# RF Interference in Lens-Based Massive MIMO Systems — An Application Note

Harsh Tataria

###### Abstract

We analyze the uplink radio frequency (RF) interference from a multiplicity of
single-antenna user equipments transmitting to a cellular base station (BS)
within the same time-frequency resource. The BS is assumed to operate with a
lens antenna array, which induces additional focusing gain for the incoming
signals. Considering line-of-sight propagation conditions, we characterize the
multiuser RF interference properties via approximation of the mainlobe
interference as well as the effective interferer probability. The results
derived in this application note are foundational to more general multiuser
interference analysis across different propagation conditions, which we
present in a follow up paper.

Notation. Boldface upper and lower case symbols are used to denote matrices
and vectors, while lightface upper and lower case symbols denote scalar
quantities. Note that the Hermitian transpose is denoted by
$\left(\cdot\right)^{\textrm{H}}$. The scalar norm is denoted as
$\left|\cdot\right|$ and the floor, ceiling and indicator functions are
expressed as $\lfloor\cdot\rfloor$, $\lceil\cdot\rceil$, and
$\mathbbm{1}\left(\cdot\right)$, respectively. The “sinc" function is given by
$\textrm{sinc}\left(x\right)=\sin\left(x\right)/x$, while the “maximum" and
“minimum" functions are given by $\max\left(\cdot\right)$ and
$\min\left(\cdot\right)$. Finally, $\mathcal{O}\left(\cdot\right)$ denotes the
“order" of a mathematical term.

##  I System Model

We consider the uplink of a single-cell system, where the base station (BS) is
equipped with a large lens antenna array containing $M$ elements. Multiuser
operation is assumed, where the lens array receives uplink data streams from
$L$ single-antenna user terminals within the same time-frequency interval. The
user terminals are located with a uniform random distribution in area covering
a net sector of $2\pi/3$ radians. The array at the BS consists of a flat
electromagnetic (EM) lens with elements that are located on the focal arc of
the lens. Without loss of generality, we consider azimuth direction-of-
arrivals (DOAs) and assume that the flat lens is employed with negligible
thickness.111For simplicity, the elevation DOAs are assumed to be zero, which
is practically valid if the relative height difference between the transmitter
and the receiver is much smaller than their separation distance. The EM lens
array has a total aperture of $D_{y}\times{}D_{z}$, as the array is located on
the $y-z$ plane and is centered at the origin. The focal arc of the lens is
defined as a semi-circle around the lens’s center in the azimuth plane ($x-y$
plane) with radius $F$. Here $F$ physically represents the focal length of the
lens. According to this, each element’s locations with respect to the lens’s
phase center can be written as $B_{m}$
($x_{m}=F\cos{}\left(\theta_{m}\right),y_{m}=-F\sin\left(\theta_{m}\right),z_{m}=0$),
where $\theta_{m}\in{}\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ is the angle
of the $m$-th antenna element with respect to the $x$-axis, where
$m\in{}\mathcal{M}$. Note that
$\mathcal{M}=\left\\{0,\pm{}1,\dots,\pm{}\frac{M-1}{2}\right\\}$ denotes the
set of antenna indices in the lens array.222For simplicity, it is worth noting
that $M$ is assumed to be an odd integer in this study. The antenna elements
are deployed on the focal arc such that
$\tilde{\theta}_{m}=\sin\left(\theta_{m}\right)$ are equally spaced in the
interval $\left[-1,1\right]$ as indicated in [1]. Doing this yields

$\tilde{\theta}_{m}=\frac{m}{\tilde{D}},\hskip 5.0ptm\in{}\mathcal{M},$ (1)

where $\tilde{D}=\frac{D_{y}}{\lambda}$ is the lens dimension along the
azimuth plane normalized by the carrier wavelength, $\lambda$. According to
this formulation, more elements are deployed in the center of the array than
those on either sides. The relationship between $M$ and $\tilde{D}$ can be
observed from (1) as $M=1+\lfloor 2\tilde{D}\rfloor$. As in [1], when the lens
array receives an uplink signal in the form of a uniform plane wave from
terminal $\ell$, with an azimuth DOA $\phi_{\ell}$, the resultant signal
received by the $m-$th element of the array can be written as

$a_{m}\left(\phi_{\ell}\right)\approx{}e^{-j\Phi_{0}}\sqrt{A}\hskip
2.0pt\left[\textrm{sinc}\left(m-\tilde{D}\tilde{\phi}_{\ell}\right)\right],\hskip
5.0ptm\in{}\mathcal{M},$ (2)

where $A=\frac{D_{y}D_{z}}{\lambda^{2}}$ is the normalized aperture,
$\Phi_{0}$ is a common phase shift from the lens’s aperture to the array, and
$\tilde{\phi}_{\ell}=\sin{}\left(\phi_{\ell}\right)\in{}\left[-1,1\right]$ is
referred to as the spatial frequency corresponding to $\phi_{\ell}$. In line
with [1, 3, 4, 5], we assume that the insertion loss of the lens, as well as
its boundary effects are negligible. The expression in (2) across all $M$
antenna elements yields a $M\times{}1$ propagation channel vector from user
$\ell$ to the BS, which we denote as $\bm{h}_{\ell}^{\textrm{LOS}}$.

##  II RF Interference Characteristics at the BS

Considering line-of-sight (LOS) propagation conditions, the received signal
strength remains deterministic under maximum-ratio combining processing, while
the only uncertainty is contained in the interference power via transmission
from undesired users to the BS. We denote the total interference power at the
BS by $I^{\textrm{LOS}}$ and analyze its form subsequently.

Theorem 1. Given $M$ antenna elements at the flat lens array, the interference
exerted by the $L-1$ interferers on to the desired user $\ell$ can be
expressed as
$I^{\textrm{LOS}}=\sum\nolimits_{\ell=1}^{L}I_{\ell}^{\textrm{LOS}}$, where
$I_{\ell}^{\textrm{LOS}}$ is the LOS interference to terminal $\ell$ given by

$\displaystyle I_{\ell}^{\textrm{LOS}}$
$\displaystyle=\frac{1}{M}\left\\{\left|\bm{g}_{\ell}^{\textrm{LOS}}\bm{h}_{k}^{\textrm{LOS}}\right|^{2}\right\\}$
(3) $\displaystyle=\frac{1}{M}\left\\{\left|\sum\limits_{m=0}^{M-1}\hskip
4.0pt\frac{\frac{A}{2}\left[\cos{}\left(m-\tilde{D}\Delta_{\ell,k}\right)-\cos\left(m-\tilde{D}\tilde{\Delta}_{\ell,k}\right)\right]}{m\left(m-\tilde{D}\Delta_{\ell,k}\right)+\tilde{D}^{2}\left(\tilde{\phi}_{\ell}\hskip
1.0pt\tilde{\phi_{k}}\right)}\right|^{2}\right\\},$

where $\bm{g}_{\ell}$ and $\bm{h}_{k}$ are the maximum-ratio combining vector
of user $\ell$ and the propagation channel vector from user $k$ to the BS.
Note that $\Delta_{\ell,k}=\tilde{\phi}_{\ell}-\tilde{\phi}_{k}$, and
$\tilde{\Delta}_{\ell,k}=\tilde{\phi}_{\ell}-\tilde{\phi}_{k}$.

Figure 1: LOS Interference power [dB] to terminal $\ell$,
$I_{\ell}^{\textrm{LOS}}$ vs. Angular separation of terminals $\ell$ and $k$,
$\tilde{\phi}_{\ell}-\tilde{\phi}_{k}$ in [Radians].

_Proof:_ Applying the definition of $\bm{g}_{\ell}^{\textrm{LOS}}$ and
$\bm{h}_{k}^{\textrm{LOS}}$, one can recognize that $I_{\ell}^{\textrm{LOS}}$
can be written as in equation (4), shown on the top of the next page for space
reasons.

