ON SPECTRAL CANTOR MEASURES

IZABELLA ŁABA AND YANG WANG

Abstract. A probability measure in ℝ^{d} is called a spectral measure if it has an orthonormal basis consisting of exponentials. In this paper we study spectral Cantor measures. We establish a large class of such measures, and give a necessary and sufficient condition on the spectrum of a spectral Cantor measure. These results extend the studies by Jorgensen and Pedersen [9] and Strichartz [20].

1. Introduction

It is known that certain Cantor measures in ℝ^{d} have an orthonormal basis consisting of complex exponentials. This was first observed in Jorgensen and Pedersen [9] and studied further in Strichartz [20]. Let μ be a probability measure in ℝ^{d}. We call μ a spectral measure if there exists a Λ ⊂ ℝ^{d} such that the set of complex exponentials {e(λt) : λ ∈ Λ} forms an orthonormal basis for L^{2}(μ) (we use e(t) to denote e^{2πit} throughout the paper). The set Λ is called a spectrum for μ; we also say that (μ, Λ) is a spectral pair. It should be pointed out that a spectral measure often has more than one spectrum.

In this paper we study spectral Cantor measures in ℝ. Our Cantor measures are selfsimilar measures associated with iterated function systems (IFS). Consider the iterated functions system (IFS) {ϕ_{j}}_{j=1}^{q} given by

ϕ_{j}(x) = ρ(x + a_{j}), (1.1)

where a_{j} ∈ ℝ and |ρ| < 1. It is well known (see e.g. Falconer [1]) that for any given probability weights p_{1}, …, p_{q} > 0 with ∑_{j=1}^{q} p_{j} = 1 there exists a unique probability measure μ satisfying

μ = ∑_{j=1}^{q} p_{j}μ ∘ ϕ_{j}^{−1}. (1.2)

We ask the following question: Under what conditions is μ a spectral measure?

The familiar middle 3rd Cantor measure given by ρ = 1/3 and a_{1} = 0, a_{1} = 2 with p_{1} = p_{2} = 1/2 is not a spectral measure, see Jorgensen and Pedersen [9]. The first known example of a spectral measure whose support has non-integer dimension was given by the same authors in that paper, who showed that the measure μ corresponding to ρ = 1/4, a_{1} = 0, a_{2} = 1 and p_{1} = p_{2} = 1/2 is spectral. A spectrum of μ is

Date: September 8, 2001.

The second author is supported in part by the National Science Fundation grant DMS 0070586.

Λ = {∑_{k=0}^{m} d_{k}4^{k} : m ≥ 0, d_{k} = 0 or 2}.

Strichartz [19] gave an alternative proof of this result, and found other examples of spectral Cantor measures. Later Strichartz [20] applied his method to a more general setting to show that a class of measures having a self-similar type of structure (but not necessarily self-similar in the more traditional sense) are spectral measures, provided that an implicit condition on the zero set of certain trigonometric polynomials is satisfied. This condition, however, is not necessary and is somewhat difficult to check.

Spectral measures are a natural generalization of spectral sets. A measurable set Ω in ℝ^{d} with positive and finite measure is called spectral if L^{2}(Ω) has an orthogonal basis consisting of complex exponentials. Spectral sets have been studied rather extensively, particularly in recent years. (A partial list of these studies is in the reference of the paper.) The major unsolved problem concerning spectral sets is the following conjecture of Fuglede [2]:

Fuglede’s Spectral Set Conjecture: Let Ω be a set in ℝ^{d} with positive and finite Lebesgue measure. Then Ω is a spectral set if and only if Ω tiles ℝ^{d} by translation.

The conjecture remains open in either direction, even in dimension one and for sets that are unions of unit intervals. As we shall see, the spectral measures–tiling connection seems to be equally compelling.

