INTERVAL EXCHANGE TRANSFORMATIONS
By WILLIAM A. VEECH*
00. Introduction
Denote by A,. C_ R m the cone of positive vectors
A =(A1,...,Am), A~>0, alli.
Let I A I = E Ai, and partition [0, I A I) into m subintervals [/3i_1,/3~), 1 =< i -< m, where /3, =/3,(A) is
= ~ 0 i=0, /3, (,x) t AI+-'-+A~ l<=i<--m.
~,, is the permutation group on {1, 2,..., m}. Given A E Am, ~r ~ ~,, set
X "/I" = (A 'rg 11, A .fi---12, " * * , a ,rf--lrtl)~
Finally define T = T(~.~), the (A, 7r) interval exchange by the formula
(0.1) T.,~)x = x -/3,_,(x) +/3~,_,(;t ~) (x ~ [/3,_1(x),/3, (x))).
For typographical reasons we shall, in places where no ambiguity might arise, speak of (A, 7r) or T instead of T(~.,~). The interval exchanges are precisely the piecewise order preserving isometries (with finitely many discontinuities) of intervals [0, ~'), ~- > 0 (z = I A I). The inverse of (A, 7r) is (A ~, ~'-~) (see [8]). Formally, T in (0.1) has m - 1 discontinuities, but there may actually be fewer depending upon zr. In any case, T is right continuous at all points.
9 Research supported by NSF-MCS75-O5577 and an Institute for Advanced Study grant.
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JOURNAL D'ANALYSE MATHI~MATIQUE, Vol. 33 (1978) INTERVAL EXCHANGE TRANSFORMATIONS 223
M. Keane [8] made the first basic study of interval exchanges, and it will be convenient to recall here certain of his results. Call T minimal if for each x E [0, I A I) the orbit r = {T"x t n ~ Z} is dense in [0, I A I). T satisfies the infinite distinct orbit condition (i.d.o.c.) if (a) r (A)) is an infinite set, 1 _-< i _-- (b) C(/3, (;~)) n r (X)) = ;~, 1 =< i ~ j _-__m - 1. 0.2. Theorem (Keane [8]). If (A, or) satisfies the i.d.o.c., then (A, zr) is minimal. An obvious necessary condition for (A, ~') to be minimal is that 7r not map dny segment {1, 2,... ,j}, j < m, into itself (for then [0,/3j(A)) would be an invariant set). Such a ~" is irreducible. If 7r is irreducible, Keane observes that a sufficient condition for the i.d.o.c, to be satisfied is that A be irrational; that is, that I AI, A1,'-',Am-1 be linearly independent over Q, or, equivalently that all rational relations between A~/IAI,..-,Am/IA I be consequences of EA,/IAI= 1. In particular, 0.3. Corollary. If Ir ~ ~, is irreducible, then for almost all A E Am the (A, ~) interval exchange is minimal. To each interval exchange (A, 7r) we associate the cone E (A, ~-) of (nonnegative) nonzero invariant measures for (A, ~r). E(A, ~r) contains Lebesgue measure on [0, IA I) (and all its positive scalar multiples). We say (A, zr) is uniquely ergodic if E(Jt, ~r) is one-dimensional; that is, if every invariant measure for (A, 7r) is a multiple of Lebesgue measure. In case m = 2 and 7r = (12----~ 21), the (A, ~r) interval exchange is simply addition by A2 modulo I A I. It is well known that minimality, which here is equivalent to the irrationality of A2/IA I, implies unique ergodicity. When m = 3, there is only one permutation, ~" = (123--~ 321), such that ~ is irreducible and (A, 7r) has 2 points of discontinuity. In this case, Keane notes that minimality is equivalent to the irrationality of (IAI-A3)/(IAI-A1), and again minimality implies unique ergodicity. The extremals of E(A, It) are well known to be the ergodic invariant measures, and any two ergodic invariant measures are either mutually singular or multiples of one another. Keane, extending a result of Oseledets [13], proves a minimal interval exchange has at most m + 1 pairwise mutually singular ergodic measures, and notes his proof can be altered to obtain the bound m. As ~(A, or) is spanned by its 224 w.A. VEECH extremals, dim 2(A, or) =< m. Notice that because any collection of nonzero, pairwise mutually singular measures is linearly independent over R, (0.4) v(A, ~r) = dim X(A, Ir) is the number of ergodic invariant measures for (A, or). In Section 2 we shall associate an m x m matrix (of O's, and • l's) L" to r and prove that dimE(A, or)_-<1 rank(L') if (A, It) is minimal. 0.5. Theorem. I[ (A, lr) is a minimal interval exchange, then 1 (0.6) v(A, ~r)<~ rank(L'). In particular, v(A, r < [m/2]. Keane conjectured that minimality implies unique ergodicity, as apparently did a number of mathematicians out of print, but Keynes-Newton [11] produced examples with m = 5 and v(A, r 2. In turn, to negate a conjecture by Keynes-Newton, Keane [9] gave examples with m = 4, v = 2. By (0.6) and the remark following, these examples maximize v(A, or) for m = 4,5. Keane's construction employed the ergodic theoretic device of inducing a transformation on a subset (this will be defined below), and as was stated in [15], his procedure suggested a formal study of this process as a possible way to settle his conjecture 0.7. Conjecture (Keane [9]). If ~r E ~m is irreducible, then for almost all A E Am the (A, 7r) interval exchange is uniquely ergodic. The purpose of this paper is to begin such a formal study of the inducing process. Our ultimate hope is that this will not only lead to a solution to Keane's conjecture but, more importantly, to a usable criterion for the unique ergodicity of (A, or). This could be of use, for example, in the study of "rational billiards" (see [10]). We shall define a set ~+~ _C ~,. and show that if ~r E ~, and if (A, It) satisfies the i.d.o.c., then for a certain collection of "admissible" intervals [~, 1,/] C_ [0, IA I), the INTERVAL EXCHANGE TRANSFORMATIONS 225 induced or first return time transformation is a (translate of) [A ct), 11"1) interval exchange, I Atl)l = ~ - ~, rr~ ~ @~, and there exists a nonnegative integer matrix A, det A = --- 1, such that A = AA t" (compare with [9]). Repeating this procedure, i.e. using a decreasing sequence F~ D F2 D 9 9 9of such admissible intervals, one obtains an "expansion" (0.8) A = A (~ (27... A (")A (") and it will develop that if F, ~O or a point (0.9) S(A, rr) = f'l A (''" "A<")A,- n is naturally isomorphic with X(A, ~r). In particular, if there is a matrix B with positive entries such that B occurs infinitely often as a product A (')A 0+~)... A o) in (0.8), then (0.9) is one dimensional, and (A, 70 is uniquely ergodic. It is in this fact that one might expect to find a usable criterion for unique ergodicity. One may specialize F,, setting F1 = [0, A~) and in general F,+I = [0, A]"~). Here one obtains a map if0:~2 • Am--~ ~2 x A= and an associated "projective" map ~:~+Xam_X'-*~++Z~m_,, Am-l={~, EArn [[,A[= 1}, where f AO) '/7"1) An application of the Poincar6 recurrence theorem, in conjunction with the criterion of the preceding paragraph will show that if ~r possesses a finite invariant measure equivalent to Lebesgue measure, then Keane's conjecture is true. In case m = 2, ~ = {(12---~ 21)}, there is an infinite invariant measure (viewing A~ as (0, 1), it is dx/(1-x)). In case m ---3, ~ = {(123---~321)}, we are able, after a rather lengthy argument, to assert that there is an invariant measure equivalent to Lebesgue, but it remains to be seen whether it is finite or infinite. In any case, we do obtain that if B~, 9 "" , Bk are matrices which can occur in (0.8) in the case of ~*3~ then for almost all A, the block B1B2. 9 9B~, occurs infinitely often as A (,§ A 0+~) in (0.8). 226 w.A. VEECH 01. The cone of invariant measures Fix m > 1, (A, 7r) E ~,, x Am, and assume (A, 7r) is minimal. As before ~(A, 7r) is the cone of nonnegative nonzero T invariant Borel measures, T = Tr We shall in this section identify ~(A, 7r) with a cone S(A, 7r)C_ Am. Let/z E E(A, 7r). Being minimal, T can have no periodic points, and therefore/z has no atoms. Also, if ~7~ O is an open set, /z(~7)>O. It follows that ~(x)= ~[0,x), 0~x is continuous and one-to-one. Define )t (tt)E A,. by x, (g) = g (/3,_,(x ). /3, (x)) = ~ (/3,) - ,p~ 03,-1). Next, define T. on [0, [A (g)l) by (1.1) T,,p~ (x)= ,p, (Tx). T. is, as we shall soon prove, the (A (/z), zr) interval exchange. Notice that by (1.1) (1.2) T~o,(x)= ~o,(T"x) (n EZ). 