\documentclass[12pt,a4paper]{article}
\usepackage{amssymb,amsthm,amsmath,amsxtra}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem*{TA}{Theorem A}
\newtheorem*{TB}{Theorem B}
\newcommand{\sgn}{\operatorname{sgn}}
\newcommand{\supp}{\operatorname{supp}}
\begin{document}
\title{Topological Groups\\Part III, Spring 2008}
\author{T.~W.~K\"{o}rner}
\maketitle
\begin{footnotesize}
\noindent
{\bf Small print}
This is just a first draft for the course.
The content of the course will be what I say, not what
these notes say.
Experience shows that skeleton notes (at least
when I write them) are very error prone so
use these notes with care.
I should {\bf very much} appreciate being told
of any corrections or possible improvements
and might even part with a small reward to the
first finder of particular errors.
This document
is written in \LaTeX2e and available in tex, dvi, ps
and pdf form from my home page \verb+http://www.dpmms.cam.ac.uk/~twk/+.
My e-mail address is \verb+twk@dpmms.cam.ac.uk+.
In the middle of the 20th century it was realised
that classical Fourier Analysis could be extended
to locally compact Hausdorff Abelian groups. The object
of this course (which may not be completely
achieved) is to show how this is done.
(Specifically we wish to get as far as the
first two chapters of the book of Rudin~\cite{Rudin}.)
The main topics will thus be
topological groups in general, Haar measure,
Fourier Analysis on locally compact Hausdorff
Abelian groups,
Pontryagin duality and the principal structure theorem.
Although we will not need deep results, we will
use elementary functional analysis, measure theory
and the elementary theory of commutative
Banach algebras. (If you know two out of three
you should have no problems, if only one out
of three then the course is probably a bridge
too far.) Preliminary reading is not expected
but the book by Deitmar~\cite{Deitmar} is a good introduction.
\end{footnotesize}
\tableofcontents
\section{Prelude}\label{S, Prelude} We start with the
following observations\footnote{I shall assume
a lot of notation and results, not because I assume
that my audience will know it all but
because I assume they will know much of it.
If something seems strange \emph{just ask}.}.
\begin{lemma}\label{L, first characters}
Consider ${\mathbb R}$,
${\mathbb T}={\mathbb R}/{\mathbb Z}$,
$S^{1}=\{\lambda\in{\mathbb C}\,:\,|\lambda|=1\}$
and ${\mathbb Z}$ with their usual (Euclidean) metrics.
Then ${\mathbb R}$, ${\mathbb T}$ and ${\mathbb Z}$
are Abelian groups under addition and $S^{1}$ is
a group under multiplication.
(i) The continuous homomorphisms
$\chi:{\mathbb R}\rightarrow S^{1}$ are precisely
the maps $\chi_{a}(t)=\exp(iat)$ $[t\in {\mathbb R}]$
with $a\in{\mathbb R}$.
(ii) The continuous homomorphisms
$\chi:{\mathbb T}\rightarrow S^{1}$ are precisely
the maps $\chi_{a}(t)=\exp(2\pi iat)$ $[t\in {\mathbb T}]$
with $a\in{\mathbb Z}$.
(iii) The continuous homomorphisms
$\chi:{\mathbb Z}\rightarrow S^{1}$ are precisely
the maps $\chi_{a}(t)=\exp(2\pi iat)$ $[t\in {\mathbb Z}]$
with $a\in{\mathbb T}$.
\end{lemma}
\begin{exercise}\label{Exercise, first Haar} We use the notation of
Lemma~\ref{L, first characters}.
(i) Show that the non-zero Borel measures $\mu$ on
${\mathbb R}$ such that
\[\int_{\mathbb R}f(x-y)\,d\mu(x)=
\int_{\mathbb R}f(x)\,d\mu(x)\]
for all continuous functions
$f:{\mathbb R}\rightarrow{\mathbb R}$
of compact support
and all $y\in{\mathbb R}$
are precisely the
non-zero multiples of $m$ the Lebesgue
measure on ${\mathbb R}$. In other words
\[\int_{\mathbb R}f(x)\,d\mu(x)
=A\int_{\mathbb R}f(x)\,dx\]
for all continuous functions
$f:{\mathbb R}\rightarrow{\mathbb C}$
of compact support and some constant $A\neq 0$.
(ii) Show that the unique measure $\mu$ on
${\mathbb T}$ such that
\[\int_{\mathbb T}f(x-y)\,d\mu(x)=
\int_{\mathbb T}f(x)\,d\mu(x)\]
for all continuous functions
$f:{\mathbb T}\rightarrow{\mathbb R}$
and all $y\in{\mathbb T}$
and such that
\[\int_{\mathbb T}1\,d\mu(x)=1\]
is $(2\pi)^{-1}m$ where $m$ is the Lebesgue
measure on ${\mathbb T}$. In other words
\[\int_{\mathbb T}f(x)\,d\mu(x)
=\frac{1}{2\pi}\int_{\mathbb T}f(x)\,dx\]
for all continuous functions
$f:{\mathbb T}\rightarrow{\mathbb C}$.
(iii) Show that the unique measure $\mu$ on
${\mathbb Z}$ such that
\[\int_{\mathbb Z}f(x-y)\,d\mu(x)=
\int_{\mathbb Z}f(x)\,d\mu(x)\]
for all continuous functions
$f:{\mathbb Z}\rightarrow{\mathbb C}$
of compact support
and all $y\in{\mathbb Z}$
and such that
\[\int_{\mathbb Z}I_{\{0\}}(x)\,d\mu(x)=1\]
(where $I_{\{0\}}(0)=1$ and $I_{\{0\}}(x)=0$
when $x\neq 0$) is the counting measure.
In other words
\[\int_{\mathbb Z}f(x)\,d\mu(x)
=\sum_{x\in{\mathbb Z}}f(x)\]
for all continuous functions
$f:{\mathbb Z}\rightarrow{\mathbb C}$.
\end{exercise}
\begin{exercise}\label{E, First convolution}
We continue with the notation
and ideas of Lemma~\ref{L, first characters}
and Exercise~\ref{Exercise, first Haar}. We write
$\chi_{a}(t)=\langle t,a\rangle$. We let
$\mu_{{\mathbb R}}=(2\pi)^{-1/2}m$ where $m$
is Lebesgue measure on ${\mathbb R}$,
$\mu_{{\mathbb T}}=(2\pi)^{-1}m$ where $m$
is Lebesgue measure on ${\mathbb T}$
and $\mu_{{\mathbb Z}}$ be the counting measure
on ${\mathbb Z}$. Let $(G,H)$ be one of the
pairs $({\mathbb R},{\mathbb R})$, $({\mathbb T},{\mathbb Z})$
or $({\mathbb Z},{\mathbb T})$. Then
(i) If $f\in L^{1}(G,\mu_{G})$, then
\[\hat{f}(a)=\int_{G}f(t)\langle -t,a\rangle\,d\mu_{G}(t)\]
is well defined for all $a\in H$.
(ii) If $f,\, g\in L^{1}(G,\mu_{G})$, then the convolution
\[f*g(s)=\int_{G}f(s-t)g(t)d\mu_{G}(t)\]
is defined $\mu_{G}$ almost everywhere and $f*g\in L^{1}(G,\mu_{G})$.
(iii) If $f,\, g\in L^{1}(G,\mu_{G})$, then
\[\widehat{f*g}(a)=\hat{f}(a)\hat{g}(a)\]
for all $a\in H$.
\end{exercise}
(Part~(ii) uses Fubini's theorem to establish the
existence of $f*g$. It is not necessary
to know the results of Exercise~\ref{E, First convolution}
in the generality given but you ought to know some version
of these results.)
\begin{theorem}\label{T, First inversion}
We continue with the notation
and ideas of Lemma~\ref{L, first characters}
and Exercises~\ref{Exercise, first Haar}
and~\ref{E, First convolution}.
If $f\in L^{1}(G,\mu_{G})$ is sufficiently well
behaved, then
\[f(x)=\int_{H}\hat{f}(a)\langle x,a\rangle\,d\mu_{H}(a).\]
\end{theorem}
\begin{exercise}\label{E, easy inversion 1}
Show that Theorem~\ref{T, First inversion}
is true for $(G,H)=({\mathbb Z},{\mathbb T})$ without
any extra conditions on $f$ (beyond that $f\in L^{1}(G,\mu_{G})$).
\end{exercise}
As it stands Theorem~\ref{T, First inversion} is
more of an aspiration than a theorem. It
is only useful if the `sufficiently well
behaved functions' form a `sufficiently large class'.
For example, if $(G,H)=({\mathbb T},{\mathbb Z})$
any twice continuously differentiable function is
sufficiently well behaved
and if $(G,H)=({\mathbb R},{\mathbb R})$, then
any twice differentiable function $f$ with
$x^{-2}f(x),\,x^{-2}f'(x),\,x^{-2}f''(x)\rightarrow 0$
as $|x|\rightarrow\infty$ is
sufficiently well behaved. (We can do much better
but it is well known that, in both cases, the result
is not true for all continuous $f$.)
The object of this course is to illustrate
the group theoretic content of classical
Fourier analysis by extending the ideas and results
of this section to the general context of
locally compact groups.
\begin{exercise} (i) Go through the results of this
section with ${\mathbb R}^{2}$, ${\mathbb T}^{2}$
and ${\mathbb Z}^{2}$ in place of
${\mathbb R}$, ${\mathbb T}$
and ${\mathbb Z}$ making the appropriate
changes.
(ii) Go through the results of this
section with the finite cyclic group $C_{n}$.
\end{exercise}
\section{Topological groups} The following method
of marrying algebraic and topological structures
will be routine for many readers and should strike
the rest as natural.
\begin{definition} We say that $(G,\times,\tau)$ is a
topological group if $(G,\times)$ is a group
and $(G,\tau)$ a topological space such that,
writing $M(x,y)=x\times y$ and $Jx=x^{-1}$
the multiplication map $M:G^{2}\rightarrow G$
and the inversion map $J:G\rightarrow G$ are continuous.
