Fourier bases on the skewed Sierpinski gasket

Transcription

1 Graduate Theses and Dissertations Iowa State University Capstones, Theses and Dissertations 2017 Fourier bases on the skewed Sierpinski gasket Calvin Francis Hotchkiss Iowa State University Follow this and additional works at: Part of the Mathematics Commons Recommended Citation Hotchkiss, Calvin Francis, "Fourier bases on the skewed Sierpinski gasket" (2017). Graduate Theses and Dissertations This Dissertation is brought to you for free and open access by the Iowa State University Capstones, Theses and Dissertations at Iowa State University Digital Repository. It has been accepted for inclusion in Graduate Theses and Dissertations by an authorized administrator of Iowa State University Digital Repository. For more information, please contact

2 Fourier bases on the skewed Sierpinski gasket by Calvin Francis Hotchkiss A dissertation submitted to the graduate faculty in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Major: Mathematics Program of Study Committee: Eric S. Weber, Major Professor Tathagata Basak Fritz Keinert Yiu Tung Poon Justin Peters Iowa State University Ames, Iowa 2017 Copyright c Calvin Francis Hotchkiss, All rights reserved.

3 ii DEDICATION For Dr. Lynne Butler, who invited me back, and Drs. Ruth Haas and Jim Henle, who brought me here. And for Dr. Charles V Stephenson, in whose footsteps I belatedly follow.

4 iii TABLE OF CONTENTS LIST OF FIGURES v ACKNOWLEDGEMENTS ABSTRACT vi vii CHAPTER 1. OVERVIEW Introduction CHAPTER 2. REVIEW OF LITERATURE The Fast Fourier Transform Diţǎ s construction Hutchinson s fractal measures In which µ 4 is spectral but µ is not Hadamard duality CHAPTER. A FAST FOURIER TRANSFORM ON S N A Fast Fourier Transform on S N Computational Complexity of Theorems.1.7 and The Diagonal Matrices CHAPTER 4. FOURIER BASES ON THE SKEWED SIERPINSKI GAS- KET The Invariant Set S Our orthonormal basis The Transversality of the Zeros Condition A different proof

5 iv 4..1 The Cuntz algebra representation The function h X Argument by rectangles Aside: a note on the zeros of m Bigger rectangles A Second Spectrum Frequencies A Third Spectrum Frequencies CHAPTER 5. RELATED FRACTALS AND SPECTRA The Upper Left fractal The Lower Right fractal The Upper Right fractal Another shared spectrum BIBLIOGRAPHY

6 v LIST OF FIGURES Figure 4.1 A fifth iterative approximation of S starting at zero Figure 4.2 Plot of the fifth iteration of 1/(x, y) + L Figure 4. Plot of l (0, 0) l ( 1, 1/2) l ( 2, 1) Figure 4.4 Plot of r 5 (0, 0) r 5 ( 1/2, 1) r 5 ( 1, 1/2) Figure 4.5 Plot of the fifth iteration of {Ϝ 0, Ϝ 1, Ϝ 2 } Figure 4.6 Plot of s 5 (0, 0) Figure 4.7 Plot of s 5 ( 1, 2) Figure 4.8 Plot of s 5 ( 2, 1) Figure 5.1 A fifth iterative approximation of F (S) starting at zero Figure 5.2 A fifth iterative approximation of F (S) starting at zero Figure 5. A fifth iterative approximation of F (S) starting at zero

7 vi ACKNOWLEDGEMENTS Many thanks to my excellent committee members, Drs. Peters, Poon, Basak, and Keinart. Without your assistance and patience this would not have been possible. The results in this thesis rely on those of many other researchers, but most especially to Drs. Palle Jorgensen and Dorin Dutkay. Extra thanks are due to Dr. Eric S. Weber, without whose guidance and assistance none of this would have been possible, and whose work is in the background of so many of these proofs. Thanks to my future mathematical sister, Anna Seitz, for working through these results, and correcting errors in the process. May you solve the problems I never could. And most importantly, my deep gratitude to Melanie Erickson, whose administrative competence brought me to Iowa State in the first place, and without whom nothing.

8 vii ABSTRACT First, we demonstrate the construction of Fourier bases for fractal approximations via a construction analogous to the Fast Fourier Transformation. Then, we construct several full Fourier bases for the skewed Sierpinski gasket and related fractals.

9 1 CHAPTER 1. OVERVIEW 1.1 Introduction We begin our investigations with the fractal S, the invariant set of the iterated function system ψ 0 (x, y) 1 (x, y), ψ 1(x, y) 1 (x + 2, y), ψ 2(x, y) 1 (x, y + 2). By Hutchinson s theorem, there exists an invariant measure ν supported on the invariant set S. The set S takes the form of a skewed Sierpinski gasket; it is the set of points (x, y) R 2 for which x, y, and x + y are all in the standard middle-thirds Cantor set. This was the reason for the choice of fractal; while it has been shown that the middle-thirds Cantor set does not have an orthonormal Fourier basis, the question of whether it has a Fourier frame is still open, and we hope that investigations on related sets such as S will lead to such a frame. It is a difficult open question, but also one that has a close connection to undergraduate analysis topics via the Cantor set. For S, we first choose a dual iterated function system ρ 0 (x, y) (x, y), ρ 1 (x, y) (x, y)+ (1, 2), ρ 2 (x, y) (x, y) + (2, 1). This iterated function system generates the frequencies R n of an orthogonal basis of exponentials on finite approximations S n of S. We proved that the matrix M n, where (M n ) j,k exp (2πiR j S k ) with (R j ) n 1 j0 and (S k) n 1 k0 orderings of R n and S n respectively, is a block matrix in the form of Diţǎ s construction for generating large Hadamard matrices. In particular, it is an orthogonal basis matrix of exponentials for S n. The blocks of M n can be expressed in terms of M 1, M n 1 and diagonal matrices. The rows and columns of M 1 n can be permuted into a block matrix of the same form. These block form allow a Fast Fourier Transform-type algorithm to quickly transform a function on S n, written as a column vector v, into its representation in terms of the orthogonal basis of exponentials that

10 2 are the rows of M n. The block form reduces the matrix computation complexity to O(N log N), as in the case of the classical Fast Fourier Transformation. This construction extends to a Fractal Fast Fourier Transformation in the general case of finite approximations of a fractal generated by an iterated function system of affine transformations on R d. With certain restrictions on the choice of affine transformations for the fractal and signal approximations, the construction generalizes fully with the same reduction in complexity. After this success, I returned to S, aiming to demonstrate that the infinite set R generated by {ρ j } was the frequency set for a Fourier basis on L 2 (ν ) of the form {e r (x) e 2πir x r R}. I determined that the exponentials were orthogonal; however, the completeness failed. In fact, I could show that the set R was not a complete Fourier basis. As it turned out there was a simple fix: following the work of Dorin Ervin Dutkay and Palle E. T. Jorgensen in (7), we took two of the bad points and added them into the set R. Our original set of frequencies was the simple application of {ρ j } to the origin; to construct a set of Fourier frequencies, it was enough to apply them to two more points: in this case ( 1, 1/2) and ( 1/2, 1). In that way we constructed our first full Fourier basis for S. Once we had that construction, I was able to quickly construct two more Fourier bases for S, based on different choices of dual iterated function systems for {ψ j }. The most interesting such basis was the one constructed using ρ 0 (x, y), ρ 1 (x, y) + (2, 1) and ρ 2 (x, y) (x, y) + (4, 2). The vectors (2, 1) and (4, 2) are linearly dependent, and in fact, the Fourier basis generated by these vectors has the form {e (u,u/2) (x) u Z}. We then extended our results to fractals related to S by rotations and reflections. It turns out that S shares some of its spectra with its with its reflection across the line y x, suggesting possible future work on the union of these two fractals. The question also remains of how many possible Fourier bases there are for S; so far, there have been the three we constructed with our Hadamard duals, as well as one constructed by Dutkay and Jorgensen in (9).

