Renormalizing an infinite rational IET

We study an interval exchange transformation of [0,1] formed by cutting the interval at the points 1/n and reversing the order of the intervals. We find that the transformation is periodic away from a Cantor set of Hausdorff dimension zero. On the Cantor set, the dynamics are nearly conjugate to the 2-adic odometer.


I
We study variations of the following interval exchange transformation: Consider the interval [0, 1) and cut it into subintervals of the form [1 − 1 k , 1 − 1 k+1 ) for integers k ≥ 1. We are interested in the dynamical system T 1 : [0, 1) → [0, 1) that reverses the order of the intervals, see Figure 1.
To study this map T 1 , we are also interested in similar maps T N on particular subintervals X N ⊂ [0, 1). For this, let N be a positive integer and let X N denote the half-open interval [0, 1 N ). Now consider the dynamical system T N : Reversing the order of these intervals can be described by applying a translation by 1 k + 1 k+1 − 1 N to each such interval. More formally, the map T N : X N → X N is de ned by Here denotes the greatest integer less than or equal to . e map T N is nearly a bijection: it is one-to-one and its image is the open interval (0, 1 N ). Following notation that is standard in the theory of dynamical systems, we use T j N (x) to indicate the point that is obtained by applying this map j times to the point x ∈ X N . A point x is called periodic under T N if there exists an integer j > 0 such that T j N (x) = x. We will show: eorem 1. For each positive integer N , there is a Cantor setΛ N ⊂ [0, 1 N ] of Hausdor dimension zero such that x is periodic under T N if and only if there exists an > 0 such that (x, x+ )∩Λ N = ∅. In particular, x is periodic if x ∈Λ N , so the vast majority of points are periodic under the map T N .
Let Λ N denote the set of points which are aperiodic (not periodic) under T N . e dynamics of the restriction of T N to Λ N turn out to be related to the 2-adic odometer which we now de ne. (1) k∈N α k 2 k with α k ∈ A for all k.
e 2-adic integers form an abelian group with the operation of addition allowing carrying of the form 1 · 2 k + 1 · 2 k = 1 · 2 k+1 . e addition-by-one map is given by adding 1·2 0 to a 2-adic integer. In terms of sequences, the addition-by-one map is the map f : is map is also called the 2-adic odometer. It is a homeomorphism when we equip A with the discrete topology and A N with the product topology. It is well known that f is minimal (all orbits are dense) and uniquely ergodic (there is only one invariant Borel probability measure) [Pyt02, §1.6.2].
Let N be the set of all 2-adic integers α ∈ A N which end in an in nite sequence of ones, i.e., Another characterization of this set is as the set of 2-adic integers α such that there exists an n > 0 for which f n (α) = 0, where 0 ∈ A N is the zero element de ned by 0 k = 0 for all k.
We show that the restriction of T N to the aperiodic set Λ N mirrors the action of the 2-adic odometer: eorem 2. For each positive integer N and T = T N , there is a continuous bijection h = h N from We give an explicit description of the aperiodic set Λ N and an explicit description of the map h in § 4. e least period of a periodic point x ∈ X N under T N is the smallest k > 0 such that T k N (x) = x. An interesting question this work leaves open is (see also Remark 7): estion 3. Which integers p > 0 appear as least periods of periodic points under T N ? For each such p what is the Lebesgue measure of the set of periodic points of least period p?
is map turns out to be semi-conjugate to the restriction of T N to Λ N as described in eorem 2.
Polygon and polytope exchange transformations (PETs) are higher dimensional analogs of IETs. ere are numerous examples in the literature of such maps admi ing an open dense set of periodic points but with interesting dynamics on the complimentary sets. See for example [AH13], [Goe00], [Goe03], [Hoo13], [Sch14], [Yi18]. is sort of behaviour is impossible for IETs formed by permuting nitely many intervals [MT02, eorem 6.6]. Part of the purpose of this article is to illustrate that this phenomenon arises in natural in nite IETs.
It is not the case that every in nite IET has a minimal component where the restriction of the map to this component is conjugate to an odometer. For example, there exists an in nite minimal IET of [0, 1] with positive entropy such that all lengths are 2-adic rationals (see [DHV,§4]) but odometers have entropy zero.

