properly fixes $C$, then by Lemma
5.1 we know that $\check{f}$ must be basic, that is $\check{f}=T^{-}\circ
J_{3}\circ(T^{+})^{-1}$ for some linear maps $T^{\pm}$. When $n$ is not
divisible by both $2$ and $3$, due to Lemma 5.3 one can choose the rotation
$\tau$ of invariant lines based on the permutation $\sigma$ in the orbit data
as follows:

  * •

if $\sigma=id$ or $(i,j)$ a transpose, then $\tau=\sigma$

  * •

if $n\equiv 1$ (mod $3$) and $\sigma$ is a cyclic permutation, then
$\tau=\sigma$

  * •

if $n\equiv 2$ (mod $3$) and $\sigma$ is a cyclic permutation, then
$\tau=\sigma^{2}$

For any $\tau\in S_{3}$ and $j\geq 1$,
$\\{\tau^{j}(i),i=1,2,3\\}=\\{1,2,3\\}$. Thus the base locus

$\\{(t_{j}+1,\tau^{j}(i)),i=1,2,3,j=1,\dots,n\\}=\\{(t_{j}+1,i),i=1,2,3,j=1,\dots,n\\}.$

Therefore this Theorem is the immediate consequence of Theorem 5.4 and
Proposition 5.5. ∎

######  Remark 5.7.

Let $\check{f}=T^{-}\circ J_{3}\circ(T^{+})^{-1}$ be a basic quadratic map in
the previous Proposition 5.5. Since the automorphism on $\mathbf{P}^{2}$ is
given by a $3\times 3$ non-singular matrix and
$p_{i}^{-}=\check{f}\,T^{+}(\\{x_{i}=0\\})=T^{-}e_{i}$, we see the each column
of $T^{-}$ is determined by $p_{i}^{-}$s. Similarly, $T^{+}$ is determined by
$p_{i}^{+}$. With the condition that $\check{f}$ fixes the singular point (the
common intersection point of three lines), one can get the explicit formula
for $\check{f}$. The computer codes written in SageMath and Mathematica are
available for download at the author’s website https://www.math.fsu.edu/
kim/publication.html.

##  6\. A rational surface $X$ with $15$ blowups

By Theorem 5.6, we see that if the common orbit length $n=5$, then all six
permutations can occur in the orbit data. Let us suppose that

$n_{1}=n_{2}=n_{3}=5.$

With the $t_{i}$’s defined in Proposition 5.5, let us define a set of $15$
distinct points in the invariant cubic $C$.

(6.1) $B=\\{p_{i,j}:=(1+t_{j},i)\in C:i=1,2,3,j=1,2,\dots,5\\}$

Let $X=B\ell_{B}(\mathbf{P}^{2})$ be a rational surface obtained by blowing up
$15$ distinct points in $B\subset\mathbf{P}^{2}$.

######  Theorem 6.1.

Let $C$ be a cubic defined in (5.3) and let $n=5$. Then for each $\sigma\in
S_{3}$, there is the unique basic birational map $\check{f}_{\sigma}$ properly
a cubic $C$ with orbit data $n,n,n$ and $\sigma$. Furthermore,
$\check{f}_{\sigma}$ lift to an automorphism $f_{\sigma}:X\to X$ on a rational
surface $X$ defined above.

###### Proof.

The existence and the uniqueness of such birational map $\check{f}_{\sigma}$
is clear from Proposition 5.5 and Theorem 5.4. From Lemma 5.3, we see that the
permutation $\tau$ of three lines must be equal to $\sigma^{-1}$. It follows
that the union of the orbit of exceptional lines is given by
$\\{(t_{j}+1,\sigma^{-j}(i)),i=1,2,3,j=1,\dots,5\\}$. Since $t_{j}$ does not
depend on $i$ and $\sigma\in S_{3}$ is a permutation, we see that

$\\{(t_{j}+1,\sigma^{-j}(i)),i=1,2,3,j=1,\dots,5\\}=\\{(t_{j}+1,i),i=1,2,3,j=1,\dots,5\\}\quad\text{for
all }\sigma\in S_{3}.$

And thus, for all $\sigma\in S_{3}$, $\check{f}_{\sigma}$ lifts to an
automorphism on $X=B\ell_{B}(\mathbf{P}^{2})$. ∎

