Spectral Hutchinson-3 Measures and Their Associated Operator Fractals

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1 University of Colorado, Boulder CU Scholar Mathematics Graduate Theses & Dissertations Mathematics Spring Spectral Hutchinson- Measures and Their Associated Operator Fractals Ian Long University of Colorado at Boulder, Follow this and additional works at: Part of the Mathematics Commons Recommended Citation Long, Ian, "Spectral Hutchinson- Measures and Their Associated Operator Fractals" (2017. Mathematics Graduate Theses & Dissertations This Dissertation is brought to you for free and open access by Mathematics at CU Scholar. It has been accepted for inclusion in Mathematics Graduate Theses & Dissertations by an authorized administrator of CU Scholar. For more information, please contact

2 Spectral Hutchinson- Measures and Their Associated Operator Fractals by: Ian Long B.A. Vassar College, 2011 M.A. University of Colorado at Boulder, 2015 A thesis submitted to the Faculty of the Graduate School of the University of Colorado in partial fulfillment of the requirement for the degree of Doctor of Philosophy Department of Mathematics 2017 i

3 ii This thesis entitled: Spectral Hutchinson- Measures and Their Associated Operator Fractals written by Ian Long has been approved for the Department of Mathematics Dr. Judith Packer Dr. Keri Kornelson Date The final copy of this thesis has been examined by the signatories, and we find that both the content and the form meet acceptable presentation standards of scholarly work in the above mentioned discipline.

4 iii Long, Ian (Ph.D. Mathematics Spectral Hutchinson- Measures and Their Associated Operator Fractals Thesis directed by Professor Judith Packer Abstract In 181, Hutchinson showed that for each iterated function system on the real line, there exists a unique probability measure, called a Hutchinson measure or invariant measure, whose support is the attractor of the of the iterated function system. A common question that has been asked about measures among this class is whether or not they are spectral. That is, for a given Hutchinson measure µ does there exist a set of complex exponential functions which forms an orthonormal basis for L 2 (µ? It has since been observed that such measures are capable of having infinitely many spectra. These multiple spectra give rise to canonically-defined unitary operators on the L 2 -spaces of Hutchinson measures. Moreover, one of these operators found on the L 2 -space of a Hutchinson measure known as the 1 - Cantor measure was shown to have a unique fractal-like construction and was dubbed the Operator Fractal by Jorgensen, Kornelson, and Shuman in In this publication, we wish to determine if other operator fractals exist on the L 2 -spaces of Hutchinson measures. Furthermore, acknowledging that it is not known in general precisely which Hutchinson measures are spectral, we divulge an if and only if condition which determines whether or not a particular class of Hutchinson measures associated to iterated function systems containing elements is spectral. We demonstrate that such a measure is spectral if and only if its components are well-spaced. We then find an explicit spectrum of each such spectral measure.

5 iv Dedication To my parents. Thanks for making sure that I always had good math teachers.

6 v Acknowledgements I would like to thank my advisor Judy Packer for the immense amount of guidance, help, and support that she has provided over the last several years. In particular, I am very grateful that she introduced me to the subject of fractal measures, which I have truly enjoyed studying during my time at the University of Colorado. I would also like to thank Keri Kornelson for several helpful discussions which contributed greatly to my understanding of fractal measures.

7 vi Contents 1 Introduction Hutchinson Measures and Their Spectra Interspectral Operators, the Operator Fractal, and Cuntz Algebras Operator Fractals Associated with Bernoulli Convolution Measures 1 The Structure of L 2 (µ 1 and B(L 2 (µ 1 2 For Which p N Is pγ(l a Spectrum for µ 1? 0 5 The Structure of L 2 (µ 1 and B(L 2 (µ 1 (a Generalization of Section 5 6 Operator Fractals Associated with Certain Hutchinson- Measures 1 7 Which Hutchinson- Measures Are Spectral? A Necessary Condition on R for the Resulting Measure to Be Spectral Well-Spacedness About the Origin and a Canonical Spectrum Well-Spacedness About the Origin and an Alternative Spectrum µ R,B Is Spectral If and Only If (R, B Is Well-Spaced The Structure of L 2 (µ R,B and B(L 2 (µ R,B (General Case 68 Additive Spectra of µ 1 10 Future Work A Generalization of Section 5 to Cases Where B > Is Prime 8 References 5 80

8 1 1 Introduction 1.1 Hutchinson Measures and Their Spectra For over three decades, there has been a great deal of attention paid to a family of fractal measures first discovered by Hutchinson [Hut81]. The definition of these measures depends on that of an iterated function system: Definition 1.1. Let (X, d be a complete metric space. An iterated function system or IFS is a set of contraction mappings {τ 1,, τ n } on X. Given an IFS I = {τ 1,, τ n } on a complete metric space it is well-known (see the previously given reference that there is a unique compact subset A X which is invariant with respect to I. That is, there is such an A that satisfies n A = τ i (A. i=1 This A is referred to as the attractor of the IFS. To each IFS I = {τ 1,, τ n } on R, Hutchinson showed that there exists a unique Borel regular probability measure µ which satisfies a self-similarity property on Borel subsets of R: µ( = 1 n n j=1 µ(τ 1 j (. Furthermore, this measure is supported on the attractor of I. In virtually all known cases of interest, this attractor is a Cantor set. Thus, the vast majority of these Hutchinson measures are singular with respect to Lebesgue measure. When integrating over such a measure, the self-similarity property equates to the following

9 2 identity: R fdµ = 1 n n f τdµ. j=1 R One of the best known and most oft-studied of these measures is the 1-Cantor measure µ 1, which is the Hutchinson measure associated to the IFS { x 1 x, x 1 x + 1 }. 2 The support of this measure is a Cantor set formed by removing the second and fourth fourths at each iteration of its creation (as opposed to removing the second third at each iteration to create the classic Cantor set. Initial enthusiasm for µ 1 that Jorgensen and Pedersen showed that it was a spectral measure [JP8]. was a result of the fact Definition 1.2. A measure µ on X R is said to be spectral if there exists Λ R such that {e λ : X C, e λ (x = e 2πiλx λ Λ} is an orthonormal basis for L 2 (µ. The set Λ is referred to as a spectrum for µ. Jorgensen and Pedersen were able to show that the 1 -Cantor measure has spectrum Λ = { m } k l k : l k {0, 1}, m N 0 k=0 = {0, 1,, 5, 16, 17, 20, 21, }. They also demonstrated that there exist a wide variety of Hutchinson measures which are not spectral. For example, the 1-Cantor measure µ 1, associated with the IFS { x 1 x, x 1 x + 2 },

10 is not spectral. In fact, Jorgensen and Pedersen were able to show that a maximal orthogonal set of exponentials in L 2 (µ 1 contains at most two elements. The method they used to determine that a given Hutchinson measure is spectral involves finding Hadamard triples: Definition 1.. Let R Z with R 2 and B, L R with 0 < B = L = N <. Assume further that 0 L and R n b l Z for all b B, l L. Then we say that (R, B, L is a Hadamard triple if the N N matrix H R,B,L = 1 N [ e 2πib l R ] b B,l L is unitary. We can relate the definition of a Hadamard triple to the definition of a Hutchinson measure in the following way: choose R Z with R 2 and B = {b 1,, b N } Z. Since R 2, we can define an IFS I R,B = {τ i : R R, τ i (x = R 1 (x + b i, 1 i N}. Therefore there is a unique Hutchinson measure µ R,B on [0, 1] satisfying the self-similarity property µ R,B ( = 1 N N µ R,B (R b i. If R and B are chosen correctly, then µ R,B is spectral: i=1 Theorem 1.. [LW02] Let R Z with R 2 and let B Z be a finite set of integers. If there exists L Z with 0 L such that (R, B, L is a Hadamard triple, then µ R,B is spectral. It is worth noting that higher-dimensional analogues to the Hutchinson measures defined here exist. In particular, if one chooses R to be an N N integer matrix which is expansive

