# The Banach–Tarski paradox for some subsets of finite-dimensional normed
spaces over non-Archimedean valued fields

Kamil Orzechowski
University of Rzeszów
The Doctoral School
Rejtana 16c
35-959 Rzeszów, Poland

###### Abstract

We show some results related to the classical Banach–Tarski paradox in the
setting of finite-dimensional normed spaces over a non-Archimedean valued
field $K$. For instance, all balls and spheres in $K^{n}$, and the whole space
$K^{n}$ (for $n\geq 2$) are paradoxical with respect to certain groups of
isometries of $K^{n}$. If $K$ is locally compact (e.g., $K$ is the field
$\mathbb{Q}_{p}$ of $p$-adic numbers for any prime number $p$), any two
bounded subsets of $K^{n}$ with nonempty interiors are equidecomposable (and
paradoxical) with respect to a certain group of isometries of $K^{n}$ (for
$n\geq 2$).

††2020 _Mathematics Subject Classification_ : Primary 47S10; Secondary 46S10,
12J25, 26E30, 20E05, 20H20, 46B04, 05A18, 03E25.†† _Key words and phrases_ :
Banach–Tarski paradox, paradoxical decomposition, non-Archimedean valued
field, non-Archimedean normed space, free group.

##  1\. Introduction

In 1924, Banach and Tarski (inspired by a result of Hausdorff [5, p. 469])
proved one of the most shocking theorems in mathematics, called the
Banach–Tarski paradox [1, Theorem 24]:

Any two bounded subsets $A$ and $B$ of $\mathbb{R}^{3}$ with nonempty
interiors can be partitioned into finite number of pieces $A_{1},\ldots,A_{n}$
and $B_{1},\ldots,B_{n}$, respectively, in such a way that for some isometries
$\tau_{1},\ldots,\tau_{n}$ of $\mathbb{R}^{3}$ we have $\tau_{k}(A_{k})=B_{k}$
for any $1\leq k\leq n.$

In modern terms, we can then say that $A$ and $B$ are $G$-equidecomposable,
where $G$ is the group of all isometries of $\mathbb{R}^{3}$. It follows that
any ball in $\mathbb{R}^{3}$ can be partitioned into finite number of pieces
in such a way that the pieces can be transformed using isometries of
$\mathbb{R}^{3}$ to obtain two balls identical with the initial one [1, Lemma
21], i.e., any ball in $\mathbb{R}^{3}$ is $G$-paradoxical. Those results are
called paradoxes as they seem counterintuitive. Clearly, the pieces involved
cannot all be Lebesgue-measurable. Note that the Banach–Tarski paradox is a
theorem in ZFC, but not in ZF [19, Corollary 15.3].

There have been many refinements, generalizations and inspired results since
the work of Banach and Tarski was published. They include minimizing the
number of pieces needed for a paradoxical decomposition (five is optimal for a
ball in $\mathbb{R}^{3}$), generalization for balls and spheres in
$\mathbb{R}^{n}$, $n\geq 3$, as well as similar results in hyperbolic spaces.
The monograph [19] is an excellent source covering most of these topics.

In this paper, we prove an analog of the Banach–Tarski paradox in the setting
of finite-dimensional normed spaces $K^{n}$ over a non-Archimedean valued
field $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ instead of
$(\mathbb{R},\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$, e.g.,
$(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ may be the field
$(\mathbb{Q}_{p},\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{p})$ of
$p$-adic numbers or the field
$(\mathbb{C}_{p},\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{p})$ of
$p$-adic complex numbers for any prime number $p$. We are especially
interested in paradoxical decompositions of balls and spheres in $K^{n}$, the
whole space $K^{n}$, and the set $K^{n}\setminus\\{\mathbf{0}\\}$, $n\geq 2$.

The paper is organized as follows. Section 2 provides basic definitions of the
theory of non-Archimedean valued fields and normed spaces over them. A
characterization of linear and affine isometries of $K^{n}$ is given and some
notation is introduced. In Section 3, for a group $G$ acting on a set $X$ we
recall the notion of a G-paradoxical set $E\subseteq X$, and some basic
results; in particular: If a group $G$ contains a non-Abelian free subgroup
and acts without nontrivial fixed points on a set $X$, then $X$ is
$G$-paradoxical using four pieces. We state and prove some auxiliary results,
especially Theorem 3.2 and Lemma 3.3. Finally, we recall the relationship
between paradoxical decompositions and amenability of groups.

Section 4 is focused on finding non-Abelian free subgroups in special linear
groups $\operatorname{SL}(n,D)$ and special affine groups
$\operatorname{SA}(n,D)$ over the ring of integers $D$ of $K$ (i.e.,
$D=\\{\alpha\in K:\lvert\alpha\rvert\leq 1\\}$). Subsection 4.1 deals with the
case of $K$ of characteristic zero; its main result is Theorem 4.9. In
Subsection 4.2, we investigate the case when $K$ is a transcendental extension
of its prime field. We introduce polynomials describing the trace of matrices
in $\operatorname{SL}(2,K)$. We prove Theorem 4.15, which provides a new tool
for finding non-Abelian free subgroups of $\operatorname{SL}(2,D)$ acting
without nontrivial fixed points on
$K^{2}\setminus\\{\mathbf{0}\\},D^{2}\setminus\\{\mathbf{0}\\}$, and any
sphere $S[\mathbf{0},r]$, $r>0$, (Theorem 4.18) and non-Abelian free subgroups
of $\operatorname{SA}(n,D)$ acting without nontrivial fixed points on balls
and spheres in $K^{n}$ (Theorem 4.19).

Section 5 contains our main results. Assume that
$(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ is a non-Archimedean
nontrivially valued field and $n\geq 2$. We prove that all balls and spheres
in $K^{n}$ not containing $\mathbf{0}$ are paradoxical with respect to certain
groups of linear isometries of $K^{n}$ using four pieces for balls, and
$5-(-1)^{n}$ pieces for spheres (Theorems 5.1 and 5.2),
$K^{n}\setminus\\{\mathbf{0}\\}$ and $D^{n}\setminus\\{\mathbf{0}\\}$
(Corollary 5.3). We also prove that all balls and spheres containing
$\mathbf{0}$ are paradoxical using four pieces with respect to certain groups
of affine isometries of $K^{n}$ (Theorem 5.4). Next, we prove a strong result
analogical to the classical Banach–Tarski paradox:

If $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ is a non-Archimedean
discretely valued field with finite residue field (in particular, if
$(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ is a non-Archimedean
locally compact field) and $n\geq 2$, then any two bounded subsets of $K^{n}$
with nonempty interiors are $G$-equidecomposable (and $G$-paradoxical), where
$G$ is the group of special affine isometries of $K^{n}$ (see Theorem 5.5).

Finally, we prove that the whole space $K^{n}$ is paradoxical using five
pieces with respect to a certain group of affine isometries (Theorem 5.7).

##  2\. Non-Archimedean valued fields and normed spaces over them

###  2.1. Preliminaries

Throughout this paper, by a field we mean a commutative division ring.

A valuation on a field $K$ is any function
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert\colon K\to[0,\infty)$
satisfying for all $x,y\in K$ the following conditions:

(1) $\lvert x\rvert=0\Leftrightarrow x=0$, (2) $\lvert xy\rvert=\lvert
x\rvert\lvert y\rvert$, (3) $\lvert x+y\rvert\leq\lvert x\rvert+\lvert
y\rvert$.

The pair $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ is then called
a valued field. If we replace the condition (3) with a stronger one (3’)
$\lvert x+y\rvert\leq\max\\{\lvert x\rvert,\lvert y\rvert\\},$ the valuation
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ will be called non-
Archimedean.

If $(X,d)$ is a metric space, $x_{0}\in X$ and $r>0$, we put
$B[x_{0},r]:=\\{x\in X:d(x,x_{0})\leq r\\}$, $B(x_{0},r):=\\{x\in
X:d(x,x_{0})<r\\}$ and $S[x_{0},r]:=\\{x\in X:d(x,x_{0})=r\\}.$

A metric $d$ on a set $X$ is called an ultrametric if it satisfies the strong
triangle inequality: $d(x,z)\leq\max\\{d(x,y),d(y,z)\\}$ for all $x,y,z\in X$.
The pair $(X,d)$ is then called an ultrametric space. It is noteworthy that,
in an ultrametric space, any point belonging to a ball (open or closed) of a
given radius may play the role of its center.

Any ultrametric space $(X,d)$ has the isosceles property: If $x,y,z\in X$ and
$d(x,y)\neq d(y,z)$, then $d(x,z)=\max\\{d(x,y),d(y,z)\\}$.

Each valuation $\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ on a field
$K$ induces a metric $d\colon K\times K\to[0,+\infty)$, $d(x,y):=\lvert
x-y\rvert$. If $\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ is non-
Archimedean, $d$ is an ultrametric. If $d_{1}$, $d_{2}$ are metrics on $K$
induced by valuations $\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{1}$,
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{2}$ respectively, they
induce the same topology on $K$ if and only if
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{1}$ and
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{2}$ are equivalent, i.e.,
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{2}=(\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{1})^{c}$
for some $c>0$ [11, Theorem 1.2.2].

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a fixed non-
Archimedean valued field. We introduce the following notation:
$D=B[0,1],\mathfrak{m}=B(0,1)$ and $D^{*}=S[0,1].$ The subset $D$ is a subring
of $K$ called the ring of integers of $K$, $\mathfrak{m}$ is its unique
maximal ideal and $D^{*}$ is the group of units (i.e., invertible elements) of
$D$ (cf. [16, Lemma 1.2]). The quotient ring $k:=D/\mathfrak{m}$ is called the
residue field of $K$.

We denote by $K^{*}$ the multiplicative group of $K$. The set $\lvert
K^{*}\rvert:=\\{\lvert x\rvert\colon x\in K^{*}\\}$ is a subgroup of the
multiplicative group $\mathbb{R}_{+}$. If $\lvert K^{*}\rvert=\\{1\\}$, the
valuation $\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ is called trivial.
If there exists $t:=\max\lvert K^{*}\rvert\cap(0,1)$, then $\lvert
K^{*}\rvert=\\{t^{n}\colon n\in\mathbb{Z}\\}$ and the valuation is called
discrete. In the remaining case $\lvert K^{*}\rvert$ is a dense subset of
$[0,\infty)$ and the valuation is called dense.

Assume that $K$ is equipped with a discrete valuation. Fix an arbitrary
$\pi\in K$ satisfying $\lvert\pi\rvert=t:=\max\lvert K^{*}\rvert\cap(0,1)$.
Such an element $\pi$ will be called a uniformizer. Then $\lvert a\rvert\leq
t^{n}$ holds if and only if $a\in\pi^{n}D$. Therefore we have
$B[0,t^{n}]=B(0,t^{n-1})=\pi^{n}D$ for all $n\in\mathbb{Z}$. Each nonzero
ideal in $D$ is principal and of the form $\pi^{n}D$ for some $n\geq 0$, in
particular $\mathfrak{m}=\pi D$.

If $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ is a complete non-
Archimedean valued field and $L$ is an algebraic extension of $K$, then there
is a unique valuation on $L$ that extends
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ [15, Theorem 14.2]. It
follows that if $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ is a
trivially valued field (e.g., a finite one), the only possible valuation on an
algebraic extension $L$ of $K$ that extends
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ is trivial.

A non-Archimedean nontrivially valued field
$(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ is locally compact if
and only if it is complete, the valuation
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ is discrete, and the residue
field $k=D/\mathfrak{m}$ is finite [15, Corollary 12.2]. A locally compact
nontrivially valued field
$(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ is called a local field.

Every nontrivial non-Archimedean valuation on $\mathbb{Q}$ is equivalent to
the $p$-adic valuation $\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{p}$,
for a prime number $p$, defined as follows

$\lvert 0\rvert_{p}:=0,\;\left\lvert
p^{k}\frac{m}{n}\right\rvert_{p}:=p^{-k}\quad\text{for
}k,m,n\in\mathbb{Z},\quad p\nmid m,p\nmid n.$ (2.1)

The completion of
$(\mathbb{Q},\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{p})$ is called
the field of $p$-adic numbers [11, Definition 1.2.7] and denoted by
$(\mathbb{Q}_{p},\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{p})$. It is
a standard example of a non-Archimedean local field. The valued field obtained
by uniquely extending the valuation
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{p}$ from $\mathbb{Q}_{p}$ to
its algebraic closure and then taking the metric completion is called the
field of p-adic complex numbers and denoted by
$(\mathbb{C}_{p},\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{p})$ [11,
Example 1.2.11]. This field is algebraically closed (unlike $\mathbb{Q}_{p}$)
but not locally compact [15, Corollary 17.2]. Remarkably, $\mathbb{C}_{p}$ and
$\mathbb{C}$ are isomorphic as fields [13, p. 83].

###  2.2. Groups of linear and affine isometries of normed spaces
$(K^{n},\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert)$

Let $R$ be a commutative ring with the identity element $1$. We write $R^{*}$
for the group of units (i.e., invertible elements) of $R$. We denote by
$M_{n}(R)$ the ring of square matrices of order $n$ over $R$, and by $I_{n}$
the identity matrix of order $n$. The group
$\operatorname{GL}(n,R)=\left(M_{n}(R)\right)^{*}$ of invertible elements of
$M_{n}(R)$ is called the ($n$-th) general linear group over $R$.

It is well known that $A\in\operatorname{GL}(n,R)$ if and only if the
determinant of $A$ is invertible in $R$, i.e.,

$\operatorname{GL}(n,R)=\\{A\in M_{n}(R)\colon\det(A)\in R^{*}\\}.$

The kernel of the homomorphism $\det\colon\operatorname{GL}(n,R)\to
R^{*},A\mapsto\det(A),$ is called the ($n$-th) special linear group over $R$
and is denoted by $\operatorname{SL}(n,R)$, i.e.,

$\operatorname{SL}(n,R)=\\{A\in\operatorname{GL}(n,R)\colon\det(A)=1\\}=\\{A\in
M_{n}(R)\colon\det(A)=1\\}.$

We adopt the convention identifying an $R$-linear mapping $U\colon R^{n}\to
R^{n}$ with its matrix in the canonical basis $\\{e_{1},\dots,e_{n}\\}$ of the
free $R$-module $R^{n}$. We will treat vectors in $R^{n}$ as columns on which
matrices operate from the left.