Let $e_{0}$ be the pre-image of a generic line in $\mathbf{P}^{2}$ and $e_{i}$
be the class of the exceptional divisor over the point $(t_{j}+1,k)\in C$ with
$i=(k-1)5+j$. Let $\\{e_{0},e_{1},\dots,e_{15}\\}$ be an ordered basis of
$Pic(X)$

$Pic(X)=\langle e_{0},e_{1},\dots,e_{15}\rangle.$

Recall that $C$ given in (5.3) consists of three lines $L_{1},L_{2},$ and
$L_{3}$ joining at one point. Each line $L_{i}$ contains five centers of
blowups $(t_{j}+1,i),j=1,2,\dots,5$. Thus the classes of their strict
transformations are given by

(6.2) $[C]=3e_{0}-\sum_{i=1}^{15}e_{i},\qquad\text{ and
}\quad[L_{i}]=e_{0}-\sum_{j=1}^{5}e_{j+(i-1)5},\ \ i=1,2,3.$

Let us use $C$ for both the invariant cubic in $\mathbf{P}^{2}$ and its strict
transformation in $X$.

###  6.1. Six Quadratic Automorphisms

The coordinate functions of those six basic maps are given by polynomials in
$\mathbb{Z}(\alpha)$ where $\alpha$ is a root of the Salem factor

(6.3) $\varphi(t)=t^{4}-2t^{3}+t^{2}-2t+1$

of the polynomial $\chi_{5}(t)$ in (5.10). Let us set

$\mathit{s}\ =\ \frac{2\alpha^{5}+1}{\alpha^{5}-1},\qquad\text{and
}\qquad\mathit{r}\ =\ \frac{2\alpha^{5}+1}{2\alpha^{5}-3\alpha+1}.$

Let us define automorphisms on $\mathbf{P}^{2}$ as follows:

$\mathcal{S}=\frac{1}{3}\begin{bmatrix}\ 1&\ 1&\ 1\\\ \
\mathit{s}&-\mathit{s}&\ 0\\\ \ \mathit{s}&\
0&-\mathit{s}\end{bmatrix},\quad\mathcal{T}_{id}=\frac{1}{3}\begin{bmatrix}\
1&\ 1&\ 1\\\ -\mathit{r}&\ \mathit{r}&\ 0\\\ -\mathit{r}&\ 0&\
\mathit{r}\end{bmatrix},\quad\mathcal{T}_{(12)}=\frac{1}{3}\begin{bmatrix}\
1&\ 1&\ 1\\\ \ \mathit{r}&-\mathit{r}&\ 0\\\ \ 0&-\mathit{r}&\
\mathit{r}\end{bmatrix},$ $\mathcal{T}_{(13)}=\frac{1}{3}\begin{bmatrix}\ 1&\
1&\ 1\\\ \ 0&\ \mathit{r}&-\mathit{r}\\\ \ \mathit{r}&\
0&-\mathit{r}\end{bmatrix},\qquad\quad\mathcal{T}_{(23)}=\frac{1}{3}\begin{bmatrix}\
1&\ 1&\ 1\\\ -\mathit{r}&\ 0&\mathit{r}\\\ -\mathit{r}&\ \mathit{r}&\
0\end{bmatrix},$ $\mathcal{T}_{(123)}=\frac{1}{3}\begin{bmatrix}\ 1&\ 1&\ 1\\\
\ 0&-\mathit{r}&\ \mathit{r}\\\ \mathit{r}&\ -\mathit{r}&\
0\end{bmatrix},\quad\text{and}\quad\mathcal{T}_{(132)}=\frac{1}{3}\begin{bmatrix}\
1&\ 1&\ 1\\\ \ \mathit{r}&\ 0&-\mathit{r}\\\ \ 0&\
\mathit{r}&-\mathit{r}\end{bmatrix}.$

Here we are using the the cycle notation for a permutation $\sigma\in
S_{n},n\in\mathbb{N}_{>0}$:

$\sigma=(a_{i_{1}},a_{i_{2}},\dots,a_{i_{j}})\in S_{n}\ \ \Leftrightarrow\ \ \
\left\\{\begin{aligned} &\sigma(a_{i_{k}})=a_{i_{k+1}},\ \ \text{for
}k=1,\dots,j-1,\\\ &\sigma(a_{i_{j}})=a_{i_{1}},\ \ \text{and}\\\
&\sigma(a_{s})=a_{s},\ \ \text{if }s\neq i_{1},i_{2},\dots,i_{j}\\\
\end{aligned}\right..$

