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% Brian C. Hall, bhall@math.ucsd.edu
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\begin{document}
\author{Brian C. Hall}
\address{University of California, San Diego\\
Department of Mathematics-0112\\
La Jolla, CA 92093-0112 U.S.A.\\
bhall@math.ucsd.edu}
\title{Yang-Mills theory and the Segal-Bargmann transform}
\date{November 18, 1997}
%\maketitle
\begin{abstract}
Motivated by the results of Landsman and Wren, I consider in this paper the
canonical quantization of Yang-Mills theory on a spacetime cylinder, with
the goal of obtaining an appropriate Segal-Bargmann transform. Using a
Gaussian measure with large variance to approximate the non-existent
Lebesgue measure on the space of connections, I consider a variant of the
classical Segal-Bargmann transform. A simple but non-rigorous argument shows
that this classical transform, restricted to the gauge-invariant subspace,
becomes a variant of the generalized Segal-Bargmann transform \cite{H1} for
the structure group $K.$ In the infinite-variance limit this yields the
invariant form of the Segal-Bargmann transform for $K,$ which may therefore
be regarded as the natural Segal-Bargmann transform for the space of
connections modulo gauge transformations.
As a consequence of this calculation, I obtain a new version of the
generalized Segal-Bargmann transform for Lie groups of compact type, which
interpolates between the two previously known versions. Although the
infinite-dimensional calculation is non-rigorous, the properties of the
predicted finite-dimensional transform may be verified rigorously by
finite-dimensional means.
\end{abstract}
\maketitle
\newtheorem{theorem}{Theorem}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{conjecture}[theorem]{Conjecture}
\let\Bbb = \mathbb
\let\frak = \mathfrak
\let\func = \text
\section{Introduction\label{intro.sec}}
The Segal-Bargmann transform was developed independently in the early 1960's
by Segal \cite{S1,S2,S3} in the infinite-dimensional context of scalar
quantum field theories and by Bargmann \cite{B} in the finite-dimensional
context of quantum mechanics on $\Bbb{R}^{n}.$ In \cite{H1} I introduced an
analog of the Segal-Bargmann transform in the context of quantum mechanics
on a compact Lie group. A natural next step is to attempt to combine the
compact group with the field theory in order to obtain a transform in the
context of quantum gauge theories. One such transform has already been
obtained by Ashtekar, \textit{et al} \cite{A}, with application to quantum
gravity.
In this paper I consider the canonical quantization of Yang-Mills theory on
a spacetime cylinder. The space of connections (on the spatial circle) is an
infinite-dimensional linear space, and in this special case the set of
connections modulo gauge transformations is a finite-dimensional set,
essentially the structure group $K.$ I argue heuristically that (a variant
of) the \textit{classical} Segal-Bargmann transform for the space of
connections, when restricted to the gauge-invariant subspace, becomes the
\textit{generalized} Segal-Bargmann transform for $K.$ Thus the transform
for $K$ is not just analogous to the classical transform, it \textit{is} the
classical transform on the gauge-invariant subspace. A forthcoming paper
\cite{AHS} with Albeverio and Sengupta will consider a Segal-Bargmann
transform for the Euclidean approach to the quantization of two-dimensional
Yang-Mills theory.
The motivation for this work comes from two sources. The first is the work
of Landsman and Wren on the quantization of Yang-Mills on a cylinder, by the
method of ``Rieffel induction.'' Their calculations show that the classical
coherent states for the space of connections correspond after reduction to
the coherent states for the generalized transform on $K.$ My approach to the
quantization is quite different from theirs, but yields similar conclusions.
The other source of motivation is the work of Gross and Malliavin, who also
obtain a version of the generalized Segal-Bargmann transform for $K$ from
the infinite-dimensional classical transform. Although the case they
consider does not admit an interpretation in terms of Yang-Mills theory, the
calculations in this paper are similar to the ones there. See also \cite{HS}
for an extension of the Gross-Malliavin method in a different direction.
The infinite-dimensional calculations yield not only a new interpretation of
the Segal-Bargmann transform for $K$ (as the transform for the space of
connections modulo gauge transformations), but also a new version of that
transform. The properties of this new transform that are redicted by the
infinite-dimensional argument may then be proved rigorously by
finite-dimensional means.
The Segal-Bargmann transform for $\Bbb{R}^{n}$ may be described as follows.
For $t>0,$ consider the density
\[
\rho _{t}\left( x\right) =\left( 2\pi t\right) ^{-n/2}e^{-x^{2}/2t},
\]
where $x^{2}=x_{1}^{2}+\cdots +x_{n}^{2}.$ Then define
\[
B_{t}:L^{2}\left( \Bbb{R}^{n},\rho _{t}\left( x\right) \,dx\right)
\rightarrow \mathcal{H}\left( \Bbb{C}^{n}\right)
\]
by
\[
B_{t}f\left( z\right) =\int_{\Bbb{R}^{n}}\rho _{t}\left( z-x\right) f\left(
x\right) \,dx,\quad z\in \Bbb{C}^{n}.
\]
Here $\mathcal{H}\left( \Bbb{C}^{n}\right) $ denotes the space of
holomorphic functions on $\Bbb{C}^{n}.$ Note that since $\rho _{t}$ admits
an analytic continuation to $\Bbb{C}^{n},$ it makes sense to allow $z\in
\Bbb{C}^{n}.$ Note also that $\rho _{t}$ is the fundamental solution at the
identity of the heat equation $\partial u/\partial t=1/2\,\Delta u.$ Thus
\[
B_{t}f=\text{ analytic continuation of }e^{t\Delta /2}\left( f\right) .
\]
There is another version $C_{t}$ of the Segal-Bargmann transform whose
domain is the Hilbert space $L^{2}\left( \Bbb{R}^{n},dx\right) ,$ but which
is given by the same formula as $B_{t}$:
\[
C_{t}f=\text{ analytic continuation of }e^{t\Delta /2}\left( f\right) .
\]
Because we will be concerned with isometry properties of these transforms,
it is desirable to give them different names to remind us of the different
domain Hilbert spaces.
Both versions of the Segal-Bargmann transform are isometric maps onto
certain $L^{2}$-spaces of holomorphic functions, as proved in a slightly
different form by Bargmann \cite{B}. Segal independently obtained similar
results for the infinite-dimensional theory \cite{S1,S2,S3}.
\begin{theorem}[Bargmann, 1961]
For each $t>0$, the map $B_{t}$ is an isometric isomorphism of $L^{2}\left(
\Bbb{R}^{n},\rho _{t}\left( x\right) \,dx\right) $ onto $\mathcal{H}\left(
\Bbb{C}^{n}\right) \cap L^{2}\left( \Bbb{C}^{n},\mu _{t}\left( z\right)
\,dz\right) $, where
\[
\mu _{t}\left( z\right) =\left( \pi t\right) ^{-n}\exp \left( -\left|
z\right| ^{2}/t\right) ,
\]
and where $dz$ denotes $2n$-dimensional Lebesgue measure on $\Bbb{C}^{n}$.
For each $t>0$, the map $C_{t}$ is an isometric isomorphism of $L^{2}\left(
\Bbb{R}^{n},dx\right) $ onto $\mathcal{H}\left( \Bbb{C}^{n}\right) \cap
L^{2}\left( \Bbb{C}^{n},\nu _{t}\left( z\right) \,dz\right) $, where
\[
\nu _{t}\left( z\right) =\left( \pi t\right) ^{-n/2}\exp \left( -\left(
\func{Im}z\right) ^{2}/t\right) .
\]
\end{theorem}
The properties of either version of the transform may be obtained from the
other by simple changes of variable. Thus in this the $\Bbb{R}^{n}$ case,
the maps $B_{t}$ and $C_{t}$ should not be regarded as different transforms,
but rather as different normalizations of the same transform. The difference
between the two may be regarded as coming from different normalizations of
the ``coherent states.'' (See the appendix of \cite{H1}.) The $B_{t}$
normalization allows easy passage to the infinite-dimensional ($n\rightarrow
\infty $) limit.
Both versions of the Segal-Bargmann transform may be generalized to the
setting of compact Lie groups \cite{H1}. (See Section \ref{main.sec} for
details.) Let $K$ be a connected compact Lie group. Let $\rho _{t}\left(
x\right) $ denote the fundamental solution at the identity of the heat
equation $\partial u/\partial t=1/2\,\Delta _{K}u$. Let $K_{\Bbb{C}}$ denote
the complexification of $K,$ which is a certain complex Lie group containing
$K$ as a subgroup. It turns out that $\rho _{t}$ has a unique analytic
continuation from $K$ to $K_{\Bbb{C}}.$ Thus we may define, by analogy to
the classical case
\[
B_{t}:L^{2}\left( K,\rho _{t}\left( x\right) \,dx\right) \rightarrow
\mathcal{H}\left( K_{\Bbb{C}}\right)
\]
by
\[
B_{t}f\left( g\right) =\int_{K}\rho _{t}\left( gx^{-1}\right) f\left(
x\right) \,dx,\quad g\in K_{\Bbb{C}}.
