# On arithmetic properties of solvable Baumslag-Solitar groups

Laurent Hayez  Université de NeuchâtelNeuchâtelSwitzerland
<EMAIL_ADDRESS>,  Tom Kaiser  Université de
NeuchâtelNeuchâtelSwitzerland<EMAIL_ADDRESS>and  Alain Valette
Université de NeuchâtelNeuchâtelSwitzerland<EMAIL_ADDRESS>
###### Abstract.

For $0<\alpha\leq 1$, we say that a sequence $(X_{k})_{k>0}$ of $d$-regular
graphs has property $D_{\alpha}$ if there exists a constant $C>0$ such that
$\operatorname{diam}(X_{k})\geq C\cdot|X_{k}|^{\alpha}$. We investigate
property $D_{\alpha}$ for arithmetic box spaces of the solvable Baumslag-
Solitar groups $BS(1,m)$ (with $m\geq 2$): those are box spaces obtained by
embedding $BS(1,m)$ into the upper triangular matrices in
$GL_{2}(\mathbb{Z}[1/m])$ and intersecting with a family $M_{N_{k}}$ of
congruence subgroups of $GL_{2}(\mathbb{Z}[1/m])$, where the levels $N_{k}$
are coprime with $m$ and $N_{k}|N_{k+1}$. We prove:

  * •

if an arithmetic box space has $D_{\alpha}$, then $\alpha\leq\frac{1}{2}$ ;

  * •

if the family $(N_{k})_{k}$ of levels is supported on finitely many primes,
the corresponding arithmetic box space has $D_{1/2}$ ;

  * •

if the family $(N_{k})_{k}$ of levels is supported on a family of primes with
positive analytic primitive density, then the corresponding arithmetic box
space does not have $D_{\alpha}$, for every $\alpha>0$.

Moreover, we prove that if we embed $BS(1,m)$ in the group of invertible
upper-triangular matrices $T_{n}(\mathbb{Z}[1/m])$, then every finite index
subgroup of the embedding contains a congruence subgroup. This is a version of
the congruence subgroup property (CSP).

††copyright: none

##  1\. Introduction

Let $G$ be a finitely generated, residually finite group. If $(H_{k})_{k>0}$
is a decreasing sequence of finite index normal subgroups of $G$, with trivial
intersection, and $S$ is a finite generating set of $G$, then the box space
$\square_{(H_{k})}G$ is the disjoint union of finite Cayley graphs

$\square_{(H_{k})}G=\coprod_{k>0}\operatorname{Cay}(G/H_{k},S);$

here by abuse of notation we identify $S$ with its image in $G/H_{k}$.
Changing the generating set $S$ does not change the coarse geometry of the box
space111In the sense that the two families of graphs are quasi-isometrically
equivalent, by a family of quasi-isometries with uniform constants. so we omit
$S$ from the notation.

In the dictionary between group-theoretical properties of $G$ and metric
properties of $\square_{(H_{k})}G$ (see e.g. [Khukhro 2018]), it is natural to
look at the behaviour of the diameter of the Cayley graphs
$\operatorname{Cay}(G/H_{k},S)$. Let $0<\alpha\leq 1$. The box space
$\square_{(H_{k})}G$ satisfies property $D_{\alpha}$ if there is some constant
$C>0$ such that for every $k>0$:

(1) $\operatorname{diam}(\operatorname{Cay}(G/H_{k},S))\geq
C|G/H_{k}|^{\alpha}.$

Note that property $D_{\alpha}$ is a coarse geometry invariant of the box
space. The following is known.

###### Theorem 1.1.

Let $G$ be a finitely generated, residually finite group.

  1. (1)

(see Cor. 1.7 and Lemma 5.1 in [Breuillard and Tointon 2016]) If some box
space of $G$ has property $D_{\alpha}$, for some $\alpha>0$, then $G$
virtually maps onto $\mathbb{Z}$.

  2. (2)

(see Theorem 3 in [Khukhro and Valette 2017]) If $G$ maps onto $\mathbb{Z}$,
then for every $0<\alpha<1$, there exists a box space of $G$ with property
$D_{\alpha}$.

  3. (3)

(see Proposition 5 in [Khukhro and Valette 2017]) The group $G$ is virtually
cyclic if and only if some (hence any) box space of $G$ has property $D_{1}$.

$\square$

This paper considers the Baumslag-Solitar groups $BS(n,m)\;(m,n>0)$ with
presentation

$BS(n,m)=\langle a,t|ta^{n}t^{-1}=t^{m}\rangle.$

They all map onto $\mathbb{Z}$, by $a\mapsto 0,t\mapsto 1$. On the other hand
they are known to be residually finite if and only if $n=1$ or $n=m$: see
Theorem C in [Meskin 1972]. It turns out that the solvable Baumslag-Solitar
groups $BS(1,m)$, with $m\geq 2$, have interesting box spaces. Indeed it is
well-known that $BS(1,m)$ may be viewed as a semi-direct product

$BS(1,m)=\mathbb{Z}[1/m]\rtimes\mathbb{Z}$

where the factor $\mathbb{Z}$ corresponds to the subgroup $\langle t\rangle$
acting by powers of $m$. We may identify this semi-direct product with the
following subgroup $G_{m}$ of upper triangular matrices in
$GL_{2}(\mathbb{Z}[1/m])$:

$G_{m}=\left\\{\left(\begin{array}[]{cc}m^{k}&r\\\
0&1\end{array}\right):k\in\mathbb{Z},\;r\in\mathbb{Z}[1/m]\right\\}.$

The isomorphism is obtained by mapping $a$ to
$A=\left(\begin{array}[]{cc}1&1\\\ 0&1\end{array}\right)$ and $t$ to
$T=\left(\begin{array}[]{cc}m&0\\\ 0&1\end{array}\right)$. The associated
embedding of $BS(1,m)$ into $GL_{2}(\mathbb{Z}[1/m])$ is called the standard
embedding.

In $GL_{n}(\mathbb{Z}[1/m])$ we may define congruence subgroups. Let $N>0$ be
coprime with $m$. The principal congruence subgroup of level $N$ is the kernel
$M_{N}$ of the reduction modulo $N$:

$M_{N}=\ker\left[GL_{n}(\mathbb{Z}[1/m])\rightarrow
GL_{n}(\mathbb{Z}/N\mathbb{Z})\right].$

###### Definition 1.2.

If $G$ is any subgroup of $GL_{n}(\mathbb{Z}[1/m])$, and $N>0$ is coprime to
$m$, then the congruence subgroup $G(N)$ in $G$ is

$G(N):=G\cap M_{N}.$

For a sequence of integers such that each one divides the next one, one
obtains a sequence of nested congruence subgroups, and thus a box space of
$BS(1,m)$. Such box spaces deserve to be called arithmetic box spaces. We will
study property $D_{\alpha}$ for the arithmetic box spaces of $BS(1,m)$ through
the standard embedding. From Theorem 1.1, we know that for every $0<\alpha<1$,
there exists a box space of $BS(1,m)$ with property $D_{\alpha}$, but what
about arithmetic box spaces? We will prove that box spaces with $D_{\alpha}$,
for $\alpha>\frac{1}{2}$, can be distinguished from arithmetic box spaces by
coarse geometry. More precisely:

###### Theorem 1.3.

For any $m\geq 2$, the following statements are true :

  1. (1)

if an arithmetic box space $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ has property
$D_{\alpha}$, then $\alpha\leq\frac{1}{2}$,

  2. (2)

there exists an arithmetic box space with property $D_{1/2}$,

  3. (3)

there exists an arithmetic box space of $G_{m}$ without property $D_{\alpha}$
for any $\alpha\in]0,1/2]$.

If $\square_{(H_{k})}G,\,\square_{(H_{k}^{\prime})}G$ are two box spaces of
the same residually finite group $G$, we say that
$\square_{(H_{k}^{\prime})}G$ covers $\square_{(H_{k})}G$ if
$H_{k}^{\prime}\subset H_{k}$ for every $k>0$. In this case
$\operatorname{Cay}(G/H_{k}^{\prime},S)$ is a Galois covering of
$\operatorname{Cay}(G/H_{k},S)$.

The following proposition bridges Theorem 1.3 with part (2) of Theorem 1.1.

###### Proposition 1.4.

Fix $\alpha<1$. Any arithmetic box space of $G_{m}$ is covered by some box
space with $D_{\alpha}$.

In addition, we study how property $D_{\alpha}$ for an arithmetic box space
depends on the prime factors of the $N$’s in the sequence of congruence
subgroups $(M_{N})_{N}$. In fact, if we denote by $D^{\prime}(P)$ the analytic
density of the prime factors (see Section 3.2), we prove the following.

###### Theorem 1.5.

Let $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ be an arithmetic box space, and let
$P$ be the set of prime factors of the sequence $(N_{k})_{k}$.

  1. (1)

If $|P|<+\infty$, then $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ has $D_{1/2}$ ;

  2. (2)

If $D^{\prime}(P)>0$, then $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ does not have
$D_{\alpha}$, for every $\alpha>0$.

Of course the choice of the standard embedding raises a natural question: how
do the congruence subgroups in $BS(1,m)$ depend on the choice of the embedding
$\rho$? Since congruence subgroups have finite index, this is related to a
form of the congruence subgroup property (CSP), which we now recall.

###### Definition 1.6.

Let $G$ be a subgroup of $GL_{n}(\mathbb{Z}[1/m])$. We say that $G$ has the
CSP if every finite index subgroup in $G$ contains $G(N)$ for some $N>0$.

Let $T_{n}(\mathbb{Z}[1/m])$ denote the group of upper triangular matrices in
$GL_{n}(\mathbb{Z}[1/m])$. We shall prove:

###### Theorem 1.7.

For every embedding $\rho:BS(1,m)\rightarrow T_{n}(\mathbb{Z}[1/m])$, the
group $\rho(BS(1,m))$ has the CSP.

This result can be reformulated as follows. Taking the congruence subgroups
$\rho(G_{m})(N)$ as a neighbourhood basis for the identity gives a topology on
$\rho(G_{m})$ (namely the congruence topology relative to $\rho$) and the
completion $\overline{\rho(G_{m})}$ with respect to this topology is a
profinite group called the congruence completion of $\rho(G_{m})$. As $G_{m}$
is residually finite, it also embeds in its profinite completion
$\widehat{G_{m}}$ which maps onto $\overline{\rho(G_{m})}$. The CSP for
$\rho(G_{m})$ says that the map
$\widehat{G_{m}}\rightarrow\overline{\rho(G_{m})}$ is an isomorphism.

