Posted by: matheuscmss | February 19, 2009

Furstenberg’s 2x, 3x (mod 1) problem

During my conversations with my friend Artur Avila around the 2x, 3x mod 1 problem and its applications, Artur explained me a cute argument (using the Fourier transform of measures) to show a weak version of Rudolph’s theorem (which is a partial answer to this problem). Of course, I believe it is a good idea to share it here, but, in order to keep this post more self-contained, I’ll not proceed directly to the argument. In fact, I’ll divide the exposition into 3 sections: the first one is a general introduction to Furstenberg problem, Rudolph’s theorem and the weak version of Rudolph’s theorem, the second section contains the proof of this weak version of Rudolph result and the third section briefly discusses Einsiedler, Katok and Lindenstrauss’ application of a (highly non-trivial) analog of Rudolph’s theorem to the so-called Littlewood conjecture (about simultaneous Diophantine approximations).

Disclaimer. While Artur’s explanation was clear and correct, any possible mistakes and errors appearing below are my responsability (of course).

Furstenberg’s problem: measure rigidity of expanding rank-two semi-group actions

Consider the following (uniformly expanding) dynamical system on the circle \mathbb{T}=\mathbb{R}/\mathbb{Z}:

f_d(x)=dx (\textrm{mod} \, 1),

where d\geq 2 is an integer. Despite its simple definition, f_d is rich from the dynamical point of view: f_d possesses plenty of ergodic invariant measures, e.g., the Lebesgue measure and the Dirac measures supported on its periodic points (in fact, due to the expanding features of f_d, one can use the so-called thermodynamical formalism [explained in Bowen’s book] to construct a whole family of ergodic invariant measure “interpolating” the Lebesgue measure and the Dirac measures); since, by Birkhoff’s theorem, the ergodic invariant measures captures the statistical features of most orbits in its support, we see that the existence of several distinct ergodic measures for f_d indicates the presence of several distinct statistical behaviours of the orbits of f_d (which justifies the adjective “rich”). Of course, the dynamics of an individual uniformly expanding (and, more generally, hyperbolic) dynamical systems is well-understood nowadays (after the works of Anosov, Smale, Sinai, Ruelle and Bowen among several other authors) and we do not plan to discuss further this issue today. Instead, let us investigate the dynamics of the joint action of two expanding maps, say f_2 and f_3. In other words, denoting by \Sigma_d:=\{d^n: n\in\mathbb{N}\} and \Sigma_{2,3}:=\{2^n3^m: n,m\in\mathbb{N}\}, we are replacing the action

F_d:\Sigma_d\times \mathbb{T}\to\mathbb{T}, \, F_d(k,x)=k x (\textrm{mod}\, 1)

of the rank-one semi-group \Sigma_d (i.e., the individual action of f_d) by the action

F_{2,3}:\Sigma_{2,3}\times \mathbb{T}\to \mathbb{T}, \, F_{2,3}(k,x) = k x (\textrm{mod}\, 1)

of the rank-two semi-group \Sigma_{2,3} (i.e., the joint action of f_2 and f_3).

Remark. In general, a similar action by a rank-two semi-group can be obtained replacing 2, 3 by any pair n,m of multiplicatively independent non-negative integers.

The basic topological result about the dynamics of F_{2,3} is:

Theorem (H. Furstenberg). The sole infinite closed F_{2,3}-invariant subset of the circle \mathbb{T} is the circle itself.

This theorem should be constrasted with the rank-one situation: while F_{2,3} has only one closed invariant set, it is not hard to see that F_d has a lot of invariant Cantor sets. In fact, since f_d is semi-conjugated to a full Bernoulli shift with d symbols (via a codification of the orbits using the Markov partition \left[ i/d,(i+1)/d\right], i=0,\dots, k-1), any subshift gives you an invariant Cantor set. More concretely, you can take a finite collection of disjoint closed intervals I_1,\dots, I_\ell whose union doesn’t coincide with the whole circle and define K the set of points of the circle whose f_d-orbit never enters \bigcup_{1\leq j\leq\ell} \textrm{int}(I_j). Of course, K is F_d-invariant and, from the expanding features of f_d, it follows that K is a Cantor set (exercise).

