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# On the Uncertainty Principle
for Metaplectic Transformations

Nicolas Lerner  N. Lerner, Institut de Mathématiques de Jussieu, Sorbonne
Université (formerly Paris VI), Campus Pierre et Marie Curie, 4 Place Jussieu,
75252 Paris cedex 05, France<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We explore the new proofs and extensions of the Heisenberg Uncertainty
Principle introduced by A. Widgerson & Y. Widgerson in [27], developed in [7]
by N.C. Dias, F. Luef and J.N. Prata and also in [24] by Y. Tang. In
particular we give here a proof of the Uncertainty Principle for operators in
the Metaplectic group in any dimension.

###### Key words and phrases:

Uncertainty principle, Wigner distribution, Metaplectic group

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###### Contents

  1. 1 Introduction
    1. 1.1 Foreword
    2. 1.2 The main result: uncertainty principle for metaplectic transformations
  2. 2 Symplectic and metaplectic groups, Weyl quantization
    1. 2.1 The symplectic group
    2. 2.2 The metaplectic group
      1. 2.2.1 Basics
      2. 2.2.2 Another set of generators for the metaplectic group
    3. 2.3 Weyl quantization
      1. 2.3.1 Wigner distribution
      2. 2.3.2 The Weyl quantization
      3. 2.3.3 Symplectic covariance of the Weyl calculus
  3. 3 Proofs of the main results
    1. 3.1 First cases: $\Xi_{12}$ invertible
    2. 3.2 Using the symplectic covariance
    3. 3.3 The singular case
  4. 4 Appendix
    1. 4.1 Classical analysis
      1. 4.1.1 Fourier transform.
      2. 4.1.2 Schatten norms, Polar decomposition
    2. 4.2 Classical phase space analysis
      1. 4.2.1 The Classical Heisenberg Uncertainty Principle
      2. 4.2.2 On some Gaussian functions
      3. 4.2.3 Warm-up in one dimension

##  1\. Introduction

###  1.1. Foreword

The Heisenberg Uncertainty Principle, a cornerstone of Quantum Mechanics,
could somehow be summarized by the fact that _Operators do not usually
commute_ and the calculation of the commutator of two operators $A,B,$ say
from a Hilbert space into itself,

(1.1.1) $[A,B]=AB-BA,$

can unravel important properties linked to stability of matter. In particular
the fact that, in one dimension,

(1.1.2) $\Bigl{[}\frac{d}{dx},x\Bigr{]}=1,$

implies readily that

(1.1.3)
$\Bigl{\|}{\frac{du}{dx}}\Bigr{\|}_{L^{2}(\mathbb{R})}\|xu\|_{L^{2}(\mathbb{R})}\geq\operatorname{Re}\langle\frac{du}{idx},ixu\rangle_{L^{2}(\mathbb{R})}=\frac{1}{2}\operatorname{Re}\langle[\frac{d}{dx},x]u,u\rangle_{L^{2}(\mathbb{R})}\\\
=\frac{1}{2}\|u\|_{L^{2}(\mathbb{R})}^{2},$

and is closely related to the fact that the Harmonic Oscillator

$\mathcal{H}=-\hbar^{2}\frac{d^{2}}{dx^{2}}+x^{2},$

is bounded from below by the positive constant $\hbar$, whereas the
Hamiltonian function $\xi^{2}+x^{2}$ (_quantized_ by $\mathcal{H}$) is
actually vanishing at the origin. Similarly, we have in $n$ dimensions with
$n\geq 2$, $\kappa>0$,

$\mathcal{L}_{\kappa}=-\frac{\hbar^{2}\Delta}{2}-\frac{\kappa}{|x|}\geq-\frac{2\kappa^{2}}{(n-1)^{2}\hbar^{2}}>-\infty,$

whereas the Hamiltonian function $\frac{|\xi|^{2}}{2}-\frac{\kappa}{|x|}$
(quantized by $\mathcal{L}_{\kappa}$) is unbounded from below. The _Heisenberg
inequality_ (1.1.3) can be written in $n$ dimensions as

(1.1.4)
$\bigl{\|}|D|u\bigr{\|}_{L^{2}(\mathbb{R}^{n})}\bigl{\|}|x|u\bigr{\|}_{L^{2}(\mathbb{R}^{n})}\geq\frac{n}{4\pi}\bigl{\|}u\bigr{\|}_{L^{2}(\mathbb{R}^{n})}^{2},$

and follows easily from the one-dimensional (1.1.3). (see e.g. Section 4.2.1
in our Appendix.) One could say as well as for (1.1.3) that (1.1.4) is a
consequence of the non-commutation of the Fourier multiplier $|D|$ with the
operator of multiplication by $|x|$. It is also interesting to notice the
following result (see e.g. Lemma 1.2.29 in [20]).

######  Lemma 1.1.

Let $\mathbb{H}$ be a Hilbert space and let $J,K\in\mathcal{B}(\mathbb{H})$;
then the commutator $[J,K]\not=\operatorname{Id}$.

That simple result, whose proof is elementary, is showing that Quantum
Mechanics must deal with unbounded operators to have properties like (1.1.2)
or creation-annihilation properties such as

$\frac{1}{2}\Bigl{[}\frac{d}{idx}-ix,\frac{d}{idx}+ix\Bigr{]}=1.$

E. Lieb’s article [21] and E. Lieb & R. Seiringer’ book [22] are providing a
whole theory of stability of matter based upon the non-commutation of
operators quantifying the kinetic energy with the operators of multiplication
by a potential depending on the configuration variables.

The Uncertainty Principle can be also seen as a _limitation on the sharp
localization of both $u$ and its Fourier transform111See Section 4.1.1 in the
Appendix for our normalization in the definition of the Fourier
transformation. $\widehat{u}$_ in the sense that, for $u$ normalized in
$L^{2}(\mathbb{R}^{n})$,

(1.1.5)
$\inf_{(y,\eta)\in\mathbb{R}^{2n}}\bigl{\|}|\eta-\xi|\widehat{u}(\xi)\bigr{\|}_{L^{2}(\mathbb{R}^{n}_{\xi})}\bigl{\|}|y-x|u(x)\bigr{\|}_{L^{2}(\mathbb{R}^{n}_{x})}\geq\frac{n}{4\pi},$

which is an immediate consequence of (1.1.4). The paper [4] by A. Bonami, B.
Demange and P. Jaming, (see also [3] by the first two previous authors),
generalizing the article [16] of L. Hörmander, proved in particular that

(1.1.6)
$\iint_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{|u(x)||\widehat{u}(\xi)|}{(1+|x|+|\xi|)^{n}}e^{2\pi|\langle
x,\xi\rangle|}dxd\xi<+{\infty}\Longrightarrow u=0.$

Also the papers [2] by M. Benedicks, [1] by W.O. Amrein & A.M. Berthier,
proved that both supports of $u,\widehat{u}$ cannot have a finite Lebesgue
measure unless $u=0$. That result was extended to more general operators than
the Fourier transformation in the article [13] by S. Ghobber & P. Jaming.

We must also quote the wide and sophisticated program around the Uncertainty
Principle given in the article [10] by C. Fefferman & D.H. Phong and in the C.
Fefferman’s survey paper [11], where the authors were able to describe an
explicit procedure to _obtain a sharp lowerbound for large classes of pseudo-
differential operators_. In fact, given a Hamiltonian function $a(x,\xi)$, a
smooth real-valued function verifying some rather natural estimates, they
would consider the set

(1.1.7) $\mathcal{N}_{a}=\\{(x,\xi)\in\mathbb{R}^{2n},a(x,\xi)<0\\},$

and evaluate its symplectic importance by counting the number of disjoint
symplectic images of the unit ball contained in it. For instance, if
$\mathcal{N}_{a}$ does not contain any symplectic image of the unit ball of
$\mathbb{R}^{2n}$, then the operator ${\text{{O}p}_{\text{{w}}}(a)}$ (the Weyl
quantization of the Hamiltonian function $a$, see e.g. Section 2.3) should be
bounded from below. Note that there is a striking link with the results of
symplectic rigidity of M. Gromov [15] (see also the articles by I. Ekeland &
H. Hofer [8], [9]) where it is proven that the very existence of a symplectic
mapping $\chi$, such that

$\chi:\underbrace{\\{(x,\xi)\in\mathbb{R}^{2n},{|x|}^{2}+{|\xi|}^{2}\leq
1\\}}_{\text{the unit
ball}}\longrightarrow\underbrace{\\{(x,\xi)\in\mathbb{R}^{2n},x_{1}^{2}+\xi_{1}^{2}\leq
r^{2}\\}}_{\text{vertical cylinder with radius $r$}}$

implies that $r$ should be $\geq 1$. Of course that obstruction is not due to
the volume when $n\geq 2$, since the cylinder has an infinite volume. In
particular, the Fefferman-Phong procedure should lead to the fact that the
operator

$\mathcal{U}^{*}\bigl{[}{\text{{O}p}_{\text{{w}}}(x_{1}^{2}+\xi_{1}^{2})}\bigr{]}\mathcal{U},$

should be bounded from below by a positive constant, for a very large class of
non-classical elliptic Fourier Integral Operators $\mathcal{U}$, quantifying a
canonical transformation. The explicit mathematical link between the results
of M. Gromov, I. Ekeland & H. Hofer mentioned above and the Fefferman-Phong
procedure is not so easy to derive rigourously, since it does not seem obvious
to quantize any smooth canonical transformation $\chi$ by a Fourier Integral
Operator without having some explicit bounds on the derivatives of $\chi$. We
may nevertheless point out that in the article [5], J.-M. Bony is providing
quite general formulas for the quantization of canonical transformations into
tractable Fourier Integral Operators, using some partitions of unity related
to metrics on the phase space.

More recently, the paper [27] by A. Wigderson and Y. Wigderson provided a
renewed point of view on uncertainty principles along with new proofs. Their
paper was developed in [7] by N.C. Dias, F. Luef and J.N. Prata and also in
[24] by Y. Tang.

This article is organized in four sections: in Section 1, after the above
Foreword, we formulate the main result of the article, a version of the
uncertainty principle for metaplectic transformations. We hope that having a
statement at the beginning of the article could serve as a motivation for the
reader to go on. However, it seems necessary to spend some time at reminding a
couple of classical facts on the symplectic and metaplectic groups: this is
what we do in Section 2, in particular with the perspective to using a special
set of generators for the metaplectic group. We end that section with some
basic elements on the Weyl quantization. In Section 3, we give the proof of
the main theorem of the paper, using the symplectic covariance of the Weyl
quantization as a key tool; we split our discussion in two different cases,
related to the invertibility of a submatrix of the symplectic transformation
linked to the metaplectic mapping under scope. Our article contains also an
Appendix as Section 4, essentially devised to ensure a reasonable degree of
self-containedness for our arguments.

###  1.2. The main result: uncertainty principle for metaplectic
transformations

######  Definition 1.2.

Let $f$ be a function in $L^{2}_{\textrm{loc}}(\mathbb{R}^{n})$. We define the
_variance_ of $f$ as

(1.2.1) $\mathcal{V}(f)=\int_{\mathbb{R}^{n}}{|x|}^{2}{|f(x)|}^{2}dx.$

######  Remark 1.3.

Inequality (1.1.4) implies that, for $u\in\mathscr{S}(\mathbb{R}^{n})$,

(1.2.2)
$\mathcal{V}(\mathcal{F}u)\mathcal{V}(u)\geq\frac{n^{2}}{2^{4}\pi^{2}}\|u\|_{L^{2}(\mathbb{R}^{n})}^{4},$

where $\mathcal{F}$ stands for the Fourier transform. We may wonder if there
are other operators satisfying this type of inequality, besides the Fourier
transform. Obviously the Identity does not satisfy any inequality of type
(1.2.2): we have

(1.2.3) $\inf_{\begin{subarray}{c}u\in\mathscr{S}(\mathbb{R}^{n})\\\
\|u\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\mathcal{V}(u)=0.$

To prove this, we just need to concentrate a function near the origin: take
with $\varepsilon>0$,
$u_{\varepsilon}(x)=\phi(x\varepsilon^{-1})\varepsilon^{-n/2},\
\phi\in\mathscr{S}(\mathbb{R}^{n}),\|\phi\|_{L^{2}(\mathbb{R}^{n})}=1.$ We
obtain $\|u_{\varepsilon}\|_{L^{2}(\mathbb{R}^{n})}=1$ and

$\mathcal{V}(u_{\varepsilon})=\int|x|^{2}{|\phi\bigl{(}\frac{x}{\varepsilon}\bigr{)}|}^{2}\varepsilon^{-n}dx=\varepsilon^{2}\int{|y|}^{2}{|\phi(y)|}^{2}dy,$

implying (1.2.3).

######  Definition 1.4.

Let $M$ be a bounded operator on $L^{2}(\mathbb{R}^{n})$, sending
$\mathscr{S}(\mathbb{R}^{n})$ into itself. We define

(1.2.4) $\mu(M)=\inf_{\begin{subarray}{c}u\in\mathscr{S}(\mathbb{R}^{n})\\\
\|u\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\sqrt{\mathcal{V}(Mu)\mathcal{V}(u)}.$

We have proven above222The constant $\frac{n^{2}}{4\pi^{2}}$ in (1.2.2) is
sharp as it is seen by checking the variances of $u,\hat{u}$ with
$u(x)=2^{n/4}e^{-\pi|x|^{2}}$, as in Section 4.2.1 of our Appendix. that

(1.2.5) $\mu(\mathcal{F})=\frac{n}{4\pi},\qquad\mu(\operatorname{Id})=0.$

We would like to introduce the symplectic group $\text{\sl Sp}(2n,\mathbb{R})$
(which is a subgroup of $\text{\sl Sl}(2n,\mathbb{R})$) and its two-fold
covering the metaplectic group $\text{\sl Mp}(n,\mathbb{R})$, with

(1.2.6) $\Psi:\text{\sl Mp}(n,\mathbb{R})\longrightarrow\text{\sl
Sp}(2n,\mathbb{R}),$

as the covering map. We shall provide in the next sections a couple of
reminders about these two groups, but at the moment, we need only to say that
with the canonical symplectic form $\sigma$ being given on
$\mathbb{R}^{n}\times\mathbb{R}^{n}$ by

(1.2.7) $\langle\sigma X,Y\rangle=\bigl{[}X,Y\bigr{]}=\xi\cdot y-\eta\cdot
x,\quad\text{with $X=(x,\xi),Y=(y,\eta)$,}$

the symplectic group $\text{\sl Sp}(2n,\mathbb{R})$ is the subgroup of
$S\in\text{\it Gl}(2n,\mathbb{R})$ such that333We shall see below that the
determinant of symplectic matrices is actually $1$.

(1.2.8) $\forall
X,Y\in\mathbb{R}^{2n},\quad[SX,SY]=[X,Y],\qquad\text{i.e.\quad}S^{*}\sigma
S=\sigma,$

where $S^{*}$ is the transpose and

(1.2.9) $\sigma=\begin{pmatrix}0&I_{n}\\\ -I_{n}&0\end{pmatrix}.$

On the other hand, the metaplectic group is a subgroup of the unitary group of
$L^{2}(\mathbb{R}^{n})$, whose elements are also isomorphisms of
$\mathscr{S}(\mathbb{R}^{n})$ and $\mathscr{S}^{\prime}(\mathbb{R}^{n})$. We
can then formulate without further postponing the main result of this paper.

######  Theorem 1.5.

Let $\mathcal{M}$ be an element of the metaplectic group $\text{\sl
Mp}(n,\mathbb{R})$ such that $\Psi(\mathcal{M})=\Xi,$ where the
$(2n)\times(2n)$ matrix $\Xi\in\text{\sl Sp}(2n,\mathbb{R})$ is written in
four $(n\times n)$ blocks as

(1.2.10) $\Xi=\begin{pmatrix}\Xi_{11}&\Xi_{12}\\\
\Xi_{21}&\Xi_{22}\end{pmatrix}.$

Then we have, with $\mu$ defined in (1.2.4),

(1.2.11)
$\mu(\mathcal{M})=\frac{1}{4\pi}\operatorname{trace}\bigl{(}(\Xi_{12}^{*}\Xi_{12})^{1/2}\bigr{)}=\frac{1}{4\pi}|\hskip-1.0pt\|\Xi_{12}\|\hskip-1.0pt|_{{\mathtt{S}^{1}}},$

where $|\hskip-1.0pt\|\Xi_{12}\|\hskip-1.0pt|_{{\mathtt{S}^{1}}}$ is the
Schatten norm of index 1.

######  N.B.

For $n=1$, the inequality $\mu(\mathcal{M})\geq|\Xi_{12}|/(4\pi)$ is already
proven for a dense subset of $\text{\sl Mp}(1,\mathbb{R})$ in the paper [27]
by A. Wigderson and Y. Wigderson. We provide here a multi-dimensional version
along with an algebraic identification of the lower bound $\mu(\mathcal{M})$.

######  Remark 1.6.

The equalities (1.2.5) are coherent with the above theorem since we have

$\displaystyle\Psi(e^{-\frac{i\pi n}{4}}\mathcal{F})$
$\displaystyle=\begin{pmatrix}0&I_{n}\\\
-I_{n}&0\end{pmatrix},\quad|\hskip-1.0pt\|I_{n}\|\hskip-1.0pt|_{{\mathtt{S}^{1}}}=n,\quad\mu(z\mathcal{F})=\mu(\mathcal{F})\quad\text{for
$|z|=1$},$ $\displaystyle\Psi(\operatorname{Id}_{L^{2}(\mathbb{R}^{n})})$
$\displaystyle=I_{2n}=\begin{pmatrix}I_{n}&0\\\ 0&I_{n}\end{pmatrix}.$

##  2\. Symplectic and metaplectic groups, Weyl quantization

We recall in this section some classical facts about the symplectic and
metaplectic groups as well as various matters on the Weyl quantization. We
rely essentially on our Memoir [20], where complete proofs are given.The
statements and examples which follow will be hopefully useful for the reader,
but we shall of course omit the proofs appearing in [20]. We have used E.P.
Wigner’s works [28], J. Leray’s book [18] and other Lecture Notes of this
author at the Collège de France such as [17], L. Hörmander’s four-volume
treatise, The Analysis of Linear Partial Differential operators and in
particular Volume III, as well as K. Gröchenig’s [14] Foundations of time-
frequency analysis, along with G.B. Folland’s [12], A. Unterberger’s [25], and
N. Lerner’s [19].

###  2.1. The symplectic group

The canonical symplectic form $\sigma$ on $\mathbb{R}^{n}\times\mathbb{R}^{n}$
is defined in (1.2.7), the symplectic group $\text{\sl Sp}(2n,\mathbb{R})$ is
defined in (1.2.8). It is easy to prove directly from (1.2.8) that $\text{\sl
Sp}(2,\mathbb{R})=Sl(2,\mathbb{R}).$ In the sequel we shall denote by

(2.1.1) $\text{\sl Sym}(n,\mathbb{R})=\\{(n\times n)\text{ real symmetric
matrices}\\}.$

######  Theorem 2.1.

Let $n$ be an integer $\geq 1$. The group $\text{\sl Sp}(2n,\mathbb{R})$ is
included in $Sl(2n,\mathbb{R})$ and generated by the following mappings

(2.1.2) $\displaystyle\begin{pmatrix}I_{n}&0\\\ A&I_{n}\end{pmatrix}$
$\displaystyle,\ \text{where $A\in\text{\sl Sym}(n,\mathbb{R})$,}$ (2.1.3)
$\displaystyle\begin{pmatrix}B^{-1}&0\\\ 0&B^{*}\end{pmatrix}$
$\displaystyle,\quad\text{where }B\in\text{\it Gl}(n,\mathbb{R}),$ (2.1.4)
$\displaystyle\begin{pmatrix}I_{n}&-C\\\ 0&I_{n}\end{pmatrix}$
$\displaystyle,\ \text{where $C\in\text{\sl Sym}(n,\mathbb{R})$.}$

For $A,B,C$ as above, the mapping

(2.1.5) $\Xi_{A,B,C}={\begin{pmatrix}[l]{B^{-1}}&{-B^{-1}C}\\\
{AB^{-1}}&{B^{*}-AB^{-1}C}\end{pmatrix}}=\begin{pmatrix}I_{n}&0\\\
A&I_{n}\end{pmatrix}\begin{pmatrix}B^{-1}&0\\\
0&B^{*}\end{pmatrix}\begin{pmatrix}I_{n}&-C\\\ 0&I_{n}\end{pmatrix}.$

belongs to $\text{\sl Sp}(2n,\mathbb{R})$. Moreover, we define on
$\mathbb{R}^{n}\times\mathbb{R}^{n}$ the generating function $S$ of the
symplectic mapping $\Xi_{A,B,C}$ by the identity

(2.1.6) $S(x,\eta)=\frac{1}{2}\bigl{(}\langle Ax,x\rangle+2\langle
Bx,\eta\rangle+\langle C\eta,\eta\rangle\bigr{)}\ \text{so that}{\hskip
15.0pt}\Xi\Bigl{(}\frac{\partial
S}{\partial\eta}\oplus\eta\Bigr{)}=x\oplus\frac{\partial S}{\partial x}.$

For a symplectic mapping $\Xi$, to be of the form (2.1.5) is equivalent to the
assumption that the mapping
$x\mapsto\pi_{\mathbb{R}^{n}\times\\{0\\}}\Xi(x\oplus 0)$ is invertible from
$\mathbb{R}^{n}$ to $\mathbb{R}^{n}$; moreover, if this mapping is not
invertible, the symplectic mapping $\Xi$ is the product of two mappings of the
type $\Xi_{A,B,C}$.

This result is proven in Theorem 1.2.6 of [20].

######  Remark 2.2.

Let us consider $\Xi$ in $\text{\sl Sp}(2n,\mathbb{R})$: we have

(2.1.7) $\Xi=\begin{pmatrix}P&Q\\\ R&S\end{pmatrix},\quad\text{where
$P,Q,R,S,$ are $n\times n$ matrices. }$

The equation

(2.1.8) ${\Xi^{*}}\sigma\Xi=\sigma$

is satisfied with $\sigma=\begin{pmatrix}0&I_{n}\\\ -I_{n}&0\end{pmatrix},$
which means

(2.1.9)
${P^{*}}R={({P^{*}}R)}^{*},\quad{Q^{*}}S={({Q^{*}S})}^{*},\quad{P^{*}}S-{R^{*}}Q=I_{n}.$

The following claim is proven in (1.2.23-24) of [20].

######  Claim 2.3.

The mapping $\Xi\mapsto(\Xi^{*})^{-1}$ is an isomorphism of the group
$\text{\sl Sp}(2n,\mathbb{R})$. As a consequence,

(2.1.10) $\displaystyle\Xi=\begin{pmatrix}P&Q\\\ R&S\end{pmatrix}\in\text{\sl
Sp}(2n,\mathbb{R})\quad\text{ is also equivalent to}$ (2.1.11) $\displaystyle
PQ^{*}={(PQ^{*})}^{*},\quad RS^{*}={(RS^{*})}^{*},\quad PS^{*}-QR^{*}=I_{n}.$

######  Examples 2.4.

The $4\times 4$ matrix

(2.1.12) $\begin{pmatrix}\begin{matrix}0&0\\\
0&1\end{matrix}&\begin{matrix}1&0\\\ 0&0\end{matrix}\\\ \begin{matrix}-1&0\\\
0&0\end{matrix}&\begin{matrix}0&0\\\
0&1\end{matrix}\end{pmatrix}=\begin{pmatrix}P&Q\\\ R&S\end{pmatrix},$

belongs to $Sp(4,\mathbb{R})$ although the block $2\times 2$ matrices
$P,Q,R,S,$ are all singular (with rank 1). The symplectic matrix

(2.1.13) $\begin{pmatrix}0&I_{n}\\\
-I_{n}&0\end{pmatrix}=2^{-1/2}\begin{pmatrix}I_{n}&I_{n}\\\
-I_{n}&I_{n}\end{pmatrix}2^{-1/2}\begin{pmatrix}I_{n}&I_{n}\\\
-I_{n}&I_{n}\end{pmatrix}=\Xi_{-I_{n},2^{1/2}I_{n},-I_{n}}^{2},$

is not of the form $\Xi_{A,B,C}$ but444Note that for a symplectic mapping to
be of type $\Xi_{A,B,C}$ requires (in fact is equivalent) to the fact that
$\Xi_{11}$ is invertible, using the notation (1.2.10). is the square of such a
matrix. It is also the case of all the mappings
$(x_{k},\xi_{k})\mapsto(\xi_{k},-x_{k}),$ with the other coordinates fixed.
Similarly the symplectic matrix

(2.1.14) $\begin{pmatrix}0&-I_{n}\\\
I_{n}&I_{n}\end{pmatrix}=\begin{pmatrix}I_{n}&-I_{n}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&0\\\ I_{n}&I_{n}\end{pmatrix},$

is not of the form $\Xi_{A,B,C}$ but is the product $\Xi_{0,I,I}\Xi_{I,I,0}$.