$\displaystyle I_{\ell}^{\textrm{LOS}}$
$\displaystyle=\frac{1}{M}\left\\{\left|{\left(\bm{h}_{\ell}^{\textrm{LOS}}\right)}^{\textrm{H}}\bm{h}_{k}^{\textrm{LOS}}\right|^{2}\right\\}=\frac{1}{M}\left\\{\hskip
1.0pt\left|\sum\limits_{m=0}^{M-1}\sqrt{A}\hskip
2.0pt\textrm{sinc}\left(m-\tilde{D}\sin\left(\phi_{\ell}\right)\right)\sqrt{A}\hskip
2.0pt\textrm{sinc}\left(m-\tilde{D}\sin\left(\phi_{k}\right)\right)\right|^{2}\right\\}$
$\displaystyle\overset{\left(a\right)}{=}\frac{1}{M}\left\\{\hskip
1.0pt\left|\hskip 2.0pt\sum\limits_{m=0}^{M-1}\frac{\overbrace{\sqrt{A}\hskip
1.0pt\sin\left(m-\tilde{D}\sin\left(\phi_{\ell}\right)\right)\sqrt{A}\sin\left(m-\tilde{D}\sin\left(\phi_{k}\right)\right)}^{=N_{\ell}^{\textrm{LOS}}}}{\underbrace{\left(m-\tilde{D}\sin\left(\phi_{\ell}\right)\right)\left(m-\tilde{D}\sin\left(\phi_{k}\right)\right)}_{=D_{\ell}^{\textrm{LOS}}}}\right|^{2}\right\\}.$
(4)

Note that $\left(a\right)$ in (4) is obtained by using the fact that
$\textrm{sinc}\left(x\right)=\frac{\sin\left(x\right)}{x}$. The numerator and
the denominator of (4), in $N_{\ell}^{\textrm{LOS}}$ and
$D_{\ell}^{\textrm{LOS}}$, can then be simplified by invoking the
trigonometric identity
$\sin\left(x\right)\sin\left(y\right)=\frac{1}{2}\left[\cos\left(x-y\right)-\cos\left(x+y\right)\right]$,
where $x=m-\tilde{D}\left(\sin\left(\phi_{\ell}\right)\right)$ and
$y=m-\tilde{D}\left(\sin\left(\phi_{k}\right)\right)$, allowing us to write

$N_{\ell}^{\textrm{LOS}}=\frac{A}{2}\left[\cos{}\left(m-\tilde{D}\left(\tilde{\phi}_{\ell}-\tilde{\phi}_{k}\right)\right)-\cos\left(m-\tilde{D}\left(\tilde{\phi}_{\ell}-\left(-\tilde{\phi}_{k}\right)\right)\right)\right],$
(5)

where $\tilde{\phi}_{\ell}=\sin\left(\phi_{\ell}\right)$ and
$\tilde{\phi}_{k}=\sin\left(\phi_{k}\right)$, respectively. Using the fact
that $\Delta_{\ell,k}=\tilde{\phi}_{\ell}-\tilde{\phi}_{k}$ and
$\tilde{\Delta}_{\ell,k}=\tilde{\phi}_{\ell}-\left(-\tilde{\phi}_{k}\right)$,
$N_{\ell}^{\textrm{LOS}}$ in (5) can be re-written as

$N_{\ell}^{\textrm{LOS}}=\frac{A}{2}\left[\cos\left(m-\tilde{D}\Delta_{\ell,k}\right)-\cos\left(m-\tilde{D}\tilde{\Delta}_{\ell,k}\right)\right].$
(6)

Similarly, after some straightforward algebraic manipulations,
$D_{\ell}^{\textrm{LOS}}$ can be expressed as

$\displaystyle D_{\ell}^{\textrm{LOS}}$
$\displaystyle=m\left(m-\tilde{D}\left(\tilde{\phi}_{\ell}-\tilde{\phi}_{k}\right)\right)+\tilde{D}^{2}\left(\tilde{\phi}_{\ell}\hskip
2.0pt\tilde{\phi_{k}}\right)$ $\displaystyle=m\left(m-\tilde{D}\hskip
1.0pt\Delta_{\ell,k}\right)+\tilde{D}^{2}\left(\tilde{\phi}_{\ell}\hskip
2.0pt\tilde{\phi}_{k}\right).$ (7)

Substituting (5) and (7) into (4) yields the desired result and concludes the
proof. ∎

Remark 1. By inspecting $I_{\ell}^{\textrm{LOS}}$ in (3), one can observe that
both the numerator and the denominator contains terms which are expressed as
the difference of the $\sin\left(\cdot\right)$ two DOA angles in
$\tilde{\phi}_{\ell}$ and $\tilde{\phi}_{k}$. We denote this as the _angular
separation_ between the DOAs of interferer $k$, and the desired terminal
$\ell$. Naturally, when $\tilde{\phi}_{\ell}$ and $\tilde{\phi}_{k}$ are
aligned, $I_{\ell}^{\textrm{LOS}}$ will reach its maximum value, inducing
maximum spatial correlation in the multiuser channel. In contrast to this,
when $\tilde{\phi}_{\ell}$ and $\tilde{\phi}_{k}$ are vastly different, a
lower value of interference is expected. Furthermore, the RF interference
power is also a function of the critical inter-element spacing, $\tilde{D}$
for which the lens array is designed. Rather interestingly, as demonstrated in
[2], the result for $I_{\ell}^{\textrm{LOS}}$ with a half-wavelength spaced
uniform linear array yields a very similar conclusion where
$N_{\ell}^{\textrm{LOS}}$ and $D_{\ell}^{\textrm{LOS}}$ are both proportional
to $\cos(2\pi{}\tilde{D}(\tilde{\phi}_{\ell}-\tilde{\phi}_{k}))$. Note that in
that case, the LOS propagation channel has a substantially different
form/structure in that it is constructed out of phase shifted exponential
functions instead of sinc functions (as for the lens arrays).

Remark 2. According to (4), Fig. 1 depicts $I_{\ell}^{\textrm{LOS}}$ as a
function of $\tilde{\phi}_{\ell}-\tilde{\phi}_{k}$. One can observe:

  1. 1.

$I_{\ell}^{\textrm{LOS}}$ is non-monotonic and non-periodic in nature, making
its probability density and cumulative density mathematically intractable to
analyze. This is true irrespective of the number of interfering sources
present in the system.

  2. 2.

With an increase in the number of elements at the lens array, majority of the
interference will lie in the mainlobe (defined from the peak to the first
null) of the interference pattern, while the relative sidelobe interference
levels are significantly lower. The power ratio between the peak of the main
lobe relative to the first sidelobe is approximately 13 dB. For further
discussions, see [6] and references therein.

  3. 3.

From the result in Theorem 1, the first nulls on either side of the mainlobe
appear at $\phi_{\ell}-\phi_{m}=\pm{}\frac{1}{Md}$. As a result, the mainlobe
width can be written as $\frac{2}{Md}$.

Considering the fact that any further analysis of the instantaneous LOS
interference power is intractable, to understand the fundamental nature of its
probability and cumulative densities, we approximate it by recognizing that
the RF interference, $I_{\ell}^{\textrm{LOS}}$, is composed of a mainlobe
surrounded by many smaller sidelobes, and the shape is determined by $M$, the
number of antenna elements at the lens array. As the relative sidelobe levels
are negligible particularly for moderate and larger arrays, the “effective" RF
interference can be approximated by the mainlobe only. The details of this
approximation are as follows, based on which an effective interferer
probability is derived.

Approximation 1. The mainlobe (effective) RF interference can be approximated
as

$I_{\ell}^{\textrm{LOS}}\approx{}J_{\ell}^{\textrm{LOS}}.\hskip
3.0pt\mathbbm{1}\left(\left|\Theta_{\ell,k}\right|\leq{}1\right),$ (8)

where
$\Theta_{\ell,k}=\tilde{D}\left(\tilde{\phi}_{\ell}-\tilde{\phi}_{k}\right)$
is the normalized angular separation of terminals $\ell$ and $k$ over half the
mainlobe width, such that when $\Theta_{\ell,k}\in{}\left[-1,1\right]$,
interferer $k$ falls within the mainlobe of the interference pattern.
Moreover, the effective interference from interferer $k$, given $M$ elements
is

$\displaystyle
J_{\ell}^{\textrm{LOS}}=\frac{1}{M}\left\\{\left|\sum\limits_{m=0}^{M-1}\frac{\frac{A}{2}\left[\cos\left(m-\Theta_{\ell,k}\right)-\cos\left(m-\tilde{\Theta}_{\ell,k}\right)\right]}{m\left(m-\Theta_{\ell,k}\right)+\tilde{D}^{2}\left(\tilde{\phi}_{\ell}\hskip
1.0pt\tilde{\phi}_{k}\right)}\right|^{2}\right\\},$ (9)

with
$\tilde{\Theta}_{\ell,k}=\tilde{D}\left(\tilde{\phi}_{\ell}-\frac{\tilde{\phi}_{k}}{-1}\right)$.

To obtain the density of the interference, we require the probability that an
arbitraray interferer (terminal $k$) is an “effective" interferer, as well as
the distribution of the effective interference. Since we are interested in a
large (yet naturally a finite element) lens array, we focus on the behavior of
the two aforementioned quantities when the number of antenna elements $M$ is
large, which leads to the following proposition on the effective interferer
probability.