In this paper we study self-similar measures satisfying

μ = ∑_{j=1}^{q} \frac{1}{q}μ ∘ ϕ_{j}^{−1} (1.3)

where ϕ_{j}(x) = \frac{1}{N} (x + d_{j}), N ∈ ℤ and |N| > 1, and D = \{d_{j}\} ⊂ ℤ. We use μ_{N,D} to denote the unique probability measure satisfying (1.3). In addition, for each finite subset A of ℝ we use T (N, A) to denote the set

T (N, A) := {∑_{j=1}^{∞} a_{j}N^{−j} : a_{j} ∈ A},

which is in fact the attractor of the IFS \{ϕ_{a}(x) := \frac{1}{N} (x + a) : a ∈ A\}, see [1]. Finally, denote

Λ(N, A) := {∑_{j=0}^{k} a_{j}N^{j} : k ≥ 1 and a_{j} ∈ A}.

Two finite sets A = {a_{j}} and S = {s_{j}} of cardinality q in ℝ form a compatible pair, following the terminology of [20], if the matrix M = [\frac{1}{\sqrt{q}} e(a_{j}s_{k})] is a unitary matrix. In other words (δ_{A}, S) is a spectral pair, where

δ_{A} := ∑_{a∈A} \frac{1}{q}δ(x − a).

For each finite set A in ℝ define its symbol by

m_{A}(ξ) := \frac{1}{|A|} ∑_{a∈A} e(−aξ).

Strichartz [20] proves the following theorem:

Theorem 1.1 (Strichartz). Let N ∈ ℤ with |N| > 1 and D be a finite set of integers. Let S ⊂ ℤ such that 0 ∈ S and (\frac{1}{N} D, S) is a compatible pair. Suppose that m_{D/N}(ξ) does not vanish on T (N, S). Then the self-similar measure μ_{N,D} is a spectral measure with spectrum Λ(N, A).

Unfortunately the condition that m_{D/N} does not vanish on T (N, S) is not a necessary condition, and can be very difficult to check, even when both D and S are simple. In general we know very little about the zeros m_{D/N}. Our objective here is to remove the above condition. We prove that a compatible pair automatically yields a spectral measure. We also give a necessary and sufficient condition for Λ(N, A) to be a spectrum.

Theorem 1.2. Let N ∈ ℤ with |N| > 1 and let D be a finite set of integers. Let S ⊂ ℤ such that 0 ∈ S and (\frac{1}{N} D, S) is a compatible pair. Then the self-similar measure μ_{N,D} is a spectral measure. If moreover gcd(D − D) = 1, 0 ∈ S and S ⊆ [2 − |N|, |N| − 2], then Λ(N, S) is a spectrum for μ_{N,D}.

In Lemma 2.2 we prove that if (\frac{1}{N} D, S) is a compatible pair for some S ⊂ ℤ, then there is also a set Ŝ ⊂ ℤ satisfying the additional conditions of Theorem 1.2 (i.e., 0 ∈ Ŝ and Ŝ ⊆ [2 − |N|, |N| − 2]) such that (\frac{1}{N} D, Ŝ) is a compatible pair.

The following theorem gives a necessary and sufficient condition for Λ(N, A) to be a spectrum. It also leads to a simple algorithm, see §3.

Theorem 1.3. Let N ∈ ℤ with |N| > 1 and D ⊂ ℤ with 0 ∈ D and gcd(D) = 1. Let S ⊂ ℤ with 0 ∈ S such that (\frac{1}{N} D, S) is a compatible pair. Then (μ_{N,D}, Λ(N, S)) is NOT a spectral pair if and only if there exist s_{j}^{∗} ∈ S and nonzero integers η_{j}, 0 ≤ j ≤ m − 1, such that η_{j+1} = N^{−1}(η_{j} + s_{j}^{∗}) for all 0 ≤ j ≤ m − 1 (with η_{m} := η_{0} and s_{m}^{∗} := s_{0}^{∗}).

The proof of Theorem 1.3 depends on the analysis of the extreme values of the eigenfunctions of the Ruelle transfer operator. The Ruelle transfer operator was studied in [9].

