text:
\section{Introduction}

It is well known that the set of exponential functions $\{e^{2\pi in x}:~n\in\mathbb{Z}\}$ is an orthonormal basis for $L^2(I)$ where $I=[0,1]$. An interval is not the only set on which an orthogonal basis consisting of exponential functions exists for $L^2(\Omega)$. For example, for the set $\Omega=[0,1]\cup [2,3]$ the set of exponential functions $\{e^{2\pi i\lambda x}:~\lambda\in\mathbb{Z} \cup \mathbb{Z}+\frac{1}{4}\}$ is an orthogonal basis for $L^2(\Omega)$. It is an interesting question to ask what sets $\Omega$ admit an orthogonal basis consisting of exponential functions. In 1974 Fuglede, in a study \cite{Fug74} that is quite well known today, proposed the following infamous conjecture, widely known today as Fuglede's Conjecture:

\medskip

\noindent
{\bf Fuglede's Conjecture:}~~A measurable set $K$ is a spectral set in ${\mathbb R}^d$ if and only if $K$ tiles ${\mathbb R}^d$ by translation.

\medskip

Fuglede coined the term {\em spectral set} to denote sets $\Omega$ that admit admit an orthogonal basis of exponential functions. Fuglede's Conjecture has been studied by many investigators, including the authors of this paper, Jorgensen, Pedersen, Lagarias, {\mathcal{L}}aba, Kolountzakis, Matolcsi, Iosevich, Tao, and others (\cite{JP98,MR1700084,IKT1,IKT2,Ko1,MR2237932,La,LRW,LaWa96,LaWa97,Tao04}), but it had baffled experts for 30 years until Terence Tao \cite{Tao04} gave the first counterexample of a spectral set which is not a tile in ${\mathbb R}^d$, $d\geq5$. The example and technique were refined later to disprove the conjecture in both directions on ${\mathbb R}^d$ for $d\geq3$ \cite{MR2159781,MR2237932,MR2264214}. It has remained open in dimensions $d=1$ and $d=2$.

\medskip

Although the Fuglede's Conjecture in its original form has been disproved, there is a clear connection between spectral sets and tiling that has remained a mystery. Furthermore, spectral sets are apparently only one of the problems among an extremely broad class of problems involving complex exponential functions as either bases or more generally, frames.

\medskip

The rest of this section will focus on Fourier bases and frames for measures. Basic concepts and some of the important recent results will be introduced and reviewed. They will be the gateway to further discussions on the subject in later sections, which include open questions and more recent progresses.

\subsection{ Fourier bases and spectral measures.}

One extension of spectral sets concerns the more general spectral measures. A {\it spectral measure} is a bounded Borel measure $\mu$ such that there exists a set of complex exponentials  $E(\Lambda): =\{e_\lambda\}_{\lambda\in\Lambda}$, where $e_\lambda:=e^{2\pi i \langle \lambda,x\rangle}$, that forms an orthogonal basis of $L^2(\mu)$. If such $\Lambda$ exists, it is called a \textit{spectrum} for $\mu$. A spectral set can be viewed as a special case of spectral measure by considering the Lebesgue measure $\chi_K dx$ on $K$. Of particular intrigue are measures that are singular, especially self-similar measures that are closely tied to the field of fractal geometry and self-affine tiles.

The study of Fourier bases on singular measures started with the Jorgensen-Pedersen paper \cite{JP98} in which they proved the following surprising result: consider the construction of a Cantor set with scale 4: take the unit interval, divide it into 4 equal pieces, keep the first and the third piece. Then for the two remaining intervals, perform the same procedure and keep repeating the procedure for the remaining intervals ad infinitum. The remaining set is the Cantor-4 set in the Jorgensen-Pedersen example. On this set consider the appropriate Hausdorff measure of dimension $\frac12$, which we denote by $\mu_4$.

Jorgensen and Pedersen proved that the Hilbert space $L^2(\mu_4)$ has an orthonormal family of exponential functions, i.e., a Fourier basis, namely
$$
   \left\{e_\lambda : \lambda=\sum_{k=0}^n4^kl_k, l_k\in\{0,1\}, n\in\mathbb{N}\right\}.
$$
Moreover, they also proved that the Middle-Third-Cantor set, with its corresponding Hausdorff measure cannot have more than two mutually orthogonal exponential functions.

\begin{definition}\label{defsp}
Let $\mu$ be a compactly supported Borel probability measure on
${\mathbb{R}}^d$ and $\langle\cdot,\cdot\rangle$ denote the standard inner product. We say that $\mu$ is a {\it spectral measure} if there
exists a countable set $\Lambda\subset {\mathbb R}^d$, which we call a {\it spectrum}, such  that
$E(\Lambda): = \{e^{2\pi i \langle\lambda,x\rangle}:
\lambda\in\Lambda\}$ is an orthonormal basis for $L^2(\mu)$. The Fourier transform of $\mu$ is defined to be
 $$
\widehat{\mu}(\xi)= \int e^{-2\pi i \langle\xi,x\rangle}d\mu(x).
 $$
 It is easy to verify that a measure is a spectral measure with spectrum $\Lambda$ if and only if the following two conditions are satisfied:
 \begin{enumerate}
\item (Orthogonality) $ \widehat{\mu}(\lambda-\lambda')=0$ for all distinct $\lambda,\lambda'\in\Lambda$ and
\item (Completeness) If for $f\in L^2(\mu)$, $\int f(x)e^{-2\pi i \langle\lambda,x\rangle}d\mu(x)=0$ for all $\lambda\in\Lambda$, then  $f=0$.
    \end{enumerate}
\end{definition}

The Jorgensen-Pedersen Cantor-4 set and the measure $\mu_4$ can be seen as the attractor and invariant measure of the iterated function system
$$\tau_0(x)=\frac{x}{4},\quad \tau_2(x)=\frac{x+2}{4}.$$

\begin{definition}\label{defifs}
For a given expansive $d\times d$ integer matrix $R$ and a finite set of integer vectors $B$ in $\mathbb{Z}^d$, with $\#B =: N$, we define the {\it affine iterated function system} (IFS) $\tau_b(x) = R^{-1}(x+b)$, $x\in \mathbb{R}^d, b\in B$. The {\it self-affine measure}, also called the {\it invariant measure}, (with equal weights) is the unique probability measure $\mu = \mu(R,B)$ satisfying
\begin{equation}\label{self-affine}
\mu(E) = \sum_{b\in B} \frac1N \mu (\tau_b^{-1} (E)),\mbox{ for all Borel subsets $E$ of $\mathbb{R}^d$.}
\end{equation}
This measure is supported on the {\it attractor} $T(R,B)$ which is the unique compact set that satisfies
$$
T(R,B)= \bigcup_{b\in B} \tau_b(T(R,B)).
$$
The set $T(R,B)$ is also called the {\it self-affine set} associated with the IFS. See \cite{Hut81} for details.
\end{definition}

The starting idea in Jorgensen-Pedersen's construction is to find first orthogonal exponential functions at a finite level and then iterate by rescaling and translations. If we start with the atomic measure
$\delta_{\frac14 B}:=\delta_{\frac14\{0,2\}}=\frac{1}{2}(\delta_0+\delta_2)$, we find that $\{e_{\ell} : \ell\in L:=\{0,1\}\}$ is an orthonormal basis for $L^2(\delta_{\frac14 B})$. In other words, $(R=4, B=\{0,2\}, L=\{0,1\})$ form a Hadamard triple.

\begin{definition}\label{hada}
Let $R\in M_d({\mathbb Z})$ be an $d\times d$ expansive matrix (all eigenvalues have modulus strictly greater than 1) with integer entries. Let $B, L\subset{\mathbb Z}^d $ be a finite set of integer vectors with $N= \#B=\#L$ ($\#$ means the cardinality). We assume without loss of generality that $0\in B$ and $0\in L$. We say that the system $(R,B,L)$ forms a {\it Hadamard triple} (or $(R^{-1}B, L)$ forms a {\it compatible pair} in \cite{LaWa02} ) if the matrix
\begin{equation}\label{Hadamard triples}
H=\frac{1}{\sqrt{N}}\left[e^{2\pi i \langle R^{-1}b,\ell\rangle}\right]_{\ell\in L, b\in B}
\end{equation}
is unitary, i.e., $H^*H = I$.

\end{definition}

The system $(R,B,L)$ forms a  Hadamard triple if and only if the Dirac measure $\delta_{R^{-1}B} = \frac{1}{\#B}\sum_{b\in B}\delta_{R^{-1}b}$ is a spectral measure with  spectrum $L$. Moreover, this property is a key property in producing examples of singular spectral measures, in particular spectral self-affine measures.

\medskip
Denote by
$$
B_n = B+RB+\dots+R^{n-1}B, \ \Lambda_n:=L+R^TL+\dots+(R^T)^{n-1}L.
$$
If $(R,B,L)$ form a Hadamard triple, then a simple computation (see e.g. \cite{JP98}) shows that $(R^n, B_n, \Lambda_n)$ is also a Hadamard triple, for all $n\in\mathbb{N}$. In other words, the measures
$$\nu_n = \delta_{R^{-n}(B_n)}=\delta_{R^{-1}B}\ast\delta_{R^{-2}B}
\ast...\ast\delta_{R^{-n}B}$$
have a spectrum $\Lambda_n$
These measures approximate the singular measure $\mu(R,B)$ in the following sense,
\begin{equation}\label{conv prod}
\mu(R,B) = \delta_{R^{-1}B}\ast\delta_{R^{-2}B}
\ast\delta_{R^{-3}B}... = \mu_n\ast\mu_{>n}
\end{equation}
where, by self-similarity, the measure $\mu_{>n}(\cdot) = \mu(R^n\cdot) $. One should expect that, under the right conditions, the set $\Lambda = \bigcup_{n=1}^{\infty}\Lambda_n$ forms a spectrum for the measure $\mu(R,B)$ and this is the case for the Jorgensen-Pedersen example. However, this is not true in general, even though it always yields an orthonormal set, but, in some cases, this set can be incomplete. Here is a simple counterexample: consider the scale $R=2$, and the digits $B=\{0,1\}$. It is easy to see that $\mu(R,B)$ is the Lebesgue measure on $[0,1]$. One can pick $L=\{0,1\}$ to make a Hadamard triple. But when we construct the set $\Lambda$ we notice that it is actually $\Lambda=\mathbb{N}\cup\{0\}$ and we know that the classical Fourier series are indexed by all the integers $\mathbb{Z}$, so the exponential functions with frequencies in $\Lambda$ are orthonormal, but not complete.

\medskip

The natural conjecture was raised to see {\it if for any Hadamard triple $(R,B,L)$ the measure $\mu(R,B)$ is spectral}. This conjecture  was settled on ${\mathbb R}^1$ \cite{LaWa02,DJ06}. The situation becomes more complicated on high dimension. Dutkay and Jorgensen showed that the conjecture  is true if $(R,B,L)$ satisfies a technical condition called {\it reducibility condition} \cite{DJ07d}. There are some other additional assumptions proposed by Strichartz guaranteeing the conjecture is true \cite{Str98,Str00}. Some low-dimensional special cases were also considered by Li \cite{MR3163581,MR3302160}. We eventually proved this conjecture in \cite{DL15,DHL15}.

\begin{theorem}\label{th1}
If $(R,B,L)$ is a Hadamard triple, then the self-affine measure $\mu=\mu(R,B)$ is a spectral measure.
\end{theorem}

The proof of this result highlights some of the useful techniques for the study of spectral self-affine measures. We will give a sketch of proof of this result as part of this review.

\bigskip

\subsection{  Fourier frames} A natural generalization of orthonormal basis is the notion of frames. It allows expansion of functions in a non-unique way, but is robust to perturbation of frequencies \cite{DHSW11}. Recall that a {\it frame} is a family of vectors $\{e_i: i \in I\}$ in a Hilbert space $H$ with the property that there exist constants $A,B>0$ (called {\it the frame bounds}) such that
\begin{equation}
A\|f\|^2\leq \sum_{i\in I}|\ip{f}{e_i}|^2\leq B\|f\|^2,\quad(f\in H).
\label{eqfr}
\end{equation}
A Borel measure $\mu$ is called a {\it frame-spectral} measure if there exists a family of exponential functions $\{e_\lambda : \lambda\in\Lambda\}$ forming a frame for $L^2(\mu)$.
\medskip

Soon after Jorgensen and Pedersen showed in \cite{JP98} that the Middle-Third-Cantor measure $\mu_3$ is not a spectral measure, a natural question was proposed by Strichartz \cite{Str00} who asked   whether the measure $\mu_3$  is frame-spectral. The question is still open.

\medskip

We proved in \cite{DL14} that if a measure has a Fourier frame, then it must have a certain homogeneity under local translations (so it must look the same locally at every point). This condition excludes the possibility of Fourier frames on self-affine measures with unequal weights, but not for the Middle-Third-Cantor measures. Some fairy large Bessel sequences of exponential functions (i.e., only the upper bound holds in \eqref{eqfr}) were constructed in \cite{MR2826404} for the Middle-Third-Cantor set. Some weighted Fourier frames were built by Picioroaga and Weber for the Cantor-4 set in \cite{PiWe15}.

\medskip

The following condition generalizing Hadamard triples was introduced in \cite{DL15} and \cite{LaiW}. We state the definition on ${\mathbb R}^1$, but it can be defined on any dimension.