Proposition 1. As $M$ grows, the probability that terminal $k$ is an effective
interferer is given by

$p_{k}\approx{}\frac{9}{\tilde{D}\pi^{2}}\left[\tanh^{-1}\left(\frac{\sqrt{3}}{2}\right)\right].$
(10)

_Proof:_ The probability of terminal $k$ being an effective interferer can be
written as

$p_{k}\approx\mathbb{P}\left(-1\leq{}\Theta_{\ell,k}\leq{}1\right).$ (11)

We recall that all terminals are placed uniformly randomly in a single sector
of $\frac{2\pi}{3}$ radians. As such, their DOAs, $\phi_{\ell}$ and $\phi_{k}$
are independently, and uniformly drawn from
$\left[\frac{-\pi}{3},\frac{\pi}{3}\right]$. From (8), the probability density
function of $\Theta_{\ell,k}$ with $M$ antennas at the lens array can be
written as

$f_{\Theta_{\ell,k}}^{\left(M\right)}\left(z\right)\approx\int\nolimits_{y_{L}}^{y_{U}}\frac{1}{\tilde{D}}\left(\frac{3}{2\pi}\right)^{2}\frac{1}{\sqrt{1-\left(\frac{z}{\tilde{D}}+y^{2}\right)}}\frac{1}{\sqrt{1-y^{2}}}\hskip
5.0ptdy;\hskip 15.0pt\left|z\right|\leq{}\sqrt{3}\tilde{D},$ (12)

where the bounds on the integral are defined given by
$y_{L}=\max\left(-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}-\frac{z}{\tilde{D}}\right)$,
and
$y_{U}=\min\left(\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}-\frac{z}{\tilde{D}}\right)$.
Then, following (11), the probability can be written as

$p_{k}\approx\int\nolimits_{-1}^{1}f_{\Theta_{\ell,k}}^{\left(M\right)}\left(z\right)\hskip
5.0ptdz.$ (13)

Applying the standard Taylor series expansion of (11) and performing some
straightforward simplifications yields

$\displaystyle p_{k}$
$\displaystyle\approx\frac{9}{\tilde{D}\pi^{2}}\left[\tanh^{-1}\left(\frac{\sqrt{3}}{2}\right)+\mathcal{O}\left(\frac{1}{M^{2}}\right)\right]$
$\displaystyle\approx\frac{9}{\tilde{D}\pi^{2}}\left[\tanh^{-1}\left(\frac{\sqrt{3}}{2}\right)\right].$
(14)

This yields the desired expression in (10), and concludes the proof. ∎

## References

  * [1] Y. Zeng and R. Zhang, “Millimeter wave MIMO with lens antenna array: A new path division multiplexing paradigm," _IEEE Trans. Commun._ , vol. 64, no. 4, pp. 1557-1571, Apr. 2016.
  * [2] D. Tse and P. Viswanath, _Fundamentals of Wireless Communication_ , Cambridge University Press, 2005.
  * [3] Y. Zeng, R. Zhang, and Z. N. Chen, “Electromagnetic lens-focsing antenna enabled massive MIMO: Performance improvement and cost reduction", _IEEE J. Sel. Areas in Commun._ , vol. 32, no. 6, pp. 1194-1206, Jun. 2014.
  * [4] Y. Zeng, L. Yang and R. Zhang, “Multi-user millimeter-wave MIMO with single-sided full-dimensional lens antenna array", in _Proc. of the IEEE Int. Conf. on Commun. (ICC)_ , May 2016, pp. 1-6.
  * [5] T. Xie, L. Dai, X. Gao, H. Yao, and X. Wang, “On the power leakage problem in beamspace MIMO systems with lens antenna array", in _Proc. IEEE 86th Veh. Technol. Conf. (VTC-Fall)_ , Sep. 2017, pp. 1-6.
  * [6] H. Tataria, M. Matthaiou, P. J. Smith, G. C. Alexandropoulos and V. F. Fusco, “Uplink interference analysis with RF switching for lens-based millimeter-wave systems," in _Proc. of the IEEE Int. Conf. on Commun. (ICC)_ , May 2018, pp. 1-7.


# Distortion in groups of generalized piecewise-linear transformations

Leonardo Dinamarca Opazo

Abstract. For each natural number $n$, we consider the subgroup
$\mathcal{R}_{n}$ of $\mathrm{Homeo}_{+}[0,1]$ made by the elements that are
linear except for a subset whose Cantor-Bendixson rank is less than or equal
to $n$. These groups of generalized piecewise-linear transformations yield an
ascending chain of groups as we increase $n$. We study how the notion of
distorted element changes along this chain. Our main result establishes that
for each natural number $n$, there exits an element that is undistorted of
$\mathcal{R}_{n}$ yet distorted in $\mathcal{R}_{n+1}$. Actually, such an
element is explicitly constructed.

Keywords: Interval homeomorphisms, piecewise-linear, Cantor-Bendixson rank,
distorted element, distortion.

2010 Mathematics Subject Classification: Primary 20F99. Secondary 37C85,
37E05.

##  1 Introduction

The notion of distorted element was introduced by Mikhail Gromov [14]. Given a
finitely generated group $\Gamma$, we fix a finite system of generators $S$,
and we denote $l_{S}(\cdot)$ the corresponding word-length. An element
$f\in\Gamma$ is said to be distorted if

$\lim_{n\to\infty}\frac{l_{S}(f^{n})}{n}=0.$

(Note that this condition does not depend on the choice of the finite
generating system.) Given an arbitrary group $G$, an element $f\in G$ is said
to be distorted if there exists a finitely generated subgroup $\Gamma\subset
G$ containing $f$ so that $f$ is distorted in $\Gamma$ in the sense above.

This notion has been studied in different groups, for example in the
automorphisms group of a shift [5, 8], in the Cremona group [6] and mainly in
groups of transformations [15, 16, 17, 23]. We will focus in the last
direction. In [13], John Franks and Michael Handel asked whether irrational
rotations are distorted in $\mathrm{Diff}^{1}_{+}(\mathbb{S}^{1})$. Danny
Calegari and Michael Freedman (see [4]) gave a positive answer to this.

######  Theorem 1.1.

Irrational rotations are distorted in $\mathrm{Diff}^{1}_{+}(\mathbb{S}^{1})$.

Later, Arthur Avila extended this result to infinite differentiability (see
[1]). To state his result in full generality, recall that $f$ is said
recurrent if $\liminf_{n\rightarrow\infty}d(f^{n},id)=0,$ where $d$ is a
metric inducing the $C^{\infty}$ topology on
$\mathrm{Diff}_{+}^{\infty}(\mathbb{S}^{1}).$

######  Theorem 1.2.

In $\mathrm{Diff}_{+}^{\infty}(\mathbb{S}^{1})$, every recurrent element is
distorted.

Emmanuel Militon generalized this result to any compact manifold (see [18]).
The following related question remains open.

Question 1. Are irrational rotations on the circle distorted in the group of
real-analytic difeomorphisms $\mathrm{Diff}_{+}^{\omega}(\mathbb{S}^{1})$?

In [20], Andrés Navas gave a list of concrete questions about group actions on
$1$-dimensional manifolds. One of these questions was the existence of
distorted elements in $\mathrm{PL}_{+}(\mathbb{S}^{1})$, the group of
piecewise-linear homeomorphisms of the circle. This question was answered
affirmatively by Juliusz Banecki and Tomasz Szarek in [2]; also, their proof
can be adapted to obtain an alternative proof of Avila’s result (Theorem 1.2).

######  Theorem 1.3.

All irrational rotations are distorted in $\mathrm{PL}_{+}(\mathbb{S}^{1})$.

There exist many interesting countable subgroups of
$\mathrm{PL}_{+}(\mathbb{S}^{1})$ for which it is unkknow whether they contain
distorted elements or not. For instance, let $F_{\tau}$ be the Thompson group
(introduced by Sean Cleary in [7]) associated to the golden number $\tau$,
defined as the set of elements in $\mathrm{PL}_{+}(\mathbb{S}^{1})$ with break
points in $\mathbb{Z}[\tau]$ and slopes that are powers of $\tau$. The next
question was suggested by Yash Lodha.

Question 2. Does $F_{\tau}$ contain distorted elements?

The situation in $\mathrm{PL}_{+}(I)$, the group of piecewise-linear
homeomorphisms of the interval $I=[0,1]$ is different, since the unique
distorted element is the identity. To see this, let
$L:\mathrm{PL}_{+}(I)\rightarrow\mathbb{R}$ be the function defined as

$L(f):=\log\max_{x\in I}\\{D_{R}f(x),D_{R}f^{-1}(x)\\}\mbox{ for all
}f\in\mathrm{PL}_{+}(I).$

Let us note that for $f,g\in\mathrm{PL}_{+}(\mathbb{S}^{1})$ the chain rule
implies that

$L(fg)\leq L(f)+L(g).$

If $f$ is a nontrivial element, then we have that $L(f^{n})\geq C\cdot n$ for
some positive constant $C.$ This easily implies that $f$ is undistorted (see
the Criterion in Section 2.1).