There appears to be a strong link between compatible pairs, tiling of integers and Fuglede’s Conjecture. All examples suggest that if D is a finite set of integers and is part of a compatible pair then D tiles ℤ. A finite set D ⊂ ℤ is called a complementing set (mod N) if there exists a E ⊂ ℤ such that D ⊕ E is a complete residue system (mod N). It is known that D tiles ℤ if and only if it is a complementing set (mod N) for some N. We prove:

Theorem 1.4. Let D ⊂ ℤ be a complementing set (mod N) with |N| > 1. Suppose that |D| has no more than two distinct prime factors. Then μ_{N,D} is a spectral measure.

4 I. ŁABA AND Y. WANG

In the next section we shall prove the results just stated. Later in §3 we give some examples and state some open problems.

We are indebted to Bob Strichartz for very helpful comments.

2. Proofs of Theorems

We first state several lemmas, many of which have been proved in Jorgensen and Pedersen [9] or Strichartz [20].

Lemma 2.1. Let A, B ⊂ ℝ be finite sets of the same cardinality. Then the following are equivalent:

(a) (A, B) is a compatible pair.

(b) m_{A}(b_{1} − b_{2}) = 0 for any distinct b_{1}, b_{2} ∈ B.

(c) ∑_{b∈B} |m_{A}(ξ + b)|^{2} ≡ 1.

Proof. Note that the condition (b) says precisely that the rows of the matrix M = [\frac{1}{\sqrt{|A|}}e(a_{j}b_{k})] are orthonormal. So (a) and (b) are clearly equivalent. To see (a) and (c) are equivalent, let δ_{A} = \frac{1}{|A|} ∑_{a∈A} δ(x − a). Then δ_{A} is a probability measure with δ̂_{A}(ξ) = m_{A}(ξ). Furthermore, (A, B) is a compatible pair if and only if (δ_{A}, B) is a spectral pair, see [20]. The equivalence of (a) and (c) follows immediately from Lemma 2.3 of [20].

Lemma 2.2. Let D, S ⊂ ℤ and N ∈ ℤ, |N| > 1 such that (\frac{1}{N} D, S) is a compatible pair. Then

(a) (\frac{1}{N} D + a, S + b) is a compatible pair for any a, b ∈ ℝ.

(b) Suppose that Ŝ ⊂ ℤ such that Ŝ ≡ S (mod N). Then (\frac{1}{N} D, Ŝ) is a compatible pair.

(c) The elements in both D and S are distinct modulo N.

(d) Suppose that |N| > 2. Then there exists an Ŝ with 0 ∈ Ŝ and Ŝ ⊆ [2 − |N|, |N| − 2] such that (\frac{1}{N} D, Ŝ) is a compatible pair.

(e) Denote D_{k} = D+ND+⋯+N^{k−1}D and S_{k} = S+NS+⋯+N^{k−1}S. Then (\frac{1}{N^{k}} D_{k}, S_{k}) is a compatible pair.

Proof. (a) is essentially trivial from Lemma 2.1. It is also well known from the fact that any translate of a spectrum is also a spectrum, and any translate of a spectral measure is also a spectral measure with the same spectra.

For (b), observe that if s ≡ ŝ (mod N) then m_{D/N}(ξ + s) = m_{D/N}(ξ + ŝ). Therefore

∑_{ŝ∈Ŝ}|m_{D/N}(ξ + ŝ)|^{2} = ∑_{s∈S} |m_{D/N}(ξ + s)|^{2} = 1.

This proves (b).

ON SPECTRAL CANTOR MEASURES 5

For (c), assume that S = {s_{j}} has s_{1} ≡ s_{2} (mod N). Then we may replace s_{2} by s_{1} in S and still have a compatible pair by (b). This means the matrix M used for defining compatible pairs has two identical columns, so it cannot be unitary, a contradiction. So elements in S are distinct modulo N. Similarly M will have two identical rows if elements in D are not distinct modulo N, again a contradiction.

