Qualitative uncertainty principle for Gabor transform on certain locally compact groups

TL;DR

This paper investigates the qualitative uncertainty principle for the Gabor transform across specific classes of locally compact groups, including Moore groups, Heisenberg groups, and certain nilpotent Lie groups.

Contribution

It extends the understanding of the qualitative uncertainty principle for Gabor transforms to new classes of locally compact groups, broadening its theoretical scope.

Findings

Uncertainty principle holds for Moore groups.

Results established for Heisenberg groups and their products.

Applicable to certain low-dimensional nilpotent Lie groups.

Abstract

Classes of locally compact groups having qualitative uncertainty principle for Gabor transform have been investigated. These include Moore groups, Heisenberg Group $\mathbb{H}_n, \mathbb{H}_{n} \times D,$ where $D$ is discrete group and other low dimensional nilpotent Lie groups.

Equations

$${\mathcal{H}}_{(x,\pi )}=\pi (x)\text{HS}({\mathcal{H}}_{\pi })$$

$${⟨\pi (x)T,\pi (x)S⟩}_{{\mathcal{H}}_{(x,\pi )}}=tr(S^{\ast }T)=⟨T,S{⟩}_{{\mathcal{H}}_{(x,\pi )}}$$

$$||F|{|}_{H^2(G\times \hat{G})}^2={\int }_{G\times \hat{G}}||F(x,\pi )|{|}_{(x,\pi )}^{2}dxd\pi <\infty .$$

$$⟨F,K{⟩}_{H^2(G\times \hat{G})}={\int }_{G\times \hat{G}}tr[F(x,\pi )K(x,\pi {)}^{\ast }]dxd\pi .$$

$$G_{\psi }f(x,\pi )={\int }_{G}f(y)\overline{\psi (x^{-1}y)}\pi (y{)}^{\ast }dm(y)$$

$$⟨G_{\psi }f(x,\pi )\xi ,\eta ⟩={\int }_{G}f(y)\overline{\psi (x^{-1}y)}⟨\pi (x{)}^{\ast }\xi ,\eta ⟩dm(x)d\mu (\pi ).$$

$$f_x^{\psi }(y)=f(y)\overline{\psi (x^{-1}y)}.$$

$$\widehat{f_x^{\psi }}(\pi )={\int }_G{f}_x^{\psi }(y)\pi (y{)}^{\ast }dy={\int }_{G}f(y)\overline{\psi (x^{-1}y)}\pi (y{)}^{\ast }dy=G_{\psi }f(x,\pi ).$$

$$||G_{\psi }f|{|}_{H^2(G\times \hat{G})}=||\psi |{|}_2||f|{|}_2.$$

$$m\times \mu \{(x,\pi ):G_{\psi }f(x,\pi )\ne 0\}<\infty ,\text{ then }f=0\text{ a.e.}$$

$$m\times \mu \{(x,\pi )\in H\times \hat{H}:G_{\psi }f(x,\pi )\ne 0\}<\infty .$$

$$\mu \{\pi :G_{\psi }f(x,\pi )\ne 0\}<\infty .$$

$$G_{\Psi }F(x,\pi )={\int }_{H}f(y)\overline{\psi (x^{-1}y)}{\chi }_H(x^{-1}y)\pi (y{)}^{\ast }dy.$$

$${\int }_{\hat{G}}{\chi }_{\{\pi \in \hat{G}:G_{\Psi }F(x,\pi )\ne 0\}}(\pi )d\pi \le [G:H{]}^3{\int }_{\hat{H}}{\chi }_{\{\pi \in \hat{H}:G_{\psi }f(x,\pi )\ne 0\}}(\pi )d\pi .$$

$${\int }_H{\int }_{\hat{G}}{\chi }_{\{\pi \in \hat{G}:G_{\Psi }F(x,\pi )\ne 0\}}(x,\pi )dxd\pi$$

$$\le [G:H{]}^3{\int }_H{\int }_{\hat{H}}{\chi }_{\{\pi \in \hat{H}:G_{\psi }f(x,\pi )\ne 0\}}(x,\pi )dxd\pi <\infty .$$

$${\int }_H{\int }_{\hat{G}}{\chi }_{\{(x,\pi )\in H\times \hat{G}:G_{\Psi }F(x,\pi )\ne 0\}}(x,\pi )dxd\pi <\infty ,$$

$$\begin{gathered} {\int }_G{\int }_{\hat{G}}{\chi }_{\{(x,\pi )\in G\times \hat{G}:G_{\Psi }F(x,\pi )\ne 0\}}(x,\pi )dxd\pi \\ <\infty . \end{gathered}$$

$$W(L^2,l^1)(G)=\{f:G\to \mathbb{C}|f\text{ is measurable, }\sum _{t\in D}{\left(\underset{H}{\int }|f(x,t){|}^{2}dx\right)}^{\frac{1}{2}}<\infty \}.$$

$$f(x,n)=\{\begin{array}{ll}\frac{1}{n},& x\in [0,1]\text{ and }n\ge 1\\ =0,& \text{ elsewhere }.\end{array}$$

$$m_G\times {\mu }_G\{(x,\pi ):G_{\psi }f(x,\pi )\ne 0\}<\infty$$

$$m_G\times {\mu }_G\{(x,\pi ):G_{\psi }f(x,\pi )\ne 0\}<\infty .$$

$${\int }_H{\left(\sum _{t\in D}|\psi (x,t)|\right)}^{2}dx\le {\left(\sum _{t\in D}{\left(\underset{H}{\int }|\psi (x){|}^{2}dx\right)}^{\frac{1}{2}}\right)}^2<\infty .$$

$$G_{\varphi }f(x,t,\pi ,\delta )$$

$$={\int }_H\sum _{s\in D}f(y,s)\overline{\psi (x^{-1}y,t^{-1}s)}\overline{\gamma (t^{-1}s)}\pi (y{)}^{\ast }\overline{\delta (s)}dy$$

