<EMAIL_ADDRESS><EMAIL_ADDRESS>Resources & Activities 00 0 2022

# Per aspera ad astra simul: Through difficulties to the stars together

David Jones  Instituto de Astrofísica de Canarias, E-38205 La Laguna, Spain
Petr Kabáth  Astronomical Institute, Czech Academy of Sciences, Fričova 298,
25165, Ondřejov, Czech Republic  Jorge García-Rojas  Instituto de Astrofísica
de Canarias, E-38205 La Laguna, Spain  Departamento de Astrofísica,
Universidad de La Laguna, E-38206 La Laguna, Spain  Josef Hanuš  Charles
University, Faculty of Mathematics and Physics, Institute of Astronomy, V
Holešovičkách 2, CZ-18000, Prague 8, Czech Republic  Marian Jakubík
Astronomical Institute, Slovak Academy of Sciences, 05960 Tatranská Lomnica,
Slovakia  Jan Janík  Department of Theoretical Physics and Astrophysics,
Masaryk University, Kotlářská 2, CZ-611 37 Brno, Czech Republic  Roman Nagy
Faculty of Mathematics, Physics and Informatics, Comenius University,
Bratislava, Slovakia  Juraj Tóth  Faculty of Mathematics, Physics and
Informatics, Comenius University, Bratislava, Slovakia

###### Abstract

In this article, we detail the strategic partnerships “Per Aspera Ad Astra
Simul” and “European Collaborating Astronomers Project: España-Czechia-
Slovakia”. These strategic partnerships were conceived to foment international
collaboration for educational activities (aimed at all levels) as well as to
support the development and growth of early career researchers. The
activities, carried out under the auspices of these strategic partnerships,
demonstrate that Key Action 2 of the Erasmus+ programme can be an extremely
valuable resource for supporting international educational projects, as well
as the great impact that such projects can have on the general public and on
the continued development of early career researchers. We strongly encourage
other educators to make use of the opportunities offered by the Erasmus+
scheme.

###### keywords:

Erasmus+; early career researchers; outreach.

††journal: aej

##  1 Introduction

Many will be familiar with the Erasmus+ programme of the European Commission
through its scheme to support students spending a semester/year abroad during
their undergraduate degrees (generally as part of its Key Action 1). However,
Key Action 2 of the Erasmus+ programme provides funding for multinational
strategic partnerships between institutions with the aim of promoting
educational, professional and personal development. These strategic
partnerships are applied for and managed on the national level with, for
example, the Czech national agency having budgets of approximately 1.5 M€ and
3 M€ in the 2017 and 2020 calls, respectively. “Per Aspera Ad Astra Simul”
(2017-2020, 2017-1-CZ01-KA203-035562) and its (currently active) successor
“European Collaborating Astronomers Project: España-Czechia-Slovakia”
(2020-2023, 2020-1-CZ01-KA203-078200), the subjects of this article, are
excellent examples of how this scheme can support astronomy education and the
professional development of early career researchers (ECRs).

##  2 Background

The Erasmus+ program is designed to support education, training and youth
through international cooperation. In such a competitive environment, the
development of ECRs in astronomy rely heavily on the most modern techniques
(be they observational or theoretical) and international collaboration.
Therefore, it is crucial for young astronomers to acquire as much wide-ranging
expertise and international experience as possible during their studies.
Ultimately, this makes ECRs better scientists with improved career prospects.

Although Spain and the Czech Republic joined the European Southern Observatory
(ESO – Europe’s premier ground-based astronomy consortium) almost
simultaneously111At the time of publication Slovakia is not an ESO member
state. (late 2006/early 2007), the Spanish astronomical community is
objectively larger and more developed [2, 8]. Spain has contributed more staff
to the ESO observatories (indeed, the current Director General is Spanish),
won more industrial ESO contracts, and been awarded more observing time on ESO
facilities. Spain is also home to several major astronomical facilities,
including the World’s largest optical-infrared telescope – the Gran Telescopio
Canarias (GTC).

The Czech and Slovak astronomical communities are smaller, but with a rich
history particularly in the fields of theoretical astrophysics, variable stars
and minor solar system bodies. More recently, both communities have expanded
significantly in the theoretical and observational study of exoplanets.
Several 1–2m class telescopes have been built in Czechia and Slovakia, but the
continental weather impacts significantly upon their usage.

The Erasmus+ strategic partnerships “Per Aspera Ad Astra Simul” and “European
Collaborating Astronomer Projects: España-Czechia-Slovakia” were conceived as
mutually beneficial agreements between institutes in Spain, the Czech Republic
and Slovakia. The significant Spanish expertise in large observing facilities
offers a clear opportunity for learning and growth for ECRs from the Czech and
Slovak partners. Similarly, Spanish ECRs would greatly benefit from extended
research stays at the Czech and Slovak partners, developing long-lasting
collaborations and gaining valuable international experience. Furthermore, the
Erasmus+ partnerships fosters collaboration also among senior staff, ensuring
continued benefit to further generations of ECRs beyond the duration of the
project.

##  3 Partners and their roles

The partners of the aforementioned Erasmus+ strategic partnerships are listed
in Table 1. The lead institute of the project is the Astronomical Institute of
the Academy of Sciences of the Czech Republic (AI ASCR), the largest
astronomical institute in the Czech Republic. The Department of theoretical
physics and astrophysics of Masaryk University and the Astronomical Institute
of Charles University (AI CU, who joined in 2020) are the other Czech partners
in the project and comprise two of the longest established and important
astronomical institutes in the Czech Republic.

In Slovakia, the Astronomical Institute of the Slovak Academy of Sciences (AI
SAS) and Comenius University Bratislava are both partners. Spain is
represented in the partnership by the Instituto de Astrofísica de Canarias
(IAC), the largest astronomical institute in the country and the operators of
the Teide and Roque de los Muchachos Observatories, the latter of which is
home to the GTC. Indeed, GranTeCan (the public company responsible for the
GTC’s design, construction, and continued operation) was an associate partner
from 2017–2020.

As lead institute, AI ASCR has been responsible for the administration and
coordination of the project and the associated funds. AI SAS and Masaryk
University have organised summer schools for ECRs in 2019 and 2020,
respectively. Further summer schools, organised by the IAC and Comenius
University Bratislava are planned for the coming years. All partners have
participated in mobilities of ECRs and more senior researchers (see Section
4), and all have been involved in local outreach and educational activities
(see Section 6). Similarly, senior researchers from all partner institutions
contributed to a book of thirteen detailed reviews, designed to be useful
graduate-level introductions to the topics ranging from meteors and space
debris through to the formation of galaxies and the standard cosmological
model, and which was published in 2020 [5].

Table 1: List of partner institutions Institution | Country | Notes
---|---|---
Astronomical Institute of the Academy of Sciences of the Czech Republic (AI ASCR) | Cz | Lead institute
Instituto de Astrofísica de Canarias (IAC) | Es | Summer school organiser 2022
Comenius University Bratislava | Sk | Summer school organiser 2023
Masaryk University | Cz | Summer school organiser 2020
Astronomical Institute of the Slovak Academy of Sciences (AI SAS) | Sk | Summer school organiser 2019
Astronomical Institute of Charles University (AI CU) | Cz | Partner since 2020
GranTeCan | Es | Associate partner until 2020

##  4 Mobilities

The strategic partnerships offered significant financial support for
mobilities of researchers between the partner institutions. Short-term
($\sim$a few weeks) exchanges of more senior researchers designed to foment
collaborations between the institutions which can then be used to achieve the
objectives of the Erasmus+ programme222For further details of these
objectives, please see What are the objectives of the Erasmus+ programme?, as
well as longer-term (2–6 months) exchanges of ECRs with the aim of aiding
their personal and professional development. At the end of the stay, the
researchers were encouraged to summarise their activities in the form of a
blog post on the project’s webpages.

In total more than 30 international exchanges have already been completed (in
spite of the obvious problems posed by the COVID-19 pandemic striking part way
through the execution of the first strategic partnership), with further
exchanges planned for 2022 and 2023. In many cases, the benefits of these
exchanges have been tangible and obvious [for example, resulting in peer-
reviewed publications in high impact journals; 4, 3, 9], while in other cases
the benefits are more “soft” – leading to long-term collaborations, personal
and professional development, improved career prospects, etc.

###  4.1 Gran Telescopio Canarias

A special case among the ECR exchanges, where the majority were conceived as
collaborations on blue skies astronomy research under the supervision of a
senior researcher at the receiving institution, was the long-term mobility of
one ECR to GranTeCan. During their stay, the PhD student combined continued
research towards their doctoral thesis with support astronomer activities at
the World’s largest optical-infrared telescope – the 10.4-m Gran Telescopio
Canarias (GTC). This offered the student a unique opportunity to obtain in-
depth experience of the tasks involved in the day-to-day operations of a
world-class astronomical observatory – experience which has undoubtedly been
extremely valuable in their career development, with the student going on to
take up highly competitive positions at the European Space Agency (ESA) and
European Southern Observatory (ESO) following the completion of their mobility
and thesis defense.

##  5 Summer schools

In June 2019, the summer school “Observational astrophysics: from proposals to
publication” was organised in Tatranská Lomnica (Slovakia) under the auspices
of the Per Aspera Ad Astra Simul grant and in collaboration with OPTICON. The
school comprised two parts: hands-on projects with archival data and
lectures/activities to teach the participants about observing time proposal
evaluation procedures.

The hands-on projects were undertaken in groups of approximately four students
(with a total of approximately 40 students in attendance) under the
supervision of an experienced tutor. The projects themselves covered topics
ranging from near-Earth asteroids [7] through to binary quasars [1] all based
on publicly available archive data. Upon completion of these projects, the
participating students presented their results in the form of a mini-
conference and later reported their analyses and results in papers published
in the Contributions of the Astronomical Observatory Skalnaté Pleso journal
[6].

A second summer school, “GAIA and TESS: Tools for understanding the Local
Universe”, was organised in 2020, originally to be hosted in Brno (Czech
Republic) but ultimately held remotely due to the COVID-19 pandemic (with 17
students attending via video conference). The school featured talks on the
capabilities and applications of the Gaia and TESS missions by highly
experienced experts, as well as hands-on sessions and research projects making
use of the public data products of these missions [10].

A third school, “Eclipsing Binaries and Asteroseismology: Precise fundamental
stellar parameters in the golden age of time-domain astronomy", will be hosted
on the Spanish island of La Palma in September of 2022. This school will be
run in a hybrid format with 15 in-person students and up to 200 attending via
video conference. A final school is planned to be held in Bratislava
(Slovakia) in 2023.

##  6 Outreach

Thus far, we have discussed the activities of the strategic partnerships
targeted towards the education and professional development of ECRs. However,
the project has always had a significant component geared towards the
education of a younger and/or wider audience, principally in the form of
activities in local schools and public talks.

We stress the importance of early education and thus target our effort to work
together with the local schools and preschool classes. In Ondřejov, activities
were focused towards children of age 5–6 years 333Similar activities for older
children were planned, but have been significantly impacted by the COVID-19
pandemic., preparing for school which begins at the age of 6 in the Czech
Republic. We invite the preschoolers to observatory and to the 2-m telescope
dome, where they can see the telescope. In the dome, we present a short talk
about a selected topic, ranging from the solar system to the life of stars,
which the children will then later build upon in class with their teachers.
After the visit, the children prepare art work or a similar project inspired
by the selected topic, which is then presented in a small gathering with their
parents invited. These activities not only engage the children with their
first experience of space and science, but also help them to develop creative
and presentation skills. In Spain, activities were organised for local 4th
grade students (ages 9–10) which focused on the importance of astronomy in the
Canary islands throughout history, from the ancient aborigines through to the
current day. These included hands-on workshops on the solar system and the
timescales of the Universe, offering many students their first contact with a
professional astronomer. Activities for high school students were also
organized, focusing on different aspects of technological development related
to astronomy and on a general view of the Universe through the different
ranges of the electromagnetic spectrum. Similarly, members of our team
contributed to the long-running “Nuestros Alumnos en el Roque de Los
Muchachos” programme, giving guided tours of the Roque de Los Muchachos
Observatory (including going inside the GTC dome) to local high school
students in the 4th grade of compulsory secondary education (4o ESO, age 15).

For the general public, several popular lectures were delivered by members of
the team in planetariums and museums across partner countries. Talks for
citizen scientists and amateur astronomers were also organised, in particular
outlining the potential for observations of stellar occultations by asteroids.
Similarly, a YouTube channel was created for the project, hosting educational
videos on the history of astronomy, astronomical techniques and principles,
and interviews. Furthermore, the partners at Comenius University in Bratislava
contributed to their pre-existing YouTube channel as part of the project.
These open talks and YouTube videos serve not only to educate but also to
strengthen the connection between the public and the scientists whom are at
least partially supported by public funding.

##  7 Conclusions

Key Action 2 of the European Union’s Erasmus+ programme offers substantial
grant funding for international collaboration on educational projects. The
(on-going) outputs of two such grants, awarded to a consortium of Czech,
Slovak and Spanish research institutes and universities, have been briefly
presented here. These grants have supported educational activities ranging
from outreach events for young children through to summer schools for graduate
students, and from public lectures and YouTube videos through to financial
support for extended international research visits to support the professional
development of ECRs.

The impact of the projects, particularly for the professional development of
ECRs, has been clear. A number of summer school participants have now
completed their Masters or PhDs (stating that the skills acquired as part of
the schools were useful for their research), while many of the ECRs to have
research trips funded by the projects have since gone on to obtain prestigious
fellowships (e.g. ESA) or permanent research/teaching positions (e.g. through
the Mexican “Càtedra CONACYT para jóvenes investigadores” scheme).

The work undertaken as part of “Per Aspera Ad Astra Simul" and “European
Collaborating Astronomers Project: España-Czechia-Slovakia" clearly
demonstrates the potential impact that Erasmus+ Key Action 2 funding can
facilitate. We strongly encourage other astronomy educators to consider how
they might make similarly good use of the available funding in future calls.

##  8 Declarations

###  8.1 List of abbreviations

  * AI ASCR

Astronomical Institute of the Academy of Sciences of the Czech Republic

  * AI CU

Astronomical Institute of Charles University

  * AI SAS

Astronomical Institute of the Slovak Academy of Sciences

  * ECR

Early career researchers

  * ESA

European Space Agency

  * ESO

European Southern Observatory

  * GTC

Gran Telescopio Canarias

###  8.2 Ethical Approval

Not applicable.

###  8.3 Consent for publication

Not applicable.

###  8.4 Competing Interests

The author(s) declare that they have no competing interests.

###  8.5 Funding

This work was supported by the Erasmus+ programme of the European Union under
grant numbers 2017-1-CZ01-KA203-035562 and 2020-1-CZ01-KA203-078200 (PI P.
Kabáth).

