Pseudo-differential Operators on Non-isotropic Heisenberg Groups with Multi-dimensional Centers

Abstract

We introduce non-isotropic Heisenberg groups with multi-dimensional centers and the corresponding Schrödinger representations. The Wigner and Weyl transforms are then defined. We prove the Stone-von Neumann theorem for the non-isotropic Heisenebrg group by means of Stone-von Neumann theorem for the ordinary Heisenebrg group. Using this theorem, the Fourier transform is defined in terms of these representations and the Fourier inversion formula is given. Pseudo-differential operators with operator-valued symbols are introduced and can be thought of as non-commutative quantization. We give necessary and sufficient conditions on the symbols for which these operators are in the Hilbert-Schmidt class. We also give a characterization of trace class pseudo-differential operators and a trace formula for these trace class operators.

1 Introduction

The Heisenberg group is the simplest non-commutative nilpotent Lie group. It is actually the first locally compact group whose infinite-dimensional, irreducible representations were classified. Harmonic analysis on the Heisenberg group is a subject of constant interest in various areas of mathematics, from Partial Differential Equations to Geometry and Number Theory.

We fix the vector (a ₁, a ₂, ⋯ , a _n ) in $\mathbb{R}^{n}$. The non-isotropic Heisenberg group on $\mathbb{R}^{n} \times \mathbb{R}^{n} \times \mathbb{R}$ is defined by the group law

$$\displaystyle{(z,t) \cdot (z',t') = \left (z + z',t + t' + \frac{1} {2}\sum _{j=1}^{n}a_{ j}(x_{j}y_{j}^{{\prime}}- x_{ j}^{{\prime}}y_{ j})\right ),}$$

for all z = (x, y), z′ = (x′, y′) in $\mathbb{R}^{n} \times \mathbb{R}^{n}$ and t, t′ are in $\mathbb{R}$. If we let a _j  = 1, for all 1 ≤ j ≤ n, then we get the ordinary Heisenberg group $\mathbb{H}^{n}$ see [4]. The center of the non-isotropic Heisenberg group $\mathbb{H}^{n}$ is the 1-dimensional subgroup Z given by

$$\displaystyle{Z = \left \{(0,0,t) \in \mathbb{R}^{n} \times \mathbb{R}^{n} \times \mathbb{R} : t \in \mathbb{R}\right \}.}$$

In the non-isotropic Heisenberg group the terms x _k y _l ^′ for l ≠ k, do not appear in the group law. In other words we do not consider these directions in the group law. We want to generalize this group to a group that has changes in other directions as well. Moreover, we want to look at a group with a multi-dimensional center which is of interest in Geometry. To do this, we consider n × n orthogonal matrices B ₁, B ₂, …, B _m such that

$$\displaystyle{ B_{j}^{-1}B_{ k} = -B_{k}^{-1}B_{ j},\quad j\not =k. }$$

(1)

Example 1.1.

Let m = 2, then

$B_{1} = \left [\begin{array}{cc} 1& 0\\ 0 & -1 \end{array} \right ]$ and $B_{2} = \left [\begin{array}{cc} 0&1\\ 1 &0 \end{array} \right ]$ satisfy the above conditions.

Then we define the non-isotropic Heisenberg group with multi-dimensional center $\mathbb{G}$ on $\mathbb{R}^{n} \times \mathbb{R}^{n} \times \mathbb{R}^{m}$ by

$$\displaystyle{(z,t) \cdot (z',t') = \left (z + z',t + t' + \frac{1} {2}[z,z']\right ),}$$

for (z, t) and (z′, t′) in $\mathbb{R}^{n} \times \mathbb{R}^{n} \times \mathbb{R}^{m}$ where z = (x, y), z′ = (x′, y′) in $\mathbb{R}^{n} \times \mathbb{R}^{n}$, $t,t' \in \mathbb{R}^{m}$ and $[z,z'] \in \mathbb{R}^{m}$ is defined by

$$\displaystyle{[z,z']_{j} = x' \cdot B_{j}y - x \cdot B_{j}y',\quad j = 1,2,\ldots ,m.}$$

The center of the non-isotropic Heisenberg group with multi-dimensional center is of dimension m and of the form (0, 0, t), $t \in \mathbb{R}^{m}$. To see this, we denote the center of $\mathbb{G}$ by $C(\mathbb{G})$. Let (z ₀, t ₀) be in $C(\mathbb{G})$, then for all $(z,t) \in \mathbb{G}$

$$\displaystyle{(z,t) \cdot (z_{0},t_{0}) = (z_{0},t_{0}) \cdot (z,t).}$$

Hence, [z, z ₀] = 0. Therefore, for all $x,y \in \mathbb{R}^{n}$

$$\displaystyle{x_{0}B_{j}y - xB_{j}y_{0} = 0,\quad 1 \leq j \leq n.}$$

In particular for x = x ₀, and for all $y \in \mathbb{R}^{n}$

$$\displaystyle{\left (x_{0},B_{j}(y - y_{0})\right ) = 0.}$$

So, B _j ⁻¹ x ₀ = 0, which implies x ₀ = 0. Similarly we get y ₀ = 0.

In fact, $\mathbb{G}$ is a unimodular Lie group on which the Haar measure is just the ordinary Lebesgue measure dzdt. Moreover, this is a special case of the Heisenberg type group. The Heisenberg type group first was introduced by A. Kaplan [6]. The geometric properties of the H-type group is studied in e.g. [7].

Note that if we let m = 1 and B ₁ = −I _n where I _n is the n × n identity matrix. Then we get the ordinary Heisenberg group $\mathbb{H}^{n}$.

It is well-known from [9, 10, 13] that Weyl transforms have intimate connections with analysis on the Heisenberg group and with the so-called twisted Laplacian studied in, e.g., [1, 11, 12]. We begin with a recall of the basic definitions and properties of Weyl transforms and Wigner transforms in, for instance, the book [13]. Let $\sigma \in L^{2}(\mathbb{R}^{n} \times \mathbb{R}^{n})$. Then the Weyl transform $W_{\sigma } : L^{2}(\mathbb{R}^{n}) \rightarrow L^{2}(\mathbb{R}^{n})$ is defined by

$$\displaystyle{(W_{\sigma }f,g)_{L^{2}(\mathbb{R}^{n})} = (2\pi )^{-n/2}\int _{ \mathbb{R}^{n}}\int _{\mathbb{R}^{n}}\sigma (x,\xi )W(\,f,g)(x,\xi )\,dx\,d\xi ,\quad f,g \in L^{2}(\mathbb{R}^{n}),}$$

where W( f, g) is the Wigner transform of f and g defined by

$$\displaystyle{W(\,f,g)(x,\xi ) = (2\pi )^{-n/2}\int _{ \mathbb{R}^{n}}e^{-i\xi \cdot p}f\left (x + \frac{p} {2}\right )\overline{g\left (x -\frac{p} {2}\right )}dp,\quad x,\xi \in \mathbb{R}^{n}.}$$

Closely related to the Wigner transform W( f, g) of f and g in $L^{2}(\mathbb{R}^{n})$ is the Fourier–Wigner transform V ( f, g) given by

$$\displaystyle{V (\,f,g)(q,p) = (2\pi )^{-n/2}\int _{ \mathbb{R}^{n}}e^{iq\cdot y}f\left (y + \frac{p} {2}\right )\overline{g\left (y -\frac{p} {2}\right )}dy,\quad q,p \in \mathbb{R}^{n}.}$$

It is easy to see that

$$\displaystyle{W(\,f,g) = V (\,f,g)^{\wedge }}$$

for all f and g in $L^{2}(\mathbb{R}^{n})$, where ∧ denotes the Fourier transform given by

$$\displaystyle{\widehat{F}(\xi ) = (2\pi )^{-n/2}\int _{ \mathbb{R}^{n}}e^{-ix\cdot \xi }F(x)\,dx,\quad \xi \in \mathbb{R}^{n},}$$

for all F in $L^{1}(\mathbb{R}^{n})$.

