This first post discusses some uniform distribution results of the orbits of on . It is a classical result that they are all uniformly distributed. Moreover, Denjoy-Koksma’s inequality states that, for iterates equal to the denominators of the continued fraction expansion of , the orbits are highly equidistributed. We give a proof of this result for the classical assumption of bounded variation and its strengthening for differentiable functions. The last part is devoted to describe the basic properties of a cylinder flow and how its ergodic properties are related to the uniform distribution of orbits of the respective irrational rotation. More specifically, we want to convince the reader that the ergodicity of a cylinder flow over implies – on the contrary to Denjoy-Koksma’s inequality – that the orbits of the irrational rotation have discrepancy, in the sense that uniform distribution is not regular. This gives another reason, beyond their intrinsic ergodic theoretical importance, to investigate cylinder flows.

**1. Uniform distribution **

It is a classical result that the only -invariant probability measure is the Lebesgue measure on . This is equivalent to

for any continuous function . By approximation, this is equivalent to, for any interval , the frequency of visits to of any orbit of being equal to :

Definition 1A sequence in isuniformly distributedorequidistributedif

The unique ergodicity of thus says that, for any , the sequence is *uniformly distributed*. We can identify equidistribution in a quantitative fashion by considering the discrepancy of the sequence.

Definition 2Let be a finite sequence of real numbers. The number

is called thediscrepancyof the sequence. For an infinite sequence, the discrepancy is equal to the discrepancy of the first terms.

It is direct, and we leave the proof as an exercise to the reader, that

Proposition 3The sequence is uniformly distributed iff

In other words, the theorem states the following (in principle stronger yet equivalent) interesting fact: whenever the sequence is equidistributed, then the convergence in (1) is uniform along all intervals .

The distinction between different types of equidistribution is made by the analysis of the non-averaged expressions

If a sequence has good equidistribution, one would hope for boundedness of . It turns out that, by the following result of W.M. Schmidt, this is never the case: is always infinite.

Theorem 4 (W.M. Schmidt)There is an absolute constant with the following property: for any infinite sequence of , we have the inequality

for infinitely many positive integers .

The interested reader may check the proof in the book Uniform distribution of sequences of Kuipers and Niederreiter.

**2. Denjoy-Koksma’s inequality **

The above theorem guarantees the existence of infinitely many times for which the uniform distribution behaves in an irregular fashion. Nevertheless, for the special case of the irrational rotation, there are indeed iterates for which the sequence is highly equidistributed in the sense that is bounded, and this is given by Denjoy-Koksma’s inequality. The iterates exhibiting such behavior are the denominators of the continued fraction expansion of .

Given irrational, consider its *continued fraction expansion*

whose –*convergent*is

The are called the *partial quotient denominators* or just *denominators* of . They give the best rational approximations of . More precisely, the approximation is equal to

and is always at most . On these denominators, the dynamics of the rotation is partially described by the rational rotation , and that’s why the sequence is highly equidistributed. This is the content of the next

Lemma 5For each , there is one and exactly one integer for which

*Proof:* The argument is based on the article Sur la conjugaison différentiable des difféomorphismes du cercle à des rotations of Michel Herman. Let . Firstly, assume

Then, for ,

and so

where is the residue of modulo . As runs overs all residues modulo , every such interval contains at least (and then exactly) one element of the set .

The case is similar, in which case .

Exercise 1Show that the above lemma implies that for every .

Translating to the setup of functions, we arrive at the

Theorem 6 (Denjoy-Koksma’s inequality)If has bounded variation, then

for every .

*Proof:* Let and . By Lemma 5, each interval

has exactly one element of the set . Thus

Denjoy-Koksma’s inequality can be improved if is of class . Let denote the -norm.

*Proof:* Again, we can assume that belongs to the set of functions of zero mean. The main property we’ll use is that the -closure of the set

of *coboundary functions* is equal to . For this, recall the set of trigonometric polynomials

of zero mean is -dense in . Hence, it suffices to prove that each trigonometric polynomial is a coboundary. This is a simple task: given ,

satisfies .

The theorem now follows: given and , let be a coboundary for which . In particular, . Then

where in the second equality we used Denjoy-Koksma’s inequality and in the third the mean value inequality. For large , the last expression is smaller than .

Remark 1Given any , the trigonometric polynomials are also dense in and then the same argument above proves that the -closure of the set

is equal to , the set of functions of zero mean.

