text:
\begin{document}

\title[Uncertainty Principles for Quaternion Windowed Offset Linear Canonical Transform of Two Dimensional Signals]
{Uncertainty Principles for Quaternion Windowed Offset Linear Canonical Transform of Two Dimensional Signals}

\author[Aajaz A. Teali]{Aajaz A. Teali}

\address{ Department of Mathematics, University of Kashmir, South Campus, Anantnag-192101 Jammu and Kashmir, India.}
\email{aajaz.math@gmail.com}

\keywords{ Quaternion algebra; quaternion-valued functions; window function, uncertainty principles; quaternion Fourier transform; quaternion linear canonical transform; quaternion offset linear canonical transform }

\begin{abstract}
The offset linear canonical transform encompassing the numerous integral transforms, is a promising tool for analyzing non-stationary signals with more degrees of freedom. In this paper, we generalize the windowed offset linear canonical transform for quaternion-valued signals, by introducing a novel time-frequency transform namely the quaternion windowed offset linear canonical transform of 2D quaternion-valued signals. We initiate our investigation by studying some fundamental properties of the proposed transform including inner product relation, energy conservation, and reproducing formula by employing the machinery of quaternion offset linear canonical transforms. Some uncertainty principles such as Heisenberg-Weyl, logarithmic and local uncertainty principle are also derived for quaternion windowed offset linear canonical transform. Finally, we gave an example of quaternion windowed offset linear canonical transform.
\end{abstract}

\parindent=0mm \vspace{.0in}
\maketitle
\section{Introduction}

\parindent=0mm \vspace{.0in}
During the culminating years of last century, a reasonably flexible integral transform associated with a general inhomogeneous lossless linear mapping in phase-space was introduced, namely, the offset linear canonical transform(OLCT)  \cite{AS,Cai}. The OLCT is a six-parameter class of linear integral transform which encompasses a number of well known unitary transforms including the classical Fourier transform, fractional Fourier transform, Fresnel transform, Laplace transform, Gauss-Weierstrass transform, Bargmann transform and the linear canonical transform \cite{DS1,Al,HKOS}. Due to the extra degrees of freedom, OLCT has attained a respectable status within a short span and is being broadly employed across several disciplines of science and engineering including signal and image processing, optical and radar systems, electrical and communication systems, pattern recognition, sampling theory, shift-invariant theory and quantum mechanics \cite{Qzy,Z,Xhch,ULT,BZ}. Recently, in \cite{Shah} we introduce a hybrid integral transform namely, windowed special affine Fourier transform which is capable of providing a joint time and frequency localization of non-stationary signals with more degrees of freedom. In view of its numerous applications, one is particularly interested in its higher dimensional analogues.

\parindent=8mm \vspace{.1in}
In the meantime, quaternion algebra has become an active area of research as it offers a refinement of classical harmonic analysis and is used to generalize the classical theory of holomorphic functions of one complex variable onto the multidimensional situation. It gives a simple and profound representation of signals wherein several components are to be controlled simultaneously. The development of integral transforms for quaternion valued signals has found numerous applications in 3D computer graphics, aerospace engineering, artificial intelligence and colour image processing. The extension of classical Fourier transform to quaternion algebra has been introduced in \cite{Ell} and its efficient implication was given by Pei et al. in \cite{Ding}. The QFTs play a vital role in the representation of quaternion signals. They transform a 2D quaternion-valued signals into the quaternion-valued frequency domain signals. Many applications of the QFTs have been found in color image processing, especially in color-sensitive smoothing, speech recognition, edge detection and data compression \cite{Bas,Bayro,Dubey,Gri}. Moreover, some authors have also generalized the classical linear canonical transform to quaternion-valued signals, known as the quaternion linear canonical transform (QLCT). The QLCT was firstly studied in \cite{Kou2} including prolate spheroidal wave signals and uncertainty principles \cite{Kou}. Some useful properties and applications of the QLCT such as linearity, reconstruction formula, continuity, boundedness, positivity inversion formula and the uncertainty principle were established in \cite{Asym,MR,Zhang} and \cite{BahApp}. Recently, based on the (two-sided) QLCT \cite{Kou2}, the quaternion windowed linear canonical transform of 2D quaternion signals has been introduced by Gao and Li in \cite{Wen}, which generalizes the quaternion windowed Fourier transform \cite{MawW}.  In \cite{QOLCT}, Haoui et al. introduced the quaternion offset linear canonical transforms(QOLCT) as a generalization of QLCTs and QFTs. But it fails in obtaining the local features of non-transient signals due to its global kernel, it is therefore interesting and worthwhile to investigate the localization of quaternion offset linear canonical transforms.

\parindent=8mm \vspace{.1in}
In the present work, we study the generalization of the windowed offset linear canonical transform for quaternion-valued signals, we introduce a novel time-frequency transform namely the quaternion windowed offset linear canonical transform of 2D quaternion-valued signals. It can reveal the local QOLCT-frequency contents and enjoys high concentrations and eliminates the cross term. Some important properties are analyzed. Moreover, inner product relation, energy conservation and inversion formula are establised. Furthermore, some uncertainty principles such as Heisenberg-Weyl, logarithmic and local uncertainty principle are also derived for quaternion windowed offset linear canonical transform. Finally, we gave an example of quaternion windowed offset linear canonical transform.

\parindent=8mm \vspace{.1in}
The organization of the article is as follows: We begin in Section 2 by presenting the brief recall of quaternion algebra and different types of quaternion offset linear canonical transforms. In Section 3, we introduce the concept of quaternion windowed offset linear canonical transform and obtain the expected properties of the QWOLCT including Parseval's formula, energy conservation, isometry and inversion formula. Moreover, the uncertainty principles such as Heisenberg-Pauli-Weyl inequality, logarithmic uncertainty principle and local uncertainty principle are generalized in the quaternion windowed offset linear canonical domains in Section 4. Finally, an example of quaternion windowed offset linear canonical transform is given in Section 5.

\section{Quaternion Algebra and Quaternion Offset Linear Canonical Transform}

\subsection{ Quaternion Algebra}
\parindent=0mm \vspace{.1in}
The theory of quaternions was initiated by the Irish mathematician Sir W.R. Hamilton in 1843 and is denoted by $\mathbb H$ in his honour.
The quaternion algebra provides an extension of the complex number system to an associative non-commutative four-dimensional algebra.
The quaternion algebra $\mathbb H$ over $\mathbb R$ is given by
\begin{align*}
\mathbb H=\Big\{f=a_{0}+i\,a_{1}+j\,a_{2}+k\,a_{3}\,:\,a_{0},a_{1},a_{2},a_{3}\in\mathbb R\Big\},
\end{align*}
where $i,j,k$ denote the three imaginary units, obeying the Hamilton's multiplication rules
\begin{align*}
ij=k=-ji,~jk=i=-kj\,,\,ki=j=-ik\,,\, {\text {and}}\,\, {i}^{2}={j}^{2}={k}^{2}={ijk}=-1.
\end{align*}
For  quaternions $f_{1}=a_{0}+i\,a_{1}+j\,a_{2}+k\,a_{3}$ and $f_{2}=b_{0}+i\,b_{1}+ j\,b_{2}+k\,b_{3}$, the addition  is defined componentwise
and the multiplication is defined as
\begin{align*}
f_{1}f_{2}&=(a_{0}b_{0}-a_{1}b_{1}-a_{2}b_{2}-a_{3}b_{3})+i\,(a_{1}b_{0}+a_{0}b_{1}+a_{2}b_{3}-a_{3}b_{2})\\
&~~~~~~~~~~~~~~+ j(a_{0}b_{2}+a_{2}b_{0}+a_{3}b_{1}-a_{1}b_{3})+k\,(a_{0}b_{3}+a_{3}b_{0}+a_{1}b_{2}-a_{2}b_{1}).
\end{align*}
The conjugate and norm of a quaternion  $f=a_{0}+i\,a_{1}+j\,a_{2}+k\,a_{3},$ are given by $\overline{f}=a_{0}-i\,a_{1}-j\,a_{2}-k\,a_{3}$ and ${\|f\|}_{\mathbb H}=\sqrt{{a_{0}}^{2}+{a_{1}}^{2}+{a_{2}}^{2}+{a_{3}}^{2}}$, respectively. We also note that an  arbitrary quaternion $f$ can be represented by two complex numbers  as $f=(a_{0}+i\,a_{1})+j\,(a_{2}-i\,a_{3})=f_1+j\,f_2$, where $f_1,f_2\in\mathbb C$, and hence, ${\overline{f}}={\overline f_1}-j\,f_2$, with $\overline{f_1}$ denoting the complex conjugate of $f_1$. Moreover, the inner product of any two quaternions  $f=f_{1}+\,jf_{2},$ and $g=g_{1}+j\,g_{2}$ in ${\mathbb H}$ is defined by
\begin{align*}
{\big\langle f,g\big\rangle}_{\mathbb H}&= f\overline{g}=(f_{1}{\overline g_{1}}+{\overline f_{2}}g_{2})+j(f_{2}{\overline g_{1}}-{\overline f_{1}}g_{2}).
\end{align*}

By virtue of the complex domain representation, a quaternion-valued function $f:\mathbb R^2 \to \mathbb H$ can be decomposed as $f(x)=f_{1}+j\,f_{2}$,  where $f_{1},f_{2}$ are both complex valued functions.

\parindent=8mm \vspace{.1in}
Let us denote $L^2(\mathbb R^2,\mathbb H)$, the space of all quaternion valued functions $f$ satisfying
\begin{align*}
 {\big\|f\big\|}_{2}=\left\{\int_{\mathbb R^{2}}\Big(|f_{1}(x)|^{2}+|f_{2}(x)|^{2}\Big)\,dx\right\}^{1/2}< \infty.
\end{align*}
The norm on $L^2(\mathbb R^2,\mathbb H)$ is obtained from the inner product of the quaternion valued functions $f=f_{1}+j\,f_{2},$ and $g=g_{1}+j\,g_{2}$  as
\begin{align*}
{\big\langle f,\,g\big\rangle}_{2}&= \int_{\mathbb R^{2}}{\big\langle f,\,g\big\rangle}_{\mathbb H}\,dx\\
&=\int_{\mathbb R^{2}}\bigg\{\Big(f_{1}(x)\,\overline{g_{1}}(x)+\overline{f_{2}}(x)\,g_{2}(x)\Big)+j \Big(f_{2}(x)\,\overline{g_{1}}(x)
-\overline{f_{1}}(x)\,g_{2}(x)\Big)\bigg\}\,dx.
\end{align*}
An easy computation shows that $L^2(\mathbb R^2,\mathbb H)$ equipped with above defined inner product is a Hilbert space.

\parindent=8mm \vspace{.1in}
Before we proceed to establish the main results of this paper, we shall recall the basic theory of quaternion offset linear canonical transform

\subsection{ Quaternion Offset Linear Canonical Transform }
Due to non-commutativity of quaternion multiplication, there are three types of the quaternion offset linear canonical transform, the left-sided QOLCT, the right-sided QOLCT, and two-sided QOLCT.  In this paper, we will extend the theory of two-sided QOLCT introduced by Haoui et al. \cite{QOLCT} as follows:

\parindent=0mm \vspace{.1in}
{\bf Definition 2.1. } Let $A_s =\left[\begin{array}{cccc}a_s & b_s &| & p_s\\c_s & d_s &| & q_s \\\end{array}\right] $, be a matrix parameter such that $a_s$, $b_s$, $c_s$, $d_s$, $p_s$, $q_s \in \mathbb R$ and $ a_sd_s-b_sc_s=1,$ for $s=1,2.$ The two-sided quaternion offset linear canonical transform of any quaternion valued function $f\in L^2(\mathbb R^2,\mathbb H)$, is given by
\begin{align*}
\mathcal O_{A_1,A_2}^{i,j}\big[f\big](w)=\left\{\begin{array}{cc} \int_{\mathbb R^2} K_{A_1}^i(t_1,w_1) \,f(t)\,K_{A_2}^j(t_2,w_2)dt, &b_1,b_2 \neq 0,\\ \sqrt{d_1}\,e^{i\big[\frac{c_1d_1}{2}(w_1-p_1)^2+w_1p_1\big]} f(d_1w_1-d_1p_1,\,t_2) K_{A_2}^j(t_2,w_2), &b_1=0,b_2\neq0,\\
\sqrt{d_2}K_{A_1}^i(t_1,w_1) f(t_1,d_2w_2-d_2p_2) \,e^{j\big[\frac{c_2d_2}{2}(w_2-p_2)^2+w_2p_2\big]}, &b_1\neq0,b_2=0,\\
\sqrt{d_1d_2}f(d_1w_1-d_1p_1,d_2w_2-d_2p_2)\,e^{i\big[\frac{c_1d_1}{2}(w_1-p_1)^2+w_1p_1\big]}& \\ \qquad\times\,e^{j\big[\frac{c_2d_2}{2}(w_2-p_2)^2+w_2p_2\big]}, &b_1=0,b_2=0,
\end{array}\right.
\end{align*}
where $t=(t_{1},t_{2}),\, w=(w_{1},w_{2})$ and the quaternion kernels $K_{A_1}^i(t_1,w_1)$ and $K_{A_2}^j(t_2,w_2)$ are respectively given by
\begin{align*}
K_{A_1}^i(t_1,w_1)=\dfrac{1}{\sqrt{2\pi b_1}} \,e^{\frac{i}{2b_1}\big[a_1t_1^2-2t_1(w_1-p_1)-2w_1(d_1p_1-b_1q_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]},b_1\neq0\,\tag{2.1}
\end{align*}

\begin{align*}
K_{A_2}^j(t_2,w_2)= \dfrac{1}{\sqrt{2\pi b_2}} \,e^{\frac{j}{2b_2}\big[a_2t_2^2-2t_2(w_2-p_2)-2w_2(d_2p_2-b_2q_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]},\,b_2\neq0\,\tag{2.2}
\end{align*}

{\bf Remark 2.2.} The left-sided and right-sided QOLCT can be defined correspondingly by placing the two above kernels both on the left or on the right, respectively.

The corresponding inversion formula for two-sided QOLCT is given by
\begin{align*}
f(x) &=\int_{\mathbb R^2}\overline{K_{A_1}^i(t_1,w_1)} \, \mathcal O_{A_1,A_2}^{i,j}\big[f\big](w)\,\overline{K_{A_2}^j(t_2,w_2)} \,d\xi.\tag{2.3}
\end{align*}

\parindent=0mm \vspace{.1in}
{\bf Theorem 2.3 (Plancherel for QOLCT).} {\it Every two dimensional quaternion valued function $f \in L^2(\mathbb R^2,\mathbb H)$ and its two-sided QOLCT are related to the Plancherel identity in the following way:}
\begin{align*}
\big\|\mathcal O_{A_1,A_2}^{i,j}\big[f\big]\big\|_{L^2(\mathbb R^2,\mathbb H)}=\left\|f\right\|_{L^2(\mathbb R^2,\mathbb H)}.\,\tag{2.4}
\end{align*}

\parindent=0mm \vspace{.0in}

\section{  Quaternion Windowed Offset Linear Canonical Transform}

In this section, we shall formally introduce the notion of the two-sided quaternion windowed offset linear canonical transform of two dimensional signals and then establish the fundamental properties of the proposed transform.

\parindent=0mm \vspace{.1in}
{\bf Definition 3.1.} Let $A_s =\left[\begin{array}{cccc}a_s & b_s &| & p_s\\c_s & d_s &| & q_s \\\end{array}\right] $, be a matrix parameter such that $a_s$, $b_s$, $c_s$, $d_s$, $p_s$, $q_s \in \mathbb R$ and $ a_sd_s-b_sc_s=1,$ for $s=1,2.$ The two-sided quaternion windowed offset linear canonical transform of any quaternion valued function $f\in L^2(\mathbb R^2,\mathbb H)$, with respect to a non-zero window function $g\in  L^2(\mathbb R^2,\mathbb H)$ is given by  \begin{align*}\label{3.1}
\mathcal O_{g}^{A_1,A_2}\big[f\big](u,w)=\left\{\begin{array}{cc} \int_{\mathbb R^2} K_{A_1}^i(t_1,w_1) \,f(t)\overline{g(t-u)}\,K_{A_2}^j(t_2,w_2)dt, &b_1,b_2 \neq 0,\\ \sqrt{d_1} \,e^{i\big[\frac{c_1d_1}{2}(w_1-p_1)^2+w_1p_1\big]} f(d_1w_1-d_1p_1,t_2) K_{A_2}^j(t_2,w_2), &b_1=0,b_2\neq0,\\
\sqrt{d_2} K_{A_1}^i(t_1,w_1) f(t_1,d_2w_2-d_2p_2) \quad\qquad\qquad\qquad\qquad & \\ \times\overline{g(t_1-u_1,d_2w_2-d_2p_2)}\,e^{j\big[\frac{c_2d_2}{2}(w_2-p_2)^2+w_2p_2\big]}, &b_1\neq0,b_2=0,\\
\sqrt{d_1d_2} f(d_1w_1-d_1p_1,d_2w_2-d_2p_2)\quad\qquad\qquad\qquad\qquad& \\ \times\overline{g (d_1w_1-d_1p_1,d_2w_2-d_2p_2) }\,e^{i\big[\frac{c_1d_1}{2}(w_1-p_1)^2+w_1p_1\big]}& \\ \times e^{j\big[\frac{c_2d_2}{2}(w_2-p_2)^2+w_2p_2\big]}, &b_1=0,b_2=0,
\end{array}\right.\,\tag{3.1}
\end{align*}
where $t=(t_{1},t_{2}),\, w=(w_{1},w_{2}),\, u= (u_1,u_2)$ and the quaternion kernels $K_{A_1}^i(t_1,w_1)$ and $K_{A_2}^j(t_2,w_2)$ are given by (2.1) and (2.2), respectively.