$$=\gamma (t)G_{\psi }f(x,t,\pi ,\gamma \delta ).$$

$$\{(x,t,\pi ,\delta ):G_{\varphi }f(x,t,\pi ,\delta )\ne 0\}=\{(x,t,\pi ,\overline{\gamma }\delta ):G_{\psi }f(x,t,\pi ,\delta )\ne 0\}.$$

$$m_G\times m_{\hat{G}}\{(x,t,\pi ,\delta ):G_{\varphi }f(x,t,\pi ,\delta )\ne 0\}<\infty .$$

$$\underset{H\hat{H}}{\iint }\sum _{s\in D}{\chi }_{\{(x,t,\pi ):G_{\varphi }f(x,t,\pi ,\delta )\ne 0\}}dxd\pi <\infty .$$

$$G_{\tilde{\psi }}\tilde{f}(x,\pi )$$

$$=\underset{H}{\int }\sum _{s\in D}f(x,s)\delta (s)\sum _{t\in D}\overline{\psi (x^{-1}y,t)}\overline{\gamma (t)}\pi (y{)}^{\ast }dy$$

$$=\sum _{t\in D}\underset{H}{\int }\sum _{s\in D}f(x,s)\delta (s)\overline{\psi (x^{-1}y,t^{-1}s)}\overline{\gamma (t^{-1}s)}\pi (y{)}^{\ast }dy$$

$$=\sum _{t\in D}{G}_{\varphi }f(x,t,\pi ,\delta ).$$

Full text

1. Introduction

For a second countable, unimodular group G of type I, let $m$ be the left Haar measure and $\mu$ the Plancherel measure on the dual object $\widehat{G}.$ Let $H$ be closed normal subgroup of $G$, we identify a representation $\pi$ of $G/H$ with its pullback to $G$, so $\widehat{G/H}$ will always considered as a closed subset of $\widehat{G}$. Let

[TABLE]

where $\pi(x)\text{HS}(\mathcal{H}_{\pi})={\{\pi(x)T:T\in\text{HS}(H_{\pi})\}}$, $\mathcal{H}_{(x,\pi)}$ is a Hilbert Space with the inner product given by

[TABLE]

Also, $\mathcal{H}_{(x,\pi)}=\text{HS}(\mathcal{H}_{\pi})$ for all $(x,\pi)\in G\times\widehat{G}.$ The family $\{\mathcal{H}_{(x,\pi)}\}_{(x,\pi)\in G\times\widehat{G}}$ of Hilbert spaces is a field of Hilbert spaces over $G\times\widehat{G}$. Let $\mathcal{H}^{2}(G\times\widehat{G})$ denote the direct integral of $\{\mathcal{H}_{(x,\pi)}\}_{(x,\pi)\in G\times\widehat{G}}$, that is, the space of all measurable vector Fields $F\text{ on }G\times\widehat{G}$ such that

[TABLE]

$H^{2}(G\times\widehat{G})$ is a Hilbert space with the inner product given by

[TABLE]

For $f\in C_{c}(G)$, the set of all continuous functions on $G$ with compact supports and $\psi$ a fixed non-zero function in $L^{2}(G),$ usually called window function. The continuous Gabor transform of $f$ with respect to $\psi$ can be defined as a measurable field of operators on $G\times\widehat{G}$ given by

[TABLE]

The operator valued integral (1) is considered in the weak sense, i.e., for each $(x,\pi)\in G\times\widehat{G}$ and $\xi,\eta\in H_{\pi},$ we have that

[TABLE]

For each $x\in G,$ let $f_{x}^{\psi}:G\to\mathbb{C}$ by

[TABLE]

Since $f\in C_{c}(G)$ and $\psi\in L^{2}(G),\text{ Therefore }f_{x}^{\psi}\in L^{1}(G)\cap L^{2}(G)\text{ for all }x\in G$. The Fourier transform of $f^{\psi}_{x}$ is given by

[TABLE]

Also, using the Plancherel Formula [8], it follows that $\widehat{f_{x}^{\psi}}$is a Hilbert Schmidt operator for almost all $\pi\in\widehat{G}.$ Therefore, $G_{\psi}f(x,\pi)$ is a Hilbert Schmidt operator for all $x\in G$. Also for almost all $\pi\in\widehat{G}\text{ and }f,\psi\in L^{2}(G)$ [3], we have

[TABLE]

We investigate several classes of locally compact groups for the following so called Qualitative Uncertainty Principle (QUP) for Gabor transform :

If $f\in L^{2}(G)\text{ and }\psi$ is a window function satisfying

[TABLE]

In [2], QUP was shown to hold for various class of locally compact groups like abelian groups, compact extensions of groups having QUP.

In section 2 we shall prove QUP for Gabor transform holds for groups of the form $H\times D,D$ being a discrete group and $H$ is a group for which QUP holds. In section 3 we show that QUP hold for certain classes of nilpotent Lie groups including Heisenberg group and other low dimensional nilpotent Lie group. In section 4, a weaker version of QUP has been discussed for Moore groups. In last section we give necessary condition for QUP to hold for Gabor transform for multiplier extension of $\mathbb{T}$ by abelian group.

2. QUP for Gabor Transform

In this section, $G$ will be second countable, unimodular, locally compact group of type I. For $\pi\in\widehat{G},\mathcal{H_{\pi}}$ will denote the Hilbert space of $\pi.$

Theorem 2.1**.**

Let $H$ be a subgroup of finite index in $G$ then $H$ has QUP for Gabor transform if and only if $G$ has QUP for Gabor transform.

Proof.

