Connes integration formula for the noncommutative plane

TL;DR

This paper proves the Connes integration formula for the noncommutative Moyal plane, establishing a way to integrate functions using singular traces in noncommutative geometry.

Contribution

It extends Connes' integration formula to the noncommutative Moyal plane, providing a new tool for analysis in noncommutative geometry.

Findings

Established the Connes integration formula for the noncommutative plane

Demonstrated the use of singular traces in noncommutative integration

Provided a framework for future analysis on noncommutative spaces

Abstract

Our aim is to prove the integration formula on the noncommutative (Moyal) plane in terms of singular traces {\it a la} Connes.

Equations

$${\mathrm{Tr}}_{\omega }(M_f(1-\Delta {)}^{-\frac{d}{2}})={\int }_{M}fd\mathrm{vol},f\in C^{\infty }(M).$$

$$\phi (x(1-\Delta {)}^{-\frac{d}{2}})={\tau }_{\theta }(x)$$

$$\phi (x(1-\Delta {)}^{-\frac{d}{2}})=c_{\phi }{\tau }_{\theta }(x),x\in W^{d,1}({\mathbb{R}}_{\theta }^d),$$

$${\mathcal{L}}_p=\{T\in \mathcal{L}(H):\mathrm{Tr}(|T{|}^p)<\infty \}.$$

$$\|T\|_{p}=\big{(}{\rm Tr}(|T|^{p})\big{)}^{\frac{1}{p}},\quad T\in\mathcal{L}_{p}.$$

$${\mathcal{L}}_{p,\infty }=\{T\in \mathcal{L}(H):\mu (k,T)=O((k+1{)}^{-1/p})\}.$$

$$\Vert T{\Vert }_{p,\infty }=\underset{k\ge 0}{\sup }(k+1{)}^{1/p}\mu (k,T),T\in {\mathcal{L}}_{p,\infty }.$$

$$\phi (TS)=\phi (ST),T\in \mathcal{I},S\in \mathcal{L}(H).$$

$${\mathrm{Tr}}_{\omega }:A\to \underset{n\to \omega }{\lim }\frac{1}{\log (2+n)}\sum _{k=0}^n\mu (k,A),0\le A,$$

$$U(t+s)=\exp (-\frac{i}{2}⟨t,\theta s⟩)U(t)U(s),t,s\in {\mathbb{R}}^d,$$

$$(U(t)\xi )(u)=e^{-\frac{i}{2}⟨t,\theta u⟩}\xi (u-t),\xi \in L_2({\mathbb{R}}^d),u,t\in {\mathbb{R}}^d.$$

$$U_1(t_1)U_2(t_2)=e^{i\alpha t_1{t}_2}{U}_2(t_2)U_1(t_1),t_1,t_2\in \mathbb{R}.$$

$$L_{p}(\mathbb{R}^{d}_{\theta})=\Big{\{}x\in L_{\infty}(\mathbb{R}^{d}_{\theta}):\ \tau_{\theta}(|x|^{p})<\infty\Big{\}}.$$

$$x=\underset{N\to \infty }{\lim }\frac{1}{(2\pi {)}^{d/4}}{\int }_{[-N,N{]}^d}f(s)U(s)ds,$$

$$x=\mathrm{Op}(f)\stackrel{def}{=}\frac{1}{(2\pi {)}^{d/4}}{\int }_{{\mathbb{R}}^d}f(s)U(s)ds$$

$$(D_k\xi )(t)=t_k\xi (t),\xi \in L_2({\mathbb{R}}^d).$$

$$[D_k,U(s)]=s_{k}U(s),s\in {\mathbb{R}}^d.$$

$$e^{i⟨t,\nabla ⟩}U(s)e^{-i⟨t,\nabla ⟩}=e^{i⟨t,s⟩}U(s),s,t\in {\mathbb{R}}^d.$$

$${\partial }^{\alpha }x=[D_{{\alpha }_1},[D_{{\alpha }_2},\cdots ,[D_{{\alpha }_n},x]]].$$

$$\Vert x{\Vert }_{W^{m,p}}=\sum _{|\alpha |\le m}\Vert {\partial }^{\alpha }x{\Vert }_p,x\in W^{m,p}({\mathbb{R}}_{\theta }^d).$$

$$\Vert x(1-\Delta {)}^{-\frac{d+1}{2}}{\Vert }_1\le c_d\Vert x{\Vert }_{W^{d,1}}.$$

$$\Vert x(1-\Delta {)}^{-\frac{d}{2}}{\Vert }_{1,\infty }\le c_d\Vert x{\Vert }_{W^{d,1}}.$$

$$T_{\varphi }:{\int }_{{\mathbb{R}}^d}f(s)U(s)ds\to {\int }_{{\mathbb{R}}^d}f(s)\varphi (s)U(s)ds,f\in L_2({\mathbb{R}}^d).$$

$$T_{\varphi }x={\int }_{{\mathbb{R}}^d}(\mathcal{F}\varphi )(u)U(-{\theta }^{-1}u)xU({\theta }^{-1}u)du,x\in L_2({\mathbb{R}}_{\theta }^d).$$