11 (all of its eigenvalues have modulus greater than 1 and B R N with B <, one can define a Hutchinson measure µ R,B which satisfies an analogous self-similarity condition. Furthermore, if L R N exists such that L = B and the matrix H R,B,L = 1 ] [e 2πi R 1 b,l N b B,l L is unitary, Dutkay and Lai proved a result analogous to Theorem 1. [DHL15]. Other recent efforts have been made to determine which conditions one can place on R and B to obtain necessary and sufficient conditions for the spectrality of a higher-dimensional Hutchinson measure µ R,B, as well as to find explicit spectra of specific higher dimensional Hutchinson measures [Li07], [DJ07]. Additionally, there have been a number of publications in recent years dedicated to finding Fourier frames (as opposed to bases for Hutchinson measures, see e.g. [DR16],[PW15], [DHW11]. While these related areas of study will not be considered again in this thesis, they are mentioned to demonstrate the vastness of the portion of this field which still consists of open problems. Example 1.5. Switching notation, we may denote the 1 -Cantor measure µ 1 as µ R,B if we choose R =, B = {0, 2}. If we also make the choice L = {0, 1} we have that the matrix H R,B,L is given by H R,B,L = 1 2 e(2πi( e (2πi( e (2πi( e (2πi( = , which is a unitary matrix. Thus, by Theorem 1. we have that µ 1 choice of L = {0, 1} and the spectrum is spectral. Note our { m } Λ = k l k : l k {0, 1}, m N 0 k=0

12 5 for µ 1 given above appear to be related. explained in Example This relation is not a coincidence and will be Additionally, one may construct µ 1 as µ R,B when R =, B = {0, 2}. One may notice with relative ease that choosing L 0 such that (R, B, L is a Hadamard triple is not possible in this case. A great deal of information about these measures can be gained by examining their Fourier transforms: given a Hutchinson measure µ R,B, define its Fourier transform by µ R,B (t = R e 2πitx dµ R,B (x for a given t R. When we define Hutchinson measures by choosing an R and a B (as we shall do from now on, we are able to conveniently express the Fourier transform of µ R,B as an absolutely convergent infinite product in terms of R and B by using the self-similarity condition of the Hutchinson measure: µ R,B (t = ( d=1 b B e 2πibt R ( 1 = B b B ( 1 = B e 2πibt R b B ( t µ R,B R ( 1 B e 2πibt R d. b B e 2πibt R 2 ( t µ R,B R 2 Lemma 1.6. The infinite product expression for µ R,B above is absolutely convergent for all t R, thus is zero if and only if one of the terms in the product is zero. Proof. It is known (for example, see [Ahl7] pg. 12 that the infinite product above will absolutely converge if and only if the series ( 1 B d=1 b B R d 1 e 2πibt

13 6 converges. Noting that 0 ( 1 B d=1 b B B R d 1 1 B e 2πibt j=1 1 + e 2πbj t R d, d=1 showing that this product converges will follow immediately if we can show that 1 + e 2πbt R d converges for each b B. Since R 2, this follows by the Ratio Test: lim d 1 + e 2πibt R d e 2πibt R = d d=1 lim e πibt R d+1 d e πibt R lim d d = lim sin ( πbt R d+1 d sin ( πbt R d = lim cos ( π d = 1 R < 1. e πibt R d+1 e πibt R d e πibt R d+1 e πibt R d R d+1 πbtr d 1 ln(r cos ( π R d πbtr d ln(r Theorem 1. can tell us that a Hutchinson measure has a spectrum, but does not explicitly find a specific spectrum for the measure. As was implied above, the spectrum of a Hutchinson measure µ R,B defined in terms of a Hadamard triple (R, B, L is likely to depend on L. Definition 1.7. Choose R Z and L a finite subset of R. We define the following sets: X(L = Γ(L = { } R k l k : l k L k=1 { m } R k l k : l k L, m N 0 k=0 Given µ R,B chosen along with a set L such that (R, B, L is a Hadamard triple, the set Γ(L

14 7 arises as a likely candidate for a spectrum. Definition 1.8. Let (R, B, L be a Hadamard triple with R Z, B, L R, and B = L = N. Define X(L as above and define the IFS {σ l : l L} on R such that σ l (x = R 1 (x + l for all l L. Then an L-cycle C is a finite set {x 1,, x n } X(L such that there exist l x1,, l xn L (not necessarily distinct satisfying σ lxi (x i = x i+1 if 1 i n 1 and σ lxn (x n = x 1. C is called B-extreme if χ B (x = 1 for all x C, where χ B (x = 1 N e 2πibx. b B Theorem 1.. ([DJ12] Let (R, B, L be a Hadamard triple with 0 B L. Define Γ(L as above. Then Γ(L is a spectrum for µ R,B if and only if the only B-extreme L-cycle is {0}. The following lemma is quite useful when attempting to find B-extreme L-cycles: Lemma ([DJ12] If C is a B-extreme L-cycle (where (R, B, L is a Hadamard triple then for all x C and all b B it must be the case that bx Z. Example Let R =, B = {0, 2} so that µ R,B is the 1 -Cantor measure. Then if we choose L = {0, 1} we know from the previous example that (R, B, L is a Hadamard triple. Our possible source of a B-extreme L-cycle is then the set [ 1 2 Z k 0, k=1 ] k 1 = 12 [ Z 0, 1 ] = {0}. k=1 It is immediate then that {0} is the only B-extreme L-cycle and thus Γ(L is a spectrum for µ R,B (as we previously claimed but did not show. Example Let R =, B = { 8, 0, 8} and define µ R,B to be the invariant measure associated to the IFS I R,B = { τ 1, τ 0, τ 1 : [0, 1] [0, 1], τ k (x = 1 x + 8k }.

15 8 Then if we choose L = {0,, 6}, one can check that (R, B, L is a Hadamard triple. Define the IFS in terms of L as well I R,L = {σ l0, σ l1, σ l2 : [0, 1] [0, 1], σ lk (x = 1 (x + k }. Our possible source of a B-extreme L-cycle is the set [ 1 8 Z k 0, k=1 ] k 6 = 18 [ Z 0, ] = k=1 { 0, 1 8, 1, 8, 1 2, 5 8, }. We have that σ l2 ( =. Thus, { } is a B-extreme L-cycle and Γ(L is not a spectrum for µ R,B. One can show (as we will in Section 7. that 1 8 Γ(L is a spectrum for µ R,B. None of the language used thus far would suggest that a given Hutchinson measure µ R,B only has one spectrum, as one could theoretically choose multiple sets L for which (R, B, L a Hadamard triple. For example, Jorgensen, Kornselson, and Shuman show using extreme cycles that 5Γ(L (where Γ(L is the spectrum found above for the 1 -Cantor measure is also a spectrum for µ 1 ([JKS1b]. We will reference this fact later in the next section (after the statement of Theorem 1.16 to introduce a remarkable operator on L 2 (µ 1. Subse- quently, Dutkay and Haussermann were able to show in [DH16] that p n Γ(L is a spectrum for the 1 -Cantor measure for any odd prime p > (notably, Γ(L is not a spectrum for the 1 -Cantor measure and Li and Xing proved similar results in [LX17]. They also gave conditions under which other odd composite numbers n could be found such that nγ(l is a spectrum for the 1 -Cantor measure. Dutkay and Kraus [DK16] later gave conditions under which nγ(l is a spectrum for µ R,B, where R is even, B = { 0, R 2 }, L = {0, 1}, and n is odd (for such measures, one can show that nγ(l is not a spectrum for µ R,B if n is even.

