# Irreversible and dissipative systems

J. Beck  Department of Mathematics, Hill Center for the Mathematical Sciences,
Rutgers University, Piscataway NJ 08854, USA<EMAIL_ADDRESS>,  W.W.L.
Chen  School of Mathematical and Physical Sciences, Faculty of Science and
Engineering, Macquarie University, Sydney NSW 2109, Australia
<EMAIL_ADDRESS>and  Y. Yang  School of Science, Beijing University of
Posts and Telecommunications, Beijing 100876, China<EMAIL_ADDRESS>
###### Abstract.

We study some new dynamical systems where the corresponding piecewise linear
flow is neither time reversible nor measure preserving. We create a
dissipative system by starting with a finite polysquare translation surface,
and then modifying it by including a one-sided barrier on a common vertical
edge of two adjacent atomic squares, in the form of a union of finitely many
intervals. The line flow in this system partitions the system into a transient
set and a recurrent set. We are interested in the geometry of these two sets.

###### Key words and phrases:

flow lines, dissipative, transient, recurrent

###### 2010 Mathematics Subject Classification:

11K38, 37E35

##  1\. Introduction

We continue our research into non-integrable systems concerning billiard flow
in polygons and solids as well as the related problem of geodesic flow on flat
surfaces and in flat manifolds. In the cases that we have studied thus far,
the flow is piecewise linear, time reversible and measure preserving. In
particular, the last property permits the application of the Birkhoff ergodic
theorem, and this leads to uniformity of the relevant orbits inside flow-
invariant sets.

In this paper, we study some new dynamical systems where the corresponding
piecewise linear flow is neither time reversible nor measure preserving. Here
the work of Veech [9] concerning the $2$-circle problem serves as our
motivation. Related to the $2$-circle problem is a polysquare translation
surface comprising $2$ atomic squares, modified by the introduction of some
partial barriers on some edges of the squares. The question then arises as to
what may happen if these barriers are one-sided, permitting flow in one
direction but not the other. We show that these one-sided barriers are
responsible for the violation of both the time reversible and the measure
preserving properties.

Dynamical systems with such long term behaviour are said to be dissipative.
Intuitively, it means that the time evolution of the flow contracts the space-
volume.

The Lorentz attractor, motivated by weather prediction and claiming that the
time evolution contracts the space-volume exponentially fast, is perhaps
famous and notorious with equal measure. It is the quintessential mathematical
model of chaos theory which has a huge literature. Thus it is somewhat
unsatisfactory that little has been proved with any mathematical rigour
concerning the Lorentz attractor.

Our flat dissipative systems here are quite different, and exhibit slow
contraction of the space-volume, so we may use the term slow chaos. However,
we are able to carry out a rigorous mathematical discussion on these systems
with precise proofs.

Time irreversibility is a very interesting property that brings one to the
well known Loschmidt paradox, or reversibility-irreversibility paradox, which
serves as a criticism of the so-called H-theorem of Boltzmann. Here a
fundamental problem of the proof of Boltzmann is circular reasoning and
concerns an element of time asymmetry artifically injected into an argument to
establish time asymmetry.

The Loschmidt paradox is still fresh after $150$ years, perhaps due to the
lack of any time irreversible mathematical model with a detailed theory
providing theorems and proofs. We hope our work here will bring a little
insight into this very intriguing question.

We introduce a dissipative system $\mathcal{P}$ by starting first with a
finite polysquare translation surface, and then modifying it by including a
one-sided barrier $B$ on a common vertical edge of two adjacent atomic
squares, in the form of a union of finitely many vertical intervals, and where
a set $A$ corresponding to $B$ is chosen on a different vertical edge on the
same horizontal street that contains $B$. We consider a flow moving from left
to right. When a flow line with irrational slope hits the barrier $B$ from the
left, it continues from the corresponding point of the set $A$. We then
observe that there are points on $\mathcal{P}$ that stay away from the flow
line, apart possibly from the early stages, and there are points on
$\mathcal{P}$ with arbitrarily small neighbourhoods that are visited by the
flow line infinitely often as time goes to infinity. The system $\mathcal{P}$
is thus divided into two subsets, and we are interested in the structure of
these two subsets which we call respectively the transient set and the
recurrent set.

We begin our investigation by considering in Section 2 the special case of the
$2$-square torus, modified by the inclusion of a one-sided barrier on the
common vertical edge. We show that there is a simple reverse flow recipe which
helps us investigate this question, and that both the transient set and the
recurrent set are finite unions of polygons with total area $1$ and with
boundary edges that are either vertical or parallel to the direction of the
flow. We also explain how this problem also possesses a form of cyclic
symmetry.

We then proceed in Section 3 to investigate the corresponding problem with the
$n$-square torus modified in a similar way. Here we may lose the cyclic
symmetry which is crucial in the earlier investigation. Nevertheless we are
able to establish some partial results. Then in Section 4, we consider the
more general problem of starting with a polysquare translation surface, and
use simple surfaces related to the L-surface to illustrate some properties of
the transient sets and the recurrent sets. We show that each can take up a
very large proportion of the system or a very small proportion of the system,
and that the recurrent set can be split into disjoint components. We then make
a thorough study of the case of the modified L-surface.

In Section 5, we study the problem of modified polysquare translation surfaces
in greater detail. Here we develop an extension process which is iterative and
helps us study more closely the recurrent set. We also use some important
results of Kakutani and Kac. However, the main question is whether this
extension process is finite, in which case we can conclude that both the
transient set and the recurrent set are finite unions of polygons and with
boundary edges that are either vertical or parallel to the direction of the
flow.

Section 6 is devoted to the study of this question. We begin this
investigation by studying two special cases of the problem, one by making some
assumptions on the one-sided barrier and the other by making some assumption
on the recurrent set. We conclude our investigation by finally showing that
the extension process is indeed always finite, and this leads to the very nice
description of the transient sets and the recurrent sets.

##  2\. Modification of the Veech model

Consider a modified billiard table comprising $2$ atomic squares, with a one-
sided partial barrier of length $b$, where $0<b<1$, on the common edge
separating the atomic squares, as shown in Figure 1. This barrier is one-sided
in the sense that when the billiard hits it from the left, as shown in the
picture on the left, it rebounds like normal billiard, whereas when the
billiard hits it from the right, as shown in the picture on the right, it
passes straight through. For simplicity, we have put this one-sided barrier
against the bottom edge, but it can be anywhere on the vertical edge
separating the atomic squares.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-1.pdf}\\\ \mbox{Figure
1: a modified billiard table}\end{array}$

Note that the billiard flow in Figure 1, if not entirely horizontal or
vertical, is a $4$-direction flow. We can convert it into a $1$-direction flow
on some modified system using the simple idea of König and Szücs [6] in 1913
that involves reflection across a horizontal axis and across a vertical axis.
This results in $1$-direction flow on the modified system as shown in Figure
2. However, we need to describe this $1$-direction flow very carefully.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-2.pdf}\\\ \mbox{Figure
2: a modified system corresponding to Figure~{}1}\end{array}$

Figure 3 describes the flow on the modified system. The left side of the
vertical barrier on the left is identified with the right side of the vertical
barrier on the right. Thus a flow line that hits the left side of the vertical
barrier on the left continues in the same direction from the corresponding
point on the right side of the vertical barrier on the right. On the other
hand, the left side of the vertical barrier on the right is also identified
with the right side of the vertical barrier on the right. Thus a flow line
that hits the left side of the vertical barrier on the right continues
straight through in the same direction. Indeed, the $1$-direction flows
indicated in Figure 3 correspond precisely to the billiard flows in Figure 1.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-3.pdf}\\\ \mbox{Figure
3: geodesic flow on the system in Figure~{}2}\end{array}$

Note that the system in Figure 3 is modified from a surface, but is not a
surface. The left side of both the vertical barrier on the left and the
vertical barrier on the right are identified with the right side of the
vertical barrier on the right, so there is no pairwise identification of
edges. The important point is that we can still study $1$-direction flow on
it, although this flow is not time reversible.

The systems in Figures 3 and 4 are equivalent. The latter dynamical system
satisfies the following convention which we now adopt.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-4.pdf}\\\ \mbox{Figure
4: a system equivalent to Figures 2 and~{}3}\end{array}$

######  Convention.

Henceforth we consider dissipative systems involving $1$-direction flow from
left to right and with irrational slope $\alpha>0$. When the dissipative flow
line hits a one-sided barrier, it then continues from the corresponding point
on the left vertical edge of the system and in the same direction.

Consider the dissipative system $\mathcal{P}$ obtained from the $2$-square
torus modified by the inclusion of a one-sided barrier which may include both,
one or neither of the endpoints $(1,a)$ and $(1,b)$, as shown in Figure 5. We
see that the two distinct flow line segments below the dotted line both
continue along the bold flow line segment. Thus any measure preserving
property of the flow is violated by a factor of $2$, and so we cannot directly
apply ergodic theory.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-5.pdf}\\\ \mbox{Figure
5: lack of measure preserving property}\end{array}$

For the system $\mathcal{P}$ shown in Figure 5, the dissipative flow of fixed
irrational slope $\alpha>0$ can be described in terms of an operator
$\mathbf{T}_{\alpha}^{(t)}$, where $t\geqslant 0$ denotes time. Roughly
speaking, this operator denotes advancing geodesic flow of the fixed slope
$\alpha$ by time $t$. More precisely, for any dissipative flow line
$\mathcal{L}(t)$, $t\geqslant 0$, of the fixed slope $\alpha$, we write
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\mathbf{T}_{\alpha}^{(t)}(\mathcal{L}(t_{1}))=\mathcal{L}(t_{1}+t),\quad
t_{1}\geqslant 0.$

In other words, the operator $\mathbf{T}_{\alpha}^{(t)}$ advances the point
$\mathcal{L}(t_{1})$ on the flow line to the point $\mathcal{L}(t_{1}+t)$
after a further time $t$ has elapsed.

Using this operator $\mathbf{T}_{\alpha}^{(t)}$, we can partition
$\mathcal{P}$ into two subsets. To define these two subsets, we use some
concepts introduced by Birkhoff [2] for a very general class of dynamical
systems in 1927.

We say that a point $P\in\mathcal{P}$ is a wandering point of $\mathcal{P}$ if
there exists an open neighbourhood $D=D(P)$ of the point $P$ and a finite
threshold $t_{0}\geqslant 0$ such that the intersection
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$D\cap\mathbf{T}_{\alpha}^{(t)}(D)=\emptyset\quad\mbox{for every $t>t_{0}$};$
(2.1)

so that a dissipative flow line of slope $\alpha$ and starting from $P$ is
bounded away from $P$ after time $t_{0}$. Furthermore, we say that a point
$P\in\mathcal{P}$ is a non-wandering point of $\mathcal{P}$ if it is not a
wandering point of $\mathcal{P}$.

The set $\mathcal{W}(\mathcal{P};\alpha)$ of all wandering points of
$\mathcal{P}$ is called the transient set of $\mathcal{P}$, and the set
$\mathcal{R}(\mathcal{P};\alpha)$ of all non-wandering points of $\mathcal{P}$
is called the recurrent set or attractor of $\mathcal{P}$. These two subsets
of $\mathcal{P}$ are complements of each other, and give a partition of
$\mathcal{P}$.

It follows easily from the definition that the transient set
$\mathcal{W}(\mathcal{P};\alpha)$ is open and that the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ is closed and invariant under dissipative
flow of slope $\alpha$. These two subsets are illustrated in Figure 6 in the
case of the modified $2$-square torus in Figure 5. We emphasize in particular
their dependence on the fixed slope $\alpha$ of the flow under consideration.
It is also worth noting that the definition of wandering points and non-
wandering points is given in terms of open neighbourhoods of the points and
their images under the operator $\mathbf{T}_{\alpha}^{(t)}$, and not in terms
of the points and dissipative flow lines that start from them.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-6.pdf}\\\ \mbox{Figure
6: transient set and recurrent set}\end{array}$

Figure 7 illustrates that any dissipative flow line of slope $\alpha$ that
starts from a point $P\in\mathcal{R}(\mathcal{P};\alpha)$ remains in the
recurrent set $\mathcal{R}(\mathcal{P};\alpha)$, so that the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ is invariant under dissipative flow of slope
$\alpha$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-7.pdf}\\\ \mbox{Figure
7: invariance of the recurrent set under dissipative flow}\end{array}$

Meanwhile, Figure 8 illustrates that any dissipative flow line of slope
$\alpha$ that starts from a point $P\in\mathcal{W}(\mathcal{P};\alpha)$
eventually moves into the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$.
Once there, it clearly never returns to the transient set
$\mathcal{W}(\mathcal{P};\alpha)$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-8.pdf}\\\ \mbox{Figure
8: leaving the transient set to join the recurrent set}\end{array}$

We comment here that the above definitions for
$\mathcal{W}(\mathcal{P};\alpha)$ and $\mathcal{R}(\mathcal{P};\alpha)$ remain
valid for any system $\mathcal{P}$ that arises when a finite polysquare
translation surface is modified by the addition of a one-sided barrier on the
common vertical edge of $2$ atomic squares. However, we need to pay some
attention to the existence of pathological points. These are points
$P\in\mathcal{P}$ where a dissipative flow line of slope $\alpha$ that starts
from the point $P$ hits a singularity of $\mathcal{P}$ and becomes undefined.
For the modified $2$-square torus in Figures 5 and 6, provided that we clearly
indicate whether the one-sided barrier includes either endpoint, then while
these endpoints are singularities, they are not pathological points because we
have clear indication on how the flow line continues after hitting either of
them. Pathological points do not affect our argument, as they form a set of
$2$-dimensional Lebesgue measure $0$.

######  Theorem 1.

Consider dissipative flow of slope $\alpha$, where $\alpha>0$ is irrational,
on a system $\mathcal{P}$ such as those shown in Figures 4–8, where the
$2$-square torus has been modified with the inclusion of a one-sided barrier
on the common vertical edge of the atomic squares, in the form of a union of
finitely many vertical intervals. For the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ and the transient set
$\mathcal{W}(\mathcal{P};\alpha)$, the following hold:

_(i)_ Both sets are finite unions of polygons with total area $1$ and with
boundary edges that are vertical or of slope $\alpha$.

_(ii)_ The recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ is invariant under
dissipative flow of slope $\alpha$, and every half-infinite dissipative flow
line of slope $\alpha$ in the set $\mathcal{R}(\mathcal{P};\alpha)$ is
uniformly distributed in this set.

_(iii)_ Every half-infinite dissipative flow line of slope $\alpha$ that
starts from a point in the transient set $\mathcal{W}(\mathcal{P};\alpha)$
eventually leaves the set, moves to the set $\mathcal{R}(\mathcal{P};\alpha)$
and is uniformly distributed there.

###### Proof.

We proceed in a number of steps.

######  Step 1.

The pointwise definition of the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ does not provide any hint on how we may
explicitly construct and describe the set. Indeed, in the general situation,
this is a rather difficult problem. However, for the modified $2$-square torus
under consideration, we consider a no-go zone
$\mathcal{N}(\mathcal{P};\alpha)$ illustrated in Figure 9. This is an open
subset of $\mathcal{P}$ obtained by sweeping the barrier $B$ without its
endpoints by the dissipative flow of slope $\alpha$ towards the right vertical
edge of $\mathcal{P}$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-9.pdf}\\\ \mbox{Figure
9: the $2$-square torus modified with one-sided barriers}\end{array}$

We claim that
$\mathcal{N}(\mathcal{P};\alpha)\subset\mathcal{W}(\mathcal{P};\alpha)$; in
other words, that every point of the set $\mathcal{N}(\mathcal{P};\alpha)$ is
a wandering point of $\mathcal{P}$. Since $\mathcal{N}(\mathcal{P};\alpha)$ is
open, every point of $\mathcal{N}(\mathcal{P};\alpha)$ is contained in an open
subset $D\subset\mathcal{N}(\mathcal{P};\alpha)$. To justify this claim, it
suffices to show that for every open subset
$D\subset\mathcal{N}(\mathcal{P};\alpha)$, there exists $t_{0}>0$ such that
(2.1) holds. Now let $D^{\prime}$ denote the image of $D$ on the right side of
the one-sided barrier $B$ under projection in the direction $(-1,-\alpha)$, as
illustrated in the picture on the right in Figure 9. Then $D^{\prime}$ is an
open subinterval of $B$ and

$D^{\prime}\cap\mathbf{T}_{\alpha}^{(t)}(D)=\emptyset\quad\mbox{for every
$t>0$},$

so that for every $t_{0}$ satisfying $0\leqslant
t_{0}\leqslant(1+\alpha^{2})^{1/2}$,

$\mathbf{T}_{\alpha}^{(t_{0})}(D^{\prime})\cap\mathbf{T}_{\alpha}^{(t)}(D)=\emptyset\quad\mbox{for
every $t>(1+\alpha^{2})^{1/2}$}.$

It is not difficult to see that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$D\subset\bigcup_{0\leqslant
t_{0}\leqslant(1+\alpha^{2})^{1/2}}\mathbf{T}_{\alpha}^{(t_{0})}(D^{\prime}).$

It then follows that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$D\cap\mathbf{T}_{\alpha}^{(t)}(D)=\emptyset\quad\mbox{for every
$t>(1+\alpha^{2})^{1/2}$},$ (2.2)

justifying our claim.

######  Step 2.

We define the open set $\mathfrak{M}(\mathcal{P};B;\alpha)$ to be the largest
possible extension of the no-go zone $\mathcal{N}(\mathcal{P};\alpha)$. To
find this set, let $A$ be an interval on the left vertical edge of
$\mathcal{P}$, identical to the interval $B$ that represents the one-sided
barrier, as shown in Figure 6. Consistent with the fact that the transient set
$\mathcal{W}(\mathcal{P};\alpha)$ is open, we assume that the interval $B$ is
open. This assumption is valid, since any deviation from it affects only sets
of $2$-dimensional Lebesgue measure $0$.

######  Reverse flow recipe.

A given point $P\in\mathcal{P}$ satisfies
$P\in\mathfrak{M}(\mathcal{P};B;\alpha)$ if and only if starting from $P$, an
ordinary geodesic on the $2$-square torus in the direction $(-1,-\alpha)$ hits
the open interval $B$ before hitting the corresponding interval $A$ on the
left vertical edge of $\mathcal{P}$.

