DIMENSION OF SLICES THROUGH THE SIERPINSKI CARPET

Transcription

1 DIMENSION OF SLICES THROUGH THE SIERPINSKI CARPET ANTHONY MANNING AND KÁROLY SIMON Abstract For Lebesgue typical (θ, a), the intersection of the Sierpinski carpet F with a line y = x tan θ + a has (if non-empty) dimension s, where s = log 8/ log 3 = dim H F Fix the slope tan θ Q Then we shall show on the one hand that this dimension is strictly less than s for Lebesgue almost every a On the other hand, for almost every a according to the angle θ-projection ν θ of the natural measure ν on F, this dimension is at least s For any θ we find a connection between the box dimension of this intersection and the local dimension of ν θ at a Introduction Let F denote the Sierpinski carpet [4, p8] in the unit square [0, ] [0, ] = I 2 in the plane R 2 and let E θ,a := {(x, y) F : y x tan θ = a} denote its intersection with the line segment Line θ,a across I 2 of slope θ through (0, a) (We only consider θ [ ) [ 0, π 2 because the case θ π, π) 2 is equivalent under the transformation (x, y) (x, y)) We shall study the dimension of E θ,a for a I θ := [ tan θ, ], as a subset of the y-axis The angle θ projection of the unit square [0, ] 2 to the y-axis is () proj θ (x, y) := ( tan θ, ) (x, y) When tan θ Q, we shall show as our main result, in Theorem 9, that for Lebesgue almost every a dim H E θ,a < log 8/ log Mathematics Subject Classification Primary 28A80 Secondary 37H5, 37C45, 37B0, 37A30 Key words and phrases Self-similar sets, Hausdorff dimension, dimension of fibres The research was supported by the Royal Society grant 2006/R4-IJP and the research of Simon by the OTKA Foundation #T 7693

2 2 ANTHONY MANNING AND KÁROLY SIMON a a a Figure The intersection of the Sierpinski carpet with the line y = 2 x + a for some a [0, ] 5 This behavior for all rational slopes is atypical because, as an easy consequence of some results of Marstrand (see [2, Chapter 0]), we shall prove in Lemma 2 that, for Lebesgue almost all (θ, a), dim(e θ,a ) = s, where s = log 8/ log 3 = dim H F When tan θ Q the opposite behavior also occurs because, for ν θ - almost all a I θ, we shall show dim(e θ,a ) s Here ν θ is defined as follows We order the vectors (u, v) {0,, 2} {0,, 2} \ {(, )} in lexicographic order and write t i for the i-th vector, i =,, 8 The Sierpinski carpet F is the attractor of the IFS { G := g i (x, y) = 3 (x, y) + } 8 3 t i i= Let Σ 8 := {,, 8} N and write σ : Σ 8 Σ 8 for the left shift We write Π : Σ 8 F for the natural projection Π(i) := lim n g i i n (0) and ν := Π µ 8 for the natural measure on F, where µ 8 is the Bernoulli measure on Σ 8 given by { 8,, 8} N Then ν θ := proj θ (ν) Feng and Hu proved [4, Theorem 22] that every self-similar measure η is exact dimensional That is, the local dimension of the measure η, given by (2) d(η, x) = lim r 0 log η(b(x, r)) log r is defined and constant for η-almost all x It was shown by Young [9] that this constant must be the Hausdorff dimension of the measure η That is (3) for η aa x, d(η, x) = dim H (η) := inf {dim H (U) : η(u) = }

3 SLICES OF THE SIERPINSKI CARPET 3 We will apply this result to for the measure ν θ, which is a self similar measure for the IFS { Φ := ϕ θ i (t) = 3 t + } 8 3 projθ (t i ) i= with equal weights That is, for every Borel set B, 8 ν θ ( (ϕ ) (B) = 8 νθ θ ) k (B) k= Since ν θ is a self-similar measure on I θ R, (3) gives (4) for ν θ aa x, d(ν θ, x) 2 Statement of results 2 Behavior for typical slope A special case of a Theorem of Marstrand [3] (also see [2, Theorem 0]) is Proposition (Marstrand) For ν-almost all z F and for Lebesgue almost all θ [0, π) we have dim ( F (z + W θ ) ) = s and H s (F (z + W θ )) <, where W θ is the straight line of angle θ through the origin Lemma 2 For Lebesgue almost all (θ, a), a I θ, θ [0, π/2) (5) dim (E θ,a ) = s, where dim denotes either dim H or dim B Proof This follows from [2, Theorem 00] using a density point argument 22 The general case We shall prove that for all θ the various dimensions are ν θ -almost everywhere constant functions Proposition 3 Fix an arbitrary θ [0, π/2) Then there exist nonnegative numbers d θ H and dθ B, dθ B such that for νθ -almost all a I θ we have (6) dim H (E θ,a ) = d θ H, dim B (E θ,a ) = d θ B and dim B (E θ,a ) = d θ B Proposition 4 For all θ [0, π/2) and a I θ if either of the two limits (7) log N θ,a (n) dim B (E θ,a ) = lim, d(ν θ log(ν θ [a δ, a + δ]), a) = lim n log 3 n δ 0 log δ

4 4 ANTHONY MANNING AND KÁROLY SIMON exists then the other limit also exists, and, in this case, (8) dim B (E θ,a ) + d(ν θ, a) = s Theorem 5 For every θ [0, π/2) and for ν θ -almost all a I θ we have dim B (E θ,a ) = s dim H (ν θ ) s The assertion includes that the box dimension exists 23 The case of absolutely continuous ν θ It is well known (see [2, Theorem 97]) that ν θ Leb for Lebesgue almost all θ [0, π/2) Theorem 6 Suppose that θ satisfies ν θ Leb (a): For Leb-almost all a I θ, there exist 0 < c 3 (θ, a) < c 4 (θ, a) < such that ( ) n ( ) n 8 8 (9) n, c 3 (θ, a) < N θ,a(n) < c 4 (θ, a) 3 3 (b): In particular, for Lebesgue almost all a I θ, dim B (E θ,a ) exists and is equal to s 24 Behavior for rational slope Recently Liu, Xi and Zhao proved Theorem 7 [] Let tan θ Q For Lebesgue almost all a I θ we have dim B (E θ,a ) = dim H (E θ,a ) = d θ (Leb) log 8 log 3, where d θ (Leb) is a constant depending only on θ They conjectured the strict inequality If tan θ Q then the dimensions in Proposition 3 are equal Proposition 8 If tan θ Q then there is a constant d θ (ν θ ) such that (0) d θ (ν θ ) := d θ H = d θ B = d θ B Our main theorem asserts the strict inequality Theorem 9 If tan θ Q then, for Lebesgue almost all a I θ, we have () d θ (Leb) := dim B (E θ,a ) = dim H (E θ,a ) < log 8 log 3 It now follows from Proposition 4 that the local dimension of the selfsimilar measure ν θ is Lebesgue almost everywhere equal to a constant which is bigger than