Let QUP for Gabor transform hold for $G$. $H$ contains a subgroup $N$ of finite index which is normal in $G$. Since $H/N$ is finite, so by [2] we can assume $H$ to be normal. Let $f\in L^{2}(H)$ and $\psi$ be a window function such that,

[TABLE]

So, there exist a zero set $K\text{ in }H$ such that for all $x\in H\setminus K,$

[TABLE]

Define $F(x)=f(x)\chi_{H}(x)\text{ and }\Psi(x)=\psi(x)\chi_{H}(x),\text{ for all }x\in G.$ Then

[TABLE]

Clearly if $x\notin H,\text{ then }G_{\Psi}F(x,\pi)=0.$ Now for $x\in H,\text{ define }F_{x}^{\Psi}(y)=f_{x}^{\psi}(y)\chi_{H}(y),\text{for all }y\in G$. Fix $x\in H\setminus K$ and using the Plancherel formula [4], it follows

[TABLE]

Equation (3) holds for every $x\in H\setminus K.$ Therefore, we have

[TABLE]

So, it follows

[TABLE]

and therefore we have

[TABLE]

Since $G$ has QUP for Gabor transform Therefore $F=0$ a.e. and consequently $f=0$ a.e.

Conversely let $H$ be a subgroup of finite index and has QUP for Gabor transform. Also $H$ has a subgroup $N$ of finite index normal in $G$. Since $H$ has QUP for Gabor transform and $N$ has finite index in $H$, therefore $N$ has QUP for Gabor transform. As $G/N$ is compact, So by [2], $G$ has QUP for Gabor transform. ∎

Remark 2.2**.**

If $G$ is a group having all irreducible representation bounded i.e. there exist $M>0$ such that dim $\pi\in\widehat{G}$ then by [5], $G$ has an abelian subgroup H of finite. Thus it follows from above theorem that $G$ has QUP for Gabor transform if and only if $G_{0}$ is non compact.

Next, we consider groups of the form $G=H\times D,$ where $H$ be a second countable, unimodular, locally compact group of type I and $D$ be a discrete type I group. We now consider Wiener amalgam space [9], which is defined as follows:

[TABLE]

Note that $W(L^{2},l^{1})(G)=L^{2}(G)$ if $D$ is finite.

Lemma 2.3**.**

$W(L^{2},l^{1})(G)\subseteq L^{2}(G)$. However the containment may be strict.

Proof.

For $f\in W(L^{2},l^{1})(G)$ and $t\in D,$ consider $b_{t}=\left(\int\limits_{H}|f(x,t)|^{2}dx\right)^{\frac{1}{2}}$. Now $0\leq b_{t}<\infty$ for all $t\in D.\text{ Let }S=\{t:b_{t}\geq 1\}\text{ and }|S|=\sum\limits_{t\in S}1\leq\sum\limits_{t\in S}b_{t}\leq\sum\limits_{t\in D}b_{t}<\infty$ which imply $b_{t}\geq 1$ for only finitely many $t$. Also $b_{t}>0$ for at most countably many $t$. So we have $\sum\limits_{t\in D}\int\limits_{H}|f(x,t)|^{2}dx=\sum\limits_{t\in D}b_{t}^{2}=\sum\limits_{t\in S}b_{t}^{2}+\sum\limits_{t\in S^{\prime}}b_{t}^{2}\leq\sum\limits_{t\in S}b_{t}^{2}+\sum\limits_{t\in S^{\prime}}b_{t}<\sum\limits_{t\in S}b_{t}^{2}+\sum\limits_{t\in D}b_{t}<\infty$ which implies that $f\in L^{2}(G).$ ∎

For the strict inclusion, consider $G=\mathbb{R}\times\mathbb{Z}$ and

[TABLE]

Clearly $f\in L^{2}(G)\text{ but }f\notin W(L^{2},l^{1})(G).$

We will say $G\text{ has }(QUP)^{\prime}$ for Gabor transform if $f,\psi(\neq 0)\in W(L^{2},l^{1})(G)$ such that

[TABLE]

then $f=0$ a.e.

It may be noted that theorem 2.1 remain valid for $(QUP)^{\prime}.$

Theorem 2.4**.**

If $H$ has QUP for Gabor transform then $G$ has $(\text{QUP})^{\prime}$ for Gabor transform.

Proof.

First of all, we assume $D$ is abelian. Let $f,\psi(\neq 0)\in W(L^{2},l^{1})(G)$ be such that

[TABLE]

By Minkowski inequality, we obtain

[TABLE]

So, for almost every $x\in H,\sum\limits_{t\in D}|\psi(x,t)|<\infty.\text{ For }\gamma\in\widehat{D},\text{ define }\tilde{\psi}(y)=\sum\limits_{t\in D}\psi(y,t)\gamma(t),\text{for all }y\in H.$ Again by Minkowski inequality, it follows that $\tilde{\psi}\in L^{2}(G)$. Since $\psi\neq 0$, therefore there exist $\gamma\in\widehat{D}$ such that $\tilde{\psi}\neq 0.$ For such $\gamma\in\widehat{D}$, define $\phi(x,t)=\psi(x,t)\gamma(t),\text{ for }(x,t)\in H\times D$. Then $\phi\in W(L^{2},l^{1})(G)$ and

[TABLE]

Thus, we have

[TABLE]

which implies that

[TABLE]

Therefore there exist a zero set $K\text{ in }\widehat{D}\text{ such that for all }\delta\in\widehat{D}\setminus K,$

[TABLE]

Now for each $\delta\in\widehat{D}\setminus K$, define $\tilde{f}(x)=\sum_{t\in D}f(x,t)\delta(t)$, for all $x\in$ H. Then $\tilde{f}\in L^{2}(H)$ also

[TABLE]

If $G_{\tilde{\psi}}\tilde{f}(x,\pi)\neq 0\text{ then }G_{\phi}f(x,t,\pi,\delta)\neq 0\text{ for all }t\in M\subset D,\text{ where }|M|\geq 1$ and

[TABLE]

Since $H$ has QUP for Gabor Transform, it follows that $\tilde{f}(x)=\sum\limits_{t\in D}f(x,t)\delta(t)=0\text{ a.e. for all }\delta\in\widehat{D}\setminus K$ which implies that $f=0$ a.e.