###  8.6 Author’s Contributions

David Jones - Conceptualization, Funding acquisition, Project administration,
Writing – original draft, Writing – review & editing
Petr Kabáth - Conceptualization, Funding acquisition, Project administration,
Writing – review & editing, Project administration Jorge García-Rojas -
Project administration, Writing - review & editing Josef Hanuš - Project
administration, Writing - review & editing Marian Jakubík - Project
administration, Writing - review & editing Jan Janík - Project administration,
Writing - review & editing Roman Nagy - Project administration, Writing -
review & editing Juraj Tóth - Project administration, Writing - review &
editing

##  9 Acknowledgements

We would like to acknowledge the role of the researchers at the participating
institutions without whom these partnerships would not have been a success –
the ECRs for their eagerness to participate in mobilities and schools, and
senior researchers for their willingness to host ECR mobilities and to
contribute to Kabáth et al., [5].

## References

  * Altamura et al., [2019] Altamura, E., Brennan, S., Leśniewska, A., Pintėr, V., dos Reis, S. N., Geier, S., and Fynbo, J. P. U. (2019).  Discovery of a binary quasar at z = 1.76.  Contributions of the Astronomical Observatory Skalnate Pleso, 49(3):528–531.
  * Barcons, [2007] Barcons, X. (2007).  Astronomy in Spain.  The Messenger, 127:4.
  * González Manrique et al., [2020] González Manrique, S. J., Kuckein, C., Pastor Yabar, A., Diercke, A., Collados, M., Gömöry, P., Zhong, S., Hou, Y., and Denker, C. (2020).  Tracking Downflows from the Chromosphere to the Photosphere in a Solar Arch Filament System.  ApJ, 890(1):82.
  * Jones et al., [2019] Jones, D., Pejcha, O., and Corradi, R. L. M. (2019).  On the triple-star origin of the planetary nebula Sh 2-71.  MNRAS, 489(2):2195–2203.
  * Kabáth et al., [2020] Kabáth, P., Jones, D., and Skarka, M. (2020).  Reviews in Frontiers of Modern Astrophysics; From Space Debris to Cosmology.  Springer.
  * Kabáth et al., [2019] Kabáth, P., Korhonen, H., and Jones, D. (2019).  Observational astrophysics: from proposals to publication.  Contributions of the Astronomical Observatory Skalnate Pleso, 49(3):522–527.
  * Kučáková et al., [2019] Kučáková, H., Mikhalchenko, O., Popescu, M., Ransome, C., and Sharma, A. (2019).  Optical spectra of near-Earth asteroids (381906) 2010 CL19 and (453778) 2011 JK.  Contributions of the Astronomical Observatory Skalnate Pleso, 49(3):532–538.
  * Palouš and Hadrava, [2007] Palouš, J. and Hadrava, P. (2007).  Astronomy in the Czech Republic.  The Messenger, 128:3.
  * Paunzen et al., [2021] Paunzen, E., Hümmerich, S., Fedurco, M., Bernhard, K., Komžík, R., and Vaňko, M. (2021).  V680 Mon - a young mercury-manganese star in an eclipsing heartbeat system.  MNRAS, 504(3):3749–3757.
  * Skarka et al., [2021] Skarka, M., Janík, J., Paunzen, E., and Glos, V. (2021).  The GATE summer school.  Contributions of the Astronomical Observatory Skalnate Pleso, 51(1):41–44.


Two groups are orbit equivalent if they both admit an action
on a same probability space that share the same orbits. In particular the Ornstein-Weiss theorem implies that all infinite
amenable groups are orbit equivalent to the group of integers. To refine this
notion between infinite amenable groups Delabie,
Koivisto, Le Maître and Tessera introduced a quantitative version of orbit
equivalence. They furthermore obtained obstructions to
the existence of such equivalence using the isoperimetric profile.

In this article we offer to answer the inverse problem (find a group being
orbit equivalent to a prescribed group with prescribed
quantification) in the case of the group of integers using the so called
Følner tiling shifts introduced by Delabie et al. To do so we use the
diagonal products defined by Brieussel and Zheng
giving groups with prescribed isoperimetric profile.

Classification 37A20

Keywords orbit equivalence, lamplighter group, inverse problem,
isoperimetric profile, diagonal products

§ INTRODUCTION

Two groups are orbit equivalent if they admit free measure-preserving actions on a same
standard probability space $(X, \mu)$ which share the same orbits. This notion
—emerging from the seminal work of Dye [5, 6]— can be seen as the
ergodic version of the famous measure equivalence introduced by
Gromov [8]. A famous result of
Ornstein and Weiss (see <ref>) implies that all amenable groups are
orbit equivalent. In particular —unlike quasi-isometry— orbit equivalence
does not preserve coarse geometric invariants.

To overcome this issue it is therefore natural to look for some refinements of this
orbit equivalence notion.
Assume for example that $G$ and $H$ are two finitely
generated orbit equivalent groups over a probability space $(X,\mu)$. Recall
that we can consider the Schreier graph associated to the action of $G$ (resp. $H$)
on $X$ and equip it with the usual metric $d_{S_G}$ (resp. $d_{S_H}$), fixing the
length of an edge to one.
A first way to refine the measure equivalence is to quantify how close the two
actions are by studying for all $g\in G$ and $h\in H$ the integrability of the
two following maps

\begin{equation*}
x\mapsto d_{S_G}(x,h\cdot x) \qquad x\mapsto d_{S_H}(x,g\cdot x).
\end{equation*}

When these two maps are $L^p$ we say that the groups are $L^p$-orbit
equivalent (see [2] for more details). In this refined
framework a famous result of Bader, Furman and Sauer [2]
implies that
any group $L^1$-orbit equivalent to a lattice in $SO(n, 1)$ for some $n\geq 2$ is
virtually a lattice in $SO(n,1)$. This refinement also lead Bowen to prove in
the appendix of [1] that volume growth was invariant under
$L^1$-orbit equivalence.

Delabie, Koivisto, Le Maître and Tessera offered in [4] to
extend this quantification to a family of functions larger than $\{x\mapsto x^p,
\ p\in [0,+\infty]\}$ (see <ref>).
They furthermore showed the monotonicity of the isoperimetric profile under this
quantified measure equivalence definition (see <ref>). In
[3] Brieussel and Zheng managed to construct amenable groups with
prescribed isoperimetric profile called diagonal product. Considering the
monotonicity of the isoperimetric profile, the striking result of Brieussel and
Zheng thus triggers a new  question: instead of trying to quantify the
equivalence relation between two given groups, can one find a group that is
orbit equivalent to a prescribed group with a prescribed quantification?

This is the problem we address in this article. Using Brieussel-Zheng’s construction we
exhibit a group that is orbit equivalent to $\bZ$ with a prescribed
quantification (see <ref>). Comparing the obtained coupling
to the constraints given by <ref>
we show that our coupling is close to being optimal for a sense of “optimal”
that we make precise in <ref>.

§.§ Quantitative orbit equivalence

Let us recall some material from [4]. A measure-preserving
actionMeasure!Preserving action
of a discrete countable group $G$ on a measured space $(X,\mu)$ is an action of
$G$ on $X$ such that the map $(g,x)\mapsto g\cdot x$ is a Borel map and
$\mu(E)=\mu(g\cdot E)$ for all $E\subseteq \mathcal{B}(X)$ and all $g\in G$. We
will say that a measure-preserving action of $G$ on $(X,\mu)$ is
freeFree action if for almost every $x\in X$ we have $g\cdot x =
x$ if and only if $g=\neutre_G$.

We recall below the definition of orbit equivalence and the
quantified version as introduced by Delabie, Koivisto, Le Maître and Tessera
[4]. We conclude this section by studying the relation between isoperimetric profile
and orbit equivalence.

Let $G$ and $H$ be two finitely generated groups. We say that $G$
and $H$ are orbit equivalent if there exists a probability space
$(X,\mu)$ and a measure-preserving free action of $G$ (resp. $H$) on $(X,\mu)$
such that for almost every $x \in X$ we have $G\cdot x = H\cdot
We call $(X,\mu)$ an orbit equivalence coupling from $G$ to $H$.

By the Ornstein Weiss theorem <cit.> below, all infinite amenable
groups are in the same equivalence class.
All infinite amenable groups are orbit equivalent to $\bZ$.

To refine this equivalence relation and “distinguish” amenable groups we
introduce the quantified version of orbit equivalence.

Recall that if a finitely generated group $G$ acts on a space $X$ and if $S_G$
is a finite generating set of $G$, we can define the Schreier graph associated to this
action as being the graph whose set of vertices is $X$ and set of edges is
$\{(x,s\cdot x) \, | \, s\in S_K\}$. This graph is endowed with a natural metric
$d_{S_G}$ fixing the length of an edge to one. Remark that if $S^\prime_G$ is another
generating set of $G$ then there exists $C>0$ such that for all $x\in X$ and
$g\in G$

\begin{equation*}
\frac{1}{C}d_{S_G}(x,g\cdot x) \leq d_{S^{\prime}_G}(x,g\cdot x) \leq Cd_{S_G}(x,g\cdot x).
\end{equation*}

We say that an orbit equivalence coupling
$(X,\mu)$ from $G$ to $H$ is $(\varphi,\psi)$-integrable if
for all $g\in G$ (resp. $h\in H$) there exists $c_g>0$ (resp. $c_h>0$) such

\begin{equation*}
\int_{X} \varphi\left( \frac{1}{c_g} d_{S_H}(g\cdot x, x)
\right)\mathrm{d}\mu(x)<+\infty \quad \text{and} \quad
\int_{X}\psi\left( \frac{1}{c_h} d_{S_G}(h\cdot x, x)\right)
\mathrm{d}\mu(x)<+\infty.
\end{equation*}

We introduce the constants $c_g$ and $c_h$ in the definition for the
integrability to be independent of the choice of generating sets $S_G$ and
$S_H$. If $\varphi(x)=x^p$ we will sometimes talk of $(L^p,\psi)$-integrability
instead of $(\varphi,\psi)$-integrability. In particular $L^0$ means
that no integrability assumption is made.
Finally, note that every $(L^\infty,\psi)$-integrable coupling is
$(\varphi,\psi)$-integrable for any increasing map $\varphi: \bR^+\rightarrow
\bR^+$.  When $\varphi=\psi$ we will say that the coupling is
$\varphi$-integrable instead of $(\varphi, \varphi)$-integrable.

* There exists an orbit
equivalence coupling between $\bZ^4$ and the Heisenberg
group $\mathrm{Heis}(\bZ)$ that is $L^p$-integrable for all $p<1$.

* Let $k\in \bN^{*}$. Their exists an $(L^{\infty},\exp)$-integrable orbit
equivalence coupling from the lamplighter group to the Baumslag-Solitar
group $BS(1,k)$.

More examples will be given in <ref>. Let us conclude on the
quantification by a remark. We chose to refine orbit
equivalence using the integrable point of view. But it is not the only
possible sharpening. For example Kerr and Li [9] defined
Shannon orbit equivalence: instead of looking at the integrability of
distance maps they consider the Shannon entropy of partitions associated
to the coupling.

§.§ Isoperimetric profile

As stated before, the orbit equivalence does not preserve the coarse geometric
invariants. But the quantified version defined above allowed Delabie
et al. [4] to get a relation between the isoperimetric profiles of two
orbit equivalent groups which we describe below.

Recall that if $G$ is generated by a finite set $S$, the isoperimetric
profile of $G$ is defined as[We chose to adopt the convention of
[4]. Note that in [3], the isoperimetric profile is defined as

\begin{equation*}
\profile_{G}(n):= \sup_{|A|\leq n} \frac{|A|}{|\partial A|}.
\end{equation*}

For example the isoperimetric profile of $\bZ$ verifies $\profile_{\bZ}(x) \simeq x$.
Remark that due to Følner criterion, a group is amenable if and only if its
isoperimetric profile is unbounded. Hence we can see the isoperimetric profile
as a way to measure the amenability of a group: the faster $\profile_G$ tends to
infinity, the more amenable $G$ is.

The behaviour of the isoperimetric profile under measure equivalence
coupling is given by the theorem below. If $f$ and $g$ are two real functions we
denote $f \preccurlyeq g$ if there exists some constant $C>0$ such that
$f(x)=\mathcal{O}\big(g(Cx)\big)$ as $x$ tends to infinity. We write $f\simeq g$
if $f \preccurlyeq g$ and $g\preccurlyeq f$.

Let $G$ and $H$ be two finitely generated groups admitting a
$(\varphi,L^0)$-integrable orbit equivalence coupling. If $\varphi$ and
$t/\varphi(t)$ are non-decreasing then

\begin{equation*}
\varphi \circ \profile_{H} \preccurlyeq \profile_G.
\end{equation*}

This theorem provides an obstruction for finding $\varphi$-integrable couplings
with certain functions $\varphi$ between two amenable groups. For example
for a coupling with $H=\bZ$ the integrability has to verify $\varphi \preccurlyeq
This lead the authors of [4] to ask the following question.
Given an amenable finitely generated group $G$, does there exist a
$(I_G,L^0)$-integrable orbit equivalence coupling from $G$ to $\bZ$?

We answer the above question for a large family of maps
$\varphi$ in <ref>. We will see that the coupling
we build to proof the aforementioned theorem answers <ref>
up to a logarithmic error.

§.§ Main results

In this paper we show the following main theorem and its corollary below.

For all non-decreasing function $\rho: [1,+\infty[
\rightarrow [1,+\infty[$ such that $\rho(1)=1$ and $x/\rho(x)$ is
non-decreasing, there exists a group $G$ such that

* $I_{G}\simeq \rho \circ \log$;

* there exists an orbit equivalence coupling from $G$ to $\bZ$ that is
$(\varphi_{\varepsilon},\exp \circ \rho)$-integrable for all
$\varepsilon>0$, where $\varphi_{\varepsilon}(x):={\rho\circ
\log(x)}/{\left( \log\circ\rho\circ\log(x) \right)^{1+\varepsilon}}$.

Let us discuss the optimality of this result. Consider a
$(\varphi,L^0)$-integrable orbit equivalence coupling from some group $G$ to
$\bZ$. By <ref> it verifies $\varphi\circ I_{\bZ} \preccurlyeq
I_G$. In particular since $I_{\bZ}(x)\simeq x$, we can not have a better
integrability than $\varphi(x)\simeq I_G$. Since $I_{G}\simeq \rho \circ
\log$ our above theorem is optimal up to a logarithmic error. We discuss this in
more length in <ref>.

