%------------------------------------------------------------------------------ % Here please write the date of submission of paper or its revisions: 31st of July 2014 %------------------------------------------------------------------------------ % \documentclass[12pt, reqno]{amsart} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color, mathrsfs} \usepackage[bookmarksnumbered, colorlinks, plainpages]{hyperref} \hypersetup{colorlinks=true,linkcolor=red, anchorcolor=green, citecolor=cyan, urlcolor=red, filecolor=magenta, pdftoolbar=true} %\usepackage{draftwatermark} \usepackage{lineno} \textheight 22.5truecm \textwidth 14.5truecm \setlength{\oddsidemargin}{0.35in}\setlength{\evensidemargin}{0.35in} \setlength{\topmargin}{-.5cm} \def\N{{\mathbb N}} \def\Q{{\mathbb Q}} \def\R{{\mathbb R}} \def\ve{\varepsilon} \def\Z{{\mathbb Z}} \newtheorem{Theo}{Theorem}[section] \newtheorem{Lem}[Theo]{Lemma} \newtheorem{proposition}[Theo]{Proposition} \newtheorem{Cor}[Theo]{Corollary} \theoremstyle{definition} \newtheorem{definition}[Theo]{Definition} \newtheorem{example}[Theo]{Example} %\newtheorem{exercise}[theorem]{Exercise} %\newtheorem{conclusion}[theorem]{Conclusion} %\newtheorem{conjecture}[theorem]{Conjecture} %\newtheorem{criterion}[theorem]{Criterion} %\newtheorem{summary}[theorem]{Summary} %\newtheorem{axiom}[theorem]{Axiom} %\newtheorem{problem}[theorem]{Problem} %\theoremstyle{Rem} \newtheorem{Rem}[Theo]{Remark} %\numberwithin{equation}{section} \usepackage{amscd} %\SetWatermarkText{Galley Proof}\SetWatermarkScale{4} \begin{document} %\linenumbers \setcounter{page}{36} \begin{center}{\footnotesize Khayyam J. Math. 1 (2015), no. 1, 36--44}\\\end{center} \noindent\parbox{2.85cm}{\includegraphics*[keepaspectratio=true,scale=0.24]{KJM.jpg}} \vspace{0.5cm} \title[Invariant means on CHART groups]{INVARIANT MEANS ON CHART GROUPS} \author[W.B. Moors]{Warren B. Moors} \address{Department of Mathematics, The University of Auckland, Private Bag 92019, Auckland, New Zealand.} \email{moors@math.auckland.ac.nz} %\dedicatory{This paper is dedicated to Professor ABCD} \dedicatory{{\rm Communicated by A.R. Mirmostafaee}} \subjclass[2010]{Primary 37B05; Secondary 22C05, 37B40.} \keywords{Topological group, invariant mean, Furstenberg's fixed point theorem.} \date{Received: 24 July 2014; Accepted: 01 August 2014.} \begin{abstract} The purpose of this paper is to give a stream-lined proof of the existence and uniqueness of a right-invariant mean on a CHART group. A CHART group is a slight generalisation of a compact topological group. The existence of an invariant mean on a CHART group can be used to prove Furstenberg's fixed point theorem. \end{abstract} \maketitle \section{Introduction and preliminaries} Given a nonempty set $X$ and a linear subspace $S$ of $\R^X$ that contains all the constant functions we say that a linear functional $m:S \to \R$ is a {\it mean on $S$} if: \begin{enumerate} \item[{\rm (i)}] $m(f) \geq 0$ for all $f \in S$ that satisfy $f(x) \geq 0$ for all $x \in X$; \item[{\rm (ii)}] $m({\boldsymbol 1}) = 1$, where $\boldsymbol 1$ is the function that is identically equal to $1$. \end{enumerate} If all the functions in $S$ are bounded on $X$ then this definition is equivalent to the following: $$1 = m(\boldsymbol{1}) = \|m\|$$ where, $\|m\| := \sup \{m(f): f \in S \mbox{ and } \|f\|_\infty \leq 1\}$.