ISOMORPHIC TO THEIR INVERSES
PRZEMYSAW BERK AND KRZYSZTOF FRCZEK
Abstract. In this paper we give a criterion for special ow to be not iso-morphic to its inverse which is a re ne of a result of [5]. We apply this crite-rion to special ows Tf built over ergodic interval exchange transformations T : [0, 1) → [0, 1)(IET) and under piecewise absolutely continuous roof func-tions f : [0, 1) → R+. We show that for almost every IET T if f is absolutely continuous over exchanged intervals and has non-zero sum of jumps the the special ow Tf is not isomorphic to its inverse. The same conclusion is valid for a typical piecewise constant roof function.
1. Introduction
The problem of isomorphism of probability measure-preserving systems to their own inverse was already stated by Halmos-von Neumann in their seminal paper [10]. In [10] the authors found a complete invariant for ergodic systems with dis-crete spectrum and then they applied it to prove that any ergodic measure pre-serving transformation with pure point spectrum is isomorphic to its own inverse. Moreover, Halmos-von Neumann conjectured that the same result is valid for an arbitrary measure preserving transformation. The rst counter-example to this conjecture was given by Anzai in [1], it was so called Anzai skew product. Anzai counter-example gave the impetus for further research on the problem isomorphism of measure-preserving systems to their inverse. As shown in [4] (for automorphisms) and in [3], the property of being isomorphic to its inverse is not a typical property. For a fairly detailed introduction to the problem we refer also to [5].
Recall that a measurable ow T = {Tt}t∈R on a standard probability Borel
space (X, B, µ) is isomorphic to its inverse if there exists measure-preserving auto-morphism S : X → X, such that
Tt◦ S = S ◦ T−t for all t ∈ R.
For any ergodic measure-preserving automorphism T : X → X and a positive measurable roof function f : X → R+ we consider a space Xf := {(x, r) ∈ X ×
R, 0 ≤ r < f (x)}. On Xf we deal with the special ow Tf (see e.g. [2], Ch.11), that is the ow which moves points vertically upwards with unit speed and we identify the point (x, f(x)) with (T x, 0). If T is an IET then the ows Tf arise naturally
as special representation of ows on compact surfaces.
In [5] the authors developed techniques to prove non-isomorphism of a ow Tf to its inverse that based on studying the weak closure of o-diagonal
3-self-joinings. The idea of detecting non-isomorphism of a dynamical system and its inverse by studying the weak closure of o-diagonal 3-self-joinings was already used by Ryzhikov in [15]. The tools developed in [5] were applied to the special ow built over irrational rotations and under piecewise absolutely continuous roof functions.
Date: October 3, 2013.
2000 Mathematics Subject Classi cation. 37A10, 37E35.
Research partially supported by the Narodowe Centrum Nauki Grant DEC-2011/03/B/ST1/00407.
The main aim of this paper is to extend the techniques of [5] to special ows over ergodic IETs. The paper is motivated by the desire to understand the problem of isomorphism of translation ows on translation surfaces to their own inverse. For the background material concerning translation surfaces, see [20]. Recall that for every translation surface from any hyperelliptic component in the moduli space the vertical ow is isomorphic to its inverse. We conjecture that for a typical translation surface from any non-hyperelliptic component the vertical ow is not isomorphic to its inverse. The result of Section 7 can be regarded as a step toward this conjecture. In Section 2 we give general background on special ows and joinings. We also recall so called o-diagonal joinings of higher rank, which serve as a main tool in latter calculations and constructions.
In Section 3 we state the conditions under which a sequence of 3-o-diagonal joinings converges weakly in the space J3(Tf) of all 3-self-joinings and how does
the limit looks like, see Theorem 3.10. In [5] the authors give an explicit formula for the whole limit measure. Now under weaker assumptions the limit measure is controlled only partially. Nevertheless, it is enough for our purpose. The proofs are based on ideas drawn from [5] and [6].
In Section 4 (using results of Section 3) we give a su cient condition for special ow built over partially rigid automorphisms to be not isomorphic to its inverse, see Theorem 4.4. This result reduces the problem of non-isomorphism of Tf and
its inverse to establishing that a probability measure ξ∗P on R does not satis es a
symmetry condition.
In the remainder of the paper we apply newly obtained criterion to special ows over IETs. In Section 5 we state some general background concerning IETs and Rauzy-Veech induction. In Section 6 for a almost every IET T we construct a sequence of Rokhlin towers, based on Rauzy-Veech induction. This combined with Theorem 4.4 leads to proving that special ow over a. e. IET under piecewise linear roof function (over exchanged intervals) with constant non-zero slope is not isomorphic to its inverse, see Theorem 6.4.
In Section 7 we modify the construction of Rokhlin tower and again using The-orem 4.4 we prove that special ows over a. e. IET T and under piecewise constant roof function f with discontinuity points β1, . . . , βr, where r ≥ 3, are not isomrphic
to their inverses, for almost every choice of β1, . . . , βr and if f has no jumps with
opposite value, see Theorem 7.3.
Finally, in Section 8 we prove that the results from Sections 6 and 7 can be extended to piecewise absolutely roof function f : [0, 1) → R, see Theorem 8.2.
2. Preliminaries
2.1. Special ows. Let (X, B, µ) be a standard probability Borel space. Let T : (X, B, µ) → (X, B, µ) be an ergodic automorphism. We will denote by B(R) the standard Borel σ-algebra on R, while by Leb we will denote the Lebesgue measure on R or [0, 1) according to the context. Let {Vn}n∈Nbe a sequence of measurable
subsets of X. We say that T is rigid along a sequence {Vn}n∈N if there exists an
increasing sequence of natural numbers {qn}n∈N, such that µ((T−qnA4A)∩Vn) → 0
for every measurable A ⊂ X. Then {qn}n∈N is called a rigidity sequence for T
along {Vn}n∈N. Assume that f ∈ L1(X, B, µ) is a strictly positive function. By
Tf = (Tf
t)t∈R we will denote the corresponding special ow under f acting on
(Xf, Bf, νf), where Xf = {(x, r) ∈ X × R; 0 ≤ r < f (x)} and Bf and µf are restrictions of B ⊗ B(R) and µ ⊗ Leb to Xf.
Moreover, if X is endowed with a metric d generating B, then we may consider Xf as a metric space, where as a metric df we may take the restriction of product
metric on X × R to Xf. Then Bf is generated by df.
2.2. Self-joinings. Let T = (Tt)t∈R be an ergodic ow on (X, B, µ). Let k ≥ 2.
By a k-self-joining of T we call any probability (Tt× . . . × Tt)t∈R-invariant measure
λon (Xk, B⊗k)which projects as µ on each coordinate. We will denote by J k(T )
the set of all k-self-joinings of T . We say that k-joining λ is ergodic, if the ow (Tt× . . . × Tt)t∈Ris ergodic on (Xk, λ). For any (Tt× . . . × Tt)t∈R-invariant measure
σ we may consider its unique ergodic decomposition σ = RM
eνdκ(ν), where Me
stands for the set of (Tt× . . . × Tt)t∈R-invariant ergodic measures on Xk and κ is
some probability measure on Me.
Remark 2.2. If λ = RMeνdκ(ν)is the ergodic decomposition of k-joining λ, then
the set of measures ν ∈ Me which are k-joinings is of full measure in measure κ.
Let {Bn; n ∈ N} be a countable family in B which i s dense in B for pseudo-metric
dµ(A, B) = µ(A4B). Then on Jk(T )we may consider a metric ρ such that
ρ(λ, λ0) = X n1,...,nk∈N 1 2n1+...+nk|λ(Bn1× . . . × Bnk) − λ 0(B n1× . . . × Bnk)|.
The set Jk(T )endowed with the weak topology derived from this metric is
com-pact. Moreover, the sequence of joinings (λn)n∈Nis convergent to λ in this metric,
whenever λn(A1× . . . × Ak) → λ(A1× . . . × Ak)for all A1, . . . , Ak∈ B.
Let t1, . . . , tk−1∈ R. Then we may consider the k-joining determined in following
way
µt1,...,tk−1(A1× . . . × Ak−1× Ak) = µ(T−t1A1∩ . . . ∩ T−tk−1Ak−1∩ Ak),
for A1, . . . , Ak∈ B. Such joining is called o-diagonal. As the image of the measure
µvia the map x 7→ (Tt1x, . . . , Ttk−1x, x), the joining µt1,...,tk−1 is ergodic.
Lemma 2.3. Suppose that (Tt)t∈R is ergodic and aperiodic. Then for any natural
k ≥ 2, the set A ⊂ Jk(T )of all k-o-diagonal joinings is Borel in Jk(T ). Moreover
h : Rk−1→ Agiven by h(t
1, . . . , tk−1) = µt1,...,tk−1 is a measurable isomorphism.
Proof. The map h is continuous and by aperiodicity of (Tt)t∈R it is an bijection
between Rk−1 and A. By Souslin's theorem (see [9]), for any measurable set A ∈
Rk−1, h(A) is a Borel set in Jk(T ). Hence h is a measurable isomorphism. In
particular A = h(Rk−1)is a Borel set.
For any probability measure P ∈ P(Rk−1)we also consider integral k-self joining
R Rk−1µt1,...,tk−1dP (t1, . . . , tk−1)such that Z Rk−1 µt1,...,tk−1dP (t1, . . . , tk−1) (A) := Z Rk−1 µt1,...,tk−1(A)dP (t1, . . . , tk−1),
for any A ∈ B⊗k. In this paper we will deal with such joinings as partial limits of
some sequences of o-diagonal joinings. For additional information about joinings see [5].
