46

The idea to study what we call irrational rotation of a torus indeed belongs to Nicole Oresme, at least he clearly understood the density of the trajectories (which is not the same as ergodicity or equidistribution! One thing is to say that a trajectory visits every interval infinitely many times, and another thing to say HOW OFTEN. Oresme has absolutely no ...

answered May 16 '17 at 13:26

Alexandre Eremenko

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28

the proof goes back to Nicole Oresme in his paper De commensurabilitate vel incommensurabilitate motuum celi [On the Commensurability or Incommensurability of the Motions of the Heavens], dated around 1360, see
Nicole Oresme and the commensurability or incommensurability of celestial motions (contains an annotated English translation of Oresme's Latin text)
...

answered May 16 '17 at 7:29

Carlo Beenakker

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27

It's indeed unbounded for every irrational $x$.
Let me identify points of $\mathbb{R}/\mathbb{Z}$ with their representatives on $[0,1)$, and order it by the usual order $<$ of $\mathbb{R}$ applied to the representatives.
Replace $x$ by $y = x/2$, and the question becomes whether for all $m$ there exists $k$ such that the orbit of $0$ in the irrational ...

26

$$\sqrt{2+\sqrt{3}}-\sqrt{2- \sqrt{3}}=\sqrt{2}$$

answered Mar 15 '17 at 15:01

Fedor Petrov

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24

There are several puzzling things about the question: Firstly of course $\theta$ must be irrational, and it is intended for $\{ x\}$ to denote the Bernoulli polynomial $x-[x]-1/2$ rather than the more usual fractional part. Secondly, where is the result of Hardy from? I did find this statement in the Cambridge ICM paper of Hardy and Littlewood where they ...

21

Edit: In a comment below, coudy asks for the following source to be included as the first in English:
Anosov, D. V. Geodesic flows on closed Riemann manifolds with negative curvature. Proceedings of the Steklov Institute of Mathematics, No. 90 (1967). Translated from the Russian by S. Feder American Mathematical Society, Providence, R.I. 1969 iv+235 pp.
...

answered Dec 26 '13 at 6:20

Benjamin Dickman

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21

The result depends on the approximation properties of $\alpha$.
Of course one has to assume $\int_{S^1} f(z)dz=0$. A rotation by $\alpha$ has the effect that the $k$-th Fourier coefficient of $f$ is multiplied by $\exp(2\pi i \cdot k \alpha)$. Hence, the $k$-th Fourier coefficient (for $k \neq 0$) of $T_n(f)$ is just
$$ \frac1{\sqrt{n}} \sum_{l=1}^n \exp(2\...

20

There is a good reason you were having difficulties in proving this.
This was an old question of Rohlin (MR0126526) which was first disproved in the topological setting by Goodwyn (MR0314023) and later independently by Thouvenot and Ornstein. An explicit example appears in the paper by Ornstein and Weiss (MR0910005; page 133); this paper should be relevant ...

17

The key words here are "large deviations"; large deviations theory addresses exactly this question. The answer depends quite a bit on the specific measure and system in question, but roughly speaking one may say the following: if the system displays a sufficient amount of hyperbolic behaviour (for example, an Axiom A system, or a system with the ...

answered Aug 25 '13 at 20:43

Vaughn Climenhaga

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17

Let $f(t) = 1$ if $t \in A$ and $f(t) = 0$ otherwise. Suppose that $a > 0$. Then
$$ \begin{aligned}
\int_a^{2 a} \operatorname{card} \{n : n t \in A\} dt & = \int_a^\infty \biggl(\sum_{n = 1}^\infty f(n t) \biggr) dt \\
& = \sum_{n = 1}^\infty \int_a^{2 a} f(n t) dt \\
& = \sum_{n = 1}^\infty \frac{1}{n} \int_{n a}^{2 n a} f(s) ds \\
& = \...

answered Nov 30 '19 at 21:25

Mateusz Kwaśnicki

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16

The sequence $\sum a_n$ is unbounded.
This is a consequence of a general result from Kesten,
On a conjecture of Erdös and Szüsz related to uniform distribution mod 1, Acta Arithmetica (1966). The proof is not very long but quite computational, using properties of continued fractions.
Theorem
Let $\xi \in [0,1]$, $0\leq a < b \leq 1$, denote by $N(M, \...

15

There's no estimate that works in general. Krengel, "On the speed of convergence in the ergodic theorem", shows that for any ergodic transformation of $[0,1]$ and any sequence $(a_n)$ converging to $0$, no matter how slowly, there is a set $A$ such that
$$\limsup_{N\rightarrow\infty}\frac{1}{a_N}|A_N\chi_A(x)-\mu(A)|=\infty$$
for almost every $x$, where $...

15

Yes. This follows from Theorem 3.2 of my paper with Michael Boshernitzan, Gregori Kolesnik and Máté Wierdl, 'Ergodic Averaging Sequences'.

15

The classical Birkhoff ergodic theorem considers a map $T$ acting on a probability space $(X,\mathcal{F},\mu)$. This could alternatively be thought of as an action of $\mathbb{N}$ on $(X,\mathcal{F},\mu)$ by the transformations $T,T^2,T^3,\ldots,T^n,$ et cetera, where $n \in \mathbb{N}$ is identified with the transformation $T^n$. Since at least the 1960s ...