Let σ be a measurable function on $\mathbb{R}^{n} \times \mathbb{R}^{n}$. Then the classical pseudo-differential operator T _σ associated to the symbol σ is defined by

$$\displaystyle{\left (T_{\sigma }\varphi \right )(x) = (2\pi )^{-n/2}\int _{ \mathbb{R}^{n}}e^{ix\cdot \xi }\sigma (x,\xi )\hat{\varphi }(x)\,d\xi ,\quad x \in \mathbb{R}^{n},}$$

for all φ in the Schwartz space $\mathcal{S}(\mathbb{R}^{n})$, provided that the integral exists. Once the Fourier inversion formula is in place, a symbol σ defined on the phase space $\mathbb{R}^{n} \times \mathbb{R}^{n}$ is inserted into the integral for the purpose of localization and a pseudo-differential operator is obtained. Another basic ingredient of pseudo-differential operators on $\mathbb{R}^{n}$ in the genesis is the phase space $\mathbb{R}^{n} \times \mathbb{R}^{n}$, which we can look at as the Cartesian product of the additive group $\mathbb{R}^{n}$ and its dual that is also the additive group $\mathbb{R}^{n}$. These observations allow in principle extensions of pseudo-differential operators to other groups G provided that we have an explicit formula for the dual of G and an explicit Fourier inversion formula for the Fourier transform on the group G. This program has been carried out in, e.g., [2, 3, 8, 14]. The aim of this paper is to look at pseudo-differential operators on the non-isotropic Heisenberg group with multi-dimensional center.

In Sect. 2, We define the Schrödinger representation corresponding to the non-isotropic Heisenberg group. Using the representation, we define the λ-Wigner and λ-Weyl transform related the non-isotropic Heisenberg group. The Moyal identity for the λ-Wigner transform and Hilbert-Schmidt properties of the λ-Weyl transform are proved. In Sect. 3, Using the Schrödinger represenation for the ordinary Heisenberg group we prove the Stone-von Neumann theorem on $\mathbb{G}$. Using the Von-Neumann theorem for the non-isotropic group with multi-dimensional center, we define the operator-valued Fourier transform of $\mathbb{G}$ in Sect. 4. Then, in Sect. 5, we define pseudo-differential operators corresponding to the operator-valued symbols. Then the L ²-boundedness and the Hilbert-Schmidt properties of pseudo-differential operators on the group $\mathbb{G}$ are given. Trace class pseudo-differential operators on the group $\mathbb{G}$ are given and a trace formula is given for them.

2 Schrödiner Representations for Non-isotropic Heisenberg Groups with Multi-dimensional Centers

Let

$$\displaystyle{\mathbb{R}^{m}{}^{{\ast}} = \mathbb{R}^{m}\setminus \{0\}}$$

and let $\lambda \in \mathbb{R}^{m}{}^{{\ast}}$. We define the Schrödinger representation of $\mathbb{G}$ on $L^{2}(\mathbb{R}^{n})$ by

$$\displaystyle{\left (\pi _{\lambda }(q,p,t)\varphi \right )(x) = e^{i\lambda \cdot t}e^{iq\cdot B_{\lambda }(x+p/2)}\varphi (x + p),\quad x \in \mathbb{R}^{n}}$$

for all $\varphi \in L^{2}(\mathbb{R}^{n})$ and $(q,p,t) \in \mathbb{G}$, where $z = (q,p) \in \mathbb{R}^{n} \times \mathbb{R}^{n}$ and B _λ  = ∑ _j = 1 ^m λ _j B _j . If we let

$$\displaystyle{\left (\pi _{\lambda }(q,p)\varphi \right )(x) = e^{iq\cdot B_{\lambda }(x+p/2)}\varphi (x + p).}$$

Then

$$\displaystyle{\pi _{\lambda }(q,p,t) = e^{i\lambda \cdot t}\pi _{ \lambda }(q,p).}$$

To prove that π _λ is a group homomorphism, we need the following easy lemma.

Lemma 2.1.

For all $z,z' \in \mathbb{R}^{n} \times \mathbb{R}^{n}$ and $\lambda \in \mathbb{R}^{m}{}^{{\ast}}$ we have

$$\displaystyle{\pi _{\lambda }(z)\pi _{\lambda }(z') = e^{\frac{i} {2} \lambda \cdot [z,z']}\pi _{\lambda }(z + z').}$$

The following theorem tells us that π _λ is in fact a unitary group representation of $\mathbb{G}$ on $L^{2}(\mathbb{R}^{n})$.

Theorem 2.2.

π _λ is a unitary group representation of $\mathbb{G}$ on $L^{2}(\mathbb{R}^{n})$ .

Proof.

By Lemma 2.1, it is easy to see that for all (z, t) and (z′, t′) in $\mathbb{G}$,

$$\displaystyle{\pi _{\lambda }((z,t) \cdot (z',t')) =\pi _{\lambda }(z,t)\pi _{\lambda }(z',t').}$$

Now let $\varphi ,\psi \in L^{2}(\mathbb{R}^{n})$. Then for all $(q,p,t) \in \mathbb{G}$,

$$\displaystyle\begin{array}{rcl} \left (\pi _{\lambda }(q,p,t)\varphi ,\psi \right )& =& \int _{\mathbb{R}^{n}}e^{i\lambda \cdot t}e^{iq\cdot B_{\lambda }(x+p/2)}\varphi (x + p)\overline{\psi (x)}\,dx {}\\ & =& \int _{\mathbb{R}^{n}}\varphi (y)\overline{e^{-i\lambda \cdot t}e^{-iq\cdot B_{\lambda }(y-p/2)}\psi (y - p)}\,dy {}\\ & =& \int _{\mathbb{R}^{n}}\varphi (y)\overline{\left (\pi _{\lambda }(-z,-t)\psi \right )(y)}\,dy {}\\ & =& \left (\varphi ,\pi _{\lambda }(-z,-t)\psi \right ). {}\\ \end{array}$$

Hence π _λ (z, t)^∗ = π _λ ((z, t)⁻¹). □ 

In fact π _λ is an irreducible representation of $\mathbb{G}$ on $L^{2}(\mathbb{R}^{n})$. To prove this we need some preparation. Let $f,g \in L^{2}(\mathbb{R}^{n})$. We define the λ-Fourier Wigner transform of f and g on $\mathbb{R}^{n} \times \mathbb{R}^{n}$ by

$$\displaystyle{V _{\lambda }(\,f,g)(q,p) = (2\pi )^{-n/2}\left (\pi _{\lambda }(q,p)f,g\right ).}$$

In fact,

$$\displaystyle{V ^{\lambda }(\,f,g)(q,p) = (2\pi )^{-n/2}\int _{ \mathbb{R}^{n}}e^{iB_{\lambda }^{t}q\cdot x }f(x + \frac{p} {2})\overline{g(x -\frac{p} {2})}\,dx.}$$

Therefore, the λ-Fourier Wigner transform is related to the ordinary Fourier Wigner transform by

$$\displaystyle{ V ^{\lambda }(\,f,g)(q,p) = V (\,f,g)(B_{\lambda }^{t}q,p). }$$

(2)

Note that

$$\displaystyle{V ^{\lambda }(\,f,g)(q,-p) = \overline{V ^{\lambda }(g,f)}(q,p),\quad q,p \in \mathbb{R}^{n}.}$$

Now, we define the λ-Wigner transform of $f,g \in L^{2}(\mathbb{R}^{n})$ by

$$\displaystyle{W^{\lambda }(\,f,g) =\widehat{ V _{\lambda }(\,f,g)}.}$$

In fact, λ-Wigner transform has the form

$$\displaystyle{W^{\lambda }\left (\,f,g\right )(x,\xi ) = \vert \lambda \vert ^{-n}(2\pi )^{-n/2}\int _{ \mathbb{R}^{n}}e^{-ip\cdot \xi }f(\frac{B_{\lambda }^{t}x} {\vert \lambda \vert ^{2}} + \frac{p} {2})\overline{g(\frac{B_{\lambda }^{t}x} {\vert \lambda \vert ^{2}} -\frac{p} {2})}\,dp}$$

and it is related to the ordinary Wigner trasform by

$$\displaystyle{W^{\lambda }(\,f,g)(x,\xi ) = \vert \lambda \vert ^{-n}W(\,f,g)(\frac{B_{\lambda }^{t}x} {\vert \lambda \vert ^{2}} ,\xi )}$$

for all x, ξ in $\mathbb{R}^{n}$. Moreover,

$$\displaystyle{W^{\lambda }(\,f,g) = \overline{W^{\lambda }(g,f)}.}$$

By using (1) and the fact that B _j , 1 ≤ j ≤ n are orthogonal matrices, we get the following result.