11 CHAPTER 2. REVIEW OF LITERATURE 2.1 The Fast Fourier Transform The standard Fast Fourier Transform is a method of reducing the complexity of the Discrete Fourier Transform on 2 N equally-spaced data points on [0, 1). A function f on this discrete space can be viewed as a vector of length 2 N, and the Discrete Fourier Transform multiplies this vector by the change of basis matrix: jk 2πi F N (e 2 N ) jk, 0 j, k < 2 N. In general, multiplying a vector of length 2 N by a 2 N 2 N matrix requires O(2 2N ) operations. The Fast Fourier Transform was developed first by Gauss in 1805 and then generalized by Cooley and Tukey in 1965, in order to reduce the complexity of this multiplication by permuting the Discrete Fourier Transform matrix into blocks and making use of symmetries in the complex matrix entries (), (16). In particular, the Fast Fourier Transform permutes the columns of F N by the permutation: 2k 0 k < 2 N 1, σ(k) 2k N 1 k < 2 N. This gives F N the block form: F N P F N 1 F N 1 DF N 1 DF N 1 (2.1) where D is the 2 N 1 2 N 1 diagonal matrix with kth diagonal entry e 2πik/2N (21).

12 4 Multiplying our vector f by this block form is significantly easier than multiplying by the full matrix F N. And since F N 1 can be permuted in the same manner, and so forth down to F 1, the overall computational complexity is reduced down to O(N 2 N ) Diţǎ s construction A block matrix form strikingly similar to that of the Fast Fourier Transform shows up in the work of P. Diţǎ, who studies complex Hadamard matrices. (4; 22). A complex Hadamard matrix is an N N matrix H, all of whose entries have norm one, and with the property that H H NI N. The Discrete Fourier Transform and Fast Fourier Transform Matrices are examples of Hadamard Matrices. Diţǎ developed a way of constructing large Hadamard matrices from smaller ones that is, in a sense, the reverse of the Fast Fourier Transform. In particular, if A is a K K Hadamard matrix, B is an M M Hadamard matrix, and E 1,..., E K 1 are M M unitary diagonal matrices, then the KM KM block matrix H is a Hadamard matrix: H a 00 B a 01 E 1 B... a 0(K 1) E K 1 B a 10 B a 11 E 1 B... a 1(K 1) E K 1 B a (K 1)0 B a (K 1)1 E 1 B... a (K 1)(K 1) E K 1 B. (2.2) To show that H is Hadamard, first notice that, for C A, H c 00 B c 01 B... c 0(K 1) B c 10 B E1 c 11 B E1... c 1(K 1) B E c (K 1)0 B EK 1 c 1(K 1) B EK 1... c (K 1)(K 1) B EK 1. (2.) Proposition H H (KM)I KM. Proof. Let G be the block matrix in Equation (2.), and let E 0 I M. Note that the product of H and G will have a block form. Multiplying the j-th row of H with the l-th column of G,

13 5 we obtain that the j, l block of HG is: K 1 K 1 (a jk E k B)(c kl B Ek ) a jk c kl (M)I M. k0 k0 Since K 1 k0 a jkc kl Kδ j,l, we obtain HG (KM)I KM. Let v be a vector of length KM. Consider H v where H is the block matrix as in Equation (2.2). We divide the vector v into K vectors of length M as follows: v 0 v 1 v.. v K 1 Then the matrix multiplication H v can be reduced in complexity, since K 1 j0 a 0jE j B v 0 K 1 j0 H v a 1jE j B v 1.. K 1 j0 a (K 1)jE j B v K 1 Let O M be the number of operations required to multiply a vector w of length M by the matrix B. The total number of operations required for each component of H v is O M + M(K 1) + MK multiplications and M(K 1) additions. The total number of operations for H v is then KO M + MK 2 2MK. We have just established the following proposition. Proposition The product H v requires at most KO M + MK 2 2MK operations. Since O M O(M 2 ), we obtain that the computational complexity of H is O(M 2 K +MK 2 ), whereas for a generic KM KM matrix, the computational complexity is O(K 2 M 2 ). Thus, the block form of H reduces the computational complexity of the matrix multiplication. Notice that the Fast Fourier Transform matrix is an example of Diţǎ s construction, with B F N 1, E 1 D, and A which is also the two-by-two Discrete Fourier Transform Matrix F 2.

14 6 2.2 Hutchinson s fractal measures In 1981, John E. Hutchinson defined the general framework for the type of fractal we discuss here, that is, the compact invariant set of a set of contraction maps on R n. Definition (18) For any metric space X with distance function d, F : X X is a contraction if d(f (x), F (y)) sup < 1 x y d(x, y) Definition (18) The compact set K R n is invariant if there exists a finite set S {S 1,..., S N } of contraction maps on R n such that K N i1 S ik. In the context of this thesis, we will discuss contraction maps of the form S(x) Ax + b, with A an n n matrix with A < 1 and b a vector in R n. Hutchinson s results apply to any finite set of contraction maps. Theorem (18) If S {S 1,..., S N } is a finite set of contraction maps on a complete metric space X, then there exists a unique, closed, bounded set K such that K N i1 S ik. Also, K is compact and the closure of the set of fixed points of finite compositions S i1 S ip of members of S. (18) In addition to a unique closed bounded set, the set S determines a set of unique Borel regular probability measures: Theorem (18) Let S as above and ρ 1,..., ρ N (0, 1) with N i1 ρ i 1. Then there exists a unique Borel regular measure µ with µ(x) 1 such that for any measurable set A, µ(a) N i1 ρ iµ(s 1 (A)). i Consequently, for any continuous function f on X, fdµ N (ρ i i1 ) f(s i (x)) dµ(x). (2.4) In particular, this applies when ρ 1 ρ N 1/N, which is the usual fractal measure considered in this context.

15 In which µ 4 is spectral but µ is not Later on, Jorgensen and Pedersen in (19) looked at the case where the contraction maps are of the form S j x R 1 x + b j, x R N, (2.5) where R is a real matrix with eigenvalues ξ i all satisfying ξ i > 1. In particular, they discuss the case of µ 4, which is the unique probability measure of compact support on R such that: fdµ 4 1 ( 2 ( x ) f 4 dµ 4 + f ( x ) ) dµ 4. 2 In this case the compact support of µ 4 is the Cantor-4 set, the Cantor set obtained by dividing I [0, 1] into four equal subintervals and retaining only the first and third; then repeating the process with each remaining interval. Similarly, there is also µ, the unique probability measure of compact support on R such that: fdµ 1 2 ( ( x ) f dµ + f ( x + 2 ) ) dµ, which is supported on the standard middle thirds Cantor set. Each of these fractal measures has a Fourier transform: µ(t) e i2πt x dµ(x) (2.6) for any t R d. Because of Equation 2.4, we can write: µ(t) χ B (t) µ(r 1 t) (2.7) where χ B (t) 1 e i2πb t. (2.8) N By applying Equation 2.7 repeatedly, Jorgensen and Pedersen obtained: µ 4 (t) n0 b B 1 ) (1 + e i πt 4 2 n 2t iπ e n0 ( ) πt cos 2 4 n, (2.9)

16 8 and µ (t) n0 1 ) (1 ( ) + e i 4πt 2πt 2 n e iπt cos n. (2.10) n0 Jorgensen and Pedersen would show that µ 4 has a Fourier basis, that is, there is at least one set of frequencies P R for which {e 2πix λ : λ P } is an orthonormal basis for L 2 (µ 4 ). The space L 2 (µ ), however, can have no such Fourier basis, since there can be no more than two orthogonal exponential functions in L 2 (µ ). Theorem (19) Any set of µ orthogonal exponentials contains at most two elements. Proof. From Equation 2.10, the set of t R with µ(t) 0 is: { n Z(µ ) 4 } (1 + 2Z) : n 1, 2,,... (2.11) For three exponentials, e γ1, e γ2, e γ to be mutually orthogonal in L 2 (µ), we must have µ (γ i γ j ) 0 for every pair i, j 1, 2,, i j. That is, γ i γ j Z(µ ). Let λ 1 γ 1 γ 2, λ 2 γ 2 γ, λ 0 γ 1 γ, and let z j Z be such that: Since λ 1 + λ 2 λ 0, we have that: λ j n j 4 (1 + 2z j). n 1 (1 + 2z 1 ) + n 2 (1 + 2z 2 ) n 0 (1 + 2z 0 ). However, the left-hand side of the equation must be even, while the right-hand side must be odd, so we have a contradiction Hadamard duality Definition (7) A set of vectors B {b 0,..., b n 1 } R d and a set of vectors L {l 0,..., l n 1 } R d are called a Hadamard pair if the matrix: ) H : (e 2πib j l k 0 j,k n 1 (2.12) is Hadamard, that is, 1 n H is unitary.

