(Underlying job: I am trying to construct an adic representation of a rotation.)

The question involves an iterative construction. At step $n$, one constructs a partition $P_n$ of $(0,1)$ and a map $\beta_n \colon (0,1) \to (0,1)$ (it could be clearer to use some indices: $\beta_n \colon (0,1)_n \to (0,1)_{n+1}$). Roughly speaking this construction will allow to define a sequence of labels $P_n(x_n)$ of a point $x \in (0,1)$, and the problem is to retrieve $x$ from this sequence.

(n=1) Let $T_1$ be a rotation with irrational step $\theta \in (0,\frac12)$ and define the interval $J=(1-\theta,1)$. The integer $a= \lfloor \frac{1}{\theta} \rfloor -1 \geq 1$ is the smallest integer such that $(0,1)$ is covered by $J$, $T_1(J)$, $\ldots$, $T_1^{a+1}(J)$.

One has the periodic approximation of $T_1$ shown by the following tower decomposition with base $B=(a\theta, 1) = J \cup T_1^{a+1}(J)$:

enter image description here

Denote by $P_1$ the partition of $(0,1)$ whose blocks can be seen in the picture: $$ P_1 = \bigl\{J, T_1(J), \ldots, T_1^a(J), (a\theta, 1-\theta)\bigr\}. $$

The orbits of this periodic approximation define an equivalence relation: each point in $x\in J$ is glued with $T_1(x)$, $\ldots$, $T_1^a(x)$, and the class of a point $x \in (a\theta, 1-\theta)$ is the singleton $\{x\}$. We take the base $B$ as the set of representatives, and for every $x \in (0,1)$, we denote by $d_1(x) \in B$ the representative of the class of $x$.

Now, the transformation induced by $T_1$ on $B$ is isomorphic to a rotation $T_2$ with irrational step $$ \theta'=\frac{\theta}{1-a\theta}=\frac{1}{1+G(\theta)} > \frac{1}{2} $$ where $G$ is the Gauss function. The isomorphism $g_1\colon B \to (0,1)$ between the induced transformation and $T_2$ is the natural affine map. The map $\boxed{\beta_1=g_1 \circ d_1}$ sends the action space of $T_1$ to the action space of $T_2$.

(n=2) We proceed as before with $T_2$ and $J = (0, 1-\theta')$. This is the similar construction for the case of a step ($\theta'$) in $(\frac12, 1)$. This time, setting $\phi=1-\theta'$, one has a tower decomposition with base $B = (0, 1-a\phi)$, where $a = \lfloor \frac{1}{1-\theta} \rfloor -1$.

enter image description here

And we have the partition of $(0,1)$: $$ P_2 = \bigl\{J, T_2(J), \ldots, T_2^a(J), (\phi, 1-a\phi)\bigr\}, $$ and the representatives $d_2(x) \in B$.

Now, the transformation induced by $T_2$ on $B$ is isomorphic to the rotation $T_3$ with irrational step $$ \frac{1-(a+1)\phi}{1-a\phi}=\frac{G(\phi)}{1+G(\phi)} < \frac{1}{2}, $$ we denote by $g_2$ the isomorphism (affine map) and we set $\boxed{\beta_2=g_2 \circ d_2}$.

(n) We can continue so on. We get the partitions $P_n$ and the maps $\beta_n$.

My question is the following one. Denote $x_1=x$, $x_2= \beta_1(x_1)$, $\ldots$, $x_{n+1}=\beta_n(x_n)$, $\ldots$. Do we know (almost) every $x \in (0,1)$ if we know the block $P_n(x_n)$ of the partition $P_n$ to which $x_n$ belongs for every $n$ ?

  • $\begingroup$ I believe this is possibly proved in this paper (I don't have the English version). $\endgroup$ Dec 28 '15 at 22:31
  • $\begingroup$ It's almost midnight here. I'm afraid it is easy actually. Because by combining the first two towers one gets a new tower decomposition for the first rotation. $x_1$ and $x_2$ give the stack to which $x$ belong. And so on. The lengths of the stacks become smaller and smaller, they go to $0$. I'll continue tomorrow... $\endgroup$ Dec 28 '15 at 22:47

Yes, this is how we got it - but then it was left out because it was too bulky if my memory serves me. A short English version can be found in Section 3 of my survey http://www.maths.manchester.ac.uk/~nikita/ad.pdf

The following papers may be also useful: http://link.springer.com/article/10.1007/BF02788235 http://citeseerx.ist.psu.edu/viewdoc/download?doi=

  • $\begingroup$ Thank you. I didn't know these papers except yours. I also wrote that on my blog. $\endgroup$ Feb 27 '16 at 22:32

I get the idea. Apply the second cut-and-stack to the first tower decomposition. This is shown on this picture (where I denote $x_n$ instead of $P_n(x_n)$):

enter image description here

An important point: $(3\theta, 1-\theta)$ goes to $(0, \phi)$ by the affine map $g_1$.

The value of $P_2(x_2)$ corresponds to one of the black pieces surrounded by a red line. If it is $1$, one necessarily has $P_1(x_1)=0$. Otherwise, one necessarily has $P_1(x_1)\neq 0$, and the value of $P_1(x_1)$ corresponds to one of the intervals in the black piece selected by $P_2(x_2)$.

Thus, knowing $P_1(x_1)$, one knows an interval $I_1(x)$ to which $x$ belongs. Then, knowing in addition $P_2(x_2)$, one knows an interval $I_2(x) \subset I_1(x)$ to which $x$ belongs. It is possible that $I_2(x)=I_1(x)$ (when $x \in (3\theta, 1-\theta)$) but in this case one has a strict inclusion $I_3(x) \subset I_2(x)$ when we continue at step $n=3$. Continuing so on, one gets intervals $I_n(x) \ni x$ whose lengths go to $0$, and $I_n(x)$ is given by $P_1(x_1)$, $\ldots$, $P_n(x_n)$. Finally one knows $x$ if one knows all the $P_n(x_n)$.

  • $\begingroup$ @NikitaSidorov, Am I right to suppose this construction is well-known to you ? Isn't it the arithmetic expansion of $x$ based on the continued fraction of $\theta$ (the integers $a$ are the digits of the continued fraction expansion) and if so, is it the idea of the proof ? $\endgroup$ Dec 29 '15 at 12:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.