To prove (d), we first translate S so that 0 ∈ S_{1} := S + a for some a ∈ ℤ. (\frac{1}{N} D, S_{1}) is still a compatible pair. Now [2 − |N|, |N| − 2] contains a complete set of residues (mod N) because it contains at least |N| consecutive integers. Choose Ŝ ⊆ [2 − |N|, |N| − 2] so that 0 ∈ Ŝ and Ŝ ≡ S_{1} (mod N). Then (\frac{1}{N} D, Ŝ) is a compatible pair.

Finally (e) is a special case of Lemma 2.5 in [20].

Lemma 2.3. Under the assumptions of Theorem 1.2, let Q(ξ) := ∑_{λ∈Λ(N,S)} |μ̂(ξ + λ)|^{2}. Then

(a) The set of exponentials {e(λξ) : λ ∈ Λ(N, S)} is orthonormal in L^{2}(μ).

(b) Q(ξ) ≤ 1 for all ξ ∈ ℝ and {e(λξ) : λ ∈ Λ(N, S)} is an orthonormal basis for L^{2}(μ) if and only if Q(ξ) ≡ 1.

(c) The function Q(ξ) is the restriction of an entire function of exponential type to the real line. Furthermore it satisfies

Q(ξ) = ∑_{s∈S} |m_{D}(N^{−1}(ξ + s))|^{2}Q(N^{−1}(ξ + s)). (2.1)

Proof. See Jorgensen and Pedersen [9]. The right hand side of (2.1) is known as the Ruelle transfer operator (operated on Q).

Proof of Theorem 1.3. Denote Λ := Λ(N, S) and μ := μ_{N,D}.

(⇐) We prove that (μ, Λ) is not a spectral pair by proving that Q(ξ) ≢ 1, where Q(ξ) is defined in Lemma 2.3. In fact, we prove that Q(η_{0}) = 0.

Observe that m_{D}(η_{j}) = 1 because η_{j} ∈ ℤ. Now ∑_{s∈S} |m_{D}(N^{−1}(η_{j} + s))|^{2} = 1 by (c) of Lemma 2.1. Since m_{D}(N^{−1}(η_{j} + s_{j}^{∗})) = m_{D}(η_{j+1}) = 1, it follows that m_{D}(N^{−1}(η_{j} + s)) = 0 for all s ≠ s_{j}^{∗} in S.

Take any λ ∈ Λ and write λ = ∑_{k=0}^{∞} s_{k}N^{k} where s_{k} ∈ S and of course only finitely many s_{k} ≠ 0. We have

μ̂(η_{0} + λ) = ∏_{j=1}^{∞} m_{D}(N^{−j}(η_{0} + λ)).

Note that

m_{D}(N^{−1}(η_{0} + λ)) = m_{D}(N^{−1}(η_{0} + s_{0}) + ∑_{k=0}^{∞} s_{k+1}N^{k}) = m_{D}(N^{−1}(η_{0} + s_{0})).

Hence m_{D}(N^{−1}(η_{0} +λ)) = 1 for s_{0} = s_{0}^{∗} and m_{D}(N^{−1}(η_{0} +λ)) = 0 otherwise. Suppose s_{0} = s_{0}^{∗}. Then the same argument together with the fact η_{1} = N^{−1}(η_{0} + s_{0}^{∗}) yield m_{D}(N^{−2}(η_{0} +

6 I. ŁABA AND Y. WANG

λ)) = 1 for s_{1} = s_{1}^{∗} and m_{D}(N^{−2}(η_{0} + λ)) = 0 otherwise. By induction we easily obtain m_{D}(N^{−j}(η_{0} + λ)) ≠ 0 if and only if s_{j} = s_{j (mod m)}^{∗}. Therefore μ̂(η_{0} + λ) ≠ 0 only if s_{j} = s_{j (mod m)}^{∗} for all j ≥ 0. But this is impossible since s_{j} = 0 for all sufficiently large j. Thus μ̂(η_{0} + λ) = 0 and Q(η_{0}) = 0.