\parindent=0mm \vspace{.1in}
{\bf Remark:} It is worth to note that the quaternion windowed OLCT (3.1), boils down to various linear integral transforms such as:
\begin{itemize}
\item Windowed versions of quaternion linear canonical transform when matrices parameters $A_s =\left[\begin{array}{cccc}a_s & b_s &| & 0\\c_s & d_s &| & 0 \\\end{array}\right] $,\\

\item Quaternion windowed fractional Fourier transform when $A_s =\left[\begin{array}{cccc}\cos\theta & \sin\theta &| & 0\\-\sin\theta & \cos\theta &| & 0 \\\end{array}\right] $,\\

\item Quaternion windowed Fourier transform when $A_s =\left[\begin{array}{cccc}1 & 0 &| & 0\\0 & 1 &| & 0 \\\end{array}\right]$.

\end{itemize}
\parindent=0mm \vspace{.1in}
For brevity, in this paper we focus for the case $b_s\neq 0,\, s=1,2,$ as in other cases proposed transform reduces to a chrip multiplications.

\parindent=0mm \vspace{.1in}
In the sequel, our intention is to study the fundamental properties of the proposed quaternion windowed OLCT (3.1).

\parindent=0mm \vspace{.1in}
{\bf Property 1.} {\it (Linearity) For any given quaternion function $f_n \in L^2(\mathbb R^2,\mathbb H), n\in \mathbb N,$ the following relationship is true:}
\begin{align*}
\mathcal O_{g}^{A_1,A_2}\Big[\sum_{n\in \mathbb N}\alpha_n f_n\Big](u,w)= \sum_{n\in \mathbb N}\alpha_n \mathcal O_{g}^{A_1,A_2}\big[ f_n\big](u,w),\,\alpha_n\in \mathbb H.
\end{align*}
\parindent=0mm \vspace{.1in}
{\bf Proof.} Since quaternion windowed OLCT is one of the linear integral transforms, Property 1 is directly obtained form the definition 3.1.

\parindent=0mm \vspace{.1in}
{\bf Property 2.} {\it (Time-Shift) For any given quaternion function $f \in L^2(\mathbb R^2,\mathbb H),$ and window function $g \in L^2(\mathbb R^2,\mathbb H),$ we have:}
\begin{align*}
{\mathcal{O}}^{A_1,A_2}_g\Big[f(t-k)\Big](u,w)&=\,e^{i\big[a_1k_1q_1-c_1k_1p_1+c_1k_1w_1-\frac{a_1c_1k_1^2}{2}\big]}\, {\mathcal{O}}^{A_1,A_2}_g  \big[f(t)\big](u-k,w-ak)\\
&\qquad\qquad\qquad\times\,e^{j\big[a_2k_2q_2-c_2k_2p_2+c_2k_2w_2-\frac{a_2c_2k_2^2}{2}\big]},
\end{align*}
{\it where $u=(u_1,u_2),\,k=(k_1,k_2),\,w=(w_1,w_2),\, a=(a_1,a_2).$}

\parindent=0mm \vspace{.1in}
{\bf Proof.} By Definition 3.1, we have
\begin{align*}
&{\mathcal{O}}^{A_1,A_2}_g\big[f(t-k)\big](u,w)\\
&\qquad=\int_{\mathbb R^2} K_{A_1}^i(t_1,w_1) \,f(t-k)\overline{g(t-u)}\,K_{A_2}^j(t_2,w_2)dt\\
&\qquad=\int_{\mathbb R^2} K_{A_1}^i(x_1+k_1,w_1) \,f(x)\overline{g(x+k-u)}\,K_{A_2}^j(x_2+k,w_2)dx\\
&\qquad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\int_{\mathbb R^2} \,e^{\frac{i}{2b_1}\big[a_1(x_1+k_1)^2-2(x_1+k_1)(w_1-p_1)-2w_1(d_1p_1-b_1q_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]}\\
&\qquad\qquad\times f(x)\overline{g\big(x-(u-k)\big)}\\
&\qquad\qquad\times\,e^{\frac{j}{2b_2}\big[a_2(x_2+k_2)^2-2(x_2+k_2)(w_2-p_2)-2w_2(d_2p_2-b_2q_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]}dx\\
&\qquad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\int_{\mathbb R^2} \,e^{\frac{i}{2b_1}\big[a_1k_1^2-2k_1(w_1-p_1)-2a_1k_1(d_1p_1-b_1q_1)-d_1(a_1^2k_1^2+2w_1a_1k_1)\big]}\\
&\qquad\qquad\times\,e^{\frac{i}{2b_1}\big[a_1x_1^2-2x_1(w_1-p_1-a_1k_1)-2(w_1-a_1k_1)(d_1p_1-b_1q_1)+d_1\big((w_1-a_1k_1)^2+p_1^2\big)-\frac{\pi b_1}{2}\big]}\\
&\qquad\qquad\times f(x)\overline{g\big(x-(u-k)\big)}\\
&\qquad\qquad\times\,e^{\frac{j}{2b_2}\big[a_2x_2^2-2x_2(w_2-p_2-a_2k_2)-2(w_2-a_2k_2)(d_2p_2-b_2q_2)+d_2\big((w_2-a_2k_2)^2+p_2^2\big)-\frac{\pi b_2}{2}\big]}dx\\
&\qquad\qquad\times\,e^{\frac{j}{2b_2}\big[a_2k_2^2-2k_2(w_2-p_2)-2a_2k_2(d_2p_2-b_2q_2)-d_2(a_2^2k_2^2+2w_2a_2k_2)\big]}\\
&\qquad= \,e^{\frac{i}{2b_1}\big[a_1k_1^2-2k_1(w_1-p_1)-2a_1k_1(d_1p_1-b_1q_1)-d_1(a_1^2k_1^2+2w_1a_1k_1)\big]}\\
&\qquad\qquad\times\int_{\mathbb R^2} K_{A_1}^i(x_1,w_1-a_1k_1)f(x)\overline{g\big(x-(u-k)\big)}K_{A_2}^j(x_2,w_2-a_2k_2)\, dx\\
&\qquad\qquad\times\,e^{\frac{j}{2b_2}\big[a_2k_2^2-2k_2(w_2-p_2)-2a_2k_2(d_2p_2-b_2q_2)-d_2(a_2^2k_2^2+2w_2a_2k_2)\big]}\\

&\qquad=\,e^{i\big[a_1k_1q_1-c_1k_1p_1+c_1k_1w_1-\frac{a_1c_1k_1^2}{2}\big]}\, {\mathcal{O}}^{A_1,A_2}_g  \big[f(t)\big](u-k,w-ak)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\,e^{j\big[a_2k_2q_2-c_2k_2p_2+c_2k_2w_2-\frac{a_2c_2k_2^2}{2}\big]},
\end{align*}
where we used the relation $ a_sd_s=1+b_sc_s,$ for $s=1,2,$ in getting last equality.

\parindent=0mm \vspace{.1in}
{\bf Property 3.} {\it (Modulation) For any given quaternion function $f \in L^2(\mathbb R^2,\mathbb H),$ and window function $g \in L^2(\mathbb R^2,\mathbb H),$ then if $\mathcal M_w f(t)= e^{it_1w_1}f(t)e^{jt_2w_2},$  we have:}
\begin{align*}
&{\mathcal{O}}^{A_1,A_2}_g\Big[\mathcal M_w f(t)\Big](u,w)\\
&\quad=\,e^{iw_1(b_1q_1-d_1p_1)-i\frac{d_1b_1w_1^2}{2}+id_1w_1^2}\, {\mathcal{O}}^{A_1,A_2}_g  \big[f(t)\big]\big(u,(1-b)w\big)\,e^{jw_2(b_2q_2-d_2p_2)-j\frac{d_2b_2w_2^2}{2}+jd_2w_2^2},
\end{align*}
{\it where $u=(u_1,u_2),\,w=(w_1,w_2),\, a=(a_1,a_2).$}

\parindent=0mm \vspace{.1in}
{\bf Proof.} We have

\begin{align*}
&{\mathcal{O}}^{A_1,A_2}_g\Big[\mathcal M_w f(t)\Big](u,w)\\
&\qquad= \int_{\mathbb R^2} K_{A_1}^i(t_1,w_1) \,e^{it_1w_1}f(t)e^{jt_2w_2}\,\overline{g(t-u)}\,K_{A_2}^j(t_2,w_2)dt\\
&\qquad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\int_{\mathbb R^2} \,e^{\frac{i}{2b_1}\big[a_1x_1^2-2x_1(w_1-p_1)-2w_1(d_1p_1-b_1q_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]}\\
&\qquad\qquad\qquad\times \,e^{it_1w_1}f(t)e^{jt_2w_2}\,\overline{g(t-u)}\,e^{\frac{j}{2b_2}\big[a_2x_2^2-2x_2(w_2-p_2)-2w_2(d_2p_2-b_2q_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]}dx\\
&\qquad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\int_{\mathbb R^2} \,e^{\frac{i}{2b_1}\big[a_1x_1^2-2x_1(w_1-b_1w_1-p_1)-2w_1(d_1p_1-b_1q_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]}\\
&\qquad\qquad\qquad\qquad\times \,f(t)\overline{e^{-jt_2w_2}}\,\overline{g(t-u)}\,\overline{\,e^{\frac{-j}{2b_2}\big[a_2x_2^2-2x_2(w_2-p_2)-2w_2(d_2p_2-b_2q_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]}}dx\\
&\qquad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\int_{\mathbb R^2} \,e^{\frac{i}{2b_1}\big[a_1x_1^2-2x_1(w_1-b_1w_1-p_1)-2w_1(d_1p_1-b_1q_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]}\\
&\qquad\qquad\qquad\qquad\times \,f(t)\overline{\,e^{\frac{-j}{2b_2}\big[a_2x_2^2-2x_2(w_2-b_2w_2-p_2)-2w_2(d_2p_2-b_2q_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]}\,g(t-u)}dx\\
&\qquad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\int_{\mathbb R^2} \,e^{\frac{i}{2b_1}\big[a_1x_1^2-2x_1(w_1-b_1w_1-p_1)-2w_1(d_1p_1-b_1q_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]}\\
&\qquad\qquad\qquad\qquad\times \,f(t)\overline{g(t-u)}\,e^{\frac{j}{2b_2}\big[a_2x_2^2-2x_2(w_2-b_2w_2-p_2)-2w_2(d_2p_2-b_2q_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]}\,dx\\
&\qquad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\,e^{\frac{i}{2b_1}\big[-2b_1w_1(d_1p_1-b_1q_1)-d_1b_1^2w_1^2+2d_1b_1w_1^2\big]}\\
&\qquad\quad\times\int_{\mathbb R^2} \,e^{\frac{i}{2b_1}\big[a_1x_1^2-2x_1(w_1-b_1w_1-p_1)-2(w_1-b_1w_1)(d_1p_1-b_1q_1)+d_1\big((w_1-b_1w_1)^2+p_1^2\big)-\frac{\pi b_1}{2}\big]}\\
&\qquad\quad\times f(t)\overline{g(t-u)}\,e^{\frac{j}{2b_2}\big[a_2x_2^2-2x_2(w_2-b_2w_2-p_2)-2(w_2-b_2w_2)(d_2p_2-b_2q_2)+d_2\big((w_2-b_2w_2)^2+p_2^2\big)-\frac{\pi b_2}{2}\big]}\,dx\\
&\qquad\quad\times\,e^{\frac{j}{2b_2}\big[-2b_2w_2(d_2p_2-b_2q_2)-d_2b_2^2w_2^2+2d_2b_2w_2^2\big]}\\
&\qquad=\,e^{iw_1(b_1q_1-d_1p_1)-i\frac{d_1b_1w_1^2}{2}+id_1w_1^2}\int_{\mathbb R^2} K_{A_1}^i(t_1,w_1-b_1w_1)\,f(t)\overline{g(t-u)}\, K_{A_1}^j(t_2,w_2-b_2w_2)\,dx\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\,e^{jw_2(b_2q_2-d_2p_2)-j\frac{d_2b_2w_2^2}{2}+jd_2w_2^2}\\

&=\,e^{iw_1(b_1q_1-d_1p_1)-i\frac{d_1b_1w_1^2}{2}+id_1w_1^2}\, {\mathcal{O}}^{A_1,A_2}_g  \big[f(t)\big]\big(u,(1-b)w\big) \,e^{jw_2(b_2q_2-d_2p_2)-j\frac{d_2b_2w_2^2}{2}+jd_2w_2^2}.
\end{align*}

\parindent=0mm \vspace{.1in}
{\bf Property 4.} {\it (Parity) Let $g \in L^2(\mathbb R^2,\mathbb H),$ be a quaternion window function and $f \in L^2(\mathbb R^2,\mathbb H),$ then we have:}
\begin{align*}
{\mathcal{O}}^{A_1,A_2}_g\Big[P f(t)\Big](u,w)=-{\mathcal{O}}^{A_1^\prime,A_2^\prime}_{Pg}\big[f(t)\big](-u,-w),
\end{align*}
{\it where $Pf(t)=f(-t),$ and $ A_s^\prime =\left[\begin{array}{cccc}a_s & b_s &| & -p_s\\c_s & d_s &| & -q_s \\\end{array}\right],\,s=1,2. $}

\parindent=0mm \vspace{.1in}
{\bf Proof.} We have

\begin{align*}
&{\mathcal{O}}^{A_1,A_2}_g\Big[P f(t)\Big](u,w)\\
&\qquad= \int_{\mathbb R^2} K_{A_1}^i(t_1,w_1) f(-t)\,\overline{g(t-u)}\,K_{A_2}^j(t_2,w_2)dt\\
&\qquad= \int_{\mathbb R^2} K_{A_1}^i(-x_1,w_1) f(x)\,\overline{g(-x-u)}\,K_{A_2}^j(-x_2,w_2)(-dx)\\
&\qquad= \int_{\mathbb R^2} K_{A_1}^i(-x_1,w_1) f(x)\,\overline{g(-x-u)}\,K_{A_2}^j(-x_2,w_2)(-dx)\\
&\qquad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\int_{\mathbb R^2} \,e^{\frac{i}{2b_1}\big[a_1(-x_1)^2-2(-x_1)(w_1-p_1)-2w_1(d_1p_1-b_1q_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]}\\
&\qquad\qquad\times f(x)\,e^{\frac{j}{2b_2}\big[a_2(-x_2)^2-2(-x_2+k_2)(w_2-p_2)-2w_2(d_2p_2-b_2q_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]}\,(-dx)\\
&\qquad=\dfrac{(-1)}{2\pi \sqrt{b_1b_2}}\int_{\mathbb R^2} \,e^{\frac{i}{2b_1}\big[a_1x_1^2-2(x_1)\big(-w_1-(-p_1)\big)-2(-w_1)\big(d_1(-p_1)-b_1(-q_1)\big)+d_1\big(w_1^2+(-p_1)^2\big)\big]}\\
&\qquad\qquad\qquad\qquad\times e^{-i\frac{\pi}{4}}\,f(x)\overline{g\big(-(x+u)\big)}e^{-j\frac{\pi}{4}}\\
&\qquad\quad\times \,e^{\frac{j}{2b_2}\big[a_2x_2^2-2(x_2)\big(-w_2-(-p_2)\big)-2(-w_2)\big(d_2(-p_2)-b_2(-q_2)\big)+d_2\big(w_2^2+(-p_2)^2\big)\big]}\,dx\\
&\qquad= -\int_{\mathbb R^2} K_{A^\prime_1}^i(x_1,-w_1) f(x)\,\overline{Pg\big(x-(-u)\big)}\,K_{A^\prime_2}^j(x_2,-w_2)\,dx\\

&\qquad= -{\mathcal{O}}^{A_1^\prime,A_2^\prime}_{Pg}\big[f(t)\big](-u,-w),\,\, \text{where}\quad A_s^\prime =\left[\begin{array}{cccc}a_s & b_s &| & -p_s\\c_s & d_s &| & -q_s \\\end{array}\right],\,s=1,2.
\end{align*}

\parindent=0mm \vspace{.1in}
In particular, it can easily be inferred that
\begin{align*}
{\mathcal{O}}^{A_1,A_2}_{Pg}\Big[P f(t)\Big](u,w)= -{\mathcal{O}}^{A_1^\prime,A_2^\prime}_{g}\big[f(t)\big](-u,-w).
\end{align*}

\parindent=0mm \vspace{.1in}
{\bf Property 5.} {\it (Anti-Linearity) For given quaternion window functions $g_n \in L^2(\mathbb R^2,\mathbb H), n\in \mathbb N,$ and $f \in L^2(\mathbb R^2,\mathbb H)$, we have:}
\begin{align*}
\mathcal O_{\sum \alpha_n g_n}^{A_1,A_2}[f](u,w)= \sum_{n\in \mathbb N}\, \mathcal O_{g_n}^{A_1,A_2}[f](u,w) \,\overline{\alpha}_n,\,\alpha_n\in \mathbb H.
\end{align*}

\parindent=0mm \vspace{.1in}
{\bf Proof.} This property follows similarly as the Property 1.

\parindent=0mm \vspace{.1in}
{\bf Property 6.} {\it (Phase-Shift) For any given quaternion function $f \in L^2(\mathbb R^2,\mathbb H),$ and window function $g \in L^2(\mathbb R^2,\mathbb H),$ we have:}
\begin{align*}
{\mathcal{O}}^{A_1,A_2}_{g(t-k)}[f](u,w)&={\mathcal{O}}^{A_1,A_2}_g  \big[f(t)\big](u+k,w)
\end{align*}
{\it where $u=(u_1,u_2),\,k=(k_1,k_2),\,w=(w_1,w_2).$}

\parindent=0mm \vspace{.1in}
{\bf Proof.} From Definition 3.1, we have
\begin{align*}
{\mathcal{O}}^{A_1,A_2}_{g(t-k)}[f](u,w)&= \int_{\mathbb R^2} K_{A_1}^i(t_1,w_1) f(t)\,\overline{g(t-k-u)}\,K_{A_2}^j(t_2,w_2)dt\\
&= \int_{\mathbb R^2} K_{A_1}^i(t_1,w_1) f(t)\,\overline{g\big(t-(u+k)\big)}\,K_{A_2}^j(t_2,w_2)dt\\
&={\mathcal{O}}^{A_1,A_2}_g  \big[f(t)\big](u+k,w)
\end{align*}

In the remaining part of this section, we will establish an inner product relation between two quaternion valued signals and their respective quaternion windowed OLCTs. As a consequence of this relation, we can deduce the Plancheral identity for the quaternion windowed OLCT as defined by (3.1).