Recall also that an affine mapping $F\colon R^{n}\to R^{n}$ is a mapping of
the form

$F(x)=L(x)+\tau,\quad x\in R^{n},$ (2.2)

where $L\colon R^{n}\to R^{n}$ is linear and $\tau\in R^{n}$ is a fixed
vector; $L$ and $\tau$ are called the linear and translational part of $F$,
respectively. Notice that $F$ is invertible if and only if so is $L$. The
group of all invertible affine transformations of $R^{n}$ is called the
($n$-th) general affine group over $R$ and denoted by
$\operatorname{GA}(n,R)$. The subgroup of $\operatorname{GA}(n,R)$ consisting
of all transformations with the linear part belonging to
$\operatorname{SL}(n,R)$ is called the ($n$-th) special affine group over $R$
and is denoted by $\operatorname{SA}(n,R)$.

If $1\leq n<m$, we can treat $R^{m}$ as the product $R^{m}=R^{n}\times
R^{m-n}$. We then define the canonical embedding
$\iota_{n,m}\colon\operatorname{GA}(n,R)\to\operatorname{GA}(m,R)$ by the
formula

$\iota_{n,m}(F)(x,y):=(F(x),y)\quad\textrm{for }(x,y)\in R^{n}\times
R^{m-n},\,F\in\operatorname{GA}(n,R).$ (2.3)

This embedding is obviously an injective group homomorphism.

If $F$ has the form (2.2), then $\iota_{m,n}(F)$ has the linear part equal to
$L\times I_{m-n}$ and the translational part equal to $(\tau,0_{m-n})$.

Let $X$ be a linear space over a non-Archimedean valued field
$(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$. A function
$\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert\colon X\to[0,\infty)$ is
called a non-Archimedean norm (briefly: norm) on $X$ if for all $x,y\in X$ and
$\alpha\in K$ we have the following conditions:

(1) $\lVert x\rVert=0\Leftrightarrow x=0$, (2) $\lVert\alpha
x\rVert=\lvert\alpha\rvert\lVert x\rVert$, (3) $\lVert
x+y\rVert\leq\max\\{\lVert x\rVert,\lVert y\rVert\\}$.

Then the pair $(X,\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert)$ is called
a non-Archimedean normed space over the non-Archimedean valued field
$(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$. The norm
$\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert$ induces an ultrametric $d$
on $X$, namely $d(x,y):=\lVert x-y\rVert$.

For any $n\in\mathbb{N}$, the function

$\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert\colon
K^{n}\to[0,\infty),\quad\lVert x\rVert:=\max\\{\lvert
x_{1}\rvert,\ldots,\lvert x_{n}\rvert\\},$

is a non-Archimedean norm on $K^{n}$; it is called the standard norm. Further
in the paper, we will denote by
$\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert$ the standard norm on
$K^{n}$, other norms will not be considered.

Let us state a useful lemma saying that if
$(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ is a non-Archimedean
discretely valued field with finite residue field, then, for any set $A$
contained in a ball $B$ in
$(K^{n},\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert)$ with
$\operatorname{Int}(A)\neq\emptyset$, the ball $B$ can be covered by a finite
number of translates of $A$. This result will be used in the proof of Theorem
5.5.

######  Lemma 2.1.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-Archimedean
discretely valued field with finite residue field $k$. If $r>0$,
$n\in\mathbb{N}$, $A\subseteq B[\mathbf{0},r]\subseteq K^{n}$ and $A$ has
nonempty interior, then there exist $m\in\mathbb{N}$ and $a_{1},\dots,a_{m}\in
B[\mathbf{0},r]$ such that
$B[\mathbf{0},r]\subseteq(a_{1}+A)\cup\dots\cup(a_{m}+A)$.

###### Proof.

Let us fix a uniformizer $\pi\in K$. We then have
$B[\mathbf{0},r]=B[\mathbf{0},\lvert\pi\rvert^{i}]$ for some $i\in\mathbb{Z}$.
Since $\operatorname{Int}(A)\neq\emptyset$, there exist $a\in A$ and $j\geq i$
such that $a+B[\mathbf{0},\lvert\pi\rvert^{j}]\subseteq A$. It will be
sufficient to show that the subgroup $B[\mathbf{0},\lvert\pi\rvert^{j}]$ of
$B[\mathbf{0},\lvert\pi\rvert^{i}]$ has finite index.

Let $q$ be the cardinality of $k$. Notice that $\pi^{j}D$ has index $q^{j-i}$
in $\pi^{i}D$. Indeed, if $T$ is a set of coset representatives in $D$ modulo
$\pi D$, then the set of all sums $\sum_{s=0}^{j-i-1}\pi^{i+s}t_{s}$, where
$t_{0},\dots,t_{j-i-1}\in T,$ is a set of coset representatives in $\pi^{i}D$
modulo $\pi^{j}D$. Since any $x=(x_{1},\dots,x_{n})$ and
$y=(y_{1},\dots,y_{n})$ in $B[\mathbf{0},\lvert\pi\rvert^{i}]$ are congruent
modulo $B[\mathbf{0},\lvert\pi\rvert^{j}]$ if and only if
$x_{s}-y_{s}\in\pi^{j}D$ for all $1\leq s\leq n$, it follows that the index of
$B[\mathbf{0},\lvert\pi\rvert^{j}]$ in $B[\mathbf{0},\lvert\pi\rvert^{i}]$
equals $q^{n(j-i)}$. ∎

The following theorem gives a characterization of linear isometries of the
normed space $(K^{n},\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert)$ for a
non-Archimedean valued field
$(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$. A similar result with
an additional assumption of completeness of the field
$(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ was proved in [14,
Corollary 1.7]. Our proof is simpler and direct.

######  Theorem 2.2.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-Archimedean
valued field and $n\in\mathbb{N}$. Let $U:K^{n}\to K^{n}$ be a linear map with
the matrix $[u_{i,j}]$ (also denoted by $U$) in the canonical basis of
$K^{n}$.

Then the following conditions are equivalent:

  1. (1)

$U$ is a linear isometry of the normed space
$(K^{n},\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert)$, i.e., $\lVert
Ux\rVert=\lVert x\rVert$ for any $x\in K^{n}$.

  2. (2)

$\lvert\det(U)\rvert=1$ and $\max_{1\leq i\leq n}\lvert u_{i,j}\rvert=1$ for
any $1\leq j\leq n$.

  3. (3)

$\lvert\det(U)\rvert=1$ and $\lvert u_{i,j}\rvert\leq 1$ for all $1\leq
i,j\leq n$.

  4. (4)

$U$ is invertible and $\lvert u_{i,j}\rvert,\lvert v_{i,j}\rvert\leq 1$ for
all $1\leq i,j\leq n,$ where $[v_{i,j}]$ is the matrix of the linear map
$V=U^{-1}$.

  5. (5)

$U\in\operatorname{GL}(n,D)$.

###### Proof.

$(1)\Rightarrow(2)$. Let $1\leq j\leq n.$ Then $\max_{1\leq i\leq n}\lvert
u_{i,j}\rvert=1$ since $\lVert Ue_{j}\rVert=\lVert e_{j}\rVert.$ Hence,

$\lvert\det(U)\rvert=\left\lvert\sum_{\sigma\in
S_{n}}\mathrm{sgn}(\sigma)\,u_{1,\sigma(1)}\dots
u_{n,\sigma(n)}\right\rvert\leq\max_{\sigma\in S_{n}}\lvert
u_{1,\sigma(1)}\rvert\dots\lvert u_{n,\sigma(n)}\rvert\leq 1.$

Clearly, $U$ is invertible and $U^{-1}$ is a linear isometry, so
$\lvert\det(U^{-1})\rvert\leq 1.$ Thus
$1\geq\lvert\det(U)\rvert=\lvert\det(U^{-1})\rvert^{-1}\geq 1,$ so
$\lvert\det(U)\rvert=1.$

$(2)\Rightarrow(3)$ is obvious.

$(3)\Rightarrow(4)$. $U$ is invertible, since $\det(U)\neq 0$. By Cramer’s
formula we have $\lvert
v_{i,j}\rvert=\left\lvert\frac{1}{\det(U)}(-1)^{i+j}\det(U_{j,i})\right\rvert=\lvert\det(U_{j,i})\rvert$,
where $U_{j,i}$ is the matrix obtained from $U$ by deleting the $j$-th row and
the $i$-th column. Estimating $\lvert\det(U_{j,i})\rvert$ as above, we get
$\lvert\det(U_{j,i})\rvert\leq 1,$ so $\lvert v_{i,j}\rvert\leq 1$ for all
$1\leq i,j\leq n.$

$(4)\Rightarrow(1)$. For $1\leq j\leq n$ we have

$\lVert
Ue_{j}\rVert=\left\lVert\sum_{i=1}^{n}u_{i,j}e_{i}\right\rVert=\max_{1\leq
i\leq n}\lvert u_{i,j}\rvert\leq 1,$

and for any $x=(x_{1},\ldots,x_{n})\in K^{n}$ we have

$\lVert
Ux\rVert=\left\lVert\sum_{j=1}^{n}x_{j}Ue_{j}\right\rVert\leq\max_{1\leq j\leq
n}\lvert x_{j}\rvert\lVert Ue_{j}\rVert\leq\max_{1\leq j\leq n}\lvert
x_{j}\rvert=\lVert x\rVert.$

Similarly, we get $\lVert Ve_{j}\rVert\leq 1$ for $1\leq j\leq n$, and $\lVert
Vx\rVert\leq\lVert x\lvert$ for any $x\in K^{n}.$ Hence, we get $\rVert
x\lVert=\rVert VUx\lVert\leq\rVert Ux\lVert\leq\rVert x\lVert,$ so $\rVert
Ux\lVert=\rVert x\ \rvert$ for any $x\in K^{n}.$

$(4)\Leftrightarrow(5)$ is obvious. ∎

######  Remark 2.3.

Thus, we can treat $\operatorname{GL}(n,D)$ as the group of all linear
isometries of the normed space
$(K^{n},\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert)$. The isometries
represented by elements of its subgroup $\operatorname{SL}(n,D)$ will be
called special linear isometries.

There is an analogy with the classical case of linear isometries of the
Euclidean space $\mathbb{R}^{3}$. All such isometries form the orthogonal
group

$\mathrm{O}(3):=\\{U\in M_{3}(\mathbb{R})\colon U^{T}U=I_{3}\\},$

whose elements necessarily satisfy $\lvert\det{U}\rvert=1.$ The special
orthogonal group

$\mathrm{SO}(3):=\\{U\in\mathrm{O}(3)\colon\det(U)=1\\}$

corresponds to the group of orientation-preserving linear isometries of
$\mathbb{R}^{3}$, which are precisely rotations around an axis containing the
origin.

An affine mapping $F\colon K^{n}\to K^{n}$ of the form (2.2) is an isometry of
the normed space $(K^{n},\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert)$ if
and only if so is its linear part $L$. By $\operatorname{GA}(n,D,K)$ we denote
the group of all affine isometries of
$(K^{n},\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert)$, and by
$\operatorname{SA}(n,D,K)$ we denote the intersection
$\operatorname{GA}(n,D,K)\cap\operatorname{SA}(n,K)$, whose elements will be
called special affine isometries. Thus,

$\operatorname{SA}(n,D,K)=\\{x\mapsto L(x)+\tau\colon
L\in\operatorname{SL}(n,D),\tau\in K^{n}\\}.$

Notice also that $\operatorname{SA}(n,D)$ can be regarded as the subgroup of
$\operatorname{SA}(n,D,K)$ consisting of all elements with translational parts
$\tau\in D^{n}$.

We will also need two related families of subgroups of
$\operatorname{SL}(n,D)$ and $\operatorname{SA}(n,D)$, respectively. Namely,
for $\varepsilon\in(0,1]$ and a subring $R\subseteq D$ containing $1$, let us
define

$\displaystyle\operatorname{SL}(n,R,\varepsilon):=\\{[a_{ij}]\in\operatorname{SL}(n,R)\colon\lvert
a_{ii}-1\rvert<\varepsilon,\lvert a_{ij}\rvert<\varepsilon\;\textrm{for }i\neq
j,1\leq i,j\leq n\\},$
$\displaystyle\operatorname{SA}(n,D,\varepsilon):=\\{x\mapsto L(x)+\tau\colon
L\in\operatorname{SL}(n,D,\varepsilon),\tau\in D^{n}\\}.$

##  3\. Paradoxical sets with respect to group actions

Most of the following definitions and facts are taken from the monograph [19]
and generalized in some cases. All group actions are assumed to be left. A set
$X$ together with a specified action of a group $G$ on it will be called a
$G$-set. For the union of (a family of) sets, the symbol “$\sqcup$” (or
“$\bigsqcup$”) will indicate disjointness of the components.

We say that a nonempty subset $E$ of a $G$-set $X$ is $G$-paradoxical [19,
Def. 1.1], if for some positive integers $m$, $n$ there exist pairwise
disjoint subsets $A_{1},\dots,A_{m},B_{1},\dots,B_{n}$ of $E$ and elements
$g_{1},\dots,g_{m},h_{1},\dots,h_{n}\in G$ such that

$E=\bigcup_{i=1}^{m}g_{i}(A_{i})=\bigcup_{j=1}^{n}h_{j}(B_{j}).$ (3.1)

Then there exist positive integers $m$, $n$, subsets
$A_{1},\dots,A_{m},B_{1},\dots,B_{n}$ of $E$ and elements
$g_{1},\dots,g_{m},h_{1},\dots,h_{n}\in G$ [7, Lemma A.12] such that

$E=A_{1}\sqcup\dots\sqcup A_{m}\sqcup B_{1}\sqcup\dots\sqcup
B_{n}=\bigsqcup_{i=1}^{m}g_{i}(A_{i})=\bigsqcup_{j=1}^{n}h_{j}(B_{j}).$ (3.2)

In such a case the decomposition (3.2) of $E$ is called a $G$-paradoxical
decomposition.

Additionally, if the subset $E$ of $X$ is $G$-invariant, we can assume that
$g_{1}=h_{1}=1_{G}$ and the $g_{i}$’s, $h_{j}$’s, $m$, $n$ from (3.2) are the
same as those from (3.1) (cf. [7, Lemma A.12]).

