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\title{Pseudofinite-dimensional Schr\"odinger representations \\ PRELIMINARY
NOTES}
\author{M. Bays \& B. Hart}
\maketitle

\abstract{
    Following Zilber \cite{Z-CCR}, we obtain the Schr\"odinger representation of
    the (3-dimensional) Heisenberg algebra on the tempered distributions as an
    ultralimit of finite-dimensional representations of certain subgroups of
    the Heisenberg group. Moreover, we see that the corresponding Weil
    representation can be obtained this way.}

\section{Introduction}
In \cite{Z-CCR}, Zilber considers a certain ultralimit of finite-dimensional
Hilbert spaces equipped with certain linear structure, and proposes it as a
suitable replacement for the traditional functional analytic domains of
quantum mechanics. Working in such a limit of finite-dimensional spaces allows
one to skirt the analytic subtleties involved in traditional techniques. Zilber
confirms that many calculations performed in his limit space agree with
standard results.

Here, we give further confirmation that such an ultralimit of finite
dimensional structures can replicate the functional analytic setup.
Specifically, we show that the Schr\"odinger representation of the 3-dimensional
Heisenberg algebra and the Weil representation of the double cover $\Mp_2(\R)$
of $\SL_2(\R)$ on the space $\S'$ of tempered distributions on $\R$ can be
obtained as a certain ultralimit of finite dimensional subspaces of $\S'$
equipped with certain traces of these representations.

Although we will treat this entirely mathematically, to draw the connection
with \cite{Z-CCR}, we briefly explain here the connection to quantum
mechanics: the representation of the Heisenberg algebra provides the
position and momentum observables $Q$ and $P$ for a particle on the line, and
the Weil representation then yields the dynamics for a quadratic homogeneous
Hamiltonian - such a Hamiltonian corresponds to an element of the lie algebra
$\sl_2(\R)$ of $\Mp_2(\R)$, and the corresponding 1-parameter subgroup gives
the evolution in time of the system. These representations can be taken to be
representations on the Hilbert space $L^2(\R)$, but in the ``rigged Hilbert
space'' presentation one extends them to representations on $\S'$, which is
what we consider here.

In short, we argue as follows: the subspaces $\S'_m$ of $m$-periodic linear
combinations of $\delta$-functions supported on $\frac 1m \Z$ have limit $\S'$
if one restricts to bounded sets; the Schr\"odinger representation of subgroups
of the Heisenberg group restricts to $\S'_m$, and a formal differentiation
process allows one to obtain the representation of the algebra in the limit;
meanwhile, restrictions of rational points of $\Mp_2$ restrict to partially
defined operators on $\S'_m$ whose limits along sufficiently divisible
ultrafilters are the correct totally defined operators on $\S'$, and the
operators for real points can be approximated by the restrictions of the
operators of approximate rational points. The rest of the paper provides the
details.

The finite-dimensional structures we consider are essentially those considered
by Zilber, but he takes quite a different ultralimit in order to have the
inner product survive to the limit. Here we ignore the inner product, which
can not be globally defined on $\S'$, and take a more simple-minded
ultralimit. We leave a full understanding of the precise relation between the
two constructions to future work.

\section{Preliminaries}

\subsection{Schwartz space and tempered distributions}
Let $\S$ be the Schwartz space of infinitely differentiable complex-valued
functions on $\R$ with ``rapidly decreasing absolute value'', namely
  \[ \S = \{ \phi(x) \in C^\infty(\R,\C) \;|\; \forall n,m.\; p_{n,m}(\phi) < \infty \} \]
where for $n,m \in \N$,
  \[ p_{n,m}(\phi) = \sup_x \;|\; x^n \delta^m \phi(x) \;|\; ,\]
and the topology on $\S$ is that defined by this collection of seminorms.

Recall also that the subspace $\D \subseteq  \S$ of compactly supported smooth
functions is dense in $\S$.

Let $\S'$ be the space of tempered distributions, the continuous dual space
of $\S$. Write the corresponding sesquilinear pairing as $\left<{T,\phi}\right> :=
T(\overline{\phi})$ for $T \in \S'$ and $\phi \in \S$. We have the seminorms
    \[ \| T \|_\phi := \;|\; \left<{T,\phi}\right> \;|\; \]
for $\phi \in \S$. A subset $B$ of $\S'$ is \underline{bounded} iff each $\|\cdot\|_\phi$ is
bounded on $B$. We consider bounded subsets with the topology induced from the
$\|\cdot\|_\phi$. In fact this topology coincides with that induced from the
strong topology on $\S'$, but we will not explicitly use or define the strong
topology. Closed bounded subsets of $\S'$ are compact
\cite[Chap.III,Thm.XII]{Schwartz-distributions}.

Recall [Rudin 7.12(d)] that every measurable function $f:\R \rightarrow  \C$ of
polynomial growth, i.e. bounded in absolute value by some polynomial, induces
a tempered distribution 
  \[ \left<{f,\phi}\right> := \int_\R f(x)\overline{\phi}(x) dx .\]
In particular this induces embeddings of $\S$ and $\D$ into $\S'$, which are
dense by \cite[Chap.III Thm.XV]{Schwartz-distributions}.