\end{definition}
\begin{exercise} Show that $(G,\times,\tau)$ is a
topological group if $(G,\times)$ is a group
and $(G,\tau)$ a topological space such that
the map $K:G^{2}\rightarrow G$ given by
$K(x,y)=xy^{-1}$ is continuous.
\end{exercise}
The systems ${\mathbb R}$, $S^{1}$, ${\mathbb T}$
and ${\mathbb Z}$ discussed in Section~\ref{S, Prelude}
are readily seen to be examples of topological
groups. Any group $G$ becomes a topological
group when equipped with the discrete or the
indiscrete topology. (Note that this shows that
the mere fact that something is a topological group
tells us little unless we know more about the topology.)
\begin{exercise} Verify the statements just made.
\end{exercise}
There is a natural definition of isomorphism.
\begin{definition} If
$(G,\times_G,\tau_{G})$ and $(H,\times_{H},\tau_{H})$
are topological groups we say that $\theta:G\rightarrow H$
is an isomorphism if it is a group isomorphism
and a topological homeomorphism.
\end{definition}
\begin{exercise} Show that ${\mathbb T}$ and $S^{1}$
are isomorphic as topological groups. Find all the
isomorphisms.
\end{exercise}
The next result gives us a source of interesting
non-commutative topological groups.
\begin{lemma}\label{L, general linear}
Let ${\mathbb F}$ be ${\mathbb R}$ or
${\mathbb C}$. The multiplicative group
$GL_{n}({\mathbb F})$
of invertible $n\times n$ matrices with the topology
induced by the usual norm is a topological group.
\end{lemma}
\begin{exercise} Prove the following generalisation
of Lemma~\ref{L, general linear}. Let $U$ be a Banach space.
If we give the multiplicative group
$GL(U)$
of continuous linear maps $T:U\rightarrow U$
with continuous inverses the topology
induced by the operator norm then we
obtain a topological group.
\end{exercise}
It is, perhaps, worth noting that we can not
replace joint continuity of multiplication
by left and right continuity (this contrasts
with the Banach algebra case).
\begin{exercise}\label{Discontinuous 1}
Let $\tau$ be the usual topology
on ${\mathbb R}$ and let $\tau_{1}$ be the set
of $U\in \tau$ such that there exists a $K>0$ such that
\[U\supseteq {\mathbb R}\setminus(-K,K)\]
together with the empty set.
Then $\tau_{1}$ is a topology on ${\mathbb R}$
such that the map $x\mapsto -x$ and the
map $x\mapsto x+a$ are continuous (as maps
$({\mathbb R},\tau_{1})\rightarrow ({\mathbb R},\tau_{1})$)
for all $a\in {\mathbb R}$
but the map $(x,y)\mapsto x+y$ is not continuous
(as a map
$({\mathbb R}^{2},\tau_{1}^{2})\rightarrow ({\mathbb R},\tau_{1})$).
\end{exercise}
Similarly, joint continuity of multiplication
does not imply continuity of the inverse.
\begin{exercise}\label{Discontinuous 2}
Let $\tau$ be the usual topology
on ${\mathbb R}$ and let $\tau_{2}$ be the set
of $U\in \tau$ such that there exists a $K$ such that
\[U\supseteq (K,\infty)\]
together with the empty set.
Then $\tau_{2}$ is a topology on ${\mathbb R}$
such that the map $(x,y)\mapsto x+y$ is continuous
(as a map
$({\mathbb R}^{2},\tau_{1}^{2})\rightarrow ({\mathbb R},\tau_{1})$)
but the map $x\rightarrow -x$ is not continuous (as a map
$({\mathbb R},\tau_{1})\rightarrow ({\mathbb R},\tau_{1})$).
\end{exercise}
Returning to more central matters we make the following
simple but basic observations.
\begin{lemma} Let $(G,\times,\tau)$ be a topological group.
Then
(i) $xU=\{xu\,:\,u\in U\}$ is open if and only if $U$ is.
(ii) $V$ is a neighbourhood of $x$ if and only if $x^{-1}V$
is a neighbourhood of $e$.
\end{lemma}
(Here and elsewhere we will use $e$ to denote the
unit of a multiplicative group and $0$ to denote the
unit of an additive one.)
\begin{lemma}\label{L, neighbourhood basis 1}
Suppose that $(G,\times)$ is a group
and ${\mathcal N}_{e}$ is a collection of sets $N$
with $e\in N\subseteq G$. Then there exists a topology
$\tau$ on $G$ having ${\mathcal N}_{e}$ as a neighbourhood
basis of $e$ and making $(G,\times,\tau)$ into a topological group
if and only if
(1) If $N,\, M\in{\mathcal N}_{e}$, then there exists
a $P\in{\mathcal N}_{e}$ with $P\subseteq N\cap M$.
(2) If $N\in{\mathcal N}_{e}$, then there exists
an $M\in{\mathcal N}_{e}$ with $M^{2}\subseteq N$.
(3) If $N\in{\mathcal N}_{e}$, then there exists
an $M\in{\mathcal N}_{e}$ with $M\subseteq N^{-1}$.
(4) If $N\in{\mathcal N}_{e}$ and $a\in G$, then there exists
an $M\in{\mathcal N}_{e}$ with $M\subseteq aNa^{-1}$.
\end{lemma}
\begin{exercise}\label{E, neighbourhood basis 2}
Suppose that $(G,\times)$ is a group
and $(G,\tau)$ a topological space. Then $(G,\times,\tau)$
is a topological group if and only if, writing ${\mathcal N}_{a}$
for the set of open neighbourhoods of $a$, we have
(1) Let $a\in G$. Then $N\in{\mathcal N}_{e}$ if and only if
$aN\in{\mathcal N}_{a}$.
(2) If $N\in{\mathcal N}_{e}$, then there exists
an $M\in{\mathcal N}_{e}$ with $M^{2}\subseteq N$.
(3) If $N\in{\mathcal N}_{e}$, then there exists
an $M\in{\mathcal N}_{e}$ with $M\subseteq N^{-1}$.
(4) If $N\in{\mathcal N}_{e}$ and $a\in G$, then there exists
an $M\in{\mathcal N}_{e}$ with $M\subseteq aNa^{-1}$.
\end{exercise}
From time to time it is useful to have neighbourhood bases
with further properties.
\begin{lemma} If $(G,\times,\tau)$ is a topological
group we can find a neighbourhood basis ${\mathcal N}_{e}$
for $e$ consisting of open sets $N$ with $N^{-1}=N$.
\end{lemma}
The following example shows that we can not choose
$M$ in condition~(4) of Lemma~\ref{L, neighbourhood basis 1}
and Exercise~\ref{E, neighbourhood basis 2} independently
of $a$.
\begin{example}\label{conjugacy and metric}
Given any $\epsilon>0$ and any $K>0$ we
can find $A,\, B\in GL_{2}({\mathbb R})$ such that
$\|I-B\|<\epsilon$ but $\|I-ABA^{-1}\|>K$.
\end{example}
In some sense a topological group is an object
were the the statement `the neighbourhood of every
point looks the same' is (a) meaningful and
(b) true. (There is a more general notion of a
uniform topology.)
\begin{exercise} Let $(G,\times,\tau)$ be a topological group.
What ought it to mean to say that $f:G\rightarrow{\mathbb C}$
is right (or left) uniform continuous.
If $(G,\times,\tau)$ is a compact topological group,
show that any continuous function $f:G\rightarrow{\mathbb C}$
is right uniformly continuous.
\end{exercise}
\section{Subgroups and quotients} It is easy to define
topological subgroups and quotient groups
along the lines given in the next lemma.
\begin{lemma} If $(G,\times,\tau)$ is a topological group
and $H$ is a subgroup of $G$, then $H$ equipped with
the standard subspace topology is a topological group.
If $H$ is a normal subgroup of $G$ then $G/H$ equipped
with the standard quotient space topology
(formally, the finest topology on $G/H$ which makes the map
$G\rightarrow G/H$ given by $x\mapsto xH$ continuous)
is a topological group.
\end{lemma}
However, it is important to realise that, without
further conditions quotient topological groups
may not behave well.
\begin{exercise} Consider $({\mathbb R},+,\tau)$ with
the usual Euclidean topology $\tau$. Show that ${\mathbb Q}$
is a normal subgroup and ${\mathbb R}/{\mathbb Q}$
is uncountable but has the indiscrete topology.
\end{exercise}
Let us look a little closer at how a subgroup can sit
in a topological group.
\begin{lemma} Let $(G,\times,\tau)$ be a topological group
and $H$ a subgroup of $G$.
(i) The (topological) closure $\bar{H}$ of $H$ is a
subgroup.
(ii) If $H$ is normal, so is $\bar{H}$.
(iii) If $H$ contains an open set, then $H$ is open.
(iv) If $H$ is open, then $H$ is closed.
(v) If $H$ is closed and of finite index in $G$,
then $H$ is open.
\end{lemma}
\begin{lemma} Let $(G,\times,\tau)$ be a topological group.
The connected component $L$ containing $e$ is an open normal subgroup.
Any open subgroup of $G$ contains $L$.
\end{lemma}
\begin{lemma} If $(G,\times,\tau)$ is a topological group
then $I=\overline{\{e\}}$ is a closed normal subgroup.
\end{lemma}
\begin{lemma} Let $(G,\times,\tau)$ is a topological group.
The following conditions are equivalent.
(i) $\{e\}$ is closed.
(ii) $G$ is Hausdorff.
(iii) $G$ is $T_{0}$.
\end{lemma}
(Part~(iii) is just an observation. Note that the spaces
in Exercises~\ref{Discontinuous 1} and~\ref{Discontinuous 2}
are $T_{1}$ but not Hausdorff.)
\begin{lemma}\label{L,Hausdorff 1} Let $(G,\times,\tau)$ be a
topological group. If $H$ is a closed normal subgroup,
then $G/H$ is Hausdorff.
\end{lemma}
\begin{lemma}\label{L,Hausdorff 2} Let $(G,\times,\tau)$ be a
topological group. Then $G/\overline{\{e\}}$ is Hausdorff.