Using time-quantitative equidistribution of geodesic flow on the $2$-square
torus in the form of the Koksma–Erdős–Turán version of the Weyl criterion, it
follows that for any given irrational number $\alpha>0$, there is a finite
threshold $T_{0}(\alpha)$ such that starting from any point $P$ and moving in
the direction $(-1,-\alpha)$, this geodesic segment hits $B$ or $A$ in time
less than $T_{0}(\alpha)$. Hence the set $\mathfrak{M}(\mathcal{P};B;\alpha)$
is a finite union of polygons with boundary edges that are vertical or of
slope $\alpha$. We shall comment further on the finiteness of $T_{0}(\alpha)$
in the Remark after the completion of the proof of Theorem 1.

We claim that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\mathfrak{M}(\mathcal{P};B;\alpha)\subset\mathcal{W}(\mathcal{P};\alpha);$
(2.3)

in other words, that every point of the set
$\mathfrak{M}(\mathcal{P};B;\alpha)$ is a wandering point of $\mathcal{P}$.
Since $\mathfrak{M}(\mathcal{P};B;\alpha)$ is open, every point of
$\mathfrak{M}(\mathcal{P};B;\alpha)$ is contained in an open subset
$D\subset\mathfrak{M}(\mathcal{P};B;\alpha)$. To justify this claim, it
suffices to show that for every open subset
$D\subset\mathfrak{M}(\mathcal{P};B;\alpha)$, there exists $t_{0}>0$ such that
(2.1) holds. An argument similar to that in Part 1 leads to an analogue of
(2.2), in the form

$D\cap\mathbf{T}_{\alpha}^{(t)}(D)=\emptyset\quad\mbox{for every
$t>T_{0}(\alpha)$},$

justifying the assertion (2.3).

Let
$\mathfrak{M}^{c}(\mathcal{P};B;\alpha)=\mathcal{P}\setminus\mathfrak{M}(\mathcal{P};B;\alpha)$
denote the set of all points $P\in\mathcal{P}$ that are not in
$\mathfrak{M}(\mathcal{P};B;\alpha)$. Then xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\mathcal{R}(\mathcal{P};\alpha)\subset\mathfrak{M}^{c}(\mathcal{P};B;\alpha).$
(2.4)

The $2$-cyclic symmetry of the system $\mathcal{P}$ then ensures that the sets
$\mathfrak{M}(\mathcal{P};B;\alpha)$ and
$\mathfrak{M}^{c}(\mathcal{P};B;\alpha)$ have essentially the same projection
on the unit torus $[0,1)^{2}$. It follows that both sets have $2$-dimensional
Lebesgue measure $1$.

######  Step 3.

Let $\mathcal{L}(t)$, $t\geqslant 0$, denote any half-infinite dissipative
flow line on $\mathcal{P}$. We say that a point $P\in\mathcal{P}$ is an
infinite-time limit point of $\mathcal{L}$ if there exists a sequence of time
instances $t_{1}<t_{2}<t_{3}<\ldots$ such that $t_{i}\to\infty$ and
$\mathcal{L}(t_{i})\to P$ as $i\to\infty$. Furthermore, we say that the
collection of all infinite-time limit points of $\mathcal{L}$ is the infinite-
time limit set of $\mathcal{L}$. By definition, this set is invariant under
dissipative flow of slope $\alpha$ and so has $2$-dimensional Lebesgue measure
$0$, $1$ or $2$. This cannot be equal to $0$, since its projection to the unit
torus $[0,1)^{2}$ has $2$-dimensional Lebesgue measure $1$. This cannot be
equal to $2$, since it does not contain points of
$\mathcal{N}(\mathcal{P};\alpha)$. Hence it has $2$-dimensional Lebesgue
measure $1$.

Suppose now that $\mathcal{L}(t)$, $t\geqslant 0$, denotes any such half-
infinite dissipative flow line on $\mathcal{P}$ that starts from a point in
the set $\mathfrak{M}^{c}(\mathcal{P};B;\alpha)$, so that
$\mathcal{L}(0)\in\mathfrak{M}^{c}(\mathcal{P};B;\alpha)$. We claim that the
entire dissipative flow line is disjoint from the set
$\mathfrak{M}(\mathcal{P};B;\alpha)$, so that the set
$\mathfrak{M}^{c}(\mathcal{P};B;\alpha)$ is invariant under dissipative flow
of slope $\alpha$. Suppose, on the contrary, that there exists $t_{1}>0$ such
that $\mathcal{L}(t_{1})\in\mathfrak{M}(\mathcal{P};B;\alpha)$. Then it
follows from the reverse flow recipe that there exists $t_{0}<t_{1}$ such that
$\mathcal{L}(t_{0})\in B$. There are two possibilities: (i) If $t_{0}>0$, then
the dissipative flow of slope $\alpha$ cannot take $\mathcal{L}(0)$ to
$\mathcal{L}(t_{0})$, a contradiction. (ii) If $t_{0}\leqslant 0$, then it
follows that $\mathcal{L}(0)\in\mathfrak{M}(\mathcal{P};B;\alpha)$, also a
contradiction. Thus the infinite-time limit set of $\mathcal{L}$ is a subset
of $\mathfrak{M}^{c}(\mathcal{P};B;\alpha)$. Since both sets have
$2$-dimensional Lebesgue measure $1$, they are essentially equal.

######  Step 4.

The essentially $2$-cyclic symmetry of the system $\mathcal{P}$ guarantees
that the sets $\mathfrak{M}(\mathcal{P};B;\alpha)$ and
$\mathfrak{M}^{c}(\mathcal{P};B;\alpha)$ split up the system $\mathcal{P}$ in
an essentially fair way. More precisely, for every $(x,y)\in[0,1)^{2}$, apart
from a set of $2$-dimensional Lebesgue measure $0$, precisely one of the two
points

$(x,y)\quad\mbox{and}\quad(x+1,y)$

lies in $\mathfrak{M}(\mathcal{P};B;\alpha)$ while the other lies in
$\mathfrak{M}^{c}(\mathcal{P};B;\alpha)$. We next use Figure 6 to guide our
discussion.

Starting from anywhere within the set
$\mathfrak{M}^{c}(\mathcal{P};B;\alpha)$, the dissipative flow of slope
$\alpha$ moves within the set $\mathfrak{M}^{c}(\mathcal{P};B;\alpha)$ which
is the white part in Figure 6. When it reaches the white part immediately to
the left of $B$ or the white part immediately to the left of $A$ on the right
vertical edge, it then jumps back to $A$. The fair split of $\mathcal{P}$
mentioned above implies that this jump is $1$-to-$1$ and measure preserving.
Thus the dissipative flow of slope $\alpha$ restricted to the set
$\mathfrak{M}^{c}(\mathcal{P};B;\alpha)$ is both invertible and measure
preserving.

Suppose that we start from $A$ and sweep it along by the dissipative flow of
slope $\alpha$. Then this certainly contains in the left atomic square the
twin of $\mathcal{N}(\mathcal{P};\alpha)$ in the right atomic square, as
illustrated in Figure 10.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-10.pdf}\\\
\mbox{Figure 10: no-go zone and its twin}\end{array}$

Furthermore, this flow returns to $A$ infinitely many times in a $1$-to-$1$
and measure preserving way. Let $\mathfrak{R}(\mathcal{P};A;\alpha)$ be the
resulting subset of $\mathcal{P}$, the construction of which requires only
finitely many zig-zaggings; see the Remark after the completion of the proof
of Theorem 1. It is also clear that

$\mathfrak{R}(\mathcal{P};A;\alpha)\subset\mathfrak{M}^{c}(\mathcal{P};B;\alpha).$
(2.5)

Observe next that $\mathcal{P}$ is a square-covering system, so the projection
of the dissipative flow of slope $\alpha$ to the unit torus $[0,1)^{2}$ is
ordinary integrable geodesic flow of slope $\alpha$. Since $\alpha$ is
irrational, the dissipative flow of slope $\alpha$ modulo one is ergodic. It
then follows that any subset of $\mathcal{P}$ of positive $2$-dimensional
Lebesgue measure that is invariant under the dissipative flow of slope
$\alpha$ must have $2$-dimensional Lebesgue measure $1$ or $2$. The inclusion
(2.5) then clearly implies that $\mathfrak{R}(\mathcal{P};A;\alpha)$ must have
$2$-dimensional Lebesgue measure $1$. This means that

$\mathfrak{R}(\mathcal{P};A;\alpha)=\mathfrak{M}^{c}(\mathcal{P};B;\alpha),$
(2.6)

apart possibly from an exceptional set of $2$-dimensional Lebesgue measure
$0$.

The measure preserving aspect then allows us to use the well known Poincare
recurrence theorem, and conclude that apart possibly from an exceptional set
of $2$-dimensional Lebesgue measure $0$, every point in
$\mathfrak{R}(\mathcal{P};A;\alpha)$ is a non-wandering point and so lies in
$\mathcal{R}(\mathcal{P};\alpha)$. Combining this observation with (2.4) and
(2.6), we conclude that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\mathfrak{R}(\mathcal{P};A;\alpha)=\mathfrak{M}^{c}(\mathcal{P};B;\alpha)=\mathcal{R}(\mathcal{P};\alpha),$

apart possibly from exceptional sets of $2$-dimensional Lebesgue measure $0$.

On the other hand, the dissipative flow of slope $\alpha$ moves $B$ first to
the gray part in Figure 6 which is $\mathcal{N}(\mathcal{P};\alpha)$. When it
reaches the gray part immediately to the left of $B$ or the gray part
immediately to the left of $A$ on right vertical edge, it then jumps to $A$.
Thus every point of $\mathfrak{M}(\mathcal{P};B;\alpha)$ is eventually moved
to $A$ by the flow. Since both sets $\mathcal{W}(\mathcal{P};\alpha)$ and
$\mathfrak{M}(\mathcal{P};B;\alpha)$ have $2$-dimensional Lebesgue measure
$1$, it follows that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\mathcal{W}(\mathcal{P};\alpha)=\mathfrak{M}(\mathcal{P};B;\alpha),$

apart possibly from an exceptional set of $2$-dimensional Lebesgue measure
$0$.

######  Step 5.

Uniformity in the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ is now
established by projecting to the unit torus $[0,1)^{2}$ and applying the
results such as the Birkhoff ergodic theorem and also unique ergodicity.

This completes the proof of Theorem 1. ∎

######  Remark.

The finiteness of $T_{0}(\alpha)$ in Step 2 of the preceding proof follows
from the fact that we are essentially considering geodesic flow in the unit
torus $[0,1)^{2}$ in the direction $(-1,-\alpha)$, and
$\mathcal{N}(\mathcal{P};\alpha)$ is extended to
$\mathfrak{M}(\mathcal{P};B;\alpha)$ in a finite number of zig-zaggings; i.e.
a finite number of traverses between the opposite vertical edges of atomic
squares. We also use the following fact from diophantine approximation.

######  Irrational rotation lemma.

Consider the irrational rotation sequence

$s_{0}+n\alpha,\quad n=0,1,2,3,\ldots,$

in the unit torus $[0,1)$, where $\alpha$ is irrational. For any fixed
$\varepsilon>0$, there is a finite threshold $N=N(\alpha;\varepsilon)$ such
that the union

$\bigcup_{n=0}^{N-1}(s_{0}+n\alpha-\varepsilon,s_{0}+n\alpha+\varepsilon)=[0,1)$

for any starting point $s_{0}$.

###### Proof.

A fundamental result in diophantine approximation says that

$\left|\alpha-\frac{p_{k}}{q_{k}}\right|<\frac{1}{q_{k}q_{k+1}},$

where $p_{k}/q_{k}$, $k=0,1,2,3,\ldots,$ is the sequence of convergents to
$\alpha$. Then

$\left|\ell\alpha-\frac{\ell
p_{k}}{q_{k}}\right|<\frac{\ell}{q_{k}q_{k+1}}\leqslant\frac{1}{q_{k+1}}<\frac{1}{q_{k}},\quad\ell=0,1,2,3,\ldots,q_{k}.$
(2.7)

Furthermore, $p_{k}$ and $q_{k}$ are coprime, so that as $\ell$ runs through
the numbers $1,\ldots,q_{k}$, $\ell p_{k}$ runs through a complete set of
residues modulo $q_{k}$. Choosing a positive integer $k$ such that the
denominator $q_{k}>1/\varepsilon$, we see that the threshold
$N=N(\alpha;\varepsilon)=q_{k}$ satisfies the requirements. ∎

##  3\. Extension of the modified Veech model

Theorem 1 can be extended to the case of the $n$-square torus modified by
identical one-sided barriers on the vertical edges of adjoining atomic
squares, in the form of a union of finitely many vertical intervals. Denote
this system by $\mathcal{P}$. Here Figure 11 illustrates $\mathcal{P}$ in the
special case $n=4$ and where the union consists of a single vertical interval.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-11.pdf}\\\
\mbox{Figure 11: the $4$-square torus modified with one-sided
barriers}\end{array}$

Consider a dissipative flow line on $\mathcal{P}$ with irrational slope
$\alpha>0$. When this flow line hits the left side of any one-sided barrier
$B_{1},\ldots,B_{n-1}$, it then continues in the same direction from the
corresponding point on the vertical interval $A$ on the left vertical edge.

We can prove that the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ has area
$1$ and is a finite union of polygons with boundary edges that are vertical or
of slope $\alpha$, and also that the transient set
$\mathcal{W}(\mathcal{P};\alpha)$ consists of $n-1$ parts that are, apart from
exceptional sets of $2$-dimensional Lebesgue measure $0$, all equivalent to
$\mathcal{R}(\mathcal{P};\alpha)$ modulo one. Furthermore, every dissipative
flow line with slope $\alpha$ in the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ is uniformly distributed in this set.
Moreover, every dissipative flow line with slope $\alpha$ that starts from any
point in the transient set $\mathcal{W}(\mathcal{P};\alpha)$ eventually leaves
the set, moves to the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ and is
uniformly distributed there.

While we have considered one-sided barriers, there is still some sort of
cyclic symmetry, of great help so far. However, the situation becomes
considerably more complicated if this cyclic symmetry is also violated, as in
the system illustrated in Figure 12 concerning a modified $n$-square torus in
the case $n=4$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-12.pdf}\\\
\mbox{Figure 12: a dissipative system while cyclic symmetry is
violated}\end{array}$

In this section, we establish the following simpler result.

######  Theorem 2 (weak area theorem).

Given any $\varepsilon>0$, there exists an irrational number $\alpha>0$ and a
positive integer $n$ such that for some $n$-square torus which is modified
with the addition of a one-sided barrier, the time evolution of the
dissipative flow of slope $\alpha$ contracts the initial space-volume to less
than $\varepsilon$-part. In other words, denoting the modified $n$-square
torus by $\mathcal{P}$, the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$
satisfies

$\frac{\lambda_{2}(\mathcal{R}(\mathcal{P};\alpha))}{n}<\varepsilon.$

As usual, $\lambda_{2}$ denotes the $2$-dimensional Lebesgue measure.

In fact, we establish the following result from which Theorem 2 follows easily
from the details of the construction by the application of a linear
transformation.

######  Theorem 3 (small attractor theorem).

Suppose that for an irrational number $\alpha>0$, the continued fraction
digits are unbounded. Then there exists an infinite sequence of vertical
intervals $B_{\sigma}$, $\sigma=1,2,3,\ldots,$ with the following property.
For every $\sigma=1,2,3,\ldots,$ let $\mathcal{P}_{\sigma}$ denote the unit
torus modified by the addition of a vertical one-sided barrier $B_{\sigma}$ as
illustrated in Figure 13, and consider the recurrent set
$\mathcal{R}(\mathcal{P}_{\sigma};\alpha)$ under dissipative flow of slope
$\alpha$ on $\mathcal{P}_{\sigma}$. Then

$\lambda_{2}(\mathcal{R}(\mathcal{P}_{\sigma};\alpha))\to 0\quad\mbox{as
$\sigma\to\infty$}.$

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-13.pdf}\\\
\mbox{Figure 13: a one-sided barrier $B_{\sigma}$ and its counterpart
$A_{\sigma}$ on the left vertical edge}\end{array}$

###### Proof.

For a one-sided vertical barrier $B_{\sigma}$ in the torus $[0,1)^{2}$, let
$A_{\sigma}=A_{\sigma}(B_{\sigma})$ denote its counterpart on the left
vertical edge of the torus, as shown in Figure 13.

We partition the unit torus into two polygonal regions
$\mathfrak{M}(A_{\sigma})$ and $\mathfrak{M}(B_{\sigma})$ using the reverse
flow recipe involving the flow in the direction $(-1,-\alpha)$. This flow
moves every point of $\mathfrak{M}(A_{\sigma})$ to $A_{\sigma}$ without
hitting $B_{\sigma}$, and moves every point of $\mathfrak{M}(B_{\sigma})$ to
$B_{\sigma}$ without hitting $A_{\sigma}$. It is easy to see that
$\mathfrak{M}(A_{\sigma})$ is invariant under dissipative flow of slope
$\alpha$, and that the dissipative flow of slope $\alpha$ moves
$\mathfrak{M}(B_{\sigma})$ to $\mathfrak{M}(A_{\sigma})$. Here the last point
follows from the observation that the barrier-free geodesic flow of slope
$\alpha$ moves $\mathfrak{M}(B_{\sigma})$ to $A_{\sigma}\cup B_{\sigma}$, and
the dissipative flow of slope $\alpha$ jumps from $B_{\sigma}$ to
$A_{\sigma}$, after which it will never return to $\mathfrak{M}(B_{\sigma})$.
Hence

$\mathcal{R}(\mathcal{P}_{\sigma};\alpha)\subset\mathfrak{M}(A_{\sigma}),$

if we ignore a possible exceptional set of $2$-dimensional Lebesgue measure
$0$. Thus to establish the small attractor theorem, it remains to show that
$\mathfrak{M}(A_{\sigma})$ is less than $\varepsilon$-part of the unit torus
$[0,1)^{2}$ provided that the index $\sigma$ is chosen appropriately.

