Asymptotic Pairs for Interval Exchange Transformations
aa r X i v : . [ m a t h . D S ] O c t ASYMPTOTIC PAIRS FOR INTERVAL EXCHANGETRANSFORMATIONS
ETHAN AKIN, ALFONSO ARTIGUE AND LUIS FERRARI
Abstract.
We provide a simple description of asymptotic pairsin the subshift associated with an interval exchange transforma-tion and show that, under reasonably general conditions, doublyasymptotic pairs do not occur.
Contents
1. Introduction 12. T Intervals 23. Interval Exchange Transformations 44. The Associated Subshift and Its Asymptotic Pairs 8References 131.
Introduction
Interval exchange transformations were introduced by Keane [1], seealso [3] for a survey. Keane begins with a partition of [0 ,
1) by left-closed intervals { J , . . . , J n } and defines a bijection T on [0 ,
1) whichis a translation on each interval. The itinerary map I : [0 , → Ω = { , . . . , n } Z is defined by I ( x ) k = i when T k x ∈ J i for k ∈ Z (We willuse, as usual, Z for the set of integers and N for the set of positiveintegers). The cocountable set R ⊂ (0 ,
1) consisting of the complementof the set of orbits of the endpoints is called the set of regular points.At points of R the iterates of T and the itinerary map I are continuous.The function I maps the homeomorphism T on R to the shift map S on Ω where S ( α ) i = α i +1 . Taking the closure of I ( R ) in Ω we obtain aclosed shift invariant subset X .Under conditions which have become known as the Keane Conditionand irreducibility, Keane showed that the subshift ( X, S ) is minimal,i.e. all orbits are dense.
Date : October, 2020 . . It will be convenient to consider at the same time the dual intervalexchange map ˜ T . This is defined using the right-closed partition of(0 , { ˜ J , . . . , ˜ J n } which have the same end-points as before and whichuses the same translation maps. The itinerary map ˜ I : (0 , → Ω isdefined as before. On R we have I = ˜ I .By using both we characterize the space X as the union I ([0 , ∪ ˜ I ((0 , F is a homeomorphism on a compact metric space K then points x, y ∈ K are positively asymptotic when lim i →∞ d ( F i x, F i y ) = 0 and negatively asymptotic when they are positively asymptotic for F − . Apair is called doubly asymptotic when it is both positively and nega-tively asymptotic.Two points α, β ∈ Ω are positively asymptotic (or negatively asymp-totic) if and only if there exists N ∈ N such that α i = β i for all i ≥ N (resp. for all i ≤ − N ).From the characterization we obtain our main result. Theorem 1.1.
If an interval exchange transformation is irreducible,fully split and satisfies the Keane Condition, then no pair of distinctpoints is doubly asymptotic for the associated subshift.
Using a different procedure, in [2] King gives an example of a subshiftwithout doubly asymptotic pairs.2. T Intervals
By an interval I we mean a bounded, nonempty subset of R whichis connected, or, equivalently, x < x ∈ I implies y ⊂ I for all y with x < y < x (no holes). We call I a positive interval whenits length ℓ ( I ) is positive. An interval which is not positive is just asingleton in R . There are four types of intervals. With a < b theintervals [ a, b ) , ( a, b ] , ( a, b ) , [ a, b ] are said to be of type Lr,lR,lr and LR,respectively. A singleton is of type LR. For any x , x ∈ R we write[ x , x ] for the closed interval spanned by the two points.Let K be a bounded subset of R and T be a transformation on K ,i.e. T : K → K a set map. An interval I is a T interval when it isa subinterval of K such that the restriction T | I is a translation, i.e. T x − x is a constant c for x in I (constant translation number) so, inparticular, T ( I ) = I + c and ℓ ( T ( I )) = ℓ ( I ). Clearly, T ( I ) has thesame type as I . SYMPTOTIC PAIRS 3
Notice that if I has end-points a < b and c > I + c \ I containsthe positive interval ( d, b + c ) with d = max( b, a + c ). Similarly, if c < I + c \ I contains a positive interval. It follows that(2.1) I a T interval with T ( I ) ⊂ I = ⇒ T x = x for all x ∈ I. For A ⊂ N I is a T A interval when it is a T n interval for every n ∈ A .In particular, I is a T N interval when it is a T n interval for everypositive n .Clearly, any subinterval of a T A interval is a T A interval.For any x ∈ K the singleton { x } is a T N interval. Lemma 2.1.
