2 Weyl transform

Recall that if $X \in S^1({\mathcal {H}})$ , the Fourier–Wigner transform of X is the function on ${\mathbb {R}}^{2n}$ defined by

$$ \begin{align*} \alpha(X)(x,y)= {\mathrm{tr}}(\rho(-x,-y,1)X) \end{align*} $$

and if $f\in L^1({\mathbb {R}}^{2n})$ , the Weyl transform of f is the operator $W(f) \in {\mathcal {B}}({\mathcal {H}})$ defined by

$$ \begin{align*} W(f) = \int_{{\mathbb{R}}^n} \int_{{\mathbb{R}}^n} f(x,y) \rho(x,y,1) \,dx \,dy, \end{align*} $$

and that these are mutual inverses (see [Reference Folland2]).

We now discuss the relation between the quantum translates and the Weyl transform (see [Reference Thangavelu11] for a detailed discussion). Recall that the Weyl correspondence of a function f on ${\mathbb {R}}^{2n}$ is defined by

$$ \begin{align*} \operatorname{Weyl}(f)= W(\mathcal{F}^{-1}(f)), \end{align*} $$

that is, the Weyl transform of the inverse Fourier transform of f. It can be shown that

$$ \begin{align*} \operatorname{Weyl}(T_{(x,y)}f)= (x,y)\cdot \operatorname{Weyl}(f), \quad (x,y)\in {\mathbb{R}}^{2n}. \end{align*} $$

The definition of the Weyl transform has been extended to tempered distributions (see, for example, [Reference Keyl, Kiukas and Werner4, Reference Maillard5]) and we briefly recall this now. Let ${\mathcal {S}}({\mathbb {R}}^{2n})$ denote the Schwartz class functions on ${\mathbb {R}}^{2n}$ and let ${\mathcal {S}}'({\mathbb {R}}^{2n})$ denote the space of tempered distributions on ${\mathcal {S}}({\mathbb {R}}^{2n})$ . If $k \in {\mathcal {S}}({\mathbb {R}}^{2n})$ , the integral operator $X \in {\mathcal {B}}({\mathcal {H}})$ defined by

$$ \begin{align*} (X\varphi)(x) = \int_{{\mathbb{R}}^n} k(x,y) \varphi(y) \, dy \end{align*} $$

is called a Schwartz operator on ${\mathcal {H}}$ . The class of Schwartz operators on ${\mathcal {H}}$ is denoted by ${\mathcal {S}}({\mathcal {H}})$ and is a Fréchet space with respect to an appropriate sequence of seminorms. Let ${\mathcal {S}}'({\mathcal {H}})$ denote the topological dual of ${\mathcal {S}}({\mathcal {H}})$ . It is well known that the Weyl transform is a topological isomorphism from ${\mathcal {S}}({\mathbb {R}}^{2n})$ to ${\mathcal {S}}({\mathcal {H}})$ . If $T \in {\mathcal {S}}'({\mathbb {R}}^{2n})$ , we define its Weyl transform $W(T) \in {\mathcal {S}}'({\mathcal {H}})$ by

$$ \begin{align*} W(T)(Y)=T(\alpha(Y)^\sim), \quad Y\in {\mathcal{S}}({\mathcal{H}}), \end{align*} $$

where $\alpha (Y)^\sim (x,y)=\alpha (Y)(-x,-y)$ . The distributional Weyl transform is a topological isomorphism from ${\mathcal {S}}'({\mathbb {R}}^{2n})$ to ${\mathcal {S}}'({\mathcal {H}})$ , and its inverse is the distributional Fourier–Wigner transform, which is defined by

$$ \begin{align*} \alpha(X)(\varphi) = X(W(\varphi^\sim)), \quad X\in{\mathcal{S}}'({\mathcal{H}}), \varphi\in{\mathcal{S}}({\mathbb{R}}^{2n}). \end{align*} $$

The following result was proved in [Reference Mishra and Vemuri7].

Theorem 2.1. Let T be a compactly supported distribution on ${\mathbb {R}}^{2n}$ . Let $\widehat {T}$ denote the Fourier transform of T. For $1 \leq p \leq \infty $ , $W(T) \in S^p({\mathcal {H}})$ if and only if $\widehat {T} \in L^p({\mathbb {R}}^{2n})$ . Moreover, if K is a compact set in ${\mathbb {R}}^{2n}$ , then there exists a constant $C_K$ such that

$$ \begin{align*} C_K^{-1} \|\widehat{T}\|_p \leq \| {W(T)}_{S^p}\| \le C_K \|\widehat{T}\|_p, \end{align*} $$

whenever ${\mathrm {supp}}(T)\subseteq K$ . Furthermore, $W(T)$ is compact if and only if $\widehat {T} \in \mathcal {C}_0({\mathbb {R}}^{2n})$ .