3. Limit theorem for off-diagonal joinings Let (Tf
t)t∈R be an ergodic special ow on the space Xf, where T : (X, B, µ) →
(X, B, µ) is an ergodic automorphism and f ∈ L2(X, B, µ)is a roof function such
that f ≥ c for some c > 0. For any measurable subset W ⊂ X with µ(W ) > 0 we will denote by µW the conditional measure given by µW(A) = µ(A|W ) for
X, increasing sequences {qn}n∈N, {qn0}n∈N of natural numbers and real sequences
{an}n∈N, {a0n}n∈N such that following conditions are satis ed:
(1) µ(Wn) → αwith 0 < α ≤ 1,
(2) µ(Wn4T−1Wn) → 0,
(3) {qn}n∈N is a rigidity sequence for T along {Wn}n∈N,
(4) {q0
n}n∈N is a rigidity sequence for T along {Wn}n∈N,
(5) nZ Wn |fn(x)|2dµ(x) o n∈N is bounded for fn= f (qn)− a n, (6) nZ Wn |fn0(x)|2dµ(x)o n∈N is bounded for f 0 n= f (q0n)− a0 n, (7) (fn0, fn)∗(µWn) → P weakly in P(R 2).
Similar conditions were stated as assumptions of Theorem 6 in [8]. By the de nition of weak convergence, we know that for any bounded uniformly continuous function φ : R2→ R, we have (8) Z Wn φ(fn0(x), fn(x))dµ(x) → α Z R2 φ(t, u)dP (t, u). We will now need the following series of auxiliary lemmas.
Lemma 3.1. If hn→ 0 in measure and {hn}n∈N is bounded in L∞(X, B, µ) then
hn→ 0in L1(X, B, µ). 2
Lemma 3.2. A sequence {qn}n∈N is rigid for T along {Wn}n∈N if and only if for
every f ∈ L1(X, B, µ)we have χ
Wn(f ◦ T
qn− f ) → 0 in measure.
Proof. Note that µ((T−qnA4A) ∩ W
n) → 0is equivalent to
Z
Wn
|χA◦ Tqn− χA|dµ → 0.
By passing to simple functions and by density of simple functions we have that χWn(f ◦ T
qn − f ) → 0 in L1 for all f ∈ L1(X, B, µ). By Markov's inequality,
χWn(f ◦ T
qn− f ) → 0in measure.
Conversely, suppose that χWn(f ◦T
qn−f ) → 0in measure for any f ∈ L1(X, B, µ).
If f is additionally bounded then, by Lemma 3.1, χWn(f ◦ T
qn− f ) → 0 in L1.
Taking f = χA we obtain χWn|χA◦ T qn− χ
A| → 0 in L1 for every A ∈ B. This
gives the rigidity of the sequence {qn}n∈Nalong {Wn}n∈N.
Lemma 3.3. Suppose that (X, B, µ) is endowed with a metric d generating the σ-algebra B. If supx∈Wnd(Tqnx, x) → 0, then {q
n}n∈N is a rigidity sequence for T
along {Wn}n∈N .
Proof. Let h ∈ L1(X, B, µ)and let ε > 0 and a > 0 be arbitrary. Then, by Lusin's
theorem, there exists a compact set Bε⊂ X such that µ(Bεc) < ε
2 and h : Bε→ R
is uniformly continuous. Therefore, there exists δ > 0 such that d(x, y) < δ implies |h(x) − h(y)| < a for all x, y ∈ Bε. By assumption, there exists n0∈ N such that
n ≥ n0 and x ∈ Wn⇒ d(x, Tqnx) < δ.
Hence, x ∈ Wn∩ Bε∩ T−qnBεimplies |h(x) − h(Tqnx)| < afor n > n0. Therefore
µ({x ∈ Wn; |h(x) − h(Tqnx)| ≥ a)}) ≤ µ(Wn∩ Bεc∩ T −qnBc
ε) ≤ 2µ(B c ε) < ε.
Since ε > 0 and a > 0 were arbitrary, then using Lemma 3.2 we complete the
proof.
Lemma 3.4. If {hn}n∈N and {gn}n∈N are bounded sequences in L∞(X, B, µ)such
that hn→ 0in measure, then hn· gn → 0in L1.
Proof. Let M ≥ 0 be such that kgnk∞≤ M for every n ∈ N. Then
khn· gnkL1 ≤ kgnk∞khnkL1 ≤ M khnkL1.
Since, by Lemma 3.1, hn→ 0in L1, it follows that hn· gn→ 0in L1.
Theorem 3.5. Suppose that (1)-(7) hold. Let h, h0
: X → R be measurable func-tions. Let g ∈ L∞(X, B, µ)and φ : R2
→ R be bounded and uniformly continuous. Then Z Wn φ(fn0(x) + h0(x), fn(x) + h(x))g(x) dµ(x) → α Z X Z R2 φ(t + h0(x), u + h(x))g(x) dP (t, u) dµ(x). (9)
Proof. We will divide the proof into steps according to the complexity of the func-tions h and h0.
Step 1. Assume that h = h0 = 0. If g is constant then (9) follows from (8)
directly. Hence we can assume that g ∈ L1
0(X, B, µ), i.e. has zero mean. By the
proof of von Neumann ergodic theorem, coboundaries, i.e. functions of the form g = ξ − ξ ◦ T with ξ ∈ L2(X, B, µ)are dense in L10(X, B, µ). Therefore, it su ces
to consider g = ξ − ξ ◦ T for ξ ∈ L∞(X, B, µ), as they are also dense in L1
0(X, B, µ).
Note that the RHS of (9) is equal to α Z R2 φ(t, u)dP (t, u) Z X g(x) dµ(x) = 0, whenever g ∈ L1
0(X, B, µ). As g = ξ − ξ ◦ T , we need to prove that
(10) Z Wn φ(fn0(x), fn(x))ξ(x) dµ(x) − Z Wn φ(fn0(x), fn(x))ξ(T x) dµ(x) → 0. However, by the T -invariance of µ, we have
Z Wn φ(fn0(x), fn(x))ξ(x) dµ(x) − Z Wn φ(fn0(x), fn(x))ξ(T x) dµ(x) = Z T−1W n φ(fn0(T x), fn(T x))ξ(T x) dµ(x) − Z Wn φ(fn0(x), fn(x))ξ(T x) dµ(x) ≤ Z Wn |φ(fn0(T x), fn(T x)) − φ(fn0(x), fn(x))||ξ(T x)| dµ(x) + kφk∞ Z T−1W n4Wn |ξ(T x)| dµ(x). By (2), we have µ(T−1W n4Wn) → 0. Thus, RT−1W n4Wn|ξ(T x)| dµ(x) → 0 as n → ∞.
Now we use the uniform continuity of φ. By the de nition of fn and fn0, we have
(fn0(T x), fn(T x)) − (fn0(x), fn(x)) = (f (Tq 0
nx) − f (x), f (Tqnx) − f (x)).
By Lemma 3.2, we have that χWn(x)(f (T
qn0x) − f (x)) → 0and χ
Wn(x)(f (T
in measure and thus χWn(x) (fn0(T x), fn(T x)) − (fn0(x), fn(x)) → 0 in measure. Since φ is uniformly continuous, we also have
χWn
φ(fn0 ◦ T, fn◦ T ) − φ(fn0, fn)
→ 0 in measure. By Lemma 3.4, we get that
Z
Wn
kφ(fn0(T x), fn(T x)) − φ(fn0(x), fn(x))||ξ(T x)| dµ(x) → 0.
This concludes the proof of (10), which also completes the proof of (9) for h = h0 =
0.
Step 2. Now let h0 =Pk
i=1hiχAi and h = P l j=1h
0
jχBj be simple functions,
where Ai and Bj for i = 1, . . . , k and j = 1, . . . , l make two measurable disjoint
partitions of X. Then Z Wn φ(fn0(x) + h0(x), fn(x) + h(x))g(x) dµ(x) = k X i=1 l X j=1 Z Wn φ(fn0(x) + h0i, fn(x) + hj)g(x)χAi(x)χBj(x) dµ(x) → k X i=1 l X j=1 α Z X Z R2 φ(t + h0i, u + hj)g(x)χAi(x)χBj(x)dP (t, u) dµ(x) = α Z X Z R2 φ(t + h0(x), u + h(x))g(x)dP (t, u) dµ(x),
where the convergence follows from the rst step of the proof applied to functions (t, u) 7→ φ(t + h0i, u + hj). This gives (9) whenever h and h0 are simple functions.
Step 3. All we need to show now is that for arbitrary measurable functions h and h0, we can nd sequences {h
m}m∈N, {h0m}m∈N of simple functions such that
Z Wn φ(fn0(x) + h0m(x), fn(x) + hm(x))g(x) dµ(x) → Z Wn φ(fn0(x) + h0(x), fn(x) + h(x))g(x) dµ(x) and Z X Z R2 φ(t + h0m(x), u + hm(x))g(x)dP (t, u) dµ(x) → Z X Z R2 φ(t + h0(x), u + h(x))g(x)dP (t, u) dµ(x),
as m → +∞. Take hmand h0msimple, such that hm→ hand h0m→ h0in measure.