######  Remark 2.5.

We have seen in (2.1.11), (2.1.9) some equivalent conditions for a matrix

(2.1.15) $\Xi=\begin{pmatrix}P&Q\\\ R&S\end{pmatrix}\quad\text{where $P,Q,R,S$
are $n\times n$ real matrices,}$

to be symplectic. We note here that when $\Xi\in\text{\sl Sp}(2n,\mathbb{R})$,
we have

(2.1.16) $\Xi^{-1}=\begin{pmatrix}S^{*}&-Q^{*}\\\ -R^{*}&P^{*}\end{pmatrix},$

as it is easily checked from (2.1.11), (2.1.9). When $\det P\not=0$, we can
prove that $\Xi=\Xi_{A,B,C}$ as555If $\Xi$ is given by (2.1.15) with $P$
invertible, we define $B=P^{-1},C=-P^{-1}Q,A=RP^{-1}$ and we get from (2.1.9)
that $\Xi=\Xi_{A,B,C}$.  defined in (2.1.5). Also from (2.1.16), we get that
if $\det S\not=0$ we have

$\Xi^{-1}=\Xi_{A,B,C},$

so that

(2.1.17) $\Xi=\begin{pmatrix}I_{n}&C\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}B&0\\\
0&B^{*-1}\end{pmatrix}\begin{pmatrix}I_{n}&0\\\ -A&I_{n}\end{pmatrix}.$

Some other properties of the same type are available when $\det Q$ or $\det R$
are different from $0$. Indeed we have for $\Xi\in\text{\sl
Sp}(2n,\mathbb{R})$ and $\sigma$ given by (1.2.9),

(2.1.18) $\Xi\sigma=\begin{pmatrix}P&Q\\\
R&S\end{pmatrix}\sigma=\begin{pmatrix}-Q&P\\\
-S&R\end{pmatrix}\underbrace{=}_{\text{if $\det Q\not=0$}}\Xi_{A,B,C},$

so that

(2.1.19) $\Xi=-\Xi_{A,B,C}\sigma=\begin{pmatrix}I_{n}&0\\\
A&I_{n}\end{pmatrix}\begin{pmatrix}B^{-1}&0\\\
0&B^{*}\end{pmatrix}\begin{pmatrix}I_{n}&-C\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}0&-I_{n}\\\ I_{n}&0\end{pmatrix}.$

If we have $\det R\not=0$, using the two first equalities in (2.1.18), we get
that $(\Xi\sigma)^{-1}=\Xi_{A,B,C}$ which gives

(2.1.20) $\Xi=\begin{pmatrix}I_{n}&C\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}B&0\\\
0&B^{*-1}\end{pmatrix}\begin{pmatrix}I_{n}&0\\\
-A&I_{n}\end{pmatrix}\begin{pmatrix}0&-I_{n}\\\ I_{n}&0\end{pmatrix}.$

However, it is indeed possible when $n\geq 2$ to have a symplectic matrix in
$\text{\sl Sp}(2n,\mathbb{R})$ in the form (2.1.15) such that all the blocks
are singular, as shown by the example (2.1.12).

###  2.2. The metaplectic group

####  2.2.1. Basics

With $\mathbb{Z}_{4}$ standing for $\mathbb{Z}/4\mathbb{Z}$, we consider the
mapping

(2.2.1) $\text{\sl Sym}(n,\mathbb{R})\times\text{\it
Gl}(n,\mathbb{R})\times\text{\sl
Sym}(n,\mathbb{R})\times\mathbb{Z}_{4}\ni(A,B,C,m)\mapsto\omega(A,B,C,m),$

where $\omega(A,B,C,m)$ is a function defined on
$\mathbb{R}^{n}\times\mathbb{R}^{n}$ by

(2.2.2) $\omega(A,B,C,m)(x,\eta)=e^{\frac{i\pi m}{2}}{|\det
B|}^{1/2}e^{i\pi\\{\langle Ax,x\rangle+2\langle Bx,\eta\rangle+\langle
C\eta,\eta\rangle\\}}.$

We note in particular that $\mathbb{Z}_{4}\ni m\mapsto e^{\frac{i\pi m}{2}}$
is well-defined. We shall always require the following link between $m$ and
$B$: we want that $m$ be _compatible_ with $B$, which means

(2.2.3) $e^{i\pi m}=\operatorname{sign}(\det B),$

in such a way that we have $\bigl{(}e^{\frac{i\pi m}{2}}{|\det
B|}^{1/2}\bigr{)}^{2}=\det B.$ In fact the above mapping (2.2.1) is defined
for

$(B,m)\in\bigl{(}\text{\it
Gl}_{+}(n,\mathbb{R})\times\\{0,2\\}\bigr{)}\cup\bigl{(}\text{\it
Gl}_{-}(n,\mathbb{R})\times\\{1,3\\}\bigr{)},\quad(A,C)\in\text{\sl
Sym}(n,\mathbb{R})^{2},$

where we consider $\\{0,2\\}$ and $\\{1,3\\}$ as subsets of $\mathbb{Z}_{4}$.
We shall denote

(2.2.4) $m(B)=\\{m\in\mathbb{Z}_{4},e^{i\pi m}=\operatorname{sign}(\det
B)\\}=\begin{cases}\\{0,2\\}&\text{if $\det B>0$,}\\\ \\{1,3\\}&\text{if $\det
B<0$.}\end{cases}$

The next proposition is Proposition 1.2.11 in [20].

######  Proposition 2.6.

Let $A,B,C$ be as in Theorem 2.1, let $S$ be the generating function of
$\Xi_{A,B,C}$ (cf. (2.1.6)) and let $m\in\mathbb{Z}_{4}$ such that (2.2.3) is
satisfied. We define the operator $M_{A,B,C}^{\scriptscriptstyle\\{m\\}}$ on
$\mathscr{S}(\mathbb{R}^{n})$ by

(2.2.5) $(M_{A,B,C}^{\scriptscriptstyle\\{m\\}}v)(x)=e^{\frac{i\pi
m}{2}}{|\det B|}^{1/2}\int_{\mathbb{R}^{n}}e^{2i\pi
S(x,\eta)}\hat{v}(\eta)d\eta.$

This operator is an automorphism of $\mathscr{S}^{\prime}(\mathbb{R}^{n})$ and
of $\mathscr{S}(\mathbb{R}^{n})$ which is unitary on $L^{2}(\mathbb{R}^{n})$,
and such that, for all $a\in\mathscr{S}^{\prime}(\mathbb{R}^{2n})$,

(2.2.6) $\bigl{(}M_{A,B,C}^{\scriptscriptstyle\\{m\\}}\bigr{)}^{\\!*}\
{\text{{O}p}_{\text{{w}}}(a)}M_{A,B,C}^{\scriptscriptstyle\\{m\\}}={\text{{O}p}_{\text{{w}}}(a\circ\Xi_{A,B,C})},$

where $\Xi_{A,B,C}$ is defined in Theorem 2.1.

######  Remark 2.7.

We have for $A,B,C$ as above,

(2.2.7) $\displaystyle(M_{A,I,0}^{\scriptscriptstyle\\{0\\}}v)(x)$
$\displaystyle=e^{i\pi\langle Ax,x\rangle}v(x),$ (2.2.8)
$\displaystyle(M_{0,B,0}^{\scriptscriptstyle\\{m\\}}v)(x)$
$\displaystyle=e^{\frac{i\pi m}{2}}{|\det B|}^{1/2}v(Bx),\ m\text{ satisfying
\eqref{compat}},$ (2.2.9)
$\displaystyle(M_{0,I,C}^{\scriptscriptstyle\\{0\\}}v)(x)$
$\displaystyle=\bigl{(}e^{i\pi\langle CD_{x},D_{x}\rangle}v\bigr{)}(x),$

three operators which are obviously automorphisms of
$\mathscr{S}(\mathbb{R}^{n})$ and of $\mathscr{S}^{\prime}(\mathbb{R}^{n})$ as
well as unitary operators in $L^{2}(\mathbb{R}^{n})$. In particular we find
that, keeping in mind (2.2.18),

(2.2.10)
$\displaystyle\bigl{(}M_{A,I,0}^{\scriptscriptstyle\\{m\\}}\bigr{)}^{-1}=M_{-A,I,0}^{\scriptscriptstyle\\{-m\\}},\quad\bigl{(}M_{0,I,C}^{\scriptscriptstyle\\{m\\}}\bigr{)}^{-1}=M_{0,I,-C}^{\scriptscriptstyle\\{-m\\}},\quad
m\in m(I_{n})=\\{0,2\\},$ (2.2.11)
$\displaystyle\bigl{(}M_{0,B,0}^{\scriptscriptstyle\\{m\\}}\bigr{)}^{-1}=M_{0,B^{-1},0}^{\scriptscriptstyle\\{-m\\}},\quad
m\in m(B).$

###### Proof.

Formula (2.2.6) is easily checked for each operator (2.2.7), (2.2.8) and
(2.2.9). We have also

(2.2.12)
$\displaystyle\mathcal{F}^{*}{\text{{O}p}_{\text{{w}}}(a)}\mathcal{F}$
$\displaystyle={\text{{O}p}_{\text{{w}}}(a(\xi,-x))},$ (2.2.13) $\displaystyle
e^{-i\pi Ax^{2}}{\text{{O}p}_{\text{{w}}}(a)}e^{i\pi Ax^{2}}$
$\displaystyle={\text{{O}p}_{\text{{w}}}(a(x,Ax+\xi))},\
\text{\footnotesize$A$ real symmetric $n\times n$ matrix,}$ (2.2.14)
$\displaystyle e^{-i\pi CD^{2}}{\text{{O}p}_{\text{{w}}}(a)}e^{i\pi CD^{2}}$
$\displaystyle={\text{{O}p}_{\text{{w}}}(a(x-C\xi,\xi))},\
\text{\footnotesize$C$ real symmetric $n\times n$ matrix,}$ (2.2.15)
$\displaystyle\mathcal{N}_{B}^{*}{\text{{O}p}_{\text{{w}}}(a)}\mathcal{N}_{B}$
$\displaystyle={\text{{O}p}_{\text{{w}}}(a(B^{-1}x,B^{*}\xi))},\quad
B\in\text{\it Gl}(n,\mathbb{R}),$

where

(2.2.16) $\bigl{(}\mathcal{N}_{B}v\bigr{)}(x)={|\det B|}^{1/2}v(Bx).$

Since we have $\Xi_{A,B,C}=\Xi_{A,I,0}\ \Xi_{0,B,0}\ \Xi_{0,I,C}$ and

(2.2.17) $\displaystyle
M_{A,B,C}^{\scriptscriptstyle\\{m\\}}=M_{A,I,0}^{\scriptscriptstyle\\{0\\}}M_{0,B,0}^{\scriptscriptstyle\\{m\\}}M_{0,I,C}^{\scriptscriptstyle\\{0\\}},$

we get (2.2.6) and the proposition. ∎

######  Remark 2.8.

With the set $m(B)$ defined in (2.2.4) and with obvious notations we have

(2.2.18) $m(-B)=m(B)+n,\quad m(B^{*})=m(B),\quad m(B^{-1})=-m(B)=m(B).$

Indeed, $m\in m(-B)$ is equivalent to having

$e^{i\pi m}=\operatorname{sign}(\det(-B))=(-1)^{n}\operatorname{sign}(\det
B)=e^{i\pi(n+m^{\prime})},\ m^{\prime}\in m(B),$

which is equivalent to $(m-n-m^{\prime})\in\\{0,2\\}$ modulo 4, proving the
first result, since $m(B)+\\{0,2\\}=m(B)$. On the other hand, $m\in m(B^{*})$
is equivalent to

$e^{i\pi m}=\operatorname{sign}(\det(B^{*}))=\operatorname{sign}(\det
B)=e^{i\pi m^{\prime}},\ m^{\prime}\in m(B),$

which means $m(B^{*})=m(B)+\\{0,2\\}=m(B).$ Also the equality
$\operatorname{sign}(\det B)=\operatorname{sign}(\det(B^{-1}))$ implies that
$m(B)=m(B^{-1})$ and moreover since the mapping

$\mathbb{Z}_{4}\ni m\mapsto-m\in\mathbb{Z}_{4},$

leaves invariant $\\{0,2\\}$ and stable $\\{1,3\\}$, we get the last equality
in (2.2.18). We note in particular that we have

(2.2.19)
$M_{0,I_{n},0}^{\scriptscriptstyle\\{0\\}}=\operatorname{Id}_{L^{2}(\mathbb{R}^{n})},\
M_{0,I_{n},0}^{\scriptscriptstyle\\{2\\}}=-\operatorname{Id}_{L^{2}(\mathbb{R}^{n})},$

and also with the notation (2.3.19), using the now proven (2.2.18),

(2.2.20) $M_{0,-I_{n},0}^{\scriptscriptstyle\\{n\\}}=e^{\frac{i\pi
n}{2}}\sigma_{0},\ M_{0,-I_{n},0}^{\scriptscriptstyle\\{n+2\\}}=-e^{\frac{i\pi
n}{2}}\sigma_{0},$

noting that $(\pm e^{\frac{i\pi n}{2}})^{2}=(-1)^{n}=\det(-I_{n})$ so that
(2.2.3) is satisfied. More generally, we have

$\text{for $\det B>0$,
}M_{A,B,C}^{\scriptscriptstyle\\{0\\}}=-M_{A,B,C}^{\scriptscriptstyle\\{2\\}},\qquad\text{for
$\det B<0$,
}M_{A,B,C}^{\scriptscriptstyle\\{1\\}}=-M_{A,B,C}^{\scriptscriptstyle\\{3\\}}.$

######  Examples 2.9.

In (2.1.13), we have seen in particular that, for $n=1$,

$\begin{pmatrix}0&1\\\ -1&0\end{pmatrix}=\left\\{2^{-1/2}\begin{pmatrix}1&1\\\
-1&1\end{pmatrix}\right\\}^{2},\quad 2^{-1/2}\begin{pmatrix}1&1\\\
-1&1\end{pmatrix}=\Xi_{-1,2^{1/2},-1}.$

We have also with the notations of Theorem 2.1,

$(M_{-1,2^{1/2},-1}^{\scriptscriptstyle\\{0\\}}v)(x)=\int_{\mathbb{R}}e^{2i\pi\frac{1}{2}(-x^{2}+2^{3/2}x\eta-\eta^{2})}\hat{v}(\eta)d\eta
2^{1/4},$

so that the kernel $k_{1}(x,y)$ of the operator
$M_{-1,2^{1/2},-1}^{\scriptscriptstyle\\{0\\}}$ is

$k_{1}(x,y)=2^{1/4}\int e^{i\pi(-x^{2}+2^{3/2}x\eta-\eta^{2})}e^{-2i\pi
y\eta}d\eta\underbrace{=}_{\text{use
\eqref{foimga}}}2^{1/4}e^{-i\pi/4}e^{i\pi(x^{2}+y^{2})}e^{-2^{3/2}i\pi xy},$

so that the kernel $k_{2}$ of the operator
$(M_{-1,2^{1/2},-1}^{\scriptscriptstyle\\{0\\}})^{2}$ is (using again
(4.1.7)),

$k_{2}(x,y)=\int k_{1}(x,z)k_{1}(z,y)dz\\\
=2^{1/2}e^{-i\pi/2}e^{i\pi(x^{2}+y^{2})}\int e^{2i\pi z^{2}}e^{-2i\pi
z2^{1/2}(x+y)}dz=e^{-i\pi/4}e^{-2i\pi xy},$

so that

(2.2.21)
$(M_{-1,2^{1/2},-1}^{\scriptscriptstyle\\{0\\}})^{2}=e^{-i\pi/4}\mathcal{F}_{1},$

with $\mathcal{F}_{1}$ standing for the 1D Fourier transformation. We get by
tensorisation that in $n$ dimensions,

(2.2.22)
$(M_{-I_{n},2^{1/2}I_{n},-I_{n}}^{\scriptscriptstyle\\{0\\}})^{2}=e^{-i\pi
n/4}\mathcal{F}_{n},$

with $\mathcal{F}_{n}$ standing for the Fourier transformation in $n$
dimensions. Similar expressions can be obtained for $\mathcal{F}_{[k;n]}$, the
Fourier transformation with respect to the variable $x_{k}$ in $n$ dimensions,
$k\in\llbracket 1,n\rrbracket$ with

(2.2.23)
$(M_{A_{k},B_{k},C_{k}}^{\scriptscriptstyle\\{0\\}})^{2}=e^{-i\pi/4}\mathcal{F}_{[k;n]},$

where $B_{k}$ is the $n\times n$ diagonal matrix with diagonal entries equal
to $1$ except for the $k$th equal to $2^{1/2}$, the $n\times n$ diagonal
matrices $A_{k}=C_{k}$ with diagonal entries equal to 0, except for the $k$th
equal to $-1$.

######  Definition 2.10.

The metaplectic group $\text{\sl Mp}(n)$ is defined as the subgroup of the
group of unitary operators on $L^{2}(\mathbb{R}^{n})$ generated by

(2.2.24) $\displaystyle
M_{A,I,0}^{\scriptscriptstyle\\{m^{\prime}\\}},\text{\footnotesize where
$A\in\text{\sl Sym}(n,\mathbb{R})$, $m^{\prime}\in\\{0,2\\}=m(I)$, cf.
\eqref{mety01}},$ (2.2.25) $\displaystyle
M_{0,B,0}^{\scriptscriptstyle\\{m\\}},\text{\footnotesize with $B\in\text{\it
Gl}(n,\mathbb{R})$, $m\in m(B)$, cf. \eqref{mety02}, }$ (2.2.26)
$\displaystyle
M_{0,I,C}^{\scriptscriptstyle\\{m^{\prime\prime}\\}},\text{\footnotesize where
$C\in\text{\sl Sym}(n,\mathbb{R})$, $m^{\prime\prime}\in\\{0,2\\}=m(I)$, cf.
\eqref{mety03}}.$

######  Remark 2.11.

The elements of $\text{\sl Mp}(n)$ are also automorphisms of
$\mathscr{S}(\mathbb{R}^{n})$ and of $\mathscr{S}^{\prime}(\mathbb{R}^{n}).$
Moreover, from (2.2.10),(2.2.11), the elements of $\text{\sl Mp}(n)$ are
products of operators of type (2.2.24),(2.2.25),(2.2.26).

######  Claim 2.12.

If $M$ belongs to $\text{\sl Mp}(n)$, then $-M$ belongs to $\text{\sl Mp}(n)$.

###### Proof.

According to Remark 2.8, we have
$M_{0,I_{n},0}^{\scriptscriptstyle\\{2\\}}=-M_{0,I_{n},0}^{\scriptscriptstyle\\{0\\}}=-\operatorname{Id}_{L^{2}(\mathbb{R}^{n})}$
so that $-\operatorname{Id}_{L^{2}(\mathbb{R}^{n})}$ belongs to $\text{\sl
Mp}(n)$, proving the claim. ∎

The next proposition is Proposition 1.2.15 in [20].

######  Proposition 2.13.

The metaplectic group $\text{\sl Mp}(n)$ is generated by

(2.2.27) $\displaystyle
M_{A,I,0}^{\scriptscriptstyle\\{m^{\prime}\\}},\text{\footnotesize where
$A\in\text{\sl Sym}(n,\mathbb{R})$, $m^{\prime}\in\\{0,2\\}$, cf.
\eqref{mety01}},$ (2.2.28) $\displaystyle
M_{0,B,0}^{\scriptscriptstyle\\{m\\}},\text{\footnotesize with $B\in\text{\it
Gl}(n,\mathbb{R})$, $m\in m(B)$, cf. \eqref{mety02}, }$ (2.2.29)
$\displaystyle e^{-\frac{i\pi n}{4}}\mathcal{F},\text{\footnotesize where
$\mathcal{F}$ is the Fourier transformation.}$

######  Remark 2.14.

We have, with $\sigma_{0}$ defined in (2.3.19),

(2.2.30)
$\sigma_{0}\mathcal{F}^{2}=\operatorname{Id},\quad[\mathcal{F},\sigma_{0}]=0,\quad{\text{so
that}\quad\mathcal{F}^{*}=\mathcal{F}^{-1}=\sigma_{0}\mathcal{F}=\mathcal{F}\sigma_{0}.}$

######  Remark 2.15.

According to (2.2.30) and to Remark 2.8, we find

(2.2.31) $(e^{-i\pi n/4}\mathcal{F})^{*}=e^{i\pi
n/4}\mathcal{F}\sigma_{0}=e^{-i\pi n/4}\mathcal{F}e^{i\pi
n/2}\sigma_{0}=e^{-i\pi
n/4}\mathcal{F}M_{0,-I_{n},0}^{\scriptscriptstyle\\{n\\}}.$

As a consequence, $e^{-i\pi n/4}\mathcal{F},e^{-i\pi n/2}\sigma_{0},e^{i\pi
n/2}\sigma_{0}$ belong to the metaplectic group.

######  Lemma 2.16.

For $Y\in\mathbb{R}^{2n}$, we define the linear form $L_{Y}$ on
$\mathbb{R}^{2n}$ by

(2.2.32) $L_{Y}(X)=\langle\sigma Y,X\rangle=[Y,X].$

For any $M\in\text{\sl Mp}(n)$ there exists a unique $\chi\in\text{\sl
Sp}(2n,\mathbb{R})$ such that

(2.2.33) $\forall Y\in\mathbb{R}^{2n},\quad
M^{*}{\text{{O}p}_{\text{{w}}}(L_{Y})}M={\text{{O}p}_{\text{{w}}}(L_{\chi^{-1}Y})}.$

This is Lemma 1.2.17 in [20]. We can thus define a mapping

(2.2.34) $\Psi:\text{\sl Mp}(n)\rightarrow\text{\sl
Sp}(2n,\mathbb{R}),\quad\text{with $\Psi(M)=\chi$ satisfying \eqref{hg54rr}.}$

In particular we have from (2.2.6) in Proposition 2.6 and (2.2.22), that

(2.2.35) $\Psi(M_{A,B,C})=\Xi_{A,B,C},\qquad\Psi\bigl{(}e^{-\frac{i\pi
n}{4}}\mathcal{F}\bigr{)}=\sigma=\begin{pmatrix}0&I_{n}\\\
-I_{n}&0\end{pmatrix}.$

The following result is Theorem 1.2.18 in [20].

######  Theorem 2.17.

The mapping $\Psi$ defined in (2.2.34) is a surjective homomorphism of groups
with kernel $\\{\pm\operatorname{Id}_{L^{2}(\mathbb{R}^{n})}\\}$.

######  Remark 2.18.

As a consequence, the mapping $\Psi$ is a two-fold covering and $\text{\sl
Mp}(n)$ appears as $Sp_{2}(2n,\mathbb{R})$, the two-fold covering of
$\text{\sl Sp}(2n,\mathbb{R})$. In fact, our results related to the
uncertainty principle, such as Theorem 1.5, can be applied to the subgroup of
the unitary group of $L^{2}(\mathbb{R}^{n})$ generated by (2.2.24), (2.2.25),
(2.2.26) and the operators of multiplication by a complex number of modulus 1
($L^{2}(\mathbb{R}^{n})\ni u\mapsto zu\in L^{2}(\mathbb{R}^{n})$ where
$|z|=1$). Indeed, for a metaplectic operator $\mathcal{M}$ and
$z\in{\mathbb{C}}$, we have obviously

(2.2.36) $\mu\bigl{(}z\mathcal{M}\bigr{)}=|z|\mu(\mathcal{M}).$

####  2.2.2. Another set of generators for the metaplectic group

######  Definition 2.19.

Let $P,L,Q$ be $n\times n$ real matrices such that $P=P^{*},Q=Q^{*}$, $\det
L\not=0$ and $m\in\mathbb{Z}_{4}$ compatible with $L$ (cf.(2.2.3)). We define
the operator $\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$ by the formula

(2.2.37) $(\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}u)(x)=e^{-\frac{i\pi
n}{4}}e^{\frac{i\pi m}{2}}{|\det
L|}^{1/2}\int_{\mathbb{R}^{n}}e^{i\pi\\{\langle Px,x\rangle-2\langle
Lx,y\rangle+\langle Qy,y\rangle\\}}u(y)dy.$

######  N.B.

Let us note that for $m$ compatible with $L$, we have

$\left(e^{\frac{i\pi m}{2}}|\det L|^{1/2}\right)^{2}=\det L.$

The next result is Proposition A.2.2 in the Appendix of [20].