\medskip

\begin{definition}\label{APFT}
 Let $\epsilon_j$ be such that $0\le\epsilon_j<1$ and $\sum_{j=1}^{\infty}\epsilon_j<\infty$. We say that $\{(N_j,B_j)\}$ is an {\it almost-Parseval-frame tower} associated to $\{\epsilon_j\}$ if
\begin{enumerate}
\item $N_j$ are integers and $N_j\geq2$ for all $j$;
\item  $B_j\subset \{0,1,...,N_j-1\}$ and $ 0 \in B_j$ for all $j$;
\item  Let $M_j := \#B_j$. There exists $L_j\subset {\mathbb Z}$ (with $0\in L_j$) such that for all $j$,
 \begin{equation}\label{ALmost}
(1-\epsilon_j)^2\sum_{b\in B_j}|w_b|^2\leq \sum_{\lambda\in L_j}\left|\frac{1}{\sqrt{M_j}}\sum_{b\in B_j}w_be^{-2\pi i b \lambda/N_j}\right|^2\leq (1+\epsilon_j)^2\sum_{b\in B_j}|w_b|^2
\end{equation}
 for all ${\bf w} = (w_b)_{b\in{B_j}}\in{\mathbb C}^{M_j}$.
 Letting the matrix ${\mathcal F}_j = \frac{1}{\sqrt{M_j}}\left[e^{2\pi i b \lambda/N_j}\right]_{\lambda\in L_j,b\in B_j}$ and $\|\cdot\|$ the standard Euclidean norm, (\ref{ALmost}) is equivalent to
\begin{equation}\label{ALmost1}
(1-\epsilon_j)\|{\bf w}\|\leq \|{\mathcal F}_j{\bf w}\|\leq (1+\epsilon_j)\|{\bf w}\|
\end{equation} for all  ${\bf w}\in {\mathbb C}^{M_j}$.
 \end{enumerate}
  Whenever  $\{L_j\}_{j\in{\mathbb Z}}$ exists, we call $\{L_j\}_{j\in{\mathbb Z}}$ a {\it pre-spectrum} for the almost-Parseval-frame tower.
We define the following measures associated to an almost-Parseval-frame tower.
$$
\nu_j = \frac{1}{M_{j}}\sum_{b\in B_{j}}\delta_{b/N_1N_2...N_j}
$$
and
\begin{equation}\label{measure}
\mu = \nu_1\ast\nu_2\ast.... : = \mu_n\ast\mu_{>n}
\end{equation}
where $\mu_n$ is the convolution of the first $n$ discrete measures and $\mu_{>n}$ is the remaining part.

\end{definition}

\medskip

When all $\epsilon_j=0$, $(N_j,B_j,L_j)$ forms a Hadamard triple.   We note that this class of measures includes self-similar measures because if we are given an integer $N\ge2$ and a set $B\subset\{0,1,...,N-1\}$ such that
$$
N_j = N^{n_j}, B_j= B+NB+...+N^{n_j-1}B,
$$
 then the associated measure $\mu$ is the self-affine measure $\mu(N,B)$. In particular if $N=3$ and $B = \{0,2\}$, $\mu$ is the standard Middle Third Cantor measure. In such situation, the almost-Parseval-frame tower is called {\it self-similar}. We prove that

 \begin{theorem} \label{th2}

 (i) If the self-similar almost-Parseval-frame tower as in Definition \ref{APFT} exists, then the associated self-similar measure is frame-spectral.

 \medskip

 (ii)  There exists almost-Parseval-frame tower with $\epsilon_j>0$ and the associated fractal measure is frame-spectral but not spectral.
 \end{theorem}

In the rest of the paper we go into more details on spectral self-affine measures as well as frame-spectral measures. We consider the explicit construction of self-affine frame-spectral measures that are not spectral. It is our hope that this survey summarizes not only recent results on the subject, but also some of the key techniques used to tackle problems. The open questions we discuss here should serve to quickly lead people into this area.

\medskip

\section{Spectral Self-Affine Measures}

 We begin with the following definitions.

\begin{definition}
We call a finite set $B\subset\mathbb{Z}^d$, a {\it simple digit set for $R$}, if distinct elements of $B$ are not congruent $(\mod R(\mathbb{Z}^d))$. We define $\mathbb{Z}[R,B]$ to be the smallest lattice in $\mathbb{R}^d$ that contains the set $B$ and is invariant under $R$, i.e., $R(\mathbb{Z}[R,B])\subset \mathbb{Z}[R,B]$.

\medskip

We say that $\mu(R,B)$ satisfies the {\it no-overlap} condition if
$$
\mu(\tau_b(T(R,B))\cap \tau_{b'}(T(R,B))) = 0, \ \forall b\neq b'\in B.
$$
\end{definition}

We have the following preliminary reductions that we can do to solve the problem.

\begin{enumerate}
\item If $(R,B,L)$ is a Hadamard triple, then $B$ is a simple digit set for $R$ and $L$ is a simple digit set for $R^T$.
\item We can assume without loss of generality that $\mathbb{Z}[R,B] = \mathbb{Z}^d$.
\item If $B$ is a simple digit for $R$, then $\mu(R,B)$ satisfies the no-overlap condition.
\end{enumerate}

Property (i) is a simple consequence of mutually orthogonality. If $\mathbb{Z}[R,B] \ne \mathbb{Z}^d$, we can conjugate some matrix to produce another Hadamard triple which satisfies with the desired property \cite[Proposition 4.1]{DL15}. The proof of the no-overlap condition of the self-affine measure can be referred to \cite[Section 2]{DL15}.

\medskip

The following proposition perhaps gives us the main idea on how to prove the completeness of an orthogonal set of exponentials and the proof are readily generalized to give our various results.

\begin{proposition}\label{prop_main}
Suppose that $(R,B,L)$ is a Hadadmard triple and $
\Lambda = \bigcup_{n=1}^{\infty}\Lambda_n $ where $\Lambda_n = L+R^TL+...+(R^T)^{n-1}L.$ Assume that
$$
\delta(\Lambda): = \inf_{n\ge1}\inf_{\lambda\in\Lambda_n}|\widehat{\mu}((R^T)^{-n}\lambda)|^2>0
$$
Then $\Lambda$ is a spectrum for $L^2(\mu(R,B))$.

\end{proposition}

\begin{proof}
The proof of mutual orthogonality follows directly from the fact that
$$
H_n = \frac{1}{\sqrt{N^n}}\left(e^{-2\pi i \ip{{ R}^{-n}\bf b}{\lambda}}\right)_{\lambda\in\Lambda_n,{\bf b}\in {B}_n}.
$$
is a unitary matrix. We now show the completeness by showing that the following frame bounds hold: for any $f\in L^2(\mu)$,
\begin{equation}\label{eq2.1i}
\delta(\Lambda)\|f\|^2\leq \sum_{\lambda\in\Lambda}\left|\int f(x)e^{-2\pi i \langle\lambda,x\rangle}d\mu(x)\right|^2\leq \|f\|^2.
\end{equation}
The positive lower bound implies the completeness. To prove (\ref{eq2.1i}), we just need to check it for a dense set of functions in $L^2(\mu)$. Let ${\bf 1}_E$ be the indicator function of the set $E$ and
$$
{\mathcal S}_n = \left\{\sum_{{\bf b}\in {B}_n}w_{\bf b}{\bf 1}_{\tau_{\bf b}(R,B)}: w_{\bf b}\in{\mathbb C}\right\}.
$$
Thus, ${\mathcal S}_n$ denotes the collection of all $n^{th}$ level step functions on $T(R,B)$. It forms an increasing union of sets. Let also ${\mathcal S} = \bigcup_{n=1}^{\infty}{\mathcal S}_n$. This ${\mathcal S}$ forms a dense set in  $L^2(\mu)$. Now for any $f = \sum_{{\bf b}\in {B}_n}w_{\bf b}{\bf 1}_{\tau_{\bf b}(R,B)}\in {\mathcal S}$, a direct computation shows that
\begin{equation}\label{eq3.1}
\int|f|^2d\mu = \frac{1}{N^n}\sum_{{\bf b}\in {B}_n}|w_{\bf b}|^2 =  \frac{1}{N^n}\|{\bf w}\|^2
\end{equation}
where ${\bf w} = (w_{\bf b})_{{\bf b}\in B_n}$ and
\begin{equation}\label{eq3.2}
\int f(x)e^{-2\pi i \ip{\lambda}{x}}d\mu(x) = \frac{1}{N^n}\widehat{\mu}((R^T)^{-n}\lambda)\sum_{{\bf b}\in {B}_n}w_{\bf b} e^{-2\pi i  \ip{R^{-n}{\bf b}}{ \lambda}}
\end{equation}
which means that
\begin{equation}\label{eq3.3}
\sum_{\lambda\in\Lambda_n}\left|\int f(x)e^{-2\pi i \ip{\lambda}{x}}d\mu(x)\right|^2 = \frac{1}{N^n} \sum_{\lambda\in\Lambda_n}|\widehat{\mu}((R^T)^{-n}\lambda)|^2\left|\sum_{{\bf b}\in {B}_n}\frac{1}{\sqrt{N^n}}w_{\bf b}e^{-2\pi i  \ip{R^{-n}{\bf b}}{\lambda}}\right|^2
\end{equation}
As $\delta(\Lambda)\le|\widehat{\mu}((R^T)^{-n}\lambda)|^2\le 1$, we obtain
$$
 \frac{1}{N^n}\delta(\Lambda)\|H_n{\bf w}\|^2\le \sum_{\lambda\in\Lambda_n}\left|\int f(x)e^{-2\pi i \ip{\lambda}{x}}d\mu(x)\right|^2 \le  \frac{1}{N^n}\|H_n{\bf w}\|^2 .
$$ But $H_n$ is a Hadamard matrix, so we have $\|H_n{\bf w}\| = \|{\bf w}\|$ and hence
$$
\delta(\Lambda)\int|f|^2d\mu\le \sum_{\lambda\in\Lambda_n}\left|\int f(x)e^{-2\pi i \ip{\lambda}{x}}d\mu(x)\right|^2 \le \int|f|^2d\mu .
$$
As ${\mathcal S}_n\subset{\mathcal S}_m$ for any $n<m$, we have
$$
\delta(\Lambda)\int|f|^2d\mu\le \sum_{\lambda\in\Lambda_m}\left|\int f(x)e^{-2\pi i \ip{\lambda}{x}}d\mu(x)\right|^2 \le \int|f|^2d\mu .
$$
By taking $m$ to infinity, we have (\ref{eq2.1i}).
\end{proof}

As we have discussed in the introduction, we cannot expect the standard orthogonal set to be always a spectrum. We have to look for some other alternatives. The main important observation is to distinguish two cases depending on whether the periodic zero set
\begin{equation}
\mathcal Z:=\{\xi\in\mathbb{R}^d : \widehat\mu(\xi+k)=0,\mbox{ for all }k\in\mathbb{Z}^d\}
\label{eqz}
\end{equation}
of $\mu(R,B)$ is empty or not.

For the case $\mathcal Z=\emptyset$, we will see that Theorem \ref{th1} follows along the same lines as Proposition \ref{prop_main}.  The case $\mathcal Z\neq\emptyset$ is more subtle but it can appear in higher dimensions $d$, but, in this case, the system $(R,B,L)$ has a special quasi-product form structure.

\subsection{The case $\mathcal Z=\emptyset$}

Given a sequence of integers $n_k$ and let $m_k = n_1+...+n_k$. The self-affine measure can be rewritten as
$$
\mu(R,B) = \delta_{R^{-n_1}B_{n_1}}\ast\delta_{R^{-m_2}B_{n_2}}\ast...\ast\delta_{R^{-m_k}B_{n_k}}\ast...
$$
Then note that if we have another set $J_{n_k}$ of integer vectors, with $J_{n_k}\equiv L_{n_k}$ (mod $(R^T)^{n_k}({\mathbb Z}^d)$), then it is easy to see that $(R^{n_k}, B_{n_k},J_{n_k})$ still form Hadamard triples. In this sequel, we produce many other mutually orthogonal sets, by:
\begin{equation}\label{eqLambda_k}
\Lambda_k = J_{n_1}+ (R^T)^{m_1}J_{n_2}+ (R^T)^{m_2} J_{n_3}+...+ (R^T)^{m_{k-1}} J_{n_{k}},
\end{equation}
\begin{equation}\label{eqlambda}
\ \Lambda = \bigcup_{k=1}^{\infty}\Lambda_k.
\end{equation}
Repeating the argument in Proposition \ref{prop_main}, the following holds true.

\begin{proposition}\label{prop_main1}
Suppose that $(R,B,L)$ is a Hadadmard triple and $
\Lambda$ if given in (\ref{eqlambda}). Assume that
\begin{equation}\label{eqdelta}
\delta(\Lambda): = \inf_{k\ge1}\inf_{\lambda\in\Lambda_k}|\widehat{\mu}((R^T)^{-m_k}\lambda)|^2>0
\end{equation}
Then $\Lambda$ is a spectrum for $L^2(\mu(R,B))$
\end{proposition}

\medskip

The following proposition guarantees some $\Lambda$ will satisfy $\delta(\Lambda)>0$.

\begin{proposition}\label{prop_main2}
Suppose that ${\mathcal Z} = \emptyset$. Then there exists $\Lambda$ built as in \eqref{eqLambda_k} and \eqref{eqlambda} such that $\delta(\Lambda)>0$.
\end{proposition}

We now give the proof of this proposition. We start with a lemma.