Proposition 2.1.

B _λ B _λ ^t = |λ| ² I, where I is the identity n × n matrix. In particular det B _λ = |λ| ⁿ .

The following proposition gives us the relation between the dimesion of the center of the non-isotropic Heisenebrg group and its phase space.

Proposition 2.2.

Let $\mathbb{G}$ be the non-isotropic Heisenberg group on $\mathbb{R}^{n} \times \mathbb{R}^{n} \times \mathbb{R}^{m}$ . Then m ≤ n ² .

Proof.

For all 1 ≤ k ≤ m and 1 ≤ i, j ≤ n, let (B _k ) _ij be the entry of the matrix B _k in the i-th row and j-th column. Then the n ² × m matrix

$$\displaystyle{C = \left [\begin{array}{lllll} (B_{1})_{11} & (B_{2})_{11} & \ldots & (B_{m})_{11} \\ (B_{1})_{12} & (B_{2})_{12} & \ldots & (B_{m})_{12}\\ \vdots & \vdots & \ddots & \vdots \\ (B_{1})_{1n} &(B_{2})_{1n} &\ldots &(B_{m})_{1n} \\ (B_{1})_{21} & (B_{2})_{21} & \ldots & (B_{m})_{21} \\ (B_{1})_{22} & (B_{2})_{22} & \ldots & (B_{m})_{22}\\ \vdots & \vdots & \vdots & \vdots \\ (B_{1})_{nn}&(B_{2})_{nn}&\ldots &(B_{m})_{nn}\\ \end{array} \right ]}$$

has rank m. To prove this, it is enough to show that the columns of C are linearly independent. Let C ⁱ be the i-th column of C and let $\lambda \in \mathbb{R}^{m}$ be such that

$$\displaystyle{\sum _{i=1}^{m}\lambda _{ i}C^{i} = 0.}$$

It follows that B _λ  = 0. Therefore by Proposition 2.1, we get λ = 0. □ 

Let $\sigma \in \mathcal{S}(\mathbb{R}^{n} \times \mathbb{R}^{n})$ and $f \in \mathcal{S}(\mathbb{R}^{n})$, then we define the λ-Weyl transform W _σ ^λ f of f corresponding to the symbol σ by

$$\displaystyle{\left (W_{\sigma }^{\lambda }f,g\right )_{L^{2}(\mathbb{R}^{n})} = (2\pi )^{-n/2}\int _{ \mathbb{R}^{n}}\int _{\mathbb{R}^{n}}\sigma (x,\xi )W^{\lambda }(\,f,g)(x,\xi )\,dx\,d\xi ,}$$

for all $g \in \mathcal{S}(\mathbb{R}^{n})$. Therefore, using the Parseval’s identity, we have

$$\displaystyle{\left (W_{\sigma }^{\lambda }f,g\right )_{L^{2}(\mathbb{R}^{n})} = (2\pi )^{-n/2}\int _{ \mathbb{R}^{n}}\int _{\mathbb{R}^{n}}\hat{\sigma }(q,p)V ^{\lambda }(\,f,g)(q,p)\,dq\,dp.}$$

Hence, formally we can write,

$$\displaystyle{\left (W_{\sigma }^{\lambda }f\right )(x) = (2\pi )^{-n}\int _{ \mathbb{R}^{n}}\int _{\mathbb{R}^{n}}\hat{\sigma }(q,p)\left (\pi _{\lambda }(q,p)f\right )(x)\,dq\,dp.}$$

Proposition 2.3.

Let $\sigma \in \mathcal{S}(\mathbb{R}^{n} \times \mathbb{R}^{n})$ . Then the λ-Weyl transform W _σ ^λ is given by

$$\displaystyle{W_{\sigma }^{\lambda } = W_{\sigma _{\lambda }},}$$

where $W_{\sigma _{\lambda }}$ is the ordinary Weyl transform corresponding to the symbol

$$\displaystyle{\sigma _{\lambda }(x,\xi ) =\sigma (B_{\lambda }x,\xi ).}$$

Proposition 2.4.

Let $\sigma \in \mathcal{S}(\mathbb{R}^{n} \times \mathbb{R}^{n})$ . Then the λ-Weyl transform W _σ ^λ is a Hilber-Schmidt operator with kernel

$$\displaystyle{k_{\sigma }^{\lambda }(x,p) = \left (\mathcal{F}_{2}\sigma \right )\left (B_{\lambda }(\frac{x + p} {2} ),p - x\right ),}$$

where $\mathcal{F}_{2}\sigma$ is the ordinary Fourier transform of σ with respect to the second variable, i.e.,

$$\displaystyle{\left (\mathcal{F}_{2}\sigma \right )(x,p) = (2\pi )^{-n/2}\int _{ \mathbb{R}^{n}}e^{-i\xi \cdot p}\sigma (x,\xi )\,d\xi .}$$

Moreover,

$$\displaystyle{\|W_{\sigma }^{\lambda }\|_{HS} = \vert \lambda \vert ^{-n/2}\|\sigma \|_{ L^{2}(\mathbb{R}^{n}\times \mathbb{R}^{n})}}$$

Proof.

By Proposition 2.4 and the kernel of the ordinary Weyl transform (see [13] for details), we have

$$\displaystyle\begin{array}{rcl} k_{\sigma }^{\lambda }(x,p)& =& \left (\mathcal{F}_{2}\sigma _{\lambda }\right )\left (\frac{x + p} {2} ,p - x\right ) {}\\ & =& \left (\mathcal{F}_{2}\sigma \right )\left (B_{\lambda }(\frac{x + p} {2} ),p - x\right ). {}\\ \end{array}$$

Hence,

$$\displaystyle\begin{array}{rcl} \|W_{\sigma }^{\lambda }\|_{HS}^{2}& =& \int _{ \mathbb{R}^{n}}\int _{\mathbb{R}^{n}}\vert k_{\sigma }^{\lambda }(x,p)\vert ^{2}\,dx\,dp {}\\ & =& \int _{\mathbb{R}^{n}}\int _{\mathbb{R}^{n}}\left \vert \left (\mathcal{F}_{2}\sigma \right )\left (B_{\lambda }(\frac{x + p} {2} ),p - x\right )\right \vert ^{2}\,dx\,dp {}\\ & =& \vert \lambda \vert ^{-n}\int _{ \mathbb{R}^{n}}\int _{\mathbb{R}^{n}}\left \vert \left (\mathcal{F}_{2}\sigma \right )\left (x,p\right )\right \vert ^{2}\,dx\,dp {}\\ & =& \vert \lambda \vert ^{-n}\|\sigma \|_{ L^{2}(\mathbb{R}^{n}\times \mathbb{R}^{n})}^{2}, {}\\ \end{array}$$

which completes the proof. □ 

Let F and G be functions in $L^{2}(\mathbb{R}^{2n})$. The λ-twisted convolution of F and G denoted by F ∗ _λ G on $\mathbb{R}^{2n}$ is defined by

$$\displaystyle{(F {\ast}_{\lambda }G)(z) =\int _{\mathbb{R}^{2n}}F(z - w)G(w)e^{\frac{i} {2} \lambda .[z,w]}\,dw.}$$

By Lemma 2.1 we get the following theorem.

Theorem 2.3.

Let σ and τ be in $L^{2}(\mathbb{R}^{2n})$ . Then

$$\displaystyle{W_{\sigma }^{\lambda }W_{\tau }^{\lambda } = W_{\omega }^{\lambda },}$$

where $\hat{\omega }= (2\pi )^{-n}(\hat{\sigma }{\ast}_{\lambda }\hat{\tau })$ .

Using the Moyal identity for the ordinary Wigner transform we have the following Moyal identity for the λ-Wigner transform and λ-Fourier Wigner transform.

Proposition 2.5.