$$x={\int }_{{\mathbb{R}}^d}f(s)U(s)ds,f\in \mathcal{S}({\mathbb{R}}^d).$$

$$\varphi (s)={\int }_{{\mathbb{R}}^d}(\mathcal{F}\varphi )(u)e^{i⟨u,s⟩}du,s\in {\mathbb{R}}^d.$$

$$T_{\varphi }x={\iint }_{{\mathbb{R}}^d\times {\mathbb{R}}^d}f(s)(\mathcal{F}\varphi )(u)e^{i⟨u,s⟩}U(s)dsdu.$$

$$e^{i⟨u,s⟩}U(s)=U(-{\theta }^{-1}u)U(s)U({\theta }^{-1}u).$$

$$T_{\phi}x=\int_{\mathbb{R}^{d}}(\mathcal{F}\phi)(u)\Big{(}\int_{\mathbb{R}^{d}}f(s)U(-\theta^{-1}u)U(s)U(\theta^{-1}u)ds\Big{)}du.$$

$${\int }_{{\mathbb{R}}^d}f(s)U(-{\theta }^{-1}u)U(s)U({\theta }^{-1}u)ds=U(-{\theta }^{-1}u)xU({\theta }^{-1}u).$$

Full text

1. Introduction

Let $M$ be a compact Riemannian manifold. The following formula can be found in p. 34 in [1] and in Corollary 7.21 in [9].

[TABLE]

Here, $M_{f}$ is the multiplication operator, $\Delta$ is the Hodge-Laplacian operator on $L_{2}(M,{\rm vol})$ and ${\rm Tr}_{\omega}$ is the Dixmier trace on the ideal $\mathcal{L}_{1,\infty}$ (see Section 2). Also, Corollary 7.22 in [9] wrongly extends this result to $f\in L_{1}(M,{\rm vol})$ (in fact, $f\in L_{2}(M,{\rm vol})$ is the necessary and sufficient condition for this formula to hold; see [14] or the book [15] for detailed proofs).

According to [1], formula (1) “led Connes to introduce the Dixmier trace as the correct operator theoretical substitute for integration of infinitesimals of order one in non-commutative geometry.” It appears suitable to refer to (1) and similar results as the “Connes Integration Formula”.

Compactness of the (resolvent of the) Hodge-Dirac operator plays a crucial role in the proofs of Connes Integration Formula for unital spectral triples (see [1] and [9]). For non-unital spectral triples (including non-compact manifolds), the proofs become radically harder. Even the case of the simplest non-compact manifold $\mathbb{R}^{d}$ required a substantial effort and the first reasonable answer was very recently given in [11] (see the book [15] for detailed proofs).

In this paper, we investigate the validity of Connes Integration Formula for the noncommutative (Moyal) plane $\mathbb{R}^{d}_{\theta}$ (here, $\theta$ is a non-degenerate antisymmetric matrix). Earlier attempts in this direction can be found in [8] (see Proposition 4.17 there), [2] and [3]. We substantially strengthen corresponding results from these papers and present a completely different approach to Connes Integration Formula. The novelty of our approach is in the consistent use of Cwikel estimates for the noncommutative plane (obtained in a recent paper [12]) — see Section 2.

Our main result is the following theorem.

Theorem 1.1**.**

If $x\in W^{d,1}(\mathbb{R}^{d}_{\theta}),$ then $x(1-\Delta)^{-\frac{d}{2}}\in\mathcal{L}_{1,\infty}$ and

[TABLE]

for every normalised continuous trace $\varphi$ on $\mathcal{L}_{1,\infty}.$

Here, $W^{d,1}(\mathbb{R}^{d}_{\theta})$ is a Sobolev space on $\mathbb{R}^{d}_{\theta}$ and $\tau_{\theta}$ is the faithful normal semifinite trace on $L_{\infty}(\mathbb{R}^{d}_{\theta}).$

Section 2 involves the preliminaries necessary to prove Theorem 1.1. In Section 3, we prove that

[TABLE]

for every normalised trace on $\mathcal{L}_{1,\infty}.$ In Section 4, we construct one particular $x\in W^{d,1}(\mathbb{R}^{d}_{\theta})$ such that $\varphi(x(1-\Delta)^{-\frac{d}{2}})$ does not depend on the choice of a normalised continuous trace $\varphi.$ The combination of these results yield Theorem 1.1.

2. Preliminaries

2.1. General notation

Fix throughout a separable infinite dimensional Hilbert space $H.$ We let $\mathcal{L}(H)$ denote the algebra of all bounded operators on $H.$ For a compact operator $T$ on $H,$ let $\mu(k,T)$ denote $k-$th largest singular value (these are the eigenvalues of $|T|$). The sequence $\mu(T)=\{\mu(k,T)\}_{k\geq 0}$ is referred to as to the singular value sequence of the operator $T.$ The standard trace on $\mathcal{L}(H)$ is denoted by ${\rm Tr}.$

Fix an orthonormal basis in $H$ (the particular choice of a basis is inessential). We identify the algebra $l_{\infty}$ of bounded sequences with the subalgebra of all diagonal operators with respect to the chosen basis. For a given sequence $\alpha\in l_{\infty},$ we denote the corresponding diagonal operator by ${\rm diag}(\alpha).$