We say that subsets $A$ and $B$ of a $G$-set $X$ are $G$-equidecomposable [19,
Def. 3.4] if for some $n\in\mathbb{N}$ there exist decompositions

$A=\bigsqcup_{i=1}^{n}A_{i},\;B=\bigsqcup_{i=1}^{n}B_{i},$ (3.3)

and elements $g_{1},\dots,g_{n}\in G$ such that $g_{i}(A_{i})=B_{i}$ for any
$1\leq i\leq n.$ Then we write $A\sim_{G}B$ or $A\sim B$. If we want to
emphasize that the decompositions (3.3) can be realized with $n$ pieces, we
write $A\sim_{G,n}B$ or $A\sim_{n}B$.

Clearly, a subset $E$ of a $G$-set $X$ is $G$-paradoxical if and only if $E$
is the union of disjoint subsets $A$ and $B$ such that $A\sim_{m}E\sim_{n}B$
for some $m,n\geq 1$. We then say that $E$ has a $G$-paradoxical decomposition
using $r=m+n$ pieces. Obviously, if $E\subseteq X$ is $H$-paradoxical (using
$r$ pieces) for some $H\leq G$, then $E$ is $G$-paradoxical (using $r$
pieces).

Let $X$ be a $G$-set and $H\leq G$. It is not hard to check that if
$E\subseteq X$ is $H$-paradoxical using $r$ pieces, then $g(E)$ is
$H$-paradoxical using $r$ pieces for any $g\in G$ with $gHg^{-1}\leq H$.

If not stated otherwise, in the case $X=G$ we mean the action of $G$ on itself
by left multiplication. If a group $G$ is $G$-paradoxical, the least number
$r$ for which there exists a paradoxical decomposition of $G$ using $r$ pieces
is called the Tarski number of $G$ [3]. In the same way we talk about the
Tarski number of a $G$-paradoxical subset $E$ of a $G$-set $X$; clearly it is
greater or equal to four.

Since all free groups of a given rank are isomorphic, we will sometimes abuse
notation and write $F_{n}$ for any free group of rank $n$. By the free group
$F_{n}$ we will denote the model group of reduced words over an alphabet
consisting of $n$ symbols and their formal inverses.

The free group $F_{2}$ is the most important example of a paradoxical group.
Its Tarski number is four [19, Theorem 5.2]. Indeed, let $\\{\tau,\sigma\\}$
be a basis of $F_{2}$. We will denote by $W(\xi)$ the set of all elements
(reduced words) of $F_{2}$ which begin with the letter
$\xi\in\\{\tau,\tau^{-1},\sigma,\sigma^{-1}\\}$, and $1_{F_{2}}$ will stand
for the empty word. Let $A_{1}:=W(\tau),A_{2}:=W(\tau^{-1}),$
$B_{1}:=W(\sigma)\cup\\{1_{F_{2}}\\}\cup\\{\sigma^{-n}\colon n\geq 1\\}$ and
$B_{2}:=W(\sigma^{-1})\setminus\\{\sigma^{-n}\colon n\geq 1\\}$. It is easy to
check that $F_{2}=A_{1}\sqcup A_{2}\sqcup B_{1}\sqcup
B_{2}=A_{1}\sqcup\tau(A_{2})=B_{1}\sqcup\sigma(B_{2});$ hence,
$A\sim_{2}F_{2}\sim_{2}B,$ where $A:=A_{1}\sqcup A_{2}$ and $B:=B_{1}\sqcup
B_{2}$.

If $G$ acts on $X$ and $x\in X$, the subgroup $\mathrm{Stab}_{G}(x):=\\{g\in
G\colon g(x)=x\\}$ of $G$ is called the stabilizer of $x$. When
$\mathrm{Stab}_{G}(x)=\\{1_{G}\\}$ for all $x\in X$, then we say that $G$ acts
on $X$ without nontrivial fixed points. If $\mathrm{Stab}_{G}(x)$ is an
Abelian group for every $x\in X$, we say that $G$ acts on $X$ locally
commutatively.

If a group $G$ is $G$-paradoxical using $r$ pieces and $G$ acts on $X$ without
nontrivial fixed points, then the $G$-set $X$ is $G$-paradoxical using $r$
pieces [19, Proposition 1.10]. Hence, if a group $G$ acts on $X$ without
nontrivial fixed points and $G$ contains a free subgroup $F_{2}$, then the set
$X$ is $G$-paradoxical using $4$ pieces.

Under the axiom of choice, we have the following characterization of
$G$-paradoxicality using $4$ pieces. A $G$-set $X$ is $G$-paradoxical using
$4$ pieces if and only if $G$ contains a free subgroup $F_{2}$ whose action on
$X$ is locally commutative. More precisely, $X$ has a decomposition

$X=A_{1}\sqcup A_{2}\sqcup A_{3}\sqcup
A_{4}=A_{1}\sqcup\sigma(A_{2})=A_{3}\sqcup\tau(A_{4})$ (3.4)

for $\sigma,\tau\in G$ if and only if $\\{\sigma,\tau\\}$ is a basis of a free
subgroup $F_{2}\leq G$ whose action on $X$ is locally commutative [19,
Theorems 5.5 and 5.8].

######  Theorem 3.1.

Let $X$ be a $G$-set, $A,B,C,Z\subseteq X$ be pairwise disjoint and $E=A\sqcup
B\sqcup C$. Let $H$ be a subgroup of $G$. If $A\sim_{H,m}E\sim_{H,n}B$ and
$C\sim_{G,l}Z$, the set $E\sqcup Z$ is $G$-paradoxical using $m+n+l+1$ pieces.

If, additionally, $A$, $B$ or $E$ is $H$-invariant, then $E\sqcup Z$ is
$G$-paradoxical using $m+n+l$ pieces.

###### Proof.

From the assumption we have

$E\sqcup Z=(A\sqcup Z)\sqcup(B\sqcup C).$ (3.5)

From the relations $B\sim_{H,n}E$ and $C\sim_{G,l}Z$, having in mind the
disjointness of $B$ with $C$ and $E$ with $Z$, we conclude that $(B\sqcup
C)\sim_{G,n+l}(E\sqcup Z)$. Similarly we justify $(A\sqcup
Z)\sim_{H,m+1}(E\sqcup Z)$, which finishes the proof of the first part of our
theorem.

If $A$ or $E$ is $H$-invariant, the relation $A\sim_{H,m}E$ can be realized by
decompositions

$A=A_{1}\sqcup\left(\bigsqcup_{i=2}^{m}A_{i}\right),\quad
E=A_{1}\sqcup\left(\bigsqcup_{i=2}^{m}\alpha_{i}(A_{i})\right),$

for some $\alpha_{2},\ldots,\alpha_{m}\in H$. That means, $A_{1}$ and $Z$ can
be joined together into $A_{1}\sqcup Z$ so that we obtain the relation
$(A\sqcup Z)\sim_{H,m}(E\sqcup Z)$ and diminish by one the number of pieces
sufficient to a $G$-paradoxical decomposition of $E\sqcup Z$. If $B$ is
$H$-invariant, one needs to repeat the reasoning, interchanging the roles of
$A$ and $B$ in (3.5). ∎

An elementary example of the situation $A\sim_{G}E\sim_{G}B$, $E=A\sqcup
B\sqcup C$ with a nonempty $C$ is the action of the free group $F_{2}$ over
the alphabet $\\{\sigma,\tau\,\sigma^{-1},\tau^{-1}\\}$ on itself. Let
$A_{1}=W(\sigma)$, $A_{2}=W(\sigma^{-1})$, $B_{1}=W(\tau)$,
$B_{2}=W(\tau^{-1})$, $C=\\{1_{F_{2}}\\}$ and $A=A_{1}\sqcup A_{2}$,
$B=B_{1}\sqcup B_{2}$, where $W(\xi)$ denotes the set of all elements of
$F_{2}$ which begin with a letter
$\xi\in\\{\sigma,\tau\,\sigma^{-1},\tau^{-1}\\}$. It is easy to check that

$F_{2}=A_{1}\sqcup\sigma(A_{2})=B_{1}\sqcup\tau(B_{2}),$

so $A\sim_{2}F_{2}\sim_{2}B$. We will make us of this in a more general
situation of a locally commutative action of $F_{2}$ on $X$.

######  Theorem 3.2.

Let $X$ be a $G$-set. Assume that $G$ contains a free subgroup $F=F_{2}$ which
acts locally commutatively on a subset $E$ of $X$. Let $C$ be a subset of $E$
such that $\mathrm{Stab}_{F}(x)=\\{1_{F}\\}$ and $Fx\cap C=\\{x\\}$ for every
$x\in C.$ Then $E$ has a decomposition $E=A_{1}\sqcup A_{2}\sqcup B_{1}\sqcup
B_{2}\sqcup C$ such that $E=A_{1}\sqcup\sigma(A_{2})=B_{1}\sqcup\tau(B_{2}),$
where $\\{\sigma,\tau\\}$ is a basis of $F$.

If $Z\subseteq X\setminus E$ and $Z\sim_{G,l}C,$ then $Z\sqcup E$ is
$G$-paradoxical using $4+l$ pieces. If $\alpha(C)\subseteq X\setminus E$ for
some $\alpha\in G$ and $Y:=\alpha(C)\sqcup E,$ then

$Y=(\alpha(C)\sqcup A_{1})\sqcup A_{2}\sqcup B_{1}\sqcup B_{2}\sqcup
C=(\alpha(C)\sqcup
A_{1})\sqcup\sigma(A_{2})=B_{1}\sqcup\tau(B_{2})\sqcup\alpha(C),$

so $\alpha(C)\sqcup E$ is $H$-paradoxical using $5$ pieces, where $H$ is the
subgroup of $G$ generated by $\sigma,\tau$ and $\alpha$.

###### Proof.

For any disjoint subsets $U,V$ of $F$ and $x\in C$ we have $Vx\cap
Ux=\emptyset,$ since $\mathrm{Stab}_{F}(x)=\\{1_{F}\\}.$ Moreover, $Fx\cap
Fy=\emptyset$ for any different $x,y\in C$, since $Fx\cap C=\\{x\\}.$ The
group $F$ acts locally commutatively on $E$ and the set
$E_{1}:=\bigsqcup_{x\in C}Fx$ is $F$-invariant, so $F=F_{2}$ acts locally
commutatively on the set $E_{2}:=E\setminus E_{1}.$ Thus, $E_{2}$ has a
decomposition

$E_{2}=A^{2}_{1}\sqcup A^{2}_{2}\sqcup B^{2}_{1}\sqcup B^{2}_{2}$

such that

$E_{2}=A^{2}_{1}\sqcup\sigma(A^{2}_{2})=B^{2}_{1}\sqcup\tau(B^{2}_{2}).$

The sets $A^{1}_{1}:=\bigsqcup_{x\in C}W(\sigma)x,A^{1}_{2}:=\bigsqcup_{x\in
C}W(\sigma^{-1})x,B^{1}_{1}:=\bigsqcup_{x\in C}W(\tau)x$ and
$B^{1}_{2}:=\bigsqcup_{x\in C}W(\tau^{-1})x$ are pairwise disjoint and

$E_{1}=A^{1}_{1}\sqcup A^{1}_{2}\sqcup B^{1}_{1}\sqcup B^{1}_{2}\sqcup
C=A^{1}_{1}\sqcup\sigma(A^{1}_{2})=B^{1}_{1}\sqcup\tau(B^{1}_{2}).$

It follows that

$E=A_{1}\sqcup A_{2}\sqcup B_{1}\sqcup B_{2}\sqcup
C=A_{1}\sqcup\sigma(A_{2})=B_{1}\sqcup\tau(B_{2}),$

where $A_{1}=A^{1}_{1}\sqcup A^{2}_{1},A_{2}=A^{1}_{2}\sqcup
A^{2}_{2},B_{1}=B^{1}_{1}\sqcup B^{2}_{1}$ and $B_{2}=B^{1}_{2}\sqcup
B^{2}_{2}.$

Let $Z\subseteq X\setminus E$ with $Z\sim_{G,l}C.$ Then $C$ has a
decomposition $C=\bigsqcup_{i=1}^{l}C_{i}$ such that
$Z=\bigsqcup_{i=1}^{l}\alpha_{i}(C_{i})$ for some
$\alpha_{1},\ldots,\alpha_{l}\in G$. Thus,

$Z\sqcup E=(Z\sqcup A_{1})\sqcup A_{2}\sqcup B_{1}\sqcup
B_{2}\sqcup\bigsqcup_{i=1}^{l}C_{i}=(Z\sqcup
A_{1})\sqcup\sigma(A_{2})=B_{1}\sqcup\tau(B_{2})\sqcup\bigsqcup_{i=1}^{l}\alpha_{i}(C_{i}),$

so $Z\sqcup E$ is $G$-paradoxical using $4+l$ pieces.

Hence, for $Z:=\alpha(C)$ we get the last part of the theorem. ∎

In particular, we have obtained a tool for proving paradoxicality of some sets
using $5$ pieces. It is a general statement of the method used originally by
Robinson [12] (see also[19, Theorem 5.7]) for getting a paradoxical
decomposition of a ball in $\mathbb{R}^{3}$ using $5$ pieces, based on an
action of a free subgroup of the group of rotations $\mathrm{SO}(3)$.

Now we will prove a simple but important lemma, which will be used several
times in our paper. It will serve as a main tool to “lift” some paradoxical
decompositions of subsets of $K^{2}$ to higher dimensions.

######  Lemma 3.3.

Let $X$ be a $G$-set, $X^{\prime}$ a $G^{\prime}$-set and $\iota\colon G\to
G^{\prime}$ a group homomorphism. Assume that $E\subseteq X$,
$E^{\prime}\subseteq X^{\prime}$, $E^{\prime}$ is $\iota(G)$-invariant and a
function $f\colon E^{\prime}\to E$ satisfies

$g\circ f=f\circ(\iota(g))\quad\textrm{for all }\;g\in G.$ (3.6)

If $A\subseteq E$ and $A\sim_{G}E$ using $m$ pieces, then
$f^{-1}(A)\sim_{\iota(G)}E^{\prime}$ using $m$ pieces. In particular, if $E$
is $G$-paradoxical using $r$ pieces, then $E^{\prime}$ is
$\iota(G)$-paradoxical using $r$ pieces.