Main Ingredients The main tools of the proof of
<ref> are Brieussel-Zheng’s
diagonal products (see <ref>) and Følner tiling shifts (see
<ref>). We show that a diagonal product $\BZ$ admits a coupling with
$\bZ$ satisfying <ref>. To prove it we use the integrability
criterion given by <ref> and involving Følner tiling

Therefore we compute in <ref> a Følner tiling shift
$(\Sigma_n)_n$ for $\BZ$. We also estimate the tiles’ diameter and the
proportion of elements in the boundary. We construct a Følner tiling shift for
$\bZ$ in <ref> and show that these two tiling shifts verify

Let us now consider the possible generalisations of this result to other
groups than the group of integers.
To do so we can use the composition of couplings described in

Given the above theorem, once we have a measure equivalence coupling from $\bZ$
to a group $H$ we can compose the two couplings to obtain a measure equivalence
from $G$ to $H$.
If the growth of the isoperimetric profile of $H$ is close to the one of $\bZ$, the
integrability of the obtained coupling will be close to the optimal one given by
<ref>. It is for example the case when $H=\bZ^d$.

Let $d\in \bN^{*}$ and $\varepsilon>0$.
For all non-decreasing function $\rho: [1,+\infty[
\rightarrow [1,+\infty[$ such that $\rho(1)=1$ and $x/\rho(x)$ is
if the map $\varphi_\varepsilon$ defined in <ref>
is subadditive and concave, then there exists a group $G$ such that

* $I_{G}\simeq \rho \circ \log$ ;

* there exists a $(\varphi_{\varepsilon},L^0)$-integrable orbit
equivalence coupling from $G$ to $\bZ^d$.

Structure of the paper In <ref> we present the
diagonal products introduced by Brieussel and Zheng. We recall some of the
properties shown in [3] and compute Følner sequences.
<Ref> is devoted to Følner tiling shifts. These tools built by Delabie
et al. [4] allow us to construct and quantify an orbit equivalence coupling
between two groups. In this section we also construct Følner tiling shifts for
diagonal products $\BZ$.
We show our main theorem in <ref> combining the results of the
two previous sections. Finally we discuss the limits of this construction and
some open problems in <ref>.

I would like to thank Romain Tessera and Jérémie Brieussel, under
whose supervision the work presented in this article was carried out.
I thank them for suggesting the topic, sharing their precious insights and for
their many useful advice. I also thank the anonymous referee for their remarks
and corrections.

§ DIAGONAL PRODUCTS OF LAMPLIGHTER GROUPS

We recall here necessary material from [3] concerning the definition of
Brieussel-Zheng’s diagonal products. We give the definition of such
a group, recall and prove some results concerning the range (see
<ref>) of an element and use it to identify a Følner
sequence. Finally we present in <ref> the tools needed to
recover such a diagonal product starting with a prescribed isoperimetric

§.§ Definition of diagonal products

Recall that the wreath product of a group $G$ with $\bZ$ denoted $G \wr \bZ$ is
defined as $G\wr \bZ:= \oplus_{m \in \bZ} G \rtimes \bZ$. An element of $G\wr
\bZ$ is a pair $(f,t)$ where $f$ is a map from $\bZ$ to $G$ with finite support
and $t$ belongs to $\bZ$. We refer to $f$ as the lamp configuration
and $t$ as the cursor. Finally we denote by $\supp(f)$ the
support of $f$ which is defined as
$\supp(f):=\left\{x\in \bZ \ | \ f(x)\neq e_G \right\}$.

§.§.§ General definition

Let $A$ and $B$ be two finite groups. Let $(\Gamma_m)_{m\in \bN}$ be a sequence
of finite groups such that each $\Gamma_m$ admits a generating set of the form $A_m \cup B_m$
where $A_m$ and $B_m$ are finite subgroups of $\Gamma_m$ isomorphic respectively
to $A$ and $B$. For $a \in A$ we denote $a_m$ the copy of $a$ in $A_m$ and
similarly for $B_m$.

Finally let $(k_m)_{m \in \bN}$ be a sequence of integers such that $k_{m+1} \geq 2 k_m$
for all $m$. We define $\Ds=\WDs$ and endow it with the generating set

\begin{equation*}
\Big\{ (\mathrm{id},1 )\Big\} \cup \Big\{ \big(a_m \delta_0,0 \big) \ | \ a_m \in A_m\Big\}
\cup \Big\{ \big(b_m \delta_{k_m},0 \big) \ | \ b_m \in B_m\Big\}.
\end{equation*}

Diagonal product Brieussel-Zheng’s
diagonal product
The Brieussel-Zheng diagonal product associated to
$(\Gamma_m)_{m\in \bN}$ and $(k_m)_{m\in \bN}$ is the subgroup $\BZ$ of $ \left(\prod_m
\Gamma_m\right)\wr \bZ$ generated by

\begin{equation*}
\Big\{ \Big({ {\big(\mathrm{id}\big)}_m,1 }\Big)\Big\} \cup
\Big\{ \big( {\left(a_m \delta_0  \right)}_m,0 \big) \ | \ a \in A\Big\}
\cup \Big\{ \big( {\left( b_m \delta_{k_m} \right)}_m,0 \big) \ | \ b \in B\Big\}.
\end{equation*}

The group $\BZ$ is uniquely determined by the sequences $(\Gs)_{m\in \bN}$ and
$(k_m)_{m\in \bN}$. Let us give an illustration of what an element in such a
group looks like. We will denote by $\mbfg$ the sequence $(g_m)_{m\in \bN}$.

represent in <ref> the element $\big(\mbfg,t \big)$ of $\BZ$ verifying

\begin{equation*}
(\mbfg,t)=\big( (g_m)_{m\in \bN},t \big):=\big({\left(a_m\delta_0\right)}_m,0 \big)
\big( {\left( b_m \delta_{k_m}\right)}_m,0\big) (0,3),
\end{equation*}

when $k_m=2^m$. The cursor is represented by the blue arrow at the bottom of the
figure. The only value of $g_0$ different from the identity is $g_0(0)=(a_0,b_0)$.
Now if $m>0$ then the only values of $g_m$ different from the identity are
$g_m(0)=a_m$ and $g_m(k_m)=b_m$.

Representation of $\big(\mbfg,t\big)=\big((a_m\delta_0)_m,0\big)%
\big((b_m\delta_{k_m})_m,0\big)(0,3)$ when $k_m=2^m$.

§.§.§ The expanders case

In this article we will restrict ourselves to a particular familiy of groups
$(\Gamma_m)_{m\in \bN}$ called expanders.  Recall that $(\Gamma_m)_{m\in \bN}$
is said to be a sequence of expandersExpander if the sequence
of diameters $(\diam{\Gamma_m})_{m\in \bN}$ is unbounded and if there exists
$c_0>0$ such that for all $m\in \bN$ and all $n\leq |\Gamma_m|/2$ the
isoperimetric profile verifies $I_{\Gamma_m}(n)\leq c_0$.

When talking about diagonal products we will always make the following
assumptions. We refer to <cit.> for an explicit example of
diagonal product verifying (H).

Hypothesis (H)

* $(k_m)_m$ and $(l_m)_m$ are sub-sequences of geometric

* $k_{m+1}\geq 2 k_m$ for all $m\in \bN$;

* $(\Gamma_m)_{m\in \bN}$ is a sequence of expanders such that $\Gamma_m$
is a quotient of $A*B$ and there exists $\Clm >0$ such that
$1/\Clm l_m \leq \diam{\Gamma_m}\leq \Clm l_m$ for all $m\in \bN$;

* $k_0=0$ and $\Gamma_0=A_0\times B_0$;

* the natural quotient map $A_m\times B_m\rightarrow
\langle{ \langle{\left[A_m,B_m\right]}}  \rangle \rangle \backslash \Gamma_m$
is an isomorphism,  where $\langle{\langle{\left[A_m,B_m\right]}\rangle}\rangle $
is the normal
closure of $\left[A_m,B_m\right]$.

Recall (see <cit.>) that in this case there exist $c_1$, $c_2>0$ such
that, for all $m$

\begin{equation}
\label{eq:encadrementlngammap}
c_1 l_m -c_2 \leq \ln \left\vert {\Gamma_m}\right\vert \leq c_1l_m + c_2.
\end{equation}

Finally we adopt the convention of <cit.> and allow $(k_m)_{m\in \bN}$
to take the value $+\infty$. In this case $\BZ_s$ is the trivial group. In
particular when $k_1=+\infty$ the diagonal product $\BZ$ corresponds to the
usual lamplighter $(A\times B)\wr \bZ$.

§.§.§ Relative commutators subgroups

Let $\theta_m : \Gs \rightarrow
\langle{ \langle{\left[A_m,B_m\right]}}  \rangle \rangle \backslash \Gs \simeq
A_m \times B_m$ be the natural projection, for all $m\in \bN$. Let $\theta^A_m$ and $\theta^B_m$
denote the composition of $\theta_m$ with the projection to $A_m$ and $B_m$
respectively. Now let $m\in \bN$ and define $\Gsp:= \langle{ \langle{
\left[A_m,B_m\right]}} \rangle \rangle$. If $(g_m,t)$ belongs to $\Delta_m$
then there exists a unique $g^{\prime}_m \ : \ \bZ \rightarrow \Gsp$ such that
$g_m(x)=g^{\prime}_m(x)\theta_m\big(g_m(x)\big)$ for all $x\in \bZ$.

Let $(\mbfg,3)$ be the element described in <ref>. Then
the only non-trivial value of $\theta_0(g_0)$ is
$\theta_0(g_0(0))=(a_0,b_0)$. If $m>0$ then the only non trivial values of
$\theta_m(g_m)$ are $\theta_m(g_m(0))=(a_m,\neutre)$ and
$\theta_m(g_m(k_m))=(\neutre,b_m)$. Finally for all $m$ we have
$g^{\prime}_m=\mathrm{id}$ since there are no commutators appearing in the
decomposition of $(\mbfg,0)$.

Assume that $k_m=2^m$ and consider first the element $(\mbff,0)$ of $\BZ$
defined by $(\mbff,0):=(0,-k_1)\big((a_m\delta_0)_{m},0 \big)(0,k_1)$. Now
define the commutator

\begin{equation*}
(\mbfg,0)=(\mbff,0)\cdot \big((b_m\delta_{k_m})_{m},0\big)\cdot
(\mbff,0)^{-1} \cdot \big((b^{-1}_m\delta_{k_m})_{m},0\big)
\end{equation*}

and let us describe the values taken by $\mbfg$ and the induced maps
$\theta_m(g_m)$ and $g^{\prime}_m$ (see <ref> for a
representation of $\mbfg$). The only non-trivial commutator appearing in the
values taken by $\mbfg$ is $g_1(k_1)$ which is equal to $
a_1b_1a^{-1}_1b^{-1}_1$. In other words
$g_0$ is the identity, thus
$\theta_0=\mathrm{id}$. Moreover when $m=1$
we have $\theta_1=\mathrm{id}$ and the only value of $g^{\prime}_1(x)$
different from $\neutre$ is $g^{\prime}_1(k_1)= a_1b_1a^{-1}_1b^{-1}_1$ (on a
blue background in <ref>). Finally
if $m>1$ then $g_m$ is the identity thus $\theta_m=\mathrm{id}$ and

[<ref>]Representation of $(\mbfg,0)$ defined in

Let us study the behaviour of this decomposition under product of
lamp configurations.

If $g_m,f_m \ : \ \bZ \rightarrow \Gamma_m$ then
$(g_mf_m)^{\prime}=g^{\prime}_m\theta_m(g_m)f^{\prime}_m\Big(\theta_m(g_m) \Big)^{-1}$.

Since  $g_m=\theta_m(g_m) g^{\prime}_m$ and  $f_m=\theta_m(f_m) f^{\prime}_m$
we can write

\begin{align*}
g_mf_m \
= \ g^{\prime}_m\theta_m(g_m) \cdot f^{\prime}_m\theta_m(f_m)\
= \ g^{\prime}_m\theta_m(g_m)f^{\prime}_m\theta_m(g_m)^{-1}\theta_m(g_m)\theta_m(f_m).
\end{align*}

But $\theta_m(g_m)\theta_m(f_m)$ takes values in $A_m\times B_m$ and
$\Gsp$ is a normal subgroup of $\Gamma_m$ thus the map
$g^{\prime}_m\theta_m(g_m)f^{\prime}_m\theta_m(g_m)^{-1}$ takes
values in $\Gsp$. Hence the claim.

Combining Lemma 2.7 and Fact 2.9 of
[3], we get the following result.

Let $(\mbfg, t) \in \BZ$. For all $m\in \bN$
and $x\in \bZ$

\begin{align*}
g_m(x)&=g^\prime_m(x) \theta^A_m\big(g_m(x)\big)\theta^B_m\big(g_m(x)\big)\\
&=g^\prime_m(x) \theta^A_m\big(g_0(x)\big)\theta^B_m\big(g_0(x-k_m)\big).
\end{align*}

In particular the sequence $\mbfg=\left( g_m \right)_{m\in \bN}$ is uniquely determined
by $g_0$ and $\left( g^\prime_m \right)_{m\in \bN}$.

In the next subsection we are going to see that we actually need only a
finite number of elements of the sequence $(g^{\prime}_m)_{m \in \bN}$ to
characterise $\mbfg$.

§.§ Range and support

In this subsection we introduce the notion of range of an element $(\mbfg,t)$ in $\BZ$
and link it to the supports of the lamp configurations $(g_m)_{m\in \bN}$.

§.§.§ Range

We denote by $\pi_{2} : \Delta \rightarrow \bZ$ the projection on the
second factor and for all $n\in \bN$ denote by $\landing(n)$ the integer such
that $k_{\landing(n)}\leq n< k_{\landing(n)+1} $.

If $w=s_1\ldots s_m$ is a word over $S_\BZ$ we define its
rangeRange!Of a word as

\begin{equation*}
\range(w) := \left\{\pi_2\left(\prod^{i}_{j=1}s_j \right) \,
| \, i=0,\ldots,n \right\}.
\end{equation*}

The range is a finite subinterval of $\bZ$. It represents the set of sites
visited by the cursor.
Range!Of an element
The range of an element $\delta \in \Delta$ is defined as the
diameter of a minimal range interval of a word over $S_{\BZ}$ representing $\delta$.

In what follows we will consider elements that can be written as a word with range
in an interval of the form $[0,n]$, where $n$ belongs to $\bN$. Therefore,
when there is no ambiguity we will denote $\range(\delta)$ this interval,
namely $\range(\delta)=[0,n]$.