(⇒) Assume that (μ, Λ) is NOT a spectral pair. Then Q(ξ) ≢ 1. Note that this can not happen if |D| = |S| = 1, in which case μ = δ_{0} and Λ = {0}. So we may assume that q = |D| = |S| > 1. It is well known that in this case T := T (N, S) is a compact set with infinite cardinality.

Since Q(ξ) ≢ 1, Q(ξ) ≢ 1 for ξ ∈ T, because Q is extendable to an entire function on the complex plane and T is an infinite compact set. Denote X^{−} := {ξ ∈ T : Q(ξ) = min_{η∈T} Q(η)}. It follows from Q(0) = 1 that 0 ∉ X^{−}.We apply the Ruelle transfer operator to derive a contradiction.

For any s ∈ ℝ denote ϕ_{s}(ξ) = N^{−1}(ξ + s). Then T = ⋃_{s∈S} ϕ_{s}(T). Hence ϕ_{s}(ξ) ∈ T for all ξ ∈ T. Now choose any ξ_{0} ∈ X^{−} and set Y_{0} = {ξ_{0}}. Define recursively

Y_{n+1} = {ϕ_{s}(ξ) : s ∈ S, ξ ∈ Y_{n}, ϕ_{s}(ξ) ∈ X^{−}} (counting multiplicity).

Claim 1: We have |Y_{n+1}| ≥ |Y_{n}| (counting multiplicity).

Proof of Claim 1. Let ξ^{∗} ∈ X^{−}. By (c) of Lemma 2.3

min_{η∈T} Q(η) = Q(ξ^{∗}) = ∑_{s∈S} |m_{D}(ϕ_{s}(ξ^{∗}))|^{2}Q(ϕ_{s}(ξ^{∗})).

But ∑_{s∈S} |m_{D}(ϕ_{s}(ξ^{∗}))|^{2} = 1 by Lemma 2.3 and Q(ϕ_{s}(ξ^{∗})) ≥ min_{η∈T} Q(η). Thus Q(ϕ_{s}(ξ^{∗})) = min_{η∈T} Q(η) whenever m_{D}(ϕ_{s}(ξ^{∗})) ≠ 0. In other words,

(2.2) ϕ_{s}(ξ^{∗}) ∈ X^{−} whenever m_{D}(ϕ_{s}(ξ^{∗})) ≠ 0.

Hence for each ξ ∈ Y_{n} there exists at least one s ∈ S such that ϕ_{s}(ξ) ∈ Y_{n+1}, proving Claim 1. □

Claim 2: The elements of Y_{n} in fact all have multiplicity one.

Proof of Claim 2. It is easy to see that elements in Y_{n} have the form ϕ_{s_{n}} ∘ ⋯ ∘ ϕ_{s_{1}}(ξ_{0}). If some element in Y_{n} has multiplicity more than one, then there are two distinct sequences (s_{1}, …, s_{n}) and (t_{1}, …, t_{n}) in S such that

ϕ_{s_{n}} ∘ ⋯ ∘ ϕ_{s_{1}}(ξ_{0}) = ϕ_{t_{n}} ∘ ⋯ ∘ ϕ_{t_{1}}(ξ_{0}).

Expanding the two expressions yields

\frac{1}{N^{n}} (ξ_{0} + s_{1} + Ns_{2} + ⋯ + N^{n−1}s_{n}) = \frac{1}{N^{n}} (ξ_{0} + t_{1} + Nt_{2} + ⋯ + N^{n−1}t_{n}).