2.2. Schatten ideals $\mathcal{L}_{p}$ and $\mathcal{L}_{p,\infty},$ $p>0$

For every $p>0,$ we set

[TABLE]

We set

[TABLE]

For every $p>0,$ $\|\cdot\|_{p}$ is a quasi-norm111A quasinorm satisfies the norm axioms, except that the triangle inequality is replaced by $||x+y||\leq K(||x||+||y||)$ for some uniform constant $K>1.$ and $(\mathcal{L}_{p},\|\cdot\|_{p})$ is a quasi-Banach space. For $p\geq 1,$ $\|\cdot\|_{p}$ is a norm. For $p<1,$ the space $(\mathcal{L}_{p},\|\cdot\|_{p})$ is not Banach — that is, its quasi-norm is not equivalent to any norm.

For a given $0<p\leq\infty,$ we let $\mathcal{L}_{p,\infty}$ denote the principal ideal in $\mathcal{L}(H)$ generated by the operator ${\rm diag}(\{(k+1)^{-\frac{1}{p}}\}_{k\geq 0}).$ Equivalently,

[TABLE]

We set

[TABLE]

For every $p>0,$ $\|\cdot\|_{p,\infty}$ is a quasi-norm and $(\mathcal{L}_{p,\infty},\|\cdot\|_{p,\infty})$ is a quasi-Banach space. For $p>1,$ $\|\cdot\|_{p,\infty}$ is equivalent to a (unitarily invariant Banach) norm. For $p\leq 1,$ the space $(\mathcal{L}_{p,\infty},\|\cdot\|_{p,\infty})$ is not Banach — that is, its quasi-norm is not equivalent to any norm. In [17], the Banach envelope of $\mathcal{L}_{1,\infty}$ was thoroughly investigated.

2.3. Traces on $\mathcal{L}_{1,\infty}$

Definition 2.1**.**

If $\mathcal{I}$ is an ideal in $\mathcal{L}(H),$ then a unitarily invariant linear functional $\varphi:\mathcal{I}\to\mathbb{C}$ is said to be a trace.

Since $U^{-1}TU-T=[U^{-1},TU]$ for all $T\in\mathcal{I}$ and for all unitaries $U\in\mathcal{L}(H),$ and since the unitaries span $\mathcal{L}(H),$ it follows that traces are precisely the linear functionals on $\mathcal{I}$ satisfying the condition

[TABLE]

The latter may be reinterpreted as the vanishing of the linear functional $\varphi$ on the commutator subspace which is denoted $[\mathcal{I},\mathcal{L}(H)]$ and defined to be the linear span of all commutators $[T,S]:\ T\in\mathcal{I},$ $S\in\mathcal{L}(H).$ It is shown in Lemma 5.2.2 in [15] that $\varphi(T_{1})=\varphi(T_{2})$ whenever $0\leq T_{1},T_{2}\in\mathcal{I}$ are such that the singular value sequences $\mu(T_{1})$ and $\mu(T_{2})$ coincide.

For $p>1,$ the ideal $\mathcal{L}_{p,\infty}$ does not admit a non-zero trace [7], while for $p=1,$ there exists a plethora of traces on $\mathcal{L}_{1,\infty}$ (see e.g. [18] or [15]). A standard example of a trace on $\mathcal{L}_{1,\infty}$ is a Dixmier trace introduced in [6] that we now explain.

Definition 2.2**.**

Let $\omega$ be a free ultrafilter on $\mathbb{Z}_{+}.$ The functional

[TABLE]

is finite and additive on the positive cone of $\mathcal{L}_{1,\infty}.$ Therefore, it extends to a trace on $\mathcal{L}_{1,\infty}.$ We call such traces Dixmier traces.

These traces clearly depend on the choice of the ultrafilter $\omega$ on $\mathbb{Z}_{+}.$ Using a slightly different definition, this notion of trace was applied by Connes [4] in noncommutative geometry.

An extensive discussion of traces, and more recent developments in the theory, may be found in [15] including a discussion of the following facts. We refer the reader to an alternative approach to the theory of traces on $\mathcal{L}_{1,\infty}$ suggested in [18] (based on the fundamental paper [16] by Pietsch).

1. (1)

All Dixmier traces on $\mathcal{L}_{1,\infty}$ are positive. 2. (2)

All positive traces on $\mathcal{L}_{1,\infty}$ are continuous in the quasi-norm topology. 3. (3)

There exist positive traces on $\mathcal{L}_{1,\infty}$ which are not Dixmier traces (see [18]). 4. (4)

There exist traces on $\mathcal{L}_{1,\infty}$ which fail to be continuous (see [15]).

Definition 2.3**.**

We say that an operator $A\in\mathcal{L}_{1,\infty}$ is measurable if $\varphi(A)$ does not depend on the choice of the continuous normalised trace $\varphi$ on $\mathcal{L}_{1,\infty}.$

2.4. Noncommutative plane: algebra

Each assertion in this subsection is rigorously established in Section 6 in [12].