Since there is no nontrivial distorted element in $\mathrm{PL}_{+}(I)$ we will
relax the condition on the quantity of break points to consider larger groups.

Let $n$ be a natural number. Following a suggestion of Lodha, we define the
group of generalized piecewise-linear transformations, that we will denote by
$\mathcal{R}_{n}$ (see Definition 3.1), as the subgroup of
$\mathrm{Homeo}_{+}[0,1]$ made by the elements that are linear except for a
subset of $[0,1]$ (that we call set of break points) whose Cantor-Bendixson
rank is smaller than or equal to $n$. The fact that this is actually a group
is not totally obvious (see Proposition 3.2). Once this established, we get an
ascending chain of groups:

$\mathrm{PL}_{+}(I)=\mathcal{R}_{0}\leq\mathcal{R}_{1}\leq\mathcal{R}_{2}\leq\mathcal{R}_{3}\leq\dots$

A general problem is to study how the notion of distortion change along a
filtration of groups. In [19], He addressed the following question:

Question 3. Given $1\leq r<s$, does there exist an undistorted element
$f\in\mathrm{Diff}^{s}_{+}(M)$ that is distorted when considered as an element
of $\mathrm{Diff}^{r}_{+}(M)$ ?

He proved that this is the case for $M=[0,1]$, $r=1$ and $s=2$. Actually, in
his example, undistortion holds in the larger group
$\mathrm{Diff}^{1+bv}_{+}([0,1])$ of $C^{1}$ diffeomorphisms with derivative
of bounded variation. Moreover, in [9], finer computations concerning
intermediate regularity are made and. As an outcome, it is established that
that there exists a diffeomorphism that is distorted in
$\mathrm{Diff}_{+}^{1+\alpha}[0,1]$ but undistorted in
$\mathrm{Diff}_{+}^{2}[0,1]$. However, for the case of the closed interval,
the question above remains open in regularity larger than $2$ and it is
completely open for higher dimensional manifolds.

In the same spirit, Lodha asked the following question: given $n$ a natural
number, does there exist an undistorted element in $\mathcal{R}_{n}$ that is
distorted in $\mathcal{R}_{n+1}$? The main result of this work answers this
question positively.

Theorem A For each natural number $n$, there exists $f\in Homeo_{+}[0,1]$ that
is undistorted in $\mathcal{R}_{n}$ but distorted in $\mathcal{R}_{n+1}$.

The strategy to prove this theorem is as follows: given a natural number $n$,
we define a length function in $\mathcal{R}_{n}$ that we will use to detect
undistorted elements. Then, we build an element that we force to be
undistorted (using the length function) and then, we establish distortion in
$\mathcal{R}_{n+1}$ via a combination of the diagonal trick and the Mather’s
argument.

The groups $\mathcal{R}_{n}$ remain mysterious to us. A study from the
perspective of large scale geometry in the sense of Rosendal [24] would be a
good starting point.

##  2 Preliminaries

In this section, we provide the definition of distorted elements in groups,
discuss some of their properties, and present several basic examples.

######  Definition 2.1.

Let $\Gamma=\langle S\rangle$ be a finitely generated group. Given an element
$f\in\Gamma$, we define the word-length of $f$ with respect to $S$ as the
minimum integer $n$ such that $f$ can be written as product of $n$ elements of
$S.$ We use the following notation:

$l_{S}(f)=||f||:=\min\\{n\,|\,\,f=s_{1}\cdot s_{2}\cdot...\cdot s_{n},\,\mbox{
with }s_{1},...,s_{n}\in S\\}.$

An important property satisfied by the word-length function
$l_{S}:\Gamma\rightarrow\mathbb{N}$ is the subadditive inequality: for every
pair of elements $f,g\in\Gamma$, we have

$l_{S}(gh)\leq l_{S}(g)+l_{S}(h).$

######  Remark 2.2.

The word metric is given by $d_{S}(f,g):=l_{S}(f^{-1}g)$. The topology induced
by the word metric is the discrete topology for any generating set.

Observe that if we take two generating sets of the group, we obtain a
lipschitz relation between the induced word-length functions. Indeed, if
$\Gamma=\langle S\rangle=\langle T\rangle$, then for each $g\in\Gamma$, the
following inequality holds

$\frac{1}{C^{\prime}}\cdot l_{S}(g)\leq l_{T}(g)\leq C\cdot l_{S}(g),$ (1)

where $C=\max_{s\in S}l_{T}(s)$ and $C^{\prime}=\max_{t\in T}l_{S}(t)$.

We next define an important limit for the rest of the text. Let
$f\in\Gamma=\langle S\rangle$ be any element. By the subadditive inequality,
we have

$l_{S}(f^{n+m})\leq l_{S}(f^{n})+l_{S}(f^{m}),$

for all natural numbers $n$ and $m$. This means that the sequence
$(l_{S}(f^{n}))$ is subadditive, therefore by Fekete’s Lemma [11], the
following limit exists:

$\lim_{n\rightarrow\infty}\frac{l_{S}(f^{n})}{n}=\inf_{n\geq
1}\frac{l_{S}(f^{n})}{n}.$

######  Remark 2.3.

Note that if the limit is equal to $0$, then the value does not depend on the
choice of the finite generating system (because of (1)).

Now we define the concept of distortion in groups, which is the crucial notion
of this work.

######  Definition 2.4.

Let $\Gamma$ be a finitely generated group. We say that $f\in\Gamma$ is
distorted if the limit above is equal to $0.$ If the limit is different from
$0$ we say that $f$ is undistorted. Note that being a distorted element is a
property of the group and the element.

Given a non-torsion distorted element $f$ in a group $G=\langle S\rangle$, we
will pay attention to the growth rate of $l_{S}(f^{n})$ as a function of $n$
to analyze different behavior. The distortion function
$D_{S,f}:\mathbb{N}\rightarrow\mathbb{N}$ is defined by

$\displaystyle D_{S,f}(n):=\max\\{k:l_{S}(f^{k})\leq n\\}.$

We have seen that the notion of distorted element does not depend of $S$. In
order to have the same property for the distortion function, we consider an
equivalence relation. Given two nondecresing functions

$\displaystyle h,g:\mathbb{N}\rightarrow\mathbb{N},$

we write $h\preceq g$ if there exists $k\geq 1$ such that

$\displaystyle h(n)\leq k\cdot g(k\cdot n+k)+k\mbox{ for all }n\in\mathbb{N}$

and we write $h\sim g$ if $h\preceq g$ and $g\preceq h$. We have that $\sim$
is an equivalence relation. With this definition, if $S^{\prime}$ is another
generating system of $G$, we obtain that $D_{S,f}\sim D_{S^{\prime},f}$ and
then we simply denote the distortion function of $f$ as $D_{f}.$

######  Definition 2.5.

For a general group $G$, we say that an element is distorted if there exists a
finitely generated subgroup of $G$ such that the element is distorted inside
this finitely generated subgroup.

###  2.1 Examples of distorted and undistorted elements

In this section we will give examples of distorted and undistorted elements in
groups. First, we recall the notion of a length function, and we define those
that we will use to state the criterion that we apply to show that an element
is undistorted. Such length functions arise from geometric group theory,
dynamical systems, algebraic geometry and many other branches of mathematics
(see [25] for a survey). After this, we present several examples of distorted
and undistorted elements

######  Definition 2.6.

Let $G$ be a group. A length function on $G$ is a function
$L:G\rightarrow\mathbb{R}\cup\\{0\\}$ such that $L(Id)=0$, $L(g)=L(g^{-1})$
and $L(gh)\leq L(g)+L(h)$ for all $g,h\in G$.

Given a length function $L$ on a group $G$ and an element $f\in G$, we define
the stabilazed limit as

$\lim_{n\rightarrow\infty}\frac{L(f^{n})}{n}.$

Again, existence of this limit follows from Fekete’s lemma.

######  Lemma 2.7.

Let $G$ be a group and $L$ a length function on $G$. If $g\in G$ is distorted,
then the stabilazed limit is equal zero

Proof. Let $g$ be distorted in $\Gamma=\langle S\rangle$, a finitely generated
generated subgroup of $G$. We let $C:=\max\\{L(s):\,s\in S\\}$. Then, by
subadditivity of $L$, we have

$\lim_{n\rightarrow\infty}\frac{L(f^{n})}{n}\leq
C\cdot\lim_{n\rightarrow\infty}\frac{l_{S}(f^{n})}{n}=0.$

We use the previous lemma to test that an element is undistorted.

Criterion: If there exists a length function $L:G\rightarrow\mathbb{R}$ such
that for an element $f\in G$ the stabilazed limit is positive, then $f$ is
undistorted.