In resume, Furstenberg’s result says that F_{2,3} is more rigid than F_d (dynamically speaking), which is quite unexpected because F_{2,3} is the combination of the dynamics of two strongly chaotic endomorphisms, namely, f_2 and f_3. Taking this theorem as a motivation, Furstenberg posed the following problem of the measure (i.e., ergodic-theoretical) rigidity of F_{2,3}:

Furstenberg’s 2x,3x (mod 1) problem. Is it true that the Lebesgue measure is the sole non-atomic ergodic F_{2,3}-invariant measure? Equivalently, is it true that the Lebesgue measure is the unique non-atomic ergodic measure which is both f_2 and f_3-invariant?

In other words, this problem asks whether the intersection of two enormous sets of measures (the set f_2-invariant probabilities and the set of f_3-invariant probabilities) is relatively small (namely, the Lebesgue measure and the Dirac measures supported on periodic orbits).

While Furstenberg’s problem is still open (to the best of my knowledge), some important partial results are available nowadays. In particular, let me quote the following theorem of D. Rudolph:

Theorem (D. Rudolph). Let \mu be a f_2 and f_3 invariant ergodic measure such that


Then, \mu is the Lebesgue measure. Equivalently, if a F_{2,3} invariant measure \mu is not the Lebesgue measure, then the semigroup

\Sigma_{2,3}(x):=\{2^n3^mx (\textrm{mod}\, 1): n,m\in\mathbb{N}\}

is a group for \mu almost every x\in\mathbb{T}.

In this statement, h_\mu(f) denotes the metric (Kolmogorov-Sinai) entropy of the f-invariant measure \mu.

Remark. The motivation of the metric entropy condition in Rudolph’s theorem comes from the fact that the metric entropy of Dirac measures supported on periodic orbits is zero. Of course, Rudolph theorem is just a partial answer to Furstenberg question because non-atomic invariant measures can have zero entropy.

While we do not pretend to give a complete proof of Rudolph’s theorem (which is not very hard but involves some amount of abstract ergodic theory), we do plan to show in the next section the following fact:

Theorem 1 (weak version of Rudolph’s theorem). Let \mu be a f_2 and f_3 invariant measure. Assume that h_\mu(f_2)>0 and \mu is f_3 ergodic. Then, \mu is the Lebesgue measure.

Weak Rudolph theorem: proof of theorem 1

Take \mu a probability measure verifying the assumptions of theorem 1 and let T(x)=x+1/2 be the translation of 1/2 on the circle \mathbb{T}.

Note that f_3 and T are commuting maps. In particular, since \mu is f_3 invariant, it follows that T_*\mu is f_3 invariant. Moreover, the assumption of ergodicity of \mu with respect to f_3 implies that \mu and T_*\mu are f_3-ergodic.

On the other hand, the positive entropy assumption h_\mu(f_2)>0 implies that the restriction of f_2 to the support of \mu is not invertible. Equivalently, we can always find some pairs of \mu generic points with the same image under f_2. However, such a pair of points is always permuted by the translation T. In particular, we have that the probability measures \mu and T_*\mu aren’t mutually singular.

Putting all these facts together, we have two f_3 invariant ergodic measures (\mu and T_*\mu) such that \mu and T_*\mu aren’t mutually singular. It follows that T_*\mu=\mu, i.e., \mu is T-invariant.

At this stage, one has enough information to conclude that \mu is the Lebesgue measure. Indeed, we claim that all of the non-zero Fourier modes of \mu vanish, that is,

(1) \widehat{\mu}(k):=\int_\mathbb{T} e^{2\pi ikx}d\mu(x)=0 for any k\in\mathbb{Z}-\{0\}.

In fact, let us start with the odd Fourier modes (i.e., \widehat{\mu}(k) where k\notin 2\mathbb{Z}). Using the T-invariance of \mu, we obtain

\widehat{\mu}(k) = \widehat{T_*\mu}(k) = \int e^{2\pi ikx}dT_*\mu(x) = \int e^{2\pi ikT(x)} d\mu(x) .