5 SLICES OF THE SIERPINSKI CARPET 5 Corollary 0 If tan θ Q then, for Lebesgue almost all a I θ, we have (2) d(ν θ, a) = s d θ (Leb) > Comparing (4) with (2) shows that, when tan θ Q, the measure ν θ on I θ is singular wrt Lebesgue measure, in contrast to the absolute continuity in section 23 It turns out that there are many slices which do not have the small dimension of Theorem 9 Theorem If tan θ Q then, for ν θ -almost all a I θ, dim H (E θ,a ) = s dim H (ν θ ) s A result related to Corollary 0 was recently proved by Feng and Sidorov, see [5] Their Proposition 4 says that, if ρ is the reciprocal of some Pisot number in (, 2), the value taken at Lebesgue almost every point by the local dimension of the Bernoulli convolution µ ρ is strictly greater than Organization of the paper: In Section 3 we develop our method of symbolic dynamics and prove Theorem 9 Then in Section 4 we prove the remaining results by using some notation and Proposition 8 from Section 3 3 The proof of our main result We make the standing hypothesis that tan θ = M N numbers M, N are coprime 0 where the natural If 3 N then 3 M and, using the isometry (x, y) (y, x), we may consider tan θ = N/M instead; thus we may assume that 3 N Because of the isometry (x, y) ( x, y) we do not need to consider a [ M N, 0) The idea of the proof of Theorem 7 was as follows: The authors found three non-negative integer matrices A 0, A, A 2 such that corresponding to the first n digits (a,, a n ) {0,, 2} n of the base 3 expansion of a [0, ] they could approximate the minimal number of squares of size /3 n one needs to cover the n-th approximation of F by the norm of A a A an Our method is similar but we use different matrices A 0, A, A 2 which carry more geometric meaning In Proposition 2 the subadditive ergodic theorem is used to express Hausdorff dimension in terms of the average of the logarithm of the norm of products of A 0, A, A 2 In Proposition 8 we show that one of these products B has each row

6 6 ANTHONY MANNING AND KÁROLY SIMON S 3 S 2 S S 0 Figure 2 Tiling S when (N, M) = (5, 2), K = either all zero or all positive (which requires extra care in 33 if N is even) In 34 we consider the action of A 0, A, A 2 on the right on the simplex of positive vectors as an Iterated Function System and, after studying powers of B, show that this IFS is contracting on average In 35 our Theorem 9 is proved using the property that there is a positive measure subset of the invariant set of the IFS where the various products do not all expand by the same amount and a theorem of Furstenberg about the integral representation of the Lyapunov exponent of a random matrix product 3 Our transition matrices Now the interior of the strip S := {(x, y) R 2 : x [0, N], y x tan θ [0, ]} meets the 2N + M (= K, say), unit squares I 2 + z for z in {( [ q, q M ]) ( [, q, q M ] ) } {([ + : 0 q < N r N ] ), r + N N M In the tiling of S by its intersection with these squares we number the tiles (3) Q i := ((q, r) + I 2 ) S, i K in increasing order of q and, for given q, in increasing order of r Let us call the (q, r) in (3) as (q i, r i ) That is (4) Q i = ((q i, r i ) + I 2 ) S, and Q i int(s) Figure 2 illustrates the case M/N = 2/5, K = Consider the three parallel narrower infinite strips S 0, S, S 2 with S t := {(x, y) R 2 : y x tan θ t/3 [0, /3]} and the expanding maps ψ t : S 0 S S 2 S t, ψ t (x, y) := (x/3, (y + t)/3), t = 0,, 2 } : 0 < r < M

7 SLICES OF THE SIERPINSKI CARPET 7 Then E θ,a = F ψ an ψ a S an n= expresses E θ,a as the intersection with F of strips of vertical height 3 n chosen according to the expansion 0a a 2 a 3 of a in base 3 Consider the intersection of S t with the eight squares of side 3 that cover F and, correspondingly, of 3S t with (q, r ) + I 2 where q, r are not both congruent to mod 3 We define, for t {0,, 2}, the K K transition matrix A t, with entries 0 and so that its (i, j)-th entry is if and only if 3(Q i S t ) contains (ln, t+lm)+q j for some l {0,, 2} with q j + ln, r j + t + lm not both congruent to mod 3 That is (5) { l {0,, 2}, Qi S A t (i, j) = t 3 ((ln, t + lm) + Q j ) and either q j + ln mod 3 or r j + t + lm mod 3 } In figure 3 the label j is marked in 3 ((ln, t + lm) + Q j ) and this illustrates, for example, that the first three rows of these transition matrices are A 0 = A = A 2 = ,, The involution (x, y) (N, M +) (x, y) sends S to itself, exchanges S 0 and S 2, and sends Q i to Q K+ i Therefore these matrices exhibit the symmetries (6) (7) i, j {,, K} A 0 (i, j) = A 2 (K + i, K + j), A (i, j) = A (K + i, K + j) Consider transitions from Q i = ((q, r) + I 2 ) S to Q j = ((q, r ) + I 2 ) (S + (0, t)) For the nine cases l, t {0,, 2}, A t (i, j) = when Q i = (([(q + ln)/3], [(r + t + lm)/3]) + I 2 ) S except for any case

8 8 ANTHONY MANNING AND KÁROLY SIMON S S S 0 3 where Figure 3 Tiling each strip S t when (N, M) = (5, 2) (8) q + ln r + t + lm mod 3, which corresponds to a square deleted from F Because 3 N, given j, (8) determines (l, t), and so each of A 0, A, A 2 has each column sum equal to 2 or 3 and (9) A s := A 0 + A + A 2 has each column sum 8 If a 0, a, a 2 and denote the row vectors in R K for which each entry of is and each A l has the row vector of its column sums equal to 3 a t, then each entry of a t is 0 or and a 0 + a + a 2 = Now A t (i, j) = whenever Q i S t contains a 3 -sized copy of Q j, and A u (j, k) = when Q j S u contains a 3 -sized copy of Q k, so the (i, k)- th entry of A t A u is the number of the 8 2 squares of side 3 2 covering F that meet Q i S t ψ t (S u ) in a 3 2 -sized copy of Q k By induction, the (, k)-th entry in the product matrix A a A a2 A an is the number of the 8 n squares of side 3 n covering F that meet Q S a ψ a (S a2 ) (ψ a ψ an )(S an ) in a 3 n -sized copy of Q k Thus the first row of A a A a2 A an counts the elements of a (3 n 2)-cover of E θ,a (In the cases (N, M) = (, ) and (2, ), certain matrix entries are 2, and the elements of the cover are still counted correctly) Thus we get: Proposition 2 For every a [0, ], a = a i 3 i we have (20) dim B E θ,a lim sup log 3 n n log A a a n, where A a a n denotes A a A an and denotes the sum of the moduli of the entries For almost all a = a i 3 i the right hand side i= i=