######  Proposition 2.20.

The operator $\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$ given in
Definition 2.19 is an automorphism of $\mathscr{S}(\mathbb{R}^{n})$ and of
$\mathscr{S}^{\prime}(\mathbb{R}^{n})$ which is a unitary operator on
$L^{2}(\mathbb{R}^{n})$ belonging to the metaplectic group (cf. Definition
2.10). Moreover the metaplectic group is generated by the set
$\\{\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}\\}_{\begin{subarray}{c}P=P^{*},\hskip
0.7ptQ=Q^{*}\\\ \det L\not=0,\hskip 0.7ptm\in m(L)\end{subarray}}$.

######  Remark 2.21.

Using the Notation (2.2.5) and (2.2.18), we see that666 We note that
$m(B)+n\in\\{m(-B),m(-B)+2\\}$ modulo 4: indeed we have modulo 4
$\begin{cases}\text{for $n$ even,}&\underbrace{\\{0,2\\}}_{\det
B>0}+n=\underbrace{\\{0,2\\}}_{\det(-B)>0},\qquad\underbrace{\\{1,3\\}}_{\det
B<0}+n=\underbrace{\\{1,3\\}}_{\det(-B)<0},\\\ \text{for $n$
odd,}&\underbrace{\\{0,2\\}}_{\det
B>0}+n=\underbrace{\\{1,3\\}}_{\det(-B)<0},\qquad\underbrace{\\{1,3\\}}_{\det
B<0}+n=\underbrace{\\{0,2\\}}_{\det(-B)>0}.\end{cases}$ We have also
$m(L)-n\in\\{m(-L),m(-L)+2\\}$ since we know already (from the above in that
footnote) that $m(L)-n\in\\{m(-L),m(-L)+2\\}-2n$, which gives
$m(L)-n\in\\{m(-L),m(-L)+2\\}$ for $n$ even; for $n=2l+1$ odd we get the same
result since
$m(L)-n\in\\{m(-L),m(-L)+2\\}-4l-2=\\{m(-L)-2,m(-L)\\}=\\{m(-L)+2,m(-L)\\}.$

(2.2.38)
$M_{A,B,C}^{\scriptscriptstyle\\{m(B)\\}}=\mathcal{M}_{A,-B,C}^{\scriptscriptstyle\\{m(B)+n\\}}\mathcal{F}e^{-i\pi
n/4},\quad\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m(L)\\}}=M_{P,-L,Q}^{\scriptscriptstyle\\{m(L)-n\\}}\left(\mathcal{F}e^{-i\pi
n/4}\right)^{-1},$

The following lemma is Lemma A.2.4 in [20].

######  Lemma 2.22.

With the homomorphism $\Psi$ defined in (2.2.34) and defining

(2.2.39)
$\Lambda_{P,L,Q}=\Psi(\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}),$

we find that

(2.2.40) $\Lambda_{P,L,Q}=\begin{pmatrix}L^{-1}Q&{\hskip 15.0pt}L^{-1}\\\
PL^{-1}Q-L^{*}&{\hskip 15.0pt}PL^{-1}\end{pmatrix}.$

######  Claim 2.23.

Let $P,L,Q,m$ be as in Definition 2.19. Then we have

(2.2.41)
$(\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}})^{-1}=\mathcal{M}_{-Q,-L^{*},-P}^{\scriptscriptstyle\\{n-m\\}},$

and moreover $n-m\in m(-L^{*})$.

Indeed, calculating the kernel $\kappa$ of
$\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}\mathcal{M}_{-Q,-L^{*},-P}^{\scriptscriptstyle\\{n-m\\}}$,
we get

$\kappa(x,y)=e^{\frac{i\pi}{2}(m+n-m-n)}|\det L|\int e^{i\pi\\{Px^{2}-2Lx\cdot
z+Qz^{2}-Qz^{2}+2L^{*}z\cdot y-Py^{2}\\}}dz\\\ =|\det
L|e^{i\pi\\{Px^{2}-Py^{2}\\}}\delta_{0}(Lx-Ly)=\delta_{0}(x-y),$

so that
$\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}\mathcal{M}_{-Q,-L^{*},-P}^{\scriptscriptstyle\\{n-m\\}}=\operatorname{Id}_{L^{2}(\mathbb{R}^{n})}$
and since $\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$ is unitary, this
proves (2.2.41). The last assertion is equivalent to $n-m\in m(L)+n,$ i.e. to
$m\in-m(L)$. Since the mapping

$\mathbb{Z}/4\mathbb{Z}\ni x\mapsto-x\in\mathbb{Z}/4\mathbb{Z},$

leaves stable the sets $\\{0,2\\},\\{1,3\\}$, we obtain the sought result,
concluding the proof of the claim.∎

######  Remark 2.24.

We may also note that since (2.2.41) holds true, having

$(\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}})^{-1}=\mathcal{M}_{P^{\prime},L^{\prime},Q^{\prime}}^{\scriptscriptstyle\\{m^{\prime}\\}}$

would imply that the kernels of the above operator should coincide, i.e.

$e^{-\frac{i\pi
n}{4}}e^{\frac{i\pi}{2}\\{n-m\\}}e^{i\pi\\{-Qx^{2}+2L^{*}x\cdot
y-Py^{2}\\}}=e^{-\frac{i\pi
n}{4}}e^{\frac{i\pi}{2}\\{m^{\prime}\\}}e^{i\pi\\{P^{\prime}x^{2}-2L^{\prime}x\cdot
y+Q^{\prime}y^{2}\\}},$

implying readily that

$P^{\prime}=-Q,L^{\prime}=-L^{*},Q^{\prime}=-P,\quad\pi\frac{(n-m-m^{\prime})}{2}\in
2\pi\mathbb{Z},$

and thus $n-m=m^{\prime}\mod 4$. More generally the mapping (2.2.1)

$\text{\sl Sym}(n,\mathbb{R})\times\text{\it Gl}(n,\mathbb{R})\times\text{\sl
Sym}(n,\mathbb{R})\times\mathbb{Z}_{4}\ni(P,L,Q,m)\mapsto\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}\in\mathcal{U}\bigl{(}L^{2}(\mathbb{R}^{n})\bigr{)},$

is injective since for a symmetric $(2n)\times(2n)$ matrix $\mathcal{A}$ ,

(2.2.42) $\forall
X\in\mathbb{R}^{2n},e^{\frac{i\pi}{2}\\{\mathcal{A}X^{2}+\mu\\}}=1\overbrace{\Longrightarrow}^{\text{\tiny
differentiation}}\forall
X\in\mathbb{R}^{2n},e^{\frac{i\pi}{2}\\{\mathcal{A}X^{2}+\mu\\}}\mathcal{A}X=0\\\
\Longrightarrow\forall X\in\mathbb{R}^{2n},\mathcal{A}X=0\quad\text{i.e.
}\mathcal{A}=0,\quad\text{and thus $\mu\in 4\mathbb{Z}$.}$

The next proposition is Proposition A.2.10 in [20].

######  Proposition 2.25.

The metaplectic group $\text{\sl Mp}(2n)$ is equal to the set

(2.2.43)
$\left\\{\mathcal{M}_{P_{1},L_{1},Q_{1}}^{\scriptscriptstyle\\{m_{1}\\}}\mathcal{M}_{P_{2},L_{2},Q_{2}}^{\scriptscriptstyle\\{m_{2}\\}}\right\\}_{\begin{subarray}{c}P_{j}=P_{j}^{*},\hskip
0.6ptQ_{j}=Q_{j}^{*}\\\ m_{j}\in m(L_{j}),\hskip 0.6pt\det
L_{j}\not=0\end{subarray}}.$

In other words, every metaplectic operator of $\text{\sl Mp}(2n)$ is the
product of two operators of type
$\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$ as given by Definition 2.19.

The next theorem is Theorem A.2.11 in [20].

######  Theorem 2.26.

Let $M$ be an element of $\text{\sl Mp}(n)$ such that
$M=e^{i\phi}\operatorname{Id}_{L^{2}(\mathbb{R}^{n})},\phi\in\mathbb{R}.$ Then
$e^{i\phi}$ belongs to the set $\\{-1,1\\}$. In other words, the intersection
of the metaplectic group with the unit circle (identified to the unitary
operators in $L^{2}(\mathbb{R}^{n})$ defined by the mappings $v\mapsto zv$
where $z\in\mathbb{S}^{1}\subset{\mathbb{C}}$) is reduced to the set
$\\{-1,1\\}$.

We may go back to the description given by Proposition 2.6 and Definition
2.10. The next proposition is Proposition A.2.12 in [20].

######  Proposition 2.27.

The metaplectic group $\text{\sl Mp}(n)$ is equal to the set

(2.2.44)
$\left\\{M_{A_{1},B_{1},C_{1}}^{\scriptscriptstyle\\{m_{1}\\}}M_{A_{2},B_{2},C_{2}}^{\scriptscriptstyle\\{m_{2}\\}}\right\\}_{\begin{subarray}{c}A_{j}=A_{j}^{*},C_{j}=C_{j}^{*}\\\
m_{j}\in m(B_{j}),\ \det B_{j}\not=0\end{subarray}}$

In other words, every metaplectic operator of $\text{\sl Mp}(n)$ is the
product of two operators of type $M_{A,B,C}^{\scriptscriptstyle\\{m\\}}$ as
given by Proposition 2.6.

###  2.3. Weyl quantization

####  2.3.1. Wigner distribution

Let $u,v$ be given functions in $L^{2}(\mathbb{R}^{n})$. The function
$\Omega$, defined on $\mathbb{R}^{n}\times\mathbb{R}^{n}$ by

(2.3.1) $\mathbb{R}^{n}\times\mathbb{R}^{n}\ni(z,x)\mapsto
u(x+\frac{z}{2})\bar{v}(x-\frac{z}{2})=\Omega(u,v)(x,z),$

belongs to $L^{2}(\mathbb{R}^{2n})$ from the identity

(2.3.2)
$\int_{\mathbb{R}^{2n}}{|\Omega(u,v)(x,z)|}^{2}dxdz=\|u\|^{2}_{L^{2}(\mathbb{R}^{n})}\|v\|^{2}_{L^{2}(\mathbb{R}^{n})}.$

We have also

(2.3.3) $\sup_{x\in\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}|\Omega(x,z)|dz\leq
2^{n}\|u\|_{L^{2}(\mathbb{R}^{n})}\|v\|_{L^{2}(\mathbb{R}^{n})}.$

We may then give the following definition.

######  Definition 2.28.

Let $u,v$ be given functions in $L^{2}(\mathbb{R}^{n})$. We define the joint
Wigner distribution $\mathcal{W}(u,v)$ as the partial Fourier transform with
respect to $z$ of the function $\Omega$ defined in (2.3.1). We have for
$(x,\xi)\in\mathbb{R}^{n}_{x}\times\mathbb{R}^{n}_{\xi}$, using (2.3.3),

(2.3.4) $\mathcal{W}(u,v)(x,\xi)=\int_{\mathbb{R}^{n}}e^{-2i\pi
z\cdot\xi}u(x+\frac{z}{2})\bar{v}(x-\frac{z}{2})dz.$

The Wigner distribution of $u$ is defined as $\mathcal{W}(u,u)$.

######  N.B.

By inverse Fourier transformation we get, in a weak sense,

(2.3.5)
$u(x_{1})\otimes\bar{v}(x_{2})=\int\mathcal{W}(u,v)(\frac{x_{1}+x_{2}}{2},\xi)\
e^{2i\pi(x_{1}-x_{2})\cdot\xi}d\xi.$

######  Lemma 2.29.

Let $u,v$ be given functions in $L^{2}(\mathbb{R}^{n})$. The function
$\mathcal{W}(u,v)$ belongs to $L^{2}(\mathbb{R}^{2n})$ and we have

(2.3.6)
$\|\mathcal{W}(u,v)\|_{L^{2}(\mathbb{R}^{2n})}=\|u\|_{L^{2}(\mathbb{R}^{n})}\|v\|_{L^{2}(\mathbb{R}^{n})}.$

We have also

(2.3.7) $\overline{\mathcal{W}(u,v)(x,\xi)}=\mathcal{W}(v,u)(x,\xi),$

so that $\mathcal{W}(u,u)$ is real-valued.

###### Proof.

Note that the function $\mathcal{W}(u,v)$ is in $L^{2}(\mathbb{R}^{2n})$ and
satisfies (2.3.6) from (2.3.2) and the definition of $\mathcal{W}$ as the
partial Fourier transform of $\Omega$. Property (2.3.7) is immediate and
entails that $\mathcal{W}(u,u)$ is real-valued. ∎

######  Remark 2.30.

We note also that the real-valued function $\mathcal{W}(u,u)$ can take
negative values, choosing for instance $u_{1}(x)=xe^{-\pi x^{2}}$ on the real
line, we get

$\mathcal{W}(u_{1},u_{1})(x,\xi)=2^{1/2}e^{-2\pi(x^{2}+\xi^{2})}\Bigl{(}x^{2}+\xi^{2}-\frac{1}{4\pi}\Bigr{)}.$

####  2.3.2. The Weyl quantization

The main goal of Hermann Weyl in his seminal paper [26] was to give a simple
formula, also providing symplectic covariance, ensuring that real-valued
Hamiltonians $a(x,\xi)$ get quantized by formally self-adjoint operators. The
standard way of dealing with differential operators does not achieve that goal
since for instance the standard quantization of the Hamiltonian $x\xi$ (indeed
real-valued) is the operator $xD_{x}$, which is not symmetric
($D_{x}=\frac{1}{2i\pi}\frac{\partial}{\partial x})$; H. Weyl’s choice in that
case was

$x\xi\ \text{should be quantized by the
operator\quad}\frac{1}{2}(xD_{x}+D_{x}x),\quad\text{(indeed symmetric),}$

and more generally, say for
$a\in\mathscr{S}(\mathbb{R}^{2n}),u\in\mathscr{S}(\mathbb{R}^{n})$, the
quantization of the Hamiltonian $a(x,\xi)$, denoted by
${\text{{O}p}_{\text{{w}}}(a)}$, should be given by the formula

(2.3.8) $({\text{{O}p}_{\text{{w}}}(a)}u)(x)=\iint
e^{2i\pi(x-y)\cdot\xi}a\Bigl{(}\frac{x+y}{2},\xi\Bigr{)}u(y)dyd\xi.$

For $v\in\mathscr{S}(\mathbb{R}^{n})$, we may consider

$\langle{\text{{O}p}_{\text{{w}}}(a)}u,v\rangle_{L^{2}(\mathbb{R}^{n})}=\iiint
a(x,\xi)e^{-2i\pi z\cdot\xi}u(x+\frac{z}{2})\bar{v}(x-\frac{z}{2})dzdxd\xi\\\
=\iint_{\mathbb{R}^{n}\times\mathbb{R}^{n}}a(x,\xi)\mathcal{W}(u,v)(x,\xi)dxd\xi,$

and the latter formula allows us to give the following definition.

######  Definition 2.31.

Let $a\in\mathscr{S}^{\prime}(\mathbb{R}^{2n})$. We define the Weyl
quantization ${\text{{O}p}_{\text{{w}}}(a)}$ of the Hamiltonian $a$, by the
formula

(2.3.9) $({\text{{O}p}_{\text{{w}}}(a)}u)(x)=\iint
e^{2i\pi(x-y)\cdot\xi}a\Bigl{(}\frac{x+y}{2},\xi\Bigr{)}u(y)dyd\xi,$

to be understood weakly as

(2.3.10)
$\langle{\text{{O}p}_{\text{{w}}}(a)}u,\bar{v}\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{n}),\mathscr{S}(\mathbb{R}^{n})}=\langle
a,\mathcal{W}(u,v)\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{2n}),\mathscr{S}(\mathbb{R}^{2n})}.$

We note that the sesquilinear mapping

$\mathscr{S}(\mathbb{R}^{n})\times\mathscr{S}(\mathbb{R}^{n})\ni(u,v)\mapsto\mathcal{W}(u,v)\in\mathscr{S}(\mathbb{R}^{2n}),$

is continuous so that the above bracket of duality $\langle
a,\mathcal{W}(u,v)\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{2n}),\mathscr{S}(\mathbb{R}^{2n})}$
makes sense. We note as well that a temperate distribution
$a\in\mathscr{S}^{\prime}(\mathbb{R}^{2n})$ gets quantized by a continuous
operator ${\text{{O}p}_{\text{{w}}}(a)}$ from $\mathscr{S}(\mathbb{R}^{n})$
into $\mathscr{S}^{\prime}(\mathbb{R}^{n})$. This very general framework is
not really useful since we want to compose our operators
${\text{{O}p}_{\text{{w}}}(a)}{\text{{O}p}_{\text{{w}}}(b)}$. A first step in
this direction is to look for sufficient conditions ensuring that the operator
${\text{{O}p}_{\text{{w}}}(a)}$ is bounded on $L^{2}(\mathbb{R}^{n})$.
Moreover, for $a\in\mathscr{S}^{\prime}(\mathbb{R}^{2n})$ and $b$ a
_polynomial_ in ${\mathbb{C}}[x,\xi]$, we have the composition formula,

(2.3.11)
$\displaystyle{\text{{O}p}_{\text{{w}}}(a)}{\text{{O}p}_{\text{{w}}}(b)}$
$\displaystyle={\text{{O}p}_{\text{{w}}}(a\sharp b)},$ (2.3.12)
$\displaystyle(a\sharp b)(x,\xi)$ $\displaystyle=\sum_{k\geq
0}\frac{1}{(4i\pi)^{k}}\sum_{|\alpha|+|\beta|=k}\frac{(-1)^{|\beta|}}{\alpha!\beta!}(\partial_{\xi}^{\alpha}\partial_{x}^{\beta}a)(x,\xi)(\partial_{x}^{\alpha}\partial_{\xi}^{\beta}b)(x,\xi),$

which involves here a finite sum. This follows from (2.1.26) in [19] where
several generalizations can be found (see in particular in that reference the
integral formula (2.1.18) which can be given a meaning for quite general
classes of symbols). We shall use the following notation

(2.3.13)
$\omega_{k}(a,b)=\frac{1}{(4i\pi)^{k}}\sum_{|\alpha|+|\beta|=k}\frac{(-1)^{|\beta|}}{\alpha!\beta!}(\partial_{\xi}^{\alpha}\partial_{x}^{\beta}a)(x,\xi)(\partial_{x}^{\alpha}\partial_{\xi}^{\beta}b)(x,\xi).$

As a consequence of (2.3.12), we get that

(2.3.14) $\displaystyle(a\sharp b)$ $\displaystyle=\sum_{k\geq
0}\omega_{k}(a,b),\quad\omega_{0}(a,b)=ab,\quad\omega_{1}(a,b)=\frac{1}{4i\pi}\left\\{a,b\right\\},$
(2.3.15) $\displaystyle\left\\{a,b\right\\}$
$\displaystyle=\partial_{\xi}a\cdot\partial_{x}b-\partial_{x}a\partial_{\xi}b,$

where $\left\\{a,b\right\\}$ is called the _Poisson bracket_ of $a$ and $b$.

######  Claim 2.32.

Let $L$ be a linear form on $\mathbb{R}^{2n}$ and let $N\geq 1$ be an integer.
Then we have

(2.3.16) $\underbrace{L\sharp\dots\sharp L}_{\text{\rm$N$ factors}}=L^{N}.$

###### Proof of the claim.

The formula is true for $N=1,2,$ from Formulas (2.3.11), (2.3.12) since
$\left\\{L,L\right\\}=0$. Let $N\geq 2$ such that (2.3.16) holds true. Then we
calculate

$\underbrace{L\sharp\dots\sharp L}_{\text{\rm$N$ factors}}\sharp L=L^{N}\sharp
L=L^{N}L+\frac{1}{4i\pi}\left\\{L^{N},L\right\\}=L^{N+1},$

concluding the induction proof. ∎

The next proposition is Proposition 1.2.2 in [20].

######  Proposition 2.33.

Let $a$ be a tempered distribution on $\mathbb{R}^{2n}$. Then we have

(2.3.17)
$\|{\text{{O}p}_{\text{{w}}}(a)}\|_{\mathcal{B}(L^{2}(\mathbb{R}^{n}))}\leq\min\bigl{(}2^{n}\|a\|_{L^{1}(\mathbb{R}^{2n})},\|\hat{a}\|_{L^{1}(\mathbb{R}^{2n})}\bigr{)}.$

The next statement is Claim 1.2.3 in [20].

######  Claim 2.34.

Let $(x,\xi)$ be in $\mathbb{R}^{n}\times\mathbb{R}^{n}$. We define the
operator $\sigma_{x,\xi}$ (called _phase symmetry_), on
$\mathscr{S}^{\prime}(\mathbb{R}^{n})$ by

(2.3.18) $(\sigma_{x,\xi}u)(y)=u(2x-y)e^{-4i\pi(x-y)\cdot\xi}.$

Then $\sigma_{x,\xi}$ is unitary and self-adjoint on $L^{2}(\mathbb{R}^{n})$.
We have in particular

(2.3.19) $(\sigma_{0,0}w)(y)=w(-y).$

The operator $\sigma_{0,0}$ will be denoted by $\sigma_{0}$.

######  Proposition 2.35.

Let $a\in\mathscr{S}^{\prime}(\mathbb{R}^{2n})$. The distribution kernel
$k_{a}(x,y)$ of the operator ${\text{{O}p}_{\text{{w}}}(a)}$ is

(2.3.20) $k_{a}(x,y)=\widehat{a}^{[2]}(\frac{x+y}{2},y-x),$

where $a^{[2]}$ stands for the Fourier transform of $a$ with respect to the
second variable. Let $k\in\mathscr{S}^{\prime}(\mathbb{R}^{2n})$ be the
distribution kernel of a continuous operator $A$ from
$\mathscr{S}(\mathbb{R}^{n})$ into $\mathscr{S}^{\prime}(\mathbb{R}^{n})$.
Then the Weyl symbol $a$ of $A$ is

(2.3.21) $a(x,\xi)=\int e^{-2\pi
it\cdot\xi}k\bigl{(}x+\frac{t}{2},x-\frac{t}{2}\bigr{)}dt,$

where the integral sign means that we take the Fourier transform with respect
to $t$ of the distribution $k(x+\frac{t}{2},x-\frac{t}{2})$ on
$\mathbb{R}^{2n}$.

###### Proof.

With $u,v\in\mathscr{S}(\mathbb{R}^{n})$, we have defined
${\text{{O}p}_{\text{{w}}}(a)}$ via Formula (2.3.10) and using Definition
2.28, we get

$\displaystyle\langle{\text{{O}p}_{\text{{w}}}(a)}u,\bar{v}\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{n}),\mathscr{S}(\mathbb{R}^{n})}$
$\displaystyle=\langle
a(x,\xi),\widehat{\Omega}^{[2]}(x,\xi)\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{2n}),\mathscr{S}(\mathbb{R}^{2n})}$
$\displaystyle=\langle\widehat{a}^{[2]}(t,z),u(t+\frac{z}{2})\bar{v}(t-\frac{z}{2})\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{2n}),\mathscr{S}(\mathbb{R}^{2n})}$
$\displaystyle=\langle\widehat{a}^{[2]}(\frac{x+y}{2},y-x),u(y)\bar{v}(x)\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{2n}),\mathscr{S}(\mathbb{R}^{2n})},$

proving (2.3.20). As a consequence, we find that
$k_{a}(x+\frac{t}{2},x-\frac{t}{2})={\widehat{a}^{[2]}}(x,-t),$ and by Fourier
inversion, this entails

(2.3.22)
$a(x,\xi)=\text{Fourier}_{t}\bigl{(}k_{a}(x+\frac{t}{2},x-\frac{t}{2})\bigr{)}(\xi)=\int
e^{-2\pi it\cdot\xi}k_{a}(x+\frac{t}{2},x-\frac{t}{2})dt,$

where the integral sign means that we perform a Fourier transformation with
respect to the variable $t$. ∎

####  2.3.3. Symplectic covariance of the Weyl calculus

######  Theorem 2.36 (Symplectic covariance of the Weyl calculus).

Let $a$ be in $\mathscr{S}^{\prime}(\mathbb{R}^{2n})$ and let $\chi$ be in
$\text{\sl Sp}(2n,\mathbb{R})$. Then for a metaplectic operator $M$ such that
$\Psi(M)=\chi$, we have

(2.3.23) $M^{*}\
{\text{{O}p}_{\text{{w}}}(a)}M={\text{{O}p}_{\text{{w}}}(a\circ\chi)}.$

For $u,v\in\mathscr{S}(\mathbb{R}^{n}),$ we have

(2.3.24) $\mathcal{W}\left(Mu,Mv\right)=\mathcal{W}(u,v)\circ\chi^{-1},$

where $\mathcal{W}$ is the Wigner distribution given in (2.3.4).