\begin{lemma}\label{lem2.1}
Suppose that ${\mathcal Z} = \emptyset$ and let $X$ be any compact set on ${\mathbb R}^d$. Then there exist $\epsilon_0>0$, $\delta_0>0$ such that for all $x\in X$, there exists $k_x\in{\mathbb Z}^d$ such that for all $y\in\mathbb{R}^d$ with $\|y\|<\epsilon_0$, we have $|\widehat\mu(x+y+k_x)|^2\geq \delta_0$. In addition, we can choose $k_0=0$ if $0\in X$.
\end{lemma}

\begin{proof}
As ${\mathcal Z} = \emptyset$, for all $x\in X$ there exists $k_x\in\Gamma$ such that $\widehat\mu(x+k_x)\neq 0$. Since $\widehat\mu$ is continuous, there exists an open ball $B(x,\epsilon_x)$ and $\delta_x>0$ such that $|\widehat\mu(y+k_x)|^2\geq\delta_x$ for all $y\in B(x,\epsilon_x)$. Since $X$ is compact, there exist $x_1,\dots,x_m\in X$ such that
$$X\subset\bigcup_{i=1}^mB(x_i,\frac{\epsilon_{x_i}}2).$$
Let $\delta:=\min_i\delta_{x_i}$ and $\epsilon:=\min_i\frac{\epsilon_{x_i}}{2}$. Then, for any $x\in X$, there exists $i$ such that $x\in B(x_i,\frac{\epsilon_{x_i}}2)$. If $\|y\|<\epsilon$, then $x+y\in B(x_i,\epsilon_{x_i})$, so $|\widehat \mu(x+y+k_{x_i})|^2\geq \delta$, we can redefine $k_x$ to be $k_{x_i}$ to obtain the conclusion. Clearly, we can choose $k_0=0$ if $0\in X$ since $\widehat{\mu}(0)=1$.
\end{proof}

\medskip

\noindent{\it Proof of Proposition \ref{prop_main2}.}  Suppose that $(R,B,L)$ is a Hadamard triple $(R,B,L)$. Then we take $X = T(R^T,L)$, the self-affine set generated by $R^T$ and digit set $L$. Define $J_{n} = L+R^{T}L+...+(R^T)^{n-1}L$
 By the definition of self-affine sets,
$$
(R^{T})^{-(n+p)}J_{n}\subset X , \quad(n\in{\mathbb N},p\geq 0).
$$

Fix the $\epsilon_0$ and $\delta_0$ in Lemma \ref{lem2.1}. We now construct the sets $\Lambda_k$ and $\Lambda$ as in \eqref{eqLambda_k} and \eqref{eqlambda}, by replacing the sets $J_{n_k}$ by some sets $\widehat J_{n_k}$ to guarantee that the number $\delta(\Lambda)$ in \eqref{eqdelta} is positive.

 We first start with $\Lambda_{0}: = \{0\} $ and $m_0=n_0=0$. Assuming that $\Lambda_k$ has been constructed, we first choose our $n_{k+1}>n_k$ so that
\begin{equation}\label{eq4.5}
\|(R^T)^{-(n_{k+1}+p)}\lambda\|<\epsilon_0 , \ \forall \ \lambda\in\Lambda_k, p\geq0.
\end{equation}
We then define $m_{k+1} = m_k+n_{k+1}$ and
$$
\Lambda_{k+1} = \Lambda_k+(R^T)^{m_{k}}\widehat{J_{n_{k+1}}}
$$
where
$$
\widehat{J_{n_{k+1}}} = \{j+(R^T)^{n_{k+1}}k(j): j\in J_{n_{k+1}}, \ k(j)\in{\mathbb Z}^d  \}
$$
where $k(j)$ is chosen to be $k_x$ from Lemma \ref{lem2.1}, with $x = (R^{T})^{-n_{k+1}}j  \in X$. As $0\in J_{n_{k}}$ and $k_0=0$ for all $k$, the sets $\Lambda_k$ are of the form (\ref{eqLambda_k}) and  form an increasing sequence. For these sets $\Lambda_k$, we claim that the associated $\Lambda$ in $(\ref{eqlambda})$ satisfies $\delta(\Lambda)>0$.

\medskip

To justify the claim, we note that if $\lambda\in \Lambda_{k}$, then
$$
\lambda = \lambda'+(R^T)^{m_{k-1}}j  +(R^T)^{m_{k}}k(j),
$$
where $\lambda'\in \Lambda_{k-1}$, $j\in J_{n_{k}}$. This means that
$$
(R^T)^{-m_k}\lambda = (R^T)^{-m_k}\lambda'+(R^T)^{-n_k}j+ k(j).
$$
By (\ref{eq4.5}), $\|(R^T)^{-m_k}\lambda'\|<\epsilon_0$. From Lemma \ref{lem2.1}, since $(R^T)^{-n_k}j\in X$, we must have $|\widehat{\mu}((R^T)^{-m_k}\lambda)|^2\geq\delta_0>0$. As $\delta_0$ is independent of $k$, the claim is justified and hence this completes the proof of the proposition.
\qquad$\Box$.

Combining Proposition \ref{prop_main1} and Proposition \ref{prop_main2}, we settle the case ${\mathcal Z}=\emptyset$.

\begin{theorem}\label{th_Z}
Suppose that ${\mathcal Z} = \emptyset$ and $(R,B,L)$ is a Hadamard triple. Then the self-affine measure $\mu(R,B)$ is a spectral measure.
\end{theorem}

\medskip

\subsection{The case $\mathcal Z\neq \emptyset$} When $\mathcal Z\neq \emptyset$, it means that there is an exponential function $e_{\xi}$ such that it is orthogonal to every exponential function with integer frequencies. This implies that none of the subsets of integers can be complete. Therefore, any construction of orthogonal sets within integers must fail to be spectral. We illustrate the situation through a simple example.

\begin{example}\label{example5.3}
Let $R = \left[
           \begin{array}{cc}
             4 & 0 \\
             1 & 2 \\
           \end{array}
         \right]
$, $$B= \left\{\left[
                \begin{array}{c}
                  0 \\
                  0 \\
                \end{array}
              \right], \left[
                \begin{array}{c}
                  0 \\
                  3\\
                \end{array}
              \right], \left[
                \begin{array}{c}
                  1 \\
                  0 \\
                \end{array}
              \right], \left[
                \begin{array}{c}
                  1 \\
                  3 \\
                \end{array}
              \right]
\right\} \ \mbox{and} \
 L= \left\{\left[
                \begin{array}{c}
                  0 \\
                  0 \\
                \end{array}
              \right], \left[
                \begin{array}{c}
                  2 \\
                  0\\
                \end{array}
              \right],  \left[
                \begin{array}{c}
                  0 \\
                  1 \\
                \end{array}
              \right], \left[
                \begin{array}{c}
                  2 \\
                  1 \\
                \end{array}
              \right]
\right\}.$$ Then $(R,B,L)$ forms a  Hadamard triple and ${\mathbb Z}[R,B] = {\mathbb Z}^2$. However, the set  defined in \eqref{eqz} ${\mathcal Z}\neq \emptyset$ for the measure $\mu = \mu(R,B)$.
\end{example}

\begin{proof}
It is a direct check to see $(R,B,L)$ forms a  Hadamard triple and ${\mathbb Z}[R,B] = {\mathbb Z}^2$.  As $M_B(\xi_1,\xi_2) = \frac{1}{4}(1+e^{2\pi i \xi_1})(1+e^{2\pi i 3\xi_2})$. It follows that the zero set of $M_B$, denoted by $Z(M_B)$, is equal to
$$
Z(M_B) = \left\{\left[
                \begin{array}{c}
                 \frac12+n \\
                  y \\
                \end{array}
              \right]: n\in{\mathbb Z}, y\in {\mathbb R}\right\}\cup \left\{\left[
                \begin{array}{c}
                  x \\
                  \frac16+\frac13 n  \\
                \end{array}
              \right]:x\in {\mathbb R}, n\in{\mathbb Z}\right\}.
$$
Let $(R^T)^j = \left[
           \begin{array}{cc}
             4^j & a_j \\
              0& 2^j \\
           \end{array}
         \right]$, for some $a_j\in{\mathbb Z}$. As $\widehat{\mu}(\xi) = \prod_{j=1}^{\infty}M_B((R^T)^{-j}(\xi))$, the zero set of $\widehat{\mu}$, denoted by $Z(\widehat{\mu})$, is equal to
$$\begin{aligned}
Z(\widehat{\mu}) =& \bigcup_{j=1}^{\infty}(R^{T})^{j} Z(M_B) \nonumber\\
=&\bigcup_{j=1}^{\infty}\left\{\left[
                \begin{array}{c}
                 4^j(\frac12+n)+a_jy \\
                  2^jy \\
                \end{array}
              \right]: n\in{\mathbb Z}, y\in {\mathbb R}\right\}\cup \left\{\left[
                \begin{array}{c}
                  4^jx+a_j(\frac16+\frac13 n) \\
                  2^{j}\left(\frac16+\frac13 n\right)  \\
                \end{array}
              \right]:x\in {\mathbb R}, n\in{\mathbb Z}\right\}.
\end{aligned}
$$
We claim that the points in  $\left[
                \begin{array}{c}
                 0 \\
                  \frac13 \\
                \end{array}
              \right]+{\mathbb Z}^2$ are in $Z(\widehat{\mu})$ which shows ${\mathcal Z}\neq \emptyset$. Indeed, for any $\left[
                \begin{array}{c}
                 m \\
                  \frac13+n \\
                \end{array}
              \right]$, $m,n\in{\mathbb Z}$, we can write it as $\left[
                \begin{array}{c}
                 m \\
                  \frac{1+3n}{3} \\
                \end{array}
              \right]$. We now rewrite the second term in the union in $Z(\widehat{\mu})$ as ${\mathbb R}\times \{\frac{2^{j-1}(1+2n)}{3}\}$. As any integer can be written as $2^j p$, for some $j\ge 0$ and odd number $p$, this means that $\left[
                \begin{array}{c}
                 m \\
                  \frac{1+3n}{3} \\
                \end{array}
              \right]\in Z(\widehat{\mu})$, justifying the claim. As ${\mathcal Z}\neq\emptyset,$ this shows that there is no spectrum in ${\mathbb Z}^2$ for this measure.  In fact,
              $$
T(R,B) = \bigcup_{x\in K_1}\{x\}\times ([0,3]+g(x)),
$$
where $K_1$ is the Cantor set of $1/4$ contraction ratio and digit $\{0,1\}$ and $g:[0,1]\rightarrow{\mathbb R}$ is a measurable function obtaining from the off-diagonal entries.

\end{proof}
\medskip

To overcome this obstruction, as we will see, we prove that ${\mathcal Z}$ has a dynamical structure and, from this structure, we obtain that such Hadamard triples have to have a special quasi-product form, and moreover, in one of the factors, the digits form complete set of representatives.

\begin{proposition}\label{pr1.14}
Suppose that $(R,B,L)$ forms a Hadamard triple and ${\mathbb Z}[R,B] = {\mathbb Z}^d$ and let $\mu = \mu(R,B)$ be the associated self-affine measure $\mu=\mu(R,B)$. Suppose that the set

$$\mathcal Z:=\left\{x\in \mathbb{R}^d : \widehat\mu (x+k)=0\mbox{ for all }k\in\mathbb{Z}^d\right\},$$
is non-empty. Then there exists an integer matrix $M$ with $\det M=1$ such that the following assertions hold:

\begin{enumerate}
	\item The matrix $\tilde R:=MRM^{-1}$ is of the form
	\begin{equation}
	\tilde R=\begin{bmatrix} R_1&0\\ C &R_2\end{bmatrix},
	\label{eq1.14.1}
	\end{equation}
	with $R_1\in M_r(\mathbb{Z})$, $R_2\in M_{d-r}(\mathbb{Z})$ expansive integer matrices and  $C\in M_{(d-r)\times r}(\mathbb{Z})$.
	\item If $\tilde B=MB$ and $\tilde L=(M^T)^{-1}L$, then $(\tilde R, \tilde B,\tilde L)$ is a Hadamard triple.
	\item The measure $\mu(R,B)$ is spectral with spectrum $\Lambda$ if and only if the measure $\mu(\tilde R,\tilde B)$ is spectral with spectrum $(M^T)^{-1}\Lambda$.
	\item There exists $y_0\in\mathbb{R}^{d-r}$ such that $(R_2^T)^my_0\equiv y_0(\mod (R_2^T)\mathbb{Z}^d)$ for some integer $m\geq 1$ such that the union
	$$\tilde {\mathcal S}=\bigcup_{k=0}^{m-1}(\mathbb{R}^r\times \{(R_2^T)^ky_0\}+\mathbb{Z}^d)$$
	is contained in the set
	$$\tilde{\mathcal Z}:=\left\{x\in \mathbb{R}^d : \widehat{\tilde \mu} (x+k)=0\mbox{ for all }k\in\mathbb{Z}^d\right\},$$
	where $\tilde \mu=\mu(\tilde R,\tilde B)$. The set $\tilde{\mathcal S}$ is invariant (with respect to the system $(m_{\tilde B},\tilde R^T,\tilde {\cj L},)$, see the definition below and Definition \ref{definv}) where $\tilde {\cj L}$ is a complete set of representatives $(\mod \tilde R^T\mathbb{Z}^d)$. In addition, all possible transitions from a point in $\mathbb{R}^r\times \{(R_2^T)^ky_0\}+\mathbb{Z}^d$, $1\leq k\leq m$ lead to a point in $\mathbb{R}^r\times \{(R_2^T)^{k-1}y_0\}+\mathbb{Z}^d$.
\end{enumerate}
\end{proposition}

The key fact in the proof of this proposition is that the set $\mathcal Z$ is invariant in the sense defined by Conze et al. in \cite{CCR}. That means that if $x\in\mathcal Z$, $k\in \mathbb{Z}^d$ and $m_B((R^T)^{-1}(x+k))\neq 0$ then $(R^T)^{-1}(x+k)$ is in $\mathcal Z$,  where
$$m_B(x)=\frac{1}{N}\sum_{b\in B}e^{2\pi i \ip{b}{x}}.$$
Then, the results from \cite{CCR} show that $\mathcal Z$ must have a very special form, and this implies the proposition.

