On the Spectral Properties of a Class of Planar Sierpinski Self-Affine Measures

TL;DR

This paper studies the spectral properties of certain self-affine measures based on Sierpinski structures, providing conditions for spectrality and orthogonality of exponential functions.

Contribution

It offers new necessary and sufficient conditions for these measures to be spectral, especially distinguishing cases where the matrix is isotropic or anisotropic.

Findings

Derived conditions for spectrality when ho_1 = ho_2.

Quantified maximum orthogonal exponential functions in absence of spectrality.

Established criteria for spectral measures with ho_1 eq ho_2.

Abstract

We investigate the spectral properties of a class of Sierpinski-type self-affine measures defined by \[ \mu_{M,D}(\cdot) = p^{-1} \sum_{d \in D} \mu_{M,D}(M(\cdot) - d), \] where \( p \) is a prime number, \( M = \begin{bmatrix} \rho_1^{-1} & c 0 & \rho_2^{-1} \end{bmatrix} \) is a real upper triangular expanding matrix, and \( D = \{d_0, d_1, \cdots, d_{p-1}\} \subset \mathbb{Z}^2 \) satisfying \( \mathcal{Z}(\widehat{\delta}_{D}) = \cup_{j=1}^{p-1} \left( \frac{j \bm{a}}{p} + \mathbb{Z}^2 \right) \) for some \( \bm{a} \in \mathcal{E}_{p}= \{ (i_1, i_2)^* : i_1, i_2 \in [1, p-1] \cap \mathbb{Z} \} \), where \( \mathcal{Z}(\widehat{\delta}_{D}) \) denotes the set of zeros of \( \widehat{\delta}_{D} \) with \( \delta_{D} = \frac{1}{\# D} \sum_{d \in D} \delta_d \). When $\rho_1 = \rho_2$, we derive necessary and sufficient conditions for $\mu_{M,D}$ to both: $(i)$ possess an infinite orthogonal set of exponential functions, and $(ii)$ be a spectral measure. When no infinite orthogonal exponential system exists in $L^{2}(\mu_{M,D})$, we quantify the maximum number of orthogonal exponentials and provide precise estimates. For $\rho_1 \neq \rho_2$, with restricted digit sets $D$, we obtain a necessary and sufficient condition for $\mu_{M,D}$ to be a spectral measure.

Equations

$${\mu }_{M,D}(\cdot )=p^{-1}\sum _{d\in D}{\mu }_{M,D}(M(\cdot )-d),$$

$$E_{\Lambda }=\left\{e^{-2\pi i⟨\lambda ,x⟩}:\lambda \in \Lambda \right\}$$

$${\phi }_d(x)=M^{-1}(x+d),x\in {\mathbb{R}}^n,d\in D,$$

$$T:=T(M,D)=\left\{\sum _{k=1}^{\infty }{M}^{-k}{d}_k:d_k\in D\right\}=\sum _{k=1}^{\infty }{M}^{-k}D$$

$$\mu =\frac{1}{\#D}\sum _{d\in D}\mu \circ {\phi }_d^{-1},$$

$$M=\left[\begin{array}{cc}{\rho }_1^{-1}& c\\ 0& {\rho }_2^{-1}\end{array}\right]\subset M_2(\mathbb{R}),D=\left\{d_0,d_1,\cdots ,d_{p-1}\right\}\subset {\mathbb{Z}}^2,$$

$$\mathcal{Z}({\hat{\delta }}_D)=\bigcup _{j=1}^{p-1}\left(\frac{j\mathbf{a}}{p}+{\mathbb{Z}}^2\right),\mathbf{a}\in {\mathcal{E}}_p=\{(i_1,i_2{)}^{\ast }:i_1,i_2\in [1,p-1]\cap \mathbb{Z}\}.$$

$$D=\{d_0,d_1,\cdots ,d_{p-1}\}=\left\{\left(\begin{array}{c}0\\ 0\end{array}\right),\left(\begin{array}{c}{d}_{1,1}\\ 0\end{array}\right),\left(\begin{array}{c}{d}_{2,1}\\ d_{2,2}\end{array}\right),\cdots ,\left(\begin{array}{c}{d}_{p-1,1}\\ d_{p-1,2}\end{array}\right)\right\}\subset {\mathbb{Z}}^2.$$

$$R:=\left[\begin{array}{cc}\frac{1}{d_{1,1}}& \frac{c}{d_{1,1}({\rho }_1^{-1}-{\rho }_2^{-1})}\\ 0& 1\end{array}\right],$$

$$M^{\prime }=RM{R}^{-1}=\left[\begin{array}{cc}{\rho }_1^{-1}& 0\\ 0& {\rho }_2^{-1}\end{array}\right],D^{\prime }=RD.$$

$$E_{\mathbf{a}}:=\left\{\frac{h}{l}:l{a}_2-h{a}_1\in \mathbb{Z}\setminus p\mathbb{Z},\gcd (h,l)=1\right\}\cup \{0\}.$$

$$\hat{\mu }(\xi )={\int }_{{\mathbb{R}}^2}{e}^{-2\pi i⟨\xi ,x⟩}d\mu (x),\xi \in {\mathbb{R}}^2.$$

$$\hat{\mu }(\lambda -{\lambda }^{\prime })=0\text{for any}\lambda \ne {\lambda }^{\prime }\in \Lambda .$$

$${\mu }_{M,D}$$

$$\mathcal{Z}({\hat{\mu }}_{M,D})=\bigcup _{k=1}^{\infty }(M^{\ast }{)}^k\mathcal{Z}({\hat{\delta }}_D)=\bigcup _{k=1}^{\infty }(M^{\ast }{)}^k\bigcup _{j=1}^{p-1}\left(\frac{j\mathbf{a}}{p}+{\mathbb{Z}}^2\right).$$

$$R=\left[\begin{array}{cc}{d_{1,1}}^{-1}& c^{\prime }\\ 0& 1\end{array}\right]\text{and}{D}^{\prime }=RD,$$

$$M^{\prime }=RM{R}^{-1}=\left[\begin{array}{cc}{\rho }_1^{-1}& 0\\ 0& {\rho }_2^{-1}\end{array}\right]\text{and}\mathcal{Z}({\hat{\delta }}_{D^{\prime }})=(R^{\ast }{)}^{-1}\bigcup _{j=1}^{p-1}\left(\frac{j\mathbf{a}}{p}+{\mathbb{Z}}^2\right).$$

$$\mathcal{Z}(\widehat{\mu}_{M^{\prime},D^{\prime}})=\bigcup_{k=1}^{\infty}\bigcup_{j=1}^{p-1}\left\{{\left({\begin{array}[]{*{20}{c}}\rho_{1}^{-k}d_{1,1}\left(\frac{ja_{1}}{p}+k_{1}\right)\\ \rho_{2}^{-k}\left(\frac{ja_{2}-c^{\prime}d_{1,1}ja_{1}}{p}+k_{2}-c^{\prime}d_{1,1}k_{1}\right)\\ \end{array}}\right)}:k_{1},k_{2}\in\mathbb{Z}\right\},$$

$$Q_{\mu ,\Lambda }(\xi )=\sum _{\lambda \in \Lambda }{\left|\hat{\mu }(\xi +\lambda )\right|}^2.$$

$$H=\frac{1}{\sqrt{N}}{\left[e^{2\pi i⟨M^{-1}b,\ell ⟩}\right]}_{b\in B,\ell \in L}$$

$${\mu }_n={\delta }_{M^{-1}D}\ast {\delta }_{M^{-2}D}\ast \cdots \ast {\delta }_{M^{-n}D},$$

$${\mu }_{M,D}={\mu }_n\ast {\mu }_{>n}.$$

$$\underset{\lambda \in {\Lambda }_{{\alpha }_{n+l}}\setminus {\Lambda }_{{\alpha }_n}}{\inf }|{\mu }_{>{\alpha }_{n+l}}(\lambda +\xi ){|}^2\ge \beta >0$$

$$\underset{\lambda \in {\Lambda }_{{\alpha }_{n+1}}\setminus {\Lambda }_{{\alpha }_n}}{\sup }|{\mu }_{>{\alpha }_{n+1}}(\lambda +\xi ){|}^2\le {\beta }_n$$

$$\prod _{n=1}^{\infty }|f_n({\rho }^{n}x)|\le c_0(\ln |x|{)}^{-\beta },|x|>1,$$

$$f(x_1,x_2,\cdots ,x_n)=f(x_1+k_1,x_2+k_2,\cdots ,x_n+k_n),(x_1,x_2,\cdots ,x_n)\in {\mathbb{R}}^n.$$

$$\left\{(x_1,\cdots ,x_n{)}^{\ast }:f(x_1,\cdots ,x_n)=1\right\}\subset \left\{{\left(\frac{\ell }{\gamma },x_2,\cdots ,x_n\right)}^{\ast }:\ell \in \mathbb{Z},x_2,\cdots ,x_n\in \mathbb{R}\right\}.$$

$$\prod _{n=1}^{\infty }f\left({\rho }^n{x}_1,x_{l,2},\cdots ,x_{l,n}\right)\le c_0(\ln |x_1|{)}^{-\beta },|x_1|>1,$$

$$\mathcal{B}=\{0,1,\cdots ,p-1\}(\mathrm{mod}p)\text{and}\gcd (\mathcal{B})=1.$$

$$a_1\left(d_{i,1}-d_{j,1}\right)+a_2\left(d_{i,2}-d_{j,2}\right)\equiv 0(\mathrm{mod}p).$$

Full text

[TABLE]

where $p$ is a prime number, $M=\begin{bmatrix}\rho_{1}^{-1}&c\\ 0&\rho_{2}^{-1}\end{bmatrix}$ is a real upper triangular expanding matrix, and $D=\{d_{0},d_{1},\cdots,d_{p-1}\}\subset\mathbb{Z}^{2}$ satisfying $\mathcal{Z}(\widehat{\delta}_{D})=\cup_{j=1}^{p-1}\left(\frac{j\bm{a}}{p}+\mathbb{Z}^{2}\right)$ for some $\bm{a}\in\mathcal{E}_{p}=\{(i_{1},i_{2})^{*}:i_{1},i_{2}\in[1,p-1]\cap\mathbb{Z}\}$, where $\mathcal{Z}(\widehat{\delta}_{D})$ denotes the set of zeros of $\widehat{\delta}_{D}$ with $\delta_{D}=\frac{1}{\#D}\sum_{d\in D}\delta_{d}$. When $\rho_{1}=\rho_{2}$, we derive necessary and sufficient conditions for $\mu_{M,D}$ to both: $(i)$ possess an infinite orthogonal set of exponential functions, and $(ii)$ be a spectral measure. When no infinite orthogonal exponential system exists in $L^{2}(\mu_{M,D})$, we quantify the maximum number of orthogonal exponentials and provide precise estimates. For $\rho_{1}\neq\rho_{2}$, with restricted digit sets $D$, we obtain a necessary and sufficient condition for $\mu_{M,D}$ to be a spectral measure.

The research is supported in part by the NNSF of China (Nos. 12201206 and 12371072), the Hunan Provincial NSF (No. 2024JJ6301).

∗Corresponding author.

1. Introduction

Let $\mu$ be a Borel probability measure on $\mathbb{R}^{n}$ with compact support, and let $L^{2}(\mu)$ denote the Hilbert space associated with $\mu$. The problem of approximating functions in $L^{2}(\mu)$ by “well-behaved” functions has a long-standing history. This area connects various fields such as analysis, geometry, and topology. The most effective approximation occurs when $L^{2}(\mu)$ possesses a basis consisting of complex exponentials, forming what is known as the Fourier basis. This work investigates the existence conditions for Fourier bases in $L^{2}(\mu)$, with special attention to singular measures $\mu$ on $\mathbb{R}^{n}$. More precisely, we say that $L^{2}(\mu)$ has a Fourier basis if there exists a countable subset $\Lambda\subset\mathbb{R}^{n}$ such that the set of complex exponentials

[TABLE]

forms an orthonormal basis for $L^{2}(\mu)$. In this case, we refer to $\mu$ as a spectral measure, $\Lambda$ as the spectrum of $\mu$, and $(\mu,\Lambda)$ as a spectral pair. Specifically, if the normalized Lebesgue measure restricted to a set $\Omega$ is a spectral measure, we refer to $\Omega$ as a spectral set.