But this is clearly not possible, since all elements of S are in different residue classes (mod N). □

ON SPECTRAL CANTOR MEASURES 7

Now X^{−} is finite. It follows that all Y_{n} have the same cardinality for sufficiently large n, say n ≥ n_{0}. Therefore for n ≥ n_{0} any ξ^{∗} ∈ Y_{n} has a unique offspring ϕ_{s}(ξ^{∗}) ∈ Y_{n+1}. Furthermore, for any t ∈ S and t ≠ s we must have m_{D}(ϕ_{t}(ξ^{∗})) = 0, because otherwise by (2.2) we would have ϕ_{t}(ξ^{∗}) ∈ Y_{n+1} as another offspring, a contradiction. Thus starting with a ξ_{n_{0}} ∈ Y_{n_{0}} we obtain a sequence {ξ_{n}}_{n≥n_{0}} in which ξ_{n} ∈ Y_{n} is the unique offspring of ξ_{n−1} ∈ Y_{n−1}, ξ_{n} = ϕ_{s_{n−1}}(ξ_{n−1}) for some s_{n−1} ∈ S. It follows from the finiteness of X^{−} that there exist k > n_{0} and m > 0 such that

ξ_{k} = ξ_{k+m} = \frac{1}{N^{m}} (ξ_{k} + s_{k} + s_{k+1}N + ⋯ + s_{k+m}N^{m−1}).

In particular ξ_{k} ∈ ℚ. Set η_{j} = ξ_{j+k} and s_{j}^{∗} = s_{j+k} for 0 ≤ j ≤ m − 1 (η_{m} := η_{0} and s_{m}^{∗} := s_{0}^{∗}. Only the case j = m−1 needs to be check). . Then η_{j} ∈ ℚ for all j. Furthermore η_{j+1} = ϕ_{s_{j}^{∗}}(η_{j}) = N^{−1}(η_{j} + s_{j}^{∗}) for all 0 ≤ j ≤ m − 1

Note that m_{D}(ϕ_{s}(η_{j})) = 0 for all 0 ≤ j ≤ m − 1 and s ≠ s_{j}^{∗}, as shown above. This yields

|m_{D}(η_{j})|^{2} = ∑_{s∈S} |m_{D}(ϕ_{s_{j−1}^{∗}}(η_{j−1}))|^{2} = 1, 1 ≤ j ≤ m.

But m_{D}(η_{j}) = \frac{1}{|D|} ∑_{d∈D} e(dη_{j}) and 0 ∈ D. The only way |m_{D}(η_{j})| = 1 can hold is e(d·η_{j}) = 1 for all d ∈ D. Hence η_{j} · d ∈ ℤ for all d ∈ D. Since all η_{j} ≠ 0 and gcd(D) = 1, it follows that all η_{j} ∈ ℤ. The theorem is proved.

Proof of Theorem 1.2. Suppose that μ_{N,D} is a spectral measure with spectrum Λ then μ_{N,aD} is a spectral measure with spectrum a^{−1}Λ. Therefore to prove that μ_{N,D} is spectral we may without loss of generality assume that 0 ∈ D and gcd(D) = 1.

If |N| = 2 then D = {0, 1} or D = {0, −1}. The corresponding self-similar measure μ_{N,D} is the Lebesgue measure supported on a unit interval. Clearly in this case μ_{N,D} is a spectral measure with spectrum ℤ.

For |N| ≥ 3, by Lemma 2.2 we may replace S by Ŝ ⊆ [2 − |N|, |N| − 2] such that 0 ∈ Ŝ and (\frac{1}{N} D, Ŝ) is a compatible pair. Now for any such Ŝ, we have T (N, Ŝ) ⊆ [−\frac{N−2}{N−1}, \frac{N−2}{N−1}] for N > 0 or T (N, Ŝ) ⊆ [−\frac{N^{2}+N−2}{N^{2}−1}, \frac{N^{2}+N−2}{N^{2}−1}] for N < 0. In either case T (N, Ŝ) contains no integer other than 0. Now suppose that (μ_{N,D}, Λ(N, Ŝ)) is not a spectral pair. Then there exist s_{j}^{∗} in S and nonzero integers η_{j} satisfying the condition of Theorem 1.3. Starting with T_{0} = {η_{j}} we see that T_{0} ⊆ ⋃_{s∈S} ϕ_{s}(T_{0}). This yields T_{0} ⊆ T (N, S). But this is a contradiction since T (N, Ŝ) contains no integer other than 0. Therefore (μ_{N,D}, Λ(N, Ŝ)) is a spectral pair by Theorem 1.3.