\medskip If $(X, \cdot)$ is a semigroup then we can define, for each $g \in X$, $L_g:\R^X \to \R^X$ and $R_g:\R^X \to \R^X$ by, $$L_g(f)(x) := f(gx) \mbox{ for all $x \in X$ \quad and \quad } R_g(f)(x) := f(xg) \mbox{ for all $x \in X$.}$$ Note that for all $g, h \in X$, $L_g \circ L_h = L_{hg}$, $R_g \circ R_h = R_{gh}$ and $L_g \circ R_h = R_h \circ L_g$. \\[6.0pt] If $S$ is a subspace of $\R^X$ that contains all the constant functions and $L_g(S) \subseteq S$ [$R_g(S) \subseteq S$] for all $g \in X$ then we call a mean $m$ on $S$ {\it left-invariant} [{\it right-invariant}] if, $$m(L_g(f)) = m(f) \quad [m(R_g(f)) = m(f)] \mbox{\quad for all $g \in X$ and all $f \in S$.}$$ We now need to consider some notions from topology. Suppose that $X$ and $Y$ are compact Hausdorff spaces and $\pi:X \to Y$ is a continuous surjection. Then $\pi^{\#}:C(Y) \to C(X)$ defined by, $\pi^{\#}(f) := f \circ \pi$ is an isometric algebra isomorphism into $C(X)$. Moreover, we know (from topology/functional analysis) that $f \in \pi^{\#}(C(Y))$ if, and only if, $f \in C(X)$ and $f$ is constant on the fibers of $\pi$ (i.e., $f$ is constant on $\pi^{-1}(y)$ for each $y \in Y$). \medskip The final notion that we need for this section is that of a right topological group (left topological group). We shall call a triple $(G, \cdot, \tau)$ a {\it right topological group} ({\it left topological group}) if $(G, \cdot)$ is a group, $(G, \tau)$ is a topological space and, for each $g \in G$, the mapping $x \mapsto x\cdot g$ ($x \mapsto g \cdot x$) is continuous on $G$. If $(G, \cdot, \tau)$ is both a right topological group and a left topological group then we call it a {\it semitopological group}. \medskip If $(G, \cdot, \tau)$ and $(H, \cdot, \tau')$ are compact Hausdorff right topological groups and $\pi:G \to H$ is a continuous homomorphism then it easy to check that $$R_g(\pi^{\#}(f)) = \pi^{\#}(R_{\pi(g)}(f)) \mbox{\quad for all $f \in C(H)$ and $g \in G$.}$$ If $\pi:X \to Y$ is surjective then $(\pi^{\#})^{-1}: \pi^{\#}(C(H)) \to C(H)$ exists. Therefore, $$(\pi^{\#})^{-1}(R_g(h)) = R_{\pi(g)}((\pi^{\#})^{-1}(h)) \mbox{\quad for all $h \in \pi^{\#}(C(H))$ and $g \in G$.}$$ From these equations we can easily establish our first result. \begin{proposition} \label{Prop1} Let $(G, \cdot, \tau)$ and $(H, \cdot, \tau')$ be compact Hausdorff right topological groups and let $\pi:G \to H$ be a continuous epimorphism (i.e., a surjective homomorphism). If $m$ is a right-invariant mean on $C(H)$ then $m^*:\pi^{\#}(C(H)) \to \R$ defined by, $m^*(f) := m((\pi^{\#})^{-1}(f))$ for all $f \in \pi^{\#}(C(H))$ is a right-invariant mean on $\pi^{\#}(C(H))$. If $C(H)$ has a unique right-invariant mean then $\pi^{\#}(C(H))$ has a unique right-invariant mean. \end{proposition} We can now state and prove our main theorem for this section. \begin{Theo} \label{Thm1} Let $(G, \cdot, \tau)$ and $(H, \cdot, \tau')$ be compact Hausdorff right topological groups and let $\pi:G \to H$ be a continuous epimorphism. If the mapping $$m: G \times \mbox{ker}(\pi) \to G \mbox{ defined by, } m(x,y) := x \cdot y \mbox{ for all $(x,y) \in G \times \mbox{ker}(\pi)$}$$ is continuous and $C(H)$ has a right-invariant mean then $C(G)$ has a right-invariant mean. Furthermore, if $C(H)$ has a unique right-invariant mean then so does $C(G)$. \end{Theo} \begin{proof} Let $L := \mbox{ker}(\pi)$. Then from the hypotheses and \cite[Theorem 2]{E:1957} $(L, \cdot, \tau_L)$ (here $\tau_L$ is the relative $\tau$-topology on $L$) is a compact topological group. Thus $(L, \cdot, \tau_L)$ admits a unique Borel probability measure $\lambda$ (called the {\it Haar measure} on $L$) such that $$\int_L L_g(f)(t) {\rm \ d}\lambda(t) = \int_L R_g(f)(t) {\rm \ d}\lambda(t) = \int_L f(t) {\rm \ d} \lambda(t) \mbox{ for all $g \in L$ and $f \in C(L)$}.$$ Let $P: C(G) \to \pi^{\#}(C(H))$ be defined by, $$P(f)(g) := \int_Lf(g\cdot t) {\rm \ d}\lambda(t) \mbox{\ i.e., $P(f)(g)$ is the ``average'' of $f$ over the coset $gL$.}$$ Firstly, since $m$ is continuous on $G \times L$ (and $L$ is compact) $P(f) \in C(G)$ for each $f \in C(G)$. Secondly, since $\lambda$ is invariant on $L$ it is routine to check that $P(f)$ is constant on the fibers of $\pi$. Hence, $P(f) \in \pi^{\#}(C(H))$. We now show that for each $g \in G$ and $f \in C(G)$, $$\int_L L_g(f)(t) {\rm \ d}\lambda(t) = \int_L f(g\cdot t) {\rm \ d}\lambda(t) = \int_L f(t\cdot g) {\rm \ d}\lambda(t) = \int_L R_g(f)(t) {\rm \ d}\lambda(t). \mbox{\quad $(*)$}$$ To this end, fixed $g \in G$ and define $G:C(L) \to C(L)$ by, $G(f)(t) := f(g^{-1} \cdot t \cdot g)$. Since $m$ is continuous, $t \mapsto (g^{-1}\cdot t)\cdot g$ is continuous and so $G$ is well-defined, i.e., $G(f) \in C(L)$ for each $f \in C(L)$. We claim that $$f \mapsto \int_L G(f)(t) {\rm \ d}\lambda(t)$$ is a right-invariant mean on $C(L)$. Clearly, this mapping is a mean so it remains to show that it is right-invariant. To see this, let $l \in L$. Then $g \cdot l \cdot g^{-1} \in L$ and \begin{eqnarray*} \int_L G(R_l(f))(t) {\rm \ d}\lambda(t) &=& \int_L R_l(f)(g^{-1} \cdot t \cdot g) {\rm \ d}\lambda(t) \\ &=& \int_L f(g^{-1} \cdot t \cdot g \cdot l){\rm \ d}\lambda(t) \\ &=& \int_L f(g^{-1} \cdot [t \cdot (g \cdot l \cdot g^{-1} )]\cdot g) {\rm \ d}\lambda(t) \\ &=& \int_L G(f)(t \cdot (g \cdot l \cdot g^{-1})) {\rm \ d}\lambda(t) \\ &=& \int_L R_{g\cdot l\cdot g^{-1}}(G(f))(t) {\rm \ d}\lambda(t) \\ &=& \int_L G(f)(t) {\rm \ d}\lambda(t) \mbox{ \quad since $\lambda$ is right-invariant.} \end{eqnarray*} Now, since there is only one right-invariant mean on $C(L)$ we must have that $$\int_L G(f)(t) {\rm \ d}\lambda(t) = \int_L f(g^{-1}\cdot t \cdot g) {\rm \ d}\lambda(t) = \int_L f(t) {\rm \ d}\lambda(t) \mbox{ \quad for all $f \in C(L)$.