16 1.2 Interspectral Operators, the Operator Fractal, and Cuntz Algebras The contents of this subsection are heavily based on work of Jorgenson, Kornelson, and Shuman (most notably [JKS12], [JKS1a], [JKS1b] and one may reference their publications for further details. Given a Hadamard triple (R, B, L such that µ R,B is spectral with spectrum Γ(L, one is able to construct a representation of the Cuntz algebra O N in B(L 2 (µ R,B, where N = B (see [Cun77] for the definition of a Cuntz algebra. This representation arises as a direct consequence of the fractal-like nature of Γ(L. Specifically, given that by definition we define Γ(L as the set of all finite sums { m } R k l k : l k L, m N 0, k=0 one may observe that Γ(L may be written as the following union (which is disjoint, see [DJ12]: Γ(L = l L(RΓ(L + l, which is reminiscent of the structure of the attractor of an IFS. Then for each l k L (where we canonically choose l 0 = 0, we can define the map S k 1 B(L 2 (µ R,B by S k 1 e γ = e Rγ+lk, γ Γ(L. Then each of the maps S 0, S 1,, S k 1 is an isometry and the set of these maps satisfies the

17 10 Cuntz-Krieger relations: (i N 1 k=0 S k S k = I, (ii S ks j = δ kj I for all k, j {0,, N 1}. Furthermore, for each N N, some Hutchinson measure µ R,B (and in fact many Hutchinson measures can be found such that B(L 2 (µ R,B supports a representation of O N ([DJ11]. In addition to an associated representation of a Cuntz algebra, those Hutchinson measures µ R,B which have spectra Γ(L and nγ(l for some n 1 give rise to a canonical unitary operator in L 2 (µ R,B. The best understood of these operators is one contained in B(L 2 (µ R,B, where R =, B = { 1, 1}, L = {0, 1}. While this is not quite the 1 -Cantor measure de- scribed above (where we chose B = {0, 2}, for spectrum-related intents and purposes they are the same. The next three lemmas demonstrate this. Lemma 1.1. (Slight Alteration of [JP8] Lemma. Let R Z, R 2 and B Z be arbitrary. Let Γ R be chosen such that {e γ : γ Γ} is orthonormal in L 2 (µ R,B and furthermore define Q R,B,Γ (t = γ Γ µ R,B (t γ 2, t R. Then {e γ : γ Γ} is an orthonormal basis for L 2 (µ R,B if and only if Q R,B,Γ (t 1. Proof. If we assume that {e γ : γ Γ} is an orthonormal basis for L 2 (µ R,B, then the fact that Q R,B,Γ (t 1 for arbitrary t R follows immediately from Parseval s identity applied to e t : 1 = e t 2 = e γ, e t 2 = µ R,B (t γ 2 = Q R,B,Γ (t. γ Γ γ Γ We are of course defining e t e t L 2 (µ R,B and e γ, e t e γ, e t L 2 (µ R,B above, which we will continue to do for notational simplicity throughout the proof.

18 11 Conversely, let us assume that Q R,B,Γ (t = 1 for all t R and {e γ : γ Γ} is orthogonal in L 2 (µ R,B. We claim then that for any t R, e t is contained in the span of {e γ : γ Γ}. Indeed, we have that e t 2 e t, e γ e γ = e t e t, e γ e γ, e t e t, e γ e γ γ Γ γ Γ γ Γ = e t, e t e t, e t, e γ e γ e t, e γ e γ, e t γ Γ γ Γ + t, e γ e γ, γ Γ e e t, e γ e γ γ Γ = e t, e t e t, e γ e t, e γ t, e γ e t, e γ γ Γ γ Γ e + e t, e γ e t, e γ e γ, e γ γ Γ γ Γ = 1 2Q R,B,Γ (t + γ Γ e t, e γ e t, e γ = 0. Thus, if f {e γ : γ Γ}, we must have that f e t for all t R. By the Stone-Weierstrass Theorem (applied to the compact support of µ R,B we must have that this implies f 0 and therefore that {e γ : γ Γ} is an orthonormal basis for L 2 (µ R,B as claimed. Lemma 1.1. Let R Z, R 2, and B Z be an arbitrary finite set and b Z be arbitrary. Let Z( µ R,B be the set of all zeroes of µ R,B and define Z( µ R,B+b analogously. Then Z( µ R,B = Z( µ R,B+b. Proof. The proof of this lemma follows from the infinite product representation of these

19 12 Fourier transforms and Lemma 1.6. Specifically: ( 1 µ R,B (t = B d=1 b B ( e 2πib t 1 R d B µ R,B+b (t = d=1 e 2πibt R d b B e 2πibt R d Since either product is zero if and only if one of its terms is zero, the result follows. Lemma Let R Z, R 2, and B Z be a finite set, and choose b Z. Let Γ R be chosen such that {e λ : γ Γ} is orthonormal in both L 2 (µ R,B and L 2 (µ R,B+b. Then {e γ : γ Γ} is an orthonormal basis in L 2 (µ R,B if and only if it is an orthonormal basis in L 2 (µ R,B+b. Proof. By Lemma 1.1, the set {e γ : γ Γ} is orthogonal in L 2 (µ R,B+b so by Lemma 1.1 it will suffice to show that µ R,B (t = µr,b+b (t for an arbitrary t R. To this end, we make use of the fact that µ R,B, µ R,B+b can be represented as an absolutely convergent infinite product by Lemma 1.6. Letting χ B (t = 1 B b B e2πibt, we have that µ R,B+b (t = d=1 = lim ( e 2πib t t R d χ B R d q q d=1 = lim q d=1 ( e 2πib t t R d χ B q e 2πib R d q = lim q χ B d=1 q = lim q d=1 q = lim q d=1 = µ R,B (t. t R d χ B ( t R d ( t R d ( t χ B R d ( t χ B R d