\parindent=0mm \vspace{.1in}
{\bf Theorem 3.2( Inner Product Relation).} {\it Let ${\mathcal{O}}^{A_1,A_2}_{g_1} [f_1]$  and ${\mathcal{O}}^{A_1,A_2}_{g_2} [f_2]$ be the quaternion windowed OLCTs of 2D quaternion-valued signals $f_{1}$ and $f_{2}$, respectively. Then, we have}
\begin{align*}
\Big\langle {\mathcal{O}}^{A_1,A_2}_{g_1} [f_1],\,{\mathcal{O}}^{A_1,A_2}_{g_2} [f_2]\Big\rangle_{L^2(\mathbb R^2\times\mathbb R^2,\mathbb H)}=\big[\big\langle f_1,f_2 \big\rangle\overline{\big\langle g_1,g_2 \big\rangle}\big]_{\mathbb H}.\tag{3.2}
\end{align*}

\parindent=0mm \vspace{.0in}
{\bf Proof.} From definition 3.1, we have
\begin{align*}
&\Big\langle {\mathcal{O}}^{A_1,A_2}_{g_1} [f_1],\,{\mathcal{O}}^{A_1,A_2}_{g_2} [f_2]\Big\rangle_{L^2(\mathbb R^2\times\mathbb R^2,\mathbb H)}\\
&\qquad\quad=\int_{\mathbb R^2}\int_{\mathbb R^2} \left[{\mathcal{O}}^{A_1,A_2}_{g_1} [f_1] (u,w)\, \overline{{\mathcal{O}}^{A_1,A_2}_{g_2} [f_2](u,w)}\,\right]_{\mathbb H}\,du\,dw\\
&\qquad\quad=\int_{\mathbb R^4} \left[{\mathcal{O}}^{A_1,A_2}_{g_1} [f_1] (u,w)\,\int_{\mathbb R^2} \overline{ K_{A_1}^i(x_1,w_1) \,f_2(x)\overline{g_2(x-u)}\,K_{A_2}^j(x_2,w_2)dx}\,\right]_{\mathbb H}\,du\,dw\\
&\qquad\quad=\int_{\mathbb R^6} \left[{\mathcal{O}}^{A_1,A_2}_{g_1} [f_1] (u,w)\, \overline{K_{A_2}^j(x_2,w_2)} \,g_2(x-u)\,\overline{f_2(x)}\,\overline{ K_{A_1}^i(x_1,w_1)}\right]_{\mathbb H}dx\,du\,dw\\
&\qquad\quad=\int_{\mathbb R^8} \left[ K_{A_1}^i(y_1,w_1) \,f_1(y)\overline{g_1(y-u)}\,K_{A_2}^j(y_2,w_2) \overline{K_{A_2}^j(x_2,w_2)} \,g_2(x-u)\,\overline{f_2(x)}\,\overline{ K_{A_1}^i(x_1,w_1)}\,\right]_{\mathbb H}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\,dy\,dx\,du\,dw\\
&\qquad\quad=\int_{\mathbb R^6} \left[ \left(\int_{\mathbb R^2}\overline{ K_{A_1}^i(x_1,w_1)} K_{A_1}^i(y_1,w_1)dw_1\right) \,f_1(y)\overline{g_1(y-u)}\right.\\
&\qquad\qquad\qquad\qquad\qquad\qquad\times\left.\left(\int_{\mathbb R^2}K_{A_2}^j(y_2,w_2) \overline{K_{A_2}^j(x_2,w_2)}dw_2 \right)\,g_2(x-u)\,\overline{f_2(x)}\,\right]_{\mathbb H}dxdydu\\
&\qquad\quad=\int_{\mathbb R^6} \left[ \delta(x_1-y_1) \,f_1(y)\overline{g_1(y-u)}\, \delta(x_2-y_2)\,g_2(x-u)\,\overline{f_2(x)}\,\right]_{\mathbb H}dxdydu\\
&\qquad\quad=\int_{\mathbb R^4} \left[ \,f_1(x)\overline{g_1(x-u)}\,g_2(x-u)\,\overline{f_2(x)}\,\right]_{\mathbb H}dxdu\\
&\qquad\quad=\int_{\mathbb R^4} \left[ \overline{f_2(x)}\,f_1(x)\overline{g_1(x-u)}\,g_2(x-u)\,\right]_{\mathbb H}dxdu\\
&\qquad\quad= \left[ \int_{\mathbb R^2}\overline{f_2(x)}\,f_1(x)dx\,\int_{\mathbb R^2}\overline{g_1(x-u)}\,g_2(x-u)\,du\,\right]_{\mathbb H}\\
&\qquad\quad= \left[\big\langle f_1,f_2 \big\rangle\overline{\big\langle g_1,g_2 \big\rangle}\right]_{\mathbb H}.
\end{align*}

This completes the proof of Theorem 3.2.\quad \fbox

\parindent=8mm \vspace{.1in}

The following corollary follows directly from Theorem 3.2.

\parindent=0mm \vspace{.1in}

{\bf Corollary 3.3(Energy Conservation).} {\it If $f,\,g\in L^2(\mathbb R^2,\mathbb H)$, then}
\begin{align*}
\int_{\mathbb R^2}\int_{\mathbb R^2}\Big|\left[{\mathcal{O}}^{A_1,A_2}_g f\right](u,w)\Big|^2 du\,dw=\big\|f\big\|^2_{L^2(\mathbb R^2,\mathbb H)}\big\|g\big\|^2_{L^2(\mathbb R^2,\mathbb H)},\tag{3.3}
\end{align*}

\parindent=0mm \vspace{.1in}
{\it Remarks:} For any normalized window function $g\in L^2(\mathbb R^2,\mathbb H)$,  the quaternion windowed OLCT becomes an isometry from the space of signals to the space of transforms or more precisely from $L^2(\mathbb R^2,\mathbb H)$ to $L^2(\mathbb R^4,\mathbb H)$.

\parindent=8mm \vspace{.1in}
Our next theorem guarantees the reconstruction of the input signal from the corresponding quaternion windowed OLCT.

\parindent=0mm \vspace{.1in}
{\bf Theorem 3.4. (Inversion Formula).} {\it Given any two quaternion-valued window functions $g_{1}$ and $g_{2}$ such that $\big\langle g_1,g_2 \big\rangle\ne 0$. Then, any $f\in L^2(\mathbb R^2,\mathbb H)$ can be reconstructed by the formula:}
\begin{align*}
f(t)=\dfrac{1}{\big\langle g_2,g_1 \big\rangle}\int_{\mathbb R^2}\int_{\mathbb R^2} \Big[K_{A_1}^{-i}(t_1,w_1){\mathcal{O}}^{A_1,A_2}_{g_1} [f](u,w)\,\overline{{\mathcal K}_{A_2}(t_2,w_2)}\cdot g_{2}(t-u)\Big]_{\mathbb H}\,dw\,du.\tag{3.4}
\end{align*}

\parindent=0mm \vspace{.0in}
{\it Proof.} As ${\mathcal{O}}^{A_1,A_2}_{g_1} [f]\in L^2(\mathbb R^4,\,\mathbb H)$, so we can easily deduce that the integral
\begin{align*}
\tilde f(t)=\dfrac{1}{\big\langle g_2,g_1 \big\rangle}\int_{\mathbb R^2}\int_{\mathbb R^2}\left[K_{A_1}^{-i}(t_1,w_1){\mathcal{O}}^{A_1,A_2}_{g_1} [f](u,w)\,K_{A_2}^{-j}(t_2,w_2)\, g_{2}(t-u)\right]_{\mathbb H}\,dw\,du
\end{align*}
is well defined in $L^2(\mathbb R^2,\,\mathbb H)$. Therefore, by virtue of the orthogonality relation (3.2), we observe that

\begin{align*}
\big\langle \tilde  f,\,h \big\rangle &=\dfrac{1}{\big\langle g_2,g_1 \big\rangle}\int_{\mathbb R^2}\int_{\mathbb R^2}\int_{\mathbb R^2}\left[K_{A_1}^{-i}(t_1,w_1){\mathcal{O}}^{A_1,A_2}_{g_1} [f](u,w)\,K_{A_2}^{-j}(t_2,w_2)\, g_{2}(t-u)\,\overline{h(x)}\,\right]_{\mathbb H}\,dw\,du\, dx\\
&=\dfrac{1}{\big\langle g_2,g_1 \big\rangle}\int_{\mathbb R^2}\int_{\mathbb R^2}\int_{\mathbb R^2}\left[{\mathcal{O}}^{A_1,A_2}_{g_1} [f](u,w)\,K_{A_2}^{-j}(t_2,w_2)\, g_{2}(t-u)\,\overline{h(x)}\,K_{A_1}^{-i}(t_1,w_1)\,\right]_{\mathbb H}\,dw\,du\, dx\\
&=\dfrac{1}{\big\langle g_2,g_1 \big\rangle}\int_{\mathbb R^2}\int_{\mathbb R^2}\left[{\mathcal{O}}^{A_1,A_2}_{g_1} [f](u,w)\,\int_{\mathbb R^2}\,K_{A_2}^{-j}(t_2,w_2)\, g_{2}(t-u)\,\overline{h(x)}\,K_{A_1}^{-i}(t_1,w_1)\, dx\,\right]_{\mathbb H}\,dw\,du\\
&=\dfrac{1}{\big\langle g_2,g_1 \big\rangle}\int_{\mathbb R^2}\int_{\mathbb R^2}\left[{\mathcal{O}}^{A_1,A_2}_{g_1} [f](u,w)\,\int_{\mathbb R^2}\overline{K_{A_1}^{i}(t_1,w_1)\,h(x)\,\overline{g_{2}(t-u)}\,K_{A_2}^{j}(t_2,w_2)}\, dx\,\right]_{\mathbb H}\,dw\,du\\
&=\dfrac{1}{\big\langle g_2,g_1 \big\rangle}\int_{\mathbb R^2}\int_{\mathbb R^2}\left[{\mathcal{O}}^{A_1,A_2}_{g_1} [f](u,w)\,\overline{{\mathcal{O}}^{A_1,A_2}_{g_2} [h](u,w)}\,\right]_{\mathbb H}\,dw\,du\\
&=\dfrac{1}{\big\langle g_2,g_1 \big\rangle}\left[\big\langle f_1,f_2 \big\rangle\overline{\big\langle g_1,g_2 \big\rangle}\right]_{\mathbb H}\\
&=\langle f_1,f_2 \big\rangle.
\end{align*}
Thus, $\tilde f=f$. This completes the proof of Theorem 3.4 \qquad \fbox

\parindent=0mm \vspace{.1in}

{\bf Corollary 3.5.} If we take $g_{1}=g_{2}=g$, then equation (3.4) reduces to
\begin{align*}
f(t)=\dfrac{1}{\| g\|^2} \int_{\mathbb R^2}\int_{\mathbb R^2} \Big[K_{A_1}^{-i}(t_1,w_1){\mathcal{O}}^{A_1,A_2}_{g} [f](u,w)\,{\mathcal K}^{-j}_{A_2}(t_2,w_2)\cdot g(t-u)\Big]_{\mathbb H}\,dw\,du.\tag{3.5}
\end{align*}

\section{Uncertainty Principles for Quaternion WOLCT}

\parindent=0mm \vspace{.1in}
The uncertainty principles in harmonic analysis are of central importance as they provide a lower bound for optimal simultaneous resolution in the time and frequency domains (see \cite{Fol}). The most famous of them is Heisenberg-Pauli-Weyl inequality, this has been extended to different time-frequency transforms and several other versions of the uncertainty principle have been investigated from time to time. For instance, Beckner \cite{Bec} obtained a logarithmic version of the uncertainty principle by using a sharp form of Pitt's inequality and showed that this version yields the classical Heisenberg's inequality by virtue of Jensen's inequality. Local version of uncertainty states that if a function is concentrated, then not only is its transformation spread out, but that cannot be localized in a subset of finite measure (see \cite{Lian}).  In this Section, we shall establish an analogue of the well-known Heisenberg's uncertainty inequality and the corresponding logarithmic version and the local uncertainty principle for the quaternion windowed offset linear canonical transform as defined by (3.1). First, we prove the following lemma

\parindent=0mm \vspace{.1in}

{\bf Lemma 4.1.} {\it Let $g\in L^2(\mathbb R^2,\,\mathbb H)\backslash \{0\}$ be a quaternion window function, then for every $f \in L^2(\mathbb R^2,\,\mathbb H),$ we have }

\begin{align*}
\|g\|^2_{L^2(\mathbb R^2,\,\mathbb H)} \, \int_{\mathbb R^2} t_k^2\,\big|f(t)\big|_{\mathbb H}^2\, dt = \int_{\mathbb R^2} t_k^2 \,\Big|\mathcal{O}^{-1}_{A_1,A_2} {\Big[\mathcal O_g^{A_1,A_2} [f](u,\xi)\Big]}(t)\Big|_{\mathbb H}^2  dt\,du\,\tag{4.1}
\end{align*}

\parindent=0mm \vspace{.0in}
{\it Proof.} Using Corollary 3.5, we observe that
\begin{align*}
&\left\|g\right\|^2_{L^2(\mathbb R^2,\,\mathbb H)} \,\int_{\mathbb R^2} t_k^2\big|f(t)\big|_{\mathbb H}^2 dt
\\&\qquad= \int_{\mathbb R^2} t_k^2\big|f(t)\big|_{\mathbb H}^2 dt \int_{\mathbb R^2}\big|g(t-b)\big|_{\mathbb H}^2 du
\\&\qquad=\int_{\mathbb R^2}\int_{\mathbb R^2} t_k^2\big|f(t)\big|_{\mathbb H}^2 \big|g(t-b)\big|_{\mathbb H}^2 dt\,du
\\&\qquad=\int_{\mathbb R^2}\int_{\mathbb R^2} t_k^2\big|f(t)\,\overline{g(t-b)}\big|_{\mathbb H}^2 dt\,du
\\&\qquad=\int_{\mathbb R^4} t_k^2\left|\dfrac{1}{\| g\|^2} \int_{\mathbb R^4} \Big[K_{A_1}^{-i}(t_1,\xi_1){\mathcal{O}}^{A_1,A_2}_{g} [f](u,\xi)\,{\mathcal K}^{-j}_{A_2}(t_2,\xi_2)\cdot g(t-u)\overline{g(t-u)}\Big]_{\mathbb H}\,d\xi\,du\right|_{\mathbb H}^2 dtdb
\\&\qquad=\int_{\mathbb R^4} t_k^2\left|\dfrac{1}{\| g\|^2} \int_{\mathbb R^2} \Big[K_{A_1}^{-i}(t_1,\xi_1){\mathcal{O}}^{A_1,A_2}_{g} [f](u,\xi)\,{\mathcal K}^{-j}_{A_2}(t_2,\xi_2)\int_{\mathbb R^2}|g(t-u)|_{\mathbb H}^2\,du\,\Big]_{\mathbb H}\,d\xi\,\right|_{\mathbb H}^2 dt\,db
\\&\qquad=\int_{\mathbb R^4} t_k^2\left| \int_{\mathbb R^2} \Big[K_{A_1}^{-i}(t_1,\xi_1){\mathcal{O}}^{A_1,A_2}_{g} [f](u,\xi)\,{\mathcal K}^{-j}_{A_2}(t_2,\xi_2)\Big]_{\mathbb H}\,d\xi\,\right|_{\mathbb H}^2 dt\,db
\\&\qquad= \int_{\mathbb R^2} t_k^2 \,\Big|\mathcal{O}^{-1}_{A_1,A_2} {\Big[\mathcal O_g^{A_1,A_2} [f](u,\xi)\Big]}(t)\Big|_{\mathbb H}^2  dt\,du
\end{align*}
This completes the proof of Lemma 4.1.\quad \fbox

\parindent=8mm \vspace{.2in}
We are now ready to establish the Heisenberg-type inequalities for the proposed quaternion windowed offset linear canonical transform as defined by (3.1).