\end{lemma}
If $G$ is a topological group, then any continuous function
$f:G\rightarrow{\mathbb F}$ must be constant on any
coset $x+\overline{\{e\}}$ so it is clear that, if we
are interested in continuous functions, nothing is lost
by confining ourselves to Hausdorff topological groups.
Rudin defines a topological group to be what we would call
a Hausdorff topological group and Bourbaki defines a compact
space to be what we would call a Hausdorff compact space.
Unless we explicitly state otherwise all our topological
groups will be Hausdorff topological groups and we will
only quotient by closed normal subgroups.
\section{Products} We shall be interested in two
different kinds of products for topological groups.
\begin{definition} Suppose $A$ is non-empty and
$(G_{\alpha},\times_{\alpha},\tau_{\alpha})$ is
a topological group for each $\alpha\in A$.
Their {\bf complete direct product} is $G=\prod_{\alpha}G_{\alpha}$
equipped with the usual product topology $\tau_{G}$
and with multiplication given by
$({\mathbf x}\times_{G}{\mathbf y})_{\alpha}=
x_{\alpha}\times_{\alpha}y_{\alpha}$.
\end{definition}
\begin{definition} Suppose $A$ is non-empty and
$(G_{\alpha},\times_{\alpha})$
is a group for each $\alpha\in A$.
Their {\bf direct product} is the subgroup of the group
$G=\prod_{\alpha}G_{\alpha}$
with multiplication given by
$({\mathbf x}\times_{G}{\mathbf y})_{\alpha}=
x_{\alpha}\times_{\alpha}y_{\alpha}$
consisting of those ${\mathbf x}\in G$ such that
$x_{\alpha}=e$ for all but finitely many $\alpha\in A$
\end{definition}
Our definition of direct product does not involve
topology but we shall usually give the direct product
the discrete topology.
\begin{exercise} (i) Verify that the complete direct
product of topological groups is indeed a topological group.
(ii) Verify that the direct
product of topological groups is indeed a group.
(iii) Show that the complete direct
product of Abelian topological groups is Abelian.
(iv) Show that the direct
product of Abelian groups is Abelian.
(v) Show that the complete direct
product of Hausdorff topological groups is Hausdorff.
(vi) Explain why the complete direct
product of compact topological groups is compact.
\end{exercise}
We illustrate these ideas by looking at $D^{\infty}$
the complete direct product of $G_{1}$, $G_{2}$, \dots
where each $G_{i}$ is a copy of $D_{2}$ the additive
group of two elements $0$ and $1$
and $D_{0}^{\infty}$ the direct product of
$H_{1}$, $H_{2}$, \dots,
where each $H_{i}$ is a copy of $C_{2}$ the multiplicative
group of two elements $-1$ and $1$,
equipped with the discrete topology.
(Of course, $D_{2}$ and $C_{2}$ are isomorphic
as groups.)
\begin{lemma} $D^{\infty}$ is homeomorphic as a topological
space to the Cantor set. It is compact, Hausdorff,
totally disconnected
and has no isolated points.
\end{lemma}
\begin{lemma}\label{L, second characters}
(i) The continuous homomorphisms
$\chi:D^{\infty}\rightarrow S^{1}$ are precisely
the maps $\chi_{{\mathbf a}}({\mathbf t})=
\prod_{j=1}^{\infty}a_{j}^{t_{j}}$ $[{\mathbf t}\in D^{\infty}]$
with ${\mathbf a}\in D^{\infty}_{0}$.
(ii) The continuous homomorphisms
$\chi:D_{0}^{\infty}\rightarrow S^{1}$ are precisely
the maps $\chi_{{\mathbf t}}({\mathbf a})=
\prod_{j=1}^{\infty}a_{j}^{t_{j}}$ $[{\mathbf a}\in D_{0}^{\infty}]$
with ${\mathbf t}\in D^{\infty}$.
\end{lemma}
\begin{exercise}\label{Exercise, second Haar}
There is a unique Borel measure $\mu$ on
$D^{\infty}$ such that
\[\int_{D^{\infty}}f({\mathbf x}-{\mathbf y})\,d\mu({\mathbf x})=
\int_{D^{\infty}}f({\mathbf x})\,d\mu({\mathbf x})\]
for all continuous functions
$f:D^{\infty}\rightarrow{\mathbb R}$
and all ${\mathbf y}\in D^{\infty}$
and $\int_{D^{\infty}}1\,d\mu({\mathbf x})=1$.
If we write
$K_{m}=\{{\mathbf x}\in D^{\infty}\,:\, x_{1}=x_{2}=\dots=x_{m}=1\}$,
then $\mu(K_{m})=2^{-m}$.
\end{exercise}
\begin{exercise}\label{E, second convolution}
We continue with the notation
of this section. We write
$\langle {\mathbf t},{\mathbf a}\rangle=
\prod_{j=1}^{\infty}a_{j}^{t_{j}}$ when ${\mathbf t}\in D^{\infty}$
with ${\mathbf a}\in D^{\infty}_{0}$. We let
$\mu$ be the measure of Exercise~\ref{Exercise, second Haar}
and $\mu_{0}$ be the counting measure on $D_{0}^{\infty}$.
Then
(i) If $f\in L^{1}(D^{\infty},\mu)$, then
\[\hat{f}({\mathbf a})=\int_{D^{\infty}}f({\mathbf t})
\langle -{\mathbf t},{\mathbf a}\rangle\,d\mu({\mathbf t})\]
is well defined for all ${\mathbf a}\in D_{0}^{\infty}$.
(ii) If $f,\, g\in L^{1}(D^{\infty},\mu)$, then the convolution
\[f*g({\mathbf s})=\int_{D^{\infty}}
f({\mathbf s}-{\mathbf t})g({\mathbf t})d\mu({\mathbf t})\]
is defined $\mu$ almost everywhere and $f*g\in L^{1}(D^{\infty},\mu)$.
(iii) If $f,\, g\in L^{1}(D^{\infty},\mu)$, then
\[\widehat{f*g}({\mathbf a})=\hat{f}({\mathbf a})\hat{g}({\mathbf a})\]
for all ${\mathbf a}\in D_{0}^{\infty}$.
\end{exercise}
\begin{theorem}\label{T, Second inversion}
{\bf (Inversion theorem for $D^{\infty}$)}
We continue with the notation
of this section.
If $f:D^{\infty}\rightarrow{\mathbb C}$
has the form
\[f=\sum_{m=1}^{M}a_{m}{\mathbb I}_{x_{m}+K_{n(m)}}\]
(that is, $f$ is a step function constant on cosets
of subgroups of the type $K_{m}$ described
in Exercise~\ref{Exercise, second Haar}),
then
\[f({\mathbf x})=\int_{D_{0}^{\infty}}\hat{f}({\mathbf a})
\langle {\mathbf x},{\mathbf a}\rangle\,d\mu_{0}({\mathbf a}).\]
\end{theorem}
Theorem~\ref{T, Second inversion} asserts that
an inversion theorem is true for functions
of a certain well behaved type. At first sight the
reader might feel that the class involved is too small to
be useful but the next result shows that this is not the case.
\begin{theorem}\label{T, Plancherel 1}
{\bf (A Plancherel theorem for $D^{\infty}$)}
If $f\in L^{2}(D^{\infty},\mu)$,
then $\hat{f}\in L^{2}(D_{0}^{\infty},\mu_{0})$ and
\[\int_{D_{0}^{\infty}}|\hat{f}({\mathbf a})|^{2}
d\mu_{0}({\mathbf a})
=\int_{D^{\infty}}|f({\mathbf x})|^{2}
d\mu({\mathbf x}).\]
\end{theorem}
\begin{exercise}\label{E, Parseval 2}
{\bf (Parseval's theorem for $D^{\infty}$.)}
If $f,\, g\in L^{2}(D^{\infty},\mu)$, then
$fg^{*}\in L^{1}(D^{\infty},\mu)$,
$\hat{f}(\hat{g})^{*}\in L^{1}(D_{0}^{\infty},\mu_{0})$
and
\[\int_{D_{0}^{\infty}}(\hat{f})({\mathbf a})
(\hat{g})({\mathbf a})^{*}
d\mu_{0}({\mathbf a})
=\int_{D^{\infty}}f({\mathbf x})g({\mathbf x})^{*}
d\mu({\mathbf x}).\]
\end{exercise}
In many ways $D^{\infty}$ is the simplest non-trivial
Hausdorff compact topological group and, if a result is
hard for ${\mathbb T}$ it may well be easier to prove
or understand for $D^{\infty}$.
\begin{exercise} If
$f\in L^{1}(D_{0}^{\infty},\mu_{0})\cap L^{2}(D_{0}^{\infty},\mu_{0})$,
then $\hat{f}\in L^{2}(D^{\infty},\mu)$ and
\[\int_{D^{\infty}}|\hat{f}({\mathbf a})|^{2}
d\mu({\mathbf a})
=\int_{D_{0}^{\infty}}|f({\mathbf x})|^{2}
d\mu_{0}({\mathbf x}).\]
(This is not
very deep. Essentially we repeat the the easy
proof of exercise~\ref{E, easy inversion 1}
but remember that
$L^{2}(D_{0}^{\infty},\mu_{0})$ is not a subset of
$L^{1}(D_{0}^{\infty},\mu_{0})$.)
\end{exercise}
The reader is probably familiar with the classical
version of the next step.
\begin{theorem}\label{T, Plancherel 2}
There exists a linear isometry
${\mathcal F}:L^{2}(D_{0}^{\infty},\mu_{0})
\rightarrow L^{2}(D^{\infty},\mu)$
such that ${\mathcal F}(f)=\hat{f}$ whenever
$f\in L^{1}(D_{0}^{\infty},\mu_{0})\cap L^{2}(D_{0}^{\infty},\mu_{0})$.