Let { I p : p ∈ P } be a family of subintervals of K with T p I p = ∅ . The subsets T p I p and S p I p are subintervals of K . If P isfinite and all the I p ’s are of the same type then both the intersectionand the overlapping union are of the same type as well.If for A ⊂ N all of the I p ’s are T A intervals, then S p I p is a T A interval.Proof. From the no holes condition it is clear that T p I p is an interval.Fix z ∈ T p I p . If x , x ∈ S p I p then [ x , x ] ⊂ [ x , z ] ∪ [ x , z ] ⊂ S p I p and so S p I p is an interval. If all of the I p ’s are T intervals then T x − x = T z − z = T x − x and so the points of S p I p have a constanttranslation number. The result for T A is obvious from the result for T .The types result is easy to check when the cardinality of P is two.The general result follows by induction. (cid:3) Lemma 2.2.
Let
T, S be transformations on K and I be a subintervalof K . Any two of the following implies the third: (i) I is a T interval. (ii) I is an S ◦ T interval. (iii) T ( I ) is an S interval.Proof. For x ∈ I , ST x − x = ( ST x − T x ) + (
T x − x ). It is thus clearthat any two of the following implies the third:(i) T x − x is constant for all x ∈ I .(ii) ST x − x is constant for all x ∈ I .(iii) ST x − T x is constant for all x ∈ I . (cid:3) Definition 2.3.
For T a transformation on K , a bounded subset of R ,and A ⊂ N the equivalence relation E AT ⊂ K × K is defined by (2.2) E AT = { ( x, y ) : there exists a T A interval I with x, y ∈ I } . When A = N we will omit the superscript. ETHAN AKIN, ALFONSO ARTIGUE AND LUIS FERRARI
Since { x } is a T N interval, the relation is reflexive. It is obviouslysymmetric and transitivity follows from Lemma 2.1. By Lemma 2.1again, the equivalence class E AT ( x ) is the union of all the T A intervalswhich contain x and so it is the maximum T A interval which contains x . From Lemma 2.2 it follows that T n ( E T ( x )) is a T N interval containing T n x . So for all x ∈ K, n ∈ N (2.3) T n ( E T ( x )) ⊂ E T ( T n x ) . Theorem 2.4.
If for T a transformation on K there exists a posi-tive T N interval, then there exists a positive subinterval I of K and apositive integer n such that T n x = x for all x ∈ I .Proof. This is a baby version of the Poincar´e Recurrence Theorem.The hypothesis says that there exists x ∈ X such that E T ( x ) is apositive interval and so has ℓ = ℓ ( E T ( x )) >
0. From (2.3) we have(2.4) ℓ = ℓ ( E T ( x )) = ℓ ( T n ( E T ( x ))) ≤ ℓ ( E T ( T n x )) . Since K is bounded, the sequence { E T ( T n x ) } cannot be pairwisedisjoint. So there exist n, k ∈ N such that E T ( T k x ) and E T ( T n + k x )intersect and so are equal. By (2.3) again the positive T N interval I = E T ( T k x ) is mapped into itself by T n . It follows from (2.1) that T n x = x for all x ∈ I . (cid:3) Interval Exchange Transformations
We review from [1] the definition of an interval exchange transfor-mation. Let J = [0 ,