Then for each $\sigma\in S_{3}$, with the Cremona involution
$J_{3}:\mathbf{P}^{2}\dasharrow\mathbf{P}^{2}$ we have

(6.4) $\check{f}_{\sigma}=\mathcal{T}_{\sigma}\circ
J_{3}\circ\mathcal{S}^{-1}.$

Notice that in terms of $t_{i}$’s in (5.11), we have $s=-1/(t_{5}+1)$ and
$r=1/(t_{1}+1)$. Thus it is easy to check that each $\check{f}_{\sigma}$ is
the one constructed in the subsection 5.2. All six birational maps lift to
automorphisms $f_{\sigma}:X\to X$.

###  6.2. Induced Actions on $Pic(X)$

Those six birational maps $\check{f}_{\sigma},\sigma\in\Sigma_{3}$ share the
same base locus and satisfy (5.6). To describe their actions on $Pic(X)$, let
$\mathbf{s}_{\kappa}$ be the Cremona involution through the vector
$\mathbf{e}-\mathbf{e}_{5}-\mathbf{e}_{10}-\mathbf{e}_{15}$:

$\mathbf{s}_{\kappa}\ \ :\ \ \left\\{\ \begin{aligned} &\mathbf{e}_{0}\mapsto
2\mathbf{e}_{0}-\mathbf{e}_{5}-\mathbf{e}_{10}-\mathbf{e}_{15}\\\
&\mathbf{e}_{i}\mapsto\mathbf{e}_{0}-\mathbf{e}_{5}-\mathbf{e}_{10}-\mathbf{e}_{15}+\mathbf{e}_{i},\qquad\text{if}\quad
i=5.10,15\\\ &\mathbf{e}_{j}\mapsto\mathbf{e}_{j},\qquad\qquad\text{ if }\quad
i\neq 0,5,10,15\\\ \end{aligned}\right.$

With the cycle notation of permutations in $S_{15}$ we have

(6.5) $\displaystyle f_{id}^{*}=\mathbf{s}_{\kappa}\,(5\ 4\ 3\ 2\ 1)\,(10\ 9\
8\ 7\ 6)\,(15\ 14\ 13\ 12\ 11)$ $\displaystyle
f_{(12)}^{*}=\mathbf{s}_{\kappa}\,(5\ 9\ 3\ 7\ 1\ 10\ 9\ 8\ 7\ 6)\,(15\ 14\
13\ 12\ 11)$ $\displaystyle f_{(13)}^{*}=\mathbf{s}_{\kappa}\,(5\ 14\ 3\ 12\
1\ 15\ 4\ 13\ 2\ 11)\,(10\ 9\ 8\ 7\ 6)$ $\displaystyle
f_{(23)}^{*}=\mathbf{s}_{\kappa}\,(5\ 4\ 3\ 2\ 1)\,(10\ 14\ 8\ 12\ 6\ 15\ 9\
13\ 7\ 11)$ $\displaystyle f_{(123)}^{*}=\mathbf{s}_{\kappa}\,(5\ 9\ 13\ 2\ 6\
15\ 4\ 8\ 12\ 1\ 10\ 14\ 3\ 7\ 11)$ $\displaystyle
f_{(132)}^{*}=\mathbf{s}_{\kappa}\,(5\ 14\ 8\ 2\ 11\ 10\ 4\ 13\ 7\ 1\ 15\ 9\
3\ 12\ 6).$

These actions are identified with elements in $W_{15}$. Let $G\subset W_{15}$
be a subgroup generated by those six actions in (6.5):

(6.6) $G=\langle\ f_{\sigma}^{*},\sigma\in S_{3}\ \rangle$

##  7\. Automorphism group $\text{Aut}(X)$ with six generators

To understand the automorphism group $\text{Aut}(X)$, let us start discussing
the properties of automorphisms on $X$.

######  Proposition 7.1.

If $\check{f}:\mathbf{P}^{2}\to\mathbf{P}^{2}$ be a birational map which lifts
to an automorphism $f\in\text{Aut}(X)$, then $\check{f}$ properly fixes the
invariant cubic $C$.

