PIECEWISE ISOMETRIES, UNIFORM DISTRIBUTION AND πmath.sfsu.edu/cheung/cgq.pdf · PIECEWISE...

PIECEWISE ISOMETRIES, UNIFORM DISTRIBUTION

AND 3 log 2− π2/8

YITWAH CHEUNG, AREK GOETZ AND ANTHONY QUAS

Abstract. We use analytic tools to study a simple family ofpiecewise isometries of the plane parameterized by an angle param-eter. In previous work we showed the existence of large numbersof periodic points, each surrounded by a ‘periodic island’. We alsoproved conservativity of the systems as infinite measure-preservingtransformations. In experiments it is observed that the periodic is-lands fill up a large part of the phase space and it has been askedwhether the periodic islands form a set of full measure. In thispaper we study the periodic islands around an important familyof periodic orbits and and demonstrate that for all angle parame-ters that are irrational multiples of π the islands have asymptoticdensity in the plane of 3 log 2− π2/8 ≈ 0.846.

1. Introduction

We consider a simple family of piecewise isometries studied previ-ously by Goetz and Quas [11]. The maps have a single parameter θand can be conveniently expressed in terms of a complex variable:

Tθ(z) =

e2πiθ(z + 1) if Im (z) ≥ 0;

e2πiθ(z − 1) if Im (z) < 0.

Systematic studies of piecewise isometries began in [10] and con-tinued in many publications by a variety of authors. One reason forstudying them is that as the pieces are simply isometries, any complex-ity that appears in a piecewise isometric dynamical system is there asa result of the discontinuity. Piecewise isometries therefore form a testcase for the study of discontinuous dynamical systems. They also havebeen applied as simple models for behaviour of electronic circuits [6, 7].The family (Tθ) that we study was introduced in [9] and shown in [11]

to have interesting behaviour that can be understood using elementarytechniques (see Figure 1 for an example of a phase portrait). In [11] itwas shown that for all values of θ the map Tθ has an infinite collection ofperiodic orbits that accumulate at infinity. It was also shown that eventhough Lebesgue measure is an infinite invariant measure under Tθ for

1

2 YITWAH CHEUNG, AREK GOETZ AND ANTHONY QUAS

Figure 1. A typical phase portrait for Tθ for an irra-tional θ value: Many different orbits are plotted

each θ, the transformation is in fact conservative (so that a PoincareRecurrence Theorem is satisfied).In this paper we restrict our attention to the case where θ is irrational

as this will allow us to make use of uniform distribution techniques fromergodic theory. The rational case, both for this map and others is alsointeresting. Since the techniques used in the rational case are differentthat those used in the irrational case we plan to make a comprehensivestudy of density in the rational case in a subsequent article. It shouldbe mentioned that the rational case is closely related to polygonal dualbilliards maps on regular polygons (see work of Vivaldi and Shaidenko[15]) as well as to some rational piecewise affine maps studied by Adler,Kitchens and Tresser in [1].The dominant feature in the phase portrait (see Figure 1) is the

large number of discs that one sees. These have a simple explana-tion: Suppose y is a periodic point of period q for Tθ and let r =min|Im T n

θ y| : n ∈ N. Then an open disc of radius r about T jθ y is

mapped rigidly by Tθ to an open disc of radius r about T j+1θ y and

none of these discs intersects the discontinuity. Suppose now that zlies outside the closed disk of radius r about some point T j

θ y on the

periodic orbit. Let R = d(z, T jθ y). We claim that there exists an n

such that T nz and T j+nθ y lie on opposite sides of the discontinuity. To

see this we start by assuming that T jθ y is exactly at a distance r from

the discontinuity. Then assuming that z and T jθ y lie on the same side

of the discontinuity for all n, we have in particular that T qnθ z is at a

PIECEWISE ISOMETRIES, UNIFORM DISTRIBUTION AND 3 log 2− π2/8 3

distance R from T jθ y for each n. Since T q

θ acts as an irrational rotation

by 2πqθ about T jθ y on the domain of continuity containing T j

θ y therewill eventually exist (by denseness of orbits on the circle under irra-tional rotations) an n such that T qn

θ z and T jθ y lie on opposite sides of

the discontinuity.Hence we have shown that for each periodic orbit there is an r such

that the open discs of radius r around each point on the orbit are rigidlypermuted (with rotation) and stay for all time on the same size of thediscontinuity as the periodic point whereas points outside the closeddisc of size r eventually fall on the opposite side of the discontinuity.These discs will be referred to as periodic islands.Given a point z in the plane we refer to its itinerary as the sequence of

elements of the partition (into pieces on which the map acts as an isom-etry) that its orbit belongs to. That is we define I(z) = (s(T n

θ z))n∈Zwhere s(z) is 1 if Im (z) ≥ 0 and −1 otherwise. It is straightforwardto see that a point has a periodic itinerary if and only if it belongs toa periodic island. If a point has an aperiodic itinerary we refer to it asaperiodically coded.A helpful heuristic for understanding the action of Tθ is that given a

point with absolute value r and argument 2πφ its image has approxi-mately the same radius but argument approximately 2π(φ+ θ−C/r).To understand this notice that the map Tθ is the composition of a ro-tation by 2πθ with a ‘shear’ sliding the two parts of the plane relativeto each other. The effect of the shear part is to diminish the argumentwhether the point lies in the upper or lower half plane. The amountthat the argument is reduced depends on the radius, as well as on thedistance of the point from the discontinuity. The second dependencetends to average out and so Tθ can be thought of as approximately atwist map (r, 2πφ) 7→ (r, 2π(φ+ θ−C/r)). If r satisfies θ−C/r = p/qthat one might expect Tθ acting on a neighbourhood of a circle of ra-dius r about the origin to behave as a rotation by 2πp/q and hence tohave periodic orbits with period q. Our previous paper [11] developedthis idea and showed rigorously the existence of orbits arising in thisway.A key concept in that paper is that of a rotationally-coded periodic

orbit. Throughout the paper we will use the notation e(θ) = e2πiθ. Wesay that a point z has a rotationally-coded orbit for Tθ if there exist arational p/q (in lowest terms) and a point y whose Rp/q-orbit does notintersect the discontinuity such that s(T n

θ z) = s(Rnp/qy) for all n ∈ Z,

where Rθ(y) = e(θ)y is the rotation of the plane by 2πθ about the


origin. In this case we say that orbit of z is p/q-coded and the rotationnumber of z will be p/q.In order to avoid repeating ourselves too much we make the fol-

lowing natural convention: when speaking about rationals p/q we willalways assume that they are expressed in lowest terms even if we avoidexplicitly mentioning it.Notice that for the map Rp/q, if the plane is divided up by the preim-

ages of the real axis under Rp/q into conical regions then two pointshave the same itinerary if and only if they lie in the same cone. If q iseven there are q cones whereas if q is odd there are 2q cones (the dif-ference arises because the negative real axis and positive real axis havedisjoint orbits in the odd case). If q is even Rp/q permutes the conestransitively whereas if q is odd the cones alternate between two familiesand the rotation sends each family to itself. For q even there is thereforeexactly one p/q-coded itinerary up to shifts (namely (s(Rn

p/qe(12q)))n∈Z)

whereas for q odd there are exactly two p/q-coded itineraries up toshifts (namely (s(Rn

p/qe(14q)))n∈Z and (s(Rn

p/qe(−14q)))n∈Z). The first of

these spends (q + 1)/2 steps per period in the upper half plane and(q− 1)/2 steps in the lower half plane whereas for the second orbit thesituation is reversed.A key question for the family (Tθ) as well as for other piecewise

isometries has been to rigorously prove quantitative theorems about thesize of the set of aperiodically coded points. In our setting, letting Adenote the set of aperiodically coded points (a measurable set becauseit is the complement of a countable family of discs), one is interestedin the density of A in the plane:

ρ(A) = limR→∞

Leb(A ∩ BR)

Leb(BR).

If the limit in the above expression fails to exist one may instead talkabout the upper density ρ(A) in which the limit is replaced by a limitsuperior and the lower density ρ(A) in which the limit is replaced bya limit inferior.Ashwin has performed computer experiments suggesting that ρ(A) is

positive ([3, 4]) but other authors have conjectured that ρ(A) is 0. Wedo not resolve this conjecture here, but instead we study quantitativeproperties of Prot, the set of points with rotationally-coded orbits. Seealso recent work of Lowenstein and Vivaldi [13] for an approach to arelated problem involving density estimates for a PWI on a compactregion with parameter values θ close to 1/4.


