\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 260, pp. 1--28.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2015 Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/260\hfil A system of Schr\"odinger equations] {A system of Schr\"odinger equations and the oscillator representation} \author[M. Hunziker, M. R. Sepanski, R. J. Stanke \hfil EJDE-2015/260\hfilneg] {Markus Hunziker, Mark R. Sepanski, Ronald J. Stanke} \address{Markus Hunziker \newline Department of Mathematics, Baylor University, One Bear Place 97328, Waco, TX 76798-7328, USA} \email{Markus\_Hunziker@baylor.edu} \address{Mark R. Sepanski\newline Department of Mathematics, Baylor University, One Bear Place 97328, Waco, TX 76798-7328, USA} \email{Mark\_Sepanski@baylor.edu} \address{Ronald J. Stanke \newline Department of Mathematics, Baylor University, One Bear Place 97328, Waco, TX 76798-7328, USA} \email{Ronald\_Stanke@baylor.edu} \thanks{Submitted October 4, 2013. Published October 7, 2015.} \subjclass[2010]{22E30, 35A30, 35J10, 58J70} \keywords{Oscillator representation; Schr\"odinger equation; \hfill\break\indent metaplectic representation; Segal-Shale-Weil representation; Jacobi group} \begin{abstract} We construct a copy of the oscillator representation of the metaplectic group $Mp(n) $ in the space of solutions to a system of Schr\"odinger type equations on $\mathbb{R}^{n}\times\operatorname*{Sym}(n,\mathbb{R})$ that has very simple intertwining maps to the realizations given by Kashiwara and Vergne. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{corollary}[theorem]{Corollary} \allowdisplaybreaks \section{Introduction} Generalizing results from \cite{SS2010,SS2005} and using techniques similar to those found in \cite{HSS2012}, this paper uses Lie symmetry analysis to study the system of partial differential equations \begin{equation} \begin{gathered} 4s\partial_{t_{ii}}f(x,t) +\partial_{x_i}^2f( x,t) =0, \quad 1\leq i\leq n,\\ 2s\partial_{t_{ij}}f(x,t) +\partial_{x_i}\partial_{x_{j} }f(x,t) =0, \quad 1\leq i0$. Turning to our realization, with the parameter choice of $r=-1/2$ and $s=-2\pi^2i$, we have a commutative diagram \[ \begin{array} [c]{ccccc} & & \mathcal{D}_{+}' & & \\ & \overset{\mathcal{G}}{\swarrow} & & \overset{\mathcal{E}}{\searrow} & \\ \mathcal{I}_{+}' & \longleftarrow & \overset{\mathcal{H} }{\longleftarrow} & \longleftarrow & \mathcal{S}_{+}(\mathbb{R}^{n}) \end{array} \] where \[ \mathcal{D}_{+}'\subseteq\mathcal{C}^{\infty}(\mathbb{R} ^{n}\times\operatorname*{Sym}(n,\mathbb{R}) ) \] is the set of smooth solutions, $f$, satisfying the system of partial differential equations \begin{equation} \begin{gathered} i\partial_{t_{i,j}}f =\frac{1}{4\pi^2}\partial_{x_i}\partial_{x_{j} }f\quad (\text{for } i\neq j) \\ i\partial_{t_{ii}}f =\frac{1}{8\pi^2}\partial_{x_i}^2f \end{gathered}\label{eqn:pdes with KV s} \end{equation} with $f(\cdot,t) \in\mathcal{S}_{+}(\mathbb{R}^{n}) $ for each $t\in\operatorname*{Sym}(n,\mathbb{R}) $ and \[ \mathcal{I}_{+}'\subseteq\mathcal{C}^{\infty}( \operatorname*{Sym}(n,\mathbb{R}) ) \] is a subspace of the noncompact picture of a certain principal series representation, see \S \ref{section: induced reps}, that essentially consists of the set of Fourier transforms of Schwartz functions pulled back as measures on $\left\{ -y^{T}y: y\in\mathbb{R}^{n}\right\} \subseteq \operatorname*{Sym}(n,\mathbb{R}) $ (see Corollary \ref{cor: Iprime identification}). The maps $\mathcal{E}$ and $\mathcal{G}$ are given by the particularly simple maps \[ (\mathcal{E}f) (x) =\widehat{f}(x,0) \] (with the Fourier transform given by $\widehat{f}(\xi) =\int_{\mathbb{R}^{n}}f(x) e^{-2\pi i\xi x^{T}}\,dx$) and \[ (\mathcal{G}f) (t) =f(0,t) . \] There is an explicit integral formula for $\mathcal{E}^{-1}$ given by \[ (\mathcal{E}^{-1}\psi) (x,t) =\int_{\mathbb{R} ^{n}}f(\xi) e^{\frac{i}{2}\xi t\xi^{T}}e^{2\pi i\xi x^{T}}\,d\xi \] which gives rise to a formula for $\mathcal{H}=\mathcal{F}_{1}$. An inverse for $\mathcal{G}$ can be given by viewing elements of $\mathcal{I}_{+}'$ as tempered distributions on $\operatorname*{Sym}(n,\mathbb{R}) $, applying a Fourier transform, and taking a limit using approximations to a $\delta$-function (see the proof of Theorem \ref{thm:x=0 injection}). The highest weight vector in $\mathcal{D}_{+}'$ is given by the function $f_{+}\in\mathcal{C}^{\infty}(\mathbb{R}^{n}\times \operatorname*{Sym}(n,\mathbb{R}) ) $ defined as \[ f_{+}(x,t) =\det(I_{n}-it) ^{-1/2} e^{-2\pi^2x(I_{n}-it) ^{-1}x^{T}} \] (Theorem \ref{thm:K finite in D}). The corresponding vector in $\mathcal{I} _{+}'$ is \[ f_{+}(0,t) =\det(I_{n}-it) ^{-1/2} \] and in $\mathcal{S}_{+}(\mathbb{R}^{n}) $ is \[ \widehat{f_{+}}(\xi,0) =(2\pi) ^{-\frac{n}{2} }e^{-\frac{1}{2}\| \xi\| ^2}. \] Note that the choice of, say, $s=2\pi^2i$ gives rise to the dual representation and Schr\"odinger-like partial differential operators with lowest weight representations. The above commutative diagram fits on top of the Kashiwara-Vergne picture to give the following commutative diagram. \[ \begin{array} [c]{ccccc} & & \mathcal{D}_{+}' & & \\ & \overset{\mathcal{G}}{\swarrow} & & \overset{\mathcal{E}}{\searrow} & \\ \mathcal{I}_{+}' & \longleftarrow & \overset{\mathcal{H} =\mathcal{F}_{1}}{\longleftarrow} & \longleftarrow & \mathcal{S}_{+}( \mathbb{R}^{n}) \\ & \underset{\operatorname*{BV}}{\nwarrow} & & \underset{\mathcal{F}_{0} }{\swarrow} & \\ & & \mathcal{O}(\mathfrak{H}_{n}) & & \end{array} \] There is a similar picture for the odd part of the oscillator representation that fits in with the Kashiwara-Vergne realization in an analogous way. There our diagram looks like \[ \begin{array} [c]{ccccc} & & \mathcal{D}_{-}' & & \\ & \overset{\mathcal{G}_{n}}{\swarrow} & & \overset{\mathcal{E}}{\searrow} & \\ \mathcal{I}_{-}' & \longleftarrow & \overset{\mathcal{H}_{n} }{\longleftarrow} & \longleftarrow & \mathcal{S}_{-}(\mathbb{R} ^{n}) \end{array} \] $\mathcal{S}_{-}(\mathbb{R}^{n}) $ denotes the odd Schwartz functions, \[ \mathcal{D}_{-}'\subseteq\mathcal{C}^{\infty}(\mathbb{R} ^{n}\times\operatorname*{Sym}(n,\mathbb{R}) ) \] is the set of smooth solutions, $f$, satisfying the system of partial differential equations from Equation \ref{eqn:pdes with KV s} with $f( \cdot,t) \in\mathcal{S}_{-}(\mathbb{R}^{n}) $ for each $t\in\operatorname*{Sym}(n,\mathbb{R}) $ and \[ \mathcal{I}_{-}'\subseteq\mathcal{C}^{\infty}( \operatorname*{Sym}(n,\mathbb{R}) ,\mathbb{R}^{n}) \] is a subspace of the noncompact picture of a certain principal series representation, see \S \ref{section: induced reps} and Corollary \ref{cor: Iprime identification}. Here the maps are given by the same $\mathcal{E}$, \[ (\mathcal{E}f) (x) =\widehat{f}(x,0), \] and the related gradient to $\mathcal{G}$, \[ (\mathcal{G}_{n}f) (t) =\nabla_{\mathbb{R}^{n}}f(0,t) . \] In this case, \begin{align*} (\mathcal{H}_{n}f) (t) & =\nabla\Big( \int_{\mathbb{R}^{n}}f(\xi) e^{\frac{i}{2}\xi t\xi^{T}}e^{2\pi i\xi x^{T}}\,d\xi\Big) |_{x=0}\\ & =2\pi i\Big(\int_{\mathbb{R}^{n}}\xi_{1}f(\xi) e^{\frac {i}{2}\xi t\xi^{T}}\,d\xi,\ldots,\int_{\mathbb{R}^{n}}\xi_{n}f( \xi) e^{\frac{i}{2}\xi t\xi^{T}}\,d\xi\Big) \end{align*} and $\mathcal{G}_{n}^{-1}$ can be recovered from certain Fourier transforms (Theorem \ref{thm:x=0 injection}). The highest $K$-finite vectors of $\mathcal{D}_{-}'$ consist of the functions $f_{a}$ given by \[ f_{a}(x,t) =\det(I_{n}-it) ^{-1/2}\big( x(I_{n}-it) ^{-1}a^{T}\big) e^{-2\pi^2x( I_{n}-it) ^{-1}x^{T}} \] where $a\in\mathbb{C}^{n}$ (Theorem \ref{thm:KV intertwine}). The corresponding vector in $\mathcal{I}_{-}'$ is \[ \nabla f_{a}(0,t) =\det(I_{n}-it) ^{-\frac{1}{2} }\big(a(I_{n}-it) ^{-1}\big) . \] and in $\mathcal{S}_{+}(\mathbb{R}^{n}) $ is \[ \widehat{f_{a}}(\xi,0) =(2\pi) ^{-\frac{n}{2} +1}i(\xi a^{T}) e^{-\frac{1}{2}\| \xi\| ^2}. \] \section{Notation} \subsection{A Double Cover of the Jacobi Group} With respect to the standard symplectic form $J_{n+1}= \begin{pmatrix} 0 & -I_{n+1}\\ I_{n+1} & 0 \end{pmatrix}$, let \begin{align*} \mathfrak{g} & =\mathfrak{sp}(n+1,\mathbb{R}) \cap\Big\{ \begin{pmatrix} \ast\\ 0_{1\times(2n+2) } \end{pmatrix} \Big\} \\ & \cong\mathfrak{sp}(n,\mathbb{R}) \ltimes\mathfrak{h}_{2n+1} \end{align*} where $\mathfrak{h}_{2n+1}$ is the $2n+1$ dimensional real Heisenberg Lie algebra. This is the Lie algebra to the \emph{Jacobi group} \begin{align*} G^{J} & =Sp(n+1,\mathbb{R}) \cap\Big\{ \begin{pmatrix} \ast & \ast\\ 0_{1\times(2n+1) } & 1 \end{pmatrix} \Big\} \\ & \cong Sp(n,\mathbb{R}) \ltimes H_{2n+1} \end{align*} where $H_{2n+1}$ is the $2n+1$ dimensional real Heisenberg Lie group. Of course, written in $n\times1\times n\times1$ block form, $Sp( n,\mathbb{R}) $ is embedded in $G^{J}$ as \[ \bigg\{ \begin{pmatrix} A & 0 & B & 0\\ 0 & 1 & 0 & 0\\ C & 0 & D & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} : C^{T}A=A^{T}C\text{, }D^{T}B=B^{T}D\text{, }A^{T}D-C^{T}B=I_{n}\bigg\} \] and $H_{2n+1}$ is embedded as \[ \left\{ \begin{pmatrix} I_{n} & 0 & 0 & x^{T}\\ y & 1 & x & z\\ 0 & 0 & I_{n} & -y^{T}\\ 