Another immediate consequence of CSP is that, if $\rho:G_{m}\rightarrow
T_{n}(\mathbb{Z}[1/m])$ is an embedding, then any box space of
$G_{m}\simeq\rho(G_{m})$ is covered by some arithmetic box space of
$\rho(G_{m})$.

When $m=p^{\ell}$ is a prime power, we also propose another approach to CSP
for $BS(1,m)$: we view $BS(1,p^{\ell})$ as a subgroup of the affine group
$\mathbb{G}(\mathbb{Q})$, and this subgroup is commensurable to the group
$\mathbb{G}(\mathcal{O}_{S})$ of $S$-integer points, with $S=\\{p,\infty\\}$.
We prove:

###### Proposition 1.8.

Let $\mathbb{H}$ be a $\mathbb{Q}$-algebraic subgroup of $GL_{n}$, and let
$\rho:\mathbb{G}\rightarrow\mathbb{H}$ be a $\mathbb{Q}$-isomorphism. Then
$\rho(BS(1,p^{\ell}))$ has CSP.

As for the structure of the paper, we will start in Section 2 by recalling
some well-known facts about elementary number theory, especially concerning
the multiplicative order of $m$ in $\mathbb{Z}/N\mathbb{Z}$.

We then study property $D_{\alpha}$ for solvable Baumslag-Solitar groups in
Section 3. We begin with some metric aspects, where we estimate the diameter
of the arithmetic box spaces of $BS(1,m)$, which is what we use to prove
Theorem 1.3 and Proposition 1.4 (see Theorem 3.6). To conclude Section 3, we
investigate the role of prime numbers in property $D_{\alpha}$ and we prove
Theorem 1.5 (see Theorem 3.8).

Finally, in Section 4, we prove Theorem 1.7.

We will use Landau’s notations $O,o,\Omega,\Theta$.

##  2\. Some elementary number theory

We gather some technical lemmas. First we recall some facts about greatest
common divisors and lowest common multiples, which will be respectively
denoted by $\gcd$ and $\operatorname{lcm}$.

###### Proposition 2.1.

Let $\mu,a_{1},\ldots,a_{n}\in\mathbb{N}^{\ast}$,

  1. (1)

$\gcd(\mu\cdot a_{1},\mu\cdot a_{2})=\mu\cdot\gcd(a_{1},a_{2})$,

  2. (2)

$\gcd(a_{1},\ldots,a_{n})=\gcd(\gcd(a_{1},\ldots,a_{n-1}),a_{n})$,

  3. (3)

$\gcd(a_{1},\gcd(a_{2},a_{3}))=\gcd(\gcd(a_{1},a_{2}),a_{3})$,

  4. (4)

$\gcd(a_{1},a_{2})\cdot\operatorname{lcm}(a_{1},a_{2})=a_{1}\cdot a_{2}$,

  5. (5)

$\operatorname{lcm}(a_{1},...,a_{n})=\operatorname{lcm}(\operatorname{lcm}(a_{1},\ldots,a_{n-1}),a_{n})$.
$\square$

The following lemma generalises Proposition 2.1.(4).

###### Lemma 2.2.

Let $a_{1},\ldots,a_{n}\in\mathbb{N}$, then

(2) $\operatorname{lcm}(a_{1},\ldots,a_{n})=\frac{a_{1}\cdots
a_{n}}{\gcd(a_{1}\cdots a_{n-1},a_{1}\cdots a_{n-2}a_{n},\ldots,a_{2}\cdots
a_{n})}.$

###### Proof.

We use induction to show that Eq. (2) is valid. If $n=1$, the formula holds.
Thus assume that the formula is true for $n\in\mathbb{N}$, and denote by
$\Pi_{n}$ the set $\\{a_{1}\cdots a_{n-1},a_{1}\cdots
a_{n-2}a_{n},\ldots,a_{2}\cdots a_{n}\\}$. Using that

$\gcd\left(\operatorname{lcm}(a_{1},\ldots,a_{n}),a_{n+1}\right)=\gcd\left(\frac{a_{1}\cdots
a_{n}}{\gcd(\Pi_{n})},a_{n+1}\right),$

we obtain by a direct computation that

$\displaystyle\operatorname{lcm}(a_{1},\ldots,a_{n+1})$
$\displaystyle=\operatorname{lcm}(\operatorname{lcm}(a_{1},\ldots,a_{n}),a_{n+1})$
$\displaystyle=\frac{\operatorname{lcm}(a_{1},\ldots,a_{n})\cdot
a_{n+1}}{\gcd\left(\operatorname{lcm}(a_{1},\ldots,a_{n}),a_{n+1}\right)}$
$\displaystyle=\frac{a_{1}\cdots a_{n}\cdot
a_{n+1}}{\gcd(\Pi_{n})\gcd\left(\frac{a_{1}\cdots
a_{n}}{\gcd(\Pi_{n})},a_{n+1}\right)}$ $\displaystyle=\frac{a_{1}\cdots
a_{n+1}}{\gcd(a_{1}\cdots a_{n},\gcd(\Pi_{n})\cdot a_{n+1})}$
$\displaystyle=\frac{a_{1}\cdots a_{n+1}}{\gcd(a_{1}\cdots
a_{n},\gcd(a_{1}a_{3}\cdots a_{n}a_{n+1},\ldots,a_{2}\cdots a_{n}a_{n+1}))}$
$\displaystyle=\frac{a_{1}\cdots a_{n+1}}{\gcd(a_{1}\cdots
a_{n},a_{1}a_{3}\cdots a_{n}a_{n+1},\ldots,a_{2}\cdots a_{n}a_{n+1})}.\qed$

We now recall the ideal structure of the ring $\mathbb{Z}[1/m]$.

###### Lemma 2.3.

Let $I$ be a proper subgroup in $\mathbb{Z}[1/m]$. TFAE:

  1. (1)

$I$ is an ideal in $\mathbb{Z}[1/m]$;

  2. (2)

There exists $N>1$ such that $I=N\mathbb{Z}[1/m]$.

Moreover $\mathbb{Z}[1/m]/N\mathbb{Z}[1/m]\simeq\mathbb{Z}/N\mathbb{Z}$

###### Proof.

The non-trivial direction follows from general results about localisations of
rings, viewing $\mathbb{Z}[1/m]$ as the localisation of $\mathbb{Z}$ with
respect to powers of $m$: the map $I\rightarrow I\cap\mathbb{Z}$ provides a
bijection between ideals of $\mathbb{Z}[1/m]$ and ideals $J$ in $\mathbb{Z}$
such that $m$ is not a zero-divisor in $\mathbb{Z}/J$ (see Proposition 2 in
section 11.3 of [Cohn 1977]). Finally observing that
$\mathbb{Z}+N\mathbb{Z}[1/m]=\mathbb{Z}[1/m]$, we have by a classical
isomorphism theorem:

$\mathbb{Z}[1/m]/N\mathbb{Z}[1/m]=(\mathbb{Z}+N\mathbb{Z}[1/m])/N\mathbb{Z}[1/m]\simeq\mathbb{Z}/(\mathbb{Z}\cap
N\mathbb{Z}[1/m])=\mathbb{Z}/N\mathbb{Z}.$

∎

Lemma 2.3 allows us to work with $\mathbb{Z}/N\mathbb{Z}$, which has a
familiar ring structure. We will write $\mathbb{Z}/N\mathbb{Z}^{\times}$ for
the multiplicative group of $\mathbb{Z}/N\mathbb{Z}$. We denote by
$\operatorname{ord}_{m}(N)$ the multiplicative order of $m$ in
$\mathbb{Z}/N\mathbb{Z}^{\times}$. We define the following function.

###### Definition 2.4.

Let $m,N\in\mathbb{N}$ be such that $\gcd(m,N)=1$. Write
$m^{\operatorname{ord}_{m}(N)}=\mu N+1$ for some $\mu\in\mathbb{N}$ and let
the function $\eta_{N}\colon\mathbb{N}^{\ast}\to\mathbb{N}$ be defined by

$\eta_{N}(k)=\begin{cases}1&\text{if }k=1\\\
\frac{N^{k-1}}{\gcd(\mu,N)}&\text{if }k\geq 2.\end{cases}$

###### Lemma 2.5.

Let $m,N\in\mathbb{N}$ be such that $\gcd(m,N)=1$. Write
$m^{\operatorname{ord}_{m}(N)}=\mu N+1$ for some $\mu\in\mathbb{N}$. Then
$\operatorname{ord}_{m}(N^{2})=\operatorname{ord}_{m}(N)\cdot\frac{N}{\gcd(\mu,N)}$,
and more generally

(3)
$\operatorname{ord}_{m}(N^{k})=\operatorname{ord}_{m}(N)\cdot\eta_{N}(k),\quad\forall
k\geq 1.$

###### Proof.

The case $k=1$ being obvious, let us consider $k=2$, and set
$\beta=\operatorname{ord}_{m}(N)$. We show that the smallest positive integer
$\lambda$ that satisfies $m^{\lambda}\equiv 1\pmod{N^{2}}$ is
$\lambda=\beta\frac{N}{\gcd(\mu,N)}$. Note that $\beta\mid\lambda$ : indeed if
$m^{\lambda}\equiv 1\pmod{N^{2}}$, then $m^{\lambda}\equiv 1\pmod{N}$ so that
$\beta$ must divide $\lambda$ (see [Gallian 2012, Cor. 2 p. 79]). Thus
$\lambda=\beta\tilde{\lambda}$ for some $\tilde{\lambda}\in\mathbb{N}$ and we
only have to show that $\tilde{\lambda}=\frac{N}{\gcd(\mu,N)}$. We have that

$\displaystyle(m^{\beta})^{\tilde{\lambda}}$ $\displaystyle=(\mu
N+1)^{\tilde{\lambda}}$
$\displaystyle=\sum_{i=0}^{\tilde{\lambda}}\binom{\tilde{\lambda}}{i}(\mu
N)^{i}$ $\displaystyle\equiv 1+\tilde{\lambda}\mu N\pmod{N^{2}}.$

The last line shows that the smallest $\tilde{\lambda}$ we can take to have
$m^{\beta\tilde{\lambda}}\equiv 1\pmod{N^{2}}$ must be
$\tilde{\lambda}=\frac{N}{\gcd(\mu,N)}$, thus demonstrating that
$\operatorname{ord}_{m}(N^{2})=\beta\cdot\frac{N}{\gcd(\mu,N)}$.