20 1 With this previous result in mind, we will refer to µ 1 interchangeably as the Hutchinson measure corresponding to R =, B = { 1, 1} and that corresponding to R =, B = {0, 2} as these measures will have the same collection of spectra. In fact, we have just proven the following theorem, which will be very useful in Section 7 when we consider which Hutchinson- measures (see Definition 1.22 are spectral: Theorem Let R Z, R 2, B Z, and Γ R be chosen such that Γ is a spectrum for µ R,B. Then for every b Z, Γ is a spectrum for µ R,B+b. Proof. This follows immediately from Lemma Returning to our thought process from prior to Lemma 1.1: we stated prior to this lemma that µ 1 and has both Γ(L = 5Γ(L = { m } R k l k : l k {0, 1}, m N 0 k=0 { m } R k l k : l k {0, 5}, m N 0 k=0 as spectra. The reader may justify this to himself or herself using Theorem 1. in a fashion analogous to the fashion it was used in Example As a consequence of fact that µ 1 operator U B(L 2 (µ 1 defined by Ue γ = e 5γ, γ Γ(L = has both Γ(L and 5Γ(L as spectra, one obtains a unitary { m } k l k : l k {0, 1}, m N 0. k=0

21 1 Definition Let µ R,B be a Hutchinson measure with L chosen such that (R, B, L is a Hadamard triple. Assume that both Γ(L and nγ(l are spectra for µ R,B for some n N. Define the unitary operator U B(L 2 (µ R,B such that Ue λ = e nλ for any λ Γ(L. Then U is called an interspectral operator for µ R,B. Much of that which has been observed about U B(L 2 (µ 1 by the aforementioned authors was observed by describing its interplay with the operators S 0, S 1 B(L 2 (µ 1 which satisfy S 0 e γ = e γ S 1 e γ = e γ+1, γ Γ(L and which form a representation of O 2 in B(L 2 (µ 1. By the definitions of these operators and the structure of Γ(L, the authors were able to show that one may orthogonally decompose L 2 (µ 1 as span{e 0 } k=0 S0 k S 1 (L 2 (µ 1. The authors showed that with respect to this decomposition, the matrix representation of U is given by U P e0 M 1 U, k=0 where P e0 is the orthogonal projection onto the space span{e 0 } and M k f = e k f for all f L 2 (µ 1, k R. This inherent self-similarly prompted Jorgensen, Kornelson, and Shu- man to dub U the Operator Fractal. Based on their work and findings, I will show in the subsequent chapter that a similar operator fractal matrix representation exists for some other interspectral operators for Hutchinson measure µ R,B.

22 15 Apart from matrix representations, these authors have made all significant headway in determining the behavior and structure of interspectral operators in the case where µ R,B is a Bernoulli convolution measure. Definition A Bernoulli convolution measure is a Hutchinson measure µ R,B such that B = { 1, 1}. These measures are of particular interest because they can be realized as the distribution of a random variable which simulates an infinite sequence of coin tosses [JKS1b]. It has been known for several years which Bernoulli convolution measures are spectral. Theorem 1.1. [Dai12] A Bernoulli convolution measure µ R,B is spectral if and only if R is an even integer. A Bernoulli convolution measure µ R,B with R = 2n has canonical spectrum Γ(L = { m (2n k l k : l k k=0 } { 0, n }, m N 0 2 When n is even, this implies that the canonical spectrum is contained in Z (as opposed to 1Z. In either case, one can show (with a proof similar to that of upcoming Proposition.2 2 that an integer p such that pγ(l is a spectrum for µ R,B must be odd. Given such a spectral Bernoulli convolution measure µ R,B, the generators of a representation of S 0, S 1 B(L 2 (µ R,B of O 2 are given by S 0 e λ = e 2nλ S 1 e λ = e 2nλ+ n 2. When n is even, one obtains an extremely useful fact: U and S 0 commute. This is not true when n is odd, and it is never true that U and S 1 commute [JKS12].

23 16 In the case of the 1 -Cantor measure and the Operator Fractal U of Jorgensen, Kornelson, and Shuman, this relationship between U and S 0 allowed these authors to make several notable insights about the spectral behavior of U. For example: Theorem [JKS1] Let µ 1 Γ(L = be the 1 -Cantor measure with canonical spectrum { m } j l j : l j {0, 1}, m N 0. j=0 Then if p is an odd integer, the set Γ(L (Γ(L + p is also a spectrum for µ 1. Theorem [JKS1a] Let µ 1 Γ(L = { m be the 1 -Cantor measure with canonical spectrum } j l j : l j {0, 1}, m N 0 j=0 and let U be the interspectral operator defined such that Ue γ = e 5γ for γ Γ(L. Let P U be the projection-valued measure (whose existence is guaranteed by the spectral theorem which satisfies U = zp U (dz σ(u and for each v L 2 (µ 1 define the real-valued Borel measure m v by m v ( = P U ( v, v L 2 (µ 1. Then if we define the operator S 0 B(L 2 (µ 1 such that S 0 γ = e γ for all γ Γ(L, we have that m S0 v m v for all v L 2 (µ 1. However, much about the behavior of the Operator Fractal U is not known. For example, while it is known to have exactly one eigenvalue, e 0 = 1, no other element of its spectrum is known [JKS1b]. It is also not generally known given λ, γ Γ(L whether or

24 17 not m λ m γ. This information could be useful in the sense that if one were able to find the unique v L 2 (µ 1 such that m v m v for all v L 2 (µ 1, one could determine the eigenvalues of U (see [Que87], Ch. 2. Additionally, that which is known about interspectral operators is only currently known about those which arise in the L 2 spaces of Bernoulli convolution measures. Even in the very similar case where we examine µ R,B where R = 2n and B Z with B = 2, it is not yet known which interspectral operators U are operator fractals or when U and the operator S 0 commute. If one chooses to examine the Hutchinson measures µ R,B such that B Z and B > 2, these properties become even more difficult to check as at present it is not known for which combinations of R Z and B Z with B > 2 one obtains a spectral measure µ R,B. This publication is an attempt to make headway in determining which such measures are spectral. For the sake of efficiency throughout the remainder of the publication, we introduce the following definition now: Definition We say that a Hutchinson measure µ R,B is a Hutchinson-N measure if R Z and B Z with B = N. This brings us to the subjects of this thesis. First, we will examine in Section 2 which Bernoulli convolution measures have interspectral operators which are operator fractals. Next, we will shift our focus to Hutchinson- measures, first examining a specific Hutchinson- measure (which we will dub the 1 -Cantor measure and a specific associated interspectral operator U in Section. We will show that for this measure, U and the operator S 0 commute and will go on to show that this relationship exists between analogous operators associated with members of a class of Hutchinson- measures in Section 5. In Section we will attempt to determine which integer multiples of the canonical spectrum for the 1 -Cantor measure are also spectra for that measure, and in Section 6 we will attempt to determine which interspectral operators associated with Hutchinson- measures possess the operator fractal

25 18 property. We will then determine precisely which Hutchinson- measures are spectral in Section 7. We will conclude by first showing that for any spectral Hutchinson- measure with an associated interspectral operator, that operator and the associated operator S 0 commute in Section 8. We will then pursue corollaries that arise from this commutativity in Section 8 and Section. Finally, in Section 10 we will generalize several previous results to certain Hutchinson-p measures, where p is an odd prime greater than.