\]
Here $dx$ denotes Haar measure on $K.$ As in $\Bbb{R}^{n},$ $B_{t}f$ is the
analytic continuation of $e^{t\Delta _{K}/2}\left( f\right) .$ We define $%
C_{t}$ by the same formula, but on the Hilbert space $L^{2}\left(
K,dx\right) .$
\begin{theorem}[Hall, 1994]
For each $t>0,$ the map $B_{t}$ is an isometric isomorphism of $L^{2}\left(
K,\rho _{t}\left( x\right) \,dx\right) $ onto $\mathcal{H}\left( K_{\Bbb{C}%
}\right) \cap L^{2}\left( K_{\Bbb{C}},\mu _{t}\left( g\right) \,dg\right) ,$
where $\mu _{t}$ is a suitably defined heat kernel on $K_{\Bbb{C}},$ and
where $dg$ denotes Haar measure on $K_{\Bbb{C}}.$
For each $t>0,$ the map $C_{t}$ is an isometric isomorphism of $L^{2}\left(
K,\,dx\right) $ onto $\mathcal{H}\left( K_{\Bbb{C}}\right) \cap L^{2}\left(
K_{\Bbb{C}},\nu _{t}\left( g\right) \,dg\right) ,$ where $\nu _{t}\left(
g\right) $ is given by $\nu _{t}\left( g\right) =\int_{K}\mu _{t}\left(
gx\right) \,dx.$
\end{theorem}
One difference between the situation for $\Bbb{R}^{n}$ and for $K$ is that
on $K$ the two transforms are inequivalent. That is, there is no simple
change of variables that will convert one form to the other; if you prefer,
they are defined in terms of inequivalent sets of coherent states. (See
again the appendix of \cite{H1}.) This generalized Segal-Bargmann transform
may be regarded as providing a phase space representation for quantum
mechanics of a particle with configuration space $K,$ a view developed in
\cite{H2,H3}. See \cite{H4} for exposition and a survey of related works.
This paper introduces a new version $B_{s,t}$ of the Segal-Bargmann
transform for $K,$ which interpolates continuously between the two
previously known versions. In all cases the transform $B_{s,t}$ itself
consists of $e^{t\Delta _{K}/2}$ followed by analytic continuation, but the
domain will be an $s$-dependent Hilbert space, namely, $L^{2}\left( K,\rho
_{s}\left( x\right) \,dx\right) .$ The result is the following.
\begin{theorem}
\label{intro.thm}For each $s,t>0$ satisfying $s>t/2,$ the map $B_{s,t}$
given by
\[
B_{s,t}f=\text{ analytic continuation to }K_{\Bbb{C}}\text{ of }e^{t\Delta
_{K}/2}\left( f\right)
\]
is an isometric isomorphism of $L^{2}\left( K,\rho _{s}\left( x\right)
\,dx\right) $ onto $\mathcal{H}\left( K_{\Bbb{C}}\right) \cap L^{2}\left( K_{%
\Bbb{C}},\mu _{s,t}\left( g\right) \,dg\right) ,$ where $\mu _{s,t}\left(
g\right) $ is a suitable heat kernel on $K_{\Bbb{C}},$ and where $dg$
denotes Haar measure on $K_{\Bbb{C}}.$
\end{theorem}
In Section \ref{ym.sec} I consider the quantization of Yang-Mills on a
cylinder, and explain how this yields, at least heuristically, the transform
$B_{s,t}$ on the gauge-invariant subspace. Then in Section \ref{main.sec} I
give a rigorous finite-dimensional proof of Theorem \ref{intro.thm}. See
Theorem \ref{main.thm} for a more precise statement.
In the case $s=t,$ the heat kernel $\mu _{s,t}=\mu _{t,t}$ coincides with
the heat kernel $\mu _{t}$ of \cite{H1}, and so in this case we have simply
the transform $B_{t}.$ In the limit $s\rightarrow \infty ,$ $\rho _{s}\left(
x\right) $ converges to the constant function 1, and the heat kernel $\mu
_{s,t}$ converges to the ``$K$-averaged'' heat kernel $\nu _{t}.$ So we
obtain in this limit the transform $C_{t}.$ I will also consider the
opposite extreme, $s\rightarrow t/2.$ The transform obtained in this case
may be regarded as a generalization of the Fourier-Wiener transform. See
Section \ref{limit.sec} for the limiting cases.
Theorem \ref{intro.thm} can (and will) be extended to the case of Lie groups
of compact type, a class which includes both compact groups and $\Bbb{R}^{n}$
(Section \ref{main.sec}). This extension allows the classical case and the
compact case to be treated in a unified way. For the results of \cite{H1},
this extension was given by Driver \cite{D}. In the case $K=\Bbb{R}^{n},$
the transform $B_{s,t}$ may be converted into the standard Segal-Bargmann
transform $B_{t}$ by simple changes of variable. See Case 2 of the proof of
Theorem \ref{main.thm}.
It is a pleasure to acknowledge useful discussions with Bruce Driver, Klaas
Landsman, and Ken Wren.
\section{Canonical quantization of Yang-Mills on a cylinder\label{ym.sec}}
In \cite{LW,W1,W2}, N. P. Landsman and K. K. Wren take on the problem of
canonically quantizing Yang-Mills theory on a cylinder, using the method of
``Rieffel induction.'' They impose gauge symmetry after quantization, by
means of a certain integral over the gauge group with respect to Wiener
measure. The unreduced Hilbert space is just an $L^{2}$ space over the set
of Lie algebra-valued 1-forms with a Gaussian (white noise) measure. Wren's
calculations \cite{W1,W2} identify the reduced Hilbert space with $%
L^{2}\left( K,dx\right) .$ (Wren considers at first only ``based'' gauge
transformations; if all gauge transformations are considered, then one would
restrict to the space of Ad-$K$-invariant functions in $L^{2}\left(
K,dx\right) .$ )
Wren further shows that under the reduction map the usual Gaussian-measure
``coherent states'' map to those coherent states in $L^{2}\left( K,dx\right)
$ that are associated to the transform $C_{t}.$ Here ``coherent state'' may
be understood as the linear functional that maps an element $f$ of a Hilbert
space to the Segal-Bargmann transform of $f$ evaluated at a point. Usually
this is a continuous linear functional, which may therefore be identified
with an element of the original Hilbert space. In the case of the transform $%
C_{t}$ the coherent states are the functions $\psi _{g}\left( x\right) =%
\overline{\rho _{t}\left( gx^{-1}\right) },$ which satisfy $C_{t}f\left(
g\right) =\left\langle \psi _{g},f\right\rangle _{L^{2}\left( K,dx\right) }.$
Thus we may say that under Landsman and Wren's reduction map, the classical
Segal-Bargmann transform for an infinite-dimensional linear space reduces to
the transform $C_{t}$ for the finite-dimensional non-linear manifold $K.$
These results raise several interesting questions, among them: 1) Is there a
geometric explanation for the occurrence of the heat equation on $K,$ and 2)
Can the measure $\nu _{t}$ on $K_{\Bbb{C}}$ be ``predicted'' by means of
infinite-dimensional analysis? Note in regard to 2) that it is not clear
what happens to the isometry properties of the Segal-Bargmann transform
after reduction. So the isometricity of $C_{t}$ appears not to be a
consequence of the calculations in \cite{W1,W2}.
In this section I take a different approach to the quantization problem, but
with the goal of obtaining similar results. I use a Gaussian measure with
large variance $s$ to approximate the non-existent Lebesgue measure on the
space of 1-forms. I then consider an appropriate version of the classical
Segal-Bargmann transform, given (of course) by the heat equation for an
infinite-dimensional linear space, followed by analytic continuation. I show
heuristically that when restricted to the gauge-invariant subspace, this
reduces to the heat equation for $K,$ followed by analytic continuation. If
the heuristics are correct, then the transform $B_{s,t}$ and its isometry
properties follow from properties of the classical transform. Thus in the
physically-relevant limit $s\rightarrow \infty ,$ we obtain the isometric
transform $C_{t}$. This way of thinking about the transform $C_{t}$ is
similar to the work of Gross and Malliavin \cite{GM} and Hall and Sengupta
\cite{HS}, except that those papers consider only the $B_{t}$ version of the
transform, that is, only the case $s=t.$
Let $K$ be a compact connected Lie group with Lie algebra $\frak{k},$ and
fix an Ad-$K$-invariant inner product on $\frak{k}.$ Think of the circle as
the interval $\left[ 0,1\right] $ with ends identified. The circle is to be
thought of as ``space,'' so that ``spacetime'' is a cylinder. The
unconstrained configuration space is the space $\mathcal{A}$ of $\frak{k}$%
-valued, square-integrable functions on $\left[ 0,1\right] .$ These are to
be thought of as $\frak{k}$-valued 1-forms. Let $W\left( K\right) $ denote
the space of continuous $K$-valued paths on $\left[ 0,1\right] ,$ starting
at the identity. The It\^{o} map $\theta $ maps $\mathcal{A}$ into $W\left(
K\right) $ by ``integrating'' a $\frak{k}$-valued 1-form $A$ from 0 to $t.$
More precisely, given $A\in \mathcal{A},$ we define $\theta \!\left(
A\right) _{t}\in W\left( K\right) $ to be the solution to the $K$-valued
differential equation
\begin{eqnarray}
\frac{d\theta \!\left( A\right) _{t}}{dt} &=&\theta \!\left( A\right)
_{t}\,A_{t}. \\
\theta \!\left( A\right) _{0} &=&e.
\end{eqnarray}
I am adopting matrix group notation. Geometrically $\theta \!\left( A\right)
_{t}$ is the parallel transport from 0 to $t$ and $\theta \!\left( A\right)
_{1}$ is the holonomy around the circle.