It is well known that if $\mu $ is a smooth measure on a hypersurface in ${\mathbb {R}}^n$ , $n \geq 2$ , whose Gaussian curvature is nonzero everywhere, then

$$ \begin{align*} |{\widehat{\mu}(\xi)}| \leq A |{\xi}|^{(1-n)/2}, \end{align*} $$

where A is a constant independent of $\xi $ , and hence, $\widehat {\mu } \in L^p({\mathbb {R}}^n)$ if $p> 2n/(n-1)$ (see, for example, [Reference Stein10, page 348, Theorem 1] and [Reference Hörmander3, Theorem 7.7.14]). The following is an immediate consequence of Theorem 2.1.

3 Proof of the main result

We prove Theorem 1.1 in this section.

Let $A \in S^p({\mathcal {H}})$ , $1 \leq p \leq \infty $ , and let $T_A \in {\mathcal {S}}'({\mathcal {H}})$ be the map

$$ \begin{align*} T_A(X) ={\mathrm{tr}}(XA). \end{align*} $$

The Fourier–Wigner transform $\alpha (A)$ is defined in a distributional sense, that is, $\alpha (A)= \alpha (T_A)$ .

Let $e: {\mathbb {R}}^{2n} \times {\mathbb {R}}^{2n} \to {\mathbb {C}}$ be the map

$$ \begin{align*} e((x',y'),(x,y))=e^{2\pi i(x'\cdot y- y'\cdot x)}. \end{align*} $$

Let $(x_1,y_1), \ldots , (x_k,y_k)$ be distinct elements of ${\mathbb {R}}^{2n}$ and let $c_1,\ldots , c_k$ be nonzero scalars. Define the difference operator D on $S^p({\mathcal {H}})$ , $1\leq p\leq \infty $ , by

$$ \begin{align*} DA = c_1 (x_1,y_1) \cdot A + \cdots + c_k (x_k,y_k) \cdot A, \quad A\in S^p({\mathcal{H}}). \end{align*} $$

Let $\psi \in {\mathcal {S}}({\mathbb {R}}^{2n})$ . Then,

$$ \begin{align*} \alpha&((x_1,y_1) \cdot A)(\psi) = T_{(x_1,y_1) \cdot A} (W({\psi}^\sim))\\ & = {\mathrm{tr}}(W({\psi}^\sim)(x_1,y_1) \cdot A)\\ & = {\mathrm{tr}}(W({\psi}^\sim)\rho (x_1,y_1,1) A {\rho (x_1,y_1,1)}^{-1})\\ & = {\mathrm{tr}} \bigg(\!\iint \psi(-x,-y) \rho(x,y,1) \rho (x_1,y_1,1) A {\rho (x_1,y_1,1)}^{-1}\,dx \,dy \bigg)\\ & = {\mathrm{tr}} \bigg(\!\iint e((x,y),(x_1,y_1)) \psi(-x,-y)\rho (x_1,y_1,1) \rho(x,y,1) A {\rho (x_1,y_1,1)}^{-1}\,dx \,dy \bigg)\\ & = {\mathrm{tr}} \bigg(\!\iint e((x,y),(x_1,y_1)) \psi(-x,-y) \rho(x,y,1) A \,dx \,dy \bigg)\\ & = {\mathrm{tr}} (W(e((x_1,y_1), \cdot)\psi)^\sim A)\\ & = \alpha(T_A)(e((x_1,y_1),\cdot) \psi)\\ & = e((x_1,y_1),\cdot) \alpha(A)(\psi). \end{align*} $$

Therefore,

(3.1) $$ \begin{align} \alpha(DA)=(c_1 e((x_1,y_1),\cdot)+ \cdots + c_k e((x_k,y_k),\cdot)) \alpha(A). \end{align} $$

The trigonometric polynomial

$$ \begin{align*} c_1 e((x_1,y_1),\cdot) + \cdots + c_k e((x_k,y_k),\cdot) \end{align*} $$

is the characteristic trigonometric polynomial of the difference equation $DA = 0$ .

We will first give the proof of Theorem 1.1 for the case $p \in [1,2]$ .