Let $n\geqslant 9$ be an integer. For every integer $i=1,2,3,\ldots,n$, let

$V_{i}=\left\\{\left(\frac{i}{n},y\right):y\in[0,1)\right\\}$

denote the vertical line segment of length $1$ and $x$-coordinate equal to
$i/n$ in the unit torus $[0,1)^{2}$. We consider a geodesic
$\mathcal{L}_{\alpha}$ on $[0,1)^{2}$, with irrational slope $\alpha>0$ and
starting from the point $(0,\beta)$ on the left vertical edge of $[0,1)^{2}$.
Then $\mathcal{L}_{\alpha}$ intersects $V_{i}$ for the first time at the point
$(i/n,\\{\beta+i\alpha/n\\})$, and the collection of the intersection points
of $\mathcal{L}_{\alpha}$ with $V_{i}$ is given by

$\left(\frac{i}{n},\left\\{\beta+\frac{(i+jn)\alpha}{n}\right\\}\right),\quad
j=0,1,2,3,\ldots.$

These points partition $V_{i}$ in a fairly regular fashion. More precisely,
let $q_{k}=q_{k}(\alpha)$ denote the denominator of the $k$-th convergent to
the number $\alpha$, where the integer $k$ is assumed to be sufficiently
large. Consider the finite collection

$\left(\frac{i}{n},\left\\{\beta+\frac{(i+jn)\alpha}{n}\right\\}\right),\quad
j=0,1,2,3,\ldots,q_{k}-1,$ (3.1)

of the first $q_{k}$ intersection points of $\mathcal{L}_{\alpha}$ with
$V_{i}$. Using the inequality (2.7) and the inequality
$\|\gamma_{1}-\gamma_{2}\|\leqslant|\\{\gamma_{1}\\}-\\{\gamma_{2}\\}|$, where
$\|\gamma\|$ denotes the distance of $\gamma$ from the nearest integer, we see
that any two distinct points in (3.1) represented by $j_{1}$ and $j_{2}$
satisfying $0\leqslant j_{1}<j_{2}\leqslant q_{k}-1$ have gap

$\displaystyle\left|\left\\{\beta+\frac{(i+j_{1}n)\alpha}{n}\right\\}-\left\\{\beta+\frac{(i+j_{2}n)\alpha}{n}\right\\}\right|$
$\displaystyle\quad\geqslant\|(j_{1}-j_{2})\alpha\|\geqslant\left\|\frac{(j_{1}-j_{2})p_{k}}{q_{k}}\right\|-\frac{1}{q_{k+1}}\geqslant\frac{1}{q_{k}}-\frac{1}{q_{k+1}},$
(3.2)

since $|j_{1}-j_{2}|<q_{k}$ and the pair $p_{k},q_{k}$ are coprime, so that
$(j_{1}-j_{2})p_{k}$ is not divisible by $q_{k}$. Thus the $y$-coordinates of
the points in the set (3.1) closely mimic an arithmetic progression with gap
$1/q_{k}$.

Consider next a subinterval $I$ on the left vertical edge of $[0,1)^{2}$, with
lower and upper endpoints given respectively by

$(0,\beta)\quad\mbox{and}\quad\left(0,\beta+\frac{1}{q_{k}}-\frac{1}{q_{k+1}}\right).$

We move $I$ by using the splitting-free geodesic flow of slope $\alpha$ on
$[0,1)^{2}$ so that the horizontal distance covered is $q_{k}$, and let $S(I)$
denote the resulting strip in $[0,1)^{2}$. In view of the inequality (3), we
see that this strip $S(I)$ is self-avoiding, and its area is precisely
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$q_{k}\left(\frac{1}{q_{k}}-\frac{1}{q_{k+1}}\right)=1-\frac{q_{k}}{q_{k+1}}.$

Furthermore, let xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$I_{\nu},\quad\nu=1,2,3,\ldots,nq_{k},$ (3.3)

denote the successive vertical intervals of intersection of $S(I)$ with the
line segments $V_{i}$, $i=1,2,3,\ldots,n$, so that if $\nu=i+jn$, where the
indices $i=1,2,3,\ldots,n$ and $j=0,1,2,3,\ldots,q_{k}-1$, then the interval
$I_{\nu}$ has lower endpoint given by (3.1).

Consider next a subcollection

$I_{\nu},\quad\nu=1,2,3,\ldots,[n^{1/2}q_{k}],$ (3.4)

of the collection of intervals (3.3), obtained in the early stage of the
process in the construction of the set $S(I)$. This gives time-closeness but
not space-closeness. The total length of these intervals is at least

$(n^{1/2}-1)q_{k}\left(\frac{1}{q_{k}}-\frac{1}{q_{k+1}}\right)>n^{1/2}-2,$

provided that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$q_{k+1}>n^{1/2}q_{k}.$ (3.5)

Hence there is some $y_{0}\in(0,1)$ such that at least $n^{1/2}-2$ intervals
in (3.4) contain points with $y$-coordinates equal to $y_{0}$. The
$y$-coordinates of the lower endpoints of these intervals are clearly
contained in an interval of length less than $1/q_{k}$, so it follows
immediately from the pigeonhole principle that the $y$-coordinates of two of
these intervals $I_{\nu_{1}}$ and $I_{\nu_{2}}$ differ by at most
$1/(n^{1/2}-2)q_{k}$. Thus there are subintervals $I^{\prime}_{\nu_{1}}\subset
I_{\nu_{1}}$ and $I^{\prime}_{\nu_{2}}\subset I_{\nu_{2}}$ that contain
precisely the same $y$-coordinates, with common length satisfying

$|I^{\prime}_{\nu_{1}}|=|I^{\prime}_{\nu_{2}}|\geqslant|I|-\frac{1}{(n^{1/2}-2)q_{k}}=\frac{1}{q_{k}}-\frac{1}{q_{k+1}}-\frac{1}{(n^{1/2}-2)q_{k}}>\frac{1}{q_{k}}-\frac{4}{n^{1/2}q_{k}},$

provided that (3.5) holds and $n\geqslant 9$.

We may assume, without loss of generality, that $\nu_{1}<\nu_{2}$. Clearly the
reverse $\alpha$-flow, starting from the far end of $S(I)$, namely
$I_{nq_{k}}$, hits $I_{\nu_{2}}$ before hitting $I_{\nu_{1}}$. We now let
$B=I^{\prime}_{\nu_{2}}$ and $A=I^{\prime}_{\nu_{1}}$, and let $\mathcal{P}$
denote the unit torus $[0,1)^{2}$ modified by the inclusion of the one-sided
barrier $B$. Then clearly the transient set has measure

$\lambda_{2}(\mathcal{W}(\mathcal{P};\alpha))\geqslant\frac{1}{n}(nq_{k}-n^{1/2}q_{k})\left(\frac{1}{q_{k}}-\frac{4}{n^{1/2}q_{k}}\right)>1-\frac{4}{n^{1/2}},$

so that the recurrent set has measure

$\lambda_{2}(\mathcal{R}(\mathcal{P};\alpha))<\frac{4}{n^{1/2}}<\varepsilon,$

provided that $n\geqslant(4/\varepsilon)^{2}$. Naturally, since
$q_{k+1}>a_{k+1}q_{k}$ and the continued fraction digits $a_{k}$ of $\alpha$
are unbounded, we can always find a sufficiently large integer $k$ so that
(3.5) is satisfied.

Finally we can determine that the vertical line segment $V_{i}$,
$i=1,2,3,\ldots,n$, that contains the edge $A$ to be the left vertical edge of
the unit torus, and this completes the proof. ∎

##  4\. Modification of polysquare translation surfaces

We next consider modification of a polysquare translation surface by the
inclusion of a one-sided vertical barrier on part of a vertical edge. We look
at some examples.

######  Example 4.1.

Consider the horizontally reversed L-surface that is modified by the inclusion
of a one-sided barrier $B$ of length $b$, where $0<b<1$, as shown in the
picture on the left in Figure 14. Here we make the assumption that the slope
$\alpha>0$ of the dissipative flow satisfies $\alpha<b$. Using the reverse
flow recipe, we see that the transient set is essentially the part of the
system that is coloured gray, and includes the entire top atomic square.
Furthermore, if we modify the original reverse L-surface by including extra
atomic squares on top of the top atomic square, as illustrated in the picture
on the right in Figure 14, we see that these extra atomic squares are all part
of the transient set. This illustrates the possibility that the transient set
may make up the overwhelming majority of the system.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-14.pdf}\\\
\mbox{Figure 14: horizontally reversed L-surface modified by the inclusion}\\\
\mbox{of a one-sided barrier}\end{array}$

######  Example 4.2.

Consider the vertically reversed L-surface that is modified by the inclusion
of a one-sided barrier $B$ of length $b$, where $0<b<1$, as shown in the
picture on the left in Figure 15. Again we make the assumption that the slope
$\alpha>0$ of the dissipative flow satisfies $\alpha<b$. Using the reverse
flow recipe, we see that the transient set is essentially the part of the
system that is coloured gray, and leaves out the entire bottom atomic square.
Furthermore, if we modify the original reverse L-surface by including extra
atomic squares below the bottom atomic square, as illustrated in the picture
on the right in Figure 15, we see that these extra atomic squares do not form
any part of the transient set. This illustrates the possibility that the
transient set may make up the overwhelming minority of the system.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-15.pdf}\\\
\mbox{Figure 15: vertically reversed L-surface modified by the inclusion}\\\
\mbox{of a one-sided barrier}\end{array}$

######  Example 4.3.

Consider two separate copies of the modified horizontally reversed L-surface
in Example 4.1, with identically placed one-sided barriers $B_{1}$ and $B_{2}$
of length $b$, where $0<b<1$, and glued together one on top of the other as
shown in Figure 16. Again we make the assumption that the slope $\alpha>0$ of
the dissipative flow satisfies $\alpha<b$. Using the reverse flow recipe, we
see that the transient set is essentially the part of the system that is
coloured gray. Clearly the recurrent set in white is made up of two non-
trivial subsets, each of which is invariant under dissipative flow of slope
$\alpha$. Naturally the dissipative flow cannot take a point in one of the
invariant subsets of the recurrent set to anywhere in the other invariant
subset, so it follows that the dissipative flow on the recurrent set cannot
possibly be ergodic.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-16.pdf}\\\
\mbox{Figure 16: L-surfaces glued together and modified by the inclusion}\\\
\mbox{of identical one-sided barriers}\end{array}$

Consider the special case where we start with the L-surface and modify it with
the inclusion of a one-sided barrier on the vertical edge separating the two
bottom atomic squares, as shown in Figure 17.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-17.pdf}\\\
\mbox{Figure 17: L-surface modified by the inclusion of a one-sided
barrier}\end{array}$

Denote such a modified system by $\mathcal{P}$. Computer-generated pictures
indicate that the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ of
$\mathcal{P}$ has area $1$ or $2$, and is a finite union of polygons with
boundary edges that are vertical or of slope $\alpha$. The pictures can be
surprisingly complicated, particularly if the barrier is short, or if it
breaks into finitely many pieces. This makes the problem rather interesting.

Meanwhile, the dissipative system $\mathcal{P}$ modulo one is the ordinary
geodesic flow of slope $\alpha$ in the unit torus $[0,1)^{2}$, and this is
uniquely ergodic. This implies that if the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ of $\mathcal{P}$ has area $1$, then the
dissipative flow of slope $\alpha$ there is uniquely ergodic. On the other
hand, if the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ of $\mathcal{P}$
has area $2$, then we do not know immediately whether the dissipative flow of
slope $\alpha$ there is also uniquely ergodic. In principle, the set
$\mathcal{R}(\mathcal{P};\alpha)$ may contain an invariant subset of area $1$.
This raises an interesting uniformity question.

We next illustrate a class of dissipative, or non-conservative, dynamical
systems. The restriction of the flow to the recurrent set gives rise to a nice
conservative measure preserving system, to which we may apply classical
ergodic theory.

Consider the system $\mathcal{P}$ shown in Figure 18, where we start with the
L-surface and then modify it by including a one-sided barrier of length $b$ as
shown, with lower endpoint $(1,0)$ and upper endpoint $(1,b)$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-18.pdf}\\\
\mbox{Figure 18: L-surface modified by the inclusion of a one-sided
barrier}\end{array}$

We consider dissipative flow of slope $\alpha$ on $\mathcal{P}$, and make the
further restrictions

$\frac{2}{3}<\alpha<\frac{3}{4}\quad\mbox{and}\quad 0<b<2\alpha-1,$ (4.1)

so that Figure 18 represents reasonable ranges of the parameters $b$ and
$\alpha$.

It is straightforward to check that the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ is represented by the part of $\mathcal{P}$
in white, and it has area $2$. The transient set
$\mathcal{W}(\mathcal{P};\alpha)$ is represented by the part of $\mathcal{P}$
in gray, and it has area $1$.

We have highlighted in Figure 18 some critical points and line segments which
help us construct the relevant interval exchange transformation of the
dissipative flow of slope $\alpha$ on $\mathcal{P}$.

Let $\mathcal{E}$ denote the union of the left vertical edges of the atomic
squares of $\mathcal{P}$. It is not difficult to see that the intersection of
$\mathcal{E}$ with the transient set $\mathcal{W}(\mathcal{P};\alpha)$ modulo
one is the unit interval $[0,1)$, while the intersection

$\mathcal{V}=\mathcal{E}\cap\mathcal{R}(\mathcal{P};\alpha)$

of $\mathcal{E}$ with the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$
modulo one is also the unit interval $[0,1)$, but with multiplicity $2$.

As shown in Figures 18 and 19, the set $\mathcal{V}$ is the union of four
vertical line segments $L_{1},L_{2},L_{3},L_{4}$, where

(1) $L_{1}$ has endpoints $(0,0)$ and $(0,\alpha)$;

(2) $L_{2}$ has endpoints $(0,1)$ and $(0,2\alpha)$;

(3) $L_{3}$ has endpoints $(0,1+\alpha)$ and $(0,2)$; and

(4) $L_{4}$ has endpoints $(1,2\alpha-1)$ and $(1,1)$.

We can make further subdivisions, as shown in Figures 18 and 19, where

(5) $L_{1}$ is split into $L_{1,1},\ldots,L_{1,4}$ at the points
$(0,3\alpha-2)$, $(0,1-\alpha)$ and $(0,2\alpha-1)$;

(6) $L_{2}$ is split into $L_{2,1},\ldots,L_{2,3}$ at the points
$(0,3\alpha-1)$ and $(0,2-\alpha)$; and

(7) $L_{4}$ is split into $L_{4,1},L_{4,2}$ at the point $(1,\alpha)$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-19.pdf}\\\
\mbox{Figure 19: L-surface modified by the inclusion of a one-sided
barrier}\end{array}$

Consider the discrete transformation

$\mathbf{T}_{\alpha}:\mathcal{V}\to\mathcal{V}$

which is the dissipative shift by the vector $(1,\alpha)$. It follows from
Figure 19 that

$\begin{array}[]{rl}\mathbf{T}_{\alpha}(L_{1,3})=L_{2,1},&\quad\mathbf{T}_{\alpha}(L_{2,3})=L_{1,1},\\\
\mathbf{T}_{\alpha}(L_{1,4})=L_{2,2}\cup
L_{2,3},&\quad\mathbf{T}_{\alpha}(L_{4,1})=L_{1,2}\cup L_{1,3},\\\
\mathbf{T}_{\alpha}(L_{3})=L_{4,1},&\quad\mathbf{T}_{\alpha}(L_{4,2})=L_{1,4},\\\
\mathbf{T}_{\alpha}(L_{1,1}\cup
L_{1,2})=L_{4,2},&\quad\mathbf{T}_{\alpha}(L_{2,1}\cup
L_{2,2})=L_{3}.\end{array}$ (4.2)

There is a good reason for listing the information in (4.2) in this particular
way, as it is not difficult to see that each column in (4.2) leads to the
projections

$\begin{array}[]{c}\mathbf{T}^{*}([1-\alpha,2\alpha-1))=[0,3\alpha-2),\\\
\mathbf{T}^{*}([2\alpha-1,\alpha))=[3\alpha-2,2\alpha-1),\\\
\mathbf{T}^{*}([\alpha,1))=[2\alpha-1,\alpha),\\\
\mathbf{T}^{*}([0,1-\alpha))=[\alpha,1),\end{array}$ (4.3)

where $\mathbf{T}^{*}$ represents the modulo one projection of the second
coordinates of $\mathbf{T}_{\alpha}$. It is clear from (4.3) that the images
in each column covers the unit interval $[0,1)$ precisely once, so the unit
interval $[0,1)$ is covered twice.

We can also describe the interval exchange transformation
$\mathbf{T}_{\alpha}$ by the directed graph in the picture on the left in
Figure 20 or by the undirected graph in the picture on the right in Figure 20,
each constructed using the information given by (4.2). In both graphs, the
$10$ vertices are the intervals

$L_{1,1},L_{1,2},L_{1,3},L_{1,4},L_{2,1},L_{2,2},L_{2,3},L_{3},L_{4,1},L_{4,2},$
(4.4)

For the graph on the left, a directed edge $L^{\prime}\to L^{\prime\prime}$
from a vertex $L^{\prime}$ to a vertex $L^{\prime\prime}$ indicates that there
exist points $P^{\prime}\in L^{\prime}$ and $P^{\prime\prime}\in
L^{\prime\prime}$ such that
$P^{\prime\prime}=\mathbf{T}_{\alpha}(P^{\prime})$. For the graph on the
right, an undirected edge $L^{\prime}L^{\prime\prime}$ between vertices
$L^{\prime}$ and $L^{\prime\prime}$ indicates that there exist points
$P^{\prime}\in L^{\prime}$ and $P^{\prime\prime}\in L^{\prime\prime}$ such
that $P^{\prime\prime}=\mathbf{T}_{\alpha}(P^{\prime})$ or
$P^{\prime}=\mathbf{T}_{\alpha}(P^{\prime\prime})$. This undirected graph is
sometimes known as the overlapping graph of the interval exchange
transformation. Clearly this is a connected graph.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-20.pdf}\\\
\mbox{Figure 20: a directed graph and the overlapping graph}\end{array}$

######  Remark.

Figure 21 shows the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ as a
parallelogram after we have performed some fake surgery on the system
$\mathcal{P}$, where we simply take the various pieces of the set and build a
parallelogram like a jigsaw puzzle. Note, however, that the two sides of the
dotted line segment between the points $(1,1)$ and $(2,1+\alpha)$ are not
neighbours.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-21.pdf}\\\
\mbox{Figure 21: visualization of the recurrent set as a
parallelogram}\end{array}$

On the other hand, let us return to Figure 19. Elementary calculations show
that the part of the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ in the
right atomic square has area equal to $2-2\alpha$, the part of the recurrent
set $\mathcal{R}(\mathcal{P};\alpha)$ in the top atomic square has area equal
to $2-\alpha^{-1}$, while the part of the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ in the bottom left atomic square has area
equal to $2\alpha+\alpha^{-1}-2$. In the special case when
$\alpha=1/\sqrt{2}$, these three numbers are respectively roughly equal to
$0.586$, $0.586$ and $0.828$, none of which is close to $2/3$. This means that
if the dissipative flow of slope $\alpha$ on the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ is uniformly distributed, then the right
atomic square and the top atomic square are under-visited, whereas the bottom
left atomic square is over-visited.

