We prove the conjecture that an interval exchange transformation on 3-intervals with corresponding permutation (1; 2; 3)
! (3; 2; 1); and rationally independent discontinuity points, is never measure theoretically isomorphic to an irrational rotation

Dajani, K.

We prove the conjecture that an interval exchange transformation on 3-intervals with corresponding permutation (1; 2; 3)

! (3; 2; 1); and rationally independent discontinuity points, is never measure theoretically isomorphic to an irrational rotation

NARCIS 

A note on rotations and interval exchange transformations on 3-intervals

Utrecht University Repository

A NOTE ON ROTATIONS AND INTERVAL EXCHANGETRANSFORMATIONS ON 3-INTERVALSKARMA DAJANIAbstract. We prove the conjecture that an interval exchange transforma-tion on 3-intervals with corresponding permutation (1;2; 3) ! (3;2;1); andrationally independent discontinuity points, is never measure theoretically iso-morphic to an irrational rotation.1. IntroductionInterval exchange transformations were rst introduced by Keane in [K1], andare dened as follows. Let I = [0; 1); n  2 and  = (1;    ; n) a proba-bility vector with i> 0: Dene 0= 0 and i=Pik=1k; and set Ii=[i 1; i): Let  be a permutation of f1; 2;    ; ng; and consider the probabilityvector = ( 1(1);    ;  1(n)): Note that  1(i)> 0 for all i: Let 0= 0 andi=Pik=1 1(k); and set Ii= [i 1; i):Dene T : I ! I byTx = x  i 1+ (i) 1if x 2 Ii: T is called an (;  ) interval exchange transformation on n intervals. Itis clear that T is invertible, Ti 1= (i) 1and T maps Iiisometrically ontoI(i): Further, T is continuous except possibly at f1;    ; n 1g: At these pointsT is right continuous. Note that T is continuous at iif and only if  (i + 1) = (i) + 1: In other words, T is discontinuous at iif and only if Ti 1; Tido notappear in this order as consecutive terms in the ordered set fn0;    ; n2ng: Wesay T is in standard form if T is discontinuous at ifor all i = 1; 2;    ; n   1or equivalently, if  (i + 1) 6=  (i) + 1 for all i = 1; 2;    ; n   1: Notice that anyinterval exchange transformation on n intervals can be written in standard form asan interval exchange transformation on m intervals with m  n: Since if T is notin standard form, then T is continuous at ifor some i, then  (i + 1) =  (i) + 1;and so T maps the interval [i 1; i+1) isometrically onto [(i) 1; (i)+1): Thus,we can redene T on intervals with end pointsf0;    ; i 1; i+1;    ; ng:We repeat this process until all the remaining 's are discontinuity points of T:1991 Mathematics Subject Classication. 28D05.Key words and phrases. Rotations, interval exchange.12 KARMA DAJANIThe permutation  corresponding to T is said to be irreducible if (f1; 2;    ; kg) 6= f1; 2;    ; kg; for all k = 0; 1;    ; n  1:Note that if  is reducible, then T can be decomposed into two interval exchangetransformations, one on [0; k) and the other on [k; 1):We assume throughout thispaper that T is irreducible.Interval exchange transformations have been studied by several authors. Herewe mention few of the known results. In [K1], Keane studied the minimality ofsuch transformations, and in [K2] questions concerning unique ergodicity were in-vestigated. It is easy to see that if n = 2; T corresponds to a rotation and if n = 3,then T can be seen as an induced transformation of a rotation. Thus, if the 'sare rationally independent, then in both cases T is uniquely ergodic. Keynes andNewton [KN], and also Keane [K2] gave examples of interval exchange transforma-tions that are not uniquely ergodic. Masur [M], and independently Veech [V1, V2,V3, V4, V5] showed that almost every minimal interval exchange transformation isuniquely ergodic. Later Boshernitzan [B] gave another proof of this result by moreelementary means. Some of the spectral properties were studied by Veech in a seriesof papers [V3,V4,V5]. Oseledets [O] and Goodson [G] constructed ergodic inter-val exchange