###### Proof.

Notice that if $\check{f}$ has a point of indeterminacy in $\mathbf{P}^{2}$,
then this point must be blown up to get an automorphism on $X$. Thus all
points of indeterminacy are in the base locus, which is in the set of regular
points of $C$. Since $C$ is an anti-canonical curve, so is $f(C)$, i.e.,
$f(C)$ is a degree $3$ curve passing through all exceptional divisors. It
follows that $\check{f}(C)$ is a degree $3$ curve passing through all $15$
points in the base locus. The cubic $C$ consists of three lines such that each
line contains exactly 5 (distinct) points in the base locus. There exists a
unique such cubic. Thus we have $\check{f}(C)=C$. ∎

######  Proposition 7.2.

If $f\in\text{Aut}(X)$ then $f$ fixes the cubic $C$.

###### Proof.

Since $f$ covers a birational map $\check{f}$ on $\mathbf{P}^{2}$, this
Proposition is the immediate consequence of the previous Proposition. ∎

Also, we have

######  Lemma 7.3.

There is no curve $W\subset\mathbf{P}^{2}$ such that $W\cap C\subset B$.

###### Proof.

Let $W$ be a curve in $\mathbf{P}^{2}$ and suppose $W\cap C\subset B$. By the
group law, there are non-negative integers $m_{1},\dots,m_{5}$ such that

(7.1)
$\sum_{i=1}^{5}m_{i}(1+t_{i})=\frac{1}{1+2\alpha^{5}}\,\left(\sum_{i=1}^{5}m_{i}(1+2\alpha^{5}-3\alpha^{i})\right)=0$

where $\alpha$ is a root of the Salem polynomial $t^{4}-2t^{3}+t^{2}-2t+1$.
Since $1+2\alpha^{5}\neq 0$, (7.1) satisfies if and only if

$\displaystyle\sum_{i=1}^{5}$ $\displaystyle m_{i}(1+2\alpha^{5}-3\alpha^{i})$
$\displaystyle=(2\sum
m_{i}-3m_{5})\alpha^{5}-3m_{4}\alpha^{4}-3m_{3}\alpha^{3}-3m_{2}\alpha^{2}-3m_{1}\alpha+\sum
m_{i}\ =\ 0$

Since the minimal polynomial of $\alpha$ is of degree $4$, there are integers
$A,B$ such that

$\sum_{i=1}^{5}m_{i}(1+2\alpha^{5}-3\alpha^{i})=(\alpha^{4}-2\alpha^{3}+\alpha^{2}-2\alpha+1)(A\alpha+B).$

By comparing the coefficients, we get $m_{1}=m_{3}=m_{5}=-m_{2}=-m_{4}$. Since
$m_{i}$ are non-negative integers, the only possible solution is $m_{i}=0$ for
all $i$. ∎

######  Proposition 7.4.

If $f\in\text{Aut}(X)$ is not linear, then the cubic $C$ does not contain a
curve of fixed points.

###### Proof.

Suppose one component, say $L_{1}$, is a curve of fixed points. Since $L_{1}$
intersects exceptional curves $E_{j},j=1,\dots,5$ over $p_{1,j}$, for each
$j=1,\dots 5$ either $E_{j}$ is fixed or $p_{1,j}$ is a point of
indeterminacy. If all $E_{j}$’s are fixed, then we have $f^{*}e_{i}=e_{i}$ for
all $i=1,\dots,5$ and $f^{*}e_{0}=f^{*}([L_{1}]-e_{1}-\cdots e_{5})=e_{0}$. It
follows that $f$ is linear. Suppose one of $p_{1,j}$ is a point of
indeterminacy. Since $L_{1}$ is a curve of fixed points, the exceptional curve
$E_{j}$ maps to a curve $fE_{j}$ passing through $L_{1}\cap E_{j}$. Since $f$
preserves the intersection number, $fE_{j}$ can not intersect $C$. It follows
that the $fE_{j}$ descends to a curve intersecting $C$ only at the base locus
$B$. This is a contradiction by Lemma 7.3. ∎