The same arguments can be applied to show that
$\operatorname{ord}_{m}(N^{k})=\beta\frac{N^{k-1}}{\gcd(\mu,N)}$ for $k\geq
2$. ∎

###### Lemma 2.6.

Let $k,N\in\mathbb{N}^{\ast}$. Then $\eta_{N}(k)\geq N^{k-2}$.

###### Proof.

If $k=1$, then $1\geq N^{-1}$. If $k\geq 2$, observe that $\gcd(\mu,N)\leq N$
implies

$\frac{N^{k-1}}{\gcd(\mu,N)}\geq N^{k-2}.$

∎

Denote by $\mathcal{P}\subset\mathbb{N}$ the set of prime numbers. The
following lemma gives us a formula to compute the order of $m$ in
$\mathbb{Z}/N\mathbb{Z}$ for any $N\in\mathbb{N}$. It is an immediate
consequence of the Chinese remainder theorem.

###### Lemma 2.7.

For every $N\in\mathbb{N}$, which we write
$N=p_{1}^{\beta_{1}}p_{2}^{\beta_{2}}\cdots p_{n}^{\beta_{n}}$ with
$p_{i}\in\mathcal{P}$ and $\beta_{i}\in\mathbb{N}$ for every
$i\in\\{1,\ldots,n\\}$, we have

(4)
$\operatorname{ord}_{m}(N)=\operatorname{ord}_{m}\left(p_{1}^{\beta_{1}}p_{2}^{\beta_{2}}\cdots
p_{n}^{\beta_{n}}\right)=\operatorname{lcm}\left(\operatorname{ord}_{m}\left(p_{1}^{\beta_{1}}\right),\operatorname{ord}_{m}\left(p_{2}^{\beta_{2}}\right),\ldots,\operatorname{ord}_{m}\left(p_{n}^{\beta_{n}}\right)\right).$

###### Lemma 2.8.

Let $P$ be a finite set of primes, not dividing $m$. There exists a constant
$C(m,P)>0$ such that, for every integer $N$ with all prime factors in $P$, we
have:

$\frac{\operatorname{ord}_{m}(N)}{N}\geq C(m,P).$

###### Proof.

Write $N=p_{1}^{\beta_{1}}\ldots p_{k}^{\beta_{k}}$, with $p_{i}\in P$, all
different, $\beta_{i}>0$ and $\eta_{p_{i}}$ defined as in Definition 2.4. In
addition, we define the set

$\displaystyle\Pi_{k}:=\\{$
$\displaystyle\operatorname{ord}_{m}(p_{1})\eta_{p_{1}}(\beta_{1})\cdots\operatorname{ord}_{m}(p_{k-1})\eta_{p_{k-1}}(\beta_{k-1}),$
$\displaystyle\operatorname{ord}_{m}(p_{1})\eta_{p_{1}}(\beta_{1})\cdots\operatorname{ord}_{m}(p_{k-2})\eta_{p_{k-2}}(\beta_{k-2})\operatorname{ord}_{m}(p_{k})\eta_{p_{k}}(\beta_{k}),$
$\displaystyle\ldots$
$\displaystyle\operatorname{ord}_{m}(p_{2})\eta_{p_{2}}(\beta_{2})\cdots\operatorname{ord}_{m}(p_{k})\eta_{p_{k}}(\beta_{k})\\}$

which contains all possible products with $k-1$ factors, each one of
$\operatorname{ord}_{m}(p_{i})\eta_{p_{i}}(\beta_{i})$, $i=1,\ldots,k$. Using
Lemmas 2.2, 2.5, and 2.7, we obtain

$\displaystyle\operatorname{ord}_{m}(N)$
$\displaystyle=\operatorname{lcm}\left(\operatorname{ord}_{m}\left(p_{1}^{\beta_{1}}\right),\operatorname{ord}_{m}\left(p_{2}^{\beta_{2}}\right),\ldots,\operatorname{ord}_{m}\left(p_{k}^{\beta_{k}}\right)\right)$
$\displaystyle=\frac{\operatorname{ord}_{m}(p_{1})\eta_{p_{1}}(\beta_{1})\cdots\operatorname{ord}_{m}(p_{k})\eta_{p_{k}}(\beta_{k})}{\gcd(\Pi_{k})},$

thus

(5)
$\frac{\operatorname{ord}_{m}(N)}{N}=\frac{\operatorname{ord}_{m}(p_{1})\eta_{p_{1}}(\beta_{1})\cdots\operatorname{ord}_{m}(p_{k})\eta_{p_{k}}(\beta_{k})}{N\cdot\gcd(\Pi_{k})}.$

Moreover, $\operatorname{ord}_{m}(p_{i})\geq 1$ for every $i$, and using Lemma
2.6 on each $\eta_{p_{i}}(\beta_{i})$, we obtain from Eq. (5)

(6) $\frac{\operatorname{ord}_{m}(N)}{N}\geq\frac{p_{1}^{\beta_{1}-2}\cdots
p_{{k}}^{\beta_{{k}}-2}}{p_{1}^{\beta_{1}}\cdots
p_{{k}}^{\beta_{{k}}}\cdot\gcd(\Pi_{k})}=\frac{1}{p_{1}^{2}\ldots
p_{{k}}^{2}\cdot\gcd(\Pi_{k})}>0.$

So we may take $C(m,P)$ as the minimum of the $\frac{1}{p_{1}^{2}\ldots
p_{k}^{2}\cdot\gcd(\Pi_{k})}$’s taken over all subsets $\\{p_{1},...,p_{k}\\}$
of $P$.

∎

The following material will be used in the proof of Proposition 4.10

###### Definition 2.9.

Let $N\in\mathbb{N}$ and $N=\Pi_{i=1}^{r}p_{i}^{\beta_{i}}$ its decomposition
in prime factors. The dominant prime of $N$ is the factor $p_{I}$ in $P$ such
that $p_{I}^{\beta_{I}}\geq p_{i}^{\beta_{i}}$ for $\forall i$. The factor
$p_{I}^{\beta_{I}}$ is called the dominating factor.

###### Lemma 2.10.

Fix $s\in(\mathbb{Z}[1/m])^{\times}$, with $s>0$. There exist infinitely many
integers $N>0$, coprime to $m$, such that $s$ has odd order in the
multiplicative group
$(\mathbb{Z}[1/m]/N\mathbb{Z}[1/m])^{\times}\simeq(\mathbb{Z}/N\mathbb{Z})^{\times}$.

###### Proof.

If $s=1$, take any $N$ coprime with $m$. So we assume $s\neq 1$ and, replacing
$s$ by $s^{-1}$ if necessary, we assume $s>1$.

Let $P$ be the set of primes dividing $m$, set $r=|P|$. Write
$s=\frac{a_{1}}{a_{2}}$ with $a_{1}>a_{2}>0$, coprime, and all their prime
factors in $P$. For $k\in\mathbb{N}$, set $a_{1}^{k}-a_{2}^{k}=N_{k}q_{k}$,
where $N_{k}$ is the maximal factor coprime to $m$. Observe that if a prime in
$P$ divides $q_{k}$, it cannot simultaneously divide $a_{1}$ and $a_{2}$,
since they are coprime.

It is clear that, for odd $k$, $s$ will also have odd order in
$(\mathbb{Z}[1/m]/(a_{1}^{k}-a_{2}^{k})\mathbb{Z}[1/m])^{\times}$, hence also
in $(\mathbb{Z}[1/m]/N_{k}\mathbb{Z}[1/m])^{\times}$. It is therefore enough
to prove that the map $k\mapsto N_{k}$ takes infinitely many values on odd
integers. This will follow immediately from the following.

Claim: There is an infinite family of odd integers $k$ such that
$q_{k}\leq(a_{1}^{2r}-a_{2}^{2r})^{r}$.

To prove the claim, take $k$ odd with $k>2r$. By the pigeonhole principle
there are $i,j\in\\{0,1\dots,r\\}$ with $i<j$ such that the dominant primes in
$q_{k-2i}$ and $q_{k-2j}$ are the same. We have that

$\displaystyle(s^{k-2i}-1)-(s^{k-2j}-1)$
$\displaystyle=s^{k-2j}(s^{2(j-i)}-1)$
$\displaystyle=\frac{a_{1}^{k-2j}}{a_{2}^{k-2i}}(a_{1}^{2(j-i)}-a_{2}^{2(j-i)}).$

Write $p_{i}^{\beta_{i}}$ for the dominating factor of $q_{k-2i}$. Set
$\beta=\min\\{\beta_{i},\beta_{j}\\}$ and $p=p_{i}$; say that
$\beta=\beta_{i}$ (otherwise replace $i$ by $j$). We see that $p^{\beta}$
divides the numerator on the left, hence it divides the numerator on the
right. We note that $p$ does not divide $a_{1}$, nor $a_{2}$, hence it must
divide $a_{1}^{2(j-i)}-a_{2}^{2(j-i)}$, which is bounded by
$(a_{1}^{2r}-a_{2}^{2r})$. So we get $p^{\beta}\leq(a_{1}^{2r}-a_{2}^{2r})$
and it follows that $q_{k-2i}\leq(a_{1}^{2r}-a_{2}^{2r})^{r}$. ∎

##  3\. Property $D_{\alpha}$ for solvable Baumslag-Solitar groups

###  3.1. Metric aspects of solvable Baumslag-Solitar groups

We study the diameter of arithmetic box spaces of $BS(1,m)$ according to Eq.
(1). In this section, we will always assume that $\gcd(m,N)=1$. We recall that
every element of $BS(1,m)$ ($m>1$) admits a unique normal form of the type
$t^{-i}a^{\ell}t^{j}$ with $i,j\geq 0,\ell\in\mathbb{Z}$ and $\ell$ can be a
multiple of $m$ only if either $i$ or $j$ is zero. Indeed, one can rewrite
$ta$ as $a^{m}t$, $ta^{-1}$ as $a^{-m}t$, $at^{-1}$ as $t^{-1}a^{m}$, and
$a^{-1}t^{-1}$ as $t^{-1}a^{-m}$ and the result follows.

The normal form of a word is usually not the geodesic form, and we want to
estimate how well the normal form approximates the geodesic form.