Now if $D$ is discrete type-I, then $D$ has an abelian subgroup $A$ of finite index. So, $H\times A$ has $(QUP)^{\prime}$ for Gabor transform and $H\times A$ has finite index in $H\times D$. So, using Theorem 2.1 $H\times D$ has $(QUP)^{\prime}.$ ∎

Remark 2.5**.**

$(i)$* If $G=H\times D$ where $D$ is discrete group such that $G$ has QUP for Gabor transform then so does $H$. Consider $f\in L^{2}(H),\text{ and }\psi$ be a window function such that*

[TABLE]

Define $g(x,t)=f(x)\chi_{\{e\}}(t)\text{ and }\phi(x,t)=\psi(x)\chi_{\{e\}}(t)$. Then $g,\phi\in L^{2}(G)$ and

[TABLE]

Therefore,

[TABLE]

which implies that $g=0$ a.e. and hence $f=0$ a.e.

$(ii)$* If $G=H\times D,\text{ where }D$ is a finite group then by Theorem 2.4 one can see if $H$ has QUP for Gabor transform then so does $G.$*

Theorem 2.6**.**

Let $G$ be a noncompact, nondiscrete, unimodular type-I group. If $G$ has a compact open subgroup $H$, Then QUP for Gabor transform does not hold for $G.$

Proof.

Let $\alpha=m_{G}(H),\text{ then }0<\alpha<\infty\text{ and }m_{H}=\alpha^{-1}(m_{G}|_{H})$ is a Haar measure for $H$. Also $H$ is non-discrete. So, $H$ is non-trivial and $A(\widehat{G},H)\subset\widehat{G}.$ Define $f=\chi_{H}\text{ and }\psi=\chi_{H}$. Then $f$ and  $\psi$ are non-zero function of $L^{2}(G).$ If $\pi\in A(G,\widehat{H})$ then $G_{\psi}f(x,\pi)=\chi_{H}(x)\alpha I.$

Let $\pi\notin A(\widehat{G},H),a\in H\text{ and }\zeta,\eta\in\mathcal{H}_{\pi}$ it follows,

[TABLE]

which means, $G_{\psi}f(x,\pi)(I-\pi(a)^{*})\zeta=0\text{for all }\zeta\in\mathcal{H}_{\pi},a\in H.$ Therefore $G_{\psi}f(x,\pi)\xi=0\text{ for all }\xi\in V$ where $V$ is the smallest closed set containing $\bigcup\limits_{a\in H}(I-\pi(a)^{*})(\mathcal{H}_{\pi}),a\in H.$

Now for all $g\in G,a\in H,\xi\in\mathcal{H}_{\pi}$, We have

[TABLE]

which implies that $V$ is a closed invariant subspace, as $\pi$ is irreducible so $V=\mathcal{H}_{\pi}.$Thus it follows

[TABLE]

Consequently, using ([hogan], Theorem 2.4)we have

[TABLE]

∎

Remark 2.7**.**

A locally compact group is said to be Plancherel if dual object $\widehat{G}$ can be equipped with a measure $\mu_{G}\ ($the Plancherel measure$)$ such that

[TABLE]

Theorem 2.6 is true for noncompact, nondiscrete Plancherel group. Maurtner Groups [8] are example of Plancherel group, unimodular not type I.

Remark 2.8**.**

Clearly, QUP for Gabor transform does not hold for groups of the type $D\times K\text{ where }D$ is discrete Maurtner group, $K$ is compact group and Moore group with compact component of identity.

3. Nilpotent Group

For a locally compact unimodular group G of type I having center $Z$, suppose that there exist a zero set $E$ in $\widehat{Z}$ such that for every $\lambda\in\widehat{Z}\setminus E,$ the induced representation ind${}_{Z}^{G}\lambda$ is a multiple of an irreducible $\pi_{\lambda}$. Then according the Plancherel formula ([4], Theorem 8.1), the Plancherel measure on $\widehat{G}$ is given by

[TABLE]

where $W\subseteq\widehat{G}\text{ and }q$ is a positive measurable function.

Proposition 3.1**.**

In the above situation, suppose that $Z$ has the following property:

If $f\in L^{2}(Z)\text{ and }\psi$ is a window function in $L^{2}(Z)$ satisfies

[TABLE]

implies $f=0$ a.e. Then QUP for Gabor transform hold for $G$.

Proof.

Let $g\in L^{2}(G)\text{ and }\phi$ be a window function such that,

[TABLE]

By Weil’s formula, it follows

[TABLE]

Now there exist a zero set $K$ in $G$ such that for all $x\in G\setminus K,$

[TABLE]

Also there exist a zero set $M$ such that for all $y\in G\setminus M,\text{ we have }(\phi_{y}|_{Z})\text{ and }(g_{y}|_{Z})\in L^{2}(Z)$ where $(g_{y}|_{Z})(h)=g(hy),\text{for all }h\in Z.$ Fix $y\in G\setminus M$ and define $k(x)=\phi(x^{-1}y),$ for all $x\in$ G. Then $k\in L^{2}(G).$ Again by Weil’s formula, there exist a zero set $N_{y}$ such that for all $x\text{ in }G\setminus N_{y}$,

[TABLE]

Since $\phi\neq 0$, Therefore we can choose $x\in G\setminus(K\cup N_{y})$ such that $0\neq({}_{x}\phi)_{y}|_{Z}\in L^{2}(Z),\text{ where }({}_{x}\phi)_{y}|_{Z}(v)=\phi(x^{-1}vy).$ Fix such $x\in G\setminus(K\cup N_{y})$ and for $h\in Z,\text{ define }g_{hx}^{\phi}(z)=g(z)\overline{\phi(x^{-1}h^{-1}z)}.$ Also $(g_{hx}^{\phi})_{y}|_{Z}=(g_{y}|_{Z})_{h}^{({}_{x}\phi)_{y}|_{Z}}.$

Now using the Plancherel formula we have,

[TABLE]

On integration it follows,

[TABLE]

which implies that $g_{y}|_{Z}=0$ a.e. for all $y\in G\setminus M$. Since $y$ was arbitrary fixed, Therefore, $g=0$ a.e. ∎

For $f\in L^{1}(G),$ let us define

[TABLE]

The idea of the proof of following lemma emerges from [13] and [11].