Recall that the $2$-cyclic symmetry of the system is a crucial ingredient in
our argument in Section 2 where we consider the modified $2$-square torus. For
the dissipative system $\mathcal{P}$ obtained from the L-surface by the
inclusion of a one-sided barrier on the common vertical edge of the bottom
atomic squares, one cannot expect to have any such symmetry property. However,
in the special case when the barrier $B$ and the slope $\alpha$ of the flow
satisfy the special conditions (4.1), which accounts for uncountably many
cases, we can nevertheless recover some form of symmetry which proves to be
very useful.

In Figure 22, the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ of area $2$
is in white and the transient set $\mathcal{W}(\mathcal{P};\alpha)$ of area
$1$ is in gray. Like in Figure 6 for the modified $2$-square torus, they split
the left side of the barrier $B$ and the left side of the corresponding
interval $A$ in an essentially fair way.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-22.pdf}\\\
\mbox{Figure 22: some form of symmetry}\end{array}$

The dissipative flow of slope $\alpha$ moves $B$ first to the gray part in
Figure 22 which is $\mathcal{W}(\mathcal{P};\alpha)$. When it reaches the gray
part immediately to the left of $B$ or the gray part immediately to the left
of $A$ on right vertical edge, it then jumps to $A$. Thus every point of
$\mathcal{W}(\mathcal{P};\alpha)$ is eventually moved to $A$ by the flow.

Starting from anywhere within the set $\mathcal{R}(\mathcal{P};\alpha)$, the
dissipative flow of slope $\alpha$ moves within the white part in Figure 22
which is the set $\mathcal{R}(\mathcal{P};\alpha)$. When it reaches the white
part immediately to the left of $B$ or the white part immediately to the left
of $A$ on the right vertical edge, it then jumps back to $A$. The fair split
of left side of the barrier $B$ and the left side of the corresponding
interval $A$ implies that this jump is $1$-to-$1$ and measure preserving. Thus
the dissipative flow of slope $\alpha$ restricted to the set
$\mathcal{R}(\mathcal{P};\alpha)$ is both invertible and measure preserving.
Starting from $A$ and sweeping it along by the dissipative flow of slope
$\alpha$, this flow returns to $A$ infinitely many times in a $1$-to-$1$ and
measure preserving way.

However, since the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ has area
$2$, we cannot simply project to the unit torus $[0,1)^{2}$ and make use of
the integrable geodesic flow there of irrational slope $\alpha$, as in the
case of the modified $2$-square torus.

Observe from (4.2) and (4.3) that the $y$-coordinates of the endpoints of the
$10$ subintervals in (4.4) modulo $1$ are
$\\{-\alpha\\},0,\\{\alpha\\},\\{2\alpha\\},\\{3\alpha\\}$.

For any irrational number $\alpha>0$ and integer $s\geqslant 2$, suppose that
the endpoints of the participating intervals modulo $1$ of the interval
exchange transformation $\mathbf{T}_{\alpha}:[0,s)\to[0,s)$ have values
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\\{-\alpha\\},0,\\{\alpha\\},\\{2\alpha\\},\ldots,\\{\ell_{0}\alpha\\},$
(4.5)

where $\ell_{0}$ is a fixed positive integer, and that $\mathbf{T}_{\alpha}$
modulo one acts as the $\alpha$-shift in the unit torus $[0,1)$. Then the
$\ell_{0}+2$ points in (4.5) divide the unit torus $[0,1)$ into $\ell_{0}+2$
critical subintervals. Lifting these critical subintervals in $[0,1)$ to
$[0,s)$, we arrive at $s(\ell_{0}+2)$ critical subintervals in $[0,s)$. If
$S_{0}\subset[0,s)$ is a non-trivial measurable
$\mathbf{T}_{\alpha}$-invariant subset, then $S_{0}$ must be a union of some
of these $s(\ell_{0}+2)$ critical subintervals.

We now use the case $s=2$ and $\ell_{0}=3$ of the following result which can
be proved by a straightforward application of the method of [1, Lemma 4.1].

######  Lemma 4.1.

If the overlapping graph of the $s(\ell_{0}+2)$ critical subintervals of the
interval exchange transformation $\mathbf{T}_{\alpha}:[0,s)\to[0,s)$ is
connected, then $\mathbf{T}_{\alpha}$ is ergodic. Furthermore,
$\mathbf{T}_{\alpha}$ is uniquely ergodic.

The special case (4.1) of the modified L-surface turns out to be a lucky case
where we can determine the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ and
the transient set $\mathcal{W}(\mathcal{P};\alpha)$ without too much
difficulty. The situation may be significantly different if the conditions
(4.1) are not satisfied. Consider the case of the system $\mathcal{P}$ as
illustrated in Figure 23, where $b=0.8$ and $\alpha=1/\sqrt{2}$. Here we start
from $B$ and sweep out a no-go zone which is the gray polygon with a vertical
side on $B$. We then extend it by the strip with left edge between $(0,b)$ and
$(0,1)$, noting that any flow that reaches any point between $(1,b)$ and
$(1,1)$ eventually reaches $A$. We can then further extend the no-go zone to
the part of $\mathcal{P}$ in gray.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-23.pdf}\\\
\mbox{Figure 23: the L-surface modified by the inclusion of a one-sided
barrier}\end{array}$

It remains true that any dissipative flow of slope $\alpha$ that reaches the
gray part immediately to the left of $B$ or the gray part immediately to the
left of $A$ on the right vertical edge continues from immediately to the right
of $A$ on the left vertical edge, but there is no guarantee that this is part
of the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$. Figure 24 shows a
further extension of the no-go zone. There is a subinterval of $B$ which is
gray immediately to the left and it is also gray immediately to the left of
the corresponding subinterval of $A$. This leads to an extension of the no-go
zone which is shown in dark gray. Clearly the set
$\mathfrak{M}(\mathcal{P};B;\alpha)$ is a proper subset of the transient set
$\mathcal{W}(\mathcal{P};\alpha)$, and the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ is a proper subset of the set
$\mathfrak{R}(\mathcal{P};A;\alpha)$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-24.pdf}\\\
\mbox{Figure 24: a further extension of the no-go zone}\end{array}$

Another interesting observation is illustrated by Figure 25 which highlights
part of the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-25.pdf}\\\
\mbox{Figure 25: part of the recurrent set}\end{array}$

We see from Figure 24 that both trapezoids in dark gray in Figure 25 are part
of the set $\mathfrak{M}(\mathcal{P};B;\alpha)$, hence part of the transient
set $\mathcal{W}(\mathcal{P};\alpha)$. These two trapezoids have the same
projection on to the unit torus $[0,1)^{2}$. Since the projection of the
dissipative flow of slope $\alpha$ to the unit torus $[0,1)^{2}$ is integrable
geodesic flow of slope $\alpha$ which is uniformly distributed if $\alpha$ is
irrational, it follows that the white trapezoid in the bottom left atomic
square in Figure 25 must be part of the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$.

It is clear from Figure 24 that the reverse flow recipe does not work in this
case and does not lead to the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$.
However, Figure 25 clearly gives a non-trivial open subset of the recurrent
set $\mathcal{R}(\mathcal{P};\alpha)$, so the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ can be obtained by sweeping this open subset
along by the dissipative flow of slope $\alpha$, and the construction is
complete in a finite number of zig-zaggings; see the Remark after the proof of
Theorem 1. The recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ is then a
finite union of polygons with boundary edges that are vertical or of slope
$\alpha$.

Again, we are lucky in this case, as the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ has area $1$, and uniformity follows from
the uniformity of geodesic flow of slope $\alpha$ in the unit torus
$[0,1)^{2}$ if $\alpha$ is irrational.

We emphasize again that in this case, the set
$\mathfrak{M}(\mathcal{P};B;\alpha)$ is a proper subset of the transient set
$\mathcal{W}(\mathcal{P};\alpha)$, and the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ is a proper subset of the set
$\mathfrak{R}(\mathcal{P};A;\alpha)$.

As for determining the transient set $\mathcal{W}(\mathcal{P};\alpha)$, Figure
24 gives the first step of an extension process through which we can grow the
dark gray strip. We discuss this in the next section where we attempt to
obtain a generalization of Theorem 1 to modifications of arbitrary finite
polysquare translation surfaces.

Whereas for the $2$-square torus or the L-surface, the choice of the interval
$A$ follows naturally from the one-sided barrier $B$, this is no longer the
case if we start with a polysquare translation surface.

Figure 26 shows that in the case of a translation surface comprising $5$
atomic squares in the form of a cross and where edge identification comes from
perpendicular translation, there is more than one choice for the interval $A$.
The only requirement is that $A$ lies on a different vertical edge on the same
horizontal street that contains $B$. If some of the edge identifications do
not come from perpendicular translation, the situation looks rather more
complicated. Figure 27 shows two equivalent dissipative systems with the
interval $A$ on the vertical edge $v_{2}$, while Figure 28 shows two
equivalent dissipative systems with the interval $A$ on the vertical edge
$v_{4}$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-26.pdf}\\\
\mbox{Figure 26: dissipative systems modified from polysquare translation
surfaces}\end{array}$

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-27.pdf}\\\
\mbox{Figure 27: dissipative systems modified from polysquare translation
surfaces}\end{array}$

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-28.pdf}\\\
\mbox{Figure 28: dissipative systems modified from polysquare translation
surfaces}\end{array}$

In the next section, we establish the following result.

######  Theorem 4.

Consider dissipative flow of slope $\alpha$, where $\alpha>0$ is irrational,
on a system $\mathcal{P}$, where a finite polysquare translation surface has
beem modified with the inclusion of a one-sided barrier $B$ on the common
vertical edge of two neighbouring atomic squares, in the form of a union of
finitely many vertical intervals, and where a set $A$ corresponding to $B$ has
been chosen on a different vertical edge on the same horizontal street that
contains $B$. Then the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ and the
transient set $\mathcal{W}(\mathcal{P};\alpha)$ can be constructed by an
extension process on the no-go zone $\mathcal{N}(\mathcal{P};\alpha)$.
Furthermore, if this process is finite, then both sets
$\mathcal{R}(\mathcal{P};\alpha)$ and $\mathcal{W}(\mathcal{P};\alpha)$ are
finite unions of polygons with boundary edges that are vertical or of slope
$\alpha$.

##  5\. Extension process and polygons

We make use of two well known results in ergodic theory. Suppose that
$(X,\mathcal{B},\mu)$ is a probability space, so that $\mu(X)=1$, and
$\mathbf{T}$ is a measure preserving transformation on $X$. For any measurable
subset $S\subset X$ with positive measure $\mu(S)>0$, since the Poincare
recurrence theorem guarantees infinite recurrence for $\mu$-almost every $x\in
S$ to the set $S$, it is meaningful to define the first return time for any
such $x\in S$. This is given by the integer xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$n_{S}(x)=\inf\\{n\geqslant 1:\mathbf{T}^{n}(x)\in S\\},$

and clearly it is finite for $\mu$-almost every $x\in S$.

The first result is due to Kakutani [5] in 1943, and concerns an induced map
of $\mathbf{T}$ on $S$.

######  Lemma 5.1.

Suppose that $(X,\mathcal{B},\mu)$ is a probability space, so that $\mu(X)=1$,
that $\mathbf{T}$ is an invertible measure preserving transformation on $X$,
and that $S\subset X$ is a measurable subset with positive measure $\mu(S)>0$.
Then the induced map

$\mathbf{T}|_{S}:S\to S:x\mapsto\mathbf{T}^{n_{S}(x)}(x),$

valid for $\mu$-almost every $x\in S$, is a measure preserving transformation
on the induced measure space $(S,\mathcal{B}\cap S,\mu_{S})$.

The second result is due to Kac [4] in 1947, and concerns the average value of
the first return time to $S$.

######  Lemma 5.2.

Suppose that $(X,\mathcal{B},\mu)$ is a probability space, so that $\mu(X)=1$,
that $\mathbf{T}$ is a measure preserving transformation on $X$, and that
$S\subset X$ is a measurable subset with positive measure $\mu(S)>0$. Suppose
further that $\mathbf{T}$ is ergodic in the space $(X,\mathcal{B},\mu)$. Then
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\int_{S}n_{S}(x)\,\mathrm{d}\mu=1.$ (5.1)

In other words, the average value of the first return time to the set $S$ is
equal to the reciprocal of the measure $\mu(S)$ of $S$.

###### Proof of Lemma 5.1.

Note first of all that both $n_{S}:S\to\mathbb{N}$ and $\mathbf{T}|_{S}:S\to
S$ are well defined and measurable mappings. For every integer $n\geqslant 1$,
let

$S_{n}=\\{x\in S:n_{S}(x)=n\\}$ (5.2)

denote the collection of points $x\in S$ that return to $S$ for the first time
at the $n$-th application of the transformation $\mathbf{T}$. By definition,
the sets $S_{n}$, $n=1,2,3,\ldots,$ are pairwise disjoint. Furthermore, we
have

$S_{1}=S\cap\mathbf{T}^{-1}S,\quad S_{2}=(S\cap\mathbf{T}^{-2}S)\setminus
S_{1},\quad S_{3}=(S\cap\mathbf{T}^{-3}S)\setminus(S_{1}\cup S_{2}),$

and in general xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$S_{n}=(S\cap\mathbf{T}^{-n}S)\setminus\bigcup_{1\leqslant m<n}S_{m},\quad
n=1,2,3,\ldots.$

Any measurable subset $A\subset S$ can be expressed as a disjoint union in the
form

$A=\bigcup_{n=1}^{\infty}(A\cap S_{n}),$

so it follows immediately that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\mu(A)=\sum_{n=1}^{\infty}\mu(A\cap S_{n}).$ (5.3)

On the other hand, since $\mathbf{T}$ is invertible and so $1$-to-$1$, the
image $\mathbf{T}|_{S}A$ can be expressed as a disjoint union in the form

$\mathbf{T}|_{S}A=\bigcup_{n=1}^{\infty}\mathbf{T}|_{S}(A\cap S_{n}),$

so it follows immediately that

$\mu(\mathbf{T}|_{S}A)=\sum_{n=1}^{\infty}\mu(\mathbf{T}|_{S}(A\cap
S_{n}))=\sum_{n=1}^{\infty}\mu(\mathbf{T}^{n}(A\cap S_{n})),$ (5.4)

where in the last step, we use the key fact that for every fixed
$n=1,2,3,\ldots,$ if a subset $B\subset S_{n}$ is measurable, then
$\mathbf{T}|_{S}B=\mathbf{T}^{n}B$. Finally, since $\mathbf{T}$ is invertible
and preserves $\mu$, it follows that

$\sum_{n=1}^{\infty}\mu(\mathbf{T}^{n}(A\cap
S_{n}))=\sum_{n=1}^{\infty}\mu(A\cap S_{n}).$ (5.5)

It now follows on combining (5.3)–(5.5) that $\mu(\mathbf{T}|_{S}A)=\mu(A)$,
and completes the proof. ∎

###### Proof of Lemma 5.2.

For every $n=1,2,3,\ldots,$ the set $S_{n}$ defined by (5.2) clearly satisfies
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$S_{n}=S_{n,0}=S\cap\mathbf{T}^{-1}S^{c}\cap\ldots\cap\mathbf{T}^{-(n-1)}S^{c}\cap\mathbf{T}^{-n}S,$

where $S^{c}=X\setminus S$ denotes the complement of the set $S$ in $X$.
Furthermore, for every $n=1,2,3,\ldots$ and every $k=1,\ldots,n-1$, write

$S_{n,k}=\mathbf{T}^{k}S_{n}=\mathbf{T}^{k}S\cap\mathbf{T}^{k-1}S^{c}\cap\ldots\cap\mathbf{T}^{k-n+1}S^{c}\cap\mathbf{T}^{k-n}S.$
(5.6)

We first show that the sets

$S_{n,k},\quad n=1,2,3,\ldots,\quad k=0,1,\ldots,n-1,$ (5.7)

are pairwise disjoint. To justify this claim, consider first the sets
$S_{n,k^{\prime}}$ and $S_{n,k^{\prime\prime}}$, where $0\leqslant
k^{\prime}<k^{\prime\prime}\leqslant n-1$. Then it follows from (5.6) that

$S_{n,k^{\prime}}\subset\mathbf{T}^{k^{\prime}}S\quad\mbox{and}\quad
S_{n,k^{\prime\prime}}\subset\mathbf{T}^{k^{\prime}}S^{c},$

and so xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$S_{n,k^{\prime}}\cap
S_{n,k^{\prime\prime}}\subset\mathbf{T}^{k^{\prime}}S\cap\mathbf{T}^{k^{\prime}}S^{c}=\emptyset.$

Suppose next that $n^{\prime}<n^{\prime\prime}$,
$k^{\prime}=0,1,\ldots,n^{\prime}-1$ and
$k^{\prime\prime}=0,1,\ldots,n^{\prime\prime}-1$. Then it follows from (5.6)
that

$S_{n^{\prime},k^{\prime}}\subset\mathbf{T}^{k^{\prime}}S\cap\mathbf{T}^{k^{\prime}-n^{\prime}}S\quad\mbox{and}\quad
S_{n^{\prime\prime},k^{\prime\prime}}\subset\mathbf{T}^{k^{\prime\prime}-1}S^{c}\cap\ldots\cap\mathbf{T}^{k^{\prime\prime}-n^{\prime\prime}+1}S^{c}.$

If $k^{\prime}\leqslant k^{\prime\prime}-1$, then clearly
$S_{n^{\prime\prime},k^{\prime\prime}}\subset\mathbf{T}^{k^{\prime}}S^{c}$,
and so

$S_{n^{\prime},k^{\prime}}\cap
S_{n^{\prime\prime},k^{\prime\prime}}\subset\mathbf{T}^{k^{\prime}}S\cap\mathbf{T}^{k^{\prime}}S^{c}=\emptyset.$

