Scaling of spectra

# Number theoretic considerations related to the scaling of spectra of Cantor-type measures

Dorin Ervin Dutkay [Dorin Ervin Dutkay] University of Central Florida
Department of Mathematics
4000 Central Florida Blvd.
P.O. Box 161364
Orlando, FL 32816-1364
U.S.A.
and  Isabelle Kraus [Isabelle Kraus] University of Central Florida
Department of Mathematics
4393 Andromeda Loop N.
Orlando, FL 32816-1364
U.S.A.
###### Abstract.

We investigate some relations between number theory and spectral measures related to the harmonic analysis of a Cantor set. Specifically, we explore ways to determine when an odd natural number generates a complete or incomplete Fourier basis for a Cantor-type measure with scale .

###### Key words and phrases:
Cantor set, Fourier basis, prime decomposition, spectral measure, base decomposition
###### 2010 Mathematics Subject Classification:
11A07,11A51,42C30

## 1. Introduction

In [JP98], Jorgensen and Pedersen constructed the first example of a singular fractal measure on a Cantor set, which has an orthonormal Fourier series. This Cantor set is obtained from the interval , dividing it into four equal intervals and keeping the first and the third, and , and repeating the procedure infinitely many times. The measure on this Cantor set associates measure 1 to , measure to and , measure to the four intervals in the next step of the construction and so on. It is the Hausdorff measure of dimension on this Cantor set, and it is also the invariant measure of the iterated function system , (see [Hut81] or [JP98] for details).

Jorgensen and Pedersen proved the surprising result that the Hilbert space has an orthonormal basis formed with exponential functions, i.e., a Fourier basis, where

 (1.1) Γ0:={n∑k=04klk:lk∈{0,1},n∈N}.

A set in is called a spectrum for a Borel probability measure on if the corresponding exponential functions form an orthonormal basis for .

Jorgensen and Pedersen’s example opened up a new area of research and many other examples of singular measures which admit orthonormal Fourier series have been constructed since, see e.g., [Str00, ŁW02, DJ06, DJ07, DHL13, Li07].

In [DJ12], it was proved also that the set is a spectrum for the measure , for any ! This means that the operator on which maps into is actually unitary, for all , which means that there are some hidden symmetries, a certain scaling by 5 in the geometry of this Cantor set. These operators were further investigated in [JKS12, JKS14a, JKS14b].

Later, Dutkay and Haussermann [DH16] studied for what digits , with odd, the set

 Γ(m):=mΓ0={n∑k=04klk:lk∈{0,m},n∈N}

is a spectrum for . Among other things, they proved that, for any prime number , the set is a spectrum for , and there are some interesting number theoretic considerations that are required to solve this problem.

Now, we will generalize these results.

Consider the iterated function system generated by a scale , with even, and the digits ,

 τ0(x)=xg,τg/2(x)=x+g/2g.

Let be the invariant measure for this iterated function system. This is the unique Borel probability measure on which satisfies the invariance equation

 μ(E)=12(μ(τ−10(E))+μ(τ−1g/2(E))), for all Borel sets E,

(see [Hut81]).

We want to find the answer to the following question:

###### Question 1.1.

For what digits is the set

 (1.2) Γ(m):=mΓ(1)={n∑k=0gklk:lk∈{0,m},n∈N}

a spectrum for ?

As in [DJ06], we look for Hadamard triples of the form with . That means that the matrix

 1√2(e2πiblR)b∈B,l∈L

is unitary, so . This means that is odd.

It was shown [DJ06] that the numbers m that give spectra can be characterized in terms of extreme cycles, i.e., we want to find the even integers for which there exist , not all equal to 0, such that

 (1.3) x1=x0+l0g, x2=x1+l1g, …,xr−1=xr−2+lr−2g, x0=xr−1+lr−1g,

and

 (1.4) ∣∣ ∣∣1+e2πig2xk2∣∣ ∣∣=1,(k∈{0,…,r−1})

where the finite set is the extreme cycle for , and are the extreme cycle points. If such an extreme cycle exists then the set of exponential functions corresponding to is incomplete but orthonormal, if no such extreme cycle exists then the set of exponential functions corresponding to is an orthonormal basis, i.e., is a spectrum.

We note that points have to be integers. Indeed, equation (1.4) implies that , for some . Assume that is not divisible by . We have

 k0g/2+l0g=x0+l0g=x1=k1g/2

for some integer .

Then so has to be divisible by , contradiction. Thus, is in so all the points in the extreme cycle have to be integers.