The based gauge group $\mathcal{G}_{b}$ is the set of absolutely continuous
maps $g:\left[ 0,1\right] \rightarrow K$ such that $g_{0}=g_{1}=e$ and such
that $\int_{0}^{1}\left| g_{t}^{-1}\,dg/dt\right| ^{2}dt<\infty .$ A gauge
transformation $g\in \mathcal{G}_{b}$ acts on $\mathcal{A}$ as follows:
\begin{equation}
\left( g\cdot A\right) _{t}=\mathrm{Ad}g_{t}\left( A_{t}\right) -\frac{dg}{dt%
}g_{t}^{-1}. \label{g.action}
\end{equation}
In terms of the parallel translation $\theta $ this action has the simple
form
\begin{equation}
\theta \!\left( g\cdot A\right) _{t}=\theta \!\left( A\right) _{t}g_{t}^{-1},
\label{g.action2}
\end{equation}
as is easily verified.
Formally, we would like to quantize $\mathcal{A}$ by passing to the Hilbert
space $L^{2}\left( \mathcal{A}\right) $ with respect to Lebesgue measure.
Since $\mathcal{A}$ is infinite-dimensional, there is no Lebesgue measure.
So instead I consider a Gaussian measure with variance $s,$ and will allow $%
s $ to tend to infinity. Although I am interested primarily in the
Segal-Bargmann transform, I will also discuss the Hamiltonian and the Wilson
loop observables.
Now, even for a Gaussian measure we must enlarge $\mathcal{A}$ to a space $%
\overline{\mathcal{A}}$ of $\frak{k}$-valued distributions. The space $%
\overline{\mathcal{A}}$ may be taken to be the space of derivatives of
continuous $\frak{k}$-valued functions. I then consider the Gaussian measure
$P_{s}$ on $\overline{\mathcal{A}}$ satisfying
\[
\int_{\overline{\mathcal{A}}}e^{i\left( \phi ,A\right) }\,dP_{s}\left(
A\right) =e^{-s\left\| \phi \right\| ^{2}/2}
\]
for all continuous linear functionals $\phi $ on $\overline{\mathcal{A}},$
where $\left\| \phi \right\| $ denotes the norm of $\phi $ restricted to $%
\mathcal{A}.$ The distribution of $A$ with respect to $P_{s}$ is a scaled
white noise, namely, $s$ times the derivative of standard $\frak{k}$-valued
Brownian motion.
The It\^{o} map may be ``extended'' to an almost-everywhere-defined map of $%
\overline{\mathcal{A}}$ into $W\left( K\right) ,$ provided that the equation
is interpreted as a Stratonovich stochastic differential equation.
Alternatively, the It\^{o} map on $\overline{\mathcal{A}}$ may be defined by
an It\^{o} s.d.e. with a ``correction term'' as in \cite[Eq. (36)]{Di}. The
distribution of $\theta \!\left( A\right) $ is then that of Brownian motion
with values in $K$ with the time variable scaled by $s.$ In particular, the
distribution of $\theta \!\left( A\right) _{1}$ is the heat kernel measure $%
\rho _{s}$ on $K.$ The action (\ref{g.action}) of the based gauge group $%
\mathcal{G}_{b}$ on $\mathcal{A}$ extends continuously to an action on $%
\overline{\mathcal{A}}$ and (\ref{g.action2}) still holds.
We now consider the Hilbert space $L^{2}\left( \overline{\mathcal{A}}%
,P_{s}\right) .$ At a heuristic level, Lebesgue measure on $\overline{%
\mathcal{A}}$ is an infinite constant times the $s\rightarrow \infty $ limit
of $P_{s}.$ We require, then, a version of the Segal-Bargmann transform
which will survive the limit $s\rightarrow \infty ,$ at least on the
gauge-invariant subspace. So we may not take the standard transform, which
involves $e^{s\Delta /2}.$ Instead we take our transform, denoted $%
\widetilde{B}_{s,t},$ to be $e^{t\Delta /2}$ followed by analytic
continuation, where $\Delta =\sum \partial ^{2}/\partial x_{k}^{2}$ with the
sum ranging over an orthonormal basis for $\mathcal{A}.$ Here $t$ is an
arbitrary parameter, which may be interpreted as Planck's constant. Let me
emphasize that in this the classical setting, the transform $\widetilde{B}%
_{s,t}$ is not truly new; it is just a different normalization of the
transform considered by Segal. See Case 2 of the proof of Theorem \ref
{main.thm}.
A cylinder function on $\overline{\mathcal{A}}$ is defined to be a function
of the form
\[
f\left( A\right) =g\left( I\left( A\right) _{t_{1}},\cdots ,I\left( A\right)
_{t_{n}}\right) ,
\]
where $I\left( A\right) _{t}$ is the $\frak{k}$-valued integral of the white
noise $A$ from zero to $t.$ If $s>t/2,$ then $e^{t\Delta /2}f$ makes sense
for all cylinder functions $f.$ As the target Hilbert space we take $%
L^{2}\left( \overline{\mathcal{A}}_{\Bbb{C}},M_{s,t}\right) .$ Here $%
\overline{\mathcal{A}}_{\Bbb{C}}$ is the space of derivatives of continuous $%
\frak{k}_{\Bbb{C}}$-valued functions, and $M_{s,t}$ satisfies
\[
\int_{\overline{\mathcal{A}}_{\Bbb{C}}}e^{i\left( \phi ,A\right)
}\,dM_{s,t}\left( A\right) =e^{\left( s-t/2\right) \left\| \phi _{1}\right\|
^{2}/2}\,e^{t\left\| \phi _{2}\right\| ^{2}/4}
\]
where $\phi _{1}$ is the restriction of $\phi $ to $\mathcal{A}$ and $\phi
_{2}$ is the restriction of $\phi $ to $i\mathcal{A}.$ Case 2 of the proof
of Theorem \ref{main.thm} shows that $M_{s,t}$ is the right Gaussian measure
to make $\widetilde{B}_{s,t}$ an isometry. The proof is by reduction to the
standard Segal-Bargmann transform. Since cylinder functions are dense, we
may extend by continuity to get an isometric map of $L^{2}\left( \overline{%
\mathcal{A}},P_{s}\right) $ into $L^{2}\left( \overline{\mathcal{A}}_{\Bbb{C}%
},M_{s,t}\right) .$ The image of $\widetilde{B}_{s,t}$ is the closure of the
space of $L^{2}$ holomorphic cylinder functions, which we may optimistically
call the ``holomorphic'' subspace of $L^{2}\left( \overline{\mathcal{A}}_{%
\Bbb{C}},M_{s,t}\right) .$
Gauge symmetry means that we should restrict our attention to
gauge-invariant functions, i.e., those such that for all $g\in \mathcal{G}%
_{b},$ $f\left( g\cdot A\right) =f\left( A\right) $ for almost every $A\in
\overline{\mathcal{A}}.$ Note that I am deliberately not ``unitarizing'' the
action of the gauge group as Wren and Landsman do; the point of letting $%
s\rightarrow \infty $ is to avoid having to do this. So although the map $%
f\left( A\right) \rightarrow f\left( g\cdot A\right) $ is not unitary for
any fixed $s,$ it formally becomes so as $s$ tends to infinity. At a
rigorous level, it is easily verified that for functions of the form $%
f\left( A\right) =\psi \left( \theta \!\left( A\right) _{t_{1}},\cdots
,\theta \!\left( A\right) _{t_{n}}\right) $ the gauge action becomes unitary
in the limit $s\rightarrow \infty .$
Now, as a consequence of the ergodicity result of \cite[Thm. 2.5]{G}
(together with (\ref{g.action2})), the functions invariant under based gauge
transformations are precisely those of the form $f\left( A\right) =\phi
\left( \theta \!\left( A\right) _{1}\right) ,$ where $\phi $ is a function
on $K.$ That is, $f$ should be a function of the holonomy around the circle.
(For invariance under all gauge transformations, not just based gauge
transformations, $\phi $ would be required to be a class function.) We now
wish to apply the Segal-Bargmann transform $\widetilde{B}_{s,t}$ to
functions of this form, which means roughly that we should compute $%
e^{t\Delta /2}f$ and then analytically continue. The key to this computation
is the following simple and well-known fact.
\begin{proposition}
If $A$ is a white-noise random variable with values in $\overline{\mathcal{A}%
},$ and $B$ is a fixed element of $\mathcal{A},$ then $\theta \!\left(
A+B\right) $ has the same distribution as $\theta \!\left( A\right) \theta
\!\left( B\right) .$
\end{proposition}
\textit{Proof.} Direct calculation shows
that for $A$
and $B$ smooth enough,
\[
\theta \!\left( A\right) _{t}\theta \!\left( B\right) _{t}=\theta \!\left(
\text{Ad}\theta \!\left( B\right) _{t}^{-1}\left( A_{t}\right) +B_{t}\right)
.