\end{theorem}
\begin{exercise}
{\bf (Parseval's theorem for $D_{0}^{\infty}$.)}
If $f,\, g\in L^{2}(D_{0}^{\infty},\mu_{0})$, then
$fg^{*}\in L^{1}(D_{0}^{\infty},\mu_{0})$,
${\mathcal F}f{\mathcal F}g^{*}\in L^{1}(D^{\infty},\mu)$
and
\[\int_{D^{\infty}}({\mathcal F}f)({\mathbf a})
({\mathcal F}g)({\mathbf a})^{*}
d\mu({\mathbf a})
=\int_{D_{0}^{\infty}}f({\mathbf x})g({\mathbf x})^{*}
d\mu_{0}({\mathbf x}).\]
\end{exercise}
\section{Metrisability}
The following easy result
is well known.
\begin{exercise}\label{E, metric one}
If $(X,d)$ is a metric space and
$\tau$ is the topology derived from the metric,
then $(X,\tau)$ is Hausdorff and given any $x\in X$
we can find open neighbourhoods $N_{j}$ of $x$ such
that $\bigcap_{j=1}^{\infty}N_{j}=\{x\}$.
\end{exercise}
The converse is false.
\begin{example} We work in ${\mathbb R}$. Let us set
\begin{alignat*}{2}
{\mathcal N}_{x}&=\{(x-\delta,x+\delta)\cap{\mathbb Q}
\,:\,\delta>0\}&&\qquad\text{if $x\in{\mathbb Q}$,}\\
{\mathcal N}_{x}&=\{(x-\delta,x+\delta)
\,:\,\delta>0\}&&\qquad\text{otherwise.}
\end{alignat*}
Then there is a unique topology $\tau$ on ${\mathbb R}$
with the ${\mathcal N}_{x}$ as neighbourhood bases.
The topology $\tau$ is Hausdorff and
given any $x\in {\mathbb R}$
we can find open neighbourhoods $N_{j}$ of $x$ such
that $\bigcap_{j=1}^{\infty}N_{j}=\{x\}$.
However there is no metric on ${\mathbb R}$
which will induce the topology.
\end{example}
The homogeneity imposed by the group structure
means that the necessary condition introduced in
Exercise~\ref{E, metric one} is actually
sufficient. Indeed, we have
an even stronger result.
\begin{theorem}\label{Metric sufficient}
Let $(G,\times,\tau)$ be a topological group.
If we can find a base of neighbourhoods of $N_{j}$ of $e$
such that $\bigcap_{j=1}^{\infty}N_{j}=\{e\}$, then
there exists a metric $d$ on $G$ which induces
$\tau$. Moreover, we can take $d$ left invariant,
that is to say $d(gx,gy)=d(x,y)$ for all
$x,\, y,\, g\in G$.
\end{theorem}
The method of proof is illustrated by the following exercise.
\begin{exercise}
We work on ${\mathbb R}$.
(i) Suppose that $|x|>2^{-k}$ for some integer $k\geq 0$.
Show, by induction on $N$, or otherwise, that if
\[x=\sum_{j=1}^{N}x_{j}\ \text{with $|x_{j}|\leq 4^{-n(j)}$
for some integer $n(j)\geq 0$ $[1\leq j\leq N]$},\]
then $\sum_{j=1}^{N}2^{-n(j)}\geq 2^{-k}$.
(ii) Show that
\begin{align*}
d(x,y)=\inf\{\sum_{j=1}^{N}2^{-n(j)}
\,:&\,x-y=\sum_{j=1}^{N}x_{j}\ \text{with $|x_{j}|\leq 4^{-n(j)}$}\\
&\text{for some integer $n(j)\geq 0$ $[1\leq j\leq N]$}
\}
\end{align*}
defines a metric on ${\mathbb R}$ such that
$d(x+a,y+a)=d(x,y)$ for all $x,\,y,\,a\in {\mathbb R}$.
(iii) Show that we can find a constant $K>1$ such that
\[K|x-y|^{1/2}\geq d(x,y)\geq K^{-1}|x-y|^{1/2}\]
for all $x,\,y\in {\mathbb R}$.
(iv) Sketch the graph of the function $x\mapsto d(0,x)$.
\end{exercise}
We make the following remarks (recall Example~\ref{conjugacy and metric}).
\begin{exercise} (i) There exist topological groups
with a topology induced by a left invariant metric
where the topology is not induced by a left and right
invariant metric. (Briefly, a metrisable topological
group
may not have a metric which is both left and right invariant.)
(ii) The metric of Theorem~\ref{Metric sufficient}
is not unique (unless $G$ is the trivial group $\{e\}$).
\end{exercise}
\begin{exercise} Let $A$ be uncountable and,
for each $\alpha\in A$ let
$G_{\alpha}$ be a copy of $D_{2}$ (the additive
group of two elements $0$ and $1$)
equipped with the discrete topology.
Then the complete direct product of the $G_{\alpha}$
is a non-metrisable compact Abelian group.
\end{exercise}
In the rest of this section we introduce a
simple but useful definition.
\begin{definition} (i) A topological space $(X,\tau)$
is locally compact
if every point of $X$ has a compact neighbourhood.
(ii) A topological space $(X,\tau)$ is $\sigma$-compact
if it is the countable union of compact sets.
\end{definition}
\begin{lemma}\label{L, one point compact}
Suppose that $(X,\tau)$
is locally compact and Hausdorff.
If we consider $X\cup\{\infty\}$ with
a topology $\tau_{\infty}$ in which the open sets are the
open sets of $\tau$ together with the sets
$\{\infty\}\cup(X\setminus K)$ with $K$ compact
in $\tau$ then $(X\cup\{\infty\},\tau_{\infty})$
(the `one point compactification' of $X$)
is a compact Hausdorff space.
\end{lemma}
Combined with Urysohn's lemma which
we quote without proof as Theorem~\ref{Urysohn}
this gives us a plentiful supply of continuous functions.
\begin{theorem}{\bf (Urysohn's lemma.)}\label{Urysohn}
If $(X,\tau)$ is a compact Hausdorff space
and $E_{1}$ and $E_{2}$ are disjoint closed sets
we can find a continuous function $f:X\rightarrow{\mathbb R}$
such that $1\geq f(x)\geq 0$ for all $x\in X$, $f(e)=1$
for all $e\in E_{1}$ and $f(e)=0$
for all $e\in E_{0}$.
\end{theorem}
\begin{lemma} Every point in a locally compact space
has a basis of compact neighbourhoods.
\end{lemma}
Since the discrete topology is locally compact,
locally compact groups can be very big indeed.
However for many purposes we can restrict ourselves
to $\sigma$-compact topological groups.
\begin{lemma}\label{Only need sigma compact}
Let $(G,\times,\tau)$ be a locally compact Hausdorff
topological group.
If $K_{j}$ is a compact subset of $G$ for $j\geq 1$,
then there exists an open (so closed) subgroup
$H$ of $G$ which is $\sigma$-compact and such that
$H\supseteq\bigcup_{j=1}^{\infty}K_{j}$.
\end{lemma}
We may also be able to restrict ourselves
to metrisable locally compact groups.
\begin{lemma}\label{Only need metrisable}
Let $(G,\times,\tau)$ be a
$\sigma$-compact Hausdorff topological group. If
$N_{j}$ is an open neighbourhood of $e$
for $j\geq 1$, then we can find a closed
normal subgroup $H$ of $G$ with
$H\subseteq\bigcap_{j=1}^{\infty}N_{j}$
and $G/H$ metrisable and $\sigma$-compact.
\end{lemma}
\section{The Haar integral} The natural (certainly,
a natural) requirement for a topology for which
we wish to develop a theory
of continuous functions and integrals
is that it should be Hausdorff and locally
compact.
\begin{definition}\label{Definition Haar}
Let $(G,\times,\tau)$ be
a Hausdorff locally compact group. Write
$C_{00}(G)$ for the collection of
continuous functions $f:G\rightarrow{\mathbb R}$
with compact support
and $C^{+}_{00}(G)$ for the set of $f\in C_{00}(G)$
such that $f(x)\geq 0$ for all $x\in G$.
If $f\in C_{00}(G)$ and $y\in G$ we write
$f_{y}(x)=f(y^{-1}x)$ for all $x\in G$.
A non-zero linear map $I:C_{00}(G)\rightarrow{\mathbb R}$
such that $If\geq 0$ when $f\in C^{+}_{00}(G)$
and $If_{y}=If$ for all $y\in G$ and $f\in C_{00}(G)$
is called a left invariant Haar integral.
\end{definition}
Measure theory gives us
much more powerful weapons than those
developed in the next exercise but for the
moment we do not need them.
\begin{exercise} Let $(G,\times,\tau)$ be
a Hausdorff locally compact group and let
$I,\, J:C_{00}(G)\rightarrow{\mathbb R}$ be
non-zero linear maps
such that $If\geq 0$ and $Jf\geq 0$
when $f\in C^{+}_{00}(G)$.
(i) Given any compact set $K$ we can find
a constant $\gamma(K)$ such that
$If\leq \gamma(K)\|f\|_{\infty}$ for
all $f\in C^{+}_{00}(G)$ with support in $K$.
(ii)\footnote{In the course I proved a slightly harder result,
but this is all we need for~(iii) and~(iv).} Suppose that $K$
is a compact set in $G$ and $H$ is
a compact neighbourhood of $e$. If
$F:G\times G\rightarrow{\mathbb R}$ is a continuous
function with support in $K\times K$, then given $\epsilon>0$
we can find $u_{j},\ v_{j}\in C_{00}(G)$
with support in $KH$ such that
\[|F(x,y)-\sum_{j=1}^{N}u_{j}(x)v_{j}(y)|\leq\epsilon\]
for all $x,\,y\in G$.
(iii) If $F:G\times G\rightarrow{\mathbb R}$ is a continuous
function with compact support
and $y\in G$, then (using dummy variable notation)
$x\mapsto J_{y}F(x,y)$
is a continuous function of compact support.
Thus
$I_{x}J_{y}F(xy)$ exists.
(iv) $F:G\times G\rightarrow{\mathbb R}$ is a continuous
function with compact support, then
\[I_{x}J_{y}F(x,y)=J_{y}I_{x}F(x,y)\]
(v) If $I$ is a Haar measure $g\in C_{00}^{+}(G)$
and $Ig=0$ then $g=0$.