Hence, Question 1.1 becomes a purely number theoretical question:

###### Question 1.2.

Given an even number , for what odd numbers , are there non-trivial extreme cycles, i.e., finite sets of integers and digits such that

 x1=x0+l0g, x2=x1+l1g, …,xr−1=xr−2+lr−2g, x0=xr−1+lr−1g?

The extreme cycle corresponding to the digit , is called the trivial extreme cycle.

###### Definition 1.3.

We say that is complete if the only extreme cycle for the digit set is the trivial one . Otherwise, is incomplete. In the paper, when we refer to an extreme cycle, we will assume it is not trivial.

As we mentioned above , if is complete then the set in (1.2) is a complete orthonormal basis, and if is incomplete then is an incomplete orthonormal set in (see [DJ06]).

According to Lemma 2.5, any odd multiple of an incomplete number is also incomplete. This justifies the next definition.

###### Definition 1.4.

We say that an odd number is if is incomplete and, for all proper divisors of , is complete. In other words, there exist non-trivial extreme cycles for the digits and there are no non-trivial extreme cycles for the digits for any proper divisor of . We say that a primitive number is non-trivial if .

In Theorems 2.10, 2.11 and 2.13 we present some large classes of numbers which are complete (such as prime powers, in most cases). On the other hand, in Theorem 2.17, we show that there are infinitely many primitive numbers.

Finding explicit formulas for primitive numbers (and thus for incomplete or complete numbers) seems to be a difficult task, even for particular choices of number .

In Section 2.4 we study the connection between primitive numbers and their order relative to (see Definition 2.8). The key technical tool is Proposition 2.23. In Theorems 2.24 and 2.25 we study primitive numbers of small order. In Theorem 2.26 we present the form of a primitive number in terms of its order and the digits corresponding to its extreme cycle. Theorem 2.31 presents a way to locate primitive numbers and in Theorem 2.37 we give an explicit example of a non-trivial primitive number.

Since, from Theorem 2.13, we see that, in most cases, prime powers are complete, in Section 2.6, we study classes of composite numbers which are complete. The results are based on some important technical lemmas : Lemma 2.3, 2.44 and 2.45. In Section 2.6, we use these lemmas in various ways to obtain new composite numbers which are complete, from simpler complete composite numbers.

## 2. Main results

For the rest of the paper will be an even integer and will be an odd integer .

### 2.1. Some preliminary lemmas

###### Lemma 2.1.

If is an extreme cycle point with digits as in (1.3), then has a periodic base expansion,

 (2.1) x0=lr−1g+lr−2g2+⋯+l0gr+lr−1gr+1+⋯+l0g2r+…,

and . We write this as , the underline indicates the infinite repetition of the digits in the base expansion of .

Hence

 x0=gr−1lr−1+gr−2lr−2+⋯+gl1+l0gr−1.

Moreover,

 {x0:x0 is an extreme cycle point }=XL∩Z,

where is the attractor of the iterated function system

 σ0(x)=xg,σm(x)=x+mg,

so

 XL=∪l∈{0,m}σl(XL),
 (2.2) XL={∞∑n=1lngn:ln∈{0,m} for all n∈N}.
###### Proof.

Recall that a finite set is an extreme cycle for digits if there exist such that

 x1=x0+l0g, x2=x1+l1g, …,xr−1=xr−2+lr−2g, x0=xr−1+lr−1g.
 x0=xr−1g+lr−1g=xr−2g2+lr−2g2+lr−1g=⋯=x0gr+l0gr+l1gr−1+⋯+lr−1g.

Iterating this equality to infinity we obtain the base decomposition of . Also

 0

From above, we know that . Therefore, is contained in . Conversely, if then, if , we have that there exists such that and we get that . If then there exists such that . Then . By induction, we obtain in and digits in such that . Since the set is finite it follows that there exists and , , such that . That means that form a cycle. We will show that actually we can start the cycle with .

We have that . Also . This means that is divisible by , and since the only digits we use are and and (odd) is not divisible by (even), it follows that and therefore . By induction, we get that must be in the same cycle.

###### Lemma 2.2.

Assume is odd and is an extreme cycle point for the digit set . Then or .

###### Proof.

We have where . Then If is 0, we get that Otherwise, Considering these modulo , we have or . Thus or . ∎

###### Lemma 2.3.

Let be an odd number not divisible by and be the largest extreme cycle point in the non-trivial extreme cycle for the digit set . Then is divisible by .