\]
Standard stochastic techniques show that this remains true almost surely if $%
A$ is a white noise and $B$ is in $\mathcal{A}.$ But the white noise measure
is invariant under the pointwise adjoint action, which is just a
``rotation'' of $\mathcal{A}.$%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}
%BeginExpansion
\endproof%
%EndExpansion
Using the proposition, we may formally calculate $e^{t\Delta /2}f.$ The
measure $P_{t}$ is the fundamental solution at the identity of the heat
equation, which means that $e^{t\Delta /2}$ should be given by convolution
with $P_{t}.$ So if $f\left( A\right) =\phi \left( \theta \!\left( A\right)
_{1}\right) $ then
\begin{eqnarray*}
e^{t\Delta /2}f\left( B\right) &=&\int_{\overline{\mathcal{A}}}\phi \left(
\theta \!\left( A+B\right) _{1}\right) \,dP_{t}\left( A\right) \\
&=&\int_{\overline{\mathcal{A}}}\phi \left( \theta \!\left( A\right)
_{1}\theta \!\left( B\right) _{1}\right) \,dP_{t}\left( A\right) \\
&=&\int_{K}\phi \left( x\theta \!\left( B\right) _{1}\right) \,d\rho
_{t}\left( x\right) \\
&=&e^{t\Delta _{K}/2}\phi \left( \theta \!\left( B\right) _{1}\right) .
\end{eqnarray*}
Here we have used the proposition and the fact that $\theta \!\left(
A\right) _{1}$ is distributed as the heat kernel on $K.$ Analytically
continuing, we arrive at the following conjecture.
\begin{conjecture}
\label{bst.conj}Suppose $f\in L^{2}\left( \overline{\mathcal{A}}%
,p_{s}\right) $ is of the form $f\left( A\right) =\phi \left( \theta
\!\left( A\right) _{1}\right) ,$ where $\phi $ is a function on $K.$ Then
\[
\widetilde{B}_{s,t}f\left( A\right) =\Phi \left( \theta \!\left( A\right)
_{1}\right)
\]
where $\Phi $ is the analytic continuation to $K_{\Bbb{C}}$ of $e^{t\Delta
_{K}/2}\phi .$
\end{conjecture}
Now, if $A$ is distributed according to the measure $P_{s},$ then $\theta
\!\left( A\right) _{1}$ is distributed according to the time $s$ heat kernel
measure $\rho _{s}.$ Thus the space of gauge invariant functions in $%
L^{2}\left( \overline{\mathcal{A}},P_{s}\right) $ may be identified with the
Hilbert space $L^{2}\left( K,\rho _{s}\right) .$ Similarly, the space of
functions in $L^{2}\left( \overline{\mathcal{A}}_{\Bbb{C}},M_{s,t}\right) $
of the form $F\left( A\right) =\Phi \left( \theta \!\left( A\right)
_{1}\right) $ with $\Phi $ holomorphic may be identified with $\mathcal{H}%
L^{2}\left( K_{\Bbb{C}},\mu _{s,t}\right) .$ If Conjecture \ref{bst.conj} is
correct, then the transform $\widetilde{B}_{s,t}$ for the
infinite-dimensional linear space $\overline{\mathcal{A}}$ reduces, when
applied to gauge-invariant functions, to the transform $B_{s,t}$ of Section
\ref{main.sec}. Thus the isometricity of $B_{s,t}$ is predicted by means of
infinite-dimensional analysis.
We may even argue for the surjectivity of $B_{s,t}$ as follows. Consider $%
F\in L^{2}\left( \overline{\mathcal{A}}_{\Bbb{C}},M_{s,t}\right) $ of the
form $F\left( A\right) =\Phi \left( \theta \!\left( A\right) _{1}\right) ,$
with $\Phi $ holomorphic. Then $F$ ought to be ``holomorphic'' as well, and
thus in the image of $B_{s,t}.$ But $F$ is certainly gauge-invariant, and $%
B_{s,t}$ commutes with the action of the gauge group, since the gauge action
is a combination of rotations and translations, and the Laplacian commutes
with such. Thus $B_{s,t}^{-1}F$ should be gauge-invariant as well, hence of
the form $f\left( A\right) =\phi \left( \theta \!\left( A\right) _{1}\right)
.$ But then Conjecture \ref{bst.conj} would tell us that $\Phi $ is the
analytic continuation of $e^{t\Delta _{K}/2}\phi ,$ which shows that every $%
\Phi $ comes from a $\phi .$ This argument is precisely in the spirit of
\cite[Cor. 7.9]{GM}.
Unfortunately, none of these arguments is rigorous. While the conjecture is
known to be true \cite[Thm. 17]{HS} if $s=t,$ even in that case the proof
uses some of the results we are trying to derive, namely the ``density
theorems'' of \cite{H1,D}. Similarly, Corollary 7.12 of \cite{GM} is not
quite the $s=t$ case of the conjecture, because the authors consider the
It\^{o} map only for $\mathcal{A}_{\Bbb{C}}$ and not for $\overline{\mathcal{%
A}}_{\Bbb{C}}.$ (Both \cite{HS} and \cite{GM} express things in terms of the
Wiener measure instead of the white noise measure, but it is merely a
difference of convention.) What is lacking in either case is a direct proof
that for $\Phi $ holomorphic on $K_{\Bbb{C}},$ the function $F\left(
A\right) =\Phi \left( \theta \!\left( A\right) _{1}\right) $ is in the
``holomorphic'' subspace of $L^{2}\left( \overline{\mathcal{A}}_{\Bbb{C}%
},M_{t,t}\right) ,$ that is, the closure of the holomorphic cylinder
functions. See the discussion in Section 2.5 of \cite{HS}.
In the case $s\neq t,$ things are even worse, since then $f$ is in $%
L^{2}\left( \overline{\mathcal{A}},P_{s}\right) $ but you are trying to
integrate $f\left( A+B\right) $ with respect to $P_{t}.$ Since $P_{s}$ and $%
P_{t}$ are mutually singular, this does not make sense for general $f.$ To
do things rigorously, one would have to approximate by cylinder functions,
and this is non-trivial. Furthermore, when $s\neq t,$ it is even less clear
how to show that $\Phi \left( \theta \!\left( A\right) _{1}\right) $ is in
the closure of the holomorphic cylinder functions. Despite these
difficulties, Conjecture \ref{bst.conj} provides motivation for Theorem \ref
{main.thm} and also suggests that $C_{t}$ is the natural Segal-Bargmann
transform for $K$ from the point of view of Yang-Mills theory.
Finally, let me consider very briefly the Hamiltonian and the Wilson loop
observables. If we could take our Hilbert space to be $L^{2}\left( \mathcal{A%
},\text{ Lebesgue}\right) ,$ then the Yang-Mills Hamiltonian should be $%
-\Delta $ \cite[Sec. 2]{Di}, the curvature term being zero since the circle
is one-dimensional. I would like to take $-\Delta $ as the Hamiltonian even
in $L^{2}\left( \overline{\mathcal{A}},P_{s}\right) .$ This operator is not
self-adjoint, but should become so as $s\rightarrow \infty .$ We have argued
formally that $e^{t\Delta /2}$ becomes $e^{t\Delta _{K}/2}$ on
gauge-invariant functions, and so by differentiating with respect to $t,$ $%
\Delta $ becomes $\Delta _{K}$ on gauge-invariant functions. Thus $-\Delta
_{K}$ should be our Hamiltonian in $L^{2}\left( K,\rho _{s}\right) .$ This
operator is self-adjoint in the limiting Hilbert space $L^{2}\left(
K,dx\right) .$ This conclusion is consistent with \cite{W1,Di}. Similarly,
if $\chi $ is a character of $K,$ we may consider the Wilson loop $\chi
\left( \theta \!\left( A\right) _{1}\right) $ as a multiplication operator
on $L^{2}\left( \overline{\mathcal{A}},P_{s}\right) .$ This corresponds to
multiplication by $\chi $ on the gauge-invariant subspace, for any $s$ and
so also when $s\rightarrow \infty .$ So as in \cite{Di} we conclude that the
Wilson loop should become multiplication by $\chi $ on $L^{2}\left(
K,dx\right) .$ The conclusions of this paragraph are also consistent with
those of Rajeev \cite{Ra}, who imposes gauge symmetry before quantization.
See \cite{H4} for a discussion of how these observables act in the
transformed Hilbert space $\mathcal{H}L^{2}\left( K_{\Bbb{C}},\nu
_{t}\right) .$
\section{Rigorous finite-dimensional results\label{main.sec}}
In this section we will prove the results concerning the transform $B_{s,t}$
which were predicted in the previous section. I will consider Lie groups of
compact type, a class which contains both compact groups and $\Bbb{R}^{n}.$
\begin{definition}
A Lie group is of \textbf{compact type} if it is locally isomorphic to some
compact group.
\end{definition}
Thus $\Bbb{R}^{n}$ is of compact type, since it is locally isomorphic to an $%
n$-torus. Of course, compact groups are also of compact type. It can be
shown \cite[Cor. 2.2]{D} that every connected Lie group of compact type is
the product of a compact group and $\Bbb{R}^{n}.$ See, for example, \cite
{He, V} for standard results about Lie groups of compact type.
Let $K$ be a connected Lie group of compact type, and let $\frak{k}$ be the
Lie algebra of $K.$ Fix once and for all an inner product on $\frak{k}$
which is invariant under the adjoint action of $K.$ A Lie group admits such
an inner product if and only if it is of compact type. Let $X_{1},\cdots
,X_{n}$ be an orthonormal basis for $\frak{k}$, and view each $X_{k}$ as a
left-invariant vector field. Let
\[
\Delta _{K}=\sum_{k=1}^{n}X_{k}^{2}.
\]
This is a bi-invariant differential operator on $K$ and is independent of
the choice of orthonormal basis.
Let $\rho _{s}\left( x\right) $ be the solution to the equation
\[
\frac{d\rho }{ds}=\frac{1}{2}\Delta _{K}\rho _{s},
\]
subject to the initial condition
\[
\rho _{s}\rightarrow \delta _{e}\;\text{as }s\rightarrow 0.