0 & 0 & 0 & 1 \end{pmatrix} \right\} . \] We write $\mathfrak{H}_{n}$ for the \emph{Siegel upper half-space} \[ \mathfrak{H}_{n}=\left\{ Z=X+iY: X,Y\in\operatorname*{Sym}(n,\mathbb{R} )\text{ with }Y>0\text{ (positive definite)}\right\} . \] The Siegel upper half-space carries a transitive action by $Sp(n,\mathbb{R})$ by linear fractional transformations, \[ g\cdot Z=(AZ+B) (CZ+D) ^{-1}. \] Note that the stabilizer of $iI_{n}$ in $Sp(n,\mathbb{R})$ is the maximal compact subgroup, $U(n) $, embedded in $Sp(n,\mathbb{R})$ by $A+iB\in U(n) \to \begin{pmatrix} A & B\\ -B & A \end{pmatrix} $. The main object of study is the double cover of $G^{J}$, \[ G=Mp(n) \ltimes H_{2n+1}. \] Here the action of $Mp(n) $ on $H_{2n+1}$ factors through its projection to $Sp(n,\mathbb{R})$ and we realize the metaplectic group as \begin{align*} Mp(n) &=\Big\{ \Big(g= \begin{pmatrix} A & B\\ C & D \end{pmatrix} ,\varepsilon\Big) : g\in Sp(n,\mathbb{R}) \text{ with smooth }\varepsilon:\mathfrak{H}_{n}\to \mathbb{C}\\ &\quad \text{satisfying }\varepsilon(Z) ^2=\det(CZ+D)\Big\}. \end{align*} The group law on $Mp(n) $ is given by \[ (g_{1},\varepsilon_{1}) \cdot(g_{2},\varepsilon _{2}) =(g_{1}g_{2},Z\to \varepsilon_{1}(g_{2}\cdot Z) \varepsilon_{2}(Z) ) . \] Note that the identity element is $(I_{n},Z\to 1) $ and $(g,\varepsilon) ^{-1}=(g^{-1},Z\to \varepsilon(g^{-1}\cdot Z) ^{-1}) $. To be explicit, the group law on $Mp(n) \ltimes H_{2n+1}$ is given by \[ ((g_{1},\varepsilon_{1}) ,h_{1}) \cdot( (g_{2},\varepsilon_{2}) ,h_{2}) =(( g_{1},\varepsilon_{1}) \cdot(g_{2},\varepsilon_{2}) ,g_{2}^{-1}h_{1}g_{2}h_{2}) . \] \subsection{Parabolic Subgroup\label{section: parabolic subgroups}} Consider the subalgebra of $\mathfrak{g}$ given, written in $n\times1\times n\times1$ block form, by \[ \overline{\mathfrak{p}}=\bigg\{ \begin{pmatrix} a & 0 & 0 & 0\\ y & 0 & 0 & z\\ c & 0 & -a^{T} & -y^{T}\\ 0 & 0 & 0 & 0 \end{pmatrix} : c^{T}=c\bigg\} . \] Then $\overline{\mathfrak{p}}$ is the semidirect product of the maximal parabolic subalgebra \[ \overline{\mathfrak{p}}_{\mathfrak{sp}}=\Big\{ \begin{pmatrix} a & 0\\ c & -a^{T} \end{pmatrix} : c^{T}=c\Big\} \] of $\mathfrak{sp}(n,\mathbb{R}) $ and a copy of $\mathbb{R}^{n+1}$ given by \[ \mathfrak{w}=\bigg\{ \begin{pmatrix} 0 & 0 & 0 & 0\\ y & 0 & 0 & z\\ 0 & 0 & 0 & -y^{T}\\ 0 & 0 & 0 & 0 \end{pmatrix} \bigg\} . \] The Langlands decomposition for $\overline{\mathfrak{p}}_{\mathfrak{sp}}$ is $\overline{\mathfrak{p}}_{\mathfrak{sp}}=\mathfrak{ma}\overline{\mathfrak{n}}$ where \begin{gather*} \mathfrak{a}=\Big\{ \begin{pmatrix} \lambda I_{n} & 0\\ 0 & -\lambda I_{n} \end{pmatrix} :\lambda\in\mathbb{R}\Big\} \\ \mathfrak{m}=\Big\{ \begin{pmatrix} a & 0\\ 0 & -a^{T} \end{pmatrix} : a\in\mathfrak{sl}(n,\mathbb{R}) \Big\} \\ \overline{\mathfrak{n}}=\Big\{ \begin{pmatrix} 0 & 0\\ c & 0 \end{pmatrix} : c^{T}=c\Big\} . \end{gather*} Before turning to the group, first note that the Lie algebra of the maximal compact subgroup of $Sp(n,\mathbb{R})$ is \[ \mathfrak{k}=\Big\{ \begin{pmatrix} a & b\\ -b & a \end{pmatrix} : b^{T}=b\text{, }a^{T}=-a\Big\} \cong\mathfrak{u}(n) \] and the corresponding maximal compact in $Mp(n) $ is \[ K=\Big\{ (k_{A,B}= \begin{pmatrix} A & B\\ -B & A \end{pmatrix} ,\varepsilon) : A+iB\in U(n) \text{, }\varepsilon ^2(Z) =\det(-BZ+A) \Big\} . \] We turn now to the group. Writing $A=\exp\mathfrak{a}$, we see \[ A=\Big\{ a_{t}=\bigg( \begin{pmatrix} e^{t}I_{n} & 0\\ 0 & e^{-t}I_{n} \end{pmatrix} ,Z\to e^{-\frac{n}{2}t}\bigg) \Big\} \] and $\overline{N}=\exp\mathfrak{n}$ is \[ \overline{N}=\Big\{ \overline{n}_{C}=\bigg( \begin{pmatrix} I_{n} & 0\\ C & I_{n} \end{pmatrix} ,\varepsilon_{C}\bigg) : C^{T}=C\Big\} \] where $\varepsilon_{C}$ is the unique smooth function \[ \varepsilon_{C}:\mathfrak{H}_{n}\to \mathbb{C} \] satisfying $\varepsilon_{C}(Z) ^2=\det(CZ+I_{n})$ determined by the condition that $\varepsilon_{C}(Z) =\sqrt{\det(CZ+I_{n}) }$ for sufficiently small $Z\in \mathfrak{H}_{n}$ (where $\sqrt{\cdot}$ denotes the principal square root). Now it is easy to check that the centralizer of $A$ in $K$ is \[ \Big\{ \bigg( \begin{pmatrix} A & 0\\ 0 & A \end{pmatrix} ,Z\to c\bigg) : A\in O(n,\mathbb{R}) ,\text{ } c^2=\det A\Big\} \] which has the structure of $SO(n) \times\mathbb{Z}_{4}$ when $n$ is odd and $SO(n) \rtimes \mathbb{Z}_{4}$ when $n$ is even. The subgroup $M$ is then defined to be the group generated by this centralizer and $\exp\mathfrak{m}$ so (using the subscript $0$ to denote the connected component) \begin{gather*} M_{0}=\Big\{ \bigg( \begin{pmatrix} A & 0\\ 0 & A^{-1,T} \end{pmatrix} ,Z\to 1\bigg) : A\in SL(n,\mathbb{R}) \Big\}, \\ \begin{aligned} M & =\Big\{m_{A,c} =\bigg( \begin{pmatrix} A & 0\\ 0 & A^{-1,T} \end{pmatrix} ,Z\to c\bigg): \\ &\quad A\in GL(n,\mathbb{R}) \text{, }\det A\in\{\pm1\} ,\; c^2=\det A^{-1}\Big\}. \end{aligned} \end{gather*} Thus the component group, $M/M_{0}$, is isomorphic to $\mathbb{Z}_{4}$. Finally, writing $W=\exp\mathfrak{w}$, we see \[ W=\Big\{ w_{y,z}= \begin{pmatrix} I_{n} & 0 & 0 & 0\\ y & 1 & 0 & z\\ 0 & 0 & I_{n} & -y^{T}\\ 0 & 0 & 0 & 1 \end{pmatrix} \Big\} . \] We let $\overline{P}$ be given by \[ \overline{P}=MA\overline{N}\ltimes W. \] \subsection{Induced Representations\label{section: induced reps}} For $q\in\mathbb{Z}$ (determined only up to $\operatorname{mod}$ $4$ or $\operatorname{mod}2$ depending on $n$), $r\in\mathbb{C}$, and $s\in \mathbb{C}$, we define a character \[ \chi_{q,r,s}:\overline{P}\to \mathbb{C} \] by \[ \chi_{q,r,s}(m_{A,c}a_{t}\overline{n}_{C}w_{y,z}) =c^{q} e^{rnt}e^{sz}. \] Note that for $n=1$, the choice of $q$ in \cite{SS2010} is the negative of the choice here. We study the induced representation \begin{align*} I(q,r,s) &=\operatorname*{Ind}\nolimits_{\overline{P}}^{G} \chi_{q,r,s}\\ &=\Big\{\text{smooth }\phi: G\to \mathbb{C} : \phi(gp) =\chi_{q,r,s}(p) ^{-1}\phi(g) \text{ for }g\in G\text{, }p\in\overline{P}\Big\} \end{align*} with action group action $(g\cdot\phi) (g') =\phi(g^{-1}g') $. We will also have occasion to use two related induced representations of $Mp(n) $. To this end, define a character and an $n$-dimensional representation of $MA\overline{N}$ \begin{gather*} \chi_{q,r} :MA\overline{N}\to \mathbb{C},\\ \pi_{q,r} :MA\overline{N}\to GL(n,\mathbb{C}) \end{gather*} by \begin{gather*} \chi_{q,r}(m_{A,c}a_{t}\overline{n}_{C}) =c^{q}e^{rnt},\\ \pi_{q,r}(m_{A,c}a_{t}\overline{n}_{C}) \cdot v =c^{q}e^{rnt}vA^{-1} \end{gather*} for $v\in\mathbb{C}^{n}$ given as a row vector. The associated induced representations are \begin{align*} I(q,r) & =\operatorname*{Ind}\nolimits_{MA\overline{N} }^{Mp(n) }\chi_{q,r}\\ & =\big\{ \mathcal{C}^{\infty}\text{ }\phi:G\to \mathbb{C}: \phi(gp) =\chi_{q,r}(p) ^{-1}\phi(g) \text{ for }g\in Mp(n) \text{, }p\in MA\overline{N}\big\} \\ I_{n}(q,r) & =\operatorname*{Ind}\nolimits_{MA\overline{N} }^{Mp(n) }\pi_{q,r}\\ & =\big\{ \mathcal{C}^{\infty}\text{ }\phi:G\to \mathbb{C}^{n} :\phi(gp) =\pi_{q,r}(p) ^{-1}\cdot\phi( g) \text{ for }g\in Mp(n) \text{, }p\in MA\overline {N}\big\} \end{align*} with action group action $(g\cdot\phi) (g') =\phi(g^{-1}g') $. \section{Boundary Values of $\varepsilon$} Recall elements of $Mp(n) $ are given by pairs $(g,\varepsilon) $ with $g=\begin{pmatrix} A & B\\ C & D \end{pmatrix} \in Sp(n,\mathbb{R}) $ and smooth $\varepsilon:\mathfrak{H}_{n}\to \mathbb{C}$ satisfying $\varepsilon(Z) ^2=\det(CZ+D) $. If we are in the special case of $\det D\neq0$, then $\det(CZ+D) =\operatorname*{sgn}(\det D) | \det D| \det(D^{-1}CZ+I_{n}) $. In particular, for all sufficiently small $Z$, \begin{align*} \varepsilon(Z) & =i^{p}| \det D| ^{\frac {1}{2}}\sqrt{\det(D^{-1}CZ+I_{n}) }\\ & =i^{p}| \det D| ^{1/2}\varepsilon_{D^{-1} C}(Z) \end{align*} where $\sqrt{\cdot}$ denotes the principal square root and $p=p( \varepsilon) $ is one of the two choices (determined precisely by $\varepsilon$) of $p\in\mathbb{Z}_{4}$ for which $(-1)^{p}=\operatorname*{sgn}(\det D) $. Note that the identity \[ \varepsilon=i^{p}| \det D| ^{1/2}\varepsilon_{D^{-1}C} \] then holds for all $Z$ since the functions are analytic. We need to extend the definition of $\varepsilon$ from $\mathfrak{H}_{n}$ to $\operatorname*{Sym}(n,\mathbb{R})$ almost everywhere. For this, let $\varepsilon:\operatorname*{Sym}(n,\mathbb{R})\to \mathbb{C}$ be given by \[ \varepsilon(X) =\lim_{Y\to 0^{+}}\,\varepsilon( X+iY) \] (here $Y\to 0^{+}$ denotes $Y\to 0$ with $Y>0$) which will be defined when $\det(CX+D) \neq0$. To see this limit exists when $\det(CX+D) \neq0$, observe that, for $Z$ with sufficiently small $\operatorname{Im}(Z) $, we can write $\varepsilon( Z) =i^{l}\sqrt{\operatorname*{sgn}(\det(CX+D) ) \det(CZ+D) }$ where $\sqrt{\cdot}$ denotes the principal square root and $l=l(\varepsilon,X) $ is one of the two choices (determined precisely by $\varepsilon$ and $X$) of $l\in \mathbb{Z}_{4}$ for which $(-1) ^{l}=\operatorname*{sgn}( \det(CX+D) ) $. In particular, we see $\varepsilon(X) $ exists and is given by \begin{equation} \varepsilon(X) =i^{l}\sqrt{| \det(CX+D)| }. \label{eqn:epsilonofX} \end{equation} In