Dimension and measure of baker-like skew-products of $\beta$-transformations

We consider a generalisation of the baker's transformation, consisting of a skew-product of contractions and a $\beta$-transformation. The Hausdorff dimension and Lebesgue measure of the attractor is calculated for a set of parameters with positive measure. The proofs use a new transverality lemma similar to Solomyak's [Solomyak, 1995]. This transversality, which is applicable to the considered class of maps holds for a larger set of parameters than Solomyak's transversality.

Together with Erdős' result [3], this implies that if λ is the inverse of a Pisot-number, then the srb-measure is singular with respect to the Lebesgue measure on [0, 1) × [0, 1). In [11], Solomyak proved that for almost all λ ∈ ( 1 2 , 1), the distribution of the corresponding Bernoulli convolution ∞ k=1 ±λ k is absolutely continuous with respect to Lebesgue measure. Hence this implies that the srb-measure of the fat baker's transformation is absolutely continuous for almost all λ ∈ ( 1 2 , 1). Solomyak's proof used a transversality property of power series of the form g(x) = 1 + ∞ k=1 a k x k , where a k ∈ {−1, 0, 1}. More precisely, Solomyak proved that there exists a δ > 0 such that if x ∈ (0, 0.64) then (1) |g This property ensures that if the graph of g(x) intersects the x-axis it does so at an angle which is bounded away from 0, thereby the name transversality. The constant 0.64 is an approximation of a root to a power series and cannot be improved to something larger than this root. A simplified version of Solomyak's proof appeared in the paper [6], by Peres and Solomyak. We will make use of the method from this simpler version. In this paper we consider maps of the form (2) (x, y) → (λx, βy) if y < 1/β (λx + 1 − λ, βy − 1) if y ≥ 1/β , where 0 < λ < 1 and 1 < β < 2, see Figure 2. Using the above mentioned transversality of Solomyak one can prove that for almost all λ ∈ (0, 0.64) and β ∈ (1, 2) the srb-measure is absolutely continuous with respect to Lebesgue measure provided λβ > 1, and the Hausdorff dimension of the srb-measure is 1 + log β log 1/λ provided λβ < 1. A problem with this approach is that the condition λ < 0.64 is very restrictive when β is close to 1. Then the above method yields no λ for which the srb-measure is absolutely continuous, and it does not give the dimension of the srb-measure for any λ ∈ (0.64, 1/β).
We prove that these results about absolute continuity and dimension of the srb-measure hold for sets of (β, λ) of positive Lebesgue measure, even when λ > 0.64. This is done by extending the interval on which the transversality property (1) holds. This can be done in our setting, since in our class of maps, not every sequence (a k ) ∞ k=1 with a k ∈ {−1, 0, 1} occurs in the power series g(x) = 1 + ∞ k=1 a k x k that we need to consider in the proof. To control which sequences that occur, we will use some results of Brown and Yin [2] and Kwon [4] on natural extensions of β-shifts.
The paper is organised as follows. In Section 2 we recall some facts about β-transformations and β-shifts. We then present the results of Brown and Yin, and Kwon in Section 3. In Section 4 we state our results, and give the proofs in Section 6. The transversality property is stated and proved in Section 5.

β-shifts
Let β > 1 and define f β : This representation, among others, of real numbers was studied by Rényi [8]. He proved that there is a unique probability measure µ β on [0, 1] invariant under f β and equivalent to Lebesgue measure. We will use this measure in Section 6.
We let S + β denote the closure in the product topology of the set { d(x, β) : x ∈ [0, 1) }. The compact symbolic space S + β together with the left shift σ is called a β-shift. If we define d − (1, β) to be the limit in the product topology of d(x, β) as x approaches 1 from the left, we have the equality where σ is the left-shift. This was proved by Parry in [5], where he studied the β-shifts and their invariant measures. Note that d − (1, β) = d(1, β) if and only if d(1, β) contains infinitely many non-zero digits. A particularly useful property of the β-shift is that β < β ′ implies S + β ⊂ S + β ′ . The map φ β : S + β → [0, 1] is not necessarily injective, but we have d(·, β) • f β = σ • d(·, β).

