Irrational rotations are rank 2 by intervals without spacers

Let $$\alpha$$ be an irrational number, and $$R_\alpha$$ be the rotation by $$\alpha$$, that is $$R_\alpha(x)=x+\alpha\bmod 1$$.

S. Ferenczi in his survey [Systems of finite rank. Colloq. Math. 73 (1997), no. 1, 35--65. MR1436950] states (Thm. 5): Every irrational rotation is of rank at most two by intervals (without spaces). For a proof he refers to a more general result (Thm. 7) stating that An ergodic exchange of $$s$$ intervals is of rank at most $$s$$ by intervals, without spacers.

As far as I understand, it means that I can find a nested family of intervals $$J_1\supset J_2\supset \ldots$$ such that $$J_n=F_n\cup G_n$$, where $$F_n=[a_n,b_n)$$ and $$G_n= [b_n,c_n)$$ are left-closed, right-open intervals on the unit circle (interpreted here as $$[0,1)$$), and for each $$n=1,2,\ldots$$ there are positive integers $$h_n^F$$ and $$h^G_n$$ such that $$[0,1)=\bigcup_{j=0}^{h_n^F-1}R_{\alpha}^j(F_n) \cup \bigcup_{j=0}^{h_n^G-1}R_{\alpha}^j(G_n)$$ is a disjoint union, and hence $$\{R_{\alpha}^j(F_n):j=0,1,\ldots,h_n^F-1\} \cup \{R_{\alpha}^j(G_n):j=0,1,\ldots,h_n^G-1\}$$ is a partition refining the previous one $$\{R_{\alpha}^j(F_{n-1}):j=0,1,\ldots,h_{n-1}^F-1\} \cup \{R_{\alpha}^j(G_{n-1}):j=0,1,\ldots,h_{n-1}^G-1\}.$$ So here are my questions:

1. Is the above interpretation correct? (Hope it is, but due to a certain vaguesness in defining rank by intervals'' I am not 100% sure.)
2. What can be said about $$h_n^F$$ and $$h^G_n$$ and their relation to each other as $$n\to\infty$$?
• I believe what you wrote is stronger than what Sebastien is claiming; but it's still true. I believe the $h_n^F$ and $h_n^G$ come from the partial quotients of the continued fraction expansion of $\alpha$. I will try and write down (or locate some more details) later. Jul 20, 2021 at 16:25
• Corinna Ulcigrai has some lecture notes on this exact topic from a course at ICTP: indico.ictp.it/event/8325/session/2/contribution/14/material/0/… Jul 20, 2021 at 20:31

As was pointed out, the answer is related to the continued fraction expansion. If $$\alpha=\frac{1}{c_1+\frac1{c_2}...}$$, we fix rational approximations $$\frac{p_k}{q_k}=\frac{1}{c_1+\frac1{c_2+...\frac1{c_k}}}$$. $$q_k+q_{k-1}$$ iterated preimages ( or images) of a point decompose the unit circle in the wished intervals that form two towers of heights $$q_k$$ and $$q_{k-1}$$. For example, $$T^i(0)$$, $$i=0..(q_k+q_{k-1}-1)$$, decompose unit circle into interval
• $$[T^i(0),T^{i+q_{k-1}}(0))$$, $$i=0..q_k-1$$,
• $$[T^{i+q_k}(0),T^{i}(0))$$, $$i=0..q_{k-1}-1$$.
The orientation switches with the parity of $$k$$. But it is likely not important for you. So, the heights are $$q_k$$ and $$q_{k-1}$$, possibly the best sequence related with $$\alpha$$ you can play with. Let me recall, that $$q_{k+1}=c_{k+1}q_k+q_{k-1}$$. The ratio can be as close to 1 as possible, along the subsequence, or unbouded in other case, or constant e.g. for Fibonnacci.