Our main theorem in this paper is the following (see Section 5 for aprecise statement).

Theorem. Let θ be irrational. Then the set Prot has upper density atleast 3 log 2 − π2/8. Moreover there is a naturally defined subset P0 ofProt such that ρ(P0) = 3 log 2− π2/8.

Further we present a conjecture that we have verified numerically ina large number of cases that would ensure that P0 = Prot.The principal ingredients of the paper are some elementary geometric

and combinatorial facts about the periodic orbits established in ourprevious paper [11] together with a number of deductions from uniformdistribution. The result is established by first showing that the densityof periodic islands is given by the sum of the limits as R approaches∞ of Godd(θ, R) +Geven(θ, R), where these are defined by

Godd(θ, R) =1

2R2

∑

q

π2R<qθ< 1

2

gcd(q,⌊qθ⌋)=1q odd

q cot2(πqθ); and

Geven(θ, R) =1

4R2

∑

qθ> q

π2R

gcd(q,⌊qθ⌋)=1q even

q tan2

(

πqθ

2

)

.

These are the approximate total densities in a disc of radius R aboutthe origin of the periodic islands with odd period and the periodicislands with even period. Our previous paper gives exact informationon the radii of the periodic islands. In this paper we use uniformdistribution techniques to approximately locate the periodic islands(so that we are able to decide those terms that should be included andthose that should be excluded). This allows us to compare the densityρ(P0) with limR→∞(Godd(θ, R) +Geven(θ, R)) in Section 5.In computing the limits of the above quantities we use uniform dis-

tribution two further times: firstly (in Section 4) we find the density ofS = q : q and ⌊qθ⌋ are coprime. It turns out S has density 4/π2 inthe even integers and 8/π2 in the odd integers. This is a mild general-ization of a theorem of Estermann [8], which in turn is closely related toa well-known theorem of Mertens [14] on the density of visible pointsin the square lattice. Secondly (in Section 3) we develop some the-orems about summation in the presence of uniform distribution thatessentially allow us to replace all the terms in the summation by their


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%PiecewiseIsometrylemmas

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%Locatingperiodic

points neardiscontinuity

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%Comparison of

densityand sum

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%Evaluation of

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DistributionTheorems

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Lemma 3Lemma 2 Lemma 8 Section 4 Lemma7

Lemma 22 Lemma 20,21 Theorem14

Lemma 23

Figure 2. Schematic diagram of the proof

averages provided that the functions being summed are Riemann inte-grable.Since in our case the functions being summed have singularities we

need to deal with the contributions from the singularities separately(Section 5).A schematic diagram of the proof is shown in Figure 2.

2. Piecewise Isometry Lemmas

We collect in this section a number of lemmas from our previouspaper [11] concerning periodic orbits of maps of the form Tθ. We shalluse the notation e(x) to denote e2πix. If ǫ ∈ ±1Z we write σ(ǫ) forthe shift of ǫ: (σ(ǫ))n = ǫn+1.We will write S±1

θ (z) = e(θ)(z ± 1) and let P+1 = z : Im z ≥ 0and P−1 = z : Im z < 0. If ǫ is a periodic ±1-valued sequence with


period q we set

F (θ, ǫ) =1

e(−qθ)− 1

q−1∑

j=0

ǫje(−jθ).

One can check that Sǫ0θ (F (θ, ǫ)) = F (θ, σ(ǫ)).

Lemma 1 (Criterion for existence of periodic points: Lemma 9 [11]).Let ǫ be a periodic word with period q in ±1Z. Then Tθ has a periodicpoint with itinerary ǫ if and only if F (θ, σjǫ) ∈ Pǫj for each 0 ≤ j < q.If this condition is satisfied then the periodic point is given by F (θ, ǫ).

We refer to F (θ, ǫ) as a potential periodic orbit with itinerary θ. Ifthe conditions of Lemma 1 are satisfied then it is a true periodic orbitwith itinerary θ.As described above for p/q ∈ Q, there are two periodic p/q-coded

itineraries if q is odd and one itinerary if q is even. For q odd and letZq = 1

4, 34, . . . , q − 1

4 whereas if q is even let Zq = 1

2, 32, . . . , q − 1

2.

These elements represent the central angle of each of the 2q conesdiscussed earlier for the map Rp/q in the case where q is odd and thecentral angle of each of the q cones for Rp/q in the case where q is even.

In either case we let(

ǫ(a)p/q

)

n= s(Rn

p/q(e(a/q))) and let z(a)p/q = F (θ, ǫ

(a)p/q),

so that provided the conditions of the lemma above are satisfied z(a)p/q

behaves under Tθ like the a point with angle 2πa/q under the map Rp/q.If gcd(p, q) = 1 and p/q < θ < (p+ 1

4)/q then we showed in [11] that

the conditions of Lemma 1 are satisfied and hence there is a p/q-codedorbit (or a pair of such orbits if q is odd). In the case where there isa pair of itineraries, the orbit corresponding to one is sent to the orbitcorresponding to the other by rotation by π so that they are eitherboth present or both absent.When the conditions of the lemma are satisfied the points of the

orbit satisfy Tθ(z(a)p/q) = z

(a+p mod q)p/q for any a in the set of allowable a’s

listed above. In the case where q is odd one orbit consists of the points

z(n+1/4)p/q : 0 ≤ n < q and the other orbit consists of z

(n−1/4)p/q : 0 ≤ n <

q. We call the first of these the +-orbit and the second the −-orbit.We showed in [11] that if p/q < θ < (p+ 1/4)/q then the p/q-coded

periodic orbit(s) ‘look like’ orbits of the rotation by 2πp/q in the sensethat the arguments of z(a) increase monotonically as a goes from 0 toq along the appropriate sequence and the periodic orbit(s) are mappedin an order-preserving way to itself/themselves by Tθ. The periodicorbits have another property in this case that will be of importance tous.


1

4

3

4

5

47

4

9

4

11

4

13

4

15

4

17

4

19

4

21

4 23

4

25

4

27

4

Figure 3. The 2/7-coded periodic islands for Tθ whereθ = 0.3219 labeled with their indices in Z7.

Let θ and p/q be given. If q is odd and there are a pair of p/q-codedorbits, the orbits are said to be well-behaved if the closest points on the+-orbit to the discontinuity are z(1/4) and z(q/2−1/4). These two pointsare shown in [11] to be the same distance from the discontinuity. Inthis case of course the closest points on the −-orbit to the discontinuityare z(−1/4) and z(q/2+1/4). In the case where q is even the four pointsz(±1/2) and z(q/2±1/2) are all at the same distance from the discontinuity.If these points are the closest points on the orbit to the discontinuitythe orbit is again said to be well-behaved. It is shown in [11] that ifp/q < θ < (p + 1/4)/q then the p/q-coded orbit is well-behaved. Thisis illustrated in Figure 3.The significance of this condition is that it is possible using an anal-

ysis of the codes ǫ(a) one can obtain exact expressions for the distanceof these four points to the discontinuity. By the earlier observation thisdetermines the radius of the periodic islands surrounding the points onthe periodic orbit(s) and hence their area.

Lemma 2 (Size of periodic islands: Lemma 9 [11]). Let gcd(p, q) = 1and θ be irrational.If q is odd then z

(1/4)p/q and z

(q/2−1/4)p/q have imaginary component cot(πqθ)/2;

while z(−1/4)p/q and z

(q/2+1/4)p/q have imaginary component − cot(πqθ)/2.

If q is even then z(q/2−1/2)p/q and z

(1/2)p/q have imaginary component

− tan(πqθ/2)/2; while z(−1/2)p/q and z

(q/2+1/2)p/q have imaginary component

tan(πqθ/2)/2.

In particular if q is odd and qθ ∈ (1/2, 1) then z(1/4)p/q lies in the

wrong half-plane and so there is no p/q-coded orbit. Notice that if qis even then by assumption p is odd so that if p/q < θ < (p + 1)/q


then πqθ/2 ∈ (πp/2, π(p + 1)/2) which ensures that − tan(πqθ/2) ispositive.