###### Proposition 3.1 ([Burillo and Elder 2015, Prop. 2.1]).

There exist constants $C_{1},C_{2},D_{1},D_{2}>0$ such that for any
$\omega=t^{-i}a^{\ell}t^{j}\in BS(1,m)$ with $\ell\neq 0$, we have

$C_{1}(i+j+\log|\ell|)-D_{1}\leq\|\omega\|\leq C_{2}(i+j+\log|\ell|)+D_{2}$

where $\|\cdot\|$ is the word metric with respect to $\\{a^{\pm 1},t^{\pm
1}\\}$. Moreover we may take $C_{2}=D_{2}=m$.

Let $A=\left(\begin{smallmatrix}1&1\\\ 0&1\end{smallmatrix}\right)$,
$T=\left(\begin{smallmatrix}m&0\\\ 0&1\end{smallmatrix}\right)\in
GL_{2}\left(\mathbb{Z}\left[1/m\right]\right)$, and denote by $G_{m}$ the
subgroup of $GL_{2}\left(\mathbb{Z}\left[1/m\right]\right)$ generated by $A$
and $T$. In the previous section, we saw that $BS(1,m)\simeq G_{m}\leq
GL_{2}\left(\mathbb{Z}\left[1/m\right]\right)$.

As mentioned before, $BS(1,m)\simeq G_{m}$ is a finitely generated, residually
finite group that surjects onto $\mathbb{Z}$ so that Theorem 1.1 applies and
we know that for every $0<\alpha<1$, there exists a box space of $G_{m}$ with
property $D_{\alpha}$. However, we are interested in specific box spaces of
$G_{m}$, namely the arithmetic box spaces, or in other words, box spaces of
the form $\square_{(G_{m}(N_{k}))_{k}}G_{m}$. To this end, we start by
studying the quotients $G_{m}/G_{m}(N)$, and then explore how the diameters
evolve.

###### Proposition 3.2.

Let $N\in\mathbb{N}$ be such that $\gcd(m,N)=1$. Then

(7)
$G_{m}/G_{m}(N)\simeq\mathbb{Z}/N\mathbb{Z}\rtimes_{m}\mathbb{Z}/\operatorname{ord}_{m}(N)\mathbb{Z},\quad\text{
and }\quad\left|G_{m}/G_{m}(N)\right|=N\cdot\operatorname{ord}_{m}(N),$

where $\mathbb{Z}/\operatorname{ord}_{m}(N)\mathbb{Z}$ acts on
$\mathbb{Z}/N\mathbb{Z}$ by multiplication by $m$.

###### Proof.

Consider reduction modulo $N$:

$\begin{array}[]{cccl}\varphi\colon&G_{m}&\to&GL_{2}(\mathbb{Z}/N\mathbb{Z})\\\
&w=\begin{pmatrix}m^{k}&x\\\
0&1\end{pmatrix}&\mapsto&\begin{pmatrix}[m^{k}]&[x]\\\
[0]&[1]\end{pmatrix}.\end{array}$

The image of $\varphi$ is clearly isomorphic to
$\mathbb{Z}/N\mathbb{Z}\rtimes_{m}\mathbb{Z}/\operatorname{ord}_{m}(N)\mathbb{Z}$,
and of order $N\cdot\operatorname{ord}_{m}(N)$. Moreover we have

$w=\begin{pmatrix}m^{k}&x\\\ 0&1\end{pmatrix}\in G_{m}(N)\iff m^{k}\equiv
1\pmod{N}\text{ and }x\in
N\mathbb{Z}\left[\tfrac{1}{m}\right]\iff\varphi(w)=1.$

So $\ker(\varphi)=G_{m}(N)$ and the result follows from the first isomorphism
theorem. ∎

###### Example 3.3.

Consider $BS(1,2)$ and $N=5$. Then $\operatorname{ord}_{2}(5)=4$, and
$G_{2}/G_{2}(5)\simeq\mathbb{Z}/5\mathbb{Z}\rtimes_{2}\mathbb{Z}/4\mathbb{Z}$.
The Cayley graph of the quotient is the graph drawn below, using $a=(1,0)$ and
$t=(0,1)$ as generators. Note that one still needs to identify the bottom line
with the upper line, and the line to the left with the line to the right.

$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$1$$a$$t$

Thanks to the familiar structure of the quotient $G_{m}/G_{m}(N)$ and
Proposition 2.1 from [Burillo and Elder 2015], we are able to estimate the
diameter of arithmetic box spaces of $BS(1,m)$.

###### Lemma 3.4.

Let $N\geq 2$. Then

(8)
$\operatorname{diam}(\operatorname{Cay}(G_{m}/G_{m}(N)))=\Theta(\operatorname{ord}_{m}(N)).$

More precisely, there exists a constant $C_{m}>0$ such that

(9)
$\frac{1}{3}\cdot\operatorname{ord}_{m}(N)\leq\operatorname{diam}(\operatorname{Cay}(G_{m}/G_{m}(N)))\leq
C_{m}\cdot\operatorname{ord}_{m}(N).$

###### Proof.

Let $m\geq 2$ and consider $BS(1,m)\simeq
G_{m}\subset\mathsf{GL}_{2}(\mathbb{Z}[1/m])$. Recall that
$G_{m}/G_{m}(N)\simeq\mathbb{Z}/N\mathbb{Z}\rtimes_{m}\mathbb{Z}/\operatorname{ord}_{m}(N)\mathbb{Z}$
so that
$\operatorname{diam}(G_{m}/G_{m}(N))=\operatorname{diam}(\mathbb{Z}/N\mathbb{Z}\rtimes_{m}\mathbb{Z}/\operatorname{ord}_{m}(N)\mathbb{Z})\geq\operatorname{diam}(\mathbb{Z}/\operatorname{ord}_{m}(N)\mathbb{Z})$.
Since the Cayley graph of $\mathbb{Z}/\operatorname{ord}_{m}(N)\mathbb{Z}$ is
a cycle, we can roughly estimate the diameter to obtain

(10)
$\operatorname{diam}(G_{m}/G_{m}(N))\geq\frac{1}{3}\operatorname{ord}_{m}(N).$

For the second inequality, let
$([x],[k])\in\mathbb{Z}/N\mathbb{Z}\rtimes_{m}\mathbb{Z}/\operatorname{ord}_{m}(N)\mathbb{Z}$
be an element realising the diameter. We rewrite $([x],[k])$ as
$\big{(}\begin{smallmatrix}m^{k}&x\\\ 0&1\end{smallmatrix}\big{)}G_{m}(N)$.
The induced metrics are always smaller in a quotient, thus

$\left\|\begin{pmatrix}m^{k}&x\\\
0&1\end{pmatrix}G_{m}(N)\right\|_{G_{m}/G_{m}(N)}\leq\left\|\begin{pmatrix}m^{k}&x\\\
0&1\end{pmatrix}\right\|_{G_{m}}.$

Recall that any word $\omega\in BS(1,m)$ can be written in normal form as
$\omega=t^{-i}a^{\ell}t^{j}$. In the quotient $G_{m}/G_{m}(N)$, the situation
is even simpler, as by the semi-direct product stucture every word can be
written as $A^{\ell}T^{j}$ with $0\leq\ell<N$ and $0\leq
j<\operatorname{ord}_{m}(N)$. With $\ell=x$ and $j=k$, we identify
$\big{(}\begin{smallmatrix}m^{k}&x\\\ 0&1\end{smallmatrix}\big{)}$ with the
element $A^{x}T^{k}$ in normal form in $G_{m}$. If $x=0$ we get $T^{k}$ and

$\|T^{k}\|_{G_{m}}=k<\operatorname{ord}_{m}(N).$

Assume that $x\neq 0$. From Proposition 3.1, we obtain

(11) $\left\|\begin{pmatrix}m^{k}&x\\\
0&1\end{pmatrix}\right\|_{G_{m}}=\|A^{x}T^{k}\|_{G_{m}}\leq 2m(k+\log x+1).$

Note that since $m^{\operatorname{ord}_{m}(N)}\geq N$ (equivalently
$\log(N)\leq\operatorname{ord}_{m}(N)\cdot\log(m)$) and moreover $\log
x\leq\log(N)$, Eq. (11) becomes

(12) $\left\|\begin{pmatrix}m^{k}&x\\\ 0&1\end{pmatrix}\right\|_{G_{m}}\leq
2m(2+\log(m))\operatorname{ord}_{m}(N).$

Setting $C_{m}:=2m(2+\log(m))$, we obtain

(13) $\operatorname{diam}(\operatorname{Cay}(G_{m}/G_{m}(N)))\leq
C_{m}\cdot\operatorname{ord}_{m}(N).$

∎

###### Proposition 3.5.

An arithmetic box space $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ has property
$D_{\alpha}$ if and only if

$\operatorname{ord}_{m}(N_{k})=\Omega(N_{k}^{\frac{\alpha}{1-\alpha}}).$

###### Proof.

Using Lemma 3.4:

$\square_{(G_{m}(N_{k}))_{k}}G_{m}\,\mbox{has}\,D_{\alpha}\iff\operatorname{diam}(G_{m}/G_{m}(N_{k}))=\Omega(|G_{m}/G_{m}(N_{k})|^{\alpha})$
$\iff\operatorname{ord}_{m}(N_{k})=\Omega(N_{k}^{\alpha}\cdot\operatorname{ord}_{m}(N_{k})^{\alpha})\iff\operatorname{ord}_{m}(N_{k})=\Omega(N_{k}^{\frac{\alpha}{1-\alpha}}).$

∎

We present here the main structure theorem for the arithmetic box spaces of
$BS(1,m)$.

###### Theorem 3.6.

For any $m\geq 2$, the following statements hold :

  1. (1)

if an arithmetic box space $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ has property
$D_{\alpha}$, then $\alpha\leq\frac{1}{2}$,

  2. (2)

there exists an arithmetic box space with property $D_{1/2}$,

  3. (3)

there exists an arithmetic box space of $G_{m}$ without property $D_{\alpha}$
for any $\alpha\in]0,1/2]$.

  4. (4)

Fix $\alpha<1$. Every arithmetic box space of $G_{m}$ is covered by some box
space with $D_{\alpha}$.

###### Proof.

  1. (1)

If $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ has property $D_{\alpha}$, using
$N_{k}\geq\operatorname{ord}_{m}(N_{k})$ and Proposition 3.5, we get
$N_{k}=\Omega(N_{k}^{\frac{\alpha}{1-\alpha}})$, which forces
$\alpha\leq\frac{1}{2}$.