Then by the uniform continuity of φ we obtain that
φ(fn0(x) + h0m(x), fn(x) + hm(x)) − φ(fn0(x) + h0(x), fn(x) + h(x)) → 0,
in measure on (X, µ) and
φ(t + h0m(x), u + hm(x)) − φ(t + h0(x), u + h(x)) → 0
in measure on (R2× X, P ⊗ µ). By Lemma 3.4 and Step 2, this completes the proof
Lemma 3.6. Let the assumptions (1)-(7) hold. Furthermore, assume that g, ξ, ξ0∈ L∞(X, B, µ). Then Z Wn φ(fn0(x) + h0(x), fn(x) + h(x)) g(x) ξ(Tqnx) ξ0(Tq 0 nx) dµ(x) → α Z X Z R2 φ(t + h0(x), u + h(x)) g(x) ξ(x) ξ0(x) dP (t, u) dµ(x). Proof. By Lemma 3.2, χWn(x)(ξ(x) − ξ(T qnx)) → 0and χ Wn(x)(ξ 0(x) − ξ0(Tqn0x)) → 0,
in measure. Then, by Lemma 3.4, it follows that | Z Wn φ(fn0(x) + h0(x), fn(x) + h(x))g(x)ξ(Tqnx)ξ0(Tq 0 nx) dµ(x) − Z Wn φ(fn0(x) + h0(x), fn(x) + h(x))g(x)ξ(x)ξ0(Tq 0 nx) dµ(x)| ≤ Z X |φ(fn0(x) + h0(x), fn(x) + h(x))||g(x)||ξ0(Tq 0 nx)| |χWn(x)(ξ(T qnx) − ξ(x))| dµ(x) → 0, and | Z Wn φ(fn0(x) + h0(x), fn(x) + h(x))g(x)ξ(x)ξ0(Tq 0 nx) dµ(x) − Z Wn φ(fn0(x) + h0(x), fn(x) + h(x))g(x)ξ(x)ξ0(x) dµ(x)| ≤ Z X |φ(fn0(x) + h0(x), fn(x) + h(x))||g(x)||ξ(x)| |χWn(x)(ξ 0(Tq0nx) − ξ0(x))| dµ(x) → 0.
Hence to conclude the proof of the lemma, we need to show that Z Wn φ(fn0(x) + h0(x), fn(x) + h(x))g(x)ξ(x)ξ0(x) dµ(x) → α Z X Z R2 φ(t + h0(x), u + h(x))g(x)ξ(x)ξ0(x)dP (t, u) dµ(x).
However, that is a direct consequence of Theorem 3.5, taking g(x)ξ(x)ξ0(x)in place
of g(x).
The following auxiliary lemma is well-known and we state it without any proof. Lemma 3.7. Let gn: X → Rm, n ∈ N be measurable maps such that (gn)∗µ → P
weakly in P(Rm). Let h
n : X → Rn, n ∈ N be measurable such that hn → 0 in
measure. Then (gn+ hn)∗µ → P weakly in P(Rm). 2
Let T : (X, B, µ) → (X, B, µ) be an ergodic automorphism, f : X → R be an L2 function such that f ≥ c > 0. Denote by T
−f : X × R → X × R the skew
product T−f(x, r) = (T x, r − f (x)). Then for every n ∈ Z we have T−fn (x, r) =
(Tnx, r − f(n)(x)), where f(n)(x) =
(
f (x) + f (T x) + . . . + f (Tn−1x) for n ≥ 0
−(f (T−1x) + . . . + f (Tnx)) for n < 0.
Denote by (σt)t∈R the ow on X × R de ned by σt(x, r) = (x, r + t).
Lemma 3.8. For all t, s ∈ R and all measurable sets A, B, C ⊂ Xf we have
µf((Tf)tA ∩ (Tf)sB ∩ C) =
X
k,l∈Z
µ ⊗ λ((T−f)kσtA ∩ (T−f)lσsB ∩ C).
Moreover, the sets that appear on the RHS of the above equation are pairwise dis-joint.
Lemma 3.9 (see Lemma 3.4 in [5]). Suppose that A = A1× A2, B = B1× B2,
C = C1× C2 are measurable rectangles in X × R. Then
µ ⊗ Leb(((T−f)k1A) ∩ ((T−f)k2B) ∩ C)
= Z
(Tk1A1)∩(Tk2B1)∩C1
Leb((A2+ f(−k1)(x)) ∩ (B2+ f(−k2)(X)) ∩ C2) dµ(x).
The following theorem was inspired by Theorem 6. in [8] and by Proposition 3.7 in [5].
Theorem 3.10. Suppose that (1)-(7) hold. Then (11) µfa0 n,an→ ρ = α Z R µf−t,−udP (t, u) + (1 − α)ν in J3(Tf), where ν ∈ J3(Tf).
Proof. By the compactness of J3(Tf), and by passing to a subsequence, if necessary,
we have µf a0
n,an → ρ in J3(T
f). First we will prove that for measurable rectangles
in Xf
A = A1× A2, B = B1× B2, C = C1× C2,
with A2, B2, C2⊂ R bounded, we have
µf (Tf)−a0 n(A ∩ (T qn0W n× R)) ∩ (Tf)−an(B ∩ (T qnW n× R)) ∩ C → α Z R2 µf (Tf)tA ∩ (Tf)uB ∩ C dP (t, u). (12) By Lemma 3.8, µf((Tf)−a0 n((A1∩ T qn0W n) × A2) ∩ (Tf)−an((B1∩ T qnW n) × B2) ∩ C1× C2) =X k∈Z X l∈Z µ ⊗ Leb((T−f)−k(T−f)−q 0 nσ −a0 n((A1∩ T qn0W n) × A2) ∩ (T−f)−l(T−f)−qnσ−an((B1∩ T qnW n) × B2) ∩ C1× C2).
Moreover, in view of Lemma 3.9, ank,l:=µ ⊗ Leb(T−f)−k(T−f)−q 0 nσ −a0 n((A1∩ T qn0W n) × A2) ∩ (T−f)−l(T−f)−qnσ−an((B1∩ T qnW n) × B2) ∩ C1× C2 = Z Un Leb(A2− a0n+ f (q0n+k)(x)) ∩ (B 2− an+ f(qn+l)(x)) ∩ C2 dµ(x), where Un := T−q 0 n−k(A 1∩ Tq 0 nW n) ∩ T−qn−l(B1∩ TqnWn) ∩ C1 = T−q0n−k(A1) ∩ T−kWn∩ T−qn−l(B 1) ∩ T−lWn∩ C1.
Fix l ∈ Z. Using Lemma 3.8 and then Lemma 3.9 for A := X × R we obtain X k∈Z ank,l≤ µ ⊗ Leb((T−f)−l(T−f)−qnσ−an((B1∩ T qnW n) × B2) ∩ C1× C2) ≤ Z T−lW n Leb((B2− an+ f(l+qn)(x)) ∩ C2) dµ(x) = Z T−lW n Leb((B2+ fn(Tlx) + f(k)(x)) ∩ C2) dµ(x),
where in the last equality we used the fact that f(l+qn)(x) − a
n = f(l)(x) + f(qn)(Tlx) − an = f(l)(x) + fn(Tlx).
Let now s = diam(B2∪ C2)and Vn= {x ∈ T−lWn; |fn(Tlx) + f(l)(x)| ≤ s}. Then
Z T−lW n Leb((B2+ fn(Tlx) + f(l)(x)) ∩ C2) dµ(x) = Z Vn Leb((B2+ fn(Tlx) + f(l)(x)) ∩ C2) dµ(x) ≤ sµ(Vn) ≤ sµ({x ∈ Wn; |fn(x)| ≥ c|l| − s}) ≤ s D (c|l| − s)2, (13) where D = supn∈NR Wn|fn(x)|
2dµ(x) and the last inequality follows from
Cheby-shev's inequality (similar calculation was used in proof of the Lemma [7]). Let Dl = s(c|k|−s)D 2 for integer l with |l| >
s
c and let Dk = 1 otherwise. Then
P
k∈Za n
k,l ≤ Dk and Pl∈ZDl < ∞. Similarly, we can nd a sequence {Dk0}k∈Z
such that Pk∈ZDk0 < ∞ and Pl∈Zan
k,l≤ D0k. Hence, for every ε > 0 there exists
M > 0such that X |k|≥M X l∈Z ank,l< X |k|≥M D0k< ε 8 and X |l|≥M X k∈Z ank,l< X |l|≥M Dl< ε 8. Thus (14) X max(|k|,|l|)≥M ank,l< X |k|≥M X l∈Z ank,l+ X |l|≥M X k∈Z ank,l< ε 4. We are now going to prove that for every pair (k, l) ∈ Z2sequence
ank,l= Z T−kW n∩T−lWn χC1(x)χT−kA 1(T q0 nx)χ T−lB 1(T qnx) Leb (A2+ fn0(x) + f (k)(Tq0nx)) ∩ (B 2+ fn(x) + f(l)(Tqnx)) ∩ C2 dµ(x)
converges. Since, by assumption (2), µ((T−kW
n∩ T−lWn)4Wn) → 0, it is enough
to check the convergence of the sequence bnk,l:= Z Wn Leb (A2+ fn0(x) + f (k)(Tq0nx)) ∩ (B 2+ fn(x) + f(l)(Tqnx)) ∩ C2 χC1(x)χT−kA1(T q0nx)χ T−lB 1(T qnx) dµ(x). Let F0 n, Fn : X → R be given by Fn0(x) = fn0(x) + f(k)(Tq0nx) − f(k)(x)and Fn(x) = fn(x) + f(l)(Tqnx) − f(l)(x). Then bnk,l= Z Wn Leb (A2+ Fn0(x) + f(k)(x)) ∩ (B2+ Fn(x) + f(l)(x)) ∩ C2 χC1(x)χT−kA 1(T q0nx)χ T−lB 1(T qnx) dµ(x).