Lemma 3 (Relative Positions of Periodic Points: Lemma 7 [11]). Letθ = (p+h)/q with 0 < h < 1 and let (z(a))a∈Zq

be the points constructedabove.If q is even, there is a collection of points y(a) (for a ∈ Zq) lying

on a circle centred at the origin with Arg(y(a)) = 2π aqso that for each

a ∈ Zq, there exists an r satisfying |r| ≤ q/4 such that

(1) z(a) − z(a−1) = e(r hq)(y(a) − y(a−1)).

If q is odd, there is a collection of points y(n) (for a ∈ Zq) lying on acircle centred at the origin with Arg(y(a)) = 2π a

qso that for each a ∈ Zq

there exists an r satisfying |r| ≤ q/2 such that

(2) z(a) − z(a−1/2) = e(r hq)(y(a) − y(a−1/2)).

3. Results on Uniform Distribution

In this section we develop a number of basic results about uniformdistribution modulo 1. While the results that we prove are tailored tothe particular applications that we study in this paper there is a generalmeta-principle at work here: in the presence of uniform distributionmodulo 1 one can replace each term in a sum by its average.A sequence of numbers (xn) is said to be uniformly distributed modulo

1 if for each interval [a, b] ⊂ [0, 1], |n ≤ N : xn ∈ [a, b]|/N → b− aas N → ∞.We shall need to make use of Weyl’s Criterion [16].

Theorem 4. (Weyl’s Criterion) Let (xn) be a sequence of real numbers.Then the following are equivalent:

(1) (xn) is uniformly distributed mod 1;(2) For each Riemann-integrable function f ,

limN→∞

1

N

N∑

n=1

f(xn) =

∫ 1

0

f ;

(3) For each non-zero integer k,

limN→∞

1

N

N∑

n=1

e2πikxn = 0.

Lemma 5 (Uniform distribution and weighted means). Suppose that(xn) is a sequence of numbers in [0, 1) and that S is a subset of N with


density α. Suppose further that (xn)n∈S is uniformly distributed mod1. Let f be a Riemann integrable function. Then

1

R2

∑

n≤R, n∈S

nf(xn) →α

2

∫ 1

0

f as R → ∞.

Proof. Let ǫ > 0. We have for all T exceeding some T0,∣

∣

∣

∣

∣

∑

n≤T, n∈S

f(xn)− αT

∫ 1

0

f

∣

∣

∣

∣

∣

< ǫT.

Summing this for T up to R and using the triangle inequality, we obtain∣

∣

∣

∣

∣

∑

n≤R,n∈S

(R− (n− 1))f(xn)− αR(R + 1)

2

∫ 1

0

f

∣

∣

∣

∣

∣

< T 20 ‖f‖+ǫ

R(R + 1)

2.

Dividing by R2 and taking a limit superior as R → ∞ we see

lim sup

∣

∣

∣

∣

∣

1

R2

∑

n≤R, n∈S

(R + 1− n)f(xn)−α

2

∫ 1

0

f

∣

∣

∣

∣

∣

≤ ǫ.

Since ǫ is arbitrary we obtain

limR→∞

1

R2

∑

n≤R, n∈S

(R + 1− n)f(xn) =α

2

∫ 1

0

f.

By simple algebraic manipulation (using limR→∞(R+ 1)/R = 1) wehave

limR→∞

1

R2

∑

n≤R, n∈S

(R + 1)f(xn) = α

∫ 1

0

f.

Taking the difference of the last two equations gives the required result.

Lemma 6 (Moving weighted averages and uniform distribution). Let0 < β < γ and let (xn) and S be as in the previous lemma. Then

limR→∞

1

R2

∑

βR≤n≤γR, n∈S

nf(xn) =γ2 − β2

2α

∫ 1

0

f.

Proof. From Lemma 5 (with R replaced by γR) we obtain

limR→∞

1

R2

∑

n≤γR, n∈S

nf(xn) =γ2

2α

∫ 1

0

f.

Obtaining a similar expression with β and subtracting gives the result.


Lemma 7 (Weighted averages of slowly varying functions). Let S bea subset of the integers of density α. Let (xn) be such that (xn)n∈Sis uniformly distributed on [0,1) and let f be a Riemann integrablefunction on [0, 1). Then

limR→∞

1

R2

∑

n≤R, n∈S

nf(xn)1[n/R,1)(xn) =α

2

∫ 1

0

x2f(x) dx.

Proof. We note that it is sufficient to prove the lemma in the case wheref is a continuous function, as a Riemann integrable function may beapproximated above and below by continuous functions whose integralsdiffer by an arbitrarily small amount.We assume without loss of generality that f is a non-negative func-

tion. Let ǫ > 0 be given and let M > ‖f‖/ǫ satisfy |f(x) − f(y)| ≤ ǫwhenever |x− y| ≤ 1/M .Let f0, f1, . . . , fM be the sequence of Riemann integrable functions

defined by

fi(x) =

f(x) if x ≥ i/M ;

0 if x < i/M.

Using Lemma 6 we can choose R0 such that for all R ≥ R0, all i andj,

∣

∣

∣

∣

∣

∣

∣

∣

1

R2

∑

(i−1)R/M<n≤iR/Mn∈S

nfj(xn)−(i2 − (i− 1)2)

2M2α

∫ 1

0

fj

∣

∣

∣

∣

∣

∣

∣

∣

≤ ǫ/M.

For n between (i− 1)R/M and iR/M , we have the inequality

nfi(xn) ≤ nf(xn)1[n/R,1)(xn) ≤ nfi−1(xn).

Summing we have

αi2 − (i− 1)2

2M2

∫ 1

0

fi − ǫ/M ≤1

R2

∑

(i−1)R/M<n≤iR/Mn∈S

nf(xn)1[n/R,1)(xn)

≤ αi2 − (i− 1)2

2M2

∫ 1

0

fi−1 + ǫ/M.

Note that∫ 1

0fi =

∫ 1

i/Mf for each i.


Summing this over i shows thatM∑

i=1

αi2 − (i− 1)2

2M2

∫ 1

i/M

f − ǫ ≤1

R2

∑

n≤R, n∈S

nf(xn)1[n/R,1)(xn)

≤M∑

i=1

αi2 − (i− 1)2

2M2

∫ 1

(i−1)/M

f + ǫ.

These lower and upper bounds may be rewritten as

α

2M2

(

12∫ 2/M

1/M

f + 22∫ 3/M

2/M

f + . . .+ (M − 1)2∫ 1

(M−1)/M

f

)

− ǫ; and

α

2M2

(

12∫ 1/M

0

f + 22∫ 2/M

1/M

f + . . .+M2

∫ 1

(M−1)/M

f

)

+ ǫ.

The upper and lower bounds differ by at most α(2M+1)‖f‖/(2M2)+2ǫ < 5ǫ.Since

∫ (i+1)/M

i/Mf differs from f(i/M)/M by at most ǫ/M , the upper

and lower bounds are within constant multiples of ǫ of the quantity

α

2

M∑

i=1

(i/M)2f(i/M)/M.

This is a Riemann sum approximation of (α/2)∫ 1

0x2f(x) dx. As ǫ is

shrunk to 0 and M grows to ∞, the two bounds converge giving therequired limiting value.

Remark. To see that Lemma 7 is an instance of the meta-principlestated at the beginning of the section, applying this principle to thequantity in Lemma 7, we expect the left hand side to be replaced by

limR→∞

(1/R2) dens(S)∑

n≤R

n

∫ 1

n/R

f = α limR→∞

R∑

n=1

(1/R)(n/R)

∫ 1

n/R

f

= α

∫ 1

0

(

x

∫ 1

x

f(t) dt

)

dx = α

∫ 1

0

(

f(t)

∫ t

0

x dx

)

dt

=α

2

∫ 1

0

t2f(t) dt

which is confirmed by Lemma 7.

For the next lemma we regard [0, 1) as the unit circle, so that additionand subtraction are interpreted modulo 1. We will write Ta for thetranslation operator on functions so that Taf(x) = f(x− a).


Lemma 8 (Averages of slowly varying functions). Let f and g be Rie-mann integrable functions on the circle. Let A > 0 and let (xn) beuniformly distributed modulo 1. Then the quantity

1

N

N−1∑

n=0

f(xn + an/N)g(xn)−1

N

N−1∑

n=0

∫ 1

0

f(t+ an/N)g(t) dt

converges to 0 uniformly for a ∈ [−A,A] as N → ∞.