Symmetric β-shifts
Let β > 1 and consider S + β . The natural extension of (S + β , σ) can be realised as (S β , σ), with S β = { (. . . , a −1 , a 0 , a 1 , . . .) : (a n , a n+1 , . . .) ∈ S + β ∀n ∈ Z }, where σ is the left shift on bi-infinite sequences. We will use the concept of cylinder sets only in S β . A cylinder set is a subset of S β of the form We will be interested in the set S of β for which S + β = S − β . This set was considered by Brown and Yin in [2]. We now describe the properties of S that we will use later on.
Consider a sequence of the digits a and b. Any such sequence can be written in the form (a n 1 , b, a n 2 , b, . . .), where each n k is a non-negative integer or ∞. We say that such a sequence is allowable if a ∈ N, b = a − 1, and n 1 ≥ 1. If the sequence (n 1 , n 2 , . . .) is also allowable, we say that (a n 1 , b, a n 2 , b, . . .) is derivable, and we call (n 1 , n 2 , . . .) the derived sequence of (a n 1 , b, a n 2 , b, . . .). For some sequences, this operation can be carried out over and over again, generating derived sequences out of derived sequences. We have the following theorem.
The "only if"-part was proved by Brown and Yin in [2] and the "if"-part was proved by Kwon in [4]. Using this characterisation of S, Brown and Yin proved that S has the cardinality of the continuum, but its Hausdorff dimension is zero.
There is a connection between numbers in S and Sturmian sequences. We will not make any use of the connection in this paper, but refer the interested reader to Kwon's paper [4] for details.
For our main results in the next section, it is nice to know whether S contains numbers arbirarily close to 1. The following proposition is easily proved using Theorem 1.
Proof. We prove this statement by explicitely choosing sequences d(1, β) corresponding to numbers β ∈ S arbitrarily close to 1. We do this by first finding some sequences that are infinitely derivable, and then we find the corresponding β by solving the equation 1 = φ β (d(1, β)). Let us first remark that the sequence (1, 0, 0, . . .) is its own derived sequence.

Results
Let 0 < λ < 1 and 1 < β < 2. Put Q = [0, 1) × [0, 1) and define T β,λ : Q → Q by  Denote by ν the 2-dimensional Lebesgue measure on Q. For any n ∈ N we define the measure The srb-measure (it is unique as noted below) of T β,λ is the weak limit of ν n as n → ∞.
The srb-measures are characterised by the property that their conditional measures along unstable manifolds are equivalent to Lebesgue measure. The existence of such measures was established for invertible maps by Pesin [7] and extended to non-invertible maps by Schmeling and Troubetzkoy [10]. We denote the srb-measure of T β,λ by µ srb . Using the Hopf-argument used by Sataev in [9] one proves that the srb-measure is unique. (Sataev's paper is about a somewhat different map, but the argument goes through without changes.) The support of µ srb is the set of which we have examples in Figure 3. One can estimate the dimension from above by covering the set Λ with the natural covers, consisting of the pieces of T n β,λ (Q). This gives us the upper bound, that the Hausdorff dimension of Λ is at most 1 + log β log 1/λ . If λβ > 1 this is a trivial estimate, since then 1 + log β log 1/λ > 2.
The following theorem states that in the case when λβ < 1, there is a set of parameters of positive Lebesgue measure for which the estimate above is optimal.
In the area-expanding case, when λβ > 1, we have the following theorem.
Since inf S = 1 by Proposition 1, there are β arbitrarily close to 1 for which we have a set of λ of positive Lebesgue measure, where the srbmeasure is absolutely continuous. In particular, this means that for these parameters, the set Λ has positive 2-dimensional Lebesgue measure.
Let us comment on the relation between Theorem 3 and the results of Brown and Yin in [2]. Brown and Yin considers any β > 1. In the case 1 < β < 2 their result is the following. They consider the map Hence their map is similar to ours when λ = 1 β . They proved that the Lebesgue measure restricted to the set Λ is invariant if β ∈ S.

