periodicity of an interval exchange transformation(IET)

Question

Let $T$ be an IET. That is, $T:[0,1] \rightarrow [0,1]$ is a piecewise orientation-preserving isometry. Let $D$ be the set of points whose entire forward iterates are well-defined. I have the following questions:

(1) If for every $x \in D$ there exists some $N$ such that $T^N x=x$ then is there some uniform $M$ such that $T^M x =x $ for all $x \in D$?

(2) Does above hold if and only if all the intervals defining $T$ have rational endpoints?

I would like to have the simplest proofs possible! Thanks.

Answer 1

(2) is false. Pick an irrational number $r \in (0,1/2)$, cut the interval $[0,1]$ at $r$ and $1-r$ into three subintervals, and define $T$ to fix the middle interval and swap the first and third intervals.

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(1) is indeed true. This is essentially a special case of the "Poincare recurrence theorem" but you ask for a simple proof. I'll give a detailed elementary proof; feel free to simplify it further by ignoring details.

Let me fix some notation. Let $I_1,…,I_K$ be the intervals on which $T$ is defined. Let $J_k = T(I_k)$, $k=1,…,K$. Let $\partial I$ denote the union of the endpoints of the $I_k$'s and let $\partial J$ be the union of the endpoints of the $J_k$'s. Note that $\partial I$, $\partial J$ have the same cardinality $K+1$.

If $x \in D$ then its forward orbit $x_i \in D$ is inductively defined for integers $i \in [0,\infty)$ by setting $x=x_0$ and $x_{i+1}=T(x_{i})$, and it follows that:

• $x_i \not\in \partial I$ for all $i \in [0,\infty)$
• $x_i \not\in \partial J$ for all $i \in [1,\infty)$ (because $x_{i-1} \not\in \partial I$).
• $x_i$ is periodic with some period $N$, that is $x_{i+N}=x_i$ (by hypothesis).
• $x_i$ can be bi-infinitely extended in a unique way so that $x_{i+1}=T(x_i)$ for all $i \in (-\infty,+\infty)$ (because $x_0=x_N \not\in \partial J$ has a unique pre-image $x_{-1} = x_{N-1}$, etc.)

Now list the points of $\partial J$ from left to right as $b_0,b_1,…,b_K$. Since the backward orbit of $b_k$ is not infinite (it cannot be extended backward even a single unit), it follows that its forward orbit is not infinite, and the only way that can happen is if its forward orbit stops at some point of $\partial I$. So its forward orbit forms a set of the form $$B_k = \{b_k=x_{k,0}, \,\, x_{k,1}, \,\, …,\,\, x_{k,i_k}=a_k\} $$ for some $a_k \in \partial I$. Note that $b_k \mapsto a_k$ is a well-defined function $\partial J \mapsto \partial I$, it is obviously one-to-one, and therefore it is onto.

The set $B_0 \cup \cdots \cup B_K$ cuts the interval $[0,1]$ into finitely many subintervals, and on the interior of each of these intervals the period is continuous and is therefore constant. Thus there are only finitely many periods.