The space of distributions $\D'$ is the continuous dual of
$\D$. We have topological embeddings
    \[ \D \subseteq  \S \subseteq  \S' \subseteq  \D' .\]

Recall that a bilinear map $u : X\times Y \rightarrow  Z$ of topological vector spaces
is {\em hypocontinuous} if each $u(x,\cdot)$ and $u(\cdot,y)$ are continuous, and
$\{u(x,\cdot) \;|\; x \in B\}$ is equicontinuous for any bounded $B \subseteq  X$, so in
particular the restriction of the pairing to $B\times Y$ is continuous, and
similarly for $B' \subseteq  Y$ bounded.

By \cite[Chap.III,Thm.XI;Chap.VII,Sect.4]{Schwartz-distributions}, the pairings $\D \times
\D' \rightarrow  \C$ and $\S \times \S' \rightarrow  \C$ are hypocontinuous.

Let $\E$ be the space of infinitely differentiable functions $\R \rightarrow  \C$ with the
topology of compact uniform convergence of all derivatives.
By \cite[Chap.V,Thm.III]{Schwartz-distributions}, multiplication of functions and
distributions is hypocontinuous as a function $\E \times \D' \rightarrow  \D'$.

\subsection{Metric logic structure on $\S'$}
\label{ultralims+prods}
We view $\S'$ as the union of its closed bounded subsets (forming a
``bornology''). $\S'$ is not covered by any countable set of bounded
sets, nor is $\S'$ the union of a chain of bounded sets.

Each closed bounded $B \subseteq  \S'$ is compact, and moreover is a complete metric
space. In fact, a single metric suffices to metrise any $B$.
Indeed, $\S$ is separable, so let $(\psi^0_i)_{i<\omega}$ be a countable
topological basis. Concretely, we can take $\psi^0_i(x) := x^i e^{-x^2}$.
Then define on $\S'$ the metric
  \[ d(T,T') := \sum_i 2^{-i} \min(1, |\left<{T-T',\psi^0_i}\right>| ). \]
This does not metrise any relevant topology on $\S'$ itself, but it does
metrise the topology on any bounded subset:

\begin{lemma}
    Let $B \subseteq  \S'$ be closed and bounded.
    Then the restriction of $d$ to $B$ is a complete metric on $B$, metrising
    the topology of $B$.
\end{lemma}
\begin{proof}
    The topology on $B$ is induced by the pseudonorms $\|\cdot\|_\phi$ and is
    complete with respect to them, so it suffices to see
    that for any $\phi \in \S$ and any $\epsilon$, there exists $\delta$ such that
    for $T,T' \in B$,
      \[ d(T,T') < \delta  \Rightarrow   |\left<{T-T',\phi}\right>| < \epsilon .\]

    But indeed, the pairing is hypocontinuous, so $\left<{T,.}\right>$ is
    equicontinuous on $B-B$; hence for a good enough approximation to $\phi$ as
    a finite linear combination $\phi_0$ of the $\psi^0_i$,
    $|\left<{T-T',\phi-\phi_0}\right>| < \epsilon/2$ for all $T,T' \in B$. Then take
    $\delta$ such that
      \[ d(T,T') < \delta \Rightarrow   |\left<{T-T',\phi_0}\right>| < \epsilon/2 .\]
\end{proof}

Compact metric spaces are the analogue in continuous logic of finite structures
in discrete logic, and most notions from continuous logic degenerate in such
spaces. In particular, the ultraproduct of subspaces may be identified with
the topological ultralimit, as we now explain.

Consider a set $X$ and a set $\B$ of subsets which covers $X$ and is closed
under finite unions,
and suppose $d$ is a metric on $X$ such that equipping each $B \in \B$ with the
restriction of $d$ makes it a compact metric space.
The example we have in mind is $X=\S'$, with $\B$ the set of its closed
bounded subsets, and $d$ defined as above.

Call $Y \subseteq  X$ \underline{bounded} if $Y \subseteq  B$ for some $B \in \B$. 
For $Y_i \subseteq  B \in \B$ and an ultrafilter $\U$, recall $\lim_{i \rightarrow  \U} Y_i = \{
\lim_{i \rightarrow  \U} y_i \;|\; y_i \in Y_i \} \subseteq  B$. Now for $Y_i \subseteq  X$, define
$\blim_{i \rightarrow  \U} Y_i = \bigcup_{B \in \B} \lim_{i \rightarrow  \U} (Y_i \cap B)$.

For $X=\S'$, this notion of ``bounded ultralimit'' does not agree with the
usual topological ultralimit for even the strong topology on $X$, since
unbounded ultrafilters in $X$ can converge; for our purposes such convergence
is pathological, so we restrict to bounded sets.

% To see it doesn't agree: take \phi_n supported near n, and consider linear
% combinations T_i of delta functions supported on $\N$; then for an appropriate
% ultrafilter and appropriate T_i, we can have for all n that T_i(\phi_n)
% is unbounded but is bounded \U-often, and if we do it right there's no
% infinite set of i's for which T_i(\phi_n) is bounded for all n. Working in
% the strong topology helps not at all, since the \phi_n are not bounded in
% \S.