Our approach to the noncommutative plane is to introduce the von Neumann algebra generated by a strongly continuous family of unitary operators $\{U(t)\}_{t\in\mathbb{R}^{d}}$, $d\in\mathbb{N},$ satisfying the commutation relation

[TABLE]

where $\theta$ is a fixed antisymmetric real $d\times d$ matrix. Namely, we set

[TABLE]

Definition 2.4**.**

Let $d\in\mathbb{N}$ and let $\theta$ be a fixed non-degenerate222A non-degenerate antisymmetric matrix is automatically of even order. antisymmetric real $d\times d$ matrix. The von Neumann subalgebra in $\mathcal{L}(L_{2}(\mathbb{R}^{d}))$ generated by $\{U(t)\}_{t\in\mathbb{R}^{d}}$, introduced in (3), is called the noncommutative plane and denoted by $L_{\infty}(\mathbb{R}^{d}_{\theta}).$

Example 2.5**.**

If $d=2,$ then $L_{\infty}(\mathbb{R}^{d}_{\theta})$ is generated by $2$ unitary groups $t\to U_{1}(t),$ $t\to U_{2}(t),$ $t\in\mathbb{R}$ satisfying the condition

[TABLE]

Here, $U_{1}(t_{1})=U((t_{1},0))$ and $U_{2}(t_{2})=U((0,t_{2})).$

The following assertion is well-known. In [12], a spatial isomorphism is constructed.

Theorem 2.6**.**

For every non-degenerate antisymmetric real matrix $\theta,$ the algebra $L_{\infty}(\mathbb{R}^{d}_{\theta})$ is isomorphic to $\mathcal{L}(L_{2}(\mathbb{R}^{\frac{d}{2}})).$

Having established the isomorphism between $r:L_{\infty}(\mathbb{R}^{d}_{\theta})\to\mathcal{L}(L_{2}(\mathbb{R}^{\frac{d}{2}}))$ we now equip $L_{\infty}(\mathbb{R}^{d}_{\theta})$ with a faithful normal semifinite trace $\tau_{\theta}={\rm Tr}\circ r.$

We can now define $L_{p}-$spaces on $L_{\infty}(\mathbb{R}^{d}_{\theta}).$

[TABLE]

Lemma 2.7**.**

*An operator $x\in L_{\infty}(\mathbb{R}^{d}_{\theta})$ is in $L_{2}(\mathbb{R}^{d}_{\theta})$ if and only if333To be precise,

$x=\lim_{N\to\infty}\frac{1}{(2\pi)^{d/4}}\int_{[-N,N]^{d}}f(s)U(s)ds,$

where the limit is taken in $L_{2}(\mathbb{R}^{d}_{\theta}).$ In what follows, we write the integral over $\mathbb{R}^{d}$ instead of the limit in order to lighten the notations. *

[TABLE]

for some unique $f\in L_{2}(\mathbb{R}^{d})$ with $\|x\|_{2}=\|f\|_{2}.$

Note that our picture is the Fourier dual of the one considered in [8]. More precisely, the paper [8] deals with operators of the form ${\rm Op}(\mathcal{F}f),$ where $f$ is Schwartz (in [8], these operators are written simply as $f$).

2.5. Noncommutative plane: calculus

Each assertion in this subsection is rigorously established in Section 6 in [12].

Let $D_{k},$ $1\leq k\leq d$ be multiplication operators on $L_{2}(\mathbb{R}^{d})$

[TABLE]

For brevity, we denote $\nabla=(D_{1},\cdots,D_{d}).$ For every $1\leq k\leq d,$ we have

[TABLE]

Moreover, we have

[TABLE]

If $[D_{k},x]\in\mathcal{L}(L_{2}(\mathbb{R}^{d}))$ for some $x\in L_{\infty}(\mathbb{R}^{d}_{\theta}),$ then $[D_{k},x]\in L_{\infty}(\mathbb{R}^{d}_{\theta}).$ This crucial fact allows us to introduce mixed partial derivative $\partial^{\alpha}x$ of $x\in L_{\infty}(\mathbb{R}^{d}_{\theta}).$

Definition 2.8**.**

Let $\alpha$ be a multiindex and let $x\in L_{\infty}(\mathbb{R}^{d}_{\theta}).$ If every repeated commutator $[D_{\alpha_{j}},[D_{\alpha_{j}+1},\cdots,[D_{\alpha_{n}},x]]],$ $1\leq j\leq n,$ is a bounded operator on $L_{2}(\mathbb{R}^{d}),$ then the mixed partial derivative $\partial^{\alpha}x$ of $x$ is defined as

[TABLE]

In this case, we have that $\partial^{\alpha}x\in L_{\infty}(\mathbb{R}_{\theta}^{d}).$ As usual, $\partial^{0}x=x.$

Therefore, we can introduce the Sobolev space $W^{m,p}(\mathbb{R}_{\theta}^{d})$ associated with the noncommutative plane in the following way.

Definition 2.9**.**

For $m\in\mathbb{Z}_{+}$ and $p\geq 1,$ the space $W^{m,p}(\mathbb{R}^{d}_{\theta})$ is the space of $x\in L_{p}(\mathbb{R}^{d}_{\theta})$ such that every partial derivative of $x$ up to order $m$ is also in $L_{p}(\mathbb{R}^{d}_{\theta}).$ This space is equipped with the norm,

[TABLE]

The following assertion is one of the main results in [12].

