LANGUAGES OF k-INTERVAL EXCHANGE TRANSFORMATIONS
S´EBASTIEN FERENCZI AND LUCA Q. ZAMBONI
Abstract. This paper gives a complete characterization of those sequences of subword complexity (k−1)n+ 1 which are natural codings of orbits ofk-interval exchange transfor- mations, thereby answering an old question of Rauzy.
Interval exchange transformations were originally introduced by Oseledec [17], following an idea of Arnold [1], see also [9]; an exchange of k intervals, denoted throughout this paper by I, is given by a probability vector ofk lengths (α1, . . . , αk) together with two permutations (π0, π1) onk letters. The unit interval is partitioned into k subintervals of lengthsα1, . . . , αk which are ordered according toπ−10 and then rearranged byI according toπ1−1. It was Rauzy [18] who first saw interval exchange transformations as a possible framework for generalizing the well-known interaction between circle rotations on one hand, and Sturmian sequences [16] on the other; thus he asked the question “how to describe the symbolic sequences which are natural codings of k-interval exchange transformations”, a natural coding meaning a sequence (xn) taking the value 1, 2, ... k when the n-th iterate of some point x lies in the first, second, ... k-th interval. This question was of huge interest to word combinatorists who are attempting to classify the systems of low complexity: they tried to generalize the Sturmian sequences by defining several classes of sequences and systems, see [3], but, up to now, a complete correspondance between sequences and systems has not been found in any of these classes - except the class of interval exchanges. Several partial answers to Rauzy’s question were given, for k = 3 in [6][19], and for allk in [12].
A complete answer to Rauzy’s question was given by the present authors plus Holton [7]
in the particular case when k = 3, π0 is the identity and π1 the symmetric permutation 17→3,27→2,37→1, and the discontinuities βi of I−1 and γi of I satisfy β1 < γ1 < β2 < γ2. In the present paper, we give a complete answer for any k, any couple of permutations, and any position of the discontinuities, extending and completing the work of [7]. Our main theorem provides a combinatorial characterization of those infinite sequences which arise as the natural codings of the orbits of points under an exchange of k intervals:
Theorem 1. A minimal sequence u is the natural coding of a k-interval exchange transfor- mation, defined by permutations (π0, π1) such that π0−1({1, ...j}) 6= π1−1({1, ...j}) for every 1 ≤ j ≤ k−1, and satisfying the i.d.o.c. condition, if and only if the words of length one occurring in u are L1 = {1, . . . , k} and it satisfies the following conditions (see Figure 2 below):
• ifw is any word occurring in u, A(w), resp. D(w)), the set of all letters x such that xw, resp. wx, occurs in u, is an interval for the order of π1, resp. π0,
Date: April 12, 2008.
1991Mathematics Subject Classification. Primary 37B10; Secondary 68R15.
1
• ifx∈A(w), y∈A(w), x≤y for the order of π1, z ∈D(xw), t ∈D(yw), then z ≤t for the order of π0,
• if x ∈ A(w) and y ∈ A(w) are consecutive in the order of π1, D(xw)∩D(yw) is a singleton.
This theorem is also conveniently expressed forlanguagesrather than sequences, and using bispcial words, see Theorem 2 below; essentially, it says that the straightforward necessary conditions on the extensions (backwards and forwards) of words arising from the minimality and the dimension 1 of the system are indeed sufficient. Its conditions have other possible expressions, in particular they imply that the complexity function is (k−1)n+ 1; they are simpler in the important particular case of symmetric interval exchanges studied in section 3.
As in the Sturmian case, the nontrivial part in the proof of our theorems is to retrieve the lengths of the intervals from the given symbolic sequence. While in [8], for symmetric interval exchanges, we use the techniques of Rauzy (an approximation algorithm or an induc- tion process), here in the general case the lengths are given by an invariant measure, which generalizes the use of frequencies in the original paper of Morse and Hedlund on Sturmian sequences.
During the submission process, the authors learned that another answer to Rauzy’s ques- tion had just been given independently by Belov and Chernyat’ev [2]; their main result is close to Theorem 1, though their methods look quite different.
Acknowledgments: the authors wish to thank T. Monteil for drawing the pictures.