Proof. Clearly it is sufficient to prove the lemma in the case when fand g are non-negative functions. We further claim that it is sufficientto prove the lemma in the case where f and g are continuous functions.To see this we note that if f and g are Riemann integrable then we mayapproximate them arbitrarily closely above and below by continuousfunctions. This then gives arbitrarily close lower and upper bounds for(1/N)

∑

n≤N f(xn + an/N)g(xn) completing the proof of the lemma.We make one further simplifying assumption for the proof, namely

that xn = nθ mod 1 for a fixed irrational θ. This is not essential for theproof, but provides a simplification at one point. Further the lemmawill only be applied in this case in the sequel.Now let ǫ > 0. There exists δ > 0 such that if |c − d| ≤ δ then

‖Tcf − Tdf‖ < ǫ/(4‖g‖). Let M = ⌈A/δ⌉ and for −M ≤ j ≤ Mlet hj(x) = (Tjδf)(x)g(x). By unique ergodicity of the rotation by θapplied to each of the continuous functions hj, there exists a K > 0such that for any N ≥ K, any x ∈ [0, 1) and any |j| ≤ M , we have

|(1/N)∑

n≤N hj(x+ nθ)−∫ 1

0hj | < ǫ/4.

Now let N be such that ⌊Nδ⌋ > K and K‖f‖‖g‖/N < ǫ/4. Assumewithout loss of generality that a > 0 and let ni = min(⌊Niδ/a⌋, N).Then for nj−1 ≤ n < nj we have |an/N − jδ| < δ. Let s = ⌈a/δ⌉. Wethen have

1

N

∣

∣

∣

∣

∣

N−1∑

n=0

f(nθ + an/N)g(nθ)−N∑

n=1

∫ 1

0

f(t + an/N)g(t) dt

∣

∣

∣

∣

∣

≤1

N

s∑

j=1

∣

∣

∣

∣

∣

∣

nj−1∑

n=nj−1

(

f(nθ + an/N)g(nθ)−

∫ 1

0

f(t+ an/N)g(t) dt

)

∣

∣

∣

∣

∣

∣

≤ǫ

2+

1

N

s∑

j=1

∣

∣

∣

∣

∣

∣

nj−1∑

n=nj−1

(

hj(nθ)−

∫ 1

0

hj(t) dt

)

∣

∣

∣

∣

∣

∣

≤ǫ

2+

Nǫ/4 +K‖f‖‖g‖

N≤ ǫ,


where the K‖f‖‖g‖ is an upper bound for the error that may arise ifthe final block is of length less than K.

4. Visible Points

We will in later sections be summing over rotationally coded orbits,that is over pairs (p, q) of coprime integers. More specifically given anirrational slope θ we need some analytic information about the densityof the points (p, q) with coprime coordinates in the strip (m,n) : θm−1 < n ≤ θm. We collect in this section the results that we shall need.It is well-known that the set of visible points in the plane (those

points in Z2 satisfying gcd(p, q) = 1) is of density 6/π2. That is, in adisc of size R about the origin the ratio of the number of visible pointsto the area of the disc tends to 6/π2 as R approaches ∞. Indeed thisresult was originally proved by Mertens [14] in the 19th century.A generalization of this result was obtained by Estermann [8] using

techniques of analytic number theory where the density 6/π2 of visi-ble points was shown to hold for any infinite strip in the plane withirrational slope. More precisely

Theorem 9 (Estermann). Let θ be irrational and let a < b. Then wehave(3)

limN→∞

1

N#(q, r) : gcd(q, r) = 1; q ≤ N, θq−b ≤ r < θq−a =

6(b− a)

π2

We make use of the following convention: We identify the circle with[0, 1) and if g is a function defined on the unit circle when we write g(x),we mean g(x mod 1). A corollary of Estermann’s theorem, convenientfor our purposes, is the following.

Theorem 10. Let θ be irrational and let g be a Riemann integrablefunction on the unit circle. Then we have

(4) limN→∞

1

N

∑

q≤Ngcd(q,⌊qθ⌋)=1

g(qθ) =6

π2

∫ 1

0

g.

We shall give a proof of this theorem using uniform distribution andsupplement it with an extension that we need in the remainder of thepaper. We initially consider the strip Sθ = (x, y) : θx−1 < y ≤ θx ofheight 1 and slope θ. For an irrational number θ, define fθ : N → 0, 1by

fθ(q) =

1 if gcd(q, ⌊qθ⌋) = 1;

0 otherwise.


This function records whether or not Sθ contains a primitive latticepoint with x-coordinate q.We remark that Theorem 10 can be informally described as saying

that the value of the fractional part of qθ is independent of whether qand ⌊qθ⌋ are coprime.Our approach to proving Theorem 10 will be to split fθ up into a

regular part and a small part. The two lemmas that follow then controlthe average values of these two parts. For r ≥ 1, define the regular part,

f(r)θ as follows:

f(r)θ (q) =

0 if there exists p < 2r such that p|q and p|⌊qθ⌋;

1 otherwise.

Notice that f(r)θ dominates fθ and the sequence (f

(r)θ )r decreases

pointwise to fθ.

Lemma 11. Let θ be irrational and let k ∈ Z. Then

limQ→∞

1

Q

Q∑

q=1

f(r)θ (q)e(kqθ) =

∏

p<2r

(

1− 1p2

)

if k = 0

0 otherwise.

(In the above formula the product is taken over primes p less than 2r.)

Lemma 12. Let θ be irrational. Then for all ǫ > 0 there exists an r0such that for all r ≥ r0 we have

(5) lim supQ→∞

1

Q

Q∑

q=1

(f(r)θ (q)− fθ(q)) < ǫ.

We now give a proof of Theorem 10 assuming the lemmas.

Proof of Theorem 10. Fix an irrational θ and let ǫ > 0. There exists anr0 satisfying the conditions of Lemma 12. Since

∏

p prime(1 − 1/p2) =

6/π2 (see Hardy and Wright [12] for a proof that∏

p prime(1−1/p2)−1 =

ζ(2) and Apostol [2] for a proof that ζ(2) = π2/6), there exists r ≥ r0such that

∏

p<2r(1− 1/p2) < 6/π2 + ǫ.By Weyl’s criterion it is sufficient to show that for any k ∈ Z,

limQ→∞

1

Q

Q∑

q=1

fθ(q)e(kqθ) =

6/π2 if k = 0;

0 otherwise.


Let k ∈ Z be fixed and let A be 6/π2 if k = 0 and 0 otherwise. Wehave

lim supQ→∞

∣

∣

∣

∣

∣

1

Q

Q∑

q=1

fθ(q)e(kqθ)−1

Q

Q∑

q=1

f (r)(q)e(kqθ)

∣

∣

∣

∣

∣

≤ lim supQ→∞

1

Q

Q∑

q=1

(f(r)θ (q)− fθ(q)) < ǫ,

(6)

where the last inequality follows from Lemma 12.By Lemma 11 we have

(7)

∣

∣

∣

∣

∣

limQ→∞

1

Q

Q∑

q=1

f(r)θ (q)e(kqθ)−A

∣

∣

∣

∣

∣

< ǫ.

Combining (6) and (7) we get

lim supQ→∞

∣

∣

∣

∣

∣

1

Q

Q∑

q=1

fθ(q)e(kqθ)− A

∣

∣

∣

∣

∣

< 2ǫ.

Since ǫ is arbitrary the proof is complete.

Proof of Lemma 11. List the primes less than 2r as p1, p2, . . . , pk andlet P = p1p2 · · ·pk.We then define a dynamical system on 0, 1, . . . , P − 1 × [0, P ) by

T (n, x) = (n+ 1 mod P, x+ θ mod P ). Define the function g : X → Cby

g(n, x) =

0 if there exists p < 2r such that p|n and p|⌊x⌋;

e(kx) otherwise.

We then claim that f(r)θ (q)e(kqθ) can be dynamically defined, namely

that f(r)θ (q)e(kqθ) = g(T q(0, 0)). To see this, note that the following

statements are pairwise equivalent

• f(r)θ (q) = 0.

• There exists p < 2r such that p|q and p|⌊qθ⌋.• There exists p < 2r such that p|(q mod P ) and p|⌊qθ mod P ⌋.• g(q mod P, qθ mod P ) = 0.• g(T q(0, 0)) = 0.