Lemma 3.2**.**

Let $f\in L^{1}(\mathbb{R}^{n}),n\geq 2\text{ be such that for some }\delta\in\mathbb{R}^{n},$

[TABLE]

where $p(x)=p(x_{1},\dots x_{n})=x_{1},{x_{1}}^{2},x_{1}x_{2},{x_{1}}^{2}+{x_{2}}^{2}.$ Then $f=0$ a.e.

Proof.

Replacing $f$ by a suitable dilate we can assume that

[TABLE]

where $\mathbb{T}^{n}=\mathbb{R}^{n}/\mathbb{Z}^{n}\text{ and }d\tilde{\mu}_{\mathbb{T}^{2}}(y)=p_{\delta}(y)dm_{\mathbb{T}^{2}}(y),$

$p_{\delta}(y)=\text{min}\{y_{1},(1-y_{1}),(y_{1}+\lceil{\delta_{1}}\rceil-\delta_{1})\},\text{ when }p(x)=x_{1},\\ p_{\delta}(y)=\text{min}\{y_{1}^{2},(1-y_{1})^{2},(y_{1}+\lceil{\delta_{1}}\rceil-\delta_{1})^{2}\},\text{ when }p(x)={x_{1}}^{2},$

[TABLE]

[TABLE]

$\lceil.\rceil$ being the greatest integer function. Now define $\phi:\mathbb{R}^{n}\rightarrow\mathbb{R}\cup\{\infty\}$ by

[TABLE]

and let $K=\{\xi\in\mathbb{R}^{n}:0<\xi_{i}<1,i=1,2\dots n\}.$

Then $\int\limits_{K}\phi(\xi)d\xi=\int\limits_{\mathbb{R}^{n}}\chi_{B_{f}}(\xi)|p(\xi)|d\xi<\infty.$ Therefore, $\phi(\xi)<\infty$ a.e.

Now for almost all $\xi=(\xi_{1},\dots\xi_{n})\in K$ and $m\in\mathbb{Z}^{n}$,

[TABLE]

and $p_{\delta}(\xi_{1},\xi_{2})\sum\limits_{m\in\mathbb{Z}^{n}}\chi_{B_{f}}(\xi+m)\leq\phi(\xi)<\infty.$

Consequently, for almost all $\xi\in\mathbb{R}^{n},\chi_{B_{f}}(\xi+m)\neq 0$ only for finitely many $m\in\mathbb{Z}^{n}.$ Fix $\xi\in\mathbb{R}^{n}$ and define $\tilde{f}_{\xi}(x)=\sum\limits_{m\in Z^{n}}e^{-\iota\langle\xi,x-m\rangle}f(x-m).$ Then $\tilde{f}_{\xi}\in L^{1}(\mathbb{T}^{n}).$ Also, $\widehat{(\tilde{f}_{\xi})}(k)=\widehat{f}(\xi+k).$

Now,

[TABLE]

Thus $\tilde{\mu}_{\mathbb{T}^{2}}\times m_{\mathbb{T}^{n-2}}\{x:\tilde{f_{\xi}}(x)=0\}>0$ which implies that $m_{\mathbb{T}^{n}}\{x:\tilde{f_{\xi}}=0\}>0.$ But for almost all $\xi\in\mathbb{R},\tilde{f}_{\xi}$ is a trigonometric polynomial. Therefore $\widehat{f}=0$ a.e. So, by Fourier inversion theorem, $f=0$ a.e. ∎

Lemma 3.3**.**

Let $f\in L^{2}(\mathbb{R}^{n})\text{ and }\psi$ be a window function such that

[TABLE]

where $p(\xi)=p(\xi_{1},\dots\xi_{n})=\xi_{1},\xi^{2},\xi_{1}\xi_{2},{\xi_{1}}^{2}+{\xi_{2}}^{2}$, then $f=0$ a.e.

Proof.

For each $(y,\delta)\in\mathbb{R}^{n}\times\mathbb{R}^{n},$ define

[TABLE]

where $T_{y},M_{\delta}$ are translation and modulation operator respectively. Then it is easy to prove that $\widehat{F_{(y,\delta)}}(\xi,z)=F_{(y,\delta)}(-z,\xi).$

Moreover $F_{(y,\delta)}$ is continuous and

[TABLE]

So it follows, $\chi_{A_{F}}(x,\xi)\leq\chi_{\{(x,\xi):G_{\psi}(M_{\delta}T_{y}f)(x,\xi)\neq 0\}}(x-y,\xi-\delta)$ and hence,

[TABLE]

Moreover, $\int\limits_{\mathbb{R}^{2n}}\chi_{A_{F}}(x,\xi)|p(\xi-\delta)|dxd\xi=\int\limits_{\mathbb{R}^{2n}}\chi_{B_{F}}(\xi,x)|p(\xi-\delta)|dxd\xi<\infty.$Now using lemma 3.2  $F_{(y,\delta)}=0$ for every $(y,\delta)\in\mathbb{R}^{2}.$

So, $F_{(y,\delta)}(0,0)=(G_{\psi}f(-y,-\delta))^{2}=0$ for every $(y,\delta)\in\mathbb{R}^{2}$ which implies that $f=0$ a.e. ∎

For a simply connected nilpotent Lie group G with Lie algebra $\mathcal{G},\widehat{G}$ can be parametreized by the set of coadjoint orbits of $G$ in the vector space dual $\mathcal{G}^{*}\text{ of }\mathcal{G}$[1]. Let $f\in\mathcal{G}^{*}$, and $\pi_{f}$ denote the irreducible representation associated to $f$ and by $O_{f}$ the coadjoint orbit of $f$. Moreover, let Z and $\mathcal{Z}$ denote the center of $G\text{ and }\mathcal{G},$respectively. Applying Proposition 3.1 to $G$, we obtain QUP for G if,

(i) For allmost all $f\in\mathcal{G}^{*},\pi_{f}\sim\text{ind}_{Z}^{G}(\pi_{f}|_{Z}).$

(ii) dim $\mathcal{Z}\leq 2,q(\lambda)\text{ of Proposition }\ref{prop1}\text{ is of the form }q(\lambda)=|p(\lambda)|,$ where $p$ is homogeneous polynomial of degree $\leq 2.$

Now, from the data presented in [12] for low dimensional groups and using lemma lemma 3.3the following groups have QUP for Gabor transform: $G_{3},G_{5,1},G_{5,3},G_{5,6},$ for $n=1$ and $G_{6,16},G_{6,17},G_{6,19},G_{6,20},G_{6,21},G_{6,22},G_{6,23},G_{6,24}$ for $n=2$ have QUP for Gabor transform.