1. Definitions 1.1. Word combinatorics.
Definition 1. Let A be a finite set called the alphabet. By a language L over A we mean always a factorial extendable language: a language is a collection of sets (Ln)n≥0 where the only element of L0 is the empty word and each Ln, n ≥ 1 consists of words of the form a1a2· · ·an withai ∈ Aand such that for eachv ∈Ln there existsa, b∈A with av, vb∈Ln+1 and for all v ∈Ln+1 if v =au=u′b with a, b∈A then u, u′ ∈Ln.
The complexity function pL : IN→IN is defined bypL(n) = #(Ln).
A word v = v1...vr occurs at place i in a word w = w1...ws or a sequence v = w1w2... if v1 =wi, ...vr =wi+r−1. We say that v is a factor of w.
A language Lis minimal if for each v ∈L there existsn such thatv is a factor of each word w∈Ln.
The language of an infinite sequence u is the language where Ln is made with all the factors of u of length n.
Definition 2. For a wordw in L, we call arrival set of w and denote by A(w) the set of all letters x such that xw is in L, and call departure set ofw and denote by D(w) the set of all letters x such that wx is in L.
A word w in L is called right special, resp. left special if #D(w)>1, resp. #A(w)>1. If w∈L is both right special and left special, thenw is called bispecial. If#L1 >1, the empty word ε is bispecial, with A(ε) =D(ε) =L1.
Definition 3. Given a word w =a1a2· · ·an with ai ∈ A, let w¯ denote the retrograde word of w, that is w=anan−1· · ·a1.
The retrograde language L is made of {w, w∈L}.
A language L is symmetric (or invariant by mirror image) if L=L.
Definition 4. The symbolic dynamical system associated to a minimal language L is the one-sided shiftS(x0x1x2...) =x1x2...on the subsetXLofAINmade with the infinite sequences such that for every r, s, xr...xr+s−1 is in Ls.
For a word w=w0...wt, the cylinder [w] is the set {x∈ X;x0 =w0, ...xt=wt}.
1.2. Permutations.
Definition 5. For a permutation π of {1, ...k}, we define the π-order by a <π b whenever π(a)< π(b).
A π-interval is a nonempty set of consecutive integers in the π-order; for π =Id or π =σ it may be denoted by [x, y].
1.3. Interval exchanges. A k-interval exchange transformationI is given by a probability vector
(α1, α2, . . . , αk),
0< αi <1, together with two permutations (π0, π1) of{1,2, . . . , k}.The unit interval [0,1[ is partitioned into k sub-intervals of lengths α1, α2, . . . , αk which are ordered according to the permutation π0−1 and then rearranged according to the permutation π1−1 (Figure 1). Note that this definition is due to Kerckhoff [11] and was revived recently in [13], we use it as it is well adapted to our results; in the classical definition, there is only one permutation π, the intervals are ordered according to π0−1 = Id and rearranged according to π1−1 =π. For any permutation χ of {1,2, . . . , k}, if we replace π0 by π0 ◦χ, π1 by π1 ◦χ, and the αi by αχ(i) in the Kerkhoff definition of I, we get the same transformation with a different coding.
0 0
1 1 [π0−1(1)] [π−01(2)] [π0−1(k)]
(π−11(1)) (π−11(2)) (π−11(k))
γ1 γ2 γk−1
β1 β2 βk−1
. . . . . .
Figure 1. I sends the interval labeled [x] onto the interval labeled (x)
Definition 6. Ak-interval exchange transformationIwith probability vector(α1, α2, . . . , αk), and permutations (π0, π1) is defined by
(1) Ix=x+ X
π1(j)<π1(i)
αj − X
π0(j)<π0(i)
αj.
when x is in the interval
X
π0(j)<π0(i)
αj, X
π0(j)≤π0(i)
αj
,
and this interval is denoted by [a] where a=π0−1i.
We denote by βi, 1 ≤ i ≤k−1, the i-th discontinuity of I−1, namely βi =P
π1(j)≤π1(i)αj, while γi is the i-th discontinuity of I, namely γi = P
π0(j)≤π0(i)αj. Then [π−10 (i)] is the interval [γi−1, γi[ if 2≤i≤k−1, while [π0−1(1)] = [0, γ1[ and [π−10 (k)] = [γk−1,1[.