G
Interval exchanges. For us, an interval exchange transformation (IET) is a one-to-one piecewise translation T : X → X where X ⊂ R is a bounded interval. at is, we have a partition of X into countably many subintervals X = j∈J I j and a choice of translations τ j ∈ R for j ∈ J such that the map We call T rational if each τ j lies in Q. e following is a classical observation: Proposition 4. If T is a rational IET and τ j takes only nitely many values, then every orbit of T is periodic. More generally, if T : X → X is a rational IET and x ∈ X, then x has a periodic orbit unless Proof. Since each τ j ∈ Q and there are only nitely many translations τ j , there is a d ∈ Q such that τ j d ∈ Z for all j. Observe that T permutes the nitely many points in (x + dZ) ∩ X.
When we were working on this project, we wondered how common it is to have a dense set of periodic points for a rational IET which is in nite in the sense that {τ j } is in nite. Some experimental work of Anna Tao (undergraduate, CCNY) seems to suggest that this sort of periodicity is rare. However, we still wonder if there are natural classes of in nite rational IETs in which having a dense set of periodic points is typical.
At this point, there are a number of in nite rational IETs in the literature. Equation (3) gives an in nite rational IET without periodic orbits, and there are other examples corresponding to p-adic odometers and the Chacon middle third transformation [Dow05, §3] [LT16]. One way to get such a rational IET is from straight-line ows in directions of rational slope on an in nite-type translation surface all of whose saddle connections have holonomy in Q 2 . Symmetric surfaces of this form have been described in [Cha04], [Bow13] and [LT16].
Assuming r < +∞ on Y , we de ne the rst return mapT : Y → Y to be the map If T is an IET in the sense above, then so isT . Furthermore,T is rational whenever T is rational.

B
Here we prove some basic results about the maps T N : X N → X N de ned in the introduction. First we fully describe the return map to X N (N +1) .
Lemma 5. For any N , the rst return map of and in particular, no point has least period 1. We have that . We get periodic points as a consequence: , because T N reverses the order of intervals and we already know . Moreover, there are only nitely many distinct translations occurring on this interval, namely the translations associated to ( 1 k+1 , 1 k ] for values of k satisfying N + 1 ≤ k < N (N + 1). Proposition 4 then guarantees that every point in Remark 7. In the case N = 2, every point in the interval [ 1 6 , 1 3 ) has a periodic orbit under T 2 that has least period 10. In general, however, there may be points in [ 1 N (N +1) , 1 N +1 ) that do not have the same least period. For example, for N = 3, points may have least period either 920 or 930 under T 3 .
To describe more examples for larger N , we de ne another family of IETs for all m > n > 0 by breaking this interval into subintervals of the form [ 1 k+1 , 1 k ) for m > k ≥ n and reversing the order of the subintervals. Note that the restriction of In fact, there are many subintervals in X 1 of the form [ 1 m , 1 n ) that are preserved by a power of T 1 and where the rst return map is R m,n .
For example, the interval [ 1 42 , 1 7 ) is sent to itself by T 6 . e restriction of T 6 to [ In all these cases, every possible least period has to be a divisor of the least common multiple of the denominators n, n + 1, . . . , m. It would be interesting to classify for which pairs (m, n) every point in the interval [ 1 m , 1 n ) has the same least period under R m,n .

C
In this section we work through a general construction of a Cantor set. We will see later in the article that the setΛ N arises as such a Cantor set. e free monoid on the alphabet A = {0, 1} is the set A * of all nite sequences in A equipped with the binary operation of concatenation. An element w ∈ A is called a word and has a length |w| ∈ N representing the number of elements strung together. We write A k to denote the collection of all w ∈ A * with length k. e unique element ε ∈ A * with length zero is called the empty word and is the unique identity element of the monoid.
More formally, xy is de ned to be the nite sequence z of length |x| + |y| such that We use exponential notation for repeated concatenation so that w k denotes the concatenation of k copies of w. For example 0 9 denotes the word w where |w| = 9 and w i = 0 for i = 0, . . . , 8.
We now informally describe the Cantor sets that we are interested in. We use a variant of the standard construction of the Cantor ternary set in R, where the Cantor set is obtained by removing the middle third interval of [0, 1], then removing the middle third intervals of the remaining segments, and so on. Our Cantor set is similarly de ned as the intersection k≥0 C k and each C k is a nite union of closed intervals. e sets C k are de ned inductively starting with a single interval C 0 = [a 0 , b 0 ] and the set C k+1 is formed by removing middle intervals of equal length from each of the intervals making up C k . In contrary to the construction of the Cantor ternary set, the ratio of the lengths of intervals making up C k+1 to the lengths of intervals making up C k is not necessarily the same for all k. We denote these ratios by numbers s k .