If the functions do not take the value 0, then they both take thevalue e(kqθ).


1

Q

Q∑

q=1

f(r)θ (q) =

1

Q

Q∑

q=1

g(T q(0, 0)).

By unique ergodicity the right hand side converges to∫

g dµ. Ifk 6= 0, the integral is clearly 0. To evaluate this integral when k = 0,we use the Chinese Remainder Theorem which states that there is aring isomorphism between ZP and Zp1 × . . .Zpk given by n → (n modp1, . . . , n mod pk). Under this isomorphism the elements divisible bypi correspond to those elements of the right side with a 0 in the Zpi

coordinate.Since [0, P ) is in measure-preserving bijection with 0, . . . , P − 1×

[0, 1) (by t 7→ (⌊t⌋, t)), we see that 0, 1, . . . , P − 1 × [0, P ) is inmeasure-preserving bijection with

∏

i≤k Z2pi×[0, 1). Under this bijection

the support of g is mapped to∏

i≤k(Z2pi\ (0, 0))× [0, 1). Since the

measure of this set is∏k

i=1(1− 1/p2i ) the lemma is proven.

Lemma 13 (Counting visible points in parallelograms). Consider aparallelogram of the form S = (x, y) : a ≤ x < 2a, θx − c < y ≤ θx.Then S contains at most ⌈4Area(S)⌉ primitive lattice points.

Proof. We consider the slopes of the lines joining the origin to the prim-itive lattice points in S. These are all distinct and have denominatorsbounded above by 2a. From the simple inequality |p/q−p′/q′| ≥ 1/(qq′)when p/q and p′/q′ are distinct, we see that each pair of slopes differsby at least 1/(4a2). Since the slopes all lie between θ − c/a and θ, thenumber of distinct slopes (and hence the number of primitive latticepoints) is bounded above by ⌈(c/a)/(1/4a2)⌉ = ⌈4ac⌉.

Proof of Lemma 12. We start the proof by making two simplifications.Firstly given ǫ > 0 it is sufficient to show the existence of a single

r > 0 such that lim supQ→∞(1/Q)∑

q≤Q(f(r)θ (q)− fθ(q)) < ǫ since the

functions f(r)θ decrease pointwise to fθ.

Secondly we observe that in order to prove the lemma it is sufficientto ensure that for any ǫ, there is an r > 0 such that the followingcondition holds for all sufficiently large t

(8)1

2t

2t+1−1∑

q=2t

(f(r)θ (q)− fθ(q)) < ǫ.

To see the sufficiency, suppose that for a fixed value of r, (8) issatisfied for all t ≥ t0 with ǫ replaced by ǫ/4. Given Q, let s satisfy


2s ≤ Q < 2s+1. We then have

∑

q≤Q

(f(r)θ (q)− fθ(q)) ≤

∑

t≤s

2t+1−1∑

q=2t

(f(r)θ (q)− fθ(q))

≤ 2t0 +∑

t≤s

(ǫ/4)2t

≤ 2t0 + ǫ2s−1.

Since Q ≥ 2s we see

1

Q

∑

q≤Q

(f(r)θ (q)− fθ(q)) ≤ 2t0/Q+ ǫ/2

So that the desired conclusion (5) follows.Since

∑

p prime 1/p = ∞, we can pick an r such that∏

p<2r(1−1/p) <

ǫ. By increasing r if necessary we may assume that 2−r < ǫ. Fix a tsuch that 2t/2 >

∏

p<2r p.We let Ar denote the set of integers greater than 1 that can be formed

as products of primes greater than 2r. Notice that for n > 1, n ∈ Ar

if and only if n mod p 6= 0 for each prime p < 2r. Using the ChineseRemainder Theorem, we see that in each block of integers of lengthP =

∏

p<2r p, Ar has exactly

∏

p<2r(p − 1) elements so that Ar has

density∏

p<2r(1 − 1/p) in any such block. Now for s ≥ t/2, have 2s+1

exceeds P so that we have

(9) |Ar ∩ [0, 2s+1)| < 2s+2ǫ.

We note that f(r)θ (q) − fθ(q) takes the value 1 if and only if there

exists p > 2r such that p divides both q and ⌊qθ⌋ but no p < 2r withthis property. This is the case if and only if gcd(q, ⌊qθ⌋) ∈ Ar. In

other words we have f(r)θ (q) − fθ(q) takes the value 1 if and only if

(q, ⌊qθ⌋) ∈ B =⋃

n∈Ar nV where V is the set of primitive lattice pointsin Z2.Let S denote the strip in R2 given by

S = (x, y) : 2t ≤ x < 2t+1, θx− 1 < y ≤ θx.

By the above, we have

(10)1

2t

2t+1−1∑

q=2t

(f(r)θ (q)− fθ(q)) = |B ∩ S|/2t.

In order to prove the lemma we need an upper bound on |B ∩ S|.If n ∈ Ar, (u, v) ∈ V and n(u, v) ∈ S, then we see that 2t/n ≤

u < 2t+1/n and θu − 1/n < v ≤ θu. If 2t−k ≤ u < 2t−k+1 then


S

Sk

Sk+1

Figure 4. The parallelograms Sj

2t/n ≤ u < 2t−k+1 so that 1/n < 2−k+1. The point (u, v) then satisfiesthe inequalities

2t−k ≤ u < 2t−k+1

θu− 2−k+1 < v ≤ θu.

We define smaller parallelograms as follows:

Sk = (x, y) : 2t−k ≤ x < 2t−k+1, θx− 1/2k−1 < y ≤ θx.

If n · (u, v) ∈ B ∩ S for n ∈ Ar, then since n > 2r, we have 1 ≤ u <2t+1−r. The above shows that (u, v) ∈

⋃tk=r Sk ∩ V .

We have

B ∩ S ⊂

t⋃

k=r

(Ar ∩ [2k−1, 2k+1)) · (Sk ∩ V ).

so that in particular

(11) |B ∩ S| ≤t∑

k=r

|Ar ∩ [2k−1, 2k+1)| · |Sk ∩ V |.

For k ≤ t/2 we use the trivial bound |Ar∩[2k−1, 2k+1)| ≤ 2k+1. UsingLemma 13, we have |Sk ∩ V | ≤ 1 + 4(2t+1−2k) < 2t−2k+4 so that we see

(12)

⌊t/2⌋∑

k=r

|Ar ∩ [2k−1, 2k+1)| · |Sk ∩ V | ≤

⌊t/2⌋∑

k=r

2t−k+5 ≤ 2t−r+6.


Notice that the rays joining the origin to the primitive lattice pointsin⋃

t/2<k≤t Sk have distinct slopes between θ − 2−(t−1) and θ and de-

nominators not exceeding 2t/2+1. As in the proof of Lemma 13, thedifference between the slopes of any two primitive points in the regiondiffers by at least 2−(t+2). It follows that there can be at most 9 pointsin⋃

k>t/2 Sk ∩ V . Given a point in Sk ∩ V , it gives rise to at most

|Ar ∩ [2k−1, 2k+1)| points of B ∩ S. Since k > t/2, this is at most 2k+2ǫby (9). Since there are at most 9 points it follows that

t∑

k=⌊t/2⌋+1

|Ar ∩ [2k−1, 2k+1)| · |Sk ∩ V | ≤ 36ǫ2t.

Combining this with (12) we see that |B ∩ S| ≤ 100ǫ2t. Since ǫ wasarbitrary, this is sufficient to complete the proof by (10).

The following is an extension of Theorem 10 that we shall need laterin the paper.

Theorem 14 (Estermann along an arithmetic progression). Let θ beirrational and let 0 ≤ a < d be integers. Let g be a Riemann integrablefunction on the unit circle. We then have

limN→∞

1

N

∑

q≤Ngcd(q,⌊qθ⌋)=1q≡a mod d

g(qθ) =1

d

∏

p|gcd(a,d)

(

1−1

p

)

∏

p∤d

(

1−1

p2

)∫ 1

0

g.

In fact all we shall need from this extension is the following corollary.

Corollary 15 (Estermann along the even and odd subsequences). Letθ be irrational and let g be a Riemann integrable function on the unitcircle. We have

limN→∞

1

N

∑


q odd

g(qθ) =4

π2

∫ 1

0

g.

and

limN→∞

1

N

∑


q even

g(qθ) =2

π2

∫ 1

0

g.