}$$ It now follows that equation $(*)$ holds. Next, we show that $R_g(P(f)) = P(R_g(f))$ for all $g \in G$ and $f \in C(G)$. To this end, let $g \in G$ and $f \in C(G)$. Then for any $x \in G$, \begin{eqnarray*} R_g(P(f))(x) &=& P(f)(x\cdot g) = \int_L f(x \cdot g \cdot t) {\rm \ d}\lambda(t) = \int_L f(x \cdot t \cdot g) {\rm \ d}\lambda(t) \mbox{ \quad by $(*)$} \\ &=& \int_L R_g(f)(x \cdot t) {\rm \ d}\lambda(t) = P(R_g(f))(x). \end{eqnarray*} Let $\mu$ be the unique right-invariant mean on $\pi^{\#}(C(H))$, given to us by Proposition~\ref{Prop1}. Let $\mu^*:C(G) \to \R$ be defined by, $\mu^*(f) := \mu(P(f))$. It is now easy to verify that $\mu^*$ is a right-invariant mean on $C(G)$. \medskip So it remains to prove uniqueness. Suppose that $\mu^*$ and $\nu^*$ are right-invariant means on $C(G)$. Since, by Proposition~\ref{Prop1}, we know that $\mu^*|_{\pi^{\#}(C(H))} = \nu^*|_{\pi^{\#}(C(H))}$ it will be sufficient to show that $\mu^*(f) = \mu^*(P(f))$ and $\nu^*(f) = \nu^*(P(f))$ for each $f \in C(G)$. We shall apply Riesz's representation theorem along with Fubini's theorem. Let $\mu$ be the probability measure on $G$ that represents $\mu^*$ and let $f \in C(G)$. Then \begin{eqnarray*} \mu^*(f) &=& \int_G f(s) {\rm \ d}\mu(s) = \int_L \int_G f(s \cdot t) {\rm \ d}\mu(s) {\rm \ d}\lambda(t) \\ &=& \int_G \int_L f(s \cdot t) {\rm \ d}\lambda(t) {\rm \ d}\mu(s) \\ &=& \int_G P(f)(s) {\rm \ d}\mu(s) = \mu^*(P(f)). \end{eqnarray*} A similar argument show that $\nu^*(f) = \nu^*(P(f))$. This completes the proof. \end{proof} This paper is the culmination of work done many people, starting with the work of H.~Furstenberg in \cite{F:1963} on the existence of invariant measures on distal flows. This work was later simplified and phrased in terms of CHART groups by I.~Namioka in \cite{N:1972}. The results of Namioka were further generalised by R. Ellis, \cite{E:1978}. In 1992, P. Milnes and J. Pym, \cite{MP:1992} showed that every CHART group (that satisfies some countability condition) admits a unique right-invariant mean (unique right-invariant measure) called the Haar mean (Haar measure). Later, in \cite{MP:1992a}, Milne and Pym managed to remove the countability condition from the proof contained in \cite{MP:1992} by appealing to a result from \cite{E:1978}. Finally, in \cite{Moors}, a direct proof of the existence and uniqueness of a right-invariant mean on a CHART group was given, however, this proof still relied upon the results from \cite{MP:1992}. \medskip In the present paper we give a stream-lined proof (that does not require knowledge from topological dynamics) of the existence and uniqueness of a right-invariant mean on a CHART group. \section{Groups} Let $(G,\cdot, \tau)$ be a right topological group and let $H$ be a subgroup of $G$. We shall denote by $(H,\tau_H)$ the set $H$ equipped with the relative $\tau$-topology. It is easy to see that $(H, \cdot, \tau_H)$ is also a right topological group. \medskip Now let $G/H$ be the set $\{xH:x\in G\}$ of all left cosets of $H$ in $G$ and give $G/H$ the quotient topology $q(\tau)$ induced from $(G,\tau)$ by the map $\pi : G \rightarrow G/H$ defined by $\pi(x) := xH$. \medskip Note that $\pi$ is an open mapping because, if $U$ is an open subset of $G$ then $$\pi^{-1}(\pi (U)) = UH=\mbox{$\bigcup$}\{Ux:x\in H\}$$ and this last set is open since right multiplication is a homeomorphism