Definition 7. I satisfies the infinite distinct orbit condition (or i.d.o.c. for short) of Keane [10] if the k−1 negative trajectories {I−n(γi)}n≥0 ,1≤i≤k−1, of the discontinuities of I are infinite disjoint sets.
The i.d.o.c. condition forIor (αi,1≤i≤k) is implied, under the conditionπ0−1({1, ...j})6=
π1−1({1, ...j}) for every 1 ≤j ≤k−1, by the total irrationality, where the only rational re- lation satisfied by αi,1≤i≤k−1, is Pk
i=1αi = 1. The i.d.o.c. condition implies that I is minimal (every orbit is dense).
Definition 8. For every point x in [0,1[, we define an infinite sequence (xn)n∈IN by putting xn=i if Inx falls into [i], 1≤i≤k; we call it the trajectory of x.
If I is minimal, all trajectories have the same language, which we call the language of I, and denote by L(I).
Definition 9. A symmetric k-interval exchange transformation is a k-interval exchange transformation I with probability vector (α1, α2, . . . , αk), and permutations (Id, σ)where Id is the identity and σj =k+ 1−j.
Let I be a symmetric k-interval exchange transformation; it has alternate discontinuities if βi < γi for each 1≤i≤k−1 and γi < βi+1 for each 1≤i≤k−2.
2. Characterization theorem
We give a slightly different form of Theorem 1 (we replace the sequence by the language, and check our properties only on bispecial words).
Theorem 2. A languageLis the language of ak-interval exchange transformationI, defined by permutations (π0, π1) such that π−10 ({1, ...j}) 6= π1−1({1, ...j}) for every 1 ≤ j ≤ k −1, and satisfying the i.d.o.c. condition, if and only if L satisfies
(H0) L1 ={1, . . . , k}, (H1) L is minimal,
(H2) if w is a bispecial word, A(w) is a π1-interval,
(H3) if w is a bispecial word and x∈A(w), D(xw) is a π0-interval,
(H4) if x∈A(w), y∈A(w), x≤π1 y, z ∈D(xw), t∈D(yw), then z ≤π0 t,
(H5) if x ∈ A(w) and y ∈ A(w) are consecutive in the π1 order, D(xw)∩ D(yw) is a singleton.
Note that (H0) depends onk, and (H2) to (H5) on π0 and π1, but these will generally be clear from the context (even if they are not, there is only a finite set to try).
We shall now prove Theorem 2.
Proposition 3.For any interval exchange transformationI defined by permutations(π0, π1), the language L(I) satisfies (H0), (H2), (H3), (H4).
If I is minimal, L(I) satisfies (H1); if I satisfies the i.d.o.c. condition, L(I) satisfies (H5).
Proof
(H0) is straightforward.
When w is bispecial, the interval [w] is cut into subintervals [wbi], 1 ≤ i ≤ s, where D(w) = {b1 <π0 b2 <π0 ...bs}and [wbj] is to the right of [wbi] whenever bi <π0 bj; [w] is also cut into subintervals I[aiw], 1≤i≤r, where A(w) ={a1 <π1 a2 <π1 ...ar}and I[ajw] is to the right of I[aiw] whenever ai <π1 aj; as these are all semi-open intervals, we have (H2) to (H4); (H4) implies also a weaker version of (H5): ifx∈A(w) andy ∈A(w) are consecutive in the π1 order,D(xw)∩D(yw) is either empty or a singleton.
Minimality implies (H1), and the i.d.o.c. condition ensures that the discontinuities sep- arating the [wb] are not the same as the discontinuities separating the I[aw], thus we have
(H5).
[wb1] [wb2] [wb3]
. . . . . .
[w]
I[a1w] I[a2w] I[a3w]
Figure 2. Bispecial intervals in the general case
The following proposition is the essential part of the proof; it relies on the fact that, for an interval exchange, the measure of the cylinders given by letters determines everything, including the language, and the measure of the cylinders associated to words, while the combinatorial conditions given are exactly the ones which allow to mimick this process for the symbolic system.