By Lemma 3.2, χWn(x)(f (k)(Tqnx) − f(k)(x)) → 0and χ Wn(x)(f (l)(Tq0 nx) − f(l)(x)) → 0 in measure. Since (f0 n, fn)∗(µWn) → P, by Lemma 3.7, it implies (Fn0, Fn)∗(µWn) → P weakly in P(R 2 ). Let us apply Lemma 3.6 to
φ(t, u) := Leb((A2+ t) ∩ (B2+ u) ∩ C2), (h0, h) := (f(k), f(l)), (fn0, fn) := (Fn0, Fn), g := χC1, ξ := χT−kA1, ξ 0:= χ T−lB 1. We obtain that bnk,l→ ck,l:= α Z X Z R2 Leb (A2+ t + f(k)(x)) ∩ (B2+ u + f(l)(x)) ∩ C2 χC1(x)χT−kA1(x)χT−lB1(x) dP (t, u) dµ(x).
By Fubini's theorem and Lemma 3.9, ck,l is equal to
α Z R2 Z T−kA 1∩T−lB1∩C1 Leb (A2+ t + f(k)(x)) ∩ (B2+ u + f(l)(x)) ∩ C2 dµ(x)dP (t, u) = α Z R2 (µ ⊗ Leb) (T−f)−kσt(A1× A2) ∩ (T−f)−lσu(B1× B2) ∩ (C1× C2) dP (t, u). By Lemma 3.8, (15) X k∈Z X l∈Z ck,l= α Z R2 µf((Tf)tA ∩ (Tf)uB ∩ C)dP (t, u) < ∞.
Hence, by enlarging M if necessary, we have
(16) X max{|k|,|l|}>M ck,l< ε 4. We know that an
k,l→ ck,l for all k, l ∈ Z. Choose N > 0 so that for all n ≥ N and
k, l ∈ Z with max{|k|, |l|} ≤ M , we have |an
k,l− ck,l| <
ε 2(2M + 1)2.
By (14) and (16), it follows that X k,l∈Z ank,l− X k,l∈Z ck,l ≤ X max{|k|,|l|}>M ank,l+ X max{|k|,|l|}>M ck,l+ X max{|k|,|l|}≤M |an k,l− ck,l| ≤ ε. Therefore, Pk,l∈Zank,l→P
k,l∈Zck,land in view of (15) this proves the convergence
(12).
Since ρ is the weak limit of µf a0 n,an, by (12), we get ρ(A × B × C) ≥ α Z R2 µf−t,−u(A × B × C)dP (t, u),
for measurable A, B, C ⊂ Xf. If α = 1 this implies (11). If 0 < α < 1 let us
consider ν := 1 1 − α ρ − α Z R2 µf−t,−udP (t, u).
Then ν is a signed measure on (Xf)3 taking non-negative values on product sets
and ν((Xf)3) = 1. All we have to prove is that ν is a 3-joining. First we need
to check that ν is positive. Let K ⊂ (Xf)3 be measurable and ε > 0. Consider
the measure |ν|, i.e. the variation of the signed measure ν. Since |ν| is a measure de ned on metric space, it is regular and hence there exists an open set U and a compact set D such that D ⊂ K ⊂ U and |ν|(U) − |ν|(D) < ε. Let us choose a nite open covering V0
1, . . . , Vm0 of D by product sets. We can also assume that
their union is contained in U. By a re ning of this covering, one can obtain nite covering (usually not open) V1, . . . , Vrof D by pairwise disjoint product measurable
sets. Take V := Si=1...rVi. We have ν(V ) ≥ 0 as V is the disjoint union of product
sets. Note that
|ν(V ) − ν(K)| ≤ |ν|(U \ D) < ε. Thus
ν(K) > ν(V ) − ε ≥ −ε for every ε > 0. It follows that ν(K) ≥ 0, so ν is a positive probability measure.
To complete the proof we have to show that ν is (Tf t × T
f t × T
f
t)t∈R-invariant and
that it projects on each coordinate as µf. That however follows immediately from
the de nition of ν and the fact that ρ and RR2µ f
−t,−udP (t, u)are 3-joinings.
4. Non-reversibility criteria for special flows Let T : (X, B, µ) → (X, B, µ) be ergodic automorphism. Let (Tf
t)t∈R be special
ow built over T under a measurable roof function f : X → R. Suppose that conditions (1)-(7) are satis ed for sequence {Wn}n∈N of measurable sets, integer
sequences {qn}n∈N, {q0n}n∈N, real sequences {an}n∈N, {a0n}n∈N, real number 0 <
α ≤ 1and measure P ∈ P(R2). Recall that
(fn0, fn)∗(µWn) → P weakly in P(R 2).
Moreover, suppose that q0
n = 2qn and a0n= 2an. Then, by Theorem 3.10, we obtain
the following.
Corollary 4.1. Let P, an, αbe de ned as above. Then
(17) µf2a
n,an→ α
Z
R
µf−t,−udP (t, u) + (1 − α)ν, where ν is a 3-joining of the special ow Tf.
We are going to show a necessary condition for Tf to be isomorphic to its inverse.
Assume then, that there exists a measure preserving isomorphism S : Xf → Xf
such that
(18) STtf = T−tf S for t ∈ R. Then we have the following lemma.
Lemma 4.2. Let (17) and (18) be satis ed. Let us consider the function θ : R2→
R2 given by θ(t, u) = (t, t − u). Then (19) α Z R2 µft,udP (t, u) + (1 − α)ρ1= α Z R2 µt,udθ∗P (t, u) + (1 − α)ρ2, where ρ1, ρ2∈ J3(Tf).
Proof. Denote by S∗ the map which pushes forward any measure on (Xf)3 by the
map S × S × S : (Xf)3 → (Xf)3. Note that S
∗ maps 3-joinings into 3-joinings.
Indeed let ρ ∈ J3(Tf). Let A, B, C ⊆ Xf be measurable sets. We have
S∗ρ(T−tf A × T f −tB × T f −tC) = ρ(S−1T f −tC × S−1T f −tB × S−1T f tA) = ρ(TtfS−1C × TtfS−1B × TtfS−1A) = ρ(S−1C × S−1B × S−1A) = S∗ρ(A × B × C).
Moreover if we take B = C = Xf we obtain
S∗ρ(A × Xf× Xf) = ν(Xf× Xf× S−1A) = µf(S−1A) = µf(A)
Additionally S∗: J3(Tf) → J3(Tf)is continuous.
Let A, B, C ⊆ Xf be measurable. Then
S∗µft,u(A × B × C) = µ(T f −tS−1A ∩ T f −uS−1B ∩ S−1C) = µ(S−1TtfA ∩ S−1TufB ∩ S−1C) = µ(TtfA ∩ TufB ∩ C) = µf−t,−u(A × B × C). By the continuity of S∗ and by (17), we have
S∗µf2an,an→ α Z R2 S∗µf−t,−udP (t, u) + (1 − α)S∗ν = α Z R2 µft,udP (t, u) + (1 − α)S∗ν. (20)
However by a direct computation, using again (17) we obtain S∗µf2an,an(A × B × C) = µ f −2an,−an(A × B × C) = µ f(Tf 2anA ∩ T f anB ∩ C) = µf(A ∩ T−af nB ∩ T−2af nC) = µ f 2an,an(C × B × A) → α Z R2 µf−t,−u(C × B × A)dP (t, u) + (1 − α)ν(C × B × A) = α Z R µ(A ∩ TufB ∩ TtfC)dP (t, u) + (1 − α)ν(C × B × A) = α Z R2 µ(T−tf A ∩ Tu−tf B ∩ C)dP (t, u) + (1 − α)ν(C × B × A) = α Z R2 µt,t−u(A × B × C)dP (t, u) + (1 − α)ν(C × B × A) = α Z R2 µt,u(A × B × C) dθ∗P (t, u) + (1 − α)ν(C × B × A).
Let ρ1:= S∗νand let ρ2= Q∗ν, where Q : (Xf)3→ (Xf)3is given by Q(x, y, z) :=
(z, y, x). Since Q permutes coordinates, Q∗ maps 3-joinings into 3-joinings. Recall
that S∗ν is also a joining. Thus the proof is concluded.
Let A ⊂ J3(Tf)be the set of all 3-o-diagonal joinings and let
h : R2→ A; h(t, u) = µt,u.
By Lemma 2.3 A is measurable. Hence we can write Z R2 µft,udP (t, u) = Z A ηd(h∗P )(η), and analogously Z R2 µft,udθ∗P (t, u) = Z A ηd(h∗θ∗P )(η).
Thus we can prove the following theorem.
Theorem 4.3. Let (Tf)t∈R be a special ow on Xf isomorphic to its inverse.
Assume that (17) is satis ed. Then
(21) αP + β(1 − α)η1= αθ∗P + β(1 − α)η2,
for some measures η1, η2∈ P(R2)and 0 ≤ β ≤ 1.