1) and ˜ J = (0 , n ≥ n ] = { , . . . , n } .Given a = ( a , . . . , a n ) a positive probability vector, i.e. a i > i ∈ [ n ] and P nj =1 a j = 1, we define b = 0 and for i ∈ [ n ](3.1) b i = i X j =1 a j , J i = [ b i − , b i ) , ˜ J i = ( b i − , b i ] . Thus, J , . . . , J n is a partition of J by Lr intervals and ˜ J , . . . , ˜ J n is apartition of ˜ J by lR intervals. Let D = { b , . . . , b n − } .If τ is a permutation of [ n ], then a τ = ( a τ − (1) , a τ − (2) , . . . , a τ − ( n ) ) isa positive probability vector and we form the corresponding b τi , J τi and˜ J τi for i ∈ [ n ]. Thus, b ττ ( i ) − b ττ ( i ) − = a ττ ( i ) = a i = b i − b i − for all i ∈ [ n ]and D τ = { b τ , . . . , b τn − } .We define the transformations T on J and ˜ T on ˜ J so that for i ∈ [ n ], T : J i → J ττ ( i ) and ˜ T : ˜ J i → ˜ J ττ ( i ) are the translations given by(3.2) x x − b i − + b ττ ( i ) − . SYMPTOTIC PAIRS 5
We call T the Lr ( a , τ ) interval exchange transformation and ˜ T thelR ( a , τ ) interval exchange transformation dual to T .These possess the following properties. • T and ˜ T are invertible and their inverses are the corresponding( a τ , τ − ) transformations. • T is everywhere continuous from the right and ˜ T is everywherecontinuous from the left. • T = ˜ T on [0 , \ ( D ∪ { , } ) and is continuous on this open set.So we have for i ∈ [ n ] ˜ T b i = lim x ↑ b i T x = b ττ ( i ) ,T b i − = lim x ↓ b i − ˜ T x = b ττ ( i ) − . (3.3)Notice that if { x i } is a monotone increasing sequence in (0 ,
1) then itis eventually contained in some J i and so eventually { T x i } is monotoneincreasing. It follows that, inductively, for all k ∈ Z , T k is everywherecontinuous from the right and similarly ˜ T k is everywhere continuousfrom the left and both of these preserve eventually monotonicity ofsequences.If τ ( { , . . . , j } ) = { , . . . , j } for some j = 1 , . . . , n − T canbe decomposed into an interval exchange transformation on [0 , b j ) andone on [ b j ,
1) (and similarly for ˜ T ). Keane calls τ irreducible when(3.4) τ ( { , . . . , j } ) = { , . . . , j } for j = 1 , . . . , n − T b τ = a ττ − (1) = a . That is, when τ (1) = 1. So if τ is irreducible, then T = 0 and so 0 ∈ T ( D ). Similarly,˜ T τ ( n ) = n and so τ ( { , . . . , n − } ) = { , . . . , n − } . Thus,irreducibility implies ˜ T = 1 and 1 ∈ ˜ T ( D ).If for some j = 1 , . . . , n − b ττ ( j ) = b ττ ( j +1) − or, equivalently, τ ( j + 1) = τ ( j ) + 1, then T is continuous at b j . In fact, the interval J j ∪ J j +1 is a T interval and T is an interval exchange transformationwith the n − a , . . . , a j + a j +1 , . . . a n ). We will call τ split when(3.5) τ ( j + 1) = τ ( j ) + 1 for j = 1 , . . . , n − . If τ is split, then T b = ˜ T b for b ∈ D and so D is the set of points atwhich T and ˜ T are not continuous.We will call τ fully split if, in addition,(3.6) τ (1) = τ ( n ) + 1 and τ − (1) = τ − ( n ) + 1 ETHAN AKIN, ALFONSO ARTIGUE AND LUIS FERRARI
These conditions exclude the possibilities T T T − T −
1, respectively.If τ is irreducible, split or fully split then τ − satisfies the corre-sponding property. Lemma 3.1. If x ∈ (0 , and k ∈ N , then { x, T x, . . . , T k − x } ∩ ( { } ∪ D ) = ∅ ⇐⇒{ x, ˜ T x, . . . , ˜ T k − x } ∩ ( { } ∪ D ) = ∅ , (3.7) in which case, T j x = ˜ T j x for j = 1 , . . . k , and (3.8) T j x ∈ J i ⇐⇒ ˜ T j x ∈ ˜ J i for j = 0 , , . . . , k − and i ∈ [ n ] .Proof. This is obvious for k = 1 and then follows easily by induction. (cid:3) Proposition 3.2.