\begin{theorem}\label{th_quasi}
Suppose that $$R=\begin{bmatrix}
R_1& 0\\ C& R_2
\end{bmatrix},
$$$(R,B,L_0)$ is a Hadamard triple and $\mu = \mu(R,B)$ is the associated self-affine measure and ${\mathcal Z}\neq \emptyset$. Then the set $B$ has the following quasi-product form:
\begin{equation}
B=\left\{(u_i,v_i+Qc_{i,j})^T : 1\leq i\leq N_1, 1\leq j\leq|\det R_2|\right\},
\label{eq1.18.1}
\end{equation}
where
 \begin{enumerate}\item $N_1 = N/|\det R_2|$, \item $Q$ is a $(d-r)\times (d-r)$ integer matrix with $|\det Q|\geq 2$ and $R_2Q=Q\tilde R_2$ for some $(d-r)\times(d-r)$ integer matrix $\widetilde{R_2}$, \item the set $\{Qc_{i,j}: 1\leq j\leq |\det R_2|\}$ is a complete set of representatives $(\mod R_2(\mathbb{Z}^{d-r}))$, for all $1\leq i\leq N_1$.
\end{enumerate}
\medskip

Moreover, one can find some $L\equiv L_0 (\mod \ R^T(\mathbb Z^d))$ so that $(R,B,L)$ is a Hadamard triple and $(R_1,\pi_1(B), L_1(\ell_2))$ and $(R_2,B_2(b_1),\pi_2(L))$ are Hadamard triples on ${\mathbb R}^r$ and ${\mathbb R}^{d-r}$ respectively for all $b_1\in\pi_1(B)$ and $l_2\in\pi_2(L)$.
\end{theorem}
Here $\pi_1,\pi_2$ are the projections onto the first and second components in $\mathbb{R}^d=\mathbb{R}^r\times\mathbb{R}^{d-r}$, and for $b_1\in\pi_1(B)$, $B_2(b_1):=\{b_2 : (b_1,b_2)\in B\}$ and
for $l_2\in\pi_2(L)$, $L_1(l_2):=\{l_1: (l_1,l_2)\in L\}$.

\medskip

In Example \ref{example5.3}, it is easy to see that the digit set is in a quasi-product form with $Q = 3$. Suppose now the pair $(R,B)$ is in the quasi-product form
\begin{equation}\label{R_4.1}R=\begin{bmatrix}
R_1&0\\
C&R_2
\end{bmatrix}\end{equation}
\begin{equation}
B=\left\{(u_i,d_{i,j})^T : 1\leq i\leq N_1, 1\leq j\leq N_2:=|\det R_2|\right\},
\label{eq1.23.1}
\end{equation}
and $\{d_{i,j}:1\leq j\leq N_2\}$ ($d_{i,j} = v_i +Qc_{i,j}$ as in Theorem \ref{th_quasi}) is a complete set of representatives $(\mod R_2\mathbb{Z}^{d-r})$.
We will show that the measure $\mu=\mu(R,B)$ has a quasi-product structure.

\medskip

Note that we have
$$R^{-1}=\begin{bmatrix}
R_1^{-1}&0\\
-R_2^{-1}CR_1^{-1}&R_2^{-1}
\end{bmatrix}$$
and, by induction,
$$R^{-k}= \begin{bmatrix}
R_1^{-k}&0\\
D_k&R_2^{-k}
\end{bmatrix},\mbox{ where }D_k:=-\sum_{l=0}^{k-1}R_2^{-(l+1)}CR_1^{-(k-l)}.$$
 For the invariant set $T(R,B)$, we can express it as a set of infinite sums,
 $$
 T(R,B)=\left\{\sum_{k=1}^\infty R^{-k}b_k : b_k\in B\right\}.
 $$
Therefore any element $(x,y)^T\in T(R,B)$ can be written in the following form
$$x=\sum_{k=1}^\infty R_1^{-k}a_{i_k}, \quad y=\sum_{k=1}^\infty D_ka_{i_k}+\sum_{k=1}^\infty R_2^{-k}d_{i_k,j_k}.$$
Let $X_1$ be the attractor (in $\mathbb{R}^r)$ associated to the IFS defined by the pair $(R_1,\pi_1(B)=\{u_i :1\leq i\leq N_1\})$ (i.e. $X_1=T(R_1,\pi_1(B))$). Let $\mu_1$ be the (equal-weight) invariant measure associated to this pair.

\medskip

For each sequence $\omega=(i_1i_2\dots)\in\{1,\dots,N_1\}^{\mathbb{N}} = \{1,\dots, N_1\}\times\{1,\dots, N_1\}\times...$, define
\begin{equation}\label{eqxomega}
x(\omega)=\sum_{k=1}^\infty R_1^{-k}u_{i_k}.
\end{equation}
 As $(R_1,\pi_1(B))$ forms Hadamard triple with some $L_1(\ell_2)$, the measure $\mu(R_1,\pi_1(B))$ has the no-overlap property. It implies that for $\mu_1$-a.e. $x\in X_1$, there is a unique $\omega$ such that $x(\omega)=x$. We define this as $\omega(x)$. This establishes a bijective correspondence, up to measure zero, between the set $\Omega_1:=\{1,\dots,N_1\}^{\mathbb{N}}$ and $X_1$. The measure $\mu_1$ on $X_1$ is the pull-back of the product measure which assigns equal probabilities $\frac1{N_1}$ to each digit.

\medskip

For $\omega=(i_1i_2\dots)$ in $\Omega_1$, define
$$\Omega_2(\omega):=\{(d_{i_1,j_1}d_{i_2,j_2}\dots d_{i_n,j_n}\dots) : j_k\in \{1,\dots,N_2\}\mbox{ for all }k\in\mathbb{N}\}.$$
For $\omega\in\Omega_1$, define $g(\omega):=\sum_{k=1}^\infty D_ka_{i_k}$ and $g(x):=g(\omega(x))$, for $x\in X_1$.  Also $\Omega_2(x):=\Omega_2(\omega(x))$.
For $x\in X_1$, define
$$X_2(x):=X_2(\omega(x)):=\left\{\sum_{k=1}^\infty R_2^{-k}d_{i_k,j_k}: j_k\in\{1,\dots,N_2\}\mbox{ for all }k\in\mathbb{N}\right\}.$$
Note that the attractor $T(R,B)$ has the following form

$$T(R,B)=\{(x,g(x)+y)^T: x\in X_1,y\in X_2(x)\}.$$

\medskip

For $\omega\in \Omega_1$,  consider the product probability measure $\mu_\omega$, on $\Omega_2(\omega)$, which assigns equal probabilities $\frac{1}{N_2}$ to each digit $d_{i_k,j_k}$ at level $k$.
Next, we define the measure $\mu_\omega^2$ on $X_2(\omega)$. Let
$r_\omega:\Omega_2(\omega)\rightarrow X_2(\omega)$,
$$r_\omega(d_{i_1,j_1}d_{i_2,j_2}\dots)=\sum_{k=1}^\infty R_2^{-k}d_{i_k,j_k}.$$
Define $\mu_x^2:=\mu_{\omega(x)}^2:=\mu_{\omega(x)}\circ r_{\omega(x)}^{-1}$.

\medskip

Note that the measure $\mu_x^2$ is the infinite convolution product $\delta_{R_2^{-1}B_2(i_1)}\ast\delta_{R_2^{-2}B_2(i_2)}\ast\dots$, where $\omega(x)=(i_1i_2\dots)$, $B_2(i_k):=\{d_{i_k,j} : 1\leq j\leq N_2\}$ and $\delta_A:=\frac{1}{\#A}\sum_{a\in A}\delta_a$, for a subset $A$ of $\mathbb{R}^{d-r}$.

The following lemmas were proved in \cite{DJ07d}.

\begin{lemma}\label{lem1.23}\cite[Lemma 4.4]{DJ07d}
For any bounded Borel functions on $\mathbb{R}^d$,
$$\int_{T(R,B)}f\,d\mu=\int_{X_1}\int_{X_2(x)}f(x,y+g(x))\,d\mu_x^2(y)\,d\mu_1(x).$$
\end{lemma}

\begin{lemma}\label{lem1.24}\cite[Lemma 4.5]{DJ07d}
If $\Lambda_1$ is a spectrum for the measure $\mu_1$, then
$$F(y):=\sum_{\lambda_1\in\Lambda_1}|\widehat\mu(x+\lambda_1,y)|^2=\int_{X_1}|\widehat\mu_s^2(y)|^2\,d\mu_1(s),\quad(x\in\mathbb{R}^r,y\in\mathbb{R}^{d-r}).$$
\end{lemma}

The two lemmas lead to the following proposition.

\begin{proposition}\cite{DHL15}\label{lem1.25}
For the quasi-product form given in (\ref{R_4.1}) and (\ref{eq1.23.1}), there exists a lattice $\Gamma_2$ such that for $\mu_1$-almost every $x\in X_1$, the set $\Gamma_2$ is a spectrum for the measure $\mu_x^2$.
\end{proposition}

Finally, the proof of the Theorem \ref{th1} follows by induction on the dimension $d$: we know it is true for $d=1$ from \cite{DJ06}. Then assume it is true for dimensions up to $d-1$. The case $\mathcal Z=\emptyset$ was treated before; if $\mathcal Z\neq\emptyset$, then the measure $\mu$ is in the quasi-product described above. With Proposition \ref{lem1.25}, and using induction, the measure $\mu_1$ has a spectrum $\Gamma_1$ and then the measure $\mu$ has the spectrum $\Gamma_1\times\Gamma_2$.

\medskip
\section{Non-spectral singular measures with Fourier frames}

Suppose that instead of the Hadamard triple, we are given the almost-Parseval-frame tower in Definition \ref{APFT}. A similar approach in Proposition \ref{prop_main} (for details see \cite{LaiW})  allows us to prove the following:

\begin{proposition}
Suppose that $\{(N_j,B_j)\}$ is an {\it almost-Parseval-frame tower} associated to $\{\epsilon_j\}$ with $\{L_j\}$ as its pre-spectrum. Let $
\Lambda = \bigcup_{n=1}^{\infty}\Lambda_n $ where $\Lambda_n = L_1+N_1L_2+...+(N_1...N_{n-1})L_n$ and let $\mu$ be the measure defined in (\ref{measure}). Assume that
$$
\delta(\Lambda): = \inf_{n\ge1}\inf_{\lambda\in\Lambda_n}|\widehat{\mu_{>n}}(\lambda)|^2>0
$$
Then $\Lambda$ is a frame spectrum for $L^2(\mu)$ and for any $f\in L^2(\mu)$,
\begin{equation}\label{eq3.1i}
\delta(\Lambda)\prod_{j=1}^{\infty}(1-\epsilon_j)^2\|f\|^2\leq \sum_{\lambda\in\Lambda}\left|\int f(x)e^{-2\pi i \langle\lambda,x\rangle}d\mu(x)\right|^2\leq \prod_{j=1}^{\infty}(1+\epsilon_j)^2\|f\|^2.
\end{equation}
\end{proposition}

\medskip

If we have a self-similar almost-Parseval-frame tower (all $N_j = N$), then the measure $\mu_{>n}(\cdot)$ can be written as $\mu(N^n\cdot)$. In this case, we can produce new candidates of frame spectra as in (\ref{eqLambda_k}) and (\ref{eqlambda}) and the consideration for ${\mathcal Z}= \emptyset$ works in a similar way as in Proposition \ref{prop_main2}. We have

\begin{theorem} \label{theorem 3.1}
Suppose that $(N_j,B_j)$ is a self-similar almost-Parseval-frame tower and the associated measure $\mu$ satisfies ${\mathcal Z} = \emptyset$. Then $\mu$ is a frame-spectral measure.
\end{theorem}

 For a self-similar measure on ${\mathbb R}^1$ as defined in Definition \ref{APFT}, ${\mathcal Z} = \emptyset$ can be obtained without additional assumption.