\]
This function is called the heat kernel (at the identity) for $K.$ It is
known that $\rho _{s}$ exists and is unique, and is a strictly positive, $%
C^{\infty }$ function on $K$. With little danger of confusion, we will also
let $\rho _{s}$ denote the associated measure
\[
d\rho _{s}=\rho _{s}\left( x\right) \,dx,
\]
where $dx$ denotes Haar measure on $K.$
Let $K_{\Bbb{C}}$ be the complexification of $K,$ which is a certain
connected complex Lie group whose Lie algebra $\frak{k}_{\Bbb{C}}$ is the
complexification of $\frak{k},$ and which contains $K$ as a subgroup. (See
\cite[Sec. 3]{H1} or \cite{Ho} for the definition.) For example, if $%
K=SU\left( 2\right) ,$ then $K_{\Bbb{C}}=SL\left( 2;\Bbb{C}\right) .$ The
key property of the complexified group $K_{\Bbb{C}}$ is that every
finite-dimensional representation of $K$ has a unique analytic continuation
to a holomorphic representation of $K_{\Bbb{C}}.$ As a consequence of this
property, it may be shown \cite{H1, D} that the heat kernel function $\rho
_{t}\left( x\right) $ has a unique analytic continuation to $K_{\Bbb{C}}.$
Consider now the Hilbert space $L^{2}\left( K,\rho _{s}\right) .$ For $t<2s$
define
\[
B_{s,t}:L^{2}\left( K,\rho _{s}\left( x\right) \,dx\right) \rightarrow
\mathcal{H}\left( K_{\Bbb{C}}\right)
\]
by
\begin{equation}
B_{s,t}f\left( g\right) =\int_{K}\rho _{t}\left( gx^{-1}\right) f\left(
x\right) \,dx,\quad g\in K_{\Bbb{C}}. \label{brt}
\end{equation}
Here $\rho _{t}$ refers to the analytically continued heat kernel and $%
\mathcal{H}\left( K_{\Bbb{C}}\right) $ to the space of holomorphic functions
on $K_{\Bbb{C}}.$ Note that the formula for $B_{s,t}$ involves the time $t$
heat kernel, whereas the domain Hilbert space involves the time $s$ heat
kernel. We will prove that the integral is convergent and that $B_{s,t}f$ is
in fact holomorphic. Unless $K$ is compact, convergence fails when $t\ge 2s.$
Now consider the operator $A_{s,t}$ on $K_{\Bbb{C}}$ given by
\begin{equation}
A_{s,t}=\left( s-\frac{t}{2}\right) \sum_{k=1}^{n}X_{k}^{2}+\frac{t}{2}%
\sum_{k=1}^{n}JX_{k}^{2}. \label{art}
\end{equation}
Here $J$ denotes the complex structure on $\frak{k}_{\Bbb{C}},$ and the $%
X_{k}$'s and $JX_{k}$'s are viewed as left-invariant vector fields on $K_{%
\Bbb{C}}.$ For example, if $K=\Bbb{R}^{n},$ $K_{\Bbb{C}}=\Bbb{C}^{n},$ then $%
X_{k}=\partial /\partial x_{k}$ and $JX_{k}=\partial /\partial y_{k}.$ For $%
t<2s$ (i.e., for $s>t/2$) $A_{s,t}$ is a second-order, left-invariant,
elliptic differential operator on $K_{\Bbb{C}}.$ Let $\mu _{s,t}$ be the
function on $K_{\Bbb{C}}$ given formally by
\[
\mu _{s,t}=e^{A_{s,t}/2}\left( \delta _{e}\right) .
\]
More precisely, let $\mu _{s,t,r}$ be the solution to the equation
\begin{eqnarray*}
\frac{d\mu _{s,t,r}}{dr} &=&\frac{1}{2}A_{s,t}\mu _{s,t,r} \\
\lim_{r\rightarrow 0}\mu _{s,t,r} &=&\delta _{e}.
\end{eqnarray*}
Then $\mu _{s,t}=\mu _{s,t,1}$. Existence and properties are shown, for
example, in \cite{Ro}. In particular $\mu _{s,t}$ is a $C^{\infty },$
strictly positive function with rapid decay at infinity.
We will let $\mu _{s,t}$ also denote the associated measure, $d\mu
_{s,t}=\mu _{s,t}\left( g\right) \,dg,$ where $dg$ is Haar measure on $K_{%
\Bbb{C}}.$ Finally we will set
\[
\mathcal{H}L^{2}\left( K_{\Bbb{C}},\mu _{s,t}\right) =\mathcal{H}\left( K_{%
\Bbb{C}}\right) \cap L^{2}\left( K_{\Bbb{C}},\mu _{s,t}\right) .
\]
\noindent We are now ready to state the main result.
\begin{theorem}
\label{main.thm}Let $s$ and $t$ be positive numbers with $s>t/2.$ Then for
all $f\in L^{2}\left( K,\rho _{s}\right) ,$ the integral
\[
B_{s,t}f\left( g\right) =\int_{K}\rho _{t}\left( gx^{-1}\right) f\left(
x\right) \,dx
\]
is absolutely convergent for all $g\in K_{\Bbb{C}}$ and depends
holomorphically on $g.$ The map $B_{s,t}$ is an isometric isomorphism of $%
L^{2}\left( K,\rho _{s}\right) $ onto $\mathcal{H}L^{2}\left( K_{\Bbb{C}%
},\mu _{s,t}\right) .$
\end{theorem}
\textit{Proof}. We will consider separately the compact and $\Bbb{R}^{n}$
cases. The general case follows by reduction to those cases.
\textit{Case 1. }$K$ \textit{is compact}. Although the method of \cite{H1}
can be used with little change, I wish to give a different argument which at
least formally does not involve matrix entries. However, the matrix entries
still play an important technical role. According to \cite[Prop. 1]{H1}, the
heat kernel $\rho _{t}$ admits a unique analytic continuation to $K_{\Bbb{C}%
}.$ Although it is possible to use the transform itself to deduce this, as
in \cite{D}, it is not difficult to obtain the result directly, as in \cite
{H1}. In this the compact case there is no trouble with the convergence of
the integral that defines $B_{s,t}$ or with the holomorphicity of $B_{s,t}f.$
So we need only address isometricity and surjectivity.
Let us first give a heuristic argument for the isometricity of $B_{s,t}$,
and then show that this can be made rigorous. Since $\rho _{s}\left(
x^{-1}\right) =\rho _{s}\left( x\right) $ \cite{H1}, integrating a function $%
\phi $ with respect to the measure $\rho _{s}\left( x\right) \,dx$ gives the
same result as computing $e^{s\Delta _{K}/2}\left( \phi \right) $ and then
evaluating at the identity. Similarly, integrating a function on $K_{\Bbb{C}%
} $ against $\mu _{s,t}\left( g\right) \,dg$ gives the same result as
applying $e^{A_{s,t}/2}$ and then evaluating at the identity. Thus the
isometricity of $B_{s,t}$ amounts to the statement
\begin{equation}
e^{s\Delta _{K}/2}\left( \overline{f_{1}}f_{2}\right) \left( e\right)
=e^{A_{s,t}/2}\left( \overline{e^{t\Delta _{K}/2}f_{1}}\,e^{t\Delta
_{K}/2}f_{2}\right) \left( e\right) . \label{norm1}
\end{equation}
Let us assume that $f_{1}$ and $f_{2}$ themselves admit an analytic
continuation to $K_{\Bbb{C}}.$ Now, $\Delta _{K}=\sum X_{k}^{2}$, regarded
as a left-invariant differential operator on $K_{\Bbb{C}},$ commutes with
complex conjugation and with analytic continuation. Thus
\[
\overline{e^{t\Delta _{K}/2}f_{1}}=e^{t\Delta _{K}/2}\overline{f_{1}}.
\]
Note that on the left, we are first applying $e^{t\Delta _{K}/2},$ then
analytically continuing, and then taking the complex conjugate. On the right
we are first analytically continuing, then taking the complex conjugate and
then applying $e^{t\Delta _{K}/2}.$
Next consider the operators
\begin{eqnarray*}
Z_{k} &=&\frac{1}{2}\left( X_{k}-iJX_{k}\right) \\
\overline{Z}_{k} &=&\frac{1}{2}\left( X_{k}+iJX_{k}\right) ,
\end{eqnarray*}
which reduce in the case $K=\Bbb{R}^{n},$ $K_{\Bbb{C}}=\Bbb{C}^{n}$ to $%
\partial /\partial z_{k}$ and $\partial /\partial \overline{z}_{k}.$ On the
holomorphic function $f_{2}$ we have $Z_{k}f_{2}=X_{k}f_{2}$ and $\overline{Z%
}_{k}f_{2}=0,$ and on the anti-holomorphic function $\overline{f_{1}},$ $%
Z_{k}\overline{f_{1}}=0$ and $\overline{Z}_{k}f_{1}=X_{k}f_{1}$. It follows
that
\[
\overline{e^{t\Delta _{K}/2}f_{1}}\,e^{t\Delta _{K}/2}f_{2}=e^{t\sum
Z_{k}^{2}/2}e^{t\sum \overline{Z}_{k}^{2}/2}\left( \overline{f_{1}}%
f_{2}\right) .