26 1 2 Operator Fractals Associated with Bernoulli Convolution Measures It turns out that there are a number of interspectral operators related to Bernoulli convolution measures which have the operator fractal property in addition to the original operator fractal of Jorgensen, Kornelson, and Shuman. To show this, we first recall their results: Theorem 2.1. ([JKS12] Let µ 1 2n be a spectral Bernoulli convolution measure with canonical spectrum Γ(L = { m (2n j l j : l j j=0 and let S 0, S 1, U L 2 (µ 1 be defined such that } { 0, n }, m N 0 2 S 0 e γ = e 2nγ S 1 e γ = e 2nγ+ n 2 Ue γ = e pγ, γ Γ(L, where p 2N + 1 is chosen such that pγ(l is also a spectrum for µ 1. Define the subspace 2n W k = S0 k S 1 (L 2 (µ 1 2n and define Γ k = ( {γ Γ(L : γ = (2n k γ + n 2 } for some γ Γ(L for each k 0. Then L 2 (µ 1 = span{e 0 } 2n k=0 W k, each W k is invariant under U, and the matrix representation for U Wk is the same as that for U W0. That is, for each γ, λ Γ k there are unique γ, λ Γ 0 such that γ = (2n k γ, λ = (2n k λ and if U Wk (γ, λ is the γ, λ entry in the matrix representation of U Wk, then U Wk (γ, λ = U W0 (γ, λ. Thus, we may

27 20 write U P e0 U Wk P e0 U W0, k=0 k=0 where P e0 is the projection onto the space span{e 0 } L 2 (µ 1 and is defined to mean 2n is matrix equivalent to. In other words, the matrix of U has the following block diagonal form with respect to this decomposition. 0 W 0 W 1 W 2 W U W 0 0 U W W U W0 0 0 W U W0 0 W U W The authors then go on to show that in the case where n = 2 (and the measure is the 1 -Cantor measure and p = 5, U W 0 is unitarily equivalent to M 1 U, where M k L 2 (µ R,B is the multiplication operator for the function e k L 2 (µ R,B. Indeed, we have that S 1 is a surjective isometry from L 2 (µ 1 onto S 1 (L 2 (µ 1 with inverse S1. With this in mind, we have for λ Γ(L that M 1 Ue λ = γ Γ(L µ 1 (1 + 5λ γe γ. Additionally if λ = λ + 1, we have since W 0 is invariant under U that U W0 e λ = γ Γ(L+1 e 5λ, e γ e γ = γ Γ(L = γ Γ(L = γ Γ(L e 5 λ+5, e γ+1 e γ+1 µ 1 ( + 5 λ γe γ+1 µ 1 (1 + 5λ γe γ+1.

28 21 In particular, for each γ, λ Γ(L + 1 with γ = γ + 1, λ = λ + 1, and γ, λ Γ(L, we have that U W0 (γ, λ = M 1 U(γ, λ. This implies that for U has the form U P e0 M 1 U. That is, we have for each k 0 that U Wk is matrix equivalent to U W0 which is in turn unitarily equivalent to M 1 U and this implies that U has the operator fractal property. We will now show that a similar property is possessed by any such interspectral operator U L 2 (µ 1 2n provided that the canonical spectrum of the Bernoulli convolution measure µ 1 2n integers. k=0 is a set of Theorem 2.2. Let µ 1 2n canonical spectrum be a spectral Bernoulli convolution measure with n even that has Γ(L = { m (2n j l j : l j j=0 } { 0, n }, m N 0 2 and let S 0, S 1, U L 2 (µ 1 be defined such that S 0 e γ = e 2nγ S 1 e γ = e 2nγ+ n 2 Ue γ = e pγ, γ Γ(L, where p 2N + 1 is chosen such that pγ(l is also a spectrum for µ 1 2n exist by [JKS11] Theorem.6. Then with respect to the decomposition L 2 (µ 1 = P e0 2n k=0 S0 k S 1 (L 2 (µ 1, 2n (such a p will always

29 22 the matrix representation of U has the form U P e0 ( I p 1 k=0 and thus U has the operator fractal property. 2 M p 1 U Proof. By Theorem 2.1, we have that U P e0 U W0, k=0 thus the proof will be complete once we are able to show that U W0 is unitarily equivalent to ( I p 1 2 M p 1 U. We have that S 1 is a surjective isometry from L 2 (µ 1 onto S 1 (L 2 (µ 1 2n 2n with inverse S 1. With this in mind, if we define λ = 2nλ + n 2 assumption that n is even, we have since W 0 is invariant under U that for some λ Γ(L( Z by our U W0 e λ = Ue λ, e γ e γ γ 2nΓ(L+ n 2 = γ Γ(L = γ Γ(L = γ Γ(L = γ Γ(L ( µ 1 p2nλ (p 1n + 2n 2 ( p 1 cos + pλ γ ( 1 p pλ+2γ µ 1 2n ( 1 p 1 2 µ 1 2n = S 1 ( ( I p 1 2 M p 1 2nγ e 2nγ+ n 2 ( p 1 µ 1 2n ( p 1 + pλ γ e 2nγ+ n 2 ( p 1 + pλ γ Ue λ + pλ γ e 2nγ+ n 2 e 2nγ+ n 2 Thus, for each γ, λ 2nΓ(L + n 2 with γ = 2nγ + n, 2 λ = 2nλ + n, and γ, λ Γ(L, we 2 have that U W0 (γ, λ = ( I p 1 2 M p 1 U(γ, λ and the theorem is proved.

30 2 Note that the fact that Γ(L Z is paramount in the proof above, as otherwise the power on the I term varies depending on our choices of λ, γ Γ(L even if λ, γ Γ k for some k 0. In other words, Bernoulli convolution measures µ 1 2n have associated operator fractals. with n odd do not necessarily

31 2 The Structure of L 2 (µ1 and B(L 2 (µ1 Using the previous section as motivation, we would like to determine whether or not operator fractals arising from Hutchinson- measures exist. To do so, we attempt to gain some initial familiarity by examining a specfic -dimensional analogue to the 1 -Cantor measure. Specifically, we will let R =, B = { 1, 0, 1}, and we dub the resulting measure µ 1 1-Cantor measure. We will demonstrate shortly that this measure has multiple spectra and make note of several properties possessed by the associated interspectral operators, as well as examine the structure of the spaces L 2 (µ 1 and B(L 2 (µ 1. the To this end, we begin by noting that µ 1 Γ(L = { m is a spectral measure with canonical spectrum } j l j : l j L, m N 0. j=0 Indeed, it was demonstrated in [DJ12] that if R = and B = {0, 1, 2} then Γ(L is a spectrum for µ R,B, thus we have by Theorem 1.16 that Γ(L is a spectrum for µ 1. One can additionally verify with relative ease that 5Γ(L is also a spectrum for µ R,B with R =, B = {0, 1, 2}: indeed, any element in a B-extreme cycle in X(5L (the attractor of the IFS {σ lk : l k 5L} must be contained in both Z and [ 0, ] [ ] 0 k=1 = 0, 15 k. However, we have that σ l1 (1 = 1 (1 + 0 Z σ l1 (2 = 1 (2 + 0 Z σ l1 ( = 1 ( + 0 Z σ l2 (1 = 1 ( Z

32 25 σ l2 (2 = 1 ( Z σ l2 ( = 1 ( + 15 = 2 σ l (1 = 1 (1 + 0 Z σ l (2 = 1 (2 + 0 Z σ l ( = 1 ( + 0 Z. Thus, there are no B-extreme 5L-cycles in X(5L besides the trivial extreme cycle. By Theorem 1. and Theorem 1.16, this implies that 5Γ is a spectrum for µ R,B, thus µ 1. Now, we define the following operators in B(L 2 (µ 1 by S 0 (e λ = e λ S 1 (e λ = e λ+ S 2 (e λ = e λ+6 U(e λ = e 5λ for a given λ Γ. It is well-known that S 0, S 1, S 2 are isometries which generate a representation of O (the Cuntz algebra on generators, see e.g. [DJ11] and furthermore it is clear that U is unitary. Lemma.1. With µ 1 defined as above, we have for all m Z that µ 1 (m = µ 1 (m.

