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Hillel Furstenberg conjectured that the only $2$-$3$-invariant probability measure on the circle without atoms is the Lebesgue measure. More precisely:

Question: (Furstenberg) Let $\mu$ be a continuous probability measure on the circle such that $$\int_{S^1} f(z) d\mu = \int_{S^1} f(z^2) d\mu = \int_{S^1} f(z^3) d\mu, \quad \forall f \in C(S^1).$$ Is $\mu$ the Lebesgue measure?

The assumption implies that the Fourier coefficients $\hat\mu(n)$ satisfy $$\hat\mu(n) = \hat \mu(2^k3^ln), \quad \forall k,l \in \mathbb N, n \in \mathbb Z.$$ Furstenberg's question is known to have an affirmative answer if one makes additional assumptions on the entropy of the measure.

The basic strategy is usually to show that a non-vanishing (non-trivial) Fourier coefficient implies the existence of an atom. The standard tool to construct atoms in a measure on the circle is Wiener's Lemma, which says that as soon as there exists $\delta>0$ such that the set $\lbrace n \in \mathbb Z \mid |\hat\mu(n)| \geq \delta \rbrace$ has positive density in $\mathbb Z$, $\mu$ has an atom. More precisely, the following identity holds: $$\sum_{x \in S^1} \mu(\lbrace x \rbrace)^2 = \lim_{n \to \infty} \frac1{2n+1} \sum_{k=-n}^n |\hat \mu(k)|^2.$$

Clearly, $$|\lbrace 2^k3^l \mid k,l \in \mathbb N \rbrace \cap [-n,n] | \sim (\log n)^2 $$ so that Wiener's Lemma can not be applied directly. My question is basically, whether this problem can be overcome for any other subsemigroup of $\mathbb N$.

Question: Is there any subsemigroup $S \subset \mathbb N$ of zero density known, such that every $S$-invariant continuous probability measure on $S^1$ is the Lebesgue measure? What about the subsemigroup generated by $2,3$ and $5$?

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I think you missed ergodicity in Furstenberg's conjecture: Furstenberg's conjecture is about 2 -3-invariant $\text{ergodic}$ probability measure. – Li Jingyang Jun 20 at 3:54
    
@LiJingyang: The only other ergodic 2-3-invariant measures are conjectured to be concentrated on periodic orbits. Since there are only countable many periodic orbits, this and absense of atoms already implies that the measure must be Lebesgue measure (and hence be ergodic). – Andreas Thom Jun 20 at 6:15
    
Hi, Andreas. sorry, i can not understand your reply of my comment. I only want to say that Furstenberg's conjecture is about 2-3-invariant $\text{ergodic}$ probability measure but not 2-3-invariant probability measure. – Li Jingyang Jun 20 at 6:30
    
Right, as I said, it is equivalent. – Andreas Thom Jun 20 at 7:33
up vote 17 down vote accepted

Manfred Einsiedler and Alexander Fish have a paper (arxiv.org/abs/0804.3586) showing that a multiplicative subsemigroup of $\mathbb N$ which is not too sparse satisfies the desired measure classification. The semigroups they consider are still somewhat large, and in particular not contained in finitely generated semigroups.

One can check that the set of perfect squares $E$ satisfies measure classification as follows: since $\frac{1}{N}\sum_{n=1}^N \exp(2\pi i n^2 \alpha)\to 0$ for every irrational $\alpha,$ one can conclude as in Wiener's lemma that an atomless measure $\mu$ on $S^1$ satisfies $\lim_{N\to \infty} \frac{1}{N}\sum_{n=1}^N|\widehat{\mu}(n^2)|=0.$ As in the original post, it follows that an atomless $E$-invariant measure is Lebesgue.

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Thanks. I did not know about this article. – Andreas Thom Oct 31 '10 at 6:23
    
Welcome to MO, John! – Kevin O'Bryant Nov 2 '10 at 20:35

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