Remark 16. In other words the corollary states that the set of q’s suchthat q is odd and gcd(q, ⌊qθ⌋) = 1 has density 4/π2 and the fractionalparts of the qθ along this sequence are uniformly distributed. Similarly


the set of q’s such that q is even and gcd(q, ⌊qθ⌋) = 1 has density 2/π2

and the fractional parts of the qθ are uniformly distributed along thissequence.

The proof of Theorem 14 is entirely analogous to the proof of The-orem 10 except that Lemma 11 has to be replaced by the followinglemma.

Lemma 17. Let θ be irrational and let k ∈ Z. Let d > 0 be given andlet 0 ≤ a < d. Further let r be such that all prime factors of d are lessthan 2r. Then we have the following:

limQ→∞

1

Q

Q∑

q=1q≡a mod d

f(r)θ (q) =

1

d

∏

p<2r

p∤d

(

1−1

p2

)

∏

p|gcd(a,d)

(

1−1

p

)

.

For k 6= 0 we have

limQ→∞

1

Q

Q∑

q=1q≡a mod d

f(r)θ (q)e(kqθ) = 0.

The proof of this lemma is essentially the same as the proof of Lemma11. The point is to show that the quantity being summed is again equalto a dynamical sequence given by evaluating a Riemann integrablefunction along the orbit of a dense rotation. Defining R = d ·

∏

p∤d;p<2r p

and P =∏

p<2r p, the dynamics are defined on the group(

0, 1, . . . , R−

1 mod R)

×(

[0, P ) mod P)

.We remark that the right hand side is very intuitive: we include

only every dth term, accounting for the factor of 1/d. Given thatq ≡ a mod d, if p divides a and d then there is a 1 − 1/p probabilitythat q and ⌊qθ⌋ do not have p as a common factor whereas if p divides dbut not a then the probability that q and ⌊qθ⌋ always have no commonfactor of p. If p does not divide d then there is a 1 − 1/p2 probabilitythat q and ⌊qθ⌋ have a common factor of p.

5. Application to Piecewise Isometries

For θ > 0 fixed we shall study p/q-coded orbits with p/q < θ <(p+ 1)/q.According to the following conjecture we believe that this exhausts

the full set of rotationally-coded orbits but we have not yet been ableto give a complete proof of this.


Conjecture 18 (Exact Ranges of Parameters for existence of each ro-tationally-coded periodic point). Let θ be irrational and let gcd(p, q) =1.If q is odd then Tθ has a pair of p/q-coded orbits if and only if p/q <

θ < (p+ 12)/q.

If q is even then Tθ has a p/q-coded orbit if and only if p/q < θ <(p+ 1)/q.In the case where these orbits exist, they are well-behaved.

Theorem 19 (Main Theorem: Explicit density of nice rotationallycoded points). Let θ be irrational and let R be the set of points inC with rotationally-coded orbits with rotation numbers p/q satisfyingp/q < θ < (p+ 1)/q. Then ρ(R) = 3 log 2− π2/8.

Remark. Notice that if Conjecture 18 holds then the conclusion ofTheorem 19 applies to all rotationally-coded orbits; not just those sat-isfying the condition p/q < θ < (p+ 1)/q.

Recall from before that for a periodic sequence ǫ ∈ ±1Z, F (ǫ, θ)is the location of a potential periodic orbit with itinerary ǫ. Now letθ be a fixed irrational. The following lemma will be used to estimateF (θ, ǫ) for various rotationally-coded itineraries ǫ.

We define the periodic itinerary ǫ(a)p/q by

(

ǫ(a)p/q

)

j= s(e((a + jp)/q))

(where the variable a will run over the set Zq = 1/4, . . . , q − 1/4if q is odd or Zq = 1/2, . . . , q − 1/2 if q is even). We then define

z(a)p/q(θ) = F (θ, ǫ

(a)p/q) so that

z(a)p/q(θ) =

1

e(−qθ)− 1

q−1∑

j=0

(

ǫ(a)p/q

)

je(−jθ).

Lemma 20 (Control of location of periodic points away from the dis-continuity (q odd)). Let θ be a fixed irrational. Then for q satisfyingqθ < 1/2, let p = ⌊qθ⌋. Provided that gcd(p, q) = 1 we have

z(a)p/q =

qe(a/q)

π2qθ(1 + o(1)),

as q → ∞ where the convergence to 0 is uniform in a running over Zq.

Proof. Let θ be as in the statement of the lemma. Now given a q suchthat gcd(⌊qθ⌋, q) = 1, let p = ⌊qθ⌋ and let h = qθ − p. Assume as in


the statement of the lemma that h < 1/2. Let a ∈ Zq. We then have

z(a)p/q = F (θ, ǫ

(a)p/q)

=1

e(−qθ)− 1

q−1∑

j=0

(

ǫ(a)p/q

)

je(−jθ)

=1

e(−h)− 1

q−1∑

j=0

s(e((a + jp)/q))e(−jθ)

=1

e(−h)− 1

q−1∑

j=0

s(e(aq+ jθ − j h

q))e(−jθ).

Now we have from Lemma 8 and writing g(t) for s(e(t)):

q−1∑

j=0

s(e(aq+ jθ − j h

q))e(−jθ)

=

q−1∑

j=0

g(aq+ jθ − j h

q)e(−jθ)

=

q−1∑

j=0

∫ 1

0

g(aq+ t− j h

q)e(−t) dt+ o(q)

=

q−1∑

j=0

∫ 1

0

g(t)e(−t+ aq− j h

q) dt+ o(q)

= e(aq)

(

q−1∑

j=0

e(−j hq)

)

∫ 1

0

g(t)e(−t) dt+ o(q)

= e(aq)e(−h)− 1

e(−hq)− 1

·−2i

π+ o(q).

Hence using the fact that h ≤ 1/2 we obtain

z(a)p/q =

−2ie(aq)

π(e(−hq)− 1)

+ o(q

h)

=qe(a

q)

π2h+ o( q

h) =

qe(aq)

π2h(1 + o(1)).


The next lemma has a similar proof. It gives information on thelocation of p/q-coded periodic orbits when q is even and (p+(1−δ))/q <θ < (p+1)/q, this being the remaining case not covered by Lemma 20.

Lemma 21 (Control of location of periodic points far from the discon-tinuity (q even)). Let θ be a fixed irrational. Then for even q satisfying0 < qθ < 1, let p = ⌊qθ⌋. Provided that gcd(p, q) = 1 we have

z(a)p/q =

qe(a/q)

π2qθ(1 + o(1)),

where the convergence to 0 is uniform in a running over Zq.

The difference between the proof of this lemma and the previous oneis that it is necessary to regroup the summation prior to applying theuniform distribution results as there can be substantial cancellation inthe summation which could otherwise result in a situation where theerrors dominate the quantity being bounded.

Proof. The case 0 < qθ ≤ 1/2 is covered by the previous lemmaso we assume that 1/2 < qθ < 1. Since the proof is very similarto the proof of the previous lemma, we abbreviate, emphasizing thefew significant differences. Let θ, p and q be as in the statement ofthe lemma. As before, let g(t) = s(e(t)). Let h = qθ − p so that1/2 < h < 1.We have

F (θ, ǫ(a)p/q) =

1

e(−qθ)− 1

q−1∑

j=0

g((a+ jp)/q)e(−jθ)

=1

e(−qθ)− 1(1− e(− q

2θ))

q

2−1∑

j=0

g(aq+ j p

q)e(−jθ)

=−1

1 + e(− q2θ)

q

2−1∑

j=0

g(a+ j pq)e(−jθ),


where for the second equality we used the fact that g(a+ (j + q2)pq) =

−g(a+ j pq). As before we have

q

2−1∑

j=0

g(aq+ j p

q)e(−jθ) =

q

2−1∑

j=0

g(aq+ jθ − j h

q)e(−jθ) + o(q)

= e(aq)(−2i/π)

q

2−1∑

j=0

e(−j hq) + o(q)

= e(aq)(−2i/π)

1 + e(− q2θ)

1− e(−hq)+ o(q).

Combining the two and using the fact that 1 + e(− q2θ) = Ω(1) we

have

F (θ, ǫ(a)p/q) =

e(aq)q

π2qθ(1 + o(1))

as required.