4. Moore Group

A locally compact group $G$ is said to be Moore if dim $\pi<\infty$ for all $\pi\in\widehat{G}.$ This class contain finite extension of abelian groups and compact groups. Let $G_{F}$ denote the subgroup consisting of all elements of G with relatively compact conjugacy classes. It is an open normal subgroup of finite index.

A locally compact $G$ has weak QUP for Gabor Transform if $f\in L^{2}(G),\psi$ be a window function satisfying

[TABLE]

Proposition 4.1**.**

Let $G$ be a locally compact Plancherel Group and $K$ is a compact normal subgroup of $G$. If $G$ satisfies QUP $($the weak QUP $)$ of Gabor transform then $G/K$ also satisfy QUP $(\text{weak QUP})$ of Gabor transform.

Proof.

Let $\dot{f}\in L^{2}(G/K)$ and $\dot{\psi}$ be a window function such that

[TABLE]

Define $f(x)=\dot{f}(q(x))\text{ and }\psi(x)=\dot{\psi}(q(x)),$ where $q$ is the quotient map.

For $\xi,\eta\in\mathcal{H}_{\pi}$ consider,

[TABLE]

So, it follows that,

[TABLE]

which means that $G_{\psi}f(x,\pi)$ is either zero a.e. on cosets of $K$ or non zero a.e. Now consider,

[TABLE]

So $f=0$ a.e. Hence $\dot{f}=0$ a.e. ∎

Remark 4.2**.**

A discrete Moore group $G$, which satisfy weak QUP for Gabor transform, is abelian. By considering $f=\chi_{\{e\}},\psi=\chi_{\{e\}},$ we see that

[TABLE]

So we have $m\times\mu\{(x,\pi):G_{\psi}f(x,\pi)\neq 0\}=\mu(\widehat{G})\text{ which mean }\mu(\widehat{G})\geq 1\text{ as }f\neq 0$, $\psi\neq 0.$ Also $\mu(\widehat{G})\leq[G:G_{F}]^{-1}.$ Therefore $\mu(\widehat{G})=1$ and $[G:G_{F}]=1.$ Now as proved in $($[6], Lemma 3.2$)$, $G$ is abelian.

Proposition 4.3**.**

Let $G$ be a Moore group with compact identity component $G_{0}$. If $G$ satisfy weak QUP for Gabor transform then $G/G_{0}$ is abelian.

Proof.

By structure Theorem [5], $G/G_{0}$ is projective limit of discrete groups. Also $G_{0}$ is compact so there exist compact open normal subgroup $H_{\alpha}$ in G such that $\cap H_{\alpha}=G_{0}$. Now for each $\alpha$ the weak QUP hold for $G/H_{\alpha}$ and hence $G/H_{\alpha}$ is abelian, which implies the commutator $[G,G]\subset\cap H_{\alpha}=G_{0}.$ ∎

Proposition 4.4**.**

Let $G$ be a Lie Moore group with compact component of identity $G_{0}$ such that $G/G_{0}$ is abelian. If $0\neq f\in L^{1}(G)\cap L^{2}(G)\text{ and }\psi$ be nonzero square integrable which is constant on cosets of $G_{0}$. Then

[TABLE]

Proof.

Define $\tilde{f}(\dot{x})=\int\limits_{G_{0}}f(xk)dk\text{ on }G/G_{0}.$ Now $\tilde{f}\in L^{1}(G/G_{0}).\text{ Since }G_{0}$ is open, therefore $\widehat{G/G_{0}}$ is compact and $\mu({\widehat{G/G_{0}}})<\infty.$ Moreover $G/G_{0}$ is abelian, therefore $\tilde{f}\in L^{2}(G/G_{0}).$ Now consider,

[TABLE]

Thus we have,

[TABLE]

Since $G/G_{0}$ is abelian and $\tilde{f},\psi$ are nonzero square integrable functions. Hence it follows that

[TABLE]

Lemma 4.5**.**

Let G be a Moore group and $f,\psi\in L^{2}(G_{F})$.

$(1)$* If g,$\phi\in L^{2}(G)$ is such that $g|_{G_{F}}=f$ and $\phi|_{G_{F}}=\psi$, then*

[TABLE]

$(2)$Let $g=\tilde{f},\phi=\tilde{\psi},$ are the trivial extension of $f,\psi$ to all of G. Then

[TABLE]

Proof.

$(1)$ We can assume that $m_{G}\times\mu_{G}\{(x,\pi):G_{\phi}g(x,\pi)\neq 0\}<\infty$, For $x\in G_{F},\ g_{\phi}^{x}|_{G_{F}}=f_{\psi}^{x}$ then by using ([7],lemma2.2) we get that

[TABLE]

i.e.$\int\limits_{\widehat{G_{F}}}\chi_{\{\pi:G_{\psi}f(x,\pi)\neq 0\}}(\pi)d\pi\leq[G:G_{F}]\int\limits_{\widehat{G}}\chi_{\{\pi:G_{\phi}g(x,\pi)\neq 0\}}(\pi)d\pi,$

On integrating both sides we get,

[TABLE]

or equivalently, $m_{G_{F}}\times\mu_{\widehat{G_{F}}}\{(x,\pi):G_{\psi}f(x,\pi)\neq 0\}\leq[G:G_{F}]m_{G}\times\mu_{\widehat{G}}\{(x,\pi):G_{\phi}g(x,\pi)\neq 0\}$