\parindent=0mm \vspace{.1in}
{\bf Theorem 4.2.} {\it Let $g\in L^2(\mathbb R^2,\,\mathbb H)\backslash \{0\}$ be a quaternion window function $\mathcal{O}^{A_1,A_2}_g [f]$ be the quaternion windowed offset linear canonical transform of any signal $f \in L^2(\mathbb R^2,\,\mathbb H)$, then we have:}
\begin{align*}
\left\{\int_{\mathbb R}t_k^2 \big|f(t)\big|_{\mathbb H}^2 dt\right\}^{1/2} \left\{\int_{\mathbb R^2} \int_{\mathbb R^2}\xi_k^2 \Big|\mathcal O_g^{A_1,A_2} [f](u,\xi)\Big|_{\mathbb H}^2 dud\xi \right\}^{1/2} \ge \dfrac{b_k}{2}\big\|f\big\|^2_{L^2(\mathbb R^2,\,\mathbb H)}\big\|g\big\|^2_{L^2(\mathbb R^2,\,\mathbb H)} \tag{4.2}
\end{align*}

\parindent=0mm \vspace{.1in}
{\it Proof.} By virtue of the Heisenberg's  inequality for the quaternion offset linear canonical transform \cite{QOLCT}, we can write
\begin{align*}
\int_{\mathbb R^2} t^2\big|f(t)\big|_{\mathbb H}^2dt \int_{\mathbb R^2} \xi_k^2 \Big|\mathcal{O}^{A_1,A_2} [f](\xi)\Big|_{\mathbb H}^2 d\xi\ge \left\{\dfrac{b_k}{2} \int_{\mathbb R}\big|f(t)\big|_{\mathbb H}^2dt \right\}^2
\end{align*}

By implementing the inversion formula and the orthogonality relation for quaternion offset linear canonical transform, we can rewrite the above inequality as

\begin{align*}
\int_{\mathbb R^2} t^2\left|\mathcal{O}^{-1}_{A_1,A_2} \Big[\mathcal{O}^{A_1,A_2} [f](\xi)\Big](t)\right|_{\mathbb H}^2dt \int_{\mathbb R^2} \xi_k^2 \Big|\left[\mathcal{O}^{A_1,A_2} f\right](\xi)\Big|_{\mathbb H}^2 d\xi \ge \left\{\dfrac{b_k}{2} \int_{\mathbb R^2}\Big|\mathcal{O}^{A_1,A_2} [f](\xi)\Big|_{\mathbb H}^2d\xi \right\}^2
\end{align*}

Since  $\mathcal{O}^M_g \left[f\right]\in L^2(\mathbb R,\,\mathbb H)$, therefore we can replace $\mathcal{O}^M [f]$ by $\mathcal{O}^M_g [f]$ to obtain

\begin{align*}
&\int_{\mathbb R^2} t^2\left|\mathcal{O}^{-1}_{A_1,A_2} \Big[\mathcal{O}^{A_1,A_2}_{g} [f](u,\xi)\Big](t)\right|_{\mathbb H}^2dt \int_{\mathbb R^2} \xi_k^2 \Big|\left[\mathcal{O}^{A_1,A_2}_{g} f\right](\xi)\Big|_{\mathbb H}^2 d\xi\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ge \left\{\dfrac{b_k}{2} \int_{\mathbb R^2}\Big|\mathcal{O}^{A_1,A_2}_{g} [f](u,\xi)\Big|_{\mathbb H}^2d\xi \right\}^2
\end{align*}

Taking square root on both sides of above equation and integrating with respect to measure $du$, we obtain

\begin{align*}
&\int_{\mathbb R^2} \Bigg(\left\{\int_{\mathbb R^2}t^2\left|\mathcal{O}^{-1}_{A_1,A_2} \Big[\mathcal{O}^{A_1,A_2}_{g} [f](u,\xi)\Big](t)\right|_{\mathbb H}^2dt\right\}^{1/2} \left\{\int_{\mathbb R^2} \xi_k^2 \Big|\left[\mathcal{O}^{A_1,A_2}_{g} f\right](\xi)\Big|_{\mathbb H}^2 d\xi\,du\right\}^{1/2}\Bigg)\,du\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ge \dfrac{b_k}{2} \int_{\mathbb R^2}\int_{\mathbb R^2}\Big|\mathcal{O}^{A_1,A_2}_{g} [f](u,\xi)\Big|_{\mathbb H}^2d\xi\,du
\end{align*}

Furthermore, an implication of the well known Cauchy-Schwarz inequality yields

\begin{align*}
&\left\{\int_{\mathbb R^2} \int_{\mathbb R^2}t_k^2\left|\mathcal{O}^{-1}_{A_1,A_2} \Big[\mathcal{O}^{A_1,A_2}_{g} [f](u,\xi)\Big](t)\right|_{\mathbb H}^2dt\,du\right\}^{1/2} \left\{\int_{\mathbb R^2}\int_{\mathbb R^2} \xi_k^2 \Big|\left[\mathcal{O}^{A_1,A_2}_{g} f\right](\xi)\Big|_{\mathbb H}^2 d\xi\,du\right\}^{1/2}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ge \dfrac{b_k}{2} \int_{\mathbb R^2}\int_{\mathbb R^2}\Big|\mathcal{O}^{A_1,A_2}_{g} [f](u,\xi)\Big|_{\mathbb H}^2d\xi\,du
\end{align*}

Finally, on implementing the Lemma 4.1 on L.H.S, and Corollary 3.3 on R.H.S of the above inequality we have

\begin{align*}
&\left\{\left\|g\right\|^2_{L^2(\mathbb R^2,\,\mathbb H)} \,\int_{\mathbb R^2} t_k^2\big|f(t)\big|_{\mathbb H}^2 dt \right\}^{1/2} \left\{\int_{\mathbb R^2}\int_{\mathbb R^2} \xi_k^2 \Big|\left[\mathcal{O}^{A_1,A_2}_{g} f\right](\xi)\Big|_{\mathbb H}^2 d\xi\,du\right\}^{1/2}\qquad\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ge \dfrac{b_k}{2}\big\|f\big\|^2_{L^2(\mathbb R^2,\mathbb H)}\big\|g\big\|^2_{L^2(\mathbb R^2,\mathbb H)}
\end{align*}

The desired result is obtained by dividing $\big\|g\big\|^2_{L^2(\mathbb R^2,\mathbb H)}$ on both sides of above inequality. \qquad\fbox

\parindent=8mm \vspace{.1in}

We now establish the logarithmic uncertainty principle for the quaternion windowed offset linear canonical transform $\mathcal{O}^M_g [f]$ as defined by (3.1). First, we remind the following definition of space of rapidly decreasing smooth quaternion functions(see \cite{Asym}).

\parindent=0mm \vspace{.1in}
{\bf Definition 4.3.} For a multi-index $\alpha =(\alpha_1,\alpha_2)\in \mathbb R^{+}\times \mathbb R^{+},$  the Schwartz space in $ L^2(\mathbb R^2,\mathbb H)$ is defined as

\begin{align*}
\mathcal S (\mathbb R^2,\mathbb H)= \left\{f\in\mathbb C^{\infty} (\mathbb R^2,\mathbb H); \sup_{t\in\mathbb R^2}\big(1+|t|^k\big)\Big|\dfrac{\partial^{\alpha_1+\alpha_2}[f(t)]}{\partial^{\alpha_1}_{t_1}\partial^{\alpha_2}_{t_2}}\Big|\,< \infty \right\},
\end{align*}
where $\mathbb C^{\infty}(\mathbb R^2,\mathbb H)$ is the set of smooth functions from $\mathbb R^2$ to $\mathbb H$.

\parindent=0mm \vspace{.1in}

{\bf Theorem 4.4.}  {\it Given a quaternion window function $g\in \mathcal S (\mathbb R^2,\mathbb H)$ and a signal $f\in \mathcal S(\mathbb R^2,\mathbb H)$, the quaternion windowed offset linear canonical transform $\mathcal{O}^{A_1,A_2}_g [f](u,\xi)$ satisfies the following logarithmic estimate of the uncertainty inequality:}
\begin{align*}
&\int_{\mathbb R^{2}}\int_{\mathbb R^{2}}\ln\Big|\frac{\xi}{b}\Big|\, \Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|_{\mathbb H}^{2} dud\xi +\|g\|^2_{L^2(\mathbb R^2,\mathbb H)} \int_{\mathbb R^2}\ln|t|\,|f(t)|_{\mathbb H}^{2}dt\qquad\qquad\qquad\qquad\qquad\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\geq\,\mathcal D\,\|g\|^2_{L^2(\mathbb R^2,\mathbb H)}\,\|f\|^2_{L^2(\mathbb R^2,\mathbb H)} \tag{4.3}
\end{align*}
{\it where $\mathcal D=\left(\dfrac{\Gamma^{\prime}(1/2)}{\Gamma(1/2)}-\ln 2\right)$ and $\Gamma$ is a Gamma function.}

\parindent=0mm \vspace{.1in}
{\it Proof.} For the quaternion-valued function $f\in L^2(\mathbb R^2,\mathbb H)$, the time and frequency spreads satisfy the inequality \cite{QOLCT}
\begin{align*}
\int_{\mathbb R^{2}}\ln|t|\,\big|f(t)\big|^{2}dt+\int_{\mathbb R^{2}}\ln\Big|\frac{\xi}{b}\Big|\,{\Big|\mathcal{O}^{A_1,A_2} [f](u,\xi)\Big|_{\mathbb H}^{2}}\,d\xi\geq\,\mathcal D\int_{\mathbb R^{2}}\Big|f(t)\Big|_{\mathbb H}^{2}\,dt.
\end{align*}

Invoking the inversion formula of QOLCT on the L.H.S and Parseval's formula for QOLCT on R.H.S, we obtain
\begin{align*}
\int_{\mathbb R^{2}}\ln|t|\left|\mathcal{O}^{-1}_{A_1,A_2} \Big[\mathcal{O}^{A_1,A_2} [f](\xi)\Big](t)\right|_{\mathbb H}^{2} dt+\int_{\mathbb R^{2}}\ln\Big|\frac{\xi}{b}\Big|{\Big|\mathcal{O}^{A_1,A_2} [f](\xi)\Big|_{\mathbb H}^{2}}d\xi\geq\mathcal D\int_{\mathbb R^{2}}\Big|\mathcal O^{A_1,A_2}[f](\xi)\Big|_{\mathbb H}^{2}d\xi
\end{align*}

Replacing $\mathcal O^{A_1,A_2}[f](\xi)$ by  $\mathcal{O}^{A_1,A_2}_g [f](u,\xi)$ in the above inequality, we obtain
\begin{align*}
&\int_{\mathbb R^{2}}\ln|t|\,\left|\mathcal{O}^{-1}_{A_1,A_2} \Big[\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big](t)\right|_{\mathbb H}^{2} dt+\int_{\mathbb R^{2}}\ln\Big|\frac{\xi}{b}\Big|\,{\Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|_{\mathbb H}^{2}}\,d\xi\qquad\qquad\qquad\qquad\qquad\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\geq\,\mathcal D\int_{\mathbb R^{2}}\Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|_{\mathbb H}^{2}\,d\xi.
\end{align*}
Integrating above equation with respect to measure $du$ and then applying the  Fubini theorem, we obtain
\begin{align*}
&\int_{\mathbb R^{2}}\int_{\mathbb R^{2}}\ln|t|\,\left|\mathcal{O}^{-1}_{A_1,A_2} \Big[\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big](t)\right|_{\mathbb H}^{2} dtdu+\int_{\mathbb R^{2}}\int_{\mathbb R^{2}}\ln\Big|\frac{\xi}{b}\Big|\,{\Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|_{\mathbb H}^{2}}\,d\xi\,du\qquad\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\geq\,\mathcal D\int_{\mathbb R^{2}}\int_{\mathbb R^{2}}\Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|_{\mathbb H}^{2}\,d\xi\,du.
\end{align*}

Applying Lemma 4.1 on L.H.S and Corollary 3.6 on R.H.S, we obtain the desired result
\begin{align*}
&\int_{\mathbb R^{2}}\int_{\mathbb R^{2}}\ln\Big|\frac{\xi}{b}\Big|\, \Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|_{\mathbb H}^{2} dud\xi +\|g\|^2_{L^2(\mathbb R^2,\mathbb H)} \int_{\mathbb R^2}\ln|t|\,|f(t)|_{\mathbb H}^{2}dt\qquad\qquad\qquad\qquad\qquad\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\geq\,\mathcal D\,\|g\|^2_{L^2(\mathbb R^2,\mathbb H)}\,\|f\|^2_{L^2(\mathbb R^2,\mathbb H)}
\end{align*}

This completes the proof of Theorem 4.4. \quad\fbox

\parindent=8mm \vspace{.1in}

In the following, we establish a local type uncertainty principle for quaternion windowed offset linear canonical transform.

\parindent=0mm \vspace{.1in}

{\bf Theorem 4.5.}  {\it Given a quaternion window function $g\in L(\mathbb R^2,\mathbb H)$  and a signal
$f\in L(\mathbb R^2,\mathbb H)$, with $\|g\|^2_{L^2(\mathbb R^2,\mathbb H)}=1=\|f\|^2_{L^2(\mathbb R^2,\mathbb H)},$
such that for a measurable set $E\subset \mathbb R^2\times\mathbb R^2
,\, \epsilon\geq 0,$ and}
\begin{align*}
\int \int_{E} \Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|_{\mathbb H}^{2} dud\xi\,\geq\,1-\epsilon.\tag{4.5}
\end{align*}
We have $2\pi (1-\epsilon)\big| b_1 b_2 \big|^{1/2}\leq \,\mu(E),$ where $\mu(E)$ is Lebesgue measure of $E$.

\parindent=0mm \vspace{.1in}
{\it Proof.} From Definition 3.1, we have
\begin{align*}
\Big|\mathcal O^{A_1,A_2}_{g}[f](u,\xi) \Big|_{\mathbb H}&=\Big| \int_{\mathbb R^2} K_{A_1}^i(t_1,\xi_1) \,f(t)\overline{g(t-u)}\,K_{A_2}^j(t_2,\xi_2)dt\Big|_{\mathbb H}
\\&\leq\,\dfrac{1}{2\pi \big| b_1 b_2 \big|^{1/2}}\int_{\mathbb R^2}|f(t)|\,\big|\overline{g(t-u)}\big|\,dx.
\end{align*}

Taking Sup-norm on L.H.S and implementing well known Holders inequality on R.H.S, we obtain

\begin{align*}
\Big\|\mathcal O^{A_1,A_2}_{g}[f](u,\xi) \Big\|_{L^{\infty}(\mathbb R^2,\,\mathbb H)}
\leq\,\dfrac{1}{2\pi \big| b_1 b_2 \big|^{1/2}}\,\big\|f\big\|_{L^{2}(\mathbb R^2,\,\mathbb H)}\big\|g\big\|_{L^{2}(\mathbb R^2,\,\mathbb H)}\,\tag{4.6}
\end{align*}

Plugging inequality (4.6) in (4.5), we get
\begin{align*}
1-\epsilon\,&\leq\,\int \int_{E} \Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|_{\mathbb H}^{2} dud\xi\\
&\leq\, \mu(E)\cdot \Big\|\mathcal O^{A_1,A_2}_{g}[f](u,\xi) \Big\|_{L^{\infty}(\mathbb R^2,\,\mathbb H)}\\
&\leq\, \mu(E)\dfrac{1}{2\pi \big| b_1 b_2 \big|^{1/2}}\,\big\|f\big\|_{L^{2}(\mathbb R^2,\,\mathbb H)}\big\|g\big\|_{L^{2}(\mathbb R^2,\,\mathbb H)}\\
&\leq\, \mu(E)\dfrac{1}{2\pi \big| b_1 b_2 \big|^{1/2}}.
\end{align*}

This completes the proof of Theorem 4.5. \quad\fbox

\parindent=0mm \vspace{.1in}

{\bf Theorem 4.6(Local uncertainty inequality).}  {\it Let $E$ be a measurable subset of $\mathbb R^2\times\mathbb R^2$, such that $0<\mu(E)<1,$ then for every
$f\in L(\mathbb R^2,\mathbb H)$ and a quaternion window function $g\in L (\mathbb R^2,\mathbb H)$ , we have}

\begin{align*}
\big\|f\big\|_{L^{2}(\mathbb R,\,\mathbb H)}\big\|g\big\|_{L^{2}(\mathbb R,\,\mathbb H)}\leq \dfrac{1}{\sqrt{1-\mu(E)}}
\Big\|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big\|_{L^2(E^c,\,\mathbb H)}\,\tag{4.7}
\end{align*}

{\it Moreover, for every $\alpha>0,$ there exist $C(\alpha)>0,$ such that}

\begin{align*}
\big\|f\big\|_{L^{2}(\mathbb R,\,\mathbb H)}\big\|g\big\|_{L^{2}(\mathbb R,\,\mathbb H)}\leq C(\alpha)\,\Big[\int_{\mathbb R^2}\int_{\mathbb R^2}\big|(u,\xi)\big|^{2\alpha}\,
\Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|^2_{\mathbb H}\,dud\xi\Big]^{1/2}.\,\tag{4.8}
\end{align*}

\parindent=0mm \vspace{.1in}
{\it Proof.} By invoking the Corollary 3.3 and Theorem 4.5, we have

\begin{align*}
\Big\|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big\|^2_{L^2(\mathbb R^2\times\mathbb R^2,\,\mathbb H)}
&=\int_{\mathbb R^2}\int_{\mathbb R^2} \Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|^2_{\mathbb H}dud\xi\\
&=\int \int_{E} \Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|^2_{\mathbb H}dud\xi+\int \int_{E^c} \Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|^2_{\mathbb H}dud\xi\\
&\leq \,\mu(E) \big\|f\big\|_{L^{2}(\mathbb R,\,\mathbb H)}\big\|g\big\|_{L^{2}(\mathbb R,\,\mathbb H)}+\Big\|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big\|^2_{L^2(E^c,\,\mathbb H)}
\end{align*}

Equivalently,
\begin{align*}
\big\|f\big\|^2_{L^{2}(\mathbb R,\,\mathbb H)}\big\|g\big\|^2_{L^{2}(\mathbb R,\,\mathbb H)}(1-\mu(E))\leq \Big\|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big\|^2_{L^2(E^c,\,\mathbb H)}
\end{align*}

Taking square root on both sides and then dividing both sides by $\sqrt{1-\mu(E)},$ we get

\begin{align*}
\big\|f\big\|_{L^{2}(\mathbb R,\,\mathbb H)}\big\|g\big\|_{L^{2}(\mathbb R,\,\mathbb H)}\leq \dfrac{1}{\sqrt{1-\mu(E)}}\Big\|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big\|_{L^2(E^c,\,\mathbb H)}
\end{align*}

which proves our first assertion.

\parindent=0mm \vspace{.0in}
Now, we fix $r_0\in(0,1]$ small enough such that $\mu\big(B_{r_0}\big)<1,$ where
$$B_{r_0}=\left\{(u,\xi)\in \mathbb R^2\times\mathbb R^2;\,\big|(u,\xi)\big|<\,r_0 \right\},$$
the ball of radius $r_0$ centered at origin, we have from inequality (4.7),

\begin{align*}
\big\|f\big\|_{L^{2}(\mathbb R,\,\mathbb H)}\big\|g\big\|_{L^{2}(\mathbb R,\,\mathbb H)}
&\leq\, \dfrac{1}{\sqrt{1-\mu(B_{r_0})}}\left[\int \int_{\big|(u,\xi)\big|>\,r_0 }\Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|^2_{\mathbb H}dud\xi\right]^{1/2}\\
&\leq\, \dfrac{1}{r^{\alpha}_0 \sqrt{1-\mu(B_{r_0})}}\left[\int \int_{\big|(u,\xi)\big|>\,r_0 }r^{2\alpha}_0\Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|^2_{\mathbb H}dud\xi\right]^{1/2}\\
&\leq\, \dfrac{1}{r^{\alpha}_0 \sqrt{1-\mu(B_{r_0})}}\left[\int \int_{\big|(u,\xi)\big|>\,r_0} \,\big|(u,\xi)\big|^{2\alpha}\Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|^2_{\mathbb H}dud\xi\right]^{1/2}\\
&\leq\, C(\alpha)\left[\int \int_{\big|(u,\xi)\big|>\,r_0} \,\big|(u,\xi)\big|^{2\alpha}\Big|\mathcal{O}^{A_1,A_2}_g [f](u,\xi)\Big|^2_{\mathbb H}dud\xi\right]^{1/2}
\end{align*}
where $C(\alpha)=\dfrac{1}{r^{\alpha}_{0} \sqrt{1-\mu(B_{r_0})}}.$ This completes the proof of  Theorem 4.6. \quad\fbox

\section{Example of Quaternion Windowed Offset Linear canonical Transform}

\parindent=0mm \vspace{.0in}
For the demonstration of the quaternion windowed offset linear canonical transform (3.1), in this section, we shall present an illustrative example.