Proof. Since $S^p({\mathcal {H}}) \subseteq S^2({\mathcal {H}})$ for $1\le p \le 2$ , it suffices to prove the result for $p=2$ .

Let $c_1,\ldots , c_k \in {\mathbb {C}}$ be such that

$$ \begin{align*} DA= c_1 (x_1,y_1) \cdot A + \cdots + c_k (x_k,y_k) \cdot A = 0. \end{align*} $$

By the Plancherel theorem for W, we may write $A=W(f)$ for some nonzero $f\in L^2({\mathbb {R}}^{2n})$ . By (3.1),

$$ \begin{align*} DA = W((c_1 e((x_1,y_1),\cdot)+ \cdots + c_k e((x_k,y_k),\cdot)) f). \end{align*} $$

Therefore, $(c_1 e((x_1,y_1),\cdot )+ \cdots + c_k e((x_k,y_k),\cdot )) f=0$ . Since f is a nonzero function in $L^2({\mathbb {R}}^{2n})$ , there exists $\varepsilon>0$ such that $\lvert f\rvert> \varepsilon $ on a set of positive measure. Therefore, the analytic function

$$ \begin{align*} c_1 e((x_1,y_1),(x,y))+ \cdots + c_k e((x_k,y_k),(x,y)) = 0 \end{align*} $$

on a set of positive measure and thus it is identically zero. Since

$$ \begin{align*} \{e((x_1,y_1),\cdot), \ldots, e((x_k,y_k),\cdot)\} \end{align*} $$

is linearly independent, it follows that $c_1 = \cdots = c_k = 0$ .

Recall that the symplectic Fourier transform of a function $f\in L^1({\mathbb {R}}^{2n})$ is the function on ${\mathbb {R}}^{2n}$ defined by

$$ \begin{align*} \breve{f}(\xi, \eta)= \int_{{\mathbb{R}}^n}\int_{{\mathbb{R}}^n} f(x,y) e^{2\pi i(\xi \cdot y - \eta \cdot x)} \,dx\,dy. \end{align*} $$

More generally (see [Reference Maillard5]), if T is a tempered distribution on ${\mathbb {R}}^{2n}$ , the symplectic Fourier transform of T is given by

$$ \begin{align*} \breve{T}(\varphi)= T((\breve\varphi)^\sim), \quad \varphi\in \mathcal{S}({\mathbb{R}}^{2n}). \end{align*} $$

Let $g(x,y)= e^{-{\pi }(\lvert x\rvert ^2+\lvert y\rvert ^2)/2}$ , $(x,y) \in {\mathbb {R}}^{2n}$ . For $X \in {\mathcal {S}}'({\mathcal {H}})$ , let $\beta (X)=g\alpha (X)$ and $\breve {\beta }(X)$ denote the symplectic Fourier transform of $\beta (X)$ . It was shown in [Reference Mishra and Vemuri7] that $\breve {\beta }$ is a bounded linear map from $S^p({\mathcal {H}})$ to $L^p({\mathbb {R}}^{2n})$ , $1 \leq p \leq \infty $ . We will use this fact in the proof of the main result.

Rosenblatt proved that the linear translates of a nonzero function in $L^p({\mathbb {R}}^n)$ , $p= 2n/(n-1)$ , are linearly independent by using the following two results (see [Reference Rosenblatt9]).

Lemma 3.2. If $f\in L^p({\mathbb {R}}^n)$ , where $2<p \leq 2n/(n-1)$ and $n>1$ , then

$$ \begin{align*} \int_{\{R< \lvert x\rvert<2R\}}\lvert f(x)\rvert^2\, dx= o(R) \quad \text{as}\ R \to \infty. \end{align*} $$

Theorem 3.3. Let u be a tempered distribution supported on a real analytic set K of codimension $k>0$ and assume that $\hat {u}\in L^2_{\operatorname {loc}}({\mathbb {R}}^n)$ . Assume that for every analytic manifold M contained in A and every $x_0\in M$ , there is an open cone $\Lambda \subset {\mathbb {R}}^n$ that contains some normal of M at $x_0$ and such that on $\Lambda _R=\{x\in \Lambda \mid R<\lvert x\rvert <2R\}$ ,

$$ \begin{align*} \lim_{R \to \infty}R^{-k}\int_{\Lambda_R} \lvert\hat{u}(x)\rvert^2\,dx=0. \end{align*} $$

Then, $u=0$ .