On the other hand, if $k^{\prime}>k^{\prime\prime}-1$, then
$-k^{\prime}\leqslant-k^{\prime\prime}$. Since $n^{\prime}<n^{\prime\prime}$,
it follows that $n^{\prime}-k^{\prime}<n^{\prime\prime}-k^{\prime\prime}$, and
so $k^{\prime}-n^{\prime}\geqslant k^{\prime\prime}-n^{\prime\prime}+1$. Then
clearly
$S_{n^{\prime\prime},k^{\prime\prime}}\subset\mathbf{T}^{k^{\prime}-n^{\prime}}S^{c}$,
and so xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$S_{n^{\prime},k^{\prime}}\cap
S_{n^{\prime\prime},k^{\prime\prime}}\subset\mathbf{T}^{k^{\prime}-n^{\prime}}S\cap\mathbf{T}^{k^{\prime}-n^{\prime}}S^{c}=\emptyset.$

Thus the sets (5.7) are pairwise disjoint as claimed. We therefore have

$\mu\left(\bigcup_{n=1}^{\infty}\bigcup_{k=0}^{n-1}S_{n.k}\right)=\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\mu(S_{n,k}).$
(5.8)

Since $\mathbf{T}$ is measure preserving, it follows from (5.6) that
$\mu(S_{n,k})=\mu(S_{n})$ for every $k=0,1,\ldots,n-1$. Combining this with
(5.2), we deduce that

$\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\mu(S_{n,k})=\sum_{n=1}^{\infty}n\mu(S_{n})=\sum_{n=1}^{\infty}\int_{S_{n}}n_{S}(x)\,\mathrm{d}\mu.$
(5.9)

The Poincare recurrence theorem now gives

$\bigcup_{n=1}^{\infty}S_{n}=S.$

Since this is clearly a pairwise disjoint union, it follows that

$\sum_{n=1}^{\infty}\int_{S_{n}}n_{S}(x)\,\mathrm{d}\mu=\int_{S}n_{S}(x)\,\mathrm{d}\mu.$
(5.10)

Finally, the pairwise disjoint union xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\bigcup_{n=1}^{\infty}\bigcup_{k=0}^{n-1}S_{n.k}$

is $\mathbf{T}$-invariant, so it follows from the ergodicity of $\mathbf{T}$
that

$\mu\left(\bigcup_{n=1}^{\infty}\bigcup_{k=0}^{n-1}S_{n.k}\right)=\mu(X)=1.$
(5.11)

The desired identity (5.1) now follows on combining (5.8)–(5.11). ∎

###### Proof of Theorem 4.

Let $\mathcal{P}$ be a dissipative system described in the statement of the
theorem. Let $X$ denote the union of the left vertical edges of the
constituent atomic squares of the underlying polysquare translation surface.
Furthermore, let

$\mathbf{T}=\mathbf{T}_{\alpha}:X\to X$

denote the discretization of the ordinary geodesic flow on the underlying
polysquare translation surface. For every $x\in X$, let $n(x)=n_{A\cup B}(x)$
denote the smallest integer $n\geqslant 1$ such that $\mathbf{T}^{n}(x)\in
A\cup B$. Then the time-quantitative version of the Gutkin–Veech theorem gives
rise to a threshold $N_{0}=N_{0}(\mathcal{P};A\cup B;\alpha)$ such that
$n(x)\leqslant N_{0}$ for $\lambda_{1}$-almost every $x\in X$.

Consider the induced map

$\mathbf{T}^{*}=\mathbf{T}|_{A\cup B}:A\cup B\to A\cup
B:x\mapsto\mathbf{T}^{n(x)}(x).$

Then it follows from the existence of $N_{0}=N_{0}(\mathcal{P};A\cup
B;\alpha)$ and Lemma 5.1 that $\mathbf{T}^{*}$ is a finite interval exchange
transformation on the set $A\cup B$. Indeed, the set $A\cup B$ can essentially
be decomposed into a finite union of disjoint intervals such that ordinary
geodesic flow of slope $\alpha$ moves each interval splitting-free either by
length $N_{0}(1+\alpha^{2})^{1/2}$ without hitting $A\cup B$ or by length at
most $N_{0}(1+\alpha^{2})^{1/2}$ when returning to $A\cup B$ for the first
time.

Let $\mathcal{F}(A\cup B)$ denote the first return strip, where
$\lambda_{1}$-almost every point of $A\cup B$ is swept along by the flow of
slope $\alpha$ until it hits the set $A\cup B$ again for the first time, and
let $\mathcal{F}(A)$ and $\mathcal{F}(B)$ denote respectively the
corresponding first return strip when we start with points of $A$ only and
with points of $B$ only.

We now apply Lemma 5.2 with $S=A\cup B$ and $\mu=\lambda_{1}/s$, where $s$
denotes the number of distinct atomic squares of the underlying polysquare
translation surface, so that $\mu(X)=1$, and deduce that

$\frac{1}{s}\lambda_{2}(\mathcal{F}(A\cup B))=\frac{1}{s}\int_{A\cup
B}n(x)\,\mathrm{d}\lambda_{1}=1.$

Hence $\lambda_{2}(\mathcal{F}(A\cup B))=s$, and so

$\mathcal{F}(A)\cup\mathcal{F}(B)=\mathcal{F}(A\cup B)=\mathcal{P},$

apart from exceptional sets of $2$-dimensional Lebesgue measure $0$. The
reader may wish to check whether this is the same split as obtained by the
reverse flow recipe.

Observe that $\lambda_{2}(\mathcal{F}(B))\geqslant 1$. To see this, we simply
observe that the continuous analogue of Lemma 5.2 applied to the projection of
$\mathcal{F}(B)$ to the unit torus $[0,1)^{2}$ shows that the image is the
whole unit torus and so has area $1$.

At this point, we recall that the Kakutani–Kac approach using Lemmas 5.1 and
5.2 gives rise to a measure preserving induced map. We now make use of this.
For every subset $H\subset\mathcal{P}$, let

$y(H)=\\{y:(x,y)\in\mathcal{P}\mbox{ and }(x,y)\in H\mbox{ for some }x\\}$

denote the collection of the $y$-coordinates of the points of $H$. We have two
cases, motivated respectively by Figures 22 and 24.

######  Case 1.

We have perfect balance, in the sense that

$\lambda_{1}(y(\mathbf{T}^{*}(A)\cap A)\cap y(\mathbf{T}^{*}(A)\cap B))=0,$
(5.12)

so that the sets $\mathbf{T}^{*}(A)\cap A$ and $\mathbf{T}^{*}(A)\cap B$ have
essentially no common $y$-coordinates. Note that

$\displaystyle\lambda_{1}((\mathbf{T}^{*}(A)\cap A)\cup(\mathbf{T}^{*}(A)\cap
B))$ $\displaystyle=\lambda_{1}(\mathbf{T}^{*}(A)\cap(A\cup B))$
$\displaystyle=\lambda_{1}(\mathbf{T}^{*}(A))=\lambda_{1}(A)=\lambda_{1}(B),$
(5.13)

where the last step is a consequence of the measure preserving property of
$\mathbf{T}^{*}$. It follows that if (5.12) holds, then the projection of
$\mathbf{T}^{*}(A)\cap A$ to the unit torus $[0,1)^{2}$ and the projection of
$\mathbf{T}^{*}(A)\cap B$ to the unit torus $[0,1)^{2}$ together form the
projection of $A$ or $B$ to the unit torus $[0,1)^{2}$, as illustrated in
Figure 29, where the parts of $\mathbf{T}^{*}(A)$ in $A$ and $B$ are indicated
in bold under $A$ and $B$ respectively. See also Figure 22 for a concrete
example.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-29.pdf}\\\
\mbox{Figure 29: $\mathbf{T}^{*}(A)\cap A$ and $\mathbf{T}^{*}(A)\cap B$ are
perfectly balanced}\end{array}$

Using the measure preserving property of $\mathbf{T}^{*}$ again, we conclude
that the sets $\mathbf{T}^{*}(A)\cap A$ and $\mathbf{T}^{*}(A)\cap B$ cover
$\lambda_{1}$-almost every $y$-coordinate in $A$ or $B$ precisely once. This
also means that the sets $\mathbf{T}^{*}(B)\cap A$ and $\mathbf{T}^{*}(B)\cap
B$ cover $\lambda_{1}$-almost every $y$-coordinate in $A$ or $B$ precisely
once. The dissipative flow of slope $\alpha$ jumps from $\mathbf{T}^{*}(B)$ to
$A$ in a $1$-to-$1$ and measure preserving way, and jumps from
$\mathbf{T}^{*}(A)$ to $A$ in a $1$-to-$1$ and measure preserving way.

Starting from $A$, we can extend it by the dissipative flow of slope $\alpha$
forever in a $1$-to-$1$ and measure preserving way. Then the resulting set
$\mathfrak{R}(\mathcal{P};A;\alpha)\subset\mathcal{P}$ is clearly invariant
under dissipative flow of slope $\alpha$. It then follows from the Poincare
recurrence theorem that $\lambda_{2}$-almost every point of
$\mathfrak{R}(\mathcal{P};A;\alpha)$ is recurrent under dissipative flow of
slope $\alpha$. The two sets $\mathfrak{R}(\mathcal{P};A;\alpha)$ and
$\mathcal{F}(A)$ have some common properties. They both contain $A$ and are
invariant under dissipative flow of slope $\alpha$. They are also both
disjoint from $\mathcal{F}(B)$. Thus we conclude that

$\mathfrak{R}(\mathcal{P};A;\alpha)=\mathcal{F}(A)=\mathcal{R}(\mathcal{P};\alpha)\quad\mbox{and}\quad\mathcal{F}(B)=\mathcal{W}(\mathcal{P};\alpha),$
(5.14)

apart from exceptional sets of $2$-dimensional Lebesgue measure $0$. If
$\mathcal{R}(\mathcal{P};\alpha)$ has area $1$, then we have uniformity of
dissipative flow of slope $\alpha$ in the set. If
$\mathcal{R}(\mathcal{P};\alpha)$ has area greater than $1$, then the set may
decompose into a union of minimal subsets that are invariant under dissipative
flow of slope $\alpha$ and with integer valued areas. It then follows from the
Birkhoff ergodic theorem that there is uniformity of dissipative flow of slope
$\alpha$ in each of these sets.

######  Case 2.

We do not have perfect balance, in the sense that

$\lambda_{1}(y(\mathbf{T}^{*}(A)\cap A)\cap y(\mathbf{T}^{*}(A)\cap B))>0,$
(5.15)

so that the sets $\mathbf{T}^{*}(A)\cap A$ and $\mathbf{T}^{*}(A)\cap B$ have
non-trivial overlapping of $y$-coordinates. Using the observation (Case 1), we
conclude that if (5.15) holds, then the projection of $\mathbf{T}^{*}(A)\cap
A$ to the unit torus $[0,1)^{2}$ and the projection of $\mathbf{T}^{*}(A)\cap
B$ to the unit torus $[0,1)^{2}$ together leave out part of the projection of
$A$ or $B$ to the unit torus $[0,1)^{2}$, as illustrated in Figure 30, where
the parts of $\mathbf{T}^{*}(A)$ in $A$ and $B$ are indicated in bold under
$A$ and $B$ respectively. See also Figure 24 for a concrete example.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-30.pdf}\\\
\mbox{Figure 30: $\mathbf{T}^{*}(A)\cap A$ and $\mathbf{T}^{*}(A)\cap B$ are
perfectly balanced}\end{array}$

In particular, the dissipative flow of slope $\alpha$ does not jump from
$\mathbf{T}^{*}(A)$ to $A$ in a $1$-to-$1$ and measure preserving way. It
follows that removing $\mathcal{F}(B)$ is insufficient to generate dissipative
flow of slope $\alpha$ that is measure preserving in the remaining set. The
extension process that we discuss next is now guided by Figure 24. This is a
process that grows the transient set in steps.

For the first step, let $A_{1}\subset A$ be defined by

$y(A_{1})=y(\mathbf{T}^{*}(A)\cap(A\cup B)).$

This means that $A_{1}$ already captures essentially all the $y$-coordinates
of $\mathbf{T}^{*}(A)$, and we throw away the complement $A\setminus A_{1}$.
More precisely, we expand the set $\mathcal{F}(B)$ by including the extra set
$\mathcal{F}(A\setminus A_{1})$. For the second step, let $A_{2}\subset A_{1}$
be defined by

$y(A_{2})=y(\mathbf{T}^{*}(A_{1})\cap(A\cup B)).$

This means that $A_{2}$ already captures essentially all the $y$-coordinates
of $\mathbf{T}^{*}(A_{1})$, and we throw away the complement $A_{1}\setminus
A_{2}$. More precisely, we expand the set $\mathcal{F}(B)$ further by
including the extra set $\mathcal{F}(A_{1}\setminus A_{2})$. And so on. We
therefore have a decreasing sequence xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$A\supset A_{1}\supset A_{2}\supset\ldots\supset A_{i}\supset\ldots$

of subsets of $A$. At the $i$-th step, we expand the set $\mathcal{F}(B)$
further by including the extra set $\mathcal{F}(A_{i-1}\setminus A_{i})$.

Of course, expanding the set $\mathcal{F}(B)$ means contracting the set
$\mathcal{F}(A)$ at the same time. Accordingly, let
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$A_{\infty}=\bigcap_{i=1}^{\infty}A_{i}.$

Starting from $A_{\infty}$, we can extend it by the dissipative flow of slope
$\alpha$ forever in a $1$-to-$1$ and measure preserving way. Then the
resulting set $\mathfrak{R}(\mathcal{P};A_{\infty};\alpha)\subset\mathcal{P}$
is clearly invariant under dissipative flow of slope $\alpha$. It then follows
from the Poincare recurrence theorem that $\lambda_{2}$-almost every point of
$\mathfrak{R}(\mathcal{P};A_{\infty};\alpha)$ is recurrent under dissipative
flow of slope $\alpha$. The two sets
$\mathfrak{R}(\mathcal{P};A_{\infty};\alpha)$ and $\mathcal{F}(A_{\infty})$
have some common properties. They both contain $A_{\infty}$ and are invariant
under dissipative flow of slope $\alpha$. They are also both disjoint from

$\mathcal{F}(B)\cup\bigcup_{i=1}^{\infty}\mathcal{F}(A_{i-1}\setminus A_{i}),$

with the convention that $A_{0}=A$. Thus we claim that

$\mathfrak{R}(\mathcal{P};A_{\infty};\alpha)=\mathcal{F}(A_{\infty})=\mathcal{R}(\mathcal{P};\alpha)$
(5.16)

and xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\mathcal{F}(B)\cup\bigcup_{i=1}^{\infty}\mathcal{F}(A_{i-1}\setminus
A_{i})=\mathcal{W}(\mathcal{P};\alpha),$

apart from exceptional sets of $2$-dimensional Lebesgue measure $0$.

We need to elaborate on (5.16). In Case 1, the sets in (5.14) are finite
unions of polygons with boundary edges that are vertical or of slope $\alpha$.
In this case, we can clearly draw the same conclusion if the extension process
is finite. However, the extension process may possibly be infinite. In this
case, the set $\mathfrak{R}(\mathcal{P};A_{\infty};\alpha)$ may be closed
which is much more complicated than a union of polygons. We then need to show
that almost every point in $\mathfrak{R}(\mathcal{P};A_{\infty};\alpha)$ is a
non-wandering point.

Note that the set $\mathfrak{R}(\mathcal{P};A_{\infty};\alpha)$ is measurable
and invariant under dissipative and measure preserving flow of slope $\alpha$,
so almost every point $P\in\mathfrak{R}(\mathcal{P};A_{\infty};\alpha)$
satisfies the Lebesgue density theorem, so that for every $\varepsilon>0$, $P$
has an $\varepsilon$-neighbourhood which intersects
$\mathfrak{R}(\mathcal{P};A_{\infty};\alpha)$ in a set of positive measure. It
then follows from the Poincare recurrence theorem that this
$\varepsilon$-neighbourhood of $P$ returns via the flow infinitely many times.
Hence $P$ is a non-wandering point.

This completes the proof. ∎

Theorem 4 leads immediately to the following question on the finiteness of the
extension process.

######  Question.

What conditions will ensure that the extension process in Theorem 4 terminates
after a finite number of steps, so that the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ is a finite union of polygons with boundary
edges that are vertical or of slope $\alpha$? Is it true that the extension
process always terminates after a finite number of steps?

In the case when the underlying polysquare translation surface is the
L-surface, we have already verified this in the affirmative in the special
case of Figure 18 where the conditions (4.1) are satisfied and in the special
case of Figure 23 with $b=0.8$ and $\alpha=1/\sqrt{2}$.

For any typical dissipative system when the underlying polysquare translation
surface is the L-surface, we can also establish this in the affirmative by
following the proof of Theorem 4. Here, in Case 1, the conclusion is clear. In
Case 2, there must be a non-empty subinterval of $B$ and corresponding
subinterval of $A$ that are gray immediately to the left, leading to a dark
gray strip that originates from the subinterval in $A$ analogous to that in
Figure 24. Clearly this dark gray strip has a twin that originates from the
subinterval in $B$, and they are both in the transient set
$\mathcal{W}(\mathcal{P};\alpha)$. This means that the third twin that
originates from the left vertical edge of the atomic square that does not
contain $A$ or $B$ must be an open set in the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$. The recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ has area $1$, and its projection to the unit
torus $[0,1)^{2}$ is the whole unit torus. The desired result now follows from
the Remark after the proof of Theorem 1 concerning a finite number of zig-
zaggings.

##  6\. More on the extension process

We continue our investigation of dissipative systems $\mathcal{P}$ where the
underlying surface is a finite polysquare translation surface, as introduced
in Section 4. We have shown in Section 5 that the extension process, starting
from the no-go strip $\mathcal{N}(\mathcal{P};\alpha)$, constructs the
transient set $\mathcal{W}(\mathcal{P};\alpha)$ in a finite or infinite number
of steps. When the number of steps is finite, then we know that both the
transient set $\mathcal{W}(\mathcal{P};\alpha)$ and the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ are finite unions of polygons with boundary
edges that are vertical or of slope $\alpha$. In this section, we establish
some partial results concerning the finiteness of the extension process. A
first result is the following.