###  7.1. Dynamical degrees and Determinants

Note that $C\subset X$ is an anticanonical curve. If $f\in\text{Aut}(X)$
preserves the invariant cubic $C$, then $f^{*}\eta=d(f)\,\eta$ where $\eta$ is
the unique (up to constant multiple) meromorphic form on $X$ with a simple
pole along $C$. We call $d(f)$ the determinant of $f$ as defined in [17]. If a
birational map $\check{f}$ properly fixing $C$ lifts to $f$ then the
restriction $\check{f}|_{C}\,(x,i)=(d(f)x+b,\tau(i))$ for a constant $b$ and a
permutation $\tau$. The determinant $d(f)$ is, in fact, the determinant of the
differential $\text{det}Df_{p}$ at the fixed point $p\notin C$. If the
determinant $d(f)$ is not a root of unity, then $d(f)$ is a Galois conjugate
of the dynamical degree $\delta(f)$. And we have

######  Theorem 7.5 ([17, 22]).

If $f\in\text{Aut}(X)$ is an automorphism on $X$, then the determinant $d(f)$
of $f$ is an eigenvalue of $f^{*}|_{Pic(X)}$.

Clearly, we have

######  Lemma 7.6.

If $f:X\to X$ is an automorphism on $X$, then we have
$d(f^{-1})=\frac{1}{d(f)}.$

######  Lemma 7.7.

If $f,g\in\text{Aut}(X)$ then the determinant of $f\circ g\in\text{Aut}(X)$ is
the product $d(f)d(g)$ of determinants of $f$ and $g$.

###### Proof.

From Proposition 7.2, we see that $f,g$ and $f\circ g$ fix the cubic $C$. Thus
the restriction maps satisfy $(f\circ g)|_{C}=f|_{C}\circ g|_{C}$ and
$d(f\circ g)=d(f)d(g)$. ∎

Since the characteristic polynomial of $f^{*}|_{Pic(X)}$ is given by a product
of a Salem polynomial and cyclotomic polynomials, it follows that

######  Lemma 7.8.

Suppose $f$ and $g$ are automorphisms on $X$ with dynamical degrees
$\delta(f)\geq\delta(g)>1$. If determinants of $f$ and $g$ are $d(f)$ and
$d(g)$ respectively, then the dynamical degree $\delta(f\circ g)$ of $f\circ
g$ is either $\delta(f)\delta(g)$, or $\delta(f)/\delta(g)$. Furthermore, by
considering $g^{-1}$ we can get both dynamical degree $\delta(f)\delta(g)$,
and $\delta(f)/\delta(g)$.

###### Proof.

Since the determinant $d(f)$ is a Galois conjugate of the dynamical degree
$\delta(f)$ and the dynamical degree is a Salem number, the minimal polynomial
$P_{f}(t)$ of $d(f)$ is given by a reciprocal poly
$P_{f}(t)=\prod(t-\alpha_{i})$ with all but two $\alpha_{i}$ are non-real
complex numbers with modulus $1$. Similarly, the minimal polynomial of $d(g)$
is given by $P_{g}(t)=\prod(t-\beta_{j})$ with only two $\beta_{j}$ are real.
The minimal polynomial $P_{f\circ g}$ of the product $d(f)d(g)$ must divide
the polynomial $\prod(t-\alpha_{i}\beta_{j})$

$P_{f\circ g}(t)|\prod_{i,j}(t-\alpha_{i}\beta_{j}).$

Since only four real numbers in $\\{\alpha_{i}\beta_{j}\\}$ are
$\delta(f)\delta(g)$, $\delta(f)/\delta(g)$, and their reciprocals. By Lemma
7.7, we know that the product $d(f)d(g)$ must be a Galois conjugate of the
dynamical degree $\delta(f\circ g)$ of the composition. Since the dynamical
degree is a real number $\geq 1$, we see that the dynamical degree
$\delta(f\circ g)$ of $f\circ g$ is either $\delta(f)\delta(g)$ or
$\delta(f)/\delta(g)$. If $\delta(f\circ g)=\delta(f)\delta(g)$ then both
$f^{*}$ and $g^{*}$ have the same eigenvector $v_{+}$ corresponding the
largest eigenvalues $\delta(f),\delta(g)$. It follows that $f$ and $g^{-1}$
will not share the same eigenvector corresponding to the largest eigenvalues,
and thus $\delta(f\circ g^{-1})\neq\delta(f)\delta(g^{-1})$. Since the only
possible options are $\delta(f)\delta(g^{-1})$ and $\delta(f)/\delta(g^{-1})$
and $\delta(g)=\delta(g^{-1})$, we have that $\delta(f\circ
g^{-1})=\delta(f)/\delta(g)$. ∎