Theorem 2.10**.**

If $x\in W^{d,1}(\mathbb{R}^{d}_{\theta}),$ then

1. (a)

$x(1-\Delta)^{-\frac{d+1}{2}}\in\mathcal{L}_{1}$* and*

[TABLE] 2. (b)

$x(1-\Delta)^{-\frac{d}{2}}\in\mathcal{L}_{1,\infty}$* and*

[TABLE]

3. Integration formula modulo a constant factor

For every $\phi\in L_{\infty}(\mathbb{R}^{d}),$ we define a bounded operator $T_{\phi}:L_{2}(\mathbb{R}^{d}_{\theta})\to L_{2}(\mathbb{R}^{d}_{\theta})$ by the formula

[TABLE]

Lemma 3.1**.**

If $\phi$ is a Schwartz function, then $T_{\phi}:L_{1}(\mathbb{R}^{d}_{\theta})\to L_{1}(\mathbb{R}^{d}_{\theta}).$

Proof.

We claim that

[TABLE]

Since both sides above define bounded operators on $L_{2}(\mathbb{R}^{d}_{\theta})$ and since the set $\{{\rm Op}(f):\ f\mbox{ is Schwartz}\}$ is dense in $L_{2}(\mathbb{R}^{d}_{\theta}),$ it suffices to establish the claim for

[TABLE]

Using the inverse Fourier transform, we write

[TABLE]

Since both $f$ and $\mathcal{F}\phi$ are Schwartz functions, it follows that

[TABLE]

It follows from (2) that

[TABLE]

Therefore,

[TABLE]

Using the definition of $x,$ we obtain

[TABLE]

This proves the claim.

Now, we prove the assertion of the lemma as follows.

[TABLE]

∎

Lemma 3.2**.**

For every $x\in W^{d,1}(\mathbb{R}^{d}_{\theta}),$ the mapping

[TABLE]

is a continuous $W^{d,1}(\mathbb{R}^{d}_{\theta})-$valued function. Moreover,

[TABLE]

Proof.

It follows from Leibniz rule that

[TABLE]

[TABLE]

Iterating the latter inequality, we obtain

[TABLE]

Thus,

[TABLE]

[TABLE]

We now establish the continuity. For every $y\in\mathcal{L}_{1},$ the mapping

[TABLE]

is continuous in the $\mathcal{L}_{1}-$norm whenever the mapping $t\to V(t)$ is strongly continuous. Recall that $(L_{\infty}(\mathbb{R}^{d}_{\theta}),\tau_{\theta})$ is $*-$isomorphic (so that trace is preserved) to $(\mathcal{L}(L_{2}(\mathbb{R}^{\frac{d}{2}})),{\rm Tr}).$ Thus, the mapping

[TABLE]

is continuous in $L_{1}-$norm. This completes the proof. ∎

Lemma 3.3**.**

1. (a)

If $f$ is Schwartz, then ${\rm Op}(f)\in W^{d,1}(\mathbb{R}^{d}_{\theta}).$ 2. (b)

The set $\{{\rm Op}(f):\ f\mbox{ is Schwartz}\}$ is dense in $L_{1}(\mathbb{R}^{d}_{\theta}).$ In particular, $W^{d,1}(\mathbb{R}^{d}_{\theta})$ is dense in $L_{1}(\mathbb{R}^{d}_{\theta}).$

Proof.

There exists a sequence $\{e_{kl}\}_{k,l\geq 0}\subset L_{\infty}(\mathbb{R}^{d}_{\theta})$ such that

1. (i)

$e_{k_{1}l_{1}}e_{k_{2}l_{2}}=\delta_{l_{1},k_{2}}e_{k_{1}l_{2}}$ and $e_{kl}^{*}=e_{lk}.$ 2. (ii)

$\tau_{\theta}(e_{kk})=1.$ 3. (iii)

$\sum_{k\geq 0}e_{kk}=1$ in strong operator topology. 4. (iv)

for every $k,l\geq 0,$ there exists a Schwartz function $f_{kl}$ such that $e_{kl}={\rm Op}(f_{kl}).$

The existence of such a sequence is established in Lemma 2.4 in [8] (see also additional references therein). A particular formula for $f_{kl}$ can be found on p. 618 in [8] in terms of Laguerre polynomials.

We prove (a). Let $f$ be a Schwartz function. By Proposition 2.5 in [8], one can write $f$ as

[TABLE]

Thus,

[TABLE]

where the series converges in $L_{1}-$norm. Thus, ${\rm Op}(f)\in L_{1}(\mathbb{R}^{d}_{\theta}).$ Let $f_{\alpha}(t)=t^{\alpha}f(t),$ $t\in\mathbb{R}^{d}.$ By (4), $\partial^{\alpha}({\rm Op}(f))={\rm Op}(f_{\alpha}).$ Since $f_{\alpha}$ is also a Schwartz function, it follows that $\partial^{\alpha}({\rm Op}(f))\in L_{1}(\mathbb{R}^{d}_{\theta}).$ This proves (a).