In 1974, Fuglede established a significant connection between the existence of commuting self-adjoint partial differential operators and spectrality, leading to the formulation of the renowned spectral set conjecture in [14]. This conjecture is known to be false in dimensions $n\geq 3$[21, 18, 19]. However interest in this conjecture is still alive.

Let $\{\varphi_{d}(x)\}_{d\in D}$ be an iterated function system (IFS) defined by

[TABLE]

where $M\in M_{n}(\mathbb{R})$ is an $n\times n$ expanding real matrix (i.e., all eigenvalues of $M$ have absolute values greater than 1), and $D\subset\mathbb{R}^{n}$ is a finite set of digits. It is well known that there exists a unique non-empty compact set

[TABLE]

such that $T=\cup_{d\in D}\varphi_{d}(T)$ [15]. Additionally, there exists a unique probability measure $\mu:=\mu_{M,D}$ supported on $T$ that satisfies

[TABLE]

where $\#D$ denotes the cardinality of $D$. The set $T$ and the measure $\mu_{M,D}$ are called the self-affine set (or attractor) and the self-affine measure, respectively. In particular, if $M$ is a scalar multiple of an orthonormal matrix, then $T$ and $\mu_{M,D}$ are called the self-similar set and self-similar measure, respectively.

In 1998, Jorgensen and Pedersen [17] discovered the first singular and non-atomic spectral measure. They proved that the one fourth Cantor measure $\mu_{4}$ supported on $T=\{\sum_{i=1}^{\infty}4^{-i}d_{i}:\;d_{i}\in\{0,2\}\}$, is a spectral measure. Since then, the study of singular spectral measures has flourished in the field of fractal analysis. For the spectrality of Bernoulli convolution in $\mathbb{R}$, it was fully resolved, see [16, 10, 8, 11]. In $\mathbb{R}^{2}$, the most work of the spectra measure was concentrated on the self-affine Sierpinski-type measures, see [9, 12, 20]. In [20], Lu et al. investigated the self-affine measure generated by the upper triangular matrices and some ternary digit sets. Recently, Yan [23] extends the results of [12] to the Moran measure generated by an expanding real diagonal matrix and some sets of $p$-ary numbers satisfying a certain zero condition, where $p$ is a prime. Fractal measures associated with this zero set have been extensively studied, yielding many significant results; see [5, 6, 7, 22] for further details.

Their results have inspired us to investigate the spectral properties of the self-affine measure $\mu_{M,D}$ with

[TABLE]

where $0<\rho_{1},\rho_{2}<1$ and $p$ is a prime number. Let $\widehat{\mu}$ be the Fourier transform of $\mu$, and $\mathcal{Z}(f)$ the zeros of $f$. The technique is to make use of some explicit expressions of $\mathcal{Z}(\widehat{\mu}_{M,D})$. In this paper, we consider the set $D$ satisfying

[TABLE]

Here $(i_{1},i_{2})^{*}$ denotes the transposed conjugate of $(i_{1},i_{2})$, $\delta_{D}=\frac{1}{\#D}\sum_{d\in D}\delta_{d}$ and $\delta_{d}$ represents the Dirac measure at $d$. Actually, many digit sets satisfy the form in equation (1.3). For example, let $D_{1}=\{0,1,\cdots,N-1\}$ and $D_{2}=\{(0,0)^{*},(1,0)^{*},(0,1)^{*}\}$, by direct calculation, we have $\mathcal{Z}(\widehat{\delta}_{D_{1}})=\{\frac{j}{N}+\mathbb{Z}:j=1,2,\cdots,N-1\}$, $\mathcal{Z}(\widehat{\delta}_{D_{2}})=\{\pm(\frac{1}{3},-\frac{1}{3})^{*}+\mathbb{Z}^{2}\}$.

Suppose that $\rho_{1}=\rho_{2}$. We first establish the following two results, which provide necessary and sufficient conditions for $L^{2}(\mu_{M,D})$ to admit an infinite set of orthogonal exponential functions. If $L^{2}(\mu_{M,D})$ does not possess an infinite orthogonal set, we can estimate the maximal number of orthogonal exponential functions and determine its exact value.

Theorem 1.1**.**

Suppose that $\rho_{1}=\rho_{2}=\rho$. Let $\mu_{M,D}$ be given by (1.1), (1.2)and (1.3). Then $L^{2}(\mu_{M,D})$ admits an infinite orthogonal set of exponential functions if and only if $\rho^{-1}=(\frac{t}{s})^{\frac{1}{r}}$ and $c=\kappa\rho^{-1}$ for some $s,t,r\in\mathbb{N}$ with $t\in p\mathbb{Z}$, $\gcd(s,t)=1$ and $\kappa\in\mathbb{Q}$.

Theorem 1.2**.**

Suppose that $\rho_{1}=\rho_{2}=\rho$. Let $\mu_{M,D}$ be given by (1.1) (1.2)and (1.3). If $\rho^{-1}=(\frac{t}{s})^{\frac{1}{r}}$ for some $s,t,r\in\mathbb{N}$ with $\gcd(s,t)=1$. Then the following statements hold:

1. $(i)$

If $c\neq\kappa\rho^{-1}$ for any $\kappa\in\mathbb{Q}$, then there exist at most $p$ mutually orthogonal exponential functions in $L^{2}(\mu_{M,D})$, and the number $p$ is the best. 2. $(ii)$

If $c=\kappa\rho^{-1}$ for some $\kappa\in\mathbb{Q}$ and $t\notin p\mathbb{Z}$, then

1. $(a)$

If $s\notin p\mathbb{Z}$, then there exist at most $p$ mutually orthogonal exponential functions in $L^{2}(\mu_{M,D})$, and the number $p$ is the best. 2. $(b)$

If $s\in p\mathbb{Z}$, then there are any number of orthogonal exponential functions in $L^{2}(\mu_{M,D})$.

The following result establishes a necessary and sufficient condition for $\mu_{M,D}$ to be a spectral measure when $\rho_{1}=\rho_{2}$. Consequently, [3, Conjecture 4.2] is incorrect, see the counterexample 6.1.

Theorem 1.3**.**

Suppose that $\rho_{1}=\rho_{2}=\rho$. Let $\mu_{M,D}$ be given by (1.1) (1.2)and (1.3). Then, $\mu_{M,D}$ is a spectral measure if and only if $\rho^{-1}\in p\mathbb{Z}$ and $c\in\{\frac{t}{s}:t\in p\mathbb{Z},\gcd(t,s)=1\}\cup\{0\}$.

For the case $\rho_{1}\neq\rho_{2}$, let $d_{0}=\bm{0}$, $d_{1}=(d_{1,1},0)^{*}\neq\bm{0}$, then

[TABLE]

Define

[TABLE]

and

[TABLE]

It is easy to check that $\mu_{M,D}$ and $\mu_{M^{\prime},D^{\prime}}$ have the same spectrality. Let $c^{\prime}=\frac{c}{d_{1,1}(\rho_{1}^{-1}-\rho_{2}^{-1})}$, $c^{\prime\prime}=d_{1,1}c^{\prime}$. For $\bm{a}=(a_{1},a_{2})^{*}\in\mathcal{E}_{p}=\{(i_{1},i_{2})^{*}:i_{1},i_{2}\in[1,p-1]\cap\mathbb{Z}\}$, define

[TABLE]

With this definition, we establish the following result.

Theorem 1.4**.**

Suppose that $\rho_{1}\neq\rho_{2}$. Let $\mu_{M,D}$ be given by (1.1), (1.2), (1.3) and (1.4). If $c^{\prime\prime}\notin\mathbb{Q}$, then $\mu_{M,D}$ is not a spectral measure.

This result also indicates that if $\rho_{1}\neq\rho_{2}$, a necessary condition for $\mu_{M,D}$ to be a spectral measure is $c^{\prime\prime}\in\mathbb{Q}$. Consequently, when examining the necessary conditions, we can restrict our attention to the case where $c^{\prime\prime}\in\mathbb{Q}$. Divide $\mathbb{Q}$ into $\mathbb{Q}\setminus E_{\bm{a}}$ and $E_{\bm{a}}$, yielding the following two results.

Theorem 1.5**.**

Suppose that $\rho_{1}\neq\rho_{2}$. Let $\mu_{M,D}$ be given by (1.1), (1.2), (1.3) and (1.4). Let $E_{\bm{a}}$ be given by (1.5) and $c^{\prime\prime}\in\mathbb{Q}\setminus E_{\bm{a}}$. Then $\mu_{M,D}$ is a spectral measure if and only if $\rho_{1}^{-1}\in p\mathbb{Z}$.

It is noteworthy that although $\rho_{1}\neq\rho_{2}$, we can conclude that when $c^{\prime\prime}\in\mathbb{Q}\setminus E_{\bm{a}}$, the spectrality of $\mu_{M,D}$ is independent of $\rho_{2}$. This is an intriguing phenomenon.

Theorem 1.6**.**

Suppose that $\rho_{1}\neq\rho_{2}$. Let $\mu_{M,D}$ be given by (1.1), (1.2), (1.3) and (1.4). Let $E_{\bm{a}}$ be given by (1.5) and $c^{\prime\prime}=\frac{c_{1}}{c_{2}}\in E_{\bm{a}}$, where $\gcd(c_{1},c_{2})=1$, $c_{2}\in p^{\ell}(\mathbb{Z}\setminus p\mathbb{Z})$ for some $\ell\in\mathbb{N}$. Then $\mu_{M,D}$ is a spectral measure if and only if $\rho_{1}^{-1},\rho_{2}^{-1}\in p\mathbb{Z}$ and $p^{\ell+1}\mid(\rho_{1}^{-1}-\rho_{2}^{-1})$.

By synthesizing the results of Theorems 1.3 to 1.6, we obtain the following theorem, which establishes the necessary and sufficient conditions for $\mu_{M,D}$ to be a spectral measure.

Theorem 1.7**.**

Let $\mu_{M,D}$ be given by (1.1), (1.2), (1.3) and (1.4), and let $E_{\bm{a}}$ be given by (1.5). Then $\mu_{M,D}$ is a spectral measure if and only if $M$ and $D$ satisfy one of the following conditions:

1. $(i)$

$\rho_{1}\neq\rho_{2}$, $\rho_{1}^{-1}\in p\mathbb{Z}$, and $c^{\prime\prime}\in\mathbb{Q}\setminus E_{\bm{a}}$. 2. $(ii)$

$\rho_{1}\neq\rho_{2}$, $\rho_{1}^{-1},\rho_{2}^{-1}\in p\mathbb{Z}$, and $c^{\prime\prime}=\frac{s}{t}\in E_{\bm{a}}$, $t\in p^{\ell}(\mathbb{Z}\setminus p\mathbb{Z})$, $p^{\ell+1}\mid(\rho_{1}^{-1}-\rho_{2}^{-1})$ for some $\ell\in\mathbb{N}$. 3. $(iii)$

$\rho_{1}=\rho_{2}\in p\mathbb{Z}$* and $c\in\{\frac{t}{s}:t\in p\mathbb{Z},\gcd(s,t)=1\}\cup\{0\}$.*

The paper is organized as follows. In Section 2, we introduce key definitions and results that are essential for proving our main theorems and provide a foundation for the study of spectral measures. In Section 3, we establish the necessary conditions for $\mu_{M,D}$ to be a spectral measure and prove Theorem 1.4. In Section 4, we analyze the case $\rho_{1}=\rho_{2}$, considering both the spectrality and non-spectrality of $\mu_{M,D}$, i.e., we give the proofs of Theorems 1.1, 1.2 and 1.3. In Section 5, we examine the case $\rho_{1}\neq\rho_{2}$, with a focus on proving Theorems 1.5 and 1.6. Finally, in Section 6, we offer some examples.