17 9 Notation 1. Let e λ (x) : e 2πiλ x Definition (7) Fix some d N. (B, L, R) is a system in Hadamard duality if: B and L are subsets of R d such that #B #L : N, R is a fixed d d invertible matrix over R with all eigenvalues λ satisfying λ > 1, the sets (R 1 B, L) form a Hadamard pair. Then let: τ b (x) R 1 (b + x), x R d ; X B be the unique compact subset of R d (see Definition 2.2.2) such that X B τ b (X B ); µ B the measure guaranteed in Theorem 2.2.4, with ρ 1 ρ N 1/N. b B τ l (x) : (R T ) 1 (l + x), x R d. X L be the unique compact subset of R d such that X L l L τ l(x L ). In this context, Jorgensen and Pedersen proved an important theorem: Theorem (19) For B, L, R as in Definition 2.2.7, with RB Z d, 0 B, and (B, L, R) in Hadamard duality, let h X (t) : λ E µ ˆ B (t λ) 2, t R d, λ E {l 0 + R l 1 + : l i L, finite sums} (2.1) Then {e λ : λ E} is an orthonormal basis for L 2 (µ B ) if and only if h X 1 on R d. Later, Dutkay and Jorgensen extended this in the following construction. Definition (7) For B, N as in Definition 2.2.7, let: W B (x) 1 N 2 2 e 2πib x b B

18 10 Lemma For B, L, N, τ l as in Definition 2.2.7, Proof. l L W B (τ l x) l L 1 N 2 1 N 2 l L W B (τ l x) 1, l L b B ( b B 1 N 2 1 N 2 1 N 2 1 N 2 1 N 2 l L b B b B l L b B b B 2 e 2πib τ lx e 2πib (RT ) 1 (l+x) x R d ) ( b B e 2πib (R T ) 1 (l+x) e 2πib (RT ) 1 (l+x) e 2πib (RT ) 1 (l+x) e 2πiR 1b (l+x) e 2πiR 1 b (l+x) e 2πiR 1b l e 2πiR 1b x e 2πiR 1 b l e 2πiR 1 b x l L b B b B e 2πiR 1b l e 2πiR 1b x e 2πiR 1 b l e 2πiR 1 b x b B b B l L b B b B e 2πiR 1b x e 2πiR 1b x l L ) e 2πiR 1 b l e 2πiR 1 b l (2.14) (2.15) (2.16) (2.17) (2.18) (2.19) (2.20) (2.21) The inner sum is the row sum of HH, for H as in Equation 2.12, therefore it is equal to N for b b and 0 otherwise. So: W B (τ l x) 1 N 2 e 2πiR 1b x e 2πiR 1b x N (2.22) l L b B 1 N 2 N 1 N 2 N 2 1 (2.2) b B Definition (7) A point x X L is a cycle if for some l 1,..., l k L, not necessarily distinct or in any particular order, τ l1 τ l2 τ lk (x) x. If in addition W B (x) 1, x is called a W B -cycle. Definition (Transversality of the zeros (7)). A function W on X satisfies the transversality of the zeros condition if:

19 11 (a) If x X is not a cycle, then there exists k x 0 such that, for k k x, {τ l1 τ l2 τ lk (x) : l 1,... l n k L} does not contain any zeros of W ; (b) If {x 0, x 1,..., x p } are on a cycle with x 1 τ l (x 0 ) for some l L, then for every y τ l (x 0 ), y x 1 is either not on a cycle or W (y) 0. Theorem (7) Suppose that: {R 1 B, L} form a Hadamard pair, #B #L : N, R n b l Z, for b B, l L, n 0, 0 B, 0 L. W B satisfies the transversality of the zeros condition. Let Λ R d be the smallest set that contains C for every W B -cycle C, and such that R T Λ + L Λ. Then {e 2πiλ x λ Λ} is an orthonormal basis for L 2 (µ B ).

20 12 CHAPTER. A FAST FOURIER TRANSFORM ON S N.1 A Fast Fourier Transform on S N We consider an iterated function system generated by contractions {ψ 0, ψ 1,..., ψ K 1 } on R d of the following form: ψ j (x) A(x + b j ) where A is a d d invertible matrix with A < 1. We require A 1 to have integer entries, the vectors b j Z d, and without loss of generality we suppose b 0 0. We then choose a second iterated function system generated by {ρ 0, ρ 1,..., ρ K 1 } of the form ρ j (x) Bx + c j where B (A T ) 1, with c j Z d, and c 0 0. We require the matrix M 1 (e 2πi c j A b k ) j,k (.1) be invertible (or Hadamard). Note that depending on A and { b 0, b 1,..., b K 1 }, there may not be any choice { c 0, c 1,..., c K 1 } so that M 1 is invertible. However, for many IFS s there is a choice: Proposition.1.1. If the set { b 0, b 1,..., b K 1 } is such that for every pair (j k), A b j A b k / Z d, then there exists { c 0, c 1,..., c K 1 } such that the matrix M 1 is invertible. Proof. The mappings φ 1 : x e 2πi x A b j and φ 2 : x e 2πi x A b k are characters on G Z d /BZ d. Since A b j A b k / Z d, the characters are distinct. Thus, by Schur orthogonality, x G φ 1(x)φ 2 (x) 0. Therefore, the matrix M (e 2πi x k A b j ) j,k, where { x k } is any enumeration of G, has orthogonal columns. Thus, there is a choice of a square submatrix of M which is invertible.

21 1 Even under the hypotheses of Proposition.1.1 there is not always a choice of c s so that M 1 is Hadamard; this is the case for the middle-third Cantor set, which is the attractor set for the IFS generated by ψ 0 (x) x, ψ 1(x) x+2 (and is a reflection of the fact that µ is not spectral). Notation 2. We define our notation for compositions of the IFS s using two distinct orderings. Let N N. For j {0, 1,..., K N 1}, write j j 0 +j 1 K + +j N 1 K N 1 with j 0,..., j N 1 {0, 1,... K 1}. We define Ψ j,n : ψ j0 ψ j1 ψ jn 1 R j,n : ρ j0 ρ j1 ρ jn 1. These give rise to enumerations of S N and T N as follows: S N {Ψ j,n (0) : j 0, 1,... K N 1} T N {R j,n (0) : j 0, 1,... K N 1}. We call these the obverse orderings of S N and T N. Likewise, we define Ψ j,n : ψ jn 1 ψ jn 2 ψ j0 R j,n : ρ jn 1 ρ jn 2 ρ j0 which also enumerate S N and T N. We call these the reverse orderings. Remark Note that for N 1, Ψ j,1 Ψ j,1 and R j,1 R j,1. We define the matrices M N and M N as follows: [M N ] jk e 2πiR j,n (0) Ψ k,n (0) and [ M N ] jk e 2πi R j,n (0) Ψ k,n (0). Both of these are the matrix representations of the exponential functions with frequencies given by T N on the data points given by S N. The matrix M N corresponds to the obverse ordering on

22 14 both T N and S N, whereas the matrix M N corresponds to the reverse ordering on both. Since these matrices arise from different orderings of the same sets, there exist permutation matrices P and Q such that Q M N P M N. (.2) Indeed, define for j {0,..., K N 1} a conjugate as follows: if j j 0 + j 1 K + + j N 1 K N 1, let j j N 1 + j N 2 K + + j 0 K N 1. Note then that j j, and Ψ k,n Ψ k,n Rk,N R k,n. (.) Now, define a K N K N permutation matrix P by [P ] mn 1 if n m, and 0 otherwise. Lemma.1.2. For P defined above, P M N P M N. Proof. We calculate [P M N P ] mn k [P ] mk [ M N ] kl [P ] ln l [P ] m m [ M N ] mñ [P ]ñn e 2πi R m,n (0) Ψñ,N (0) e 2πiR m,n (0) Ψ n,n (0) [M N ] mn by virtue of Equation (.). Proposition.1.. For scale N 1, M 1 M 1 exp(2πi c 1 A b 1 )... exp(2πi c 1 A b K 1 ) exp(2πi c K 1 A b 1 )... exp(2πi c K 1 A b K 1 ). Proof. The proof follows from Remark.1. Lemma.1.4. For N N, 0 j < K N, and x, y R d,