9 SLICES OF THE SIERPINSKI CARPET 9 gives the same value and we can replace lim sup by lim Consider the random product of the matrices A 0, A, A 2 each taken with probability /3 independently in every step Then the Lyapunov exponent γ of this random matrix product is the almost sure value of the limit above That is (2) γ := lim n log A a a n, for aa (a, a 2, ) Similarly, (22) γ = lim n n 3 log A n i i n i i n Further, n (23) γ [log 2, log 3] As an easy consequence of Theorem 7, the definition of γ and the translation invariance of the Lebesgue measure we obtain the following proposition Proposition 3 If tan θ Q then for Lebesgue almost all a I θ we have (24) dim B (E θ,a ) = dim H (E θ,a ) = γ log 3 We will use the following definitions Definition 4 The symbolic space to code the translation parameter a [0, ] is Σ := {0,, 2} N Let π y : Σ [0, ] be defined by π y (i) := i k 3 k, i = (i, i 2, ) k= We denote the uniform distribution on Σ by P := { 3, 3, 3} N The push down measure (π y ) of P by π y is the Lebesgue measure Leb on [0, ] The measure P is ergodic with respect to (Σ, σ), where σ is the left shift on Σ As an easy case consider first the intersection of F with horizontal lines Proposition 5 For almost every a dim H (E 0,a ) = log 8/ log 3 < 3 log 8/ log 3 Proof If tan θ = 0/ then the above construction gives K = 2N +M = and matrices A 0 = A 2 = (3), A = (2) Then E 0,a is covered by A a A an squares of side 3 n This leads us to study the function on Σ := {0,, 2} N taking the value log 2 where the first symbol is and log 3 elsewhere, whose integral with respect to P is log 8 3

10 0 ANTHONY MANNING AND KÁROLY SIMON So, from now on we may assume that M/N 0 By symmetry, without loss of generality we may assume that M/N > 0 To prove Proposition 2 we need the following simple observation which will also be used later Fact 6 Consider the non-negative K K matrices A, B Let K K (25) c A (j) := A(i, j) and r B (i) := B(i, j) i= be the j-th column sum and the i-th row sum of the matrices A, B respectively Then (26) A B = i j= c A (i) r B (i) Proof The proof of the Fact is a simple calculation Now we turn to the proof of Proposition 2 Proof of Proposition 2 The inequality in (20) immediately follows from the discussion right above the proposition The fact that we can replace lim sup by lim in (20) is an immediate corollary of the sub-additive ergodic theorem (see [8, p 23]) For non-negative matrices it is easy to check that is submultiplicative Indeed A B = i c A (i) r B (i) K i= K c A (i) r B (j) = A B This implies that the sequence of the bounded functions f k : Σ R j= (27) f k (i) := log A T i k A T i = log A i i k is sub-additive Using the ergodicity of P and the sub-additive ergodic theorem [8, p 23] we obtain that the limit (28) γ := lim n n f n(i) = lim n n log A i A in exists for P-almost all (a, a 2, ) Σ and gives the same value Further, using n f n (i)dp(i) = n i i n 3 n log A i i n, the sub-additive ergodic theorem implies that (22) holds To verify (23) we make the following observation: each factor in the matrix product A i A in has each column sum either 2 or 3 This

11 SLICES OF THE SIERPINSKI CARPET implies that all the column sums of the product matrix are between 2 n and 3 n This yields (29) [i,, i n ] we have: K 2 n A i A in K 3 n The inequality (30) below is an immediate corollary of (20) and (2) Corollary 7 The following holds: (30) dim B E θ,a γ log 3 32 Positive rows in some products of our matrices Recall from Section that (3) 3 N In this section and the next we prove Proposition 8 There exists n 0 and (a a n0 ) {0,, 2} n 0 such that the rows of the matrix A a a n0 are vectors with either all positive or all zero elements Our approach is as follows A row will be positive when labels to K appear in the intersection of each square with the strip in the n 0 th version of Figure 3 The labelling depends on the relation between N and 3 When M/N < we shall find the labels near the left or right edge of the square in positions corresponding to squares in the n 0 th stage in the construction of F When M/N the strip may not be near either edge, so we shall search instead in small rectangles with diagonal of slope near M/N and carefully chosen horizontal and vertical coordinates 32 An IFS leaving S invariant First we introduce some notation For t, l {0,, 2} we define the contraction: ψt(x, l y) := [x + l N, y + l M + t] 3 To compute the n-fold compositions we introduce the following notation: (32) l,n := l 3 n + + l n 3 + l n and (33) a,n := a 3 n + + a n 3 + a n If we write ψ l l n a a n for ψ l a ψa ln n have and S l l n a a n for ψ l l n a a n (S) then we (34) ψ l l n a a n (x, y) = 3 n [x + N l,n, y + M l,n + a,n ]

12 2 ANTHONY MANNING AND KÁROLY SIMON and (35) S = Sa lln a n, (a a n),(l l n) {0,,2} n where the sets on the right hand side have disjoint interior The set S l l n a a n is the intersection of the slope θ level n strip { S a a n := (x, y) : 0 x N and 0 y (x tan θ + a a n 3 ) } n 3, n with the level n vertical strip { ( V l l l n := (x, y) : N l ) n 3 n x < N for all (l l n ), (a a n ) {0,, 2} n See Figure 4 ( l l n 3 + )}, n 3 n Earlier in (4) we defined the K = 2N + M different level zero squares which together cover S Similarly, here we define the level n square of shape j in S l l n a a n by (36) Q l l n a a n (j) := ψ l l n a a n (Q j ) 3 V 0 V 0 V 2 S 2 S 0 2 Q 2 (3) S 0 S 0 0 V 2 (0, 0) Figure 4 The subsets defined in (35) and (36) when M/N = 2/5 3

13 SLICES OF THE SIERPINSKI CARPET The n-th approximation of the translated copies of the Sierpinski carpet Definition 9 Let F be the union of the translated copies of the Sierpinski carpet to those unit squares that intersect S That is K F := ((q i, r i ) + F ) i= Let F n be the level n approximation of F Put { U n i = V n i = { u [q i, q i + ) : (u,, u n ) {0,, 2} m st u = q i + v [r i, r i + ) : (v,, v n ) {0,, 2} m st v = r i + } n u m 3 m m= } n u m 3 m m= and say that u m, v m are the m-th ternary (that is base 3) digits of u, v respectively For some n n 2 and for u U n i and v V n 2 i we define the so called level (n, n 2 ) grid rectangle in the square (q i, r i ) + I 2 : (37) ( n n 2 R i (u, v) := (q i, r i )+ u m 3 m, v m 3 )+ [ m 0, 3 ] n [ 0, 3 ] n 2 m= m= The collection of all level (n, n 2 ) grid rectangles in (q i, r i )+I 2 is called R i (n, n 2 ) That is R i (n, n 2 ) := {R i (u, v) : u U n i and v V n 2 i } When n = n 2 then the elements of C i (n) := R i (n, n 2 ), n = n = n 2 are called level n grid squares in (q i, r i )+I 2 Those level n grid squares that are contained in F n are called n-cylinder squares For a given level n grid square we can decide if it is an n-cylinder square using the following Fact, whose proof follows immediately from the observation that all the elements (x, y) of the Sierpinski carpet F can be represented as (x, y) = ( uk ), v 3 k k 3 such that for all k either k u k {0, 2} or v k {0, 2} k= Fact 20 The level n grid square R i (u, v) C i (n) with u = q i + n 3 m and v = r i + n m= m= v m 3 m is an n-cylinder square if and only if (38) p n, either u p {0, 2} or v p {0, 2} u m