###### Proof.

The first property follows immediately from (2.2.6) and Definition 2.10,
whereas (2.3.24) is a consequence of (2.3.10) and (2.3.23). ∎

We note also that for $Y=(y,\eta)\in\mathbb{R}^{2n}$, the symmetry $S_{Y}$ is
defined by $S_{Y}(X)=2Y-X$ and is quantized by the phase symmetry $\sigma_{Y}$
as defined by (2.3.18) with the formula

(2.3.25) ${\text{{O}p}_{\text{{w}}}(a\circ
S_{Y})}=\sigma_{Y}^{*}{\text{{O}p}_{\text{{w}}}(a)}\sigma_{Y}=\sigma_{Y}{\text{{O}p}_{\text{{w}}}(a)}\sigma_{Y}.$

Similarly, the translation $T_{Y}$ is defined on the phase space by
$T_{Y}(X)=X+Y$ and is quantized by the phase translation $\tau_{Y}$,

(2.3.26) $(\tau_{(y,\eta)}u)(x)=u(x-y)e^{2i\pi(x-\frac{y}{2})\cdot\eta},$

and we have

(2.3.27) ${\text{{O}p}_{\text{{w}}}(a\circ
T_{Y})}=\tau_{Y}^{*}{\text{{O}p}_{\text{{w}}}(a)}\tau_{Y}=\tau_{-Y}{\text{{O}p}_{\text{{w}}}(a)}\tau_{Y}.$

##  3\. Proofs of the main results

###  3.1. First cases: $\Xi_{12}$ invertible

######  Proposition 3.1.

Let $\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$ be a metaplectic
operator as given by Definition 2.19. Then with the variance defined by
(1.2.1), we have for all $u\in\mathscr{S}(\mathbb{R}^{n})$,

(3.1.1)
$\sqrt{\mathcal{V}(\mathcal{M}u)\mathcal{V}(u)}\geq\frac{\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}}{4\pi}\operatorname{trace}\left((L^{*-1}L^{-1})^{1/2}\right)=\frac{\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}}{4\pi}|\hskip-1.0pt\|L^{-1}\|\hskip-1.0pt|_{{\mathtt{S}^{1}}},$

where $|\hskip-1.0pt\|L^{-1}\|\hskip-1.0pt|_{{\mathtt{S}^{1}}}$ is the trace-
norm of $L^{-1}$, that is the sum of the eigenvalues of $|L^{-1}|$ (see
Section 4.1.2 in our Appendix).

###### Proof of the Proposition.

We start with a lemma.

######  Lemma 3.2.

Let $\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$ be a metaplectic
operator as given by Definition 2.19. Let $a,b$ be linear forms on the phase
space given by (3.1.6). Then we have, with $\Lambda_{P,L,Q}$ defined in
(2.2.40),

(3.1.2) $2\operatorname{Re}\left\langle(\zeta\cdot y-z\cdot
D_{y})(\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}u)(y),i\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}\bigl{(}\tau\cdot
x-t\cdot D_{x})u\bigr{)}\right\rangle_{L^{2}(\mathbb{R}^{n})}\\\
=2\operatorname{Re}\left\langle{\text{{O}p}_{\text{{w}}}(a)}\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}u,i\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}{\text{{O}p}_{\text{{w}}}(b)}u\right\rangle\\\
=\frac{1}{2\pi}\Bigl{\\{}a\circ\Lambda_{P,L,Q},b\Bigr{\\}}\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}\\\
=\frac{1}{2\pi}\Bigl{[}(\zeta-Pz)\cdot\bigl{(}L^{-1}(\tau+Qt)\bigr{)}+Lz\cdot
t\Bigr{]}\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}.$

######  N.B.

We know from Proposition 2.25 that every element in the metaplectic group is
the product of two operators of type
$\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$ as given in Definition 2.19
. Of course it is _not_ true that all metaplectic operators are of type
$\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$ since the Identity itself
has the singular kernel $\delta(x-y)$ which is different from all the
imaginary Gaussian functions serving as kernels of the operators
$\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$.

###### Proof of the lemma.

Let $\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$ be a metaplectic
operator as given by Definition 2.19. We would like to calculate for
$u\in\mathscr{S}(\mathbb{R}^{n})$, $a,b$ real valued linear forms on
$\mathbb{R}^{n}_{x}\times\mathbb{R}^{n}_{\xi}$, with $L^{2}(\mathbb{R}^{n})$
dot-products,

$\displaystyle 2\operatorname{Re}$
$\displaystyle\left\langle{\text{{O}p}_{\text{{w}}}(a)}\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}u,i\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}{\text{{O}p}_{\text{{w}}}(b)}u\right\rangle$
$\displaystyle=2\operatorname{Re}\left\langle(\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}})^{\hskip-1.0pt*}{\text{{O}p}_{\text{{w}}}(a)}\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}u,i{\text{{O}p}_{\text{{w}}}(b)}u\right\rangle$
(from Lemma 2.22)
$\displaystyle=2\operatorname{Re}\left\langle{\text{{O}p}_{\text{{w}}}(a\circ\Lambda_{P,L,Q})}u,i{\text{{O}p}_{\text{{w}}}(b)}u\right\rangle$
$\displaystyle=\left\langle\bigl{[}{\text{{O}p}_{\text{{w}}}(a\circ\Lambda_{P,L,Q})},i{\text{{O}p}_{\text{{w}}}(b)}\bigr{]}u,u\right\rangle$
(3.1.3)
$\displaystyle=\frac{1}{2\pi}\left\langle{\text{{O}p}_{\text{{w}}}\bigl{(}\left\\{a\circ\Lambda_{P,L,Q},b\right\\}\bigr{)}}u,u\right\rangle.$

We may assume777We may keep in mind that the phase space is $\Phi=E\oplus
E^{*}$ (here $E$ is the _configuration space_ , a real finite dimensional
vector space) and that the symplectic form $\sigma$ is bilinear alternate on
$\Phi$ with  (3.1.4) $\langle\sigma
X,Y\rangle_{\Phi^{*},\Phi}=[X,Y]=\langle\xi,y\rangle_{E^{*},E}-\langle\eta,x\rangle_{E^{*},E},\quad
X=x\oplus\xi,Y=y\oplus\eta,$ in such a way that our notation $\xi\cdot
y-\eta\cdot x$ is the difference of two brackets of duality. We have also that
the kernel of the operator $\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$,
given by (2.2.37) is  $e^{-\frac{i\pi n}{4}}e^{\frac{i\pi m(L)}{2}}{|\det
L|}^{1/2}e^{i\pi\left\\{\langle Px,x\rangle_{E^{*},E}-2\langle
Lx,y\rangle_{E^{*},E}+\langle Qy,y\rangle_{E^{*},E}\right\\}},$ with  (3.1.5)
$P:E\longrightarrow E^{*},\ P=P^{*},\qquad Q:E\longrightarrow E^{*},\
Q=Q^{*},\qquad L:E\longrightarrow E^{*}\text{ invertible.}$ that, with
$Y=(y,\eta)$, $Z=(z,\zeta)$, $T=(t,\tau)$, $X=(x,\xi)$, in
$\mathbb{R}^{n}\times\mathbb{R}^{n}$,

(3.1.6) $a(Y)=[Z,Y]=\zeta\cdot y-\eta\cdot z,\qquad b(X)=[T,X]=\tau\cdot
x-\xi\cdot t.$

We have

(3.1.7) $(a\circ\Lambda)(X)=[Z,\Lambda X]=[\Lambda^{-1}Z,X],$

and since $\Lambda$ is a symplectic mapping we have
$\Lambda^{*}\sigma\Lambda=\sigma$, that is from (2.2.40) and (3.1.5)

$\displaystyle\Lambda^{-1}=\sigma^{-1}\Lambda^{*}\sigma$
$\displaystyle=\begin{pmatrix}0&-I\\\
I&0\end{pmatrix}\begin{pmatrix}QL^{*-1}&QL^{*-1}P-L\\\ {\hskip
15.0pt}L^{*-1}&{\hskip 15.0pt}L^{*-1}P\end{pmatrix}\begin{pmatrix}0&I\\\
-I&0\end{pmatrix}$ $\displaystyle=\begin{pmatrix}0&-I\\\
I&0\end{pmatrix}\begin{pmatrix}L-QL^{*-1}P&{\hskip 15.0pt}QL^{*-1}\\\
-L^{*-1}P&L^{*-1}\end{pmatrix}$ $\displaystyle=\begin{pmatrix}L^{*-1}P&{\hskip
15.0pt}-L^{*-1}\\\ L-QL^{*-1}P&QL^{*-1}\end{pmatrix},$

and thus (3.1.7) gives

(3.1.8)
$(a\circ\Lambda)(X)=\Bigl{[}\Bigl{(}L^{*-1}Pz-L^{*-1}\zeta,(L-QL^{*-1}P)z+QL^{*-1}\zeta\Bigr{)},(x,\xi)\Bigr{]}\\\
=\Bigl{(}(L-QL^{*-1}P)z+QL^{*-1}\zeta\Bigr{)}\cdot
x-\xi\cdot\Bigl{(}L^{*-1}Pz-L^{*-1}\zeta\Bigr{)}.$

We calculate now the Poisson bracket, using (3.1.6), (3.1.8),

(3.1.9) $\displaystyle\Bigl{\\{}a\circ\Lambda,b\Bigr{\\}}$
$\displaystyle=\bigl{(}-L^{*-1}Pz+L^{*-1}\zeta\bigr{)}\cdot\tau+\bigl{(}(L-QL^{*-1}P)z+QL^{*-1}\zeta\bigr{)}\cdot
t$ $\displaystyle=\tau\cdot L^{*-1}(\zeta-Pz)+\bigl{(}Lz+QL^{*-1}(\zeta-
Pz)\bigr{)}\cdot t$ $\displaystyle=\underbrace{(\zeta-
Pz)}_{E^{*}}\cdot\bigl{(}\underbrace{L^{-1}\underbrace{(\tau+Qt)}_{E^{*}}}_{E}\bigr{)}+\underbrace{Lz}_{E^{*}}\cdot\underbrace{t}_{E},$

which does not depend on the variable $X$ as the Poisson bracket of two linear
forms in $X$. We get thus from (3.1.3)

(3.1.10)
$2\operatorname{Re}\left\langle{\text{{O}p}_{\text{{w}}}(a)}\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}u,i\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}{\text{{O}p}_{\text{{w}}}(b)}u\right\rangle_{L^{2}(\mathbb{R}^{n})}\\\
=\frac{1}{2\pi}\Bigl{[}(\zeta-Pz)\cdot\bigl{(}L^{-1}(\tau+Qt)\bigr{)}+Lz\cdot
t\Bigr{]}\|u\|_{L^{2}(\mathbb{R}^{n})}^{2},$

that is

(3.1.11) $2\operatorname{Re}\left\langle(\zeta\cdot y-z\cdot
D_{y})(\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}u)(y),i\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}\bigl{(}\tau\cdot
x-t\cdot D_{x})u\bigr{)}\right\rangle_{L^{2}(\mathbb{R}^{n})}\\\
=\frac{1}{2\pi}\Bigl{[}(\zeta-Pz)\cdot\bigl{(}L^{-1}(\tau+Qt)\bigr{)}+Lz\cdot
t\Bigr{]}\|u\|_{L^{2}(\mathbb{R}^{n})}^{2},$

concluding the proof of Lemma 3.2. ∎

Let us go on with the proof of Proposition 3.1. Choosing $z=t=0$ in Lemma 3.2,
we get

$2\operatorname{Re}\left\langle(\zeta\cdot
y)(\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}u)(y),i\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}\bigl{(}(\tau\cdot
x)u\bigr{)}\right\rangle_{L^{2}(\mathbb{R}^{n})}=\frac{1}{2\pi}(\zeta\cdot
L^{-1}\tau)\|u\|_{L^{2}(\mathbb{R}^{n})}^{2},$

which implies that

(3.1.12) $\|(\zeta\cdot
y)(\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}u)(y)\|_{L^{2}(\mathbb{R}^{n})}\|(\tau\cdot
x)u\|_{L^{2}(\mathbb{R}^{n})}\geq\frac{1}{4\pi}|(\zeta\cdot
L^{-1}\tau)|\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}.$

######  N.B.

We note at this stage of the proof, that Inequality (3.1.12) is already
providing a non-trivial estimate by choosing $\zeta=L^{-1}\tau$: we get

$\|L^{-1}\tau\|\|\tau\|\||y|(\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}u)(y)\|_{L^{2}(\mathbb{R}^{n})}\||x|u\|_{L^{2}(\mathbb{R}^{n})}\geq\frac{1}{4\pi}\|L^{-1}\tau\|^{2}\|u\|_{L^{2}(\mathbb{R}^{n})}^{2},$

implying for any $\tau$,

$\sqrt{\mathcal{V}\bigl{(}\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}u\bigr{)}\mathcal{V}\bigl{(}u\bigr{)}}\geq\frac{1}{4\pi}\frac{\|L^{-1}\tau\|}{\|\tau\|}\|u\|_{L^{2}(\mathbb{R}^{n})}^{2},$

and thus

(3.1.13)
$\mu\bigl{(}\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}\bigr{)}\geq\frac{1}{4\pi}\|L^{-1}\|=\frac{1}{4\pi}|\hskip-1.0pt\|L^{-1}\|\hskip-1.0pt|_{{\mathtt{S}^{{\infty}}}}.$

However, we are looking for a better estimate, and we want to replace
$|\hskip-1.0pt\|L^{-1}\|\hskip-1.0pt|_{{\mathtt{S}^{{\infty}}}}$ in the right-
hand-side (just the operator-norm) by the much larger Schatten norm of index
1, $|\hskip-1.0pt\|L^{-1}\|\hskip-1.0pt|_{{\mathtt{S}^{1}}}$ (see (4.1.12) in
our Appendix); as a matter of fact, we shall see below that this will give the
equality

$\mu\bigl{(}\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}\bigr{)}=\frac{1}{4\pi}|\hskip-1.0pt\|L^{-1}\|\hskip-1.0pt|_{{\mathtt{S}^{1}}}.$

Let $(\zeta_{j})_{1\leq j\leq n}$, $(\tau_{j})_{1\leq j\leq n}$ be two
orthonormal basis of $\mathbb{R}^{n}$ for the canonical Euclidean structure on
$\mathbb{R}^{n}$. We find from (3.1.12) that for each $j\in\llbracket
1,n\rrbracket$,

$\left(\int{|\zeta_{j}\cdot
y|}^{2}{|(\mathcal{M}u)(y)|}^{2}dy\right)^{1/2}\left(\int|\tau_{j}\cdot
x|^{2}{|u(x)|}^{2}dx\right)^{1/2}\geq\frac{\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}}{4\pi}|\langle\zeta_{j},L^{-1}\tau_{j}\rangle|,$

so that

$\sum_{1\leq j\leq n}\left(\int{|\zeta_{j}\cdot
y|}^{2}{|(\mathcal{M}u)(y)|}^{2}dy\right)^{1/2}\left(\int{|\tau_{j}\cdot
x|}^{2}{|u(x)|}^{2}dx\right)^{1/2}\\\
\geq\frac{\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}}{4\pi}\sum_{1\leq j\leq
n}|\langle\zeta_{j},L^{-1}\tau_{j}\rangle|,$

which implies that

$\left(\sum_{1\leq j\leq n}\int{|\zeta_{j}\cdot
y|}^{2}{|(\mathcal{M}u)(y)|}^{2}dy\right)^{1/2}\left(\sum_{1\leq j\leq
n}\int{|\tau_{j}\cdot x|}^{2}{|u(x)|}^{2}dx\right)^{1/2}\\\
\geq\frac{\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}}{4\pi}\sum_{1\leq j\leq
n}|\langle\zeta_{j},L^{-1}\tau_{j}\rangle|,$

i.e.

$\left(\int{|y|}^{2}{|(\mathcal{M}u)(y)|}^{2}dy\right)^{1/2}\left(\int{|x|}^{2}{|u(x)|}^{2}dx\right)^{1/2}\geq\frac{\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}}{4\pi}\sum_{1\leq
j\leq n}|\langle\zeta_{j},L^{-1}\tau_{j}\rangle|,$

so that

(3.1.14)
$\mathcal{V}(\mathcal{M}u)\mathcal{V}(u)\geq\frac{\|u\|_{L^{2}(\mathbb{R}^{n})}^{4}}{4^{2}\pi^{2}}\left(\sum_{1\leq
j\leq n}|\langle\zeta_{j},L^{-1}\tau_{j}\rangle|\right)^{2}.$

Thanks to our Proposition 4.5 in our Appendix we have with a unitary matrix
$W$ the identity

$\langle\zeta_{j},L^{-1}\tau_{j}\rangle=\langle\zeta_{j},W|L^{-1}|\tau_{j}\rangle=\langle
W^{*}\zeta_{j},|L^{-1}|\tau_{j}\rangle.$

The set $(W^{*}\zeta_{j})_{1\leq j\leq n}$ is also an orthonormal basis and we
can choose $\tau_{j}=W^{*}\zeta_{j}$ so that we get

$\sum_{1\leq j\leq n}|\langle\zeta_{j},L^{-1}\tau_{j}\rangle|=\sum_{1\leq
j\leq n}\langle|L^{-1}|\tau_{j},\tau_{j}\rangle=\operatorname{trace}|L^{-1}|,$

and this, together with (3.1.14), gives the result of Proposition 3.1. ∎

Let us now take a look at a simple example.

######  Example 3.3.

Let $n\geq 2$ and let $\mathcal{F}_{1}$ be the partial Fourier transformation
with respect to the variable $x_{1}$. Then the operator
$e^{-i\pi/4}\mathcal{F}_{1}$ belongs to the metaplectic group
$\textsl{Mp}(n)$. Indeed, let us consider, with obvious notations, the
symmetric $n\times n$ matrix,

(3.1.15) $C=\begin{pmatrix}-1&0_{1,n-1}\\\
0_{n-1,1}&I_{n-1}\end{pmatrix},\quad e^{i\pi CD^{2}}=e^{-i\pi
D_{1}^{2}}e^{i\pi D_{2}^{2}},\quad D_{2}=(D_{x_{2}},\dots,D_{x_{n}}).$

Using the notations (2.2.5), (2.2.38), we have

(3.1.16) $e^{i\pi
CD^{2}}=M_{0,I_{n},C}^{\scriptscriptstyle\\{0\\}}=\mathcal{M}_{0,-I_{n},C}^{\scriptscriptstyle\\{n\\}}\mathcal{F}e^{-i\pi
n/4}\in\textsl{Mp}(n),$

where $\mathcal{F}$ stands for the (full) Fourier transform. We calculate the
kernel $\kappa$ of the operator $e^{i\pi CD^{2}}e^{-i\pi/4}\mathcal{F}_{1}$
and we find that

$\displaystyle\kappa(x,y)$ $\displaystyle=\int
e^{2i\pi(x-z)\cdot\eta}e^{i\pi(-\eta_{1}^{2}+\eta_{2}^{2})}d\eta_{1}d\eta_{2}e^{-2i\pi
z_{1}y_{1}}\delta(z_{2}-y_{2})dz_{1}dz_{2}e^{-i\pi/4}$ $\displaystyle=\int
e^{2i\pi[(x_{1}-z_{1})\eta_{1}+(x_{2}-y_{2})\cdot\eta_{2}]}e^{i\pi(-\eta_{1}^{2}+\eta_{2}^{2})}d\eta_{1}d\eta_{2}e^{-2i\pi
z_{1}y_{1}}dz_{1}e^{-i\pi/4}$
$\displaystyle=e^{i\pi[-x_{2}^{2}-2(x_{1}y_{1}-x_{2}y_{2})-y_{1}^{2}-y_{2}^{2}]}e^{\frac{i\pi(n-2)}{4}}.$

With

$L=\begin{pmatrix}1&0\\\ 0&-I_{n-1}\end{pmatrix},\quad\text{we have }(\det
L)^{1/2}=\pm e^{\frac{i\pi(n-1)}{2}},$

so that

$e^{i\pi
CD^{2}}e^{-i\pi/4}\mathcal{F}_{1}=\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}},\quad
P=\begin{pmatrix}0&0\\\ 0&-I_{n-1}\end{pmatrix},Q=-I_{n},$

since

$e^{-\frac{i\pi n}{4}}e^{\frac{i\pi
m}{2}}=e^{\frac{i\pi(n-2)}{4}}\Longleftrightarrow
m-\frac{n}{2}=\frac{n}{2}-1,\mod 4\Longleftrightarrow m=n-1,\mod 4.$

Eventually we have $e^{i\pi(n-1)}=\operatorname{sign}(\det L)$

$e^{i\pi
CD^{2}}e^{-i\pi/4}\mathcal{F}_{1}=\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m_{L}\\}},\quad
m_{L}\in m(L),$

proving that $e^{-i\pi/4}\mathcal{F}_{1}$ belongs to $\textsl{Mp}(n)$,
although it is not888Indeed, the kernel of $e^{-i\pi/4}\mathcal{F}_{1}$ is
$e^{-i\pi/4}e^{-2i\pi x_{1}y_{1}}\delta_{0,n-1}(x_{2}-y_{2}),$ which is
supported in a vector subspace with dimension $n+1$ in $\mathbb{R}^{2n}$,
whereas the kernel of $\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m(L)\\}}$ is
the Gaussian function  $e^{-\frac{i\pi n}{4}}e^{\frac{i\pi
m(L)}{2}}e^{i\pi(Px^{2}-2Lx\cdot y+Qy^{2})},$ whose support is the whole
$\mathbb{R}^{2n}$.  of the type
$\mathcal{M}_{P,L,Q}^{\scriptscriptstyle\\{m\\}}$. We check now,

$\displaystyle\frac{1}{2\pi}\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}=2\operatorname{Re}\langle(D_{1}u)(x_{1},x_{2}),ix_{1}u(x_{1},x_{2})\rangle_{L^{2}(\mathbb{R}^{n})}$
$\displaystyle=2\operatorname{Re}\langle\xi_{1}e^{-i\pi/4}\mathcal{F}_{1}u,ie^{-i\pi/4}\mathcal{F}_{1}(x_{1}u)\rangle_{L^{2}(\mathbb{R}^{n})}$
$\displaystyle\leq
2\left(\iint{|\xi_{1}|}^{2}\left|(e^{-i\pi/4}\mathcal{F}_{1}u)(\xi_{1},x_{2})\right|^{2}d\xi_{1}dx_{2}\right)^{1/2}\left(\iint{|x_{1}|}^{2}\left|u(x_{1},x_{2})\right|^{2}dx_{1}dx_{2}\right)^{1/2}$
$\displaystyle\leq
2\mathcal{V}\bigl{(}e^{-i\pi/4}\mathcal{F}_{1}u\bigr{)}^{1/2}\mathcal{V}(u)^{1/2},$

so that, according to the notation (1.2.4),

(3.1.17) $\mu(e^{-i\pi/4}\mathcal{F}_{1})\geq\frac{1}{4\pi}.$

######  Lemma 3.4.

Let $n$ be an integer $\geq 1$ and let $n_{1},n_{2}$ integers such that
$n_{1}\geq 1$ and $n=n_{1}+n_{2}$. We define $\mathcal{F}_{n_{1}}$ as the
Fourier transformation with respect to $n_{1}$ variables amongst
$\\{x_{1},\dots,x_{n}\\}$. Then we have, according to the notation (1.2.4),

(3.1.18) $\mu(e^{-i\pi n_{1}/4}\mathcal{F}_{n_{1}})=\frac{n_{1}}{4\pi}.$

###### Proof.

The proof of the inequality $\mu(e^{-i\pi
n_{1}/4}\mathcal{F}_{n_{1}})\geq\frac{n_{1}}{4\pi}$ follows the same lines as
in Example 3.3. We do have with the notation

$x=(\underbrace{x_{1},\dots,x_{n_{1}}}_{x_{[n_{1}]}},\underbrace{x_{n_{1}+1},\dots
x_{n_{1}+n_{2}}}_{x_{[n_{2}]}}),$
$\displaystyle\frac{n_{1}}{2\pi}\|u\|_{L^{2}(\mathbb{R}^{n})}^{2}=\sum_{1\leq
j\leq
n_{1}}2\operatorname{Re}\langle(D_{j}u)(x),ix_{j}u(x)\rangle_{L^{2}(\mathbb{R}^{n})}$
$\displaystyle=\sum_{1\leq j\leq
n_{1}}2\operatorname{Re}\langle\xi_{j}\mathcal{F}_{n_{1}}u,i\mathcal{F}_{n_{1}}(x_{j}u)\rangle_{L^{2}(\mathbb{R}^{n})}$
$\displaystyle\leq 2\sum_{1\leq j\leq
n_{1}}\left(\iint{|\xi_{j}|}^{2}\left|(\mathcal{F}_{n_{1}}u)(\xi_{[n_{1}]},x_{[n_{2}]})\right|^{2}d\xi_{[n_{1}]}dx_{[n_{2}]}\right)^{1/2}\left(\int{|x_{j}|}^{2}\left|u(x)\right|^{2}dx\right)^{1/2}$
$\displaystyle\leq 2\left[\sum_{1\leq j\leq
n_{1}}\iint{|\xi_{j}|}^{2}\left|(\mathcal{F}_{n_{1}}u)(\xi_{[n_{1}]},x_{[n_{2}]})\right|^{2}d\xi_{[n_{1}]}dx_{[n_{2}]}\right]^{1/2}\hskip-15.0pt\times\left[\sum_{1\leq
j\leq n_{1}}\int{|x_{j}|}^{2}\left|u(x)\right|^{2}dx\right]^{1/2}$
$\displaystyle\leq
2\mathcal{V}\bigl{(}\mathcal{F}_{n_{1}}u\bigr{)}^{1/2}\mathcal{V}(u)^{1/2},$

proving the sought inequality. To prove the equality (3.1.18), we choose $u$
as a tensor product