23 15 1. Ψ j,n ( x + y) Ψ j,n ( x) + A N y 2. Ψ j,n ( x + y) Ψ j,n ( x) + A N y. R j,n ( x + y) R j,n ( x) + B N y 4. Rj,N ( x + y) R j,n ( x) + B N y. Proof. 1. We prove by induction on N. Base case, n 1, we have j 0, 1,... K 1 and Ψ n,j Ψ n,j ψ j. j 0: ψ 0 ( x + y) A ( x + y) A( x) + A( y) ψ 0 ( x) + A y j 1,... K 1: ψ j ( x + y) A ( x + y + b j ) A( x) + A( y) + Ab j ψ 0 ( x) + A y Assume the equality in Item 1. holds for N 1. For j j 0 + j 1 K + + j N 1 K N 1, let l j j N 1 K N 1. We have: Ψ j,n ( x + y) Ψ l,n 1 (ψ jn 1 ( x + y)) Ψ l,n 1 (ψ jn 1 ( x) + A y) Ψ l,n 1 (ψ jn 1 ( x)) + A N 1 A y Ψ j,n ( x) + A N y 2. Proof is the same as for 1.. Base case: For N 1, R j,1 ρ j. ρ 0 ( x + y) B x + B y, and for j 1,..., K 1, ρ j ( x + y) B( x + y) + c j ρ j ( x) + B 6. Let R j,n 1 ρ j1 ρ jn 1 be the composition of N 1 ρ k s with the property that ρ j0 R j,n 1 R j,n.

24 16 Assume true for n < N. Then: R j,n ( x + y) ρ j0 ρ j1 ρ jn 1 ( x + y) ( ρ j0 Rj,N 1 ( x + y) ) ( ρ j0 Rj,N 1( x) + B N 1 y ) ρ j0 (R j,n 1 x) + B(B N 1 y) R j,n x + B N y 4. Proof is the same as for. Lemma.1.5. For N N and 0 j < K N, 1. Ψ j,n (0) A N z for some z Z d, 2. Ψ j,n (0) A N z for some z Z d,. R j,n (0) Z d, 4. Rj,N (0) Z d. Proof. 1. We prove by induction on N. Base case, N 1, we have j 0, 1,..., K 1 and Ψ j,n Ψ j,n ψ j. For j 0, ψ 0 ( 0) A 0 0, and for j 1,..., K 1, ψ j ( 0) A( 0 + b j ) A b j, so the lemma is satisfied for z b j, which is in Z d by definition. Assume the equality in Item 1. holds for N 1. For j j 0 + j 1 K + + j N 1 K N 1, let q j j j N 1 K N 1. We have: ( Ψ j,n (0) ψ jn 1 Ψqj,N 1(0) ) A (A N 1 z + ) b j A N ( z + A (N 1) bj )

25 17 Since A 1 is an integer matrix, so is A (N 1) and thus z + A (N 1) bj Z d. Item 2. is analogous. For Item., note first that ρ j (Z d ) Z d, so by induction, ρ j0 ρ jn 1 (0) Z d. Likewise for Item 4. Lemma.1.6. Assume N 2, let l be an integer between 0 and K 1, and suppose l K N 1 j < (l + 1)K N 1. Then, 1. Ψ j,n (0) Ψ j l K N 1,N 1(0) + A N bl, 2. Ψ j,n (0) A Ψ j l K N 1,N 1(0) + A b l,. R j,n (0) R j l K N 1,N 1(0) + B N 1 c l, 4. Rj,N (0) B R j l K N 1,N 1(0) + c l. Proof. For l K N 1 j < (l + 1)K N 1, j N 1 l, so we have: Ψ j,n (0) ψ j0 ψ j1 ψ jn 2 ψ l (0) ψ j0 ψ j1 ψ jn 2 (A(0 + ) b l ) Ψ j l K N 1,N 1(0 + A b l ). Applying Lemma.1.4 Item i) to Ψ j l K N 1,N 1: Ψ j l K N 1,N 1(0 + A b l ) Ψ j l K N 1,N 1(0) + A N 1 A b l. The proof of Item iii) is similar to Item i) with one crucial distinction, so we include the proof here. We have: R j,n (0) ρ j0 ρ j1 ρ jn 2 ρ l (0) ρ j0 ρ j1 ρ jn 2 (B0 + c l ) R j l K N 1,N 1(0 + c l ). Applying Lemma.1.4 Item iii) to R j l K N 1,N 1: R j l K N 1,N 1(0 + c l ) R j l K N 1,N 1(0) + B N 1 c l.

26 18 For Item ii), we have Ψ j,n (0) ψ l ( Ψ j l K N 1,N 1(0)) A Ψ j l K N 1,N 1(0) + A b l. The proof of Item iv) is analogous. Note that in Item i), the extra term involves A N, whereas in Item iii) the extra term involves B N 1. We are now in a position to prove our main theorem. Theorem.1.7. The matrix M N representing the exponentials with frequencies given by T N on the fractal approximation S N, when both are endowed with the obverse ordering, has the form M N m 00 M N 1 m 01 D N,1 M N 1... m 0(K 1) D N,K 1 M N 1 m 10 M N 1 m 11 D N,1 M N 1... m 1(K 1) D N,K 1 M N m (K 1)0 M N 1 m (K 1)1 D N,1 M N 1... m (K 1)(K 1) D N,K 1 M N 1. (.4) Here, D N,m are diagonal matrices with [D N,m ] pp e 2πiR p,n 1(0) A N b m, and m jk [M 1 ] jk. Proof. Let us first subdivide M N into blocks B lm of size K N 1 K N 1, so that B B 0(K 1) M N B (K 1)0... B (K 1)(K 1) Fix 0 j, k < K N and suppose lk N 1 j < (l + 1)K N 1 and mk N 1 k < (m + 1)K N 1 with 0 l, m < K. Let q j j lk N 1 and q k k mk N 1. Observe that [M N ] jk [B lm ] qj q k. (.5) Using Lemma.1.6 Items ii) and iv), we calculate R j,n (0) Ψ k,n (0) ( R qj,n 1(0) + B N 1 ) ) c l (Ψ qk,n 1(0) + A N bm. By Lemma.1.5 Item i), for some z Z d, B N 1 c l Ψ qk,n 1(0) B N 1 c l A N 1 z c l z Z.

27 19 Note that Therefore, combining the above, we obtain B N 1 c l A N bm c l A b m. [M N ] jk e 2πiR j,n (0) Ψ k,n (0) e 2πiR q j,n 1(0) Ψ qk,n 1(0) e 2πiR q j,n 1(0) A N b m e 2πi c l A b m [M N 1 ] qj q k e 2πiR q j,n 1(0) A N b m [M 1 ] lm. (.6) Letting j vary between lk N 1 and (l+1)k N 1 and k vary between mk N 1 and (m+1)k N 1 corresponds to q j and q k varying between 0 and K N 1. Therefore, we obtain from Equations (.5) and (.6) the matrix equation B lm [M 1 ] lm D N,m M N 1 where [D N,m ] pp e 2πiR p,n 1(0) A N b m as claimed. Corollary.1.8. The matrix M N is invertible. If M 1 is Hadamard, then M N is also Hadamard. Proof. If M 1 is invertible, then by induction, M N is invertible via Proposition If M 1 is Hadamard, then again by induction, M N is Hadamard by Diţǎ s construction. Theorem.1.9. The matrix M N representing the exponentials with frequencies given by T N on the fractal approximation S N, when both are endowed with the reverse ordering, has the form M N m 00 MN 1 m 01 MN 1... m 0(K 1) MN 1 m 10 MN 1 DN,1 m 11 MN 1 DN,1... m 1(K 1) MN 1 DN, m (K 1)0 MN 1 DN,K 1 m (K 1)1 MN 1 DN,K 1... m (K 1)(K 1) MN 1 DN,K 1 (.7) Here, D N,q is a diagonal matrix with [ D N,l ] pp e 2πic l A( Ψ p,n 1 (0)), and m jk [M 1 ] jk.