14 4 ANTHONY MANNING AND KÁROLY SIMON Fact 2 If Q l l n a a n (j) (q i, r i ) + I 2 then the inclusion (39) Q l l n a a n (j) F n ((q i, r i ) + I 2 ) is equivalent to the following assertion: (Assertion): Let (u,, u n ), (v,, v n ) {0,, 2} n be defined by ( n ) (40) ψ l l n a a n (q j, r j ) = (q i, r i ) + n 3 n l u 3 n l, 3 n l v l l= Then for every p n we have (4) either u p {0, 2} or v p {0, 2} Lemma 22 There is m 0 such that { 3 m m } km N is a full k= residue system modulo N Proof Clearly we can find k < l such that 3 k 3 l mod N Let m 0 := l k Then (42) 3 m 0 mod N holds (since we assumed that 3 N) Thus 3 m m0 + 3 km 0 k mod N Definition 23 (a): First we define k 0 as the smallest non-negative integer satisfying M/N < 3 k 0 That is if M/N then (43) 3 k 0 M/N < 3 k 0 On the other hand if M/N < then k 0 := 0 (b): We fix m 0 which satisfies (42) (c): Finally, we introduce the equivalence relation on {0,, N } as follows: if N is odd: then k l holds for all k, l if N is even: then k l holds iff either both k and l are even or both of them are odd (d): Assume that M/N < We shall argue later in proving Proposition 8 that for all shapes Q i, i K, the region in Q i with first coordinate in the interval (44) J n 0 (i) = [q i, q i +(3 m 0 N+ ) 3 n ) or J n 2 (i) = [q i + ( 3 m 0 N+ + ) 3 n, q i + 3 n ) contains an image of Q j by ψ l l n a a n l l n l= for an appropriately chosen The definition of the intervals in (44) will be much more complicated when M/N

15 SLICES OF THE SIERPINSKI CARPET The definition of the intervals J n 0 (i), J n 2 (i) in the general case First we have to place some restrictions on (a,, a n ) {0,, 2} n for the strip S a,,a n considered To do that we divide the set {, 2,, n} into four regions: I := {,, 2k 0 }, I 2 := {2k 0 +,, n }, where n := n (m 0 N + ) 4N3 2k 0 and I 3 := {n +,, n }, I 4 := {n +,, n}, where n := n + 4N3 2k 0 = n (m 0 N + ) When M/N < then k 0 = 0 and in that case I = Assumption (A): (a): a n = 0 (b): i, u U 2k 0 i, v V k 0 i we assume that both the top left and bottom right corners of the rectangle R i (u, v) are farther from S a a n than 3 n This can be arranged by excluding NM3 3k 0 3 n grid intervals of level n (k 0 +2) when selecting the level n grid interval determined by (a,, a n ) from the interval [0, ] So, this is a restriction implemented by excluding some intervals whose indices are from I I 2 and their total length is less than NM3 3k0 2 3 n 3 k0+2 so now we assume that n is large enough that this is (c): Let us denote the bottom edge of S a a n by Bottom a a n See Figure 5 We define y α (y α ) for α N3 2k 0 (0 α N3 2k 0 ) as the second coordinate of the intersection of the line Bottom a a n with the vertical line x α := α 3 2k (n +4α ) (x α := α 3 2k (n +4α+) ) respectively Note that all these x α, x α lie in [0, N] We assume that (a,, a n ) is chosen in such a way that c: for all α N3 2k 0, both the (n + 4α )-th, and the (n + 4α)-th digits of the ternary expansion of y α are zero and c2: for all 0 α N3 2k 0, both the (n + 4α + )-th and (n + 4α + 2)-th digits of the ternary expansion of y α are zero Now we prove that there is a positive proportion (independent of n) of all possible (a,, a n ) {0,, 2} n for which assumptions (c) and (c2) hold First, for having a more convenient notation we write b α := n +4α, α N3 2k 0 and f α := n +4α+, 0 α N3 2k 0 (Here f α refers to forward and b α refers to backward relative to α3 2k 0 see Figure 5)

16 6 ANTHONY MANNING AND KÁROLY SIMON Fact 24 There are 3 n possible choices of (a,, a n ) {0,, 2} n such that the following holds: (45) 0 = y α b α = y α b α+, α {,, N3 2k 0 } and 0 = y α f α = y α f α+, α { 0,, N3 2k 0 }, where y α k (yα k ) is the k-th ternary digit of yα (y α ) respectively Proof Observe that (46) y α = where n a k 3 k + z α and y α = k= n a k 3 k + z α, k= z α := M N xα and z α := M N xα We prove that there is a way to choose the elements a k {0,, 2}, for k I 3 I 4 such that (45) holds for all possible choice of a k, k I I 2 We construct these values a k for k I 3 I 4 by mathematical induction starting from k = n and moving towards to smaller values of k Namely, we define a k := 0 for all k I 4 Fix an arbitrary k I 3 We assume that we have already defined a k for all k k n Clearly, we can either find an α N3 2k 0 such that k {b α, b α + } or we can find an 0 α N3 2k 0 such that k {f α, f α + } For symmetry without loss of generality we may assume that we are in the latter case and k = f α + Then we compute the overflow o k from the (k +)-th n ternary place to the k -th ternary place when adding up a k 3 k and k=k + z α k 3 k That is if n k=k + a k 3 k + k=k + z α k k=k + 3 k > 3 k then there is an overflow to the k -th ternary place and then o k := otherwise there is no overflow and o k := 0 Observe that the value of o k depends only on the ternary digits a k for k < k n (which have been determined at this stage of the mathematical induction) and zk α for k < k which are given numbers So, we can compute the number o k It follows from (46) that y α k = a k + zα k + o k Then for a k := (z α k + o k ) mod 3 we obtain that yα k = 0 We continue this process with doing the same first for k then k 2 and so on for all k n +