2. Preliminaries

In this section, we present some preliminary lemmas and notations that will be used throughout the paper. Let $\mu$ be a probability measure with compact support in $\mathbb{R}^{2}$. The Fourier transform of $\mu$ is defined in the usual manner:

[TABLE]

It is straightforward to show that $\Lambda$ is an orthogonal set with respect to $\mu$ if and only if

[TABLE]

In other words, the family $E_{\Lambda}=\{e^{-2\pi i\langle\lambda,x\rangle}:\lambda\in\Lambda\}$ is an orthogonal family in $L^{2}(\mu)$ if and only if $(\Lambda-\Lambda)\setminus\{\bm{0}\}\subset\mathcal{Z}(\widehat{\mu}),$ where $\mathcal{Z}(\widehat{\mu})$ denotes the zero set of the Fourier transform $\widehat{\mu}$. We also say that $\Lambda$ is a bi-zero set of $\mu$. Based on (1.1), we can express $\mu_{M,D}$ as an infinite convolution of the following form:

[TABLE]

where $\delta_{E}=\frac{1}{\#E}\sum\limits_{e\in E}\delta_{e}$, $\delta_{e}$ is the Dirac measure at the point $e\in E$ and the convergence is in weak sense.

From (1.2) and (2.1), it follows that

[TABLE]

When $\rho_{1}\neq\rho_{2}$, recall that

[TABLE]

where $c^{\prime}=\frac{c}{d_{1,1}(\rho_{1}^{-1}-\rho_{2}^{-1})}$, and $D$ satisfies (1.4). Then,

[TABLE]

Since similarity transformations preserve the spectral properties of a self-affine measure [13], the spectral characteristics of $\mu_{M,D}$ and $\mu_{M^{\prime},D^{\prime}}$ are identical. By combining (2.2) and (2.3), it is straightforward to verify that

[TABLE]

where $\bm{a}=(a_{1},a_{2})^{*}\in\mathcal{E}_{p}.$ Define the function

[TABLE]

The following theorem establishes a fundamental criterion for determining the spectrality of the measure $\mu$ [17].

Lemma 2.1** ([17]).**

Let $\mu$ be a Borel probability measure with compact support in $\mathbb{R}^{n}$, and let $\Lambda\subset\mathbb{R}^{n}$ be a countable subset. Then

• $(i)$

$E_{\Lambda}$* is an orthogonal family of $L^{2}(\mu)$ if and only if $Q_{\mu,\Lambda}(\xi)\leq 1$ for $\xi\in\mathbb{R}^{n}$.*

• $(ii)$

$E_{\Lambda}$* is an orthogonal basis for $L^{2}(\mu)$ if and only if $Q_{\mu,\Lambda}(\xi)=1$ for $\xi\in\mathbb{R}^{n}$.*

• $(iii)$

$Q_{\mu,\Lambda}(\xi)$* has an entire analytic extension to $\mathbb{C}^{n}$ if $\Lambda$ is an orthogonal set of $\mu$.*

As an immediate consequence of Lemma 2.1, we obtain the following useful result.

Lemma 2.2** ([11]).**

Let $\mu=\mu_{0}*\mu_{1}$ be the convolution of two probability measures $\mu_{i}$, $i=0,1$, and they are not Dirac measures. Suppose that $\Lambda$ is a bi-zero set of $\mu_{0}$, then $\Lambda$ is also a bi-zero of $\mu$, but cannot be a spectrum of $\mu$.

The following lemma is well-known.

Lemma 2.3**.**

([1, Lemma 2.2])* Let $I$ be a compact set in $\mathbb{R}^{2}$. Then, the set $F(I)$, which denotes the Fourier transforms of all Borel probability measures supported on the compact set $I\subset\mathbb{R}^{2}$, is equi-continuous.*

Next, we give the definition of compatible pair, which is essential for the construction of the spectrum.

Definition 2.4** ([17]).**

Let $M\in M_{n}(\mathbb{Z})$ be an $n\times n$ expansive matrix with integer entries. Let $B,L\subset\mathbb{Z}^{n}$ be a finite set of integer vectors with $\#B=\#L=N$ and $\bm{0}\in B\cap L$. We say that $(M^{-1}B,L)$ forms a compatible pair (or $(M,B)$ is admissible, or $(M,B,L)$ forms a Hadamard triple) if the matrix

[TABLE]

is unitary, i.e., $HH^{*}=I$, where $H^{*}$ denotes the transposed conjugate of $H$.

It is highly convenient to construct a family of orthogonal exponential functions for $L^{2}(\mu)$ using a Hadamard triple. However, the primary challenge lies in verifying whether the constructed set $\Lambda$ forms an orthonormal basis for $L^{2}(\mu)$. For self-affine measures, we have the following well-known result.

Lemma 2.5** ([13]).**

Let $(M,B,L)$ be a Hadamard triple. Then the self-affine measure $\mu_{M,B}$ is spectral.

For $n\geq 1$, define

[TABLE]

and

[TABLE]

Let $\Lambda_{n}$ be a bi-zero set of $\mu_{n}$ with $\bm{0}\in\Lambda_{n}$ and $\Lambda_{n}\subset\Lambda_{n+1}$ for $n\geq 1$. Then $\Lambda=\cup_{n=1}^{\infty}\Lambda_{n}$ is an orthogonal set of $\mu_{M,D}$. The following proposition determines whether $\Lambda$ is a spectrum of $\mu_{M,D}$. Since the proof follows a similar approach as [2, Theorem 2.3], we omit it.

Lemma 2.6** ([2]).**

With the above notations, suppose that $\xi=(\xi_{1},\xi_{2}),|\xi_{i}|\leq\frac{1}{3}$ and $\{\alpha_{n}\}_{n=1}^{\infty}$ is an increasing sequence of integers.

• $(i)$

If $\Lambda_{\alpha_{n}}$ is a spectrum of $\mu_{\alpha_{n}}$ and

[TABLE]

for $n,l\geq 1$, then $\Lambda$ is a spectrum of $\mu_{M,D}$.

• $(ii)$

If

[TABLE]

for some sequence $\{\beta_{n}\}_{n=1}^{\infty}$ with $\sum_{n=1}^{\infty}\beta_{n}<\infty$, then $\Lambda$ is not a spectrum of $\mu_{M,D}$.

In [20], Lu et al. established the following two lemmas, which are instrumental in our proof.

Lemma 2.7** ([20, Lemma 2.5]).**

Let $\rho=\frac{s}{t}\in(0,1)$ with $s>1$ and $\{f_{n}(x)\}_{n=1}^{\infty}$ satisfy $|f_{n}(x)|\leq 1$ for any $x\in\mathbb{R}$. Suppose that there exists a positive constant $b_{0}<1$ such that $|f_{n}(x)|\leq b_{0}$ for any $|<x>|\geq\frac{1}{2t}$, where $<x>\in(-\frac{1}{2},\frac{1}{2}]$ and $x-<x>\in\mathbb{Z}$. Then there is a constant $c_{0}>0$ such that

[TABLE]

where $\beta=\log_{a}b_{0}>0$ with $a=\log_{t}s\in(0,1)$.

Lemma 2.8** ([20, Lemma 2.6]).**

Let $\rho=\frac{s}{t}\in(0,1)$ with $s>1$ and $\gcd(s,t)=1$. Suppose that a continuous non-negative real-valued function $f(x_{1},x_{2},\cdots,x_{n})$ with $f(x_{1},x_{2},\cdots,x_{n})\leq 1$ satisfies the following conditions:

1. $(i)$

There exist non-zero integers $k_{1},k_{2},\cdots,k_{n}$ such that

[TABLE] 2. $(ii)$

There exists an integer $\gamma\in\mathbb{N}^{+}$ such that

[TABLE]

Then, there exist two constants $c_{0},\beta>0$ such that

[TABLE]

for any sequence $\{x_{l,i}:2\leq i\leq n,1\leq l\}\subset\mathbb{R}$.

At the end of this section, we present a conclusion that reflects the structure of the digit set satisfying condition (1.2).

Lemma 2.9**.**

Let $D=\{d_{0},d_{1},\cdots,d_{p-1}\}\subset\mathbb{Z}^{2}$, and let $E_{\bm{a}}$ be defined by (1.2) and (1.5), respectively. Set $d_{i}=(d_{i,1},d_{i,2})$, $c^{\prime\prime}=\frac{c_{1}}{c_{2}}\in\mathbb{Q}\setminus E_{\bm{a}}$ with $\gcd(c_{1},c_{2})=1$ and $\mathcal{B}=\{c_{2}d_{0,1}+c_{1}d_{0,2},\cdots,c_{2}d_{p-1,1}+c_{1}d_{p-1,2}\}$, then

[TABLE]

Proof.

Assume that $\mathcal{B}\neq\{0,1,\cdots,p-1\}\pmod{p}$. Since $\#B=p$, there exist $i\neq j\in\{0,1,\cdots,p-1\}$ such that $c_{2}d_{i,1}+c_{1}d_{i,2}\equiv c_{2}d_{j,1}+c_{1}d_{j,2}\pmod{p}.$ This implies that $c_{2}(d_{i,1}-d_{j,1})+c_{1}(d_{i,2}-d_{j,2})\equiv 0\pmod{p}.$ By $c^{\prime\prime}\in\mathbb{Q}\backslash E_{\bm{a}}$, we have $c_{2}a_{2}-c_{1}a_{1}\in p\mathbb{Z}$. Since $p$ is prime, $\gcd(c_{1},c_{2})=1$ and $1\leq a_{i}\leq p-1$, we have $\gcd(p,a_{i})=\gcd(p,c_{i})=1$ and

[TABLE]

By $\mathcal{Z}\left(m_{D}\right)=\cup_{j=1}^{p-1}(\frac{j\bm{a}}{p}+\mathbb{Z}^{2})$, we obtain that $\sum_{k=0}^{p-1}e^{-2\pi i\langle\frac{m\bm{a}}{p},d_{k}\rangle}=0$ for any $m=1,2,\cdots,p-1$. Hence

[TABLE]

By $\gcd(p,a_{i})=1$ and $p$ is prime, it is easy to see

[TABLE]

Then for $k\neq i,j$, $\sum_{m=0}^{p-1}e^{-2\pi i\frac{m}{p}(a_{1}\left(d_{k,1}-d_{j,1}\right)+a_{2}\left(d_{k,2}-d_{j,2}\right))}=0$ or $p$, while the right-hand side of (2) is equal to $p$, resulting in a contradiction. Hence, $\mathcal{B}=\{0,1,\cdots,p-1\}\pmod{p}$.

By the fundamental theorem of arithmetic, let $\gcd(\mathcal{B})=h$, where $h$ is a prime number. Now, we prove that $h=1$. Clearly, $h\neq p$. Then, we may assume that $c_{2}d_{i,1}+c_{1}d_{i,2}=ht_{i}$. From $\mathcal{B}=\{0,1,\cdots,p-1\}\pmod{p}$, we obtain $\{t_{i}\}_{i=0}^{p-1}=\{0,1,\cdots,p-1\}\pmod{p}$. Furthermore,

[TABLE]

which implies that $\frac{(c_{2},c_{1})^{*}}{ph}\in\mathcal{Z}(m_{D})=\cup_{j=1}^{p-1}(\frac{j\bm{a}}{p}+\mathbb{Z}^{2})$. Therefore, $(c_{2},c_{1})^{*}\in h\mathbb{Z}^{2}$. Combining with $\gcd(c_{1},c_{2})=1$, we have $h=1$. ∎

3. Necessary conditions

In this section, we first establish that if $\mu_{M,D}$ is a spectral measure, then $\rho_{1}$ can be expressed as $\frac{s}{t}$ for some $s,t\in\mathbb{N}$ with $\gcd(s,t)=1$ and $p\mid t$. Subsequently, under the assumptions that $\mu_{M,D}$ satisfies conditions (1.1), (1.2), (1.3) and (1.4) with $\rho_{1}\neq\rho_{2}$, we further prove that if $\mu_{M,D}$ is a spectral measure, then $\rho_{1}^{-1}\in p\mathbb{Z}$ and $c^{\prime\prime}\in\mathbb{Q}$ (Theorem 1.4). Prior to delving into the proofs, we introduce the concept of the selection map and present several lemmas that will be instrumental in the subsequent arguments of this section.