1.3. Lemma. With notations as above T~ is the (A (tt), or) interval exchange. ProoL If x E [13~-~(A),/3~ (A)), then by definition Tx = x -/3,_1(A) +/3~,_,(A"), and ,p,.(Tx) = ~[0,x - t3,-,(X) + t3~,-,(X ~)) = ~.L [0, ~Iri--l(/~ ~r)) "4- [.~ [~r,--l(/~ ~r), ~Tri--l(/~ "n') "JI- X -- ~i--l(/~)). Now use the invariance of/z for both terms on the right to obtain ~ (ix) = t3~,_l(x (~)~) + ~ [~,-,(x), x) (1.4) = 9, (x)- t3,_,(x (~)) + t3,-dx (~)"). INTERVAL EXCHANGE TRANSFORMATIONS 227 Of course (1.4) says simply that T~ is the (A(/z), zr) interval exchange. 1.5. Lemma. The map I~--->A(p.) defined above is an affine isomorphism between E(A, ~r) and its image, denoted S(A, 7r) C_ Am. Proof. The map is obviously affine. We must prove it is injective. To this end note that for each /~, ~, (0)= 0, and therefore by (1.2) (1.6) TT~0 =/.t [0, Tn0). T~ is now known to be (A(p~), ~'), and therefore T~0 is determined by A(/~) (and 7r). As n varies, T"0 runs through a dense subset of [0, [A I), and therefore by (1.6) /~[0, x) is determined for all x ~ [0,1A I) by A(/~) (and 7r). That is, /~---> A(/~) is injective. 1.7. Lemma. Let T be a minimal (A, ~r) interval exchange. Then dim S (A, It) is the number of pairwise distinct ergodic invariant probability measures [or T. Proof. This follows from Lemma 1.5 and the remarks in the introduction. Remark. Of course dimS(A, Ir)_-- Remark. Suppose given minimal interval exchanges (A, It) and (A', or') and a homeomorphism q~ : [0, t A I)'-> [0, I A'[) which intertwines the two. If/~ is the pull back on [0, ]A I) of Lebesgue measure on [0, I A'[), then ~ = ~ and so Tt~, ,~,) = T~. This does not say A'= A(/x), ~-'= 1r because, e.g., A' and A may have a different number of components. If however (A, 7r) has m - 1 discontinuities and A, A' are m dimensional, then A'= A (/x). We have 1.8. Proposition. Fix r E ~m irreducible, and let ~t 7. C Am be the set of A such that (A, ~r) is minimal. The sets {S(A, ~')IA E J,t ~,} partition Jl,g~,. If (A, or) has m - 1 dicontinuities, then A'~ S(A, cr) if and only if (A, 7r) and (A',cr) are continuously isomorphic. 228 W.A. VEECH {}2. A bound for dimS(A, Tr) Associate to 1r E ~= the permutation matrix 7r,j = B.m). Let E= = E be the m x m matrix of O's and l's which has 0 on and below the diagonal and 1 above. Finally, define an (alternating) matrix L" by (2.1) L ~ = E - rr'Err. 2.2. Lemma. I[ A E A,~ is regarded as a column vector, and it" T is (A, ,r), then (2.3) Tx =x + (L"A), (x E [/3,_,(A ), /3, (A ))). Proof. By a simple computation (2.4) Therefore, (2.5) (L"A), = ~ A,- Z A,. j>i jwi The first sum on the right contains those summands in fl,,-l(A ") which do not occur in /3,_,(A), while the second sum on the right contains those summands in /3~_t(A) which do not occur in /3,,_ffA*). In other words, by (2.5) (L'A), = fl.,_,(A ~)- ~,_,(A). The lemma follows from (0.1). 2.6. Remark. Note that (2.4) implies L" is an alternating matrix. With the aid of L" we introduce an alternating bilinear pairing on R m by (2.7) (u,v).=u'L'v (u,v E R"). The pairing is possibly degenerate (always so if m is odd). Notation -l-,r denotes perpendicularity relative to ( , ),~. INTERVAL EXCHANGE TRANSFORMATIONS 229 2.8. Lemma. If (A, 7r) is a minimal interval exchange, then S(A, ~') J_,, s(x, 9). Proof. If ~ E E(A, zr), then by the invariance of /~ and (2.3) we have Ixl o = f (Tx - x)g(dx) o =- ~ (L ~A ),A, (/x) i=l = (x (#), x).. This computation implies the lemma because S()~,~r)=S(A',~) for all x'E s(x, ~r). It is convenient to isolate those zr ~ ~m which are irreducible and which have the property that ~r(j + 1)~ (rq)+ 1 for all j. The latter condition ensures that (A, rr) will have m -1 actual points of discontinuity. ~*~ denotes the set of irreducible permutations with the property that 1r(j + 1)~ it(j)+ 1, all j. It is also convenient to introduce notation for the purpose of expressing T'x. Define for each x E[O, IAI), n > 1, and i, l= n-1 (2.9) M(n, A, zr, x, i) = ~'~ Xt~,_~.a,)(Tix). j=O That is, M(. ) is the number of visits (by x) to [/3,_5,/3,) under T j, 0 =< j < n. Now it is clear that (2.10) T"x = x + ~ M(n, A, zr, x, i)(L'A),. n=l (For example, if n = 1, (2.9)-(2.10) reduce to (2.3).) Define V(A, It) = S(A, zr)- S(A, 7r) to be the subspace generated by S (dim V = dim S). Also, let N(1r) be the null space of L", and note that L ~ and its transpose have the same nullspace. 2.11. Lemma. Assume ~r ~ ~*.. and that (A, ~r) is minimal. Then S(A, It) O NOr) = {0}. 230 w.A. VEECH Proof. Fix w E N(Tr), and suppose w = A'- A", where )t', A" E S(A, ~'). We are to prove A' = A". Because (A ', 7r), (A", ~r) are continuously isomorphic and have the maximal number of discontinuities,* we have M(n, ~', A ', O, i) = M(n, "n', A", O, i) for all n >0 and i. Therefore, because A'= w +A" (and consequently L '~X'= L '~X "), Tr = ~ M(n, ~', A ', 0, i)(L'~A '), = ~ M(n,'n',h",O,i)(L'~A"), i=l = Trn ,~)0. Minimality of (A',~r) and (A",Ir) now implies Ttx,~)= Tt~..,,), and therefore A' = )t" 2.12. Theorem. If 7r E 6*, and if (A, It) is minimal, then dim V(A, ~) _--< 89(Rank L ~). Proof. Because V• there is a well defined linear map L : V---~ W*, the dual space of W = Rm/(V+N(Ir)), given by (L")'. (Lu(v + V + N(zr)) = u'L~v,) If Lu = 0, then u'L~v = 0 for all v ~ R m, meaning u'L "~ =0 or L"u =0. Thus u =0 by Lemma 2.11. The same lemma implies dim W = (Rank L'~)- dim V, and therefore 2 dim V =< Rank L'~. The theorem is proved. w Transformations on the body of interval exchanges In this section we shall elaborate on the transformations mentioned in [15], which in turn were suggested by Keane's construction [9]. Fix m > 1, (A, ~r) E Am • ~*, and let T = T(~.,~). Let F be an interval of the form r = 0_- * Set A, = tA" + (1 - t)A', 0 --_ The Poincar6 recurrence theorem implies that for almost all a ~ F there exists n > 0 such that T"a E F. Taking n = n (a) > 0 small as possible with this property (or 0o if there is no such n), define (3.1) Ua = T"ta)a (n(a)< oo). U is called the transformation induced on F by T. It is known U is measure preserving and, because T is one-to-one, one-to-one ([7]). If a E F is such that n(a) < 0% the right continuity of T implies there exists b > a such that n(x) = n(a), x ~ [a, b), (3.2) T n is continuous on [a, b), 0<= n <- n(a). If (a, b) is an open interval on which n(x) is constant (finite) and such that the second line of (3.2) holds for (a, b), then by right continuity all of (3.2) holds for [a, b). It follows then that every x E F for which n(x) < ~ is contained in a maximal interval [a, b) satisfying (3.2) (hereafter referred to as a maximal interval). Suppose [a, b), b < r/, is a maximal interval. We claim there exists j = ](b) (chosen as small as possible) such that 0=<] < n(a) and TJb is one of the m + 1 points ~, 71,/3,,. 