\]
So the desired norm equality becomes
\begin{equation}
e^{s\Delta _{K}/2}\left( \overline{f_{1}}f_{2}\right) \left( e\right)
=e^{A_{s,t}/2}e^{t\sum Z_{k}^{2}/2}e^{t\sum \overline{Z}_{k}^{2}/2}\left(
\overline{f_{1}}f_{2}\right) \left( e\right) . \label{norm2}
\end{equation}
Now, a holomorphic vector field $Z_{k}$ automatically commutes with an
anti-holomorphic vector field $\overline{Z}_{l}.$ Thus the second and third
exponents on the right of (\ref{norm2}) may be combined. The exponent that
results is
\[
\frac{t}{2}\sum_{k=1}^{n}\left( Z_{k}^{2}+\overline{Z}_{k}^{2}\right) =\frac{%
t}{4}\sum_{k=1}^{n}\left( X_{k}^{2}-JX_{k}^{2}\right) .
\]
This is a constant times the Casimir operator for $K_{\Bbb{C}},$ which is
bi-invariant and therefore commutes with the left-invariant operator $%
A_{s,t}.$ So in the end all three exponents on the right in (\ref{norm2})
may be combined. It thus suffices to have the exponents be equal, and so,
multiplying by two and recalling the definition (\ref{art}) of $A_{s,t}$ it
suffices to have
\begin{equation}
s\sum X_{k}^{2}=\left( s-\frac{t}{2}\right) \sum X_{k}^{2}+\frac{t}{2}\sum
JX_{k}^{2}+\frac{t}{2}\sum \left( X_{k}^{2}-JX_{k}^{2}\right) ,
\label{norm3}
\end{equation}
which is true!
Let us now make this argument rigorous. Let $\mathcal{F}$ denote the space
of finite linear combinations of matrix entries for finite-dimensional
irreducible representations of $K.$ Since $K$ is compact, $\rho _{s}\left(
x\right) $ is bounded and bounded away from zero. Thus $L^{2}\left( K,\rho
_{s}\right) $ is the same space of functions as $L^{2}\left( K,dx\right) ,$
with a different but equivalent norm. So by the Peter-Weyl theorem, $%
\mathcal{F}$ is dense in $L^{2}\left( K,\rho _{s}\right) .$ On $\mathcal{F}$
there is little trouble in justifying the above arguments. Indeed,
\cite[Lem. 2, Lem. 8]{H1} shows that when integrating matrix entries against
heat kernels, one may expand everything in power series. Thus $B_{s,t}$ is
isometric on $\mathcal{F}$ and so also on $L^{2}\left( K,\rho _{s}\right) .$
The proof of surjectivity of \cite{H1} applies essentially without change.
In particular, the proof of the ``averaging lemma'' \cite[Lem. 11]{H1}
applies with $\mu _{t}$ replaced by $\mu _{s,t}.$ This lemma is the key to
surjectivity.
\textit{Case 2: }$K=\Bbb{R}^{n}.$ Convergence is an easy calculation, and
isometricity can be proved by the method of Case 1. But it is not evident
how to prove surjectivity directly. In particular, the arguments of \cite{B}
or \cite{D} do not apply if $s\neq t.$ Instead, I will reduce everything to
properties of the ``classical'' Segal-Bargmann transform $B_{t},$ thus
showing that $B_{s,t}$ is, in the $\Bbb{R}^{n}$ case, just a disguised form
of $B_{t}.$ Because the calculations rely on the explicit Gaussian forms for
the heat kernels, no such calculations are possible for general groups.
Let $r=2\left( s-t/2\right) ,$ so that $r>0.$ Recalling the definition of $%
A_{s,t}$ we may compute $\mu _{s,t}$ explicitly:
\[
\mu _{s,t}\left( x+iy\right) =\left( \pi r\right) ^{-n/2}\left( \pi t\right)
^{-n/2}e^{-x^{2}/r}e^{-y^{2}/t}.
\]
Now consider the map $U$ defined on $\mathcal{H}L^{2}\left( K_{\Bbb{C}},\mu
_{s,t}\right) $ given by
\[
UF\left( z\right) =e^{az^{2}}F\left( z\right) .
\]
Since $\exp az^{2}$ is holomorphic and nowhere zero, $U$ is a unitary map of
$\mathcal{H}L^{2}\left( K_{\Bbb{C}},\mu _{s,t}\right) $ onto the Hilbert
space
\[
\mathcal{H}L^{2}\left( \Bbb{C}^{n},\left| e^{az^{2}}\right| ^{-2}\mu
_{s,t}\right) .
\]
Choosing $a=\left( r-t\right) /4rt$ gives
\[
\left| e^{az^{2}}\right| ^{-2}\mu _{s,t}=c_{1}e^{-\left| z\right| ^{2}/\tau
}
\]
where $\tau =2rt/\left( r+t\right) $ and the value of $c_{1}$ is immaterial.
The composite transform $UB_{s,t}$ is
\[
UB_{s,t}f\left( z\right) =c\int_{\Bbb{R}^{n}}e^{az^{2}}e^{-\left( z-x\right)
^{2}/2t}f\left( x\right) \,dx.
\]
By combining the exponents and completing the square we obtain
\begin{eqnarray}
UB_{s,t}f\left( z\right) &=&c\int_{\Bbb{R}^{n}}\exp \left\{ -\frac{1}{2\tau }%
\left( z-\frac{2r}{r+t}\,x\right) ^{2}\right\} \nonumber \\
&&\times f\left( x\right) \exp \left\{ \frac{r-t}{2t\left( r+t\right) }%
\,x^{2}\right\} \,dx. \label{ubst}
\end{eqnarray}
This is just the time-$\tau $ heat operator, combined with a dilation and
applied to the function
\[
g\left( x\right) =f\left( x\right) \exp \left\{ \frac{r-t}{2t\left(
r+t\right) }\,x^{2}\right\} .
\]
Now let $\widetilde{g}\left( x\right) =g\left( 2rx/\left( r+t\right) \right)
.$ Making the change-of-variable $y=2rx/\left( r+t\right) $ in (\ref{ubst})
gives
\[
UB_{s,t}f\left( z\right) =c_{2}\int_{\Bbb{R}^{n}}\exp \left\{ -\frac{1}{%
2\tau }\left( z-y\right) ^{2}\right\} \widetilde{g}\left( y\right) \,dy.
\]
Furthermore, calculation shows that the norm of $f$ in $L^{2}\left( \Bbb{R}%
^{n},\rho _{s}\right) $ is the same (up to a constant) as the norm of $%
\widetilde{g}$ in $L^{2}\left( \Bbb{R}^{n},\rho _{\tau }\right) .$ Note that
the map $\widetilde{g}\rightarrow UB_{s,t}f$ is just the standard
Segal-Bargmann transform $B_{\tau },$ which is isometric from $L^{2}\left(
\Bbb{R}^{n},\rho _{\tau }\right) $ onto $\mathcal{H}L^{2}\left( \Bbb{C}%
^{n},\mu _{\tau }\right) .$ Undoing all of these steps shows that $B_{s,t}$
is, up to a constant, an isometric isomorphism of $L^{2}\left( \Bbb{R}%
^{n},\rho _{s}\right) $ onto $\mathcal{H}L^{2}\left( \Bbb{C}^{n},\mu
_{s,t}\right) .$ By considering the case $f=1$ we see that the constant is
one.
\textit{Case 3: }$K=\Bbb{R}^{n}\times G,$\textit{\ with }$G$\textit{\
compact and semisimple}. Then $K_{\Bbb{C}}=\Bbb{C}^{n}\times G_{\Bbb{C}}$
and the Lie algebra decomposition $\frak{k}=\Bbb{R}^{n}\oplus \frak{g}$ is
automatically orthogonal with respect to any Ad-invariant inner product. It
follows that the heat kernels on both the real side and the complex side
will factor. Now (suppressing the measures) we have the standard result
\[
L^{2}\left( \Bbb{R}^{n}\right) \otimes L^{2}\left( G\right) \cong
L^{2}\left( \Bbb{R}^{n}\times G\right) ,
\]
where the isomorphism takes $f_{1}\otimes f_{2}$ to the product function $%
f_{1}\left( x\right) f_{2}\left( y\right) .$ Meanwhile on the complex side,
we have
\[
\mathcal{H}L^{2}\left( \Bbb{C}^{n}\right) \otimes \mathcal{H}L^{2}\left( G_{%
\Bbb{C}}\right) \cong \mathcal{H}L^{2}\left( \Bbb{C}^{n}\times G_{\Bbb{C}%
}\right) ,
\]
with $F_{1}\otimes F_{2}$ mapping to $F_{1}\left( z\right) F_{2}\left(
g\right) .$ This result requires a small proof which is given in the
appendix.
Now, since the heat kernel on $\Bbb{R}^{n}\times G$ factors, the transform
applied to a product function will be just the product of the separate
transforms. That is, the transform for the product group is (under the above
isomorphisms) just the tensor product of the separate transforms. So all the
desired properties of the transform for $K$ follow from the corresponding
properties of the transforms for $\Bbb{R}^{n}$ and for $G.$
\textit{Case 4: the general case}. Although every connected Lie group of
compact type is a product of a compact group with $\Bbb{R}^{n}$
\cite[Cor. 2.2]{D}, the Lie algebras of the two factors need not be
orthogonal, since the compact factor need not be semisimple. Thus the method
of Case 3 is not sufficiently general. But for any $K$ which is connected
and of compact type, the universal cover of $K$ is of the form $\widetilde{K}%
=\Bbb{R}^{n}\times G,$ where $G$ is compact and simply-connected, hence
semisimple. (See \cite[Chap. II, Cor. 6.5 and Thm. 6.9]{He}.) Thus by Case
3, the transform for $\widetilde{K}$ is well-defined, isometric, and
surjective.