$(2)$ For almost every $x\in G_{F},\ g_{\phi}^{x}$ is trivial extension of $f_{\phi}^{x}.$

Then by ([7],Lemma 3.7)it follows that

[TABLE]

Thus

[TABLE]

As $G_{F}$ is a subgroup of finite index, so QUP of Gabor transform holds for G if and only if it holds for $G_{F}$, but it may not be true for weak QUP of Gabor transform. Every nonabelian discrete group for which $G_{F}$ is abelian will serve our purpose $e.g.(Z\rtimes\{1,-1\})$. But if $G$ has weak QUP of Gabor transform then (ii) of Lemma 4.5 we can conclude that $G_{F}$ has weak QUP of Gabor transform. Moreover if $G_{F}$ has weak QUP of Gabor transform then we at least have the following,

[TABLE]

Next we consider bounded representation dimension group.Define d(G) = sup{dim $\pi:\pi\in\widehat{G}\}.$

Lemma 4.6**.**

Let $G$ be a group with bounded representation dimension. Then

[TABLE]

Proof.

We can assume that $m\times\mu\{(x,\pi)\in G\times\widehat{G}:G_{\psi}f(x,\pi)\neq 0\}<\infty.$ For almost every $(x,\pi)\in G\times\widehat{G},G_{\psi}f(x,\pi)=f_{\psi}^{x}(\pi)$ and $G_{\psi}f(x,\pi)^{*}=f_{\psi}^{x}(\pi)^{*}.$ Now $\|G_{\psi}f(x,\pi)\|^{2}_{HS}=tr[G_{\psi}f(x,\pi)^{*}G_{\psi}f(x,\pi)]=tr[\pi({f_{\psi}^{x}})^{*}\pi(f_{\psi}^{x})]$ for almost every $(x,\pi)\in G\times\widehat{G}.$

Let $\{\xi_{\pi,j}1\leq j\leq d_{\pi}\}$ be an orthonormal basis of $\mathcal{H(\pi)}.$ Then,

[TABLE]

Consequently, $tr(\pi(f_{\psi}^{x})^{*}\pi(f_{\psi}^{x}))\leq d_{G}\|f\|_{2}^{2}\|\psi\|_{2}^{2}.$ Now

[TABLE]

or equivalently $\|f\|^{2}\|\psi\|^{2}\leq m\times\mu\{(x,\pi):G_{\psi}f(x,\pi)\neq 0\}d_{G}\|f\|_{2}^{2}\|\psi\|_{2}^{2},$ Hence $\frac{1}{d(G)}\leq m\times\mu\{(x,\pi):G_{\psi}f(x,\pi)\neq 0\}.$ ∎

Remark 4.7**.**

$(1)$* Weak QUP for Gabor transform always hold for abelian groups as $M=1$.*

$(2)$* A discrete moore group satisfy weak QUP of Gabor Transform if and only if it is abelian.*

If $G$ is a group of bounded representation dimension with connected component of identity non compact then $m\times\mu\{(x,\pi):G_{\psi}f(x,\pi)\neq 0\}=\infty$ for every $0\neq f,\psi\in L^{2}(G)$. So, the functions for which equality hold in lemma 4.6 can only exist when $G_{0}$ is compact.

Let $N$ be a closed normal subgroup of $G$. Then a $G-$invariant character $\gamma$ of $N$ is a continuous homomorphism from $N$ to the circle Group $\mathbb{T}$ satisfying $\gamma(y^{-1}xy)=\gamma(x)$ for all $x\in N$ and $y\in G$.

\widehat{G}_{\gamma}=\{\pi\in\widehat{G}:\pi(x)=\gamma(x)I_{\mathbb{H_{\pi}}}\text{ for all x\in N}\}

Lemma 4.8**.**

Let $G$ be a locally compact group of bounded representation dimension. If there exist a compact open normal subgroup $N$ of $G$ and a $G-$invariant character $\gamma$ of $N$ such that $d_{\pi}$ = d$($G$)$ for all most all $\pi\in\widehat{G}_{\chi}.$ Then there exist non-zero function $f,\psi\in L^{2}(G)$ satisfying,

[TABLE]

Proof.

Define $f=\overline{\gamma(x)}\chi_{N}(x)$ and $\psi(x)=\chi_{N}(x).$ Then

[TABLE]

Thus, we have $\{(x,\pi):G_{\psi}f(x,\pi)\neq 0\}=N\times\widehat{G_{\chi}}.$ Thus by $($[7], Lemma 1.2$)$ it follows,

[TABLE]

For existence of such groups see [7].

Now every pair $(f,\psi)$ of nonzero functions in $L^{2}(G)$ such that $m\times\mu\{(x,\pi)\in G\times\widehat{G}:G_{\psi}f(x,\pi)\neq 0\}=\frac{1}{d(G)}$ is defined to be minimizing function.

Proposition 4.9**.**

Let $K$ be a compact normal subgroup of $G$ such that d$(\text{G/K})=d(G)$, and suppose that there exists minimizing functions for $G/K$. Then there exist minimizing functions for $G$.

Proof.

Let Haar measures on $G,K\text{ and }G/K$ be normalised so that Weil’s formula hold and $m_{K}(K)=1$.Then $\mu_{G/K}$ equals the measure induced from $\mu_{G}$ on $\widehat{G/K}\subset\widehat{G}.$ Now Let $g,\phi\in L^{2}(G/K)$ satisfy,

$m_{G/K}\times\mu_{G/K}{\{(\dot{x},\dot{\pi}):G_{\phi}g(\dot{x},\dot{\pi})\neq 0\}}$

and define $f,\psi\in L^{2}(G)$ by $f(x)=g(\dot{x}),\psi(x)=\phi(\dot{x}).$ Then by Proposition 4.1 we have,

[TABLE]