Let k ∈ N and let j p ∈ [ n ] for ≤ p ≤ k − . Define (3.9) I = k − \ p =0 T − p ( J i p ) , ˜ I = k − \ p =0 ˜ T − p ( ˜ J i p ) . If either I or ˜ I is nonempty, then for some a < b ∈ [0 , I = [ a, b ) , ˜ I =( a, b ] and for all p = 0 , . . . , k T p is a translation on I and ˜ T p is atranslation on ˜ I with the same translation constant. For all p = 0 , . . . , k on ( a, b ) T p = ˜ T p and these are continuous there. The endpoints satisfy (3.10) a ∈ { } ∪ k − [ p =0 T − p D, b ∈ { } ∪ k − [ p =0 ˜ T − p D As we vary the sequence { j p } ∈ [ n ] k , the nonempty intervals I forman Lr partition of [0 , , which we will call the k -level partition , withthe intervals ˜ I the corresponding lR partition of (0 , , which we willcall the dual k -level partition Proof.
Again the result is clear for k = 1. Assuming the result for k we prove it for k + 1.By induction hypothesis T k is a translation on I = [ a, b ). If T k ( I )meets J j k = [ b j k − , b j k ) then I ∩ T − k ( J j k ) = [ a ′ , b ′ ) with(3.11) a ′ = ( a if T k a > b j k − ,T − k b j k − otherwise b ′ = ( b if T k b < b j k ,T − k b j k otherwiseand on it T ◦ T k is a translation. By induction hypothesis, ˜ T k hasthe same translation constant on ( a, b ] as T k does on [ a, b ). So ˜ I ∩ SYMPTOTIC PAIRS 7 ˜ T − k ( ˜ J j k ) = ( a ′ , b ′ ] and on it ˜ T ◦ ˜ T k has the same translation constantas T ◦ T k does on [ a ′ , b ′ ). Finally, T ◦ T k = ˜ T ◦ ˜ T k on ( a ′ , b ′ ).The endpoint result (3.10) follows inductively from (3.11).Distinct sequences yield disjoint intervals I and every point of (0 , I . Hence, they form an Lr partition. (cid:3) Corollary 3.3.
If the Lr ( a , τ ) interval exchange transformation T or its dual ˜ T has a periodic point of period k , then there is a positivesubinterval of (0 , on which T k = ˜ T k = id .Proof. Assume that T k x = x for some x ∈ [0 , I = [ a, b ) is theelement of the k level partition which contains x then since T k is atranslation on I and so T k y − y = T k x − x = 0 for y ∈ [ a, b ). Since˜ T k = T k on ( a, b ) we see that ˜ T k = T k = id on ( a, b ). Similarly,˜ T k x = x for x ∈ ˜ I = ( a, b ] implies ˜ T k = T k = id on ( a, b ). (cid:3) Definition 3.4.
We say that the interval exchange transformation T satisfies the Keane Condition when for all k ∈ N D ∩ T k ( D ) = ∅ . Of course, D ∩ T k ( D ) = ∅ if and only if T − k ( D ) ∩ D = ∅ . The KeaneCondition says exactly that the T orbits of the points of D are infiniteand distinct. Lemma 3.5. D ∩ T k ( D ) = ∅ for all k ∈ N if and only if D ∩ ˜ T k ( D ) = ∅ for all k ∈ N .Proof. Assume that a ∈ D and T k a = b i − ∈ D so that 1 < i < n .Let I be an interval of the k -level Lr interval described in Proposition3.2 with a ∈ I . From equation (3.11) I has a as left end point. Itis translated by T k to an interval with b i − as left endpoint. So someinterval ˜ K = ( c, d ] of the dual k -level lR partition is translated by˜ T k so that ˜ T k d = b i − . By (3.10) d ∈ { } ∪ S k − p =0 ˜ T − p D . That is,˜ T k − p a = b i − for some a ∈ D and some p < k or else ˜ T k b i − . Inthe latter case, 1 is not fixed and so 1 ∈ ˜ T ( D ) and b i − ∈ ˜ T k +1 ( D ).Thus, D meets ˜ T j ( D ) for some j with 1 ≤ j ≤ k + 1.Similarly, D ∩ ˜ T k ( D ) = ∅ implies D ∩ T j ( D ) = ∅ for some j ∈ N . (cid:3) The major application of this condition is Keane’s Theorem:
Theorem 3.6.