Now suppose $Y_i \subseteq  X$ are subsets such that $\blim_{i \rightarrow  \U} Y_i = X$. If
the $Y_i$ are equipped with relations
$R^{Y_i} \subseteq  Y_i^n$, define the induced relation on $X$ by $R^X := \blim_{i \rightarrow 
\U} R^{Y_i} \subseteq  X^n$. Then this notion of bounded ultralimit of structures agrees with
the notion of ultraproduct of metric structures, in the following sense.
Consider $(Y_i,R^{Y_i})$ as a metric structure in the language with a sort for
each $B \in \B$, interpreted in $Y_i$ as $Y_i \cap B$, and a real-valued
metric predicate $R'_B$ on each sort, interpreted in $Y_i$ as the distance
predicate $R'_B(x) = d(x,R^{Y_i} \cap B^n) = \inf_{y \in Y_i \cap B^n} d(x,y)$
(where we define $d$ on $X^n$ by $d(x,y) = \max_i d(x_i,y_i)$, say). Then the metric
ultraproduct is isomorphic to $(X,R^X)$ considered as a multisorted metric
structure in the same way.

We apply this to (partial) operators $X\rightarrow X$ by considering their graphs, so we define
$F^X(\lim_{i \rightarrow  \U} y_i) := \lim_{i\rightarrow \U} F^{Y_i}(y_i)$
whenever $\{y_i\}_i$ is bounded, $F^{Y_i}(y_i)$ are defined, and
$\{F^{Y_i}(y_i)\}_i$ is bounded,
if this definition yields a well-defined partial operator.

Say $f^{Y_i}$ are \underline{provably bounded} if $\{F^{Y_i}(y_i)\}_i$ is bounded
whenever $\{y_i\}_i$ is bounded and the $F^{Y_i}(y_i)$ are defined. Note that
the composition of provably bounded operators is provably bounded.

Metric logic admits a \Los theorem, but the only consequence of this we will
need is the following weak statement for exact equality of compositions of
operators, which follows directly from the above definitions.

\begin{lemma} \label{LosComp}
    Suppose $F_j^{Y_i}$ are partial operators, and $F_1^{Y_i}
    \o F_2^{Y_i} = F_3^{Y_i}$ holds $\U$-often, and suppose that the
    ultralimits $F_1^X$ and $F_2^X$ are well-defined partial operators.
    Suppose also that the $F_2^{Y_i}$ are provably bounded. Then
    $F_3^X$ is a well-defined partial operator, and $F_1^X \o F_2^X = F_3^X$.
\end{lemma}
\begin{proof}
    Restricting the index set, we may assume the equality holds for all $i$.
    Let $\{y_i\}_i$ be bounded with $y_i \in \dom(F_3^{Y_i})$. Then
    $\{F_2^{Y_i}(y_i)\}_i$ is bounded by provable boundedness, so $\lim_{i\rightarrow \U}
    F_1^{Y_i}(F_2^{Y_i}(y_i)) = F_1^X(\lim_{i\rightarrow \U}F_2^{Y_i}(y_i)) =
    F_1^X(F_2^X(\lim_{i\rightarrow \U}y_i))$.
\end{proof}

\subsection{Schr\"odinger representations}
The 3-dimensional real Heisenberg Lie algebra $\eta$ is the central extension of
the trivial Lie algebra on $\R^2$ via the canonical symplectic structure on
$\R^2$; it can be defined by saying it has basis $(P,Q,E)$, with $E$ being
central and $[Q,P]=E$. The infinitesimal Schr\"odinger representation (with
respect to $Q$) of $\eta$ on the Schwartz space $\S \subseteq  L^2(\R)$ is given by
$P\phi(x) = -\phi'(x)$, $Q\phi(x) = 2\pi i x\phi(x)$, $E\phi(x) = 2\pi i\phi(x)$.
% Not sure what technically accurate comment along these lines to make:
%Note that $\S \subseteq  L^2(\R)$ is precisely the largest subspace
%for which this representation can be defined with $P$ and $Q$ invertible
The dual representation on $\S'$ is defined by
$\left<{XT,\phi}\right> := \left<{T,-X\phi}\right>$; note that this is
consistent with the embedding $\S \subseteq  \S'$, since $\eta$ acts (formally)
skew-adjointly.

The corresponding real Heisenberg Lie group $N$ is the corresponding central
extension of $\R^2$, and can be defined as triples $(u,v,e)$ with group
structure $(u,v,e)*(u',v',e') = (u+u',v+v',e+e' + (uv'-u'v)/2)$. The corresponding
exponential map is then $\exp(pP+qQ+tE) = (p,q,t)$.
We have $[\exp(X),\exp(Y)] = \exp([X,Y])$.
Let $\chi : \R \rightarrow  \C$ be the character $\chi(x) := e^{2\pi ix}$.
The corresponding Schr\"odinger unitary representation of $N$ on $\S$ (and even
on $L^2(\R)$) is given by
\begin{align*} \exp(pP)f(x) &= f(x-p) \\
      \exp(qQ)f(x) &= \chi(qx)f(x) \\
      \exp(tE)f(x) &= \chi(t)f(x) . \end{align*}
This can be considered as the exponential of the infinitesimal
representation of $\eta$ defined above; in particular,
$\frac d{dt} \exp(tX)f(x)|_{t=0} = Xf(x)$ for $X \in \eta$ and $f \in \S$.
The dual representation on $\S'$ can be defined by
$\left<{XT,g}\right> := \left<{T,X^{-1}g}\right>$.

\section{Finite-dimensional approximations}

\subsection{Cyclic Schr\"odinger representations}
Let $N_m$ be the finite subgroup of $N$ generated by $U_m := \exp(\frac Qm)$
and $V_m := \exp(\frac Pm)$. This group is presented by the relations
$[U_m,[U_m,V_m]]=1=[V_m,[U_m,V_m]]$ expressing that $[U_m,V_m]$ is central.