######  Theorem 5.

In the setting and terminology of Theorem 4, suppose further that the
endpoints of the vertical intervals that make up the one-sided barrier $B$ all
have rational $y$-coordinates. Then the extension process terminates after a
finite number of steps, and both the transient set
$\mathcal{W}(\mathcal{P};\alpha)$ and the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ are finite unions of polygons with boundary
edges that are vertical or of slope $\alpha$.

###### Proof.

We follow the proof of Theorem 4 and use the Kakutani–Kac approach. In Case 1,
the extension process does not start at all, so is clearly finite. In Case 2,
we have the situation illustrated in Figure 31.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-31.pdf}\\\
\mbox{Figure 31: a step of the extension process}\end{array}$

At the start of the $i$-th step of the extension process, we have the growing
no-go zone at the end of the $(i-1)$-th step, coloured in gray, immediately to
the left of $A_{i-1}\setminus A_{i}$ as well as immediately to the left of the
corresponding subintervals of $B$. We then construct the set
$\mathcal{F}(A_{i-1}\setminus A_{i})$, which may contain one or more strips
depending on the number of intervals $A^{\dagger}$ that make up the set
$A_{i-1}\setminus A_{i}$. This is done by starting from each interval
$A^{\dagger}$ and sweeping it along by the flow of slope $\alpha$ until it
hits the set $A\cup B$, as illustrated in dark gray in Figure 31.

This flow from each interval $A^{\dagger}$ that makes up the set
$A_{i-1}\setminus A_{i}$ may split at a singularity of the system
$\mathcal{P}$, as illustrated in Figure 24. However, any singularity of
$\mathcal{P}$ which causes splitting in the $i$-th step is then eliminated, in
the sense that it will not feature in any subsequent step of the extension
process. Thus all the singularities will be eliminated after finitely many
steps of the extension process, assuming that the extension process has not
actually terminated before that. Hence by ignoring a finite number of steps at
the start of the extension process, we may therefore assume without loss of
generality that the flow corresponding to $\mathcal{F}(A^{\dagger})$ from each
interval $A^{\dagger}$ of $A_{i-1}\setminus A_{i}$ is splitting-free, and its
image $\mathbf{T}^{*}(A^{\dagger})$ on $A\cup B$ is an interval. We then need
to investigate the location of $\mathbf{T}^{*}(A^{\dagger})$. The part of
$\mathcal{P}$ immediately to the left of $\mathbf{T}^{*}(A^{\dagger})$ at the
beginning of the $i$-th step is not yet part of the growing no-go zone and is
therefore coloured white. We study its intersection with $A\cup B$. For the
purpose of our discussion, we assume that $\mathbf{T}^{*}(A^{\dagger})$ lies
on $A$, as the case if it lies on $B$ is essentially the same, with reference
to $A$ and to $B$ interchanged.

For each subset $A_{i}\subset A$, we also consider the corresponding subset
$B_{i}\subset B$. We say that an interval $A^{\dagger}\subset A_{i}$ is a ww-
subinterval (resp. gg-subinterval) of $A_{i}$ if it is maximal in the sense of
inclusion with the property that $\mathcal{P}$ is coloured white (resp. gray)
immediately to the left of $A^{\dagger}$ and is also coloured white (resp.
gray) immediately to the left of the corresponding subinterval
$B^{\dagger}\subset B_{i}$. We say that an interval $A^{\dagger}\subset A_{i}$
is a wg-subinterval (resp. gw-subinterval) of $A_{i}$ if it is maximal in the
sense of inclusion with the property that $\mathcal{P}$ is coloured white
(resp. gray) immediately to the left of $A^{\dagger}$ and is coloured gray
(resp. white) immediately to the left of the corresponding subinterval
$B^{\dagger}\subset B_{i}$. Furthermore, we define the special sum at the
start of the $i$-th step of the extension process by

$S(i)=|\\{A^{\dagger}\subset A_{i-1}:\mbox{$A^{\dagger}$ is a ww-subinterval
or a gg-subinterval of $A_{i-1}$}\\}|,$

and this represents the total number of ww-subintervals and gg-subintervals of
$A_{i-1}$ at the start of the $i$-th step of the extension process.

We need the following intermediate result.

######  Lemma 6.1.

Suppose that there is no splitting in the $i$-th step of the extension process
where the flow takes the set $A_{i-1}\setminus A_{i}$ to the image
$\mathbf{T}^{*}(A_{i-1}\setminus A_{i})$. Then the inequality $S(i+1)\leqslant
S(i)$ holds. In particular, there exists a positive constant
$c_{1}=c_{1}(\mathcal{P};B;\alpha)\geqslant 1$ such that $S(i)\leqslant c_{1}$
for every $i=1,2,3,\ldots.$

###### Proof.

The proof breaks down into a number of cases.

Note first of all that the set $A_{i-1}\setminus A_{i}$ is the union of all
the gg-subintervals $A^{\dagger}$ of $A_{i-1}$, and recall that at the start
of the $i$-th step of the extension process, the part of $\mathcal{P}$
immediately to the left of each image $\mathbf{T}^{*}(A^{\dagger})$ is
coloured white. It follows that each image $\mathbf{T}^{*}(A^{\dagger})$ is
contained in a union of finitely many ww-subintervals and wg-subintervals of
$A_{i-1}$.

######  Case 1.

Suppose that the image $\mathbf{T}^{*}(A^{\dagger})$ of a gg-subinterval
$A^{\dagger}$ of $A_{i-1}$ lies in the interior of a ww-subinterval
$A^{\ddagger}$ of $A_{i-1}$, as illustrated in the picture on the left in
Figure 32. Then after $\mathcal{F}(A^{\dagger})$ has been added to the growing
no-go zone, this ww-subinterval $A^{\ddagger}$ of $A_{i-1}$ is converted to
$2$ ww-subintervals and a gw-subinterval of $A_{i}$. It follows that going
from $S(i)$ to $S(i+1)$, we gain $2$ new ww-subintervals, lose the ww-
subinterval $A^{\ddagger}$ and lose the gg-subinterval $A^{\dagger}$. Thus
$S(i+1)=S(i)$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-32.pdf}\\\
\mbox{Figure 32: the splitting-free image $\mathbf{T}^{*}(A^{\dagger})$ and
consequences}\end{array}$

######  Case 2.

Suppose that the image $\mathbf{T}^{*}(A^{\dagger})$ of a gg-subinterval
$A^{\dagger}$ of $A_{i-1}$ lies in the interior of a union of consecutive ww-
subintervals and wg-intervals of $A_{i-1}$, with a ww-subinterval of $A_{i-1}$
at each end of the union, as illustrated in the picture on the right in Figure
32. Then after $\mathcal{F}(A^{\dagger})$ has been added to the growing no-go
zone, the ww-subinterval of $A_{i-1}$ at the each end is converted to a ww-
subinterval and a gw-subinterval of $A_{i}$, each ww-subinterval of $A_{i-1}$
in the middle is converted to a gw-subinterval of $A_{i}$, and each wg-
subinterval of $A_{i-1}$ is converted to a gg-subinterval of $A_{i}$. It
follows that going from $S(i)$ to $S(i+1)$, we gain one more gg-subinterval of
$A_{i}$ than ww-subinterval of $A_{i-1}$ that we lose, but we also lose the
gg-subinterval $A^{\dagger}$. Thus $S(i+1)=S(i)$.

######  Case 3.

Suppose that the image $\mathbf{T}^{*}(A^{\dagger})$ of a gg-subinterval
$A^{\dagger}$ of $A_{i-1}$ lies in the interior of a union of consecutive wg-
subintervals and ww-intervals of $A_{i-1}$, with a wg-subinterval of $A_{i-1}$
at each end of the union, as illustrated in the picture on the left in Figure
33. Then after $\mathcal{F}(A^{\dagger})$ has been added to the growing no-go
zone, the wg-subinterval of $A_{i-1}$ at the each end is converted to a gg-
subinterval and a wg-subinterval of $A_{i}$, each wg-subinterval of $A_{i-1}$
in the middle is converted to a gg-subinterval of $A_{i}$, and each ww-
subinterval of $A_{i-1}$ is converted to a gw-subinterval of $A_{i}$. It
follows that going from $S(i)$ to $S(i+1)$, we gain one more gg-subinterval of
$A_{i}$ than ww-subinterval of $A_{i-1}$ that we lose, but we also lose the
gg-subinterval $A^{\dagger}$. Thus $S(i+1)=S(i)$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-33.pdf}\\\
\mbox{Figure 33: the splitting-free image $\mathbf{T}^{*}(A^{\dagger})$ and
consequences}\end{array}$

######  Case 4.

Suppose that the image $\mathbf{T}^{*}(A^{\dagger})$ of a gg-subinterval
$A^{\dagger}$ of $A_{i-1}$ lies in the interior of a union of consecutive wg-
subintervals and ww-intervals of $A_{i-1}$, with a wg-subinterval of $A_{i-1}$
at one end of the union and a ww-subinterval of $A_{i-1}$ at the other end of
the union, as illustrated in the picture on the right in Figure 33. Then after
$\mathcal{F}(A^{\dagger})$ has been added to the growing no-go zone, the wg-
subinterval of $A_{i-1}$ at one end is converted to a gg-subinterval and a wg-
subinterval of $A_{i}$, the ww-subinterval of $A_{i-1}$ at the other end is
converted to a ww-subinterval and a gw-subinterval of $A_{i}$, each wg-
subinterval of $A_{i-1}$ in the middle is converted to a gg-subinterval of
$A_{i}$, and each ww-subinterval of $A_{i-1}$ in the middle is converted to a
gw-subinterval of $A_{i}$. It follows that going from $S(i)$ to $S(i+1)$, we
gain one more gg-subinterval of $A_{i}$ than ww-subinterval of $A_{i-1}$ that
we lose, but we also lose the gg-subinterval $A^{\dagger}$. Thus
$S(i+1)=S(i)$.

We next investigate those instances when an endpoint of the image
$\mathbf{T}^{*}(A^{\dagger})$ coincides with an endpoint of the union of
consecutive wg-subintervals and ww-intervals of $A_{i-1}$.

######  Case 5.

Suppose that the extreme endpoint of the wg-subinterval of $A_{i-1}$ at one
end of the union of consecutive wg-subintervals and ww-intervals of $A_{i-1}$
coincides with an endpoint of the interval $\mathbf{T}^{*}(A^{\dagger})$, as
illustrated in the picture on the right in Figure 34. Compared to the picture
on the left, we see that the only difference is that there is now one fewer
wg-subinterval of $A_{i}$, and this does not contribute to $S(i+1)$. Thus
$S(i+1)=S(i)$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-34.pdf}\\\
\mbox{Figure 34: the splitting-free image $\mathbf{T}^{*}(A^{\dagger})$ and
consequences}\end{array}$

######  Case 6.

Suppose that the extreme endpoint of the ww-subinterval of $A_{i-1}$ at one
end of the union of consecutive wg-subintervals and ww-intervals of $A_{i-1}$
coincides with an endpoint of the interval $\mathbf{T}^{*}(A^{\dagger})$, as
illustrated in the picture on the right in Figure 35. Compared to the picture
on the left, we see that the only difference is that there is now one fewer
ww-subinterval of $A_{i}$, and this makes a difference to $S(i+1)$. Thus
$S(i+1)<S(i)$.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-35.pdf}\\\
\mbox{Figure 35: the splitting-free image $\mathbf{T}^{*}(A^{\dagger})$ and
consequences}\end{array}$

This completes the proof. ∎

Assume, on the contrary, that the extension process does not terminate after a
finite number of steps. That means that for every $i=1,2,3,\ldots,$ at the
beginning of the $i$-th step, there are gg-subintervals of $A_{i-1}$ to work
with. On the other hand, we see from the proof of Lemma 6.1 that each gg-
subinterval $A^{\dagger}$ of $A_{i-1}$ originates from a gg-subinterval of
$A_{i-2}$, which in turn originates from a gg-subinterval of $A_{i-3}$, and so
on, and so has its origin from a gg-subinterval of $A$. Thus we can examine
their combined horizontal flow. More precisely, for each gg-subinterval
$A^{\dagger}$ of $A_{i-1}$, we consider the total length $L(A^{\dagger})$ of
the horizontal flow, starting from a gg-subinterval of $A$ and culminating in
the image $A^{\dagger}$, as illustrated in Figure 36 where the flow is
stretched out on the plane.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-36.pdf}\\\
\mbox{Figure 36: illustration of $L(A^{\dagger})$ on the plane}\end{array}$

Summing $L(A^{\dagger})$ over the collection $\mathcal{H}(i)$ of all the
individual gg-subintervals $A^{\dagger}$ of $A_{i-1}$ gives rise to the sum
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$L(i)=\sum_{A^{\dagger}\in\mathcal{H}(i)}L(A^{\dagger}).$

Then the sequence $L(i)$, $i=1,2,3,\ldots,$ is increasing and not bounded
above.

We need a well known number theoretic result.

######  Lemma 6.2.

Let $q_{k}=q_{k}(\alpha)$ denote the denominator of the $k$-convergent of the
irrational number $\alpha>0$. Then for every integer $n=1,2,3,\ldots,q_{k}-1$,
we have

$\|n\alpha\|>\frac{1}{2q_{k}},$

where $\|\gamma\|$ denotes the distance of $\gamma$ from the nearest integer.

Using the time-quantitative version of the Gutkin–Veech theorem and Lemma 6.1,
we see that there exists a sufficiently large positive constant
$c_{2}=c_{2}(\mathcal{P};B;\alpha)$ such that any interval of positive real
numbers of length at least $c_{2}$ contains at least one member of the
sequence $L(i)$, $i=1,2,3,\ldots.$

Let $q_{k_{1}},q_{k_{2}},q_{k_{3}},\ldots,$ where $k_{1}<k_{2}<k_{3}<\ldots,$
be a sequence of sufficiently large denominators of the convergents of
$\alpha$ such that

$q_{k_{\nu}}>2c_{2},\quad q_{k_{\nu+1}}\geqslant
4c_{1}q_{k_{\nu}},\quad\nu=1,2,3,\ldots.$ (6.1)

Then there exists an increasing sequence of integers $i_{\nu}$,
$\nu=1,2,3,\ldots,$ such that

$L(i_{\nu})\in\left(\frac{q_{k_{\nu}}}{2},q_{k_{\nu}}\right),\quad\nu=1,2,3,\ldots,$

and the intervals are pairwise disjoint, since $c_{1}\geqslant 1$. We may
further assume that there is no splitting in the extension process from step
$i_{1}$ onwards.

For every $i=1,2,3,\ldots,$ let $\mathscr{A}(i)$ denote the subset of the
transient set $\mathcal{W}(\mathcal{P};\alpha)$ that the extension process
gives after $i$ steps.

We wish to give a lower bound to the the quantities

$\lambda_{2}(\mathscr{A}(i_{\nu})\setminus\mathscr{A}(i_{\nu-1})),\quad\nu=2,3,4,\ldots.$

In view of Lemma 6.1, there exists $A_{\nu}^{\dagger}\in\mathcal{H}(i_{\nu})$
such that

$L(A_{\nu}^{\dagger})\geqslant\frac{q_{k_{\nu}}}{2c_{1}}.$ (6.2)

To determine the length of the interval $A_{\nu}^{\dagger}$, we use the
assumption that all of the vertical intervals that make up the one-sided
barrier $B$ have endpoints with rational $y$-coordinates. Suppose that
$Q=Q(\mathcal{P};B;\alpha)$ denotes the lowest common multiple of the
denominators of these $y$-coordinates. On magnifying the dissipative system
$\mathcal{P}$ by a factor $Q$ in each direction, we arrive at a dissipative
system $\mathcal{P}_{Q}$ where the one-sided barrier corresponding to $B$
consists of whole vertical edges of atomic squares of $\mathcal{P}_{Q}$. Thus
the magnified analogue of the interval $A_{\nu}^{\dagger}$ in
$\mathcal{P}_{Q}$ must have endpoints with $y$-coordinates of the form
$\\{b+m\alpha\\}$ for some $m=0,1,2,3,\ldots,q_{k_{\nu}}-1$, where
$b\in\mathcal{R}$ is fixed, and length

$|\\{b+m_{1}\alpha\\}-\\{b+m_{2}\alpha\\}|\geqslant\|(m_{1}-m_{2})\alpha\|>\frac{1}{2q_{k_{\nu}}},$

where $0\leqslant m_{1}<m_{2}<q_{k_{\nu}}$, in view of Lemma 6.2. Thus the
length of the interval $A_{\nu}^{\dagger}$ is at least
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\frac{1}{2q_{k_{\nu}}Q}.$

It follows from (6.1) and (6.2) that

$\lambda_{2}(\mathscr{A}(i_{\nu})\setminus\mathscr{A}(i_{\nu-1}))\geqslant\frac{L(A_{\nu}^{\dagger})-L(i_{\nu-1})}{2q_{k_{\nu}}Q}\geqslant\frac{1}{2q_{k_{\nu}}Q}\left(\frac{q_{k_{\nu}}}{2c_{1}}-q_{k_{\nu-1}}\right)\geqslant\frac{1}{4c_{1}Q}.$

The disjointness of the sets
$\mathscr{A}(i_{\nu})\setminus\mathscr{A}(i_{\nu-1})$, $\nu=2,3,4,\ldots,$ now
implies infinite total area, and this gives the desired contradiction. ∎

Under the hypotheses of Theorem 5, the restriction of the dissipative flow of
irrational slope $\alpha$ to the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ modulo one is integrable geodesic flow on
the unit torus $[0,1)^{2}$. The ergodic theorem then implies that this
restricted flow is a $k$-fold covering of geodesic flow on the unit torus
$[0,1)^{2}$ for some integer $k$ satisfying $1<k<s$, where $s$ is the number
of atomic squares of the underlying finite polysquare translation surface of
the dissipative system $\mathcal{P}$. Since the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ is a finite union of polygons with boundary
edges that are vertical or of slope $\alpha$, the usual discretization of the
flow restricted to $\mathcal{R}(\mathcal{P};\alpha)$ defines a bijective
interval exchange transformation from the interval $[0,k)$ to itself. If the
overlapping graph of this bijection is connected, then Lemma 4.1 guarantees
equidistribution of the orbits inside $\mathcal{R}(\mathcal{P};\alpha)$. In
this case, we say that the attractor $\mathcal{R}(\mathcal{P};\alpha)$ is
minimal.

We also establish a second partial result concerning the finiteness of the
extension process.

