Dimension of Slices Through Fractals with Initial Cubic Pattern?

Lifeng XIWen WUYing XIONG

1 Introduction

1.1 Dimension of slices

The intersections of Borel sets inwith(n?m)-dimensional subspace in random directions are studied in many publications.The following Marstrand’s theorem(see[12–13])is well known:Suppose that A?Rnis a Borel set with<∞and m

in Rn,and obtained that for a fixed(n?m)-dimensional subspace V,there are constantssuch that for Hm-almost all a∈ΛV,

Figure 1 Example 1.1.

Again,the equality(1.3)does not hold for k ∈ Q.Recently,B′ara′any,Ferguson and Simon proved in[1]that for all k∈Q,

Remark 1.1 When the slices of a fractal take the value in Marstrand’s theorem?The above examples show us that this is not an easy question.

1.2 Slices of self-similar fractals

Suppose thatand the self-similar set satisfies

Theorem 1.1 Suppose that E is a self-similar set satisfying(1.4),and V is an(n?m)-dimensional subspace satisfying(1.6).Then for Hm-almost all a∈Λ,

where λ is the Lyapunov exponent for the symmetric independent random product ofdefined in(1.8).

Remark 1.2 The box dimension of the intersection of the Sierpinski carpet with lines of rational slopes can be computed using the nature cover of the Sierpinski carpet directly,since if the intersection of the square[0,1]×[0,1]with a line is not empty,then the intersection of the Sierpinski carpet with the same line is not empty(see Figure 2).However,this is not true for general self-similar sets.Figure 2 shows that there exist lines that pass through the unit square but do not meet any point of the self-similar set.Therefore,we can not use the nature cover of the self-similar set directly,which complicates the computation of the box dimension of the slices.

Figure 2

1.3 The case of absolute continuousμV

Figure 3 The intersection of the limit set generated by a rule satisfying the 2-star condition with the line?3x+4y=b for some b∈[0,1].

Remark 1.5 In fact,the converse of the first part of Theorem 1.2 is true.That is,in case of self-similar sets satisfying(1.4),the slices take the typical value in Marstrand’s theorem if and only if the projection measure is absolutely continuous with respect to the Lebesgue measure.This result is proved by Feng[4]in a more general case recently.When we turn to the Sierpinski carpet,which does not satisfy the s-star condition,Niu and Xi proved in[14]that the projection of the self-similar measure on Sierpinski carpet onto a line with rational slope is singular.

The s-star condition is easy to verify.However,the following example shows that it is not a necessary condition to ensure the continuity of the projection measure.

Example 1.5 Consider a planar set intersect with a line,i.e.,n=2 and m=1.Let p=6, α =(5,1),V⊥=span{α}and ? ={(0,0),(0,1),(0,3),(0,4),(1,1),(1,2),(1,4),(1,5),(2,2),(2,3),(3,0),(3,1)}.Then

which implies that ? does not satisfy the s-star condition.We will prove in Section 6 that μVis absolutely continuous with respect to the 1-dimensional Lebesgue measure L1.Figure 4 is the first two steps in generating the self-similar set satisfying(1.4).

Figure 4 The case that the s-star condition does not hold.

1.4 Fractals generated by rules

Let p ∈ N with p ≥ 2 and Q be the unit cubeas a “rule” for defining a subset of Q.

Remark 1.7 If the rule sequence ω is chosen to be a constant sequence,say ?? ···,then the limit set Eωis a self-similar set satisfying the open set condition.In this case,for the sake of simplicity,we denote the setsby E and Ekrespectively.

Example 1.6 Let ? ={0,2}2and ω = ?? ···,then the limit set Eωis the self-similar set C×C,where C is the Cantor ternary set(see Figure 5).

Figure 5 C×C.

Example 1.7 Let ?={0,1,2}2{(1,1)}.The Sierpinski carpet E?R2is generated by only one rule ?(see Figure 6).

Example 1.8 Let ? ={0,2}2, ?0={(0,1),(1,0),(2,2)}and ω = ??0? ···.Figure 7 illustrates the first two steps in the construction of Eω.

1.5 Slices of fractals generated by multi-rules

Figure 6 The Sierpinski Carpet.

Figure 7 The limit set generated by the rule sequence ??0?···.

Further,we discuss the box dimension for the following fractal sets which are generated by multi-rules.For any fixed s∈ [1,pn?1],let

Figure 8 Example 1.9.

Remark 1.8 The projection of the Sierpinski carpet E in any direction is an interval.This good property ensures that if the intersection of the Ekwith a line,say L,is not empty,then the intersection of E with the line is also not empty.However,the limit set F generated by rules may not hold this good property.That is to say,even if L,we can not deduce that.In fact,for some line L and the limit set F,although,the intersection L∩Fk+1may be empty(see Figure 9).