Let $C_{m^2}^*$ be the space of complex-valued functions on $C_{m^2} :=
\Z/m^2\Z$. We identify $C_{m^2}^*$ with the space of $m^2$-periodic functions on
$\Z$ in the obvious way.
Define a finite dimensional ``cyclic'' unitary representation of $N_m$ on
$C_{m^2}^*$ by
 \begin{align*} U_mf(k) &:= q^kf(k) \\
    V_mf(k) &:= f(k-1) , \end{align*}
where $q := \chi(\frac 1{m^2})$.
So $[U_m,V_m]$ acts centrally as multiplication by $q$.

We will show that the infinitesimal Schr\"odinger representation of $\eta$ can
be obtained as an ultralimit of these $C_{m^2}^*$.

We identify $C_{m^2}^*$ with the subspace $\S'_m$ of $\S'$ of $m$-periodic
linear combinations of $\delta$ functions supported at $\frac 1m \Z$, via the
embedding $\theta_m$ defined by
    \[ \theta_m(f) := \sum_{k \in \Z} f(k) \delta_{\frac km} .\]

This is easily verified to be an embedding of the representation of $N_m$ on
$C_{m^2}^*$ into the Schr\"odinger representation of $N_m \leq  N$ on $\S'$ defined in
the previous section.

Define the discrete Fourier transform on $C_{m^2}^*$ by
\[ \F_m(f)(p) = \frac 1m \sum_{0 \leq  k < m^2} f(k) \chi(\frac{pk}{m^2}) .\]

Meanwhile, let $\F$ be the Fourier transform on $\S'$ 
defined by $\left<{\F T, \F \phi}\right> = \left<{T, \phi}\right>$,
where $\F$ on $\S$ is the usual Fourier transform defined by
$(\F f)(p) = \int_\R f(x) \chi(px) dx$.

\begin{lemma} \label{FourierRestrict}
    $\F_m$ agrees with the restriction of $\F$ to $\S'_m$;
    i.e. $\F(\theta_m(f)) = \theta_m(\F_m f)$.
\end{lemma}
\begin{proof}
    Let $\comb_m := \sum_{l \in \Z} \delta_{ml}$ be the Dirac comb, consider
    the shifts by convolutions with $\delta$ functions
    $\delta_{\frac km} * \comb_m = \sum_{l \in \Z} \delta_{ml+\frac km}$,
    and note that
        \[ \theta_m(f) = \sum_{0 \leq  k < m^2} f(k) (\delta_{\frac km} * \comb_m) .\]
    Meanwhile, $\F \comb_1 = \comb_1$ by Poisson summation, and similarly
    $\F \comb_m = \frac 1m \comb_{\frac 1m}$. Hence
    \begin{align*} \F \theta_m(f)
    &= \sum_{0 \leq  k < m^2} f(k) \F(\delta_{\frac km} * \comb_m) \\
    &= \sum_{0 \leq  k < m^2} f(k) \left({\left({\F \delta_{\frac km}}\right) \cdot \left({\F \comb_m}\right)}\right) \\
    &= \sum_{0 \leq  k < m^2} f(k) \left({\left({\chi \o (\frac km \cdot)}\right) \cdot \left({\frac 1m \comb_{\frac 1m}}\right)}\right) \\
    &= \frac 1m \sum_{0 \leq  k < m^2} f(k) \left({\left({\chi \o (\frac km \cdot)}\right) \cdot \left({\sum_{0 \leq  l < m^2} (\delta_{\frac lm} * \comb_m)}\right)}\right) \\
    &= \sum_{0 \leq  l < m^2} \frac 1m \sum_{0 \leq  k < m^2} f(k) \left({\left({\chi \o (\frac km \cdot)}\right) \cdot \left({\delta_{\frac lm} * \comb_m}\right)}\right) \\
    &= \sum_{0 \leq  l < m^2} \frac 1m \sum_{0 \leq  k < m^2} f(k) \left({\chi(\frac {lk}{m^2}) \cdot \left({\delta_{\frac lm} * \comb_m}\right)}\right) \\
    &= \sum_{0 \leq  l < m^2} \F_m(f)(l) \cdot \left({\delta_{\frac lm} * \comb_m}\right) \\
    &= \theta_m (\F_m f) . \end{align*}
\end{proof}

Define on $C_{m^2}^*$ approximations to the derivatives of $\exp(qQ)$ and
$\exp(pP)$:
\[ Q_m := \frac m 2 (U_m - U_m^{-1}), \]
\[ P_m := \frac m 2 (V_m - V_m^{-1}). \]

Meanwhile, consider $P$ and $Q$ as operators on $\S'$ via the
infinitesimal Schr\"odinger representation defined above.