28 20 Proof. The proof proceeds similarly to the proof Theorem.1.7. Let us first subdivide M N into K N 1 K N 1 blocks B lm, so that B B0(K 1) M N B (K 1)0... B(K 1)(K 1) Fix 0 j, k < K N and suppose lk N 1 j < (l + 1)K N 1 and mk N 1 k < (m + 1)K N 1 with 0 l, m < K. Let q j j lk N 1 and q k k mk N 1. Observe that [ M N ] jk [ B lm ] qj q k. (.8) We calculate using Lemma.1.6 items ii) and iv): R j,n (0) Ψ k,n (0) (B R qj,n 1(0) + c l ) (A Ψ qk,n 1(0) + A b m ) R qj,n 1(0) Ψ qk,n 1(0) + c l A Ψ qk,n 1(0) + R qj,n 1(0) b m + c l A b m. By Lemma.1.5 Item iv), Rqj,N 1(0) b m Z. Thus, [ B lm ] qj q k [M N 1 ] qj q k e 2πi c l A Ψ qk,n 1(0) [M 1 ] lm and as in the proof of Theorem.1.7, we have B lm [M 1 ] lm MN 1 DN,l..1.1 Computational Complexity of Theorems.1.7 and.1.9 As a consequence of Proposition 2.1.2, the matrix M N can be multiplied by a vector of dimension K N in at most KP N 1 + K N+1 2K N operations, where P N 1 is the number of operations required by the matrix multiplication for M N 1. Since M N 1 has the same block form as M N, P N 1 can be determined by P N 2, etc. The proof of the following proposition is a standard induction argument, which we omit. Note that this says that the computational complexity for M N is comparable to that for the FFT (recognizing the difference in the number of generators for the respective IFS s).

29 21 Proposition The number of operations to calculate the matrix multiplication M N v is P N K N 1 P 1 + (N 1)K N+1 2(N 1)K N. Consequently, P N O(N K N ). The significance of Theorem.1.9 concerns the inverse of M N. If P is the permutation matrix as in Lemma.1.2, then M 1 N 1 1 P M P. By Proposition 2.1.1, M construction, and so the computational complexity of N M 1 N N has the form of Diţǎ s is the same as M N. Thus, modulo multiplication by the permutation matrices P, the computational complexity of multiplication by M 1 N is the same as that for M N..1.2 The Diagonal Matrices The matrices M N and M N have the form of Diţǎ s construction as shown in Theorems.1.7 and.1.9. The block form of Diţǎ s construction involves diagonal matrices, which in Equations (.4) and (.7) are determined by the IFS s used to generate the matrices M N and M N. As such, the diagonal matrices satisfy certain recurrence relations. Theorem The diagonal matrices which appear in the block form of M N (Equation (.4)) satisfy the recurrence relation D N,m D N 1,m E N,m, where E N,m is the K K diagonal matrix with [E N,m ] uu e 2πicu AN b m. That is: [D N,m ] pp [D N 1,m ] p p e 2πi(cp 0 AN b m) where p (p p 0 )/K. Likewise, the diagonal matrices which appear in the block form of M N (Equation (.7)) satisfy the recurrence relation D N,l D N 1,l ẼN,l, where ẼN,l is the K K diagonal matrix with [ẼN,l] uu e 2πi c l A N b u. That is: [ D N,l ] pp [D N 1,l ] p p e 2πi c l A N b p0. Proof. As demonstrated in Theorem.1.7, for p 0, 1,..., K N 1, [D N,m ] pp e 2πiR p,n 1(0) A N b m. Note that p N 1 0, and ρ 0 (0) 0. We want to cancel one power of A in A N bm, so we factor out a B from R p,n 1 (0): R p,n 1 (0) ρ p0 ρ p1 ρ pn 2 (0) B ( ρ p1 ρ pn 2 (0) ) + c p0.

30 22 Since p p 1 + p 2 K + + p N 2 K N, R p,n 1 (0) BR p,n 2 (0) + c p0. Thus, [D N,m ] pp e 2πiR p,n 1(0) A N b m e 2πi(BR p,n 2(0) A(AN 1 b m)) e 2πi( cp 0 AN b m) e 2πi(R p,n 2(0) (AN 1 b m)) e 2πi( cp 0 AN b m) [D N 1,m ] p p e 2πi( cp 0 AN b m). Similarly, as demonstrated in Theorem.1.9, [ D N,l ] pp e 2πi c l A( Ψ p,n 1 (0)). We write: Ψ p,n 1 (0) ψ pn 2 ψ pn ψ p1 ψ p0 (0) ψ pn 2 ψ pn ψ p1 (0 + A b p0 ) Ψ p,n 2 (0 + A b p0 ) Ψ p,n 2 (0) + A N 1 bp0. where in the last equality we use Lemma.1.4 item ii). Therefore: [ D N,l ] pp e 2πic l A( Ψ p,n 1 (0)) e 2πi c l A( Ψ p,n 2 (0)+A N 1 b p0 ) e 2πi c l A( Ψ p,n 2 (0)+A N b p0 ) e 2πi c l A( Ψ p,n 2 (0)) e 2πi c l A N b p0 [ D N 1,l ] p p e 2πi c l A N b p0.

31 2 CHAPTER 4. FOURIER BASES ON THE SKEWED SIERPINSKI GASKET 4.1 The Invariant Set S Now, instead of looking at fractal approximations as in Chapter, we construct an orthonormal basis of exponentials on L 2 (ν ), the unique Borel regular measure supported on the complete closed bounded set S guaranteed by Hutchinson in (18). Definition Let S 2 j0 ψ j(s), with: ψ 0 (x, y) 1 (x, y) (4.1) ψ 1 (x, y) 1 (x + 2, y) (4.2) ψ 2 (x, y) 1 (x, y + 2) (4.) By (18), there exists a unique Borel regular measure, which we call ν, supported on S and with the property that, for any continuous function f on S, First, we examine the set S. fd ν 2 j0 1 f ψ j d ν (4.4) Proposition Let A {(x, y) [0, 1] [0, 1] : (x, y) j1 (x j, y j ) j, (x j, y j ) {(0, 0), (0, 2), (2, 0)}} Then A is the closed invariant set for (4.1), (4.2), (4.), that is, A S. S {(x, y) [0, 1] [0, 1] : (x, y) j1 (x j, y j ) j, (x j, y j ) {(0, 0), (0, 2), (2, 0)}}. Proof. First, we show that A ψ 0 (A) ψ 1 (A) ψ 2 (A). Claim For all (x, y) A, ψ 0 (x, y) A.