###### Proof.

From the assumption we have

$A=\bigsqcup_{i=1}^{m}A_{i},\quad E=\bigsqcup_{i=1}^{m}g_{i}(A_{i}),\quad
g_{1},\dots,g_{m}\in G;$

therefore also

$f^{-1}(A)=\bigsqcup_{i=1}^{m}f^{-1}(A_{i}),\quad
E^{\prime}=\bigsqcup_{i=1}^{m}f^{-1}(g_{i}(A_{i})),\quad g_{1},\dots,g_{m}\in
G.$

It follows from (3.6) that, for any $i\in\\{1,\dots,m\\}$, we have

$\displaystyle f^{-1}(g_{i}(A_{i}))$ $\displaystyle=\\{x\in E^{\prime}\colon
f(x)\in g_{i}(A_{i})\\}=\\{x\in E^{\prime}\colon(g_{i}^{-1}\circ f)(x)\in
A_{i}\\}$ $\displaystyle=\\{x\in
E^{\prime}\colon(f\circ\iota(g_{i}^{-1}))(x)\in A_{i}\\}=\\{x\in
E^{\prime}\colon\iota(g_{i}^{-1})(x)\in f^{-1}(A_{i})\\}$
$\displaystyle=\iota(g_{i})(f^{-1}(A_{i})),$

so $E^{\prime}=\bigsqcup\limits_{i=1}^{m}\iota(g_{i})(f^{-1}(A_{i}))$, which
shows that $f^{-1}(A)\sim_{\iota(G),m}E^{\prime}$. ∎

If a group $G$ acts on $X$ and $E\subseteq X$ is $G$-paradoxical, then $E$ is
$G$-equidecomposable with any subset $A$ of $E$ such that

$E\subseteq g_{1}A\cup\dots\cup g_{n}A\quad\textrm{for some
}n\in\mathbb{N},\;g_{1},\dots,g_{n}\in G.$ (3.7)

Hence, any two such subsets $A^{\prime},A^{\prime\prime}$ of $E$ are
$G$-equidecomposable and both $G$-paradoxical [19, Corollary 10.22].

###  3.1. Amenability and nonexistence of paradoxical decompositions

The concept of amenability (i.e., admitting some invariant measure by a group
$G$) is closely connected to nonexistence of paradoxical decompositions of
$G$-sets and their subsets.

Let $X$ be a $G$-set. A finitely additive measure
$\mu\colon\mathcal{P}(X)\to[0,\infty]$ is called $G$-invariant if it satisfies

$\mu(gA)=\mu(A)\quad\textrm{for any }\;g\in G,\,A\subseteq X.$ (3.8)

If $X=G$, we say that such $\mu$ is left-invariant.

A group $G$ is amenable if there exists a left-invariant, finitely additive
measure $\mu\colon\mathcal{P}(G)\to[0,1]$ with $\mu(G)=1$.

If $\mu\colon\mathcal{P}(X)\to[0,\infty]$ is a $G$-invariant, finitely
additive measure on a $G$-set $X$ and $E\subseteq X$ is $G$-paradoxical with
$\mu(E)<\infty$, then $\mu(E)=2\mu(E)$ and so $\mu(E)=0$. Thus the existence
of a $G$-invariant measure on $X$ excludes $G$-paradoxical decomposition of
any $E\subseteq X$ with $0<\mu(E)<\infty$.

In fact, the converse also holds (if the axiom of choice is assumed). The
following fact is known as Tarski’s theorem (see [17] and a modern approach
[19, Corollary 11.2]): A subset $E$ of a $G$-set $X$ is $G$-paradoxical if and
only if there is no $G$-invariant measure $\mu$ on $X$ with $\mu(E)=1$.

It follows that a group $G$ is amenable if and only if $G$ is not
$G$-paradoxical [7, Theorem A.13].

If an amenable group $G$ acts on a set $X$, then $X$ is not $G$-paradoxical;
in fact, any subset of $X$ containing a nonempty $G$-invariant subset is not
$G$-paradoxical (see [19, Theorem 12.3] and its proof).

Let us recall a few basic facts about amenable groups [19, Theorem 12.4], [7,
Theorem 3.2]:

  1. (1)

All finite and Abelian groups are amenable.

  2. (2)

A subgroup of an amenable group is amenable.

  3. (3)

If $H$ is a normal subgroup of a group $G$, then $G$ is amenable if and only
$H$ and $G/H$ are both amenable.

  4. (4)

If $G$ is the direct union of a directed system $(G_{\alpha})_{\alpha\in I}$
of amenable groups, then $G$ is amenable.

  5. (5)

All solvable groups are amenable.

It follows from (4) that any locally finite group (i.e., such that its every
finite subset generates a finite subgroup) is amenable.

If a group $G$ contains a free subgroup $F_{2}$ of rank two, then $G$ is not
amenable, since $F_{2}$ is $F_{2}$-paradoxical, so nonamenable. We will see
that for an important class of groups the converse also holds.

There is a famous work of Tits [18] classifying linear groups, i.e., subgroups
of the group $\operatorname{GL}(V)$ of all linear automorphisms of a linear
space $V$ over a field $K$. Its main results can be reformulated in the
context of amenability. The next statement is a conclusion from [21, Theorem
1] and [18, Theorem 2] (see also Theorem 12.6 in [19] and the discussion
following it).

######  Proposition 3.4.

Let $K$ be a field and $n\in\mathbb{N}$. Let $G$ be a subgroup of
$\operatorname{GL}(n,K)$. Then $G$ is amenable or has a free subgroup of rank
two.

###### Proof.

Suppose that $G$ has no subgroup of rank two. If $\operatorname{char}{K}=0$,
then G has a normal solvable subgroup of finite index [21, Theorem 1], so $G$
is amenable. If $\operatorname{char}{K}>0$, then $G$ has a normal solvable
subgroup $N$ such that $G/N$ is locally finite [18, Theorem 2], so $G$ is
amenable. ∎

Let us present a nontrivial example of an amenable linear group.

######  Example 3.5.

Let $K$ be a finite field and $L$ an algebraic extension of $K$. It is known
that $L$ can be expressed as the union $L=\bigcup_{\alpha\in I}K_{\alpha}$ of
all subfields $K\subseteq K_{\alpha}\subseteq L$ such that $K_{\alpha}$ is a
finite extension of $K$ (in particular, also a finite field). The system
$(K_{\alpha})_{\alpha\in I}$ is directed by inclusion. Therefore, for any
$n\in\mathbb{N}$, there is a decomposition

$\operatorname{GL}(n,L)=\bigcup_{\alpha\in I}\operatorname{GL}(n,K_{\alpha})$

of $\operatorname{GL}(n,L)$ as the direct union of a directed system of finite
groups, so $\operatorname{GL}(n,L)$ is amenable.

The following simple lemma will be used in the proof of Theorem 5.8.

######  Lemma 3.6.

Let $K$ be a field and $n\in\mathbb{N}$. If $\operatorname{GL}(n,K)$ is
amenable, then so is $\operatorname{GA}(n,K)$.

###### Proof.

Let $\Phi\colon\operatorname{GA}(n,K)\to\operatorname{GL}(n,K)$ be the
epimorphism sending an affine transformation to its linear part. Clearly, its
kernel is the group of all translations $x\mapsto x+\tau$, $\tau\in K^{n}$.
Hence, $\operatorname{Ker}{\Phi}\cong(K^{n},+)$ is amenable as an Abelian
group. Since
$\operatorname{GA}(n,K)/\operatorname{Ker}{\Phi}\cong\operatorname{GL}(n,K)$
is amenable by the assumption, so is $\operatorname{GA}(n,K)$. ∎

##  4\. Some free subgroups of special linear groups $\operatorname{SL}(n,D)$
and special affine groups $\operatorname{SA}(n,D)$

Let us begin with a simple observation from linear algebra.

######  Lemma 4.1.

Let $K$ be a field and $M\in\operatorname{SL}(2,K)$. Then
$\operatorname{tr}{M}=2$ if and only if the linear map $M\colon K^{2}\to
K^{2}$ has a nonzero fixed point.

###### Proof.

The characteristic polynomial of $M$ has the form

$f(\lambda)=\det(M-\lambda I_{2})=\lambda^{2}-(\operatorname{tr}M)\lambda+1,$
(4.1)

so $\det(M-I_{2})=f(1)=0$ if and only if $\operatorname{tr}M=2$. ∎

The following lemma gives some conditions under which a group of isometries of
$K^{n}$ acts on certain balls and spheres.

######  Lemma 4.2.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-Archimedean
valued field and $n\in\mathbb{N}$. Let $G$ be a group of isometries of the
normed space $(K^{n},\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert)$ and
$\Sigma\subseteq G$ a generating set of $G$. Let $x\in K^{n}$ and $r>0$ be
fixed. If $\lVert g(x)-x\rVert<r$ for all $g\in\Sigma$, then the balls
$B[x,r]$, $B(x,r)$, and the sphere $S[x,r]$ are $G$-invariant.

###### Proof.

Let $g\in\Sigma$. Since $g$ is a surjective isometry, the images of $B[x,r]$,
$B(x,r)$, $S[x,r]$ under $g$ are equal to $B[g(x),r]$, $B(g(x),r)$,
$S[g(x),r]$, respectively. Since $\lVert g(x)-x\rVert<r$, we obtain
$B[g(x),r]=B[x,r]$ and $B(g(x),r)=B(x,r)$ (by the strong triangle inequality),
as well as $S[g(x),r]=S[x,r]$ (by the isosceles property). Thus, the sets
under consideration are invariant with respect to all $g\in\Sigma$, hence also
$G$-invariant. ∎

The proposition below provides a tool for finding explicit bases of some free
groups of affine mappings acting on certain balls and spheres in $K^{n}$
without nontrivial fixed points.

######  Proposition 4.3.

Let $K$ be a field and $n\in\mathbb{N}$. Assume that a family
$\\{A_{i}\\}_{i\in S}$ is a basis of a free subgroup
$F\leq\operatorname{GL}(n,K)$ such that $\det(M-I_{n})\neq 0$ for all $M\in
F\setminus\\{I_{n}\\}$. Let $\hat{x}\in K^{n}$ and
$u_{i}:=(I_{n}-A_{i})\,\hat{x}$ for $i\in S$. Then, the family
$\\{T_{i}\\}_{i\in S}$, where

$T_{i}\colon K^{n}\to K^{n},\quad T_{i}(x):=A_{i}(x)+u_{i},$ (4.2)

is a basis of a free subgroup $F^{\prime}$ of $\operatorname{GA}(n,K)$ acting
on $K^{n}\setminus\\{\hat{x}\\}$ without nontrivial fixed points.

If, additionally, $K$ is equipped with a non-Archimedean valuation
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$, $x\in
K^{n}\setminus\\{\hat{x}\\}$, $0<r<\lVert x-\hat{x}\rVert$ and $F$ is a
subgroup of $\operatorname{SL}(n,D,\varepsilon)$ for some
$0<\varepsilon\leq\varepsilon_{0}$, where

$\varepsilon_{0}:=\begin{cases}\min\left(\frac{r}{\lVert
x-\hat{x}\rVert},\frac{1}{\lVert\hat{x}\rVert}\right)&\textrm{if
}\hat{x}\neq\mathbf{0},\\\ \frac{r}{\lVert x\rVert}&\textrm{if
}\hat{x}=\mathbf{0},\end{cases}$ (4.3)

then $F^{\prime}\leq\operatorname{SA}(n,D,\varepsilon)$ and $F^{\prime}$ acts
on the balls $B[x,r]$, $B(x,r)$, and the sphere $S[x,r]$ without nontrivial
fixed points.

###### Proof.

Denote by $F^{\prime}$ the subgroup of $\operatorname{GA}(n,K)$ generated by
the family $\\{T_{i}\\}_{i\in S}$. Let
$\Phi\colon\operatorname{GA}(n,K)\to\operatorname{GL}(n,K)$ be the
homomorphism sending an affine transformation to its linear part. Since $\Phi$
maps $\\{T_{i}\\}_{i\in S}$ one-to-one onto the basis $\\{A_{i}\\}_{i\in S}$
of $F$, the group $F^{\prime}$ is free with $\\{T_{i}\\}_{i\in S}$ as basis
[8, Proposition 1.8] and $\phi:=\Phi|_{F^{\prime}}\colon F^{\prime}\to F$ is
an isomorphism.

If $T=M+u\in F^{\prime}$ and $T\neq 1_{F^{\prime}}$, then $M=\phi(T)\neq
I_{n}$, so $\det(I_{n}-M)\neq 0$ and $(I_{n}-M)^{-1}\,u$ is a unique fixed
point of $M$ in $K^{n}$. On the other hand, $\hat{x}$ is a fixed point of
$T_{i}$ for all $i\in S$ and, as a consequence, for all $T\in F^{\prime}$.
Therefore, $F^{\prime}$ acts on $K^{n}\setminus\\{\hat{x}\\}$ without
nontrivial fixed points.

Suppose additionally that the assumptions of the second part of the
proposition are satisfied. Clearly, the sets $B[x,r]$, $B(x,r)$, $S[x,r]$ are
contained in $K^{n}\setminus\\{\hat{x}\\}$. Since
$F\leq\operatorname{SL}(n,D,\varepsilon)$ and $\lVert
u_{i}\rVert\leq\varepsilon\lVert\hat{x}\rVert\leq\varepsilon_{0}\lVert\hat{x}\rVert\leq
1$ for all $i\in S$, we have
$F^{\prime}\leq\operatorname{SA}(n,D,\varepsilon)$. The choice of
$\varepsilon_{0}$ guarantees that

$\lVert
T_{i}(x)-x\rVert=\lVert(A_{i}-I_{n})\,x+u_{i}\rVert=\lVert(A_{i}-I_{n})(x-\hat{x})\rVert<\varepsilon\lVert
x-\hat{x}\rVert\leq\varepsilon_{0}\lVert x-\hat{x}\rVert\leq r$

for all $i\in S$. Therefore, by Lemma 4.2, $F^{\prime}$ acts on $B[x,r]$,
$B(x,r)$ and $S[x,r]$ without nontrivial fixed points. ∎

######  Remark 4.4.

To ensure that the group $F^{\prime}$ from Proposition 4.3 acts without
nontrivial fixed points on the open ball $B(x,r)$, a weaker assumption on $r$
is sufficient, namely $0<r\leq\lVert x-\hat{x}\rVert$.

###  4.1. The case when the characteristic of $K$ is zero

Let $K$ be a field of characteristic $0$. We will identify its prime field
with the field $\mathbb{Q}$ of rational numbers. If $K$ is equipped with a
non-Archimedean valuation and $D$ is the ring of integers of $K$, then
$\mathbb{Z}\subseteq D$ and
$\operatorname{SL}(2,\mathbb{Z})\leq\operatorname{SL}(2,D)$.

We will rely on the following remarkable result due to Neumann [10] and Magnus
[9].