**3. Cylinder flows **

We now focus our attention on cylinder flows. As defined in CF0, for any irrational number and function , where or , we define the cylinder flow by

The dynamics of is intimately connected to the Birkhoff sums of under : denoting these by , we have

By Birkhoff’s theorem, converges to almost surely. In particular, if has *drift*, i.e. , then

and so almost every orbit of diverges to infinity. From now on, we assume has no drift.

There is a natural invariant measure for : it is , where and are the Lebesgue measure of and , respectively. It is an infinite sigma-finite measure. Measures of this type often behave in very strange ways and fail to comply with rules forming the basis of finite ergodic theory but, regardless this, are worth studying and open the possibility for very different statistical behaviors.

Let be a *measure-preserving system*: is a measure space, a (possibly infinite) -finite measure and is a measurable map on invariant under .

Definition 8We say isconservativeif whenever is such that are pairwise disjoint. We say isergodicif it has only trivial invariant sets, that is, if or whenever is a measurable set invariant under .

If is finite, then it is necessarily conservative. This is Poincaré’s recurrence theorem (see ERT1). In the infinite measure situation, this is not generally the case. For example, the translation , , is -invariant but non-conservative. Even more weird, it is ergodic! This is a pathological example that indeed is global in the sense of the

Exercise 2Let be an invertible measure-preserving system. If it is ergodic and non-conservative, then it is isomorphic to the translation , .

In the non-invertible situation, there are many other examples. See the end of this paper of J. Aaronson and T. Meyerovitch. As cylinder flows are invertible and clearly not isomorphic to the translation in , conservativity follows if we prove ergodicity.

Coming back to our specific skew product, the mere assumption of zero drift guarantees that is conservative, according to the following

Theorem 9 (Atkinson)Let be an ergodic probability measure-preserving system and a measurable function such that . Then -almost every has the following property: for any set containing with and any , the set

is infinite.

The proof is in the paper Recurrence of co-cycles and random walks. A classical result of W. Gottschalk and G. Hedlund characterizes transitivity.

Theorem 10 (Gottschalk – Hedlund)Let be a compact metric space, a minimal homeomorphism and continuous. Then defined by

is transitive if and only if

- and
- .

Another theorem of Gottschalk and Hedlund indeed shows that, under the assumptions of Theorem 10, the two conditions above are equivalent to not being a coboundary. See the book Topological dynamics for a proof.

Not much can be obtained in the topological perspective. By this we mean that, when is continuous, is never minimal, whose proof is the content of this paper of P. Le Calvez and J.-C. Yoccoz.

Theorem 11 (Le Calvez – Yoccoz)There are no minimal homeomorphisms on an open annulus.

The next post will discuss ergodicity of cylinder flows. Before finishing the present one, let’s emphasize the relation between ergodicity and irregularity of equidistribution: Denjoy-Koksma’s inequality only gives information on the partial quotient denominators of . It might happen that, for different iterates, the Birkhoff sums become unbounded. The analysis of this, at least in the qualitative perspective, can be made via cylinder flows: if is ergodic, then must assume arbitrarily large values, because the orbit of a generic point has to visit sets far from the fiber . In particular no such high equidistribution as that in Denjoy-Koksma’s inequality might happen. It is because of this that

**“Ergodicity of cylinder flows are related to the irregularity of equidistribution”.**

**Previous posts:** CF0.

Hi Yuri,

As a side remark, it is worth to mention that there are no good analogues of the Denjoy-Koksma inequality in “higher dimensions”: for instance, in “Appendice 1” of this article here (http://www.ams.org/mathscinet-getitem?mr=1367354), Yoccoz shows that, given an increasing non-bounded function from to itself, there are a minimal rotation of the two-torus , a real-analytic function of zero mean, and a full (Lebesgue) measure subset such that, for every , one has .

Best, Matheus

By:

matheuscmsson February 17, 2012at 2:14 pm

Hi,

A remark: you should assume that phi is not a coboundary in theorem 10. As otherwise, the map F is bounded on the fibers and you do not have transitivity… Moreover, it enlights your last remark as Birkhoff sum of a coboundary is definitely regular!

Theorem 11 is definitely geometric. I suspect that there exists a theorem which says that a skew-product with fiber Z is never minimal. Do you know if this is true ?

By:

vdelecroixon March 26, 2012at 5:38 pm

Hi Vincent,

You’re absolutely right. As it is, this is just equivalent to conservativity. I forgot to add that

1. the basis transformation is minimal, and

2. the Birkhoff sums are unbounded:

.