The following lemma states that provided qθ is bounded away fromthe endpoints of the range in which there are ⌊qθ⌋/q-coded orbits then

the periodic orbit is ‘locally well-behaved’ in the sense that the z(a)p/q lie

above z(1/4)p/q or z

(1/2)p/q for small values of a with similar results near the

other periodic points just above and below the discontinuity.

Lemma 22 (Control of locations of periodic points near the disconti-nuity). Let ǫ > 0 and let θ be irrational. Let p = ⌊qθ⌋, let h = qθ − p

and suppose that gcd(p, q) = 1. If q is odd let b = Im z(1/4)p/q otherwise

let b = Im z(1/2)p/q .

If q is odd and h < 12−ǫ or if q is even and h < 1−ǫ then Im (z

(a)p/q) ≥

b for a ∈ Zq ∩ ((0, ǫq/2) ∪ (q/2 − ǫq/2, q/2). Similarly Im (z(a)p/q) ≤ −b

for a ∈ Zq ∩ ((q/2, q/2 + ǫq/2) ∪ (q − ǫq/2, q)).

Proof. We give the proof only for the odd case in the first quadrant, theeven case and the other quadrants being precisely analogous. If q is odd

then we have by Lemma 3, arg(z(a+1/4)p/q −z

(a−1/4)p/q ) = 2π(a/q+1/4+rh/q)

with |r| ≤ q/2. It follows that | arg(z(a+1/4)p/q −z

(a−1/4)p/q )−π/2| ≤ π/2(1−

2ǫ + 4a/q). Provided a < ǫq/2 the difference is less than π/2 which

ensures that z(a+1/4)p/q − z

(a−1/4)p/q has positive imaginary part. The result

follows.


For θ irrational, let Podd(θ) denote the set of z such that z is rota-tionally coded with rotation number some p/q (in lowest terms) with qodd and p/q < θ < (p+ 1)/q. In fact by the remark following Lemma2 this can only happen if p/q < θ < (p+ 1

2)/q. Also let Peven(θ) denote

the set of z such that z has rotation number some p/q with q even andp/q < θ < (p+ 1)/q.Define

Godd(θ, R) =1

2R2

∑

q

π2R<qθ< 1

2


q cot2(πqθ); and

Geven(θ, R) =1

4R2

∑

qθ> q

π2R

gcd(q,⌊qθ⌋)=1q even

q tan2

(

πqθ

2

)

Similarly let Hodd(θ, R) = |DR ∩ Podd(θ)|/|DR| and Heven(θ, R) =|DR ∩ Peven(θ)|/|DR|.

Lemma 23 (Comparison of density with averages). Let θ be irrational.Then Godd(θ, R) and Hodd(θ, R) have identical limits superior and in-ferior as R → ∞. Similarly for the even versions.

Proof. We deal with the odd case first. Let an irrational θ be fixed andsuppose ǫ > 0 is given. We may suppose that ǫ is small enough so thattan(πǫ) < 4ǫ. By Lemma 20 let q0 satisfy the following conditions:

• 1/q0 < ǫ/(2π);• If q ≥ q0, let p = ⌊qθ⌋. If qθ < 1/2 and gcd(p, q) = 1 one has

|z(a)p/q − qe(a/q)/(π2qθ)| < ǫq/(4π2qθ).

Let R be so large that DR contains all periodic islands surroundingperiodic points with rotation number whose denominator is less thanq0.We first give an upper bound for |DR∩Podd(θ)|. Suppose the intersec-

tion meets a periodic island of a periodic point z(a)p/q with q odd. If q ≥ q0

then the periodic point is at a distance at least (1−ǫ/4)q/(π2qθ) fromthe origin. The periodic island has size cot(πqθ)/2 < 1/(2πqθ) <(ǫ/2)q/(π2qθ). In particular the entire periodic island is at a dis-tance at least (1− ǫ)q/(π2qθ) from the origin. If the island intersectsDR we must have (1−ǫ)q/(π2qθ) < R so that (1−ǫ)q/(π2R) < qθ.In order to exist we must have qθ < 1/2.


This establishes the inequality

|DR ∩ Podd(θ)| ≤∑

(1−ǫ)q/(π2R)<qθ< 1

2

q oddgcd(q,⌊qθ⌋)=1

(2q)π cot2(πqθ)/4.

We then see that

(13) Hodd(θ, R) ≤ (1− ǫ)2Godd(θ, R/(1− ǫ)).

For a comparison in the opposite direction we will make a comparisonbetween terms appearing in Godd(θ, R) and areas of periodic islands inD(1+ǫ)R. Suppose q ≥ q0 is odd and let p = ⌊qθ⌋. Suppose furtherthat gcd(p, q) = 1 and that q/(π2R) < qθ < 1

2for some R. We will

separate the cases where 12− ǫ ≤ qθ < 1

2and those where q/(π2R) <

qθ < 12− ǫ. In the latter case we will show that there is a p/q-coded

pair of orbits; that they are well-behaved and that they lie completelyinside D(1+ǫ)R so that the p/q-coded periodic islands in D(1+ǫ)R have acombined area of 2q(π cot2(πqθ)/4). For the former case we shall showthat the terms do not make a substantial contribution to Godd.We deal first with the case where q/(π2R) < qθ < 1

2− ǫ. If q < q0

then the disc was chosen to contain the periodic islands so we onlyneed to cover the case q ≥ q0. Using Lemma 20 as before the potential

periodic points with codes ǫ(a)p/q lie inside DR(1+ǫ/4) and the potential

islands surrounding these points are completely contained in DR(1+ǫ).We next show that the periodic orbit is well-behaved. To see this

we will verify that Im z(a)p/q ≥ Im z

(1/4)p/q for a ∈ Zq ∩ (0, q/4), the

other quadrants being similar. Lemma 22 establishes this for 1/4 <a < ǫq/8. If ǫq/8 < a < q/4 then we use Lemma 20 to say that

|z(a)p/q − qe(a

q)/(π2qθ)| < (ǫ/(4π2))q/qθ. We then have Im z

(a)p/q >

q/(π2qθ)(sin(2πǫ/8) − ǫ/4) > ǫq/(4π2qθ). Since ǫq > 2π we see

Im z(a)p/q > 1/(2πqθ) > cot(πqθ)/2 = Im z

(1/4)p/q as required.

The p/q-coded periodic islands therefore lie completely insideDR(1+ǫ)

and have combined area 2qπ cot2(πqθ)/4. Since the periodic islands aredisjoint we see that

∑

q

π2R<qθ< 1

2

gcd(q,⌊qθ⌋)=1q odd, q≥q0

qπcot2(πqθ)

2≤ |D(1+ǫ)R ∩ Podd(θ)|.


This gives us the inequality

(14)1

2R2

∑

q

π2R<qθ< 1

2−ǫ

gcd(q,⌊qθ⌋)=1q odd, q≥q0

q cot2(πqθ) <(1 + ǫ)2|D(1+ǫ)R ∩ Podd(θ)|

D(1+ǫ)R

.

We now consider the terms in Godd(θ, R) where 12− ǫ < qθ < 1

2.

For these terms cot(πqθ) < tan(πǫ) < 4ǫ. In order for the inequalityq/(π2R) < qθ < 1

2to be satisfied we must have q < π2R/2 < 5R.

The sum over those terms in Godd(θ, R) for which 12−ǫ < qθ < 1

2may

therefore be trivially bounded above by (1/(2R2))∑

q<5R 16ǫ2q < 100ǫ2

giving

(15)1

2R2

∑

q

π2R<qθ< 1

2

gcd(q,⌊qθ⌋)=1

q odd, qθ≥ 1

2−ǫ

q cot2(πqθ) < 100ǫ2.

Letting A be the sum of the terms of the sum for which q < q0 andcombining (14) and (15) we obtain

(16) Godd(θ, R) ≤ (1 + ǫ)2Hodd(θ, R) +A

2R2+ 100ǫ2.

From (16) (and using the fact that ǫ can be chosen arbitrarily) weobtain that the limits superior and inferior of Godd as R → ∞ areless than those of Hodd. Using (13) we obtain the converse inequalitiescompleting the proof of the theorem in the odd case. The even case isessentially identical except that Lemma 21 is used in place of Lemma20

To complete the proof of Theorem 19 we now see that it is sufficientto calculate the limits limR→∞ Godd(θ, R) and limR→∞Geven(θ, R).