the special case where $X=0$ and $\det D\neq0$, there is a useful formula for recovering the $p$ in the formula $\varepsilon=i^{p}| \det D| ^{1/2}\varepsilon_{D^{-1}C}$. Namely, \[ i^{p}=\frac{\varepsilon(0) }{| \det D| ^{1/2}}. \] Finally, define an almost everywhere action of $Sp(n,\mathbb{R})$ on $\operatorname*{Sym}(n,\mathbb{R})$ given by \[ g\cdot X=(AX+B) (CX+D) ^{-1} \] for $X\in\operatorname*{Sym}(n,\mathbb{R})$ when $\det(CX+D) \neq0$ so that $g\cdot X=\lim_{Y\to 0^{+}}g\cdot(X+iY)$. \section{Noncompact Pictures\label{sec: noncompact picture}} Let \[ \mathfrak{x}=\Big\{ \begin{pmatrix} 0 & 0 & 0 & x^{T}\\ 0 & 0 & x & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \Big\} \] so that $X=\exp\mathfrak{x}$ is given by \[ X=\Big\{ e_{x}= \begin{pmatrix} I_{n} & 0 & 0 & x^{T}\\ 0 & 1 & x & 0\\ 0 & 0 & I_{n} & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \Big\} \cong\mathbb{R}^{n} \] and let \[ \mathfrak{n}=\big\{ \begin{pmatrix} 0 & b\\ 0 & 0 \end{pmatrix} : b^{T}=b\big\} \] so that $N=\exp\mathfrak{n}$ is given by \[ N=\Big\{ n_{B}=\Big( \begin{pmatrix} I_{n} & B\\ 0 & I_{n} \end{pmatrix} ,Z\to 1\big ) : B^{T}=B\Big\} . \] Restriction to $XN\cong\mathbb{R}^{n}\times\operatorname*{Sym}( n,\mathbb{R}) $ gives what would be called the \emph{noncompact realization} of the induced representation if we were in the semisimple category and which we denote by \begin{align*} &I'(q,r,s) \\ & =\Big\{ f:\mathbb{R}^{n}\times \operatorname*{Sym}(n,\mathbb{R}) \to \mathbb{C}: f(x,B) =\phi(e_{x}n_{B}) \text{ for some }\phi\in I(q,r,s) \Big\}. \end{align*} We make $I'(q,r,s) $ into a $G$-module so that the restriction map $\phi\to f$ is an intertwining map. When necessary, we will coordinatize $\operatorname*{Sym}(n,\mathbb{R}) $ as $\mathbb{R}^{\frac{n(n+1) }{2}}$ by writing \[ B= \begin{pmatrix} t_{11} & t_{12} & \cdots & t_{1n}\\ t_{12} & t_{22} & \cdots & t_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ t_{1n} & t_{2n} & \cdots & t_{nn} \end{pmatrix} . \] \begin{theorem} \label{thm:noncompact picture group action} For $f\in I'(q,r,s) $, the action of $g=\Big( \begin{pmatrix} A & B\\ C & D \end{pmatrix} ,\varepsilon\Big)\in Mp(n) $ on $f$ is given by \begin{align*} ((g,\varepsilon) \cdot f) (x,t) &=i^{lq}| \det(A-tC) | ^{r}e^{-sxC( A-tC) ^{-1}x^{T}}\\ &\quad\times f(x(-C^{T}t+A^{T}) ^{-1},(A-tC)^{-1}(tD-B) ) \end{align*} when $\det(A-tC) \neq 0$ and $l\in\mathbb{Z}_{4}$ satisfies $\varepsilon(g^{-1}\cdot t) =i^{l}| \det(A-tC) | ^{-1/2}$. The action of $h= \begin{pmatrix} I_{n} & 0 & 0 & y_{0}^{T}\\ x_{0} & 1 & y_{0} & z_{0}\\ 0 & 0 & I_{n} & -x_{0}^{T}\\ 0 & 0 & 0 & 1 \end{pmatrix} \in H_{2n+1}$ on $f$ is given by \[ (h\cdot f) (x,t) =e^{s(2xx_{0}^{T} +z_{0}-x_{0}tx_{0}^{T}-y_{0}x_{0}^{T}) }\,f(x-y_{0} -x_{0}t,t) . \] \end{theorem} \begin{proof} When $\det D\neq0$, write $\varepsilon(Z) =i^{p}| \det D| ^{1/2}\sqrt{\det(D^{-1}CZ+I_{n}) }$ for all sufficiently small $Z$ and recall that $i^{p}=\varepsilon(0) | \det D| ^{-1/2}$. It is straightforward to verify that \begin{equation} (g,\varepsilon) =n_{BD^{-1}}\,m_{| \det D| ^{-\frac{1}{n}}D^{-1,T},i^{p}}\,a_{\ln(| \det D| ^{-\frac{1}{n}})}\,\overline{n}_{D^{-1}C} \label{eqn:nman} \end{equation} and \begin{equation} (g,\varepsilon) \,e_{x}=n_{BD^{-1}}\,e_{xD^{-1}}\,( \begin{pmatrix} D^{-1,T} & 0\\ C & D \end{pmatrix} ,\varepsilon)\,w_{-xD^{-1}C,-xD^{-1}Cx^{T}}. \label{eqn:g ex} \end{equation} Suppose $f\in I'(q,r,x) $ corresponds to $\varphi\in I(q,r,s) $. Then \begin{align*} ((g,\varepsilon) \cdot f) (x,t) &=\phi(g^{-1}e_{x}n_{t}) \\ & =\phi(( \begin{pmatrix} D^{T} & D^{T}t-B^{T}\\ -C^{T} & -C^{T}t+A^{T} \end{pmatrix} ,Z\to \varepsilon(g^{-1}\cdot(Z+t) ) ^{-1})e_{x}). \end{align*} Using Equations \ref{eqn:g ex} and \ref{eqn:nman} when $\det( A-tC) \neq0$, it follows that \begin{align*} &((g,\varepsilon) \cdot f) (x,t) \\ &=\Big(\frac{\varepsilon(g^{-1}\cdot t) ^{-1}}{| \det(-tC+A) | ^{1/2}}\Big) ^{-q}\big( | \det(-tC+A) | ^{-\frac{1}{n}}\big)^{-rn}\\ &\quad\times \cdot e^{-sx(-C^{T}t+A^{T}) ^{-1}C^{T}x^{T}}\phi(n_{( D^{T}t-B^{T}) (-C^{T}t+A^{T}) ^{-1}},e_{x( -C^{T}t+A^{T}) ^{-1}}). \end{align*} Finally, it is easy to see that $C(g^{-1}\cdot t) +D=(A^{T}-C^{T}t) ^{-1}$. Looking at Equation \ref{eqn:epsilonofX}, there is an $l\in\mathbb{Z}_{4}$ so that $\varepsilon(g^{-1}\cdot t) =i^{l}| \det(A^{T}-C^{T}t) | ^{-1/2}$ and the result follows. The calculation for $H_{2n+1}$ is similar and omitted. \end{proof} A straightforward calculation yields: \begin{corollary} \label{cor: lie alg action noncompact} Let $f\in I'(q,r,s) $. The element $h=(x_{0},y_{0},z_{0})\in\mathfrak{h}_{2n+1}$ acts on $f$ by \[ h\cdot f(x,t)=s(2x_{0}x^{T}+z_{0})f(x,t)-\sum_{i=1}^{n}(x_{0}t+y_{0} )_i\partial_{x_i}f(x,t). \] The element $a_{\lambda}\in\mathfrak{a}$, $\lambda\in\mathbb{R}$, acts on $f$ by \[ (a_{\lambda}\cdot f) (x,t)=nr\lambda f(x,t)-\lambda\sum _{i=1}^{n}x_i\partial_{x_i}f(x,t)-2\lambda\sum_{i\leq j}t_{i,j} \partial_{t_{i,j}}f(x,t). \] The element $n_{c}\in\overline{\mathfrak{n}}$, $c^{T}=c$, acts on $f$ by \begin{align*} (n_{c}\cdot f) (x,t) & =-rTr(tc)f(x,t)-sxcx^{T}f(x,t)+\sum _{i=1}^{n}(xct)_i\partial_{x_i}f(x,t)\\ &\quad +\sum_{i\leq j}(tct)_{i,j}\partial_{t_{i,j}}f(x,t). \end{align*} If $k_{a,b}\in\mathfrak{k}$, $b^{T}=b$, $a^{T}=-a$, then $k_{a,0}$ acts on $f$ by \[ (k_{a,0}\cdot f) (x,t)=\sum_{i=1}^{n}(xa)_i\partial_{x_i }f(x,t)+\sum_{i\leq j}(ta-at)_{i,j}\partial_{t_{i,j}}f(x,t) \] and $k_{0,b}$ acts by \begin{align*} (k_{0,b}\cdot f) (x,t) & =rTr(tb)f(x,t)+sxbx^{T}f(x,t)\\ &\quad -\sum_{i=1}^{n}(xbt)_i\partial_{x_i}f(x,t)-\sum_{i\leq j} (Tr(tb)t+b)_{i,j}\partial_{t_{i,j}}f(x,t). \end{align*} \end{corollary} In a similar fashion, we also have the noncompact realizations of $I(q,r) $ and $I_{n}(q,r) $ given by restriction to $N\cong\operatorname*{Sym}(n,\mathbb{R}) $. We denote these realizations by \begin{gather*} I'(q,r) =\big\{ f:\operatorname*{Sym}( n,\mathbb{R}) \to \mathbb{C}: f(B) =\phi( n_{B}) \text{ for some }\phi\in I(q,r) \big\} \\ I_{n}'(q,r) =\big\{ f:\operatorname*{Sym}( n,\mathbb{R}) \to \mathbb{C}^{n}: f(B) =\phi(n_{B}) \text{ for some }\phi\in I_{n}(q,r) \big\} . \end{gather*} Simple modifications of the proof Theorem \ref{thm:noncompact picture group action} give the following result. \begin{corollary} For $f\in I'(q,r) $, the action of $\big(g= \begin{pmatrix} A & B\\ C & D \end{pmatrix} ,\varepsilon\big)\in Mp(n) $ on $f$ is given by \[ ((g,\varepsilon) \cdot f) (t) =i^{lq}| \det(A-tC) | ^{r}\,f(( A-tC) ^{-1}(tD-B) ) \] when $\det(A-tC) \neq0$ and $l\in\mathbb{Z}_{4}$ satisfies $\varepsilon(g^{-1}\cdot t) =i^{l}| \det( A-tC) | ^{-1/2}$. For $f_{n}\in I_{n}'(q,r) $, the action of $\big(g= \begin{pmatrix} A & B\\ C & D \end{pmatrix} ,\varepsilon\big)\in Mp(n) $ on $f_{n}$ is given by \[ \big((g,\varepsilon) \cdot f\big) (t) =i^{lq}| \det(A-tC) | ^{r-\frac{1}{n}} \,f_{n}((A-tC) ^{-1}(tD-B) )(-tC+A) ^{-1}. \] \end{corollary} We also see that: \begin{corollary}\label{cor: x=0} There is an $Mp(n) $-intertwining map \[ \mathcal{G}:I'(q,r,s) \to I'(q,r) \] given by the mapping $f\to f(0,\cdot)$. The corresponding map from $I(q,r,s) \to I( q,r) $ is given by $\phi\to \phi|_{Mp(n) }$. There is also an $Mp(n) $-intertwining map \[ \mathcal{G}_{n}:I'(q,r,s) \to I_{n}^{\prime }(q,r-\frac{1}{n}) \] given by mapping $f\to \nabla f(0,\cdot) $. The corresponding map from $I(q,r,s) \to I_{n}( q,r-\frac{1}{n}) $ is given by $\phi\to \nabla( \phi(\cdot e_{x}) ) |_{x=0}$. \end{corollary} \begin{proof} The first statement is obvious since \[ ((g,\varepsilon) \cdot f) (0,t) =i^{lq}| \det(A-tC) | ^{r}\,f(0,( A-tC) ^{-1}(tD-B) ). \] It also follows trivially from the definitions that the map $f\to f(0,\cdot) $ on $I'(q,r,s) \to I'(q,r) $ corresponds to the map $\phi\to \phi|_{Mp(n) }$\ on $I(q,r,s) \to I(q,r) $. For the second statement, observe that \begin{align*} &\Big(\frac{\partial}{\partial x_i}((g,\varepsilon) \cdot f) \Big) (0,t)\\ & =i^{lq}| \det( A-tC) | ^{r} \sum_{j}\big((-C^{T}t+A^{T}) ^{-1}\big) _{ij} \frac{\partial f}{\partial x_{j}}(0,(A-tC) ^{-1}( tD-B) ). \end{align*} Thus \begin{align*} &\nabla\big((g,\varepsilon) \cdot f\big) ( 0,\cdot) \\ & =i^{lq}| \det(A-tC) | ^{r} \nabla f(0,(A-tC) ^{-1}(tD-B) )( -C^{T}t+A^{T}) ^{-1,T} \end{align*} and the map intertwines. Finally, we claim that the map given by $f\to \nabla f(0,\cdot) $ on $I'(q,r,s) \to I_{n}'(q,r-\frac{1}{n}) $ is induced by the map $\varphi\to \nabla(\varphi(\cdot e_{x}) ) |_{x=0}$ on $I(q,r,s) \to I_{n}(q,r-\frac{1}{n}) $. To check this, note that it is easy to verify that \[ (g,\varepsilon) e_{x} =e_{xD^{-1}}\,(g,\varepsilon) \,w_{-xD^{-1}C,-xD^{-1}Cx^{T}} \] when $D$ is invertible. Then, for $\gamma\in Mp(n) $ and $p\in MA\overline{N}$ written as $p=m_{A,c}a_{t}\overline{n}_{C}$, \begin{align*} \nabla(\phi(\gamma pe_{x}) ) |_{x=0} &=\nabla(\phi(\gamma m_{A,c}a_{t}\overline{n}_{C}e_{x}) ) |_{x=0}\\ & =\nabla(\phi(\gamma e_{e^{t}xA^{T}}m_{A,c}a_{t}\overline {n}_{C}w_{-xA^{T}A^{-1}C,-xA^{T}A^{-1}Cx^{T}}) ) |_{x=0}\\ & =\nabla(c^{-q}e^{-rnt}e^{sxA^{T}A^{-1}Cx^{T}}\phi(\gamma e_{e^{t}xA^{T}}) ) |_{x=0}\\ & =c^{-q}e^{-rnt}\nabla(\phi(\gamma e_{x}) ) |_{x=0}\,e^{t}A\\ & =c^{-q}e^{-(r-\frac{1}{n}) nt}\nabla(\phi( \gamma e_{x}) ) |_{x=0}\,A\\ & =\pi_{q,r-\frac{1}{n}}(p) ^{-1}\cdot\nabla(\phi( \gamma e_{x}) ) |_{x=0}. \end{align*} Thus $\nabla(\phi(\cdot e_{x}) ) |_{x=0}\in I_{n}(q,r-1/n) $. Moreover, noting that $n_{B}e_{x}=e_{x}n_{B}$, we have $\nabla(\phi(e_{C}e_{x}) ) |_{x=0}=\nabla f( 0,C) $ so that $\nabla(\phi(\cdot e_{x}) ) |_{x=0}\in I_{n}(q,r) $ corresponds to $\nabla f( 0,\cdot) \in I_{n}'(q,r) $. \end{proof} \section{An Invariant Subspace} \begin{theorem} \label{thm:magic r} For $r=-1/2$, the set of functions $f\in I'(q,r,s) $ satisfying the system of partial differential equations (from Equation \eqref{eqn:system of pdess}) \begin{gather*} 2s\partial_{t_{i,j}}f+\partial_{x_i}\partial_{x_{j}}f =0, \quad i\neq j\\ 4s\partial_{t_{ii}}f+\partial_{x_i}^2f =0 \end{gather*} is $G$-invariant. \end{theorem} \begin{proof} Temporarily write $D=\left\{ 2s\partial_{t_{i,j}}+\partial_{x_i} \partial_{x_{j}}, 4s\partial_{t_{ii}}+\partial_{x_i}^2:1\leq i\neq j\leq n\right\} $. First observe that the differential operators in $D$ commute with the Heisenberg group action. This is clear for $(0,y,z)\in H_{2n+1}$ since $D$ consists of constant coefficient differential operators and $((0,y,z).f) (x,t) =e^{sz}\,f(x-y,t) $ by Theorem \ref{thm:noncompact picture group action}. Checking commutivity for $(x,0,0)\in H_{2n+1}$ is a straightforward application of the chain rule and is omitted. The invariance of $D$ under $Mp(n) $ follows by a Lie algebra calculation showing that $[ X,D_i] $ lies in the $\mathcal{C}^{\infty}( \mathbb{R}^{n}\times\operatorname*{Sym}(n,\mathbb{R}) ) $-span of $D$ for any $X\in\mathfrak{g}$ and $D_i\in D$. As the details are straightforward and all similar, we give the particulars only for the element $X=E_{n+1,1}\in\mathfrak{sp}(n,\mathbb{R}) $ as representative of the most interesting case. By Corollary \ref{cor: lie alg action noncompact}, \[ E_{n+1,1}\cdot f=-rt_{11}f-sx_{1}^2f+\sum_{i=1}^{n}x_{1}t_{1,i} \partial_{x_i}f+\sum_{i\leq j}t_{1,i}t_{1,j}\partial_{t_{i,j}}f. \] Then \begin{align*} & [-rt_{11}-sx_{1}^2+\sum_{i=1}^{n}x_{1}t_{1,i}\partial_{x_i}+\sum_{i\leq j}t_{i,1}t_{1,j}\partial_{t_{i,j}},\,4s\partial_{t_{11}}+\partial_{x_{1}} ^2]\\ & =-4s(-r+x_{1}\partial_{x_{1}}+2t_{1,1}\partial_{t_{1,1}}+\sum_{j=2} ^{n}t_{1,j}\partial_{t_{1,j}})-(-2s-4sx_{1}\partial_{x_{1}}+2\sum_{i=1} ^{n}t_{1,i}\partial_{x_{1}}\partial_{x_i})\\ & =2s(1+2r) -2t_{1,1}(4s\partial_{t_{1,1}}+\partial_{x_{1}} ^2)-2\sum_{j=2}^{n}t_{1,j}(2s\partial_{t_{1,j}}+\partial_{x_{1}} \partial_{x_{j}}). \end{align*} The result follows. \end{proof} It is helpful to be able to write down explicit formulas for solutions to Equation \eqref{eqn:system of pdess}. \begin{theorem}\label{thm:pde soln} Let $s\neq0$ be purely imaginary. If $f\in\mathcal{C} ^2(\mathbb{R}^{n}\times\operatorname*{Sym}(n,\mathbb{R} ) ) $ satisfying $f(\cdot,0) ,\,\widehat{f( \cdot,0) }\in L^{1}(\mathbb{R}^{n}) $ and the system of partial differential equations from Equation \eqref{eqn:system of pdess}, then \[ f(x,t) =\int_{\mathbb{R}^{n}}\widehat{f}(\xi,0) e^{\frac{\pi^2}{s}\xi t\xi^{T}}e^{2\pi i\xi x^{T}}\,d\xi. \] \end{theorem} \begin{proof} By standard Fourier techniques, when $f(\cdot,0) $ is a tempered distribution, there is a unique solution to the Cauchy problem in the space of $\mathcal{C}(\operatorname*{Sym}(n,\mathbb{R}) ,\mathcal{S}'(\mathbb{R}^{n}) ) $--i.e., $f(x,t) $ is continuous in $t$ and takes values in the set of tempered distributions on $\mathbb{R}^{n}$. In fact, if $\int_{\mathbb{R}^{n} }(1+\| x\| ^2) f(x,0) \,dx<\infty$, the solution is classical in the sense that it has continuous derivatives with respect to each $t_{i,j}$ and continuous second order derivatives with respect to each $x_i$. Alternately, if $f( \cdot,0) \in L^2(\mathbb{R}^{n}) $, then $f\in \mathcal{C}(\operatorname*{Sym}(n,\mathbb{R}) ,L^2(\mathbb{R}^{n}) ) $ with $\| f( \cdot,t) \| _{L^2(\mathbb{R}^{n}) }=\| f(\cdot,0) \| _{L^2(\mathbb{R}^{n}) }$. The calculation goes as follows: take the Fourier transform with respect to $x$ of the partial differential equations from Equation \eqref{eqn:system of pdess} to get \begin{gather*} (2s\partial_{t_{i,j}}-4\pi^2\xi_i\xi_{j}) \widehat{f} =0, \quad i\neq j\\ (4s\partial_{t_{ii}}-4\pi^2\xi_i^2) \widehat{f} =0. \end{gather*} Thus \begin{equation} \widehat{f}(\xi,t) =\widehat{f}(\xi,0) e^{\frac{\pi^2}{s}(\sum_{i=1}^{n}\xi_i^2t_{ii}+2\sum_{i0$ with $\arctan\frac{\mu}{\lambda}<\frac{\pi}{n}$. The \emph{Cartan involution} $\theta:Mp(n) \to Mp( n) $ is the anti-involution $\theta(g,\varepsilon) =(g^{T},\varepsilon^{T}) $ where \[ (g^{T},\varepsilon^{T}) =\widetilde{J}(g,\varepsilon ) ^{-1}\widetilde{J}^{-1}. \] \end{definition} Notice that \begin{align*} (g^{T},\varepsilon^{T}) & =\widetilde{J}(g,\varepsilon) ^{-1}\widetilde{J}^{-1}\\ & =\Big(J_{n}g^{-1}J_{n}^{-1},Z\to \varepsilon_{\widetilde{J} }(g^{-1}J_{n}^{-1}\cdot Z) \varepsilon(g^{-1}J_{n} ^{-1}\cdot Z) ^{-1}\varepsilon_{\widetilde{J}}(-Z^{-1}) ^{-1}\Big) \\ & =(g^{T},Z\to \varepsilon(-(B^{T}Z+D^{T}) (A^{T}Z+C^{T}) ^{-1}) ^{-1}\\ &\quad\times \varepsilon_{\widetilde{J}}\Big(-(B^{T}Z+D^{T}) (A^{T}Z+C^{T}) ^{-1}\Big) \varepsilon_{\widetilde{J}}( -Z^{-1}) ^{-1}) \end{align*} so that \begin{align*} \varepsilon^{T}(Z) & =\varepsilon(-(B^{T} Z+D^{T}) (A^{T}Z+C^{T}) ^{-1}) ^{-1}\\ & \quad\times \varepsilon_{\widetilde{J}}(-(B^{T}Z+D^{T}) (A^{T}Z+C^{T}) ^{-1}) \varepsilon_{\widetilde{J}}( -Z^{-1}) ^{-1}. \end{align*} Of course, \begin{align*} \varepsilon^{T}(Z) ^2 & =\frac{\det(-( B^{T}Z+D^{T}) (A^{T}Z+C^{T}) ^{-1}) }{\det( -C(B^{T}Z+D^{T}) (A^{T}Z+C^{T}) ^{-1}+D) \det(-Z^{-1}) }\\ & =\frac{\det(B^{T}Z+D^{T}) }{\det(C(B^{T} Z+D^{T}) -D(A^{T}Z+C^{T}) ) \det( -Z^{-1}) }\\ & =\det(B^{T}Z+D^{T}) \end{align*} as required. \begin{theorem}\label{thm:nonzero} When $\sigma>0$ and$\ q\equiv-1$, we can define $\phi _{+},\phi_{+,\alpha}\in I(q,r,s) $ with $\alpha\in\mathbb{C}^{n}$ by \begin{gather*} \phi_{+}((g,\varepsilon) \,h_{x,y,z}) =\frac{e^{i\sigma(-z-xy^{T}+x(g^{T}\cdot iI_{n}) x^{T}) }}{\varepsilon^{T}(iI_{n}) }, \\ \phi_{+,\alpha}((g,\varepsilon) \,h_{x,y,z}) =\frac{(x(Bi+D) ^{-1}\alpha^{T}) e^{i\sigma( -z-xy^{T}+x(g^{T}\cdot iI_{n}) x^{T}) }}{\varepsilon ^{T}(iI_{n}) } \end{gather*} (recall $\varepsilon^{T}(Z) ^2=\det(ZB+D) $). The corresponding elements $f_{+},f_{+,\alpha}\in\mathcal{D}'$ are \begin{gather*} f_{+}(x,t) =\varepsilon_{t}(iI_{n}) ^{-1} e^{-\sigma x(I_{n}+it) ^{-1}x^{T}} \\ f_{+,\alpha}(x,t) =\varepsilon_{t}(iI_{n}) ^{-1}(x(I_{n}+it) ^{-1}\alpha^{T}) e^{-\sigma x(I_{n}+it) ^{-1}x^{T}} \end{gather*} where, recall, $\varepsilon_{t}(Z) $ is the analytic continuation to $Z\in\mathfrak{H}_{n}$ of the function $Z\to \sqrt {\det(I_{n}+tZ) }$ for sufficiently small $Z$. When $\sigma<0$ and$\ q\equiv1$, we can define $\phi_{-},\phi_{-,\alpha}\in I(q,r,s) $ with $\alpha\in\mathbb{C}^{n}$ by \begin{gather*} \phi_{-}((g,\varepsilon) \,h_{x,y,z}) =\frac{e^{i\sigma(-z-xy^{T}+x(g^{T}\cdot(-iI_{n}) ) x^{T}) }}{\overline{\varepsilon^{T}(iI_{n}) }} \\ \phi_{-,\alpha}((g,\varepsilon) \,h_{x,y,z}) =\frac{(x(-Bi+D) ^{-1}\alpha^{T}) e^{i\sigma (-z-xy^{T}+x(g^{T}\cdot(-iI_{n}) ) x^{T}) }}{\overline{\varepsilon^{T}(iI_{n}) }}. \end{gather*} The corresponding elements $f_{-},f_{-,\alpha}\in\mathcal{D}_{+}'$ are \begin{gather*} f_{-}(x,t) =\overline{\varepsilon_{t}(iI_{n}) }^{-1}e^{\sigma x(I_{n}-it) ^{-1}x^{T}} \\ f_{-,\alpha}(x,t) =\overline{\varepsilon_{t}( iI_{n}) }^{-1}(x(I_{n}-it) ^{-1}\alpha^{T}) e^{\sigma x(I_{n}-it) ^{-1}x^{T}}. \end{gather*} \end{theorem} \begin{proof} To determine when $\phi_{+}\in I(q,r,s) $, first write $\overline{p}=m_{A_{0},c_{0}}a_{t_{0}}\overline{n}_{C_{0}}=( \overline{p}_{0},\varepsilon_{p_{0}}) $ so that \begin{gather*} \overline{p}_{0}= \begin{pmatrix} e^{t_{0}}A_{0} & 0\\ e^{-t_{0}}A_{0}^{-1,T}C_{0} & e^{-t_{0}}A_{0}^{-1,T} \end{pmatrix}, \\ \varepsilon_{p_{0}}(Z) =c_{0}e^{-\frac{n}{2}t_{0}} \varepsilon_{C_{0}}(Z) . \end{gather*} Since $\varepsilon_{p_{0}}^{T}(Z) ^2=\det(e^{-t_{0} }A_{0}^{-1}) =e^{-nt_{0}}\det A_{0}^{-1}$ and $c_{0}^2=\det A_{0}^{-1}$, it follows that $\varepsilon_{p_{0}}^{T}(Z) =\pm c_{0}e^{-\frac{n}{2}t_{0}}$. The exact answer can be determined by using the continuity of the Cartan involution and its evaluation on the central elements, $Z=(\pm I_{n},c) $ with $c^2=(\pm1) ^{-n}$: \[ \varepsilon^{T}(Z) =c^{-1}\varepsilon_{\widetilde{J}}( -Z^{-1}) \varepsilon_{\widetilde{J}}(-Z^{-1}) ^{-1}=c^{-1}. \] It follows that \[ \varepsilon_{p_{0}}^{T}(Z) =c_{0}^{-1}e^{-\frac{n}{2}t_{0}}. \] In particular, it we see that \[ ((g,\varepsilon) \overline{p}) ^{T}=( \overline{p}_{0}^{T}g^{T},c_{0}^{-1}e^{-\frac{n}{2}t_{0}}\varepsilon ^{T}) . \] Turning to $\phi_{+}$, a straightforward calculation shows that \begin{align*} & \phi_{+}((g,\varepsilon)h_{x,y,z}\,\overline{p}w_{y_{0},z_{0}})\\ & =\phi_{+}((g,\varepsilon)\overline{p}h_{e^{-t_{0}}xA_{0}^{-1,T},e^{t_{0} }yA_{0}+e^{-t_{0}}xA_{0}^{-1,T}C_{0},z}w_{y_{0},z_{0}})\\ & =\phi_{+}((g,\varepsilon)\overline{p}h_{e^{-t_{0}}xA_{0}^{-1,T},e^{t_{0} }yA_{0}+e^{-t_{0}}xA_{0}^{-1,T}C_{0}+y_{0},z+z_{0}-e^{-t_{0}}xA_{0} ^{-1,T}y_{0}^{T}})\\ & =e^{-i\sigma(z+z_{0}-e^{-t_{0}}xA_{0}^{-1,T}y_{0}^{T}) }e^{-i\sigma e^{-t_{0}}xA_{0}^{-1,T}(e^{t_{0}}yA_{0}+e^{-t_{0}} xA_{0}^{-1,T}C_{0}+y_{0}) ^{T}}\\ &\quad\times e^{i\sigma(e^{-t_{0}}xA_{0}^{-1,T}) (e^{2t_{0} }A_{0}^{T}(g^{T}\cdot iI_{n}) A_{0}+C_{0}) ( e^{-t_{0}}A_{0}^{-1}x^{T}) }/[c_{0}^{-1}e^{-\frac{n}{2}t_{0} }\varepsilon^{T}(iI_{n}) ]\\ & =c_{0}e^{\frac{n}{2}t_{0}}e^{-i\sigma z_{0}}\phi_{+}(( g,\varepsilon) \,h_{x,y,z}) . \end{align*} It follows that $\phi_{+}\in I(-1,-1/2,\iota\sigma) $. Next observe that the $\varepsilon$ for \[ n_{t}^{T}=\Big( \begin{pmatrix} I_{n} & 0\\ t & I_{n} \end{pmatrix} ,Z\to \varepsilon_{\widetilde{J}}(-(tZ+I_{n}) Z^{-1}) \varepsilon_{\widetilde{J}}(-Z^{-1}) ^{-1}\big). \] Now for $Z=\rho e^{i\theta}I_{n}$, $\det(-Z^{-1}) =\rho^{-n}e^{in(\pi-\theta) }$ so that $\varepsilon_{\widetilde{J} }(-Z^{-1}) =\rho^{-\frac{n}{2}}e^{i\frac{n(\pi -\theta) }{2}}$ for $\pi-\theta$ sufficiently positively small and $\rho>0$. Therefore $\varepsilon_{\widetilde{J}}(-Z^{-1}) =\rho^{-\frac{n}{2}}e^{i\frac{n(\pi-\theta) }{2}}$ for all $0<\theta<\pi$. Similarly, $\det(-(tZ+I_{n}) Z^{-1}) =\det(tZ+I_{n}) \rho^{-n}e^{in(\pi -\theta) }$ so that $\varepsilon_{\widetilde{J}}(-( tZ+I_{n}) Z^{-1}) =\sqrt{\det(tZ+I_{n}) } \rho^{-\frac{n}{2}}e^{i\frac{n(\pi-\theta) }{2}}$ for $\pi-\theta$ and $\rho$ sufficiently positively small. It follows that $\varepsilon_{\widetilde{J}}(-(tZ+I_{n}) Z^{-1}) \varepsilon_{\widetilde{J}}(-Z^{-1}) ^{-1}=\varepsilon _{t}(Z) $ for all $Z\in\mathfrak{H}_{n}$. In particular, we see that $n_{t}^{T}=\overline{n}_{t}$. Thus \[ \phi_{+}(n_{t} h_{x,0,0}) =\frac{e^{i\sigma x\frac {\varepsilon_{\sigma}i}{\varepsilon_{\sigma}it+I_{n}}x^{T}}}{\varepsilon _{t}(\varepsilon_{\sigma}iI_{n}) }=\varepsilon_{t}( \varepsilon_{\sigma}iI_{n}) ^{-1}e^{-\sigma x(I_{n}+it) ^{-1}x^{T}}. \] Finally, we must show $f_{+}\in\mathcal{D}_{+}$. As $f_{+}( \cdot,0) $ is clearly Schwartz when $\sigma>0$, it remains only to show that $f_{+}$ satisfies the system given in Equation \eqref{eqn:system of pdess}. For the sake of brevity, we will only show $4s\partial_{t_{ii}}f_{+}+\partial_{x_i}^2f_{+}=0$ and omit the similar calculation that $2s\partial_{t_{i,j}}f+\partial_{x_i}\partial_{x_{j}}f=0$, $i\neq j$. For $X\in M_{n}(\mathbb{C}) $, write $X_{( i,j) }$ for the $(i,j) $ minor of $X$. Then \begin{align*} \partial_{t_{i,i}}f_{+} & =-i\frac{1}{2}\det(I_{n}+it) ^{-1}\det(I_{n}+it) _{(i,i)}f_{+}\\ &\quad +i\sigma x(I_{n}+it) ^{-1}E_{i,i}(I_{n}+it) ^{-1}x^{T}f_{+}\\ & =-i\frac{1}{2}((I_{n}+it) ^{-1}) _{i,i} f_{+}+i\sigma x(I_{n}+it) ^{-1}E_{i,i}(I_{n}+it) ^{-1}x^{T}f_{+} \end{align*} while \begin{align*} \partial_{x_i}^2f_{+} & =\partial_{x_i}\big(-2\sigma e_i( I_{n}+it) ^{-1}x^{T}f_{+}\big) \\ & =-2\sigma e_i(I_{n}+it) ^{-1}e_i^{T}f_{+} +4\sigma ^2\big(e_i(I_{n}+it) ^{-1}x^{T}\big) ^2f_{+}\\ & =-2\sigma\big((I_{n}+it) ^{-1}\big) _{i,i}f_{+} +4\sigma^2x(I_{n}+it) ^{-1}e_i^{T}e_i( I_{n}+it) ^{-1}x^{T}f_{+}\\ & =-2\sigma\big((I_{n}+it) ^{-1}\big) _{i,i}f_{+} +4\sigma^2x(I_{n}+it) ^{-1}E_{i,i}(I_{n}+it) ^{-1}x^{T}f_{+} \end{align*} which finishes the claim. Turn now to the second part of the Theorem. Taking conjugates, it follows that $\phi_{-}=\overline{\phi}_{+}\in I(1,-1/2,-\iota\sigma) $, $f_{-}(\cdot,0) $ is Schwartz, and $f_{-}$ satisfies the system given in Equation \eqref{eqn:system of pdess} (with $\sigma$ replaced by $-\sigma$). Renaming $\sigma$, the result follows. The calculations for $\phi_{\alpha}$ are trivial modifications of the above argument. \end{proof} \begin{corollary} For $q=-\operatorname*{sgn}\sigma$, $\mathcal{D}_{\pm}'$ is nonzero. \end{corollary} \section{Restriction to $t=0$} By Theorem \ref{thm:pde soln}, the map from $\mathcal{D}'$ to $\mathcal{S}(\mathbb{R}^{n}) $ given by restriction to $t=0$ is injective. Following this map by the Fourier transform gives the following injective map. Recall that $\mathcal{D}_{\pm}'$ is nonzero when $q=-\operatorname*{sgn}\sigma$ and we assume this is so for the rest of the paper. \begin{definition} \rm Let $\mathcal{E}:\mathcal{D}'\to \mathcal{S}(\mathbb{R}^{n})$ be given by \[ (\mathcal{E}f) (x) =\widehat{f}(x,0). \] We also write $\mathcal{S}=\operatorname{Im}(\mathcal{E})$ and $\mathcal{S}_{+}$ and $\mathcal{S}_{-}$ for the images of $\mathcal{D} _{+}'$ and $\mathcal{D}_{-}'$, respectively. We make $\mathcal{S}$ into a $G$-module by requiring $\mathcal{E}$ to be an intertwining isomorphism \[ \mathcal{E}:\mathcal{D}'\to \mathcal{S}. \] \end{definition} \begin{theorem} \label{thm:S-action}For $f\in\mathcal{S}$ and $(g,\varepsilon) \in Mp(n) $, $((g,\varepsilon) \cdot f) (x) $ is given by \begin{itemize} \item[(1)] For $m_{A,a}=\Big( \begin{pmatrix} A & 0\\ 0 & A^{-1,T} \end{pmatrix} ,Z\to a\Big)$ with $a^2=\det A^{-1}$ (so $(a| \det A| ^{1/2}) ^2=\operatorname*{sgn}(\det A) $), \[ (m_{A,a}\cdot f) (x) =\big(a| \det A| ^{1/2}\big) ^{q}| \det A| ^{1/2} f(xA) . \] \item[(2)] For $n_{B,\varepsilon}=\Big( \begin{pmatrix} I_{n} & B\\ 0 & I_{n} \end{pmatrix} ,Z\to \varepsilon\Big)$ with $\varepsilon^2=1$, \[ (n_{B,\varepsilon}\cdot f) (x) =\varepsilon ^{q}e^{-\frac{\pi^2}{s}xBx^{T}}\,f(x) . \] \item[(3)] For $\overline{n}_{C}=\Big( \begin{pmatrix} I_{n} & 0\\ C & I_{n} \end{pmatrix} ,\varepsilon_{C}(Z) \Big)$, \[ (\overline{n}_{C}\cdot f) (x) =\big( e^{-s(\cdot) C(\cdot) ^{T}}\,f^{\vee}( \cdot) \big) ^{\wedge}(x) =(\widehat{e^{-s(\cdot) C(\cdot) ^{T}}}\ast f)(x) . \] \item[(4)] Let $\omega= \begin{pmatrix} 0 & -I_{n}\\ I_{n} & 0 \end{pmatrix} $ and $\varepsilon_{\omega}(Z) $ satisfy $\varepsilon_{\omega }(Z) ^2=\det(Z) $ with $\varepsilon_{\omega }((\lambda+i\mu) I_{n}) =\sqrt{( \lambda+i\mu) ^{n}}$ for $\lambda,\mu\in\mathbb{R}^{+}$ with $\arctan(\frac{\mu}{\lambda}) <\frac{\pi}{n}$. Then \[ ((\omega,\varepsilon_{\omega}) \cdot f) ( x) =e^{-\varepsilon_{\sigma}\frac{i\pi n}{4}}| \frac{\pi }{\sigma}| ^{n/2}\widehat{f}(\frac{\pi}{\sigma }x). \] \end{itemize} \end{theorem} \begin{proof} For $f\in\mathcal{S}$ and $(g,\varepsilon) \in Mp(n) $, \[ \big((g,\varepsilon) \cdot f\big) (x) =\mathcal{E}((g,\varepsilon) \cdot(\mathcal{E} ^{-1}(f) ) ) (x) =(( g,\varepsilon) \cdot(\mathcal{E}^{-1}(f) ) ) ^{\wedge}(x,0) . \] Since \[ (\mathcal{E}^{-1}(f) ) (x,t) =\int_{\mathbb{R}^{n}}f(\xi) e^{\frac{\pi^2}{s}\xi t\xi^{T} }e^{2\pi i\xi x^{T}}\,d\xi, \] we use Theorem \ref{thm:noncompact picture group action} to calculate the new action. In the first case, $(m_{A,a}\cdot f) (x,t) =i^{lq}| \det A| ^{r}\,f(xA^{-1,T},A^{-1}tA^{-1,T})$ with $i^{l}=a| \det A| ^{1/2}$. Therefore \begin{align*} (m_{A,a}\cdot(\mathcal{E}^{-1}(f) )) ^{\vee}(x,0) & =i^{lq}| \det A| ^{r}\,(\mathcal{E}^{-1}(f) ) (xA^{-1,T},0)\\ & =i^{lq}| \det A| ^{r}\,f^{\vee}(xA^{-1,T}) \end{align*} so that \[ (m_{A,a}\cdot(\mathcal{E}^{-1}(f) ) ) (x,0) =i^{lq}| \det A|^{r+1}\,f(xA) . \] In the second case, $(n_{B}\cdot f) (x,t) =\varepsilon^{q}\,f(x,t-B)$ so \begin{align*} (n_{B}\cdot(\mathcal{E}^{-1}(f) ) ) ^{\vee}(x,0) & =\varepsilon^{q}\,(\mathcal{E} ^{-1}(f) ) (x,-B)\\ & =\varepsilon^{q}\,\int_{\mathbb{R}^{n}}f(\xi) e^{-\frac {\pi^2}{s}\xi B\xi^{T}}e^{2\pi i\xi x^{T}}\,d\xi \end{align*} so that \[ (n_{B}\cdot(\mathcal{E}^{-1}(f) ) ) (x,0) =\varepsilon^{q}e^{-\frac{\pi^2}{s}xBx^{T}}\,f( x) . \] For the third case, \begin{align*} (\overline{n}_{C}\cdot f) (x,t) & =i^{lq} | \det(I_{n}-tC) | ^{r}e^{-sxC( I_{n}-tC) ^{-1}x^{T}}\\ & \quad\times f(x(-Ct+I_{n}) ^{-1},(I_{n}-tC) ^{-1}t) \end{align*} with $i^{l}| \det(I_{n}-tC) | ^{-\frac{1}{2} }=\sqrt{\det(I_{n}-tC) ^{-1}}$ for small $t$. Therefore \begin{align*} (\overline{n}_{C}\cdot(\mathcal{E}^{-1}(f) ) ) ^{\vee}(x,0) & =e^{-sxCx^{T}}(\mathcal{E}^{-1}(f) ) (x,0)\\ & =e^{-sxCx^{T}}\int_{\mathbb{R}^{n}}f(\xi) e^{2\pi i\xi x^{T}}\,d\xi \end{align*} so \begin{align*} (\overline{n}_{C}\cdot(\mathcal{E}^{-1}(f)) ) (x,0) & =(e^{-s(\cdot) C(\cdot) ^{T}}\,f^{\vee}(\cdot) ) ^{\wedge }(x) \\ & =(\widehat{e^{-s(\cdot) C(\cdot) ^{T}}}\ast f)(x) . \end{align*} Finally, when $t$ is invertible, \[ ((\omega,\varepsilon_{\omega}) \cdot f) ( x,t) =i^{lq}| \det t| ^{r}e^{sxt^{-1}x^{T}}\,f(-xt^{-1},-t^{-1}) \] where $\varepsilon_{\omega}(-t^{-1}) =i^{l}| \det t| ^{-1/2}$. In the case of $t=\lambda I_{n}$ with $\lambda<0$, \[ \varepsilon_{\omega}(-t^{-1}) =\lim_{\mu\to 0^{+} }\varepsilon_{\omega}((-\lambda^{-1}+i\mu) I_{n}) =\sqrt{(-\lambda^{-1}+i\mu) ^{n}}=| \lambda| ^{-n/2} \] so that $i^{l}=1$ and $((\omega,\varepsilon_{\omega}) \cdot f) (x,\lambda I_{n}) =| \lambda| ^{nr}e^{s\lambda^{-1}\| x\| ^2}\,f(-\lambda^{-1} x,-\lambda^{-1}I_{n})$. We now will calculate the action of $(\omega,\varepsilon_{\omega}) $ on $\mathcal{S}(\mathbb{R}^{n}) $ using \begin{align*} ((\omega,\varepsilon_{\omega}) \cdot f) ( x) & =((\omega,\varepsilon_{\omega}) \cdot(\mathcal{E}^{-1}(f) ) ) ^{\wedge }(x,0) \\ & =\lim_{\lambda\to 0^{-}}((\omega,\varepsilon_{\omega }) \cdot(\mathcal{E}^{-1}(f) ) ) ^{\wedge}(x,\lambda I_{n}) . \end{align*} Now \[ ((\omega,\varepsilon_{\omega}) \cdot( \mathcal{E}^{-1}(f) ) ) ^{\vee}(x,\lambda I_{n}) =| \lambda| ^{nr}e^{s\lambda^{-1}\| x\| ^2}\,(\mathcal{E}^{-1}(f) ) (-\lambda^{-1}x,-\lambda^{-1}I_{n}). \] We first rewrite $(\mathcal{E}^{-1}(f) ) (w,-\lambda^{-1}I_{n})$ using the identity \[ \int_{\mathbb{R}^{n}}e^{-2\pi i\xi x^{T}}e^{-\pi\alpha\| \xi\| ^2}\,d\xi=\alpha^{-n/2}e^{-\frac{\pi}{\alpha}\| x\| ^2} \] for $\operatorname{Re}\alpha>0$. We get (taking $\alpha=\varepsilon +\pi/(s\lambda) $), using Dominated Convergence and Fubini, \begin{align*} (\mathcal{E}^{-1}(f) ) (w,-\lambda^{-1}I_{n}) &=\int_{\mathbb{R}^{n}}f(\xi) e^{-\frac{\pi^2}{s\lambda }\| \xi\| ^2}e^{2\pi i\xi w^{T}}\,d\xi\\ & =\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\widehat{f}(y) e^{2\pi i\xi y^{T}}e^{-\frac{\pi^2}{s\lambda}\| \xi\| ^2 }e^{2\pi i\xi w^{T}}\,dyd\xi\\ & =\lim_{\epsilon\to 0^{+}}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n} }\widehat{f}(y) e^{-\pi(\varepsilon+\pi/s\lambda) \| \xi\| ^2}e^{2\pi i\xi(y+w) ^{T}}\,dyd\xi\\ & =\lim_{\epsilon\to 0^{+}}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n} }\widehat{f}(y) e^{-\pi(\varepsilon+\pi/s\lambda) \| \xi\| ^2}e^{-2\pi i\xi(-y-w) ^{T}}\,d\xi dy\\ & =\lim_{\epsilon\to 0^{+}}(\varepsilon+\pi/s\lambda) ^{-n/2}\int_{\mathbb{R}^{n}}\widehat{f}(y) e^{-\frac{\pi }{\varepsilon+\pi/s\lambda}\| y+w\| ^2}\,dy. \end{align*} Now write $s=i\sigma$ (and recall $\lambda<0$) so that analytic continuation of $\alpha^{-n/2}$ on $\mathbb{R}^{+}$ gives \[ \lim_{\epsilon\to 0^{+}}(\varepsilon+\pi/s\lambda) ^{-n/2}=\begin{cases} | \frac{\pi}{s\lambda}| ^{-\frac{n}{2}}e^{-\frac{i\pi n} {4}}, & \sigma>0\\[4pt] | \frac{\pi}{s\lambda}| ^{-\frac{n}{2}}e^{\frac{i\pi n}{4} }, & \sigma<0. \end{cases} \] Thus \[ (\mathcal{E}^{-1}(f) ) (w,-\lambda^{-1} I_{n})=| \frac{\pi}{s\lambda}| ^{-\frac{n}{2} }e^{-\varepsilon_{\sigma}\frac{i\pi n}{4}}\int_{\mathbb{R}^{n}}\widehat {f}(y) e^{-s\lambda\| y+w\| ^2}\,dy. \] Therefore, \begin{align*} &((\omega,\varepsilon_{\omega}) \cdot( \mathcal{E}^{-1}(f) ) ) ^{\vee}(x,\lambda I_{n}) \\ & =| \lambda| ^{nr}e^{s\lambda ^{-1}\| x\| ^2}\,(\mathcal{E}^{-1}(f) ) (-\lambda^{-1}x,-\lambda^{-1}I_{n})\\ & =| \lambda| ^{nr}| \frac{\pi}{s\lambda }| ^{-\frac{n}{2}}e^{-\varepsilon_{\sigma}\frac{i\pi n}{4} }e^{s\lambda^{-1}\| x\| ^2} \int_{\mathbb{R}^{n}}\widehat{f}(y) e^{-s\lambda \| y-\lambda^{-1}x\| ^2}\,dy\\ & =| \frac{\pi}{s}| ^{-\frac{n}{2}}e^{-\varepsilon _{\sigma}\frac{i\pi n}{4}}e^{s\lambda^{-1}\| x\| ^2} \,\int_{\mathbb{R}^{n}}\widehat{f}(y) e^{-s\lambda\| y-\lambda^{-1}x\| ^2}\,dy\\ & =| \frac{\pi}{s}| ^{-\frac{n}{2}}e^{-\varepsilon _{\sigma}\frac{i\pi n}{4}}\,\int_{\mathbb{R}^{n}}\widehat{f}(y) e^{-s\lambda\| y\| ^2}e^{2syx^{T}}\,dy\\ & =| \frac{\pi}{\sigma}| ^{-\frac{n}{2}}e^{-\varepsilon _{\sigma}\frac{i\pi n}{4}}\,\int_{\mathbb{R}^{n}}\widehat{f}(y) e^{-s\lambda\| y\| ^2}e^{2\pi i\frac{\sigma}{\pi}yx^{T} }\,dy\\ & =| \frac{\pi}{\sigma}| ^{n/2}e^{-\varepsilon _{\sigma}\frac{i\pi n}{4}}\,\int_{\mathbb{R}^{n}}\widehat{f}(\frac{\pi }{\sigma}y) e^{-\frac{i\lambda\pi^2}{\sigma}\| y\| ^2}e^{2\pi iyx^{T}}\,dy\\ & =| \frac{\pi}{\sigma}| ^{-\frac{n}{2}}e^{-\varepsilon _{\sigma}\frac{i\pi n}{4}} \int_{\mathbb{R}^{n}}\widehat{f\circ M_{\frac{\sigma}{\pi}}}( y) e^{-\frac{i\lambda\pi^2}{\sigma}\| y\| ^2 }e^{2\pi iyx^{T}}\,dy \end{align*} where $M_{\sigma/\pi}$ is the multiplication map given by $M_{\sigma/\pi }(x) =\sigma x/\pi$. As a result, \begin{align*} &((\omega,\varepsilon_{\omega}) \cdot f) (x) \\ & =\lim_{\lambda\to 0^{-}}((\omega ,\varepsilon_{\omega}) \cdot(\mathcal{E}^{-1}(f) ) ) ^{\wedge}(x,\lambda I_{n}) \\ & =\lim_{\lambda\to 0^{-}}\int_{\mathbb{R}^{n}}(( \omega,\varepsilon_{\omega}) \cdot(\mathcal{E}^{-1}( f) ) ) (\xi,\lambda I_{n}) e^{-2\pi i\xi x^{T}}\,d\xi\\ & =e^{-\varepsilon_{\sigma}\frac{i\pi n}{4}}| \frac{\pi}{\sigma }| ^{-\frac{n}{2}} \lim_{\lambda\to 0^{-}}\int_{\mathbb{R}^{n}}\int_{\mathbb{R} ^{n}}\widehat{f\circ M_{\frac{\sigma}{\pi}}}(y) e^{-s\lambda \| y\| ^2}e^{2sy\xi^{T}}e^{-2\pi i\xi x^{T}}\,dyd\xi\\ & =e^{-\varepsilon_{\sigma}\frac{i\pi n}{4}}| \frac{\pi}{\sigma }| ^{-\frac{n}{2}}\lim_{\lambda\to 0^{-}}\widehat{f\circ M_{\frac{\sigma}{\pi}}}(x) e^{-s\lambda\| x\| ^2}\\ & =e^{-\varepsilon_{\sigma}\frac{i\pi n}{4}}| \frac{\pi}{\sigma }| ^{-\frac{n}{2}}\widehat{f\circ M_{\frac{\sigma}{\pi}}}( x) \\ & =e^{-\varepsilon_{\sigma}\frac{i\pi n}{4}}| \frac{\pi}{\sigma }| ^{n/2}\widehat{f}(\frac{\pi}{\sigma}x) . \end{align*} \end{proof} To match these formulas with the realization of the oscillator representation in, say, Kashiwara and Vergne, consider the dilation operator defined by \[ (Tf) (x) =f(\frac{| \sigma| ^{1/2}}{\pi\sqrt{2}}x). \] Making $T$ into an intertwining map, Theorem \ref{thm:S-action} gives an equivalent action on $T(\mathcal{S}) \subseteq\mathcal{S}( \mathbb{R}^{n}) $. Note, of course, that the map $T$ can be modified by multiplying by the scalar $(| \sigma| ^{1/2}/( \pi\sqrt{2}) )^{n/2}$ to make it a unitary map with respect to $L^2(\mathbb{R}^{n}) $. This modification will not change the theorem below. \begin{theorem} \label{thm:KV intertwine} The action of $Mp(n) $ on $T( \mathcal{S}) $ is given by \begin{gather*} (m_{A,a}\cdot f) (x) =| \det A| ^{1/2}\,f(xA) ,\text{ for }a>0\\ (n_{B}\cdot f) (x) =e^{\varepsilon_{\sigma }\frac{i}{2}xBx^{T}}\,f(x),\\ (\overline{n}_{C}\cdot f) (x) =(\widehat {e^{-\varepsilon_{\sigma}2i\pi^2(\cdot) C(\cdot) ^{T}}}\ast f)(x) \\ ((\omega,\varepsilon_{\omega}) \cdot f) ( x) =e^{-\varepsilon_{\sigma}\frac{i\pi n}{4}}(\frac{1}{2\pi }) ^{n/2}\,\int_{\mathbb{R}^{n}}f(\xi) e^{-\varepsilon_{\sigma}i\xi x^{T}}\,d\xi. \end{gather*} In particular, when $s=i\sigma$ with $\sigma<0$, this is a dense $Mp( n) $-invariant subspace in the oscillator representation. When $\sigma>0$, this representation is isomorphic to the dual to the oscillator representation. In either case, this action completes to a unitary representation on $L^2(\mathbb{R}^{n}) $ and decomposes as a direct sum of irreducible representation via the set of odd and even function, \[ L^2(\mathbb{R}^{n}) =L^2(\mathbb{R}^{n}) _{+}\oplus L^2(\mathbb{R}^{n}) _{-}. \] \end{theorem} \begin{proof} For $a>0$, \begin{align*} (m_{A,a}\cdot f) (x) & =(T(m_{A,a}\cdot T^{-1}f) ) (x) \\ & =(m_{A,a}\cdot T^{-1}f) (\frac{| \sigma|^{1/2}}{\pi\sqrt{2}}x)\\ & =| \det A| ^{1/2}(T^{-1}f) ( \frac{| \sigma| ^{1/2}}{\pi\sqrt{2}}xA) \\ & =| \det A| ^{1/2}\,f(xA) , \end{align*} and \begin{align*} (n_{B}\cdot f) (x) & =(T(n_{B}\cdot T^{-1}f) ) (x) \\ & =(n_{B}\cdot T^{-1}f) (\frac{| \sigma| ^{1/2}}{\pi\sqrt{2}}x)\\ & =e^{-\frac{\pi^2}{i\sigma}\frac{| \sigma| }{2\pi^2 }xBx^{T}}\,(T^{-1}f) (\frac{| \sigma| ^{1/2} }{\pi\sqrt{2}}x)\\ & =e^{(\varepsilon_{\sigma}) \frac{i}{2}xBx^{T}}\,f(x), \end{align*} and \begin{align*} (\overline{n}_{C}\cdot f) (x) & =( T(\overline{n}_{C}\cdot T^{-1}f) ) (x) \\ & =(\overline{n}_{C}\cdot T^{-1}f) (\frac{| \sigma| ^{1/2}}{\pi\sqrt{2}}x)\\ & =(\widehat{e^{-s(\cdot) C(\cdot) ^{T}}}\ast T^{-1}f)(x) (\frac{| \sigma| ^{1/2}}{\pi \sqrt{2}}x)\\ & =(\frac{| \sigma| ^{1/2}}{\pi\sqrt{2}})^{n} (T\widehat{e^{-s(\cdot) C(\cdot) ^{T}}}\ast f)(x) \\ & =(\widehat{T^{-1}e^{-s(\cdot) C(\cdot) ^{T}} }\ast f)(x) \\ & =(\widehat{e^{-\frac{2\pi^2s}{| \sigma| }( \cdot) C(\cdot) ^{T}}}\ast f)(x) \end{align*} and \begin{align*} ((\omega,\varepsilon_{\omega}) \cdot f) ( x) & =(T((\omega,\varepsilon_{\omega}) \cdot T^{-1}f) ) (x) \\ & =((\omega,\varepsilon_{\omega}) \cdot T^{-1}f) (\frac{| \sigma| ^{1/2}}{\pi\sqrt{2}}x)\\ & =e^{-\varepsilon_{\sigma}\frac{i\pi n}{4}}| \frac{\pi}{\sigma }| ^{n/2}\widehat{f\circ M_{\frac{\pi\sqrt{2}}{| \sigma| ^{1/2}}}}(\frac{\pi}{\sigma}\frac{| \sigma | ^{1/2}}{\pi\sqrt{2}}x)\\ & =e^{-\varepsilon_{\sigma}\frac{i\pi n}{4}}| \frac{\pi}{\sigma }| ^{n/2}| \frac{| \sigma| ^{1/2}}{\pi\sqrt{2}}| ^{n}\widehat{f}(\frac{\pi}{\sigma} \frac{| \sigma| }{2\pi^2}x)\\ & =e^{-\varepsilon_{\sigma}\frac{i\pi n}{4}}(\frac{1}{2\pi}) ^{n/2}\widehat{f}(\frac{\varepsilon_{\sigma}}{2\pi}x)\\ & =e^{-\varepsilon_{\sigma}\frac{i\pi n}{4}}(\frac{1}{2\pi}) ^{n/2}\int_{\mathbb{R}^{n}}f(\xi) e^{-\varepsilon _{\sigma}i\xi x^{T}}\,d\xi. \end{align*} \end{proof} \section{Restriction to $x=0$} Recall from Corollary \ref{cor: x=0} that there is an $Mp(n) $-intertwining map $\mathcal{G}:I'(q,r,s) \to I'(q,r) $ given by \[ (\mathcal{G}f) (t) =f(0,t) \] and an intertwining map $\mathcal{G}_{n}:I'(q,r,s) \to I_{n}'(q,r-\frac{1}{n}) $ given by \[ (\mathcal{G}_{n}f) (t) =\nabla f( 0,t) . \] By the definitions and Theorem \ref{thm:pde soln}, restricting to $\mathcal{D}'$ and pre-composing with $\mathcal{E}^{-1}$ gives $Mp(n) $-maps $\mathcal{H}:\mathcal{S}\to I^{\prime }(q,r) $ and $\mathcal{H}_{n}:\mathcal{S}\to I_{n}'(q,r-\frac{1}{n}) $ given by \[ (\mathcal{H}f) (t) =\int_{\mathbb{R}^{n}}f( \xi) e^{\frac{\pi^2}{s}\xi t\xi^{T}}\,d\xi \] and \begin{align*} (\mathcal{H}_{n}f) (t) & =\nabla\Big( \int_{\mathbb{R}^{n}}f(\xi) e^{\frac{\pi^2}{s}\xi