Let $(\mbfg,0) \in \BZ$ such that
$\range(\mbfg,0)=[0,6]$, that is to say: the cursor can only
visit sites between $0$ and $6$. Then the map $g_m$ can “write” elements of
$A_m$ only on sites visited by the cursor, that is to say from $0$ to $6$, and
it can write elements of $B_m$ only from $k_m$ to $6+k_m$. Thus $g_0$ is
supported on $[0,6]$, since $k_0=0$. Moreover, commutators (and hence elements
of $\Gsp$) can only appear between $k_m$ and $6$, thus $\supp(g^\prime_m)
\subseteq [k_m,6]$. In particular $\supp(g^\prime_m)$ is empty when $k_m>6$.

Such a $(\mbfg,0)$ is represented in <ref> for $k_m=2^m$.

An element of $\BZ$

Recall that $g_m$ : $\bZ \rightarrow
\Gamma_m$. If $m\leq \landing(6)$, then $g_m(x)$ belongs to $A_m$ if $x \in
[0,k_m-1]$, it belongs to $\Gs$ if $x\in [k_m,6]$ and to $B_m$ if $x\in [7,6+k_m]$ and
equals $\neutre$ elsewhere. If $m>\landing(6)$ then $g_m(x)$ belongs to $A_m$ if $x \in
[0,6]$ and to $B_m$ if $x\in [k_m,6+k_m]$ and equals $\neutre$ elsewhere.

Let us now recall a useful fact proved in [3].
An element $(\mbfg,t) \in \BZ$ is uniquely determined by $t$, $g_0$ and the sequence
$ (g^{\prime}_m)_{m\leq \landing(\range(\mbfg,t))}$.

Consider again $(\mbfg,0) \in \BZ$ such that
$\range(\mbfg,0)=[0,6]$, which was illustrated in <ref>.
Since $k_3=8>6$, the element $(\mbfg,0)$ is uniquely determined by the data
$g_0$ (that is to say, the values read in the bottom line) and the values of
$g^\prime_i$ for $i=1,2$ (namely, the value taken in the blue area).
<Ref> represents the aforementioned characterizing data.

Data needed to characterise $\mbfg$ such that $\range(\mbfg)\subset
[0,6]$ when $k_m=2^m$

§.§.§ Relation between range and support

Recall that for all $m\in \bN$ we can write
$g_m(x)=g^\prime_m(x)\theta^A_m\big(g_0(x)\big)\theta^B_m\big(g_0(x-k_m)\big)$ and
that $\landing(n)$ denotes the integer such that $k_{\landing(n)}\leq n <

To work with the Følner sequence we compute in <ref> and
deduce a Følner tiling shift from it, we will need to link the range
of $(\mbfg,t)$ in $\BZ$ with the support of $g_0$ and the sequence of supports of
$({g^\prime}_m)_{m \in \bN}$. This is what the following lemma formalises.

Let $n\in \bN$ and take $(\mbfg,t) \in \Delta$.
Then $\range(\mbfg,t)$ is included in $ [0,n] $ if and only if

\begin{equation*}
\begin{cases}
t\in [0,n] \\
\supp(g_0) \subset [0, n]\\
\supp({g^\prime}_m) \subseteq [k_m,  n] \quad &\forall 1\leq m \leq \landing(n)  \\
{g^\prime}_m\equiv \neutre \quad &\forall m > \landing(n).
\end{cases}
\end{equation*}

Let $n \in \bN$ and first assume that $\range(\mbfg,t)\subseteq [0,n]$, that
is to say: the cursor
can only visit sites between $0$ and $n$. Let $(\mbfg,t)=\prod^{l}_{i=0} s_i$
be a decomposition in a product of elements of $S_{\BZ}$ with range of minimal length.
Let $m\in \bN$, then by definition of
$S_{\BZ}$, an element $s_i$ can “write” elements of $A_m$ only between $0$ and
$n$, and it can write elements of $B_m$ only between $k_m$ and $n+k_m$. Thus $g_0$ is
supported on $[0,n]$, since $k_0=0$. And commutators can only appear between
$k_m$ and $n$, hence $\supp(g^\prime_m)\subseteq [k_m,n]$. In particular if
$k_m>n$ then $g^\prime_m \equiv \neutre$. Finally we obtain that $t$ belongs
to $[0,n]$ by noting that $t=\pi_2\left(\prod^{l}_{j=1}s_j \right)$.

Now let us prove the other way round. Consider $m\in [1,\landing(n)]$ then
$g^{\prime}_m(x)\in \Gsp$. It is therefore a product of conjugates of
commutators of the form $[a_m,b_m]$, where $a_m\in A_m$ and $b_m\in B_m$.
Applying <ref> with $x$ instead of $k_1$ we can show that we
can write $[a_m,b_m]$ at $g_m(x)$ without changing any other entry in $\mbfg$
(see also <ref>). In a similar way, we can write a conjugate
of $[a_m,b_m]$ at $g_m(x)$ without changing any other entry in $\mbfg$.
Finally writing $(a_0,b_0)$ at the entry $g_0(x)$ writes $a_m$ at $g_m(0)$ and
$b_m$ at $g_m(k_m)$ (see also <ref>). Therefore using
<ref> we can obtain
$(\mbfg,0)$ by first considering the word in $S_\BZ$ that writes
all the values of $g_0$, then multiplying it on the left by a word that writes
the value of $g^{\prime}_1$, and continue this process to write all
$g^{\prime}_m$ for $m\leq \landing(n)$.

Let us now check that the cursor
remains in $[0,n]$ when writing $g_0$ and $g^{\prime}_m$.
Take $m\in [1,\landing(n)]$, then $k_m\leq n$ and $\supp(g^\prime_m)$ is
contained in $[k_m,n]$. Now let $x \in \supp(g^\prime_m)\subseteq [k_m,n]$.
Since $\Gsp \subseteq \Gamma_m$, which is generated by $A_m \times
B_m$, we can decompose $g^\prime_m(x)$ as a product of elements in $A_m$ and
$B_m$. To write some $a_m\in A_m$ at the position $x$ the cursor needs to
visit sites in $[0,x]$. To write some $b_m\in B_m$ it needs to visit sites in
$[0,x-k_m]$. Therefore, the cursor remains in $[0,n]$ when writing $g_m(x)$ at
position $x$.
Finally, for all $x$
the cursor needs only to visit position $x$ in order to write $g_0(x)$. Since
$\supp(g_0)$ is contained in $[0,n]$ then the cursor needs only to visit sites
between $0$ and $n$.

Combining what precedes with <ref> and the hypothesis that $t\in
[0,n]$, we get that the cursor needs only
to visit cites between $[0,n]$ to write $(\mbfg,t)$.
Hence the lemma.

§.§.§ Følner sequence

In this subsection we describe a Følner sequence $(F_n)_{n\in \bN}$ for $\BZ$.
Recall that $\landing(n)$ denotes the integer such that $k_{\landing(n)}\leq n <

The following sequence is a Følner sequence of $\BZ$

\begin{equation*}
F_n:= \left\{ \left(\mbff, t \right) \mid
\range \left( \mbff,t\right) \subseteq \{0, \ldots, n-1\} \right\}.
\end{equation*}

Let $n\in \bN$ and $\delta:=(\mbff,t)\in F_n$. Remark that since
$\delta$ belongs to $ F_n$, <ref> implies that $t$ belongs to $\{0,\ldots,n-1\}$. Now let
$s_1,\ldots,s_l \in S_{\BZ}$ such
that $\delta=s_1\cdots s_l$ and take $s_{l+1}\in S_{\BZ}$. If
$s_{l+1}=\big((a_m\delta_0),0 \big)$ for some $a \in A$ or if
$s_{l+1}=\big((b_m\delta_{k_m}),0 \big)$ for some $b\in B$ then
since the cursor of $s_{l+1}$ equals $0$,

\begin{equation*}
\range (\delta s_{l+1}) =\left\{\pi_2\left(\prod^{i}_{j=1}s_j \right) \,
| \, i=1,\ldots,l+1 \right\} = \range(\delta).
\end{equation*}

Thus $\delta s_{l+1}\in F_n$.
Finally denote by $[x,y]$ the range of $\delta$. Using the same formula
as above we get

\begin{align*}
\range (\delta\cdot (\mathrm{id},1))\subseteq [x,y+1] &\quad \text{if} \ t=y,\\
\range (\delta\cdot (\mathrm{id},1))\subseteq [x,y] &\quad \text{if} \ t<y.
\end{align*}

Hence for all $t<n-1$ we have $\range(\delta\cdot (\mathrm{id},1)) \subseteq
[0,n-1]$. Now if $t=n-1$ then the cursor of $\delta
(\mathrm{id},1)$ visits the site $n$, thus $\range(\delta\cdot
(\mathrm{id},1))$ is not included in $[0,n-1]$ and therefore
$\delta(\mathrm{id},1)$ does not belong to $F_n$.

A similar argument shows that $\delta(0,-1)$ belongs to $F_n$ if and only if
$t \neq 0$.
Hence $\partial F_n= \left\{\left(\mbff , t \right)\in F_n \mid
t\in \{0,n\} \right\}$ and thus

\begin{equation*}
{\left\vert \partial F_n \right\vert}/{\left\vert F_n \right\vert}=
{2}/{n} \underset{n \rightarrow \infty}{\longrightarrow} 0.
\end{equation*}

§.§ From the isoperimetric profile to the

We saw how to define a diagonal product from two sequences $(k_m)_m$ and
$(l_m)_m$. In this section we recall the definition given in <cit.> of a Brieussel-Zheng group from its isoperimetric profile.
We conclude with some useful results concerning the metric of these groups.

§.§.§ Definition of the
diagonal product

Recall that in the particular case of expanders (see <ref>) a
Brieussel-Zheng group is uniquely determined by the sequences
$(k_m)_{m\in \bN}$ and $(l_m)_{m\in \bN}$ (where $l_m$ corresponds to the
diameter of $\Gs$).
Thus, starting from a prescribed function $\rho$, we will define sequences
$(k_m)_{m\in \bN}$ and $(l_m)_{m\in \bN}$ such that the corresponding $\BZ$
verifies $\profile_{\BZ} \simeq \rho \circ \log$.

\begin{equation*}
\calC:=\left\{ \zeta :[1,+\infty) \rightarrow [1,+\infty) \ \left\vert
\begin{matrix}
\zeta \ \text{continuous}, \ \ \zeta(1)=1 \\
\zeta \ \text{and} \ x\mapsto x/\zeta(x) \text{non-decreasing}
\end{matrix} \right.
\right\}.
\end{equation*}

Equivalently this is the set of functions $\zeta$ satisfying $\zeta(1)=1$ and

\begin{equation}
\label{Rq:IneqTildeRho}
\left( \forall x,c\geq 1 \right) \quad \zeta(x)\leq \zeta(cx) \leq c\zeta(x).
\end{equation}

So let $\rho \in \calC$.
Combining <cit.> we can show the following result (remember that with our
convention the isoperimetric profile considered in [3] corresponds to

Let $\kappa, \lambda \geq 2$. For any $\rho \in \calC$ there exists a
subsequence $(k_m)_{m\in \bN}$ of $(\kappa^n)_{n\in \bN}$ and a subsequence
$(l_m)_{m\in \bN}$ of $(\lambda^n)_{n\in \bN}$ such that the group $\BZ$ defined in
<ref> verifies $\profile_{\BZ}(x) \simeq \rho \circ \log$.

[<cit.>] Let $\alpha >0$. If
$\rho(x):=x^{1/(1+\alpha)}$ then the diagonal
product $\BZ$ defined by $k_m=\kappa^m$ and $l_m=\kappa^{\alpha m}$ verifies
$I_{\BZ}\simeq \rho \circ \log$.

§.§.§ Technical tools

We recall the intermediate functions
defined in <cit.> and some of their properties.

Let $\rho \in \calC$ and let $f$ such that $\rho(x)=x/f(x)$. The construction of
a group corresponding to the given isoperimetric profile $\rho \circ \log$ is
based on the approximation of $f$ by a piecewise linear function $\bar{f}$.
For the quantification of orbit equivalence, many of our computations will use
$\bar{f}$ and some of its properties. We recall below all the
needed results, beginning with the definition of $\bar{f}$.

Let $\rho\in \calC$ and $f$ such that $\rho(x)=x/f(x)$.
Let $(k_m)$ and $(l_m)$ given by <ref> above and $\BZ$ the
corresponding diagonal product. The function
$\bar{f}$ defined by

\begin{equation}\label{Def:Tilde}
\bar{f}(x):= \begin{cases}
l_m & \text{if} \ x\in [k_ml_m,k_{m+1}l_m],\\
\frac{x}{k_{m+1}} & \text{if} \ x\in [k_{m+1}l_m,k_{m+1}l_{m+1}],
\end{cases}
\end{equation}

verifies $\bar{f} \simeq f$. In particular the map $\rhoaff$ defined by
$\rhoaff(x)= x/\bar{f}(x)$ verifies $\rhoaff\simeq \rho$.

If $\rho(x)=x$ then $f(x)=1$ leads to $l_m=1$ for all $m$ and
$k_m=+\infty$ for all $m\geq 1$. In this case $\BZ=(A\times B)\wr \bZ$.

Remark that both $\bar{f}$ and $\rhoaff$ belong to $\calC$. In particular
they verify <ref>, which is only true when $c$ and
$x$ are greater than $1$. When $c<1$ we get the following inequality.

If $0< c^\prime <1$ and $x^\prime \geq 1/c^\prime$ then $c^\prime\rhoaff(x^\prime)
\leq\rhoaff(c^\prime x^\prime )$.

If $0< c^\prime <1$ then $1/c^\prime >1$, thus we can apply <ref>
with $c=1/c^\prime$ and $x=c^\prime x$ to obtain

\begin{equation*}
\rhoaff(x^\prime) =\rhoaff\left(\frac{1}{c^\prime} c^\prime x^\prime\right)
=\rhoaff(cx) \leq c\rhoaff(x)
= \frac{1}{c^\prime}\rhoaff(c^\prime x^\prime).
\end{equation*}

§.§.§ Metric

We recall here some useful material about the metric of $\BZ$ and refer to
<cit.> for more details.
First, let $(x)_+:=\max\{x,0\}$.

For $j\in \bZ$ and $m\in \bN$ let $I^m_{j}:=[jk_m/2,(j+1)k_m/2-1]$.

Let $f_m$ : $\bZ \rightarrow
\Gamma_m$. The essential contribution of $f_m$ is defined as

\begin{equation*}
E_m(f_m):= k_m \sum_{j: \range(f_m,t)\cap I^m_j \neq \emptyset}
\max_{x\in I^m_j} \left( |f_m(x)|_{\Gamma_m}-1 \right)_{+}.
\end{equation*}

The following proposition sums up <cit.>.