We now give a more formal construction of our Cantor set. Fix an initial interval [a 0 , b 0 ] and a sequence s = {s k } k∈N of real numbers satisfying (5) 0 < s k ≤ 1 2 for all k ∈ N and lim sup s k < 1 2 .
F 2. e intervals I w0 and I w1 produced from I w when s |w| = 1 4 .
We inductively de ne an interval I w for each w ∈ A * . We de ne I ε = [a 0 , b 0 ]. Assuming I w is de ned to be [a, b], we de ne see Figure 2. Observe that if s |w| < 1 2 then I w0 ∪ I w1 is the interval I w with the middle open interval removed whose length is 1 − 2s |w| times the length of the whole interval. On the other hand, if s |w| = 1 2 the intervals I w0 and I w1 are formed by cu ing I w at the midpoint. In particular, the length of the interval I w only depends on |w| and the xed sequence s. e length is given by |w| where We de ne the Cantor set C = C s, [a 0 , b 0 ] by de ning It is a standard observation that as long as the sequence s satis es the conditions in (5) that C is a Cantor set: it is compact, totally disconnected and perfect. e following is a standard result on the Hausdor dimension of C (compare [Mat95, §4.10-11]).
Proposition 8. If lim k→∞ s k = 0 then the Hausdor dimension of C is zero.
Proof. Recall that the d-dimensional Hausdor content of C is there is a covering of C by balls of radius r i > 0 .
e Hausdor dimension of C is inf{d : C d H (C) = 0}. Fix d > 0. Now consider an integer k > 0 and consider that w∈A k I w contains C. Each interval in the union has length k and there are 2 k words in A k , so for this covering i r d i yields 2 k ( k /2) d . Observe from (7) that 2s d j and since s j → 0, this limit is zero. is shows that the d-dimensional Hausdor content is zero for any d > 0 and so the Hausdor dimension is zero.
We can now de ne the map h that was announced in eorem 2 to give a continuous bijection from A N N to the aperiodic set of T N . Recall that A N is the set of 2-adic integers, consisting of all sequences α = (α 0 , α 1 , . . .) with each α k ∈ A = {0, 1}. De ne the map h : A N → R depending on a sequence s as in (5) and on an interval [a 0 , b 0 ] by We will see in Lemma 10 that the function h is closely related to our construction of the Cantor set C(s, [a 0 , b 0 ] . We will also see in § 4 that h can be used to give a 2-adic in nite address to every point in the aperiodic set of T N , and that h describes a semi-conjugacy to the 2-adic odometer. But rst we observe that h can be used to describe the endpoints of the intervals I w used in the construction of the Cantor set C(s, [a 0 , b 0 ] . Proposition 9. For each w ∈ A * , we have I w = h(w0), h(w1) , where w0 and w1 denote the elements of A N whose rst |w| entries are given by w and whose remaining entries are all zeros or all ones respectively.
Proof. Fix w and let k = |w|. Observe that the lengths of I w and h(w0), h(w1) match since the length of I w is k and since lim j→∞ j = 0. It follows that checking I w = h(w0), h(w1) is equivalent to checking that the le endpoint of I w is h(w0) or checking that the right endpoint of I w is h(w1). We proceed by induction on the length of the word w. Observe that h(0) = a 0 and hence I ε = h(0), h(1) . Now suppose that I w = [a, b], h(w0) = a and h(w1) = b. We have to check that I w0 = h(w00), h(w01) and I w1 = h(w10), h(w11) . e statement for I w0 holds because the le endpoint of I w0 coincides with the le endpoint of I w by de nition in (6), and by hypothesis we have a = h(w0) = h(w00).
e statement for I w1 holds because the right endpoint of I w1 coincides with the right endpoint of I w by (6), and by hypothesis we have b = h(w1) = h(w11).
Lemma 10. e image h(A N ) is the Cantor set C = C(s, [a 0 , b 0 ] . Furthermore, h is one-to-one at all x ∈ C except at those x of the form x = h(w01) = h(w10) for some w ∈ A k with s k = 1 2 . e la er case happens only nitely o en and in this case, h is two-to-one at x.