Proof of Theorem 1.4. It is known that if D is a complementing set (mod N) then there exists an L|N whose prime factors are precisely those of |D| such that D is a complementing set (mod L). We prove that there exists an S ∈ ℤ such that (\frac{1}{N} D, S) is a compatible pair, using a theorem of Coven and Meyerowitz [3]. The argument below is essentially a repetition of the proof of Theorem 1.5(i) in [13] specialized to the case of two prime factors.

8 I. ŁABA AND Y. WANG

Let Φ_{n}(z) denote the n-th cyclotomic polynomial. Let also D(z) = ∑_{d∈D} z^{d} so that m_{D}(ξ) = D(e^{2πiξ}). Assume that |D| = p^{α}q^{β} and L = p^{α′}q^{β′}, where p, q are distinct primes. (If |D| is a prime power, the proof below works and is simpler.) Let

P := {p^{k} : Φ_{p^{k}}(z) | D(z), k ≤ α′}, Q := {q^{k} : Φ_{q^{k}}(z) | D(z), k ≤ β′}.

Coven and Meyerowitz [3] prove that |P| = α, |Q| = β, and that Φ_{p^{k}q^{l}}(z) | m_{D}(z) for all k, l such that p^{k} ∈ P, q^{l} ∈ Q.

We construct the set S. Write P = {p^{k_{j}}} and Q = {q^{l_{j}}}, where k_{1} < k_{2} < ⋯ < k_{α} and l_{1} < l_{2} < ⋯ l_{β}. Define

E = {∑_{j=1}^{α} a_{j}p^{−k_{j}} + ∑_{j=1}^{β} b_{j}q^{−l_{j}} : 0 ≤ a_{j} < p, 0 ≤ b_{j} < q},

and let S = NE. Clearly S ⊂ ℤ. To prove that (\frac{1}{N} D, S) is a compatible pair it suffices to prove that m_{D/N}(s_{1} − s_{2}) = 0 for any distinct s_{1}, s_{2} ∈ S. Equivalently, we only need to show that m_{D}(λ_{1} − λ_{2}) = 0 for any distinct λ_{1}, λ_{2} ∈ E. Note that

λ_{1} − λ_{2} = ∑_{j=1}^{α} c_{j}p^{−k_{j}} + ∑_{j=1}^{β} d_{j}q^{−l_{j}}, −p < c_{j} < p, −q < d_{j} < q.

If all c_{j} = 0 then not all d_{j} = 0. So λ_{1} − λ_{2} = r/q^{l_{j^{∗}}} for some r with gcd(r, q) = 1, where j^{∗} is the largest j such that d_{j} ≠ 0. Therefore Φ_{q^{l_{j^{∗}}}} (e^{2πi(λ_{1}−λ_{2})}) = 0 and hence m_{D}(λ_{1} − λ_{2}) = 0. Similarly m_{D}(λ_{1} − λ_{2}) = 0 if all d_{j} = 0. Finally, assume none of the above is true. Let j_{1} be the largest j such that c_{j} ≠ 0 and j_{2} be the largest j such that d_{j} ≠ 0. Then λ_{1} − λ_{2} = r_{1}/p^{k_{j_{1}}} + r_{2}/q^{l_{j_{2}}} with gcd(r_{1}, p) = gcd(r_{2}, q) = 1. This yields

λ_{1} − λ_{2} = \frac{r_{1}q^{l_{j_{2}}} + r_{2}q^{k_{j_{1}}}}{p^{k_{j_{1}}} q^{l_{j_{2}}}} .

The numerator is clearly coprime with the denominator. Hence Φ_{p^{k_{j_{1}}} q^{l_{j_{2}}}} (e^{2πi(λ_{1}−λ_{2})}) = 0 and therefore m_{D}((λ_{1} − λ_{2}) = 0.

So (D/N, S) is a compatible pair, and hence μ_{N,D} is a spectral measure.