\parindent=0mm \vspace{.1in}
{\bf Example 5.1.} Consider a two-dimensional quaternion valued Gaussian  function $$f(t)=\beta \exp\left\{-
(\alpha_1 t_1^2+\alpha_2 t_2^2)\right\}, \alpha_1,\alpha_2\,\in \mathbb C, \,\beta\in L^2(\mathbb R^2,\,\mathbb H)$$ Then, the quaternion windowed offset linear canonical transform of $f(t)$ with respect to
the unimodular matrices $A_s=\left(a_s,b_s,c_s,d_s, \,|\, p_s, q_s\right),\,s=1,2,$ and the rectangular window function
\begin{align*}
g(t)=\left\{\begin{array}{cc} 1,\, &\text{if}\,\, |t_1|<a, |t_2|<a, \,a>0,\\ 0,\, &\text{elsewhere}\\ \end{array}\right.
\end{align*}
is given by
\begin{align*}
&{\mathcal{O}}_g^{A_1,A_2} [f](u,w)\\
&\quad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\int_{\mathbb R^2} \,e^{\frac{i}{2b_1}\big[a_1t_1^2-2t_1(w_1-p_1)-2w_1(d_1p_1-b_1q_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]} \,\beta \,e^{-(\alpha_1 t_1^2+\alpha_2t_2^2)}\\
&\qquad\qquad\times\overline{g(t-u)} \,e^{\frac{j}{2b_2}\big[a_2t_2^2-2t_2(w_2-p_2)-2w_2(d_2p_2-b_2q_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]}dt\\
&\quad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\int_{u_1-a}^{u_1+a}\int_{u_2-a}^{u_2+a} \,e^{\frac{i}{2b_1}\big[a_1t_1^2-2t_1(w_1-p_1)-2w_1(d_1p_1-b_1q_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]-\alpha_1 t_1^2} \\
&\qquad\qquad\times\,\beta \,e^{\frac{j}{2b_2}\big[a_2t_2^2-2t_2(w_2-p_2)-2w_2(d_2p_2-b_2q_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]-\alpha_2t_2^2}dt\\
&\quad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\,e^{\frac{i}{2b_1}\big[-2w_1(d_1p_1-b_1q_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]}\\
&\quad\times\int_{u_1-a}^{u_1+a} \,e^{\frac{i}{2b_1}\big[t_1^2(a_1-2ib_1\alpha_1)-2t_1(w_1-p_1)\big] }dt_1 \beta \int_{u_2-a}^{u_2+a}\,e^{\frac{j}{2b_2}\big[t_2^2(a_2-2jb_2\alpha_2)-2t_2(w_2-p_2)\big]}\,dt_2  \\
&\qquad\qquad\times \,e^{\frac{j}{2b_2}\big[-2w_2(d_2p_2-b_2q_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]}\\
\end{align*}
For simplicity, choose $\alpha_1=\frac{-ia_1}{2b_1}$ and $\alpha_2=\frac{-j a_2}{2b_2},$ we have
\begin{align*}
&{\mathcal{O}}_g^{A_1,A_2} [f](u,w)\\
&\quad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\,e^{\frac{i}{2b_1}\big[-2w_1(d_1p_1-b_1q_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]}\,\int_{u_1-a}^{u_1+a} \,e^{\frac{i(p_1-w_1)}{b_1}t_1 }\,dt_1   \\
&\qquad\qquad\qquad\times \,\beta \int_{u_2-a}^{u_2+a}\,e^{\frac{j(p_2-w_2)}{b_2}t_2}\,dt_2\,e^{\frac{j}{2b_2}\big[-2w_2(d_2p_2-b_2q_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]}\\
&\quad=\dfrac{1}{2\pi \sqrt{b_1b_2}}\,e^{\frac{i}{2b_1}\big[2w_1(b_1q_1-d_1p_1)+d_1(w_1^2+p_1^2)-\frac{\pi b_1}{2}\big]}\,\dfrac{b_1}{i(p_1-w_1)}\left[ e^{\frac{i(p_1-w_1)(u_1+a)}{b_1}}-e^{\frac{i(p_1-w_1)(u_1-a)}{b_1}}\right]  \\
&\qquad\qquad\times \,\beta \,\dfrac{b_2}{j(p_2-w_2)}\left[ e^{\frac{j(p_2-w_2)(u_2+a)}{b_1}}-e^{\frac{j(p_2-w_2)(u_2-a)}{b_2}}\right] \,e^{\frac{j}{2b_2}\big[2w_2(b_2q_2-d_2p_2)+d_2(w_2^2+p_2^2)-\frac{\pi b_2}{2}\big]}\\
&\quad=\dfrac{\sqrt{b_1b_2}}{2\pi(w_1-p_1)}\,e^{\frac{i}{2b_1}\big[2w_1(b_1q_1-d_1p_1)+d_1(w_1^2+p_1^2)+\frac{\pi b_1}{2}\big]}\left[ e^{\frac{i(p_1-w_1)(u_1+a)}{b_1}}-e^{\frac{i(p_1-w_1)(u_1-a)}{b_1}} \right]  \\
&\qquad\times \dfrac{\beta}{(w_2-p_2)} \left[ e^{\frac{j(p_2-w_2)(u_2+a)}{b_1}}-e^{\frac{j(p_2-w_2)(u_2-a)}{b_2}} \right] \,e^{\frac{j}{2b_2}\big[2w_2(b_2q_2-d_2p_2)+d_2(w_2^2+p_2^2)+\frac{\pi b_2}{2}\big]}.\\
\end{align*}

config: arxiv


text:
\begin{document}

\title{Simple mixing actions with uncountably many prime factors}

\author[A. I. Danilenko]{Alexandre I. Danilenko}
\address{Institute for Low Temperature Physics \& Engineering of the National Academy of Sciences of Ukraine, 47 Lenin Ave., Kharkiv, 61103, UKRAINE}
\email{alexandre.danilenko@gmail.com}

\author[A. V. Solomko]{Anton V. Solomko}
\email{solomko.anton@gmail.com}

\subjclass[2010]{37A05, 37A10}

\begin{abstract}
Via $(C,F)$-construction we produce a 2-fold simple mixing transformation which has uncountably many non-trivial proper factors and all of them are prime.
\end{abstract}

\maketitle

\section{Introduction}
This paper is about prime factors of simple probability preserving actions.
We first recall these and related definitions from the theory of joinings.

Let $T=(T_g)_{g\in \Gamma}$ be an ergodic action of a locally compact second countable group $\Gamma$ on a standard probability space $(X,\mathfrak B,\mu)$.
The main interest for us lies in $\Z$ and $\R$-actions.
A measure $\lambda$ on $X\times X$ is called a \emph{2-fold self-joining} of $T$ if it is $(T_g\times T_g)_{g\in\Gamma}$-invariant and it projects onto $\mu$ on both coordinates.
Denote by $J_2^e(T)$ the set of all ergodic 2-fold self-joinings of $T$.
Let $C(T)$ stand for the centralizer of $T$, i.e.\ the set of all $\mu$-preserving invertible transformations of $X$ commuting with $T_g$ for each $g\in \Gamma$.
Given a transformation $S\in C(T)$, we denote by $\mu_S$ the corresponding \emph{off-diagonal measure} on $X\times X$ defined by $\mu_S(A\times B) := \mu(A\cap S^{-1}B)$ for all $A,B\in\mathfrak B$.
In other words, $\mu_S$ is the image of the measure $\mu$ under the map $x\mapsto(x,Sx)$.
Of course, $\mu_S\in J_2^e(T)$ for every $S\in C(T)$.
If $T$ is weakly mixing, $\mu\times\mu$ is also an ergodic self-joining.
If $J_2^e(T) \subset \{\mu_S\mid S\in C(T)\}\cup\{\mu\times\mu\}$ then $T$ is called \emph{2-fold simple} \cite{Ve}, \cite{dJR}.
By a \emph{factor} of $T$ we mean a non-trivial proper $T$-invariant sub-$\sigma$-algebra of $\mathfrak B$.
If $T$ has no non-trivial proper factors then $T$ is called \emph{prime}.
In \cite{Ve} it was shown that if $T$ is 2-fold simple then for each non-trivial factor $\mathfrak{F}$ of $T$ there exists a compact (in the strong operator topology) subgroup $K_{\fr F}\subset C(T)$ such that $\fr F=\fr F_{K_{\fr F}}$, where
$$
\mathfrak{F}_K = \{ A\in\mathfrak B \mid \mu(kA\bigtriangleup A)=0 \text{ for all } k\in K \}
$$
is the fixed algebra of $K$.
In particular, $\mathfrak{F}$ (or, more precisely, the restriction of $T$ to $\mathfrak{F}$) is prime if and only if $K_\mathfrak{F}$ is a maximal compact subgroup of $C(T)$.

One of the natural questions arising after the general theory of simple actions was developed in \cite{dJR} is: are there simple maps with non-unique prime factors?
The first example of such maps was constructed by Glasner and Weiss \cite{GlW} as an inverse limit of certain horocycle flows.
For that they used some subtle facts from Ratner's theory of joinings for horocycle flows and properties of lattices in $SL_2(\Bbb R)$.
The authors of a later paper \cite{DdJ} utilized a more elementary cutting-and-stacking technique  to construct a weakly mixing 2-fold simple transformation which has countably many factors, all of which are prime.
Our purpose in the present paper is to use a similar cutting-and-stacking technique to produce a {\it mixing} transformation which has {\it uncountably} many factors, all of which are prime.

Via $(C,F)$-construction we produce a measure preserving action $T$ of an auxiliary group $G=\mathbb{Z}\times(\mathbb{R}\rtimes\Z_2)$ such that the transformation $T_{(1,0,0)}$ is mixing 2-fold simple and $C(T_{(1,0,0)}) = \{T_g\mid g\in G\}$.
Since all non-trivial compact subgroups of $G$ are $G_b=\{(0,0,0),(0,b,1)\}$, $0\neq b\in\R$, and all of them are maximal, this gives an example of 2-fold simple transformation with uncountably many prime factors.
All these factors are 2-to-1 and pairwise isomorphic.

We also correct a gap in the proof of \cite[Lemma~2.3(ii)]{DdJ} (see Remark~\ref{gap_rem}).

The skeleton of the proof of the main result is basically the same as in \cite{DdJ}, where the ``discrete case'' (i.e.\ the auxiliary group is discrete) was under consideration.
To work with the $(C,F)$-construction for actions of continuous (i.e.\ non-discrete) groups we use the approximation techniques from \cite{Da_H}.

\section{$(C,F)$-construction}

We now briefly outline the $(C,F)$-construction of measure preserving actions for locally compact groups.
For details see \cite{Da_CF} and references therein.

Let $G$ be a unimodular locally compact second countable (l.c.s.c.)\ amenable group.
Fix a ($\sigma$-finite) Haar measure $\lambda$ on it.
Given two subsets $E,F\subset G$, by $EF$ we mean their algebraic product, i.e. $EF=\{ef \mid e\in E, f\in F\}$.
The set $\{e^{-1} \mid e\in E\}$ is denoted by $E^{-1}$.
If $E$ is a singleton, say $E=\{e\}$, then we will write $eF$ for $EF$.
Given a finite set $A$, $|A|$ will denote the cardinality of $A$.
Given a subset $F\subset G$ of finite Haar measure, $\lambda_F$ will denote the probability on $F$ given by $\lambda_F(A) := \lambda(A)/\lambda(F)$ for each measurable $A\subset F$.
If $D$ is finite, then $\kappa_D$ will denote the equidistributed probability on $D$, that is $\kappa_D(A) := |A|/|D|$ for each $A\subset D$.

To define a $(C,F)$-action of $G$ we need two sequences $(F_n)_{n=0}^\infty$ and $(C_n)_{n=1}^\infty$ of subsets in $G$ such that the following are satisfied:
\begin{align}
& (F_n)_{n=0}^\infty \text{ is a F{\o}lner sequence in $G$}, \label{CF1}\\
& C_n \text{ is finite and } |C_n| > 1,  \label{CF2}\\
& F_n C_{n+1} \subset F_{n+1},  \label{CF3}\\
& F_n c \cap F_n c' = \varnothing \text{ for all } c\neq c'\in C_{n+1}.  \label{CF4}
\end{align}
This means that $F_n C_{n+1}$ consists of $|C_{n+1}|$ mutually disjoint `copies' $F_n c$,
$c\in C_{n+1}$, of $F_n$ and all these copies are contained in $F_{n+1}$.

First, we define a probability space $(X,\mu)$ in the following way.
We equip $F_n$ with the measure $(|C_1|\cdots|C_n|)^{-1}\lambda\upharpoonright F_n$ and endow $C_n$ with the equidistributed probability measure.
Let $X_n := F_n \times \prod_{k>n} C_k$ stand for the product of measure spaces.
Define an embedding $X_n \to X_{n+1}$ by setting
$$
(f_n,c_{n+1},c_{n+2},\ldots) \mapsto (f_n c_{n+1},c_{n+2},\ldots).
$$
It is easy to see that this embedding is measure preserving.
Then $X_1\subset X_2\subset\cdots$.
Let $X:=\bigcup_{n=0}^\infty X_n$ denote the inductive limit of the sequence of measure spaces $X_n$ and let $\goth B$ and $\mu$ denote the corresponding Borel $\sigma$-algebra and measure on $X$ respectively.
Then $X$ is a standard Borel space and $\mu$ is $\sigma$-finite.
It is easy to check that $\mu$ is finite if and only if
\begin{equation} \label{fin}
\lim_{n\to\infty}\frac{\lambda(F_n)}{|C_1|\cdots|C_n|} < \infty.
\end{equation}
If (\ref{fin}) is satisfied then we choose (i.e., normalize) $\lambda$ in such a way that $\mu(X)=1$.

Now we define a $\mu$-preserving action of $G$ on $X$.
Suppose that the following is satisfied:
\begin{equation} \label{CFgap}
\text{for any $g\in G$, there is $m\geqslant 0$ with $gF_n C_{n+1}\subset F_{n+1}$ for all $n\geqslant m$.}
\end{equation}
For such $n$, take $x\in X_n\subset X$ and write the expansion $x=(f_n,c_{n+1},c_{n+2},\ldots)$ with $f_n\in F_n$ and $c_i\in C_i$, $i>n$.
Then we let
$$
T_g x := (gf_nc_{n+1},c_{n+2},\ldots)\in X_{n+1}\subset X.
$$
It follows from (\ref{CFgap}) that $T_g$ is a well defined $\mu$-preserving transformation of $X$.
Moreover, $T_gT_h=T_{gh}$, i.e. $T:=(T_g)_{g\in G}$ is a $\mu$-preserving Borel action of $G$ on $X$.
$T$ is called the \emph{$(C,F)$-action of $G$ associated with $(C_{n+1},F_n)_{n=0}^\infty$}.

We now recall some basic properties of $(X, \goth B, \mu, T)$.
Given a Borel subset $A\subset F_n$, we put
$$
[A]_n := \{x\in X \mid x=(f_n,c_{n+1},c_{n+2},\ldots)\in X_n \text{ and } f_n\in A\}
$$
and call this set an \emph{$n$-cylinder}.
It is clear that the $\sigma$-algebra $\goth B$ is generated by the family of all cylinders.
Given Borel subsets $A,B\subset F_n$, we have
\begin{align}
& [A\cap B]_n = [A]_n \cap [B]_n, [A\cup B]_n = [A]_n \cup [B]_n, \\
& [A]_n = [AC_{n+1}]_{n+1} = \bigsqcup_{c\in C_{n+1}} [Ac]_{n+1}, \\
& \mu([A]_n) = |C_{n+1}| \mu([Ac]_{n+1}) \text{ for every } c\in C_{n+1}, \\
& \mu([A]_n) = \mu(X_n) \lambda_{F_n}(A), \\
& T_g[A]_n = [gA]_n \text{ if } gA\subset F_n, \\
& T_g[A]_n = T_h[h^{-1}gA]_n \text{ if } h^{-1}gA\subset F_n. \label{CFhg}
\end{align}
Each $(C,F)$-action is of funny rank one (for the definition see \cite{Fe} for the case of $\Z$-actions and \cite{So} for the general case) and hence ergodic.
It also follows from (\ref{CF2}) that $T$ is conservative.

\section{Main result}

By $\Z_n$ we denote a cyclic group of order $n$, i.e. $\Z_n=\Z/n\Z=\{0,1,\ldots,n-1\}$.
Let $G:=\Z\times(\R\rtimes\Z_2)$ with multiplication law as follows
$$
(x,a,n)(y,b,m) := (x+y,a+(-1)^n b,n+m).
$$
Then the center $C(G)$ of $G$ is $\Z\times\{0\}\times\{0\}$.
Each compact subgroup of $G$ coincides with $G_b=\{(0,0,0),(0,b,1)\}$ for some $b\in\R$.
Notice that $G_b$ is a maximal compact subgroup of $G$ if $b\neq 0$.