Proof. First we show that for any α ∈ A N we have h(α) ∈ C. We must show h(α) ∈ C k for every k. Fix a k and set w = α 0 α 1 . . . α k−1 . en observe that Now suppose x ∈ C. We study the number of preimages of x under h. Observe that for each k ≥ 0 there exists a w ∈ A k such that x ∈ I w . We break into two cases. First suppose that for each k there exists a unique w ∈ A k such that x ∈ I w . Denote each such word by w k . Observe that w is an initial word of w k if and only if I w ⊃ I w k . It follows that for j < k, w j is the initial subword of w k of length j. en we can unambiguously de ne α ∈ A N by α i = w k i for some k > i. Now observe that h(α) ∈ I w k for each k. Since the length of I w k tends to zero as k → ∞, we see that h(α) = x. Finally, suppose β ∈ A N is distinct from α. en there is a k such that the initial word of length k of β di ers from w k . We see that h(β) ∈ I β 0 ...β k−1 but x is not in this interval, so h(β) = x. us h is one-to-one at x.
If we are not in the rst case, then there is a smallest k such that there are two words in A k for which x lies in both the corresponding intervals. From the argument about initial words in the previous paragraph, we see that because k is smallest, the two words have the same initial words. at is, the two words must have the form w0 and w1. us we have x ∈ I w0 ∩ I w1 . By (6) we see that I w0 ∩ I w1 = ∅ if and only if s k = 1 2 . And if this intersection is non-empty then the intersection just consists of the midpoint of I w . In this case, x is the right endpoint of I w0 and the le endpoint of I w1 . So by Proposition 9 we see x = h(w01) = h(w10). Furthermore, it can be deduced by an inductive application of (6) that for any j > 0, we have w ∈ A k+j and x ∈ I w if and only if w ∈ {w01 j , w10 j }. en if β ∈ A N {w01, w10}, there is some initial word w of β of length k + j such that w ∈ {w01 j , w10 j } and we have h(β) ∈ I w but x is not in this interval, so h(β) = x. is shows that h is two-to-one at x. Furthermore, there are only nitely many k > 0 such that s k = 1 2 because of (5), so this case only appears nitely o en. Recall from the introduction that N = {w1 : w ∈ A * } ⊂ A N . is is an important set for us, and we prove the following.

Proposition 11.
(1) e restriction of h to A N N is injective.
(2) e Cantor set C is the closure of h(A N N ). ( Proof. Statement (1) is a consequence of Lemma 10 since h is one-to-one at all points except that it is possible that x = h(w01) = h(w10). But we have w01 ∈ N . Since C = h(A N ) and C is closed by construction, to prove statement (2) we just need to nd for each α ∈ N a sequence α k ∈ A N N such that h(α k ) converges to h(α). For each k, let w k = α 0 . . . α k−1 ∈ A k and de ne α k = w k 0. en both h(α) and h(α k ) lie in I w k for each k and the length of I w k tends to zero so we see that h(α) = lim h(α k ) as desired.
Finally consider statement (3). First suppose that α ∈ A N N . en there exists a sequence k j → ∞ such that α k j = 0. For j ≥ 0, de ne β j ∈ A so that the sequence agrees with α except that β j k j = 1. Observe that by de nition of h, we have h(β j ) > h(α) and lim h(β j ) = h(α). is proves that h(α), h(α) + intersects C = h(A N ) for all > 0.
On the other hand, suppose that x ∈ C h(A N N ). We need to show that there exists an > 0 such that (x, x+ )∩C = ∅. If x = h(1) then this is clearly true since h(1) is the right endpoint of I ε by Lemma 13 and C ⊂ I ε . Otherwise there exists a w ∈ A * such that x = h(w01). Furthermore, h is one-to-one at x since otherwise we would have x = h(w10) as well which would contradict that x ∈ h(A N N ). Se ing k = |w| we see therefore that s k < 1 2 by Lemma 10. Since x = h(w01) we see that x is the right endpoint of I w0 . Let [a, b] = I w . en we see in the notation of (6) that x = a + s k (b − a) and the removed interval gives an interval of positive length not intersecting C as required.
4. T Fix a positive integer N and extend it to a sequence inductively by de ning N 0 = N and N k+1 = N k (1 + N k ) for all k ≥ 0.
By an inductive application of Lemma 5 we see: Corollary 12. For each k, the rst return map of T N to X N k is T N k .