Proof. Let ρ1and ρ2 be measures satisfying (19). Then let
ρ1= Z J3(Tf) ηdκ1(η) = β Z A ηd(κ1)A(η) + (1 − β) Z Ac ηd(κ1)Ac(η) ρ2= Z J3(Tf) ηdκ2(η) = β0 Z A ηd(κ2)A+ (1 − β0) Z Ac ηd(κ2)Ac(η),
be ergodic decompositions, where κ1, κ2 are Borel measures on J3(Tf). Note that
ergodic decomposition of 3-joining consists of ergodic 3-joinings. By the uniqueness of ergodic decomposition and by (19) we obtain that
α Z A ηdh∗P (η) + β0(1 − α) Z A ηdκ02(η) = α Z A ηdh∗θ∗P (η) + β(1 − α) Z A ηdκ01(η), and hence β = β0. Again, by uniqueness of ergodic decomposition, we obtain that
(22) αh∗P + β(1 − α)κ02= αh∗θ∗P + β(1 − α)κ01.
And thus, by pushing forward with h−1 (recall that h was a measure theoretic
isomorphism), we conclude the proof of the theorem. Let us consider ξ : R2→ R; ξ(x, y) = x − 2y. Recall that P = lim n→∞(f 0 n, fn)∗µWn, where fn= f(qn)− an and fn0 = f (2qn)− 2a
n. Then for any x ∈ X we have
ξ ◦ (fn0, fn)(x) = fn0(x) − 2fn(x) = 2qn−1 X i=0 f (Tix) − 2an− 2( qn−1 X i=0 f (Tix) − an) = 2qn−1 X i=qn f (Tix) − qn−1 X i=0 f (Tix) = f(qn)(Tqnx) − f(qn)(x). (23) It follows that (24) ξ∗P = lim n→∞(ξ ◦ (f 0 n, fn))∗µWn= limn→∞(f (qn)◦ Tqn− f(qn)) ∗µWn.
Theorem 4.4. Suppose that ξ∗P =Pmi=0ciδdi is a discrete measure with d0= 0.
Assume that Pm i=1ci >
1−α
α , where 0 < α ≤ 1 is given by (1), and di 6= −dj for
i 6= j. Then the special ow Tf is not isomorphic to its inverse.
Proof. Note that
ξ ◦ θ(t, u) = 2u − t = −ξ(t, u). Hence we also have
ξ∗(θ∗P ) = m
X
i=0
Combining this with (21) we get α m X i=0 ciδdi+ β(1 − α)ξ∗η1= α m X i=0 ciδ−di+ β(1 − α)ξ∗η2
with 0 ≤ β ≤ 1. After normalization this gives α α + β(1 − α) m X i=0 ciδdi+ β(1 − α) α + β(1 − α)ξ∗η1 = α α + β(1 − α) m X i=0 ciδ−di+ β(1 − α) α + β(1 − α)ξ∗η2.
The LHS probability measure has non-zero atoms at di, i = 1, . . . , m and the sum
of their measures is equal to α α + β(1 − α) m X i=1 ci> α α + β(1 − α) 1 − α α = 1 − α α + β(1 − α) ≥ β(1 − α) α + β(1 − α). The RHS probability measure has atoms at −di, i = 0, . . . , m and the sum of their
measures is greater or equal to 1 − β(1−α)
α+β(1−α). By assumption, the union of these
atoms is disjoint from the set of atoms {di: i = 1, . . . , m}. It follows that the LHS
measure with at least 2m + 1 dierent atoms with total measure greater than 1. This yields a contradiction, and hence the proof is complete.
5. Interval Exchange Transformations
On the interval I = [0, 1) we consider the standard Lebesgue measure Leb. Let Sd(d ≥ 2) be the set of all permutations of d elements. Let Λd= {λ ∈ Rd+; |λ| = 1}
be the standard unit simplex, where |λ| = Pd
i=1λi is the length of a vector λ. Let
Ik = [Pj<kλk,Pj≤kλk) for k = 1, . . . , d. The interval exchange transformation
Tπ,λ is a map that rearranges intervals Ik according to permutation π. In other
words it is an automorphism of I such that Tπ,λacts as a translation on Ik for k =
1, . . . , d. Note that Tπ,λpreserves Leb (see [16], [17] or [20] for detailed information
about IETs).
In this section we will construct a sequence of sets Wn for interval exchange
transformation (IET) such that the assumptions of Theorem 3.10 are met. The construction is based on the construction made by Veech in [19] (see also [11]). Let Tπ,λ : [0, 1) → [0, 1)be an ergodic interval exchange transformation of d intervals
given by a permutation π ∈ Sd, and a length vector λ = (λ1, . . . , λd) ∈ Λd. Then π
is irreducible, i.e.
π({1, . . . , k}) = {1, . . . , k}for some k ∈ {1, . . . , d} ⇒ k = d. Denote by S0
d the set of all irreducible permutations. Let (Ij)j=1...d be the
ex-changed intervals and by ∂Ij we will denote the left endpoint of the interval Ij.
Note that Ij = [Pi<jλi,Pi≤jλi)and ∂Ij =Pi<jλi. We say that Tπ,λ satis es
Keane's condition, if
Tπ,λk (∂Ii) = ∂Ij for some k ∈ N and i, j ∈ {1 . . . d} ⇒ k = 1 and j = 1.
Theorem 5.1 (see Proposition 3.2. in [20]). If π ∈ S0
d then for a.e. λ ∈ Λd the
interval exchange transformation Tπ,λ is ergodic and satis es Keane's condition.
From now on for any interval J we will denote by |J| the length of this interval. Let s = min{|Id|, |Iπ−1(d)|}. If |Id| 6= |Iπ−1(d)| then we can consider the rst
return map to the interval [0, 1 − s). Note that newly obtained automorphism is also an interval exchange of d subintervals of [0, 1 − s) which is given by a pair (π1, λ1) ∈ Sd0× (0, +∞)d. The map (π, λ) 7→ (π1, λ1) is called the Rauzy-Veech
induction and we will denote it by ˆR. If possible we can iterate Rauzy-Veech induction.
Proposition 5.2 (see Veech in [18]). Let Tπ,λ be an interval exchange
transfor-mation satisfying Keane's condition. Then for each n ∈ N the n-th Rauzy-Veech induction is well de ned.
Let (πn, λn) = ˆR(π, λ). By In we will denote the interval obtained after n-th
step of Rauzy-Veech inductions. Denote by An(π, λ) = An = [An
ij]i,j=1...d the n-th
Rauzy-Veech induction matrix (or shortly Rauzy's matrix), i.e. An
ij is the number
of times that the interval In
j visits the interval Iiunder iterations of Tπ,λbefore its
rst return to In.
Remark 5.3. Let T = Tπ,λ : I → I be an interval exchange transformation
of d intervals satisfying Keane's condition. Then for every n ∈ N (see ...), the interval I is decomposed into d Rokhlin towers of the form Ssnj−1
i=0 T iIn
j (i.e. TiIjn
for i = 0, . . . , sn
j− 1are pairwise disjoint intervals), where s n j := Pd i=1A n ij(π, λ)for j = 1, . . . , d.
The following theorem states some properties of Rauzy's matrices. Theorem 5.4 (see [16]). Let (π, λ) ∈ S0
d× Λd be such that Tπ,λ satis es Keane's
condition. Then 1. An(π, λ)λn= λ;
2. An(π, λ) = A1(π, λ) · . . . A1(πn−1, λn−1) · A1(πn, λn);
3. There exists n ∈ N such that An(π, λ)is strictly positive matrix and πn= π.
For any positive d × d matrix B let ρ(B) = max 1≤i,k,l≤d Bij Bik . Set bj =P d
i=1Bij and let A be any nonnegative nonsingular d × d matrix. The
following properties are easy to prove
(25) bj≤ ρ(B)bk for any 1 ≤ j, k ≤ d,
(26) ρ(AB) ≤ ρ(B).
By normalized Rauzy-Veech induction we call the map R : Sd0× Λd→ Sd0× Λ
d, R(π, λ) =π1, λ1
|λ1|
. The set of permutations S0
d splits into subsets called Rauzy graphs G ⊂ Sd0 such
that the product G × Λdis R-invariant (see [20] for details). The following theorem
was proven by Veech in [18].
Theorem 5.5. For every Rauzy graph G ⊂ S0
d there exists a σ- nite R-invariant
measure ζG on G × Λd equivalent to the product of the counting measure on G and
the Lebesgue measure on Λd such that the normalized Rauzy-Veech induction R is
ergodic and recurrent on (G × Λd, ζ G).
6. Non-reversibility of special flows over interval exchange transformation
6.1. Piecewise linear roof functions. Let I = [0, 1) be the unit interval. Let (I, B, Leb, T )be an interval exchange transformation of d intervals given by a pair (π, λ), where π is a non-reducible permutation of d elements, and λ is a length vector. Let (Tf
f : [0, 1) → R which is linear over each exchanged interval with the same slope r 6= 0. Suppose that T is ergodic and satis es Keane condition.
Now we will perform the main construction. Lemma 6.1. For every π ∈ S0
d and a.e. λ ∈ Λ
d there exists sequence {W n}n∈N
of measurable sets, increasing sequences {qn}n∈N, {q0n}n∈N of integer numbers such
that q0
n= 2qn and conditions (1)-(4) are satis ed.
Proof. Fix (π0, λ0) ∈ Sd0× Λ
d and let π
0 ∈ Gbe the corresponding Rauzy graph.