To prove (b), note that, for every $x\in L_{1}(\mathbb{R}^{d}_{\theta}),$

[TABLE]

in $\mathcal{L}_{1}-$norm as $N\to\infty.$ Note that $e_{kk}xe_{ll}$ is a scalar multiple of $e_{kl}={\rm Op}(f_{kl}).$ Since a linear combination of Schwartz functions is again a Schwartz function, it follows that

[TABLE]

This proves (b). ∎

Lemma 3.4**.**

If $F$ is a continuous functional on $W^{d,1}(\mathbb{R}^{d}_{\theta})$ such that

[TABLE]

then $F=\tau_{\theta}$ (up to a constant factor).

Proof.

Let $T:W^{d,1}(\mathbb{R}^{d}_{\theta})\to W^{d,1}(\mathbb{R}^{d}_{\theta})$ be defined by setting

[TABLE]

The integral is understood as a Bochner integral of a continuous $W^{d,1}(\mathbb{R}^{d}_{\theta})-$valued function (the continuity and convergence of the integral follow from Lemma 3.2).

For every $x\in W^{d,1}(\mathbb{R}^{d}_{\theta}),$ we have

[TABLE]

Thus,

[TABLE]

We claim that $\|Tx\|_{W^{d,1}}\leq c_{d}\|x\|_{1}$ for every $x\in W^{d,1}(\mathbb{R}^{d}_{\theta}).$ To see this, let

[TABLE]

If, in the proof of Lemma 3.1, we select $\phi(t)=e^{-\frac{1}{2}|t|^{2}},$ $t\in\mathbb{R}^{d},$ then the argument given there yields

[TABLE]

By (4), we have

[TABLE]

Let $\phi_{\alpha}(s)=s^{\alpha}e^{-\frac{1}{2}|s|^{2}},$ $s\in\mathbb{R}^{d}.$ We have that $\partial^{\alpha}\circ T=T_{\phi_{\alpha}}.$ By Lemma 3.1, $T_{\phi_{\alpha}}:L_{1}(\mathbb{R}^{d}_{\theta})\to L_{1}(\mathbb{R}^{d}_{\theta})$ is a bounded operator. This proves the claim.

For every $x\in W^{d,1}(\mathbb{R}^{d}_{\theta}),$ we have

[TABLE]

Thus, a functional $F$ on $W^{d,1}(\mathbb{R}^{d}_{\theta})$ is bounded in $\|\cdot\|_{1}-$norm. By the Hahn-Banach Theorem, $F$ extends to a bounded functional on $L_{1}(\mathbb{R}^{d}_{\theta}).$ Hence, there exists $y\in L_{\infty}(\mathbb{R}^{d}_{\theta})$ such that

[TABLE]

Clearly,

[TABLE]

Comparing the last 2 equalities, we obtain

[TABLE]

Since $W^{d,1}(\mathbb{R}^{d}_{\theta})$ is dense in $L_{1}(\mathbb{R}^{d}_{\theta}),$ it follows that $y=U(t)yU(-t)$ for every $t\in\mathbb{R}^{d}.$ In other words, $y$ commutes with every $U(t)$ and, therefore, with every element in $L_{\infty}(\mathbb{R}^{d}_{\theta}).$ Since $L_{\infty}(\mathbb{R}^{d}_{\theta})$ is a factor (see Theorem 2.6), it follows that $y$ is a scalar operator. This completes the proof. ∎

The following proposition is a light version of Theorem 1.1.

Proposition 3.5**.**

If $x\in W^{d,1}(\mathbb{R}^{d}_{\theta}),$ then $x(1-\Delta)^{-\frac{d}{2}}\in\mathcal{L}_{1,\infty}$ and

[TABLE]

for every continuous trace on $\mathcal{L}_{1,\infty}$ and for some constant $c_{\varphi}.$

Proof.

By Theorem 2.10 (b), the functional

[TABLE]

is a well defined bounded linear functional on $W^{d,1}(\mathbb{R}^{d}_{\theta}).$

Since $\varphi$ is unitarily invariant, it follows that

[TABLE]

By the Spectral Theorem, we have

[TABLE]

and so

[TABLE]

For every $s\in\mathbb{R}^{d},$ we have (see (5))

[TABLE]

On the other hand, it follows from (2) that

[TABLE]

Comparing preceding equalities, we arrive at

[TABLE]

It follows that

[TABLE]

Combining the preceding paragraphs, we obtain

[TABLE]

Applying Lemma 3.4 to our functional $F,$ we conclude the argument. ∎

4. Proof of measurability

Lemma 4.1**.**

If $K\in W^{2d+2,1}([0,1]^{d}\times[0,1]^{d})$ and if $T:L_{2}((0,1)^{d})\to L_{2}((0,1)^{d})$ is an integral operator with integral kernel $K,$ then $T\in\mathcal{L}_{1}$ and $\|T\|_{1}\leq c_{d}\|K\|_{W^{2d+2,1}}.$

Proof.