9 For if it were otherwise, there would exist e > 0 such that [a,b +e) satisfies (3.2), contradicting maximality. Denote by w(b) the point w = T'b, j = j(b). Suppose now that [a',b') is a second maximal interval and that b' TJ(b- e,b) =Tr(b '- e,b'). (One may take 0 < e < min(b - a, b'- a').) If one of j, ]' = 0, say j = 0, then as ]' < n(b' - e), it must be that/' = 0, and b = b', a = a'. If both j and/' are positive, then n(a)-j=n(a')-j', and the definition of U implies U(b-e,b)= U(b'- e, b'). But U is one-to-one, and therefore b = b', a = a'. We have now proved that U has at most m + 1 discontinuities. An additional hypothesis on F will reduce this number to m -1. In what follows (A, ~r) is assumed to satisfy the i.d.o.c. (Section 0), and ~, 7/ are to have the form = Tk/3, (1 _-- 3.4. Lemma. If (A, or, I') is admissible, the induced transformation U has at most m - 1 points of discontinuity on F. Proof. We shall prove that two of the four values 6, 7/,/3,/3, are omitted by {w(b)[ [a, b) maximal, b < 17}. Thus, U has at most m maximal intervals and at most (m- 1) points of discontinuity. Fix a maximal interval [a, b), b < ~/, and suppose w(b) =/3, with (first) j(b) = 0. We shall prove that if [a', b'), b'< vl, is a maximal interval, then w(b')~ 6. For if w (b')= sr, then j'=/(b') > 0, and Trb '= = TE/3,. Because b =/3,, we have k < 0 or k > 0. But admissibility does not, because/3, E F, allow k <0. Therefore k >0. Now Tk-r/3, = b'E F, and therefore by admissibility k-j'_-<0. In fact k-j'<0 because /3,~ b'. Finally, from 0< j'- k < j' and Tr-kb ' =/3,, we obtain contradiction to the definition of j (b') = ]'. A similar argument shows that if w(b)=/3, with j(b)= O, then w(b')~ 71 for any maximal [a', b'), b' < 7/. (If j(b) = 0, it is not possible that w(b) = ~ or ~/, i.e. that b = sr or 7/ because ~ 0. We shall show that if k -> 0, then w (b) ~ 6, while if k < 0, w(b) ~/3,. Only the proof for k _-->0 will be given because they are so similar. ThuS, suppose k _-> 0, j(b) >0, TJb = 6. As TE/3, = (;, and hence T~-J/3, = b, admissibility requires k - j _-< 0. Again,/3, ~ b so k - j < 0. Therefore, TJ-~b =/3,, and 0 < j - k < j, contradicting the choice of j(b). Entirely analogous arguments prove that if /(b)>0 and l->0, then w(b)~vt, while if l<0, then w(b)#/3t. The function b ~ w(b) is one-to-one from maximal intervals (with b < v/), and therefore there are at most m maximal intervals (m - 1 with b < ~/). The lemma is proved. Remark. With notations as above, suppose [a, b) is maximal and w(b) = ~: or 71. Then U is certainly discontinuous at b (in case w(b) = 6, use the right continuity of T). However, if w(b)=/3j for some 1 <] _-< m- 1, it can happen that U is continuous at b. For example, if r ~ ~*~,/is may be a point of continuity for T. In fact, there are other instances in which there will in general be fewer than m - 1 discontinuities for U. The discussion to follow has the purpose of determining these instances. Example. m =3, w=(123-,321). Keane observes that if a3 0.5) For some j r w(j + 1) = 1 Irm = (w 1)- 1 (3.6) For some ] or] = m r = r + 1)- 1 (3.7) For some j r + 1) = 1 then F can be chosen so that U has at most m- 2 discontinuities. Notice that (123--> 321) satisfies (3.6) with j = 1, m = 3. ~ denotes the set of irreducible permutations. 3.8. Lemma. Assume r E G~ and let A be irrational. (Hence the i.d.o.c. holds.) If 1 < j <- m - 1 is such that T is discontinuous at [3j, and if there exists n > 0 such that T" is continuous at {3i, then r satisfies (3.5) and T 2 is continuous at [3j. Proof. Choose the smallest v > 0 such that T ~ is continuous at /3j. Certainly v > 1, and the i.d.o.c, implies T~-~/3~ is a point of continuity for T. If T~-~/Sj >0, then T -~ is continuous at TV[3j, and so T v-~= T-~T v is continuous at /~j. This contradicts our choice of v, and therefore TV-~flj = 0. As Tfl~ = 0 for some k, the i.d.o.c, implies v = 2, T/Sj = 0. In particular, r + 1) = 1. Now let w = xT~lim Tx. We claim w =/3, for some t _-< m. For if not, T is continuous at w. As w > 0, T -~ is continuous at Tw -- T2/3;, and then w = T/ij, a contradiction. Finally, supposing w = fit for some t, the definitions of w and T imply /3, =/3j + (L'rA)~. 234 w.A. VEECH Since a is irrational, it must be that t = m and /3, = fxf, I ko, k>j. That is, lrj = m. We also have T(O) = lim Tx, x11,~f and therefore (L "A )1 = lit I + (L"A ),~, or by irrationality L~'l= 1 + L,~, (1_-< I-< m). Thus, if rrl < ~rl (resp. lrl > ~rl) it must be that lrm- r (resp. ~rm < ~rl). It follows that ~rm = (It 1)- 1, and rr satisfies (3.5), as claimed. 3.9. Lemma. Assume lr ~ ~ and A is irrational. Let j, 1 <-_ j <- m - 1, be such that T is discontinuous at/3j and T/3j # O. Let w = lim Tx, w*= lim T2x. xT~aj x .-"~j We have : (i) If w < I A I, then T~w ~ Tq/3j for all p, q. (ii) If w = [A I, and if TPw * = Tq/3i for some p, q, then either ~r satisfies (3.5) or else 7r satisfies (3.6) and T/3j = w*. Proof. Assume first w T/3j =/3j + (L"A)j§ (3.10) w * =/3i + (L "A )j + (L'~A).,. By irrationality, (3.11) LT§ (l= As 7rj = m (because w = [hi), (2.4) implies L~={ 0 k<=l' 1 k>j. Set k = ] + 1 in (3.8), and use the antisymmetry of L" to conclude that L~'+l.m = 1; that is, 7rm < ~r(j + 1). If for some p, 1rm < ~rp < ~'(j + 1), then L~ = LT§ and L T~p = -1. Then (3.8) must be false for k =p, and we conclude that ~rm = rr(/" + 1)-1. That is, (3.5) holds. The lemma just proved excludes the possibility that T/3s = 0. Suppose now that T/3~ = 0. If w = I h [, the fact T -~ is the (A ~,1r -~) interval exchange implies w* =/3~,,. If T~'w * = Tq/3s = Tq-~O = Tq-2(T(O)) the i.d.o.c, for T -~ implies p = q - 2 and/3 ;,~ = T(0). Thus, w * = T2/3j, a contradiction if Ir does not satisfy (3.5). If T/3j = 0, and if w < [ A I, arguments analogous to those in Lemma 3.9 show that if Taw = T'~/3j for certain p, q, then w = T2/3j, which entails 236 w.A. VEECH /3~ + (L"A)j =/3j + (L'A)t +1 + (L"A)I and L; = Li+l.k + L[k. These relations imply in turn that r satisfies (3.7). 3.12. Definition. Let ~* be the set of ~r ~ ~ such that for all ] ar(j + 1) ~ (~'j)+ 1. Let @+~ be the set of lr E ~* which do not satisfy any of the conditions (3.5)-(3.7). Remark. It is easily verified that ~+~ = (~2)-1. If lr E @*~ satisfies one of conditions (3.5)-(3.7), it is possible for any A to choose F admissible so that the induced transformation U has at most m - 2 discontinuities. It is possible to argue that if "almost all" minimal interval exchanges on fewer than m intervals are uniquely ergodic, then for almost all A, the (A, ~r) interval exchange is uniquely ergodic. We turn our attention now to ~. If 7r E @2, and if A E Am is irrational, then for any admissible interval F the transformation U induced by (A, Ir) will have precisely m - 1 points of disconIinuity. It has no more than this number because F is admissible. On the other hand, if 1 _-< ] _-< m - 1, and if ~ < T-"/3j < 7/ with n _-> 0 as small as possible, then T-"/3j visits flj before returning to F. If/3j = ~ or r/, T-"flj is surely a discontinuity of U. Otherwise, if T-"ISj is not a discontinuity, it must either be that w = limx.,o~ < [A [ and Taw = Tqflj for certain p, q _-> 0 or else w = [A [ and Taw * = Tql3j for certain p, q _-> 0. We have seen that these are possibilities only if zr E ~+~. If we translate F = [~, r/) to F-~ = [0, 17- ~), then U is a (A', lr') interval exchange for some A' E Am, [A '1 = ~/- ~:. Now U is minimal because (A, ~r) is, and therefore zr'~ @~ Because U has m - I discontinuities 1r'E @*~. Finally, we claim 1r'E @2. For if this were not so, there would exist F'_C F, admissible for both (A ', lr') and (A, It), such that the transformation U' induced by (A ', It') on F' has at most m - 2 discontinuities. But then the same is true for (A, It) induced on F' (which is the same as U'). We have 3.13. Proposition. Let (A, ~r) E Am • ~+~ be such that A is irrational, and let INTERVAL EXCHANGE TRANSFORMATIONS 237 F = [Tk/3,, TZ[3,) be admissible. If U is the induced transformation on F, then U is (effectively) a (;t', or') interval exchange with (A', r Am x @2. Let F = [~, 7/) be admissible, and suppose the induced map has m - 1 discon- tinuities with separations a~,- 9 a,~, al + 9 9 