\begin{lemma} \label{QPFlim}
    For any non-principal ultrafilter $\U$, identifying $C_{m^2}^*$ with $\S'_m$
    via $\theta_m$,
    \[ \blim_{m\rightarrow \U} (C_{m^2},P_m,Q_m,\F_m) = (\S',P,Q,\F) .\]
\end{lemma}
\begin{proof}
    \begin{itemize}\item We first show that $\blim_{m\rightarrow \U} \S'_m = \S'$. So let $T \in \S'$. By
      \cite[Chap.VI~Sect.4]{Schwartz-distributions}, $T$ is the limit of a
      sequence of smooth functions with compact support, say
      $T = \lim_{m \rightarrow  \infty} g_m$ with $g_m \in \D \subseteq  \S \subseteq  \S'$.
      Define $m$-periodic approximations $T_m \in \S'_m$ to
      $g_m$ by sampling around 0, $T_m(\frac km) := \frac 1m g_m(\frac km)$ for
      $\floor{\frac{-m^2}{2}} \leq  k < \floor{\frac{m^2}{2}}$. Then for any
      $\phi\in\S$, we have $\left<{g_m-T_m,\phi}\right> \rightarrow _{m \rightarrow  \infty} 0$. So
      $\lim_{m \rightarrow  \infty} T_m = T$. Hence $\{T_m\}_m$ is bounded, and the
      ultralimit along any non-principal ultrafilter is $T$.

    \item Next, we verify that $\blim_{m\rightarrow \U} \F_m = \F$.
      $\F$ is provably bounded and is uniformly continuous on $\S'$, since
      $\|\F T\|_\phi = \|T\|_{\F^{-1}\phi}$. So if $T_m \in \S'_m$ are bounded,
      so are $\F T_m$, and by Lemma~\ref{FourierRestrict},
      $\lim_{m\rightarrow \U}\F_mT_m = \lim_{m\rightarrow \U}\F T_m = \F \lim_{m\rightarrow \U} T_m$.

    \item To show $\blim_{m\rightarrow \U} Q_m = Q$,
      we must show that if $T = \lim_{m \rightarrow  \U} T_m$ with
      $\{T_m\}_m$ bounded, then $\{Q_mT_m\}_m$ is
      bounded and $\lim_{m \rightarrow  \U} Q_mT_m = (2\pi ix)\cdot T$,
      i.e. for any $\phi = \phi(x) \in \S$,
        \[ \lim_{m \rightarrow  \U} \left<{Q_mT_m, \phi}\right> =
          \lim_{m \rightarrow  \U} \left<{T_m, -2\pi ix\phi}\right> .\]

      From the definition of $Q_m$, we have
      $\left<{Q_mT_m, \phi}\right> = \left<{T_m, -mi\sin(2\pi x/m)\phi}\right>$.
      Now $\lim_{m \rightarrow  \infty} -mi\sin(2\pi x/m)\phi = -2\pi ix\phi$ in $\S$,
      since $\lim_{m \rightarrow  \infty} -mi\sin(2\pi x/m) = -2\pi ix$ in $\E$ and multiplication
      is hypocontinuous.
      So by hypocontinuity of the pairing,
      $\lim_{m \rightarrow  \infty} \left({\left<{Q_mT_m, \phi}\right> - \left<{T_m, -2\pi ix\phi}\right>}\right) = 0$.
      This implies provable boundedness and that the ultralimit is as required.

    \item We verify $\blim_{m\rightarrow \U} P_m = P$ by appeal to Lemma~\ref{LosComp}.
        One can check by direct calculation that on each $\S'_m$ we have
        $V_m = \F_m^{-1} \o U_m \o \F_m$, and hence by linearity
        $P_m = (\F_m^{-1} \o Q_m \o \F_m)$.

        Meanwhile, on $\S'$ we have $\ddx = \F^{-1} \o (-2\pi i x \cdot) \o \F$.

        We have already verified provable boundedness of the $\F_m$ and $Q_m$,
        so we conclude by Lemma~\ref{LosComp}.
    \end{itemize}
\end{proof}

\subsection{Weil representations}

$\SL_2(\R) \cong  \Sp(2,\R)$ acts by automorphisms on $\eta$ and on $N$, acting
trivially on the centre and via the canonical action on $\R^2$. We will write
the action in exponential notation, $X^g$.

Via the Stone-von Neumann theorem, this induces a projective
unitary representation on $S$, which turns out to yield a ``metaplectic''
unitary representation of the double cover $\Mp_2(\R)$ of $\SL_2(\R)$ on $\S$,
such that $X^g(\phi) = (\widetilde{g} \o X \o \widetilde{g}^{-1})(\phi)$ for $\widetilde{g} \in \Mp_2(\R)$, $g$
the image in $\SL_2(\R)$, $X \in \eta$, and $\phi \in \S$.
The dual representation on $\S'$ is then defined by demanding formal
unitarity, $\left<{\widetilde{g}T,\widetilde{g}\phi}\right> := \left<{T,\phi}\right>$.

$\SL_2(\R)$ is generated by $F := \minimat{0}{-1}{1}{0}$ and $G^b :=
\minimat{1}{b}{0}{1}$ for $b\in\R$, so to define the metaplectic
representation it suffices to define it on the lifts of these generators to
$\Mp_2(\R)$. Following \cite[p.179]{Folland-phase}, we have for $\phi \in \S$
\begin{align*}
  \widetilde{F} \phi(x) &= \pm \chi(\frac 18)\F \phi(x) \\
  \widetilde{G}^b \phi(x) &= \pm \chi(-b\frac{x^2}2)\cdot \phi(x) ,\end{align*}
where the signs depend on the choice of lift.