Denote $\Xi=\{0,1,\cdots,p-1\}$, $\Xi^{k}=\{\textbf{I}=i_{1}i_{2}\cdots i_{k}:\;i_{k}\in\Xi\}$ for $k\geq 1$ and $\Xi^{*}=\cup_{k=0}^{\infty}\Xi^{k}$, where $\Xi^{0}=\{\emptyset\}$. Notice that $\Xi^{*}$ can be thought of as a tree: the root is the empty set $\emptyset$, and for each node I, the nodes following it are $\textbf{I}\Xi$. In particular, $\emptyset\textbf{I}=\textbf{I}$, $\textbf{I}0^{\infty}=\textbf{I}00\cdots$, $\;0^{k}=0\cdots 0\in\Xi^{k}.$

Definition 3.1** ([11]).**

If $p\mid t$, we call a map $\gamma:\Xi^{*}\to\{-1,0,\cdots,t-2\}$ a selection mapping if

1. $(i)$

$\gamma(\emptyset)=\gamma(0^{n})=0$* for all $n\geq 1$.* 2. $(ii)$

For any $\mathbf{I}=i_{1}\cdots i_{k}\in\Xi^{k}$, $\gamma(\mathbf{I})\in i_{k}+p\mathbb{Z}\cap\mathcal{C}$, where $\mathcal{C}=\{-1,0,\cdots,t-2\}$. 3. $(iii)$

For any $\mathbf{I}=i_{1}\cdots i_{k}\in\Xi^{*}$, there exists $\mathbf{J}\in\Xi^{*}$ such that $\gamma$ vanishes eventually on $\mathbf{IJ}0^{\infty}$, i.e., $\gamma(\mathbf{IJ}0^{k})=0$ for sufficiently large $k$.

Let

[TABLE]

where $\textbf{I}|_{k}$ denotes the prefix of I with length $k$. Furthermore, we can define a mapping $\gamma^{*}$ from $\Xi^{\gamma}$ to $\mathbb{Z}$ by

[TABLE]

Lemma 3.2** ([8]).**

Let $(\Lambda-\Lambda)\backslash\{0\}\subset\{\rho^{-k}\ell:k\geq 1,\ell\in\cup_{i=1}^{p-1}(\frac{i}{p}+\mathbb{Z})\}$ for a real number $0<\rho<1$. Then, $\Lambda$ is an infinite set if and only if $\rho$ is the $r$-th root of a fraction $\frac{s}{t}$, where $\gcd(s,t)=1$, $p\mid t$.

Dai $et\;al.$ [11] obtained the following result, which illustrates the relationship between maximal orthogonal sets and selection maps.

Lemma 3.3**.**

([11, Theorem 4.5] )* Suppose that $\rho=\frac{s}{t}\in(0,1)$ with $p\mid t$, and*

[TABLE]

with $0\in\Lambda$. Then $\Lambda$ is maximal if and only if there exists $j_{0}\geq 1$ and a selection map $\gamma$ such that $\Lambda=\rho^{-j_{0}}p^{-1}(\gamma^{*}(\Xi^{\gamma})).$

For convenience, for any set $\Lambda\subset\mathbb{R}^{2}$, we define

[TABLE]

where $i=1,2$. Similar to [20, Proposition 3.3], we give the structure of the spectrum.

Proposition 3.4**.**

Let $\nu=\delta_{1}*\delta_{2}*\cdots*\delta_{n}*\cdots$ be a spectral measure satisfying $\#\operatorname{spt}\nu=\infty$ and

[TABLE]

where $0<\rho<1$ and $\eta\neq 0$. If $\bm{0}\in\Lambda$ is a spectrum of $\nu$, then $\rho=\frac{s}{t}$ with $t\in p\mathbb{Z}$, and there exist a selection map $\gamma$ and a subset $\Xi^{\prime}\subset\Xi^{\gamma}$ such that

[TABLE]

Moreover, the spectrum $\Lambda$ can be decomposed as

[TABLE]

where $\Lambda_{n}:=\{\lambda(\mathbf{I}):|\mathbf{I}|\leq n,\mathbf{I}\in\Xi^{\prime}\}$ is a bi-zero set of $\nu_{n}=\delta_{1}*\delta_{2}*\cdots*\delta_{n}$.

Proof.

By adapting the methodology used in the proof of Proposition 3.3 in [20] and leveraging Lemmas 2.2, 3.2, and 3.3, the proof of this proposition follows straightforwardly. ∎

Remark 3.5**.**

For any $\lambda=(\lambda_{1},\lambda_{2})^{*}\in\Lambda$, by the above proposition, there exists a unique word $\mathbf{I}\in\Xi^{\prime}$ such that $\lambda_{1}=\eta\rho^{-1}p^{-1}\gamma^{*}(\mathbf{I}).$ We can denote $\lambda_{1}=\lambda_{1}(\mathbf{I})$ and $\lambda=\lambda(\mathbf{I})$.

It is straightforward to verify that the measure $\mu_{M,D}$, defined by conditions (1.1), (1.2) and (1.3), satisfies the requirements of Proposition 3.4. Thus, we conclude that if $\mu_{M,D}$ is a spectral measure, then $\rho_{1}=\frac{s}{t}$ with $t\in p\mathbb{Z}$. In the remainder of this section, we will assume that the digit set $D$ satisfies conditions (1.2), (1.3)and (1.4). We first prove that if $\mu_{M,D}$ is a spectral measure, then $\rho_{1}\in p\mathbb{Z}$, followed by the proof of Theorem 1.4.

Let $Q=\begin{bmatrix}d^{-1}_{1,1}&0\\ 0&1\end{bmatrix}$. Then,

[TABLE]

and $\mu_{M^{\prime\prime},D^{\prime\prime}}$ shares the same spectral property as $\mu_{M,D}$. Write

[TABLE]

It is readily verified that

[TABLE]

fulfilling the conditions of Proposition 3.4. Consequently, we establish the following lemma.

Proposition 3.6**.**

Let $\rho_{1}=\frac{s}{t}$ with $s>1$ and $t\in p\mathbb{Z}$. Suppose there exist $\alpha,\beta>0$ such that for all $n\geq 1$ and $\xi=(\xi_{1},\xi_{2})^{*}\in\mathbb{R}^{2}$ with $|\rho_{1}^{n}\xi_{1}|>1$,

[TABLE]

Then $\mu_{M,D}$ is not a spectral measure.

Proof.

Suppose that $\mu_{M,D}$ is a spectral measure and that $\bm{0}\in\Lambda$ is a spectrum of $\mu_{M^{\prime\prime},D^{\prime\prime}}$. From Proposition 3.4, we know that there exists a selection map $\gamma$ and a subset $\Xi^{{}^{\prime}}\subset\Xi^{\gamma}$ such that $\Lambda^{(1)}=d_{1,1}\rho_{1}^{-1}p^{-1}(\gamma^{*}(\Xi^{{}^{\prime}}))\subset d_{1,1}\rho_{1}^{-1}p^{-1}(\gamma^{*}(\Xi^{\gamma}))$. Additionally, $\Lambda=\cup_{n=1}^{\infty}\Lambda_{n}$, where $\Lambda_{n}:=\{\lambda(\textbf{I}):|\textbf{I}|\leq n,\textbf{I}\in\Xi^{{}^{\prime}}\}$. For any $n\geq 1$ and $\lambda=(\lambda_{1},\lambda_{2})^{*}\in\Lambda\setminus\Lambda_{n}$, there exists $\textbf{I}\in\Xi^{{}^{\prime}}$ with $|\textbf{I}|\geq n+1$ such that $\lambda_{1}=\lambda_{1}(\textbf{I})=d_{1,1}\rho_{1}^{-1}p^{-1}\gamma^{*}(\textbf{I}).$ This implies that

[TABLE]

Let $K>\beta^{-1}$ and $\alpha_{n}=n^{K}$. Then, there exists a positive integer $n_{0}$ such that $(n+1)^{K}\leq\left(1+\frac{1}{2}\log_{t}^{s}\right)n^{K}\;\text{for any }n\geq n_{0}.$ For any $\xi=(\xi_{1},\xi_{2})^{*}\in[0,|\frac{d_{1,1}}{ps}|]^{2}$ and $n^{K}<|\textbf{I}|\leq(n+1)^{K}$, it follows from (3.3) that

[TABLE]

for some constants $C,N_{0}>0$, and for all $n\geq N_{0}\geq n_{0}$. Since $\sum_{n=1}^{\infty}\frac{C^{2}}{n^{2}}<\infty$ and by applying Lemma 2.6 (ii), we conclude that $\Lambda$ cannot be a spectrum of $\mu_{M^{\prime\prime},D^{\prime\prime}}$, which leads to a contradiction. This completes the proof that $\mu_{M,D}$ is not a spectral measure. ∎

Lemma 3.7**.**

Let $\mu_{M,D}$ be given by (1.1), (1.2), (1.3)and (1.4). If $\mu_{M,D}$ is a spectral measure, then $\rho_{1}^{-1}\in p\mathbb{Z}$.

Proof.

If $\mu_{M,D}$ is a spectral measure, $\mu_{M^{\prime\prime},D^{\prime\prime}}$ is also a spectral measure. From Proposition 3.4, it follows that $\rho_{1}^{-1}=\frac{t_{1}}{s_{1}}$, where $t_{1}\in p\mathbb{Z}$ and $\gcd(s_{1},t_{1})=1$. Therefore, we only need to prove that $s_{1}=1$. By contradiction, assume that $s_{1}>1$. Let $\xi=(\xi_{1},\xi_{2})^{*}$, and define $f_{n}(\xi_{1}):=|m_{{M^{\prime\prime}}^{-n}D^{\prime\prime}}((\rho_{1}^{-n}\xi_{1},\xi_{2})^{*})|$. Then, it follows that $\widehat{\mu}_{M^{\prime\prime},D^{\prime\prime}}(\xi)=\Pi_{j=1}^{\infty}f_{j}(\rho_{1}^{j}\xi_{1}).$ Note that

[TABLE]

For convenience, we can denote

[TABLE]

where $\vartheta_{n}$ is determined by $\rho_{1}$, $\rho_{2}$, and $c{d_{1,1}}^{-1}$. Furthermore, we can estimate $f_{n}(\xi_{1})$ as

[TABLE]

It is easy to see that if $|\langle\xi_{1}\rangle|\geq\frac{1}{2t_{1}}$, then $f_{n}(\xi_{1})<1.$ For each $n>1$, by Lemma 2.7, there exist $\alpha,\beta>0$ such that

[TABLE]

for any $|\rho_{1}^{n}\xi_{1}|>1$. It follows from Proposition 3.6 that $\mu_{M^{\prime\prime},D^{\prime\prime}}$ is not a spectral measure, a contradiction. Hence $\rho_{1}^{-1}\in p\mathbb{Z}$. ∎

At the end of this section, we will prove that when $\rho_{1}\neq\rho_{2}$ and $c^{\prime\prime}\notin\mathbb{Q}$, $\mu_{M,D}$ is not a spectral measure. From (2.2), we define $\mathcal{Z}_{k}:={M^{\prime}}^{*^{k}}\mathcal{Z}(m_{D^{\prime}})$, which will be used frequently later.

Lemma 3.8**.**

Let $\rho_{1}^{-1}=t_{1}\in p\mathbb{Z}$, and let $\mu_{M^{\prime},D^{\prime}}$ be a spectral measure with spectrum $\Lambda$. Then, there exists $\lambda\in\Lambda$ such that $(\Lambda-\lambda)\cap\mathcal{Z}_{k}\neq\emptyset$ for all $k=1,2,3$, where $\mathcal{Z}_{k}={M^{\prime}}^{*^{k}}\mathcal{Z}(m_{D^{\prime}})$.

Proof.

From Proposition 3.4, we obtain the decomposition $\Lambda=\cup_{n=1}^{\infty}\Lambda_{n}$, where each $\Lambda_{n}:=\{\lambda(\mathbf{I}):|\mathbf{I}|\leq n,\mathbf{I}\in\Xi^{\prime}\}$ forms a bi-zero set for the measure $\mu^{\prime}_{n}$. If $(\Lambda-\Lambda)\cap\mathcal{Z}_{3}=\emptyset$, then $(\Lambda-\Lambda)\setminus\{\bm{0}\}\subseteq\cup_{k\neq 3}\mathcal{Z}_{k}$. By Lemma 2.2, it follows that $\Lambda$ cannot be a spectrum of $\mu_{M^{\prime},D^{\prime}}$. This contradiction enables us to choose an element $\lambda\in\Lambda$ such that $(\Lambda-\lambda)\cap\mathcal{Z}_{3}\neq\emptyset$. We claim that $\lambda(\mathbf{I})-\lambda(\mathbf{J})\in\mathcal{Z}_{s}\;\text{for any }\mathbf{I},\mathbf{J}\in\Xi^{\prime}$, where $s=\min\left\{i:\mathbf{I}|_{i}\neq\mathbf{J}|_{i}\right\}$. In fact

[TABLE]

This shows that the claim is true.