9+ am = 17 - ~. Because of (3.2), each point of [~ +/3k-t(a), ~ +/3k (a)) spends the same amount of time in the interval [/3i-l()t ), /3~ (A )), before returning to F. Let A(i, k) denote this number. Since T is minimal, we must have x, = ~ A (i, k)a~, k~l or what is the same, (3.14) A = Aa. (See [9].) On the other hand, each division point has the form T~/3j for certain v, j. Now TVl3s is an integer combination of the Aj's, and therefore each ak is an integer combination of At's. Thus, there is an integer matrix B such that (3.15) a = B)t. Putting (3.14)-(3.15) together, we have (3.16) ABA = ,L If rank A < m, the set of ;t satisfying (3.16) has measure 0. The set of integer matrices being countable, it must be for almost all ;t that rank A = m and B = A -~. In particular, det A = det B = +_ 1. It is possible to express T in terms of U by the procedure of "stacking". With notations as above, let Rk = ~ A(i,k), i=l and let h = [r + jsk_,(o~), ~ + ~k (,~)), 1 --- k _-< m. 238 w.A. VEECH Set up a space X'= {(Ik, p)ll =< k _-< m, 0-< p < Rk}, and define T':X'---} X' by I (x,p + 1) xEh, 0_---p T'(x, p) = [(Ux, O) XEIk, p=Rk--1. With the natural measure (of mass Y. Rkak = I A I) T' is measure preserving, and on F'= [-J(h,0), T' induces U'~ U. Now define q~ : X'---~ [0,1A I) by requiring q~'--- 1 on each (h,p) and (3.17) q~ ((h, p)) = TVlk. Evidently q~T' = Tq~. Note that q~ merely determines an "order of appearance" for the intervals (h, p) in [0, I)t I), and this prescription for order of appearance may be used for any a E A,, and the fixed permutation which corresponds to U. As a varies in A,. (and X' varies in total mass but not number of "floors" above (h, 0)) (3.17) may be used to define T on [0, E Rka ;,). Evidently, the number (and position) of the discontinuities for T varies continuously, and so T = T ca'~ is an interval exchange on m intervals a', )t'= Aa'. The permutation associated to T ta'} also varies continuously and hence is always zr. We have 3.18. Proposition. With notations and assumptions as above, assume U is, for fixed (A, ~r, F) the (a, It') interval exchange, and let ;t ', a ' E A., be related by A' = Aa'. Then if F = [r r/) = [T~.,o/3, (A), TI~.,,)/3,(X)), and if we set F' = [ T~,..~)fl. (A '), T~A,,.,fl,(A ')), the transformation U' induced on F' by T~,.,o is the (a', zr') interval exchange. In particular, there is a local map (3.19) 8-(k.~.s.,) : Am X ~+~---> A,, x ~+., given by (3.20) gr(k.~,)(A, It) = (A-'A, qr') INTERVAL EXCHANGE TRANSFORMATIONS 239 and realizing the process of inducing on an admissible interval F = [Tk[3,, T*I3,). Remark. (3.19) may not be defined a.e. because F, for fixed k, l, s, t may not be admissible a.e. If we choose F carefully, e.g., F= [/3H(A ), /3, (A )), then F is admissible, and (3.19) is defined a.e. If (A, ~-), (A', ~')EAm x ~+~ are continuously isomorphic, there is a natural correspondence between admissible intervals F for (h, zr) and F' for (A', zr) (i.e. the same 4-tuples (k, I, s, t) give rise to admissible F's for each). For each such F, F', the induced maps U, U' are also continuously isomorphic. It therefore follows from the discussion in Section 2 that if(s(x, ~)) = s(if(x, ~)) for any (A, ~r) E Am x ~+~ which satisfies i.d.o.c. In terms of A in (3.20) S(A, ,r) = AIf(S(A, ~')) = AS(If(A, rr)), where if = if(k,~,,,,). Let F ") D F (2) _D 9 9 9be admissible intervals decreasing to 0 or a point, and iterate the above to obtain for all n > 1 (3.21) S(A, ~r) = A r"'A r '2," 9 9A r,-,S((A ("), *r("))), where (x ("~, ,r("5 = if~,.,(X ~"-', Ir ~"-"), (X(% ~r ~~ = (X, ,r). 3.22. Proposition. Let (A, It) E Am x ~+~ satisfy the i.d.o.c., and let l~(1)_~ F(2) ~_ .. 9 be a sequence of admissible intervals shrinking to 0 or a point. Then (3.23) S(A,~r)= fi (Ar,,,Ar~2,'"Ar,.,A.). n=l (Compare with [9].) Proof. The inclusion C_ is immediate from (3.21). On the other hand, the cone on the right in (3.23) is spanned by the set of cluster points of the images of the extremal rays of Am under the sequence 240 W.A. VEECH A (') = A r")" " 9A re-), n -> 1. Each column of A ('), when normalized, gives the relative frequency of visits of a specific maximal interval in F t') to the intervals [13,_~(A),/3, (A)) before returning to F in). Therefore, any cluster point of the sequence of normalized columns corre- sponds to an invariant measure for T(~.~) (see [9]). The inclusion _D in (3.23) now follows. In [3] Furstenberg defines a pseudometric on A,, with respect to which each nonnegative matrix A acts contractively. More precisely, if D(-,-) is the pseudometric, and if A is a nonnegative matrix having no row identically 0, then (3.24) D(Au, Av)<=D(u,v) (u,v E A,). D(., .) restricted to A,_] x &m-1 is a metric, but in general (3.25) D(su, tv)= D(u,v), s,t >O. In other words, D (.,.) identifies each ray to a point. Following [3], call a matrix B with positive entries 6-bounded, 6 > 0, if for all i, j, k (3.26) Bij ----<8Bi~, Bj, _-< 8Bk~. (There is a corresponding notion for matrices with some entries zero, but we shall not need it.) The main feature of ~5-boundedness is the inequality (3.27) D(Bu, By) <- D(u, v) + log (1 + 8 exp(- D(u, v ))~ \ 1 + 8 exp(D(u, v)) / valid for B 8-bounded and u, v E Am. 3.28. Lemma. Let B be 6-bounded, and let UI, U2,"" be a sequence of matrices with nonnegative entries not all zero. Then (B UI B U2. . " UnB )A,~ = II is one dimensional. INTERVAL EXCHANGE TRANSFORMATIONS 241 Proof. Given u, v E ll, we are to prove D(u, v) = 0. Let ~k = ~ (BUk BUk+I " " U.B )Am, nick and notice that BUk-ftk = flk-l. Now choose u~, vk EfI, inductively, so that BUk-luk = uk-1, BUk-lvk = vk-l, k > 2 (ul = u, vl = v). By (3.24) the sequence D(uk, vk) is monotone nondecreasing. But also, uk-1, vk-1 ~ BUk Am C BArn for each k, and on (B A,) x (B Am); which corresponds to a relatively compact set in Am-l, D(., .) is bounded. Thus, limkD(uk, Vk) = o" exists and is finite. Now let wk = Uk-lUk, Zk = Uk-~v~, so that uk-1 = Bwk, vk-a = Bzk, and D(wk, zk ) <= D(uk, v~ ). Now by (3.27) + 6 exp( l D(wk, Zk )) ~ (3.29) D(uk-,, vk-1) <- D(wk, zk ) + log / 1 \ 1 + 8 exp(D(wk, zk)) } " If D (ul, vl)= a > 0, there exists e = e (a, 8)> 0 such that the second term on the right in (3.29) is =< - e for each k. We get that tr = lim D (uk-t, vk-1) -< - e + lim D (wk, z~) = -e+o" which is absurd. Thus, D(u, v)= 0, and [1 is one dimensional. 3.30. Proposition. Let (A, ~r) E A, x ~+~ satisfy i.d.o.c., and let F ~ Fr --. be a sequence of admissible intervals which shrink to 0 or a point. If there'exists a matrix B with positive entries such that for infinitely many i, j, A r~,)A r"+')" " " A rO) = B, then (A, zr) is uniquely ergodic. 242 w. A. VEECH Proof. Clear from (3.23) and Lemma 3.28. Remark. If (A, 7r) satisfies i.d.o.c., and if F ~') are as above, there exists n > 0 such that B = A r"~" 9 9 A i~"~ has all entries positive. For otherwise there would be a vector A' in (3.23) having a component 0, contradicting the fact that A'= A(/z) for some /z E ~(h, zr). 