  \medskip

 \noindent{\it Proof of Theorem \ref{th2} (i).} By Theorem \ref{theorem 3.1}, it suffices to show that ${\mathcal Z} = \emptyset$ for self-similar measures $\mu(N,B)$ defined by the almost-Parseval-frame tower in Definition \ref{APFT}. In fact, as $B\subset \{0,1,...,N-1\}$, the self-similar set $T(N,B)$ is a compact set inside $[0,1]$. By the Stone-Weierstrass theorem, the linear span of exponentials $e_n$ with integer frequencies is complete in the space of continuous functions on $T(N,B)$. This shows that ${\mathcal Z} = \emptyset$, completing the proof. \qquad$\Box$

\medskip

In the end of this section, we demonstrate the existence of an almost-Parseval-frame tower with $\epsilon_j>0$, which gives a proof of Theorem \ref{th2}(ii). More precisely, we prove

\begin{theorem}\label{th0.1} Let $N_n$ and $M_n$ be positive integers satisfying
\begin{equation}\label{eq0.2}
N_j = M_jK_j+\alpha_j
\end{equation}
for some integer $K_j$ and $0\leq\alpha_j<M_j$ with
\begin{equation}\label{eq0.3}
\sum_{j=1}^{\infty}\frac{\alpha_j\sqrt{M_j}}{K_j}<\infty.
\end{equation}
Define
\begin{equation}\label{eq0.4}
B_j = \{0, K_j,...,(M_j-1)K_j\}, \ L_j = \{0,1,...,M_j-1\}.
\end{equation}
Then   $(N_{j}, B_{j})$ forms an almost-Parseval-frame tower associated with
$$
\epsilon_j=\frac{2\pi\alpha_j\sqrt{M_j}}{{K_j}}
$$ and its pre-spectrum is $\{L_j\}$.
\end{theorem}

\begin{proof}

Let
$$
{\mathcal F}_j= \frac{1}{\sqrt{M_j}}\left[e^{2\pi i b \lambda/N_j}\right]_{\lambda\in L_j,b\in B_j}, \ {\mathcal H}_j= \frac{1}{\sqrt{M_j}}\left[e^{2\pi i b \lambda/M_jK_j}\right]_{\lambda\in L_j,b\in B_j}.
$$
Then ${\mathcal H}_j$ is a unitary matrix (in fact Hadamard matrices). We first show that for any $j>0$, the operator norm ($\|A\|: = \max_{\|x\|=1}\|Ax\|$)
\begin{equation}\label{eq2.0}
\|{\mathcal F}_j-{\mathcal H}_j\|\leq\frac{2\pi \alpha_j\sqrt{M_j}}{K_j}.
\end{equation}
To see this, We note that by Cauchy-Schwarz inequality,
\begin{equation}\label{eq2.1}
\|{\mathcal F}_j-{\mathcal H}_j\|^2\leq  \frac{1}{M_j}\sum_{b\in B_j}\sum_{\lambda\in L_j}\left|e^{2\pi i b\lambda/N_j}-e^{2\pi i b\lambda/M_jK_j}\right|^2.
\end{equation}
We now estimate the difference of the exponentials inside the summation using an elementary estimate
$$
|e^{i\theta_1}-e^{i\theta_2}| = |e^{i(\theta_1-\theta_2)}-1|\leq |\theta_1-\theta_2|.
$$
This implies that
$$
\begin{aligned}
\left|e^{2\pi i b\lambda/N_j}-e^{2\pi i b\lambda/M_jK_j}\right|^2\leq& \left|\frac{2\pi b\lambda}{N_j}-\frac{2\pi b\lambda}{M_jK_j}\right|^2\nonumber\\
=& 4\pi^2\frac{b^2\lambda^2 \alpha_j^2}{M_j^2K_j^2N_j^2}  \ \ \  \ \mbox{(by $N_j = M_jK_j+\alpha_j$)}\nonumber\\
\leq &4\pi^2\frac{M_j^2\alpha_j^2}{N_j^2}  \ \ \ \ \mbox{(by $b\leq M_jK_j$ and $\lambda\leq M_j$)}\\
\end{aligned}
$$
Hence, from (\ref{eq2.1}),
\begin{eqnarray}
\|{\mathcal F}_j-{\mathcal H}_j\|^2&\leq& \frac{1}{M_j}\sum_{b\in B_j}\sum_{\lambda\in L_j}4\pi^2\frac{M_j^2\alpha_j^2}{N_j^2} \nonumber\\
 &=& 4\pi^2\frac{M_j^3\alpha_j^2}{N_j^2} \nonumber \\
 &= & 4\pi^2\frac{M_j\alpha_j^2}{\left( K_j+\alpha_j/M_j\right)^2}
\end{eqnarray}
As $\alpha_j\ge 0$, $\|{\mathcal F}_j-{\mathcal H}_j\|^2\leq  4\pi^2{\alpha_j^2 M_j}/{K_j^2}$ and thus  (\ref{eq2.0}) follows by taking square root.

\medskip

 We now show that $\{(N_{j}, B_{j})\}$ forms an almost-Parseval-frame tower with pre-spectrum $L_j$. The first two conditions for the almost-Parseval-frame tower are clearly satisfied. To see the last condition, we recall that $\epsilon_j = 2\pi \sqrt{M_j}\alpha_j/{K_j} $.  From the triangle inequality and (\ref{eq2.0}), we have
$$
\begin{aligned}
\|{\mathcal F}_j{\bf w}\|\leq&\|{\mathcal H}_j{\bf w}\| + \|{\mathcal F}_j-{\mathcal H}_j\|\|{\bf w}\|\\
\leq&  \left(1+\frac{2\pi \alpha_j\sqrt{M_j}}{K_j}\right)\|{\bf w}\| = (1+\epsilon_j)\|{\bf w}\|.\\
\end{aligned}
$$
Similarly, for the lower bound,
$$
\begin{aligned}
\|{\mathcal F}_{j}{\bf w}\|\geq &\|{\mathcal H}_{j}{\bf w}\|-\|{\mathcal F}_{j}-{\mathcal H}_{j}\|\|{\bf w}\|\\
\geq& \left(1-\frac{2\pi \alpha_j\sqrt{M_j}}{K_j}\right)\|{\bf w}\| = (1-\epsilon_j)\|{\bf w}\|.
\end{aligned}
$$
Thus, from (\ref{ALmost1}), the last condition follows and $(N_j,B_j)$ satisfies the almost-Parseval-frame condition associated with $\{\epsilon_j\}$ and $\sum_{j=1}^{\infty}\epsilon_j<\infty$ is guaranteed by (\ref{eq0.3}) in the assumption.
\end{proof}

Some of the fractal measures induced by the almost-Parseval-frame tower were found to be non-spectral, one can refer to \cite{LaiW} for detail.

\medskip

\section{Explicit construction of spectrum}
In general, the canonical orthogonal set in Proposition \ref{prop_main} is not necessarily a spectrum. But in some cases, we can complete this set by adding some more points, and, in some cases, it is possible to gives an explicit formula for the spectrum of the measure $\mu(R,B)$. Such a description can be given in the following definition. This is always true in dimension one, as explained in \cite{DJ06}.
\begin{definition}\label{definv}
Let $(R,B,L)$ be a Hadamard triple. We define the function
$$m_B(x)=\frac{1}{\#B}\sum_{b\in B}e^{2\pi i \ip{b}{x}},\quad (\xi\in\mathbb{R}^d).$$
The Hadamard triple condition implies that $\delta_{R^{-1}B}$ is a spectral measure with a spectrum $L$ and $m_{R^{-1}B}$ is the Fourier transform of this measure.  The Hadamard condition implies that
\begin{equation}\label{M_B}
\sum_{\ell\in L}|m_B((R^T)^{-1}(x+\ell))|^2=1, \mbox{or} \ \sum_{\ell\in L}|m_B(\tau_{\ell}(x))|^2=1,
\end{equation}
where we define the maps
$$\tau_{\ell}(x)=(R^T)^{-1}(x+\ell),\quad(x\in\mathbb{R}^d,\ell\in L), \ \mbox{and} \ \tau_{\ell_1...\ell_m} = \tau_{\ell_1}\circ...\circ\tau_{\ell_m}.
$$
A closed set $K$ in $\mathbb{R}^d$ is called {\it invariant (with respect to the system $(R,B,L)$)} if, for all $x\in K$ and all $\ell\in L$
$$
 |m_B(\tau_{\ell}(x))|>0 \ \Longrightarrow \ \tau_{\ell}(x)\in K.
 $$
We say that {\it the transition, using $\ell$, from $x$ to $\tau_\ell( x)$ is possible}, if $\ell\in  L$ and $m_B(\tau_{\ell} (x))>0$. A compact invariant set is called {\it minimal} if it does not contain any proper compact invariant subset.

For $\ell_1,\dots,\ell_m\in L$, the cycle $\mathcal C(\ell_1,\dots,\ell_m)$ is the set
$$\mathcal C(\ell_1,\dots,\ell_m)=\{x_0,\tau_{\ell_m}(x_0),\tau_{\ell_{m-1}\ell_m}(x_0),\dots, \tau_{\ell_2\dots\ell_m}(x_0)\},$$
where $x_0:=\wp(\ell_1,\dots, \ell_m)$ is the fixed point of the map $\tau_{\ell_1...\ell_m}$. i.e. $\tau_{\ell_1...{\ell_m}}(x_0)=x_0$.
The cycle $\mathcal C(\ell_1,\dots, \ell_m)$ is called an {\it extreme cycle for $(R,B,L)$} if $|m_B(x)|=1$ for all $x\in \mathcal C(\ell_1,\dots, \ell_m)$.
\end{definition}

\begin{definition}\label{definv2}
We say that the Hadamard triple $(R,B,L)$ is {\it dynamically simple} if the only minimal compact invariant set are extreme cycles. For a Hadamard triple $(R,B,L)$, the {\it orthonormal set $\Lambda$ generated by extreme cycles} is the smallest set such that
 \begin{enumerate}
 \item it contains $-{\mathcal C}$ for all extreme cycles ${\mathcal C}$ for $(R,B,L)$
 \item it satisfies $R^T\Lambda+L\subset \Lambda$.
 \end{enumerate}
When this set $\Lambda$ is a spectrum (see Theorem \ref{th_dynamic} below), we call it {\it the dynamically simple spectrum}.

More generally, the {\it set generated by an invariant subset $A$} of $\mathbb{R}^d$, is the smallest set which contains $-A$ and satisfies (ii).
\end{definition}

\medskip

\begin{theorem}\label{th_dynamic}\cite{DL15c}
Let  $(R,B,L)$ be a dynamically simple Hadamard triple. Then the orthonormal set $\Lambda$ generated by extreme cycles is a spectrum for the self-affine measure $\mu_{R,B}$ and $\Lambda$ is explicitly given by
$$\Lambda=\{\ell_0+R^T\ell_1+\dots (R^T)^{n-1}\ell_{n-1}+(R^T)^n(-c) : \ell_0,\dots,\ell_{n-1}\in L, n\geq0, \ c\mbox{ are extreme cycle points}\}.
$$
Moreover, if $(R,B,L)$ is a Hadamard triple on ${\mathbb R}^1$, it must be dynamically simple.
\end{theorem}

\begin{example}
There are Hadamard triples which are not dynamically-simple. For example let
$$R=\begin{bmatrix}2&1\\
0&2\end{bmatrix},\quad B=\left\{\begin{bmatrix}0\{\iota_0}\end{bmatrix},\begin{bmatrix}3\{\iota_0}\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix},\begin{bmatrix}3\\1\end{bmatrix}\right\},
L=\left\{\begin{bmatrix}0\{\iota_0}\end{bmatrix},\begin{bmatrix}1\{\iota_0}\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix},\begin{bmatrix}1\\1\end{bmatrix}\right\}.$$

\begin{figure}[h]
\centering
\begin{minipage}{.5\textwidth}
\centering
\includegraphics[width=\linewidth]{example2_3xb}%
\caption{$T(R,B)$}
\label{a}
\end{minipage}%
\begin{minipage}{.5\textwidth}
\centering
\includegraphics[width=\linewidth]{example2_3}%
\caption{$T(R^T,L)$}
\label{b}
\end{minipage}
\end{figure}

Note that $B$ is a complete set of representatives $\mod R\mathbb{Z}^2$ and $L$ is a complete set of representatives $\mod R^T\mathbb{Z}^2$; thus we have a Hadamard triple. As shown in \cite[Example 2.3]{LW1}, the measure $\mu(R,B)$ is the normalized Lebesgue measure on the attractor $T(R,B)$ which tiles $\mathbb{R}^2$ with $3\mathbb{Z}\times\mathbb{Z}$. Hence $\frac13\mathbb{Z}\times\mathbb{Z}$ is a spectrum for $\mu(R,B)$ (by \cite{Fug74}).

We look for the extreme cycles: we have
$$m_B(x,y)=\frac{1}{4}(1+e^{2\pi i3x}+e^{2\pi iy}+e^{2\pi i(3x+y)}).$$
If we want $|m_B(x,y)|=1$, then we must have equality in the triangle inequality and we get that $3x,y,(3x+y)\in\mathbb{Z}$. So $(x,y)\in \frac13\mathbb{Z}\times\mathbb{Z}$.
For an extreme cycle $(x,y)$, we must also have that $(x,y)$ is in the attractor of the IFS $(R^T,L)$, and this is contained in $[0,1]\times [-1,1]$ (on can check the invariance of this rectangle for the IFS). So, one can check which of the points in $\frac13\mathbb{Z}\times\mathbb{Z}\cap [0,1]\times[-1,1]$ are in an extreme cycle. The only extreme cycles are $\left\{\begin{bmatrix}0\{\iota_0}\end{bmatrix}\right\}$,$\left\{\begin{bmatrix}1\{\iota_0}\end{bmatrix}\right\}$,$\left\{\begin{bmatrix}1\{\iota_0}\end{bmatrix}\right\}$,$\left\{\begin{bmatrix}0\\1\end{bmatrix}\right\}$,$\left\{\begin{bmatrix}1\\-1\end{bmatrix}\right\}$. Note that these form a complete set of representatives $\mod R^T\mathbb{Z}^2$. The set generated by the extreme cycles is then just $\mathbb{Z}^2$ which is a proper subset of the spectrum $\frac13\mathbb{Z}\times\mathbb{Z}$, so the Hadamard triple is not dynamically simple.
\end{example}

\medskip

\section{Open problems}

The major open problem in the study of Fourier analysis on fractals is to see whether the non-spectral self-affine measures are still frame-spectral. The idea of almost-Parseval-frame towers turns this problem into a problem of matrix analysis. Given an integral expanding matrix $R$ and a set of simple digits $B$ with $N=\#B<|\det R|$, the condition of  almost-Parseval-frame towers can be reformulated equivalently as {\it for any $\epsilon>0$, there exists $n\in{\mathbb N}$ and a set of $L_n\subset {\mathbb Z}^d$ such that the matrix}
$$
{\mathcal F}_n (B_n,L_n) = \frac{1}{\sqrt{N^n}}\left(e^{2\pi i \langle R^{-n} b,\ell\rangle}\right)_{\ell\in L_n,b\in B_n}
$$
{\it satisfies}
$$
(1-\epsilon)\|{\bf w}\|^2\le \|{\mathcal F}_n{\bf w}\|^2\le(1+\epsilon)\|{\bf w}\|^2
$$
{\it for any vectors ${\bf w}\in {\mathbb C}^{N^n}$. (Recall that $B_n = B+RB+...+R^{n-1}B$)}

\medskip

We observe that if we let ${\overline{B_n}}$ and $\overline{L_n}$ be respectively the complete representative class  (mod $R^n({\mathbb Z}^d))$ and  (mod $(R^T)^n({\mathbb Z}^d))$. Then the matrix
$$
\overline{{\mathcal F}_n}: = \frac{1}{\sqrt{|\det R|^n}}\left(e^{2\pi i \langle R^{-n} b,\ell\rangle}\right)_{\ell\in \overline{L_n},b\in \overline{B_n}}
$$
forms a unitary matrix. i.e.
$$
\|\overline{{\mathcal F}_n}{\bf w}\| = \|{\bf w}\|, \ \forall {\bf w}\in{\mathbb C}^{|\det R|^n}
$$
 As $B_n\subset {\overline{B_n}}$, we can take the vectors ${\bf w}$ such that they are zero on the coordinates which are not in $B_n$. This implies that
 $$
\| {\mathcal F}_n(B_n,\overline{L_n}){\bf w}\| = \frac{|\det R|^n}{N^n}\|{\bf w}\|.
 $$
 In other words,
 $$
 \sum_{\lambda\in \overline{L_n}}\left|\sum_{b\in B_n} w_b \frac{1}{\sqrt{N^n}}e^{-2\pi i \langle R^{-n}b,\lambda\rangle}\right|^2 = \frac{|\det R|^n}{N^n}\sum_{b\in B_n}|w_b|^2.
 $$
 This shows that the collection of vectors $\{ \left(\frac{1}{\sqrt{N^n}}e^{-2\pi i \langle R^{-n}b,\lambda\rangle}\right)_{b\in  B_n}:\lambda\in \overline{L_n}\}$ forms a tight frame for ${\mathbb C}^{N^n}$ with frame bound $\frac{|\det R|^n}{N^n}$. Our problem is to extract a subset $L_n$ from $\overline{L_n}$ such that we have an almost tight frame with frame constant nearly 1. This reminds us about the Kadison-Singer problem that was open for over 50 years and solved recently in \cite{MSS}.