\end{exercise}
\begin{theorem} If a left invariant Haar integral
exists it is unique up to multiplication by a strictly positive constant.
\end{theorem}
Here are some examples. (Note that if $G$ is Abelian a left invariant
Haar integral must be right invariant.)
\begin{exercise}\label{First Haar examples}
(i) Show that the set $G$ of
all matrices of the form
\[\left(\begin{matrix}x&y\\0&x\end{matrix}\right)\]
with $x$ and $y$ real and $x>0$ with the usual norm
gives rise to a locally compact metrisable Abelian group.
By informal but reasonably coherent arguments show that
a Haar integral must be a multiple of
\[\int_{G}f\,d\mu=\int_{-\infty}^{\infty}\int_{0}^{\infty}
f\left(\begin{matrix}x&y\\0&x\end{matrix}\right)\frac{1}{x^{2}}\,dx\,dy.\]
Verify that this is indeed a Haar integral.
(ii) Show that the set $G$ of
all matrices of the form
\[\left(\begin{matrix}x&y\\0&1\end{matrix}\right)\]
with $x$ and $y$ real and $x>0$
gives rise to a locally compact metrisable non-Abelian group.
By informal but reasonably coherent arguments show that
the left invariant Haar integral must be a multiple of
\[\int_{G}f\,d\mu=\int_{-\infty}^{\infty}\int_{0}^{\infty}
f\left(\begin{matrix}x&y\\0&1\end{matrix}\right)\frac{1}{x^{2}}\,dx\,dy.\]
Verify that this is indeed a left invariant Haar integral.
By informal but reasonably coherent arguments show that
the right invariant Haar integral must be a multiple of
\[\int_{G}f\,d\mu=\int_{-\infty}^{\infty}\int_{0}^{\infty}
f\left(\begin{matrix}x&y\\0&1\end{matrix}\right)\frac{1}{x}\,dx\,dy.\]
Verify that this is indeed a right invariant Haar integral.
\end{exercise}
Once the ideas of Exercise~\ref{First Haar examples}
have been understood it is fairly easy to extend them
to more complicated situations.
\begin{exercise}\label{Second Haar examples}
(i) Show that $GL({\mathbb R}^{n})$ is a locally compact
group which is non-Abelian if $n\geq 2$.
If we identify $GL({\mathbb R}^{n})$ as a subset of ${\mathbb R}^{n^{2}}$
in the usual manner and let $m$ be Lebesgue measure on ${\mathbb R}^{n^{2}}$
show by informal but reasonably coherent arguments that
a left invariant Haar integral must be a multiple of
\[\int_{G}f(A)\,d\mu=\int_{{\mathbb R}^{n^{2}}}f(A)|\det A|^{-n}dm\]
where we define $f(A)=0$ when $A\notin GL({\mathbb R}^{n})$.
Verify that this is both a left and a right invariant Haar integral.
(ii) Consider the set $GA({\mathbb R}^{n})$ of invertible
affine transformations ${\mathbf x}\rightarrow{\mathbf t}+A{\mathbf x}$
with ${\mathbf t}\in{\mathbb R}^{n}$ and $A\in GL({\mathbb R}^{n})$.
If we identify $GA({\mathbb R}^{n})$ as a subset of ${\mathbb R}^{n+n^{2}}$
in the usual manner and let $m$ be Lebesgue measure on ${\mathbb R}^{n+n^{2}}$
show by informal but reasonably coherent arguments that
the left invariant Haar integral must be a multiple of
\[\int_{G}f(A)\,d\mu_{L}=
\int_{{\mathbb R}^{n+n^{2}}}f({\mathbf t},A)|\det A|^{-n}dm\]
and the right invariant Haar integral must be a multiple of
\[\int_{G}f(A)\,d\mu_{R}=\int_{{\mathbb R}^{n+n^{2}}}f(A)|\det A|^{-n-1}dm\]
(with natural conventions). Verify that we have indeed left invariant and
right invariant Haar integrals.
\end{exercise}
The following remarks range from the obvious to
the fairly obvious.
\begin{lemma} (i) Let $(G,\times,\tau)$ be
a Hausdorff locally compact group. If $G$ has a
left invariant Haar integral it has a right invariant
Haar integral.
(ii) Suppose $I$ is a left invariant Haar integral for $G$.
If $t\in G$ there exists a $\Delta(t)>0$ such that
\[I_{x}f(xt)=\Delta(t)I_{x}f(x)\]
for all $f\in C_{00}(G)$.
(iii) The function $\Delta:G\rightarrow (0,\infty)$
depends only on $G$ (and not on the particular choice of $I$).
\end{lemma}
The function $\Delta$ is called the modular function.
(Although we discuss it briefly, we do not use it.)
The statement that $\Delta=1$ is equivalent to the
statement that left invariant Haar integrals are
right invariant and vice versa. A group with this
property is called unimodular.
\begin{exercise} Let $(G,\times,\tau)$ be
a Hausdorff locally compact group
with a left invariant Haar measure.
(i) Give $(0,\infty)$ its standard multiplicative structure
and topology. The map $\Delta:G\rightarrow (0,\infty)$
is a continuous homomorphism.
(ii) If $G$ is Abelian
or compact or discrete, then it is unimodular.
(iii) If $I$ is a left invariant Haar integral for
$G$, then
\[I_{x}(f(x^{-1})\Delta(x^{-1}))=I_{x}f(x).\]
\end{exercise}
\begin{exercise} Calculate the modular function
for the groups of Exercises~\ref{First Haar examples}
and~\ref{Second Haar examples}.
\end{exercise}
\section{Existence of the Haar integral} The main
business of this section is the proof of the
following theorem.
\begin{theorem}\label{Haar for metric}
Any metrisable $\sigma$-compact
topological group $(G,\times,\tau)$ has a
left invariant Haar integral.
\end{theorem}
We need a preliminary result.
\begin{lemma} (i) Any $\sigma$-compact metric space has
a countable dense subset.
(ii) If $G$ is a $\sigma$-compact metric group then
we can find a countable collection of compactly
supported, everywhere non-negative, continuous
functions $g_{i}:X\rightarrow{\mathbb R}$
with the following property. If $\epsilon>0$,
$H$ is a compact neighbourhood of $e$ and
$g:X\rightarrow{\mathbb R}$ is a non-negative continuous
function vanishing outside a compact set $K$ we
can find $i\geq 1$ such that
$g_{i}$ vanishes outside $KH$ and
\[|g(x)-g_{i}(x)|<\epsilon\]
for all $x\in X$
\end{lemma}
From now on until the completion of the proof of
Theorem~\ref{Haar for metric}, $(G,\times,\tau)$
will be a $\sigma$-compact topological group
with topology derived from a metric $d$.
If $u:G\rightarrow {\mathbb R}$ we write $(T_{y}u)(x)=u(xy)$
for $x,\, y\in G$.
Although the discovery of the Haar integral appears to
have been unexpected, much of the construction follows
very natural lines.
If $f$ and $\phi$ are members of $C_{00}^{+}(G)$
and $\phi$ is non zero we define
\[(f;\phi)=\inf\{\sum_{j=1}^{n}c_{j}
\,:\,\sum_{j=1}^{n}c_{j}\phi(y_{j}x)\geq f(x)\ \text{for all $x\in G$},
\ c_{j}\geq 0, y_{j}\in G, in\geq 1\}.\]
The following results are routine.
\begin{lemma}\label{First upper integral}
Let $f,\, f_{1},\, f_{2}$ be members of $C_{00}^{+}(G)$
and let $\phi, \, \psi$ be non zero members of $C_{00}^{+}(G)$.
Let $y\in G$ and $\lambda>0$. Then
(i) $(f;\phi)$ is well defined.
(ii) $(f_{1}+f_{2};\phi)\leq (f_{1};\phi)+(f_{2};\phi)$.
(iii) $(\lambda f;\phi)=\lambda (f;\phi)$.
(iv) If $f_{1}(x)\leq f_{2}(x)$ for all $x\in G$ then
$(f_{1};\phi)\leq (f_{2};\phi)$.
(v) $(T_{y}f;\phi)=(f;\phi)$.
(vi) $(f;\psi)\leq (f;\phi)(\phi;\psi)$.
(vii) $(f;\phi)\geq \|f\|_{\infty}/\|\phi\|_{\infty}$.
\end{lemma}
We need to normalise the `upper approximation' $(f;\phi)$.
To do this, fix, once and for all, $f_{0}$ as
a particular non-zero element of $C_{00}^{+}(G)$
and set
\[I_{\phi}f=\frac{(f;\phi)}{(f_{0};\phi)}\]
\begin{exercise} Interpret the results of
Lemma~\ref{First upper integral} in terms
of $I_{\phi}$ and $I_{\psi}$. In particular
note that
\[\frac{1}{(f_{0},f)}\leq I_{\phi}f\leq(f;f_{0})\]
whenever $f$ is non-zero.
\end{exercise}
As might be hoped the `quality' of the `upper approximation'
is improved by taking $\phi$ of small support. (Note, however,
that the approximation is not uniform.)
\begin{lemma}\label{Near additive}
Given $f_{1},\, f_{2}\in C_{00}^{+}(G)$
and $\epsilon>0$,
we can find a neighbourhood $V$ of $e$ such that
if $\phi\in C_{00}^{+}(G)$ is non-zero and $\supp \phi\subseteq V$
then
\[I_{\phi}f_{1}+I_{\phi}f_{2}\leq I_{\phi}(f_{1}+f_{2})+\epsilon.\]
\end{lemma}
A sequential compactness argument now gives Theorem~\ref{Haar for metric}
(the existence of a left invariant Haar integral on
a $\sigma$-compact metrisable group).
Using Lemmas~\ref{Only need metrisable}
and~\ref{Only need sigma compact} we obtain
first Lemma~\ref{Haar for sigma compact}
and then the general
Theorem~\ref{Haar for all}
\begin{lemma}\label{Haar for sigma compact}
Any $\sigma$-compact Hausdorff
topological group $(G,\times,\tau)$ has a
left invariant Haar integral.