######  Theorem 7.9.

If $f\in\text{Aut}(X)$ is an automorphism on $X$, then the dynamical degree
$\delta(f)$ of $f$ is $\delta^{n}$ for some $n\geq 0$ where $\delta$ is the
largest real root of the Salem polynomial

$\chi_{5}(t)=t^{4}-2t^{3}+t^{2}-2t+1$

###### Proof.

Recall that there are six automorphisms $f_{\sigma}:X\to X,\sigma\in S_{3}$
constructed in Theorem 6.1 and $\delta(f_{\sigma})=\delta$. Suppose there is
an automorphism $g\in\text{Aut}(X)$ such that $\delta(g)\neq\delta^{n}$ for
all $n\geq 0$. The composition $f_{\sigma}\circ g\in\text{Aut}(X)$.
Furthermore due to Lemma 7.8, the dynamical degree $\delta(f_{\sigma}\circ g)$
is either $\delta\cdot\delta(g)$, $\delta/\delta(g)$, or $\delta(g)/\delta$.
If there is no Salem number $\tau$ such that $\delta=\tau^{p}$ and
$\delta(g)=\tau^{q}$ for some $p,q\geq 1$, then none of them is Salem number
due to Proposition 4.3. This is a contradiction since the dynamical degree of
an automorphism of a rational surface is either $1$ or a Salem number.

Now suppose there is a Salem number $\tau$ such that $\delta=\tau^{p}$ and
$\delta(g)=\tau^{q}$ for some $p,q\geq 1$. If $p=1$, then
$\delta(g)=\delta^{q}$. If $p>1$, applying the Lemma 7.8 repeatedly, we get an
automorphism $h\in\text{Aut}(X)$ such that $\delta(h)=\tau$. Note that since
$h$ is an automorphism on $X$, $h^{*}\in W_{15}$ and thus
$\delta(h)\geq\lambda_{\star}$, the Lehmer’s number [16]. Direct computation
shows that $\delta^{1/4}\approx 1.17145<\lambda_{\star}$ and the minimal
polynomials for $\delta^{1/2}$ and $\delta^{1/3}$ have two or more real roots
bigger than $1$. It follows that $p=1$ and therefore $\delta(g)=\delta^{n}$
for some $n\geq 0$. ∎

###  7.2. The Automorphism group

######  Proposition 7.10.

Suppose $f\in\text{Aut}(X)$. If the induced action $f^{*}:Pic(X)\to Pic(X)$ is
the identity, then $f$ itself is an identity map.

$f^{*}\ =\ Id\quad\Leftrightarrow\quad f\ =\ Id.$

###### Proof.

If $f^{*}=Id$, then the corresponding birational map $\check{f}$ on
$\mathbf{P}^{2}$ must fix all $15$ distinct base points and the three lines
joining at one point. Since each line contains $5$ base points, those three
lines are, in fact, lines of fixed points for $\check{f}$. Thus, we have
$\check{f}=Id$ and $f=Id$. ∎

######  Proposition 7.11.

Let $\delta$ is the largest real root of $\varphi(t)$ defined in (6.3). If an
automorphism $f$ on $X$ has the dynamical degree $\delta(f)=\delta$ then

$f\in\\{f_{\sigma},f^{-1}_{\sigma}:\sigma\in S_{3}\\}$

where $f_{\sigma}$ is the lift $\check{f}_{\sigma}$ defined in (6.4)

###### Proof.

Suppose $f\in\text{Aut}(X)$ is an automorphism with $\delta(f)=\delta>1$. Then
the determinant $d(f)$ is a root of $\varphi(t)$. Due to the construction, for
each $\sigma\in S_{3}$, automorphisms $f_{\sigma}$ have the same determinant.
Furthermore, $\varphi(t)$ has two real roots $\delta,1/\delta$, and two non-
real roots $\omega,\bar{\omega}$ on the unit circle. We have two cases:

  * •

If $d(f_{\sigma})$ is real, then $d(f)$ has to be real. Otherwise, we have
$d(f\circ f_{\sigma})=d(f)\cdot d(f)$, which is non-real outside unit circle.
It follows that the determinant $d(f\circ f_{\sigma})$ of composition is not a
Galois conjugate of a Salem number, and thus $f\circ f_{\sigma}$ is not an
automorphism, which is a contradiction. Thus the determinant $d(f)$ has to be
real, that is, $d(f)$ is either $\delta$ or $1/\delta$.