######  Theorem 4.5 ([9, Theorems 3 and 4]).

The matrices

$A_{0}:=\left(\begin{array}[]{cc}1&1\\\ 1&2\end{array}\right),\quad
A_{n}:=\left(\begin{array}[]{cc}4n^{2}+1&2n\\\ 2n&1\end{array}\right),\quad
n=1,2,\dots$ (4.4)

form a basis of a free subgroup $F$ of $\operatorname{SL}(2,\mathbb{Z})$.
Moreover, every element of $F\setminus\\{I_{2}\\}$ is nonparabolic, i.e., its
trace is different from $\pm 2$.

######  Remark 4.6.

The first part of the above-stated result corresponds to [9, Theorem 3], which
is directly based on Neumann’s paper [10]. The second part is contained in [9,
Theorem 4] and proved thereafter.

Taking into account Theorem 4.5 and Lemma 4.1, we obtain an immediate
conclusion.

######  Corollary 4.7.

If $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ is a non-Archimedean
valued field and $\operatorname{char}{K}=0$, the group
$\operatorname{SL}(2,\mathbb{Z})\leq\operatorname{SL}(2,D)$ contains a free
subgroup of infinite rank acting on $K^{2}\setminus\\{\mathbf{0}\\}$, and (for
any $r>0$) on $B[\mathbf{0},r]\setminus\\{\mathbf{0}\\}$ and $S[\mathbf{0},r]$
without nontrivial fixed points.

If $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ is a non-Archimedean
valued field of characteristic $0$, we distinguish two cases depending on the
characteristic of $k$, the residue field of $K$. We either have
$\operatorname{char}{k}=0$, in which case the valuation
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ restricted to the prime
field $\mathbb{Q}\subseteq K$ is trivial, or $\operatorname{char}{k}=p$ for a
prime number $p$. In the latter case, the valuation restricted to $\mathbb{Q}$
is equivalent to the $p$-adic valuation
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{p}$, given by (2.1).

The following results provide the existence of some free groups of linear and
affine isometries of the normed space
$(K^{n},\lVert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rVert)$, in the case when
$\operatorname{char}{K}\neq\operatorname{char}{k}$, acting on certain sets
without nontrivial fixed points. They will be used in Section $5$.

######  Proposition 4.8.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-Archimedean
valued field and $k$ its residue field. Assume that $\operatorname{char}{K}=0$
and $\operatorname{char}{k}>0$. Then, for any $\varepsilon\in(0,1]$, there
exists an infinite set $S(\varepsilon)\subseteq\mathbb{N}$ such that the
family $\\{A_{i}\\}_{i\in S(\varepsilon)}$, where $A_{i}$ are as in (4.4), is
a basis of a free subgroup
$F(\varepsilon)\leq\operatorname{SL}(2,\mathbb{Z},\varepsilon)$.

###### Proof.

We can see from (4.4) that, for $n\geq 1$, all entries of $(A_{n}-I_{2})$ have
valuation bounded from above by $\lvert n\rvert$. Let
$p:=\operatorname{char}{k}$. Since $\lvert p\rvert<1$, we have
$A_{p^{m}}\in\operatorname{SL}(2,\mathbb{Z},\varepsilon)$ for sufficiently
large $m\in\mathbb{N}$. ∎

######  Theorem 4.9.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-Archimedean
valued field and $k$ its residue field. Assume that $\operatorname{char}{K}=0$
and $\operatorname{char}{k}>0$. Let $n\geq 2$,
$\hat{x}=(\hat{x}_{1},\hat{x}_{2},\dots,\hat{x}_{n})\in K^{n}$,
$x=(x_{1},x_{2},\dots,x_{n})\in K^{n}$,
$(x_{1},x_{2})\neq(\hat{x}_{1},\hat{x}_{2})$,
$0<r<\lVert(x_{1},x_{2})-(\hat{x}_{1},\hat{x}_{2})\rVert$ and
$\varepsilon\in(0,1]$. Then, there exists a free subgroup
$F^{\prime}\leq\operatorname{SA}(n,D,\varepsilon)$ of infinite rank acting
without nontrivial fixed points on
$K^{n}\setminus\left(\\{(\hat{x}_{1},\hat{x}_{2})\\}\times K^{n-2}\right)$,
the balls $B[x,r]$, $B(x,r)$, and the sphere $S[x,r]$.

Moreover, $F^{\prime}$ can be chosen so that
$F^{\prime}\leq\operatorname{SA}(n,\mathbb{Z})$ if
$(\hat{x}_{1},\hat{x}_{2})\in\mathbb{Z}^{2}$, and
$F^{\prime}\leq\operatorname{SL}(n,\mathbb{Z},\varepsilon)$ if
$(\hat{x}_{1},\hat{x}_{2})=(0,0)$.

###### Proof.

First, let us consider the case $n=2$. Let $\varepsilon_{0}$ be defined as in
(4.3) and $\varepsilon_{1}:=\min(\varepsilon,\varepsilon_{0})$. Applying
Proposition 4.8, we obtain a free subgroup
$F(\varepsilon_{1})\leq\operatorname{SL}(2,\mathbb{Z},\varepsilon_{1})$ with
$\\{A_{i}\\}_{i\in S(\varepsilon_{1})}$ as basis. Notice that
$\det(M-I_{2})=2-\operatorname{tr}M\neq 0$ for all nonidentity $M\in
F(\varepsilon_{1})$. A direct application of Proposition 4.3 yields a desired
free subgroup
$F^{\prime}\leq\operatorname{SA}(2,D,\varepsilon_{1})\leq\operatorname{SA}(2,D,\varepsilon)$
with a basis $\\{T_{i}\\}_{i\in S(\varepsilon_{1})}$ as in (4.2).

Assume now that $n>2$. Let us again apply Proposition 4.3, but with $2$ in
place of $n$, the vectors $\hat{x}$, $x$ replaced by their projections on the
first two coordinates, and with $\varepsilon_{0}$, $\varepsilon_{1}$ changed
appropriately. We obtain a free subgroup
$F^{\prime\prime}\leq\operatorname{SA}(2,D,\varepsilon_{1})\leq\operatorname{SA}(2,D,\varepsilon)$
acting on $K^{2}\setminus\\{(\hat{x}_{1},\hat{x}_{2})\\}$ without nontrivial
fixed points. As before, $\\{T_{i}\\}_{i\in S(\varepsilon_{1})}$ from (4.2) is
a basis of $F^{\prime\prime}$. Let
$F^{\prime}:=\iota_{2,n}(F^{\prime\prime})$, where $\iota_{2,n}$ is defined as
in (2.3). Clearly, $F^{\prime}$ is a free subgroup of infinite rank of
$\operatorname{SA}(n,D,\varepsilon)$. We will show that $F^{\prime}$ satisfies
the claim of our theorem. It is not hard to check (by examining the equation
(2.3)) that $F^{\prime}$ acts without nontrivial fixed points on the set
$Y:=K^{n}\setminus\left(\\{(\hat{x}_{1},\hat{x}_{2})\\}\times K^{n-2}\right)$,
which contains $B[x,r]$, $B(x,r)$ and $S[x,r]$. As in the proof of Proposition
4.3, we have

$\lVert\iota_{2,n}(T_{i})(x)-x\rVert=\lVert
T_{i}(x_{1},x_{2})-(x_{1},x_{2})\rVert<\varepsilon_{0}\lVert(x_{1},x_{2})-(\hat{x}_{1},\hat{x}_{2})\rVert\leq
r$

for all $i\in S(\varepsilon_{1})$. Therefore, by Lemma 4.2, $F^{\prime}$ acts
on $B[x,r]$, $B(x,r)$ and $S[x,r]$.

The final assertion of our theorem follows easily from the definition of
$u_{i}$ in Proposition 4.3 and the fact that the linear part of any element of
$F^{\prime}$ belongs (in any case) to
$\operatorname{SL}(n,\mathbb{Z},\varepsilon)$. ∎

###  4.2. The case when $K$ is a transcendental extension of its prime field

Let $F_{2}$ be the free group of reduced words over the alphabet
$\\{a,b,a^{-1},b^{-1}\\}$ and $R$ some commutative ring with identity. For any
pair $A,B\in\operatorname{SL}(2,R)$, the mapping $a\mapsto A,b\mapsto B$
extends uniquely to a homomorphism $\rho_{A,B}\colon
F_{2}\to\operatorname{SL}(2,R)$, called a representation of $F_{2}$. If such a
homomorphism is injective, the representation is called faithful. The function
$\chi_{A,B}\colon F_{2}\to R$,
$\chi_{A,B}(g):=\operatorname{tr}(\rho_{A,B}(g))$, where $\operatorname{tr}$
stands for the matrix trace, is called the character of a representation
$\rho_{A,B}$. It follows from well-known properties of the trace that
$\chi_{A,B}(ghg^{-1})=\chi_{A,B}(h)$ for $g,h\in F_{2}$, so $\chi_{A,B}$ is
constant on conjugacy classes in $F_{2}$. Hence, it induces a function
$\tilde{\chi}_{A,B}\colon\operatorname{Conj}{F_{2}}\to R$, where
$\operatorname{Conj}{F_{2}}$ is the set of all conjugacy classes in $F_{2}$.

Obviously, if $\chi_{A,B}(g)\neq 2$ for $g\neq 1_{F_{2}}$, then $\rho_{A,B}$
has trivial kernel, so it must be faithful. It turns out that if $R\subseteq
K$ for a field $K$ with $\operatorname{char}{K}>0$, the converse also holds,
as we briefly explain in the following remark.

######  Remark 4.10.

Let $K$ be a field with $\operatorname{char}{K}=p>0$ and
$A,B\in\operatorname{SL}(2,K)$ be such that $\rho_{A,B}$ is faithful. We claim
that $\chi_{A,B}(g)\neq 2$ for $g\neq 1_{F_{2}}$. Indeed, if
$\chi_{A,B}(g)=2$, we can see from (4.1) that $1$ is an eigenvalue of
$\rho_{A,B}(g)$. Hence, $\rho_{A,B}(g)$ has the matrix
$\left(\begin{array}[]{cc}1&a\\\ 0&1\end{array}\right)$, for some $a\in K$, in
an appropriate basis of $K^{2}$. By direct multiplication and using the
assumption on characteristic, we obtain $\rho_{A,B}(g^{p})=I_{2}$. Since
$\rho_{A,B}$ is injective, $g^{p}=1_{F_{2}}$, so $g=1_{F_{2}}$ because $F_{2}$
is torsion-free.

######  Corollary 4.11.

If $\operatorname{char}{K}>0$, then any free subgroup of rank two in
$\operatorname{SL}(2,K)$ acts on $K^{2}\setminus\\{\mathbf{0}\\}$ without
nontrivial fixed points.

It is known that, for a given $w\in F_{2}$, the trace of $\rho_{A,B}(w)$
depends in a polynomial manner on the three values: $\operatorname{tr}A$,
$\operatorname{tr}B$, $\operatorname{tr}AB$. This fact was first discovered by
Fricke and Klein in [4] for matrices with real entries, using reasoning based
on arguments in non-Euclidean geometry. Horowitz gave a constructive,
algebraic proof of a stronger result [6, Theorem 3.1] for any free group of
finite rank and for any commutative ring $R$ of characteristic zero with
identity. Traina obtained a more explicit formula describing the polynomial
relation between $\operatorname{tr}(\rho_{A,B}(w))$ and $\operatorname{tr}A$,
$\operatorname{tr}B$, $\operatorname{tr}AB$. However, his paper [20] concerns
only matrices in $\operatorname{SL}(2,\mathbb{C})$ and some proofs are
omitted.

We are going to formulate and prove an analogical result in the case of
representations of $F_{2}$ in $\operatorname{SL}(2,K)$ for any field $K$ (of
arbitrary characteristic). We will need an explicit construction of some
polynomials of three variables $X,Y,Z$ over $\mathbb{P}$, the minimal subring
of $K$ containing $1$, called also the prime ring of $K$. If
$\operatorname{char}{K}=0$, we identify $\mathbb{P}$ with $\mathbb{Z}$,
otherwise $\mathbb{P}=\mathbb{F}_{p}:=\\{0,1,\dots,p-1\\}$ is the prime field
of $K$ and is isomorphic to $\mathbb{Z}/p\mathbb{Z}$, where
$p=\operatorname{char}{K}$.

We are going to define a function
$\Phi\colon\operatorname{Conj}{F_{2}}\to\mathbb{P}[X,Y,Z]$. For this purpose,
we need some canonical way of choosing a representative of a given conjugacy
class in $F_{2}$ (see [8, Proposition 2.14] for a solution of the conjugacy
problem for free groups). After conjugating by a suitable element, we can find
in any class $W\in\operatorname{Conj}{F_{2}}$ a representative $w\in F_{2}$ of
one the following forms:

$w=1_{F_{2}},w=a^{n_{1}},w=b^{m_{1}}\;\textrm{or }w=a^{n_{1}}b^{m_{1}}\dots
a^{n_{k}}b^{m_{k}}\;\textrm{for }k\geq 1,$ (4.5)

where $n_{i},m_{i}\in\mathbb{Z}\setminus\\{0\\}$ for $i=1,\dots,k$. If $W$ is
not one of $[1_{F_{2}}]$, $[a^{n_{1}}]$, $[b^{m_{1}}]$, let us call each
subword $s_{i}:=a^{n_{i}}b^{m_{i}}$, $i=1,\dots,k$, of $w$ a syllable. Then, a
representative $w\in W$ of the form (4.5) is determined up to a cyclic
permutation of syllables (the possible representatives vary only in the choice
of an initial syllable). To avoid ambiguity, let us choose $w$ so that the
vector $(\max\\{\lvert n_{1}\rvert,\lvert m_{1}\rvert\\},\dots,\max\\{\lvert
n_{k}\rvert,\lvert m_{k}\rvert\\})$ is maximal with respect to the
lexicographic ordering in $\mathbb{N}^{k}$. If there are several such choices,
then let us additionally maximize lexicographically
$(n_{1},m_{1},\dots,n_{k},m_{k})\in\mathbb{Z}^{2k}$ among them. This
particular $w\in W$ will be called the canonical representative of a conjugacy
class $W$. The numbers

$l_{a}(W):=\sum_{i=1}^{k}\lvert n_{i}\rvert,\quad
l_{b}(W):=\sum_{i=1}^{k}\lvert m_{i}\rvert,\quad l(W):=l_{a}(W)+l_{b}(W)$
(4.6)

will be called the $a$-length, $b$-length and length of $W$, respectively. We
extend $l_{a}$, $l_{b}$ and $l$ to the cases $W=[1_{F_{2}}]$, $W=[a^{n_{1}}]$
or $W=[b^{m_{1}}]$ in the obvious way.