% TODO: could try to be more explicit? Need a nice way to understand the group
% structure of $\Mp_2(\R)$\ldots

We have seen that $\F$, and hence the action of $\widetilde{F}$ on $\S'$, restricts to an
operator on each $\S'_m$. The action of $\widetilde{G}^b$ on $\S'$ is by multiplication
by $\pm \chi(b\frac{x^2}2)$. For $b=\frac cd$ a rational number expressed in
lowest terms, this restricts via $\theta_m$ to a partial operator $\widetilde{G}^b_m$
on $\S'_m$ if $d$ divides $m^2$, with domain $D_{m,b}$ consisting of
$m$-periodic linear combinations of $\delta$ functions supported on $\frac dm
\Z$.

In terms of $C_{m^2}^*$, $\widetilde{G}^b_m$ acts by
$\widetilde{G}^b_m(f)(k) = \pm \chi(b\frac{k^2}{2m^2})\cdot f(k)$, defined on
$d\Z / m^2\Z \subseteq  \Z / m^2\Z$.

Call an ultrafilter $\U$ on $\N$ \underline{ultradivisible} if for all $d \in \N$, $d\N
\in \U$.

\begin{lemma} \label{domLimAll}
    For $\U$ ultradivisible,
    $\blim_{m\rightarrow \U} D_{m,b} = \S'$.
\end{lemma}
\begin{proof}
    If $d|m$,
    \[ D_{m,d} \supseteq  \S'_{m/d} ,\]
    so the result follows from ultradivisibility and Lemma~\ref{QPFlim}.
\end{proof}

It follows fairly immediately that for $b \in \Q$ and $\U$ ultradivisible,
$\blim_{m\rightarrow \U} \widetilde{G}^b_m = \widetilde{G}^b$.
But for $b \in \R \setminus \Q$, simply restricting to $\S'_m$ does not yield a
nontrivial partial operator. Rather, we must simultaneously approximate $b$ by
rationals.

So given $m\in\N$, write $m = e^2f$ where $e \in \N$ and $f$ is a squarefree
integer, and define $b_m := \frac {\floor{be}}e$. So for $\U$ ultradivisible,
$\lim_{m\rightarrow \U} b_m = b$. Then define $\widetilde{G}^b_m := \widetilde{G}^{b_m}_m$.

\begin{lemma} \label{metaRepG}
    Let $b \in \R$. Then the $\widetilde{G}^b_m$ are provably bounded on the $\S'_m$, and
    for $\U$ ultradivisible, $\blim_{m\rightarrow \U} \widetilde{G}^{b_m} = \widetilde{G}^b$.
\end{lemma}
\begin{proof}
    First we prove provable boundedness. So suppose $T_m \in D_{m,b_m}$ are
    bounded, let $\phi \in \S$, and let $\epsilon > 0$. Then by hypocontinuity
    of the pairing, for all but finitely many values of $b_m$,
    \begin{align*} \left|{\left<{\widetilde{G}^{b_m}T_m, \phi}\right> - \left<{T_m, \widetilde{G}^{-b}\phi}\right>}\right|
      &= \left|{\left<{T_m, \widetilde{G}^{-b_m}\phi - \widetilde{G}^{-b}\phi}\right>}\right| \\
      &= \left|{\left<{T_m, \pm \chi((b-b_m)\frac{x^2}2)\phi(x)}\right>}\right| \\
      &< \epsilon .\end{align*}
    So since $\left<{T_m, \widetilde{G}^{-b}\phi}\right>$ is bounded,
    also $\left<{\widetilde{G}^{b_m}T_m, \phi}\right>$ is bounded.

    Now we consider the ultralimit.

    If $e^2|m$, we have $D_{m,b_m} = \S'_e$. So as in
    Lemma~\ref{domLimAll}, $\blim_{m\rightarrow \U} D_{m,b_m} = \S'$.

    So let $T \in \S'$; then we may write $T = \lim_{m\rightarrow \U} T_m$ with
    $T_m \in D_{m,b_m}$ and with the $T_m$ bounded.

    Now since $\lim_{m\rightarrow \U} b_m = b$,
    $\lim_{m\rightarrow \U} \chi(b_m\frac{x^2}2) = \chi(b\frac{x^2}2)$ in $\E$.

    Then by hypocontinuity of multiplication,
    \[ \lim_{m\rightarrow \U} \widetilde{G}^b_m T_m
      = \lim_{m\rightarrow \U} \pm \chi(b_m\frac{x^2}2)\cdot T_m
      = \pm \chi(b\frac{x^2}2)\cdot T
      = \widetilde{G}^b T ,\]
    as required.
\end{proof}
    
We have already seen the analogous result for $\F$, and hence for $\widetilde{F}$.
Now for $\widetilde{h} \in \Mp_2(\R)$, write $\widetilde{h}$ as a word in the generators
$\widetilde{h}=w(\widetilde{F},\widetilde{G}^{b_1},\ldots,\widetilde{G}^{b_n})$ and define $\widetilde{h}_m$ as the corresponding word
$\widetilde{h}_m=w(\widetilde{F}_m,\widetilde{G}^{b_1}_m,\ldots,\widetilde{G}^{b_n}_m)$.
Then by the provable boundedness and Lemma~\ref{LosComp},
$\blim_{m\rightarrow \U} \widetilde{h}_m = \widetilde{h}$. Adding this to Lemma~\ref{QPFlim}, we conclude:

\begin{theorem} \label{main}
    For $\U$ ultradivisible,
    \[ \blim_{m\rightarrow \U}(\S'_m,Q_m,P_m,(\widetilde{h}_m)_{h\in\Mp_2(\R)}) = 
    (\S',Q,P,(\widetilde{h})_{\widetilde{h}\in\Mp_2(\R)}) ,\]
    where the latter structure encodes the standard Schr\"odinger and Weil
    representations on $\S'$ defined above.
\end{theorem}

