Self-Joinings for 3-IETs

We show that typical interval exchange transformations on three intervals are not 2-simple answering a question of Veech. Moreover, the set of self-joinings of almost every 3-IET is a Paulsen simplex.


Introduction
Defintion 1. Let (X, B, µ, T ) be a probability measure preserving system. A selfjoining is a T × T invariant measure on X × X with marginals µ.

Defintion 3.
A Poulsen simplex is a metrizable simplex where the extreme points are dense.
Lindenstrauss, Olsen and Sternfeld proved that a Poulsen simplex is unique up to affine homeomorphism [10].

Question 1. Is almost every 3-IET prime?
It is also natural to wonder what happens for IETs with other permutations and flows on translation surfaces. It is likely that our techniques can show that residual sets of interval exchange transformations on more intervals, and flows on translation surfaces of genus greater than 1 are not simple, but we do not see how they can be applied to almost every flow on translation surface or IET with different permutation.
To prove Theorem 1.1 we define in Section 2 a distiguished class of self-joinings called "shifted power joinings." In Section 2 we also show that a special type of transformations called "rigid rank 1 by intervals" (which includes IET's by [16,Part 1,Theorem 1.4 ]) have the property that linear combinations of shifted power joinings are dense in their self-joinings. M. Lemanczyk brought to our attention that this result was proved in an unpublished paper of J. King [9]. We then prove that almost every 3-IET has the property that its ergodic self-joinings are dense in linear combinations of the shifted power joinings. We do this by having an abstract criterion (Section 3) and showing 3-IETs verify this criterion (Section 4). Context of our results: Before Veech's work, D. Rudolph introduced the notion of minimal self joinings, using it as a fruitful class of examples, including examples of prime systems [11]. The property of 2-simple generalizes minimal self joinings and in particular, no rigid system has minimal self joinings. The typical IET is rigid [16, Part 1, Theorem 1.4], so the typical IET does not have minimal self joinings, but there are rigid 2 simple systems. Ageev proved that the set of measure preserving transformations which are not 2 simple contains a dense G δ , i.e. it is a residual set, (with the topology being the so called weak topology) [1]. Our construction can be modified to give a new proof of this fact.
Our result that the self-joinings form a Paulsen simplex is also perhaps a little unexpected. Many examples of systems whose set of invariant measures form a Paulsen simplex are well known, but typically these systems are high complexity, satisfying some form of specification. In contrast, our examples have very low complexity, as T × T has quadratic block growth. Since systems of linear block growth have only finitely many ergodic measures [2], such a system can not have that the set of its invariant measures form a Paulsen simplex (though as our examples show its Cartesian product could). We remark that in the previously mentioned unpublished work, J. King proved a residual set of measure preserving transformations (which therefore must include rank 1 transformations) have that their set of their self-joinings form a Paulsen simplex [9], giving many (non-explicit) entropy zero examples. Our result is perhaps still surprising, because we treat a previously considered family of examples and we show typicality in a metric, rather than topological setting.
Two key steps are showing that the typical 3-IET admit (n j , n j + 1) approximation (see the proof of Proposition 4.5) and that this implies the existence of all sorts of ergodic joinings (see Proposition 3.1). Some consequences of transformations with (n j , n j + 1) approximation were studied by Ryzhikov [13] and as a result we get some spectral consequences for T n and T × ... × T n times , see Remark 2.
Acknowledgments: We thank El Abdaloui, M. Lemaczyk and V. Ryzhikov for numerous illuminating discussions on connections of other results with this work. Defintion 5. We say T is rigid rank 1 by intervals if there exists a sequence of intervals I 1 , . . . and natural numbers n 1 , . . . so that This is a condition saying that our transformation is well approximated by periodic transformations. A similar condition, admiting cyclic approximation by periodic transformations was considered in [6]. Let Then, R k is the Rokhlin tower over I k ,R k is the Rokhlin tower over I k ∩T −n k I k ∩T n k I k , andR k is the Rokhlin tower over 2 Heuristically one can think of R k as the set of points we can control.R k andR k let us control the points for long orbit segments, which is necessary for some of our arguments.