∎

5. Multipliers Extension of $\mathbb{T}$

For a locally compact abelian group $G$ and normalised multiplier $\omega$ on $G$, $h_{\omega}:G\to\widehat{G}$ is a continuous homomorphism where, $h_{\omega}(x)(y)=\omega(x,y)\omega(x,y)^{-1}.$ for every $x,y\in G.$ Also, $S_{\omega}=\{x\in G:h_{\omega}(x,y)=1,\text{ for all }y\in G\}$ is a closed subgroup of $G$ and $h_{\omega}(G)\text{ is a dense subgroup of }\widehat{G/S_{\omega}}\subset\widehat{G}$. For a normalised multiplier $\omega\text{ on }G,G(\omega)$ denote the central extension of $G$ defined by $\omega,i.e.\ G(\omega)=G\times\mathbb{T}$ with Weil topology and multiplication $(x,s)(y,t)=(xy,stw(x,y))$ for $x,y\in G$ and $s,t\in\mathbb{T}$. $G(\omega)$ is a locally compact group in which $\mathbb{T}$ is a central subgroup and $G(\omega)/\mathbb{T}\text{\ is isomorphic to G}$. . The center of $G(\omega)\ \text{is }S_{\omega}\times\mathbb{T}$.It is shown in $\omega$ is type 1 if and only if $h_{\omega}$ is an open homeomorphism of G onto $\widehat{G/S_{\omega}}$.

Proposition 5.1**.**

Let $\omega$ be a type-I multiplier on an arbitrary abelian group $G$ and QUP for Gabor transform holds for $G(\omega)$ then, $(S_{\omega})_{0}$ is non-compact.

Proof.

Let us assume that $(S_{\omega})_{0}$ is compact, which means that $S_{\omega}$ contains a compact open subgroup $K$ and let $H=K\times\mathbb{T}$. So $H$ is compact open subgroup of $Z=S_{\omega}\times\mathbb{T}$, the center of G($\omega$). Define $\chi\in\widehat{H}$ by$\chi(x,t)=t,x\in K,t\in\mathbb{T},\widehat{Z_{\chi}}=\{\alpha\in\widehat{Z}:\alpha|_{H}=\chi\}\text{ and }\widehat{G(\omega)_{\chi}}=\{\alpha\in\widehat{G(\omega)}:\alpha|H\text{ is a multiple of }\chi\}.$ For every $\alpha\in\widehat{Z_{\chi}}\text{ there is a unique }\pi_{\alpha}\in\widehat{G(\omega)}$such that $\pi_{\alpha}|Z\text{ is a multiple of }\alpha\text{ and }\alpha\rightarrow\pi_{\alpha}$ is a homeomorphism between $\widehat{Z_{\chi}}\text{ and }\widehat{G(\omega)}_{\chi}$[10].

Since $H$ is open in $Z$ it follows that $\widehat{G(\omega)}_{\chi}$ is compact and hence has finite Plancherel mesure. Define $f,\psi:G(\omega)\rightarrow\mathbb{C}\text{ where }f(x,t)=t\chi_{{K}}(x)$ and $\psi(x,t)=\chi_{K}(x)$, then $f,\psi\in L^{2}(G{(\omega}))$ and we have

[TABLE]

If $\ x\notin K\text{ then for all }y\in K,x^{-1}y\notin K.$ Also if $\pi\in\widehat{G(\omega)}/\widehat{G(\omega)_{\chi}}\text{ then }\pi|_{K\times\mathbb{T}}$ is not a multiple of $\chi.$ So it follows that,

[TABLE]

Thus,

[TABLE]

which is a contradiction to the hypothesis. ∎

Proposition 5.2**.**

Let $G$ be a locally compact, unimodular group of type-I and $\omega$ is type-I multiplier on $G$. If $G(\omega)$ has QUP for Gabor transform then $(G,\widehat{G}^{\omega})$ has QUP for Gabor transform where $\widehat{G}^{\omega}$ is the set of equivalence classes of irreducible $\omega$ representation.

Proof.

Let $f\in L^{2}(G)\text{ and}\ 0\neq\psi\in L^{2}(G)$ be a window function such that

[TABLE]

Extend $f\text{ and }\psi$ on $G(\omega)$ as $F(x,t)=tf(x)\text{ and }\Psi(x,t)=\psi(x)$ and we have

[TABLE]

Now if $\tau\in\widehat{G}^{\omega}$ then we can define $\rho_{\tau}\in\widehat{G(\omega)}$ by $\rho_{\tau}(x,t)=t\tau(x),x\in G,t\in\mathbb{T}$

Then $\tau\rightarrow\rho_{\tau}$ is a homeomorphism between $\widehat{G}^{\omega}$ and

[TABLE]

and this maps the projective Plancherel measure on $\widehat{G}^{\omega}$ onto the restriction to $\widehat{G(\omega)}_{1}$ of the Plancherel measure on $\widehat{G(\omega)}$. Now if $\pi\in\widehat{G(\omega)}_{1}$, then $\pi=\pi_{\tau}$ for some $\tau\in\widehat{G}^{\omega}$ and $\pi_{\tau}(y,s)=s\tau(y).$

Then(6) becomes

[TABLE]

and if $\pi\notin\widehat{G(\omega)}_{1}\text{ then }\pi(x,t)=t^{k}\pi(x,1)\text{ for some }k\in\mathbb{Z},k\ \neq 1$ and hence (6) becomes

[TABLE]

So, we have

[TABLE]

which implies that $m\times\mu\{(x,t,\pi):G_{\Psi}F(x,t,\pi)\neq 0\}<\infty.$ So $F=0$ a.e. and it follows that $f=0$ a.e. ∎

Remark 5.3**.**

If we take $G=\mathbb{R}^{2},\text{ and }\omega((x_{1},y_{1}),(x_{2},y_{2}))=e^{-\iota(y_{2}x_{1}-x_{2}y_{1})/2},$ Then $\mathbb{R}^{2}(\omega)$ is isomorphic to three dimensional nilpotent Lie group [4], which is Heisenberg group. Hence, $(G,G^{\omega})$ has QUP for Gabor transform.

acknowledgements

The first author is supported by the Junior Research Fellowship of Council of Scientific and Industrial Research, India (Ref. No:21/12/2014(ii)EU-V).