Proof of Theorem 19. The results of Section 3 do not immediately ap-ply since the cotangent function has a singularity at 0 (and the tangentfunction has a singularity at π/2). The functions therefore fail to beRiemann integrable. We split the function into two parts: a boundedpart and the singularity. Lemma 7 will apply to the bounded part.We argue that the part near the singularity makes an arbitrarily smallcontribution.


We deal first with limR→∞Godd(θ, R). Let r > 0 and write f(x) =cot2(πx)1[2−r ,1/2)(x) and g(x) = cot2(πx)1[0,2−r](x). We then have(17)

Godd(θ, R) =1

2R2

∑

q

π2R<qθ


q · f(qθ) +1

2R2

∑

q

π2R<qθ


q · g(qθ)

Clearly f is Riemann integrable. We apply Remark 16. We arestudying the first term of (17):

1

2R2

∑

q

π2R<qθ


q · f(qθ).

Notice that in order to get a contribution, q must be smaller than π2R(otherwise the condition q

π2R< qθ cannot be satisfied). This first

term can then be rewritten:

π4

2(π2R)2

∑

q<π2Rgcd(q,⌊qθ⌋)=1

q odd

q · f(qθ)1(q/(π2R),1)(qθ).

We now apply Lemma 7 (with π2R replacing R) to show that thefirst term in (17) converges as R → ∞ to

π4

2·4

π2·1

2

∫ 1/2

2−r

x2 cot2(πx) dx

As r approaches ∞, this tends to

(18) π2

∫ 1/2

0

x2 cot2 x dx = log 2−π2

24.

We then need to control the second term of (17). We will use theestimate g(x) ≤ 22ℓ on the interval [2−(ℓ+1), 2−ℓ] for ℓ ≥ r. It suffices tocontrol the second term for R’s of the form 2t/π2 for large t. Let Ns,ℓ

be the number of q’s in the range (2s−1, 2s] such that gcd(q, ⌊qθ⌋) = 1and qθ ∈ [2−(ℓ+1), 2−ℓ).The quantity that we need to control is then overestimated by

(19)1

22t

t∑

s=1

t−s∑

ℓ=r

Ns,ℓ2s+2ℓ =

1

22t

t−r∑

s=1

t−s∑

ℓ=r

Ns,ℓ2s+2ℓ.

We separate this sum into three ranges: s ≤ ℓ ≤ t− r, ℓ > t− r andℓ < s. These ranges are illustrated in Figure 5.


ℓ

s

r

Figure 5. Regions of Summation

We start by controlling the range ℓ < s. By Lemma 13 provided thatℓ < s, we have Ns,ℓ ≤ 2s−ℓ+1. This part of the sum is then estimatedby

1

22t

t−r∑

s=r

min(s,t−s)∑

ℓ=r

2s−ℓ+1 · 2s+2ℓ =2

22t

t−r∑

s=r

min(s,t−s)∑

ℓ=r

22s+ℓ

≤2

22t

t−r∑

s=r

t−s∑

ℓ=r

22s+ℓ

≤4

22t

t−r∑

s=r

2t+s

≤8

22t22t−r = 8 · 2−r.

Outside this range we have ℓ ≥ s so that Lemma 13 ensures thatNs,ℓ ≤ 1. The part of (19) corresponding to the range s ≤ ℓ ≤ t− r isbounded above by

1

22t

⌊t/2⌋∑

s=1

min(t−s,t−r)∑

ℓ=max(r,s)

2s+2ℓ ≤2

22t

r∑

s=1

2s+2t−2r +

t/2∑

s=r+1

22t−s

≤4

22t22t−r = 4 · 2−r.


The final part of (19) remaining to be bounded is the range ℓ > t−r.We assume that t > 2r. The contribution is then given by

1

22t

r∑

s=1

t−s∑

ℓ=t−r

Ns,ℓ22ℓ.

Let δ = minq≤2rqθ. If 2−(t−r+1) < δ then all of the Ns,ℓ appearingin this sum are 0, so that for sufficiently large t, this range makes nocontribution.We have therefore shown that as t → ∞, the contribution to (17)

coming from the second term is bounded above by 12 · 2−r. By takingr large, this term can be made arbitrarily small.It follows that limR→∞Godd(θ, R) = log 2− π2/24.A similar argument works to compute limR→∞Geven(θ, R). We recall

the definition:

Geven(θ, R) =1

4R2

∑

qθ>q/(π2R)gcd(q,⌊qθ⌋)=1

q even

q tan2

(

πqθ

2

)

.

Notice that the summand has period 2 in the variable qθ rather than1 so that a priori the earlier results do not apply. The ambiguity isresolved by observing that in order for gcd(q, ⌊qθ⌋) to be 1 when q iseven, one must have that ⌊qθ⌋ is odd so that qθ/2 must lie in [1

2, 1).

Notice that in this case tan(πqθ/2) = − cot(πqθ/2). We thereforere-express the sum as

Geven(θ, R) =1

4R2

∑

qθ>q/(π2R)gcd(q,⌊qθ⌋)=1

q even

q cot2(

πqθ

2

)

,

thereby restoring the periodicity of the summand.From this point the calculations are exactly similar to the odd case

(using Remark 16, Lemma 7 and the splitting into singular and non-singular parts) yielding:

limR→∞

Geven(θ, R) =π4

4·2

π2·1

2

∫ 1

0

x2 cot2(πx

2

)

dx

= 2 log 2−π2

12= 2 lim

R→∞Godd(θ, R).

The combined asymptotic density is therefore given by 3 log 2−π2/8.


6. Open Questions

Question 1. Does the set of aperiodically-coded points have positive(upper) density?

Question 2. It is known that there periodic points whose codings do notcome from any rational rotation. Is there a general description of theseperiodic irrotational codings? What is the density of the correspondingislands?

Note that an answer to Question 2 together with establishing Con-jecture 18 would answer Question 1.

Question 3. For a fixed θ, let Lθ denote the itinerary language: the setof all finite words that can appear in the itinerary of a point. What canbe said about the growth rate of the number of words of length n? Buzzi[5] proves that piecewise isometries have sub-exponential numbers of‘names’ but his proof relies on compactness of the ambient space. Fornow it is not clear whether his proof extends to the general non-compactcase.

Question 4. Prove or disprove Conjecture 18.

References

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[2] T. Apostol. A proof that Euler missed: Finding ζ(2) the easy way. Math.

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232:409–416, 1997.[5] J. Buzzi. Piecewise isometries have zero topological entropy. Ergodic Theory

and Dynamical Systems, 21(5):1371–1377, 2001.[6] L. O. Chua and T. Lin. Chaos in digital filters. IEEE Trans. Circuits and

Systems, 35(6):648–658, 1988.[7] L. O. Chua and T. Lin. Chaos and fractals from third-order digital filters.

Internat. J. Circuit Theory Appl., 18(3):241–255, 1990.[8] T. Estermann. Lattice points in a parallelogram. Canad. J. Math., 5:456–459,

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4(4):593–608, 1998.[10] A. Goetz. Dynamics of piecewise isometries. Illinois Journal of Mathematics,

44(3):465–478, 2000.[11] A. Goetz and A. Quas. Global properties of piecewise isometries. Ergodic The-

ory Dynamical Systems, 29:545–568, 2009.


[12] G. H. Hardy and E. M. Wright. An introduction to the theory of numbers.Oxford, 1938.

[13] J. Lowenstein and F. Vivaldi. Approach to a rational rotation number in apiecewise isometric system. Nonlinearity, 23:2677–2721, 2010.

[14] F. Mertens. Uber einige asymptotische Gesetze der Zahlentheorie. Journal furreine und angewandte Mathematik, 77:289–338, 1874.

[15] F. Vivaldi and A. Shaidenko. Global stability of a class of discontinuous dualbilliards. Commun. Math. Phys., 110:625–640, 1987.

[16] H. Weyl. Uber die Gleichverteilung von Zahlen mod Eins. Math. Ann., 77:313–352, 1916.

(Cheung and Goetz) Department of Mathematics, San Francisco State

University, 1600 Holloway Avenue, San Francisco, CA 94132, U.S.A.

(Quas)Department of Mathematics and Statistics, University of Vic-

toria, Victoria BC, V8W 3R4, Canada

E-mail address : cheung(a)math.sfsu.eduE-mail address : goetz(a)sfsu.eduE-mail address : aquas(a)uvic.ca

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