on $G$. \medskip If $H$ is a normal subgroup of a right(left)[semi] topological group $(G, \cdot, \tau)$ then one can check that $(G/H, \cdot, q(\tau))$ is also a right(left)[semi] topological group. \medskip In order to continue our investigations further we need to introduce a new topology. \subsection{The $\sigma$-topology} Let $(G, \cdot, \tau)$ be a right topological group and let $\varphi : G\times G \rightarrow G$ be the map defined by $$\varphi (x,y) := x^{-1}\cdot y.$$ Then the quotient topology on $G$ induced from $(G\times G, \tau\times\tau)$ by the map $\varphi$ is called the $\sigma(G,\tau)$-topology or $\sigma$-topology. \medskip The proof of the next result can be found in \cite[Theorem~1.1,Theorem~1.3]{N:1972} or~\cite[Lemma~4.3]{N:2011}. \begin{Lem} \label{th:1.3} Let $(G, \cdot, \tau)$ be a right topological group. Then, \begin{enumerate} \item[{\rm (i)}] $(G,\sigma)$ is a semitopological group. \item[{\rm (ii)}] $\sigma \subseteq \tau$. \item[{\rm (iii)}] $(G/H,q(\tau))$ is Hausdorff provided the subgroup $H$ is closed with respect to the $\sigma$-topology on $G$. \end{enumerate} \end{Lem} \subsection{Admissibility and CHART groups} Let $(G,\cdot, \tau)$ be a right topological group and let $\Lambda(G,\tau)$ be the set of all $x\in G$ such that the map $y\mapsto x\cdot y$ is $\tau$ continuous. If $\Lambda(G, \tau)$ is $\tau$-dense in $G$ then $(G,\tau)$ is said to be {\it admissible}. \medskip The proof for the following proposition may be found in \cite[Theorem 1.2, Corollary 1.1]{N:1972} or \cite[Proposition 4.4, Proposition 4.5]{N:2011}. \begin{proposition}\label{th:1.5b} Let $(G, \cdot, \tau)$ be an admissible right topological group. \begin{enumerate} \item[{\rm (i)}] If $\mathcal{U}$ is the family of all $\tau$-open neighbuorhoods of $e$ in $G$ then \\ $\{U^{-1}U:U\in\mathcal{U}\}$ is a base of open neighbuorhoods of $e$ in $(G,\sigma)$. \item[{\rm (ii)}] If $N(G, \tau) :=\bigcap\{U^{-1}U:U\in\mathcal{U}\}$ then $N(G, \tau) = \overline{\{e\}}^\sigma$. \end{enumerate} \end{proposition} A compact Hausdorff admissible right topological group $(G,\cdot, \tau)$ is called a {\it CHART group}. \medskip The proof for the following result may be found \cite[Proposition 2.1]{N:1972} or \cite[Proposition 4.6]{N:2011}. \begin{proposition}\label{th:1.7a} \label{th:1.7b} Let $(G,\cdot, \tau)$ be a CHART group. Then the following hold: \begin{enumerate} \item[{\rm (i)}] If $L$ is a $\sigma$-closed normal subgroup of $G$, then so is $N(L, \sigma_L)$. \item[{\rm (ii)}] If $m:(G/N(L,\sigma_L),q(\tau))\times(L/N(L,\sigma_L),q(\tau))\rightarrow(G/N(L, \sigma_L),q(\tau))$ is defined by $$m(xN(L, \sigma_L),yN(L,\sigma_L)) := x\cdot yN(L,\sigma_L) \mbox{\quad for all $(x,y) \in G \times L$}$$ then $m$ is well-defined and continuous. \end{enumerate} \end{proposition} \begin{Rem}\label{Rem} By considering the mapping $\pi:G/N(L, \sigma_L) \to G/L$, Theorem~\ref{Thm1} and Proposition~\ref{th:1.7a} we see that if $(G/L, q(\tau))$ admits a unique right-invariant mean then so does $(G/N(L, \sigma_L), q(\tau))$. Hence if $N(L, \sigma_L)$ is a proper subset of $L$ then we