3.31. Proposition. Letg-o:A, x ~+~--~A, x~+~bethemapcorrespondingto F = [0,ill(A)). Define J-: A~_~ x ~2--*A,_~ x ~ by letting where (A ', ~r') = 9-o(A, 7r). If there exists a finite ineasure on A~-I x ~, equivalent to Lebesgue measure and invariant under 9-, then almost every (A, ~), 1r E ~, is uniquely ergodic. Proof. Let A satisfy i.d.o.c., and suppose B = Ar,~"" Arc-, is such that A E BArn. Now BA,. n Am-1 consists of vectors A' such that the expansion (3.21) for (A', rr) agrees to time n with that for (A, ~r). The Poincar6 recurrence theorem implies almost all such A' are such that 9-k(h', 7r) = (h", ~'), A"EBA,, nAm-1 for infinitely many k. The result now follows from Proposition 3.30 and the remark which follows. Remark. Since this paper was written Rauzy (Echanges d'intervalles et transformationes induites, preprint) has studied another transformation (say ~') which induces on the longest interval of the form [0,1-) which is admissible. By his analysis ~2 divides into "Rauzy classes" R~ "), 1 _- w The groups O(L "~) and SO(L ~) Let ~o be the irreducible elements of ~,,. The argument involving the ranks of A, B in (3.16) applies here to show that for almost all A E A,, and admissible (A, 7r, F), F will contain m maximal intervals for (3.2)*, although the induced transformation U may actually have fewer discontinuities. Assuming (A,~r) is minimal, the new interval exchange, (A', ~") will also be minimal, and therefore 7r'~ ~. Recall that A(i, k), 1 <= i, k <= m, denotes the number of visits by the interval [~ +/3~_1(A '), sr +/3~ (A ')) = J~, F = [sr, r/), to the interval [/3,_2(A),/3, (A)), before returning to F. It is evident that the induced transformation U is given by (see (2.9)) (4.1) Ux = x + (A 'L'~A )k (x E Jk ). On the other hand, it is also given by Ux = x + (L'~'A ')k (4.2) = x + (L "'A -1A )k. Since (4.1) and (4.2) hold for all A E AA,,, a set of positive measure, we have A 'L" = L '~'A -~, or (4.3) L ~'= A 'L"A. Of course, also (4.4) det A = - 1. * Add to (3.2) the requirement that /3j~ T"[a, b), 1 -<] _- Recall that F = [/3H(~t ), /3~ (a )) is always admissible, and therefore the above considerations apply to F. Suppose we induce (a, 1r) on F = [/3,-t(a), I a I) and that Am > 1/2 I a I- Then because TF O F = [/3,,_~(A),/3,_~(A) + ~'), I- > 0, the new (a ', It') will satisfy 1r'm = 1. Let A be the associated matrix satisfying (4.3)--(4.4), and let us now begin afresh with It'. If ~' satisfies A,>~, 1 IA,I, A" >~IX'I,1 and if F' = [0, A I), the induced transformation U will be of the form (A", ~r"), where lr"l = m, ~"m = 1. Indeed, Tr' n r' = [x~- z', xl) and Tr' O [/3._1(A '), 1) = [p,,,-1(/~ '), /3m--1(/~ ') "~ T'), 7" > 0. On the next application of T, the latter interval is sent to [0, r"). Let B be the associated matrix, B'L"'B = L"'. Now (4.5) B'A'L'AB =L", det AB = +--1, and because zr"l = m, ~'"m = 1, L"" has the form (4.6) L ~" = i71.... i) Here * is of the form L ~, y E @m-2, but y need not be irreducible. However, y decomposes into irreducible components, and L" has a block form L ~' 0 ) L ~'2 (4.7) L ~ = (0 ""L ~, INTERVAL EXCHANGE TRANSFORMATIONS 245 where each % is irreducible. One checks easily that det L"'= det L ~, and so by (4.5)-(4.7) P (4.8) det L" = 1-I det L vJ. If L ~ is nondegenerate, m is necessarily even (because L" is alternating), and then each L ~j is nondegenerate. In case m = 2, has determinant 1 and is the only nondegenerate possibility. Arguing by induction in (4.8), we see that if L '~ is nondegenerate, then det L" = 1. (From the fact L" is an alternating integer matrix, one may infer, when m is even, that det L" = ! 2 for some 0 _-< l ~ Z.) In what follows we shall use Jk.m, 2k _-< m, to denote the matrix 2k A Jk,. = I 0 m 0 0 We shall prove 4.9. Proposition. If 7r E ~,,, there exists an integer matrix A, det A = - 1, such that (4.10) A 'L'~A = Jk. m, where 2k = rank(L"). Proof. Notice that L", being alternating, has even rank. If rr is reducible, L" decomposes into blocks as in (4.7). In that case an induction assumption enables one to find A0, det Ao = -1, such that 246 w.A. VEECH Jk.... 0 ) (4.11) A 'oL ~Ao = Jk2, m2 (0 "" "Jk.m, ml+".+ rn~ = m. From here it is easy to find A~ with At 1AoLt ~rAoA1 --- Jk , ~, 2k = 2(k1+...+ kp)= rankL". If 1r is irreducible, there exists A such that A 'L'~A = L "" has the form (4.6) with L" the inner (m - 2) x (m - 2) submatrix. Using an induction hypothesis one finds A0, detA0 = • 1, such that 0 1 ... 1 -1 1 (4.12) A ~L'~'A0 = J~m-2 1 -1 .... 1 0 where 21 = rankLL From the form of (4.12) it is evident that rank L '~" = 21 + 2 = rank L". It is an elemetary problem to show that an integer matrix A1, det A1 = • 1, exists so that if M is the matrix (4.12), A~MA1 = J,+l.m. Details are omitted. For each 7r E ~o, the orthogonal group, O(L~), of L '~ is the group of real invertible matrices A such that (4.13) a 'L ~A = L ~ Proposition 4.9 implies that there exists an invertible integer matrix A such that AO(L~)A -1 = O(Jk.m), 2k = rank L". The Lie algebra of the latter is the set of real m x m matrices B such that BJk, m + Jk. mB = O, INTERVAL EXCHANGE TRANSFORMATIONS 247 or what is the same, such that BJk.,. is symmetric. If A E O(Jk.,,,), write A(: :) where a is 2k x2k, 6 is (m -2k)x(m -2k), etc. The condition that A'Jk.~,A = Lk.,~ translates into a'Jk ~ka = A~k, /3%.:~/3 = 0. Thus, a ~ Sp(2k, R) and fl = 0 (because Jk. 2k is nondegenerate). It follows det a = 1, and detA = det & We have 4.14. Proposition. Let N(~r) be the nullspace for L". If A E O(L"), then det A = det A I~,t,,). Example. If N(~) is one dimensional, let v span N(Ir). If A E O(L'~), then Av = Av, and A = det A. In case m = 3, ~r = (123--> 321), we have L~ = (011 10 li) and v = (1, - 1,1). 4.15. Proposition. IrA E O(L~), lr as above (m = 3), and ifthefirst row of A is (1,1,1) then detA = 1. (We write A ESO(L').) Proof. The first component of Av=(detA)v is 1-1+1=1=1.1. 05. Projective transformations If A is an m x m matrix with nonnegative entries and determinant _+ 1, we define a transformation ~,~ : Am-l--> A,~_I by 248 W.A. VEECH Au (5.1) ~?A~ = l Au I" Such transformations will be important in sections to follow. 5.2. Proposition. If A has nonnegative entries and determinant +-1, the jacobian (determinant) of ~A on Am_l is given by 1 JA(u) = iAulm . ProoL Fix u E A,,_I, and let U be a relatively compact neighborhood of u in Am_,. Set V = l,.Jo<, b'( U~ vol v = f = J m o As A preserves volume, vol(AV) = vol(V), and an approximation to vol(AV) is IAul vol*(AV)= f t"-lu(~AU)dt o (5.3) = ]Au I" v(.~,~U). m The error in using (5.3) for vol(AV) is dominated by R" -r m vGSgAU)--, R = sup IAwl, r = inf 1Awl. m w~u wEU This is o(v(.~,~U)) as U shrinks to u, and so lira v(~AU)_ 1 ux. v(U) IAu l" as claimed. 5.4. Proposition. If A is as in Proposition 5.2, define INTERVAL EXCHANGE TRANSFORMATIONS 249 Rj = 2 A~s, l < j <<- m. i=l Then (5.5) ,,(~e,, a._,) = m l R1R2 . . . R,, " Proof. Let F,,(RI,---,R,~), R1,'-.,R~, >0, be the integral dul 9 9 9dum_l r-(R1,'',R,) = f (Rxul + 9 9 9+ R,,,Um)" " Am-- 1 Use the fact u., = 1-ut ..... urn_, and integrate first over um-~ (i.e.u.,-l_--- 1 - ul ..... urn-2) to find r,(R,,...,R,)= (5.6) 1 m (R. - R.-1) {F._,(R1,..., R~_,)- F._,(R1,..., R.-1, R.)}, where ^ denotes omission. Assume inductively that Fk(S1,---, Sk) is given by k Using (5.6), the same relation holds for m. A direct verification for m = 2 establishes the general formula for Fro. The proposition follows from this and Proposition 5.2. w Description of 5 r in case m = 3 In this section we shall suppose m = 3, in which case ~;~ = {(123--* 321)} is a singleton. 3-o is the map of A3 x ~;~ -~ A3 to itself given by inducing (A, lr) on the first interval [0, hi). Now restrict attention to A2, and normalize the vector component of J-o(A) = gro(h, zr) to obtain h---, W(A)E A2. There exists a certain 250 w.A. VEECH collection ,~ = {A } of matrices A ~ O (L "~) such that on ~A A2, ff is given by (~A)-1 (Section 3). Since (A, ~r) is induced on [0, a~), and since the first row of A tells the number of visits of the various maximal intervals to [0, A~) before returning to [0, A1), the first row of A is (1, 1, 1). By Proposition 4.15, det A = 1, and A E SO(L'~), L "~ = (01 0li) . 