Transversality
The main results of this paper, Theorem 2 and Theorem 3, only deal with 1 < β < 2. However, the arguments in this section work just as well for larger β, so for the rest of this section we will be working with a fixed β > 1.
Proof. Let and assume that no such δ exists. We will show that if ε is too small, then we get a contradiction.

If we have
[β]ε (1−1/β−ε)(1−1/β) ≥ x 0 , then let k = 0. Otherwise, let k be the largest integer such that x k 0 > [β]ε (1−1/β−ε)(1−1/β) . Since h 1 (x) is of the form (8) and all its terms are non-negative we must have a i = 0 for i ≤ k. This implies that By the maximality of k, we have x k+1 (10) and (5) implies To estimate h ′ 2 (x 0 ) from below, we note that since h 2 (x) is of the form (8), we must have h ′′ 2 (x) ≥ 0 for all x. We also have h 2 (x 0 ) ≥ 1 since we get a contradiction to the fact that g ′ (x 0 ) ≥ 0. By (11) and (12) we see that it is enough to choose ε so small that So, by (5) it is sufficient to choose To get a bound on k recall that by definition, either k = 0 or it satisfies . .
Inserting this estimate into (13), we get the sufficient condition (14) But ε log ε → 0 as ε shrinks to 0, so it is clear that we can find an ε > 0 satisfing (14).