$u_{\varepsilon}(x)=2^{n_{1}/4}e^{-\pi\|x_{[n_{1}]}\|^{2}}\phi\bigl{(}\frac{x_{[n_{2}]}}{\varepsilon}\bigr{)}\varepsilon^{-n_{2}/2},\quad\phi\in\mathscr{S}(\mathbb{R}^{n_{2}}),\|\phi\|_{L^{2}(\mathbb{R}^{n_{2}})}=1,\quad\varepsilon>0.$

We check easily that

$\displaystyle\mathcal{V}(\mathcal{F}_{n_{1}}u_{\varepsilon})$
$\displaystyle=\int{|\xi_{[n_{1}]}|}^{2}2^{n_{1}/2}e^{-2\pi\|\xi_{[n_{1}]}\|^{2}}d\xi_{[n_{1}]}+O(\varepsilon^{2}),$
$\displaystyle\mathcal{V}(u_{\varepsilon})$
$\displaystyle=\int{|x_{[n_{1}]}|}^{2}2^{n_{1}/2}e^{-2\pi\|x_{[n_{1}]}\|^{2}}dx_{[n_{1}]}+O(\varepsilon^{2}),$

and we already know from (4.2.11) that

$\int{|\xi_{[n_{1}]}|}^{2}2^{n_{1}/2}e^{-2\pi\|\xi_{n_{1}}\|^{2}}d\xi_{[n_{1}]}\int{|x_{[n_{1}]}|}^{2}2^{n_{1}/2}e^{-2\pi\|x_{n_{1}}\|^{2}}dx_{[n_{1}]}=\frac{n_{1}^{2}}{2^{4}\pi^{2}},$

proving the lemma. ∎

###  3.2. Using the symplectic covariance

######  Theorem 3.5.

Let $\mathcal{M}$ be an element of the metaplectic group and let
$\Xi=\Psi(\mathcal{M}),$ where $\Psi$ is the homomorphism defined in (2.2.34).
Then for $a,b$ real-valued linear forms on
$\mathbb{R}^{n}\times\mathbb{R}^{n}$, we have, for all
$u\in\mathscr{S}(\mathbb{R}^{n})$,

(3.2.1) $\displaystyle 2\operatorname{Re}$
$\displaystyle\left\langle{\text{{O}p}_{\text{{w}}}(a)}\mathcal{M}u,i\mathcal{M}{\text{{O}p}_{\text{{w}}}(b)}u\right\rangle_{L^{2}(\mathbb{R}^{n})}=\frac{\left\\{a\circ\Xi,b\right\\}}{2\pi}\|u\|^{2}_{L^{2}(\mathbb{R}^{n})}.$

######  N.B.

The Poisson bracket $\left\\{a\circ\Xi,b\right\\}$ is a constant scalar
function since $a\circ\Xi$ and $b$ are both linear forms on $\mathbb{R}^{2n}$.

###### Proof.

The proof is obvious and follows indeed the same lines as the proof of (3.1.3)
above. ∎

######  Corollary 3.6.

Let $\mathcal{M}$, $\Xi$ be as in Theorem 3.5 and let $b$ be the linear form
on $\mathbb{R}^{2n}$ given by $b(X)=[T,X]$ for some $T\in\mathbb{R}^{2n}$.
Then there exists a real-valued linear form $a$ on $\mathbb{R}^{2n}$ such
that, for all $u\in\mathscr{S}(\mathbb{R}^{n})$,

(3.2.2)
$\|{\text{{O}p}_{\text{{w}}}(a)}\mathcal{M}u\|_{L^{2}(\mathbb{R}^{n})}\|{\text{{O}p}_{\text{{w}}}(b)}u\|_{L^{2}(\mathbb{R}^{n})}\geq\frac{\|\Xi
T\|^{2}}{4\pi}\|u\|^{2}_{L^{2}(\mathbb{R}^{n})}.$

We may choose $a(Y)=[Y,\sigma\Xi T]$.

######  N.B.

Assuming that $b$ is not the zero linear form, with the linear form
$\tilde{a}$ defined by

(3.2.3) $\tilde{a}(Y)=\frac{[Y,\sigma\Xi T]}{\|\Xi T\|^{2}},$

we get

(3.2.4)
$\|{\text{{O}p}_{\text{{w}}}(\tilde{a})}\mathcal{M}u\|_{L^{2}(\mathbb{R}^{n})}\|{\text{{O}p}_{\text{{w}}}(b)}u\|_{L^{2}(\mathbb{R}^{n})}\geq\frac{1}{4\pi}\|u\|^{2}_{L^{2}(\mathbb{R}^{n})},$

a somewhat general uncertainty principle for metaplectic transformations.

###### Proof of the corollary.

Let $Z\in\mathbb{R}^{2n}$ and let $a$ be the linear form given by
$a(Y)=[Z,Y]$. We have seen that $(a\circ\Xi)(X)=[\Xi^{-1}Z,X]$. According to
(3.2.1) and the Cauchy-Schwarz inequality, it is enough to find $a$ such that
$|\left\\{a\circ\Xi,b\right\\}|=\|\Xi T\|^{2},$ that is to find
$Z\in\mathbb{R}^{2n}$ such that

(3.2.5) $[\Xi^{-1}Z,T]=\|\Xi T\|^{2}.$

We choose $Z=-\sigma\Xi T$, and this gives $[\Xi^{-1}Z,T]=[Z,\Xi
T]=\langle\sigma Z,\Xi T\rangle=\|\Xi T\|^{2},$ proving the corollary. ∎

###### Proof of Theorem 1.5.

We assume first that $\Xi_{12}$ is invertible. The result follows almost
immediately from Proposition 3.1: indeed, using (2.2.40), (2.1.9) and
(2.1.11), we choose

$L=\Xi_{12}^{-1},\quad P=\Xi_{22}\Xi_{12}^{-1},\quad Q=\Xi_{12}^{-1}\Xi_{11},$

and using the fact that $\Xi$ is a symplectic matrix, which is expressed by
the equalities

(3.2.6)
$\displaystyle\bigl{(}\Xi_{11}^{*}\Xi_{21}\bigr{)}^{*}=\Xi_{11}^{*}\Xi_{21},$
(3.2.7)
$\displaystyle\bigl{(}\Xi_{12}^{*}\Xi_{22}\bigr{)}^{*}=\Xi_{12}^{*}\Xi_{22},$
(3.2.8) $\displaystyle\Xi_{11}^{*}\Xi_{22}-\Xi_{21}^{*}\Xi_{12}=I_{n},$

as well as999Note that the three conditions (3.2.6), (3.2.7), (3.2.8) are
equivalent to $\Xi\in\textsl{Sp}(2n,\mathbb{R})$, which is equivalent to
$\Xi^{*-1}\in\textsl{Sp}(2n,\mathbb{R})$, itself equivalent to (3.2.9),
(3.2.10), (3.2.11). As a result we can freely use these six consequences of
the assumption $\Xi$ symplectic.

(3.2.9)
$\displaystyle\bigl{(}\Xi_{11}\Xi_{12}^{*}\bigr{)}^{*}=\Xi_{11}\Xi_{12}^{*},$
(3.2.10)
$\displaystyle\bigl{(}\Xi_{21}\Xi_{22}^{*}\bigr{)}^{*}=\Xi_{21}\Xi_{22}^{*},$
(3.2.11) $\displaystyle\Xi_{11}\Xi_{22}^{*}-\Xi_{12}\Xi_{21}^{*}=I_{n},$

we get that $P,Q$ are symmetric since

$P^{*}-P=\Xi_{12}^{*-1}\Xi_{22}^{*}-\Xi_{22}\Xi_{12}^{-1}=\Xi_{12}^{*-1}\underbrace{\bigl{(}\Xi_{22}^{*}\Xi_{12}-\Xi_{12}^{*}\Xi_{22}\bigr{)}}_{=0\text{
from \eqref{symp02}}}\Xi_{12}^{-1},$
$Q^{*}-Q=\Xi_{11}^{*}\Xi_{12}^{*-1}-\Xi_{12}^{-1}\Xi_{11}=\Xi_{12}^{-1}\underbrace{\bigl{(}\Xi_{12}\Xi_{11}^{*}-\Xi_{11}\Xi_{12}^{*}\bigr{)}}_{=0\text{
from \eqref{symp04}}}\Xi_{12}^{*-1},$

and we have moreover

$\displaystyle
PL^{-1}Q-L^{*}-\Xi_{21}=\Xi_{22}\Xi_{12}^{-1}\Xi_{12}\Xi_{12}^{-1}\Xi_{11}-\Xi_{12}^{*-1}-\Xi_{21}$
$\displaystyle=\Xi_{22}\Xi_{12}^{-1}\Xi_{11}-\Xi_{12}^{*-1}-\Xi_{21}$
$\displaystyle\hskip-9.0pt\underbrace{=}_{\eqref{symp06}}\Xi_{22}\Xi_{12}^{-1}\Xi_{11}-\Xi_{12}^{*-1}-\bigl{(}\Xi_{22}\Xi_{11}^{*}-I_{n}\bigr{)}\Xi_{12}^{*-1}$
$\displaystyle=\Xi_{22}\Xi_{12}^{-1}\Xi_{11}-\Xi_{12}^{*-1}-\Xi_{22}\Xi_{11}^{*}\Xi_{12}^{*-1}+\Xi_{12}^{*-1}$
$\displaystyle=\Xi_{22}\bigl{(}\Xi_{12}^{-1}\Xi_{11}-\Xi_{11}^{*}\Xi_{12}^{*-1}\bigr{)}$
$\displaystyle=\Xi_{22}\Xi_{12}^{-1}\underbrace{\bigl{(}\Xi_{11}\Xi_{12}^{*}-\Xi_{12}\Xi_{11}^{*}\bigr{)}\Xi_{12}^{*-1}}_{=0\text{
from }\eqref{symp04}},$

and $PL^{-1}=\Xi_{22}\Xi_{12}^{-1}\Xi_{12}=\Xi_{22}$, so that
$\Xi=\Lambda_{P,L,Q}$ and we may apply Proposition 3.1 to get the inequality

(3.2.12)
$\mu(\mathcal{M})\geq\frac{1}{4\pi}|\hskip-1.0pt\|\Xi_{12}\|\hskip-1.0pt|_{\texttt{S}^{1}}.$

We would like now to prove that the above inequality can be improved into an
equality, while we still assume the invertibility of $\Xi_{12}$. To start
with, we could take, with $\lambda>0$,

$v_{\lambda}(x)=\phi(x\lambda)\lambda^{n/2},\quad\phi\in\mathscr{S}(\mathbb{R}^{n}),\quad\|\phi\|_{L^{2}(\mathbb{R}^{n})}=1=\|v_{\lambda}\|_{L^{2}(\mathbb{R}^{n})}.$

To handle $\mathcal{M}_{P,L,Q},$ it is better to define
$w_{\lambda,Q}(x)=v_{\lambda}(x)e^{-i\pi Qx^{2}},$ which has $L^{2}$ norm 1
and is such that

$\bigl{(}\mathcal{M}_{P,L,Q}w_{\lambda,Q})(x)=e^{-\frac{i\pi n}{4}+\frac{i\pi
m}{2}}e^{i\pi Px^{2}}{|\det L|}^{1/2}\int e^{-2i\pi Lx\cdot
y}v_{\lambda}(y)dy\\\ =e^{-\frac{i\pi n}{4}+\frac{i\pi m}{2}}e^{i\pi
Px^{2}}{|\det L|}^{1/2}(\mathcal{F}v_{\lambda})(Lx)\\\ =e^{-\frac{i\pi
n}{4}+\frac{i\pi m}{2}}e^{i\pi Px^{2}}{|\det
L|}^{1/2}\widehat{\phi}(Lx\lambda^{-1})\lambda^{-n/2},$

entailing that

(3.2.13)
$\displaystyle\mathcal{V}\bigl{(}\mathcal{M}_{P,L,Q}w_{\lambda,Q}\bigr{)}=|\det
L|\int|x|^{2}{|\widehat{\phi}(Lx\lambda^{-1})|}^{2}\lambda^{-n}dx=\lambda^{2}\int{|L^{-1}\xi|}^{2}{|\widehat{\phi}(\xi)|}^{2}d\xi$
(3.2.14)
$\displaystyle\mathcal{V}\bigl{(}w_{\lambda,Q}\bigr{)}=\lambda^{-2}\int{|x|}^{2}{|\phi(x)|}^{2}dx.$

We obtain thus with $\Xi_{12}=L^{-1}$,

(3.2.15)
$\sqrt{\mathcal{V}\bigl{(}\mathcal{M}_{P,L,Q}w_{\lambda,Q}\bigr{)}\mathcal{V}\bigl{(}w_{\lambda,Q}\bigr{)}}\\\
=\left(\int{|\Xi_{12}\xi|}^{2}{|\widehat{\phi}(\xi)|}^{2}d\xi\right)^{1/2}\left(\int{|x|}^{2}{|\phi(x)|}^{2}dx\right)^{1/2}\\\
\leq\frac{1}{2}\Bigl{\langle}{\text{{O}p}_{\text{{w}}}\bigl{(}{|\Xi_{12}\xi|}^{2}+{|x|}^{2}\bigr{)}}\phi,\phi\Bigr{\rangle}_{L^{2}(\mathbb{R}^{n})}.$

Using the polar decomposition of Proposition 4.5 in our Appendix, we have

$\Xi_{12}=U|\Xi_{12}|,\quad\text{where $U$ is a unitary matrix},$

so that

(3.2.16)
${\text{{O}p}_{\text{{w}}}\bigl{(}{|\Xi_{12}\xi|}^{2}+{|x|}^{2}\bigr{)}}={\text{{O}p}_{\text{{w}}}\bigl{(}\bigl{|}|\Xi_{12}|\xi\bigr{|}^{2}+{|x|}^{2}\bigr{)}},$

and the symplectic covariance of the Weyl calculus (see Theorem 2.36) shows
that (3.2.16) is unitarily equivalent to the operator

(3.2.17)
${\text{{O}p}_{\text{{w}}}\bigl{(}\bigl{|}|\Xi_{12}|B^{*}\xi\bigr{|}^{2}+{|B^{-1}x|}^{2}\bigr{)}},\quad\text{where
$B$ belongs to $\textsl{Gl}(n,\mathbb{R})$}.$

We may choose

(3.2.18) $B={|\Xi_{12}|}^{-1/2}=B^{*},$

and we get that (3.2.16) is unitarily equivalent to the operator

(3.2.19)
${\text{{O}p}_{\text{{w}}}\bigl{(}\bigl{|}{|\Xi_{12}|}^{1/2}\xi\bigr{|}^{2}+\bigl{|}{|\Xi_{12}|}^{1/2}x\bigr{|}^{2}\bigr{)}}={\text{{O}p}_{\text{{w}}}\bigl{(}\langle|\Xi_{12}|\xi,\xi\rangle+\langle|\Xi_{12}|x,x\rangle\bigr{)}}\\\
={\text{{O}p}_{\text{{w}}}\bigl{(}\sum_{1\leq j\leq
n}\lambda_{j}\bigl{(}|\Xi_{12}|\bigr{)}(x_{j}^{2}+\xi_{j}^{2})\bigr{)}},$

where $\\{\lambda_{j}\bigl{(}|\Xi_{12}|\bigr{)}\\}_{1\leq j\leq n}$ are the
eigenvalues of $|\Xi_{12}|$, that is the singular values of $\Xi_{12}$. A
consequence of the inequality (3.2.15) and the identity (3.2.19) is that

(3.2.20)
$\mu(\mathcal{M}_{P,L,Q})\leq\inf_{\begin{subarray}{c}\phi\in\mathscr{S}(\mathbb{R}^{n})\\\
\|\phi\|_{L^{2}(\mathbb{R}^{n})}=1\end{subarray}}\sqrt{\mathcal{V}\bigl{(}\mathcal{M}_{P,L,Q}w_{\lambda,Q}\bigr{)}\mathcal{V}\bigl{(}w_{\lambda,Q}\bigr{)}}\\\
\leq\frac{1}{2}\inf\Bigl{[}{\text{{O}p}_{\text{{w}}}\bigl{(}\sum_{1\leq j\leq
n}\lambda_{j}\bigl{(}|\Xi_{12}|\bigr{)}(x_{j}^{2}+\xi_{j}^{2})\bigr{)}}\Bigr{]}.$

On the other hand, its is easy to calculate the right-hand-side of (3.2.20),
which is

(3.2.21) $\frac{1}{4\pi}\sum_{1\leq j\leq
n}\lambda_{j}\bigl{(}|\Xi_{12}|\bigr{)}=\frac{1}{4\pi}|\hskip-1.0pt\|\Xi_{12}\|\hskip-1.0pt|_{{\mathtt{S}^{1}}},$

proving that

(3.2.22)
$\mu(\mathcal{M}_{P,L,Q})\leq\frac{1}{4\pi}|\hskip-1.0pt\|\Xi_{12}\|\hskip-1.0pt|_{{\mathtt{S}^{1}}},$

and eventually with (3.2.12) the sought equality in of Theorem 1.5 in the case
where $\Xi_{12}$ is invertible. We are thus left with proving Theorem 1.5 in
the case where $\Xi_{12}$ is a singular matrix. Example 2.4 shows that
$\Xi_{12}$ could be singular and, even worse, it could happen that the four
blocks $\\{\Xi_{jk}\\}_{1\leq j,k\leq 2}$ in the decomposition of $\Xi$ in
Theorem 1.5 are all singular $n\times n$ matrices when $n\geq 2$.

######  Remark 3.7.

Before tackling the general singular case, it is interesting to check first
that we have $\mu(\mathcal{M})=0$ when $\Xi_{12}=0$: indeed in that case we
find that, thanks to (3.2.8),

$\Xi_{11}^{*}\Xi_{22}=I_{n},$

and thus $\Xi_{11}$ is invertible proving that

$\Xi=\Xi_{A,B,C}={\begin{pmatrix}[l]{B^{-1}}&{-B^{-1}C}\\\
{AB^{-1}}&{B^{*}-AB^{-1}C}\end{pmatrix}}=\begin{pmatrix}I_{n}&0\\\
A&I_{n}\end{pmatrix}\begin{pmatrix}B^{-1}&0\\\
0&B^{*}\end{pmatrix}\begin{pmatrix}I_{n}&-C\\\ 0&I_{n}\end{pmatrix},$

where $B$ is invertible and $A,C$ symmetric. The condition $\Xi_{12}=0$ forces
$C=0$ and we find

$\Xi=\begin{pmatrix}I_{n}&0\\\ A&I_{n}\end{pmatrix}\begin{pmatrix}B^{-1}&0\\\
0&B^{*}\end{pmatrix},$

and $M=M_{A,B,0}^{\scriptscriptstyle\\{m\\}}$ with

$(M_{A,B,0}^{\scriptscriptstyle\\{m\\}}v)(x)=e^{\frac{i\pi m}{2}}{|\det
B|}^{1/2}e^{i\pi
Ax^{2}}\int_{\mathbb{R}^{n}}e^{i\pi(2Bx\cdot\eta)}\hat{v}(\eta)d\eta\\\
=e^{i\pi Ax^{2}}(\det B)^{1/2}v(Bx).$

In particular we have

$\mathcal{V}(Mu)\mathcal{V}(u)=\iint|\det
B|{|x|}^{2}{|v(Bx)|}^{2}{|y|}^{2}{|v(y)|}^{2}dxdy\\\
=\iint{|B^{-1}x|}^{2}{|v(x)|}^{2}{|y|}^{2}{|v(y)|}^{2}dxdy\leq\|B^{-1}\|^{2}\mathcal{V}(u)\mathcal{V}(u).$

We already know that

$\inf_{\begin{subarray}{c}u\in\mathscr{S}(\mathbb{R}^{n})\\\
\|u\|_{L^{2}(\mathbb{R}^{n})}=1\end{subarray}}\mathcal{V}(u)=0,$

and this gives $\mu(M)=0$.

###  3.3. The singular case

We need now to tackle the case where $\Xi_{12}$ is a singular matrix with
positive rank. We may assume that $n\geq 2$, since the singular case in one
dimension is treated by Remark 3.7. According to Theorem 3.5, we find that,
choosing

$a(y,\eta)=\zeta\cdot y,\quad b(x,\xi)=\tau\cdot x,$

we have with a “vertical” $Z=(0,\zeta)$

$\Xi X=(\Xi_{11}x+\Xi_{12}\xi,\Xi_{21}x+\Xi_{22}\xi),\quad a(\Xi X)=[Z,\Xi
X]=\zeta\cdot(\Xi_{11}x+\Xi_{12}\xi),$

and thus, with a unitary operator $\mathcal{U}$ (thanks to Proposition 4.5 in
our Appendix),

$\left\\{a\circ\Xi,b\right\\}=\langle\Xi_{12}^{*}\zeta,\tau\rangle=\langle\zeta,\mathcal{U}|\Xi_{12}|\tau\rangle.$

We get then

$\sum_{1\leq j\leq n}\|{\text{{O}p}_{\text{{w}}}(\zeta_{j}\cdot
y)}\mathcal{M}v\|\|{\text{{O}p}_{\text{{w}}}(\tau_{j}\cdot
x)}v\|\geq\frac{1}{4\pi}\|v\|^{2}\sum_{1\leq j\leq
n}|\langle\mathcal{U}^{*}\zeta_{j},|\Xi_{12}|\tau_{j}\rangle|.$

Let $(\tau_{j})_{1\leq j\leq n}$ be an orthonormal basis of $\mathbb{R}^{n}$.
We choose $\zeta_{j}=\mathcal{U}\tau_{j}$ and we obtain that
$(\zeta_{j})_{1\leq j\leq n}$ is also an orthonormal basis. Moreover, we have

$\frac{1}{4\pi}\sum_{1\leq j\leq
n}|\langle\tau_{j},|\Xi_{12}|\tau_{j}\rangle|\|v\|^{2}\\\
=\frac{1}{4\pi}\sum_{1\leq j\leq
n}|\langle\mathcal{U}^{*}\mathcal{U}\tau_{j},|\Xi_{12}|\tau_{j}\rangle|\|v\|^{2}=\frac{1}{4\pi}\sum_{1\leq
j\leq n}|\langle U^{*}\zeta_{j},|\Xi_{12}|\tau_{j}\rangle|\|v\|^{2}\\\
\leq\left(\sum_{1\leq j\leq n}\|{\zeta_{j}\cdot
y}\mathcal{M}v\|^{2}\right)^{1/2}\left(\sum_{1\leq j\leq n}\|{\tau_{j}\cdot
x}v\|^{2}\right)^{1/2}\leq\mathcal{V}(\mathcal{M}v)^{1/2}\mathcal{V}(v)^{1/2},$

proving indeed that, even in the case where $\Xi_{12}$ is a singular matrix,
we have

(3.3.1)
$\mu(\mathcal{M})\geq\frac{1}{4\pi}|\hskip-1.0pt\|\Xi_{12}\|\hskip-1.0pt|_{{\mathtt{S}^{1}}}.$

We are now left with proving that the reverse inequality of (3.3.1) holds
true, when $\Xi_{12}$ is a singular matrix with positive rank.  It is helpful
to begin with a simple particular case.

######  Lemma 3.8.

Let $n\geq 2$ be an integer, let $r\in\llbracket 1,n-1\rrbracket$ and let $C$
be a real symmetric $n\times n$ matrix with rank $r$. Then with
$M=e^{i\pi\langle CD,D\rangle}$, we have

$\mu(M)=\frac{1}{4\pi}\sum_{1\leq j\leq n}|\lambda_{j}(C)|,$

where $\\{\lambda_{j}(C)\\}_{1\leq j\leq n}$ are the eigenvalues of $C$
(listed with their multiplicities).