The first example of distortion concerns the Baumslag-Solitar group and its
realization as a group of homeomorphisms of the real line.

######  Example 2.8.

Let $f$ and $g$ be the elements in $\mbox{Homeo}_{+}(\mathbb{R})$ defined by
$f(x)=x+1$ and $g(x)=2x$. Consider the group $\Gamma=\langle S\rangle$ with
$S:=\\{f^{\pm 1},g^{\pm 1}\\}$. An explicit computation shows that

$g^{n}fg^{-n}(x)=f^{2^{n}}(x)\mbox{ for all }n\geq 1.$

Hence, $f$ is distorted in $\langle S\rangle$, since

$\lim_{n\rightarrow\infty}\frac{l_{S}(f^{2^{n}})}{2^{n}}\leq\lim_{n\rightarrow\infty}\frac{2n+1}{2^{n}}=0.$

In the next example, we exhibit distorted and undistorted element in a finite
chain of groups. We will see explicitly how an element is undistorted in a
group but distorted in a larger one. We consider a group of upper triangle
matrices of $5\times 5$ with integer coefficients and a chain of three
subgroups.

######  Example 2.9.

Let us consider the group

$\displaystyle\mathcal{H}_{5}:=\left\\{\begin{pmatrix}1&a_{1,2}&a_{1,3}&a_{1,4}&a_{1,5}\\\
0&1&a_{2,3}&a_{2,4}&a_{2,5}\\\ 0&0&1&a_{3,4}&a_{3,5}\\\ 0&0&0&1&a_{4,5}\\\
0&0&0&0&1\end{pmatrix}:a_{i,j}\in\mathbb{Z}\right\\}.$

Also, we consider the subgroups of $\mathcal{H}_{5}$ defined by

$\displaystyle\Gamma_{1}$
$\displaystyle:=\\{A\in\mathcal{H}_{5}:a_{1,2}=a_{1,3}=a_{1,4}=a_{1,5}=a_{2,3}=a_{2,4}=0\\},$
$\displaystyle\Gamma_{2}$
$\displaystyle:=\\{A\in\mathcal{H}_{5}:a_{1,2}=a_{1,3}=a_{1,4}=0\\}.$

We have a finite chain of groups

$\Gamma_{1}\leq\Gamma_{2}\leq\mathcal{H}_{5}.$

Let us define the matrices in $\mathcal{H}_{5}$ given by the formula

$E_{i,j}:=e_{i,j}+Id,$

where $2\leq j\leq 5$, $1\leq i<j$ and $e_{i,j}$ are the canonical matrices.
Observe that the set $S$ made by the matrices $E_{i,j}$ is a finite generating
system of $\mathcal{H}_{5}$. Moreover, for $k\in\\{1,2\\}$, we have that
$S_{k}=S\cap\Gamma_{k}$ is a finite generating system of $\Gamma_{k}$. An
explicit computation shows that

$E_{3,5}^{n^{2}}=[E_{3,4}^{n},E_{4,5}^{n}],$

so

$\lim_{n\rightarrow\infty}\frac{l_{S_{1}}(E_{3,5}^{n^{2}})}{n^{2}}\leq\lim_{n\rightarrow\infty}\frac{4n}{n^{2}}=0.$

This means that $E_{3,5}$ is distorted in $\Gamma_{1}.$ On the other hand, the
element $E_{2,5}$ is undistorted in $\Gamma_{1}$. To show this, it suffices to
consider $L_{1}:\Gamma_{1}\rightarrow\mathbb{N}$ defined as

$L_{1}(A):=|(A)_{2,5}|$

for each $A\in\Gamma_{1}$. Note that $L_{1}$ is a length function and
$L_{1}(E_{2,5}^{n})=n$. Then, our Criterion assert that $E_{2,5}$ is
undistorted in $\Gamma_{1}$. However, $E_{2,5}$ is distorted in $\Gamma_{2}$.
Actually, we have that

$E_{2,5}^{n^{2}}=[E_{2,3},E_{3,5}^{n^{2}}]=[E_{2,3},[E_{3,4}^{n},E_{4,5}^{n}]],$

so

$\lim_{n\rightarrow\infty}\frac{l_{S_{2}}(E_{2,5}^{n^{2}})}{n^{2}}\leq\lim_{n\rightarrow\infty}\frac{8n+2}{n^{2}}=0.$

On the other hand, we have that $E_{1,5}$ is undistorted in $\Gamma_{2}$. To
check this, we just repeat the argument before with the length function
$L_{2}:\Gamma_{2}\rightarrow\mathbb{N}$ defined as $L_{2}(A):=|(A)_{1,5}|$ for
each $A\in\Gamma_{2}$. Nevertheless, we claim that $E_{1,5}$ is distorted in
$\mathcal{H}_{5}$. Indeed, we have that

$E_{1,5}^{n^{4}}=[E_{1,3}^{n^{2}},E_{3,5}^{n^{2}}]=[[E_{1,2}^{n},E_{2,3}^{n}],[E_{3,4}^{n},E_{4,5}^{n}]],$

and so

$\lim_{n\rightarrow\infty}\frac{l_{S}(E_{1,5}^{n^{4}})}{n^{4}}\leq\lim_{n\rightarrow\infty}\frac{16n}{n^{4}}=0.$

######  Remark 2.10.

In Example 2.9, we have that the elements $E_{3,5}$ and $E_{2,5}$ have
distortion function of order $g(n)=n^{2}$ and the element $E_{1,5}$ has
distortion function of order $g(n)=n^{4}$. One can increase the size of the
matrix to produce a longer finite chain of subgroups with distorted elements
with polynomial distortion function of degree as large as we want. In Example
2.8, the element $f$ has distortion function of order $g(n)=2^{n}.$

Let $f:\mathbb{N}\rightarrow\mathbb{R}$ be a function. We say that $f$ is
superpolynomial if

$\displaystyle\lim_{n\rightarrow\infty}\frac{\log(f(n))}{\log(n)}=\infty.$

We say that $f$ is subexponential if

$\displaystyle\lim_{n\rightarrow\infty}\frac{\log(f(n))}{n}=0.$

Finally, $f$ is said to have intermediate growth if $f$ is both subexponential
and superpolinomial. We are not able to exhibit an exmaple of an element with
distortion function of intermediate growth, eventhouth for specialists it must
be known.

The next example can be considered as an affirmative answer to Question $3$
for the case of $M=\mathbb{S}^{1}$ and regularity $C^{0}$ versus $C^{1}.$

######  Example 2.11.

In the group Homeo${}_{+}(\mathbb{S}^{1})$, Calegari and Freedman prove that
every element is distorted (see [4]). If we go to $C^{1}$ regularity, we
observe that the existence of an hyperbolic fixed point is an obstruction to
$C^{1}$-distortion.

Indeed, define
$L:\mathrm{Diff}_{+}^{1}(\mathbb{S}^{1})\rightarrow\mathbb{R}_{+}$ by

$L(f):=\log\max\\{||Df||_{\infty},||Df^{-1}||_{\infty}\\}.$

Note that $L$ is a length function. Now, if
$f\in\mathrm{Diff}_{+}^{1}(\mathbb{S}^{1})$ has an hyperbolic fixed point $x$,
i.e, $f(x)=x$ and $\lambda:=\log Df(x)\neq 0$, then, by the chain rule,

$L(f^{n})\geq n|\lambda|.$

This implies that $f$ is undistorted in
$\mathrm{Diff}_{+}^{1}(\mathbb{S}^{1})$.

###  2.2 Cantor-Bendixson rank

In this section we recall the notion of Cantor-Bendixson rank and state the
properties that will be useful for us. We will use this rank to define the
groups $\mathcal{R}_{n}$ (see Definition 3.1). Given a subset $A$ of the
interval $[0,1]$, we denote by $A^{\prime}$ its derived set, i.e, the set of
accumulation points of $A$.

######  Definition 2.12.

Let $X\subset[0,1]$ be a set. The Cantor-Bendixson rank, denoted by $r(X)$, is
defined by induction as follows:

  * •

If $X$ is a finite set, then $r(X)=0$.

  * •

$r(X)=r(X^{\prime})+1.$

In this work, we only consider subset of the interval with finite Cantor-
Bendixson.

######  Proposition 2.13.

Let $X,Y\subseteq[0,1]$ and let $g\in$ Homeo${}_{+}[0,1]$. Then

  * •

$r(X)=r(g(X))$.