If $\{i\mathbf{I}\in\Xi^{\prime}:i=1,2,\cdots,p-1\}=\emptyset$, then it follows that $\lambda(\mathbf{I})-\lambda(\mathbf{J})\in\cup_{k=2}^{\infty}\mathcal{Z}_{k}$ for any $\mathbf{I},\mathbf{J}\in\Xi^{\prime}$. Using Lemma 2.2 again, we conclude that $\Lambda$ is not a spectrum of $\mu_{M^{\prime},D^{\prime}}$, which leads to a contradiction. Thus, $\{i\mathbf{I}\in\Xi^{\prime}:i=1,2,\cdots,p-1\}\neq\emptyset$. Since $\bm{0}\in\Lambda$ and by the claim, we have $\lambda(\mathbf{I})\in\mathcal{Z}_{1}$ for any $\mathbf{I}\in\Xi^{\prime}$ with $\mathbf{I}|_{1}\neq 0$. Therefore,

[TABLE]

Let $\xi_{0}:=d_{1,1}p^{-1}t_{1}\left(t_{1},-\rho_{2}^{-1}c^{\prime}\right)^{*}$, then

[TABLE]

Furthermore, we have the following equation:

[TABLE]

where $\mu^{\prime}_{>1}$ is defined similarly to (3.2). We assert that $\Lambda_{\textbf{I}}:=\{\lambda(\textbf{IJ}):\textbf{IJ}\in\Xi^{\prime}\}$ is a bi-zero set of $\mu^{\prime}_{>n}$ for any $\textbf{I}\in\Xi^{n}$. For any $\textbf{IJ}=\textbf{IJ}^{\prime}\in\Lambda_{\textbf{I}}$, by the first Claim, it follows that

[TABLE]

where $s=n+\min\{j:\textbf{IJ}|_{j}\neq\textbf{IJ}^{\prime}|_{j}\}$. Therefore, $\Lambda_{\textbf{I}}$ is a bi-zero set of $\mu^{\prime}_{>n}$, and

[TABLE]

for any $\xi\in\mathbb{R}^{2}$. Assume that $\{0i\textbf{I}\in\Xi^{\prime}:i=1,2,\cdots,p-1\}=\emptyset$. Then, from (3.4), it follows that

[TABLE]

From (3.5), the expression above is bounded as follows

[TABLE]

According to Lemma 2.9, we obtain $p\nmid d_{1,1}$, and thus the last inequality holds. From Lemma 2.1, $\Lambda$ is not a spectrum of $\mu_{M^{\prime},D^{\prime}}$, which leads to a contradiction. ∎

In the following, we will prove Theorem 1.4.

Proof of Theorem 1.4.

Suppose that $\mu_{M^{\prime},D^{\prime}}$ is a spectral measure. By Lemma 3.7, $\rho^{-1}_{1}=t_{1}\in p\mathbb{Z}$. According to Lemma 3.8, we can assume that $\bm{0}\in\Lambda$ is a spectrum with $\Lambda\cap\mathcal{Z}_{j}\neq\emptyset$ for $j=1,2,3$. For each $n=1,2,3$ and $\lambda_{n}\in\mathcal{Z}_{n}\cap\Lambda$, from (2.4), let

[TABLE]

where $s_{n}=(s_{n,1},s_{n,2})^{*}\in\mathcal{E}_{p}$ and $k_{n}=(k_{n,1},k_{n,2})^{*}\in\mathbb{Z}^{2}$. For any $n>\ell$, the difference $\lambda_{n}-\lambda_{\ell}$ can be written as

[TABLE]

Since $\lambda_{n}-\lambda_{\ell}\in\cup_{j=1}^{\infty}\mathcal{Z}_{j}$, it follows from the first component that $\lambda_{n}-\lambda_{\ell}\in\mathcal{Z}_{\ell}={M^{\prime*}}^{\ell}{R^{*}}^{-1}\mathcal{Z}(m_{D})$. Therefore, the second component satisfies the following relationship

[TABLE]

We can then derive the following equation

[TABLE]

Note that $\frac{s_{n\ell,2}+pk_{n\ell,2}+s_{\ell,2}+pk_{\ell,2}}{t_{1}^{n-\ell}}\in\mathbb{Q}$. This implies that for any $\ell_{1}<\ell_{2}<n$,

[TABLE]

Set $\ell_{1}=1$, $\ell_{2}=2$, and $n=3$. We obtain

[TABLE]

Returning to (3), we obtain $c^{\prime},c^{\prime\prime}\in\mathbb{Q}$, which leads to a contradiction of the initial assumption. Hence, $\mu_{M^{\prime},D^{\prime}}$ is not a spectral measure. ∎

4. The case $\rho_{1}=\rho_{2}$

In this section, we explore the scenario where $\rho_{1}=\rho_{2}:=\rho$ and present the proofs of Theorems 1.1, 1.2, and 1.3. Initially, we establish a necessary and sufficient condition for the existence of an infinite orthogonal set of exponential functions in $L^{2}(\mu_{M,D})$. When such an infinite orthogonal set does not exist in $L^{2}(\mu_{M,D})$, we not only provide an estimate for the number of orthogonal exponential functions but also determine its maximal cardinality. Finally, we obtain a necessary and sufficient condition for $\mu_{M,D}$ to be a spectral measure.

Lemma 4.1**.**

Suppose that $\sigma\in\{(\frac{s}{t})^{\frac{1}{r}}:s,t,r\in\mathbb{N},\gcd(s,t)=1\}$ satisfies the equation

[TABLE]

where $i,j,l$ are nonnegative integers. Then the following statements hold.

• $(i)$

If $k_{1},k_{2},k_{3}\in\mathbb{Z}\setminus\{0\}$, then $i\equiv j\equiv l\pmod{r}$.

• $(ii)$

If $k_{1},k_{2},k_{3}\in\mathbb{Z}\setminus p\mathbb{Z}$, and $s\in p\mathbb{Z}$ or $t\in p\mathbb{Z}$, where $p$ is a prime, then at least two of the integers $i,j,l$ must be equal.

Proof.

$(i)$ This result is an immediate consequence of [8, Lemma 2.5]. $(ii)$ Suppose, for the sake of contradiction, that $i,j,l$ are distinct nonnegative integers. Without loss of generality, we assume that $i>j>l$. By $(i)$, we can express $i$, $j$, and $l$ as follows: $i=i_{1}r+k,\;j=j_{1}r+k,\;l=l_{1}r+k,$ where $i_{1}>j_{1}>l_{1}$ and $0\leq k\leq r-1$. From $k_{1}\sigma^{i}+k_{2}\sigma^{j}=k_{3}\sigma^{l}$, we obtain the equation

[TABLE]

which can be easily verified to be impossible, thus leading to a contradiction and completing the proof. ∎

Proof of Theorem 1.1.

$"\Rightarrow".$ Suppose that $L^{2}(\mu_{M,D})$ admits an infinite orthogonal set $E_{\Lambda}$. Then

[TABLE]

where

[TABLE]

We claim that $\Lambda^{(1)}$ is infinite. Suppose on the contrary that $\Lambda^{(1)}$ is finite, then $\Lambda^{(2)}$ must be infinite. By the pigeonhole principle, there exist $\lambda_{1}\neq\lambda_{2}\in\Lambda\setminus\{\bm{0}\}$ such that $\lambda_{1,1}=\lambda_{2,1}$, where $\lambda_{i}=(\lambda_{i,1},\lambda_{i,2})^{*}$ for $i=1,2$. Thus,

[TABLE]

which leads to a contradiction. Thus, $\Lambda^{(1)}$ is an infinite set satisfying

[TABLE]

According to Lemma 3.2, we have $\rho^{-1}=(\frac{t}{s})^{\frac{1}{r}}$ for some $s,t,r\in\mathbb{N}$ with $t\in p\mathbb{Z}$ and $\gcd(s,t)=1$.

In the following, we will prove that $c=\kappa\rho^{-1}$, where $\kappa\in\mathbb{Q}$. Since $\#\Lambda=\infty$, without loss of generality, there exist $\lambda_{1}\neq\lambda_{2}\in\Lambda\setminus\{\bm{0}\}$ such that $\lambda_{1},\lambda_{2}\in\mathcal{H}_{1}$. From the definition of $\mathcal{H}_{1}$, we can write

[TABLE]

and

[TABLE]

where $i\in\{1,2,\cdots,p-1\}$, $k_{j_{s}}=(k_{j_{s},1},k_{j_{s},2})^{*}\in\mathbb{Z}^{2}$ for $s=1,2,3$, and $\bm{a}=(a_{1},a_{2})^{*}\in\mathcal{E}_{p}$. By (4.1), we obtain

[TABLE]

and

[TABLE]

Note that $\rho^{-1}=(\frac{t}{s})^{\frac{1}{r}}$ and $t\in p\mathbb{Z}$. By Lemma 4.1, we conclude that $j_{1}\equiv j_{2}\equiv j_{3}\pmod{r}$, and at least two of $j_{1},j_{2},j_{3}$ are equal. Since $p(k_{j_{1},1}-k_{j_{2},1})\in p\mathbb{Z}$ and $ia_{1}+pk_{j_{3},1}\notin p\mathbb{Z}$, it follows from (4.2) that $j_{1},j_{2},j_{3}$ are not all equal. Without loss of generality, we assume that $j_{1}\neq j_{2}=j_{3}$. Thus, combining with (4.2) and (4), we obtain that

[TABLE]

The necessity follows.

$"\Leftarrow".$ Suppose that $\rho^{-1}=(\frac{t}{s})^{\frac{1}{r}}$ and $c=\kappa\rho^{-1}$, where $s,t,r\in\mathbb{N}$, $t\in p\mathbb{Z}$, $\gcd(s,t)=1$, and $\kappa\in\mathbb{Q}$. Then $\gcd(s,p)=1$. Let $\kappa=\frac{v}{u}$, where $u\in\mathbb{N}$, $v\in\mathbb{Z}$, and $\gcd(u,v)=1$. Define

[TABLE]

Note that $t=s\rho^{-r}$, we have $\frac{t^{pu\ell}\bm{a}}{p}=\rho^{-rpu\ell}\frac{s^{pu\ell}\bm{a}}{p}$. Furthermore, since

[TABLE]

we get

[TABLE]

It follows from (4.4), $\gcd(s,p)=1$, and $\frac{(a_{2}-rpu\ell\kappa a_{1})s^{pu\ell}}{p}\in\frac{a_{2}s^{pu\ell}}{p}+\mathbb{Z}$ that $\Lambda\setminus\{\bm{0}\}\subset\mathcal{Z}(\widehat{\mu}_{M,D})$. For any $\lambda_{1},\lambda_{2}\in\Lambda\setminus\{\bm{0}\}$ with $\lambda_{1}\neq\lambda_{2}$, we can write

[TABLE]

for two positive integers $\ell_{1}>\ell_{2}$. Similar to (4.4),

[TABLE]

which implies that $(\Lambda-\Lambda)\setminus\{\bm{0}\}\subset\mathcal{Z}(\widehat{\mu}_{M,D})$. Therefore, $E_{\Lambda}$ forms an infinite orthogonal set in $L^{2}(\mu_{M,D})$. In summary, we have thus established the proof of the theorem. ∎

Proof of Theorem 1.2.