17 SLICES OF THE SIERPINSKI CARPET 7 Fact 25 Let J [0, ] be a non empty interval Whenever n is big enough we can choose (a,, a n ) {0,, 2} n which satisfies the requirements of Assumption (A) and [ n ] n (47) a k 3 k, a k 3 k + 3 n J k= k= Proof The three parts of the assumption posed restrictions for indices in different regions, so these restrictions cannot conflict Let G be the biggest grid interval, say level g which is contained in J This means that we need to fix the first g ternary digits In parts (a) and (c) of assumption (A), we fixed the last 4N3 2k 0 + m 0 N + ternary digits of (a,, a n ) So, from now we need to fix g +4N3 2k 0 +m 0 N + ternary digits to provide that (47) and parts (a), (c) of assumption (A) hold In this way we restrict ourselves to a set of level n grid intervals with a total length of at least 3 (g+4n32k 0 +m 0 N+) (which does not depend on n) among which only an amount of total length of NM3 3k0 3 (n (k 0+2)) (which tends to zero as n ) is lost for part (b) So, if n is big enough then we find an (a, a n ) satisfying the assumptions (A) and (47) The reason for part (c) of assumption (A) is as follows: Remark 26 We consider α 3 2k 0 as the end point of two level 2k 0 grid intervals: (48) I L (α) := [ (α )3 2k 0, α3 2k 0 ] and I R := [ α3 2k 0, (α + )3 2k 0 ] Assume that the corresponding ternary digits of these intervals are (u L,, u L 2k 0 ) and (u R,, u R 2k 0 ) Let (49) n α,l := # { l 2k 0 : u L l = }, n α,r := # { l 2k 0 : u R l = } Then we define the level n grid intervals (50) { { L (α), if n J L (α) := α,l is odd; R (α), if n J L 2 (α), otherwise R (α) := α,l is odd; R 0 (α), otherwise where the level n grid intervals L (α), L 2 (α), R 0 (α), R (α) are defined in Figure 5 In this way J L (α) and J R (α) are level n grid intervals contained in I L (α) and I R (α) respectively For V {L, R} we obtain the ternary digits of J V (α) as the concatenation of (u V,, u V 2k 0 ) and a vector of n 2k 0 components of all zeros or twos if n α,v is an even number If n α,v is an odd number then the ternary digits of J V (α) are obtained in the same way with the difference that we have digit one in the b α -th place (f α -th place) if V = L (V=R) respectively In this way,

18 8 ANTHONY MANNING AND KÁROLY SIMON for both J L (α) and J R (α) the number of ones among the ternary digits is an even number y Bottom aa n y α 3 fα 3 bα 3 bα 3 fα 3 fα y α 3 bα a a n 3 n L 2 (α) R 0 (α) L (α) α 3 2k0 R (α) x Figure 5 L (α), L 2 (α), R 0 (α), R (α) are level n 2k 0 grid intervals Definition 27 We say that S a a n is an n-good strip if (a,, a n ) satisfies Assumption (A) From now on we fix n and an n-good strip S a a n For this strip Q i = ((q i, r i ) + I 2 ) S is a relevant shape if int(s a a n Q i ) We remark that, in the case when M/N <, we do not use part (c) of Assumption (A) We recall that in Definition 23 we have already defined the intervals J n 0 (i) = [q i, q i + 3 (n ) ) and [q i + 3 (n ) 3 n, q + 3 n ) Now we extend this definition to the case when M/N and Q i is a relevant shape for the strip S a a n The idea of the construction is as follows: Using the notation of Definition 9, for every i K we shall define, by an inductive procedure, R i (u k 0+l, v l ), for 0 l k 0 such that (C): S a a n R i (u k 0+l, v l )

19 (C2): u k 0+l U k 0+l i (5) u k 0+l = q i + SLICES OF THE SIERPINSKI CARPET 9 k 0 +l m= and v l V l i with u m 3 m, v l = r i + l v m 3 m m= (C3): u k0 +,, u k0 +l {0, 2} and v,, v l {0, 2} In this way we will obtain (52) R i := R i (u 2k 0, v k 0 ) R(2k 0, k 0 ) It is clear by the definition that the following two assertions hold: Remark 28 () Using Fact 20, all the 3 k 0 level 2k 0 grid squares contained in Ri are in F 2k 0 (2) The slope of the increasing diagonal of the rectangle Ri is 3 k 0 which is slightly bigger than M/N Thus every line of slope M/N that enters the rectangle via the bottom horizontal line will leave Ri through its Eastern side The rectangle Ri corresponds, in the simpler case M/N <, k 0 = 0, to the whole square (q i, r i ) + I 2 for which every line of slope M/N that meets it must cross its Western or Eastern side Definition 29 Using the previous notation of this section especially (5) and (50) we define the intervals J n 0 (i) = J R (u 2k 0 3 2k 0 ), J n 2 (i) = J L (u 2k 0 3 2k 0 + ) The properties of these intervals, which are summarized in the following lemma, are immediate consequences of the definition Lemma 30 (a): The orthogonal projection of S a a n Ri (the rectangle Ri was defined in (52)) to the x-axis contains at least one of the level n grid intervals J0 n (i) or J2 n (i) If the line Bottom a a n enters the rectangle Ri on its Western side then J0 n (i) is such an interval If the line Bottom a a n enters the rectangle Ri on its Southern side then J2 n (i) is such an interval (b): Using the notation of (5), by Fact 20, all the 3 n k 0 level n grid squares both in J0 n (i) [v k 0, v k k 0 ] and in J2 n (i) [v k 0, v k k 0 ] are contained in F n Now we present the inductive construction of rectangle Ri Fix an i K such that Q i is a relevant shape for S a a n We recall that the line Bottom a a n was defined as the bottom edge of the strip S a a n Using the notation of Definition 9 we construct a nested sequence of rectangles { R i (u k0+l, v l ) } k 0 satisfying the conditions (C), (C2) and (C3) on page 8 To construct R i (u k 0 l=0, v 0 ) note that it follows from (43) that we can find an 0 m 3 k 0

20 20 ANTHONY MANNING AND KÁROLY SIMON { such that the line Bottom a a n intersects the vertical line segments (x, y) : x = qi + m 3 k 0, q i < y < q i + } Let us call this segment l 0 and the point where l 0 intersects Bottom a a n A 0 See Figure 6 Observe that q i + m 3 k 0 is the (left or right) end point of two level k 0 grid intervals At least one of these two intervals lies in [q i, q i + ] We call this level k 0 grid interval I 0 (if there are two such intervals then we pick one) Note that the left end point of I 0 (which is either q i +m 3 k 0 or q i + m 3 k 0 3 k 0 ) is defined as u k 0 = q i + k 0 u m 3 m Let v 0 := r i Note that A 0 lies on one of the vertical sides of Box 0 := R i (u k 0, v 0 ) If the first ternary digit of the y-coordinate of A 0 is either 0 or 2 then we call it v and define Box Box 0 as that /3 scaled copy of Box 0 which contains A 0 on one of its vertical sides Clearly, the orthogonal projection I of Box to the x-axis is a level k 0 + grid interval which has a level k 0 end point therefore the last digit, called u k0 +, of the ternary representation of I is either 0 or 2 In this way we defined R i (u k0+, v ) = Box On the other hand, if the first ternary digit of the y-coordinate of A 0 is equal to then the definition of u k0 +, v is more complicated Namely, in this case we define Box as follows: Without lost of generality we may assume that Box 0 is on the left hand side of l 0 (as in Figure 6) In this case we define v := 0 (otherwise we would have chosen v = 2) We divide the bottom third part of Box 0 into three equal vertical strips corresponding to the level k 0 + grid intervals contained in I 0 (Figure 6) It follows from (43) that one of the vertical sides of one of these three 3 (k0+) 3 rectangles, which is different from the middle one, intersects Bottom a a n Let us call this point A and the non-middle positioned 3 (k0+) 3 grid rectangle which contains A on one of its vertical sides is called Box and the projection of Box to the x-axis is called I Clearly, I I 0 is a level 3 k0+ grid interval and its k 0 + ternary digit is different from This follows from the non-middle position of Box as mentioned above So, the rectangle R i (u k0+, v ) := Box satisfies the requirements (C)- (C3) on page 8 We continue the construction R i (u k0+l, v l ) := Box l for all l k 0 exactly in the same way m= 324 The proof of Proposition 8 Level n shapes are labeled using N l,n and which of these appear in F n is affected by u,n so we compare these, using (34) and Fact 2, in the following lemma, which is the key step in proving Proposition 8 Lemma 3 Fix any i, j {,, K} satisfying q i m 0 N + Assume that we are given q j Let n (53) u,, u n (m0 N+) {0,, 2}