Proposition 4. Let L be a language satisfying (H0) to (H5), and let L′ be a language satisfying(H0), (H2), (H3), (H4), for the same permutations π0 andπ1; if there exist shift- invariant probabilities on the associated symbolic dynamical systems,µonXL andµ′ onXL′, such that µ′[i] =µ[i] for every i∈ {1, . . . , k}, then L=L′.
Proof
By minimality, µ[w]>0 for any w∈L, as for anyN there is at least one wordw1 of length N withµ[w1]>0, and, aswis a factor ofw1ifN is large enough, [w1] is included in∪Ni=0Si[w].
We prove inductively on nthat w∈Ln iffw∈L′n, and for these wordsµ[w] = µ′[w]. This is true for n = 1 by hypothesis. We denote by AL, AL′,DL, DL′ the arrival and departure sets in L and L′.
Let the hypothesis be proved for n, and take a word w in L of lengthn−1 (possibly the empty one ifn= 1). Because of the induction hypothesis, AL(w) =AL′(w),DL(w) =DL′(w) (as they depend only on Ln and L′n).
Ifwis not right special,DL(w) is made with a single lettera, and every wordxw is always followed bya, thusxwais in Lif and only ifxw is inL, withµ[xwa] =µ[xw]; because of our induction hypothesis, all this remains true with “prime” signs added. Thus our hypothesis
is carried over to all awb, for a and b letters. And similarly if w is not left special.
Suppose now that w is bispecial in L (hence also in L′); note that, by looking at the possible xwy, from (H2) to (H4) we get also
(H4R) if x∈D(w),y∈D(w), x≤π0 y, z ∈A(wx),t∈ A(wy), then z ≤π1 t.
LetAL(w) = [a1, ...ap], in the π1-order, andDL(w) = [b1, ...bq] (see Figure 2 above) in the π0-order.
We show now that if (i, j)6= (p, q) then
i
X
r=1
µ[arw]6=
j
X
s=1
µ[wbs];
the equality should not occur because of the i.d.o.c. condition, and indeed it cannot occur under hypotheses (H2) to (H5): if i=p, this implies all [wbs] are of µ-measure 0 for s > j, hence wbs is not in L for s > j, hence j = q; if i < p, then both [aiwbj+1] and [ai+1wbj] are of µ-measure 0, hence neither aiwbj+1 nor ai+1wbj is in L and DL(aiw)∩DL(ai+1w) is empty, which contradicts (H5).
Note that this result will force L′ to satisfy also (H5), and that it precludes the existence of weak bispecial factors (see [5] for a definition).
Starting from the left of A(w), DL(a1w) must contain b1 by (H4). Then a1wb1 is in L, and
• if µ[a1w] < µ[wb1], then DL(a1w) is reduced to the element b1 and thus µ[a1wb1] = µ[a1w],
• ifµ[a1w]> µ[wb1], thenDL(a1w) contains also another element thanb1, thus contains b2 by (H3); hence A(wb2) contains a1, hence by (H4R) A(wb1) is reduced to the elementa1 and µ[a1wb1] = µ[wb1].
Thus µ[a1wb1] = min(µ[a1w], µ[wb1]) > 0; similarly, because of (H3), (H4) and (H4R), a1wb1 is always inL′, and, using our induction hypothesis, we get
µ′[a1wb1] = min(µ′[a1w], µ′[wb1]) = min(µ[a1w], µ[wb1]) = µ[a1wb1].
Suppose we have proved that arwbs is in L iff it is in L′, and µ[arwbs] = µ′[arwbs], for every r ≤i, and s≤j; suppose that aiwbj is in L and L′, with i < por j < q.
Suppose first thatPi
r=1µ[arw]>Pj
s=1µ[wbs]; then∪ir=1D(arw) contains strictly{b1, ...bj}, thus by (H3) D(aiw) contains bj+1, thusaiwbj+1 is in L while ai+1wbj is not in L. And we have again two cases:
• ifPi
r=1µ[arw]>Pj+1
s=1µ[wbs], then by (H4R)A(wbj+1) is reduced to the element ai, thusµ[aiwbj+1] =µ[wbj+1],
• if Pi
r=1µ[arw]<Pj+1
s=1µ[wbs], then by (H4) D(aiw) is reduced to the two elements bj and bj+1, and µ[aiwbj+1] =Pi
r=1µ[arw]−Pj
s=1µ[wbs].