Proof. By the third condition in the definition of rigid rank 1 by intervals we have lim k→∞ λ(R k ) = 1. By (2.1), and thus by the fourth condition of the definition of rigid rank 1 by intervals, lim Defintion 6 (Shifted Power Joining). Let (X, T, µ) be a measure preserving dynamical system. A self-joining of (X, T, µ) that gives full measure to {(x, T a x)} for some a ∈ Z with a = 0 is called a shifted power joining.
These have also been called off diagonal joinings.
The operator A σ and convergence in the strong operator topology. Let σ be a self-joining of (T, λ). Let σ x be the corresponding measure on [0, 1] coming from disintegrating along σ on the fiber {x} Recall that one calls the strong operator topology the topology of pointwise convergence on L 2 (λ). That is A 1 , ... converges to A ∞ in the strong operator topology if and only if lim Theorem 2.2. Assume ([0, 1], T, λ) is rigid rank 1 by intervals and σ is a self-joining of ([0, 1], T, λ). Then A σ is the strong operator topology (SOT) limit of linear combinations, with non-negative coefficients, of powers of U T , where U T : Corollary 2.3. (J. King) Any self-joining of a rigid rank 1 by intervals transformation is a weak-* limit of linear combinations of shifted power joinings.
These results (or very closely related results) were established earlier by J. King [9] using a different proof. In fact he shows that if the joining in Corollary 2.3 is ergodic then there is no need to take a linear combination. See also [5, Remark. Note that n k is roughly λ(T j I k ) −1 .
Proof. Suppose 0 ≤ j < n k , and suppose x ∈ T j I k . From (2.3) we have T i R c k ⊂R c k for all −n k < i < n k . We claim that Since we can choose in (2.6) so that j + takes any value in [0, n k − 1] ∩ Z, we get (2.7) where the last estimate uses that σ has projections λ. So we obtain Now the estimate (2.4) follows from (2.7) and (2.8).
We want to guess coefficients c j so that σ is close to n k −1 j=0 c j (id × T i ) * µ. The next lemma comes up with a candidate pointwise version. Theorem 2.2 and Corollary 2.3 follow because by Egoroff's theorem this choice is almost constant on most of the T I k and the lemma after this (Lemma 2.6), which shows that they are almost T invariant.
where 0 ≤ a < n k and i + j ≡ a (mod n k ). For all 1-Lipschitz f we have Morally c j (x) is the σ x measure of the level in R k that is j levels above the level x is on. Because j + can be bigger than n k the definition is slightly more complicated. Note that the c j (x) are non-negative.
Proof. Suppose x ∈R k ∩ T j I k . First notice that if y, z ∈ T i I k for some 0 ≤ i < n k we have that d(y, z) < diam(I k ). So if j + < n k we have (2.9) .
T i I k for all 0 ≤ < n k . By summing over the j in (2.10) we obtain In view of the fact that we obtain the lemma.
Lemma 2.6. Suppose 0 ≤ < n k . If x ∈ T I k and − ≤ i < n k − then Proof. Suppose 0 ≤ < n k , 0 ≤ j < n k , and − ≤ i < n k − . First note that if 0 ≤ m < n k and z ∈ T m I k ∩R k then by (2.1), we have T s z ∈ T m+s I k ∩R k for all −m ≤ s < n k − m. Thus, if j + < n k and i + j + < n k , we have . This gives c j (x) = c j (T i x) if j + < n k and i + j + < n k . By similar reasoning we have that c j (x) = c j (T i x) if j + ≥ n k and i + j + ≥ n k . Now lets assume that j + < n k and i + j + ≥ n k . Then, , and z ∈ T j+ I k ∩R k . Therefore, the symmetric difference between T j+ −n k I k ∩ T −iR k and T j+ I k ∩R k is contained in the union of T i+j+ −n k I k ∩R c k and T j+ I k ∩R c k . Thus, in view of (2.13), and (2.14), . The last case, where j + ≥ n k and 0 ≤ i + j + < n k gives analogous bounds. So and obtain the lemma.
Let d KR denote the Kantorovich-Rubinstein metric on measures. That is The next lemma is an immediate consequence of this definition.
We say 0 ≤ j < n k is k-good if there exists y j in T j I k so that at least 1− proportion of the points in T j I k have their disintegration close to y j . That is Proof. By Lusin's Theorem there exists a compact set K of measure at least 1 − 2 4 so that the map y → σ y is continuous with respect to the usual metric on [0, 1] and the metric d KR on measures. Because K is compact this map is uniformly continuous and so there exists δ > 0 so that x, y ∈ K and |x − y| < δ then Then, because the T j I k are disjoint and of equal size and n k −1 and thus η < /2. This completes the proof of the lemma.