For any $\delta=(\mbff,t) \in \BZ$ we have

\begin{align*}
\left\vert {(\mbff,t)}\right\vert_{\BZ}
& \leq 500 \sum^{\landing\left( \range(\delta) \right)}_{m=0} |(f_m,t)|_{\BZ_m},\\
|(f_m,t)|_{\BZ_m} &\leq 9 \big( \range(f_m,t) + E_m(f_m)  \big).
\end{align*}

§ FOLNER TILING SHIFTS

We start by recalling some material of [4] about Følner tiling
shifts and then construct such a tiling for diagonal products.

§.§ Følner tiling shifts

The tools we are going to use to build orbit equivalence are Følner tiling
shifts[Delabie et al. [4] use the term “Folner tiling
sequence”. We chose to call $(\Sigma_n)_n$ a tiling shift in
order to avoid confusion with usual Følner sequences.]. These
sequences lead to Følner sequences defined recursively: the term of rank $(n+1)$ is
composed of a finite number of translates of the $n$-th term of the sequence.

Følner tiling shift
Let $G$ be an amenable group and $(\Sigma_n)_{n\in \bN}$ be a sequence of finite subsets
of $G$. Define by induction the sequence $(T_n)_{n\in \bN}$ by $T_0:=\Sigma_0$ and
$T_{n+1}:=T_n\Sigma_{n+1}$. We say that $(\Sigma_n)_{n\in \bN}$ is a (left) Følner
tiling shift if

* $(T_n)_{n\in \bN}$ is a left Følner sequence, viz. $\lim_{n \rightarrow \infty} {|gT_n \backslash T_n|}/{|T_n|} = 0$ for all $g\in G$;

* $T_{n+1}= \sqcup_{\sigma \in \Sigma_{n+1}} \sigma T_n$.

We call $\Sigma_n$ the set of shiftsShifts and $(T_n)_{n\in
\bN}$ the tilesTiles.

We can also consider right Følner tiling shifts, that is to say sequences
$(\Sigma_n)_n$ such that $T_{n+1}:=\Sigma_{n+1}T_n$ defines a right Følner

Let $S$ be a generating part of $G$. We say that $(\Sigma_n)_{n\in \bN}$ is a
$(R_n,\varepsilon_n)$-Folner tiling shift if for all $n$ we have

\begin{equation*}
\diam{T_n} \leq R_n, \qquad \left\vert sT_n\backslash T_n \right\vert
\leq \varepsilon_n |T_n| \quad (\forall s\in S).
\end{equation*}

Delabie et al. obtained in [4] the two following examples.

If $G=\bZ$ the sequence defined by $\Sigma_{n+1}:= \left\{ 0,2^n
\right\}$ is a $(2^n,2^{1-n})$-Følner tiling shift and the sequence $(T_n)$
thus defined verifies $T_n=[0,2^{n}-1]$.

If $G=(\bZ/2\bZ) \wr \bZ$ then the sequence $(\Sigma_n)_{n\in \bN}$ defined by

\begin{equation*}
\begin{cases}
\Sigma_{0}&:=\left\{ (f,0)\in G \mid \supp(f)\subseteq \{0,1\} \right\},\\
\Sigma_{n+1}&:= \left\{ (f,0)\in G \mid \supp(f)\subseteq [2^n,2^{n+1}-1] \right\}\\
&\phantom{:=}\cup \big\{ (f,2^n)\in G \mid \supp(f)\subseteq [0,2^{n}-1] \big\},
\end{cases}
\end{equation*}

is a right $(3\cdot 2^{n},2^{-n})$-Følner tiling shift. Moreover the
tiling $(T_n)_{n\in \bN}$ thus defined verifies $T_n=\left\{(f,m)\in G \mid
\supp(f)\subseteq [0,2^{n}-1], \ m \in [0,2^{n}-1] \right\}$.

In [4] the authors used Følner tiling shifts to build an explicit
orbit equivalence coupling between two amenable groups and quantify its
integrability. Indeed if $G$ admits a
Følner tiling shift $(\Sigma_n)_{n\in \bN}$ then we can define $X:=\prod_{n\in \bN}
\Sigma_n$ and endow it with an action of $G$. Up to measure zero, two elements of
$X$ will be in the same orbit under that action if and only if they differ by
a finite number of indices. The equivalence relation thus induced is called the
cofinite equivalence relation. Now if $G^\prime$ admits a Følner
tiling shift $(\Sigma^{\prime}_n)_{n\in \bN}$ verifying
$|\Sigma_n|=|\Sigma^{\prime}_n|$ for all integer $n$, then there exists a
natural bijection between $X$ and $X^{\prime}:=\prod_{n\in
\bN}\Sigma^{\prime}_n$ which preserves the cofinite equivalence relation. That
is to say $G$ and $H$ are orbit equivalent.
Furthermore they showed that if we know
the diameter and the ratio of elements in the boundary of each tile, then we can deduce
the integrability of the coupling. This is what the following proposition sums up.

[<cit.>]  Let $G$ and
$G^\prime$ be two discrete amenable groups and let $(\Sigma_n)_n$ be an
$(\varepsilon_n,R_n)$-Følner tiling shift for $G$ and
$(\Sigma^\prime_n)_n$ be an $({\varepsilon^\prime}_n,{R^\prime}_n)$-Følner
tiling shift for ${G^\prime}$.

If $|\Sigma_n|=|{\Sigma^\prime}_n|$, then the groups are orbit equivalent over
$X=\prod_{n\in \bN} \Sigma_n$. Moreover if $\varphi:\bR_+ \rightarrow \bR_+$
is a non-decreasing map such that the sequence $\big( \varphi(2
{R^\prime}_n)\left( \varepsilon_{n-1} -\varepsilon_n \right) \big)_{n\in
\bN}$ is summable, then the coupling from $G$ to $G^\prime$ is

Using this tiling technique and the above theorem, Delabie et al.
[4] obtained the first point of <ref> and the two
following quantifications.

For all $n$ and $m$
there exists an orbit equivalence coupling from $\bZ^m$ to $\bZ^n$ which is
$(\varphi_{\varepsilon},\psi_{\epsilon})$-integrable for every $\varepsilon
>0$ where

\begin{equation*}
\varphi_{\varepsilon}(x)=\frac{x^{n/m}}{\log(x)^{1+\varepsilon}} \quad
\psi_{\varepsilon}(x)=\frac{x^{m/n}}{\log(x)^{1+\varepsilon}}.
\end{equation*}

Remark that in particular
for all $p<n/m$ and $q<m/n$ there exists a $(L^p,L^{q})$-orbit equivalence
coupling from $\bZ^m$ to $\bZ^n$.

Let $m\geq 2$. There exists an orbit equivalence coupling from $\bZ$ to
$\bZ/m\bZ\wr \bZ$ that is $(\exp, \varphi_{\varepsilon})$-integrable for all
$\varepsilon>0$ where

\begin{equation*}
\varphi_{\varepsilon}(x)=\frac{\log(x)}{\log(\log(x))^{1+\varepsilon}}.
\end{equation*}

Note that the above example corresponds to the case when $\rho(x)=x$ in our

§.§ Følner tiling shifts of diagonal products

Let $(k_m)_m$ and $(l_m)_m$ be two sequences verifying the conditions of
(H) and consider $\BZ$ the associated diagonal product (see
<ref>). We define below a Følner tiling shift for $\BZ$. Our goal
is to obtain a tiling verifying $T_n=F_{\kappa^n}$. After defining the shifts sets
$\Sigma_n$ we prove that the sequence $(\Sigma_n)_{n\in \bN}$ is actually a Følner
tiling shift. Finally we make this last statement precise by
computing $(R_n)_{n\in \bN}$ and $(\varepsilon_n)_{n\in \bN}$ such that
$(\Sigma_n)_{n\in \bN}$ is a
$(R_n,\varepsilon_n)$-Følner tiling shift (see <ref>).

§.§.§ Definition of the shifts

For any $n\in \bN$, let $\Landing(n)=\landing(\kappa^n-1)$, that is to say $\Landing(n)$ is the
integer such that $k_{\Landing(n)}\leq {\kappa}^{n}-1 <k_{\Landing(n)+1}$. For
example if $k_n:=\kappa^n$ for all $n\in \bN$, then $\Landing(n)=n-1$.

Before defining our sequence $(\Sigma_n)_{n\in \bN}$, let us show some
practical results on $\Landing$. First remark that
since $(k_n)_{n\in \bN}$ is a subsequence of $(\kappa^n)_{n\in \bN}$, it
verifies $k_n\geq \kappa^n$ for all $n\in \bN$. Thus $\Landing(n)\leq n$ and

\begin{equation*}
k_{\Landing (n)} < \kappa^n \leq k_{\Landing(n)+1}.
\end{equation*}

Let $n\geq 0$, then either $\Landing(n+1)=\Landing (n)$ or
$\Landing(n+1)=\Landing (n)+1$. Moreover in this second case

Recall that by definition $\Landing (m)=\max
\left\{i\in \bN\, | \, k_i \leq \kappa^m-1 \right\}$ for all $m\in \bN$.

$n\in \bN$, then $\Landing(n+1)\geq \Landing (n)$. Moreover if $k_{\Landing
(n)+1}\geq \kappa^{n+1}$ then $\Landing(n+1)< \Landing(n)+1$. That is to say
$\Landing(n+1)\leq \Landing(n)$ and thus $\Landing(n+1)=\Landing (n)$.

On the contrary, if $k_{\Landing(n)+1}<\kappa^{n+1}$ then $\Landing(n+1)\geq
\Landing (n)+1$. But, by definition of $\Landing (n)$ it verifies $k_{\Landing
(n)+1}\geq \kappa^{n}$ and by
construction of $(k_m)_{m\in \bN}$ we also have $k_{\Landing(n)+2} \geq \kappa
k_{\Landing (n)+1}$ thus $k_{\Landing(n)+2} \geq \kappa^{n+1}$. Hence
$\Landing(n+1)<\Landing(n)+2 $ and the first assertion.

Finally if $\Landing(n+1)=\Landing(n)+1$ then by definition of $\Landing$

\begin{equation*}
k_{\Landing(n)} < \kappa^n \leq k_{\Landing(n)+1}=k_{\Landing(n+1)} \leq \kappa^{n+1}-1.
\end{equation*}

But $(k_m)_{m\in \bN}$ is a subsequence of $\kappa^m$ thus the above
inequality implies $k_{\Landing(n+1)}=\kappa^n$.

Now, let us define the shifts. First let $\Sigma_0:=F_0$, then if $n\geq 0$
we distinguish two cases depending on whether $\Landing(n+1)=\Landing (n)$ or
$\Landing(n+1)=\Landing (n)+1$ and in both cases
we split the set of shifts $\Sigma_{n+1}$ in $\kappa$ parts.

If $\Landing(n+1)=\Landing (n)$, let for all $j \in \{0, \ldots,\kappa -1\}$

\begin{equation*}
\Sigma^j_{n+1}
:=\left\{  \left( \mbfg, j\kappa^n \right)\in \BZ \ \left\vert \
\supp \left(g_0\right) &\subseteq
\left[ 0, j \kappa^n-1 \right] \cup \left[ (j+1)\kappa^n, \kappa^{n+1}-1 \right],\\
\forall m \in [1,\Landing (n)] & \\
\supp \left(g^\prime_m\right)& \subseteq
\big[k_m, j \kappa^n + k_m -1\big] \cup \big[(j+1)\kappa^n, \kappa^{n+1}-1 \big],\\
\forall m \notin [0,\Landing (n)]& \\
\supp \left(g^\prime_{m}\right)
& =  \emptyset.
\end{aligned}}
\right. \right\}.
\end{equation*}

Now if $\Landing(n+1)=\Landing (n)+1$, we add the condition that $g^\prime_{\Landing(n)+1}$ has support
contained in $[k_{\Landing(n+1)},\kappa^{n+1}-1]$, namely for all $j \in \{0, \ldots,\kappa -1\}$

\begin{equation*}
\Sigma^j_{n+1}
:=\left\{  \left( \mbfg, j\kappa^n \right)\in \BZ \ \left\vert \
\supp \left(g_0\right)& \subseteq
\left[ 0, j \kappa^n-1 \right]
\cup \left[ (j+1)\kappa^n, \kappa^{n+1}-1 \right]\\
\forall m \in [1,\Landing (n)]&\\
\supp \left(g^\prime_m\right)&\subseteq
\big[k_m, j \kappa^n + k_m -1\big]%\\
\cup \big[(j+1)\kappa^n, \kappa^{n+1}-1 \big],\\
\supp \left(g^\prime_{\Landing (n)+1}\right)
&\subseteq\left[k_{\Landing (n)+1}, \kappa^{n+1}-1 \right], \\
\forall m \notin [0,\Landing (n+1)] & \\%\\
\ \supp \left(g^\prime_{m}\right)
&= \emptyset.
\end{aligned}}
\right. \right\}.
\end{equation*}

Finally, in both cases we define $\Sigma_{n+1}:= \cup^{\kappa -1}_{j=0} \Sigma^j_{n+1}$.

Let $(\mbfg,t)$ be an element of some $\Sigma^j_{n+1}$. We represent in
<ref> the supports and the sets where the maps $g_0, {g^\prime}_1,
\ldots, {g^\prime}_{\Landing (n+1)}$ take their
values. The light-blue rectangle with dotted outline is in
$\Sigma^j_{n+1}$ if and only if $\Landing(n+1)=\Landing(n)+1$.

Support and values taken by $(\mbfg,t) \in \Sigma^j_{n}$

Now that we have the shifts sequence, let us turn to the definition of the

§.§.§ Tiling

Recall that $(F_n)_{n\in \bN}$ denotes the Følner sequence of $\BZ$ defined in
<ref>. The aim of this section is to show the theorem below.

The sequence $(\Sigma_n)_{n\in\bN}$ defined in <ref> is a
Følner tiling shift of $\BZ$.

Before showing that the sequence of tiles $(T_n)_{n\in \bN}$ thus induced
verifies indeed the conditions of <ref>, let us show the following lemma.

The sequence $(T_n)_{n\in \bN}$ defined by $T_0:= F_0$ and $T_{n+1}:=
\Sigma_{n+1} T_n$ for all $n>0$  verifies

\begin{equation*} \big(\forall n \in \bN \big) \quad
\end{equation*}

Let us discuss the idea of the proof. We proceed by induction and use a double
inclusion argument to prove the induction step. To
show that $\Sigma_{n+1}T_n$ is included in $F_{\kappa^{n+1}}$ we rely on
<ref>, that is to say we verify that every element of
$\Sigma_{n+1}T_n$ has range included in $[0,\kappa^{n+1}-1]$. For the reversed
inclusion we consider an element $(\mbfh,t)$ of $F_{\kappa^{n+1}}$ and make the
elements $(\mbfg,j\kappa^n)$ of $\Sigma_{n+1}$ and $(\mbff,t^{\prime})$ of $T_n$ explicit
such that $(\mbfh,t)=(\mbfg,j\kappa^n)(\mbff,t^{\prime})$.