  * •

If $r(X)\leq n$ and $r(Y)\leq n$ for a certain $n\in\mathbb{N}$, then $r(X\cup
g(Y))\leq n.$

Proof. We prove both statements by induction. For the first statement, the
base case $(r(X)=0)$ holds because a homeomorphism sends a finite set into a
finite set. Suppose that $r(X)=n$. By the induction hypothesis,
$r(g(X^{\prime}))=r(X^{\prime})=n-1,$ hence

$r(g(X))=r(g(X)^{\prime})+1=r(g(X^{\prime}))+1=(n-1)+1=n.$

For the second statement, the case $n=0$ means that both $X$ and $Y$ are
finite sets, hence $X\cup g(Y)$ is a finite set, therefore, $r(X\cup g(Y))=0$.
Suppose $r(X)\leq n$ and $r(Y)\leq n$. Observe that

$(X\cup g(Y))^{\prime}=X^{\prime}\cup g(Y)^{\prime}=X^{\prime}\cup
g(Y^{\prime}).$

Thus, by the induction hypothesis, it follows that $r((X\cup
g(Y))^{\prime})=r(X^{\prime}\cup g(Y^{\prime}))\leq n-1$ because
$r(X^{\prime})\leq n-1$ and $r(g(Y^{\prime}))=r(Y^{\prime})\leq n-1$.
Therefore, $r(X\cup g(Y))\leq n$.

##  3  Distortion in groups of generalized piecewise-linear transformations

This is the main section of this work. Here we introduce the groups we are
going to deal with and we develop the necesary tools to prove Theorem A.

######  Definition 3.1.

Let $n\in\mathbb{N}$ be any natural number. We define $\mathcal{R}_{n}$ as the
set of all $f\in Homeo_{+}[0,1]$, such that there exist a sets (which we will
call the set of break points and denote by $BP(f)$) such that $f$ is linear on
each component of the complement of $BP(f)$ and $r(BP(f))\leq n$. Note that
$\mathcal{R}_{0}=\mathrm{PL}_{+}(I).$

######  Proposition 3.2.

For each $n\in\mathbb{N}$, the set $\mathcal{R}_{n}$ is a group.

Proof. Let $f$ and $g$ be any elements of the set $\mathcal{R}_{n}$. Let $x$
be a point that belongs to both the complement of $g^{-1}(BP(f))$ and the
complement of $BP(g)$. By the chain rule, we have

$\displaystyle D(f\circ g)(x)=Df(g(x))\cdot Dg(x).$

We this conclude that $x$ belongs to the complement of $BP(f\circ g)$. This
means that

$BP(f\circ g)\subset g^{-1}(BP(f))\cup BP(g).$

Since $r(BP(f))\leq n$ and $r(BP(g))\leq n$, Proposition 2.13 implies

$\displaystyle r(BP(f\circ g))\leq r(g^{-1}(BP(f))\cup BP(g))\leq n.$

This shows that $f\circ g$ belongs to $\mathcal{R}_{n}.$ Also, since
$BP(f^{-1})=f^{-1}(BP(f))$, using Proposition 2.13 we have that
$r(BP(f^{-1}))=r(BP(f))\leq n$. Therefore, $\mathcal{R}_{n}$ is a group.

We will refer to $\mathcal{R}_{n}$ as group of generalized piecewise-linear
transformations. By definition, we have that the groups $\mathcal{R}_{n}$
defined above give the following ascending chain of groups:

$\displaystyle\mathrm{PL}_{+}(I)=\mathcal{R}_{0}\subset\mathcal{R}_{1}\subset\mathcal{R}_{2}\subset\mathcal{R}_{3}\subset\dots$

In the next proposition, we build a length function in $\mathcal{R}_{n}$ that
we will use to test that an element is undistorted. To do this, we will use
the following notation: given a subset $A$ of the interval $[0,1]$, we denote
by $A^{(n)}$ the subset that we obtain by applying derivation $n$ times to the
set $A$ (i.e. $A^{(n)}$ is the $n^{\mathrm{th}}$ derived subset of $A$).

######  Proposition 3.3.

Given $n\in\mathbb{N}$, the function
$L_{n}:\mathcal{R}_{n}\rightarrow\mathbb{N}$ defined as
$L_{n}(f):=|BP(f)^{(n)}|$ is a length function.

Proof. It is clear that $L_{n}(id)=0$ because the identity has no break point.
Also, again by the chain rule, we have that

$\displaystyle BP(f)=f^{-1}(BP(f^{-1})).$

It thus follows from Proposition 2.13 that $L_{n}(f)=L_{n}(f^{-1})$.

It remains to prove the subadditive inequality. Let $f$ and $g$ be any
elements in the group $\mathcal{R}_{n}$. We already know that

$\displaystyle BP(f\circ g)\subseteq g^{-1}(BP(f))\cup BP(g).$

Taking derived sets $n$ times we obtain finite sets, and

$\displaystyle L_{n}(f\circ g)$ $\displaystyle=|BP(f\circ g)^{(n)}|$
$\displaystyle\leq|(g^{-1}(BP(f))\cup BP(g))^{(n)}|$
$\displaystyle=|g^{-1}(BP(f)^{(n)})\cup BP(g)^{(n)}|$
$\displaystyle=|g^{-1}(BP(f)^{(n)})|+|BP(g)^{(n)}|-|g^{-1}(BP(f)^{(n)})\cap
BP(g)^{(n)}|$ $\displaystyle\leq|g^{-1}(BP(f)^{(n)})|+|BP(g)^{(n)}|$
$\displaystyle=|(BP(f)^{(n)})|+|BP(g)^{(n)}|$
$\displaystyle=L_{n}(f)+L_{n}(g).$

We conclude that $L_{n}$ is a length function.

######  Definition 3.4.

Let $n\in\mathbb{N}$ and let $f\in\mathcal{R}_{n}$ be any element. Since
$(L_{n}(f^{m}))_{m\geq 0}$ is a subadditive sequence, by Fekete´s lemma, the
following limit exists:

$\displaystyle l_{n}(f):=\lim_{m\rightarrow\infty}\frac{L_{n}(f^{m})}{m}.$

As before, this gives a tool to establish that an element in the group
$\mathcal{R}_{n}$ is undistorted. The criterion is given by the proposition
below.

######  Proposition 3.5.

Let $n\in\mathbb{N}$ and let $f\in\mathcal{R}_{n}$ be any element. If
$l_{n}(f)>0$, then $f$ is undistorted in $\mathcal{R}_{n}$.

Let us recall the endpoint homomorphism
$\eta:\mathrm{PL}_{+}[0,1]\rightarrow\mathbb{R}\oplus\mathbb{R}$ defined as
$\eta(f)=(\log Df_{+}(0),\log Df_{-}(1)).$ To prove Theorem A, we will work
with elements in the kernel of $\eta.$

Theorem A Let $n$ be a natural number. There exists $f\in Homeo_{+}[0,1]$ that
is undistorted in $\mathcal{R}_{n}$ but distorted in $\mathcal{R}_{n+1}$.

Outline of the proof: We will consider $f$ as a commutator in
$\mathcal{R}_{n}\setminus\mathcal{R}_{n-1}$ that satisfies our criterion
(Proposition 3.5). Then, we will deal with the distortion of $f$ in
$\mathcal{R}_{n+1}$ with a combination of two techniques:

  * •

(Diagonal trick) The powers of the commutator $f$ can be written as the
product of a bounded quantity of commutators.

  * •

(Mather’s argument) This is a recipe to build functions (belonging to
$\mathcal{R}_{n+1}$) allowing to write sequences of commutators with lengths
in a prescribed sequence of natural numbers (see [1]).

Building the element $f$. We fix $n\in\mathbb{N}$. Consider a sufficiently
small interval $I$ that is contained $(0,1)$. Let $f_{1}$ satisfy the
following properties:

  * •

We have that $supp(f_{1})\subseteq I$ and $f_{1}(x)\geq x$ for all
$x\in[0,1]$.

  * •

There exist $x_{0}$ and $x_{1}$ in $Supp(f_{1})$ such that $x_{0}<x_{1}$,
$f_{1}(x_{0})=x_{1}$ and $f_{1}$ is linear outside $[x_{0},x_{1}]$.

  * •

$f_{1}\in\mathcal{R}_{n}\setminus\mathcal{R}_{n-1}$.

Observe that $f_{1}$ cannot be distorted in $\mathcal{R}_{n}$. Indeed, for
every $k\in\mathbb{N}$ we have
$|BP(f_{1}^{k})^{(n)}|=k\cdot|BP(f_{1})^{(n)}|$, and since
$0<|BP(f_{1})^{(n)}|<\infty$, this implies that
$l_{n}(f_{1})=|BP(f_{1})^{(n)}|>0,$ which by Proposition 3.5 shows that
$f_{1}$ is undistorted in $\mathcal{R}_{n}.$

We want to build $f$ as a commutator. To do this, consider $t\in Ker(\eta)$
such that:

  * •

$t(x)\geq x$ for all $x\in[0,1].$

  * •

$t(I)\cap I=\emptyset.$

Let $f_{2}:=tf_{1}t^{-1}$. We define $f:=[f_{1},t]=f_{1}f_{2}^{-1}$. Note that
$f_{1}$ and $f_{2}$ have disjoint support. Since $f_{1}$ is undistorted in
$\mathcal{R}_{n}$, the element $f$ is also undistorted.