Since \int e^{2\pi ikT(x)} d\mu(x) = \int e^{2\pi ik(x+1/2)} d\mu(x) = e^{2\pi ik/2}\widehat{\mu}(k), we get

\widehat{\mu}(k)=e^{2\pi ik/2}\widehat{\mu}(k).

In particular, it follows that \widehat{\mu}(k)=0 whenever k\in\mathbb{Z} is odd. Finally, we can use the f_2-invariance of \mu (with a similar argument) to obtain

\widehat{\mu}(k)=\widehat{\mu}(2k) for every k\in\mathbb{Z},

that is, the even Fourier modes can be deduced from the odd ones (more precisely, writing a non-zero even number k as k=2^n m where m\neq 0 is odd, we have \widehat{\mu}(k)=\widehat{\mu}(m)). Consequently, our claim (1) about the vanishing of the Fourier modes of \mu is proved. As we know (1) forces \mu to be the Lebesgue measure, so that the proof of theorem 1 is complete.

Measure rigidity of higher-rank hyperbolic actions: Einsiedler-Katok-Lindenstrauss theorem

In this final section, we briefly outline a recent application of the measure rigidity of higher-rank hyperbolic actions to a partial solution of a number-theoretical problem.

A well-known fruitful interaction in Mathematics occurs between Dynamical Systems and Number Theory. Among the several applications of dynamical ideas to number-theoretical problems, one finds Margulis’ solution to Oppenheim conjecture (see e.g. this blog post of Terence Tao). Here, the basic idea is to convert the study of the values of indefinite quadratic forms in k variables (k\geq 3) into the study of the dynamics of an action of a higher-rank group H in the space SL(k,\mathbb{R})/SL(k,\mathbb{Z}) of unimodular lattices in \mathbb{R}^k. Another (more recent) application of these ideas was performed by M. Einsiedler, A. Katok and E. Lindenstrauss in the study of the so-called Littlewood conjecture:

Littlewood conjecture. For every \alpha,\beta\in\mathbb{R}, it holds

\liminf\limits_{n\to\infty} n\langle n\alpha\rangle \langle n\beta\rangle=0.

Here \langle x\rangle denotes the fractional part of x.

Again, the basic idea is to convert this problem into the study of the dynamics of the action of the group A of positive diagonal SL(k,\mathbb{R}) matrices in SL(k,\mathbb{R})/SL(k,\mathbb{R}).

However, although the basic strategy of Margulis and Einsiedler-Katok-Lindenstrauss is the same, there is a subtle and important difference: while in Margulis’ context the acting group H contains only unipotent (parabolic) elements, in Einsiedler-Katok-Lindenstrauss’s setting the acting group A contains only hyperbolic elements.

In particular, at the present moment, we have the following picture: one can completely classify the invariant measures of the H-action (by Ratner’s theorems), but one can say something interesting about the invariant measures of the A-action only when they have positive entropy. In other terms, it is “morally” more easy to show measure rigidity statements (in the spirit of Furstenberg’s problem) for the H-action than A-action because the individual elements of the group H are already parabolic, i.e., rigid. Here, the positive entropy condition for the measure rigidity statement (for the A action) is certainly motivated by Rudolph’s result.

In any case, the moral philosophy here is the following: the number-theoretical problems quoted above (Oppenheim and Littlewood conjectures) can be reduced to the complete classification of the invariant measures of adequate higher-rank group actions. Since such a complete classification is possible for the H-action, Margulis completely solved Oppenheim conjecture; but because our current technology only allows us to classify A invariant measures of positive entropy, Einsiedler-Katok-Lindenstrauss partially (almost) solved Littlewood conjecture. More precisely, they proved the following result:

Theorem (Einsiedler, A. Katok and E. Lindenstrauss). The set of (possible) exceptions (\alpha,\beta) to Littlewood conjecture has Hausdorff dimension 0.