Mind the involved maps here: we study the values of $g_m$ and $f_m$ instead of the
“derived” functions $g^{\prime}_m$, $f^{\prime}_m$ usually considered.

The assertion is true for $T_0$. Now let $n\geq 0$ and assume that
$T_{n}=F_{\kappa^n}$. We show the induction step by double inclusion.

First inclusion

Let us prove that  $\Sigma_{n+1}T_{n} \subseteq F_{\kappa^{n+1}}$. Recall that
\Sigma^j_{n+1}$.

Let $\left( \mbff , t \right) \in T_n$ and $j \in \{0,
\ldots, \kappa-1\}$. Take $ \left(\mbfg , j\kappa^{n} \right) \in
\Sigma^j_{n+1} $, then the following product

\begin{equation*}
\left(\mbfg ,  j\kappa^n\right)\left(\mbff, t \right)
=  \Big( \big(g_mf_m\left(\cdot -j\kappa^n\right)\big)_m,
t+ j\kappa^n \Big)
\end{equation*}

verifies $t + j\kappa^n \in
\left[j \kappa^n,\kappa^{n}-1 + j\kappa^n \right]$ which is contained in $\left[
0,\kappa^{n+1}-1 \right]$ since $j\leq \kappa-1$. Moreover

\begin{equation*}
\begin{cases}
g_0(x)& \text{if} \ x \in  \left[ 0, j \kappa^n \right]
\cup \left[ (j+1)\kappa^n, \kappa^{n+1}-1 \right]\\
f_0(x-j\kappa^n)& \text{if} \ x \in \left[j\kappa^n,(j+1)\kappa^n-1 \right] \\
0 & \text{else}.
\end{cases}
\end{equation*}

Thus $\supp(g_0f_0(\cdot- j\kappa^n)) \subseteq \left[0,\kappa^{n+1}-1\right]$.

Furthermore, for all $m \in \{1, \ldots,\Landing (n)\}$

\begin{align*}
\supp(g^\prime_m)
&\subset \left[k_m, j \kappa^n + k_m -1\right]
\cup \left[(j+1)\kappa^n, \kappa^{n+1}-1 \right]\\
\supp\left(f^\prime_m\left(\cdot-j\kappa^n \right) \right)
&\subseteq [j\kappa^n +k_m,(j+1)\kappa^{n}-1],
\end{align*}

hence by <ref> the support of ${\left(
g_mf_{m}(\cdot-j\kappa^m) \right)}^\prime$ is contained in $[k_m,\kappa^{n+1}-1]$.

Now if $\Landing(n+1)=\Landing (n)+1$ consider $m=\Landing
(n)+1$. In that case $f^{\prime}_m\equiv \neutre$ since $m>\Landing(n)$.
Thus ${\left(
g_{m}f_{m}(\cdot-j\kappa^m) \right)}^\prime = g^\prime_n$ whose support is
contained in $[k_{\Landing(n)+1}, \kappa^{n+1}-1]$.

Finally ${\left( g_{m}f_{m}(\cdot-j\kappa^m)\right)}^\prime \equiv 0$ for all $m \notin [0,
\Landing(n+1)]$.
Hence by <ref> the product $\left(\mbfg,j\kappa^n
\right)(\mbff,t)$ has range included in $\left[0,\kappa^{n+1}-1\right]$ and thus belongs to $

Second Inclusion

Let us show that $F_{\kappa^{n+1}}$ is contained in $\Sigma_{n+1}T_{n}$. So take
$\left(\mbfh,t \right) $ in $ F_{\kappa^{n+1}}$. We want to define $\big(\mbff,
t^\prime \big) \in T_{n}$ and $\left(
\mbfg, j\kappa^n \right) \in \Sigma_{n+1}$ such that
$\left(\mbfg, j\kappa^n\right)\left(\mbff, t^\prime\right) =
\left(\mbfh,t \right)$.
First remark that $t<\kappa^{n+1}$ since $\left(\mbfh,t \right)$ belongs to
$F_{\kappa^{n+1}}$. Thus there exists $t_0, \ldots, t_n$ in $[0,\kappa-1]$
such that $t=\sum^n_{i=0}t_i \kappa^i$. Let $j=t_n$ and $t^\prime =
\sum^{n-1}_{i=0}t_i \kappa^i$. Then $j$ does belong to $[0,\kappa-1]$ and
$t^\prime$ to $[0,\kappa^n-1]$. We now have to define $\mbff$ and $\mbfg$ such

\begin{equation*}
\left( {\left( g_m f_m\left( \cdot -j\kappa^n \right) \right)}_m
, t^\prime +j\kappa^n  \right)
= \big(\mbfh,t \big).
\end{equation*}

We refer to <ref> for an illustration of the different
supports. Let

\begin{align*}
f_0(x)&:= \begin{cases}
h_0(x+j\kappa^n) &\text{if} \ x\in [0,\kappa^n-1],\\
\neutre & \text{else,}
\end{cases} \\
g_0(x)&:= \begin{cases}
h_0(x) &\text{if} \ x\in [0,j\kappa^n-1] \cup [(j+1)\kappa^n, \kappa^{n+1}-1],\\
\neutre & \text{else.}
\end{cases}
\end{align*}

One can verify immediately that $ g_0 f_0\left( \cdot -j\kappa^n \right)=h_0$.
Then take $m \in [1,\Landing (n)]$ and let

\begin{align*}
&:= {\begin{cases}
{h^\prime}_m(x+j\kappa^n) &\text{if} \ x\in [k_m,\kappa^n-1],\\
\neutre & \text{else,}
\end{cases} }\\
&:= {\begin{cases}
{h^\prime}_m(x) & \text{if}\ x \in [k_m,j\kappa^n+k_m-1]
\cup [(j+1)\kappa^n,\kappa^{n+1}-1]\\
\neutre &\text{else.}
\end{cases}}
\end{align*}

Now if $\Landing(n+1)=\Landing(n)+1$ then $k_{\Landing(n+1)}\geq
\kappa^n$ and in that case define
Finally let ${f^\prime}_{\Landing(n+1)}\equiv \neutre$ and if
$m> \Landing(n+1)$ let $ {g^\prime}_m\equiv \neutre \equiv {f^\prime}_m$.

With the above definitions $\mbff$ and $\mbfg$ are uniquely defined. Moreover,
by definition $(\mbfg,j\kappa ^n)$ belongs to
$\Sigma^j_{n+1}$ and by <ref> we have $\range(\mbff,t) \subseteq
[0,\kappa^n-1]$ thus $(\mbff,t^\prime)$ belongs to $T_{n}$.
Now, using <ref> we verify that $g_mf_m(\cdot
-j\kappa^n)=h_m$ thus $(\mbfh,t) \in \Sigma_{n+1}T_{n}$.

Hence, combining the first and second inclusion we get $F_{\kappa^{n+1}}=T_{n}$.

We now know that $(T_n)_{n\in \bN}$ is a Følner sequence. To prove <ref> we have
to show that $(\Sigma_n)_{n\in \bN}$ a Følner tiling shift.

The sequence $(T_n)_{n\in \bN}$ is a Følner sequence, by the last lemma. Thus we
only have to show that for all $\sigma \neq \tilde{\sigma} \in \Sigma_{n+1}$,
$\sigma T_{n}  \cap \tilde{\sigma} T_{n}=\emptyset$. So let us denote by
$(\mbfh,t)$ an element of $\sigma T_{n}  \cap \tilde{\sigma} T_{n}$. We
distinguish two cases.

First if $\sigma \in \Sigma^j_{n+1}$ and $\tilde{\sigma} \in
\Sigma^i_{n+1}$ for some $i \neq j$, then the cursor of $\sigma$ is equal to
$j\kappa^n$ and the one of $\tilde{\sigma}$ to $i\kappa^n$. Thus

\begin{align*}
\left(\mbfh, t\right) \in \sigma T_{n} & \Rightarrow t \in [j \kappa^n, (j+1)\kappa^n-1],\\
\left(\mbfh, t\right) \in \tilde{\sigma} T_{n} & \Rightarrow t \in [i\kappa^n,(i+1)\kappa^{n}-1].
\end{align*}

But since $i \neq j$ these two intervals are disjoint, thus $\sigma T_{n} \cap
\tilde{\sigma} T_{n}=\emptyset$.
Now fix $j\in \{0, \ldots, \kappa-1\}$ and take $\sigma, \tilde{\sigma} \in
\Sigma^{j}_{n+1}$. Let $\sigma:=\left(\mbfg ,j \kappa^n\right)$ and
$\tilde{\sigma}:=\left(\tilde{\mbfg} ,j\kappa^n\right)$. Assume
that there exists $\left( \mbff,t \right), \left( \tilde{\mbff}, \tilde{t}
\right) \in T_{n}$ such that $(\mbfg,j\kappa^n) \left( \mbff,t \right)= (\tilde{\mbfg},j\kappa^n)
\left( \tilde{\mbff}, \tilde{t}\right)$. Then

\begin{equation} \label{eq:fgtiling}
\forall m \in \bN,\ \forall x \in \bZ
\quad g_mf_m(x- j\kappa^n)=\tilde{g}_m(x)\tilde{f}_m(x - j\kappa^n).
\end{equation}

First remark that

\begin{align*}
\sigma, \tilde{\sigma} \in \Sigma^j_{n+1}
\quad  &\Longrightarrow  \quad \supp(g_0), \, \supp(\tilde{g}_0)
\subseteq [0,j\kappa^n-1] \cup [(j+1)\kappa^n, \kappa^{n+1}-1]\\
\left( \mbff,t \right), \left( \tilde{\mbff}, \tilde{t}\right) \in T_{n}
\quad &\Longrightarrow \quad  \supp({f_0(\cdot - j\kappa^n)}),\,
\supp(\tilde{f}_0(\cdot - j\kappa^n)) \subseteq [j\kappa^n, (j+1)\kappa^n-1].
\end{align*}

In other word the support of $g_0$ (resp $\tilde{g}_0$) is disjoint from
the one of $f_0(\cdot - j\kappa^n)$ (resp ${\tilde{f}}_0(\cdot - j\kappa^n)$).
Combining this with <ref> we obtain that $g_0=\tilde{g}_0$ and

Now let $m>0$ and let us show that $g_m=\tilde{g}_m$. Due to supports overlap
(see <ref>) we need to decompose
$[0,\kappa^{n+1}-1]$ in five subintervals, namely

\begin{align*}
\Big[ 0, \kappa^{n+1}-1 \Big] =
&\Big[ 0, j\kappa^n-1 \Big]\sqcup
\Big[ j\kappa^n, j\kappa^n + k_m -1 \Big] \sqcup
\Big[ j\kappa^n + k_m, (j+1)\kappa^n -1 \Big],\\
\Big[(j+1)\kappa^n,(j+1)\kappa^n + k_m -1 \Big]\sqcup
\Big[(j+1)\kappa^n + k_m,\kappa^{n+1} -1 \Big].
\end{align*}

If $x\leq j\kappa^n-1$ or $x\geq (j+1)\kappa^n+k_m$, then
$f_m(x-j\kappa^n)=\neutre = \tilde{f}_m(x-j \kappa^n)$
and thus $g_m(x)=\tilde{g}_m(x)$ by <ref>.

If $x\in [j\kappa^n, j\kappa^n + k_m -1]$ then using <ref> and the fact
that on that subinterval $f_0=\tilde{f}_0$, we get

\begin{equation*}
f_m(x-j\kappa^n) = \theta^A_0\left( f_0\left( x-j\kappa^n\right) \right)
= \theta^A_0\left( {\tilde{f}}_0\left( x-j\kappa^n\right) \right)={\tilde{f}}_m(x-j\kappa^n).
\end{equation*}

Hence by <ref> we get $g_m(x)=\tilde{g}_m(x)$.

If $x$ belongs to $ [j\kappa^n+k_m, (j+1) \kappa^n -1]$ then
$g_m(x)=\tilde{g}_m(x)=\neutre$ and thus <ref> implies that
$f_m(x-j\kappa^n)=\tilde{f}_m(x-j\kappa^n)$, that is to say $f_m$ and
${\tilde{f}}_m$ coincide on $[k_m, \kappa^n-1]$.

Finally if $x\in [(j+1)\kappa^n,(j+1)\kappa^n+k_m-1]$ then using <ref> and the fact
that $f_0=\tilde{f}_0$ on that subinterval, we get

\begin{equation*}
f_m(x-j\kappa^n) = \theta^B_0\left( f_0\left( x-j\kappa^n-k_m \right) \right)
= \theta^B_0\left( {\tilde{f}}_0\left( x-j\kappa^n-k_m \right) \right)={\tilde{f}}_m(x).
\end{equation*}

Hence by <ref>, we have $g_m(x)=\tilde{g}_m(x)$.

Thus $\mbfg=\tilde{\mbfg}$ and then $\sigma=\tilde{\sigma}$. Which concludes
the proof of the theorem.

Supports overlap

§.§.§ Diameter and boundary

Let us now quantify our shifts sequence.

The sequence $(\Sigma_n)_{n\in \bN}$ defined in <ref> is a
$(R_n,\varepsilon_n)$-Følner tiling shift where

\begin{equation*}
R_n=\Cdiam \kappa^n l_{\Landing (n)} \qquad \varepsilon_n=\frac{2}{\kappa^{n}},
\end{equation*}

for some strictly positive constant $\Cdiam$.

First we prove the following lemma.

There exists $\Cdiam>0$ depending only on $\BZ$ such that
$\diam{F_n} \leq \Cdiam n l_{\landing(n-1)}$  for all $n\in \bN$.

To show this result, we use <ref>.