33 26 Proof. This follows immediately from the inherent self-similarity of µ 1 : µ 1 (m = e 2πimx dµ 1 (x ( = 1 2πim e e 2πimx dµ 1 (x + = 1 ( 2πim e 1 + e 2πim + e πim e 2πim(x+1 dµ 1 e 2πimx dµ 1 = 1 ( ( 2πim e 1 + e 2πim + e πim m µ 1. (x (x + e 2πim(x+2 dµ 1 (x This equation holds for all integers m (and moreover for all real numbers, thus by instead plugging in m we have that µ 1 (m = 1 µ 1 ( m = µ 1 (m as desired. Lemma.2. Defining Z( µ 1 to be the set of all zeroes of µ 1, we have that ( m } Z( µ 1 = { n + k : m {1, 2}, n 1, k Z. Proof. Again, we use the self-similarity of µ 1 : µ 1 (t = 1 ( 2πit e 1 + e 2πit + e πit µ 1 = 1 ( 2πit e 1 + e 2πit + e πit 1 = n=1 1 ( 2πit e n Φ e 2πit n, ( t 2πit e 81 ( ( 1 + e 2πit 81 + e πit t 81 µ 1 81 where Φ (x = 1 + x + x 2 is the rd cyclotomic polynomial. We have by Lemma 1.6 that this product is zero if and only if one of the terms in the product is zero, i.e., if and only if ( Φ e 2πit n = 0 for some n N and t R. Since all roots of Φ are of the form e 2πi+6πik or for k Z, we must have that µ 1 (t = 0 if and only if there is some k Z, n N e πi+6πik such that e 2πit n = e 2πi+6πik or e πi+6πik or t = n ( 2 + k as claimed.. Equivalently, µ 1 (t = 0 if and only if t = n ( 1 + k

34 27 Lemma.. Noting that by construction L 2 (µ 1 = span{e 0 } k=0 S0 k S 1 (L 2 (µ 1 k=0 S0 k S 2 (L 2 (µ 1, we have that S0 m (L is invariant under U for all m 0, where L = k=0 Sk 0 S 1 (L 2 (µ 1 k=0 Sk 0 S 2 (L 2 (µ 1. Proof. We note that each S m (L = k=m Sk 0 S 1 (L 2 (µ 1 k=m Sk 0 S 2 (L 2 (µ 1 is the closed linear span of the set of all e λ, 0 λ Γ(L such that the first nonzero term in the series λ = m i=0 i l i is l n for some n m. We will make use of this fact during this proof and the next proof. We assume first that m = 0. We choose λ Γ(L and will show that Ue λ span{e 0 } if and only if λ 0. By the characterization in the paragraph above, this will suffice to prove the lemma in this case. If λ = 0, then Ue λ = e 0 span{e 0 }. If λ 0, then Ue λ = e 5λ, thus e 5λ e 0, since 5Γ(L is a spectrum for µ 1 which contains 0. Now, we assume that m 1 and choose λ Γ(L\{0}. Then we have that US m 0 (e λ = e 5 m λ = γ Γ(L\{0} µ 1 (5 m λ γe γ, where we need not include 0 in this sum since 5Γ(L is a spectrum for µ 1 which contains 0 and 5 m λ 0. Furthermore, we claim that µ 1 (5 m λ γ = 0 if γ m Γ(L. Indeed, if γ m Γ(L, then the first nonzero entry in the series which defines γ occurs at or before the m 1 entry, which implies that γ = i 0 ( + γ or i 0 (6 + γ for some i 0 < m and

35 28 γ Γ(L. Thus, for such a γ we have that µ 1 (5 m λ γ = µ 1 (5 m λ i 0 ( + γ = µ 1 ( i0+1 (5 m i0 1 λ 1 γ = 0 or µ 1 (5 m λ γ = µ 1 (5 m λ i 0 (6 + γ = µ 1 ( i0+1 (5 m i0 1 λ 2 γ = 0 by Lemma.2. Thus, we have that US m 0 (e λ = γ m Γ(L\{0} µ 1 (5 m λ γe γ S0 m (L which gives the desired result. Theorem.. Given U and S 0 defined as before Lemma.1, we have that US 0 = S 0 U. Proof. First, we note that it is clear that US 0 (e 0 = S 0 U(e 0 = e 0. Now, for a given nonzero λ Γ: S 0 Ue λ = S 0 e 5λ = S 0 e 5λ, e γ e γ = e 5λ, e γ e γ = γ Γ(L γ Γ(L γ Γ(L µ 1 (5λ γe γ.

36 2 On the other hand US 0 e λ = e 5λ = = γ Γ(L γ Γ(L µ 1 ( 5λ γe γ µ 1 ( 5λ γe γ + γ Γ(L+ µ 1 ( 5λ γe γ + γ Γ(L+6 µ 1 ( 5λ γe γ. We now claim that µ 1 ( 5λ γ = 0 if γ Γ(L+ or γ Γ(L+6. Indeed, λ 0 implies that e λ S0 k S 1 (L 2 (µ 1 or e λ S0 k S 2 (L 2 (µ 1 for some k 0. Thus, we have that e λ L so that S 0 e λ S 0 L. By Lemma. we have that US 0 e λ S 0 (L, which implies that US 0 e λ is contained in the closed linear span of the set of e γ, γ Γ(L such that γ = m i=0 i l i with l 0 = 0, l i 0 for some i. Thus, in particular, US 0 e λ span{e λ : λ Γ(L}, and therefore is orthogonal to any γ Γ(L + or Γ(L + 6. This implies then that US 0 e λ = γ Γ(L µ 1 ( 5λ γe γ = γ Γ(L µ 1 ( 5λ γe γ = µ 1 (5λ γe γ = S 0 Ue λ γ Γ(L (using Lemma.1 as desired.

37 0 For Which p N Is pγ(l a Spectrum for µ1? Having demonstrated that µ 1 it has more spectra. As we shall soon see, µ 1 has two spectra, one is naturally curious about whether or not has infinitely many spectra, and this fact relies heavily on the fact that is a perfect square (see Theorem.6. We begin by reiterating the definition of µ 1 : throughout the entirety of this chapter, µ 1 refers to the 1 -Cantor measure, i.e., the Hutchinson measure µ R,B with R =, B = { 1, 0, 1}, L = {0,, 6}. Section tells us that Γ(L is a spectrum for µ 1. Lemma.1. If λ, γ Γ(L, then we may write where each l k { 6,, 0,, 6}. m λ γ = k l k, k=0 Proof. By definition, we must have that λ = m 1 k=0 k l k and γ = m 2 k=0 k l k, where each l k, l k {0,, 6}. In fact, we may let m = max{m 1, m 2 } and write λ = m k=0 k l k and γ = m k=0 k l k with each l k, l k {0,, 6}. Then for any k with 0 k m, it is clear that l k l k { 6,, 0,, 6} and the lemma is proved. Proposition.2. If n, then nγ(l is not a spectrum for µ 1. Proof. First, we wish to show that if n, then {e j : j nγ(l} is an orthonormal set. Indeed, we know by Lemma.2 that ( m } Z( µ 1 = { d + k : m {1, 2}, d N, k Z. So if we choose γ, λ nγ(l with λ γ, we have by Lemma.1 that there exists m N 0 such that m λ γ = k l k, k=0