  2. (2)

Let $(N_{k})_{k}\subset\mathbb{N}$ be the sequence defined by
$N_{k}=(m^{2}-1)^{k}$. Clearly, $N_{k}\mid N_{k+1}$ for every $k$. We apply
Lemma 2.5 with $N=m^{2}-1$, so that $\operatorname{ord}_{m}(m^{2}-1)=2$ and
$\mu=1$. We thus obtain:

(14) $\operatorname{ord}_{m}(N_{k})=2\cdot(m^{2}-1)^{k-1},\forall k\geq 1.$

i.e. $\operatorname{ord}_{m}(N_{k})=\Omega(N_{k})$. By Proposition 3.5 the box
space $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ has property $D_{1/2}$.

  3. (3)

We consider the sequence $(N_{k})_{k}$ defined by $N_{k}=m^{2^{k}}-1$ and
prove that the arithmetic box space $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ does
not have property $D_{\alpha}$ for any $\alpha\in]0,1/2]$. It is
straightforward that $N_{k}\mid N_{k+1}$ for every $k$, and
$\operatorname{ord}_{m}(N_{k})=2^{k}$. We have

$\lim_{k\to\infty}\frac{\operatorname{ord}_{m}(N_{k})}{N_{k}^{\frac{\alpha}{1-\alpha}}}=\lim_{k\to\infty}\frac{2^{k}}{(m^{2^{k}}-1)^{\frac{\alpha}{1-\alpha}}}=0,$

i.e. $\operatorname{ord}_{m}(N_{k})=o(N_{k}^{\frac{\alpha}{1-\alpha}})$. By
Proposition 3.5 this shows that the arithmetic box space
$\square_{(G_{m}(m^{2^{k}}-1))_{k}}G_{m}$ does not have property $D_{\alpha}$
for any $\alpha\in]0,1/2]$.

  4. (4)

We adapt the proof of Theorem 3 in [Khukhro and Valette 2017]. Pick an integer
$D>0$ with $\frac{D}{D+1}\geq\alpha$. Let $\square_{G_{m}(N_{k})}G_{m}$ be any
arithmetic box space of $G_{m}$. Define

$n_{k}=\operatorname{ord}_{m}(N_{k})\cdot N_{k}^{D}$

and

$M_{k}=\left\\{\left(\begin{array}[]{cc}m^{\ell n_{k}}&r\\\
0&1\end{array}\right):\ell\in\mathbb{Z},\;r\in
N_{k}\mathbb{Z}[\tfrac{1}{m}]\right\\}.$

It is readily checked that $M_{k}$ is a subgroup and, because
$\operatorname{ord}_{m}(N_{k})|n_{k}$, that $M_{k}$ is normal in $G_{m}$ and
is contained in $G(N_{k})$. As $n_{k}|n_{k+1}$, we have that
$\square_{(M_{k})}G_{m}$ is a box space which covers
$\square_{(G_{m}(N_{k}))}G_{m}$.

It remains to check that $\square_{(M_{k})}G_{m}$ has property $D_{\alpha}$.
But $G/M_{k}$ maps onto the cyclic group $\mathbb{Z}/n_{k}\mathbb{Z}$ so we
have

$\operatorname{diam}(G/M_{k})\geq\operatorname{diam}(\mathbb{Z}/n_{k}\mathbb{Z})\geq\frac{n_{k}}{3}.$

On the other hand

$|G/M_{k}|^{\alpha}=n_{k}^{\alpha}N_{k}^{\alpha}=\operatorname{ord}_{m}(N_{k})^{\alpha}N_{k}^{\alpha(D+1)}\leq\operatorname{ord}_{m}(N_{k})N_{k}^{D}=n_{k}.$

This concludes the proof.

∎

###  3.2. Density results

A natural question after encountering the constructions of Theorem 3.6.2 and
3.6.3 is “how many arithmetic box spaces of $BS(1,m)$ have $D_{1/2}$”? In the
following paragraphs, we give a partial answer to this question.

Let $(N_{k})_{k}\subset\mathbb{N}$ be such that $N_{k}\mid N_{k+1}$ for every
$k>0$, and denote by $P_{k}$ the set of prime factors of $N_{k}$. Moreover, we
define the set of prime factors of the sequence $(N_{k})_{k}$ by

(15) $P:=\bigcup_{k=1}^{+\infty}P_{k}.$

Before stating our main result from this section, we need to introduce some
definitions about the density of prime numbers. We follow Powell [Powell 1980]
for the terminology.

###### Definition 3.7.

Let $P\subset\mathcal{P}$ be a subset of the prime numbers. The natural
primitive density of $P$ is (if the limit exists)

$d^{\prime}(P):=\lim_{N\to\infty}\frac{\left|\\{p\leq N\mid p\in
P\\}\right|}{\left|\\{p\leq N\mid p\in\mathcal{P}\\}\right|}.$

The analytic primitive density of $P$ is (if the limit exists)

$D^{\prime}(P)=\lim_{s\to 1^{+}}\frac{\sum_{p\in
P}\frac{1}{p^{s}}}{\sum_{p\in\mathcal{P}}\frac{1}{p^{s}}}.$

If $P$ is finite then $d^{\prime}(P)=D^{\prime}(P)=0$. Suppose now that
$D^{\prime}(P)>0$. In this case, we see that $\sum_{p\in
P}\frac{1}{p}=+\infty$, otherwise $D^{\prime}(P)$ would be equal to $0$.
Observe that

$\prod_{p\in P}\left(1-\frac{1}{p}\right)=0\iff\sum_{p\in
P}\ln\left(1-\frac{1}{p}\right)=-\infty.$

But using that $\ln(1+x)\leq x$ for $x>-1$

$\sum_{p\in P}\ln\left(1-\frac{1}{p}\right)\leq-\sum_{p\in
P}\frac{1}{p}=-\infty.$

Therefore, we obtain

(16) $\prod_{p\in P}\left(1-\frac{1}{p}\right)=0$

if $D^{\prime}(P)>0$.

###### Theorem 3.8.

Let $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ be an arithmetic box space, and let
$P$ be the set of prime factors of the sequence $(N_{k})_{k}$.

  1. (1)

If $|P|<+\infty$, then $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ has $D_{1/2}$ ;

  2. (2)

If $D^{\prime}(P)>0$, then $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ does not have
$D_{\alpha}$ for every $\alpha>0$.

###### Proof.

In view of Proposition 3.5, we must study the asymptotics of the quotient
$\frac{\operatorname{ord}_{m}(N_{k})}{N_{k}}$.

  1. (1)

By Lemma 2.8, there exists a constant $C(m,P)$ such that
$\frac{\operatorname{ord}_{m}(N_{k})}{N_{k}}\geq C(m,P)$, i.e.
$\operatorname{ord}_{m}(N_{k})=\Omega(N_{k})$. Proposition 3.5 applies to show
that $\square_{(G_{m}(N_{k}))_{k}}G_{m}$ has $D_{1/2}$.

  2. (2)

Assume now $D^{\prime}(P)>0$, pick $N=p_{1}^{\beta_{1}}...p_{k}^{\beta_{k}}$
with $p_{1},...,p_{k}\in P$. We have
$\operatorname{ord}_{m}(N)\leq\varphi(N)=|(\mathbb{Z}/N\mathbb{Z})^{\times}|$,
where $\varphi$ denotes Euler’s totient function. Then

$\frac{\operatorname{ord}_{m}(N)}{N}\leq\prod_{i=1}^{k}\frac{\varphi(p_{i}^{\beta_{i}})}{p_{i}^{\beta_{i}}}=\prod_{i=1}^{k}\left(1-\frac{1}{p_{i}}\right).$

In view of Eq. (16) we then get $\operatorname{ord}_{m}(N_{k})=o(N_{k})$,
which proves the second part of the theorem thanks to Proposition 3.5.

∎

We state here a few open questions related to the previous theorem.

  * •

If we assume that $d^{\prime}(P)>0$, does the associated arithmetic box space
have $D_{\alpha}$ for some $\alpha\in]0,1/2[$ or not?

  * •

What happens in the case $|P|=+\infty$ and $D^{\prime}(P)=0$?

  * •

Given $\alpha\in]0,1/2]$, can we create an arithmetic box space with exactly
$D_{\alpha}$?

##  4\. CSP for $BS(1,m)$

For a subring $A$ of $\mathbb{Q}$, we define three subgroups of $GL_{n}(A)$:

  * •

$T_{n}(A)$, the subgroup of upper triangular matrices;

  * •

$U_{n}(A)$, the unipotent subgroup, i.e. the subgroup of $T_{n}(A)$ consisting
of matrices with 1’s down the diagonal;

  * •

$D_{n}(A)$, the subgroup of diagonal matrices.

The map $\Delta:T_{n}(A)\rightarrow D_{n}(A)$ taking a matrix to its diagonal,
is a surjective group homomorphism with kernel $U_{n}(A)$.

We will also need the set $N_{n}(A)$ of upper triangular nilpotent matrices,
i.e. upper triangular matrices with 0’s down the diagonal. Note that
$T_{n}(A)={\bf 1}_{n}+N_{n}(A)$.

###  4.1. Representations of $A$ into $U_{n}(A)$

We recall Chernikov’s theorem (see Theorem 4.10 in [Warfield 1976]): if $N$ is
a torsion-free nilpotent group, for every $k\geq 1$ the map $N\rightarrow
N:x\mapsto x^{k}$ is injective.

The first lemma discusses one-parameter subgroups in $U_{n}(A)$.

###### Lemma 4.1.

For every $g\in U_{n}(A)$, there exists a unique homomorphism
$\alpha:A\rightarrow U_{n}(A)$ such that $\alpha(1)=g$.

###### Proof.

The existence follows from Cor. 10.25 in [Warfield 1976]: if $g={\bf
1}_{n}+X$, with $X\in N_{n}(A)$, then for $r\in A$:

$\alpha(r)=({\bf 1}_{n}+X)^{r}=\sum_{k=0}^{\infty}\binom{r}{k}X^{k}={\bf
1}_{n}+rX+\frac{r(r-1)}{2}X^{2}+...$

(note that the sum is finite as $X^{n}=0$). For the uniqueness, let $\beta$ be
another homomorphism with $\beta(1)=g$; for $r\in A$, write $r=\frac{a}{b}$
with $a,b\in\mathbb{Z},b>0$ and $a,b$ coprime. Then

$\alpha(r)^{b}=\alpha(a)=g^{a}=\beta(a)=\beta(r)^{b},$

so $\alpha(r)=\beta(r)$ by Chernikov’s theorem. ∎

###### Definition 4.2.

For a subgroup $H$ of $U_{n}(A)$, the isolator of $H$ is:

$I(H)=\\{g\in U_{n}(A):g^{k}\in H\,\mbox{for some}\,k\geq 1\\}.$

By Theorem 3.25 in [Warfield 1976], $I(H)$ is a subgroup of $U_{n}(A)$.
Clearly $H\subset I(H)$.