Notation. If j is k-good let i.e. G j is the set of points that are almost continuity points of the map Lemma 2.9. For all > 0 there exists k 1 so that for all k > k 1 there exists 0 ≤ < n k and y k ∈ T I k ∩R k so that σ y k (R c k ) < and and so for all large enough k (so that λ(R k ) is close to 1 and Lemma 2.8 holds) we have Therefore, the measure of the set of y k satisfying (2.15) (for some ) is at least 1/2. Recalling that by Lemma 2.1 we have lim k→∞ λ(R c k ) = 0 and so for k large enough, Thus, we can pick y k satisfying the conditions of the lemma.
Proof of Theorem 2.2. For each k large enough so that Lemmas 2.8 and 2.9 hold and diam(I k ) < and λ(R c k ) < , let y k be as in the statement of Lemma 2.9 and assume it is in T I k for some 0 ≤ < n k .
Step 1: We show that for all 1-Lipschitz functions f with f sup ≤ 1 we have First, observe that by Lemma 2.5 and the fact that f sup ≤ 1, By our assumptions that diam(I k ) < and σ y k (R c k ) < we have Let V denote the set of x satisfying (2.16) and such that Recalling that by assumption diam(I k ) < and invoking Lemma 2.6 we have Since y k satisfies the assumptions of Lemma 2.9 and λ(R c k ) < we have that Estimating trivially on V c we have Since f sup ≤ 1 and is arbitrary this establishes Step 1.
Step 2: Completing the proof. The idea of the proof is that by step 1 and linearity we have the limit on a dense set in L 2 . Since the functions on L 2 we consider have operator norm uniformly bounded (by 1) they are an equicontinuous family and so convergence on a dense set implies convergence.
To complete the formal proof of the theorem, observe that for any z we have Therefore since we have shown lim This is the definition of strong operator convergence.
Proof of Corollary 2.3. Letδ p denote the point mass at p. By the proof of the theorem that there exists y k so that for all x ∈ V . By (2.17) we may assume λ(V c ) is as small as we want. The corollary follows.
3. An abstract criterion Let (S, Y, λ) be a uniquely ergodic topological dynamical system. Letδ p denote a point mass at p. Note we will consider the metric d KR on the Borel probability measures on Y ×Y (which is a weak-* closed set since Y is compact) and the measureŝ Motivated by Corollary 2.3 we wish to build ergodic joinings that are close to finite linear combinations of shifted power joinings. For example we wish to have ergodic measures with d KR distance from the joining that gives measure 1 2 to {(x, x)} and measure 1 2 to {(x, Sx)}. Naively, one wants to find a sequence of shifted power joinings that spend half their time close to {(x, x)} and half their time shadowing {(x, Sx)}. Taking a weak-* limit of these we wish to have a measure close to the joining that gives measure 1 2 to {(x, x)} and measure 1 2 to {(x, Sx)}. Our approach will be to do this inductively, to have a sequence of measure ν i and µ i so that ν 0 is the shifted power joining supported on {(x, x)} and µ 0 is the joining supported on {(x, Sx)}. Inductively, µ i+1 spends a definite proportion of its time near µ i and a definite proportion near ν i and similarly for ν i+1 . That is, we want to have sets A i+1 and B i+1 so that when Clearly we want the union of A i and B i to have almost full measure and it is helpful that they each have measure at least c > 0. This isn't quite good enough, in particular if A i and B i were constant sequences. We now make the next technical proposition to overcome these issues and additionally guarantee that limiting joining is ergodic.
Of course we want to consider the case of a linear combination of d off diagonal joinings. That is, if we are given a finite number of shifted power joinings ν We do this analogously to the previous case. Indeed, we have A 1 , B 1 and {ν We repeat this and obtain {ν We continue repeating to approximate 1 Proposition 3.1 makes this precise. Conditions (a)-(e) are basic setup, Condition (A) gives the inductive switching as above and Condition (B) lets us rule out a previously mentioned issue to show that the weak-* limit of the ν i and µ i is close to 1 2 (µ 0 + ν 0 ) and moreover that it is ergodic.