For all f ₁ ,f ₂ ,g ₁ ,g ₂ in $L^{2}(\mathbb{R}^{n})$

$$\displaystyle{\left (W_{\lambda }(\,f_{1},g_{1}),W_{\lambda }(\,f_{2},g_{2})\right ) = \vert \lambda \vert ^{-n}\left (\,f_{ 1},f_{2}\right )\overline{\left (g_{1},g_{2}\right )},}$$

and

$$\displaystyle{\left (V _{\lambda }(\,f_{1},g_{1}),V _{\lambda }(\,f_{2},g_{2})\right ) = \vert \lambda \vert ^{-n}\left (\,f_{ 1},f_{2}\right )\overline{\left (g_{1},g_{2}\right )}.}$$

Now, we are ready to prove the following theorem.

Theorem 2.4.

For all $\lambda \in \mathbb{R}^{m}{}^{{\ast}}$ , π _λ is a unitary irreducible representation of $\mathbb{G}$ on $L^{2}(\mathbb{R}^{n})$ .

Proof.

suppose $M \subset L^{2}(\mathbb{R}^{n})$ is a nonzero closed invariant subspace of π _λ and f ∈ M \{0}. Then

$$\displaystyle{\pi _{\lambda }(q,p,t)M \subset M,\quad (q,p,t) \in \mathbb{G}.}$$

If $M\not =L^{2}(\mathbb{R}^{n})$, then we can find $g \in L^{2}(\mathbb{R}^{n})$ such that

$$\displaystyle{\left (\pi _{\lambda }(q,p,t)f,g\right ) = 0,\quad (q,p,t) \in \mathbb{G}.}$$

But,

$$\displaystyle\begin{array}{rcl} \left (\pi _{\lambda }(q,p,t)f,g\right )& =& e^{i\lambda \cdot t}\left (\pi _{\lambda }(q,p)f,g\right ) {}\\ & =& e^{i\lambda \cdot t}(2\pi )^{n/2}V _{\lambda }(\,f,g)(p,q). {}\\ \end{array}$$

So,

$$\displaystyle{V _{\lambda }(\,f,g)(q,p) = 0}$$

for all $(p,q) \in \mathbb{R}^{n} \times \mathbb{R}^{n}$. By the Moyal identity,

$$\displaystyle{\|V _{\lambda }(\,f,g)\|_{L^{2}(\mathbb{R}^{n}\times \mathbb{R}^{n})}^{2} = \vert \lambda \vert ^{-n}\|\,f\|_{ L^{2}(\mathbb{R}^{n})}^{2}\|g\|_{ L^{2}(\mathbb{R}^{n})}^{2} = 0.}$$

So, f = 0 or g = 0 which is a contradiction. □ 

3 Stone-Von Neumann Theorem on $\mathbb{G}$

Let $U(L^{2}(\mathbb{R}^{n}))$ be the space of unitary operators on $L^{2}(\mathbb{R}^{n})$. Let $h \in \mathbb{R}^{{\ast}}$, then the Schrödinger representation $\rho _{h} : \mathbb{H}^{n} \rightarrow U(L^{2}(\mathbb{R}^{n}))$ on the ordinary Heisenebrg group is defined by

$$\displaystyle{\left (\rho _{h}(q,p,t)\varphi \right )(x) = e^{iht}e^{iq\cdot (x+hp/2)}f(x + hp),\quad x \in \mathbb{R}^{n},}$$

for all $f \in L^{2}(\mathbb{R}^{n})$. Then ρ _h is an irreducible unitary representation of $\mathbb{H}^{n}$ on $L^{2}(\mathbb{R}^{n})$. By the Stone-von Neumann theorem, any irreducible unitary representation of $\mathbb{H}^{n}$ on a Hilbert space that is non-trivial on the center is equivalent to some ρ _h . More precisely we have

Theorem 3.1.

Let π be an irreducible unitary represenatation of $\mathbb{H}^{n}$ on a Hilbert space $\mathcal{H}$ , such that π(0,0,t) = e ^iht I for some $h \in \mathbb{R}^{{\ast}}$ . Then π is unitarily equivalent to ρ _h .

Similarly, we prove the Stone-von Neumann theorem for the non-isotropic Heisenberg group $\mathbb{G}$. To prove we use the following lemma.

Lemma 3.2.

Let $\lambda \in \mathbb{R}^{m}{}^{{\ast}}$ . The mapping $\alpha _{\lambda } : \mathbb{G} \rightarrow \mathbb{H}^{n}$ defined by

$$\displaystyle{\alpha _{\lambda }(q,p,t) = (B_{\lambda }^{t}q, \frac{p} {\vert \lambda \vert }, \frac{\lambda \cdot t} {\vert \lambda \vert }),\quad (q,p,t) \in \mathbb{G}}$$

is a surjective homomorphism of Lie groups. In particular, G∕ker α _λ is isomorphic to $\mathbb{H}^{n}$ where

$$\displaystyle{\ker \alpha _{\lambda } =\{ (0,0,t) : (t,\lambda ) = 0\}.}$$

Proof.

To prove α _λ is a group homomorphism, let $(q,p,t),(q',p'.t') \in \mathbb{G}$. Then

$$\displaystyle\begin{array}{rcl} & & \alpha _{\lambda }((q,p,t) \cdot _{\mathbb{G}}(q',p',t')) =\alpha _{\lambda }(q + q',p + p',t + t' + \frac{1} {2}[z,z']) {}\\ & =& \left (B_{\lambda }^{t}(q + q'), \frac{p + p'} {\vert \lambda \vert } ,\lambda \cdot (t + t' + \frac{1} {2}[z,z'])/\vert \lambda \vert \right ) {}\\ \end{array}$$

Since λ ⋅ [z, z′] = (q′, B _λ p) − (q, B _λ p′), therefore

$$\displaystyle\begin{array}{rcl} & & \alpha _{\lambda }((q,p,t) \cdot _{\mathbb{G}}(q',p',t')) \\ & & = (B_{\lambda }^{t}q, \frac{p} {\vert \lambda \vert }, \frac{\lambda \cdot t} {\vert \lambda \vert }) \cdot _{\mathbb{H}^{n}}(B_{\lambda }^{t}q', \frac{p'} {\vert \lambda \vert }, \frac{\lambda \cdot t'} {\vert \lambda \vert }) \\ & =& \alpha _{\lambda }((q,p,t) \cdot _{\mathbb{H}^{n}}\alpha _{\lambda }(q',p',t')). {}\end{array}$$

(3)

Surjectivity is easy to see, since B _λ is invertible. □ 

The following lemma gives the connection between the Schrödinger representation on the ordinary Heisenberg group $\mathbb{H}^{n}$ and the representations π _λ on the non-isotropic Heisenberg group $\mathbb{G}$.

Lemma 3.3.

For all $\lambda \in \mathbb{R}^{m}{}^{{\ast}}$ ,

$$\displaystyle{\pi _{\lambda } =\rho _{\vert \lambda \vert }\circ \alpha _{\lambda }.}$$

Now, we are ready to prove the Stone von-Neumann theorem for the non-isotropic Heiseneberg group.

Theorem 3.4.

Let $\Pi _{\lambda }$ be an irreducible unitary group representation of $\mathbb{G}$ on a Hilbert space $\mathcal{H}$ such that $\Pi _{\lambda }(0,0,t) = e^{i\lambda \cdot t}I$ , for some $\lambda \in \mathbb{R}^{m}$ . Then $\Pi _{\lambda }$ is unitarily equivalent to π _λ

Proof.