###### Lemma 4.3.

For $m\geq 2$, let $\alpha:\mathbb{Z}[1/m]\rightarrow U_{n}(\mathbb{Z}[1/m])$
be an injective homomorphism. Then $I(\alpha(\mathbb{Z}[1/m]))$ is abelian and
the exponent of the group $I(\alpha(\mathbb{Z}[1/m]))/\alpha(\mathbb{Z}[1/m])$
is finite.

###### Proof.

The proof is in three steps.

  1. (1)

By Lemma 4.1, there exists a unique homomorphism
$\tilde{\alpha}:\mathbb{Q}\rightarrow U_{n}(\mathbb{Q})$ that extends
$\alpha$. We show that
$I(\alpha(\mathbb{Z}[1/m])\subset\tilde{\alpha}(\mathbb{Q})$, from which the
first statement will follow. For $g\in I(\alpha(\mathbb{Z}[1/m]))$, there
exists $k\geq 1$ and $r\in\mathbb{Z}[1/m]$ such that $g^{k}=\alpha(r)$. Then

$\tilde{\alpha}\left(\frac{r}{k}\right)^{k}=\tilde{\alpha}(r)=\alpha(r)=g^{k},$

hence $g=\tilde{\alpha}(\frac{r}{k})$ by Chernikov’s theorem.

  2. (2)

Let us show that there exists some non-zero $x_{0}\in\mathbb{Z}[1/m]$ such
that, for $g\in I(\alpha(\mathbb{Z}[1/m])),k\geq 1,r\in\mathbb{Z}[1/m]$:

$g^{k}=\alpha(r)\Longrightarrow\frac{r}{k}x_{0}\in\mathbb{Z}[1/m].$

Write $\alpha(1)={\bf 1}_{n}+X$, with $X\in N_{n}(\mathbb{Z}[1/m])$, as in
Lemma 4.1. Note that $X\neq 0$ as $\alpha$ is injective. Then

$g=\tilde{\alpha}\left(\frac{r}{k}\right)=({\bf 1}_{n}+X)^{\frac{r}{k}}={\bf
1}_{n}+\frac{r}{k}X+...$

For $1\leq i<n$ and an upper triangular matrix $Y$ of size $n\times n$, we
denote by $Y_{(i)}$ the $i$-th parallel to the diagonal (moving upwards from
the diagonal). Let $i$ be the smallest index such that $X_{(i)}\neq 0$. Since
$(X^{k})_{(i)}=0$ for $k\geq 2$, we have $g_{(i)}=\frac{r}{k}X_{(i)}$. Let
$x_{0}$ be any non-zero coefficient of $X_{(i)}$. Since $g\in
U_{n}(\mathbb{Z}[1/m])$, we have $\frac{r}{k}x_{0}\in\mathbb{Z}[1/m]$ as
desired.

  3. (3)

Let $\pi(m)$ be the set of primes dividing $m$. An integer is a
$\pi(m)$-number if all its prime divisors are in $\pi(m)$. Write
$x_{0}=\frac{b}{t}$, with $t$ a $\pi(m)$-number, and $b\in\mathbb{Z}$ is
coprime to $t$. For $g\in I(\alpha(\mathbb{Z}[1/m]))$, with $g^{k}=\alpha(r)$
as above, write $\frac{r}{k}=\frac{a}{s\ell}$, where $a,s,\ell$ are pairwise
coprime, $s$ is a $\pi(m)$-number and $\ell$ is coprime with $m$. By the
previous step $\frac{r}{k}x_{0}=\frac{ab}{st\ell}\in\mathbb{Z}[1/m]$. Since
$a$ and $\ell$ are coprime, this may happen only if $\ell$ divides $b$.
Finally

$g^{b}=\tilde{\alpha}\left(\frac{r}{k}\right)^{b}=\tilde{\alpha}\left(\frac{br}{k}\right)=\tilde{\alpha}\left(\frac{ab}{s\ell}\right).$

But $\frac{ab}{s\ell}\in\mathbb{Z}[1/m]$ as $\ell$ divides $b$. This implies
that $g^{b}\in\alpha(\mathbb{Z}[1/m])$, hence the exponent of
$I(\alpha(\mathbb{Z}[1/m]))/\alpha(\mathbb{Z}[1/m])$ divides $b$.

∎

###  4.2. Special representations of $BS(1,m)$

For $m\geq 2$, set $G_{m}=BS(1,m)=\mathbb{Z}[1/m]\rtimes\mathbb{Z}$. We will
write $A_{m}$ for $\mathbb{Z}[1/m]$ when viewed as a normal subgroup of
$G_{m}$.

###### Definition 4.4.

A  special representation of $G_{m}$ is an injective homomorphism
$\rho:G_{m}\rightarrow T_{n}(\mathbb{Z}[1/m])$ such that $\rho(A_{m})\subset
U_{n}(\mathbb{Z}[1/m])$.

We note that the standard embedding $G_{m}\rightarrow T_{2}(\mathbb{Z}[1/m])$,
is a special representation.

###### Lemma 4.5.

  1. (1)

If $\rho$ is a special representation, then
$\rho^{-1}(U_{n}(\mathbb{Z}[1/m]))=A_{m}$.

  2. (2)

If $m$ is even, then any injective homomorphism $\rho:G_{m}\rightarrow
T_{n}(\mathbb{Z}[1/m])$ is a special representation.

###### Proof.

We work with the presentation $G_{m}=\langle a,t|tat^{-1}=a^{m}\rangle$,
observing that the normal subgroup $A_{m}$ coincides with the normal subgroup
generated by $a$.

  1. (1)

Suppose by contradiction that $A_{m}$ is strictly contained in
$\rho^{-1}(U_{n}(\mathbb{Z}[1/m]))$. Then there exists $k>0$ such that
$\rho(t^{k})\in U_{n}(\mathbb{Z}[1/m])$. Consider the subgroup $H$ of $G_{m}$
generated by $A_{m}\cup\\{t^{k}\\}$, so that $\rho(H)\subset
U_{n}(\mathbb{Z}[1/m])$. Since $\rho$ is injective, we see that $H$ is
nilpotent. As $H$ also has finite index in $G_{m}$, we deduce that $G_{m}$ is
virtually nilpotent, which is a contradiction.

  2. (2)

Suppose that $m$ is even and $\rho:G_{m}\rightarrow T_{n}(\mathbb{Z}[1/m])$ is
an injective homomorphism. It is enough to see that $\rho(a)$ belongs to
$U_{n}(\mathbb{Z}[1/m])=\ker(\Delta)$. But we have

$\rho(a^{m-1})=\rho([a,t])\in[T_{n}(\mathbb{Z}[1/m]),T_{n}(\mathbb{Z}[1/m])]\subset\ker(\Delta).$

Now the image of $\Delta$, namely
$D_{n}(\mathbb{Z}[1/m])\simeq(\mathbb{Z}[1/m]^{\times})^{n}$, contains only
2-torsion; since $\Delta(\rho(a))^{m-1}={\bf 1}_{n}$ and $m-1$ is odd, we have
$\Delta(\rho(a))={\bf 1}_{n}$ as desired.

∎

We now head towards CSP for special representations of $BS(1,m)$. The next
lemma is proved exactly as Lemma 4 in Formanek [Formanek 1976], using the same
ingredient, namely a number-theoretical result by Chevalley [Chevalley 1951].

###### Lemma 4.6.

Let $R$ be a subring of a number field with $R^{\times}$ finitely generated,
let $G$ be a subgroup of $D_{n}(R)$. There exists a function
$\varphi:\mathbb{N}\rightarrow\mathbb{N}$ such that, if $g\in G$ satisfies
$g\equiv{\bf 1_{n}}\mod\varphi(r)$, then $g$ is an $r$-th power in $G$.
$\square$

In Theorem 5 of [Formanek 1976], Formanek proved that, if $R$ is the ring of
integers of a number field and $G$ is a subgroup of $T_{n}(R)$, then $G$ has
CSP. The proof of the next result is inspired by Formanek’s proof.

###### Proposition 4.7.

Let $\rho:G_{m}\rightarrow T_{n}(\mathbb{Z}[1/m])$ be a special representation
of $BS(1,m)$. Then $\rho(G_{m})$ has CSP.

###### Proof.

Let $N\triangleleft G_{m}$ be a normal subgroup of finite index $r$. Denote by
$G_{m}^{r}$ the subgroup generated by $r$-th powers in $G_{m}$. Then
$G_{m}^{r}\subset N$ by Lagrange’s theorem. Let $b$ be the exponent of
$I(\rho(A_{m}))/\rho(A_{m})$ (which is finite by Lemma 4.3), and let $e$ be
the exponent of the finite group
$T_{n}(\mathbb{Z}[1/m]/(br)^{2}\mathbb{Z}[1/m])$. Define then

$M=(br)^{2}\varphi(re),$

where $\varphi$ comes from Lemma 4.6 applied to $\Delta(\rho(G_{m}))$. We will
show that if $\rho(g)\equiv{\bf 1}_{n}\mod M$, then $g\in G_{n}^{r}$, so that
$\rho(N)$ contains the congruence subgroup $\rho(G_{m})(M)$.