###### Proof.

We note first of all that the eigenvalues of $\sqrt{C^{*}C}=|C|$ are
$\\{|\lambda_{j}(C)|\\}_{1\leq j\leq n}$ so that
$|\hskip-1.0pt\|C\|\hskip-1.0pt|_{\texttt{S}^{1}}=\sum_{1\leq j\leq
n}|\lambda_{j}(C)|.$ As a consequence, from (3.3.1), we get the inequality

(3.3.2) $\mu(M)\geq\frac{1}{4\pi}\sum_{1\leq j\leq n}|\lambda_{j}(C)|,$

since $e^{i\pi\langle CD,D\rangle}$ is a metaplectic transformation such that

$\Psi(e^{i\pi\langle CD,D\rangle})=\begin{pmatrix}I_{n}&-C\\\
0&I_{n}\end{pmatrix}.$

Since $C$ is real symmetric, there exists an orthogonal matrix $\Omega$ such
that

$C={{}^{t}}\Omega\Delta\Omega,\quad\text{$\Delta$ diagonal
$(\mu_{1},\dots,\mu_{r},\underbrace{0,\dots,0}_{\text{$(n-r)$ zeros}})$},$

where the $(\mu_{l})_{1\leq l\leq r}$ are the non-zero eigenvalues of $C$
(counted with their multiplicities). As a consequence, we have
$M=e^{i\pi\langle\Delta\Omega D,\Omega D\rangle}$ and thus

$\displaystyle(Mv)(x)$ $\displaystyle=\int_{\mathbb{R}^{n}}e^{2i\pi
x\cdot\xi}e^{i\pi\langle\Delta\Omega\xi,\Omega\xi\rangle}\hat{v}(\xi)d\xi$
$\displaystyle=\int_{\mathbb{R}^{n}}e^{2i\pi\Omega
x\cdot\eta}e^{i\pi\langle\Delta\eta,\eta\rangle}\hat{v}(\Omega^{-1}\eta)d\eta=\bigl{(}e^{i\pi\langle\Delta
D_{y},D_{y}\rangle}(v\circ\Omega^{-1})\bigr{)}(\Omega x)$ (using the notation
(2.2.5))
$\displaystyle=\bigl{(}M_{0,\Omega,0}^{\scriptscriptstyle\\{m\\}}e^{i\pi\langle\Delta
D_{y},D_{y}\rangle}M_{0,\Omega^{-1},0}^{\scriptscriptstyle\\{-m\\}}v\bigr{)}(x),$

implying that

(3.3.3)
$MM_{0,\Omega,0}^{\scriptscriptstyle\\{m\\}}=M_{0,\Omega,0}^{\scriptscriptstyle\\{m\\}}e^{i\pi\langle\Delta
D,D\rangle}.$

We check now the operator $e^{i\pi\langle\Delta D,D\rangle}=e^{i\pi\sum_{1\leq
j\leq r}\mu_{j}D_{x_{j}}^{2}}$. With the notation

$y=(x_{1},\dots,x_{r}),\quad z=(x_{r+1},\dots,x_{n}),$

we consider for
$v\in\mathscr{S}(\mathbb{R}^{r}),\phi\in\mathscr{S}(\mathbb{R}^{n-r})$,

$u(y,z)=v(y)\phi(z/\varepsilon)\varepsilon^{-\frac{(n-r)}{2}},\quad\varepsilon>0,\quad\|\phi\|_{L^{2}(\mathbb{R}^{n-r})}=1=\|v\|_{L^{2}(\mathbb{R}^{r})}.$

We have

$\displaystyle\mathcal{V}(u)$
$\displaystyle=\iint_{\mathbb{R}^{r}\times\mathbb{R}^{n-r}}\bigl{(}{|y|}^{2}+\varepsilon^{2}{|t|}^{2}\bigr{)}{|v(y)|}^{2}{|\phi(t)|}^{2}dydt,$
$\displaystyle\mathcal{V}(e^{i\pi\langle\Delta D,D\rangle}u)$
$\displaystyle=\iint_{\mathbb{R}^{r}\times\mathbb{R}^{n-r}}\bigl{(}{|y|}^{2}+\varepsilon^{2}{|t|}^{2}\bigr{)}{|e^{i\pi\sum_{1\leq
j\leq r}\mu_{j}D_{x_{j}}^{2}}v(y)|}^{2}{|\phi(t)|}^{2}dydt,$

so that, with obvious notations,

$\mathcal{V}_{n}(e^{i\pi\langle\Delta D,D\rangle}u)\mathcal{V}_{n}(u)=\\\
\Bigl{(}\int{|y|}^{2}{|e^{i\pi\sum_{1\leq j\leq
r}\mu_{j}D_{x_{j}}^{2}}v(y)|}^{2}+\varepsilon^{2}\int{|t|}^{2}|\phi(t)|^{2}dt\Bigr{)}\times\Bigl{(}\int|y|^{2}{|v(y)|}^{2}+\varepsilon^{2}\int{|t|}^{2}{|\phi(t)|}^{2}dt\Bigr{)}\\\
=\left(\mathcal{V}_{r}(e^{i\pi\langle\Delta
D,D\rangle}v)+\varepsilon^{2}\int{|t|}^{2}{|\phi(t)|}^{2}dt\right)\times\left(\mathcal{V}_{r}(v)+\varepsilon^{2}\int{|t|}^{2}|\phi(t)|^{2}dt\right),$

which means with

$\gamma_{\phi}=\left(\int{|t|}^{2}{|\phi(t)|}^{2}dt\right)^{1/2},$
$\mathcal{V}_{n}(e^{i\pi\langle\Delta
D,D\rangle}u)\mathcal{V}_{n}(u)=\mathcal{V}_{r}(e^{i\pi\langle\Delta
D,D\rangle}v)\mathcal{V}_{r}(v)\\\
+\varepsilon^{2}\gamma_{\phi}^{2}\bigl{(}\mathcal{V}_{r}(e^{i\pi\langle\Delta
D,D\rangle}v)+\mathcal{V}_{r}(v)\bigr{)}+\varepsilon^{4}\gamma_{\phi}^{4}.$

This proves that for any
$\varepsilon>0,v_{0}\in\mathscr{S}(\mathbb{R}^{r}),\phi_{0}\in\mathscr{S}(\mathbb{R}^{n-r}),$

$\inf_{\begin{subarray}{c}u\in\mathscr{S}(\mathbb{R}^{n})\\\
\|u\|_{L^{2}(\mathbb{R}^{n})}=1\end{subarray}}\mathcal{V}_{n}(e^{i\pi\langle\Delta
D,D\rangle}u)\mathcal{V}_{n}(u)\leq\inf_{\begin{subarray}{c}v\in\mathscr{S}(\mathbb{R}^{r})\\\
\|v\|_{L^{2}(\mathbb{R}^{r})}=1\end{subarray}}\mathcal{V}_{r}(e^{i\pi\langle\Delta
D,D\rangle}v)\mathcal{V}_{r}(v)\\\
+\varepsilon^{2}\gamma_{\phi_{0}}^{2}\bigl{(}\mathcal{V}_{r}(e^{i\pi\langle\Delta
D,D\rangle}v_{0})+\mathcal{V}_{r}(v_{0})\bigr{)}+\varepsilon^{4}\gamma_{\phi_{0}}^{4},$

and thus that

(3.3.4) $\inf_{\begin{subarray}{c}u\in\mathscr{S}(\mathbb{R}^{n})\\\
\|u\|_{L^{2}(\mathbb{R}^{n})}=1\end{subarray}}\mathcal{V}_{n}(e^{i\pi\langle\Delta
D,D\rangle}u)\mathcal{V}_{n}(u)\leq\inf_{\begin{subarray}{c}v\in\mathscr{S}(\mathbb{R}^{r})\\\
\|v\|_{L^{2}(\mathbb{R}^{r})}=1\end{subarray}}\mathcal{V}_{r}(e^{i\pi\langle\Delta
D,D\rangle}v)\mathcal{V}_{r}(v)\\\ =\mu_{r}\bigl{(}e^{i\pi\langle\Delta
D,D\rangle}\bigr{)}=\frac{1}{4\pi}|\hskip-1.0pt\|\Delta\|\hskip-1.0pt|_{\texttt{S}^{1}}=\frac{1}{4\pi}\sum_{1\leq
j\leq r}|\mu_{j}|=\frac{1}{4\pi}\sum_{1\leq j\leq r}|\lambda_{j}(C)|,$

proving that, from (3.3.3)

(3.3.5)
$\mu_{n}\bigl{(}M_{0,\Omega^{-1},0}^{\scriptscriptstyle\\{-m\\}}MM_{0,\Omega,0}^{\scriptscriptstyle\\{m\\}}\bigr{)}=\frac{1}{4\pi}\sum_{1\leq
j\leq r}|\lambda_{j}(C)|.$

######  Claim 3.9.

Let $\Omega\in O(n)$. Then we have for $u\in\mathscr{S}(\mathbb{R}^{n})$,

$\mathcal{V}(M_{0,\Omega,0}^{\scriptscriptstyle\\{m\\}}u)=\mathcal{V}(u).$

We have indeed

$\mathcal{V}(M_{0,\Omega,0}^{\scriptscriptstyle\\{m\\}}u)=\int_{\mathbb{R}^{n}}{|x|}^{2}{|u(\Omega
x)|}^{2}dx=\int_{\mathbb{R}^{n}}{|y|}^{2}{|u(y)|}^{2}dy=\mathcal{V}(u),$

proving the claim. As a consequence we get that for
$u\in\mathscr{S}(\mathbb{R}^{n})$,

$\mathcal{V}(M_{0,\Omega^{-1},0}^{\scriptscriptstyle\\{-m\\}}MM_{0,\Omega,0}^{\scriptscriptstyle\\{m\\}}u)\mathcal{V}(u)=\mathcal{V}(MM_{0,\Omega,0}^{\scriptscriptstyle\\{m\\}}u)\mathcal{V}(u),$

and thus

$\inf_{\begin{subarray}{c}u\in\mathscr{S}(\mathbb{R}^{n})\\\
\|u\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\mathcal{V}(M_{0,\Omega^{-1},0}^{\scriptscriptstyle\\{-m\\}}MM_{0,\Omega,0}^{\scriptscriptstyle\\{m\\}}u)\mathcal{V}(u)=\inf_{\begin{subarray}{c}u\in\mathscr{S}(\mathbb{R}^{n})\\\
\|u\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\mathcal{V}(MM_{0,\Omega,0}^{\scriptscriptstyle\\{m\\}}u)\mathcal{V}(u)\\\
=\inf_{\begin{subarray}{c}v\in\mathscr{S}(\mathbb{R}^{n})\\\
\|v\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\mathcal{V}(Mv)\mathcal{V}(M_{0,\Omega^{-1},0}^{\scriptscriptstyle\\{-m\\}}v)=\inf_{\begin{subarray}{c}v\in\mathscr{S}(\mathbb{R}^{n})\\\
\|v\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\mathcal{V}(Mv)\mathcal{V}(v)=\mu(M)$

which eventually proves the lemma, using (3.3.5). ∎

We want finally to deal with the general case where $\Xi_{12}$ is singular
with rank $r\in\llbracket 1,n-1\rrbracket$. According to Proposition 2.25, it
is enough to tackle the case where

(3.3.6)
$\mathcal{M}=\mathcal{M}_{P_{1},L_{1},Q_{1}}^{\scriptscriptstyle\\{m_{1}\\}}\mathcal{M}_{P_{2},L_{2},Q_{2}}^{\scriptscriptstyle\\{m_{2}\\}},$

where $P_{j},Q_{j}$ are symmetric $n\times n$ matrices and $L_{j}$ are
invertible $n\times n$ matrices.

######  Lemma 3.10.

Let $\mathcal{M}$ be given by (3.3.6). We have, using the notations (2.2.5),

(3.3.7)
$\displaystyle\mathcal{M}=M_{P_{1},I_{n},0}M_{0,-I_{n},0}^{\scriptscriptstyle\\{-n\\}}M_{0,B,0}^{\scriptscriptstyle\\{m_{B}\\}}M_{0,I_{n},C}^{\scriptscriptstyle\\{0\\}}M_{Q_{2},I_{n},0}^{\scriptscriptstyle\\{0\\}},$
(3.3.8) $\displaystyle B=L_{2}^{*-1}L_{1}\in\textsl{Gl}(n,\mathbb{R}),\quad
C=L_{2}^{*-1}(Q_{1}+P_{2})L_{2}^{-1}\in\textsl{Sym}(n,\mathbb{R}).$

Moreover, defining

(3.3.9)
$\widetilde{\mathcal{M}}=M_{0,B,0}^{\scriptscriptstyle\\{m_{B}\\}}M_{0,I_{n},C}^{\scriptscriptstyle\\{0\\}}=M_{0,B,C}^{\scriptscriptstyle\\{m_{B}\\}},$

we have

(3.3.10) $\displaystyle\Psi(\widetilde{\mathcal{M}})$
$\displaystyle=\widetilde{\Xi}=\begin{pmatrix}B^{-1}&-B^{-1}C\\\
0&B^{*}\end{pmatrix},$ (3.3.11) $\displaystyle\Psi({\mathcal{M}})$
$\displaystyle=\Xi=\begin{pmatrix}[l]{B^{-1}CQ_{2}-B^{-1}}&{B^{-1}C}\\\
{P_{1}(B^{-1}CQ_{2}-B^{-1})-B^{*}Q_{2}}{\hskip
15.0pt}&{P_{1}B^{-1}C-B^{*}}\end{pmatrix},$ (3.3.12)
$\displaystyle\mu(\mathcal{M})$ $\displaystyle=\mu(\widetilde{\mathcal{M}}),$

where $\mu$ is defined in (1.2.4).

###### Proof.

We have

$\displaystyle(\mathcal{M}v)(x)=\overbrace{e^{i\pi\frac{m_{1}+m_{2}}{2}}e^{-\frac{i\pi
n}{2}}{|\det L_{1}|}^{1/2}{|\det L_{2}|}^{1/2}}^{\omega_{12}}$
$\displaystyle\hskip 75.0pt\times\iint
e^{i\pi(P_{1}x^{2}-2L_{1}x\cdot\xi+Q_{1}\xi^{2})}e^{i\pi(P_{2}\xi^{2}-2L_{2}\xi\cdot
y+Q_{2}y^{2})}v(y)d\xi dy$ $\displaystyle=\omega_{12}e^{i\pi P_{1}x^{2}}\iiint
e^{i\pi(Q_{1}+P_{2})\xi^{2}}e^{-2i\pi\xi\cdot(L_{1}x+L_{2}^{*}y)}\mathcal{F}(e^{i\pi
Q_{2}y^{2}}v(y))(\eta)e^{2i\pi y\cdot\eta}d\eta dyd\xi$
$\displaystyle=\omega_{12}e^{i\pi P_{1}x^{2}}\iint\delta_{0}(\eta-
L_{2}\xi)e^{i\pi(Q_{1}+P_{2})\xi^{2}}\mathcal{F}(e^{i\pi
Q_{2}y^{2}}v(y))(\eta)e^{-2i\pi\xi\cdot L_{1}x}d\eta d\xi$
$\displaystyle=\omega_{12}e^{i\pi P_{1}x^{2}}\int
e^{i\pi(Q_{1}+P_{2})\xi\cdot\xi}\mathcal{F}(e^{i\pi
Q_{2}y^{2}}v(y))(L_{2}\xi)e^{-2i\pi\xi\cdot L_{1}x}d\xi$
$\displaystyle=\omega_{12}{|\det L_{2}|}^{-1}e^{i\pi P_{1}x^{2}}\int e^{i\pi
L_{2}^{*-1}(Q_{1}+P_{2})L_{2}^{-1}\xi^{2}}\mathcal{F}(\underbrace{e^{i\pi
Q_{2}y^{2}}v(y)}_{v_{Q_{2}}(y)})(\xi)e^{-2i\pi\xi\cdot L_{2}^{*-1}L_{1}x}d\xi$
$\displaystyle=e^{i\pi\frac{m_{1}+m_{2}}{2}}e^{-\frac{i\pi n}{2}}{|\det
L_{1}|}^{1/2}{|\det L_{2}|}^{-1/2}e^{i\pi P_{1}x^{2}}\bigl{(}e^{i\pi
L_{2}^{*-1}(Q_{1}+P_{2})L_{2}^{-1}D^{2}}v_{Q_{2}}\bigr{)}(-L_{2}^{*-1}L_{1}x).$

We have thus, using the notations (2.2.5), (2.3.19), (2.2.19),

(3.3.13) $\displaystyle\mathcal{M}$
$\displaystyle=M_{P_{1},I_{n},0}M_{0,-I_{n},0}^{\scriptscriptstyle\\{-n\\}}M_{0,L_{2}^{*-1}L_{1},0}^{\scriptscriptstyle\\{m_{1}+m_{2}\\}}e^{i\pi
L_{2}^{*-1}(Q_{1}+P_{2})L_{2}^{-1}D^{2}}M_{Q_{2},I_{n},0}^{\scriptscriptstyle\\{0\\}}$
(3.3.14)
$\displaystyle=M_{P_{1},I_{n},0}M_{0,-I_{n},0}^{\scriptscriptstyle\\{-n\\}}M_{0,B,0}^{\scriptscriptstyle\\{m_{B}\\}}e^{i\pi
CD^{2}}M_{Q_{2},I_{n},0}^{\scriptscriptstyle\\{0\\}},$ (3.3.15)
$\displaystyle\text{with }B$ $\displaystyle=L_{2}^{*-1}L_{1},\quad
C=L_{2}^{*-1}(Q_{1}+P_{2})L_{2}^{-1},$

proving (3.3.7), (3.3.8). Moreover we have

$\displaystyle\Psi(\widetilde{\mathcal{M}})=\begin{pmatrix}B^{-1}&0\\\
0&B^{*}\end{pmatrix}\begin{pmatrix}I_{n}&-C\\\
0&I_{n}\end{pmatrix}=\begin{pmatrix}B^{-1}&-B^{-1}C\\\
0&B^{*}\end{pmatrix}=\widetilde{\Xi}$
$\displaystyle\widetilde{\Xi}_{11}=B^{-1},\quad\widetilde{\Xi}_{12}=-B^{-1}C,\quad\widetilde{\Xi}_{21}=0,\quad\widetilde{\Xi}_{22}=B^{*},$

proving (3.3.10); we calculate now

$\Psi(\mathcal{M})=\begin{pmatrix}I_{n}&0\\\
P_{1}&I_{n}\end{pmatrix}\begin{pmatrix}-I_{n}&0\\\
0&-I_{n}\end{pmatrix}\widetilde{\Xi}\begin{pmatrix}I_{n}&0\\\
Q_{2}&I_{n}\end{pmatrix},$

and we get101010The most important point in (3.3.10), (3.3.11) is the fact
that $\Xi_{12}=-\widetilde{\Xi}_{12}$, which is compatible with (3.3.12) and
the statement of Theorem 1.5.  easily (3.3.11). We note also that

$\mathcal{V}\bigl{(}\mathcal{M}v\bigr{)}=\mathcal{V}\bigl{(}M_{P_{1},I_{n},0}M_{0,-I_{n},0}^{\scriptscriptstyle\\{-n\\}}\widetilde{\mathcal{M}}M_{Q_{2},I_{n},0}^{\scriptscriptstyle\\{0\\}}v\bigr{)}=\mathcal{V}\bigl{(}\widetilde{\mathcal{M}}M_{Q_{2},I_{n},0}^{\scriptscriptstyle\\{0\\}}v\bigr{)},$

so that we find

$\mu(\mathcal{M})^{2}=\inf_{\begin{subarray}{c}v\in\mathscr{S}(\mathbb{R}^{n})\\\
\|v\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\mathcal{V}(\mathcal{M}v)\mathcal{V}(v)=\inf_{\begin{subarray}{c}v\in\mathscr{S}(\mathbb{R}^{n})\\\
\|v\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\mathcal{V}(\widetilde{\mathcal{M}}M_{Q_{2},I_{n},0}^{\scriptscriptstyle\\{0\\}}v)\mathcal{V}(v)\\\
=\inf_{\begin{subarray}{c}u\in\mathscr{S}(\mathbb{R}^{n})\\\
\|u\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\mathcal{V}(\widetilde{\mathcal{M}}u)\mathcal{V}(M_{-Q_{2},I_{n},0}^{\scriptscriptstyle\\{0\\}}u)=\inf_{\begin{subarray}{c}u\in\mathscr{S}(\mathbb{R}^{n})\\\
\|u\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\mathcal{V}(\widetilde{\mathcal{M}}u)\mathcal{V}(u)=\mu(\widetilde{\mathcal{M}})^{2},$

proving (3.3.12) and the lemma. ∎

######  Claim 3.11.

Let $\mathcal{M},\widetilde{\mathcal{M}},B,C$ be given by Lemma 3.10. To
complete the proof of Theorem 1.5, it is enough to prove that

(3.3.16)
$\mu\bigl{(}\widetilde{\mathcal{M}}\bigr{)}\leq\frac{1}{4\pi}|\hskip-1.0pt\|B^{-1}C\|\hskip-1.0pt|_{\texttt{S}^{1}}.$

Moreover, we have with $v\in\mathscr{S}(\mathbb{R}^{n})$, $w=e^{i\pi
CD^{2}}v$, the inequality

(3.3.17)
$\sqrt{\mathcal{V}(\widetilde{\mathcal{M}}v)\mathcal{V}(v)}\leq\frac{1}{2}\langle{\text{{O}p}_{\text{{w}}}(\|B^{-1}x\|^{2}+\|x+C\xi\|^{2})}w,w\rangle_{L^{2}(\mathbb{R}^{n})}.$

###### Proof of the claim.

The first statement in the above claim is obvious, since we already know from
(3.3.1), (3.3.12) that

$\frac{1}{4\pi}|\hskip-1.0pt\|\Xi_{12}=B^{-1}C\|\hskip-1.0pt|_{\texttt{S}^{1}}\leq\mu(\mathcal{M})=\mu(\widetilde{\mathcal{M}}).$

On the other hand, we have

$\displaystyle\sqrt{\mathcal{V}(\widetilde{\mathcal{M}}v)\mathcal{V}(v)}\leq\frac{1}{2}\left(\int\|B^{-1}x\|^{2}{|(e^{i\pi
CD^{2}}v)(x)|}^{2}dx+\int\|x\|^{2}{|v(x)|}^{2}dx\right)$
$\displaystyle=\frac{1}{2}\left(\int\|B^{-1}x\|^{2}{|w(x)|}^{2}dx+\int\|x\|^{2}{|(e^{-i\pi
CD^{2}}w)(x)|}^{2}dx\right)$
$\displaystyle=\frac{1}{2}\left(\int\|B^{-1}x\|^{2}{|w(x)|}^{2}dx+\sum_{1\leq
j\leq n}\int{x_{j}}^{2}{|(e^{-i\pi CD^{2}}w)(x)|}^{2}dx\right)$
$\displaystyle=\frac{1}{2}\left(\int\|B^{-1}x\|^{2}{|w(x)|}^{2}dx+\sum_{1\leq
j\leq n}\|x_{j}e^{-i\pi CD^{2}}w\|_{L^{2}(\mathbb{R}^{n})}^{2}\right)$
$\displaystyle=\frac{1}{2}\left(\int\|B^{-1}x\|^{2}{|w(x)|}^{2}dx+\sum_{1\leq
j\leq n}\|e^{i\pi CD^{2}}x_{j}e^{-i\pi
CD^{2}}w\|_{L^{2}(\mathbb{R}^{n})}^{2}\right),$

and applying the symplectic covariance of the Weyl calculus as given by
Theorem 2.36 and Claim 2.32, we find that, with $(\mathbf{e}_{j})_{1\leq j\leq
n}$ the canonical basis of $\mathbb{R}^{n}$,

(3.3.18)
$\sqrt{\mathcal{V}(\widetilde{\mathcal{M}}v)\mathcal{V}(v)}\leq\frac{1}{2}\langle{\text{{O}p}_{\text{{w}}}\bigl{(}\|B^{-1}x\|^{2}+\sum_{1\leq
j\leq n}\langle
x+C\xi,\mathbf{e}_{j}\rangle_{\mathbb{R}^{n}}^{2}\bigr{)}}w,w\rangle_{L^{2}(\mathbb{R}^{n})}\\\
=\frac{1}{2}\langle{\text{{O}p}_{\text{{w}}}\bigl{(}\||B^{-1}|x\|^{2}+\|x+C\xi\|^{2}\bigr{)}}w,w\rangle_{L^{2}(\mathbb{R}^{n})},$

proving the claim. ∎

######  Claim 3.12.

Let $\mathcal{M},\widetilde{\mathcal{M}},B,C$ be given by Lemma 3.10. We
define the following quadratic form on $\mathbb{R}^{2n}$,

(3.3.19) $Q(x,\xi)=\||B^{-1}|x\|^{2}+\|x+C\xi\|^{2}.$

We have then

(3.3.20)
$\mu\bigl{(}\widetilde{\mathcal{M}}\bigr{)}\leq\frac{1}{2}\inf\Bigl{[}\operatorname{Spectrum}\bigl{\\{}{\text{{O}p}_{\text{{w}}}(Q)}\bigr{\\}}\Bigr{]}.$

###### Proof.

Inequality (3.3.20) follows immediately from (3.3.18) and the definition of
$w$ in Claim 3.11. ∎

######  Lemma 3.13.

Let $C$ be a symmetric real $n\times n$ matrix and let $\Omega$ be a symmetric
positive-definite $n\times n$ matrix. We define the operator

(3.3.21) $\mathcal{H}_{C,\Omega}={\text{{O}p}_{\text{{w}}}\bigl{(}\|\Omega
x\|^{2}+\|C\xi+x\|^{2}\bigr{)}},$

where $\|\cdot\|$ stands for the canonical Euclidean norm on $\mathbb{R}^{n}$.
Then we have

(3.3.22) $\inf_{\begin{subarray}{c}v\in\mathscr{S}(\mathbb{R}^{n})\\\
\|v\|_{L^{2}(\mathbb{R}^{n})}=1\end{subarray}}\langle\mathcal{H}_{C,\Omega}v,v\rangle_{L^{2}(\mathbb{R}^{n})}\leq\frac{1}{2\pi}\operatorname{trace}{\sqrt{C\Omega^{2}C}}.$

######  N.B.

The lemma is already proven when $C$ is $0$ from Remark 1.3, and also when
$n=1$ from Lemma 4.12 in our Appendix.

