Consider the torus $\mathbb T^2:=\mathbb{R}^2/\mathbb{Z}^2$, with,for each $\alpha\in \mathbb R$, the discrete dynamical system $f_\alpha:\mathbb T^2\to\mathbb T^2$ defined by $f_\alpha(x,y)=(x+\alpha\mod 1,x+y\mod 1)$. I am trying to show that if $\alpha$ is irrational, and $A$ is a non-empty $f$ invariant open subset of $\mathbb T^2$, then $A$ is in fact dense in the torus.

It is fairly easy to show, for any $n\in\mathbb N$, that $f_\alpha^n(x,y)=(x+n\alpha\mod 1,y+nx+\frac{n(n-1)}{2}\alpha \mod 1)$. Now as translation is a homeomorphism we can assume w.l.o.g that $(0,0)\in A$. I aim to show that the $f_\alpha$ orbit of $(0,0)$ is dense for some fixed irrational $\alpha$ (we thus define $f:=f_\alpha$). Using the pigeonhole principle we can see that this orbit is "dense in each argument". This meaning that for any $t\in [0,1)$ and $\varepsilon>0$ there exists points $(u,v),(u',v')\in \mathcal O_f(0,0)$ such that $\min\{|u-t|,1-|u-t|\} 1<\varepsilon$ and $\min\{|v'-t|,1-|v'-t|\}<\varepsilon$. Sadly, this is not sufficient. I need to show that for any $(t,s)\in \mathbb T^2$ that there exists a point $(u,v)\in\mathcal O_f(0,0)$ such that $\min\{|u-t|,1-|u-t|\} 1<\varepsilon$ and $\min\{|v-s|,1-|v-t|\}<\varepsilon$.

As I have not used the opennes of $A$ in my above arguments I think that I may need to somehow consider the union of all neighbourhoods of all points in some neighbourhood. I am not sure how to do this, because at the moment I can't even see what prevents $A$ being some sort of open band spiralling around the torus, and not intercepting entire connected swathes of the torus. Any help would be much appreciated.

As an aside, am I correct in thinking that if the claim is true, then any such $A$ is necessarily the whole of $\mathbb T^2$?

I am aware that this question has been asked before here and here, but neither have been answered in a way that helps my understanding at all.

  • $\begingroup$ Do you know some Ergodic theory? I have a sketch of proof using ergodic theory, my approach is completely different than yours.. $\endgroup$ – Yanko Oct 7 '17 at 22:31
  • $\begingroup$ @yanko unfortunately not, we are only supposed to do that later in the course. $\endgroup$ – K.Power Oct 7 '17 at 22:32

I see nobody is answering this question, so I will give a proof using Ergodic theory Ideas (In fact I do not use Ergodic theorems only some Fourier analysis and measure theory).

Lemma: Let $g:\mathbb{T}^2\rightarrow \mathbb{C}$ be a function in $L^2(\mathbb{T}^2)$ such that $g\circ f = g$ then $g$ is constant (up to measure zero).

Proof: Let $g(x,y)=\sum_{n,m\in\mathbb{Z}}a_{n,m}x^ny^m$ be the fourier series for $g$ we have that $$g\circ f = \sum_{n,m\in\mathbb{Z}} a_{n,m}\alpha^ny^mx^{n+m}=\sum_{n,m\in\mathbb{Z}}a_{n-m,m}\alpha^{n-m} y^mx^n=g$$

some use the notation $e^{2\pi i n x}$ instead of $x^n$

As the Fourier series is unique, we have that for every $n,m\in\mathbb{N}$ that $a_{n-m,m}\alpha^{n-m}=a_{n,m}$. Hence for every $m\not=0$ we have that $|a_{n,m}|=|a_{n-m,m}|=...=|a_{n-km,m}|$ (because each is a constant multiplier of the other by an element in $\mathbb{T}$, it follows that the fourier series is convergence only if each of these terms is zero. (you can get that by using parseval identity which gives that $\|g\|^2 = \sum_{n,m}|a_{n,m}|^2$)

Now if $m=0$, then $a_{n-m,m}\alpha^{n-m}=a_{n,m}$ implies that $a_{n,0}\alpha^n=a_{n,0}$ hence $a_{n,0}=0$ for every $n\not=0$. It follows that $a_{n,m}=0$ whenever $n$ or $m$ is not zero, as the fourier series is unique we have that $g$ is a constant. This completes the proof of the Lemma.

Back to your question, the set $A$ is invariant under $f$ and is open, let $1_A$ be the characteristic function for $A$ (i.e gets $1$ is $x\in A$ and $0$ otherwise), this is obviously a function in $L^2(\mathbb{T}^2)$ and is satisfying $1_{A}\circ f = 1_{A}$ because $x\in A$ if and only if $f(x)\in A$. It follows that $1_A$ is constant up to measure zero, hence $A$ is of full measure in $\mathbb{T}^2$.

Now let $U$ be an open set in $\mathbb{T}^2$, be so that $U\cap A=\emptyset$ then $U$ must be of measure zero because $\mu(A\cup U)=\mu(A)+\mu(U)$. This is a contradiction to the fact that every open set is of positive measure, it follows that $A$ intersects any open set, hence $A$ is dense.

  • $\begingroup$ Thanks for the answer! It is a very elegant solution. I won't accept it just yet, because I'm interested if anyone else has a topological/combinatorial approach. $\endgroup$ – K.Power Oct 8 '17 at 11:57
  • $\begingroup$ sure no problem. $\endgroup$ – Yanko Oct 8 '17 at 15:45
  • $\begingroup$ You won't happen to know why any forward orbit is dense would you? My book claims that it is, but my approach runs into the same issue that I mentioned in the post, and I can't see how to adapt your approach, as it was necessary that $A$ had nonzero measure. $\endgroup$ – K.Power Oct 10 '17 at 15:15
  • $\begingroup$ Please could you expand on why $|a_{n,m}|=|a_{n-km,m}|$ for all $k\in \mathbb Z$? I can't see why that follows. $\endgroup$ – K.Power Oct 10 '17 at 21:02
  • 1
    $\begingroup$ because $a_{n,m}=\alpha^{n-m}a_{n-m,n}$ hence $|a_{n,m}|=|a_{n-m,n}|$, and $a_{n-m,n}=\alpha^{n-2m}a_{n-2m,n}$... and the rest follows by induction on $k$... $\endgroup$ – Yanko Oct 10 '17 at 22:03

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