 \medskip

\begin{theorem}\cite[Corollary 1.5]{MSS}
Let $r$ be a positive integer and let $u_1,...,u_m\in{\mathbb C}^d$ such that
$$
\sum_{i=1}^m|\langle w, u_i\rangle|^2 = \|w\|^2 \ \forall w\in {\mathbb C}^d
$$
and $\|u_i\|\leq \delta$ for all $i$. Then there exists a partition $S_1,...,S_r$ of $\{1,...,m\}$ such that
$$
\sum_{i\in S_j}|\langle w, u_i\rangle|^2 \le  \left(\frac{1}{\sqrt{r}}+\sqrt{\delta}\right)^2\|w\|^2 \ \forall w\in {\mathbb C}^d.
$$
 \end{theorem}

 \medskip

 This statement says that we can partition a tight frame into $r$ subsets such that the frame constant of each partition is almost $1/r$. Iterating this process allowed Nitzan et al \cite{NOU} to establish the existence of Fourier frames on any unbounded sets of finite measure. One of their lemmas states:

\begin{lemma}\label{lem6.1}
\cite[Lemma 3]{NOU} Let $A$ be an $K\times L$ matrix and $J\subset \{1,...,K\}$, we denote by $A(J)$ the sub-matrix of $A$ whose rows belong to the index $J$. Then there exist universal constants $c_0,C_0>0$ such that whenever $A$ is a $K\times L$ matrix, which is a sub-matrix of some $K\times K$ orthonormal matrix, such that all of its rows have equal $\ell^2$-norm, one can find a subset $J\subset\{1,...,K\}$ such that
$$
c_0\frac{L}{K}\|{\bf w}\|^2\leq \|A(J){\bf w}\|^2\leq C_0\frac{L}{K}\|{\bf w}\|^2 ,\ \forall {\bf w}\in {\mathbb C}^n.
$$
\end{lemma}

\medskip
This lemma leads naturally to the following:

\begin{proposition}\label{prop1.3}
With $(R,B)$ as in Definition \ref{defifs}, there exist  universal constants $0< c_0< C_0<\infty$ such that for all $n$, there exists $J_n$ such that
$$
c_0\sum_{b\in B_n}|w_b|^2\leq \sum_{\lambda\in J_n}\left|\frac{1}{\sqrt{N^n}}\sum_{b\in B_n}w_be^{-2\pi i \langle R^{-n}b, \lambda\rangle}\right|^2\leq C_0\sum_{b\in B_n}|w_b|^2
$$
for all $(w_b)_{b\in B_n}\in{\mathbb C}^{N^n}$.
\end{proposition}

\begin{proof} Let
$$
{\mathcal F}_n = \frac{1}{|\det R|^{n/2}}\left[e^{2\pi i \langle R^{-n}b,\ell\rangle}\right]_{\ell\in \overline{L}_n,b\in\overline{B}_n}
$$
where $\overline{B}_n$ is a complete coset representative (mod $R({\mathbb Z}^d)$) containing $B_n$ and  $\overline{L}_n$ is a complete coset representative (mod $R^T({\mathbb Z}^d)$). It is well known that ${\mathcal F}_n$ is an orthonormal matrix. Let $K = |\det R|^n$ and
$$
A_n = \frac{1}{|\det R|^{n/2}}\left[e^{2\pi i \langle R^{-n}b,\ell\rangle}\right]_{\ell\in \overline{L}_n,b\in B_n}.
$$
Then $A_n$ is a sub-matrix of ${\mathcal F}_n$ whose columns are exactly the ones with index in $B_n$ so that the size $L$ is $L = N^n$. By Lemma \ref{lem6.1}, we can find universal constants $c_0,C_0$, independent of $n$, such that for some $J_n\subset \overline{L}_n$, we have
$$
c_0\frac{N^n}{|\det R|^n}\|{\bf w}\|^2\leq \|A(J_n){\bf w}\|^2\leq C_0\frac{N^n}{|\det R|^n}\|{\bf w}\|^2 ,\ \forall {\bf w}\in {\mathbb C}^{N^n}.
$$
As $\frac{|\det R|^{n/2}}{N^{n/2}}A(J_n) = \frac{1}{|\det R|^{n/2}}\left[e^{2\pi i \langle R^{-n}b,\ell\rangle}\right]_{\ell\in J_n,b\in B_n}: = F_n$, this shows
$$
c_0\|{\bf w}\|^2\leq \|F_n{\bf w}\|^2\leq C_0\|{\bf w}\|^2 ,\ \forall {\bf w}\in {\mathbb C}^{N^n}.
$$
This is equivalent to the inequality we stated.
\end{proof}

This proposition shows that there always exists some subsets $J_n$ in which the norm  of ${\mathcal F}(B_n,J_n)$ is uniformly bounded by universal constants $c_0, C_0$, this indicates that the existence of almost-Parseval-frame pairs $B_n,L_n$ is possible.

 \begin{acknowledgements}
This work was partially supported by a grant from the Simons Foundation (\#228539 to Dorin Dutkay).
\end{acknowledgements}

\bibliographystyle{alpha}

redpajama_set_name: RedPajamaArXiv


I) The Distance from Sun – The farther a planet is from the Sun, the higher its IP. This is attributed to the fact that planets close to the sun start becoming invisible due to the sheer brightness of the sun and hence lose their own 'lusture'. The farther a planet is from the sun, the brighter its own appearance in the sky. This is most clearly visible for the Moon for obvious reasons. Exactly opposite to the Sun, the Moon is Full and bright and conjunct to the Sun is the New Moon or Amavasya where the moon is fully dark.
Likewise, for other planets (except Rahu-Ketu) the farther they are from the Sun, the brighter they will be.
Within the same context, safe to mention here without getting into details, is the fact that the outer planets – Mars, Saturn and Jupiter get retrograde when far away from the Sun and hence when Retrograde planets will generally have a higher IP.
II) Distance from Point of Exaltation – The closer a planet is to its point of deep exaltation, the higher its IP and lower its KP. Vice-versa is also true.
Now when you add the points (I) & (II) an important point to note is that a low IP DOES NOT automatically imply a high KP. It might, but not necessarily.
As an example assume Saturn at 15 degrees in Libra. This Saturn is within 5 degrees of its deepest exaltation point at 20 degrees of Libra and hence will have a very low KP and a high IP by default. Now assume Sun is at 10 degrees of Libra. Now Saturn being so close to Sun will drastically reduce the IP of Saturn however will NOT increase the KP. Hence such a Saturn will have a very low KP and a low IP. On the other hand notice how Sun is at exact debilitation point. Hence Sun will have a very high KP and almost NIL IP.
Another example – Moon at 3 degrees Taurus with Sun in Scorpio at 3 degrees will have a full IP (100%) and zero KP as its fully bright and also in deep exaltation.
4) High IP planets tend to give their promised results easily in their dasas as opposed to high KP planets. A planet with a high KP will tend to throw many obstacles in the native's path during its dasa. However the final result is not dictated by IP/ KP. Its dicated by the functional nature, dignity, Yogas etc the planet is involved in.
7) More than the planet itself, the IP of the Rashi, Nakshatra and Navamsa depositor of a planet is very crucial. This is because these planets act like a source for nourishment for the deposited planet from where the planet get its 'Poshaak' or Food. If a planet is deposited in a Rashi and Nakshatra of a planet with a very high IP and low KP, such a planet will have a lot of good energy to feed from and will give good/ benefic results.


text:
'Thuggish' police acted like 'schoolboys from Lord of Flies' when they tasered student high on LSD who later died
Coroner Mary Jerram, from New South Wales, said it would be impossible to find the exact cause of death, but that if the police hadn't attacked the student he would still be alive. She explained to Skynews.com that it was "impossible to believe that he would have died but for the actions of police".
Jerram recommended that their be a review on the way the police use their taser guns.
Commissioner Andrew Scipione from the police force supported the use of the stun guns, explaining that they "save lives."
It's believed that the police perused the young Brazilian student when he was acting irrationally due to the effects of an LSD tab. The police allegedly mistook him for a suspect, who had committed armed robbery.
The police shocked him 14 times and blasted him with pepper spray.
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UK Trip: Munro bagging in Scotland

redpajama_set_name: RedPajamaCommonCrawl


text:
DA, LGU Valencia break ground for P7.1-M seed storage facility
The groundbreaking ceremony led by DA-10 and LGU Valencia signal the start of the construction of the seed storage facility in barangay Pinatilan.
VALENCIA CITY, BUKIDNON, September 9 – The Department of Agriculture – Regional Field Office 10 (DA-RFO 10) led the groundbreaking ceremony, signaling the start of the construction of the seed storage facility at Nutriland Purok 1, in barangay Pinatilan, this city.
The soon-to-rise storage facility, which costs P7.11 million has a total area of 320 square meters, with cold storage of 160 square meters and ambient of 160 square meters.
Funded by DA's Rice Program, it has a 99 ton-capacity for the cold storage of rice seeds, while 156 tons for the ambient.
Particularly, cold storage is intended to prolong the shelf life and improve the quality of the seeds up to about two years.
Valencia City Vice Mayor Policarpo P. Murillo noted, that the city has 30,000 hectares of agricultural area, of which 11,000 hectares are devoted to rice production.
He also expressed his heartfelt gratitude to DA-10 Regional Executive Carlene C. Collado for considering Valencia City as the recipient of the said project among other towns in Bukidnon province.
In early April of this year, the department has also turned over one unit warehouse in Nutriland, which costs more than P4.3 million.
Meanwhile, DA-10 Bukidnon Provincial Operations Center Chief Dr. Janet D. Lopez, who represented RED Collado, reiterated his recurring message, that the provision of projects to beneficiaries relies on the diligence and enthusiasm of an agriculture's office at the local level.
She emphasized, that DA welcomes any requests for agri-assistance to make farming communities become more productive.
The completion and inauguration of the project are set sometime in February 2022.
Attending the ceremony are Valencia City Councilor Glenn P. Peduche, City Agriculturist Connie L. Dalangan, DA-RFO 10 Regional Agricultural Engineering Division Agri-Infra Section Chief Engr. Junel W. Ablanque, Sta. Ana Raycon, Corp. Authorized Managing Officer Raymund Casiño, and other staff from the local government of Valencia City.# (EMBentulan)


I knew right away which picture I wanted to enter for this…. and the timing could not have been more perfect (today's his birthday!). This is a picture of my husband and those of you who know him would agree that this is so totally him –a big goofball.
We were at the pool and the kids were taking turns jumping into the water. Chris decided to join in the fun. He took a running jump, holding on to his shorts, telling the kids that he's afraid that his pants might fall off when he hits the water. Of course, that got the kids giggling and laughing hard.
To the special guy who always knows how to make me laugh and who inspires me to be a better me, who is a most wonderful father to our children, and the one whom I love and trust most in the world –happy birthday.


Does anyone now the new "character"??? Please tell me who will create she or he???
how to enter that codes ?
My favorite song is " Let it Go"
Thanks for the codes! You're the best!!
That last code didn't work but thanks!
what clan am I in.
Sorry but I do not know any promo codes.But can you tell me the code for the second book of spirit animals please and thank you.
Thanks I love this game so much. How do get Space how do you get a unique weapon quickly.
Hey peeps. Check out the Spirit Animals books in your schools' libraries for codes if that was any help.
beware am coming to kill you all.
Does anybody know the code to book three?