(i) We prove the conclusion by contradiction. Suppose, for the sake of argument, that there exists an orthogonal set $E_{\Lambda}$ such that $\#\Lambda>p$. Without loss of generality, we assume that $\#\Lambda=p+1$, and let $\Lambda=\{\lambda_{0}=\bm{0},\lambda_{1},\lambda_{2},\cdots,\lambda_{p}\}$. For any distinct $i,j\in\{0,1,2,\cdots,p\}$, $\lambda_{i}-\lambda_{j}\in\mathcal{Z}(\widehat{\mu}_{M,D})=\cup_{j=1}^{p-1}\mathcal{H}_{j}$. Let $c=\kappa\rho^{-1}$ where $\kappa\notin\mathbb{Q}$. We can express $\lambda_{i}$ as

[TABLE]

where $\{j_{i}\}_{i=1}^{p}\subset\mathbb{N}$ is a monotonically decreasing sequence, $k_{j_{i}}=(k_{j_{i},1},k_{j_{i},2})^{*}\in\mathbb{Z}^{2}$, $l_{i}\in\{1,2,\cdots,p-1\}$ for $i=1,2,\cdots,p$, and $\bm{a}=(a_{1},a_{2})^{*}\in\mathcal{E}_{p}$. Considering the orthogonality of $\Lambda$ and its first component, we have $j_{1}\equiv j_{2}\equiv\cdots\equiv j_{p}\pmod{r}$, which implies that there exists $0\leq b\leq r-1$ such that

[TABLE]

Clearly, $\{h_{i}\}_{i=1}^{p}$ is a monotonically decreasing sequence. We will divide the proof into two cases as follows.

Case I. $s,t\notin p\mathbb{Z}$. By the pigeonhole principle, and since $s,t,l_{i}a_{1}+pk_{j_{i},1}\notin p\mathbb{Z}$ for $i=1,2,\cdots,p$, it is clear that at least two of the following set

[TABLE]

are congruent modulo $p$. We can assume, without loss of generality, that

[TABLE]

Since $\lambda_{1}-\lambda_{2}\in\cup_{j=1}^{p-1}\mathcal{H}_{j}$, there exist $j_{p+1}\in\mathbb{N}$, $k_{j_{p+1}}=(k_{j_{p+1},1},k_{j_{p+1},2})^{*}\in\mathbb{Z}^{2}$, and $l_{p+1}\in\{1,2,\cdots,p-1\}$ such that

[TABLE]

Using Lemma 4.1, $j_{1}\equiv j_{2}\equiv j_{p+1}\pmod{r}$. Let $j_{p+1}=h_{p+1}r+b$. Then, the equation

[TABLE]

implies that

[TABLE]

One can easily derive that

[TABLE]

The left-hand side of the above equation belongs to $p\mathbb{Z}$ as indicated in (4.6), while the right-hand side does not. This leads to a contradiction.

Case II. $s\in p\mathbb{Z},\ t\notin p\mathbb{Z}$ or $s\notin p\mathbb{Z},\ t\in p\mathbb{Z}.$ By the pigeonhole principle, at least two of $\lambda_{1},\lambda_{2},\cdots,\lambda_{p}$ must fall into the same $\mathcal{H}_{i}$. We may assume that $\lambda_{1},\lambda_{2}\in\mathcal{H}_{1}$, and similarly to (4.5), $(a_{1}+pk_{j_{1},1})\equiv(a_{1}+pk_{j_{2},1})\pmod{p}$. Since $\lambda_{1}-\lambda_{2}\in\cup_{j=1}^{p-1}\mathcal{H}_{j}$, there exist $j_{p+2}\in\mathbb{N}$, $k_{j_{p+2}}=(k_{j_{p+2},1},k_{j_{p+2},2})^{*}\in\mathbb{Z}^{2}$, and $l_{p+2}\in\{1,2,\cdots,p-1\}$ such that

[TABLE]

From (4.5) and (4.7), it follows that

[TABLE]

and

[TABLE]

Given that either $s$ or $t$ is an element of $p\mathbb{Z}$, it follows from Lemma 4.1 that $j_{1}\equiv j_{2}\equiv j_{p+2}\pmod{r}$, with at least two of the indices $j_{1},j_{2},j_{p+2}$ being identical. Note that $(a_{1}+pk_{j_{1},1})\equiv(a_{1}+pk_{j_{2},1})\pmod{p}$, and $l_{p+2}a_{1}+pk_{j_{p+2},1}\notin p\mathbb{Z}$. It follows from (4.8) that $j_{1}$, $j_{2}$, and $j_{p+2}$ cannot all be equal. If $j_{1}=j_{2}$, then $j_{1}\neq j_{p+2}$. Since $j_{1}\equiv j_{p+2}\pmod{r}$, there exists $h_{p+2}\in\mathbb{Z}\setminus\{0\}$ such that $j_{1}-j_{p+2}=h_{p+2}r$. Using $\rho^{-1}=(\frac{t}{s})^{\frac{1}{r}}$ in conjunction with (4.8) and (4), we arrive at the conclusion that

[TABLE]

This leads to a contradiction since $\kappa\notin\mathbb{Q}$. Similarly, if $j_{1}=j_{p+2}$ or $j_{2}=j_{p+2}$, we can also derive a contradiction. Therefore, based on the above reasoning, there can be at most $p$ mutually orthogonal exponential functions in $L^{2}(\mu_{M,D})$.

Let $k\in\mathbb{N}$, and let

[TABLE]

It is straightforward to verify that $(\Lambda-\Lambda)\setminus\{\bm{0}\}\subseteq\mathcal{Z}(\widehat{\mu}_{M,D})$. Therefore, having $p$ elements is optimal, and the result follows.

(ii) Using a proof similar to that in Case I of $(i)$, we can easily complete the proof of $(a)$, and will omit the details here. We will now prove $(b)$. Since $\kappa\in\mathbb{Q}$ and $s\in p\mathbb{Z}$, we express $\kappa=\frac{v}{u}$ for some $u\in\mathbb{N}$, $v\in\mathbb{Z}$ with $\gcd(u,v)=1$, and let $s=p^{\ell}s_{1}$ where $\gcd(s_{1},p)=1$. For any $N\in\mathbb{N}$ and $n\leq N$, we define

[TABLE]

Note that $t,s_{1}\notin p\mathbb{Z}$. It follows easily that

[TABLE]

For any $\lambda_{1}\neq\lambda_{2}\in\Lambda_{n}^{N}$, we define

[TABLE]

where $n_{1}>n_{2}$. From $s=p^{\ell}s_{1}$ and $t,s_{1}\notin p\mathbb{Z}$, it follows that

[TABLE]

which implies that

[TABLE]

Combining with (4.10) and (4.11), we can conclude that for any $N$, $E_{\Lambda_{n}^{N}}$ is the orthogonal set of $L^{2}(\mu_{M,D})$, thus completing the proof. ∎

Next, we proceed to prove Theorem 1.3. From Proposition 3.4, it follows that when $\mu_{M,D}$ is a spectral measure, $\rho=\frac{s}{t}$ with $\gcd(s,t)=1$ and $p\mid t$. Furthermore, we can establish the following proposition.

Proposition 4.2**.**

Let $\mu_{M,D}$ be given by (1.1), (1.2) and (1.3), where $\rho_{1}=\rho_{2}=\rho=\frac{s}{t}$ and $c=\kappa\rho^{-1}$ for some $s,t\in\mathbb{N}$ with $\gcd(s,t)=1$, $p\mid t$, and $\kappa\in\mathbb{Q}$. If $s>1$, then $\mu_{M,D}$ is not a spectral measure.

Proof.

The proof of this proposition follows by a straightforward adaptation of the proof of Proposition 3.2 in [4], and we omit the details here. ∎

Lemma 4.3**.**

Let $\mu_{M,D}$ be defined by (1.1), (1.2) and (1.3), assume that $\rho_{1}=\rho_{2}=\rho^{-1}=t\in p\mathbb{Z}$, $c=a^{\prime}t$, and $a^{\prime}=\frac{v}{u}\in\left\{\frac{k}{l}:\gcd(k,l)=1\text{ and }k\in\mathbb{Z}\right\}$, where $\gcd(v,u)=1$ and $u=p^{\ell_{1}}u^{\prime}\in p^{\ell_{1}}(\mathbb{Z}\setminus p\mathbb{Z})$ for some $\ell_{1}\in\mathbb{N}$. If $p^{\ell_{1}+1}\mid t$, then $\mu_{M,D}$ is a spectral measure.

Proof.

Let $P=\begin{bmatrix}u^{\prime}&0\\ 0&1\end{bmatrix}$. Then, we can write $M_{1}=PMP^{-1}=\begin{bmatrix}t&\frac{vt}{p^{\ell_{1}}}\\ 0&t\end{bmatrix}\;\text{and}\;D_{1}=PD.$ The spectral properties of $\mu_{M,D}$ and $\mu_{M_{1},D_{1}}$ are identical, and

[TABLE]

where $\bm{a}=(a_{1},a_{2})^{*}.$ Define

[TABLE]

Since $p^{\ell_{1}+1}\mid t$, it follows that $L_{1}\subset\mathbb{Z}^{2}$. By calculation, we find that $(M_{1},D_{1},L_{1})$ forms a Hadamard triple. Therefore, by Lemma 2.5, it follows that $\mu_{M_{1},D_{1}}$ is a spectral measure. This completes the proof. ∎

Lemma 4.4**.**

Suppose that $\rho_{1}^{-1}=\rho_{2}^{-1}=t\in p\mathbb{Z}$ and $\lambda_{j}\in\mathcal{L}_{i_{j}}$ for $j=1,2$. If $\lambda_{2}-\lambda_{1}\in\mathcal{Z}(\widehat{\mu}_{M,D})$, then $\lambda_{2}-\lambda_{1}\in\mathcal{L}_{i}:=\mathcal{Z}(\widehat{\delta}_{M^{-i}D})$, where $i=\min\{i_{1},i_{2}\}$ if $i_{1}\neq i_{2}$, and $i\geq i_{1}$ if $i_{1}=i_{2}$.

Proof.

Based on the structure of $\mathcal{L}_{j}$, we can assume

[TABLE]

for some $k_{j}=(k_{j,1},k_{j,2})^{*},s_{j}=(s_{j,1},s_{j,2})^{*}\in\mathbb{Z}^{2}$. Then,

[TABLE]

If $i_{1}=i_{2}$ and $i<i_{1}$, the first component of the above vector is given by

[TABLE]

then

[TABLE]

Since $p\mid t$, the left-hand side is divisible by $p$, while the right-hand side belongs to $\mathbb{Z}\setminus p\mathbb{Z}$, leading to a contradiction. Therefore, we conclude that $i\geq i_{1}$. Similarly, we assume that $i_{2}>i_{1}$, so $\lambda_{2}-\lambda_{1}\in\mathcal{L}_{i}$ for some $i\geq i_{1}$. If $i>i_{1}$, then

[TABLE]

which is a contradiction. Therefore, we must have $i=i_{1}$. ∎

Before proving Theorem 1.3, we define $\mathcal{L}_{j}=\mathcal{Z}(\widehat{\delta}_{M^{-j}D})$ in the same manner as in Lemma 4.4. Thus, $\mathcal{Z}(\widehat{\mu}_{M,D})=\cup_{j=1}^{\infty}\mathcal{L}_{j}$.

Proof of Theorem 1.3.