If the Lr ( a , τ ) interval exchange transformation T isirreducible and satisfies the Keane Condition, then the maps T and ˜ T are minimal. That is, for every x ∈ [0 , the orbit { T k x : k ∈ Z } isdense in [0 , and for every x ∈ (0 , the orbit { ˜ T k x : k ∈ Z } is densein (0 , . ETHAN AKIN, ALFONSO ARTIGUE AND LUIS FERRARI
Proof.
In [1] Keane proves this for T . The results for ˜ T follow becausethe ˜ T lR interval exchange transformation is conjugate to an Lr intervalexchange transformation via the map x − x . From Lemma 3.5 itfollows that the conjugate also satisfies the Keane Condition.Irreducibility says that no interval [0 , b i ) is invariant for T and no(0 , b i ] is invariant for ˜ T . Thus, the conjugate Lr transformation isirreducible and Keane’s Theorem applies to it. Minimality is preservedby conjugation. (cid:3) For constructing examples the following Irrationality Theorem from[1] is useful.
Theorem 3.7. If τ is irreducible and { a , . . . , a n − , a n } is linearly inde-pendent over the field of rationals Q (or, equivalently, { a , . . . , a n − , } is linearly independent over Q ) then the ( a , τ ) interval exchange trans-formation satisfies the Keane Condition. The Associated Subshift and Its Asymptotic Pairs
Let T, ˜ T be the interval exchange transformations associated with( a , τ ) as described in the previous section.We define the itinerary functions I : [0 , → Ω , ˜ I : (0 , → Ω by I ( x ) k = i ⇐⇒ T k x ∈ J i for k ∈ Z , i ∈ [ n ] , x ∈ [0 , , ˜ I ( x ) k = i ⇐⇒ ˜ T k x ∈ ˜ J i for k ∈ Z , i ∈ [ n ] , x ∈ (0 , . (4.1)Clearly we have the following commutative diagrams showing that I maps T and ˜ I maps ˜ T to the shift S on Ω.Ω S / / Ω[0 , I O O T / / [0 , I O O Ω S / / Ω(0 , ˜ I O O ˜ T / / (0 , ˜ I O O Define(4.2) D ∞ = { , } ∪ [ k ∈ Z T k ( D ) = { , } ∪ [ k ∈ Z ˜ T k ( D ) . Notice that either 0 ∈ T ( D ) or else T ∈ ˜ T ( D ) orelse ˜ T D τ ⊂ D ∞ .It follows that R = [0 , \ D ∞ is a dense G δ subset of [0 ,
1] onwhich T = ˜ T and the set is invariant, i.e. T ( R ) = ˜ T ( R ) = R . Hence, T − = ˜ T − on R as well. The points of R are called the regular points SYMPTOTIC PAIRS 9 for the transformation. Since R is disjoint from D ∪ D τ , it follows that T and its inverse are continuous at the points of R . Proposition 4.1.
The map I is everywhere continuous from the rightand ˜ I is everywhere continuous from the left. On R the maps I and ˜ I are equal and are continuous.Proof. On each interval J i = [ b i − , b i ) the function I ( · ) is constantly i as is ˜ I ( · ) on ˜ J i . So I ( · ) is continuous from the right and ˜ I ( · ) iscontinuous from the left. They agree and are continuous on the openinterval ( b i − , b i ).For every k ∈ Z I ( T k x ) = ( S k I ( x )) = I ( x ) k . That is, I ( · ) k = I ( · ) ◦ T k . Since T k and I ( · ) are continuous from the right and T k preserves eventually monotonicity, it follows from the definition of theproduct topology on Ω that I is continuous from the right. Similarly, ˜ I is continuous from the left. Furthermore, since R is invariant, it followsthat I = ˜ I on R and so at points of R the map is continuous from bothdirections. (cid:3) Define(4.3) X = I ( R ) = ˜ I ( R ) and X = X . The set X is an S invariant subset of Ω, i.e. S ( X ) = X , and sothe closure X is a closed, invariant subset. That is, ( X, S ) is a subshiftof the full shift (Ω , S ) on n symbols.Keane provides an explicit description of X as follows. Let J =([0 , × { } ) ∪ (( D ∞ \ { } ) × {− } ). Order J lexicographically and usethe order topology. Since every nonempty subset has a supremum andinfimum, it follows that J is a compact Hausdorff space. Identify each x ∈ [0 ,
1) with ( x,
0) in J and for x ∈ D ∞ \ { } write x − for ( x, − x − , x with x ∈ D ∞ \ { , } is a gap pair, i.e. x − < x and there are no points between them. If D ∞ is dense, as in the minimalcase, then J is homeomorphic to the Cantor Set. The projection to thefirst coordinate defines a continuous, order-preserving surjection from J onto [0 ,
1] which is one-to-one over the points of R ∪ { , } and whichmaps x and x − to x .Let J i = [ b i − , b i − ] for i ∈ [ n ]. This is a partition of J by n clopenintervals. Define the homeomorphism T on J by mapping J i to theclopen interval J ττ ( i ) in the order-preserving way which extends T on J i . Now the itinerary map is a continuous map from J onto X , mapping T to S and extending I on the dense set R .Our alternative description of X is the following Theorem 4.2. X equals the union I ([0 , ∪ ˜ I ((0 , . Proof.