Figure 9 The intersection of the limit set E generated by the rule ? with the linewhere ?={(0,1),(1,0),(1,1),(1,2),(2,1),(2,2),(2,3),(3,2)}.

Combining Theorem 1.1 and Theorem 1.3,we have the following corollary which tells us that when the s-star condition holds,then the Hausdorff dimension of slices of self-similar sets take the Mastrand’s value.

Corollary 1.1 Suppose that E is a self-similar set satisfying(1.4)and V is an(n?m)-dimensional subspace satisfying(1.6).If ? in(1.4)satisfies the s-star condition,then λ =logs and for Hm-almost all a∈Λ,

This paper is organized as follows.Section 2 gives some preliminary information such as Definitions and basic lemmas.The box dimension of slices is discussed in Section 3.The equivalence of the box dimension and the Hausdorff dimension of the slices and Theorem 1.1 are proved in Section 4.In Section 5,we discuss slices of the fractal generated by multi-rules satisfying the s-star condition,and prove Theorem 1.3.In the last section,we focus on the case that the projection measure is absolutely continuous with respect to the Lebesgue measure,and prove Theorem 1.2 and other remaining results.

2 Preliminaries

Fix a V such thatsatisfying(1.7).LetWrite

where

When∈ J,we have

and we can list the elements of Γb,i.e.,

3 Box Dimension of Slices

due to(2.1)–(2.2).

Moreover,we have the following auxiliary lemma about Uk(b)which will be used in estimating the upper bound of the box dimension of the slices.

Lemma 3.2 Nk(b)≤3nUk(b).

Proof Given a p-adic cube of side lengthintersecting,we denote it by B,then

According to the sub-additive ergodic theorem(see[17,Theorem 10.1]),there exists a constant λ such that

We also assume that the Lyapunov exponent

For notational convenience,we also write

Proposition 3.1 For Hm-almost all b∈(0,1)m,we have

is ergodic.Applying ergodic theorem to(3.8),we obtain that for Hm-almost all bthere is a sequence

Taking a subsequencethen we have for every k=mq,

Claim 3.1 Every entry of Tqin the row respect to i/∈Ξ is zero.

Therefore,by(3.10),(3.12)and(3.14),we have

4 Sections in Torus

Letbe the n-dimensional torus and P:the map defined by

wherethe fractional part of xifor every i.,the metric d onis

Now,we will show that the Hausdorff dimension of the slice equals its box dimension almost everywhere.During the proof,we will use the following result provided by Ledrappier(see[8,Proposition 2.6]).

Lemma 4.3(Ledrappier)Let Tpdenote the endomorphism Tpx=px(mod 1)of the(n?m)-dimensional torus,and let S be a continuous transformation of a metric space Y.

Assume thatis compact and invariant under the map Tp× S,and that ν is an S-invariant probability measure on Y.Then for ν-a.e.y,we have

5 Sections of Fractal Generated by Multi-rules

In this section,we will prove Theorem 1.3.

Fix a sequencesatisfying s-star condition.We will discuss the slices of following sets

due to(2.2)and(2.1).

The following proposition shows us how to compute

Lemma 5.4 For any integers k,t≥1,we have

Corollary 5.1 For any

It follows from(5.7)that=0.However,by Proposition 5.1,k,tptbis a contradiction.

Proof of Theorem 1.3 Theorem 1.3 follows from Propositions 5.2–5.3.

6 The Case of Continuous Projection Measure

6.1 Proof of Proposition 1.1

6.1.1 The local dimension ofμV

6.2 Proof of Theorem 1.2

Proof The necessity.By(6.7),we have

6.3 Proof of Example 1.5

To prove Example 1.5,we need some notations and results in[15].

A tree is said to be the tree of order pif it is constructed as follows:

1.Put 2π at the root;

2.Put the numberat the vertices on the first level;

3.Inductively,let a number α be associated to a vertex on the l-th level,then the numbers

Definition 6.1(see[15])A subset A of vertices of the tree of order p is a blocking set if the following conditions are satis fi ed:

(a)

(b)

(c)every in finite path starting at the root of the tree includes exactly one element of A.

From[15,Theorem 1],we have the following lemma.

Lemma 6.3(see[15])The solution of(6.1)is absolutely continuous if and only if there is a blocking set that consists of roots of(6.6).

Proof of Example 1.5 According to Lemma 6.3,we only need to show that there is a blocking set consisting of the roots of the following equation:

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