The sufficiency follows directly from Lemma 4.3, and we now need to consider the necessity. If $\mu_{M,D}$ is a spectral measure, then by Proposition 4.2, $\rho^{-1}=t\in p^{\ell_{1}}(\mathbb{Z}\setminus p\mathbb{Z})$, where $\ell_{1}\geq 1$. If the necessity does not hold, then either $a^{\prime}\notin\mathbb{Q}$ or $a^{\prime}=\frac{v}{u}$, where $u\in p^{\ell_{1}}\mathbb{Z}$ and $\gcd(v,u)=1$. Let $\Lambda$ be the spectrum of $\mu_{M,D}$. From Lemma 2.2, it follows that the intersection of $(\Lambda-\Lambda)$ and $\mathcal{L}_{2}$ is non-empty, i.e., $(\Lambda-\Lambda)\cap\mathcal{L}_{2}\neq\emptyset$. Without loss of generality, we assume that $\bm{0}\in\Lambda$ and $\Lambda\cap\mathcal{L}_{2}\neq\emptyset$. Lemma 4.4 then imply that $\Lambda\cap\mathcal{L}_{1}\neq\emptyset$. Furthermore, since $(\Lambda-\Lambda)\setminus\{\bm{0}\}\subset\cup_{j=1}^{\infty}\mathcal{L}_{j}$, it follows that there exist $\lambda_{1}\in\mathcal{L}_{1}\cap\Lambda$ and $\lambda_{2}\in\mathcal{L}_{2}\cap\Lambda$ such that $\lambda_{2}-\lambda_{1}\in\mathcal{L}_{1}$. Denote

[TABLE]

for some $k_{i}=(k_{i,1},k_{i,2})^{*},s_{i}=(s_{i,1},s_{i,2})^{*}\in\mathbb{Z}^{2}$, $i=1,2$. Then,

[TABLE]

By examining the second component of the above expression and the structure of $\mathcal{L}_{1}$, we conclude that

[TABLE]

for some $s_{3,2},k_{3,2}\in\mathbb{Z}.$ However, since $s_{3,2}+s_{1,2}\equiv 0\pmod{p}$, it follows that $t\frac{v}{u}(s_{2,1}+pk_{2,1})\in p\mathbb{Z}$. Thus, $p^{\ell_{1}+1}\mid t$, leading to a contradiction. Therefore, the proof of necessity is complete. ∎

5. The Case $\rho_{1}\neq\rho_{2}$

In this section, we focus on proving Theorems 1.5 and 1.6, which investigate the spectral properties of the measure $\mu_{M,D}$ defined by conditions (1.1), (1.2), (1.3) and (1.4) when $\rho_{1}\neq\rho_{2}$. As established by Theorem 1.4, the spectrality of $\mu_{M,D}$ necessarily implies that $c^{\prime\prime}$ must be a rational number, i.e., $c^{\prime\prime}\in\mathbb{Q}$. Recall that

[TABLE]

and

[TABLE]

Let $R^{\prime}=\begin{bmatrix}1&c^{\prime\prime}\\ 0&1\end{bmatrix}$, $\widetilde{D}=R^{\prime}D$. Then

[TABLE]

Consequently, $\mu_{M,D}$ and $\mu_{\widetilde{M},\widetilde{D}}$ share the same spectral properties. Define

[TABLE]

and $\widetilde{\mathcal{Z}}_{j}={\widetilde{M}}^{*^{j}}\mathcal{Z}(m_{\widetilde{D}}).$

Proof of Theorem 1.5.

The necessity is directly established by Lemma 3.7. We now prove the sufficiency. Let $\rho_{1}^{-1}=t\in p\mathbb{Z}$, $c^{\prime\prime}=\frac{c_{1}}{c_{2}}\in\mathbb{Q}\backslash E_{\bm{a}}$ with $\gcd(c_{1},c_{2})=1$. Define

[TABLE]

From Lemma 2.9, we know that $\mathcal{B}=\{0,1,\cdots,p-1\}\pmod{p}.$ Let

[TABLE]

Then the triplet $(t,\mathcal{B},L)$ forms a Hadamard triple. By [13, Theorem 5.4], we know the associated self-similar measure $\mu_{t,\mathcal{B}}$ is a spectral with a spectrum in $\mathbb{Z}$. Denote $L_{n}=L+tL+t^{2}L+\cdots+t^{n-1}L$ and

[TABLE]

where $0\in J_{n}\equiv L_{n}\pmod{t^{n}}$ and $\ell_{k}=n_{1}+n_{2}+\cdots+n_{k}$. By [13, Proposition 4.6], we obtain a spectrum $\Lambda$ of $\mu_{t,\mathcal{B}}$ and

[TABLE]

This shows that

[TABLE]

Define $\Lambda^{\prime}_{k}=\{(c_{2}\lambda,0)^{*}:\lambda\in\Lambda_{k}\}$ and $\Lambda^{\prime}=\cup_{k=1}^{\infty}\Lambda^{\prime}_{k}$. For any distinct $\lambda_{1}^{\prime},\lambda_{2}^{\prime}\in\Lambda^{\prime}_{k}$, since $\lambda_{1}-\lambda_{2}\in\frac{1}{p}t^{\ell}(\mathbb{Z}\setminus p\mathbb{Z})$ with $\ell\leq\ell_{k}$, it can be concluded that

[TABLE]

Note that

[TABLE]

for any $z=(z_{1},z_{2})^{*}\subset\mathbb{Z}^{2}$. Given that $\gcd(c_{2},p)=1$ and $c^{\prime\prime}\in\mathbb{Q}\backslash E_{\bm{a}}$, it follows that $\lambda_{1}^{\prime}-\lambda_{2}^{\prime}\subset\mathcal{Z}(\widehat{\widetilde{\mu}}_{\ell_{k}}).$ Therefore, it is not hard to conclude that $\Lambda^{\prime}_{k}$ is a spectrum of $\widetilde{\mu}_{\ell_{k}}$. Moreover, since for any $\lambda^{\prime}=(c_{2}\lambda,0)^{*}\in\Lambda^{\prime}_{k}$, we have

[TABLE]

the equicontinuity of the sequence $\{\widehat{\widetilde{\mu}}_{>\ell_{k}}\}_{k=1}^{\infty}$ implies that

[TABLE]

Thus, $\Lambda^{\prime}$ is the spectrum of $\mu_{\widetilde{M},\widetilde{D}}$, as established in Lemma 2.6 (ii). ∎

Next, we consider the case $c^{\prime\prime}\in E_{\bm{a}}$.

Proposition 5.1**.**

If $\mu_{\widetilde{M},\widetilde{D}}$ is a spectral measure and $c^{\prime\prime}\in E_{\bm{a}}$, then $\rho_{1}^{-1},\rho_{2}^{-1}\in p\mathbb{Z}$.

Proof.

$\rho_{1}^{-1}\in p\mathbb{Z}$ follows from Lemma 3.7. Let $c^{\prime\prime}=\frac{c_{1}}{c_{2}}\in E_{\bm{a}}$ with $\gcd(c_{1},c_{2})=1.$ Since

[TABLE]

similar to the proof of Proposition 3.4, we can conclude that $\rho_{2}^{-1}=\frac{t}{s}$ for some $t\in p\mathbb{Z}$ with $\gcd(s,t)=1$. If $s>1$, let $\xi=(\xi_{1},\xi_{2})^{*}\in\mathbb{R}^{2}$ and

[TABLE]

Clearly, $f(\xi_{1}+c_{2},\xi_{2}+1)=f(\xi_{1},\xi_{2})$ and

[TABLE]

From Lemma 2.7, for any $n\geq 1$ and $\xi=(\xi_{1},\xi_{2})^{*}$ with $|\rho_{2}^{-n}\xi_{2}|>1$, there exist positive constants $\alpha,\beta$ such that

[TABLE]

Thus, by Proposition 3.6, $\mu_{\widetilde{M},\widetilde{D}}$ cannot be a spectral measure, leading to a contradiction. Hence, $s=1$, which implies $\rho_{2}^{-1}\in p\mathbb{Z}$. ∎

Proposition 5.2**.**

Suppose that $\rho_{1}^{-1},\rho_{2}^{-1}\in p\mathbb{Z}$ and $c^{\prime\prime}=\frac{c_{1}}{c_{2}}\in E_{\bm{a}}$, where $\gcd(c_{1},c_{2})=1$ and $c_{2}\in p^{\ell}(\mathbb{Z}\setminus p\mathbb{Z})$ for some $\ell\in\mathbb{N}$. If $p^{\ell+1}\nmid(\rho_{1}^{-1}-\rho_{2}^{-1})$, then $\mu_{\widetilde{M},\widetilde{D}}$ is not a spectral measure.

Proof.

Assuming that $\mu_{\widetilde{M},\widetilde{D}}$ is spectral measure with a spectrum $\bm{0}\in\Lambda$. From $\rho_{1}^{-1},\rho_{2}^{-1}\in p\mathbb{Z}$ and Lemma 3.8, it follows that $\Lambda\cap\widetilde{\mathcal{Z}}_{j}\neq\emptyset$ for $j=1,2$. Denote

[TABLE]

for some $k=(k_{j,1},k_{j,2})^{*}\in\mathbb{Z}^{2}$, $i_{j}\in\{1,2,\cdots,p-1\}$, $j=1,2$. Then, $\lambda_{2}-\lambda_{1}\in\widetilde{\mathcal{Z}}_{1}$ follows as

[TABLE]

Furthermore, we get

[TABLE]

From the expression $i_{3}a_{2}+pk_{3,2}+i_{1}a_{2}+pk_{1,2}\in p\mathbb{Z}$, it follows that $(\rho_{1}^{-1}-\rho_{2}^{-1})c^{\prime\prime}(i_{2}a_{1}+pk_{2,1})\in p\mathbb{Z}$. This implies that $p^{\ell+1}\mid(\rho_{1}^{-1}-\rho_{2}^{-1})$, which leads to a contradiction. Therefore, the proof is complete. ∎

Combining with Proposition 5.1 and Proposition 5.2, we can derive the necessary part of Theorem 1.6. For the sufficiency, it requires several additional technical lemmas. For convenience, we define

[TABLE]

and the integral periodic zero set of $P^{\prime}$ as

[TABLE]

Lemma 5.3**.**

Suppose that $\rho_{1}^{-1}=t_{1},\;\rho_{2}^{-1}=t_{2}\in p\mathbb{Z}$ and $c^{\prime\prime}=\frac{c_{1}}{c_{2}},\;\text{where }\gcd(c_{1},c_{2})=1\text{ and }c_{2}=p^{u}c_{2}^{\prime}\in p^{u}(\mathbb{Z}\setminus p\mathbb{Z})$ for some $u\in\mathbb{N}$. If

[TABLE]

then $\mu_{\widetilde{M},\widetilde{D}}$ is a spectral measure.

Proof.

According to Lemma 2.3, $\widehat{\mu}_{\widetilde{M},\widetilde{D}}$ is uniformly continuous. This implies that there exists $\delta>0$ such that for any $\xi\in G$, there exists $k_{\xi}\in\mathbb{Z}^{2}$ satisfying

[TABLE]

whenever $\parallel y\parallel\leq\delta.$ Construct the sequence $\{\Lambda_{n_{\ell}}\}$ with $n_{0}=0$ and $n_{\ell+1}>n_{\ell}$ such that

[TABLE]

For any $\bm{w}_{\ell}\in\{0,1,2,\cdots,p-1\}^{n_{\ell+1}-n_{\ell}}:=\Xi^{n_{\ell+1}-n_{\ell}}$, let

[TABLE]

and

[TABLE]

Along with (5.1) and (5.2), we have

[TABLE]

Furthermore, we will prove that $\Lambda=\cup_{\ell=1}^{\infty}\Lambda_{n_{\ell}}$ is the spectrum of $\mu_{\widetilde{M},\widetilde{D}}$. For any $\lambda_{1}\neq\lambda_{2}\in\Lambda_{n_{\ell}}$, there exist two distinct sequences $\{w_{1}^{(i)}\}_{i=1}^{n_{\ell}}$,$\{w_{2}^{(i)}\}_{i=1}^{n_{\ell}}\subset\{0,1,2,\cdots,p-1\}^{n_{\ell}}$, such that

[TABLE]

where $k_{x_{w_{i}^{(j)}}}=(k_{j_{i},1},k_{j_{i},2})^{*}$ and $l=\min\{j:w_{1}^{(j)}\neq w_{2}^{(j)}\}\in(n_{s},n_{s+1}]$ for some $0\leq s\leq\ell-1$. The expression for $\lambda_{1}-\lambda_{2}$ can be rewritten as follows:

[TABLE]

It follows from $p^{u+1}\mid(t_{1}-t_{2})$ that

[TABLE]

Therefore, we conclude that $\widetilde{\mu}_{n_{\ell}}(\lambda_{1}-\lambda_{2})=0$. Since $\#\text{spt}\,\widetilde{\mu}_{n_{\ell}}=p^{n_{\ell}}$, it immediately follows that $\Lambda_{n_{\ell}}$ is a spectrum of $\widetilde{\mu}_{n_{\ell}}$ for all $\ell\geq 1$. By Lemma 2.3, the sequence $\{\widehat{\widetilde{\mu}}_{>n_{\ell}}\}_{\ell=1}^{\infty}$ is equicontinuous. Therefore, for $\lambda\in\Lambda_{n_{\ell+1}}\setminus\Lambda_{n_{\ell}},\,\xi\in[0,r]^{2}$ with some $r>0$, using (5.1), we obtain that

[TABLE]

Moreover, by Lemma 2.6, $\Lambda$ is a spectrum of $\mu_{\widetilde{M},\widetilde{D}}$. ∎

Lemma 5.4**.**

Suppose that $\rho_{1}^{-1}=t_{1},\;\rho_{2}^{-1}=t_{2}\in p\mathbb{Z}$ and $c^{\prime\prime}=\frac{c_{1}}{c_{2}},\;\text{where }\gcd(c_{1},c_{2})=1\text{ and }c_{2}=p^{u}c_{2}^{\prime}\in p^{u}(\mathbb{Z}\setminus p\mathbb{Z})$ for some $u\in\mathbb{N}$. If $Z(\mu_{\widetilde{M},\widetilde{D}})=\emptyset$, then

[TABLE]

Proof.