32 24 Proof. We know that ψ 0 ( j1 (x j, y j ) j ) j1 (x j, y j ) j 1. Let ψ 0 (x, y) : ( x, ỹ) j1 (x j, y j ) j. Show that ( x, ỹ) A. Check: for j 1, x 1 ỹ 1 0. For j 2, x j x j+1 and ỹ j ỹ j+1, so ( x j, ỹ j ) {(0, 0), (0, 2), (2, 0)}. Therefore, ( x, ỹ) A. Claim For all (x, y) A, ψ 1 (x, y) A. Proof. We know that ψ 1 ( j1 (x j, y j ) j ) (2 1, 0 1 ) + j1 (x j, y j ) j 1. Let ψ 1 (x, y) : ( x, ỹ) j1 (x j, y j ) j. Show that ( x, ỹ) A. Check: for j 1, x 1 2, ỹ 1 0; for j 2, x j x j+1 and ỹ j ỹ j+1, so ( x j, ỹ j ) {(0, 0), (0, 2), (2, 0)}. Therefore, ( x, ỹ) A. Claim For all (x, y) A, ψ 2 (x, y) A. Proof. We know that ψ 2 ( j1 x j, y j ) j (0 1, 2 1 ) + j0 (x j, y j ) j 1. Check: for j 1, x 1 0, ỹ 1 2; for j 2, x j x j+1 and ỹ j ỹ j+1, so ( x j, ỹ j ) {(0, 0), (0, 2), (2, 0)}. Therefore, ( x, ỹ) A. Now show that A ψ 0 (A) ψ 1 (A) ψ 2 (A). Again, let (x, y) j1 (x j, y j ) j. Now let (ˆx, ŷ) ( j2 x j j+1, j2 y j j+1 ). Notice that ˆx j x j+1, ŷ j y j+1, so (ˆx, ŷ) A. We will show that, for all (x, y) A, ψ k (ˆx, ŷ) (x, y) for some k {0, 1, 2}. There are three cases: (x 1, y 1 ) (0, 0), (x 1, y 1 ) (2, 0), and (x 1, y 1 ) (0, 2). Claim When (x 1, y 1 ) (0, 0), ψ 0 (ˆx, ŷ) (x, y). Proof. ψ 0 ( j2 (x j, y j ) j+1 ) j2 (x j, y j ) j (x, y), since x 1 y 1 0. Claim When (x 1, y 1 ) (2, 0), ψ 1 (ˆx, ŷ) (x, y).

33 25 Proof. ψ 1 ( j2 (x j, y j ) j+1 ) (2 1, 0 1 ) + j2 (x j, y j ) j (x, y), since x 1 2 and y 1 0. Claim When (x 1, y 1 ) (0, 2), ψ 2 (ˆx, ŷ) (x, y). Proof. ψ 2 ( j2 (x j, y j ) j+1 ) (0, 2 1 ) + ( j2 (x j, y j ) j (x, y), since x 1 0 and y 1 2. Now, show that A is compact. It is obvious that A is bounded in R 2, we need only show that it is closed. Let {(x k, y k )} k1 A be a Cauchy sequence. Notice that by construction, x k j1 x k j j, x j {0, 2} is contained in the Cantor- set C, a well-know closed set. The same is true for y k. Therefore, x k x, with x C, and y k y, with y C. Therefore we have that (x k, y k ) (x, y), with (x, y) C C. We need to show that (x, y) A, that is, that for (x, y) j1 (x j, y j ) j, (x j, y j ) (2, 2) for any j. Suppose that (x j, y j ) (2, 2) for some j 1. By assumption, (x kj, y kj ) (2, 2) for any j. Consider: (x, y) (x k, y k ) 2 (x x k ) 2 + (y y k ) 2 2 (x j x kj ) j + (y j y kj ) j j0 j0 If (x j, y j ) (2, 2), then for every k, we have (x j x kj ) 2 or (j j y kj ) 2, or both. So for every k: 2 (x, y) (x k, y k ) 2 (2) j Therefore, (x k, y k ) cannot converge to (x, y) and we have a contradiction. So (x, y) A and A is closed, thus compact. Therefore, since A is compact and A ψ 0 (A) ψ 1 (A) ψ 2 (A), we must have that A S. Another characterization of S:

34 26 Figure 4.1 A fifth iterative approximation of S starting at zero. Proposition S {(x, y) R 2 : (x, y) C C and x + y C } where C is the standard middle-thirds Cantor- set mentioned earlier, the compact attractor set of τ 0 (x) x/, τ 1 (x) x/ + 2/. Proof. Let A {(x, y) R 2 x C, y C, and x + y C }. Again, it is sufficient to show that A ψ 0 (A) ψ 1 (A) ψ 2 (A). First, we show that ψ 0 (A) ψ 1 (A) ψ 2 (A) A. Claim ψ j (x, y) C C for j 0, 1, 2 and (x, y) A. Proof. Let (x, y) A ψ 0 (x, y) (x/, y/) (τ 0 (x), τ 0 (y)). ψ 1 (x, y) (x/ + 2/, y/) (τ 1 (x), τ 0 (y)). ψ 2 (x, y) (x/, y/ + 2/) (τ 0 (x), τ 1 (y)).

35 27 Since x C and y C, ψ j (x, y) C C for j 0, 1, 2. Claim Let ψ j (x, y) ( x, ỹ), for j 0, 1, 2, (x, y) A. We will show that x + ỹ C. Proof. ψ 0 (x, y) (x/, y/), x/ + y/ 1/(x + y) τ 0 (x + y)/ C. ψ 1 (x, y) (x/ + 2/, y/), x/ + 2/ + y/ (x + y)/ + 2/ τ 1 (x + y) C. ψ 2 (x, y) (x/, y/ + 2/), x/ + y/ + 2/ (x + y)/ + 2/τ 1 (x + y) C. ψ 2 (A): These two claims give us ψ 0 (A) ψ 1 (A) ψ 2 (A) A. Now, we show A ψ 0 (A) ψ 1 (A) Let (x, y) A. There are three cases: 1. x 1/ and y 1/. 2. x 2/ and y 1/.. x 1/ and y 2/. Since x, y C, neither x nor y can be in (1/, 2/). Since x + y C, x and y can t both be greater than 2/. So the three cases given cover all of A. Claim (Case 1). For (x, y) A, x 1/, y 1/, and (x + y) 1/, (x, y) ψ 0 (A). Proof. Let ( x, ỹ) (x, y). Since ψ 0 (x, y) (x, y), we will get (x, y) ψ 0 (A) if (x, y) A. For (x, y) A we need to show three things: x C : Since x C, x τ 0 ( x) or x τ 1 ( x) for x C. Since x C x 0, τ 1 ( x) x/ + 2/ 2/; since x 1/ < 2/, we cannot have τ 1 ( x) x. Therefore, we must have τ 0 ( x) x; thus x x C. y C : Since y C, y < 1/, y < 1. And, τ 0 (ỹ) 1 (y) y. Thus, ỹ y C. x + y C : We need to show that x + y 1/. Suppose not. Since x 1/, and y 1/, x + y 2/. Since x + y C, this means that x + y 1/. Therefore, x + y < 1. Then τ 0 (x + y) x + y C by assumption. Therefore, x + y C.

36 28 Claim (Case 2). For (x, y) A, x 2/ and y 1/, (x, y) ψ 1 (A). Proof. Let ( x, ỹ) (x 2, y). Since ψ 1 (x 2, y) 1 (x 2 + 2, y) (x, y), if ( x, ỹ) A, (x, y) ψ 1 (A). For (x 2, y) A we need to show three things: x 2 C : Since x C, x τ 0 ( x) or x τ 1 ( x) for x C. Since x 2/ and x 1, we must have that x τ 1 ( x); thus x x 2 C. y C : Since y C, y < 1/, y < 1. And, τ 0 (ỹ) 1 (y) y. Thus, ỹ y C. x 2 + y C : τ 1 (x 2 + y) 1 (x 2 + y) 2/ x + y C by assumption, so x + ỹ C. Claim (Case ). For (x, y) A, x 1/ and y 2/, (x, y) ψ 2 (A). Proof. Take ( x, ỹ) (x, y 2). Since ψ 2 (x, y 2) 1 (x, y 2+2) (x, y), if ( x, ỹ) A, (x, y) ψ 2 (A). Show that (x, y 2) A: For (x, y 2) A we need to show three things: x C : Since x C, x τ 0 ( x) or x τ 1 ( x) for x C. Since x C x 0, τ 1 ( x) x/ + 2/ 2/; since x 1/ < 2/, we cannot have τ 1 ( x) x. Therefore, we must have τ 0 ( x) x; thus x x C. y 2 C : Since y C, y τ 0 (ỹ) or y τ 1 (ỹ) for ỹ C. Since y 2/ and ỹ 1, we must have that y τ 1 (ỹ) 1 (y 2) + 2/ y; thus ỹ y 2 C. x + y 2 C : Since x 2/ and y 1/, 0 < x + y 2 < 1, and τ 1 (x 2 + y) 1 (x 2 + y) 2/ x + y C by assumption, so x + ỹ C. Now, we show that A is compact. A is obviously bounded, so we need only show it is closed.