By Theorem 5.4 there exists m > 1 such that B := Am
(π0,λ0)is positive matrix and
π0= π0m. Let ε > 0 and let δ > 0 be such that
6δ < εand (1 − 3δ)(1 − ρ(B) δ
1 − δ) > 1 − ε.
Let Y ⊆ {π0} × Λd be the set of (π0, λ) such that λ1 > (1 − δ)|λ|and λj > 2dδ|λ|
for 2 ≤ j ≤ d. Moreover, let V = {(π0, Bλ); (π0, λ) ∈ Y }. Since ζ(V ) > 0, by
the ergodicity and recurrence of R, for a.e. (π, λ) ∈ Gd× Λd, there are in nitely
many times rn for n ∈ N such that Rrn(π, λ) ∈ V. Since π0 = πm, we have that
Rm+rn(π, λ) ∈ Y, as by the de nition of V , we have that RmV ⊆ Y. Moreover, by
the Theorem 5.4
Am+rn(π, λ) = Arn(π, λ)B.
Hence, by (26), we obtain that
(27) ρ(Am+rn(π, λ)) ≤ ρ(B). Since Rm+rn(π, λ) = (πm+rn, λm+rn |λm+rn|) ∈ Y, we have (28) λm+rn 1 > (1 − δ)|λ m+rn| (29) λm+rn j > δ 2d|λ m+rn|for 2 ≤ j ≤ d. Let T := Tπ,λ and snj = Pd i=1A m+rn
ij . Note that snj is the rst return time of
interval Im+rn j to I m+rn under T . Let Jn:= Im+rn 1 ∩ T −sn 1Im+rn 1 ∩ T −2sn 1Im+rn 1 . Since Tsn1(Im+rn
1 ) ⊂ Im+rn and Leb is Tπ,λ-invariant, we have
Leb(Jn) = Leb(Im+rn∩ Tsn1Im+rn 1 ∩ T 2sn1Im+rn 1 ), and Tsn1Im+rn 1 ∩ T2s n 1Im+rn 1 ⊂ Im+rn. Moreover, by (28) Leb(Tsn1Im+rn 1 ∩ T 2sn1Im+rn 1 ) = Leb(I m+rn 1 ∩ T sn1Im+rn 1 ) ≥ (1 − 2δ)|I m+rn|. Thus Leb(Jn) ≥ |Im+rn| − Leb(Im+rn\ Im+rn 1 ) − Leb(Im+rn\ (Tsn1Im+rn 1 ∩ T 2sn 1Im+rn 1 ))
> |Im+rn| − δ|Im+rn| − 2δ|Im+rn| = (1 − 3δ)|Im+rn|
(30)
Since Ssn1−1 i=0 TiI
m+rn
1 is a Rokhlin tower (see Remark 5.3), we have
Jn∩ TlJn= ∅for 1 ≤ l < sn1, and by the fact that Tsn1J ⊆ Im+rn we have that
Leb(Jn∩ Tsn1Jn) ≥ |Im+rn| − Leb(Im+rn\ Jn) − Leb(Im+rn\ Tsn1Jn)
By Remark 5.3, we have that |λ| = Pd j=1λ m+rn j snj and by (25), (27) and (28) we obtain that |λ| − sn 1λ m+rn 1 = d X j=2 snjλm+rn j ≤ ρ(A m+rn)sn 1δ|λ m+rn| ≤ ρ(B)sn 1δ|λ m+rn| ≤ ρ(B)sn 1 δ 1 − δλ m+rn 1 ≤ ρ(B) δ 1 − δ|λ|. Therefore sn1λ m+rn 1 = Leb( sn 1−1 [ l=0 TlIm+rn 1 ) > 1 − ρ(B) δ 1 − δ |λ|, which, by (30), implies that
Leb( sn1−1 [ l=0 TlJn) = sn1Leb(Jn) ≥ sn1|Im+rn| > (1 − 3δ)1 − ρ(B) δ 1 − δ |λ| > (1 − ε)|λ|. Note that nal inequality does not depend on n.
Let (31) Wn = sn1−1 [ l=0 TlJn. By passing to a subsequence if necessary, we get
(32) lim
n→∞Leb(Wn) = αfor some 1 ≥ α ≥ 1 − ε.
Set
(33) qn:= sn1.
Since TlJn∩ TkJn = ∅for 0 ≤ k < l < q
n and k 6= l, we get that
(34) |Wn| = sn1|J
n| → α.
Thus condition (1) is satis ed.
Let x ∈ Wn, and let 0 ≤ l < qn be such that x ∈ TlJn. By the de nition of Jn
we have that
(35) Tqn(x) ∈ Tqn(TlJn) ⊂ TlIm+rn
1 .
By the properties of Rauzy-Veech induction |Im+rn| → 0. Hence
(36) lim sup
n→∞
|Tqnx − x| = 0.
Analogously we can prove that in (35), we can replace qn by 2qn and as a result
(37) lim sup
n→∞
|T2qnx − x| = 0.
Be Lemma 3.3, (36) and (37) imply (3) and (4). Moreover (38) Leb(Wn4T−1Wn) ≤ 2Leb(Jn) → 0with n → ∞,
which veri es condition (2).
Lemma 6.2. Assume that sequences {Wn}n∈N, {qn}n∈N and {qn0}n∈N are as in
Lemma 6.1 and (29) is satis ed. Then for every x ∈ Wn and 0 ≤ l < qn the points
Tlxand Tqn+lxbelong to the same exchanged interval and the sequence de ned by
γn:= qn−1
X
l=0
converges, up to taking a subsequence, to some γ > 0.
Proof. Let x ∈ Wn, and let 0 ≤ l < qn be such that x ∈ TlJn. By (35) we have
that Tqnx ∈ TlIm+rn
1 . By the non-reducibility of considered interval exchange
transformation there is 2 ≤ p ≤ d such that πm+rn(p) < πm+rn(1). Hence by (29)
we get Tqn(x) − x > |Im+rn p | > δ 2d|I m+rn|.
Since |Im+rn| > Leb(Jn), we get
γn = qn−1 X l=0 (Tqn+l(x) − Tl(x)) ≥ q n δ dLeb(J n) = δ dLeb(Wn) → δα d > 0, for x ∈ Wn. Note that γn does not depend on x ∈ Wn, because T acts as a linear
function on TlJn for each 0 ≤ l < q
n. Hence, by passing to a subsequence if
necessary, we obtain that
(39) γn→ γ for some γ > 0.
Lemma 6.3 (see also [11]). Let T : I → I be an IET and let f : I → R be a function of bounded variation. Let {TiJ }q−1
i=0 be a Rokhlin tower such that T iJ,
i = 0, . . . , q − 1 are intervals. Then there exists a ∈ R such that |f(q)(x) − a| ≤ V ar [0,1)f for x ∈ q−1 [ i=0 Ti(J ∩ T−qJ ). Moreover, |f(2q)(x) − 2a| ≤ 2V ar [0,1)f for x ∈ q−1 [ i=0 Ti(J ∩ T−qJ ∩ T−2qJ ). Proof. Let (40) a := 1 |J | Z Sq−1 l=0TlJ f (t)dt. Then for x ∈ Tk(J ∩ T−qJ )we have
|f(q)(x) − a| ≤ X k≤i<q 1 |J | Z TiJ |f (Ti−kx) − f (t)|dt + X 0≤i<k 1 |J | Z TiJ |f (Tq+i−kx) − f (t)|dt ≤ X 0≤i<q V arTiJf ≤ V ar[0,1]f. If x ∈ Tk(J ∩ T−qJ ∩ T−2qJ )then Tqx ∈ Tk(J ∩ T−qJ ). Hence |f(2q)(x) − 2a| ≤ |f(q)(x) − a| + |f(q)(Tqx) − a| ≤ 2V ar [0,1]f,
which concludes the proof of the lemma.
Theorem 6.4. Let π ∈ S0
d. For a.e. λ ∈ Λ
d if f : [0, 1) → R is a roof function
which is linear over each exchanged interval with the same slope r 6= 0 then the special ow Tf
Proof. Let Wn be the tower obtained in (31) and let {qn}n∈N be the sequence of
integer numbers de ned in (33). Take q0
n = 2qn. Then by Lemma 6.1 conditions
(1)−(4) are satis ed. Moreover, by Lemma 6.3, we obtain a real sequence {an}n∈N
such that sequences nZ Wn |f(qn)(x) − a n|2dx o n∈N and nZ Wn |f(2qn)(x) − 2a n|2dx o n∈N
are bounded, that is (5) and (6) are satis ed. Also, by Lemma 6.3 and by Prokhorov's theorem, passing to a subsequence if necessary, we can assume that
P = lim n→∞(f 0 n, fn)∗LebWn weakly in P(R 2 ), where fn= f(qn)− an and fn0 = f (2qn)− 2a n.
By (35) and (39), for x ∈ Wn we have
f(qn)◦ Tqn(x) − f(qn)(x) = qn−1 X i=0 (f (Tqn+ix) − f (Tix)) = r qn−1 X i=0 (Tqn+ix − Tix) = rγ n→ rγ,
for some γ > 0. Therefore, by (24), ξ∗P = δrγ. Since, by (32), we can take
lim Leb(Wn) = α > 0arbitrary close to 1 (in this case it is enough to take α > 12).