To construct the required $(C,F)$-action of $G$ we will determine a sequence $(C_{n+1},F_n)_{n=0}^\infty$.
Let $(r_n)_{n=0}^\infty$ be an increasing sequence of positive integers such that
\begin{equation}
\lim_{n\to\infty}\frac{n^4}{r_n}=0.
\end{equation}
Below --- just after Lemma~\ref{dJlemma} --- one more restriction on the growth of $(r_n)_{n=0}^\infty$ will be imposed and we will assume that $r_n$ is large so that (\ref{dJlem}) is satisfied.
We define recurrently three other sequences $(\widetilde{a}_n)_{n=0}^\infty$, $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ of positive integers by setting
\begin{align*}
\widetilde{a}_0 &:= 1, \\
a_n &:= (2r_{n-1}+1)\widetilde{a}_{n-1} \text{ for $n\geqslant 1$}, \\
b_n &:= (2n-1)\widetilde{a}_{n-1} \text{ for $n\geqslant 1$}, \\
\widetilde{a}_n &:= a_n + b_n + n \text{ for $n\geqslant 1$}.
\end{align*}
For each $n\in\N$, we let
\begin{align*}
I_n &:= \{-n,\ldots,n\}^2\subset\Z^2, \\
H_n &:= \{-r_n,\ldots,r_n\}^2\subset\Z^2, \\
F_n &:= (-a_n,a_n]_\Z \times (-a_n,a_n]_\R \times \Z_2 , \\
S_n &:= (-b_n,b_n]_\Z \times (-b_n,b_n]_\R \times \Z_2 , \\
\widetilde{F}_n &:= (-\widetilde{a}_n,\widetilde{a}_n]_\Z \times (-\widetilde{a}_n,\widetilde{a}_n]_\R \times \Z_2 .
\end{align*}
We also consider a homomorphism $\phi_n\colon \Z^2 \to G$ given by
$$
\phi_n(i,j) := (2i\widetilde{a}_n, 2j\widetilde{a}_n, 0).
$$
We then have
\begin{equation}
S_n\subset F_n, \quad F_n S_n = S_n F_n \subset \widetilde{F}_n \subset G,
\end{equation}
\begin{equation}
S_{n+1} = \widetilde{F}_n \phi_n(I_n) = \bigsqcup_{h\in I_n} \widetilde{F}_n \phi_n(h) = \bigsqcup_{h\in I_n} \phi_n(h) \widetilde{F}_n,
\end{equation}
\begin{equation}
F_{n+1} = \widetilde{F}_n \phi_n(H_n) = \bigsqcup_{h\in H_n} \widetilde{F}_n \phi_n(h) = \bigsqcup_{h\in H_n} \phi_n(h) \widetilde{F}_n,
\end{equation}

Suppose also that $F_n$ is equipped with a finite partition $\xi_n$ such that the following are satisfied:
\begin{enumerate}
\item[(i)] the diameter of each atom of $\xi_n$ is less than $\tfrac 1 n$,
\item[(ii)] for each atom $A\in\xi_{n-1}$ and each element $c\in C_n$, the subset $Ac\subset F_n$ is $\xi_n$-measurable and
\item[(iii)] $\xi_n$ is symmetric, that is $A^{-1}\in\xi_n$ whenever $A\in\xi_n$.
\end{enumerate}
It follows that for each measurable subset $A\subset F_n$, any $\varepsilon>0$ and for all $k$ large enough, there is a $\xi_k$-measurable subset $B\subset F_k$ such that $\mu([A]_n \bigtriangleup [B]_k)<\varepsilon$.
We will denote by $\sigma(\xi_n)$ the $\sigma$-algebra on $F_n$ generated by $\xi_n$.

For a finite subset $D$ in $S_n$, we denote by $\kappa_D$ the corresponding normalized Dirac comb,
i.e.\ a measure on $S_n$ given by $\kappa_D(A):=\dfrac{|A\cap D|}{|D|}$ for each subset $A\subset S_n$.
Given two subsets $A,B\subset F_n$ define a function $f_{A,B}\colon S_n\times S_n \to \R$ by setting $f_{A,B}(x,y) := \dfrac{\lambda(Ax\cap By)}{\lambda(F_n)}$, $x,y\in S_n$.
Choose a finite subset $D_n$ in $S_n$ such that
\begin{equation} \label{discr_meas}
\left| \int_{S_n\times S_n} f_{Ag,Bh} d\kappa_{D_n}d\kappa_{D_n} - \frac{1}{\lambda(S_n)^2}\int_{S_n\times S_n} f_{Ag,Bh} d\lambda d\lambda \right| < \frac{1}{n}
\end{equation}
for each $\xi_n$-measurable subsets $A,B\subset F_n$ and any $g,h\in F_n$ with $AgS_n,BhS_n\subset F_n$.
For instance, let $\xi_n$ consists of `rectangles' $\{a\}\times \Delta\times \{m\} \subset G$, where $a\in (-a_n,a_n]_{\Z}$, $m\in\Z_2$ and $\Delta\subset (-a_n,a_n]_{\R}$ is a subinterval of length $n^{-1}$, and set $D_n:=\{ (a,kn^{-2},m) \mid a\in(-b_n,b_n]_{\Z}, k\in (-n^2 b_n,n^2 b_n]_{\Z}, m\in\Z_2 \}$.
It is an easy exercise to check that (\ref{discr_meas}) is satisfied for such $\xi_n$ and $D_n$.
We notice also that $|D_n^0| = |D_n^1|$.

Given a finite (signed) measure $\nu$ on a finite set $D$, we let $\|\nu\|_1 := \sum_{d\in D}|\nu(d)|$.
If $\pi\colon D\to E$ then clearly $\|\nu\circ\pi^{-1}\|_1 \leqslant \|\nu\|_1$.
Given a finite set $Y$ and a mapping $s\colon Y\to D$, let $\dist_{y\in Y}s(y)$ denote the image of the equidistribution on $Y$ under $s$:
$$
\dist_{y\in Y}s(y) := \frac{1}{|Y|}\sum_{y\in Y} \delta_{s(y)} = \kappa_D\circ s^{-1}.
$$

The following lemma easily follows from \cite[Lemma~2.1]{dJ} (cf. \cite[Lemma~3.2]{Da_H}).
\begin{lemma}\label{dJlemma}
Let $D$ be a finite set.
Then given $\varepsilon>0$ and $\delta>0$, there is $R\in\N$ such that for each $r>R$, there exists a map $s\colon \{-r,\ldots,r\}^2 \to D$ such that
$$
\left\| \dist_{0\leqslant t <N} (s_n(h+(t,0)),s_n(h'+(t,0))) - \kappa_D\times \kappa_D \right\|_1 < \varepsilon
$$
for each $N>\delta r$ and $h\neq h'\in \{-r,\ldots,r\}^2$ with $h_1+N<r$ and $h'_1+N<r$.\footnote{Here and below by $a\neq b\in A$ we denote two different elements $a$, $b$ of a set $A$.}
\end{lemma}

Applying this Lemma with $\varepsilon=\frac{1}{n}$ and $\delta = \frac{1}{n^2}$ we get the following.
If $r_n$ is large enough then there is a mapping $s_n\colon H_n \to D_n$ such that for any $N>\dfrac{r_n}{n^2}$ and $h\neq h'\in H_n\cap (H_n - (N-1,0))$ we have
\begin{equation} \label{dJlem}
\left\| \dist_{0\leqslant t <N} (s_n(h+(t,0)),s_n(h'+(t,0))) - \kappa_{D_n}\times \kappa_{D_n} \right\|_1 < \frac 1 n.
\end{equation}
From now on we will assume that $r_n$ is large so that this condition is satisfied and for each $n$ fix $s_n\colon H_n \to D_n$ satisfying (\ref{dJlem}).

Now we define a map $c_{n+1}\colon H_n \to G$ by setting $c_{n+1}(h) := s_n(h)\phi_n(h)$.
We put $C_{n+1} := c_{n+1}(H_n)$.

The reader should have the following picture in mind.
The set $F_{n+1}$ is exactly tiled with the sets $\widetilde{F}_n \phi_n(h)$, $h\in H_n$, which may be thought of as `windows'.
Each $F_n$ has a `natural' translate $F_n \phi_n(h)$ in $\widetilde{F}_n \phi_n(h)$ but the translate we actually choose is the natural translate perturbed by a further translation $s_n(h)$ which is chosen in a `random' way and does not move $F_n \phi_n(h)$ out of its window.

It is easy to derive that (\ref{CF1})--(\ref{CFgap}) are satisfied for the sequence $(F_n,C_{n+1})_{n=0}^\infty$.
Hence the associated $(C,F)$-action $T=(T_g)_{g\in G}$ of $G$ is well defined on a standard probability space $(X,\fr B,\mu)$.

We now state the main result.

\begin{theorem} \label{MainTh}
The transformation $T_{(1,0,0)}$ is mixing and 2-fold simple.
All non-trivial proper factors of $T_{(1,0,0)}$ are of the form $\fr F_{G_b}$, $0\neq b\in\R$.
All these factors are 2-to-1, prime and pairwise isomorphic.
\end{theorem}

We first prove some technical lemmata.
After that in Proposition~\ref{mixingProp} we show mixing for $T_{(1,0,0)}$ and in Proposition~\ref{simpl} we prove simplicity and describe the centralizer of $T_{(1,0,0)}$.
The structure of factors follows then from Veech's theorem.

Denote by $G^0$ the subgroup $\Z\times\R\times\{0\}$ of index 2 in $G$.
Given any subset $A$ in $G$ we set $A^0 := A\cap G^0$ and $A^1 := A\setminus A^0$.
We will refer to $A^0$ and $A^1$ as `\emph{levels}' of $A$.
We will say that a subset $A\subset G$ is \emph{$\varepsilon$-balanced} if $$|\lambda(A^0) - \lambda(A^1)| < \varepsilon \lambda(A).$$
Denote by $\pi_3\colon G\to\Z_2 $ a natural projection on the third coordinate.
Since $\kappa_{D_n}\circ\pi_3^{-1} = \kappa_{\Z_2 }$, it follows from (\ref{dJlem}) that
\begin{equation} \label{balanced}
\| \dist_{h\in H_n} \pi_3\circ s_n(h) - \kappa_{\Z_2 } \|_1 < \frac{1}{n}.
\end{equation}
In particular, for any $A^* \subset F_n$ the set $A=A^*C_{n+1}$ is $\frac 1 n$-balanced:
\begin{equation} \label{balanced}
|\lambda(A^0) - \lambda(A^1)| < \frac{1}{n} \lambda(A).
\end{equation}
Indeed, since $$A^0 = \bigsqcup_{h\in s_n^{-1}(G^0)} A^{*0} c_n(h) \sqcup \bigsqcup_{h\in s_n^{-1}(G^1)} A^{*1} c_n(h),$$
$$\lambda(A^0) = \lambda(A^{*0})|s_n^{-1}(G^0)| + \lambda(A^{*1})|s_n^{-1}(G^1)|,$$
and similarly
$$\lambda(A^1) = \lambda(A^{*1})|s_n^{-1}(G^0)| + \lambda(A^{*0})|s_n^{-1}(G^1)|.$$
Hence
\begin{align*}
|\lambda(A^0) - \lambda(A^1)| &= |\lambda(A^{*0}) - \lambda(A^{*1})| \left| |s_n^{-1}(G^0)| - |s_n^{-1}(G^1)| \right| \\
&\leqslant \frac{1}{|H_n|} \lambda(A) \left| |s_n^{-1}(G^0)| - |s_n^{-1}(G^1)| \right|.
\end{align*}
It remains to notice that
\begin{align*}
\frac{1}{|H_n|} \left| |s_n^{-1}(G^0)| - |s_n^{-1}(G^1)| \right| &\leqslant \left| \frac{|s_n^{-1}(G^0)|}{|H_n|} - \frac{1}{2} \right| + \left| \frac{|s_n^{-1}(G^1)|}{|H_n|} - \frac{1}{2} \right| \\
&= \| \dist_{h\in H_n} \pi_3\circ s_n(h) - \kappa_{\Z_2 } \|_1 < \frac{1}{n}
\end{align*}
by (\ref{balanced}).
It follows that $A=A^*C_{n+1}$ is $\frac 1 n$-balanced for each $A^* \subset F_n$.

Given $h=(h_1,h_2)\in\Z^2$, we let $h^* := (h_1,-h_2)$.

\begin{lemma} \label{mainLem}
Let $f=f'\phi_{n-1}(h)$ with $f'\in\widetilde{F}_{n-1}$ and $h\in\Z^2$.
\begin{enumerate}
\item[(i)]
Suppose $f\in G^\alpha$ and let $\beta:=1-\alpha$.
Let
\begin{align*}
L^-_n &:= \widetilde{F}_{n-1}^\alpha \phi_{n-1}(I_{n-2} + h) \sqcup \widetilde{F}_{n-1}^\beta \phi_{n-1}(I_{n-2} + h^*) \text{ and} \\
L^+_n &:= \widetilde{F}_{n-1}^\alpha \phi_{n-1}(I_{n} + h) \sqcup \widetilde{F}_{n-1}^\beta \phi_{n-1}(I_{n} + h^*).
\end{align*}
Then $L^-_n \subset fS_n \subset L^+_n$.
Hence
$$
\frac{\lambda(fS_n\bigtriangleup L^-_n)}{\lambda(S_n)} = \overline{o}(1).
$$

\item[(ii)]
If, in addition, $fS_n\subset F_n$ then for any subset $A=A^*C_{n-1}$ with $A^*\subset F_{n-2}$ we have
$$
\frac{\lambda(AC_n\cap fS_n)}{\lambda(S_n)} = \lambda_{F_{n-1}}(A) + \overline{o}(1).
$$
\end{enumerate}
\end{lemma}

Here $\overline{o}(1)$ means a sequence that goes to 0 as $n\to\infty$ and does not depend on the choice of $A^*$ in $F_{n-2}$.

\begin{proof}
(i) Suppose $f\in G^0$ (the case $f\in G^1$ is considered in a similar way).
We have
\begin{align*}
fS_n &= f'\phi_{n-1}(h) \widetilde{F}_{n-1} \phi_{n-1}(I_{n-1}) \\
&= f'\widetilde{F}_{n-1}^0 \phi_{n-1}(h+I_{n-1}) \sqcup f'\widetilde{F}_{n-1}^1 \phi_{n-1}(h^*+I_{n-1}).
\end{align*}
Since $\widetilde{F}_{n-1}^0 \widetilde{F}_{n-1}^\alpha \subset \bigsqcup_{u\in I_1} \widetilde{F}_{n-1}^\alpha \phi_{n-1}(u)$, there exists a partition of $\widetilde{F}_{n-1}^\alpha$ into subsets $A_u^\alpha$, $u\in I_1$, such that $f'A_u^\alpha \subset \widetilde{F}_{n-1}^\alpha \phi_{n-1}(u)$ for any $u$ and $\alpha=0,1$.
Therefore
$$
fS_n = \bigsqcup_{u\in I_1} (f'A_u^0 \phi_{n-1}(u)^{-1}\phi_{n-1}(u+h+I_{n-1}) \sqcup f'A_u^1 \phi_{n-1}(u)^{-1}\phi_{n-1}(u+h^*+I_{n-1})).
$$
It remains to notice that $\bigsqcup_{u\in I_1} f'A_u^\alpha \phi_{n-1}(u)^{-1} = \widetilde{F}_{n-1}^\alpha$.

(ii) Since $fS_n \subset F_n$ and $F_n = \widetilde{F}_{n-1} \phi_{n-1}(H_{n-1})$, it follows from (i) that the subsets $K:=I_{n-1} + h$ and $K^*:=I_{n-1} + h^*$ are contained in $H_{n-1}$.
Therefore
\begin{align*}
&\frac{\lambda(AC_n\cap fS_n)}{\lambda(S_n)}
= \sum_{k\in H_{n-1}} \frac{\lambda(Ac_n(k)\cap fS_n)}{\lambda(S_n)} \\
&= \sum_{k\in H_{n-1}} \frac{\lambda(Ac_n(k)\cap L^-_n)}{\lambda(S_n)} + \overline{o}(1) \\
&= \frac{1}{\lambda(S_n)} \sum_{k\in H_{n-1}} \lambda(As_{n-1}(k)\phi_{n-1}(k)\cap \widetilde{F}_{n-1}^\alpha \phi_{n-1}(K) \sqcup \widetilde{F}_{n-1}^\beta \phi_{n-1}(K^*)) + \overline{o}(1) \\
&= \frac{1}{\lambda(S_n)} \sum_{k\in K} \lambda(As_{n-1}(k) \cap \widetilde{F}_{n-1}^\alpha) +
\frac{1}{\lambda(S_n)} \sum_{k\in K^*} \lambda(As_{n-1}(k) \cap \widetilde{F}_{n-1}^\beta) + \overline{o}(1).
\end{align*}
Notice that
$$\lambda(As_{n-1}(k) \cap \widetilde{F}_{n-1}^{\alpha}) =
\begin{cases}
\lambda(A^\alpha), \text{ if $s_{n-1}(k) \in G^0$;} \\
\lambda(A^\beta), \text{ if $s_{n-1}(k) \in G^1$.}
\end{cases}
$$
In any case, since $A=A'C_{n-1}$ is $\frac{1}{n-2}$-balanced, we conclude from (\ref{balanced}) that $$\lambda(As_{n-1}(k) \cap \widetilde{F}_{n-1}^{\alpha}) = (\frac 1 2 + \overline{o}(1)) \lambda(A).$$
In a similar way $$\lambda(As_{n-1}(k) \cap \widetilde{F}_{n-1}^{\beta}) = (\frac 1 2 + \overline{o}(1)) \lambda(A).$$
Hence
\begin{multline*}
\frac{\lambda(AC_n\cap fS_n)}{\lambda(S_n)}
= \frac{\lambda(A)|K|(1+\overline{o}(1))}{\lambda(S_n)}+ \overline{o}(1) \\
= \frac{\lambda(A)}{\lambda(F_{n-1})} \cdot \frac{\lambda(F_{n-1}) |K|}{\lambda(S_n)} \cdot (1+\overline{o}(1)) + \overline{o}(1) \\
= \lambda_{F_{n-1}}(A) \cdot \frac{\lambda(F_{n-1}) (2n-1)^2}{(2n+1)^2 \lambda(\widetilde{F}_{n-1})} \cdot (1+\overline{o}(1)) + \overline{o}(1)
= \lambda_{F_{n-1}}(A) + \overline{o}(1).
\end{multline*}
\end{proof}

\begin{remark} \label{gap_rem}
We note that there is a gap in \cite[Lemma~2.3(ii)]{DdJ}.
It was stated there that the claim (ii) is true for each subset
$A\subset F_{n-1}$.
This is not true.
However --- as was shown in Lemma~\ref{mainLem}(ii) above --- the
claim is true if $A=A^*C_{n-1}$ for an arbitrary subset $A^*\subset
F_{n-2}$.
This corrected  version of the claim suffices to apply it in the proof
of \cite[Theorem~2.5]{DdJ} which is the only place in that paper where
\cite[Lemma~2.3(ii)]{DdJ} was used.
\end{remark}