Models and musicians go together like cookies and milk. And given how frequenty celebrity chefs are being compared to rock stars these days, perhaps it won't be such a random stretch to see the likes of Bill Telepan, Emeril Lagasse and Cat Cora installed at Bon Appétit's pop-up cafe come Fashion Week.
Today, Bon Appétit announced that from September 12-16, the magazine will host a pop-up cafe inside Lincoln Center's Avery Fisher Hall. Breakfast offerings will include coffee and pastries from city hotspots like Momofuku Milk Bar, Le Bernardin, and Baked NYC, while lunch will feature a rotating roster of mostly grab-and-go dishes created by Laurent Tourondel of BLT market, Marc Murphy of Landmarc, Missy Robbins of A Voce, and other notable NYC chefs. At night the focus shfts to wine—clearly Bon Appétit knows the fashion crowd well—and small plates.
We'll be thrilled to have such sweet food options under the same roof as the shows, but we should note—and we'll say it with our mouths full—that between Fashion's Night Out, an event spearheaded by Vogue (by way of Anna Wintour), and the cafe, Condé Nast is really staking a sizable claim on Fashion Week.


A separation with no conflicts is one of the main goals most upcoming Kentucky divorcees have. The proceedings are easy and the couple and does not suffer as much of an emotional roller coaster if they easily agree on how the asset division goes.
However, the circumstances of your divorce should determine if an uncontested divorce is the right path to take. Only the courtroom can handle certain unavoidable marital problems. If you plan on separating from your spouse, you should know some factors which can make uncontested divorce too difficult in Kentucky.
Uncontested divorces can only work if the couple is cooperative and willing to hear each other's ideas to come to an agreement. If you cannot be in the same room as your spouse, then you likely do not want to spend hours alone with them talking about who gets what. You cannot let your emotions overpower you as these proceedings have a large impact on your future.
If you are separating because your spouse cheated on you, it is important to know that Kentucky is a "no fault" divorce state. It means that most of the time, the court does not consider factors like adultery to be a major ground for divorce. While it can be brought up in the proceedings to get you potentially more assets or alimony, there is no guarantee that the court will take it into consideration.
However, there are some faults that the court does allow to impact the case. For example, if you and your child are the victims of domestic abuse, you can apply for emergency protective orders in Kentucky to keep your aggressive spouse away from you and the court does allow this to impact child custody and asset division. They do this by determining how much the violence has affected the child's relationship to the parents.
There are a lot of aspects to consider when separating property and determining child custody. Many couples take the uncontested route to get the divorce over as soon as possible and move on with their lives. As emotionally difficult as the process is, you want to avoid rushing so you do not overlook any potential issues you might have with the agreement. Even if you take the uncontested route, you still have to deal with Kentucky's 60 day waiting period before the final divorce decree, so there is even less purpose in getting it done too quickly.
Divorce can be an emotionally complex period that could cause you to make rash decisions when you and your spouse divide your assets. To lessen some of the stress, you can contact a family law attorney that has experience in helping divorcees get the most out of property division and spousal support.


I enjoy bargain hunting for furniture whether it's furniture someone is throwing out, garage sales, Goodwill or Craigslist. If you are moving into your first place and don't have any furniture. Refurbishing old furniture is a great way to fill up your house without spending tons of $$$. There are so many awesome pieces out there that just need a little TLC like this coffee table that I found for ONLY $5 on Craigslist. The legs were a dark green and the top was a light wood color, now its an off white/cream with a dark stained top.
I am very pleased with how it turned out. Now I just need an area rug for the room!


text:
Q: Resolution and scaling issues on dual monitor with nvidia (one 4k HiDPI, one VGA with converter) I am trying Kubuntu 18.10 and I have a nvidia GTX 1060 (driver 418), a LG 27" 4k monitor (27UD58-B ) and an older HP 22" 1680x1050 monitor (HPw2207) with a VGA cord and a VGA-> DP converter.
My issues are as follow:
1)  I can't get my vga screen with converter to use a resolution above 1280*1024
2)  I'd like to use screen scaling because everything is so damn small on the 4k display.  I'd want the scaling to only be done on the high-dpi display if possible.
Current setup:
I am using Nvidia drivers 418.56.
$ nvidia-smi
Mon Apr 15 15:07:47 2019
+-----------------------------------------------------------------------------+
| NVIDIA-SMI 418.56       Driver Version: 418.56       CUDA Version: 10.1     |
|-------------------------------+----------------------+----------------------+
| GPU  Name        Persistence-M| Bus-Id        Disp.A | Volatile Uncorr. ECC |
| Fan  Temp  Perf  Pwr:Usage/Cap|         Memory-Usage | GPU-Util  Compute M. |
|===============================+======================+======================|
|   0  GeForce GTX 106...  Off  | 00000000:23:00.0  On |                  N/A |
| 10%   64C    P0    25W / 120W |    739MiB /  6076MiB |      3%      Default |
+-------------------------------+----------------------+----------------------+

+-----------------------------------------------------------------------------+
| Processes:                                                       GPU Memory |
|  GPU       PID   Type   Process name                             Usage      |
|=============================================================================|
|    0       958      G   /usr/lib/xorg/Xorg                           386MiB |
|    0      1413      G   /usr/bin/kwin_x11                            192MiB |
|    0      1432      G   /usr/bin/krunner                               1MiB |
|    0      1434      G   /usr/bin/plasmashell                         111MiB |
|    0      1509      G   /usr/bin/latte-dock                           29MiB |
|    0      9387      G   /usr/lib/firefox/firefox                       1MiB |
+-----------------------------------------------------------------------------+

and normal higher resolutions are not available:
$ xrandr
Screen 0: minimum 8 x 8, current 4864 x 2160, maximum 32767 x 32767
DVI-D-0 disconnected (normal left inverted right x axis y axis)
HDMI-0 disconnected (normal left inverted right x axis y axis)
DP-0 connected primary 3840x2160+0+0 (normal left inverted right x axis y axis) 600mm x 340mm
   3840x2160     60.00*+  30.00
   2560x1440     59.95
   1920x1080     60.00    59.94
   1600x900      60.00
   1280x1024     60.02
   1280x800      59.81
   1280x720      60.00    59.94
   1152x864      59.96
   1024x768      60.00
   800x600       60.32
   720x480       59.94
   640x480       59.94    59.93
DP-1 disconnected (normal left inverted right x axis y axis)
DP-2 disconnected (normal left inverted right x axis y axis)
DP-3 disconnected (normal left inverted right x axis y axis)
DP-4 connected 1024x1280+3840+832 left (normal left inverted right x axis y axis) 376mm x 301mm
   1280x1024     60.02*+
   1280x960      60.00
   1152x864      75.00
   1024x768      85.00    75.03    70.07    60.00
   800x600       85.06    75.00    72.19
   640x480       85.01    59.94
DP-5 disconnected (normal left inverted right x axis y axis)

First issue: fixing resolution
I have tried to "addmode" using xrandr, but I get a BadMatch error. This is apparently normal with nvidia video cards.
$ cvt 1680 1050 60
# 1680x1050 59.95 Hz (CVT 1.76MA) hsync: 65.29 kHz; pclk: 146.25 MHz
Modeline "1680x1050_60.00"  146.25  1680 1784 1960 2240  1050 1053 1059 1089 -hsync +vsync
$ xrandr --newmode  "1680x1050_60.00"  146.25  1680 1784 1960 2240  1050 1053 1059 1089 -hsync +vsync

$ xrandr --addmode DP-4 1680x1050_60.00
X Error of failed request:  BadMatch (invalid parameter attributes)
  Major opcode of failed request:  140 (RANDR)
  Minor opcode of failed request:  18 (RRAddOutputMode)
  Serial number of failed request:  45
  Current serial number in output stream:  46

My /etc/X11/xorg.conf file has been generated by nvidia usind sudo nvidia-xconfig :
$ sudo nvidia-xconfig

Using X configuration file: "/etc/X11/xorg.conf".
Backed up file '/etc/X11/xorg.conf' as '/etc/X11/xorg.conf.backup'
New X configuration file written to '/etc/X11/xorg.conf'

The xorg.conf only contains one monitor, is this normal.  I have tried to add a "Modeline" line under the Monitor0, but it didnt change anything.
# nvidia-xconfig: X configuration file generated by nvidia-xconfig
# nvidia-xconfig:  version 418.56

Section "ServerLayout"
    Identifier     "Layout0"
    Screen      0  "Screen0" 0 0
    InputDevice    "Keyboard0" "CoreKeyboard"
    InputDevice    "Mouse0" "CorePointer"
EndSection

Section "Files"
EndSection

Section "InputDevice"

    # generated from default
    Identifier     "Mouse0"
    Driver         "mouse"
    Option         "Protocol" "auto"
    Option         "Device" "/dev/psaux"
    Option         "Emulate3Buttons" "no"
    Option         "ZAxisMapping" "4 5"
EndSection

Section "InputDevice"

    # generated from default
    Identifier     "Keyboard0"
    Driver         "kbd"
EndSection

Section "Monitor"
    Identifier     "Monitor0"
    VendorName     "Unknown"
    ModelName      "Unknown"
    HorizSync       28.0 - 33.0
    VertRefresh     43.0 - 72.0
    Option         "DPMS"
EndSection

Section "Device"
    Identifier     "Device0"
    Driver         "nvidia"
    VendorName     "NVIDIA Corporation"
EndSection

Section "Screen"
    Identifier     "Screen0"
    Device         "Device0"
    Monitor        "Monitor0"
    DefaultDepth    24
    SubSection     "Display"
        Depth       24
    EndSubSection
EndSection

Second issue : scaling
I have taken two steps to allow the HiDPI screen to display text and GUI in a more readable size:
1)  System settings - fonts - Force Font DPI 144 (instead of 96)
2)  system settings- display and monitors - scale display 1.5 (instead of 1).
I would lke these settings to only apply to the 4k monitor, is this possible at all?

A: I found a workaround: I connected my display directly to an old tower that has a VGA and exported the edid using
sudo apt install read-edid
sudo get-edid -m 0 > edid.bin

then I erased the old xorg.conf on the machine and created a new one where I forced the edid for the monitor on display port 4:
sudo rm /etc/X11/xorg.conf
nvidia-xconfig --custom-edid="GPU-0.DP-4:/home/MYNAME/edid.bin"

Ta-da, the higher resolution now works.   When I look at my xorg.conf file, I see the following line has been added to the "Screen" section:
   Option         "CustomEDID" "GPU-0.DP-4:/home/MYNAME/Downloads/edid.bin"

here is what I ended up doing regarding the scaling with a high-dpi monitor and a standard-dpi monitor:

Ok, mais là tout est trop gros sur le 22" .. il faut que je rescale de 1.4 dans la direction inverse.
Ça me fait un écran de 3840 + 1.4 * 1680 = 6192 pixels de large, que je dois indiquer dans le frame buffer.
Ensuite, j'indique la position des écrans avec --pos, où le 0x0 est le coin haut-gauche de l'écran.
https://askubuntu.com/questions/393400/is-it-possible-to-have-different-dpi-configurations-for-two-different-screens
xrandr --output DP-0 --scale 1x1 --pos 0x0 --fb 6192x2160
xrandr --output DP-4 --scale 1.4x1.4 --pos 3840x690

redpajama_set_name: RedPajamaStackExchange


BT chairman Sir Christopher Bland is the sixth most powerful individual in the UK's media sector, according to Monday's edition of The Guardian.
This year's Media Guardian 100 list puts Bland behind only Rupert Murdoch, BBC director-general Greg Dyke, Tony Blair, Tony Ball -- the chief executive of BSkyB -- and advertising titan Sir Martin Sorrell when it comes to cultural influence, economic clout and political power.
Several other senior figures from the new media scene appeared in the Media Guardian 100, although none as high as Bland.
Bland was credited for lowering BT's debt mountain, and for making the company give broadband a much higher priority. Even though chief executive Ben Verwaayen is generally seen as the real driving force behind BT broadband push, Bland shares the credit for having chosen Verwaayen to succeed Sir Peter Bonfield.
It is now seen as very unlikely that BT will attempt to become a broadcasting company that generates its own content, despite Bland's past experience as chairman of the BBC. However, the company is still important to the media world because of broadband. BT is aiming for five million broadband connections by 2006 -- such an audience could help the Internet to become a major platform for broadcasting content.
Richard Parsons, chief executive of AOL Time Warner, was ranked 9th on the list, one behind Bill Gates. The Guardian felt that the success of Microsoft's MSN Web sites, on top of the company's grip on the operating system market, makes Gates a significant beast in the UK's media jungle.
Sir Christopher Gent, Adam Singer, Barclay Knapp and John Pluthero -- chief executives of Vodafone, Telewest, ntl and Freeserve respectively -- were all also included, illustrating the important role that technology and the Internet play in the world of the media.
Cabinet ministers Patricia Hewitt and Tessa Jowell were also included in the list thanks to their involvement in the Communications Bill, which will change the way telecoms, TV and radio are regulated.