We will now use these results to prove that quantum translates of a nonzero operator in $S^p({\mathcal {H}})$ , where $2< p \leq 4n/(2n-1)$ , are linearly independent.

Let $A\in S^p({\mathcal {H}})$ , $2< p \leq 4n/(2n-1)$ , be a nonzero operator such that $DA=0$ . It follows that

(3.2) $$ \begin{align} 0= \alpha(DA) =(c_1 e((x_1,y_1),\cdot)+ \cdots + c_k e((x_k,y_k),\cdot))\alpha(A). \end{align} $$

Let $Z_k$ denote the zero set of the characteristic trigonometric polynomial

$$ \begin{align*}c_1 e((x_1,y_1),\cdot)+ \cdots + c_k e((x_k,y_k),\cdot).\end{align*} $$

Case 1: $Z_k$ is of positive measure. Since the analytic function $c_1 e((x_1,y_1),(x,y))+ \cdots + c_k e((x_k,y_k),(x,y))$ vanishes on a set of positive measure, it follows that it is identically zero. Since the set

$$ \begin{align*} \{e((x_1,y_1),\cdot), \ldots, e((x_k,y_k),\cdot)\} \end{align*} $$

is linearly independent, it follows that $c_1 = \cdots = c_k = 0$ . Therefore, the quantum translates of A are linearly independent.

Case 2: $Z_k$ is of measure zero. Then, $Z_k$ is an analytic submanifold of codimension one (see, for example, [Reference Whitney13]). By (3.2), $\alpha (A)$ is supported in $Z_k$ and hence, $\beta (A)$ is supported in $Z_k$ . Take $u=\beta (A)$ in Theorem 3.3. Since $A\in S^p({\mathcal {H}})$ , it follows that $\breve {u}=\breve {\beta }(A) \in L^p({\mathbb {R}}^{2n})$ . Therefore, $\hat {u}\in L^2_{\operatorname {loc}}({\mathbb {R}}^{2n})$ . By Lemma 3.2, u satisfies the growth condition required in Theorem 3.3 for every cone $\Lambda $ . Therefore, $u=0$ and so $\alpha (A)=0$ . It follows that $A=0$ .

4 Example

We will now find an explicit example of a nonzero operator $A \in S^p({\mathcal {H}})$ , $p>4n/(2n-1)$ , whose quantum translates are linearly dependent. The argument used here is similar to that given by Edgar and Rosenblatt in [Reference Edgar and Rosenblatt1, Example 2.4]. The same example was given in [Reference Mishra and Vemuri6] to prove that there exists a nonzero operator in $S^p({\mathcal {H}})$ , $p>n \geq 6$ , whose quantum translates are linearly dependent, and in [Reference Mishra and Vemuri8] to prove that there exists a nonzero compact operator on ${\mathcal {H}}$ whose quantum translates are linearly dependent.

Consider the difference equation given by

(4.1) $$ \begin{align} 2(2n-1) A = \sum_{j=1}^n [(e_j, \mathbf{0})\cdot A + (-e_j, \mathbf{0})\cdot A + (\mathbf{0}, e_j)\cdot A +(\mathbf{0}, -e_j)\cdot A], \end{align} $$

where $\mathbf {0}=(0, \ldots , 0) \in {\mathbb {R}}^n$ and $\{e_1, \ldots , e_n\}$ is the standard basis of ${\mathbb {R}}^n$ .

Then, the characteristic trigonometric polynomial

$$ \begin{align*} p(x_1, \ldots, x_n, y_1, \ldots, y_n) = 2(2n-1) - 2\sum_{j=1}^n (\cos(2 \pi x_j) + \cos(2\pi y_j)) \end{align*} $$

has a zero set which is a disjoint union of compact $(2n-1)$ -dimensional surfaces of positive Gaussian curvature. Let $S_{2n}$ be the connected component of the zero set containing the points with all coordinates zero except for one which is $\pm 1/4$ . Let $\sigma $ be the measure on $S_{2n}$ induced by the Lebesgue measure on ${\mathbb {R}}^{2n}$ . Then, by Theorem 2.2, $W(\sigma ) \in S^p({\mathcal {H}})$ for $p> 4n/(2n-1)$ .

It follows from (3.1) that $A=W(\sigma )$ is a nonzero solution of the difference equation (4.1). Hence, for $p> 4n/(2n-1)$ , there exists an operator in $S^p({\mathcal {H}})$ whose quantum translates are linearly dependent.