######  Theorem 6.

In the setting and terminology of Theorem 4, suppose further that a component
of the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ contains an open subset
$G$. Then this component of the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ can be constructed by a finite number of
zig-zaggings and is a finite union of polygons with boundary edges that are
vertical or of slope $\alpha$.

Note that we have no information on the structure of the part of the recurrent
set $\mathcal{R}(\mathcal{P};\alpha)$ outside this special component, so we
are unable to draw any conclusions about the recurrent set
$\mathcal{R}(\mathcal{P};\alpha)$ as a whole.

We begin with some general facts.

Suppose that the underlying polysquare translation surface of the dissipative
system $\mathcal{P}$ has $s$ atomic squares. Then the dissipative flow of
irrational slope $\alpha>0$ on $\mathcal{P}$ defines a non-bijective
transformation

$f=f_{\alpha}:[0,s)\to[0,s),$ (6.3)

where the interval $[i-1,i)$ represents the left vertical edge of the $i$-th
atomic square.

######  Remark.

Strictly speaking, $f=f_{\alpha}$ is not an interval exchange transformation
as dissipative flow is in general not reversible.

Denote modulo one projection by

$\pi:[0,s)\to[0,1).$ (6.4)

The function $\pi$ relates the discretization of dissipative flow of slope
$\alpha$ on $\mathcal{P}$ to the discretization of geodesic flow of slope
$\alpha$ on the unit torus $[0,1)^{2}$.

Let $\mathscr{S}\subset[0,s)$ be a subset of multiplicity $1$. In other words,
the restriction

$\pi_{\mathscr{S}}:\mathscr{S}\to[0,1):y\mapsto\pi(y)$

of $\pi$ to the set $\mathscr{S}$ is injective. We further assume that
$\mathscr{S}$ is a union of finitely many intervals, so likewise for the set

$S=\pi(\mathscr{S}).$

Let the modulo one projection of $f=f_{\alpha}$ be denoted by

$g=g_{\alpha}:[0,1)\to[0,1),$ (6.5)

so that xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$g\circ\pi=\pi\circ f.$ (6.6)

Then $g=g_{\alpha}$ describes rotation by $\alpha$ on the unit circle $[0,1)$,
and is the interval exchange transformation corresponding to the
discretization of geodesic flow of slope $\alpha$ on the unit torus
$[0,1)^{2}$.

For any $x\in S\subset[0,1)$, let

$n(x)=n_{S}(x)=\min\\{n\geqslant 1:g^{n}(x)\in S\\}.$

Then it follows from Lemma 5.2 due to Kac that there exists a common threshold
$N^{*}(S)=N^{*}(S;\alpha)$ such that

$n(x)=n_{S}(x)\leqslant N^{*}(S)\quad\mbox{for all $x\in S$}.$

As a consequence of (5.11) in the proof of Lemma 5.2, we have

$\\!\\!\\![0,1)=\bigcup_{n=1}^{N^{*}(S)}\bigcup_{k=0}^{n-1}g^{k}(\\{x\in
S:n_{S}(x)=n\\})=\bigcup_{n=1}^{N^{*}(S)}\bigcup_{k=1}^{n}g^{k}(\\{x\in
S:n_{S}(x)=n\\}),$ (6.7)

apart from an exceptional set of measure $0$ and where each union is pairwise
disjoint. We then lift this up to $[0,m)$ and obtain a quantity

$\Psi(\mathscr{S})=\bigcup_{n=1}^{N^{*}(S)}\bigcup_{k=1}^{n}f^{k}(\\{y\in\mathscr{S}:n_{S}(\pi(y))=n\\}),$
(6.8)

and of course $\pi(\Psi(\mathscr{S}))=[0,1)$.

The assumption that $\mathscr{S}$ is a set of multiplicity $1$ implies that
$\pi$ is injective on $\Psi(\mathscr{S})$; in other words, $\Psi(\mathscr{S})$
is a copy of the unit torus $[0,1)$ generated from $\mathscr{S}$ by the
dissipative flow of slope $\alpha$. More importantly, $\Psi(\mathscr{S})$ is a
set of multiplicity $1$.

The set xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\phi(\mathscr{S})=\bigcup_{n=1}^{N^{*}(S)}f^{n}(\\{y\in\mathscr{S}:n_{S}(\pi(y))=n\\})$

is essentially a copy $\mathscr{S}$, since

$\displaystyle\pi(\phi(\mathscr{S}))$
$\displaystyle=\bigcup_{n=1}^{N^{*}(S)}\pi(f^{n}(\\{y\in\mathscr{S}:n_{S}(\pi(y))=n\\}))$
$\displaystyle=\bigcup_{n=1}^{N^{*}(S)}g^{n}(\pi(\\{y\in\mathscr{S}:n_{S}(\pi(y))=n\\}))$
$\displaystyle=\bigcup_{n=1}^{N^{*}(S)}g^{n}(\\{x\in S:n_{S}(x)=n\\})$
$\displaystyle=S=\pi(\mathscr{S}),$

apart from an exceptional set of measure $0$, as it follows from (6.7) that

$\bigcup_{n=1}^{N^{*}(S)}g^{n}(\\{x\in
S:n_{S}(x)=n\\})=\bigcup_{n=1}^{N^{*}(S)}\\{x\in S:n_{S}(x)=n\\}=S,$

apart from an exceptional set of measure $0$.

Clearly $\phi(\mathscr{S})\subset\Psi(\mathscr{S})$, so $\phi(\mathscr{S})$ is
a set of multiplicity $1$.

We next set up the crucial step in the proof of Theorem 6.

The open subset $G$ of $\mathcal{P}$ contains a disc of positive radius. We
may assume that this disc is sufficiently small that it is contained in a
single atomic square of the underlying polysquare translation surface of
$\mathcal{P}$ and the reverse flow in the direction $(-1,-\alpha)$ projects it
on to the left vertical edge of a single atomic square, as illustrated in
Figure 37.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-37.pdf}\\\
\mbox{Figure 37: constructing the interval $\mathscr{I}_{0}$}\end{array}$

This gives rise to an interval $\mathscr{I}_{0}\subset[0,s)$ of positive
length and multiplicity $1$ on the left vertical edge of an atomic square of
the underlying polysquare translation surface of $\mathcal{P}$.

Define the sets $\mathscr{I}_{1},\mathscr{I}_{2},\mathscr{I}_{3},\ldots$ in
terms of $\mathscr{I}_{0}$ inductively by

$\mathscr{I}_{1}=\phi(\mathscr{I}_{0})\setminus\mathscr{I}_{0},\quad\mathscr{I}_{j}=\phi(\mathscr{I}_{j-1})\setminus\mathscr{I}_{0},\quad
j=2,3,4,\ldots,$ (6.9)

assuming that $\mathscr{I}_{j-1}$ has positive measure. Note that each set
$\mathscr{I}_{1},\mathscr{I}_{2},\mathscr{I}_{3},\ldots$ is of multiplicity
$1$, and each $\Psi(\mathscr{I}_{j})$ is a copy of $[0,1)$ if
$\mathscr{I}_{j}$ has positive measure.

######  Lemma 6.3.

For any non-negative integers $j^{\prime}\neq j^{\prime\prime}$, the sets
$\Psi(\mathscr{I}_{j^{\prime}})$ and $\Psi(\mathscr{I}_{j^{\prime\prime}})$
are disjoint.

###### Proof of Theorem 6.

By Lemma 6.3, the $s+1$ sets

$\Psi(\mathscr{I}_{j})\subset[0,s),\quad j=0,1,2,3,\ldots,s,$

are pairwise disjoint. Note next that $\Psi(\mathscr{I}_{j})$ is a copy of
$[0,1)$, and so has measure $1$, if $\mathscr{I}_{j}$ has positive measure.
This implies that $\Psi(\mathscr{I}_{j^{*}})$ must have measure $0$ for some
$j^{*}=0,1,2,3,\ldots,s$. Hence the set xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\bigcup_{j=0}^{j^{*}-1}\Psi(\mathscr{I}_{j})$ (6.10)

is generated from $\mathscr{I}_{0}$ in at most xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\sum_{j=0}^{j^{*}-1}N^{*}(\pi(\mathscr{I}_{j}))$

zig-zaggings. Clearly the set (6.10) is invariant under $f$. Note that each
set $\Psi(\mathscr{I}_{j})$ for any integer $j\geqslant j^{*}$ has measure $0$
and so can be ignored. ∎

######  Remark.

Figure 38 illustrates how we set up an induction process and explains the
definitions in (6.9). It also explains the purpose of Lemma 6.3.

$\begin{array}[]{c}\includegraphics[scale={0.8}]{figure-38.pdf}\\\
\mbox{Figure 38: the early steps of the induction process}\end{array}$

The idea is as follows. Starting from the set $\mathscr{I}_{0}$, we generate
the set $\Psi(\mathscr{I}_{0})$ by the flow which is represented in discrete
form by $f$. This set $\Psi(\mathscr{I}_{0})$ has an interesting subset
$\phi(\mathscr{I}_{0})$ which collects together all the points corresponding
to first returns to $\mathscr{I}_{0}$ modulo one. In general,
$\phi(\mathscr{I}_{0})$ has a part that lies in $\mathscr{I}_{0}$ and a part
that lies outside $\mathscr{I}_{0}$. Since the part in $\mathscr{I}_{0}$ is
already accounted for, we remove it from $\phi(\mathscr{I}_{0})$ and consider
the subset $\mathscr{I}_{1}=\phi(\mathscr{I}_{0})\setminus\mathscr{I}_{0}$.
Next, we generate a set $\Psi(\mathscr{I}_{1})$ which has an interesting
subset $\phi(\mathscr{I}_{1})$. Again, $\phi(\mathscr{I}_{1})$ has a part that
lies in $\mathscr{I}_{0}$ and a part that lies outside $\mathscr{I}_{0}$.
Hence we consider the subset
$\mathscr{I}_{2}=\phi(\mathscr{I}_{1})\setminus\mathscr{I}_{0}$. And so on. In
this way, we ensure that the sets
$\Psi(\mathscr{I}_{0}),\Psi(\mathscr{I}_{1}),\Psi(\mathscr{I}_{2}),\ldots$ are
pairwise disjoint. Together they take up all the points in $[0,s)$
corresponding to the component of the recurrent set under consideration.

###### Proof of Lemma 6.3.

Note that the lemma follows immediately from the uniqueness property of
Assertion A below.

######  Assertion A.

For every integer $\ell\geqslant 0$ and $y^{*}\in\Psi(\mathscr{I}_{\ell})$,
there is a unique sequence

$(y_{0},h_{0}),\quad(y_{1},h_{1}),\quad(y_{2},h_{2}),\quad\ldots,\quad(y_{\ell},h_{\ell}),$

where $y_{j}\in\mathscr{I}_{j}$ and $h_{j}=1,2,3,\ldots$ for every
$j=0,1,\ldots,\ell$, such that

$y_{1}=f^{h_{0}}(y_{0}),\quad y_{2}=f^{h_{1}}(y_{1}),\quad\ldots,\quad
y_{\ell}=f^{h_{\ell-1}}(y_{\ell-1}),\quad y^{*}=f^{h_{\ell}}(y_{\ell}).$

Assertion A follows from Assertion B below by backtracking from $y^{*}$ to
$y_{0}$, since the set $\mathscr{I}_{0}$ is never visited in between.

######  Assertion B.

For every integer $\ell\geqslant 0$ and $y^{*}\in\Psi(\mathscr{I}_{\ell})$,
there is a sequence

$(y_{0},h_{0}),\quad(y_{1},h_{1}),\quad(y_{2},h_{2}),\quad\ldots,\quad(y_{\ell},h_{\ell}),$

where $y_{j}\in\mathscr{I}_{j}$ and $h_{j}=1,2,3,\ldots$ for every
$j=0,1,\ldots,\ell$, such that the numbers

$\begin{array}[]{lllll}y_{0},&\quad f(y_{0}),&\quad
f^{2}(y_{0}),&\quad\ldots,&\quad f^{h_{0}-1}(y_{0}),\vspace{3pt}\\\
y_{1}=f^{h_{0}}(y_{0}),&\quad f(y_{1}),&\quad f^{2}(y_{1}),&\quad\ldots,&\quad
f^{h_{1}-1}(y_{1}),\vspace{3pt}\\\ y_{2}=f^{h_{1}}(y_{1}),&\quad
f(y_{2}),&\quad f^{2}(y_{2}),&\quad\ldots,&\quad
f^{h_{2}-1}(y_{2}),\vspace{3pt}\\\ &&\quad\ldots,\vspace{3pt}\\\
y_{\ell}=f^{h_{\ell-1}}(y_{\ell-1}),&\quad f(y_{\ell}),&\quad
f^{2}(y_{\ell}),&\quad\ldots,&\quad f^{h_{\ell}-1}(y_{\ell}),\vspace{3pt}\\\
y^{*}=f^{h_{\ell}}(y_{\ell}),\end{array}$ (6.11)

apart possibly from $y_{0}$ and $y^{*}$, do not belong to $\mathscr{I}_{0}$.

######  Remark.

In Assertion A and Assertion B, for every $j=0,1,\ldots,\ell-1,$ we have

$f^{h_{j}}(y_{j})=y_{j+1}\in\mathscr{I}_{j+1}\subset\phi(\mathscr{I}_{j}),\quad\mbox{so
that}\quad h_{j}=n_{\pi(\mathscr{I}_{j})}(\pi(y_{j})).$

On the other hand, we have

$f^{h_{\ell}}(y_{\ell})\in\Psi(\mathscr{I}_{\ell}),\quad\mbox{so that}\quad
h_{\ell}\leqslant n_{\pi(\mathscr{I}_{\ell})}(\pi(y_{\ell})).$

It remains to establish Assertion B. We proceed by induction on $\ell$.

Consider first the initial case $\ell=0$, and let
$y^{*}\in\Psi(\mathscr{I}_{0})$. It follows from (6.8) that there exist
$n=1,\ldots,N^{*}(\pi(\mathscr{I}_{0}))$ and $k=1,\ldots,n$ such that

$y^{*}\in
f^{k}(\\{y\in\mathscr{I}_{0}:n_{\pi(\mathscr{I}_{0})}(\pi(y))=n\\}),$

so that $y^{*}=f^{k}(y_{0})$ for some $y_{0}\in\mathscr{I}_{0}$ satisfying
$n_{\pi(\mathscr{I}_{0})}(\pi(y_{0}))=n$, and so (6.11) for $\ell=0$ is
satisfied with $h_{0}=k\leqslant n$ and none of
$f(y_{0}),f^{2}(y_{0}),\ldots,f^{k-1}(y_{0})$ belongs to $\mathscr{I}_{0}$.

Suppose next that $i>0$ and Assertion B holds for every $\ell<i$. Let
$y^{*}\in\Psi(\mathscr{I}_{i})$. It follows from (6.8) that there exist
$n=1,\ldots,N^{*}(\pi(\mathscr{I}_{i}))$ and $k=1,\ldots,n$ such that
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$y^{*}\in
f^{k}(\\{y\in\mathscr{I}_{i}:n_{\pi(\mathscr{I}_{i})}(\pi(y))=n\\}),$

so that $y^{*}=f^{k}(y_{i})$ for some $y_{i}\in\mathscr{I}_{i}$ satisfying
$n_{\pi(\mathscr{I}_{i})}(\pi(y_{i}))=n$. Write $h_{i}=k\leqslant n$. Then we
have a pair $(y_{i},h_{i})$ where $y_{i}\in\mathscr{I}_{i}$ and
$h_{i}=1,2,3,\ldots,$ leading to the numbers xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\begin{array}[]{lllll}y_{i},&\quad f(y_{i}),&\quad
f^{2}(y_{i}),&\quad\ldots,&\quad f^{h_{i}-1}(y_{i}),\vspace{3pt}\\\
y^{*}=f^{h_{i}}(y_{i}).\end{array}$ (6.12)

Since $y_{i}\in\mathscr{I}_{i}\subset\Psi(\mathscr{I}_{i-1})$, using the case
$\ell=i-1$, there is a sequence

$(y_{0},h_{0}),\quad(y_{1},h_{1}),\quad(y_{2},h_{2}),\quad\ldots,\quad(y_{i-1},h_{i-1}),$

where $y_{j}\in\mathscr{I}_{j}$ and $h_{j}=1,2,3,\ldots$ for every
$j=0,1,2,\ldots,i-1$, such that the numbers

$\begin{array}[]{lllll}y_{0},&\quad f(y_{0}),&\quad
f^{2}(y_{0}),&\quad\ldots,&\quad f^{h_{0}-1}(y_{0}),\vspace{3pt}\\\
y_{1}=f^{h_{0}}(y_{0}),&\quad f(y_{1}),&\quad f^{2}(y_{1}),&\quad\ldots,&\quad
f^{h_{1}-1}(y_{1}),\vspace{3pt}\\\ y_{2}=f^{h_{1}}(y_{1}),&\quad
f(y_{2}),&\quad f^{2}(y_{2}),&\quad\ldots,&\quad
f^{h_{2}-1}(y_{2}),\vspace{3pt}\\\ &&\quad\ldots,\vspace{3pt}\\\
y_{i-1}=f^{h_{i-2}}(y_{i-2}),&\quad f(y_{i-1}),&\quad
f^{2}(y_{i-1}),&\quad\ldots,&\quad f^{h_{i-1}-1}(y_{i-1}),\vspace{3pt}\\\
y_{i}=f^{h_{i-1}}(y_{i-1}),\end{array}$ (6.13)

apart possibly from $y_{0}$ and $y_{i}$, do not belong to $\mathscr{I}_{0}$.
Clearly (6.13) combines with (6.12) to give the existence aspect of (6.11) for
the case $\ell=i$. It remains to show that the numbers in (6.12), apart
possibly from $y^{*}$, do not belong to $\mathscr{I}_{0}$.