Proofs
Before we give the proofs of Theorems 2 and 3, we make some preparations that will be used in both proofs.
For any point (x, y) ∈ Λ, there is sequence (x (k) , y (k) ) of points from ∞ n=0 T n β,λ (Q) that converges to (x, y). But each of the points (x (k) , y (k) ) is of the form (π 1 (a (k) , λ), π 2 (a (k) , β)) for some a (k) ∈ S β . Since the space S β is closed we conclude that (x, y) ∈ Λ is also of this form.
Consider a sequence a ∈ S β and the corresponding point p = π(a, β, λ). In the symbolic space, T β,λ acts as the left-shift, so the local unstable manifold of p corresponds to the set of sequences b such that a k = b k for k ≤ 0. For λ ≤ 1/2, π is injective on S β so the local unstable manifold of p is unique. If λ > 1/2, then π need not be injective on S β , so the local unstable manifold of p need not be unique. Indeed, when π is not injective there are a = b such that p = π(a, β, λ) = π(b, β, λ), giving rise to different unstable manifolds.
Because of the description (3) we have that π(b, β, λ) is in the unstable manifold of π(a, β, λ) if (b 1 , b 2 , . . .) ≤ (a 1 , a 2 , . . .). Hence for the unstable manifold of π(a, β, λ), there is a maximal c, with c k = a k for all k ≤ 0, such that π(c, β, λ) is contained in the unstable manifold. For this c we have that the unstable manifold is the set i.e. a vertical line. So, if a is such that (a 1 , a 2 . . . ) does not end with a sequence of zeros, then the unstable manifold has positive length. Since Λ is a union of unstable manifolds, we conclude that Λ is the union of linesegments of the form { (x, y) : x fixed, 0 ≤ y ≤ c }.
We will be using the symbolic representation of Λ given by (15), so we transfer the measure µ srb to a measure η on S β by η = µ srb • π(·, β, λ). We take a closer look at this measure η before we start the proofs. Recall, from Section 2, the probability measure µ β on [0, 1] that is invariant under f β and equivalent to Lebesgue measure. We get a shift-invariant measure on S + β by taking µ β • φ β and it can be extended in the natural way to a shift-invariant measure η β on S β .
Since µ srb and µ β are the unique srb-measures for T β,λ and f β respectively, we conclude that µ β is the projection of µ srb to the second coordinate. Thus η and η β coincide on sets of the form { a : a k = b k , k = 1, . . . , n }. By invariance η and η β will coincide. Since η β does not depend on λ by construction, η does not depend on λ. We now get the following estimates using the relation between η and µ β .
where K < ∞ is a constant. It follows from (17) that for η almost all a ∈ S β , the sequence (a 1 , a 2 , . . . ) does not end with a sequence of zeros. As already noted, this means that the unstable manifold is a vertical line segment of positive length. Hence for η almost all a the corresponding unstable manifold is of positive length. We will use this fact in the proofs that follow.
Proof of Theorem 2. Let β > 1 and pick any β ′ ≥ β such that β ′ ∈ S. For η almost every sequence a, the local unstable manifold of π(a, β, λ) corresponding to a, contains a vertical line segment of positive length. Note that this length does not depend on λ. Let ω δ be the set of sequences a, such that the corresponding local unstable manifold of π(a, β, λ) has a length of at least δ > 0. Take δ > 0 so that ω δ has positive η-measure. Then the set Ω δ = π(ω δ , β, λ) has the same positive µ srb -measure. Consider the restriction of µ srb to Ω δ and project this measure to [0, 1) × {0}. Let µ s srb denote this projection.
Take an interval I = (c, d) with 0 < c < d < 1/β ′ . Let t be a number in (0, 1). We estimate the quantity If this integral converges, then for Lebesgue almost every λ ∈ I, the dimension of µ s srb is at least t, and so the dimension of µ srb is at least 1 + t. Writing J(t) as an integral over the symbolic space we have that Since η does not depend on λ we can change order of integration and write Now, a, b ∈ S β ⊂ S β ′ , so for a and b with a j = b j for j = −k + 1, . . . , 0 and a −k = b −k , we have where g is of the form (4). Since I = [c, d] ⊂ [0, 1/β ′ ], we can use the transversality from Lemma 1 to conclude that Since η(a) = 0 for all a ∈ S β , we can replace ω δ × ω δ by A in the estimates, so after using (18) we get by (17) and the fact that η is a probability measure. This series converges provided that t < log β log 1/c . We have now proved that for a.e. λ in I = (c, d), the dimension of the srb-measure is at least 1 + log β log 1/c . To get the result of the theorem, we let ε > 0 and write I = (0, 1/β ′ ) as a union of intervals I n = (c n , d n ) such that log β log 1/cn > log β log 1/dn −ε. Then the dimension is at least 1+ log β log cn ≥ 1+ log β log 1/λ −ε for a.e. λ ∈ I. Since ε and β ′ was arbitrary this proves the theorem.
Proof of Theorem 3. In [6], Peres and Solomyak gave a simplified proof of Solomyak's result from [11], about the absolute continuity of the Bernoulli convolution ∞ k=1 ±λ k . The proof that follows uses the method from [6] and we refer to that paper for omitted details.
Let γ ∈ S, pick ε according to Lemma 1 and let β be such that 1/β ∈ [1/γ, 1/γ + ε). Let µ s srb be the projection of µ srb to [0, 1] × {0}. We form D(µ s srb , x) = lim inf r→0 µ s srb (B r (x)) 2r , where B r (x) = (x−r, x+r), and note that µ s srb is absolutely continuous with respect to Lebesgue measure if D(µ s srb , x) < ∞ for µ s srb almost all x. Since we already have absolute continuity in the vertical direction, it would then follow that µ srb is absolutely continuous with respect to the two-dimensional Lebesgue measure. If where µ Leb is the one-dimensional Lebesgue measure. Now, a, b ∈ S γ , so for a and b with a j = b j for j = −k + 1, . . . , 0 and a −k = b −k , we have where g is of the form (4). Since I = [c, 1/γ +ε] we can use the transversality from Lemma 1 and we get which converges since cβ > 1. Since c > 1/β was arbitrary, we are done.