Any x ∈ [0 ,
1) is a limit of a decreasing sequence { x i } in R . So I ( x ) = lim I ( x i ) is in the closure of X . Similarly, ˜ I ((0 , ⊂ X .Now suppose { α i = I ( x i ) = ˜ I ( x i ) } is a sequence in X which con-verges to β ∈ Ω. By going to a subsequence we can assume that { x i } converges monotonically to a point y ∈ [0 , β = I ( y ). If the sequence is increasing, then β = ˜ I ( y ). (cid:3) Corollary 4.3.
If the Lr ( a , τ ) interval exchange transformation T isirreducible and satisfies the Keane Condition, then ( X, S ) is a minimalsubshift.Proof. By Keane’s Theorem 3.6 the closure of every T orbit and every˜ T orbit contains R . By Theorem 4.2 and continuity of I = ˜ I at pointsof R , it follows that the closure of every S orbit in X contains X andso equals X . (cid:3) Theorem 4.4.
Assume that the Lr ( a , τ ) interval exchange transfor-mation T has no periodic points. (i) If x, y are distinct points of [0 , , then the pair I ( x ) , I ( y ) isneither positively, nor negatively asymptotic. (ii) If x, y are distinct points of (0 , then the pair ˜ I ( x ) , ˜ I ( y ) isneither positively, nor negatively asymptotic. (iii) If α, β are distinct points of X and α ∈ X , then the pair isneither positively nor negatively asymptotic. (iv) The maps I and ˜ I are injective.Proof. (i) Assume that i k = I ( x ) k = I ( y ) k for all k ≥
0. By Proposition3.2 { I k = T k − p =0 T − p ( J i p ) } is a decreasing sequence of intervals each ofwhich contains x and y . For k > p it is a T p interval. By Lemma 2.1the intersection I = T k> I k = T k>p I k is a T p interval for all p . Sinceit contains x and y , it is a positive T N interval if x = y . Theorem 2.4would then imply the there is an interval on which T is periodic. Since T is minimal, it does not admit periodic points and so x = y .Now if I ( x ) k = I ( y ) k for k ≥ N , then I ( T N x ) k = I ( T N y ) k for k ≥ T N x = T N y . Because T is injective. it follows that x = y .Apply the result to T − to see that the pair is negatively asymptoticonly when x = y .(ii) As in (i) ˜ I ( x ) k = ˜ I ( y ) k for k ≥ N with x = y implies that thereis a positive interval on which ˜ T is periodic. Such an interval meets R and the intersection would consist of periodic points for T . Thecontradiction shows that x = y .(iii) Assume that α = I ( x ) = ˜ I ( x ) for some x ∈ R . If β = I ( y ), then x = y and so α = I ( x ) and β = I ( y ) are not asymptotic. Similarly, if SYMPTOTIC PAIRS 11 β = ˜ I ( y ), then x = y and so α = ˜ I ( x ) and β = ˜ I ( y ) are not asymptotic.By Theorem 4.2 one of these two cases applies.(iv) If for x = y in [0 , I ( x ) = I ( y ) then the “pair” I ( x ) , I ( y ) isasymptotic contradicting (i). Similarly, for ˜ I . (cid:3) The result (i) above is equivalent to a comment at the end of Section5 of [1].The main result of this paper is the description of asymptotic pairs.