Let J k be a sequence of intervals, U k be a sequence of measurable sets, r k be a sequence of natural numbers, n ( ) k be sequences of natural numbers for ∈ {1, ..., d} and j > 0 be a sequence of real numbers. Let (e) i are non-increasing and j < ∞. If and any ∈ {1, ..., d}. 2 Then the weak-* limit of any ν ( ) k (as k goes to infinity) is the same as the weak star limit of 1 d d =1 ν ( ) k as k goes to infinity. In particular these limits exist. Call this measure µ. It is ergodic and there exists C so that d KR (µ, 1 To connect this to the remarks above, consider the case that the ν and this is the measure we approximate as above. This lets us treat general linear combinations of shifted power joinings. Remark 1. One can drop the assumption that (S, Y, λ) is uniquely ergodic. In this case one replaces (B) by This requires some straightforward changes to the estimates in the proof of for all k ≥ 0, i > k and s ∈ {1, ..., d}.
Proof. Letγ The second term is at most δ i−1 1−δ i−1 + δ i−1 and using this we inductively see that To see this note thatγ where 1 ≥ c ,s > ζ > 0 for some fixed ζ depending only on c and d. Consider the matrix A i which has ( , s) entry equal to c ,s . This matrix is a definite contraction in the Hilbert projective metric. Indeed, for every ζ there exists θ > 0 so that if M is a positive matrix where the ratio of every pair of entries is at most ζ and v, w are any vectors in the positive cone then D HP (M v, M w) < θD HP (v, w) where D HP denotes the Hilbert Projective metric. Nowγ ( ) k+rd is the th entry of A k A k+d ...A k+(r−1)dγ whereγ is the vector whose i th entry isγ (k) i . Since each A i+jd is a definite contraction in the Hilbert projective metric, we see that |γ Remark. Corollary 3.3 establishes all the conclusions of Proposition 3.1 except the ergodicity of µ.
Proof of Corollary 3.3. First notice that by (A) we have that We now claim that for all , Indeed, for f 1-Lipschitz with f sup ≤ 1 we have and by (c) (i.e. the size estimate on U j ),

Then (3.3) follows because
So for any 1-Lipschitz function, f , with f sup ≤ 1, we claim that we may apply Lemma 3.2 to γ

Then, by (3.3) and (3.4),
This completes the verification of (3.1), and, in view of Lemma 3.2, the proof of Corollary 3.3.
To complete the proof of Proposition 3.1, we need to prove that µ is ergodic. We start with the following:  Proof.μ has an ergodic decompositionμ = Ỹμy dμ whereμ y is an ergodic probability measure withμ y ({z :μ z =μ y }) = 1 forμ-almost every y. For each y, let Because there is a countable · sup -dense subset of C c (Y ), by the Birkhoff Ergodic Theorem, we have thatμ y (Z y ) = 1 for all y. ∪Z y has fullμ measure and satisfies the conclusion of the lemma.
Proof of Lemma 3.4. By our assumptions, a positive µ measure set of (x, y) have that µ is a weak-* limit point (in particular the set lim sup of the G for a choice of going to 0). Throwing out a set of µ measure zero where the limit may not exist, Lemma 3.5 implies this is the unique weak-* limit point and it is ergodic.
We now identify a set of full measure for µ. As a preliminary, by the assumptions (e) and (A) of Proposition 3.1 we have that if x / ∈ ∞ n=1 ∞ k=n U k (this is a full measure condition) then there exists p 1 (x), ..., p d (x) so that Proof. It is straightforward to see that for any f (x, p (x))dλ.

By Corollary 3.3, the left hand side is
for every , establishing the lemma.
For all large enough k we have that λ(G k ) ≥ 1 2 . This is a straightforward measure estimate using Assumptions (c), (d) and (e). Suppose x ∈ G k and y ∈ supp(µ x ). The next claim shows that there exists j so that lim(ν j ) x is the point mass at y.