We next apply the diagonal trick to the element $f.$ This trick is well-known;
we refers to [3], where in his study of the stable conmutator length in
subgroups of $\mathrm{PL}_{+}(I)$, Calegari applies the diagonal trick in the
group $\mathrm{PL}_{+}(I)$.

Diagonal trick: We will prove that, for $m\geq 1$, the element $f^{m+1}$ can
be written as a product of two commutators.

Let us take an interval $J$ contained in $(0,1)$ that contains the support of
$f$. Let $h\in Ker(\eta)$ be such that:

  * •

$h(x)\geq x$ for all $x\in[0,1].$

  * •

$h(J)\cap J=\emptyset.$

Observe that the last property implies that for every
$i\in\mathbb{Z}\setminus\\{0\\}$, one has $h^{i}(J)\cap J=\emptyset.$ Consider
the two finitely generated groups $G_{0}:=\langle f_{1},t\rangle$ and
$G_{1}:=\langle f_{1},t,h\rangle$. The map $\Delta_{m}:G_{0}\rightarrow G_{1}$
defined as

$\displaystyle\Delta_{m}(g):=\prod_{i=0}^{m}h^{i}gh^{-i},$

is an injective homomorphism.

Let $f^{\prime}\in G_{1}$ be defined as

$\displaystyle f^{\prime}=\prod_{i=0}^{m}h^{i}f^{i+1}h^{-i}.$

We compute the commutator between $f^{\prime}$ and $h$,

$\displaystyle[f^{\prime},h]$
$\displaystyle=f^{\prime}\cdot(hf^{\prime-1}h^{-1})$
$\displaystyle=\left(\prod_{i=0}^{m}h^{i}f^{i+1}h^{-i}\right)\cdot\left(\prod_{i=1}^{m+1}h^{i}f^{-i}h^{-i}\right)$
$\displaystyle=\left(\prod_{i=0}^{m}h^{i}fh^{-i}\right)\cdot
h^{m+1}f^{-(m+1)}h^{-(m+1)},$

where the last equality is justified because the maps in each product have
support contained in one of the sets $I,h(I),...,h^{m}(I)$ or $h^{m+1}(I)$,
and maps with disjoint support commute. Then, we have the following relation:

$\displaystyle[f^{\prime},h]=\Delta_{m}(f)h^{m+1}f^{-(m+1)}h^{-(m+1)}.$

Finally, we can write $f^{m+1}$ as a product of two commutators:

$\displaystyle f^{m+1}$
$\displaystyle=h^{-(m+1)}(\Delta_{m}(f)\cdot[f^{\prime},h]^{-1})h^{(m+1)}$
$\displaystyle=h^{-(m+1)}[\Delta_{m}(f_{1}),\Delta_{m}(t)]\,\,h^{(m+1)}\cdot
h^{-(m+1)}[h,f^{\prime}]\,\,h^{(m+1)}$
$\displaystyle=[h^{-(m+1)}\Delta_{m}(f_{1})\,h^{(m+1)},h^{-(m+1)}\Delta_{m}(t)\,h^{(m+1)}]\cdot[h,h^{-(m+1)}f^{\prime}h^{(m+1)}],$

where we use that the inverse of a commutator is a commutator, a homomorphism
sends a commutator to a commutator, and conjugating a commutator gives us
another commutator.

Mather’s argument: The next lemma is Mather’s argument stated for our
situation. This classical trick can be found (in different versions) in [1],
[4], [12], [17], [18] and probably elsewhere. Let us point out that the
statement is inspired by Avila’s work (see [1]).

######  Lemma 3.6.

Let $[a,b]$ be an interval properly contained in (0,1) and let
$n\in\mathbb{N}$. There exists a (linear) sequence of natural numbers
$(k_{m})$ such that for any sequences $(f_{m})$ and $(g_{m})$ of
$\mathcal{R}_{n}\setminus\mathcal{R}_{n-1}$, such that
$supp(f_{m})\subseteq[a,b]$ and $supp(g_{m})\subseteq[a,b]$, there exists a
finite set $\mathcal{G}\subset\mathcal{R}_{n+1}$ such that in the subgroup of
$\mathcal{R}_{n+1}$ generated by $\mathcal{G}$ one has
$l_{\mathcal{G}}[f_{m},g_{m}]\leq k_{m}.$

Proof of the Lemma: After conjugating by an element in $\mathrm{PL}_{+}[0,1]$,
we can assume that $1/2$ belongs to $(a,b)$ and $0<b-a<1/2$. First we define
three homeomorphisms in $\mathrm{PL}_{+}[0,1]$ that will be used along the
proof. Let $0<a^{\prime}<a<b<b^{\prime}<1$. Let us define
$h\in\mathrm{PL}_{+}[0,1]$ as $h^{-1}(x)=\frac{x}{2}+\frac{1}{4}$ on
$[a^{\prime},b^{\prime}]$ and being linear outside $[a^{\prime},b^{\prime}].$
Let $r\in$ $\mathrm{PL}_{+}[0,1]$ be defined as $r(x)=2x$ on $[0,\frac{3}{8}]$
and being linear on $[\frac{3}{8},1]$.

Let $\alpha\in(0,1)$ be sufficiently small such that $\alpha+b<1$. We let
$\widetilde{f}\in\ker(\eta)$ be such that

  * •

$\widetilde{f}(x)=x+\alpha$ for each $x\in[a,b]$.

  * •

$supp(\widetilde{f})\subseteq[a^{\prime},b^{\prime}]$ and
$0<b^{\prime}-a^{\prime}<1/2$.

We will define the sequence that we will use to compute commutators. Consider
the sequence $(F_{m})$ defined as

$\displaystyle F_{m}=r^{-m}h^{-m}\widetilde{f}h^{m}r^{m}.$

For each $m\geq 1$, we denote $T_{m}$ the interval
$T_{m}:=[r^{-m}h^{-m}(a^{\prime}),r^{-m}h^{-m}(b^{\prime})]$. Note that
$Supp(F_{m})\subseteq T_{m}$.

Let us prove that, for all $m\geq 1$, we have that $T_{m+1}\cap
T_{m}=\emptyset.$ By construction, it is enough to show that
$r^{-(m+1)}h^{-(m+1)}(b^{\prime})<r^{-m}h^{-m}(a^{\prime})$. An explicit
computation shows that

$\displaystyle T_{m}$
$\displaystyle=[r^{-m}h^{-m}(a^{\prime}),r^{-m}h^{-m}(b^{\prime})]$
$\displaystyle=\left[\frac{a^{\prime}}{2^{2m}}+\frac{1}{2^{2m+1}}+\frac{1}{2^{2m-1}}+\cdots+\frac{1}{2^{m+2}},\frac{b^{\prime}}{2^{2m}}+\frac{1}{2^{2m+1}}+\frac{1}{2^{2m-1}}+\cdots+\frac{1}{2^{m+2}}\right].$

Since $0<b^{\prime}-a^{\prime}<1/2$, it follows that, for all $m\geq 1$, one
has

$0<(2^{m+1}-3)-2(b^{\prime}-a^{\prime})+6a^{\prime}.$

This inequality implies that

$\displaystyle r^{-(m+1)}h^{-(m+1)}(b^{\prime})$
$\displaystyle=\frac{b^{\prime}}{2^{2m+2}}+\frac{1}{2^{2m+3}}+\frac{1}{2^{2m+2}}+\cdots+\frac{1}{2^{m+3}}$
$\displaystyle<\frac{a^{\prime}}{2^{2m}}+\frac{1}{2^{2m+1}}+\frac{1}{2^{2m-1}}+\cdots+\frac{1}{2^{m+2}}$
$\displaystyle=r^{-m}h^{-m}(a^{\prime}),$

and, therefore, $T_{m+1}$ and $T_{m}$ are disjoint sets.

We will build two elements $\widetilde{F}$ and $\widetilde{G}$ in
$\mathcal{R}_{n+1}$ with support contained in $\cup_{m\geq 1}T_{m}$. First, we
define the set where the maps $\widetilde{F}$ and $\widetilde{G}$ will be
supported. Let us fix $m_{0}\in\mathbb{N}$ such that

$2^{m_{0}}\alpha>b-a.$ (2)

For each $m\in\mathbb{N}$, we define the interval
$T_{m}^{\prime}:=[r^{-m}h^{-m_{0}-m}(a),r^{-m}h^{-m_{0}-m}(b)]$. Let us
justify the choice of $m_{0}$ because it implies that $T_{m}^{\prime}$ and
$F_{m}(T_{m}^{\prime})$ are disjoint. Indeed, since for all $x\in[0,1]$ we
have $F_{m}(x)\geq x$, to check this it is enough to prove that

$r^{-m}h^{-m_{0}-m}(b)<F_{m}(r^{-m}h^{-m_{0}-m}(a)).$

Using the definition of $F_{m}$, the previous inequality is equivalent to

$h^{-m_{0}}(b)<\widetilde{f}h^{m}(a).$

The last inequality is exactly the one in (2).