Another theorem of Gottschalk and Hedlund guarantees that is not a coboundary, as you remarked.

With respect to your second remark, I don’t know and, up to my knowledge, there aren’t results in this direction (but I have to check with more care).

Thanks, Yuri.

By:

yglimaon March 27, 2012at 9:33 am

Hi Matheus,

what is the precise (original) reference for Theorem (Remark ?) 7 ?

Is it due to Koksma ?

This is closely related to my work with M. Triestino http://arxiv.org/abs/1207.1481, where we extend this to any C^2 circle diffeomorphism of irrational rotation number as the base dynamics. Indeed, if you read our proof carefully in the context of an irrational rotation, it is the very same as yours…

Best,

Andrés

By:

Andrés Navason September 18, 2012at 11:14 am

Hi Andrés,

as far as I know, the original reference for “Theorem 7” in this post is Michel Herman’s article “Sur la conjugaison différentiable des difféomorphismes du cercle à des rotations”.

Indeed, by looking in this article at the proposition at the paragraph (3.2) at page 74 (after the proof of Denjoy-Koksma inequality), you can find a slightly more general version of “Theorem 7” based on the modulus of continuity of .

Best,

Matheus

By:

matheuscmsson September 18, 2012at 11:38 am

Nice, many thanks Matheus.

Here is a closely related question for which affirmative answer would trivialize everything (though I’m afraid the answer should be negative): assume you have a C^2 circle diffeomorphism of irrational rotation number, as well as a C^1 test function u on the circle. Is it true that the derivatives of the Birkhoff means of u at the approximating times of the rotation number uniformly converge to zero ? Is this true at least for a subsequence of these times ?

All the best,

Andrés

By:

Andrés Navason September 18, 2012at 11:47 am

Hi Andrés,

I’ve not thought seriously about your question, but I think that the following remark gives some reason to think that a positive answer in some cases is not completely impossible.

Let me oversimplify the situation by considering a smooth circle diffeomorphism with Diophantine rotation number and let the test function be a coboundary of a function , i.e., . Then, we know that is smooth () conjugated to the rotation , i.e., with a diffeomorphism. On the other hand, the th Birkhoff sum of is , and hence its first derivative is . Now, we note that , so that is close to and is close to the identity when . In particular, in this very special case ( smooth with Diophantine rotation number and a coboundary of a function, the question has a positive answer (if I’m not mistaken).

Best,

Matheus

By:

matheuscmsson September 20, 2012at 8:10 am

Hi again Matheus,

yes, actually, if you assume only that the differ is C^1 conjugate to the rotation one can still show the desired convergence without the assumption on u. The problem are Liouvillean rotation numbers. I’m afraid that this shouldn’t be true in such case. Anyway, some kind of convergence can be established (ion L^1, if I remember well).

Best,

Andrés

By:

Andrés Navason September 20, 2012at 8:43 am

Hi Andrés,

Ah, ok, now I see your point! I’ll try to keep your question in mind and if I find something interesting to tell you, I will add it as a reply here.

Best,

Matheus

By:

matheuscmsson September 20, 2012at 10:50 am

Hi Matheus. Very quickly just to say that the answer to the question above is negative. This arose in a discussion with S. Crovisier who carried out the construction of a counter-example using the Anosov-Katok method (counter-examples are indeed generic…)

By:

Andrés Navason March 21, 2013at 5:25 pm

Hi,

If I am not mistaken, the argument in the proof of Denjoy-Koksma does not work: you can not necessarily find a value y such that the sum is zero (zero mean does not imply the existence of a value 0… unless you assume that the function is continuous). The proof is still valid for continuous with bounded variation. It is not hard to fix to make it work for general BV functions but need some Abel summation.

Vincent

By:

vdelecroixon July 30, 2013at 2:42 am

Hi Vincent,

You’re absolutely right. I will ask the owner of the post (Yuri) to make the necessary changes.

Best,

Matheus

By:

matheuscmsson August 1, 2013at 11:04 am

Hi Vincent,

Thanks for the observation. I just posted a different proof (without the use of Abel summation) which follows Herman’s paper Sur la conjugaison différentiable des difféomorphismes du cercle à des rotations. I hope this time the proof is correct.

Best, Yuri.

By:

yglimaon August 5, 2013at 2:38 pm