\end{lemma}
\begin{theorem}\label{Haar for all}
Any locally compact
Hausdorff
topological group $(G,\times,\tau)$ has a
left invariant Haar integral.
\end{theorem}
\noindent{\bf Note:} The standard argument for obtaining
the Haar integral on a general Hausdorff
locally compact group from Lemma~\ref{Near additive} uses
Tychonoff's theorem and so the axiom of choice
(see~\cite{Nachbin}). Our treatment uses countable
choice. Cartan produced a proof which avoids any use
of the axiom of choice (see~\cite{Nachbin} again).
The reader probably does not need to be told how remarkable
and useful Theorem~\ref{Haar for all} is. The absence
of such a result makes possible the following phenomenon.
\begin{example} If $G$ is the group freely generated
by two generators then we can find a function
$f:G\rightarrow{\mathbb R}$ such that $1\geq f(x)\geq 0$
for all $x\in G$, together with $y_{1},\, y_{2},\, y_{3},\, y_{4}\in G$
such that
\[f(y_{1}x)+f(y_{2}x)-f(y_{3}x)-f(y_{4}x)\leq -1\]
for all $x\in G$.
\end{example}
\section{The space $L^{1}(G)$}
Once we have a Haar integral $I$
on a locally compact Hausdorff topological
group we can develop the
standard theory of integration. The quickest way
is simply to define $L^{1}(G)$ as the completion
of the space $C_{00}(G)$ normed by
\[\|f\|_{1}=I(|f|).\]
This would cover all the integration theory that
we need but does not connect the integral
with the underlying space. The natural path
is to define
\[\lambda(K)=
\inf\{If\,:\, f\in C_{00}(G),\ f(x)\geq 1\ \text{for $x\in K$}\}\]
whenever $K$ is a compact set and show that $\lambda$ satisfies
the appropriate consistency conditions which allow
it to be extended to a measure $\mu$ on all Borel sets
such that
\[If=\int_{G}f\,d\mu\]
for all $f\in C_{00}(G)$.
(For details, see for example,~\cite{Halmos}~Chapter~X.)
An important consequence of Lemma~\ref{Only need sigma compact}
is that, although the measure $\mu$ itself may
not be a $\sigma$-finite measure, the pathologies
(failure of Fubini's theorem etc) associated with
non-$\sigma$-finite measures cannot occur.
We shall not give a specific proof of the following result.
\begin{lemma} Let $I$ be a left invariant Haar integral
on a Hausdorff locally compact topological group
$(G,\times,\tau)$.
Then $C_{00}(G)$ is dense in $(L^{1}(G),\|\ \|)$.
If $f\in L^{1}(G,m)$ then, setting $f_{y}(x)=f(yx)$,
we have $f_{y}\in L^{1}(G,m)$ and
\[\int_{G}f\,dm=\int_{G}f_{y}\,dm.\]
\end{lemma}
\begin{exercise}
If $f\in L^{1}(G,m)$, then the map $y\mapsto f_{y}$
from $G$ to $L^{1}(G)$ is continuous.
\end {exercise}
\begin{exercise} $\int_{G}1\,dm<\infty$ if and only if $G$ is compact.
\end{exercise}
Unless specifically stated we normalise Haar measure on compact
groups to give $\int_{G}1\,dm=1$ and on discrete groups so that
$m(\{e\})=1$.
\begin{lemma} If $f,\, g\in L^{1}(G,m)$ then
\[f*g(x)=\int_{G}f(xy)g(y^{-1})\,dm(y)\]
is well defined $m$ almost everywhere
and $f*g\in L^{1}(G,m)$ with $\|f*g\|_{1}\leq \|f\|_{1}\|g\|_{1}$.
\end{lemma}
From now on we only deal with Hausdorff locally compact
{\bf Abelian} groups (although some of the results
carry over to the non-Abelian case).
\begin{theorem}\label{T, L one Banach}
If $(G,+,\tau)$ is a locally compact
Hausdorff Abelian group with Haar measure $m$. then
the Banach space $L^{1}(G,m)$ equipped with convolution
$*$ as multiplication is a commutative Banach algebra.
Further, the map $f\mapsto f^{*}$ is an involution.
(That is to say $f^{**}=f$, $(f+g)^{*}=f^{*}+g^{*}$,
$(\lambda f)^{*}=\lambda^{*}f^{*}$ and $(f*g)^{*}=f^{*}*g^{*}$.)
\end{theorem}
\begin{exercise} If
$(G,+,\tau)$ is a locally compact
Hausdorff Abelian group the map $f\mapsto \tilde{f}$ with
$\tilde{f}(x)=f^{*}(-x)$ is an involution on
$L^{1}(G,m)$ equipped with convolution.
\end{exercise}
\begin{exercise} Convolution is commutative if and only
if $G$ is commutative.
\end{exercise}
\section{Characters} Recall that a character $\theta$ on
a commutative Banach algebra $B$ is non-zero linear functional
$\theta:B\rightarrow{\mathbb C}$ such that
$\theta(ab)=\theta(a)\theta(b)$ for all $a,\, b\in B$.
We give the collection ${\mathcal M}$ the weak-* topology
in which $\theta_{1}$ has neighbourhood basis formed
by sets of the form
\[\{\theta\,:\,|\theta(a_{j})-\theta_{1}(a_{j})|<\epsilon_{j}
\ [1\leq j\leq n]\}.\]
We define $\hat{a}(\theta)=\theta(a)$. The Gelfand
transform $a\mapsto \hat{a}$ is a continuous
algebra homomorphism
from $B$ to $C({\mathcal M})$ with $\|\hat{a}\|_{\infty}
=\rho(a)$.
How does this fit with the Banach algebra of Theorem~\ref{T, L one Banach}
and the classical Fourier analysis of section~\ref{S, Prelude}?
Following the pattern of section~\ref{S, Prelude} we
make the following definition.
\begin{definition} If $G$ is a Hausdorff locally compact
Abelian group we say that a continuous group homomorphism
$\chi:G\rightarrow S^{1}$ is a character of the group.
We write $\hat{G}$ for the set of such characters
and $\langle x,\chi\rangle=\chi(x)$ for all
$x\in G$ and $\chi\in \hat{G}$.
\end{definition}
\begin{theorem} If $G$ is a Hausdorff locally compact
Abelian group with Haar measure $m$ then if we fix
$\chi$ and write
\[(\chi)(f)=\int_{G}\langle -x,\chi\rangle f(x)\,dm(x),\]
the map $L^{1}\rightarrow{\mathbb C}$ is a non-zero
multiplicative linear functional. Every
multiplicative linear functional arises in this way
and distinct $\chi$ give rise to distinct multiplicative
linear functionals.
\end{theorem}
Thus if we give $\hat{G}$ the topology in which a character
$\chi_{1}$ has neighbourhood basis formed
by sets of the form
\[\{\chi\,:\,|\chi(f_{j})-\chi_{1}(f_{j})|<\epsilon_{j}
\ [1\leq j\leq n]\}\]
(with $\epsilon_{j}>0$, $f_{j}\in L^{1}(G)$, $1\leq j\leq n$, $n\geq 1$)
we can identify $\hat{G}$ with ${\mathcal M}$
and obtain
\[\hat{f}(\chi)=\int_{G}\langle -x,\chi\rangle f(x)\,dm(x)\]
satisfactorily uniting the Gelfand and the Fourier transform.
Note that we have extended the classical slogan
`Fourier transformation is a way to convert convolution
into multiplication' by showing that it is the only
(reasonable) way.
The nature of the Gelfand topology on $\hat{G}$ is illuminated
by the following lemma.
\begin{lemma}\label{First steps dual}
If $G$ is a Hausdorff locally compact
Abelian group then
(i) The map $G\times \hat{G}\rightarrow{\mathbb C}$ given
by $(x,\chi)\mapsto\langle x,\chi \rangle$ is continuous.
(ii)$_{a}$ Sets of the form
\[\{\chi\in\hat{G}\,:
\,|\langle x,\chi\rangle-\langle x,\chi_{1}\rangle|<\epsilon
\ \text{for all $x\in K$}\}\]
with $K$ compact in $G$ and $\epsilon>0$ are open
in $\hat{G}$.
(ii)$_{b}$ Sets of the form
\[\{x\in G\,:
\,|\langle x,\chi\rangle-\langle x_{1},\chi\rangle|<\epsilon
\ \text{for all $\chi\in K$}\}\]
with $K$ compact in $\hat{G}$ and $\epsilon>0$ are open
in $G$.
(iii)$_{a}$ Sets of the form
\[\{\chi\in\hat{G}\,:
\,|\langle x,\chi\rangle-\langle x,\chi_{1}\rangle|<\epsilon
\ \text{for all $x\in K$}\}\]
with $K$ compact in $G$ and $\epsilon>0$
form neighbourhood bases
at each $\chi_{1}\in\hat{G}$.
(iv) $\hat G$ is a locally compact Hausdorff Abelian group.
\end{lemma}
Note the absence (for the time being) of any part~(iii)$_{b}$.
\begin{exercise}\label{Dual discrete}
(i) If $G$ is compact, $\hat{G}$ is discrete.
(ii)If $G$ is discrete, $\hat{G}$ is compact.
\end{exercise}
\section{Fourier transforms of measures} Although we want an
inversion theorem stated in terms of Haar measures alone,
the standard treatments (see~\cite{Rudin} which
we follow closely and~\cite{Loomis}) require us to look at
more general measures. This is not surprising since
$L^{1}(G)$ is not weak-* closed in the appropriate
space of measures.
\begin{definition} If $X$ is a locally compact space
then $M(X)$ is the set of measures $\mu$ on the Borel sets
of $X$ such that $\|\mu\|$ is finite
and $\mu$ is \emph{regular} that is to say:-
Given any $\epsilon>0$ we can find a compact set $K$
such that $|\mu|(X\setminus K)<\epsilon$.