Let $n \in \bN$ and $(\mbff,t) \in F_n$.
First, take $m\leq \landing(n-1)$ and let us bound $E_m$ by above. Recall that
$I^m_j=[jk_m/2,(j+1)k_m/2-1]$. Since $(f,t)$ belongs to $F_{n}$ its range is
included in $[0,n-1]$, thus

\begin{align*}
&\left\vert\left\{j \in \bZ: \range(f_m,t)\cap I^m_j \neq \emptyset\right\} \right\vert\\
\leq&
\left\vert \big\{j \in \bZ:\, [0,n-1]\cap [jk_m/2,(j+1)k_m/2-1]
\neq \emptyset\big\} \right\vert,\\
\leq& \left\vert \big\{j \in \bZ:\, jk_m/2\leq n -1\
\text{and}\ (j+1)k_m/2\geq 1 \big\} \right\vert,\\
\leq& \frac{2(n-2)}{k_m} + 1.
\end{align*}

Moreover remark that $|f_m(x)|_{\Gamma_m} \leq \diam{\Gamma_m} \leq \Clm l_m$
for all $x$, thus

\begin{align*}
&=  k_m \sum_{j: \range(f_m,t)\cap I^m_j \neq \emptyset}
\max_{x\in I^m_j} \left( |f_m(x)|_{\Gamma_m}-1 \right)_{+},\\
&\leq k_m  \sum_{j: \range(f_m,t)\cap I^m_j \neq \emptyset} l_m,\\
&\leq k_m l_m \left(\frac{2(n-2)}{k_m}+1\right)=l_m(2(n-2)+k_m).
\end{align*}

Thus, applying the second part of <ref> we get

\begin{align*}
|(f_m,t)|_{\BZ_m}\ &\leq\  9 \big( \range(f_m,t) + E_m(f_m)  \big) %,\\
\  \leq\  9\left(n +l_m\big(2(n-2)+k_m\big)\right).
\end{align*}

But if $m\leq\landing(n-1)$ then $k_m\leq n-1 \leq n$ thus we can bound $
|(f_m,t)|_{\BZ_m}$ by above by $9n(3l_m+1)$. Now remark that
$\landing\left(\range(\mbff,t) \right)\leq \landing(n-1)$. Thus,
using the preceding inequality and the first part of <ref>, we

\begin{align*}
\left\vert {(\mbff,t)}\right\vert_{\BZ}
\leq 500 \sum^{\landing\left( \range(\mbff,t) \right)}_{m=0} |(f_m,t)|_{\BZ_m}%,\\
& \leq 500 \sum^{\landing (n-1)}_{m=0} 9n\left(3l_m+1\right),\\
& \leq 4500 n \sum^{\landing (n-1)}_{m=0}\left(3l_m+1  \right)
\end{align*}

Finally, since $l_m$ is a subsequence of a geometric sequence, there exists
$C_l>0$ such that $ \sum^{\landing (n-1)}_{m=0}\left(3l_m+1  \right)\leq C_l
l_{\landing (n-1)}$. Denoting $\Cdiam:=4500 C_l$ we get the lemma.

Let us now show the wanted proposition.

First remark that by the proof of <ref> we have

\begin{equation*}
\varepsilon_n=\frac{|\partial T_n |}{|T_n|} =
\frac{|\partial F_{\kappa^{n}} |}{|F_{\kappa^{n}}|}= \frac{2}{\kappa^n}.
\end{equation*}

Now by <ref> we have $\diam{T_n}=\diam{F_{\kappa^n}}\leq \Cdiam
\kappa^n l_{\Landing(n)}$.

§ COUPLING WITH THE GROUP OF

Our aim in this section is to show <ref>.
What we actually show is that a diagonal product $\BZ$ admits a coupling with
$\bZ$ satisfying <ref>. We start by defining a Følner tiling
shift for $\bZ$ in <ref>. We compute in <ref>
an estimate of the diameter of such tiles, namely the cardinal $|T_n|$.
We conclude by showing the integrability of the coupling
using the criterion given by <ref>. And then show that
$\BZ$ thus considered satisfies <ref>.

§.§ Tiles for the integers

We will denote by $(\SZ_n)_{n\in \bN}$ a Følner tiling shift of $\bZ$ and by
$(\TZ_n)_n$ the corresponding tiles.

Consider $(\Sigma_n)_n$ and $(T_n)_n$ as defined in
<ref> respectively. In order to use
<ref> to get an orbit equivalence coupling between $\bZ$
and $\BZ$ we need $\Sigma_{n+1}$ and ${\SZ}_{n+1}$ to have the same number of
elements. We thus define

\begin{equation}\label{eq:TZn}
\begin{cases}
&\SZ_0=\big[0,|T_0|-1\big] \\
\forall n \in \bN &
\SZ_{n+1}:=\big\{ 0, |T_n|, 2|T_n|, \ldots,
\left( \left\vert {\Sigma_{n+1}}\right\vert - 1 \right)|T_n|\big\}.
\end{cases}
\end{equation}

It induces a sequence $({\TZ}_n)_{n\in \bN}$ defined by $\TZ_0=\SZ_0$ and
$\TZ_{n+1}=\SZ_{n+1}\TZ_n$ for all $n\geq 0$. We are going to prove that
$(\SZ_n)_{n\in \bN}$ is a Følner tiling shift for $\bZ$.

The sequence $(\SZ_n)_{n\in \bN}$ defined by <ref> is a $(\RZ_n, \epsZ_n)$-Følner
tiling shifts for $\bZ$ with

\begin{equation*}
\RZ_n= \left\vert{T_n}\right\vert \qquad \epsZ_n={2}/{\vert T_n \vert}.
\end{equation*}

Moreover the induced sequence $(\TZ_n)_{n\in \bN}$ verifies $\TZ_n=\left[ 0,
\left\vert{T_n}\right\vert -1 \right]$ for all $n \in \bN$.

Let $(\SZ_n)_{n\in \bN}$ be as defined by <ref> and recall that the
induced tiling $(\TZ_n)_{n\in \bN}$ is the sequence defined by $\TZ_0:=\SZ_0$ and
$\TZ_{n+1}=\SZ_{n+1} \TZ_n$ for all $n \in \bN$.
One can easily prove that for all $n\geq 0$

\begin{equation}
\TZ_n= \left[ 0, \left\vert{T_n}\right\vert -1 \right].
\end{equation}

It is now immediate to check that $\diam{\TZ_n}=|T_n|$ and $|\partial \TZ_n
|/|\TZ_n|=2/|T_n|$. Furthermore note that if $\sigma, \sigma^\prime \in
\SZ_{n+1}$ such that $\sigma\neq \sigma^{\prime}$ then
$d_{\bZ}(\sigma,\sigma^\prime)\geq |T_n|=\diam{\TZ_n}$. Thus for such $\sigma$
$\sigma^\prime$ we get $\sigma \TZ_n \cap \sigma^\prime T_n = \emptyset$.
Therefore $(\Sigma_n)_{n\in \bN}$ is a Følner tiling shift and the proposition follows
from the above quantifications on $T_n$.

§.§ Estimates: diameter and boundary

The integrability of the coupling between $\bZ$ and $\BZ$ depends on
$(R_n,\varepsilon_n)$ and $(\RZ_n,\epsZ_n)$ but by the above proposition, that last
couple depends on the value of the cardinality of the tiles $(T_n)_{n\in \bN}$.
The aim of this section is to give
estimates of $|T_n|$ involving only terms of $(k_m)_{m\in \bN}$ and
$(l_m)_{m\in \bN}$.
First let us make the value of $|T_n|$ precise.

The sequence $(T_n)_n$ defined in <ref> verifies

\begin{equation*}
|T_n|=\kappa^n \big( {|A| |B|} \big)^{\kappa^n}
\prod^{\Landing(n)}_{m=1} {\left\vert {\Gamma^\prime}_m \right\vert}^{\kappa^{n}-k_m}.
\end{equation*}

Recall that $T_n=F_{\kappa^n}=\left\{(\mbff, t) \mid \range
\left(\mbff,t\right) \subseteq \{0, \ldots, \kappa^n-1\} \right\}$ for all
$n\in \bN$. We use here <ref> linking range and supports.
Let $n\in \bN$ and take $(\mbff,t)\in T_n$, then there are exactly $\kappa^n$
values of $t$ possible. Moreover $\mbff$ is uniquely determined by $f_0$ and
$f^\prime_1, \ldots, f^\prime_{\Landing(n)}$ (see <ref>). But $f_0$
is supported on $[0,\kappa^n-1]$ which is set of cardinal $\kappa^n$ so
there are exactly $\big( {|A| |B|} \big)^{\kappa^n}$ possible values for
$f_0$. Moreover if $m>0$ then remark that $f^\prime_m$ is supported on
$[k_m,\kappa^n-1]$ which has
$\kappa^n-k_m$ elements so there are exactly
${\left\vert \Gamma^\prime_m \right\vert}^{\kappa^n-k_m}$ possible values for
$f^\prime_m$. Thus the number of elements in $T_n$ is

\begin{equation*}
{\kappa}^{n} \big( {|A| |B|} \big)^{\kappa^n}
\prod^{\Landing(n)}_{m=1} {\left\vert \Gamma^\prime_m \right\vert}^{\kappa^n-k_m}.
\end{equation*}

Now let us bound $|T_n|$ so that the bounds depend only on $(\kappa^m)_{m\in \bN}$ and
$(l_m)_{m\in \bN}$.

There exists two constants $C_2, \CTn >0$
such that for all $n \in \bN$,

\begin{equation*}
C_2 \kappa^{n-1}l_{\Landing(n)}\leq \ln |T_n| \leq \CTn \kappa^nl_{\Landing(n)}.
\end{equation*}

Before showing the above proposition let us give an estimate of the right
factor of the expression of $|T_n|$.

There exists two constants $C_1, C_2>0$
such that for all $n \in \bN$,

\begin{equation*}
C_2 \kappa^{n-1} l_{\Landing(n)}
\leq \ln \left(
\prod^{\Landing(n)}_{m=1} {\left\vert {\Gamma^\prime}_m \right\vert}^{\kappa^n-k_m} \right)
\leq C_1\kappa^n l_{\Landing(n)}.
\end{equation*}

Recall that by <ref> there exists $c_1$, $c_2>0$ such that, for all $m$

\begin{equation*}
c_1 l_m -c_2 \leq \ln \left\vert {\Gamma_m}\right\vert \leq c_1l_m + c_2.
\end{equation*}

Since $\Gsp \leq \Gamma_m$ we thus have

\begin{align*}
\ln \left(
\prod^{\Landing(n)}_{m=1} {\left\vert {\Gamma^\prime}_m \right\vert}^{\kappa^n-k_m} \right)
&\leq \sum^{\Landing(n)}_{m=1} \left({\kappa^n-k_m}\right) \ln |\Gamma_m |,\\
&\leq \sum^{\Landing(n)}_{m=1} \left({\kappa^n-k_m}\right)\left( c_1l_m+c_2 \right).%,\\
\end{align*}

But we can bound $\kappa^n-k_m$ from above by $\kappa^n$ and since
$(l_m)_{m\in \bN}$ is a subsequence of a sequence having geometric growth, the sum
$\sum^{\Landing(n)}_{m=1} \left( c_1l_m+c_2 \right)$ is bounded from above by
its last term up to a multiplicative constant. That is to say: there exists
$C_1>0$ such that

\begin{equation*}
\ln \left(
\prod^{\Landing(n)}_{m=1} {\left\vert {\Gamma^\prime}_m \right\vert}^{\kappa^n-k_m} \right)
\leq C_1\kappa^n l_{\Landing(n)}.
\end{equation*}

Hence the upper bound.
Now, using that $[\Gamma_m:\Gsp]=|A||B|$ we have

\begin{equation*}
\ln \left(
\prod^{\Landing(n)}_{m=1} {\left\vert {\Gamma^\prime}_m \right\vert}^{\kappa^n-k_m} \right)
= \sum^{\Landing(n)}_{m=1} \left({\kappa^n-k_m}\right)
\ln|\Gamma^\prime_m|
= \sum^{\Landing(n)}_{m=1} \left({\kappa^n-k_m}\right)
\ln \left(\frac{|\Gamma_{m} |}{|A||B|}\right).
\end{equation*}

Bounding the sum from below by its last term and using once more
<ref>, we get

\begin{align*}
\ln \left(
\prod^{\Landing(n)}_{m=1} {\left\vert {\Gamma^\prime}_m \right\vert}^{\kappa^n-k_m} \right)
&\geq (\kappa^n-k_{\Landing(n)})\ln \left( \frac{|\Gamma_{\Landing(n)} |}{|A||B|} \right) ,\\
&\geq (\kappa^n-k_{\Landing(n)})\left( c_1l_{\Landing(n)}-c_2 -\ln \left( |A||B| \right) \right),\\
&\geq C_2(\kappa^n - k_{\Landing(n)})l_{\Landing(n)},
\end{align*}

for some $C_2>0$. We get the wanted inequality by noting that
$\kappa^n - k_{\Landing(n)}\geq \kappa^{n-1}$.

Applying <ref> to the cardinal of $T_n$ given by
<ref> we obtain that there exists $\CTn>0$ such that $\ln|T_n|
\leq \CTn  \kappa^n l_{\Landing(n)}$. Hence the upper bound.
The minoration comes imediately from <ref>.

Equipped with these bounds on $|T_n|$ we can now show the wanted integrability
for the coupling.

§.§ Integrability of the coupling

We will show that $\BZ$ is the group satisfying <ref>, but
first let us quantify the integrability of the orbit equivalence coupling with
$\bZ$ induced by the Følner tiling shifts we built.
Recall that $\calC$ denotes the set of non-decreasing functions $\rho\colon
[1,+\infty[ \rightarrow [1,+\infty[$ such that $x/\rho(x)$ is non-decreasing.

Let $\rho \in \calC$ and take $\BZ$ to be the Brieussel-Zheng diagonal
product defined from $\rho$.
Let $\varepsilon >0$ and $\Psi := \exp \circ \rho$ and let

\begin{equation*}
\varphi_{\varepsilon}(x):=\frac{\rho\circ \ln(x)}%
{\left( \ln\circ \rho\circ \ln (x) \right)^{1+\varepsilon}}.
\end{equation*}

There exists an orbit equivalence coupling from $\BZ$ to $\bZ$ that is

Let us discuss the strategy of the proof. The demonstration is based on
<ref>, thus we first prove that
$(\Psi(2R_n)\epsZ_{n-1})_n$ is summable and then that
is. In both cases we use <ref> to get upper bounds.
So far, we have the following quantifications.