If $\rho(g)\equiv{\bf 1}_{n}\mod M$, in particular $\Delta(\rho(g))\equiv{\bf
1}_{n}\mod\varphi(er)$. By Lemma 4.6, the matrix $\Delta(\rho(g))$ is an
$(er)$-th power in $\Delta(\rho(G_{m}))$, i.e. there exists $z\in G_{m}$ such
that

$\Delta(\rho(g))=\Delta(\rho(z^{er}))=\Delta(\rho((z^{e})^{r})).$

But $\rho(z^{e})\equiv{\bf 1}_{n}\mod(br)^{2}$, by definition of $e$, so also
$\rho(z^{er})\equiv{\bf 1}_{n}\mod(br)^{2}$. On the other hand by definition
of $M$ we have $\rho(g)\equiv{\bf 1}_{n}\mod(br)^{2}$, so
$\rho(g^{-1}z^{er})\equiv{\bf 1}_{n}\mod(br)^{2}$. Since
$g^{-1}z^{er}\in\ker(\Delta)=U_{n}(\mathbb{Z}[1/m])$, Lemma 1 of [Formanek
1976]222Formanek states this lemma for the ring of integers of a number field,
but the proof only uses that the ring is commutative with characteristic 0.
applies to guarantee that $g^{-1}z^{er}$ is a $(br)$-th power in
$U_{n}(\mathbb{Z}[1/m])$, so we find $h\in U_{n}(\mathbb{Z}[1/m])$ such that

$\rho(g^{-1}z^{er})=h^{br}.$

Hence $g^{-1}z^{er}\in\rho^{-1}(U_{n}(\mathbb{Z}[1/m]))=A_{m}$ (the equality
follows from the first part of Lemma 4.5). This means that $h\in
I(\rho(A_{m}))$, so by definition of $b$ we have $h^{b}\in\rho(A_{m})$, say
$h^{b}=\rho(y)$ for $y\in A_{m}$. Then $h^{br}=\rho(y^{r})$ and
$\rho(g^{-1}z^{er})=\rho(y^{r})$. As $\rho$ is injective $g^{-1}z^{er}=y^{r}$,
i.e $g=(z^{e})^{r}y^{-r}$, so $g\in G_{m}^{r}$. ∎

###  4.3. From special representations to all injective representations in
$T_{n}(\mathbb{Z}[1/m])$

In this section we show that for any injective representation $\rho$ of
$G_{m}$ into $T_{n}(\mathbb{Z}[1/m])$, the group $\rho(G_{m})$ has CSP. The
main idea is to pass to subrepresentations which are special and extract
information about $\rho$.

###### Lemma 4.8.

Let $G$ be a subgroup of $GL_{n}(\mathbb{Z}[1/m])$.

  1. (1)

Let $H$ be a finite index subgroup of $G$, where $H$ has CSP. If $H$ contains
a congruence subgroup of $G$, then $G$ also has CSP.

  2. (2)

For $1\leq i\leq k$, let $H_{i}$ be a finite index subgroup of $G$. Suppose
$G=\cup_{i=1}^{k}H_{i}$, then if all $H_{i}$ have CSP, $G$ also has CSP.

###### Proof.

  1. (1)

Let $N$ be a finite index subgroup of $G$ , then $N\cap H$ is a finite index
subgroup of $H$ and thus contains some congruence subgroup of the form $H(M)$.
By assumption there is some $\tilde{M}$ such that $G(\tilde{M})\subset H$,
hence we have that

$G(M\tilde{M})\subset G(M)\cap G(\tilde{M})\subset H(M)\subset N\cap H\subset
N.$

  2. (2)

Suppose by contradiction that $G$ does not have CSP, then there is some finite
index subgroup $N$ which does not contain any congruence subgroup. However
each $N\cap H_{i}$ contains a congruence subgroup of the form $H(M_{i})$. Set
$\overline{M}=\Pi_{i}M_{i}$, then there is some $g\in
G(\overline{M})\backslash N$. Note that this $g$ is necessarily in one of the
$H_{i}$’s. So it is necessarily in $H(M_{i})$ since $H(\overline{M})\subset
H(M_{i})$. However it is not in $N\cap H_{i}$: a contradiction.

∎

###### Lemma 4.9.

Let $\rho:G_{m}\rightarrow T_{n}(\mathbb{Z}[1/m])$ be a monomorphism. Suppose
$\Delta(\rho(a))\neq{\bf 1}_{n}$ and the diagonal of $\rho(t)$ is strictly
positive, then $\rho(G_{m})$ has CSP.

###### Proof.

Note that the subgroup $B=\langle a^{2},t\rangle$ is also a Baumslag-Solitar
group isomorphic to $BS(1,m)$. The restriction $\rho|_{B}$ is a special
representation of $BS(1,m)$; indeed by the proof of Lemma 4.5, we have
$\Delta(\rho(a))^{m-1}={\bf 1}_{n}$, so $\Delta(\rho(a))$ has finite order in
$D_{n}(\mathbb{Z}[\frac{1}{m}])$; but the torsion subgroup of
$D_{m}(\mathbb{Z}[\frac{1}{m}])$ is $\\{\pm 1\\}^{n}$, so
$\Delta(\rho(a^{2}))={\bf 1}_{n}$. By Proposition 4.7, the group $\rho(B)$ has
the CSP.

We show there is some congruence subgroup of $\rho(G_{m})$ that is fully
contained inside $\rho(B)$. Since $\Delta(\rho(a))\neq{\bf 1}_{n}$, there is
some $-1$ on the diagonal, suppose it appears at the $i^{\text{th}}$ position.
Let $s$ be the element of $(\mathbb{Z}[1/m])^{\times}$ in the $i^{\text{th}}$
position in $t$. By Lemma 2.10, we find $M>1$, coprime with $m$, such that $s$
has odd order in $(\mathbb{Z}[1/m]/M\mathbb{Z}[1/m])^{\times}$.

We note that if $\omega$ is a word with letters in $\\{a,t\\}$ representing an
element of $G_{m}\setminus B$, then it contains an odd number of $a$’s. Hence,
the element on the $i^{\text{th}}$ position on the diagonal of $\rho(\omega)$
is of the form $-s^{r}$ for some $r\in\mathbb{Z}$. If we suppose that
$\rho(\omega)\in\rho(G_{m})(M)$, then $s^{r}=-1\mod M$. This is impossible
since $s$ has odd order in
$(\mathbb{Z}[\frac{1}{m}]/M\mathbb{Z}[1/m])^{\times}$ and thus cannot have a
power equal to $-1$. So $\rho(\omega)$ cannot be in the corresponding
congruence subgroup, i.e. $\rho(G_{m})(M)\subset\rho(B)$. By Lemma 4.8, the
result follows. ∎

We are ready to complement Proposition 4.7.

###### Proposition 4.10.

For any injective homomorphism $\rho:G_{m}\rightarrow T_{n}(\mathbb{Z}[1/m])$,
the group $\rho(G_{m})$ has CSP.

###### Proof.

Since we already know the result for $m$ even (by Proposition 4.7 and Lemma
4.5), we may and will assume that $m$ is odd.

The group $G_{m}$ has three subgroups $B_{1}=\langle a^{2},t\rangle$,
$B_{2}=\langle a^{2},at\rangle$ and $B_{3}=\langle a,t^{2}\rangle$, which are
all isomorphic to $BS(1,m)$ or $BS(1,m^{2})$. We show that they cover $G_{m}$.
Any element $\omega\in G_{m}$ can be described by a word of the form
$t^{-k}a^{r}t^{\ell}$, with $k,\ell\in\mathbb{N}$ and $r\in\mathbb{Z}$.
Replacing $\omega$ by $\omega^{-1}$ if necessary, we may assume $k\leq\ell$.

  * •

If $r$ is even, then $\omega\in B_{1}$;

  * •

If $k$ and $l$ are both even or odd, then $\omega\in B_{3}$; indeed
$\omega=t^{-2k}(t^{k}a^{r}t^{-k})t^{k+\ell}=t^{-2k}a^{rm^{k}}t^{k+\ell}\in
B_{3}$ as $k+\ell$ is even.

  * •

If $r$ is odd and $k,\ell$ do not have the same parity, then $\omega\in
B_{2}$. Indeed

$t^{-k}a^{r}t^{\ell}=(t^{-1}a^{-1})^{k}a^{\sum_{i=0}^{k-1}m^{i}}a^{r}a^{-\sum_{i=0}^{\ell-1}m^{i}}(at)^{l}=(at)^{-k}a^{r-\sum_{i=k}^{\ell-1}m^{i}}(at)^{\ell}.$

Since $r,\ell-k$ and $m$ are odd, the exponent of $a$ in the latter expression
is even, so $\omega\in B_{2}$.

Note that if $\Delta(\rho(a))={\bf 1}_{n}$, then it is a special
representation and Proposition 4.7 applies. If it is not then the restrictions
of $\rho$ to $B_{1}$ and $B_{2}$ are special representations, and the
restriction of $\rho$ to $B_{3}$ satisfies the assumptions of Lemma 4.9. Hence
the $\rho(B_{i})$’s all have CSP. Now simply applying Lemma 4.8 finishes the
argument. ∎

###  4.4. An alternative approach to CSP for $BS(1,p^{\ell})$ ($p$ prime)

We use the language of $\mathbb{Q}$-algebraic groups. Denoting by
$\mathbb{G}_{a}$ (resp. $\mathbb{G}_{m}$) the additive group (resp. the
multiplicative group), we set
$\mathbb{G}=\mathbb{G}_{a}\rtimes\mathbb{G}_{m}$, the affine group viewed as a
subgroup of $GL_{2}$ via the standard embedding.

Let $p_{1},...,p_{r}$ be distinct primes, and let
$S=\\{p_{1},...,p_{r},\infty\\}$ be viewed as a set of places of $\mathbb{Q}$.
Then the ring $\mathcal{O}_{S}$ of $S$-integers in $\mathbb{Q}$ is precisely

$\mathcal{O}_{S}=\mathbb{Z}[1/p_{1},...,1/p_{r}],$

so that $\mathcal{O}_{S}^{\times}=\\{\pm
p_{1}^{k_{1}}...p_{r}^{k_{r}}:k_{1},...,k_{r}\in\mathbb{Z}\\}$ and

$\mathbb{G}(\mathcal{O}_{S})=\left\\{\left(\begin{array}[]{cc}\pm
p_{1}^{k_{1}}...p_{r}^{k_{r}}&a\\\
0&1\end{array}\right):k_{1},...k_{r}\in\mathbb{Z},\;a\in\mathbb{Z}[1/p_{1},...,1/p_{r}]\right\\}.$

It is known that $\mathbb{G}(\mathcal{O}_{S})$ satisfies CSP: see formula
$(\ast\ast)$ on page 108 of [Raghunathan 1976].

Note that taking $r=1$, i.e $S=\\{p.\infty\\}$ and
$\mathcal{O}_{S}=\mathbb{Z}[\frac{1}{p}]$, makes $BS(1,p^{\ell})$ appear as a
finite index subgroup in $\mathbb{G}(\mathcal{O}_{S})$, namely
$[\mathbb{G}(\mathcal{O}_{S}):BS(1,p^{\ell})]=2\ell$.