Then, by Theorem 4.4, this concludes the proof of the theorem. 7. Piecewise constant roof functions
In this section we will consider special ows built over interval exchange trans-formations and under piecewise constant roof functions. For some class of these functions we will prove the non-reversibility of such ows, by using similar argu-ments as in previous sections.
We will need the following general lemma.
Lemma 7.1 (cf. [12]). Let T be an ergodic automorphism of a standard proba-bility space (X, B, µ). Let {Wn}n∈N be the sequence of Rokhlin towers such that
lim infn→∞µ(Wn) > 0 and such that µ(Jn) → 0, where Jn is the basis of tower
Wn. Then almost every x ∈ X belongs to Wn for in nitely many n ∈ N.
We will apply a construction and arguments similar to those shown in the proof of Lemma 6.1 as well as some notation. We will also need the following fact considering interval exchange transformations.
Lemma 7.2 (see [14]). Let Tπ,λ : [0, 1) → [0, 1) be an interval exchange
trans-formation of d intervals satisfying Keane's condition. Let Gπ be a Rauzy graph
associated with π. Then there exists π0∈ Gsuch that π0(1) = dand π0(d) = 1.
Theorem 7.3. Let π ∈ S0
d. Suppose that T : [0, 1) → [0, 1) is an interval exchange
transformation of d intervals de ned by (π, λ). Assume that f : [0, 1) → R+ is a
piecewise constant function with r ≥ 3 discontinuity points β1, . . . , βr and such that
f has no jumps with opposite value. Then for almost every λ ∈ Λd and almost
every choice of discontinuity points of the roof function f the special ow Tf on
[0, 1)f is non-reversible.
Proof. Fix π0 ∈ Gπ such that π0(1) = d and π0(d) = 1. By Theorem 5.4, there
exists m ∈ N such that B := Am(π
Let 0 < ε < min (1/ρ(B)20, 1/8(2r + 1) and let ε/3 < δ0 < δ < ε/2be such that
(41) δ − δ0 < ε
4ρ(B). Let Y ⊆ {π0} × Λd stand for the set of (π0, λ)such that
1 2− δ < λ1< 1 2 − δ + δ − δ0 4 , 1 2 + δ 0< λ d< 1 2 + δ 0+δ − δ0 4 . Then λd− λ1> δ0+ δ − δ − δ0 4 > 2 3ε − 1 16ε > ε 2.
Note that Y is an open set and hence it is of positive measure. By the arguments used in the proof of Lemma 6.1 we have that for almost every λ ∈ Λd there exists
an increasing sequence of natural numbers {rn}n∈N, such that Rm+rn(π, λ) ∈ Y.
Therefore λm+rn 1 > 1 2− δ |λm+rn|, λm+rn d > 1 2+ δ 0|λm+rn| and (42) λm+rn d − λ m+rn 1 > ε 2|λ m+rn|. (43) Set sn j = Pd i=1A m+rn ij and de ne Jn:= Im+rn 1 ∩ T −sn 1−s n dIm+rn 1 ∩ T 2(−sn1−sn d)Im+rn 1 . Recall that sn
j is the rst return time of interval I m+rn j to I m+rn under T and, by (25) and (27), we have (44) snj ≤ ρ(B)sn k for 1 ≤ j, k ≤ d.
Since π0(1) = d and π0(d) = 1, the interval I1m+rn is translated by T sn
1 to the
interval ending at the end of Im+rn and Im+rn
d is translated by T
snd to the interval
starting at 0. Moreover, in view of (42), Tsn1Im+rn
1 ⊂ I m+rn d . It follows that T sn1+snd acts on Im+rn 1 by the translation by |I m+rn d | − |I m+rn 1 | = λ m+rn d − λ m+rn 1 > 0. Moreover, since ε < 1/20, by (42), (45) λm+rn d − λ m+rn 1 < |λ m+rn| − 2λm+rn 1 ≤ 2δ|I m+rn| ≤ ε|Im+rn| ≤ 1 4|I m+rn 1 |.
Therefore Jn is an interval starting from 0 and
|Jn| = λm+rn 1 − 2(λ m+rn d − λ m+rn 1 ) > λ m+rn 1 − 4δ|I m+rn| > (1 − 4δ)|Im+rn| >1 2|I m+rn|. (46)
Note also that by the Remark 5.3 and by the fact that Tsn 1Im+rn 1 ⊂ I m+rn d , we have that (47) Jn∩ TiJn= ∅for i = 1, . . . , sn 1+ s n d − 1. Let now Wn =S sn 1+snd−1
i=0 TiJn. Let x ∈ TiJn for some 0 ≤ i < sn1 + snd. We have
T−ix ∈ Jn and thus
Tsn1+sndx = Ti◦ Ts1n+snd ◦ T−ix = Ti(T−ix + λm+rn d − λ m+rn 1 ) = x + λm+rn d − λ m+rn 1 , (48)
where the last equality follows from the fact that T−ix, T−ix + λm+rn d − λ m+rn 1 ∈ I m+rn 1 and Ti acts on Im+rn
1 as a translation. Hence we obtain
(49) sup x∈Wn d(x, Tsn1+sndx) = λm+rn d − λ m+rn 1 < |I m+rn| → 0.
Similar argument shows also that (50) sup x∈Wn d(x, T2(sn1+s n d)x) = 2(λm+rn d − λ m+rn 1 ) < 2|I m+rn| → 0.
Note that the points that do not belong to Wn come from three sources, namely:
the tower of height sn
1 built over I m+rn
1 \ Jn, the towers built of height snj over
the intervals Im+rn
j for j = 2, . . . , d − 1 and the tower of height s n d built over Im+rn d \ T sn 1Jn. Since, by (46), Leb(Im+rn 1 \ J n) = |Im+rn 1 | − |J n | = 2(λm+rn d − λ m+rn 1 ), Leb(Im+rn d \ T sn 1Jn) = |Im+rn d | − |J n | = 3(λm+rn d − λ m+rn 1 )
and by (42) the sum of lengths of intervals Im+rn
j , j = 2, . . . , d − 1 is |Im+rn| − λm+rn d − λ m+rn 1 ≤ (δ − δ 0)|Im+rn|, it follows that Leb([0, 1) \ Wn) < (λm+rd n− λ m+rn 1 )(2s n 1+ 3snd) + (δ − δ0)|Im+rn| max 1<j<d(s n j). (51)
Let us consider r disjoint segments Jln=h(2l − 1)|J n| 2r + 1 , 2l|Jn| 2r + 1 ⊂ Jn for l = 1, . . . , r. In view of (47), Wl n= Ssn1+snd−1 i=0 T iJn
l are also Rokhlin towers for 1 ≤ l ≤ r. Note
that, by (44), (52) sn1|Im+rn| ≥ d X j=1 sn j ρ(B)|I m+rn j | = 1/ρ(B). Hence, by (46) Leb(Wnl) = (sn1 + snd) 1 2r + 1|J n| > sn 1 1 2(2r + 1)|I m+rn| ≥ 1 2(2r + 1)ρ(B) > 0. By Lemma 7.1, for almost every choice (β1, . . . βr) ∈ [0, 1)r we have βl ∈ Wln for
in nitely many n ∈ N for all 1 ≤ l ≤ r. Consider now the sets
Vln= sn 1+snd−1 [ i=0 T−iβl− (λm+rd n− λ m+rn 1 ), βl for l = 1, . . . , r. Note that Vn
l are Rokhlin towers. Indeed, since βl ∈ Wnl, there exists 0 ≤ k =
k(l) < sn
1 + snd such that T −kβ
l∈ Jn for l = 1, . . . , r. Note also that, by (45) and
(46), (53) λm+rn d − λ m+rn 1 < ε|I m+rn| < 1 8(2r + 1)|I m+rn| < 1 2(2r + 1)|J n|. Hence [βl− 2(λm+rd n− λm+r1 n), βl) ⊂ Tk h2l − 2 2r + 1|J n|, 2l 2r + 1|J n|⊂ W n. Since Tsn
1+snd acts on Wn as the translation by λm+rn
d − λ m+rn 1 , we have Tsn1+snd[β l− 2(λm+rd n− λ m+rn 1 ), βl− (λm+rd n− λ m+rn 1 )) = [βl− (λm+rd n− λ m+rn 1 ), βl). It follows that T−sn1−snd[β l− (λm+rd n− λ m+rn 1 ), βl) ⊂ Tk h2l − 2 2r + 1|J n|, 2l 2r + 1|J n|.
Thus for 0 ≤ i ≤ k we have T−i[βl− (λm+rd n− λm+r1 n), βl) ⊂ Tk−i h2l − 2 2r + 1|J n|, 2l 2r + 1|J n|⊂ W n and for k < i < sn 1+ snd we have T−i[βl− (λm+rd n− λ m+rn 1 ), βl) ⊂ Ts n 1+snd+k−i h2l − 2 2r + 1|J n|, 2l 2r + 1|J n|⊂ W n.
By (47), this implies that T−i[β
l− (λm+rd n− λm+r1 n), βl)are pairwise disjoint for
i = 0, . . . , sn
1 + snd − 1 and l = 1, . . . , r and all these sets are subsets of Wn. It
follows that Vn
l , l = 1, . . . , r are pairwise disjoint Rokhlin towers all included in
Wn. Hence (54) Leb( r [ l=1 Vln) = r(λm+rn d − λ m+rn 1 )(s n 1+ s n d) ≥ rε 2 s n 1|I m+rn|.