###### Proof.

Taking into account the above Nota Bene, we may assume that $C$ is a symmetric
matrix with rank $r\in\llbracket 1,n\rrbracket$ with $n\geq 2$. We may as well
suppose that

$C=\begin{pmatrix}[l]\Lambda&0_{r,n-r}\\\
0_{r,n-r}&0_{n-r,n-r}\end{pmatrix},\qquad\Omega=\begin{pmatrix}[l]\beta_{11}^{r,r}&\beta_{12}^{r,n-r}\\\
{\beta_{12}^{*}}^{\\!\\!n-r,r}&\beta_{22}^{n-r,n-r}\end{pmatrix},$

with $\Lambda=\operatorname{diag}(\lambda_{1},\dots,\lambda_{r})$ with all
$\lambda_{j}\not=0$, $\Omega$ symmetric positive-definite, which implies in
particular that $\beta_{11}$ is positive-definite. Now the Weyl symbol of
$\mathcal{H}_{C,\Omega}$ is, with obvious notations111111 The norm
$\|\cdot\|_{r}$ stands for an Euclidean norm on $\mathbb{R}^{r}$.

$a(x_{1},x_{2},\xi_{1})=\|\beta_{11}x_{1}+\beta_{12}x_{2}\|^{2}_{r}+\|\beta_{12}^{*}x_{1}+\beta_{22}x_{2}\|_{n-r}^{2}+\|\Lambda\xi_{1}+x_{1}\|^{2}_{r}+\|x_{2}\|^{2}_{n-r}.$

Since it does not depend on $\xi_{2}$, we may apply Proposition 4.9 and take
advantage of the fact that

(3.3.23)
$\inf\bigl{[}\operatorname{Spectrum}\bigl{(}\mathcal{H}_{C,\Omega}\bigr{)}\bigr{]}\leq\inf\bigl{[}\operatorname{Spectrum}\bigl{(}{\text{Op}_{w,r}(a(x_{1},0,\xi_{1}))}\bigr{)}\bigr{]}.$

We have in fact

$a(x_{1},0,\xi_{1})=\|\beta_{11}x_{1}\|^{2}_{r}+\|\beta_{12}^{*}x_{1}\|_{n-r}^{2}+\|\Lambda\xi_{1}+x_{1}\|^{2}_{r}\\\
=\langle(\beta_{11}^{2}+\beta_{12}\beta_{12}^{*})x_{1},x_{1}\rangle_{r}+\|\Lambda\xi_{1}+x_{1}\|^{2}_{r}.$

We note that $(\beta_{11}^{2}+\beta_{12}\beta_{12}^{*})$ is positive-definite,
so applying the result in dimension $r<n$, we get that

(3.3.24)
$\inf\bigl{[}\operatorname{Spectrum}\bigl{(}{\text{Op}_{w,r}(a(x_{1},0,\xi_{1}))}\bigr{)}\bigr{]}\leq\frac{1}{2\pi}\operatorname{trace}\sqrt{\Lambda(\beta_{11}^{2}+\beta_{12}\beta_{12}^{*})\Lambda}\\\
=\frac{1}{2\pi}\operatorname{trace}\sqrt{C\Omega^{2}C},$

and thanks to (3.3.23), this gives the proof of the lemma. ∎

We can now apply this lemma to the case where $\Omega=|B^{-1}|$, where $B$ is
the invertible matrix given in Lemma 3.10 and we find that, with $Q$ defined
in Claim 3.12, we have

$\inf\Bigl{[}\operatorname{Spectrum}\bigl{\\{}{\text{{O}p}_{\text{{w}}}(Q)}\bigr{\\}}\Bigr{]}\leq\frac{1}{2\pi}\operatorname{trace}{\sqrt{C{|B^{-1}|}^{2}C}}=\frac{1}{2\pi}\operatorname{trace}{\sqrt{CB^{*-1}B^{-1}C}}\\\
=\frac{1}{2\pi}|\hskip-1.0pt\|B^{-1}C\|\hskip-1.0pt|_{\mathtt{S}^{1}},$

which implies, thanks to Claim 3.12, that

$\mu\bigl{(}\widetilde{\mathcal{M}}\bigr{)}\leq\frac{1}{4\pi}|\hskip-1.0pt\|B^{-1}C\|\hskip-1.0pt|_{\mathtt{S}^{1}},$

and according to Claim 3.11, this finally gives the proof of Theorem 1.5. ∎

##  4\. Appendix

###  4.1. Classical analysis

####  4.1.1. Fourier transform.

We use in this paper the following normalization for the Fourier transform and
inversion formula: for $u\in\mathscr{S}(\mathbb{R}^{n})$,

(4.1.1) $\hat{u}(\xi)=\int_{\mathbb{R}^{n}}e^{-2i\pi x\cdot\xi}u(x)dx,\quad
u(x)=\int_{\mathbb{R}^{n}}e^{2i\pi x\cdot\xi}\hat{u}(\xi)d\xi,$

a formula that can be extended to $u\in\mathscr{S}^{\prime}(\mathbb{R}^{n})$,
with defining the distribution $\hat{u}$ by the duality bracket

(4.1.2)
$\langle\hat{u},\phi\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{n}),\mathscr{S}(\mathbb{R}^{n})}=\langle
u,\hat{\phi}\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{n}),\mathscr{S}(\mathbb{R}^{n})}.$

Checking (4.1.1) for $u\in\mathscr{S}^{\prime}(\mathbb{R}^{n})$ is then easy,
that is

(4.1.3) $\check{\hat{\hat{u}}}=u,$

where the distribution $\check{u}$ is defined by

(4.1.4)
$\langle\check{u},\phi\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{n}),\mathscr{S}(\mathbb{R}^{n})}=\langle
u,\check{\phi}\rangle_{\mathscr{S}^{\prime}(\mathbb{R}^{n}),\mathscr{S}(\mathbb{R}^{n})},\quad\text{with}\
\check{\phi}(x)=\phi(-x).$

It is useful to notice that for $u\in\mathscr{S}^{\prime}(\mathbb{R}^{n})$,

(4.1.5) $\check{\hat{u}}=\hat{\check{u}}.$

Our normalization yields simple formulas for the Fourier transform of Gaussian
functions: for $A$ a real-valued symmetric positive definite $n\times n$
matrix, we define the function $v_{A}$ in the Schwartz space by

(4.1.6) $v_{A}(x)=e^{-\pi\langle Ax,x\rangle},\quad\text{and we have
}\quad\widehat{v_{A}}(\xi)=(\det A)^{-1/2}e^{-\pi\langle
A^{-1}\xi,\xi\rangle}.$

Similarly when $B$ is a real-valued symmetric non-singular $n\times n$ matrix,
the function $w_{B}$ defined by

$w_{B}(x)=e^{i\pi\langle Bx,x\rangle}$

is in $L^{{\infty}}(\mathbb{R}^{n})$ and thus a tempered distribution and we
have

(4.1.7) $\widehat{w_{B}}(\xi)={|\det
B|}^{-1/2}e^{\frac{i\pi}{4}\operatorname{sign}B}e^{-i\pi\langle
B^{-1}\xi,\xi\rangle},$

where $\operatorname{sign}B$ stands for the signature of $B$ that is, with
$\mathtt{E}$ the set of eigenvalues of $B$ (which are real and non-zero),

(4.1.8)
$\operatorname{sign}B=\underbrace{\operatorname{Card}(\mathtt{E}\cap\mathbb{R}_{+})}_{\nu_{+}(B)}-\underbrace{\operatorname{Card}(\mathtt{E}\cap\mathbb{R}_{-})}_{\nu_{-}(B)=\operatorname{index}{(B)}}.$

The integer $\nu_{-}(B)$ is called the _index_ of $B$, noted
$\operatorname{index}{(B)}$; Formula (4.1.7) can be written as

(4.1.9) $e^{-i\pi n/4}\mathcal{F}\left(e^{i\pi\langle
Bx,x\rangle}\right)=i^{-{\operatorname{index}{B}}}{|\det
B|}^{-1/2}e^{-i\pi\langle B^{-1}\xi,\xi\rangle},$

since $\nu_{+}+\nu_{-}=n$ (as $B$ is non-singular),

$e^{\frac{i\pi
n}{4}}e^{-\frac{i\pi\nu_{-}}{2}}=e^{\frac{i\pi}{4}(\nu_{+}+\nu_{-}-2\nu_{-})}=e^{\frac{i\pi}{4}\operatorname{sign}(B)}.$

We note also that

(4.1.10) $\operatorname{sign}(\det B)=(-1)^{\operatorname{index}{B}},$

so that

$\left(i^{-{\operatorname{index}{B}}}{|\det
B|}^{-1/2}\right)^{2}=(-1)^{\nu_{-}}{|\det B|}^{-1}=\operatorname{sign}(\det
B){|\det B|}^{-1}=(\det B)^{-1},$

and thus the prefactor $i^{-{\operatorname{index}{B}}}{|\det B|}^{-1/2}$ in
the rhs of (4.1.9) is a square root of $1/\det B$.

####  4.1.2. Schatten norms, Polar decomposition

######  Definition 4.1.

Let $\mathbb{H}$ be a (real or complex) Hilbert space, let
$\mathcal{B}(\mathbb{H})$ be the Banach space of bounded linear endomorphisms
of $\mathbb{H}$ and let $T\in\mathcal{B}(\mathbb{H})$. Using the fact that
$T^{*}T$ is a self-adjoint non-negative bounded operator, we may define, using
the spectral theorem,

(4.1.11) $|T|=\sqrt{T^{*}T}.$

Elements of the spectrum of $|T|$ will be called the _singular values_ of $T$.
Let $p\in[1,+{\infty}]$. We define the _Schatten norm of index $p$ of the
operator $T$_, as

(4.1.12)
$|\hskip-1.0pt\|T\|\hskip-1.0pt|_{\mathtt{S}^{p}}=\bigl{(}\operatorname{trace}(|T|^{p})\bigr{)}^{1/p}\text{
for $p$
finite},\quad|\hskip-1.0pt\|T\|\hskip-1.0pt|_{\mathtt{S}^{{\infty}}}=\||T|\|_{\mathcal{B}(\mathbb{H})}=\|T\|_{\mathcal{B}(\mathbb{H})}.$

######  Lemma 4.2.

Let $\mathbb{H}$ and $T$ be as in Definition 4.1. Then we have

(4.1.13) $\ker T=\ker|T|.$

###### Proof.

We have,

$Tx=0\Longrightarrow T^{*}Tx=0\Longrightarrow 0=\langle
T^{*}Tx,x\rangle=\||T|x\|^{2}\Longrightarrow|T|x=0,$ $|T|x=0\Longrightarrow
0=|T|^{2}x=T^{*}Tx\Longrightarrow 0=\langle
T^{*}Tx,x\rangle=\|Tx\|^{2}\Longrightarrow Tx=0,$

concluding the proof. ∎

######  Remark 4.3.

The notation $|T|$ could be misleading. In particular we may have
$|T|\not=|T^{*}|$: take for instance the right shift operator $S_{r}$ defined
on $\ell^{2}(\mathbb{N})$ by

$x=(x_{0},x_{1},\dots,x_{k},\dots)\mapsto(0,x_{0},x_{1},\dots,x_{k},\dots)=S_{r}x.$

The operator $S_{r}$ is isometric since we have obviously
$\|S_{r}x\|_{\ell^{2}(\mathbb{N})}=\|x\|_{\ell^{2}(\mathbb{N})}$ and thus

$S_{r}^{*}S_{r}=I,\quad\text{which implies }|S_{r}|=I.$

On the other hand we have with $S_{l}=S_{r}^{*}$

$\langle
S_{l}\mathbf{e}_{j},\mathbf{e}_{k}\rangle_{\ell^{2}(\mathbb{N})}=\langle\mathbf{e}_{j},\mathbf{e}_{k+1}\rangle_{\ell^{2}(\mathbb{N})}=\delta_{j,k+1},$

so that $S_{l}\mathbf{e}_{0}=0$, and $\forall j\geq
1,S_{l}\mathbf{e}_{j}=\mathbf{e}_{j-1},$ i.e.

$x=(x_{0},x_{1},\dots,x_{k},\dots)\mapsto(x_{1},\dots,x_{k},\dots)=S_{l}x.$

We have thus $S_{l}^{*}S_{l}=S_{r}S_{r}^{*}$ and $S_{r}S_{r}^{*}$ is the
orthogonal projection onto $\ell^{2}(\mathbb{N}^{*})$, since

$\forall j\geq 1,\
S_{r}S_{r}^{*}(\mathbf{e}_{j})=S_{r}(\mathbf{e}_{j-1})=\mathbf{e}_{j},\quad
S_{r}S_{r}^{*}(\mathbf{e}_{0})=0,$

implying that

$|S_{r}^{*}|=I-\mathbb{P}_{0},\quad\text{where $\mathbb{P}_{0}$ is the
orthogonal projection onto $\mathcal{V}(\mathbf{e}_{0}).$}$

More generally if $T:\mathbb{H}\longrightarrow\mathbb{H}_{0}$ is isometric
onto a proper subspace $\mathbb{H}_{0}$ of $\mathbb{H}$, we have

$T^{*}T=\operatorname{Id}_{\mathbb{H}},\quad\text{implying
}|T|=\operatorname{Id}_{\mathbb{H}},$

whereas $TT^{*}$ is the orthogonal projection onto $\mathbb{H}_{0}$: indeed,
it is an orthogonal projection since $TT^{*}TT^{*}=TT^{*}$ and its range is
included in the range of $T$ (which is $\mathbb{H}_{0}$); also if
$y\in\mathbb{H}_{0}$, we have $y=Tx$ with $x\in\mathbb{H}$ and thus

$TT^{*}Tx=Tx=y,\text{ implying that
$\mathbb{H}_{0}\subset\operatorname{ran}(TT^{*})\subset\operatorname{ran}T=\mathbb{H}_{0}$,}$

proving that $TT^{*}=\mathbb{P}_{\mathbb{H}_{0}}$ and
$|T^{*}|=\mathbb{P}_{\mathbb{H}_{0}}$.

Another caveat about the notation $|T|$ is that the commutator $[|T|,T]$ can
be different from $0$, even in two dimensions. We consider the matrix

$T=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix},\quad\text{so that
}T^{*}T=\begin{pmatrix}1&1\\\ 1&2\end{pmatrix}.$

Then $[|T|,T]\not=0$, otherwise we would have $[T^{*}T,T]=[|T|^{2},T]=0$ and
in fact

$[T^{*}T,T]=\begin{pmatrix}1&1\\\ 1&2\end{pmatrix}\begin{pmatrix}1&1\\\
0&1\end{pmatrix}-\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}\begin{pmatrix}1&1\\\
1&2\end{pmatrix}=\begin{pmatrix}-1&-1\\\ 0&1\end{pmatrix}\not=0.$

Moreover the operators $|T|$ and $|T^{*}|$ could be different, even in two
dimensions: using the previous example, we get

$|T|^{2}=T^{*}T=\begin{pmatrix}1&1\\\
1&2\end{pmatrix},\quad|T^{*}|^{2}=TT^{*}=\begin{pmatrix}2&1\\\
1&1\end{pmatrix},$

which are two different matrices which are not even commuting since

$\begin{pmatrix}1&1\\\ 1&2\end{pmatrix}\begin{pmatrix}2&1\\\
1&1\end{pmatrix}=\begin{pmatrix}3&2\\\
4&3\end{pmatrix},\quad\begin{pmatrix}2&1\\\
1&1\end{pmatrix}\begin{pmatrix}1&1\\\ 1&2\end{pmatrix}=\begin{pmatrix}3&4\\\
2&3\end{pmatrix}.$

######  N.B.

Nevertheless, we can note that the singular values of $T$ and $T^{*}$ are the
same (cf. Corollary 4.6).

######  Lemma 4.4.

Let $\mathbb{H}$, $\mathcal{B}(\mathbb{H})$ be as in Definition 4.1. Let $T$
be an invertible operator in $\mathcal{B}(\mathbb{H})$. Then the operator
$|T|$ is invertible and

$U=T{|T|}^{-1}\quad\text{is a unitary operator and we have $T=U|T|.$}$

###### Proof of the lemma.

Since $T$ is invertible, we find that $T^{*}$ is invertible as well as
$T^{*}T$: we have that

$(T^{*}T)^{-1}=T^{-1}(T^{*})^{-1}=T^{-1}(T^{-1})^{*}.$

As a consequence with $|T|=\sqrt{T^{*}T}$, we find that there exists
$\alpha>0$ such that for all $x\in\mathbb{H}$, we have

$\||T|x\|^{2}=\langle T^{*}Tx,x\rangle\geq\alpha\|x\|^{2},$

and thus $|T|$ is invertible. Now we get also that

$U^{*}U={|T|}^{-1}T^{*}T{|T|}^{-1}=(T^{*}T)^{-1/2}T^{*}T(T^{*}T)^{-1/2}=I,$

and
$UU^{*}=T{|T|}^{-1}{|T|}^{-1}T^{*}=T(T^{*}T)^{-1}T^{*}=TT^{-1}T^{*-1}T^{*}=I,$
so that $U$ is indeed unitary. ∎

######  Proposition 4.5.

Let $\mathbb{H}$ be a finite-dimensional Hilbert space. Let $T$ be in
$\mathcal{B}(\mathbb{H})$. Then there exists a unitary operator $\mathcal{U}$
such that

(4.1.14) $T=\mathcal{U}|T|.$

###### Proof.

According to Lemma 4.4, we may assume that the kernel of $T$ has dimension
$r\in\llbracket 1,n-1\rrbracket$, where $n=\dim\mathbb{H}$. Then we have

$\mathbb{H}=\underbrace{\ker T}_{\text{dim }r}\oplus\underbrace{(\ker
T)^{\perp}}_{\text{dim
}n-r}=\underbrace{(\operatorname{ran}T)^{\perp}}_{\text{dim
}r}\oplus\underbrace{\operatorname{ran}T}_{\text{dim }n-r},$

and thus

$\begin{matrix}[r]\tilde{T}:(\ker
T)^{\perp}&\longrightarrow&\operatorname{ran}T\\\
x&\mapsto&Tx\end{matrix}\quad\text{ is an isomorphism.}$

We have thus, with $P$ standing for the orthogonal projection on $(\ker
T)^{\perp}$ (note that $P^{*}$ is the canonical injection121212Indeed we have
for $x_{2}\in(\ker T)^{\perp}$, $x_{1}\in\mathbb{H}$,  $\langle
P^{*}x_{2},x_{1}\rangle_{\mathbb{H}}=\langle x_{2},Px_{1}\rangle_{(\ker
T)^{\perp}}=\langle x_{2},x_{1}\rangle_{\mathbb{H}},\quad\text{since
$x_{2}\in(\ker T)^{\perp}.$}$ of $(\ker T)^{\perp}$ into $\mathbb{H}$) and $J$
for the canonical injection of $\operatorname{ran}T$ into $\mathbb{H}$ (note
that $J^{*}$ is the orthogonal projection from $\mathbb{H}$ onto
$\operatorname{ran}T$),

(4.1.15) $T=J\tilde{T}P,$

so that
$T^{*}T=P^{*}\tilde{T}^{*}\underbrace{J^{*}J}_{\operatorname{Id}_{\operatorname{ran}T}}\tilde{T}P,$
and for $x=x^{\prime}\oplus x^{\prime\prime},x^{\prime\prime}=Px$,

$T^{*}Tx=P^{*}\tilde{T}^{*}J^{*}J\underbrace{\tilde{T}Px}_{\in\operatorname{ran}T}=P^{*}{|\tilde{T}|}^{2}P.$

We note also that $P^{*}|\tilde{T}|PP^{*}\underbrace{|\tilde{T}|Px}_{\in(\ker
T)^{\perp}}=P^{*}|\tilde{T}||\tilde{T}|Px,$ proving that
$|T|=P^{*}|\tilde{T}|P.$ We have in fact, thanks to Lemma 4.4, with a unitary
$W$, the following commutative diagram:

---
$\textstyle{\mathbb{H}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{T}$$\scriptstyle{P}$$\textstyle{\mathbb{H}}$$\textstyle{(\ker
T)^{\perp}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{|\tilde{T}|}$$\scriptstyle{\tilde{T}}$$\textstyle{\operatorname{ran}T\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{J}$$\textstyle{(\ker
T)^{\perp}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{W}$

We have also from (4.1.15) and the above diagram

$T=JW|\tilde{T}|P,$

and defining with $R_{0}$ isometric from $\ker T$ onto
$(\operatorname{ran}T)^{\perp}$ (which are finite-dimensional spaces with the
same dimension),

(4.1.16) $\mathcal{U}x=\mathcal{U}((I-P)x\oplus
Px)=\underbrace{R_{0}\overbrace{(I-P)x}^{\in\ker
T}}_{\in(\operatorname{ran}T)^{\perp}}+\underbrace{W\hskip-7.0pt\overbrace{Px}^{\in(\ker
T)^{\perp}}}_{\in\operatorname{ran}T},$

we get

$\mathcal{U}|T|x=\mathcal{U}\overbrace{P^{*}\underbrace{|\tilde{T}|Px}_{\in(\ker
T)^{\perp}}}^{\in(\ker
T)^{\perp}}=\underbrace{W|\tilde{T}|\overbrace{Px}^{\in(\ker
T)^{\perp}}}_{\in\operatorname{ran}T}=JW|\tilde{T}|Px=Tx,$

so that

$T=\mathcal{U}|T|,$

and $\mathcal{U}$ is unitary since, using (4.1.16) and the isometries $R_{0}$
and $W$, we get

$\|\mathcal{U}x\|^{2}=\|R_{0}(I-P)x\|^{2}+\|WPx\|^{2}=\|(I-P)x\|^{2}+\|Px\|^{2}=\|x\|^{2},$

proving that $\mathcal{U}^{*}\mathcal{U}=I,$ providing the result since
$\mathbb{H}$ is finite dimensional. ∎

######  Corollary 4.6.

Let $\mathbb{H}$, $T$ be as in Proposition 4.5. Then $|T|$ and $|T^{*}|$ are
unitarily conjugate and in particular have the same spectrum, so that $T$ and
$T^{*}$ have the same singular values and for all $p\in[1,+{\infty}]$, we have

(4.1.17)
$|\hskip-1.0pt\|T^{*}\|\hskip-1.0pt|_{\mathtt{S}^{p}}=|\hskip-1.0pt\|T\|\hskip-1.0pt|_{\mathtt{S}^{p}}.$

###### Proof.

We have
$TT^{*}=\mathcal{U}|T||T|\mathcal{U}^{*}=\mathcal{U}{|T|}^{2}\mathcal{U}^{*}$
and

$\mathcal{U}|T|\mathcal{U}^{*}\mathcal{U}|T|\mathcal{U}^{*}=\mathcal{U}{|T|}^{2}\mathcal{U}^{*},$

so that $|T^{*}|=\mathcal{U}|T|\mathcal{U}^{*},$ proving the sought result. ∎

######  Remark 4.7.

As it is apparent from the proof of Proposition 4.5, the decomposition
(4.1.14) is not unique, since we have many possible choices for the unitary
operator $\mathcal{U}$, in particular with the choice of $R_{0}$ in (4.1.16).
As a consequence, except in the case where $T$ is invertible, it does not
coincide with the so-called _polar decomposition_ (see e.g. Theorem VI.10 in
the B. Simon & M. Reed’ book [23] and Section 3.9 in J. Conway’s book [6])
providing a decomposition $T=U|T|$ where $U$ is a partial isometry. However,
the fact that $\mathcal{U}$ in (4.1.14) can be chosen as unitary is helpful
for us at several places, in particular for the precise computation of the
lower bound of an Harmonic Oscillator.