######  Definition 4.12.

Let $\Phi\colon\operatorname{Conj}{F_{2}}\to\mathbb{P}[X,Y,Z]$ be defined as
follows. At the beginning, we specify $\Phi(W)$ for so called special classes
$W$, namely we put

$\begin{array}[]{l}\Phi([1_{F_{2}}]):=2,\quad\Phi([a])=\Phi([a^{-1}]):=X,\quad\Phi([b])=\Phi([b^{-1}]):=Y,\\\
\Phi([ab])=\Phi([a^{-1}b^{-1}]):=Z,\quad\Phi([a^{-1}b])=\Phi([ab^{-1}]):=XY-Z,\\\
\Phi([aba^{-1}b^{-1}])=\Phi([ab^{-1}a^{-1}b]):=X^{2}+Y^{2}+Z^{2}-XYZ-2,\\\
\Phi([aba^{-1}b])=\Phi([ab^{-1}a^{-1}b^{-1}]):=-X^{2}-Z^{2}+XYZ+2.\end{array}$
(4.7)

We will use the recursion on $l(W)$, $W\in\operatorname{Conj}{F_{2}}$. Notice
that the cases $l(W)\in\\{0,1\\}$ are contained in (4.7). Assume that $n\geq
2$ and we have already defined $\Phi$ for all the classes of length less than
$n$. Let $l(W)=n$ and $w$ be the canonical representative of $W$ of the form
(4.5). If $w$ satisfies $\lvert n_{1}\rvert\geq 2$, we can write
$w=a^{n_{1}}u$, where $u=b^{m_{1}}\dots a^{n_{k}}b^{m_{k}}$ or $u=1_{F_{2}}$.
Let us define

$\Phi(W)=\Phi([a^{n_{1}}u]):=X\,\Phi([a^{n_{1}-\mathrm{sgn}(n_{1})}u])-\Phi([a^{n_{1}-2\,\mathrm{sgn}(n_{1})}u]).$
(4.8)

If $\lvert m_{1}\rvert\geq 2$, but $\lvert n_{1}\rvert=1$ or $w=b^{m_{1}}$, we
similarly write $[w]=[b^{m_{1}}u]$, where $u=a^{n_{2}}b^{n_{2}}\dots
a^{n_{k}}b^{m_{k}}a^{n_{1}}$ or $u=1_{F_{2}}$, and define

$\Phi(W)=\Phi([b^{m_{1}}u]):=Y\,\Phi([b^{m_{1}-\mathrm{sgn}(m_{1})}u])\\\
-\Phi([b^{m_{1}-2\mathrm{sgn}(m_{1})}u]).$ (4.9)

Since the classes appearing on the right-hand side of the formula (4.8) or
(4.9) are of lengths less than $l(W)$, the definition of $\Phi(W)$ is correct
in both cases.

We have one remaining case to deal with, namely, all the exponents $n_{i}$,
$m_{i}$ of $w$ in $\eqref{postkan}$ have absolute value $1$ and $W$ is not one
of the special classes described by (4.7). In this case, there exist $1\leq
i<j\leq k$ such that $n_{i}=n_{j}$. Let us assume that such a pair $(i,j)$ is
chosen to be as small as possible with respect to the lexicographic ordering
in $\mathbb{N}^{2}$. We can thus write $[w]=[uv]$ for $u=s_{i}\dots s_{j-1}$,
$v=s_{j}\dots s_{k}s_{1}\dots s_{i-1}$. Let us define

$\Phi(W)=\Phi([uv]):=\Phi([u])\,\Phi([v])-\Phi([u^{-1}v]).$ (4.10)

The definition is correct since $l([u])<l(W)$, $l([v])<l(W)$ and, by virtue of
$n_{i}=n_{j}$, there is a reduction in the product $u^{-1}v$, hence
$l([u^{-1}v])<l(W)$. The definition of
$\Phi\colon\operatorname{Conj}{F_{2}}\to\mathbb{P}[X,Y,Z]$ is now complete.

Let us write down some elementary properties of the trace of matrices in
$\operatorname{SL}(2,K)$. They can be verified by a direct calculation [4, p.
338] or by applying the Cayley–Hamilton theorem [2, Chapter V: Lemma 18].

######  Lemma 4.13.

For any field $K$ and any $A,B\in\operatorname{SL}(2,K)$, the following
identities hold:

  1. (1)

$\operatorname{tr}A^{-1}=\operatorname{tr}A$.

  2. (2)

$\operatorname{tr}{A^{-1}B}=\operatorname{tr}A\cdot\operatorname{tr}B-\operatorname{tr}AB$.

Using the above-stated lemma and the definition of $\Phi$, we can inductively
prove the following theorem.

######  Theorem 4.14.

Let $K$ be a field. For any $A,B\in\operatorname{SL}(2,K)$ and $w\in F_{2}$,
we have the following equality:

$\chi_{A,B}(w)=\operatorname{tr}(\rho_{A,B}(w))=\Phi([w])(\operatorname{tr}A,\operatorname{tr}B,\operatorname{tr}AB).$
(4.11)

###### Proof.

First, we will show (4.11) for $w$ such that $[w]$ is one of the special
classes from (4.7). The cases $w=1_{F_{2}}$, $w=a$, $w=b$, $w=ab$ are obvious.
Using the part (1) of Lemma 4.13, we get (4.11) for $w=a^{-1}$ and $w=b^{-1}$.
Since
$\operatorname{tr}A^{-1}B^{-1}=\operatorname{tr}B^{-1}A^{-1}=\operatorname{tr}AB$,
the result follows for $w=a^{-1}b^{-1}$. From the part (2) of Lemma 4.13, we
obtain

$\operatorname{tr}A^{-1}B=\operatorname{tr}A\cdot\operatorname{tr}B-\operatorname{tr}AB=\Phi([a^{-1}b])(\operatorname{tr}A,\operatorname{tr}B,\operatorname{tr}AB).$

Moreover,
$\operatorname{tr}AB^{-1}=\operatorname{tr}B^{-1}A=\operatorname{tr}A^{-1}B$,
which yields the result for $w=a^{-1}b$ and $w=ab^{-1}$. For the case of
$w=aba^{-1}b^{-1}$, we calculate

$\begin{split}\operatorname{tr}ABA^{-1}B^{-1}&=\operatorname{tr}A\cdot\operatorname{tr}BA^{-1}B^{-1}-\operatorname{tr}A^{-1}BA^{-1}B^{-1}\\\
&=(\operatorname{tr}A)^{2}-(\operatorname{tr}A^{-1}B\cdot\operatorname{tr}A^{-1}B^{-1}-\operatorname{tr}B^{-2})\\\
&=(\operatorname{tr}A)^{2}+\operatorname{tr}B^{2}-(\operatorname{tr}A\cdot\operatorname{tr}B-\operatorname{tr}AB)\cdot\operatorname{tr}AB\\\
&=(\operatorname{tr}A)^{2}+(\operatorname{tr}B)^{2}-2+(\operatorname{tr}AB)^{2}-\operatorname{tr}A\cdot\operatorname{tr}B\cdot\operatorname{tr}AB,\end{split}$

hence
$\operatorname{tr}ABA^{-1}B^{-1}=\Phi([aba^{-1}b^{-1}])(\operatorname{tr}A,\operatorname{tr}B,\operatorname{tr}AB)$.
Moreover,
$\operatorname{tr}AB^{-1}A^{-1}B=\operatorname{tr}BAB^{-1}A^{-1}=\operatorname{tr}ABA^{-1}B^{-1}$,
so (4.11) holds for $w=ab^{-1}a^{-1}b$ as well. To deal with the remaining
two-syllable cases simultaneously, let $\varepsilon\in\\{1,-1\\}$. We
calculate

$\begin{split}\operatorname{tr}AB^{\varepsilon}A^{-1}B^{\varepsilon}&=\operatorname{tr}AB^{\varepsilon}A^{-1}\cdot\operatorname{tr}B^{\varepsilon}-\operatorname{tr}AB^{-\varepsilon}A^{-1}B^{\varepsilon}\\\
&=(\operatorname{tr}B)^{2}-\operatorname{tr}ABA^{-1}B^{-1}\\\
&=-(\operatorname{tr}A)^{2}-(\operatorname{tr}AB)^{2}+\operatorname{tr}A\cdot\operatorname{tr}B\cdot\operatorname{tr}AB+2,\end{split}$

hence
$\operatorname{tr}AB^{\varepsilon}A^{-1}B^{\varepsilon}=\Phi([ab^{\varepsilon}a^{-1}b^{\varepsilon}])(\operatorname{tr}A,\operatorname{tr}B,\operatorname{tr}AB)$.

We complete the proof by induction on $l([w])$. Assume that $n\geq 2$ and
(4.11) holds for any $v\in F_{2}$ with $l([v])<n$. Let $l([w])=n$ and
$W:=[w]$. We may assume that $w$ is the canonical representative of $W$ of the
form (4.5). If $\lvert n_{1}\rvert\geq 2$, we denote $U:=\rho_{A,B}(u)$, where
$u$ is such as in (4.8). Then, by (4.8) and the inductive assumption, we
obtain the relation

$\Phi([w])(\operatorname{tr}A,\operatorname{tr}B,\operatorname{tr}AB)=\operatorname{tr}A\cdot\operatorname{tr}A^{n_{1}-\mathrm{sgn}(n_{1})}U-\operatorname{tr}A^{n_{1}-2\mathrm{sgn}(n_{1})}U.$
(4.12)

Using Lemma 4.13, we conclude that the right-hand side of (4.12) is equal to
$\operatorname{tr}A^{n_{1}}U=\operatorname{tr}(\rho_{A,B}(w))$.

If $\lvert m_{1}\rvert\geq 2$, but $\lvert n_{1}\rvert=1$ or $w=b^{m_{1}}$, we
proceed similarly using (4.9), the inductive assumption and Lemma 4.13.

If $\Phi(W)$ is defined by (4.10), we put $U:=\rho_{A,B}(u)$,
$V:=\rho_{A,B}(v)$. As before, we obtain

$\begin{split}\Phi([w])(\operatorname{tr}A,\operatorname{tr}B,\operatorname{tr}AB)&=\operatorname{tr}U\cdot\operatorname{tr}V-\operatorname{tr}U^{-1}V\\\
&=\operatorname{tr}UV=\operatorname{tr}(\rho_{A,B}(w)),\end{split}$

which finishes the proof. ∎

In our next theorem, we prove an interesting property of polynomials obtained
from $\Phi([w])$ after the substitution of three rational functions satisfying
certain conditions. We use an auxiliary valuation on $K(X)$.

######  Theorem 4.15.

Let $K$ be a field and $K(X)$ the field of rational functions over $K$. Let
$K(X)$ be equipped with a non-Archimedean valuation
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ whose restriction to the
subfield $K\subseteq K(X)$ is trivial. If $f,g,h\in K(X)$ satisfy

$\lvert f\rvert>1,\quad\lvert g\rvert>1,\quad\lvert h\rvert=\lvert
fg\rvert=\lvert fg-h\rvert,$ (4.13)

then, for any $W\in\operatorname{Conj}{F_{2}}$, $W\neq[1_{F_{2}}]$, the
rational function $\Psi(W):=\Phi(W)(f,g,h)\in K(X)$ satisfies

$\lvert\Psi(W)\rvert=\lvert f\rvert^{l_{a}(W)}\cdot\lvert
g\rvert^{l_{b}(W)}>1,$ (4.14)

where $l_{a}$, $l_{b}$ are defined by (4.6). In particular, $\Psi(W)\not\in K$
for $W\neq[1_{F_{2}}]$.

###### Proof.

First, we will show (4.14) for special classes $W\neq[1_{F_{2}}]$. It is clear
for $W\in\\{[a],[a^{-1}],[b],[b^{-1}]\\}$. By (4.13), we also have

$\displaystyle\lvert\Psi([ab])\rvert$
$\displaystyle=\lvert\Psi([a^{-1}b^{-1}])\rvert=\lvert h\rvert=\lvert
f\rvert^{1}\cdot\lvert g\rvert^{1},$
$\displaystyle\lvert\Psi([a^{-1}b])\rvert$
$\displaystyle=\lvert\Psi([ab^{-1}])\rvert=\lvert fg-h\rvert=\lvert
f\rvert^{1}\cdot\lvert g\rvert^{1},$
$\displaystyle\lvert\Psi([aba^{-1}b^{-1}])\rvert$
$\displaystyle=\lvert\Psi([ab^{-1}a^{-1}b])\rvert=\lvert
f^{2}+g^{2}+h^{2}-fgh-2\rvert$ $\displaystyle=\lvert
f^{2}+g^{2}-2-(fg-h)h\rvert,$ $\displaystyle\lvert\Psi([aba^{-1}b])\rvert$
$\displaystyle=\lvert\Psi([ab^{-1}a^{-1}b^{-1}])\rvert=\lvert-f^{2}-h^{2}+fgh+2\rvert$
$\displaystyle=\lvert(fg-h)h-(f^{2}-2)\rvert.$

Since

$\lvert f^{2}+g^{2}-2\rvert\leq\max\\{\lvert f\rvert^{2},\lvert
g\rvert^{2}\\}<\lvert f\rvert^{2}\cdot\lvert g\rvert^{2}=\lvert
fg\rvert^{2}=\lvert fg-h\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert
h\rvert=\lvert(fg-h)h\rvert$

and $\lvert f^{2}-2\rvert=\lvert f\rvert^{2}<\lvert(fg-h)h\rvert$, we obtain

$\displaystyle\lvert\Psi([aba^{-1}b^{-1}])\rvert$
$\displaystyle=\lvert\Psi([ab^{-1}a^{-1}b])\rvert=\lvert\Psi([aba^{-1}b])\rvert=\lvert\Psi([ab^{-1}a^{-1}b^{-1}])\rvert$
$\displaystyle=\lvert(fg-h)h\rvert=\lvert f\rvert^{2}\cdot\lvert g\rvert^{2},$

as desired.