% ## Further directions
% 
% # Inner product?
% 
% Ill-defined in general (e.g. delta paired with heaviside), but
% sometimes well-defined. No one choice of normalisation will suffice if we want
% finite values (e.g. delta with delta vs. delta with $\chi(px)$).
% Boris: restrict to lagrangian cross lagrangian, with appropriate
% normalisation.
% Worth considering generalisations of this?
% 
% Mention [HH], who seem to want to get a global inner product.
% 
% # Trace
% 
% If the trace of the representation of $\widetilde{h}\in\Mp_2(\R)$ is defined as
% in mathematical physics (namely what??), can we confirm that the trace is
% preserved by the ultraproduct, generalising last section of [Z]?
% 
% # Feynmann integrals
% 
% Can we? Should we?
%
% # Algebra to analysis
%
% Fourier inversion.
%
% Gaussian integrals via Gauss sums?

\bibliography{ultrarigging}
\end{document}

\section{TODO: Dirac calculus}
One can try to define an inner product on $\S'$ extending that on $L^2(\R)$,
induced by the standard inner products on the $\S'_m$. Is there a way to make
sense of this?

\section{Old stuff and offcuts}

\subsection{Uniform logic}

{\it All we actually need of uniform logic is the definition of the
ultraproduct and the fact that the ultraproduct commutes with composition of
provably uniformly continuous provably bounded maps.
Possibly we should just explain that rather than treating uniform logic in
full generality.}

Syntax: pseudometrics OR system of entourages; functions and real-valued
relations, coming with proofs of uniform continuity and boundedness. Fix a
basepoint. The uniform structure induces a system of (closed) bounded sets
("bornology"). Quantifiers are over bounded sets (presumably?).

Ultraproducts: on each bounded, take discrete ultraproduct then quotient by
all pseudometrics being infinitesimal, i.e. by the $\Wedge$-definable
equivalence defined by the entourages; take union over the boundeds.

Remark: if a single pseudometric suffices to define the uniformity (equiv,
countably many do), then we can equivalently work on the whole space, but keep
only those points in the ultraproduct with finite distance from the base
point.

\begin{lemma}
   \L{}os.
\end{lemma}
\begin{proof}
    TODO
\end{proof}

The following lemma gives the analogue of finite structures in this setting.
\begin{lemma} \label{HB}
    Suppose $X$ is Hausdorff and has the Heine-Borel property that every
    closed bounded subset is compact. Then any ultrapower of $X$ is isomorphic
    to $X$.
\end{lemma}
\begin{proof}
    If $B \subseteq  X$ is closed and bounded, any ultrafilter on $B$ converges
    uniquely in $B$. This defines a map from the product to $X$ which induces
    a well-defined homeomorphism between the ultraproduct and $X$.

    Then one must verify that the functions and relations behave properly,
    which they will if everything's defined right\ldots TODO.
\end{proof}

\subsection{blah}
Fix $h \in \R$, and let $\hbar := 2\pi h$.

For $m\in \N$, let $\S'_m \subseteq  \S'$ be the space of $m$-periodic
linear combinations of Dirac delta distributions supported on $\frac \Z m$.
Explicitly, $\S'_m$ has $\C$-basis
$\{ \sum_{k\in\Z} \delta_{mk+\frac i m} \;|\; i \in \{1,\ldots,m^2\} \}$.

When $f\in\S'_m$, we sometimes (abusively) write $f(x)$ for the complex
number which is the multiple of $\delta_x$ appearing in $f$.

Let $^*\S'$ be a non-principal continuous ultraproduct of the
$(\S'_m)_{m\in\N}$ with respect to the seminorms $\|\cdot\|_\phi$.

By definition, this may be obtained by taking the union over bounded subsets
$B$ of the discrete ultraproduct on $B$, and then quotienting by the
equivalence relation
    \[ \forall \phi \in \S. \st(\|f\|_\phi - \|g\|_\phi) = 0, \]
where $\st$ is the usual standard part map on finite non-standard reals.

\begin{lemma}
    $^*\S'$ is isomorphic to $S'$ as a uniform structure in the language of
    the $\|\cdot\|_\phi$.
\end{lemma}
\begin{proof}
    By Lemma~\ref{HB}, the embeddings of $\S'_m$ into $\S'$ induce an
    embedding of $^*\S'$ into $\S'$. It remains to show that this embedding is
    surjective.

    Let $f \in \S'$. By \cite[Chap.VI~Sect.4]{Schwartz-distributions}, $f$ is the limit of a
    sequence of smooth functions with compact support, say
    $f = \lim_{m \rightarrow  \infty} g_m$. Define approximations $f_m\in\S'_m$ to
    $g_m$ by $f_m(\frac km) := \int_{km}^{(k+1)m} g_m(x)\dx$ for
    $-\floor{\frac{m^2}{2}} \leq  k < \floor{\frac{m^2}{2}})$. Then for any
    $\phi\in\S$, we have $\left<{g_m-f_m,\phi}\right> \rightarrow _{m \rightarrow  \infty} 0$. So
    $\lim_{m \rightarrow  \infty} f_m = f$. Hence $\{f_m\}_m$ is bounded, and the
    ultralimit along any non-principal ultrafilter is $f$.
\end{proof}

Our aim now is to prove that some of the conventional structure on $\S'$
used in quantum mechanics arises as the ultraproduct of corresponding
structure on the $\S'_m$.