38 1 where each l k { 6n, n, 0, n, 6n}. Choosing k to be the first nonzero term in this series, we get that λ γ = k +1 m k=k k k 1 l k Z( µ 1 by construction. Thus, since γ, λ were arbitrary, this implies that {e j : j nγ(l} is orthonormal as desired. Case 1: n but 2 n In this case, we claim that no two elements of {e j : j nγ(l} are orthogonal. Indeed, let n = r where r N and r and let γ, λ nγ(l be arbitrary. Then we have by Lemma.1 that there exists m N 0 such that m λ γ = k l k, k=0 where each l k { 18r, r, 0, r, 18r}. Equivalently, m λ γ = k l k, k=1 where each l k { 2r, r, 0, r, 2r}. For each N N 0, we have that m k=1 k N l k 1 Z\Z since by assumption r. Thus, ( m λ γ = N k N l k Z( µ 1 k=1 and thus no two elements of {e j : j nγ(l} are orthogonal. This necessarily implies that nγ(l is not a spectrum for µ 1. Case 2: q n for some q > 1

39 2 In this case, we may write n = q k, where q N and k either satisfies k or satisfies k but 2 k. In the former case, we have that {e j : j nγ(l} is a proper subset of an orthonormal subset in L 2 (µ 1 (by the argument preceding Case 1 and thus cannot be an orthonormal basis for L 2 (µ 1. In the latter case, we have that {e j : j nγ(l} is a proper subset of a set no two of whose elements are pairwise orthogonal (by Case 1. Either way, nγ(l is not a spectrum for µ R,B as claimed. Proposition.. If n, then nγ(l is not a spectrum for µ 1. Proof. Let n = k for some k Z where k (the case where k was dealt with by Proposition.2. Then one can easily verify that R =, B = {0, 1, 2}, kl = {0, 12k, 2k} forms a Hadamard triple. We also have that j=1 j 2k = k Z, and furthermore one has that 1 (k + 2k = k. Thus, {k} is a B-extreme L-cycle in this case and the proposition is proved. Definition.. For a given n N such that gcd{, n} = 1, define G n = { j (mod n : j N}. Lemma.5. Choose n N such that n and n. Then if one of 1,, is contained in G n, we have that nγ(l is a spectrum for µ 1. Proof. Assume toward a contradiction that one of 1,, is contained in G n but nγ(l is not a spectrum for µ 1. Given the assumptions about our choice of n we have that (R, B, nl is a Hadamard triple by construction, thus there must exists a B-extreme L-cycle of nonzero numbers C = {x 0,, x r 1 }. We claim first that for each j with 0 j r 1, it is the case that x j. Indeed, since 1 B, we have that x j Z and x j+1 (or x 0 in the case that j = r 1 is also an integer. This implies that one of x ( j, xj + n ( xj, + 2n Z.

40 This can only be the case if x j as claimed. With this in mind, we note that for each j such that 0 j r 1 (and with the convention that x r = x 0, we have that x j+1 = x j + l j where l j {0, n, 6n}. Without loss of generality, we can therefore assume that x j Z and x j+1 = x j+l j where l j {0, n, 2n} for each j. We may then note that this implies x j+1 x j (mod n and r x j x j x j C, k x j x j for some x j C. (mod n for each j. This in turn implies that for any k N and Before proceeding to the next step of the proof, we note that in addition to our assumption that x j Z for all j, we have that 0 < x j < n for all j. This follows from the fact that n and if each l d = 2n, then d=1 d l d = n. With this in mind, we denote the maximal element of C by x N. Now, assume that 1 G n. Then there is some j such that x j x N (mod n. However, by our assumption that 0 < x N < n, we have that x j > n, which is a contradiction. If we assume instead that G n, then there is some j such that x j x N (mod n. However, by our assumption that 0 < x N < n, we have that x j > n, which is a contradiction. Finally, if we assume that G n, then there is some j such that x j x N (mod n. But since 0 < x N < n, we have that x N < x j < qn for some integer q 1 by construction, which contradicts the assumed maximality of x N. Thus, the desired result has been proved. Theorem.6. If p > is an odd prime and n N is arbitrary, then p n Γ(L is a spectrum for µ 1.

41 Proof. It is a well-known result (see, for example, [IR0] pg. 5 that for a given b, x 2 b (mod p n has either zero or two solutions. With this in mind, we let a N be the smallest positive integer such that a 1 (mod p n. If a is even, then ( a 2 a 2 1 so that 2 1 (mod p n by the minimality of a. If this is the case, we have by Lemma.5 that p n Γ(L is a spectrum for µ 1. If a is odd, then we have that ( a (mod p n so that a+1 2 ± (mod p n, which implies either way that p n Γ(L is a spectrum for µ 1 by Lemma.5.

42 5 5 The Structure of L 2 (µ 1 and B(L 2 (µ 1 (a Generalization of Section We will now attempt to generalize the results from Section to include a larger class of Hutchinson- measures. By Section 8, we will have demonstrated that most of the results below in fact generalize for arbitrary spectral Hutchinson- measure, but in the meantime these results will be helpful for Section 6, in which we will find operator fractals in the L 2 - spaces of Hutchinson- measures. Prior to this, such operators had yet to be observed. Let R =, q 2, B = {0, 1, 2}, L = {0, q, 2q}. Then µ R,B is a spectral measure with canonical spectrum Γ(L = { m } ( k l k : l k L, m N 0 k=0 according to [DJ12] and thus we have if R = and B = { 1, 0, 1} that µ R,B = µ 1 spectrum Γ(L by Theorem Assume that pγ(l is also a spectrum for µ 1 make sure to note that p is not necessarily odd. has (where we Now, we define the following operators in B(L 2 (µ 1 by S 0 (e λ = e λ S 1 (e λ = e λ+q S 2 (e λ = e λ+2q U(e λ = e pλ for a given λ Γ(L. It is well-known that S 0, S 1, S 2 are isometries which generate a representation of O (the Cuntz-Krieger algebra on generators, see e.g. [DJ11] and furthermore it is clear that U is unitary.