Proof of Claim 3.8. We first state the following straightforward consequence of the condition (A) of Proposition 3.1 (by considering if x ∈ A k or x ∈ B k ): By iterating we obtain: Corollary 3.10. For all j > k, ∈ {1, ..., d} and x ∈ G k there exists so that Note that if L ≥ r k+1 9 by condition (B) of the proposition we obtain By Corollary 3.10 there exists so that if S i x / ∈ ∞ =k+1 U then for some we have In this section, we will verify the conditions of Proposition 3.1. Before beginning the proof we set up a geometric context connected to our situation. A 3-IET with lengths 1 , 2 and 3 is a rescaling of the Poincaré first return map of rotation by 2 + 3 1 +2 2 + 3 to the interval [0, 1 + 2 + 3 1 +2 2 + 3 ) ⊂ [0, 1) [6, Section 8]. If ω sq denotes the area one square torus oriented horizontally and vertically, observe that rotation by α corresponds to the first return map of the vertical flow on 1 −α 0 1 ω sq to a horizontal side, which is also the time one map of that flow. To set up the geometric context, let M 1,2 denote the moduli space of area 1 tori with two marked points. Note that M 1,2 is isomorphic to (SL(2, R) R 2 )/(SL(2, Z) Z 2 ). For ω ∈ M 1,2 let F t ω denote the vertical flow on ω, which corresponds to left multiplication by the element 1 0 0 1 0 t . Letω ∈ M 1,2 be the square torus with two marked points distance 1 2 apart on the same horizontal line segment. Let S ⊂ M 1,2 be the set of surfaces ω so that F 1 ω p is on the same horizontal as ω and its distance along this horizontal is at most 1 2 . That is, if p is one marked point the other marked point is at 1 0 0 1 s 0 where s ≤ 1 2 . Let T be a 3-IET. It arises as the first return map of a rotation R α to an interval K. Let Then, for any x ∈ K so that R M α x ∈ K, Let ω T ∈ M 1,2 be the torus defined by taking the torus 1 −α 0 1 ω sq and marking two points on the bottom horizontal line that are |K| apart. When it is convenient, in what follows we will consider K as being embedded in ω T and g t K as being embedded Here we are identifying g t with the matrix e t 0 0 e −t 0 0 and think of g t as acting on M 1,2 ∼ = SL(2, R) R 2 /SL(2, Z) Z 2 by left multiplication. Thus, for any M ∈ N we can identify ψ M (x) as the intersection number between K and a vertical line of length M on ω T starting at a x, see Figure 1.
Using this as a definition, we can make sense of ψ M for all M ∈ R + . If we embed K in ω T , then for x ∈ K and M ∈ N, we have   Proof. Let U + denote the subgroup 1 * 0 1 * 0 of SL(2, R) R 2 . Then, U + is the expanding horospherical subgroup with respect to the action of g t , or in other words, the orbits of U + are the unstable manifolds for the flow g t . By construction, map T → ω T projects to a positive measure subset D of a single U + orbit on M 1,2 . Moreover, the pushforward of the Lebesgue measure on the space of 3-IET's to D is absolutely continuous with respect to the pushforward of the Haar measure on U + to D. The lemma then follows from the ergodicity of g t . Proof. Sinceω is square, for p ∈ω, F 1 ω p = p. Therefore, for ω ∈ B(ω, δ ) and p ∈ ω , v 2 ), and note that for δ > 0 sufficiently small and for small s ∈ R, for p ∈ g s ω , we have . Therefore, given ω ∈ B(ω, δ), we can choose s ∈ R, with |s| < c 2 (δ ) where c 2 (δ ) → 0 as δ → 0, such that 1 − e s + e s v 2 = 0, i.e. g s ω ∈ S. We have g s ω ∈ B(ω, c 3 (δ )) with c 3 (δ ) → 0 as δ → 0.
Suppose T is such thatω is a limit point of {g t ω T } t≥0 . Choose δ > 0 such that c 3 (δ ) < δ and choose t such that g t ω T ∈ B(ω, δ ) and then let t = t + s where s is as in the previous paragraph. Then g t ω T ∈ B(ω, δ) ∩ S as required.
We now apply g t to Figure 1, with t = log M . Note that ψ M (x) is also the intersection number between a vertical segment γ 1 of length 1 and a horizonal slit γ 2 of length M |K| (see Figure 2). From now on, we assume that g t ω T ∈ B(ω, δ) ∩ S for some δ 1. The following lemma references Figure 3.  . Closing the curves. We complete the vertical segment γ 1 to a closed curveγ 1 by adding a horizontal segment ζ 1 (drawn in green). Note that since g t ω T ∈ B(ω, δ)∩S, the length of ζ 1 is at most 1/2+O(δ). Similarly, we close up the horizontal slit γ 2 to obtain a closed curveγ 2 by adding in a horizontal segment ζ 2 and a vertical segment ζ 2 (drawn in purple).