Now, $K$ itself is $\widetilde{K}/N,$ where $N$ is a discrete central
subgroup, and
\begin{equation}
\rho _{s}\left( x\right) =\sum_{n\in N}\widetilde{\rho }_{s}\left(
\widetilde{x}n\right) \label{periodize}
\end{equation}
where $\rho _{s}$ is the heat kernel for $K,$ $\widetilde{\rho }_{s}$ is the
heat kernel for $\widetilde{K},$ and $\widetilde{x}$ is any point in $%
\widetilde{K}$ which projects to $x.$ Note that $N$ is finitely-generated
(since it is $\pi _{1}\left( K\right) $) and abelian, hence a product of
cyclic groups, and each generator of an infinite cyclic factor must have
non-zero $\Bbb{R}^{n}$-component. So there is no problem with convergence of
the sum. It follows that
\begin{equation}
L^{2}\left( K,\rho _{s}\right) =L^{2}\left( \widetilde{K},\widetilde{\rho }%
_{s}\right) ^{N}. \label{n1}
\end{equation}
Here a superscript $N$ indicates the functions which are invariant under the
right (or equivalently left) action of $N.$
The complexification of $K$ will be $K_{\Bbb{C}}=\widetilde{K}_{\Bbb{C}}/N,$
where $\widetilde{K}_{\Bbb{C}}$ means the complexification of the universal
cover of $K,$ which is the same as the universal cover of the
complexification. We have, by analogy to (\ref{n1})
\begin{equation}
\mathcal{H}L^{2}\left( K_{\Bbb{C}},\mu _{s,t}\right) =\mathcal{H}L^{2}\left(
\widetilde{K}_{\Bbb{C}},\widetilde{\mu }_{s,t}\right) ^{N}. \label{n2}
\end{equation}
By construction, the transform for $\widetilde{K}$ commutes with the action
of $N,$ and so it maps $L^{2}\left( \widetilde{K},\widetilde{\rho }%
_{s}\right) ^{N}$ into $\mathcal{H}L^{2}\left( \widetilde{K}_{\Bbb{C}},%
\widetilde{\mu }_{s,t}\right) ^{N}.$ Furthermore, in light of (\ref
{periodize}), the transform for $\widetilde{K},$ restricted to the $N$%
-invariant subspace, coincides with the transform for $K.$ So convergence
and isometricity of the transform for $K$ follow from the corresponding
properties for $\widetilde{K}.$
To establish surjectivity for the transform for $K,$ we must prove that the
transform for $\widetilde{K}$ maps onto $\mathcal{H}L^{2}\left( \widetilde{K}%
_{\Bbb{C}},\widetilde{\mu }_{s,t}\right) ^{N}.$ So suppose $F\in \mathcal{H}%
L^{2}\left( \widetilde{K}_{\Bbb{C}},\widetilde{\mu }_{s,t}\right) ^{N}.$
Then by the surjectivity of the transform for $\widetilde{K},$ there exists $%
f\in L^{2}\left( \widetilde{K},\widetilde{\rho }_{s}\right) $ such that
\[
F\left( g\right) =\int_{\widetilde{K}}\widetilde{\rho }_{t}\left(
gx^{-1}\right) f\left( x\right) \,dx.
\]
Since $F\left( g\right) =F\left( gn\right) ,$ we have
\begin{eqnarray}
\int_{\widetilde{K}}\widetilde{\rho }_{t}\left( gx^{-1}\right) f\left(
x\right) \,dx &=&\int_{\widetilde{K}}\widetilde{\rho }_{t}\left(
gnx^{-1}\right) f\left( x\right) \,dx \label{change} \\
&=&\int_{\widetilde{K}}\widetilde{\rho }\left( gx^{-1}\right) f\left(
xn\right) \,dx.
\end{eqnarray}
Now, if $f\left( x\right) \in L^{2}\left( \widetilde{K},\widetilde{\rho }%
_{s}\right) ,$ then both $f\left( x\right) $ and $f\left( xn\right) $ are in
$L^{2}\left( \widetilde{K},\widetilde{\rho }_{r}\right) $ for all $rt$%
\[
\mu _{s,t}=e^{\left( s-t\right) \sum X_{k}^{2}}\,e^{t/2\sum
JX_{k}^{2}}\,e^{t/2\sum X_{k}^{2}}\left( \delta _{e}\right) .
\]
Thus
\[
\mu _{s,t}\left( g\right) =\int_{K}\mu _{t,t}\left( gx^{-1}\right) \rho
_{s-t}\left( x\right) \,dx,
\]
from which it follows that $\lim_{s\rightarrow \infty }\mu _{s,t}\left(
g\right) =\nu _{t}\left( g\right) $ for all $g.$ Furthermore, applying the
averaging lemma to $\mu _{t,t}$ we see that for all $s>t,$ $\mu _{s,t}\left(
g\right) $ is dominated by a constant (independent of $s$) times $\nu
_{t}\left( g\right) .$ So Dominated Convergence gives the second limit in
the theorem. The methods of \cite{H1} are sufficient to make all of this
rigorous.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}
%BeginExpansion
\endproof%
%EndExpansion
Thus we obtain the unitary transform $C_{t}$ as the $s\rightarrow \infty $
limit of $B_{s,t}.$ There is almost certainly a theorem of this sort for
general $K$ of compact type, but in the non-compact case one would have to
multiply the measures on $K$ and $K_{\Bbb{C}}$ be a suitable function of $s$
in order for the measures to converge, for example, $\left( 2\pi s\right)
^{n/2}$ if $K=\Bbb{R}^{n}.$ Furthermore, in the non-compact case one no
longer has the same space of functions for different values of $s,$ so one
would have to be careful about how the theorem is formulated.
We now turn to the opposite extreme, the limit $s\rightarrow t/2.$ In the
case $K=\Bbb{R}^{n}$ the measure $\mu _{s,t}$ collapses as $s\rightarrow t/2$
onto the imaginary axis, and our transform becomes a finite-dimensional
version of the Fourier-Wiener transform. In the case $K$ compact, we still
have when $s=t/2$ an isometric map \textit{into} an $L^{2}$-space of
holomorphic functions, but at least in the simplest case the map is not onto.
\begin{theorem}
Let $K=\Bbb{R}^{n}$ so that $K_{\Bbb{C}}=\Bbb{C}^{n}.$ As $s$ tends to $t/2,$
the measures $\mu _{s,t}$ on $\Bbb{C}^{n}$ converge in the weak-$*$ topology
to the Gaussian measure $\left( \pi t\right) ^{-n/2}\exp \left(
-y^{2}/t\right) \,dy=\rho _{t/2}\left( y\right) \,dy$ on the ``imaginary
axis'' $i\Bbb{R}^{n}.$ The transform $B_{t/2,t}$ makes sense, say, on
polynomials and is given by
\begin{eqnarray*}
B_{t/2,t}f\left( iy\right) =\int_{\Bbb{R}^{n}}\rho _{t}\left( iy-x\right)
f\left( x\right) \,dx \\
=\left( 2\pi t\right) ^{-n/2}e^{y^{2}/2t}\int_{\Bbb{R}%
^{n}}e^{iyx/t}e^{-x^{2}/2t}f\left( x\right) \,dx.
\end{eqnarray*}
This transform maps the space of polynomials isometrically into $L^{2}\left(
i\Bbb{R}^{n},\rho _{t/2}\right) .$ Extending continuously we obtain an
isometric isomorphism of $L^{2}\left( \Bbb{R}^{n},\rho _{t/2}\right) $ onto $%
L^{2}\left( i\Bbb{R}^{n},\rho _{t/2}\right) .$
\end{theorem}
Note that $B_{t/2,t}$ is essentially the Fourier transform, unitary from $%
L^{2}\left( \Bbb{R}^{n},dx\right) $ to itself, disguised by conversion to
Gaussian measure on both sides. Note also that in the limit $s\rightarrow
t/2,$ the analyticity on the range is lost. That is, the range of $B_{t/2,t}$
is not an $L^{2}$-space of holomorphic functions, but an ordinary $L^{2}$%
-space, the elements of which need not even be continuous. This loss of
analyticity coincides with a loss of convergence in the integral that
defines $B_{t/2,t}.$
\textit{Proof}. Convergence of the measure follows from the explicit formula
\[
\mu _{s,t}\left( x+iy\right) =\left( 2\pi \left( s-t/2\right) \right)
^{n/2}e^{-x^{2}/2\left( s-t/2\right) }\left( \pi t\right)
^{-n/2}e^{-y^{2}/t}.
\]
If $f$ is a polynomial, then $F=e^{t\Delta /2}f$ is also a polynomial. So
there is no difficulty in letting $s\rightarrow t/2$ to obtain
\[
\left\| f\right\| _{L^{2}\left( \Bbb{R}^{n},\rho _{t/2}\right)
}=\lim_{s\rightarrow t/2}\left\| f\right\| _{L^{2}\left( \Bbb{R}^{n},\rho
_{s}\right) }=\lim_{s\rightarrow t/2}\left\| F\right\| _{L^{2}\left( \Bbb{C}%
^{n},\mu _{s,t}\right) }=\left\| F\right\| _{L^{2}\left( i\Bbb{R}^{n},\rho
_{t/2}\right) }.