have made some progress towards showing that $G \cong G/\{e\}$ admits a unique right-invariant mean. \end{Rem} \section{$N(L, \sigma_L) \not= L$} In this section we will show that if $L$ is a nontrivial $\sigma$-closed normal subgroup of a CHART group $(G, \cdot, \tau)$ then $N(L, \sigma_L)$ is a proper subset of $L$. \begin{Lem} \label{Lem1} Let $(H, \cdot)$ be a group and $X$ be a nonempty set. Then for any $f :H \to X$, $S := \{s \in H: f(hs) = f(h) \mbox{ for all $h \in H$}\}$ is a subgroup of $H$. \end{Lem} \begin{proof} Clearly, $e \in S$. Now suppose that, $s_1, s_2 \in S$. Let $h$ be any element of $H$ then $$f(h(s_1s_2)) = f((hs_1)s_2) = f(hs_1) = f(h)$$ Therefore, $s_1s_2 \in S$. Next, let $s$ be any element of $S$ and $h$ be any element of $H$ then $$f(h) = f(h(s^{-1}s)) = f((hs^{-1})s) = f(hs^{-1}) .$$ Therefore, $s^{-1} \in S$. \end{proof} \begin{Lem}\label{th:3.1} Let $(G,\cdot, \tau)$ be a compact right topological group and let $\sigma$ be a topology on $G$ weaker than $\tau$ such that $(G,\cdot, \sigma)$ is also a right topological group. If $U$ is a dense open subset of $(G,\sigma)$ then $U$ is also a dense subset of $(G, \tau)$. \end{Lem} \begin{proof} Let $C :=G \backslash U$. Then $C$ is a $\sigma$-closed (hence $\tau$-closed) nowhere-dense subset of $G$. If $U$ is not $\tau$-dense in $G$ then $C$ contains a nonempty $\tau$-open subset. By the compactness of $(G,\tau)$ there exists a finite subset $F$ of $G$ such that $G=\bigcup \{Cg: g\in F\}$. Now each $Cg$ is nowhere dense in $(G,\sigma)$ since each right multiplication is a homeomorphism. This forms a contradiction since a nonempty topological space can never be the union of a finite number of nowhere dense subsets. \end{proof} \begin{Lem}\label{th:3.2} Let $(G,\cdot, \tau)$ be a CHART group and let $\Lambda=\Lambda(G, \tau)$. If $A$ and $B$ are nonempty open subsets of $(G,\tau)$, then $A^{-1}B=(A\cap\Lambda)^{-1}B$. \end{Lem} \begin{proof} Let $x\in A^{-1}B$. Then for some $a\in A, ax\in B$. Since $B$ is open and $A\cap\Lambda$ is dense in $A$ there is a $c\in A\cap\Lambda$ such that $cx\in B$. Hence $x \in c^{-1}B \subseteq (A \cap \Lambda)^{-1}B$. Thus, $A^{-1}B \subseteq (A \cap \Lambda)^{-1}B$. The reverse inclusion is obvious. \end{proof} \begin{Lem}\label{th:3.3} Let $(G,\cdot, \tau)$ be a compact Hausdorff right topological group. If S is a nonempty subsemigroup of $\Lambda(G,\tau)$ then $\overline{S}$ is a subgroup of $G$. \end{Lem} \begin{proof} In this proof we shall repeatedly use the following fact, \cite[Lemma 1]{E:1958} ``Every nonempty compact right topological semigroup admits an idempotent element (i.e., an element $u$ such that $u\cdot u = u$). Firstly, it is easy to see that $\overline{S}$ is a subsemigroup of $G$. Hence, $(\overline{S}, \cdot)$ is a nonempty compact right topological semigroup and so has an idempotent element $u$. However, since $G$ is a group it has only one idempotent element, namely $e$. Therefore, $e = u \in \overline{S}$. Next, let $s$ be any element of $\overline{S}$. Then $\overline{S} \cdot s$ is a nonempty compact right topological semigroup of $\overline{S}$. Therefore, there exists an element $s' \in \overline{S}$ such that $(s' \cdot s)\cdot (s' \cdot s) = (s' \cdot s)$. Again, since $G$ is a group, $s' \cdot s =e$. By multiplying both sides of this equation by $s^{-1}$ we see that $s^{-1} = s' \in \overline{S}$. \end{proof} The following lemma is a simplified form of the structure theorem found in \cite{Moors}. \begin{Lem} \label{th:3.4} Let $(G,\cdot, \tau)$ be a CHART group and let $\sigma$ denote its $\sigma$-topology. Suppose $L$ is a nontrivial $\sigma$-closed subgroup of $G$. Then $N(L, \sigma_L)$ is a proper subset of $L$. \end{Lem} \begin{proof} Let $\mathcal{U}$ denote the family of all open neighbuorhoods of $e$ in $(G,\tau)$. Then it follows from Proposition \ref{th:1.5b} that $\mathcal{V}=\{U^{-1}U:U\in \mathcal{U} \}$ is a base for the system of open neighbourhoods of $e$ in $(G,\sigma)$. Then $\{V \cap L:V\in \mathcal{V}\}$ is a basis for the system of neighbourhoods of $e$ in $(L,\sigma_L)$. From the definition of $N(L, \sigma_L)$ (see Proposition \ref{th:1.5b} part (ii)) it follows that $$N(L,\sigma_L)=\mbox{$\bigcap$} \{(V\cap L)^{-1}(V\cap L):V\in \mathcal{V}\}.$$ The proof is by contradiction. So assume that $N(L, \sigma_L) = L$. Then $$L=\mbox{$\bigcap$} \{(V\cap L)^{-1}(V\cap L):V\in \mathcal{V}\}.$$ Hence, for each $V\in \mathcal{V}$, $(V\cap L)^{-1}(V\cap L)=L$, or equivalently, for each $V\in \mathcal{V}$, $(V\cap L)$ is dense in $(L,\sigma_L)$. That is, for each $U\in \mathcal{U}$, $(U^{-1}U\cap L)$ is open and dense in $(L,\sigma_L)$ and hence, by Lemma \ref{th:3.1}, dense in $(L,\tau_L)$. \medskip Since $L\neq \{e\}$, there exists a point $a\in L$ such that $a \neq e$. Note that since $(G,\tau)$ is compact and Hausdorff there is a continuous function $f$ on $(G,\tau)$ such that $f(e)=0$ and $f \equiv 1$ on a $\tau$-neighbuorhood of $a$. \medskip For the rest of the proof, the topology always refers to $\tau$ and we shall denote $\Lambda(G,\tau)$ by $\Lambda$. By induction on $n$, we construct a sequence $\{U_n: n\in\N\}$ in $\mathcal U$, a sequence $\{V_n:n\in\N\}$ of nonempty open subsets of $G$, each of which intersects $L$ and sequences $\{u_n:n\in\N\}$ and $\{v_n:n\in\N\}$ in $G$ which satisfy the following conditions: \begin{enumerate} \item[(i)] $v_n\in U_{n-1}^{-1}U_{n-1}\cap (V_{n-1}\cap \Lambda) = (U_{n-1}\cap\Lambda)^{-1} U_{n-1}\cap(V_{n-1}\cap \Lambda)$; by Lemma~\ref{th:3.2}. \item[(ii)] $u_n\in U_{n-1}\cap\Lambda$; \item[(iii)] $V_n\subset\overline{V_n}\subset V_{n-1}\subset f^{-1}(1)$ \ and \ $V_n \cap L \not= \emptyset$; \item[(iv)] $u_n V_n\subset U_{n-1}$; \item[(v)] if $H_n$ denotes the semigroup generated by $\{u_1, v_1,u_2, v_2,\ldots,u_n, v_n\}$; which we enumerate as: $H_n := \{h^n_j:j\in\N\}$ and $$U_n:= \{t \in G: |f(h^i_j\,t)-f(h^i_j)| < 1/n\quad \text{for}\quad 1\le i,j\le n\} $$ then $H_n \subset \Lambda$ and $e\in U_n\subset \overline{U_n}\subset U_{n-1}$. \end{enumerate} \noindent {\bf Construction.} We let $U_0 :=G$ and let $V_0$ be the interior of $f^{-1}(1)$ and $u_0,\,v_0$ are not defined. Assume that $n\in\N$ and that $U_k,\,V_k$ are defined for $0\le k