1 -1 Denote elements of A2 by triples u=(x,y,z), x,y,z >0, x+y+z =1. Introduce new coordinates p(u), s(u) by o(u) = z + y , x+y s(u)= x x+y 6.1. Proposition. The map q~(u)= (p(u),s(u)) is a diffeomorphism of A2 onto the region X C R 2, X = {(O,u)lO< O <~, max(O, 1-p)< s < 1}. Also, dpds q~ *(dxay) = (p + s) 3 . Proof. Clearly p takes on all values 0 < p < ~, and s > O. Also, X--Z X 1-p = x+y <--=S.x+y For fixed p, s assumes all values between max(O, 1- p) and 1. As for the second statement, the jacobian of ~ is INTERVAL EXCHANGE TRANSFORMATIONS 251 -y-1 x -1 (x + y)2 (x + y)2 a(x,y) y --X ix + y)2 (x + y)2 which has determinant (x + y)-3 = (p + s)3. The Proposition follows. It will be useful below to recall that if A E SO(L~), then A( 1 1 That is, if (6.2) A = ~3 s then a-b+c = 1, (6.3) a-/3+y=-l, r-s+t= 1. 6.4. Lemma. Let A E SO(L~) have nonnegative entries. For every u E A2, we have p(~u) = TAp, where TA is a fractional linear transformation with matrix y+t a+r) (6.5) O(A ) = \y + c a + a " Proof. Since 252 w.A. VEECH Au ~Au = Imul ' _ (Au)2 + (Au)3 p(*LPAU) -- (AU)l + (Au)2 (6.6) = (~ + r)x +.(~ + s)y + (r + t)z (a + a)x +(/3 + b)y +(3" + c)z " By (6.3),/3 + s = (a + r)+ (3' + t) and (/3 + b) = (a + a)+ (3' + c), and therefore by (6.6) (a + O(x + y)+(r + t)(z + y) (6.7) ~ = (a + a)(x + y)+ (3' + c)(z + y) " The desired result now follows by dividing numerator and denominator of (6.7) by x+y. 6.8. Lemma. If A ~ SO(L'~), then tr(A ) ~ SL(2, R). Proof. Write a a+c-1 c A = a a+3'+l 3' lr r+t-1 t and note l=detA=det 1 -1 a ot c ; 3'/ = det : 01 \r+a 0 t+3"] = det 0 (A). A fractional linear transformation which maps [0, 0o) into itself determines a unique nonnegative 2 x 2 matrix with determinant 1 (cf. [14]). Clearly, if A, B E SO(L") have nonnegative entries, then TAB = TATB, and O(AB), O(A)O(B) are INTERVAL EXCHANGE TRANSFORMATIONS 253 nonnegative elements of SL(2, R) determining the same fractional linear transfor- mation. Thus, O(AB)= O(A )O(B). Remark. In fact, A---,O(A) defines a homomorphism of SO(L ~) onto SL(2, R). The kernel is a two-dimensional vector group. In order to study ~ we shall determine the elements of SO(L ~) which have the form A = a+y+l (a, y,r,t >O), r+t-1 Actually, it will be more convenient to write 1 1 1 ) (6.9) A = a-1 a+y-1 y-1 r+l r+t+l t+l with a, y > 1 and a priori r, t ___> - 1. In this form we see that O(A)=(y+t a+r) y ot ' and a straightforward computation shows any element of SL(2,R) gives rise to A ~ SO(L ~) by inverting this correspondence. 6.10. Lemma. If the matrix A in (6.9) belongs to SO(L'). Proof. Direct computation. Given 254 w.A. VEECH a necessary condition that A in (6.9) have nonnegative entries is that. a, 3' _-> 1, r,t_->-l. Ifr= -1, then t => 0 because we must have r + t + l => 0. NowdetA0= at + 3i, and as a, 3' --> 1, t = 0 and 3' = 1. That is, if r = - 1, A has the form (6.11) A = Ot 1 O~ 0 If t=-1, we know r_->0, and therefore detAo=-a-r3'_-<0. Thus, t~0. Collecting results, 6.12. Lemma. A~ (~ t3')E SL(2, Z) defines a nonnegative element A ~ SO(L "~) by (6.9) if and only if a, 3' >= 1, t _-> 0, and r >= - 1. When r = - 1, A has the form (6.11). It is useful now to compute s(~Au), u E A2, A ~ SO(L") having nonnegative entries. 6.13. Lemma. If A E SO(L ~) has nonnegative entries (6.2), then for all U E A2, a-l+cp+s (6.14) s(~Au) = (c + 3')p + (a + a) where on the right, p = p(u), s = s(u). Proof. By definition, (Au)l s(~Au) = (Au)l + (Au)2 = ax + by + cz (a + a)x + (b + fl)y + (c + 3')z and using (6.3) this is INTERVAL EXCHANGE TRANSFORMATIONS 255 a(x + y)- y + c(z + y) (6.15) s(~au)= (a + a)(x + y)+(c + 3,)(z + y)" Now s-1 = ---L--x 1- -y , x+y x+y and therefore (6.14) follows upon dividing numerator and denominator in (6.15) by (x +y). 6.16. Definition. If A E SO(L") has nonnegative entries, define .I/la :X--~X by ./~A~0(u) = ~0(LeAu). By Lemmas 6.4 and 6.3 we have a-l+cp+s '~ (6.17) ~A (p, s) = TAp, (c + y)p + (a + a)] " Thus, .//.~ is fractional linear in the first variable and affine in the second, the atfine map depending on p. Remark. The maps ~A and q~ are projective, and therefore ~/~a is projective. In particular, ~A takes lines to lines and triangles to triangles. Notice that ./~AX is a triangle with vertices ~A(0,1)= +a 'a+a~ ' {7+t+a+r c+a-1 ] u/,gA (1, 0) = ' ~A (~, , ) = (r+t r \y+c ' y+c]" The Jacobian of MA is computed to be ((c + 3')0 + (a + a)) -3, and therefore ~AX has Euclidean area 256 W.A. VEECH I ~l~x I = f pdp + f dp ((c + 3')0 + (a + a))3 ((c + 3')0 + (a + a))3 o I = 11 + 12, where 1 1 1 L =-~ (c + 3")(a + a)(c + 3" + a + a) 2(c + 3')(c + 3" + a + o~)2' (6.18) I2- 2(c + 3')(c + 3' + a + c~)2" That is, I /J vl 1 (6.19) r'~aA J = 2(c + y)(a + c~)(c + 3' + a + ~) I For later reference we record that if X 1 = {(p, S) ~ X I P > 1}, then by the second line in (6.18) 1 (6.20) [JL~X~[ = 2(c + 3')(c + 3, + a + a) 2 " When A has the form (6.9), the vertices of AAAX are (o+r 1) (6.21) ~A(1,0)= (~ +t+a+r 1) \ 3'+or y+a ' and the corresponding area (6.22) I MAX I = 2a3"(a + y) " Because INTERVAL EXCHANGE TRANSFORMATIONS 257 the pairs (a+r o+t) (a+r..... a+r+3"+t) and (a+r+3"+t 3"+9 ' 3" ' a ' a+3" ' a+3" ' 3" are neighbors in the Farey series of orders max(a, 7), a + 3', and a + 3' respectively (see [6]). Call the corresponding intervals Farey intervals. Farey intervals have the property that if two of them intersect, then one is contained in the other. Based on this fact we see that if two triangles defined by vertices (6.21) have a nonempty open set in their intersection, then they are identical. For if (: defines the second triangle, the "long" intervals (o r t)and (o r a ' T -a-7 ' T' will surely intersect, and therefore one, say the second, is contained in the other. If they are identical, the three vertices must agree, and the triangles are identical. If, say, the second is properly contained in the first, then it must also be contained in one of , or , . a a+ T a+T T But this, by definition of Farey series, implies a', y'_-> a + 7, and the second triangle lies below the lowest vertex of the first. That is, they do not intersect in a nonempty open set. It follows now that as A runs through the elements of SO(L~) in Lemma 6.12, the (open) triangles A/AX are pairwise disjoint. On the other hand, their union must contain almost all (p, s) E X because ff is defined a.e. (This may also be verified directly without difficulty.) We conclude that (6.23) if(P, $) = E x.aAx(p,S)./,~A'(p, S), AE~ 258 w.A. VEECH where ~ is the set of matrices which arise in (6.9) from with a,y = 1, t_>0, r=> -1. Remark. If a, 3,>0, (a, 3,)=l, there exist r,t with t=>0, r= -1 such that (6.21) defines a triangle contained in the square 1 _-< p _---2, 0 _--- ay(a + y) 2. ~,y>o (c~,~,)= 1 Professor Selberg has pointed out that for any k => 2 =k! 9 9 (~1,...,~k)=l 0~10/2 " O~k (a,+ " 9+ at) ~1,-- ',ak >O The proof is analytic, but perhaps there is a geometric proof as above. Remark. 3 can be defined "directly" as follows. Given (p,s), p irrational, there will exist for all but countably many s a Farey interval (b/d, a/c) containing p and such that (p, s) is in the triangle with vertices .... +d 'c+ " Define el =dp - b, er = a - co. Then it may be seen that ~r " We have no need of this representation, and so details are omitted. Suppose A E SO(L") has nonnegative entries. We shall estimate the Jacobian of //'r9 with respect to the measure/z = dpds/(p + s) 3. Write (15, g) = A,~a (p, s). Then by (6.17) INTERVAL EXCHANGE TRANSFORMATIONS 259 dpds d~dg = ((y + c)p + (a + a))3 , ~+g= (y+t+c)p+(a+r+a)+s-1 (y + c)o + (~ + s) On MaX we have d~dg (p + s) 3 dpds (6.24) (~+g)3 ((y+t+c)p+(a+r+a)+s_l)3 (p+s)3- Thus, if we define p+s )3 JA(P'S)= (y+t+c)p+(a+r+a)+s--1 (6.24) says Ja is the Jacobian of Ma. We introduce two quantities: (6.25) M(A) = sup Ja (p, s) r162 JA (p ', s') ' (6.26) M,(A) = sup J~ (p's) ~o.,.