aa r X i v : . [ m a t h . D S ] A p r THERE EXISTS A TOPOLOGICALLY MIXING INTERVALEXCHANGE TRANSFORMATION
JON CHAIKA
Abstract.
We prove the existence of a topologically mixing interval exchangetransformation and that no interval exchange is topologically mixing of allorders. Introduction
The main result of this paper is that there exists a topologically mixing IETanswering a question of Boshernitzan. The construction is based on the work ofKeane [11]. The plan of the paper is as follows: First, we state the results ofthe paper and introduce terminology. Second, we describe the Keane constructionpresented in [11], which is used to prove topological mixing. Third, we provetopological mixing. Last, we prove that no IET is topologically mixing of all orders.
Definition 1.
Let ∆ n − = { ( l , ..., l n ) : l i > , l + ... + l n = 1 } be the ( n − -dimensional simplex. Given L = ( l , l , ..., l n ) ∈ ∆ n − we can obtain n subintervalsof the unit interval: I = [0 , l ) , I = [ l , l + l ) , ..., I n = [ l + ... + l n − , . Ifwe are also given a permutation on n letters π we obtain an n- Interval ExchangeTransformation T : [0 , → [0 , which exchanges the intervals I i according to π .That is, if x ∈ I j then T ( x ) = x − P k Remark . The various conventions for naming the permutation of IETs can be asource of confusion. The naming convention this paper uses says that if an IEThas permutation (4213) then the 4 th interval is placed first by T , the 2 nd intervalis placed second the 1 st is placed third and the 3 rd is placed last. It does not saythat the 4 th interval gets sent to the 2 nd . Definition 2. Let X be a topological space. A dynamical system T : X → X issaid to be topologically mixing if for any two nonempty open sets U , V there exists N U,V := N such that T n ( U ) ∩ V = ∅ for all n ≥ N . Theorem 1. There exists a topologically mixing 4-IET.Remark . It has been shown that no IET satisfies the stronger property of beingmeasure theoretically mixing [9]. It was shown [8] that minimal, uniquely ergodicdynamical systems of linear block growth can not be measure theoretically mixing.No 3-IET can be topologically mixing [3]. However, almost every IET that is not ofrotation type satisfies the weaker property of being topologically weak mixing [12].In fact, almost every IET that is not of rotation type is measure theoretically weakmixing [2]. This was earlier shown for almost every 3-IET [10], almost every type W IET [13] and every type W IET satisfying an explicit Diophantine condition(Boshernitzan type, which almost every IET satisfies) [4]. For type W IETs it hasbeen shown that idoc (the Keane condition) implies topological weak mixing [5].The formulation of type W used in the two previous results is in [6]. Remark . Since writing this paper we were able able to show that a residual (dense G δ ) set of IETs with permutation ( n, n − , ..., 1) are topologically mixing for any n ≥ 4. The methods of this proof are different. Remark . The IET exhibited will be topologically mixing with respect to thestandard topology on [0,1) and also the finer topology given by considering theIET as a symbolic dynamical system. As a symbolic dynamical system the IET iscontinuous.A complementary result to Theorem 1 is: Theorem 2. No IET is topologically mixing of all orders.Remark . Another result along these lines is proved in [7], which shows thatno substitution dynamical system can be measure theoretically mixing, but thatthere exists a topologically mixing substitution dynamical system. However, nosubstitution dynamical system can be topologically mixing of all orders. Somesubstitution dynamical systems have IETs that are obviously conjugate to them.However, the topologically mixing substitution dynamical system in [7] is a veryparticular substitution on two letters, which has a closely related substitution thatis not topologically mixing. The IETs obviously corresponding to substitutions