t\xi^{T} }e^{2\pi i\xi x^{T}}\,d\xi\Big) \Big|_{x=0}\\ & =2\pi i\Big(\int_{\mathbb{R}^{n}}\xi_{1}f(\xi) e^{\frac{\pi^2}{s}\xi t\xi^{T}}\,d\xi,\ldots,\int_{\mathbb{R}^{n}}\xi _{n}f(\xi) e^{\frac{\pi^2}{s}\xi t\xi^{T}}\,d\xi\Big) . \end{align*} Clearly $\mathcal{S}_{-}\subseteq\ker\mathcal{H}$ and $\mathcal{S} _{+}\subseteq\ker\mathcal{H}_{n}$ (equivalently, $\mathcal{D}_{-}^{\prime }\subseteq\ker\mathcal{G}$ and $\mathcal{D}_{+}'\subseteq \ker\mathcal{G}_{n}$). To show these are the entire kernels involves inverting $\mathcal{H}|_{\mathcal{S}_{+}}$ and $\mathcal{H}_{n}|_{\mathcal{S}_{-}}$ (equivalently, $\mathcal{G}|_{\mathcal{D}_{+}'}$ and $\mathcal{G} _{n}|_{\mathcal{D}_{-}'}$). Straightforward Fourier analysis requires a bit more care due to the fact that the images usually do not have sufficient decay properties to be $L^{1}$ or $L^2$ functions (unless $n=1$, see \cite{SS2010}). In fact, if we could view $f\in\mathcal{D}'\subseteq I'(q,r,s) $ as a tempered distribution $f( x,\cdot) \in\mathcal{S}'(\operatorname*{Sym}( n,\mathbb{R}) \cong\mathbb{R}^{n(n+1) /2}) $ and writing $\mathcal{F}$ for the Fourier transform on $\mathcal{S}( \operatorname*{Sym}(n,\mathbb{R}) ) $ given by \[ (\mathcal{F}f) (\tau) =\int_{\operatorname*{Sym} (n,\mathbb{R}) }f(t) e^{-2\pi i\operatorname*{tr} (t\tau) }\,dt, \] we would have \begin{gather*} 8\pi is\tau_{i,j}\mathcal{F}f+\partial_{x_i}\partial_{x_{j}}\mathcal{F}f =0, \quad i\neq j,\\ 8\pi is\tau_{i,i}\mathcal{F}f+\partial_{x_i}^2\mathcal{F}f =0. \end{gather*} Looking at $\partial_{x_i}^2\partial_{x_{j}}^2\mathcal{F}f$ written in two ways for $i\neq j$, we would get \[ (\tau_{i,i}\tau_{j,j}-\tau_{i,j}^2) \mathcal{F}f=0 \] so that $\mathcal{F}f$ would be supported on $\{\tau\in\operatorname*{Sym} (n,\mathbb{R}) :\tau_{i,i}\tau_{j,j}=\tau_{i,j}^2$ all $i\neq j\}$. This is, of course a rank of at most one condition on $\operatorname*{Sym}(n,\mathbb{R}) $. As a result, it will be useful to consider the cone defined by the function $\theta:\mathbb{R} ^{n}\to \operatorname*{Sym}(n,\mathbb{R}) $ given by \[ \theta(y) =\frac{\pi}{2\sigma}\,y^{T}y. \] \begin{lemma}\label{lem:FT of x=0} \textbf{(1)} For $f\in\mathcal{D}'\subseteq I'(q,r,s) $ and each $x\in\mathbb{R}^{n}$, $f( x,\cdot) $ may be viewed as a tempered distribution on $\operatorname*{Sym}(n,\mathbb{R}) $ given by \[ \left\langle f(x,\cdot) ,\phi\right\rangle =\int _{\operatorname*{Sym}(n,\mathbb{R}) }f(x,t) \phi(t) \,dt \] for each $\phi\in\mathcal{S}(\operatorname*{Sym}(n,\mathbb{R} ) ) $. Its Fourier transform $\mathcal{F}f(x,\cdot ) \in\mathcal{S}'(\operatorname*{Sym}( n,\mathbb{R}) ) $ is given by \[ \left\langle \mathcal{F}f(x,\cdot) ,\phi\right\rangle =\int_{\mathbb{R}^{n}}\widehat{f}(\xi,0) (\phi\circ \theta) (\xi) e^{2\pi i\xi x^{T}}\,d\xi =(f(\cdot,0) \ast(\phi\circ\theta) ) (x) \] and is supported on $\operatorname{Im}\theta$. \textbf{(2)} For each $1\leq j\leq n$, $\partial_{x_{j}}f( x,\cdot) $ may be viewed as a tempered distribution on $\operatorname*{Sym}(n,\mathbb{R}) $ given by \[ \left\langle \partial_{x_{j}}f(x,\cdot) ,\phi\right\rangle =\int_{\operatorname*{Sym}(n,\mathbb{R}) }\partial_{x_{j} }f(x,t) \phi(t) \,dt \] for each $\phi\in\mathcal{S}(\operatorname*{Sym}(n,\mathbb{R} ) ) $. Its Fourier transform $\mathcal{F}(\partial _{x_{j}}f) (x,\cdot) \in\mathcal{S}'( \operatorname*{Sym}(n,\mathbb{R}) ) $ is given by \begin{align*} \left\langle \mathcal{F}(\partial_{x_{j}}f) ( x,\cdot) ,\phi\right\rangle & =2\pi i\int_{\mathbb{R}^{n}}\xi _{j}\widehat{f}(\xi,0) (\phi\circ\theta) ( \xi) e^{2\pi i\xi x^{T}}\,d\xi\\ & =2\pi i((\partial_{x_{j}}f) (\cdot,0) \ast(\phi\circ\theta) ) (x) \end{align*} and is supported on $\operatorname{Im}\theta$. \end{lemma} \begin{proof} First of all, since \[ | f(x,t) | \leq\int_{\mathbb{R}^{n} }| \widehat{f}(\xi,0) e^{\frac{\pi^2}{s}\xi t\xi^{T} }e^{2\pi i\xi x^{T}}| \,d\xi=\| \widehat{f}( \cdot,0) \| _{L^{1}(\mathbb{R}^{n}) }<\infty, \] $f(x,\cdot) $ is bounded. As it is also continuous, it is clearly locally integrable and therefore gives rise to an element of $\mathcal{S}'(\operatorname*{Sym}(n,\mathbb{R}) ) $. To calculate its Fourier transform, use Fubini to see that \begin{align*} \left\langle \mathcal{F}f(x,\cdot) ,\phi\right\rangle & =\left\langle f(x,\cdot) ,\mathcal{F}\phi\right\rangle \\ & =\int_{\operatorname*{Sym}(n,\mathbb{R}) }f( x,t) \mathcal{F}\phi(t) \,dt\\ & =\int_{\operatorname*{Sym}(n,\mathbb{R}) }\int_{\mathbb{R} ^{n}}\widehat{f}(\xi,0) e^{\frac{\pi^2}{s}\xi t\xi^{T}}e^{2\pi i\xi x^{T}}\mathcal{F}\phi(t) \,d\xi dt\\ & =\int_{\mathbb{R}^{n}}\int_{\operatorname*{Sym}(n,\mathbb{R}) }\widehat{f}(\xi,0) e^{2\pi i\xi x^{T}}\mathcal{F}\phi( t) e^{\frac{\pi^2}{s}\xi t\xi^{T}}\,dtd\xi\\ & =\int_{\mathbb{R}^{n}}\int_{\operatorname*{Sym}(n,\mathbb{R}) }\widehat{f}(\xi,0) e^{2\pi i\xi x^{T}}\mathcal{F}\phi( t) e^{2\pi i(-\frac{\pi}{2\sigma}) \operatorname*{tr} (t\xi^{T}\xi) }\,dtd\xi\\ & =\int_{\mathbb{R}^{n}}\widehat{f}(\xi,0) e^{2\pi i\xi x^{T} }\mathcal{F}^2\phi(-\frac{\pi}{2\sigma}\xi^{T}\xi) \,d\xi\\ & =\int_{\mathbb{R}^{n}}\widehat{f}(\xi,0) e^{2\pi i\xi x^{T} }\phi(\theta(\xi) ) \,d\xi. \end{align*} Finally, \begin{align*} \left\langle \mathcal{F}f(x,\cdot) ,\phi\right\rangle & =\int_{\mathbb{R}^{n}}\widehat{f}(\xi,0) (\phi\circ \theta) (\xi) e^{2\pi i\xi x^{T}}\,d\xi\\ & =(\widehat{f}(\cdot,0) (\phi\circ\theta) (\cdot) ) ^{\vee}(x) \\ & =(f(\cdot,0) \ast(\phi\circ\theta) ) (x) . \end{align*} Turning to $\partial_{x_{j}}f$, \[ | \partial_{x_{j}}f(x,t) | \leq \int_{\mathbb{R}^{n}}| 2\pi i\xi_{j}\widehat{f}(\xi,0) e^{\frac{\pi^2}{s}\xi t\xi^{T}}e^{2\pi i\xi x^{T}}| \,d\xi =2\pi\| (\cdot) _{j}\widehat{f}(\cdot,0) \| _{L^{1}(\mathbb{R}^{n}) }<\infty \] so that $\partial_{x_{j}}f(x,\cdot) $ gives rise to an element of $\mathcal{S}'(\operatorname*{Sym}(n,\mathbb{R} ) ) $. The rest of the Lemma is a simple modification of the above argument and is omitted. \end{proof} \begin{theorem}\label{thm:x=0 injection} $\mathcal{H}|_{\mathcal{S}_{+}}$ is injective and $\mathcal{H}_{n}|_{\mathcal{S}_{-}}$ is injective. Equivalently, $\mathcal{G}|_{\mathcal{D}_{+}'}$ is injective and $\mathcal{G} _{n}|_{\mathcal{D}_{-}'}$ is injective. \end{theorem} \begin{proof} We show how to construct the inverse maps. Let $f\in\mathcal{S}$. By the definitions and Lemma \ref{lem:FT of x=0}, \[ \left\langle \mathcal{FH}f,\phi\right\rangle =(f^{\vee}\ast( \phi\circ\theta) ) (0) \] for $\phi\in\mathcal{S}(\operatorname*{Sym}(n,\mathbb{R}) ) $. Fix $\psi\in\mathcal{S}(\operatorname*{Sym}( n,\mathbb{R}) ) $ with $\int_{\operatorname*{Sym}( n,\mathbb{R}) }\psi(t) \,dt=1$ and let $\psi_{\epsilon }(t) =\varepsilon^{-n(n+1) /2}\psi( \varepsilon^{-1}t) $ for $\varepsilon>0$ so that $\psi_{\epsilon }\to \delta_{0}$ as an element of $\mathcal{S}'( \operatorname*{Sym}(n,\mathbb{R}) ) $\ as $\epsilon \to 0^{+}$. Then, for any $x\in\mathbb{R}^{n}$, $\tau_{\theta( x) }\psi_{\epsilon}\to \delta_{\theta(x) }$ as $\epsilon\to 0^{+}$. As $\theta(y) =\frac{\pi}{2\sigma }y^{T}y$, it is trivial to check that $(\tau_{\theta(x) }\psi_{\epsilon}) \circ\theta\to \delta_{x}+\delta_{-x}$ as elements of $\mathcal{S}'(\mathbb{R}^{n}) $\ as $\epsilon\to 0^{+}$. If $f\in\mathcal{S}_{+}$, then \[ \lim_{\epsilon\to 0^{+}}\left\langle \mathcal{FH}f,\tau_{\theta( x) }\psi_{\epsilon}\right\rangle =\lim_{\epsilon\to 0^{+} }(f^{\vee}\ast((\tau_{\theta(x) } \psi_{\epsilon}) \circ\theta) ) (0) =f^{\vee}(x) +f^{\vee}(-x) =2f^{\vee}( x) . \] In particular, $f^{\vee}\in\mathcal{S}_{+}$ (and therefore $f$) can be recovered from $\mathcal{H}f$ by taking the Fourier transform and looking at approximations to translations of the delta distribution. Next, view the image of $\mathcal{H}_{n}$ as landing in $\oplus_{j=1} ^{n}\mathcal{S}'(\operatorname*{Sym}(n,\mathbb{R} ) ) $. Evaluating via the diagonal map (so viewing the image as landing in $\mathcal{S}'(\operatorname*{Sym}( n,\mathbb{R}) ,\mathbb{R}^{n}) $) and applying the Fourier transform in each coordinate, it follows that \[ \left\langle \mathcal{FH}_{n}f,\phi\right\rangle =2\pi i(( \partial_{x_{1}}f^{\vee}\ast(\phi\circ\theta) ) ( 0) ,\ldots,(\partial_{x_{n}}f^{\vee}\ast(\phi\circ \theta) ) (0) ). \] As above, when $f\in\mathcal{S}_{-}$, \[ \lim_{\epsilon\to 0^{+}}\left\langle \mathcal{FH}_{n}f^{\vee} ,\tau_{\theta(x) }\psi_{\epsilon}\right\rangle =4\pi i(\partial_{x_{1}}f^{\vee}(x) ,\ldots,\partial_{x_{n}}f^{\vee }(x) ). \] In particular $f^{\vee}\in\mathcal{S}(\mathbb{R}^{n}) _{-}$ (and therefore $f^{\vee}$) can also be recovered from $\mathcal{H}_{n}f$ by taking the Fourier transform and looking at approximations to translations of the delta distribution. \end{proof} \begin{definition} \rm Let $\mathcal{I}_{\pm}'$ be the image of $\mathcal{D}_{\pm}'$ under $\mathcal{G}$ and $\mathcal{G}_{n}$, respectively (alternately, the image of $\mathcal{S}_{\pm}$ under $\mathcal{H}$ and $\mathcal{H}_{n}$, respectively). \end{definition} From Corollary \ref{cor: x=0} and Theorem \ref{thm:x=0 injection}, we see $\mathcal{I}_{\pm}'$ is