But then Pi
r=1µ′[arw]>Pj
s=1µ′[wbs], and the same analysis applies toL′, thus we get the same conclusions with “prime” signs added; as all estimates depend only on properties of words of length n, we use our induction hypothesis to get µ′[aiwbj+1] =µ[aiwbj+1].
In the opposite case where Pi
r=1µ[arw] < Pj
s=1µ[wbs], then aiwbj+1 is not in L while ai+1wbj is in L, and a similar reasoning applies.
Thus our induction hypothesis is again carried over to all awb, for a and b letters, and is
now proved for all words of length n+ 1
Proof of Theorem 2
We prove the “if” part, the “only if” part having been disposed of by Proposition 3. Let L be a language satisfying (H0) to (H5). Let µ be an invariant probability measure on the symbolic system (XL, S) associated to L; we can build one as in [4]. Let I be the k-interval exchange transformation defined by the vector (µ[1], ..., µ[k]) and the permutations (π0, π1), L′ = L(I), µ′ be the shift-invariant measure on XL′ defined by the Lebesgue measure on [0,1[. ThenL=L′ by Proposition 3 and Proposition 4, and this implies thatI satisfies the i.d.o.c. condition, as otherwise we can find a word w and lettersa,b such that DL(aw) and DL(bw) are two adjacentπ0-intervals with an empty intersection, which contradicts (H5).
Given a language L, there may be several k-interval exchange transformations such that L(I) = L. The solution I is unique if and only if I is uniquely ergodic (it has a unique invariant probability measure); a famous result of Veech [20] and Masur [14] states that the set of (α1, . . . , αk) in IR+k for which I defined by the vector (α1+...αα1 k, . . . ,α αk
1+...αk) is uniquely ergodic has full Lebesgue measure.
We have stated a minimal number of properties characterizing L, but these have more consequences:
Proposition 5. A language L satisfying (H0) to (H5) satisfies also (H6) the complexity ofL is p(n) = (k−1)n+ 1 for all n,
(H7) if k ≥ 2, L is aperiodic: L does not consist of all the factors of an infinite word of the form ww...
If L satisfies(H1) to (H5) so does its retrograde language L.
IfL satisfies(H1)to(H5), then the conclusions of (H2)to (H5)are satisfied by all words w.
Proof
Letw be a bispecial word (possibly empty); from (H2) to (H5) we get
k
X
x=1
(#D(xw)−1) = #D(w)−1;
this allows us to compute the complexity [5], and, as (H0) implies p(1) = k, we get (H6).
This implies (H7) through a famous result of Morse and Hedlund [15].
The assertion about Lis straightforward.
If L satisfies (H1) to (H5), the conclusions of (H2) to (H5) are trivially satisfied if w is neither left nor right special; if w is left but not right special, they reduce to the fact that A(w) is a π1-interval, and this is true as A(w) = A(w′) where w′ is the shortest bispecial word beginning byw, and this exists becauseLis minimal and aperiodic (#L1 ≥2 otherwise there is no left special word); if w is right but not left special, they reduce to the fact that D(w) is a π0-interval, and this is true as D(w) =D(w”) where w” is the shortest bispecial
word ending by w, and this exists because L is minimal and aperiodic.
The last result of Proposition 5 allows us to deduce Theorem 1 from Theorem 2.
The sets D(x), 1 ≤x≤k, determine the language L2; they satisfy (H2) to (H5) applied to the empty bispecial wordw, withA(w) = D(w) ={1, . . . , k}intoπ0-intervals, but not all sets satisfying these conditions are possible. For example, for k = 3, π0 = Id, π1 = σ (the symmetric permutation, see Definition 9), there are four possibilities for {D(3), D(2), D(1)}
which can actually occur for interval exchanges, namely
{(1),(1),(1,2,3)},{(1),(1,2),(2,3)},{(1,2),(2,3),(3)},{(1,2,3),(3),(3)}.
The conditions above are also satisfied by{(1,2),(2),(2,3)}and{(1),(1,2,3),(3)}, but these last two possibilities are forbidden by considerations of measure: for example D(x) cannot contain both x and its predecessor and successor in theπ0-order as this would imply µ[x]>
µ[x] for any invariant measure µ.