Let $\Pi _{\vert \lambda \vert } : \mathbb{H}^{n} \rightarrow U(\mathcal{H})$ be defined by $\Pi _{\vert \lambda \vert } = \Pi _{\lambda }PT$ where T is the isomorphism of $\mathbb{H}^{n}$ onto G∕kerα _λ (see Lemma 3.2) and P is the projection from $\mathbb{G}/\ker \alpha _{\lambda }$ onto $\mathbb{G}$. Then $\Pi _{\vert \lambda \vert }(0,0,t_{0}) = e^{i\vert \lambda \vert t_{o}}I$, for all $t_{0} \in \mathbb{R}$. Moreover, $\Pi _{\vert \lambda \vert }$ is an irreducible unitary representation of $\mathbb{H}^{n}$ on the Hilbert space $\mathcal{H}$. This can be easily seen by using the fact that $\Pi _{\lambda }$ is an irreducible unitary representation of $\mathbb{G}$ on $\mathcal{H}$. □ 

4 Fourier Transforms and the Fourier Inversion Formula on $\mathbb{G}$

By the Stone-von Neumann theorem every irreducible unitary representation of $\mathbb{G}$ which acts non-trivially on the center is in fact unitarily equivalent to exactly one of π _λ , $\lambda \in \mathbb{R}^{m}{}^{{\ast}}$. Hence, the identification of $\{\pi _{\lambda } :\lambda \in \mathbb{R}^{m}{}^{{\ast}}\}$ with $\mathbb{R}^{m}{}^{{\ast}}$ will be used. Let $f \in L^{1}(\mathbb{G})$ and $\lambda \in \mathbb{R}^{m}{}^{{\ast}}$. We define the Fourier transform of f at λ to be the bounded linear operator $\hat{f}(\lambda )$ from $L^{2}(\mathbb{R}^{n})$ into $L^{2}(\mathbb{R}^{n})$ given by

$$\displaystyle{\hat{f}(\lambda )\varphi =\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}f(z,t)\left (\pi _{\lambda }(z,t)\varphi \right )\,dz\,dt,\quad \varphi \in L^{2}(\mathbb{R}^{n}).}$$

To see the boundedness of $\hat{f}(\lambda )$, let $\varphi ,\psi \in L^{2}(\mathbb{R}^{n})$. Then By Schwarz inequality

$$\displaystyle\begin{array}{rcl} \left \vert \left (\hat{f}(\lambda )\varphi ,\psi \right )\right \vert & \leq & \int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\vert \,f(z,t)\vert \ \vert \left (\pi _{\lambda }(z,t)\varphi ,\psi \right )\vert \,dz\,dt {}\\ & \leq & \int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\vert \,f(z,t)\vert \ \|\pi _{\lambda }(z,t)\varphi \|_{L^{2}(\mathbb{R}^{n})}\|\psi \|_{L^{2}(\mathbb{R}^{n})}\,dz\,dt. {}\\ & \leq & \|\,f\|_{L^{1}(\mathbb{G})}\|\varphi \|_{L^{2}(\mathbb{R}^{n})}\|\psi \|_{L^{2}(\mathbb{R}^{n})}. {}\\ \end{array}$$

Set

$$\displaystyle{f^{\lambda }(z) = (2\pi )^{-m/2}\int _{ \mathbb{R}^{m}}e^{i\lambda \cdot t}f(z,t)\,dt.}$$

Then $\hat{f}(\lambda )\varphi$ has the form

$$\displaystyle{\hat{f}(\lambda )\varphi = (2\pi )^{m/2}\int _{ \mathbb{R}^{2n}}f^{\lambda }(z)\left (\pi _{\lambda }(z)\varphi \right )\,dz.}$$

Therefore we have following proposition relating the Fourier transform $\hat{f}(\lambda )$ to the λ-Weyl transform.

Proposition 4.1.

Let $f \in L^{1}(\mathbb{G})$ . Then for all $\lambda \in \mathbb{R}^{m}{}^{{\ast}}$

$$\displaystyle{\hat{f}(\lambda ) = (2\pi )^{(2n+m)/2}W_{ (\,f^{\lambda })^{\vee }}^{\lambda },}$$

where ( f ^λ ) ^∨ is the inverse Fourier transform of f ^λ on $\mathbb{R}^{2n}$ .

We have the following Plancheral’s formula for the Fourier transform on the non-isotropic Heisenberg group with multi-dimensional center.

Theorem 4.1.

Let $f \in L^{2}(\mathbb{G})$ and $\lambda \in \mathbb{R}^{m}{}^{{\ast}}$ . Then $\hat{f}(\lambda ) : L^{2}(\mathbb{R}^{n}) \rightarrow L^{2}(\mathbb{R}^{n})$ is a Hilbert-Schmidt operator. In fact we have

• (i) The kernel of $\hat{f}(\lambda )$ is given by

  $$\displaystyle{k_{\lambda }(x,p) = (2\pi )^{(n+m)/2}\left (\mathcal{F}_{ 1}^{-1}f^{\lambda }\right )\left (B_{\lambda }(\frac{x + p} {2} ),p - x\right )}$$

  where $\mathcal{F}_{1}^{-1}f^{\lambda }$ is the ordinary inverse Fourier transform of f ^λ with respect to the first variable, i.e.,

  $$\displaystyle{\left (\mathcal{F}_{1}^{-1}f^{\lambda }\right )(x,p) = (2\pi )^{-n/2}\int _{ \mathbb{R}^{n}}e^{ix\cdot q}f^{\lambda }(q,p)\,dq.\quad (x,p) \in \mathbb{R}^{n} \times \mathbb{R}^{n}.}$$

• (ii) The Hilbert-Schmidt norm of $\hat{f}(\lambda )$ is given by

  $$\displaystyle{\|\,\hat{f}(\lambda )\|_{HS}^{2} = (2\pi )^{m+n}\vert \lambda \vert ^{-n}\|\,f^{\lambda }\|_{ L^{2}(\mathbb{R}^{2n})}^{2}.}$$

• (iii) Let dμ(λ) = (2π) ^−(n+m) |λ| ⁿ  dλ. We have the following Plancheral’s formula

  $$\displaystyle{\int _{\mathbb{R}^{m}}\|\,\hat{f}(\lambda )\|_{HS}^{2}\,d\mu (\lambda ) =\|\, f\|_{ L^{2}(\mathbb{G})}^{2}.}$$

Proof.

Let φ be in $L^{2}(\mathbb{R}^{n})$. Then for all $x \in \mathbb{R}^{n}$,

$$\displaystyle\begin{array}{rcl} \left (\hat{f}(\lambda )\varphi \right )(x)& =& (2\pi )^{m/2}\int _{ \mathbb{R}^{2n}}f^{\lambda }(q,p)\left (\pi _{\lambda }(q,p)\varphi \right )(x)\,dq\,dp {}\\ & =& (2\pi )^{m/2}\int _{ \mathbb{R}^{2n}}f^{\lambda }(q,p)e^{iq\cdot B_{\lambda }(x+\frac{p} {2} )}\varphi (x + p)\,dq\,dp {}\\ & =& \int _{\mathbb{R}^{n}}\left ((2\pi )^{m/2}\int _{ \mathbb{R}^{n}}e^{iq\cdot B_{\lambda }(\frac{x+p} {2} )}f^{\lambda }(q,p - x)\,dq\right )\varphi (p)\,dp {}\\ & =& \int _{\mathbb{R}^{n}}k_{\lambda }(x,p)\varphi (p)\,dp {}\\ \end{array}$$

where

$$\displaystyle{k_{\lambda }(x,p) = (2\pi )^{(n+m)/2}\left (\mathcal{F}_{ 1}^{-1}f^{\lambda }\right )\left (B_{\lambda }(\frac{x + p} {2} ),p - x\right ).}$$

Hence the Hilbert-Schmidt norm of $\hat{f}(\lambda )$ is given by

$$\displaystyle\begin{array}{rcl} \|\,\hat{f}(\lambda )\vert \|_{HS}^{2}& =& \|k_{\lambda }\|_{ L^{2}(\mathbb{R}^{n}\times \mathbb{R}^{n})}^{2} \\ & =& (2\pi )^{(n+m)}\int _{ \mathbb{R}^{n}}\int _{\mathbb{R}^{n}}\left \vert \left (\mathcal{F}_{1}^{-1}f^{\lambda }\right )\left (B_{\lambda }(\frac{x + p} {2} ),p - x\right )\right \vert ^{2}\,dx\,dp \\ & =& (2\pi )^{(n+m)}\int _{ \mathbb{R}^{n}}\int _{\mathbb{R}^{n}}\left \vert \left (\mathcal{F}_{1}^{-1}f^{\lambda }\right )\left (x,p\right )\right \vert ^{2}\vert \lambda \vert ^{-n}\,dx\,dp \\ & =& \vert \lambda \vert ^{-n}(2\pi )^{(n+m)}\|\,f^{\lambda }\|_{ L^{2}(\mathbb{R}^{2n})}^{2} {}\end{array}$$

(4)

where in (4) we used the Parseval’s identity for the ordinary Fourier transform. □ 

Now we are ready to prove the inversion formula for the non-isotropic group Fourier transform.