Clearly
$y_{i}\in\mathscr{I}_{i}=\phi(\mathscr{I}_{i-1})\setminus\mathscr{I}_{0}$ does
not belong to $\mathscr{I}_{0}$. Suppose on the contrary that some other
number in (6.12), apart from $y_{i}$ and $y^{*}$, belongs to
$\mathscr{I}_{0}$. Then it is of the form xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$f^{h}(y_{i})\quad\mbox{for some $h=1,\ldots,h_{i}-1$},$

and is clearly in $\Psi(\mathscr{I}_{i})$ but not in $\phi(\mathscr{I}_{i})$.
Let $y^{\prime}\in\mathscr{I}_{0}$ denote the one with minimal value of $h$.
Then xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$y^{\prime}\in\mathscr{I}_{0}\quad\mbox{and}\quad
y^{\prime}\not\in\phi(\mathscr{I}_{i}).$ (6.14)

We can next truncate (6.12) and replace it with

$\begin{array}[]{lllll}y_{i},&\quad f(y_{i}),&\quad
f^{2}(y_{i}),&\quad\ldots,&\quad f^{h-1}(y_{i}),\vspace{3pt}\\\
y^{\prime}=f^{h}(y_{i}).\end{array}$ (6.15)

Combining (6.13) and (6.15) leads to

$\begin{array}[]{lllll}y_{0},&\quad f(y_{0}),&\quad
f^{2}(y_{0}),&\quad\ldots,&\quad f^{h_{0}-1}(y_{0}),\vspace{3pt}\\\
y_{1}=f^{h_{0}}(y_{0}),&\quad f(y_{1}),&\quad f^{2}(y_{1}),&\quad\ldots,&\quad
f^{h_{1}-1}(y_{1}),\vspace{3pt}\\\ y_{2}=f^{h_{1}}(y_{1}),&\quad
f(y_{2}),&\quad f^{2}(y_{2}),&\quad\ldots,&\quad
f^{h_{2}-1}(y_{2}),\vspace{3pt}\\\ &&\quad\ldots,\vspace{3pt}\\\
y_{i-1}=f^{h_{i-2}}(y_{i-2}),&\quad f(y_{i-1}),&\quad
f^{2}(y_{i-1}),&\quad\ldots,&\quad f^{h_{i-1}-1}(y_{i-1}),\vspace{3pt}\\\
y_{i}=f^{h_{i-1}}(y_{i-1}),&\quad f(y_{i}),&\quad
f^{2}(y_{i}),&\quad\ldots,&\quad f^{h-1}(y_{i}),\vspace{3pt}\\\
y^{\prime}=f^{h}(y_{i}),\end{array}$ (6.16)

where, apart possibly from $y_{0}$ and $y^{\prime}$, the terms do not belong
to $\mathscr{I}_{0}$.

Meanwhile, it follows from (6.14) that

$\pi(y^{\prime})\in\pi(\mathscr{I}_{0})\setminus\pi(\phi(\mathscr{I}_{i}))=\pi(\mathscr{I}_{0})\setminus\pi(\mathscr{I}_{i}).$

It is easy to check that $\pi(\mathscr{I}_{j})\subset\mathscr{I}_{j-1}$ for
every $j=1,2,3,\ldots,$ so there exists some $\kappa<i$ such that
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\pi(y^{\prime})\in\pi(\mathscr{I}_{\kappa})\quad\mbox{and}\quad\pi(y^{\prime})\not\in\pi(\mathscr{I}_{\kappa+1}).$

Now xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\displaystyle\pi(\mathscr{I}_{\kappa+1})$
$\displaystyle=\pi(\phi(\mathscr{I}_{\kappa})\setminus\mathscr{I}_{0})=\pi(\phi(\mathscr{I}_{\kappa}))\setminus\pi(\phi(\mathscr{I}_{\kappa})\cap\mathscr{I}_{0})$
$\displaystyle=\pi(\mathscr{I}_{\kappa})\setminus\pi(\phi(\mathscr{I}_{\kappa})\cap\mathscr{I}_{0}),$

so that
$\pi(y^{\prime})\in\pi(\phi(\mathscr{I}_{\kappa})\cap\mathscr{I}_{0})$.
Together with the assumption $y^{\prime}\in\mathscr{I}_{0}$, this leads to

$y^{\prime}\in\phi(\mathscr{I}_{\kappa})\subset\Psi(\mathscr{I}_{\kappa}).$

The case $\ell=\kappa$ of the assertion now gives a sequence

$(y^{\star}_{0},h^{\star}_{0}),\quad(y^{\star}_{1},h^{\star}_{1}),\quad(y^{\star}_{2},h^{\star}_{2}),\quad\ldots,\quad(y^{\star}_{\kappa},h^{\star}_{\kappa}),$

where $y^{\star}_{k}\in\mathscr{I}_{k}$ and $h^{\star}_{k}=1,2,3,\ldots$ for
every $k=0,1,2,\ldots,\kappa$, and the numbers

$\begin{array}[]{lllll}y^{\star}_{0},&\quad f(y^{\star}_{0}),&\quad
f^{2}(y^{\star}_{0}),&\quad\ldots,&\quad
f^{h^{\star}_{0}-1}(y^{\star}_{0}),\vspace{3pt}\\\
y^{\star}_{1}=f^{h^{\star}_{0}}(y^{\star}_{0}),&\quad f(y^{\star}_{1}),&\quad
f^{2}(y^{\star}_{1}),&\quad\ldots,&\quad
f^{h^{\star}_{1}-1}(y^{\star}_{1}),\vspace{3pt}\\\
y^{\star}_{2}=f^{h^{\star}_{1}}(y^{\star}_{1}),&\quad f(y^{\star}_{2}),&\quad
f^{2}(y^{\star}_{2}),&\quad\ldots,&\quad
f^{h^{\star}_{2}-1}(y^{\star}_{2}),\vspace{3pt}\\\
&&\quad\ldots,\vspace{3pt}\\\
y^{\star}_{\kappa}=f^{h^{\star}_{\kappa-1}}(y^{\star}_{\kappa-1}),&\quad
f(y^{\star}_{\kappa}),&\quad f^{2}(y^{\star}_{\kappa}),&\quad\ldots,&\quad
f^{h^{\star}_{\kappa}-1}(y^{\star}_{\kappa}),\vspace{3pt}\\\
y^{\prime}=f^{h^{\star}_{\kappa}}(y^{\star}_{\kappa}),\end{array}$ (6.17)

where, apart possibly from $y^{\star}_{0}$ and $y^{\prime}$, the terms do not
belong to $\mathscr{I}_{0}$.

Note that the two sequences (6.16) and (6.17) have the same endpoint
$y^{\prime}$. Working backward along (6.16) and (6.17) one term at a time, we
may reach $y_{0}$ or $y^{\star}_{0}$ first, or at the same step.

If this process reaches $y_{0}$ first, then $y_{0}$ must be one of the terms
in (6.17) that is different from $y^{\star}_{0}$ and $y^{\prime}$, so that
$y_{0}\not\in\mathscr{I}_{0}$, a contradiction.

If this process reaches $y^{\star}_{0}$ first, then $y^{\star}_{0}$ must be
one of the terms in (6.16) that is different from $y_{0}$ and $y^{\prime}$, so
that $y^{\star}_{0}\not\in\mathscr{I}_{0}$, a contradiction.

If the process reaches $y_{0}$ and $y^{\star}_{0}$ at the same step, then
clearly $y^{\star}_{0}=y_{0}$. Then successively we have
$y^{\star}_{1}=y_{1}$, $y^{\star}_{2}=y_{2}$, and so on, and eventually
$y^{\prime}=y_{\kappa+1}$. But $y^{\prime}\in\mathscr{I}_{0}$, and
$y_{\kappa+1}\in\mathscr{I}_{\kappa+1}=\phi(\mathscr{I}_{\kappa})\setminus\mathscr{I}_{0}$
does not belong to $\mathscr{I}_{0}$, thus leading to a contradiction also.

This completes the proof. ∎

We complete this section by giving a complete answer to the question of the
finiteness of the extension process.

######  Theorem 7.

Consider dissipative flow of slope $\alpha$, where $\alpha>0$ is irrational,
on a system $\mathcal{P}$, where a finite polysquare translation surface has
beem modified with the inclusion of a one-sided barrier $B$ on the common
vertical edge of two neighbouring atomic squares, in the form of a union of
finitely many vertical intervals, and where a set $A$ corresponding to $B$ has
been chosen on a different vertical edge on the same horizontal street that
contains $B$. Then the recurrent set $\mathcal{R}(\mathcal{P};\alpha)$ and the
transient set $\mathcal{W}(\mathcal{P};\alpha)$ can be constructed by an
extension process on the no-go zone $\mathcal{N}(\mathcal{P};\alpha)$.
Furthermore, this process is finite, and both sets
$\mathcal{R}(\mathcal{P};\alpha)$ and $\mathcal{W}(\mathcal{P};\alpha)$ are
finite unions of polygons with boundary edges that are vertical or of slope
$\alpha$.

We first make some preliminary discussion where we introduce the new ideas.

As before, we make use of functions $f$, $\pi$ and $g$ satisfying (6.3)–(6.6)
first introduced earlier in this section, where we use the interval $[0,s)$ to
describe the left vertical edges of the $s$ atomic squares of the finite
polysquare translation surface, with $[i,i+1)$ representing the left vertical
edge of the $i$-th atomic square, where the index
$i\in\mathscr{M}=\\{0,1,\ldots,s-1\\}$.

The one-sided barrier $B$ on a common vertical edge of two atomic squares is a
union of finitely many intervals, so that

$B\subset[i,i+1)\subset[0,s)$

for some $i\in\mathscr{M}=\\{0,1,\ldots,s-1\\}$, and the boundary $\partial B$
is a finite set. Denote the set of splitting points by
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\mathcal{Q}=\pi(\partial B)\cup\\{0\\}.$

To avoid these splitting points, we say that an interval $I=(a,b)\subset[0,1)$
is $k$-stable if xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$g^{j}(I)\cap Q=\emptyset\quad\mbox{for every $j=0,1,\ldots,k$}.$

It is easy to check that the following hold:

(S1) If $I^{\prime}\subset I^{\prime\prime}$, then $I^{\prime}$ is $k$-stable
if $I^{\prime\prime}$ is $k$-stable.

(S2) If $I$ is $k$-stable, then $I$ is $j$-stable for every $j=0,1,\ldots,k$.

(S3) If $k\geqslant 2$ and $I$ is $k$-stable, then $g(I)$ is $(k-1)$-stable.

(S4) If $I=(a,b)$ is not $k$-stable, then there exists a finite partition

$a=x_{0}<x_{1}<\ldots<x_{n}=b$

of $I$ such that $(x_{\ell-1},x_{\ell})$ is $k$-stable for every
$\ell=1,\ldots,n$.

(S5) If $I$ is $1$-stable, then there exists a mapping
$\sigma_{I}:\mathscr{M}\to\mathscr{M}$ such that

$f(I+i)=g(I)+\sigma_{I}(i),\quad i\in\mathscr{M}.$ (6.18)

To see this, note that if $I$ is $1$-stable, then it follows from (6.6) that

$\pi(f(I+i))=g(\pi(I+i))=g(\pi(I))=g(I)$

does not split, and so it follows that
$f(I+i)\subset[i^{\prime},i^{\prime}+1)$ for some $i^{\prime}\in\mathscr{M}$.
We now simply take $i^{\prime}=\sigma_{I}(i)$.

The assumption that $I$ is $k$-stable leads to a sequence

$I\longrightarrow g(I)\longrightarrow
g^{2}(I)\longrightarrow\ldots\longrightarrow g^{k}(I)$ (6.19)

of splitting-free intervals. In view of (S2), (S3) and (S5), this gives rise
to a sequence

$\mathscr{M}\mathop{\longrightarrow}^{\sigma_{I}}\mathscr{M}\mathop{\makebox[24.18501pt]{\rightarrowfill}}^{\sigma_{g(I)}}\mathscr{M}\longrightarrow\ldots\longrightarrow\mathscr{M}\mathop{\makebox[36.98866pt]{\rightarrowfill}}^{\sigma_{g^{k-1}(I)}}\mathscr{M}$

of mappings xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\sigma_{g^{j}(I)}:\mathscr{M}\to\mathscr{M},\quad j=0,1,\ldots,k-1.$

Denote their composition mapping $\sigma_{k,I}:\mathscr{M}\to\mathscr{M}$ by

$\sigma_{k,I}=\sigma_{g^{k-1}(I)}\circ\ldots\circ\sigma_{g(I)}\circ\sigma_{I}.$

Then corresponding to (S5) and (6.18), we have

$f^{k}(I+i)=g^{k}(I)+\sigma_{k,I}(i),\quad i\in\mathscr{M}.$ (6.20)

######  Remark.

Note that (6.18) and (6.20) represent not only equality of sets but crucially
also mappings of intervals.

Next, to avoid self intersection of the sets in the sequence (6.19), we say
that an interval $I=(a,b)\subset[0,1)$ is $k$-clear if $I$ is $k$-stable and
the sets

$I,g(I),g^{2}(I),\ldots,g^{k}(I)$

are pairwise disjoint. It is easy to check that the following hold:

(C1) If $I$ is $k$-stable, then there exists a $k$-clear subinterval
$I^{\prime}\subset I$. To see this, take $x\in I$ to be the midpoint of $I$,
let

$r=\min_{\begin{subarray}{c}{j_{1},j_{2}\in\\{0,1,\ldots,k\\}}\\\ {j_{1}\neq
j_{2}}\end{subarray}}|g^{j_{1}}(x)-g^{j_{2}}(x)|,$

and consider the interval $(x-r/2,x+r/2)$. We now simply take a sufficiently
small $r^{\prime}<r/2$ to ensure that
$I^{\prime}=(x-r^{\prime},x+r^{\prime})\subset I$. Here the choice of $x$ is
justified by the Remark above, and the value of $r$ is independent of the
choice of $x\in I$.

(C2) If $I$ is $k$-clear and $i^{\prime}\not\in\sigma_{k,I}(\mathscr{M})$,
then for every $i\in\mathscr{M}$ and integer $j\geqslant 0$,

$f^{j}(I+i)\cap(g^{k}(I)+i^{\prime})=\emptyset.$ (6.21)

To see this, suppose on the contrary that

$x_{1}\in f^{j}(I+i)\cap(g^{k}(I)+i^{\prime}),\quad\mbox{so
that}\quad\pi(x_{1})\in g^{j}(I)\cap g^{k}(I).$

This means that $g^{j}(I)\cap g^{k}(I)\neq\emptyset$, so that we must have
$j\geqslant k$. There exists $x_{2}\in I+i$ such that $x_{1}=f^{j}(x_{2})$.
Then the number $x_{3}=f^{j-k}(x_{2})$ satisfies

$f^{k}(x_{3})=f^{j}(x_{2})=x_{1}\in g^{k}(I)+i^{\prime},$

and it follows from (6.20) that $i^{\prime}\in\sigma_{k,I}(\mathscr{M})$, a
contradiction.

(C3) If $I$ is $k$-clear, $i^{\prime}\in\sigma_{k,I}(\mathscr{M})$ and
$i^{\prime\prime}\not\in\sigma_{k,I}(\mathscr{M})$, then for every integer
$j\geqslant 0$,

$f^{j}(g^{k}(I)+i^{\prime})\cap(g^{k}(I)+i^{\prime\prime})=\emptyset.$

To see this, suppose on the contrary that

$x_{1}\in f^{j}(g^{k}(I)+i^{\prime})\cap(g^{k}(I)+i^{\prime\prime}).$

Then there exists $x_{2}\in g^{k}(I)+i^{\prime}$ such that
$f^{j}(x_{2})=x_{1}$. Meanwhile, it follows from (6.20) that there exists
$i\in\mathscr{M}$ such that

$f^{k}(I+i)=g^{k}(I)+i^{\prime},$

so there exists $x_{3}\in I+i$ such that $f^{k}(x_{3})=x_{2}$. Thus
$x_{1}=f^{j+k}(x_{3})$, so that
$f^{j+k}(I+i)\cap(g^{k}(I)+i^{\prime\prime})\neq\emptyset$, contradicting
(6.21) and (C2).

The proof of Theorem 7 is based on the following critical intermediate result.

######  Lemma 6.4.

If $I=(a,b)\subset[0,1)$ is $k$-clear, then one of the following holds:

_(i)_ The set xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\bigcup_{i\in\sigma_{k,I}(\mathscr{M})}(g^{k}(I)+i)$ (6.22)

generates a recurrent region in a finite number of steps.

_(ii)_ There exists a subinterval $I^{*}\subset I$ and an integer $k^{*}>k$
such that $I^{*}$ is $k^{*}$-clear and xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$|\sigma_{k^{*},I^{*}}(\mathscr{M})|\leqslant|\sigma_{k,I}(\mathscr{M})|-1.$

Here a recurrent region is a set of recurrent points that is invariant under
the dissipative flow.

######  Remark.

Suppose that a recurrent region $\mathcal{R}$ of a dissipative system
$\mathcal{P}$ has non-trivial intersection with every other recurrent region
of $\mathcal{P}$. Then apart possibly from exceptional sets of measure $0$,
$\mathcal{R}$ contains every other recurrent region of $\mathcal{P}$. To see
this, note that if $\mathcal{R}^{\prime}$ is another recurrent region of
$\mathcal{P}$, then so is $\mathcal{R}^{\prime}\setminus\mathcal{R}$. If
$\mathcal{R}^{\prime}\setminus\mathcal{R}$ is non-trivial, then $\mathcal{R}$
intersects non-trivially with $\mathcal{R}^{\prime}\setminus\mathcal{R}$,
clearly a contradiction. Hence apart possibly from exceptional sets of measure
$0$, we must have $\mathcal{R}^{\prime}\subset\mathcal{R}$.

Here we have used the fact that if $\mathcal{R}$ and $\mathcal{R}^{\prime}$
are recurrent regions of $\mathcal{P}$, then so are the region
$\mathcal{R}\cup\mathcal{R}^{\prime}$ and its three pairwise disjoint
constituent parts $\mathcal{R}\cap\mathcal{R}^{\prime}$,
$\mathcal{R}\setminus\mathcal{R}^{\prime}$ and
$\mathcal{R}^{\prime}\setminus\mathcal{R}$. To see this, note that it is
straightforward that $\mathcal{R}\cup\mathcal{R}^{\prime}$ and
$\mathcal{R}\cap\mathcal{R}^{\prime}$ are recurrent regions of $\mathcal{P}$.
Clearly $\mathcal{R}^{\prime}\setminus\mathcal{R}\subset\mathcal{R}^{\prime}$,
so no point of $\mathcal{R}^{\prime}\setminus\mathcal{R}$ can move into
$\mathcal{R}\setminus\mathcal{R}^{\prime}$ under the dissipative flow. To show
that $\mathcal{R}^{\prime}\setminus\mathcal{R}$ is a recurrent region, suppose