On $\S'_m$ define the unitary operators
    \[ ^U(f)(x) = \chi(\frac x m) f(x) \]
    \[ ^V(f)(x) = f(x - \frac 1m), \]
and the self-adjoint operators
    \[ Q := \frac m {2i} (^U - ^U^{-1}) \]
    \[ P := h \frac m {2i} (^V - ^V^{-1}). \]

Note that ${^U}{^V}{^U}^{-1}{^V}^{-1} = \chi(\frac 1{m^2}) =: q_m$.

Define the (discrete) Fourier transform on $\S'_m$ by
\[ \F(f)(p) = \frac 1m \sum_{n=1}^{m^2} \chi(p \frac nm) f(\frac nm) .
\]

Note that $\F ^V = ^U \F$, $\F ^U = ^V^{-1} \F$, and $\F^2 = \I$ where
$\I(f)(x) := f(-x)$.

\begin{lemma}
    On the ultraproduct identified with $\S'$,
    \[ Qf(x) = 2\pi xf(x) .\]
\end{lemma}
\begin{proof}
    By definitions, we must check that if $f = \lim_{m \rightarrow  \U} f_m$ with
    $\{f_m\}_m$ bounded, and $\phi = \phi(x) \in \S$, then 
    $\lim_{m \rightarrow  \U} \left<{Qf_m, \phi}\right> = \lim_{m \rightarrow  \U} \left<{f_m, 2\pi x\phi}\right>$.

    From the definition of $Q$ on $\S'_m$, we have
    $\left<{Qf_m, \phi}\right> = \left<{f_m, m\sin(2\pi x/m)\phi}\right>$.
    Now $\lim_{m \rightarrow  \infty} m\sin(2\pi x/m)\phi = 2\pi x\phi$ in $\S$,
    since $\lim_{m \rightarrow  \infty} m\sin(2\pi x/m) = 2\pi x$ in $\E$ and multiplication
    is hypocontinuous.
%    So by boundedness of $(f_m)_m$ and hypocontinuity of the pairing,
%    $\left<{f_m, m\sin(2\pi x/m)\phi - x\phi}\right>$ converges uniformly to 0.
%    Hence $\left<{Qf_m, \phi}\right> - \left<{f_m, x\phi}\right>$ is bounded with $m$.
%    Hence $(Qf_m)_m$ is bounded.
    So we conclude by hypocontinuity of the pairing.
\end{proof}


\begin{lemma}
    \begin{enumerate}[(i)]\item $Pf(x) = -i\hbar f'(x)$;
    \item $[P,Q] = i\hbar$
    \end{enumerate}
\end{lemma}
\begin{proof}
    These follow from the previous Lemma and the \L{}os theorem,
    since $P$ is defined as $P = h \F^{-1} Q \F$, and $[P,Q] = PQ - QP$.
\end{proof}

\subsection{Feynman integrals}


\subsection{Metaplectic nonsense}

The metaplectic group $\Mp_2(\R)$ can be represented as M\"obius transformations
with a choice of square root, as follows: elements are of the form
$(\frac{az+b}{cz+d},\sqrt(cz+d))$ where $\minimat{a}{b}{c}{d} \in \SL_2(\R)$
and $\sqrt(cz+d)$ is a choice of holomorphic square root on the upper half
plane, and the product is given by $(m,s)*(m',s') = (m \o m', (s \o m')*s')$.

Now $\SL_2(\R)$ is generated by $F := \minimat{0}{1}{-1}{0} \approx \frac{1}{-z}$
and $G_b := \minimat{1}{b}{0}{1} \approx \frac{z+b}{1}$ for $b\in\R$, so to define
the metaplectic representation it suffices to define it on the lifts of these
generators to $\Mp_2(\R)$. Following \cite[p.179]{Folland-phase}, and
choosing $+\sqrt{-z}$ such that $+\sqrt{-i} = e^{\frac{-pi i}{4}}$, we define
this as follows:
\begin{align*} (F,\pm \sqrt{-z}) f(x) &= \pm \chi(\frac 18)\F^{-1}f(x) \setminus
    (G_b,\pm 1) f(x) &= \pm  e^{-bx^2}\cdot f(x) ,\end{align*}

Hold on - this representation of $\Mp_2(\R)$ (lifted from wikipedia) appears
to be nonsense.

TODO: check, and fix wikipedia as necessary!


\subsection{TODO:}
    What structure on $\S'$ is pseudofinite?

    Feynmann integrals as ultralimits of finite (Gauss) sums.

    Generalities on pseudometric structures, including a \L{}os theorem.

%Note:
%    Came across this paper
%    \url{http://iopscience.iop.org/1742-6596/306/1/012026/pdf/1742-6596_306_1_012026.pdf}
%    whose basic setup is similar to the one here (but the definition of $Q$ is
%    a bit different, so the finite-dimensional structures aren't
%    quite representations of the Weyl algebra).
\end{document}