15

Where to begin!
The ergodicity of non-compact subgroups (singular tori) was used by Margulis to prove that higher rank lattices $\Gamma$ are arithmetic.
Once you have that $\Gamma $ is arithmetic, this has the following consequences: (1) if $Comm (\Gamma)$ is the abstract commensurator, then $Comm (\Gamma)/\Gamma$ is infinite.
(2) The cohomology groups $...

14

[Edited because I had misread $c_b(k)$ to be the number of non-zero digits in the base $b$ representation, rather than the number of digit transitions. The argument works for both variants. -T]
The claim is true, and one can in fact argue by purely Archimedean methods (with the only number theoretic input being the irrationality of $\log 2 / \log 3$).
If the ...

answered Nov 24 '20 at 17:06

Terry Tao

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13

Why not just prove the result from scratch? It's only a couple of pages and involves only basic Fourier analysis for locally compact abelian groups.
Let
$$\mathbb{T}^n = \left\{(z_1,\ldots,z_n) \in \mathbb{C}^n : |z_l| = 1 \text{ for all $1 \leq l \leq n$}\right\}$$
be the $n$-torus. Let $t_1, \ldots, t_n$ be arbitrary real numbers, and let $H$ be the ...

13

Yes. That is the same thing as ergodicity. To explain it, you have probably seen somewhere that the way to understand random variables formally is functions from a (hidden) underlying space $\Omega$ to $\mathbb R$. That is: knowing the point $\omega$ of $\Omega$, one can recover $X_n(\omega)$ for each $n\in \mathbb Z$.
In fact, a standard $\Omega$ to use ...

13

We have
Theorem. Let $\psi(x)$ and $\varphi(x)$ be positive increasing functions such that
$$\int_1^\infty \frac{dx}{\psi(x)}=+\infty,\qquad \int_1^\infty \frac{dx}{\varphi(x)}<+\infty.$$
Then for almost all $\alpha\in(0,1)$ we have
$$\Omega(\log N\cdot \psi(\log\log N))\le\sup_{n\le N}\sum_{j=1}^n(-1)^{\lfloor j\alpha\rfloor}=O(\log N\cdot \varphi(\...

12

There are very precise results on the repartition of orbits of $SL(2,\mathbb{Z})$ of irrational points in $\mathbb{R}^2$, such as this one.
In particular these orbits are dense, and this can be seen easily like this.
If $\theta$ is irrational, there are coprime $(a,b)$ such that $a+b\theta$ is less than $\epsilon$ for any $\epsilon>0$. Then there are $(...

12

Suppose that $(S-n)\cap S$ has upper density 0 for all $n\leq N$. Then the density of the set of integers $x$, such that $|[xN, (x+1)N]\cap S|\geq 2$ is also 0, hence the upper density of $S$ is $\leq \frac{1}{N}$. Taking $N\rightarrow\infty$ gives a positive answer to your question.

answered Apr 17 '16 at 12:35

Jan-Christoph Schlage-Puchta

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12

It was proved by Donnay that any compact orientable surface can be given a Riemannian metric for which the geodesic flow is ergodic. See Theorem 1 of this article. On the other hand, there are no negatively curved metrics on a sphere or a torus.

12

Let $\mu=(\delta_A+\delta_B)/2$. Then the claim of Arnold - Krylov is the weak convergence of the convolutions $\mu^{*n}*\delta_x$ to the rotation invariant probability measure on the sphere (where $\mu^{*n}$ is the $n$-th convolution power of the probability measure $\mu$ on the group of rotations). A general answer to this question had been given by ...

12

The points on the spiral with $\theta=$
$$
3.4535999354657\\
15.1248305526170\\
22.0370015553781\\
16.3950081067565
$$ form a square.
The numbers can be generated from this Mathematica code, if anyone wants to play with the algorithm and the initial conditions:
start = {{r, 15}, {s, 16}};
curve[r_] := {r Cos[r], r Sin[r]};
rotate[{x_, y_}] := {-y, x};
...

11

Let $X$ be a compact metric space and $T\colon X\to X$ be a continuous map. The set of $T$-invariant Borel probability measures $\mathcal{M}_T(X)$ is well known to be non-empty, convex, compact, and metrizable. Moreover, its extreme points $\mathcal{M}^e_T(X)$ coincide with ergodic invariant measures and every invariant measure can be represented as a ...

answered Sep 12 '13 at 18:58

Dominik Kwietniak

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11

I think the brief answer is Yes; the answer seems to be well-known to the grand master of billiards, Bunimovich himself. See http://www.scholarpedia.org/article/Dynamical_billiards and
the references quoted therein.
The above example of rectangular billiard with round corners has several ergodic components in the invariant measure so it is not uniquely ...

11

Topological mixing is called "permanent regional transitivity" (and demonstrated) by Gustav Hedlund in his 1939 article on the dynamics of the geodesic flow in constant negative curvature (third page). I think this is the first occurrence of the concept.

11

You want to find an $s$ such that $s, \sqrt{2} s, \sqrt{3} s, \sqrt{5} s$ are all close to integer. Your $t$ is then given by $2\pi s.$ The first question is a problem in simultaneous Diophantine approximation, an algorithm for which (using lattice reduction) is given by W.Bosma (probably among others).

answered May 12 '15 at 19:52

Igor Rivin

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11

See the paper "Le poisson n'a pas d'arêtes" by Thierry Bousch. This is a joke that was explained to me much later. The set is considered to resemble a fish. The French word arête means both bone and edge (of a polygon); so the English title of the paper would be "The fish has no bones/The fish has no edges". The paper proves that every point on the boundary ...

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