We use this information to de ne the intervals I w as before.
Recall the de nition of the 2-adic odometer f : A N → A N in (2). We want to extend this addition-by-one map to A * = ∪ k≥0 A k . At words of the form 1 k for some k ≥ 0, we leave the map f unde ned. We de ne f : For this section, if I is a closed interval, we write I to denote I with its right endpoint removed. e key to the results announced in the introduction is the following: Lemma 13. For any w ∈ A k {1 k }, the restriction of T N to I w is a translation carrying I w to I f (w) . If w = 1 j 0 for some j ≥ 0 then this is a translation by − 1 N 0 + 1 N j + 1 1+N j . Proof. First we prove this for the special case when w = 1 j 0. By Proposition 9, the endpoints of I w are us if x ∈ I w then we see by de nition of T N that e word f (w) is a string of j zeros followed by a one. us, we see that the endpoints of I f (w) are Observe that these new endpoints di er from the endpoints of I w found earlier by a translation by − 1 N 0 + 1 N j + 1 1+N j which is exactly how T N acts. is proves the second statement of the lemma. Now suppose that w ∈ A * is a word which has at least one zero. As in (9), we can then de ne j = min {i : w i = 0}. Hence w = w 0 . . . w j is a word consisting of j ones followed by a zero, so the previous paragraph implies that T N restricted to I w is a translation by − 1 N 0 + 1 N j + 1 1+N j . Recall that I w ⊂ I w which implies that the restriction of T N to I w also acts by the same translation. e intervals I w and I f (w ) have the same length and their le endpoints di er by N N ). en either x is not contained in the closed set C or we can apply statement (3) of Proposition 11. In both cases, there is an such that (x, x + ) ∩ C = ∅. Since X N = I ε , the interval (x, x + ) must lie in one of the gaps of the Cantor set, i.e., there is a w ∈ A k such that (x, x + ) ⊂ I w (I w0 ∪ I w1 ). It follows that x ∈ I w (I w0 ∪I w1 ). en I 0 k has the same length as I w and so we have I w = τ +I 0 k for some τ ∈ R acting by translation. Set x 0 = x − τ ∈ I 0 k . Observe that there exists an m ≥ 0 such that f m (0 k ) = w, where f is as in (9). By Lemma 13, we know that T m N restricted to I 0 k is a translation carrying this interval I 0 k to I w . us T m N (x 0 ) = x. It also follows that T m N (I 0 k+1 ) = I w0 and T m N (I 0 k 1 ) = I w1 and in particular x 0 ∈ I 0 k+1 ∪ I 0 k 1 . Now observe that I 0 k = X N k = [0, 1/N k ) by Proposition 9, and by Corollary 12 the rst return map of T N to this interval is T N k . Since I 0 k+1 = [0, 1/N k+1 ) and I 0 k 1 = [1/N k − 1/N k+1 , 1/N k ), Corollary 6 tells us that x 0 is periodic under T N k and therefore also periodic under T N . Since x = T m N (x 0 ), x is also periodic. eorem 15. For any α ∈ A N N , we have T N • h(α) = h • f (α). In particular, no point in h(A N N ) has a periodic orbit.
Proof. Fix α ∈ A N N . De ne j = min ({k : α k = 0} ∪ {+∞}) as in (2). Since α ∈ N we have j < +∞. e initial word of α then has the form 1 j 0 and the initial word of f (α) is 0 j 1. e rest of the sequence f (α) agrees with α. erefore we have Let x = h(α). en x ∈ I 1 j 0 and T N acts as a translation by − 1 N 0 + 1 N j + 1 1+N j on I 1 j 0 ; see Lemma 13. us by equation (11) we see that h • f (α) = T N • h(α). Since f has no periodic orbits and h restricted to A N N is injective, we see that T N has no periodic orbits in h(A N N ).
We nish by proving the rst two theorems of our article.
Proof of eorems 1 and 2. Recall that Λ N denoted the set of points in X N with aperiodic orbits under T N . Together eorem 14 and eorem 15 guarantee that Λ N = h(A N N ). eorem 15 then directly implies eorem 2. Statement (2) in Proposition 11 shows that the closureΛ N is the Cantor set C. Further by statement (3) of Proposition 11 we see that Λ N has the form claimed in eorem 1. e fact that Λ N has Hausdor dimension zero follows from Proposition 8.