Theorem 4.2.

Let f be a Schwartz function on $\mathbb{G}$ . Then we have

$$\displaystyle{f(z,t) =\int _{\mathbb{R}^{m}}tr\left (\pi _{\lambda }(z,t)^{{\ast}}\hat{f}(\lambda )\right )\,d\mu (\lambda ),\quad (z,t) \in \mathbb{G}.}$$

Proof.

For all $(z,t) \in \mathbb{G}$,

$$\displaystyle\begin{array}{rcl} \pi _{\lambda }(z,t)^{{\ast}}\hat{f}(\lambda )& =& \pi _{\lambda }(-z,-t)\int _{ \mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}f(\tilde{z},\tilde{t})\ \pi _{\lambda }(\tilde{z},\tilde{t})\,d\tilde{z}\,d\tilde{t} {}\\ & =& \int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}f(\tilde{z},\tilde{t})\ \pi _{\lambda }\left ((-z,-t)) \cdot (\tilde{z},\tilde{t})\right )\,d\tilde{z}\,d\tilde{t} {}\\ & =& \int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}f(\tilde{z},\tilde{t})\ \pi _{\lambda }\left (-z +\tilde{ z},-t +\tilde{ t} + \frac{1} {2}[-z,\tilde{z}]\right )\,d\tilde{z}\,d\tilde{t} {}\\ & =& \int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}f(\tilde{z},\tilde{t})e^{i \frac{\lambda }{2} \cdot [-z,\tilde{z}]}\pi _{ \lambda }\left (-z +\tilde{ z},-t +\tilde{ t}\right )\,d\tilde{z}\,d\tilde{t}. {}\\ \end{array}$$

Now, we let $z' = -z +\tilde{ z}$ and $t' = -t +\tilde{ t}$. W get

$$\displaystyle\begin{array}{rcl} \pi _{\lambda }(z,t)^{{\ast}}\hat{f}(\lambda )& =& \int _{ \mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}g(z',t')\pi _{\lambda }(z',t')\,dz'\,dt', {}\\ \end{array}$$

where

$$\displaystyle{g(z',t') = e^{-i \frac{\lambda }{2} \cdot [z,z']}f(z' + z,t' + t).}$$

Hence,

$$\displaystyle{\pi _{\lambda }(z,t)^{{\ast}}\hat{f}(\lambda ) =\hat{ g}(\lambda ).}$$

By Theorem 4.1, the kernel of $\hat{g}(\lambda )$ is given by

$$\displaystyle{k_{\lambda }(x,p) = (2\pi )^{(n+m)/2}\left (\mathcal{F}_{ 1}^{-1}g^{\lambda }\right )\left (B_{\lambda }(\frac{x + p} {2} ),p - x\right ).}$$

Therefore,

$$\displaystyle{tr\left (\pi _{\lambda }(z,t)^{{\ast}}\hat{f}(\lambda )\right ) =\int _{ \mathbb{R}^{n}}k_{\lambda }(x,x)\,dx.}$$

So, for $z = (u,v) \in \mathbb{R}^{n} \times \mathbb{R}^{n}$,

$$\displaystyle\begin{array}{rcl} k_{\lambda }(x,x)& =& (2\pi )^{(n+m)/2}\left (\mathcal{F}_{ 1}^{-1}g^{\lambda }\right )\left (B_{\lambda }x,0\right ) {}\\ & =& (2\pi )^{m/2}\int _{ \mathbb{R}^{n}}e^{iB_{\lambda }x\cdot \xi }g^{\lambda }(\xi ,0)\,d\xi . {}\\ \end{array}$$

On the other hand, it is easy to see that

$$\displaystyle{g^{\lambda }(z') = e^{-i \frac{\lambda }{2} \cdot [z,z']}e^{-i\lambda \cdot t}f^{\lambda }(z + z').}$$

So, for $z = (u,v) \in \mathbb{R}^{n} \times \mathbb{R}^{n}$, and z′ = (ξ, 0), we get

$$\displaystyle{g^{\lambda }(\xi ,0) = e^{\frac{-i} {2} B_{\lambda }v\cdot \xi }e^{-i\lambda \cdot t}f^{\lambda }(\xi +u,v).}$$

Hence,

$$\displaystyle\begin{array}{rcl} k_{\lambda }(x,x)& =& (2\pi )^{m/2}\int _{ \mathbb{R}^{n}}e^{iB_{\lambda }x\cdot \xi }e^{\frac{-i} {2} B_{\lambda }v\cdot \xi }e^{-i\lambda \cdot t}f^{\lambda }(\xi +u,v)\,d\xi \\ & =& (2\pi )^{m/2}e^{-i\lambda \cdot t}e^{i(-B_{\lambda }x+B_{\lambda }v/2)\cdot u}\int _{ \mathbb{R}^{n}}e^{i(B_{\lambda }x-B_{\lambda }v/2)\cdot \xi }f^{\lambda }(\xi ,v)\,d\xi {}\end{array}$$

(5)

Therefore,

$$\displaystyle\begin{array}{rcl} & & tr\left (\pi _{\lambda }(z,t)^{{\ast}}\hat{f}(\lambda )\right ) {}\\ & =& (2\pi )^{m/2}e^{-i\lambda \cdot t}e^{iB_{\lambda }v/2\cdot u}\int _{ \mathbb{R}^{n}}e^{-ix\cdot B_{\lambda }^{t}u }\left \{\int _{\mathbb{R}^{n}}e^{i\xi \cdot (-B_{\lambda }v/2+B_{\lambda }x)}f^{\lambda }(\xi ,v)\,d\xi \right \}\,dx {}\\ & =& (2\pi )^{(m+n)/2}e^{-i\lambda \cdot t}e^{iB_{\lambda }v/2\cdot u}\int _{ \mathbb{R}^{n}}e^{-ix\cdot B_{\lambda }^{t}u }\left (\mathcal{F}_{1}^{-1}f^{\lambda }\right )(-B_{\lambda }v/2 + B_{\lambda }x,v)\,dx {}\\ & =& (2\pi )^{(m+n)/2}e^{-i\lambda \cdot t}\vert \lambda \vert ^{-n}\int _{ \mathbb{R}^{n}}e^{-ix\cdot u}\left (\mathcal{F}_{ 1}^{-1}f^{\lambda }\right )(x,v)\,dx {}\\ & =& (2\pi )^{m/2+n}e^{-i\lambda \cdot t}\vert \lambda \vert ^{-n}f^{\lambda }(u,v). {}\\ \end{array}$$

By integrating both sides of

$$\displaystyle{tr\left (\pi _{\lambda }(z,t)^{{\ast}}\hat{f}(\lambda )\right )(2\pi )^{-(n+m)}\vert \lambda \vert ^{n} = (2\pi )^{-m/2}e^{-i\lambda \cdot t}f^{\lambda }(z)}$$

with respect to λ, we get the Fourier inversion formula. □ 

5 Pseudo-differential Operators on Non-isotropic Heisenberg Groups with Multi-dimensional Centers

Let $B(L^{2}(\mathbb{R}^{n}))$ be the C ^∗-algebra of all bounded linear operators on $L^{2}(\mathbb{R}^{n})$. Then consider the operator valued symbol

$$\displaystyle{\sigma : \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}}\rightarrow B(L^{2}(\mathbb{R}^{n})).}$$

We define the pseudo-differential operator $T_{\sigma } : L^{2}(\mathbb{G}) \rightarrow L^{2}(\mathbb{G})$ corresponding to the symbol σ by

$$\displaystyle{\left (T_{\sigma }f\right )(z,t) =\int _{\mathbb{R}^{m}}tr\left (\pi _{\lambda }(z,t)^{{\ast}}\sigma (z,t,\lambda )\hat{f}(\lambda )\right )\,d\mu (\lambda ),\quad (z,t) \in \mathbb{G}}$$

for all $f \in L^{2}(\mathbb{G})$. Let $HS(L^{2}(\mathbb{R}^{n}))$ be the space of Hilbert-Schmidt operators on $L^{2}(\mathbb{R}^{n})$. We have the following theorem on L ²-boundedness of pseudo-differential operators.

Theorem 5.1.