transformations with simple spectrum. Recently, Berthe, Chekhovaand Ferenczi [BCF] proved that every ergodic interval exchange transformation onthree intervals has simple spectrum. The rst interval exchange transformationwith continuous spectrum was given by Katok and Stepin [KS], their example isalso an exchange on three intervals. In [BCF], the authors gave other examplesof exchanges on three intervals with continuous spectrum, and they conjecturedthat no non-trivial exchange on three intervals is measure theoretically isomorphicto an irrational rotation. In section 2 we prove this conjecture as a corollary of arecent result by Simin Li [S], where he gave necessary and sucient conditions foran interval exchange transformation to be conjugate to an irrational rotation.2. Non-trivial exchanges on 3-intervalsLet 0 < l < m < 1 with 1; l;m rationally independent. Consider the intervalexchange transformation T given byTx =8<:x+ 1  l x 2 [0; l);x+ 1  l  m x 2 [l;m);x m x 2 [m; 1):T corresponds to the permutation (1; 2; 3) ! (3; 2; 1): Notice that T is the onlyinterval exchange transformation on 3-intervals which is irreducible and in standardform. Moreover, by a result of Keane [K1], T is minimal. We call T a non-trivialexchange transformation on 3-intervals. It is well known that T is an inducedtransformation of the interval exchange transformation S dened on [0; 1  l +m)bySx =x+ 1  l x 2 [0;m);x m x 2 [m; 1  l +m):Since after normalization S is isomorphic to an irrational rotation, S is minimaland uniquely ergodic, and hence so is T:A NOTE ON ROTATIONS AND INTERVAL EXCHANGE TRANSFORMATIONS 3Let  =1 l1 l+mand  =11 l+m. In [KS], the authors proved that if  hasunbounded partial quotients and if for some subsequence qnof denominators ofconvergents of ; we havej pnqnj < o(1q2n); and j  rqnj >cqnfor all r and some constant c > 0; then T is not measure theoretically isomorphic toan irrational rotation. In [BCF], it is proved that when  has bounded partial quo-tients, and  2 K() for some Cantor set K(); then T is not measure theoreticallyisomorphic to an irrational rotation.Simin Li [Li] gave recently necessary and sucient conditions for an intervalexchange transformation to be conjugate to an irrational rotation.Theorem 1 (Li). Let T be an interval exchange transformation, and let d(Tn) bethe number of discontinuities of Tn: Then, T is conjugate to an irrational rotationif and only if (i) Tnis minimal for all n  1; (ii) fd(Tn)g is bounded by someinteger N > 0 and (iii) there exists k > 0 and M  2N3+3N2such that d(Tk) =d(T2k) =    = d(TMk):Since a non-trivial interval exchange on 3-intervals is uniquely ergodic, to showthat it is not measure theoretically isomorphic to an irrational rotation, we provethat fd(Tn)g is an unbounded sequence.Theorem 2. Let T be a non-trivial interval exchange transformation on 3-intervalswith rationally independent discontinuity points. Let D(Tn) be the set of disconti-nuity points of Tn; and let d(Tn) denote the cardinality of D(Tn): ThenD(Tn) = fT il; T jm : 0  i; j  n  1g;and hence, d(Tn) = 2n:Proof: The proof is done by induction on n: The result is true for n = 1: SupposeD(Tk) = fT il; T jm : 0  i; j  k   1g;for k = 1; 2;    ; n: We prove the result for k = n + 1: Let0 < 1< 2<    < 2n< 1be the discontinuities of Tnwritten in increasing order. By the induction hypoth-esis,D(Tn) = fi: 1  i  2ng = fT il; T jm : 0  i; j  n  1g:Let 0= 0 and 2n+1= 1: The underlying partition of Tnis given byP(Tn) = f[i; i+1) : i = 0; 1;   2ng:Let nbe the permutation corresponding to Tn(notice that Tnis an intervalexchange transformation). Then,Tnf0; 1;    ; 2ng = fn0; n1;    ; n2ngwith 0= n0= 0; and Tni= nn(i+1) 1for i = 0; 1;    ; 2n: Furthermore, since1; l and m are rationally independent, and each niis a linear combination of 1; land m with integer coecients, it follows that l;m =2 fn0; n1;    ; n2ng: Nowinvertibility of T implies that Tn0;    ; Tn2n; Tm; T l