###  4.2. Classical phase space analysis

####  4.2.1. The Classical Heisenberg Uncertainty Principle

Let $u\in\mathscr{S}(\mathbb{R})$. We have, with
$D_{t}=\frac{\partial}{2i\pi\partial t}$,

(4.2.1) $2\operatorname{Re}\langle
D_{t}u,itu\rangle_{L^{2}(\mathbb{R})}=\langle D_{t}u,itu\rangle+\langle
itu,D_{t}u\rangle=\langle D_{t}itu-itD_{t}u,u\rangle\\\
=\langle[D_{t},it]u,u\rangle_{L^{2}(\mathbb{R})}=\frac{1}{2\pi}\|u\|^{2}_{L^{2}(\mathbb{R})},$

implying in particular

(4.2.2)
$\|D_{t}u\|_{L^{2}(\mathbb{R})}\|tu\|_{L^{2}(\mathbb{R})}\geq\frac{1}{4\pi}\|u\|^{2}_{L^{2}(\mathbb{R})},$

which is an equality for $u(t)=e^{-\pi t^{2}}2^{1/4}$; moreover we infer also
from (4.2.1) that

(4.2.3) $\displaystyle\langle\pi(D_{t}^{2}+t^{2})u,u\rangle$
$\displaystyle\geq\frac{1}{2}\|u\|^{2}_{L^{2}(\mathbb{R})},$

proving that

(4.2.4)
$\min\bigl{[}\operatorname{Spectrum}{\pi(D_{t}^{2}+t^{2})}\bigr{]}=\frac{1}{2}.$

The core argument in the proof of (4.2.2) is the integration by parts given by
the third equality, taking advantage of the non-commutation of the operators
of momentum $D_{t}$ and of position $t$. We can infer a non-trivial
generalization of (4.2.2) by noticing that for
$u\in\mathscr{S}(\mathbb{R}^{n})$, we have

$2\operatorname{Re}\langle
D_{x_{j}}u,ix_{j}u\rangle_{L^{2}(\mathbb{R}^{n})}=\frac{1}{2\pi}\|u\|^{2}_{L^{2}(\mathbb{R}^{n})},$

which implies for $p\in[1,+{\infty}]$, $\frac{1}{p}+\frac{1}{p^{\prime}}=1$,

$\displaystyle\frac{n}{4\pi}\|u\|^{2}_{L^{2}(\mathbb{R}^{n})}=\sum_{1\leq
j\leq n}\operatorname{Re}\langle
D_{x_{j}}u,ix_{j}u\rangle_{L^{2}(\mathbb{R}^{n})}\leq\sum_{1\leq j\leq
n}\|D_{x_{j}}u\|_{L^{p}(\mathbb{R}^{n})}\|x_{j}u\|_{L^{p^{\prime}}(\mathbb{R}^{n})}$
$\displaystyle\leq\Bigl{(}\sum_{1\leq j\leq
n}\|D_{x_{j}}u\|^{p}_{L^{p}(\mathbb{R}^{n})}\Bigr{)}^{1/p}\Bigl{(}\sum_{1\leq
j\leq
n}\|x_{j}u\|^{{p^{\prime}}}_{L^{p^{\prime}}(\mathbb{R}^{n})}\Bigr{)}^{1/{p^{\prime}}}=\|Du\|_{L^{p}(\mathbb{R}^{n})}\|xu\|_{L^{p^{\prime}}(\mathbb{R}^{n})},$

where for a function $\mathbb{R}^{n}\ni x\mapsto
v(x)=\bigl{(}v_{j}(x)\bigr{)}_{1\leq j\leq n}\in\mathbb{R}^{n}$, we have
defined

(4.2.5) $\|v\|_{L^{p}(\mathbb{R}^{n})}=\left(\int\sum_{1\leq j\leq
n}{|v_{j}(x)|}^{p}dx\right)^{1/p}=\left(\int\|v(x)\|_{p}^{p}dx\right)^{1/p}.$

We get thus

(4.2.6)
$\bigl{\|}Du\bigr{\|}_{L^{p}(\mathbb{R}^{n})}\bigl{\|}xu(x)\bigr{\|}_{L^{p^{\prime}}(\mathbb{R}^{n})}\geq\frac{n}{4\pi}\|u\|^{2}_{L^{2}(\mathbb{R}^{n})},$

and in particular for $p=2$, since we have with the standard Euclidean norm on
$\mathbb{R}^{n}$, and the vector
$(Du)(x)=\bigl{(}(D_{x_{j}}u)(x)\bigr{)}_{1\leq j\leq n}$,

$\bigl{\|}Du\bigr{\|}_{L^{2}(\mathbb{R}^{n})}^{2}=\int\sum_{1\leq j\leq
n}{|(D_{{x_{j}}}u)(x)|}^{2}dx\\\
=\int\langle(Du)(x),(Du)(x)\rangle_{\mathbb{R}^{n}}dx=\bigl{\|}|Du|\bigr{\|}_{L^{2}(\mathbb{R}^{n})}^{2},$

we get

(4.2.7)
$\bigl{\|}|D|u\bigr{\|}_{L^{2}(\mathbb{R}^{n})}\bigl{\|}|x|u(x)\bigr{\|}_{L^{2}(\mathbb{R}^{n})}\geq\frac{n}{4\pi}\|u\|^{2}_{L^{2}(\mathbb{R}^{n})},$

with the Fourier multiplier

(4.2.8) $|D|=\bigl{(}\sum_{1\leq j\leq n}D_{x_{j}}^{2}\bigr{)}^{1/2}.$

The inequality (4.2.7) can be written as

(4.2.9)
$\Bigl{(}\int{|\xi|}^{2}{|\hat{u}(\xi)|}^{2}d\xi\Bigr{)}\Bigl{(}\int{|x|}^{2}{|u(x)|}^{2}dx\Bigr{)}\geq\frac{n^{2}}{2^{4}\pi^{2}}\|u\|^{2}_{L^{2}(\mathbb{R}^{n})},$

and is an equality for the Gaussian function

(4.2.10) $u(x)=2^{n/4}e^{-\pi\|x\|_{2}^{2}}.$

Indeed we have $\hat{u}(\xi)=2^{n/4}e^{-\pi\|\xi\|_{2}^{2}},$ and

$\displaystyle 2^{n}\Bigl{(}\int{|y|}^{2}e^{-2\pi|y|^{2}}\Bigr{)}^{2}$
$\displaystyle=2^{n}\Bigl{(}\int_{0}^{+{\infty}}r^{n+1}e^{-2\pi
r^{2}}dr\Bigr{)}^{2}\frac{4\pi^{n}}{\Gamma(n/2)^{2}}$ (using
$r=s^{1/2}(2\pi)^{-1/2}$)
$\displaystyle=2^{n}\frac{4\pi^{n}}{\Gamma(n/2)^{2}}\frac{1}{(2\pi)^{n+1}}\Bigl{(}\int_{0}^{+{\infty}}s^{n/2}e^{-s}\frac{ds}{2^{3/2}\sqrt{\pi}}\Bigr{)}^{2}$
(4.2.11)
$\displaystyle=\frac{1}{4\pi^{2}\Gamma(n/2)^{2}}\Gamma(\frac{n}{2}+1)^{2}=\frac{1}{4\pi^{2}}\frac{n^{2}}{4}=\frac{n^{2}}{2^{4}\pi^{2}}.$

######  N.B.

We note that the proof of (4.2.7) in dimensions larger than $2$ does not
follow immediately from an integration by parts since we have to deal with
proving the inequality

$\bigl{\|}|D|u\bigr{\|}_{L^{2}(\mathbb{R}^{n})}\bigl{\|}|x|u\bigr{\|}_{L^{2}(\mathbb{R}^{n})}\geq\frac{n}{4\pi}\bigl{\|}u\bigr{\|}_{L^{2}(\mathbb{R}^{n})}^{2}.$

When writing

$2\operatorname{Re}\langle|D|u,i|x|u\rangle=\langle[|D|,i|x|]u,u\rangle,$

the bracket $[|D|,i|x|]$ is not so easy to calculate and we are not really
able to find a direct $n$-dimensional argument to handle Inequality (4.2.7).

######  Remark 4.8.

If $M$ is a unitary operator in $L^{2}(\mathbb{R}^{n})$, then we have

(4.2.12) $\mu(M)=\mu(M^{*}),$

where $\mu$ is defined by (1.2.4). Indeed, we have

$\displaystyle\mu(M^{*})^{2}$ $\displaystyle=\inf_{\begin{subarray}{c}u\in
L^{2}(\mathbb{R}^{n})\\\
\|u\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\bigl{(}\mathcal{V}(M^{*}u)\mathcal{V}(u)\bigr{)}$
$\displaystyle=\inf_{\begin{subarray}{c}v\in L^{2}(\mathbb{R}^{n})\\\
\|v\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\bigl{(}\mathcal{V}(M^{*}Mv)\mathcal{V}(Mv)\bigr{)}=\inf_{\begin{subarray}{c}v\in
L^{2}(\mathbb{R}^{n})\\\
\|v\|_{L^{2}(\mathbb{R}^{n})=1}\end{subarray}}\bigl{(}\mathcal{V}(v)\mathcal{V}(Mv)\bigr{)}=\mu(M)^{2}.$

######  Proposition 4.9.

Let $n_{1},n_{2}\in\mathbb{N}^{*}$ and let $n=n_{1}+n_{2}$. Let
$Q(x_{1},x_{2},\xi_{2})$ be a non-negative quadratic form on
$\mathbb{R}^{n_{1}}\times\mathbb{R}^{n_{2}}\times\mathbb{R}^{n_{2}}$. We
define

(4.2.13)
$\lambda_{0}(x_{1})=\inf\Bigl{[}\operatorname{Spectrum}\bigl{\\{}{\text{Op}_{w,n_{2}}\bigl{(}Q(x_{1},\cdot,\cdot)\bigr{)}}\bigr{\\}}\Bigr{]},$

where ${\text{Op}_{w,n_{2}}(b)}$ stands for the Weyl quantization of the
Hamiltonian $b$ defined on $\mathbb{R}^{n_{2}}\times\mathbb{R}^{n_{2}}$. Then
we have

(4.2.14)
$\inf\Bigl{[}\operatorname{Spectrum}\bigl{\\{}{\text{Op}_{w,n}\bigl{(}Q\bigr{)}}\bigr{\\}}\Bigr{]}=\inf_{x_{1}\in\mathbb{R}}\lambda_{0}(x_{1}).$

###### Proof.

Let $v\in\mathscr{S}(\mathbb{R}^{n_{1}+n_{2}})$: we have

$\displaystyle\langle{\text{Op}_{w,n}\bigl{(}Q\bigr{)}}v,v\rangle_{L^{2}(\mathbb{R}^{n})}=\iiint\hskip-5.0pt\iiint
e^{2i\pi(\langle
x_{1}-y_{1},\xi_{1}\rangle+\langle(x_{2}-y_{2}),\xi_{2}\rangle)}$
$\displaystyle\hskip 108.0pt\times
Q\bigl{(}\frac{x_{1}+y_{1}}{2},\frac{x_{2}+y_{2}}{2},\xi_{2}\bigr{)}v(y_{1},y_{2})\overline{v(x_{1},x_{2})}dy_{1}dy_{2}d\xi_{1}d\xi_{2}dx_{1}dx_{2}$
$\displaystyle=\iint\hskip-5.0pt\iint e^{2i\pi\langle
x_{2}-y_{2},\xi_{2}\rangle}Q\bigl{(}x_{1},\frac{x_{2}+y_{2}}{2},\xi_{2}\bigr{)}v(x_{1},y_{2})\overline{v(x_{1},x_{2})}dy_{2}dx_{2}d\xi_{2}dx_{1}$
$\displaystyle=\int_{\mathbb{R}^{n_{1}}}\langle{\text{Op}_{w,n_{2}}\bigl{(}Q(x_{1},\cdot,\cdot)\bigr{)}}v_{x_{1}},v_{x_{1}}\rangle_{L^{2}(\mathbb{R}^{n_{2}})}dx_{1},$

with $v_{x_{1}}(x_{2})=v(x_{1},x_{2}).$ As a consequence, we have

$\langle{\text{Op}_{w,n}\bigl{(}Q\bigr{)}}v,v\rangle_{L^{2}(\mathbb{R}^{n})}\geq\int_{\mathbb{R}^{n_{1}}}\lambda_{0}(x_{1})\|v_{x_{1}}\|_{L^{2}(\mathbb{R}^{n_{2}})}^{2}dx_{1}\\\
\geq\inf_{x_{1}\in\mathbb{R}}\lambda_{0}(x_{1})\int_{\mathbb{R}^{n_{1}}}\|v_{x_{1}}\|_{L^{2}(\mathbb{R}^{n_{2}})}^{2}dx_{1}=\bigl{(}\inf_{x_{1}\in\mathbb{R}}\lambda_{0}(x_{1})\bigr{)}\|v\|_{L^{2}(\mathbb{R}^{n})}^{2},$

proving that

(4.2.15)
$\inf\Bigl{[}\operatorname{Spectrum}\bigl{\\{}{\text{Op}_{w,n}\bigl{(}Q\bigr{)}}\bigr{\\}}\Bigr{]}\geq\inf_{x_{1}\in\mathbb{R}}\lambda_{0}(x_{1}).$

Let $z_{1}\in\mathbb{R}^{n_{1}}$. We check with $\varepsilon>0$,

$v_{1}(z_{1},\varepsilon,x_{1})=\phi\bigl{(}(x_{1}-z_{1})/\varepsilon\bigr{)}\varepsilon^{-n_{1}/2},\quad\phi\in\mathscr{S}(\mathbb{R}^{n_{1}}),\quad\|\phi\|_{L^{2}(\mathbb{R}^{n_{1}})}=1,$

using the first computation in this proof, for
$v_{2}\in\mathscr{S}(\mathbb{R}^{n_{2}})$,

$\langle{\text{Op}_{w,n}\bigl{(}Q\bigr{)}}(v_{1}(z_{1},\varepsilon,\cdot)\otimes
v_{2}),(v_{1}(z_{1},\varepsilon,\cdot)\otimes
v_{2})\rangle_{L^{2}(\mathbb{R}^{n})}\\\
=\int_{\mathbb{R}^{n_{1}}}\langle{\text{Op}_{w,n_{2}}\bigl{(}Q(x_{1},\cdot,\cdot)\bigr{)}}v_{2},v_{2}\rangle_{L^{2}(\mathbb{R}^{n_{2}})}{|\phi\bigl{(}(x_{1}-z_{1})/\varepsilon\bigr{)}|}^{2}\varepsilon^{-n_{1}}dx_{1}\\\
=\int_{\mathbb{R}^{n_{1}}}\langle{\text{Op}_{w,n_{2}}\bigl{(}Q(z_{1}+\varepsilon
t_{1},\cdot,\cdot)\bigr{)}}v_{2},v_{2}\rangle_{L^{2}(\mathbb{R}^{n_{2}})}{|\phi\bigl{(}t_{1}\bigr{)}|}^{2}dt_{1},$

so that

$\lim_{\varepsilon\rightarrow
0_{+}}\langle{\text{Op}_{w,n}\bigl{(}Q\bigr{)}}(v_{1}\otimes
v_{2}),(v_{1}\otimes v_{2})\rangle_{L^{2}(\mathbb{R}^{n})}\\\
=\int_{\mathbb{R}^{n_{1}}}\langle{\text{Op}_{w,n_{2}}\bigl{(}Q(z_{1},\cdot,\cdot)\bigr{)}}v_{2},v_{2}\rangle_{L^{2}(\mathbb{R}^{n_{2}})}{|\phi\bigl{(}t_{1}\bigr{)}|}^{2}dt_{1}\\\
=\langle{\text{Op}_{w,n_{2}}\bigl{(}Q(z_{1},\cdot,\cdot)\bigr{)}}v_{2},v_{2}\rangle_{L^{2}(\mathbb{R}^{n_{2}})}.$

Let $\gamma>0$ be given. We may choose $v_{2}(x_{2})=w(z_{1},\gamma,x_{2})$
such that

$\|v_{2}(z_{1},\gamma,\cdot)\|_{L^{2}(\mathbb{R}^{n_{2}})}=1,$

and verifying

$\langle{\text{Op}_{w,n_{2}}\bigl{(}Q(z_{1},\cdot,\cdot)\bigr{)}}v_{2},v_{2}\rangle_{L^{2}(\mathbb{R}^{n_{2}})}\leq\gamma+\lambda_{0}(z_{1}).$

As a consequence, we find that

$1=\|v_{1}(z_{1},\varepsilon,\cdot)\otimes
v_{2}(z_{1},\gamma,\cdot)\|_{L^{2}(\mathbb{R}^{n})},$

and

$\lim_{\varepsilon\rightarrow
0_{+}}\langle{\text{Op}_{w,n}\bigl{(}Q\bigr{)}}(v_{1}(z_{1},\varepsilon,\cdot)\otimes
w(z_{1},\gamma,\cdot)),v_{1}(z_{1},\varepsilon,\cdot)\otimes
w(z_{1},\gamma,\cdot)\rangle_{L^{2}(\mathbb{R}^{n})}\\\
\leq\gamma+\lambda_{0}(z_{1}),$

so that

$\langle{\text{Op}_{w,n}\bigl{(}Q\bigr{)}}(v_{1}(z_{1},\varepsilon,\cdot)\otimes
w(z_{1},\gamma,\cdot)),v_{1}(z_{1},\varepsilon,\cdot)\otimes
w(z_{1},\gamma,\cdot)\rangle_{L^{2}(\mathbb{R}^{n})}\\\
\leq\gamma+\lambda_{0}(z_{1})+\beta(z_{1},\gamma,\varepsilon),\quad\lim_{\varepsilon\rightarrow
0_{+}}\beta(z_{1},\gamma,\varepsilon)=0.$

The previous inequality implies that for
$\gamma>0,\varepsilon>0,z_{1}\in\mathbb{R}^{n_{1}}$,

(4.2.16)
$\inf\Bigl{[}\operatorname{Spectrum}\bigl{\\{}{\text{Op}_{w,n}\bigl{(}Q\bigr{)}}\bigr{\\}}\Bigr{]}\leq\gamma+\lambda_{0}(z_{1})+\beta(z_{1},\gamma,\varepsilon).$

Taking the $\mathtt{\limsup}$ with respect to $\varepsilon\rightarrow 0_{+}$
on both sides of (4.2.16), we obtain for
$\gamma>0,z_{1}\in\mathbb{R}^{n_{1}}$,

(4.2.17)
$\inf\Bigl{[}\operatorname{Spectrum}\bigl{\\{}{\text{Op}_{w,n}\bigl{(}Q\bigr{)}}\bigr{\\}}\Bigr{]}\leq\gamma+\lambda_{0}(z_{1}).$

Taking then the $\mathtt{\limsup}$ with respect to $\gamma\rightarrow 0_{+}$
on both sides of (4.2.17), we get for $z_{1}\in\mathbb{R}^{n_{1}}$,

(4.2.18)
$\inf\Bigl{[}\operatorname{Spectrum}\bigl{\\{}{\text{Op}_{w,n}\bigl{(}Q\bigr{)}}\bigr{\\}}\Bigr{]}\leq\lambda_{0}(z_{1}).$

proving the converse inequality of (4.2.15) and the proposition. ∎

######  Corollary 4.10.

Let $C$ be a symmetric real $n\times n$ matrix and let $\Omega$ be a symmetric
positive-definite $n\times n$ matrix. We define the operator

(4.2.19) $\mathcal{H}_{C,\Omega}={\text{{O}p}_{\text{{w}}}\bigl{(}\|\Omega
x\|^{2}+\|C\xi+x\|^{2}\bigr{)}},$

where $\|\cdot\|$ stands for the canonical Euclidean norm on $\mathbb{R}^{n}$.
Then we have

(4.2.20) $\inf_{\begin{subarray}{c}v\in\mathscr{S}(\mathbb{R}^{n})\\\
\|v\|_{L^{2}(\mathbb{R}^{n})}=1\end{subarray}}\langle\mathcal{H}_{C,\Omega}v,v\rangle_{L^{2}(\mathbb{R}^{n})}\leq\frac{1}{2\pi}\operatorname{trace}{\sqrt{C\Omega^{2}C}}.$

###### Proof.

We may assume that $C$ is a symmetric matrix with rank $r\in\llbracket
1,n\rrbracket$. We may as well suppose that

$C=\begin{pmatrix}[l]\Lambda&0_{r,n-r}\\\
0_{r,n-r}&0_{n-r,n-r}\end{pmatrix},\qquad\Omega=\begin{pmatrix}[l]\beta_{11}^{r,r}&\beta_{12}^{r,n-r}\\\
{\beta_{12}^{*}}^{\\!\\!n-r,r}&\beta_{22}^{n-r,n-r}\end{pmatrix},$

with $\Lambda=\operatorname{diag}(\lambda_{1},\dots,\lambda_{r})$ with all
$\lambda_{j}\not=0$, $\Omega$ symmetric positive-definite, which implies in
particular that $\beta_{11}$ is positive-definite. Now the Weyl symbol of
$\mathcal{H}_{C,\Omega}$ is

$a(x_{1},x_{2},\xi_{1})=\|\beta_{11}x_{1}+\beta_{12}x_{2}\|^{2}_{r}+\|\beta_{12}^{*}x_{1}+\beta_{22}x_{2}\|_{n-r}^{2}+\|\Lambda\xi_{1}+x_{1}\|^{2}_{r}+\|x_{2}\|^{2}_{n-r}.$

Since it is a quadratic form independent of $\xi_{2}$, we may apply
Proposition 4.9. We have for instance

$a(x_{1},0,\xi_{1})=\|\beta_{11}x_{1}\|^{2}_{r}+\|\beta_{12}^{*}x_{1}\|_{n-r}^{2}+\|\Lambda\xi_{1}+x_{1}\|^{2}_{r}\\\
=\langle(\beta_{11}^{2}+\beta_{12}\beta_{12}^{*})x_{1},x_{1}\rangle_{r}+\|\Lambda\xi_{1}+x_{1}\|^{2}_{r}.$

We note that $(\beta_{11}^{2}+\beta_{12}\beta_{12}^{*})$ is positive-definite,
so applying the result when $C$ is invertible we get that

$\inf\bigl{[}\operatorname{Spectrum}(\mathcal{H}_{C,\Omega})\bigr{]}\leq\frac{1}{2\pi}\operatorname{trace}\sqrt{\Lambda(\beta_{11}^{2}+\beta_{12}\beta_{12}^{*})\Lambda}$

which is indeed equal to
$\frac{1}{2\pi}\operatorname{trace}\sqrt{C\Omega^{2}C},$ proving the sought
result. ∎

####  4.2.2. On some Gaussian functions

We consider for $x\in\mathbb{R},t>0$,

(4.2.21) $v_{t}(x)=e^{-\pi tx^{2}}(2t)^{1/4},\quad\text{so that
}\|v_{t}\|_{L^{2}(\mathbb{R})}=1,$

and

(4.2.22) $(\mathcal{M}_{p,l,q}v_{t})(x)=e^{-i\pi/4}e^{i\pi
m/2}{|l|}^{1/2}\int_{\mathbb{R}}e^{i\pi(px^{2}-2lxy+qy^{2})}e^{-\pi
ty^{2}}(2t)^{1/4}dy,$

with $l\not=0$, $e^{i\pi m}=\operatorname{sign}l$. We find that

$\displaystyle{|(\mathcal{M}_{p,l,q}v_{t})(x)|}^{2}$
$\displaystyle=(2t)^{1/2}|l|\iint
e^{i\pi(px^{2}-2lxy)}e^{-\pi(t-iq)y^{2}}e^{-i\pi(px^{2}-2lxz)}e^{-\pi(t+iq)z^{2}}dydz$
$\displaystyle=(2t)^{1/2}{|l|}{\left|\int e^{-2i\pi
lxy}e^{-\pi(t-iq)y^{2}}dy\right|}^{2}$
$\displaystyle=(2t)^{1/2}{|l|}{|e^{-\pi(t-iq)^{-1}l^{2}x^{2}}|}^{2}(t^{2}+q^{2})^{-1/2},$

and thus we have

$\displaystyle\mathcal{V}\bigl{(}\mathcal{M}_{p,l,q}v_{t}\bigr{)}$