Let $K\in W^{2d+2,1}([-\pi,\pi]^{d}\times[-\pi,\pi]^{d})$ be an extension of $K$ such that

[TABLE]

and such that $K$ vanishes on and near the boundary. Thus, $K\in W^{2d+2,1}(\mathbb{T}^{d}\times\mathbb{T}^{d}).$ Let $S:L_{2}(\mathbb{T}^{d})\to L_{2}(\mathbb{T}^{d})$ be an integral operator with integral kernel $K.$ We have $T=M_{\chi_{(0,1)^{d}}}SM_{\chi_{(0,1)^{d}}}.$ Thus, $\|T\|_{1}\leq\|S\|_{1}.$

Let us write Fourier series

[TABLE]

Set

[TABLE]

It is an integral operator on $L_{2}(\mathbb{T}^{d})$ with the integral kernel $(t,s)\to e_{m_{1}}(t)e_{m_{2}}(s).$ Hence,

[TABLE]

By triangle inequality, we have

[TABLE]

[TABLE]

Observe that $(1+|m_{1}|^{2}+|m_{2}|^{2})^{d+1}c_{m_{1},m_{2}}$ is the $(m_{1},m_{2})-$th Fourier coefficient of the function $(1-\Delta_{\mathbb{T}^{2d}})^{d+1}(K)$ (here, $\Delta_{\mathbb{T}^{2d}}$ is the Laplacian on the torus $\mathbb{T}^{2d}$). Taking into account that Fourier coefficients do not exceed the $L_{1}-$norm, we infer that

[TABLE]

Here, the last inequality follows from the definition of a Sobolev space. ∎

In what follows, we consider the tensor product of 2 bounded operators on a Hilbert space $H$ as a bounded operator on the Hilbert space $H\bar{\otimes}H.$

Lemma 4.2**.**

If $T\in\mathcal{L}_{1,\infty}$ and $S\in\mathcal{L}_{1},$ then $S\otimes T\in\mathcal{L}_{1,\infty}$ and

[TABLE]

for every continuous trace $\varphi$ on $\mathcal{L}_{1,\infty}.$

Proof.

Firstly, we show that $S\otimes T\in\mathcal{L}_{1,\infty}.$ Let $z(t)=t^{-1},$ $t>0.$ By definition, we have $\mu(T)\leq\|T\|_{1,\infty}z.$ The crucial fact that $\mu(S\otimes z)=\|S\|_{1}z$ is proved on p. 211 in [13]. Thus,

[TABLE]

We now turn to the proof of (6). If $S$ is a rank one projection, then there is nothing to prove. If $S$ is a positive finite rank operator, then the assertion follows by linearity. If $S$ is an arbitrary finite rank operator, then the assertion again follows by linearity.

Let $S\in\mathcal{L}_{1}$ be arbitrary. Fix $\epsilon>0$ and choose $S_{1},S_{2}\in\mathcal{L}_{1}$ such that $S=S_{1}+S_{2},$ $S_{1}$ is finite rank and $\|S_{2}\|_{1}\leq\epsilon.$ Clearly,

[TABLE]

[TABLE]

By the preceding paragraph, the summand in the first bracket vanishes. Thus,

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Hence,

[TABLE]

[TABLE]

By the norm estimate in the first paragraph and by the assumption on $S_{2},$ we have

[TABLE]

Since $\epsilon>0$ is arbitrarily small, the assertion follows. ∎

In the following lemma, we consider the direct sum of bounded operators on a Hilbert space $H$ as a bounded operator on a Hilbert space $\bigoplus_{m\geq 0}H.$

Lemma 4.3**.**

If the operators $\{T_{m}\}_{m\geq 0}$ are pairwise orthogonal, i.e. $T_{m_{1}}T_{m_{2}}=T_{m_{1}}^{*}T_{m_{2}}=0$ for $m_{1}\neq m_{2},$ then $\sum_{m\geq 0}T_{m}$ is unitarily equivalent444To be pedantic, $\sum_{m\geq 0}T_{m}$ is unitarily equivalent to the direct sum $\bigoplus_{m\geq 0}T_{m}|_{r_{m}(H)\to r_{m}(H)},$ where $r_{m}$ is the projection defined in the proof of Lemma 4.3. Clearly, $T_{m}$ is unitarily equivalent to the direct sum $T_{m}|_{r_{m}(H)\to r_{m}(H)}\bigoplus 0_{(1-r_{m})(H)\to(1-r_{m})(H)}.$ Thus, a direct sum $\bigoplus_{m\geq 0}T_{m}$ is unitarily equivalent to $(\sum_{m\geq 0}T_{m})\bigoplus 0.$ In what follows, we ignore this subtle difference and write unitary equivalence as stated in Lemma 4.3. to $\bigoplus_{m\geq 0}T_{m}.$ Here, the sums are taken in the weak operator topology.

Proof.

Let $p_{1}$ and $p_{2}$ be projections on $H.$ Since $t\to t^{\frac{1}{n}},$ $t>0,$ is an operator monotone function for every $n\geq 1,$ it follows that

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Similarly, $p_{2}\leq{\rm supp}(p_{1}+p_{2})$ and, therefore,

[TABLE]

This simple fact can be also found in Proposition 2.5.14 in [10].