Assume that $n\geq 2$ and (4.14) holds for all classes of length less than
$n$. Let $l(W)=n$ and $w$ be the canonical representative of $W$ of the form
(4.5). Assume that $\lvert n_{1}\rvert\geq 2$ and let us abbreviate the
elements of $F_{2}$ on the right-hand side of (4.8) by $w_{1}$ and $w_{2}$.
Thus we have $\Psi(W)=f\,\Psi([w_{1}])-\Psi([w_{2}])$. Examining the
$a$-length and $b$-length of $[w_{1}]$ and $[w_{2}]$, we get
$l_{a}([w_{1}])=l_{a}(W)-1$, $l_{b}([w_{1}])=l_{b}(W)$ and $l_{a}([w_{2}])\leq
l_{a}(W)-2$, $l_{b}([w_{2}])\leq l_{b}(W)$. We deduce that $l([w_{1}])<l(W)$
and $l([w_{2}])<l(W)$. By the inductive assumption,

$\displaystyle\lvert\Psi([w_{1}])\rvert$ $\displaystyle=\lvert
f\rvert^{l_{a}([w_{1}])}\cdot\lvert g\rvert^{l_{b}([w_{1}])}=\lvert
f\rvert^{l_{a}(W)-1}\cdot\lvert g\rvert^{l_{b}(W)},$
$\displaystyle\lvert\Psi([w_{2}])\rvert$ $\displaystyle=\lvert
f\rvert^{l_{a}([w_{2}])}\cdot\lvert g\rvert^{l_{b}([w_{2}])}\leq\lvert
f\rvert^{l_{a}(W)-2}\cdot\lvert g\rvert^{l_{b}(W)}.$

Therefore,

$\lvert f\,\Psi([w_{1}])\rvert=\lvert f\rvert^{l_{a}(W)}\cdot\lvert
g\rvert^{l_{b}(W)}>\lvert\Psi([w_{2}])\rvert,$

so $\lvert\Psi(W)\rvert=\lvert f\rvert^{l_{a}(W)}\cdot\lvert
g\rvert^{l_{b}(W)}$.

If $\lvert m_{1}\rvert\geq 2$, but $\lvert n_{1}\rvert=1$ or $w=b^{m_{1}}$, we
proceed analogically using (4.9).

If $\Phi(W)$ is defined by (4.10), we have $l_{a}(W)=l_{a}([u])+l_{a}([v])$,
$l_{b}(W)=l_{b}([u])+l_{b}([v])$. Hence, by the inductive assumption,

$\lvert\Psi([u])\,\Psi([v])\rvert=\lvert f\rvert^{l_{a}([u])}\cdot\lvert
g\rvert^{l_{b}([u])}\cdot\lvert f\rvert^{l_{a}([v])}\cdot\lvert
g\rvert^{l_{b}([v])}=\lvert f\rvert^{l_{a}(W)}\cdot\lvert g\rvert^{l_{b}(W)}.$

Because of a reduction in the product $u^{-1}v$, we have
$l_{a}([u^{-1}v])<l_{a}(W)$, $l_{b}([u^{-1}v])\leq l_{b}(W)$, so

$\lvert\Psi([u^{-1}v])\rvert=\lvert f\rvert^{l_{a}([u^{-1}v])}\cdot\lvert
g\rvert^{l_{b}([u^{-1}v])}<\lvert f\rvert^{l_{a}(W)}\cdot\lvert
g\rvert^{l_{b}(W)}=\lvert\Psi([u])\,\Psi([v])\rvert.$

We finally get $\lvert\Psi(W)\rvert=\lvert\Psi([u])\,\Psi([v])\rvert$, from
which (4.14) follows. ∎

Let us formulate an important corollary from Theorem 4.15.

######  Corollary 4.16.

Let $K$ be a field and $K_{0}$ its prime field. Let $K_{0}(X)$ be equipped
with a non-Archimedean valuation
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ trivial on $K_{0}$ and let
$f,g,h\in K_{0}(X)$ satisfy (4.13). If $t\in K$ is transcendental over
$K_{0}$, then $\Phi(W)(f(t),g(t),h(t))\neq 2$ for all classes
$W\in\operatorname{Conj}{F_{2}}$ distinct from $[1_{F_{2}}]$.

###### Proof.

Suppose that $\Phi(W)(f(t),g(t),h(t))=2$ for some
$W\in\operatorname{Conj}{F_{2}}$, $W\neq[1_{F_{2}}]$, and let $\Psi(W)\in
K_{0}(X)$ be as in Theorem 4.15. Then $t$ would satisfy the equation
$\Psi(W)(t)-2=0$. However, by virtue of Theorem 4.15, the rational function
$\Psi(W)$ is not constant, neither is $\Psi(W)-2$. Therefore $t$ would be
algebraic over $K_{0}$, which leads to a contradiction. ∎

Let us introduce a family of nontrivial non-Archimedean valuations on $K(X)$,
each being trivial when restricted to $K$. Namely, for $q$ being an
irreducible polynomial in $K[X]$, we define
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{q}$ on $K(X)$ by putting
$\lvert 0\rvert_{q}:=0$ and

$\left\lvert q^{k}\frac{f_{1}}{f_{2}}\right\rvert_{q}:=2^{-k}\quad\text{for
}k\in\mathbb{Z},\;q\nmid f_{1},q\nmid f_{2},\;f_{1},f_{2}\in K[X],\;f_{2}\neq
0.$ (4.15)

######  Lemma 4.17.

Let $K_{0}$ be a prime field and $q\in K_{0}[X]$ be the irreducible polynomial
specified as below:

$q:=\begin{cases}2X+1&\text{if }K_{0}\neq\mathbb{F}_{2},\\\ X^{2}+X+1&\text{if
}K_{0}=\mathbb{F}_{2}.\end{cases}$ (4.16)

Then, the conditions (4.13) are satisfied for the valuation
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{q}$ on $K_{0}(X)$ and
$f,g,h\in K_{0}(X)$ defined by $f=g:=q+q^{-1}$, $h:=q^{-2}(q^{4}+X+1)$.

###### Proof.

We have $\lvert f\rvert_{q}=\lvert g\rvert_{q}=2$ and $\lvert h\rvert_{q}=4$
since $q\nmid X+1$. We also have $\lvert fg-h\rvert_{q}=\lvert
2-q^{-2}X\rvert_{q}=\lvert q^{-2}X\rvert_{q}=4$ since $q\nmid X$. ∎

Now we are ready to prove results analogical to those in Proposition 4.8 and
Theorem 4.9, but in the case of $K$ being a transcendental extension of its
prime field $K_{0}$.

######  Theorem 4.18.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-Archimedean
valued field, $D$ its ring of integers and $K_{0}$ its prime field. Let $q\in
K_{0}[X]$ be chosen according to (4.16). If $K$ is a transcendental extension
of $K_{0}$, then there exists $t\in D$ such that the matrices $A,B$ given by

$A:=\left(\begin{array}[]{cc}q(t^{2})&t\\\
0&(q(t^{2}))^{-1}\end{array}\right),\quad
B:=\left(\begin{array}[]{cc}q(t^{2})&0\\\ (q(t^{2}))^{-2}\cdot
t&(q(t^{2}))^{-1}\end{array}\right)$ (4.17)

form a basis of a free subgroup $\left<A,B\right>\leq\operatorname{SL}(2,D)$
that acts on on $K^{2}\setminus\\{\mathbf{0}\\}$, and (for any $r>0$) on
$B[\mathbf{0},r]\setminus\\{\mathbf{0}\\}$ and $S[\mathbf{0},r]$ without
nontrivial fixed points.

Moreover, if the valuation $\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$
is nontrivial, then, for any $\varepsilon\in(0,1]$, $t$ can be chosen so that
$A,B\in\operatorname{SL}(2,D,\varepsilon)$.

###### Proof.

If the valuation $\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ is trivial,
let $t\in K$ be any element transcendental over $K_{0}$; hence, $q(t^{2})\neq
0$ and $A,B$ are well-defined elements of
$\operatorname{SL}(2,K)=\operatorname{SL}(2,D)$. If
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ is nontrivial, then, for any
$\varepsilon\in(0,1]$, we can always find an element $t\in K$ transcendental
over $K_{0}$ and satisfying $\lvert t\rvert<\varepsilon$. We then have $\lvert
q(t^{2})-1\rvert\leq\lvert t^{2}\rvert<\varepsilon$ and $\lvert
q(t^{2})\rvert=1$. Since $q(t^{2})$ belongs to the coset $1+B(0,\varepsilon)$
modulo the ideal $B(0,\varepsilon)$ in $D$, so does $(q(t^{2}))^{-1}$.
Therefore, $A,B\in\operatorname{SL}(2,D,\varepsilon)$.

Let $f,g,h\in K_{0}[X]$ be defined as in Lemma 4.17. By a direct calculation,
$\operatorname{tr}A=f(t^{2})$, $\operatorname{tr}B=g(t^{2})$ and
$\operatorname{tr}AB=(q(t^{2}))^{2}+(q(t^{2}))^{-2}+(q(t^{2}))^{-2}\,t^{2}=h(t^{2})$.

Let $\rho_{A,B}\colon F_{2}\to\operatorname{SL}(2,D)$ be the representation of
$F_{2}$ specified by $A,B$ from (4.17). Assume that $w\in F_{2}$, $w\neq
1_{F_{2}}$. From Theorem 4.14 we get

$\operatorname{tr}(\rho_{A,B}(w))=\Phi([w])(\operatorname{tr}A,\operatorname{tr}B,\operatorname{tr}AB)=\Phi([w])(f(t^{2}),g(t^{2}),h(t^{2})).$

Since $t^{2}$ is transcendental over $K_{0}$, we deduce from Corollary 4.16
that $\operatorname{tr}(\rho_{A,B}(w))\neq 2$, which implies that
$\rho_{A,B}(w)\neq I_{2}$ and (by virtue of Lemma 4.1) $\rho_{A,B}(w)$ has no
fixed points in $K^{2}\setminus\\{\mathbf{0}\\}$. ∎

######  Theorem 4.19.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-Archimedean
nontrivially valued field that is a transcendental extension of its prime
field $K_{0}$. Let $n\geq 2$,
$\hat{x}=(\hat{x}_{1},\hat{x}_{2},\dots,\hat{x}_{n})\in K^{n}$,
$x=(x_{1},x_{2},\dots,x_{n})\in K^{n}$,
$(x_{1},x_{2})\neq(\hat{x}_{1},\hat{x}_{2})$,
$0<r<\lVert(x_{1},x_{2})-(\hat{x}_{1},\hat{x}_{2})\rVert$ and
$\varepsilon\in(0,1]$. Then, there exists a free subgroup
$F^{\prime}\leq\operatorname{SA}(n,D,\varepsilon)$ of rank two acting on
$K^{n}\setminus\left(\\{(\hat{x}_{1},\hat{x}_{2})\\}\times K^{n-2}\right)$,
the balls $B[x,r]$, $B(x,r)$, and the sphere $S[x,r]$ without nontrivial fixed
points.

Moreover, $F^{\prime}$ can be chosen so that
$F^{\prime}\leq\operatorname{SL}(n,D,\varepsilon)$ if
$(\hat{x}_{1},\hat{x}_{2})=(0,0)$.

###### Proof.

The proof is essentially the same as that of Theorem 4.9. The only significant
difference is that we take for $F(\varepsilon_{1})$ the free group with basis
$\\{A,B\\}$, where $A$, $B$ are the matrices (4.17) with $t$ chosen so that
$A,B\in\operatorname{SL}(2,D,\varepsilon_{1})$. ∎

##  5\. The Banach–Tarski paradox for subsets of $K^{n}$

In this section, by $R$ we will mean $\mathbb{Z}$ if
$\operatorname{char}{K}\neq\operatorname{char}{k}$, and $D$ otherwise.

Clearly, any subset $A$ of $K^{n}$ containing $\mathbf{0}$ is not paradoxical
with respect to any group of linear isometries of $K^{n}$. For balls in
$K^{n}$ not containing $\mathbf{0}$, we obtain the following positive result.

######  Theorem 5.1.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-Archimedean
nontrivially valued field, $k$ its residue field, $\varepsilon\in(0,1]$ and
$n\geq 2$. Then any ball $B$ in $K^{n}$ not containing $\mathbf{0}$ is
$\operatorname{SL}(n,R,\varepsilon)$-paradoxical using $4$ pieces.

###### Proof.

We either have $\operatorname{char}{K}=0\neq\operatorname{char}{k}$, or
$\operatorname{char}{K}=\operatorname{char}{k}$. In the latter case,
$\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert$ is trivial on the prime
field $K_{0}$ of $K$ and any element $t\in K^{*}$ with $\lvert t\rvert\neq 1$
is transcendental over $K_{0}$, so $K$ is a transcendental extension of
$K_{0}.$

Assume that $B$ is a closed ball $B[x,r]$ in $K^{n}$ not containing
$\mathbf{0}$. Then $\lVert x\rVert>r$, so $\lvert x_{i}\rvert>r$ for some
$1\leq i\leq n.$

If $i=1$, we put $\hat{x}:=\mathbf{0}$. From Theorem 4.9 or 4.19, it follows
that there exists a non-Abelian free subgroup
$F^{\prime}\leq\operatorname{SL}(n,R,\varepsilon)$ acting on $B[x,r]$ without
nontrivial fixed points, which yields
$\operatorname{SL}(n,R,\varepsilon)$-paradoxicality of $B[x,r]$ using $4$
pieces.

Assume that $i>1$. Let $\gamma$ be a permutation of the set $\\{1,\ldots,n\\}$
with $\gamma(1)=i$ and $P\in\operatorname{GL}(n,R)$ with
$P(e_{j})=e_{\gamma(j)}$ for any $1\leq j\leq n.$ If
$V\in\operatorname{SL}(n,R,\varepsilon)$ and $M=PVP^{-1}$, then
$m_{k,l}=v_{\gamma^{-1}(k),\gamma^{-1}(l)}$ for all $1\leq k,l\leq n$, so
$M\in\operatorname{SL}(n,R,\varepsilon)$. Put $z=P^{-1}x.$ Then $\lvert
z_{1}\rvert=\lvert x_{i}\rvert>r$ and $P(B[z,r])=B[x,r]$, so
$\mathbf{0}\not\in B[z,r].$ Thus, $B[z,r]$ is
$SL(n,R,\varepsilon)$-paradoxical using $4$ pieces, so $B[x,r]=P(B[z,r])$ is
$SL(n,R,\varepsilon)$-paradoxical using $4$ pieces too.

The same holds for any open ball $B=B(x,r)$ in $K^{n}$ not containing
$\mathbf{0}$ because the assumption $\lVert x\rVert\geq r$ is then sufficient
by Remark 4.4. ∎

As in the case of balls, spheres not containing $\mathbf{0}$ are also
$\operatorname{SL}(n,R,\varepsilon)$-paradoxical, but the number of pieces
involved is expressed in a more complicated manner.