Theorem 4.5.
Assume that the Lr ( a , τ ) interval exchange transfor-mation T is irreducible and satisfies the Keane Condition. Let α, β ∈ X . (i) If α = β then the pair { α, β } is positively asymptotic if and onlyif there exist i ∈ [ n ] and k ∈ Z such that with x = b τi ∈ D τ , { α, β } = { S k I ( x ) , S k ˜ I ( x ) } . (ii) If α = β then the pair { α, β } is negatively asymptotic if andonly if there exist i ∈ [ n ] and k ∈ Z such that with x = b i ∈ D , { α, β } = { S k I ( x ) , S k ˜ I ( x ) } . (iii) If I ( x ) = ˜ I ( y ) for some x ∈ [0 , , y ∈ (0 , , then x = y and thecommon point lies in R . (iv) Assume, in addition, that τ is fully split. If the pair { α, β } isdoubly asymptotic, then α = β . Conversely, if τ is not split,then there exist doubly asymptotic pairs of distinct points.Proof. By the Keane Condition every T orbit or ˜ T of a point of D meets D just once. Because τ is irreducible 0 ∈ T ( D ) and 1 ∈ ˜ T ( D ).(i) First, assume that x ∈ D τ . Let b i − = T − x , the left endpoint ofsome J i and, b j = ˜ T − x the right end-point of some ˜ J j . Furthermore, i = j since the endpoints b i − and b i are translated by T and ˜ T to theendpoints of the interval J ττ ( i ) . If τ is split, then b i − = b j . But in anycase, I ( x ) − = i = j = ˜ I ( x ) − . Thus, letting α = I ( x ) , β = ˜ I ( x ). Wehave α = β .By the Keane Condition and irreducibility, T k x
6∈ { } ∪ D for k ≥ T k x = ˜ T k x and I ( x ) k = ˜ I ( x ) k for all k ≥ α, β is positively asymptotic.Conversely, assume that α = I ( x ) and β = ˜ I ( y ) are positively as-ymptotic. By shifting down we may assume that I ( x ) k = ˜ I ( y ) k for all k ≥ T k x
6∈ { } ∪ D for all k ≥
0. Again Lemma 3.1 impliesthat T k x = ˜ T k x for all k ≥
0. It follows that ˜ I ( x ) k = I ( x ) k = ˜ I ( y ) k forall k ≥
0. From Theorem 4.4 (ii) it follows that x = y . So α = I ( x )and β = ˜ I ( x ). If α = β , x R and so the T orbit of x hits { } ∪ D . Now we can move in the negative direction and shift so that we mayassume that T − x ∈ { } ∪ D and T k x
6∈ { } ∪ D for all k ≥
0. Soagain T k x = ˜ T k x for all k ≥
0. Note that if T − x = 0 then T − x ∈ D .Thus, x ∈ T ( { } ∪ D ) \ { } = D τ .(ii) Apply (i) to T − which is also irreducible and satisfies the KeaneCondition.(iii) If I ( x ) = ˜ I ( y ), then the pair is trivially positively asymptotic.The proof of (i) essentially shows that T N x = T N y for some N ∈ N and so x = y . If the orbit of x hits D then the proof of (i) shows that I ( x ) = ˜ I ( x ). It follows that a point x with I ( x ) = ˜ I ( x ) must lie in R .(iv) We use the following(a) If d ∈ D , then T − k d
6∈ { }∪ D τ for k ≥ T − k d
6∈ { }∪ D for k ≥ T − k d = ˜ T − k d for k ≥ I ( d ) − k = ˜ I ( d ) − k for k ≥ T − ∈ D so by (a) T − k ( T −
0) = ˜ T − k ( T − k ≥ I ( T − − k = ˜ I ( T − − k . Similarly, for k ≥ T − k ( ˜ T −
1) = ˜ T − k ( ˜ T −
1) and for k ≥ I ( ˜ T − − k = ˜ I ( ˜ T − − k .Now suppose that τ is not split. Thus, for suitable x ∈ D τ we have b i − = T − x and b j = ˜ T − x with j = i − d = T − x = ˜ T − x ∈ D . Thus, T k x = ˜ T k x for all k ∈ Z . We have i = I ( x ) − , j = i − I ( x ) − . On the other hand, the proof of (i) and (a) above imply that I ( x ) k = ˜ I ( x ) k for all k = − Z . That is, I ( x ) and ˜ I ( x ) is a pair ofdistinct doubly asymptotic points.Also, if T T x and T − T −