Closing this post, let me say that this last result deserves further explanation and I plan to discuss it a little bit more in a future post (probably after the series of posts around Trieste’s conference). By the way, the curious reader may consult (besides the original article) these expository notes of Akshay Venkatesh for a nice introduction.  Ciao!


  1. This is a very nice summary of the `easy case’ of
    Rudolph’s argument. I think that this argument is probably close to that of Russel Lyons. Dan Rudolph based his argument upon the Fourier analytic argument of Russ.

  2. Dear Michael Lacey,

    you’re completely right: indeed the argument is very close to Lyons’ one! Thanks for the reference!

    By the way, let me say that I like this argument of Lyons because it makes clear the utilization of two “hyperbolic” dynamics 2x, 3x and a third “central” element T(x) = x+1/2, which is the starting point of Katok-Spatizer and Eisiendler-Katok-Lindenstrauss’ results (namely, the utilization of “hyperbolic” dynamics and “central” [unipotent] dynamics).

  3. Find this this post when googling ‘Furstenberg conjecture’. A very short argument! Just the definition of ‘generic point’ is a little bit vague. Also could you explain a few more words how to derive \mu and T_\ast \mu are not mutually singular?

    • Hi Pengfei,

      In fact, I was using the term ‘generic point’ in a vague way to mean ‘a point in some “adequate” set of full measure’. Of course, the precise meaning of ‘generic point’ depends on the context: if you are interested in topological properties, you can consider the points in the support of the measure as ‘generic points’, while if you are interested in ergodic averages of an ergodic measure, you can say that the points satisfying the conclusion of Birkhoff’s theorem are ‘generic points’.

      In the context of this post, we mentioned ‘generic point’ in a vague manner as a way to keep an informal discussion. However, it is not difficult to convert what was said into a formal argument as follows.

      As it was mentioned in the post, the assumption h_{\mu}(f_2)>0 means that there exists a subset A of positive \mu-measure of the support of \mu (our set of ‘generic points’) such that f_2 is not invertible at points in A, or, equivalently, T^{-1}(A) also has positive \mu-measure. Since the support of \mu (‘generic points’) has full \mu-measure, we can (and do) also assume that T^{-1}(A) is contained in the support of \mu. It follows that A is a set of positive \mu and T_*\mu measures contained in the intersection of the supports of \mu and T_*\mu.

      In other words, the support of \mu (a set of full \mu-measure) intersects the support of T_*\mu (a set of full T_*\mu-measure) in a set whose both \mu-measure and T_*\mu-measure are positive. Therefore, \mu and T_*\mu can’t be mutually singular.



      • Oi Carlos, I really like the short argument you present here. It took me a while to fully understand the part about positive entropy implying that the measures \mu and T_*\mu could not be mutually singular so I decided to write here the details of that step (I hope you don’t mind) for future readers who may face similar difficulties.

        It helped me to think of a concrete set as the `support’ of \mu; I used the set A of generic points for the transformation f_3 (i.e. points which are good for the pointwise ergodic theorem for every continuous function). Since T and f_3 commute, TA consists of the generic points for ([0,1),T_*\mu,f_3), so either \mu=T_*\mu (which is what we want to conclude) or A\cap TA=\emptyset. In the second case, (A,f_2) is an invertible system, which is isomorphic as measure preserving system to ([0,1),f_2).

        Now this part confused me because there are, of course, invertible systems with positive entropy (think of a two-sided Bernoulli system). The contradiction arises from the fact that the system ([0,1),\mu,f_2) has a one-sided generator (namely, the partition \big\{[0,1/2),[1/2,1)\big\}). The fact that invertible systems with a one sided generator must have 0 entropy (or equivalently, as used in the post, that a positive entropy system with a one sided generator can not be invertible) is well known and can be found, for instance, as Corollary 1.23 in the upcoming book of Einsiedler, Lindenstrauss, Ward (available online for now). This proves that \mu=T_*\mu.

  4. Are the Fourier modes missing a multiple of 2 \pi in the exponential? [Corrected; Thanks, Matheus]

    • No worries. You can simplify exp(2 i \pi k / 2) as well.

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