# Iterative renormalizations of a rotation

(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 $B=(a\theta, 1) = J \cup T_1^{a+1}(J)$:

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 $a = \lfloor \frac{1}{1-\theta} \rfloor -1$.

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 $n$ ?

• I believe this is possibly proved in this paper (I don't have the English version). Dec 28 '15 at 22:31
• 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... 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

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 $P_n(x_n)$):
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 $P_n(x_n)$.
• @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 ? Dec 29 '15 at 12:05