In other words, for the set of points x whose green segment does not cross the purple segment, φ M (x) is m if its red segment crosses the purple segment (where φ M is as in (4.2)) and φ M (x) = m + 1 if it does not.
Note that because Figure 3 is of g log(M ) ω T , vertical trajectories of length 1 in Figure  3 correspond to vertical trajectories of length M on ω T .
Proof. Indeed, the family of curves we define are all homotopic and so their intersection withγ 2 is all the same. So for such curves, if the green and purple segments have intersection number zero then the intersection of the red segment and blue segment depends only on the intersection of the purple segment and the red segment, which by construction is either 0 or 1.

Lemma 4.4.
For all > 0 there exists δ > 0 so that if ω ∈ B(ω, δ) ∩ S and the flow F s ω is minimal then there exists ρ < and L ∈ N so that for any interval J with |J| = ρ we have Proof. Suppose p is a point in ω, and ω ∈ S. Then, F 1 p is horizontally adjacent to p. For all > 0 there exists δ > 0 so that if ω is in S ∩ B(ω, δ) then F 1 p is translated by less than 9 . Since the vertical flow on ω is minimal, F 1 p = p. Therefore, F 1 p is translated horizontally by some amount ρ > 0. Let J be a horizontal interval of length ρ. We choose L = min{s > 0 : F s J ∩ J = ∅}. We have that λ Moreover, Remark 2. Specializing to the case where a = 0 and b = k, we see that 1 2 (Id + T k ) is in the weak closure of the powers of T . Veech showed that almost every 3-IET has simple spectrum [16,Theorem 1.3]. Combining these two facts with Ryzhikov's [13, Theorem 6.1 (3) and (4) Proof. In view of Lemma 4.4, we can choose δ so small that for any ω ∈ B(ω, δ), (i) The horizontal purple line has length between 1 2 − 4 and 1 2 + 4 .
Because T × T is uniquely ergodic on the support of ν (a) and ν (b) there exists L 0 so that if d(p i , T a+i y) < for all 0 ≤ i ≤ L then for all L ≥ L 0 and y ∈ K. Indeed, T × T is uniquely ergodic on supp(ν (a) ) and supp(ν (b) ) and uniquely ergodic systems have uniform convergence of Birkhoff averages of continuous functions (see for example [3, Proposition 4.7.1]). We choose t 0 so large that any vertical trajectory of length e t 0 on ω T crosses K at least L 0 times. We further assume L 0 > max{|a|, |b|}.
We now set about defining J and A. Let V be the horizontal purple line segment. Let ρ be as in the previous lemma for g t ω T . For any horizontal interval I on g t ω T of length ρ we have one of the following mutually exclusive possibilities:   Let ρ > 0 be given by Lemma 4.4 and I be an interval of length ρ inÃ ∩ g t K so that F −1 I ⊂Ã. Now F 1 I is horizontally adjacent to I, and so F j I is horizontally jρ over from I. So by our assumption on the length of I, we have (4.5) F j I ⊂Ã, for all 0 ≤ j ≤ |V | ρ − 2(2 + |a + b|) − 3 ≡p.
(Note that by (ii) and the fact that |V | > 1 2 − we havep ≥ 1.) We now use what we have done for the flow on g t ω T to establish some of our claims about the IET, T . Let r be the cardinality of the set of intervals of length ρ in s∈[0,p) F s I ∩ K. Note that because in our setÂ a vertical trajectory of length 1 crosses g t K ⊂ g t ω T exactly m times, r = mp.
Let A = g −tÂ ∩ K ⊂ ω T ∩ K, which we can consider as a subset of the domain of T as well (because it is contained in K). Note that we have We also have for all x ∈ A , because when we apply g −t to pull back our dynamics from g t ω T back to ω T we contract horizontal distances by e −t . It follows from (4.2), (4.6) and (4.7) that Let J denote the interval corresponding to I in the domain of our IET, T . That is, we consider J = g −t I ⊂ K ⊂ ω T , which since it is in K we consider as an interval in the domain of T . Let A = r−1 i=0 T i J, which we can consider as a subset of K ⊂ ω T . We now claim that (4.9) A ⊂ g −tÃ ∩ K.