~'.,'~ J., (O', s') " p,p'~l If is readily verified that for some constant D < ~, independent of A, {(~+r+a] 3 (~ + t + c )3} (6.27) M(A)<=Dmax +-i~c/ ' +r+a " If y+t a+r~ O(A)= y+c ~+a/ is such that __a + r ~,+t ot+a "y+c c [/,/+ 11, then one checks that 260 W. A. VEECH 1 (a+a]< a+r+a A similar inequality holds if [ a+r , y+t] Finally, if a +r=0, then a = 1 and M(A)<-D(y+c) 3. In all cases there is a uniform constant D (possibly larger) such that (6.28) M(A ) <- D max \\-~-~--~/ , \'ff--~a / / " It the case of M~(A) the estimate looks like (6.29) MI(A)<--_max(D,D(~+r+o~3~ +~u as one sees from (6.24) and (6.26). In terms of a + a, y + c, it reads (with D possibly larger) [ (6.30) M~(A ) <_- max ~D, D \~--~-c-/ / w Canonical products and continued fractions Let If (a has nonnegative entries with c, d > 0, the continued fraction expansions of a/c, b/d may be put in the forms [ao; a~,. 9 ak-~] and [a0; a~,- 9 ak], the longer expansion INTERVAL EXCHANGE TRANSFORMATIONS 261 being for whichever of a/c or b/d has the larger denominator. Indeed, b/d < a/c are neighbors in the Farey series of order max(c,d) and therefore occur as successive convergents in some continued fraction expansion ([6]). Now we claim it follows that where ~ = tr or ~" as k is even or odd. First, note p p' p p'+ap) (q q')~ra = (q q'+aq/' p p' p+ap' p' If po = ao, qo = 1, pm= 1 + aoa~, q~ = a~, an inductive argument shows the (2]) 'h partial product, j => 1, is of the form p2,-~ p2,~ (q2j-l q2j/ ' while the (2j- 1)s' is of the form (p2,-1 p2,-2~ q2i-1 q2i-2/ ' where for all l pt+l = az§ + pH, ql+l = at+lq~ + ql-1. The same recursion is satisfied by the convergence to a/c,b/d, with the same "initial data", and so our claim is established. Notice that the quantity max(c/d, d/c) from (7.1) (with c, d# 0) is on the order of ak. Using (6.29) and (6.30) we see that if 0(A)= (a b) in (7.1), then 262 w.A. VEECH M(A ) <-- Da 3 while = ID k odd, (7.2) M1(A ) t Da 3 k even. The set of matrices ~ C SO(L ~) generates a free semigroup in SO(L ~) because the set of images d,tAX, A E ~, are pairwise disjoint. Let :~n be the set of n-fold products of elements of ~. The set 0(~) not being free, there will exist for "most" Ao ~ 0(~n) many representations Ao --- O(A), A ~ ~;n. For each of these, d~AX is a triangle whose Euclidean area is the same (determined by Ao). The ratios M(A), M,(A) are also determined by Ao. For the purposes of counting the number of representations of Ao as O(A), A ~ ~,, it is necessary to identify the elements 0(A), A ~ .~. If A = (1u-1 uli) , 0 0 rthen (I ~ U+I = "/'Oru" Otherwise, 0(A)= (: b) maps [0,oo] into [1, oo), and so 0 (A) = o-%~"a, *, where aoa, > 0 and * denotes an arbitrary product of ~'s and z's. Now given a positive element Ao E SL(2, Z), ~t o ---- 0-%,{- a, . . . ~ ak, INTERVAL EXCHANGE TRANSFORMATIONS 263 one can in principle determine how many times Ao = 0(A), A ~ ~,, by counting how many different ways A0 is a product of n matrices ~" of the form fGra~ a~* aoai > O, (7.3) / L r o'~ a=>O. A good procedure for making this count seems to be essential if one is to determine whether the measure whose existence is proved in Section 8 is finite or infinite (unless one can compute the measure explicitly). w Invariant measures tor ~3r in cases m = 2, 3 To begin with let m = 2 or 3, and 9- : Am_,---~ A,,_I correspond to - 1 0 or = 1 0 1 -1 as m = 2 or 3. As before, v denotes Lebesgue measure on A,~_, and if .~ is the set of (2 • 2 or 3 • 3) matrices which are used to define 9- (i.e. 9-u = ~L~'?~tu for an appropriate A = A (u), a.e. u ), there is an operator 9-* : L t(v)__~ L ~(v) given by 9-*f(u) = ~_~ JA(u)f(Au) A E~; (8.1) = ~ IAul-~'f(Au). A EJ; The measure fv with density f ~ L1(v) is 9- invariant precisely when 9-*f = f. In fact, (8.1) is defined for certain locally integrable f, and so infinite (absolutely continuous) invariant measures arise the same way. For example, in case m = 2, ~= A= k+l = and if A, = {(x, 1 - x)10 < x < 1}, v(dx) = dx, f(x) = 1/(1 - x), then 264 W. A. VEECH ff*f(x)= ~o. (k +2-x)1 1 k+2-x =~, 1 = 1 ~o. (k+2-x)(k+l-x) 1-x = f(x). Thus, f(x) dx defines an invariant measure which is infinite. In the case m = 3 we have been unable to verify directly that 3 has an invariant measure equivalent to Lebesgue measure, but we are able to argue that one must exist. This is the purpose of what follows. Recall the definition X~={(p,s)~X [p >_ 1}, and below consider ~r as a mapping of X to X. 8.2. Lemma. For almost all (p,s)EX1 there exists n = n(p,s) such that er"(p,s)~ X1. Proof. If if(p, s) ~ XI there is nothing to prove. Therefore, it will be sufficient to prove for almost all (p, s) E X~ that ffk(p, s) E XI for some k > 0. It is true for almost all (p,s) that there exists for each n At")=A~')...A~')EJ;, with (p, s)E ~A,-)X. To say ~-k(p, s)E X; for all k is to say O(A~ ")) = ~'(rbj for certain integers bj > 0. (Of course, A~ ") = A~ "+~) .... .) For any matrix Ao = (ca db)~SL(2,Z) with nonnegative entries and a0 = 0 in (7.1) there exists precisely one value of n and one representation of A0 as a product of n matrices of the form ~'trbj, b~ = 0. The value of n is clearly al + a3 + " " " + a2z+l, 21 + 1 --<_ k < 21 + 3. If A E ~, is the corresponding element, ~AX has Euclidean area l/2cd(c + d). It follows that the sum of the measures of these triangles is dominated by 1 (c,d)ffi~" 1 2cd(c + d) = 1 c,d>0 INTERVAL EXCHANGE TRANSFORMATIONS 265 and by the Borel-Cantelli lemma, the set of (p, s) which belong to infinitely many of them has measure 0. In other words, for almost all (p, s) E X~, ffk(p, s) ~ Xt for some k > 0. The lemma is proved. The proof that 8- possesses an invariant measure will be based on the fact that the induced transformation 3rl : Xt ~ X~, which by Lemma 8.2 is defined a.e., has a finite invariant measure which is equivalent to Lebesgue measure (i.e. to/z Ix,). The latter follows from a simple criterion of Hajian-Kakutani [4] and 8.3. Proposition. For every e > 0 there exists 8 > 0 such that if E C XI has /z(E)< 8, then for all k >0 /~(~kE)< e. Remark. Here ffl is not one-to-one, but the criterion, that Proposition 8.3 implies the existence of a finite absolutely continuous measure, is still valid. Proof of the proposition shall be deferred until later in the section. For now we shall assume it to be true. Let/xo < ~ Ix, be a finite nonzero invariant measure. We shall derive some of the ergodic properties of (if1,/~o), as well as the equivalence of /~o and /x Ix,. Write /Zo = fo~, and let P = {.fo> 0}, N = {/o = 0}. Let ~o~ be the set of matrices A ~ SO(L ~) involved in the definition of ff~. That is, for a.e. (#, s) E X~ there exists a unique A E ~o~ such that ffl(p, s) = d~)(p, s). Let Q = Q C~, be the partition of X~, and then set n-1 Q(")--- V ff~kQO). k=0 Applying the martingale theorem to the join n=o we have for any h E L I(/~) Lim E(h I Q '"~) = E(h ]~) (a.e. /~). 266 w.A. VEECH For/z-almost all x = (p, s) there exist elements A v), Ate>,..., e~<1) such that for each n > 0 x belongs to ./~BX~, B = Att). 9 9 A t"). As we have seen, there exists a.e. x an n such that B has positive entries. As in Proposition 3.31, the Poincar6 recurrence theorem implies that for/zo-almost all x {x} ~ 2.