21 SLICES OF THE SIERPINSKI CARPET 2 y l 0 Box 0 A 0 Bottom a a n 2 3 Box A 2 A Box 2 u k 0 3 I 0 I x in such a way that Figure 6 The inductive definition of Box l (54) u + + u n (m0 N+) is an even number We can choose u n m0 N,, u n {0,, 2} and (l,, l n ) {0,, 2} n such that n (55) q j + N l,n = 3 n q i + 3 n k u k Proof Clearly we can choose (l,, l n) {0,, 2} n such that for p := N(l 3 n + +l n) 3 n q i +q j ( ) 3 n u m 0 N+ u n (m0 N+) we have k= (56) 0 p N Now we distinguish two cases based on the parity of N N is odd: Then it follows from (42) that { 2 ( 3 m m km 0 )} N is a complete residue system modulo N since {2k} N k= is such a system So we can find integers k N and v N such k=

22 22 ANTHONY MANNING AND KÁROLY SIMON that (57) 2 ( 3 m km 0 ) = v N + p Choose (l,, l n ) {0,, 2} n such that N(l 3 n + + l n) + vn = N(l 3 n + + l n ) Then (58) N(l 3 n + +l n ) 3 n q i +q j ( ) 3 n u m 0 N+ u n (m0 N+) = v N + p = 2 ( 3 m km 0 ), which immediately implies the assertion of the lemma N is even: In this case { 2 ( 3 m m km 0 )} N is not a complete residue system but contains (actually twice) all the even number residues On the other hand, using q i q j and (54), we see that p is an even number This follows from (50) and Definition 29 So, as above, we can find integers k N and v N such that (57) holds The rest of the proof is the same as in the case when N is odd k= This implies Corollary 32 Fix an arbitrary (a,, a n ) {0,, 2} n and also fix i, j K such that q i q j For any u {0, 2} we can find (l,, l n ) {0,, 2} n such that for the first component function (ψ l l n a a n ) of ψ l l n a a n we have (a): (ψ l l n a a n ) (q j ) Ju n (i) and (b): Q l l n a a n (j) F n Proof Using Fact 2 and Lemma 3 the proof immediately follows from the observation that for u = 0, 2 choosing (l,, l n ) as in Lemma 3 we get that the first coordinate of ψ l l n a a n (q j, r j ) is (ψ l l n a a n (q j, r j )) = 3 n ( qj + N(l 3 n + + l n ) ) = q i + u ( m 0 N+ n ) n ( 3 m km 0 ), where 0 k N

23 SLICES OF THE SIERPINSKI CARPET 23 Lemma 33 Assume that S a a n is an n-good strip For each i, j K let r i be the i-th row vector of the matrix A a a n and write r i (j) for the j-th element of r i Then (59) r i 0 and q j q i imply A a a n (i, j) = r i (j) > 0 Proof Recall (60) r i (j) = # { (l,, l n ) {0,, 2} n Q l l n a a n (j) Q i F n } Assume that int(q i S a a n ) It is enough to prove that (6) j K, q i q j (l,, l n ) {0,, 2} n s t Q l l n a a n (j) Q i F n Namely, by definition, Q l l n a a n (j) S l l n a a n S a a n To verify (6) we fix j K such that q i q j Let u {0, 2} be chosen such that assumption (A2) holds for Ju n (i) For this u, i, j and (a,, a n ) we choose an (l,, l n ) {0,, 2} n which satisfies Corollary 32 This implies that we have Q l l n ( a a n (j) Q i F n Namely, ψ l l n a a n ) (q j) J k (i) Corollary 34 We assume that S a a n is an n-good strip and we write r i for the i-th row vector of the matrix A a a n Assume that r i 0 (a): If N is odd then all the elements of r i are positive, (b): If N is even then for all j satisfying q j q i, we have r i (j) > 0 Note that for a shape Q i which is relevant for the n-good strip S a a n we have r i 0 33 The case when N is an even number The above argument shows that in the case when 3 N we can find a,, a n such that all the rows of some matrices A a a n are either all-positive or all-zero Now we would like to add to Corollary 34 (b) and prove the same in the case when N is even In this section we always assume that N is even We fix n which is large enough (It will be specified later how large n has to be) We always assume that a = (a,, a n ) {0,, 2} n is chosen in such a way that a n = 0 and S a a n is an n-good strip For every q {0,, N } we can find a unique i = i(q, a) {,, K} such that (a): q = q i (b): Q i is a relevant shape for S a a n

24 24 ANTHONY MANNING AND KÁROLY SIMON (c): for J(q) := [ ] q +, q n 3 we have n (62) J(q) J 0 (i) (q, q + ) π (e (i, a,, a n )) π (e 2 (i, a,, a n )), where π is the projection to the first axis Observe that (63) [ n ] n J(q) = q+ u k 3 k, u k 3 k + 3 n k= k= where u = = u n = 0, u n = One of the motivations to consider the intervals J(q) is as follows Fact 35 For some 0 q N let C = J(q) J, where [ n ] n (64) J := m + v k 3 k, v k 3 k +, 3 n k= for some 0 m M Then k= (65) C (q, m) + F n if and only if v n, where F n is the n-th approximation of F The proof of this Fact is an immediate corollary of Fact 2 and (63) Given n and q define l,, l n {0,, 2} n as follows 3 n q + = l 3 n + + l n 3 + l n N Then we have (66) J(q) [ ( l N l ) ( n l, N 3 n l n 3 + )] n 3 n Thus there is a unique z = z (q) {0,, N } such that ( (67) z (q) := 3 n q + N 3 n l l ) n 3 n (Thus z (q) is the number of the interval J(q) when we count modulo N the horizontal intervals of width 3 n ) It is easy to see that q z (q) is a bijection on {0, N } Since N is even we obtain (68) q {0,, N }, q z (q) Further, since 3 N, the map z : {0,, N } {0,, N } is a bijection Now we define z 2 := z 2 (q) {0,, M } as follows: (69) z 2 (q) := min { k ( (z (q), k) + I 2) int(s), k + M l n (q) mod 3 }