\end{definition}
(If we give the group ${\mathbb T}$ the discrete
topology and define $\mu(E)=0$ if $E$ is countable and
$\mu(E)=1$ if ${\mathbb T}\setminus E$ is countable
then $\mu$ is not regular.)
\begin{exercise} Consider the following statements about
a measure $\mu$ on a locally compact Hausdorff group $G$.
(i) $\mu$ is $\sigma$-finite.
(ii) $G$ is $\sigma$-compact.
(iii) $\mu$ is regular.
(iv) $\mu$ is finite.
What relations, if any, hold between these concepts.
Show that if $\mu$ is regular its support lies inside
a $\sigma$-compact normal open subgroup of $G$.
\end{exercise}
\begin{lemma} Let $(G,+,\tau)$ be a locally compact
Hausdorff Abelian group.
(i) If $E$ is a Borel set in $G$ then
\[\tilde{E}=\{(x,y)\in G^{2}\,:\,x+y\in E\}\]
is Borel in $G^{2}$.
(ii) If $\mu_{1},\, \mu_{2}\in M(G)$ then writing
\[\mu_{1}*\mu_{2}(E)=\mu_{1}\times\mu_{2}(\tilde{E})\]
gives us $\mu_{1}*\mu_{2}\in M(G)$.
\end{lemma}
\begin{exercise} If $(G,+,\tau)$ is a locally compact
Hausdorff Abelian group then $(M(G),+,*,\|\ \|)$
is a commutative Banach algebra with unit.
\end{exercise}
If $\mu\in M(G)$ we can define the Fourier transform
\[\hat{\mu}(\chi)=\int_{G}\langle -x,\chi \rangle d\mu(x).\]
\begin{exercise} (i) If $\mu\in M(G)$ then $\hat{\mu}$
is a bounded uniformly continuous function $G\rightarrow{\mathbb C}$.
(ii) If $\mu_{1}\, \mu_{2}\in M(G)$
then $\widehat{\mu_{1}*\mu_{2}}(\chi)=\hat{\mu}_{1}(\chi)\hat{\mu}_{2}(\chi)$.
\end{exercise}
We can now prove our first uniqueness theorem.
\begin{theorem} Let $(G,+,\tau)$ be a locally compact
Hausdorff Abelian group. If $\mu\in M(\hat{G})$ and
\[\int_{\hat{G}}\langle x, \chi\rangle\,\mu(\chi)=0\]
for all $x\in G$, then $\mu=0$.
\end{theorem}
\section{Discussion of the inversion theorem}
We want a theorem of the form
\begin{equation*}
\int_{\hat{G}}\langle \chi, x\rangle\hat{f}(\chi)
\,dm_{\hat{G}}(\chi)
\stackrel{?}{=}Af(x),
\tag*{$\bigstar$}
\end{equation*}
but we know from the classical case that such a result
does not hold without restriction.
\begin{example} (i) Consider $f:{\mathbb R}\rightarrow{\mathbb R}$
defined by $f(x)=1$ for $|x|\leq 1$, $f(x)=0$ otherwise.
Then $f\in L^{1}$ but $\hat{f}\notin L^{1}$.
(ii) There exists a continuous function
$g:{\mathbb R}\rightarrow{\mathbb R}$ of compact support
such that $\hat{g}\notin L^{1}$.
\end{example}
It is trivial that if $\bigstar$ is to hold
we must have $f(x)=\hat{\mu}(x)$ for some $\mu\in M(\hat{G})$.
Remarkably the converse holds.
\begin{TA} If $f\in L^{1}(G)$ and $f=\hat{\mu}$
for some $\mu\in M(\hat{G})$ then
\[\int_{\hat{G}}\langle \chi, x\rangle\hat{f}(\chi)
\,dm_{\hat{G}}\chi
=Af(x)\]
$m_{G}$ almost everywhere for some $A$ independent of $f$.
\end{TA}
How can recognise that $f=\hat{\mu}$? In general,
we can not but, remarkably, we can characterise
$\hat{\mu}$ when $\mu$ is a positive measure in $M(G)$.
(Since every member $\mu$ of $M(\hat{G})$ an be written
$\mu=\mu_{1}-\mu_{2}+i\mu_{3}-i\mu_{4}$ with
$\mu_{j}\in M^{+}(\hat{G})$ this also gives
a handle on general elements of $M(\hat{G})$.
One again we start with a near triviality.
\begin{lemma}
If $\mu\in M^{+}(\hat{G})$ then $\hat{\mu}$ is continuous and
\[\sum_{1\leq j,k\leq n}c_{j}c_{k}^{*}\hat{\mu}(x_{j}-x_{k})\geq 0\]
for all $x_{j}\in{\mathbb C}$ $c_{j}\in{\mathbb C}$
$n\geq 1$.
\end{lemma}
Remarkably the converse holds
(this was first observed by Herglotz
for $G={\mathbb Z}$ and then by Bochner for
the deeper case
$G={\mathbb R}$.)
\begin{definition} A function $\phi:G\rightarrow{\mathbb C}$
is called positive definite if
\[\sum_{1\leq j,k\leq n}c_{j}c_{k}^{*}\phi(x_{j}-x_{k})\geq 0\]
for all $x_{j}\in G$ $c_{j}\in{\mathbb C}$
$n\geq 1$.
\end{definition}
\begin{TB} {\bf (Bochner's theorem.)}
A continuous function $f:G\rightarrow{\mathbb C}$
is positive definite if and only if $f=\hat{\mu}$
for some $\mu\in M^{+}(\hat{G})$.
\end{TB}
Although we introduced Theorems $A$ and $B$ in alphabetical order
we shall first prove $B$ and then $A$.
As an indication of their utility we make the following observation.
\begin{lemma} (i) If $f,\, g\in L^{2}(G)$ then
$f*g$ is a well defined continuous function.
(ii) If $f\in L^{2}(G)$ and we set $\tilde{f}(x)=f(-x)^{*}$,
then $f*\tilde{f}$ is positive
definite.
\end{lemma}
\section{The inversion theorem} Before proving Bochner's theorem
we need some simple results on positive definite functions.
\begin{lemma} Suppose that $(G,+,\tau)$ is a locally compact
Abelian Hausdorff group and $\phi:G\rightarrow{\mathbb C}$ is
positive definite. Then $\phi(0)$ is real and positive and
(i) $\phi(-x)=\phi^{*}(x)$,
(ii) $|\phi(x)|\leq \phi(0)$
(iii) $|\phi(x)-\phi(y)|^{2}\leq 2\phi(0)\Re(\phi(0)-\phi(x-y))$
\noindent for all $x,\, y\in G$.
In particular, if $\phi$ is continuous at $0$, then $\phi$ is
uniformly continuous.
\end{lemma}
\begin{lemma}
Suppose that $(G,+,\tau)$ is a locally compact Abelian
Hausdorff group and $\phi:G\rightarrow{\mathbb C}$ is
positive definite and continuous. Then
\[\int_{G}\int_{G}f(x)f^{*}(y)\phi(x-y)\,dm_{G}(x)\,dm_{G}(y)\geq 0\]
for all $f\in L^{1}(G)$.
\end{lemma}
We now prove Bochner's theorem.
\begin{theorem}\label{Bochner theorem}
Suppose that $(G,+,\tau)$ is a locally compact Abelian
Hausdorff group. Then
a continuous function $\phi:G\rightarrow{\mathbb C}$
is positive definite if and only if $\phi=\hat{\mu}$
for some $\mu\in M^{+}(\hat{G})$.
\end{theorem}
Now we can prove our main inversion theorem.
\begin{theorem}\label{Main inversion theorem}
Suppose that $(G,+,\tau)$ is a locally compact Abelian
Hausdorff group. Let $m_{G}$ be a Haar measure on $G$.
Then there exists a Haar measure $m_{\hat{G}}$ on $\hat{G}$
such that,
if $f\in L^{1}(G)$ and $f=\hat{\mu}$
for some $\mu\in M(\hat{G})$, then $\hat{f}\in L^{1}(\hat{G})$ and
\[\int_{\hat{G}}\langle \chi, x\rangle\hat{f}(\chi)
\,dm_{\hat{G}}(\chi)=f(x)\]
$m_{G}$ almost everywhere.
\end{theorem}
Henceforward we shall assume that our Haar measures on $G$ and $\hat{G}$
are chosen so that the formula of Theorem~\ref{Main inversion theorem}
holds.
\begin{exercise} Recall Exercise~\ref{Dual discrete}.
Check that our new convention for Haar measure
is consistent with our
previous conventions for choosing Haar measures for
discrete and compact groups at least if the groups are
not both compact and discrete.
What happens if our group
is both compact and discrete?
\end{exercise}
\begin{theorem}\label{T, Plancherel 3}
{\bf (Plancherel.)}
Suppose that $(G,+,\tau)$ is a locally compact
Hausdorff group.
If $f\in L^{1}(G)\cap L^{2}(G)$,
then $\hat{f}\in L^{2}(\hat{G})$ and
\[\int_{\hat{G}}|\hat{f}(\chi)|^{2}
\,dm_{\hat{G}}(\chi)
=\int_{G}|f(x)|^{2}
\,dm_{G}(x).\]
\end{theorem}
\begin{theorem}\label{T, Plancherel 4}
Suppose that $(G,+,\tau)$ is a locally compact
Hausdorff group.
There exists a linear isometry
${\mathcal F}:L^{2}(G)\rightarrow L^{2}(G)$
such that ${\mathcal F}(f)=\hat{f}$ whenever
$f\in L^{1}(G)\cap L^{2}(G)$.
\end{theorem}
\begin{lemma} If $F,\,G\in L^{2}(\hat{G})$ then we can find
a $k\in L^{1}{G}$ with $\hat{f}=F*G$.
\end{lemma}
\section{Pontryagin duality} Using the inversion theorem
we can increase the symmetry between $G$ and $\hat{G}$.