* That is (8.4) Lim E(h I Q <")) = h (a.e./~o). There exists a uniform constant D such that whenever B, C E SO(L "~) have nonnegative entries with all entries in B positive, then M(BCB) <- DM(B), (8.5) MI(BCB) <-_ DMI(B). (For example, if B is 8-bounded (Section 3), then BCB is also 8-bounded.) We conclude that for /zo-almost all (p,s)~ X~ there exists M(p,s) with (p, s) E f~k and 1 (8.6) JA'k,(X ) => M(p, s) Ja"'(Y) for all x,y ~ X~. Integrate (8.6) with respect to g Ix1 (and normalize so that (X0 = 1) to obtain (8.7) Ja,~,(x) _-> /~(f~Q M(p, s) " If E C_ X1 is any measurable set, then using a change of variables and (8.7), we have * In fact, this statement is true for p -almost all x because, e.g., it is already known that p.-almost all x are uniquely ergodic (m = 3). This fact is not necessary in the present argument (and is a consequence of it). INTERVAL EXCHANGE TRANSFORMATIONS 267 X1 = f x.~(~t.,*,z)J~,.,(z)~(az) X1 (8.8) = f x~ (z).r,,,~,(z)ix (dz) XI > Ix (E)Ix (~) M(p, s) Now specialize h in (8.4), taking h = XP and specialize E in (8.8), taking E = N. In this case, (8.1) and the fact 3 *if0 = )Co implies ~,~,,,N tq P has Ix-measure 0. Now if Ix (N)> 0, (8.8) implies that for Ix0-almost all (p, s) lira inf E(X~ ] Or")) < 1. n~ This contradicts (8.4), and therefore Ix (N)= 0. That is, 8.9. Proposition. If Ixo < tx is a finite nonzero nonnegative invariant measure for 3-1, then Ix Ix1 < Ixo. 8.10. Definition. Define the tail g-field for (3"1, QO)) by ~= f~ V 3-;kQ <'. n=l k~n One has E ~ ~ if and only if there exists for each k a measurable set Ek such that E = 3";kEk. One says (X1, Ixo, 3-1) is an exact endomorphism ([12]) if for each B ~ ~| Ix0(B) = 0 or IXo(Bc) = 0. As Ixo < > IX, it will suffice to verify the condition for IX. 8.11. Proposition. (X, Ix0, 3-~) is an exact endomorphism. ProoL Let E E ~| E = 3--IEk, k > 0. If E has positive measure, let h = X~ in (8.4). For almost all (p, s) there will exist k, nk as in (8.8) such that if in that equation we let E = Ek, E n f~ #(E O ~(~h >/~ (E~)~ (1~'~)) "= M(p, s) " Now if tz (E) > 0, then tzo(E) > 0 and inf/a, (E~) > 0. Thus, )t'~ = 1 a.e. and Iz(E') = 0. Remark. It is undoubtedly the case that (ff~,/~0) is "Bernoulli" in the sense that the natural extension to a K-automorphism [12] is isomorphic to a Bernoulli shift. We do not presently know whether (~',, ~o) has finite or infinite entropy. Remark. Let Q C X~ be a measurable set. Then almost all x E Q visit Q infinitely often, and therefore the tail set Q- = {x E X1 [ ~-~x E O infinitely often} has measure at least that Of Q. If/z (Q) > 0, it must be that/~ (Q~-) = 0. In particular, if A1,'.. ,A, E J-t" are arbitrary, there will exist for/~-almost all (O,s)EX1 an expansion (p, s) -- (A (1)(p, s), A (2,(p, s),"" ) with the property that infinitely often A ('+')(p, s), 9 9 9 A ('+"J(p, s) = A,, " 9",A,. Using the fact that almost all x E X visit X1, one obtains the corresponding fact for the expansion of x in terms of ff and ~1. It is now left for us to prove Proposition 8.3. Recall that if A E SO(L'~) has the form (6.2), and if y + c > (or + a), then by (6.20) 1 1 1 8 (~ + c) 3 = = 2(~ + c) 3 If 0(a, with q > q', this becomes INTERVAL EXCHANGE TRANSFORMATIONS 269 1 1 (8.12) 8q3 =< I ddaX, [ =< 2q3 . Let A1, 9 9 9 A~ E :~1 be such that (i) AI~:T (u. (ii) Exactly n - 1 of A2,...,A~ belong to ~(1). (iii) The representation (7.1) of O(A~) has k odd. Condition (iii) guarantees that if O(A~... A,) = ( pq P')q' , then q > q'. Conditions (i) and (ii) guarantee that J-7(~,...A,x) = x (x ~ x,). For each integer a > 0, there exists a unique matrix A ~., E $;~ such that o(a,,,,) = 0(a,)trL Indeed, use the correspondence given in Lemma 6.10. We have from (6.20) that if A (") = A~ 9 9 9At.,, then 1 (8.13) I~ta,-,X,I = 2q(q'+ (a + 1)q) 2 " If T >0 is any integer, the integral test and (8.13) combine to give ~Yr d,r A,-,X~ = 2q (q' + (a + 1)q)2 1 1 = 2q2(q'+ Tq) = 2Tq'" By hypothesis, q > q', and therefore by (8.12) I U .aA,.,xl] <__!__1 o~r = 2Tq" (8.14) = 4 I~AXlJ T 270 W.A. VEECH Now by (7.2) we have (8.15) M,(A ")) <- Da 3 for all a >0 (M(A) (8.16) 0~' = {f~ e 0 ~"' I M,(tO > DT'} then the total area of those 1~ E Q~) which lie in the square 1 -< p _-< 2, 0 = s -< 1, is dominated, using (8.14), by (4/T)Y.ld~,~XI I, the sum extending over those A with A/tAX1E Qt") and such that (iii) above holds. As these are pairwise disjoint triangles, the area is dominated by 4/T. Now the same argument may be applied to each square j < p _- (8.17) Now let e >0 be given. We are to choose 8 >0 so that if /z(E)<8, then /., (~--"E)< e for all n > 0. To this end let T be so large that 4c~'(3)/T < 89e, and then choose 8 < e/2DT 3. Now suppose/z (E)< 8. If t'/E O (") does not belong to O~ ), then arguing as in (8.8) (but using M(l)) for upper bounds) we have for l) = ~tAX1, < DT 3. 8 9I~ (A/lAX,) < If l-/~ O~), then at least/z (.~tAE) -- Y, NzM(AX1EO ("~ INTERVAL EXCHANGE TRANSFORMATIONS 271 8 <2+~=~. Proposition 8.3 is thereby proved. The essential facts used about Xz in the existence proof are (a) that J-i is defined, and (b) the inequalities (8.14)--(8.15). One may extend these and thus the existence proof to larger open sets X~, so long as the point (1,0) is not in the closure of the larger set. For example, let Ao 1 e SO(L"), (ix!)0 and let x~ = (~AoX)*, 9 9 -, x. = (~,,~x)', 9 9 9 At each stage one obtains ff.:X~--~ X. and a measure /~.-/z Ix. such that (J-~, X.,/z.) is exact. If ~r is induced on X~-,, one obtains ft. -1, and therefore up to a scalar multiple, ~[~n-I ~ ~l'n IXn-1- If we normalize so that /z,(X1)= 1 for all n, it will then be true that /~. is an extension of P..-1 for each n. In this way a measure tt| is obtained on X,/.t| Ix. =/z,, which is equivalent to ~, and such that (if,/z| is exact (even though/z| may not be finite). Remark. Let n(p,s) be the least n, if any, such that J'"(p,s)~X~. If f n(p,s)iz| XI then one sees that/z| is finite and conversely. Now for #a-almost all (p, s) E X,, (8.16) lim ~ n(~r~(p, s)) = n(p, S)l~| s)) X| 272 w.A. VEECH and it should be possible to estimate the left-hand side of (8.16) using the remark at the end of Section 7. We hope to have more to say on this at a later time. Problem. What is /z~? REFERENCES 1. N. Bourbaki, AIg~.bre, Ch. IX, Paris, Hermann, 1959. 2. N. A. Friedman, Introduction to Ergodic Theory, New York, Van Nostrand-Rheinhold, 1970. 3. H. Furstenberg, Stationary Processes and Prediction Theory, Annals of Math. Studies, Princeton, 1960. 4. A. Hajian and S. Kakutani, Weakly wandering sets and invariant measures, Trans. Amer. Math. Soc. 110 (1964), 136--151. 5. P. R. Halmos, Lectures on Ergodic Theory, Math. Soc. of Japan, 1956. 6. G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford, Clarendon Press, 1938. 7. S. Kakutani, Induced measure preserving transformations, Proc. Imp. Acad. Tokyo 19 (1943), 635-641. 8. M. Keane, Interval exchange transformations, Math. Z. 141 (1975), 25-31. 9. M. Keane, Non-ergodic interval exchange transformations, Israel J. Math. 26 (1977), 188-196. 10. C. Boldrighini, M. Keane and F. Marchetti, Billiards in polygons, preprint, 1977. 11. H. Keynes and D. Newton, A minimal non-uniquely ergodic interval exchange transformation, Math. Z. 148 (1976), 101-105. 12. V. A. Rohlin, Exact endomorphisms of a Lebesgue space, Amer. Math. Soc. Transl. (2) 49 (1966), 171-240. 13. Ya. G. Sinai, Introduction to Ergodic Theory, Princeton Lecture Notes Series, Princeton University Press, 1977. 14. W. A. Veech, A Second Course in Complex Analysis, New York, Benjamin, I967. 15. W. A. Veech, Topological dynamics, to appear in Bull. Amer. Math. Soc. DEPARTMENT OF MATHEMATICS RICE UNIVERSITY HOUSTON, TEXAS 77001 USA (Received January 12, 1978) 0, there is no j, 0 < j < k, such that T~/3, E F; if k < 0, there is no ], 0 _-> ] > k, such that TJ/3, E F. The triple (A, ~r, F) is admissible if the requirements of this paragraph are met.