###### Proof.

Assume by contradiction that is not divisible by . Then, we know that the next cycle point is

 xt+1=xt+mg.

Since is the largest cycle point in this cycle, we have that . If then so is divisible by , contradiction. Otherwise, we get that , which contradicts Lemma 2.1. ∎

###### Lemma 2.4.

If , then it has the extreme cycle with digits .

###### Proof.

Letting and , we get that . So, , and since , is indeed an extreme cycle for the digit

###### Lemma 2.5.

If is incomplete, then any odd multiple of is also incomplete.

###### Proof.

The number is complete if and only if the only extreme cycle for the digit set is the trivial one . Suppose that is incomplete, so has the non-trivial extreme cycle , where

 x1=x0+l0g, x2=x1+l1g, …,xr−1=xr−2+lr−2g, x0=xr−1+lr−1g.

Consider the extreme cycles of , where is an odd number. Multiplying the previous expression by , we get that,

 kx1=kx0+kl0g, kx2=kx1+kl1g, …,kxr−1=kxr−2+klr−2g, kx0=kxr−1+klr−1g.

Thus, we get an extreme cycle for the digit .

Hence, odd multiples of are incomplete whenever is incomplete. ∎

###### Lemma 2.6.

All of the odd numbers between and are complete.

###### Proof.

Let be an odd number, . Suppose is incomplete. Then, by Lemma 2.1, the set contains a cycle point so it is non-empty. But , and this is a contradiction. ∎

###### Lemma 2.7.

Let be a cycle point, i.e., it has the form in (2.1). Suppose is an integer. Then is an extreme cycle point, i.e., all the other points in the cycle are integers.

###### Proof.

Suppose that is a cycle for , and that . Since is a cycle point, we know that . Then , so . By induction, all points in the cycle are integers. ∎

###### Definition 2.8.

Let be an odd natural number. We will denote by the finite ring of integers modulo . We denote by the multiplicative group of elements in that have a multiplicative inverse, i.e., the elements in which are relatively prime with . For , we denote by the order of the element in the group . We also say that has order (with respect to ). We denote by (or ) the group generated by in , that is

###### Proposition 2.9.

Assume is odd. If a coset of in has the property that for all , , then is an extreme cycle for the digit set .

###### Proof.

Let be such a coset. Label the elements in such that , and if is the number of elements in , then . Then, since , we have that . Now, since , we have that , where . Consider the following possibilities for the value of .

If , then , a contradiction.

If , then , a contradiction.

So, , and it follows that, for , and similarly for and . Rearranging, we find that

 xj+ljg=xj+1

for , and similarly for and . Since contains only integers, is an extreme cycle. ∎

### 2.2. Some complete numbers

###### Theorem 2.10.

Let be an odd number not divisible by . If any of the numbers or is in , then is complete.

###### Proof.

Assume by contradiction that is incomplete. Then there is a non-trivial extreme cycle for the digit set . From the relation between the cycle points,

 xj+1=xj+ljg,

where , we have that . Thus,

 gr−kx0≡xk(modm), with k∈{0,…,r},xr:=x0.

So, for all , the number is congruent modulo with an element of the extreme cycle . If, as in the hypothesis, there is a number in such that the number is congruent modulo with an element in , and since is arbitrary in the cycle, we get that is congruent to an element in for any .

In the following arguments, we use the fact that since is not divisible by , the condition on cycle points implies .

If , then, since , we have so

 cx0(modm)=m+cx0>m+cmg−1=m(g+c−1)g−1>mg−1,

a contradiction with the fact that is a cycle point.

For the second set , by a similar argument, we have that for some in this set, for all . Let be the largest element of the extreme cycle. Since , so , a contradiction to the maximality of . ∎

###### Theorem 2.11.

Let be an odd number not divisible by . If any of the numbers or is in , then is complete.

###### Proof.

Assume by contradiction that is incomplete. Then there is a non-trivial extreme cycle for the digit set . As in the proof of Theorem 2.10, for all , the number is congruent modulo with an element of the extreme cycle . But then, the hypothesis implies that there is a number in such that the number is congruent modulo with an element in , and since is arbitrary in the cycle, we get that is congruent to an element in for any .

In the following arguments we use the fact that since is not divisible by , the condition on the cycle points implies . Let be the largest element in the extreme cycle. We have

 0

By Lemma 2.3, is divisible by . Therefore, dividing by , we get the next element in the extreme cycle, called , and we have

 xN

For , , so is a point in bigger than , a contradiction to the maximality of . ∎

###### Corollary 2.12.

For , the numbers are complete. For , the numbers are complete. For , the numbers are complete.