######  Theorem 5.2.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-Archimedean
nontrivially valued field, $k$ its residue field, $\varepsilon\in(0,1]$,
$r\in\lvert K^{*}\rvert$ and $n\geq 2$. Then any sphere $S[x,r]$ in $K^{n}$
not containing $\mathbf{0}$ is
$\operatorname{SL}(n,R,\varepsilon)$-paradoxical using $4$ pieces if $\lVert
x\rVert>r$, and $5-(-1)^{n}$ pieces if $\lVert x\rVert<r$.

###### Proof.

First, notice that if $\lVert x\rVert>r$, we can proceed exactly as in the
proof of Theorem 5.1. As before, Theorem 4.9 or 4.19 provides the existence of
a non-Abelian free subgroup $F^{\prime}\leq\operatorname{SL}(n,R,\varepsilon)$
acting on $S[x,r]$ without nontrivial fixed points.

In the case $\lVert x\rVert<r$, we have $S[x,r]=S[\mathbf{0},r]$ by the
isosceles property. Hence, we assume from now on that $x=\mathbf{0}$. Let us
denote by $F(\varepsilon)$ a free subgroup of
$\operatorname{SL}(2,R,\varepsilon)$ obtained from Proposition 4.8 or Theorem
4.18, depending on whether $\operatorname{char}{K}$ and
$\operatorname{char}{k}$ are distinct or equal.

Assume that $n=2k$, $k\geq 1$. We can treat $K^{n}$ as the product of $k$
copies of $K^{2}$. Let us define an embedding $\varphi\colon
F(\varepsilon)\to\operatorname{SL}(n,R,\varepsilon)$ by putting

$\varphi(g)(y_{1},\dots,y_{2k}):=\left(g(y_{1},y_{2}),\dots,g(y_{2k-1},y_{2k})\right)\in(K^{2})^{k}\cong
K^{n}$

for $g\in F(\varepsilon)$, $(y_{1},\dots,y_{2k})\in K^{n}$, i.e., $\varphi(g)$
acts like $g$ on each copy of $K^{2}$. Since $F(\varepsilon)$ acts without
nontrivial fixed points on $K^{2}\setminus\\{0_{2}\\}$, so does
$\varphi(F(\varepsilon))$ on $K^{n}\setminus\\{\mathbf{0}\\}$, hence also on
$S[\mathbf{0},r]$. Therefore, $S[\mathbf{0},r]$ is
$\operatorname{SL}(n,R,\varepsilon)$-paradoxical using $4$ pieces.

Assume now that $n=2k+1$, $k\geq 1$. Let us partition $S[\mathbf{0},r]$,
regarded as a subset of $K^{2k}\times K\cong K^{n}$, into two disjoint sets.
Namely, we put

$A_{1}:=S[0_{2k},r]\times B[0_{1},r],\qquad A_{2}:=B(0_{2k},r)\times
S[0_{1},r].$

It is easy to see that $A_{1}$ and $A_{2}$ are
$\operatorname{SL}(n,R,\varepsilon)$-invariant.

Let us define two embeddings $\varphi_{1},\varphi_{2}\colon
F(\varepsilon)\to\operatorname{SL}(n,R,\varepsilon)$ as follows

$\begin{split}\varphi_{1}(g)(y_{1},\dots,y_{n}):=\left(g(y_{1},y_{2}),\dots,g(y_{2k-1},y_{2k}),y_{n}\right)\in(K^{2})^{k}\times
K\cong K^{n},\\\
\varphi_{2}(g)(y_{1},\dots,y_{n}):=\left(y_{1},g(y_{2},y_{3}),\dots,g(y_{2k},y_{2k+1})\right)\in
K\times(K^{2})^{k}\cong K^{n},\end{split}$

for $g\in F(\varepsilon)$, $(y_{1},\dots,y_{n})\in K^{n}$. Our choice of
$\varphi_{1},\varphi_{2}$ guarantees that, for each $i\in\\{1,2\\}$, any
nonidentity element of $\varphi_{i}(F(\varepsilon))$ has no fixed points in
$A_{i}$. Let $\\{\sigma,\tau\\}$ be a basis of $F(\varepsilon)$. As in
$\eqref{characterization_four_pieces}$, we have the decompositions

$\begin{split}A_{1}=A_{11}\sqcup A_{12}\sqcup A_{13}\sqcup
A_{14}=A_{11}\sqcup\varphi_{1}(\sigma)(A_{12})=A_{13}\sqcup\varphi_{1}(\tau)(A_{14}),\\\
A_{2}=A_{21}\sqcup A_{22}\sqcup A_{23}\sqcup
A_{24}=A_{21}\sqcup\varphi_{2}(\sigma)(A_{22})=A_{23}\sqcup\varphi_{2}(\tau)(A_{24}).\end{split}$
(5.1)

Let us abbreviate $E:=A_{11}\sqcup A_{21}$, $E^{\prime}:=A_{13}\sqcup A_{23}$.
Combining the equations (5.1), we obtain

$\begin{split}S[\mathbf{0},r]&=E\sqcup A_{12}\sqcup A_{22}\sqcup
E^{\prime}\sqcup A_{14}\sqcup A_{24}\\\
&=E\sqcup\varphi_{1}(\sigma)(A_{12})\sqcup\varphi_{2}(\sigma)(A_{22})\\\
&=E^{\prime}\sqcup\varphi_{1}(\tau)(A_{14})\sqcup\varphi_{2}(\tau)(A_{24}),\end{split}$
(5.2)

which yields an $\operatorname{SL}(n,R,\varepsilon)$-paradoxical decomposition
of $S[\mathbf{0},r]$ using $6$ pieces. ∎

######  Corollary 5.3.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-
Archimedean, nontrivially valued field, $\varepsilon\in(0,1]$ and $n\geq 2$.
Then, the union of any family of nonempty spheres in $K^{n}$ centered at
$\mathbf{0}$ is $\operatorname{SL}(n,R,\varepsilon)$-paradoxical using
$5-(-1)^{n}$ pieces. In particular, the claim holds for
$K^{n}\setminus\\{\mathbf{0}\\}$ and $D^{n}\setminus\\{\mathbf{0}\\}$.

###### Proof.

Let $Z\subseteq\lvert K^{*}\rvert$. Since the group elements witnessing the
$\operatorname{SL}(n,R,\varepsilon)$-paradoxical decomposition of
$S[\mathbf{0},r]$ in the proof of Theorem 5.2 do not depend on $r\in Z$, the
corresponding pieces from those decompositions can be gathered together,
yielding a desired $\operatorname{SL}(n,R,\varepsilon)$-paradoxical
decomposition of $\bigsqcup_{r\in Z}S[\mathbf{0},r]$. ∎

Now we will prove that balls and spheres containing $\mathbf{0}$ admit a
paradoxical decomposition with respect to a certain group of affine isometries
of $K^{n}$.

######  Theorem 5.4.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-
Archimedean, nontrivially valued field, $\varepsilon\in(0,1]$ and $n\geq 2.$
Then any ball and any sphere in $K^{n}$ containing $\mathbf{0}$ is
$\operatorname{SA}(n,D,\varepsilon)$-paradoxical using $4$ pieces.

###### Proof.

Let $x\in K^{n}$ and $r>0$. Let us choose a point $\hat{x}\in K^{n}$
satisfying $\lVert(x_{1},x_{2})-(\hat{x}_{1},\hat{x}_{2})\rVert>r$. Then, by
Theorem 4.9 or 4.19, there exists a non-Abelian free subgroup
$F^{\prime}\leq\operatorname{SA}(n,D,\varepsilon)$ acting on $B[x,r]$,
$B(x,r)$ and $S[x,r]$ without nontrivial fixed points. It follows that
$B[x,r]$, $B(x,r)$, and $S[x,r]$ (if $r\in\lvert K^{*}\rvert$) are
$\operatorname{SA}(n,D,\varepsilon)$-paradoxical using $4$ pieces. ∎

Our next theorem is an analog of the strong result of Banach and Tarski [1,
Theorem 24]. It follows that if $K$ is a non-Archimedean discretely valued
field with finite residue field (in particular, if $K$ is a locally compact
non-Archimedean valued field, e.g., $K=\mathbb{Q}_{p}$), then any two balls in
$K^{n}$ are equidecomposable with respect to a certain group of isometries.

######  Theorem 5.5.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-Archimedean
discretely valued field with finite residue field (e.g., a locally compact
non-Archimedean valued field) and $n\geq 2$. Then any two bounded subsets of
$K^{n}$ with nonempty interiors are
$\operatorname{SA}(n,D,K)$-equidecomposable.

###### Proof.

Let $G:=\operatorname{SA}(n,D,K)$ and $A,A^{\prime}\subseteq K^{n}$ be bounded
sets with nonempty interiors. There exists a closed ball
$E:=B[\mathbf{0},r]\subseteq K^{n}$ containing both $A$ and $A^{\prime}$. By
Lemma 2.1, we have $E\subseteq g_{1}A\cup\dots\cup g_{m}A$ for some
$m\in\mathbb{N}$ and translations $g_{1},\dots,g_{m}\in G$.

Since $E$ is $G$-paradoxical (by Theorem 5.4), an application of [19,
Corollary 10.22] yields that $A\sim_{G}E\sim_{G}A^{\prime}$; hence,
$A\sim_{G}A^{\prime}$. ∎

######  Corollary 5.6.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-Archimedean
discretely valued field with finite residue field, e.g.,
$(\mathbb{Q},\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert_{p})$ or
$(\mathbb{Q}_{p},\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert p)$, and
$n\geq 2$. Then any bounded subset of $K^{n}$ with nonempty interior is
$\operatorname{SA}(n,D,K)$-paradoxical.

###### Proof.

By the previous theorem, any bounded subset of $K^{n}$ with nonempty interior
is $\operatorname{SA}(n,D,K)$-equidecomposable with the ball
$B[\mathbf{0},1]$, which is $\operatorname{SA}(n,D,K)$-paradoxical by Theorem
5.4. The assertion follows from [19, Proposition 3.5]. ∎

It turns out that the whole space $K^{n}$ is also paradoxical with respect to
a certain group of affine isometries.

######  Theorem 5.7.

Let $(K,\lvert{\mspace{1.0mu}\cdot\mspace{1.0mu}}\rvert)$ be a non-
Archimedean, nontrivially valued field and $n\geq 2$. Let $L:=\mathbb{Z}$ if
$\operatorname{char}{K}=0$, and $L:=D$ if $\operatorname{char}{K}>0$. Then
$K^{n}$ is $\operatorname{SA}(n,L)$-paradoxical using $5$ pieces.

###### Proof.

First, we will prove the claim for $n=2$. If $\operatorname{char}{K}=0$, let
$F\leq\operatorname{SL}(2,\mathbb{Z})$ be a free group from Corollary 4.7. If
$\operatorname{char}{K}=p>0$, then (by [15, Theorem 14.2]) $K$ is necessarily
a transcendental extension of its prime field $\mathbb{F}_{p}$, so let
$F\leq\operatorname{SL}(2,D)$ be the free group from Theorem 4.18. Let us
choose any point $c\in L^{2}\setminus\\{\mathbf{0}\\}$, we then have
$\mathrm{Stab}_{F}(c)=\\{I_{2}\\}$. Let $C:=\\{c\\}\subseteq
E:=K^{2}\setminus\\{\mathbf{0}\\}$. If $\alpha$ denotes the translation along
$-c$, then $\alpha\in\operatorname{SA}(2,L)$ and $\alpha(C)=\\{\mathbf{0}\\}$.
By virtue of Theorem 3.2, the space $K^{2}=E\sqcup\alpha(C)$ is
$\operatorname{SA}(2,L)$-paradoxical using $5$ pieces.

Assume now that $n>2$. Let
$\iota\colon\operatorname{SA}(2,L)\to\operatorname{SA}(n,L)$ be the
restriction of the canonical embedding $\iota_{2,n}$, defined by (2.3), to
$\operatorname{SA}(2,L)$. Let also $f\colon K^{n}\to K^{2}$ be the projection
onto the first two coordinates. Since $\iota$ and $f$ satisfy the condition
(3.6), we infer from Lemma 3.3 that $K^{n}$ is
$\operatorname{SA}(n,L)$-paradoxical using $5$ pieces. ∎

Let us briefly discuss the case when the valuation on $K$ is trivial. Then
$D=K$ and the linear (or affine) isometries of $K^{n}$ are just the linear (or
affine) bijections.

######  Theorem 5.8.

Let $K$ be a field and $n\geq 2$.

If $\operatorname{char}{K}=0$, then $K^{n}$ is
$\operatorname{SA}(n,\mathbb{Z})$-paradoxical using $5$ pieces.

If $\operatorname{char}{K}=p>0$ and $t\in K$ is transcendental over
$\mathbb{F}_{p}$, then $K^{n}$ is
$\operatorname{SA}(n,\mathbb{F}_{p}(t))$-paradoxical using $5$ pieces.

If $K$ is an algebraic extension of a finite subfield, then $K^{n}$ is not
$\operatorname{GA}(n,K)$-paradoxical and $K^{n}\setminus\\{\mathbf{0}\\}$ is
not $\operatorname{GL}(n,K)$-paradoxical.

###### Proof.

Let $n\geq 2$. If $\operatorname{char}{K}=0$, let
$F\leq\operatorname{SL}(2,\mathbb{Z})$ be a free group from Corollary 4.7. If
$\operatorname{char}{K}=p$ and $t\in K$ is transcendental over
$\mathbb{F}_{p}$, let $F\leq\operatorname{SL}(2,\mathbb{F}_{p}(t))$ be the
free group from Theorem 4.18. We further proceed exactly as in the proof of
Theorem 5.7.

If $K$ is an algebraic extension of a finite field, then (see Example 3.5) the
group $\operatorname{GL}(n,K)$ is amenable, which excludes the existence of a
$\operatorname{GL}(n,K)$-paradoxical decomposition of a
$\operatorname{GL}(n,K)$-invariant set $K^{n}\setminus\\{\mathbf{0}$}. The
group $\operatorname{GA}(n,K)$ is amenable by Lemma 3.6, so $K^{n}$ cannot be
$\operatorname{GA}(n,K)$-paradoxical. ∎

## Acknowledgements

The author would like to express his gratitude to Professor Wiesław Śliwa for
many inspiring discussions and helpful suggestions.

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