37 29 Let {(x j, y j )} k1 A be a Cauchy sequence. Since x k C, y k C, x k x C and y k y C, we have that {x k, y k } (x, y) C C. Moreover, by continuity, x k + y k x + y, and since x k + y k C for all k, x + y C and (x, y) A. Therefore, A is compact, and A S. 4.2 Our orthonormal basis This section will use Theorem to show that {e t,t/2 t Z} is an orthonormal basis for L 2 (ν ). Following Dutkay and Jorgensen (9), we begin with a dual iterated function system of the form R(x, y) + L, for R I and L {(0, 0), (2, 1), (4, 2)}. Check first that (R, B, L) are in Hadamard duality (Definition 2.2.7), that is, that the ( ) matrix M 1 e 2πil R 1 b is Hadamard: Claim M 1 M 1 I. l L,b B Proof. We calculate: M 1 1 e 2πi(1/) e 2πi(2/) 1 e 2πi(2/) e 2πi(4/) (4.5) Proof is by basic matrix multiplication. Recall the Definition For the fractal S, B {(0, 0), (2, 0), (0, 2)}, therefore: W B (x, y) e 4πix + e 4πiy ( + 2 cos (4πx) + 2 cos (4πy) + 2 cos (4π (x y))). W B (x, y) 1 if and only if e 4πix e 4πiy 1, that is, when (x, y) Z/2 Z/2. W B (x, y) 0 if and only if 1 + e 4πix + e 4πiy 0; that is, when e 4πix + e 4πiy 1; or when (x, y) (1/ + Z/2, 1/6 + Z/2) (1/6 + Z/2, 1/ + Z/2). Proposition X X L {(2t, t) : t [0, 1]}, that is, the straight line between (0, 0), and (2, 1).

38 0 Proof. Let: l 0 (x, y) (x/, y/) τ (0,0) (x, y) (4.6) l 1 (x, y) (x/ + 2/, y/ + 1/) τ (2,1) (x, y) (4.7) l 2 (x, y) (x/ + 4/, y/ + 2/) τ (1,2) (x, y) (4.8) so that X L is invariant under {l 0, l 1, l 2 }. It is clear that T {(2t, t) : t [0, 1]} is compact, so we need only show that T l 0 (T ) l 1 (T ) l 2 (T ). Let (x, y) (2t, t) for some t [0, 1]. Then: l 0 (2t, t) (2t/, t/) (2s, s) with s t/ [0, 1/] [0, 1]. l 1 (2t, t) (2t/ + 2/, t/ + 1/) (2s, s) with s t/ + 1/; since t [0, 1], t/ [0, 1/ so t/ [1/, 2/] [0, 1]. l 2 (2t, t) (2t/ + 4/, t/ + 2/) (2s, s) with s t/ + 2/; since t [0, 1], t/ [0, 1/ so t/ [2/, 1] [0, 1]. Now for (x, y) (2t, t) find some ( x, ỹ) (2s, s), j {0, 1, 2} with l j ( x, ỹ) (x, y). Case 1: y t [0, 1/). Take s t; then s [0, 1] and l 0 (2s, s) l 0 (6t, t) (2t, t) (x, y). Case 2: y t [1/, 2/), so that t [1, 2). Take s t 1; then s [0, 1] and l 1 (2s, s) l 1 (6t 2, t 1) (2t, t) (x, y). Case : y t [2/, 1], so that t [2, ]. Take s t 2; then s [0, 1] and l 2 (2s, s) l 1 (6t 4, t 2) (2t, t) (x, y). in X L. To find W B -cycles on X L, first we find which of the points in the lattice 1 2 Z 1 2Z are also By construction, each y [0, 1] has at most one x [0, 2] with (x, y) X L ; so we really only need to check three possibilities: y 0 x 2(0) 0: (0, 0) X L. y 1/2 x 1(1/2) 1: (1, 1/2) X L. y 1 x 2(1) 2: (1, 1/2) X L.

39 1 Figure 4.2 Plot of the fifth iteration of 1/(x, y) + L.

40 2 So the only three points in 1 2 Z 1 2 Z X L are (0, 0), (1, 1/2), and (2, 1). All of these are cycles, in fact, they are fixed points: l 0 (0, 0) (0, 0) (4.9) l 1 (1/2, 1) 1 (1/2, 1) + (1/, 2/) (1/2, 1) (4.10) l 2 (1, 1/2) 1 (1, 1/2) + (2/, 1/) (1, 1/2). (4.11) Therefore, the W B -cycles are exactly (0, 0), (1, 1/2) and (1/2, 1). Proposition Let: ρ 0 (x, y) (x, y) (4.12) ρ 1 (x, y) (x, y) + (2, 1) (4.1) ρ 2 (x, y) (x, y) + (4, 2) (4.14) so that {ρ j (x, y)} j0,1,2 R(x, y) + B. Let l 1 (x, y) {ρ 0 (x, y), ρ 1 (x, y), ρ 2 (x, y)}, then for n 2, l n (x, y) {ρ 0 (s, t) (s, t) l n 1 (x, y)} {ρ 1 (s, t) (s, t) l n 1 (x, y)} {ρ 2 (s, t) (s, t) l n 1 (x, y)}. Then let L(x, y) n N l n(x, y). We will show that: L(0, 0) L( 1/2, 1) L( 1, 1/2) {(t, t/2) t Z}. The proof is in the form of three Claims. Claim L(0, 0) {(2t, t) t Z, t 0}. Proof. ( ): Let L {(2t, t) t Z, t 0}. We have: l 1 (0, 0) {(0, 0), (2, 1), (4, 2)} L. By construction, we need now only show that if (x, y) L, each of ρ 0 (x, y), ρ 1 (x, y), ρ 2 (x, y) is in L : ρ 0 (2t, t) (2t, t) (6t, t); with t 0, t 0 and ρ 0 (2t, t) (2(t), t) L. ρ 1 (2t, t) (2t, t) + (2, 1) (6t + 2, t + 1); with t 0, t and ρ 1 (2t, t) (2(t + 1), t + 1) L.

41 Figure 4. Plot of l (0, 0) l ( 1, 1/2) l ( 2, 1).

42 4 ρ 1 (2t, t) (2t, t) + (4, 2) (6t + 4, t + 2); with t 0, t and ρ 2 (2t, t) (2(t + 2), t + 2) L. ( ): To show that each (2t, t) with t 0 is in L(0, 0), we write t N j0 t j j, t j {0, 1, 2}, then induct on N: Base case, N 0, t 0, 1, 2: t 0: ρ 0 (0, 0) (0, 0); t 1: ρ 1 (0, 0) (2, 1); t 2: ρ 2 (0, 0) (4, 2). Induction step: Suppose (2t, t) L(0, 0) for all t n j0 t j j, t j {0, 1, 2}, n < N; let t N j0 t j j. Case 1: t 0 0. Then t/ N 1 j0 t j+1 j Z, t/ 0; so (2t/, t/) L(0, 0) by inductive assumption; then (2t, t) ρ 0 (2t/, t/) L(0, 0). Case 2: t 0 1. Then (t 1)/ N 1 j0 t j+1 j Z, (t 1)/ 0, so (2(t 1)/, (t 1)/) L(0, 0) by inductive assumption; therefore (2t, t) ρ 1 ((2(t 1)/, (t 1)/) L(0, 0). Case : t 0 2. Then (t 2)/ N 1 j0 t j+1 j Z, (t 2)/ 0, so (2(t 2)/, (t 2)/) L(0, 0) by inductive assumption; therefore (2t, t) ρ 1 ((2(t 2)/, (t 2)/) L(0, 0). Claim L( 2, 1) {(2t, t) t Z, t 1}. Proof. ( ): Let L {(2t, t) t Z, t 1}. We have: l 1 ( 2, 1) {( 6, ), ( 4, 2), ( 2, 1)} L. By construction, we need now only show that if (x, y) L, each of ρ 0 (x, y), ρ 1 (x, y), ρ 2 (x, y) is in L :