Let $p_{m}={\rm supp}(T_{m})$ and $q_{m}={\rm supp}(T_{m}^{*}).$ It follows from the assumption that $p_{m_{1}}p_{m_{2}}=p_{m_{1}}q_{m_{2}}=q_{m_{1}}q_{m_{2}}=0,$ $m_{1}\neq m_{2}.$ Set $r_{m}=p_{m}\vee q_{m}.$ We have

[TABLE]

Thus,

[TABLE]

By the preceding paragraph, we have $r_{m_{1}}r_{m_{2}}=0,$ $m_{1}\neq m_{2}.$

If $T=\sum_{m\geq 0}T_{m},$ then $r_{m}T=T_{m}$ and $Tr_{m}=T_{m}$ for every $m\geq 0.$ Thus, $T=\bigoplus_{m\geq 0}T_{m},$ where $T_{m}$ acts on the Hilbert space $r_{m}(H).$ ∎

Let

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The following proposition yields a special case of Theorem 1.1.

Proposition 4.4**.**

If $f$ is a Schwartz function supported on $[-1,1]^{d}$ and if $x={\rm Op}(f),$ then $xh(\nabla)$ is measurable.

Proof.

Step 1: We have that $xh(\nabla)$ is an integral operator with the kernel

[TABLE]

By assumption on $f,$ we have that

[TABLE]

Thus,

[TABLE]

where $T_{m,l_{1}}$ is an integral operator whose integral kernel is given by the formula

[TABLE]

Step 2: We claim that $T_{l_{1},l_{2}}\in\mathcal{L}_{1,\infty}$ and is measurable.

Note that the operators $\{T_{m,l_{1}}\}_{\begin{subarray}{c}m\in\mathbb{Z}^{d}\\ m=l_{2}{\rm mod}3\end{subarray}}$ are pairwise orthogonal. Therefore, we have ($\sim$ denotes unitary equivalence)

[TABLE]

By definition, $T_{m,l_{1}}:L_{2}(m+[-1,2]^{d})\to L_{2}(m+[-1,2]^{d}).$ Define a unitary operator

[TABLE]

by setting

[TABLE]

Define an operator $S_{l_{1}}:L_{2}([-1,2]^{d})\to L_{2}([-1,2]^{d})$ to be an integral operator with the integral kernel

[TABLE]

A direct computational argument shows that555Indeed,

$(U_{m}^{-1}\xi)(t)=e^{-\frac{i}{2}\langle m,\theta t\rangle}\xi(t+m),\quad\xi\in L_{2}(m+[-1,2]^{d}),\quad t\in[-1,2]^{d}.$

Thus,

$(S_{l_{1}}U_{m}^{-1}\xi)(t)=\chi_{l_{1}+[0,1]^{d}}(t)\cdot\int_{[0,1]^{d}}f(t-s)e^{\frac{i}{2}\langle s,\theta(t+m)\rangle}\xi(s+m)ds.$

Thus,

$(U_{m}S_{l_{1}}U_{m}^{-1}\xi)(t)=\chi_{l_{1}+[0,1]^{d}}(t-m)\cdot\int_{[0,1]^{d}}e^{\frac{i}{2}\langle m,\theta t\rangle}f(t-s-m)e^{\frac{i}{2}\langle s,\theta t\rangle}\xi(s+m)ds=$

$=\chi_{m+l_{1}+[0,1]^{d}}(t)\cdot\int_{m+[0,1]^{d}}f(t-s)e^{\frac{i}{2}\langle s,\theta t\rangle}\xi(s)ds.$

[TABLE]

Hence,

[TABLE]

By Lemma 4.1, $S_{l_{1}}\in\mathcal{L}_{1}.$ The claim follows now from Lemma 4.2. ∎

Proof of Theorem 1.1.

Choose a Schwartz function $f_{0}$ supported on $[-1,1]^{d}$ such that $f_{0}(0)\neq 0$ and let $x_{0}={\rm Op}(f_{0}).$ Set

[TABLE]

Clearly, $k$ is a bounded function on $\mathbb{R}^{d}.$

By Lemma 3.3 (a), we have $x_{0}\in W^{d,1}(\mathbb{R}^{d}_{\theta}).$ Using the obvious equality

[TABLE]

and Theorem 2.10 (a), we infer that

[TABLE]

By Proposition 4.4, we have that $x_{0}h(\nabla)$ is measurable and, hence, so is the operator $x_{0}(1-\Delta)^{-\frac{d}{2}}.$

Let now $x\in W^{d,1}(\mathbb{R}^{d}_{\theta})$ be arbitrary. Since $f_{0}$ is a Schwartz function, it follows that

[TABLE]

Without loss of generality, $\tau_{\theta}(x_{0})=1.$ Let $z=x-\tau_{\theta}(x)x_{0}\in W^{d,1}(\mathbb{R}^{d}_{\theta}).$ Clearly, $\tau_{\theta}(z)=0.$ We have

[TABLE]

By Proposition 3.5, the first summand vanishes. By the preceding paragraph, the second summand does not depend on $\varphi.$ This completes the proof. ∎