ontwo (or even three) letters are never topologically mixing. Definition 3. Given an IET T : [0 , → [0 , and a subinterval J , let T | J : J → J denote the first return map induced by T on J . That is, if x ∈ J let r ( x ) = min { n > T n ( x ) ∈ J } and T | J ( x ) = T r ( x ) ( x ) if r ( x ) is finite.A tower over J is the consecutive images of a subinterval before it returns to J .We call the individual images levels.Remark . The discontinuities for the induced map of an IET on an interval [ a, b )come from the preimages of discontinuities the IET and preimages of the endpoint.Therefore, if one chooses the endpoints carefully one gets an IET on at most thesame number of intervals. Definition 4. I (1) j denotes the j th subinterval of the IET given by inducing on the th interval. I ( k +1) j denotes the j th subinterval of the IET given by inducing on I ( k )4 . Definition 5. Let O ( I ( k ) j ) denote the disjoint images under T of I ( k ) j before firstreturn to I ( k ) .Remark . This is the tower over I ( k ) given by images of I ( k ) j . Definition 6. Set e = , e = , e = , e = .Remark . Some of the above terminology is nonstandard, but convenient for ourpurposes. HERE EXISTS A TOPOLOGICALLY MIXING INTERVAL EXCHANGE TRANSFORMATION3 The Keane construction This construction is given in [11] and it is included here for completeness. Con-sider IETs with permutation (4213). Observe that the second interval gets shiftedby l − l . If this difference is small relative to l then much of I gets sent to itself.At the same time, pieces of I do not reach I until they have first reached I .This is the heart of the Keane construction. The details of the Keane constructionare centered around iterating this procedure. Keane showed that by choosing thelengths appropriately one could ensure that T | I (1)4 is a (2413) IET. Name the inter-vals in reverse order and we once again get a (4213) IET. Moreover, Keane showedthat for any choice m, n ∈ N one can find an IET whose landing pattern of I (1) j isgiven by the following matrix. A m,n = m − m n n n − n ; m, n ∈ N = { , , ... } .In order to see this, pick lengths for I (1) and write them as a column vector. Nowassign lengths to the original IET by multiplying this column vector by A m,n . Theinduced map will travel according to this matrix by construction. For instance, ifone chooses lengths [ , , , ] for I (1)4 , ..., I (1)1 one gets lengths of[ 22 + 2 m − n − , m − m + 4 n + 4 , n − m + 4 n + 4 , m + 4 n + 4 ]for the original IET (after renormalizing). For any finite collection of matrices onecan just iterate this construction (assign lengths for I ( k ) by multiplying the lengthsof I ( k +1) by A m k ,n k , multiply the resulting column vector by A m k − ,n k − , ... ). Com-pactness (of ∆ , which can be thought of as the parameterizing space of (4213)IETs) ensures that we can pass to an infinite sequence of these matrices.Since the intervals are named in reverse order, the discontinuity (under the in-duced map) between I (1)2 and I (1)3 is given by T − ( δ ) where δ denotes the disconti-nuity between I and I . As the first row of the matrix suggests I = T ( I (1)4 ∪ I (1)3 ).The discontinuity (under the induced map) between I (1)1 and I (1)2 is given by T − m ( δ ) where δ denotes the discontinuity between I and I . As the secondrow of the matrix suggests I = T ( I (1)2 ∪ I (1)1 ) ∪ T ( I (1)2 ∪ I (1)1 ) ∪ ... ∪ T m − ( I (1)2 ∪ I (1)1 ) ∪ T m ( I ) . The discontinuity (under the induced map) between I (1)3 and I (1)4 is given by T − n − ( δ ) where δ denotes the discontinuity between I and I . As the thirdrow of the matrix suggests I = T m ( I (1)1 ) ∪ T m +1 ( I (1)2 ) ∪ T ( I (1)4 ∪ I (1)3 ) ∪ T m +1 ( I (1)1 ) ∪ T m +2 ( I (1)2 ) ∪ T ( I (1)4 ∪ I (1)3 ) ∪ ... ∪ T m + n − ( I (1)1 ) ∪ T m + n ( I (1)2 ) ∪ T n ( I (1)4 ∪ I (1)3 ) ∪ T m + n ( I (1)1 ) ∪ T m + n +1 ( I (1)2 ) ∪ T n +1 ( I (1)4 ) .I = I (1)4 ∪ I (1)3 ∪ I (1)2 ∪ I (1)1 . As the columns of the matrix suggest, this is also I = T n +1 ( I (1)3 ) ∪ T m + n +1 ( I (1)2 ) ∪ T m + n ( I (1)1 ) ∪ T n +2 ( I (1)4 ) . To summarize, the composition of I j can be given by the j th row of the matrix.The travel before first return of I (1) j can be given by the j th column. Additionally, J. CHAIKA because the intervals were named in reverse order, the permutation of the inducedmap is once again (4213).It is important for this construction that everything be iterated. The compositionof I ( k ) j in pieces of I ( k + r ) is given by e τj A m k ,n k ...A m k + r − ,n k + r − (where e τj denotesthe transpose of e j ). Likewise, the travel of I ( k + r ) j under T | I ( k ) before first returnto I ( k + r ) is given by A m k ,n k ...A m k + r − ,n k + r − e j .Now for some explicit statements about the travel of subintervals of I ( k ) underthe induced map T | I ( k ) . When I ( k )3 returns to I ( k ) it entirely covers I ( k )4 . It is asubset of I ( k )3 ∪ I ( k )4 . When I ( k )4 returns to I ( k ) it entirely covers I ( k )1 . It intersects I ( k )2 . Moreover part of this intersection will stay in O ( I ( k )2 ) for the next m k b k, images (the other part ( m k − b k, images). When I ( k )2 returns to I ( k ) it intersects I ( k )3 . Moreover this piece of intersection will stay in O ( I ( k )3 ) for the next n k b k, images. Definition 7. Let b k,i be the first return time of I ( k ) i to I ( k ) .Remark . b k, = m k − b k − , + n k − b k − , + b k − , and b k, = b k − , + ( n k − − b k − , + b k − , .Some facts to keep in mind:(1) The choice of n k has no effect on b i, for i ≤ k .(2) The choice of n k has no effect on b i, for i ≤ k .(3) The choice of m k has no effect on b i, for i ≤ k .(4) The choice of m k has no effect on b i, for i ≤ k + 1.3. There exists a topologically mixing Keane IET Conditions for b k, and b k, to ensure topological mixing:(1) b k, is prime for all k .(2) b i, b k, for all i < k .(3) The group of multiplicative units mod k Π i =1 b i, has more than k Π i =1 b i, ele-ments.(4) b k, b k +1 , + b k +1 , + b k, + b k − , < m k b k, . (5) b k, b k, + b k, < n k b k, . Theorem 1 will be proven by showing that any Keane IET chosen in this way istopologically mixing. We first show that the set of such IETs is nonempty. Lemma 1. We can choose b k, and b k, to fulfill these conditions.Proof. By induction. Assume we have chosen n , m , n , m , ..., n k − , m k − ; wedescribe how to choose n k and then given this n k how to choose m k . Considercongruence modulo g := k Π i =1 b i, . Choose a congruence class [ f ] that is in the groupof multiplicative units and so that [ f + b k, − b k, ] is in the multiplicative groupof units. This can be done by pigeon hole principle (by condition 3). Pick n k sothat b k +1 , ∈ [ f ] and so that n k > b k, b k, + b k, b k, . This can be done because b k, is relatively prime to the b i, . Next we pick m k so that b k +1 , is prime, m k > b k, b k +1 , + b k, + b k +1 , b k, and condition 3 is satisfied. This can be done because we wishto find a prime in the arithmetic progression n k b k, + b k, + b k, N and the starting HERE EXISTS A TOPOLOGICALLY MIXING INTERVAL EXCHANGE TRANSFORMATION5 point and the increment are relatively prime and the other conditions merely requirechoosing m k large enough. This is Dirichlet’s Theorem (see for example [1, Chapter7]). (cid:3) Let c k = b k +1 , b k +2 , + b k +2 , + b k +2 , + b k +1 , and d k = b k +2 , b k +2 , + b k +2 , .Let J be a subinterval in I containing at least one level of a tower over I ( k ) .This means that it contains at least 1 level from each of the 4 towers over I ( k +2) .For all j > k , i > c j , T i ( J ) intersects every level of every tower over I ( j − . Thisis proved in the following lemmas. In these arguments it will be important to pickout a level from O ( I ( k +2)2 ) and O ( I ( k +2)3 ). These will be denoted