Let $\sigma : \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}}\rightarrow HS(L^{2}(\mathbb{R}^{n}))$ be such that

$$\displaystyle{C_{\sigma }^{2} =\int _{ \mathbb{R}^{m}}\int _{\mathbb{G}}\|\sigma (z,t,\lambda )\|_{HS}^{2}\,dz\,dt\,d\mu (\lambda ) <\infty .}$$

Then $T_{\sigma } : L^{2}(\mathbb{G}) \rightarrow L^{2}(\mathbb{G})$ is a bounded linear operator and

$$\displaystyle{\|T_{\sigma }\|_{op} \leq C_{\sigma },}$$

where ∥⋅∥ _op is the operator norm on the C ^∗ -algebra of bounded linear operators on $L^{2}(\mathbb{G})$ .

Proof.

Let $f \in L^{2}(\mathbb{G})$. Then by Minkowski’s inequality we have

$$\displaystyle\begin{array}{rcl} & & \|T_{\sigma }f\|_{L^{2}(\mathbb{G})} = \\ & & \left \{\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\left \vert \int _{\mathbb{R}^{m}}tr\left (\pi _{\lambda }(z,t)^{{\ast}}\sigma (z,t,\lambda )\hat{f}(\lambda )\right )\,d\mu (\lambda )\right \vert ^{2}\,dz\,dt\right \}^{1/2} \\ & & \leq \int _{\mathbb{R}^{m}}\left \{\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\left \vert tr\left (\pi _{\lambda }(z,t)^{{\ast}}\sigma (z,t,\lambda )\hat{f}(\lambda )\right )\right \vert ^{2}\,dz\,dt\right \}^{1/2}\,d\mu (\lambda ) \\ & & \leq \int _{\mathbb{R}^{m}}\left \{\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\|\sigma (z,t,\lambda )\|_{HS}^{2}\|\,\hat{f}(\lambda )\|_{ HS}^{2}\,dz\,dt\right \}^{1/2}\,d\mu (\lambda ) \\ & & =\int _{\mathbb{R}^{m}}\|\,\hat{f}(\lambda )\|_{HS}\left \{\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\|\sigma (z,t,\lambda )\|_{HS}^{2}\,dz\,dt\right \}^{1/2}\,d\mu (\lambda ). \\ & & \leq C_{\sigma }\|\,f\|_{L^{2}(\mathbb{G})} {}\end{array}$$

(6)

where in (6), we used Hölder’s inequality. □ 

The following result tells us that under suitable conditions, two symbols of the same pseudo-differential operator are equal.

Proposition 5.1.

Let $\sigma : \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}}\rightarrow HS(L^{2}(\mathbb{R}^{n}))$ be such that

$$\displaystyle{\int _{\mathbb{R}^{m}}\int _{\mathbb{G}}\|\sigma (z,t,\lambda )\|_{HS}^{2}\,dz\,dt\,d\mu (\lambda ) <\infty .}$$

Furthermore suppose that

$$\displaystyle\begin{array}{rcl} \int _{\mathbb{R}^{m}}\|\sigma (z,t,\lambda )\|_{HS}\,d\mu (\lambda ) <\infty ,\quad (z,t) \in \mathbb{G},& &{}\end{array}$$

(7)

$$\displaystyle\begin{array}{rcl} \sup _{(z,t,\lambda )\in \mathbb{G}\times \mathbb{R}^{m}{}^{{\ast}}}\|\sigma (z,t,\lambda )\|_{HS} <\infty ,& &{}\end{array}$$

(8)

and the mapping

$$\displaystyle{ \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}}\ni (z,t,\lambda )\mapsto \pi _{\lambda }(z,t)^{{\ast}}\sigma (z,t,\lambda ) \in HS(L^{2}(\mathbb{R}^{n})) }$$

(9)

is weakly continuous. Then T _σ f = 0 for all f only if

$$\displaystyle{\sigma (z,t,\lambda ) = 0}$$

for almost all $(z,t,\lambda ) \in \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}}$ .

Proof.

For all $(z,t) \in \mathbb{G}$, we define $f_{z,t} \in L^{2}(\mathbb{G})$ by

$$\displaystyle{\widehat{f_{z,t}}(\lambda ) =\sigma (z,t,\lambda )^{{\ast}}\pi _{ \lambda }(z,t).}$$

Then, for all $(w,s) \in \mathbb{G}$

$$\displaystyle{\left (T_{\sigma }f_{z,t}\right )(w,s) =\int _{\mathbb{R}^{m}}A_{z,t}^{\lambda }(w,s)\,d\mu (\lambda ),}$$

where

$$\displaystyle{A_{z,t}^{\lambda }(w,s) = tr\left (\pi _{\lambda }(w,s)^{{\ast}}\sigma (w,s,\lambda )\sigma (z,t,\lambda )^{{\ast}}\pi _{ \lambda }(z,t))\right ).}$$

Let $(z_{0},w_{0}) \in \mathbb{G}$. Then by the weak-continuity of the mapping (9),

$$\displaystyle{A_{z,t}^{\lambda }(w,s) \rightarrow A_{ z,t}^{\lambda }(z_{ 0},t_{0})}$$

as (w, s) → (z ₀, t ₀). Moreover, by (8), there exits C > 0 such that

$$\displaystyle{\vert A_{z,t}^{\lambda }(w,s)\vert \leq C\|\sigma (z,t,\lambda )\|_{ HS}}$$

Therefore, by (7) and Lebesgue’s dominated convergence theorem,

$$\displaystyle{\left (T_{\sigma }f_{z,t}\right )(w,s) \rightarrow \left (T_{\sigma }f_{z,t}\right )(z_{0},t_{0})}$$

as (w, s) → (z ₀, t ₀). Therefore T _σ f _z, t is continuous on $\mathbb{G}$ and since by the assumption of the proposition T _σ f _z, t  = 0 almost every where, hence

$$\displaystyle{\left (T_{\sigma }f_{z,t}\right )(z,t) = 0.}$$

But

$$\displaystyle\begin{array}{rcl} \left (T_{\sigma }f_{z,t}\right )(z,t)& =& \int _{\mathbb{R}^{m}}tr\left (\pi _{\lambda }(z,t)^{{\ast}}\sigma (z,t,\lambda )\sigma (z,t,\lambda )^{{\ast}}\pi _{ \lambda }(z,t)\right )\,d\mu (\lambda ) {}\\ & =& \int _{\mathbb{R}^{m}}tr\left (\sigma (z,t,\lambda )^{{\ast}}\sigma (z,t,\lambda )\right )\,d\mu (\lambda ) {}\\ & =& \int _{\mathbb{R}^{m}}\|\sigma (z,t,\lambda )\|_{HS}^{2}\,d\mu (\lambda ) = 0 {}\\ \end{array}$$

Hence, ∥ σ(z, t, λ) ∥  _HS  = 0 for almost all $\lambda \in \mathbb{R}^{m}{}^{{\ast}}$ and therefore,

$$\displaystyle{\sigma (z,t,\lambda ) = 0}$$

for almost all $(z,t,\lambda ) \in \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}}$ □ 

The following theorem gives necessary and sufficient conditions on a symbol σ for $T_{\sigma } : L^{2}(\mathbb{G}) \rightarrow L^{2}(\mathbb{G})$ to be a Hilbert-Schmidt operator.

Theorem 5.2.

Let $\sigma : \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}}\rightarrow HS\left (L^{2}(\mathbb{R}^{n})\right )$ be a symbol satisfying the hypothesis of Proposition  5.1 . Then $T_{\sigma } : L^{2}(\mathbb{G}) \rightarrow L^{2}(\mathbb{G})$ is a Hilbert-Schmidt operator if and only if

$$\displaystyle{\sigma (z,t,\lambda ) =\pi _{\lambda }(z,t)W_{(\alpha (z,t)^{-\lambda })^{\wedge }}^{\lambda },\quad (z,t,\lambda ) \in \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}},}$$

where $\alpha : \mathbb{G} \rightarrow L^{2}(\mathbb{G})$ is weakly continuous mapping for which

$$\displaystyle\begin{array}{rcl} & & \quad \quad \int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\|\alpha (z,t)\|_{L^{2}(\mathbb{G})}^{2}\,dz\,dt <\infty , {}\\ & & \sup _{(z,t,\lambda )\in \mathbb{G}\times \mathbb{R}^{m}{}^{{\ast}}}\vert \lambda \vert ^{-n/2}\|(\alpha (z,t))^{-\lambda }\|_{ L^{2}(\mathbb{R}^{2n})} <\infty {}\\ \end{array}$$

and

$$\displaystyle{\int _{\mathbb{R}^{m}}\vert \lambda \vert ^{n/2}\|(\alpha (z,t))^{-\lambda }\|_{ L^{2}(\mathbb{R}^{2n})}\,d\lambda <\infty .}$$

Proof.