43 6 Lemma 5.1. With µ 1 defined as above, we have for all m Z that µ 1 (m = µ 1 (m. Proof. This follows immediately from the inherent self-similarity of µ 1 : µ 1 (m = e 2πimx dµ 1 (x ( = 1 2πim e e 2πimx dµ 1 (x + e 2πim(x+1 dµ 1 (x + e 2πim(x+2 dµ 1 (x = 1 2πim e (1 + e 2πim = 1 2πim e (1 + e 2πim + e πim + e πim e 2πimx dµ 1 µ 1 ( m. (x This equation holds for all m Z (and for all t R, thus if we instead plug in m we get ( m that µ 1 (m = 1 µ 1 = µ 1 (m as desired. Lemma 5.2. Defining Z( µ 1 to be the set of all zeroes of µ 1, we have that ( m } Z( µ 1 = {( n + k : m {1, 2}, n 1, k Z. Proof. Again, we use the self-similarity of µ 1 : µ 1 (t = 1 2πit e (1 + e 2πit = 1 2πit e (1 + e 2πit 1 ( 2πit ( n Φ = e n=1 + e πit µ 1 + e πit 1 e 2πit ( n, ( t ( 2πit e ( e 2πit ( 2 ( + e πit ( t 2 µ 1 ( 2 where Φ (x = 1 + x + x 2 is the rd cyclotomic polynomial. It is a well-known result that this product is zero if and only if one of the terms in the product is zero, i.e., if and only if ( Φ e 2πit ( n = 0 for some n N and t R. Since all roots of Φ are of the form e 2πi+6πik or e πi+6πik for k Z, we must have that µ 1 (t = 0 if and only if there is some k Z, n N such that e 2πit ( n = e 2πi+6πik or e πi+6πik. Equivalently, µ 1 (t = 0 if and only if t = ( ( n 1 + k or

44 7 t = ( n ( 2 + k as claimed. Proposition 5.. Given S 0, S 1, and S 2 defined as above, we have that the adjoints of these operators are defined by Sj e λ = e λ jq if λ Γ(L + jq 0 if λ Γ(L\(Γ(L + jq (1 Proof. We consider the case where j = 0, noting that the j = 1, 2 cases follow mutatis mutandis. We assume first that λ Γ(L, so that there is some λ Γ(L Z such that λ = λ. Then if γ Γ(L is arbitrary, we have that e γ, S0e λ = S 0 e γ, e λ = µ 1 ((γ λ = e γ, e λ by Lemma 5.1. If we assume instead that λ Γ(L + jq for j = 1 or j = 2, so that λ = λ + jq for some λ Γ(L, we have for arbitrary γ Γ(L that e γ, S0e λ = S 0 e γ, e λ = µ 1 (γ λ jq = µ 1 ( ( γ λ j = 0 by Lemma 5.1. Thus, S 0 is defined as is given in the statement of the proposition. Lemma 5.. Noting that by construction L 2 (µ 1 = span{e 0 } k=0 S0 k S 1 (L 2 (µ 1 k=0 S0 k S 2 (L 2 (µ 1, we have that S0 m (L is invariant under U for all m 0, where L = k=0 Sk 0 S 1 (L 2 (µ 1 k=0 Sk 0 S 2 (L 2 (µ 1. Proof. We note that each for each m 0, S m 0 (L = k=m S0 k S 1 (L 2 (µ 1 k=m S0 k S 2 (L 2 (µ 1

45 8 is the closed linear span of the set of all e λ, 0 λ Γ such that the first nonzero term in the series λ = m i=0 (i l i is l n for some n m. We will make use of this fact during this proof and the next proof. We assume first that m = 0. We choose λ Γ(L and will show that Ue λ = e pλ span{e 0 } if and only if λ 0. By the characterization in the paragraph above this will suffice to prove the lemma in this case. If λ = 0, then Ue λ = e 0 span{e 0 }. If λ 0, then Ue λ = e pλ, thus e pλ e 0, since pγ(l is a spectrum for µ 1 which contains 0. Now, we assume that m 1 and choose λ Γ(L\{0}. Then we have that US m 0 (e λ = e p ( m λ = γ Γ\{0} µ 1 (p ( m λ γe γ, where we need not include 0 in this sum since pγ(l is a spectrum for µ 1 which contains 0 and p ( m λ 0. Furthermore, we claim that µ 1 (p ( m λ γ = 0 if γ ( m Γ(L. Indeed, if γ ( m Γ(L, then the first nonzero entry in the series which defines γ occurs at or before the m 1 entry, which implies that γ = ( i 0 (q + γ or ( i 0 (2q + γ for some i 0 < m and γ Γ(L. Thus, for such a γ we have that µ 1 (p ( m λ γ = µ 1 (p ( m λ ( i 0 (q + γ = µ 1 (( i0+1 (p ( m i0 1 λ 1 γ = 0

46 or µ 1 (p ( m λ γ = µ 1 (p ( m λ ( i 0 (2q + γ = µ 1 (( i0+1 (p ( m i0 1 λ 2 γ = 0 by Lemma 5.2. Thus, we have that US m 0 (e λ = γ ( m Γ(L\{0} µ 1 (p ( m λ γe γ S0 m (L which gives the desired result. Theorem 5.5. Given U and S 0 defined as before Lemma 5.1, we have that US 0 = S 0 U. Proof. First, we note that it is clear that US 0 (e 0 = S 0 U(e 0 = e 0. Now, for a given nonzero λ Γ(L: S 0 Ue λ = S 0 e pλ = S 0 e pλ, e γ e γ = e pλ, e γ e γ = γ Γ(L γ Γ(L γ Γ(L µ 1 (pλ γe γ. On the other hand US 0 e λ = e pλ = = γ Γ(L γ Γ(L µ 1 ( pλ γe γ µ 1 ( pλ γe γ + γ Γ(L+q µ 1 ( pλ γe γ + γ Γ(L+2q µ 1 ( pλ γe γ. We now claim that µ 1 ( pλ γ = 0 if γ Γ(L + q or γ Γ(L + 2q. Indeed, λ 0 implies that e λ S0 k S 1 (L 2 (µ 1 or e λ S0 k S 2 (L 2 (µ 1 for some k 0. Thus, we have that e λ L so that S 0 e λ S 0 L. By Lemma 5. we have that US 0 e λ S 0 (L, which implies that US 0 e λ is contained in the closed linear span of the set of e γ, γ Γ(L such that γ = m i=0 (i l i with l 0 = 0, l i 0 for some i. Thus, in particular, US 0 e λ span{e λ : λ

47 0 Γ(L}, and therefore is orthogonal to any γ Γ(L + q or Γ(L + 2q. This implies then that US 0 e λ = γ Γ(L µ 1 ( pλ γe γ = γ Γ(L = γ Γ(L µ 1 ( pλ γe γ µ 1 (pλ γe γ = S 0 Ue λ (using Lemma 5.1 as desired.

48 1 6 Operator Fractals Associated with Certain Hutchinson- Measures As was noted in the previous section, one finds a number of interspectral operators in the L 2 -spaces of Hutchinson- measures which possess the operator fractal property themselves, and some such operators which do not possess the operator fractal property but nevertheless exhibit very interesting behavior. In this section, we will observe these phenomena and share several interesting results. Let R = for q Z and let B = { 1, 0, 1} so that by Section 5, µ R,B = µ 1 has spectrum Γ(L = { m } R j l j : l j {0, q, 2q}, m N 0 Z. j=0 Let S 0, S 1, S 2 L 2 (µ R,B be defined as in Section 5 and define U L 2 (µ 1 by U γ = U pγ where p N is chosen such that pγ(l is also a spectrum for µ 1. By generalizing the argument from the proof of Proposition.2, one can note that it must be the case that p. We also adopt the following notation for the remainder of this section (all k 0: Γ k = {γ Γ(L : γ = ( k (γ + q for some γ Γ(L} Λ k = {γ Γ(L : γ = ( k (γ + 2q for some γ Γ(L} W k = S0 k S 1 (L 2 (µ 1 V k = S0 k S 2 (L 2 (µ 1 We also note that it is a simple exercise to demonstrate that W k = span{e γ : γ Γ k } V k = span{e λ : λ Λ k }.