###### Proof.

Let and . Then so , so so is not divisible by . Then . Since , by Theorem 2.10, , so is complete. Similarly for .

Let and . Then so is not divisible by . Also . Since , by Theorem 2.10, , so is complete. Similarly for .

Let and . Then so is not divisible by . Also . Since , by Theorem 2.11 , so is complete. Similarly for . ∎

###### Theorem 2.13.

If is a prime number, and , then is complete whenever the order of , is even. Otherwise, is complete provided that is a perfect square.

###### Proof.

If is even, then is even for all , see Proposition 2.21 below. Since is prime and greater than , we have that and are relatively prime. It is well known that the equation has zero or two solutions. Let . If is even, then we have so . Since we get that . The result follows from Theorem 2.10. If is a perfect square and is odd, then . Therefore . If is a perfect square, or is in and the result again follows from Theorem 2.10.

###### Remark 2.14.

There are prime numbers which are not complete. Consider and the prime number . Then , so the order of in , is odd. An extreme cycle for this digit set is

 {311,9383,10895,11147,11189,11196,1866},

so we see that is incomplete.

### 2.3. Primitive numbers

###### Proposition 2.15.

A number is incomplete if and only if it is divisible by a primitive number.

###### Proof.

Suppose that is incomplete. Then either is primitive, and hence divisible by a primitive number, or is not primitive. If is incomplete and not primitive, then a proper divisor of must be incomplete. Similarly, either is primitive, or a proper divisor of is incomplete. Continuing this process until we run out of proper divisors, we find that a proper divisor of must be primitive.

On the other hand, suppose that is divisible by a primitive number . Since is incomplete, by Lemma 2.5, all odd multiples of are also incomplete, so is incomplete. ∎

###### Lemma 2.16.

If is a primitive number for , then and are relatively prime.

###### Proof.

Suppose that is a primitive number and that , with . We know by Lemma 2.1 that there is an extreme cycle point in , , with . Since each is either or , where is divisible by , and since is not divisible by any of the prime factors of , we have that is also divisible by . Dividing everything by we get that is an extreme cycle for . But is complete, because is primitive, a contradiction. Thus and are relatively prime. ∎

###### Theorem 2.17.

There are infinitely many primitive numbers.

###### Proof.

Suppose there are only finitely many primitive numbers and let be the ones bigger than . By Lemma 2.16, the numbers are relatively prime with so the order of in is well defined. Let be a common multiple of , , larger than .

Then . Let . This is an odd number. We have that is not divisible by or , otherwise is divisible by , or . Consider the cycle point with digits , as in Lemma 2.1. Then

 x0=m(1+g+⋯+gg−2)gn+1−1=1+g+⋯+gg−2g−1

But so . So . With Lemma 2.7, it follows that is incomplete, so it is divisible by a primitive number, contradiction. ∎

### 2.4. Properties of the order of a number

###### Definition 2.18.

For a prime number , we denote by the largest number such that . We say that is if , i.e., .

###### Proposition 2.19.

Let and be relatively prime odd integers. Then

 og(mn)=lcm(og(m),og(n)).
###### Proof.

We have that is the smallest integer such that . So is the smallest integer such that and , which means that is the smallest integer that is divisible by and so it is the lowest common multiple of these two numbers. ∎

###### Lemma 2.20.

Let be an odd prime number relatively prime with . Then .

###### Proof.

Suppose to the contrary that . Let and , with . Then we have that and , so and . Since , we also have that . Thus , which means that divides . This contradicts the fact that , so we have that . ∎

###### Proposition 2.21.

Let be an odd prime number relatively prime with . Then for and for all .

###### Proof.

For , the statement follows from Lemma 2.20. Assume by induction that for , and . Then there exists not divisible by such that . Raise this to power using the binomial formula:

 gpak=1+p⋅qpk+q′pk+2,

for some integer . This implies that divides and also that is not . Since we have also that so divides . Thus is a number that divides and is divisible by , and by the induction hypothesis . Thus . Also, so . Using induction we obtain the result. ∎

###### Proposition 2.22.

Let be distinct odd primes relatively prime with and . For , let be the largest integer such that divides . Then

 (2.3) og(pk11…pkrr)=(r∏i=1pmax{ki−ji−ιg(pi),0}i)lcm(og(p1),…,og(pr)).
###### Proof.

We have that by Proposition 2.19. By Proposition 2.21,