are all distinct.4 KARMA DAJANISuppose l 2 (nr 1; nr); and m 2 (ns 1; ns): We consider three cases.Case 1. If r = s; then T nl; T nm 2 (p 1; p) where p = n 1(r): Since Tis an order preserving isometry on [p 1; p); it follows that T nl < T nm: Theunderlying partition of Tn+1is then given byP1(Tn+1) = f[0; 1);    ; [p 2; p 1); [p 1; T nl); [T nl; T nm); [T nm;p);[p; p+1);    ; [2n; 2n+1)g:To prove the result, we need to show thatf1;    ; p 1; T nl; T nm;p;    ; 2ngis the set of discontinuity points of Tn+1: LetD1= f0;    ; p 1; T nl; T nm;p;    ; 2ngandE1= fn0;    ; nr 1; l;m; nr;    ; n2ng;both considered as ordered sets. Then TD1= E1; and by discontinuity of Tnatp; we have Tnp6= nr: Further, ni2 (0; l) for 1  i  r   1; and ni2 (m; l)for r  i  2n; : Hence,Tn+1D1= TE1= fTm = 0; Tnr;    ; Tn2n; T l = 1 m;Tn0= 1  l; Tn1;    ; Tnr 1g:Here, the elements of TE1are listed in increasing order.We rst show that Tn+1is discontinuous at ifor i 6= p: To do this, we needto prove that Tn+1i 1and Tn+1ido not appear in this order as consecutiveterms in TE1: By assumption, Tnis discontinuous at i; hence Tni 1and Tnido not appear as consecutive terms of the form nj; nj+1in E1: Let I0= [0; l);I1= [l;m) and I2= [m; 1): If Tni 1; Tni2 Ijfor some j = 0; 2; then sinceT maps Ijisometrically onto TIj; it follows that Tn+1i 1and Tn+1icannotappear as consecutive terms in TE1: If Tni 12 Ijand Tni2 Ikfor j 6= k;then either Tni 12 I0and Tni2 I2; or Tni 12 I2and Tni2 I0: In therst case we get Tn+1i< 1   m < Tni 1; and in the second case, we getTn+1i 1< 1 m < Tn+1i: Hence, Tni 1and Tnido not appear as consecutiveterms of the form nj; nj+1in E1; and so Tn+1is discontinuous at i:Now, the discontinuity of Tnat pimplies that Tn+1p6= Tnr; andTm = 0 < Tnr< Tn+1p:Hence Tn+1(T nm) = Tm = 0 and Tn+1pdo not appear as consecutive terms inTE1: So Tn+1is discontinuous at p:The discontinuity of Tn+1at T nl follows from the fact that Tn+1p 1is an inte-rior point of TI2; while Tn+1(T nl) = 1  m is the left end-point of TI1: Finally,Tn+1(T nm) = 0 < Tnr< 1 m = Tn+1(T nl) implies that Tn+1is discontinu-ous at T nm: Therefore, D1= D(Tn+1):Case 2: If r < s and p =  1nr <  1ns = q; then T nl 2 (p 1; p) and T nm 2(q 1; q): The discontinuity of Tnat pand qimplies Tnp6= nrand Tnq6=ns: The underlying partition of Tn+1is easily seen to beP2(Tn+1) = f[0; 1);    ; [p 2; p 1); [p 1; T nl); [T nl; p);A NOTE ON ROTATIONS AND INTERVAL EXCHANGE TRANSFORMATIONS 5[p; p+1);    [q 1; T nm); [T nm;q); [q; q+1);    ; [2n; 1)g:To show the discontinuity of Tn+1at 1;    ; 2n; T nl; T nm; we consider theordered setsD2= f0;    ; p 1; T nl; p;    ; q 1; T nm;q;    ; 2ngandE2= fn0;    ; nr 1; l; nr;    ; ns 1;m; ns;    ; n2ng:Then, TnD2= E2:Notice that n1;    ; nr 1are interior points of I0; nr;    ; ns 1are interior points of I1and ns;    ; n2nare interior points of I2: Thus,Tn+1D2= TE2= fTm = 0; Tns;    ; Tn2n; T l = 1 m;Tnr;    ; Tns 1; Tn0= 1  l;    ; Tnr 1g:Here, the elements of TE2are listed in increasing order. We rst prove that Tn+1is discontinuous at ifor i 6= p; q: If Tni 1; Tni2 Ij; then since Tni 1; Tnido not appear as consecutive terms in E1and since T is an isometry on Ij; wehave that Tn+1i 1and Tn+1iare not consecutive terms of E2; and thus Tn+1is discontinuous at i: If Tni2 Ijand Tni 12 Ikfor k 6= j; then we considerseveral cases. If Tni2 I2and Tni 12 I0or I1, then since Tni 16= l we have Tn+1i<1 m < Tn+1i 1: If Tni2 I1and Tni 12 I2; then since Tni6= l it follows that Tn+1i 1<1 m < Tn+1i: If Tni2 I1and Tni 12 I0; then Tn+1i< Tn+1i 1: If Tni2 I0and Tni 12 I1; then since i 6= q we have Tn+1i 1< Tns 1<Tn0 Tn+1i: If Tni2 I0and Tni 12 I2; then Tn+1i 1< 1 m < Tn+1i:In all the above cases we see that Tn+1is not continuous at i:The discontinuity of Tn+1at pand qfollows from the fact that Tn+1p6= Tnrand Tn+1q6= Tns; so that neither Tn+1pand Tn+1(T nl) nor Tn+1qandTn+1(T nm) appear as consecutive terms in TE2: Finally, from Tn+1(T