###### Proposition 4.11.

Let $\mathbb{H}\subset GL_{n}$ be a $\mathbb{Q}$-subgroup, and let
$\rho:\mathbb{G}\rightarrow\mathbb{H}$ be a $\mathbb{Q}$-isomorphism. Then
$\rho(BS(1,p^{\ell}))$ satisfies the CSP.

###### Proof.

Let $K$ be a finite index subgroup in $\rho(BS(1,p^{\ell}))$. Since
$\mathbb{G}(\mathcal{O}_{S})$ satisfies the CSP, we find $N>0$ such that
$\mathbb{G}(\mathcal{O}_{S})(N)\subset\rho^{-1}(K)$. By Lemma 3.1.1.(ii) in
Chapter I of [Margulis 1991], the subgroup
$\rho(\mathbb{G}(\mathcal{O}_{S})(N))$ is an $S$-congruence subgroup in
$\mathbb{H}(\mathcal{O}_{S})$, i.e. it contains
$\mathbb{H}(\mathcal{O}_{S})(M)$ for some $M>1$. Then
$(\rho(BS(1,p^{\ell})))(M)\subset\mathbb{H}(\mathcal{O}_{S})(M)\subset\rho(\mathbb{G}(\mathcal{O}_{S})(N))\subset
K$. ∎

## References

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is the descendant of a sibling on the right of $\nodi'$ and the result follows;

* or $\noda$ it is the parent of $\nodj$, treated symmetrically.

We define:

\begin{equation*}
p \defined \mu(w[\maxe{\nodi'}{\forest}{+}1{:}\mine{\nodj'}{\forest}{-}1]).
\end{equation*}

It is always clear that $\mu(w[\maxe{\nodi}{\forest}{+}1{:}{\mine{\nodj}{\forest}{-1}}])  = e_1 p e_2$.
The other equations can be checked by a (rather long) case disjunction.
Let us detail one case.

Assume that $\nodi$ is a middle child of $\basis{\nodi}{\forest} = \basis{\nodj}{\forest}$,
i.e. $\nodi = \mimi{\nodi}{\forest}$.
Then $m_1 = n_1  = \mu(\epsilon)$ and $e_1 = e$.
Let $\nodi, \nodi', \nodi'', \nodi'''$ be the successive right siblings below $\basis{\nodi}{\forest}$
and assume that $\nodj$ is a descendant of $\nodi'''$.
Then $p = em_2 e_2$ or $p = em_2$ (depending on whether $\nodj'$
is a middle child or not).
Then $m_1 e_1 p e_2 n_2 = e m_2 e_2 n_2 = e$; $em_1e_1p e_2= em_2e_2$;
and  $e_1p e_2 n_2 e=  e m_2 e_2 n_2 e = e= e_1n_1e$.

Finally we get $\bitype{\nodi, \nodj} = em_1 e_1 \cro{\uu} e_1 p e_2 \cro{\ud} e_2 n_2 e$.
To conclude the proof, we use the $K$ of Definition <ref>
whose conditions are met. Switching $\tau_1$ and $\tau_2$ gives the same $K$,
using the second case of Definition <ref>.

§.§.§ Proof of Lemma <ref>

Let $\nodi < \nodj$ and $(\nodi, \nodj) \in I(F)$.
Let $\{i_1, \dots, i_{\ell}\} \defined \fr{\nodi}{\forest}$
and $\{j_1, \dots, j_{\ell'}\} \defined \fr{\nodj}{\forest}$.
Hence $i_{\ell} = \maxe{\nodi}{\forest} < \mine{\nodi}{\forest} = j_1$.

\begin{equation*}
\begin{aligned}
\pro{\nodi, \nodj} &= \sum_{\substack{1 \le k \le \ell \\ 1 \le k' \le \ell'}} \pro{i_k,j_{k'}}
\end{aligned}
\end{equation*}

Let $q \defined \mu([1{:}i_1{-}1])$,  $q' \defined \mu(w[i_{\ell}{+}1{:}j_{1}{-}1]) $
and $q'' \defined \mu(w[j_{\ell'}{+}1{:}|w|])$.
Let $u \defined w[\fr{\nodi}{\forest}]$ and $u' \defined w[\fr{\nodj}{\forest}]$
be the words described by their frontiers.

By definition, $\bitype{\nodi, \nodj} = q\cro{u} q' \cro{u'} q''$.

\begin{equation}
\begin{aligned}
&\pro{\bitype{\nodi, \nodj}} = \sum_{\substack{1 \le  k \le \ell  \\ 1 \le k' \le \ell'}} P(k,k') \text{ where}\\
&P(k,k') \defined q \mu(u[1{:}{k}{-}1])\cro{u[k]}
\mu(u[{k}{+}1{:}{\ell}] )
q' \mu(u'[{1}{:}{k'}{-}1] )\cro{u'[{k'}]} \mu(u'[{k'}{+}1{:}\ell']) q''.
\end{aligned}
\label{eq:Pkk}
\end{equation}

Hence to show Lemma  <ref>,
is is enough to show $\forall k,k'$ $\pro{i_k, i_k'} = P(k,k')$. And:

\begin{equation}
\begin{aligned}
\pro{i_k, i_k'}&=  \pro{\mu(w[1{:}i_{k}{-}1])\cro{w[i_k]} \mu(w[i_{k}{+}1{:}j_{k'}{-}1] )\cro{w[j_{k'}]} \mu(w[j_{k'}{+}1{:}|w|])}\\
&=  \pro{\mu(w[1{:}i_{k}{-}1])\cro{u[k]} \mu(w[i_{k}{+}1{:}j_{k'}{-}1] )\cro{u'[{k'}]} \mu(w[j_{k'}{+}1{:}|w|])}\\
\end{aligned}
\label{eq:prodkk}
\end{equation}

Comparing Equations <ref> and <ref>, we see that we only need to show that $\forall k,k'$:

\begin{equation*}
\left\{
\begin{array}{l}
\mu(u[1{:}k{-}1]) = \mu(w[i_1{:}i_{k}{-}1])\\
\mu(u[k{+}1{:}\ell]) = \mu(w[i_{k}{+}1{:}i_{\ell}])\\
\mu(u'[1{:}k'{-}1]) = \mu(w[j_{1}{:}j_{k'}{-}1])\\
\mu(u'[k'{+}1{:}\ell']) = \mu(w[j_{k'}{+}1{:}j_{\ell'}])\\
\end{array}
\right.
\end{equation*}

We only treat the first case, since the others
are similar. It follows by applying the claim
below to the node $\nodi \in \facto{\mu}{w[i_1{:}i_{\ell}]}$
seen as a factorization.

Let $\forest \in \facto{\mu}{x}$,
$y \defined x[\fr{\forest}{\forest}]$ and
$\{1 = i_1 < \cdots < i_{\ell} = |x|\} \defined \fr{\forest}{\forest}$.

Then $\forall 1 \le k \le \ell$, $\mu(x[1{:}i_k{-1}]) = \mu(y[1{:}k{-}1])$.

The result is shown by induction on the factorization.
The only interesting case is when
$\forest = (\forest_1) \dots (\forest_n)$ starts with an idempotent node.
Let $x = x_1 \cdots x_n$ be such
that $\forest_i$ is a factorization of $x_i$. Then $ e \defined \mu(x_i)$ is an idempotent.

Consider some $1 \le k \le \ell$.
By definition of the dependency and Lemma <ref>,
there exists $1 \le p \le k$ such that $i_{p} = |x_1|$ and $i_{p+1} = |x| - |x_n| +1$.

\begin{equation*}
y = x_1[i_1] \cdots x_1[i_p] x_n[i_{p+1} {-} |x|{+}|x_n|] \cdots x_n[i_{\ell} {-} |x|{+}|x_n|]
\end{equation*}

Let $1 \le k \le \ell$, we assume that $p+1 \le k$
(the case $k \le p$ is easier).

\begin{equation*}
y[1{:}k{-}1] = x_1[i_1] \cdots x_n[i_p] x_n[i_{p+1} {-} |x|{+}|x_n|] \cdots x_n[i_{k{-}1} {-} |x|{+}|x_n|].
\end{equation*}

By Lemma <ref>:

\begin{equation*}
\mu(x_1[i_1] \cdots x_1[i_{p-1}]x_1[i_p]) = \mu(x_1) = e
\end{equation*}

And by induction hypothesis:

\begin{equation*}
\mu(x_n[i_{p+1}{-}|x|{+}|x_n|] \cdots x_n[i_{k-1} {-} |x|{+}|x_n|]) = \mu(x_n[i_{p+1}{-}|x|{+}|x_n|{:}i_{k} {-} 1 {-} |x|{+}|x_n|])
\end{equation*}

Finally since $e = \mu(x_2) = \cdots = \mu(x_{n-1}) $ is idempotent, it follows that:

\begin{equation*}
\begin{aligned}
\mu(x[1{:}i_k{-}1] ) &= e \mu(x_n[i_{p+1}{-}|x|{+}|x_n|{:}[i_{k} {-} 1 {-} |x|{+}|x_n|])\\
&= \mu(x_1[i_1] \cdots x_1[i_{p-1}]x_1[i_p]) \mu(x_n[i_{p+1}{-}|x|{+}|x_n|] \cdots x_n[i_{k-1} {-} |x|{+}|x_n|])\\
\end{aligned}
\end{equation*}

§ PROOF OF THEOREM <REF> - CHARACTERIZATION

We only show that $\ref{it:trt} \Rightarrow \ref{it:det}$.
Assume that the $1$-blind bimachine $\trans$ is symmetrical.

By Proposition <ref>, one can
build a two-way transducer (with non-unary output) that
given $w \in A^+$, computes some  factorization $\forest \in \facto{\mu}{w}$ of height at most $\mh$,
described as a word, that is $\forest \in \apar{A}^+$. We denote this function $f_{\ltm}$.

Using lemmas <ref>, <ref>, <ref>,
the functions $f_D, f_L$ and $f_I$ are computable by a $1$-blind transducer.
We want to compute their sum, in the sense of Claim <ref>.
It follows from this claim that $f_D + f_L + f_I$ can be computed
by a $1$-blind bimachine, or equivalently a $1$-blind transducer.
By Lemma <ref> and the definitions
of $f_D, f_L$ and $f_I$, it follows that:

\begin{equation*}
(f_D + f_L + f_I) \circ f_{\ltm} = f.
\end{equation*}

It follows from [11]
that (even for non-unary alphabets) $1$-blind transducers are effectively closed under
composition with two-way transducers, hence $(f_D + f_L + f_I) \circ f_{\ltm}$
can be computed by a $1$-blind transducer.