Indeed, by (4.5), we have (4.10) It follows in view of (4.6), that for x ∈ J, By (4.2), we have for x ∈ J and i ∈ N, Since for a fixed x ∈ J, the map s → φ s (x) is monotone increasing in s, for 0 < i < r we have in view of (4.11), This, together with (4.10) implies (4.9). The same argument shows that (4.12) T x ∈ A for x ∈ A and | | ≤ m(|a − b| + 1).
We now claim that for all x ∈ A we have: Indeed, by (4.12) and (4.8) we have d(T jm+a x, T (j−1)m+a x) = ρe −t for all |j| ≤ |a − b|, because |a| < m. We obtain the second inequality by (ii). We now show that for all x ∈ A d KR ( 1 m m−1 i=0 δ (T i x,T i+n x) , ν (a) ) < 2 .
By construction, if x ∈ A then T i x ∈ A for all −m ≤ i ≤ m. So we have that d(T i+n x, T i+a x) < for all |i| ≤ |m|. So by (4.3) and the fact that m ≥ L 0 we have our condition on d KR . We now show that λ(A) > 1 2 λ(K) − . This follows from the fact that by (ii) the measure of the set of x ∈ g t ω T so that F x crosses the horizontal purple strip for 0 ≤ ≤ 1 and −1 ≤ ≤ 0 and F s gtω T x does not have this property for some −1 ≤ s ≤ 1 has measure at most 2 10(|a|+|b|) . By our condition on the length of the purple horizontal strip, the measure condition on A is completed.
The fact that the return time of T to J is at most 3 2 r follows from the fact that the measure of A c is at most 1 2 λ(K) + and so the orbit of J after leaving A and before returning to J has measure at least 1 2 λ(K) − − > 1 2 λ(A). So J has at least 1 2 r images outside of A before part of it returns.
We now similarly define B ⊂ A c with the desired properties. First let B = {x ∈ g t ω T : ∪ s∈[−3−|a−b|,3+|a−b|] F s gtω T (x) ∩ V = ∅}. Similarly to before letB = ∩ s∈[−1,1] F s gtω TB and B = g −tB ∩ K ⊂ ω T , considered as a subset of the domain of T . Now as above, by Lemma 4.3 if x ∈B then we have that a vertical trajectory of length 1 or -1 emanating from x crosses g t K exactly m + 1 times. Moreover, F s gtω T x has this property for all −|a − b| ≤ s ≤ |a − b|. Since n = b + (m + 1)(a − b), for any x ∈ B and |i| ≤ m we have x, T (i−1)(m+1) x) ≤ . Thus, as above we have d KR ( 1 m m−1 i=0 δ (T i x,T i+n x) , ν (b) ) < 2 for all x ∈ B. The fact that λ(B) > λ(K) − is similar to the case of λ(A) above. Now given two number a, b we may iteratively apply Proposition 4.5 to obtain the assumptions of Proposition 3.1. Indeed, we choose i satisfying assumptions (c) and (e). We apply Proposition 4.5 to the pair of numbers (a, b) and = 1 to obtain m, A, B and r. Denote m by a 1 . We apply Proposition 4.5 to the pair of numbers (a, b) and = 1 to obtain m, r , A , B , and denote m by b 1 . We repeat this procedure with a 1 and b 1 in the place of a and b and 2 in place of and obtain a 2 , b 2 . We further request that the interval J produced by Proposition 4.5 have max{r, r }λ(J) < 2 . Iterating this we have the conditions of Proposition.
Proof of Theorem 1.1. Let µ be an invariant measure for T ×T . By Corollary 2.3 there exists n 1 , ..., n d so that ν i is the joining supported on {(x, T n i x)} and d KR (µ, 1 d d i=1 ν i ) < . For each pair n i , n i+1 and 2 we apply Proposition 4.5 to obtain δ, t 0 . We further do this for the pair n d , n 1 . We choose δ to be the smallest of these and t 0 to be the largest. We obtain t > t 0 so that g t ω T ∈ B(ω, δ). We then obtain m i , r i which we denote n (1) i and r (1) i . We now repeat this n (1) i in place of n i , 2 2 in place of 2 and max{r (1) i }. In doing this we obtain n (2) i and r (2) i . We repeat this recursively having our kth choice of be 2 k .
We are now left to prove that there is an ergodic self-joining that is neither λ × λ nor one-to-one on almost every fiber. Let ν