We first prove the sufficiently. Let $f \in \mathcal{S}(\mathbb{G})$. Then by Proposition 4.1,

$$\displaystyle{\left (T_{\sigma }f\right )(z,t) = \vert \lambda \vert ^{n}(2\pi )^{-m/2}\int _{ \mathbb{R}^{m}}tr\left (W_{(\alpha (z,t))^{-\lambda })^{\wedge }}^{\lambda }W_{(\,f^{\lambda })^{\vee }}^{\lambda }\right )\,d\lambda .}$$

By Proposition 2.3 and the trace formula in [5], we get

$$\displaystyle\begin{array}{rcl} & & tr\left (W_{(\alpha (z,t)^{-\lambda })^{\wedge }}^{\lambda }W_{(\,f^{\lambda })^{\vee }}^{\lambda }\right ) {}\\ & =& (2\pi )^{-n}\int _{ \mathbb{R}^{2n}}(\alpha (z,t)^{-\lambda })^{\wedge }(B_{\lambda }x,\xi )\ (\,f^{\lambda })^{\vee }(B_{\lambda }x,\xi )\,dx\,d\xi {}\\ & =& (2\pi )^{-n}\vert \lambda \vert ^{-n}\int _{ \mathbb{R}^{2n}}(\alpha (z,t)^{-\lambda })^{\wedge }(x,\xi )\ (\,f^{\lambda })^{\vee }(x,\xi )\,dx\,d\xi {}\\ & =& (2\pi )^{-n}\vert \lambda \vert ^{-n}\int _{ \mathbb{R}^{2n}}(\alpha (z,t)^{-\lambda })(z')\ (\,f^{\lambda })(z')\,dz'. {}\\ \end{array}$$

Hence,

$$\displaystyle\begin{array}{rcl} & & \left (T_{\sigma }f\right )(z,t) = (2\pi )^{-(m+2n)/2}\int _{ \mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}(\alpha (z,t)^{-\lambda })(z')\ (\,f^{\lambda })(z')\,dz'\,d\lambda {}\\ & & = (2\pi )^{-(m+2n)/2}\int _{ \mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\alpha (z,t)(z',\lambda )f(z',\lambda )\,dz'\,d\lambda . {}\\ \end{array}$$

So, the kernel of T _σ is a function on $\mathbb{R}^{2n+m} \times \mathbb{R}^{2n+m}$ given by

$$\displaystyle{ k(z,t,z',t') = (2\pi )^{-(m+2n)/2}\alpha (z,t)(z',\lambda ),\quad (z,t),(z',t') \in \mathbb{R}^{2n+m}. }$$

(10)

Therefore,

$$\displaystyle\begin{array}{rcl} & & \int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\vert k(z,t,z',\lambda )\vert ^{2}\,dz\,dt\,dz'\,d\lambda {}\\ & & = (2\pi )^{-(2n+m)}\int _{ \mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\vert \alpha (z,t)(z',\lambda )\vert ^{2}\,dz\,dt\,dz'\,d\lambda {}\\ & & = (2\pi )^{-(2n+m)}\int _{ \mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\|\alpha (z,t)\|_{L^{2}(\mathbb{G})}^{2}\,dz\,dt <\infty . {}\\ \end{array}$$

Thus, T _σ is a Hilbert-Schmidt operator. Conversely, suppose that $T_{\sigma } : L^{2}(\mathbb{G}) \rightarrow L^{2}(\mathbb{G})$ is a Hilbert Schmidt operator. Then there exists a function k in $L^{2}(\mathbb{R}^{2n+m} \times \mathbb{R}^{2n+m})$ such that

$$\displaystyle{\left (T_{\sigma }f\right )(z,t) =\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}k(z,t,z',\lambda )f(z',\lambda )\,dz'\,d\lambda ,\quad (z,t) \in \mathbb{G},}$$

for all $f \in L^{2}(\mathbb{G})$. We define $\alpha : \mathbb{G} \rightarrow L^{2}(\mathbb{G})$ by

$$\displaystyle{\alpha (z,t)(z',\lambda ) = (2\pi )^{(m+2n)/2}k(z,t,z',\lambda ).}$$

Then reversing the argument in the proof of the sufficiency and using Proposition 5.1, we have

$$\displaystyle{\sigma (z,t,\lambda ) =\pi _{\lambda }(z,t)W_{(\alpha (z,t)^{-\lambda })^{\wedge }}^{\lambda },\quad (z,t,\lambda ) \in \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}}.}$$

 □ 

Corollary 5.3.

Let $\beta \in L^{2}(\mathbb{G} \times \mathbb{G})$ be such that

$$\displaystyle{\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\vert \beta (z,t,z,t)\vert \,dz\,dt <\infty .}$$

Let

$$\displaystyle{\sigma (z,t,\lambda ) =\pi _{\lambda }(z,t)W_{(\alpha (z,t)^{-\lambda })^{\wedge }}^{\lambda },\quad (z,t,\lambda ) \in \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}},}$$

where

$$\displaystyle{\alpha (z,t)(z',\lambda ) =\beta (z,t,z',\lambda ),\quad (z,t),(z',\lambda ) \in \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}}.}$$

Then $T_{\sigma } : L^{2}(\mathbb{G}) \rightarrow L^{2}(\mathbb{G})$ is a trace class operator and

$$\displaystyle{tr(T_{\sigma }) = (2\pi )^{-(2n+m)}\int _{ \mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\beta (z,t,z,t)\,dz\,dt.}$$

Corollary 5.3 follows from the formula (10) on the kernel of the pseudo-differential operator in the proof of the preceding theorem.

Theorem 5.4.

Let $\sigma : \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}}\rightarrow HS\left (L^{2}(\mathbb{R}^{n})\right )$ be a symbol satisfying the hypothesis of Proposition  5.1 . Then $T_{\sigma } : L^{2}(\mathbb{G}) \rightarrow L^{2}(\mathbb{G})$ is a trace class operator if and only if

$$\displaystyle{\sigma (z,t,\lambda ) =\pi _{\lambda }(z,t)W_{(\alpha (z,t)^{-\lambda })^{\wedge }}^{\lambda },\quad (z,t,\lambda ) \in \mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}},}$$

where $\alpha : \mathbb{G} \rightarrow L^{2}(\mathbb{G})$ is a mapping such that the conditions of Theorem  5.2 are satisfied and

$$\displaystyle{\alpha (z,t)(z',\lambda ) =\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\alpha _{1}(z,t)(w,s)\alpha _{2}(w,s)(z',\lambda )\,dw\,ds}$$

for all (z,t) and (z′,λ) in $\mathbb{G} \times \mathbb{R}^{m}{}^{{\ast}}$ , where $\alpha _{1} : \mathbb{G} \rightarrow L^{2}(\mathbb{G})$ and $\alpha _{2} : \mathbb{G} \rightarrow L^{2}(\mathbb{G})$ are such that

$$\displaystyle{\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\|\alpha _{j}(z,t)\|_{L^{2}(\mathbb{G})}^{2}\,dz\,dt <\infty ,\quad j = 1,2.}$$

Moreover, the trace of T _σ is given by

$$\displaystyle\begin{array}{rcl} tr(T_{\sigma })& =& \int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\alpha (z,t)(z,t)\,dz\,dt {}\\ & =& \int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\int _{\mathbb{R}^{m}}\int _{\mathbb{R}^{2n}}\alpha _{1}(z,t)(w,s)\alpha _{2}(w,s)(z,t)\,dw\,ds\,dz\,dt. {}\\ \end{array}$$

Theorem 5.4 follows from Theorem 5.2 and the fact that every trace class operator is a product of two Hilbert-Schmidt operators.