\]
So $B_{t/2,t}$ is isometric on polynomials. But $e^{t\Delta /2}$ is
invertible on the space of polynomials of degree at most $k,$ and so every
polynomial is in the image of $B_{t/2,t}.$ Since polynomials are dense in $%
L^{2}\left( \Bbb{R}^{n},\rho _{t/2}\right) ,$ $B_{t/2,t}$ is densely defined
and has dense image.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}
%BeginExpansion
\endproof%
%EndExpansion
\begin{theorem}
\label{cpct.t2}Let $K$ be compact. Then as $s\rightarrow t/2$ the measures $%
\mu _{s,t}$ on $K_{\Bbb{C}}$ converge in the weak-$*$ topology to a
probability measure, denoted $\mu _{t/2,t}.$ If $K$ is semisimple, $\mu
_{t/2,t}$ is absolutely continuous with respect to Haar measure on $K_{\Bbb{C%
}}.$ For any compact $K,$ the transform $B_{s,t}$ makes sense with $s=t/2,$
and $B_{t/2,t}$ is an isometry of $L^{2}\left( K,\rho _{t/2}\right) $ \emph{%
into\ }$\mathcal{H}L^{2}\left( K_{\Bbb{C}},\mu _{t/2,t}\right) .$
\end{theorem}
\textit{Proof}. Every connected compact Lie group is of the form $K=\left(
G\times T\right) /N,$ where $G$ is compact and semisimple, $T$ is a torus,
and $N$ is a finite central subgroup. In the semisimple case, the function $%
\mu _{s,t}$ on $G_{\Bbb{C}}$ exists and is strictly positive \cite[IV.4]{Ro}
even when $s=t/2,$ because in that case the operator $\sum JX_{k}^{2}$ is
sub-elliptic and therefore hypo-elliptic. Furthermore, as in the proof of
Theorem \ref{inf.thm}, we have for $s>t/2$%
\[
\mu _{s,t}\left( g\right) =\int_{G}\mu _{t/2,t}\left( gx^{-1}\right) \rho
_{s-t/2}\left( x\right) \,dx.
\]
It follows that the weak-$*$ limit of $\mu _{s,t}$ as $s$ approaches $t/2$
is $\mu _{t/2,t.}$ Furthermore,
\begin{equation}
\int_{G}\mu _{t/2,t}\left( xg\right) dx=\nu _{t}\left( g\right) =\int_{G}\mu
_{s,t}\left( xg\right) dx \label{mut2.nut}
\end{equation}
where $\nu _{t}\left( g\right) $ is as in Theorem \ref{inf.thm} and decays
rapidly at infinity. If $f$ is in the space $\mathcal{F}$ (as in proof of
Theorem \ref{main.thm}), then the function $F=$ analytic continuation of $%
e^{t\Delta _{K}/2}f$ grows only exponentially at infinity. Using (\ref
{mut2.nut}) we may control the integral of $F$ near infinity uniformly in $%
s, $ giving
\[
\lim_{s\rightarrow t/2}\left\| F\right\| _{L^{2}\left( K_{\Bbb{C}},\mu
_{s,t}\right) }=\left\| F\right\| _{L^{2}\left( K_{\Bbb{C}},\mu
_{t/2,t}\right) .}
\]
Thus, $B_{t/2,t}$ is isometric on $\mathcal{F}$ and so also on $L^{2}\left(
G,\rho _{t/2}\right) .$
In the torus case $T$, we have $T_{\Bbb{C}}=T\times \Bbb{R}^{n},$ and the
measure $\mu _{s,t}$ is the product of the heat kernel measure $\rho
_{\left( s-t/2\right) /2}$ on $T$ and the Gaussian measure $\rho _{t/2}$ on $%
\Bbb{R}^{n}.$ So again there is no trouble in letting $s$ tend to $t/2.$
Taking products and periodizing over $N$ pose no problem.
Note, though, that in neither case is there any obvious way to show that
square-integrability with respect to $\mu _{t/2,t}$ implies
square-integrability with respect to $\mu _{s,t}.$ So we cannot obtain the
surjectivity of $B_{t/2,t}$ from that of $B_{s,t}.$ Compare this to the
situation in Theorem \ref{inf.thm}.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}
%BeginExpansion
\endproof%
%EndExpansion
I will address the surjectivity of $B_{t/2,t}$ only in the simplest case. I
conjecture that if $K$ is compact and semisimple, then $B_{t/2,t}$ maps onto
$\mathcal{H}L^{2}\left( K_{\Bbb{C}},\mu _{t/2,t}\right) .$
\begin{theorem}
\label{s1.t2}Let $K=S^{1}=\Bbb{R}/2\pi \Bbb{Z},$ so that $K_{\Bbb{C}}=\Bbb{C}%
/2\pi \Bbb{Z}.$ Then the image of the map $B_{t/2,t}$ is precisely the space
of functions $F$ satisfying:
\begin{enumerate}
\item $F$ is holomorphic on $\Bbb{C}$ and satisfies $F\left( z+2\pi \right)
=F\left( z\right) $ for all $z\in \Bbb{C}.$
\item $F$ is square-integrable with respect to $\mu _{t/2,t};$ that is,
\[
\left( \pi t\right) ^{-1/2}\int_{\Bbb{R}}\left| F\left( iy\right) \right|
^{2}e^{-y^{2}/t}\,dy<\infty .
\]
\item There exists constants $a,$ $b,$ and $c,$ with $c<1/2$ such that
\[
\left| F\left( x+iy\right) \right| \le a\exp \left( be^{c\left| y\right|
}\right)
\]
for all $x,y.$
\end{enumerate}
If $F$ is in the image of $B_{t/2,t},$ then for some constant $d$%
\[
\left| F\left( x+iy\right) \right| \le d\,e^{y^{2}/2t}
\]
for all $x,y.$
\end{theorem}
We may not take $c=1/2$ in Condition 3, as the function
\[
F\left( z\right) =\exp \left( ie^{iz/2}\right) +\exp \left(
-ie^{iz/2}\right)
\]
demonstrates. (Observe that $F$ is $2\pi $-periodic, even though neither
term is separately.) After all, $\left| F\left( iy\right) \right| \le 2,$
which means that $F$ is square-integrable, but $F\left( \pi +iy\right) =\exp
\left( -e^{-y/2}\right) +\exp \left( e^{-y/2}\right) $ which grows too
rapidly as $y\rightarrow -\infty $ for $F$ to be in the image of $B_{t/2,t}.$
Note that in the presence of Conditions 1 and 2 of the theorem, the very
weak bounds of Condition 3 imply the stronger bounds given in the last part
of the theorem. This is a consequence of the Phragmen-Lindel\"{o}f method
(e.g., \cite[Chap. 12]{Ru}), which allows the values of a holomorphic
function in the strip $0<\func{Re}z<2\pi $ to be controlled by its values on
the boundary, given Condition 3.
\textit{Proof}. If $F$ is $B_{t/2,t}f$ for some $f,$ then it is clear from
the definition of $B_{t/2,t}$ that $F$ satisfies Conditions 1 and 2.
Meanwhile,
\[
B_{t/2,t}f\left( x+iy\right) =\int_{0}^{2\pi }\rho _{t}\left( x+iy-x^{\prime
}\right) f\left( x^{\prime }\right) \,dx^{\prime }
\]
where
\begin{eqnarray}
\rho _{t}\left( x+iy-x^{\prime }\right) &=&\frac{1}{\sqrt{2\pi t}}%
\sum_{n=-\infty }^{\infty }\exp \left\{ -\left( x-x^{\prime }-2\pi
n+iy\right) ^{2}/2t\right\} \nonumber \\
&=&\frac{1}{\sqrt{2\pi t}}e^{y^{2}/2t}\sum_{n=-\infty }^{\infty }e^{-i\left(
x-x^{\prime }-2\pi n\right) y/t}e^{-\left( x-x^{\prime }-2\pi n\right)
^{2}/2t}. \label{rho.t}
\end{eqnarray}
But the sum in (\ref{rho.t}) is easily seen to be bounded uniformly in $x,$ $%
x^{\prime },$ and $y$ which gives the estimate in the last part of the
theorem, and so certainly Condition 3.
Conversely, suppose we have a function $F$ which satisfies Conditions 1, 2,
and 3. Then choose $\alpha $ with $c<\alpha <1/2,$ and let
\[
h_{\varepsilon }\left( z\right) =\exp \left\{ -\varepsilon \cos \alpha
\left( z-\pi \right) \right\} .
\]
For $x\in \left[ 0,2\pi \right] ,$ $h_{\varepsilon }\left( x+iy\right) $
tends to zero very rapidly as $y$ tends to infinity. Let us apply Cauchy's
Formula to the function $F\left( z\right) e^{z^{2}/2t}h_{\varepsilon }\left(
z\right) ,$ over the rectangle $0<\func{Re}z<2\pi ,$ $-A<\func{Im}zt,$ there exists $f_{r}$ such that $F$ is
(the analytic continuation of) $e^{r\Delta /2}f_{r},$ or $f_{r}=e^{-r\Delta
/2}F.$ Now, the $L^{2}$ norm on $S^{1}$ with respect to Lebesgue measure is
bounded by a constant times the $L^{2}$ norm with respect to heat kernel
measure. Furthermore, a function on the imaginary axis which is
square-integrable with respect to $\rho _{t/2}$ will also be
square-integrable with respect to $\rho _{r/2}$ for all $r