25 We write SLICES OF THE SIERPINSKI CARPET 25 (70) C(q, a) := ψ l (q)l n(q) a a n ((z (q), z 2 (q)) + I 2 ) Using Fact 2 and a n = 0, by (34) we obtain (7) C(q, a) Q i(q,a) F n and C(q, a) S a a n Let Q(q) := ((z (q), z 2 (q)) + I 2 ) S Note that we could choose Q(q) independent of a only because we always assume here that we restrict our attention to those n-good strips S a a n for which a n = 0 Then (72) ψ l l n a a n (Q(q)) = C(q, a) S a a n Summarizing what we have proved above: Lemma 36 Assuming that S a a n is an n-good strip and a n = 0, for every m {0,, N } we can find i, j K such that q i q j, q j = m and A a a n (i, j) > 0 Proof With q := z (m), i := i(q, a) and Q j := Q(q) the assertion of the lemma follows Lemma 37 We can find n and a = (a,, a n ) {0,, 2} n with a n = 0 and q q {0,, N } such that S a,,a n is an n-good strip and both Q(q ) and Q(q ) are relevant shapes for S a,,a n Proof First suppose that M/N Then Q(z (0)) is either Q = I 2 S or Q 2 = ((0, ) + I 2 ) S Each is relevant for any n-good strip S a,,a n By Fact Fact 25 we can now choose any n-good strip S a,,a n with a n = 0 for which Q(z ()) is relevant and it is also relevant for Q(z (0)) As 0 we have z (0) z () as required Now suppose that M/N < Then Q(z (0)) is either Q or Q 2 Clearly, if Q(z (0)) = Q then whichever way we choose Q(z ()) ([, 2] R) S we can find a narrow enough strip of slope M/N through this shape and Q This would immediately give us by Fact 25 that q = z (0), q = z () satisfy the requirements of the lemma So, we may assume that Q(z (0)) = Q 2 By symmetry, for the same reason we can assume that Q(z (N )) = Q K = ((N, M )+I 2 ) S If M/N > /2 then M/N < (M )/(N ) and so we can find a narrow enough strip of slope M/N which traverses both Q 2 and Q K So, by Fact 25, we can choose q = z (0), q = z (N ) Now assume that 0 < M/N < /2 First suppose that (73) Q(z ()) = Q 3 = ((, 0) + I 2 ) S and Q(z (N 2)) = Q K 2 = ((N 2, M) + I 2 ) S

26 26 ANTHONY MANNING AND KÁROLY SIMON Then by elementary geometry one can find a narrow strip which intersects the interior of both Q 3 and Q K 2 So, by Fact 25, we can choose q = z (), q = z (N 2) If (73) does not hold then either (74) Q(z ()) = Q 4 = ((, ) + I 2 ) S or (75) Q(z (N 2)) = Q K 3 = ((N 2, M ) + I 2 ) S If (74) holds we put q = z (0), q = z () and if (75) holds we put q = z (N 2), q = z (N ) Now we are ready to prove Proposition 8 Proof of Proposition 8 If N is odd then the assertion of the Proposition follows from Lemma 33 So, may assume that N is even Let us fix n and ã = (ã,, ã n ), and q, q whose existence is guaranteed by Lemma 37 Using the notation of Lemma 36 we substitute m = q to define i, j as in Lemma 36 Further, we also substitute m = q in Lemma 36 to define i, j As a shorthand notation we write in this proof A := Aã,,ã n Then by definition we have (76) A(i, j ), A(i, j ) > 0 Without loss of generality may assume that q = q j and q i are even and q = q j and q i are odd We will use the following observation: it follows from Corollary 34 (b) that for any k, l K we have (77) A 2 (k, l) A(k, k) A(k, l) A(k, l), since all the non-zero entries of A are at least one First we prove that (78) A 2 (i, k) > 0 and A 2 (i, k) > 0 for every k K To see that the first inequality holds observe that it follows from (76) and Corollary 34 (b) that whenever q k is even we have A 2 (i, k) A(i, j ) A(j, k) > 0 Similarly if q k is an odd number then using (78) and Corollary 34 (b) we get A 2 (i, k) A(i, k) > 0, which completes the proof of the first half of (78) By symmetry, the second inequality in (78) can be proved in the same way

27 SLICES OF THE SIERPINSKI CARPET 27 Now let u {,, K} be arbitrary such that the u-th row of A is a non-all-zero row To prove Proposition 8 it is enough to show that for every v {,, K} we have (79) A 3 (u, v) > 0 If q u, q v have the same parity then this follows from Corollary 34 (b) and (77) If q u and q v have different parity then without loss of generality we assume that q u is odd Since we assumed that q i is also odd we get from Corollary 34 (b) that A(u, i ) > 0 So, (78) yields that (80) A 3 (u, v) A(u, i ) A 2 (i, v) > 0 Thus we have verified Proposition 8 with n 0 := 3n and (a,, a n0 ) = (ã,, ã n, ã,, ã n, ã,, ã n ) 34 A corollary of the Perron-Frobenius theorem We fix n 0 and A a a n0 which satisfies Proposition 8 We consider the matrices We write T := 3 n 0 Put Since B := { A i i n0 }(i i n0 ) {0,,2} n 0 and B = {B,, B T }, where B s = B := A a a n0 B s := T B k k= using (9) one immediately gets that i i n0 A i i n0 = (A 0 + A + A 2 ) n0 (8) B s has each column sum 8 n 0 Further, (82) k T, each column sum of B k [2 n 0, 3 n 0 ] In particular all the matrices B k are column allowable (every column contains a non-zero element) non-negative integer matrices We define the Lyapunov exponent for the random product of the matrices {B i } T i=, where for each i in every step we choose B i independently with probability /T Using (2) the Lyapunov exponent (83) γ B := lim n n log B i i n, for aa (i, i 2, )

28 28 ANTHONY MANNING AND KÁROLY SIMON Then clearly we have (84) γ = n 0 γ B where we recall that T = 3 n 0 Note that it follows from (22) that (85) lim k k T log B k j j k = γ B j j k Let B be the matrix that we obtain from B when we replace all the column vectors of B that correspond to an all-zero row by all-zero columns That is if the column and row vectors of B are B = [c,, c K ] and B = [r,, r K ] then the matrix B is defined by its column vectors as follows: (86) B = [c,, c K], where c i = Note that for any k we have { ci, if r i > 0; 0, if r i = 0 (87) B k+ = B k B Let (88) l := # {i : c i > 0} We choose an orthogonal matrix Q that corresponds to a change of the order of the basis vectors in the natural basis such that [ ] (89) Q B C 0 Q T =, 0 0 where C is a positive (all the elements are positive) l l matrix It follows from the Perron-Frobenius theorem [, p 85], [7, p 9] that for the leading eigenvalue ρ and normalized left and right leading eigenvectors u, v R l of C we have (90) l l u T C = ρ u T, C v = ρ v, u > 0, v > 0, u i v i =, v i = Furthermore, there exists 0 < ρ 0 < ρ and l l matrices R k such that for the l l matrix P := v u T we have (9) k C k = ρ k P + R k, i= i=