Our first step is to obtain the missing
part~(iii)$_{b}$ of Lemma~\ref{First steps dual}.
\begin{lemma}\label{Second steps dual}
If $G$ is a Hausdorff locally compact
Abelian group then
sets of the form
\[\{x\in G\,:
\,|\langle x,\chi\rangle-\langle x_{1},\chi\rangle|<\epsilon
\ \text{for all $\chi\in K$}\}\]
with $K$ compact in $\hat{G}$ and $\epsilon>0$
are open
and form neighbourhood bases
at each $x_{1}\in G$.
\end{lemma}
\begin{lemma} If $\gamma\in\hat{G}$ and $K$ is a compact
neighbourhood of $\gamma$ we can find an $f\in L^{1}(G)(G)$
with $\hat{f}(\gamma)>0$, $\hat{f}(\chi)\geq 0$ for all $\chi\in\hat{G}$
and $\hat{f}(\chi)=0$ for all $\chi\notin K$.
\end{lemma}
\begin{theorem} {\bf (Pontryagin's duality theorem.)}
If $G$ is a Hausdorff locally compact
Abelian group, then
the map $\Phi:G\rightarrow \Hat{\Hat{G}}$
defined by
\[\Phi(x)(\chi)=\chi(x)\]
for $x\in G$ and $\chi\in \hat{G}$
is an algebraic isomorphism and a topological homeomorphism.
\end{theorem}
Thus we can identify $G$ with $\Hat{\Hat{G}}$
in a natural manner.
Here are some immediate corollaries.
\begin{lemma}
Let $G$ be a Hausdorff locally compact
Abelian group.
(i) If $\mu\in M(G)$ and $\hat{\mu}(\chi)=0$ for
all $\chi\in\hat{G}$ then $\mu=0$.
(ii) If $\mu\in M(G)$ and $\hat{\mu}\in L^{1}(\hat{G})$
then there exists an $f\in L^{1}(G)$ such that
$\mu=fm_{G}$ and
\[f(x)=\int_{\hat{G}}\hat{\mu}(\chi)\langle x,\chi\rangle
\,dm_{\hat{G}}(\chi).\]
(iii) If $G$ is not discrete then $L^{1}(G)$
has no unit. Thus $L^{1}(G)=M(G)$ if and only
if $G$ is discrete.
\end{lemma}
It is worth noting that since the dual of a compact Hausdorff
Abelian group $G$ is a discrete group $\hat{G}$ and the
dual of $\hat{G}$ is $G$ all the topological
information about $G$ is encoded as algebraic information
about $\hat{G}$.
\section{Structure theorems} We conclude by deriving
a substantial amount of information about the
structure of a general locally compact Hausdorff
Abelian group.
Our first results are not unexpected.
\begin{definition} Let $G$ be a Hausdorff locally compact
Abelian group and let $H$ be a closed subgroup of $G$.
We write
\[H^{\perp}=\{\chi\in\hat{G}\,:\,\langle x,\chi\rangle=1
\ \text{for all $x\in H$}\}\]
and call $H^{\perp}$ the annihilator of $H$.
\end{definition}
\begin{lemma} Let $G$ be a Hausdorff locally compact
Abelian group and let $H$ be a closed subgroup of $G$.
(i) $H^{\perp}$ is a closed subgroup of $\hat{G}$.
(ii) $H^{\perp\perp}=H$.
(iii) $H^{\perp}$ is isomorphic as a topological
group with $(G/H)\hat{\ }$.
(iv) $\hat{G}/H^{\perp}$ is isomorphic as a topological
group with $\hat{H}$.
\end{lemma}
\begin{theorem}\label{Extend}
Let $G$ be a Hausdorff locally compact
Abelian group and let $H$ be a closed subgroup of $G$.
Then any character of $H$ can be extended to a character
of $G$.
\end{theorem}
The next results are also expected. Let us write $G\oplus H$
for the direct sum of two Abelian topological groups $G$ and $H$.
\begin{lemma}
Let $G_{j}$ be a Hausdorff locally compact
Abelian group for each $1\leq j\leq n$
Then
\[(G_{1}\oplus G_{2}\oplus G_{3} \dots \oplus G_{n})\hat{\ }
=\hat{G}_{1}\oplus \hat{G}_{2}\oplus \hat{G}_{3} \dots \oplus \hat{G}_{n}.\]
\end{lemma}
The next exercise echos Lemma~\ref{L, second characters}.
\begin{exercise} The dual of the complete direct sum of
a collection $G_{\alpha}$ $[\alpha\in A]$
of compact Hausdorff Abelian groups is the
direct sum of the dual groups $\hat{G}_{\alpha}$ $[\alpha\in A]$.
\end{exercise}
We can now state the structure theorem we wish to prove.
\begin{theorem} {\bf (Principal structure theorem.)}\label{T, principal}
If $G$ is a Hausdorff locally compact
Abelian group then we can find an open
(so closed) subgroup $H$ such that
$H=W\oplus {\mathbb R}^{n}$ with $W$ a compact group.
\end{theorem}
The lemmas that follow are directed towards the proof
of the principal structure theorem.
\begin{lemma} Suppose that $G$ is a locally compact
Hausdorff topological group containing a dense
cyclic group. If $G$ is not compact then $G$
is (topological group isomorphic to) ${\mathbb Z}$.
\end{lemma}
\begin{lemma} Suppose that $G$ is a locally compact
Hausdorff topological group generated by a compact
neighbourhood $V$
of $0$. Then $G$ contains a closed
subgroup (topological group isomorphic to) ${\mathbb Z}^{n}$
such that $G/{\mathbb Z}^{n}$ is compact
and $V\cap{\mathbb Z}^{n}=\{0\}$.
\end{lemma}
The next two lemmas concern groups like $D^{\infty}$.
\begin{lemma} Suppose $E$ is a compact open set
in a locally compact
Hausdorff topological group $G$.
(i) There exists a neighbourhood $W$ of $0$
with $W=-W$ and $E+W=E$.
(ii) If $0\in E$ then $E$ contains a compact open subgroup of $G$.
(iii) $E$ is the finite union of open cosets in $G$.
\end{lemma}
Recall that a totally disconnected topological space is
one in which the connected components are singletons.
\begin{lemma} IF $G$ is
a totally disconnected locally compact
Hausdorff topological group then every neighbourhood of $0$
contains a compact open subgroup of $G$.
\end{lemma}
Our final preliminary lemma tells us that what
looks like ${\mathbb R}^{k}$ is actually ${\mathbb R}^{k}$.
\begin{lemma} Suppose $G$ is a connected locally compact
Hausdorff topological group containing no infinite compact subgroup
and locally isomorphic to ${\mathbb R}^{k}$
in the sense that there exists a neighbourhood $V$ of $0$
and a homeomorphism $\phi$ from the unit ball $B({\mathbf 0},1)$ of
${\mathbb R}^{k}$ to $V$ such that
\[\phi({\mathbf x}+{\mathbf y})=\phi({\mathbf x})+\phi({\mathbf y})\]
for all
${\mathbf x},\,{\mathbf y},\,{\mathbf x}+{\mathbf y}\in B({\mathbf 0},1)$.
Then $G$ is (topological group isomorphic) to ${\mathbb R}^{k}$.
\end{lemma}
\begin{lemma} If $f$ is a continuous open map
of a locally compact Hausdorff space $X$ onto a
Hausdorff space $Y$ and if $K$ is a compact subset
of $Y$ then we can find a compact subset $C$ of $X$
with $f(C)=K$.
\end{lemma}
We can now prove Theorem~\ref{T, principal}.
Our proof is summarised in the following two lemmas.
\begin{lemma} If $G$ is a Hausdorff locally compact
Abelian group then we can find an open
(so closed) subgroup $H$ such that $H$ is
compactly generated and contains no open
subgroup of infinite index.
\end{lemma}
\begin{lemma} If $H$ is a compactly generated
Hausdorff Abelian group
with no open subgroup of infinite order
then
$H=W\oplus {\mathbb R}^{n}$ with $W$ a compact group.
\end{lemma}
It may be helpful to run throught the proofs in the particular
cases $G={\mathbb R}$, $G={\mathbb R}\times{\mathbb T}\times{\mathbb Z}$
and $G={0,1}^{\mathbb R}$.
\section{Final remarks} A different approach which reflects the
original way in these results were discovered is given in
the book of Hewitt and Ross~\cite{Hewitt}. First one
obtains a more powerful version of the structure theorem.
\begin{theorem} If $G$ is a compactly generated Hausdorff
topological group
then
\[G\cong F\times W\times{\mathbb Z}^{l}\times {\mathbb R}^{k}\]
where $W$ is a compact group and $F$ is a discrete group.
\end{theorem}
Then one establishes Pontryagin duality for compact and discrete
groups and uses this to obtain the full duality theorem.
The book of Hewitt and Ross contains proper attributions
of the various theorems.
(In particular, what we call Pontryagin duality should really be called
Van Kampen-Pontryagin duality.)
As a research tool the theory presented in this course was
perhaps too successful for its own good.
It shows that, for many purposes,
${\mathbb R}$, ${\mathbb T}$, $D^{\infty}$ and their
duals are typical locally compact Abelian groups and
if we understand these cases it is not hard to extend
results to the general case.
\begin{thebibliography}{9}
\bibitem{Deitmar} A. Deitmar \emph{A First Course in Harmonic Analysis}
Universitext.
Springer-Verlag, New York, 2002.
\bibitem{Loomis} L.~H.~Loomis \emph{Abstract Harmonic Analysis}
Van Nostrand, Princeton, 1953.
\bibitem{Halmos} P.~R.~Halmos \emph{Measure Theory}
Van Nostrand, Princeton, 1950.
\bibitem{Hewitt} E.~Hewitt and K.~A.~Ross \emph{Abstract Harmonic Analysis}
(Vol I) Springer, Berlin 1963.
\bibitem{Nachbin} L.~Nachbin \emph{The Haar Integral}
(translated from the Portuguese by L.~Bechtolsheim)
Van Nostrand, Princeton, 1964.
\bibitem{Rudin}
W. Rudin \emph{Fourier analysis on groups}
Reprint of the 1962 original.
Wiley Classics Library. A Wiley-Interscience Publication. John Wiley
and Sons, Inc., New York, 1990.
\end{thebibliography}
\end{document}