  * •

Similarly, if $d(f_{\sigma})$ is non-real, then $d(f)$ is either $\omega$ or
$\bar{\omega}$.

Either case, we have either $d(f)=d(f_{\sigma})$ or $d(f)=1/d(f_{\sigma})$ for
all $\sigma\in S_{3}$. It follows that there is $\sigma\in S_{3}$ such that
the composition $f\circ f_{\sigma}$ (or $f\circ f_{\sigma}^{-1}$) has
determinant $1$. It follows that there is a positive integer $n$ such that
$(f\circ f_{\sigma})^{n}=Id$. If the restriction map $f\circ
f_{\sigma}|C:t\mapsto t+\beta$, then $t+n\beta=t$. It follows that $\beta=0$
and thus $C$ is a curve of fixed points. By Proposition 7.4, we see that
$f\circ f_{\sigma}$ is linear and thus $(f\circ f\sigma)\circ f_{\sigma}^{-1}$
is quadratic. The statement follows from Theorem 5.6 ∎

######  Lemma 7.12.

Let us set six linear maps

(7.2) $L_{\sigma}:=f_{\sigma}\circ f_{id}^{-1}\qquad\text{ for }\sigma\in
S_{3}.$

Let $\mathcal{L}$ be the group generated by six linear maps above, then
$\mathcal{L}$ is equivalent to the dihedral group $D_{3}$

$\mathcal{L}=\langle L_{(12)},L_{(13)},L_{(23)},L_{(123)},L_{(132)}\rangle\
\cong D_{3}.$

###### Proof.

The the formulae for $\check{f}_{\sigma}=\mathcal{T}_{\sigma}\circ
J_{3}\circ\mathcal{S}^{-1}$ in (6.4), it is easy to see $L_{\sigma}$ is a
linear. In fact we have

$L_{id}=L_{(12)}^{2}=L_{(23)}^{2}=L_{(13)}^{2}=L_{(123)}^{3}=L_{(132)}^{3}=Id,\quad\text{and}\quad
L_{(123)}^{-1}=L_{(132)}.$

∎

The linear maps (7.2) defined in the previous Proposition permute three
concurrent lines. If $\sigma\in S_{3}$, $L_{\sigma}$ maps the line
$L_{\sigma(i)}$ to the line $L_{i}$.

######  Proposition 7.13.

Suppose $f\in\text{Aut}(X)$ is an automorphism on $X$. If the dynamical degree
of $f$, $\delta(f)=1$, then $f\in\mathcal{L}$

###### Proof.

Let $P(t)=\prod(t-\alpha_{i})$ be the minimal polynomial of the dynamical
degree $\delta>1$ of $f_{\sigma}$. If $\delta(f)=1$, then the determinant of
$f$ is a root of unity. Thus, the minimal polynomial of $d(f\circ
f_{\sigma})=d(f)d(f_{\sigma})$ divides $\prod(t-\alpha_{i}\beta_{j})$ where
$\beta_{j}$’s are roots of unity. For any root of unity $\beta$, Since
$\alpha_{i}\beta$ is not a root of unity and $|\alpha_{i}\beta|=|\alpha_{i}|$.
Thus we see that the dynamical degree of $f\circ f_{\sigma}=\delta$. On the
other hand, the determinant $d(f\circ f_{\sigma})=d(f)d(f_{\sigma})$ is a
Galois conjugate of $\delta$ and thus $d(f)=1$. With the same argument in the
proof of Proposition 7.11 $\check{f}$ is linear and thus
$\check{f}\circ\check{f}_{\sigma}$ is a quadratic map with dynamical degree
$\delta$. Thus, by Theorem 5.6, we have $f\circ f_{\sigma}=f_{\tau}$ for some
$\tau\in S_{3}$. Thus