We will also use the following simple lemma.

\begin{lemma} \label{techLem}
Let $A$, $B$ and $S$ be subsets of finite Haar measure in $G$.
Then
$$
\int_{S\times S} \lambda(Ax\cap By) d\lambda(x)d\lambda(y) = \int_{A\times B} \lambda(aS\cap bS) d\lambda(a)d\lambda(b).
$$
\end{lemma}

\begin{proof}
Notice that $G$ is unimodular.
Consider two subsets in $G^3$:
\begin{align*}
\Omega_1 :=& \{ (a,x,y) \mid x\in S, y\in S, a\in A\cap Byx^{-1} \} \\
=& \{ (a,x,y) \mid a\in A, y\in S, x\in a^{-1}By\cap S \} \text{ and} \\
\Omega_2 :=& \{ (a,b,y) \mid a\in A, b\in B, y\in b^{-1}aS\cap S \} \\
=& \{ (a,b,y) \mid a\in A, y\in S, b\in B\cap aSy^{-1} \}.
\end{align*}
It is clear that the maping $\Omega_1\ni(a,x,y)\mapsto (a,axy^{-1},y)\in\Omega_2$ is 1-to-1 and $\lambda^3$-preserving.
Applying Fubini theorem we obtain that
$$
\int_{S\times S} \lambda(Ax\cap By) d\lambda(x)d\lambda(y) = \lambda^3(\Omega_1) = \lambda^3(\Omega_2) = \int_{A\times B} \lambda(aS\cap bS) d\lambda(a)d\lambda(b).
$$
\end{proof}

The following lemma is the first step to prove mixing for $T_{(1,0,0)}$.
Let $h_0 := (1,0) \in \Z^2$.
Then $\phi_n(h_0) = (1,0,0)^{2\widetilde{a}_n}$.

\begin{lemma} \label{LemWM}
Given a sequence of subsets $H_n^*\subset H_n$ such that $\frac{|H_n^*|}{|H_n|}\to\delta$ for some $\delta\geqslant 0$, we let $C_n^*:=c_n(H_{n-1}^*)$.
Then
\begin{equation} \label{WMlem}
\sup_{A^*,B^*\in\sigma(\xi_{n-1})} \left| \mu(T_{\phi_n(h_0)} [A^*C_n^*]_n \cap [B^*]_{n-1}) - \mu([A^*C_n^*]_n) \mu([B^*]_{n-1}) \right| \to 0.
\end{equation}
\end{lemma}

\begin{proof}
Let $A,B\subset F_n$ be $\xi_n$-measurable.
We set $F^\circ_n := \{f\in F_n\mid fS_nS_n^{-1}\subset F_n\},$
$A^\circ:= A\cap F^\circ_n$, $B^\circ:=B\cap F^\circ_n$, $H'_n:=H_n\cap (H_n - h_0)$.
It is clear that $\mu(F_n\setminus F^\circ_n)\to 0$ and $\frac{|H'_n|}{|H_n|}\to 1$ as $n\to\infty$.
Since $\phi_n(h_0)\in C(G)$ for all $n\in\N$, we have
$$
\phi_n(h_0) A c_{n+1}(h) = A s_n(h) \phi_n(h_0 + h) = A s_n(h) s_n(h_0 + h)^{-1} c_{n+1} (h_0 + h).
$$
whenever $h\in H'_n$.
In particular, $\phi_n(h_0) A^\circ c_{n+1}(h)\subset F_{n+1}$ for all $h\in H'_n$.
Then
\begin{align*}
\mu(T_{\phi_n(h_0)}[A]_n \cap &[B]_n)
= \mu(T_{\phi_n(h_0)}[A^\circ]_n \cap [B^\circ]_n) + \overline{o}(1) \\
&= \sum_{h\in H_n} \mu(T_{\phi_n(h_0)}[A^\circ c_{n+1}(h)]_{n+1} \cap [B^\circ]_n) + \overline{o}(1) \\
&= \sum_{h\in H'_n} \mu(T_{\phi_n(h_0)}[A^\circ c_{n+1}(h)]_{n+1} \cap [B^\circ]_n) + \overline{o}(1) \\
&= \sum_{h\in H'_n} \mu([A^\circ s_n(h) s_n(h_0 + h)^{-1} c_{n+1} (h_0 + h)]_{n+1} \cap [B^\circ]_n) + \overline{o}(1) \\
&= \sum_{h\in H'_n} \mu([(A^\circ s_n(h) s_n(h_0 + h)^{-1} \cap B^\circ) c_{n+1} (h_0 + h)]_{n+1}) + \overline{o}(1) \\
&= \frac{1}{|H_n|}\sum_{h\in H'_n} \mu([A^\circ s_n(h) s_n(h_0 + h)^{-1} \cap B^\circ]_n) + \overline{o}(1) \\
&= \frac{1}{|H_n|}\sum_{h\in H'_n} \lambda_{F_n}(A^\circ s_n(h) \cap B^\circ s_n(h_0 + h))\mu(X_n) + \overline{o}(1) \\
&= \frac{1}{|H'_n|}\sum_{h\in H'_n} \lambda_{F_n}(A^\circ s_n(h) \cap B^\circ s_n(h_0 + h)) + \overline{o}(1) \\
&= \frac{1}{|H'_n|}\sum_{h\in H'_n} \lambda_{F_n}(As_n(h) \cap B s_n(h_0 + h)) + \overline{o}(1).
\end{align*}
Let $\nu_n := \dist_{h\in H'_n}(s_n(h),s_n(h+h_0))$.
Set $f_{A,B}(x,y) := \lambda_{F_n}(Ax\cap By) = \frac{\lambda(Ax\cap By)}{\lambda(F_n)}$.
Notice that
$$
\nu_n = \frac{1}{2r_n-1} \sum_{i=-r_n}^{r_n} \dist_{-r_n \leqslant t < r_n}(s_n(t,i),s_n(t+1,i)).
$$
It follows from (\ref{dJlem}) that $\| \nu_n - \kappa_{D_n}\times \kappa_{D_n} \|_1 < \frac{1}{n}$.
Then by (\ref{discr_meas})
\begin{multline*}
\mu(T_{\phi_n(h_0)}[A]_n \cap [B]_n) = \int_{S_n\times S_n} f_{A,B}d\nu_n + \overline{o}(1) \\
= \int_{S_n\times S_n} f_{A,B}d\kappa_{D_n}d\kappa_{D_n} + \overline{o}(1)
= \frac{1}{\lambda(S_n)^2} \int_{S_n\times S_n} f_{A,B}d\lambda d\lambda + \overline{o}(1),
\end{multline*}
Now take $A:=A^* C^*_n$ and $B:=B^* C_n$ for some $\xi_{n-1}$-measurable subsets $A^*,B^*\subset F_{n-1}$.
We say that elements $c$ and $c'$ of $C_n$ are \emph{partners} if $F_{n-1} c S_n \cap F_{n-1} c' S_n \neq \varnothing$.
We then write $c \bowtie c'$.
Since $A^*c x \cap B^*c' y = \varnothing$ for $c \not\bowtie c'$, it follows that
\begin{align*}
\int_{S_n\times S_n} f_{A,B}d\lambda d\lambda &=
\int_{S_n\times S_n} \lambda_{F_n} (A^*C_n x \cap B^*C_n y) d\lambda(x) d\lambda(y) \\
&= \frac{1}{\lambda(F_n)} \int_{S_n\times S_n} \sum_{C^*_n\ni c\bowtie c'\in C_n} \lambda (A^*c x \cap B^*c' y) d\lambda(x) d\lambda(y).
\end{align*}
Applying Lemma~\ref{techLem} we now obtain that
$$
\int_{S_n\times S_n} f_{A,B}d\lambda d\lambda =
\frac{1}{\lambda(F_n)} \sum_{C^*_n\ni c\bowtie c'\in C_n} \int_{A^* \times B^*} \lambda (ac S_n \cap bc' S_n) d\lambda(a) d\lambda(b).
$$
Next, we note that
$$
|\lambda (ac S_n \cap bc' S_n) - \lambda (c S_n \cap c' S_n)| \leqslant 8n\lambda(\widetilde{F}_{n-1}) = \overline{o}(1) \lambda(S_n).
$$
Each $c\in C_n$ has no more than $2(4n+1)^2$ partners.
Therefore
\begin{align*}
&\mu(T_{\phi_n(h_0)}[A^*C^*_n]_n \cap [B^*]_{n-1}) \\
&= \frac{1}{\lambda(S_n)^2} \sum_{C^*_n\ni c\bowtie c'\in C_n} \int_{A^* \times B^*} \frac{\lambda (c S_n \cap c' S_n) + \lambda(S_n)\overline{o}(1)}{\lambda(F_n)} d\lambda(a) d\lambda(b) + \overline{o}(1) \\
&= \frac{\lambda(A^*)\lambda(B^*)}{\lambda(F_{n-1})^2}
\frac{\lambda(F_{n-1})^2}{\lambda(S_n)^2 \lambda(F_n)}
\sum_{C^*_n\ni c\bowtie c'\in C_n} \left( \lambda (c S_n \cap c' S_n) + \lambda(S_n)\overline{o}(1) \right) + \overline{o}(1) \\
&= \lambda_{F_{n-1}}(A^*) \lambda_{F_{n-1}}(B^*) \theta_n
\pm \frac{\lambda(F_{n-1})^2 |H^*_n| 2(4n+1)^2 \lambda(S_n) \overline{o}(1)}{\lambda(S_n)^2 \lambda(F_n)}
+ \overline{o}(1) \\
&= \lambda_{F_{n-1}}(A^*) \lambda_{F_{n-1}}(B^*) \theta_n
\pm \frac{\lambda(F_{n-1})^2 |H^*_n| 2(4n+1)^2 \overline{o}(1)}{\lambda(\widetilde{F}_{n-1})^2 (2n-1)^2 |H_n|}
+ \overline{o}(1) \\
&= \lambda_{F_{n-1}}(A^*) \lambda_{F_{n-1}}(B^*) \theta_n
+ \overline{o}(1),
\end{align*}
where $\theta_n = \frac{\lambda(F_{n-1})^2}{\lambda(S_n)^2 \lambda(F_n)} \sum_{C^*_n\ni c\bowtie c'\in C_n} \lambda (c S_n \cap c' S_n)$.
Substituting $A^* = B^* = F_{n-1}$ and passing to the limit we obtain that $\theta_n\to \delta$ as $n\to\infty$.
Hence
$$
\mu(T_{\phi_n(h_0)}[A^*C^*_n]_n \cap [B^*]_{n-1}) = \mu([A^*C^*_n]_n)\mu([B^*]_{n-1}) + \overline{o}(1).
$$
Since $\overline{o}(1)$ does not depend on the choice of $A^*$ and $B^*$ inside $F_{n-1}$, the claim is proven.
\end{proof}

\begin{cor}
The transformation $T_{(1,0,0)}$ is weakly mixing.
\end{cor}

\begin{proof}
Substituting $H_n^* := H_n$ to (\ref{WMlem}) we obtain that
$$
\sup_{A^*,B^*\in\sigma(\xi_{n-1})} \left| \mu(T_{\phi_n(h_0)} [A^*]_{n-1} \cap [B^*]_{n-1}) - \mu([A^*]_{n-1}) \mu([B^*]_{n-1}) \right| \to 0.
$$
Since each measurable subset of $X$ can be approximated by $[A^*]_{n-1}$ for large $n$ and $\xi_{n-1}$-measurable subset $A^*\subset F_{n-1}$, it follows that the sequence $(\phi_n(h_0))_{n=1}^\infty$ is mixing for $T$, that is $\mu(T_{\phi_n(h_0)}A\cap B) \to \mu(A)\mu(B)$ for every pair of measurable subsets $A,B\subset X$.
\end{proof}

\begin{prop} \label{mixingProp}
The transformation $T_{(1,0,0)}$ is mixing.
\end{prop}

\begin{proof}
We have to show that
$$
\lim_{n\to\infty} \mu(T_{g_n}A\cap B) = \mu(A)\mu(B)
$$
for any sequence $(g_n)_{n=1}^\infty$ that goes to infinity in $C(G)$ and every pair of measurable subsets $A,B\subset X$.
Let $g_n\in F_{n+1}\setminus F_n$.
It suffices to show that a subsequence of $(g_n)_{n=1}^\infty$ is mixing for $T$.
We write $g_n = f_n \phi_n(h_n)$ for some $f_n\in \widetilde{F}_n\cap C(G)$ and $h_n\in H_n$.
Denote by $z\colon \Z\to C(G)$ the natural embedding $z(x):=(x,0,0)$.
We may assume that $f_n\in z(\Z_+)$ for all $n$
(the case $f_n\in z(\Z_-)$ is considered in a similar way).
Let $H'_n:= H_n\cap (H_n-h_n)$ and $F'_n:= F_n\cap (f_n^{-1}F_n)$.
Passing to a subsequence, if necessary, we also may assume without loss of generality that
$$
\frac{|H'_n|}{|H_n|}\to \delta_1 \quad\text{and}\quad \frac{\lambda(F'_n)}{\lambda(F_n)}\to \delta_2
$$
for some $\delta_1,\delta_2\geqslant 0$.
Partition $H_n$ into three subsets $H_n^1$, $H_n^2$ and $H_n^3$ as follows
\begin{align*}
& H_n^1 := \{ h\in H_n\mid g_nF_nc_{n+1}(h)\subset F_{n+1}\phi_{n+1}(h_0) \}, \\
& H_n^2 := \{ h\in H_n\mid g_nF_nc_{n+1}(h)\subset F_{n+1} \}, \\
& H_n^3 := H_n\setminus (H_n^1\sqcup H_n^2).
\end{align*}
As before $h_0=(1,0)\in\Z^2$.
Let $C^i_{n+1} := \phi_{n+1}(H_n^i)$.
It is clear that $|H_n^3|\leqslant 4(n+1)(2r_n+1)$ and $\left|H_n^2 \bigtriangleup H'_n\right|\leqslant 2r_n+1$.
Since $|H_n| = (2r_n+1)^2$, it follows that
$$
\frac{|H^1_n|}{|H_n|}\to 1-\delta_1, \quad \frac{|H^2_n|}{|H_n|}\to \delta_1, \quad \frac{|H^3_n|}{|H_n|}\to 0.
$$

Take two $\xi_n$-measurable subsets $A,B\subset F_n$.
Since $$\mu([AC^3_{n+1}]_{n+1})=\frac{|C^3_{n+1}|}{|C_{n+1}|} \mu([A]_n)\leqslant \frac{1}{2r_n+1}\to 0,$$ we have
\begin{equation} \label{first}
\left| \mu(T_{g_n}[AC^3_{n+1}]_{n+1} \cap [B]_n) - \mu([AC^3_{n+1}]_{n+1}) \mu([B]_n) \right| \to 0,
\end{equation}
so $[F_nC^3_{n+1}]_{n+1}$ is negligible.
It suffices to show mixing separately on each of the remaining subsets $[F_nC^1_{n+1}]_{n+1}$ and $[F_nC^2_{n+1}]_{n+1}$.

First, we note that $\phi_{n+1}(h_0)^{-1} g_n F_n C^1_{n+1} \subset F_{n+1}$.
Thus, by (\ref{CFhg}),
$$
T_{g_n}[AC^1_{n+1}]_{n+1} = T_{\phi_{n+1}(h_0)} [\phi_{n+1}(h_0)^{-1}g_n AC^1_{n+1}]_{n+1}.
$$
By Lemma~\ref{LemWM} (with $C^*_{n+1}:=\phi_{n+1}(h_0)^{-1}\phi_n(h_n) C^1_{n+1}$ and $A^*:=f_n A$) we obtain that
\begin{equation}
\left| \mu(T_{g_n}[AC^1_{n+1}]_{n+1} \cap [B]_n) - \mu([AC^1_{n+1}]_{n+1}) \mu([B]_n) \right| \to 0.
\end{equation}

It remains to consider the second case involving $C^2_{n+1}$.
If $\delta_1=0$, then obviously
\begin{equation}
\mu([AC^2_{n+1}]_{n+1}) \to 0.
\end{equation}
Suppose now that $\delta_1 > 0$.
Partition $A$ into three subsets $A_1$, $A_2$ and $A_3$ in the following way: $A_1:=A\cap f_n^{-1}F_n$, $A_2:=A\cap f_n^{-1}F_n \phi_n(h_0)$ and $A_3:=A\setminus (A_1\sqcup A_2)$.
In other words, $f_n A_1\subset F_n$, $f_n A_2\subset F_n \phi_n(h_0)$, $f_n A_3\cap(F_n \sqcup F_n \phi_n(h_0)) = \varnothing$.