By (52) and by passing to a subsequence, if necessary, we get
(55) lim n→∞Leb [r l=1 Vln= Γ ≥ rε 2ρ(B) > 0, and in particular (56) lim n→∞Leb(Wn) = α > 0.
Moreover, by (51), (54), (43), (44), (41) and (52), we have
Leb r [ l=1 Vln− Leb([0, 1) \ Wn) > (λm+rn d − λ m+rn 1 ) (r − 2)sn1 + (r − 3)snd− (δ − δ0) max 1<j<d(s n j)|I m+rn| > ε 2s n 1|I m+rn| − (δ − δ0) max 1<j<d(s n j)|I m+rn| >ε 2 − (δ − δ 0)ρ(B)sn 1|I m+rn| > ε 4s n 1|I m+rn| > ε 4ρ(B). (57) Therefore (58) Γ − (1 − α) ≥ ε 4ρ(B) > 0.
Let f : [0, 1) → R+be a piecewise constant roof function for which β1, . . . , βrare
all discontinuities and with jumps equal to d1, . . . , dr respectively. By assumption,
dj 6= −dk for all 1 ≤ j, k ≤ r.
Let qn = sn1 + snd and qn0 = 2qn. By the de nition of {Wn}n∈N, (49), (50) and
Lemma 6.3, there exist {an}n∈N, {a0n}n∈Nthat meet the conditions (1)-(6). Passing
to a subsequence, if necessary, we can assume that P = lim(f0
n, fn)∗LebWn and let
us consider the measure ξ∗P = lim n→∞(ξ ◦ (f 0 n, fn))∗µWn= limn→∞(f (qn)◦ Tqn− f(qn)) ∗µWn. Since f(qn)(Tqnx) − f(qn)(x) = Pqn−1 i=0 (f (T qn+ix) − f (Tix)), by the de nition of sets Vn
l ⊂ Wn, for every x ∈ Wn we have
f(qn)(Tqnx) − f(qn)(x) =
(
dl if x ∈ Vln for l = 1, . . . , r
Thus, by (55) and (58), it follows that the measure ξ∗P has r non-zero atoms at dl
for l = 1, . . . , r with total mass Γ α >
1 − α α .
In view of Theorem 4.4, this gives that Tf is not isomorphic to its inverse.
8. Piecewise absolutely continuous roof functions
We will use the results of the previous section to prove the non-reversibility of special ows built over IETs under piecewise absolutely continuous functions (AC functions). Let T := Tπ,λ : [0, 1) → [0, 1) be an uniquely ergodic IET. Let
f : [0, 1) → R be a positive piecewise absolutely continuous roof function. Then the derivative Df is well de ned almost everywhere, Df ∈ L1([0, 1), Leb) and we
can de ne the sum of jumps of f as S(f ) =
Z 1
0
Df (x)dx.
We can decompose the function f into the sum of functions fpl and fac, where fpl
is a piecewise linear function with the slope S(f) and fac given by the formula
(59) fac(x) =
Z x
0
Df (t)dt − S(f )x, is an absolutely continuous function. Note that R1
0 Dfac(t)dt = fac(1) − fac(0) = 0.
The proof of the following result is partially based on the proof of Lemma 4.8 in [13].
Lemma 8.1. Let g : [0, 1) → R be a function of bounded variation. Let {Ti∆ n}
qn−1 i=0
be a sequence of Rokhlin towers such that Ti∆
n for i = 0, . . . , qn− 1 and n ∈ N are
intervals and qn → +∞. Let Jn:= ∆n∩T−qn∆n∩T−2qn∆nand Wn=S qn−1 i=0 TiJn.
Suppose that lim inf Leb(Wn) = α > 0. Then there exists a sequence of real numbers
{an}n∈N, such that for x ∈ Wn the sequences of functions {g(qn)− an}n∈N and
{g(2qn)− 2a
n}n∈Nare uniformly bounded for n ∈ N. Moreover, there exists measure
Q ∈ P(R2) such that g(2qn)− 2a n, g(qn)− an ∗LebWn→ Qweakly in P(R 2).
up to taking a subsequence. If additionally g is absolutely continuous function with R1
0 Dg(t)dt = 0then
(g(qn)◦ Tqn− g(qn))χ
Wn→ 0 uniformly.
Proof. By Lemma 6.3, we immediately obtain a real sequence {an}n∈N such that
|g(qn)(x) − a
n| ≤ V ar(g)and |g(2qn)(x) − 2an| < 2V ar(g)for x ∈ Wn and n ∈ N.
Moreover, by Prokhorov's theorem, the weak limit measure of the sequence { g(2qn)− 2a
n, g(qn)− an
∗LebWn}n∈N
in P(R2) exists up to taking a subsequence. Denote the limit measure by Q ∈
P(R2).
Assume that g is absolutely continuous function with R1
0 Dg(t)dt = 0. Since g is
absolutely continuous, for every ε > 0 there exists a function gε ∈ C1([0, 1)) such
that V ar(gε− g) = kDg − DgεkL1 < εand gε(0) = g(0). Then
Z 1 0 Dgε(t)dt = Z 1 0 (Dgε(t) − Dg(t))dt ≤ Z 1 0 |Dgε(t) − Dg(t)|dt < ε.
By the unique ergodicity of T , we have 1 qn qn−1 X i=0 Dgε◦ Ti → Z 1 0 Dgε(t)dt uniformly. For su ciently large n and for x ∈ Wn, we have
qn−1 X i=0 (gε(Tqn+ix) − gε(Tix)) = qn−1 X i=0 Z Tqn+ix Tix Dgε(t)dt ≤ Z Tqnx x qn−1 X i=0 Dgε(Tit)dt dt < qnε|T qnx − x|. Since |Tqnx − x| ≤ Leb(∆
n) and qnLeb(∆n) = Leb(S qn−1 i=0 Ti∆n) ≤ 1, we have qn|Tqnx − x| ≤ 1, and thus (60) qn−1 X i=0 (gε(Tqn+ix) − gε(Tix)) < ε. Since [Tix, Tqn+ix) for 0 ≤ i < q
n are included in dierent levels of the tower
{Ti∆ n}
qn−1 i=0 , [T
ix, Tqn+ix)for 0 ≤ i < q
n are pairwise disjoint. Hence for x ∈ Wn
we get qn−1 X i=0 (gε(Tqn+ix) − gε(Tix)) − qn−1 X i=0 (g(Tqn+ix) − g(Tix)) = qn−1 X i=0 (gε− g)(T qn+ix) − (g ε− g)(Tix) ≤ qn−1 X i=0 V ar[Tix,Tqn+ix)(gε− g) ≤ V ar[0,1)(gε− g) < ε. (61)
Combining (60) and (61), for su ciently large n > 0 and x ∈ Wn we have
|g(qn)(Tqnx) − g(qn)(x)| = qn−1 X i=0 (g(Tqn+ix) − g(Tix)) < 2ε,
which completes the proof.
Now we will state a more general version of Theorems 6.4 and 7.3.
Theorem 8.2. Let f : [0, 1) → R+ be a piecewise absolutely continuous function
with β1, . . . , βrits discontinuity points. Then for almost every (π, λ) ∈ Sd0× Λd we
have:
(1) if S(f) 6= 0 and f is absolutely continuous over exchanged intervals, or (2) for almost every choice of (β1, . . . , βr) ∈ [0, 1)r with r ≥ 3 if S(f) = 0 and
f has no jumps with opposite value, then Tf
π,λ is not isomorphic to its inverse.
Proof. Let f = fpl+ fac be the decomposition of f into its piecewise linear part
with slope S(f) and the absolutely continuous part satisfying R1
0 Dfac(t)dt = 0. If
fac= 0, then the assertion of the theorem follows straightforwardly from Theorems
7.3 and 6.4. We will show that the result remains unchanged when facis non-zero.
Let the sequence of Rokhlin towers {Wn}n∈N, the sequence of integer numbers
{qn}n∈N and real sequence {an}n∈N be as in the proof of Theorems 6.4 and 7.3,
where the roof function is equal to fpl. Then
|f(qn)
pl (x)−an| ≤ V ar(fpl)and |f (2qn)
and lim n→∞(f (2qn) pl − 2an, f (qn) pl − an)∗LebWn = P in P(R 2).
In both cases we obtain that the measure ξ∗P ∈ P(R) satis es the assumption of
Theorem 4.4. By applying Lemma 6.3 to the function f and using Prokhorov's theorem we obtain a sequence {bn}n∈N such that
|f(qn)(x) − b
n| ≤ V ar(f )and |f(2qn)(x) − 2bn| < 2V ar(f )for x ∈ Wn and n ∈ N
and there exists the weak limit of measures lim
n→∞(f
(2qn)− 2b
n, f(qn)− bn)∗LebWn= Q ∈ P(R 2),
up to taking a subsequence. We will show that ξ∗Q = ξ∗P, that is the measure Q
also satis es the assumption of Theorem 4.4. Indeed, by (24) we have ξ∗Q ← (f(qn)◦ Tqn− f(qn))∗LebWn = (f(qn) pl ◦ T qn− f(qn) pl + f (qn) ac ◦ T qn− f(qn) ac )∗LebWn. By Lemma 8.1 (f(qn)
ac ◦Tqn−fac(qn))χWn→ 0uniformly, as n → ∞. Using Lemma 3.7
we obtain that ξ∗Q = lim n→∞(f (qn) pl ◦ T qn− f(qn) pl )∗LebWn= ξ∗P.
This completes the proof.
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