As seasons change we switch out our closets for warmer clothes, and we need to change our skincare during this time too! During the falls months breakouts can be more common with weather fluctuations going from warm to cold, and many experience more dryness and start to loose that summer glow. Here are some tips to keep your skin it's best this fall!
No surprise here, exfoliating keeps fresh live skin cells visible which allows you to keep a glow year round. I love using my [Cranberry Scrub] twice a week, and my [Cranberry Peel] once a week. Using a scrub removes surface skin cells, keeps skin soft and smooth and helps your products penetrate better. [Cranberry Peel] uses natural enzymes that digests dead skin cells and acids lactic acid to increase the natural moisture factor of the skin. Cranberry contains vaccinium which is natures benzoyl peroxide and helps fight breakouts too!
Keeping hydration balanced in the fall is tricky! One day it's warm and maybe you're oily, the next day it's cooler and maybe you feel dry! The best thing to do during this time is layering products! Maybe on a warm day use a light product like [Hydrotant Gel] or [Hydro-Lane] mixed with [Z-Plus] for SPF protection. On a cooler day add a product like [Brightening Lotion] for more moisture! [Hydrotant Gel] and [Hydro-Lane] contain hyaluronic acid which binds 1,000 times it's weight in moisture to the skin without being heavy. This is what helps give a GLOW! [Hydro-Lane] also contains 20% peptides so if you are looking for more plump and firm skin, this is a great daytime choice. [Brightening Lotion] contains daisy flower extract, lemon peel and L-Arbutin to fight discolorations, while [Glucana Crème] is a rich moisturizer with a lightly matte finish for hydration without the heaviness and oiliness.
Masques are the bonus category of skincare. Adding a masque during fall months can help keep your skin balanced and beautiful. I love the [Hydrotant Marine Masque] which is a creamy masque for all skin types with anti-aging properties and leaves your skin silky smooth! [Mineral Enzyme Masque] is another great masque if you're looking for water hydration on your skin and something lightweight with a little zing to it.
Fall is a great time for repairing and renewing your skin. [Cellular Repair Concentrate] or [Repair Serum] are perfect for your morning regimen, and [Chiral A Discoloration] or [Retinoxyl Elixir] in the evening. [Cellular Repair] is vitamin C which brightens and evens skin tone. This combined with [Brightening Lotion] I call my glow in a jar combo! [Repair Serum] contains stem cells and peptides for anti-aging along with protecting your skin from future damage. [Chiral A Discoloration] uses L-arbutin and retinol to reverse discolorations and summer sun damage. [Retinoxyl Elixir] is your fix all for all skin types! Anti-aging, pore size, fine lines, discolorations. All the results you want without redness, inflammation and irritation!
Fall is also the perfect time to get your skin ready for the holidays with a [Cabernet Peel]! A peel is a quick way to reduce discolorations, fine lines and wrinkles and refresh your skin for a huge glow! It also boosts collage and elastin production for more firm and toned skin. Talk to your esthetician about if this treatment is right for you!
Lastly, we all love our pumpkin products. Yum!
Don't forget we have [Pumpkin Scrub], [Foaming Face Wash], [P.P.E. Nutritional Solution], [Pumpkin Moisturizer] and 5% (retail), 10%, 20% and 35% (professional) [Pumpkin Peels]!


Big thanks to all the kind and encouraging comments. 20 has to be a record. Even if the last one was JK commenting on the fact the entry had attracted 19 comments.
Last week was a whirlwind of excitement and (for others) shock. Despite receiving numerous ignorant comments along the lines of "you'll have given up running then?" - which were generally reciprocated with "sorry, I said pregnant. Not disabled" - OK, I appreciate people are concerned, but I'm big and ugly enough to know my own body. And, more importantly, know my own limits. I have been keeping generally quite active. I know there hasn't been much blogging, but I haven't resigned myself to slippers, soaps and giant bars of Cadbury...YET! So twos-up to the clueless and here's to a fit and healthy pregnancy.
My running - which is now jogging - has been confined within the barriers of 4-7 miles at 9-10 m/m pace. Last week I did 2 x 5 miles and 7 miler at an average of 9m/m. I've spoken to my gym fitness instructor and he's happy for me to continue with pilates and yoga for the time being. I'll be given alternatives for anything that he deems inappropriate. I've also been using the elliptical trainer for 30 mins a few times a week. And last Friday - despite being adverse to most water-based sports - did a bit of swimming and aqua jogging.
This is the book that I've been avidly reading. In summary: I can run as long as it doesn't hurt and I don't get too breathless. What an education ;-) I think I'll write my own version at the end of this adventure. Joking aside it was comforting to read other runner's stories and generally get a little bit more information.
So, that's the boring updates out of the way.
Sonic and I decided to escape on a caravan trip to do a spot of hillwalking for a few days. We headed up to Ullapool on Saturday. The day was spent travelling, so we took our time. Stopping frequently for tea breaks, shopping stops and view appreciation. And off course I needed to pee every 15 minutes. Hey, it's all part of the course.
Setting up for the night, the weather didn't look too promising. Howling winds and bouts of torrential rain. And on Sunday morning, it was pretty much the same. As we had travelled so far we could hardly bow out, so we braved the elements and headed on the course for the day - the ascent of Beinn Dearg, Cona Mheall and Meall nan Caepraichean. Surprisingly enough the weather wasn't actually that bad. OK, it rained a lot and the view was non-existent at most times, but I have experienced much worse. Back at the caravan with we celebrated a successful trip with tomato soap and spaghetti hoops on toast. Gawd we know how to live.
Monday was morning was a whole different story. Blue skies and sunshine. Surprising considering Sunday was supposed to be the best day. Thus confirming my opinion on never believing a weather forecast. Due to the long walk-in we only managed one Munro - Eididh nan Clach Geala. Sonic wanted to push on for another, but that would have involved finishing about 10pm. I managed to talk him out of it with the promise of fish 'n' chips. The way to Sonic's heart. We went to the Seaforth in Ullapool, which is billed as the UK's best fish pub restaurant. I think we were the only people in the town with English as a first language, but I think it's a lovely wee place. And the food was definitely worth missing another Munro for. The fish was more like a whale in batter. Despite my crazy appetite, I barely made a dent in it. I was more interested in the chips and mushy peas. Sacrilege, I know.
Tuesday was another cracking day. We couldn't believe out luck. Probably just a well, as the route to An Foachagach involved crossing two rivers. We had spoken to a walker on Sunday, who told us he ended chest-deep. I think Sonic has been watching too many Bear Grylis programmes, as he was looking forward to a spot of skinny-dipping. Thankfully the countryside was spared, as we made it over thigh-deep with shoes and socks in hand. The steep grassy slopes were pretty relentless. As lame as it is, I'm a big fan of a nice trail or path. The views from the top were breathtaking. Views for miles. We even managed to have lunch at the summit - without risk of exposure or being blown off. We were less cautious on the descent and wadded through the rivers fully clothed. I think my boots might have drained by next month.
All in all, a fabulous trip with limited drama. Only a few stumbles and bruises. We only tipped the caravan once. However, we did manage to lose one of the sunroof shutters though, as we forgot to close it. It flew off on the A9 and hit an American tourist's car - which they informed us they had borrowed from the minister. Oops. And last night Sonic drove off with the wheel lock on though. RIP the wheel lock. One day we'll grasp this caravan malarkey. In the meantime, it keeps us highly amused.


On September 13th, 2018, a significant modification to Law No. 19,496 Consumer Protection Act ("CPA") was enacted through its publication in the Official Gazette. This publication ends a process which lasted over four years.
This reform process commenced on June 2nd, 2014, when the Government filed a bill before the Congress which intended to modify the CPA, informally known as "Bill to strengthen the National Consumers Bureau ("SERNAC")" (hereinafter, the "Bill") since its main modifications introduced to the CPA are aimed to strengthen SERNAC's institutionality, granting it more abilities.
On October 25th, 2017, the Bill was approved by Congress. After this approval, on January of 2018, the Bill was significantly altered by the Constitutional Court, in order to ensure its constitutionality. In particular, the Court rejected the proposed sanctioning abilities of SERNAC.
Afterwards, and following an extensive administrative procedure involving the Executive Branch, the General Comptrollership of the Republic, and the Constitutional Court (process which lasted almost a year), the reform to the CPA (hereinafter the "Reform") was published in the Official Gazette on September 13th, 2018, thus officially becoming Law No. 21,081.
The Reform will enter into force progressively, depending on the provision and the region of the country in which it is implemented.
Hereunder is an outline on the Reform's principal guidelines and main changes to the CPA.


text:
Forsaken Bankzy
Everything posted by Forsaken Bankzy
Pay attention to the Deepfake queens speech today.
Forsaken Bankzy replied to Tetragrammaton's topic in Over 14 days old
In the new Matrix the machines philosophical equation to human behavior went from choice to people 'wanting to be controlled'. That is what I see in the fakery scripted Queens speech broadcasted to millions of brits lol.
The Keys To Religion
Forsaken Bankzy replied to pi3141's topic in Religion & Secret Societies
https://www.youtube.com/watch?v=X9fR1vSxNEQ
Polish Government prohibits abortions (Polish People Protest in 100 Towns / Cities)
Forsaken Bankzy replied to a topic in Over 14 days old
No, i'm not happy because i've seen what's happened in defective newborns in vietnam and I don't have answers for such controversy. You are just blind to the fact that the controversial topic of abortion in the liberal world is to do with 'WOMEN's rights more so than to do with defective babies so I doubt you have any answers yourself. My post was not about Polish Laws, but more to do so about people who have abortions not because deformed fetuses or emotional implications from rape but because they were irresponsible on a night out or any other situation where sexual tensions collided. As the fella said, this is an excuse so they can have freedom to fornicate like rabbits.
I was focused on whether abortions is right or wrong. Polish laws cannot give man the wisdom on whether it's right or wong.
Let's be honest with ourselves. Most abortions happen because two individuals were irresponsible both in their actions and thought processes, this far outweighs abortions happening because an individual was brutally raped or because of deformed fetus; but somehow these become the core argument when defending abortions in a over sexualized society, where even young fragile minds cannot escape its imagery. We need to draw a line of right and wrong and whilst liberalism tries to find common grounds for both, God has already given us what is right from wrong.
Liverpool Cathedral - Mark of the Beast Exercise - Thank God I'm not a Christian
Forsaken Bankzy replied to DannyUK's topic in Over 14 days old
They don't sound like Christians. Sure it wasn't a Catholic Church?
Forsaken Bankzy replied to Kais_1's topic in Religion & Secret Societies
The Flat Earth/Globe Earth Discussion Thread
Forsaken Bankzy replied to bflat's topic in Space & Universe
Good footage. This was in the comment section of your video.
Kanye West running for president 2020
Forsaken Bankzy replied to Jack's topic in Over 14 days old
I thought Trump was a joke at first then he actually became president.
BERMUDA-TRIANGLE USOs & UFOs
Forsaken Bankzy replied to THESTRONG-THEVALIANT's topic in UFOs & Aliens
I used to think the triangle was some sort of portal and aliens came from it. Now I just think its the landing site for space shuttles.
Post your rig, I mean computer build...
Forsaken Bankzy replied to motleyhoo's topic in Technology, Transhumanism and AI
i7 9700K NZXT Kraken x52 CPU cooler GTX 1080 MSI Z390 Gaming Edge 8gb Corsair DDR4 Samsung SSD 850 EVO 500gb Toshiba HDD 2TB 34" Ultrawide LG Monitor 144hz 2 x NZXT RGB fans, black/white cable mods, Model O Glorious mouse, HyperX keyboard. I kickass in cs:go, cod:mw, aoe:de
The Rainbow
Forsaken Bankzy replied to alexa's topic in General Chat
There's something about the rainbow that hints to the Divine presence guarding the stormy world. The sign isn't something tangible, like a stone or tree — it's more faint and ethereal. It can be seen only briefly and you might miss it if you're not looking for it. No one can get to close to it, and each person sees it a little differently. It's faint but it hints to a light beyond the darkness, even if all one can see is its reflection. Yechezkel was dealing with a traumatic period for the Jewish people, when they were going into exile in Babylonia. His vision is filled with frightening forces — a stormy wind, a flashing fire — people don't know what will happen. But at the end of his vision, he sees a rainbow — a sign of the Divine, something beyond that will guide his people. In difficult times, from floods to exile, sometimes all that's needed is this faint sign.
"they". Well, that's if you believe in that sort of thing.
There have been numbers footages at 100,000 ft clearly showing a flat horizon. However, youtube has been on a censoring rampage in the last two years.
Until someone refutes it with a fisheye lens ?
A dark circular anomaly is evident that it's the moon's shadow. Not quite.
What i observe during a solar eclipse phenomena is beyond my understanding. There is no observable evidence that this is the moon's shadow other than just believing that it is.
Coronavirus Mega-Thread.
Forsaken Bankzy replied to numnuts's topic in Covid-19 & NWO
Israel Cohen - A Racial Program for the Twentieth Century, 1912. "We must realize that our Party's most powerful weapon is racial tension. By propounding into the consciousness of the dark races that for centuries they have been oppressed by the Whites, we can mould them to the program of the Communist Party. In America, we will aim for subtle victory. While inflaming the Negro minority against the Whites, we will instill in the Whites a guilt complex for their exploitation of the Negroes. We will aid the Negroes to rise to prominence in every walk of life, in the professions and in the world of sports and entertainment. With this prestige, the Negro will be able to intermarry with the Whites and begin a process which will deliver America to our cause." This has been refuted by the MSM stating that Israel Cohen had never made such remarks but to me this is very Zionist like and I wouldn't be surprised if the elders were attempting to cover their tracks.
122,847 replies
Actually you was quite clear on it, how else do you interpret what you said without 'twisting it'? But now you have cleared up that the movement was created by a bunch of uneducated people who can't deal with the simplicity of pseudoscience. No, i've acknowledge it, i've believed it, and now I don't. I remember years ago when i believed aliens had infiltrated US military bases too. Why was my belief in aliens and distant planets so strong at one point? because I loved science fiction and still do. You are wrong here. You are speaking for the deep state about their agenda. When youtube removes all flat earth videos, that is not promoting flat earth is it. Flat earth has been censored in the same manner as false flag terrorist attacks.
So the flat earth conspiracy is masonic move to prevent the people from realising the truth of John Lear's work? That is a ludacris claim. More people believe and are exposed to the idea of Aliens from outer space than people exposed to the flat earth conspiracy. I can't see no reasoning behind your claim. To believe in aliens and alien UFO's requires the same type of faith to believe in any religion and my guess would be that you strongly believe in aliens with the likelihood that their intellect is far superior to humans. Btw, I don't deny that the flat earth movement is not somewhat controlled., because I believe they have always given us pinch of truth for whatever reasons, and this is one of them.