We define $\widetilde{F}$ as the map whose restriction to each
$T_{m}^{\prime}$ is given by

$\displaystyle\widetilde{F}|_{T_{m}^{\prime}}=r^{-m}h^{-m_{0}-m}f_{m}h^{m_{0}+m}r^{m},$

and the identity elsewhere. Analogously, we build $\widetilde{G}$ using the
sequence $(g_{m})$.

Observe that since the elements of each sequence $(f_{m})$ and $(g_{m})$
belong to $\mathcal{R}_{n}$, after taking the $n^{\mathrm{th}}$ derived
subsets of $BP(\widetilde{F})$ and $BP(\widetilde{G})$, we obtain finite sets
on each $T_{m}^{\prime}.$ Since the intervals $T_{m}^{\prime}$ accumulate at
$0$, after taking the derived set once more, we obtain that
$|BP(\widetilde{F})^{(n+1)}|=|BP(\widetilde{G})^{(n+1)}|=1$. Therefore,
$\widetilde{F}$ and $\widetilde{G}$ belong to $\mathcal{R}_{n+1}$.

We define
$\mathcal{G}:=\\{\widetilde{F},\widetilde{G},h,r,\widetilde{f}\\}\subset\mathcal{R}_{n+1}$.
For each $m\geq 1$, we consider the following elements in
$\langle\mathcal{G}\rangle$:

$\displaystyle
A_{m}:=[\widetilde{F},F_{m}],\quad\,B_{m}:=[\widetilde{G},F_{m}],\quad\,C_{m}:=[\widetilde{F}^{-1}\widetilde{G}^{-1},F_{m}].$

We note that each of these homeomorphisms has support in the disjoint
intervals $T_{m}^{\prime}$ and $F_{m}(T_{m}^{\prime})$. The maps $A_{m},B_{m}$
and $C_{m}$ restricted to $T_{m}^{\prime}$ are equal to $\widetilde{F}$,
$\widetilde{G}$ and $\widetilde{F}^{-1}\widetilde{G}^{-1}$, respectively, and
restricted to $F_{m}(T_{m}^{\prime})$ are equal to
$F_{m}\widetilde{F}^{-1}F_{m}^{-1}$, $F_{m}\widetilde{G}^{-1}F_{m}^{-1}$ and
$F_{m}\widetilde{G}\widetilde{F}F_{m}^{-1}$, respectively.

Hence, $A_{m}B_{m}C_{m}$ has support in $T_{m}^{\prime}$, and

$\displaystyle
A_{m}B_{m}C_{m}=r^{-m}h^{-m_{0}-m}f_{m}g_{m}f_{m}^{-1}g_{m}^{-1}h^{m_{0}+m}r^{m}.$

Then,
$H_{m}:=h^{m_{0}+m}r^{m}A_{m}B_{m}C_{m}r^{-m}h^{-m_{0}-m}\in\langle\mathcal{G}\rangle$
satisfies that $H_{m}=[f_{m},g_{m}]$ and $l_{\mathcal{G}}(H_{m})\leq
k_{m}=28m+2m_{0}+14$.

Proof of Theorem A: We consider the element $f$ built above. As we mentioned
before, this element is undistorted in $\mathcal{R}_{n}.$

Given the sequence $(k_{m})$ of Lemma 3.6, we fix a sequence $(i_{m})$ of
natural numbers such that

$\lim_{m\rightarrow\infty}\frac{k_{m}}{i_{m}+1}=0.$

By the Diagonal Trick, the powers of $f$ can be written as products of two
commutators $f^{i_{m}+1}=[a_{i_{m}},b_{i_{m}}][c_{i_{m}},d_{i_{m}}]$ where

$a_{i_{m}}:=h^{-(i_{m}+1)}\Delta_{i_{m}}(f_{1})h^{(i_{m}+1)},\,\,\,b_{i_{m}}:=h^{-(i_{m}+1)}\Delta_{i_{m}}(t)\,h^{(i_{m}+1)},\,\,\,c_{i_{m}}:=h,$
$d_{i_{m}}:=h^{-(i_{m}+1)}f^{\prime}h^{(i_{m}+1)}.$

Also, from the construction we know that all the supports of the elements
$a_{i_{m}},b_{i_{m}},c_{i_{m}}$ and $d_{i_{m}}$ are contained in the support
of $h$, so there exists an interval properly contained in $(0,1)$ that
contains the supports of $a_{i_{m}},b_{i_{m}},c_{i_{m}}$ and $d_{i_{m}}$.
Then, applying Lemma 3.6 to the sequences of commutators, we obtain that there
exist finite subsets $\mathcal{G}$ and $\mathcal{G^{\prime}}$ of
$\mathcal{R}_{n+1}$ such that $l_{\mathcal{G}}([a_{i_{m}},b_{i_{m}}])\leq
k_{m}$ and $l_{\mathcal{G^{\prime}}}([c_{i_{m}},d_{i_{m}}])\leq k_{m}.$ Hence,
by the triangular inequality,

$\displaystyle l_{\mathcal{G}\cup\mathcal{G^{\prime}}}(f^{i_{m}+1})\leq
2k_{m},$

and therefore $f$ is distorted in $\mathcal{R}_{n+1}.$

Question 4. Is every element $f\in$ $\mathrm{PL}_{+}(I)$ distorted in
$\mathcal{R}_{1}$? More generally, is every element of $\mathcal{R}_{n}$
distorted in $\mathcal{R}_{n+1}$?

######  Observation 3.7.

Let us define for each $n\in\mathbb{N}$ the following set

$\widetilde{\mathcal{R}_{n}}:=\\{f\in\mathcal{R}_{n}:f\mbox{ is
bilipchitz}\\}.$

With this definition, again we obtain a chain of groups as before

$\displaystyle\mathrm{PL}_{+}(I)\leq\widetilde{\mathcal{R}_{1}}\leq\widetilde{\mathcal{R}_{2}}\leq\widetilde{\mathcal{R}_{3}}\leq\dots$

Similarly, we could ask whether there exist a distorted elements in
$\widetilde{\mathcal{R}_{n+1}}$ that is undistorted in
$\widetilde{\mathcal{R}_{n}}$. However, the situation here is different,
because there is no distorted elements in the groups
$\widetilde{\mathcal{R}_{n}}$. Indeed, we can consider the function
$L:\widetilde{\mathcal{R}_{n}}\rightarrow\mathbb{R}$ defined as

$\displaystyle L(f):=\log(||f||_{Lip}),$

where $||\cdot||_{Lip}$ is taken in the complement of the set of break points.
We see that $L$ is a length function and if $f\in\widetilde{\mathcal{R}_{n}}$
is different of the identity, then we have that $L(f)>0$ and $L(f^{n})\geq
n\cdot L(f).$ Therefore, $f$ is undistorted.

######  Remark 3.8.

Let us explain why the combination of the Diagonal Trick and the Mather’s
argument does not work for the groups $\widetilde{\mathcal{R}_{n}}$. Indeed,
as $m$ increases, the commutators of the diagonal arguments take account
arbitrarily large powers of the element $f$, and so the elements
($\widetilde{F}$ and $\widetilde{G}$) created by Mather’s trick cannot belong
to $\widetilde{\mathcal{R}_{n+1}}$ because they are not bilipchitz.

Acknowledgments. I would like to gratefully acknowledge to Yash Lodha for
suggesting the problem and Andrés Navas for proposing it to me as well as for
the guidance and many encouragements during the work. Also, I thank to Hélène
Eynard-Bontemps, Cristóbal Rivas and Michelle Triestino for the comments on
previous versions of this article. I thank the hospitality of IHP during the
trimester ”Around the Zimmer program”, I acknowledge support of the Institut
Henri Poincaré (UAR 839 CNRS-Sorbonne Université) and LabEx CARMIN
(ANR-10-LABX-59-01). I was partially funded by Fondecyt regular 1220032.

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Leonardo Dinamarca Opazo

Dpto. de Matemática y Ciencias de la Computación, Univ. de Santiago de Chile
(USACH)

Alameda Lib. Bdo O’Higgins 3363, Estación Central, Santiago, Chile

Emails<EMAIL_ADDRESS>