3. The symmetric case
We look now at the important particular cases defined in Definition 9. We remark that, in general, ifI is defined by the vectorα and the permutationsπ0 and π1, the tranformation I−1 is the interval exchange defined by a reordering of α and the permutations π1 and π0. Thus in this particular case, I is defined by α, Id and σ, while I−1 will be defined in the same way after reversing the orientation, as σ is an involution. Thus I is conjugate to I−1 by x→1−x, and it is not surprising that
Lemma 6. If L satisfies (H0) to (H5) for π0 =Id, π1 =σ, L is a symmetric language.
Proof
Because π−10 = π0 and π1−1 = π1, we check that L satisfies also (H0) to (H5) for the same permutations, and we can apply Proposition 4 toL′ =L, using any invariant probability µ,
and the probability µ′[w] =µ[w].
The extra assumption of alternate discontinuities simplifies pleasantly the statements:
Proposition 7. A language L is the language of a symmetric k-interval exchange trans- formation with alternate discontinuities, satisfying the i.d.o.c. condition, if and only if it satisfies (H1), and
(H8) if w is a bispecial word, A(w) = {a, b} with a and b consecutive in the π1-order, D(w) ={c, d}, with c and d consecutive in the π0-order, there are three words of the form xwy, x∈ A, y∈ A and these are awc, bwd and either bwc or awd,
(H9) L2 ={1k,1(k−1),2(k−1),2(k−2), . . . ,(k−1)2,(k−1)1, k1}.
Proof
Let L be as in the hypothesis: the condition on discontinuities implies (H9) directly. It implies that for every w bispecial #A(w) = #D(w) = 2, and (H8) is just a translation of (H2) to (H5) in this particular case.
Suppose L satisfies (H1), (H8) and (H9). (H9) implies (H0), and also that for every w bispecial #A(w) = #D(w) = 2; thus (H8) implies (H2) to (H5), and we can apply Theorem 2 to getL=L(I), with permutations Id andσ; the condition of alternate discontnuities for
I is a translation of (H9).
I[aw] I[bw] I[aw] I[bw]
[wc]
[wc] [wd] [wd]
[w]
[w]
Figure 3. The two possibilities for bispecial intervals in this case
The case of alternate discontinuities seems to be a very particular case, but it can be considered as fairly typical, as
Lemma 8. IfLis anyL(I)as in Theorem 2, for every wordwlong enough,#A(w)≤2and
#D(w) ≤ 2, and thus the conclusions of (H8) are satisfied by every long enough bispecial word w.
Proof
Whenn is large enough, words of lengthn correspond to small intervals because of minimal- ity, and these cannot contain two different βi or the image by the same power of I of two
different γi, which implies the lemma.
We finish by showing an equivalent form of Proposition 7, which is a direct generalization (fromk = 3 to every k) of Theorem 1.1 of [7].
Proposition 9. A language L is the language of a symmetric k-interval exchange trans- formation with alternate discontinuities, satisfying the i.d.o.c. condition, if and only if it satisfies (H1), (H9) and
(H10) for every n there are exactly k−1 left special words of length n, one beginning in i for each 1≤i≤k−1,
(H11) L is symmetric,
(H12) if w is a bispecial word beginning in i and ending in j, w(k−j) is left special if and only if w(k−i) is left special.
Proof
Let L be as in the hypothesis: Proposition 7 implies that L satisfies (H8) and (H9); (H9) ensures that (H10) is true forn = 1 and (H8) that it extends fromnton+ 1. (H11) comes from Lemma 6, and (H12) from (H8) and (H11).
Suppose L satisfies (H9) to (H12). Let w be bispecial; then in Lthere are at least two xw and two wy, hence there are exactly two of each because of (H9); there are at least two and at most four xwy; if there are two or four of them this contradicts (H10) (see [5] for a general theory). Thus either wc or wd is left special but not both; if wc is left special, then by (H12) wb is left special, bw is right special, thus bwc, awc, bwdare in L; similarly, if wd is left special, bwd, awd, awc are in L, and we have proved (H8). Thus we can apply
Proposition 7 to get L=L(I) for the required I.
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