For every $\xi\in G$, the condition $Z(\mu_{\widetilde{M},\widetilde{D}})=\emptyset$ implies the existence of an integer vector $k_{\xi}\in\mathbb{Z}^{2}$ such that

[TABLE]

By the continuity of $\widehat{\mu}_{\widetilde{M},\widetilde{D}}$, we obtain

[TABLE]

Define $B(\xi,\delta_{\xi}):=\{x\in\mathbb{R}^{2}:\text{dist}(x,\xi)<\delta_{\xi}\}.$ Since $G$ is a compact set and $G\subset\cup_{\xi\in G}B(\xi,\delta_{\xi})$, by the finite covering theorem, there exists a finite collection of balls $\{B(\xi_{i},\delta_{\xi_{i}})\}_{i=1}^{\ell}$ such that $\cup_{i=1}^{\ell}B(\xi_{i},\delta_{\xi_{i}})\supset G.$ For any $\xi\in G$, there exists a $j$ with $1\leq j\leq\ell$ such that $\xi\in B(\xi_{j},\delta_{\xi_{j}})$. Thus,

[TABLE]

∎

According to Lemma 5.3 and Lemma 5.4, for the sufficiency of Theorem 1.6, we only need to prove that $Z(\mu_{\widetilde{M},\widetilde{D}})=\emptyset$. However, before proceeding, we require a key lemma.

Lemma 5.5**.**

Suppose that $\rho_{1}^{-1}=t_{1},\;\rho_{2}^{-1}=t_{2}\in p\mathbb{Z}$ and $c^{\prime\prime}=\frac{c_{1}}{c_{2}},\;\text{where }\gcd(c_{1},c_{2})=1\text{ and }c_{2}=p^{u}c_{2}^{\prime}\in p^{u}(\mathbb{Z}\setminus p\mathbb{Z})$ for some $u\in\mathbb{N}$. For any $k=(k_{1},k_{2})^{*}\in\mathbb{Z}^{2}$, if $\widehat{\mu}_{\widetilde{M},\widetilde{D}}(\xi+P^{\prime}k)=0$, then there exists $j_{k}$ such that $m_{\widetilde{D}}({\widetilde{M}}^{-j_{k}}(\xi+P^{\prime}k))=0.$

Proof.

From the inequalities $|z|\geq|\text{Re}(z)|$ and $\cos(2\pi x)\geq\cos(2\pi r)$ for all $|x|\leq r<\frac{1}{4}$, it follows that

[TABLE]

We can choose an appropriate $J$ such that for all $i\in\{0,1,\cdots,p-1\}$, the following inequality holds:

[TABLE]

Hence,

[TABLE]

Furthermore, we know that

[TABLE]

Together with the assumption that $\widehat{\mu}_{\widetilde{M},\widetilde{D}}(\xi+P^{\prime}k)=0,$ this implies that $\widehat{\widetilde{\mu}}_{J}(\xi+P^{\prime}k)=0.$ Consequently, we have $m_{\widetilde{D}}({\widetilde{M}}^{-j_{k}}(\xi+P^{\prime}k))=0$ for some $j_{k}\leq J$. ∎

We now proceed to demonstrate that $Z(\mu_{\widetilde{M},\widetilde{D}})=\emptyset$.

Lemma 5.6**.**

Suppose that $\rho_{1}^{-1}=t_{1},\;\rho_{2}^{-1}=t_{2}\in p\mathbb{Z}$ and $c^{\prime\prime}=\frac{c_{1}}{c_{2}},\;\text{where }\gcd(c_{1},c_{2})=1\text{ and }c_{2}=p^{u}c_{2}^{\prime}\in p^{u}(\mathbb{Z}\setminus p\mathbb{Z})$ for some $u\in\mathbb{N}$, then $Z(\mu_{\widetilde{M},\widetilde{D}})=\emptyset$.

Proof.

If not, we let $\xi=(\xi_{1},\xi_{2})^{*}\in Z(\mu_{\widetilde{M},\widetilde{D}})$. For any $k=(n,\ell)^{*}\in\mathbb{Z}^{2}$, the definition of $Z(\mu_{\widetilde{M},\widetilde{D}})$ ensures that $\xi+P^{\prime}k\in\mathcal{Z}(\widehat{\mu}_{\widetilde{M},\widetilde{D}})$. By Lemma 5.5, there exists $j_{k}$ such that $\xi+P^{\prime}k\in\mathcal{Z}(m_{{\widetilde{M}}^{-j_{k}}\widetilde{D}})$, i.e.,

[TABLE]

for some $k_{1},k_{2}\in\mathbb{Z}$, $i_{k}\in\{1,2,\cdots,p-1\}$. For the convenience of subsequent discussion, we define $j_{k}=j_{n,\ell_{n}},k_{i}=k_{i}^{(n,\ell_{n})}$ and $i_{k}=i_{n,\ell_{n}}.$ Take $n=0$ and $\ell_{0}\in\mathbb{Z}$. By equation (5.4), there exist two positive integers $j_{0,\ell_{0}}$ and $k_{1}^{(0,\ell_{0})}$ such that

[TABLE]

Substituting this into equation (5.4), we obtain

[TABLE]

We claim that for any $(n,\ell_{n})^{*}\in\{0,1,\cdots,p-1\}\times\mathbb{Z}$, there are exactly $p-1$ elements in $\{j_{n,\ell_{n}}\}_{n=0}^{p-1}$ that are equal to $1$, and the remaining one is greater than $1$.

Let $n\in\{1,2,\cdots,p-1\}$. If $j_{n,\ell_{n}}\leq j_{0,\ell_{0}}$, then $j_{n,\ell_{n}}=1$ and $t_{1}\in p(\mathbb{Z}\setminus p\mathbb{Z})$. Otherwise, using equation (5.5), we obtain

[TABLE]

which leads to a contradiction since $c^{\prime}_{2}n\in\mathbb{Z}\setminus p\mathbb{Z}$. If $j_{n,\ell_{n}}>j_{0,\ell_{0}}$, then $j_{n,\ell_{n}}\geq 2$ and we have

[TABLE]

This implies $j_{0,\ell_{0}}=1$ and $t_{1}\in p(\mathbb{Z}\setminus p\mathbb{Z})$. Using (5.4) again, for any distinct $n_{1},n_{2}\in\{1,2,\cdots,p-1\}$, we obtain

[TABLE]

Together with $c^{\prime}_{2}\in\mathbb{Z}\setminus p\mathbb{Z}$, this leads to the conclusion that $\min\{j_{n_{1},\ell_{n_{1}}},j_{n_{2},\ell_{n_{2}}}\}=1$. The arbitrariness of $n_{1}$ and $n_{2}$ tells us that at least $p-1$ elements in the set $\{j_{n,\ell_{n}}\}_{n=0}^{p-1}$ must be equal to 1. If they are all equal to 1, let $hc^{\prime}_{2}=-t_{1}\frac{i_{0,\ell_{0}}a_{1}}{p}\pmod{p}$ for some $h\in\{1,2,\cdots,p-1\}$. Then,

[TABLE]

which leads to a contradiction, and the claim follows.

Building on the previous claim, we assume that $j_{0,\ell_{0}}>1$ and $j_{1,\ell_{1}}=j_{2,\ell_{2}}=\cdots=j_{p-1,\ell_{p-1}}=1$ for all vectors $(\ell_{0},\ell_{1},\cdots,\ell_{p-1})^{*}\in\mathbb{Z}^{p}$. Let $n=1$ and $\ell_{1}\in\{0,1,\cdots,p-1\}$. It follows from (5.4) that

[TABLE]

Through simple calculations, we get

[TABLE]

Note that $\ell_{1}\in\{0,1,\cdots,p-1\}$ and $i_{1,\ell_{1}}\in\{1,\cdots,p-1\}$. Then, we can select two distinct integers $\ell_{1}^{\prime},\ell_{1}^{\prime\prime}\in\{0,1,\cdots,p-1\}$ such that $i_{1,\ell_{1}^{\prime}}=i_{1,\ell_{1}^{\prime\prime}}$. From (5.6),

[TABLE]

which leads to a contradiction, thus completing the proof. ∎

6. Examples

In the final section, we present some examples to emphasize our key findings. Firstly, we give a counterexample to refute [3, Conjecture 4.2].

Example 6.1**.**

Define

[TABLE]

Then

[TABLE]

It is known that $\mu_{M,D}$ and $\mu_{M_{1},D_{1}}$ have the same spectrality. Let

[TABLE]

We can check $(M_{1},D_{1},L_{1})$ forms a Hadamard triple. Thus $\mu_{M,D}$ and $\mu_{M_{1},D_{1}}$ are both spectral measure. The spectral measure $\mu_{M,D}$ can be verified to satisfy all conditions of Theorem 1.3, while $c=\frac{3}{4}=\frac{1}{8}\times 6=\kappa\beta$ and $p\kappa=2\times\frac{1}{8}\notin\mathbb{Z}$. This implies that [3, Conjecture 4.2] is false. As shown in Figs. 4 to 4, these figures also illustrate the first four approximations of the Sierpinski fractal of $\mu_{M,D}$.

Example 6.2**.**

Let

[TABLE]

Then

[TABLE]

According to Theorem 1.7, the measure $\mu_{M,D}$ is a spectral measure if and only if the matrix $M$ satisfies one of the following conditions:

1. $(i)$

$\rho_{1}\neq\rho_{2}$, $\rho_{1}^{-1}\in 5\mathbb{Z}$, and $c^{\prime\prime}\in\mathbb{Q}\setminus E_{\bm{a}}$, where $E_{\bm{a}}=E_{(1,1)^{*}}=\{\frac{s}{t}:t-s\in\mathbb{Z}\backslash 5\mathbb{Z},\;\gcd(s,t)=1\}\cup\{0\}$. 2. $(ii)$

$\rho_{1}\neq\rho_{2}$, $\rho_{1}^{-1},\rho_{2}^{-1}\in 5\mathbb{Z}$, and $c^{\prime\prime}=\frac{c_{1}}{c_{2}}\in E_{\bm{a}}$, $c_{2}\in 5^{\ell}(\mathbb{Z}\setminus 5\mathbb{Z})$, $5^{\ell+1}\mid(\rho_{1}^{-1}-\rho_{2}^{-1})$ for some $\ell\in\mathbb{N}$. 3. $(iii)$

$\rho_{1}=\rho_{2}\in 5\mathbb{Z}$* and $c\in\{\frac{s}{t}:s\in 5\mathbb{Z},\gcd(s,t)=1\}\cup\{0\}$.*

We present the following example to illustrate Theorems 1.1 and 1.2.

Example 6.3**.**

Let

[TABLE]

and

[TABLE]

Then by Theorems 1.1 and 1.2, $L^{2}(\mu_{M_{1},D})$ admits an infinite set of orthogonal exponential functions, while $L^{2}(\mu_{M_{2},D})$ contains an arbitrary number of orthogonal exponential functions. Moreover, there can be at most $5$ mutually orthogonal exponential functions in both $L^{2}(\mu_{M_{3},D})$ and $L^{2}(\mu_{M_{4},D})$, with $5$ being the maximum possible. Figs. 8 to 8 show the second iterative fractal graphs of the measure $\mu_{M_{1},D}$, $\mu_{M_{2},D}$, $\mu_{M_{3},D}$, and $\mu_{M_{4},D}$, respectively.

Finally, we propose two natural questions.

Question 6.4**.**

In Theorem 1.7, if we replace $d_{1}=(d_{1,1},0)^{*}$ with $d_{1}=(d_{1,1},d_{1,2})^{*},d_{1,2}\neq\bm{0}$ in the condition (1.4), does the conclusion still hold?

Question 6.5**.**

How to extend our results to higher dimensions or Moran measure?

Acknowledgment

Competing interests. The authors declare that they have no competing interests.

Data availability. Data availability is not applicable to this article as no new data were created or analyzed in this study.