1, then we can similarly seethat I ( x ) k = ˜ I ( x ) k for all k = − , − Z . I ( x ) and ˜ I ( x ) is a pair ofdistinct doubly asymptotic points.Now assume that τ is fully split. We begin with a positively asymp-totic pair and after shifting we can assume the pair is I ( x ) , ˜ I ( x ) with x ∈ D τ .Since τ is split, d = b i − = T − x and d = b j = ˜ T − x are not equaland d ∈ { } ∪ D, d ∈ { } ∪ D . Since τ is fully split, we do not haveboth d = 0 and d = 1.Let y = T − d , y = ˜ T − d . From (a) and (b), we have T − k y =˜ T − k y and T − k y = ˜ T − k y for k ≥ k ≥ I ( y ) − k = I ( y ) − k = I ( x ) − k − , I ( y ) − k = ˜ I ( y ) − k = ˜ I ( x ) − k − . (4.4) SYMPTOTIC PAIRS 13 If I ( x ) and ˜ I ( x ) were negatively asymptotic, then from Equation(4.4) it would follow that I ( y ) and I ( y ) are negatively asymptotic. Itwould then follow from Theorem 4.4 that y = y .Now either d ∈ D or d ∈ D . Suppose d ∈ D . then ˜ T − d = T − d = y = y = ˜ T − d . Applying ˜ T we obtain d = d which is nottrue. Similarly if d ∈ D .Thus, I ( x ) and ˜ I ( x ) are not negatively asymptotic. Consequently,pairs of distinct points of X are never doubly asymptotic. (cid:3) Thus, we have obtained our promised result.
Theorem 4.6.
If an interval exchange transformation is irreducible,fully split and satisfies the Keane Condition, then no pair of distinctpoints is doubly asymptotic for the associated subshift.
We conclude by describing the simplest examples. n = 2: The transposition τ = (1 ,
2) is irreducible and split. If a is chosen irrational, then by Theorem 3.7 the ( a , τ ) interval exchangemap on two intervals satisfies the Keane Condition. If we identify 0with 1 then the maps T and ˜ T become irrational rotations of the circle,or, equivalently, the translation by a on the quotient R / Z . The liftto ( X, S ) yields a Sturmian subshift. The permutation τ is not fullysplit. In fact, T T T − T −
1. In this case, b = a and,as is demonstrated in (iv) above, the orbit of the pair I ( b ) , ˜ I ( b ) is theunique pair of orbits which are positively or negatively asymptotic andit is doubly asymptotic. n = 3: We choose a , a so that { a , a , } is linearly independentover Q . The irreducible permutations are the three cycle (1 , ,
3) (andits inverse) and the transposition (1 , a , τ ) trans-formation satisfies the Keane Condition by Theorem 3.7 again.The three cycle (1 , ,
3) is not split. Both pairs I ( b ) , ˜ I ( b ) and I ( b ) , ˜ I ( b ) are doubly asymptotic. The maps T and ˜ T are the sameas those for the n = 2 case above with probability vector ( a + a , a ).The subshift is obtained from the translation by a + a on R / Z , bycutting at the points of two orbits instead of just one. See Keane’spoint-doubling construction of X described above.On the other hand, the transposition τ = (1 ,
3) is irreducible andfully split. For such an ( a , τ ) transformation no pair of distinct pointsis doubly asymptotic for the associated subshift ( X, S ). References
1. M. Keane,
Interval exchange transformations , Math. Z, (1975), 25-31.
2. J.L. King,
A map with topological minimal self-joinings in the sense of delJunco , Ergod. Th. & Dynam. Sys., (1990), 745-761.3. M. Viana, Ergodic theory of interval exchange maps , Rev. Mat. Complut, (2006), 7-100. Mathematics Department, The City College, 137 Street and Con-vent Avenue, New York City, NY 10031, USA
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