Note that
\begin{equation}
\mu([A_3C_{n+1}^2]_{n+1}) \leqslant \mu([A_3]_n) \leqslant \frac{2n+1}{2r_n+1} \to 0.
\end{equation}
For $A_1$ and $A_2$ we argue as in the proof of Lemma~\ref{LemWM}.
Set $F^\circ_n := \{f\in F_n\mid fS_nS_n^{-1}\subset F_n\}$,
$A_1^\circ:= A_1\cap F^\circ_n$, $B^\circ:=B\cap F^\circ_n$.
We have
\begin{align*}
\mu(T_{g_n}[A_1 C_{n+1}^2]&_{n+1} \cap [B]_n)
= \sum_{h\in H'_n} \mu([\phi_n(h_n) f_n A_1^\circ c_{n+1}(h)]_{n+1} \cap [B^\circ]_n) + \overline{o}(1) \\
&= \sum_{h\in H'_n} \mu([(f_n A_1^\circ s_n(h) s_n(h_n+h)^{-1} \cap B^\circ) c_{n+1}(h)]_{n+1}) + \overline{o}(1) \\
&= \frac{1}{|H_n|} \sum_{h\in H'_n} \mu([f_n A_1^\circ s_n(h) s_n(h_n+h)^{-1} \cap B^\circ]_n) + \overline{o}(1) \\
&= \frac{\delta_1}{|H'_n|} \sum_{h\in H'_n} \lambda_{F_n}(f_n A_1^\circ s_n(h) \cap B^\circ s_n(h_n+h)_n) + \overline{o}(1) \\
&= \delta_1 \int_{S_n\times S_n} f_{A_1 f_n,B} d\nu_n + \overline{o}(1),
\end{align*}
where $\nu_n := \dist_{h\in H'_n}(s_n(h),s_n(h_n+h))$ and $f_{A_1 f_n,B}(x,y) = \lambda_{F_n}(A_1 f_nx\cap By)$.
Write $h_n=(t_n,0)$.
Since $\frac{2r_n-t_n+1}{2r_n+1} = \frac{|H'_n|}{|H_n|}\to \delta_1>0$ and
$$
\nu_n = \frac{1}{2r_n-1} \sum_{i=-r_n}^{r_n} \dist_{-r_n \leqslant t \leqslant r_n-t_n}(s_n(t,i),s_n(t+t_n,i)),
$$
it follows from (\ref{dJlem}) and (\ref{discr_meas}) that
$$
\mu(T_{g_n}[A_1 C_{n+1}^2]_{n+1} \cap [B]_n) = \frac{\delta_1}{\lambda(S_n)^2} \int_{S_n\times S_n} f_{A_1 f_n,B} d\lambda d\lambda + \overline{o}(1).
$$
Now take $A:=A^* C^*_n$ and $B:=B^* C_n$ for some $\xi_{n-1}$-measurable subsets $A^*,B^*\subset F_{n-1}$.
Let $C'_n := C_n\cap F'_n$.
It follows that $\frac{|C'_n|}{|C_n|}\to\delta_2$ and $\mu([A_1]_n\bigtriangleup [A^* C_n]_n)=\overline{o}(1)$.
Hence $\mu([A_1]_n) = \delta_2 \mu([A^*]_{n-1}) + \overline{o}(1)$.
Arguing as in the proof of Lemma~\ref{LemWM} we obtain that
$$
\mu(T_{g_n}[A^*C'_nC^2_{n+1}]_{n+1}\cap [B^*]_{n-1}) = \delta_2 \mu([A^*]_{n-1}) \mu([B^*]_{n-1}) + \overline{o}(1).
$$
Therefore
\begin{equation}
\left| \mu(T_{g_n}[A_1C^2_{n+1}]_{n+1} \cap [B]_n) - \mu([A_1C^2_{n+1}]_{n+1}) \mu([B]_n) \right| \to 0.
\end{equation}
Since $T_{g_n}[A_2]_n = T_{\phi_n(h_n+h_0)}[\phi_n(h_0)^{-1} f_n A_2]$ with $\phi_n(h_0)^{-1} f_n A_2\subset F_n$,  a similar reasoning yields
\begin{equation} \label{last}
\left| \mu(T_{g_n}[A_2C^2_{n+1}]_{n+1} \cap [B]_n) - \mu([A_2C^2_{n+1}]_{n+1}) \mu([B]_n) \right| \to 0.
\end{equation}
Since
$$
[A^*]_{n-1} = [A^* C_n C^1_{n+1}]_{n+1} \sqcup \bigsqcup_{i=1}^3 [A_i C^2_{n+1}]_{n+1} \sqcup [A^* C_n C^3_{n+1}]_{n+1},
$$
it follows from (\ref{first})--(\ref{last}) that
$$
\lim_{n\to\infty} \sup_{A^*,B^*\in\sigma(\xi_{n-1})} \left| \mu(T_{g_n}[A^*]_{n-1}\cap[B^*]_{n-1}) - \mu([A^*]_{n-1}) \mu([B^*]_{n-1}) \right| = 0.
$$
Since $\xi_n$-measurable cylinders generate the entire $\sigma$-algebra $\fr B$ as $n\to\infty$, it follows that $(g_n)_{n=1}^\infty$ is a mixing sequence for $T$, as desired.
\end{proof}

\begin{prop} \label{simpl}
The transformation $T_{(1,0,0)}$ is 2-fold simple and $C(T_{(1,0,0)}) = \{T_g\mid g\in G\}$.
\end{prop}

\begin{proof}
Take an ergodic joining $\nu\in J_2^e(T_{(1,0,0)})$.
Let $K_n := \left[-\frac{a_n}{n^2},\frac{a_n}{n^2}\right]_\Z$, $J_n := \left[-\frac{r_n}{n^2},\frac{r_n}{n^2}\right]_\Z$ and $\Phi_n := K_n + 2\widetilde{a}_n J_n$.
We claim that $\nu$-a.e. point $(x,y)\in X\times X$ is \emph{generic} for $T_{(1,0,0)}\times T_{(1,0,0)}$, i.e. for all cylinders $A,B\subset \bigcup_{n=1}^\infty \sigma(\xi_n)$ we have
\begin{equation} \label{generic}
\nu(A\times B) = \lim_{n\to\infty} \frac{1}{|\Phi_n|} \sum_{i\in\Phi_n} \chi_A (T_{(i,0,0)}x) \chi_B (T_{(i,0,0)}y).
\end{equation}
To see this, we first note that $(\Phi_n)_{n=1}^\infty$ is a F{\o}lner sequence in $\Z$.
Since
$$
\frac{a_n}{n^2} + \frac{2\widetilde{a}_n r_n}{n^2} < \frac{\widetilde{a}_n(2r_n+1)}{n^2} < \frac{2a_{n+1}}{(n+1)^2},
$$
it follows that $\Phi_n\subset K_{n+1} + K_{n+1}$ and hence $\bigcup_{m=1}^n \Phi_m\subset K_{n+1} + K_{n+1}$.
This implies that $\left| \Phi_{n+1} + \bigcup_{m\leqslant n} \Phi_m \right| \leqslant 3\left|\Phi_{n+1}\right|$ for every $n\in\N$, i.e. Shulman's condition \cite{Li} is satisfied for $(\Phi_n)_{n=1}^\infty$.
By \cite{Li}, the pointwise ergodic theorem holds along $(\Phi_n)_{n=1}^\infty$ for any ergodic transformation.
Since $T\times T$ is $\nu$-ergodic, (\ref{generic}) holds for $\nu$-a.a.\ $(x,y)\in X\times X$ and for every pair of cylinders $A,B\subset X$ from $\bigcup_{n=1}^\infty \sigma(\xi_n)$.

Fix a generic point $(x,y)\in X\times X$.
Since $x,y\in X_n$ for all sufficiently large $n$ and we have the following expansion
\begin{align*}
x &= (f_n, c_{n+1}(h_n), c_{n+2}(h_{n+1}),\ldots),\\
y &= (f'_n, c_{n+1}(h'_n), c_{n+2}(h'_{n+1}),\ldots)
\end{align*}
with $f_n,f'_n\in F_n$, $h_i,h'_i\in H_i$, $i\geqslant n$.
We let $H_n^- := \left[ -(1-\frac{1}{n^2})r_n, (1-\frac{1}{n^2})r_n \right]_\Z^2 \subset H_n$.
Since the marginals of $\nu$ both equal to $\mu$, we may assume without loss of generality that $h_n,h'_n\in H_n^-$.
Indeed, $$\mu(\{x=(f_n, c_{n+1}(h_n), c_{n+2}(h_{n+1}),\ldots)\in X_n \mid h_n\not\in H_i^-\})<\frac{2}{i^2},$$ and hence by Borel-Cantelli lemma for $\mu$-a.e.\ $x\in X_n$ and all but finitely many $i$ we have $h_i\in H_i^-$.
Then we may replace $x=(f_n, c_{n+1}(h_n), c_{n+2}(h_{n+1}),\ldots)\in X_n$ with $x=(f_n c_{n+1}(h_n)\cdots c_m(h_{m-1}), c_{m+1}(h_m),\ldots)\in X_m$ for some $m>n$ if necessary.
Similarly, $h'_n\in H_n^-$.

This implies, in turn, that
\begin{multline} \label{H_n_incl}
f_{n+1} = f_n c_{n+1}(h_n) \in
\widetilde{F}_n \phi_n(H_n^-) \subset [ -c_n, c_n ]_\Z \times [ -c_n, c_n ]_\R \times \Z_2 ,
\end{multline}
where $c_n = \widetilde{a}_n(1+2r_n(1-\frac{1}{n^2}))$,
and, similarly, $f'_{n+1} \in [ -c_n, c_n ]_\Z \times [ -c_n, c_n ]_\R \times \Z_2 $.

Given $g\in\Phi_n$, there are some uniquely determined $k\in K_n$ and $j\in J_n$ such that $g = k + 2\widetilde{a}_n j$, i.e. $(g,0,0) = (k,0,0) \phi_n(j,0)$.
Moreover, $(j,0) + h_n \in H_n$ since $h_n\in H_n^-$.
It also follows from (\ref{H_n_incl}) that
\begin{equation} \label{b_incl}
(k,0,0)f_n S_n S_n^{\pm 1} \subset F_n.
\end{equation}

Take $g\in\Phi_n$ and calculate $T_{(g,0,0)}x$.
$$
x = (f_n, c_{n+1}(h_n),\ldots) = (f_n c_{n+1}(h_n),\ldots) = (f_n s_n(h_n) \phi_n(h_n),\ldots).
$$
\begin{align*}
(g,0,0)f_n s_n(h_n) \phi_n(h_n)
&= (k,0,0) \phi_n(j,0) f_n s_n(h_n) \phi_n(h_n)\\
&= (k,0,0) f_n s_n(h_n) \phi_n((j,0)+h_n)\\
&= (k,0,0) f_n s_n(h_n) s_n((j,0)+h_n)^{-1} c_{n+1}((j,0)+h_n)\\
&= d c_{n+1}((j,0)+h_n),
\end{align*}
where $d := (k,0,0) f_n s_n(h_n) s_n((j,0)+h_n)^{-1} \in F_n$ by (\ref{b_incl}).
This means that $T_{(g,0,0)}x = (d,\ldots)\in X_n$.
Similarly,
$$
(g,0,0)f'_n, s_n(h'_n) \phi_n(h'_n) = d' c_{n+1}((j,0)+h'_n)
$$
with $d' := (b,0,0) f'_n s_n(h'_n) s_n((t,0)+h'_n)^{-1} \in F_n$.

Now take any $\xi_{n-2}$-measurable subsets $A^*,B^*\subset F_{n-2}$ and set $A:=A^*C_{n-1}C_n$, $B:=B^*C_{n-1}C_n$.
\begin{align*}
\nu([A^*]_{n-2}&\times[B^*]_{n-2}) = \nu([A]_{n}\times[B]_{n}) \\
&= \lim_{n\to\infty} \frac{\left|\{ g\in\Phi_n \mid T_{(g,0,0)}x\in[A]_n, T_{(g,0,0)}y\in[B]_n \}\right|}{|\Phi_n|} \\
&= \lim_{n\to\infty} \frac{\left|\{ g\in\Phi_n \mid d\in A, d'\in B \}\right|}{|\Phi_n|} \\
&= \lim_{n\to\infty} \frac{1}{|K_n|} \sum_{k\in K_n}  \frac{\left|\{ j\in J_n \mid d\in A, d'\in B \}\right|}{|J_n|} \\
&= \lim_{n\to\infty} \frac{1}{|K_n|} \sum_{k\in K_n} \zeta_n (A^{-1}(k,0,0)f_n s_n(h_n) \times B^{-1}(k,0,0)f'_n s_n(h'_n)),
\end{align*}
where $\zeta_n := \dist_{j\in J_n} (s_n((j,0)+h_n),s_n((j,0)+h'_n))$.
We consider separately two cases.

\textit{First case}.
Suppose first that $h_n\neq h'_n$ for infinitely many, say \emph{bad} $n$.
Since $|J_n|\geqslant \frac{r_n}{n^2}$ it follows from (\ref{dJlem}) that $\|\zeta_n - \kappa_{D_n}\times \kappa_{D_n}\|<\frac{1}{n}$.
Moreover, it follows from (\ref{b_incl}) and (\ref{discr_meas}) (we need (\ref{discr_meas}) for $A^{-1}(k,0,0)f_ns_n(h)$) that $$\kappa_{D_n}(A^{-1}(k,0,0)f_ns_n(h)) = \lambda_{S_n}(A^{-1}(k,0,0)f_ns_n(h)) + \overline{o}(1).$$
Hence
\begin{align*}
&\frac{1}{|K_n|} \sum_{k\in K_n} \zeta_n (A^{-1}(k,0,0)f_n s_n(h_n) \times B^{-1}(k,0,0)f'_n s_n(h'_n))\\
&= \frac{1}{|K_n|} \sum_{k\in K_n} \kappa_{D_n}(A^{-1}(k,0,0)f_n s_n(h_n)) \kappa_{D_n}(B^{-1}(k,0,0)f'_n s_n(h'_n)) + \overline{o}(1) \\
&= \frac{1}{|K_n|} \sum_{k\in K_n} \lambda_{S_n}(A^{-1}(k,0,0)f_n s_n(h_n)) \lambda_{S_n}(B^{-1}(k,0,0)f'_n s_n(h'_n)) + \overline{o}(1)
\end{align*}
Now we derive from Lemma~\ref{mainLem}(ii) that
\begin{multline*}
\lambda_{S_n}\left(A^{-1}(k,0,0)f_n s_n(h_n)\right)
= \frac{\lambda\left(A^{-1}(k,0,0)f_n s_n(h_n) \cap S_n\right)}{\lambda(S_n)} \\
= \frac{\lambda\left(A \cap (k,0,0)f_n s_n(h_n) S_n\right)}{\lambda(S_n)}
= \lambda_{F_{n-2}}(A^*) + \overline{o}(1)
\end{multline*}
and, in a similar way, $\lambda_{S_n}\left(B^{-1}(b,0,0)f'_n s_n(h'_n)\right) = \lambda_{F_{n-2}}(B^*) + \overline{o}(1)$.
Hence
$$
\nu([A^*]_{n-2}\times [B^*]_{n-2}) = \lambda_{F_{n-2}}(A^*)\lambda_{F_{n-2}}(B^*) + \overline{o}(1) = \mu([A^*]_{n-2})\mu([B^*]_{n-2}) + \overline{o}(1)
$$
for all bad $n$ and all $\xi_{n-2}$-measurable subsets $A^*,B^*\subset F_{n-2}$.
Since any measurable set can be approximated by $[A^*]_{n-2}$, it follows that in this case $\nu = \mu\times\mu$.

\textit{Second case}.
Now we consider the case where $h_n = h'_n$ for all $n$ greater than some $N$.
Then it is easy to see that $y = T_k x$, where $k=f'_N f^{-1}_N\in G$ and then it follows immediately that $(x,y)$ is generic for the off-diagonal joining $\mu_{T_k}$:
\begin{multline*}
\nu([A]_{n}\times[B]_{n})
= \lim_{n\to\infty} \frac{1}{|\Phi_n|} \sum_{i\in\Phi_n} \chi_{[A]_n} (T_{(i,0,0)}x) \chi_{[B]_n} (T_{(i,0,0)}T_k x) = \\
= \lim_{n\to\infty} \frac{1}{|\Phi_n|} \sum_{i\in\Phi_n} \chi_{[A]_n \cap T_k^{-1}[B]_n} (T_{(i,0,0)} x)
= \mu([A]_n \cap T_k^{-1}[B]_n) = \mu_{T_k}([A]_{n}\times[B]_{n})
\end{multline*}
for all $A,B\in\sigma(\xi_n)$, since $\nu$ projects onto $\mu$.
Since each measurable set can be approximated by cylinder sets, we deduce that in this case $\nu = \mu_{T_k}$ with $k\in G$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{MainTh}]
follows now from Veech's theorem, Propositions~\ref{mixingProp}, \ref{simpl} and the fact that $\fr F_{G_a}$ and $\fr F_{G_b}$ are isomorphic if and only if $G_a$ and $G_b$ are conjugate in $G$ \cite[Corollary~3.3]{dJR}.
It is clear, that $G_b = h G_a h^{-1}$ with $h=(0,\frac{a+b}{2},1)$.
\end{proof}

Notice that with some additional conditions on $s_n$ in Lemma~\ref{dJlemma} (cf. \cite[Lemma~2.3]{Da_fs}) one
can show that $T_{(1,0,0)}$ is actually mixing of all orders.

\section{Concluding remarks}

If we replace $G=\Z\times(\R\rtimes\Z_2)$ with $\Gamma:=\R\times(\R\rtimes\Z_2)$ and apply the same construction (with obvious minor changes) we obtain a probability preserving $\Gamma$-action $R$ such that the flow $(R_{(t,0,0)})_{t\in\R}$ is 2-fold simple mixing and its centralizer coincides with the entire $\Gamma$-action.
This gives an example of 2-fold simple mixing \emph{flow} with uncountably many prime factors.
By \cite{Ry}, each 2-fold simple flow is simple.
For the definitions of higher order simplicity we refer to \cite{dJR}.
Moreover, since $\Z\subset\R$ is a closed cocompact subgroup, the corresponding $\Z$-subaction is also 2-fold simple and $C(R_{(1,0,0)}) = \{R_g\mid g\in \Gamma\}$ by \cite[Theorem~6.1]{dJR}.
Thus we get examples of two nonisomorphic 2-fold simple transformations with uncountably many prime factors: $R_{(1,0,0)}$ is embeddable into a flow while $T_{(1,0,0)}$ is not.


TITLE: Does $2\to 2$ particle scattering occur in a plane?
QUESTION [4 upvotes]: I am reading a physics textbook which implicitly uses this assumption in dealing with photon electron scattering. In general, I don't see why this is true. I can imagine the first two particle momenta spanning the x-y plane and the second two spanning the x-z plane, for instance, as long as the z-components of the final momenta are equal and opposite.

REPLY [0 votes]: You are right about there being no reason for 2-particle collisions to be restricted to a plane. However, if one of the particles has mass, then there is always a frame of reference where the action happens in a plane. For the electron-photon collision, if you go to the rest frame of the electron, then there are three momentum vectors: the photon's pre-collision momentum, the photon's post-collision momentum, and the electron's post-collision momentum. To conserve momentum, these three vectors have to lie in the same plane.
If you are studying Compton scattering, then the electrons in experiments are bound to atoms, making them effectively motionless in the lab frame compared to the incoming photon. This makes it valid to study the problem in two dimensions.