3. Examples and Open Questions

Theorem 1.3 leads to an algorithm for determining whether Λ(N, S) is a spectrum for μ_{N,D}. To find whether the integer sequence {η_{j}} exists we only need to check a finite number of integers. This is because we have shown in the proof of Theorem 1.3 that if the sequence {η_{j}} exists, it must be contained in T (N, S). However, T (N, S) is compact. In fact T (N, S) ⊆ [\frac{a}{N−1}, \frac{b}{N−1}] for N > 0 and T (N, S) ⊆ [\frac{a+Nb}{N^{2}−1}, \frac{b+Na}{N^{2}−1}] for N < 0, where a, b are the smallest and the largest elements in S, respectively.

Example 3.1. Our first example addresses the condition in Theorem 1.1 by Strichartz. Let N = 5 and D = {0, ±2, ±11}. Since D is a residue system (mod N), the set T (N, D) is a fundamental domain of ℤ and μ_{N,D} is simply the restriction of the Lebesgue measure to

ON SPECTRAL CANTOR MEASURES 9

T (N, D), see Gröchenig and Haas [5]. Let S = {0, ±1, ±2}. Then Λ(N, S) = ℤ is a spectrum for μ_{N,D}. However, m_{D}(ξ) has a zero in [0, \frac{1}{4}], which is contained in T (N, S) = [−\frac{1}{2}, \frac{1}{2}].

The condition can also be hard to check. Consider the same D as above, let N = 10 and S = {0, 2, 4, 6, 8}. Then (D/N, S) is a compatible pair. By Theorem 1.2 Λ(N, S) is a spectrum for μ_{N,D}. Nevertheless it is difficult to check whether T (N, S) contains a zero of m_{D}(ξ).

Example 3.2. Our next example illustrates that a spectral measure can have many spectra. Take N = 6 and D = {0, 1, 2}. Then (D/N, S) are compatible pairs for both S = {0, 2, 4} or S = {0, −2, 2}. By Theorem 1.2 Λ(N, S) are spectra of μ_{N,D} for both S.

A far more striking example is to take S′ = {0, 4, 8}. Then (D/N, S′) is a compatible pair because S′ ≡ S = {0, 2, 4} (mod N). One can check using the algorithm described earlier that Λ(N, S′) is indeed a spectrum of μ_{N,D}, as is Λ(N, S). But Λ(N, S′) = 2Λ(N, S)! This is rather striking because Λ(N, S′) is intuitively “twice as sparse as” Λ(N, S).

The study in this paper also leaves several questions unanswered. For example, do Theorem 1.2 and Theorem 1.3, or something similar, hold in higher dimensions? The difficulty is that an analytic function of two or more variables may attain its infimum at infinitely many points, even on a compact set. The technique of the Ruelle transfer operator employed in this paper has its origin in the study of wavelets and self-affine tiles, see e.g. [5], in dimension 1. It was extended to higher dimensions in Lagarias and Wang [17]. Could the techniques there be applied to higher dimensions to yield results on spectral measures?

Note that we have studied spectral Cantor measures in which the probability weights are equally distributed. Is this a general rule? We conclude this paper with the following conjecture:

Conjecture 3.1. Let μ be the self-similar measure associated with the IFS ϕ_{j}(x) = ρ(x + a_{j}), 1 ≤ j ≤ q, with probability weights p_{1}, …, p_{q} > 0, where |ρ| < 1. Suppose that μ is a spectral measure. Then

(a) ρ = \frac{1}{N} for some N ∈ ℤ.

(b) p_{1} = ⋯ = p_{q} = \frac{1}{q}.

(c) Suppose that 0 ∈ A = {a_{j}}. Then A = αD for some α ∈ ℝ and D ⊂ ℤ. Furthermore, D must be a complementing set (mod N).

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Department of Mathematics, University of British Columbia, Vancouver, B.C. V6T 1Z2, Canada

E-mail address: ilaba@math.ubc.ca

School of Mathematics, Georgia Institute of Technology, Atlanta, Georgia 30332, USA.

E-mail address: wang@math.gatech.edu