$f=f_{\tau}\circ f_{\sigma}^{-1}=L_{\tau}\circ L_{\sigma}^{-1}\in\mathcal{L}.$

∎

######  Theorem 7.14.

Let $X$ be a rational surface defined in the beginning of Section 6 and let
$f_{id}$, $f_{(12)}$, $f_{(13)},f_{(23)},f_{(123)},$ and $f_{(132)}$ be six
quadratic automorphisms on $X$ which are induced from birational maps defined
in (6.4). The automorphism group $\text{Aut}(X)$ is generated by $6$ quadratic
automorphisms properly fixing three lines meeting at a single point and

$\text{Aut}(X)=\langle
f_{id},f_{(12)},f_{(13)},f_{(23)},f_{(123)},f_{(132)}\rangle\ \cong\
D_{3}\rtimes\mathbb{Z}$

###### Proof.

Direct computation using formulae in (6.4) shows that $L_{\sigma}\circ
f_{id}=f_{id}\circ L_{\sigma}$. Since $f_{\sigma}=L_{\sigma}\circ f_{id}$, it
is clear that the second part $\langle
f_{id},f_{(12)},f_{(13)},f_{(23)},f_{(123)},f_{(132)}\rangle\ \cong\
D_{3}\rtimes\mathbb{Z}$. Now suppose $f\in\text{Aut}(X)$. By Theorem 7.9, the
dynamical degree $\delta(f)=\delta^{n}$ for some $n\geq 0$. Applying Lemma 7.8
repeatedly and using Proposition 7.13, we have the desired result. ∎

###### Proof of Theorem A.

The assertion of Theorem $A$ is the immediate consequence of the previous
theorem 7.14. ∎

##  8\. The Subgroup of $W_{15}$

######  Theorem 8.1.

Let $X=B\ell_{B}(\mathbf{P}^{2})$ be a rational surface obtained by blowing up
a set $B$ of $15$ distinct points defined in Equation 6.1 and let $G$ be a
subgroup of the Coxeter group $W_{15}$ defined in (6.6). Then $G$ is realized
by automorphisms on $X$. In fact, the subgroup $G$ of $W_{15}$ is isomorphic
to the automorphism group $\text{Aut}(X)$.

$G=\langle
f^{*}_{id},f^{*}_{(12)},f^{*}_{(13)},f^{*}_{(23)},f^{*}_{(123)},f^{*}_{(132)}\rangle\
\cong\ \text{Aut}(X)$

###### Proof.

The map sending $f\in\text{Aut}(X)$ to the actions $f^{*}$ on $Pic(X)$ is a
homomorphism. This map is, in fact, one-to-one due to Proposition 7.10. It
follows that $G\cong\text{Aut}(X)$.

∎

###### Proof of Theorem B.

The statement of Theorem $B$ is clear from the previous Theorem 8.1. ∎

Also, we have other subgroups that are realized by automorphisms on rational
surfaces. The construction is essentially the same as the subgroup $G\in
W_{15}$ in this section and the rational surface $X$ in the section 7. We will
list them below. The following automorphism groups are generated by quadratic
surface automorphisms fixing concurrent lines.

###  8.1. Other subgroups

Let $X_{3n}$ be a rational surface obtained by blowing a set of finite points
on a cubic $C$ given by three concurrent lines

$B_{n}=\\{p_{i,j}:=(1+t_{j},i)\in C:i=1,2,3,j=1,\dots,n\\}$

  1. (1)

If $n\geq 5$ is odd and $n\equiv 0$ $(\text{mod }3)$, then
$\text{Aut}(X_{3n})\cong(\mathbb{Z}/2\mathbb{Z})^{2}\rtimes\mathbb{Z}$.

  2. (2)

If $n\geq 4$ is even and $n\not\equiv 0$ $(\text{mod }3)$, then
$\text{Aut}(X_{3n})\cong(\mathbb{Z}/3\mathbb{Z})\rtimes\mathbb{Z}$.

  3. (3)

If $n\geq 4$ is even and $n\equiv 0$ $(\text{mod }3)$, then
$\text{Aut}(X_{3n})\cong\mathbb{Z}$.

## References

  * [1] Eric Bedford, Jeffery Diller, and Kyounghee Kim.  Pseudoautomorphisms with invariant curves.  Proceedings of the Abel symposium 2013, pages 1–27, 2015.
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