isomorphic to $\mathcal{D}_{\pm}'$ (and $\mathcal{S}_{\pm}$) as $Mp(n) $-representations. In particular, they complete to unitary highest ($\sigma<0$) or lowest ($\sigma>0$) weight representations isomorphic to the oscillator representation or its dual. The next corollary identifies $\mathcal{I}_{\pm}'$ by viewing the Schwartz space as tempered distributions supported on $\operatorname{Im} \theta$, taking their Fourier transform, and implicitly identifying the resulting tempered distribution with the smooth function it generates. \begin{corollary} \label{cor: Iprime identification} \textbf{(1)} Embed $\mathcal{S} \hookrightarrow\mathcal{S}^{'}(\operatorname*{Sym}( n,\mathbb{R}) ) $ via $\theta$ by mapping $\psi\to \left\langle \psi,\cdot\right\rangle $ where \[ \left\langle \psi,\phi\right\rangle =\int_{\mathbb{R}^{n}}\psi( \xi) (\phi\circ\theta) (\xi) \,d\xi \] for $\phi\in\mathcal{S}(\operatorname*{Sym}(n,\mathbb{R}) ) $. Then $\mathcal{I}_{+}'\subseteq I'( q,r) $ is given explicitly by \[ \mathcal{I}_{+}'=\left\{ \mathcal{F}\psi:\psi\in\mathcal{S} \subseteq\mathcal{S}^{'}(\operatorname*{Sym}( n,\mathbb{R}) ) \right\} . \] \textbf{(2)} Embed $\mathcal{S}\hookrightarrow\mathcal{S}^{'}( \operatorname*{Sym}(n,\mathbb{R}) ,\mathbb{R}^{n}) $ via $\theta$ by mapping $\psi\to \left\langle \psi,\cdot\right\rangle $ where \[ \left\langle \psi,\phi\right\rangle =\Big(\int_{\mathbb{R}^{n}}\xi_{1}\psi( \xi) (\phi\circ\theta) (\xi) \,d\xi ,\ldots,\int_{\mathbb{R}^{n}}\xi_{n}\psi(\xi) (\phi \circ\theta) (\xi) \,d\xi\Big) \] for $\phi\in\mathcal{S}(\operatorname*{Sym}(n,\mathbb{R}) ) $. Then $\mathcal{I}_{-}'\subseteq I_{n}'( q,r-\frac{1}{n}) $ is given explicitly by \[ \mathcal{I}_{-}'=\left\{ \mathcal{F}\psi:\psi\in\mathcal{S} \subseteq\mathcal{S}^{'}(\operatorname*{Sym}( n,\mathbb{R}) ,\mathbb{R}^{n}) \right\} . \] \end{corollary} \begin{proof} Part (1) follows immediately from the formula $\left\langle \mathcal{FH}f,\phi\right\rangle =\int_{\mathbb{R}^{n}}f(\xi) ( \phi\circ\theta) (\xi) \,d\xi$ and Lemma \ref{lem:FT of x=0} and Theorem \ref{thm:x=0 injection}. Similarly, part (2) follows from the formula $\left\langle \mathcal{FH}_{n}f,\phi\right\rangle =(\int_{\mathbb{R}^{n}}\xi_{1}\psi(\xi) (\phi\circ \theta) (\xi) \,d\xi,\ldots,\int_{\mathbb{R}^{n}}\xi _{n}\psi(\xi) (\phi\circ\theta) ( \xi) \,d\xi)$. \end{proof} \section{$K$-finite Vectors} If $M\in M_{n}(\mathbb{C}) $ and $p$ is a complex valued polynomial on $\mathbb{R}^{n}$, define $\widetilde{p}(x,M)$ by \[ \widetilde{p}(x,M)=e^{| \sigma| xMx^{T}}p( \partial_{x}) \Big(e^{-| \sigma| xMx^{T}}\Big) \] with $p(\partial_{x}) $ representing the constant coefficient differential operator obtained by replacing $x_{j}$ by $\partial_{x_{j}}$. For $p$ of the form $x^{\alpha}$, $\widetilde{p}$ defines a generalization of the Hermite polynomials. \begin{theorem} \label{thm:K finite in D} The highest ($\sigma<0$) and lowest ($\sigma>0$) $K$-finite vector of $(\mathcal{D}_{+}') _{K}$, up to a constant multiple, is given by the function $f_{-}$ and $f_{+}$, respectively (see Theorem \ref{thm:nonzero}). The highest and lowest $K$-type vectors of $(\mathcal{D}_{-}') _{K}$ consist of the functions $f_{-,a}$ and $f_{+,a}$, respectively, for $a\in\mathbb{C}^{n}$. In general, the $K$-finite vectors in $\mathcal{D}'$ consists of the functions $f_{-,p}$ and $f_{+,p}$ where \begin{gather*} f_{-,p}(x,t) =\overline{\varepsilon_{t}( iI_{n}) }^{-1}\widetilde{p}(x,(I_{n}-it) ^{-1}) e^{\sigma x(I_{n}-it) ^{-1}x^{T}}\\ f_{+,p}(x,t) =\varepsilon_{t}(iI_{n}) ^{-1}\widetilde{p}(x,(I_{n}+it) ^{-1}) e^{-\sigma x(I_{n}+it) ^{-1}x^{T}} \end{gather*} where $p$ is a complex valued polynomial on $\mathbb{R}^{n}$. \end{theorem} \begin{proof} It is well known that the $K$-finite vectors in the oscillator representation (see, e.g., \cite{KV1978} or \cite{HT1992}) are spanned by functions of the form $p(x) e^{-\| x\| ^2/2}$ with $p$ a polynomial on $\mathbb{R}^{n}$. Pulling back this standard picture by $Tf=M_{| \sigma| ^{1/2}/\pi\sqrt{2}}f$, we see that the $K$-finite vectors in the image of $\mathcal{E}$, $\mathcal{S}( \mathbb{R}^{n}) _{K}$, are spanned by functions of the form $p( x) e^{-\frac{\pi^2}{| \sigma| }\| x\| ^2}$ (a different $p$ of the same degree). Pulling these functions back to $\mathcal{D}'$ involves solving a system of partial differential equations with initial condition at $t=0$ given by the inverse Fourier transform of $p(x) e^{-\frac{\pi^2}{| \sigma| }\| x\| ^2}$, that is, functions of the form $\widetilde{p}(x) e^{-| \sigma| \| x\| ^2}$ for some polynomial $\widetilde{p}$ determined by $p$. By Theorem \ref{thm:pde soln}, the solution of this system is given by \begin{align*} f(x,t) & =\int_{\mathbb{R}^{n}}p(\xi) e^{-\frac{\pi^2}{| \sigma| }\| \xi\| ^2}e^{\frac{\pi^2}{s}\xi t\xi^{T}}e^{2\pi i\xi x^{T}}\,d\xi\\ & =\int_{\mathbb{R}^{n}}p(\xi) e^{-\frac{\pi^2}{| \sigma| }\xi(1+i\varepsilon_{\sigma}t) \xi^{T}}e^{2\pi i\xi x^{T}}\,d\xi\\ & =\Big(p(\cdot) e^{-\frac{\pi^2}{| \sigma| }(\cdot) (1+i\varepsilon_{\sigma }t) (\cdot) ^{T}}\Big) ^{\vee}(x) \\ & =p(-2\pi i\partial_{x}) \Big(e^{-\frac{\pi^2}{| \sigma| }(\cdot) (1+i\varepsilon_{\sigma }t) (\cdot) ^{T}}\Big) ^{\vee}(x) . \end{align*} As a result, the problem comes down to finding the function defined by \begin{align*} F(x,t) & =(\frac{\pi}{| \sigma| }) ^{n/2}\Big(e^{-\frac{\pi^2}{| \sigma | }(\cdot) (1+i\varepsilon_{\sigma}t) (\cdot) ^{T}}\Big) ^{\vee}(x) \\ & =(\frac{\pi}{| \sigma| }) ^{\frac{n}{2} }\int_{\mathbb{R}^{n}}e^{-\frac{\pi^2}{| \sigma| }\| \xi\| ^2}e^{\frac{\pi^2}{s}\xi t\xi^{T}}e^{2\pi i\xi x^{T}}\,d\xi. \end{align*} We claim that this function is given exactly by $F=f_{\operatorname*{sgn} \sigma}$ from Theorem \ref{thm:nonzero}. To verify this claim, note that, by definition, $F$ is the unique solution to the system given in Equation \eqref{eqn:system of pdess} satisfying the initial condition of $\widehat{F}(\xi,0) =(\pi/| \sigma| ) ^{n/2}e^{-\frac{\pi^2}{| \sigma | }\| \xi\| ^2}$ or, equivalently, that $F( x,0) =e^{-| \sigma| \| x\| ^2}$. Obviously, our proposed solution, $f_{\operatorname*{sgn}\sigma}$, satisfies that initial condition. By the proof of Theorem \ref{thm:nonzero}, it also satisfies the system of differential operators which finishes the claim. Since the highest/lowest $K$-type space in the oscillator representation is spanned by $e^{-\| x\| ^2/2}$ (for the even functions) and $x_ie^{-\| x\| ^2/2}$ (for the odd functions), the above discussion shows that the corresponding functions (up to a multiple) in $\mathcal{D}'$ are $f_{\operatorname*{sgn}\sigma}$ and $\partial _{x_i}f_{\operatorname*{sgn}\sigma}$. Since $f_{\operatorname*{sgn}\sigma}$ has been calculated, consider $\partial_{x_i}f_{\operatorname*{sgn}\sigma}$: \begin{gather*} \partial_{x_i}f_{-} =2| \sigma| \overline {\varepsilon_{t}(iI_{n}) }^{-1}(x(I_{n}-it) ^{-1}e_i) e^{\sigma x(I_{n}-it) ^{-1}x^{T}}\\ \partial_{x_i}f_{+} =-2\sigma\varepsilon_{t}(iI_{n}) ^{-1}(x(I_{n}+it) ^{-1}e_i) e^{-\sigma x( I_{n}+it) ^{-1}x^{T}}. \end{gather*} Finally, the last statement follows from the fact that the element of $\mathcal{D}'$ corresponding to the function $p(x) e^{-\frac{\pi^2}{| \sigma| }\| x\| ^2 }$ in the image of $\mathcal{E}$ is $p(-2\pi i\partial_{x}) f_{+}(x) $. \end{proof} \begin{corollary} The highest ($\sigma<0$) and lowest ($\sigma>0$), respectively, $K$-finite vector of $(\mathcal{I}_{+}') _{K}$ is spanned by the function $f_{\operatorname*{sgn}\sigma}$ given by \[ f_{-}(0,t) =\overline{\varepsilon_{t}(iI_{n}) }^{-1}, \quad f_{+}(0,t) =\varepsilon_{t}(iI_{n}) ^{-1}. \] The highest ($\sigma<0$) and lowest ($\sigma>0$), respectively, $K$-type vectors of $(\mathcal{I}_{-}') _{K}$ is given by the functions $f_{\operatorname*{sgn}\sigma,a}$ where \begin{gather*} f_{-,a}(t) =\overline{\varepsilon_{t}(iI_{n}) }^{-1}a(I_{n}-it) ^{-1}\\ f_{+,a}(t) =\varepsilon_{t}(iI_{n}) ^{-1}a(I_{n}+it) ^{-1} \end{gather*} for $a\in\mathbb{R}^{n}$. \end{corollary} It is possible to describe the general $K$-finite vector, though the details are more involved. For instance, it is straightforward to check that the $K$-finite vectors of $(\mathcal{I}_{+}') _{K}$ are spanned by functions of the form \[ f(t) =\det(I_{n}+i\varepsilon_{\sigma}t) ^{-1/2}\sum_{\sigma\in\,\widetilde{S}_{2k}}\prod_{l=1}^{k}( (I_{n}+i\varepsilon_{\sigma}t) ^{-1}) _{j_{\sigma( 2l-1) },j_{\sigma(2l) }} \] where $k\in\mathbb{N}$, $j_{1},\ldots,j_{2k}\in\left\{ 1,\ldots,n\right\} $ and$\widetilde{S}_{2k}$ denotes the set elements of the symmetric group $S_{2k}$ satisfying $\sigma(2l-1) <\sigma(2l) $ and $\sigma(1) <\sigma(3) <\cdots<\sigma( 2k-1) $. Notice that each term in the summand is the $k$-fold product of the determinant of a minor of $(I_{n}+i\varepsilon_{\sigma}t) $ divided by $\det(I_{n}+i\varepsilon_{\sigma}t) $. \begin{thebibliography}{99} \bibitem{AL} S. Ali, M. Englis; Quantization methods: a guide for physicists and analysts, \emph{Rev. Math. Phys.} 17, no. 4, 391--490, 2005. \bibitem{Berc} S. 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