nl) =1 m < 1  l < Tnr 1and Tn+1(T nm) = 0 < 1 m < Tns 1we have that Tn+1is discontinuous at Tn+1(T nl) and Tn+1(T nm): Thus, D2= D(Tn+1):Case 3: If r < s and p =  1nr >  1ns = q; then the underlying partition of Tn+1is given byP3(Tn+1) = f[0; 1);    ; [q 2; q 1); [q 1; T nm); [T nm;q); [q; q+1);    ;[p 2; p 1); [p 1; T nl); [T nlp); [p; p+1);    ; [2n; 1)g:LetD3= f0;    ; q 1; T nm;q;    ; p 1; T nl; p;    ; 2ngandE3= fn0;    ; nr 1; l; nr;    ; ns 1;m; ns;    ; n2ng:Then,Tn+1D3= TE3= fTm = 0; Tns;    ; Tn2n; T l = 1 m;Tnr;    ; Tns 1; Tn0= 1  l;    ; Tnr 1g:6 KARMA DAJANIThe elements of D3; E3and TE3are listed in increasing order. A similar argumentas in the above two cases shows that D3= D(Tn+1): Thus, the theorem is proved.Theorem 3. Any non-trivial interval exchange transformation on 3-intervals withrationally independent discontinuity points is not measure theoretically isomorphicto an irrational rotation.Proof: By theorem 2 and unique ergodicity, the result follows from Li's theorem.In [BCF], the authors proved that every ergodic interval exchange transformationon three intervals has simple spectrum. Using this result and theorem 3, we havethe following corollary.Corollary 1. Every non-trivial interval exchange transformation on three inter-vals with rationally independent discontinuity points has either rational or contin-uous spectrum.References[BCF] Berthe, V., N. Chekhova and S. Ferenczi - Covering numbers: arithmetics and dynamicsfor rotations and interval exchanges, J. D'Analyse Math. 79 (1999), 1-31.[B] Boshernitzan,M. - A condition for minimal interval exchange maps to be uniquely ergodic,Duke Math. J. 52 (1985), 723-752.[G] Goodson, G.R. - Functional equations associated with the sectral properties of compactgroup extensions, Proceedings of Conference on Ergodic Theory and its connection withHarmonic Analysis, Alexandria 1993, Cambridge University Press, 1994, 309-327.[K1] Keane, M.S. - Interval exchange transformations, Math. Z. 141 (1975), 25-31.[K2] Keane, M.S. - Non-ergodic interval exchange transformations, Israel J. Math. 26 (1977),188-196.[KN] Keynes, H. and D. Newton - A minimal non-uniquely ergodic interval exchange transfor-mation, Math. Z. 148 (1976), 101-105.[KS] Katok, A.B. and A.M. Stepin - Approximations in ergodic theory, Uspekhi Math. Nauk22, 5 (1967), 81-106 (Russian), translated in Russian Mth. Surveys 22, 5 (1967), 76-102.[Li] Li, Simin - A Criterion for an Interval Exchange Map to be Conjugate to an IrrationalRotation, J. Math. Sci. Univ. Tokyo 6 (1999), 679-690.[M] Masur, H. - Interval exchange transformations and measured foliations, Ann. of Math.115 (1982), 169-200.[O] Oseledets, V.I. - On the spectrum of ergodic automorphisms, Doklady Akad. Nauk SSSR168, 5 (1966),1009-1011 (in Russian), translated in Soviet Math. Doklady 7 (1966), 776-779.[R] Rauzy, G. - Echanges d'intervalles et transformations induites, Acta Arith. 34 (1979),315-328.[V1] Veech, W.A. - Interval exchange transformations, J. D'Analyse Math. 33 (1978), 222-272.[V2] Veech, W.A. - Gauss measures for transformations on the space of interval exchangemaps, Ann. of Math. 115 (1982), 201-242.[V3] Veech, W.A. - The metric theory of interval exchange transformations. I Generic spectralproperties, Amer. J. Math. 106 (1984), 1331-1359.[V4] Veech, W.A. - The metric theory of interval exchange transformations. II Approximationby primitive exchange, Amer. J. Math. 106 (1984), 1361-1387.[V5] Veech, W.A. - The metric theory of interval exchange transformations. III The Sah-Arnoux-Fathi invariant, Amer. J. Math. 106 (1984), 1389-1422.Universiteit Utrecht, Fac. Wiskunde en Informatica and MRI, Budapestlaan 6, P.O.Box 80.000, 3508 TA Utrecht, the NetherlandsE-mail address: dajani@math.uu.nl

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A note on rotations and interval exchange transformations on 3-intervals

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