J ′ . Lemma 2. At times c k to d k J ′ , a level in O ( I ( k +2)3 ) , intersects every level of O ( I ( k +1)2 ) .Proof. There exists 0 < i ≤ b k +2 , (it is equal to b k +2 , for I ( k +2)3 butfor pieces of the orbit it is less than) such that I ( k +2)4 ⊂ T i ( I ( k +2)3 ). So T i + b k +2 , ( I ( k +2)3 ) ∩ I ( k +2)2 = ∅ . Also T i + b k +2 , + b k +1 , ( I ( k +2)3 ) ∩ I ( k +1)2 = ∅ . In fact, T i + b k +2 , + b k +1 , + jb k +2 , ( I ( k +2)3 ) ∩ I ( k +1)2 = ∅ for j < n k +3 . Pieces of I ( k +2)3 are inserted into I ( k +1)2 with a delay of b k +2 , whichis coprime to b k +1 , . It follows that T c k ( J ′ ) intersects every level of O ( I ( k +1)2 ). Bycondition 5 it follows that T r ( J ′ ) intersects every level of O ( I ( k +1)2 ) for c k ≤ r ≤ d k . Moreover, the pieces inserted take m k +1 b k +1 , to leave O ( I ( k +2)2 ). Because m k +1 b k +1 , > b k +2 , b k +1 , (condition 4) the piece does not leave O ( I ( k +2)2 ) beforeanother is inserted into its level. (cid:3) Lemma 3. At times d k to c k +1 J ′ , a level in O ( I ( k +2)2 ) , intersects every piece of O ( I ( k +2)3 ) .Proof. There exists 0 < i ≤ b k +2 , (it is equal to b k +2 , for I ( k +2)2 but forpieces of the orbit it is less than) such that I ( k +2)3 ∩ T i ( I ( k +2)2 ) = ∅ . Also I ( k +2)3 ∩ T i + jb k +2 , = ∅ for j < m k +2 . Because b k +2 , is relatively prime to b k +2 , we have T i + jb k +2 , ( I ( k +2)2 ) intersects each level of O ( I ( k +2)3 ) for j = b k +2 , .It follows from condition 4 that T r ( I ( k +2)2 ) intersects each level of O ( I ( k +2)3 ) for d k ≤ r ≤ c k +1 . Moreover, the pieces inserted take ( n k +2 − b k +2 , to leave O ( I ( k +2)3 ). Because ( n k +2 − b k +2 , > b k +2 , b k +2 , (condition 5) the piece doesnot leave O ( I ( k +1)3 ) before another is inserted into its level. (cid:3) Proof of Theorem 1. Let J , J be any two nonempty intervals in I . Therefore thereexists k such that both contain some level of a tower over I ( k ) . This implies thatthey contain a level from each tower over I ( k ) for all k > k + 1. This implies that T n ( J ) ∩ J = ∅ for n ∈ [ c k , d k ] because J contains a level of I ( k +2)3 and J containsa level of I ( k +1)2 . Also T n ( J ) ∩ J = ∅ for n ∈ [ d k , c k +1 ] because J contains a levelof I ( k +2)2 and J contains a level of I ( k +2)3 . It follows that T n ( J ) ∩ J = ∅ for any n > c k +1 . (cid:3) J. CHAIKA No IET is topologically mixing of all orders The argument is a straightforward application of [9]. Let T be a d -IET. Observethat a topologically mixing IET must be minimal (otherwise it splits into disjointinvariant components). Let J , J ′ be any disjoint intervals bounded by discontinu-ities of T l for some l , and n , ..., n d be natural numbers. We will find a violation oftopological mixing of order d + 1 at bigger times. Pick an interval V such that allof the first returns to V are greater than max { l, n , ..., n d } . We may also choose V so that T | V is an s -IET for some s ≤ d . By our assumption that the return times to V are larger than l , each level of a tower over V is either contained in J or disjointfrom J . Let U , U , ..., U s be its subintervals. T | U i is an s i -IET for s i ≤ s . Callits intervals U i, , ..., U i,s i and their return times r i, , ..., r i,s i . If x ∈ O ( U i ) ∩ J and x ∈ O ( U i,j ) then T r i,j ( x ) ∈ J . This is because x ∈ T k ( U i ) ⊂ J for some k < r i ,in fact x ∈ T k ( U i,j ). T r i,j − k ( x ) ∈ U i . So T k ( T r i,j − k ( x )) ∈ T k ( U i ) ⊂ J . Therefore d ∩ i,j =1 T r i,j ( J ) ∩ J ′ = ∅ . 5. Acknowledgments I would like to thank my advisor, M. Boshernitzan, for posing this problem,helpful conversations and encouragement. I would also like to thank A. Bufetovand W. Veech for helpful conversations and encouragement. I would like to thankthe referee for suggestions that improved the paper. 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