$R_n$   $=C_R \kappa^n l_{\Landing(n)}$           $\RZ_n$   $= \left\vert{T_n}\right\vert$

$\varepsilon_n$   =${2}{\kappa^{-n}}$           $\epsZ_n$   $={2}/{\left\vert T_n \right\vert}$

Let $\rho \in \calC$
and take $\BZ$ to be the diagonal product defined from $\rho$ as described in

To begin, let us recall some preliminary results about $\rho$.
Remember that $\rho \simeq \rhoaff$ where $\rhoaff$ is defined below
<ref>. By definition of $\Landing(n)$ we have
$k_{\Landing(n)}l_{\Landing(n)} \leq \kappa^n l_{\Landing(n)} \leq
k_{\Landing(n)+1}l_{\Landing(n)}$, thus by <ref>

\begin{equation}\label{eq:tilderhokappan}
\rhoaff(\kappa^n l_{\Landing(n)}) =\kappa^n.
\end{equation}

Now let us show that the coupling from $\bZ$ to $\BZ$ is
$\Psi$-integrable. To do so we prove that
$\left(\Psi(2R_n)\epsZ_{n-1}  \right)$ is summable.
First note that by <ref> we have the following lower bound on $|T_{n-1}|$

\begin{equation}\label{eq:MinorationTnmoinsun}
|T_{n-1}| \geq \exp\left(C_2 \kappa^{n-2} l_{\Landing(n-1)} \right).
\end{equation}

Moreover recall that $R_n=C_R \kappa^nl_{\Landing(n)}$ and $\epsZ_{n-1}=2/|T_{n-1}|$
thus by the inequality above

\begin{align*}
\Psi(2R_n)\epsZ_{n-1}
&=\exp\Big[ \rho(2C_R\kappa^nl_{\Landing(n)}) \Big] \frac{2}{|T_{n-1}|},\\
&\leq 2 \exp
\Big[\rho \left(2C_R\kappa^nl_{\Landing(n)}\right)- C_2\kappa^{n-2}l_{\Landing(n-1)}\Big].
\end{align*}

But remember that $\rho\simeq \rhoaff$. Thus using
<ref> we get

\begin{equation} \label{eq:jaipasdidee}
\rho \left(2C_R\kappa^nl_{\Landing(n)}\right)
\simeq \rhoaff \left(2C_R\kappa^nl_{\Landing(n)}\right)
\leq 2C_R \rhoaff \left(\kappa^nl_{\Landing(n)}\right) = 2C_R \kappa^n.
\end{equation}

Combining the above result with the previous inequality, we get

\begin{align*}
\Psi(2R_n)\epsZ_{n-1}
& \preccurlyeq 2 \exp\left[2C_R\kappa^n-C_2\kappa^{n-2}l_{\Landing(n-1)} \right],\\
& = 2 \exp\left[\kappa^{n-2}\left(2C_R\kappa^2-C_2l_{\Landing(n-1)}\right)\right],
\end{align*}

which is
summable. Indeed $l_{\Landing(n)}$ tends to infinity and thus
$\left(2C_R\kappa^2-C_2l_{\Landing(n-1)}\right)<-1$ for $n$ large enough. Hence by
<ref> the orbit equivalence from $\bZ$ to $\BZ$ si $\Psi$-integrable.

Now, let us show that for all $\varepsilon>0$ the coupling from $\BZ$ to $\bZ$ is
$\varphi_{\varepsilon}$-integrable. Based on <ref> we only
have to prove that $\varphi_{\varepsilon}(2\RZ_n)\varepsilon_{n-1}$ is summable.
Recall that $\RZ_n=|T_n|$ and $\varepsilon_{n-1}=2/\kappa^{n-2}$ and remark
that by both the lower and upper bounds given in <ref>
we have

\begin{equation*}
\varphi_{\varepsilon}(2\RZ_n)\varepsilon_{n-1}
=\frac{2 \rho\circ\ln\big(2|T_n|\big)}%
{\Big( \ln\circ \rho\circ \ln\big(2|T_n| \big)\Big)^{1+\varepsilon}\kappa^{n-1}}
\leq \frac{2 \rho\big(2\CTn \kappa^nl_{\Landing(n)}\big)}%
{\Big(\ln\circ\rho\left(2C_2 \kappa^{n-1}l_{\Landing(n)}\right)\Big)^{1+\varepsilon}\kappa^{n-1}}.
\end{equation*}

Let us give a lower bound for $\rho\left(2C_2
\kappa^{n-1}l_{\Landing(n)}\right)$. Recall that $\rho \simeq \rhoaff$
furthemore if $2C_2\geq 1$ then by <ref> and since
$\rhoaff$ is non-decreasing

\begin{align*}
\kappa^{n-1}=\rhoaff\left(\kappa^{n-1}l_{\Landing(n)}\right)
\rhoaff\left(2C_2 \kappa^{n-1}l_{\Landing(n)}\right)% ,\\
\simeq \rho\left(2C_2 \kappa^{n-1}l_{\Landing(n)}\right).
\end{align*}

Now if $2C_2 <1$ using <ref> with $c^\prime=2C_2$ and
$x^\prime =\kappa^{n-1}l_{\Landing(n)}$ we get (for $n$ large enough)

\begin{equation*}
2C_2 \kappa^{n-1} =2C_2 \rhoaff(\kappa^{n-1}l_{\Landing(n)})
\leq \rhoaff(2C_2\kappa^{n-1}l_{\Landing(n)})
\simeq \rho\left( 2C_2 \kappa^{n-1}l_{\Landing(n)}\right)
\end{equation*}

Hence, in both cases $\kappa^{n-1} \preccurlyeq \rho( 2C_2
\kappa^{n-1}l_{\Landing(n)})$. Finally replacing $C_R$ by $\CTn$ in
<ref> we can show that $\rho
\left(2\CTn \kappa^nl_{\Landing(n)}\right) \leq 2\CTn \kappa^n$. Thus,
combining the two preceding results, we obtain

\begin{align*}
\varphi_{\varepsilon}(\RZ_n)\varepsilon_{n-1}
&\leq \frac{2 \rho\big(\CTn \kappa^nl_{\Landing(n)}\big)}%
{\Big( \ln\circ\rho\left(C_2 \kappa^{n-1}l_{\Landing(n)}\right)\Big)^{1+\varepsilon}\kappa^{n-1}}\\
\end{align*}

which is a summable sequence. Hence by
<ref> the orbit equivalence coupling from $\BZ$ to $\bZ$
si $\varphi_{\varepsilon}$-integrable.

This result is stated in the general case, that is to say for an abstract
$\rho$. Nonetheless, for some particular functions $\rho$ the quantification
can be improved. For example the case where $k_n= 2^n$ and $l_n=2^{\alpha n}$
corresponds to $\rho(x)\simeq x^{1/(1+\alpha)}$. In that case $\Landing(n)=n-1$ and we
can show that the coupling from $\bZ$ to $\BZ$ is $\exp$-integrable (instead of
$\exp\circ \rho$-integrable). Indeed,
let $c_{\varphi}< C_2/(C_R2^{3+\alpha})$ and $\Psi(x):=\exp(c_{\varphi} x)$,
then by <ref>

\begin{align*}
\Psi(2R_n)\epsZ_{n-1} &= \exp\left[c_{\varphi} 2C_Rk_nl_{n-1}\right] \frac{2}{|T_{n-1}|}\\
&\leq \exp \left[c_{\varphi}2C_R2^n2^{\alpha(n-1)}- C_22^{n-2}2^{\alpha(n-2)}\right]
&= 2\exp \left[2^{n-2}2^{\alpha(n-2)} \left(c_{\varphi}C_R2^{3+\alpha}- C_2  \right)\right].
\end{align*}

Which is summable by choice of $c_{\varphi}$.

We can verify that the integrability obtained for the coupling from
$\BZ$ to $\bZ$ is “almost” optimal.
Indeed if the coupling from $\BZ$ to $\bZ$ is
$\varphi$-integrable, then by <ref> we have

\begin{equation*}
\varphi \circ I_{\bZ} \preccurlyeq I_{\BZ}
\end{equation*}

where we recall that $I_{\bZ}(n) \simeq n$ and $I_{\BZ}(n)\simeq
\rho \circ\ln(n)$. Thus using the inequality above, we get $\varphi(n)
\preccurlyeq \rho \circ \ln(n)$. Hence the quantification of
<ref> is optimal up to a logarithmic factor.

It is now easy to prove our first main theorem.

Let $\rho \in \calC$
and $\BZ$ to be the
group defined in <ref>. By the aforementioned proposition it
verifies $\profile_{\BZ}\simeq \rho \circ \log$. Moreover by
<ref> there exists an orbit equivalence coupling from $\BZ$ and
$\bZ$ that is $(\varphi_{\varepsilon},\exp \circ \rho)$-integrable for all

To prove <ref> we use the composition of couplings introduced in
We recall below the proposition concerning the integrability of this composition
and refer to <cit.> for more details on the
construction of the corresponding coupling.

Let $\varphi,\psi:\bR^+\rightarrow \bR^+$ be non-decreasing subadditive maps with
$\varphi$ moreover concave.
If $(X_1,\mu_1)$ (resp. $(X_2,\mu_2)$) is a $(\varphi,L^0)$-integrable (resp.
$(\psi,L^0)$-integrable) orbit equivalence coupling from $\Gamma$ to $\Lambda$
(resp. $\Lambda$ to $\Sigma$), the composition of couplings gives a $(\varphi
\circ \psi,L^0)$-integrable orbit equivalence coupling from $\Gamma$
to $\Sigma$.

Let us now show <ref> concerning the coupling with $\bZ^d$.

Let $d \geq 1$. Let $\rho \in \calC$ and let $\BZ$ be the
group defined in <ref>, in particular it verifies
$\profile_{\BZ}\simeq \rho \circ \log$.
Assume moreover that the map $\varphi_\varepsilon$ defined by $\varphi_{\varepsilon}(x):={\rho\circ
\log(x)}/{\left( \log\circ\rho\circ\log(x) \right)^{1+\varepsilon}}$ is subadditive and concave.

Since $d=1$ is precisely the case of <ref>,
we only have to treat the case of $d\geq 2$. For such a $d$ recall (see <ref>)
that for all $p<d$ and
all $q<1/d$ there exists a $(L^p,L^{q})$-integrable orbit equivalence coupling from
$\bZ$ to $\bZ^d$. In particular taking $p=1$ and $q=0$ gives a $(L^1,L^{0})$-integrable orbit
equivalence coupling from $\bZ$ to $\bZ^d$.
Hence, using the composition of couplings described in [4]
we can deduce from <ref> and <ref> above
that there exists a
$(\varphi_{\varepsilon},L^0)$-integrable orbit equivalence coupling
from $\BZ$ to $\bZ^d$.
Hence the corollary.

We make the hypothesis that $\varphi_\varepsilon$ is subadditive and concave
only in order to use <ref> and the composition of couplings. Building directly a coupling from $\BZ$ to $\bZ^d$ (instead of transiting via $\bZ$)
might allow to remove the aforementioned assumption.

§ CONCLUSION AND OPEN PROBLEMS

Let us conclude with some questions and remarks.

§.§ Optimality and coupling building techniques

The tiling technique —though inspiring— is not always usable to get orbit
equivalence couplings. Indeed the condition that the two Følner tiling shifts
must have at each step the same cardinality is very restrictive. Furthemore this
technique does not seem to produce couplings with the best quantification: wether
it is our coupling with $\bZ$ or the one built in [4]
(<ref>) the integrability is always optimal
up to a logarithmic factor.
One can thus ask: is the optimal integrability reachable? Is the logarithmic
error due to the building technique?

§.§ Inverse problem

We studied here the inverse problem for the group of integers
(<ref>) but one can also ask the same question for other groups
than $\bZ$.

Given a function $\varphi$ and a group $H$
is there a group $G$ such that there exists a $(\varphi,L^0)$-measure
equivalent from $G$ to $H$? Can $G$ be chosen such that $\varphi\circ I_H
\simeq I_G$?

In [7] we answer this question when $H$ is a diagonal
product, in particular $H$ can be a lamplighter group. This coupling is
obtained with another building technique than the tiling process and the
integrability is optimal, answering the questions of
<ref> positively.

[1]
T. Austin.
Integrable measure equivalence for groups of polynomial growth.
Groups, Geometry, and Dynamics, 10:117–154, February 2016.
arXiv:1310.3216 [math.GR].

[2]
U. Bader, A. Furman, and R. Sauer.
Integrable measure equivalence and rigidity of hyperbolic lattices.
Inventiones mathematicae, 194:313–379, 2013.

[3]
J. Brieussel and T. Zheng.
Speed of random walks, isoperimetry and compression of finitely
generated groups.
Ann. of Math., 193:1–105, 2021.
arXiv:1510.08040 [math.GR].

[4]
T. Delabie, J. Koivisto, F. Le Maître, and R. Tessera.
Quantitative measure equivalence between amenable groups,.
Annales Henri Lebesgue, 5:1417–1487, December 2022.
arXiv:2002.00719 [math.GR].

[5]
H. A. Dye.
On groups of measure preserving transformations. i.
American Journal of Mathematics, 81(1):119–159, 1959.

[6]
H. A. Dye.
On groups of measure preserving transformations. ii.
American Journal of Mathematics, 85(4):551–576, 1963.

[7]
A. Escalier.
Sofic aproximations and optimal quantitative orbit
equivalence, 2022.
arXiv:2205.12137 [math.GR].

[8]
M. Gromov, G. Niblo, and M. Roller.
Asymptotic invariants of infinite groups.
In Geometric group theory, Vol. 2: Asymptotic invariants of
infinite groups, pages 1–295. Cambridge Univ. Press, 1993.

[9]
D. Kerr and H. Li.
Entropy, Shannon orbit equivalence, and sparse connectivity.
Mathematische Annalen, 380:1497–1562, 2021.
arXiv:1912.02764 [math.DS].

[10]
D.S. Ornstein and B. Weiss.
Ergodic theory of amenable group actions. I: The Rohlin lemma.
Bulletin of the American Mathematical Society, 2:161–164,

§ NOTATIONS INDEX

$\preccurlyeq$, $\simeq$ See above <ref>.

$|X|$ Cardinal of the set $X$.

$\partial F$ Boundary of the set $F$.

$\BZ$ See <ref>.

$\BZ_m$ See <ref>.

$F_n$ Følner sequence of $\BZ$.

$\mbfg$ The sequence of maps $(g_m)_{m\in \bN}$.

$g^\prime_m$ See <ref>.

$\Gsp$ Normal closure of $[A_m,B_m]$.

$\profile_{G}$ Isoperimetric profile of $G$.

$R_n$ Diameter of $T_n$.

$\RZ_n$ Diameter of $\TZ_n$.

$\range(\mbff,t)$ The range of $(\mbff,t)$, see <ref>.

$S_G$ A generating set of the group $G$.

$\Sigma_n$ Følner tiling shifts (of $\BZ$).

$\SZ_n$ Følner tiling shifts of $\bZ$.

$T_n$ Tile of $\BZ$ defined by $T_n=\prod^{n}_{i=0}\Sigma_i$

${\TZ}_n$ Tile of $\bZ$ defined by $\TZ_n=\prod^{n}_{i=0}\SZ_i$

$\theta^A_m(f_m)$ Natural projection of $f_m$ on $A_m$ (see

$\theta^B_m(f_m)$ Natural projection of $f_m$ on $B_m$ (see

Amandine Escalier

Mathematisches Institut,

Fachbereich Mathematik und Informatik der Universität Münster,

Orléans-Ring 12,

48149 Münster,