43 5 ρ 0 (2t, t) (2t, t) (6t, t); with t 1, t 1 and ρ 0 (2t, t) (2(t), t) L. ρ 1 (2t, t) (2t, t) + (2, 1) (6t + 2, t + 1); with t 1, t and ρ 1 (2t, t) (2(t + 1), t + 1) L. ρ 1 (2t, t) (2t, t) + (4, 2) (6t + 4, t + 2); with t 1, t and ρ 2 (2t, t) (2(t + 2), t + 2) L. ( ): To show that each (2t, t) with t 1 is in L( 2, 1), we write t N j0 t j j, t j {0, 1, 2}, then induct on N. Base case, N 0, t 1, 2 (note that t 0 is not in the set): t 1: ρ 2 ( 2, 1) ( 2, 1); t 2: ρ 1 ( 2, 1) ( 4, 2). Induction step: Suppose (2t, t) L( 2, 1) for all t n j0 t j j, t j {0, 1, 2}, n < N, t 1; let t N j0 t j j. Further, suppose that t, since we already know ( 2, 1) and ( 4, 2) L( 2, 1). Case 1: t 0 0. Then t/ N 1 j0 t j+1 j, and since t, t/ 1. So we have that (2t/, t/) L( 2, 1) by inductive assumption, therefore (2t, t) ρ 0 (2t/, t/) L( 2, 1). Case 2: t 0 1. Case : t 0 2. Then (t 1)/ N 1 j0 t j+1 j, and since t, (t 1)/ 1. So we have that (2(t 1)/, (t 1)/) L( 2, 1) by inductive assumption, therefore (2t, t) ρ 1 (2t/, t/) L( 2, 1). Then (t 2)/ N 1 j0 t j+1 j, and since t, (t 2)/ 1. So we have that (2(t 2)/, (t 2)/) L( 2, 1) by inductive assumption, therefore (2t, t) ρ 2 (2t/, t/) L( 2, 1).

44 6 Claim L( 1, 1/2) {(2t, t) t Z/2, t / Z}. Proof. ( ): Let L {(2t, t) t Z/2, t / Z}. We have: l 1 ( 2, 1) {(, /2), ( 1, 1/2), (1, 1/2)} L. By construction, we need now only show that if (x, y) L, each of ρ 0 (x, y), ρ 1 (x, y), ρ 2 (x, y) is in L : ρ 0 (2t, t) (2t, t) (6t, t); with t Z/2, t / Z, t will also be in Z2\Z, so ρ 0 (2t, t) L. ρ 1 (2t, t) (2t, t) + (2, 1) (6t + 2, t + 1); with t Z/2, t / Z, t + 1 will also be in Z2 \ Z, so ρ 1 (2t, t) L. ρ 1 (2t, t) (2t, t) + (4, 2) (6t + 4, t + 2); with t Z/2, t / Z, t + 2 will also be in ( ): Z2 \ Z, so ρ 2 (2t, t) L. One way of expressing the set {t Z/2, t / Z} is as the set {t t + 1/2 Z}. For our induction argument, we write the integer t + 1/2 in balanced ternary: t + 1/2 N j0 t j j, t j {0, 1, 1}. With this expression, we can induct on N and conveniently cover all of our set L. Base case, N 0, t 1/2, 1/2, /2: t 1/2: ρ 1 ( 1/2, 1) ( 1/2, 1); t 1/2: ρ 2 ( 1/2, 1) (1, 1/2); t /2: ρ 0 ( 1/2, 1) (, /2). Induction step: Suppose (2t, t) L( 1, 1/2) for all t + 1/2 n j0 t j j, t j {0, 1, 1}, n < N; let t + 1/2 N j0 t j j. Case 1: t 0 1. Then t + 1/2 N j0 t j j 1 + N j1 t j j, or t + /2 N j1 t j j. Therefore, t/ + 1/2 N 1 j0 t j+1 j, and by our induction assumption, (2t/, t/) L( 1, 1/2). Thus (2t, t) ρ 0 (2t/, t/) L( 1, 1/2).

45 7 Case 2: t 0 0. Then t + 1/2 N j0 t j j N j1 t j j, or (t 1) + /2 N j1 t j j. Therefore, (t 1)/ + 1/2 N 1 j0 t j+1 j, and by our induction assumption, (2(t 1)/, (t 1)/) L( 1, 1/2). Thus (2t, t) ρ 1 (2(t 1)/, (t 1)/) L( 1, 1/2). Case : t 0 1. Then t + 1/2 N j0 t j j 1 + N j1 t j j, or (t 2) + /2 N j1 t j j. Therefore, (t 2)/ + 1/2 N 1 j0 t j+1 j, and by our induction assumption, (2(t 2)/, (t 2)/) L( 1, 1/2). Thus (2t, t) ρ 2 (2(t 2)/, (t 2)/) L( 1, 1/2). Proof of Proposition Putting together Claims 4.2.4, 4.2.5, 4.2.6, we get that L(0, 0) L( 1/2, 1) L( 1, 1/2) {(t, t/2) t Z}. Claim Let Λ R 2 be the smallest set that contains C for all W B -cycles C, and such that SΛ + L Λ. Then Λ {(t, t/2) t Z}. Proof. Recall that S R T I, and that L {(0, 0), (2, 1), (4, 2)}. Let L 0 {(t, t/2) t Z}. By Claims 4.2.4, 4.2.5, 4.2.6, we know that L(0, 0) L( 2, 1) L( 1, 1/2) L 0. For Λ to contain C for all W B cycles C, and also contain SΛ + L it must contain at least these three sets, so we have L 0 Λ. For L 0 Λ, we need only show that it itself has the necessary properties; we know it contains C for all W B cycles C, so we need only show SL 0 +L L 0, that is, {ρ 0 (t, t/2), ρ 1 (t, t/2), ρ 2 (t, t/2)} L 0 for all (t, t/2) L 0. Let (t, t/2) with t Z. ρ 0 (t, t/2) (t, t/2); if t Z, t Z, so ρ 0 (t, t/2) L 0. ρ 1 (t, t/2) (t + 2, t/2 + 1) (t + 2, (t + 2)/2); if t Z, t + 2 Z, so ρ 1 (t, t/2) L 0. ρ 1 (t, t/2) (t + 4, t/2 + ) (t + 4, (t + 4)/2); if t Z, t + 4 Z, so ρ (t, t/2) L 0.

46 8 Therefore, by Theorem 2.2.1, if the transversality of the zeros condition is satisfied, {e 2πi(t,t/2) (x,y) t Z} {e 2πitx e πity t Z} is an orthonormal basis for L 2 (ν ) The Transversality of the Zeros Condition Recall the transversality of the zeros condition from Definition The function W X on X L satisfies the transversality of the zeros condition if: (a) If x X L is not a cycle, then there exists k x 0 such that, for k k x, {τ l1 τ l2 τ lk x : l 1,... l n L} does not contain any zeros of W ; (b) If {x 0, x 1,..., x p } are on a cycle with x 1 τ l (x 0 ) for some l L, then for every y τ l (x 0 ), y x 1 is either not on a cycle or W (y) 0. For the current case, B {(0, 0), (2, 0), (0, 2)}, so W B (x, y) e 4πix + e 4πiy 2 (4.15) 1 ( + 2 cos (4πx) + 2 cos (4πy) + 2 cos (4π (x y))) (4.16) 9 Therefore, W B (x, y) 1 if and only if e 4πix e 4πiy 1, that is, when (x, y) Z/2 Z/2. Furthermore, W B (x, y) 0 if and only if 1 + e 4πix + e 4πiy 0; that is, when e 4πix + e 4πiy 1; or when (x, y) (1/ + Z/2, 1/6 + Z/2) (1/6 + Z/2, 1/ + Z/2). We show our set X L satisfies Condition (b) first. Recall from Proposition that X X L {(2t, t) : t [0, 1]}. Lemma (2t, t) X L is on a cycle if and only if t k/( n 1) for some integer k, n N. Proof. By definition, (2t, t) is on a cycle if and only if (2t, t) r n (2t, t) for some n N, where r is the common right inverse of l 1, l 2, l : r(2t, t) (2s, s) with s t mod 1.