29 SLICES OF THE SIERPINSKI CARPET 29 where for all i, j K the (i, j)-th element r (k) i,j satisfies (92) r (k) i,j < c ρ k 0, of the matrix R k for some constant c > 0 Sometimes we will have to extend vectors defined in R l to R K We will do this in two different ways: namely, for a R l we write [ ] a (93) a := and a 0 := Q a The meaning of a is as follows: to provide that all the l positive rows of B become the first l rows we needed to use a permutation of the basis vectors of the natural basis This permutation was provided by multiplying by the orthogonal matrix Q on the left To permute all the coordinates to the original order we have to multiply by Q on the left Remark 38 (a): In the following definition we use the fact that by the definition of Q the last K l rows of the matrix Q B Q T are all-zero rows and the first l rows are all-positive rows That is there exists an l K matrix D such that [ ] D (94) Q B Q T = and D = [C E] with E > 0 0 (95) (b): By the definition of Q we have B Q v = Q ((C v) ) = Q ρv = Q [ C E 0 ] v = ρq v Note that Q is an orthogonal matrix that corresponds to some change of the order in the natural basis Hence Q 0 In this way we see that ρ is an eigenvalue of the matrix B Furthermore, v = Q v is a non-negative eigenvector of the eigenvalue ρ Definition 39 For every n and for every (b,, b n ) {,, 3 n 0 } n : (a): We write E b b n for the l K matrix which satisfies [ ] Eb b n = Q B 0 Q T Q B b b n Q T = Q B B b b n Q T (b): We define the positive vector of K components u T b b n := u T E b b n (c): For ε R we define the (column) vector ε R l as a vector with all components ε

30 30 ANTHONY MANNING AND KÁROLY SIMON Lemma 40 Let 0 < ε < min i l v i Then there exists k 0 such that for every n and for every (b,, b n ) {,, 3 n 0 } n we have (96) ( ρ k0 v ) ( 0 ε u T b b n < C k0 E b b n < ρ k0 v + ) 0 ε u T b b n Proof The proof follows from (92) Namely, (97) { k 0 := k 0 (ε ) is defined as k 0 := min k : c Then by (92) we have ρ k 0 r(k 0) i,j < ε ρ k 0 (v u T 0 ) ε u T < C k 0 < ρ k 0 0 min u i Hence { } } k ρ0 < ε ρ 0 min u i i l (v u T + 0 ε u T ) This completes the proof of the lemma because all the elements of all the matrices and vectors in the last inequality as well as in the assertion of the lemma are nonnegative Lemma 4 There is (i,, i n0 ) {0,, 2} n 0 (98) A a a n0 v A i i n0 v, such that where we recall that v was defined by the convention introduced in (93) Proof To get a contradiction we assume that (99) (i,, i n0 ) {0,, 2} n 0, A a a n0 v = A i i n0 v Then 3 n0 ρ = 3 n0 ρ v = 3 n0 B v = A i i n0 v = (i,,i n0 ) = A n 0 s v = 8 n 0 v = 8 n 0 (i,,i n0 ) A i i n0 v so ρ = 8 n 0 /3 n 0 However this is impossible since 8 n 0 /3 n 0 cannot be a root of the characteristic polynomial of B which is a matrix of integer coefficients We assume from now on that we numbered the elements of B such that (00) B v B 2 v Without loss of generality we may assume that B v < B 2 v

31 SLICES OF THE SIERPINSKI CARPET 3 Definition 42 Let us define 0 < ε < 20 min v i and c 0 > 0 such that (0) B (v + 0 ε ) (v 0 ε ) + c 0 < B 2 (v 0 ε ) (v + 0 ε ) Lemma 43 Here we use the notation of Lemma 40 For an arbitrary (b,, b n ) {0,, 2} n we define F b b n := B k 0+ B b b n Then for an arbitrary n and for arbitrary (b,, b n ) {0,, 2} n we have B F b b (02) n + c 0 < B 2 F b b n, F b b n F b b n where A means the sum of the modulus of the elements of the matrix Proof First we start with a simple observation that we will use at the end of this argument Let a, b 0 be vectors of K components (03) If A = a b T then A = a b T Because QAQ T is obtained from A by permuting the rows and columns A = Q A Q T Thus in order to verify (02) it is enough to estimate the ratio of the norms Q B j F b b n Q T for j =, 2 and Q F b b n Q T Using (89), (87) and Definition 39 (a) we obtain [ ] [ ] [ C (04) Q F b b n Q T k 0 0 Eb b = n C k0 E = b b n Lemma 40 asserts that ( ρ k0 v ) 0 ε u T b b n < C k0 E b b n < ρ k0 ] ( v + ) 0 ε u T b b n, where we recall that u T b b n = u T E b b n Using (03) this implies that (05) ρ k0 ( v ) 0 ε u T b b n F b b n ρ k0 ( v + ) 0 ε u T b b n Now we estimate Q B j F b b n Q T = Q B j Q T Q F b b n Q T j =, 2 Also from the assertion of Lemma 40 we obtain that (06) ρ k0 ( v ) 0 ε u T b b n Q T Q F b b n Q T ρ k0 ( v + ) 0 ε u T b b n

32 32 ANTHONY MANNING AND KÁROLY SIMON Using (03) again we get on the one hand ( (07) Q B j F b b n Q T ρ k0 B j v + ) 0 ε u T b b n and on the other hand (08) Q B j F b b n Q T ρ k0 B j ( v 0 ε ) u T b b n Putting together these last two inequalities with (05) the assertion of the Lemma immediately follows from (0) Namely, for j =, 2 (09) Bj (v ( v + 0 ε ε 0 ) ) B j F b b n B j (v + ε 0 ) F b b n ( v ε ) 0 Lemma 44 For every m and b, b m and i =,, T we have (0) B i B b b m B b b m [2 n 0, 3 n 0 ] Proof Using the notation and the assertion of Fact 6 and the fact that, for all i, every column sum of B i is between 2 n 0 and 3 n 0 we obtain K K 2 n 0 r (i) Bb bn B i B b b m 3 n 0 r (i) Bb bn i= i= The simple observation that B b b n = K r (i) Bb bn completes the proof of the lemma i= Definition 45 (a): Let } R := {(x,, x T ) [2 n 0, 3 n 0 ] T : i, j such that x i x j c 0, where c 0 > 0 is the constant defined in Lemma 43 Using the well known inequality between the arithmetic and geometry means we obtain that the continuous function T x j j= f(x,, x T ) := log log T T x x T takes only positive values on the compact set R We define () δ 0 := min f R > 0 (b): Let m > k 0 + Put I m := {,, T } m, I m := {i I m : i = = i k0 + = } and I m := I m \I m

33 SLICES OF THE SIERPINSKI CARPET 33 Fact 46 For every i I m we have (2) T T log j= B j B i B i + δ 0 log T T j= B j B i B i Proof It follows from Lemma 43 that for every i I m B B i B i + c 0 < B 2 B i B i This and Lemma 44 imply that for every i I m ( B B i,, B ) T B i R B i B i The assertion of the Fact immediately follows from () Observe that for every i I m we have (3) T j= B j B i B i = T B j B i j= B i = B s B i B i = 8n0 B i B i = 8 n 0, where in the last but one step we used (26) and (8) This and Fact 46 imply that (4) i I m : T T log j= B j B i B i n 0 log 8 3 δ 0 Lemma 47 For every m k 0 + we have (5) T m T log T B j B i n 0 log 8 B i I m j= i 3 T (k 0+) δ 0